Power System Analysis

1.

( 2 . 1 4)

wh ere N I is the number o f turns a n d L I I is the self-inductance o f coil Under the same condition of i l acting a lone the fl ux l inkages of coi l 2 are given by

( 2 .1 5) w here N2 i s t h e n umber o f t urns of c oil 2 a n d L 2 1 is the mutual i n ductance between the coils. Similar definitions a pply when i 2 acts a lone. It produces fl ux

1 due to ( 2 . 17)

. When both currents act together, the flux l i n kages add t o give

( 2 . 1 8)

The order of the subscripts of L j 2 a n d L 2 1 is not i mportant since mutual i nductance is a single reciprocal property of the coils, a n d so L 1 2 = L 2 l . The d i rection o f the currents and the orientation of the coils determine the sign of m u tu al inductance, which i s positive i n Fig. 2.4 because i l and i 2 are ta�en to m a gnetize in the same sense.

2 .2

MAGNETICALLY COUPLED COILS

49

When the flux linkages change with time, the vol tage d rops across the coils in the direction of their circulating currents are

( 2 . 19 )

( 2 .20) The positive signs of Eqs. (2. 1 9) a n d (2.20) are usually associated with a coil that is absorbing power from a source as i f the coil were a load. For instance, in Fig. 2.4 if both V 2 and i 2 h ave positive values simultaneously, then instantan eous power is being absorbed b y coi l 2. I f the voltage d rop across coi l 2 is n ow - v � , we h JVC r e v e rse d so t h a t v'2 =

( 2 .2 1 ) For positive instantaneous values of v� a n d i 2 power is b e i n g supplied b y coil 2 . T h u s t h e n e g a t i ve signs of E q . (2.2 1 ) are characteristic of a coil acting as a generator del ivering power (and energy ove r time) to an external load. I n the steady state, with ac vol tages and curre n ts i n the coils, Eqs. (2. 1 9) a n d (2.20) assume the phasor form ,

(2.22)

(2.23) H e r e we l I s e

l owe rcase

:: ij

to

i m p e d a n c e s /'j ' I n ve c t uI - I 1 1 a t r i x

d is t i ngu ish

t h e co i l

Co rm I � q s . ( 2 .22)

and

i m red a n ce s

( 2 . 2J )

f ro m

become

node

( 2 .24) that t h e V 's a re t h e vol tage d rops across the term i n a l s of the coils and the I 's are the circu l a t i ng cu rrents in the coils. The inverse o f t h e coefficient mat rix is t h e m atrix of a d m ittances d e n o t e d by We s h o u l d a l so n o t e

[Yll

Y21

( 2 .25 )

50

CHAPTER 2

Multiplying

TRAN S FO R M ERS

Eq. (2.24) by the admittance matrix gives ( 2 .2 6 )

Of course, t h e

and z parameters w i t h the s a m e subscripts are n o t si mple reciprocals of each other. I f the term i nals of c oil 2 are ope n, then setting 12 = 0 in Eq. (2.24) shows that the open-circuit input impedance to coi l 1 i s y

( 2 . 27)

If the term inals of coil 2 are closed , then short-circuit input impedance to coil 1 is

V2

=

0 and Eq. (2.26) shows that the

( 2 .28) By substituting the expressions defining the z if from Eqs . (2.22) and (2.23) into Eq. (2.28), the reader can show that t h e apparent reactance of coil 1 is r e duced b y the presence of closed coil 2. In Chap. 3 a similar result is fou n d for the synchronous m achine under s hort-circuit con ditions. An important e quivalent c ircuit for the m u t ually coupled coils is shown i n Fig. 2.5 . The current o n the coil 2 side appea rs a s 1 2 1a and t he term i n a l vol tage as a V2 , w here a is a positive constant. O n the coil 1 s i de V[ and I[ a re the same as b e fore. B y writing Kirch hoff's vol t age equa tion around the path o f each of t h e c urrents 11 and 1 2 1a in Fig. 2.5, t h e reader shou l d find t h a t Eq s . (2.22) and (2.23) are satisfied exactly. The i nductances i n b rackets in Fig. 2.5 a re t h e leakage inductances L l l a n d L 2I o f t h e coils i f we l e t a = N , IN2 . This i s shown

FI G U RE 2.S An ac equivale n t circuit for Fig. 2.4 with seco n d a ry c u rre n t a n d voltage r e d e fi n e d a n d

a =

N1 /N2 .

2.3

THE EQUI VALENT CI RCUIT OF A S I NGLE·PHASE TRANSFORMER

51

FIGURE 2.6 The equiva lent circu i t of Fig. 2.5 with i nductance paramete rs renamed.

from E qs. (2. 1 4) through (2. 17) as foll ows:

L it

�

L I I - aL 2 1

L 2 1 #::. L n

-

L 1 2/a

=

=

N I 4> I l il

-

-

N2 rP 22 12

N I N2 cP 2 1 N2 II

N2 N l cP 1 2

--

NI

[2

=

'v I . (

=

"V-,

( 2 .29)

.

q; I {

1 2 "'-

�

(

( 2 .30)

CP u

wh er e 4> 1 1 and cP 2 1 are the l eakage ftuxes of the coils. Likewise, with a Ni lN2 , t h e s h u n t indu ctance aL 2 1 i s a magnetizing inductance associated w i t h t h e m u t u a l fl u x rP 2 1 l inking t h e coi l s d u e t o i l since =

( 2 .3 1 )

Defining t h e se ries leakage reactances X l = w L l ! and x 2 w L 2 ! , a n d the shunt magnetizing susceptance IJ", ( (U a /' 2 1 ) . 1 , leads t o the e quival e n t c ircuit of F i g . 2 . 6 , \vhich i s the basis of the equivalent circu it of the practical tra nsformer in S ec. 2 . 3 . =

2.3 THE EQUIVALENT CIRCUIT OF A SIN GLE- PHASE TRANSFORMER

=

The equ ivalent circu i t of Fig. 2.6 comes close to matching the p hysical c harac­ teris t i cs of the practical transformer. However, i t has three deficie ncies: (1) I t does n o t reflect any curre n t o r voltage t ransformation, (2) i t d o e s n o t provi d e for e lectrical isol ation of t h e primary from t h e secondary, a n d (3) it does n o t accoun t for the core losses.

52

CH APTER 2

TRANSFORMERS

When a s inusoidal voltage is applied to the primary winding of a practical transformer on an iron core with the secondary winding open, a small cu rrent It: called the exciting current flows . The major component of this cur r e n t is called the magnetizing current , which corresponds to the current t hrough the magnetizing susceptance B m of Fig. 2.6. The magnetizing current produces the flux in the core. The much s m a l ler component of ]E ' which accounts for losses

in the i ro n core, leads the magnetizing cu rrent by 90° an d is not represented i n Fig. 2 . 6 . T h e core losses occur d ue , fi rst, t o t h e fact that t h e cyclic changes o f the d i r e c t i on o f t h e flux i n the iron require energy which i s d issipated a s h eat and is called hysteresis loss . The second l oss is due to the fact that circula t i n g currents a r e indu ced in t h e i ro n due to t h e changing flux, and t hese currents 2 produce an I I I R loss in the i ro n cal led eddy-current loss . Hysteresis loss is r educed by the use of c e rtain high g r a d e s of a lloy steel for t he core . E d d y- c u r ­ re nt loss is re d uc e d b y b u i l d i ng up the c o r e with l a minated s h e e t s of st e e l . W i t h the secondary open, t h e t r a n s fo r m e r primary c i rcu i t is s i mply one of very high inductance due t o the i ron core . In the equivalent c i rcuit IE i s taken fu lly i n to accou nt by a co n d u c t a n ce G( i n p a ra l l e l with the magnetizing s uscep t a nce Bill > as shown i n Fig. 2.7. In a well -designed transforme r the maximum flux density in the core occurs at the knee of the B-H or satura tion curve of the transformer. So, flux density is not linear with respect to field intensity . The magnetizin g c u r r e n t cannot b e s in usoidal if it is to produce sinusoidally varying flux required for inducing sinuso idal voltages e I an d e2 when the applied voltage is sinusoidal. The exciting current IE will h ave a t hird harmonic content as h igh as 40% a n d l esser amoun ts o f higher h armonics. Since JE is s m a l l compared to rated current, it is treated as sinusoidal for convenience, a nd so use of Gc and Bm is accept a b l e in the equivalent circuit. V ol tage and current transformation and electrical isolation of the primary from t h e secondary can be obtained by adding to Fig. 2.6 an ideal transformer

rl

1 -j

+

VI

FIG UR E 2.7

Xl

a2 X 2

a2 r 2

r

12

a:l

•

T-

�

•

+

1 j-

V2 N1

N2 Ideal

Eq u i va l e n t c i rc u i t for a s i n g l e - ph a s e t r a n sfo r m e r w i t h a n i d e a l t r a n s fo r m e r o f turns r atio a =

+

N,1 /N2.

2.3

53

TH E E Q U I VA L ENT C I R C U I T OF A S I N G LE · P H A S E T R A NSFO R M E R

FI G U RE 2.8

Transfo r m e r e qu iv a l e n t circuit w i t h magnetizing current n e ­ glected.

with t urns ratio a = N I l Nz , as shown in Fig. 2.7. The location of t he ideal transformer is not fixed. For instance, it may be moved to the left past the series ele m e n ts a Z r2 and a 2 x 2 ' w hich then become the winding resistance 'z and the leakage reactance x 2 of the seco n dary w i n d i ng . This is i n keeping with the rule e s t a b l i s h e d fo r the ideal transformer in Sec. 2 . 1 that whenever a bran ch i m p e d a n c e is referred from a g i v e n s i d e to the oppos i te side of a n ideal t r ,1n s fo r m e r , i t s i m p e d a n c e va l u e i s m u l t i p l ie d hy the sqU : HC o f t he rat i o of the t u r l l S o n t h e o p [! ( ) s i t e s i d e

all

t o t h e t lI r n s

( ) Il

t he g i v e n s i d e .

1l1 ,ly b e o m i t t e d i ll t h e e q u i v ,d e n t c i rcu i t i f w e refer e i ther t h e h i gh- or the l ow-vol tage s ide of the t ra nsformer. For i n s t a nce, in F i g . 2.6 we say that a l l vol t ages, curr ents, a n d i mped a nces are re fe rr e d to the primary cirel.! it o f t h e transformer. W i thout the i d e a l t rans­ fo r m e r , w e h ave t o he ca refu l not t o create u n n ecessary short c i rcuits w h en developing e q u iva l e n ts for m u l tiwi n d ing transformers. Ofte :1 we negl ect exc i t i n g current because it is so sma l l compared to the u c;ual load currents and to s i m p i ify the c i rcuit further, we let T h e i d e a l t ra n s fo r m e r

q u a n t i t i es t o

( 2 .32) to obt a in the equ ivalent c i rcu i t of Fig. 2 . S . A l l impedances and voltages in the part of t h e c i rcuit connected t o the seco n d a ry t erm inals must now be referred to regulation i s

t h e p r i m a ry s i d e . f/o/:(/ge

t u d e a t t h e I O (l d k r m i n (l l s

p c rc c n t u f fu l l - l o a d vo l t a g e

c q l l ;l t i O I l

d i fference between t h e voltage magni­ o f t h e t r (l Ils fo rm c r a t fu l l l o a d a n e! a t no load in w i t h i n pu t v o l t a g e h e l d co n s t a n t . I n t h e fo rm o f a n

dcflned

a s t he

Perce n t regu l a t ion

( 2 .33)

I V::

l i s t h e magni t u d e of l oad voltage V7 . NL m (l g n i t u cJ e of V2 a t fu l l load with I V I I constant.

w h e re

'

Exa mple 2 . 2 . A s i n g l e - p b ase t r a nsfor m e r h a s

at no load and I Vz

2000

a re X I

= R.O fl

I is

the

t u rn s o n t h e p r i m ary w i n d i n g

a n d 5 0 0 t u r n s on t h e s ec o n d a ry. W i n d i n g r e si s t a n c e s

0 . 1 2 5 n . Leakage r e a c t a n c e s

FL

3re r1

and x2 = 0.50

fl.

= 2.0 fl

an d

r2

=

T h e r e s i s t a nce l oa d

54

CHAPTER 2

TRANSFORMERS

j16 Q

4 Q

FIGURE 2.9 Circuit for Exa m p l e 2.2.

22

12 n . If app l ied voltage a t the terminals of the prim ary w i n d i n g is 1 200 V2 and the voltage regulation. Neglect magnet i z i ng curre n t .

is

find

V,

Solution

2000

N

a = -I = -=4 500 N2

RI

= 2 + 0 .125 ( 4 ) 2 = 4 .0 n

XI

= 8 + 0 .5( 4 ) 2 = 16 n

2�

= 12 X

LQ:

(4) 2

The equivalent circu i t is shown i n F i g .

1 = I

1 200 1 92 +

a V2 = 6 1 0/ .

V2 =

= 1 92 n

2.9, a n d we c a n calcu late

4 + j1 6 = 6 . 1 O!- 4.6r 4 6r

/

-

.

x

1 1 7 1 .6 - 4 .67 :

Voltage regulation

4

=

1 92

= 1 17 1 .6/

A

-

4 . 6r

= 292.9/ - 4 .6r

1 200/4 - 292 . 9 292 . 9

V

V

= 0 .0242 or 2.42%

The p arameters R and X o f the two-winding transformer are determined by the short-circuit test , where impedance is measured across the terminals of one winding when the other winding is short-circuited. Usually, the low-voltage side is short-circuited and just enough voltage is applied to the h igh-voltage terminals to circulate rated current. This is because the current rating of the source supplying the high-voltage side can be smaller. Voltage, current; and

2 .3

TH E EQUI VALENT CI RCU IT OF A SING LE-PHASE TRANSFO R M E R

55

power input are determined_ Since only a smal l vol tage is required, the exciting c urrent is i nsignificant, and the calculated impedance is essentially equal to

R + jX.

Exa mpJe 2.3. A single-phase tran sformer is rated 1 5 M VA, 1 1 .5/69 kY. I f the 1 1 .5 kV w i n d i ng (designated winding 2) is short-circuited, the rated current flows w h e n the vol tage a p p l i e d to wi nding 1 is 5.50 kV. T h e power i n p u t is 1 05 . 8 kW. F i n d R} a n d X I in ohms referred to t h e h i g h -vol tage wind ing. Solution.

Rated curre n t for the 69-kY winding has the magnitude 1 5 , 000 69

Th e n ,

( 2 1 7.4r R I ')

RI

IL)

=

1 05 ,800

=

2 . 24 n 5500

=

--

2 1 7 .4

=

=

2 1 7 .4

A

25 . 3 0 D

The example illustrates t h e fact that the w i n ding resistance m ay o ften be omitted i n the t ransforme r equivalent circuit. Typical ly, R is l ess t h a n 1 %. Al though exc i t i ng current may be n eglected (as in Exam ple 2.2) for most power system calculat ions, Gc jBm can be calculated for t h e equivalent circuit by a n open · circuit test R ated voltage i s applied to the l ow-voltage termin a ls, a n d the p mver input and cu rrents are measu red . This is because the voltage rating o f t he source s u p pl yi n g t h e low-vol tage s i d e c a n be sma l l e r. The measured i m p e d a nc e i n c l u des the resi st a n ce a nd leakage reacta nce of t h e wind ing, but t h e s e v a l ue s are i nsignificant when compared to l /( Ge jBn). -

_

-

2.3 the open-circuit test with 1 1 .5 kY a p p l i e d resu l t s in a power i n p u t of 66.7 kW a n d a current of 30.4 A. Find t h e v a l u e s of Gc a n d E m referred to t h e h igh-vol tage w i n d i n g 1 . W h a t i s t h e efficiency of t h e tra nsformer for a l o a d of 12 MW at 0.8 power-factor l a g g i n g at rated vol tage? Exa m p l e 2.4. For the transfor m e r o f Exa mple

The t u rn s ratio is a = N \ /N2 = 6. Measurements are made o n the low-vo ltage s i d e . To t r a nsfer shunt admittance Y G c - jAm from high-voltage 2 side 1 to l ow-vo l tage s i d e 2, m u l t i p ly by a since we wou l d divide by a 2 to tran sfer

Solution.

=

56

CHAPTER 2

TRANSFORMERS

impedance from side 1 to s i d e 2. U n d e r open-circuit t e s t con d i tions

Gc I YI

=

Bm =

Under rated

I V2 1 1 12 1

V I YI 2

X -

= 1 4 .0 X 1 0 - 6

30.4 = 1 5 a2 1 , 00 1

G}

10-6

X

S

1 7 3 .4 36 =

-/73 .42

-

X

10 6 S

1 4 . 0 2 = 72 .05

1 0 (, S '

co n d i t i o n s t h e t o t a l l oss is a p p rox i m a t e l y t h e s u m of s h o r t -c i rc u i t a n d

o p cll-circ u i t test losses, � n d s i nce c t l i c i e ncy is t h c r;l l i o o f k i l ow a tts, w e h ave

Efficiency

2.4

X

1 2 , 000

=

+ ( 1 05 .8

----

1 2 ,000

+ 66 .7)

X

t i l e u u t r HI l t o t h e i n p u t

1 00 = 9 8 . fi %

This example illustrates the f a c t that G c i s so m u c h s m a l l e r than Bm that i t m ay b e omitted. B m is a lso very small s o t h a t IE is oft e n n e g l ected e n t irely.

PER-UNIT IMPEDANCES IN SINGLE-PHASE TRANSFORMER CIRCUITS The ohmic values of resistance and l e akage re actance of a transformer depend on whether t hey are measured on the h i gh - or low-vol tage side of the trans­ former. If they are expressed in per u n i t , t h e base kilovol tamperes is understood to be the kilovoltampere rating of the transformer. The base voltage is under­ stood to b e the voltage rating of the l ow-voltage wind ing if the ohmic values of resista nce and leakage reacta nce are refe rred to the low-voltage side of the transformer. Likewise, the base volt age is taken to be the voltage rating of the high-vol tage wind ing if the ohmic values are referred to the high-vol tage side of the transformer. The per-unit impedance of a transformer is the same regard­ less of whether it is determined from ohmic values referred t o the high-voltage or low-voltage sides of the transfo rmers, as show n by the following example. 2.S. A s i n gle-p h as e t ransfo r m e r is rated 1 1 0/440 V, 2.5 k VA. Leakage reactance measured from the low-vo ltag e side i s 0.06 fl. Determine leakage reactance in per u n i t .

Example

Solution. From E q . ( 1 .46) we h ave

Low-voltage base i mp e d a nce

0 . 1 1 0 2 X 1 000 =

2.5

= 4 .84 fl

2.4

P E R - U N IT I M P ED A NCES I N S I N G L E - P HA SE TRANSFO R M E R C I R C U ITS

57

In per u n i t

X

0 .06 = -- =

4 . 84

0 .0 1 24 per u n i t

If l eakage reactance h a d been measured on t h e high-voltage side, t h e value wo u l d b e

X

I Iigh-vo l t agc b a s e i m pe d a n c e I n per unit

X

= -- = 0 . 96

7 7 .5

=

0 .0

( ) 6 - = 440 2 110

0 .440 2 =

X

1 000

2.5

0 .9

6n

= 77.5 n

0 .0 1 24 pcr u n i t

great advan tage i n m aking per-un i t computat ions i s real ized by the p roper selection o f different b ases for circuits connected t o each o t h er t hrough a transfonner. To achieve the adv an tage in a single-phase system, the voltage bases for the circuits connecred through [he transformer must have the same ratio as the turns ra tio of the transformer l1:indings. With such a selection of voltage bases and the same kilovo l t a mpere base, the per-unit value of an i mpedance will be t h e same when it is expressed on t h e b ase selected for its own side of the transformer as when i t is referred to the other side of the transform er a nd expressed on the base of that s i d e. So, the transformer i s represented completely b y i t s imped a n ce (R + jX ) i n per u ni t when magnetizing curre nt is n eglected . No per-unit volt age transfor­ m a t i o n occurs when t h i s sys t e m is used, a n d the c u rrent wi l l a l so h ave t h e same p e r - u n i t v a l u c o n b o t h s i d e s o f t h c t r a n s fo r m e r i f m a g n e t i z i n g cu rrent is A

neglected .

Exa m p l e 2.6. T h r e e p a r t s o f a s i n g l e - D h ase e l e c t r i c system arc desi g n a t e d A , B ,

u n d (' a n d a rc con n ec t e d 2 . 1 0 . T h e t ra nsform e rs a r c A -B B-C

1 0,000 kY A, 1 0 , 000 kYA,

t o e a c h o t h e r t h r o u g h t ra n s fo r m e rs, as

rated as

shown i n Fig.

[o l l ows:

13.8/ 138 kY, leakage r e a c t a n c e 1 0 % 138/69 k Y , leakage r e a c t a n ce 8%

138

I f t h e b a se i n c i rc u i t B i s c h os e n a s 1 0 , 000 kYA, k Y , fi n d t h e per-u n i t i m p e d a n ce of t h e 300-D r es i s t i v e l o a d i n c i r c u i t C r e fe r r e d to c i r c u i ts C , B , a n d A . D r aw t he im p e d a n c e diagram n e g l e c t i ng m a g n e t i z i n g current, transformer r es i s t a n c es, and l i n e i m p e d a n c e s

.

58

CHAPTER 2

TRANSFO RM E RS

1-10

2-1

A

B

A-B

300n

B--{;

FIGURE 2 . 1 0 Circuit for Examp le

2.6.

Solution Base vol tage for ci rcu i t A : Base voltage for circu i t C :

Base imped ance of circu i t C :

Per-unit impedance of load i n circ u i t C :

0.1

x

O.S X

138

=

1 3 . 8 kV

138 = 69 k V

692 X 1000 10,000 300 476

- =

=

476 n

0.63 per u n i t

Because the selection of base i n various p arts of the system is d e t e r m i n e d by the t u r n s ratio of the transformers, and b ecause the base kilovol tamperes i s t h e s a m e in all parts o f the sys tem, the per-u n i t i mpedance of t h e l o a d re ferred to a n y p a rt of the sys tem w i l l b e t h e s a m e . This i s verified a s fol l ows:

Base impedance o f c i rcui t B : Impedance o f load referred to circuit B :

Per-un i t impedance of load refe rred to B :

Base i m pedance of c ircu i t A :

Impedance of load referred t o c i rcu i t A :

Per-un i t imped ance of load referred t o A :

1382 X 1000 10,000

=

1900 n

300 X 22 1200 n =

1200 1 900

-- =

0. 63 per u n i t

13 .82 X 1 000 10,000

--- -- =

19 n

300 X 22 X 0.12 12 n 12 19

- =

=

0.63 p e r u n i t

=:=10.63+)0 j O. l

2 .5

T H R E E-PHASE TRANSFO R MERS

59

j O.08

Imped ance d i agram for Example 2.6. Impeda n ces are m a rked in per u n i t .

F I G U R E 2. 1 1

S i nce t h e chosen bases for k i l ovol ts a n d k i l ovol tamperes agree w i t h t h e tra nsformer rat ings, the t ransform e r reacta nces in pe r u n i t a re 0 . 08 and 0. 1 , r esp e ct iv ely . Figu re 2. 1 1 IS t h e r e q u i r e d i m p e d ance d i agram w i t h impedances m a rked in per unit.

Because o f t h e advantage p reviously pointed o u t , t h e p r i n c i p l e demon­ st r a t e d i n t h e preceding exa m p l e fo r se l ecting b ases in various p arts of t h e s i n g l e - p h ase system is always followed i n m a k i n g co m p u t a t i o n s i n per u n i t . That is, the k i/olJo/ta mpere base sh o u ld h e the saine in all pa rts of t h e system , a n d the

select ion of ri t e base ki/o (:o/rs il l ki/o L'o/rs

to

the system d('(ermines the base ra tios of the rTan'<'jormers , to the

Th i s p r i nc i p l e il i l o w s liS to c o m b i n e on o n e

De assign ed , acc()rdill�

of r h e system .

pa rt of

t o the OtiC

tUTIlS

d i a g r a rn t h e p c r - u n i t i m p e d a n c e s 0 1 t h e e n t i re sys t e m .

or/lcr parts

2.5

i mpedance

THR.E E - PfLA.S E TRAN SFORM ERS

Th ree i d en t ic a l s i n g l e - p h a s e t ra n sforme rs may be connected so that the t h re e w i n d i ngs o f o n e voltage rat i ng a re L\-connected and t h e t h r e e w i n d i n g s o f t h e o t h e r vol t age rcl t i n g (l IT Y - c o nn e c t e d t o form a t h r e e - p h a s e tran sformer. S uc h a transformer is s a i d to be con nected 'y'-6. o r 6 -Y . T h e o t h e r possi ble connections a r e Y- Y a n d l'l - L\ . I f e a c h o f the t h ree s i n g l e-ph as e transformers h a s t h re e w i n d ings ( a primarj, secondary, and t er t i a ry) , two sets m ig h t b e conn ected i n Y a n el o n e i n 6 , or two cou l d b e 6. - co n n ec t e d with one Y - co n n e ct e d . I nstead u s i n g t h ree i d e ;l t i c a l s i n gi e - p h a s e t r a n s fo r m e rs, a m o re usual u n i t is a three­ p h a s e t a n s fo r m e r w h e r e a l l t h ree p h a s e s a re o n t h e s a m e iron structure. The theory is t h e s a m e for a t h ree-phase t r Cl n s fo r m e r a s fo r a t h re e-ph ase b a n k of s i n io' : e - p h a s e t ra n s fo rrn c r s . T h e t h r e e - p h Cl s e u n i t has t h e a dv a n t a g e of requi r i ng

of

r

co r e ,

more econ omical t h a n t h re e s i n g le-phase u n i ts h ave t he a d v a n t a g e of r e pl a c e m e n t o f o n ly one u n i t o f t h e t h ree-phase bank i n case of a fa i l u re rather t h a n l o s i n g t h e whole th ree-ph ase bank. I f a fai l u re occurs i n a 6 - 6. b a n k c o m p o se d of t h r e e s e p a r a t e u n i t s, o n e of t h e s i n g l e - p h as e t ra nsform­ e rs c a n b e r e m ove u and t h e re m a i n i n g t w o wi l l s t i l l o p e r a t e a s a t h ree-phase trans former a t a red uced k i l ovol t a m pere . Such an ope r a t i o n is called open delta . For a s i n g l e -· p h ase t r a n s for m e r we can con t i n u e to p l ace a dot on one e n d o f e ac h w i n d i n g , o r a l t e r n a t ively, t h e dutted e n d s m a y b e m a r k e d H I for t h e h i g h -v o l t a g e w i n d i ng a n d X l for t h e low-voltage wind i n g . The oppos i t e e n d s a r e t h e n l a b eled H :. and X 2 , respect ively. F i g u r e 2 . 1 2 shows how t hree s i n g l e - phase transformers a re conn ected to form a Y - Y th r e e - p h as e t r a n sfo r m e r b a n k . In this text \\l e s h a l l u se capital less

iron

to

fo r m

the

and

is

t h e r e fo r e

s i n g l e - p h a se t1 n i t � (l n u occ u p i e s l e ss s p ' l c e . T h r e e

60

CHAPTER 2

TRANSFORM ERS

IB

A

Hl

C

H2

H3 N

-

•

-

-

-

•

n

x� b

( a ) Y-Y connection d i a g ram A �------�

B

-0--..

-

•

N

C

H3 --0-------'

X2

.-------c>-

-

-

b

,----0--- a

n

,-1 -------------� c

(b) Alternate form of c onnection d i ag ram

FI GURE 2 . 1 2 Wiring d iagrams for Y - Y tra nsformer.

l e tters A , B, and C to identify the p hases of the high-voltage wind ings and lowercase l etters a, b, and c for the low-vol tage w indings. The h igh-voltage terminals of three-phase transformers are marked H I ' H 2 ' and H 3' and the low-vol tage terminals are marked X l ' X 2 , and X3. I n Y - Y or � - � transfo rmers t h e m arkings are such that voltages t o n eutral from terminals H I ' H 2 , and H3 are in phase w ith the voltages to n e u tral from terminals X" X 2 , and X3, respective ly. Of course, the .6. w i n d ings have n o neutral, but the part of the system to w h ic h the � winding is c onnected will have a con nection to groun d . Thus, the g round c a n serve a s t h e e ffective n eutral u n d e r balanced conditions and voltages to neutral from the term inals of the � do exist. To conform with the American standard, the terminals of Y-� and � - Y transformers are l abeled so that the vol t a ges from H I ' 2 ' and H3 to n e utral lead the voltages to neutral from X l ' X 2 , and X 3' respectively, b y 30° '. W e consider t hi s p hase shift more fu l ly i n the n ext section .

H

2.5 THREE·PHASE TRANSFO R ME R S

61

'r I 1. �_�'------'

66 kV

6.6 kV

3.8 1 kV

Fl G U RE 2 . 1 3 y-y t ra n s form e r r a t e d 66/6.6 k V

Figure 2 1 2( h ) p r ov i d e s t h e S il nl C i n fo r m a t i o n as ! -' ig. 2 . 1 2( a ). W i n d i n gs of t h e p r i m a ry , \ n d s e c on d a ry , w h i c h a r c d ra w n i n p , \ r a l l c l d i r e c t i o n s i n Fig. 2 . 1 2( 6 ) , a r c for t h e same s i n g le-ph a s e t r a n s fu r m e r or o n t h e s a m e leg o f a t h re e p h a s e t r a n s for m e r . Fo r i n s t a n c e , t h e w i n d i n g from A to N is l i n ke d by t h e s a m e A u x as t h e w i n d i n g from (l t o 1 1 , , \ Il d V! N is i n p h a se w i t h V;" I . T h e d iagrams o f F i g . 2 . 1 2 ( b ) a r e wiring diagrams only. They a r e not phasor d i a

-

­

F i g u re 2 . 1 3 i s

-

g r Zl m s .

a schematic met hod of i n dicating w i n d i n g con nections o f a t h ree phase tra nsformer. Voltages are shown for a 6 6 j 6 . 6-kV, Y-Y t ransfor m e r supplying 0.6-0 re s is t o rs or impeda nces. Figure 2 . 1 3 sho ws a balanced system i n which e ach phase can b e tre a t e d s e p a ra t e l y , whether o r not t h e n e u t ra l p o i n t s a r e c o nn e c t ed . Then , i m p e cJ a n c es wou l d t ransfer f r om t h e low-voltage to t h e high-vol tage s i d e b y t h e square o f t h e ratio o f l i ne-to-n e u t ral voltages, w h i c h i s t h e s a m e a s t h e square o f t h e r a t i o of l i n e-to-line vol tages; t h a t is,

0.6

r �) , �8 1

\

_� . 8 1

2

=

0 .6

( -(i6- ) )

6.6

2 =

60 D

t h e resis to r s w i t h t i l e S;l llle () () · � V p r i m ; I I'Y, t i l e 6 w i n d i n g s wOl l l d he r;l l e ll (d) k V r(l t h e r t h an 3.8 1 kY. So fa r a s t h e vo l t age m agnitude a t t h e l uw-v o l t a g e t e r m i n a l s is c onc e rn e d , th e Y - 6 t ra n s fo r m e r cou l d t h e n b e re p l aced b y a Y-Y trans­ for m e r b a n k h avi n g a n e ffe c t ive p h a s e - t n- n e u t ra l t u rns rat i o of 38 . 1 : 6.6/ /3 , o r I f we h a d l I '; c d ; \ Y - � t r ,\ n S rl ) f fll C r t o o ht a i n (l . () � V , \ c ross

N2 j f3 ,

w o u l d b e s e e n by t h e p r i m a ry . So, we see t h a t t h e c r i t e r i on fo r the selection of base v ol t a g e i nvolves the square of t h e ratio of r ne-to- line voltages a n d not t he squ a re of t h e turns ratio of t h e i n d ividu a l windings of the Y -6 transformer. This d iscussio n l eads to the conclusion that to transfe r the ohmic v a l u e of i m pedance from the vol t age l evel on o n e side of a three-phase transfor m e r to the voltage l evel o n the oth er, the m ul t i p l ying factor is t he square of the ratio o f l i ne-to-line voltages regard l ess o f w h e ther t h e t ransformer con nection i s Y-Y o r ,

NI

:

as s h o w n i n Ta b l e 2 . 1 ,

so

that the

same

6 0 - n r e s i s t ance p e r p h a s e

TABLE 2.1 Transferring ohmic val ues of per-phase impedances

from one side of a three-phase tra nsformer to another t

I

y-y

I

y-tl

- -+­ : vl n l _ t_

'-_ _ _ _ _ _

tl -Y

t Secondary

load consists of b a l a nced Y -con n e c l e d i m peda nces Z" .

2.5

THREE-PHASE TRANSFO R M E R S

63

This is shown in Table 2. 1 , w hich summarizes the relations for the effective turns ratio of the different types of transformer connections. Therefore, in per-u nit calculations involving transformers in three-phase circuits we require the base voltages on the two sides of the transformer to have the same ratio as the rated line-to-line voltages on the two sides of the transformer. The kilovoltampere base is the same on each side . y-� .

Y-�

Three transformers, e ach rated 25 MVA, 38. 1 /3.81 kV, are connected with a bala nced load of three 0.6-.0, Y-connected resistors. Choose a base of MVA, kV for t h e hig h-voltage side of the transformer and specify the base for the low-voltage side. Determine the per-unit resistance of the load on the base for the low-vo ltage s i d e . Then, d etermine the load resistance RL i n ohms refe rred to the high-voltage side and the per-unit va lue of this resistance on t h e chosen base.

Example

2.7.

75

66

� ,

13 X 38. 1 kV e q uals 66 k V the ra t i n g of the transformer as a bank is 75 MV A, 66Y /3_8 1 kY. So, base for t he low-voltage side is 75 MY A , 3 8 1 k Y . B y Eq . (J .54) b a s e i m p e d a n ce on t h e l ow vo l t a g e s i d e i s

.

Solution. S i n c e

t h ree-phase

( base

kVL L ) ='

base M VA J a n d on

the

Jow-\'oltage side

RL B ase

i m p e d a nce on t he

=

( 3 .81 ) 2 --75

0 . 1 93 5 .0

.6 -. 0 . 1 935 0

=

h i g h -voltage

= 3 1 0 per u n i t

side

-75 = ( 66)

is

2

58 . 1 .0

-

Th e

=

res i s t a n c e referred to t h e h i gh - v o l t age side is 0.

R j-

6

=

(

66 -

3 .8 1

-1 8 ()

58 . )

)2

=

=

1 80 .0

3 . 1 0 per u n i t

The res istance R and leakage reactance X of a t hree-ph ase transformer are me asured by the short-circu i t test as discussed for single-phase transform­ ers. In a three-ph ase equivalen t circuit R and X are connected in each line to an ideal three-phase transformer. Since R and X will have the same per-u nit value whether on t he low-voltage or the high-voltage side of t he ,transformer,

64

CHAPTER 2

-

TRANSFOR.\il E R S

the per-ph ase equivalent circuit will account for the t ra n s fo r m e r by the p e r uni t impedance R + jX without the ideal tra nsformer, i f phase-shift is not i m p o r tan t in t he calculations and all qua nti ties in the circ u i t are i n per u n i t w i t h the p roper selection of base. Tab l e A . 1 in the Appendix l ists typical values of transformer impedances, which are essentially equ al to the lea kage reacta nce si nce the resista nce is usually less than 0.0 1 per unit.

Exa m p l e 2 . 8 . A t h re e - p h ase t ra n s fo r m CJ' is r a t e d 4 0 0 M YA , 2 2 0 Y k Y . The Y - e q u iva l e l l t s h o rt -circ u i t i m pe d a nce measu red o n the l ow-\,ol t
/22 6.

Solution. On i t s o w n b a s e t h e t r a n s former re a c t a n c e is

0.121

( 22 )2/400

---:;:--

= 0 . 1 0 per

A, 230

unit

On t h e chosen base t he reactance becomes

( 220 ) 2 1 00 0.1 = 0.0228 230

-

2.6

400

-

THREE-PHASE TRANSFORMERS: SHIFf AND EQUIVALENT CIRCUITS

per u nit

PHASE

As m e n t i o n e d in Sec. 2 . 5 , ; 1 p h ; l s e s h i ft occu rs in Y - 6. t r a n s ro fm e r s . W e n ow examine p h ase sh i ft i n m o r e d e t a i l , u n d t h e i m p o r t a n c e o f p h a s e s e q u e n c e becomes a p p a r e n t . L a t e r i n s t u d y i n g fa u l ts w e have to d e a l w i t h b o t h pos i t i v e ­ or ABC-s equence q uantities a n d ne g a l i ve - or A CE- s e CJ u e nc e q u a n t i t i e s . S o , we need to examine phase s h i ft for both p osi tive and n e ga tive se q ue n c e s . Posi tive­ sequence voltages and currents are identified by the sup e r s c ri p t 1 a n d negative­ sequence voltages and currents by the superscript 2. To avoid too many subscripts, we so m e t i m es w r i t e Vj l ) i nstead of vj � for t h e vol t a g e d ro p fro m terminal A to N a n d similarly iden t i fy other voltages a n d c u rr e n ts to neutra l . I n a positive-sequence set of l ine-to-neutral vol tages VA l ) lags VY ) b y 1 200 , whereas V� l) lags V,J I ) by 2400 ; i n a negative-sequence set of line-to-neutral voltages V(B 2 ) leads VA(2) by 1 200 whereas VC(2) leads VA(2) by 2400 ' La t e r on when we d iscuss u n bala nc ed currents and voltages ( in Chaps. 1 1 and 1 2), we must b e c areful to d istinguish between voltages to neutra l and voltages to ground since they can differ u n d er u nbala nced con d i tions. Figure 2. 14(a) is the schematic wiring diagram of a Y - 6. t r a ns fo F m e r, where the Y side is t h e h i gh-voltage side. We reca ll that c a p i t a l l e t t e rs a p p J y to

,

2.6

T H R EE- P H A SE T R A N SFO R M E R S : P H A S E S H I FT A N D EQU I VA L E N T C I RC U ITS

HI A ---�o----,

�------�-- a

----

----

1...

Hz • B -----o---..J --->-

18

-c

-------.-

Ibc

H3 c -----o----"

lb

C

b

( a) Wiring d i agram

al l

(l)

V/l B )

Ie

-

-

B( l )

A( l

X3 Xz

----

Ie

65

VBe

) V(I A

I)

VlJ( lCl

)

C( I

(b)

VUI A

2) VA( ll

bt l

)

l) Vc(o

)

Positive sequ ence components

C (2 2) V( (. /\

'

l)

VCi I

VC{ A

Va( �I

)

V( (.'

�l

Vl(J2 )

C

O)

)

VlJ( 2e)

B(2

)

a ( 'L 1

C c) N e g ative s e q u ence components F1 G u RE 2 . 1 4 W i r i n g d i ag r a m a n d vol t ag e p h a s ors f o r a t h r e e - p h ase i s t h e h i g h - vo l t a g e s i d e .

t r a n s fo r m e r co n n e c t ed

Y- 6..

w h e re t h e

Y side

66

CHAPTER 2

TRANSFORM ERS

the high-voltage side and that windings drawn in p a raUei are linked by the same flux. In Fig. 2.14(a) wind ing AN i s the ph ase o n the Y -connected side, which is linke d magnetically with the phase w i n d i ng a6 on the 6.-connected side. The location of dots on the windings shows that VA N i s always in phase with Vab regardless o f phase sequence. If H I i s the terminal to which line A i s con nected, it is customary to con nect p hases B a n d C to terminals H2 and H3 , respectively. The Ame rican standard for d esigna ting termi nals H I and X l o n y - � transformers requ i res t h a t thc pos i t i vc-sequence vol tage drop from H I to neutral lead the posit ive-sequence voltage d rop from Xl to neu tral by 3 0° regardless of whether the Y or the � w i n d i n g is on the high-vol tage side. Similarly, the voltage from H 2 to n e u t ra l l e a d s t h e vol t age from X 2 t o n e u t r a l b y 30° , a n d t h e v o l t a g e from /1 .1 t o n c u t r ; i l I C ; l d s t h c vo l t a g c fro m X � t o n e u t ra l by 30° . The p h a s o r d i a g ra m s for t h e [)os i t i v c - a n d n e g a t ive-scq u c nc e c o m p o ­ n en ts of voltage are shown i n Figs. 2. 1 4( 6 ) and 2. 1 4( c ), respecti\ c l y . Figure 2 . 1 4( 6 ) shows the r e lation of the voltage p h asors whe n positive­ sequence volt ages are applied to term i n a l s A , B , a n d C. The vol t ages Vj l ) ( t h a t is, Vl�) a n d Va(� ) are i n p hase because of the dots, a n d a s s o o n .as we have d rawn Vl ) in p hase with Va(; ), the other voltages for the phasor diagrams can be d etermined. For i nstance, on the high-vol tage side VA l ) l ags Vl ) b y 1 20° . These two vol tages and vt ) meet at the tips of their a r rows. L ine-to-line voltages can then be d rawn. For the low-voltage d i a gram V�� ) a n d Vc�l ) c a n b e d rawn i n p h ase w i t h V� I) and vt), respectively, and t h e n t h e line-to-neutral vol tages fol low. We see that Vl 1 ) leads Va(l ) by 30° and term inal a must be marked Xl to satisfy the American standard. Term i nals 6 and c a re ma rked X 2 and X 3 ' respectively. F igure 2 .14(c) shows the relation of the voltage phasors when negative­ sequence voltages are applied to terminals A , B, and C. We note from the dots on the w i ring d i agram that V,P) ( n ot n ecessarily in p h ase with VJ I » i s in phase with v}l). After drawing Vj 2 ) i n phase with �IS;), we complete the d i a grams simil a rl y to the posit ive-se q u ence d i agrams but keep i n g in mind t h at VJ 2) l eads Vl2) by 120° . The completed d iagrams of F i g . 2. 1 4(c) show that VY) lags VY) by 30° . If N I and N2 represent the number of t urns i n the high-voltage and l ow-vo ltage windings, respectively, of a ny phase, then F i g. 2.l4(a) shows that �( l ) = ( N l /N2)Va(I� ) and v,f) = ( Nl / N2 ) Va(l) b y transformer action. I t then follows from the geometry of Figs. 2. 1 4( b ) and 2. 14( c ) that

VAO )

=

N

/ 3 00 _I v3 Vae l )� N2

VA(2)

=

N

_1· 13 Va(2)/ - 30° N2

( 2 . 34 )

Likewise, currents i n the Y-6. transformer a re d isplaced by 30° in the d i rection of the voltages since the p hase angles of the currents with respect to their associated voltages are determined by the load i mpedance. The ratio o r the

2.6

TH R EE·PHASE TRANSFO R MERS: PHASE S H I FT AND E Q U I VALENT CI RCU ITS

67

rated line-to-line voltage of the Y w in d in g to the rated line-to-line voltage of the D. wi n d i n g equals fS N I /N2 , so that in choosing t h e line-to-line voltage bases on the two sides of the transformer in t h e same ratio, we obtain in per u n i t

JA(2}

=

] (2) a

X

1/_

( 2 .35 ) -

30°

Tran sformer imped a nce and magnetizing cur ren ts are hand led separately from t h e p hase shi ft , which can be represe n ted by an :deal tra nsformer. This explains why, a c c ord i n g to Eq. (2.35), the per-uni( magni tudes of voltage and cu rrent a re exactly the same on both sides of t h e t ransformer (for i n stance I V(l( l ) I = I VY ) I ). Usual ly, the h igh-voltage w i n d i n g i n a Y -D. t ransformer is Y -connected. I n s u l a t i o n cos t s fo r , I g i v e n s t e p u p i n v o l ta g e a re thereby red u ced si nce t h is co n n e c t i o n t a k e s a d v a n t a g e o f t h e fact t h a t t h e vo l t a g e t r a n s form a t ion from the l o w - v o l t a g e s i d e t o t h e h i g h -v o l t i\ g e s i u e o f t h e t ra n s fo r m e r i s t h e n fS ( N I / N2 ), ,

w h e r e /VI ([ n ei N2 ,I re t h e S , l m e ,IS i n Eq.

w i n u i n gs il re 8. - co n n e c t e d , t h e t ransformat ion ratio of l i n e v o l t ages i s reduced r a t h e r t h a n i n crease d . F i gu r e 2 . 1 5 is the schematic diagram for t h e 8. - Y transform e r w here t h e 6 side i s the high-voltage side. The reader should ve rify that the vol t age p hasors are exactly the same as in Fig s 2 . ] 4( 0 ) and 2. 1 4(c), an d EC]s. (2.34) and (2.35) a re therefore s t i l l v a l i d . These e q u a t i o n s s t i l l hold if W e reverse t h e d i rect ions of all c u r re n t s on the wiring diagram. Under n ormal oper a t i n g con d i t ions only positive-sequence quant I t I es are i nvolved an d then the general rule Jor a ny Y - 8. or 8.- Y transforme r is that vol t ag e is a d v a n c ed 3 0 ° when it i s s t e p p ed u p . A s already d iscussed, we can i n d icate t h is phase shift i n vol tage b y an ideal t ransformer o f c o m p l e x turns ratio 1 : EJ·; /6 . S i n c e VY ) / J� I ) = V} Ij / /,; 1 ) i n Eq. (2.35), p e r u n i t impedance If

the

( 2 . 3 4).

h i g h -vol t a g e

.

-

A

Xl

H\

--------�o�--�

•

X3

FI G u RE 2 . 1 5

W i r i n g d i a g r a m for a t h re e - p h a s e t ra n s fo r m e r side.

Xz

H:;

----j:r=-------------'

--

COIl l l c c t e u 6 · Y w h e re t h e

0.

a

------

C

]a

Ie

B C

------

Ib

------

b

s i d e is t he h i g h -vo l t a g e

68

CHAPTER 2

TRANSFO R M E RS

values are t h e same when moved from o n e side of t h e ideal transform er to the other. R e a l a n d reactive power flow i s also not a ffected b y the phase s hift because the current p hase shift compensa tes exactly for t h e voltage phase s hift as far as power values are concerned. This is easily seen by writing the per-un i t complex power for each side o f t h e Y - � ( o r � -Y) transformer from Eqs. (2.35) a s fol lows: ( 2 .3 6) Hence, i f only P and Q quantities are required, i t is not necessary t o include ideal transform ers for the phase s h i rt o [ Y -� and �-Y tra nsformers i n t h e i m pedance d i a g r a m . T h e o n l y case i n w h i c h t h e i d e a l t r a n s fo r m e r c a n n o t b e ignored i s i n a ny closed-loop portion o f a system i n which t h e prod uct o f a l l t h e act u a l transformer voltage ratios is n o t u n i ty around t h e loop. We encounter one such case i n Sec. 2.9 when parallel con nections of regulating transformers are considered. In most other situations we c a n eliminate the ideal transformers from the per-un i t impedance d iagram, a n d then the calculated currents and vol t ages are proportional to the a c tu a l currents and voltages. P hase a ngles of t h e a ctual currents and voltages can b e foun d i f n eeded by noting from the one-l i n e d iagram the positions of the Y -� a n d �-Y transformers a n d b y applying t h e rules o f E q . (2.35); namely, When stepping u p from the l ow-vol tage to the h igh-voltage side of a �-Y o r Y -� transformer, advance positive-sequence voltages and currents by 30° and retard negative-sequence voltages and currents by 30° . It i s important to note from Eq . (2.36) that ( 2 . 3 7) w h i c h shows t hat t h e current ratio of a ny transformer with phase shift is the reciprocal of the cample.:r conjugate of t h e vol tage ratio. General ly, only vol tage ratios are shown in circuit diagrams, b u t i t is always u n derstood that the curren t ratio is t h e reciprocal o f t h e complex conj ugate o f t h e voltage ratio. In Fig. 2 . 1 6( a ) the single-line d i a gram indicates Y -6 transformers to step up voltage from a generator to a high-voltage transmission l i ne and t o step down the voltage to a l ower l evel for d i stribution. In the equivalent circu i t of Fig. 2. 1 6( b ) transformer resistance and leakage reactance are in per unit a n d exci t i n g current is neglected. B locks w it h i d e a l transformers indicating phase shift are shown along with the equ iva lent circu it for the transmission line, which

i s d eveloped in Chap. 6. Figure 2 . 1 6 ( c ) is a furt her simplification where the resistances, shunt capacitors, and ideal transformers are neglected. Here we rely u p o n the single-line d iagram to rem i n d u s to account for phase shift due to t h e Y - 6 transformers. W e must remember t h a t positive-sequence voltages a n d

2.6

TH R EE-PH ASE T R A N S FO R M E R S : PHASE S H I Ff A N D EQU IVALENT CIRCU ITS

�

�t

O�E I

�

3E 3E

Transm ission l i n e

----t----+-

t. - y

Y - t.

(a)

-'

>----I-

69

to load

(b)

XL

Xb

(c )

HGl: RE 2 - 1 6 ( a ) S i n gl e - l i n e d i a gr a m ; ( b ) p e r· p h a s e e q u iv a k n t c i rc u i t s w i t h p a r a m e t e rs i n p e r u n i t ; ( c ) p e r -p h ase e q u i v a l e n t c i rc u i t w i t h r e s i s t a n c e , c a p a c i t a n c e , a n d i d e a l t ra n s formers n e g l e c t e d . T h e p e r - p h ase e q u i v a l e n t c i rc u i t for t h e t ra n s m i s s i o n l i n e i s d e v e l o p e d in C h a p . 6.

cu r r e n t s in the

higher-yol tClge tran s m ission l i ne lead the corresponding quanti­

t. i c :, i n t h e l ow e r-vo l t a ge ge n e ra t o r a n d d i s t r i h u t i o n c i rc u i t s h y Exa m p l e 2 . 9 : F i g u r e: 2 . 1 7 s h ows

gC I 1e:r'ltor

3 0° .

]()O M Y 1\, n k Y s u p p l y i ng a sys t e m l o at.! o f 2 4 () M Y 1\ , D.t) powe r - fa c t or l a g g i n g a t 230 k Y t h ro u g h a 330 - M V A 2 3 6 j230Y - k Y s t e: p - u p t r a n s for m e r o f l e a ka ge r e a c ta n c e 1 1 % . N e g l e c t i n g m a g n e t i z i n g c u r re n t a n d c h oos i n g b a s e v a l u e s a t t h e l oa d o f 1 00 M YA a n d 230 k Y , fl n d 1/1 , In , a n d Ie s u p p l i e d to t h e load i n p e r u n i t w i t h VA as

I,

a

t h r e: e: - p h ase:

t e r m i n a l vol t a g e .

m t c tl

r e fe r e n c e . S p e c i fy i n g t h e p r o p e r base fo r t h e g e n e rator c i rc u i t , d e t e r m i n e 10 ' Ih '

a n t.!

fro m t h e g e n e r a t o r a n d i t s

Solution . T h e c u r re n t s u p p l i e d to t h e l o a d i s 2 4 0 ,000

.f3

X

230

602 .45 A

70

CHAPTER 2

TRANSFORMERS

To

Or------t---� �>-------f--�LOad ( a)

t::. - y

1 : Bjrr /6

10

--

+, J

fA

--

r L

v,.

V"

Load

(b) FIGURE 2. 1 7 (a) Single-line d i a g ram;

The

base

e q u ivalent c i rc u i t for Example 2.9, a l l param eters in pe r unit.

(b) per-phase

current at the load is

1 00,000 = 251 .02 A y 3 x 230 h

The power-factor a ngle of t he load current is

0 = cos -- I 0.9 = 25 .84°

Hence, with VA load are

= 1 .0� as reference

1A = 18

Ie

=

602.45

i n Fig.

lag

2.17(6),

/ - 25 . 84° = 2.40/ - 25 .84°

25 1 .02 -

'----

2 .40/ - 25 .84° - 1 20°

= 2.40/ - 25 .84°

+

1 20°

= =

the line currents i nto the

per u n i t

2.40/ - 145 .84° 2.40/ 94 . 160

per u n i t

per unit

Low-vol tage side currents further lag by 30° , and so i n p e r u n i t lb

=

2.40/

1 75 . 84°

2.7

THE AUTOTRANSFORMER

71

The transformer reactance modified for t h e chosen base is

0 .11 and so from Fig.

2. 1 7( b )

x

1 00

1

-

330

=

-

30

per unit

the terminal voltage of the generator is

/ - 30°

=

1 .0

=

0 .9322

-

+

�

x

30

j0.45 5 1

=

2.40/ - 55 .84°

1 .0374L

- 26 .02"

per u n i t

hase g en e r a tor voltage is 2 3 k Y , w h ich m e a n s t h a t t he terminal voltage of t h e g e n erator is 23 X 1 . 0374 = 23.86 k Y . T h e re a l powe r s u p p l i e d by t h e generator is

The

R c { VJa* }

�c

1 . 0 3 74 X 2 . 4 c o s ( - 26 .02° + 5 5 . 1-\4° )

=

2 . l GO

per

unit

which corresponds t o 2 1 6 M W absorbed by t h e l oad since there are no ] 2 R losses. The i n t e re s t e d re a d e r wi l l find the same v a l u e for I VI i by o m i t t i n g the phase s h i ft of the transformer altogether or by recalculating VI w i t h the reactance j /30 per u ni t o n the high-voitage s i d e of F i g . 2. 1 7(b). 2.7

THE AUTOTRANSFORMER

a u totransformer diffe rs from the ordinary transformer in t ha t t h e windin gs of t h e a u totransforme r are e l ectrical l y connected as well as coupled by a m u t u al flux . We exa mine the a utotransformer by electrically connecting t h e windin gs of an i d eal transformer. Figure 2. 1 8( a ) is a schematic d i agram of a n ideal t rans­ fOlmer, and Fig. 2. 1 8(b) shows how the windings are connected e lectrical ly t o form a n a u t otrcDsformer. Here t h e w i n d i ngs are shown so that their vol tages a r e a d d i t ive a l t hough t h ey coul d have been connected to oppose each other. The great di sadvantage of t h e a u totransformer is t h at electrical isolation is lost, but the fol l o w i n g cxample dcmonstrates the increase i n pow er r ating obtained. pJl

\.

f.

T3

/:

-

-l

VI

.

N2

NI

Inl

r 1-

V2

12 !1

+1

VI

-l

l,n

II

j

N2

+

V2 !Ii ) I hl

FIG U RE 2 . 1 8 Sch e m a t ic d i a g r a m of a n i d e a l t ra ns-

( b ) a s a n autptransformer.

fo rmer co n n e c t e d : (a) i n the u s u a l manner;

72

CHAPTER

2

TRANSFORMERS

Example 2 .10. A 90-MVA s ingle-phase transformer rated SO/ 1 20 kV is connected as a n a u totransformer, as shown in Fig. 2 . 1 S( b ). Rated vol tage I V} I = SO k V is applied to the low-voltage winding of the transformer. Co nsider the transformer to be ideal and the load to be such that currents of rated magnitudes 1 /1 1 and 1 /2 1 flow in the windings. D e term i ne I V2 1 and the kilovoltampere rat i n g of the a u totransformer. Solution

111 1

90 ,000 =

1 /2 1 =

I V2 1

=

SO 90 ,000 1 20 80 +

=

1 1 25 A

=

750 A

1 20 = 20 0 k V

The d i rections chosen for I } and 12 in relation to the dotted terminals s how that these currents are in phase. So, the input current is I lin I = 1 1 25

+

750 = 1 S75 A

Input kilovoltamperes are

Outp u t kilovoltamperes are

The increase in the kilovoltampere rating from 90,000. to 1 50 000 kVA a n d in the o u tput voltage from 1 20 t o 200 k V demonstrates t h e adva ntage o f t h e a u t o t r a n s fo r m e r. T h e a u t o t l " a l 1 S rO fl1l e r prov i d es a higher r a t i ng for t h e s a m e cost, ,

i.Jll d

i t s e t I i c i e n cy i s g r e a t er

since

c on n ec t ioll of the same t ra l l s fo r m e r.

the

l osses a r e t h e same a s i l l

the

o rd i n a ry

Single-phase autotransformers can be co n n e c t e d for Y - Y three-ph a se operation o r a t hree-phase u n i t can be built. Three-phase autotransformers are often used to con nect two transmission l ines opera ting a t differen t voltage levels. If t h e t r a nsformer o f E x a m ple 2 . 1 0 were con n ected as o n e p h ase o f a three-ph ase y -Y autotransformer, the rating of the three-ph ase u n i t would be 450 MY A, 1 38 /345 k Y (or more exactly 1 3 8 .56/346.4 1 kY).

2.8 PER· UNIT IMPEDANCES OF THREE-WINDING TRANSFORMERS

Both t he primary and the secondary w i n d i ngs of a two-winding transfo rmer,have the same k ilovoltampere rating, but a l l three w i n d in gs of a t hree-w i n d i n g transformer may h ave d iffe r e n t k i lovu] t a m p c r e r a t i n g s . T h e i m p e d a n c e of e a c h

2.8

P

-{ -----h.

PER UN IT -

n---t---- s

,...L---+---- t

(a)

I M PEDAN CES O F THREE-W I N D I N G TRANSFO R M ER S

73

p --------,

s

t .------� (6)

FI G U RE 2 . 1 9 T h e e n ) sche m a tic d i a g r a m a n d ( b ) e q u iva l e n t c i rc u i t o f a t h r ee - w i n d i n g t r a n s former. Poi n t s p , S , a n d I l i n k t h e c i rc u i t o f t h e t r a n s fo r m e r t o t h e a p p r o p r i a t e e q u iv al e n t c i r c u i t s repre s e n t i ng p a r t s o f t h e sys t e m con n e c t e d t o t h e p r i m a ry, s e co n d a ry , :l nll t e r t i a ry w i n d i n g s .

w in d i ng of a th ree-winding t ran sformer may b e gIven I n perce n t or per u n i t b 3 s c d on t h e )",l t i n g o f i t s own w i n d i n g, o r tests m ay b e m a d e t o d eterm i ne t h e imped ances. I n a n y case, a l l t h e p e r- u n i t impedances m u s t b e expressed o n the s a m :..: k ilovo l t arn p e re base_ A single-phase thre e-win d i ng transformer is shown schematically in Fig. 2. 1 9( a ), where we designate the t hree w i n d ings as primary , secondary , a n d tertia ry . Three impedances m ay b e m e asured b y the stan dard short-circui t test, as follows:

Zps

-

Zp l

Z51

leakage impedance measured 1D primary with secondary short-circuited and terti ary open leakage impedance m easured i n pnmary w i t h tertiary short-circ u i t e d a n d secon d a ry open lea kage i mpedance measured 1 D secondary with tertiary short-circu ited a n d primary open

of t h e w i n d i n g s , t h e i m p e d a n c e s o f e �l c h s e p a r a t e w i n d i n g refe rred t o that s a m e w i n d i ng arc related to t h e m easured impe da nces so referred a s fol lows: I f t h e t h rc e i rn r c d a n c e s m c a s l I rc d i n o h m s ,Ire rlf e rr e d to t h e vo l t a g e of one

( 2 .38)

and Z are t h e i m p ed ances of the prim a ry, second a ry, and tertiary w i n d i ngs, respectively, referred to t h e primary circu it i f Zp s ' Zp I ' a n d Z SI a r e t h e m e asured impedances referred t o t h e prim ary c i rcuit. Solving ' Eqs. (2.38)

Here

ZP ' Zs '

I

74

CHAPTER 2

TRANSFORMERS

simultane ously yields

( 2 . 3 9)

The impedances of the t h re e w i n d i ngs a re connected to represent t h e equiva l e n t circuit o f t h e single-phase three-wi n d i n g transformer w i t h magnetiz­ ing c u rr e n t neglected, as s h o w n i n Pig. 2. l 9 ( h ) . T h e common rJoint is fictitioL l s a n d u n r e l a t e d to t h e ne ut r a l o r t h e s y s t e m . The p o i n t s {J , s , a n d f a re c o n n c c t c tl to the p a rts of the i mpedance diagrams repres e n t i n g the pa rts of t h e system connected to the primary, seconda ry , and tertiary windings, res p e c t i v e l y , of the t ransform er. As in two-winding transformers, conversion to per-unit im pedance requires t h e same kilovo!tam pere base for a l l t h re e c i rc uits and req u i res vol tage b ases i n t h e t h ree c irc uits that a re in the same ratio as the rated l i ne-to-line vol tages of t h e three c ircuits of t h e transformer. When t hree such t ransformers a re connected for three-phase opera tion , t h e primary and secondary windings are usually Y -connected and the tertiary w in dings are connected in t::. to provid e a p a t h for t h e t h i rd harmonic of the exciting curren t. Example

2.11.

The three-phase ratings of a three-winding

15

Primary

Y-connected, 66

Secondary

Y-connected,

13.2 kV, 1 0 MVA

Tertiary

6-connected,

2.3 kV, 5

kV,

transformer are:

MVA

MY A

Neglecting resistance, t he leakage i mp eda nces a re ZPS

=

Zp I

=

ZS I

=

7% on 15 MVA , 66 kV base 9% on 1 5 MVA, 6 6 kV base 8% on 1 0 MV A, 13.2 kV base

Find t h e per-u nit impedances of t h e per-phase equivalent circuit for a base of MY A, 6 6 kV i n the primary circ u i t .

15

With a base of 15 MVA, 66 k V i n t h e primary circuit, t h e proper bases for the per-unit impedances of the equivalent circuit are 15 MVA, 66 kV for prima ry-circuit quan tities, 1 5 M Y A, 13.2 kV for secondary-circui t quantities, and 15 MYA, 2.3 kY for tertiary-circuit quantities.

SoLution.

2.8

PER-UNIT I M P EDANCES OF TH R E E-WINDING TRANSFO R M E R S

75

Since Z ps and Zpt are m easured in the primary circui t, they are already expressed on the proper base for the equivalent circuit. No change of voltage base is required for Zst . The required change in base megavol tamperes for Z S ( is made as follows : 15 Z SI

=

8% X

10 =

12%

In per unit on the specified base

2. 1 2.

Zp

=

Zs

=

}( j0.07 + jO.09 - j0.12) = jO.02 per u n i t }- ( j0.07 + j0 . 1 2 - jO.09) jO.OS per u n i t =

Z, }- ( j0.09 j0.12 - iO.07) jO.07 p e r u n i t =

=

+

co nst ant-vo ltage source ( i n fi n i t e bus) supplies a purely resistive 5 MW, 2.3 kY three-phase load and a 7.5 MY A , 1 3.2 kY syn chronous motor having ; \ s u b t ra n si c n t r e a c t a n ce o f X " = 20 % . T h e source is con nected to t h e primary of t h e l h r e e -w i n u i n g t r a n s fo r m e r desc r i b e d i n Exa m pl e 2 . 1 1 . The motor a n d resis t ive load are connected to t h e secondary a n d tertiary of the t ransformer. D raw t h e impedance diagram of t h e system a n d mark the per-u n i t impedances for a base o f 6 6 k Y , 1 5 M Y A in t h e p r i m a ry . Neglect exci ting curre n t a n d a l l resistance except that of the resistive load. Exa m p l e

A

Solution. The constant-vol tage source can be represented by a generator having no

internal impedance. The resistance of the load i s 1 .0 per unit on a base of 5 M YA, 2.3 tertiary . Expressed o n a I S M YA , 2.3 kY base, t h e load resistance i s R

=

1 .0 X

15 5

=

3.0 per unit

The reactance of the motor on a base of 1 5 MYA, X"

kY in t h e

15

7.5

= 0 . 20 - = 0 . 4 0

per

13.2 kY is

unit

Figure 2.20 is the required d iagram . We must remember, however, the phase s h i ft which occurs between t h e Y -connected primary and the 6-connccted tertiary. j0 05

+

E""

F I G UR E 2.20

I m p e d a n c e d i agram for E x a m p l e 2. 1 1 . ,

76

CHAPTER 2

TRANSFO R M ERS

2.9 TAP-CHANGING TRANSFORM ERS

AND REGULATING

Tra nsformers which provide a sma l l adj u stment of vol tage magnitude, usually i n the range o f ± 10%, a n d others which s h i ft the p h ase a ngle o f t h e l i n e voltages are important components of a pow e r system. Some transformers regulate both the magnitude and phase a ngle. Alm ost a l l transformers provide t a ps on windings to adjust the ra tio of transformation b y changing taps when the transformer is deenergized . A change in tap can be made while the transformer is e n e rgized, and such transformers a re c a lled load-tap-changing (LTC) transformers or tap-changing-under-load (TCU L) transformers. The tap changi n g is a utomatic and ope rated by motors w h ich respond to rel ays set to h o l d t h e vol t age a t the prescribed l evel. Spec i a l circuits a l low t h e change t o b e made without i n terrupting t h e current. A type of transformer designed for small adjustments of voltage rather than l a rge changes in vol tage levels is called a regulating trallsformer. Figure 2.21 s hows a regu lating t ransforme r for control of voltage magnitude, and F ig. 2.22 s hows a regulating t ransformer for p hase-angle con t rol . The ph asor d ia gr a m of Fig. 2.23 helps to expla i n the s h i ft in phase a ngle. Each of the three w i n d in gs to w h ich taps are made is on the same magnetic core as the phase w inding whose voltage is 90° out of p hase with the voltage from neutral to the point connected to the center of the t a pped wind ing. For instance, the voltage to neutral Va n is increased by a component � Va n ' which is in phase or 1 80° out of phase with Vbc ' Figure 2.23 shows h o w the t h ree l ine voltages are sh ifted i n p h a s e angle with very l i t t l e change i n m a gnitude.

b .-------�--�--+--�

Series transformers

FIGURE 2.21 Regula t i n g t ra nsformer for con trol of voltage m a g n i t u d e .

2.9

TAP-CHA N G I N G A N D REG ULATI N G T R A N S FO R M E R S

VCII + � Vcn

77

F I G L R E 2.22

R e g u l a t i n g t r a n s fo r m e r for co n t rol of phase a n g l e . W i n d i ngs d rawn D a r a l k l to e a c h o t h e r a re on t h e s a m e i ron c o r e .

The p roce d u r e to d e t e rm i n e t h e bus a d m i ttance m a t r Lx Y bus i n p e r u n i t for a network con t ain i ng a regu l a t i n g t r ansformer is the same as t h e p rocedure to accou n t for any transform e r whose turns ratio is other than t h e ratio used to select the ratio of base vol tages on the two sides of t h e tran sformer. We d e fe r co nsider a t ion o f t h e proced ure u nt i l Ch ap. 9 . W e can, howeve r, i nvestigate t h e u se fu lness of tap-changing and regu lating t ransformers b y a simple exampl e . I f we have two buses connected by a transforme r . and i f t h e ratio of t h e line-to-line voltages of t h e transforme r i s the s a m e a s t h e r a t i o o f t h e b ase vol t ages of the two buses, the per-phase equivalent circuit (wi t h the m ag n e t izing current n eglected) is simply t h e transformer impedance i n per unit o n the chosen base connected between the buses. Figure 2 . 24C a ) is a one- p n e d i a gram of two t r a n s formers in parallel. Let us assume t h a t one of them has the voltage r a t i o l in , w h i c h i s also t h e r a t i o of base vol t ages o n the two sides o f t he t r a n s form e r , a n d t h a t the vol t age ra tio of the other i s l in'. The equivalent c i rcu i t is t h e n t h ,l( of Fig. 2 .24( b ). W e n e e d the ideal ( n o imped ance) t ra ns­ for m e r w i t h t h e r(l t i o 1 / ( i n t h e p e r - u n i t r e ,l c ta n cc d i a g ra m t o take ca re of the o fT- n o mi n ; 1 1 t u rn s r;d i o o r t h e s e c o n d t r a n s fo r m er b e c a u s e b a s e vol t
Shd(c·(j

V" ,, ·'

- O r ig i n a l V""

F I G U R E 2.23

P h a s o r d i agram for t h e r e g u l a t i n g t r a n s fo r m e r shown i n

Fig. 2 . 2 2 .

78

CHAPTER 2

TRANSFORMERS

ljn

h

CD

®

l in '

(a)

----I---�--

FIG U RE 2 . 24 Tran sformers w i t h d i ffe r i n g t u rn s ratio connected i n paral­ lel: (a) the single-line diagram; (b) t h e per-ph ase reactance d i ­ agram in p e r u n i t . T h e t u rns ratio I II i s e q u a l to n ln ' .

(b)

determined by the turns ratio o f the fi rs t transformer. Figure 2.24( b ) m ay b e interpreted a s two transmission l ines i n p a ra l le l with a reg ulating transformer i n o n e l ine. Two transformers are con nected i n para l le l to supply a n i m p e d a n c e to n eutral per phase of 0 . 8 + j O . 6 per unit ·at a vol tage of V2 1 .0ft per u n i t . Transformer T" h as a vol tage r atio e q u a l to t h e ratio of t h e base vo l tages on t h e two sides o f t h e t ransformer. T h is transformer h a s an impedance of jO. l p e r u n i t on the a ppropriate base. T h e second transformer Tb also has a n impedance of j O . 1 per u nit o n the same base b u t h as a step-up toward the load of 1 .05 t imes t ha t of Ta (seco n d a ry w i n d i n gs on 1 . 0 5 tap). Figure 2 .25 shows the equ ivale n t circuit w ith transformer Tb r epresented by its impedance and the insertion of a vol t age tl v. Find the complex power transmitted to the load t h rough each transformer. Exa mple 2 . 13.

=

Solution.

Load current

is 1 .0

---- = + jO.6

0.8

0.8

-

jO.6

per unit

2.9

TAP-CH A N G ING AND R E G U LATING TRANSFO R M E R S

79

)0. 1

s

lcirc

tN

=

O.OSlQ:.

0.8

)0.6

F1 G C RE 2.25 An equivalent c i rcuit for Exa m p l e 2 . 1 3 .

An ap proxi m a t e sol u t i o n to t h i s prob l e m i s fou n J b y recogn izing t h a I rig. 2.25 with sw i t ch 5 c l o s c u is a n e CJ l I i va l e n t c i rc u i l ror t h e p ro b l e m i f t h e vol t age � V, which is i n t he branch of the c i r c u i t equiva l e n t t o transformer Tb , i s equal to I 1 in per u n it. I n other words, if T" is p roviding a voltage r a t i o S o/c higher than Tb, I equals 1 .05 and �V equals 0 . 05 per u n i t . If we consider that the current set up by �v circulates arou nd the loop i n dicated b y Ie i r c w ith switch S open, and that with S closed only a very small fraction of that current goes through the load impedance (because it is much larger than the transformer impedance), then we can apply the supcrposition pri n c i p l e to �V and the sou rce voltage. From � V alone we ob tain -

[e i re 0.4

0.05 jO.2

=

. - -jO .2S

p er unit

and with �V short-circuited, the currcnt in each path is half the load current, o r - /0.3. Then, superimposing the ci rcul ating current gi\cs ITo

IT

/.

so that

= =

0 .4 - jO.3

.-

=

0 4 - j O .OS per unit .

0 . 4 - j O . 3 + ( -jO .25) = 0 .4 - jO.55 p e r u n i t ST

o

=

5Th =

Th is

( -) 0 .25)

0 .40

+

jO .OS

per unit

0 .40 + jO.55 per u n i t

example s hows t h a t t h e t r ansformer w i t h the higher t a p s e t t ing i s supplying most o f t h e react ive power to t h e load. The real power i s d ivid i n g e q u a l l y b e tween t h e transfo rm ers. Since both t ransformers h ave t h e same imped ance, they would sh a re both t h e real and re a c t i v e power equ ally if they h a d the same t urns ratio. In t hat case each would b e represented by the same per - u n i t react ance of jO. 1 betwe e n the t w o buses a n d wou l d c a r ry e q �l a l c u r r e n t .

80

CHAPTER 2

TRANSFORMERS

When two t r ansformers are in p arallel , we can v ary the d istribution of reactive power betw een the transformers by a dj usting the voltage-magnitude r atios. When two p aralleled t ransformers of equal kilovoltamperes do not share t he kilovoltamperes equally because their impedances differ, t h e kilovoltamperes may be more nearly equalized by a djustment of the voltage-magnitu d e r atios through tap c h a n g in g .

ft

Example 2 . 1 3 excep t t h a t Th i n c l u d e s h o t h a t r a n s fo r m e r h a ving the same turns ratio as Ta a n d a regu lating transformer w i t h a phase shift of 3° (t £:)'rr /60 = l .O ). The i m pe dance o f the two components of Tb is iO. l p e r u n i t on the base of Ta '

Exa mple 2 .14. Repe a t

=

Exa m p l e 2 . 1 3, we c a n o bt a i n an a p pr ox i m a t e so l u t i o n o f t h e problem by inserting a vol tage SOurce � V i n s e r i e s w i t h t h e i m p e d a nc e o f tra nsformer Tb . The proper per-uni t voltage i s

Solution. As in

[eire =

IT

=

IT"

=

a

So ,

0.0524

/91 .5°

-�-!./::::===- = 0 .262 + iO.0069 p e r unit 0.2 90°

0 .4 - iO.3 - (0.262

+

iO .007)

=

0 . 1 38 - iO .307 per unit

0.4 - iO.3

+

iO .007)

=

0 . 662 - iO.293 per unit

+

( 0.262

51 a 5 .,.

"

=

0 . 1 38

+

iO.307 p e r unit

= 0 . 6()2 + iO.293

per

unit

The example shows that the p hase-sh ifting transformer is usefu l to control the a m ou n t of real power flow b u t has less effect on the flow of reactive power. B oth E xa m p les 2.1 3 and 2.14 are i llustrative of two transmission l i nes in parallel with a regulating transforme r in one of the l i n es . 2.10 THE ADVANTAGES OF PER-UNIT COMPUTATIONS

W hen b ases are specified p roperly for the various p arts of a circuit connected by a t r a n sfor m er the per-unit values of impedances d etermined in their own p ar t of t h e system a r e t h e s ame w hen viewed from another part. Therefore, i t is n ec essary only to compute e ach impedance on t h e base of its own part of t h e circuit. The gre a t advantage o f using per-unit values i s t h a t no computati9ns are req u i re d t o refe r an impedance from one side of a transformer to the other. ,

2. 1 0

1.

T H E A DVANTAGES OF P E R - U N I T COM PUTATIONS

81

The fol lowing points should be kept in mind:

A base k ilovolts and base kilovoltamperes is selected in one part of the system. The base values for a t h ree-phase system are u n derstood to b e l ine-to-l i ne kilovolts and t h ree-phase k i lovoltamperes o r megavoltamperes. 2. For other parts of the system, t h a t is, on other sides of transformers, the b ase kilovolts for each part is d etermined according to the l i ne-to- l i ne volt age ratios of the transformers. The base kilovoltamperes wi l l be the same in all p arts of the sys t em. I t will be helpful to mark the base kilovolts of each p art of t h e system on the one-l i ne d i agram 3 . I mp ed ance i n formation ava ilable for t h ree-ph ase tra nsformers will usually be i n per u n i t or perce n t on the base determined b y their own ratings. 4. For t hree single-ph ase t ransform ers connected as a t h ree-phase u n i t the t hree-phase rat i ngs a re determ ined from the si ngle-phase rating of e ach i n d i v i d u a l t r a n s for m e r . I m p e d a n c e i n p e rce n t fo r t h e t h r e e p hase u n i t is the s a m e as t h a t fo r each i n divi d u a l t r a nsfo r m e r. 5 . P e r- u n j t imped ance given on a base o ther than t hat determined for the p art of the system in which the e l e m e n t is l ocated must be c h a nged to the p roper base b y Eq. ( 1 .56). .

-

�1aking compu ta tions for e lectric systems in terms of per- u n i t values simplifies the work great l y. A rea l appreci a t io n of the value of the pe r-un i t method comes throug h experience. Some of t h e advantages of the method are summarized b ri efly below: 1 . M a n u fa c ture rs

2.

usually speci fy the i m pedance of a p iece of apparatus in perce n t or per u n i t on t h e base of t h e nameplate rating. T h e per-unit impedances of machines of the same type and widely d iffe rent r a t i n g u s u a l l y l i e w i t h i n a n a rrow range a l t hough the ohmic values d iffer m a t e r i a l l y for mach ines of d i ffe re n t ratings. Fo r t h i s reason when the i m p e d a nce i s n o t k n ow n d e fi n i t e l y , i t i s g e ne r a l l y p o ssi b l e t o se lect from ; I ve r a g e

correc t . E x p e r i e n c e i n tab u l ateu

3.

w i l l be r e a so n a b ly per- u n i t v a l u e s b r i n gs fa m i l i a r i ty with t h e

p e r- u n i t

i m p e d a n ce w h i c h

working w i t h proper val ues o f per- u n i t i m p e dance for d i ffe rent t y p e s of appara tus. When impeda n ce in ohms i s specifi e d in an equivalent circuit, each imped a nc e must b e r e fe rred t o t h e same c i rc ll i t b y m u l t i p l y i n g it by the square of t he ra t i o of t h e r a t e d v o l t a g e s of t h e two s i d e s of t h e t ra nsfo r m e r connect i n g t h e reference c i rc u i t and t h e cir c u i t con t a i n i ng the i mpeda n ce . The per-unit im p edance once expressed on t h e p roper base, is t h e same referred to e i ther s i de of any tra nsformer. The way in which transformers are connected i n t h ree-ph ase circuits does not affect the per-un i t impedances of the equivale n t circ u i t a lthough the t r ansformer connection does d e termine the relation between the volt age bases on t h e two sides of the t ransformer. ,

4.

v;I i u cs a

82

CHAPTER

2.1 1

2

TRANSFORM ERS

S U M MARY

The intro duction in this chapter of t h e s implified equivalent c ircuit for t h e transfo rm e r i s o f great i mportance. Per-unit calculations a r e useu a lmost continuously throughout the chapters to fol l ow . We h av e seen h ow t h e trans­ former is e l i mi nated in the equivalen t circui t by the u s e of per-unit calculations.

It i s important to remember t h a t fJ does not enter t h e deta i l e d p er-unit computations because o f t h e specification of a base l i n e-to-line voltage a nd b a se line-to-neu tral vol lage related by fJ . The concept of proper selection of base i n the various parts of a c irc u i t linked by transformers and t h e calculation o f parameters i n per u n i t o n t h e base specified for the part or t h e circuit in w hich the parameters exist is fu ndamental in building a n equivalent c i rc u i t from a single-l ine diagram.

PROBLEMS single-phase transformer rated 7.2 kY A, 1 .2 kY / 1 20 Y h as a primary winding of 800 t urns. Determine (a) the turns ratio and t h e number of turns in the secondary winding and (b) the currents carried by the two windings when the transformer d elivers i ts rated kY A at rated voltages. Hence, verify E q. (2.7). 2.2. The transformer of Prob. 2. 1 is delivering 6 kYA at i ts rated voltages and 0.8 power-factor lagging. (a) Determine the impedance Z 2 connected across i ts sec­ ondary terminals. (b) What is the value of t h is impedance referred to the prim a ry side (i.e. Z ;)? (c) Using the value of Z; obtained in part (b), determine t he magnitude of the primary current and the kY A supplied by the source. 2.3. W i t h reference to Fig 2.2, consider that the flux density inside the center-leg of t he transformer core, as a function of time I , is B ( t ) = Bm sin(27T It ), where Bm is the peak value of the sinusoidal flux density and I is the operating frequency in2 H z . I f t h e fl ux density i s uniformly d istribu ted over t h e cross-sectional area A m of t he center-leg, determine ( a ) The instantaneous flux 4> ( 1 ) in terms of B"" f, /1 , and f . ( b ) Thc instantaneous induced-voltage e / d, according to Eq . (2. 1 ) . ( c ) Hencc, show that the rms magnitude of the induced voltagc of the primary i s given b y I E ) I = firrfN) BIII A . 100 cm 2 , f 60 Hz, Bm = 1 .5 T, and N) 1 000 turns, compute I E, I . ( d ) If A 2.4. For the pair of mutually coupled coils shown in Fig. 2.4, consider that L ' I 1.9 H, L ' 2 = L 2 , 0.9 H, L 2 2 = 0.5 H , and r , = r2 = a fl . The system is operated at 60 2.1. A

= =

=

=

=

H z. ( a ) Write the impedance form [Eq. ( 2.24)] of the system equations.

2.S.

( b ) Write the admittance form [Eq. (2.26)] of the system equ ations. ( c ) D etermine the primary voltage V, and the primary current I I whe n the . secondary is CD open-circuited and has the induced voltage V2 = 1 00 v. Oi) s hort-circu ited and carries the current 12 A. For the pair of mutually coupled coils shown in Fig. 2.4, develop an equival en t-T network i n the form of Fig. 2.5. Use the parameter values given in Prob. 2,4 and assu me that the tu rns ratio a equals 2. What are the values of the leakage reactan ces o f t h e windings and the magnetizing susceptance of the cou pled coils?

= 2�

LQ:.

P ROBLEMS

2.6.

2.7.

2.B.

83

A single-phase transformer rated 1 .2 kV / 1 20 V, 7.2 kVA has the following winding parameters: ' 1 = 0.8 fl, x I = 1 .2 n, r2 = 0.01 .0, and x2 = 0.01 n. Determine ( a ) The combined winding resistance and leakage reactance refe rred to the primary side, as shown in Fig. 2.8. ( b ) The val ues of the combined paramete rs referred to the secondary side. (c) The vol tage regulation of the transformer when it is delivering 7.5 kVA t o a load at 1 20 V and 0.8 power-factor l agging. A single-phase transformer is rated 440/220 V, 5.0 kVA. When the low-voltage side is short-circu ited and 35 V is applied to the high-voltage side, rated current flows in the windings and the power input is 1 00 W. Find the resistance and reactance of the high- and low-voltage windings if the power loss and ratio of reactance to resistance is the same in both windings. A single-phase transformer rated 1 .2 kV / 1 20 V, 7.2 kVA yields the following test results: O p e n - c i rcu i t t e s t ( p r i m a ry -o p e n ) V o l t a ge V=,

= 1 20 V ;

c ur re n t I .,

=

1 .2

rower W2 =

A;

S hort-circuit test (secondary-shorted) Vol tage

2.9.

VI

=

20 V ;

curren t

JI =

power

6.0 A ;

W,

=

4 () W

36 W

Determine 2 2 ( 0 ) The para meters R , = ' , + a r 2 , X l = X I + a x 2 , Ge , and Bm referred to the primary side, Fig. 2.7. ( b ) The values of the above parameters referred to the secondary side. ( c ) The efficiency of the transformer when it delivers 6 kVA at 1 20 V and 0.9 power factor. A single-phase transformer rated 1 .2 kV / 120 V, 7.2 kV A h as primary-referred p arame ters R I r l -+- 0 2 ' 2 1 .0 fl and X I X I + a 2 x 2 = 4.0 fl. At rated volt­ age its core loss may be assumed to be 40 W for a l l values of the load current. ( 0 ) De t e r m i ne the eA1cicncy and regu ;ation of the tra nsformer when it de livers 7.2 1 20 V a n d powe r factor of ( i ) 0.8 laggi ng and ( ii) 0.8 leading. kVA at V2 (b) for a given load vol tage and power factor it can be shown that the efficiency of ; \ t r a n s ro r m e r ;I l t a i n s i t s m ax i m u m v a l u e a t t h e k V A load level which makes t h e I :' R w i n c.J i ng l o sses e q u a l t o t l t e c o r e l o s s . U s i n g t h i s r e s u l t , d et c r m i n e the maximum e rti c i e n cy o f the above t ra nsformer a t rated voltage and 0.8 power factor, and the kV A load level at which it occurs. A s i n gle -p h a s e sys t e m s i m i l a r to that shown in Fig. 2 . 1 0 has two transformers A -B and B-C connected by a line B fec d i n g a load a t t he receiving end C. The ratings a n d p a ra m e t e r v a l u es o f the com ponents a r c : =

=

=

o c

2 . 1 0.

Transformer A -B : 500 V / 1 .5 kV, 9.6 kVA, leakage reactance

=

5%

Tra n sfo r m e r B- C :

=

4%

Line B:

Lo a d C :

1 .2 kV / 1 20 V, 7.2 k V A , leakage reactance series impeda nce

=

1 20 V , 6 kVA a t U . S

CO.5

j3.0) n

powe r-factor -1

lagging'

84

CHAPTER

2 TRANSFO R M E R S

(a)

Determine the value of the load imped ance in ohms and t he a c t u a l ohmic impedances of the two transformers referred to both their prim ary a n d secondary sides. (b) Choosing 1 .2 kV as the voltage base for circuit B and 1 0 kYA as the systemwide k VA base, express all system impedances in per unit. ( c ) What value of sending-end voltage corresponds to the given loading condi­ tions? 2 . 1 1 . A balanced �-connected resistive load of 8000 kW is connected to the low-vol tage, 6 -connccted s i d e of a Y -6 transformer rated 1 0,000 kYA, 138/ 1 3.8 kY. Find the load resistance in ohms in each p hase as measu red from line to neutral on the high-vol tage side o f the transformer. Neglect t ransformer impedance and assume rated vol tage is appl ied to the t ransformer primary . 2.12. Solve Prob. 2. 1 1 if t h e s a m c r e s i s t a n c e s a r e r ec o n n e c t e d i n Y . 2 . 1 3 . Th ree t r a n s fo rm e rs , e a c h r a t e d 5 kY A , 22() Y o n t h e s e co n d a ry s i d e , , Ir e con n e c t e d �-� and h ave b e e n su p p l y i n g a b a l a n c e d 1 5-kW p ur e l y resist ive l o a d a t 2 2 0 V . A change is made which reduces t h e load to 1 0 k W , s til l purely resistive a n d balanceu. Someone suggests that w i t h two-thirds o f t h e load , one t r a n s fo r m e r can be removed and the system can bc opera ted open � . B a l an c e d t h r ee - p h a s e voltages will still be supplied to the load since two of the I ine vol t a g e s (and thus also the third) will be u nchanged. To investigate the suggestion fur ther, ( a ) Find each of the line curre nts (magnitude and angle) with the lO-kW load and the transformer between a and c removed . (Assume Va'" = 220 V, se­ quence abc.) ( b ) Find the kilovoltamperes supplied by each of the remaining transformers. ( c ) What restriction must be placed on the load for open-6 operation with these transformers? ( d ) Thin k about why the individual t ransformer kilovoltamperc values i nclude a Q component when the load is purely resistive. 2. 1 4. A transformer rated 200 M YA, 345Y /20.56 kY connects a bala nced load rated 180 MYA, 22.5 kY, 0.8 powe r- fa c t o r lag to a t ra n s m i ssion l i n e . D e t e r m i n e ( a ) The rating of each of three single-phase transformers which when properly connected will be equivalent to t h e above t h ree-phase t ra nsfo r m e r . ( b) T h e complex impedance of the load in 'per unit i n the impedance d iagram if the base in the transmission l ine is 100 M YA, 345 kY. 2 . 1 S. A three-phase transforme r rated 5 MY A, 1 15 / 13.2 kY has per-phase series imp edance of (0.007 + jO.075) per u nit. The transformer is eonnected to a short d istribution line wh ich can be represented by a series impedance per p hase of (0.02 + jO. 1 0) per unit o n a b a s e o f 10 MV A , 1 3.2 kY . T h e l i n e supplies a balanced th ree-phase load rated 4 M VA, 1 3 . 2 kY, with lagging power factor O.�S . ( a ) D raw an equivalent circuit of the system indicating all impedances in per u n it. Choose 10 M YA, 1 3.2 kY A as the base at the load. ( b ) With the voltage at the p rimary side of the transformer held constant at 1 1 5 kY, the load at the receiving end of the line is d isconnected. Find the voltage regulation a t the load. 2 . 1 6. Three identical single-phase transformers, each rated 1.2 kY / 1 20 Y, 7.2 kY A a n d h aving a leakage reactance of 0.05 p e r unit, are connected together t o form a three-phase bank. A bala nced Y -connected load of 5 n per phase is con nected across the secondary of the b a n k . D e ter m i n e the Y-equivalent p e r- p h ase i m p e d a n ce

LQ:.

PROBLEMS

2. 17.

85

( in ohms and in per u n it ) seen from the primary side when the transformer b a n k is connected ( a ) Y-Y, ( b ) Y - � , ( c ) � -Y , and ( d ) �-�. Use Table 2. l . Figure 2. 1 7( a ) shows a t h ree-phase generator supplying a load t hrough a three­ p hase transformer rated 1 2 kVA/ 600 V Y, 600 k Y A The transformer has per-phase leakage reactance of 1 0 % . The line- to-line voltage and the line current at the generator term i nals are 1 1 .9 k Y and 2 0 A , respectively. The power factor seen b y the generator is 0. 8 l agging and the phase sequence of supply is ABC. ( a ) Determine the l ine curren t and the line-to-line volt age at t he load, and t h e per-phase C equiva lcnt-Y) i mpedance of t h e l oa d . ( b ) Using t h e li ne-to-neutral v oltag e VA a t t h e t r a n s form e r p r i m a ry a s reference, draw cO Ill p lcte per-phase phasor diagrams of all vol tages and curren ts. Show t h e co r r e ct phase re lations be tween primary and secondary q u a n t i t i e s . ( c ) Com p u t e t h e r e a l and r e a c t i v e pow e r supplied by t h e gene rator and consumed by t h e l oa d .

2 . 1 S . S o l ve P r ob . 2 . 1 7 w i t h p h a s e s e q u e n ce A CE . 2 . 1 9. A s i n g l e -p h ase t ra n s fo r m e r r a t e d 30 k Y A , 1 20 0 / 1 20 V i s

(a)

t r a n s fo r m e r t o s u p p l y 1 3 20 Y f r o m a 1 200 Y b u s .

as a n au to­

marks on each winding so

D r a w ;\ d i ag r ,l m o f t h e t ra n s fo r m e r con n e c t i o n s sh o\\ i n g t h e p o l a r i t y

the w i n d i ngs

( u ) i'.'1 ark

tha t

(d)

co n n ec t e d

and d i rections c h o se n as

t h e c u r re n t s w i l l be on t

he

in phase.

pos i

d i ag r a m t h e v a l u e s o f ra t e d

t ive

for

curren t

CU iTent

in the

in

windings and at the

i n p u t and o u tp u t .

Determine t h e rilted ki lovoltampercs of t h e u n i t as an autotransformer. If t h e efficiency o f t h e trz:nsformer con nected for 1 200/ 1 20 Y operation at r a t e d load u n i t y p owe r fl1ctor is 97%, d e t e r m i ne its e fficiency as an au totrans­ fo r m e r w i t h r a t e d c u , re n t in t he windings and o p e r a t i ng a t rated voltage to supply a l o a d at u n i ty power factor. Sol'/e P r ob . 2 . 1 9 if the t raf'.sformer is t o supply 1 080 Y h o m a 1 200 Y bus. Two b u s e s n and b a r e con nected to each other through imped ances X l = 0.1 a n d X 2 = 0 . 2 p e r u n i t i n p a r a l l e l . B u s b is a load bus s u p p l y i n g a cu rrent J = 1 .0 - 300 pcr u n i t . T he per-u n i t b u s vol t ag e Vb is 1 .0 �. Find P and Q into bus b through C n c h o f t h e pa ra l l e l b r a n c h e s ( (I ) i n the circll it described, ( b ) i f a regulating t r a n s fo r m er i s c O n lH.: C f l: d < I t b t i S Ii i n t h e l i n l: o f h i g h e r !T � l c t ;l n c c t o g i v e i t boos t of 3 % i n v o l t a g e rn a g n i t l l ci e towa rcl l i l l: l o a d ( 0 1 . 0J), a n d ( c ) i f t h e r e g u l a t i n g t r ;1 I 1 s fo r l1l e r a d v a n c e s t h e p h a s e 2° ( (I I' j;l /C)(I). Use t h e c i r c u l a l in g - c u r r e n t m e t h od fo r p;\ r t s ( h ) a n d ( c), a n d ;\ S S U l1l e t h il t ,<, is a dj u s t e d [o r each p a r t of t h e prob lem s o t h a t VI) re m a i n s con s t a n t . F i g u re 2 . 2 () i s t h e s i n g l e - l i n e d i a g r a m showing buses a 3 n d b of t h e sys t e m w i t h t h e r e g u l a t i n g t ransfo r m e r i n p l ac e . Neglect t he impeda n ce (c)

2.20.

2.21.

/

=

= ,

of t h e t r a n s fo r m e r .

,. \

\!)

x

�-,

)0.1

1

---

FI G U RE 2 . 2 6

l i r c u i t fo r P r o b . 2 . 2 1 .

86

CHAPTER 2

TRANSFO R M ERS

�

o reactances X l = 0.08 and X2 = 0 . 1 2 per u9i� _are i n parallel between two per unit, buses a and b in a power system. If Va = 1 .05 � and Vb = 1 . 0 w h a t should b e the turns ratio o f t h e re gu l a t ing transformer to be inserted in series

2.22. Tw

with X2 at bus b s o that

2 .23.

no vars flow into bus b from t he branch whose reactance i s Xl? Use t he circu la t i ng c u rren t me t hod a n d neg l ec t the reactance of the r egu la t i ng transformer. P a nd Q of t he load and Vb rem a i n constant. Two t r a n s former s e d e h r at e d 1 1 5Y / 13.2il kY, operate in parallel to supply a load of 35 MYA, 1 3.2 k Y a t O.H pow c r - fa <.: tor l a g g i n g . T r a ns fo rm<.:r 1 i s r a t e d 20 M Y A with X = 0.09 p e r uni t , an a t ransformer 2 is r a t e d 15 MY A w i t h X = 0.07 p e r unit. Find the magni t u d e o f t h e current in per u n i t through each transformer, the meg av o l ta mpere output of each transfo rmer, a n d the megavol tamperes to wh ich t h e total load m u s t be limited so that neither transformer is overloaded . If t he taps on t ra ns fo r m e r 1 a rc s c t at I I I kY to g i v e a 3 . 1l % boost in vo l t a ge t o w a r d t h e l ow v o l t a g e s i d e o f t h at t rans fo r m e r c o m p a red L O t r Ll I1 s fo rm c r 2, w h i c h r emLl i n s on t he 1 15-kY tap, find the megavoltampere output of each transformer for the origin a l 35-MYA total load and the m aximum megavoltamperes of t he t o t a l l o a d w h i c h w i l l not overload the transformers. Use a base of 35 MYA, 13.2 kY on the l o w v o l t a ge side. The circulating-curre nt method is sat isfactory for this prob l em -

,

,

-

-

.

CHAPTER

3

THE SYNCHRONOUS MACHINE

syncIHonou s m achine as an ac genera tor, driven by a turbine to convert m e c h a n i c a l e n e r gy into e l e c trical energy, is the major e l ectric power generat i n g s o u r c e throughout t h e world. A s a motor t h e machine converts e lectrical e n e rgy

Th e

mech anical energy. \Ve tOL b u t we shall give som e

are chiefly concern ed with the synchronous genera­ consideration to the synchronous motor. We c a n n ot treat the synch ronous machine fu lly, but t here are m a ny books o n the subject o f a c m a c h i n e ry w h i c h provi d e qu ite a d e q u a te a n a l y s i s o f generators a n d m otors . ! O ur i n t e r e s t i s i n t h e J pp! ica t i o n a n d opera t i o n o f the syn chronous m a c h i n e w i t h i n a l a rg e i n t e rconnected power syste m . Em ph asis i s o n principles a n d ext e r n a l b e h avior u n d er both steady-state a n d t ransie n t conditions. The w i n d i ngs of the polyph ase syn c h ronous machine constitute a group o f ind uctively coupled electric c ircuits, some of which rotate relative to others so t h a t mu t u a l i n d u ctances are variab le . The general equations developed for the flux lin k2.ges u f t h e various win d i ngs a r e appl icabl e to both steady-state a n d tra n s i e n t a n al)'sis. On l y l i n ear m agnetic circuits are con sidered, w i t h saturation n eglecte d . T his allows us, whenever convenient, to refer separately t o t h e flux

to

I

For

3

m u c h m o re d e t a i l e d discussion of s y n c h r o n o u s m a c h i n e s , co n s u l t any of

m a c h i n e ry s u c h a s A . E . F i t zge r a l d , C. K i n g s l e y ,

M c G r aw-H i l i , I n c , N e w York, 1 98 3 .

the

texts on

electric

J r. , a n d S . D . U m a ns, Electric Mach inery , 4th e d . ,

87

88

CHAPTE R

3

THE SYNCH RONOUS M AC H I N E

and flux l i nkages produced b y a compone n t m a gnetomot ive force (mmn-even though in a ny e lectric machine there exists only net physical flux due to the resultant mmf o f a l l t h e magnetizing sources. S implified equ ivalent circuits a r e developed t hrough which important p hysical relationships within the m a c h i n e can be visualized. Therefore, our t r e a t m e n t o f the synchronous machine shoul d p rovide confidence i n the equivalent circuits sufficient f o r un derstanding the role o f the generator i n our fu rther s t u d i e s of power syste m analysis. 3.1 DESCRIPTION OF THE SYN C H RO N O U S MACHINE

The two p r i n ci p a l p a rt s o r a syn c h ro n o u s m a c h i n e a rc fe r ro m a g n e t i c s t r u c t ur e s . The stationary p a r t w h i c h i s ess e n t i a l l y a h o l l ow cy l i n d e r, c a l l e d t h e .\'/(I / oJ' o r armature , has l ongitud inal slots in w h ich there a re coils -of t h e a rm a t ure wind ings. These windings carry the current supplied to a n electrical load by a generator or t he current received from a n a c supply by a motor. The rotor i s the p a r t of the machine w h ieh is mounted on the shaft and rotates i n s i d e the holl ow stator. The winding on the rotor, called the field winding , is s u p p l i e d w i t h d e current. The very high m m f produced b y this curre n t i n the fi e l d w i n ding combines with the mmf produced b y c urrents i n the a rm a t ure windings. The resultant flux across the air gap between the stator a n d rotor generates vol tages in the coils o f the armature w i n di ngs and provides the e lectromagnetic torq ue between the stator and rotor. Figure 3.1 shows t he t h reading o f a fou r-pol e cylindrical rotor into t h e stator o f a 1 52 5 - MVA generator. The dc current is supplied to the field winding by an exciter , which may be a generator mounted on the same shaft or a separate d c sou rce connected to the field winding through brushes bearing on slip r i n g s . L a r g e ae generators usually h ave exc iters consisting of an ac source with solid-state rectifiers. If the machine is a generator, the sh aft is d riven by a prime mover , w h ich is usually a steam or hydraulic turbine . The electromagnetic torque developed in the generator when it delivers power opposes the torque of the p rime m ove r. The d ifference between these two torques is due to losses in the iron core a n d friction. I n a motor the electromagnetic t orque developed in t h e machine (except for core and friction losses) is converted to the shaft torque which drives the mechanical load. Figure 3.2 shows a very element a ry three-phase generator. The fi e l d w inding, indicated b y the j-coil, gives rise to two poles N and S a s m arke d . The axis of the field poles is called the direct axis or s i mply the d-axis, while the centerline of the interpolar space is c a l l e d the quadrature a:ris or simply the q-axis. The positive direction along the d-axis leads the positive d irection along the q-axis b y 90° as shown. The generator i n Fig. 3 .2 i s called a nonsalient or round-rotor machine b eca use i t has a cyli n drical rotor l ike that of Fig. 3 . 1 . In the actual machine the winding has a la rge number of t u rns d istributed i n slots around the c ircumference of the rotor. The strong magnetic field produce � l in ks

J. I

F1 G lJRE

D ESCR I PT I O N O F THE S Y N C H R O N O U S MACH I N E

89

:U

P h otograph s ho\',ing t h e t h re a d i n g of a fo u r - p o l e cyl i nd r i c a l r o t o r i n t o t h e s t a t o r of a l S2S - M VA g e n e r a t o r . ( Cou rtesy Utilit}' POIIEr Corpora tioll , Wisconsin .) t h e s t a t o r coils t o i n u c e vol t a g e i n

d

by

t h e p r i m c m ov e r .

Th e

t h e a rm a t u r e windi ngs as the shaft is turned

s i d e s of a c o i l , w h i c h i s a l m o s t r c c t a n g u l ; l r , ;I rc i n s l o t s a a n d a' I SO° a p a r t . S i m i lar c o i l s a re i n s l ots h a n d h i ,l !ld s l o t s c , I n d ( . Co i l s i d e s i n s l o t s [{ , h , ,l n d c a re 1 200 a part. T h e c o n u u c t or s s h o w n i n t h e s l o t s i n d i c l t c a co i l o r o n l y Olle t u r n , but slIch a coil m a y h ave m a ny t u r n s a n d i s u su a l ly i n series with i d e n t ical coils i n a djacent slots to fo r m a w i n d i n g having e n d s des ignated a and a' . Wind ings w i t h ends designated b hi a n d c c' a r c t h e s a m e a s t h e a d w i nd i ng except fo r t h e i r sym m e t r i c a l l o c a t i o n at a n g l e s of 1 2 ()0 a n c i 24() 0 , respectively, a ro u n d t h e S i ll t o r

i s s h o w n i n cross s e c t i o n i n F i g . 3 . 2 . O p p o s i t e "

-

-

-

armature.

Figure 3 . 3 shows a sa lien t -pole m a c h in e w h i c h h a s four poles . O ppos i t e sides of a n arm a t u r e coil a r e 90° apart. So, t here are two coils for each p hase. Coi l sides a , b , a n d c of adj ace n t coils a r e 60° a p a r t . The two coils of each p h a se m a y be connected i n series or in parallel.

90

,,-axiS ,

CHAPTER 3

Ai r ga p

THE SYNCHRONOUS M A C H I N E

Stator

FI G U RE 3.2

Elementary three- p hase ac g e n e ra­

Rotor

d-axis

"

tor showing end view of t he two-pole cyl i n drical rotor and cross sectiOn of the stator.

mmf of de w i n d i n g de field � w i n d i ng coi l s

q-axis

-'. FIG U R E 3.3

Cross section of an element ary stator and s a l i e n t-pole rotor.

Although not shown in Fig. 3 . 3 , salient-pole machines usually h ave damper windings, whic h consist of s hort-circ uited copper b ars t h rough the pole face similar to part of a " squirrel cage" winding of an induction motor. The purpose of the d amper winding is to reduce the mechanical oscillations of t h e rotor about synchronous speed, w h ic h is d etermined by the n umber of poles of t h e machine a nd the frequency of t h e system to w h ic h the mach ine is connected. I

3.2

THREE-PHASE G E NERATION

91

I n the two-pole machine one cycle o f voltage is generated for each revolution of the two-pole rotOL I n the four-pole machine two cycles a r e generated i n e a c h coil per revolu t io n . Si nce t h e number of cycles p e r revolution equals the n umber of pairs of p oles, the frequency of the generated voltage is

f=

P N

2 60 = 2 fm

w h e re f

=

e l ectrical frequency in

P

=

number of poles

N

=

fill

=

P

(3.1)

Hz

Hz

rotor speed i n revol utions per minute ( rpm )

Nj6() ,

0. 1 )

the

mcch:tn iCid

I ILlt

freq l l e n cy i n revo l u t i o n s

rer s ec o n d ( r p s ) .

a t J GOO r p m , w h e r e a s :l fo ur - p o l e m a c h i n e ope rCl t es : I t I t) ()() r p m . U s u a l l y , foss i l - flred steam t urbogcn e rators are two- p o l e m a c h i nes, whereas hyd roge ncrat ing u n its a re slOwer machi nes w i t h many pole p a i rs . S i nce one cycle o f voltage (3 600 of t h e vol tage wave) is generated every time a pair of poles passes a co i l , we must d istinguish b e tween electrical degrees used to express vo l tage and curre n t and mechanical degrees used to express t h e pc)s i t i o n o f t h e rotor. I n a two-·pole m achine electrical a n d mecha n ical degrees are e q u a l . I n any other mach i n e the number of e lectrical degrees or radians e q u a l s P / 2 ti mes the n umber of mechanical degrees or radians, a s can be seen 2 from E q . (3. 1 ) b y multiplying both sides b y 7T . I n a fou r-pole mach ine, therefore, two cycles or 720 e lectrical degrees are produced per revolution of 360 mechanical d egrees. I n this chapter all angu l a r measurements are exp ressed in electrical d egrees u n less otherwise stated, a n d the d i rect axis always leads the quadrature ?,xis b y 90 e l e c t r i c a l d e g rees i n t h e coun terclock wise d i rect ion o f rot a t ion rcg( "Hdiess of t h e n u m ber of poles o r the type of rotor const ructio n . Eq u d l i o n

3.2

t e l ls u s

a

t w o - p o l e , ()( ) - I- I z

m ac h i ne

uperates

THREE- PHA S E G EN ERAT IO N

The R e i d and armature w i n d ings o f the synchronous mach ine described in Sec. 3 _ 1 a r e d i s t r i b u t e d i n s l o t s a ro u n d t h e p e r i p h e ry o f t h e a i r gap . Section A. I of t h e Append ix s hows t h a t t h e se d i s t r i b u ted w i n d i ngs can be replaced a long t h e i r axe s b y con cen tra ted coils w i th a p p ro p r i a t e s c J f- a n d m u t u a l i nducta nces. Figure 3.4 \hows t h ree s uch coils-a, b , a n d c-whid: represent the three armatu re w i n d i ngs on t h e stator of t h e rou n d - rotor machine, and a conce ntrated coi l f, which represents the distri b u t e d fie l d winding on the rotor. The t hree station ary a r m a t ure coi l s are identical in every respect and each has one of its two t e r m i n a l s conn ected to , I co m m o n p o i n t o. The other t h ree ter m inals a re

92

CHAPTER 3

THE SYNCH RONOUS M A C H I N E

()d

=0 a- axis

Quad rature axis

-

Direct axis

Rotation

Rotating field

ir/ f

� � v{f' """

b- axis

�

"

.-'

b

+

__

f'

\ \ \ I I I f

R , Le, .....

c

"-

-....

c - axis

FIG U RE 3.4

Idealized thr ee-p h ase generator s h owi n g i d e n t i cal a r m a t u re c oils a , b, and c . and fieJd coi l axis leads q u adrature axis by 90° in the a nt i clockwise d i re c t i on of rot a t i o n .

f. D i rect

marked a, b, and c . The axis o f coil a is chosen a t (J d = 0° , and coun terclock­ wise around the air gap are t h e axes of the b-coil at (J d = 1 200 and of the c-coil at (Jd = 240° . For the round-rotor machine it is shown in Sec. A.1 of t h e Appendix that: •

Each of the concentrated coils a , b, and c has self-inductance L which is equal to the self-inductances Lac/ > Lb1p and Lee of the d istributed a rm a ture w i n d i ngs w h ich the coils represent s o t h a t S

'

( 3 2) .

•

The mutual inductances Lab' Lbe, and Lea between each adjacent p a i r of concentrated coils are negative constants d enoted by - Ms so that

( 3 .3) •

The mutual i n ductance between t h e field coi l f and each of the stator coils v a ries w i th the rotor position (Jd as a cosinusoidal function with m ax'imum

3.2

value

Mf

T H REE·PHASE G ENE RATION

93

so that

( 3 .4)

The

field co i l has a co n s t a n t se l f- i n d u c t a n ce L ff ' T h i s i s because i n t h e ro u n d - rotor m a ch i n e ( a n d , indeed, i n t h e sal ient-pole machine also), t h e fi e l d w i n d i n g o n t h e d-axis prod uces fl u x t h ro u g h a s i m i l a r magnetic path i n the stator for a l l pos it ions o f t h e rotor (neglec ting the small effect of armature sl o t s ) .

t h e co i l s a , b ,

c u rr e n t a n d t h e cu rr e n t s i n t h re e t h e refore w r i t t e n for a l l fo u r c o i l s as fo l l ow s :

of the

Flux l i n kages with e ac h

Armature :

an d

f a r e due to

own other coils. Flux- l i nkage equations a re c,

its

..

Field :

( 3 .6 ) I f i ll ' i i"

a n d I e a rc a

S e t t i n g i ll

balan ced t h r e e - p h a s e set o f c u rr e n t s , the n

( 3 . 7)

( 3 . 8)

For n o w we are i nt erested in steady-state condi tions. We assume, there­ fore, that c u rr e n t if i s dc w i t h a constant value If and that the fiel d rotates a t ,

94

CHAPTER 3

constant

THE SYNCHRONOUS MACHIN E

a ngul ar velocity w so that for the two-pole m achine

d8d - =w dt

and

( 3 .9)

i nitial p osition of the fi e l d winding is given by the angle fJdO ' which c a n b e arb itrarily chosen a t t = O. Equations (3 .4) give the express ions for Luf' Lhf , and LeJ i n terms o f 8d• Substituting ( w I + 8d O ) for 8d and using the resu lts along with i f = If in Eqs. (3 .8), we obta i n

The

Aa

=

( Ls +

Ms ) i a + Mf If cos ( w t +

8d O ) ( 3 . 1 0)

The fi rst of t h ese equations shows t h a t A a has two fl u x - li nkage cum p o n e n t s - o n e due to t h e fi e l d current If a n d t h e o t h e r d u e t o t he armature current i a , which is flow i ng out o f the machine for generator action. If c o i l a has resistance R , then t h e voltage drop v a across t h e coil from terminal a to terminal 0 i n F i g. 3.4 i s g iven b y

The nega t ive signs apply, a s d iscussed i n S e c . 2.2, because t h e machine i s being treated as a g en erator. The last term o f Eq. (3 . 1 1 ) represents an internal emf, which we n ow call ea' . This emf can be written ( � . 1 2)

w here the rms m agnitude I Ei l , p ro p o r t i o n a l to the Reid curren t, is defined by

IE I I

=

f .. I... f..:.... _ w_ M....:.

fi

( 3 . 13)

action of the field c u r r e n t c a us e s ea' t o a p p e a r across the terminals o f t h e a-phase when i a is zero, a n d so it i s called by vari ous n ames such as the no-load voltage, the open-circuit voltage, t h e synchronous internal coltage, or generated emf of phase a . The angle ()d O ind icates the position of the field w inding (and the d-axis) rela tive t o the a-phase a t t = O . Hence, 8 � 8dO - 90° ind icates the position o f the q - axis, which i s 90° behind t h e d-axis i n Fig. 3 . 4 . For l a t e r convenience we now set 8dO = 8 + 90° , a n d then w e have The

(,3 . 14)

3.2

TH R E E - P HA S E G E N E RA TI O N

9S

where ed ' w, and 0 h ave consiste n t u n i ts o f angular measurement. Subst i tu ting from Eq. (3 . 1 4) into Eq _ (3. 1 2) a n d noting that sin(a + 900 ) = cos a , we obtain for the open-circuit voltage of p hase a

The terminal voltage

V

a

of

( 3 . 15) Eq . (3 . 1 1 ) i s t h e n g iven by ( 3 . 1 6)

This equ ation corresponds to the a - phase circu i t of Fig. 3.5 i n which the no-load vol tage c a ' i s t h e s o u r c e c t n d t h e e x t e rn a l l oad i s b a l a n ced across a l l th ree phases.

(3. 1 0) c a n h e t rea ted i n t h e same S ince t h e a rm a t u re w i n d i n g s a re identical, r e s u l t s similar to Eqs. (3. 1 5 ) and ( 3 . 1 6) ca n be fou n d for the no-load vol tages e b , a n d e c' which l ag e a' T h e fl u x l i n k a g e s A " a n d A c g iv e n h y Eq .

way as A a '

by 1 200 and 2400 , respe ctively, in Fig. 3 . 5 . Hence, e a , e b " and ec' constitute a b a l a nced th ree-phase set of emfs which give rise to balanced t h ree-phase l i n e

,--- '--�-----:-_.

+

a

- - - - r- - - - - r L-_____________________l_+ i:_'..... : L c -�- � -

_ _

Fl G G RE 3.5

Arm a t u r e e q u i va l e n t c i r c u i t

vol t a g e s ea' , E,i . a n d

ee'

of

th e

i d e a l ized

in t h e s t e a d y s t a t e .

n

t h re e - p h ase g e n e r a t o r s h ow i n g b a l a nc e d

no-load

96

CHAPTER 3

THE SYNCHRONOUS MACH I N E

currents, say, ia

=

i ll

=

ie

=

V2 l /a I cos( w I + 0

(5

n l / l cos ( w t + a

n I /u 1 cos ( w t

+ (5

-

-

-

ea ) ea ea

-

120° )

-

240° )

( 3 . 1 7)

where I fa l i s t h e rms v a l u e and Va is t h e phase a n g l e or lag o f t h e c u r re n t i ll with respect to e,l ' W h e n t h e cm rs a n d t h e c u rre n t s a re expre ssed CiS p h a s o rs , Fig . 3 .5 becomes v e ry m uc h l i ke t h e eq u i va l e n t c i r c u i t i n t rod uced i n F i g . 1 . 1 l . B e fo r e e m p l oy i n g t h e e q u iv;t l c n t c i rcu i t , l e l u s c o n s i d e r t h e n u x l i n k ,l g e s '\ f o r t h e fi e l d wi n d i n g . T h e e x p r e ss i o n s for I� I/ ! ' 1> 1>1 ' a ll d L ' I i ll Eq s . 0.4) C l ll h e s u b s t i t u t e d i n t o E:q . (3 .6) t o y i e l d

The fi r s t t e r m w i t h i n t h e b rackets c a n be exp a n d e d a cco r d i n g to Eqs .

(3. 17) a s

(3. 1 4 )

and

fol l ows :

(3 . 19)

The E: q.

t ri go n o m e t r i c i d e n t i t y

2 cos

a cos

(3. 1 9 ) yields

(3

=

cos( a

-

(3 ) +

cos( a

+ (3) a p p l i e d t o

( 3 . 20 ) T he i u a n d

Ie

terms in E q . (3 . 1 8) l e a d

I /a l

12

-

s i milar results, a n d

sin(2( wl

+

8)

-

e"

we

-

h av e

1 20° ) } ( 3 .2 1 )

(3.20) t h rough (3 . 22) a r e b a l anced s e c o n d ­ har m o n i c s in u so i d a l q u a n t i ties w h i c h sum t o z e r o a t e a c h p o i n t i n t i m e . H e n c e , addi ng t h e b racke ted t e r m s of E: q . (3. 1 8 ) toge ther, w e ob t a i n

The t e r m s i nvolving

2wt

{ - si n e"

to

i n E:q s.

3.2

TH REE-PHASE G EN E RATION

97

and t h e expression for ;\ f t a k e s on t h e s i m p l e r form

( 3 .24)

cu rren t

w h e r e de

by Eq . (3 .23)

/f [ i a cos fid

ld =

+

iI.J COS( (Jd

1 20° )

-

i / cos

+

(Jd

-

240° )] or

( 3 .25)

..

w h i c h i s u s e fu l l a t e r i n

t h i s c h a p t e r. F or n o w l e t u s observe fro m Eq_ (3 . 24) t h a t t h e fl u x l i n k a ge s w i t h t h e fi e l d w i n d i n g d u e t o t h e comb i n a t ion o f i a ' i b • a n d ie d o n o t va ry w i t h t i m e _ \Ve c a n , t h e r e fore , r e g a r d those fl u x l i n kages a s com i n g from t h e s t e a d y d c c u rre n t i " i n � l fict i t i o u s d c c i rc u i t c o i n c i d e n t w i t h t h e d-ax.i s a n d t h u s s ta ti o n a ry w i t h r e s p e c t to t h e fi e l d c i rcu i t . The t w o c i rc u i ts ro t a t e t o g e t h e r i n s y n c h ro n i s m em d h ave a m u t u a l i n d u c t a nce Mf be tw e e n t h e m , as s h o w n i n F i g . 3 . n . I n g e n e r ;t I , t h e fi e l d w i n d i n g w i t h r e s i s t (l n ce R f a nd e n t e r i n g c u r r e n t i l b as term i n a l vol t a g e UIt give n by

( /3/2 )

( 3 _26)

�:

a - axis

" /"

Di rect axis

0,

i/&6)�) j 3 :

"

""

"

/

r

I

L

winding rotati n g with ro/ lor)

Arn1;]t u r e e q U i v a l e n t

2-

""""

/: , '/;/ ( /

f

I fr

� ".. ".. ".. "

__

_ ".. - l'rr

"...­

Quad rature axis ,

..-' "..

:

M

1

M utual ,, " /

(

J /

/

/

"

/

"

"

.'

/

"

"

"

/

i n d uctance

"

Field windi n g rota t i n g

with rotor

,,,-

f'

'"

b - axis

c - axis

""-

FI G L RE 3 . 6

/3 / 2 /1f!

R e p r e se n t i n g in ductance

lhe

a r m a t u re with the

of

lhe

li e l cJ

s yn c h r o n o u s

m3chine

by

a

d i rect-axis w i n d i n g of m u t u a l synchro � ism.

w i n d i n g . B o t h w i n d i n gs r o t a t e t o g e t h e r i n

98

CHAPTE R

3

THE S Y N C H R O N O U S M AC H I N E

Because A f is not varying w i t h time i n the steady state, the field vol tage becomes vff' = R f lf and i f = If can be supplied by a dc sou rce . Equ ation (3.25) shows t h a t the n u me rical value of id depends on the magnitud e of the armature curre n t I Ia I a n d i ts phase angle of lag e a relative to the internal voltage ea" For l agging power factors ea i s positive and so id is negative, w hich means that the combined effect of the armature currents i l b ' a n d i e is d emagnetizing; t h a t is, i d opposes the magnetizing influence of the field current If ' To overcome this demagnetizing influence, If has to be i ncreased by the excitation system of t h e gen erator. At leading power fac tors 8a takes o n smaller val ues, which me ans t h a t t h e de magnetizing influence of the a rmature currents (represen ted by i d = - /3 I Ia 1 sin ea) is reduced and If can then be l owered b y the excitation sys t e m. In an actual machine the effect of the currents i Q ' ib, and i c is called a rm at u re rea c tio l l a n d the control of the field current is called excitation system con trol, w h i c h is d iscussed i n Sec. 3 . 4. a '

Example 3 . 1 . A 60- Hz three-phase synchronous generator with negligible armature

resistance has the following in ductance para meters: Laa

Lab

=

=

Ls

Ms

2 .7656

=

1 .3828

=

Mf

mH

mH

L

ff

=

=

3 l .6950

mH

433 . 65 69

mH

The machine is rated at 635 MY A, 0 . 9 0 power-factor lagging, 3600 rpm, 2 4 kY. When operating under rated load co ndi tions, the l ine-to-neutral terminal voltage and l ine current of phase a m ay be written Va =

1 9596 cos

wI

Y

in

=

2 1 603 COS ( w l

-

2 5 . 84 1 9° ) A

Determine t h e m a g n i t u u e o r t h e sync i l r o n o u s i n t e rn a l v o l t a ge , t h e fi e l u curren l if ' and t h e fl u x l in k a g e s w i t h t h e fi e l d w inding. C a l c u l a t e t h e v a l u e s o f t h e s e q u a n t i t i e s wh e n a load of 635 M YA is s e rved a t r a t e d v o l t a ge a n d unity po w e r f a c t o r . W h a t is the field current for r a t e d a r m a t u r e vo l t a g e on an open ci rcu it?

The given maximum value of Va is )2 (24,000/ [3 ) = 1 9596 Y , the maximum value of in is )2 (635,000/ fS X 2 4 ) = 2 1 603 A, and the power-factor angle 0 = cos - 1 0.9 = 25 .84 1 9° l agging. W ith R = 0, in Eq. ( 3 . 1 6) the synchronous i nternal voltage can be written

Solution.

= =

Va + ( 2.765 6

+

1 95 96 cos

- ( 4 . 1 484) 1 0 � 3

w(

1 . 3828) 1 0 - 3

di a --:it X w X

2 1 603 sin ( w t - 25 . 1)4 1 9° )

3.2

S et t i n g

w = 1 20rr,

w e ob t a i n

e a, = V2 I Ei l cos ( w l and expanding t h e seco n d

gives

+

8 ) = 1 9596 cos w i - 33785 sin ( w t - 25 _84 1 9° ) V

term

according to s i n( a - (3 )

ea, = V2 I E, l cos( w l

+

=

45855 cos ( (tJ I

a

cos

=

4 5055

-� ---

1 2 ( ) rr x 3 1 . (/Xi x I ( )

·

=

flu x l i n ka g e s w i t h t h e fi e l d w i n d i n g s are g i v e n by Eq.

behind

Va ' w hic h

l a gs

en

I fa l s in e a

4 1 .5384°

= 25 .84 1 9°

=

cos

fi I E, I = 45855

a

sin

(3 .24),

ea, .

b e h i n d e a" it fo l l ows t h a t

+

V and

S i nce In

l a gs

4 1 .5384° = 67.3803°

2 1 603 .fi s m 67 .3803° = 1 4 1 00 .6 A

a n d subst i t u t i n g i n t h e a bove e x p r e ss i o n fo r A f y i e l d s Af

= ( 433 .656()

x 10

1 ) 3838

- --�.f2=i��3 x

1 664 .38 - 948.06 = 7 1 6 .32

3 1 .695

x j0

.1

x

14 1 00 .6

Wb- t u r n s

R e p e a t i n g t h e a bove s e q u e nce o f c a l c u l a t i o n s a t u n i t y p o w e r fa ctor, we

e , = fi I E I cos( w t il

I

+

f3

3 � n � ;\

\v here ea is t h e a n gl e of J a g of in m e as u r e d w i t h respect t o

25 .84 1 9°

(3 -

+ 4 1 .5384° ) V

Bence, t h e syn c h ro n o u s i n t e r n a l v o l t a g e has m a g n i t u d e angle 8 4 1 .5384° . From Eq . 0. 1 3) w e l i n d

{(1 M,

sin

0 ) = 34323 cos w i - 30407 sin w t =

The

99

T H R E E - P H A S E G E N E RA TION

0)

=

1 9596 cos w I - 33785 s i n w I

= 39057 cos (

(u

( -f

59 .88 )4° )

ob t a i n

100

CHAPTER

3

THE SYNCH RONOUS MACH I N E

Because I Ei l is directly proportional to If' we have from previous calcu la tions

If = Current t a

is in

I Ja l sin Oa

phase =

Al = =

with

VII

3 90 5 7

45855

a n d l a gs

1 5276 sin 5 9 .8854°

(433 .()S()c) 1 4 1 7 .62

-

X

X

=

3838

e ll' by

0/)0 .43

=

3269

A

5 9 . 8854° . Ther e fo re ,

1 32 1 4 A

1 ) 3 2()c)

10

=

3 -

X

3 1 .695

./2

X

10-3 X

1 32 1 4

5 2Sl . 1 Sl Wb- L u rns

Thus, when the power factor of the load goes from 0 . 9 lagging to 1 . 0 under rated megavoltamperes load ing and voltage cond itions, the field current is reduced from 3838 to 3269 A . Also, t h e net air-gap fl u x l i nking the field winding of t h e generator is reduced along with the demagnetizing influence of armature reaction. The field current required to maintain rated terminal voltage in the machine u nder open-circuit conditions is found from Eq. ( 3 . 1 3), and Eq. ( 3 . l 6) with i a = 0 , /f

-

v'2I Ei l

-w Mf

1 9596 X 1 0 3

-

----= 120 X 3 1 .695 ..

1 640

A

3.3 SYNCHRONOUS REACTAN C E A N D EQUIVALENT CIRCUITS

The coupled-circu it model in Fig. 3 . 4 represents the idealized V-connected round-rotor synchronous machine. Le t us assume that the mach i n e is rotating at syn c h ronous speed w and that the field current Ii is steady dc. Under t hese conditions the balanced three-phase circuit of Fig. 3 . 5 gives the steady-s tate operation of the machine. The no- load voltages are the emfs e a" eh" and e c, · Choosing a-phase a s t h e refe r ence p hase for t h e m ac h i n e , we o b t a i n t h e p er-phase equivalent circuit of Fig. 3 .7 ( a ) with s teady-state sinusoidal currents a n d voltages which lead the corresponding currents and vol tages of phases b a n d c by 1 200 a n d 2400 , respectively. We recall that the phase angle of the current ia in Eq. (3. 1 7) i s chosen with respect to the no-load v oltage e a , of the a-phase. In practice, e a , cannot be m e as u red under load, and so i t is prefe rable to choose the terminal voltage Va as r e ference and to measure the phase angle of the current ia with resp'ect to

3. 3

SYNCH R O N O U S R EA CTANCE A N D E Q U I VA L E NT C I R C U ITS

101

-

1-

o

FI G U R E 3 . 7

1-

E q u i va l e n t

�------ �

o

(b)

Vo.

T he r e fore , w e

circllit

for

re fe r e n c e

p h a s e a of t h e synchronous m a ­ c h i n e s h o w i n g voltages a n d c ur ­ r e n t s a s ( n ) cos i n u so i d a l a n d ( b ) ph asor q u a n t i t ies.

deAne ta

= Ii I I 1 c o s ( w t - 8 ) a

( 3 .27)

Note that e o, cor resp o n d s to Eq. (3 . 1 5) a n d t h a t ia d iffers from Eq. (3 . 1 7) o n ly i n t h e r e s p e c t t h a t t h e p h a se a n gle 8 = 80 - 0 i s n ow the a ngle o f lag o f i a m e a s u r e d w i t h r e s p e c t t o t e r m i n a l v o l t a g e VtI • T h e p h a s o r e q u i v a l e n t s o f E q s .

(3 . 2 7 )

a rc

v11

and t hese

n rc

'-'=

/ 0° I V{J I �

'

.

m a rk e d on t h e e q u i v a l e n t c i rc u i t o f F i g . 3 .7( 6 ) fo r w h i c h

p h a s o r-vo l t ll g e e q u a t i o n I S

va

=

------fU a

E/

--­

n o load

W h e n the curre n t

a r m a t u re s i st a nc e

D u e to

GenerJted at

( 3 .28)

re

10

jw Ls Ia

----­

Due to ar rr:at ure se I f- re a c t a nce

jw MJa -------

the

( 3 . 2 9)

'--

Due to arm ature m u t u a l r e a c t a nce

l e ads Va ' t h e a ngle 8 is num erically negative; and when l a

102

CHAPTER 3

THE SYNCHRONOUS MACHINE

lags Va ' the angle () is numerically positive. S ince symmetrical cond itions apply, phasor equations corresponding to Eq. 0 . 29) c a n be written for b-phase a n d c-phase. The combined quantity w(Ls + M) o f Eq. (3.29) has the dim ensions of reactance and is customarily called t h e synchronous reactance Xd of the ma­ c hine. The synchronous impedance Zd of the machine is defined b y ( 3 .30)

and Eq. (3.29) then can be written in the more compact form ( 3 .3 1 )

from which follows the generator equivalent circuit of Fig. 3 .8( a). The equiva­ lent circuit for the synchronous motor is ide ntical to that of the generator, except that the di rection of I(2 is reversed, as shown in Fig. 3 .8( b ), which has t he

equation ( 3 . 3 2)

r

Zd '"

"\

1a

-

1+

Va 0

r

Zd A

l-

(a)

iQUO' jXd

\

1a

-

1+ •

Va

0

l..

(b)

FIG U R E 3.8

E q uiv a l e n t c i rcu i ts for ( a ) the syn c h ronous gen e r ator a n d (b) the synch rono u s mot o r w i t h const a n t sy nchro no u s i m pedance 2,, '= R +

jX.t.

J.J

S Y N C HR O NO U S R E AC TANCE A N D E O U I VALENT CIRCU ITS

103

(6) FI G U R E 3 . 9

( 0 ) ovncxc i t e d g e n e ra t or d e l ive r i n g I it g g i ' l g c u r re n ! d r awi n g l a g g i n g c u r re n t la ' P h a� o r d i a g r a m s o f :

I,, ;

( h ) u n d e re x c i t e u m o t o r

P h asor d iagrams for Eqs . 0 . 3 1 ) and (3 .32) are shown i n Fig. 3.9 for t h e c ase of l agging power-factor angle f) measured wi t h respect to the term i n a l voltage . I n Fig. 3 . 9( a ) for the generator note t h at Ei always leads Va ' and i n Fig. 3 . 9( b ) for the motor Ei always lags Va ' Except for the case of an isolated generator supplying its own load, most synchronous machi nes are con n e cted to la rge interconnected powe r systems such t h at t h e t e r m i n a l vol tage Va ( soon to b e cal led VI for e mph asis ) is not a lten.:d by machi ne loa d i ng. I n that case the poi n t of con nection is called a n infin ice bus, which means that i ts voltage remains constant a n d n o frequency ch ange occurs regardless of cha nges made in operating the sync h ro nous m a ­ chi ne . S y n c h ronolls mach i ne rarameters a n d opera ting Cl uan tities such as vol tage a n d c u r re n t lI r c n o r m a l l y rc r rcs e n t c d i n r c r u n i t or n or m a l i z e d va l u es usi ng b a s e s c o r r e s po n d i n g to t h e Il < l m c p l (l t c (l a t a o f t h e mach i ne. Such parame ters are prov i ded by the manufact u rer. Mach i nes of similar design h ave norm a lized param eters which fal l i n a very n arrow ra nge rega rd less o f size, and this is very (see Ta b l e

A.2 in

Appe n d ix ) . I n the a rm a t u re or t h e t h r e e - p h a s e m a c h i n e u s u a lly the kilo­ vo l t ampere base correspon d s to the t h ree-phase ra t i ng of the mach ine and base voltage in kilovol ts correspon ds to the rated l in e- to-li n e voltage i n kilovolts . Accordi n gly, the per-phase equivalent circuit of Fig. 3 .8 has a k Y A b ase equal to t h e k ilovol tam pere rat i n g of one p h ase a n d a voltage base eq u a l to the r a te d l i n e - to-n eutral vol t age o f t h e m achi n e . B ase arm ature i m peda nce i s t herefore c a l c u l a t e d from Eq. ( 1 .5 4 ) in t h e usual w a y. u s e fu l wh e n d a t a fo r a p Clr t i c u l a r m a c h i n e are n o t a v a i l ab l e the

104

CHAPTER 3

THE SYNCH R O N O U S MAC H I N E

Although the generated vol tage Ei is con trol led by the fiel d current, nonetheless, i t is a per-phase armature voltage which can be normal ized on the arma ture base. Equa tions (3.3 1 ) and (3 .3 2) are thus d irectly applicab l e I n per unit on the armature base. The 60-Hz synchronous gener ator described in Example 3 . 1 is serving its rated load u nder ste ady-state operating conditions. Choosing the armature base equal to t he rat i n g of t h e m ac h i n e , determ ine th e valu e of th e syn c h ronous react a nc e a n d t h e p l1 asor e x p re s s i o n s f or the stator q u a n t i t i e s V:p ' Exa mple 3.2.

a n d Ej i n per u n i t . If t h e b;lse fie l d cu rre n t e q u a l s t h a

t va l u e o r If

",

w h i c h p ro d u c es

rateu t e r m i n a l vol t a ge U IH J cr o p e n - c i rc u i t co n d i t i o n s , d e t e rm i n e the v a l u e o f If

under th e sp ec i fied op e ra t i n g cond i t i o n s .

Solution .

a p

From Ex m l e 3 . 1 w e l i n d for t h e a rma t ur e

B ase kYA B ase

k VL L

B ase cu r rent B ase i m pedance

=

=c

635 ,000 kYA 24 k Y 635 ,000

= =

that

IS X

242

-

635

24

=

= 1 5275 .726 A

0 .9071 n

Using the values g iven for the i nductan ce parameters L J and Ms of the a rmatu re, we compute Xd

=

w(L,

+

MJ

= 1 20 .. ( 2 .7656 + 1 . 3 828) 1 0 - 3

=

1 .5 639 n

w h ich i n p e r u n i t i s

X"

1 . 5639 =

0 . 907 1

=

1 .724 1

The load is to be served at rated voltage e q u a l to th e specified base, and so use the terminal vol t ag e Va as the refere nce p hasor, we obtain V,l = 1 . 0

L.2: per u ni t

if

we

The load current h as the rms m ag n i t u d e I Ia I 635 ,000j(y3 x 24) A, w h i c h is also the base arm atu re current. He nce, I Ia I = 1 . 0 per un it, a n d since the power­ factor angle of the load is e = cos 1 0.9 = 25 .84 19° lagging, the p hasor form of the lagging current fa is =

-

1 .0

/

-

25 . 84 1 9°

per

unit

Synch ro nous i n t e r n a l

=

=

E, c a n

vo l t a g e

LQ:

1 .0

1 . 75 1 5

+

+

3.4

be

cal cu l a ted

=

from Eq. (3. 3 1 )

1 .0/ 2.340/ 4 1 .53840

j 1 . 724 1 X

j 1 .55 1 7

a rm a t u r e

p roport ion a l t o If ' w e t h e s p e c i fi c d o p e rat i n

g

volt age)

with R

per u n i t

0,

1.0 pe -u

is 1 640 A . T h e r e fo r e , s i n ce I E, I i s d irec t l y cu r r e n t of 2 . 34 x 1 640 = 3 8 3 8 A u n d er

to produce

h av e a n exci t a t i o n co n

=

25 . 84 1 9°

I [1 Exa m p l c 3 . 1 t h e b a se fie l d curre n t (which i s requ i red

o p e n - c i rc u i t

1 05

R E A L AN D R EA Cf I VE P O W E R CONTR O L

r

nit

d i t i ons .

The i n terested reader may wish to d raw a ph asor d i agram for the resu l ts of t h is exa mple a 'n d comp are t h e phaso r method o f sol u t ion w i th the t i me ­ dom a i n a p p roac h of Example 3 . 1 . 3 .4

REAL A � D REACTIVE P OWER C O),;TROL

When the synch ronous m achine is con n e c t e d t o an i n fi n i te bus, its speed a n d te rm i n a l voltage a re fued and u n a l terable. Two con t rollable variables, how ever, a re the field current and the m echanical torque on t he shaft. The variation of the fi eld curr e n t If ' referred to as excita tion systenl control , i s applied t o ei t her a g e n e r a t o r or a motor to su pply or ab sorb a variable amount of rea ct ive powe r . Because t h e synchronous m achine runs a t const ant speed, the only m eans o f varying t h e real power i s through con trol of t h e torque i mposed o n t h e shaft b y either t h e prime mover i n t h e case o f a generator o r the m echan ical load in t he case of a motor. I t is conven i e n t to neglect resi sta nce a s we consider reactive power con t ro l o f the rou n d -rotor generator. Assume that t h e genera tor is del ive r ing pow e r s o t h a t a ce r t a i n a n g l e [) exists b e tween t h e t e rm i n a l v ol t a ge v: a n d t h e gener a t e d v o l t a g e E, 0 f t h e m achine [sec Fi g . 3 . J O( (l )] . T h e c o m p l ex power d e l ivered t o t h e system b y t h e generator i s g i ve n i n per u n i t by s

=

P +

jQ

=

V, f:

=

I V, I

1 1,, 1 ( cos e

+ j sin e)

(3.33)

Equ a t i ng real a n d i mag i n a ry q u a n t i t ies i n t h i s equ a t i on, we obt a i n P

=

I v: I I J ) cos

e

Q

=

I V, I I IJ s i n e

( 3 . 34 )

We note that Q is posit ive for l agg ing power factors since the a ngle e i s n u merically positive. I f w e d ecid e t o m ainta in a certa in power del ivery P fro m the generator to the constant vol t age system , it is clear from Eq. (3 .34) t h a t I la l eos e m u s t remain cons t a n t . As w e v a ry t h e d e fi e l d c ur r e n t If u n d e r t hese ,

106

C H A PTER 3

THE SY;-"CHRONOUS M ACH I N E

/ I 1- - - - - - - --,I - - - - - - - 1 _ I 1 I

:

: I

: I

Ia �

Constant power locus of E i

- - - - - - - - - - - _ _ _ _ _ _ _

I

Constant power l ocus of

"" i

8

I

j1a Xd

VI I

I Ia l Xd

E '

cos

e

(a)

�----�-+--, ---L----� - - - - - - - - - - - J I

:

f l1alcos e

4 - - - - - +_ 1 I 1 I I I I I 1

:

- - - - - - - -

o

I I I I I

:

�

I I

Ei -

l Ie Xd sin e

- - - - - - - - - - -

I Ia l Xd

cos

1I<...__ .__ ---L...l.L......__ __ _ __ _ _ _ �'1

0

(b)

_ _ _ _ _ _ _ _ _ _ _ _ _ _ _

FIGURE 3 . 1 0

P h asor d i a g r a m s show i n g c o n s t a n t - powe r l o c i o f a n

« (1 ) ove r e x c i t e d ge n e ra t o r d e l i v e r i ng r e a c t ive

power t o t il e syst e m ; ( I) ) u n d c rexc i t ed g e n c r ; t t o r receivi n g r e a c t i ve rowe r from t h e sys t e m . T h e

powe r delivered by t h e ge n e r
same

in

b o t h cases.

cond itions, the generated voltage E, varies proportionally but a lways so as to keep I fa l cos e consta nt, as shown by the loci of Fig. 3 . 1 0( a ). Normal excitation is defined as t h e cond ition w h e n

I E · I cos fj I

=

!VI I

( 3 . 35)

a n d the machine i s said to be e i t h e r ouerexcited or underexcited acco r d i ng t o w hether I Ei l cos 8 > I � I or I Ei l cos 8 < I VJ For t h e condition of F i g . 3 . 10(a) the generator is overexcited a n d supplies reactive power Q to the syste m . Thus, from the system viewpoint the mach ine is acting l i ke a capacitor. Figure J . I 0( b )

3. 4

gI

of a n

(a)

107

gI

ea) FIG U RE 3 . 1 1 Phasor d i a g r a m s

R EAL AND REA CfI V E POWER CONTROL

(6)

ovc r n c i t e d a n d

(h)

u n d c rcxci t e d sy n c h ro n o u s m o t or d r aw i n g C lI rre n t

a n d C() n s t �l l l t pow e r a t cOlls t a n t t e r l 1 1 i l l ; d vol t a g e .

fa

is for a n u n d c rexci ted gene rator supplyi ng the same a mo u n t of rea l power a n d a l e a d i n g current to the system , or i t may be considered to be d rawi ng l aggi n g c u r r e n t fro m t he syst e m . Th e underexcited gener<1 tor d raws reactive power fro m the system and i n this respect acts l i ke a n i n ductor. The reader is encouraged to explain t his a ction in terms of the a rm a tu re react ion discussed i n connect i on w i t h Eqs. (3.24) and (3.25). Figure 3 . 1 1 shows ove rexci ted and underexcited synchronous motors d raw­ i n g t h e s a me real power at t h e s a me term ina l vol t a ge. The overexcited motor d raws leading current and a cts l ike a capacitive circuit when viewed from the n e twork to which it suppli e s reactive power. The unde rexcited motor draws l agging current , absorbs rea ctive power, a nd i s acting l i ke an i n d uctive circ u i t w h e n viewed from the ne twork. B riefly t h e n , Figs. 3 . 1 0 a nd 3. 1 1 show t ha t overexcited generators and motors supply reactive power t o the system a n d underexcited generators and motors absorb reactive power from the system . Now we t u rn our a t t e n tion to re (l l powe r P, which is con t ro l l ed by ope n i n g o r c losi ng t h e valves t hrou gh which steam (or water) enters a turb i n e . I f t h e powe r i nput t o t h e gene rator is i ncrease d , t h e rotor speed w i l l s t art t o increase, a nd i f the field c urrent 1[ and hence l Ei I a rc he ld cons tant, t he a ngle 0 b e tween Ei and V, w i l l i ncrease . I ncreasing 0 resu l t s in a l a rger I f l eos (j , as may be seen b y ro tati ng the ph asor Ei cou n terclockwise i n Figs . 3 . 1 0( a ) a n d 3 . 1 O( b ) . The genera tor wit h a l a rger 0 , t he refore, del ivers more power t o t h e n etwork; exerts a higher cou n tertorq u e on the prime mover; a n d h ence, the i npu t from the prime mover i s reestablished at the speed correspon d i ng to t he frequency of the i nfinite bus. Sim ilar reason ing applies also to a motor. The dependence of P on the power angle 0 is also show n as foll ows . I f a

and

J

108

C HAPTER 3

. where � a n d

Ei are expressed in volts to n e u tral or in per unit, then I Eil� - I V;I and THE SYNCHRO N O US MACH I N E

/ a* -

- )Xd

Therefore, the complex powe r d el ivered to t h e system at t h e genera tor is given by

S = P + j'Q = VI f* a

=

of the

I V I I E -I / - o - I V, 1 2 _I _' = � ==- j'X

I VI I I E; I( cos 8

-j

_ _ _

cl

=

t e rm i n a l s

( 3 . 36)

sin

8)

-

IV/

( 3 . 3 7)

-jXd

The rea l a n d imaginary parts of Eq. ( 3.37) are

( 3 .38)

E;

W he n volts rather than per-un i t values are substituted for V; and Ei i n Eqs. (3 .38), w e must b e careful t o note t h a t V; a n d are line-to-neutral voltages and P and Q will be per-phase quan tities. However, l in e-to- l i n e voltage val u es substituted for V; and E i will y ield total three-phase values for P a n d Q . The per-unit P and Q of Eqs. (3. 38) are m u l tiplied by base three-phase m eg avol tamperes or base megavoltampe rcs per phase depend ing on whether tot a l t hree-phase power o r power per phase i s wanted . Equation (3.38) shows very clea rly the dependence of P on the power angle 8 if I E ; I and I V; I a re consta n t . Howeve r, i f P and V; a re consta nt, Eq . (3 .38) s hows that 0 must d ecrease i f I E; I is i ncreased by boosting the d c field excit ation. With P const a n t i n Eq. (3 .38), both a n increase in I E, I and a d ecrease i n 0 mean that Q will i n c reas e i f i t is a l ready posi t ive, or i t will d ecrease in magnitude and pe rhaps become pos itive i f Q is al ready negative b e fore the field excitation i s booste d . These operating characteristics of the generator are m a d e graph i c a l l y evid ent in S e c . 3 . 5 .

LQ:

The generator o f Example 3 . 1 has synchronous reactance Xd = per unit and is connected to a very large system. The terminal vol tage i s p e r u n i t a n d t h e generator i s supply i ng to t h e system a current of 0.8 p e r unit a t 0.9 power-factor l aggi n g. Al l p e r-unit values a r e on the mach i n e base . Neglecting resistance, find the m agnitude a n d a ngle of the synchronous i n te rnal voltage Ei, and and Q de livered to the i nfi n i te bus. If the real power output of the generator remains constant but t h e excitat ion of the generator is i n;:reas e d Exa mple 3.3.

1 .724 1 1 .0

P

(a)

3.4

by anu (b) bus vol r a e ,

20%

REAL AND R EACTIVE POWER CONTROL

E;

decreased by 20%, fi n d the angle 0 between Q d e l i v ere d to the bus by the generator.

g and

Solution. The power-factor angle is e

0.9 25.8419° synchro nous internal vol tage given b y Eq . (3.31) is

= =

Equ a t ions 0 .3 8 )

1 .0LQ:

+

jl .724 1

1 .6012 j l .24 1 4 +

give t h e P a n

d

=

i Vl l

- ( I E I cos 0 Xd

r

I V: 1 )

-

--

I V, I I E, I Xd

sin 0

i5 t h e:

new value

Q

-

=

of

l .0 =

=

1 .0 -1 .724 1 [ 1 .20

X

p e r u ni t

t h .: g e n e ra t o r.

--

l V; I I E, 1 XJ

--

-

sin -

X

l

.

- sin

�

c o n st a n t gives

1 .2 X 2.0261 1 .7241

( 0.72

sin

l .724 1 1 .20 2 .0261 X

X

era t o r i s

)

0.72

i5

=

=

30.701 6°

2 .0261 cospO.70 1 6° ) - 1 .0J 0.6325 per unit

( b ) Wi th e x c i t a t i o n d ecre ased -

and so

=

Q s u p p li e d by the g e n

-

lagging,

term inal

1 .0 2.0261 s i n 37.7862° 0.7200 p e r u n i t 1 . 724 1 l.0 l .7241 ( 1 . 6012 1 .0) = 0.3487 p e r u n i t

20% w i t h P a n d

exci t a t i o n b y

-

=

the

0.8! 25 .84 1 9°

2.0261/37.7862°

x

=

=

x

( a ) I n creasi n g

anu

COS - I

Q output of

p =

Q

=

a nd

0

=

o =

=

20%, we o n t a i n l . ()

X

sin - }

G.oD

)(

2.0261

1 .7241

( 0 .72

1 .7241 0 .80 X 2.0261 X

s i n i5

)

=

=

0 .72

49.98270

109

the

110

CHAPTER 3

THE SYNCH RONOUS M ACH I N E

a n d the value

Q=

of

Q

1 .0 1 .7241

now supp lied

b y the

generator is

[0 .80 X 2 .0261 cos ( 49 .9827° ) - 1 .0]

=

0.0245 per u n i t

Thus, w e sec how excitation con t ro l s the reactive powe r ou tpu t of t h e ge nerat o r. 3.5

LOADIN G CAPABILITY DIAGRAM

Al l the normal o p e r a t i n g c o nd i t i o n s 0 (" t h e ro u n d - ro t o r g e n e r a t o r co n n e c t e d to a n infi n ite bus c a n be shown on a s i n g l e d iagra m , u su a l ly cal led the loadillg capability diagram o r opera tioll chart 0 (" t h e m ac h i n e . The ch a r t is i m p o r t clll t to the pow e r - p l a n t o p e r a t o rs w h o ,I re r e s D o n s i b l e for p ro p e r l o a d i n g a n d o p e r a l i o n of the gen erator. The chart i s constructed on the a ss umption that the generator has fix ed terminal voltage V, a nd negligib l e a r m a t ure resistance . Construct ion begins with the ph asor d iagram of the mach ine havi ng V; a s the reference phasor, as s h own in Fig. 3 . 1 0( a ). The m irror image of Fig. 3 . 1 0( a ) can be rotated to give t h e p hasor d iagram of Fig. 3 . 12, which shows five l oci passing through the opera t i ng point m . These l oci cor respond to five possible opera t ing modes, in eac h of w hich one parameter of the generati ng u n i t is kept constant. The con stant excitation circle has point n a s center and a radius of length n-m equ a l t o the internal voltage magnitude I E i l , w h ich can be m ai ntained constant by hol d i ng the dc cu rrent If in the field winding constant according t o Eq. (3 . 13).

CONSTANT EXCITATION.

I Ia l .

The circl e for constant arma t u re cu rrent has point ° as center and a rad ius o f length o-m proportional to a fixed value o f l Ia I . Beca u se I Vt 1 i s fixed , the operating points on t his l ocus correspond t o consta n t megavo ltampere output ( 1 � I I Ia l ) from the generator.

CONSTANT

CONSTANT POWER. Active power output of the m a ch ine is given by P = 1 l!; I I Ia l cos (J i n per u nit. S ince I � I i s constant, vertical l ine m-p a t the fixed distance Xd l Ia l eos (J from the vertical axis n-o represents a locus of operating points fo r cons t an t P . The megawatt o utput of the generator is a lways posit ive regardless of the power factor of the ou tpu t .

The reactive power output of the m ach i n e is given b y Q = I VI I l Ia \ sin e in per u n i t w h en t h e a ngle (J i s defined posi tive for lagging power factors. When I Vt I is consta nt, horizontal l i ne q-m at the fixed d ista nce Xd l Ia l l sin e I from the hor izo ntal axis represents a locus of oper ating points for constant Q. For u nity power-factor operation the Q output ? f the

CONSTANT REACTIVE POWER.

Q

I

t r II

---------

I

- .� - - .... ...

� - - - - - - - -

- ....

.... ....

--

I Ic Xd

cos

0:

... ... "

"' ....... ...... .....

,

(a) Constant P ' .....

"

"

...

,

I! /"/

,

,�

- - - - - - -

"

-

l:

>-

- - t- - - ----------------------�-� < q

�

,,' \

, \

/

,,/

-

,

\

/

,

\ \

,

-

\

( e ) Con stant power facto r angle

/ /

/

- - - - -/ ,/ /

/

( 6 ) Constant Q

\ C c ) Co n s t a n t � Ei l \ \� \ \ \ \

,

\

'

'

\ \ \

C d ) Constant : 10 1

�� \ 1

p -------

�

Lead i n g power fact o r

n

Phasor diagram obtained from mirror image of Fi g . 3. 1 0( a ) showing five loci t h rough po i n t m correspond ing to: ( a ) const ant power P; (b) constant reactive power Q; (c) constant internal voltage l Ei I ; ( d) constant armature current I /a I ; (e) constant power-factor angle e. FIGURE 3.12

r

-------__

- .... ...

- ""-

"

"

.... ,

_

"

"

�,

""

"

Constant P

cos 8 1

"

,�, ,

- - - - - - -

,

q

I

/

\ " \ " \ , , \ , \

',

1

:1 l o _'!... t

I EY, I

p

�

---L

,, /

�,

I

Constant power factor angle

,/

,

r - - -------------------"'"-a:- -:... /m .

- - - - - - - - - - I � la

-

,

- .... ... ... .. ....

,

"

'

//

,

\\

\ \ \ \

Constant Q

'�

Constant

\

\

�

\

, ,

/'

I EYt l Xd

nstant ! VI la l

t �

Lagging power factor

� ----------------- --___ --_____ _ ____----L--+-

Real power P

-

n

Phasor di agram obtained b y m u l t i plyi ng (resca li ng) a l l dista nces i n Fig. 3 . 1 2 by I V,/Xd I .

FIGURE 3 . 1 3

Leading power factor

3 .5

LOA D I NG CA P A B I LITY D I A G R A M

LOAD I N G CAPAB I LITY CURVE

0 .8

113

r

0.7 0.6 E�

(1) U � Q.J -. U 0 o � c�

0.5 0.9 PF

0.4

0.95 PF Armature h eating limit

0.3 0.2

k

0.1 -:


C1l > ('Ij OJ Cll

Q 0. 0

E

-c ::J , 'Cll

0.1

0

0.2

0. 1

D--

0.3 '-

'- ...

"-

...

...

0.4

:S E

0.5

>- .
�


0 '-

LL


c :J

�

0 . 58 O.G

"-

0.9

\

\

1

\

l . 0 PF l.0 Prime mover l i mit

00% excitation circle

\

\

0.95 PF

\ \

m'

\ \

0.90 PF

\

n

FI C U R E 3 . 1 4 =

"-

\

Lo a d i l l g c a D a b i l i t y c u rv e ro r

Xci

0.8

'-

0.3 (1) U (lJ
0.7

0.6

I I / P e r - u n it m e gawatts P

0.2

E �

0.5

0.4

P

a

cy l i I1l1 r i c ; d - ro t o r l u rt)()g c n c r a t o r ,

1 72 . 4 % wi t h m a x i m u m t u rb i n e o u t p u t

=

6]:; MV 1\ , 24 kY, 0.9 power fa ctor,

() ] ) M W . P o i n t k

rel a t e s to Ex a mp l e 3 . 4 .

1 14

CHAPTER 3

THE SYNCHRONOUS MACH I N E

generator is zero corresponding to an operating p o i n t o n the horizo n t al axis a-p o For lagging (leading) power facto rs the Q ou tpu t is positive (negat ive) and the operating point is in the half-pl ane a bove ( below) the lin e o-p . CONSTANT POWER-FACTOR. The r a d i a l l i n e o -m corresponds to a fixed power-fa ctor a ngl e () between the arm'l ture curre n t fa and te rminal voltage � . In Fig . 3 . 1 2 t h e angle e is for a l agging power-factor load. When 0 = 0° , the power factor is un ity and t h e opera t ing poi nt is actually on the horizontal axis o-p . The half plane below the horizo n t a l axis a p p l i c s t o lea d i ng power factors. Figu re 3 . 1 2 is most usefu l w h e n t he axes arc scaled to in d i c a t e t h e P and Q load i n g of the generator. Accordi ngly, we rea rrange Eqs. (3.38) to read

p =

I Ei l l V: 1

Since sin 2 0 + cos 2 0

x" =

.

S l l1 a

( 3 . 3� )

1 , squari ng each s i d e of Eq. (3.39) and add i n g give ( 3 .40)

wh ich h a s the form of ( x - a ) 2 + ( y - b) 2 = r 2 for a circle of center (x = a , y = b ) and radius r . The locus of P a n d Q is therefore a circle of 2 radius I Ei l I � I /Xd and center (0, - I V, 1 /Xd ). This circle c a n be obtained by mul tiplying the length of each phasor i n Fig . 3 . 1 2 by I � I /Xd or, equivalen tly, by rescaling the diagram to conform to Fig. 3. 1 3 , which has axes l abeled P horizontally a n d Q vertically from the origin a t point o . On the vertical axis of F ig . 3 . 1 3 the length o-n equals I � 1 2/Xd of reactive power, w here � is the term i n a l voltage. Usu a l ly, the loading d iagram is constructed for 1 � 1 1 .0 per u nit, i n wh ich case length O-I ! represents react ive power equal to l /X" per unit. So, l ength o-n i s the key to setting the sca le for the real and reactive power on th e P and Q axes . The loa ding chart of the synchronous generator can be made more practica l by taking account of the maximum permissible heating ( I 2 R losses) i n t h e armature and t h e fi eld windings, a s wel l a s the power lim its of the prime m over a n d heating in the a rma ture core . Using the exa m pl e of a cyl i n d rical­ rotor t urbogen era t ing u n i t rated 635 MYA, 24 kY, 0.9 power factor, Xd = 172.4 1 %, let u s demonstrate the procedure for constructing the loading capabil­ i ty diagram 0 f Fig . 3 . 1 4 as follows: =

•. •

Take I � I = 1 . 0 per u n i t on the rated-vo l tage base of the machine. Using a convenient volt ampere sca le, m ark t he poin t n on the vertical axis so that length o-n equals 1 /Xd in per u n i t on the rated base of the m achipe. I n

3,5

LOAD I N G CAPAB ILITY D I A GR A M

115

our example Xd 1 .724 1 p e r u n i t , a n d s o t h e length o-n i n Fig. 3 . 1 4 corresponds t o l /Xd = 0.58 per u n i t o n the vertical Q-axis. T h e s a m e scale obviously appl ies to active power P i n per unit on the horizontal axis. Along the P-axis, m ark the d istance corresponding to the m aximum power output of the prime m over. For present p urposes t h e meg awatt l i m it of the t u rbine is assumed in Fig. 3. 1 4 to be 1 .00 per u nit on the rated m egavoltam­ pere base of the m ach i n e . D raw t h e vertical l ine for P = 1 .00 per unit. M ark the l en gt h o-m = 1 .0 per u n i t on t h e rad ial l i ne from the origin at t h e r a t e d powe r-factor a n gl e fJ , which i n t h is case equals cos - 1 0.90. With 0 a s center a n d l ength o-m a s rad ius, d raw t h e per-unit megavoltampere circular a rc correspond ing to t h e armatu re-cu rre n t l i m it. Construct the a rc m -r of m ax i m u m pe rmiss ible excitation using n as center a n d d istance t/ -m as ra d i us . Th is c i rcu l a r a rc corre sponds to t h e m axi m u m fiel d-cu rrent l i mi t . T h e co n s t a n t exci t <1 t ion c i rcl e w i t h ra d ius o f l e ngth o-n u s u a l l y d di n e s 1 0 0 % o r 1 . 0 pe r-u n i t exc i t a t i o n , a n u so rig. 3 . 1 4 s hows the fi e ld-current l im it occu r r i n g a t 2 .340 per- u n i t exc i t a t io n , t h at is, (length r-Il ) /(l e n g t h o-n ) o n the Q- axis, An und erexc i tation l i m i t also app l i es at l ow l evels of excitation when vars are he i ng imported from t h e system to t h e m ach ine. It is determ ined by the manufactu rer's design as d iscllssed below. =

•

•

•

•

I n F i g. 3.1 4 poi n t m corresponds to the megavoltam pere rating o f the gene ra tor at rated power- factor l agging, The m ach i n e designer h as t o a rrange sufficient field cu rrent to su pport overexcited ope ration of the generato r at rated point m . The l evel o f t he fie l d curre n t is l im ited to this m aximum value along the c i rcu l a r arc m -r , a n d the capab i l ity of t h e generator t o d e liver Q to t h e system is thereby reduce d . I n actu a l i ty, m ach ine satura tion decreases the v a l u e o f t h e synch ronous reacta n ce Xd , and for t h is reason most m an u facturers ' cu rves depart fro m t h e t h eoretical fi e l d - heating limits described h ere. The m i rror image of m is t h e operat i n g poi n t m' in the underexcited r e g i o n , Power-pl a n t operators try t o avoi d opera ting c o nd i t i o ns i n the u nderex­ c i ted region of the capab i l i ty cu rve fo r two d i fferent reasons. The fi rst rela tes to steady-state s t ab i lity of the system a n d t h e second rel at es to overheating of the m achine itself. Theore t ica l ly, the so-ca l l ed sIet1dy -state sta bility limit occurs when the angle 8 b e tw e e n E j and VI i n F igs. 3 . 1 2 a n d 3. 1 3 reaches 9 0° . In practice, however, system dyn a mics e n t e r i n t o the p ictu re to complicate t h e determ ina­ t i o n of the actual 'st abi l i ty l i m i t . For t h is reason power-p l a n t operators prefer to avoid u n d e r ex c i t e d mach i n e operat ion whenever possible. As the mach i n e e n ters i n to the u n d e rexcited region of operat ion eddy cu rre n ts i nduced by the syste m in i ron p a rts of � h e armature beg i n to increase. Th e accompa nying 1 2 R h e a t i n g also i ncreases in the e nd region of the a r m a­ t u re. To l i m i t such hea t i ng, t he m ach i n e manufact u re rs prepare capab i l ity

116

CHAPTER 3

TH E S Y N CH RO N O U S M A C H I N E

curves specific to t hei r own designs a n d recommend l i mits within wh ich to operate. In Fig. 3 . 1 4 t h e l i ne m' -n is t h e refore d rawn for il lust rative purposes only. To ob tai n m egawa t t and megava r values for any operating point in Fig. 3.14, the per u n i t values of P and Q as read from the chart are m ul t i p l ie d by the megavolt amp e r e r a t i n g of the m a c h i n e , w h ich i n this case i s 635 M YA . Also, distance n -m of Fig. 3 . 1 4 i s the per- u n i t mega vol tampere va l u e o f the qua n tity I Ej V, I IXd at operating poi n t m , as s hown in Fig. 3 . 1 3 . Therefore, we can calculate t h e value of I E; I in per u n i t o n the rated vol tage base (24 kV in t h is case) b y mult iply i n g l e ngth n -m ( ex pre s s e d i n per-u n i t vol tamperes) b y th e per- u n i t ratio Xd l 1 V, I , or s imply by Xd si nce I � I = 1 . 0 per u n i t i n Fig. 3 . 14. Conversi o n to k ilovolts t h e n req u i res m u l t i p l ica t ion by the vol tage rating of t h e machine i n kilovolts. I f t h e act u a l t e rm i n a l vo l t age I V; I i s n o t 1 . 0 D e r u n i t, t h e n t h e pe r - u n i t value I IXd assigned to d i st a n ce o-n of Fig . 3 . 14 h a s to be changed to I V; 1 2 IXd i n per u n it, as shown i n Fig. 3 . 1 3 . This c h a n g e a l ters the scale of Fi g . 3 . 1 4 by 1 � 1 2, a n d so the p er- u n i t P a n d Q r e a d i n g s from the cha rt must be firs t m ul t i p l i ed b y I V; 1 2 i n per u n i t and t h e n by the m egavol tampere base (635 M Y A in this case) in order to give correct m egawa t t a n d megavar va l u es for the actu a l operati n g con d itions. For instance, if t h e a c t u a l term i n a l vol tage i s 1 .05 p e r u n i t, then t h e p o i n t n on t he Q-axis of Fig. 3. 1 4 corresponds to the actual va l u e 0.58 X 0 .05)2 = 0 .63945 p e r u n i t o r 406 Mvar, a n d t h e po i n t shown a s 0 . 9 per unit on t h e P-axis has an actual va lue of 0 .9 x 0 .05 ) 2 0.99225 p e r unit or 630 MW.

=

To calculate t h e correct excitation v o l ta g e E; c or r e s p o n d i ng to a n operat­ ing poin t m when the termi n a l vol tage i s not exactly equ a l to i ts rated vol tage, we cou l d first m u l tiply l ength l1 -m o b t a i n e d d irect ly from Fig. 3 . 1 4 by I V; 1 2 i n per u n it t o correct the scale a n d t h e n b y the ra t io Xul I V; I i n per u n i t to convert to I E; I , as a lready d iscussed . The n e t res u l t is t h a t l e n g t h n - m obt a i n e d d irectly from F i g . 3 . 1 4 w h e n m u l t i p l i e d b y t h e a C l u a l p e r- u n i t v a l u e of the product Xd X I VI I y i e l ds the correct p e r- u n i t va lue of I E; ! . T h e n , i f p h ys i cal units of k i lovolts are desired, mu lt i plica t io n by the rated k ilovo l t base o f t h e machi n e follows. I t is i mport a nt to n ote t hat t h e power-fa ctor a ng l e f) a n d inter nal a ngle 8 a r e t h e same before a n d a fte r the resca l i ng si nce t h e geometry of Figs. 3 . 1 2 and 3. 13 i s preserved . The reader should note, however, that the operati n g constrain ts forming the bo u n d a ry of t h e opera ting region of the chart are p hysical limits. So, the b ou n d a ry of the operating region may be affected once the scale is a ltered . The fol lowi n g example ill ustrates t h e p rocedures. Example 3 .4. A 6 0-Hz t hree-p h as e g e n erator r a t ed a t 635 M VA, 0.90 powe r factor, 24 kV, 3 600 rpm h a s t h e operat i n g chart shown i n Fig. 3 . 1 4 . The gener a tor

is d el iveri n g 4 5 8 .47 M W a nd 1 14 . 62 Mvar at 22. 8 kV to a n i n fi ni t e bus. Calcu l a t e t h e exci t a t i o n

vol tage

E;

u sing

(a)

the equ ivale nt

circ u i t of

Fig .

3 . 8( a )

and

( b ) the I

3 .6

THE TWO-AXIS MACHINE MODEL

loading d i agram of Fig. 3 . 1 4. Th e synchronous reactance is Xd on the machine b ase and resi s t ance is n egligible.

=

1 17

1 .724 1 per unit

Solution. I n the calc u l ations which follow all per- u n i t val u es a re based o n the

megavoltamp ere a n d ki lovolt rati ngs of the m acp.. i n e .

(a) Choosing the term i n a l

� P

+

iQ

J(l

Ei

=

vol t age a s the refe re n ce p hasor, w e h ave

22 . 8 /

--L...i?: 24 .0 45 8 .47

=

(\0

+

/ 0 . 95LJ?: p er u n i t {"\O

=

=

635 0 . 722 - j O. 1 805

=

=

=

=

=

0 .95�

�

0.722 + i O . 1 805 per

j 1 1 4 .62

+ jXd I(l

=

0 .76 - jO . 1 9 p e r u n i t

0 . 95�

1 .2776 + j 1 .3 1 03

=

unit

+ j 1 .724 1 ( 0 . 7h

- jO . 1 9)

1 .830/ 45 .72390 per

u ni t

43 .920/ 45 .7239° kV

( 6 ) The poi n t k corresponding to the a c t u a l operating con d i t i o n s c a n be located on t h e ch a rt of Fig . 3 . 1 4 as fol lows :

0 . 722 + i O . 1 805 ---�-- =

0 . 95 2

+

0.8 + iO.2

per u nit

0 . 78 2 = 1 . 1 1 73 per u n i t w h e n calcu l a t e d o r The dista nce n -k equ als ../0 . 8 2 measured on t he scale of t he c h a r t o f F i g 3 . 14. T h e actu a l v a l u e o f I Ei l i s t h e n .

co m p u t e d as

l EI I 3 .6

w h i c h is t h e s a m e

as

=

( 1 . 1 1 73

x

0 . 95 : )

1 . 724 ] 0 . 95

=

o b t a i n e d a b ove . The a n g l e 0

1 .830 per u n i t

=

4 5° c a n be e a s i l y m e asured.

THE TWO-AXIS MACHINE M O D EL

The round-rotor theory already d eveloped in this chapter gives good results for the steady-state performance of the synchronous machine. However, for tran­ sien t analysis we need to consider a two-axis model . In this section we i ntrod uce the two-axis mode l by means of the equations of the salient-pole machine i n which t h e ai r gap i s much narrower along the direct axis t h a n a long the

118

CHAPTER

3

T H E SYNCH RONO US MACH I N E

The largest generating u nits are s team - tu rb ine ­ driven alternators of round-rotor construction; fossil-fired units have two pol es and nucl ear units have four p ole s for reasons of economical design and operational efficiency. Hyd roelectric generators usually have more pole-pairs and are of sal ient-pole construction. These units run a t lower speeds so as to avoid mechan ical damage due to centri fuga l forces. The three-phase salient-pole machine, l i ke i ts round-roto:- counterpa rt, has t h ree symmetrically distributed armature windings a, b, and c, and a field winding f on the rotor which produces a sinusoidal flux d istribution around the air gap? In both types of m achines the field sees, so to speak, t h e same air gap and magnetizing paths in the stator regardless of the rotor position. Conse­ quently, the field winding has constant sel f-inductance Lff o Moreover, both m ac h i n e t y p e s h ave t h e s a m e cos i n l l s o i d a l m u t u a l i n d u ct a n c e s r il l ' l ' hI ' a n d L ei with t h e a r m a t u re p h a ses a s g i v e n by Eqs . ( J .4 ) . A d d i t i l l l1 ; t l i y , t h ro u g h o u t each revolution of the rotor the self- i n ductances L a a , L b b , and L : : o f t h e s t a t o r windings, a n d t he mutual inductances L a b ' L b O and L e a between them, are not constant i n the salient-pole machi ne but also vary a s a function of the rotor angular d isplacement (Jd . The flux linkages of phases a, b, and c are related to the currents by the inductances so that quadrature axis b etween poles.

( 3 .41)

These equations look similar to Eqs. 0 . 5 ) for the round-rotor machine but all the coefficients are variable, as summarized i n Table 3.1.3 A s a result, the equations for t h e flux linkages A a , A b , and A e of the salient-pole machine are more difficult to use than their rou nd-rotor counterparts. Fortunately, the equations of the salient-pole machine can °be expressed in a simple form by transforming the a , b, and c vari ables of the stator into correspond i ng sets of new variables, c a l l e d the direct-axis , quadra ture-axis , and zero-sequence quanti­ t ie s which are distinguished by the subscripts d, q , and 0: respectively. For example, the three stator curren ts i a ' i b ' and ie can be transformed into t hree equivalent currents, called t h e direct-axis current i d ' t h e qua dra ture-axis current iq and the zero-sequence current i o . The transformation is made by the matrix P,

2 For further d iscu ssion on the sal i e n t-pole m ac h i n e , s e e P. M . And erson a n d A A. Fo uad, Power System Control and Stability , Chap. 4, The Iowa S tate U n iversity Press, Ames, Iowa, 1 977.

3

The D-

and Q-damper wind ings refe rred

to in

Table

3 . 1 are d i s cussed in SeCo 3 . 8 .

1J"lE lWO-AXIS MACHINE MODEL

3.6

119

TABLE 3.l

Expressions for the inductances of three-phase salient-pole synchronous generator with field. D-damper and Q-damper windings on the rotor.

Stator

{ {

Field windin g: Lrf Self- i n d uctances D da mpc r winding: Lf) Q-dalllJ)t'r winli i nt!: I. v

Rotor

M u t ual-i nductances

-

P'icld /D-wind ing : M, F i dd /Q wi nd i ng : 0

D-winding/Q-wind ing: 0 -

Stator-rotor mutual i n du c ta n c e s

cal led

Park IS tramformatioJl

P

=

h

® ® ©

I

where

@

cos ()tI sin ()eI I

' Ii

®

( ()eI

©

- 1 200 )

cos ( ()eI - 2400 )

s i n « ()d - 1200 )

sin ( ()d - 2400 )

I

1

cos

li

( 3 .42)

{i

which was introduced by R. H. Park in slightly d i fferent form from that shown here. The matrix P has the convenient property (called inverse P '

1

orthogonality )

that its

equals its transp([)se P T, which is found simply by interch'lnging rowS

120

CHAPTER 3

THE S Y NCH RONOUS M A C H I N E

in Eq. (3.42). This property is most i mportant as it ensures that power i n t h e a , b , and c variables is not altered by P , as d iscussed in Sec. 8 . 9 . The currents, voltages, and flux linkages of p hases a , b , and c are transformed by P to d, q, and 0 variables as follows: and columns

ld lq

-

ta

lb Ie

P

10

( 3 .43 )

The P-transformation defines a set of currents, voltages, and fl ux l i nkages for three fictitious coi ls, one of which is the stationary O-coil. The other two coils are the d-coil and t h e q-coil, which rolate i n syn c h ro n ism with the rotor. The d- and q-eoils have constant fl u x linkages \v i t h th e fleld a n d a n y other w i n d i ngs which may exist o n the rotor. Section A . 2 of the Appendix illustrates the detailed manipulations which transfo rm the currents, vol tages, and fl ux l inkages of phases a , b , and c into d-q-O quantities according to Eqs. (3.43). The resulting d, q , and 0 flux-linkage equations are

( 3 .44 )

in

which if is the actua l field curr;e nt, and the inductances are defined by

( 3 . 45 ) Ms

have the same meanings as before and L is a positive number. The i nductance Ld is c al le d the direc t-axis inductance , Lq is called the quadrature-axis in duc tance and L o is known as the zero-sequence inductance. The flux linkages of the field are still g iven by Eq. (3.24), w hich is repeated here i n the form Parameters Ls and

m

,

( 3 .46) Equations (3.44) and (3.46) have constant i nductance coefficients, and thus are quite simple to use. Physically interpreted, these simpler fl ux-l inkage equations show that L d is t h e self-indu ctance o f a n equ iv a len t d-axis armature w i'nding ..

...

,

�

.

_

_

......

..L ' _

_

_

&

_ __

_

�_ �

� ...J

�

�

4-

t... .....

.c. ..-. ' ..-J

..-.. -.. ......J

� . ,

1.-. � ....... t...

...... .,.."

..... ...� ; D {.

r- t I t"t"� n t

l'

trl

3. 6

D i rect

axis

a-axis Quadrature

axis

Rr, L rr

f' FIG U RE 3 . 1 5

121

All c oils rotate together

R otation

b- axis

THE TWO-AXIS M A C H I N E MODEL

c - axis

R e p r ese n t a t i o n of the s a l i e n t -p o l e synch r o n o u s g e n e ra t o r by a r m a t u r e - e q uival e n t d i rect-axis a n d

q u a d r a t u r e - axis coi l s ro t a t i n g i n synchro n i s m w i t h t h e fi e l d w i n d i ng on t h e ro t o r .

produce the same mmf on the d-axis as do the actual stator currents i a ' ib , and i Simil arly, Lq and i q apply to the q-axis. Accordingly, i d and i q give rise to mmfs which are stationary with respect to the rotor. The fictitious d-axis winding and the f winding representing the physical field can be considered to act l ike two coupled coils which are stationary with respect t o each other a s they rotate together sharing the mutual ind uctance kMf ( k = {3/2 ) between them, as shown by Eqs. 0.44) and 0 . 4 6 ) . Fu rthermore, the ReId and the d-axis coil do not couple magnet ically with the fkt itioliS q winding on t he q-axis, which lags t he d-axis in space by 90° . The zero-sequence inductance La is associated with a stationary fictitious armature coil with no coupling to any other coi ls. Under balanced condit ions this coil carries n o current, and therefore we omit i t from further discussion . The d-axis and q-axis coi l s representing the stator windings are shown i n Fig. 3 . 15 , w h i ch s h o u l d b e c o m p a r e d w i th t h e s i n g l e - axis d i a g r a m for t h e r ou n d - r o t o r machine in Fig. 3 .0 . c '

U n d e r s t eady-state ope r a t i n g condi tions t he a rmature o f t h e sal ient­ p o le sync h ro nous genera to r c arries symm et rical s i n u soidal thre e-phase cu rrents Ex a m p l e 3 . 5 .

122

CHAPTER 3

TH E SYN CHRO N O U S M A C H I N E

where ()d = w t + 8 + 90° , a s shown in Eq. (3. 1 4). Using the P-transformation matrix, find expressions for the corresponding d-q-O currents of the a r m a t u r e . Solution.

From Eqs. (3.42) a n d (3 .43) we h a v e ld

if

iq

iO

.J

cos () d sin

1

()d

Ii

cos( () d - 1 200 )

cos ( ed - 240° )

si n e 0d - 1 200 )

s i n e ed - 2400 )

Ii

Ii

1

1

a n d row -by-col u m n m u l t ip l i c a t i o n t h e n gives

I

q

=

Under b a lanced con d itions ia + i b + i c = 0, a n d so = O. By m e a n s of t h e trigonom etric i d e n t i ty 2 s i n a cos f3 = si n ( a + f3 ) + s i n ( a - f3), w e o b t a i n

io

Likewise, w e have

,.

'":,. ...

In t he t h re e p rece d i n g trigonometric expa n s i o n s the fi rst terms i n s i d e t h e b rackets are second harmonic si nusoidal q uant i t i e s which sum to zero a t every i n stant of

3.7

VOLTA G E EQUATIONS: SALI ENT-POLE MACHINE

123

time as in Sec. 3.2, a n d so we obtain

We reca ll from Sec. 3.3 that (Ja = (J + 8, where (J is the phase angle of lag o f m easured with respect to the term i n a l voltage and (Ja is the p ha se a ngle of lag of i a with respect to the i n ternal vol tage of the machine. Accord i ngly,

ia

We c a n

s how

in a si m i l a r m a n n e r t h a t t h e q u a d r a tu re-axis c u r re n t

i '/

=

v'J 1 /0 1 cos 0"

=

If I (, I cos( 0

+

8)

Thus, the expression for i d is exactly the same for the salient-pole a n d the rou n d-rotor machines. The flux l inkages in the field winding are given by Eq. (3 .46), which shows that the di rect-axis current i d is directly opposing the magnetizing influence of the field when ()a = 'TT' /2, and the quadrature-axis current i q is then zero.

3 .7

VOLTAGE EQUATIONS: SALIE NT-POLE

MACHINE

In Sec. 3.6 the flux-linkage equations for the salient-pole machine are remark­ ably simple when expressed in tenns of the d, q , and 0 variables. We now consider other important simplifications which occu r when the P-transformation is also applied to the vol tage equations of the armature. Using the voltage polarities and cu nent directions of Fig. 3.4, l e t us write the terminal-voltage equations for the armature wind ings of the salient-pole machine in the fonn V a = - Ri a -

J /I. 'I

'

dr '

V"

= - Ri h -

d /l. h dt '

vc =

- Ri c

d A.c -

--

dt

( 3 . 4 7)

In these equations the voltages V I P V " , and V c are the line-to-neutral terminal vol tages for the armatu re phases; the negative signs of the coefficients arise because currents i i b ' and i are directed out of the generator. While simple in format, Eqs. (3.47) are in fact very difficult to handle if left i n terms of A. a , A. b , a n d /I. e ' Again, a m u c h simpler set o f equations for the voltages V d ' vq • a n d V a i s fou n d b y e m p l o y i n g t h e P - t r a n s fo rm a t io n . The calculations l eading , to the new a'

c

124

CHAPTER 3

THE SYNCHRONOUS MAC H I N E

volt a ge equations are straightforward but tedious, as shown in Appendix, which yields

Sec.

A.2 of the

( 3 .4 8 )

is t h e rotation a l speed de,,1 dl . Equ ation (3.26) for t h e field w i n d i n g i s not subject t o P- t ra n s ro r rn a t i o n , a n d so a rr a n g i n g t h e d-q-O /l u x - l i n k a g e a n d vo l tage equations according to their axes gives d-axis: w here

w

( 3 .49)

( 3 . 50) q

-

ax l s :

vq =

- Ri q

dAq -

dr

--

+

( 3 .5 1 )

WA d

where k = /3/2 . Equations involving i o a n d A D sta nd alone and are not of interest under balanced conditions. Equa tions (3.49) through (3.5 1 ) are much sim p ler to solve than their correspondi ng voltage and flux-linkage equations in a-b-c v a ri ab les Furthermore, a set o f e q u ivalent circuits m ay b e drawn to satisfy ·. these simpler equations, as shown in Fig. 3 . 16 . The f c i rcuit represent s the actual fi eld since t h e P-transformation affects only t h e armature phases, w hich are replaced by the d- a n d q-coils. We see that the f-coil is mutually coupled to

'

.

-

the d-coil on the d-axis, and so

ft ux-linkage

and voltage equations can he

3.7

VOLTAGE EQUATIONS: SALI ENT-POLE MACH I N E

125

written to agree with Eqs. (3 .49) and (3.50). The fictitious q-coil is shown m agnetically uncoupled from the other two windings since the d-axis a n d the q-axis are spatially in quadrature with one another. However, there is interac­ tion between the two axes by means of the voltage sources - wA q and W A d' w hich are rotational emfs or speed voltages internal to the machine due to the rotation of the rotor. We note that the speed voltage i n the d-axis depends on A q , and similarly, the speed voltage in the q-axis depends on A d ' These sources represent ongoing electromechanical energy conve rsion. No such energy conver­ sion could occur at standstill ( w = 0) since the field and the other d-axis circuit would then act l ike a stationary transformer and the q-axis circuit l ike an ord inary inductance coi l . To summarize, Park's transformation repl aces the physical stationary windings of the armature by: 1 . A d i re c t - a x i s c i r c u i t w h i c h r o t a t e s w i t h t h e

AcId

c i rcu i t

and

is

mutually

coupled to it, 2 . A q u a d r a tu r e - a x i s circ u i t w h i c h i s d i s p l a c e d 90° from the d-ax is, and thus has no mutual inductance with the field or other d-axis circuits al though it rotates in synchronism with them, and 3. A stationary stand-alone a-coil with no coupling to any other circuit, and thus is not shown in Fig. 3 . 1 6 . Figure 3 . 1 6 is most useful in analyzing the performance of the synchronous machine under short-circu it conditions, which we consider in the next section.

�JV���+r - - - - - -

}-----'--O- - -

----

(b) FI G U RE 3 . 1 6 E q u iva l e n t c i rc u i t for t h e s a l i e n t - po l e syn c h ro n o u s g e n e rator:

( b ) w i t h a rm a t u re sho r t - c i rc u i t e d .

( a ) w i t h t e rm i n a l vol tages

Vd and

vq ;

126

CHAPTER 3

THE SYN CHRON OUS MACH I N E

Example 3 .6. Direct current

If

is supplied to the field winding of an un loaded salient-pole synchronous generator rotating with constant angular velocity w . Determine the form of the open-circu i t armature voltages and their d-q-O components. The armature currents open circuit, and so Solution.

l a ' i I> ,

and

lc

are zero because o f the armature

Substitut ing these va lues i n Eqs. ( 3 .49) through (3.5 1 ), we obtain fo r k

/3/2

=

0

=

0

(v A "

=

=

a n d from Eqs . (3 .48) we then fi n d

va =

- Ri "

-

dA " til

+

k w Mf lf

Thus, we see that thc co nstant flux l i n k ages A d on the d-axis give rise to t h e rota tional emf k w Mf lf o n t h e q-axis. S i n ce p - l = p T, i t foll ows fro m Eqs. (3.43) that

3. 8

TRANSIENT AND SUBTRANSIENT EFFECfS

127

Therefore, the steady-state open-circuit armature voltages in the idealized salient­ pole machine are balanced sinusoidal q u a ntities of amplitude J2 IEj l = w Mflf, as obtained previously for the rou nd -rotor machine.

3.8 TRANSIENT AND SUBTRANSIENT EFFECTS

When a fault occurs in a power network, the current flowing is determi ned by the internal emfs of the m achines in the network, by their impedances, and by t he impedances in the network between the machines and the fault. The current flowing in a synchronous machine immediately after the occurrence of a faul t differs from t h at flowing a few cycles later and from th e sustained, or steady-state, value of the fault current. This is because of the effect of the faul t current in the armature on the flux generating the voltage in the machine. The current changes re l a t ively slowly fro m it s initial v a l u e t o i ts steady-state value owin g to the changes in reactance of thc synchronous machinc. Our immed iate i nterest is in the inductance effective in the armature of the synchronous machine when a th ree-p hase short circuit suddenly occurs at its term i n a ls. Before the fault occurs, suppose that the annature voltages are V a ' V b ' and Vc and that these give rise to the voltages Vel ' v q , and V a accordin g to Eq . (3.43). Figure 3 . 1 6 ( a ) shows the vol tages v d and Vq at the terminals of the d-axis and q-axis eq uivalent circuits. The short circuit of phases a , b, and c imposes the condi tions VII = V b = Vc = 0, which lead to the conditions V d = V = O. Thus, to simulate short-circuit conditions, the terminals of the d-axis q and q-axis circuits in Fig. 3.16(a) must also be shorted. Each of these circuits has a net terminal vol tage of zero when e q ua l but opposite voltage sources are connected in series, as shown in Fig. 3 . l 6(b). In that figure the switches S should be interpreted i n a symbolic sense; namely, when the switches a re both open, the sources - V d and - V q are in the circuit, and when the switches are closed, those two sources are removed from the circuit. The principle of superposi tion can be applied to the series-connected vol tage sou rces, provided we assu m e that the rotor speed W rema ins at its p refault steady-state va lue-for Eqs. (3.49) through (3.5 1 ) are then linear. With both switches closed in Fig. 3 . 1 6( b ), we have the steady-state operation of the machine since the sources V d and Vq then match perfectly the d-axis and q-axis voltages at the terminals just before the fault occurs. Sud denly opening the swi tches S adds the voltage sou rce - V(/ in series with the source V d and - V q i n series with t h e sou rce Vq to produce t h e required short circuits. Thus, the sources - V ri and - Vq a r e those d etermining the instantaneous changes from the steady state due to the sudden short-circuit fault. By superposit ion, we can calculate the fault-induced changes of all variables by setting the external sources vjf' , Vd ' and Vq of Fig. 3 . 1 6( b ) equal to zero and suddenly applying the voltages - Vd and - v q to the u n excited rotating machine, as shown in Fig. 3 . 17. T h e in ternal speed voltages - W A q and W A d are initially zero because flux l inkages with all coils are zero i n Fig. 3 . 1 7 before applying the - � d and - vq

128

CHAPTER 3

THE SY N CHRONOUS M ACH I N E

r-

q-axis

WAd +

=

0

FIGURE 3 . 1 7

Equ ivalent c i rc u i t o f s a l i e n t - p o l e sy n c h ro n o u s g e n e r a t o r r o t a t i n g a t c o n s t a n t , p t: c d w i t h f i e l d shon-circuited. Clo s i n g sw i l c l l e s : I t ( ( ) c o r r e s p o n d s t o s l I d d c l l a p p l i C : l t i o l l ( , : S I H ) r t c i rc l l i t t o �

'

machine t e r m i n a l s .

sources. The flux-linkage changes on the d-axis of Eq. (3. 4 9), which gives

the

machine

af;:

governed by

( 3.52)

where � denotes incremental changes. Since the field winding is a closed physical winding, its flux linkages cannot change instantaneously according to the principle of constant flux linkages. Therefore, setting !:l A f equal to zero in Eq. (3 .52) gives

,and substituting for

!:li

f in the equation for !:l A d yields ( 3 . 5 3)

flux l inkage per unit current I n Eq. (3 . 5 3 ) defines the inductance I..:d , where

The

d-axis transien t

( 3 . 54)

Since (k Mf) 2 / L If is posItIve, Eq. (3 .54) shows t h a t the direc:-axis transien t W Ld is always less than the direct-axis synchronou s reac'tance rea ctance Xd =

3 .8

T R A N S I ENT A N D S U BTR ANSI ENT EFFECT'S

129

Xd = w L d ' Thus, following abrupt changes at its terminals, the synchronous machine reflects in -its a rmature the transient reactance X�, which is less than its steady-state reactance Xd . In defining X�, we assume that the field is the only physical rotor winding. In fact, most salient-pole machines of p ractical importance have damper wind­ ings consisting of shorted copper bars through the pole faces of the rotor; and even in a round-rotor machine, u n d e r short-circuit conditions eddy currents are induced in the sol id rotor as if in damper windings. The effects of the eddy-current damping circu i ts are represented by d irect-axis and quadrature-axis closed coils, which are treated in very much the same way as the field winding except that they have no applied vol tage. To account for the addition of damper wind ings, we need only add to Fig. 3. 1 6 the closed D-circuit and Q-circuit of Fig. 3 . 1 8 , which have self-ind uctances L D and L Q and mutual inductances with the other wind ings as shown. In the steady state the flux linkages are constant between all circuits on the same rotor axis. The D- and Q-circuits are then passive (baving neit her ind uced n o r applied vo l tages) and do not enter into steady-st ate analysis. Under short-ci rcu it condit ions, however, we ca n d e termine from F i g . 3 . 1 8 th e initial d - a x is flux- l i n kage changes result ing from sudden shorting of the synchronous machine with damper-winding effects. The proce­ dure is the same as already discussed. The field and D-dampe r circuits repre­ senting closed phys ical wind ings are mutually coupled to each other and to the d-coil representing the armatu re al ong the direct axis. There cannot be sudden change in the flux linkages of the closed windings, and so we can write for the flux-linkage changes along the d-axis

( 3 .55)

' a re simil a r to Eqs. (3. 52), but they have ext ra terms because of the addi tional se l f- and mutual ind ucta nces associa ted with the D-damper circu i t . Coeffi c i e n t s re flecting stator-to-rotor mutual coupling have the multi­ plier k = V3/2 . Mr re la tes to mutual coupling between rotor-based windings on t he d - a x i s a n d t h u s h a s no k m U l t i p l ier. S o l v i n g Eqs. ( 3 .55) for t1 i f a n d .j, i D in terms of D. i d yields These e q u a t io n s

130

CHAPTER 3

+

TH E SYNCHRONOUS MACHI N E

•

Field wi nding

+

kM

,-!...

d-axis armature equ ivalent w i n d i n g

f'--�--------------� in -+

Rn D - d amper wind ing

=

-I

Vn

Ln

0

+

d-axis

RQ

•

Q-damper w i n d i n g vQ 0 LQ

-I

Vq

=

Equivalent circuit

FIGURE 3.18

q-axis armat u re equivalent winding

q-axis

o f the s a l i e n t-pole synchronous g e n e r a tor w i t h o o e fi e l d wi n d i n g a n d two d am p e r

windings on the rotor.

and

substituting these results into the

!:l. A d

expression direct-axis sub transient inductance L,� , defi n ed b y

of Eq. (3 .55) yields the

( 3 .5 6 )

direct-axis subtransient rea c ta n ce X� , defined as X� = w L:� , is considerably smaller than X�, which means that X � < X � < X d. The reader should check The

data given by t he machine manufacturers in Table A.2 in the Appe ndix to confirm these inequal ities. We should note that similar reactances c an be defined for the q-axis. We have shown that the synchronous m achine h as d ifferent reactances when it is subjected to short-circuit faults at its terminals. Immediately upon occurrence of the short circuit, the armature of the machine behaves with an effective reactance X�, which combines with an effective r e s i s t a n c e d eterm i n e d by the dam ping circuits t o define a direct-axis, short-circuit sublransient timethe numerical

.

3.8

T R A N SIENT AND S U BTRANSIENT EFFECfS

131

constant T:; in the range of 0.03 s . The period over which X� is effective is called the subtransient period, and this is typically 3 to 4 cycles of system freq uency in duration . When the d a m per-winding currents decay to negligible levels, the D- and Q-circuits are no longer needed and Fig . 3 . 1 8 reverts t o Fig. 3 . 16 . The machine currents t h e n decay more slowly with a direct-axis, short-circuit transient time-constant T� d etermined by X� and a m a chine resis­ tance which depends on R f of the field . The period of effectiveness of Xd is called the transient period a nd T� is of the order of 1 s. Finally, for sustained steady-state con di tions the d- and q-axis reacta nces Xd = wLd and Xq = wLq determine t h e performance of t h e s a l i e nt-pole machine, j u s t as t h e synchronous reacta nce Xd applies to the rou n d - rotor synchronous machine in the steady state. The various reactances supp l i e d by the machine manufactu rers are usually expressed in per u n i t based on the n a m e p l a t e rating of the m a ch i n e while time co n s t a n t s a rc g i v e n i n s e co n d s . Tah l c 1\.2 i n t h e A p pe n d i x s e ts forth a summary oj" typ i c a l p a ra m d � rs fo r t h e s y nc h ro n o u s m a c h i n e s of p r a c t i c a l i m port a nce . t h e p e r - u n i t v a l u e o f X:, fo r t h e 6 0 - H z synchronous generator of Exa m p \ c 3 . 1 . Use t h e m ac h i n e rat ing of 635 MVA, 2 4 kV as base.

Exa m p l e 3 . 7 . Ca l c u la t e

Solution . Va l u e s for t h e i n d u c t a n ces o f t h e a r m a t ure and field w i n d i n gs are given in Example 3 . 1 , a n d Fig . 3 . 1 0 s hows Lt/ Ls + Ms ' Th e r e fore, L"

=

Ls + Ms

=

=

2 . 765 6

+

1 .3828 = 4 . 1 4 84 mH

The t ransi e n t i n d u c t a nce [;d is n ow c a l c u l a ted from Eq. ( 3 . 5 4 ) :

=

4 . 1 4�4

-

( /3 / 2

X 3 1 . 6950

4 3 3 . 6 5 69

r ')

= 0 . 67 3 6 m H

a n d t he t r a n s i e n t r e a c t a nce is

Xd

=

w L'd = 1 20 7T X 0 . 6736 X 1 0 - 3 = 0 . 2540 n

T h e impe d a n c e base o n t h e m ac h i n e rat i n g e q ua ls (24 2/635) n so that

Xd Thus,

X:,

0 . 2540 X 6 35 =

242

=

0 . 28 per u n it .

i s m u c h l e s s t h a n t h e syn c h ronous reactance

X"

=

1 .734 1

�er

unit.

132

CHAPTER 3

3.9

SHORT�CIRCUIT CURRENTS

THE SYNCH RONOUS MAC H I N E

When an ac voltage i s a p p l i e d s u d d e n ly across a series R - L c i rc u i t , t h e current w h i ch flows gen e r ally has two compone nts-a d c compo n e n t , w h i c h decays according to t h e t i m e con s t a n t

L /R

o f the c i r c u i t , a n d a s teady-s t a t e

sin u s o id a l l y v a ry i n g compo n e n t o f con s t a n t ampl i tu d e .

A

s i m i l a r b u t more

complex p h e n o m e n o n occurs when a short ci rcu i t a p p e ars s u d d e n ly a c ross the term i n a l s o f a syn c h ronous m a ch i n e . T h e

res u l t i n g p hase cu rrents i n

the

machine will h ave d c compon e n ts, w h i c h c a u s e them t o be offs e t o r asymmetri ­ cal w h e n p l ot t e d a s a fu n c t ion o f t i m e . I n Ch a p . 1 ( ) w e s h a l l d i sc us s how t h e symmetrical p orti o n of t he s e s h o rt -c i rc u i t c u rre n t s i s u s e d i n t h e r a t i n gs o f ci rcu i t b r e a k e rs . Fo r n ow l e t liS c o n s i d e r h ow s h o rt c i rcli i ts a ffe c t t h e reac t a n ces o f th e m a c h i n e . A good w a y to a n a lyze t h e e ffe c t o f' a t h re e - p h a s e s h o rt c i rc u i t a t t h e terminals o f a p re v i o u s l y u n l oaded g e n e r a tor i s t o t a ke a n

osci l l ogram of

the vo l ta g e s g e n e r a t e d i n the p hases o f a t h r e e - p h a s e m a c h i n e a r e d i s p l a ced

the

current i n o n e o f t h e p h ases u p o n t h e occu rrence o f such a fau l t . S i n ce

1 20

the voltage wave o f each phase. Fo r t h i s re ason t h e u n i d i r e ct i o n a l o r d c

e l e c trical d eg r ees from each other, t h e s h o r t c i rc u i t occurs a t d i ffe r� n t p o i n ts on

transient compo n e n t of current i s d i ffe r e n t i n each p h as e . 4 I f the dc compo n e n t

o f current i s elim i n a t e d from t h e c u rre n t o f e ach p hase, t h e amp l i t u d e o f t h e a c component o f e ach p hase cu rrent p l o t t e d v e rsus t i m e , s hown i n F i g . 3 . 1 9 , v a r i e s approximately a ccordi n g to

le t )

=

1

I E · I - + I E· I

w h e re ei =

I Xd

I

Ii I Ei l cos w t E q u a t i o n (3 .57)

( 1

X'd

1

Xd

- - -

)

, c - r / Td

(1 1) "

+ l E I - - - [ - [ I Tt! ( 3 . 5 7 ) I · X"d X'd

is the syn c h ro n o u s i n t e r n a l or n o - l oad vol tage of t h e

the dc remov e d , h a s t h ree compon e n t s , t w o o f w h i c h d e c a y a t d i ffe re n t r a t e s over t h e s u b t r a n s i e n t a n d tra n s i e nt p e riod s . N e g l e c t i n g t h e co m p a r a t ively s m a l l

machi n e .

c le a r ly s h ows t h a t t h e a r m a t ure p h a s e c u rr e n t , w i t h

res is tance o f t h e a r m a t u re , t h e d is t ance o -a i n Fig. 3 . 1 9 i s t h e m ax i m u m v a l u e of the s u st a i n e d s h ort-c i r c u i t current, w i t h t h e rms va l u e I I I given by

III -

( 3 .58)

o -a

If t h e e nvelope o f t h e curre n t wave is exte n d ed back t o z e ro t i me a n d t h e fi rs t few cycles w h e re the d ecre m e n t a p p e a rs t o b e very r a p i d a re neglecte d , t h e

4

1 983 a n d C h a p . 1 0 o f t h i s book.

For further d iscussion of t h e dc compon ents s e e A . E. F i t zger a l d e t a l . ,

McGraw-Hili, I nc., N ew York,

Electric Machinery ,

4th e d . ,

3. 9

S H O RT-CIRCUIT C U R RENTS

133

c

b

a

o

Time

)0

FlGURE 3.19

Curre n t a s a fu n c t i o n o f t i m e for a sy n c h ro n o u s g e n e r a t o r s h o rt -c i rc u i t e d w h i l e ru n n i n g a t n o l o a d . T h e u n i d i rect i o na l , t ra n s i e n t com pon e n t o f cu r re n t h a s b e e n e l i m i n a t e d i n red ra w i n g t h e osci l lo­ gram.

i n t e rcept is the d ist ance a -b . The rms v a l u e of the c u r rent represented by this i n t e rce p t is known as the tran sient current 1 1' 1 , defined by

11' 1 -

o -b

( 3 .59)

Ii -

The r m s va l li e of t h e cu rre n t d e t e rm i n e d b y t h e d is t a nce o-c I n Fig. ca l l e d

t h e suiJtrtlllsiel l t currel l t I !" I , g i v e n b y

I I" I -

I E, I

o -c

3.19

is

( 3 . 60 )

X d"

S u b transient cu r r e n t is oft e n ca l l e d t h e initia l symmetrical rms current , which is more d escriptive because

it

conveys t h e i d ea of n eglect i n g the d c compone n t

a n d t a k i n g t h e r m s v a l u e o f t h e a c component o f c u r r e n t i m m e d i a te l y a fter t h e · occu rrence of the fau l t . Equ a t i o n s p a r a meters Fig.

3 . l9

is

(3.59)

and

(3 . 60)

c a n be used to calcu late t h e

X; o f t he m a c h i n e w h e n an oscillographic r ecor d s u ch as ava i l ab l e . On t h e other h a n d , Eqs. 0.59) and (3 . 60) also i n d i cate the X�

and

,

134

CHAPTER 3

TIlE SYNCHRONOUS MACH I N E

(c)

(b) FI G U RE 3.20

X:i ;

XJ ; C c )

(a) subtransient Xci. Vol tage E, c h a n ges w i t h

Equ iva l e n t ci rc u i t s for a syn c h r o n o u s g e n e r a t o r w i t h i n t e r n a l vo l t a g e E; a n d

reactance

(b)

10.2.

t ra n si e n t r e a c t ance

load a s discussed i n S e c .

s ync h ro n o u s rea c t a nce

m e t h o d o f d e t e rm i n i n g t h e fa u l t c u r r e n t in a g e n c r a tor w h e n i ts re a c t a n cc s a re

kno w n .

If t h e generator i s unloaded w h e n t h e fa u l t occurs, the m a ch i n e i s r e pr e s e n t ed by t h e no-load vo l t age t o n e u t r a l i n series with the p r o p e r rc:ac­ tance . T o c a l c ulate currents for s u b t r a n s i e n t co n d i t io n s , we use r e a c t a n c e , series with the no-load vol t age

Ei,

Xd

X� ,

Xd.

as s hown i n F i g . 3 . 20( b ) . I n t h e

i s u s e d , a s shown i n Fig. 3 . 20( c} The s ub t r a ns i e nt c u r r e n t

is much l arger t h an t h e s t e a dy-state c u r r e n t than

The i n t e r n a l voltage

Ei

in

as shown i n Fig. 3 .20( a ) and fo r t r a n s i e n t

con d itions w e u s e t h e series reac t a n c e steady s t a t e

X�

III

because

X:;

1 1" 1

i s m uch s m a l l e r

i s t h e s a m e i n e a c h ci r cu i t of F i g . 3 . 20 b e c a u s e

the generator i s assumed to b e i ni t i a lly u n l oa d e d . I n C h a p . 10 we s h a l l co n s i d e r how t h e e q u iva l e n t circu i ts a r e a l tered t o acco u nt for l o a d i n g o n t h e m a c h i n e when t h e s hort c i rcu it occu rs.

Two generato rs are con n e c t e d in p a r a l l e l to the l ow-vo l t age s i d e of a t h re e-phase Q - Y t ra n sformer, as shown i n F i g . 3 .2 1 ( a ). G e n e r a t o r 1 is r a t e d 50,000 kYA, 1 3 .8 kY. G e n e rator 2 is ra t e d 25,000 k V A , 1 3 . 8 k Y . Each g e ne r a t o r \ h as a subtransi e n t reactance of 25 % o n i t s own base. The t r a n s former i s rated 75 ,000 kYA, 1 3 . 8 Q / 69Y kY, with a r e a ct a nce of 1 0% . B e fore the fau l t occu rs, t h e vol tage o n the high -vol tage side o f t h e t r a n s fo r m e r i s 6 6 k V . T h e t ra n s form e r is u nl o a d e d a n d t her e is no c i rcu l a t i n g c u r re n t b e t\V een t h e ge n e r a t o rs . Fi n d t h e subtransi e n t eurre n t i n each generator w h e n a t h re e - p h ase short c i r c u i t occurs o n t h e high-vol tage side o f t h e tran s fo r m e r .

E x a m p l e 3.8.

Solution. S e l e c t 69 kY, 75 ,000 k Y A as b ase i n t h e h i g h -vol t a ge c i rc u i t . Th e n , t h e b as e voltage o n t h e low-vol tage side i s 1 3 . 8 kY. Generator 1

X�l

=

0 .25

75 ,000 5 0 , 000

66 E; J

=

-

69

=

=

0 .375 per u n i t

0 . 957 p e r u n i t

SHO RT-CIRCUIT C U R RENTS

3.9

jO.375

jO.10

j O . 75

( a)

p

(6)

( a ) S i n g l e - l i n e d i agram; ( 6 ) react a n c e d i ag r a m for Exam p l e FIGURE 3 . 2 1

Gen erator

135

3.8.

2

X;;2

=

Ei 2

=

0 .25

75 ,000 =

25 , 000

0 . 750 per

0 .957 per

unit

XI = 0 . 1 0 p er

unit

66 69

=

unit

Transformer

Figure 3 . 2 1 ( b ) shows t h e reacta n c e d i agram before t h e fau l t. A t h re e-phase faul t at P is sim u l at e d by closing switch S. Th e i n ternal vol tages o f the two m a c h i nes may be con s i d e red to be in parallel s i nce they are i d e n tical i n m a g n it ude a n d p h ase a n d no c i r c u l a t i n g current flows betwe en t h e m . The e q uiva l e n t p a ra l l e l subtransient r e a c t an ce is

X"d

=

X"(/ 1 X"d 2

as

a

J"

=

v;

=

0 . 375

X ti" l + X "" 2

current in t h e s h o r t c i rc u i t is

Th e re fo r e ,

------ =

ph asor w i t h

£I.

J'Xd" + J'X I

X

0 .75

0 . 375 + 0 . 75

Ei

� Ei l = Ei 2

0 .957

j0 . 1 0

------

j O .25

+

as

0 . 25 per unit r e fe r e n c e ,

the

- -j2 .735 per u n i t

The vo l t age VI on the 6. s i d e o f t h e tran sformer i s 1" X

jXI

c."

( - j 2 .735 ) ( jO . l O)

=

0 . 2735 p e r unit

s u b t r a nsient

136

CHAPTER 3

THE SYNCHRONOUS MAC H I N E

In genera tors 1 and 2 1" I

I� 3.10

-

-

Ej}

-

�

0 .957 - 0 . 2735

}'Xd" l

-­

Ei 2 VI -'XIIJ .1 2

jO .375 0 . 957 - 0 .2735

-

=

-j 1 . 823 per u n i t

--

=

jO .75

-

-j O . 9 1 2 p e r

unit

SUMMARY

Simplified equivalent circuits for the synchronous genera tor a re developed i n this c hapte r for use throughout the rema i n d e r of the text. We h ave seen that the steady-state performance of the synch ro nous m ac hine rel ies on the concept of synchronous reactance Xd , which is the basis of the steady-state equivalent circuit of the machine. In steady-state operation w e h ave observed that the synchronous generator delivers an i ncreasing amoun t o f reactive power t o the system t o which i t i s con n e cted as its exci tat ion is increased. Conversely, as its excitation i s reduced, the synchronous generator furnishes less reactive power, and when u n derexci ted, it d raws reactive pow�r from t h e system. All of these normal ste ady-state operating co ndi tions of the roun d-rotor generator, connected to a l arge system as i f to a n infinite b us , are -shown by the loading capability d iagram of the machi n e . Tra nsien t analysis o f the synchronous generator requires a two-axis m a ­ chine model . W e have seen t h a t t h e correspond ing equat ions involving p hysical a-b-c phase variables can be simplified by Park's transformation, which i ntro­ duces d-q-O currents, voltages, and flux l i n kages. S i m p l ified equiva l e nt circuits which fol low from the d-q-O equa tions of the mach i n e a l low definition s of the s ubtransient reactance X:; a n d tra ns i e n t reactance X:, . S u b transi e n t reactance X� is important in calcul ating currents resulting from short-circuit fa u l ts a t or near syn ch ronous generators, a s discussed i n Ch ap. 1 0. The tra nsient reactance X� is used i n stab i l i ty s tud ies, as d emonstrated i n Cha p . 1 6 . P ROBLEMS

3 . 1 . Determ i n e the h i g h est speed at w h ie h two ge n e r a t o r s m ou n te d on the same s h a ft can be d riven so t h a t t h e frequency o f o n e generator is 60 Hz a n d t h e freq u e n cy of the o t h e r is 25 Hz. How many poles does each m a c h i n e h ave?

3.2. The t h re e-ph ase syn c h ronous generator d escribed i n Example 3 . 1 is operated at 3600 rpm and suppl ies a un i ty power-factor load. If t h e terminal voltage of t h e m achine is 2 2 kV a n d t h e fi e l d current i s 2500 A, d e te r m i n e t h e l i n e curre n t a n d t h e tot a l powe r consumption o f t h e load .

1.0&

3.3. A t h ree-phase round-rotor sync h ro n o u s g e nerator h a s n e gl i gible arm at u re r esis­ t a n ce a nd a synchronous reacta nce Xd of 1 .65 per u ni t . The mach i n e is con n ected d i r ectly to a n i nfin ite bus of vol t a ge per u n i t . Fi n d the i n tern a l voltage Ei

/

of the m achine when it del ivers a current o f (a) 1 .0

137

� per u n i t , (b) L!!:. per P R OBLEMS

1 .0

u n i t , and (c) 1 . 0 - 300 per u n i t to the i n fi n ite bus. Draw p h a so r diagrams depicting the operation of t h e m a c h i n e in each case. 3.4. A t h ree-ph ase rou n d-rotor synchronous generator, rated 1 0 k V , 5 0 MV A h as armat ure resistance R of 0 . 1 per u n i t and synchronous reacta n ce Xd of 1 .65 per u ni t . The machine operates o n a l O-kV i n fi n i te b u s deliverin g 2000 A a t 0 . 9 power-factor lead ing. ( a) Deter m i ne the i n t e r n a l vol tage E, and the power angle 8 of t h e machine. D raw a phasor diagram d epicting i t s opera t i o n . W h a t i s the open-ci rc u i t vol tage o f t h e m ac h i n e a t t h e s a m e level o f exc i t a t io n ? (c) What i s t he steady-s tate short-circu i t c u r r e n t a t t h e same l evel o f excitati o n ? Neglect a l l satu r a t i o n e ffects.

(6)

3.5. A t h ree-phase rou nd-rotor syn c h ro n ous gene rator, rated 16 kV and 200 MVA, h as

n eg l i gi b l e losses a n d syn chronous reactance of 1 . 65 per u n i t . It is operated on a n i n fl n i te b u s having a voltage o f 1 5 kV. Th e i n t e r n a l e m f E j a n d t h e powe r a n g l e 0 o f t he mac h i ne are fou nd to be 24 kV ( l i n e to l i n e ) and 27.40 , respectively. ( a ) Deter m i n e the l i ne c u r r e n t and the t h ree-phase rea l and reactive power b e i n g d e l ivered t o t h e sys t e m . ( 6 ) I f t he m echanical power i npu t a n d t h e fi e l d c u r r e n t of t h e g e n erator a r e n ow changed so that the l i n e c u rre n t of t he m achine i s reduced by 25 % at t h e power factor of part ( a ), fi n d t h e new internal e m f El a n d t h e pow e r a ng l e 8 . ( c ) W h i l e del iveri n g t h e red uced l i n e curre n t o f part ( 6 ), t h e m e c h a n i c a l pow e r i n p u t and the exci t a t i o n a rc further adjusted so t h a t the m ac h i n e operates a t u n i ty power fa ctor a t i t s terminals. Cal cu l ate t h e n e w v a l u e s o f E; a n d 8.

3.6. The t hree-phase synchronous gen erator of Prob. 3.5 i s operated o n an i n fin i te bus of voltage 1 5 k V a n d delivers 1 0 0 MVA at 0 . 8 power-factor l aggi n g . ( a ) Determine t h e i n ternal vol tage El , t he power a n gle 8 , a n d t h e l i n e curre n t o f t h e machine. ( 6 ) If the field c u r r e n t of t h e m a ch ine is reduced b y 1 0 % , w h i l e t h e mechanical power i n put to the m a c h i n e is m a i n tain ed const a n t , determine the new value o f 8 an d t h e react ive power d e l i vered to t h e syste m . ( c ) The p r i me mover power i s n e x t adjusted wit hou t changing t h e excitation so t h a t the machine d e l ivers zero re act ive power to t h e sys t e m . Determine t h e n ew power a n g l e 0 a n d the real power b e i n g d e l ivered to t h e syst e m . ( d ) Wh a t i s t h e maxi m u m reac tive powe r t h a t t h e m a c h i n e can d e l iver i f the l evel of exci tat ion is m a i n t a i ned as i n parts (b) and ( c )? D raw a ph asor d i agram for t h e operation of the mach i n e i n parts ( a ) , ( b), a n d (c). 3.7. S t a rl i n g w i t h Eq . <:1 . ::1 1 ), m od i fy Eq .

')

p =

R-

Q

=

R

2

I�I +

to show t h a t

') { I E; I ( R cos o + Xrl sin o ) - I � I R )

Xd-

1 y;, 1 +

O . ::IR)

2

Xd

.

{ Xi I E, l cos 8

-

I � I ) - R I E; l sin

{j

w h e n t h e sy nchronous generator has nonzero a r m ature resistan ce

} R.

138

CHAPTER 3

THE SYNCHRONOUS MACH I N E

3.8. The three-phase synchronous g e n e r a to r described in Exam p l e 3 . 4 is now o p er a t e d on a 25.2-kV infinite b u s . I t i s fou n d t ha t t h e i n t ernal vol t age m ag n i t u d e = 49.5 kV a n d that the power a n g l e 8 3 8 . 5° . U sing t h e loading capab i l i ty d i agram of Fig. 3 . 1 4 , determine graphically t h e real and reactive power d e l ivered to the system by t h e machine. Verify your a n swers using Eqs. ( 3 .38).

IEi l

=

3.9.

A t hree-phase salient-pole synchronous generator with n egligible armature resis­ t ance has the fol lowing values fo r t h e i n d u c t a nce p a r a m e t e rs spec i fi e d in Tab l e 3 . 1 ;

Ls = 2 .7656 mH Ms

=

1

. 3 8 28

Mf

mH

=

L II

=

3 1 .6950 m H

433 . 6569 roH

Lm

During balanced stea dy-state o p e ra t i o n t h e /I c 1 d c u r re n t curr e nt o f t h e m a ch i n e h ave t h e res p e c t i v e v a l u e s if

=

4000

A

i ll

=

20 ,000 s i n e 0d

-

0 .377 1 m H

=

a n cl (I -phase

30°

)

a r m a t ur e

A

( a)

Using Eq. (3 .41), determine t h e i ns t a n t a n eous values of t h e flux l i n kages A (I ' A b , A c , and A ! when ()d 60° . (b) Using P ark's t ransformation g iven by Eq s . ( 3 .42) a n d (3.43 ), d e t e r m i n e t h e instantaneous values of the fl u x l i n kages A d ' A (I ' a n d A o , a n d the c u r re n t s id, i q , a n d i o when a J 60° . (c) Verify resul ts using Eqs. (3.45) a n d (3. 46)

=

=

3.10. The a rmature of a t hree-phase s a l i e n t -p o l e g e n e r a to r carries t h e c u r r e n t s

(a) Using the P-transform ation m a trix o f Eq. ( 3 . 42), find t h e d i rec t - a x is c u r r e n t and t h e quadrature-axis c u r r e n t ill ' W h a t i s t h e z e ro-se q u e nc e curr e n t io? (b) Suppose t hat the armature curre n ts a rc fa

id) i q , and

=

fi

X 1 000 sin( O"

- 6;, )

id

A

io.

3.11. Calculate t h e d irect-axis synchronous reactance Xd , t h e d i rect -axis transient reac­ tance Xd, and the d irect-axis subtransient reactance X; of the 60-Hz s a l i e n t-pole synchronous m achine with the fol l o wi n g p ar a m e t ers: D etermine

L If = 433.6569 mH

LD

mH

M!

=

Mo = 3 . 1 523

0 . 377 1 mH

M,

=

L s = 2.7656 mH

Ms L ",

= =

1 . 3 8 28

3 1 . 6 9 50 mH

37.028 1 mH

= 4 .2898 mH mI l

P R O BLEMS

139

3.12. T h e single-l ine diagram of an unloaded power system is shown in Fig. 3.22. Reactances of the two sections of the transmission line are shown on the diagram. , The generators and transformers are rated as follows:

1:

Generator

G e n erator 2: Ge nerator 3: Tran sfo r m e r

20 MYA, 1 3 . 8 kY, X:; = 0.20 per u n i t 30 MY A, 1 8 k Y , X; 0 . 20 p e r unit 30 MYA, 20 kY, X:; = 0 .20 per unit 25 MYA, 220Y / 1 3.8Ll kY, X = 1 0 % s i n g l e - p h a se u n its, each r a t e d 1 0 MVA, 1 27/1 8 kY, 3 5 M VA , 220Y 122Y k V , = 10%

=

T] :

T r a n s fo r m e r T2 : T r a n s fo r m e r T3 :

X

X

=

10%

D raw t h e i m ped a n c e d i a g r a m w i t h a l l r ea c ta n c e s marked

in per unit and with e t t e rs to i n d i c a t e p o i n t s correspo n di n g to the single-line d i agram. Choose a l b a s e o f 50 MV A, 1 3 . 8 kV i n t h e c i rc u i t o f g e n e r a t o r 1 . ( b ) S u ppose t h a t t h e sys t e m is u n l oaded a nd t h a t the vol tage throughout the sy s t e m is 1 . 0 p e r u n i t o n bases c h ose n in part { / . If a three-phase short circuit o cc u rs rrom b u s C t o g ro u n d , rind t h e p h asor v a l u e of t h e s h o rt-c i rcuit current ( i n a m p e res) i f each g e n e ra to r is r e p r es e n t e d by i t s sub t r a n s i e n t r e a c t a n c e. ( c ) Fi n d t h e m e g a vo l t a m p e res su p p l i e d by e ac h g e n e r a t o r u n d e r the cond itions o f p a r t ( b ).

(a)

F I G U RE 3.22 O n e - l i n e d i a g r a m for P rob. 3 . 1 2 .

3 . 1 3 . The r a t i n gs

of

t h e g e n e ra tors, m o t o rs, a n d t r a ns fo r m e rs of Fig. 3 .23

G e n erator 1 : Gen erator 2: S y n c h ro n ous m o t o r

3:

Th re e-phase Y-Y t r a n s fo r m e rs : Thre e - phase Y-Ll t r a n s fo r m e r s :

(0) D raw

fo r

20 20 30 20 15

X; 20% X:; = 20% 1 3 .8 kY, X:; = 20% 1 38Y 120Y kY, X = 1 0% 1 3 8Y 1 1 3. 8 Ll kY, X = 1 0 %

M V A , 18 k V ,

M V A , 18 k V , MVA, MV A, M VA,

are

=

t h e p o w e r sys t e m . M a r k i m p e q a n ces i n p er

u n i t . N e g l ect r es i s t a nce and use a base o f SO MY A, 138 kY i n the 40-D l i n e . t h e i m p e d a n c e d i a gr a m

140

CHAPTER 3

TIlE SYNCHRONOUS MACH I N E

(b) Suppose that t h e system i s un loaded and t h a t t h e voltage throughout t h e system i s 1 .0 per u n i t on b ases chosen i n part ( a ). If a three-phase s h o r t circuit occurs from bus C to grou nd, find the ph asor value of the short-circu i t current (in amperes) if each generator is represented by i ts subtransient reactance. (c) Find the megavoltamperes supplied by each synchron ous machin e under the cond itions of part (b). � Y±.

- t �

�

A

__

j40 n

�-

�

..-:J:-·2_0_n£'(-_-y -' C t>

j20 0 y t:>

--.L--.---L..- c

FIGURE 3.23

One-line d iagram for Prob.

3.13.

CHAPTER

4

S ERIES IMPEDANCE OF TRANS MIS S I ON LINES

An

electric transmission l i n e has fou r parameters which affect its ability to fulfill its function as part of a power system: resistance, inductance , capacitance , a n d con ductance. In t h i s chapter we discuss t h e first two of these parame ters, and w e shall consider capacitance i n t h e n ext chapter. The fourth parameter, con d uc­ t a n c e , e x i s t s b e tw e e n con ductors or between conductors and t h e groun d . Con­ ductance accounts fo r t h e l eakage current at the insulators of ove rhead l ine� a n d t h ro u g h t h e i nsu lat ion of cables. S i ncc lea kage at insulators of overh e a d l i n es i s n e g l i g i bl e , t h e co n d u ct a n c e b e t w e e n conductors of an ove rh e a d line is u s u a l l y n e g l e c te d . Another reason for n eglecti ng conducta nce is t h a t since i t i s quite vari able, t h ere i s n o good way of t a k in g i t i n t o a ccou n t . Le a k a g e a t i n su l a tors, the princi pal sou rce of conductance, cha nges a ppreciably with atmospheric condi­ t ions and with the conducti n g pro perties of d i rt that coll ects on the insulators. Corona, which resu lts in l c akage betwee n l ines, is also quite variable with atmospheric conditions. It is fortunate that the effect of conductance is such a n egligible component of shuht admittance. Some of the properties of a n e l ectric circuit can be explained b y the elec t ric and magnetic fields which accomp a ny i ts current flow. Figure 4.1 shows a s i n g l e - ph a se line and i ts associ a ted magnetic and electric fields. The lines of mag n e t ic flux fo rm closed l oops linking t h e circu i t , and the l i nes of' electric flux ori g i n a t e on the positive charges on one con d u c tor and term i n a t e on the

142

CHAPTER 4

SERIES I M PEDANC E OF TRANSMISSION L I N ES

Magnet

FI G U R E

..U

M a g n e t i c a mI e l e c t r i c t i t: k � s a ssoc i a t e d w i t h

a two-w i r e l i n c .

negative c har ge s 0 11 t h e othcr c o n d u ct o r. V a r i ,l l i o l1 o f t h e c u r re n t i l l t h e con ductors causes a change i n t h e number of l ines o f m a g n e l l c n u x l i n k i n g t h e circu i t . Any change i n t h e ft u x l in k i ng a circui t induces a vol tage i n the circui t which i s proport iona l to the r a t e o f c h a ng e of tl u x . The i n d ucta nce of the circuit relates t h e vol tage induced b y c h a n g i n g flux t o t h e ra t e of change of curre nt. The capaci tance which exists between t h e cond uctors is defined as the c harge o n the condu ctors per u n i t of poten t i a l d iffe rence between them. The resistance and i nductance u niformly d istributed along the l i ne form the series impedance. The conductance and capaci tance exi sting b e tween con­ ductors of a single-phase l i ne or from a con d uctor to neutral of a three-phase line form the shunt admittance. Alt hough the resistance , inductance, and capacitance are distribu ted, the equivale n t circui t of a line is made up of l umped parameters, a s we shall see when w e d iscuss them. 4.1

TYPES O F C ON D U CTORS

In the early days of the transmission of e lectric power conductors were usually copper, b u t aluminum conductors have completely replaced copper for ove r­ head l i nes because of the much l ower cost and l ighter weight of an a l um i num conductor compared with a copper con d u ctor of t h e same resistance. T h e fact that an aluminum con d u c t o r h as a l a rger d i a m e t er t h a n a co p p e r c o n d u c t o r of t h e same resistance is also an a dvantage. W i t h a l a rger d iame ter, the l i nes of e l ectric fl ux originating on the cond uc tor will be farther a p a r t at the cond uctor surface for the same vol tage. This means t h e re is a l ower vol tage gradient at the conductor surface and l ess tend ency to io nize the air around the con ductor. Ionization produces the u ndesirable effect called coron a . Symbols identify ing diffe rent types of aluminum cond uctors are as fol lows: AAC AAAC ACSR

a ll -alum i num cond uctors all-aluminum- alloy cond uctors a l u m i n u m co n d uctor, stee l-rei nfo rced ., 1 " rn ;

n"

rn

.... A n

r1 " ,... r n ,-

-:l l l ",,_rp ; n fo n'prJ

4.2

RESISTANCE

143

Al u m i n u m

FIGURE 4.2

Cross section of a s t ee l - r e i n forced conductor, 7 steel

s t r a n d s , a n d 24 a l u m i n u m st rands.

Alu m i num-alloy conductors have higher tens ile stre ngth than the ordinary electrical -cond uctor grade of a l u m i n u m . ACS R consists of a central core of steel stran ds su rrou nded b y l a ye r s of a l u m i n u m st ra n ds. ACA R has a central core of hi g h e r- s t r e n g t h a l u m i n u m s u r ro u nded b y layers of elcct rica l-conductor­ gra d e a l u m i n u m . Al ternate l ayers o f w i re o f a s t randed cond uctor are spiraled i n opposite directions to preven t u nw i n d i n g and to make the outer rad i us of one l ayer coincide with the i n·n er rad ius of t h e n ex t layer. Strand ing provides flexibility for a l a rge cross-sectional area. The nu mber of strands depends on the number o f I"ayers a n d o n whether all the strands a re o f t h e s a m e d iameter. T h e total number of strands in concen t rical ly stranded cabl es, where the total a nnular space is filled with s trands of uniform d i a meter. is 7, 1 9, 37, 6 1 , 91, or more. Figure 4.2 s hows the cross section of a typical steel-reinforced alu m inum cable (ACS R). The conductor shown has 7 steel strands forming a central core, around which t here are two l ayers of aluminum strands. There a re 24 aluminum strands in the two outer layers . The c onductor stranding is specified as 2 4 Al/7 St, or simply 24/7. Various tensile strengths, cu rre nt capacities, and cond uctor sizes are obtained by using different combinations of steel and aluminum. Appendix Table A�3 gives some el ectrical characte ristics of ACS R. Cod e names, uniform t h roughout t h e a l u m i n u m industry, have been assigned t o e a c h cond uctor for easy reference. A type of conductor known as expanded ACS R has a filler such as paper separating the inner steel strands from the outer aluminum strands. The paper gives a larger d iameter (and he nce, l ower corona) for a given cond uctivity a n d tensile strengt h . Expanded ACS R i s llsed for some extra-high-voltage (EHV) lines. 4.2

RESISTA N C E

The resistance of transmission-line conductors is the most i mportant cause of power loss i n a t ransm ission line. The term " resistance," u n less specifically q u a l ified, means effective resistance. The e ffective resistance of a conductor is R =

powe r

�

l o ss i n

)J 1 2

------

conductor ----

n

(4 . 1 )

144

CHAPTER 4

S E R I ES I M PEDANCE OF TRANSMISSION LI N ES

where t h e power is i n watts a n d I is the rms cu rrent in t h e conductor i n amp eres. The effective resistance i s e q u a l to t h e dc resistance of t h e con d uctor only if the distribution of curre n t t h roughout the con ductor i s uniform. We shall discuss nonuniformity of curre n t d istribution b ri e fly after reviewing some fun d a­ mental concepts of d c resistance. Di rect-current resistance is given by the form ula

where

p = I =

A =

Ro

=

pI

- fl A

( 4 . 2)

resistivi ty of con ductor lenoth I::!

cross-sectional area

Any con s istent set of units m ay be used . I n power work in the enited S tates / is usually given in feet, A i n circular mils (cmiI), a n d p i n ohm-ci rcu l a r mils per foot, sometimes called ohms per circular m il-foot. I n S 1 u nits I i s in meters, A in square m eters and p i n ohm-m eters. l A circu l a r mil is the a rea of a circle having a d iameter of 1 m i l . A m i l is equal to 1 0 - 3 in. The cross-sectional area of a solid cyl ind rical con d uctor i n circular m ils i s equal t o the square o f t he d iameter o f the conductor expressed in mils. The n umber of circular mils m u ltiplied by 17/4 equals the n umber of square mils. Since m a n u facturers i n the U nited States identify conductors by their cross-sectional area in circular m i ls, we must use this unit occasiona lly. The a re a in square mill imeters e q uals the area i n circular m ils multiplied by 5.067 X 1 0 - 4. The i n t e rn a t i o n a l s t a n d a rd o r co n d u c t i v i t y i s l i l ; l l o f a n n e a l e d c op p e r. Com me rc ial h a r d d r a w n cop p e r w i re h a s <J 7 . YYo a n d a l u m i n u m h J S 6 1 % o r t h e con d uctivity of s t a n d a rd annea led co pper. A t 20°C fo r hard-d rawn copper p is 1 .77 X 1 0 - H fl . m 0 0.66 n c m i l / h ). Fo r a l u m i n u m a t 2 0 ° C p i s 2. 83 X 1 0 - 1) .n m ( 1 7.00 n c miljft). The dc resistan ce of s t ra nd e d cond u ctors is greater than the value com­ p uted by Eq. ( 4 .2) because s p i ra l i n g o f t h e s t r a n d s m a kes t h e m longer than the conductor i tself. For each m i le of cond uctor the current in all strands except the o n e i n t he center flows i n more than a m i l e of wire. The i ncreased resistance due to s piraling is estimated as 1 % for th ree-strand cond uctors and 2% for concentrically stranded conductors. The variation of resistance of metallic cond uctors with temperature is practically l inear �ver the norm a l range of ope ration. If temperature is p lotted o n the vertical axis and resistance on the horizontal axis, as i n Fig. 4.3, extension -

.

.

lSI

.

is

t h e o ffi c i a l des i g n a t ion for the I n tern a t i o n a l Sys t em o f

U n i ts.

4.2

R ESISTANCE

145

t

R

I T

1

/

/

/

/

/

FIG U RE 4.3

R e s i s t a nce of a m e t a l l i c con ductor as a fu nction of tem­

per ature.

of the straigh t-line portion of the graph provides a conve nient method of correcting res ista nce for changes in tem perature . The point of in tersection of the extended line with the tempera ture axis at zero resistance is a constant of the materi a l . From the geometry of Fig. 4.3

( 4 .3 ) where R l and R 2 are the resistan ces of the conductor a t t e mperatures t 1 a n d f 2 ' respectively, i n degrees Ce lsius a n d T i s the constant de termined from t h e graph . Values o f the constant T i n d egrees Celsius a r e a s fol lows : T

=

{

234 .5 241 228

fo r anne a l e d copper of 1 00% conductivity for hard-d rawn copper of 97 . 3 % conductivity fo r h a rd-drawn alum i n um of 6 1 % conductivity

Uniform distribu tion of c urre n t t h roughout the cross section of a conduc­ tor exists only for d i rect curre n t . As the frequency of a lterna ting current i ncreases, the no n u n i form i ty of d ist ribu t ion beco m e s more p ronounced. An increase in frequency causes non u n i fo rm cu rrent dens ity. This p he nomenon is cal led skin effect . I n a circular co nd uctor the curren t d ensity usually increases from the interior towa rd the surface. For cond uctors of sufficiently large radius, however, a cu rrent density oscillat ory with respect to radial d istan ce from t h e center may result. A s we shal l see when d iscussing i n d u ctance, some l i nes o f magnetic flux exist in side a conductor. Filaments on the surface of a conductor are not linked by internal flux, and the flux linking a filament near the surface is less than the fl ux l inking a filame n t in the i n te rior. The alternating flux induces higher vol t ages acting on the inte r ior fi l aments than are ind uced on fi laments near the surface of the conductor. B y Le nz's l a w the ind uced vol tage opposes the

146

CHAPTER 4

S ERIES I MPEDANCE OF TRANS M I S SION LINES

changes of current p roducing i t , and the h igher i nduced voltages acting o n the inner filame n ts cause t h e higher current d e ns i ty i n filaments nearer t h e surface, a n d therefore higher effective resistance results. Even a t power system frequencies, skin effect i s a significant factor i n large conductors. 4.3

TABU LATE D R E S I STAN C E VALUES

The dc resistance o f various types of conductors is easily found by Eq. ( 4 . 2 ) , a n d the increased resistance d u e t o spira l i n g can be estimated. Te m pe ra t u re correc­ tions a re d et ermined by Eq. ( 4 .3). The increase i n resistance caused by skin effect can be calcul ated for rou nd wires a n d tu bes or solid m a t erial, and c u rves of R /R o a re ava i lable for these simple cond uctors. 2 This information i s not necessary, however, s ince m a n ufact ure rs s upply tables of e l ect rical ch aracteris­ tics of the i r con duct ors. Ta b l e A . 3 is a n e x a m p l e o r s o m e o r t h e ( Li l a a v a i l a b l e . Example 4 . 1 . Tab l es o f e l e c lrical c h aracteri stics o f a l l - a l u m i n u m .\fa rigo!d s t r a n d e d conductor l i s t a d c resistance of 0 . 0 1 5 5 8 D p e r 1 00 0 ft a t 200e a n d a n a c r e s i s t a n c e of 0 . 0 9 5 6 Djmi a t 5 0°C. The conductor has 6 1 s t rands and i t s size is 1 , 1 1 3 ,0 0 0 c m i l . Verify t h e dc resistance a n d fi n d t h e ratio of ae to dc res i s t a n c e .

Solution. At 2 0 0 e

Ro

At a t e m p e r a t ur e

from Eq. (4.2) w i t h an increase of 2 % for s p i r a l i n g

1 7 .0 1 1 13

=

x

1 00 0

X 1 03

X 1 .02

=

0 . 0 1 558 D

per

1 0 00 ft

of 5 0 0 e from Eq. ( 4 . 3 )

Ro R Ro

=

0 .0 1 5 5 8

228 + 5 0

228 + 2 0

=

0 . 0 1746 n p e r 1 000 ft

0 .0956 ------ = 1 . 037 0 .0 1 746 x 5 . 2 8 0

Skin effect causes a 3.7% i n crease in resistance.

4.4 INDUCTANCE OF A C O N D U CTOR DUE TO I NTERNAL FLUX

The inductance of a tran smission line is calculated as flux linkages per ampere. If p ermeabi lity IL is constant, sinusoid a l current produces sinusoidally varying flux in phase with the current. The res u l ting flux lin kages can then be expressed 2 S e e T h e A l u m i n u m Asso c i a t i o n , A luminum Electrical Condu rlnr Handbouk , 2d ed., W a s h i,n g t o n ,

DC, 1 982.

4.4

I N DUcTANCE O F A CONDUcrOR D U E TO I NTERNAL FLUX

147

as a p hasor A , and

L

A

( 4 .4 )

I

I f i , the instantaneous value of current, is subst i t uted for the p hasor 1 in Eq. (4.4), then A should be the v a l u e of the i nstantaneous flux l inkages prod uced by i. Fl ux li nkages are meas u red in weber-turns, Wbt. Only flux lines external to the cond uctors are sh own in Fig. 4. 1. Some of t h e magnetic field, however, exists i n side the cond uctors, as we mentioned w he n considering skin effect. T h e changing l i ne s o f fl u x inside the conductors a lso con trib u te to the induced vol t age of t h e circuit and therefore to the i n ductance. The correct value of ind uctance d u e to i n ternal A ux can be computed as t h e ra tio o f fl ux l i n kages t o cu rre n t b y t a k i n g i n t o account the fact t h a t each l i ne o f i n t e r n a l fl u x l i n ks o n l y ,I fr<'lction o f t he tot a l c u r r e n t . To obtain a n accurate va l u e for the in ductance o f a transm ission line, i t is necessary to consider t h e fl ux inside e a c h conductor as wel l as the external fl ux. Let us consider the l ong cylind rical conductor whose cross section is shown in Fig. 4.4. We assume t h a t the return path for the current in this c onductor is s o far away that i t does not appreciably affect the magnetic field of t h e conductor shown. Then, the l ines of flUX are concentric with the conductor. By Ampere's law the magnetomotive force (mmi) i n ampere-turn s aro u n d a n y closed path i s equal t o t h e n e t c u r r e n t i n amperes enclosed b y the p a t h , a s discussed i n Sec. 2 . 1 . The mmf e q u a l s t h e l i n e integral around the closed p a t h o f the component of the magnetic field i ntensity tangent to the path a n d i s given by Eq . (2.4), now written as Eq . (4.5) : mmf

=

¢ H . ds = J At

( 4 .5 )

//;:-- - -::� � �\ - - -

I I I I I f I I I \ \ \ , ,

"

" ......

_

�x

\ \ \ \ \

x f

"'

,/

/

' -� (' - -Flux

I

...... -- - .....-- _ / ..... /

I I / ' /

/

/

FIGURE 4 . 4

Cross sect ion of a cyl i n d ri c a l con d u ctor.

'

148

CHAPTER 4

H

where

SERIES I M PEDANCE OF TRAN S M I S S I O N LINES

= magnetic field intensity , At/m

s = distance along path , m I

=

curren t enclosed , A

H and I are shown as phasors to represent s i nu soidally a lternating quantities si nce our work here appl ies e q u a l ly to alternating and d i rect curre n t . For si m pl i c i ty t h e current I could be interpreted a s a d irect cu rre nt a n d H a s a real number. We recal l that the dot b e tween H and ds i n d icates t h a t t h e value of H i s the component of the field i n t e nsity tangent to ds. Let the field intensity at a d istance x meters from the center of t h e conductor be designated Hx ' Si nce t h e fi e l d is symmetrical, Hx is constant a t a l l points equid istant from t h e ce nter o f t h e cond u ctor. I f t h e i n tegrat ion i n d icated in Eq. (4.5) is pe rformed a round a circ u l a r path co nce n t ric w i t h t h e con d u ctor at x meters from the center, fIx i s const a n t over the p a th and tange n t to i t . Equ ation (4.5) becomes

' Note t ha t

( 4 . 6) ( 4 .7)

and where

Ix is the current enclose d . The n , a ss u m i ng u niform current density, (4 .8)

I i s t h e total curre nt in t h e con d uctor. T h e n , substit ut i n g Eq . ( 4 . 8) I n Eq . (4 .7) a n d solvi ng fo r Hx ' we obtain

where

( 4 . 9) The

flux d e ns i ty x meters from the cent e r o f the con d u ctor is

where J.L

Bx

= J.L Hx

/-L xI =

2 7T" r 2

Wb/m 2

( 4 . 10 )

i s the permeability of the con ductor. 3 I n the tubular element of thickness dx t h e fl u x d¢ is B times t h e cross-sectional area of the element normal t o the flux lines, the area being dx J

ILr = J-L /

1 n SI u n i ts

J-Lo·

x

the permeabi lity of free space

is

/.Lo

�

4 7T

X

10-7

Him, and t h e r e l a t ive p e r m e a b i l i t y

is

4.5

FLUX LI N KAGES BETWEEN TWO POI NTS EXTE RNAL TO AN ISOLATED CONDUCTOR

149

t imes the axial length. The flux p e r m e t e r of length is

J.LxI

dx

2 d¢ = 2 TT'f

(4 . 1 1 )

Wb/m

The flux l inkages d A per meter o f length, which are caused by t h e flux in the tubular element, a re the p ro duct of the flux per meter o f length and the fraction of the current l i nke d . Thus, dA

=

1T X 2

d e/>

rr r 2

=

j.L lx J

2 TT r

4

dx Wbt/m

( 4. 12)

I ntegrating from the cen te r of the conductor to i ts outside edge to fi n d A inp the total flux li nkages i nside the conductor, we ob tain

A jrll

=

1

r

o

For a relat ive permeab i l i ty of

1,

A int

J.L lx 1 4 2 1T r

L int =

f..L I

=

8 rr

-

= 4 7T X 1 0 - 7

J.L

=

d'(

I '2 1

-

2

X

X

( 4. 13)

Wbt/m H i m, a n d

10 - 7

Wb t / m

( 4. 14)

10-7

Him

( 4. 15)

W e have computed the inductance per u n i t le ngth ( h e n rys p e r m e te r) of a rou n d conductor attri buted only to the flux inside the conductor. Here after, for convenience, we refe r to inductance per unit length s imply as inductance, but we m u s t b e ca refu l to use the correct d i m ensional u n i ts. The validity of com p u t ing the i nternal i n d uctance of a sol i d rou n d wire by the method of partial flux l i nkages can be demonstrated by deriving the internal i n d uctance i n a n entirely d ifferent m anner. Equating energy stored in t h e m a g netic fi e l d wit h i n the co n d uc t o r p e r u n i t length at a ny instant t o L int i 2 /2 and solving for L illt wi ll yield Eq. (4. 1 5 ). 4.5 FLU X LINKAGES B ETWE E N TWO PO INTS EXTERNAL TO AN I S O IATED C O N D UCT OR

As a step i n comp uting i n ductance d u e to flux external to a conductor, let us de r ive an expression for the flux l i n kages of an isol ated conductor due only to that portion of the extern al fl ux which lies b e tween two points at D I and D 2 m e ters from the center of the cond uctor. I n Fig. 4.5 P I and P2 are two such points. The cond uctor carries a current of J A . S ince the flux paths are J

150

CHAPTER 4

S E R IES I M P E D A N C E OF T R A N S M I S S I O N LI N ES

FI C U R E 4.5

A co nductor alld ext e rn a l poi n ts P I and P2 '

concentric circles around the con d uctor, a l l the flux between P I and P 2 l ies within the concentric cyli n drical surfaces (ind i cated by solid circular li n es) w hich pass t h rough Pl and P2 ' At the tubular element which is x meters from the cen ter of the cond u ctor the field i ntensity is Hx ' The mmf around the ele m e n t is

( 4 . 16) Solving that

for Hx a n d multiplying b y

yield t h e fl ux density

J.L

Ex

i n t h e e l e m e n t so

( 4 . 1 7) The

flux dcjJ in the tubular clement of thickness d(p =

j.L /

--

2 rr x

dx

dx

IS

Wb/m

( 4 . 18)

flux l i nkages di\. p e r meter are n u merically e qual t o t h e flux d¢ s i nce fl ux extern a l t o t h e con d u ctor l i n ks a l l t h e c u rr e n t i n t h e cond uctor only once. So, between PI and P 2 the flux l in kages are The

( 4 . 19 ) or for a rel ative p e rm e a b i l i ty

A 12

=

of

1

2

X

10

7

I

D2 I n - Wbt / m D,

( 4 .20) ,

4.6

I N D UCTANCE

OF A SINGLE-PHASE TWO-WIRE

LINE

151

The inductance d u e only to the flux included between P I and P2 i s

( 4 .21) INDUCTANCE TWO-WIRE LINE 4.6

OF

A SINGLE-PHASE

We can now d e termine the i nductance of a simple two-wire l i n e composed o f sol i d rou nd conductors. Figu re 4 .6 shows such a line having two conductors o f rad i i ' 1 and ' 2 ' One cond uctor is t h e return circuit for the other. First, consider only the fl u x l i nkages of the circu i t caused by the current in conductor 1 . A l ine of flux s e t up by cu rrent in conductor I at a distance e Cl u a l to or greater than D + ' f r o m t h e ce n t e r of cond u ctor 1 docs n o t l i n k thc c i rcu i t . A t a d i s t a nce less t h a n D r2 t h e fr" c l i o n o f t h e t o t a l c ur r e n t l i n ked by a l i n e o f fl ux is 1 .0. Therefore , it is l ogical w h e n D is much greater than ' I a n d ' 2 to assume t h a t D can b e used instead of D - ' 2 o r D + ' 2 ' I n fact, i t c a n be show n that calculations made w i t h t h i s assumption are correct even when D is smal l . We add i n ductance d u e t o i n ternal fl u x l inkages determ i n e d by Eq. ( 4. 15) to i n d uctance due to exte rnal fl ux l i nkages determ i ned by Eq. (4 .2 1 ) with ' 1 replacing D 1 and D replacing D2 to obtain "

-

( 4.22) which is t h e i n ductance of the circu i t d u e to the curr e n t i n cond u ctor

1

o n ly.

FI GURE 4.6

Con d u ctors of d i ffe r e n t r a d i i a n d t h e mag­ netic fi e l d d u e to cu r re n t In co n d u ct o r 1

only.

152

CHAPTER 4

S E R I ES I M PEDANCE Or: TRA N S M I S S I O N L I N ES

The expression for inductance may b e p u t in a more conCIse form by factoring Eq. (4.22) and by noting that In £ 1 /4 = 1 14, whence ( 4 .23)

Upon

combining terms, we ob tain ( 4 .24 )

If

we substitute r ; for r 1 f - 1 / 4 , L)

=

2

X

10 - 7

In

D ,

r

1

Him

( 4 . 25 )

The radius r 1 is that of a fictitious con d uctor assumed to have n o internal fl ux but w i th the same inductance as the actual con d uctor of radius r I ' The quantity £ - 1 / 4 is equal to 0.7788. Equation (4 .25) omi ts t h e t e rm accounting fo r i nternal flux but compensates for it by using an adjusted value for the rad ius of the conductor. T h e multiplying factor of 0.7788 , whi c h adj usts the radius in order to accou n t for i nternal fl ux, applies only to solid rou n d cond uctors. We consider other con d uctors later. Since the current i n cond u ctor 2 fl ows in the d i rection opposite to that i n conductor 1 (or i s 1800 out o f phase with it), the fl u x li nkages p roduced by current in conductor 2 considered a lo n e are i n the same d i rect i o n t h rough t h e circui t a s those p roduced b y CU IT e n t i n con d uctor 1 . T h e resul ting fl u x for the two cond uctors is determined by the sum of the mmfs of both cond uctors. For constant permeability, however, the fl ux linkages (and l ikewise the i n d ucta n ces) of the two con d uctors considered separately m ay be a d d e d . By comparison with Eq . (4.25), the i n d uctance d u e t o current in con d uctor 2 is

L2

and

for

the

=

2

X

D

10 - 7 In -, H i m r2

( 4 .2 6 )

complete circuit ( 4.27) .

If r�

= r; =

4.7

FLU X L I N KA G ES OF O N E CONDUcrOR IN A G R O U P

153

r' , the total i nductance reduces to L =

4

X

10-7

In

D -

r'

( 4.28)

Him

This value of inductance is someti mes called the inductance per loop meter or per loop mile to distinguish it from that component of the inductance of the circui t attributed to the cu rrent in one conductor only. The latter, as given by Eq . (4. 25) , is one-half the total i n d u ct ance of a single-phase l ine and is called the inductance per conductor .

4. 7

FLUX L I N KAGES O F O N E COND UCTOR IN A GROUP

A more general problem than that of the two-wire l i ne is presented by one cond uctor i n a group o f con d u ctors where the sum o f the currents i n all the con d u ctors is zero. Such a group of cond uctors is shown in Fig. 4.7. Conduc­ tors 1 , 2, 3, . , n carry the phasor currents 1 1 , 12 , 13 " " , In " The d istances of t h ese conductors from a remote point P are i ndicated on the figure as D I P ' D z p , D 3 P " ' " Dn P ' Let us dete rmine A I P I ' t h e flux l inkages of con ductor 1 due to I I i n cl u d ing internal flux l i nkages but excluding all the flux beyond the point P. By Eqs. (4. 14) and .

.

(4.20) (� + 2I l n Dlpl10-7

A I Pl =

2

I

( 4.29)

rI

( 4 .30) The fl ux linkages A I P 2 with con d u ctor 1 due 10 12 but excluding flux beyon d point P is equal to t h e fl u x produced by / 2 between t h e point P and conductor

FIGURE 4.7

Cross-sectional view of of

n

a

group

cond uctors carrying cur­

rents whose sum i s zero. Point

n

F is remote from the con d uc­ tors.

154

CHAPTER 4

SERIES I M PEDANCE OF TRANSMISSION LINES

1 (that is, within the limiting d i st a n ces

D 2 f'

an d

from cond uctor 2), a n d so

D 12

( 4 .3 1 ) The flux l i n k a g e s A I I > w i l h conductor but excluding flux beyond point P is

due

to aI/ ,lie conductors

I II

t h e g ro u p

( 4 . 32) w hich becomes, by expanding the loga rithmic terms and regrouping,

A

IP

=

2

X

( -

1 0 - 7 I I In

+ I, In

1 r�

1

I2 In - +

+

D,p

D I2

+ I, I n

D2P

1

I 3

In D 13

+ 13 I n

+

D3P +

... +

In In

. . . + In

1 DIn

In Dn p

)

( 4 . 33)

Since the sum of all the currents in the group is zero, _

and solving for

I" ,

II +

we obtain

12 + 13 +

...

+

III

=

0

Subst i tu t i ng Eq. (4.34) i n the second t e r m contammg recombining some loga r i t h m i c terms, we have

+ /1 ln -- + /2 1 n -- + /3 I n -- + . . . + DIP

D2 P

D 3P

Dn p

Dn P

Dn P

III

In

In - I I n

Eq. (4.33) and

D (n

-

I)?

Dn P

1

( 4 .35)

Now letting the point P move i n fi n itely far away so that the set of t.erms containing l og a r i th m s of ratios of d istances from P becomes infinitesimal, since

4 .8

I N D U CTA N C E OF COM POSITE-C ON DUcrOR L I N ES

155

the ratios of the distances approach 1 , we obtain

_1_ )

+ I In n D In

Wbt/m

( 4 .3 6)

B y letting point P move infin ite ly far away, we have included all the fl u x l i n kages of con ductor 1 i n o u r derivation. Therefore, Eq. (4.36) expresses all t h e fl u x l inkages of conductor 1 in a group o f cond uctors, p rovided the s u m o f a l l t h e currents is zero. ] f t h e currents a re al ternating, they must b e expressed a s instantaneous currents t o obtain i nstantaneolls flux l i nkages o r as com p l e x r m s values t o u h t a i n the r m s va l u e of n u x l i n k i t g c s as a co m p l e x n u m her.

4.8

I N D U CTAN C E O F C O M P OS IT E­

CO \l D U CTO R LI N ES

S tra nded co n d u c tors cum e u n d e r the general c lassification of compo site conduc­ tors, wh ich means con d u ctors composed of two or more elements or strands electrically in parall e l . W e l i m i t o u rselves to the case where all t h e s t r a n d s a re identical and share the current equal ly. The va lues of internal i n d ucta nce of specific cond uctors a re generally avail abi e from the various m a n ufactu rers a n d can be fou nd in ha ndbooks. The method to b e developed indicates t h e a p proach to the more com plicated problems of nonhomogeneous con d uctors and u n eq u a l division of curre nt between strands. The tnethod i s applicable to the determina­ t ion of inductance of l i nes consisting ofcircuits electri.cally i n parallel since two conductors in par all el can be treated as strands of a s i ngle composite conductor. ' Figure 4.8 shows a single-phase line composed of two conductors. In o rder to be more general, each conduc tor forming one side of the line i s s hown a s a n arbitrary arrangement o f a n indefinite n u mber o f con ductors. The o n ly restric­ tions are that the p a r a l l el filamentS are cyl indrical and share the current equally. Cond uctor X i s composed of n identical, parallel fi la m e nts, e a ch o f which carries t h e current lin . Conductor Y , which i s the return c ircu it for t h e current i n conductor X, is composed of m identical, parallel fi.l a me nts, e a c h o f which carries the current - / Im . Distances between the elements w i l l b e designated by the letter D w i t h a ppropriate subscripts. A pplying Eq. (4.36) t o

'-�---...V'_-_J Condo X

FIG U R E 4.8

'----v---" Condo Y

S i n g l e -ph ase l i n e consis t i n g of t wo compos i t e c o n ductors.

156

CHAPTER

filament a

Aa

=

S E R I ES I M PEDANCE OF TRANSM ISSION LI N ES

4

o f conductor X, we obtain for flux l in kages of filament

2 X 10

-

7

I -

n

( 1

2 X 10-7

_

r�

In

-

� m

(

+

In

I n _1_ Doa'

1 -

Da b +

+

In

1 -

D ac

I n _1_ Doh'

+

+ . . , +

I n _1_ Doc'

1

In

-

Da n

..

+

)

)

+ In

.

a

_1_ ) Da m

( 4 . 37)

from wh ich

Vlb t/ m

( 4 . 38)

Dividing Eq . (4.38 ) by the current lin , we find t h a t t h e inductance of fi lament a is L

Aa

=

a

l in

-

=

2n X 1 0 - 7 I n

VDaa' Dab, Dac' Vr�Dab Dac

l i t .---_ _ _ _

n ,---_

Dam

----

. . .

-- -.

.

Dan

.

Him

( 4 .39)

Similarly, the inductance of filament b is

( 4 . 40)

The

average inductance of the fi la ments of conductor X is La + L b + L c + .

+ Ln L av = -------

n

. .

( 4.41 )

Conductor X is composed of n fi laments electrically in p arallel. If a l l the filaments had t h e same indu ctance, the i n ductance of the con ductor would b e lin t imes t h e inductance of one filament. H e r e a l l the filaments have d ifferent inductances, but the inductance of all of them i n p aral lel is l i n times the average inductance. Thus, the inductance of conductor X is La + Lb + Lc +

' "

+ Ln

n2

Sub stituting the logarithmic expression fo r inductance

of

( 4 .42)

each filament i � Eq.

4 .R

I N D UCTA N C E OF COM POS I TE·CO N D UcrOR LI N ES

157

(4.42) and combining terms, we obtain Lx

= 2 X

10-7

( 4 .43 ) where r� , r� , and < h ave been replaced b y Dall , Dhb , and Dnn , respectively, to make the exp ression appear more symmetric a l . N ote t h a t the numera tor of t h e a rgument of the logari t h m i n Eq. (4.43) i s the mn th root o f m n terms, which are the p rod ucts of t h e distances from a l l t h e n fi l a ments of con ductor X to a l l the In filaments o f cond uctor Y . For each fi l a m e n t in cond uctor X t h e re a re 11/ d istances to filaments in cond uctor Y, a nd there a re !l fi l aments i n con d uctor X. The prod uct of m distances for each of n fi l aments results i n /7111 t e rm s. The m n t h root of the prod uct o f the mn d i s t a nces is cal led the geometric mean distance between conductor X a n d conductor Y . I t i s abb rev i a ted D", or G M D and is also called t h e mutual GMD between the two cond uctors. 2 The denominator of the argu ment of the logarithm in Eq. (4.43) i s the n root of n 2 terms. There are n filame nts, and for each filament t here a re n terms consisting of r' for that fi l am ent times the d i sta nces from that filament to every other fi l ament in conductor X . Thus, we account for n 2 terms. Sometimes r� i s called t h e d istance from fi l a ment a t o itself, especially when i t i s designated as Daa . With this in m ind, the terms u nder the radical in the denominator m ay be described a s the prod uct of the d istances from every fi l a ment in the conductor to i ts e l f and to eve ry other fi l amen t . The n2 root of these ter m s i s called the self GMD of conductor X, a nd the r' of a separate filament is ca l l e d the self G M D of t h e filament. Sel f G M D is also called geometric mean radius , or G M R . T h e correct m a t he m Cl t i c a l expression is s e l f G M D, b u t com mon practice h as m a d e G M R more prevalent. W e u s e G M R i n orJ e r to conform t o t h i s practice a n d ide n t i fy it b y D I n terms of Dil l a nd D, Eq. (4.43) becomes \ .

. Lx = 2

x

to

7

f)1Il In Ds

Hjm

( 4 . 44 )

ThL: reader should compare Eqs. (4.44) and (4.25 ). The inductance of cond uctor Y i s determined i n a similar m a nner, a n d t h e indu ctance of the l i n e i s

L

=

Lx

+

Ly

158

CHAPTER 4

SERIES I M PEDANCE OF TRANS M I SS I O N LI N ES

One ci rcuit of a single-phase tra nsmission line is composed of t h ree solid O.2S-cm-radius wires. The r eturn ci rcuit is composed o f two O.S-em-ra d i u s wires. T h e arrangement of conductors i s shown i n F i g . 4 . 9 . Fi n d t h e i n d u ct a n c e due to the cu rrent in each side of the line and t h e i ndu ctance of the com p l e te l i n e . i n h e n rys per meter (and i n millihe n rys per m i le). Example 4.2.

So/utioll.

Fi nd t he GMD b e t w ee n sides X and Y:

DIII

Then, find

the G M R

=

=

V92

6

____-----X 15 X 1 1 7 3/2

=

1 0 .743 m

for side X

V(O.25

9 ,--------- ----,.----3 X 0 . 7788 X 1 0 - 2 )

X 64

X

1 2 2 = 0 . 48 1 m

6m

6m

10

c

'--y---'

Side X

'--r--' Side Y

FIGURE 4.9 Arrangement

of con d u ctor s for Exa m p l e 4.2.

4.9

a n d for side Y

Ds Lx

= V (O . S 4

=

2

X 10

X 0 . 7788 X 1 0 - 2 )

-7

In

L y = 2 X 1 O - 7 1n

L = Lx

(L

6

=

2

62 =

,-____________________

+

1 0 . 743

6.212

X

0.153

=

X 10-7

H im

0.153

= 8 . 503 X 1 0 - 7

H im

1 0 . 743

1 0 -7

X

1 609

X

Him

103

=

2 .37

OF

TABLES

159

m

0 .4 8 1

L y = 1 4 .7 1 5 X 1 0 - 7

1 4 .7 1 5

X

THE U S E

mH/mi)

I n Exa m p l e 4 .2 t he con d u ct ors i n para ll e l on o n e side o f the l in e are separated by m, a n d t h c d i st ance between the two sides of the l i ne is 9 m . Here t he ca lcul ation of m u t u a l G M D is importa n t . For stranded con ductors the d istance between sides of a l i n e composed of one cond u ctor per side is usually so great that the mutual G M D can be taken as equal to the center-to-center d i st ance with n egl igible error. If the effect of the steel core of ACS R is neglected i n calculating i n du ct a nce, a h igh degree of accu racy results, provided the aluminum strands are in a n even n u mber of l ayers. The effect of the core i s more apparent for a n odd n u mber o f layers o f a l u m in u m strands, b u t t h e accuracy i s good w h e n t h e calcula tion s are based o n the a l u m inum strands alone. 4.9

THE U S E OF TABLES

Tables l isting values of G M R are generally ava i lable for standard conduc tors and p rovide other i n formation for calcu lating i nductive reactance as well as shunt capacitive reactance and resistance. S i nce i ndustry i n the U n ited States conti nues to use units of i nches, fe e t , and m i les, so do th ese tables. The refore, some of our exa mples will use fect and m iles, but others will use meters a n d k ilometers. I nductive reactance rather than i nductance is usually desired. The i n duc­ t ive reactance of one cond uctor of a single-phase two-conductor l i n e is

( 4 .45 )

or

( 4 . 46)

160

CHAPTER

4

S ERI ES I MPEDANCE OF TRAN SM ISS ION LINES

where Dm is the distance between conductors. Both Dm and Ds must be i n the same units, usu ally e ither meters or feet. The GMR found i n tables is a n equ ivalent Ds ' which accounts for s kin e ffect where i t is appreciable enough t o affect inductance. O f course, skin effect is greater a t h igher frequencies for a conductor of a given d iameter. Values of Ds listed i n Table A.3 of the Appendix are for a fre quency of 60 Hz. Some tables g ive values of i nductive reactance in a d dition to GMR. O n e method is to expa nd t h e loga ri thmic t e r m o f Eq . (4.46), a s fo l l ows:

XI,

=

2 .022

x

1 0 - :If In

1

D."

-

2.022

+

10

x

:If I n

Dill D j m i

( 4 .4 7 )

I f both Ds a n d Dill a r c in feet, the fi rs t term in Eq . (4.47) is t h l: inductive reactance of one conductor of a two-conductor l ine h aving a distance of 1 ft between con d uctors, as m ay b e s e e n by co m p a r i n g Eq . (4.47 / w i t h E q . ( 4 . 4 () ) ' The refore, t h e first term of Eq. ( 4 . 47) i s c a l led the inductive reactance a t l -ft spacing Xa' I t d epends on the GMR o f t h e con d uctor and the fre quency. The secon d term of Eq. (4.47) i s called t h e inductive reactance spacing factor Xd . This second term is independent of the type of cond uctor and depends o n frequency a n d spacing only. Table A.3 i ncl udes v a lues o f induc[ive reactance a t I-ft spacing, a n d Tabl e A . 4 l ists va lues o f t h e i n d u ctive reactance spacing factor. Find the inductive reactance per m i le of a single-p:"ase l ine operating at 6 0 Hz. The conductor is Partridge, a n d spacing is 2 0 ft between centers.

Exa mple 4.3.

For this conductor Table A.3 lists D s onc conductor

Solution.

XL

=

=

2 .022

X

1 0-3

X

=

0.02 1 7 ft. From Eq. (4 .46) for

20 60 In 0 . 02 1 7

--

0 .828 fl/ m i

The above calculation i s u s e d only i f Ds is known. Table ..\.3, however, l ists i nductive reactance at 1-ft spacing Xa = 0.465 fl/m i . From Table A.4 the i nductive reactance spacing factor is X" = 0.3635 O/ mi, and so the i n d '-l c t ivc reactance o f one conductor is 0 . 4 65

+ 0 .3635

=

0 .8285 fl/m i

S ince the con ductors composing t h e two s i des o f t he l i ne arc iden tical, the i nductive reactance of the line is 2 XL =

2

X

0 . 8285

=

1 . 657 fl/mi

4. 1 1

I N D U CTA N C E O F TH R E E - P HASE LI N ES WITH U N S Y M METRICAL S P A C I N G

FI GURE 4. 1 0

Cross-sectional view of t h e equ i l aterally t h ree-p h ase l i n e .

s pa ce d

161

conductors o f a

4.10 I N D UCTAN C E O F THREE- PHASE L I N E S WI T H E Q U I LATERAL S PA C I N G

So fa r in our d iscussion w e h ave considered only single-phase l i n es. The e q u a t ions we have deve loped a re qu ite easily adapted, however, to the cal cula­ tion of the ind uctance of th ree-phase l i nes. Figu re 4 . 1 0 shows the conductors o f a t h ree-phase l i ne spaced a t t h e corners o f a n equilateral triangle. If we assume that there is no neutral w i re, o r if we assume balanced t hree- p hase p hasor currents, I + Ib + Ie = O. Equ ation (4.36) dete rmines the fl ux l inkages of cond u c tor a : a

Aa

S ince Ia

Aa and L a

=

�

=

=

2

- ( Jb +

X 10-7

X

1a I n

1

+

Iii I n

Ie ), Eq. (4.48) becomes

2 X 1O - 7

2

( Ds

(

fa I n

�, -

1 0 - 7 In - Him

fa

In

�1

�

-DI

+

Ie I n

-DI I Wbt/m

2 X 1 0 - 7 fa In

D

Os

� Wbt/m

( 4 .48 )

( 4 .4 9 ) ( 4 .50 )

Eq u a tion (4.50) is the same in form as E q . (4.25) for a si ngle-phase line except that Ds repl aces r ' . Because of symmetry, the ind uctances of cond uctors b a n d c a re the ,same a s the i n d uctance o f conductor a . S i nce each phase consists o f only o n e co n d u ct o r , Eq . (4.50) gives t h e i n d u c t a n c e per phase of t h e three-p h ase line.

4. 1 1 l N D U CTANC E O F THREE-PHASE L I N E S WI T H U N SYMM ETRI CAL SPAC I N G

When t h e cond uctors o f a t hree-phase l i n e a r e n o t spaced equilateral ly, the p robl em of finding the inductance becomes more d ifficult. The flux lin kages a n d i n d ucta nce o f each p hase are not t he same. A different i n ductance i n e ach . \,

162

CHAPTER 4 SE RIES IMPEDANCE OF TRANS M ISSION LINES Condo a

Condo

Pos. 2

Condo b

Cond. a

Condo

Pos. 3

Condo c

Co n d o b

Condo a

Pos. 1

Condo b

c

c

FIGURE 4 . 1 1 Transposi t ion cycle.

p hase resul ts i n a n u nbal a nced c i rcu i t . Balance of t h e t h ree ph ases can be restored by exc h a nging the pos i tions o f t h e conductors at regular i n t ervals a long t h e line so that each conductor occ u p i es the o r i g i n a l pos ition of every o t h e r con d uctor ove r a n e q u a l d istance. S uch a n exc ha nge o f conductor posit ions i s called transposition . A complete t ra nsposition cyc l e is s hown i n F i g . The phase conductors a rc designated a , b, a nd c , a n d thc positions occ upied a r e n umbered 1, 2, and 3, respectively. Transposition results in each con ductor havin g the same average inductance ove r the whole cycle. Modern power l i nes are usu a l ly n o t transposed at regular i ntervals al­ though a n i nterchange in the positions of the con d uctors may be made at switching stations i n order to balance the i n d uctance of t h e phases more closely. Fortun ately, the dissymmetry between the phases of an un transposed line is small and n eglected in most calcu l ations of i nductance. I f the dissymmetry is neglected, the inductance of the u ntransposed l i n e is taken as equal to t h e average value o f the inductive reacta nce of one p h ase of the same l i ne correctly transposed. T he derivations to follow a re for tra nsposed l ines. To find the average i nductance of one con ductor o f a transposed line, we first determine the fl ux l i nkages of a con ductor for each p osition i t occup i es in the transposition cycle and then dete r m i ne t h e average fl ux l i nkages. Applying Eq. (4.36) t o co nd uctor ( / 0 [' r i g . 4 . 1 I l o f I n d t h e r l 1 ; l s or e x p r e s s i o n fo r t h e n u x linkages o f a i n posit ion 1 w h e n /; is i n pos i t i o n 2 a n d c i s i n posi t ion 3 , w e o btain

4.1 l.

Aa l

=

2

x

(

1O - 7 Ia

(

1 In _ Ds

+

I" In

With a in position 2, b i n position 3 , a n d

Aa 2 = 2

X

1

10 - 7 Ia I n -

Ds

+

Ib

D 12

+

1

__

c

Ie

In

1

D 23

and, with a i n position 3 , b in position 1 , and

c

+

)

Wbt / m

( 4 .5 1 )

1 Ie In - Wbt/m

( 4 . 52)

i n position

In -

D31 1

__

1,

D !2

)

i n position 2, ( 4,.53)

4. 1 1

I N D U Cf A N C E

OF TH R E E- P H A SE

The average value of the flux l in kages of

With

the rest riction that 1a A"

=

. =

a n d the

a verage

2 X IO-7 ( 3

2 X I ()

'

7

a

163

is

+ J),

= - ( Jb

----

L I N E S WITH U N SYM METRICAL SPAC I N G

1

1 1

] /Ii I n - - I Ii I n D, D 1 2 D �.' D J I

-----

.1

�2 /)2J UJI

I Ii I n

Wbt

( 4 .55)

m

ind uctance per phase is

L

a

=

2

X

10-7 In

D

�

Ds

( 4.56 )

Him

where

( 4.57)

and Ds is the GMR of the condu ctor. Deq , the geometric mean of the three d istances of the u n symmetrical l ine, is the equivale nt equ i lateral spacing, as may be seen by a com parison of Eq. (4.56) with Eq. (4.50). \Ve should note the s i m i l arity between all the equations fo r the in ductance of a conductor. I f the inductance is in henrys per meter, the factor 2 X 1 0 - 7 appears in all the e q u a t i o n s , and t h e denom i nator of t h e logari t h m ic term i s always the G M R of t h e co n d u c t o r . T h e n u m e r a t or i s t h e d i s t a n c e b e tween w i res of a two-wire l i ne, the mutual GMD b e tw e e n si des of a composi te-conductor s i ngle-phase l ine, t h e d istance b etween conductors of a n equi lateral ly spaced l ine, o r the equivalent eq u i l ateral spacing of an u nsymm e trical l i n e . Exa m p l e 4 . 4 . 1\ s i n g l e -c i rc u i t t h re e - p h a se l i n e ope r a t e cJ a t

shown in Fig. 4 . 1 2 . The cond uctors p e r m i l e p e r ph ase.

60 Hz is

a rc ACS R Drake. Fi n d the

FIGURE 4 . 1 2

a r r a nged, as inductive reactance

Arra ngem e n t of c o n d u ctors for l::: x ample 4 . 4 .

164

CHAPTER

4

Solution.

S ERIES I M PEDA N CE OF TRAN S M I S S I O N LI N ES

From Table

Ds = L

XL

Equation

D CQ

0 .0373 ft

= 2

X

1 0 - 7 In

(4.46) m ay

Xd

=

XL =

= �/20

24.8 0 . 0373

= 2 17 60 X 1 609

and by i nter(1ola tio I 1 "

A.3

X

= 1 3 .00 X 1 0 - 7

=

24.8

ft

Him

1 3 .00 X 1 0 - 7 = 0 . 788 .fl/mi per p h ase

b e useu a lso, or

fro m Ta b l e s A.J

XII for

X 20 X 3 8

=

and

A.4

O . : W'J

24.l'i n

0 .3896 0 . 399

+

0 .3896 = 0 . 7886 .fl/mi

4. 1 2 INDUCTANCE CALCULATI O N S FOR BUNDLED C ONDUCTORS

per p hase

At extra-high voltages ( EHV) , that is, voltages above 230 kV, coron a w i t h its . resultant power loss and particularly its i n terference with commu nications is excessive i f the circui t has only one c on ductor per p h ase. The high-voltage g�adient at the conductor in the EHV range is reduced considerably by hav i n g two o r more conductors p e r phase i n close p roximity compa red with the spacing . between phases. Such a l ine is said to be composed of bundled conducto rs. The bundle consists of two, three, or fou r cond uctors. Figure 4. 1 3 shows the a r rangements. The current will not d ivide exactly b e tween the conductors of the bundle unless there i s a transposi tion o f t h e conductors w i t h i n t h e bundle, b u t the diffe rence i s o f n o practical import
F I G U R E 4 . 13

B u n d l e a r r 3 n ge m c n ls.

4. 1 2

I N DUcrA N C E CALCULATIONS FOR O U N DLED CONDUcrORS

individual conductors composing the b u n d l e,

we

165

find, referring to Fig. 4 . 1 3 :

For a two-strand bundle

( 4 .5 8 ) For a t hree-strand bundle ( 4 .59) For a four-strand b u n d l e

D;>

=

V ( D,

]6

x

d X d X

12 d ) 4

L09

=

JD,

X

dJ

( 4 . 60 )

I n comput i ng i nductance using Eq . (4.56), D; of t h e b u n d l e replaces D� o f a single con ductor. To com p u t e De ll ' t h e d ista nce fro m t h e c e n te r o f o n e b u n d l e to t h e center o f another bundle i s sufficie ntly accurate for D oln D bc' a n d Dca. Obtain ing the actu al G M D between conductors of one b u n d l e and those o f a n o t h e r wou ld be a lmost i n d istinguishable from the ce nter-to-center d is t a nces for the usual spacing. Pheasan t .

Exa m p l e 4 . 5 . E a c h con d l l c t or of t h e b u n d l e d -co n d u c t o r l i n e s h own i n F i g . 4 . 1 4 i s

ACSR,

1 , 272,OOO-c m i l

Find

the

k i l o m e t e r ( a n d p e r m i l e ) p e r p h a s e for d

=

i n d uc t ive

re actance

in

ohms

per

45 c m . A l so, fi n d t h e p e r- u n i t s e r i e s

r e a c t a n c e o f t h e l i n e i f i t s l e n g t h i s 1 60 k m a n d t h e base i s 1 00 M Y A , 3 4 5 k Y .

Solution. Fro m Ta b l e A . 3

to m e t e rs .

D" "

X I.

�

,=

�

= Base

D,

= 0 . 04 6 6 f t , a n d w e m u l t i p l y [e e t by 0 . 3 04 8 t o convert

/0 .0466

x 0

2 rr o()

2 x 10

U . 365

x

.

.1048 x (l A S

x

iO

f l j k m per p h ase

0 . 3 65 X 1 . 6 0 9 .,

=

"\

In

O . OSO

X=

0 . 36 5 x 1 60 1 1 90

=

m

l O .m�

-­

(1 . OS

0 . 5 8 7 i1 / m i per p h as e

= 1 1 90 n

Z=

(345)� 1 00

7

=

0 . 0 49 p e r u n i t

166

CHAPTER

4

SERIES I M PEDANCE OF TRANSM ISSION LINES

d

4.13

=

45

FI GURE 4.14 Spacing of cond uctors of a bun­ d l ed -conductor line.

elm

SUMMARY

programs are usua lly ava i l ab l e or written rather easily for calcu l ating i n ductance of a l l k i n ds of l i n e s, some u n d erstanding of the develop­ m e n t of the equations used is rewarding from the standpoint of apprec iat ing t h e effect of variables i n designing a l in e . However, tabul a te d val ues such a s those in Tables A.3 and A. 4 make the calc u l a tions q u i te s i m p l e except for parallel-cir­ cuit lines. Tabl e A.3 a lso l ists resistance . The important equation for i nd u c t a nce p e r p hase of single-circuit three­ phase lines is given here for conve nience:

Although co m p u t e r

L

=

2

10 - 7 I n � H j m per p h ase D

X

Ds

(4.61)

I nductive reactance i n ohms p e r k i l ometer a t 60 H z i s found b y mul tiplying i nductance i n henrys per meter by 2 7T 6 0 x 1 000: XL

or

XL

=

=

0 .0754

X

In

0 .1213

x

In

Dcq Ds

-

Dc<.J Ds

-

fljkm per p hase

( 4 . 62)

fl j m i per ph ase

( 4 .6 3 )

Both D C q a n d Ds must be i n t h e same u nits, usually feet. I f the l i ne has one con ductor per phase, Ds is foun d d irectly from tables. For bundled condu ctors D:, as defined in Sec. 4. 1 2, is subst i tu t e d for Ds . For both single-conductor and b u n dled-co n ductor l ines ( 4 . 64) For bundled-conductor

lines Da b ' Dbc ' and Dca are distances between t h e centers of the bundles o f p hases a , b , a n d c . For lines with one conductor per p hase it i s convenient to determine XL from tables by a dding Xa for the con ductor as foun d in Table A.3 to ){d as fou n d in T abl e AA correspond ing to DC q .

PROBLEMS

167

PROBLEMS

The all-al u m i n um co nductor (AAC) i d e n tified by the code word Bluebell is composed o f 37 strands, e a ch h aving a d iameter of 0 . 1 672 in. Tab les o f characteris­ tics of AACs list an area o f 1 ,033,500 emil for t h is conductor ( 1 emil = (7T /4) X 1 0 - 6 in2). Are t h ese v a l u e s cons i st e n t with each other? Find t h e overall area of the strands i n square m i l l i meters.

4.1.

Determine the dc resistance in o h m s per km of Bluebell a t 20°C by Eq. (4.2) a n d t h e i nformation i n Prob . 4. 1 , a n d check t h e result against the value l is ted i n tables of 0.0 1 678 D. per ] 000 [t . Com p u t e t h e d c resistance i n ohms per kilometer at 50 ° C and compare the resu l t with t h e ac 60- H z resistance of 0. 1 024 D./mi l i sted i n tab les for this cond uctor at 50°C. Exp l a i n any d i fference in values. Assume t h a t the increase i n resistance d u e to s p i r a l i n g i s 2%. A n AAC i s composed o f 37 s t r a n ds , e a c h h a v i n g a d i a m e t e r o f 0.333 c m . Com p u t e t h e d c r e s i s t a n c e i n o h m s p e r k i l o m e t e r a t 75°C. Assume t h at t he i ncrease i n r e s i s t a n c e d u e t o s p i r;t l i n g i s 2%.

4.2.

4.3.

4 . 4 . T h e e ne rgy d e n s i ty ( t h a t i s , t h e c l l e rgy pc r L1 ll i t vo l u m e ) a t a p o i n t i n a m a g n e t ie fi e l d c a n be s h ow n

4.5.

to bl! B 2/2 /-L , w h e r e [] is t h e fl ux d e n s ity a n d /-L is t h e p e rm e a b i l i t y . U s i n g t h i s r e s u l t a n d Eq (4 . 1 0), show that t h e total magne tic fi e l d e n e rgy s t o r e d w i t h i n a u n i t l en g t h of s o l i d c i rc u l a r co n d u c t o r carrying curren t I i s given by JL / 'J. / 1 6 7T . Neglect s k i n e ffect, a n d thus ver i fy Eq . (4. 1 5). The conductor of a single -phase 60-Hz l i n e i s a sol i d rou n d a l u m i n u m w i re h avi n g a d i ameter of 0.4 1 2 cm. The conductor spacing is 3 m . Determine t h e inductance o f the l i ne i n m i l l i henrys per m i l e . How m u c h o f the indu ctance i s d u e t o i n ternal flux linkages? Assume skin effect i s negl i g i b l e .

4.6.

A singl e-phase 60-Hz overhead power l i n e is symme trically supported o n a horizon­ tal crossarm. Spacing between the cent ers o f the condu ctors (say, a and b ) i s 2.5 m . A telephone l i ne i s also sym m etrically supported o n a horizontal crossarm 1 .8 m d irectly below the power l i n e . Spacing between the centers of t hese con d u ctors (say, c and d) is 1 .0 m. (a) Using Eq. (4.36), show t h a t the m u t u a l indu ctance per unit length between circuit a -b and c ircu i t c-d i s given by

Dud [) J,e

Dl lc lJ ,,,,

(b) (c)

4.7.

where, for exam ple, a n c! d .

H e nce, co m p u t e

the

Dud d e notes t h e di stance i n meters betwe en conducto rs m u t u a l i n d uc t a n c e per

a n d t h e t e l e p h o ne l i n e . Find the

H/m

G O - H z vo l t age

kil ome ter hetween

p e r k i l o m e t e r i n d uc e d i n

pow e r l i n e c a r r i e s 1 5 0 I\.

t h e pow e r

a

line

th e telephone l i n e when t h e

I f t h e pow e r l i n e and t h e t e l e p h o n e l i n e d escribed in P r o b . 4.6 arc in the same horizontal plane and the distance b etw een the nearest conductors o f the t w o l i n e s is 18 m , usc t h e resu l t o f Prob . 4 . 6 ( a ) t o find the m u tual inductance between t h e power and tel ephone circuits. Also, fi n d t h e 60-Hz volt age p e r kilometer i n duced i n t h e tel ephone l i n e w h e n 1 5 0 A fl o w s i n t h e power l i n e .

168

CHAPTER 4

4.8. 4.9.

S E R I ES I M PEDA NCE OF TRA N S M I S S I O N L I N E S

Find t h e G M R of a three-strand conductor in terms of r of a n i ndividual strand. Find the GMR of each of the u nconventional conductors shown in Fig. 4 . l 5 in terms o f the radius r of a n i n d ividual s trand.

83

(b)

(a)

(e)

FIGURE 4. 1 5 C los�-sec li(JIl;t1

(d)

l i () l l �d C O l l u l i C l ur .'

4.12. 4. 13. 4. 14.

4.15.

4.17.

4 .18.

f

U Il C O I 1 V t: I l ­

distance between con d u ctors o r a s i n g l e - p h a se l i n e i s t o I t . E a c h o f i t s conductors is composed of s i x strands symmetrically placed around o n e center strand so that there are seven equal strands. The diameter of each strand is in. Show t h a t Ds o f each co nductor i s 2 . 1 7 7 times the r a d i us o f e ac h s t ra n d . F i nd t h e i nductance o f t h e l i n e i n m H/mi. Solve Example 4 . 2 for t he case where side Y of the single-phase line is identical to side X and the two sides a r c 9 m apart, as s hown i n Fig. 4 . 9 . Fin d t h e inductive react an ce o f ACS R Rail i n ohms per ki lometer a t I - m spacing. Which conductor listed in Table A.3 has an induct ive rea c t a n c e a t 7 -ft spacing of 0 . 65 1 Djm i? A three-phase line has t hree equilaterally spaced conductors of ACSR Dove. I f the conductors are 1 0 ft apart, determine the GO-Hz per-phase reactance of the l ine in D/km. A t h ree-phase l ine is designed with equ i l ateral spacing of fl. I t is decided to build the line w ith horizontal spacing ( D I 3 2 D l 2 2 D n). The conductors are transposed. What should be the spacing between adjacent conductors in order to obta i n the same i n d u c t a n c e a s i n t h e o r i g i n a l de s i g n ? A t h r e e - p h a se 60-Hz t ra ns m ission l i n e h a s i t s con J u c tors a r ra n ge d i n a triangu lar for mation so that two of the d istances between conductors a r e 25 ft an d t h e t hird distance is 42 ft. The conductors a rc ACS R Osprey . Determine the inductance and i nductive reactance per p hase per mile. A th ree-phase l i n e h a s fl a t h o r i zo n t a l s pa c i n g . T h e conductors have a G M R of m with 10 m between adjacent conductors. Determine the induct ive react ance per phase in ohms per kilometer. W hat is the name of this cond u ctor? For short t ransmission l ines if resistance is neglected, the maximum power which can be transmitted per phase is equal to

=

4.16.

(J

l ur ( ' ro b , 4 . l) ,

0.1

4.10. T h e

4.1 1.

\' i t: w

0.0133

=

16

60-Hz

where Vs and VR are the l ine-to-neutral voltages at the sending and receiving ends of the line and X is the i n ductive reactance of the l ine. This rel ationship will become apparent in the study of Chap. 6 . If the m a gnitUdes o f Vs and Vn are held

PROBLEMS

4.19.

4.20.

4.21.

4.22.

169

constant, and i f the cost of a conductor is p roportional to its cross-sectional area, find the conductor in Table A.3 which has the maximum power-handling capacity per cost of conductor at a given geometric mean spacing. A t hree-phase underground d istribution line is operated at 23 kY. The three conductors are i n sulated w i th 0.5-cm solid black polyethylene insulation and lie flat, side by side, d irectly next to each other in a dirt trench. The conductor is circular in cross section and has 33 strands of aluminum. The d iameter of the cond uctor is 1 .46 cm. The m a n u factu rer gives the GMR as 0.561 cm and the cross section of the conductor as 1 .267 cm 2 . The thermal rating of the line buried in normal soil whose maxi mum temperature is 30°C is 350 A. Find the dc and ac resistance at 50°C and the inductive reactance in ohms per kilometer. To decide w hether to consider skin e ffect in calcu lati ng resistance, de termine the percent skin effect at 50°C in the ACSR conductor of the size nearest that of the u nderground conductor. Note that the series i mped ance o f the distrib ution line is dominated by R rather than XI_ because of the very low inductance d u e to the close spacing of the conductors. The single-phase power line of Prob. 4.6 is replaced by :1 three-phase line on a horizontal crossarm i n t h e same posit ion as that of t he original single-ph ase line. Spacing of the conductors o f the powe r l i ne is D ' 3 = 2 D ' 2 = 2 D 2 3, and equivalent equi lateral spacing is 3 m. The telephone line re mains i n t he position described in Prob. 4.6. If the curre n t i n t h e power line is 1 50 A, find the voltage per kilometer induced in the telephone l i n e . Discuss the phase relation of the induced vol tage with respect to the power-line current. A 60-Hz three-phase line composed of one ACSR Bluejay conductor per p hase has flat horizontal spacing of 1 1 m between adjacent conductors. Compare the induc­ tive reactance i n ohms per ki lometer per p hase of this line with that of a line using a two-conductor bundle o f ACSR 26/7 conductors having the same total cross-sec­ tional area of aluminum as t h e s i ng le-conductor line and 1 1 -m spaci ng measured from the center of the bund les. The spacing between conductors in the bundle is 40 cm. Calculate the inductive reactance in ohms per kilometer of a bundled 60-Hz three-phase l ine h aving t h ree ACSR Rail conductors per bundle with 45 cm between conductors of the bundle. The spacing between bundle cen te rs is 9, 9, a n d 1 8 m.

CHAPTER

5

CAP ACITANCE OF TRANS MISS ION LINES

M

w e d iscussed briefly a t t h e beginn i n g o f C h a p . 4, t h e s h u n t admittance of a transmissio n line consists of con ductance a n d capacitive reactance. We h ave also men tion e d that conductance is usually neglected because its contribution to shunt admitt a nce is very smal l . For this reason this chapter has been given the title of capacitance rather than shunt a d m ittance. Capacitance of a transmission l i ne is the resu l t of the potential d i fference betwee n the conductors; it causes t h e m to be c h a rged i n the same m a n n e r as the p lates of a capacitor when there is a poten t i a l d ifference between t h e m . The capacitance between conductors is the charge per unit of potential d i ffe rence. Cap acitance between parallel conductors i s a constant depending on the size and spacing of the conductors. For power lines l ess than about 80 k m ( 5 0 m i ) long, the effect of capacitance can be slight a n d is often neglected . For longer lines o f h igher voltage capaci t a nce becomes i ncreasin gly impor t a n t .

An alternating voltage impressed o n a transmission line causes the c h a rge o n the cond uctors a t a ny point t o i n crease and d e c rease with the i ncrease and decrease of the instantan eous val u e of the vo ltage between conductors at t h e point. The flow o f charge i s current, a n d t h e current caused b y t h e a l t e r n a t e chargin g a n d d ischarging o f a l i n e d u e t o a n a l t er nating voltage is c a l l e d the charging current of the l i n e. Since capacitance is a s h u n t between con d uctors, charging c urrent flows in a transmission line even when it is open-circuited. It affects the voltage d rop a long the lines as well as efficiency and power fact or of the line a n d the stab i l i ty of the system of wh ich t h e 1 i n c is a D 'l f t . 170

5.1

ELECT R I C F I ELD

O F A LON G ,

STR A I G HT

CONDUCTOR

171

The basis of our analysis of capacitance is Gauss's l aw for electric fields . The l aw states that the total e l ectric charge with i n a closed surface equals the tot al electric flux emerging from t h e surface. In other words, t h e total charge within the closed surface e quals t h e i ntegral ove r t h e su rf a ce of the normal component of the electric fl ux d ensity. The l ines of electric flux originate on posit ive charges and terminate on negative charges. Charge density n ormal to a surface is d esignated Df and equals kE, where k is the perm ittivity of t h e m aterial su rrounding t h e surface and E is the electric field i n tensity. I 5.1

ELECTRIC FIELD OF A L O N G , S T RAI G HT

CON D U CTOR

If a long, s t r a i g h t cy l i n d r i c a l conductor l i es in a u n i form medium such as air and is iso l a ted from other charges so t ha t t he c h a rge is u n i formly distributed around its periphery, the flux is ra d i a l . All poi nts e q u i d i s t a n t from such a conductor are poi n ts of equipotential and have t h e same electric fl u x de nsity. Figure 5 . 1 shows sllch an i sol a ted conductor. T h e electric fl u x d e ns ity at x meters from the con ductor can be computed by imagining a cyl i n drical surface concentric with the con d uctor and x meters in radius. S ince a l l p arts of the surface are equ i d istant from the conductor, the cyl i n drical s u r face is a surface of equipoten­ tial and the electric flux density on t he surface is equal to the flux leav i ng the conductor per meter of length d ivided by t h e area of the surface in an axial length of 1 m. The electric fl ux densi ty is (5. 1)

where q i s the charge o n t h e con d u ctor i n coulombs per meter o f l e ngth and x is t he d istance in meters from t h e conductor t o t h e p oint where the electric flux density is comp uted. The el ectric fi e l d intens i ty, or the negat ive of t he potential g r a d i e n t , is e q u a l t o t h e e l e c t r i c fl u x d e nsity d ivided by the pe rm i t tivity of the medium. Therefore, t h e e l ectric field i n tens i ty is

E E

l

q =

2 TT Xk

Vim

( 5 .2)

and q bot h may be instantaneous, p hasor, or d e expressions.

of f r e e s p a c e ko is 8.85 X 1 0 - 1 2 F 1 m (fa ra d s p er m eter). Relat ive p e rm i t t iv i t y k r is t h e ratio of t h e act u a l p e rm i t t iv i t y k of a material of t h e p e r m i t tivity of free space. Th us, k , k l k Q • For d ry air k r is 1 . 00054 a n d is ass u m e d equal to 1 .0 in c a l c u l a t i o n s for overhead I n 5 1 u n i t s t h e p e r m i t t i vity =

l i nes.

172

CHAPTER

/ .,... -

5

CAPAClTANCE OF TRAN S M ISSION LINES

- ...... ,

FIG U RE 5. 1

Lines o f electric flux origi n a t i n g on the pos i t ive charges u n i ­ formly d istribu ted over t h e su rface o f an isolated cyl i n drical conductor.

S.2

THE POTENTIAL DIFFERENCE BETWEEN lWO POINTS DUE TO A CHARGE

The potential diffe rence between two points in volts is numerically equal to the work i n joules per coulomb necessary to move a coulomb of charge b etween t h e two points. The electric field intensity is a measure of t h e force o n a charge i n the field. T h e electric field intensity i n vol ts p er m eter i s equal t o t h e force i n newtons p e r coulomb on a coulomb of charge a t t h e p o i n t considered. B e tween two points the line integral of the force i n n ewtons a cting on a coulomb of positive c harge is the work done i n moving the charge from the point of lower potential to the point of higher potential and is numerically equal to t h e poten t i a l d ifference between the two poin ts. Consider a long, straight wire c arrying a positive charge of q elm, as shown in Fig. 5.2. Points P I and P 2 are located a t distances D ) and D 2 meters, respectively, from the center of t h e w i re. The wire is an equipotential surface and t h e u ni fonnly distributed c harge on the wire is equivalent to a c harge concen tr ated at t he ce n t e r of the wire fo r ca l c u l a t i n g flux exte r n al to t h e wire. The positive c h a rge on t h e w i re w i l l exe r t t I r e p e l l i n g fo rce 0 1 1 , \ pos i t iv e c h a rge placed in the field. For this reason a n d because D 2 in t h i s case i s g re a t e r t h a n

Path of i ntegration

I

I

,

�-��I P2

/

/

/

I

I I

/

I

I

I

I I I /

F IG U RE 5.2

Path of i n tegra t i o n be tween two points ext ernal to a cyl i n d rical conductor h aving a u n i formly d ist.-ibuted pos i t ive charge.

5.3

CAPACITANCE OF A TWO·WI R E L I N E

1 73

P2 to P I ' and PI is at a h igher potential than P2 . The d ifference i n potential is the amount of work done per coulomb of charge move d. On t h e other hand, i f the o n e coulomb o f charge moves from P I t o P2 , it expends energy, a n d the amoun t of work, o r energy, in newton-meters i s t h e vol tage drop f r
D I , work must be done on a posit ive charge to move i t from

(5 .3 ) w h e r e q i s the instanta neous charge on t he wire in coulombs per meter of length. Note that t he voltage d rop between two poi nts, as given by Eq. (5.3), may be posit ive or negative depending on whether the charge causing the pote n t i a l d i ffe r e n c e is positive or negative and on whether the vol t age d rop i s computed from a point n e a r t h e conductor t o a poin t farther away, o r vice versa. The sign of q may be either positive or negative, and the logarithmic term is e i ther positive or negative depending on whether D2 is greater or less than D I . 5.3

CAPACITANCE OF A TWO -WI RE LI NE

Capacitance between t he con ductors of a two-wire l ine is defined as the c harge on the cond uctors per u n i t of potential d ifference between them. In the form of an equation capacitance per u n i t l ength of the line is C

=

q

-

v

F jm

(5 .4)

the potential d i l le r e n ce b e t w e e n t h e co n d u c t ors i n vo l t s . H e re a ft e r, for c o n ve n i e n ce, we re fe r to cul)(lcit u l / cC !)(.'r i ll / it h'l I/.',r/t , I S c([pa citallce el ml i n d i c ,l t e the correct d i mens ions [or t h e e q u a t i ons d e r ived . T h e capaci t a n ce be tw e e n two condu ctors can b e fou n d by subst ituting in Eq . (SA) th e expression fo r v i n terms of q from E q . (5 . 3 ) . T h e vol t age U " h b e twe e n t h e two con d uctors of t h e two-wire line shown in F i g . 5 . 3 c a n be fo u n d by d e t e rm i n i n g t h e p o t e n t i a l d i ffe r e n ce between t h e two co n d uctors of t h e line, first by comp uting t he vol tage d rop due to the charge q a on c o n ducto r a a nd then by com puting the voltage d rop d ue to the charge q b on conductor b . By the principle of s up e rposition the vol tage d rop from cond uctor a t o conductor b due to the cha rges on both conductors is the sum of the vol t age drops caused by each c ha rge alone. The c h a rge q a on conductor a of Fig. 5 . 3 causes surfaces of equipotential i n the vicinity of conductor b , which are shown in Fig. SA. We avoid the w h e r e q i s t h e c h ,lr g e

on

t h e l i n e i n co u l o m b s p e r

me t e r

,I n d

v

is

I

174

CHAPTER 5 CAPACITANCE OF TRAN SMISSION L I N ES

FIGURE 5.3

Cross section of a paralle l-w i re l i n e .

distorted equ ipotential surfaces by i n tegrating Eq . (5.3) a long the a l te rnate rather t ha n t h e direct path o f Fig. 5 .4 . In determ i n i n g V a b due to q u ' w e fo l l ow the p a t h t hrough the u n distorted region a n d see t h a t d istance D \ o f Eq. (5.3) is the r a d i u s ra of conductor a and d istance D2 is the cen ter- to-center d i sta nce between conductors a and b . S i m i l a r ly, in d e t e rm i n ing v a /> due to q" , we fi n d that the d istances D 2 and D \ a r e t il a n d D , respectively. Conve rt i n g t o p h asor notation ( q n a nd q ,} become phasors), we obt a i n

and s ince

qa =

-

due to

qb

( 5 .5 ) due to

qa

for a two-wire l in e , Va b

=

qb

--qa ( -D - ) 2 rr k

In

ra

- In

rb

D

V

( 5 . 6)

Equipotential suriaces

Direct path of integration from a to b

FIG U RE 5.4 Eq u i po t e n t i a l s u r faces of a por­ t i o n of the e l e c t r i c fi e l d c a u s e d by

c h a rged c o n d u c t o r a ( n o t shown). Cond uctor b causes t h e equipot e n t i a l s u rfaces to become distorted. A r rows i n d ic a t e op­ tional paths of i n te g r a t i o n be­ tween a p o i n t o n the e q u i p o t e n ­ tial s u rface o f conductor b a n d t h e con d uctor a , whose c h a rge qa creates t he equ i pote n t i a l' s u rfaces show n . a

5.3

1 75

CAPACITANCE OF A nvO-W I R E L I N E

or by combining the logarithmic terms, we obtain

( 5 . 7) The capacitance between conductors is

2 7T k

----If r

=

u

rb

In( D2jra rb )

Fjm

( 5 .8 )

= r'

Cu b

=

'TT k

I n ( D jr )

Fjm

( 5 .9)

Equation (5.9) gives the capacitancc betwe en the conductors of a two-wire l in e . If the l i n e i s supp l i e d by a transformer h av ing a grou nded center tap, t h e potential difference between each conductor and ground i s h a l f t h e pote nt i a l d ifference between the two conductors a n d t h e capacitance t o ground, o r capacitance r o neutral, is

Ca n

=

Cbn

=

=

----

In( Djr )

(5 . 1 0 )

F j m to neutral

The concept of capacitance to neutral is i llustrated i n Fig. 5 . 5 . Equation (5 . 1 0) corresponds t o E q . (4.25) for inductance. O n e difference between the equations for capacitance and inductance shou l d b e noted c a re­ fully. The radius in the equation for capacita nce is the actual outside radius of the conductor and not the geometric mean ra tio (GMR) o f the con d u ctor, as in t h e i n d u c t a nce fo rm u l a . Equa t ion (5.3), from \vh ich Eqs. ( 5 . 5 ) t h rough (5. 1 0) were d erived , i s based on the assumpti on o f u n i form charge d is tribution ove r the su rface of the conductor. Whcn other charges are present, the d istribu tion of c h arge o n the surface of the conductor is not u ni fo rm a n d the equations derived from Eq. (5 . 3 ) are not strictly correct. The nonuniformity of charge distrib ution, h owever, c a n

aO

--I1 ,--( ----0 --1

_______

(a)

eM

R e p re s e n lation o f l i n e -to - l i n e c a pa c i t a n c e

b

(b)

R e p rese ntation of l i n e - to - n e utral c a p a c i t a nce

FI G U RE 5.5 R e l a t io n s h i p b e tw e e n t h e c o n ce p t s of l in e - t o - l i n e c a p a ci t a n c e a n d l i n e - t o -n e u t ra l c a p a c i t a n c e . I

176

CHAPTER 5

CAPACITANCE OF TRAN S M ISSJ O N LINES

be n eglected e n t irely in overhead l in es since t h e error in Eq . ( 5 . 10) is o n l y 0.01 %, even for such a close spacing as that where t h e ratio D Ir = SO. A question arises about the val u e to be used in the denominator o f t h e argument o f the l ogar i t h m in Eq. (5 . 10) w h e n the con ductor is a stra nded cable because the equation was derived for a sol id round conductor. Since electric flux is p e rp e n d icular to the surface of a pe rfect conductor, the electric fi e l d at t h e surface of a stranded conductor is not the s a m e a s the field at t h e su rface o f a cylin drical conductor. Therefore, t h e capaci tance calcula ted for a s t r a n d e d c o n ducto r by subst i t u t ing t h e ou t side r a d i u s of t h e conductor fo r r in Eq. ( 5 . 10) will be slightly in e rror because o f t h e d i fference b e tween the field in t h e n eighborhood o f such a cond uctor anci t he fi e l d ncar a so l id con d u c t o r fo r w h ich Eq. (5. 1 0) was d erive d . The error is ve ry small, however, si ncc only the fi e l d very close to t h e surface of the conductor is a ffected . The outside radius of the stranded conductor is used in calculating the capacitance. Afte r the capacitance to n e u t r a l h a s been determined, the capacitive reactance existing between one con d uctor a n d neutral for relative per m i t tivity k = 1 is fou n d by using t h e expression for C given in Eq. (5 . 10) to y i e l d r

Xc

-

1 2Tf f C

=

2 .862 f

X

D r

109 In

-

.n

. m to neutral

(5 . l 1 )

Since C i n Eq. (5. 1 1) is in fa rads p e r m eter, the p roper un its for Xc m u s t b e ohm-meters. W e should also note t h a t Eq. (5. 1 1) expresses t he reactance from line to n eutral for 1 m of line. S i nc e capacitance reactance is in para llel a l o ng t h e line, Xc i n ohm-meters must b e divided b y the length of the l ine i n m e ters to obtain t h e capaci tive reactance i n ohms to neutral for thc ent ire length of t h e line. When E q . ( 5 . 1 1 ) is d ividcd by 1 6 09 t o convcrt t o ohm-miles, we obtain

Xc

1 . 779 =

--

f

X

1 0 () I n

D

-

r

.n

.

O1 i

to neutral

( 5 . 1 2)

Table A.3 l ists the outside d i a meters of the most widely used s izes of ACSR. If D and r in Eq. (5 . 1 2) a re in feet, capacitive reactance at l -ft spacing X� is t h e first term a n d capacitive reactance spacing factor X� is the seco n d term w h e n t h e e q uation is expanded as fol lows : Xc

=

1 . 779 f

--

X

106 In

1 -

r

+

1 . 779 --

f

X

106 I n D n . m i to neutral ( 5 . 13 )

Table A . 3 i ncludes values o f X� for c o m m on sizes o f ACSR, a n d similar tables . a re readily available for other types and sizes o f cond uctors. Tab l e A.S i n t h e Append ix lists values o f X� which, o f course, i s different from t h e synchronous mach i n e t ransient reactance bearing the same symbol.

5.4

OF

CAPA C I T A N C E

A TH R EE - P HASE L I N E W ITH EQU I LATERAL SPACING

177

Find the capacitive susceptance per mile of a single-ph ase line operating at 60 Hz. The conductor is Partridge, and spacing is 20 ft between centers. Exa mple 5 . 1 .

Solution. For t h i s conductor

Table A.3 lists an o u t s i d e d iameter of 0.642 in, and so r =

and from

Eq.

Xc

0 .642

2

12

x

=

0 .0268 ft

(5. 12) 1 .779 "60

=

Be =

1

Xc

=

1Q6 ln

x

5.10

X

20 0 .0268

=

0 . 1961

106

x

n .

mi to neutral

1 0 - 6 Simi to neutral

o r i ll t e rm s o f c J p a c i t ivc re a c t a n ce at l - n spacing and c a p ac i t ive reactance spacing factor from Tables A.3 and A.5

X;

=

X�.

=

Xc

=

0 . 1 074

Mn

0 . 0889

MD

0 . 1 074

+

. mi

. mi

0 .0889

=

0 . 1 963

Mn

. m i per conductor

Line-to-line capacit ive reactance and susceptance are Xc

Be

= =

2

x

1 Xc

0 . 1 963 =

2 .5 5

X X

106

=

0 .3926

X

106

n

. mi

10- 6 Simi

5.4

CAPACITAN CE OF A THREE-PHASE LINE WITH EQUILATERAL S PACING

5.6 .

The t hree identical conductors of radius r of a three-phase l ine wit h equilateral s p a c i n g a re shown i n Fig. E q u a t i o n (5 . 5 ) exp r e s s e s t h e vol t a g e between two con d u c tors due to the c harges on e a ch one if the charge d istrib ution o n the cond uctors can be assu med to b e u n iform. Thus, the vol tage Vab of the t hree-phase line due only to the cha rges on conductors a and b I S

Vab

=

. -�- -

2 Tfk

(q

a

In

D r

q �)

+ b in

D

y

( 5 . 14)

178

CHAPTER

5

CAPACITANCE OF TRANSM ISS I O N LI N ES

D

a

c

FI G U R E 5 . 6 Cross sec t i o n of a t h r e e - p h ase l i ne w i t h e q u i l a t e r a l s p a c i n g .

Equation (5 .3) enab les us to i nc l u d e the e f fe c t o r q c s i nce u n i fo r m c h 'lrge d istribution ove r the surface of a con d u ctor is eq uivalent to a conce n t ra ted charge at the center of the conductor. Therefore, due only to the c h a rge CJ( ,

Vab

=

D 2Tfk D V qc

--

In

-

w h ich is zero since qc is equi distant from a and b. However, to s how t h a t we are considering all three charges, we c a n write

Vab

=

2Tfk 1

--

( qa D In

r

+

qb I n

D r

-

+

qc

In

D]V D

(5 . 15 ) ( 5 . 1 6)

Adding Eqs. (5 . 1 5) and (5 . 1 6) gives

(5 . 17)

I n deriving these equations, we have assumed t h a t gro und i s far e nough away to h ave negligible effect. Since the voltages a re assumed to be s i nusoidal and expressed a s phasors, the charges are sinusoidal and expressed a s p hasors. I f there a r e n o o ther charges i n t h e vici n i ty , t h e sum of t h e charges on t h e thre e conductors is zero and we can substitute - qa in Eq. ( 5 . 1 7) for q b + qc and obtain

( 5 . 18) Figure 5.7 is the phasor d iagram of voltages. From this figure we obtain the following relations between the line v ol t ag e s Vab and Vac and t he vol tage Va n

5.4

179

CAPA CITANCE OF A T H R E E , P H AS E L I N E WITH E Q U I LATERAL SPACI N G

F IG U R E 5 . 7 P h asor d i agram of the balanced volt ages o f a t h re e - phase l i n e .

from l ine

a

13- V:1I1�

to the neutral of the th ree-phase circuit:

V:I1>

Val = Add i ng Eqs. ( 5 . 1 9)

-

=

=

Veil = /3 V:'" I - 30 °

and

13 V:'1I( 0 .866 + )0 . 5 ) =

(5.20) gives

( 5 . 19)

13 Va,, ( 0 .866 - )0.5 )

( 5 .20 )

( 5 .2 1 ) S u hs t i t u t in g 3 Va n for Va b

+

Va c in Eq. (5 . 1 8 ), we obta i n Va n

qa =

--

2 TTk

In

D -

r

V

( 5 .22)

S ince capaci tance to neutral is the ratio of t he charge on a conductor to the vol tage b e tween that condu ctor and neutra l , 2 7T k ----

I n ( Dj r )

F j m to neutral

( 5 . 23 )

Comparison of Eq s. ( 5 .23) a n d (5. 10 ) shows t h a t t h e two a re i d e n tica l . These e q u a tions express t h e capacitance to neu t ral for single-phase a n d equi­ l a t e r a l l y spaced t h ree-phase l ines, respective ly. S i m i l a r ly, we reca l l t h a t the equ ati ons for inductance per conductor are the same for single-phase a nd e q u i l at eral ly sp ac�d t h ree-phase l ines. The t e rm cha rging current is appl ied to the current associated w i t h the c a p �l c i t a n c e o f a l i n e . F o r

p ro d u c t of p h asor,

the

a

t h e l i ne- to- l i ne

single-ph ase c i rc u i t t h e ch a rg i n g c u rre n t i s the

l i n c - t o- l i n e v o l t a g e

and

suscepta nce, or

as a

(5 .24 ) For a three-phase line t h e c h a rging current is found by multiplying the voltage to n e u t r a l by t h e capaci tive susceptance to neutral. This gives t h e charging I

180

CHAPTER

5

CAPACITANCE OF TRA NSM I SS ION L I NES

curren t per p hase and is in accord with t h e calculation of balanced three-phase circuits on t h e basis of a single p hase with n eutral return. The phasor c harging current in p h ase a is

( 5 .25) Since t h e rms voltage varies along the line, t h e charging current is not t h e same everywhere. Often the voltage used to obtain a value for charging current is the normal vol tage for which the line is designed, such as 220 or 500 kY, which is probably not t he actual vol tage at either a generating station or a load.

5.5 CAPACITANCE OF A THREE-PHASE LINE WITH UNSYMMETRICAL SPACING When the conductors o f a t hree-phase l i n e a re not equilaterally spaced, t h e problem o f calculating capacitance becomes more d i fficult. In the u s u a l u n t rans­ pose d line the capacitances of e ac h p h ase to n e u tral are unequal. In a t rans­ posed line the average capacitance to neutral of any phase for the com plete t ransposition cycle is the same as the average capacitance to neutral of any other p h ase since each phase con ductor occupies the same position as every other p h ase conductor over an equal distan ce along the transposition cycle. The dissymmetry of the u ntransposed line is slight for the usual configuration, and capacitance calculations are carried out as though all lines were transposed. For t h e line shown i n Fig. 5 . 8 three e q u ations are found for Vab for the three d ifferent parts of t he transposition cycle. With phase a in position 1 , b i n position 2 , and c i n position 3 ,

V;l /J With p h ase

a

=

( D1 2

1 q n 2 7T k " I

r-

-

+ q ,)

D21 ) V

r

In D

1

+ (j, I n D 3"\

2

i n position 2, b i n position 3 , a n d

c

in

( 5 . 26 )

position 1 , ( 5 . 27)

2

FIGURE 5.8

1

3

Cross s e c t i o n o f a t h ree -phase line with u nsy m m etrical spacing.

5.)

a n d with

a

CAPACITANCE OF

181

A TH R E E- P H A S E f . I N E WITH UNSYMMET R I CA L SPACING

in posi tion 3, b i n posi tion 1 , and

i n position 2,

c

( 5 .28) Equations (5 .26) through (5.28 ) a re sim i l a r to Eqs. (4.5 1 ) t hrough (4.53) for the magnetic flux l i nkages of o n e cond uctor of a transposed line. However, i n t h e equations for magn e tic fl ux l i n kages w e note t h a t t h e current i n a n y p h ase is the same in every part of the transposition cycle. In Eqs. (5 .26) through (5.28), if we d isregard the voltage d ro p along the l ine, the voltage to neutral of a p hase in one p a r t of a transpos i t io n cycle is e q u a l to the voltage to neutral of t h a t p h ase i n a ny part of t h e cyc l e . H e nce, t h e vol t age between any two conductors i s t h e same in a l l p a r t s of t h e t r a n sposition cyc l e . It fo l lows t h at t h e charge o n a co n d uctor m ust be d i ffe re n t w h e n the pos i t ion of the conductor changes with respect to other conductors. A tre a t m e n t of Eqs. (5 .26) th rough (5 .28) analogous to that of Eqs. (4.5 1 ) through (4.53) is not rigorous. The rigorous solu tion for capacita nces is too i nvolved to be p ractical except perhaps for fiat spacing with equal d istances bet\veen adjacent conduc­ tors. With t h e usu a l spacings a n d cond uctors, sufficient accuracy is obtained by assuming t h a t the charge p e r u n i t length on a cond uctor is the same i n every part of the transposition cycle. When the above assumption is made with rega rd to charge, the voltage between a pair of cond uctors is different for each p a rt of the transposition cycle. Then an average val u e of voltage betwee n the cond u c­ tors can b e found and the capacitance calcul ated from the average voltage. We obtain the average voltage by a d d ing Eqs. (5 .26) through (5 .28) and by dividing the resu lt by 3. The average voltage between cond uctors a and b, assum i ng the same charge on a conductor regardl ess of i ts position i n the transposit i o n 1 cyc.e, IS •

( 5 29 ) .

( 5 .30)

where S imilarly, t h e average vol tage d rop from conductor a to conductor

Va c

1

=

2 7T k

--

( Dcq qa

In

.

-

t

+ q

C

In

1 D V r

-

eq

C

IS

1

(5 . 3 )

182

CHAPTER 5

CAPACITANCE OF TRANSM ISSION LINES

Applying Eq. (5.21) to fi n d the vol tage to neutral, we h ave

3 Va n - Va b + Vac _

=

1

2k Tr

(

D eq 2 qa I n r

+

r qb I n Deq

+

qc

)

In

r - V ( 5 .32 ) De q

( 5 .3] ) e"

and

q"

=

=

---( I n D cq / r )

( 5 .34)

F / m to neutral

Equ ation (5 .34) for capacitance to neutral of a transposed t h ree-phase l i n e correspo n d s to Eq. (4.56) for the i n ductance p e r p hase of a similar l i n e . I n fi nding capacitive reactance t o neutral corresponding t o en , w e can spl i t t h e reactanc e into compone n ts o f capacitive reactance t o neutral a t I - ft spacing X� and capaci t ive reactance spacing factor X� , as defined by Eq. (5 . 13). Find the capaci tance and the capacitive reactance for 1 m i o f the line described i n Example 4.4. I f the length o f the line is 1 75 mi and the norma) operating voltage is 220 kV, find capacitive reactance to neutral for the entire leng t h of the line, the charging current per mile, and the total charging megavoltamperes.

Example S.2.

Solution

r =

1 . 1 08 2 X 12

=

0. 0462

Dcq = 24 .8 ft

ft

-

ell =

2 IT x 8 .85 x 1 0 1 2 = 8. 8466 1 n (24.8/0 .0462)

Xc =

1012 2-rr x 60 x 8 . 8466 x 1609

=

x

1 0- 1 2

0 . 1 864

F/m

X

106

.0.

. mi

or from tables

X�

Xc

=

=

0.0912

X

(0.0912

10 6

+

x�

=

0 .0953

0.0953) x 1 0 6

=

X

106

0 . 1 865

X

1 0 6 .0.

.

m i t o neutral

J

5.6

EFFECf OF EARTH ON TH E CA PACITANCE OF THREE-PHASE TRANSMISSION LINES

For a length of

175

183

mi

Capacitive reactance I fchg l

=

0 . 1 865 X 1 0 1 75

220 , 00 0 =

13

6 =

1 066 n

13

220 ,000

Xc

to neutral X

10 - 6

x 0 . 1 865

=

0 .6 8 1

Ajmi

A for the l ine. Reactive power is Q = 13 X 220 X 1 1 9 X 10-3 43.5 Mva r. This amou n t of reactive power a b s o rb e d by t he d istributed capacitance is negat ive i n keeping with the convention d iscussed in Chap. 1. In o t h e r w o r d s , p o s i t i v e re(lctive power is b e i n g [?('l I eratecl b y t h e distributed or

0 . 68 1 x 1 75 =

=

119

C
5.6

EFFECT O F EA RTH O N T H E

CAPACITAN C E OF TH REE-PHASE TRA N S M I S S I O N L I N E S

Earth affects the capacitance of a transmission line because its presence a lters t he electric field of the l i ne. I f w e assume that the earth is a perfect cond u ctor in the form of a horizontal p l a n e of i n fi nite extent, we realize that the electric field of charged conductors above t h e earth is n o t the same as it wou l d b e i f t h e equ ipotential su rface o f t h e e a r t h were not p resent. The electric fi e l d o f t h e charged conductors i s forced t o conform t o t h e presence of t h e earth's s u rface. The assumption of a flat, equ i pot e n t i a l su rface is, of course, l imited by t h e i rregularity o f terrain a n d the type o f surface o f t h e e a r th. The assumption enables us, however, to understa n d the effect o f a conducting earth on capaci­ t ance calculations. Consider a circuit consisting o f a s i ngle overhead conductor with a re turn p a t h through the earth. In c ha rging t h e cond uctor, charges come from the earth to reside on the con ductor, and a potential d i fference exists between t h e conductor and the eart h . T h e e a r t h has a charge e q u a l i n magni tude t o t h a t o n t h e conductor b u t of oppos ite sign . T h e e l ectric fl ux from the cha rges o n t h e conductor to the charges o n t h e e a r t h is perpend icu l a r to t h e earth's equipoten­ tiel l su rface since the su rface is assumed to be a perfect conductor. Let us i m agine a fictit ious con d uctor of t h e same size and shape as the ove r h e a d cond uctor lying di rectly below t h e o r i g i n a l conductor at a d istance e q u a l to t w i ce t h e d i s t a n ce o f t h e c o n d u c t o r a bove

the p l ane

o f t h e gro u n d .

The

fi c t i t i o u s

conductor is be l ow the su rface of tbe e a rth by a distance equal to the distance o f the overhead cond uctor ahove the e a rt h . If the earth is removed and a c h a rg e e q u a l a n d opposite t o that o n t h e overhead cond uctor i s assume d o n t h e fictitious conductor, t h e p l a ne m idway between the original conductor a n d t h e fictitious conductor is an equ i pot e n t i a l surface a n d occupies t h e same position a s the equipotential surface o f t h e e a r t h . The el ectric flux between the overhead con d uctor and this eq u ip ot e n t i a l surface is the same as that w � ich existe d

184

CHAPTER

5

CAPACITANCE O F TRANS M I S S I O N LINES

between the conductor and the earth. Thus, for purposes of calcu l a tion of capacitance t h e earth m ay be replaced by a fictitious charged conductor b e l ow the surface of the earth by a distance equal to that of the ove rh e ad cond u c t or above the e arth. Such a conductor h a s a charge equal i n mag n i t u d e a n d opposite i n sign t o that of the origi n a l con ductor a n d i s ca l l e d t h e image

The method of calculat ing capacit a nce by replacing th e e a r t h by t he i m age of an overhead conductor can be extended to more than one con d uctor. I f w e locate a n i m a g e cond uctor for e a c h ove rhead conductor, t h e A u x h c twc c n t h e original conductors a n d their images i s perpen d icula r t o the p l a n e w h ich repl aces the earth, and that plane is a n equ i potenti a l su rface . T h e A u x above the plane is the same as it i s when t h e earth is prese n t i n stead u r t he i m age conductors . To apply t h e met hod of i mages to t h e ca l c u la t i o n o f c a r a c i l a n c c ror a three-phase line, refe r to F i g . 5 .9. We a s s u m e that t h e l i n e is t r a n sposed a n d

conductor .

77/ 7

FIGURE 5.9 T h r e e-phase l i n e and its i m a g e .

5.6

EFFECT OF EA RTH ON T H E CA PA CITA NC E OF T H R E E - P H A SE T R A N S M IS S ION L I N ES

1 85

that conductors, a , b, a n d c c a r ry t h e c h a rges q qb ' and q c a n d occupy positions 1, 2, and 3, respec tively, in the first p a r t of the transposition cycle. The plane of the earth is show n , a n d below it a re the conductors with the i mage charges - q a ' - q b ' and - qc Equ ations for the t hree p arts of the transposition cycle can be written for the voltage drop from conductor a to conductor b as determined by the three c harged conductors a n d their images. With conductor a in position 1 , b in position 2, a n d c in pos ition 3, by Eq. (5.3) a'

(5 .35) Similar equ ations for Vall a re written for the other parts of the t ra nsposition cyc l e . Accepting t h e approxim ately correct assu mption of constant charge per unit length of each conductor t h roughout the transposit ion cycle allows u s to obta i n a n average value of the p hasor Va ll ' The equat ior. for the average value of the p h asor Vac is fou nd in a s i m i l a r m an ner, and 3 Var. is obtained by a d di ng t he average values of Vlln a n d Vac - Knowing t h a t the sum of the charges is zero, we then fi n d cn

=

In

( - ) - ..,; -DCq

r-

In

(5. 3 4)

""\

H l � HZ3 H3 1

J r-

-

yH) H2 H3

1

F1m to neutral

( 5 .36)

Comparison of Eq s . and (5.36) shows that t h e effect of the earth is to in c r e a s e th e cI [J <1 c i t il n c c o f ,I I i n c . T o ,ICCO li n t fo r t h e e a r t h , t he denominator of Eq . (5.34) m u s t !l,IVC s u b t r a c t e d fro m i t t h e term

I f the con ductors are h i g h a bove ground compared with the d istances between them, the d i agonal distances i n t he numera tor of the correction term a re nearly equal to the vertical distances in the denominator, and the term is very small. This is the usual case, a n d the effect of groun d is ge nerally neglected for

186

CHAPTER 5

CAPACITANCE OF TRANSM ISSI ON LI NES

three-phase l ines except for calculations by symmetrical components when t h e sum o f t h e three l in e currents i s n o t zero.

CAPACITANCE CALCULATIONS FOR BUNDLED CONDUCTORS Figure 5 . 1 0 shows a bu ndled-conductor l i ne for which we ca n write an equat ion for the voltage from conductor a t o con d uctor b a s w e did i n deriving Eq . ( 5 .2fl), 5.7

except t h a t now we must consider t h e c h a r g e s on a l l s i x i n d iv i d u a l conductors. The conductors of any one bundle a re in para l l el , and we can assume the charge per bundle d ivides equa l ly between the con d u ctors of the bund I e since t h e separation between bundles i s usu a l ly m o r e t h a n 15 t i mes the spaci ng between the conductors of the bundle. A l so, si nce D 1 2 is m uc h greater than d, we can use D J 2 i n p lace of the d is ta nces D I 2 d and D I 2 + d and m a ke o t h e r s i m i l a r substitutions o f b u n d l e separation d istances i nstead of u s i n g t h e m o re exact expressions that occur in fi n ding Val) ' The d i ffere nce due to this a p p roxima tion cannot be d etected i n the final res u l t for usual spacings even when the calc ulat ion is carried to five or s ix significant figures. If charge on p hase a is q a ' each of conductors a and a f has the ch arge qa/2; similar division of charge is assumed for p hases b a n d c . Th e n ,

-

1

--

2rrk

l(qa

- In

2

+

D 12 r

+ In

D

J2

--

d

�

�

a

a

'

]

+ b

- ( D21 D21 1 j qc

In

2

-"

f) .1 \

�--

c

+

b'

( 5 .37)

I n -' D -:, \

-------

c'

The l etters u nder each loga r i thmic term indicate the con ductor whose charge is accounted for by that term. Combi n i n g te rms gives

( 5 .38 )

a

1�----- D3 1 --------�� 1 0

..--- D 1 2 1....

0

I� d -l

a'

1

----· -----

b o

O o

I- d-\

D 23 -----... � c o

o c'

I- d - I

FIGURE S.lO

Cross section of a b u n d l ed-con ­ ductor t h ree-phase l i n e .

5.7

CA P A C ITA N C E CA LCU LAT I O NS F O R B U N D L E D C O N D UCTORS

1 87

Equation (5 .38) is the same as Eq. (5.26), except that VrJ has replaced r. I t therefore follows that if w e consider t h e l i n e t o b e t ransposed, w e fi n d e'l

=

--(--D)

-

In

eq

( 5 .3 9)

F 1 m to n eutral

Vrd

,

The Vrd is the same as D:' for a two-conductor bundle except t h a t r has r e p l a c ed DJ • T h is leads us to the very i mport ant conclusion that a modified geom e t ric m e a n d istance ( G M D ) m e t h od a p p l ies to t he calculation of capaci­ t a n c e of a h u n d l e d -cond u c t o r I h r e e - r h a se l i ne h a ving two conductors per h u n d l e . Thc mOll i fi c a t i o n is I h ,1 I w c a rc l I s i n g o u t s i d e r a d i u s in p l a ce o f the G M R o f a s i n g k conduct o r. I t i s l og i e
2 ... k

( )

-�n--D-�qThe n , for a two-strand bund l e

Dsc

F1 m

( S AO)

to neutral

( S Al )

-

fo r a t h r e e s t r a n d b u n d l e

( 5 .42)

a nd fo r

a

four-strand hu n d l e

1 . 09

Exa m p l e 5 . 3 . Fi n d t h e

Example 4 . 5

capac I t I ve reac t a nce

..;rd J

4

. ( 5 .43)

t o n e u t r a l of the l i n e d e s c ribed i n

i n o h m - k i l o m e t e r s ( a n d i n o h m - m i les)

per phase.

188

CHAPTER 5 CAPACITANCE OF TRANSMISSION L I N ES

Solution. Computed from the diameter given i n Table A.3 r =

D:c Deq

=

=

ell =

(

Xc

=

Xc

=

1 .382 2

X X

0 .3048 12

JO .01755 ""{8 X 8

X

0.45

3 �----=--...,. .

271

X

16

(

=

)

=

0 .01755 m 0.0889 m

=

1 0 .08 m

8 .g5 X 1 0 - J 2 1 0 . O /) In O.08�9 X

10J2 X 10-3 27160 X 1 1 .754 0.2257 X 1 0 6 1 .609

=

=

1 1

.754

0 .2257

0 . 1403

X

X

X 10

1 0 ('

f2

.

J

2 F/m

km

per

p h a s e to n e u t ra l

1 0 6 11 . m i per phase to neutral

)

5.8 PARALLEL-CIRCUIT THREE-PHASE LINES If two t hree-phase circuits that are identical i n construction and operating i n parallel a r e so close together t h a t cou pling exists between them , t h e GMD method can be used to calcul ate the inductive and capacitive reactances of t heir equ ivalent circuit. Figure 5 . 1 1 shows a typical a rrangement of parallel-circui t th ree-phase l i n es on the same tower. Although the l i ne will probably not be transposed, we obtain practical values for inductive and capaci tive reactances j f transposition is assumed. Conductors a and a ' are in p a rallel to compose phase a. Ph ases b a n d

a

O----

18· ---0 �

1 0'

8-----2 1' -----0

b

'.

J ----0 b'

1 81

FIGURE 5. 1 1 Typical arrangement o f co n d u ctors o f a p a r a l lel­ circu i t t h ree-p h a s e l i n e .

5 . l'i

PA RA LLEL-C I R C UIT TH R EE - P H A S E L I N E S

189

c are similar. We assume t h a t a and a ' take the positions of b a nd b' and then o f c and c' as those conductors are rotated similarly in the transposition cycle .

To calculate Deq the GMD method requires that we use D:b , D bc ' and D(� , where the superscript indicates that these quantities are for p a rallel l ines and where D:b means the GMD between the conductors of p hase a and those of p hase b. For inductance calculations Ds of Eq . (4.56) is replaced by D/, which is · the geometric mean of t h e GMR values of the two cond uctors occu pying fi rst the p o siti o n s of a and a' ) then the positions of b and b', and fi n a l ly the positions of c and c'. Because of the sim i l a ri ty b e t w e e n i n d uctance and capacitance calcula tions, we c a n ass u m e t h a t the D/c for ca p a c i t a nce i s t h e same as D! for i nductance, excep t that r i s used ins tead of D} of the i n d ivid ual condu ctor. Fol lowi n g each step of Ex a m p l e S A is possibly the best means of u nd er­ s t a n d i ng the procedure . Exa m p l e 5 .4 A t h re e - p h ase d o u b l e - c i rc u i t I i n e i s c o m pos e d o f 300,OOO-cm i l 2 6 / 7 Ostrich cO l l d u c t o r s a r r a ll g e d a s s h o w n i n F i g . 5 . 1 1 . F i n d t h e 6 0 - H z i n d u c t ive r e a c t a n c e and c a p a c i t i ve su sce p t a n ce in o h m s per m i l e p e r p h ase a n d s i e m e n s p e r .

.

m i l e pe r p h a s e , respe c t i v e l y .

Solution . From

Table

D i stance

A.3

for Ostrich

a to b :

D is t a n ce a to

b' :

Ds

=

0 .0229 ft

o r igi n a l pos i t ion =

/1 0 2

+ 1 .52

o r i g i n a l p os i t i o n =

/1 02

+ 1 9. 5 2 = 21 . 9 ft

1 0 . 1 ft

T h e G M D s b e t w e e l l p h a s e s a rc

1 8 . 9 7 ft

DC Q =

�1 4 .88

X

1 4 .88

X

1 8 .97

= 1 6 . l ft

For i n d u c t a n c e c a l c u l a t i o n s the G M R fo r t h e p a ral l e l -ci rcu i t l i n e is fou n d after fi rs t obta i n i n g the G M R v a l u e s for t h e t h r e e posi t i o n s . The a c t u a l d is t a nce from a

190

CHAPTER 5 to

a' is

CAPACITANCE OF TRANSM ISSION L I NES

';20 2

+

18 2 = 2 6 . 9 ft. Then, G M R o f each p hase is ../26 .9

' a - a :

I n position

../2 1

In position b - b' :

X 0 . 0229 = 0 . 785

X 0 . 0229

-./26 .9 X

In position c - c' :

= 0

ft

. 69 3 ft

0 . 0229 = 0 . 785

ft

Therefore, D;'

=

L =

XL

�O.n5 X 0 . 693 2

x

X

(�7H5

1 O - 7 1 n -16.1

0 . 753

=

0 .7 53

It

= 6.13 X 10-7

= 2 11- 60 x 1 0 9 x 6 . 1 3 x

6

10 - 7

Him per phase

= 0 . 3 -:- 2

D/ mi

per phase

For capacitive calculations Die is t h e same as t h a t of Dj' , except that the ou tside radius of the Ostrich conductor is used i nstead of i t s G M R . Th..-: o u ts i de d i ameter of Ostrich i s 0 . 680 i n : r =

Dj'e

=

0 . 680 2 x

= 0 . 0283

12

( ";26 .9

x 0 .0283 /21 x 0 . 0283 ";26 . 9 x

= /0 .0283

ell =

Be

2 11"

= 2 11"

X

= 1 1 .4 1

( 26 . 9

� . �5

111

X

ft

x

x 21 x

10

-­ O .�37

1 (> . 1

6 0 x 1 8 . � 07

X

10-6

12

x

26.9) 1 / 6 l H . H07

=

x

1 3 0 .02S3 ) /

0 . 837

ft

12

F

10 -

1m

1 609

Simi per phase to neu tral

5.9 SUMMARY The similarity between inductance a n d capacitance calculations has been em­ phasized t h roughout our discussions. As in inductance calculations, computer programs are recommended i f a large n umber of calcul ations of capacitance is required. Tables l ike A.3 a n d A.S make t h e calculations quite simple, however, except for p arallel-circuit l ines.

191

P R O B LEMS

The important equation for capacitance to n e utral for a s in g l e-circuit, three-phase l ine is

.

cn

DSL' is the outside r ad ius

=

Dcq

--=--

In

F / m to n eutral

(5 .44 )

Ds c

-

of the conductor for a l i n e consisting of one 2 cond uctor per phase. For overhead l ines k is 8.854 X 1 0 - 1 since k r for air is 1 .0 . Capacitive reactance i s ohm-meters is 1 /2 1T!C, v,,' here C is in farads per meter. So, at 60 H z XC' =

r

4 . 77

X

1 0 4 In

D

eq

n . km to n e u t ral

( 5 ,45 )

Deq n . mi to neutral Dsc

( 5 ,4 6 )

--

D se

o r upon d ivid ing by 1 .609 km / m i , wc have

Xc

=

2 . 965

X

1 0 4 In

-

Values for capacit Ive susce p t ance in siemens per kilometer a n d SIemens per m i l e are the reciprocals of Eqs. (5 ,45) a n d (5 ,46), respectively. Both Deq and Dsc must be in the same u nits, usual ly feet. For bund led conductors DsbC is substituted for Dsc, For both single- and b u n d led-conductor l ines

(5 4 7 ) ,

For bundled-conductor l i nes Da b ' Dhc ' and Dca a re d istances b e tween the centers of the bundles of phases a , h, and c . For l i nes with one conductor per phase i t i s convenient to determine X c by adding X:/ for the conductor as fou n d in Tabl e A . 3 to X:, as found in Table A . S correspond i ng to D"q ' I nd u c t ance, capaci t a n c e ,
A t h re e - p h a s e t ra n s m i s s i o n l i n e h a s fl at horizontal spacing with 2 m between adjacent conductors. At a c e r t a i n instant the charge on one of t he outside condu ctors is 60 ,u C/km, and the charge on the center conductor and on the other outside conductor is 30 J.L C/km. The radius of each conductor is 0.8 cm. Neglect the effect of the ground and find the vol tage d rop between the two identically charged conductors at the i n s t a n t s p e c i fi e d . -

192 5.2.

5.3 . 5.4.

5.5.

CHAPTER

5

CAPACITANCE OF TRANS M ISS I O N L I NES

The 60-Hz capacitive reactance to neutral of a solid conductor, which is one conductor of a single-phase line with 5-ft spacing, is 1 96.1 k O-mi. W h at value of reactance would b e specified in a table l isting the capac itive reactance i n oh m - m i l es to neutral of the conductor at I -ft spacing for 25 Hz? W hat is the cross-sectional area of the conductor in circular mils? Solve Example 5.1 for 50-Hz operation and 1 0 ft spacing . Using Eq. (5 .23), determine the capacita nce to neutral (in p, F / k m ) of a t h re e - p h ase l i ne with t h re e Cardinal ACS R c o n d u c t o r s e q u i l a t e ra l l y s pa c e d 2(j f t a p a r t . W h a t is [he c h a rg i ng current of t h e l i n e (in A / k m ) at 60 H z and 1 00 k Y l i n e t o l i ne ? A t h ree-phase 60- H z t ra ns m i ss i o n l i n e h a s i t s c o n d u c t o rs a rr a n g e d i n d t r i ,l l1 g u L l r formation so t h a t two of t h e d is t a n c e s b e t w e e n co n d u c t o rs a rc 25 :'[ a n d t h e t h i r d is 42 [t . The c o n q u c t o r s arc ACS R Os"I'ey . D e t e r m i n e t h e e a p a e i L t nee t o n e u t r a l i n m icrofa rads p e r m i l e and t h e c a p a c i t i v e r e a c t a nce t o n e u t ra l i n ll h m - l11 i l es . If t h e l i n e i s 1 50 m i l o n g , fi n d t he c a pac i t a n c e t o n e u t r a l a n d c a pa e i t i e r e a c t a n c e o t' t h e -

\'

line.

thr ee -ph a s e 60-Hz l i n e h a s fl a t h o r i zo n t a l spac i n g . T h e c o r. J u e t o r s h a v e a n out s i d e d i a m e t e r o f 3 . 2H c m w i t h J 2 III b e t w e e n c () n d u c t o r � . D e t e r m i n e t h e c a pac i t i ve r e a c t a nce t o n e u t r a l i n o h m - m e t e rs a n d t h e capac i t ivc r C tl c t d n ce o f t h e line in o h ms i f its length is 1 25 m i . 5.7. ( a ) Derive an equation for the capacitance to n e u t r a l i n far a d s p e r m et e r of a single-phase l ine, t a k i ng into account t h e e ffe ct of ground. Use t h e same n o m e n cla ­ tu re as i n the equation derived for t h e c a pac i tance of a t hree-phi:: s e l i ne where t h e effect of ground is represented b y i m a ge c harges. (b) Using the derived equation, calculate the capacitance to n e u , r a l i n farads per meter of a single-phase line c o m p o s e d o f two s o l i d c i rcu l a r co n d u c o rs , each having a d i ameter of 0.229 in . The conductors are 10 ft apart and 2 5 f t abov e grou nd. Co m pare the result with the value obtained by applying Eq. (5. 10). 5.8. Solve Proh . 5.6 while t a k i n g i n t o a c c o u n t the e ffe c t o f gro u n d . .-\ s s u m e t h a t the conductors a r c horizont a l ly p l ace d 20 m above g ro u n d . 5.9. A 60-Hz t hree-phase line composed of one ACSR Blllejay conductor per phase has flat horizontal spacing of 1 1 m b e tw e e n a dj ac e n t co n d u c t o rs . C o m p a re the c a p a c i ­ tiv e reactance in o h m - k i l o m e t e rs p e r p h ase of th i s l i n e w i t h t h a : o f ,I l i n e u s i n g a t w o-co n d uctor b undle of ACS R 26/7 conductors having t he s a m e t o t a l cross­ sec t i o n a l area of a l u m i n u m as t h e s i n g l e conduc t or l i ne a n d t he I l -m spacing measured between b u n dl es. The spacing between conducto r s in the b u n d l e is 40 c m 5.10. C a l cu l a t e the capacitive reactance i n ohm-kilometers of a bund led 60-Hz three­ phase line having three A C S R R ail cond uctors per b undle with 45 em between conductors of the bundle. The spacing between bundle cen ters is 9, 9, and 18 m. S . U . S ix conductors of ACSR D ra ke cons titute a 60-Hz double-circu i t three-phase l i n e arranged a s shown i n Fig. 5 . 1 1 . T h e vertical spacing, howeve r , i s 1 4 ft; t h e longer horizontal d istance is 32 ft ; and the shorter horizontal d i st ances are 25 ft. Find (a) The inductance pe r ph ase (in H / mi) and the inductive reactance Un D / mi) . ( b ) T h e capacitive reactance to ne u t ra l (in n · m i ) and the charging current in A / m i per phase and per con ductor a t 138 kY. 5.6. A

-

.

CHAPTER

6

CURRENT AND VOLTAGE RELATIONS · ON A TRANSMIS S I ON LINE

We h ave examined the parameters of a transmission line and a re ready to con sider the l ine as an element of a power system. Figure 6.1 s hows a 5 00-kV l in e h aving bundled conductors. I n overhead lines the conductors are suspended from the tower a n d i nsulated from it and from each other b y insulators, t h e n u mber of which i s determined b y the vol tage o f the line. Each insulator s t r i n g i n F i g. 6.1 h a s 22 insu lators. T h e two shorter arms above t h e p h ase conductors support wires usua l ly made of steel. T h ese wires, much smaller i n diameter than the phase conductors, are not visible in the p ic ture, but they are e lectrically connected to the tower and a re therefore at ground potential. These wires are referred to as shield or ground wires and shield the phase conductors from l ightning strokes. A very important problem in the d esign and operation of a power system i s t h e m a intenance of the vol tage within specified l i mits at various points i n t he system. I n this chapter we d evelop formulas by which we can calcu l a t e the voltage, current, and powe r at any point on a transmission l i ne, p rovided w e know these values at one poin t , usually at o n e e n d o f t h e l ine. The purpose of this chapter, however, is not merely to develop the pertinent equ ations, but also to p rovide an opportunity to u n d,erstand the effects of the parameters of the line on bus voltages and t he flow of power. In 193

194

CHAPTER

6

CUR R ENT AND VOLTAGE RELATIONS ON A TRANSMISSION U i\ E

500-kV transmission line. Conductors are 76/ 1 9 ACS R w i t h a l u m i n u m cross section of 2,5 1 5 ,000

FIGURE 6.1

A

em i l . Spacing between phases is 30 ft

3

i n a n d t h e two con d u ct ors per b u n d l e are 1 8 in a p a r t .

(Courtesy Carolina Power and Light Company.)

6.1

R E P R E S E NTATION OF L I N ES

195

this way we can see the importance of the design of the line and better understand the developments to come in l ater chapters. This chapter a l so p rovides an introduction to the study of transients on lossless l i nes in order to in dicate how problems a rise due to su rges caused by l ightning and switching. I n the modern power system d ata from all over the system are b e i ng fed contin uously i n to on-line compu ters for control and information purposes. Power-flow stu d i es performed by a computer readily supply a nswers to q ues­ tions concern ing the effect of switching l i nes into and out of the system or of cha nges in l ine parameters . Equa t ions derived in this chapter remain i m portant, however, i n developing an overall u nderstand i ng of what is occurring on a system and i n calcu l a t i ng efficiency of transm ission, losses, a n d I i m i ts of power flow over a line for both s te a d y-s t a t e and transient conditions.

6. 1

REPRESENTAT I O N O F L I N ES

The g e n e r a l e q u a t ions re l a t i n g vo l U t g c (l n u c u r re n t o n a t ransmission l ine recogn ize t h e f.t c t t h a t a l l fo u r of t h e parameters of a transmission l i n e d iscussed i n t h e two preced ing c h a p te rs a re u n i formly distributed along t h e l i n e . W e derive these general equations l a ter, b u t first we use lu mped paramete rs w hich give good accu racy for short l i nes and for l i nes of medium lengt h . I f a n overhead line i s classified a s short, shunt capacitance i s so small that i t c a n b e om itted entirely with l ittle loss o f accuracy, and we need to consider only t h e series resistance R and t h e series i n d uctance L for t h e total length o f t h e l i n e. A medium-length line can be represented sufficiently well by R a n d L a s I u m p e d parameters, as shown i n Fig. 6.2, w i t h h a l f the capacitance to n e u t r a l o f t h e l ine lumped a t each e n d of t h e equivalent circuit. S h unt conductance G , a s me ntioned previously, i s usually neglected i n overhead power transmission l ines when calculating voltage and current. The same circu i t represents the short l in e i f capacitors are omitted. I nsofar as the h a n dling of capacitance is concerned, open-wire l ines l ess than about 80 km (50 m i ) long are short l ines. Medium -length l i nes a re roughly betwe en 80 km (50 mi) a n d 240 km 0 50 mi) long. Lines longer than ( I SO m i) req u i re c a lcul a t i o n s i n terms of d istributed constants if a h igh degree of accu racy is req u i re d , a l t ho u g h fo r som e pu rposes a lumped-parameter representation can be used for l i nes L I P to 320 km (200 mi) long.

60-Hz

km +

J I�

R

L

240

+

FI G U RE 6.2

S i ngle-phase equ ivalent of a m e d i u m ­ l e n g t h l i n e . T h e capacitors a r e omitted fo r a s h or t l i n e .

196

CHAPTER

6

CURRENT AND VOLTAGE R ELATIONS ON A TRANSM ISSIOi' LINE

Normally, transmission l i nes are operated w i t h b a l anced three-phase l oads. Although the lines are n o t spaced equ ilaterally and not transposed, the resu l t ing dissymmetry is slight and the p hases are considered to be balanced. I n o rder to distinguish between the total series impedance of a l i n e and the s eries i m p edance per u n i t length, the following nomenclature is adopted : z = series im pedance per u n i t l ength per p hase

y I

=

=

Z =

6.2

shunt admittance per u n i t l ength per p h ase to neutral l ength of l i ne

zl

=

Y = yl

=

tot a l series i mpeda nce per pbase total shunt a d m ittance per phase to neu t r a l

THE SHORT TRANSMISSION LINE

The equivale n t circuit of a short transmission l i ne is s hown in Fig. 6.3, w here Is a n d IR a re t h e sending- and receivi ng-end currents, respectively, and Vs and VR are the sending- and receiving-end l i n e-to-neutral vol tages. The circui t is solved as a simple series ac circuit. So,

(6.1) ( 6 .2 )

where Z is zl, the total series impedance of the line. The effect of t h e variation of t h e power factor of the load on the voltage regulation of a line is most easily u nderstood for the short li ne and therefore will be considered at this time. Voltage regul ation of a t ransmission l i ne is the rise in voltage at the rece iving end, expressed i n percent of ful l -load voltage, when full load at a specified power factor is removed while the send ing-end

Z -R +j wL

Gen.

Equ iva l e n t circu i t of a short transmission l i n e w h ere t h e resis t a n c e R and i nd u c t a nce L are values

FIGURE

6.3

for the e n t i re length of the line.

6.2

( a ) Load

p.

f. = 70 %

lag

( b ) L oad

p.

f.

=

1 00 %

TH E S H O RT TRANS M I SS I O N L I N E

(c)

Load p.

1 97

f. = 70 % lead

FI G U R E 6.4

Phasor d i agrams of a s h ort tra nsmission l i n e . A l l d i agrams are d rawn for t he same m ag n i t u des of VR a n d JR '

vol tage is held constant. Correspond i ng to Eq. (2.33) we can write Percent regu l a tion

=

I Vn . N f. 1 - I VR . FI. : I VI( . 1-'1. 1

X

1 00

( 6 .3)

where I VI? N L I is the magn i t u d e of receivi ng-end volt age at no load and I VR. F L I i s the magnitude o f receiving-end vol t age at ful l load with i Vs l constant. After the load on a short t ransmission l i n e , rep resented by the circuit of F ig. 6.3, is removed, the vol tage a t the recei v i ng end is equal to the voltage at the sen d ing end. I n Fig. 6 .3, wi th the load connected , the receiving-end voltage is designated by VR , and I VR I = I VR. FL I. The send ing-end voltage is Vs ; and I Vs l = I VR . NL I. The p h asor d i agrams of Fig. 6.4 are d r awn for the same m agnitudes of the receivi ng­ end voltage and current and s how t h at a l arger value of the se nding-en d voltage i s req u i red to maintain a given receivi ng-end vol tage \vhen the receiving- e n d curre n t i s l agging t h e vol t age t h a n when t h e s a m e curre nt and vol tage a re i n p h ase. A s t i l l s m a l l e r sen d ing-end voltage i s required t o m a i n t a i n t h e g iven receiving-end voltage when the receiving-end current leads the vol t age. The voltage d rop i s the same i n the series impedance of the l ine i n all cases; becau se of the d i ffe rent power factors, however, the vol tage drop is added t o t h e r e ce i v i n g - e n d v o l t il g e il t a d i ffe r e n t a n g l e i n each c a s e . T h e r eg u l a t i o n i s grea test fo r l a g g i n g pow e r factors a n d l e a s t , o r e v e n negat ive, ro r l e a d i n g pow e r factors. The inductive re actance of ( t tran s m is s io n line i s l a rger t h a n the res ista nce, a nd t h e p rinciple of reg ula t ion i l l u s trated in F i g . 6.4 is true for any load s u p p lied by a p redominantly indu ctive c i rc u i t . The magn i t udes of the volt age drops fa R a n d lu X L for a short line have been exagge rated with respect to Va i n drawing t h e p h a s o r d i a g r a m s i n o r d e r t o i l l u s t ra t e t h e p o i n t m o r e c l e a r l y . The relat ion between power factor and regu l a tion for longer l i n e s is similar to t h a t for s hort l i n es but i s not visua lized so easily.

300-MVA 20·kY three-phase generator has a subtransient react­ ance of 20%. The generator supplies a number of synchronous motors over a 64-km transmission line having transformers a t both ends, as shown on the one-line d iagram of Fig. 6.5 . The motors, a l l rated 1 3.2 k Y , are represen ted by just

Exam p l e 6 . 1 . A

, '198

T� --o -------'?l�]--.:. , � o--y �>-l !CHAPTER

( 2 0 kV)

�

6

C U RRENT A N D VOLTA G E RELA T I O N S ON A T R A N S M ISSION L I N E

T

( 1 3 , 8 kV)

( 2 3 0 kV)

FIGURE 6.5 One-line d iagram

for

M

.,r-"r t>

<J q

M

1

Y

Example 6. 1 .

two equivalent motors. The neutral of M , i s g o u n d d through reactance. The neutral of the c ond M2 not to ground (an u nusual condition). Rated inputs to the motors are 200 M Y A and 1 00 kYA for M J and M 2' respectively. For both motors X� = 20% . The transformer Tl is 350 MY A, 230/20 kY with l k g e of 10%. T2 is composed of rated 1 27 / 1 3.2 kY, 1 00 MYA with l ea k g of 1 0 % . of t r a n s m i ss i o n is 0.5 il / km. Draw the reactance d iagram with all reactances marked i n per unit. Select t h e generator rating as base in the generator

se motooner motis or ctohnnecree-ptrehad se e rated a therreeeacstiannglcee-phase t raSernseafioeasrmrereraces,actaeanctacncehe the Transforml ierne c i rc u i t . is

Solution.

The three-phase rating of transformer T2 3

X

1 00

= 300 kYA

and its l ine-to-line voltage ratio is I3 x

220 127 = - kY 1 3 .2 1 3 .2

A base of 300 MY A, 20 kY in the generator circu i t requires a 300-MYA b ase i n all parts of the system and fol lowing vol

the

In the transmission line: In the motor circui t:

tage bases:

230 kY ( since 1 3 .2 230 220

=

Tl

is rated 230/20 kY)

1 3 .8 kY

These bases are shown in parent h eses on the one-line diagram of Fig. 6.5. The reactances of the transformers converted to t h e proper base are Transformer T1 : Transformer T2 :

X

X

= 0.1 =

X

300 350

= 0.0857 per unit

( 1 3 .2 ) 2 0 . 1 -1 3 .8

=

0.09 15 per unit

6.2

jO.0857

jO. 1 8 1 5

THE SHORT TRANSMISSION L I N E

199

j O . 09 1 5

m

j O . 5490

FIGURE 6.6 R e a c t a n c e d i agram for Example 6 . l . Reactances a re in per u n i t on the spec i fie d base.

The

base impedance

of the transm ission (230) 2 300

l i n e is 176.3f1

and the reactance of the line is 0 .5 x 64 1 76 .3 Reactance X� of motor MJ

=

Reactance X� of motor M2

=

JS t h

e MW, respectiveltyh, e

Figure 6.6 om itted.

0 . 1 8 1 5 per unit

=

0 .2

( 300 ) ( 1 3 .2 ) -

--

-

--

200

1 3 .8

2

( 300 ) ( 1 3 .2 ) 2 0.2 1 00

1 3 .8

p

=

0 .2745 e r unit

=

0 .5490 per unit

required react ance diagram when transformer

operate e e MW, p

phase shifts a re

motors M J and M 2 o f Example 6. 1 h a v inputs of 1 20 and 60 at 1 3.2 kV, a n d bot h at un ity power factor, find the vo lt age at the termin als of the generator and the vol tage regu l ation of the line. Exa m p le 6.2. I f

Solution.

Together the motors t a k 1 80 l �O

300 Therefore, with V and I

at

t

he

=

or

0.6 er unit

motors i n per unit,

IVI

x

III

=

0.6 per u n it

200

CHAPTER 6

CURR ENT AND VOLTAGE RELAT I ONS ON A TRANSMISSION L I N E

V

W ith phase-a voltage at the motor terminals as reference, we have

J

=

=

1 3 .2

1 3 .8

/ O.9565L.9.: no

=

0.6

=

0.9565

unit per unit

per

0 . 6273LQ: 0"

P h a:; e -a p e r - L1 n i t vo l t a g e s a t nt l l er po i n t s o r F i g . (d) ; \ r e

At m :

At

At

V

I: V V k:

=

0 .9565 0 . 9565

=

0.9565 0.9565

=

0 .9565 0.9565

+ + + + + +

0.6273 ( j0.09 1 5 ) jO .0574

=

0.9582/ 3 .4340

per unit / 10. 1 540 uni t 0.9826/ 13.23T uni t

0. 6273 ( J0 .0915 j0 . 1 7 1 3

=

=

j0 . 1 815)

0 .971 7

0 . 6273 ( J0 .091 5 jO.2250

+

+

per

j0 . 1 8 1 5

+

iO.08S7)

per

The voltage regulation o f the l i n e i s

0 .9826 - 0 .9582 0.9582

Percent regulation and magnitude l generator s h ow t h e pha s e s h i f t s due t o t h e t r a ns f o r m e r s , t h e an g l e s thphae se-a cur elnta ine thae lineanshoulshdoulalsdo bebe iinnccrreeaasseedd bbyy fThenrom the angle of =

the

or t h e vo t a g e a t t h e

0.9826 X 20

I f it is desired to of phase-a vo t g s t the

6.3

m

d

100

=

2.55%

t erm i n a l s i s

=

1 9 .652 kV Y

I

x

-

�

30° . 30°

0° .

THE MEDIUM-LENGTH LINE

The shunt admittan ce, usually pure c ap aci t ance, is i ncluded in the calculations for a line of medium length. I f the total s hu n t admittance of the line is d iv ided into two equal parts placed at the sendi n g and receiving e nds o f the line, the circuit is called a nominal 7r . We refe r to Fig. 6.7 to derive equations. To Qbt a i n a n ' expression for Vs) w e note t h at the c u rr e n t i n t h e capacitance a t t h e

6.3

Tl-I E M E D I U M-LENGTH L I N E

201

FIGCRE 6 . 7 Nom i n a l -;;- circu i t o f a med i u m-length

t ransmission line.

receiving end is VRY/2 and the curre n t in th e series arm is IR + VRY/2. Then,

( 6 .4) ( 6 .5 ) To derive Is , we note th a t t h e cu rre n t i n t h e sh u n t capacitance a t t h e sen d i n g end i s Vs Y/ 2 , which added t o t h e cu rren t i n the se ries a rm gives

( 6 .6) Substituting Vs , as given by Eq.

(6.5), in Eq . (6.6) yields ( 6 .7)

Equ ations

(6.5) and (6.7) may be expressed in the ge neral form ( 6 .8) ( 6 .9)

wh ere

ZY

A = D = - + l

2

( 6 . 10)

B = Z These ABeD constants are sometimes called the generalized circuit constants of the transmission line. In general, they are complex n umbers. A a n d D are d im ensionless and equ a l each other i f the l ine is the same when viewed from either end. The d i mensions of B and C a re ohms a n d m hos or Siemens,

202

CHAPTER 6

C U RRENT AND VOLTAGE R ELATI O N S ON A TR A N S M I S S I O N L I N E

respectively. The cons tants apply to any l inear, p assive, and bilateral four- termi­ nal n e twork h aving two pairs of ter m inals. S uch a netwo rk is c alled a two-port

network .

A p hysical m eaning is easily a ssigned to the constants. By letting 1,< be zero i n Eq. (6.8), we see t h at A is t h e r atio VS / VR at no load . Similarly, B is t h e ratio Vs /IR when the receivi ng end is sho rt-ci rcuited. The constan t A is u s eful in c om p u ting regul ation. I f VI?, '.L i s t h e receiving-en d vol tage at fu [ I load for a sending-end voltage of ��. , Eq. (6.3) becom es

Percent regulation

=

- I VJ?.F/J I VU, FL I

I Vs l /l A I

x

1 00

(6.1 1 )

Table A.6 in the Appendix l ists ABeD constants for v a riou s networks and combinations of n c tworks.

THE LONG TRANSMISSION LINE: SOLUTION OF THE DIFFERENTIAL EQUATIONS

6.4

150

The exact sol ution of any transmISSIon l in e and the one required for a h igh degree of accuracy in calculating 60-Hz l ines more than approximately mi long m u s t consider t h e fact that t he parameters of t he : i nes are not lumped but, rather, a re distributed uniformly throu gh o u t the length of the l ine. Figure 6.8 s hows one phase and the neut ral connection of a three-phase line. Lumped p aram e ters are n ot shown because we are ready to consi der the solution of the line with the impedance and admittance uniform ly d istributed . I n Fig. 6.8 we consider a d ifferential element of length dx i n the l in e a t a d ista nce x from the receiving end o f the line. Then z dx and y dx a re, respectively, the series i mpedance and shu n t ad m it tance of the elemental section . V and I a re ph asors which v a ry w i th x .

+

I I I

Gen.

Vs

V + dV

:

�

I

I

dx --t+-I

----

x --------I

FIGURE 6.8 Sche matic diagram of a transmission l i n e show i n g one phase a n d the neu t ra l ret u r n . Nomenc; l a t u re for the l i ne and the elemental length a re i n dica t e d .

6. 4

TH E LONG TRAN S M ISSION LI N E : SOLUTI ON OF T H E D I FFER ENTIAL E Q U ATIONS

203

Average line current in the el ement is (J + I + dJ) /2, and the increase of V i n the distance dx is quite accurately expressed as

dV =

I + 1 + dI

�---

2

z dx

=

Iz dx

( 6 . 1 2)

when products of the differential qu antitie s are neglected. Similarly, dJ

=

v

+

V + dV y d.x: 2

-----

=

Vy dx

( 6 . 13)

The n , from Eqs. (6. 1 2) and (6. 1 3) we have

dV

( 6 . 14)

- = fz dx

and Let u s differenti a te Eqs.

dI dx

-

(6. 14) and (6. 1 5) w ith respect t o d2 V dx 2

--

and

Vy

=

d2]

dx 2

-

= z

=y

( 6 . 1 5) x,

dI dx

d2V dx

and

dx 2

=

( 6 . 1 6)

-

dV dx

( 6 . 17)

­

If we subst itute the va lues of dI/dx and dV/dx: from Eqs. Eqs . (6. 1 6) and (6. 17), resp ectively, we obt a i n -2 =

and w e obtain

(6. 15) and (6. 14) in

yz V

( 6.18)

yzl

( 6 . 1 9)

Now we h ave E q . (6.18) i n which t h e o n l y vari ables are V and x and Eq. (6. 1 9) in which the only vari ab l es are I a n d x . The sol u tions of those eq uations for V and I , respectively, must be expressions which when d ifferentiated twice with respect to x yield the original expression t i m es the constant yz . For instance, the solution for V when d i fferent iated twice with respect to x must yie ld yzv. ,

204

CHAPTER

6

C U RRENT AND VOLTAG E RELATIONS ON A TRANS M I SS IO� L 1 � E

This suggests a n exponential form of solution. Assume t hat the solu tion of Eq. (6. 1 8) is ( 6 .20) Takin g the second derivative of V with respect to d2 V

dx 2

�- = yz

[AI

c

jYix

+ A

2

c

x

in Eq. (6 .20) yields

- fYl

x

]

( 6 .2 1 )

which is yz t imes the assumed solut ion for V. Therefore, E q . (6.20) i s t he solution of Eq. (6. 18). \V hen we subst itute the value given by Eq . (6.20) fo r V i n Eq. (6 . 1 4), w e obt a i n

J 1 =

vz/y

A c fYlx I

..;

1

z /y

A -, E

· Fzx

( 6 .22)

�

The constants A 1 and A 2 can b e evalu a ted by u sing the con d i tions at the receiving end o f the line; namely, when x = 0 , V = VR and J = JR' Substitu tion of these values in Eqs. ( 6.20) and (6 .22) yiel ds

and Substituting Zc = vz/y and solving for A I gIVe and Then, substituting t h e values found for A I and A 2 in Eqs. (6.20) and (6.22) and letting y fYZ, w e obtain =

( 6 . 23 )

( 6 . 24 ) where Zc

fiY

=

vz/y and is called t h e characteristic impedance of the l in e , and

and is called the propagation constant . Equations (6.23) and (6.24) g ive the rms values of V and J and their p hase angles at any specified point along the l i n e i n t e rm s of the distance x from the receiving end to t he specified poi n t , p rovided v�? , JR , and t h e parameters of the line are known.

y =

6.S

T H E LO N G T R A N S M 1 SS 1 0N Ll N E: l N TE R P R ETATION OF THE EQUATIONS

205

6.5

THE LONG TRANSMI S SI ON LINE: I NTERPRETATION OF THE EQUATIONS Both y and Ze are complex q u a ntities. The r e a l part o f the propagation constant y is called the attenuation constant a and is measu red i n nepers per unit l e ngth . The quad rat u re p a rt o f y is called the phase constant f3 and is measu red in rad ians per u nit l e ngt h . Th us, y

= a

+ jf3

( 6 . 25 )

a n d Eqs. (6.23) and (6.24) become

( 6 . 26 ) and

( 6 .27)

The properties of E ClX a n d E J{3x help to expla i n the vanatlOn of the p h asor values of vol tage and cu rrent as a fu nction of d istance along the l i ne . The term E Cl .r changes in magnitude as x c h a nges, but E J{3x ( i dent ical to cos {3x + j sin (3x) always has a magnitude of 1 a n d cau ses a shift in phase o f {3 rad ians p e r u nit length of l ine. The fi rst term in Eq. (6.26), [( Vu + fR Z)/2]E Clx£ i/3x, i ncreases in m agni­ tude and advances in phase as d i s tance x from the receiving e nd incre ases. Conversely, as progress along t h e l i ne from the sending end toward the receiving end is considered, the term d i mi nishes i n magnitude and i s retarded in phase. This is the characteristic of a t raveling wave and is simil a r to the beh avior of a wave i n water, wh ich va ries in magnitude with time at any poin t, whe reas its phase is retarded a n d its maximu m val ue dimi nishes with d istance from the origi n . The var iat ion in i ns tanta neous value is not expressed in the t e r m b u t i s u n d e rs t o o d s i n ce VI? a nd I N (I r e p h Cl sors . T h e fi rs t term i n Eq. (6.26) is c a l l e d t h e in ciden t I 'o/tage . The second term i n E q . ( 6 .2(»), [( Vu - lu Z)/2]E - Ir.I C - ii3 r , d i m i nishes in magni tude a n d i s retarded i n p hase from the rccc iving end toward the send i ng end . I t is called the reflecred voltage . At any point along the l ine the voltage is the s u m of t h e c o m p on e n t i nc i d e n t a n d reflected vol tages at that point. S i n c e t h e e q u a t i o n fo r c u r r e n t i s s i m i l a r to t h e e q u a t io n for vol t a g e , t h e

cu r r e n t may bc considered t o be com poscci of i ncident and reRected cu rrents. If a l ine is termi nated in its cha racteristic i m pedance Ze ' rece ivi ng-en d vol tage VR i s e qual to fR Ze and t h e re i s n o reflected wave of e ither vol tage o r cu rrent, as may be seen by subst i t u t i ng fR Ze for VR i n Eqs. (6.26) and (6.27). A line term inated i n its characte rist i c i mpedance is cal led a fiat line or a n infinite line. The latter term arises from the fact that a l i ne of infinite l ength c a nn ot h ave a reflected wave . Usu a l ly, power l i nes are not terminated i n t � eir charac­ te ristic impedance, but com m u n i cation l i nes are frequently so terminated i n

206

CHAPT E R 6

CURRENT AND VOLTAGE RELATIONS ON A TRAN S M I S S I O N LI N E

order to eliminate the reflected wave. A typical valu e of Zc is 400 fl for a single-circui t overhead l in e a n d 200 n for two circuits in p a ra l l e l . Th e ph ase angle of Zc is usually between 0 and 1 5° . Bundled-conductor lines have lower values of Zc since such lines have lower L and h igher C than lines with a single conductor per phase. In power system work characteristic impedance is sometimes called surge impedance. The term "surge impedance," however, is usually reserved for the special case of a loss\ess l i n e . I f a l i n e is lossless, its series resistance and shunt conductance are zero and the characteristic impedance reduces to the real number ..;L IC , which has the dimensions of ohms when L is the series inductance of the line in henrys and C is the shunt capacitance in farads. Also, the propagation constant y = {ZY for the line o f length I red uces to the im aginary number j{3 = jW'/LC' ll since the attenuation constant a res u l ting from line losses is zero. When dealing with h igh frequencies or with surges d ue to lightning, losses are often negl ected and the surge impedance becomes important. Surge-impedance loading (S I L) of a line is the power delivered by a line to a purely resistive load equal to its su rge impedance. When so load ed , the line supplies a current of -

where I VL I is the line-to-line vol tage at the load. Since the load resistance,

or with

I V" I

in

k i lovo l ts,

SIL

=

I V 12 '..;i7c MW L ie

IS

pure

( 6 .28 )

Power system engineers sometimes find it convenient to express the power transmitted by a line in terms of per unit of S I L, that is, as the ratio 'of the power tra nsmitted to the surge-impedance loading. For i nstance, t h e p ermissi­ ble loading of a transmission line may be expressed as a fraction of its S I L, and S I L provides a comparison of load-carrying capabilities of lines.I A wavelength � is the d istance along a line between two poin ts of a wave which d iffer in phase by 360° , or rad. If {3 is the phase shift in radians per

27T"

I See R. D . D u n lop R. Gu t m a n , a n d P. P. M a rc h e n k o " A n a l ytica l Developme n t of Loa'd ab i l i ty Cha racteris t ics fo r EH Y a n d U H Y Trunsm ission L i n e s , " IEEE Tra nsactions on Power Apparatus ,

and System s , vol. PAS·98,

,

nO.

2, 1 979, p p . 606- 6 1 7 .

6.6

207

THE LONG TR A N S M ISSI ON L I N E : H Y P E R BOLIC F O R \1 OF TH E E Q U AT I O NS

mile, the wavelength in miles is 2 rr

( 6 .29)

.1 = -

{3

The veloci ty of propagation of a wave i n miles per second is the product of the wavelength in miles and the frequency in hertz, or Velocity

=

A. I

For the lossless line of length I meters f3 (6.30) become I

fiL e

----= --, ---

m

2 rrI

( 6 30 )

f3

= -­ =

v e l oc i t y

.

2rr tlLC I I and Eqs . (6.29) and =

I

!LC

--

mls

values of L and C for low-loss overhead l ines are subst ituted i n these equations, it is found that the wavelengt h is approximately 3000 mi at a frequency of 60 Hz and the velocity of propagation is \'e ry nearly the speed of l ight in air (approximat ely 1 86,000 mils or 3 X 1 0 8 m/s). If there is no load on a line, IR is equal to zero, and as d etermined by Eqs. (6.26) and (6.27), the incid ent and reA ected voltages are equal i n magn itude and in phase at the receiving end. In this case the i ncident and reflected currents are equal i n magnitude bu t are 1 800 out of phase at the receiving end . Thus, the i ncident and reflected cu rrents cancel each other at the receiving end of an open line but not at any other poi nt unless the l ine is entirely lossless so t hat the atten uation a is zero. Wh e n

6.6

LONG TRANSMISSION LINE: HYPERBOLI C FORM OF T HE E Q U ATIO N S THE

Th e n i

i c de n t a n d re fl e c t e d waves of vol t a ge a re s e l d om fo u n d w h e n calculating the voltage of a power l ine. The reason for discussing the vol tage and the cu rrent of a l i n e i n terms of the i ncident an d reflected components is that such an a n alys i s i s helpful i n obt a i n i ng a be t t e r u nd e rs t a n d i ng of some of the p h e no m e n a o f t ra n sm i s s i o n l i n e s . A m o re c o n v e n i e n t fo r m o f t h e e q u a t i o n s for c o m p u t i n g c u rr e n t a n d vo ! l
2

= ----

=

2

( 6 .3 1 ) ( 6 .32)

208

CHAPTER 6

C U R RE NT AN D VOLTAGE R ELATIONS ON A T RANSMISS ION LIN E

By rearranging Eqs. (6.23) and (6.24) and substitut ing hyperbolic functions for the expone ntial terms, we find a new set of equat ions. The new equat ions, giving voltage and current anywhere along the l in e, are ( 6 .33 )

I = Iu cosh "IX Letting x = /

to

obta in

the

vo l t a g e

and

+

VR

- si n h "IX

( 6 . 34 )

Z(

t h e c u r re n t

(It

t h e s e n u i n g e n u , we

h a ve

( 6 .35 ) Is

=

If(

cosh "1 /

+

VI?

Z

sinh

c

( 6 . 36)

yl

From examinat ion of t hese equ ations we see that the general ized circui t constants for a long l ine are A

= cosh yl

B = Zc sinh yl

Solving Eqs. (6.35) and (6 .36) for VR and

sinh 1'1 c = ---

D

IR

( 6 . 37)

= cosh 1'1 in t erms of Vs and

Is ,

we obtain ( 6 .3 8 )

II?

=

Is cosh "1 / -

v:�. .

Zc

S J Il h

( () . 3 9)

"1 /

For balanced three-phase lines the currents in the above equat ions are l i ne currents and the voltages are li ne-to-neutral volt ages, that is, l ine volt ages d ivided by 13 . In order to solve t he equ a t ions, t he hyperbolic functions must be evaluated. S ince "II is usually complex, the hyperbolic functions are a lso complex and can be evaluated with t he assistance of a calculator or compu ter. For solving an occasional problem without resort ing to a compu t er t here are several choices. The following equat ions give the expansions of hyperbolic sines and cosines of complex arguments in terms of circular and hyperbolic functions of real arguments: cosh ( exl + j{3/ )

sinh ( exl + j{3l )

= =

cosh exl cos {31 + j sinh al sin {31

sinh exl cos {31 + j cosh exl sin {3i

Equa tions (6. 4 0) and (6. 4 1 ) make poss i ble th e

co m p u ta t i o n

( 6 . 40 ) ( 6 .41 )

of hype rbolk

fu nc-

6,6

209

TH E L O N G T RA N S M I SS I ON L I N E : H Y P E R B O L I C FO RM OF TH E EQUATI O N S

tions of complex arguments. The correct mat hematical unit for {3/ is the radian, and the radian is the unit found for {3 1 by computing the quadrature component of "II. Equations (6 . 40) and (6 . 4 1 ) can be verified by substituting in t hem t h e exponential forms o f t h e hyperbolic functions a n d t h e similar exponential forms of the circular functions. Another method of evaluating complex hyperbolic functions is suggested by Eqs . (6.3 1 ) and (6. 32). S ubstituting a + j{3 for 0, we obtain cosh ( a + j{3 )

c "E Jf3 +

=

-

sinh( a + j{3 )

-

£ (r E jf3

-

£ - "£ -jfJ

2 E

2

- a C - jf3

=

1

2

(E"iJ!.

+ £-"

U)

Z (eaL!i - c -aU ) 1

( 6 .42)

( 6 .43 )

370 km ( 230 m i ) l ong. The w i th fl a t ho r i zon t a l spacing and 7.25 m (23.8 ft ) b etween on the l i n e i s 1 25 MW at 215 kY w i t h 1 00% powe r factor.

cond u c to rs a re Rook cond uctors, The load F i n d t h e v o l t a g e , c u r re n t , a n d powe r at t h e se n d i ng e n d a n d the voltage reg u l a t ion of the l ine . Also, d e te rm i n e t h e wavelength a nd veloci ty of propagation of the l i n e .

60-Hz t ra n s m ission

Exa mp le 6 .3. A s i n g l e -c i rc u i t

line is

Solution.

Feet a n d m i l e s r a t h e r t h a n m e ters a n d ki lome t ers a r e chosen fo r t h e calcu l a tions i n ord e r to u s e Tab les A.3 t h rough A.5 in the Appendix : Deq

/23 .8

J �-:------:-=--::--- ---:3 X ==

=

a n d from t h e tabl e s for Rook z =

y =

"II

VR iR

=

=

0 . 1 603

+

;Y; I

=

O . 4 772

+

0 .4 1 27)

+

L84 , 5 2°

=

0 .0456

0 . 843 1

=

215 ,000 13

=

i

S mi

/, / 79 .04° + 90° 230 0 .843 1 X 5 . 105 X 1 0 - 1> �--2

5 . 1 05 X 1 0 - 6 =

30.0 ft

/ 0.843 1 79 .0r n/mi 6 L22: ] 0 . 1 009) X 1 0 - = 5 . 1 05 X 1 0 - 6

j(0,4 1 5

[ j 1 / ( 0 ,0950

2 . 8 X 47.6

=

1 24 , 1 30

j0 .4750

/ +

7 9 . 04°

2

L.2: V to neut ral

1 25 ,000,000 Li2.: = 335 .7 / 13 3 X 2 1 5 ,000 {\O

A

-

90° =

406 .4

/ - 5 . 48°

n

210

CHAPTER 6

Eqs.

From

CURRENT AND VOLTA G E R ELATIONS O N A TRANS M ISSION L I N E

( 6.42) and ( 6.43) and noting tha t 0.4750 rad

= 27.22°

= 0 .4654 + j O . 23 94 + 0 .4248 - jO.2185 =

sinh yl

0.8902 + jO .0209

= 0 .4654

+

0.0406

+

=

jO .2394 - 0.4248 j0 .4579

Then, from Eq. (6.35)

Vs =

124 , 1 30 x 0.8904 / 1 .34°

/27.77°

=

1 10,495

=

1 37 ,860

and from Is

Eq .

+ j2,585

0.H904 / 1 .34°

=

+ 1 1 ,483

+

jO.21�5

= 0 .4597/ 84 .93°

335 .7 x

+

+

406.4/ - 5 .48°

x

j61 ,656

V

( 6.36)

= 335 .7 x 0.8904/ 1 .34° = 298 .83 + j6.99 - 1 .00 = / A 332.3 1

0 .4597[84 .93°

26 .330

+ +

124,130

406 .4/ - 5 .48°

/t... ....- -

x 0 .4597 84.93°

j 1 40 .4 1

At the sending end

= fS x 137 .86 = 238.8 k V Line current = 332.3 A Power factor = co ( 27 . 77 ° - 26 .330 ) = 0.9997 1 .0 Power = fS x 238.8 x 332.3 x l .0 137 ,443 ( 6.35 ) we see that a t no load = 0) Line voltage

s

=

=

From

Eq.

( JR

V

I?

Vs

- cosh y l -

--­

kW

6 . 6 THE LONG TRANS MISSI ON L I N E: H YPER BOLIC FORM OF THE E QUATIONS

211

So, the vol tage regulation is

1 37 .86/0 .8904 - 124 . 13 ------- X 100 = 24 .7% 124 . 1 3 The wavelength and velocity of propagation are computed as follows:

0.4750 230

{3 =

Vel ocity

{3

=

fA.

=

0.002065 rad / m i

0 .002065

----

=

60

x 3043

=

=

3043 m i

1 82,580 mijs

We note particularly in this example t h at in the equations for Vs a nd Is the val ue of vol tage m ust be expressed i n vol ts and must be the l ine-to-neu tral

voltage.

Exa mple 6 . 4 .

Solve for t he send ing-en d voltage and t he current found in Example

6 .3 using per-un i t calculations.

Solution . We choose a base of 1 25 M VA, 2 1 5 kV t o achieve the simplest per-unit values and to compute base impedance and base current as follows: Base impedance B ase c u rre n t So,

Zc V

=

=

/I

For l i s e i n E q . ( 6 . 3 5 )

VI?

=

/

=

x

-

we

1 .0

c h o s e VI<

/..Q:

as

n

= f3125 ,0002 1 5 =

- 5 .48° --�===370 v0" 215/ 215 215 2 1 5 / 13 406 .4

= 370

2152 -1 25

335 . 7

/ 1 .098 - 5 .48° per u n i t 1 .0

per u n i t

t h e r e fe re nce vo l t <1 g e . So,

per unit ( as

a

l ine-Io-neutra l v o ltage )

and s ince the load is at unity power factor,

IR

=

A

/..Q: 337.S 337 .5

/..Q: l .o

212

CHAPTER 6

CURRENT A N D VOLTAGE R E LAT[ONS O N A TR A N S MISSION LI N E

If the power factor had been less than 1 00%, IR would have been greater than l.0 and would have been at an angle d etermined by the power factor. By Eq. (6.35) Vs

= 0 .8902 jO.0208 0.0923 = / per u nit

= 1 .0 X 0 .8904

+

+

1 . 1 102

1 .0

1 .098

x

+

/

+

-

1 .0

X

=

+

/

0.4597 84 .93°

27 .7SO

0 .8904 / l . 34°

= 0 .8902

X

j0 .4961

and by Eq. (6.36) Is =

5 .48°

+

1 .0

LQ:

1 .098L- 5 . 4 8 °

x

0.4597/ 84 .93°

fO .0208 - 0 . 003 1 + j 0 . 4 1 86

0 .990/ 26 .35° per u n i t

At t h e sending end Line voltage

=

1 . 1 1 02

Line current

=

0 . 990

x

X

2 15

335 .7

=

=

238.7 kV 332 . 3 A

Note that we multiply li ne-to-li ne voltage base by t h e per-unit magnitude of the voltage t o find the li ne-to-line voltage magnitude. W e could have multipl ied the line-to-neutral voltage base by the per-uni t voltage to nnd the li ne- to-neu tral voltage magni tude. The factor 13 does not enter t h e calculations a fter we h ave expressed all quanti ties in per unit. 6.7 OF

THE EQUIVALENT CIRCUIT A LONG LINE

The nominal-7T circuit does not represent a transm ission l ine exactly because i t does n o t account for the parameters o f t h e line be i ng uniformly distributed. The discrepancy between the nom inal 7T and the actual line becomes larger as the length o f l ine increases. I t is possible, however, t o find the equivalent circui t o f a l ong t r a n s m iss io n line and to rep resent t h e l i n e accu rately, in sofar as measure­ men ts at the ends of the line are concerned, by a network of lumped parame­ ters. Let us assume that a 7T circu i t similar to that of Fig. 6.7 is the equivalent circui t of a long l ine, but let us call the series a rm of our equivale nt-7T circuit Z' and the shunt arms Y' /2 to distinguish them from the arms of t he nominal-7T circuit. Equation (6.5) gives the sending-e n d voltage of a symmetrical -7T circuit in terms of its series and shunt arms and the voltage and current at the receiving end. By substituting Z' and Y ' /2 for Z a n d Y/2 in Eq. (6.5), we obtq.in t he sending-end vol tage of our equ ivalent circu it in terms of its series and shunt

6.7

THE EQUIVALENT CIRCUIT OF A LONG LI NE

( Z'Y' 1

213

arms and the voltage and current at t h e receiving end: Vs =

2- + 1

-

VR + Z ' IR

( 6 .44)

For our circuit to be equivalent to t h e long transmission l ine the coefficients of VR and IR in Eq . (6.44) must be identical, respectively, to t h e coefficients of VR and IR in Eq. (6.35). Equating the coefficients of JR in the two equations yie l ds Z'

=

Z'

=

Zc

r; -

y

z' = z

( 6 .45)

s i nh yl

sinh y l

= z/ -=--

sinh y ! \ zy

I

sinh y l

( 6 .46 )

-­

yl

where Z is equ al to zl, the total series impedance of the l ine. The term (sinh y /)/yl is t he factor by which the series impedance of the nominal 7T m ust be m u l tiplied to convert the nominal 7T t o the equ ivalent 7T . For small values of y/, both sinh y I and y I are almost i d entical, and t his fact shows t hat the nominal 7T represents the med ium-l ength transmission I ine quite accu rately, insofar as the series arm is concerned . To investigate the s hu nt arms of the equivalent-" circu it, we equ ate t h e coefficients o f VR i n Eqs. (6.35) and (6.44) and obt ain + 1 = cosh y !

Z'Y' --

2

( 6 .4 7 )

S ubstituting Zc sinh yl for Z' gives Y 'Zc

sinh y !

-----

2

+ 1

=

cosh y l

1 cosh yl

Y' 2

Zc

-

1

( 6 .48 ) ( 6.49)

sinh yL

An other form of the expression for the s hunt admittance of the equival ent circu i t can be found by substituting in Eq. (6.49) the identity

tanh

yl -

2

cosh y l =

-

sinh y l

1

( 6 .5 0 )

The i dent ity can be verified by substituting the exponential forms of Eqs. (6.3 1) ,

214

CHAPTER 6

Z'

=

C U R R ENT AND VOLTAG E R E LATI O N S ON A TRANSMISSION L I N E

Z C si n h "'l I

=

Z

sinh YI yl

1:: = J.... 2

y' 2

Zc

:=: 2

ta n h

!..!. .2

FIGURE 6.9 Eq u iva J e n t -7T ci rcu i t of a t r ans­ m ission l i n e.

y tanh rl/2

1//2

and (6.32) for the hyperbol ic fu nctions and by recal ling that tan h sinh () Icosh e . Now Y'

2

-

2

-

"II 1 - tanh -

Y' Y

Y

2

=

( 6 .5 1 )

2

Zc

()

tanh ( "1112 )

( 6 .52)

"1 112

where is equ al to yl, the total shunt admittance of the l ine. Equation (6.52) shows the correction factor u sed to convert the admittance of the shunt arms of the n ominal 7r to that of the equivalent 7r . S ince tanh( "1112) and "1112 are very n e arly equ al for small values of "I I, the nominal TT represents the medium-length transmission l ine quite accurately, for we h ave seen previously that the correc­ tion factor for the series arm is n egligible for medium-length lines. The equivalent-TT ci rcuit is shown in Fig. 6.9. An equ ivalent-T circuit can also be found for a transmission line. 6 . 5 . Find the equivalent-7T circuit for the l ine described in Example 6.3 and compare it with the nominal-7T circu it.

Example

Solution. S i nce s i n h y / a n d cos h y / a rc a l re a d y k nown fro m Exa m p l e 6 . 3 ,

( 6 . 45)

Z'

and =

/ - 5.480

(6.49)

4aC1 . 4

0 .8902

Y'

2

are now used .

+

la.a2a8

/

1 86 .82/ 79 .450

=

0 .4597 / 84 . 930

X

-

1

U36 .82/ 79.45°

/ 1 86 .82/ 79 .45° =

n

Eqs.

in series arm

0 . 1 1 1 8 169 .27°

0.000599 89 .8ZO S in each shunt arm

Using the val ues of z series impedance of

and

y

from Example 6.3,

Z = 230 X

0.84311

79 . 04°

we

=

find for

the

193.9/79.04°

nominal-rr

circuit a

6.8

and equal shunt arms

y 2 For t h i s

is

5 . 105 10 - 6 L22: of

230

---===- x

-

2

l ine the i m p e d a n c e

of t h e equiva l e n t 7T

POWER F LOW THRO UGH A TRANSMISSION LINE

by

=

0.000587 L22:

series arm

of t h e

S

i l

of t h e n o m n a 7T

3 . 8 % . The co n d u c t a nce o f t h e s h u n t a rm s of

2 . 0 % l ess t h a n t ha t o f t h e equ i v a le n t

215

TT .

exceeds t h at the n o m i n a l 7T

We conc lude from the preced ing example that the nomi nal 7T' may represent long l i nes sufficiently w e l l i f a h i gh degree of accuracy i s not requ i red. 6.8

POWER FLOW THROU G H A TRAN SMISSI O N LI N E

A l t h o u g h powe r now a t a ny po i n t

along ( \ l ra nsm ission l i n e can always he fou n d i f t h e voltage , c u r r e n t , a n d power fa ctor a r e k n own o r can b e calcu l ated, very interesting equations for powe r can be d erived i n terms of ABeD constants. Th e equations ap ply to any network of two ports or two terminal p aIrs. Repeating Eq. ( 6 .8) and solving for the receiving-end current JR yields (6 .53 )

( 6 .54)

Le tt i ng

A

=

IAI�

B =

I B ill

we obtain

( 6 .55)

Then, the complex power VR l� at t h e receiving end is ( 6 .56)

216

CHAPTER 6

CURR ENT A N D VOLTAG E R E LATIONS ON A TRA:"<SM ISSJON L I N E

var

-f------L---l-W

FI G U R E 6. JO P h asors of Eq. ( 6 . 5 6 ) p lo l l e d i n t h e c o m p k x p l a n e , w i t h

magn i t u des a n d a n g l es a s i n d i c a t e d .

and real and reactive power at the receiving end a re (6.57)

( 6 .58 )

Noting that the expression for complex power PR + iQ R is shown by Eq. (6.56) to be t h e resultant of combining two phasors expressed in polar form, we can plot th ese phasors in the complex plane whose horizontal and vertical . coordi nates a re in power units (watts and vars). Figure 6. 1 0 s h ows the two complex quantities and their diffe rence as expressed by Eq. (6.5 6). Figure 6 . 1 1 shows the same p h as o r s w i t h . t h e o r i g i n o f t h e coord i n at e ax es s h i ft e d This figure is a power d iagram wi t h t h e res u l t a n t w hose m a g n i t u d e is I Pn + )'Q u l, or I VR I I JRI, at an angle 8 R with the horizontal axis. As expected , the real and imagin a ry components of I PR + iQul a re .

( 6 .5 9 ) ( 6 .60)

where 8R is t h e p hase angle by whic h VR l eads JR , as discussed in C h ap . 1 . T h e sign of Q is consistent with t he convention w h ich assigns positive values to Q when current is l agging the voltage. Now let us determine some points on the power d iagra'm various loads with fixed values of I Vs l and I VR I . First, we notice that the position of point n is not dependen t on the current JR and wil l not ch ange so long as

6.B

p OWE R FLOW TH R O U G H A TRANSM ISS ION L I N E

217

var +

FI G U R E 6 . 1 1

Power d i agram o b t a i n e Ll by s h i fting t h e origin o f the coo rd i na t e axes of Fig. 6. 1 0 .

n

kk

i s constant. We note further t hat the d istance from point to point i s constant for fixed values o f I Vs I a n d I Vu l. Therefore, as the distance 0 to changes with changing load , the point k , since it must remain at a constant distance from the fixed point I I , i s const rai ned to move i n a c ircle whose center is at n . Any change i n P R wi l l req uire a change i n Q R t o keep k on the c i rcle. I f a d ifferent value of I Vs l is held constant for the same value of ! VR I, the location of point n is unchanged bu t a new c ircl e of radius is found. Examination of Fig. 6 . 1 1 shows that the re is a l imit to the power that can b e transmitted to the receiving end o f the l ine for specified m agnitudes of sending- and receiving-end volt ages. A n increase i n power delivered m eans that the point k will move along the ci rcle unti l the a ngle {3 - [) is zero; that i s, more power will be del ivered u nt i l (5 = {3. Further i ncreases in (5 result i n l ess power received. The maximum power is

I VR I

nk

I A I I VR I 2

-- - cos ( f3 - a ) IBI

( 6 .6 1 )

The load mu st draw a la rge l e a d i ng current to achieve t he condition of maximum power received . Usua lly, operation i s l imited by keeping (5 less t h a n about 350 and I Vs l/I VRI equal t o o r greater than 0.95. For short li nes thermal ratings l imit the loading. In Eqs. (6.53) t hrough (6.61) \ Vs \ and \ VR I are l ine-to-neutral vol tages a nd coordinates in Fig. 6. 1 1 are watts a n d vars per ph ase. However, if I Vs l and I VR I are li ne-to-line voltages, each d istance in Fig. 6. 1 1 is i ncreased by a factor o f 3 and the coord inates on t he d iagram a re total three-phase watts and vars. If t h e voltages are kilovolts, t he coordi nates are megawatts and megavars.

218 OF

6.9

CHAPTER

6

C U RRENT A ND VOLTAGE R ELAT I O NS O N A TRA N SM I SSION L I N E

REACTIVE COMPENSATION TRANSMISSION LINES

The performance of transmission l in es, especially those of med ium length and longer, can be improved by reactive compensation of a series or parallel type. Series compensation consists of a capacitor bank pl aced in series with each phase conductor of the line. Shunt compensation refers to the placement of inductors from e ach line to neutral to reduce partially or completely t h e shunt suscep­ tance of a h igh-voltage line, which is particularly important at light loads when the voltage at the receiving end may othe rwise become ve ry h igh. Series compensation re duces the se ries i mped ance of the l i ne, w h i ch i s t h e principal cause of vol tage d rop and the most important fac tor i n d etermi n i ng the maximum power which thc l ine can transmit. I n ord er to understand the effect of series impedance Z on maxim u m power transm ission , we exa m i n e Eq. (6.6 1 ) and see that maximum power transmitte d is d ependent on the reci procal of the generalized circu i t constant B , which for the nominal-1T equals Z and for the equivalent-1T equals Z (si nh ),1)/)'1. Because the A , C , and D constants a re functions of Z , t hey will also c hange in value, but these changes will be small in comparison to the change in B . T h e d esired reactance of the capacitor bank can b e determ ined by compensating for a specific amount of the total ind uctive reactance of the line. This l eads to the term " compensation factor," which is d efined by Xc iXL > w h ere Xc is the capacitive reactance of t h e series capacitor bank per phase and XL is the total inductive reactance of the l i ne per phase. When the nominal-7T circuit is u se d to represent the l ine and capacitor bank, the p hysical location of the capacitor bank along the l ine is not taken into accoun t. If only the sending- and receiving-end conditions of the line a re of interest, this will not create any significant e rror. However, when the operating conditions along the l ine are of interest, t h e physical location of t h e capacitor bank must be taken into account. This can be accomplished m ost e asily by determining ABeD constants of the porti ons of line on each side · of t h e capacitor b a n k a n d by representing the capacitor bank b y i ts ABeD constants. The e quivalent constants of the comb ination (actually referred to as a ca scaded connection) of l ine-capaci tor�l ine can then be d etermined by applying the equ ations found in Table A.6 in the Appendix. In t h e southwestern part of the United S tates series compensa tion is especially i mp ort a n t beca use l a rge generating p l a n ts a re located hun dreds of miles from load centers and l arge amounts of power must be transmitted over l ong d i stances. The lower vol tage drop in t he l ine with series compensation is a n additi onal a dvantage. S eries capacitors are also useful in balancing the voltage drop of two p arallel l ines. Example 6.6. In order to show the relative changes in the B constant with respect to the change of the A, C, and D constants of a line as series compensation is applied , find the constants for the line of Example 6.3 when uncompensated ,and for a series compensation of 70%.

6.9

R EA CT IVE COM PENSATION OF TRA N S M ISSION L I NES

The equivalent-7T circuit and quantities found in Exa m p l e s can be used with Eqs. (6.37) to find, for the uncompensated l ine

Solution.

A

D

=

B (.'

=

=

cosh y l 2'

=

sinh y'

=

= 0.00 1 1 3 1

/

/

/ - 5 .48°

0.4590/ 84 .940

406 .4

90 .420 S

of

t h e e q u i v ,l I e n t -7T

new se r i es ;I r m i m pe d ; l l1 c e i s a l so t h e ge n e ra l i z e d co n s t a n t D. So,

=

=

and by Eqs. ( 6 . 1 0) A

c

The

=

= =

60 .88 2

X

1 86.78/ 79 .46" 34 . 1 7

+

/ 55 .850

j5 0 . 3

x

R

=

0 .000599/89.81°

0.000599/ 89 .8 10 +

0.001 180/ 90.4 10

j O.7 X 230( 0.41 5

60.88/ 55.85°

-

6.5

79 .460 n

T h e se r i e s co m p e n s a t i o n a l t e rs o n l y t h e s e ri e s a rm

J3

and

0 .8 90 4

= 1 86 .78

= _____

/ 1 . 340

6.3

219

+

+

circu it. The

0 .4 1 27)

n

1

=

0.970/ 1 .240

60.88/55.85° (0.000599/ 89 .81° ) 2

S

ex ample shows t hat compe nsation has reduced the const ant B to about one - t h i rd of its value for the uncompensated line without affecting t h e A a n d C co nstants appreciably. Thus, maxi mum power which can be transmitted is i ncreased by ahout ]O() % . When a transmission l i ne , with or withou t se ries compensation, has the d es i re d l o a d t ra n s m i s s i o n c a p a b i l i ty, a t t e n t i o n i s t u r n e d t o o p e r a t i on u n d e r l i ght loads or at no load. Charg i ng cu rrent is an important factor to be considered and should not be al lowed to exceed the rated fu ll-load current of the l i ne . Equ ation (5.25) shows us that the charging current is usually d efined as Bc l V I i f Be is the tot al capacit ive susceptance of the l i ne and I V I is the rated voltage to neu traL As noted foll owing Eq. (5 .25), this calculation is not an exact determination of chargi ng cu rrent because of the variation of I V I along t he l i ne. H we co nnect inductors from l i ne to neutral at various poi nts along the l i ne so

220

CHAPTER 6

C U RRENT A N D VOLTAG E R E LATIONS ON A TRANS M I SSION LIN E

that the total inductive susceptance is B v the charging current becomes ( 6 .62)

We recognize that the charging current is reduced by the term in pare ntheses. The shunt compensation factor is B,j Be The other benefit of shunt compensation is the reduction of the receivi ng­ e n d voltage of the line which on long high-voltage l i nes tends to become too high at no load. In the discussion preceding Eq. (6 . 1 1) we noted that I Vsl /1 A I e quals I VR, N LI. We also have seen that A equals 1 .0 when shunt capacitance is neglected. In the medium-length and longer lines, howeve r, the presence of capacitance reduces A . Thus, the reduction of the shunt susceptance to the value of ( Be - B L ) can limit the rise of the no-load voltage at the receiving end of the l i n e if shunt inductors are i ntroduced as load is removed. By applying both series and shunt compensation to long transm ission li nes, we can t ransmit l a rge amou nts of power efficiently and within the desired voltage constraints. Ideally, the series and shunt elements should be placed at int ervals along the l ine. Series capacitors can be bypassed and shunt inductors can be switched off when desirabl e . As with series compensation, ABeD constants p rovide a straigh tforward method of analysis of shunt compensation . Find the voltage regu la tion of the line of Example 6.3 when a shunt inductor is connected a t t h e receiving end of the line during no-load conditions if the reactor compensates for 70% of the total shunt admittance of the l ine.

Exa mple 6.7.

Solution.

From Example 6.3 the shunt admi ttance of the line is

y = jS . l OS X 1 0 - 6

Simi

and for the e ntire line

Be = S . 1 0S X 1 0 - 6 X 230 = 0 .001174 S For 70% compensation

B L = 0 . 7 X 0 .001 1 74

= 0.000822

We know the ABCD constants of t h e l ine from ExampJe 6.6. Table A.6 of the Appendix tells us that the ind uctor a lone is represented by the generalized constants

A =D= 1

B=O

C

=

-jBL = -jO.000822 S

The equation in Table A.6 for combi ning two networks in series teJJs us that for

6.10

/ 1 .340 0 1 . 41 1 / - 0 .4

the l in e and i n d uctor A eq

= =

0.8904

+

T R A NS M IS S I ON - L I N E T R A N S I ENTS

L

/

186 .78 79 .46° ( 0 .000822

- 90

0

221

)

The vol t age reg u l a t ion w i t h t he s h u n t reac tor con ne c t e d at n o lo ad becomes

1 3 7 . 8 6/ 1 . 0 4 1 1 - 1 24 . 1 3

------ =

] 24 . 1 3

6 . 67%

w h i ch i s a consid e r a ble red u c t io n from t h e v a l u e o f 24 .7% for the regu l a t i o n o f t h e u n compe n s a ted l i ne .

6. 1 0

TRANSMISSION-LINE TRAN S I ENTS

The transient overvo l tages which occur on a powe r system a re e i ther of external ori g i n (for example, a l ightning d ischarge) or generated i nte rnally by swi tching operations. I n general , the t ransients on transmission systems a re caused by any sudden change in the operating con d i t ion or configu ration o f the syste m s . . Lightning is always a potent ial hazard to power system equipment, but switching operations can also cause equ i pment damage. At voltages up to about 230 kV, the insulation l evel of the l i nes and equi pment is d ictated by the need to protect against I ightning. On syste ms where voltages are above 230 kV but less t h an 700 kV swi tching operations as wel l as l ightning are poten tially d a m a g i ng to insulation. At vol tages above 700 kV switching surges are the main d etermi n a n t o f the level of insu lation . Of cou rse, underground cables are immune to d irect lightn i ng strokes a n d can b e protected agai nst t ransie nts o riginating on overhead lines. Howeve r, for economi c and technical reason s overhead lines at t ransmission voltage l evel s prevail except u nder unusual c i rcumstances and for short d istances such a s under a river. Overhead lines can be p rotected from d i rect st rokes of ligh t n i ng i n most cases by one or more w i re s at g round potent ial strung above the pow er-l i ne conductors as mentioned in the d escript ion of Fig. 6. 1 . These protecting wi res, cal led ground wires , or sh ield wires , are connected to ground t h rough the transm ission towers supporting the l ine. The zone of p rotection is u s ually co n s i d ered t o b e 3 o n e a c h s i d e o f v e r t i c a l b e n e a t h a g r o u n d w i r e ; t h a t i s , the power l i nes must come within this 6 sector. The ground wires, rather than the power line, rece ive the ligh tning strokes in most cases. Lightning strokes h i t ting either ground wi res or power conductors c ause an i njection of current, whi ch divides with h alf the current flow i ng in one direction and half i n the other. The crest value of current along the stru ck con ductor varies widely because of t he wide vari a tion in the intensity o f t h e strokes. Values o f 1 0,000 A and u pward a re typical. I n t h e case w h ,e r e a p ower

222

CHAPTER 6

CURRENT A N D VOLTAGE R E L AT I O N S ON A TRA N S M I S S I O N LI N E

line receives a d irect stroke the d amage to equipment at line terminals i s caused by the vol tages between the line and t h e ground resulting from the injected charges which travel along the line as curre n t . These voltages are typically above a million volts. Strokes to the ground w ires can also cause h igh-vol tage surges on the power lines by electromagnetic i n d uction .

TRANSIENT ANALYSIS: TRAVELING WAVES The study of t ra n s m i s s i o n -l i n e s u r g e s , 6.11

r e g a r d l e ss o r t h e i r o r i g i n , is very comrlex and w e can consider here only the case of a l o s s l c s s l ine.2 A lossless line is a good representation for l ines of high frequency w here wL and we become very la rge compared to R and G . For ligh tning su rges on a power t ransmission l ine the study of a lossless l i ne is a simplification that enables us to understand some of the phenomena without becoming too involved in complicated theory. Our approach to the problem is similar to that used earlier for deriving the steady-state voltage and current re lations for the long l ine with d istributed constants. We now measure the distance x along the line from the sending end (rather than from the receiving end) to t h e d ifferential element of length � x shown i n Fig. 6.12. The vol tage v and the current i are functions of both x and t so t h a t we need to use partial derivatives. The series voltage drop along the elemental l ength of line is

ai i( R tl x ) + ( L tl x ) ­ at and we can write

av tl x ax

= -(

Ri + L at

ai

)

tl x

( 6 .63 )

The negative sign i s necessary because v + (a v lax) 6.x must be less t h a n v for positive v a l ues of

i

and

ai/a t . S i m i la rly,

:� A x

�

-

(

Gv +

C

:� ) A x

( 6 .64 )

For further s tudy, see A. G reenwood, Electrical Tra nsients ill Power Systems, 2d ed., Wiley-Lnters ci­ ence, New York, 1 99 1 .

2

6. 1 1 .

�

+

TRANSIENT ANALYSIS: TRAVELING WAVES

ai 61x ax

FIGURE 6.12

u

x

1-

223

I

f-- t1 x

Schematic diagram of an elemental section of a t ransmission line showing on e phase and neu­ tral return. Voltage v and current i are func­ t ions of both x and t. The distance x is mea­ su red from the sending end of the l in e .

I

---.,

We can divide through both Eqs. (6.63) and (6.64) by � x , and s ince we are considering only a lossless line, R and G wil l equal zero to give au (Ix

=

ai

and

ai at

-L -

( 6 .65 )

au -c at

ax

( 6 .66)

Now we can e li m ina t e i by taking the partial derivative of both terms in Eq. (6.65) with respect to x and the partial derivative of both terms in Eq. (6 .66) with respect to t . This proce dure yields a 2 ijax a t in both resulting equations, and eliminating this second partial derivative of i between the two equations yi el d s ( 6 .67) Equation (6.67) is the so-ca l l ed tra veling-wave equation of a lossless v I ), a n d the

r an s m i ss ion line. A solution of the equation is a function of (x vol t age is expressed by t

-

V = J( X - Jl I )

Th e fu n c t i o n is u n d e fi n e d b u t m u s t b e s i n g l e v a l u e d . The const an t

( 6 .68)

must have the dimensions of meters per secon d if x is in meters and t is in s econ ds . W e c a n verify this sol u t i o n by s u bs t i t u t i n g this expression for v into Eq. (6.67) to d e t erm in e v . Fi rs t we make the change in variable v

,

u

=

x

-

vt

( 6 .69)

and write u ( x , t ) = f( u )

( 6 .70)

224

CHAPTE� 6

C U R RENT AND VOLTAGE R E LAT I O N S ON A TRA N S M I S S I O N

LI NE

v

FIGURE 6.13 A voltage wave which is

··1 a

fu nction o f (x

-

v i ) i s s h own for va l u es o f

I

x -

equal to

II

a nd

12,

Then,

au at and

=

af ( u ) a u au at

-

-v

af( u ) au

a 2f( u ) a2u - = v2 at 2 au2

( 6 .7 1 )

( 6 .72 )

Similarly, we obtain

( 6 .73 ) Substituting these second partial derivatives of

u

i n E q . (6.67) yie1ds

( 6 .7 4) and we see that Eq. (6.68) is a solution of Eq. (6.67) i f

v=

1

( 6 .75 )

VLC

--=-

The voltage as expressed by Eq. (6.68) i s a wave trave ling in the positive x direction. Figure 6 . 1 3 shows a function of ( x v t ), wh ich is similar to the shape of a wave of voltage traveling along a line w hich has been struck by lightning. The function is shown for two values of time t I and t , where t 2 > t 1 . An 2 nt on the wave sees observer t raveling with the wave and staying at the same poi no change in voltage at that point. To the observer -

x

.

JJ { = a

consta n t

6. 1 1

from which it follows that

dx

-

dt

= v =

TRANSI ENT ANALYSIS: TRAVEL I NG WAVES

1

/LC

--

m/s

225

( 6 .7 6 )

for L and C in henrys per meter and farads per meter, respectively. Thus, the volt age wave travels in the positive x d irection with the velocity v . A function of ex + v t) can also be shown to be a solution of Eq. (6.67) and, by similar reasoning, can be p roperly interpreted as a wave traveling in the negative x direction. The gen e ral solution of Eq. (6.67) is

( 6 .77) wh ich is a solu tion for simultaneous occurrence of forward and backward components on the l ine. I n i tial conditions and boundary (terminal) cond i tions determine the particular values fo r each component. If we ex p r e ss a forward t ra v e l i ng wave, also called an inciden t wave, as

( 6 .78) a wave of current will resul t from the moving charges and will be expresse d by

( 6 .79) which can be verified by substitut ion of these values of voltage and current i n Eq. ( 6.65) a n d b y the fact that v i s equal t o 1 / /L C . S imil arly, for a backward moving wave of voltage where

( 6 .80) the correspond ing current is l

"'"'

( 6 .81 )

From Eqs. (6.78) and (6.79) we note that

( 6 .82 )

226

CHAPTER 6

C U R R ENT AND V O LTAGE R E LATI O NS ON A T R ANSM ISSION L I N E

and fro m Eqs. (6.80) and (6. 8 1 ) that ( 6 .83 )

If w e had decided to assume the positive direction of current for i - to be in the direction of travel of the backward traveling wave the m inus signs would change to plus signs in Eqs. (6. 8 1 ) and (6.83). We choose, h owe v er to keep the p os i t i ve x dire ction as the direction for posit ive current fo r both forw a r d a n d b a c kw a r d traveling waves. The ratio of v + to i + is called the characteristic i m p e d an ce Zc of the l ine. We have encountered characteristic impedance p re v i o u s l y i n th e steady-state solution for the long l i n e w h e re Zc was d e fi n e d as Fly ) w h i c h e q u a l s ';f../C when R and G are zero. ,

6. 12

TRANSIENT ANALYSIS: REFLECTIONS

We now consider what happens when a vol tage is first applied to the sending end of a transmission l ine which is termin a ted in an impeda nce Z R ' For our very si mple treatment we consider Z R to be a pure resistance. If the termination is other than a pure resistance, we woul d resort to Laplace transforms. The transforms of voltage, current, and impedance wou ld be functions of the Laplace transform variable s . When a switch is closed applying a voltage to a l ine, a wave of vol tage v + accompanied by a wave of current i + starts to travel along the line. The ratio of the vol tage VR at the end of the l ine at any time to the current i R at the end of the line must equal the terminating resistance Z R ' Therefore, the arrival of v + and i + at the receiving end where their values are v;' and i � must result in backward traveling or reflected waves v - and i- havi ng values v ii and if;. at the receiving end such t h a t ( 6 .84 )

where v ii and i'R are the reflected waves v - and i - measured at the receiving end. If we let Zc = ,;L IC we find from Eqs. (6.82) and (6.83) that ,

( 6 .85)

(.6 .86)

6. 1 2

T R A N S IENT AN A LY S I S : REFLECTIONS

227

Then, substituting th ese values of i; and i"R i n Eq. (6.84) yields ( 6 .87 )

The voltage vii at the receiving end is evi dently the same function of t as v; (b ut with d iminishe d magnitud e u nl ess Z R is zero or infinity). The reflection coefficient P R for voltage at the receiving end of the line is d efined as vii Iv; , so, for voltage ( 6 . 88)

We note from

Eq s . ( 6. 85) a n d (6.86) t h a t =

( 6 .89)

and therefore the reflect ion coefficient for curre nt is always the negative of the reflection coefficient for voltage. If the line is terminated in its characteristic impedance Zo we see that the reflection coefficient for both voltage and current will be zero. There w i l l be no reflected waves, and the l ine will be have as though it i s i nfinitely long. O n ly when a reflected wave returns to the se nd ing end does the source sense that t he l ine is neither infini tely long nor terminate d i n Zc ' Termination in a short circui t results i n a P R for voltage of - 1 . I f t h e tennination i s a n open circuit, Z R i s i nfi nite a n d P R i s found by d ividing t h e n umerator and d e nomi nator i n Eq. (6.88) by ZR a n d by allowing Z R to approach infinity to yield P R = 1 in the l imit for voltage. We s h o u l d n o t e a t t h i s ro i n t t h eI r w av e s t rave l i n g back t o w a rd the s e n d i ng end w i l l c a u s e n e w r c l l e c t i o n s as d e t e rm i n ed by th e r e l1c c t i on c oe ffi c i e n t at t he senJing e n d P ,I ' For impedance a t t h e se n d i n g e n d eq u a l to Z \. Eq. (6.88) becomes Ps

(6.90)

W i th sendi ng-end impedance of ZS ' the value of the initial voltage impressed across the l i ne will be the sou rce volt age m ultiplied by Zc/(Zs + Z). Equation (6.82) shows that t he inci d e n t wave of vol tage experiences a l in e imped ance of Zc ' and a t t h e instant when t he source is connected to t h e l in e Zc and Zs in serie s act as a voltage d ivider.

228

CHAPTER 6

C U R RENT A N D VOLTAGE R E LA TIONS ON A TRAl\SMISSION L I N E

Zc

30 Q

=

(a)

1 20 T

1 20 V , , ,

60 2T

: 1 80 V - 60

3T

1 20 V - 30

4T -

5T

- -

-

-

30

, , , , , - - + -

90-V

- -

4.25T

1 20 V 135 V (b) 1 80

1 20 V

- - ----

90,

T

-------

3T

-

135 V

F I G U RE 6 . 1 4 Circuit d i a g r a m , l a ttice d ia g ram, a n d p l o t of v olta g e versus time for Exa m p l e 6.8, where the receiving-end

5T

resistance is 90 D.

( c ),

A dc source of 120 V with n egligible resistance is connected through a switch S to a lossless transmission l ine having Zc = 30 11 . The line is termi n a ted i n a resistance of 90 11. If the switch closes at I 0, plot V R versus time until t = ST, where T i s the time for a voltage wave to tra\·el the length of the l ine. The circuit is shown in Fig. 6.14(a).

Exa mp le 6.8.

=

Solution. When switch S is closed, the i ncident wave of vol tage starts to travel along the line and is expressed as v

= 1 20U( v t

-

x)

where U(v t x ) is the u n i t step function, which equals zero when (v t x ) is negative and equals unity when ( v t x) is positive. There can be no reflected wave u ntil the incident wave reaches the end of the l ine. With impedance to the i nc ident wave of Zc = 30 11, resistance of the source zero and + = 120 V , the reflection coefficient becomes -

-

-

v

PI<

=

90 - 30 90 + 30

1 2

i

6. 1 2

T R A NSI ENT A N A LYS I S : R E FLECfIONS

229

W h e n v + reaches t h e e n d of t h e l i ne , a reflected wave origina tes of value

v

and so

-=

VR

=

(�) 1 20 = +

1 20

60

60 V

=

1 80 V

W h e n I = 2 T, t h e reflec t e d wave a rrives a t t he s e n d i n g e n d w h e re t he sen d i n g- e n d r e fl e c t i o n coe ffi c i e n t P s is c a l cu l a t e d b y Eq. (6.90). The l i n e term i n a t i o n for t h e

r e fl e c t e d wave i s

Zs '

t h e i m p e d a n ce i n series w i t h t h e sou rce , o r z e ro i n t h is case.

So, 0 - 30

PR

=

-1

0 + 30

a n cl a rc l1 e e t e u wave o f - 60 V s t a rts t o w a r d t h e r e c e i v i n g e n d to k e e p t h e se n d i ng e n cl vo l t age e q u a l t o 1 20 V . T h i s n ew w ave r e a c h e s t h e r e c e i v i n g e n d a t t

=

-

3T a n d re.A ccts tow a r d t h e se n d i ng e n d a wave o f 1 2

( - 60 )

- 30 V

a n d t h e rece i v i n g - e n d vo l t a g e b e comes

la ttice diagra m

VR =

1 80 - 60 - 30

=

90 V

An exce l l e n t m e t h o d o f k e e p i n g t rack of t h e various r e fl e c t i o n s as t hey o c c u r is the

shown in Fig.

6.14(b).

H e r e t i me is meas ured a l o n g t h e

v e r t i c a l axis i n i n t e rva l s o f T. O n t h e s l a n t l i n e s t h e re a r e reco r d e d t h e v a l u e s of t h e i n c i d e n t a n d r e fle c t e d waves. In the space be twee n t h e slant l i n es t he re a re s h o w n t h e s u m of a l l t h e waves a b ove a n d t h e c u r r e n t or \'oltage fo r a p oi n t in t h a t a r e a o f t h e c h a r t . For i n s t a n ce , a t x e q u a l t o t h r e e- fo u r t h s o f t h e l i n e l e n g t h a n d

t

=

4 . 2 S T t h e i n terse c t i o n o f t h e d as h e d l i n e s t h ro u g h t hese po i n ts i s w i t h i n t h e

a r e a w h i c h i n d i c a t e s t h e vo l t ag e i s 9 0 V .

Figure

6 .14(c)

s h o ws t h e r e c e i v i n g - e n d vo l t a g e p l o t t e d a g a i n s t t i m e . T h e

vol t age is a ppr o a c h i n g i t s s t e a d y - s ta t e va l u e o f 1 20 V .

Lattice diagrams for current m ay also be d rawn. We must remember, however, that the reflection coefficient for current is always the negative of the

I f the resisumce at t h e end of the l i n e of Example 6.8 is reduced to 1 0 n as shown in the circuit of Fig. 6 . 1 5 ( a ), the lattice diagram and plot of voltage are as s hown in Figs. 6. 15(b) a n d 6 . 1 5( c). The resistance of 10 n giv es a negat ive value for the reflection coefficien t for voltage, which always oc curs for resistance Z R l ess than Zc- As we see by comparing Figs. 6 . 1 4 and 6 . 1 5 , the negative P R causes the receiving-end vol tage to build up graduaIIy to 120 V, while a positive P R causes an initial jump in vol tage to a value greater than that of ! he voltage orig i n a l ly applied at the sending e n d . re fl e c t i o n co e ffi c i e n t for v o l t a ge .

811 20 230

CHAPTER 6

Ps

=

Jon

C U R R ENT AND VOLTAGE R E LAT I O N S ON A TRAN S M I S S I ON L I N E

Zc

30 Q

=

(a)

-1

x�

1 20

P R ==

0

- 1-

T - 1 20 V

2T

<:-::

60 V

3 T - 1 20 V

-30

4T

90 V 30

5T 105 V

( b)

1 20 V

60

90- - - - - - - - - 1 0 5 '

- - - --- - - -

T

I

(el

3T

5T

F I G U R E 6. 1 5

Circu i t d iagra m , l a t t ice di agram, a n d p l o t of voltage versus time when the receivi n g-end res istance for Ex­ a m p l e 6.8 is changed to 10 n .

R eflections d o not necessarily a rise only at the end of a l ine. If one line is joined to a second l ine of different c h a r a c t e r i s t i c impedance, as i n the case of an overhead l ine connected to an underground cable, a wave incident to the junction will behave as though the fl.rst l ine is terminated i n the Z c of the second line. However, the part of the incident wave which is not reflected will t ravel (as a refracted wave) along the second line at whose termination a reflected wave will occur. Bifurcations of a line will also cause reflected and refracted waves. It should now be obvious t hat a t horough study of t ransm ission -line transients in general i s a complicated problem. We realize, however, t hat a voltage surge such as that shown in Fig. 6. 1 3 e ncountering an impedance at the end of a lossless line (for ins tance, at a transformer bus) will cause a voltag e w ave of t h e same shape to travel back toward the source of t h e surge. The reflected wave will be reduced in m agnitude i f the terminal impedance is other than a short or open circuit, but if Z R is greater than Z c ' our study has shown that t h e peak terminal voltage will be higher than, often close to d ouble t h e peak voltage o f t h e surge.

6. 1 3

D I RECf-CU R RENT TRANSMISSION

231

Terminal equipment is p rotected by surge arresters, which are also called lightning arresters and surge diverters. An ideal arrester connected from the l ine to a grounded neutral wou l d (1) become conducti ng at a design voltag e above the arrester rating, (2) limit the voltage across its terminals to this d esign value, and (3) become nonconducting again when the l i ne-to-neutral voltage d rops below the design value. Originally, an arrester was simply an air gap. In this application w h e n the surge vol tage reaches a value for w h ich the gap is de signed, an arc occurs that causes an ion ized path to ground, essentially a short circuit. However, when the surge ends, the 60-Hz current from the generators still flows through the a rc to ground. The a rc has to be exting u ished by the opening of circui t breake rs. Arresters capable of ext inguishing a 60-Hz current after conducting s urge current to ground were deve loped l ater. These arresters are m a d e u s i ng non l i near res istors in series with a i r gaps to which an arc-q uench ing cap ab i l ity h a s bee n a d d e d . The n o n l i n e a r r e s i s t a n c e d e c r e a s e s r a p i d l y as the voltage across it r ises. Typical resistors m ade of sil icon carbide cond uct curre n t p ropor­ tional to approxim ately the fou r t h powe r of the vol tage across the resistor. \Vhen the gaps arc over as a result of a voltage surge, a low-resistance current path to ground is provided t h rough the nonlinear resistors. After the s urge ends and the voltage across the arrester returns to the normal l ine-to-neutral level the resistance is sufficient to limit the a rc current to a value which can be quenched by the series gaps. Quenching is usually accomplished by cooling and deionizing the arc by elongating it magnetically between insulat ing p l a tes. The most recent development in surge arresters is the use of zinc oxid e in place of silicon carbide. The voltage across the zinc oxide resistor is extremely constant over a very high range of cu rrent, which means that its resistance at normal l i ne voltage is so high t h at a series a ir gap i s not necessary to l im i t t h e drain o f a 60-Hz current a t norm al voltage. 3 6 . 13

DIRECT-CURRENT TRANSMISSION

transmission of energy by d i rect cu rrent becomes economical when com­ pa red to ac t r a n s m i s s i o n o n l y w h e n t h e cxtra cost of the terminal equipment r e q u i re d for d c l i n es is o frs e t b y the l owe r cost o f h u i l d i n g t h e l i nes. Con v e r t e rs a t t h e two ends of t h e d c l i n c s ope ra t e as b o t h recti fiers to c h a nge t h e gener a ted alternating to d i rect curre n t and inverters for converting d i rect to a l ternat ing current so t h a t pow e r c a n /low i n e i t h e r d i re ct i o n .

T he

Th e yea r

1 95 4 is g e n e ral l y reco g n i zed a s t h e s t a r t i n g d a t e for m o d e rn

h igh-voltage dc t ransmission when a dc line began service at 1 00 kY from V a s t e rv i k o n t h e m a i n l a n d of S w e d e n t o V i sby on t h e i sl an d o f Gotland, a J

See E. C. S a k s h a u g , 1 . S. Kre s g e , a n d S . A M is k e , J r. , " A N e w Concept in S t a t ion A r re s t e r

D e s i g n , " IEEE

Transactions o n Power Appara tus and Systems, v o l . PAS-96, no. 2, Marchi A p r i l

1 97 7 , pp. 647 - 656.

232

CHAPTER 6

CU RRENT AND VOLTAGE R E LATIONS ON A T R A N S M ISSION L I N E

distance of 1 00 km (62.5 mi) across the Baltic Sea. Static conversion equipme nt was in ope ra tion much earlier to transfe r e nergy between systems of 25 and 60 Hz, essentially a d c transmission line of zero length. In the United S tates a d c line operating at 800 k V transfers power generated in the Pacific northwest to t h e southern part of California. As the cost of conversion equ ipme nt d ecreases with respect to t he cost of line const ruction, the economical minimum length of d c lines also decreases a n d at this time is about 600 km ( 3 75 mi). Operation of a dc l ine began in 1 977 to transmit power from a mine-mouth generating plant burning lignite at Center, North D a k o t a , to n e a r Duluth, M inn eso t a , a d i s t a n c e o r 740 k ill ( 4 ()() ll1 i ). P r c l i m i n ; l r y s t u u i e s s h ow e d t h ; l ! t h e dc l ine including terminal facilities wo u ld cost about 30% less t han the compa­ rable ac l i ne and auxiliary equipment. This l i n e o p e r a t e s a t ± 250 kV (SOO kV l in e to line ) and transmits 500 MW. Direct-current lines usually have one con d u ctor wh ich is at a positive potential with respect to ground and a second conductor operating at an equal negative potential. S uch a l i ne is said to be bipola r . The line could be ope rated with one energized conductor with the return p a t h through the earth, which has a much lower resistance to direct current than to alternating curre nt. I n this case, or with a grounded retu rn conductor, the line is said to be monopolar . I n addition to the lower cost of d c transmission over long distances, there a re other advantages. Voltage regulation is less of a problem since at zero frequency the series reactance wL is no longer a factor, whereas it is t h e chief contributor to voltage drop in an ac l ine. Another advantage of direct current is the possibility of monopolar operation in an emergency when one side of a b ipolar l i ne becomes grounded. Due to the fact that underground ac transmission is limited to about 5 km because of excessive charging current at longer d istances, direct current was chosen to transfer power under the English Channel between Gre at Britain and France. The use of direct cu rren t for this installation also avoided the d ifficulty o f synch ron izing t h e a e systems o f t h e two countries. No n e twork or d e l i n es i s poss i b l e a t t h i s t i m e b e c a ll s e no c i rcu i t b r e a k e r i s available for direct current that is com parabl e to the highly developed ac b reakers. The ac breaker can extinguish the arc which is formed whe n the breaker opens because zero cu rrent occu rs twice in each cycle. The d i rection and a mount of power in the d c line is controlled by the converters in w hich grid-controlled mercury-arc devices are being d ispl aced by the semiconductor r ectifier (SCR). A rectifier u n i t w i l l conta in pe rhaps 200 SCRs. Still another advantage of dire ct curre nt is the smal ler amount of right of w ay required. The distance between the two conductors of the North Dakota­ D ul ut h 500-kV line is 25 ft. The 500-kV ac l i ne shown i n Fig. 6.1 has 60.5 ft b etween the outside conductors. Another consideration is the peak vol t age of t h e ac l ine, which is fi x 500 707 kV. So, the line requires more insulation b etween the tower and conductors as well as greater clearance above the e arth. We conclude that dc transmission has many advantages over altqnating current, b u t dc transmission rem ains very li mited i n usage except for long l i n es =

PROBLEMS

233

since t here is no dc device which can provide the excellent switching operations and p rotection of the ac c i r c u it breaker. There i s also no simple device to change the vol tage l evel, which the transformer acco mplishes for ac systems.

6.14 SUMMARY The long-line equations given by Eqs. (6.35) and (6.36) are, of course, valid for a line of any length. The approximations for the short- and medium-length lines make analysis easier in the absence of a computer. Circle diagrams were i n t roduced because of their instructional v a l ue in showing the maximum power which can be transmitted by a l ine and also in showing the effect of the power factor o f the load or the addition o f capacitors. ABeD constants provide a straightforward means of writing equations i n a more concise form and are v e ry convenient i n probl ems involving n etwork reduction. Their usefulness is a pparent in the discussion of series and s h u n t reactive compensation. The simple discussion of transients, although confined to lossless lines a n d d c sources, should give some i d e a o f t h e complexity o f t h e study of transi e n ts which arise from lightning and swi tching in power systems.

PROBLEMS

IS-km, 60-Hz, single-circuit, th ree-phase line is composed of Partridge con d u c­ tors equilaterally spaced with 1 .6 m between cen ters. The line delivers 2500 kW a t 1 1 kV t o a balanced load. Assume a w ire temperature of 5C. (a) Determine the per-phase series impedance of the line. C b ) What must be the sending-end voltage when the power factor i s (j) SO% lagging, ( ii) unity, (iii) 90% leading? (c) Determine the percent regu lation of the line at the above power factors. Cd) Draw phasor d iagrams depicting the operation of the line in each case. A I OO-mi, single-circuit, t hree-phase t ransmission l ine del ivers 55 MVA at O.S power-factor lagging to the load at 1 32 kV (line to li ne). The l i ne is compose d of Drake conductors with flat horizontal spacing of 1 1 .9 ft between adjacent con d u c­ tors. Assume a wire temperature o f 5C. Determine (a) The s e ri es i m p ed a nc e a n d the s h u n t adm i t t ance o f t h e l i n e . ( b ) The ABeD constants of t h e l i ne. ( c ) The sending-end Voltage, current, real and reactive powers, and the power factor. Cd) The percent regul ation of t h e line. circuit having a 600-.0 resistor for t h e shunt Find t h e ABeD constants of a b ranch at the send ing end, a l - k fl resistor for the shunt branch at the receiving e n d , and an SO-o. resistor for the series branch.

6.1. An

6 .2.

6.3.

7T

234 6.4.

CHAPTER 6

CURR ENT AND VOLTAGE RELATIONS ON A TRA N SMISSION L I N E

A

=

D B C

6.5.

6.6.

6.7.

6.8.

/ 0 .98° 142/76.40 D

The ABCD constants of a three-phase tra nsmission line are = = =

0 .936 3 3.5

+

+

iO .016

i1 3 S

( - 5 .] 8

+

=

=

i914)

0 .936

x

The load a t the receiving en d is 50 M W at 220 kV with a power factor of 0.9 lagging. F i nd t h e magnitude o f the sending-end vo l lage and the vol tage reg u l �l t ion. Assume that the magn itude of the send ing-e n d voltage rem a ins consta n t. A 70-m i, single-circu it, three-phase line composed of Ostrich conductors is a r­ ranged i n flat horizontal spacing w i th 1 5 ft betwee n adjacent conductors. The line deli vers a load of 60 MW at 230 kV w i th 0 .8 power-factor l aggi ng. ( a ) Using a base of 230 kV, 1 00 M VA, de termin e the series i mpedance a n d t he shunt admittance of the line in per u n i t. Assume a wire temperat ure of 500 C. Note that the base admittance m ust be the reciprocal of base i m p e d a n c e . (b) Find the voltage, current, real and reactive power, and the power factor a t the send ing end in both per unit and a bsol u te u n its. Cc) What is the percent regulation of the line? A single-circuit, three-phase transmiss ion line is composed of Parakeet con ductors with flat horizonta l spacing of 19.85 ft between adjacent conductors. Determine the characteristic impedance and the propagation constant of th e l ine a t 60 Hz a n d 5 0° C temperatu re. Using Eqs. (6.23) and (6.24), show that if the receiving end of a line is terminate d by its characteristic i mpedance Zc ' t h e n the impedance seen a t the sen d i ng e n d of the l i n e is also Zc regardless of l i ne length. A 200-m i transmiss ion line has the fol lowing pa rameters at 60 Hz: Resistance r = 0 . 2 1 D/mi per phase

Series reacta nce

x =

Shunt susceptance b

=

0 .78 D / m i 5 .42

X

6.10.

per

phase

1 0 - 6 Simi per phase

Determine the a ttenuation consta nt a, wavelength A, and the veloc ity of propagation of the line at 60 Hz. (b) I f the line is ope n-ci rcu ited at the rece iving end and the receiving-end voltage is maintained at 100 kV line to line, use Eqs. (6.26) and (6.27) to determine the incident and reflected components o f t he sending-end voltage and current. ( c ) Hence, determ ine the sendi ng-end voltage and current of t he l ine. . Evaluate cosh e a n d s inh e for e = 0.5 Usi ng Eqs. (6. 1), (6.2), (6.10), and (6.37), show that the gene ralized circuit consta nts of all three transmission-line models satisfy the condition that (a)

6.9.

10-6 S

�

AD - BC = 1

P R O B LE M S

6. 1 1 .

6.12.

6.13. 6.14.

The sending-end voltage, current, and power factor of the line described in Example 6.3 are found to be 260 kY (line to l i n e), 300 A, and 0.9 lagging, respectively. Find the corresponding receiving-en d voltage, current, and power factor. A 60-Hz three-phase transmission line is 175 mi long. I t has a total series impedance of 35 + D. and a shunt admittance of 930 X S. It delivers 40 MW at 220 kY, with 90% power-factor lagging. Find the voltage at the sending end b y ( a ) the short-line approximation, (b) the nominal-'1T approximation, and (c) the long-line equation. Determine the voltage regulation for the line described in Prob. 6.12. Assume t h at the sendi ng-end voltage remains constant. A three-phase, 60-Hz transmission line is 25 0 mi long. The voltage at the sending c n d is 2 2 0 kY. The parameters of the line arc R 0. 2 D. / m i , X = 0 8 D./mi, a nd Y = 5.3 ,u S / m i F i n d th e se n d i n g e n d c u r r e n t w h e n t h e r e is no l o a d on the line. I f t h e l oad on t h e l i ne d es c r i b ecl i n Prob. is SO M\V a t 220 k Y , with u nity pow e r fa ct o r calculate t h e c u rre n t , vol t ag e , and powe r at t h e sending end. Assume that the sen ding-end voltage is held constant and calculate the voltage regulation of t h e line for the load specified above. A three-phase' transmission line is 300 mi long and serves a load of 400 MY A, w i t h 0.8 lagging power factor at 345 kY. The ABCD constants of the line are

j140

1O-6�

.

6.15.

6.14

-

,

6.16.

235

A B C

=

=

=

D

=

=

.

�

0 .8180

1 72 .2/ 84 .20 D.

0 .001 933

/90.40

S

( a ) Determine the sendi ng-end lin e-to-neutral voltage, the sending-end current,

and the percent voltage drop a t full load. ( b ) Determine the receiving-end l i n e-to-neutral vol tage at no load, the sending-end c u rre n t

y

6 . 1 7. J usti f

a t n o l oa d ,

;l n cl t h e

E q . ( C>.50) by s u b s t i t u t i n g

e x po n e n t i a l e x p r e ss i o ns.

6.20.

for

t he

y

h pe r bo l ic

fu nctions the equivalent

et e r m i n e t h e e q u iv a l e n t -7T c i rc u i t fo r t h e l i n c of Prob. 6. 1 2. Usc Eqs. (C>. J ) and (6.2) to s i mp l i fy Eqs. (6.57) and (6.58) for the short transmission line with (a) series r e a c ta n ce X and resistance R and (b) series reactance X a n d neg l i g i b l e r e s i s t an c e . R ig h t s of w a y fo r t r a n s m ission circuits a rc d i flicu lt to obtain in urban areas, a n d cxi�ting l ines are often u pgraded by rccond uctoring the line with larger conductors or by reinsulating the line for operation a t higher voltage. Thermal considerations and m aximum power which the line can transmit are the important considerations. A 138-kY line is 50 km long and is composed of Partridge conductors with fiat horizontal spacing of 5 m between adjacent conductors. Neglect resistance and find

6. 1 8. D 6.19.

vo l t a ge reg u l a t io n .

236

CHAPTER

6

C U R RENT A N D VOLTAGE R ELAT I O N S O N A T R A N SM I S S I O N L I N E

the perce nt i ncrease in power which can be transmitted for constant I Vsl and I VR I while 0 is limited to 45° (a) If the Partridge conductor is replaced by Osprey, which has more than twice the area of aluminum in square millimeters, (b) If a second Partridge conductor is placed in a two-conductor bundle 40 cm from the original conductor and a center-to-center d ista nce between bundles of 5 m, and ( c ) I f t h e v oltag e o r t h e o r i g i n a l l i n e is raised to 230 kV with i nc reased conductor spacing of I) m. 6.21. Construct a receivi ng-end power-circle d iagram similar t o Fig. 6. 1 1 [ r the line o f Prob. 6. 12. Lo c a t e t h e poi n t corresponding t o t h e load o f P rob. 6 . 1 2 , a d locate lhe center o f circles for v a r iou s values of I Vs l i f I Vu l 220 kY. Draw the circle passing through the load point. From the measured r a d i u s of the latter circle determine Wil l and compare this value with the val ues calculated for P r o 6 . 1 2 . 6.22. A synchronous condenser is connected i n parallel \vi t h the load described in Prob. 6. 1 2 to improve the overall power factor at the receiving end. The sending-end voltage is always adjusted so as to maintain the receiving-end vol tage fixed a t 220 kY. Using the power-circle diagram constructed for Prob. 6 . � 1 , determine the sending-end voltage and the reactive power supplied by the synchronous condenser when the overall power factor at the receiving end i s (a) unity (b) 0.9 leading. 6.23. A series capacitor bank having a reactance of 146.6 n is to be i nstalled a t the m i d point of the 300-mi line of Prob. 6. 1 6. The ABCD constants for each 1 S 0-mi portion of line are

on

=

b

.

A = D = 0 .9S34fu B = 90 .33 �t. C

=

0 .00 1 0 1 4/ 90 . 1° S

cascade combination

(a) 6.24.

Determine the e q u i v a l ent A B CD c o n s ta n ts for the l ine-capacitor-line. (See Table A.6 i n the Appendix.) (b) Solve Prob. 6.16 using these equivalent ABCD constants. The shunt admittance of a 300-m i t ransmission l i ne is

yc

6.25.

n

=

0

+

of the

}6 .87 X 1 0 - 6 Simi

o.Doni

D etermine the ABeD constants of a shunt reactor that will compensate for 60% of the total shunt admittance. 90° S is con­ A 2S0-Mvar, 34S-kV shunt reactor whose admittance is n ected to the receiving end of the 300-mi line of Prob. 6 . 1 6 at no load. (a) Determine the e quivalent ABCD constants of the line in series with the shunt reactor. (See Table A.6 in the Appendix.) (b) Rework part (b) of P rob. 6 . 1 6 using these equivalent ABCD constants and the sending-end voltage found in Prob. 6. 1 6. -

PROBLEMS

6.26.

6 . 27.

6.28. 6.29.

6.30.

237

Draw the lattice d iagram for curre n t and plot current versus time at the se n d i n g end of the line of Example 6.8 for the line terminated in (a) an open circuit (b) a short circuit. Plot voltage versus time for the line of Example 6.8 at a point d istant from the sending end equal to one-fourth of the length of the line if the line is terminated in a resistance of 10 D. Solve Example 6.8 if a resistance of 54 n is in series w ith the source. Voltage from a dc source is applied to a n overhead transmission line by closing a swi tch. The end of the overh ead line is connected to a n underground cable. Assume that both the l ine and the cable are lossl ess and that the initial voltage along the l ine is v + . I f the characteristic impedances of the line and cable are 400 and 50 fl, respectively, and the end of the cable is open-circuited, find in terms of v + ( a ) T he vol tage at the ju nction of the line and cable i m med iately after the arrival of the i n c i de n t wave a n d ( b ) The vo l t age at t h e open end of the cable immediately after arrival of the first voltage wave. A dc sou rce of voltage VI and internal resistance Rs i s connected through a switch to a l os s l c s s l i n e having ch a r a c t e r i s t i c i mp e d a nc e R c . The l i n e is terminated in a r e s i stance R R . The traveling t ime of a vo l t a ge across the line is T. The switch o. cl oses at ( a ) D raw a l a t t ic e diagram showing the voltage of the line during the period t 0 to t = 7T. Indicate the voltage components i n terms of � and the reflection coetflcien ts P R and P ( b ) D e ter m i n e the receiving-end voltage a t t 0, 2T, 4T, and 6T, and hence at ( 2 n T where ! l is any non-negative i nteger. (c) Hence, determine the steady-state voltage at the receiving end of the l i ne i n terms o f v" R s ' R R ' a n d R e C d ) Verify the result i n part (c) b y analyzing the system as a simple d c circuit i n the steady state. (Note that the line is loss less and remember how i nductances and capacitances behave as short circuits and open circuits to dc.)

(

=

=

S"

=

=

CHAPTER

7

THE AD MITTANCE MODEL AND NETWORK CALCULATIONS

The typical power transmission network spans a large geographic a rea and involves a l arge number and variety of network components. The e lectrical characteristics of the individual components a re d eveloped in previous chapters and now we are concerned with the composite representation of those compo­ nents when they are interconnected to form the network. For large-scale system analysis the network model takes on the form of a network matrix. with elements determined by the choice of parameter. There are two choices. The current flow t h rough a network component " can be related to the voltage d rop across it by either an a dmittance or an impedance parameter. This c hapter treats t h e a d m ittance representation in the form of a primitive model which describes the electrical characteristics of the network components. The primitive model neither requires nor provides any information about how the components are interconnected to form the network. The steady-state behavior of all the components acting together as a system is given by the nodal admittance matrix based on nodal analysis of the network equations. The nodal admittance matrix of the typical power system is large and sparse, and can be constructed in a syste matic building-block manner� The building-block approach provides insight for developing algori thms to account for network changes. Because t h e n e t w o r k m a t rices a r e very l a rge, !)pa rsily

7. 1

BR ANCH A N D N O D E

A DMITTANCES

239

techniques are needed to enhance the computational efficiency of computer programs employed in solving many of the power system problems described in later chapters. The particular importance of the present chapter, and also Chap. 8, which develops the nodal impedance matrix , becomes evident in the course of pow er­ flow and fault analysis of the system. 7.1

BRANCH AND NODE ADMITTANCES

In per-phase analysis the components of the power transmiSSion system are modeled and represented by passive impedances or equivalent admittan ces accompanied, where appropriate, by active voltage or current sources. In the steady state, for example, a generator can be re presented by the circuit of e ither Fig. 7 . l ( a ) or Fi g . 7 . 1 ( b ) . The circu it having the constant emf Es ' series impedance Za ' and terminal vol tage V has the vol tage equation ( 7 . 1)

D ividing across by Z a gives the current equation for Fig . 7. l(b) E !s = _s = I + VYa 2a

( 7 .2)

where Ya = 1 /2a ' Thus, the emf Es and its series impedance Za can b e interchanged with the current source Is and its shunt admittance Ya , p rovided Es 1 = ­ Za s

I

r

V Es

(a)

I

1 Ya = ­ Z

and

( 7 .3)

a

I N

N

e

w 0

Yo

V

e t

w 0

r k

r k

(b)

Fl G U RE 7 . 1 Circ u i t s i l l u strat ing t h e eq uivalence of sources when Is

=

Es/Zu a n d

Y" =

1 /Z/I'

240

CHAPTER 7

THE ADMITTANCE

M O D E L A N D N ETW O R K CALCU LAT I O N S

Sources such as Es and Is may be considered externally applied at the nodes of the transmission network, which then consists of only passive branches. In this chapter subscripts a and b distinguish branch quantities from node quantities which have subscripts m , n, p, and q or else numbers. For network modeling we may then represent the typical branch by either the branch impedance Za or the branch admittance Ya , wh ichever is more convenient. The branch i mpedance Za is often called the primitive impedance, a n d l ikewise, Y{l is called the primitive admittance. The equations characterizing the branch are vII

=

Z U Ia

( 7 .4 )

or

where Y:l is the reciprocal of Z{l and V:I is the vol tage d rop across the branch in the d i rection of the branch current I {l ' Regardl ess of how it is co nnected into the network, the typical branch has the two associated va riables V{l and fa related by Eqs. (7.4). In this chapter we concentrate on the branch a dmittance form in order to establ ish the nod al a d m ittance representation o f the power ' network, and Chap. 8 treats t he impedance form. I n Sec. 1 . 1 2 rules are given for forming the bus a dmittance m atrix of the network. Review of t hose rules is recommended since we are about to consider an a lternative method for Y hus formation. The new method is more general because i t is easily extended to networks with mutually coupled elements. Our approach first considers each branch separately and then in combination with other b ranches of the network. S uppose that only branch admittance Ya is connected between nodes ® and ® as part of a larger network of which only the reference node appears in Fig. 7.2. Current i njected into the network at any node is considered positive and c ur re n t leaving the network at a ny node is considered nega tive . In Fig. 7.2 c u rrc n l / 1/1 i s that p or t io n o r t h e t o t a l cu rr e n t i njected i nto node ® w h i c h passes through r:1 ' L i k ewise, I" is that portion of the current i njected into node ® wh ich passes t h ro u g h Y:, . The volt ages V;,/ and V;1 a re the vui t Cl g e s of nodes

1m

@ )0

I

+ I

r+

�-

Vm

fa

�

c==

Va

Ya

)o ®

=

Za 1

-

1+ !-

Vn

I

I" �

Reference node

Primi tive b ra n c h voltage drop V;" branch c u r re n t III ' i njected cu rrents JI/I and I" , node volt flges Vm a n d V,I w i t h respect to ne twork refe rence.

FIGURE 7.2

7.1

B RANCH AND NODE A D M ITTANCES

241

@J

and ® , respectively, m e asured with respect to network reference. By Kj rchhoff's l aw 1m = 1a a t node @) and 1/1 = - la at node ® . These two current equations arranged in vector form a re ( 7 .5 )

In Eq. (7 . 5) the labels or pointers @J and ® associate the d irection o f I a from node § to node ® wi t h t h e e nt ries 1 and - 1 , w h ich arc then said to be in row ® anu row "i!) , r e s p e ct i v e l y I n a s i m i l a r m a n n e r, the vo ltage d rop in the d i re c t i o n of 1 has t h e equa tion Va = V,II v, , , O r expressed in vector for m .

a

-

@)

®

- 1

[ 1

Subst i t u ting t h i s expression for �I In

the

-I]

®

( 7 .6)

]

admittance equation Ya Va

� [ 1

� I,

=

1a gives ( 7 .7 )

and premul tiplying both si des of Eq. (7.7) by the column vector of Eq. (7.5), we

obtain

®® [ l lYa -1

which,

s i m p l i fies

[

@) 1

to

@) ®

®

-1]

[ �: 1 [ �: 1

®

( 7 .8)

( 7 . 9)

Ya

and the coefficient m a t rix is the noda l admittance matrix . We note that t h e off- diagonal e le ments e q u al t he n egat ive of the branch admittance. The m atrix of Eq. (7.9) is singul a r b ecause neither node § nor node ® connects to the refe re nce. In the par ticular case w h e r e one of the two nodes, s ay, ® , is the reference node, then n � d e vol tage

This is the nodal admittance equa tion for b ranch

242

CH APTER 7

TH E ADMITTANCE MO DEL A N D N ETWO R K

Vn is zero and Eq. (7.9) reduces to the 1

x

@) ® [ Ya ] Vm correspo n d i n g t o re m ova l o r

row ®

CALCULATIONS

1 m atrix equation

=

and

1m

( 7 . 1 0)

co l u m ll ®

fro lll

t h e c o e /'n cie n t

matrix. Desp ite its s t r a i g h t forw a r d d e r i v a t i o n , Eq . (7.<) ) a n d t h e p mc e d ur e l e a d i n g t o it arc important in more general s i t u a t i o n s . W e n o t e t h a t t h e b ranch vo l t age Va is transT'ormed to the node vol tages v,1/ and V, I ' and the b r a n c h cu rrent l a is l ikewise re presented by the cu rrent injections III/ and 1/1 " The coefficient ma trix relating the node voltages a n d cu r re nts o f Eq . (7 .9) fol lows from the fact that in Eq. (7.8) ®

-1

@) ®

(7.11 )

]

This 2 X 2 matrix, also seen i n Eq. 0 .64), is an im portant bllilding block for represe nting more general n etworks, as we shall soon see. The row and column pointers identify each entry in the coe fficient m a trix by node numbers. For instance, in the first row and second column of Eq. (7 . 1 1 ) the entry - 1 is identified with nodes @) a nd ® of Fig. 7.2 and the other entries are similarly identified. Thus, the coefficient m atrices of Eqs. (7.9) and (7. 1 0) a re s i m p l y storage matrices with row and col umn l abels determ ined by t h e e n d nodes of the branch. Each branch of t he ne twork has a sim i l ar mat rix labe led according to t h e nodes of the network to w h ich t hat bra nch i s connec ted. To obtain the overall nodal admittance matrix of the e n t i re network, w e s i m p l y co m b i n e the ind ividual branch matrices by a dding together clements with identical row a n d column labels. Suc h add ition causes t h e sum of the b r a n c h currents flowing from each node of the network to equal t he total curre nt inj ected into that nod e, as required by Kirchhotl's current law. I n the overall m a t r i x , the off-d iag­ onal element 0j is the negative sum of the adm ittances con nect ed between nodes CD and CD and the d iagon al e n t ry Y,i is the algebraic sum of the admittances connected to node CD . Provided at least one of the networ k branches is connected to the refe rence node, t h e n et result is Y bu S of the system, as show n in the following example .

The single-line diagram of a small power system is shown i n Fig. 7. 3. The corresponding reactance diagram, with re actances specified in per u n i t , is per u n i t is conllected shown in Fig. 7.4. A generator with e m f equal to t hrough a t ransformer t o h igh-voltage node G) , w h i l e a motor w i t h internal voltage equal to 0 . 85 - 45° is similarly c o n n e c t e d to n o d e GG · Deve l o p t h e n o d a l Example 7.1.

!

1 .25&

7.1

B R A N C H A N D N O D E A D M ITTANCES

243

®

CD

F I G U RE 7.3

S i n g l e - l i n e d i agram of t he fou r-bus sys­ t e m of Exa m p l e 7.1. R e fe rence n o d e is n o t shown .

)0.25

)0.2 jO.25

)0.4

@ --,--+--

--+--'-

@

0

R e a c t a n c e di ag r a m fo r F i g. 7.3.

FIG U R E 7.4

Node

@ is reference, react a n ces and vo l t a ges a te in rer u n i t .

a d m ittance m atrix for e a c h o f t he network b ra n ch es and then write the n odal a d m i t tance equations of the system. The re a c t a n ce s o f t h e g e n e r a t o r and t h e m o t o r may b e combined w i t h t he i r respective step-up t ransformer reactances. Then, by t ransfo rm a tion of sou rces t h e com bincd rcactanccs a n d the gencrated e m rs a rc repl aced by t h e e quival e n t curre nt sources a n d s h u nt a d m i t ta nces shown in Fig. 7 . 5 . We wi l l t re a t t h e current sources as external injections at nodes G) and @ and name the seven pass ive b ran ches accord i n g to the subscri pts of their currents and vol tages. For example, the branch between nodes CD and G) will be called branch c . The a d mittance o f each b ranch is s i m ply the r eciprocal of the branch impedance a n d Fig. 7.5 shows the resulta nt a d m itt ance d i agram with a l l values i n per u n i t. The twp branches a a n d g connected to t h e r e ference node arc characterized by Eq. (7. 1 0), w h ile Eq. Solution .

244

CHAPTER

7

THE A D M ITTANCE M O D E L A N D N ETW O R K CA LCU LA T I O N S

-- v � +

c

--- v:(�+

_

+

1.00/- 90°

t

V}.'

t

@

FIGURE 7.5 Per- u n i t admit tance d i agram for Fig. 7 . 4 w i t h c u r r e n t sou rces r e p l a c i n g vo l t a g e sou rces. B r a n c h names a t o g correspond t o t h e subscripts o f b ra n ch voltages and c u rrents.

(7.9) applies to each of the other five branches. By setting

and n in those equations equal to the node n u mbers at the ends of the individual b ranches of Fig. 7.5, we obtain

0

0 (1)

G) [ l ] Ya

G) CD

-

1 1

G) CD

[ -: ]

-

1 1

0 G)

[ -: ] Yd

@ G)

Yb

@

0 CD

-1 1

Yc

@ CD

[ -: ]

G) CD

[ -: ]

-(1) 1 Yc 1

[ -: ]

m

@

@

[1]

Yg

-1

@ CD

1 Yf

The order in which the labels are assigned is not important here, provided t h e columns a n d rows follow the same order. Howe\ er, for consistency with later sections l et us assign the node numbers i n the directions of the branch currents of Fig. 7.5, w h i ch a l s o shows the n u merical v a lues o f the a d m ittances. Comb i n i n g together those elements of the above matrices ha\'ing identical row and column labels gives

CD G)

G)

@

CD

( � + Yd

Yd Yc - Y,

+

Yf )

Yd ( Yb + Yd - Yb - Y:,

@

G)

G)

-

+

Yc )

Yc -Y b ( Ya + Yb + � ) 0

(Y

c

- Yf Yc 0 -;- YJ + YJ

7.2

M UTUALLY CO UPLE D B RANCHES I N Y bus

245

which is t he same as Y hlls of Eq . ( 1 .02) si nce F i g s . 1 .23 and 7.5 me for the same ne twork. S u b s t i t u t i ng t he n u m e rical val ues of the branch admitta n ces i n t o this matrix, we obtain for the overall n e twork the nodal admi ttance equations -j 1 4 . S j8.0

j8 .0 -j 1 7 .0

j 4 .0 j4 .0

j2.S j5 .0

j 4 .0

j4 .0

-jS . S

0 .0

j2.5

j) .O

0 .0

-jS .3

VI V2 V]

/

/ - 900 0 .68 / - 1 350 0 0

=

�

1 .00

/

w h e re V I ' V� , VJ , and Vol
The coe l fl c i e n t m , l t r i x o o t
is

the s a m e (I S t h e b u s (t u m i t l a n c c m�lt l i x ro u n d i n S e c . 1 . 1 2 lIs i n g t h e u s u a l ru l e s fo r Y b u s fo r m a t i o n . However, t h e approach based on the build ing-block m a trix h as adv a n t ages w h e n extended to n e tworks w i t h m u t u a lly cou p l e d bra n ches, as w e n ow d e m on s t r a t e . exac t l y

M UTUALLY C O U PLED B RA N C H E S

IN Y b u s

7.2

The p roce d u re based on t h e b u i l d i ng-block m a trix is n ow extended to two m u t u a l ly cou p l e d branches w h i c h a re p a r t of a l a rger n et\vork but w h ich a r e n o t i n d u c t ively coupled t o a n y othe r b r a n c h e s . I n Sec. 2 . 2 t h e p r i m i t iv e equations of s u c h m u t u a l ly coupled branches a r e d eveloped in t h e for m of Eq. (2.24) for i m p e d a nces and Eq . (2.26) for a d m i t t a n ces. The nota t i on is d ifferent h e re b e c a u s e we a re n ow u s i n g n u m bers to i d e n t i fy n odes r a t h e r t h a n b r a n ches. Assume t h ; l t h r due t o t h e b l a n c h c u rr e n t s I " (l n cl I I ! ;l l e t h e n g i v c n b y t h e p r i m i t ive i m p e d a n ce e q u a t i o n co r r e s p o n d i n g t o E q . ( 2 .24) i n the fo r m

T
( 7 . 1 2)

i ll w h i c h t h e coe fll c i c n t m a t r i x i s sy m m e t r i c a l . As n o t e d i n S e c . 2.2, t h e m u t u a l

a n d II! e n ter the ter m i nals m a r ked w i t h clots i n Fig. 7 . 6( a ) ; t h e voltage drops Va a n d Vb t h e n h ave t h e p o l a r i t i e s show n . \1 u l t i p l y i n g E q . ( 7 . 1 2) b y t h e i nverse of t h e prim i t ive i m p e d ance m a t r i x i n1 p cc.L. l l1 cc

Z,\/

is

co n s i d e r e d

2M]Zh

1

=

pos i t iv e

when

c u r re n t s

Iii

YM] Yb

( 7 . 13 )

246

1

CHAPTER

-'"

@)

7

THE ADMITTANCE MODEL A N D NETWORK CALCULATIONS

III

Ib

---'--+-

�

-I p

+

+

®

®

Zb

Vb

®

In-----

� Iq (a)

@)

1m---

In--

®

®

III

-1p

---'--+-

VIl

Ya ------ 1q

(b)

FIG URE 7.6 Two m u tu a lly c o u p l e d branches with ( a ) i m ped a n ce p a ra m e t ers a n d ( b ) correspo n d i n g a d m i t t ances.

we obtain the admittance form of Eq. (2.26) for the two branches

( 7 . 14) which is also symme trical . The admi ttance matrix of Eq. (7.14), ca lled the

primitive admittance matrix of the two coupled branches, corresponds to Fig. 7.6(b). The primitive self-admitta nce "y;, equals Z,,/( Z,, Z,, Z;1 ) and similar expressions from Eq. (7. 13) apply t o Y,) and the primitive mutual aumitta nce YM • We may write the vol tage-drop equations V" = ��I/ v" and VI> = Vp - Vq of Fig. 7.6 in the matrix form -

-

@ ®

-1 o

o 1

(7.15)

which the first row of the coefficient matrix A is associated with branch admi tt a nc e Ya and the second row relates to branch admittance Yh. The node

ill

vo l ta g es Vm , Vn , Vp ' and � are measured with respect to the network reference.

In Fig. 7.6 the branch current fa is related to the injected currents by the two

M UTUALLY COU P LE D B RA NCHES IN Ybu .

7.2

247

node equations 1m = la a n d In = - 10 ; similarly, branch current lb is related to the currents 1p and 1q by t h e two node equations 1p = 1b and 1q = lb' These fou r current equations arranged in m atrix form become -

o

1

@) @ ® ®

-

1

o

(7. 16)

o -1

with the coefficient matrix equal to t h e transpose of that in Eq. (7. 1 5 ) we substitute for the vol tage d rops of Eq . (7. 14) to find

[

Yh

Y "

�\1 Y"

1

VI I I

V:, �,

A

[ �;: 1

v:/

(7. 1 5).

( 7 . 17)

and premultiplying both sides of t h is equation by the matrix AT of Eq. obtain

[

AT

'-.......----' .....

4 X

2

Yo

Y "

YM

Yb

1

J

2

X

2

A --

-

2 X 4

From Eq.

1m In Ip

Vm Vn Vp

(7. 1 6),

we

( 7 . 1 8)

lq

Vq

\Vhen the multipl ications ind icated in Eq. (7. 1 8) are performed, the result gives the nodal admittance equations of the two mutually coupled branches in the m atrix form

® ® ® ®

® Yo

-Y

0

YM - YM

® -

Y Ya a

- YM YM

®

YM - YM Y"

- Y"

®

- YM YM - Y" Y"

V:'I v"

�, VI]

fl/ l In =

If!

II]

J

( 7 . 1 9)

The two m utually couple d branch es a re actuall y part of a larger n etwork , a n d so the 4 X 4 m at rix of Eq. (7 . 1 9) forms part of the larger nodal admitt ance matrix of the overal l system . The pointe rs @ , @ , 0 , and ® indica te the rows and colum ns of the system matrix to which the elemen ts of Eq. (7 . 1 9) belong . Thus, for examp le, the' quanti ty e ntered in row ® and colum n ® of the system

248

CHAPTER 7 THE A D M IIT ANCE

M O D E L A N D N ETW O R K CALCU LATI O N S

nodal admi ttance matrix is - YM and similar e ntries are made from the other

elements o f Eq. (7.19). The nodal admittance matrix of the two coupled branches may be for med directly by i nspection. This becomes clear w h e n w e write t h e coefficient m atrix of Eq. ( 7 . 1 9) in the alternative form

@ ® ® ®

®

- 11 ]

®

r:l

®

1 - 1

® ®

1

YM

®

( 7 .20)

To obtain Eq. (7.20), we m ul tiply each element of the pnm l tlve admittance matrix by the 2 X 2 buildi ng�block matrix. The l abels assigned to the rows and columns of t h e multipliers in Eq. ( 7.20) are easily determined. First, we not e that t h e self-admittance Ya is measured between nodes @) and ® with t h e dot at node @j . Hence, the 2 X 2 matrix multiplying Ya in Eq. (7.20) has rows and col umns l abeled @) and ® in that order. Then, the self-admit tance Yb between nodes ® and ® is m ultiplied by the 2 X 2 matrix with labels ® and ® i n the order shown since node ® is marked with a dot. Finally, the l abels of the matri ces multiplying the m utual admittance �\r are assigne d row by row and then column by column so as to align and agree with those already give n to the self-inductances. In the nodal admi ttance m atrix of Eqs. (7. 1 9) and (7.20) the sum of the columns (and of the rows) adds up to zero. This is because none of the nodes @j, ®, ®, and ® has been considered as the refe rence nod e of the n e twork. I n the special case where one of the nodes, say, n ode ® , is in fact the reference, Vn is zero and col u m n n in Eq. (7. 1 9) does not need to appear; furthermore, III does not h ave to be expl icitly represented since the refe rence node current is not an independent quantity. Consequently, when node ® is the reference, we may eliminate the row and col umn of that node from Eqs. (7.19) a nd ( 7 .20). It is important to note that nodes ® , ® , ® , and ® are often not distinct. For i nstance, suppose that nodes ® and ® are one and t h e same nod e. In that case columns ® and ® of Eq. (7. 1 9) can be combined together since � = Vq , and the correspond ing rows can be added because In and Ip are parts of the common inj ected current. The fol lowing exam ple illustrates this situation. Examp]e

7.2. Two branches h aving i m pedances equal t o j O .25 per u n i t' are

co upled through m u t u a l i m p e d a n ce 7M

=

iO. 1 5 per u n i t , as s h own i n Fig.

7.7.

Fi n d

MUTUALLY COUPLED B R A N CHES

7.2

I N Ybu<

249

F I G U R E 7.7

two

The

(a)

mutually

cou p l e d

b r a n c h e s of Example 7 . 2 ,

their

p r i m i t ive i mp e d a n ces a n d

( b ) p r i m i t ive a d m i t t an c e s

(a)

in

pe r u n i t .

(6)

t h e n o d a l a d m i t t a n c e m a t r i x for t h e m u t u a lly c o u p l e d branches a n d w r i t e t h e correspo n d i ng n od a l a d m i t t a n c e e q u a t i o n s .

Sollllion.

The pri m i t ive

i m pe d a n ce m � l t r i x ror t h e m u t ll ,t l ly c o u p l e d b r a nc hes o f e ll t i t y t o y i e l d t h e p r i m i t ive t l cl m i t t a n c c s o f Fig.

f i g . 7 . 7( (I ) i s i n v e r t e d d S , I s i n g l e

7 . 7( h ) , I h ; 1 1 i s ,

jO . 1 5 j O . 25

[ iO . 25 jO.IS

] [ -I

=

-j6 .25

i 3 .7 5 ] · -)6 .25

j3 .75

F i r s t , t h e rows and co l u m ns of t he b u i l d i n g- b l o c k m a t r i x w h i ch m u l t i pl i e s t h e are l a b e l e d G) a n d in p r i m i t ive s e l f-ad m i t t a n ce b e twe e n nodes and t h a t o r d e r to corre spo n d to t h e dot m a r k i n g n ode Next, t h e rows a n d col u m n s of t h e 2 x 2 m a t rix m u l t i p l y i n g t h e s e l f-a d m i t t a n c e b et w e e n nodes and are l ab e l e d and i n t h e o r d e r s hown b e ca u se node G) is marked. Fi n a l ly, the

CD

G)

(1)

® G) -

CD

G)

(1)

p o i n t ers of the m a t r i ces mu l t ip l y i n g the m u t u a l a d m it t a n c e are a l igned w i t h t h ose of t h e s e l f- a d m i t t a nces to fo r m t h e 4 x 4 array s i m i l a r to Eq. (7.20) as fol lows:

G)

G) CD G) (1)

-:

CD

[ -: 1 CD

CD

[ : -: 1

( -)6 .25 )

G) I CD @ (1)

( i 3 .7 5 )

G)

W

G)

(1)

[ -: -: ]

[ -: : 1

( j 3 . 75 )

( -)6 .25 )

I

i n f i g . 7 . 7 , t i l e r eq u i re d 3 X ] m a t rix is fo u n d by a d d i n g t h e co l u m ns a n d rows o f c o m m o n node to obtain

S i n c e t h e re ; I re o n l y t h re e I l o d e s

CD (1) G)

�

CD

G)

(1)

G)

-)6 .25

) 3 .75

j3 .75

-j 6 . 25

) 6 . 2 5 -- ) 3 .75

- ) 3 .75

+

j6 .25

)6.25 - )3 .75 -

)3 7 5 .

+

-j6.25 - )6 .25

j6 .25

+

2( j3 .75)

J

250 ® •

Za

CHAPTER 7

� Ia

® •

THE ADMITTANCE MODEL A N D N ETWORK C ALCU LAT I O NS

� Ib

([)

•

Zc

Zb

'----./ ZM l ®

PC

ZM 2

®

FI G U R E 7.8 Three branc hes w i t h b ra n c l l L: s 1I a n d h a n d

7.M 2

m u t u a l co u p l i n g 2M I b c twL:L:o b c tWL:L: 11 brallchcs (I , l i l t ! ( ,

The new diagonal e l e m e n l r e prese n t i ng node CD, for instance, i s t h e s u m o f the four clements ( -j6.25 - j6.25 + j3.75 + j3.75) in rows G) and co l u mns CD of the previous matrix. The three nodal admittance equations in vector-matrix form are then written

[

-j6 .25 j3 .75 j2.50

j3 .75 -j6.25 j2.50

where VI ' V2 • and V3 are the voltages at nodes CD, (1), and G) measured with respect to reference, while 11 ' 12 , and 13 are the external currents injected at the respective nodes.

As in Sec. 7. 1 , the coctl1cient matrix or the last equation can be combined with the nodal adm ittance matrices of the other branches of the network In order to obtain the nodal admittance matrix of the entire system. For three or more coupled branches we follow the same proced ure as above. For example, the three coupled branches of Fig. 7.8 have primitive impedance and admittance matrices given by

( 7 .2 1 )

in the Z-matrix arise because branches b and c are not directly coupled. For all nonzero values of the current l a of Fig. 7 .8, branches b and c are indirectly coupled through branch a , as shown by nonzero YM 3 df the primitive admittance matrix.

The zeros

7.3

AN E Q U I V A LENT A D M ITT A N C E N ETW O R K

25 1

Therefore, to form Y b u s for a ne twork which has mutually coupled branches, we do the following in sequence: 1. I nvert the primitive impedance m atrices of the network branches to obtain the

2.

3.

corresponding p r i m i t ive admittance m a t rices. A s i ng le branch h a s a 1 x 1 matrix, two mut u a l ly coupled branches have a 2 X 2 matrix, t hree mutually coupled branches h ave a 3 x 3 matrix, and so on. M u l tiply the eleme nts o f e a c h primitive admittance matrix by the 2 X 2 b u il d i n g b l oc k m a t r i x . ] ,abe l the two rows a n d t h e two col u mns of e a c h diagonal building-block m a t r i x w i t h t h e e n ei - n o d e n u mb e r s of t h e co r re s p o n d i n g self-admittance. For I1l L I l u , t l l y c o u p l e el h r a n c i l e s i t i s i m p or t ll n t t o 1 ,I h e l in t h e o r ei e r o f t he marked -

( do / l cd ) - IIICI l - lll l ! / l(/ril. ('(/

m , l t rix w i t h

w i t h t h e row 1 , l b e l s ,Issignecl t h e c o l u m l l s co n s i s l e n l w i t h t h e co l u i l l n 1 ; l h e l s o r D ) .

4 . La b e l

I h e t w o rows o r

n u m hers a l igned lind

5.

( line/oiled ) n o d e n u m b e rs . e ; l c h ojJ'-dia.£.!,ollo/ b u i l d i n g - b l oc k

c() n s i s t e n t

in (3);

node b

then l a el

C o m b i n e . b y a d d i n g t og e t h e r , t h o s e c l e m e n t s w i t h ident ical row a n d col u m n l a b e l s t o o b t a i n t h e nod a l a d m i t t a nc e m a t r ix o f t h e overall network. If o n e of t h e n o d e s encounte reci is the refe rence node , omit i ts row and col u m n to

o b t a i n t h e sys t e m Y I "" .

7.3

A N E Q U IVALE I\'T A D M I TTA N C E

N ETWO RK

\Ve h ave d e monst r a t e d

how to write the nodal admittance equations for one b r a n c h () r a n u mber of mu t u a l ly coupled branches which are part of a l arger n e t w o r k . W e n o w s h o w t h a t such equ a t ions can be i nterpreted as representing a n e q u i va l e n t a dmi t t a n c e network with no m u t u a l l y coupled elements. This m ay b e useful w h e n fo r m i n g Y b u s for an origi nal network having m utually coupled

d e s c r i b e d i n t e r m s of n o d e vo l t a g e s a n e! ,I d m i t t a n c e s by Eq . 0 . 1 <)). Fo r e x a m p l e , t h e equation for the c u r r e n t Jill ,I t n o d e CD i s g i v e n b y t h e f i rs t row o f Eq . (7 . 1 <)) a s fol l ows:

clements.

T h e c u rr e n t s i nj e c t e d i n t o t h c n od e s o f F i g .

7.0

,I rc

( 7 .22 ) A d d i n g tl n d s u b t r a c t i n g t h e

c ( ) m b i ll i n � t e r m s w i t h c o m m o n

e q u Ll t i o n �d l

noLle CD

t h e r i g h t - h a n d s i d e of

coc fllc i c n ts , w e o b t a i n t h e

t e r m YM V:lI

on

Eq. (7.22) and K i r c h h o ff s c u r re n t '

( 7 .23) I�n p

IlIlil

"

,

:.. ;:;:

252

Aq ,

CHAPTER 7 THE ADM ITfANCE MODEL A N D N E TWORK CALCULATIONS

@)

Im p-

Yo Im n i ®

®

®

Ya

@

(a)

- YM

®

- Ill

-- 'IHII Ip Q

t

/,

FIGURE 7.9

(c)

@)

®

-Y M

1m-

In-

YM

®

@)

It ...-

In

Y

In p

- YM

q

(6)

® -

1,,

®

-

®

-+-- 1 1 1

Ya

Yb

®

t

Yr, -Y M

®

-+-- I

q

Cd) ;'

Developing the nodal admittance network of two m u tually coupled branches.

The double subscripts indicate the dire ctions of the currents 1m , Imp ' and Imq from node ® to each of the other nodes @ , ® , and ® , respectively, of Fig. 7.9(a). A similar analysis of the second and third rows of Eq. (7. 1 9) leads to equations for the currents In and Ip in the form I

I

( 7 . 24)

( 7 . 25 )

and these two equations represent the p a rtial networks of Figs. 7.9( 6 ) and ( c ) . The fourth row of Eq. (7. 1 9) does not yield a separate partial network because it is not independent of the other rows. Combining the three partial networks without duplicating branches, we obtain an equivalent circu it in the form of the lattice network connected among nodes @), @ , ® , and ® in Fig. 7. 9 ( d). This l attice network has no mutually coupled branches, but it is equivalent in every respect to the two original cou p led branches of Fig. 7.6 since it sa tisfies

7.�

AN

EOLJ I VALENT ADM rITANCE N ETWO RK

253

®

F I G U RE 7 . 1 0

0) .

T h e n o d a l a o m i l l , m ce n e t work o f two m u t u a l ly cou pled branches c o n n ected to nodes

and

([0 . ® ,

Eq . ( 7 . 1 9). Accord ingly, t h e stand ard rules of circuit analysis may b e applied to the c q u iv a l c n t . For e x a m p l e , if the two co u p l e d branches are p hy s i c a l l y con­ nect ed among three independent nodes as in Example 7.2, we m ay regard nodes ® and ® of Fig . 7.9(d) as one and the same node, which we simply join together, as shown in Fig. 7. 10. The th ree-bus equivalent circu i t of Fig. 7 . 1 0 t h e n yields the nodal equations for the original branches. Thus, each physical branch or mutually coupled pair gives rise to an equivalent admittance network to w h i ch the usual rules of circuit analysis a pply. The fol lowing example i l l ustrates the role of the equivalent circuit in forming

You s '

CD G)

Exa m p l e 7 . 3 . R e p l a c e b r a n c h es b
W- G)

-

a n d the nodal e q u a t i o n s of t he n e w ne twork. So/ulion . T h e

a d m i t t a n ce

d i agra m

or

t ile

n ew

n e twork

c ( ) l I r l i n g i s s l w w n i n f i g . 7 . 1 1 . F r o m E X l I m p l c 7.2

c o u D l e d b r a n c h e s have t h e

n o d a l a d m i t t a nce m a t r i x

m[

CV Q)

CD

-j 6 . 2 5

G)

j3 .75

we

mutual

k n ow t h a t t h e m u t u a l ly

G)

j2 5 0

j3 . 7 5

-j6 .25

j2.50

j2 .50

j 2 50

-js .OO

.

i n cl u d i ng t h e

j

w h i c h corresponds t o t h e e q u iva l e n t c i rcu i t shown e n c ir c l e d i n Fig. 7 . 1 2. The

rem a i n i n g portion of Fig. 7 . 1 2 is d r a w n from Fig. 7.5. S i n c e m u t u a l coupl i n g is n ot evi d e n t i n Fig. 7 . 1 2, w e m ay a p p l y t h e s t a n d a r d rules o f Y hu s for m a t i o n to t h e ,

254

CHAPTER 7

THE ADMITTANCE MODEL A N D NEnvORK CALCULATIONS

-j8.0

+

1 . 00/- 900

-)0.8

.

jS

-

.

O

�

+

0.68/- 135'

®

FIGURE 7. 1 1 Per-unit a d mi t t a nce d i agram [or Examp l e 7 . 3 .

@ 1 . 00 /- 90°

-jO.8 ®

0.68/- 135'"

FIGURE 7.12

Nodal admittance ne twork for Example branches connected betwee n buses CD,

@, 7.3.

The shaded portion represents two m ut u a ll y cou pled and Q) .

7.4

.•

MODIFICATION O F Ybus

255

overall network, which l eads to the n'odal admittance equations

CD -j16.75 jl 1 .75 j2.S0 j2.S0

G) (1) G) 0

@ jl 1 .75 -j19 .25 j2.S0 jS .oo

® j2 50 j2.50 -jS .80 0

yield

7.4

=

j2.50

.

N o t e t h a t the two a d m i t t a nces between Yl 2

0

jS .OO

0 -j8.30

n o d es

CD

and

VI

V2

V3 V4

(1)

0 0

/ - 900 0.68 / - 1350 1 .00

combine in p a rallel to

- ( -j:' .75 - j8.00) = j l 1 .75

MODIFI CATION OF Y bus

The b uilding-block approach and the equivalent circuits of Sec. 7.3 provid e import a nt insights into the manner in which each branch self- a n d mutual admi ttance contributes to the entries of Ybus and the corresponding equivalent nehvork of the overall system. As a result, it is clear that Ybus is merely a systematic means of combi n in g the nodal admittance matrices of the various nehvork branches. We simply form a large array with rows and columns ordered according to the sequence in which the nonrefere nce nodes of the n e twork are n u mbered, and within it we combine e ntries with matching l abels drawn from the nodal admittance matrices of the i ndividual branches. Consequently, we can easily see how to modify the system Y bus to account for branch a dditions or other changes to the system n etwork. For instance, to modi fy Ybus of the existing n etwork to reflect the addition of the branch admittance betwee n the n odes @) and ® , we simply add Ya to the elements Ymm and Ynn of Ybus and subtract Ya from the symmetrical elements Ymn and Ynm• I n other words, to i ncorporate the new branch admittance Ya into the network, we add to the existing Yh u s the change matrix L1Yhus give n by L1Y h u s =

( 7 .26)

Aga in, we recognize 6.Vhlls as a storage matrix with rows and columns marked @) a n d ® . Using Eq . (7.26), we may change the admittance value of a single branch of the network by adding a new branch between the same e nd nodes @) a n d ® such that the para l l el combination of the old and new branches yields the desired value. Moreover, to remoIJe a branch admittance Ya already con­ n ected between nodes ® a n d ® of the network, we simply add the branch admittance Ya between t h e same nodes, w hich amounts to s4btracting the -

256

CHAPTER

7

THE A D M ITTANCE MODEL AND N ETWO RK CALCULATIONS

elements of t1Ybus from the existing YbuS' Equation (7.20) shows that a pair o f mutually coupled branches ,can b e removed from the network by subtracting the entries in the change matrix

@J

Ya - �l YM - YM

@ ® t1Y bus = ® ®

®

®

-Y

®

- YM YM - Yh

l';\1 - YM Yh

a

�l - Y""I Y" ,

-

Y"

( 7 .27)

Y"

from the rows a n d co l u m ns o f Yhus co r respo n d i n g to t he e n d nodes @ , ® , ® , a n d @ . O f c o u rs e , i f o n l y o n e of t he two m u t u a l l y co u p l e d b r a n ch e s i s t o be removed from the network, we cou ld fi r st remove all t he e n t r i e s for t he mutually coupled pair from Y bus using Eq. (7 .27) and then add t he ent ries for the branch to be retained using Eq. (7.26). Other strategies for mod i fy i n g Y b u s to reflect network changes become clear from the i n sigh ts developed i n Secs. 7 . 1 through 7.3. Exa mple 7.4. Determine t h e b u s a d m i ttance matrix of the ne twork of Fig.

removi n g the effects of m u t u a l cou p l i n g from Y bus of Fig.

Solution. found

in

The Y bus for the e n tire sys t e m of Fig. Exa m ple 7 . 3 to be

CD

CD

-j1 6 .75 j 1 1 .75 CV Y bus = j2 .50 G) j2.50 @

7. 1 1

0

j 1 1 .75 - j 1 9 .25 j2.50

j5 .00

7. 1 1 .

7.5

by

i n cl u d i n g mu tual coupl i n g is

0

j2.50 j2.50 -j5 .80 0

@

j2 .50 j5 .00 0

-jo . J O

T o r e m ov e completely t h e e ffe ct o f m u t u a l cou p l i n g from the n e twork YlJus , we proceed in two steps by ( a) fi rst remov i n g the two m u t u a l l y coupled br<mches a l together a n d (b) then restori n g each of the two bra nches w i thout m u tu a l cou p l i n g betwee n t h e m . ( a ) To r e m ove the two mutually coupled branches from the network, we s u b tract fr o m the syst e m Y bus t h e e n tries i n .... " . ,

i:

. l-ll r . �. ;" �

AYb u s• I =

CD CV

0 @

CD

-j6 .25 j3 .75 j 2 .50

correspo n d i n g to the encircled portion of

cv

j3 .75 -j6 .25 j2.50 F i g . 7. 12.

G) , @

j2 .50 j2 .50 -j5 .00

7.5

(b) Now we must

257

recon nect to t h e network the u n c oupled branches, of 1 - j4.0 p e r u n i t . Accordingly, to reconnect

each

w h i c h h a s an admitta nce ( jO.25 ) the b ra n c h between nodes

6Y bus

THE N ETWO R K I NCI D ENCE MATRIX A N D "bus

')

CD

=

and

=

G), we a d d to Y CD CD

W- I C1) G) - 1 @

6Y hu s . J

=

CD W G) @

the change m a trix

G) 8)

-1

( -j4 .0)

1

a n d s i m i l a r l y fo r t h c b r a n c h b c t w e e n nodes

CD

bus

G)

and

G)

we a d d

G) @ @ -1

- 1

( -j4 .0)

1

A p prop riate l y' s u b tract i n g a n d a d d i n g t h e t h ree c hange m a trices a n d t h e o r i g i n a l

Y b u s g ive t h e new b u s a d m i t t ance m a tr i x for t he u n coupled branches

Y hus ( new)

=

CD G) G) @

CD

- j 1 4 .S

G)

@ j2.S

j8 0

j4 .0 j4 .0

jS .O

j4 .0

-j8 .8

0

j8.0

-j 1 7 . 0

j 2 .S

jS .O

j4 .0

@

.

0

-

j8 3 .

w h i c h a g r e e s w i t h E xa m r l c 7 . 1 .

T H E N ETWO R K I N CI D EN C E M ATRIX

A N D Yh u � 7.5

In S e e s . 7. 1 and 7 . 2 nodal ad mittance equations for each branch and m utually coupled pa i r of b ra n c h e s a re d e r i ved i n d e p e n d e n t ly from those of other branches in the network. The nodal admittance matrices of the individual branches are then combined together in order to bu i l d Y hu s of the overall system. Since we now understand the process, we may p roceed to the more formal approach which t reats all the equat ions of the system simultaneously rather than sepa­ rate ly. We will use the example system of Fig. 7 . 1 1 to e stablish the general procedure. Two of the seven branches in Fig. 7 . 1 1 are mutually coupled as shown. The mutually coupled p a i r is characterized by Eq. (7 . 14) and the other five ,

258 .

THE ADMI1TANCE MODEL AND N EnvORK CALCULATIONS

CHAPTER 7

by Eq. (7.4 ). Arranging the seven branch equations into an array fonnat, we ob t a i n

branches

-jO .80 -j6 .2S

Va

fa

Ve

Ie

Vb

j3 .7S

j3 .7S -j6 .25

V:,

-j8 .00 -j5 .00 -j2 . S 0 -jO .80

fb

Ia

Vc

Ie

Vf

If Ig

VR

( 7 .28)

The coefficient matrix is the p r i m i tive admittance m a trix fo rmed by in sp e c t i o n o f Fig. 7.11. Each branch of the network contributes a diagonal entry equal to the simple reciprocal of its branch impedance except for branches b and c , which are mutually coupled and h ave entries determined by Eq. (7. 1 3). For the general case Eq. '(7 .28) may be more compactly written in the form ( 7 . 29) where Vpr and I pf are the respective column vectors of branch voltages and cu rre nts , while Ypr represents the primitive admittance matrix of the network. The primitive equations do not tell how the branches are configured within the network. The geometrical configuration of the branches, called the topology , is provided by a directed graph, as s hown in Fig. 7.l3( a ) in which each branch of the network of Fig. 7. 1 1 is represented between its end nodes by a directed l ine ® c

d

f e

g

'to

•

"

ea)

®

@)

®

a ( b)

tree branch link

The linear graph for Fig. 7. 1 1 showing: (a) directed-line segments for branches; (b) branches ,a, b, 'and f define a tree while bra nches d, e, a n d R are lin ks.

, FIGURE 7.13

c,

7.5

TH E NETW O R K I NC I D ENCE MATRI X A N D Ybus

259

segment with an arrow in the direction of the b ranch current. When a branch con nects to a node, the branch and node are said to be incident . A tree of a graph is formed by those branches of the graph which i nterconnect or span all the nodes of the graph w ithou t forming any closed path. In general, t here a re m any possible t rees of a network since different combinations of branches can be chosen to span the nodes. Thus, for example, b ra nches a , b , c, and f in Fig. 7. 1 3 ( 6 ) define a tree. The remain ing branches d, e, and g are called links, and w hen a l ink is added to a tree, a closed path or loop is formed. A graph may be described in terms of a connection or incidence ma trix . Of particular interest is the branch -to-node incidence malrix A, w hich h as one row for each branch and one col umn for e a c h n o d e with an entry a ij in row i a nd co l u m n j �Icco rd i ll g to t h e fo l l o wi n g r u l e : "

(J

o

I)

-1

i f b r:l I1 c h i i s

not con nected to n od e CD if curr e n t in branch i is directed away from node CD ( 7 .30) i f c u rr e n t i n bra n c h i i s directed toward node 0

Th is r u l e for m a l izes fo r t h e network as a whole the procedure used to set up the coefficient matrices of Eqs. (7.6) and (7. 1 5 ) for the individual branches. In netwo rk calculations we usually choose a reference !lode. The column corre­ spondi ng to the reference nod e is then omitted from A and the resultant matrix is denoted by A. For example, choosing node ® as the reference in Fig. 7 . 1 3 and invoking the rule of Eq. (7.30), w e obtain the rectangular branch-to-node m atrix

CD

a A =

d c

e

1

-1

1

0 0

- 1

0

1

U

0

-1

-

I I

g

0

0 0

b

Q) G) @

I

- 1

0

J ()

0

()

0

0

0

0 1

(7.31)

1

1

ca l l ed independcn t nodes

or buses , and when we s a y t h a t the network has N b u ses, we generally mean t hat there are N independent nodes not including the reference node. The A m a trix has the row-column dimension B X N for any n e twork with B branches and N nodes excluding the reference. We note that each row of Eq. (7. 3 1 ) has two nonzero entries which add to zero except for rows a and g , each of w h ich has T h e I W ll r e fc r e n c e n o d e s o r

,\

n e t wo r k ; I r e o rt e l l

260

CHAPTER

7 THE ADM I1TANCE M O DEL AND N ETWORK CALCULATIONS

This is because branches a a nd g of Fig. 7 . 1 1 have one con nected to the reference node for which no column is shown. The voltage across each branch may be expressed as the difference in i ts end-bus voltages measured w i t h respect to the reference node. For example, i n Fig. 7. 1 1 the voltages a t buses CD , (1) , (1) , and @ with respect to reference node ® ·a re denoted by VI ' V2 , V3, and � , respectively, and so the voltage drops across the branches are given by

o"n ly one nonzero entry. end

V = V1 Vb = V3 - V2 v:: = V3 - VI Vd = V2 - V I � = V4 - V2 Vf = V - V Vg = V4 /J

V:, �) V:. V:,

•

4

or

Ve

()

=

Vg

()

0

-1

-1 0

1

-1

0 0

-1

0

0

1

0

0

0

1

-1

V I

1

()

1

0

1

0

VI

0

V3

0

1

V2

V4

in which the coefficient matrix i s the A matrix of Eq. (7.31). This is one illustration of the general res u lt for any N-bus network given by

Vp r = AV

( 7 .32 )

Vpr is the B X 1 column vector of branch voltage drops and V is the 1 column vector of bus vol tages measured w ith respect to the chosen reference node. Equations (7.6) a nd (7. 15) are p a r t icu l a r a p p l i ca t i o n s of Eq . (7.32) to i ndividual branches. We fu rther n o t e t h a t K i r c h h o ff' s cu rrent law at nodes CD to @ of Fig. 7.1 1 yields

where

N

X

fa

0

-1

1

-1 1

0

0

0 0

/

0

0

-1 1

-1

-1

0

0

1

0

0

0

0

0

0

1

1

1

/

0

III Ie Id Ie I f

Ig

0

-

0 13 14

1 .00 - 900 and 14 = 0 . 6 8 - 1350 are the external currents in­ j ected at nodes (1) and @, respectively. The coeffi cient matrix in this equation 'is AT. Again, this is illustrative of a general result applicable to every elec;rical :network since it simply states, in accordance with Kirchhoff's current law, that the sum of all the b r a n c h c u r r e n ts i nc i d e n t to a node of the network equals the where

13 =

7.5

Ti l L::

N L::T WO R K I N C l D t..:: N C t..::

M ATR I X A N D y�u,

26 1

i njected curre nt at the node. Accord i ngly, we m ay write

A I =I T

( 7 .33)

pr

where I p r i s the ' B X 1 col u m n vector of branch currents and I is the N X 1 col u m n vector w ith a nonzero entry for each bus with a n external current source . Equations (7.5) and (7. 1 6) are pa rticular examples of Eq. (7.33). The A matrix ful ly d escribes t he topology of t h e n e twork and is indepe n­ dent of t h e particular values of t h e branch parameters. The l atter a re suppli ed by t h e primitive adm ittance rn a trix. There fo re, two dlff eren t network configura­ t i o n s e m p l o y i n g the s a m e b r anc h e s w i l l have d i ffere n t A matrices but the same Y rJr" On t h e o t h e r h a n d , if c h a n g e s occ u r in b r a nc h p a r a m e t e rs w h i l e mai ntain­ i n g the s a m e n e t w o r k co n fi g u r a t i o n , o n l y Y p r i s a l t e re d b u t not A. M u l t i p l y i n g E q . ( 7 . i. lJ ) b y A , w e o b t a i n ( 7 .34 ) The rig h t - h a n d side

( 7.32), we

n nd

of Eq . (7.34)

equ als

I

and

s u b s t i t ut i n g fo r Vpr from

Eq.

�.

( 7 . 35) W e m y \v r i t e

a

where

the N

Eq. (7 . 3 5 )

X

N

in

t h e more concise form

bus admitt ance matrix Ybus

=

N x N

Y"us h as so

of

A

Ybus

fo r

( 7 . 3 6) the

A

T

o n e ro w l l m l o n e c o l u m n fo r c , \ c h o f

t hc N h uscs

t h c s t a n u a rd fo rm o f l h c fo u r i ll ll c [) c n d e n l e q u a t i o n s 0 ["

Fig.

7 . 1 1 is

YI I Y2 1 Y:; J Y<1 1

YI 2 Yn Y:;2 Y42

Yu Y2J Y:n Y4 1

YI 4 Y2<1 Y-, 4 Y4 4

V2

12

11

VI

V:; V4

( 7 . 37 )

B x N

B x R

N x R

system is given by

=

I :;

and t h c exa mple system

i n t h e n e t work,

( 7 . 38)

14

four u n knowns are the bus voltages VI ' V2 , V3 , and V4 when the bus-in­ j ected c u rre n t s I I ' 12 , '-:" a nd 14 are specified. Generally, Y pr is symmetrical, in w h i c h c a se t a k i n g the transpose of each s i de of Eq. (7.37) shows that Yhlls is also Th e

sy m m e t r i c a l .

262

CHAPTER 7

THE A DM ITTANC E

MODEL AND NETWORK CALCU LATIONS

Example 7.5. Determine the per-unit bus admittance matrix of the example system of Fig. 7 . 1 1 using the tree shown in Fig. 7.13 with reference node ® .

The (primi tive a dmittance matrix Ypr describing the branch admittances is given by Eq. 7.28) and ) the branch-to-node incidence matrix A for the specified tree is given by Eq (7.31 . Therefore, performing the row-by-column mUltiplications for AT Yp r indicated by

Solution.

.

0 0 -1

0 -1 0

0 -1 0

-1 0 0

-1

1

1

T

0 0 0 0 1

1

1 0 0 0 0

- jO.S

- j6 .25 j3 .75 j3 .75 -j6 .2S

-jS.O

-j5 .0

-j2 .S

- jO .S

we obtain the intermediate result

AT ypr

=

0 0 - jO .8 0

-j3 .75 j6 .25 - j2.5 0

j6 .25 -j3.7S - j2.5 0

which we may now p ostm u lt i pl y by

Ybus

=

T A Ypr A

=

Q) @ 0 @

A

jS.O - jS.O 0 0

jl 1 .75 -j 19 .25 j2 .50 j5 .00

0

0 - j O .S

to calculate

CD

-j 1 6 .75 j 1 1 .75 j2.50 j2.S0

Since currents are injected at o n ly buses m atrix form are wri tten

- j 1 6 . 75 j 1 1 .75 j2.50 j2.50

0

j2 .5 0 j 0 - 2 .5

0 jS .O 0 -j5 .0

j 2 .50 )2.50 - j5 .S0 0

0

j 1 1 .75 -j19 .25 j2 .50 jS .oo

G)

j2.50 )5 .00 0 - jS .30

and

G)

@

j2.50 jS .oo 0 -j8 .30

j2.S0 j2.S0 -j5 .S0 0

@,

VI V2 V3 V4

the

I

nodal e qu a tions

In

0 0

1 .00/ - 900 0.68 / - 1 350

Example 7.6. Solve the node equations of Example 7.5 to find the bus vol tages by inverting the bu s admittance mat rix.

Solution. Premultiplying

both sides of

the m atrix nodal equation

t h e bus admittance matrix (determin ed by using a standard

by the inverse of

progr a m

on a com puter

7.6

THE M ETHOD OF SUCCESS I V E ELIMI NATI O N

263

or calculator) y i e l d s

VI

V2

V3

=

V4

jO.73 1 28 jO . 69 1 4 0

j O . 6 9 1 40 j O . 7 1 966

j O . 6 1 323 jO . 60822

jO.63677 j O . 64 1 7 8

j O . 6 1 323

j O . 6 0 822

jO . 69890

j O . 55 1 1 O

jO .64 1 78 jO.55 1 1 O

jO .69890

jO .63677

! - 1 8.41 1)9° 0 . 96734 / 0 _ 99964/ - 1 5 .37 1 8° . 9 4 6�

1 .00/ - 90° 0 0

0 .68/ - 135°

Pe rform i n g the i n d i c a t e d m u l t i p l i cations, we o b t a i n the per-u n i t resu l t s

0 . 96903

I H .6()28°

0

7.6

86

2 0 . 74 ()6°

=

I lI

- j O .30859 0 . 96388 - jO . 2 6 49 9

0.91 939 - j O .3 0 6 1 8 0 . 9 1 680

0 . 887 1 5 - jO .3360S

T H E M ETHOD OF S U C C E SS IVE

ELI M I N ATI O N .

In industry-based studies of power systems the networks being solved a re geographically extensive and often e ncompass m any hundreds of substations, generating plants and load centers. The Y b u s matrices for these l arge netw orks of thousands of nodes have associated systems of nodal equations to be solved for a correspondingly large numbe r of u nknown bus voltages. For such solutions computer-based numerical techniques are required to avoid direct matrix inver­ sion, thereby minimizing com putational effort and com puter storage. The m ethod of successive elimination, called gaussian elimina tion , under! ies many o f the n umerical methods solving the equations of such large-scale power systems. We now describe this method using the nodal equations of the four-bus system ( 7 .39) ( 7 . 40 ) (7.41 ) of

( 7 .42)

reduci ng t h is system of fou r equations i n t h e fo u r u n k n ow n s V I ' V2 , V I ,
264

CHAPTER 7

THE ADMITTANCE MODEL AND NETWORK CALCULATIONS

called back substitution . Forward e limination b egins by selecting one equation and eliminating from this equation one variable whose coefficient is cal l ed the

pivot. We exemplify the overall procedure by first e liminating V I from Eqs. (7.39) t hrough (7.42) as follows: Step 1

1.

Divide Eq. (7 .39) by

t h e p ivo t Y I I l o

obt a i n 1

Y1 I

-

2.

[

( 7 .43 ) 1

Multiply Eq. (7.43) by Y2 1 , Y3 1 , and Y4 1 and s u bt ract the resu l ts from Eqs. ( 7.40) through (7.42), respect ively, to get

( 7 . 44 )

(7 .45)

( 7 .46) Equations (7.43) thro ugh (7 .46) may

be

w r i t t e n more compactly in t h e form

( 7 .4 7)

Y22(l)v2 + y(32l ) v2 +

y(23I) V3

+

y(1)V 33 3

+

( 7 .48 )

y(I)V 24 4 I(1) 2 y(1)V 34 4 -- I(3 1 ) =

( 7 .49)

' ,' ,

( 7 .50 )

',wh ere the superscript denotes the Step 1 set of derived coefficients

Yjk(l) - Yjk _

_

1] 1 Yl k Yl l

for j and k

=

2

,

3 4 ,

(7 .5 1)

265

T H E METHOD OF S U CCESS I VE E L IM I N AT I O N

7.6

and t h e modified right-hand side expressions

1� 1 )

)

=

I.

-

J

Y, �J� J

YI I

for j =

I

2, 3, 4

( 7 .52)

Note that Eqs. (7.48) through (7.50) may now be solved for V2 , V3 , and V4 since VI has been eliminated; and the coefficie n ts constitute a reduced 3 X 3 m atrix, which can be interpreted as representing a reduced equivalent network with bus CD absent. In this th ree-bus equivalent the voltages V2 , V3 , and V4 h ave exactly the same values as in the original fou r-bus system . Moreover, t he effect of the current injection l i on the network i s taken into account at buses W , ®, and @ , a s shown by Eq. (7.52). The current I I at bus CD is multiplied by the factor �- I /YI I before being distributed , so to speak, to each bus CD still in the network. \Vc nex t consider t h e e l i m ina t io n or t h e v a r ia b l e V . -

Step 2

1 . D iv i d e

Eq.

(7.48)

by

the

new p ivot Y2ci l to

obtain

-ym 1 22

1

2.

2

C)

( 7 .5 3 )

Multiply Eq. (7.53) by Y§i l and Y4�1 ) and subtract the results from Eqs. (7.49)

and

(7.50)

In a manner

t o g et

similar t

o that of

S t e p 1,

we rewrite Eqs. (i.53) through (7.55) I n

t h e fo rm

V 2

y(l)

-, y (2I ) 2 ")

+

y(l) --'.\

VJ + -----'0l V4 Y2 2

y33( 2 ) V3

+

y34(2 ) V4

y4(2) :1 V3

+

y4(42 ) V4

I

=

=

=

J( 2 I ) ( Y

22

1 ( 2) 3

](2) 4

I)

( 7 .5 6 ) ( 7 57 ) .

(7 58 ) .

266

CHAPTER 7 THE ADMITTANCE MODEL A N D N ET WO R K CALClJLATIONS

where the second set of calculated coefficients is given by

Yjk(2 )

_ -

y(1)

_

jk

YJ·2(1) y2(kl ) yii)

and the net currents injected at buses

for j and k = 3 , 4

( 7 .59 )

G) and @ are for j

=

3, 4

( 7 .60)

Equations (7.57) and (7.58) describe a fu rther r e d u c e d equ ivale n t n e twork ha vi ng only buses Q) a nd @ . V o l t ages V] a n l! V .I a re exactly t h e s a m e a s t h o s e of the original four-bus network because the c u rr e n t i nject ions If) a n d I.f 2) re p re s e n t the effects of al l t h e o r i g i n a l c u rr e n t sources. We now consider el imination of the variable V3 . Step 3

1.

Divide Eq. (7.57) by the pivot Y1P to obtain 1

y ( 2 ) 13( 33

2.

( 7 .6 1 )

2)

Multiply Eq. (7. 6 1 ) by Y4�2) and subtract t h e result from Eq. (7.58) t o obtain y44( 3 ) V4

in which we have defined y(443 )

=

y(2 44 )

_

y4( 2 ) y(342 ) 3 y(2)

=

14( 3 )

( 7 .62)

and

( 7 .6 3 )

33

Equation (7.62) describes the single equivalent branch admittance YN) with voltage V4 from bus @ to reference caused by the equivalent injected current

/(3) 4

.

The final step in the forward elimination process yields

V" .

Step 4

1.

Divide Eq. (7.62) by ��) to obtain ( 7 .64)

At this

p o i nt

we have

fou n d a va l u e ror bus

vo l t age VI which

C ; 1 1l

be s u b s t i t u t ed

267

7.6 THE METHOD OF SUCCESSIVE ELIM INA TION

back in Eq. (7.6 1 ) to obtain a value for V3• Continuing this process of back substitution using the values of V3 and � in Eq. (7.56), we obtain V2 and then solve for V I from Eq. (7.47). Thus, the gaussian-elimination procedure demonstrated h ere for a four-bus system provides a systematic means of solving large systems of equations without having to invert the coefficient matrix. This is most desirable when a largc:-scale power system is being analyzed. The following example numerically illustrates the procedure. Exa m ple 7.7. U s i n g g a uss i a n e l i m i n a t io n , solve t h e n o d a l equat ions of Exa m p l e

7 . 5 to find t h e bus vo l t ages. At e a c h step of t h e so l u t i o n h n d t h e equivalent c i r c u it

m a t rix.

of t h e r e d u c e d c o e fl i c i e n t

Solution. 1 n Exa m p l e 7 . 5 t he nodal

t o be

CD Q)

CD

-j 1 6 . 7S

] j l 1 .75 ]

G) @

W

j l 1 . 7S - · j 1 9 .2S

a cl m i t t a n c e

G)

@ j 2 .S0 jS .OO 0 .0 0

I j2 . S 0 ] j2 .S0

j2.S0

j2.S0

-jS .SO

j2.S0

jS .OO

0 . 00

e q u a t i o n s in m a t r i x form are fou n d

VI

/ - 900 0.68/ - 1350 0

1 . 00

V2

V3

-j8 .30

V4

0

S tep 1

To eliminate the variable VI from rows 2, 3, and 4, we first divide the first row by the pivot - j 1 6.75 to obtain VI - 0 . 70 1 49 V2 - 0 . 14925 V3 - 0 . 1 4925 V4

=

0

We now use this equation to eliminate the j 1 1 .75 entry in (row 2, column 1 ) of Y b u s ' and in the process all the other elements of row 2 a r e modified. Equation (7.5 1 ) shows the p roced ure. For example, to modify t he element j2.50 under­ l i n e d in (row 2, co l u m n 3 ), s u b t r ac t from i t t h e p roduct of the elements enclosed by r e c t a n g l e s d i v i d e d by t h e pivot - j 1 6.75; t h a t i s ,

Y(I) = 2.1

Y2 .1 -

Y2 I YI ''\ YI I

= J'/ � . 5() -

S i m i l a r l y, t h e o t h e r c l e m e n t s o f n e w

Yi� )

=

- j 1 9 . 25 -

y24 ( l ),

= )'5 . 00 -

j l 1 . 75

row

2

X

j2,SO

= j4 .25373 per unit

:I r e

j 1 1 .75 x j 1 1 .7 5 - - j 1 1 .00746 - j 1 6 . 75 j 1 1 .75 X j2 .50 - j 1 6 .75

=

j6.75373

per unit

per unit

268

CHAPTER

7

THE ADM ITTANCE MODEL A N D NETWORK CALCULATIONS

--,.---..... ®

�-------�--�

®

-jO.373 13

-jO.S

1.00/- 909

0.68/- 13s<'

®

FIGURE 7.14

The equivalent three-bus network following Step 1 of Example 7 . 7

Modified elements of rows 3 and 4 are likewise found to yield 1

- 0 .70149

- 0 . 1 4925

- 0 . 14925

VI

0

-

0 0 0

-j l l .00746 j4.25373 j6 .75373

j4.25 373 -j5 .42686 jO .373 1 3

current

j6 .75373 jO .373 1 3 -j7.92686

is d i s t r ib u t e d

V2

V3

-

V4

0 1 .00/ - 900 0 .68 / - 1 350

bu s Ii , a n d

t o the remaini ng buses @ , G)' and @ , and so t h e c u r re n ts Ii , Id in the right-hand-side vector have the same values before and after S tep 1 . The p a r t i t i oned system of equations involving the unknown voltages V2 , V3, and V4 corresponds to the three-bus equivalent network constructed in Fig. 7.14 from the reduced coeffi­ cient matrix. Because

II =

0,

no

CD

from

Step 2 Forward elimination applied to the partitioned 3 tion proceeds as in Step 1 to yield 1 0

- 0 . 70149 1

- 0 . 14925 - 0 .38644

- 0 . 14925 - 0 .6 1356

0 0

0 0

-j3 .78305 j2.98305

j2 .98305 -j3 .78305

X

3 system of the last equa­

VI

0

V2 V3

V4

-

a

�

1 .OO . 0 .68 - 1350

7.6 THE M ETHOD O F S UCCESS I VE ELI M I N AT I O N

®

-)2.98305

-�L......,--

1 . 00/- 900

t

269

-)0.8

0.68/- 135"

@

FI G U RE 7 . 1 5 The e q u iva l e n t two - b u s n e twork fo l lowi n g S t e p

2

of Exa m p l e 7.7.

in which the i njected currents n 2 ) and / P) of Step 2 a re also u nchange d because / � I ) = /2 = O. A t t h is stage w e have e l i minated V I and V2 from the o r i g i n a l 4 X 4 system of equations and t h ere remains the 2 X 2 syste m i n the variables V3 and V4 correspond i n g t o Y b u s of Fig. 7 . 1 5 . Note t ha t nodes CD a n d W are elim in a te d . Step 3

Con t i n u ing the forward e l i m inatio n , we fi n d

1 0 0

- 0 .70 1 49 1 0

- 0 . 1 4925 - 0 . 38644 1

- 0 . 1 4 925 - 0 .6 1 35 6 - 0 .78853

0

0

0

-j 1 .43082

G)

/ - 90° 3 . 7 R3 ()5 / - (jO°

i n w hich the e n t ry for b u s

a n d the modified current at bus

Figure

=

o 68 .

@

/ - 1 35°

1 . 35738

V3 V4

LQ:

0 0 0 .26434

-

1 .35 7 3 8 / - 1 1 0 .7466°

in the right-hand-side vector is calc u lated to be

1 .00

=

VI V2

=

O . 2()434

�

per u n i t

is given by

-

j2 .98305

/ - 1 1 0 .7466°

-j3 .78305

1 .00

- 90° L--

per u n it

7 . 1 6 ( a ) shows the s ingl e a d m i ttance resu lting from Step 3.

270

CHAPTER 7

THE ADM rITANCE M O D EL A N D N ETWO R K CALCULAT I O N S

1@ +

.--..--

1 .35 738/- 1 1 0 .7466"

j 1 . 43 0 82

-

1

@) +

)4 '-------1...---11 @ -

0.94867/- 20.7466

"--�

(a)

(b)

I

FIGURE 7.16

®

1 r

+

The equivalent circuits following ( 11 ) S t e p 3 and ( h ) Step 4 o f Exa m ple 7.7. Step 4

The forw a r d e l i m i n a t i on process term i nates with the calculation -

corresponding to the source transform ation of Fig. 7 . 1 6( 6 ). Therefore, forward elimination l eads to the triangular coefficient matrix given by 1 0 0 0

- 0 . 70149 1 0 0

- 0. 14925 - 0.38644 1 0

VI V2 V3

- 0 . 14925 - 0 .6 1356 - 0 .78853 1

=

�

0 0 0. 26434 0 .94867 / - 20 . 7466°

�

Since V4 = 0.94867/ 20 .7466° , we now begin the back-substitution p rocess to determine V3 u sing the third row entries as follows: -

. V3

- 0 .78853 V4

=

V3 - 0 .74805 /

-

20 .7466� = 0 .26434�

which yie lds V3

Back substitution for

V3

=

0 . 99965 ! - 15 .37 16°

per

unit

and V4 in the second-row equation

V2 - 0 .38644 V3 - 0 .61356V4 = 0

yields

V2

= 0 .9 6 734L - 1 8 . 6030° per unit

7.7

N O D E E L I M I NATI O N ( K R O N R E D UCfI O N )

271

When the calculated values o f V2 , V3, and V4 are substit ute d in the first-row

equ a tion

V1 we obtain

- 0 .70149 V2 - 0 . 14925 V3 - 0 . 14925 V4 = 0 VI

and s o t he p e r- u n i t b u s VI

V2 V, V-1

=

0 .96903 ! - 1 8 .41 89° per u n i t

vo l t ages a rc

= 0 .06903 / - 1 8 .4 1 890

= = =--=

O .96734L - 1 8 . 6030°

O .t)l)t)()4/ - J 5 . 3 7 1 ()0 0 . 94867/ - 20 . 7466°

=

=

= =

0 . 9 1 9 39

- j O . 3 06 1 8

0 . 9 J 6 80 - j O . 3 0 8 5 9

( ) . 963��

- j O .26499

0 . 887 1 5 - j O . 3 3605

w h ich agree a l most exactly w i th t he resu l ts fou n d in E x am p le 7. 7

7.6.

N O D E ELI M I N ATI O N ( KRO N

RE D U CT I O N )

S e c t i o n 7 .6 s h ow s t h a t gaussian e l i m i n a ti on removes the need for m a trix

-

i nve rsion when solving t h e n o d a l e q u a t ions of a l a rg e - s c a l e powe r system. At t h e s a m e t i me i t i s a l so s h own t h a t e l i m i n a t i o n o f v a riables i s identical t o network r e d uction s ince it l e a d s to a s e q u e n ce o f re d u ce d o r d e r network equivalents by node e l im i n at ion at e a c h s t e p . T h i s may b e i m p o r t a n t in an alyzing a l arge i n t e rc o n n e c t e d powe r system if t h e re is s p ec ia l i n t e rest in the vol tages a t o n ly sume of t h e b u se s o f t h e o v e r a l l sys t e m . For i ns t a nce, one e l e c tr i c u t i l i ty com pany \v i t h i n t e rc o n n e c t i o n s to o t h e r c o m r ; l I 1 i cs may w i s h to con fin e i t s s t u d y o f v u l t a g e l e v e l s t o t h o s e S L l b s t ; l t i o fl S w i t h i n i t s o w n se rv i ce t e rr itory. By se lect ive n u m b e r i n g of t h e sys t e m b u s e s , w e m a y a p p l y g a u ss i a n elimi nation so as to rc cJ u ce t h e Y illl, e q u a t i o n s o f t h e ove r a l l sys t e m t o a se t w h ich con t a i ns o n l y t h o s e b u s vu l t ;l g e s o r s p e c i ; d i n t e r e s t . T h e coe l I i c i e I l t m a t r i x i n t h e red uced-order se t o f e q u ; l t i u n s t h e n r e p r e s e n t s t h e Y " , , , fo r ;\ ll e q u iv a l e n t n e twork con t a i n i ng o l l l y t h ose b u s e s w h i c h ; l l C 1 0 he rc t ' l i n c li . A l l o t h er lnl ses Me c l i m i n a t c u i n t h e mat hematical s e n se t h a t t h e i r b u s v o l t a g e s a n d c u r r e n t injections d o n o t app e ar expli c i t l y. S u c h r e u uc t i o n in s i z e of the equ n t io n set l e a d s to efficiency of com p u t a t ion a n d h elps to foc u s more d i rectly on t h at portion of t h e overall n e twork w h i c h is of p r i m a ry i n tere s t . I n gaussian eliminn tion o n e bus-voltage variable at a t i m e is seque n t i a l l y removed from t h e original system o f N e q u a t i o n s i n N un knowns. Fo ll owi n g S t e p 1 of t h e p roce d ur e , the v a r i a b l e V I d o es not exp l icitly appear in the

272

CHAPTER 7

THE A DMITIANCE M ODEL AND NETWORK CALCULATIONS

resultant ( N 1) X (N 1) system, which fully represents the original network if the actual value of the voltage Vl at bus G) is not of direct interest. If knowledge of V2 is also not of prime in terest, we can interpret the (N 2) X (N 2) system of equations resulting from S tep 2 of the procedure as replacing the actual network by an (N 2) bus equivalent with buses CD and CV removed, and so on. Consequ ently, i n our network calculations if i t is advanta­ geous to do so, we may eliminate k nodes from the network representation by employing the first k steps of the gaussian-elimination procedure. Of course, the current i njections (if any) at the eliminated nodes are taken into account at the remaining ( N k) nodes by successive application of expressions such as that in Eq. (7.54). Current injection is a lways zero at those buses of the network to which there is no external load or generating source connected. At such buses i t is usually not necessary to calculate the voltages explicitly, and so we m ay eliminate them from our representation. For example, when 1 \ = 0 in the four-bus system, w e may write nodal admittance' equations i n the form -

-

-

-

-

-

CD @ W @

CD

W

Y2 1 Y3 1 .Y4 1

Y2 2 Y3 2

Yl l

[

CD Yij , G) Y3(i) @ Y.t�)

@

Y1 3 Y23

Y1 2

Y1 4

Y2 4 Y3 4

Y33 Y4 3

Y4 2

and following elimination of node

@

W

Y44

CD , we

W

y(l) 23

@

Y24( l '

y(l)

yell

ye ll

43

0 -

V3 V4

34

44

12

( 7 .65)

13 14

obtain the 3

y
]

VI V2

X

3 system

[�: J ��: ] =

( 7 . 66)

in which the superscripted elements of the reduced coefficient matrix are. calculated as before. Systems in which those nodes w ith zero current injections a re eliminated are said to be Kron \ reduced. Hence, the system having t he particular form of Eq. (7.65) is Kron reduced to Eq. (7.66), and for such systems node elimination and Kron reduction are synonymous terms. Of course, regardless of w hich node has the zero current injection , a system can be Kron reduced without having to rearrange the equations as in Eq. nodal equations of the N-bus system, we example, if Ip 0 in the , , (7.65). For . =

IAfter Dr. Gabriel Kron (190 1 - 1 968) of General Electric Company, Schenectady, NY, who con­ tributed greatly to power system analysis.

7 .7

N O D E ELI M I N ATION (KRON REDUCfION)

273

may d irectly calculate the elements of the n ew, reduced bus admittance matrix by choosing Ypp as the p ivot and by e liminating bus p using the formul a ( 7 . 67)

where j and k take on a l l the i nteger values from 1 to N except since row p and column p are to be eliminated . The subscript (new) d istinguishes the elements in the new Y of d imension ( N - 1 ) X (N - 1) from the elements in the origin al Y

p

bus

bu s '

Exa m p l e 7.8. Using Y2 2 as t h e in i t i l pivot, e l i m i n a t e n o d e @ corresponding voltage V, from t h e 4 x 4 syst e m o f Example 7.7

a

The i ot Y22 q u l s may e l i m i n a t e row 2 col u m

Solution. 1

pv

cl ements

Y 1 1 ( n ew) YI 3(new)

=

=

and

Y1 1 -

YI J -

YI 4( new) = YI 4 -

e a

-j 1 9.25 . With p se t e q u a l to 2 in Eq. (7.67), we n 2 from Ybus of Exa m p l e 7.7 to o b t a i n t h e new row

Yl 2 Y2 1

-j 1 6 .75

Y22

Y l 2 Y23

Y22

Y 1 2 Y2 4 Y22

= j2 .50 -

=

j2.50 -

-

( j 1 1 .75 ) ( j 1 l . 7 5 ) -

j 1 9 25

-j9 .57792

.

( j l 1 .75 ) ( j2 'sO) =

-j 1 9 .25 ( j l 1 . 7 5 ) ( j5 . 00 )

j4 .02 597

= j5 .55 1 95

-j1 9 .25

-)0.64935 -)5.55195

-j4.02597

1 .00/- 900

CD -j O.S

jO 8

-

@

Kron -reduced network of Example 7.8.

FI G l IR E 7 . 1 7

The

t

and the

.

,I,

0.68/- 1 350

274

CHAPTER 7

THE ADM I1TANCE MODEL AND NETWORK CALCU LATIONS

Similar calculations yield the other elements of the Kron-reduced matrix

Q) r

G) @

7.8

CD

-j9 .5 779 1

j4.02597

j4 .02597 j5 .55 1 95

-j5 .47532 jO .64935

G)

r

/ - 90° 0

1 .00

0 .68

/

- 1 35°

1

Because the coefficient matrix is symmetrical, the e q u iv a l e n t c i rcu i t of Fig. 7 . 1 7 applies. Further use o f Eq. (7.67) to e l i m i n a t e node CD from F i g . 7. 1 7 l e a d s t o t h e Kron-reduced equivalent circuit shown i n Fig. 7 . 1 5 . TRIANGULAR FACTORIZATI O N

In p,r actical studies the nodal admittance equations o f a given large-scale power system are solved under different operating conditions. Often in such studies the network configuration and parameters are fixed and the operating condi­ tions differ only because of changes made to the external sources connected at the system buses. In all such cases the same Y bus applies and the problem then is to solve repeatedly for the voltages corresponding to different sets of cu rrent injections. In finding repeat solutions considerable computational effort is avoided if a ll the calculations in the forward phase of the gaussian-elimination procedure do not have to be repeated. This may be accomplished by expressing Y bus as the product of two matrices L and U defined for the four-bus system by 1

Yl l

L=

Y2 1

y(l) 22

YJ I

yell 32

y (2)

Y.t l

y(1) 42

y ( 2)

U= :\3

43

Y1 2

Y! 3

Y14

Y1 1

Y1 1 yell

Y1 1 y(l) 24 y(l) 22 y (2) 34 y ( 2)

1

23

y(l) 22

1

( 7 . 68 )

33

y (3)

1

44

The matrices L and U are called the lower- and upper-triangular factors of Yb u s because t h ey have zero elements above a nd below t h e respective p r i ncipal d iagonals. These two matrices have the remarkably convenient property that their product equals Ybus (Problem 7 . 1 3). Thus, we can write LU = Y bus

( 7 .69)

The process of developing the triangular matrices L and U from Y bu s is called triangular factorization since Ybus is factored into the product LV. Once Y b us is so . factored, the calculations in the forward-elimination phase of ga � ssian

7.8

TR I A N G ULAR FACfORI Z ATION

275

e l i m ination do not h ave to be repeated s ince both L an d U are u n ique a n d do not change for a given Y b u s ' The entries i n L and U are formed by systematically recording the outcome of t h e calculations at e ach step of a single pass through the forward -elim ination p rocess. Thus, in forming L and U, no new calculations are involved . This is now demonstr ated for the four-b us system w ith coefficient matrix

Y hlls

=

CD W G)

@

CD

YI I Y2 1

Y,\ 1 Y4 1

C1)

Yl 2 Yn Y,,\ � Y4 2

G)

Y I :1

Y2:1 YJ :1

Y4:1

@

YI -1 Y�:4

Y ;.1 Y�-1

( 7 .70)

g;ll1 ss i ;l I1 e l i lll i n il t i o n i s ;t p p l i e d to t he fO l l r 110ti ;t i cql 1 <1 t ions co rrespo n d i ng to t h i s Ybli S ' w e n o t e t h e fo l l owing. Step 1 y i e l d s res u l ts g iven by Eqs. ( 7.47) t hrough ( 7 .50) in which:

When

The coefficients Y1 1 , Y2 1 ' Y3 1 , and Y4 1 a re e l i m i n a ted from t h e first column of t h e original coefficient m atrix of Eq. ( 7.70). 2. New coe ffici e n ts 1, Y I 2/Y1 1 , Y U / Y1 1 , and YI 4/YI I are generated t o replace t hose in the first row of Eq . ( 7.70 ) . 1.

The coefficients in the o t h e r rows and col u mns a re also altered, b u t we keep a separate record of only th ose specified i n ( 1 ) and (2) si nce t hese are the only resu lts from S tep 1 w hic h a re neither used nor a l tered i n Step 2 or subsequent steps o f the forwa rd -el i m i n a t ion p rocess. Col u m n 1 of L and row 1 of U i n Eqs. (7.68) show the recorded coefficients. Step 2 y iel d s resu l ts giv e n b y Eqs. (7.56) t h rough ( 7.58 ) i n w h ic h : Coc fll c i c n t s Y ii >' Y.� i >'
These coefflcients a r c n o t n e e d ed i n t he r e m a i n i n g steps o f forward e l i m i na t ion, and so w e record them as column 2 of L and row 2 of U to s h ow the Step 2 record . Con t i n u i n g t h is record-keep ing proce d ure, we fo rm col u m ns 3 and 4 of L and rows 3 and 4 of U u s i n g the res u l ts from Steps 3 and 4 of Sec. 7.6. Therefore, m a t r ix L is s im pl y a record of those col umns w hich a r e succe ssive ly e l i minated a n d m a t r ix U records t h ose row e n t ries w h i c h are s u ccessive ly generated at each step of the forw a rd stage of gaussian el imination.

276

CHAPTER 7

THE ADM ITI'ANCE MODEL AND N ETW O R K CALCU LATI ONS

We may use the triangular factors to solve the original system of equations by substituting the product LU for Y bus in Eq. ( 7.38) to obtain L

NxN

As

u

NxN

v

I

Nx l

(7.71)

Nx l

an intermediate step in the solution of Eq. (7.71), we p ro duct UV by a new voltage vector V' such that

L

V'

I

NxN Nx 1

u

and

N x 1

N x N

V

N x !

m ay replace

V'

the

( 7 .72)

.V x 1

Expressing Eq. (7.72) in full format shows that the original system of Eq. (7.38) is now replaced by two triangular systems given by Y1 1 Y2 1

y(l) 22

Y3 1

y(l) 32

y (2 ) 33

Y4 1

y(l) 42

y ( 2) 43

1

and

Y1 2 Y1 1

1

y (3 ) 44

Y1 3

YI 4

YI I

YI I

y (3l ) 2 y(l) 22

1

y24( l )

V'1

II

V'2

12

V'3

13

V'4

14

VI

V'I

yO) 22 y3( 2 ) 4 y33 (2 )

V2 V3

V'3

1

V4

V'4

-

V'2

( 7 .73 )

( 7 . 7 4)

The lower triangular system of Eq. (7.73) is readily solved by forward substitu­ -tion beginning with V;. We then use the calculated values of V; , V� , V;, and V; to solve Eq. (7.74) by back substitution for the actual unknowns VI ' V2 , V3 • � and �. . Therefore, when changes are made in the current vector I, the solution vector V is found in two sequential steps; the first involves fonvard substitution using L and the second employs back substitution using U.

7.8

Example

277

T R I A N G ULA R FACTOR I ZATION

G) 0.60/ - 1200

7.9. U s i n g t h e t r i a n g u l a r factors of Y nUS ' d e t e r m i ne t h e voltage at bus

of F i g . 7. 1 1 when the c u rre n t sou rce a t b u s

7. 1 1

per u n i t . A l l o t h e r con d i ti o n s of F i g .

Ynu s

Solution . T h e

@

is c h anged to

are u n c h a n g e d .

for t h e n e two rk of Fi g . 7 . 1 1

14

=

is given in Example

7.3.

The

correspond ing m a t r i x L may b e a ss e m b l e d col u m n by col u m n from Example

7.7

s i m ply by reco r d i n g t h e col u m n w h i c h is e l i m i n a t e d from the coe ffi c i e n t m a trix a t

e a c h step o f t h e forw a rd - e l i m i n a t i o n p rocedure. Th e n , subs t i t u t i n g for L a n d t h e n e w c u r r e n t vector

I

i n t h e e q u a t i o n LV '

=

I , w e ob t a i n

-j I 6 .7 5

j 1 1 .75

-j 1 1 .00746

j2.50

j 6 . 75 3 73

j4 .25373

j2 .50

-j3 .78305 j2

.

V;

=

V2

=

.

- 900 0.60/ - 1200

V(

1 .00/

�

yields

900 3 . 78305 / - 900 -

-

- ( j2 .9S305 ) V; -j 1 .43082

=

The m a t ri x U i s shown d i re c t l y fol l o w i n g S t e p

4

.

r/I

-j 1 .43082

0.60,[ - 1 20° =

1 0 0/

V:.l

9 83 0 5

V'' -

0;

0

V2

S o l u t i o n by forw a rd s u b st i t u t i o n b e g i n n i n g w i t h

V(

0

V I'

=

0.26434

0.93800 / - 1 2.9163° "----o f t h e forward e l i m i n a t i o n i n

Ex a m p l e 7 . 7 . Subs t i t u t i n g U a n d t h e c a l c u l a t ed e n t r i e s o f V ' i n t h e e q u a t i o n

U V = V ' g i ve s

1

- 0 .70 149

-

0 . 14 25

9

- 0 .38644

-

0

.

1 4 92

5

- 0 . 6 D5 6 - 0 .7885 3

VI

0 0 0 . 264 34 0 . 9 3 80 0 / - 1 2 9 1 630

V2 V:l V4

.

0 . 9 3 8 00/ - 12.9163°

w h ich \Ve m ay s o l ve by b a c k s u b s t i t u t i o n to o b t a i n

V4

Vl

=

V;

= 0 . 6434 - ( - 0 .7885 3 ) V4 =

2

=

0 . 99904 /

per u n i t

- 9 .5257°

per u n i t

I f d e s i r e d , w e m a y co n t i n u e t h e b a ck s u b s t i t u t ion u s i n g t h e v a l u e s for eva l u a t e unit.

V2

=

0.96 1 18/- 1 1 .555 10

p e r u n i t and

VI

=

V)

and

V4

0.96324/ - 1 1 .43880

to

per

278

CHAPTER 7

THE ADMITTANCE MODEL A N D NETW O R K CALCULATIONS

When the coefficient matrix Ybu s is symmetrical, which is almost always the case, an important simplification results. As can be seen by inspection of Eq. (7.68), when t h e first column of L is divided by Yl 1 , we obtain the first row of U; when the second column of L is divided by Yi�\ we obtain the second row of U ; and so o n for the other columns and rows of Eq. (7.68), provided �j = �' i ' Therefore, dividing the entries in each column of L by the principal diagonal element in that column yields U T whenever Y "us is symmetrical. We can then write YI I

Y2 I

L = uTn

YI I

=

Y3 1 YI I

�l

Yl l

y(l) 32 y(l) 22 yO) 42 yell 22

y(l) 22

( 7 .75)

y ( 2)

1

33

y (2 ) 43 y (2 )

1

33

y (3 ) 44

where diagonal matrix D contains the d iagonal e lements of L. Substituting i n Eq. (7.71) for L from Eq. (7.75), we obtain t h e nodal admittance equations in the form Y busV =

u Tnuv =

( 7 .76)

I

Equation (7.76) may be solved for the u nknown voltages V in three consecutive steps as follows: ( 7 .77)

nv' =

V"

( 7 . 7 8)

=

V'

( 7 .79)

UV

These equations will be recognized as an extension of Eqs. (7.72). The interme­ diate resu l t V" i s first found from Eq. (7.77) by forward substitution. Next, each entry in V' is calculated from Eq. (7.78) by dividing the corresponding element of V" by the appropriate diagonal element of n. F i na l l y, the solution V is ; ob taine d from Eq. (7.79) by b ack substitution as demonstrated in Example 7.9. Example 7.10.

Using Eqs.

(7.77)

through

(7.79),

d etermine the solution vector V of

u nknown voltages for the system a n d operating cond itions of Example

7.9.

·

7.1)

Solution. S u b st i tu t i ng i n Eq.

(7.77)

S P A R S ITY A N D N EA R - O PTI M A L O R D E R I N G

fro m the c u rre n t vector

Ex a m p l e 7.9, w e obt a i n

1 .00/ /

V�'

- 0.70 149 - 0 . 1 4925

- 0 .38644

- 0 . 1 4925

-

0

.

V":1

1

6 1 356

V" 4

- O . 7RR53

i

and

matrix U of

0

V"1

1

I

279

0

0 . 60

- 9 00 - 1 200

S t ra i g h t fo rward sol u t i o n o f t h i s sys t e m of e q u a t i o n s y i e l d s

v;'

v;'

=

V�'

O.(loL- .� :20: ' c

v;'

=

0

-I-

O . 7XX:'i l l "-;'

S u b s t i t u t i n g for Y " i n E q . ( 7 . 7 8 ) l e n c i s -

j 1 6 75

I .()O

=

=

I

. 342

9()0 p e r u n i t

rO�02 .9 1 64°

-

t o t h e d i ag o n a l sys t e m

V'1

.

-j 1 1 . 00 7 4 6

V'3

-j 3 .78305 - j 1 .43082

V'4

pe r u n i t

o o

/ 1 .342 1 0/ 1 . 00

-

- 90° 1 02 . 9 1 64°

t h e s o l u t i on of w h i c h is e x a c t l y e q u a l to V' of Example 7 . 9 , a n d so the re m a i n i n g

s t e p s o f t h i s exa m p l e m a t c h t h o s e o f Ex a m p l e 7 . 9 .

7.9

SPARSITY

A ND NEA R - O PT I M AL

O RD E RI N G

La r g e - sc a l e p o w e r con n e c t e d

to

systems

a small n u mher of transmiSSion l ines s u b s t a t i o n I n t h e n e tw o r k g r a p h fo r stich systems

h ave onl y

each h u l k - rower

.

a bou t 1 .S

and

the c o r r e s p o n d i n g Y \1us h a s 1ll ;l i n l y ze ro c l e m e n t s . 1 1 1 fa c t , i f t h e r c a re 750 h r a n c h e s i n (l S O( ) - no d e n e t w o r k ( e x c l u d i n g t h e r de rc n c e n o d e ), t h e n s i n c e each n o d e has ,I n ( l s s o c i ,l [ e d d i a g o n a l c l e m e n t a n d e a c h b r(l n c h gives r i s e t o t wo sym metr ically p l a c e d o tf- d i a gona l elements, t h e t o t a l n u m b e r of nonzero e l e ments is (500 + 2 X 750) = 2000. Th i s comp a res to a total of 250, 000 e l e m e n ts in Y h ilS ; that is, o n l y O.R% o f t h e c l e m e n t s i n Y blls ar c n o n z e r o . Bec a u s e o f t he s m a l l number o f n o n z e ro e l e m e n t s, such m a t rices a r e said to b e spa rse . F r o m the viewpoint o f computational speed, accuracy, and storage i t i s desi rab l e to process only the n o n z e ro entries in Y h u s a n d to avoid filling-in n e w nonzero elements in the co u r s e o f g a u s s i an e l i m i n a t i o n and triangular factorization . Orde r ing refe rs to t h e s e q u en c e i n w h i c h t h e e q u a t i o n s o f a s y s t e m a re p r o c e s s ed When a sparse t h e r ; l l i o 0 1' t h e n u m b e r o r b r ;l I1 c h e s to t h e n u m b e r o f n o c i e s i s

.

280

CHAPTER 7

THE ADMITTANCE MODEL AND N ETWORK CALCULATIONS

matrix is triangularized, the order in which the unknown variables are elimi­ nated affects the accumulation of new nonzero entries, called fill-ins, in the triangular matrices L and U. To help minimize such accumulations, we may use ordering schemes as described in Sec. B.1 of the Appendix. 7.10

SUMMARY

Nodal representation of the power transmission network is developed i n this chapter. The essential background for understanding the bus admittance matrix and its formation is provided. I ncorporation of mutu ally coupled branches into Y bus can be handled by the bu ilding-block approach described here. Mod i fica­ tions to Y bus to reflect network changes arc thereby facil itated. Gaussian elimination offers an alternative to matrix i nversion for solving large-scale power systems, and triangular factorization of Y bus enhances compu­ tational efficiency and reduces computer memory requirements, especially when the network matrices are symmetric. These modeling and numerical p rocedures underlie .the solution ap­ proaches now being used i n daily practice by the electric power industry for power-flow and system analysis. PROBLEM S 7.1.

Usin g t h e building-block procedure described i n Sec. 7 . 1 , d e t e r m i n e Y bus circuit of Fig. 7.18, Assume there is no m u tual coupling between any branches.

fo r t h e of t h e

CD

)0. 125

+

®

1.10LQ:.

•

®

+

0.90/- 300

F1GURE 7.18 Value's shown are voltages and impedances in per u nit. Dots represent mutual coupling between ,

branches u nless otherwise stated in problems.

P R O BLEMS

281

Y bus modification proce d u re d escribed in Sec. 7.4 and ass u m i n g no m u t u a l coupling between branches, m o d i fy t h e Y b u s obtained in Prob. 7 . 1 to reflect removal of the two b ra nc hes CD - G) a n d 0 - G) from the circuit of Fig. 7 . 1 8.

7.2. Using the

7.3. The circuit of Fig. 7 . 1 8 h as t h e l i n e a r graph shown i n Fig. 7 . 1 9, with ar rows i n dicating d irections ass u m ed for t h e branches a to h . Disregarding all m u tu al

coupling between branches: (a) Determi � e the branch-to-node incidence matrix as reference. ( 0 ) Find the circu i t Y bus u s i n g Eq. (7.37).

a

b

g

A

for the circuit with node

®

® d

c

r

e

®

®

h FI GURE 7 . 1 9

@ 7.4.

L i n e a r g r a p h for P rob. 7 . 3 .

Co n s i d e r t h at only t h e two bra nches C!) - G) and 0 - G) in the circ u i t o f Fig. 7. 1 8 are m u t ua l l y co u p l e d as ind icated by t h e d o t s besi d e them and t h a t their m u t u a l impeda nce is j 0 . 1 5 per u n i t (that is, ignore the dot on branch 0- G) ). D e t e r m i n e t h e c i rcu i t Y b u s by the p roce d u re d escribed i n Sec. 7.2.

7.5. Solve Prob. fr o m t h e

7.6. U s i ng

u s i n g E q . 0 . 3 7). D e t e rm i n e the b ra nch-to-node incidence m a trix l i n e ,l r g r a p h o f F i g . 7 . 1 9 w i t h n o d e @ as r e fe r e n c e . 7.4

A

t h e m od i fi.c a t i o n proce d u re of S c c. 7 . 4 , mod i fy the Y bus sol u t i o n of Prob. 7.4 (o r P r o b . 7 . S ) t o re flect re moval of the branch W - G) [rom t h e circ u i t . 7.7 . Mo d i fy the Yb ll S d e t e r m i n e d i n Exa m p le 7.3 to reflect r emoval of the m u tu ally co upled b r a n c h CD - G) from the c i rc u i t of Fig. 7. 1 1 . Use t he modification p rocedu re of Sec. 7.4.

7.8.

7.9.

u n i t i s a d d e d between nodes 0 a n d G) in t h e c i rc u i t o f Fig. 7 . 1 1 . M u t u a l i m p e da n ce of jD. l per u n i t couples t h is n ew b ranch to the branch a l ready exist i n g between nodes 0 and G) . M o d i fy the Y b u s o b t a i n e d in Example 7.3 to acco unt for the a d dition of the n ew branch.

A n e w b r a n c h h a v i n g a se l f- i m pe d a n ce o f

j O.2

per

S u ppo se t hat m u tual cou p l i n g exists p a i rw i se b etween bra nches CD - G) a n d W- G), and a lso between branches W - G) a n d 0 - G) of Fig. 7.18, a s shown b y t h e dots in t h a t fi g u re . T h e m u t u a l i m p e d a n ce be tween the former p a i r o f

282

CHA PTER 7

THE ADMITTANCE M O DEL A N D N ETWORK CALCU LATI ONS

branches is jO. 1 5 per unit (the same as i n Prob. 7.4) and between the latter pair is jO. l per u nit. Use the procedure of Sec. 7.2 to find Y bus for the overall circui t including the three mutually coupled br anches. 7.10. Solve for the Yb us of Prob. 7.9 u sing Eq. (7.37). Use the l in e ar graph of Fig. 7 . 1 9 with reference node ® to d etermine the branch-to-node i ncidence matrix A. 7.1 1 . Suppose that the direction of branch d in Fig. 7. 1 9 is reversed so that it is now dire ct e d from node 0 to node G) . Find the bra n c h -to-node inciden ce matrix A of th is modified graph and t h e n so lve for t h e Ybus of P rob 7.9 using Eq. (7.37). 7.12. U s i n g the Yhus nl odi tica t i o ll p roce d u r e d escri bed in Sec. 7.4, rcmove b r a n c h (1)- G) from the Y t,lIS so l u t i o n o b t a i n e d in P ro b . 7.9 (or P ro b . 7. 1 0 or P rob. 7. 1 1 ). 7.13. Write n o d a l a d m i t t a nce e q u a t i o n s for t h e c i rc u i t o r rig. 7. 1 i) d i s re g a rd i ng a l l m utual c oupling. Solve t h e r esu l t a n t e q u a t i o n s for t h e bus volt ages b y t h e me thod of gauss ian e limination. 7.14. Prove Eq. (7.69) based on Eq. (7.68). 7.1S. Using the g aussian-el i minat ion calcu l ations o f Prob . 7 . 1 3 , fi n d the t ri a n g u l a r factors of Y bus fo r the circui t of Fig. 7.18. 7.16. Use the triangular factors obtained i n Prob . 7 . 1 5 to calculate new bus vol tages for Fig. 7.18 when the voltage source at bus G) is changed to per u n i t . Follow the p rocedure of Example 7.9. 7. 17. Using the triangular factors obta ined in Example 7.9, fi n d t h e voltage a t b u s G) o f t h e circuit o f Fig. 7.1 1 w h e n an additional curre n t of 0.2/ - 1200 pe r u n i t is injected a t bus 0 . All other conditions of Fig. 7. 1 1 are unchanged. .

1 .0�

G) .

reduce Y bus of the c ircui t of Fig. to reflect elimination of node ( b ) Use t h e Y transformation o f Table to eli minate node from the 6. circuit o f Fig. and find Y b u s for the resul t i n g reduced network. Compare results of parts ( a ) and ( b ) .

7.18. (a) Kro n

7. 18 1 .2

-

7.18

7.19.

Find the L a n d

U

triangular factors o f t h e symmetric m atrix

Verify the result using Eq. (7.75).

0

CHAPTER

8

THE I MPEDANCE MODEL AND

NE'rWO R K CALCULA'rI O N S

The bus admittance matrix o f a large-scale interconnected power syste m i s typically very sparse with mainly zero elements. I n Chap. 7 w e saw how Y b u s i s constructed branch by branch from primitive admittances. I t i s conceptually simple to i n v e r t Ybu s to find the bus impedance matrix Z b u � ' but such d i rect inversion is rarely employed when the systems to be analyzed are large scale. In p r a c t i ce Z bus is r a r e l y e xp l i c i t l y required, and so the tria ngular factors of Y b u s a re u s e d to ge n e r(l t e c l e m e n t s of Z h l l S o n l y as t h ey are n e e d e d s i n ce t h is i s often ,

as b e i n g a l r e a d y constructed and a n a l y s t can de r ive a great deal of i n s i g h t .

r e g ar d i n g Z h u s

t h e m o s t C O I l 1 I Hl t i l t i o l l ; t l l y d J i c i e n t m d i l oli . U y se t t i n g c O IH r l l t a l i o n a l co n s i d e r a ­

t ions aside,

however,

and

is t h e approach t a k e n in t h i s c h a p t e r . The bus impedance matrix c a n be d i re c t l y constructed element by ele m e n t using simple algorithms to incorporate one element at a time i n to the syste m representation. T h e w o r k e n t a i l e d i n const ructing Z bus is much greater than that required to construct Y b u s ' but the informat ion content of the bus impeda nce rna trix is far greater than that of Y b us ' We shall see, for example, that each diagonal element of Z b u s reflects i mportant characteristics of the entire system in the form of the Thevenin impedance at the corresponding bus. Unlike Y bus ' the b u s impedance matrix of an interconnected system is never. sparse a n d contains z e ros only when t h e s y s t e m i s regarded a s being subdivided i nto ex p l i c i t l y a v a i l a b l e , t h e rowe r sys t e m

Th i s

283

284

CHAPTER 8

THE I M PEDANCE MODEL AN D N ETWO RK CALCULATIONS ·

independent parts by open circuits. In Chap. 1 2, for instance, such open circuits arise in the zero-sequence network of the system . The bus admittance matrix is widely used for power-flow analysis, as we shall see in Chap. 9. On the other hand, the bus impedance matrix is equally well favored for power system fault analysis. Accordingly, both Ybus and Z bus have important roles in the analysis of the power system network. I n this chapter we study how to construct Z b us directly and how to explore some of the conceptual insights which it offers into the characteristics of the power transmis­ sion network. 8.1 THE BUS ADMITTANCE AND IMPEDANCE MATRICES In Example 7.6 we inverted the bus admittance matrix resultant the bus impedance matrix Z b u s ' By definition

Y hus

and called the (8.1)

and for a network of three independen t nodes the standard form is

CD Z b us = Q) G)

CD Q) G)

2 1 1 2 12 213

2 2 1 222 223

( 8 .2)

2 3 1 2 3 2 2 33

Since Y bus is symmetrical around the principal diagonal, Z bus must also be symmetrical . The bus admittance matrix need not be determined in order to obtain Z bus > and in another section of this chapter we see how Z b us may be formulated directly. The impedance elements of Z hus on the principal diagonal are called driving-point impedances of the buses, and the off-diagonal elements are called the transfer impedances of the buses. The bus impedance matrix is important and very useful in making fault ca1cula tions, as we shall see l ater. In order to understand the physical signifi­ cance of the various i mpedances i n the m a trix, we compare them with the bus admittances. We can easily do so by looking at the equations at a particular bus. For instance, starting with the node equations expressed as I = Y busV

( 8 .3 )

we have at bus (I) of the three independent nodes '( 8 . 4)

H. I

THE

BUS ADM ITTANCE A N D I M PEDANCE M AT R I C ES

285

J

VI and V3 are reduced to zero by shorti n g buses CD a n d Q) to t h e reference n o d e , and voltage V2 is a p p l i e d a t b u s (I) s o t h a t c urren t 2 e n t e r s a t bus W , t h e self-adm i ttance at bus (I) i s If

( 8 .5 ) T h u s , the s e l f-ad m i tt a n ce of a particu l a r bus cou l d be m easured by short ing all other b u ses to the reference node and then finding the ratio of the cu rren t i njected a t t h e bus to the vol tage applied at t h a t bus. Figure 8 . 1 illustra tes t h e m e thod for a t h ree-bus reactive network. T h e result i s obviously eq u iv a l e n t t o add i n g a l l the a d m i t ta nces d i rectly connected to the bus, w h i c h is t h e p rocedure up to now w h e n m u t ua l l y coupled branches are abs e n t . Figure 8.1 a l s o s e rv e s to i l l u s t r3 t e the off-di agonitl a d m i t t a nce t e rm s of Ybll s ' At bus CD t h e equ a t ion obta i ned by expan d i n g Eq . (8.3) is

( 8 .6)

from \vh i ch w e s e c t h a t

( 8 .7) Thus, the m u t ual a d m i t t a n ce term Yl 2 i s m easu red b y shorting all b u ses exc e p t bus G) to the refe rence n o d e a n d b y app lyin g a vol tage V2 at bus Cl), a s s h ow n i n Fig. 8 . 1 . The n , YI 2 is the ratio of the n egat ive of the curre n t l eavi n g t h e n e t work i n t h e short c i rcu i t a t n o d e CD to t h e vol tage V:, . T h e n eg a t ive o f t h e curre n t l e aving t h e network a t node CD is used s i n ce I I i s d e fi n e d a s t h e c u rrent entering t h e n e twork. The resultant admitta nce i s t h e n egative

®

CD ®

..

12

+

V2

~ FI G C RE 8 . 1 Circ u i t fo r m e a s u r i n g Yn . Y1 2 • a n d

Y32 .

286

CHAPTER 8

THE I MPEDANCE MODEL A N D N ETWORK CALCULATIONS

of the adm;ttance directly connected between buses CD and @ , as we would expect since m u tu a lly coupled branches are absent. We have made this detailed examination of the bus admittances in order to differentiate them clearly from the impedances of the bus impedance matrix. Conceptually, we solve Eq. (8.3) by premultiplying both sides of the equa tion by Y;;:I� = Z hus to yield ( B .B)

and we must remember w h e n dea l i ng w i t h Z hus t h a t V a n d I a re col u m n vec t o rs of the bus voltages and the c u r r e n ts e n t e r i n g the buses from c u r r e n t s o u rc e s , respectively. Expanding Eq. (8.8) for a network of three in d ependent nodes

yields

( 8 .9) ( 8 . 1 0) (8 . 1 1 )

From Eq. (8. 1 0) we see that the driving-point impedance 222 is deter­ mined by open-circuiting the current sources at buses CD and G) and by injecting the source current 12 at bus W . Then, V2 222 = -12

Figure 8.2 shows the circuit described . Si nce 2 2 2

®

+

FIGURE 8.2

Circuit for measuring

2 22 , 21 2 ,

and

232.

( 8 . 12) IS

defined by openmg the

THEVENI N'S THEOREM A N D Z bus

8.2

287

cur re n t sources coimected to the other buses whereas Y22 is found with the oth er buses shorted, we m ust not expect any reciprocal relation between t h ese two quantities. The c i rcuit of Fig. 8.2 a lso e nables us to m easure some transfer impedances, for we see from Eq. (8.9) t h a t with cu rrent sou rces 1\ an d 1 3 open-circuited

( 8 . 1 3) a n d fro m Eq. (8. 1 1 ) Zl� .

-

=

v�I I �

�

I,

-

I,

-

II

( 8 . 1 4)

Th us, w e can meas u re t h e tran sfer i m peda nces Z l 2 and Z:l 2 by i nj ecting c u r rent at b u s W a n d by fi n d i n g t h e ratios of VI a n d V� to f; w i t h the sources open at a l l buses except bus W . We note that a m u t u a l adm i tt a n ce is measured wit h a l l b u t on e b u s s h o r t circ u i t e d and that a tra nsfe r i mpedance i s measure d wit h a l l sources open-circu ited exce p t o ne. Equ a t io n (8.9) tells us t h a t i f we i nject cu rrent into bus CD with current sources at buses @ and G) ope n , the only impedance t h rough which 1\ fl ows is Z 1 1 ' U n d e r the same con d i t ions, Eqs. (8. 1 0) a n d (8 . 1 1 ) show that 1 1 is causing vo l t ages at buses @ and G) expressed by '

-

and

( 8 .15)

I t is i m portant t o re a l ize the i mp l i ca t ions o f t h e p reced ing d iscussion, for Z bus is som e t i m e s u sed i n power-flow studies a'ld is extremely valuab l e in fau lt calcula­ tiuns.

8.2

TH EV E N I N 'S T H EO R E M A N D Z hus /

Th e b u s i m p e d a n c e m a t r i x p r ov i d e s i m p o rt a n t i n form a t i o n rega rd i n g the p ower sys t e m n e t wo r k w h i c h w e C
=

288

CHAPTER 8

THE IM PEDANCE MODEL AND N ETWO R K CALCULATIONS

superposition equation v = Z bus( I o + � I) = Z bus I o + Z bus � I �

VO

(8 .16)

t:N

where �v represents t h e changes in t h e bus voltages from their original values. Figure 8.3( a) shows a large-scale system in schematic form with a repre­ . sentative bus ® extracted along with the reference node of the system. InitiallY, we consider the circuit not to be energized so t h a t t h e bus c u r r e n t s 1 ° and voltages V O a re zero. Then, into b u s ® a curre n t of 11 ft-; a m p ( o r 11 ft-; per unit for Z hus in per unit) is injected in to the sys te m fro m a c u r r e n t s o u rce connected to the reference node. The resulting v olt a g e c h a nges at t h e buses of

- �v

3

Il @.

2 II ® �V I CD 1I

�V

--4

Original network Zbus

� Vn

® I

I

®

Reference

� Vk

cD

(a)

Original network Zbus V;.G

+

®

Z th

:0=

Zk k

®

t

+

!-

6 Ik

VII

Reference

: FIGURE 8.3 (a) Original network with bus

®

(b)

and reference node extracted . Voltage �Vn at bus by curre n t tlI k entering the network. (b) Theven i n equivalent circu it at node CD ·

®

is s:aused

8.2 THEVEN IN'S THEOREM

the

n e twork, indicated by t h e incr emental q u an tities 6 V1 to 6 VN ,

il V I il V2

CD W

CD

W

®

Z21

Z 22

Zl 1

are give n by

ZlN

0

Z2k

Z2 N

0

Zlk

Z12

®

il Vk

®

Zk L

Zk 2

Zk k

Zk N

il VN

®

ZN l

ZN 2

Z Nk

Z NN

289

AND z.,...

tJ.

Ik

0

( 8 . 1 7)

w i t h the only nonz ero e n t ry in t h e cu rre n t vcctor equal t o Ll lk i n row k . Row-by-col u m n m U l t i p l i c a t i o n i n Eq. (8. 1 7) y i e l d s the i ncreme n ta l bus vol tages

il V, Ll V

2

Ll V"

il VN

CD CV

®

Z lk

Z 2/.:

®

Zu

®

Z Nk

Ll Ik

( 8 . 1 8)

w h ich are n u merically equal to t h e ent ries in col u m n k of Z bus m u l tiplied by t h e c u r r e n t Ll Ik . Ad d i ng t hese voltage c h anges to the original vol tages at t h e b uses according t o Eq. (8. 1 6) yields a t bus ® (8.19)

Th e c i rcu i t correspo n d i n g t o t h is cq u a t ion i s shown i n Fig . 8.3( b ) from w h i c h i t i s evident that t h e Theve n i n imped a n ce Z th a t a represe n t a tive b u s ® o f t h e syst e m i s given by ( 8 . 20)

where Z k k is the d i a go n a l e n t ry in row k and col u m n k of Z bu s ' W i t h k s e t equal to 2 , t h is i s esse n t i al ly t h e s a m e resu l t obtained i n Eq. (8. 1 2) for the d riving-p oint impe dance at bus W of Fig. 8.2. In a s i m i l a r m a n n e r, we can d e te rm i n e the Theve n i n impedance b e tween any two b uses (j) and ® o f t h e n e twork. As s hown i n Fig. 8.4(a), t h e otherwise dead n etwork i s e n e rgized by t h e current i nj ections 6 Ij .a t b u s 0)

290

CHAPTER 8

THE I MPEDANCE MODEL A N D NETWORK CALCULATIONS

and I:1 Ik at bus ® . Denoting the changes in the bus voltages resulting from the combination of these two current inj ections by � Vl to � VN ' we obtain

CD

� Vl � v1.

� Vk

-

� VN

(j)

OJ

CD

ZI I Zj t

®

Z lj

Zt k

ZJJ. .

Zj k

®

ZI N

0

Z1 v

6 /1.

I

®

Zk l

Zk j

Zk k

Zk N

� lk

®

ZN I

Z Nj

ZN k

Z NN

0

� l1" + Z1"k � lk Z k j � Ij + Z kk � Ik

ZlJ"

=

( 8 . 21 )

in which the right-hand vector is nume rically equal to the product of � Ij and

column j added to the product of � lk and column k of the system Z bus ' Adding these voltage changes to the orig inal bus voltages according to Eq. (8.1 6), we obtain at buses OJ and ®

( 8 .22 )

: Adding and subtracting Zjk \ � Jj in Eq.

(8.23), give

v. 1

=

V1 O + ( Z11" "

(8.22), and l ikewise,

- Z1"k. ) � I1"

Zk j � lk in Eq.

+ Z1"k ( � I1 + � lk )

(8 .24) ( 8 .25 )

Since Z bus

is symmetrical, Zjk equals Zk j and the circuit corresponding to these two equations is shown in Fig. 8.4Cb), which represents the Thevenin equivalent circuit of the system between buses (j) an d ®. Inspection of Fig. 8.4Cb ) sp ows that the open-circuit voltage from bus ® to bus OJ is VkO Vi o , and the -

8.2

THEVEN lr-;'S THEOREM A N D Zbus

291

O r i g i n a l n etwork

l ® 2I 6 V I CD 1 I

Zbus .

LiV

®

o

Ll JJ.

R efere n c e

(a )

Jj)

O r i g i n a l n e tw o r k

+

jk

=

VkO

Zk k - Z k ;

I "

V) O

Z ) k

®

®

�

...--

Zbus

Z

®

:'J. lk

+

J)

Zjj - Z) k

0(

0

� I)

(

Ib

�

Reference

FI G C R E 8 . 4

e q u i v a l e n t c i rc u i t :

Z

b

(d)

(b)

O r i g i n a l n e t wo r k w i t h ( a ) cu r re n t so u rc e s

(

6 I} a t b u s

CD

and

([); (b) Thevenin b u s e s CD and ([) .

6 1� a t b u s

(c) s h o r t - c i rc u i t c o n n e c t i o n ; ( tI ) i m p e d :1n c e Z " b e t w e e n

i m pe d a n c e e n co u n t e re d by t h e sil ort -cif'C[ {il Cl I rre n t I I ( i n F i g . H . rJ ( ( ' ) i s e v i d e n t l y t h e T h c VL' l l i l 1 i l l l p e t \ ; I I 1l' c .

Crom

bus

®

to b u s

(j)

( 8 . 26) This

res u l t is rea d i l y con fi rmed by subs t i t u t i n g 1,.(, = 6. lj = 6. /" i n Eqs. ( 8 .24) Vk between the resu ltant e q u a ­ a n d (R.2S) a n d by s e t t i n g the d i ffe rence �. t i o n s e q u a l to z e r o. A s fa r a s exte r n a l con n ections t o buses (]) a n d ® a re concerned, Fig. 8A( b ) represe nts t h e e ffect of the o riginal syste m . From bus 0 to the refere nce node w e can t race t h e Theven i n i mpedance 2jj ( 2jj - 2j k ) + Zjk a n d t h e open-circ u i t vol tage Vj Q ; from b u s ® t o t h e refe rence node we h ave the Theve n i n i mp ed a n ce Zkk ( Zkk Z k, ) + Z kj and the open-circuit -

-

=

=

-

292

CHAPTER 8

THE IMPEDANCE MODEL AND N ETWOR K CALCULATIONS

voltage VkO; and between buses ® and Q) the Thevenin i m p ed a nce of Eq. (8.26) and the open-circuit voltage VkO - VjO is evident. Finally, when the branch impedance Zb is connected between buses (]) and ® of Fig. 8.4( d ), the resulting current Ib is given by

( 8 .27 ) We u s e this e q uat ion in Sec. 8.3 to show how to modify imp e d an ce is added between two buses of the network.

Z bus

when a branch

capacitor having a reactance o f 5.0 p e r u n i t is connected between the reference node and bus @ of the circu i t of Examples 7.5 and 7.6. The original emfs and the corresponding external current i njections a t buses CD and @ are the same as in those examples. Find the current d raw n by the capacitor. Example 8 . 1 . A

L.

Solution. The Thcvenin equ ivalent c i rc u i t at bus

@

has an emf with respect to reference given by V40 = 0 . 9 4 866 - 20 .7466° p er u n it, wh ich is the vol tage at bus @ fou n d in Example 7.6 before the capacitor is connected. The Thevenin impedance Z44 at bus @ is calculated i n Example 7.6 to be Z44 = jO.69890 per unit, and so Fig. 8.5(0) follows. Therefore, the current leap drawn b y the capacitor IS

Ieap

=

0.94866/ - 20 .7466° -j5 .0 + jO.69890

=

0.22056/ - 69 .2534°

If an additional current equal to injected into the network at bus @ of Example buses CD, (1), Q), a n d @ . Example 8.2.

per u n it

- 0.22056/69.2534: per umt IS 7.6, find the resulting voltages a t

.

Solution The voltage changes at the buses d u e to t h e additio nal i nj e cted current . .

can be calculated b y making use of the bus impedance matrix found i n Example The required impedances are in colu m n 4 of Z b uS ' The voltage changes due to the added current i njection at bus @ i n per u n i t are

7.6.

�Vl � V2 � V3 � V4

=

=

=

=

- IeapZl 4

=

-leapZ24

=

- lcapZ34

=

- lcapZ4 4

=

- 0.22056�.2534°

X

jO.63677

- O.22056�?534° x jO .64178 - 0.22056/ 69.2534° x jO .55110 -

O

.

22 0

56�?534� X jO.69890

=

= =

=

0 .14045/ - 20.7466° 0 .14155/ - 20 .7466° 0.12155/ - 20 .7466°

0.15415/ - 20.7466°

8.2 THEVENI N 'S THEOREM A N D

o I

I

V4O

293

�

/'

�

jO.69890 0.94866/- 20.7466°

Z hus

'i' -j5 .0

Reference (a )

- 2 0 . 7 4 6 G"

(b) Exam ples 8 . 1 and 8.2 s h o w i n g : ( 0 ) Theve n i n e q uivalent c i r c u i t ; ( b )

FI G L' RE 8.5

0.

Ci rc. u i t fo r bus

p h asor d i a g r a m a t

(R. 1 6)

B y s u p e rp o si t i o n t h e r e s u l t i n g vo l t ag e s a re d e t e r m i n e d from E q . a d d i n g t h e s e c h a n g e s to t h e o r i g i n a l bus vol t ag e s fo u n e! in Ex a m p l e

bus volt ages VI

V2 VJ

V4

=

=

=

=

per u n i t a re

0.96901/ in

- I R . 4 1 R9°

+ 0 . 1 4045/ - 20 . 746(')°

O .96734� P 3 .6028°

+

0 . 99964

+

0 .94866

/ - 15 .371 8°

/ - 20 .7466°

+

0.14155 O . l 2 1 55 0 . 1 54 1 5

/ - 20 .7466°

=

1 . 1 0938

7.6.

L- I R .7 1 3 5°

= 1 . ] () 8 80

� R7 64° .

�20.7466° = 1 .12071/ -- 1 5 .9539° / - 20.7466°

=

by

The new

1 . 10281/ -;- 20 .7466°

·294

CHAPTER 8

THE I MP EDANCE M O D EL A N D NETWORK CALCU l;ATIONS

S ince the changes in voltages due to the injected current are all at the same angle shown in Fig. 8.SCb) and this angle d iffers little from the angles of t h e original voltages, an approximation will often give satisfactory answers. The change in voltage magnitude at a .bus may be approximated by the product of the magnitude of the per-unit current and t he magn itude of the appropriate driving-point or transfer i mpedance. These values added to the original voltage magnitudes approximate the magni tudes of the new voltages very closely. This approximation is valid here because the network is p urely reactive, but it also provides a good estimate where reactance is consid erably larger than resistance, as is u sual in transmission systems. The last two examples illustrate the importance of the bus impedance matrix and incidentally show how adding a capac itor at a bus c a u s e s a rise i n bus voltages. The assumption t h a t t h e an gles o f voltage and current sources remain constant after connecting capacitors at a bus is not entirely valid if we are considering operation of a power system . We shall conside r such system operation in Chap. 9 using a compu ter power-flow program.

8.3

MODIFICATION OF AN EXISTING Z bU S

Z bus

In Sec. 8.2 we see how to use the Theven in equivalent circu it and the exist ing to solve for new bus voltages in the network following a branch addition without h av ing to develop the new Z b u s ' S ince Z b u s is such an i mportant tool in power system analysis we now examine how an existing Z bus m a y be modified to add n ew buses or to con nect new lines to established buses. Of cou rse, we cou ld create a new Y bu s and invert it, but direct methods of modi fying Z b u s are available and very much simpler than a matrix i nversion even for a small number of buses. Also, when we k n ow how to m o cJ i fy Z htl � ' we c a n see how to build it directly. We recognize several types of modifications in which a branch having impedance Z/J is added to a network with k nown Z b\l s ' The origi n a l bus impedance matrix is identified as Z ori!o: ' an N X N m a trix. In the notation to be used in our analysis existing buses will be identified by n umbers or the letters h, i, j, and k. The l etter p or q will designate a new bus · to be added to the network to convert Z orig to an e N + 1 ) X e N + 1 ) m atrix. A t b u s ® t h e original voltage w i l l be denoted by VI?' t h e new voltage after modifying Z bus will be Vk , and .1 Vk Vk - VkO will denote the voltage change at that bus. Four cases are considered in this section. =

CASE

1.

bus ® to the reference node. addition of the new bus ® connected to the reference node through

Adding Zb from

a

new

The Zb wit hout a connection to a ny o f the buses of the original network cannot alter the o riginal bus voltages when a current Ip is i nj ected at the new bus. The

Ik

Ip

�

8.3

® I I

®

", I I

M O D I FICATION OF AN EXISTING Z bus

295

�

Original network with

bus

Zb

®

and the

reference node extracted

f--

A d d i t ion of new bus

@

to exis t i ng bus

0

0 Z Orig

=

®

0

con­

®.

I(lZ" , Then,

VIlJ

VIS �)

®

nected t h rough i m p e d ance Zb

Reference

voltage �) a t the new b us i s e q u a l to

V.,O

FI G U RE 8.6

0

0

Z bus(ncw)

;" J

II

/2

IN

( 8 .28)

lp

We note that t h e column vector of c u rren ts m u l t i p l i e d by the new Z bu s w i l l not alter the vol t ages of the ori g i n a l n e twork and w ill result i n the correct vol tage a t t h e n ew b u s ® . CAS E 2 .

A dding

Zb

from

a

new bus

®

t o an existing bus

®

The addition of a n ew bus ® connected through Zb to a n existing b u s ® i njected at b u s ® will cause t h e c u rren t e n t e ring t he original n etwork with at bus ® to beco me t he s u m o f lJ..:. i nj e c t e d at bus ® plus t h e curren t 1p com i n g t h ro u g h Z" , a s s h ow n i n r i g . H . 6 . T h e c u r re l l t 1 " I l ow i ng i n t o t i l L: n e t wo r k ; 1 1 l H I S ® w i l l i n c r L: a s L: t h e IJ o r i g i n a l vo l t a g e V" by t h e vol t a ge lp ZkJ..:. j ust l i k e i n Eq. ( O . P »); t h a t i s ,

Ip

a n d �) w i l l be l a rg e r

( 8 . 2 9) t h a n t h e n ew

V"

by t h e v o l t a ge

I{) 2". S o , ( 8 .30)

and

su

bst i t u t i ng for V"o, w e obt a i n

( 8 .3 1 ) Vu I;

296

CHAPTER 8

THE I MPEDANCE MODEL A N D N ETWO RK CA LCULATIONS

We now see that the new row which must be a dded to � is

Z ori g I n

order to find

Since Z bus must be a square matrix around the principal diagonal, we must add a new column which is the transpose of the new row. The n ew column accounts for the increase of all bus voltages due to I", as shown in Eq. (8 . 1 7). The matrix equation is

VI V2

®

I

Vp

/2

LU -

Z o rig

VN I -

/1

Z I t.:

®

Zk l

Z Nk

Zk 2

Zk N

Zk k

+

Zb

Iv

.( 8. 3 2 )

lp

z hUs( nCw)

Note that the first N elements of the new row are the e lemen ts of row k o f Z ori g and the first N elements of the n ew column are the elements of column k of

Z orig'

CASE 3. A dding Zb from existing bus

®

to the reference node

To see how to alter Z Orig by con necting an impedance Z/J from an existing bus ® to the reference node, we add a new bus ® connected through Z b to bus ® . Then, we short-circuit bus ® to the refe rence nod� by letting Vp equal zero to yield the same matrix equat ion as Eq. (8.32) except that Vp is zero. So, for the modification we proceed to create a n ew row and new column exactly the same as in Case 2, but we then elimi nate the ( N + 1 ) row and ( N + 1 ) column b y Kron reduction, wh ich i s possible because o f the ze ro in the column matrix of voltages. We use the method developed in Eq. (7 .50) to find each element Z"i(IICW) in the new matrix, where Zh i Zhi(new) -

CASE

4.

_

Zh ( N + l ) Z ( N + l ) i Zk k

Adding Zb between two existing buses

+

CD

Zb

® buses 0

and

( 8 .33 )

and ® already To add a branch impedance Zb between established in Z on g , we examine Fig. 8.7, w hich shows these buses extracted , from the original network. The current Ib flowing from bus ® to bus CD is similar to that of Fig. 8.4. Hence, fro m Eq. (8.2 1 ) the change i n voltage a t each .

MODIFICATIO N OF AN EXISTING Z bu'

8.3

(])

I

J

)0

I I

®

�

bus

16

I I

®

1

Ij

1

+

Ib

297

)-

Orig i n al network with

Zb

1

1k

buses

([) , ® and

the reference

node extracted

Addition of i m pedance Z6 between existing buses OJ and @. FI G U R E 8.7

)0 -

1b

®

caused by the i njection

Reference

(j)

a t bus

I,)

and

-

Ih

at bus

®

is given by

( 8 .34)

which means t h a t the ve c t o r 6V of hus vol t a ge c h a n g es is found by subtracting col u m n k from col umn j of Z orig a n d by m u l t i p l yi n g the result by I,) . Based on the d efin i t ion of vol tage c h a n g e we now write some equa tions for the bus vol t ages as fo l lows: ,

( 8 .35) and using Eq . (8.34) g i v e s

(j)

Similarly, at buses v )

Zj j 1j +

Vk = Z k j l j

.

. .

+

and

®

2jj Ij

+

Zjk

I

k

+

V} O

+

.

.

.

+ Z k Jj + Z k k Ik VIiO

+

.

.

.

+

2jN J N +

( 2jj

+ Z k N IN + ( ZkJ

-

Zjk ) Ib

( 8 . 3 7)

6. V}

-

2kk ) l b ( 8 . 38 )

6. Vk

\Ve need one more equation s i n ce Ib is u nknown. Th is is supplied b y Eq. (8 . 27), which can be rearranged in to the form

( 8 .39) From Eq. (8. 3 7) we note tha t V, 0 equals the prod uct of row j of Z orig and the col u m n of bus curre nts I; l i kewise, VkO of Eq. (8.38) equals row , k of Z o rig mul tiplied by 1 . Upon subs t i t u t i n g t h e expressions for � o and VkO i n E q . (8.3 9),

298

CHAPTER 8

THE IMPEDANCE MODEL AND N ETWORK CALCULATIONS

we obtain

o = [ ( row j - row k ) of Z Orig ]

By examining thc cocHlcicnls can write the matrix equation

or Eqs .

W.3() )

( 8 .40)

t h ro u g h

(C;.3C;)

a n d £4.

VI

11

V-)

IJ,

Vk VN 0

( col . j

Z ori g

=

( row j

-

-

col . k )

(c;.40), wc

Ik

( 8 .41 )

of Z orig

row k ) of Z Orl' g

IN

Z bb

-

lb '

in which the coefficient of lb in the last row is denoted by (8 .42 )

The ncw column is column j m i n lls col u m n k 0 [" Zorig w i t h ZJ,h in t h e ( N + 1 ) row. The ncw row i s the t ra n spos e o f th e ne w column. El i m i n a t i ng t he ( N + 1 ) row and ( N + 1 ) col u m n o f t h e square matrix of Eq. (8. 4 1) i n t h e same manner as previously, we see that ea e h element Z" i(llcw) i n the new matrix is 2h i(ne w)

=

Z"i

-

2Jz( N + 1 ) 2( N + I )i 211 + 2 kk - 2 2jk + Zb

( 8 .43 )

We need not consider the case of introducing two new buses connected by Zb because we could always connect one of these new buses through an impedance to an existing bus or to the reference bus before adding the second new bus. Removing a branch. A single bra nch of impedance Z b between two nodes can be removed from the network by adding the negative of Zb between the , same terminating nodes. The reason is, of course, that the parallel combination o f the existing branch (Z b) and the added branch ( - Z b) amounts to an effective , , open circuit. Tabl e 8 . 1 summarizes the procedu res of Cases 1 to 4.

8.3

Modification of existing Z bU8

MODIFICATION OF A N EXISTING Zbwo

TABLE S. 1

Case Add branch Z b fro m

®

Refe re nce node to new bus

1

® Existing bus

®

to new bus

®

Existing bus

Zorig

3

®+

® ®

to reference node

f-+1-C==::J--ih ®

( N ode

®

®�

( Node

®

2 and

- Remove row p and column p by Kron reduction

is temporary . )

Existing b u s

Zorig

- Repeat Case

Zb

L-____________�

4

1

col . k

2

(j)

to existing bus

®

r+ H�� lZ--b --r- lh---1

® (j) l ) H,f---

@

•

®

--------'

----

is temporary . )

[

Fo r m the

matrix

Z o ,',

row j

where Z th , jk

=

-

row k

Z j + Z kk j

and

®

col . j

Z th , jk

-

col . k

+ Zb

2 Zjk

-Remove row q and column q by Kron reduction

-

-

,

-

299

300

,

CHAPTER 8

THE I M PEDANCE MODEL A N D N ETWORK CALCULATIONS

Modify the bus impedance matrix of Example 7.6 to account for the connection of a capacitor having a reactance of 5.0 per unit between bus @ and the reference node o f the circuit o f Fig. 7.9. Then, fi n d V4 using the impedances o f the new matrix a n d t h e current sources of Example 7.6. Compare this value o f V4 with t h a t foun d in Example 8.2. Example 8.3.

and recognize that Z O rig is the 4 X 4 matrix of that subscript k = 4, and that Zb = - j5.0 per unit to find

Solution. We use

Example

7.6,

�

Eq. (8.32)

=

jO . 6 3 6 77

j O . M l 7!)

jO .55 I 1 0

j ( J . ()l)!)lJO

jO.63677 jO.64178 jO.55 1 10 )0 .69890

12

-j4 .30 l l 0

'"

11

13 14

The terms i n the fi fth row and column were obtained by repeating the fourth row and column of Z orig and noting that

Z44

+

Zb

=

jO .69890 - j5 .0

=

-j4 .301 10

Then, eliminating t h e fi fth row and column, we obtain for

Zl 1(new) jO.73 1 28 =

ZZ

4{ ne w)

=

jO.64178 -

jO.63677 X jO.63677 -}4. .30110 jO.69890 X jO.64 1 78 -}4 .3011 0

-

.

Z b u S( new)

=

=

from

Eq. (8.33)

jO.82555 jO.74606

and other elements i n a similar m a n n e r to give

Zb us(nc w)

=

jO.82555 jO.78641 jO.69482 jO.74024

jO.78641 jO.81542 jO .69045 jO.74606

jO.69482 jO.69045 jO.7695 1 jO.64065

jO.74024 jO .74606 jO.64065 jO.8 1 247

The column matrix of currents by w hich the new Zb us is mUltiplied to obtai n the new bus voltages is the same as i n Example 7.6. S ince both II and 12 are zero while 13 a n d 1 4 are nonzero, we 0 bta in

�

=

=

=

jO.64065 (1 .00/ - 900 ) + jO.81247 ( 0.68/ - 1350 ) 1 .10281 L - 2 0.7466° 1 .03131 - j0.39066

as fou n d in Example

8.2.

t

"

per unit

IJ

S. 4

D I R ECT D ETE R M I NATION OF Z bw

301

=

It is of i nterest to note t h at V4 may be calculated d irectly from Eq. (8 .27) by setting node (J) equ al to t h e reference node. We then obtain for k 4 and

2 th = 2 44

since 8.4

=

1 . 1 028 1

/

-

20 .7466° p e r

Ic3P i s already calculated i n Exam p l e

unit

8.1.

D I RECT D ETERMI NATI O N O F Z bus

We could determine Z o us by fi rs t fi nd in g Y b u s a n d then i nverting i t, but this is not convenient for large-scale systems as we h ave seen. Fo rtunate ly, formulation o f Z hus u s i n g a d i rect b u i l d i n g a l go r i t h m is a straightforward p rocess on the com pu ter. At the outset we h 8ve a l i st of the branch impeda nces showing the buses to which t hey are connected . We start by writing the e q uation for one bus connected through a branch i m ped ance Zu to the refe rence as

CD

[ VI ] = G) [ Z a ] [ I I ]

(8. 44 )

a n d t his can be considered as an equation i nvolving three m a trices, e a ch of w hich h as one row and one colum n . Now we m ight a d d a new bus connected to the fi rst bus or to the reference node. For instance, if t h e second bus I S connected to the refere nce node through Z b ' we h ave the matrix equation

( 8 .4 5 ) a n d we p roceed to modify the evolving Z bus matr ix by adding other buses a n d branches following the p roced ures described i n Sec. 8.3. The combin ation of t hese procedures cons t i tutes the Z bus building a lgorithm . U s u a l ly, the buses of a network m ust be renumbered intern a ll y by t h e computer algorithm to a gree with the order in which they a re to be added to Z b u s as it is b u i l t u p . Example 8.4. D e t e rmi ne Z bu s for the n e twork shown in Fig. 8.8, where t h e impedances l abeled 1 through 6 a r e s hown i n p e r u nit. Preserve a l l buses. The branches are added i n t h e order of their l abe l s and numbered s u b s cr i p t s o n Z h us wi l l i n d i c a t e i n te rmediate steps of the so lution. 'We start by

Solution .

302

CHAPTER 8

THE IMPEDANCE MODEL AND N ETWORK CALCULATIONS

@

�-r-

j 1 .25

(4)

@

Reference

establishing bus

CD

FIGURE 8.8

Ne lwork for Examples 8 . 4 a n d 8 . 5 . B r a n c h i m ped a nc e s a rc i n pe r u n i l ; 1 I l l! b r a n c h Il u m b e rs a r c i n p ; l r C I l ­ I h eses.

with i ts impedance to the refe re nce nod e and write

CD [ VI ] = CD U l .2S ] [ 11 ] We then have the

1

x 1 bus imped a nce matrix

Zbu s , 1 To establish bus

CV

=

CD CD [ J 1 .25 ]

with i ts i mpedance to bus

[

CD i l .2S Zb us, 2 =

Cl) il .25

CD ) we i 1 .25

j l .SO

]

follow Eq. (8.32) to write

The term j 1 .50 above is the sum of j l .2S and iD.2S. The el ements j l.2S in t h e n ew row and column are the repetition o f t h e clemen ts of row 1 and co lumn 1 o f the matrix be ing modified. Bus G) with the impedance con necting it to bus CV is established by writing

[

CD CD j l .2S Zbus , 3 = CV il .2S Q) il .2S

CV i 1 .2S iI .SO i I .SO

Q) j. 1 2S iI .SO i I .90

J

Si nce the new bus Q) is being con nected to bus CV , the te rm i 1.90 above is t h e s u m o f 222 of the m atrix being modified a n d t h e imped ance 2 b of t h e branch being connected to bus CV from bus Q) , The o t he r e l e m e n ts of the new r9w a n d column are t h e repetition of row 2 a n d column 2 of the matrix be i n g modified since t he new bus is being con ne c t e d to bus CV .

8 .4

Dl R ECf D ETE R M I NATION

OF Z bua

303

If we now decide to add the impedance Zb = j l .25 from b us G) to the re ference n ode, we fol l o w Eq . ( 8 . 3 2 ) to conn ect a new bus ® t h ro u gh Z b a n d o b t a i n the i mpedance matrix

j.

Z bu s . 4 =

CD

CD @ G) 3

®

W

j l .25

j l . 25

j 1 .25

j I .50

j l .25 j l .25

+

G)

®

j. 1 25

j 1 .5 0

j 1 .50

j 1 .25

j l .90

j 1 . 90

j 1 .50

j 1 . 90

j3.15

j . I 50

w here 3 1 5 a bove i s t h e s u m o f Z :1 3 T h e o t h e r c l e m e n t s i n t h e n ew row a n d col u m n a re t h e re p e t i t io n o f row 3 a n d col u m n 3 of t h e m a t r ix b e i n g m odi fi ed s i n c e bus is be i n g con n e ct e d to t h e reference nod e t h roug h We now e l i m i n a te row p a n d co l u m n p by Kron red u c t i o n . Som e o f the c l e m e n ts o f t he n e w m a t rix fro m Eq. ( � . 3 3 ) a r e

G)

ZI

Z22(new)

Z2.\(new)

j l . 25

J ( new) =

= Z 3 2(new)

=

j 1 . 50

.

=)1

Z" .

Zb'

( j l . 25 ) ( j l . 2S ) -

-

.5 0 -

=

j3 . 1 5

( j l .50) ( jl .50) =

j3 . 1 5

( j 1 . 50) ( j 1 .90)

j3 . 1 5

j O . 75 3 97

j O .7857 1

= j O . 59524

�jOO.75Y7n jO.65477() jO .496U3 J

W h e n all the e l em e n t s are d e t e rm i n e d , we have

Z b " , . :;

=

CD

Q) j

G)

u s i n g E q . (8.32), a nd we obt a i n We

lJOw

decide

10

Z bus. 6

add

=

I he

CD @ G) @

CD

. 65 4 6

j O .7H5

CD

1

j O .59 5 2 4

j O .49 60 3

i m ped a n ce Z"

G)

W

�

jO . 5 9524

)0 .75397

jO .20 fro m b u s

@

CD

t o e st a b l is h bus

G)

@

jO.59524

jO .49603 I jO . 49603

jO. 753 97

jO.65476

j O . 65476

jO.785 7 1

jO.59524

j O.49603

jO. 59524

jO. 753 97

j O .49603

j O . 5 9524

jO.75397

I

jO . 75397 jO.95397

@

304

CHAPTER 8

THE IMPEDANCE MODEL AND N ETWORK CALCULATIONS .

The off-diagonal el ements of the new row and column are the re p e ti t i o n of row 3 and column 3 of the matrix being modified because the new bus @ is being connected to b u s ® . The new diagonal element is the sum of Z33 of the previous matrix and Zb = jO.20. Finally, we add the impedance Zb = jO.l25 between buses G) and @ . If we l et j a n d k in Eq. (8.41) equal 2 and 4 , respectively, we obtain the e l ements for row 5 and column 5 Z I5

=

ZI 2 - ZI 4

=

j O . 65476 - j0.49603

Z2 5

=

Z22 - Z24

=

jO. 7 85 7 1 - jO.59524

Z 35

=

Z45

=

Z32

- Z34 = j O . 5 95 24 - jO .75397

Z42 - Z4 4 = j O . 5 95 24 - jO .95397

=

jO. 1 5873

=

jO . 1 9047

= =

- jO. ) 5873

-jO .35873

and f r o m Eq. (8.42)

=

So, em ploy ing

j { 0 .785 7 1

Z b uS. 6

® jO. 1 5873

+

0.95397 - 2 ( 0 .5 9524 ) }

+

previously fou nd , we write the 5

jO . 1 25 = j O . 6742 1 x

5

m a t r ix

®

jO. I 5873 j O. I 9047 -jO. I 5873 -j O.35873 jO. 1 9047

-jO. 1 5 873

- jO.35873

jO.67421

a n d from Eq. (8.43) we fi nd by Kron red u ction

CD Z bu s

=

G)

® @

@

j O .60992

®

j O .53340

jO.58049

jO .60992

j O .73 1 90

jO . 64008

jO .6965 9

jO .53340

j O . 64008

j O . 7 1 660

jO.6695 I

jO.58049

jO .69659

jO .6695 1

jO .763 1O

CD

G)

j O .7 I 660

which is the bus impedance matrix to be determined. All calcu lations ha,-:e been rounded off to five decimal places.

8.4

D I R ECT DETE R M INATION OF Z bu.

305

Since we shall again refer to these results, we note here that the reactance diagram of Fig. 8.8 is derived from Fig. 7. 1 0 by omitting the sources and one of the mutually coupled branches. Also, the buses of Fig. 7. 1 0 have been renum­ bered in Fig. 8.8 because the Z b us building algorithm must begin with a bus connected t o the reference node, as previously remarked. The Z bus bu ilding procedures are simple for a computer which first must determine the types of modification involved as each branch impedance is added. However, the opera tions must follow a sequence such that we avoid connecting an impedance between two n ew buses. As a matter of interest, we can check the impedance values of Z b us by the network calculations of Sec. 8 . 1 . Exa m p l e 8 . 5 . Find

ZII

m e a s u red be t w e e n b u s

W , CD,

Solution .

a nu

Q)

ar�

o f t he

CD

c ircu it 0 r Exam p l e 8.4 by d e t e rm i n i ng the i mpeda nce t h e re ference nod e wlH ; n cu r rents i nj ected at buses

and

z e ro .

T h e e q u ati o n co r respond i n g t o E q . ( 8 . 1 2) i s

W e r e cog n i ze tw o p a ra l l e l pa t hs

8.8 w i t h

b e tw e e n

b u ses

the resu l t ing i m p ed a n c e of

+ jO . 20 ) ( j 0 .40) j ( 0 . 1 25 0 .20 + 0 .40)

( jO . 1 25

-�-----

+

This i mp e d ance in ser i es

w i t h ( j0.25

y i e ld

+

j 1 . 25 )

(1)

and

G)

of

the c ircuit

of Fig.

= j O . 1 793 1

comb i n e s in

parallel

w i th

j 1 .25 to

j 1 .25 ( j0 .25 + j l .25 + j O . 1 793 1 ) = jO . 7 1 660 = ZI I j( 1 .25 + 0 .25 + 1 .25 + 0 . 1 793 1 ) which

is i d e n t i c a l

w i t h t h e val u e

fo u n d i n

E x a m p l e 8.4 .

Although the network reduction m e t hod of Example 8.5 may appear to b e s i m p l e r b y com parison wi t h othe r methous of form i n g Z bus ' such is not the case because a different network reduction is requ ired to evaluate each element of the matrix. In Example 8.5 the n etwork reduction to find 244 , for instance, is more difficult than that for find ing 2 1 1 , The computer could make a network reduction hy node e l i m i n ation bu t wou ld have to repeat the process for each n o d e.: .

306

CHAPTER 8

THE IMPEDANCE MODEL AND N ETWO RK CALCULATIONS

CALCULATIO N OF FROM Ybus

8.5

ELEME NTS

Zbus

When the full numerical form of Z b us is not explicitly required for an applica­ tion, we can readily calculate elements of Z b us as needed if the upper- and lower-triangular factors of Y bus are available. To see how this can be done, consider postmultipiying Z h u s by a ve c to r with only one nonzero element 1 = 1 in row m and all other elements cqual to zero. When Z o u s IS an N X N matrix, we have m

CD @ @ ®

-

CD

Zl 1

Z2 1

Zm l

ZN l

@

Z 12

Z22

Zm 2

ZN 2

@)

®

Zmm

Zm N

ZIN

Zim

Z2N

Z2m

C§

-

1m

Z NN

ZNm

CD 0

0 0

®

0

Z b us

C§

Zim

Z2 m

Zmm

( 8 .46)

ZNm

-----

m) z (bus

Thus, postmultiplying Z bus by the vector shown extracts the m th column, which we have called the vector Z h�;; that is

z (m) bus £

column

of

m

Z bu s

CD 0 ® ®

-

@)

Zim

Z2 m

Zm m Z Nm

Since the product of Y b us and Z bus e qual s the unit matrix, we have

o o Y bus Z bus 1 m

o

-

) Y bus z(m bus -

o o 1

(8. 47)

m

o

I

If t h e lower-triangular matrix L and the upper-triangular matrix U of Ybus are

CA LCULATION OF Zbu. ELEMENTS FROM Ybu£

8.5

307

available, we can write Eq. ( 8 .47) in the fonn 0 0

1m

= Luz(m) bus

( 8 .48 )

0

I t i s now a pparent t ha t t h e e l e m ents i n t he col u mn ve ctor Zb':� can be fou n d from Eq . (8,48) by forw a rd e l im i n a t ion and back subs t i tu tion, as ex p l a i n e d i n S e c . 7.8. I f o n ly som e o f t h e e le m e n t s o f z b:l a re req u i re d , t h e calcu lations c a n b e r e d uc e d accord ingly. For exam p l e , su ppos e t ha t w e wish t o generate Z 3 3 a nd 243 o f Zbus for a fou r- b u s sys t e m . Us i n g conve nient notatio n for t h e e le m e n ts o f L an d U , w e have 1

III 12 t

/ 22 /3 2

/ 3 3

'4 1

[ 42

1 4 :'

'3 1

U 13

Ul2

il 2

1

3

Ul4

.2 1 3

U 24

Z2 3

U 34

Z 33

1

Z4 3

1 14 4

0 0 1 0

( 8 .49)

�

We c a n solve this equat ion for

i n two st eps as follows:

XI

'I I 12 1

1 22

1 '\ 1

/32

{4 1

1 2 4

U l2

w here

Z�Js

(3) Z bus

I 33

1 4 .1

un

u 23

1

1

1 44 U I4 U 24 U 34

1

0 0

1

X2

X3

zu ]

0

X4

223

2 33

XI

x2

=

x]

24 3 J

L

------

By forward

( 8 . 50)

(3) Z bllS

( 8.51)

X4

substitution Eq . (8.50) i m m ediately yie lds

XI

=

0

X2 -

a n d by b a c k s u b s t i t u t i o n o f

0

1

X4

-

t hese i n t e rm e d i a t e re s u l t s i n Eq. ( 8. 5 1 ) we fi n d the

308

CHAPTER 8

THE IM PEDANCE MODEL A N D N ETWORK CALCU LATIONS

required elements of column 3 of Z bus ,

If

all elements of

z i;�s

are required, we can contin ue t he calcu l ations,

The computational effort in gene rating t he required clements can be red uced by judiciously choosing the bus numbers. In l ater chapters we shall find it necessary to evaluate terms like ( Z illl Zjn) involving differences b etween columns (if!) and ® of Z b I f the elements o f Z bus are not available exp licitly, we can calculate the required differences by stS1ving a system of equations such as us '

o n Luz(m bus- ) =

( 8 .52)

- 1n o

where Zb�s- n) = Z�:l - Z���� is the vector formed by subtracting column n from column m of Z bus , and 1 /11 = 1 in row m and - I " = - 1 in row n of the vector s hown . In large-scale system calculations considerable computational efficiency can be realized by solving equations in the' triangu larized form of Eq. (8.52) while the ful l Z bus need not be developed. Such computational consider(\tions underlie many of the formal developments based on Z b in this text.

, '

The five-bus system shown in Fig . 8.9 has per-unit impedances as marked. The symmetrical bus a dmittance m atrix for the system is g iven by us

Example 8.6.

Ybus

=

CD CV ® 0 G)

CD

CV

®

0

0

G) 0

0

-)30.0

) 1 0 .0

)1 0 .0

-)26 .2

) 1 6 .0

0

) 1 6 .0

-)3 6 .0

0

0

-j20 .0

0

0

-f20 .0

0 )20.0 0

0 0

f20.0

)20 .0

)20 .0

8.5

CALCULATION OF Z tKu ELE M ENTS FROM Ybus

309

a n d i t is fou n d t ha t t h e triangu l a r factors of Y bus are

-

j 30

.

0

}1O.0

L=

0 }20 .0

-} 2 2 . 866667 } 1 6 .000000

-}24 .804666

}6 . 66 66 67

}4 .664723

0

1

0 - 0 .333333

-j3 .84579 3 j3 .761 164 -jO.195604

} 20 .000000

o - 0 .69970 8

U =

0 666667 - 0.291545 - 0. 1 88051) -

.

o o - 0 . 806300

- 0 . 977995 1

Use the t ria ngu l a r fact ors to c a l c u l a t e Z lh. 45 = ( Z 4� - 245) ( Z54 - Z55), the Thcvenin i m pedance l ooki n g i n t o t h e system between b uses @ a n d G) of F i g . -

8 .9.

Solution. S i n c e Y bus is symm e t r i ca l , t h e reader should c heck t h a t t h e row of U equ al the col u m n e l eme n ts of L d ivi ded t h e i r corresponding

elements d i a gonal elements. With l's represen ting the n u merical values of L, forward solu tion of the system of equa tions

by

'1 1 '21

'22 1 32

1 33

141

1 42

14 3

1 44

15 1

1 52

15 3

1 54

131

15 5

Xl

0

X2

0

X3

0

x4

1 -1

Xs

® ®

) 0.0625

)5.0 )0.1 FIG U RE 8.9

R e a ct a nce d ia g r a m for Exa m p l e 8.6, a l l imped a nces.

,

va l u e s

are p e r-unit

310

CHAPTER 8 THE I MPEDANCE MODEL AND NETW O R K CALCULATIONS

yie lds the intermediate

X4 =

1«\

=

values

( -j3.845793) - \ =

Backsubstituting in

1

=

jO.260024

- 1 - j3 .76 1 1 64 x j0.260024 -j0 . 1 95604

-j0. 1 12500

t h e system o f eq u at i o n s

U

12

1

UD UI4 1 5 u2) U24 u25 [ZC4 - 5 l] U 1I 45

0 0 0 jO.260024 -jO.1 12500

It

:14

I t :IS

hu�

1

where u 's represent the numerical values of U, we

fi n d

from the l ast two rows that

-jO. I l 25 per u nit

Z54 - Z55

=

Z44 - Z45

=

jO.260024 -

=

jO. 1 500 per u nit

U45 ( Z54 - Z55 )

=

jO.260024 - ( - 0 .977995 ) ( -jO. 1 125)

Th e desi red Thevenin impedance is therefore calculated as follows :

Z th . 45

=

( Z44 - Z 45 ) - ( Z54 - Z55 )

I nspection of Fig.

8.9 verifies

=

jO. 1500 - ( -jO.l 125)

=

jO.2625

per u n i t

this resu lt.

8.6 POWER INVARIANT TRANSFORMATIONS

The complex power in a network is a physical quantity with a value which should not change simply because we change the way we represent the network. For example, in Chap. 7 we see t h a t t h e n e twork cu rrents a n d v ol t a g es may be chosen as branch quantities or as bus quantities. In either case w e should expect the power in the branches of the network to be the same regardless of which quantities are used in the calculation. A transformation of network variables which preserves power is said to be For such transformations involving the bus impedance matrix certain general relationships must be satisfied, which we now establish for use in l ater chapters. Let V and I describe the set of bus voltages and currents, respectively, i n the network. The complex power associated with these variables is a

power invariant.

I

scalar

8.6

quantity ,

311

P O W E R I N V A R I ANT T R A N S FO RM A:r I O N S

w h i ch w e may rep r e s e nt b y ( 8 .53 )

o r in matrix form by

v2

'

"

VN ]

1 1*

/ 2*

( 8 .54)

1,".,. *

S u p p ose t h a t

we t r a n sfo r m t h e bus c u r r e n t s I to a new set of u s i ng the t ransformat i on m a trix C s u ch t h a t I

=

bus c u r r e n t s

I new

( 8 . 55 )

C 1 n ew

we shall s e e b e l ow s u c h a tran s fo r m a t i u n occu rs, for i n s tance, w hen the re fe r e n ce node of t he n e t work i s cha nged a n d it i s req u ired to compute t h e new bus i m ped a n ce matrix, wh ich we c a l l Z b ll S(lleW) ' The bus v ol t age s in terms of the existing and the new variab les are repre se n ted by As

,

( 8 .56)

an d

and we now seek t o establ ish the co n d i t ions to be satisfied b y Vn ew and Z bUS(neW) so t h a t the power rema ins inva r i a n t w h e n the curren ts are c h a nged ac c o r d i n g to

Eq . (8 .55).

S u b s t i t u t in g V from Eq . (8.56) in E q . (8.54) gives

where

obt a i n

Z h m i s sy m m e t r i c a l . F r o m Eq .

(K5S)

we su bs t i t u t e fo r I

( 8 .57) i n Eq . (8 .57)

to

( 8 .5 8 ) fro m

which

it

fol l ows t h a t

( 8 .5 9) Z huS(nCw )

Compa ring Eqs. (8.57) a n d (8.59), we see t h at the complex p ow er will be i nvariant in t erms of the n ew variables, p rovid e d the new bus impe 9 a nce m atrix

312

CHA PTER 8

TH E IMPEDANCE M O D EL AND NETWORK CALCU LATIONS

is calculated from the relationship ( 8 .60 )

This is a fundamental result for constructing the new bus impedance matrix. From Eqs. (8.56) and (8.59) we find that ( 8 .6 1 )

It also follows from

Eq.

(8.54) that ( 8 .62 )

and we m ay then conclude from Eqs. (8.6 1 ) and (8.62) that the new voltage variables Vnew must be related to the existing voltage variables V by the fundamental relationship ( 8 .63 )

In many transformations, especially those involving the connection matrices of the n etwork, all the entries in C are real, and in such cases we drop the complex conjugate superscript of C* . Equation (8.53) is the net sum of all the real and reactive power entering and leaving the buses of the network. Hence, SL represents the complex power loss o f the system and is a phasor quantity with real and reactive parts given by Eq. (8.59) as The complex conj ugate of the S*L

-

PL

transpose -

J'Q L

-

of Eq. (8.64)

is

I Tnew CTZT* b us C* 1 *n ew ·

( 8 .64 )

( 8 .65 )

Adding Eqs. (8.64) and (8.65) together and solving for PL yield PL

=

[

1

Z b us + Z Tb us* IT CT _ C* 1*new new

2

( 8 .66 )

When Zbus is symmetrical, which is almost always the case, we may write Z b us = R b us +

jXbus

( 8 .67)

where both R b us and X b us are symmetrical. We note that R b us and X b u s are available by inspection after Z bu s has been constructed for the ne�ork. Substituting from Eq. (8.67) in Eq. (8.66) cancels out the reactance part of Z bu S '

POWER I N VARIANT TRANSFORMATIONS

8.6

II

CD ------+-

12

®

®

Vl

Zbus

Vz

t

Initial Reference ®

FI G U RE 8.10

----:----:

a n c.!

we t he n

3 13

II n cJ

Changing

Z bus ·

I"

I)L

-

IF

n l' W

e r R e* l * bus

the

re fe re nc e

of

( 8 .68 )

new

wh ich simplifies the n u m e ri cal calcu l a t i o n of PL smce o nly the resistance p o r t io n o f Z bu s is i nvolved. An i mpor t ant a p p l i ca t i o n o f Eqs . (8.60) and (8.6 3 ) a ri ses when the re fe r­ ence n o d e for the Z bu s represe n t a t i o n of the system is change d . Of course, we cou l d again use the b u i l d i n g a lgori t h m of Sec. 8.4 to rebu i ld completely t he n ew Z bus starting with the new reference node. However, this wou l d be com p u ta­ ti o n a lly i n efficient, a n d we n ow s how h ow to mod ify the existing Z b us to accou nt for the ch ange of reference n od e . To i ll us trate , consider t h a t Z b us has already b een constructed for the five-nod e system of Fig. 8 . 1 0 based on node ® as reference . The standard b us e q u a t i o n s are t hen written as

VI

V2 V:l

V4

CD @ G) @

CD

211 Z21

212 Z 22

Z4 1

Z 32 Z42

Z :lI

a n e!

®

213 Z 23 Z 33 Z4 3

@

214 224 Z34 Z 44

II 12

( 8 .69)

I." 14

V4

a re measu red w ith respect to node as reference, a n d the c u rre n t i nj ections 1 1 , /2 , /3 , and /4 a re i ndepend ent.

i n w h i c h t he b u s vo l t a ge s VI ' V2 , V:l '

®

@

Ki rchhoff's curre n t l aw fo r Fig. 8 . 1 0 shows t hat

( 8 . 70 ) If we n ow change the refe rence from n o d e ® to nod e @ , for inst ance, t h e n 14 is no longer i n d epen d e n t s i nce i t c a n be expressed in terms of the other four

314

CHAPTER 8

TH E IMPEDANCE MODEL A N D NETWORK CALCULATIONS

node currents. That is, (8 71) .

From Eq. (8.71) it fol lows that the new independent current vecto r I new is related to t h e old vector I by

I} 12 I." 14

0 0

0

1

0 0

1

0

-1

1

-1

1

-

--

I

0 0 0 -1

I} 12

( 8 . 72)

13

1/1

-

I ncw

C

Equation (8.72) is merely a statement that II > 12 , and 13 remain as before but

that 14 is rep l aced by the inde p endent current In ' which app e a rs in the new

current vector I n ew as shown. In the tra nsformation matrix C of Eq. (8.72) all the entries are real, and so substit u tin g for C and Z b u s in Eq. (8.60); we fin d that

Zb us(n ew ) =

1 0 0 0

0 0 -1 1 0 -1 0 1 -1 0 0 -1 c

211 22 1 23 1 24 1

212 2 22 2 32 2 42

T

2 13 223 2 33 243

1 0 0

214 2 24 2 34 244

-1

0 0 1 -1

0 1 0 -1 c

Z bus

0 0 0 -1

( 8 .73 )

The matrix m u ltiplications in Eq. (8.73) arc accompl ished in two easy stages as fol lows. First, we compute

2 1 1 - 241 2 2 1 - 24 1 C T Z buS = 23 1 - Z4 1 - Z4 1

2 12 - 242 2 22 - 242 2 32 - 242 - 2 42

2 1 3 - 2 43 22 3 - 2 43 233 - 2 43 - 243

2 } 4 - 244 224 - 244 234 - 2 44 - 244

( 8 .7 4)

which for convenience we write in the form

C TZ bu S =

2� 1 2;1 2; 1 2� I

212 2 ;2 2 ;2 2 �2

2 i3 2 ;3 2 ;3 2 �3

2� 4 2 ;4 2 ;4 2'44

( 8.75)

lH,

P O W E R I N V A R I A NT T R A N S F O R M AT I O N S

3 15

I t is evi d e nt from Eqs. (8.74) a nd (8.75 ) t h at t h e e l em ents w i t h p r i m e d super­ scri p ts a re fou nd by sub tract i ng t h e exist i ng row 4 from e ach of t h e other rows of Z b u s a n d by changing t h e sign of existi ng row 4. S econd, we postm u l t i ply Eq. (8.75 ) by C to obta i n

=

CD @ G) ®

CD

Z; I Z 2' 1 Z; I L 4' -- 1

@

Z'

Z ; 2 - Z;4 Z �2 - Z 24 Z;2 - Z;4 Z�2 - Z�4

Z �4

Z ;4 Z �4

W

Z;3 - Z;4 Z;3 - Z;4 Z ;l 34 Z�3 - Z�-l

-

@

Z� 4

- Z;4

( 8 .76 )

- Z; 4 - Z� 4

w h i c h i s e q u a l l y s i m p l e t o c o m p u t e from C " Z i JUS b y s u b t r a c t i n g t h e forth colu m n o f Eq. (8.75) fro m each o f i t s o t h e r col u m ns a n d b y changing t h e s i g n o f the fou rt h colu m n . I t is worthw h i l e n o ting that the fi rst d iagonal e l e m e n t of Z b u s(ne w) ' expressed in t e rms of t h e original Z b us e nt ries, has t h e va lue ( Z ; 1 Z ; ) = ( Z I I + Z-l 4 - 2 Z I ), w h ich is the Th e ven i n i mpedance between nod es CD a n d @ J a s we wou l d expect from Eq . (8. 26). S i m i l a r rema rks a p p ly to e a c h of t h e o t h e r d iagona l e le m e n ts o f Z b u s ( ncw ) ' The bus vol tages with respec t to t he n ew reference node @ are given by Eq . ( 8 . 6 3 ) as fol lows : -

Vnew

=

VI

new

V2.

new

v.I . n e w

v'1 .

new

1 0

0

0

1 0

0

0

0 0 1

()

"

-} -1 - }

-- 1

VI V2 V3 V4

VI - V4 V2 - V4 V3 - V4 - V4

( 8 .77 )

C'

Therefore, i n the ge n e ra l c ase whe n b u s ® of a n ex is t i ng Z h u s i s c hosen as the n ew re fe re nce node , we m ay d e t e rm i ne t he n ew b u s i m pe d a n ce m a t r ix

Z b uS(n cw) i n two consecll t ilJe s t e p s :

1. S u b t ra ct the ex ist ing row k from each of the oth e r rows i n Z bu s a n d c h ange t he sign of row k. The resu l t i s C T Z bus ' T 2. S u b tract col u m n k of t h e r e s u l t a n t m a t r ix C Z bu s from e ach of i ts other c o l u m n s and c hange the s ign of col umn k. The result is C T Z b u C = Z h u s( new) with row k and col u m n k now r e p r e se n t i n g t h e node which was t h e p revi o us S

refe re nce node.

316

CHAPTER 8 TH E IMPEDANCE MODEL AN D N ETWORK CALCU LATIONS

We use these procedures, for example, when studying economic operation In Chap. 13. I N Z bus

8.7

MUTUALLY COUPLED BRANCHES

So far we have not conside red how to inco rporatc mutually couplcd c lcments of the network into Z hu s ' The proccdures for doing so are not difficult. B u t they a re somewhat unwieldy as we now d emonstrate by extend i ng the Z h us buildi ng algorithm to provide for addit ion to thc netwo rk of one pa ir of mu tua lly coupled branches. l One of the bra nchcs may be a d u cd to Z u rig using thc appropr i a te procedurc of Scc. K3, a nd thc qucstion rcma in ing i s how to add thc sccond branch so that it mutually cou p lcs with thc branch already i nc luded in Z u ri g ' W c consider bus ® already established within Z u rig in the fol lowi ng four cases, which continue the numbering system introd uced in Sec. 8.3. CASE

5.

Adding mutually coupled Zb from existing bus

®

to new bus ®

. Let u s assume that the branch impedance Z a is a lready added to the energized ·n etwork between nodes @) and ® of Fig. 8.1 1 . The bus impedance rn a trix Z orig then includes Z a and the existing buses @) , ® , a !1 d ® as shown. Be.tween bus ® and the n ew bus ® of Fig. 8. 1 1 it is required to add the branch impedance Zb ' which is mutually coupled to Za through mutu a l impedance ZM ' The voltage-drop equations for the two coupled branches are given i n Eq. (7.9) and repeated here as

( 8 .78 ) ( 8 .7 9 ) where fa is the branch current flow in Z a from bus @) to bus ® and t he c urrent fb from bus ® to bus ® equals the negative of current injection fq . Solving Eq. (8.78) for fa and substituting t he result a long w i th Ih = - Iq in Eq. (8.79) give

( 8 .80) In terms of bus voltages the voltage d rops across the branches are given by Va = ( Vm - Vn ) and Vb = ( Vp - Vq ), and substituting these relationships in Eq. I For a more general treatment, see System Analysis, McGraw-H i l l , I n c . ,

G. W.

Stagg and A. H. EI-Abiad, Computer Methods in Power 1 968 .

New York,

,

8.7

M UT U A LLY COUPLED B R A N CH ES IN Z bu ,

317

Zor ig ® -.---1----, @) •

I

,

I

® FIG L' RE 8. 1 1 A d d i n g m \l t u a l l y co u p l ed bra n c h Z" to

Reference

(8.80)

yie l ds

ZM Z

q

V = V · - - ( Vm - Vn ) p

-

a

This eq u a t i o n gives t he vo l t a g e a t n e w b us

[ Z� Z 1 n

®

- zb 1

( 8 .8 1 )

q

with

both mu tually cou p l e d b ra nches now i ncluded i n the n etwork. T h e b us i mpedance m atrix for t h e sys tem augme nt ed b y n e w b u s ® i s g iven by

®

Z lq Z 2 Q 12 ZNq Zq f',' Zqq Iq

VI

V2

VN

Vq

®

Zq l Z q 2

Z o r ig

11

IN

( 8 . 8 2)

a n d it is now req u i r e d to n n d e x p re ss i o n s ro r t he new c l e m e nts w i t h subscrip ts q i n row q and co lumn q . A typical row i of Eq. (8 .82) may be wr i t te n i n t he for m

( 8 . 83) V/

where

for conven ience we h ave

O

d e n oted

VO r

N

" L Z .I.

j� )

I} }

( 8 . 8 4)

318

CHAPTER

8

THE I MPEDANCE MODEL A N D NETWORK CALCULATIONS

Setting i equal to m, n, p , and q in Eq. (8.83) gives expressions for Vm , Vn , V , p and Vq , which can be substit uted in Eq. (8.8 1 ) to obtain

_

(

2 1� 2 II

_

)

2 b Iq

( 8 .85 )

Equation (8.85) is a general equ ation fo r the augmented network regardless of the particular values of the cu rrent inj ections. Therefore, when 1 q = 0 , it must follow from Eq. (8.85) that

Vqa

=

Va p

-

2 �

2

a

( Vam

-

Va)

( 8 .86)

n

Substituting i n Eq . (8.86) for v,�, Vno , Vpo , and VqO from Eq. (8.84), collecting terms, and eq uating coefficients o f lj on both sides of the resultant equation, we find tha t ( 8 .87)

1

for all values of j from to N but not including q. Thus, excep t for 2qq, Eq . (8.87) tells us how to calcu la te the entries in the new row q of the bus i mpedance matrix using the known val u es of 2 M , ZU , and certain el ements of Z o rig' Indeed, to obtain the entry in row q , each c lement of row n is subtrac ted from the corresponding eleme n t of row m and the difference mul tiplied by 2m /Za is t hen subtracted from the corresponding el ement of row p . Thus, only ' rows m , n , and p of Z orig enter into the calculations of new row q . Because of symmetry, the new col umn q of Eq. (8.82) is the transpose of the new row q, and so Zq J Zjq . The expression for the d iagona l element Zqq is determined by considering a l l currents except lq set equal to zero and then eq uating t h e coefficients of 1q o n both sides o f Eq. (8.85), which gives =

>

( 8 .88)

This eq uation shows that there is a seq ue nce to be fol lowed in determining the

new el ements of the bus impedance matrix. First, we calculate the elements 2qj of the new row q (and thereby 2jq of col umn q ) employing Eq. (8.87), and then we use the newly calculated quantities 2mq , Znq ' and 2pq to find 2qq froI? Eq. (8.88).

H.7

M UT U A LLY COU P L E D I3RAN C H ES I N Z bul

319

T here are three o the r cases of i nterest i nvolving mutually coupled b ra n ch es.

Adding mutually coupled

CAS E 6.

Zb

from existing bus ® to reference

For t hi s case t he procedure is b as i cally a special application of Case 5 . First, betwee n bus ® and new bus ® w e add i mped ance Zb cou p l e d by m u t u a l i m p e d ance Z M t o t he i m p e dance Za a lready i n cluded i n Z orig ' Then, we shor t-circu it bus ® t o the refe rence node by setting Vq equal to zero, w hich yie l ds the same m at r ix e q u ation as Eq. (8.82) excep t that Vq is zero. Thus, for t h e modified bus i m p e d ance matrix we proceed to cre ate a n ew row q a n d colu m n q exactly t h e s a m e as i n Case 5 . Then, w e el im inate t h e newly forme d row a n d col u m n b y t h e stand ard tech n i q u e of Kron reduction si nce VI] i s zero i n t h e column o f vol tages 0 f Eq. (8.8 2 ).

A dding mutually coupled Zh bctween cxisting buses ® and CD

CA S E 7 .

The p r oc e d u r e i n t h i s case essen t i a lly combines Cases 5 a n d 4. To begin, we follow the proce dure of Case 5 to add the mutually cou p led bra n ch i mp edance Zh from existing bus ® to a new temporary bus ® , recogn iz ing that Z a is a lready part of Z orig ' Th e resu l t i s the augmented matrix of Eq. (8 .82) w h ose q-elements are given i n Eqs. (8.87) and (8.88). N ext, we short-circu i t bus ® to bus ® by a d d i n g a b ranch of zero i mp edance between t hose buses. To do so, we apply Case 4 to Eq . (8 .82) as follows. S i nce ( Vq - Vk ) is require d to b e zero, w e fi n d a n expression for t h a t quantity b y subtracting row k from row q in Eq . (8.82) and t hen u se the resu lt to replace the existing row q of Eq. (8.82). By symmetry as in Case 4, a n ew colu m n follows d i re ctly from the t ranspose of t h e n ew row, and w e obt a i n

( co l . q - col . k ) Eq .

( row

q

-

row

of

( 8 .82)

k ) o f Eq . ( 8 . 82) ( 8 .8 9 )

wh ere Zc equals t h e su m C Z q q + Zk/.: - 2 Zqk ) of elements d rawn from Eq. (8 .82). Becau se ( Vq - V,) e quals 0, we m ay e l i m i nate the n ew row a n d n ew co l u m n of Eq. (8 .89) by Kron re d u c tion to fi n d the final form of t he N X N b u s i mp e dance mat rix. To remove a m u t u a l ly coupl e d b ranch , we must mod ify t h e above proce­ du res as now expl ai n e d . ., j

320

CHAPTER 8 THE I MPEDANCE MOD\ L A N D N ETWORK CALCU LATI ONS

CASE 8.

Removing mutually coupled Z" from

existing buses ® a nd ® . "

A single u ncoupled branch of impedance Zb can be removed from t he network model by adding the negative of imped ance Zb between the same terminating buses. When the impedance Z b to be removed is a lso mutually coupled to a second branch of impedance Za ' the rule for mod i fy i n g Z b us is to add between the end buses of Z" a bra nch, which has n egat i ve i m p eda nc e - Z h and the sam e m u t u a l co u p l i ng to Z" as t he o r ig i nal Z" . The p r oc e d u r e is i l l u str a t ed in Fig. 8. 1 2. As shown, the m u t u a l ly cou p l e d branches ZI/ a n d Z" with mutual impedance Z M a re a l re a d y i n cl u d e d i n Z orig ' I n acco r d a n c e w i t h the a bove rule, we add the impedance - Z/; having m u t u a l coupl i n g Z M w i th the branch Za ' and the voltage-drop eq u a t i on s fo r t h e three m u t u a l l y cou pled branches then are ( 8 .90 ) ( 8 .9 1 ) ( 8 .92)

where the branch Currents la' Ib , and I� are as shown in Fig. 8 . 1 2 . S u bt r a c t i n g Eq. (8.92) from Eq. (8.9 1 ) yields ( 8 .93 )

which shows that ( Ib + I� ) is zero, and substituting this result in Eq. ( 8 .90) gives

Because

Zorlg

( J"

( 8 . 94 ) + 1;,) equals zero, the net cou plin g effect of the two pa ral lel

®

@ I.

J

Za

ZM

Zb

®

•

FIG U RE 8. 1 2

Removing mutua lly cou pled branch Zb from Z orig ' Initially, the switch is considered open and the branch Zb is added from existing bus ® to tem­ -

ZM ®

1 I.

•

•

porary bus ® . The swi{ch is t hen closed to con nect bus ® to bus

®.

117

M UTUALLY CO U P LED BRANCHES IN Z bus

321

branches between buses ® and ® is zero. Consequen tly, the im pedance Z a between buses @) and ® can be made to stand alone as evide nced by Eq. (8 .94). To do so, we fo llow exactly the same proced ure as i n Case 7, except that the eleme n ts o f t h e n ew row and col umn of the temporary bus ® are calc u l a ted sequentially lI s i n g the m od i fi e d fo r m u l � s

( 8 .95 ) with index j

ranging

�

L'I'I

=

fro m 1 to N and

- - Z,H [ Y., ! YM ] [ - Z 1 - ( y" Z;, -

LillI

Z ""I z

1"1

_

z

" '1

-' k 'i

+

2/,)

( 8 .96)

Usi n g t h ese clements i n Eqs . (RR2) and (R.R9), (l n d Kron red ucing the l atter, gi v e s the des i red new b u s i m p e cJ: i n c e m a t r i x fro m w h i c h t h e m u t u a l ly coupled branch b e tw e e n b uses ® a n cl ® i s o m i t t e d . Equ a tions (8 . 95 ) and (8.96)' ( d eveloped in Prob. 8 . 1 9) h ave admittances Ya and YM , which a re cal cu l ated from the impedance pa rameters Z(I ' 2 ,> , and 2 M accord i n g to Eq. (7. 1 0). Table 8.2 summarizes the p rocedures of Cases 5 t hrough 8. E x a m p l e 8 . 7 . Betwt.:en buses CD a n d @ o f F i g . 8.8 i m pe d a nce Zb e q ua l t o jO.25 per u n i t is connected so that it cou p l es t h rough m u t u a l i m pedance j0. 1 5 p e r u n it to the branch impedance a l ready con nected between buses CD « n d 0 . Modify t h e b u s i m pe d a nce m a t rix o f Example 8.4 to i ncl u d e the a d d i ti o n of Zb to Fig . 8.8.

Zb between buses CD and @ correspo n d s to the condit ions of Case 7 above. Our calcu l a t i on s beg i n with Z orig ' t h e 4 x 4 m a trix solution of Examp l e 8.4 which i n c l u de s the br anch between buses CD and 0. T o estab l is h row q a n d col u mn q fo r temporary b u s ®, w e fol low Eq. (8.87), recog n izing t h a t subscript m = 1 , su bscri pt n = 2, s ubsc r i pt (J = 1 , a nd we find t h a t

Solution. Con n ec t i n g

/. /1{ --Z 1.1 �

The r a t i o ( Z M /Z) is g i v t.: n by ( jO . 1 5 /j O . 25 ) � · a long w i t h t h e t.: l e l11 e l 1 t s o f ro w I i l l 1 d row 2 0 1" obt a i n

Zq l

=

0 .4 Z 1 1

Z,/ 2

=

0 . 4Z 1 2

Zq)

=

0.4Z13

Z I/ 4

=

+

+

+

0 .6Z:; 1 = 0 .4 ( j O . 7 1 660)

+

)

Z 1 /·

-

I

- Z -. ZM /. (I J

-J

s u bst i t u t i ng t h is v a l u e in t h e «bove e q u a t i o n, we

0.6, a n d

Z orig

O .6 ( j0 .60992) = jO .65259

0 . 6 Z 22 = 0 . 4 ( JO.60992) + 0 .6( j 0 .73 1 90) = jO .683 1 1 0 . 6 Z 2:1 = 0 .4 ( jO.5J340) + 0.6( j 0.64008) = jO .59741

0 . 4 Z 1 4 + 0 . 6 Z 24

=-

0 .4 ( jO .58049 )

+

O .6 ( j 0 .69659) = jO . 650 1 5

ZbU8 modifications; mutual coupling

TABLE 8.2

Case

Add mutually coupled Z b from Existing bus

Z bU8(new)

® to new bus ®

•

®

®

J I

5

Existing bus ®

F or m the matrix

•

[

® Zmi,

row q

col . q

1

refe rence node

10

® T I

•

Repeat Case 5 and

•

Eliminate row

•

6

® a nd column ®

by Kron reduction

Existing bus

® to

exis ting bus

®

®

• •

Repeat Case 5

Then, form the matrix

l

col . q

7

row q

-

row k

and •

E l i m in ate last row a n d column by Kron reduction

-

col . k

1

Removal of mutually coupled line

8

zorig @·12 . Za

®

-----

•

Zb ZM

_�e-----=--t �-)c:>

I..:

@ -,-

I..::

Zb

• �-- -

Continue as in Case

7.

-- --- -----�--- - --- ---- , . -.- -------'

1l.7

323

MUTUA LLY COU PLED BRANCHES I N Zb...

These ele ments const i t u t e n ew row q and new colu mn q except for the d i agonal element which is fou nd from Eq . by s e t ting the subscripts m , n , a n d p as a bove to yield

Zqq'

(8.88)

S u bstituting in this e q u a t i o n for

Z M jO.IS, Za = jO.2S and =

Zb

=

jO.2S gives

Zq q = 0 .4Z 1 <7 + 0 .6Z21} + j0.16 = 0.4( j0 .65259) + 0 .6( j0.683 1 l ) jO. 1 6 = jO.83090 +

Assoc i a t i n g E xa m p l e

(0;.4

the

n e w l y ca l c u l a t e d c l e m e n t s of

res u l t s in t h e

5

x

5

row

m a t rix

q a n d col u m n q

Z orig

jO.7 1 660 jO.60992 jO.53340 jO.58049 ® jO.65259

jO.60992 jO.73 1 90 jO.64008 jO.69659 jO.683 1 1

jO.53340 jO.64008 jO.71 660 jO.6695 1 jO.59741

jO.5�049 jO.69659 jO.66951 jO.76310 jO.65015

with

Z ori g

from

-®

jO.65259 jO.683 1 1 jO.59741 jO.6501 5 jO.83090

At t h i s point in our sol u t i o n t h e m u t ually cou p l ed b ranch i mpedance has been i ncorpora ted i n t o t h e n e twork b e t w e e n bus CD a n d b u s T o complete the c o n n e c t i o n o f Zb to b u s 0 , w e m u s t fi n d ( v:, V4 ) and then set i t e q u a l t o zero. The fi rst of th ese steps is accom p lished by s e t t i n g subscript k equal to 4 i n Eq. subt ract ing row from row q of the a b ove x m a t rix, and u s i n g the re s u l t t o r e p l ac e t h e e x i s t i n g r o w q a n cl col u m n q to obt a i n

Zb

®.

-

(8.89),

4

5 5

Z orig

jO.072 1 0 -jO .0 1 348 Th e

new

The only

-jO.072 1 0 -jO. 1 1 295

jO.07210 1 -jO.0 1348 -jO.0721O -jO.1 1295 jO.29370

d i ag onal e l e m e n t is c a l cu l ated from

re m a I n i ng c a l c u l a t i o n s i n vo l v e

se t t i

ng

e v:,

--

V4 ) equal to' zero

In

Eq.

324

CHAPTER 8 THE I M PEDANCE MODEL AND N ETWORK CALCULATIONS

(8.89) and elim inating

Z bus

=

CD CV G) @

the new row and col u m n by Kron reduction to g i ve

CD

j O .69890

CV

jO . 6 1 32 3

G)

j O .55 1 1O

j O. 60822

j O . 6 1 323

j O .73 1 28

j O .63677

jO . 69 1 40

j O .55 1 1 0

j O . 63677

j O . 69890

j O .64 1 78

j O .60822

j O . 69 1 40

jO.64 1 78

j O .7 1 966

0

is t h e d esired bus imped ancc m a t rix. T h is res u l t is the same as t h a t shown i n Exam ple 7.6 except for t h e ch ange i n l hc b u s nu mbers of F i g . 7.9 t o t h o s e 0 f F i g . 8.8.

which

8.8

_

SUMMARY

This chapter introduces the importan t Z bus building algorithm, which starts by choosing a branch tied to the reference from a node and adding to this node a second branch connected from a new node. The starting node and the new node then have one row and one column each i n the 2 X 2 bus imped ance matrix which represents them. Next, a third branch connected to one or both of the first two chosen nodes is added to expand the evolving network and i ts Z bus representation. In this manner the bus impedance matrix is built up one row and one column at a time u ntil all b ranche s of the physical network have been i ncorporated i nto Z bu s . Whenever possible at any stage, it is compu tationally m or e efficient to select the next branch to be added between two nodes with rows a n d columns already included in the evolving Z hu S . The elements of Z hus can also be generated as n ee de d using the triangu l a r factors of Yhll s , which i s often the most attractive method computationally. The variables used to analyze a powe r network can take on many different forms. However, regardless of the particu l ar choice of representation, powe r i n the physical network must not be arbitrarily altered when representing currents and voltages in the chosen format. Power i nvariancy imposes requirements whe n transforming from one set of curren ts and vol tages to another. Mutually coupled branches, which do not generally arise except under u nb a l a n ced short-circuit (fault ) con di tions, c a n be a dd e d to and removed f r o m Z b u s b y algorithmic methods.

PROBLEMS 8.1.

Form Z bu s for the circu i t of Fig. 8. 1 3 a fter removing node G) by converting t h e voltage source t o a current source. Dete rmine t h e voltages with respect t o refe r­ ence node at each of the four other nodes when V 1 .2 and t h e load currents are IL l -j O. l , IL2 jO. I, Iu = jO.2, and IL 4 = -jO.2, all i n per u n i t . =

=

=

-

-

�

325

PROBLEMS

®

)0.6

tID

)0.5 I1A

IL3

Load 3

Load 4

F I G U R E 8.13

Load 2

Load 1

C I rc u i t d i a g ram show i n g con­ stant current l o a d s s u pp l i e d by an i d e a l vol t a g e sou rce. C i rc u i t

ILZ

IL l

8.2.

®

)0.4

CD

Reference

p a ra m e t e rs

a re i n

per u n i t .

t h e sol u t i o n o f P r o b . 8 . 1 , d raw t h e T h cve n i n e q u i v a l e n t c i r c u i t a t b u s of f i g . 8 . 1 3 lI n u u s e i t t o d c t e r m i n e t h e c ur re n t d r a w n hy i \ c a p a c i to r o f r e a c t a n c e 5 . 4

@

r

F om

@

p e r u n i t c o n n e c t e d b e tw e e n b u s

lI n u r e fe r e n ce . F o l l o w i n g t h e p roced u re o f

EX�l m p l e R . 2 , c a l c u l a t e t h e v o l tage ch anges a t e a c h o f t h e b u s es d u e t o t h e capacitor. 8 . 3 . M o d i fy Z b us o f Pro b . 8 . 1 to i n c l u d e a c a pa c i t o r of r e a c t a n ce 5 . 4 per u n i t c o n n e c t e d

@

from b u s t o refe r e nce a n d t h e n calcu l a te t h e new b u s v o l t ages u s i n g t h e m o d i fi e d Z b u s ' Check your a nswers u s i n g t h e resu l t s o f Probs. 8.1 and 8.2.

8.4. M o d i fy t he Z bu s d e t e r m i n e d i n Ex a m p l e 8.4

G)

n ew node c o nnected t o b u s

fo r t h e c i r c u i t o f Fig.

8 . 8 by ad d i ng a

t h ro u g h an i m pe d a nce of jO.S per u n i t .

8 . 5 . Modify t h e Z bus d e t e r m i ned in Exa m p l e

.' per u n i t between b u se s

CD

and

@

8.4 b y a d d i n g a b ranch o f i m p e d a n c e jO.2 of t h e circ .J i t o f Fig. 8.8.

8.6. M o d i fy the Z b u s d e t e r m i n e d i n Exa m p l e 8 . 4 by removing t h e i m pe d a n c e con n e c t e d b e t we e n buses and o f t h e c i rc u i t o f F i g . 8.8.

W

G)

8 . 7 . F i n d Z b us for the c i rcu i t of Fig. 7 . 1 8 by t h e Z O ll S bui l d i n g a l gori t h m d iscussed 1 0 S e c . 8 . 4 . Assu me t h ere i s no m u t u a l co u p l i n g between branches.

'.

8.8. For t he reactance n e twork of Fig, ( a ) Z b u s by d i rect for m u l a t io n ,

(0 ) The v o l t a g e

at

8 . 1 4 , fi n d

each bus,

( c ) T h e c u r r e n t d ra w n b y a c a p ll c i t o r I l ; \ v i n g , \ r C ll c t il l1 C C o f 5 . 0 p e r u n i t con n e c t e d f ro m b u s

G)

t o n e u t ra l ,

o

jO.5 j0 2

)2.0

jO.4

FIG U R E 8.14 Circ u i t fo r Prob. 8.8. Vol t a g e s a n d impeda nces a re in per u n i t .

326

CHAPTER 8 TH E I MPEDANCE MODEL A N D NETWO RK CALCULATIONS

(d)

The change in voltage at each bus when the capacitor is connected at bus G) , and (e) The voltage at each bus after connecting the capacitor. The magnitude and angle of each of the generated voltages may be assumed to remain constant. Find Zb us for the three-bus circuit of Fig. 8.15 by the Z b us building algorithm of Sec. 8.4.

8.9.

�-@+I-j-H�

R eference

8.10. 8.1 1 .

FIG U R E 8. 1 5 Ci rc u i l for Probs. K . () , X . I reac t a nces i l l pe r u n i l .

Find Z bus for the four-bus circuit of Fig. 7. 12, which has per-unit admittances as marked. The three-bus circuit of Fig. 8.15 h as per-unit reactances as marked. The symmetri­ cal Yb us for the circuit has triangular factors

L

=

r l

-

j6 .0 j5�0

-j21 .633333 j20.0

-j 1 .5 1 0038

8. 13.

8.14. 8.1 5 . 8.16. 8 . 1 7.

1

- 0 .833333

1

� 1 - 0.92 499

Use L and U to calculate (a) The elements 2 1 2 , 223 , and 233 of the system Z bus and (b) The Thcvcn i n i m p e d a nc e 2 th ' 1 1 looking into thc circuit of Fig. 8. 1 5 between buses CD and G) . Use the Yb us triangular factors of Prob. 8 . 1 1 to calculate the Thcvenin impedance 222 looking into the circuit o f Fig. R. 1 5 b e tween bus W a n d the reference . C h e c k your a nswer by i n spec t i o n o r rig. �. I ) . The Ybus for the circuit of Fig. 7. 1 2 has triangular factors L and U give'n in Example 7 .9. Use t he triangular factors to calculate the Thcvehin impedance Zlh 24 looking into the ci rcu i t of Fig. 7. 1 2 betwe e n buses @ and @ ' Check your ans�er using the solution of Prob. 8 . 1 0. Using the notation of Sec. 8.6, prove that the total reactive power loss is given by the formula QL = 1 T X bus I * . Calculate the total reactive power loss in the system of Fig. 8 . 1 3 using Eq. ( 8.57). Using the procedure discussed in Sec. 8.6, modify the Z bus determined in Example 8.4 to reflect the choice of bus CV of Fig. 8.8 as the reference. (a) Find Z b us for the network of Fig. 8 . 1 3 lIsing node G) as the reference. Change the reference from node G) to node @ and determine the new Zb u s of the .

8.12.

I , a n d H . 1 2 . V a l u c s s h ow i l a r c

327

PROBLEMS

n e twork using Eq. (8.60 ) . Use the n u m er i c a l values of the load currents ILi of Prob. 8 . 1 to d e t e rm i n e I n e w by E q. (8.55) a nd Vnew by E q. (8.56). ( b ) Change the Z bus refe re n c e from node b ack to node and u sing Eq. (8.63), d e t ermine t h e voltages a t buses and relative to node What are the values of these b u s vol t ages with respect to the gro u n d reference of Fig. 8 . 1 3?

@ CD

W,

@

W.

n e w b ranch havi n g a n imp e d a n c e o f iO.25 p e r u n i t is connected b etween nodes and of t h e circ u i t of Fi g . 8 . 8 i n p a ra l l e l w i t h t h e exist i n g i m p e d a nce o f jO.2 per u n it between t h e s a m e two n o d e s . These two bra n c h e s have m u tu al imp e d a nce o f jO .1 per u n i t . Modify t h e Z bus d e te r m i n e d i n Examp l e 8 . 4 to account f o r t h e a d d ition of the n e w b r a n c h .

8 .18. A

G)

8)

8.19.

D e rive E q s . ( 8 . 9 5 ) a n d (8.96).

8.20.

M od i fy t he Z bus deter m i n e d in Ex a mp l e 8.7 to remove t h e b r a n c h b e tween b u ses and c o u p l e d by m u t u a l i mp e d a n c e i O . l s per u n i t to t h e b r anch and

@ already between buses CD

8) .

CD

8 . 22.

8.23.

CD - CD

0 - CD

t h e o n l y m u t u a l ly co u p l e d b ra n c h e s ( as i n d i c a t e d by t h e dots) w i t h a mu tual i m p e d a n c e o f jD. I S p e r u n i t b e tw e e n them. Find Z bus fo r t h i s c i r c u i t by t h e Z b us bu i l d ing algori t h m .

8.2 1 . !\ s s u m e t l) ; l t t h e t wo b r a n c h e s

and

i n t h e c i rcu i t o f Fig. 7 . 1 8 are

Prob.

0 - G) ,

M od i fy t h e Z hu, ob t a i n e d i n 8 . 2 1 t o re move b ra n c h w h i c h is c ou p l e d to b r a n c h t h ro u g h a m u t u a l i mp e d a nce of i O . I S per u n i t .

CD - CD

I n Fig. 8. 1 6 a new b u s ® i s t o be c o n nected t o a n exi s t i n g b u s ® t h ro u g h a n ew bra nch c. New b r a n c h c is m u t u a l l y cou p l e d to b ra n ches a a n d b , w h i c h are a l ready m u t u a lly coupled to one a no t h e r as show n . The pri m i t ive imp e d a nce m a t rix d e fi n i n g se l f- a n d m u t ual i m pe d a n c e s o f t h ese t h ree m u t u a l l y coupled branches and its rec iproca l , the p ri m i t ive a d m i t t a nce ma trix, h ave the forms

To acco unt for the a d d i t i o n of new bus ® , prove t h at the exist i n g bus imp e d ance ma t r ix of t h e n e twork m u s t b e a u g m e n t e d by a n e w row q a n d col u m n q w i t h

a

@ i(, ® (JJ

0 --

•

-.

@ @ New bus

'--------f----..

FI G U RE 8. 1 6 New b ranch c from new

and b.

bus ®

i s m u t u a l l y coupled to b ranches

a

328

CHAPTER 8 THE IMPEDANCE MODEL AND NETWO R K CALCULATIONS elements g iven

8.24.

by

YCb ] Z .. � Z k "

[ �m i ] Zn i

JI

I

for i

=

1,

. .,N .

Note that these equations are a g e n e ra l i z a t i o n of Eqs. (8.87) and (8.88). Branch @ - G) of the circuit of Fig. 7. 1 8 is m utually coupled to two bra n ch e s CD - G) and C.V - G) through m ulual impedances o f j O . I S pe r u n i t and jO . l p e r unit, respectively, as indicated by dots. Using the formulas gi e n in Prob. 8.23, fi n d Z bus for the circuit by the Z bus bui l d i n g algorithm. v

CHAPTER

9

P OWER-FLOW S OLUTIONS

Power-flow studies a r e o f great i mportance i n p l an n i n g a n d desig n i n g t h e futu re exp a nsion of power systems as w ell as in d etermin ing the best operation o f existin g systems. The p r i n c i p a l i n formation obtained from a powe r-flow study i s t h e m agnitude a n d ph ase a n g le of the vol t age a t e ach bus a n d t h e real a n d reactive power fl owing i n e ach l i n e . Howeve r, m u c h additional i n forma t io n o f val u e i s provided by t h e p r i n tout of the so lution from computer p rograms u s e d by t h e electric u tility compani es. M o s t o f t h e s e fe atu res are made evid e n t i n o u r d i scussion o f power-flow studies i n t h i s c hapter. W e shall exami n e some of t h e meth ods upon w hich s o l utions to the pow e r - fl ow problem ar c b a s e d . Th e g rc::lt va l u e of the pow e r-flow com p u t e r prog ram i n pow e r sys t e m d e s i gn a n d o p e r a t ion w i l l become appare n t . 9. 1

THE POWER-FLOW P R O BLEM

E i t h e r the bu s s e l f- and m u t u a l a d m i t tances w h i ch com pose t he b u s a d m i t t a n c e m a t rix Y b u s or the driving-po i n t and t ra n sfer i m p e d a nces w h i ch co mpose Z hus may be used in solving t h e power-flow p rob l e m . We confine o u r study to m e thods using admittances. The s t a r t i n g poi n t in obtaining t h e d a t a w h ich m u s t b e furnished to t h e compu t e r i s the single-line d iagram o f t h e system. Tra n s m is­ sion l i n es are represented by t he i r p e r-phase nominal-Ii equiv a l e n t c ircuits l ike t h a t s hown in Fig . 6.7. For e a ch line n um e r i ca l val u es for the serie s i mped ance ( usually in terms of l i n e-charging Z and the total line-charg i n g a d m i tt a n ce megavars at nom i n a l vol tage of t h e system ) a re ne cessary so t h a t t h,e computer can d et e rm ine a l l t h e e l e m ents of t h e N X N b u s a d m i ttance m a tr ix of w h ich

Y

329

330

CHAPTER 9 POWER·FLOW SOLUTI ONS

the typical elemen t �j is

Other essential information includes tra nsformer ratings and impedances, shunt capacitor ratin gs, and transforme r tap s e tt i n gs I n a d v a n c e o f e ach power - fl o w study certain bus vol ta ges and power injections m u s t b e g iven k n ow n v a l u e s as d iscussed b elow. The vol ta ge a t a typical bus CD of the system is g i v e n in po l a r coord i n a t e s .

,

by

( 9 .2)

and the vo l t age a t a n o t h e r bus (J) i s s i m i l a rl y w r i t t e n from i to j . The net current inj ected i n to t h e netw o r k elements � n of Y bus is given by the summa tion Ii = Y; I VI + � 2 V2 + . . . + Y;N VN =

at bus

t h e s u b sc r i p t i n te rms o f t h e

b y cha n g i n g

CL

N

L t:n Vn

( 9 .3)

n= i

Let Pi and Q ; denote the net real and reactive powe r entering the n etwork at the bus CD . Then, the complex conjugate of the power i njected at b u s CD IS N

Pi - jQi = V:* L Y; n Vn = 11

( 9 .4 )

1

in which we substitute from Eqs. (9. 1 ) and (9.2) to obtain

Pi

-

jQ i

N

=

L

II

=

1

1 Y;1IV;V,, 1 / 8ill + 011 - 0 i

Expanding this equation and equating

Pi

=

N

.

Qi =

.

a n d r e a c t ive p a rt s , we obt a i n

-

+

On

-

0i)

( 9 . 6)

J

N

L I Y;n V: Vn I s i n e 8i n + On

II = 1

.

al

L 1 �,Y:· � l cos ( 8ill

n =

.

re

( 9 .5 )

-

0i)

( 9 .7)

Equations (9.6) and (9.7) constitute the polar form of the power-flow equations; they provide calculated values for the net real power Pi and reactive power Qi entering the n etwork at typical bus CD . Let Pg i denote the scheduled power being generated at bus CD and Pdi den ote the sch eduled power demand of the

9. 1

TH E POWE R-FLOW P R O B L E M

(a)

331

(b)

Notation fo r ( a ) a c t i v e a n d ( b ) react ive p ow e r a t a typ ical bus

FIGURE 9 . 1

CD

i n pow e r-ftow s t u d i e s .

0,

load at that bus. The n , Pi. sch = P i - Pdi is the net scheduled power b e i n g i nj e cted i n to the n e twork a t b u s as i l l ustrated in F i g . � . l ( a ) . D e noting t h e cal c u l ated va l u e of Pi by Pi . ca l c leads to the definit ion of mismatch 6 Pi as t h e schedu led value Pi . seh m i n u s the ca l c u l a te d val u e Pi. c�ie ' 6 P,

Li kewi s e , fo r

r eac t ive

=

Pi . seh

-

Pi. ca l c

=

( P);i

-

power a t b u s CD w e h ave

Pi/i )

-

Pi. c alc

( 9 .8)

( 9 .9 ) a s s hown i n Fig. 9. l ( b ) . M ismatches occu r i n t h e cou rse of solvin g a power-flow probl e m w he n calcu l a t e d v al u es o f Pi and Q i do not coi ncide w i t h t h e s ch e d­ u le d valu e s . If the calc u l ated values Pi. calc and Qi . OIC mat ch the s ch ed ul e d val u es Pi, sch a n d Q i. sch p erfectly, then we say that the mism atches 6 Pi a n d D. Q i are zero a t b u s CD , and we write t h e power-balance equations

( 9 . 1 0) (9.11) s h a l l s e e i n S e c . 9 . ::\ , t h e fu nct i o n s g i' a n cl gi" ,I r e conve n i e n t fo r w r i t in g c e r t a i n equat ions i nvolving t h e mismatches 6 Pi a n'd 6 Qi' I f b u s CD has n o g e n e r a t io n or load, t h e a p p ro p r iate t e rms a re se t e q u a l t o z e ro i n Eqs. (9. 1 0) a n d ( 9 . 1 1 ) . Each bus of the n e twork has two s u ch e q u a t ions, and the power-flow p rob lem is to solve Eqs. ( 9 . 6 ) and (9.7) for va l u es of t h e u n known b u s vol t ag e s w h i c h c a u s e Eqs. (9. 1 0) and (9. 1 1 ) t o be n u m erically sat isfied a t each b us. I f there i s no sc heduled val u e Pi. sell for bus CD , then the m ismatch D. Pi = Pi , sc h Pi, ca l c c a n not be defi ned a nd t he re is no requ i rement to sat isfy t h e correspond­ i n g Eq. (9. 1 0) i n t h e co urse of solv i n g the power-flow prob l e m . S i m i l ar ly, if Q i. sch is not specified at b u s CD , t h e n Eq . (9. 1 1 ) does not have to be satisfied . Four potentially u n know n q u anti ties associated with each b u s CD a r e Pi , Q i ' voltage a n g l e 8 i , a n d vol tage magnitude I v: I . At most, t h e re a r e two e q u a t ions like Eqs. (9. 1 0) and (9. 1 1 ) ava i l ab le for e ach node, and so w e m u st c o n s i de r how t h e n u m b e r of u n k n own q u a n t i t ies can be reduced to .agree w i t h As

we

-

332

CHAPTER 9

POWER· FLOW SOLUTIONS

the number o f available equations before beginning to solve the power-flow problem. The general p ractice in power-flow studies is to identify three types of buses in the network. At each bus CD two of the four quantities 8 i , I V; I , Pi ' and Qi are specified and the remaining two a re calculated. Specified quantities a re chosen accor ding to the fol lowing d iscussion:

Load buses. At each nongenerator bus, called a load bus, both PK i and Qg; are zero and the real power Pd; a nd reactive power Qdi drawn from t he system by the load (negative inputs into the system) are known from histori­ cal record, load forecast, or measu reme n t . Quite often in practice on ly reaJ power is known and the reactive powe r is then based on an assumed power factor such as 0.8 5 or higher. A load bus CD is often ca lled a P-Q bus because the scheduled val ues Pi , sell Pcli and Qi , sch = Q dj are kn own and mismatches 6. Pj and 6. Qi can be defined. The corresponding Eqs. (9. 1 0) and (9. 1 1 ) are then explicitly incl uded in the st atement of the power-flow prob lem, and the two unknown quantities to be determined for the bus are 0 i and I V; I . 2. Voltage-controlled buses. Any bus of the system at which the voltage magni­ tude is kept constant is said to be voltage con trolled . At each bus to which there is a generator connected the megawatt generation can be controlled by adj usting the prime mover, and the voltage magnitude can be controlled by adj usting the generator excitation. Therefore, at each generator bus CD we may p roperly specify Pg i and I v: I . With Pdi also known , we can define mismatch f1 Pj according to Eq. (9 .8). Generator reactive power Q g i required to support the sched u l ed voltage I �I can not be known in advance, and so mismatch 6. Qi is not d efi n e d . T h e re fore, at a ge nerator bus CD voltage angle 0i is the unknown quantity to be determ ined and Eq. (9. 1 0) for P i is the available equ ation. After the powe r-flow problem is solved, Qi can be calculated from Eq. (9.7). For obvious reasons a generator bus is usu ally called a voltage-con­ trolled or PV bus. Certain buses without genera tors may have vol tage control capability; such buses are a lso design ated vol tage-con trol led buses at which the real power genera tion is simply zero. 3. Slack bus. For convenience throughout this chapter bus CD is almost always designated as the slack bus. The voltage angle of the slack bus serves as reference for the angles of all other bus voltages. The particular angle assigned to the slack bus voltage is not important because voltage-angle differences d etermine the calculated values of Pi and Q i in Eqs. (9.6) and (9. 7). The usual practice is to set 8 ) 0° . Mismatches are not defined for the s lack bus, as explained below, and so voltage magnitude I VI I is specified as the other known quantity along with ° 1 0° . Then, there is no require­ ment to includ e either Eq. (9. 1 0) or Eq. (9. 1 1 ) for the slack bus in the power-flow probl em.

1.

=

-

-

=

=

.

.

9.1

THE POWER-FLOW P R O B L E M

333

To understand why P I and Q 1 are not sched u l ed at the slack bus, consider that at each of the N buses of the syst e m an equation similar to Eq. (9. 10) can be written by l e t ti ng i range from 1 to N. When the resul ting N e q u a t ions are a d d e d toget h e r, we obt a i n

PL

- [ Pi = N

i

�

=

I

R e a l powe r loss

N

L Pg i i I =

To t a l ge n e ra t i on

N

[, Pdi i= 1 -- -

( 9 . 1 2)

To t a l l o a d

The t e r m PI- in this equation is evi d e n t l y the tota l / 2 R loss i n the t ransmission I i n e s and t ransformers of the n e twork. The i nd iv i d u a l currents in the various t r a n s m ission l i n e s of the network cannot be calcu l a ted u n t i l a fter the vol ta ge m a g n i t u d e a n d a n gl e a r e k nown a t e ve r y b u s o r t h e sys t e m . The r e fore, PL is i n i t i a l ly u n k nown and i t i s not poss i b le to prespecify all the qua n t i t ies i n t h e s u mm ations of Eq . (9. 1 2) . I n t h e fo rmulation of t he power-flow p roblem we choose one b u s , the slack bus, at which PI; i s not sch e d u l e d or othemise p respecified . A ft e r t h e power-flow p roblem has been solve d , the d i ffe re n ce ( slack ) between the total specified P going into the system a t all the other b uses a n d t h e total output P plus / 2 R losses are assigned to the slack bus. For t his reason a generator bus must be s e lected as t h e slack bus. The difference between the total me gavars s u p p l i e d by the generators at the buses a n d the megavars received by the loads is given by N

[ Qi =

N

L Qg i

i= I

N

- L Q di i= I

( 9 . 1 3)

This equation is satisfied on an ind ividual bus basis by satisfyin g Eq .

(9. 1 1 ) at in the course of solving the power-flow p roblem. Ind iv i d u a l Q i c a n

each bus (J) be eval u ated from Eq. (9.7) aft e r t h e power-flow solution becomes ava ilable. Th us, t he q u a nt i ty on the l e ft- h a n d s i d e of Eq . (9. J 3) accounts for the c ombi n e d Jl1 c g
n s c h e u u l e d h u s- vo l t agc m a g n i t u d e s and a n g l e s i n t h e i n p ut d a t a o f t h e p o w e r - n ow s t u d y , t f e c a l l ed s l U l c l!{I ria h!cs or dependent variables s i nce t heir v a l u e s , which describe t h e sta l e of t h e system, depend on the q u a n t ities s pec i fi e d at all the buses. H e nce, the power-flow p ro b le m i s to de t e r m i n e val ues for all s t a te va riables b y so l v i n g an e q u a l n u mb e r of power-flow equ a t ions base d on the i nput data specificat i o n s . If t here a re NI: voltage-controlled buses (not cou n ti ng the sl ack bus ) i n the syste m of N buses, t here will be (2 N - N - 2) g e q u a t ions to be solved fo r (2 N - NI: - 2) state variables, as shown in T able 9 . 1 . O nce t h e state variables h ave b e e n calculated, t h e complete state of t h e system is known a n d all other qu a n t i t ies w hich depend on the state variables can b e t r : m s rn i s s i o n l i n e s . The

u

334

CHAPTER 9 POWER-FLOW SOLUTIONS

TABLE 9.1 Summary of power-flow problem

Bus

type

=

S l ack: i

=

No. of buses

1

2, . . . , Ng +

Volt age con trol led

(i (i

=

Nx

Load +

2,

N

Total:;

° 1 , I VI I I)"

1)

. .. , N)

Q uantities specified

-

N,, -

N

I V, I

P, . Q ,

2N

I Vj I state variables

available eq u a t i ons

No. of

No. of OJ,

0

0

NI:

2( N - N�

-

I)

2 N - NI: - 2

2( N

2N

N!] -

-

N" -

I)

tv',

2

-

d etermined. Quantiti es such as P I a n d Q 1 at the sl ack bus, Qi at each voltage-controlled bus, a n d the power loss PL of the system are examples of depe n d e nt functions. The functions Pi and Qi of Eqs. (9.6) and (9.7) are nonl i near functions of the s t a t e variables 8i and \ V, \ . H e nce, power-flow calculations us u ally employ i terative t echniques s uch as the Gauss-Sei d e l and Newton-Raphson proce d ures, which are d escribed in this chapter. The Newton-Raphson method solves t h e polar form of t h e power-flow equations u n t i l t h e 11 P a n d I1Q mismatches a t all buses fal l within specified t olerances. The Gauss-Seidel method solves the power-flow equations in rectangul a r (complex variable) coordi nates u n t i l d if­ ferences i n bus voltages from one i t e ration t o another are sufficiently smal l . Both met hods are b ased o n bus admitta nce equations.

Suppose that the P-Q load is known at each of the nine buses of a small power system and that synchronous generators are connected to b u ses CD, 0 , G) , a n d (j) . For a power-flow study, identify the t1P and t1Q m ismatches and the state v ariables a ssoc i a t e d with each b u s . Choose bus CD as t h e slack b us.

Example

9.1.

Solution. The

nine

buses

of t h e sys t e m

P-Q buses :

a re

categorized as follows:

G) , @ , (0 , @ , a m i ®

P-V buses: 0 , G) , and (j) Slack bu s :

CD

The mismatches corresponding to the specified P and Q arc At P-Q buses :

t::. P3 , t1 Q3; t1 P... , 6. Q4 ; 6. P6 , t::. Q 6 ; t1 P13 , t1 QI3; t1 P9 , t1 Q9

At P-V buses :

9.2

THE GAUSS·S E I D E L METHOD

335

and the s t ate variables a re

S i n c e N = 9 a n d N = 3, t h e re a re 2 N g t h e 1 3 s t a t e variables s h own .

9.2

-

Ng

-

2

= 1 3 equations to be solved for

THE GAUSS-SEIDEL METHOD

The complexi ty of ob t a i n i n g a fo rmal sol u t ion for power flow in a pow e r syst e m arises because of t he d iffer ences i n t h e type o f d a t a specined for t h e d ifferent k inds of buses. A lthough the for m u l a t ion of s u ffi c i e n t e q u a t i o n s to match the n u m b e r of u n k n o w n s t a t e v a r i a b l e s i s n o t d i Hi c u l t , a s we h a ve see n, the closed form of sol u t ion i s not practica l . D i g i tal sol u tions of the powe r-flow pro b l e m s fol l ow a n i t e r ative p rocess b y assigning estimat e d va l u es to t h e u nk nown bus vol t ages a n d by calc u l a t i n g a n ew val u e f or e a c h b u s v o l t a g e from the e st im a t e d v a l ues a t t he o t h e r b u ses a n d t h e r e a l a n d reactive pow e r s p e c i n e d . A new s e t o f v a l u e s for t h e volt age a t e a c h b u s is t h u s obt a i ne d a n d used to calcu l a t e s t i l l a n o t h e r s e t o f bus vol t ages. E a c h ca lcu l a t io n o f a n e w s e t o f voltages i f c a l l e d a n iteration . The itera t ive p rocess is r e p e a ted u n t i l t h e cha nges a t e a c h bus a r e less t h a n a specined minimum v a l u e . We d e rive equat ions for a fou r-bus system and w r i t e t h e gene ral equa tions l a t e r . With the slack bus designated as number 1 , comput a t ions start with bus W . If P2. sch a n d Q 2. SCh are the sch e d u le d real a n d reactive power, respectively, e n te r i n g t h e n e twork a t b u s @ , i t fol lows from Eq. (9.4) with i se t e qu a l t o 2 a n d N equal to 4 t h a t

( 9 . 1 4) Sol v i n g for Vz gives

( 9 . 15) us ass ume t h a t buses CD a n d @ a rc also l o a d buses w i t h re a l a n d re a c tive power specified. Express i o ns simi lar t o t h a t i n Eq. (9. 1 5 ) m a y b e wri t t e n for e a c h b u s . At bus G) w e have

For nOw

let

( 9 . 1 6) I f we were to e q u a t e r e a l a n d i m a g i n a ry p ar t s o f Eqs. (9. 1 5), (9. 1 6), a n d t h e

336

CHAPTER

9

POWER-FLOW SOLUTIONS

02 04

V21

V41 .

similar equation of bus @ , we would obtain six equations in the six state variables to and I to I However we solve for the complex voltages directly from the equations as they appear. The solution proceeds by iteration based on the scheduled real and reactive power at buses 0, ® , and @ , the scheduled slack bus voltage VI = I and initial voltage estimates 0 an d V4( ) at the other buses. Solution of Eq. ( 9 . 1 5) gives the corrected vol tage VP) calcu lated from

VJO),

el) V2

=

y22

_1_

[P2. Kh

VI I&,

j Q2 , seh.

V2(0 )* -

_

viO),

( Y2 1 V

1

+

Y

2J

V (O) + Y J

2.1

V4(O» )

]

( 9 . 17)

]

( 9 . 1 8)

]

( 9 . 1 9)

i n w hich a l l quantities in the right-hand-side express ion are either fixed s pe c ifi ­ cations or initial estimates . The calculated value V? ) and the estimated value ' will not agree. Agreement would be reached to a good degree of accu racy after several iterations and would be the correct value for V2 with the esti mated voltages but without regard to power at the other buses. This value wou ld not be the sol ution for V2 for the specific powe r-flow conditions, however, because the voltages on which this calculation for V2 depends are the estimated values and V4(0 ) at the other buses, and the actual vol tages a r e not yet know n . As the corrected voltage is found at each bus, it is used to calculate the corrected voltage at the next bus. Therefore, substituting V�- I) i n Eq. ( 9 . 1 6), we obtain for t h e first calculated value at bus G)

V�O)

vjO)

vel) 3

= _1_ y33

[ P3 , sel V

-

jQ J , seh

0* 3( )

_

31

(Y

V1

+ Y32 V2( l ) + Y34 V4(O» )

The p rocess is repeated at bus @ a n d at e ach b u s consecu tively throughout t he network (except at the slack bus) to complete the first iteration in w hich calcu lated values are found for each state variable. Then, the entire p rocess is carried out again and again until the amou nt of correction in voltage at every bus is less than some predetermined precision index. This process of solving the power-flow equations is known as the Gauss-Seidel iterative method. Convergence upon an e rroneous solution is usually avoided if the in itial values are o f reasonable magnitude and do not differ too widely in p hase. It is common p ractice to set the initial estimates of the u nknown voltages at all load buses equal to 1 . 0 per unit. Such initialization is called a fiat start because of the uniform voltage profile assumed. For a system of N buses the general equation for the calculated voltage a t a ny b u s CD where and Q a re scheduled i s

L!:. 1

P

VI ( k ) = y .

_

II

[PI, sel

-

I, seh,

'Q .

}

V (k - I )* I

_

i- J

'\" l...J

J= I

YIJ. VJ ( k )

_

�

L.J j=i+ 1

YIJV) (k - I )

The superscript ( k ) denotes the number of the iteration in which the voltage is

9.2

TH E GA U S S -S E I D E L

M ET H O D

337

c urre n tly being calcu l ated a n d ( k - 1) i n d icates the n umber of the p receding i te r a t i o n . Th us, we see t h a t the val ues for the vol t ages on the right-ha n d side of t h i s e q u at ion a re the most recently calcul ated v alues for the correspo n d i n g b uses ( or t h e estimated v o l t a g e if k is 1 a n d n o i teration h a s y et b e e n m a d e at that p a rticular bus ) . S i nce Eq. (9. 1 9) a p p l i es only a t load buses w here re al a n d reactive powe r a r e specified, a n addi tional s t e p i s n ecessary a t vol t age-contro l l e d buses w h e re vol tage magnitude is to r e m ain cons t a n t . Before investigating t h is a d d i t io n a l ste p , l e t u S look a t a n exa m p l e o f t h e cal c u l a t ions a t a load b u s . E xa m p l e 9 _ 2 . F i g u r e 9 . 2 s h ows t h e o n e - l i n e d i a g r a m o f G e n e ra t o r � ,I r e co n n e c t e d a t b u s e s

(I)

and

@

(1 s i m p l e p ow e r sys t e m

w hi le l oads arc i nd ica t e d a t a l l

fo u r h u se s . Rise v,! l lI e s fo r t h e t r , l I1 s m i s s i o n syst e m ,I re I ( ) ( ) M VA , 2 3 0 k Y . T h e

data

of

Tab l e <;.2

.

fine

g i ve p e r - u n i t s e r i e s i m p e d a n c e s ,Ill el l i n e -c h a r g i n g s l i s ce p t a nces

blls da ta

Q,

w hich each bus.

for t h e n o m i n ,t \ - 7T e q ll i " , t \ e n t s o f t h e fo u r l i n e s i cJ e n t i fl c c! by t h e b u ses at

Q values

t hey t e r m i n a t e . The

The

i n Ta b l e 9 . 3 l is t v a l u e s for P ,

the

load

b uses

(1)

and

G) .

G e n e ra t e d

Q

CD

!

Qi. SCh '

o r load (l rc c a l cu l a t e d from t h e c o r rcspo l l ci i n g

Q i; i

powe r fa c t o r o r 0 . 8 S . T h e n e t s c h e d u l e d v a l u e s , P' . ,ch a n d

Birch

is

and

P

V

at

va l u e s a s s u m i n g a a r e n e g a t ive a t

n o t spec i fi e d w h e re vo l t age

®

Elm

Maple

FI G U RE 9_2

O n e - l i n e d i ag r a m for Exam p l e s h ow i n g t he b u s n a mes a n d n u moers.

9.2

TA B L E 9 . 2

Li n e d a t a fo r Exa m p l e

Series Z

Line,

Series Y

=

Z -

I

T ot a l

Shunt Y Y / Z

per unit

II

c h a rg i n g

per unit

G

per u n i t

0 . 0 1 000

3 . 8 1 5629

- 1 9. 07 8 1 44

] 0.25

0.05 1 25

0.0074 4

0.0 504ll 0.03720

5 . 1 69561

- 2 5 . 8 4 7 809

7 . 75

0 . 03875

2-4

0 . 00744

0.03720

5 . 1 69 5 6 1

- 2 5 . 8 4 7 809

7 . 75

0. 03875

3 -4

0 . 0 1 272

0.06360

3 .023705

- 1 5 . 1 :. 8528

1 2. 7 5

0.06375

R

per unil

1 -2 1 -3

bus to bus

t B a s c 1 00 M V A , 2 3 0 k V .

tAt

9.2t

230 k V .

X

;\ l v a rt

per u n i t

338

CHAPTER 9 POWER-FLOW SOLUTIONS

data for Example 9.2

Bus

TABLE 9.3

Generation

Bus

P, MW

Q, Mvar

1

2

0

0

P, MW

Load

Q, Mvart

50

3 0 .99

1 70

1 05 . 3 5

V, per I .()()

u nit

�

I . ()O�

R emarks

Slack

bus

Load b u s ( i n d u c t ive)

3

0

0

200

1 23.94

1 .00�

Loao b u s ( i n d u c t i ve)

4

3 18

49.58

80

1 .02�

V o l t age con t ro l l e d

tThe Q values o f load a re ca lcu l a ted from t h e correspond i n g P v a l u e s assu m i n g a pow e r fa c t o r o f 0 . 8 5 .

m agnitude is constant. In the voltage colum n the values for the load buses are flat-start estimates. The slack bus voltage magnitude I VI I and angle 8 ) , and also magnitude I V4 1 at bus @ , are to be kept constant at the values listed. A power-flow study is to be made by the Gauss-Seidel method. Assuming that the iterative calculations start at bus CD , find the value of V2 for the first iteration.

In order to approach the accuracy of a digital computer, we perform the computations indicated below to six decimal places. From the line data given in Table 9.2 we construct the system Y bus of Table 9.4. For example, associated with bus @ of Fig. 9 . 2 are the nonzero off-diagonal elements Y2\ and Y24 , which are equal to the negative of the respective line admittances

Solution.

Y2 1

=

- ( 3 .8 1 5 629 - j 1 9 .078 1 44) ;

Y24 =

-

( 5 . 1 695 6 1 - j25 . 847809)

Since Y22 is the sum of all the admittances connected to bus (1) , including the

TABLE 9.4 Bus a d mitt ance m a t ri x fo r Exa mple Bus no.

CD

@

® 8)

CD

9.2"\"

(i)

8.985 1 90

- 3. 8 1 5 629

-j44.835953

+ j 1 9.078 1 44

- 3 . 8 1 5629

8.98 5 1 90

+ j I 9.078 1 4 4

-j44.835953

- 5. 1 69561

+j25. 847809

0

0

(J) - 5 . 1 69 5 6 1

+ j 2 5 . 847809

0

@ 0 - 5 . 1 6956 1 + j25.847809

8 . 1 93267

- 3.023705

-j40.863838

+ j 1 5. 1 1 8528

- 5 . 1 69 5 6 1

- 3. 023705

8 . l 93267

+j25.847809

+ j 1 5 . l 1 8528

-j40.863838

t Per-unit val ues rou nded t o s ix deci m a l p l aces

T H E GAUSS-S E I D E L M ETHOD

9.2

s h u n t susceptances for l i n e charging of l i n e s

Y22 = e Y2 1 ) + j O .05 1 25 -

Substit u t ion in ( 1)

V2

_ -

Eq. (9. 17) 1 -

Y22

[-

+

e

-

Y2 4 )

+

@ CD

and

-

jO.03875

1 .7 + j l .0535 - 1 .00 e - 3 .8 1 5 629 + j 1 9 .078144) 1 .0 + j O . O

Y"2 [ 1 .7 -

7 . 3�05S I

have

yi e l d s the per-un i t voltage

1

-

we

8 .985 1 90 - j44 .835953

=

- 1 0 2 ( - 5 . 1 69 5 6 1

=

(1)- @ ,

. 33 9

+

-

j l . 0535 +

9 .088581 - j 4 5

j44 . 3Sl)40l)

------ = (l . lJS 3 5 64 0 . 985 1 90 j44 . 8 3 5 9 5 3

- -

�-

.

+

j 2 5 .847809)

44 2 9 0 9]

1

jO . OJ23 1 6

Exper ien ce with the Gauss- S e i d e l met hod of solution o f power-flow p rob­ lems has shown t ha t the nu mber of i te r a t ions r e q u i re d m ay be reduced consid­ e rably i f the co rrection in voltage a t e a ch b u s i s m u ltiplied by some con s t a n t t h a t i ncreases t h e amou n t o f correct i o n t o b r i n g the voltage closer to t h e v a l u e i t i s approachi ng. Th e m u l t i p l i e r t h a t a ccompl ishes t h i s improved convergence is c a l l e d an acceleration jactor_ Th e d i ffe rence b e tween t h e n ewly calc u l a t e d voltage a nd t h e best previous voltage a t t h e b u s is multiplied b y the a ppropri a t e a cceleration factor to obt a i n a b etter correc tion t o b e a dded to the previous val u e . For example, a t bus W in the fi rs t i teration we h ave the acce l e r a t e d value VF1cc defined b y the stra igh t - l i n e formu l a Vel) 2 . ace

=

(1

-

a ) V2(O)

+

a V2( I )

=

V2(O)

+

a ( V)( I ) _

-

V (O» 2

In which a is the acce l e ra t ion factor. More generally, for bus i teration k the llcce lerated val u e is given by V(k)

/ . acc

=

(1

-

a ) V( k -

I . acc

1) +

a V( k ) /

=

V(k � l ) / • acc

+

( 9 .20)

)

CD

d uring

a ( V(k ) - VY - 1 » ) ( 9 .21) /

/ . acc

If a = 1 , t h e n t h e Gauss-Se i d e l compu t e d val u e of Vi is stored as t he c urrent val u e . I f 0 < a < 1, then the v a lue to be stored is a weighted average of t h e Gauss-Seidel val u e and t h e v a l ue s t o r e d from t h e p revious i teration. If 1 < a < 2, t h e n the value to be stored is esse n t i a l ly a n extrapolated val u e . I n power-flow s t u d ies a is ge nerally set a t abo u t 1 . 6 a n d c a nnot exceed 2 i f convergence i s t o occ u r .

340

CHAPTER 9

POWER· FLOW SOLUTIONS

Substituting the results of Example 9.1 and an acceleration factor of 1 .6 i n Eq. (9.20), w e find that

v?�cc

U SIng ·

V (l)

2, acc

= =

1 + 1 .6 [ ( 0 .983564 - jO .0323 1 6) - 1 ]

0. 973703 - jO .05 1706 per u n i t

In . SImI . ' 1 ar calculations for bus r:l\ (V g ives fi rst-i terat ion value

V{I:'CC

=

0 .953949 - jO .066708 per u n i t

Since bus @ i s v o l t ag e con t ro l l e d , w e m l l s t t r e ,\ t i t d i ffe r e n t l y a s e x p l a i n e d below. The acce l e r a t ion fac t o r fo r t il e r e a l co m p o n e n t o r t h c c or r ect i o n lll a y d iffer from that [or t h e i m a g i n a ry co m po n e n t . For a ny sys t c m o p t i m u m v a l u e s for acceleration factors exist, and poor choice of factors may result in less rapid convergence or make convergence i mpossible. An acceleration factor of 1 . 6 for both the real and imaginary components is usually a good choice but studies may be made to determine the best choice for a part icul ar system. Voltage-controlled buses. When vol tage magn itude rather than reactive power is specified at bus CD , the real and imaginary components of the vol tage for each i teration are found by fi rst computing a value for the reactive power. From Eq. (9.4) we have Q I.

=

{

- Im vI *

�

L j= l

y. v } IJ

J

( 9 .22 )

which h as the equivalent algorithmic expression ( 9 . 2 3) where 1m means " imaginary part of" and the superscripts indicate the relevant iteration. Reactive power Q (I k ) is eva l u ated by Eq. (9.23) for the best p. revious voltage values at the buses, and this value of Q � k ) is substi tuted in Eq. (9. 1 9) to find a new value of v:( k ) . The components of the n ew v:( k ) are then multiplied by the ratio of the specified constant magnitude , V: , t o the magnitude of v:( k ) found by Eq. (9. 19). The result is the corrected complex voltage of the specified magnit ude. In the fou r-bus example if bus @ is voltage controlled, Eq. (9.23) yields the calculated value ( 9 .24) wherein the calculated voltages of

buses 0 a n d Q) are accelerated v alues of the first iteration. Substituting Q �l) for Q 4 sch in the equ ivalent of Eq. (9.19) for

9.2 bus

@

THE GAUSS-SEIDEL M ETHOD

341

yields

( 9 .25 ) a n d a l l quantities o n t h e right-hand s i d e a re now known . S ince I V4 1 is s pecified, we correct t h e magnitude of VP ) as fol l ows:

( 9 .26) a n cl p roceed

to t h e n e x t s t e p w i t h stored v a l u e V4:�Hr u f bus @ vo l ta g e h av ing t h e speci ned magn itude i n t he rem a i n ing calculat ions of the i t e ration. As discussed in Sec. 9 . 1 , either voltage magnitude or reactive power must be specified at every bus except the slack bus, where voltage is specifi e d by b o t h vol tage magn itude and angle. At b u s e s with generation t h e vo l t a g e m a g n i t u d e is specified as well as the real power Pg supplied by the generator. The reactive power Q g entering the n e twork from the generation is then determi ned b y the comput e r in solving the power-flow p roble m. From a practical v iewpoin t the Qg output of the generator must be w i t h i n d efinite l imits given by the i ne quality

w here Q min is the m J n l m u m and Q max is t h e maximum l im i t imposed o n the react ive power output of the generator a t t h e bus. I n the cou rse of power-flow sol u t ion if the calcu l ated va l u e of Qg is outside e i t h e r l im i t , t h e n Qg is se t equal to the l imit violated, t h e origina l ly specified vol t a g e magni t u d e at the b u s is rela." ed, and the bus is then treated as a P-Q bus for which a n ew vol t a g e is calc u l a te d by t h c com p u t c r p r o g r a m . I n s u b s e q u c n t i terat i o n s t h e p ro g ram c n d eavors to sust a i n the origi n a l l y spcci flcd vo lt age at t h e bus w h i l e ensu ring t h a t Q R is w ithin the permitted ra nge of values. This cou ld w e l l be possible s i n ce other changes may occu r elsewhere in the system to support the l ocal action of the generator excita tion a s it adj usts to sat isfy the spe c i fi e d terminal volt age. 9.3. To complete the first iteration of the G a uss- S e i d e l proce d u r e , fi n d the v o l t g e a t b u s @ o f E a m p l e 9 . 2 co m put e d with the origi nally estimated voltages at buses W and G) rep l aced by t h e acce l e r a t e d v a l u e s i n d i c a t e d above. Example

a

Solution.

x

T a b l e 9.4 shows Q(l) 4

-

-:-

_

Y4 1

Im

e q u a l to zero, and so Eq.

{ V(O)* [ Y 4

42

V2,e l a) cc

+

Y Vel) 3 , acc 43

(9.24) +

gives

Y V(O)] } 44

4

342

CHAPTER 9

POWER·FLOw SOLUTIONS

{

Substituting values

Q�l) = =

This

=

_

for

}

the indicated quantities in this equation, we obtain

1 .02[( - 5 . 169561 + j25 .847809)(0.973703 - jo.05 1 706) 1m + ( - 3 .023705 + j 15 . 1 18528)(0.953949 - jO.06670t;) + (8.193267 - j40 .863838) ( 1 .02) ]

- Im{ 1 .02[ - 5 .573064 + j40.059396 + (8. 193267 - j40.863838) 1 .02 ]) 1 .654 151 per u n i t

val u e

for Q �l ) i s n ow s u bs t i t u te d i n t o

=

=

1 Y44

E(j .

(9.25) t o y i e l d

[ 2.381 .02- j1-.6541 51 jO.O - ( - 5 .573066 + j40.059398) 1

j

7.906399 - j41 .681 1 1 5. 8 . 1 93267 - }. 40. 8 63 8 38

H ence, I VP ) I equals

=

l .017874 jO.Ol0604 p e r u n it -

1 .017929, and so we must correct the magnitude

V}, 12orr = =

to

1 .02,

1 .02 ( 1 .017874 - jO.010604) 1 . 0 17929 1 .019945 - jO.010625 per u n i t

In t h is example Q�I ) is found t o be 1 .654 1 5 1 per u nit i n the first i teration. I f the reactive power generation at bus @ were l imited below 1 .654 1 5 1 per unit, then the specified l i m i t value would be used for Q�l ) and bus @ in that case would be treated as a load bus within the i te ration. The same strategy is used w i t h i n any o t h e r i teration i n which t h e generator Q -limits are violated . The G au ss-Seidel p rocedure is one method of solving the power-flow p roblem. H owever, today's industry-b ased studies generally employ the alterna­ tive Newton-Rap hso n iterative method, which is reliable in convergen ce, com­ puta tionally faster, and more economical in storage requirements. The con. verged solution of Examples 9.2 and 9.3 agrees with t he results given in Fig. 9.4 fou n d later by the Newton-Raphson method.

9.3

THE NEWfON-RAPHSON METHOD

Taylor's series expansion for a function of two or more variables is the b � sis for t he N ewto n-Raphson method of solvin g the power-flow p roblem. Our study o f

9.3

THE N EWfON-R A P H S O N METHOD

343

the method begins by a d iscussion of the �olu t:on of a p roblem i nvolvin g o nly two equations and two var iables. The n , we see h ow to extend t h e a nalysis to the sol u t io n of power-flow e qu a tions. Le t us consider the equation of a fu n ction h I of two variables x 1 and x2 equal to a constant b 1 expressed as

( 9 .27) and a second equation involving a n o t h e r funct ion h 2 such that

( 9 .2 8 ) w h ere 6 2 is also a con stan t . The symbol II re pr e se n ts an indep e n d e n t con trol, w h i c h is con s i d e re d consta n t in t h is c h a p t e r . As in E q s . (9.9) and (9. 1 0), t h e fu n c t i on s 6" 1 a n d g 2 a re i n t ro d u ce d for c o nv e n i e n c e t o a l l ow u s t o d iscuss t h e d i ffe r e n c e s b e t w e e n c alcu l a t e d v a l u e s o r h i a n d 11 2 a n c! t h e i r r e s p e c t ive speci­ fied values 6 1 a n d 6 2 • Fo r a s p e c i fl e d va l u c o f I I l e t l i S e s t i m a t e t h e sol u t ions o f t he s e equ a t ions to b e x \O) a n d x �()). T h e zero superscripts i n d icate t h at these va lues are i n itial est imates and not the act u a l sol u t ions x :i and x; . We designate the correc tions � x iO) and � x �O) as the values to be added to x \O) and x �O) to yield the correct sol ut i o n s x7 and x� . So, we can write

( 9 .29) ( 9 .30) Our problem now is to solve for � x iO) and .6. x iO), which we do by expa n d i n g E q s . ( 9 .29) a n d ( 9 .30) i n Tayl or's series about the assumed solu tion to give

( 9 .3 1 )

= 0 ( 9 .32) where the pa rti a l derivatives of ord e r greater t h a n 1 i n the series of t e rms of t h e exp a n sion have n o t been listed. T h e t e rm 3 g 1 /3x 1 1 ( 0 ) ind i ca tes t h a t t h e p a rtial d e rivative is evaluated for the estim a t e d values of x \O ) and x �O). Other such terms are eva l u ated si m i l a rl y .

344

CHAPTER 9 POWER-FLOW SOLUTIONS

If we neglect the partial derivatives of order greater than 1 , we can rewrite Eqs. (9.3 1 ) and (9.32) in matrix form_ We then h ave

0 -

o

g I ( x(O) J '

x(O) 2 ' u)

- .:, 2 ( x(O) I ' x(O) 2 ' u) (1

=

bI

- h I ( x eoI ) ' x(O) 2 , u)

b2

- h 2 ( x CI() ) ' x(O) 2 ' u)

where the square matrix of partial derivatives is c a l l e d the jacobian J or in this case J ( O) to indicate t hat the initial estim ates x �O) and x�O) have been used to compute the n u merical va lu es o f the partial derivatives. W e note that g l(X�O ), x�O), u) is the calculated value of g I based on the estimated values of x\O) and x�O), b u t t his calculated value is not the zero value specified by Eq . (9.27) unless our estimated values x\O) and x�O) are correct. As before, we designate the specified value o f g I mi n u s the calculated value of g I as the mismatch i1 g �O) and define the mismatch i1 g�O) simi l arly . We then have the linear system of mismatch equations J (O)

[ i1 X C O) ] J i1 x(O) 2

=

[ i1 g eo ] 1 )

( 9 . 34)

i1 g(O) 2

By solving the mismatch e quat i o n s , either by triangular factorization of the jacobian or (for very smal l problems) by finding its inverse, we can determine Llx\O ) and i1 x �O). However, since we tru ncated t h e series expansion, these v al ues added t o our initial guess do not determine for us the correct solution and we m ust try again by assuming new estima tes x i I ) and x �J ), where X CI I )

=

+ i1 x CO1 ) ., x (O) 1

( 9 . 35 )

We repeat the p rocess u ntil the corrections become so sma ll in magnitude that they satisfy a chosen pr eci s i o n index c > 0; that is, until l i1 x J I and l i1 x 2 1 are both less than c . The concepts und erlying the N ewt on - R a phs o n method are now exemplified n umerically_ Example 9.4. Using the Newton-Raphson method, solve for nonli n ear equations

Xl

and

x2

of the

9.3

Treat the parameter con d i t ions x iO)

=

a

u

=

.

og l

3g 1


dg2

aX I

t r e a t e d as a

=

XI

cos

u

Setting

has a fixed

0 - R2. calc

bl -

=

b2 -

=

1 .0

u =

4 u sin x 1

4 w: 2 sin X I

value equ a l

specifiab le o r cont ro l vari abl e .

ll g iO) = 0 - g 1 . catc

ll g �O )

4 uX 2

oX2

w e calculate t h e mismatches First i tera t i o n .

with respect to the x 's yie lds

aX 2

aX I

l Iere t he p a ra m e t e r

be

345

as a fixed number e q u a l to 1 .0, and choose t he initial x�O) = 1 0 . The precision i n d e x � is 1 0 - 5 .

rad and

Solution. Partial differen t iation

J

T H E N EWTON·RAPHSON M ET H OD

8x2

l .0,

to

but in

some s t u d ies i t co u l d

a n d u si ng t h e i n i t i a l estim ates

h \O) = - 0 . 6 - 4 s in ( O ) h �O)

4 11 cos X I

-

=

- 0 .3 - 4

X

=

of

and x2'

X1

-0.6

2 ( 1 .0 ) + 4 cos( 0 )

- 0 .3

which we use i n Eq . ( 9.34) to y i e l d the m ismatch e q ua t ions

[

4 si n ( 0)

4 COS( 0 ) 4 si n ( O)

8 x 2 - 4 cos ( O )

1[llX}O) 1 [ 1 L1 X �O)

-

- 0 .6 - 0 .3

I nverting this s i m ple 2 X 2 m a trix, we determine t h e initial corrections

[ 1 [ ][ 1 [ 1 � x �O)

. 6. x �O)

=

4

0

0

4

..

w h i c h p rov i d e fi r s t i t e r a t i o n v a l u e s of

X \ I ) = x \O) x ii ) :.= x �()

+ +

L1 X\()

6 X �»

=

=

- 0 . 1 50

- 0.6

0.0 1 .0

I

XI +

+

-

0.3

(tnd

=

-

X2

a s fo l l ows:

( - O . I SO)

( -- 0 .075)

= =

T h e c o r r e c t i o n s exce ed the s p e c i fi e d t o l e r a n c e , a n d so

[LlLl ggi�ll )) 1 [

S ec o n d i t e r a t i o n .

=

The n ew m ismatches are

0 .07 5

- O . l S0 rad 0 . 925

we

- 0 .6 - 4 ( 0 .925 ) s i n ( - 0 . 1 5 ) 2 - O J - 4 ( 0 .925 ) + 4 ( 0 .925 ) cos ( - 0 . 1 5 )

con t i n u e .

1 [

- 0 .047079 - 0 .064047 ,

1

346

CHAPTER 9

POWER·FLOW SOLUTIONS

[aX�a X�ll)) 1 [

][

] [-

and updating the jacobian, we compute the new corrections =

3 .658453 -

- 0 .597753

0 .55292 1

3 .44491 6

-1

-

0.047079

- 0 .064047

These corrections also exceed t h e p recision index, a n d next iteration with t h e new corrected values X �2 ) x�2 )

ax�3)

-

=

=

0 . 1 50 +

0 .925 +

( - 0 .0 1 (335 )

( - 0 .021 2 1 4 )

=

=

-

so

=

0 .0 1 6335

- 0 . 02 1 2 1 4

]

we move on t o t h e

0 . 1 66335 rad

0 . 903786

Continuing on to the th ird iteration, we find that the corrections 6. x\ J ) a nd are each smaller in magnitude than the stipulated tolerance o f 1 0 - 5 . Accordingly, we calculate the solution X \4)

=

- 0 . 1 66876 rad ;

X �4 )

=

0 . 903057

The resultant mismatches a re insignificant as may be easily checked . I n this example we have actually solved our first power-flow p roblem by the Newton-Raphson method. This is because the two nonlinear equations of t h e example are the power-flow equations for the simple system shown i n Fig. 9.3,

( 9 . 3 6)

( 9 . 37)

Pg 1

-

Qg l

CD

I Vl l !!!

--+--+-

)6.0

® )0.25 To load

FIGURE 9.3 The

system

eq u a t i o n s

with

power-flow

c o r r e s p o n d in g

t hose o f Exa m p l e 9 . 4 .

to

T H E NEWTON- RAPHSON POWER- FLOW SOLUTI O N

9.4

347

w he re x I represents t h e a n g le O 2 a n d x 2 represents the vol tage magn itude I V2 1 at bus G) . The co n t ro l u d e n o tes t h e vol tage magn i t u de I VI I of the slack bus, and by changing i ts val u e fro m the specified val u e of 1 .0 per u n i t , we m ay control the sol u t i on to t h e prob l e m . I n this textbook we do not i nvestigate this con t rol c haracteristic b u t , r a t her, concen trate on the application of the Newton-Raphson proced u re in p ower-flow stud ies. 9.4

T H E N EWTO N - RAPH S O N POWER-FLOW SOLUTI O N

T o apply t h e N ewton - R ap h son m et hod t o t h e so l u t ion o f t h e power-flow e q u a t io n s, we express bus vol tages a n d l i ne a d m i t t a n c e s in pol a r form . W hen n is s e t e q u a l t o i i n ECl S . (9.6) a n d (9.7) a n cl t he corresponding terms a re s c r ; l r a t c c..! rro m t h c S U ll1 m ; l { i o n s , wc o bt
/J;

Q;

= I �f ( ; u

=

-

+

I

/\'

II

L - iI

1/ I

I V, i :: B i i -

Vi v't Y, ,, l cos( (Ji"

N

L. I V, V,) �

11 = 1 t1

1/

+

l si n ( 8iI /

() 1/

- ('5J

+ 0 1/ - 0 ; )

( 9 .38 )

( 9 .39)

*" i

These e q u a t ions can be rea d i ly d iffe re n t iated with respect t o voltage a ngles a n d m ag n i t u d es. T h e t e rms invo l v i n g Gi; a n d Eii come from the defin ition o f }�j i n E q . ( 9 . 1 ) and t h e fact t ha t t h e a n g l e ( 0 1 / - 0 ) is zero whe n n = i . Let u s p ostpone cons i d e ra t i on o f vol tage-con trolleu buses for now a nd reg a rd a l l buses (except t h e sl a ck bus) as load b uses w it h known dema n d s Pdi a n d Qdi' The slack bus has sp ecified values for o ( and I VI I , and each of the o t h er buses in t he n etwork h as t h e two state v a r i a bles 0i and I V,I t o b e c a l c u lated i n the pow er -flow sol u t ion . T h e known values o f Pdi a n d Q d i correspond to t he n ega t ive of t h e b const a n t s shown in Eqs. (q . 27) a n d (9.28), as d e monstrated i n Ex a m p l e 9 . 4 . A t c a c h or t h e nonsl ack buses e s t i m a t e d val ues of 0/ a n d I Vi i c orrespond to t h e c s t i m a t e s X �") and x �)) i n the p rece d i n g section. Co r responde nce to t h e 6 g m i s m a t c h e s or E q . (9.34) fol lows from Eqs. (9. 8) a n d (9.9) b y writing t he powe r m ism a t ch es for t h e typical load bus CD ,

6. Q i

( 9 .40 ) =

Q' . sch - Qi, calc

( 9 .4 1 )

Fo r sim p l ic ity sake, we n ow w r i te m i s m a tch equations for a fou r-bus system, a n d i t w i l l be come obvious h ow t o exte n d t hose equ ations to systems w ith more than fou r buses.

348

CHA PTER

9

POWE R-FLOW SOLUTIONS

For real power

Pi we have

( 9 A2)

The last t h ree te rllls c a n be magnitudes without a l t e r i n g

so

m u l t i p l i e d a n d d i v i d e d hy t h e i r r e s p e c t ive v o l l cl g c

t h e i r va l u c s , a n u

wc

o b ta i n

( 9 .43 )

There are advan tages to this form o f the equation as we shall soon see. A similar m ismatch equation can be written for reactive power Qi'

( 9 .44 )

Each nonslack bus of the system has two equations like those for 6. Pi and Collecting all the mismatch equations into vector-matrix form yields

a P2 a04

a P2 a0 2 aP4 a0 2 a Q2 a0 2 I

a Q4 a0 2

JII

aP4 a 04 a Q2 a0 4

aP I V4 1 a l v2 41

aP2 V 1 2 1 a l V2 1 ap4 1 V21 3 1 V2 1 aQ I V21 al V2 21

JI2

J22

J 21

aQ 4 a0 4

I V2 1

aQ 4 a l v2 1

Jacobian

ap I V4 1 a l v4 41 aQ I � I al v2 41 a Q4 I V4 1 a l v4 1

6.0 2

6. P2

6.0 4

6. P4

6. 1 V2 1

6. Q 2

--

I V2 1

il l

V4 1 I V4 1

--

----- � ----

Co rrect ions

6. Qj.

6. Q4 ---I

M is m a t c h e s

( 9 .45)

9.4

T H E N EWTO N - R A PHSON POWER- FLOW SOLUTIO N

349

We c a nn o t i ncl ude mismatches for t he slack bus si nce 6. P I a n d 6. Q I are u nd e fi n ed when P I and Q I a re not scheduled. We a lso omit all terms involving 6. 01 and 6. 1 VI I from t h e equ ations because t hose corrections a re b o t h zero at the s lack bus. The parti tioned for m of E q (9.45) emphasizes t h e fou r d ifferent types of partial d erivatives which e n te r i n t o the j a cobian J . The elements of J 1 2 a n d J 2 2 h ave volt age-magnitude m u l t i p l i e rs because a simpler and more symmetr i c a l j acob i a n resu l ts. I n choos i n g t h is format, w e h ave used the i d e n tity

.

------- -----lVI

c7 PI,

j aI

X

VJ I

E l e me n t of J 1 2

L1 1 � 1 I VJ I

=

(7 p.

-I () Vj' I

X

( 9 .46)

�I �I

Co rrec t io n

and t h e correct ions become 6 1 � 1 / I � I a s s h ow n r a t h e r t h a n 6. I Vj I . T h e so l u t i o n of Eq. ( 9 . 4 5 ) is fo u n d by i t e ra t i o n a s fo l l ows: • •

• •

E s t i m a t e values ofO) and I � I (O) fo r the state variables, Use the estima tes to c a l c u l a t e : p/:Ojotc a n d Q ���alc from E q s . ( 9.38) a n d (9.39), mismatches 6. p/O) and 6. Q�O) from Eq s . (9.40) and (9.41), and the p a rt i a l derivat ive e l e m e n ts o f the jacobian J . Solve E q . (9.45) for the i n i t i a l corrections 6. 0�0) a n d � I v: I ( 0 ) / I v: I (0). A d d t h e solved corrections to the i n i t i a l estim a tes to obtain

I VI I ( I ) = I VI,' I ( 0 ) + L1 1 VI, I erl) •

U s e the new v a l u es

()� I ) and I � 1 ( l l

co n t i n u e .

more general v a r i a b l e s are In

t e rm s , t h e

+ 8(k I

I)

=

+

6. 8 I( k )

(

I VI I (0) 1 +

..'l l v: I (0) I VI 'i ( O )

)

as s t
u p da t e fo r m u l a s

8 ( 1.: ) I

=

( 9 .4 7 )

( 9 .48)

2 and

fo r t h e s t a r t i n g v a l u e s o f t h e s t a t e

( 9 .49 )

350

CHAPTER 9

POWER-FLOW SOLUTIONS

For the fou r-bus system subm a trix J I I has the form

JI I

a P2 a02 a p:, a0 2 a P4 a0 2

=

a P2 a 04 a p-" a 04 aP4 a 04

a P2 a03 a p:, a03 aP4 dO"

(9.51)

Expressions for the c l e m e nts of t h i s e q u a t i o n ( I re e a s i l y found by d i ffe re n t i a t i n g the appropriate n umber of t erms in Eq . ( � . 3 8 ) . W h e n the v a r i ab l e ! l e q u a l s t h e p articu l ar value j , only o n e o f t h e cos i n e terms i n t h e s u m m a t i o n o f E q . (9 .3R) contains OJ , and by partial d iffere n t i a t i n g that single t e rm with respect t o Dj , we obtain the typical off-d iagon al elem e n t of J I I '

( 9 . 52 ) On t h e other h a n d , ecery term i n the summation of so the typical d i agonal element of J I I I S

a p1aD·

L. I V; V,I Y,) s i n ( 8ill + D I I - Di ) N

I

I *i

n =" n

By comparing this expression and t h a t for Q i i n

ap a D 1. 1

- Q I.

Eq.

- I vf B -I

II

Eq. ( 9 .38) contains 0;, N a p1 - L. -

nn

I oF ;

=

d Oli

a nd

( 9 .53)

(9.3 9 ) , we ob t a in ( 9 .54)

J 2 1 as follows:

In a q uite similar manner, we can d e rive form ulas for t h e elements of submatrix

=

ao I·

- I VI VJ YIJ· i cos ( 8 IJ. .

+

0J - 0 I )

L. 1 V; � �n I cos ( 8i n + 0"

nn

N �

1

oF ;

-

( 9 .55 )

oJ

( 9.56)

9.4

THE N EWfON-RAPHSON POWER-FLOW SOLUTION

351

Comparing this equation for aQjaoi with Eq. (9.38) for Pi ' we can show that aQ i 2 - = p. - I V I G /..I a DI, I

(9.57)

I

The e lemen ts o f submatrix J 1 2 a r e easily found by first fi n d i ng the expression for t h e d er i vat i ve CJPja l Vi i and then multiplying by I Vi i to o b ta in o P' I Vj l -oI VI )

Cu mparisol1 wi t h Eq _ ( 9 . ) ) )

=

I Vj l I V; Y) cos( 81) + 0) - 01 )

( 9 .58 )

yields

( 9 .59) T h i s is a most u s e fu l resul t , for it re d u c es the compu tation i nvolve d i n forming the j acobian since the off- d i agona l c l e m e n ts o f J 1 2 are now simply the negatives of the cor r es p o n d i n g e l e m e n t s in J 2 1 - Th is fact wou ld not have become ap parent if we had not mu l t i p l i e d JP';J I Vi I by t h e ma g n itu d e I L� I i n Eq. (9.43). In an a n a logous manner, the d iago n a l e l e m e n ts of J 1 2 are found to be

( 9 .60)

a n d comparing this resu l t w i t h Eqs. (9_56) and (9.57), we arrive at the formula

I V I -71 V1 a Pr

I

(

I I

=

a Qi a oI

-

2 + 21 VI G , I

1/

=

p. + I VI I GI'I' I

2

(9.61)

Fin al l y , t h e off-diagon a l a n d d i agonal eleme nts o f submatrix J22 o f the jacobian a re d etermined t o be

( 9 . 62)

( 9 .63)

352

CHAPTER 9

J

POWER-FLOW SOLUT IONS

Let US now bring togeth er the results develo ped above In the follow ing defini­ tions: Off-diagonal elements, i

=1=

j

ap.I

M I. J �

i =

j

d OJ

aQI

NI J. �

Diagolla l c!CJ1lCIl ls ,

lVI

a Q I.

( 9 . 64 )


-

( 9.65)

j

M/ I �

D PI

ao· I

I

- Mii

apI l VI a I l V:

-

=

N/ I

- .'

? I Vi I 2 S il

2 '' VI I

+

( 9 . 66)

-

2

G / I·

( 9 . 67)

interr elation ships among the eleme n ts in the four subma trices of the jacobi an are more clearly seen if we use the definit ions to rewrite Eq. (<).45) in the follow ing form:

M 22

M 23

M 24

M32

M 33

M 34

N22 + 2 1 V2 1 2 C22 - N3 2

M4 2

M4 3

M4 4

- N4

N22

N 3 2 N3 3

N24

N3 2

N4 3

N4 2

x

N34 N44

Ll o 2

22

-M

i103 i1 0 4 i1 I V2 1 / I V2 1 i1 1 V3 1 / 1 V3 1 i1 1 � 1 / I V4 1

-

M3 2 M4 2

N33

1 2 B 22

-S

2

N 3

+ 2 1 v3 1 2 C 33 - N4

2

2 1 V2

-

3

24

- N34

N.!, :,

+

2 2 1 VI 1 C4 4

M 23

M �.!,

- M33 - 2 1 V3 1 2 B':l

M� 4

M4 3

2 - M')4 - 2 1 V4 1 B 44

Ll P2

i1 P3 i1 P4 i1 Q 2 i1 Q3 i1 Q 4

( 9 .68 )

T H E N EWTO N - R A P HSON POW E R - FLOW SOLUTION

9.4

353

So fa r we h ave reg arded a l l nons lack buses as l o a d buses. We now consider vol ta ge-cont rol led buses a lso. Voltage-con trolled b u s e s . I n the polar form o f the powe r-flow equations volt age-con trolled buses a re easily t aken into account. For example, i f bus @ of t h e fou r-bus system is vol tage con t rol l e d , then I V4 1 has a specified constant v a l u e and the voltage corr e ct ion � I V4 1 / I V4 1 must a lways be zero. Conse­ q uently, the s i xth column of the jacobian of Eq . (9. 68) i" a lways mul tiplied by zero, and so it may be removed. Fu rthermore, since Q-\ is not specified, the m i sm a tch � Q 4 cannot be d e fi n e d , a n d so we must omit the s ixth row of Eq. ( 9 . 6 8 ) corresponding to Q4 . O f cou rse , Q4 c a n be calcul ated after the power-flow sol u tion becomes ava il a b l e . I n t h e g e n e r a l case if th ere a re N,,, vol tage-contro l l e d buses besides t h e s l a c k b u s , a row a n d co l u m n fo r e a c h s u c h b u s i s omitted from t h e p o l a r form o f t h e sys t e m jacob i a n , which t h e n has (2 N - Ng - 2 ) rows a n d (2 N - NK 2) co l u m n s c o n s i s t e n t w i t h Ta h l e 9. 1 . -

Ex a m p l e 9 . 5 . T h e s m
cl a t a g i v e n i n T8 b l e s 9 . 2 a n d 9 . 3 . A power- Aow s t u d y o f t h e sys t e m i s t o b e m a d e b y t h e N e w t o n - R a p h s o n m e t hod u s i n g t h e p o l a r fo rm o f t h e e q u a t ions for

P

a n d Q.

De t e rm i n e t h e n u m b e r of rows a n d co l u m n s i n t h e j a cob i a n . C a l c u l a t e t h e i n i t i a l

t1 PjOJ

mismatch

a n d t h e i n i t i a l v a l u e s o f t h e j a cob ian e l e m e n t s o f t h e (second row,

t h i rd c ol u m n ) ; of t h e ( s e c o n d row, s e c o n d co l u m n ) ; a n d of t h e ( fi f t h row, fi ft h co l u m n ) . U s e t h e sp e c i fi e d v a l u e s a n d i n i t i a l vo l t age e s t i m at e s shown i n Ta b l e 9 . 3 . Solution . S i nce t h e s l a c k b u s h a s n o rows or col u m ns i n t h e j a c o b i a n , a 6 X 6

m a t rix wou l d be n e c e s s a ry i f

P

a n d Q w e re s p e c i fi e d for t h e rem a i n i ng t h ree

b u s e s . In fa c t , h oweve r , the v o l w g e m Cl g n i t u d e is s p e c i fi e d ( h e l d c o n s t a n t ) at bus

@,

and t h u s the j a cob i a n w i l l b e a 5

based on

x

t h e e s t i m a t ed ;1 11t1 t h e spcc i fi c Ll vol t a ges

26. 3 5969:')/ 1 0]

fo r m o f t h e o fI- d i u gon a J e n t r i e s o f Ta b l e 9 .4,

Y:l I

8 �1I)

-

(j �())

a n d t h e (kl go n
,I n c.!

c l e l l l e n t Y.1.1 ,l IT

all

. 3 0993° ; ,,�

ze r o ,

0 . \ lp, 2() 7

f ro m Eq .

( 1 . 0 ) 2 8 . 1 93 2 6 7 + ( 1 . 0 X

+ ( 1 .0

=

-

X

1 .02

P3. c alc

o r Ta h l e 9.3, we n e e d t he p o l a r

5 m a t r i x . In ord e r t o c a l c u l a t e

X

(9.30) we obt a i n -

1 .0

j 4 0 . (; () J H 3 t: . S i n c e

X

and the i n i t ial

26 . 359695 ) cos ( 1 0 1 .30993°

1 5 A l 7934 )cos( 1 0 1 .3 0993°

0 . 060 4 7 p e r u n i t

Y.12

)

)

- 354

CHAPTER 9

POWER-FLOW SOLUTIONS

Scheduled real power into the network at bus ® is 2.00 per unit, and initial mismatch which we are asked to c al c u l a t e has the v a l u e -

tl

P���alc =

- 2 . 00 -

From Eq_ (9.52) the (second

=

-

a n d from Eq. (9.53) t h e

For

1 5 .420B9B

element

1 .0

X

per

=

4 1 . 26B707 per u n it

1 5 .4 1 7934 ) s i n ( l O I .30993° ) unit

2 6 . 35 9695 ) si n ( 1 0 1 .30993° )

r o w fi ft h ,

unit

e

c leme n t is

o f t h e (second row, second

( 1 .0

the clement o f t h e (fi fth

x

per

=

X

- 1 .93953

c

1 . 02

X

=

t h i rd co l u m n ) j a o b i a n

row,

- ( 1 .0

( - 0 .06047)

so t h

co l u m n ) i s

+ 1 5 .420898

co l u m n ) E q . (9.63) yields

- 4 l .268707 - 2 ( l .0) \ - 40J) 63838)

= 40 .458969 p e r u n i t

Using t he initial i n p ut data, we c a n simil a r ly ca l c u l at e initial v a l u e s o f t he other elements of the jacobian and o f the power m i s m a tche s a t a l l the b u s es of the system. For the system of the preceding example the numerical values for i nitial­ ization o f the mismatch equations are now shown, for convenience, to only t hree

9.4

T H E N EWTO N - RA P HS O N POWER- FLOW SOLUTION

355

decimal places as fol lows:

@ G) @ G) W

G)

G)

45 A43

0

@

4 1 .269

0

- 26 . 3 65

- 1 5 .42 1

8 . 882

- 1 5. 4 2 1

4 1 . 786

- 5.273

- 9.089

0

5 . 273

44. 229

- 8. 254

- -- I

3 . 084

i1 0 2

i1 o)

0

i1 04

8 . 1 33

0

- 26.365 0

G)

G)

i1 1 V2 1

- 3.084

1 V2 1

0

i1 1 V3 1 1 V) I

40.459

0

.597

- 1 . 94U 2 .2 1 3 - 0 . 447 - 0 . 835 Th i s system of equations y i e l d s v a l u e s for t h e vol tage corrections of t h e first i t e r a t ion wh ich are needed to u p d a te the s t a t e variables a ccord ing t o E qs. (9.49) and (9.50). A t the end of the first i te ra t ion the set of upda t e d vol tages at t h e buses is: B u s no. i

=

I V, I (per u n i t )

0,

( d e g.)

CD

0 l . 00

W

- 0 .93094 0 . 98 335

G)

- l . 78790 0 .97095

CD

- l . 54383 1 .02

Th ese updated voltages a re t h e n used to reca l cu l a t e t h e jacobi a n a n d m is­ ma t ches of t h e second i t e r a t i o n , a n d so on . The i t e ra t ive proc e d u re con t i n u e s u n t i l e i t h e r t h e m is m a t ches 6. Pi a n d 6. Q, b e c o m e l ess t h a n t he i r stip u l a t e d al lowable val ues or a l l 6.8, and 6. 1 V: I a re l ess t h a n t h e chosen p recision index. When the solu t ion is comp l e t e d , w e can use Eqs. (9. 38) and (9.39) to calc u l a t e re a l and reactive powe r, P I a n d Q I ' at t h e s l ack b u s , and t h e reactive power Q 4 a t vo l t age-con t rol led bus @ . L i n e flows can also be computed from the d ifferences i n bus vol t ages a nd t he known p a ramet ers of t h e l i n es. Solved values of the b u s voltages and I ine fl ow s for the system of Example 9.5 a r e shown in F i g s . 9.4 and 9.5. The number of i t e r a t i o n s req u i re d by t h e N ew ton- Raphson m e t hod using bus a d m i t t a nces i s practic a l ly i n d e p e n d e n t of t h e n u mber of buses. The t i m e for t h e Gauss-Seidel method ( e m ploy i n g b u s a d m i ttances) in creases a l most d i rectly with t h e n u mber of b u ses. O n t h e other h a n d , com p u ting the e lements of t h e

356

CHAPTER 9

POWER-FLOW SOLUTIONS

j acobian is time-consuming, and the time per iteration is considerably longer for the Newton-Raphson method. When sparse matrix tech niques are employed, the advantage of shorter computer time for a solution of the same accu racy is in favor of the Newton-Raphson method for all but very small systems. Spars i ty features of the j acobian are d iscussed in Sec. B . 2 of the Append ix. 9.5

POWER-FLOW ST U D I ES D E S I G N AND OPERATION

IN

SYS TEM

Electric u t l l ity companies use very elaborate programs for pow e r - n o\\' studies a imed at evaluating t h e adequacy of a c om p l ex, i n t e rcon n e c t e d n e t w o r k I m por­ tant i n fo rm a t i o n i s ob t a i n e d c o n c e rn i l l g t h e d e s i g n a n d o p e ra t i o n o f sy s t e m s t h a t h a ve n o t y e t b e e n h u i l t a n d t h e e l rc e t s 0 [" c l 1 ; l l l ges o n e x i s t i n g s ys t e m s . A power-now study ["o r a sys t e m o p e r �l t i l l g LI n d e r a c t u a l m p roj c c t c u n o r m a l o perating cond itions i s ca l led a base case. The results from t he base case constitute a benchmark for comparison of changes in network R o w s and vol tages u nder abnormal or con tingency cond itions. The transmission pl anning engineer can d iscover system weaknesses such as low vol tages, line overloads, or loading con d i tions deemed excessive. These weaknesses can be removed by m a k i n g design studies i nvolving changes and/or add itions to the base case system. The system model is then subjected to computer-based contingency testing to discover whether weaknesses arise under contingency condit ions involving ab­ normal generation schedules or load levels. I nteraction between t he system designer and the compu ter-based power-flow program continues until system p erformance satisfies local and regional planning or operating criteria. A typical power-flow program is capable o f handling systems of more than 2000 buses, 3000 lines, and 500 transformers. Of course, programs can be expanded to even greater size provided the available computer faci lities are sufficiently large. Data supplied to the computer must include the n u m e r i c a l val u es of the l i n e and bus data (such as Tables 9 . 2 and 9 . 3 ) and an indication of \v h e t h e r a bus is the slack bus, a regulated bus where voltage magn itude is held constant by generation of reactive power Q , o r a l o a d bus with fixed P a n d Q. Where values a re not t o b e h e l d c o n st a n t t h e q u a n t i t i e s given in t h e t ab l e s a re interpreted as initial estimates. Lim i ts of P and Q generation usually must be specified as well as the lim its of line kilovoltamperes. Unless otherwise speci­ fied, programs usually assume a base of 1 00 MY A. Total line-charging megavars specified for each line accoun t for shunt capacitance and equal 13 times the rated line voltage in kilovolts times Ic hg ' d ivided by 1 0 3. That is, .

( 9 .6 9 )

w here

I VI

IS

the rated line-to-line voltage in k ilovolts, ell

IS

line-to-neutr al

9.5

P O W E R · FLOW STU D I ES IN S YSTEM DES I G N A N D OPERATION

357

c apacitance in fa r a d s for the e n t i re l e ngth of t h e l i ne , a n d Ichg is d e fi n e d by Eqs . (5 .24) and (5.25). The p rogram creates a n o m i n al -7T representation of t h e l i n e similar to Fig. 6 . 7 b y d iv i d i n g t h e capacitance computed from t h e g iven v a l u e of charging megavars equ a l ly between the two ends of the line. It is evi d e n t from Eq. (9.69) that l i ne-c h a rg i ng m egavars in p e r u n i t equals the per-un i t shunt susceptance of the l i n e a t 1 .0 per- u n i t vol tage. For a l o n g l i n e the comput e r could be programmed to compute the equivalent 7T for capacitance d i s t r i buted even ly along t h e l i n e . The p r i ntout o f resu lts p rov i d e d b y t h e computer consists o f a n u m b e r o f t a b u l a t i o n s . Usual ly, t h e most i mp o r t a n t i n form ation t o b e con s i d e red fi rs t is the table which l ists e ach bus n u m b e r and name, b us-voltage magn i tude in p e r u ni t and p hase a ngle, g e n e ra t i o n a n d load a t each b u s i n m egawa tts a n d megava rs, and megavars of s t a t i c c a p a c i tors or r eactors on t h e o u s . Acco m p a ny­ i n g the bus i !l format ion i s the fl ow of meg a w a t t s a n d megava rs from t h a t b u s ove r e a c h t r , t n s m i ss i o n l i n e c o n n e c t e d t o t h e b u s . T h e t o t a l s o f system ge n e ra­ tion and l oads a re l isted in megaw a t t s a n d megavars. T h e t a b u l a t ion described is shown in Fig. 9.4 for t h e fou r-bus system of Example 9. 5 . A sys t e m m a y be d i vi d e d i n t o a re a s , o r o n e s t u d y m a y i n c l u d e t h e sys t e m s u f severa l com p a n ies with each designated as a d i ffere n t a r e a . The computer p rogram wi ll exam i n e the flow between areas, and d ev i a t ions from t h e pre­ scri b e d flow will be ove rcome by c a u s i n g the approp riate change i n ge nerat ion o f C:! sel ected gene rator i n e ach a re a . In a c t u a l system operat ion i ntercha nge of power between a reas is monitored to determine whe t h e r a g iven a r e a is p rodu cing that amount of pow e r w h i ch will result i n the desired i nterc h a n g e . Among other i n for mation t h a t may be obt a i n e d is a l isting of a ll b u s e s where the pe r-unit volt age m a gn i t ud e is above 1 .05 or below 0 . 9 5 , o r o t h e r l i m i t s t h a t ma y b e spec i fi e d . A l ist of l in e l o a d i n gs i n megavoltampe res c a n b e 2 o b t (1 i n e d . Th e p r i n t o u t w i I I a l s o l i s t t h e t o t a l me g a w a t t ( \ / \ R ) l osses a n d 2 m e g av a r ( \ / \ X ) r eq u i r e men ts of t h e syst em, and b oth P a n d Q m isma tch a t e ach bus. M ismatch is a n i n d i c a t ion of t h e preciseness of the solut ion and i s the d i fference between P (and also u s uatly Q) entering a n d leaving each bus. T h e n u m c r i ct l resu l t s i n t h e p r i n t o u t o f Fig. 9.4 a rc from ,I N c w t o n ­ R a p hson powe r-fl ow s t u d y of t h e s y s t e m described i n Ex a m p l e 9 . 5 . T h e sy s t e m l i n e d a t a a n u b u s d a t a LI re p ro v i d e d i n Ta b l e s 9 . 2 .I n d 9 . 3 . Th ree N ew t o n ­ R a p hson i t e r; \ t i o n s w ere req u i re d . S i m i l a r st u d ies e m p l o y i n g t h e G a u ss - S e i d e l p ro c e d u re re q u i r e d m a ny m o r e i t e r a t i o n s , ;I n d t h i s i s a c o m m o n obs e rva t i o n i n com p a r i n g t h e two i t e ra t iv e m e thods . I n spe c t i o n o f t h e p r i n t o u t reve a l s t h a t t h e 4.81 MW. 1 1 1 2 R l osses of t h e sys t e m a r e (504 . 8 1 - 5 0 0 .0) Figure 9 . 4 m ay b e exa m i n e d for more info rm a t i o n t h a n just t h e tab u l at e d results, and t h e i n form a t ion p rovi d e d c a n b e d isplayed on a one-line d iagram showi ng the ent ire system o r a portion of t h e system such a s a t load bus G) of F i g . 9 .5. Transm ission design e n g i n e e rs and system operators usually can call for video d isplay of such selected power-flow resu l ts from computer-interactive te rm inals or workstations. The m e g aw a t t loss i n a ny of the l ines can be fou n d by com p aring the values of P at t h e two e n ds of the I i n e . A s an ex?mple, from =

x - - - - - - - - - - - - - - - - - - - - - - - - - -Bu s in fo rm ati on - - - - -

Bus

Volts

Angle (degJ

no.

Name

(p. u . )

1

Birch

1 .000

2

Elm

0.982

3 4

Pine Maple

0.969 1 . 020 Area totals

FI GURE 9.4

O. - 0.976 - 1 .872 1.523

-

X -

-

-

-

-

--

-

-

-

(MW)

(Mvar)

- -

- - - - - - - - - - -X - - - - - - - - - - - L in e fl o w

186.81

1 14.50

50.00

O.

O.

1 70.00

O. 318.00 504.81

O.

(MW)

181.43 295.93

230 kV

30.99

SL

2 3

1 05.35

PQ

49.58

500.00

To

B u s Name

123.94

80.00

Bu s

type

(Mvar)

200.00

Newton-Raphso n powe r-flow sol u t i o n for the sys tem of Example 9 . 5 . B a s e is respe c t ive ly.

-

- - G eneration - - - - X - - - - - Load - - - - X

PQ PV

Elm Pi n e

1

Birch Maple

1

- - - - - - - - - -X Line fl o w

(MW)

4 4 2 3

-

Birch Maple Elm Pine

(Mvar)

22.30 38.69 61.21 98. 12 - 38.46 - 31.24 - 131.54 - 74. 1 1 - 97.09 - 63.57 - 102. 9 1 - 60.37 74.92 133.25 104.75 56.93

309.86 and

1 00

MVA. Ta b l e s 9.2 a n d 9.3 show t h e l i n e d a t a a n d bus d a t a ,

9.5

P O W E R -FLOW STU D I ES IN SYSTEM D ES I G N A N D O P E R ATION

! t

MW flow -

98. 1 2

61.21

97.09

63. 5 7

®

359

1 t ! t

200

Mvar flow

102.91

1 04 . 75

-

-

----+--

----+--

60.37

56.93

123.94

V:I

=

0.969/- 1 . 87°

V4

=

W 1 .02/1 .52°

To IC3d

of P

FI G U RE 9.5

fl o w

Flow

anu Q at bus

G)

fo r the sys t e m of Exa m p le 9 . 5 . N u m b e rs b e s i u e t h e a rrows show t h e

o f P 3 n u Q i n me gawa t t s a ll d m e g a v a rs. T h e b u s vol t age i s s h o w n i n p e r u n i t .

Fig. 9 . 5 w e see that 98. 1 2 MW flow from b u s CD i n to l i n e (1)- G) a nd 97.09 M \V flow into bus G) from t h e same l i n e . Evidently, the 1 1 1 2 R loss i n a l l t h ree phases of the l i n e is 1 .03 M W . Consideration of t h e megavar fl o w on the l i n e b e tween b us CD and b u s ® is s l ig h tly more com p l ic a t e d because of t h e charging m e g avars. The computer con s i d e rs the distributed capacitance of the line t o be concen trated, h a l f at one e n d of the l i ne a n d half at t h e o t h e r e n d . In t h e l i n e d at a g iven i n Tab l e 9.3 the line charging for line CD - G) i s 7 .75 Mvars, b u t t h e computer recog n izes t hi s value to be t h e va l ue w h e n t h e voltage i s 1 .0 p e r u n i t . S ince charg i n g m e g avars vary as the square of the vol tage accor d i n g to Eq. (9.69), vol t ages a t b us e s CD a n d G) o f 1 . 0 p e r u n i t a n d 0.969 p e r u n i t , respec t ively, m ake t h e c h a rg i n g at those buses eq u a l t o

7 .75 -2

7 .75

-- X

2

X

( 1 .0 )

2

( 0 .969) 2

=

=

3 . 875 Mvars at b us W ----

---3 . 638 Mva rs a t b u s U)

Figure 9.5 shows 6 1 .2 1 Mva rs going from bus CD i n t o the l i n e to b u s G) a n d 63.57 Mva rs received a t b u s ® . T h e i ncrease i n m e gavars is d u e t o the l in e chargi ng. T h e t hree-ph ase fl o w o f megawatts a n d m e g avars i n t h e l i n e i s s hown in the single-line d iagram of Fig . 9.6.

360

CHAPTER 9

�

1.0

POWER-FLOW SOLUTIONS

�

98 2 98 . 2 9 7 .0 9 9 1 1 t-__ - __.--___________ _____----r-_ 61.2

1

t

FIGURE 9.6

65. 0 85

59 . 9 3 2

3.875

3 . 638

I

®

09 ;

63 . 5 7

l

Single-line diagram showing now o[ mcgawa t t s
0 969

buses

/- 1 . 8 72'

CD a n d G)

Exa m ple 9.6. From t h e l i l l e ! l ows s i 1 n w l l i l l F i g . I) . h , c a l c u la t e thL C l I rr e n t ilow i ng

in t he equivalent

of Fig.

9.2.

c i rc u i t of

from b u s

the l ine

CD

to b u s

G)

i n t h e 2 3 0 - k Y sys t e m

Usi ng t h e ca l c u l a t e d c ur re n t a n d t h e l i n e p a r a m e te rs g i ve n i n T a b l e

c a l c u l a t e the l i n e

9.2,

loss and c o m p a r e t h i s v a l u e w i t h t h e d i ffc r e n c e b e t w e e n t h e pow e r i n t h e l i n e from b u s a n d the power out a t b u s S i m i l a r l y , fi n d J 2 X

{ 2R

G) .

CD

i n t h e l i ne a n d compare w i t h t h e v a l u e w h i c h cou l d b e fou n d from t h e d a t a i n Fig.

9.6.

Solution. Figure 9 . 6 s h ows t h e megawa t t a n d m e g ava r flow e q u i v a l e n t c i rc u i t of l i ne X of

all

t h ree p h ases is

or

CD - G) .

S

=

S

= 97 . 0 9

98 . 1 2

and

or

calculated

using I I I

+

1 17.744/33.56° j59 .932 = 1 1 4 .098 / 3 1 . 69° _ 1 1 7 ,744

h

---,-----

I II

-=

I Vt

X

13

230

x

230

x

/ 2R

R

1 .0

I

-

x

/ 2 X of l i n e

the

per-ph ase

MVA

MVA

A

= 295 . 5 7 A

0 . 969

i n the s e r i e s R +

jX of l i ne

CD - G)

V3 1 / I R + JX I . T h e base i mp e dance

3

X

( 295 . 5 6 )

=3

X

( 295 .56)

=

295 . 5 6

1 1 4 ,098

(23 0 ) 2 1 00

a n d X p a r a m e te rs of T ab l e

loss

=

-----

Z base = and using t h e

=

j65 .085

III

The m ag n i t u d e of c u rren t =

+

in

The t o t a l m e g av o l t am p e r e flow t h ro u g h R a n d

=

529

9.2,

can a lso be

is

n

w e h ave

2

X

0 . 00744

X

529

X

10-6

2

X

0 .03720

X

529

X

10-6

= 1 . 03 =

MW

5 .157

Mvar

T h ese v a l u e s com p a re w i t h (98. 1 2 5 . 1 5 3 Mva r from Fig. 9 . 6 .

9.6

- 97.09)

9.6

R EG U LAT I N G TRANSFO R M ER S

1 .03

=

361

M W a n d (65.085 - 5 9 . 932)

=

REGULATING TRANSFORM ERS

As we h ave seen in Sec. 2.9, regu l at i n g transformers can be used to control the real and react ive power flows i n a circu i t . W e now d evelop the bus a dm it t ance e quations to i nc l u d e such transformers i n power- fl ow s t u d ies. Figure 9.7 i s a more d e t a i l e d repres e n t at ion of the regu l a t i n g t r a n s form e r o f Fig. 2.24( b). The a d m i t t a nce Y i n per u n i t is t h e reci procal of t h e p er-unit imped ance of the transformer which has the transform a t i on ratio 1: t as s hown. T h e a d m it tance Y is shown on t h e s i d e of the ideal t ra nsform e r nearest node CD , w h i ch is t h e tap-changing s i d e . T h i s design a t ion is i m port a n t i n u s i n g the e q u a tions which are to be d e r i ve d . I f w e a re co n s i d e r i ng a t r a n s fo r m e r with off- n o m i n a l turns rat io, t m ay be rc a l or i m a g i n a ry. s u c h a s 1 . 02 for an approxi mate 2% boost in vol t age magnitude or f irr / 6 0 for a n a pproxim a t e 3° s hift p e r phase. Figure 9.7 shows curr e n t s I i and Ij e n t e r i n g t h e two bus es, ancl the voltages are V; n nd Vj refe n' c d t o t h e refe rence node. T h e compl e x expressions for power i n to the ideal t ra n s former from bus CD and bus CD are, respective ly, 5I

=

Sj.

VJ* I

I

=

t VI /J*

( 9 . 70 )

S ince we are assu m i n g t h a t we have an i d e a l transformer w i t h n o losses, t h e power 5 i i n to t h e i d e a l transformer from bus CD m ust e q u a l t h e power - Sj out of the i de a l transformer on t h e bus CD side, and so from Eqs . (9. 7 0) w e obtain JI

The curre n t Ij can b e expressed by

=

- ( *1

( 9 .7 1)

)

( 9 . 72)

+ �

CD

1

�

I,

Ideal transformer

--1

:

t

+

VI

1

-.L •

�

/. J

t tV, t

t

(j)

�

�

�

.

+

t- -

v-j

F I G U R E 9.7 More

detailed

l a nce

d i a g r a m fo r the

pe r-u n i t

t u rn s ra t i u is 1 It.

fo r m e r o f Fig.

2.24( b )

react r answ hose

362

CHAPTER 9

Vi

-t

POWER-FLOW SOLUTIONS

tY

t ( t - l)Y

CD (1 - t ) Y

!-

Vj

FIGURE 9.8 C i rcuit h av i n g the node a d m i t t a nces o f Eq. ( 9.74) when , is real.

.---�---- --------�-

Multiplying by

Setting [ [ * we have

-

t * and substituting /I.

=

CD (JJ

=

f;

for

1 1 * Y VI

-

-

t * Ij yield

, * Y V)

2 1 / 1 and rearranging Eqs. (9.72) and (9.73 )

( 9 . 73 ) i n t o Yhu,

matrix form,

[Y;i Y:i] [�l [l t l 2 Y CD (JJ yjl.

yJJ.

VJ

(])

CD (jJ

- (Y

( 9 .74)

The equivalent-7T circuit correspon d i ng to these values of node admittances can be found only if t is real because then Yi j �' i ' Otherwise, the coefficient matrix of Eq. (9.74) and the overall system Y bus ar e not symmetrical because of the phase shifter. If the transformer is changing magnitude, not phase s h i fting, the circuit is that of Fig. 9.8. This c ircu it cannot be realized if Y has a real component, which would require a negative resistance in the cir cuit. Some texts show admittance Y on the side of t h e t r a n s fo r m e r opposite to the tap-changing side, and often the t r a n s [o r m a t i o n ratio i s c x p r e s s e d as 1 : a , as shown in Fig. 9.9( a). Analysis sim i l a r to that developed abovc shows that the bus admittance equat ions for Fig. 9 . 9( 0 ) take the form =

0J ] [�] [

(]J

�)

which c a n be verified

j

CD CD

-

CD Y

Y/a*

OJ

- Y/ a

2 Y/ i a 1

] [V:] �'

:...

[I, ]

( 9 .75 )

lj

from Eq. (9.74) by interchanging bus numbers i and j and setting t = l /a . When a is real, the equivalent circu it is that shown in Fig. 9.9( 6). Either Eq. (9.74) o r Eq. (9.75) can be used to incorporate the m odel of the tap-changing trans former into the rows and columns marked i and j in Yb u s of the overall system. Simpler equations follow if the represen ta tion o f Eq. (9.74) i s used, but the importan t factor, however, i s that w e can n ow account for magn itude, p hase shifting, and off-nominal-turns-ratio transformers in calcu­ lations to obtain Y bus and Z b u s ' I f a pa rticular transmission line i n a system is carrying too small o r too large a reactive power, a regulating trans fo rmer can b e p rovided at one end of

9.6

--

I d eal transformer l:a

-�

R E G U LATI NG T R A N S FO R M E R S

363

JJ.

- (})

y

VI

(a)

(1) +

1

l

V,

a

- 1

)'/0

-- y a

(j)

--

t

1 - a -- y

[< I C U R E 9 . 9 P n - r h ase repres e n t a t ion

t he

t-

VJ

lal2

(6)

+

regu l a t i n g

(a)

of t ra n s fo rm e r

show i n g : per- u n i t a d m itt a n ce Y oppos i t e to t h e tapc il ,u gi n g s i d e ; ( b ) per-u n i t equ Iva l e n t circu i t w h e n a i s r ea l .

t h e l i n e to make t h e l ine tran smit a larger o r smaller reactive power, as d e monstrated in Sec. 2.9. And any appreciable drop in voltage on the primary of a tra nsformer due to a change of load may make it desirable to change the t a p setting on transformers provided w i th adjustable taps in o r d e r to maintain p roper volt a ge a t the loa d . W e can i nvestigate voltage-magnitude adj u stment at a bus by means of the a u tomatic ta p-ch anging feature in the power-flow prog r a m . For i n s t ance, in t h e fou r-bus system of Example 9.5 suppose we wish to raise the load volt age at bus G) by inserting a magnitude-regu l a ti n g trans­ for mer between t h e load and t h e b us. With t real, in Eq. (9.74) we set i = 3 a n d a s s i g n t h e n u mb e r 5 to b u s 0 from w h i c h t h e load i s n ow to be served . To a c c o m m o d a t e t h e reg u l a tor in t h e p ow e r- fl ow e q u a t i o n s , Y b u s of the n e twork is exp a n d ed by one r o w and o n e col u m n for bus G) , and the e l ements o f buses G) a n d 0) i n t h e matrix of Eq. (9.74) a re then added to t h e p revious bus ad m i t t a n c e m a t r ix t o g i v e

Y b u s( n e w )

=

CD W W @ G)

W

Q)

@

W

Y2 1

Yn

U

0

Y3 1

0

Y3 3 + t 2 y

Y21 Y3 4

- tY

0

Y4 2

Y4 3

Y4 4

0

0

0

- tY

0

Y

CD Y1

1

Yl 2

Y1

:l

()

0

( 9 .76)

364

CHAPTER 9

POWER-FLOW SOLUTIONS

The �j elements correspond to the parameters already in the network before the regulato r is added. The vector of state variables depends on how bus G) is treated within the power-flow model. There a re two alternatives: e ither •

•

Tap t can be regarded as an independent parameter w ith a prespecified value before the power-flow solution begins_ Bus G) is then tre ated as a load bus with angle Os and vo l t age magn i t u d e I Vs l to be determined along with the other five state variables represented in Eq. (9.68). I n this case the state­ variable vector is

or Vol tage IllLlgnitude at b u s G) ca n b e prespeeilicd . T a p I tlH.:n rep l aces I V) ! as the state variable to be determined along with 85 at the voltage-controlled bus G) . In this case x = [ 82 , 03, 84, 85, I V2 1 , I V3 1 , t V and the jacobian changes accordingly.

VI

=

! t

1.0LQ:.

--,.--

-

CD

MW flow

98.29 69.02

Mvar flow

97.17 70.94

102.83

104.74

---

---

64.62

6 l .59

-+-+--

! t

200

200

V3

135.56 0 .966/ - 1 .85° =

V4

=

@) 1 .02/1 .52"

123.94

®

I 1 123.94 t � V5 0.976/ - 4 . 19° =

To load FIGURE 9.10 Flow of P a n d Q at bus G) of the system of Fig. 9.5 when a regu lating t ra nsformer is in t\irposed betwee n the b u s and the load.

9.6

R EG U LATING TRANSFO R M E R S

365

In some studies the variable tap t is considered as an i n dependent control variable. The reader is e n couraged to write t h e j acob i a n matrix and the m ismatch equations for e ach of t h e above a l te r n at ives (see P robs. 9.9 a n d 9 . 1 0). Because there i s a definite step between tap settings, discrete control action occurs when regulators are used to boost voltage at a bus. The results of r egu l a t in g the voltage of t h e load p reviously a t bus G) of t h e system of Fig. 9.5 a re s hown in the one-l i n e d i ag ram of Fig. 9 . 1 0 . A per- un i t reactance of 0 .02 was assumed for the load-tap- c h a n g i n g (LTC) transformer. W h e n t h e voltage a t the load is raised by se tting the LTC ratio t equal to 1 .0375 , t h e voltage at bus ® is s l i g htly lowered compared t o Fig. 9.5, resu l t i n g in slightly higher vol tage d rops across the l ines CD G) and @ - G) . The Q supplied t o t h ese l i n e s from huscs CD a n d @ is i ncreased ow i n g to t h e reactive power required by t h e r e g ul a tor, but t h e re a l power fl o w is r e l a t i v e l y u n a ffect ed. The i ncreased me gavars on t h e l i n e s cause the losses to i n c r e a s e and the Q contributed by the c h a rg i n g capac i t a n ces a t bus G) is decre ased . To d e t e r m ine t h e effect of p h ase-shi ft i ng t ransfo r m e rs, we l e t t b e complex wi t h Ll m agni t ude of u n i ty i n Eq. (9.74). -

Exa m p l e 9.7. S o l v e Ex a m the two re s u l t s .

parallel

Hence, the currents in

unit.

1(0)

jX

1 ------+-

11

----J-

+ �

Y

=

using t he Y b u s model of Eq. (9.74) for each of and co m p a re the solu t io n w i t h the approximate

2. 1 3

t r a n s fo r m e r s

Solution . T h e a d m i t t n nc e

�

ple

each

is given by 1 /jO. 1 -jlO per Fig. 9. 1 1 can be determ ined from

of t r a ns for m e r of t r a n s fo r m e r

Ta

jO.1

Ta

lIb) 1

=

1'2

1 / 1 .05

jX

,. ,

)0. 1

I(b) �

+

T" VI

F I G U RE 9. 1 1 C i rc u i t for Exa m p l e 9.7. V a l u e s ,lT e i n per u n i t .

j O .6

366

CHAPTER 9

POWER-FLOW SOLUTIONS

the bus admittance equation

CD

[ ;j:: 1 � [ �

-

(1)

� -� ] [ �: 1

CD

[

W

�

and the currents in transformer T" with ( = LOS, by Eq. (9.74), w hich takes the numerical form

CD

W

[

w

CD

-;:� 1 [�: 1

-jlO jlO

as

shown i n Fig. 9. 1 1 ,

w

CD

j 1 0 J OO

-j 1 1 .025

-j 1 0 .000

j 1 0 .500

are given

1 [VV21 1

In F i g . 9. 1 1 current 11 = ( f� a ) + I}b» , and likewise, 12 ( Jia) + lib » , w h i c h m eans t h a t t h e prece d i n g two m a t r i x e q u a t ions can be d i rect l y added ( l i k e admittances i n pani.llel) to obtain =

[ II ] 12

G) @

=

[

CD

- j21 .025 j20 .500

W j20 .500 -j20 .000

][ ]

LQ:.

VI V2

I n Example 2. 1 3 V2 is the reference vo l t a g e 1 .0 and the current 12 is ca l c u l a te d to be - 0.8 + jO.6. Therefore, [rom the second row o f t h e p r e ce d i ng e q u a t i o n we have

12 = - 0 .8 + jO.6 = j20 .5 V1 - j20( 1 .0) w h i c h g ives t h e

per-u n i t

VI =

vo l t age a t b u s

- 0 .8 + j20 .6 ---j20.5

1 .0049 + jO .0390 per unit

--

Since

VI and V2 are now both fo r t r a n s fo r m e r Ta to obt a i n

Ia

CD

known, we can return to the admittance equation

i ) = j l O V1 - j l 0 V2 = j1 0 ( 1 .0049

=

jO .0390 - 1 .0)

= - 0 .390 + jO.049 per unit

and from the admittance matrix for transformer

l� b )

+

j 1 0 . 5 VI - j10V2

= - 0 .41

+

=

T"

j 1 0 . 5 ( 1 .0049 + jO .0390) - jlO

jO.55 L pe r

unit

9.6

367

R EG U LATIN G T R A N SFO R MERS

H ence, the complex power outputs of the transformers are

= - V2 Iib)*

5Tb

=

0.41

+

j0.55 1 per u n it

Th e results fou nd by the approximate circulating c u rre n t method of Example compare favorably with the above exact solution.

2. 1 3

Solve the phase-sh ifter problem of Example 2 . 1 4 by the exact Y bus model of Eq. (9.74 ) and compare resu lts. E x a m p l e 9.8.

Solution .

The

I = E j rr / 60

=

l .O

w hi c h c a n be i n Exa m p l e

ft

(9.74)

bus a d m i ttance equation i s g i v e n by E q .

for

t h e p h ase - s h i ft i n g t r a n s for m e r

obtain

[II ] [ 12

=

w ith

as

a d d e d d i re c t l y t o t h e a d m i t t a nce e q u a t i o n fo r

9.7 to

Tn

tra nsformer

0 .5234 -: j 1 9 .9863 -}20.0

-j20 .0 0 .5234 + j 1 9 .9863

To

given

][ ] VI V2

Fol lowing the proced u re of Example 9.7, we have - 0 . 8 + jO.6 w h ic h

( - 0 .5234

=

yields the voltage a t bus

VI

=

- 0.8

- 0 . 5 234

CD ,

=

/ �h) =

=

j l O( VI I :.

j 1 9 .9863

+

- O J)7

V2 )

-

. - I i")

=

+

a !1d t h e complex powel' o u t p u t s

S Th

=

=

- O tI

j O .2 9

=

1 . 03 1

+

- j20 ( 1 .0)

jO.013 per unit

- 0 . 1 3 + j O . 3 1 per

.

+ jO.6

-

( - 0.13

unit

+

jO .3 1 )

rer u n i t

,Irc

- V2 /i h )*

A g a i n , t h e a p p r o xi m a t i o n v a l u e s o f

t h i s examp l e .

j 1 9 .9863) VI

j20 . 6

+

W e t h e n d e t e r m i n e the c u r re n t s

l�a)

+

=

0 . 67

+

j O 2 9 per u nit

.

E x a m p l e 2. 1 4 arc

close t o t h e

ac tu a l values o f

368

CHAPTER 9

POWER-FLOW SOLUTIONS

THE DECOUPLED POWER-FLOW METHOD

9.7

the s trictest u s e of the Newton-Raphson procedu re, the jacobian is calcu­ lated a n d triangularized in each itera tion in order to update the LU factors. In practice, however, the jacobian is often recalculated only every few iterations and this speeds up t h e ove ra l l sol u l i o n r:>rocess. T h e l i n a l so l u l i o n i s J C l e r m i n e cJ , of course, b y the allowable power mismatches a n d voltage tolerances a t the In

buses.

When solving egy for i m p rov i n g require m e n t s

ing t h e

is

l a rge-s c ; d e D o

•

•

co m p u t
the

a p p rox i m a t e v e r s i o n

we r

t ra n s m i s s i o n sys t e m s , an a l t e rn a t ive s t r a t ­

l

e l ic ie nc

y ancJ red ucing

dccoup/ct/ 1)()w{'/"�f!()1I'

or the

d e c o ll p l e d
w h i c h m a k e s u s e o f , I ll rHocc <.i ur e . The r r i n c i p \ c u n cl e rly­ o\)sc rv ; t l i O l l s :

m et/l O d ,

N c w t o l1 - R ,l phsO I l

h;lsed U il t wu

c o mp u t e r s t o r a g e

Change in the voltage angle 0 at a b u s primarily affects the flow of real power P in the transmission l ines and leaves the flow of reactive power Q rel a tively unchanged. Change in the voltage magnitude I V I at a bus primarily affects the flow of reactive power Q in the transmission lines and leaves the flow of real power P relatively u n c hanged.

We h ave noted both of these effects in Sec. 9.6 when studying the phase shifter and voltage-magnitude regulator. The fi rst observation states essentially that aPjaD j is m uch larger than a Qja Dj , which we now consider to be approxi­ mately zero. The second observation states that a Q ja l V; I is much larger than 8Pja l V; I , which is also considered to be approximately zero. Incorporation of these approximations into the jacobian of Eq. (9.45) makes the elements of the s ub matrices J 1 2 and J 2 1 zero. W e are then l eft w i th two separated systems of equations, a P2 a02

a P2 Ao4

6. 0 2

a P4 a84

6. 8 4

( 9 . 77 )

J11

aP4 a 02

aQ2 I V4 1 a l V4 1

a Q2 1 V2 1 a l V2 1 and

aQ4 V I 21 al v2 1

J 22

a Q4 I V4 ! \ a V4 1

6. V2

--

I V2 1

6. V4

I V4 1

--

( 9 .7 8 )

These e q u ations a re

decoupled

9.7

369

THE DECOU PLED POWER-FLOW M ETHO D

i n the sense that the vol tage-a ngle corrections

/1 P ,

/1 8 a r e c a l c u l a ted u s i n g o n l y r e a l powe r mismatches

w h i l e the voltage-m ag­

n i t u d e c o r r e c t i o n s are c a l c u l a t e d u s i n g o n ly 6. Q m i sm a tc h e s . However, th e

whereas the

co efficien t m a t ri c es J I l a n d J22 a re sti ll i nte r d e p e n d en t because the e le m e n t s of

J 1 1 d e p e n d o n t h e vo l tage ma g n i t u des b e i n g solved i n E q .

(9.78),

equ ati ons c o u l d be solved a l t e r n a t e ly, u s i n g in one set the most re c en t e l e m e n t s of J 22 depend on the a ngles of Eq.

(9 .77). Of

course, the two sets of

solutions

of t h e two c oe ffi c ie n t m a t r i c es at e a c h i te r a t ion . To avoi d such com pu t a t i o n s , we

fro m the o t h e r set. B u t this s c h e m e w ou l d s t i l l requ i re eva l u a t i o n a n d f actoring

s i o n - l i n e pow e r fl ow , as we now exp l a i n .

i n t r o d u c e fu r t h e r s i m p l i fic a t i o n s , w h i c h a re j u s t i fi e d by t h e p hysics of t r a n s m is­ I n a w e l l - d e s i g n e d a n d p ro p e rly o p e r a t e d power t ra n sm i s s i o n syst e m :

•

1 y so s m a I I t h a t

T h e a n gu l a r d i ffe re n c es ( 8 , - 8) b e tw e e n typ i c a l bu ses o f t h e system 1I S Ll a 1

are

( 9 . 79 )

•

Th e l i n e susc e p t a n c e s

C;i j so t h a t

Bli

a r e m a ny t i m e s l a rger t h a n t h e l i n e co n d uc t ances

G I ) s i n ( 8 I.

•

Qi

8j )

-

BI· ).

«

( 9 .80 )

cos ( 8 I - 8). )

(J)

o p e ra t i o n i s m uc h l es s t h a n the r e a c tive powe r w h i c h would fl ow if a l l l i nes

The r e a c ti ve pow e r

i nj e c t e d i n t o a n y bus

of t h e system d u r i n g n o rmal

fro m t h a t bus we re s h o r t - c i rc u i t e d to r e fe r e nce . Th a t i s,

( 9 .8 1 ) T h e s e a p p roxi m a t i o n s c a n b e u s e d t o s i m p l i fy t h e c l e m e n ts o f t h e j a cob i a n . I n

Eq.

(9. 62)

t h e off- d i a go n a l c l e m e n t s o f J I

ii P,

lVI

G O)

Using t he i d e n t i ty

apI

3 81·

wherc

B, ;

=

sinea

3Q .

j

+

rl Q

I J i V) I

I

a n d J 22 a re g ive n by

( 9 .82)

-

(3)

=

sin

I - l Vj I d l Vj I

a

cos

{3

+ cos

a

sin

(3

in Eq.

(9.82)

g ives u s

-

I Y;) s i n

e'i

a n d Gij

=

I �) cos eii'

T h e a p p roxi m a t i o n s l i s t e d ab ov e

370

CHAPTER 9

POWER-FLOW SOLUTIONS

then yield the off-diagonal elements apI.

JQi - J VJ ao}. J a I Vj I

�

- I VI V) I EIJ. .

( 9 . 84)

The diagonal elements of J I I and J 22 have the2 expressions shown in Eqs. (9 . 54) and ( 9 .63). Applying the inequality Qi « I V;· 1 Bii to t hose expressions yields iJ Q ; I V.I I --, a D I.

By substituting the approximations of m atrices J I I and J 22 , we obtain

==

( 9 .8 5 )

- I VI I BI. I.

Eqs. (9.84)

2

and (9.85 ) in

the

coe fficient

( 9 .86)

and

- I V2 V2 1 Bn

- I V2 V:\ I B21

- I V2 V4 1 B 24

- I V2 V3 1 B 3 2

- 1 V3 V3 1 B 33

- I V3 V4 1 B 34

- 1 V2 V4 1 B 4 2

- I V3 V4 1 B4 3

- 1 V4 V4 1 B44

� I V2 1 1 V2 1

� I V3 1

�Q2

1 V3 1

� Q3

I V4 1

L1 Q4

� I V4 1

( 9 . 87)

To show how the voltages arc removed from the entries in the coefficient matrix of Eq. ( 9 .87), let us multiply the first row by the correction vector and then d ivide the resultant equation by I V2 1 to obtain ( 9 . 88 )

The coefficients in this equation are constants e qual to the negative of the susceptances in the row of Ybu S corresponding to bus @ ' Each row of Eq. (9. 87) can be similarly treated by representing the reactive mismatch at bus CD by the quantity � Qj I V: I . All the entries in the coefficient matrix of Eq. (9.87) become constants given by the known susceptances of Y lJUS' We can also modify Eq. (9.86) by m ultiplying the first row by the vector of angle corrections and

9.7

THE D ECOUPLED POWER-FLOW M ET H O D

371

rearranging the res u lt to obtain

( 9 . 89 ) The coefficients i n t h i s equation can be m a d e the same a s those i n Eq. (9. 88) by setting I V2 1 , I V3 1 , and I � I e q u a l to 1 .0 p e r u nit in the left-hand-s i de expression. Note that the q u a ntity I1 P 2/ I V2 1 represents the real-pow e r mismatch in Eq. (9.89). Treating all t h e rows of Eq . (9.86) in a simi l ar manner leads to two d ecoupled systems of e q u ations for the fou r-bus network - B .."

- 8 24

t1 8 ..,

- /J ., -)

- B"

- 13 14

t1 [) :.

- ·- B

t1 8 -1

t1 P4

Q2 --

'

- B.

I J

�

p

_.'

. "> .'

-

B � ."

Ii

.

and

t1 P2 -I V2 1

- B"..,

.

44

- Bn

- B 23

- B2

4

6 1 V2 !

- B 32

- B 33

- B :' 4

11 1 V3 !

- B4 2

- B4

-B

11 1 V4 1

Ii

3

44

=

t1

PJ -I VJ I

( 9 . 90 )

I V4 1

I1

I V2 1

Q3 -6

=

I V3 1

( 9 .9 1 )

Q4 -I1

I V4 1

Matrix B is general ly symmetri c a l a n d s parse w i t h nonzerO elements, which a re co nstan t , real numbers exact ly e q u a l to t h e n ega tiue of the suscept a nces o f Y bus ' Consequen t ly, m a t r ix B i s easily formed a n d i ts t riangular factors, once com- ' p u ted at the beginning of t he sol u tion , d o not h ave to be recomputed, which l eads t o very fast iterations. A t voltage-co n t rolled buses Q is not specified a n d t1 1 V I is zero; t h e rows a n d c o l u m n s correspond i n g t o slI ch buses a rc omitted from Eq . ( 9 . 9 1 ) .

One typi c a l sol u t io n strategy is to:

2. 1.

3. 4.

5.

Calcu l ate the i n itial m i s matches t1 P / I VI , Solve Eq. (9.90) for 11 0 , Update the a ng les 0 a n d use t hem to calcul ate m ismatches I1 Q / I V I , Solve Eq. (9. 91 ) for 6. 1 VI a n d u pdate the m a gn itu des I VI , a n d Return to Eq_ (9 . 90) t o rep e at t h e it erat ion u ntil a l l mismatches are w ithin specified tolerances.

372

CHAPTER 9

POWER-FLOW SOLUTIONS

Using this decoupled version of the Newton- Raphson proced ure, faster power­ flow solutions may be found within a specified d egree of precision. Example

Using the deeoupled for m of the N ewton-Raphson method, d e t e r m in e the first-iteration s olu t i o n to the power-flow problem of Example 9.5. 9.9.

Solution. The B

l

Tab l e

ma t r i x can b e read directly from

9.4 and t h e m is m a tches co r re s po n d ing to t h e initial voltage estimates are a l ready calcula ted in Exa m ple 9.5 so that Eq. (9.90) becomes 44 .835953 ()

() 4() . H(,3H38

- 25 . 8471)09

- 1 5 . 1 1 H5 2H

Solv ing this equation gives t h e angle corrections in radians

6 82

683

- 0 .0205 7 ;

=

Add i ng these results to the

=

-

6 84

0 .0378 1 ;

fl at st ar t estimates of Table

=

0 .02609

updated values of 82, 83, and 84, which we then use along w ith the e lements of Y hus to calculate t he reac tiv e m is ma t c h e s

1 =

=

Ll Q 3 1 V3 1

9.3 givc:: s

.0 1 -11-

-

-

- 1 .0535

+

1

2

.0 ( - 44 .835953)

=

=

OJ)

1 9 .455965

1T/ 1 80 + 0 + 0 .02057 ) + 2 6 . 35 9695 1 .02 s i n ( 1 0 1 . 3 0993 x 1T/ 1 80 + 0 . 02609 + 0 .0205 7)

si n ( 1 0 1 .30993 x

+

X

- 0 . 80370 p e r u n it

1

=

the

2 Q 2 , SCh - [ - I V2 1 B22 - I YI2VIV2 1 sin(812 + o J . + 84 - I Y24V2V4 I sm(824 02)] I V2 1 l�

_

=

{ {

-

(3)}

3 V3 1 { Q , sCh - Q 3. calc }

_1_ {Q3. 5Ch 1

1 V3 1 1

--

1 1 .0 1

{

-

2 [ - I V3 1 S33

-

- 1 .2394

+

I YI 3 V\ V3 1 si n ( 0 I 3 + 8 \ . 8 + 4_ 34 8 03)] 1 Y34V3V4 1 SIn(

1 .0 2 ( - 4 0 . 863838)

sin ( 1 0 1 .30993 X

-

x

7T / 1 80

1 .02 sin ( 10 1 .30993

- 1 .27684 per u ni t

X

+

0

+

+

7T / 1 80

26 .359695 0 . 0378 1 ) +

+

0 .0260 9

1 5 . 417934 +

}

}

0.0378 1 ) ,

1).7

T H E D ECOU PLED POWER-FLOW M ETHOD

A r eac t i v e m is m a t c h c a l cu l a t i o n is not required for bus c o n t ro l l e d _ Accord i n g ly, in t his example Eq. ( 9 . 9 1 ) b e comes

[44.8305953

which y i elds the so l u t i o n s

t

0

40 .863838 t. I V2 1

vo l t age mag n i u d e s a t b u s e s

W

=

] [ V21 ] [ 11 1 11 1 V) l .

- 0 . 0 1 793

and

G)

a re

=

@,

- 0 .80370

- 1 .27684

and 6. I V3 1 1 V2 1 = 0 . 98207 =

373

w h i c h is vol t age

]

and I

- 0 . 0 3 1 25 . T h e n ew

V3 1

=

0 . 9 6875 ,

m i s m a tches for t h e s e c o n d i te r a t i o n o f E q . (9.90) a re c a l c u l a t e d u s i n g t h e n ew vol t a g e v a l u e s . Repe a t i ng t h e p r oce d u r e over a n u mb e r o f i t e r a t i o n s y i e l d s s a m e so l u t i o n as t a b u l a t e d i n Fig. 9 . 4 .

w h i c h c o m p l etes the fi rs t i t e r a t i o n . U p d a t e d

the

O ft e n i n industry-based programs c e r t a i n m o d i n c a t i o n s a re m a d e i n Eqs. ( 9 . 9 0 ) a n d ( tJ . 9 J ) . The m o d i fi c a t i o n s to TI i n Eq. ( 9 . 9 1 ) a re g e n e r c.l l 1 y as fol lows: •

LQ:..

O m i t the angl e-sh i ft i n g e ffects of phase s h i ft e rs from 13 by setting t = 1 . 0 W h e n rows and col u m n s for voltage-contro l l e d b uses are a lso o m it t e d as p reviously i n d icated , the res u l t i n g ma trix is cal led B" .

The coefficient matrix in Eq . (9 .90) is gen e r a l ly modified as fol l ows: •

O mit from B t hose elements that m a i n l y affect megavar flows such as shunt capaci tors and reactors, and set t a ps t of off-nominal transformers e qu a l to 1 . Also, ignore series resistances i n t h e eq uivalent-Ii circuits o f the t ra ns mission l in es in forming Y b u s from which B in Eq. (9.90) is obt a i n e d . The resul t i n g m at r ix i s ca l led B' .

n e two r k .

When B i n Eq. (9.90) is repl aced by B', t h e model becomes a l ossl ess I f, in a d d i t i o n , a l l bus vo l t ages a rc a s s u m ed to re m a i n const a n t at n o m i n a l values of 1 . 0 per u n i t , t h e so-cc.l l 1 e d de power-flow model res u lts. U nder t h e s e a d d i t i o n a l assum pt ions, Eq . (9.9 1 ) is n o t necessary (since � I � I = 0 at e a c h bus CD ) and Eq . (9 .9CJ) for the d c p ow e r flow hecomes

( 9 . 92)

B'

where it is und erstood t h a t the elements of B' are calculated ass u m i n g all l i n es a re lossless. Dc power-ftow a n a l ysis can b e used where approximate solutions are acceptable, as i n the con t i n g e n cy studies d iscussed in Chap. 1 4.

374

CHAPTER 9

9.8

SUMMARY

POWER-FLOW SOLUTIONS

This chapter explains the power-flow problem, which is the determination of the voltage magnitude and angle at each bus of the power network under specified conditions of operation. The Gauss-Seidel and Newton-Raphson iterative pro­ cedures for solving the power-flow problem are described and numerically exemplified. In addition to d iscussing how power-flow studies are undertaken, some m etho d s of real and react ive power-flow con t rol are prese n t e d . The resu l t s of paralleling two transformers when the voltage-m agn i tude ratios arc d i fferent or when one transformer provides a phase s h i ft are examined. Equations are d eveloped for the nodal adm ittances of these transformers and equivalent circuits which allow a n a lysis of reactive-power control are i n t roduced. The computer-based power-flow p rogram can b e used t o study the appli­ cation of capacitors a t a load bus simply by in corpora t i ng the sh u n t admittance of the capacitor i nto the system Y bus ' Vol tage control at a generator bus can also be i nvestigated by specifyi ng the values of the P V bus vol tages. Fast, approximate methods for solvi n g the power-flow problem are intro­ duced by means of the dc pow�r-flow model, which depends on the linkages b etwee n real power P and voltage angle 8 , and reactive power Q and vol tage m agnitude. Table 9.5 summarizes the equations for each of the methods of power-floW a n alysis.

Summary

TABLE 9.5

of power-flow equations a n d solution m ethods

Power-flow equ ations

Pi

L, N

I Y" , vi V" l co s(Oill

Pi

Qi

0" - 0, )

+

=

L N

I Y,1l V; V" I s i n (8in

+

On

0)

!1 Q i = Qi. sch - Q i . calc !1 p.I = • calc �------r- ------ ------ --- ' ------ ---' Gauss-Seidel m ethod

=

11 = 1 pI. , .seh

To obt a i n V a t

VI < k )

a

-

bus i w i t h known P a n d Q :

=

I

_

Y/ I

[ P.

V< k

I . se

h

-

jQ l , seh

-

I

To obt a i n Q at a regul ated bus

I)

_

i- I

\' L.. Y')V< ) k) j� 1

=

_

1

N

\' L.. j=i+ !

YI)V( k- l) )

1

i:

U s e 01 acceler atio n factor a a t b u s i in i t er a tio n k .

VI,( ace k)

11 0 ,

( 1 - a ) V(k - l) I. ace

+

a V(k) I

=

V
a (VY ) I

-

VCk - I ») I . ace

-

TABLE 9.5

9.8

( Continued)

il P2

Newton-Raphson met hod

a P2

a0 2

a PN

(J O N

.

.

.

.

a Q2

.

.

.

.

.

.

.

.

.

.

(J Q "2

aoz

(IO N ,) ° 2

.

..

(J Q N aDN

- I I ; 1 -; Y" I \ i r l ((J'J

,1 1 ' i

'I ,')j

.. i v

I

(lP' IVI -

vJ y. /

I V2 1 a 1 V1 1

J12

a PN

aON J 21

r

I Vl l a W2 1

a PN

a0 2

.

a P2

aON J1j

! cos( . () lJ, .

.

.

.

.

I V" I

il l V:1 .

(J Q " .

.

.

.

.

.

.

.

.

= I V I -iJ a Q,

)

l V; 1

=

I

O e co u p l c d cJ pow er-flow

sol u t i o n t e c h n i q u e

-8

+

,�

(/ 1 ',

. 6, )

.I

f)J. . /) ) •

- B :.1

I

- Ii "

-, / 1 2"

- B .'

l

- BN

- IJ N

J

Bii

2

.

.

.

.

'

. .

.

.

_

[,

II Il

-�

I

:F I

rJo

r) r'i , a p' ' --

;, , V. I ,

�Q i I V, : (" . V, I

l

IJ :' N

- B NN

- iJ 2 N - /J.' N - B NN

6Q

.

.

1/

iJ Q i ') + - + 2 I V I - CI I· QO •.

tJ. P2/

6.1 ve l

6. ° N

tJ. IV, 1

6. I VN I

iJ Pi

aa),

1 V2 1

6. PN/ I VN I tJ. Q2 / 1 V1 1

6. 0.1 / I V, I

L tJ. QN/ I VN I

a r e t h e i ma g i n a ry p a r t s o f t h e correspo n d i n g Yblls e l e m e n t s .

V o l t a ge-co n t ro l l ed b u ses arc n o t re prese n tt:d i n

6. 1 V I

'

- - - 2 1 VI I B·-I I

tJ. PJ/ I V;\ I

- n2 N -

.

il Pr

I

- BN

- 1J e."2

.

8 (k) + 6 8 ( k )

Z

- BN

.

r7 Q ,

a Pi

- Bl l

2

.

il il ,

O () }

+

.

(i( J N : v\ ' a WN i

- !l I V" 1

I

.

W,

-B D

n

.

i1 Q ",

S t a t e v a r i a h l e u pd a t e f orm u la s :

a(k + I )

.

I v, I --- - -

a VI I

j

375

SUMMARY

e q u ations.

2

376

CHAPTER 9

POWER-FLOW SOLUTIONS

PROBLEMS 9.1 .

9.2.

9.3.

In Example 9-3 suppose that the generator's maximum reactive power generat ion a t bus @ is limited to 1 25 Mvar. Recompute the first-iteration value of the voltage at bus @ using the Gauss-Seid el method. Fo r the system of Fig. 9.2, complete the second iteration of the Gauss-Se idel p rocedure using the first-iteration bus voltages obtained i n Examples 9.2 and 9 . 3 . Assume an acceleration factor of 1 .6. A synchro nous condenser, whose n ..: a c t ivc po e assumed be u n l i m i t d i s i n s t a l l e d at l o a d b u s 0) or t h c s s t c rn or Exa m p l e tJ . 2 t o hold b u s vol t a g e m g n u d e a t 0 . 99 p n u n i t . U s i n g (J ; ! u ss-Se i d c l m e t h o d , f i n d t h e v l t e s buses CD a n d CD for t h e !i rst i t e ra t i o n . Take Fig. 9.12 t e equ ivalent-7T representation of the t a s m is s i bus 0 and bus @ of the system of F i . 9 . 2 . v n in Fig. 9 . 4 , d e t e rm i n e a n d i u i c on f7ig. 9 . 1 2 t e v a l u e s o f ( a ) J> and Q l e a v i n g buses 0 a nd @ on line 0- @, ( b ) charging megavars o f t h e e qu i va l e t l i n e 0 - @, a n d ( c ) P a d Q at both ends o f t h e series part o f t h e e q u v a e t of line G) - @ .

w r capabi l i t y i s t o c y t i l e a i t t h c o ag at as h r n on l i n e bet w een g Us i n g t h e powe r f l o w s o l u t i o n gi e nn atc h ni l n of ,

-

9.4.

"

IT

®

+-----�--�---+

0 Diagram for Prob. 9.4. FIG U RE 9 . 12

9.5.

From the l i n e F g . 9 .4, determ i n e / 2 R l ss each of the f u r t ra n s m issi n l i n e s and ver i fy t h a t t e the 4.8 1 M W . o f th ese l i n e losses i s e q u l Suppose t hat a s h u t c p c i tor b a n k 1 8 Mv r is bus G) in and the reference node t he system of Example 9.5. Ybus Table 9.4 to account for this and estimate th actual megavar reactive power i nj ected into the system from the capacitor. For the system of Example 9.5 a ugmented a synchronous condenser as d escribed in Prob. 9.3, find the j acobia n calculated at the i n i al e s t i m a t e s . Hin t : i n Sec. 9.4 fol lowing Example 9.� t h a to start calculations from the Suppose t hat i n Fig. 9.7 the t a p i s on t h e side of node CD so that t h e transforma­ tion ratio is t : 1. Fin d elements of Y bus similar to those in Eq . (9.74), and d raw t h e equivalent-1T representation similar t o Fig. 9.8. In the four-bus system of Example 9.5 suppose that a magnitude-regulating transformer with 0 . 2 per-unit reactance is i nserted b e tween the load and the bus a t bus G), a s shown in Fig. 9 . 1 0 . The variable tap i s o n the load side of t h e transformer. If t h e voltage magnitude at t h e new load bus G) is prespecified, and

oflowininfoarmtato ion oftootatlhesyspowetem lro-sflooofw solution given in hi sum n a ian capacitorra,ted a coeModinnectfeydthbeetwe n given wi t h It would be simplenr to modify the jacobian matberigxinsnihownng. it ,

9.6.

9.7.

9.8.

9.9.

PROBLEMS

9.10.

therefore is not a state variable, t he variable tap t of the transformer 'should be regarded as a state variable. The Newton-Raphson method is to be applied to the so l u t i o n of the power-flow e q u atio n s . ( a ) Write mismatch equations for t h is problem in a symbolic form similar to Eq. (9.45). (b) Write e q u a ti o n s o f the j acob i a n elements of the co lumn corresponding to variable t (that is, part i a l derivatives w ith respect to t ), and evaluate them using the i n it i a l voltage estimates shown in Table 9.3 and assuming that the voltage magnitude at bus G) is specified to be 0.97. The i nitial estimate of 0 5 is O. C c ) Write equat ions of P and Q m ismatches at bus G) and evaluate the m for the first i teration. Assu me the i n i t ial estimate of variable t is 1 .0. I f the tap setting of the transformer of Prob. 9.9 is prespecified instead of the voltage magnitude at bus G) , then V5 should be regarded as a state variable. Suppose that t he tap se t t i n g ( i s spec i fied to be 1 .05. ( II ) In t h i s ease w r i t e m i s m a tch e q u a t ions i n a s y m ho l i \.' for m s i m i lar to E q . (9.45). ( b ) Write eC]uations of t h e j acob i a n c lements which Zlre partial derivatives with respect to I Vs I , and eva luate t h e m using t he i n i tial estimates. The i n i tial est imate of V5 i s 1 . 0 ( c ) Write equat ions fo r t h e P a n d Q m i s m a t c h e s at bus G) , and evaluate them for the fi rst i teration. Redo Example 9.8 for ( = 1 .0 , and compare the two results as to cha nges in real and reactive power flows. The generator at bus @ of the system of Example 9.5 is to be represented by a generator connected to bus @ t h rough a generator step-up transformer, as shown in Fig. 9. 1 3 . The reacta nce of this transformer is 0 . 02 per unit; the tap is on the h igh-voltage side of the tra nsformer with the off-nominal turns ratio of 1 .05. Evaluate the jacobian elements of the rows corresponding to buses @ and G) .

iQ:.

9. 1 1 .

9.12.

377

�

---rr --- I ..----- @

®

F I GURE 9 . 1 3 G e n er a t o r s t e p - tl iJ t r a n s fo r m e r fo r Prot.

9.13.

9. 1 4.

9.12.

For the system of Prob. 9. 12, find matrices B ' and B " for use in the decoupled power-flow method. A five-bus power system is shown i n Fig. 9.14. The line, bus, transformer, and capacitor data are g iven in Tables 9.6, 9.7, 9.8, and 9.9, respectively. Use the Gauss-Seidel method to find the bus voltages for the first iteration.

378

CHAPTER 9

POWER-FLOW SOLUTIONS

for Probs. 9 . 4

FI G U R E System 9. 1 8 .

9_ 1 4

Th e lin e

1

t h ro u g h

giv e n i n Tables 9 . 6 t h ro u g h

Line

data for the system of Figure 9 . 1 4

TABLE 9.6

Per-unit Series Y

Per· u n i t

Line

bus to bus

CD CD @ G) @

R

@ @ G) G) G)

Series Z

Charging Mvar

X

G

0.0108

0.0649

2.5

- 15

6.6

0. 0235

0.09 4 1

2.5

- 10

4.0

0.0 1 1 8

0.0471

5.0

- 20

7.0

B

0.0 1 47

0.0588

4.0

- 16

8.0

0.01 1 8

0.0529

4.0

- 18

6.0

Bus data for the system of Figure 9 . 1 4

TABLE 9.7

Q (Mvar)

Generation

Bus

CD @ ® G) CD

P (MW)

P

(MW)

Q (Mvar)

Lo a d

V

p.u.

L.Q: 1 .0LQ: 1.0La" l .olit 1 . 0 LQ:

1 .0 1

1 90

60

35

70

42

80

50

65

36

Transformer data for the system of Figure 9.14 TABLE 9.8

Transformer b u s t o bus

Per-unit reactance

Tap setting

0.04

0.975

9.9.

a n d b u s d a t a a re

Remark Slack bus

PV bus

PROBLEMS

Ca pacitor data for the system of Figure 9.14

379

TABLE 9.9

Rating i n Mvar

Bus

G) o

18

15

p o w e r fl o w s o l u t i o n o f t h e sys t e m of F i g . 9 . 1 4 , d e t e r m i n e ( (/ ) Yb"s of t h e sys t e m , ( lJ ) t h e mismatch equa tion at bus @ e v a l u a t e d a t t h e i n i t i a l v o l t a g e e s t i m ,l t e s of T a b l e 9 . 7 fo r t h e fi r s t i t e ra t i o n , a n d ( c ) w r i t e m i s m a t c h eq u a t i o n s in a for m s i m i l a r t o Eq . ( 9 . 4 5 ) . r m t h e sys t e m o r r i g . 9 . 1 4 . f i n d m ,l l r i c e s ll' a n d n " for l i s e i n t h e decou pled ro w e r I l o w m e t h od . A l s o , d e t e r m i n e t h e f i rs t - i t e r ,l t i o n I) a n d Q m i s m a t ch equa­ t i o n s ,I t b u s @ , a n d f i n d t h e vo l t age 1 l 1 ;l g n i t u li e a t b u s @ a t t h e e n d o f t h e fi rst l

9 . 1 5 . To a p p y t h e [\; c w t o n - R a p h s o n m e t ho d t o t h e

9. 1 6.

-

iteration .

9.17.

-

Suppose t h a t s h i ft e r w h e re

in

Fig.

9 . 1 4 t he t ra n s fo r m e r

G) is a phase Find Y ou s o f th is

and

t i s now a complex variable and is 1 .0 . (a) s y s t e m . ( b ) W h e n c o m p a r e d w i t h the power- flow sol u t ion of Prob. 9 . 1 5 , w i l l t h e real power flow i n t h e l i n e fro m b u s to i n c r e a se or d e cr e a s e ? What a b o u t t h e r e a c t ive power flow? E x p a i n why q u alitatively.

l

W

bus G)

9. 1 8. To a pp l y t h e d e c o u p l e d pow e r - flow m e t h o d m at r i c e s B' and B" .

to t he

sys t e m

of Prob. 9 . l 7 ,

find

an 1 8-Mvar shunt capacitor b a n k IS added to b u s G) . I n a pp l y i ng the N ew t o n - R a p h s o n m e t h o d i f th e amount of reactive power r e q u ir e d t o m a i n t a i n t h e sp e c i fi e d voltage at a PV bus ex c e e d s the m ax i m u m l i m i t o f its react ive power generat ion capability, the r e ac t i v e pow e r a t that b u s i s s e t t o t h a t l i m i t and t h e t yp e of bus b e c o m e s a load bus. Su ppo s e t h e maxi m u m r e a c t ive power g e n e r a t io n at b u s 0 is l i m i t e d to 150 Mvar in t h e system of Ex a m p l e 9.5. Usinf! the first-iteration r e s u l t g i v e n i n S e c . 9.4 fol low i n g Ex a mp l e 9.5, d e t er m in e w h e t h e r or n o t t h e t y p e o f b u s (j) s h o u l d be conve r t e d to Cl l o ad bus at t he s t a rt of t h e s e co nd i t e r a t i o n . If so, c a l c u l a t e t he react ive power mismatch at b us 0 t h a t

9 . 1 9 . Redo Exam p l e 9.20.

H

between buses (2)

9. ] 0

when

,

s h o u l d be u sed i n t h e s e c o n d - i t e r , l t i o n m i s m a t c h e q u a t i o n .

iCHAPTER

10

SYMMETRI CAL FAULTS

A fault in a circuit i s any failure which i n terferes with the normal flow of current. Most faults on transmission lines of 1 15 k V and higher are caused by lightning, which results in the fl ashover of insulators. The high voltage betw e e n a conductor and the grounded supporting tower causes ionization, which pro­ vides a path to ground for the charge induced by the l ightning stroke. Once the ionized path to ground i s established, the resultant low impedance to ground allows the flow of power current from the conductor to ground and through the ground to the grounded neutral o f a transformer or generator, thus completing the circuit. Line-to-line faults not involvi n g ground are l ess common. Open i n g Circuit b reakers to isolate the faulted portion of the l i ne from t h e rest of the system i nterrupts the flow of current i n the ionized path and allows deionization to take place. After a n i nterval of about 20 cycles to allow deionization, breakers can usually be reclosed without reestablishing the arc. Experience i n the operation o f transmission lines has shown that ul tra-high-speed reclosing breakers successfully rec10se after most faults. Of those cases where reclosure is not successful, many are caused by permanent fa ults where rec\ o s u re wou l d b e impossible regardless of the interval between opening and reclosi ng. Permanent faults are caused by lines being on the ground, by insulator strings breaking because of ice loads, b y permanent d a m age to towers, and by surge-arrester failures. Experience has shown that between 70 and 80% of transmission-line faults are single line-to-ground faults, which arise from the flashover of only one line to the tower and ground. Roughly 5 % of all faults i nvolve all three p hases. These are the so-called symmetrical three-phase faults which are considered i n this chapter. Other types of transmission-line faults are line-to-line faults, which do not i nvolve ground, and double line-la-ground faults. All the above fau l ts

381

T R AN S I ENTS IN RL S E R I ES C I RCU ITS

10.1

exc e p t t h e t h re e - ph ase type c a u s e a n i m b a l a n ce b e tween t h e p h a s e s , a n d they a r e c a l l e d

unsymmetrical faults.

T h e s e a r e c o n s i dered i n C h a p .

so

12.

T h e c u r r e n ts w h i c h fl o w i n d i ffe r e n t p a r t s o f a p ower s y s t e m i m med i a tely a ft e r t he occ u rrence of a f a u l t d i ffe r from t h ose fl owi n g a few cyc l e s l a t e r just b e fore c i rc u i t b reakers a re c a l l e d upon to o p en the line on both s i d es of the f a u l t . A n d a l l o f t h ese c u r re n ts d i ffe r w i d e l y f ro m the c u r r e n ts w h i ch wou l d

flow

u n d e r s teady-state con di t i o n s i f t h e f a u l t w e r e n ot i s o l a t e d from t h e rest of the s y s t e m by the opera t i o n of c i rcu i t b r e a k e r s . Two o f the f a c t o r s on w h i c h the p ro p e r selection of c i rc u i t b re ak e rs d e p e n d s are the c u r r e n t fl ow i ng i m m e d i ately a ft e r t h e fa ul t occ u rs and the c u rr e n t w h i c h the b r e a k e r m u s t i n terru p t . I n

fault

a n a lysis va l u es of t h e s e cu r r e n t s a re c a l c u l a t e d for t h e d iffe r e n t types of fau lts

at va r i o u s locations in the s ys t e m . T h e da ta o b t a i n ed from fa u l t c a l c u l a t i o n s a l so

s e rv e to d e t e r m i n e t h e s e t t i n gs o f r e l a y s w h i c h c o n t r o l t h e c i rc u i t b re a ke r s .

TRA \l S I ENTS I N R L S E R I E S C I R C U I TS

10.1

The s e l e c t i o n o f a c i rc u i t b r e a k e r for a p ow e r sys te m d e pe n d s not o n l y on the c ur r e n t t h e b r e a k e r i s to c a rr y u n cl e r n o r m a l o p e r a t i n g co n d i t i o n s , but a l so on

t h e m a x i m u m c u rre n t i t m ay h a v e t o c a r ry mom e n ta rily a n d t h e c u r r e n t i t may

h a ve to in terrupt at the vol t ag e of t h e l i n e i n w h i ch it is p l a c e d .

I n o r d e r t o a p p ro a c h t h e p r o b l e m o f c a l c u l a t i ng t h e i n i t i a l c u r ren t w h e n a

to a c i rc u it con tai n i n g c o n s t a n t v a l u e s of r e s is t a n ce a n d i nd u c t a n ce. Let t h e

s y s t e m i s short-ci rcu i t e d , con s i d e r w h a t h a p p e n s w h e n an a c v o l t a g e is a p p l i ed a p p l i e d vol tage b e Vmax s i n ( w t

vol t a g e . The n ,

a

+ a),

where

is zero a t the t i m e of a p p l y i n g t he

t

d e termi n e s t h e m a g n i t u d e of the vol t a g e w h en t h e c i rc u i t i s

c l o s e d . I f t he i n s t a n ta n eo u s v o l t a g e i s z e r o a n d i nc r e a s i n g i n a p o s i ti ve d i rect ion m ax i m u m i n s t a n t a n eo u s v a l u e , a is

whe n it i s a p p l i ed by c l os i n g a s w i t c h ,

Vm�lx

sin (

i s z e ro . I f the vol tage is at i t s p o s i t ive

a

11/2.

wt +

a

)

The d ifferen t i a l e q u a t i o n is

Ri + L ­

di

=

( 10.1)

dt

T h e so l u t i o n o f t h i s e q u a t i o n I S

i

w here

12 1

=

=

�;

31x

[sin(

VR2 + ( w L ) 2

T h e first t e rm of Eq. is no nperiodic

and

(ll

t

+

and

0 0.2)

d ec a y s

a

e

=

e)

- E -R

I

/ L s i n ( (l'

-

e)]

( 1 0 .2)

t a n - 1 ( w L /R ).

v a r i e s s i n u soi d a l ly w i th t i m e . T h e second t e r m

exp o n e n t i a l l y w i t h a

t im e con s t a n t

L/ R.

This

s i n u so i d a l t e rm a s t h e s t e a dy - s t a t e v a l u e o f t h e c u r r e n t i n a n RL c i rc u i t for t h e

n o n p e r i o d i c t e rm i s c a l l e d t h e

=-: 0,

dc component

o f t h e curren t . W e recognize t h e

g iv e n a p p l i e d vol tage. I f t h e v a l u e o f t h e s t e a d y - s t a t e term i s n o t z e r o w h e n

t

t h e d c com p o n e n t a p p e a r s i n t h e s ol u t i o n i n o r d e r t o s a t i s fy , t h e p h y s i c a l

382

CHAPTER 10

SYMMETRICAL FAU LTS

(al

FI G U R E 1 0 . 1

t i m e i n a n R L circu i t 0 = - rr/2, w h e re 0; (h) a l ( w L / !n. T h e vo l t a g e i s V" "" s i n ( ll) / +

Cu rre n t as a fu n c t i o n of

Time

for: (0) () = t a l l

IY )

( b)

a

-

(]

=

a p I1 1 i c d < I t / C"

-

() .

cond ition of zero current at the instant of closing the swi tch . Note t hat the dc term does not exist if the ci rcui t is closed at a point on the vol tage wave such t h a t a e = 0 or a e = rr . Figure 1 0 . I ( a ) shows the variation of curre n t w i th time according to Eq. 00.2) when a e = O. If t h e switch is closed at a point on the voltage wave such that a e = ± rr / 2, the de component has i ts maximum initial value, which is equal to t he maxim u m value of the sinusoidal component. Figure 1 0.l(b) shows curre n t versus time when a e = - rr /2. The d c component may have any value from 0 to Vmax / I Z I , depending on the instantaneous value of the vol tage when the circuit is closed and on the power factor of t he circu i t . At the instant of applying the vol tage the dc and steady-state components always have the same magnitude but are opposite in sign i n order to express the zero value of current then existing. In Chap . 3 we discussed the principles of operation of a synchronous generator consisting of a rotating magnetic field which generates a voltage in an armature winding having resistance and reactance . The current flowi ng when a generator is short-ci rcu ited is similar to that flowing when an alternating voltage is s uddenly applied to a resistance a n d an inducta nce in series. There are important differences, however, because the curren ts in the damper w indings and the armature affect the rotating field, as d iscussed i n Secs. 3 . 8 and 3.9. I f the d c component of current i s eliminated from t h e short-circuit curre n t of each a rm ature phase, the resulting plot of each p hase current versus time is t h at shown in Fig. 3 . 19. Comparison of Figs. 3 . 1 9 a nd 1 0 . l ( a ) shows the d ifference between applying a voltage to the ord i n a ry RL circuit and applying a short circuit to a synchr
-

-

-

-

,

1 0. 2

383

Xd , respectively. These re a c t a nces h ave i ncreas i n g values ( t h a t is, X; < X:t < Xd ) a n d t h e corresponding compo­ n e n t s of the shor t - c i rcu i t c u rrent have d e c r e a s i n g m a gn i t u d es ( 1 1" 1 > 1 1' 1 > I I I ). W i t h t h e d c com ponent re moved , t h e initial symmetrical rms current i s t h e rms reactance

va l u e

of

X� ,

I NTERNAL vOLTAGES OF LOADED MAC H INES UNDER FAULT CON DITIONS

a nd t h e s t e a d y-state r e a c t a nce

the ac

component of

t h e faul t

current immediately a fter the fault

occurs. In a n a l y t i cal work t h e i n t e r n a l vol t a ge to t h e m a c h i n e a n d t h e s u b t r a n ­ s i e n t , t ra n s i e n t , and steady-state c u rre n ts m a y be e x p r e s s e d a s p h a s o r s . T h e vu l t age i n d u ce d i n t h e a r m a t u re w i n d i n g s j u s t a ft e r t h e fa u l t o c c u rs d i ffe rs from

t h a t w h i c h exists a ft e r s t e ady s t a t e is rea c h e d . We acc o u n t for t h e d i ffe r e n c e s in i n d uced \'o l t a ge by u s i n g the d i ffe r e n t reactances

s t e a d y - s t a t e cond i t i o n s . 1 f

( X;, X� ,

and

Xd )

in s eries

w i t h the i n t e r n a l vo l t a g e t o c a l c u l a te c ur rents fo r s u b t ra n s i e n t , t ra n s i e n t ,

a

and

m a c h i n e i s r e p r e se n t e d hy the n o - l oa d vol t a g e t o n e u t r a l in s e r i e s w i t h the

gen e ra t o r i s u n l oa d e d w h e n t h e fau l t o cc u rs , the

r ru p e r re :l c t a n ce , ; I S s h o w \l in f7 i g , l , 20 , The r e s i st a n c e i s t a k e n i n t o a c c o u n t i f

t e r m i n a l s and t h e s h o r t c i rc u i t, t h e e x t e r n a l i m pe d a n ce m u s t b e

g rc a t e r a c c u r;ICY i s d e s i re d . H t h e r e is i m p e d a n ce e x t e r n a l t o t h e g e n e ra tor

between i t s

i n c l u d ed in t h e c i rc u i t . \V e s h a l l ex a m i n e t h e t ra n s i e n t s fo r m a c h i n e s c a rr y i n g a

l o a d in t h e n e xt sect i o n ,

Al t h o u g h mach i n e r e a c t a n c es a r e n o t t r u e co n s t a n ts o f t h e m a c h i n e a n d d e p e n d o n t h e degree o f s a t u r a t i o n o f t h e m a g n e t i c c i rc u i t , t h e i r v a l u e s u s ua l l y l i e w i t h i n ce r t a i,n l i m i t s a n d c a n b e pred i c ted fo r va r i o u s t y p e s o f m a c h i n es. Ta b l e A.2 in the Appe n d i x g ives t y p i c a l va l u es o f m a c h i n e r e a c t a n c e s t h a t a re n e e d e d i n m a k i n g fa u l t c a l c u l a t i o n s a n d i n s t a b i l i ty s t u d i e s . I n g e n e r a l , s u b t r a n ­ s i e n t reactances o f gene r a tors a n d motors a re u s e d t o d e t e r m i n e t h e i n i t i a l c u rr e n t flowing on the occu rrence o f a s h o r t c i r c u i t . For d e t e rm i n i n g t h e i n t e r ru p t i ng c a p a c i ty o f c i rcu i t b re a k e rs, exc e p t t h ose w h i c h o p e n i n s t a n t a ­ n e o u s l y , s u b t r a n s i e n t r e a c t ance i s u s e d fo r g e n e rators a n d t r a n s i e n t rea c t a n c e i s u s e d f o r synch ronous m o t o rs . In s t a b i l i ty s t u d i e s , wh ere t h e problem i s t o d e t e r m i n e whether a fa u l t w i l l c a u se a m a c h i n e to l os e syn c h ro n ism

reactances apply,

with

t he

rest of t h e system if the fa u l t is r e m oved a ft e r a cert a i n t i m e i n t e rv a l , t r a n s i e n t

10.2

I NTERNAL VO LTAGES O F LOAD E D

MAC HINES U N D E R FAU LT C O N D ITIONS Lc t

li S

co n s i d e r a g e n e r a t o r t h a l i s l o a d e d w h e n a fa u l t occ u rs . F i g ur e I O . 2 ( a ) is

I n t e r n a l \'ol t a ges a n d r e £.l c t a n c es o r t h e g e n e ra to r a re n ow i d e n t i fi e d by t h e t h e c q u i v ,l i c l1 l c i rc u i t or
has

,\

b a l a nced t h r e c - p h a s e l o a d .

s u b s c r i p t g s i n c e some o f t h e c i rcu i t s to be c on s i d e red a l so h ave m o tors. Exte r n a l i m p e d a n c e is shown b e tw e e n the g e n erator t e rm i n a l s a n d the p o i n t

w h e r e t h e fa u l t occurs. The c u r re n t f1 0w i n g b e fore t h e fau l t o c c u r s a t po i n t IL , the vo l tage at t h e fa u l t i s

Vf '

P

P is

a n d the term i n a l vo l t age of the g e n e ra t o r i s V; .

T h e s t e a dy- s t a t e e q uiva i c n t c i rc u i t o f t h e synchronous g e n e r a t o r i s i t s n o - l o a d vo l t a g e E � i n s e r i e s with i ts sy n c h ro n o u s r e a c t a nce

Xdg .

If a thre�-phase fault

384

C HAPTER 10

SYMMETRICAL FAU LTS

Zext

p "

jXdg

r

V(

(a)

L

FIG U RE 10.2

Z ext

p

+

'X" J de ZL

S

E g" (b)

V(

L

Equiva lent c i rc u i t for a generator s u pplying a b a l anced t h ree-ph ase l o a d . A p p l i c a t ion of

a

Zt.,

t h rcc­

p h ase fa u l t a t J> is s i m u lated by clos ing :iw i t cl! S : (a) usual s t eady-s t a l e g e n e r a t o r e q u i va l e n t c i rcu i t

wi t h load;

( b ) c i rc u i t for calcu lation o f 1" .

occurs at point P, we see t hat a short circuit from P to neutral i n the equivalent circuit does n o t satisfy the conditions for calculating subtransient current, for the reactance of the generator m ust be X�g i f we are calculating subtransient current 1" or X�g if we are calculating transient current J' . The circuit shown in Fig. lO.2( b ) g ives us the desired result. Here a voltage E� i n series with X�g supplies the steady-state current fL when swi tch S is open and supplies the current to the short circuit through X�g and Zext when switch S is closed. If we can determine E� , the current through X�g will be 1" . With switch S open, we see that ( 10.3)

and this equation defines E� , which is cal1ed the subtransient internal voltage. Simi larly, when calculating transient current I ' , which must be su pplied t h rough the transient reactance X:fll , the d rivi ng vo l tage is the transient internal voltage E� , where ( l O A)

Thus, the value of the load current iL d etermines the values of the vol tages E� and E� , w hi ch are both equal to the no-load voltage Eg only when IL is zero so that Eg is t hen equal to v, . At t h is point it is important to note that the particular value of E� in series with X�g represents the generator imme d iately before and immedi ately after the fau lt occurs only if the prefa u l t current in the generator has the correspondi n g value of fL ' On the other hand, Eg i n series with the syn­ chronous reactance Xd8 is the equivalent circuit of the machine under steady­ state condit ions for a ny value of the load current. The magnitude of 1;g is determined by the field current of the machine, and so for a different value of

·

IL

E�

1 0. 2

I NTERNAL V O LTA G ES OF LOA D E D M A C H I N ES U N D E R F A U LT CO N D ITI O NS

i n the c i rc u i t of Fig.

1 0.2( a )

l E/: I

385

wou l d rem a i n t h e same b u t a new v a l u e

wo u l d be req u i r e d .

of

S y n c h r o n o u s motors h ave r e a c t a n c e s o f t h e same type a s g e n e r ators.

W h e n a motor is short-c i r c u i t e d , it n o l o n g e r r e c e ives e l e ctric e n e rgy fro m t h e

p o w e r l in e , b u t i t s field r e m a i n s e n ergized a n d the i n e r t i a o f i t s r o t or a nd c o n n ec t e d l o a d k eeps i t r o t a t i n g for a s h o r t p e r i o d of t i m e . The i n t e r n a l v o l t age o f a sy n c h ro nous motor c a u ses it to co n t ri b u te c u rr e n t to the sys t e m , for it i s t h e n a c t i ng l i ke a g e n e r a t o r . By c o m p a r i s o n w i t h t h e c o rrespon d i n g form u l a s for

E;�

a n d t r a n s i e n t i nt e r n a l vol t age

" IL f/I - J'X dm

( 10 .5)

for a synchronous motor are g i v e n by

a g e n e r a t o r t h e s u b t ra n s i e n t i nt e r n a l vo l t a g e

E;"

£"! I I

l·� '

� I II

- f/I - }'X dill ' --

/I.

( 1 0 .6)

w h e re VI i s now the t e rm i n a l v o l t a g e o f the m o t o r . Fau l t c u rr e n t s in systems

by c a l c u l c H i ng t h e s u b t r a n s i e n t ( o r tra n s i e n t ) i n t e r n a l vol t a g e s o f t h e

c o n t a i n i n g g e n e r a t o rs and m o t o rs u n d e r l o a d m a y be s o h' ed in e i t h e r o n e of two

ways :

(1)

machi nes or

(2)

b y u s i n g T h eve n i n ' s t h e o r e m .

A

s i m p l e e xa m p l e w i l l i l l u s t r a t e

t h e two a p p ro a c h e s . S u p p o s e t h a t a s y n c h r o n o u s g e n e r a t o r i s c o n n e c t e d t o a s y n c h ro nous

m o t o r by a l i ne o f e x t e r n a l i m p e d a nce Zexl ' Th e m o t o r is d ra w i n g l o a d c u r r e n t f r o m t h e g e n e rator w h e n a s ym m e t r i c a l t h re e - p h ase fa u l t o c c u r s a t the m otor

Ii"

t e r m i n a ls. Figure

1 0.3

s h ows the e q u i v a l e n t c i r c u i t s and c u rr e n t fl ows of the

syst e m immed i a t e l y b e fo r e a n d i m m e d i a t e l y a ft e r the fau l t occurs. By r e p l a c ing the

synchronous

s h o w n in Fig.

r e a c t a n c e s of the m a c h i n e s by t h e i r

1O.3( a),

r e a c t a n c e s as

sub transient

we c a n c a l c u l a t e the s u b t r a n s i e n t i n t e r n a l v o l t a g e s o f the

m a. c h i n e i m m ed i a t e l y be fo r e t h e f a u l t o c c u r s by s u bs t i t u t in g the v a l u es of

( j O. 20)

+

p

(jO. l0) 1" (jO.20)

-----+-

( j O.20)

E"8

Neutral

( a ) Before the fau lt

g

jXd m

E"In

En

e

VI

p

--

1"m

p;

(jO.20)

Neut ral

jXdm E" m

(b) After the fault

FIG URE 10.3 E q u i v a l e n t c i rc u i t s a n d c u r r e n t flows b e fore a n d a ft e r a fau l t at t h e t e r m i n a l s of a synchronous motor con n e c t e d to a synchro n o u s g e n e r a t o r by l i n e i m p e d a nce 20,1 ' N u m erical va l u e s are for Exa m p l e 1 0 . 1 .

386

and

CHAPTER 10

IL

SYMMETRICAL FAULTS

i n the equations

( 10 .7) E" 'm

=

V1 - )'X"dm 1L

( 10 .8)

When t h e faul t is on the system, as s hown In Fig. 1 0 3C b ) the subtransie nt currents I; out of the generator and I;:, out of the m otor arc fou n d from the rel ations .

J�"

/"

+ !I .

E" =

==

m

-

--� ' --

'Xtil' " L --' e x l -I- J

" Enr

_ _ _

_

,

( 1 0 .9)

( 10 . 10)

" J'Xdnz

These two currents add together to give the total symmetrical fault current 1; shown in Fig. 1 0.3(b). That is, I" 1

=

I" + I" 8 m

=

Z ext + J'Xdg "

/"gf

+

jXdm

( 10 . 1 1 ) '

["m!

where 1;1 and I; I are the respective contributions of the generator and motor to the fau l t current If. Note that the fa ult current does not i nclude the prefault (load) current. The alternative approach using Thcvc n i n 's theorem i s based on the observation that Eq. (10. 1 1 ) requires a knowledge of only VI ' the prefault voltage of the fault point, and the parameters of the network with the subtran­ sient reactances rep resenting the machines. Therefore, 1; and the additional currents produced throughout the network by the fault can be found simply by applying voltage VI to the fault point P i n the dead subtransient network of the system, as shown i n Fig. 1 0.4( a). If we redraw t h a t network as shown in Fig. lO.4(b), it becomes clear that the symmetrical values of the subtransient fault curren ts c a n be found from t h e Thevenin equiva lent circuit of the subtransient network a t the fau l t point. The Thevenin equivalent circu i t is . a single generator and a single impedance terminating at the point of application of the fault. The equivalent g enerator has an internal volt age equal to VI ' the voltage at the fault point before the fault occurs. The impedance is that measured at the point of application of the fault looking back into the circui t with all the generated voltages short-circuited. Subtransient r eactances are used since the initial sym-

1 0.2

I NTERNAL VOLTAG E S OF LOADED MACHINES U N D E R FAULT CON DITIONS

Z ext

I"g (

-

P

"g J'X d

I"m (

-

jX d m

I If

Vr

(a)

C i rc u i t s i l l u s t ra t i ng t h e a d d i t i o n a l

+

Vr

+

FI G L R E 1 0.4 dead

'X" J dg

Z ext

g(

jXd m

-

[If

[If

(b) c u r re

nt

( /J )

! l ows d u e: to t h e t h re e - p h a s e fa u l t at

T h c \'t� n i n e: q u iv ;l I e n t loo k i n g i n t o t h e

m e t r i c a l f{l u l t c u r r e n t i s d e s i re d . ) n f i g .

I O. 4C h )

m(

/If

P: ( a ) a pplying VI to circu i t at po i n t P .

the T h c v e n i n i mp e d a n c e

. X' '' X" ) J. dm ( 2 " X I -+- ] dli

Zth

P

-

r ---

n e t w o r k t o s i m lJ l a t e t h e fa lI l t :

387

Z th

is

( 1 0 . 1 2)

U p o n t h e occu rrence of a t h re e - p h a se s h o r t c i rcu i t a t P , s i m u l a t e d by c l o s ing sw i t c h S , t h e s u b t ra n s i e n t c u rre n t i n t h e fau l t is

Vf [ ZCXI + j( X;;R + X;;m ) ] jX:;m ( Z exI + jX�g )

I"f

( 1 0 . 13)

T h u s , t h re e - p h a s e s ym m e t r i c a l fa u l ts o n systems containing g e n e r a t o r s a n d

vol t ag e s o r Theven i n ' s t h e o r e m , a s i l l u s t r a t e d i n the fo l l ow i n g exa m p l e s .

m o to rs u n d e r l o a d m a y b e a n a l y z e d b y e i t h e r the use of s u b t ra n s i e n t i n t e rn a l

Exa m p l e 1 0 . 1 . A sy n ch ro n o u s ge n e ra t o r a n d m o t o r a re ra t e d 3 0,000 k Y A , 1 3 . 2 kY,

a n d bo t h h ave

kW

s u b t r;\ I1 s i e n t re a c t a nc e s

o f 20%. T h e l i n e co n n e c t i ng t h e m has a

r e a c t a n ce o f 1 0 So o n t h e b a s e o f t h e m a c h i n e ra t i ngs. T h e m o t o r is d raw i n g 2 0,000 at

0 . 0 powe r - f a c t or

s Y ll l l l 1 c l r i c t i l l m: c - p i 1 : t S C f; 1 \ I I I n1 ; 1 c h i n e s .

c u rre n l s

the

kY w h e n a O C C l \ I'S , I I l i l e J l l o t o r l eI' ll 1 i l1 : t i s . r i n d t h e s u b t r ll n s i e n t Ill t l i O f , , l l l l l t h e ra l l l t h y l I s i t 1 g l he i n l e r n a l v o l t a ges o f

l e a cl i n g a n d

i n t i l l' g e n e f' : l i ( lI . t he

,\

l e rm i n ;\ 1 vo l t a ge of 1 2 . 8

SO/Illion. T h e p r c fa u l t e q u i va l e n l c i rcu i t o f t h e sys l e m curresponds t o Fig.

C hoo s i n g a base o f 3 0 , 0 0 0 k Y A , 1 3 . 2 k Y a n d u s i n g t h e v o l t age a s t h e r e ference p h asor, w e o b t a i n 1 2 .8 V = -

f

1 3 .2

0 . 970

L..2.:

per unit

Vi

10.3(a).

a t t h e fa u l t poi n t

388

CHAPTER 10

S Y MMETRICAL FAU LTS

B ase current

IL

=

=

=

30,000 13 X 1 3.2

=

1312 A

20,000/ 36.90 0.8 X fS3 X 12.8

O .�() ( O .K

+

jO J) )

=

1 1 28

O N)

=

/ 36 .9 A �---0

+

jO .52 p e r u n i t

For the g enerator

�

E; Ig"

0 .970

=

0.918

=

0 .8 14

= -

=

+ +

jO . l (0 .69 jO.069

jO.S2)

jO .2(0 .69

jO.207 = 0.69 jO.3 +

1312(0.69

Fv.

+

+

-

j2.71)

-

+

0.918

-

+

j O .52)

}· 2 . 7 1 p e r

90 5

=

=

jO.069 p e r u n i t

= 0.814 +

jO .207

per u n i t

unit

j3SS0 A

"r

�

£;;,

=

�

.

0 .970

=

+

/ 0° per u nit ') . 6 9

J

= 1 .074 - jO. l 38 1" '"

=

=

1 .074 - jO . 1 38 jO .2

+

jO .S2)

= 0 . 970

- -

jO. I 3R

+ 0 . 1 04 per u n i t

1--

- 0 .69

1 3I2( - 0.69 - is .37)

=

-

uni t

-90 5 - j7'v_

In the fault

If

= I; + I� =

-

j8.0S

=

X

0.69 - j2.7 1 - 0 .69 - jS .37 1312

=

-

jl0,600 A

Figure IO . 3(b) s hows the paths o f 1:, I�" an d 1[.

7=

-j8 .u,

10.2

I NTER N A L

E xample 1 0 .2. Solution.

VOLTAGES OF LOA D ED M ACH I N ES U N D ER F A ULT CO N DI TION S

389

Solve Example 10. 1 by the use of Thcvenin's t heorem .

The Thevenin equivalent circuit corresponds to Fig. 10.4.

Zth = VJ

In the fau l t 1 '/ f -

_

=

VJ

_

= . 1 2 per u nit iO

i O . 3 X iO . 2 ' ' J 0 .3 + J 0 .2 0 .970

L.Q:

per unit

0 . 97 + i O

Zlh

-i8 . OS

iO . 1 2

per u n i t

t h e p a r a l l e l c i rc u i ts o f t h e m ac h i n e s i n ve r s e l y s i m p l c current divis ion we o b t a i n t h c fa u l t c l l r re n t s

T h i s b u l t c u rre n t i s c1i \ i d e d b e t w ee n �\S

t h c : � i m p c ( Li n c c s By

From gener a tor: 1;r

From motor: 1�'r

=

=

-

is . 08

iO.2 X

- i3 .23 p e r

-

iO.S

iO.3

- i 8 .08

X

-i 4 .SS

-­

j O .S

N e g l e c t i n g load c u r r e n t g ives

Fau lt curren t from generator Fau l t current fro m motor

Curre nt i n fault

=

3 .23

X

131 2

=

4 .8S

x

1 3 12

=

8 .08 x 1 3 1 2

=

unit

p e r unit

4240 A 6 360 A

=

1 0 , 600

=

A

The current i n t he fa ult is the same whether or not load current is considered, but the currents in the l i nes differ. When load curre nt I L is i ncluded, we fi nd from Exa m p l e 1 0. 1 that I:

1;:,

=-

=

':1 +

1::,!

-

fL

11.

= =

-

i3 . 23 + 0 .69 + iO .5 ?

- i4 .85 - 0 . 69 - i O . 5 2

= =

0 . 69 - i 2 .7 1 p e r u n i t -

0 . 69

-

is .37 per unit

Note that I/. is i n the sa m e d ir ect i o n a s I ; b u t opposite to ,;: . T h e p e r un i t v a l u e s Ii , 1; , and 1;;, a r e t h e s a m e as i n Exa m p l e 1 0 . 1 , a n d s o t h e am pere values w i l l also be the s ame, ,

fo u n d fo r

Fault current from generator Fault curre nt from motor

= =

1 905 - j3SS0 1

1

-

=

905 - j70S0 1

-

3 600 A =

72 0 0

A

The sum of the magnitudes of the generator and motor c urrents does not e q ual t h e fa u l t current because the currents from t h e g ene r a t or and motor ar � not i n phase when load current i s included.

390

CHAPTER 10

SYMMETRICAL FAULTS

Usually, load current is omitted in determining t h e current in each l i ne upon occurrence o f a fault. I n the Thevenin method n eglect of load current means that the prefaul t current in each l ine is not added to the component of current flowing toward the fault in the l ine. The method of Example 1 0 . 1 neglects load current i f the subtransient internal voltages o f all machines are assumed equal to the voltage VI at the fau l t before the fault occurs, for such is the case if no current flows anywhere in the network prior to the fau l t . Resistances, charging capacitances, a n d off-nominal tap-changing o f transform­ ers are also usual ly omitted in fault studies since they are not l ikely to influence the level of fault current significantly. Ca l cu l a t i on o f th e fa u l t c u r re n t s i s thereby simplified s i nce t h e n etwork m o d e l b e c o m e s an in t e r co n n e c t i o n of inductive reactances and all currents throughout t he fa u l t e d sys t e m are t h e n I n phase, as demonstrated in E x a m p l e 1 0.2.

10.3

FAULT CALCULATIONS USING Z bus

Our discussion of fault calculations has been confined to s i m p l e c i rc u i t s , b u t now w e extend o u r study to general networks. W e proceed t o t h e general equations by starting with a specific network with which we are already familiar. In the circu it of Fig. 7.4 if the reactances i n series with the genera ted vol tages are ch anged from synchronous to subtransient values, and if the generated ' volt ages become subtransient interna l vol tages, we have the network show n i n Fig. 1 0.5. This n etwork can be regarde d a s the · per-phase equivalent of a balanced three-phase system. I f we choose to study a fau l t a t bus (1) , for example, we can follow the notation of Sec. 1 0.2 and designate Vf as the actual voltage at bus 0 before the fault bccurs. +

® -'--11---

--I-.L...

G) FI GURE 10.5

E"a

+

+

E'b

React a n ce d i agram obta ined from Fig. 7.4 by subst i t u t i ng subtransient values for sync h ronous re acta n!;:es and synchronous i n ternal voltages o f t he m a ch ines. Reactance values are m arked in per u n i t .

1 0. :;

®

FAU LT CALCULATIONS U S I N G

@

CD

+

+

'l E �u

Vi -

Vi F I G U R E 10.6 C i r c u i t o f Figure 1 0.5 w i t h a

E,:

+

A t h r e e - p h ase fa u l t a t b u s

w h e r e t h e sou rce vo l t a g e s

Vf

@

Cl n d

fa li l t

t i 1 r � e - p h a s t:

, i ll 1 l 1 i ;t t l: d by

on

V and I

bus -

V f

S t.:r I l: S .

i s s i m u l a t e d b y the network of Fig. -

391

Z b us

Vf

(1)

In

1 0 . 6,

i n s e r i e s con s t i t u t e a s h o r t - c i rc u i t

b ra n c h . S o u r ce v o l t a g e Vr a c t i n g a l o n e i n t h i s branch wo u l d m a t c h t h e p r e fa u l t

(1) ,

v o l t a g e a l re a d y a t b u s b ranch . W i t h

Vf

a n d t h e re fore w o u l d n o t c a u s e c u r r e n t t o fl o w i n t h e

a n d - Vf i n s e r i e s , t h e b r a n c h becomes a s h o r t c i rc u i t a n d t h e

Ii

branch current is

t h e 8. d d i t i o n of t h e -

a s s h own .

VI

It

source. The current

Ii

sys t e m from t h e refe re n ce n o d e b e fo r e fl o w i n g

E� , E;; ,

a c t a l o n e and - Ii in fO b u s

W

fro m e x t e r n a l sou rc e s . W i t h -

V;

Vf

and

is c a u s e d

by

d istributes i tsel f t h roughout the

out

of bus

W

t hrough t h e -

changes

so u rc e . I n d o i n g so, i t p r o d u ce s w h a t ev e r b u s vol t age sys t e m d u e t o t h e fa u l t . I f

Ii

i s evi d e n t , t h e r e fore, t h a t

t h a t occur i n t h e

a r e s h o r t -ci rcu i t e d , t h e n -

is t h e n t h e only c u r r e n t

Vf

entering

Vf

i s l e ft to

t h e n e tw o r k

a s t h e o n l y source, t h e n e t w o r k h a s t h e n o d al

i m p e d a n ce e q u a t i o n s i n t h e Z bl l S m a t r i x fo rm

6. V

I

6. V,

6 V, 6. V -1

6. V

t

- V J 6 V, � V.J

CD (1)

CD

CD

21t

.% t 2

Ln

2 :' 1

L 12

G) L , t

@

2 ,\

24 1

The p re fi x 6. i s chosen to i n d i c a t e t h e t o t h e c u rrent -

Ii

i nj ec t e d i n t o b u s

Zt

2

changes

W

G)

Z2

:-

�

L �l

Z . I :1

@

2 1 .1

2 2 '1

/ :\.1

7,44

() -

II

"

()

( l O . 1 4)

0

i n t h e vo l t a g e s a t t h e b u s e s d u e

b y t h e fa u l t .

Th e Z b l l s b u i l d i n g a l g o r i t h m , o r some o t h e r m e a n s such a s

Y bus

t r i a n g u l a r­

i z a t i o n a n d i n v e r s i o n , c a n be u s e d to eva l u a t e t h e b u s i m p e d a n c e m a t r ix for t h e ,

392

CHAPTER

1 0 SYMMETRICAL FAU LTS

network of Fig. 10.6. The numerical values of the elements of the matrix will be different from those in Example 7.6 because subtransient reactances are now being used for the synchronous machines. The changes in the bus vol tages due to - If a r e given by � V1

� V2

� V3

�

�

The second row of this

� V1

- VI

- Ij'

=

� V:I

� V4

e q u a t i o n s h ow s

l

Cul u m n ot

Z hll.\

21

- Z 1 2 If =

-L 22 I" /

- 2 .1 2 If - 2'1 2 1j'

( 10. 1 5)

t hat

( 1 0 . 1 6) We recognize 2 22 as the diagonal element of Z bl ls representing the Theveni n impedance o f t h e network a t bus @ . Substituting t h e expression for Ii into �q. 00.15) gives � Vl

� V2

� VJ � V4

-

,

-

-

2 12

2 22 J V f -

v

Z 32 V Z 22 f Z4 2 v Z 22 I

( 1 0 . 17)

-

-

When the genera to r vol tage V is s h o r t -ci rell i t e d in t h e n e tw o r k of Fig. 1 0.6 and the sources E�, E;; , and Vf Iare rei nserted i nto the network, the currents and vol tages everywhere in the network will be the same as those existing before the fault. By the principle of superposi t ion these prefault vo l t a ge s a d d t o the changes given by Eq. (10. 1 7) to yield the total vol tages existing after the faul t occurs. The faulted network is usually, but not a lways, assumed to be without load before the fault occurs. In t h e absen ce of loads, as remarked previously, no prefault currents flow and there are no voltage differences across t he branch impedances; all bus voltages throughout the network are then the same as Vf' the prefault voltage at the fault point. The assumption of no prefault current simplifies our work considerably, and by applying the principle of s uperpos it ion, -

lU.3

FAULT CALCULATIONS U S I N G

Z bu s

393

we obtain the bus voltages

VI

VI

6. V

V2

VI

6. V 2

+

1

V�

VI

6. V 3

V4

VI

6. V4

VI - 2 1 2 Ii VI - VI VI - 2 :12 I]' VI - 2 42 IJ

2 12 Z 22

1 0 =

VI

2 :n 1 2 22 2 42 1 2 22

( 1 0 . 1 8)

-

Thus, the vol tages at all buses of t h e ne twork can be c al c u lat e d using t h e prefa u l t vo ltage VI o f the fa u l t bus a n d the c l ements i n t h e co lumn of Z bus correspo n d i n g to t h e fa u l t bus. The calcu lated values of the bus vo l t a g e s w i l l y i e l d t h e subtransicnt cu rre n ts i n t h e b r a n c h e s o f t h e ne t w o rk i f t h e sys t e m Z bus h a s been fo rmed w i t h s u b t ra n s i e n t va l u es fo r the mac h i n e r e acta n c e s. In more ge neral terms, when the three-phase fa u l t occ urs on bus ® of a l arge-scale n e twork, we h ave V

- _I_ II" 2" "

( 10 . 1 9 )

a n d n e g lectin g prefau l t l o a d currents, we can t h e n wri te fo r t h e voltage a t a ny b u s 0 d ur i n g the fau l t

( 1 0 .20 ) w h ere 2j " a n d 2 u a re e l ements i n col umn k of the s y s t e m Z OllS . I f the p re fa ul t vo l t age of b u s (j) is not t h e same a s t h e prefa u l t vol tage of fau l t bus CD, t h e n we s i m p l y re p l a ce V o n t h e l e ft in Eq . ( I O . 2()) hy t h e a c t u a l prcfa u l t vo ltage of I bus (j) . Knowing the bus vol t ages d u ring the fau l t , we can calculate t he subtran s i e n t current I:�. from bus CD to bus (j) i n the l i n e of i mp edance 2b con n ecting t hose two buses,

( 1 0 .21 ) This e q u ation shows I:; as the fraction of t h e fau l t c u rr e nt Ii appearing as a line flow from b us CD to b u s (]) i n the fau l te d n e twork. I f b u s 0 is d i rectly con n e cted to the fa ulted bus CD by a l i n e of series i mpedance 2b, then the current contributed from bus (]) to the current i n the fau lt at b u s CD is s i m pl y Vj/ 2 b ' w here Vi i s g iv e n by Eq . 0 0 .20).

394

CHAPTER 10

SYMMETRICAL F AULTS

The d iscussion of this section s hows that only column k of Z bu s , which we now denote by Z b:�, is required to evaluate the impact on t h e sys t e m of a symmetrical three-phase fault at bus ® . I f n ecessary, the e l e m ents of Zb:� c a n b e generated from the triangular factors of Y bus ' as d emonstrated in Sec. 8.5 . Example

three-phase fau l t occurs a t bus W of the network of Fig. 10.5. Determine the initial symm etrical rms currc n t (that is, thc subt ra n s i e n t c u r r e n t ) i n the fault; the voltages a t buses CD , G) , a n d @ d u r i n g the fault; the current flow in the line from bus ® to bus CD ; and the cu rr e n t contributions to t he fault from 10.3. A

1 .0ZQ:. p e r u n i t

lines G) - W, CD - W,

e q u a l to

(l n o

@ - (1) .

Ta k e t h e p r da u l t vll l t ;\ge

a n o n e g l e c t ;1 1 1 p r da u l t c u rr e n t s .

Solution. Applying t h e Z hllS b uil d i n g a l gor i t h m t o f i g .

Z bus

(1)

CD

=

jO. 1 938 jO.22Q5 jO. 1 494 jO.1506

CD jO.2436

CD jO.1938

G) jO. 1 544 @ jO. 1 456

G)

1 0.5, we

jO. 1544 jO . 1494 jO. 1 954 jO . l 046

.� �

1 .0

-

Z22

=

1 .0 jO .2295

---

bus is

-j4 .3573 per unit

and [rom Eq. (10.18) t h e vo l ta g e s d ur i n g t h e fa u l t a r c VI

1 -

V2 =

V3

�

The current flow in

l i ne

11-

jO . 1 938 jO.2295 0 jO.1494

® - CD

jO .2295

jO.1506 jO.2295 is

0.3490 - 0 . 1 556 jO.25

0. 1556 o

per

unit

-jO.7736

pe r

0.3490

at bus

G)

that

jO . 1 456 jO .1 506 jO.1O-l6 jO. 1 954

Since l oa d currents are n eglecte d , the pr e fault voltage at each unit, the same as Vf at bus @ . W h e n the fault occu rs, [" = f

fi n el

T'J

0.3438 unit

l .Oft p e r

FAU LT CALCULATIONS USING Z bw EQUIVALENT CIRCUITS

I DA

Faul t currents contributed to bus From bus CD :

Exce p t ror

10.4

From bus

G) :

From bus

@:

395

W by the a dj a c e n t unfaulted buses are

Vj

0 . 1 55 6

Zbl VJ

0 .3490

Zb 3

jO .25

V4

Zb 4

round-off e rr o r s , the sum

-j l .2448 per unit

jO.125

=

0 .3438 =

of

jO .20

=

-j l .3960 per unit -jl .7 1 90 per unit

these current contributions equals If.

FAU LT CALC U LAT I O N S U S I N G Z blls

EQ U IVALE NT . C I R C U ITS

We

cannot devise a physica l l y re a l izable n e twork which d irec tl y i n c orpora te s all the i n d iv i d u a l e l e m e n t s of the bus i mpedance m a t r i x . However, Fig. 8.4 shows t h a t we can use the m a tr ix elements to construct the Thevenin equival ent circuit between any pair of b uses i n the n etwork that may be of interest. The Thevenin e q u ivale nt circuit is very helpfu l for illustrating the symmetrical fault equations, which are developed in Sec. 10.3. In the Theven i n e q u i val e nt c i rc u i t of Fig. lO.7( a ) b u s ® i s assumed to be the fau l t bus and bus (j) is u n f au lt ed . The i mpedances shown correspond directly to the eleme n ts of t h e n e twork Z bus and all the pre fa u l t bus voltages are the same as Vf of t he fau l t b u s if load currents a re neglected . The two points U nfaulted bus

U nfaulted bus }

Bus to be faulted

Z· · - Z k

v r = _ f

'-!... ""'"

II; (a)

(])

))�

� (])

Z · }· - Z · k

..r---- �� x

Faulted bus

® � s Zk k

�jli

(b)

a n d ® of sys t e m w i t h no p r e fa u l t 1 0
FI G U RE 1 0 . 7

Thev e n i n e q u ival e n t b e t w e e n b u s e s

CD

( a ) befor e

396

CHAPTER 10

SYMMETRICAL FAU LTS

marked x h ave the same potential, a n d so they can b e joined together to y i e ld the equivalent circuit of Fig. 1 0. 7 ( b ) w i th a single voltage source Vr as shown. If the switch S is open between bus ® a n d t h e reference node, there is n o short circuit and no current flows in any branch of the network. When S is closed to represen t the fault on bus ® , c urre n t flows in the circuit toward bus ® . This current is If = VrIZk k ' which agrees w i t h Eq . C 1 0. 1 9), and it i n d uces a voltage d rop ( Zjk l Zk k )Vr in the direclion ji'o/11 t he r e fe r e n c e node towa rd b u s CD . The voltage from bus (]) to the reference changes therefore by the amou n t - (Zjk IZ kk ) Vf s o t h a t t h e voltage a t b u s (]) d u ring the fau l t is VI (Zj k IZkk ) Vr , which is co ns i s t ent with Eq. ( 1 0 . 20). Thus, by substituting appropriate n u m e r i c a l v a l u e s for the i m p e d a n ces i n the simple e quivalent circ uit of Fig. 10.7( b ), we can calculate the bus vol tages of the system before and after the fau l t occurs. With swi tch S open in the circu i t , the voltages at b u s ® a n d the represe n tative bus (J) are equal to Vf . The s a m e u niform voltage profile occurs i n Fig. 1 0.6 i f there a r e n o prefa u l t curren ts so that E� and E� equal Vf ' If S is closed in Fig. 1 0.7(b), the circuit reflects t h e voltage of representative b u s CD w i t h respect to reference while the fau l t i s o n bus ® . Therefore, i f a three-ph as e short-circu i t fau l t occurs a t bus ® of a large-scale network, we can cal culate the c urrent i n the fau l t and the vol tage at any of the u n faulted buses simply by i nserting the proper impedance values i n to elementary circuits like t hose i n Fig. 10.7. The fol lowi ng example i l l ustrates t h e procedure.

10.4. A five-bus network has generators at buses CD and G) rated 270 and 225 MVA, respectively. The generator subtransient reactances plus the reactances of the transformers con necting them to the buses are each 0 . 3 0 per u n i t on t h e generator rating a s base. T h e t u rn s ratios o f the t ransformers are such that the voltage base in each generator circuit is e qual to t h e v o l t age rating of the generator. Line impedances in per u n it o n a 1 00-MVA system base are shown in Fig. 1 0.8. Al l resistances are neglected. U sing the bus impedance m a t r i x for the network which includes t h e g e n e r a l o r a n d t r a n s fo rm e r re a c t , I I 1 C C S , fi n d t h e

Example

1.0�

®

)0. 168 jO. 126

®

)0. 126

) 0.210

®

1 . 0�

FIGURE 10.8 Impedance diagram for Example 1 0.4. G en erator react ances i nc l u d e subtransient value6 p l u s reactances of s et-up tra nsformers. A l l values i n p e r u n i t on 3 1 00-MYA b a s e.

\ 0 .4

FAULT CALCULATIONS USING

Z bus

EQUI VALENT CIRCUITS

397

subtransient current in a t hree-phase fault at bus @ and the current coming to the faulted bus over each line. Prefault current is to be neglected and all voltages are assumed to be 1 .0 p er unit before the fault occurs. Converted to the lOO-MVA base, the combined generator and transformer reactances are

Solution.

Generator at bus

CD :

X

Generator a t bus

G) :

X

100

=

0 .30

X

-

=

0 .30

X

-

270

=

0.1 111

per unit

=

0 . 1 333

per unit

100 225

These valu es, a long with the line impedances, are marked in per unit in Fig. 10.8 from which the bus i mped an ce matrix can be determined by the Z bu s building algorithm to yic ld

Z bus

=

CD W G) @ G)

CD

W

j O .0793

W

jO .0558

0)

jO.0382

C1)

jO . 05 1 1

j O.0608

j O .0558

j O . 1 338

j O . 0664

j O .0630

j O .0605

jO .0382

j O .0664

jO .0875

jO .0720

jO .0603

j O .05 1 1

j 0 .0630

jO .0720

jO .232l

jO . 1 002

j O . 0 608

j O .0605

jO .0603

jO . 1 002

jO. 130l

Since we are to calcu late the currents from buses

G) and G) i nto the fault at bus Unfaulted bus

U nfaulted b us

Z 33 - Z 34 ®

V(

_.

l . O�O°

+

.

j O.0720 If

Z44� S

=

j O . 1601

-j4 .308

@

Vr

=

l.oLQ:,

Z4 5

=

_

Z54 jO.l 002

If

--

G).

U s e of Theve n i n e q u i v a l e n t c i rc u i t s to c a l c u l a t e voltages at

at b u s

(a)

-

Z5'

+

ea) Fl G U RE 1 0 . 9

®

-;.02�9 I Z 1)1)

bus

G)

(b) and

=

jO.1319 -j4.30B

s

--

(b) b u s

@

d u e to fault

398

CHAPTER 10 SYMMETRICAL FAULTS

need to know V3 and Vs during the fault. Visualizing equivalent circui ts like those of Fig. 10.9 helps in finding the desired currents and voltages. The subtransient current in the three-phase faul t at bus 8) can be calculated from Fig. 1 O.9(a). Simply closing switch S gives

0 , we

1 .0

V _ IfII - I Z44 _

From

j O.2321

Fig. 1O.9( a ) the voltage at bus

VJ

From Fig.

=

Vf - IlZJ4

1 O.9( b )

=

1 .0

-

G)

=

-j4 .308

per u n i t

during the fault is

( - j4 . 3 08 ) ( J0 .0720)

t he voltage a t bus G)

Vs = Vf - fl Zs4

=

@

0 . 6891-: p e r u n i t

d u r i ng t h e fa u l t is

1 . 0 - ( -j4 .3(8) ( j0 . 1 002)

Currents i nto the fault a t bus

=

=

0 .568 3 p e r u n i t

over t h e line impedances Zb are

From bus

@:

-

From bus

G) :

-

VJ Z b3 Vs

Zb S

0 . 6898 =

jO.336 0 .5683

=

jO .252

Hence , total fault current at bus

@

=

=

-j2 . 05 3

per u n i t

- j 2 .255

per unit

- j4 .308 per u n i t

Other equivalent circuits based on t h e given bus impedance matrix c a n b e developed for three-phase faults o n a n y o f the other buses o r transmission lines of the system. A specific application w i l l demonstrate how this is accomplished. Three-p hase fau lts occur more often on transmission l ines than on substa­ tion buses because of the greater exposure of the l ines to storms and acciden tal disturbances. To analyze a line fault, the point of fault on the l i ne can be assigned a new bus n umber and Z b u s for the normal configuration of the network can then be modified to accommodate the new bus. Sometimes t he circuit breakers at the two ends of the l i n e do not open simultaneously when a line fault is being cleared. If only one circ u i t breaker has opened and the fau l t is not fully c leared, shor t-circuit curre n t persi sts . T he so-ca l l e d line-end fa ult represents the p articu l a r situation where the three-phase fault occurs very close to one of the terminating buses of the l i n e , on the line side of the first breaker (nea r the fault) to open. The line breaker near the fault is called the near-end breaker and that at the end away from the faul t is called the remme-end breaker . The single-line d iagram of Fig. 1 0 . 1 0 shows a four-bus network with a line-end fau lt a t point P on the line con necting b uses CD and 0 The line has series impedance Zb o The near-en d breaker at bus @ is open a n sI the remote-end breaker is closed, leaving the fault still on at point P , which we now °

1 0 .4

CD

®

@ o

F I GLRE 1 0 . 1 0 L i n � · e nd fa u l t Fig

399

FAU LT C A LCULATIONS U S I N G Z buS EQU I VA L E N T C I RCUITS

at

I O.H.

point

P

on l i n e o f s e r i e s i m pe d ;tnce

-

Closed breaker

-

Open breaker

Z" b e t w e e n

b J se s

CD

nnd

G)

of sys t e m of

call bus ® . I n order to study t hi s fa u l t con d i t ion, we need to modify the existing b us impedance matrix Z orig for the normal con figuration of the system t o reflect the near-end breaker o p e r ation. This is accomplished in two steps: 1.

2.

Establish the new b us ® b y a d d in g a line of series i mped ance Z b b etwe e n bus CD a n d bus ®. Remove t h e line between b u s CD a n d b u s W b y add ing l i n e impe da nce Zb b etween t hose two buses i n t h e manner explained in Sec. 8.4. -

The first step fol lows the proce d u r e given for Case 2 in Table 8.1 a n d y i e l ds, i n terms o f the e lements Z ij o f Z orig , t h e first five rows and columns o f t h e

®

ZII

221 .% � I

Z4 1

211

+ Zb

( Z I I - 22 1 )

Z21

®

-

Z I I - 2 12 .1..1 1

-

-

222 2:12

24 1 - Z42 ZII

-

Z I I1 . 1 2

2 12

-

2b -

( 1 0 .22)

whe r e Z th. 1 2

=

Z I I + Z22 - 2 Z 1 2 w h e n

be accom p l i s h e d

by

fo r m i n g

Z ori g is sym met ric. The second step can row ® a mI co l u m n ® as shown a n,d then Kron

400

CHAYTER 10

SYMMETRICAL FAULTS

reducing the matrix Z to obtain the n ew 5 X 5 matrix Z bu s . new including bus ® , as explained for Case 4 of Table 8.1. However, since 2 k , n ew is t he only element required to calculate the current in the fault at bus ( point P of Fig. 1 0 . 1 0), we can save work by observing from Eq. 00.22) that the Kron reduction form glves

®

( 1 0 .23)

LQ:

Again, w e n o t e t h a t 212 2 2 1 a n d 2(h, 1 2 = 2 I I + 2 22 - 2 Z 1 2 ' By neglecting prefault currents and assigning pre fault voltage VI = 1 .0 per unit to the fault point P, we find the line-end fau l t cu rrent 1; out of b u s ® as fol l ows: =

I" f

- 1 .0 - 2 k k , ne w

1 .0

=

Z b - ( 2 1 1 - Z 21 ) 2 / ( Zth

- Zb )

------

2\ !

+

12

( 1 0 .24)

Thus, the only el em e n ts of Z orig entering into the calculation of I; are 2 1 1 , Z 12 = Z 21' and Z22 ' It is worthwhile observing that the same equation for the line-end faul t current c a n be found directly by inspection of Fig. 1 0 . l l (a ), which shows t h e Theveni n equivalent circuit between buses CD a n d W o f the prefa u l t network. The impedances Zb and - 2b are con nected as shown in accordance with steps 1 and 2 above. Circuit analysis then shows in a straightforward manner that the imp ed a n ce looking back into the circu it from the terminals of the open switch S is

( 2 1 1 - 2 1 2 ) ( 222 - 22 1 - 2b ) + 2 12 2 k k , new = 2b + 2 1 1 2 1 2 + 2 22 2 21 2 b _

_

( 1 0 .25 )

_

S ince 2 1 2 = 2 2 1 and 2(h 1 2 = 2 1 1 + 2 22 - 2 2 12 , Eq. 00.25) can be reduced to gIve

( 1 0 .26 ) Thus, simply by closing switch S as shown in Fig. 10. 1 1 ( b ) and using elementary circuit analysis, we can ca l cu l a te the line-end faul t current I; in agreement with Eq. (10.24). Of course, the circuit approach using the Thevenin equivalent , m ust yield th e same results as the matrix manipulations of Eq. 00.22)-for the same

1 0.4

FA U LT CA LCU LATIONS U S I N G :Z bU,

EQUIVALENT CIRCUITS

®

401

s

p -z

(a)

�

--' --- -----

ZH , ,, � w

=

Zi t

+

Zb

L--____

V{

=

1.0iQ:.

� -__ .:.. _ Z 21-.)_ _ l l_ _ Z � -

ZLh , 12 - Zb

___

® L I J\' ! 1 t

S

P

(b) Fl G U RE 1 0. 1 1

S i m u l ating t h e l i n e - e n d fau l t o f Fig. 1 0. 1 0 by Theve n i n equivalent c i rcu i t : ( a ) w i t h l i n e open before t h e fau l t ; ( I» d u ri n g t h e fau l t ( S close d),

CD - (1)

external connections a rc being m a d e to the l a rger system model a s to its Theve n i n equiva l e nt. Other uses of the equ ivalent circuits based on the bus impedance matrix are poss i b l e . Exa m p l e 1 0 . 5 . In the five-bus system o f F i g . 1 0.8 a l i ne-end, short-circu i t fau l t occurs on I inc CD - W , on the line side of the breaker at bus W . Neglecting p rc fa u l t c u r re n t s a n d assu m i ng r a t e d sys t e m vo l t age a t the fault poi n t , calculate the subtransicnt current into the faul t when only the ncar-end breaker a t bus @ opens.

Figu re 1 0.8 shows that the imped ance of line J) - (l) is Z b jO. 1 68 per unit a nd the required elem ents of Z hus a re given in Example 10.4. The Thevenin equivalent circuit looking into t h e intact system between buses ,CD a n d @

Solution.

=

402

CHAPTER 10 SYMMETRI CAL FAU LTS

corresponds to Fig. 1O.1 1(a). The numerical values of the im p e dan c e s shown i n parallel are calculated a s follows: Z11

222 - 2 2 1 - 2b

=

=

jO .0793 - jO .0558

=

jO .0235

jO. 1338 - jO.0558 - j0. 1 68

=

-

jO .09

The new Thcvenin impedance seen looking back i n t o t h e fa u l t e d system between the fault point P and the reference is therefore g iven by Eq . ( 1 0.25) as .

2H , n cw = 1 0 . 1 68 + =

Thus,

( jO.0235) ( -jO . ( 9 ) ( j0 .0235 jO .09) _

+

j O.255() per u n i t

jO .0558.

l he s u b t ra l l s i c l l t c u r re l l t i l l t o t l l c l i l l c -c n J h u l t is

= 1; jO . 255 6 1

=

- j3.9 1 2 per uni t

10.5 THE SELECTION OF CIRCUIT BREAKERS

The electric u ti lity company furnishes data to a customer who must determine the fault current in order to specify circu i t breakers p roperly for an i nd ustrial plant or industrial power distrib ution system connected to the u ti l i ty system at a certain point. I nstead of providing the Theve n i n i mped ance of the system at the

point of connection, usually the power company i nforms the customer of the short-circui t megavoltamperes which can be expected at nominal voltage; that IS,

S hort-ci rcuit

MYA

=

13 X ( no m i n a l kY) X I isc i

X

1 0 -- 3

( 1 0 .27)

where I Isc i in amperes is the r ms magni t u de of the short-ci rcu i t current i n a three-phase fault at the connection point. Base megavoltamperes are rel ated to base kil ovolts and base amperes I/basc i b y Base

MYA

=

13 X ( base kY)

1 1 baSe I X 1 0 - 3

( 10 .2 8 )

l I se I in per unit

( 1 0 .29)

x

If base kil ovol ts equ al nominal kilovol ts, t h e n divi d i ng Eq. 00.27) by Eq. 0 0.28) converts the former to per u n i t , and we obtain Short-circuit

MY A in per u ni t

=

�

At nomi nal vo l ta g e the Thevenin equivalent circuit l ooking back into the system from the point of connection is an emf of 1 .0 per unit in series with the

per-u n i t impedance

Zth. Therefo re , 1 .0 -- per u n i t l Is e !

1 0.5

THE SELECTION OF CIRCUIT BREAKERS

403

u n d e r short-circuit cond itions,

1 .0 =

short-circuit

MVA

per u n i t

( 1 0 .30)

O fte n resistance and shunt capacitance are n eglected, i n which case Zth = Xth . Thus, by specifying short-circu it megavoltamperes a t the customer's bus, the e l ectric u t i l i ty is effectively describing the short-ci rcu i t curre n t at nominal vol tage and the reciprocal of the Th e venin impedance of the system at the point of connection. M uch study has been g iven to ci rcu i t-b reake r ratings a n d applications, a n d our d iscuss ion h e re gives so me i n troduct ion to t h e subject. T h e p resentat ion is n o t i n t en d e d Zl S a s t u cl y of b r e a k e r a p p l i ca t i o n s b u t , ra t h e r, to i n d i c a t e t h e i m p o r t an c e o f u n d e rs t a n d i n g fa u l t c Zl l c u l a t i on s . For additional gui dance in s p e c i fy i n g b re a k e rs t h e (T i l ( i e r s h o u l d c ( ) l l s u l t t h e A N S I p u b l i c a t i o ns listed in t h e footnot es w h i c h a cc o m p a n y t h i s s e c t i on . From t h e current viewpoi n t two factors to be considered i n selecting c i rc u i t b re a k e rs a r e : •

•

The maximum instantaneous cu rr e n t which the breaker m ust carry ( withstand ) and Th e total cu rrent when the breake r contacts part to in terrupt the circui t.

Up t o this point we h ave devote d most o f our attention to t h e subtransient current call ed t h e initial symmetrical current , which does not i nclude t h e d e component. Inclusion of the dc com ponent res u l ts i n a rms val u e of c u rrent immediately aft e r the fau l t, which is h igher t h a n the subtransient current . For oi l circu i t b reakers above 5 kV t he subtran s i e nt c u rrent m u l tipl ied by 1 .6 is considered to be the rms v a l ue of the current whose d isruptive forces the bre a ke r must withstand during the fi rst half cycle a fter the fa u l t occurs. This c u r r e n t i s ca l l e d t h e momentary current , a n d for many years c i rcu i t breakers we re ra t e d i n t e rms o f t h e i r mo m e n t a ry c u r re n t as w e l l Zl S o t h e r criteri a . I T h e ill t amptill/{ ra t il/g o f a c i rc u i t b re ,l ke r w a s s I1 ec i fi e d in k i lovo l tam­ peres or m e gavol t a m pe re s . T h e i n t e r rupting k ilovoltamperes equal 13 X ( the k i l ovo l t s of the b u s to w h i ch the b rc , l k e r i s c o n n e c t e d ) X ( t h c c u rr e n t wh ich the b r e a k e r m u s t be c l ]J a b l c o r i n t e r r u p t i n g w h e n i t s c o n t a cts p a r t ) . This interrupt­ ing c u rren t i s , o f cou rse , l ow e r t h a n t he m o m e n t a ry C U ITe n t a n d d e p e n d s on t h e spe ed of the breaker, such as 8, 5, 3 , o r 2 cycl es, w hi ch is a m easu re of t h e t i me from the occu rrence o f t h e fa u l t to t h e ex tinct ion of the a rc . Breakers of different speeds are class ified by t h e i r rated interrupting times. The rated

l See

Lester, " H igh Yoltilge Ci rcu i t Bre a k e r S t a n d a rds i n t h e USA: P a s t , Prese n t, a n d IEEE Transactions on Power Apparatus a n d Systems, v o l . 93, 1 974, p p . 5 90,600.

G . N.

Fu t u re , "

404

CHAPTER 10

SYMMETRICAL FAULTS Extinction of arc on

Initiation of

primary contacts

short ci rcu it Energization of t ri p circuit

Parti ng of primary arcing

contacts

Time �

;

Inte rrupti ng time

{

•

Tripping Opening time d e l ay (

I

•

)

Arcing time �

)

Contact parting ti m e (

l

FIG U RE 1 0. 1 2 D e fi n i t io n

of

i n t e r ru p t i n g

time

give n

tion Guide for A C High VO/lage Circuit Rated Of! a Symmetrical Currell t Basis .

In

ANS I / I EEE S t a n d a r d C.37.0 1 0 - 1 979, Applica ­

Bre a k ers

interrupting time of a circuit breaker is the period between the instan t of energiz ing the trip circuit and the arc extinction on an opening operation, Fig. 10. 12. Preceding this period is the tripping delay time, which is usu ally assumed to b e � cycle for relays to pick up. The current which a breaker must inte rrupt is usually asymmetrical since it still contains some of the decaying dc component. A schedule of preferred ratings for ac high-voltage oil circuit breakers specifies the interrupting cu rrent ratings of breakers in terms of the component of the asymmetrical current which is symmetrical about the zero axis. This current is properly called the required symmetrical interrupting capability or simply the rated symmetrical short-circuit current . Often the adjective symmetrical is omitted. Selection of circuit breakers m ay also be made on the basis of total current (dc component inclu ded). 2 W e shall l im i t o u r discus s ion t o a brief treatment of the symmetrical basis of breaker selection.

a Symmetrical Current Basis, ANSI C37.06 - 1 987, a n d Guide for Calculation of Fault Currents for Application of A C High-Vo/tage Circuit Breakers Rated on a Total Current Basis, ANSI C37. � - 1 979, 2See Preferred Ratings and Related Required Capabilities for A C High-Voltage Circuit Breakers Rated

on

American National Standards Inst i t u t e, New York.

10.5 THE SELEcnON OF CIRCUIT BREAKERS

405

Breakers are identified by nominal-voltage class, such as 69 kY. Among oth er factors specified a re rated continuous current, rated maximum voltage, vol tage range factor K, and rated short-circuit current at rated maximum kilovol ts. Th e rated maximum voltage of a circuit breaker is the highest rms vol tage for wh ich th e circuit breaker is designed. The rated voltage range factor K is the rat io (rated maximum voltage -7- t he lower l imit of the range of operating voltage). K determines the range of voltage over which the product (rated short-circuit current X operating voltage) is con s t a nt . In the application of circuit breakers it is important not to exceed the short-circu it capabilities of the breakers. A br e ake r is req ui red to have a maximum symmetrical internlpting capability equa l to K x rated short -circuit current. Between the rated maximum voltage and 1/ K times the rated maximum voltage the symmetrical interrupting capability is defined as the prod uct [rated short-circu it current X (rated m axi­ mum vol tage/operating voltage)].

Exa m ple 10.6. A 69-kV circuit b reaker having a voltage range factor K of 1 .2 1 and a con ti nuous curre nt rati ng o f 1 200 A has a rated short-circuit current o f 1 9,000 A at the m aximum rat ed voltage of 72 .5 kV . Determine the maximum symmetrical i nt errupt ing capab il ity of the brea k er and expl ain its significance at lower operat ing voltages.

Solution .

The max imum symmetrical i nterrupting capab i l i ty is given by K x

rated short-circuit current

=

1 .21

x

1 9 ,000

=

22 ,990 A

This value of symmetrical i nterrupting current must not be exceeded. From the definition of K we have Lower l i mit of operating voltage

rated maximum voltage =

K

=

72.5 1 .21

-

=

60 kV

Hence, in the operating voltage range 72.5-60 kV, the symmetrical interrupting current may exceed the rated short-circu it current of 1 9,000 A, but it is lim ited to 22,990 A. For example, at 66 kV the interrupting curre n t can be 72 .5 66

-

x

1 9,000

=

20,87 1 A

B re a kers of the 1 1 S -kV class a n d h i gh e r have a K of 1 .0. A simplified procedure for calcu J at ing the symmetrical

short-circuit cur­

rent , called the E/X method, 3 d isrega rds all resistance, all stat ic load, and all :

3 S e e Application Guide for A C High- Voltage CirCLIit B reakers Rated on a Symmetrical Current Basis, is also A NS I C37.0 1 O - 1 979, A m e ri c a n N at iona l S tand ards I nstitute, New York. This publication I EEE Std. 320 - 1 979.

.

'

... - � . �

406

CHAPTER 1 0 SYMMETRICAL FAULTS

current. Subtransient reactance is used for generators in the E jX method, and for synchronous motors the recommended reactance is the X� of the motor t imes 1 .5 , which is the approximate value of the transient reactance Xd of the motor. Induction motors below 50 hp are neglected, and various multiplying factors are applied to the X� of larger i nduction motors depending o n their size. If no motors are present, symmetrical short-circuit current equals subtrans ient current. The impedan ce by which the vol tage Vf at the fau l t is divided to find short-circ u it current must be examined when the E jX method is used. In specifying a bre aker for bus ® , this impedance is Zk k of the bus impedance matrix with the proper machine reactances si nce the short-circuit current is expressed by Eq . ( 1 0. 1 9). If the radio of XjR of t h i s i mpedan ce i s 1 5 or le ss a breaker of the correct voltage and kilovoltamperes may be used if its i n terrupt­ ing current rating is eq ual to or exceeds the cal culated current. If t h e Xj R r a t i o is unknown, the calcul ated curren t should be no more than 80O/C of the allowed value for t h e bre aker at the existing bus voltage. The ANSI application g uide s pecifies 'a corrected method t o accou nt for ac and dc ti me constants for the decay o f the current amplitude i f the XjR ratio exceeds 15. The corrected method also considers breaker speed. prefault

,

10.7. A 25,000-kVA 1 3.8-kV generator with X;; = 1 5 % is con nected through a transformer to a bus which s upplies fou r identical motors, as shown in

Exam ple

Fig. 1 0 . 13 . The s ubtransient reactance X; of each motor i s 20% o n a base of 5000 kVA, 6.9 kV. The three-phase rating of the transformer is 2 5 ,000 k YA, 1 3 .8/6.9 kV, with a leakage re actance of 10%. The bus voltage at the motors is 6.9 kV when a three-phase faul t occurs at point P . For the fau l t specified , determine (a) t h e subtra nsien t current i n t he fault, ( b ) t he subtra nsicnt curre nt i n breaker A , and CC) the symmetrical short-c ircuit interru pting current ( a s defined fo r circuit-breaker appl ica tions) i n the fault and in breaker A .

(a) For a base o f 25,000 kV A, 1 3 .8 kV i n the generator circui t the base for the motors is 25,000 kVA, 6.9 kV. The subtransicnt reactance o f each motor is

Solution.

X"d

=

0 .20

25 ,000 5000

=

1 .0 per u n it

Figure 1 0. 14 is the diagram with sub transient values of reactance marked. For a

FIGURE 10. 13 One-l i n e

diagram for Exa mple 1 0,7.

THE SELECfION OF CIRCUIT BREAKERS

1 0.5

407

FIGU R E 10. 14 Reactance d i agram for Example 10.7.

fault at P

VI =

J .O�

I .()�

/" J -

The base current i n the 6 .9

jO . 1 25

circuit is

kY

i Il i

(b)

=

-jR .n per

jO . 1 25

u

per

nit

unit

25 ,000 = 2090 A 3 X 6. 9 f3

=

and so

Zlh

per u n i t

2090

8 X

=

=

16,720 A

Through breaker A comes the contribution from the generator and three of the four motors. The generator con tributes a current of - j 8 .0 X

0.25 0 .50

--

=

-j

4.0 per unit

Each motor contributes 25% of t he remaining fault current, or amperes each. breaker A J"

(c)

=

-j4 .0

To

+-

� ( - j l .O)

=

-j 7 0 .

per u

ni t

or

7

x

2090

=

-

j l .O

per-unit

1 4 ,630 A

the j 1 .5 in the motor

c o m p u t e t h e c u r re n t to be i n t e rr u p t e d by b re a k e r A . r e p l ace

c n j l .O circuits of Fig. 1 0. 1 4. Then,

s u b l r a n s i e n t re a t a c e of

Zlh

=

j

b y t h e t ra ns i e n t r e ac t a nce of

0 .375 0 . 375

X

+

0.25 0.25

The generator contributes a cu rrent of 1 .0 jO. 1 5

--

x

0.375 0.625

=

jO . l S

per unit

-j4.0 per unit

·

408

CHAYfER 10

SYMMETRICAL FAU LTS

Each motor contributes a current of 1 - X

4

1 .0 -jO.15

X

0 .25 -0 . 625

=

-jO.67

per unit

The symmetrical short-circuit curren t to he interrupted is ( 4 0 + 3 X 0 .67) X 2090 .

1 2,560 A

=

Suppose that all the breakers connected to the bus are rated on t h e basis of t h e current into a fault on the bus. In t hat case the short-circuit curren t interrupting rat ing of the breakers connected to the 6.9 kY bus must be at l east 4 + 4 X

or

0 . 67 = 6 .67 per unit

6 . 67 X 2090

1 3 ,940

=

A

.

A 14.4-kV circuit breaker has a rated maximum voltage of 1 5.5 kY and a K of 2 .67. At 15.5 kY its rat ed short-circuit interrupting current is 8900 A. This breaker is rated for a symmetrical sh ort-circuit interrupting current of 2 6 7 X 8900 23, 760 A, at a voltage of 15.5/2.67 5 8 kY. This current is the maximum that can b e interrupted even though the breaker may be in a circui t of l ower voltage. The short-circuit interrupting current rating at 6.9 kY is =

=

.

15 .5 -- X 8900 = 20 ,000 A 6 .9

The required capability of 13,940 A is well below 80% of 20,000 A, and t h e breaker i s suitabl e with respect to short-circuit current.

The short-circui t current cou l d have been found by usin g the bus impedance matrix. For this purpose two buses CD a n d @ a re identified in Fig. 10. 1 4 . Bus CD i s o n the low-vo l tage s i d e o f t h e t ra ns former and b u s 0 i s o n the high-voltage side. For motor reactance of per unit Y = \I Y1 2

The

= jl0

-

' 10 +

)

Y 22

=

j1.5 /4 1

-j l 0 -

1.5

-j 1 2

j 6 . 67

=

.

67 - j 16 .67

node admittance matrix and its inverse are

Y bus

=

CD:(l}[

CD

- j12.67

jI0 .00

@ j l0 .00 -jI 6.67

1

Z bus

CD

=

CD[j O. 150

@ j O.090

@

1

j O .090 jO . 1 1 4 ,

THE SELECTION OF CIRCUIT BREAKERS

10.5

[

1

)0. 15 - )° .09 CD 11--__-+__

Vf

=

l.O�

0 . � J.o� l l 4 - ) o. 9 __ L___�__

jO.09

Fig u re

Z !HIS o f

s,

tl 1 s,

®

__

FIG U RE 1 0. 1 5 Bus i m p e d a nce e q u i v a l e n t c i rc u i t fo r t h e

409

__ _

f'ig. 1 0. 1 4 .

1 0. 1 5 is t he n e twu r k corresponding to Z ou s a n d VI = 1 .0 per unit. Closing 5[ with 5 2 open represents a fau lt on bus CD . The symme trical sho rt-circuit i n terrupting current in a three-phase faul t at bus CD is I')c

l .0

=

--

jO . 1 5

=

-j6 .67 per unit

w h ich agrees with our previous calculati ons. The bus impedance matrix also gives us the voltage at bus Cl) w i th the fau l t on bus CD · and

V2

=

1 .0 -

1St Z 2 J

=

1 . 0 - ( -j6 .67) ( j0 .09 ) = 0 .4

since the admittance between buses the fau l t from the transformer is ( 0 .4 - 0 .0) ( -jI 0)

CD

and

W

is

- j l 0, the cur rent i nto

- j4 .0 per unit

which also agrees with ollr p revious resu l t . We a lso know immediately the sho rt-ci rcu it current in a th ree-phase fault a t b u s W , which, by referring to Fig. 1 0. 1 5 with S I open and S2 closed, is 1sc

l .0

=

jO.1 14

=

-j8.77 per unit

This simple example ill ustrates the value of the bus impedance matrix , where the

410

CHAPTER 1 0 SYMMETRICAL FAULTS

effects of a fault at a number of buses are to be studied. Matrix i nversion is not necessary, for Zbus can be generated d irectly by compute� using the Z bu s building algorithm of Sec. 8.4 or the triangular factors of Y bu s ' as explained I n Sec. 8.5.

of the network of Fig . 1 O. 1 6( a ) have s y n ch ro no u s reacta nces X" I = X" 2 = j l 7 0 r>er u n i t (as marked) a n d subtransient r e ac t a n ce s X� = X:; = j O .2S p e r u n i t . If a t h re e - phase s ho r t - c i rcu i t faul t occurs a t bus wh � n the :e is no load (al l b u s vol t ages e q u a l 1 .0 0° pe r un it), find the i n i ti a l symmetrical (sub t r a n s i e n t ) c u rr e n t ( 0 ) i n t h e fau l t l i ne and ( c) the voltage a t b u s Use t r i a n g u l a r fac tors of Y bus i n t h e

10.8. The generators at b u s es CD

Example

G)

CD G) ,

and

(1) .

/ U.0m"

@.

-

calculations.

fig. I O. I ()( h),

Solution. Fo r the give n ra u l t co n d i t i o n s t I l e n e t wo r k l I a s t he s u h t r ; l Il s i e n t re a c t ;I I1CC

diagram

shown

factors

i ll

and

-0.1

tIl(': co r re spo n d i n g

[

h a s t h e l r i �I I l g ll b r

- O.�4304 - 0 .5

1

-j7.9 j3 .5 L

Y I )II'

u

. ][X J [OJ

1

Since the faul t is at bus 0, Eqs. (10. 1 9) through 0 0.21 ) show that the calculation s involve column 3, Z�Js' o f the subtransient Z bu s ' which we n ow generate as follows: -jl O jl jS

jO.2 ® )O.3333 1.0LQ:.

...L (a) -

FIGURE 10.16

l.O�

-j7 .9 j3.S

1

.

-j3 .94937

X2

X3

=

0 1

)0.2 ® )0.3333 j O.25

j O.25

(b)

t

-

The reactance d iagram for Example 1 0.8 with g en erators represented behind Xd ; (b) e q u ivale n t cu rre n t sou rce i n parall e l w i t h X:;.

by:

(a) series voltage �ou rce

10.6

Solving, we obtain

1 = ---=j . -j3 .94937

O 25 3 20 per

X3

and so

the e l e m e n t s

of Z b�s

given by

are

- 0.5

- 0.1

(u)

2:122..11 2

t hat

Acco r d i n g t o EC] .

2� ,

u

( 2

I" _ _ V I

b

_

(c)

1_

_

jO . 2 During

I

0

=

"

jO.25320

2 33

= j O . 25 3 20 per u n i t = jO . 1 1 2 1 t{ pcr u n i t = j O . 1 3 782 p e r u n i t

jO .25 320

w ri t e

2 1 3 - 233

(

1

---

( b ) From Eq . ( 1 0.21) we c a n D

u nit

( J 0. 1 9), t h e s u b t ra n s i c n t cu r r e n t i n t h e fa u l t i s

VI " - _ J1 -

-

411

0

- 0 .44304

We And

][222I)3 ] [ ]

SUMMARY

Z �3

)

for

-

=

j3 . c)493 7 per u n i t

t h e current i n l i n e

jO . 1 3 782 - jO . 25320 j O .25320

V=, = Vr

W

( - 7."�) %::.

t h e f(l u l t t h e vo l t a ge at b u s I

)

0 - G)

= - j 2 . 27844 per u nit

(

)

is g iven by Eq . OU.20) = 1 .0

jO . 1 1 2 I H j O . 25320

as fol lows:

= 0 .55695 p e r u n i t

10.6

SUMMARY

The cu rrent flowing immed iate ly afte r a fau l t occu rs in a power network is dete rmined by the impedances of the netwo rk components and the synchronous machines. The initial symmet rical rms fault cu rrent can be determined by representing each machi ne by its subtrans ient reactance in series with its subtransient internal vo ltage . S u btransient cu rrents are larger than" the transient

,

.

412

CHAPTER 1 0

SYMMETRICAL FAU LTS

and steady-state currents. Circuit breakers h ave ratings determined by the maximu m i nstan taneous current whi ch the breaker m ust withstand and later interrupt. Interrupting currents depend on the speed of breaker operation. · Proper selection and application of circu it breakers should follow the recom­ mend ations of ANSI standards, some of wh ich are re ferenced in this chapter. S implifying assumptions usually made in i n dust ry-based fault stud ies are: •

•

•

•

•

All shunt connections from system buses to the reference node (neutral) can be n eglected in the equivalent circu its representi ng trans miss ion lines and transformers. Load ·impedances are much larger t han those of network co mponents, and so they can be neglected in system mode I i ng. All b u ses of t he sys t e m have ra ted /nom i n a l vo l tage or I .()� pe r u n i t s o t h a t no prcfaul t currents tlow in the network. Syn chronous machines can be represe nted by vol tage of l . 0 per unit behind subtransient o r transient reactance, depe nding on t h e speed of t he circu it breakers and whether the mom en tary or interrupting fau l t curren t is being calculated (ANSI standards should b e consulted). The voltage-source-plus-series-impeda nce equivalen t circu it of each syn­ c hronous machine can be transformed to an equivalent current-source-plus­ shunt-im pedance model. Then, the shunt i mpedances of the machine models represent the only shunt connections to the reference node.

�

The bus impedance matrix is most often used for fault current calcu la­ tions . The elements of Z bu s can be made available explicitly using the Z bus building algorithm or they can 'be generated from the triangular factors of Y bus ' Equivalent circuits based �:m the elements of Z b u s can simpl i fy fault-current cal culations as demonstrated in this chapter for the line-end fault.

PROBLEM S 10. 1.

A 60-Hz alternating voltage h aving a rms value of 1 00 V is applied to a series RL circui t by closing a switch. The resistance is 1 5 n and the inductance is 0.12 H. (a) Find the value of the dc component of current upon closing the switch if the instantaneous value of the voltage is 50 V at that time. (b)

W h a t i s the

i n s t a n t a n e o u s va l u e

of t h e v o l t age w h i ch w i l l p r o d uce t h e

maximum dc component of current u pon closin g the swi tch? What is the instantaneous val ue of the voltage which will result in the absence of any dc component of c urrent u p on closing the switch? (d) If the switch is closed when the i nstantaneous voltage is zero, find the instantaneous current 0.5 , 1 .5, and 5.5 cycles l ater. A generator connected through a 5-cycle circuit b reaker to a transformer is rated 1 00 MVA, 18 kV, with reactances of X� 1 9 %, Xd 26%, and Xd = 1 30 � . I t (c)

10.2.

=

=

P R OBLEMS

1 0.3.

413

" is operating a t no load and rated vol tage when a three-phase short circui t o ccurs between the breaker and the transformer. Find (a) the sustained short-circu it current in the breaker, ( b ) the initial symmetrical rms current in the breaker, and (c) the maximum possible d c component o f the short-circuit current in the breaker. The three-phase transformer connected to the generator described in Prob. 1 0.2 is rated 1 00 M YA, 240Y / 1 8 6 k Y X = 1 0%. If a three-phase short circ u i t occurs on the high-voltage side of the transformer at rated vol tage and no load, find (a) the initial symmetrical rms current in the transformer windings o n the h igh-volt­ age side and ( b) the initial symme trical rms current in the line on the low-vol tage side. A 60-Hz generator is rate d 500 M Y A , 20 k Y , w i t h X':, = 0.20 per u n i t. It supplies a purel y resist ive load of 400 M W at 20 kY. The load is connected d irectly across the terminals of t he generator. If all three phases of the load are short-circuited s i m u l t a neou sly. fi nd t h e i ni t i a l sy m m e t r i c a l rms current in t h e generator i n per u nit on a base of SOO M Y A, 20 k Y . A g e ne r a to r is c onn e c t e cJ t h r o u g h a t ra n s fo r m e r to a s y n c h ro n o u s m ot o r. Reduced to the same base, the per-unit subtransient reactances of t h e generator and motor a r e 0 . 1 5 and 0.35, respect ive l y, and the leakage reactance of the transformer is 0 . 1 0 p e r u n i t . A t h r ee p h a s e fa u l t occu rs a t t he ter m i n (l l s of the motor when the ter m i nal voltage of t h e ge n e ra t o r is 0 . 9 p e r u n i t a n d the o u t p u t current o f the ge n e r a t o r is 1 .0 per u n i t at 0 . 8 powe r factor l e a d ing. Find th e subtransient current in per u n it in t h e fau l t in t he generator, and in the motor. Use the terminal voltage of the generator as the reference phasor and obta in the sol u tion (a) by computi ng the voltages beh ind subtransient reactance in the generator and motor and (b) by using Thevenin's theore m. Two synchronous motors having subtransient reactances of 0.80 and 0 . 2 5 per u nit, respectively, on a base o f 480 Y , 2000 k Y A are connected to a b u s. Th is motor is connected by a lin e having a reactance of 0.023 fl t o a bus of a power system. At t he power system bus t h e short-circui t megavoltamperes of the power system are 9.6 M Y A for a nomi nal vol tage of 480 Y . When the vol tage at the motor bus is 440 Y , neglect load current and find the initial sym metrical rms current i n a three-phase fault at t he motor bus. T h e bus impedance m a t r i x o f a four-bus network with va lues i n p e r unit IS ,

1 0 .4.

1 0.5.

-

,

10.6.

10.7.

Z bus =

jO . 1 5 jO .08 jO .04 j O .07

jO .08 jO . 1 5 jO .06 jO .09

jO.04 jO .06 jO . 1 3 jO .OS

jO .07 jO .09 jO .05 jO . 1 2

Generators connected to buses CD and 0 have their sub transient reactances incl uded i n Z bu s ' If prefa u l t current is neglected , find the subtransient curren t in per u n i t in 19c fault for a three-phase fault on bus @ . Assume the vol t � ge at the fault is 1 .0L.2.:. per unit before the fault occurs. Find a lso the per-unit current from generator 2, whose sub transient reactance is 0.2 per u n it.

414 10.8.

CHAPTER 10 SYMMETRICAL FAULTS

-

For the network shown in Fig. 1 0 . 1 7, find the subtransient c urrent in per u nit from generator 1 and in line CD @ and the voltages at buses CD and G) for a t hree-phase fault on bus CD . Assume that n o current is flowing prior to the faul t a n d that t h e prefault voltage a t b u s @ is 1.0& per unit. Use the bus impedance matrix i n the calculations.

FIGURE 1 0. 1 7 N etwo rk for P robs. 1 0 .8 a n d 1 0 . 9 .

: . 10.9.

10.10.

10. 1 1 .

10. 12.

10.13.

For the n etwork shown in Fig. 1 0. 17, determine Y bus and its triangular factors. Use the triangular factors to generate the elements of Z b u s needed to solve Prob. 10.8. If a three-phase fault occurs at bus CD of the network of Fig. 1 0.5 when there is no load (all bus voltages equal 1 .0 per unit), find t h e subtransient current in the fault; the voltages a t buses @ , ® . and @ ; and the current from the generator connected to bus @ . Use e quivalent c ircu its based o n Z bu s of Example 1 0.3 and simil ar to those of Fig. 1 0 . 7 to illustrate you r calcu l a t ions. The network of Fig. 1 0.8 has the bus i m pe d a nce m a t rix given in Exa mple l O A . I f a · short-circuit fault occurs a t bus @ o f t h e netwo rk w h e n t h ere i s n o load (al l bus voltages equal 1 .0 per u n it), find the subtransicnt c u rrent i n t he fau l l, the voltages at buses CD and G) . and the current from the generator con nected to bus CD . Use equivalent c ircuits based on Z bu s and similar to those of Fig. 1 0.7 to illustrate your calculations. Z bus for the network of Fig. 1 0.8 i s g iven in Example 10.4 . If a l i ne-end short-circuit fault occurs on line ® G) of the network on the l ine side of the breaker at bus G) , calculate the subtransient current i n the fault when only t h e near-end breaker at bus G) has opened. Use the equivalent c ircuit approach of Fig. 1 0. 1 1 . Figure 9.2 shows the one-line d iagram o f a single power network which has the line data given in Table 9.2. Each generator connected to buses CD and @ has a s ubtransient reactance of 0.25 per unit. M aking the usual fault study assumptions, s ummarized in Sec. 1 0.6, determine for the network ( a ) Y bU5 ' ( b ) Z bus ' (c) the s ubtransient current in per unit i n a t hree-phase fault on bus ® and Cd ) the contributions to the fault current from line CD G) and from line @ G) .

&

&

-

-

-

PROBLEMS

10.14. A 625-kV generator with X'd

415

0.20 per unit is connected to a bus through a circuit breaker, as shown in Fig. 1 0. 18. Connected through circuit breakers to the same bus are three synchronous motors rated 250 hp, 2.4 kV, 1 .0 power factor,

90% efficiency, with

X'd

=

= 0.20 per unit. The motors are operating at ful l bad,

u nity power factor, and rated voltage, with the load equally divided among the machines. (a) Draw the impedance diagram with the impedances marked i n p e r unit on a base of 625 kVA, 2.4 kY. ( b) Find the symmetrical short-circuit current in amperes, which must be i nter­ rupted by breakers A and B for a three-phase fault at point P. Simplify the calcu lations by neglect ing the pre fault cu rrent. (c) Repeat part ( b ) for a three -phase fau l t at point Q. Cd) Repeat part ( b ) for a three-phase fau l t at point R .

FI G U RE 1 0. 1 8 O n e · l i n e d i a g ra m for P rob. 1 0 . 1 4 .

circuit breaker having a nominal rating of 34.5 kV and a continuous current rating of 1500 A has a voltage range factor K of 1 .65. Rated maximum vol tage is 38 kV and the rated short-circu it current at that vol tage is 22 kA Find (a) t he voltage b elow which rated short-circuit current does not increase as operat ing vol tage decreases and the value of that current and (b) rated short-circuit current at 34.5 kY.

10.15. A

CHAPTER

11

SYMMETRI CAL COMPO NENTS AND SEQUENCE NETWORKS

One of the m os t p owerful tools for dealing wi th unbalanced polyphase circui ts is the method of symmetrical components in troduced by C. L. Fortescue. ! Fortescue ' s work proves that an unbalanced system of n related phasors can be resolved into n systems of balanced phasors called the symmetrical components of the original phasors. The n phasors of each set of componen ts are equal in length , and the angles between adjacent phasors of the set are equal. Al though the m ethod is applicable to any unbalanced polyp hase sys tem , w e con fine our d iscussion to three-phase systems. I n a three- phase system which is normally balanced, unbalanced fault conditions generally cause u nbalanced c urren ts and vol tages to exist in e ach of the p hases. I f the curren ts and vol tages are related by constant impedances, the system is said to b e linear and the principle of superposi tion applies. The voltage response of the linear system to the u nbalanced cu rrents can b e determined by considering the separate responses of the individual elements to the symmetrical components of the currents . The system elements of interest are the machines, t ransformers, transmission lines, and loads connec ted to 6.. or Y configurations. I e. L . For t escue, " M ethod of Symmetrical Coord i n at es Applied to the Sol u t ion of Polyph ase Networks, " Transactions of AlEE, vo l. 37, 1 9 1 8, p p. 1 027- 1 1 40.

4H l

1 1.1

SYNTH ESIS OF UNSYMMETRICAL PHASORS FROM TH EIR SYMMETRICAL COMPONENTS

417

In this chapter we study symmetrical components and show that the response of each system element depends, in general, on its connections and the component of the cu rren t being considered. Equivalent circui ts, called sequence circuits , will be developed to refl ect the separate responses of the elements to each current component. There are three equivalent circuits for each element of the three-phase system. By organizing the individual equivalent circuits into n e tworks accord ing to the interconnections of the elements, we arrive a t the concept of three sequence net wo rks Solving the sequence networks for the fault con d i tions gives symmetrical cu rrent and voltage components which can be combined together to reflect t h e effects of the original unbalanced fault currents on t he ove ra ll system . Ana lys is by symme trical compon ents is a powerful tool which makes the calculation of unsymmetrical faults almost as easy as the calculation of t hree­ phase fau lts. Unsymm etrica l faults are stud ied in Chap. 1 2. .

S Y NTHES I S OF UNSYM M ETRI CAL PHASORS FROM THEI R S Y M M ETRI CAL COMPO NENTS

11.1

Accord ing to Fortescu e's t h eorem, th ree unba l anced phasors of a three-ph ase system can be resolved into th ree balanced systems of phasors . The balanced sets of components are: Positive-sequence components consisting of three phasors equal in magnitude, d isplaced from each other b y 1 200 in phase, and havi ng the same phase sequence as the original phasors, 2 . Negative-sequence components consisting of three phasors equal in magni­ tude, displaced from each o ther by 1200 in phase, and having the phase sequ ence opposite to that of the original phasors, and 3 . Zero-sequence components consisting o f t hree phasors equal in magn itude and with zero ph ase d ispl acemen t from each other. 1.

It is customary w h e n solvi ng a p rob lem by sym metrica l components to designate the th ree phases of t he syslem as a, b, and c i n such a manner that the phase sequ ence of the voltages and cu rrents i n the system is abc. Thus, the ph ase sequence of the pos itive-sequence components of the unbalanced p h asors is abc, and the p h ase seq u e n ce o f the n e g a t i ve-se q u e n c e comp o n e n t s is acb . I f the ori ginal phasors are voltages, they may b e designated Va ' Vh ' and VC . The three sets of symmetrical components are designated by the additional s uper­ script 1 for the positive-sequence components, 2 for the negative-sequence components, and 0 for t he zero-sequence components. Superscripts are chosen so as not to confuse bus numbers with sequence indicators later on i n this chapter. The positive-sequence components of Va ' Vh ' and Vc are Va( l ) , Vb( I ), and Vc( l ) , respectively. Simi larly, the negative-sequence components are VY), VP), ,

418

CHAPTER 1 1

SYMMETRICAL COMPONENTS A N D SEQUENCE NETWORKS

y (l)

y (2 )

a

a

VP) -------{

Zero-sequence components

N egative-seq u ence com ponents

P os itive-s equence c omponents

FIGURE 1 1 . 1 Three sets of b a lanced p h asors which a re the sym m e t r i c a l compon ents of th ree u n b a l a nced ph asors.

and Vc(2 ), and the zero-sequence components are Va(O ), V� O ), and Ve(O), respec­ tively_ Figure 1 1. 1 shows three such sets of symmetrical components. Phasors represen ting currents will be design a ted by J with superscripts as for voltages. S in ce each of the original unbalanced phasors is the sum of its compo­ nents, the original phasors expressed in terms of their components are

va

vc

=

Va(O) + Va( l ) + Va(2 )

(11 .1) ( 1 1 .2)

=

V (O) + V ( I ) + V (2) c

c

( 1 l .3)

c

The synthesis of a set of three unbalanced phasors fro m the three sets of symm etrical componen ts of F i g . 1 1 . 1 is shown in F i g . 1 1 .2_ The many advantages of analysis of power systems by t he method of symmetrical components will become apparent gradually as we apply the method to the stu dy of unsymmetrical fa ults on otherwise sym metrical systems. It is sufficient to say here that the method consists i n finding the symmetrical compon e n ts of c u rre n t a t the fa u l t . Th e n , t h e v a l u e s o f c u r r e n t a n d vol t a g e at various poi n ts in the system can be fou n d by means of the bus impedance matrix. The method is simple and leads to accu rate predictions of system b ehavior.

1 1 .2 THE SYMMETRICAL COMPONENTS

OF

UNSYMMETRICAL PHASORS

In Fig. 1 1 .2 we observe the synthesis of three u nsym metrical phasors from th ree sets of symmetrical p h a so rs . The syn t h e s i s is made in accor d a n ce w i t h EQs.

1 1 .2 THE SYMMETRICAL COMPONENTS OF U NSYMMETRICAL PHAS ORS

VCl( O l

Vc( 2)

vc( l

VCl(2

419

)

)

FI G U R E 1 1 .2

showll ill Fig. 1 1 . 1

G raphical a d d i t i o n of t he compo nents to o b t a i n three u n ­

balanced phasors.

( 1 1 . 1 ) t h rough ( 1 1 .3). Now l e t us examine t hese same equations to determ i ne h ow to resolve t h ree unsymmetrica l phasors i nto the i r sym metrical compone nts. F i rst, we note that the n u m be r of unknown quantities can b e reduce d by exp ressing each componen t of Vb and Vc as t h e p roduct of a compon e nt o f Va and some function of the ope rator a = 1 , which was i ntroduced i n Chap. 1 . Refe re nce to Fig. 1 1 . 1 verifies t he fol lowing relations:

�

V (O)

=

b

VhOl

Va(O)

V ( O) c V(l)

=

=

C

=

Va(O)

a V(I)

( 1 1 04)

11

a V (2) a

R C f) C (l t i n g Fq . ( 1 1 . 1 ) a n d s u bs t i t u t i n g Eqs. ( I J . 4 ) i n Eqs. ( 1 1 . 2 ) a n d ( 1 1 .3 ) V( I ) /I

+

yield

( 1 1 .5 ) ( 1 1 .6)

( 1 1 .7) or in m a t r ix form V a

Vb Vc

=

1

1

1

1

a2

a

Va(O ) Vel)

a2

) V (2

1

a

a

a

=

A

V eO ) a ) V(l a 2) V( a

( 1 1 J8)

420

CHAPTER 1 1

SYMMETRICAL COMPON ENTS A N D SEQUENCE NETWO R KS

where, for convenience, we let

( 1 1 .9 )

Then, as m ay be verified easi ly,

( 1 1 . 10)

a nd premul tiplying both sides Va(O) V(l) Va(2) a

1 -

3

0

f

Eq. ( 1 1 .8) by

1

1

1

1

a

a2

1

a2

a

A-

Va Vb Vc

I yic I d s

A- I

=

Va Vb

( 1 1 .11)

Vc

which shows u s how to resolve three unsymmetrical phasors into their sym metri­ cal components. These rel ations are so i mportant that w e write the separate equ ations in the expanded form V(a O) Va( l ) Va(2)

=

=

-

�3 ( V +

( 1 1 . 12 )

II

�3 ( V + a Vh + a 2 V )

( 1 1 . 13)

+ a 2 v.,) +

( 1 1 . 14)

a

� 3

(V

a

C

,/

2 I f required , the components v,(0) b , Vb( l ) ' Vb(2)' Vc( O )' Vc( l ) ' and V ( ) can b e fou nd by Eqs. ( 1 1 .4). Similar resul ts apply to l ine-to-line voltages simply by replacing Va ' Vb ' a nd � in above e q u a t ions by Vab , V"c , a n d VC t" respectively. Equ ation ( 1 1 . 12) shows that no zero-sequence components exist if the sum o f the unbal a nced ph asors i s zero. Since t h e sum of t h e line-t6-line voltage p hasors i n a three-phase system is alw ays zero, zero-sequ ence compon ents are never present i n the l ine voltages regardless of the degree of unbal ance. The s u m of the three line-to-line neutral volt age p hasors is not necessarily zero, and voltages to neutr al may cont ain zero-sequence components. The preceding equ ations could h ave been written for any set of rel ated phasors, and we m ight have written them for cu rrents instead of for vol tages. They m ay be solved either analytica l ly or grap hic(l lly. Because some of the C

1 1 .2 THE SYMMETRICAL COMPONENTS OF UNSYMMETRICAL PHASORS

421

preceding equations are so fund amental, they are summarized for currents:

Ia Ib Ic J (O) a

Ja( 1 ) J a( 2 )

=

1a(0 ) +

I� O)

+

a 2 1a( 1 )

+

aI(2) a

1a(0) +

a 1(a 1 )

+

a2I(2) a

= =

1a(2)

1 a( 1 ) +

=

1..3 ( Ia +

=

t ( la

=

1.. :I

(1

+

IJ

Ib +

alb + a2/c )

+

u

( 1 1 . 15)

( 1 1 . 1 6)

a 2 Ib + alc)

Finally, these results can be extended to phase currents of a � c ircuit [such as t h a t of Fig. 1 1 .4( a )] by rep lacing 111 ' 1,." and 1, by I,,/) , I;" . and ICIl ) respect ively. Exa m p l e 1 1 . 1 . One conductor o f

t hrec -phnse l i n e is o p e n . The current fl ow i n g to the 6.-connccled load t h ro u g h l i n e ( [ is l O A . W i t h the curre n t in l i ne (/ as rcfcn:nce and assu m i n g that l i n e c i s o pe n f i n d the sym m e t r i c a l c o m po n e n t s o f the line cu rrents. a

,

Solution. Figu re 1 1 . 3 is a di agram of the circu i t . T h e line currents are 1c

Fro m Eqs.

0 1 . 1 6)

l�O)

1� I)

= = =

1� 2 )

= =

H 10L2: H 1 0L2:

5

-

j2 .89

HlOLo 5

+

+ + =

+

j2 .89

=

1 0� 10L 1800

+

+

5 .78L - 300

10/ 1800

+

5 . 7 8�

0)

=

1 200

0

+

0)

A

2400

+

A

a --------�

c ----

=

FIGURE 1 1 .3 Circuit for Exampl e 1 1 . 1 .

0)

OA

422

CHAPTER 1 1

SYMMETRICAL COMPONENTS A N D SEQUENCE NETWORKS

From Eqs. ( 1 1 .4)

[g I )

The result I�U)

If!.)

= 5.78/ - 1500 =

S.7H� A

= J�U) = I�O) =

leO} c

=0

A

IP)

=

5.7B/ - <JO° A

0 holds for any t h ree-wi r e sys t e m .

In Example 1 1 . 1 we note t h a t components 1� ' ) and 1�2) have nonze ro values although line c is open a n d can carry no net current. As is expected , therefore, t he sum o f t h e components in l ine c i s ze ro. O f cou rse, t h e s u m o f t h e components i n line a i s 1 0� A, a n d t h e s u m o f the components i n line b is 10� A. 1 1 .3

SYMMETRICAL Y AND d CIRCUITS

In three-phase systems circuit elements are connected between l ines a , b , a n d c in either Y or fl. configuration. Relationships between the symmetrical compo­ n e nts of Y and fl. currents and volt ages can be established by referring to Fig. 1 1 .4, which shows symmetrical i mped ances connected in Y and fl. . Let us agree that the reference phase for fl. qu antities is branch a -b . The p articular choice of reference phase is arbitrary and does not affect the results. For currents we h ave

( 1 1 . 17)

a �--�----� +

Zy + C ---1_---'-______

FIGURE 11.4

(a)

-----'

Symmetrica l impedances: (a) �-co n nected; ( b) Y -co n nected.

(6)

1 1 .3 SYMMETRICAL Y AND A CIRCUITS

423

Adding all three equations together and invoking the definition o f zero­ sequence current, we obtain I�O) ( Ia + Ib + 1)/3 0 , which means that line currents into a D.. -connec(ed circuit have no zero-sequence currents. Substituting components of current in the equation for Ia yields =

=

=

( J�g)

-

o

15�»)

+

( /�b)

15�»)

-

+

( /!�)

-

11�»)

( 1 1 . 1 8)

Evidently, if a nonzero value of circulating current 1��) exists i n the D.. circu it, it cannot be determined from the l ine currents alone. Not i ng that 15� ) a"�h) and 1 ��) a 211�t), we now write Eq. ( 1 1 . 1 8) as follows: =

=

1(1) a

+

f (2) a

=

(1

-

(1) a ) 1 ab

+

(1

- a2

) laC�) �

( 1 1 . 1 9)

A s i m i l a r equat ion for phase b is 1 h( 1 ) + 1b( 2) = ( 1 - a ) Ib( eI ) + ( 1 - a 2 ) Jb(2) e , and 2 1 2 2 ( 1 1 express ing 1b( 1 ) ' 1b( ) ' /b( e) ' and 1 be ( ) in terms' of 1a ) ' 1(2) ' 1(1) a D ' and a( b) ' w e obtain a result ant equat ion which can be solved along with E q . ( 1 1 . 1 9) to yield the important results a

ja( 1 )

=

13 /

-

30°

X

1a( b1 )

( 1 1 .20 )

These results amount to equ ating cu rrents of the same sequence in Eq. ( 1 1 . 1 9). Complete sets of posi t ive- and negative-sequence components of currents are shown in the phasoT diagram of Fig. 1 1 . 5 ( a). I n a sim ilar. manner, the li ne-to-line vol tages can be written in terms of l ine-to-neutral voltages of a Y -connected system, vu f)

V" ,.

v;,· u

=

:co

=

Vo n

-

V,,,, -

v:."

-

Vb n

v: "

( 1 1 .21)

�,"

Ad d i n g together a l l three equat ions shows that V:,(j/) e V:,h + V"c + V,, )/3 = O. J n words, lille-Io-line uolluges l/{fpe 1 1 0 zero-sequence comp on e nt s . Substituting components of the vol tages in the e quation for Vu h yields

=

V0e0l )

+ V ( 2)

at>

=

( VeOl an

+

()

V(anI ) + V ez») Ill!

-

( V(O) bll

+ V ( l ) + Vb('�» on ,

424

CHAPTER I I

SYM METRICAL COMI'ON ENTS AND SEQUENCE N ETWO R KS

/ (1 ) e

/ (1 ) ea

(2 1be

)

[(2 ) a

Y-----

/ (1 ) b

[( l )

[��)

[ (' 2 )

be

ca

Positive-sequence components

Neg ative-sequence components

(a )

v(l) ab

va( n\ )

Ol Vea

VbOl

( l) ven

)

vbe( l ) 2) Va(n

Positive-sequence components

va(b2 )

Negative-sequence com pon e nts (b )

FIGURE 1 1 .5 Posit ive- a n d nega t ive-sequence components of (a) l i n e a n d d e l t a currents a n d ( b ) l i ne-to- I i n e a n d line-to- n e u tral voltages o f a t h ree-phase syst e m .

Therefore, a nonzero value df the zero-sequence vol tage v}�) can not be deter­ mined from the l ine-to-line voltages alone. Separating positive- and negative­ sequence quan tities in the manner exp l ai n ed for Eq. (1 1 . 1 9), we obta in the importa n t voltage relations

VaCI) b

=

( 1 - a 2 ) V C1 ) an

( 1J. .23 )

I I .J

SYMMETRICAL Y A N D 6 CIRCU ITS

425

Complete sets of positive- and n egative-sequence components of voltages are shown in the phasor d iagrams of Fig. 1 1 .5(b). If the voltages to neutral are in per u ni t referred to the base voltage to neu tral and the line voltages are in per unit referred to the base vol tage from l ine to l ine, the 13 multipliers must be omi tted from Eqs. ( 1 1 .2 3 ). If both vol tages are referred to the same b ase, however, the equations are correct as given. Similarly, when l ine and 6. currents are expressed in per unit, each on i ts own base, the 13 in Eqs. ( 1 1 .20) disappears since the two bases are related to one another in the ratio of f3 : l . When the currents are exp ressed on the same base, the equation i s correct a s written. From Fig. 1 1 .4 we note that �"jlab = Z 6. when there are no sources or mutual cou p l i ng inside the c. c i r cuit. When positive- and negative-sequence q u a n t i t i es are both prese n t , we have 1 (1)

VII(hI )

=

IIi>

S u bst i t u t i n g from

Eqs. ( J J .20)

a n c.J ( J J . 2 3 ),

13- Vu(nI )�

13 �

1(\)

II _

(I)

- Z6. -

Va(nl )

so t hat

Z:-,

=

---

fa

Z(). -

3

Va(i,2 )

( 1 1 .24)

1 (2) a i,

we

obta i n

13 V}�)/ 1a( 2 )

13

_

-

30°

/ - 300

5

v ( 2)

( 1 1 .2 )

an

1a( 2 )

which shows that the c. c on nec t e d i mped ances Z6. are equivalent to the per­ phase or Y -co n nected impedances Z y = Z6./3 of Fig. 1 1 .6(a) insofar as posi-

(a)

(b)

FI G U RE 1 1 .6 ( (I ) S y m m e t r i c ; 1 i CI - cO I l Il e c t e d i m r e d ; l!l ccs and t h e i r V-co n n e c t e d e q u iva l e n ts re l a t e d ( b ) Y - c o n n e c t e d i m p e d a nces w i t h n e u t ra l c o n n e c t i o n to g ro u n d .

by

'

2y

=

26/3;

426

CHAPTER 1 1

SYMMETRICA L COMPONENTS A N D S EO U EN C I : N ETWO R KS

tive- or negative-sequence currents are have been anticipated from the usu a l

Of

concerned.

course, this resul t could

/1 -Y transformations o f Tabl e 1 .2. The Z/l/3 is correct whe n the impedances Z/l and Zy are both

Zy expressed in ohms or in per unit on the same kilovo)tampere and voltage bases.

relation

=

1 1 .2. Three identical Y -con ne cted resistors form a load bank wi th a three-phase rating of 2300 V and 5 00 kVA. If the load bank has applied vol tages

Exa mple

I V',c l

=

I V:'11 I

2760 V

= 230() V

find the line voltages and currents i n per unit i n t o the load. Assume that the neutral of the load is not connecte d to the neutral o f the system a nd select a base of 2300 V, 500 kV A. The r a t i n g of the l o a d bank c o i n c i d e s w i t h the s p e c i f i e d base, and s o t he resistance values are 1 .0 per u nit. O n the same base the g ive n line vo l tages i n per unit are

Solution.

I

Vca I

=

1 .0

Assuming an angle of 1 80° for Vc a a n d using the law of cosines to fi nd of the other l ine vol tages, we find the per-u n i t values

Vab

=

/ 82 .8°

=

= =

1 (0 . 1003 + jO .7937 0 . 2792

angles

0 .8

The symmetrical components of the l ine vol tages

=

the

+

� (0 . 1 003

jO .9453 =

+

jO .7937

- 0 . 1 790 - jO .l51 7

+

0 . 2 3 72 + j l . 1 7 63

O.9t{S7/ 73 '()o

0 .2346

+

0 .5

+

jO . 8(60)

p C I' u n i t ( I i n c - to - l i n e

- 1 . 1 373 - jO . 3 8 28 =

a rc

/ 220 .30

+

O .S

--c

v o l t age base )

jO .8660)

per u n i t ( line-to-l ine voltage base )

The absence of a neutral connection means that zero-sequence currents a rc not present. Therefore, the p hase voltages at the load contain positive- and negative­ sequence components only. The p hase voltages a re fou nd from Eqs : ( 1 1 .23 ) with the {3 factor omitted since the line voltages are expressed in terms of the base voltage from line to line and the p hase voltages are desired in per u nit of the' base

1 1 .4 POWER IN TERMS OF SYMMETRICAL COMPONENTS

427

voltage to neutral . Thus,

=

==

0.9857/ 43 .6° per unit (line-to-neutral voltage base)

/ 250 .30

0 .2346

per unit (Iine-to-neutral voltage base)

S i ncc e a c h resistor has an i m p e d a nce

J�I ) / (2 ) a

=

v(l)

1 . 0/� V (2 )

=

1 .0L.Q:

�

=

=

of 1 .0LQ:.

0.9857L

pc r u n i t ,

43 . 6° p e r u n i t

O.2346i.3S0.3°

per

unit

The pos itive direction of current is chosen to be from the supply toward the load. 1 1 .4

POWER IN TERMS OF SYMMETRICAL COMPONENTS

If the sym metrical components of current and voltage are known, the power exp ended in a three-phase circuit can be computed directly from the compo­ nents. Demonstration of this statement is a good example of the mat rix manipulation of symmetrical components. The total complex power flowing into a three-phase circui t through three lines a , b, and c i s S·3
"- ,

jJ

+ JI) �: =

V" ! "*

+

V" I I>*

+

Vt · jC*

( 1 1 .26)

where Va ' Vb ' and Ve are the vol tages to reference at the terminals and la ' lb ' and Ie are the currents flowing into the circuit in the three lines. A neutral conn ection may or may not be present. If there is impedance in the neu tral con nect ion to grou n d , then the vol tages V;/ ' VI" a n d V;. must be in terpreted as vol tages fro m the l i ne to grou nd ra ther t h a n to n e u t r a l . I n matrix notation ( 1 1 .27)

428

CHAPTER 1 1

SYMMETRICAL COMPONENTS AND SEQU ENCE N ETWORKS

where the conjugate of a matrix is understood to be composed of elements that are the conjugates of the corresponding elements of the original matrix. To introduce the symmetrical components of the voltages and currents, we make use of Eq. ( 1 1 .8) to obtain ( 1 1 .28) ] (0)

va( O)

a

Va( 2 )

where

and

( 1 1 .29)

(Z )

Ia

The reversal rule of matrix al gebra states that the transpose of the prod uct of two matrices is equal to the p roduct of the transposes of the matrices in reverse order. According t o this rule,

( 1 1 . 30) ( 1 1 .3 1 )

and so Noting t h at AT s 3

=

=

A and that a

[ V (O)

and a Z are conjugates, we obtain

Va( l )

a

1

1 1 1

1

1 1

1 ( 1 1 . 32)

1

a

or smce

1

1

1

a 534>

=

3

v(O) a

o 1 o

1 1

V(l) a

V ( Z) a

1a( 0

)

U 0 1

*

( 1 1 . 33 )

So, complex power is

which shows how complex power (in voltamperes) can be computed from the symmetrical components of the voltages to reference (in volts) and line currents

1 1 .5

SEQUENCE CIRCUITS OF Y AND A IMPEDANCES

429

(in amperes) of an unbalanced three-phase circuit. It is important to note that the transformation of a-b-c voltages and currents to symmetrical components is power-invariant in the sense d iscussed in Sec. 8.6, only if each product of sequence voltage (in volts) times the complex conjugate of the corresponding sequence current (in amperes) is multiplied by 3, as shown in Eq. ( 1 1 .34). When the complex power S 3 1> is expressed in per unit of a three-phase voltampere base, however, the multiplier 3 d isappears. Example 1 1 .3 . Using symmetrical components, calcu late the power absorbed i n the load o f Example 1 1 .2 a n d check the answer. Solution.

In per unit of the three-phase 500-kYA b ase, Eq . ( 1 1 .34) becomes

Substitut ing the components

obtain S )
=

=

=

0 +

O.9S57[ 4 3 .�".

X

( 0 .9857) 2 + ( 0 .2346) 2

of

vol tages

0.9857[ . . 4 3 .6° =

cu rre nts from Example 1 1 .2 , we

,md

+

0.2346/>50 .3"

X

0.2346/ - 250 .3"

1 .02664 per unit

5 1 3 .32 kW

The per-unit ya)ue o f the resistors in each phase of the Y-connected load b a n k is 1 .0 per unit. In ohms, there fore, 2 ( 2300) = = 1 0 .58 D R Y 500 ,000 and the equivalent 6.-connected resistors are R Cl.

=

3Ry

=

3 1 .74

D

From the given line-to-line vol tages we calculate directly

=

11. 5

.

( 1 840) 2

+

(2760) 2 3 1 .74

+

(2300) 2

- -

SEQUENCE CIRCUITS OF Y AND � ' IMPEDANCES

= 5 1 3 .33 kW

impedance Zn is inserted between the neutral and ground of the Y �co nnected impedances shown in Fig. 1 1 .6( b ), then the sum of the line currents is equal to If

430

CHAPTER 11

SYMMETRICAL COMPONENTS AND S EQUENCE NETWORKS

the current /n in the return path through the neutral. That is, ( 1 1 .35 ) Expressin g the u nbalanced l ine currents in terms of their symmetrical compo­ n ents gives

( /a( 0)

=

+

/b(0 )

+

1(0» ) c

+ ( / ( 1 ) + /b( 1 ) + fI

1(1» ) c

�---- --�

o

3 1 (0)

+

( 1(2) a

+

1 (2) h

()

a

+

1 (2» ) c

(II

.36)

Since the positive-seq uence and nega tive-sequence cu rrents add sep a rately to zero a t neu tral point n, there cannot be any positive-sequence or negative­ sequence currents in the connections fro m neutral to ground regardless of the value of Zn ' Moreover, the zero-sequence currents combining together at n become 3 /�0) , which produces the vol tage drop 3 1� 0) Z /l between neutral and ground. It is important, therefore , to d ist inguish between voltages to neutral and voltages t o ground under unbal anced cond itions. Let us designate volt ages of phase a with respect to neutral and ground as Va ll and � , respectively. Thus, the voltage of phase a with respect to ground is given by Va = Van + v,, , where � = 31�0)ZTI ' Referring to Fig, 1 1 .6( b ), we can write the vol tage d rops to ground from each of the l ines a, b, and c as ( 1 1 .37) symmetrical compon ents a s fo l l ows:

The a-b-c voltages

a n d cu rr e n t s i ll

A

Va(O) V}t)

t h is equation

be

1 (0)

rep laced by t he ir

a

=

ZyA

J! I )

( 1 1 .38)

1a(2)

Va(2)

M U ltiplying across by the inverse matrix

· v
c a ll

1(2) a

A- I,

we obtain 1

+

3 1a(0)Zn A - I 1 1

1 1 .5

Postmultiplying

of A- I ,

1

A - 1 by [1

SEQU ENCE CIRCUITS OF Y AND A IMPEDANCES

' I V amounts to adding the elements in each row

v (O )

1

Va( I ) a

and so

431

Zy

=

V( 2)

+

3 1a(0)Zn 0

( 1 1 .39)

o

a

In expanded form, Eq. ( 1 1 . 39) becomes three separate or decoupled equations, v (O ) a

vel)

va( 2 ) a

( Z Y + 3Z

=

n

) 1(0) a

ZY/ ( 1)

=

a

Z Y 1 (2 )

=

a

=

=

=

Z O 1(0) a

( 1 1 AO)

Z I 1(J( 1 )

( l 1 A1 )

Z 2 /a(2)

( 1 1 .42 )

It is customary to use the symbols Z o , Z l ' and Z 2 as shown. Equations ( l 1 AO) throug h ( 1 1 .42) could have been easily developed in a less formal manner, but the matrix approach adopted here will be usefu l in developing other important rel ations in the sections wh ich follow. Equati ons ( 1 1 .24) and ( 1 1 .25) combine with Eqs. ( 1 1 .40) through ( 1 1 .42) to show that currents of one sequence cause voltage drops of only the same sequence in /:::,. ­ or Y -connected circuits with symmetrical impedances i n each phase. This most important result allows us to draw the three single-phase sequence circuits shown i n Fig. 1 1 .7. These three circuits, considered simultaneously, provide the sam e information as the actual circuit of Fig. 1 1 .6 ( 6), and they are independent of one another because Eqs . ( l 1 AO) through ( 1 1 .42) are decoupled . The c ircuit of Fig. 1 1 .7( a ) is called the zero-sequence circuit because it relates t he zero­ sequence voltage Va( O ) to the zero-sequence current 1� O), and thereby s e rv e s to define the impedance to zero-sequence curren t given by v(J«( I )

1 (0)

=

(1

a �

+Vt iOl -�

,0

J (O)

Zy

a

n

3 Z"

Zo

(a)

Reference

FI G U R E 1 1 .7

+t

VO) "

J(l)

a -----

Z'

I

Zo

Zy

=

2y +

3 2 1/

( 1 1 A3)

n

I

+

Reference

Reference

(6)

(c)

Z e r o - , posi t iv e - , a n d n e g a t i v e - s e q u e n c e c i r c u i t s for Fig.

l 1 .6(b).

432

CHAPTER 1 1

SYMMETRICAL COMPONENTS AND SEQUENCE NETWORKS

impedance to positive-sequence current , whereas Fig. 1 1 .7( c ) is t h e negath'e­ sequence circuit and Z 2 is t h e impedance to negative-sequence current . The

Likewise, Fig. 1 1 .7(b) i s called the positive-sequence circuit and Z I is called the

names of the impedances to currents of the d ifferent sequences are usually shortened t o the less descriptive terms zero-sequence impedance 2 0 , positive­

sequence impedance 2 I ' and negative-sequence impedance 22 , Here the positive­ and n e g at iv e -s e qu e nc e i m p e d a n ces Z I < t n d Z2 ' r e s r c c t i v c l y, a rc /1 o t h fou n d t o b e equal t o the usual per-phase i m p e d a n ce Z y , w h i c h i s ge n e r a l l y t h e c a s e for stationary sy mme t ric a l circuits. Each or the three seque nce c i rc u i ts re p r e sen t s one p hase of th e actual three-phase circuit when t h e l a t t e r c a r r i e s c u r r e n t o f only t h a t seq u e nce . W h e n t h e t h r e e s e q u e nce c u r r e n t s a r c s i m u l t a n e o u s l y p re s e n t , all three sequence c i rcuits a re n e e d e d t o fu l ly r e p r e se n t t h e o r i g i n a l circuit.

Voltages in the positive-sequ ence and n e g at ive -s e q u e n c e c i rc u i t s can be regarded as vol tages measured with res p e c t to e ither neutral or grou nd whet h e r o r not there i s a connection of some finite value of i mpedance Zn b e tw e e n neut r a l and grou nd. Accordingly, in t h e positive-sequence circuit there is no d ifference between V ( I ) and V ( I ) and a s i m i l a r s t a t e m e n t a p p l i e s t o V ( 2 ) and Va�) in t h e negat ive-sequ ence circu it. However, a voltage d i fference can exis t between the n eutral and the reference of the zero-sequence circuit. I n the circuit o f Fig. 1 l .7( a) th e current 1� 0) flowing through impedance 3Z n produces t h e same voltage drop from neutral to ground as the current 3 1�0) flowi ng through impedance Z n in the actual circuit of Fig. 1 1 .6(b). If the neutral of t h e Y -conn ected circuit is grounded through zero impedance, we set Zn = 0 and a zero- imped ance connection then joins the neutral point t o the reference node of t he zero-sequence circuit. If th e re i s no connection between neutral and ground, there cannot be any zero-sequence current flow, for then Zn = which is indicated by the open circuit between neutral and the reference node i n the zero-sequence ci rcui t of Fig. 1 1 .8( a). Obviously, a L1-connected circu i t cannot provide a path through n e utral, and s o line currents flowi ng into a L1-con nected load or its equivalent Y circuit cannot cont ain any zero-seq uence components. Consider t h e symmetrical L1 connected circui t of Fig. 1 1 .4 with U II

U

a

,

00 ,

( 1 1 .44 ) Adding the three prece d i n g

equations together, we obtain

( 1 1 .45) and

since the

sum of the

l i ne - t o- l i n e voltages is always zero, we therefore have

v
=

[a(0b)

=

0

( 1 1 .46)

Thus, in �-connected circuits with impedances only and no sources or m'u tual coupling t h e re cannot be any circulating c u r r e n t s . S o m e t imes s ing l e -p h a s e

1 1 .5

1( 0) �

SEQUENCE CIRCUITS OF Y AND A IMPEDANCES

433

Z

a +jl : y

I_ � Reference

z,..

FI G U R E 1 1 .8 ( a ) Un grounded Y-co nnected and (b) 6-connected circuits and their zero-sequence circ uits.

Reference

( b)

circula ting currents can be produced in the 11 circuits of transformers and generators by either induction or zero-seque nce generated voltages. A 11 circuit and its zero-sequence circuit are shown in Fig. 1 1 .8(b). Note, however, that even if zero-sequence voltages were generated in the phases of the 11, no zero­ sequence vol tage could exist between the 11 term inals, for the rise in vol tage in each phase would then be matched by the vol tage drop in the zero-sequence impedance of each phase. Three equal impedances of j21 .n are connected in .1 . Determine the sequence impedances and circuits of the combination. R epeat the solution for the case where a mutual i mpedance of j6 n exists b e tw e e n each pair of adj acent

Example 1 1 .4.

branche� in the

6. .

Vl an 1 lj2 1 1 2 j l j l V}2)

Solution. The line-to-line volt ages a rc related t o the 6. currents by VI,..-

r< "

0

0

j2 !

0

0

Transform ing to symmetrical c ompon e n ts of voltages and curren ts gives

A

VoOj,. )

2)

va(t>

=

0

0

o

j2 1 a

434

CHAPTER

11

j7

: --++0

SYMMETRICAL COMPONENTS AND SEQUENCE N ETWORKS

� �t � . Reference

+

I t -](l) a

V ( 2)1

V (l)

Positive-sequence

Zero-sequence

�t :

,

](2 ) . a

'7 J

'

Reference

Negative-seq uence

� � +t l[j] '� +t 'D +! D (a)

V(O):

�+ :

j21

j3 3

[(0)

.

V ( l ):

I (1

�t :

Refere nce

,

)

j5

j5

V ( 2 }1

Reference

Positive-sequence

Zero-sequence

II �t :

I ( 2)

,

R e fere n c e

N eg ative-seq uence

(b) FIGURE 1 1.9 Zero-, positive-, a n d negat i ve-se quence ci rcu i ts for Cl-con nected i m p e d a n ces o f Ex a m p k 1 1 . 4 .

a n d premuitiplying ea c h side by A - I ) we obtain o

j2 l o

Thc pos i t ivc a n d negat ive-se q u e nc e c i rcu i t s have p e r- p h ase i m p e d a nces Z = 22 = i7 fl, as shown i n Fig, 1 1 .9( a ), a n d s i nce V"cJ}) 0 , t h e zero-se q u e nc e c u r r e n t I��) = 0 so that the zero-sequence circuit is a n o p e n c i rc u i t . T h e j 2 1 -fl res i s t a n c e in the zero-sequence network has s i g n i A c an ce o n l y w h e n t h ere is an i n ternal sou rce in the or i g i n al t1 c i rc u i t . When there is mutual i nductance j6 n b e t w e e n phases, I

-

=

A

[V}2) l lj2 1 (1l .Vnb

V (2l ab

=

i6 j21

)' 6

j6

)' 6

[1

6 j ] [/ (0) ] j6 A 2) J'21 1a( b

I�i)

The coefficient matrix can be separated into two parts as follows: i 21 , [ j6 i6

i6 j21 i6

i6 j6 ] = j1 5 0 a i 21

o 1 a

1 1 1

1 1 :6

and substituting into

SEQUENCE CIRCUITS OF A SYMMETRICAL TRANSMISSION LINE 435 the previous equation,

( O) Vab

Va(bl )

2) v( ab

::::

j 15 A - JA

we

1 1

+ j6 A - [

obtain

1

1

1

1

1

1

1

) 1 (0 ab

A

1 j 6 r� ° lW;� l {rT 'D) n3 WJ L

�

0

=

115

0

0

jlS

�

) Ia( bI ( 2) 1ab

0

j15

o

o

j 15

0

0

0

+

0

o

o

/(1) ab / (2 ) ab

II h

/(1)

!�;) li b

The positive- and negative-seque nce i m p e d a nce s Z 1 and Z 2 now take on t h e value jS n , as shown in Fig. 1 1 .9(b), and since �il?) 1��) 0, t h e zero-sequence ci rc u i t is open. Ag(}in, we note t h a t the j33-[2 resistance in the zero-sequence network has no significance because there is no i nterna! source i n t h e o r igin a l 6 circuit, =

=

The matrix manipu lations of this exa mple are useful in the sections which fol low . 1 1.6

SEQUENCE CIRCU ITS O F A SYMMETRICAL TRAN SMISSI O N LIN E

W e are con cerned primarily with systems that are essentially symmetrically hal anced and which become unbalanced only u pon the occurre nce of an u n symme trica l fau l t. I n p ractical t ransmission systems such complete symmetry is more i d e a l t h a n realized, but s i nce the effect o f the departure from symmetry is usually smal l , perfect bala nce between ph ases is often assumed especially if the l i nes a re t ransposed a long their l e ng ths. Let us consider F i g 1 1 . 1 0, for i ns l Ll n c c , w h i c h s h ow s o n e s e c t i o n o f a t h r e e p h a se transm ission l ine with a n e u t ral conductor. The se lf-imp ed ance Z a a is the same for each phase conduc­ tor, and t h e neu t ral conductor has sel f- i mpedance Z"" . When currents la ' 1b , a n d Ie i n t h e p h a s e co n d u c t o rs a rc u n b a l a n ce d , t h e n e u t r a l c o n d u c t o r serves as a ret urn path. A l l the currents are assumed positive in the d i rections shown even though some of their numerical values may be negative under unbalanced conditions caused by faul ts. Because of mutual coup l ing, curren t flow in any one of the phases induces volt ages in each of the other adjacent p hases and i n the neu tral conductor. Similarly, In in the neutral conductor induces voltages in each of the phases. The coupl i n g between all three phase conductors is regarded as being symmetrical and m u tu a l impedance Z a b is assumed between .

-

436

CHAPTER 1 1 SYMMETRICAL COMPON ENTS AND SEQUENCE N ETWO R KS

Zo b Zo b Z o Z o b Zo I e l Vb!n +t

�+�-----

10

a'

a __

Ib

b'

b�

Ie

V• •

_

Vc n

_

_

!

�

I n>..==n= ,�

Z --C== n n�1

_ _

--------r+�

'

c

L I

Zo n

n

FIGURE 1 1. 10 Flow of u n b a lanced cu rre nts in a symmetrical t h rce- phasc - l i n e

'

�I +

±-

Ve . n .

s ec t i o n w i t h

n e utral

conductor.

each pair. Lik ewise, the m u t u a l i m pe d a nce b e t w e e n t h e n e u t r a l co n d u c t o r a n d each of t h e p hases i s taken to b e Za n " The voltages induced i n p hase a , for example, b y currents i n the other two phases and the neutral conductor are shown as sources in the loop circu i t of Fig. 1 1 . 1 1 , along with the similar voltages induced in the neutral con ductor. Applying Kirchhoff's voltage law around the loop circuit gives

( 1 1 .47) from which voltage drop across the line section is found to be v:,n - Va'n' = ( Zaa - Zan ) la

+

( Za b - Z an ) ( lb + IJ + ( Zan - Znn ) In ( 1 1 .48)

Similar e q uations can be written for p hases b and V::n - �' n ' = ( Zaa - Zan ) Ie

T �

a

10

�

Va n

_

n

In

�

Zab 1b

+

Z an Ia

+

Za n In

+

c

as follows:

( Zab - Z an ) ( la + Ib ) + ( Zan - Znn ) ln

Zo Zn ,.

( 1 1 .49)

t

T n'�

a

Va' ' n

F1 G U RE 1 Ll 1

W r i t i n g Ki rchhoff's voltage e q u a t ion a r ou n d t h e loop formed by line a and the. n eutral conductor.

1 1 .6 SEQUENCE CIRCU ITS OF A SYMMETRICAL TRANSMISSION LINE

437

Wh en the line currents la ' Ib, and Ie return together as In in the neutral

conductor of Fig. 1 1 . 10, we have

( 1 1 .50)

Let us now substitute for In in Eqs. ( 1 1 .48) and ( 1 1 .49) to o b ta i n

+

( 7" /1

+

21111 -

( 1 1 .51 )

2 211/1 ) I,.

T h e coefficients in these equations show t hat the presence

o f the n eu tral conductor changes the self- and mutual impedances of the p h ase conductors to the fol lowing effective values:

( 1 1 .52)

� 'J l

�z

Using these definitions, we can rewrite Eqs. ( 1 1 . 5 1 ) in the convenient m atrix

form

Va a

V ,

. 1> / ,

V;· r'

=

V

(i l i

V/>11

V;

"

-_

vo " n

V/! "

V;·'n' /I

J

=

2/11 .r

( 1 1 .5 3 )

Z/II

w h e re the vol tage drops across the p hase conductors a r e n o w denoted by V , ,@ V UU

(l lP

- vII " n

Vee'

£

V:·n

-

v:.,

n'

( 1 1 .54)

S ince Eq. ( 1 1 .53) does not explicitly include the neutral conductor, Z s and Zm can b e regarded as parameters of the phase conductors alone, w ithout any self-' or m u t u a l indu ctance being associated with the return path. The a -b-c vol tage d rops and currents of the line s ec t i on can be written in terms of their symmetrical components according to Eq. ( 1 1 .8) so , that with

438

.

CHAPTER

II

SYMMETRICAL COMPONENTS A N D SEQUENCE NETWO RKS

phase a as the reference phase, we h ave A

Va(0) a Va(l) a

Va(a2) ( 1 1 . 55 )

This pa rticular form of the equation m akes calcula tions easier, as demonstrated i n Example 1 1 .4. Multiplying across by A I , we obtai n -

v(O) 00

Ve l)

1 1

v (2) aa

1 1

aa

( 1 1 .5 6 )

The matrix multiplication here i s t h e s a m e as in Exa mple 1 1 .4 and yields

va(0) a

V(I)

( 1 1 .57)

aa

va(a2) Let us now define zero-, positive-, and negative-sequence impedances in terms of 2s and 2m introduced in Eqs . ( 1 1 .52), ( 1 1 . 58)

From Eqs. ( 1 1 .57) and ( 1 1 .58) t h e sequence components of the voltage drops between the two ends of the line section can be written as three simple equations of the form

v(�) aa v(�) aa va(a2')

= = =

Van(O)

-

V el) an

Va�o� 11 Va� nl �

Van( 2) - Va�2n�

= = =

1(0) 2 1(1 ) 2 1(2)

2

0

a

I

a

2

a

( 1 1 . 5 9)

Because of the assumed symmetry of the circuit of Fig. 1 1 . 1 0, once again we see

1 1 .6

+r V O)

-t

[(0)

a

a

--

SEQUE:--I C E CI RCUITS OF A SYMMETRICAL TRANSMISSION L INE

Zo

I Va��' �

a

(

a ll

+r V(l'

-t

n

a

-+

I '

n

/(1) �

ZI

-r V I

a

�

a('�)'

a ll

+tV (2)

439

n

a

n

1(2) a

-

Z2

j+ V n� I

a

(?),

on I

a II

n

FIGURE 1 1 . 1 2

Seque ncl: c i rc u i t s for t h e symm e trical l i n e section of Fig. 1 1.10.

that the zero-, posltlve-, and negat ive-sequence equa tions decouple from one another, and correspond ing zero-, posit ive-, and negative-sequence circuits can be drawn without any mutual cou pling between them, as shown in Fig. 1 1 . 12. Despite the simplicity of the line model in Fig. 1 l . 1 0, the above development has demonstrated importa n t characterist ics of the sequence impedances which apply to more elaborate and pract ical line models. We note, for instance, that the positive- and negat ive-sequence impedances a re equ al and that they do not i nclude the neu tral -con du ctor i m pedances Zn Jl and Zan ' which enter into the calcul ation of only the zero-sequ ence imped a nce Zo, as shown by Eqs. ( 1 1 . 58). In o t h e r wo r d s , i mped a n ce p a r a m e t e rs of the retu rn-pa th conductors enter i nto the values of t he zero-sequence impe dan ces of transmission lines, but they do not affect e i t h e r the p os i t i v e or negative-sequence impedance. Most aerial tra nsmission l i nes have at least two overhead condu ctors cal led ground wires, which are grou n d ed at u n i form intervals along the length of the l i ne. The ground wires combi n e with the earth re t ur n path to constitute an e ffe c tive n e u t r a l co n d u c t o r w i t h i m ped a n c e p a ra m e t e r s , l i k e ZlI n a n d Z an ' which depend on the re sistivity of the earth. The more spec i alize d literature shows, as we have demonstrated here, that the parameters of the return path are i nclu ded in the zero-sequence impedance of the line. B y regarding the neu t ral conductor of Fig. 1 1 . 1 0 as the effect ive return path for the zero­ sequence compone nts of the unbal anced currents and including its parameters in the zero-sequence impedance, we can treat the ground as an i deal conductor. The voltages of Fig. 1 1 . 1 2 are then i nterpreted as being measured with respect -

I

440

CHAPTER 1 1

S Y MMETRICAL COMPONENTS AND S EQ U ENCE N ETWORKS

to perfectly conducting ground, and we can write VCO)

V CO) - V CO) '

=

aa

a

V ( l)

V ( l ) - V ef l )

=

aa

V ( 2) oa

a

a

a

V (2 ) - V ('2 ) a Il

=

=

=

=

Z /a( 0)

0

Z I /a( 1 )

( 1 1 .60 )

Z 2 [( 2) a

where the sequence com ponents or the vol tages V:, a n d Vu' arc n ow w i t h respect to ideal ground. In derivi ng the equations fo r ind uctance and capacitance of tra nsposed transmission lines, we assumed balanced th ree-phase currents and did not specify p hase order. The resu l t i ng p a r a m e t e r s are therefore val i d fo r both positive- and negative-seque nce impeda nces. When only ze ro-sequence current flows in a transmission line, the cu rrent in each phase is identical. The current returns through the ground, through over head ground wires, or t hrough both. Because zero-sequence current is identical in e a ch phase condu ctor ( r a th e r than equal only in magnitude and d isplaced in p h a s e by 120 from other phase currents), the magnetic field due to zero-sequence current is very different from the magnetic field caused by either positive- or negat ive-sequence current. The d i fference in magn etic field results in the zero-sequence i n d u c t ive react ance of ove rhead t ransmission lines b e in g 2 to 3 .5 times as large as the posi tive­ sequence reactance. The ratio i s toward the h igher portion of the specified range for double-circuit l ines and l i nes without ground wires. Example 1 1.5. In Fig. the terminal voltages at the left-hand and right-hand e nds of the line are given by

11.10 Val +.170.0 Vbl 170.24 Zaa j60 2ab j20 fa' Ib' Zo Za 2Zab 32 -6Za ZI Z2 Zao -Zab j60 -j20 = 1 82 . 0 kY = 72 . 24 - j32.62 k Y

V::n

-

=

Va'n' 154.0 + Vb'I ' Ve'd + Znn

= 44 .24 - j 74 . 62 k Y

=

+ j88 .6 2 kY

The line impedances in ohms are

=

j28 .0 kY

=

=

- 1 98 . 24

= j80

Z

j46 .62 kY

a ll

=0

D etermine the line currents a nd Ie using symmetrical components. Repeat the solution without using symmetri cal components. Solution.

The sequence impedances have calculated values

=

=

+

=

+

nn

=

n

=

=

+

0 j40 + j240 -j180 j1 60 j40

j6 '

n

=

n

1 1.6 SEQUENCE C I RCUITS OF A SYMMETRICAL TRANSMISS I ON LINE

[ VaVa��/1 l [VaVbnn -Vb-Va''nn'' l V};I v:.n -v:.'n' [228.8.00 ++ jj442.2.00 1

441

[ (1(72.82.204 --44.154.204)) +- j(j(730.2.062-28.-74.0)62) 1 -(170.24 - 198.24) + j(88.62 -46.62) [28.0 +0j42.0 1

The sequence components of the voltage drops in the line a re

= A-

=

A-

= A-I

)

I

28.0 j42.0 +

=

kV

0

S u bs t i t u t i n g i n Eq. ( 1 1 .59), we obt a i n

V«a (l�) = 20 , 000 + ;'42 , 000 = j' 1 6 0 !(/(O)

j40/( l )

()

=

()

= j40 ! (2)

a

II

from which we d e te r m i n e t h e symmet rical components of t h e

I� O)

=

262.5 -j1 75

currents in phase

a,

A

a b 262.(1 1.552)-j175 + Znn -2Z(1l j60 + j80 -j60 j80 Z"" - 2ZIlf + j60 j40 (1 1.53) j 10) [j80j40 j40j80 j40 j40 j40j40 j - l [2828 +j+ j4422 j 103 [262.262.55 -j175 j -j175

The l i n e c u rrents are therefore

I

=I

=

Ie

=

A

The s elf- a n d m u t u al impeda n ces of E q .

Zs Zm

= Za a =

[V(1V"(1'' j [ 2828 . 28

h ave val u e s

=

= j20

Zil " +

JRO -

�e'

n

=

n

without symmetrical

a n d so l i n e curren ts can be c al c u l a ted from Eq, components as fol l ows:

=

=

+ j42 +

j42

+ j42

X

=

j80 28 + j42

X

=

262.5 -j175

A

442

CHAPTER I I

S Y MM ETRI CAL COMPONENTS AND SEQUENCE NE1WORKS

--

a .,....-

Z"

______

I"

FIG U RE 1 1 . 1 3

b

c

C i rc l l i t d i a g r < l m o f

Eall , E"n , t h ro u g h

a

�!

g e n e r a t or g ro u n d e d

Ecn arc

reac t a n c e .

and

The

r h asc

e m fs

po� i t ivc seq u e n c e .

11.7

SEQUENCE CIRCUITS OF THE SYNCHRONOUS MACHINE A synchronous generator, grounded through a reactor, is shown in Fig. 1 1 . 1 3 .

Whe n a fau l t (not indicated in t h e figure) occurs a t t h e term inals o f t h e generator, currents la ' lb ' and Ie flow in t h e lines. If the fa u l t invulves ground, the current flowing i nto the neutral of the generator is designated In and the line currents c a n b e resolved into their symm etrical com ponents regard less of how u nbal anced they m ay be. The equations developed in Sec. 3.2 for the idealized synchronous ma­ chine are all based on t h e assumption of b alanced i nstantaneous armature currents. In Eq. (3 .7) we assumed that i a + i b + ie 0, and then set i a - (i b + i) i n Eq. (3.5) i n order to arrive a t E q . (3. 1 ] ) for the terminal voltage of p hase a i n t h e form =

=

( 1 1 .61) The steady-state counterpart of t h is equation i s given i n Eq. (3 .24) as

( 1 1 .62) where Eo n is the synchronous internal voltage of the machine. The subscripts of the voltages in Eqs. (1 1 .61) and (1 1 .62) d i ffer slightly from those of Chap. 3 so as to emph asize that t h e vol tages are with respect to neutral. If the subs t it u tion La = - (ib + iJ had not b een made as i nd icated, then we would h ave found Va n

=

�

Ri a

-

di a

LS - +

dt

d

M - ( ib 5 dt

+ ie) +

ean

( 1 l .63 )

Assuming for now t ha t steady-state s i n usoidal curren ts and voltages of nominal

l i .7

443

SEOUENCE C[ RCUITS O F THE SYNCH RONOUS MACH INE

system frequency w continue to exist i n the armature, we can write Eq. (1 1 .63) i n the phasor form

( 1 1 .64 ) where E a n again designates the phasor equivalent of e a n . The armature p h ases b a n d c of the idealized machine h ave similar equations

V[ I/ I 1

( 1 1 .65)

We can a rrange �>n

=

-

�

Eq s .

[lu 1 [1

( 1 1 .64) and 0 1 .65) in vector-matrix form as follows:

[ R + j w ( L , + MJ ]

Ih �.

+ j w Ms 1 1

1 1 1

( 1 1 .66)

Followi ng the procedure demonstrated in the two preceding sections, we now express the a -b-c q u a n t i t i e s of the m achine in t er m s of symmetrical components of p h a s e a of the armature

VV( I) (2)

vall( O) all all

=

- [ R + jw ( L s + MJ ]

+

j(u M 1 A - 1

1

1

1

1

1

E'III

1 (0)

A I II

1/

(I)

1 II( 2 )

2£ + A- I a .

( 1 1 .67)

1111

a E O II

S ince the synch ronous g e n e r a tor is designed to su pply balanced three-phase vol tages, we have shown t h e generated vol tages Ea ll • E'm ' and Ec n as a posit ive-sequence set of phasors i n Eq . ( 1 1 .67), w h e r e the operator a = 1 /1 200 anu

a2

V ( 2)

=,

/2400 . T h e

m ;l t r i x

of Eqs . ( 1 1 .56), anu so we

va(nO) V(I

an an

)

=

In

li l t i p l i c a t i o n s o f Eq .

o b ta i n

- [ R + jw ( Ls + Ms ) ]

1 (2) /aCO)

I a(

a

I)

3

+ jw Ms

( 1 1 . 67)

ar e

s i m i l a r t o t h ose

1a(0 ) 1a( 1 ) 1a(2 )

0

+

Ean

0

( 1 1 . 68)

,.

444 '

CHAPTER 1 1

SYMMETRICAL COMPONENTS AND SEQUENCE NETWORKS

The zero-, positive-, a n d negative-sequence equations decouple to give

v(O an ) Va(nl ) v ( 2) all

=

= =

-

RIa( O) - J'w( L s - 2 Ms ) 1a(0)

- RIaO ) - J'w ( L s + M ) 1a( 1 ) + Ea n - Ri 1Ia)

- J· w ( / .

.\"

+

M.,. ) /,(, :!)

Drawing the corresponding s e q u e n c e c ircu i ts is ( 1 1 .69) in the form

a(n0 )

vel) - E

v

=

Gil

vall(2 )

=

(l ll

( 1 1 .69)

5

m ade simple by w r i t i n g

--- E a ll

-

j (O ) Z

-

1a( 1 ) ,c' 7l

(l

Eqs.

!; o

- 1a(2)2 2:

( 1 1 .70)

where Zg O ' Z I ' and Z2 are the zero-, posltlve-, and negative-sequence impedances, respectively, of the genera tor. The sequence circuits shown in Fig. 1 1 .14 are the single-phase equivalent circuits of the balanced three-phase m a ch i n e t h rough which the symmetrical components of the unbalanced currents are considere d to flow. The sequence components of current a re flowing through i mped ances of their own sequence o nly, as ind icated b y the appropriate subscripts on the impedances shown in the figure. This is because the machine is symmetrical with respect to phases a , b, and c . The positive-sequence circuit is comp osed o f an emf in series w ith t he positive-sequence impedance of t he generator. The negative- and zero-sequ ence circuits con tain no emfs b u t i nclude the i m p e d a nces of t h e g ene r ato r to negative- a n d zero-sequence c u rren ts, respectively. The reference node for the posi tive- and negative-sequence circ u i ts is the neutral o f the gene rator. So fa r as p os itive- and negat ive-seque nce com ponents a r c co n c e rn e d , t h e n e u t r a l o f t h e g e n e r a t o r is a t g ro u n d p o t e n t i a l i f t h er e is a connection between neutral a n d ground h aving a finite or zero i mpedance s ince the connection will carry no positive- or negative-sequence curre nt. Once again, w e see that there is no essential difference betwee n V} l ) and Va� ) i n the positive-sequence circui t or between VP) and Va<'� ) in the nega tive-sequence circuit. This explains w hy the posit ive- and negative-sequence voltages v}l ) and of Fig. 1 1 . 14 are written without subscript n . The current flowing i n the impedance Zn between neutral and ground is 3I�O), By referring to Fig. 1 1 . 1 4( e ), we see that the voltage drop of zero sequence from point a to ground is - 3 /�O) Zn - I�O )Zg O ' where Zg O is the zero-sequence impedance per phase of the generator. The �ero-sequence cir­ cuit, which is a single-phase circui t assumed to carry only the zero-sequence

VY)

1 I .7

( a) Positive-sequence current paths a

SEQUENCE CI RCUITS OF THE SYNCHRONOUS MACHINE

44S

Reference ( b ) Positive-sequence network

.------'/ (2) a

n

c

(c) N e g ative-seque nce current paths

Reference ( d ) Negative-sequence network

'-_--.. ---, n

Reference ( e ) Zero-sequ ence current paths

(n Zero-s equence network

FI G U R E 1 1 . 1 4 P a t h s f o r c u r r e n t o f each sequence i n a g e n e r a t o r and t h e correspo n d i n g s eq u e nce n e tworks.

cu rrent of one phase, m u s t t here fore have an impedance of 3 Zn + Zg a , as shown in Fig. 1 1 . 1 4( / ). The total ze ro-sequence impedance through which J�a) fl ows is ( 1 1 . 71) U s ually, the components o f current and vol tage for phase a are fou n d from equations determined b y t h e sequence circu its. The equa ti ons for the components of vol tage drop from point a of phase a to t he reference node (or

446

CHAPTER I I

S Y M M ET R I CA L CO M PO N ENTS A N D S EO U FN ( ' I ;. N I ;:rWO I { K S

ground) a re written from Fig. 1 1 . 1 4 as

v(O) - Q

V (I ) a

Va( 2 )

=

= =

£ a ll

-

/(l)Z a

1

( 1 1 . 72)

where £ /1/1 is the positive - seq u e nce voltage to n e u t ra l , Z 1 a n d Z 2 arc the positive- and negative-sequence i m p e d a nces of t h e generator, respectively, and 2 0 is d e fi n e d by Eq. ( I 1 .7 1 ). The equations d eveloped to this p o i n t are b a s e d o n a s i m p l e m J c h i n e model which assumes the cxistence o r o l 1 l y rU I l lL l I ll c n l a l c o m p o n e n t s o r c u r ­ rents; o n this basis the positive- and negative-sequence impedances are fou n d to be equal to one another b u t qu ite different from the zero-sequence impedance. I n fact, h owever, the impedances of rotating machi nes to currents of the th ree sequences will generally be different for each sequence. The m m f produced by negative-sequence armatu re current rotates in the direction opposite to that of the rotor wh ich has the dc field winding. U n l ike the flux produced by pos i t ive­ sequence current, which is stationary with respect to the rotor, the flux pro­ duced by the negative-sequence current is sweeping rapidly over the face of the rotor. The currents induced in the field and damper windings cou nteract the rotating mmf of the armature and thereby reduce the flux penetrating the rotor. This condition is similar to the rapidly changing fl ux immediately u pon the occurrence of a short circu i t a t the terminals of a machine. The flux paths are the same as t hose encou ntered in eva luating subtransient reactance. So, in a cylindrical-rotor m achine subtra nsient and negative-sequence re act ances are equal. V a l ues given in Table A.2 in the Appendix confirm this statement. The reactances in both the positive- and negative-sequence circuits are often taken to be equal to the subtransient or transient reactance, depending on whether subtransient o r transient conditions a re being studied. When only zero-sequence c u rrent flows in the armature wind ing of a three-phase m ach ine, the current and m m f of one phase are a maxi mum at the same t i m e as t h e current and mmf of each of the other phases. The windings are so distributed around the ci rcu m ference of the armature that the point of maxi m u m m m f produced by o n e phase is d i s p l ac e d 1 20 e l e ctrical d eg r ee s i n space from the point of maximu m mmf of each of the other phases. If the m m f p roduced b y t h e current of each p h ase had a perfectly s i nusoidal distribution i n space, a plot o f m m f around t h e armature wou ld result in three s in usoidal curves whose sum would be zero a t every point. No flux would be produced across the air gap, and the only reactance of any phase winding woul d be that due to l eakage and end turns. In an actual machine the winding is not distributed to p roduce perfectly s i nusoida l mmf. The flux resulting from the,sum o f the m m fs is very small, wh ich makes the zero-sequence reactance the smallest

SEOUENCE CIRcu i TS OF THE SYNCHRONOUS MACHINE

1 1 .7

447

of the m achine's reactances-just somewhat higher than zero of the ideal case where there is no air-gap flux d u e to zero-sequence current. Equations (1 1 .72), wh ich apply to any generator carrying unbalanced currents, are the starting points f or the derivation of equations for the compo­ nents of current for different types of faults. As we shall see, they apply to the Th6venin equivalent circuits a t any bus of the system as well as to the case of a loaded generator under steady-state conditions. When computing transient or subtransient conditions, the equations apply to a loaded generator if £' or E" is substituted for

Eo n '

salient-pole generator without dampers is rated 20 MVA, 1 3.8 kY and has a direct-axis subtransient reactance of 0.25 per unit. The negative­ and zero-sequence reactances are, respectively, 0.35 and 0. 1 0 per unit. The neutral of the generator is sol idly grounded. With the generator operating unloaded at rated vol tage with E a n l .Oft per u nit, a single l i ne-ta-ground fault occurs at the m achine terminals, which then have per-unit vol tages to ground, Example

1 1 .6. A

=

v"

=

0

v"

=

1 .01 3/

-

1 02 .25°

Determine the subtransient current in the generator and the line-to-line vol tages for sub transient conditions due to the fault. Figure 1 1 . 1 5 shows the line-to-ground fault on phase In rectangular coordinates Vb and Vc are

Solution.

Vb JJ;..

=

-

=

-

0 . 2 1 5 - j O .990 0 .2 1 5

a

of the machine.

per unit

+ jO.990 per unit

a .------,

I

z"

FIGURE 1 1 . 1 5

Circu i t gro u nd nals of neutral tance.

d iagram for a single l i n e-to­ fau l t on p hase a at the termi­ an u n loaded generator whose is grounded through a reac­

448

CHAPTER 1 1

S Y M M ETRICAL CO M PO N ENTS A N D S EQ U ENCE N ETWO R KS

] [-0..1433 ]

The sym m etrical components of the voltages at the fault poi n t are

[V}OV(l)) ] V a ( 2) a

From Eqs.

1a( 0)

/(/( 1 ) 1a( 2)

13 [11 1 1 ][ 1 ( 1 . 7 2) =

0 - 0 . 2 1 5 - }0 .990

a2

-

- 0 .215 +

a

and Fig. 1 l . 1 4 w i t h 2 "

=

=

=

V eil) a - --

ZgO

VZ(,2)

0

=

=

we ca l c u l a t e

( - 0 . 1 43 + }O) )0 . 1 0

=

E a " - V11( I )

( 1 .0

ZI

a - --

}0 .990

+ }O 0 64 + }O per u n i t - 0 .500 + }O

+

- j l '+ 3 .

+

(0 .643 jO .25

}O)

-

-j l . · n pcr

jO)

( - 0 .500 + }O ) jO. 5

=

per u ni t

=

3

unit

-j 1 . 4 3 per u nit

Therefore, the fau l t current into the ground is I

a

=

1a( 0

)

+

13

1 (1) a

The base current is 20,000/( i n l ine a is fa

+ [a(2) X

3 / ( 0) = a

13.8) 837 837 =

-}4 .29

=

=

=

X

-)'4 .29 per u n i t

A, and so the subtransient curre n t

-j3 ,590 A

Li ne-to-line vol tages during the fau l t arc vti ll

V"C �'1I

=

OC�,

=

Va - V"

VI,

V

0.2 1 5

=

V c -

=

=

- 0.21 5

+

1 .0 1 / 77.T per u n i t

1 . 980�

- j l .980

0

v:, = c

+ jO.990

1 .0 1 i..l2 . r

jO.990

per

unit

per u n i t

Since t h e generated voltage-to-neutral E a ll was taken as l .0 per u nit, the above l ine-to-l i ne voltages are expressed i n per unit of the base vol tage to neutral. Expressed i n vo l ts , t h e pos t fa u l t l i n e vo l t ages a r c

V . 1 1 3 .8 / 77.70 8 / 77.70 kY 1 3 .8 1 . 1 1 3 .8 8 k Y a

.

b

Vbc

=

l 0

=

1 .980

Vc a =

0

X

X

X

13

= . 05

_

/ .. 13 LE!!. ., .., ,,0

13

/

1 02 .30

=

=

/ 15 .78LE!!. .. " .., ,...., 0

.05 L/_ 102 .30

kY

SEQUENCE CI RCUITS OF Y-A TRANSFORMERS

1 1 .8

449

b

b

a �---+

c

( a ) Prefa u l t

Postf a u l t

FIG U RE 1 1 . 1 6

P h asor d iagrams of the l i n e voltages of Example 1 1 .6 be fo re a n d after the fau l t .

t h e ra u l t t h e l i n e vo l t a g e s w e re b " la n c e d l i n d e q u a l t o

B e fore

l�l fl

(b)

c

c o m p a r i s o n wi l h l h e l i n e vo l l a ges a fte r t h e f,l ll l t occu rs, t h e prcfa u l t =

F.i/ I/

�,"

=

as

rdc re nce ,

! J .8! 3 (jO

; I IT

= 13.8�

g ivCI! as

kV

v" , .

kY

v:"

=

1 3.8 kY. For vol tages, with

13.R�

kY

F i g u re 1 1 . 1 6 s h ows p h ll s o r d i ag r a m s o f p rc fa u l t a n d pos t fa u l t voltages.

The p reced ing example s hows that I(�(J) I�l) I�2) in the case of a single l ine-to-ground fault. This is a general result, which is established in Sec. 1 2.2. =

=

1 1 .8

S EQUENCE CIRCUITS OF V-A TRANS FORM ERS

The sequ ence equivalent circu its of three-phase transformers depend on the con nections of the primary and seco n d a ry wind ings. The d ifferent combi nations of L1 and Y windings determine the configurat ions of the zero-sequence circuits and the phase shift in the positive- and n egative-sequence circu its. On this acco unt the re ader may wish to review some port ions of Chap. 2, notably Secs. 2.5 a n d 2 . 6 . We remember t h a t n o cu rre n t flows in the primary of a transformer u n ­ less cu rrent fl ows in the seco nd a ry , i f we n eglect the relatively s m a l l m agnetizing curre n t . We know also that the primary current is determined by t h e secondary current and the turns ratio of the windings, again with magnetizing current n e g l e c t e d . These p r i n c i p l e s guide us in the a n alysis of i n d ividual cases. Five possible connections of two-w i n d ing tra nsfo rmers will be d iscussed. These connections are summarize d , along w i t h t heir zero-sequence circuits, in Fig. 1 1 . 17. The arrows on the connection d i agrams of the figures to fol l ow show the poss ible paths for the flow of zero-seq u ence current. Absence of an a rrow ind icates that the transformer connection is such that zero-sequence current cannot flow. The zero-sequence equival ent circuits are a pproximate a s shown since resist ance and the magnetizing-cu rren t p a t h a re omitted from each circuit. The l e t t ers P and Q ide n t i fy correspo n d i ng points on the connection d iagram f

450 CASE

1

2

3

CHAPTER 1 1

SYMMETRICAL COM PONENTS AND SEQUENCE N ETWORKS

SYMBOLS

� �.

CONNECTION DIAG RAMS

.FY �

ZERO-SEQUENCE EQUIVALENT C I R C U ITS

pi I I I

•

Reference bus

Zo

p

�

�y

•

Reference b u s -- --

��

I

I

:Q I

6

p

--

•

Zo

n

Reference bus

Zo

� p

4

•

p

•

Reference b u s

Zo

�

5

Q

.�----�r-t�----�.

�� 6

Zo

p

I

•

Reference b u s

•

Q •

•

Q

-

•

Q •

•

Q •

•

FIGURE 1 1 . 1 7 Zero-seq u e n ce e q u iv a l e n t c i rcuits o f t h ree-phase t r a n s form e r b a n k s , toge t h e r w i t h d i ag r a m s o f con n ections a n d t h e symbols for o n e - l i n e d i a g r a m s . I m pe d ance 2 0 accou n t s for t h e l ea k a ge impe d a n ce 2 a n d t h e n eu t ral i mpedances 3 2 N and 3 2" where a pp l i cab l e .

and equivalent circuit. The reaso n i ng to justify the equivalen t c i rc u i t for each connection follows. CASE

1. Y-Y

Ban k , Both Neutrals Grounded

Figure 1 1 . 1 8(a ) shows the ne u t ra l s of a Y - Y b a n k gro u n d e d t hrou g h

imped an ce ZN on t h e high-vol t age s i d e a n d 211 on t h e low-vol tage s i d e . The arrows o n the diagram show the d i rections chosen for the cu rrents. W e first treat the trans formers as ideal and add series leakage impeda nce later when the shunt m agnetizing current can also be i ncluded i f necessary . We continue to designate voltages with respect to gro u n d by a single subscript such a s VA ' VN , and Va' Voltages with respect to neutral h ave two subscripts such as VA N and �n' Capital letters are assigned to t h e hi gh-voltage and lowercase l etters are assigned on the other side. As before, windings that are d rawn in paral lel

1 1 .8 SEQU ENCE CiRCUITS OF Y-L\ TRANSFORM ERS ]

A

]

B

]C

=

A =

B

=

c

] ( 0 ) + ] ( 1 ) + ] ( 2) A

A

A

] b

-,

_ __ __ __ __ _ _ _ _

] (0) + ] ( 1 ) + ] ( 2 )

B

B

�

[ (0) c

+

] (0) + ] ( 1 ) + ] (2) b b b ,

�-----

]

B

a

•

] ( 1 ) + ](2) c

=

]

C

c

-------'

=

=

451

b

::;:::=)oo� a

] ( 0 ) + ] ( 1 ) + ] ( 2) a

a

a

](0) + ]( 1 ) + 1 (2 ) c

,

c

c

'----- c

(a)

----�

i\ -,-------.--, + +

.-----.----,-- a + +

NI : N 2 3

[ (0) "

N

I

3 ] a( 0

ZN Zn

) •

Va n

(b) FIGURE 1 1 . 1 8 (a)

Y-Y

connected

tra nsfo r m er

bank

wi t h

bot h

n e u trals

grounded

t h rough

impeda nces;

(b) a pair o f t h e m a g n e t ical ly l i n k e d wi n d i n g s .

d i rections are t h ose l i n ked m ag n e t i c a l l y o n the same core. Two such windings t a ke n fro m Fig. 1 1 . 1 8( a ) a rc s h ow n i n Fig. 1 1 . 1 8(b ). The vol tage m easured w i t h respect t o g r o u n d o n t h e h i g h -v o l t a g e s i d e i s g i v t.: n b y ( 1 1 .73)

S ubstit uting the symmetrical compon e n ts of each vol tage g ives ( 1 1 .74)

and equating quantities o f again confirms the fact t h a t a r e equal t o posit ive- a n d sequence voltage difference

t h e s a m e sequence, as explained for Eq. ( 1 1 . 1 9) , positive- and negative-sequence voltages to gro u n d negative-sequence vol t ages to n eutral. The zero­ between neu tral and ground is equal t o (3ZN ) I�O). J

452

CHAPTER I I

SYMMETRICAL COMPONENTS AND SEQUENCE N ETWO RKS

., Similarly, o n the low-voltage side we h ave

V CO) a

+

V C l ) + V( 2 ) = ( vCO) a

a

all

V C I ) + V ( 2 ) ) - 32

+

an

a ll

fl

] (0)

( 1 1 .75 )

a

There is a minus sign in this equation because the d i rection of 1 (0) is out of the transformer a n d into the l ines on the low-voltage side. Voltages ; n d cu rre n ts on both sides of t h e transformer are related by turns ratio N I /N2 so that

Multiplying across by N, /N2 gives

N -l ( V (O) N2

a

+

Vo) a

+

v (2 » ) = ( VACO)N + VAe l ) + V ( 2 » ) N AN

-

a

3 21 1

(N

1

_

N2

J

, 2

JA( O ) ( 1 1 . 77)

and substituting for ( Vl� + Vl� + Vl� ) fro m Eq. ( 1 1 .74), we u b t a i n

N

l ( VCO) + V(l) + V ( 2» ) = ( V CO) + V e l ) + V (2)) A A A

N2

a

a

a

_

3 2 N JA(O)

_

( )

N j 3 2n N2

2 1A(0)

( 1 1 .78)

By equating voltages of the same sequence, we can write NI ( ) _ V I = N2
( ) VA I

N V CO ) = VA(O) N2 a

_ 1

NI N2

_ _

_

[

V

( 2)

=

V

3 2N + 3 2

(2)

(NI )2] ] «() A

11

II

N2

A

( 1 1 .7 9 )

( 1 1 , 80)

The positive- and negative-sequence relations of Eqs. ( 1 1 .79) are exactly the same a s i n Chap. 2 , a n d the usual p e r-phase e q u iva l e n t circuit o f the trans­ former therefore applies when posi tive- or negative-sequence voltages a n d currents a r e present. The zero-seq uence equiva l e n t ci rcuit representing Eq. ( 1 1 .80) is d rawn in Fig. 1 1 . 19. We h ave a d ded the l eakage impedance 2 of the transformer i n series on t h e high-voltage s i d e as s hown so t h a t the total impedance to zero-sequence current is now 2 + 3 2 N + 3(Nl/N2 )22n referred to the h igh-voltage side. It is appa rent t h a t the s h u n t magnetizing impedance could also be added to the circuit of Fig. 1 1 . 1 9 if so d esired. When voltages , on both sides of the transformer are expressed i n per u ni t on kilovolt line-to- l i n e

1 1 .8

O) VA( '

SEQUENCE CI RCUITS OF

y-� TRANSFORMERS

VA( Ol

FIG l'RE 1 1 . 1 9 Zero - s e q u e nce c i rc u i t of Y- Y co n n e c t e d t r a n sform e rs of Fig . 1 1 . I R . I mpedance re a c t a nc e

bases

453

,IS

measl i red

Oil

the

h i g h-vo l t ;l g e side or

t i l e t r; l I 1 s fo r m e r .

Z is t h e

le akage

in accorda nce with the rated voltages, the t u rns ratio in Fig. 1 1 . 1 9 unity a n d N t /N2 d i s a p p ears, a n d we arrive a t the zero-seq uence circu it shown under Case 1 in Fig. 1 1 . 1 7, where chosen

becomes

Zo

=

Z

+

3 ZN + 3Z n per u n i t

( 1 1 .8 1 )

Aga in, we note that impedances connected from neutral t o ground i n t he actual circu it are multiplied b y 3 i n the zero-sequence circui t . Where both neutrals of a y -Y bank are grounded d irectly or through impedance a path t hrough t h e t ransformer exists for zero-sequence currents i n b oth wi ndings. Provided t h e zero-sequence cu rrent can fol l ow a complete circuit o utside t h e transformer o n both sides, it can flow in both windings of the transformer. I n t h e zero-sequ ence circu it points on the two sides of the transformer are connected b y t h e zero-sequence impedance of the transform er i n the same manner as in t h e positive- a n d negative-sequence n etworks.

CASE

2. Y-Y

Bank , One

Neutral Grounded

If either one of the neutrals of a Y Y bank is u ngrounded, zero-sequence current cannot flow in either winding. Th is can b e seen b y setting either Z N or Z n equal to in Fig. 1 1 . 1 9 . The absence of a path t h rough one winding prevents cu rren t i n the other and a n open ci rcu it exists for zero-sequence current between the two p a rts of the system connected b y the transformer, a s shown in F i g . 1 1 . 1 7. -

cc

CASE

3.

6-6 Bank

The phasor sum o f the l i n e-to-line voltages e q u a l s zero on each s i d e of t h e 6-6 t ransformer of Fig. 1 1 . 20, and so V,J<]i = vaCE) O. Appl ying t h e rules o f t h e =

,

454 II. A

CHAPTER I I

=

11°)

�

+

111)

+

S Y MMETRICAL COM PONENTS A N D SEQUENCE N ETWORKS

I

112)

=

a

1(0) + 1(l) + 1(2) a a a

� r------ a

+

B -.1.....----411'--. -..

In

Ie

1�0) + Ij/) + I�2) O = I� ) + Ig) + I�2)

�

=

.

�

FIGURE 1 1 . 2 0 Wiring d i a �raJ1l of l i ll:

C -------�

-

�------ b

I l l rl: l: · p l l ; I � C � . � C U I 1 1 I C l' l c d I l' a 1 1 S I( 1 I 1 l 1 C I.

conventional dot notation for coupled coils to that figure, we have N]

N2

VA B = - V" ....,.

V (l) + AB

VA(2)B

=

N1

_

N2

( V(l) + ab

( 1 1 .82) V ( 2» ab

The l ine-to-line voltages can be written as l i n e-to-neutral voltages according to Eqs. ( 1 1 .23) giving

and so

VAON)

=

I v( l ) N ""

N

_

2

v ( 2) AN

=

N

I v (2) N :>. {/II

_

( 1 1 .84)

Thus, the positive- and negative-sequ ence equivalent circu its fo r the 6. - 6. tra nsformer, like those for the Y Y con nection, correspond exactly to the usual per-phase equivalent circuit o f Chap. 2. Since a 6. circuit provides no return p at h for zero-sequ ence current, no zero-sequ ence current can flow i nto either side of a Ll-Ll bank a lthough i t can somet imes circulate within the Ll windings. Hence, I�O) = J�O) = 0 in Fig. 1 1 .20, a n d we obtain the zero-sequence equivalent circuit shown i n Fig. 1 1 . 1 7. -

CASE 4. Y-� Bank, Grounded Y

I f the n eutral of a Y-Ll bank is grounded, zero-sequence currents h,ave a p at h to ground through the Y b ecause corresponding i nduced currents can

1 1 .8

I

I

1 ( 0) + 1 (1 ) + 1 ( 2 )

=

A

A

A

SEQUENCE CI RCU ITS OF Y·I!> TRANS FORMERS

A

�

c

Ie

=

Ib

-

•

=

] (0)

c -----

+

](1) c

+

+

1a( 1 )

,----- a

A

•

Ic

1a( 0)

=

a

](2) c

455

+ 1d:(2)

le(1) + 1�2 )

I�O)

+

�

C

1b(0)

+

1b(1)

�----- b

+

1(2) b

FI G U RE 1 1 .2 1

W i r i n g d i agram i m pe cJ a nc e ZN '

Y-O

for a t h ree-p hase

transfo r m e r bank wit h

neutral

gr ounded t h rough

c i rcu l a t e in t h e 6. . The ze ro-seque nce cu rrent c i rcu l a t i ng in t h e 6. magnetically b Cl l d n c e s t h e z e r o - se q u e n ce curre n t i n the Y b u t c a n n o t flow i n the l ines

connected to the 6. . Hence, l�O) = 0 in Fig. 1 1 . 2 1 . P h a se-A voltage on the Y side can be written t h e same a s in Eq. ( 1 1 .74 ) from w h ich we obtain y eo) + V ( I ) + V ( 2 ) A A A

gIves

=

N

I 2

_

N

y ebo ) n

+

N

1

_

N2

Vn(1) b +

N

1 ( 2) 2 Vab

_

N

+

3 2 1 (0) N A

( 1 1 .85)

Equating correspond ing sequence components, as expla ined for Eq. ( 1 1 . 1 9),

VA((l)

-

3 2 1 (0) N A

=

N

I

_

N2

VaeO) h

-

N

V ( I ) = _I V ( I ) = ,

,.

VA(2)

N2

NI

( :2 ) :2 Vtil!

= _

N

ill>

=

0

( 1 1 . 86)

N _1 /3 / 3(f N:;>,

N

1

_

N2

13/

y .)

_

-

X

30°

V(I) 1I

X .

vtI( 2 )

( 1 l .87)

Eq uation ( 1 l .8 6 ) enables us to draw t h e zero-sequence circuit s hown i n F i g . 1 1 .22( a ) i n w h i c h Zu = 2 + 3 2 N when l eakage impedance 2 is referred t o

the high side of the transformer. T h e equ ivalent circuit p rovides for a zero­ sequence cu rrent path from t h e line on t h e Y side through t h e equivalent resistance and leakage reacta nce of the transformer to the reference node . An open circuit must exist between t h e line and the reference node o n the 6. s i de. W hen the connection from neutral to grou n d conta i n s impedance 2 N as s h own,

456

CHAPTER

SYMMETRICAL COMPONENTS A N D SEQUENCE N ETWORKS

II

] (0) A

z

-

1 (0)

N

= 0

a -

( a) 1---------'

I

z +

I I

13 _1 /30" 1 N N

2

I

II

:

Ideal

:

+

I

Z

--�c===J

t\

I

Vf" : I

-

- - - - ,

- /- 30 N2 Nj

I

0

:

I I

11

] (2)

t+

1

a _

:

V(2)

:

� � _____-�r�i _�1 1 -�i--�!-1

I

J

'-

I V3

1

I

I

_ _ _ _ _ _ _ _ _ _

(b)

- - - - -

I I

,_

FI G U RE 1 1.22

�-4 1--�

�

(c)

Ideal

_ _ _ _ __ _ _ _ _

I I J

( a ) T h e zero-sequence circ u i t of Y -� t r a nsfo r m e r b a n k w i t h g r o u n d i n g i m p e d a n c e Z N a n d correspo ndi ng (b) pos i t ive-sequence a n d ( c ) n e g a t ive-sequ e n ce c i r c u i t s .

t h e zero-sequence equivalent circ u i t m u s t have impeda nce 3 ZI\. in series with t h e equivalent resistance and leakage reactance of the t ransformer to con­ n ect the l i n e on the Y side to grou n u . CASE 5. Y-il Ban k , Ungrounded

Y

u ngrounded Y is a special case where the impedance Z N between ne utral a n d ground is infinite. The impedance 3Zn in the zero-sequence equivalent circui t of Case 4 becomes i n fi nite and zero-sequ ence current cannot flow in the transformer windings. An

T h e positive- and negative-sequence equival en t circuits of the Y- Ll trans­ former shown i n Figs. 1 1 .22(b) a n d 1 l .22( c ) are b ased on Eqs . ( 1 1 .87). We recall from Sec. 2.6 that the multiplier 13 Ni lN2 in Eqs. ( 1 1 .87) is the ratio of the two rated line-to-line (and a lso line-to-neutraU voltages of the Y-Ll transformer. In p er-unit calcu l ations, therefore, Eqs. ( 1 1 .87) become exactly the same as Eqs. (2.35), a n d again we have the rules V(I) A

=

l) V( a

X 1� / 300

I�I )

I�2)

=

=

I� ! ) I� 2 )

X X

1� 1/ - 300

( 1 1 .88)

1 1 .8 SEQUE NCE CIRCUITS O F Y - A TRANSFO RMERS

457

That is, When stepping up from the low-voltage side to the high-voltage side of a � -y or Y - � transformer, advance the positive-sequence voltages (and currents) by 3 0 ° and retard the negative-sequence voltages (and currents) by 3 0 ° . The next example shows the numerical application of Eqs. (1 1 .88).

1 1 .2

The resistive Y -connected load bank of Example i s supplied from the low-voltage Y-side of a Y-tl transformer. The voltages a t the load are the same as i n that example. Find the line vol tages and currents i n per u n i t on the high-voltage s i d e o f t h e t ra n s form e r .

E x a m p l e 1 1 .7.

Sollliion . In Exa m p l e 1 1 .2 w e

c urrents flowing

fo u n d

t h e pos i t i v e - a n d n e g a t i v e - se q u e n c e

l hat

0.9857/ 43. °

t o w a r d t h e re s i s t ive load a re

f,� ' )

I(�!.)

= =

6

0 . 2346

/ 250 .3°

per u nit per

unit

while the correspond ing vol tages on the low-voltage Y side of the transformer are

/ 43 .6°

per unit ( I ine-to-neutral voltage base )

0.9857 0.2346/250 ,3° 30° 0.9857�6 + 30° 0.9857//73.6° 0.2783 + }0.9456 0.2346/ 250 .3 -30° 0.2346 2 0 .3° -0.1789 -}0 . 15 1 7 0.0994 jO.7939 0.9857! -46.4° 0.6798 -}0.7138 O.2346i..=- 19.7° 0.2 09 -}0.0791 -0.9581 -}0.2318 -0.0419 + }0.2318 -1.0 +}O 1.0�

Va� )

=

Va\�)

=

p e r unit ( line-to-neutral voltage base )

Advancing the phase angle of t h e low-voltage positive-sequence vol tage by 30° and retarding the negative-sequence voltage b y give on the high-voltage side

Vj I )

vY) VA

=

=

=

=

V� I )

=

VII =

+

Vc

=

lTvy) V,� I )

V�I)

=

=

vY)

+

= ,

VJ I ) = a !' VJ I ) = Vf)

=

+

+

=

=

v,F)

VP)

=

=

=

=

0 . 9 007

-

jO .7929

=

=

(U�/ 82 .80

per

1 .20i..=- 4 1 .4°

per u n i t

unit

per u n i t

458

CHAPTER

II

SYM M ETRICAL COM PON ENTS A N D S EO U E N C' I : N ETWO R K S

Note that the line-ta-neutral vol tages on the h igh -voltage a rc equal in p er unit to h e lillc-to-lillc voltages low-vol tage Y side. The line-to-l i n e vol ta ges are

t found insi of for 0.09 4 jO.7939 -0.9007 jO .7929 -0.8013 jl .5868 1 .78/ 1 16.8" p ( i n a voltage base) 1 .7S 0.90 7 -jO.7939 2.2.0066//-22 .7° per ni / in n ra vo b s -22 .7° l.1 9 -22.7° per unit (line-line voltage base) -1.0 -0.0994 -jO.7939 l.09 4 - jO .7939 pe (line-neutral voltage base) / 1.356 / 0.783/ 215 ,8" per unit ( line-line voltage base) pha s e i s a r e s i s t a nc e of 1 . 0 / Q: per uni t , and p i va l u e s i n t h i s pr o bl e m. Li k e w i s e , and in p Therefore, must be identical to VA expres ed in

VA B VA - VB =

=

=

Vn c

= =

=

Vc 'A

= =

=

r:; vJ

/

=

I I I () X) I .02XL -----

=

f3

+

+

the transformer 1 1 .2 the

Exam p l e

=

+

pc r u n i t ( I i n c - l i ne vo l L l g c h a s e )

I .n =

u t ( l e - eut l

1 .9007 -

l t age a

jO .7l)]l) e)

=

Ve - VA

f3

er u n it l e - neu t r l

L L 6 .H." =

Vn - Vc

1 .35 6

+

de

6.

=

2 1 5 .80

2 1 5 .8"

=

r unit

=

I� l ) If-)

Since the load impedance in each VY> are fou n d to h ave ident ical er - u n t VY> are i dentical er unit. IA per un it. Thus,

in

Ie

=

=

1.20/

-

4 1 4° .

p

er

unit

1.0� per ni

u t

example emphasizes t h e fact t h a t i n going from one side of the 6. - Y o r Y - 6. transformer to the other, the posi t ive-sequence components of curr e n ts a nd voltages on one side must be phase shifted separately from the negative­ sequence components on the same side before combining them toget her to fo rm the actual voltage on the other side. Remarks on Phase Shzft. The American National Standards I nstitute (ANSI) requires connection of y-� and � -y transformers so that the positive­ sequence voltage to neutral, VH N ' on the h igh-voltage side leads the positivesequence vol tage to neutral, VX )n' on the low-voltage side by 30° . The wiring This

I

-�'

1 1 .9

A

8 HI

X1

B

8 H2

X2

c

-O H3

X30

(a)

V)l)

(} ()

a

A

C HI

XI

UNSYMM ETRICAL SERIES I MPEDANCES

�

b

B

-o

H2

X2 8

c

C

8 H3

X3 0

le ads Va( l ) by 300

459

b c a

FIGURE 1 1.23

Labeling of l ines connected to a t h re e - p h a s e Y -� t r a n s ­ former.

di agram of Fig. 1 1 .2 1 and the con nection diagram of Fig. 1 1 .23( a ) both satisfy the ANSI requi rement; and b ecause the connections of the p hases to the transformer term inals H I ' H 2 , H3 - X I ' X2 , X3 are respectively m a rked A , B, C a, b, c as shown in those fi gures, we find that the positive-sequence vol tage to neutral Vl� leads the positive-sequence vol tage to neutral Va�) b y 3 0° . I t is not abso l u tely necessa ry, however, to l abe l l i nes attached to t h e transformer terminals Xl ' X2 , a n d X 3 b y t h e letters a , b, a n d c , respectively, a s w e have done, since n o standards have been adopted for such labeling. I n fact, in ca lcu l a t ions the designa tion of l i nes could be chosen as show n in Fig. 1 l .23( b ), \\'h ich shows the letters b, c, and a associated, respectively, with X l ' X 2 ; and X:; . If the scheme of Fig. 1 1 .23( b ) is preferred, i t is n ecessary o n l y t o exchange b for a , C for b , and a for c in the wiring and phasor di agrams of Fig . 1 l .2 1 , which wou ld then show that VaS: ) l e ads V)� b y 90° a n d t h a t V};) l ags V);� by 90° . It is easy to show that similar statements also apply to t h e correspond ing currents. We shall continue to fol low the l abel ing scheme of Fig. 1 1 .23( a ), a n d Eqs. ( 1 l .88) then become identical to the ANSI requirement. When problems i nvolv­ ing u n symmetrical fau lts a re solved, positive- and nega tive-sequence compo­ nents are found sep a rately and phase shift is taken into account, if n ecessary, by applying Eqs. ( 1 1 .88). Computer programs can be written to i n corporate the effect s of phase shift. A transformer in a three-phase circu it m ay co nsist of three indivi d u a l s i n g l e - p h a s e u n i ts , o r i t m a y b e a t h re e - p h ase t ra n sfo r m e r . A l t ho u gh t h e zero-sequence series imped ances of three-phase units may d i ffer slightly from the positive- and negative-sequence v a lues, it is customary to assume that series i m p e dances of a l l se q u e n c e s are e q u a l r e g a rd l ess o f t h e type o f t ransformer. Tab l e A . 1 in the Appendix l ists tranformer reactances. Reactance and i mpedance are almost equal fo r transformers o[ 1 000 kYA or l arger. For s i m p l i c i ty in our c a l c u l a t i o n s we o m i t the s h u n t a d m i t t a nce which a cc o u n t s for excit i ng current. -

1 1 .9

U N SYM M ETRI CAL SERIES

IMPEDANCES

In the p revious sections we have been concerned particularly with systems t h a t are normally balanced. Let us l ook, however, a t the equa t ions o f a three-phase ,

460 a

CHAPTER

1 1 SYMMETRICAL COMPONENTS AND S EOUENCE NEnVO RKS

-

10

Za

Ib

Zb

b

-

c

-

Ie

a' b'

Zc

c'

flGURE 1 1 .24

Portion of a three-phase system showing th ree u n equal series i mped a n ces.

circuit when the series impedances are unequal. We shall reach a conclusion that is important in analysis by symmetrical components. Figure 1 1 .24 shows the unsymmetrical part of a system with three u nequal series impedances Z a ' Z b ' and Zc ' If we assume no mutual inductance (no coupling) among the three impedances, the voltage drops across the part of the system shown are given by the matrix equation

( 1 1 .89 )

and i n

t enns of the symmetrical components of voltage and current vaa' ( O)

A Vaa'(I ) where A

equation

Za

0

0

0

Zb

0

0

0

Zc

(2 ) vaa'

( 1 1 .90 )

a

I

1a(2)

is the matrix defined by Eq. ( 1 1 .9). Premuitiplying both sides of the by A - I yields the matrix equation from which w e obtain vaa( O)

=

13 1a(0) ( Za

+

Z b + Z c ) + 1.3 1a( 1 ) ( Z a

( + aZ b + 11(2 3 a ) Za

vaael) V{2} ad

I f the

A

1a(0) 1(1)

=

=

11( 0) ( Z a 3 a

) ( Za l[(O 3 a

+

+

a2Zb + aZ ) c

a 2 Zc )

+

aZ b + a2 Z c ) + 1.[ 3 a( 1 ) ( Za + Zb + Z )

+

a2Zb

c

+

aZ c )

+

i mpedances are made equal (that

1 )( Za + aZb 11( 3 a

IS,

if Za

=

Zb

+

=

( 1 1 .91)

a 2Zc )

ZJ, Eqs. ( 1 1 . 9 1 )

1 1.10 SEQUENCE NETWORKS

461

reduce to v(O) aa

=

1(0)Z a

a

( 1 1 .92)

If t h e imped ances are

un e qu a l , however, Eqs. ( 1 1 .91) s h ow that the voltage drop of a ny one sequence is dependent on the currents of all three sequences. Thus, we conclude that the symmetrical components of unbalanced currents flowing in a balanced load or in balanced series impedances produce voltage d rops of like sequence only. If asymmetric coupling ( such as unequal mutual inductances ) existed among the three impedances of Fig. 1 1 .24, the square matrix of Eqs. ( 1 1 . 89) and ( 1 1 .90) would contain off-diagonal elements and Eqs. (1 1 .91) would h ave additional terms. A l though current i n any conductor of a th ree-phase transmission line ind uces a voltage in the other phases, the way in which reactance is calculated eliminates consideration of coupling. The self-ind uctance calculated on the basis of complete transposition i ncludes the effect of mutual reactance. Th e assumption of transposi tion yields equal scries impedances. Thus, the compo­ nent currents of any one sequence produce only voltage drops of l i ke sequence in a transmission line; that is, positive-sequence currents produce positive­ sequence voltage drops only. Likewise, negative-sequence currents produce negative-sequence voltage drops only and zero-sequence currents produce zero­ sequence voltage drops only. Equations 0 1 .9 1 ) apply to u nbalanced Y l oads because points a' , b', and c' may be connected to fonn a neutral. We could study variations of these equations for special cases such as single-phase loads where Zb = Zc = 0, but we continue to confine our discussion to systems that are balanced before a fault occurs. 11.10

SEQUENCE N ElWORKS

In the preceding sections of this chapter we have developed single-phase equivalent circuits in the form of zero-, positive-, and negative-sequence circuits for load impedances, transformers, transmission l ines, and synchronous ma­ chines, all of which constitute the main parts of the three-phase power transmis­ sion network. Except for rotating machines, all parts of the network are static and w ithout sources. Each individual part is assumed to be l inear and t hree­ phase symmetrical when connected in Y or fl configuration. On the basis of these assumptions, we have found that: •

•

n any part of the network voltage drop caused by current of a certain sequence depends on only the impedance of that part of the network to current flow of that sequence. The impedances to positive- a nd negative-sequence currents, Z I and Z2' are equal in any static circuit and m ay be considered approximately equal i n synchronous machines u nder subtransient conditions. I

462

•

•

•

•

•

CHAPTER 1 1

SYMMETRICAL CO M PO N ENTS AND SEQU ENCE N ETWO RKS

I n any part of the network impedance to zero-sequence current, 2 o , is generally different from 21 and 22. Only positive-sequence circuits of rotating machines contain sources which are of positive-sequence voltages. Neutral is the reference for voltages in posltlve- and negative-sequence circuits, and such voltages to neutra l a re the same as vol tages to grou nd if a p hysical connection of z e ro or o t h e r li n i t e i m ped a nce e xists between n eutral and ground in the actual circuit. No positive- or negative-sequence c u r r e n ts How b e tw e e n n e u t r a l p o i n t s a n d ground. Impedances 2 n in the physical connections between neutral and ground are not included in positive- an d negative-sequence circu its but arc represen ted b y impedances 3 2" b e tw ee n t h e p o i n t s fo r n e u t r a l a n d g ro u n d in t h e z c r o ­ sequence circuits only.

These characteristics of individual sequence circuits guide the construction of corresponding sequence networks. The object of obtaining t he values of the sequence i mpedances of the various p a rts of a power system is to enable us to construct the sequence networks for the complete system. The network o f a particular sequence-constructed by joining together all the correspond ing sequence circuits of the separate parts-shows all the paths for the flow of current of that sequence in one phase of the actual syste m . In a balanced t hree-phase system t h e currents flowing in the three p hases under normal operating conditions constitute a symmetrical positive-sequence set. These positive-sequence currents cause voltage drops of the same sequence only. Because currents of only one sequence occurred in the preced ing chapters, we considered them to flow in an i ndependent per-phase network which combined the positive-sequence emfs of rotating m a c h i n e s a n d t h e impedances of other static circuits to positive-sequence currents only. That same per-phase equivalent network is now called the positive-sequence network i n orde ' to d istinguish it from the networks of the other two sequences. We have discussed the construction of impedance and admittance rep re­ sentations of some rather complex positive-sequence networks in e a rl i e r ch ap­ ters. Generally, we have not included the phase s h i ft ass o c i a t e d w i t h 6. - Y and y-� transformers in positive-sequence networks since p ractical systems are d esigned with such p h ase s h i ft s su m m i ng to z e ro a r o u n d all loops. I n detailed calculations, however, we must remember to advance all positive-sequ ence voltages and currents by 30° when stepping up from the low-voltage side to the high-voltage side of a !:J.-Y or Y -� transformer. The transition from a positive-sequence network to a nega tive-sequence network is simple. Three-phase synchronous generators and motors have inter­ nal voltages of positive sequence only because they are designed to generate balanced voltages. Since the positive- and negative-sequence impedances are the same in a static symmetrical system, conversion of a positive-sequence

I LID SEQUENCE NETWORKS

463

network to a negative-sequence network is accomplished by changing, if n eces­ sary, only the impedances that represent rotating m a c h i ne ry and by omitting the emfs. Electromotive forces are omitted on the assumption of balanced gener­ ated voltages and the absence of negative-sequence voltages induced from o u tsi d e sources. Of course, in using the negative-sequence network for detailed calculations, we must also remember to retard the negative-sequence voltages and currents by 30° when stepping up from the low-voltage side to the

high-voltage side of a D.-Y or Y -Do transformer. Since all the neutral points of a symmetrical three-phase system are at the same potential when b a l a nced t h ree-phase currents are flowing, all the neutral points must be at the same potential for either positive- or negative-sequence currents. Therefore, the neutral of a symmetrical three-phase system is the logical reference potential for specifying positive- and negative-sequence voltage drops and is the reference node of the positive- and negative-sequence n et­ works. Impedance connected between the neutral of a machine and ground is not a p a rt of either the positive- or negative-sequence network because neither positive- nor negative-sequence current can flow in an impedance so connected. Negative-sequence networks, like the positive-sequence networks of previ­ ous chapters, may con tain the exact equivalent circui ts of parts of the system or be s imp l ified by omitting series resistance and shunt admittance.

Draw the negative-sequence network for the system described i n Example 6 . l . Assume t hat t he negat ive-seque nce reactance o f each m achine is equal to its sub transient reactance. Omit resistance and phase shifts associated with the transformer connections.

Exa m p l e 1 1 .8.

Since all the negative-sequence reactances of the system are equal to the positive-sequence reactances, t he negative-sequence network is identical to the positive-sequence network of Fig. 6.6 except for the omission of emfs from t he negative-sequence network. The required n etwork is drawn without tra nsformer ph ase sh i fts i n F i g . 1 1 .25 .

Solution.

d e t e r m i n e d fo r t h e

various separate parts of the system are readily com bined to form the complete zero-sequence network . A three-phase system operates single phase insofar as the zeroZe ro-se q u e n ce e q u iva l e n t c i rc u i t s

) 0 . 08 5 7

l

jO. 1 8 1 5

j O.20

m

j O . 09 1 5

jO. 5490

Reference

FIGURE 1 1 .25 Negative-sequence network for Exa mple 1 1 . 8.

464

CHAPTER 1 1

R

SYM METRICAL COMPON ENTS A N D SEQUENCE N ETWORKS

T

T s

M

N

-

-

Reference

FIGURE 1 1.26 One-l i n e d iagram of a sma l l power sys t e m a n d t h e correspon d i ng zero- s e q u e n c e n e twork.

sequence currents are concerned, for the zero-sequence currents are the same in magnitude and phase at any point in all the phases of the system. Therefore, zero-sequence currents will flow only if a return path exists through which a completed circuit is provided. The reference for zero-sequence voltages is the potential of the ground at the point in the system at which any particular voltage is specified. Since zero-sequence cu rrents may be flow i n g i n the ground, the ground is not necessarily at the same potential at all points and the reference node of the zero-sequence network does not represent a ground of unifonn potential. We have already d iscussed the fact that the impedance of the ground and ground wires is included in the zero-sequence impedance of the transmission line, and the return circuit of the zero-sequence network is a conductor of zero impedance, which is the reference node of the system. It is because the impedance of the ground is included in the zero-sequence impedance that vol tages measured t o the r eference node of the zero-sequence network give the correct voltage to equivalent ideal ground. Figures 1 1 .26 and 1 1 .27 show one-line diagrams of two small power systems and their correspond­ ing zero-sequence networks simplified by omitting resistances and shunt admit­ tances. The analysis of an unsymmetrical fault on a symmetrical system consists i n finding the symmetrical components o f the u nbalanced currents that a r e flow­ ing. Therefore, to calculate the effect of a fault by the method of symmetrical

1 1 .10

g�

N

:�

�

8.

8. 8.

N

Q

p

46S

z

� y

g� � fi J: ut � � 8. r Tu 6�

SEQUENCE NElWORKS

Y

X

� �

w

z

[] _� R T

S

Cfj

Reference FI GURE 1 1 .27 O n e - l i n e d iagram of a s m a l l power s ys t e m and t he corres p o n d i n g zero-sequence network.

components, it is essential to determine the sequence impedances and to combine them to form the sequence networks. The sequence networks carrying the symmetrical-component currents l�O), I� l ), and 1�2) are then interconnected to represent various unbalanced fault conditions, as described in Chap. 1 2. Draw the zero-sequence network for the system d escribed in Assu me zero-sequence reactances for the generator a n d motors of

Exa mple 1 1 .9.

Example

6. 1 .

The

ze ro-se q u e n c c r e a c t a n c e

0 . 0 5 p e r u n i t . A c u r re n t - l i m i t i n g r e a c t o r o f 0 . 4 n i s in each o f t h c ncutrals of the

ge ne r a t or a n o the l a rg e r motor. l i n e is 1 .5 n/km.

of the

transmission

The zero-sequence l eakage reactance of transformers is equal to the positive-sequence reactance. So, for the transformers X 0 = 0.0857 per unit and 0.09 1 5 per unit, as i n Example 6 . 1 . Zero-sequence reactances of the generator and

Solution.

466

CHAPTER I I

S YMMETRICAL COM PONENTS A N D SEQ UENCE NE1W O R KS

motors are Generator : Motor 1 :

Motor 2 :

I n t h e generator circui t

Xo

0.05 0.05 ( -200300 ) ( --1 3.3 .82 ) 0.0686 D.05 ( 300 ) ( 11 33 ..28 ) 0. 1372 (32000)2 1.3 3 300 3Zn 3 ( 1.0.333 ) 0.900 3Zn 3 ( 0.0.6435 ) 1.890 176.3 1 .2�. =

per u n i t

Xo = Xu

1

=

2

1 00

=

'2

-

--

'-=

Base Z =

per u n i t

per

n

a n d i n t h e motor circuit

Base Z

=

( 1 3 K) 2

=

0 . 635 H

I n t h e i mpedance network for the ge nerator

=

and for the motor

=

4

--

=

per u nit

per u ni t

--

For the transmission line

1 . 5 x ()4

�

0 .5445 per

u ni t

T h e zero-sequence network i s shown i n F i g . k

jO.0857

l

jO.5445

m

jO.0915

n

r

,-_L-_ r

jO.06B6 )1.890

Reference FlGURE 1 1.28 Zero-s eque n ce n etwork for Exam p l e

1 1 .9.

jO. 1372

unit

PROBLEMS

11.11

467

SUMMARY

Unbalanced voltages and currents can be resolved into their symmetrical components. Problems are solved b y treating each set of components separately and superimposing the results. I n balanced n etworks h aving strictly symmetrical coupling between phases the currents of one phase sequence induce voltage drops of like sequence only. Impedances of circuit elements to currents of different sequences are not necessarily equal. A knowledge of the positive-sequence network is necessary for power-flow stu d i e s, faul t calculations, and stabil i ty studies. I f the fau lt calculations or s t a b i l ity studies involve unsymmetrical fau l ts on otherwise symmetrical systems, the negat ive- and zero-sequence networks are also needed. Synthesis of the zero-sequence network requires particular care because the zero-sequence network may diffe r from t h e others considerably. PRO BLEMS

1 1 .2.

1 1.3. 1 1 .4 .

1 1 .5.

&

�

10�

, V2) = 20 , and V};;) = V , determine analytica l ly Va� ) = 50 the vol tages to neutral Van ' Vb" , and Veil ' and also show graphically the sum of the given symmetrical components which determine the l ine-to-neutral voltages. When a generator has term inal a open and the other two terminals are connected to each othcr with a short c i rcuit from this connection to ground, typical values for the symmetrical com ponents of current i n phase a are 1� 1) I� 2) = and I� O) = A. Find the cu rrent into the ground a n d the current in each phase of the generator. Determine the symmetrical components of the three currents I = 10&, Ib and Ie = A. The currents flowing in the l i nes toward a balanced l oad connected i n � are 10 = Ib = a nd Ie Find the symmetrical compo­ n e n ts of the g i v en l i n c c urre n t s and d raw ph asor diagrams of the positive- and

1 1 .1. If

250L22:., 350� 10/230 , 10� 100&, 141 .4/2250 , 100�. Thc voltagcs there is 100&, .sO.SL-1 2 1 .4 ° , 1O-D.

=

=

a

=

n e g a t i v e - s e q u e n ce l i m: (l n d p h a se c u r r e n t s . W h a t is luI, i n a m pe res?

at the t e r m i n a l s of a bal anced load consisting of t h r e e 1 O-D. resistors in Y are Va l> = and v;.a = V/Jc = V. Assuming t h a t no connection to the neutral of the l oad, find the l ine cu rrents from the symmetrical components of the given l i ne voltages. Find the power expended in the t h ree resistors of Prob. 1 1 .5 from the symmetrical com p o n e n t s o f curre n t s and voltages. Cheek t h e answer. If therc is impedance in t h e neutral connection to ground of a Y -connected load, then show that the vol t ages Va ' Vb ' and v;. of Eq. ( 1 1 . 26) must be interpreted as voltages with respect to grou n d . A bal anced three-phase l o a d consists of �-connected impedances Z 6 i n p a rallel with solidly grounded Y -con nected i mpedances Z y . ( a ) Express the currents la ' 1b, a n d Ie flowing i n t h e lines from the supply source toward the l oad in terms of the source vol tages Va ' Vb) and �.

90�

con ne c t e d

1 1 . 6. 1 1 .7 .

1 1 .8.

60 /-90°,

,

468

CHAPTER

11

SYMMETRICAL COM PO NENTS AND S EQ UENCE N ETWORKS

,

,

(b) Transform the expressions of part ( a ) i nto their symmetrical component equivalents , and thus express 1(0) and [(a2 ) in terms of Va(O) Vel) , and a , 1(1) a a

vy).

(C) Hence, draw the sequence circuit for the combined load.

11.9. The Y-connected impedances in parallel with the t.-connected impedances

Ze:,. of Prob. 1 1 .8 are now grounded thro ugh a n i m pedance Zg . ( a ) Express the currents la ' a n d Ie flowing i n the lines from the supply source t ow ard the load in te rms of the source voltages Va ' Vb ' a nd � and the voltage � of the neutra l point. ( b) Expressing Vn in terms of I�() , 1� 1), 1�2), and Z}; , fi nc.l the e q u a t i o n s fo r these currents in terms of V(O) a , and Va( 2 ) · a , V(I) (c) Hence, draw the seq u en c e circuit for the co mb in e d l oa d

Ib,

1 1. 10.

.

impedances Zo , Z ] ,

1 1. 1 1 .

11.12.

( b ) Hence, determine t h e line curren t IL i n amperes. (c) D etermine the open-circuit vol tages to n eu tra l of phases b and c a t the receiving end. ( d ) Verify your answer to part (c) w i thout using symmetrical compon ents. and Z2 o f t h e l i n e .

Solve Prob . 1 1 .10 if the same 420-D i nductive load is connected b etwe e n phases a a nd b at the receiving end. In part ( c ) find the open-circu i t voltage of phase c only.

A Y-connected synchronous generator has sequence rea c tan ces Xo = 0.09, X I = 0.22, and X2 = 0.36, all in per u n i t. The neutral p oi n t of the machine is grounded through a reactance of 0.09 per unit. The machine is running on no load with rated terminal voltage when i t suffers a n unbal anced fault. The faul t currents out of the machine are la 0, = 3.75 , and Ie = 3.75� , all in per u n i t with respect to p h a s e a l i n e to -n e u t r a l voltage. Determine ( a ) The terminal voltages i n each p hase of the machine with respect to ground, ( b ) The voltage o f the neutral point of the machine with respect to ground, and ( c ) The nature (type) o f the fault from the r esu l ts of part ( a). Solve Prob. 1 1 . 1 2 i f the fault currents i n per u n i t are la 0, Ib = - 2.986 , =

1 1.13.

and

1 1.14.

1 1.15.

-

Suppose t ha t the l ine-to-neutral voltages at the send ing end of the l i n e described in Example 1 1 .5 can be maintained constant at 20()-kY and t h a t a s ingl e p h a s e inductive load of 420 n is con nected b e tw e e n p h as e a and n eu t r a l at t he receiving end. (a) Use Eqs. ( 1 1 .5 1 ) to express n u merically the receiving-end sequence voltages VY;/., Va�� , and V}�l in terms of the load current IL and the sequence

Ie

=

2.986

&.

Ib-

�

=

&

Assume that the currents specified i n P r o b . 1 1 .4 a r e flowi ng toward a load from lines connected to the Y s i d e of a Ll-Y t ransformer rated 10 MY A, 1 3 . 2 6.j66Y kY. Determine the currents flowi ng in the l i n e s on the t. s i de by converting the symmetrical components of the currents to per unit on the base of the trans­ former rating and by shiftin g the components according to Eq. ( 1 1 .88). Check the results by computing the current s in each p h ase of the t. windings i n amperes d irectly from the currents on the Y side by mUltiplying by the turns ratio of the windings. Complete the check by co mp u ting the line currents from the phase currents on the � side.

Three single-phase transformers a re connected as shown in Fig. 1 1.29 to f� r m a Y-Ll transfo rmer. Th e hig h -v o l t a g e w i n d ings are Y-connected with polarity marks

PROBLEMS

as ind icated. Magnetically coupled wind ings are d r awn

469

in parallel directions.

Determine the correct placement of polarity marks on the low-voltage windings. Identify the numbered terminals on the low-voltage side (a) with the letters a, b, and c , w he re l� l ) l e a d s 1;1) by 30, and ( b) wi th the l etters a', b', and c ' so t hat I�}) is 90 out of p hase with I�l) .

d

- -

FIGURE 1 1 .29

Ie

Circu it for Prob. 1 1 . 1 5 .

- -----+-

1 1 . 1 6.

1 1 . 17 .

1 1 . 1 8.

1 1 .19.

Bal anced three-phase voltages of 1 00 V l i n e t o l i n e are applied t o a Y-conn ected load consisting of three resistors. The neutral of t h e load is not grounded. The resistance in p hase a is 10 D , in p h ase b is 20 D , and in p hase c is 3 0 D . Select voltage to neutral of the three-phase l in e as reference and determine the current in phase a and the volt age Va n . Draw the negative- and zero-sequence impedance networks for the power system of Prob. 3 . 1 2 . Mark t h e values of all reactances i n pe r unit on a base of 50 MVA, 13.8 kV in the circui t of generator 1 . Letter the networks to correspond to the single-line diagram. The neutrals of generators 1 and 3 are connected to ground through current-limiting reactors h aving a reactance of 5%, e ach on the b ase of the machine to which it is connected. Each generator has negative- and zero­ sequence reactances of 20 and 5 % , respectively, on i ts own rating as base. The z e ro - s e q u ence r e a ct a n c e of the transm ission line is 2 1 0 D fro m B to C and 250 D from C to E. Draw the negative- and zero-sequence impedance networks for t he power system of Prob. 3 . 1 3 . Choose a base of 50 MVA, 1 38 kV in the 40-fl transmission line and mark all reactances in per unit. The negative-sequence reactance of each synchronous machine is equal to i ts subtransient reactance. The zero-sequence reactance of each m a c h i ne is 8% based on its own rati ng. The neutrals of the m a c h i n e s a re con n e c t e d to ground t hrough c u r re n t -l i m i t i n g reactors h aving a r e a c t a n c e of 5 % , each on the base of the machine to w hich i t is connected. Assume that the zero-sequence reactances of the transmission l i n e s are 300% of their positive-sequence reactances. Determine the zero-sequence Theven i n impedance seen at bus © of the system ' described in Prob. 1 1 . 1 7 if t ransformer T) has (a) one u ngrounded a n d one solidly grounded neutral, as s hown 10 Fig. 3.23, and (b) both neutrals are solidly grounded.

CHAPTER

12

UNSYMMETRICAL FAULTS

M ost of the faults that occur on power systems are u nsymmetrical faults, which may consist of unsymmetrical s hort circuits, unsymmetricar faults through impedances, or open conductors. Unsymmetrical faults occur as single line-to­ ground faults, line-to-line faults, or double line-to-ground faults. The path of the fault current from line to line or line to ground may or may not contain impedance. One or two open conductors result in unsymmet rical faults, through either the breaking of one or two conductors or the action of fuses and other d evices that may not open the three phases simultaneously. S ince any u nsym­ m etrical fault causes unbalanced currents to flow in the system, the method of symmetrical components is very useful in an analysis to determine the currents and voltages in all parts of the system after the ocCurrence of the fault. We consider faul ts on a power system by applying Thevenin's theorem, which allows us to find the current i n the fault by replacing the entire system by a single generator and series impedance, and we show how the bus impedance matrix is applied to the analysis of unsymmetrical faults.

12.1

U NSYMMETRICAL FAULTS O N POWER SYSTEMS J

In the derivation of equations for the symmetrical components of currents and voltages if! a general network the curren t s flowing o u t of the original balanced

1 2. 1

UNSYMMETRICAL FAU LTS ON POWER SYSTEMS

-----.-1�Irll b------�i-F-rb------r Ij

a

c

471

­

------------------

-

If'

F I G U RE 12.1 Three conductors of a t h ree- p h a se system. The stubs ca rrying currents '/0 ' I[h ' and l[e may be i n t erconnected to represe n t d i ffer­ ent types of fa u l t s .

system from p h ases a, b, and c at the fau l t point wil l be designated a s If lib ' a n d l[c ' respect ive ly. We can visualize these cu rrents by refe rring to Fig. 1 2 . 1 , wh ich sh ows the th ree l i n es a , b , and c o f t h e t h ree-phase system a t t h e p a r t o f t h e network where the fa u l t occurs. The flow of current from e a c h l i n e i nto the fa u l t is indicated by arrows sh own on the d i agram beside hypothetical stubs con nected to each l i ne at t h e fau l t l ocation. Appropriate connections of the s t u bs represent the various types of fa u l t . For instance, d i rect con n e ction of stubs b and c produces a l i ne-to-l ine fau l t t h rough z ero impedance. The current in stub a is then zero, and lib equ a l s l[c ' The l i ne-to-grou nd vol tages at any bus (J) of the sys tem Juring t h e fau l t will be designated V}a ' �h' a n d � c ; and w e s h a l l con tinue to u s e s u perscripts 1 , 2 , a n d 0 , respect ively, to d e note positive-, negative-, and zero-seque nce quanti­ t ies. Thus, for example, Vj.�l ), V}�12 ), and Vj�O) will denote , respective ly, t h e posi tive-, negat ive-, a n d zero-seque nce components of the l i ne-to-gro u n d volt­ age Vj·a at bus CD during t h e fau l t . The l ine -to-neutral vol tage o f p hase a at t h e fau l t point before t h e fau l t occurs w i l l be designated s imply by VI ' w hich i s a positive-sequence voltage si nce the system is ba lanced. We m e t t h e p re faul t voltage VI previously i n Sec. 1 0.3 when calcu l at ing the curre n t s i n a p ower system with a symmetrical th ree-phase fau l t app lied. A singl e-l ine d iagram o f a power system conta ining two synchronous Ill � l c h i il c s i s s h ow n i ll Fi g . 1 2 . 2 . Such ,l sys t e m i s s u ffi c i e n t l y ge ne ral for e q u a t i o ll s de rived t h e re fro m t o be , l p p l i c ( l b l e t o , I ll Y b;I I Cl llccd sys t e m regardless o f the com plexity. F i g ur e 1 2 . 2 a l so s h o w s t he seq u e nce n e t w o r k s o f the system. Th e poi n t wh ere a fa u l t i s a s s u m e d to occ u r is m a r k e d P , a n d in this particular exam ple it is cal led bus CD on the single-l ine di agram and in the s equence ne tworks. Machines are represented b y their subtransient i n te r n a l voltages in series with t h e i r subtrans ient reactances when subt ransient fau l t condit ions a re b e ing studied. I n Sec. 1 0.3 we u s e d the bus impedance matrix composed o f positive­ seque nce impedances to d e t e r m ine currents and vol tages upon t h e occurre nce of a symmet rical th ree-phase fau l t . The method can be eas i ly extended to apply to unsymmetrical fau l ts by realizing t h at the negative- and zero-sequ ence ne tworks also can be represented by bus impedance matrices. The bus imped ance m a t rix w i l l now be wri tten symbolica lly for t h e posit ive-sequence n e twork i n t h e a,

-

,

472

CHAPTER

12

UNSYMM ETRICAL FAULTS

following form:

CD

Z (buI )s -

W

®

(l)

( l)

®

CD W

) Z(l

11 (I) Z 21

Z 12

®

(I)

Zk l

(I)

Zk 2

Zk( Ik)

Z k( IN)

®

) ZN( I I

Z N( I 2)

( \k) ZN

(I) Z NN

(\) Z 22

Z lk

Z 2( Ik)

) Z I( IN ) Z 2( IN

( 12. 1 )

Similarly, the bus impedance matrices for the negative- and zero-sequence

n etworks will be written

) Z (b2us

O Z (bu s )

Thus,

CD W

CD

@

®

) Z (I 2I ) Z (2

) Z 1( 22 ( 2) Z 22

) Z (I 2k Z 2( 2k)

21

®

) Z (I2N ) Z 2( 2N ( 1 2 .2)

=

®

) Zk( 21

Zk( 22)

Z k( 2k)

Z k( 2N)

®

Z N( 2)I

Z N(22)

Z N( 2k)

( 2) Z NN

CD

W

®

®

CD W

Z (I OI ) Z 2( O1 )

Z (1O2) ( O) Z 22

Z (I Ok) Z ( O)

®

) Z k( Ol

Z k(O2)

Z k( Ok)

Z k( ON)

@

( Ol ZN

) Z N( O2

(O Z Nk

)

(O ) Z NN

2k

) Z (I ON ) Z 2(ON

=

)

ZW, zg>, and ZfJ) denote representatiye elements of the bus impedance

matrices for the positive-, negative-, and zero-sequence networks, respectively. I f s o desired, each of the networks can be replaced by i ts Thevenin equivalen t b etween a ny one of the buses and the reference node.

12.1

UNSYMM ETRICAL FAULTS ON POWER SYSTEMS

473

y (a) Single-line diagram of balanced three-phase system

p

( b ) Positive-seq uence n etwo rk

+

®

( e ) The venin equivalent of the

positive-seq uence network fa -

1(2)

Reference

Cc) Negative-sequence n etwork

Reference ( d ) Zero-sequence n etwork

p

+0

( f) The venin equivalent of the

negative-sequence n etwork

( g ) Th eveni n eq uivalent of the zero-seq uence network

FlGURE 1 2.2

S i n g l e - l i n e d i a g r a m of a t h re e-phase sys t e m , the t h ree sequence networks of the system, a n d the Thev e n i n e q uivalent of each n e twork fo r a fa u lt at P , whieh is cal led bus CD.

The Thevenin equivalent circuit between t h e fault point P and the reference node in each sequence network is shown adjacent to the diagram of the corresponding n etwork in Fig. 1 2.2. As in Chap. 10, the v o l t ag e source in the positive-sequence network and i ts Thevenin equivalent circuit is VI ' the prefault voltage to neutral at the fault point P, which happens to be bus ® in this illustration. The Thevenin impedance measured between point P and the reference node of the positive-sequence network i s zW, and its value depends

474

CHAPTER

12

UNSYMMETRICAL FAU LTS

on the values of the reactances used i n the network. We recall from Chap. 1 0 that subtrans ient reactances of generato rs and 1 .5 t imes the subtrans ient reactances (or else the t ransient reactances) o f synchronous motors a re the values used i n c a l cu l a t i n g the sym metrical current t o b e interrupted. There a re no negative- or zero-sequence curre nts flowing before the fau l t occurs, and the prefa u l t voltages are zero a t a l l buses o f t h e n e g a t i ve a n d zero -sequence n e tworks . Therefore, t h e prefa u l t voltage between p o i n t P a n d the reference node is zero i n t h e ne ga t i ve - a n d zero-se q u e n c e �e tw o rks a n d n o electromotive for c e s (emfs) a ppear i n t h e i r Thcv e n i n e q u iv a l e n t s . T h e n e g a t ive­ a n d zero-s equence i mped ances be t w e e n p o i n t P at bus CD J n cJ the re fe re n c e n o de i n the respective networks are r e p r e s e n t e d by t h e Thc\ e n i n i m p e d a n c e s Zk� and Z fOJ d i a go n a l e lements o f Z \��s and Z \��S ) r e s p c c t i v c : y . Since Ifl1 i s t h e c u r re n t n ow i n g froll/ t h e sys t e m il / to t h e fa u l t , i t s symmetrical components 1fa ( 1 ) ) 1fa( 2 ) ' ' ll1 d 1I(0) How O l l l o f t h e r c '-' p c c t i v e s e Cl u c n e e networks and t h e i r equivalent circu i ts a t point P , as s h own i n F i g . 1 2 . 2 . T h u s , t h e currents I}�), - Ij;), an d - I}�) represent injected currems ill lO the fa u l ted bus ® of the positive-, negative-, and zero-sequence networks due to the fa u l t . These curren t i nj ections cause voltage changes a t t h e b u ::, c ::, u f t h e p o s i t i v c - , negative-, a n d zero-sequence networks, which can b e calcul ated from t h e b u s impedance m a t rices in t h e manner d emonstrated in Sec. 10.3. For instance, d u e t o t h e inj ection I}�) i n t o b u s ®, t h e voltage changes in the pos itive-seq uence network o f the N-bus system are given i n general terms by -

-

(

iJ

�

-

-

®

Ll V1Oa ) 6. Vk( al )

zf\) zW Z � II ) zW =

®

Z k( Il)

) Z k( !2

) Z ( I,\' I I Z( )

2 1\'

o o

..- I ( aI ) f

(I) ZH (I) Z Nk

®

®

(I) Z NN

o

- 2 ( 1 )1 ( 1 ) l k fa

2 k fa

- Z ( I )1 ( 1 )

k k fa

- Z ( I )1 ( 1 )

fa

- 2(1) 1 (1) Nk

( 12.3)

1 2.1

UNSYMMETRICAL FAULTS ON POWER SYSTEMS

475

This equation is quite similar to Eq. ( 1 0. 1 5) for symmetrical faults. Note t h at only column k of ZblJs enters into the calcu lations. I n industry practice, it is customary to regard all p refault currents as being zero and to designate the voltage Vf as the positive-sequence voltage at all buses of the system before the fault occurs. Superimposing the changes of Eq. ( 1 2.3) on the prefault volt ages then yields the total positive-sequence voltage of phase a at each bus during the fau l t, Vl(a\ )

Vf Vf

V2( a\ )

=

Vk( lIl )

VI'

.1 V2( aI ) +

�(

VNil (I)

- Z ( k1 ) 1f( a1 ) I (1) Vf L 2( \k) 1fa Vf

.1 Vl(al )

-

.1 VA( llI )

V/

-

'kli. fa

7(1)[(I)

( 12 .4)

6. VNil (I)

Th is equation is similar to Eq . 0 0. 1 8) for symmetrical faults, the only difference

being the added superscripts and subscripts denoting the positive-sequence components of the phase a quanti ties. Equations for the negative- a nd zero-sequence voltage changes due to the fau l t at bus ® of the N-bus system are similarly written with the superscripts in Eq. 0 2. 3 ) changed from 1 to 2 and from 1 to 0, respectively. Because the prefault voltages are zero in the negative- and zero-sequence networks, the voltage changes express the total negat ive- and zero-sequence voltages during the fau lt, and so we have V ICIIO )

V I(2) {I

V"( a2 )

( 2) VNa

=

- Z(0) (0 ) I k 1fa

V2(O)11

- _ 7 (O) j ( O} ' 2k fa

- Z k( 2k) Jfa( 2 )

Vk(O) a

- Z(O) kk 1(0) fa

( 2") [fa (2) ZN

vNil (O )

- Z Nk (0) ( O ) 1fa

_

( 12 .5)

When the fault is at bus ® , note that only the entries in col umns k of Zb2Js and ZhoJs are involved in the calcul ations of negative- and zero-sequence voltages. Thus, knowing the symmetrical components 1}� ) , 1}�), and I};) of the fault curre n ts at bus ®, we can determine the sequence voltages at any bus CD of the system from the jth rows of Eqs. 0 2.4) and ( 1 2.5). That is, during the fault

476

CHAPTER

12

UNSYMMETRICAL FAU LTS

at bus ® the voltages at any bus vJa(O )

(]) are =

(O ) - Zj(Ok )/fa

( 12 .6) V(2) Ja

=

I f the prefault voltage at bus CD is not Vf ' then we simply replace Vf in Eq. 02. 6 ) b y the actual value of the prefault (posi tive-sequence) vol tage at th at bus. Since Vf is by definition the actual prcfault voltage at the faulted b u s CD , we

always have at that bus

V(O) ku

V(I) ka

= =

V(2) ka -

_

Z(O) kk 1fa(0)

Vf - Z(l) kk /(1) fa

( 1 2 .7)

- Z (2) / (2) kk fa

and these are the terminal voltage equations for the Thevenin equivalents of the sequence networks shown in Fig. 12.2. It is important to remember that the currents Ij�), Jj�), and I};) are . symmetrical-component currents i n the stubs hypothetically attached to the system at the fault point. These currents take on values determined by the particular type of fault being studied, and once they have been calculated, they can b e regarded as negative injections into the corresponding sequence net­ works. I f the system has .1 -Y transfo rmers, some of the sequence voltages calculated from Eqs. (1 2.6) may have to be shifted in phase angle before being combined with other components to form the new bus voltages of the fau l ted system. There are no phase shifts involved in E q . 02.7) when the voltage Vr at the fault point is chosen as reference, which is customary. In a system with .1-Y transfo rmers the open circuits encountered in the zero-sequence network require carefu l consideration in computer appl ications of the Z b us building algorithm. Consider, for instance, the solidly grounded Y-tJ. transformer con nected between buses § and ® of Fig. 12.3(a). The positive­ and zero-sequence circuits are shown in Figs. 12.3( b ) and 1 2.3( c), respectively. The negative-sequence circuit is the same as the positive-sequence circuit. It is straightforward to include these sequence circuits in the bus impedance matri­ ces Z�Js ' Z��s ' and Zb2�s using the pictorial representations shown in the figures. This will be done in the sections which follow when Y-/l transformers are present. Suppose, however, that we wish to represen t removal of the trans­ former connections from bus ® i n a computer algorithm which cannot avai l of pictorial representations. We can easily undo the con nections to bus ® in the positive- and n egative-sequence networks by applying the building algorit nm to

1 2. 1

z 'OOlJ'

?

�

Reference

Z

(b)

Reference

Z

UNSYMMETRICAL FAULTS O N POWER SYSTEMS

'-

Reference

477

Z

(c) Z

� Reference

Cd)

(e)

FIGURE 12.3

( a ) 6 -Y grou nded transformer w i t h I e a kage impedance Z ; ( b ) posItIve-sequence circu it; (c) ze ro-seq uence circuit; (d) posit ive-sequ ence c i rcuit w i t h i nternal node; ( e ) zero-seq u e n ce circ u i t w i t h i n ternal node.

the matrices ZbIJs and zb22s in the usual manner-that is, by adding the negative of the leakage impedance Z between buses @) and ® in the positive- and negative-sequence networks. However, a similar strategy does not apply to t h e zero-sequence matrix ZboJs if i t has been formed di rectly from t h e pictorial represcntation shown i n Fig. 12.3(c). Adding - Z between buses @) and ® d o c s n ot re move t h e zero-sequence connect ion from bus ® . To permit uniform procedures fo r all sequence networks, one strategy is to include an internal node ® , as shown in Figs. 1 2.3(d) and 1 2.3(e ). 1 Note that the leakage i m p e d a n c e is now s u b d i v i d e d into two parts between node ® and t h e o t h e r n o d e s as s h o w n . Con n e c t i n g - Z /2 between b u s e s ® and ® in each of the s e q u e n c e c i rcu i t s o f Figs. 1 2. 3 ( d ) a n d 1 2.3( c ) will o p e n the transformer con nec­ t ion s t o bus ® . Also, t h e o p e n circuits c a n b e re p resen ted w i t h i n t h e com p u t e r algori thm by br a nches of arbitrarily large impedances (say, 106 per u n i t). Intcrn al nodes of transformers can be useful in p ractical computer applications

ISee New

H. E. York,

Brown, Solution of Large Networks by Matrix Methods , 2 d e d., J o h n Wi ley & Sons, I nc., 1 9R5.

478

CHAPTER

12

U N SYMM ETRICAL FAU LTS

of the Z bus building algorithm. The reader is re fe r red to the reference cited in footnote 1 for further guidance i n handlin g open-circui t and short-circuit (bus tie) branches.

The faults to be discussed in succeedi ng sections may involve impedance Zt between lines and from one or two l i n es to ground. When Zj = 0, we have a d irect short circuit, which is called a bolted fault . Although such direct short c ircuits result in the highest value of fault current and are therefore the most conservative values to use when determ i n i n g the effects of anticipated faults, the faul t impeda n ce is seldom zero . Most fa u l ts a re the result o f i n su l a t o r flashovers, where the impedance between the l i n e and g ro u n d d e pe n d s o n t h e re s i s ta nce o f Ir a

a

b

Ifb

c

Ire

t

j �

a

Zr

O z/

b

Irb

c

Zr

Ira

b

Ir b

c

Ire

t

t

�

a

r

I

T

t

j

l£l ,

\ -

S i n g l e l i n e-to- g r o u n d fau l l

------------��-------------Ira

t

�

r

b ------------� • --------------

Zr

(c) Llne-to- line fau l t

FIGURE 1 2.4

Ire

(6)

(a) Three-phase fau lt

a

Ira

irb

-� Zr

c --I----�.-- ----Cd) Double

l i n e-lo- g ro u n d

Connection d iagrams of the hypot hetical stubs for various fau lts t hrough i m p e d an ce.

fault

1 2. 1

UNSYM M ETR ICAL FAU LTS ON POWER SYSTEMS

479

the arc, of the tower itself, and of t h e tower footing if ground wires are not used. Tower-footing resistances form the m ajor part of the resistance between line and ground and depend on t h e soil conditions. The resistance of dry earth is 10 to 100 times the resistance of swampy ground. Connections of the hypothetical stubs for faults through impedance Z f are shown in Fig. 1 2.4. A balanced system remains symmetrical after the occurrence of a three­ phase fault having the same impedance between each line and a common point. Only positive-sequence currents flow. With t h e fault impedance Zf equal in all phases, as shown in Fig. 12.4(a), we simply add impedance Zf to the usual (positive-sequence) Theve n i n equivalent circuit of the system at the faul t bus ® and calculate the fault curre n t from the equation ( 1 2 .8)

Fur each of the other fau l ts shown in Fig. 1 2.4, formal derivations of the equations for the symmetrical-compone nt currents I}�), I};>' and If;) are pro­ v i d e d i n the sections wh ich follow. I n e ach case the fault point P is designated as bus CD . Two synchronous mach ines are connected t hrough three-phase transformers to the transmission l i ne shown in Fig. 1 2.5. The ratings and reactances of the machi nes and transformers are

Example 12. 1 .

Machines

Transformers On

T)

1

and

and

2:

T2 :

1 00 M V A , 20 kV ;

XI

=

X2

networks a nd find t h e b u i lding algorithm.

=

15%

Xo

1 00 M V A, 20�/345Y kY ; a n d X()

z e r o - s eq u e n c e

FI G U RE 12.5 S i n g l e - l i n e d i agram of the system of Exa m p l e 1 2. 1 .

=

in

=

=

XI

X2

=

4%, Xn X

=

=

=

8%

20% , 5%

circuit the line D raw c
a c h o s e n b a se o f 1 00 M Y A , l 4 5 k Y

r e a c t a n ce s a r c

x;;

t h e t r a ns m i s s i o n - l i n e

50%.

480

CHAPTER

12 UNSYMMETRICAL FAU LTS

Reference

(a)

jO.04 (2)

® �1 (6)

Reference

7 jO. 15

(b) FIGURE 12.6 (a ) Positive-sequence and ( b ) zero-sequence n e t works o f the system of Fig. 1 2 . 5 . Buses are internal nodes of t h e transformers.

0

and

®

The given per-unit impedance values correspond to the chosen base, and so they can be used directly to form the sequence networks. Figure 1 2.6( a ) shows the positive-sequence network, which is identical to the negative-sequence network when the emfs are short-circuited; Fig. 1 2.6( b) shows the zero-sequence network with reactance 3 Xn = per u n i t in the neutral connection of each machine. Note that each transformer is assigned a n i n t ernal node-bus W for transformer T J and bus ® for transformer T2 • These i nt ernal nodes do not have an active role in the analysis of the system. In order to apply the Z bus building algorithm, w h ich is particularly simple in this example, let us l abel the zero-sequence branches from 1 to 7 as shown.

Solution.

0.15

Step 1

Add b r a n c h

1 t o t h c r c fc rencc node

CD

CD

[jO. 1 9]

Step 2

Add branch 2 to the reference node

CD G)

CD

[jO.019

W

0jO.04 1

UNSYM METRICAL FAULTS ON POWER SYSTEMS

12.1

Step 3

G)

A d d branch 3 between buses

[ G)

CD (])

Step 4 A d d branch

4

between buses

[0 CD

CD

CD G) W

[0

®

and

0 jO.04 jO . 04

Q)

O jO O4 jO.os

[

W

jO '()4

i

G)

0 iO .04 iO.OS

iO 04 iO .OS

iO .OS

iO .5S

G) and ® G)

j �

0

0 jO . 04 iO .04

0

! 0

0

J-

I G)

@

®

0 jO.04 jO .04 jO .04

0 j O .04 jO .08 j O .08

0 jO .04 jO .08 jO .58

0 j O.04 jO .08 jO.58

0

j O . 04

jO .08

jO .58

jO.66

Add branch 6 from bus

CD

@ to the reference

CD - jO . 1 9 G) 0 G) 0 ® @

@

I @

jO . 1 9 0 0 0

Step 6

0

@

jO . 1 9 0 0

jO . 1 9 0 ()

w

Add branch 5 between buses

CD

CD

CD

Step 5

f

and

G)

G)

G)

©

@

jO . 1 9

0 0

0 j O . 04 j O . 04 jO .04 jO .04

0 jO .04 jO .08 jO .08 jO .08

0 jO.04 jO .08 jO .58 jO.58

0 jO .04 jO .08 jO .58 jO.66

0

0

0

0

0

0 0 0 0 0

481

482

CHAPTER

12

UNSYMMETRICAL FAULTS

Buses G) and ® are the fictitious internal nodes of the transformers which facilitate computer application of the Z bu S building algorithm. We have not shown calculations for the very high impedance branches representing the open circuits. Let us remove the rows and columns for buses G) and ® from the matrix to obtain the effective working matrix

CD Z(O) - W G) @ tHIS -

CD

jO . 1 <)

W

G)

@

0

j O . O�

jO .Oc;

0

0

j O . OS

j O .58

0

()

()

n

()

0

0

jO . 1 9

The zeros i n Z ��s show that z e r o - se q u c n ce c ur re n t i nj ec t c c.J i n t l ! b u s CD o r b u s @ of Fig. 1 2.6( b ) cannot cause vol tages at the other b u s e s b C G.l u S C o f the o pen circuits i n t roduced by the 6-Y transfo rmers. Note also that the jO.08 per-unit reactance in series with the open circu it between buses ® a n d @ d o e s n o t affect Z��s s in ce it cannot carry current. By applying the Z bus building algorithm to the positive- and negative­ sequence n e tworks in a similar manner, we obtain

Z (bIu)s

=

Z(b2u)s

=

CD W ® @

CD

j O . 1 437 j O. 1 2 1 1 j O .0789 j O .0563

W

jO.121 1 j O . 1 696 j O . 1 1 04 jO .0789

G)

j O .0789 jO . 1 1 04 jO . 1 696 j O . 1 21 1

@

jO .0563 jO .0789 jO . 1 21 1 jO . l 437

We use the above matrices i n the examples which ful low. 12.2

S IN GLE LINE-TO - G RO U N D FAU LTS

The single l ine-to-ground fau lt, the most common type, is caused by lightning or by conductors making contact with grounded structures. For a single l ine-to­ ground fau l t through impedance Zf the hypothetical stubs on the three l i lies are connected, as shown in Fig. 1 2.7, w here phase a is the one on which the faul t occurs. The relations to be developed for t h i s type of fau l t will apply o nly when t h e fau l t is on phase a, but this should cause no difficu lty since the phases are lab eled arbitrarily and any phase can be designated as phase a . The con d itions at t h e fau l t bus ® are expressed by the fol lowing equations: ( 12.9)

12.2 SINGLE L1NE-TO-GROUND FAULTS

a

b

483

® �� �-----------------

I,o t ® r Irb{ -'-

--__ __ __ __ __ __

�

�

-

®

� I

c --------------�. __ Ir,

With

by

If b

=

ffe

-=-

----

--

Con nection diagram of the hypothetical st u bs for a s i ngle l i n e-to-gro u n d fau l t . The fa u l t poi n t is ca lled bus

CD .

0, t h e sym metrical c ompon e nts o f the stub c u r rents a r e given

=

1 (0 fa

)

(1)

1fil

1 (2 fa and

FI G U R E 1 2.7

--__ __ __ ___

1

3

=

)

1

1

1

1

a

a

1

a2

a

Ifa

2

0

0

p e rforming the m u l t i p l ication yields 1 (0 ) fa

Substituting I}:�)

for

ollLl i n

=

1( 1 ) Ja

=

1 (2) fa

I}I; ) a n d 1/(;) shows t h a t

Vk(2) a

=

H

V f

-

=

V1..(O) 0

+

Vkeal )

+ V ( 2) /; a

=

V

f

( 12 .10 )

3 1}�:), and from Eqs. 0 2.7) we

=

fa

( 0) Z k( lk) 1fa

fa

( 12.11)

( 2) / ( 0 ) - Z kk

=

S u m ming these equa tions a n d n o t i n g t h a t vk a

I a f

Ifa 3

- Z (0) I (0)

v (O) ka -

V1..e 11l )

=

-

Vk o

=

3Zf IX?

g ive

1(0) ( Z{O) + Z { l ) + 2(2)) kk fa kk

kk

=

3 2f ffa rO}

484

C HAPTER

12

UNSYMMETRICAL FAULTS

Solving for /J�) and combining the result with Eq. 0 2.10), we obtain /f(0)a

=

/f(a1 )

=

/f(a2 )

=

v

f

'

( 12 . 12)

Zk( Ik) + Z(2) kk + Z(O) k k + 3 Zf

Equations 02. 12) are t h e fau l t current equations particular to t h e single l i n e to groun d fault through impe d a nce Zf ' a n d they are used with t h e symm et­ rica l compon ent relations to d e te rm i n e all the voltages and currents at t h e fau l t point P . I f the Thevenin equivalent c ircu its o f t h e t hree sequ ence ne tworks o f the system a r e connected in series , as shown i n F i g . 12.8, we s e e t h a t the -

-

-

current s a n d voltages resulting therefrom satisfy t he above e q u a tions-for the Theve n i n impedances looking into the three sequence networks a t fau l t bus CD a re t he n in series w i t h t h e fau l t i m p e d a nce 3Zf a n d t h e p rc fa u l l vul t age sou rce Vf ' W i t h the equivalent circuits so connected, the volt age across each sequence network is the corresponding sym metrical component of the vol tage Vk a at the fault bus ® , and the current i nj ected into e ach sequence n etwork at bus ® is the negative of the cOfn;spondin g sequence current i n the fZlUl t . T h e s eries connection of the Thevenin equivalents of the sequence networks, as shown i n Fig. 12.8, i s a convenient m eans of remembering the equations for the solution of the single line-to-ground fault, for a l l the necessa ry equa tions for t h e fault ] ( 1) fa

zi�

Zi�

•

]( 2 ) fa

,

t+

®

1 (0) fa

t

Vk(2) a

1 CD )

fa

,

®

+

Vk(O) a

=

](\) fa

=

]('2) fa

3Zr

FlGURE 12.8 Con n ection of the Thevenin equivalents of the sequence networks to simulate a single l i n e-to1ground fault on phase a at bus ® of the syste m .

12.2

SIN G LE L1NE·TO·G ROU N D FAULTS

485

point can be determined from the sequence-network connection. Once the currents I}�), I}�), and I};) are known, the components of voltages at all other buses of the system can be determined from the bus impedance matrices of the sequence networks according to Eqs. (12.6). Two synchronous machines are connected t h rough three-phase transformers to t he transmission line shown in Fig. 1 2.9(a). The rati ngs and reactances of the m achines and t ra nsformers are Example 1 2 . 2.

Machines Tra nsfo r m ers T,

1

and 2 :

a n d T.... :

1 00

100

MVA ,

20

kV;

MVA, 20Y /34SY

Bo t h t r a n s fo r m e rs a r c s o l i d l y g ro u n d e d

on

x;;

=

XI

=

X2

=

X"

20% , =

5%

X = R%

kV;

e b ase o f 1 00

two s i d e s . O n (\ c h os n

:145 kY i n t h e t r a n s m i s s i o n - l i n e c i r c u i t t h e l i ne r e a c t a n ces are X I = X2 = 1 5 % a n d XI) = 50%. T h e sys t e m i s o p e ra t i n g a t n o m i n a l v o l t a g e w i t h o u t p r c fa u l t curre n ts w h e n a bo l t e d ( Zr = 0 ) s i n g l e l ine -to -grou n d fau l t occ u rs o n phase A a t

MV 1\ ,

bus G) . Using t h e b u s i mpcdance m a t r i x for e a c h o f t h e three sequence networks, determine the subtransient curre n t to ground at the fault, the li ne-to-g round voltages at the tcrminals of machine 2 , and the subtransient current out of p h ase c of machine 2.

The system is the same as in Example 1 2. 1 , except that the transformers are now Y Y connected. Therefore, we can conti nue to use Zb1Js and Z\,2Js corresponding to Fig. 1 2.6( a), as given in Example 1 2 . 1 . However, because the

Solution.

-

Mac;:�

oH�>---+---< -+-I 1 �Ho � CD

TI

®

@

n Y ·.I.

( a)

T2

@

n �

ne 2

)0. 15

Reference (b) FI G U RE 1 2.9 ( a ) T h e s i n gl e · l i n e

d iagram

and ( b ) z e r o - se q u e nc e n e tw o r k of t h e sys t e m o f Exam p l e

1 2. 2 .

486

CHAPTER 12

UNSYMMETRICAL FAU LTS

I (l) fA

V{

= 1.0�

+

®

I

j O . 1696

It

j O. 1696

+

V3( 2,1)

®

[ ( 0) fA

jO. 1 999

,

--

FIGURE 12. 1 0 Series connecti on o f the Thevenin equiva len ts o f the sequence ne tworks for the single l i n e-to-gro u n d fault o f Exa mple 1 2.2.

transformers are solidly grounded on both sides, the zero-sequ ence network is fu lly connected, as shown in Fig. 1 2.9( b ), and has the bus impedance matrix

CD

Z(O) - CV hll:-i

-

G) G)

CD

(1)

iO . l S S 3

i O . 1 407 j O . 1 999 j O . 07 0 I jO .049]

jO . 1 407 jO .0493 iO .O]47

G)

j O 04 9 3 jO .070 1 j O . 1 099 j O . 1 407 .

@

j O 0 3 47 jO .0493 j O . 1 407 i O . 1 SS ] .

Since the l ine-to-ground fau l t is at bus G), we must connect t h e Thevenin equivalent circuits of the sequence n etworks i n series, as shown in Fig. 1 2. 1 0. From this figure we can calculate the symmetrical components of the current IfA out of the system and into the fault, ] (0) - 1 ( 1 ) fA - fA

-

1 (2 ) fA Z( I ) -

1 .0 j(0 . 1 696

L2Q: 33

+

+

0 . 1 696

Zm 33

+

+

Z «l) 33

0 . 1 999)

- - j 1 .8549 per u n i t

487

12.2 SINGLE LlNE-TO-GROUND FAULTS

The total current in the faul t is =

=

-j5 .5643 1 0 , 0 00/ 45 6 5 -j5 .5643 6 5 93 1/ 270 2 , ( 1 2.6) j 4, -( J O . 1 407 ) ( -J· I .8549) -0.26 0 - 1 - ( J·0. 1 2 1 ) ( -J· 1 .8549) 0.7754 (j0 . 1 2 11 ) ( -j1 .8549) -0.2246 IjA

3 I}?J

per unit

/3

and since the base cu rre nt in the high-voltage transmission line is 3 = 1 7 3 A, we have .

IrA

X

=

1 73 .

- Z� 'I((J)� /J(IJ)/ 1

V-I(III )

=0

V J

i n vo l v e d .

=

- zf:, ) /J:i

=

(/

unit

per unit

d e n o t e v o l t ages a n d c u r re n l s i n t h e h i g h -volta g e

r e s p e ct i v e l y , From t h e a b o ve

l i n e - t o -grou n d v o l t a g e s a t bus

per

=

-

are

1 pe r u ni t

=

low-vo l t a ge c i rc u i ts ,

is

=

the terminals of machi ne

=

J /1

� o t e t h a t s u bsc r i p t s /1 a n d s h i ft

@,

=

Z ( 1 )/ ( I ) -1�

A

=

The phasc-a sequence voltages at bus calcu lated from Eqs. with k = 3 a nd

X

@

and Y-Y con nected transformer. No phase sym m e t r i c a l compo n e n t s w e can calculate a -b -c of the

as fo l l ows:

-0.0.72675104 -0.0.52386984 -jjOO..O36 0 -0.2246 -0.5364 jO.36 0 0.1.20819887 -121.3° L . 1

+

a

+

=

"

a-

a

[Q: L

I () un

20/

Ii ,

V-l iI

=

1 2 I �o

--

To e x p r e ss t h e l i n e - t o - g r o u n d vo l t a g e s

3.346L.Q:

which gives kV ·

V4 h = 1 1 . 763

of

/

m ach i n e i n t h e

Ollf

-

1 2 1 .8°

m a ch i n e

V4IOl u

--

=

=

=

1 1 .763

�.8Q

kY

must first calculate t h e b r anch e s repre senting the t h e zero-sequence current

we

t h e p h ase-a c u r re n t i n t h e

s e q u e n c e networks. From Fig. o f the machine is

-

V4 c

kV

21 2.9,(b) 0.jO.261040 -j6 .525

To d e t e r m i n e p h a se -c c u r re n t o u t of

sy m m e t r i c a l c o m p o n e n t s

of m achine 2 in k i l o vo l t s, we mult iply by

per

unit

488

CHAPTER 1 2 UNSYMMETRICAL FAULTS

and from Fig. 12.6(a) the other s equence curr ents /a( 1 ) =

/a( 2 ) =

f

v

-

-

V4(la )

" jX

1 .0 - 0 . 7754 =

v4(a2 ) jX2

--

jO . 2

0 2246

0

are calculated as

-j 1 . 1 23 per u n i t

- -jl . 1 23 per u n i t

.

j O . 20

Note that the mach i ne currents a rc s h o w n w i t h o u t subscript J , w h i ch i s rese rved exclusively for the (stub) currents a n d voltages of th e fa u l t poi n t . The phJse-c currents in machine 2 are now easily calcul ated, =

-j 6 . 5 2 5

+

a ( -j 1 . 1 23 ) + a 2 e -j1 . 1 23 )

= -j5 .402

per

unit

The base current i n the machine circu i t s i s 100,OOO/({3 X 20) = 288 6 . 7 5 1 A , a n d so I Ie l = 1 5,594 A . Other voltages a n d c u rrents i n t h e system can b e calculated s imi la rly .

12.3

LINE-TO-LI NE FAULTS

To represent a l i n e-to-lin e fau l t through i m pedance Zf ' the hypothetical stubs on t he t h re e l i n es at the fault a re con n ected, as s hown in Fig. 12. 1 1 . Bus ® is aga i n the fault point P, and without a n y loss of generality, t h e li ne-to-l i ne fa u l t i s regarded a s being o n phases b and satisfied a t t h e fau l t point

c.

The

following rel a tions must b e

( 1 2 . 1 3)

12.3 LlNE-TO-LINE FAULTS

Vr

®

z( l) kk

+

®

Zr

+

Vk( al )

2) Zk( k

Vk( 2a)

Reference

489

FIGURE 1 2 . 1 2

Con n ection of t h e Theven i n equ ivalents of t he positive- and negative-sequence networks for line-to-line fa u l t between phases b a n d c at bus ® o f the system.

Since

If b

=

- Ife

and

Ifa �

0, the sym metrical com ponen ts of curre n t are

0)

If( li If( a

I)

J (2) fa

a

1

a

1

3

1

a

2

a

2

a

0 Ifb

- lfb

and m u l tiplying t h rough i n t h i s equation shows t h a t

( 1 2 .14) 1(1) fa

=

-

( 12.1 5 )

/ (2) fa

The voltages t h roughout t h e zero-sequence n e twork must b e zero since there are no zero-sequence sourc es, a n d because 1J�) 0, cu rrent is not being i nj ected into t h a t network d u e to t h e fault- Hence, line-to-line faul t c alcula tions do not i nvolve the zero-seq uence n e twork, which remains the same as before the fa u l t - a dead network. T o satisfy the req u i re m e n t t h a t I}(�) = - l}�>' let us connect the Thevenin e q u ivalents of the pos i t ivc- and negat ive-seque nce networks in parallel, as shown i n Fi g. ] 2 . ] 2 . To show t h a t t h is con nection of the networks a lso satisfies t h e voltage e q u a t ion V" h - Vkc 111> /, f , we now exp a nd e ach side of that e q u ation separately as fol lows: =

� ,

1 Z fb f

= ( 1f(b1 ) + 1fb(2» ) Zf -- ( a 21fa( 1 ) + a lfa(2» ) Zf

490

CHAPTER 12

Equating

UNSYMMETRICAL FAU LTS

both terms and setting I};)

=

- IJ�) as in Fig.

( a 2 - a ) ( Vk(al ) - V
or

=

Vk( al ) - Vk( a2 )

=

1 2 . 1 2, we ob tain

( I )Z f ( a 2 - a ) Ifa Ifa( I ) Zf

( 1 2 . 1 6)

which is p recisely the vol tage-drop e q u a t ion for i m p e d a n c e Zf i n F i g . 1 2 . 1 2. Thus, all the fa u l t cond i t i o n s o f Eq s . ( 1 2. 1 3 ) a r c s a t i s fi e d by co n Il e c t i ng the positive- a n d negat ive-seq u e n c e n e t w o r k s il l pam llel l h ro u g h i m p c J ; l l1 c e Zr ' as s how n in Fig. 1 2. 1 2. T h e z e ro - se q ll c l l c � I l c l w o r k i s i n a c l i v e il n u dm:s n o l e n L e r into t h e l i n e - to- l i ne fau l t c a l c u l a t i o n s . T h e e q u a l i o n fo r t h e p os i t i ve - s e q u e n c e cu r r e n t in t h e fa u l t ca n b e d e t e r m i n e d d i r L' L" t l y f ro m F i g . 1 2 . 1 2

1fa( 1 )

For

=

- 1f(2) a

so

t h:lt

( 1 2 . 1 7)

_

a bolted line-to- l i n e fau l t we set Lf O . Equations ( 1 2 .1 7) are the fau l t c u r re n t e q u a t i o n s fo r a l i n e - t o - l i n e fau l t t h rough impedance Zf ' Once I}�) and Ii;) a re know n , they can h e t re a t e d as i nj ections I}2 and -: IJ�) into the p o s i t ive- and negative-sequence networks, r e sp e ctively, and the changes i n t he s eq uence vol tages at the buses o f the sys tem due to t h e fau l t can be ob tained from the bus impedance m a t rices, as p r e v i o u s l y demonstrated. When 6. - Y transformers a re present, p h as e s h i ft o f the pos i tive­ and negative-sequence currents and voltages must be t a ken i n t o acco u n t in t h e calculations. The following example shows how t h i s i s accomplish ed . =---'

-

Example 12.3. The same system as i n Example 1 2 . 1 is opera ting at nominal sys tem voltage without prefault cu rrents when a bolted l i ne-to-li ne fault occ urs at bus G) . Using the bus impedancc matrices 0 f the seque nce networks for subtransient conditions, determine the currents ill the rau l t, t h e l i n c - t o - l i n c Yo l t (l g e s a t t h e Ll u l t bus, a n d t h e line-to-line vol tages a t t h e t er minals o r machine 2 .

z�)12s a n d Z\;J, a rc a l re a d y s e t fo r t h i n Example 1 2 . 1 . A l t h o u g h Z\:�s is also given, we arc not concerned w i th the zero-sequence ne twork in this solu t ion since the fault is line to line. To simulate the fault, the Thcve n i n equivalent circuits at bus G) of the positive- and negative-sequence networks of Example 1 2. 1 arc connected in parallel, as shown in Fig. 1 2. l 3 . From this ligure the seque nce curre n ts arc calculated as follows:

Solution.

1 + jO

j O . 1 696

+ jO . 1 696

- j2.948 1 p e r

unit

Uppercase A is'used here because the fau l t is in the high-voltage t ransmission-line

1 2.3

j O. 1696 v(

=

1 . 0�

®

(1) 1(A

�

+

+

Fl C U RE 1 2 . 1 3

1f( 2A)

t+

491

-

jO. 1696

) V3(2A

)

Vel

-l

®

L1NE-TO-L1 NE FAU LTS

3A

I

Reference

Conn ection of t h e Theve n i n equiva l e n t c i rcuits for t h e l i ne - t o - l i ne fa u l t of Example 12.3.

circuit. Since

If

A

Ifll

=

0, the com po ne n ts of currents in the fault are calculated from

Ij�) + IJ�) =

=

=

Ij�)

a 2 IJ�) + alJ;; =

-

j 2. 948 1

+

j2.9481 = 0

-j2 . 948 1 ( - 0 .5 - jO .866) + j2 .9481 ( - 0 .5 + jO . 866) - 5 . 1 06 1 +

jO per u n i t

5 . 1 06 1 + j O per unit

As in Example 1 2 .2, base current in the transmission line

ffll

Irc

=

=

Vj�)

=

=

1 67.35 A , and so

�A - 5 . 1 06 1 X 1 67 .35 = 8ssLQ: A

- 5 . 1 061

X

1 67 .35

= 855

Symmet rical compon ents o f phase-A vol t age to ground

vJ(O)A

is

at

bus

®

are

0

Vj�) =

1

Zi� /}�)

-

=

1 - ( j0 . 1 696)( -j 2 . 948 1 ) = 0 . 5 + jO per u n i t

Line -to-ground vol tages a t fa ult b u s G) are V3 A - V3(O) 1'1 + -

V3 C = V3 8

,;"

vj�)

0 .5

+

�

vj�)

= 0 +

per unit

0 .5 + D .S

=

LQ: per unit

1 .0

492

CHAPTER 12

UNSYMMETRICAL FAU LTS

G)

Line-to-line voltages at fault bus

V3, A B V3A - V3B =

V3 , BC = V3B - V3C V3, CA

( - 0 .5 + jO) - ( - 0.50 + jO )

=

V3C - V3 A =

=

1 .0 + jO) - ( - 0.50 + jO)

(

=

( - 0 .5 + j O) - (

l .0 + jO )

Expressed i n volts, these line-to-line vol tages a re V3 . A ll

=

V3 , CA

=

1 .5

LQ:

1 .5/

1 .5L.! p er unit

are

X f3

= 299

345

1 80'" x

345 ",'

v3

=

=

0

=

1 .5

=

L..Q:

�

pe r u n i t

kY

299L 1 800

kY

._

For the moment, let us avoid phase shifts due to the Ll-Y transformer connected to m achine 2 and proceed to calcu late the sequence voltages of phase A at bus @ using the bus impedance matrices of Example 1 2 . 1 and Eqs. ( 1 2.6) with k = 3 a nd j = 4 V4(AO)

V4�)

V4�)

=

=

=

-

2 (0)1 3 (0)

4

fA

VI - Zg>lj� -

=

=

0 1 - ( j0 . 1 2 1 1 ) ( -j 2 . 94 8 1 )

ZWI};J =

- ( j0 . 1 2 1 1 ) ( j2 .948 1 )

=

0 . 643 p e r u n i t

= 0 .357 per u n i t

T o account for phase shifts i n stepp i n g down from t h e high-vol tage t ra nsmission line to the low-voltage termi nals of machine 2, we must retard the posi tive-sequence voltage and advance the negative-sequence voltage by 30° . At machine 2 terminals, ind icated by lowercase a, t h e voltages are

V4';) ��)�

=

=

V4 a

=

��) + �(� )

+

�

0 .3 5 7

V4(;)

= =

0

+

=

0 . 3092

+ jO . 1 785 per unit

(0 . 5569 - j O . 3 2 1 5 )

0 .866 1 - jO . 1 430

+ (0 .3092 + j0 . 1 785)

= 0.87781 - 9 .40

per u n i t

1 2.3 LINE-TO-UNE FAULTS

493

Phase-b voltages at terminals o f machine 2 are now calculated,

V4b

and for v4((l) c

V4\1 )

V4�) + V4(� )

=

ph ase

=

=

V4(O) "

=

=

=

+

+

+

-

-

=

=

=

=

=

=

(1/ 2400

�(j )

+

V2)

0 .357L

=

+

=

+

V4 a - V4b

70 60 per u n i t

V4b - V4c

V4(

_...

V4 0

per unit

=

-

+

per unit

are

=

=

-

0 . 8661

= 0 . 9665

V4 , ca

=

at

+

per unit

pcr unit

tcrm i n als arc

x

X

153 .65°

jO . 1 43)

=

machinc

�

0 , 9665

=

-

- ( 0 .866 1

+

v o l t s , l i n c - t o - l i n c v o l t ages

V4 , bc

+

( - 0 .866 1 -

=

� . a h = 1 . 7322

per unit

+

=

In

jO. 1 785)

per u n i t

=

=

V4. CiJ

-

(1� )(0.643/ -300 ) 0.643� )(0 .357� ) -90 0 (j0.643) ( 2jO.357) 0 jO.286 (1.07.8322661 -jjOO. 143) -(0.8661 -jO. 143) j0.143) - (0 jO.286) j0.429 0.96 5/ -153 .650 (0 jO.286) -CLr;661 j0.429 0.96 5/ 153.65" 2 20 / -1 5 3. 65° 20 11. 2 / -1 5 3. 65° 1 1 .2/153.650 / =

V4 be

0.8778/

+

2

Line-to-line voltages at the terminals of m achine V4. ab

per u n i t

+

0

a V}� ) =

��)

V4<;)

c of machine

V2) = a 2 V4(;) V4c

+

-0.3092 j0.1785 0 ( 0.5569 -jO_3215) ( - 0.3092 0.8661 -jO. 143 1 .

X

fS

-

=

13

=

20 13

=

20

�

kY

kV

kV

494

CHAPTER 12

U NS YMMETRICAL FAULTS

Thus, from the currents Ijr;t, Ij!), and Ij;; of the faul t and the bus impedance matrices of the sequence networks we can determine the unbalanced bus voltages and branch currents throughout the system due to th e l i ne-to- line fau l t . 12.4

DOUBLE LINE-TO-GROUND FAULTS

For a double line-to-ground fau l t the hypothetical stubs are connecte d , as shown in Fig. 12.1 4 . Again, the fa u l t is taken to be on phases b and c, and t h e relations now existing a l t he fa u l t b u s ® a rc Since Ifa i s zero, the ze ro - s eque nce cu rre n t is given by t h e vol tages of Eq. 0 2 . 18 ) t h e n become

Ij�)

=

( Jjb +

( 12. 10)

If) 1 3, a n d ( 1 2 . 19)

Substituting Vk b for Vk c in the symmetrical-component transformation, we fi n d that

Vk( aO ) 1 1 Vka( l ) - - 1 (2)

vka

a

b

c

Ir
Irbt Ire

t

3

1

1

1

a

a

a

2

2

a

Vk a Vk b Vk b

( 12 .20)

®

i

® •

®

FIGU�E 12. 14

Connection diagram for the hypot h e t i c a l stubs for a double line-to-grou nd fau l t . The fau l t po int i s called b u s

®.

1 2.4

bOUBLE LlNE-TO-G ROUND FAULTS

495

The second and third rows of this equation show that v: ( l )

=

ka

Vk(2) a

( 12.21)

while t h e first row a n d Eq. (12.19) show that

Collecting zero-sequence terms on o n e side, setting Vk(;), we obtain

Vel) ka

VJ;)

=

Vf�),

and s o l v i n g for

Vk(O)u - 3 Zf 1f(0a )

=

Bringing together Eqs. 0 2 . 2 1 ) a n d 0 2 . 22), and again noting t hat arrive at the resul t s V(I) ka

=

Vk( a2 )

+

1f(a0 )

V (O)

=

ka

1f(a1 )

+

-

1 (2) fu

Ifa

( 1 2 .22 ) =

0, w e

3 Zf 1fll(0 ) =

0

( 1 2 .23 )

These characterizing equations o f the double l in e- to-ground faul t are satisfied when all t hree of the sequence n etworks are con nected in parallel, as shown in Fig. 12.15. The d iagram of n e twork c onnections shows t ha t the positive­ sequence current I}�) is determ i n e d by applying p re fault voltage Vf ac ross the total impedance consisting of Zk� in series with the parallel comb i nation of zi� and (Zko; + 3Zf ). That is, 1f(a1 )

] ( 1)

9

�

Z(I)

kk

fa

®

-! + vkOa}

�-

=

Z k( Ik)

+

[

Z k( 2k) ( Z k(O)k + 3 Zf Z k( 2k) + Z k( Ok) + 3 Z J.

)

fa ® -] ( 2)

�

Z( 2 )

kk

-t + vk(2) a

�-

1

� ] (0)

-c::::J-

Zk(O) k

( 1 2. 24 )

®

-\ +

Vk(O) a I•

FIG U RE 1 2 . 1 5

Co n nection of t h e Theve n i n equivalents of t h e sequence networks for a double l i n e-to-ground fault on phases b and c at bus ® of t h e system.

496

CHAPTER 12

UNSYMMETRICAL FAULTS

The negative- and zero-sequence currents out of the system and into the fault can be determined from Fig. 12. 15 by simple current division so that

[(III2) _

_ /(1) III

fa

1 (0) _ 1(1) = la

[ Z(O) [

3Z

Z (2) + Z (O) + 3 2 k /.; kk f kk

+

Z [�

2 k( 2k) + L k(Ok)

I

1

( 1 2 .2 5 )

( 1 2 .26)

+

For a bolted fault Zf is set e q ua l to 0 in the above e q u a t i o n s . \\' h e n LI = the zero-sequence circuit b e c o m e s a n opc n c i rc u i t ; 1 1 0 z e ro - se q u e nc e c u rr e n t c a n then flow and the equations rcvc r L back t o t h o s e fo r t h e l i n c - t o - l i n e fa u l t discussed in t h e preceding section. Again, we observe that the sequence currents I}�), I}�l, a n d I}�l, once calcu lated , can be treated as negative inj ections into the sequ e nce n e tw o r k s at the fault bus ® and the sequence voltage changes at all buses of the system can then be calcul ated from the bus i mpedance matrices, as we have done i n p receding sections. 0::- ,

12.4. Find the subtransie n t currents and the l ine-to-l ine voltages at the fa u l t under subtransient conditions when a doublL: line-to-ground fault with Z 1 = 0

Example

occurs at the term inals of machine

i n the syste m of Fig. 1 2 . 5 . Assume that the system is u nloaded and operating a t rated voltage when the fault occurs. Use the bus impedance matrices and neglect resistance. 2

. . . SoIutwn. · Th e b us Impe. d ance matnces Z
VI

=

/,., l . OL..9�

+ ,--j O . 1437 V e l ) -{=::Jr-----t-t:--:-'+ \ -

40

j O . 19

FIGURE 12.16 Connection of the Theve n i n e quivalents of t h e s e q u e nce n e t works for t h e double Example 1 2.4.

l i n e -lo- I i ne fault o f

497

1 2.4 DOUBLE LlNE-TO-GROUND FAULTS

obtain

I (a! ) f

--

r-V _f___..,,-

_ _ _

z(l) 44

+

[ Z(2+)Z(0) 1 . 44

ZUl

44

Z��)

j O . l 437

[

+

1 +

jO

( j0 . 1 437) ( j0 . 1 9 ) ( j 0 . 1 43 7

+

jO . 19)

-j4 .4342 per unit

Therefore, the sequence voltages a t the fault are V(I) 4

a

=

V Ol � II

=

V (O) 4 (/

=

V j

- If(aI )Z ( I ) 4-1

-

1

=

( -JA 4342 ) ( J' 0 . 1 437 ) •

Cu r re n t i nj e c t i o n s i n t o t h e n e ga t iv e - a n d ze ro-se q u e n c e arc

networks

c a l c u l a t e d hy c u rre n t d i v i s i o n a s fo l l ow s :

/ (0) fa

=

_ f( l ) fa

[

Z�) 1 Z�) + Z��)

=

-

'4 434 ?

J .

[

jO. l 437

j(0. 1437

0 . 1 9)

+

1

=

The currents out of the system at the fa u l t point are

IIG Ij�) Ij�) + Ij;) =

=

=

+

j 1 . 9095

j 1 .9095 - j4 .434 2

(1/240'" )(4.4342/

- 90"

+

- 6 .0266 j 2 . 8642 0,6726/ 1 54 .6"

= j 1 .9095 =

+

=

+

+

=

and the current

=

6 .6726

/ 25 .40

If i n t o t he ground is If Ifb + fie =

=

3 Ij�)

=

=

0 3 6 2 8 per •

unit

a t t h e fa u l t bus

j 1 . 9095 per u n it

0

) + ( 1 LE2: ) ( 2 .5247� )

(l� ) (4 .4342/ - 90CJ )

6.0266 + j 2 . 8 6 42

j2 . 5247

=

per u n i t

+

( !� ) ( 2 . 5247L22: )

p e r unit

j5.7285 per unit

498

CHAPTER 12

UNS Y M M ETRICAL FAU LTS

Calculating a-b-c voltages at the fault bus, we find that

V4 • co

=

V4 C - V4 0

Base current equals and s o w e find that

lib

Ifc

If

=

1 00

X

2887

lO:l/(!X

=

X

-

1 .088 4 per u n i t 20)

X

=

6 .6726/ 154 .6°

2887

=

= 2887 X 6.6726/ 25 .40 =

=

2887

X

5 .7285LiQ:

=

The base line-to-neutral voltage in machine 2 is

V4 • ub

V4• cu

= =

1 .0884

X

- 1 .0884

f3

20

X

f3

20

=

=

A i n t h e circuit o f m a c h m e

2,

19,262/ 154 .6° A

/

1 9 ,262

25 .4° A

16 ,538LiQ: 20/ 13

-A

kV, and so

1 2 .568LQ: kV

1 2 .568� kV

Examples 12.3 and 12. 4 show that phase shifts due to /l- Y transformers do not enter into the calculations of seq uence cu rrents and vol tages in that part of the system where the fault occurs, provided Vf at the fault point is chosen as the reference voltage for the calculations. However, for those parts of the system which are separated by /l - Y transformers from the faul t point, the sequence currents,

and voltages calculated by bus impedance matrix must be shifted in

phase before being combined to form the actual voltages. This is because the bus impedance matrices of the sequence networks are formed without consider­ a t i o n of phase shifts, and so they consist of per-unit impedances referred .to the part of the network which includes the fault point.

1 2.4

DOUBLE L1NE-TO-GROUND FAULTS

499

@, the end of the transmission l ine remote from the double l ine-to-ground fault, in the system of Example 1 2.4. Example 12.5. Solve for the sub transient voltages to ground at bus

Numerical values of the faul t-current components are given in the solution of Example 1 2.4 and the elements of zb11s, Z �Js ' and zbols are provided in the solution of Exampl e 1 2. 1 . Neglecting phase shift of the �-y transformers for the moment and substi tuting the appropriate values in Eq. ( 1 2 .6) , we obtain for the voltages at bus @ due to the fault at bus 0 ,

Solution.

Vi�) Vi�)

- I}�)Z��)

=

=

Vi�) =

v, - IJ�)Z��)

- Ig)z W

- ( j l .9095 ) (0)

=

=

1

=

� V2(� = Vi�)/ - 300 =

Vi� )

=

=

0

- ( -j4 .4342) ( j0. 0789) = 0 . 65 0 1 per unit

- ( -j2.S 247 ) ( j O .0789)

Account i n g for ph ase s h i ft in stepp i ng fau l t at b u s G) , we have

Vi �

=

up

=

0 . 1 992 per unit

to the transmission-line circu it from the .

� 0 . 1 992/ - 300

0 .6501

=

0 .5630

+

jO.325 1 per unit

= 0 . 1 725 - jO . 0996 per unit

The required vol tages can now be calculated: v2 A

V2 B

=

=

V2(O) A + V2( AJ )

V2A ( O)

+

V2( 2A)

+

a 2 V2( AJ )

+

a V2(2A)

=

(0 .5630 + j O .325 1 )

= 0.7355 =

(O V2C - V2 A)

+

aV(I) 2A

+

a 2 V2(2A) -

=

0 .7693

- jO .0996 )

/ 17 .0° per unit

( 0 . 1 725

/ 30° ) ( 1 / 2400 )( 0 . 653 1 �

�) - 0 . 1 725 - j O .5535 = 0 .5798/ 1 07 .30 +

=

+ jO.2255

+

( 1 LE2: ) (

0 . 1 992

per unit

(1/ ] 20(> )(0 . 6531/30C' )

( 1 / 2400 )( 0 . 1 992� ) - 0 .5656 j O . l 274 0 .5798/ 1 67 .30 _

+

+

=

per unit

These per-unit values can be converted to volts by multiplying by the line-to-neutr al base vol tage 345 / f3 k V of the transmission 1 ine.

SOO 12.5

CHA PTER 12

UNSYMMETRICAL FAU LTS

DEMONSTRATION PROBLEMS

Large-scale computer programs b ased on t h e bus i mpedance matrices of the sequence networks are generally used to analyze faults o n electric u til ity transmission systems. Three-phase and single l ine-to-ground fau l ts are usually the only types of faul t studied. S ince circuit-breaker applications are made according to the symmetrical short-circuit current that m ust be i n terru pted, this current i s calculated for the two types o f fault. The p r i n tout includes the total fault current and the contributions from each l ine. The results a lso l ist those quantities when each l ine co n n ecte d to the fau l ted b u s is opened i n t u rn w h i l e a l l others are i n opera tion. The program uses the impedances for the l i nes as provided i n t h e l i n e d a t a for t h e power-flow program and inc l u d e s the appropriate r e a c t a n c e for each m achine in forming the positive- and zero-sequence bus impedance matrices. As far as impedances are concerned , the n egative-sequence network is t a ke n to be the same as the positive-sequence network. So, for a single line-to-ground fault a t b us ®, I}�) i s calculated in p er unit as 1 .0 d ivided b y the sum (2 Zk� + Zk� + 3 Zr). T he bus voltages are included in the compu ter printout , i f called fOf, as well as the current in l ines other than t ho s e connected to the faul ted bus since this information can easily be found from the bus imped ance matrices. The fol lowing numerical examples show the a nalysis of a single l ine-to­ ground fault on (1) a n industrial power system and (2) a small electri� u ti l i t y system. Both of these systems are quite small in extent compared to the large-scale systems normally enco untere d . The calculations are presented with­ out m atrices in order to emphasize t h e circuit concepts wh ich underlie fau l t analysis. The presentation should allow t h e r e ader to become more fam i l i a r wit h the sequence networks and how t h ey a r e used to analyze f aul ts. The principles demonstrated here are essentially t h e same as those employed within the l arge-scale compu ter programs used by industry. The same examples are to be solved by the bus impedance matrices in the pro blems at the end of t his chapter. A group of identical syn chronous motors is con nected t h rough a transformer to a 4 . l 6-kV bus a t a loca tion remote from the generating pl ants of a power system. The motors are rated 600 V a n d operate at 89.5% efficiency when carrying a ful l load at u n ity p ower factor and rated voltage. The sum of their output ratings i s 4476 kW (6000 h p ) . T h e r e actances i n per u n i t of e a c h motor based on its own input kiJovoltampere rat i n g are X� = X I 0.20, X 2 = 0.20, Xo 0.04, and each is grounded through a reactance of 0.02 per unit. The motors are con nected to the 4.l6-kV bus .t hrough a transformer bank composed of t hree single-phase units, each of which is rated 2400/600 V, 2500 kVA. T h e 600-V windings are connected in 6. to the m otors and the 2400-V windings are connected in Y. The leakage reactance of e ach transformer is 10% . The power system which supplies t h e 4.16-kV b u s is represented by a Thevenin equivalent generator rated 7500 kVA, 4 . 1 6 kV with reactan<;es of Example 12.6.

=

=

1 2.5

DEMONSTRATION PROBLEMS

501

CD

F IGURE 1 2 . 1 7

Single-line d iagram of the system of Example 1 2.6.

Motors

x;;

X2 = 0. 1 0 per u n i t , Xo = 0.05 per u n i t a n d Xn from neutral to grou nd eq u a l to D.05 p e r u n i t . Each o f t h e i d e n t i c a l m o t o rs i s s u pp l y i n g
,

subtransien t l i ne currents in all parts of the system with prefault current n e gl ected. The single-l i ne diagram of t h e system i s shown in Fig. 1 2 . 1 7 . The 600-V bus and the 4 . l 6- k Y bus are nu mbered CD an d (1), respectively. Choose the rating of the equivale n t generator a s base: 7500 k Y A , 4 . 1 6 kY at the system bus. Since

Solution.

f3

X 2 4 00

=

3

4 1 60 Y

X

2500

7500

=

kYA

the three-phase rating of the transformer is 7500 kYA, 41 60Y /600� V. base for the motor circuit is 7500 kVA, 600 Y. The input rating of the single equivalent motor is 6000 X 0 .746 0 .895

=

5000

So,

the

kYA

re;lctances or t i l C e q u i v a l e n t motor in perce n t arc t h e s a m e on t h e base of t h e com b i n e d ra t i ng a s t h e reac t a nc e s of t h e i n d i v i d u a l motors on the base of the ra t i n g o r an i n d i v i d u ;t 1 motor. The re actanccs of t h e e q u i v a l e n t motor in p e r unit on t h e s e l e c t e d base a rc
X:;

=

Xl

=

In t h e z e ro-se q u e nc e

X2

=

0 .2

7500

5000

n e tw o r k

the

3 XII

X

equ ival ent motor is =

3

Xo

0.3 =

=

0.04

7500 --

5000

=

0 .06

reactance between neutral and ground of the

0 .02

7500 5000

=

0 .09 p e r u nit

502

CHAPTER 12

UNSYMMETRICAL FAULTS

and for the equivalent generator the reactance from neutral to ground is

3Xn 3 0. 05 0. 15 X

=

=

per unit

Fi gure 1 2.18 shows the series connection o f the sequence networks. S ince the motors are operating at rated vol tage equal to the base voltage of t he motor circuit, t h e prcfault volt a ge of ph ase a at the fault bus CD is

,

1.0 7,500,060000 Vf

=

per u n i t

B ase current for the motor circu i t is

= 72 1 7 A

13 X

---=---

and the actual motor current is

746 5000 0.88 600 0.85 0.85 0.665 '0 .565 X

13 X

-----;0=------ =

X

X

Current drawn by the motor through line

--I 7217 L------

4798

-

cos - 1

! 1-

=

a

before the faul t occurs is

3 1 .80

-

4798 A

=

-

jO.350 per unit

1.0&

If prefaul t current is neglected, E; and E;� are made equal to in Fig. 12.1 8. Thevenin imped ances a re compu ted at bus CD in each sequence network as follows:

(i O . l + j(O. l +

- jO0..1l )( j0 .3) jO.12 1. 0 jO. 12 jO. 1 2 1(2) 1(1 ) -)'2 564 3 1}�) 3( -j2.564) -j7.692 -j2.5jO.6450 jO.30 -j1.538

Z( I ) 11

=

Z (2) 11

_

+

0.3)

Fau lt curre n t i n t he series con nection V,_ ..;; _ ____.. If(a1 ) _ - zg) + zg) + z \7)

---=-

fa

=

1 (0a) f

=

fa

Curr ent i n the faul t

=

the sequcnCL: networks is

+

+

per u n i t

sequence network the portion of =

of

-------

.

=

found by current division to be

X

I}�)

z\�) = jO. 15 per unit

per unit

=

jO . 1 5

=

1 .0 jO.39

----

=

-j2.564

per unit. In the posltlve­ flowing toward P from the t r ansformer is =

per unit

12.5

®

CD

0.565 - j 1 .SSS

-

j O. 3 0

Positive-seq uence network

CD

-j 1 .538 rv-"'><><-..

-

j O. 30

N eg ative-sequence netwo rk

®

jO.IO

CD

p

1

-j l . 026

p

jO. IO

503

0.565 - jO. 676

p

jO.lO

DEMONSTRATI ON PRO B LEMS

-j2.S64

-j 2.564

) 0 .05

jO.06 Zero-sequence network jO.09

jO. I S

FI G U RE 1 2 . 1 8 Con n e c t i o n of t h e s e q u e n c e n e twor k s of Exam p l e 1 2 .6. S ub t ra n s i e n t currents a re m a rked i n per u n i t

f o r a si n g l e l i ne - t o- g r o u n d faul t a t

P.

P refa u l t c u r r e n t i s i nc l ud e d .

P

and the portion of flowing from the motor ward is - j 2.5j6O4.50 jO.20 -j 1.026 per unit Sicfroommpone imlarly,nmottthofeoporr. tifornomofthe motfroomr ist h-je t1r.a0n26sroperrmerunit. of perflouniwst,toandward Cur r e nt s a t t h e f a u l t , s h own wi t h out s u b s c r i p t ar e : To from the t ransformer in per unit: 7 0 . 6 0 1l 1 [- 1 I}�)

to

X

Ii;')

t he

P

is

I};)

in th e fines

1

a

a1 a2

j 1 .538

- j 1 .538

-

=

- 1 .538 j

All

3

J}�)

jj1 .538 j 1 .538

j,

the P

S04

CHAPTER

12

UNSYMMETRICAL FAULTS

To P from the motors in per unit:

a

1 ][ -j2.-jl.506426 ] [-j4 .616 -jl.026 lL23(a )

a a2

-j 1 .538

=

-jL538

Our method of l abeli ng the l ines is the same as in Fig. s u c h t h a t c u rre n ts 2 I and � ) in the lines on the high-voltage side of t h e transformer are related to the currents 1� 1 ) and I�2) in the l ines o n t h e l ow-vo l t a g e side by

I�l)

Hence,

I�l) = I�2) =

( -J l .538) �

- j 1 . 3 32

3 / -60° 0.769 ( -jL538)/ -30° 1 .538/ -1 2 0° 12.18 0 (0 .769 -Jl.332) ( -0.769 -jl.332) j2.6b4 (1/2400 )(1 .538/ -60° ) -1. 5 31) )(1.538/ -1200) 1.538 0 = 0. 769 j1. 3 32 ( 1� j2.664 .

= 1 5

8

=

=

=

- 0 .769 -

j 1 .332

and from Fig. we note that I�O) = in the zero-sequence network. S i nce there are no zero-sequence currents on the high-voltage side of t h e t ra n sform e r , we have

fA I�l ) If) I8

IP)

It2)

Ie

= =

=

I�I) + I�2)

a 2 I�1 )

8

= a =

=

+

l�l )

a 2 I�2)

/ 8(2 )

Itl) + 12) =

per u n i t

+

) ( 1 .538/ - 600 )

( 1 / 2400 ) ( 1 .538/ - 1 200 )

-

+ jO

=

-

=

=

+ jO

=

=

a I�2 ) = ( 1�

= /(\)

=

+

=

=

- 0 . 769

+

j1 .332

per unit

If voltages throughout t h e system a re to be fou n d by circuit analysis, their components at any point can be calculated from the currents and reacta n ces of the sequence networks. Components of voltages o n the high-voltage side of the transfo rmer are found first without regard for phase shift. Then, the effect of p hase shift must be determined . B y evaluating the base currents o n the two sides of the transformer, we can convert the above per-unit currents to amperes . Base current for the motor circu it

12.5

was found previously a n d equals 721 7 IS

A.

13 X 4 1 60

Curre nts in

50S

Base current for the h igh-voltage circuit

7 ,500,000

Current in the fa ul t

DEMONSTRATION PROBLEMS

---=---

=

7 . 6 9 2 x 72 1 7

=

1 04 1 A

is

55 ,500 A

the l i n es bet wee n the transformer a n d the 3 .070

X

72 1 7

In line h :

1 .538

721 7

In l ine

X

1 .5 38 x

In line

a:

c:

721 7

= 22,200 A

=

=

1 1 , l OO A

1 1 , 1 00

Currents in the lines between the motor and the fault

In l i n e

a :

In line b : In line

c:

4 . () l(j

l .538

l .538

X

x x

727 1

72 1 7

fa u l t a re

=

=

A

are

:r� , 3rn \

1 1 , 1 00 A

72 1 7 = 1 1 , 1 00 A

Cu rrents in the lines between the 4 . 1 6 kY bus and the transformer are In line A : 2 . 664

x

1 04 1 = 2773 A

In l ine B : I n line

C:

0

2 . 664

x

1 04 1

=

2773

A

The cu rre n t s we h ave calculated i n the above example a re those which w o u l d fl ow u po n the occu r r e n c e of a s i n g l e l i n c - t o- g ro u n cl fau l t whe n t here is no load on the motors. These c u rrents a re correct o n ly if the motors are drawing no current whatsoever. The statement of the problem specifies the l oad condi­ tions at the time of the fa u l t , however, a n d the load can b e considered. To acco u n t fo r t h e load, we a d d t h e per-unit cu rre n t d rawn by t h e motor through l ine a before the fa u l t occu rs to the port ion of IJ�) flowi ng toward P from the transformer and subtract t he same cu rrent from t h e portion of I};) flowing from the motor to P. The n ew va l u e of posit ive-sequence current from the trans­ forme r to the fau lt in phase a i s 0 .5 65 - jO . 350 - j1 .538

=

0 .565 - j1 .888

a n d the n ew val u e of posit ive-sequence curren t from the motor to t h e fault in

506

CHAPTER 1 2

UNSYMMETRICAL FAULTS

- j 2 .664

j 1 . 538

,

)

.

)

} 1 . 538

)

- j 3 . 07 6

, +j 2 . 6 6 4 FIGURE 1 2. 1 9 Per-u nit va lues

o f subt ransient l i ne currents i n a l l pa rts of

current neglected.

phase

t h e sys t e m of Ex a m p l e 1 2 . 6 , p refa u l t

a IS

- 0 .565

+

jO.350 - j l .026

=

- 0 .5 65 - j O . 676

These values are shown in Fig. 1 2 . 1 8 . The remainder of the ca lc u la t hese n ew values, proceeds as in the example. Figure 12. 1 9 gives the per-unit values of subtransient line currents i n all parts of the system when the fault occurs at no load. Figu re 12.20 shows the values for the fault occurring on the system when the load specified i n the exam p l e is considered. I n a larger system where the fault current is much higher in comparison with the load current the e ffect of neglecting t h e load current is less tha n is indicated by comparing Figs. 12. 1 9 and 1 2.20. In the la however, the prefault currents determined b y a power- fl ow study cou ld simply be added to the fault current found with the load neglected.

0 . 6 6 5 -j 2 . 68 5

-0. 586 +j l . 2 2 4 -

>

---:==�=-_-----' - 0.35 1 - j 0. 565 -

....__

•

- 0.314 + j 3.250

O.0 2 1 +j 2 . 20 2 •

-- �

0. 565-j3.42 6

+

-n .692

FlGURE 1 2.20 Per-unit values of subtransient line currents i n a l l p a rts of the sys tem of Example 12.6, prefa u l t current considered.

1 2.5

507

DEMONSTRATION PROBLEMS

The single-l ine di agram of a small power system is shown in Fig. 1 2.21 . A bolted single line-to-ground fault a t point P is to be analyzed. The ratings and reactances the generator and the transformers are Example

12.7.

of

G ene r a

tor

:

Transformers T I and T2 :

On a choseanr of reactances

base e

X"

1 00 MVA , 20 kV ;

X2

=

1 00 M VA , 206. /345Y k V ;

From T2

t to

o P:

=

4% ,

1 0%

kY n

i the transmission-l ine circuit the l ine

1 00 MV A , 345

From T I

x

Xo =

= 20% ,

XI

=

X�

= 20 % ,

X() = 5 0 %

P:

Xo =

r i c s , as s h ow n i n F i g .

30%

the with the valures of e c r n flows maapplrkyed. in the transformers vaoreltage ofeWip thedae ntcce poiseopeennt lno,oking eirnettaoaktehnecaussertqheuencentrseafeerreneezntecwreorovokasnltdaatgtehte per t The sequence cur e t of phase t

reactances marked the l i g u l e a n u d r aw a co m p l e t e t h r e e - p h ase c i rcu i t diag a m with per u n i t . Ass u m e that are lettered so

To s i m u l a t e t h e fa u l t , t h e s e q u e n c e n e t wo r ks o f

in

rer

u n i t : l rc con n e c t e d i n s

all u e t t h at Eqs. 0 1 .88)

c u r re n t s

Solu tion.

u n i t . Th

S l lOW Il in the

sw i h S h as A at i mp s

P

Z pp (I) =

1(1)

-

A - fA -

�

T CD \ ®

-

open-ci rcuit

1.0 + jO.O per

( j0 .6) ( j0 .4) jO.6

+

jO.4

2 Z"" ( )

/f{2} A

-

1 .0

=

jO .24

= jO.S

s ub

per u ni

A

+ jO .O

j O . S + jO .5 + j O . 24

unit

=

at P are

.

-jO.8065 p e r u n i

® Tz @)

l3f'---lI ---------� ---t- +33E -tE r/_ t- S -

VDE .6 h

G enerator n

t he p f ult can b

n t s i n t h e hypot h e t i c a l

-

1 2 . 2 2 . Ve r i fy

h e faul t point

Z�� =

l}()

system

0 .6

Switch open

FI G U RE 1 2.2 1

Single-l i n e diagr a m of t he sys t e m of Ex a mple 1 2 .7. S i ngle l i n e - to-ground fault

is

at poi nt P.

508

CHAPTER 12

U NSYMMETRICAL FAU LTS

t Ii� = jO.2 + tp jO.l -�

t Ii�

@)

@ Tz

@)

-

�......--.

t

VP(2) A

1

jO.B065

=

jO.2 + P jO. l

] (0 ) fA

=

IrA

3

= -jO .B065 -j O . 3226 j 0.48�

@

Reference

® T2

V(l) PA

Reference

Reference

j O.B065

-

t li�

.

jO.5 + P jO.3 _

)·0 . 1

-

t

VP(O) A

J

@

G) -J

Tz

J'0 . 1

FIGURE 12.22 Con nection of t h e sequ ence networks for t h e system of Fig . 1 2. 2 1 to s i m u l a t e si ngle l i n e-to-ground fault a t poi n t P .

The total c urrent in the fau lt is

If

A

= 3 1}�)

In the stub of phase B at point P

-) 2 . 4 1 95 per unit

=

have

we

I}V = a 21}:.J = 0.8065 / - 900 I}�) = aI};/ = 0 . 8065 /

-

+ 2400 = 0 . 8065�

90° + 1 20°

- 1f(0) 1f(0) B A

= =

-

Likewise, i n the stub of phase C a t point

IIe

-

�

1(0) Ie

+

P

0.8065 0. 8065

we have

1 ( 1 ) + 1 (2) Ie Ie

=

0

� /

-

90°

1 2.5

509

D E MONSTRATION PROBLEMS

In the zero-sequence n etwork the cu rrents are: Toward P from T]

1A(0) = = In the

jO.4 ( 0 .8065 / jO .6 + JO A O . 3226

L - 900

t r a n s m i s sion

Tow a r d P from T I

O .J22Cl

90°

L...-_ _

)

P from T2

jO.6 ( 0 .8065 . j O .6 + j O .4

I�Ol = .

per unit

=

0 .4839

/

/ - 90° )

L...-__

900 per unit

-

l ine the currents arc:

/-

I n l ine

A:

In l ine

B:

0.3226/ - 9 00

+

I n line

C:

0 .3226

+

Toward P from

�

Toward

90

0

/ - 9 00

/ 90 O .8005 � 0 .8065 L2Q:

O . K065

+

-

0

L 90 0.806S L2Q: + 0.8065 �

+

O .S06S

-

+

0

= -j I .93S6 per uni� = jO .4839 per unit = j0 .4839 per unit

T�

IA

In l ine A : In l i n e B :

=

-

j0 .4839 per u n i t

I B = -j0 .4839 per u n i t

Tn line C:

Ie

=

-j0.4839 pe r u n i t

Note that POS ] t lve -, negat ive-, a n d zero-sequence components o f cu rrent flow i n l ines A , B , and C from T ] but o n l y zero-seq uence components flow i n these l ines from T2 • Kirchhoff's curre n t l aw i s fu l fil led, howeve r. In the generator t he cu rrents arc

= - j l . l 969

1,

=

1a(0 )

+ a

J( I ) (1

= j 1 .3909 +

0 2 Ja( 2 )

=

0

+

0 .8065/ -. 1 200 _

+

1 200

+-

0.8065/ - 600 _

+ 2400

=0 Th e three-phase circui t diagram of Fig. 1 2.23 shows all the current flows

in

per

Ia

=

-j 1 .3969 a

c

P

B

Ib

FIG URE 1 2.23

Current flows i n

=

j 1 .3969

b

j0 .4839 j0.4839 -

Ira

Ire

P

=

I

1

0

=

a

IrA

= -j 2 .4 1 95 .

-j0.4839 -j0.4839

b

0

the system of Fi:

1 2.5

DEMONSTRATION PROBLEMS

511

u nit. From this diagram Dote that: • • •

•

• •

•

10

Lines are lettered and pol a rity marks are pl aced so that Eqs. ( 1 1 .88) are valid. Stubs are connected to each l in e at the fault. For a single line-to-ground fault stub curre nts IB = Ie = 0, b u t IhO), Ii:?) , It) , and It2) in the stubs all have n onzero values . Fa u l t current flows out of s tub A , then partly to T) and partly to T2 . I n the generator only positive- and n egative-sequence currents are flowing . In the Ll windi ngs of T2 only zero-sequence currents are flow ing I n the Ll windings of T\ each phase wind ing contains positive-, n egative-, and zero-sequence com ponents of cu rrent. These components are shown i n Fig. .

�

=

1 . 3969/- 90c

Ie

n�) l��) l��) lea

0

=

=

=

=

0. 1 863/- 90°

0. 4656�

0 . 4 656�

•

0.2 794� c

I6�) 0 . 1 863/ - 90° I6�) 0.4656/ 1500 I6�) 0.4656/ 30c Ibe O.2794� 1.3969/900 --=

-..

-----

a ------.... =

=

=

16

c

•

=

=

•

Fl G U RE 12.24

I��)

I!V I!i) la b

=

=

=

=

0 . 1863/- 90" 0 .4656/ gOo 0.4656/ 90° -

-

1 . 1 1 75/ - 900

b

------::..

3 I�O)

• =

B

0 .9678/- 900

IE

=

0.4839/ 900

Sym m e t rica l components of c u rre n t s i n t ra n s fo r m e r 7, of Fig . 1 2.23.

512

CHAPTER

12

UNSY M M ET R I CA L r:'A U LTS

12.24 and yield

12.6

OPEN-CONDUCTOR FAULTS

When one ph ase of a bal created and asymmetrical currents flow. A s im i ! w hen any two of the three phases a re opened while t h e t h i rd p h a s e re m a i n s closed. These u nbalanced conditions are caused, fo r example, \v h e n o n e - or two-phase conductors of a transmiss ion l i n e are p hy s i cal storm. I n other circu its, due to current overl oad, fuses o r o t h e r switching d evices may operate in one or two conductors and fail to conductors. Such open-conductor faults can be analyzed by means of tbe bus impedance matrices of the sequence networks, as we now demonstrate. Figure 1 2.25 depicts a section of a t hree-phase c i rcu i t i n w h i c h t h e l i n e currents in the respective p h ases are la ' Ib , and Ie ' with pos i t i bus @ to bus ® as s hown. Phase a is o p e n b etw e e n p o i n ts p and p i i n Fig . 12.25(a), whereas phases b and c are open b etw e e n t h e sam e t\VO po i nts i n Fig. 12.25(b ). The same open-cond uctor faul t phases are fi rst opened between points p and p i , and short circuits are then applied in those phases which are s hown to be closed i n F i g . 1 2.25. The e ns u i ng developmcnt fo ll ows t h i s reason i n g .

8 - ® a l toge t h e r @ a n d ® t o t h e po i n ts

O pe n i n g t h e t h re e p h ases i s t h e s a m e a� r e m o v i n g l i n c:

and then adding approp riate i mpedances from b u s es p and p ' . I f Ii simulate the opening of the three phases b y a d d i ng t h e n ega t ive i mp e d a nces - Zo , - Z I ' and - Z 2 between buses @ a n d ® i n t h e co rrespond i ng Thevenin equivalents of t h e t h ree s e q u e nce n e tworks of t h e intact sys t e m . T o exem p l i fy, , consider Fig. 12.26(a), which s hows the connection of Z 1 to sequence Thevenin equivalent between b uses @J a n d ® . The im p e d a nc es shown are the elements Z m( I m) ' Z n( ln) ' and Z m( I n) Zn( lm) of t h e p os i t ive- s e q u e nce b us ZJ� � + Z,� � - 2Z��� impedance m atrix Zb1Js of the intact system, and Z(��mn is the corresponding Thevenin impedance between buses @) a n d ® . Vol tages Vm and Vn are the normal (positive-seque nce) voltages of phase a at b uses ® and @ before the open-conduc tor faults occur. The p os i t i v e -s eq ue n ce imp edances kZ I a n d ( 1 k )Z l ' where 0 � k � 1 , a re a d d e d as s h own t o -

=

=

-

12.6 OPEN-CONDUCTOR FAULTS

@ a

10 1 -

C

@) Ie 1 -----

p

P

I

@ Ib b 1 -----

@) a I

fa --

C

® I ® I ® 1

(a) pi

p

® 1

®

@) f6 6 1 @) 1

513

I

Ie

(6 )

® 1-------11

FIGURE 1 2.25 Open-cond uctor fau l t s on a sect ion of a t h ree-phase syst e m between

buses § a n d ® : (a) conductor a open ; ( b ) can ductors b and c ope n between points p a n d p'.

represent the fractional lengths of the b roken line @ - ® from bus @) to point p and bus ® to point p ' , respectively. To u s e a convenien t notation , let the vol tage VY ) denote the phase-a positive-sequence component o f th e voltage drops Vp P '. a ' Vpp' b ' and Vp P'. from p to p' in the phase conducto rs. We shall , soon see that Va(l ); and the corresponding negative- and zero-seque nce comp o­ nents Va(2) and Va(O ), take on d ifferent values depending on which one of the open -conductor fau It By sou rce transformation we can replace the voltage drop Va( l) in series wi t h t h e i m p e d a n c e [ kZ 1 + ( 1 - k ) Z I ] in Fi in p a r a l l e l w i t h t h e i m pe d a n ce Z I ' as s h o w n in Fig. 1 2.26( 0 ). I n t h i s latter figure t h e p a r a l l e l comb i n a t i o n o f - Z 1 a n d Z 1 can be canceled, as shown in Fig. 1 2. 2GC c ). The above considerations for the positive-sequence network also apply d i rect l y to the n e g a t ive- and zero-sequence n etworks, but we must remember t h a t the l a t t er ne tworks do not contain any i nternal sou rces of their own. I n drawi understood that the currents VY)jZ and Va( O )jZo , l ike the current Va( l)j Z l of 2 Fig. 1 2. 26C c ), owe their origin to the open-cond uctor fault between points p a n d p ' i n the system. I f there is no open conductor, the voltages V} l), VP), a n d Va(O) are al each of the sequence currents V} O)jZo , V} I )jZ \ , and �(2)jZ2 can be regarded in turn as a pair of injections into b uses ® and ® of the corresponding c

z(l)

mm

-

Z( 1 )

mn

®

p kZ 1

+

(

Vm

) Z( l th ,

z( l )

nm

=

ZmO)n

Z(l) nn

_

z( 1)

nm

-Z

mn

z ( 1 ),

®

(1

pp

-

k )Z I

r

VaO

0(

l

p

I

)

1-

.... N

(a) ( l) - z( 1 ) Zm m mn +

Z mC I m) - Z m( I rt) @

I

+

Z ( l) th o

Vn + Zn( l) n

Z n( I) m

=

-

Z ( l)

nm

( I) Zm n

t

mn

V(!( l

)

® I

2(1)

rl r11

(b) FlGURE 1 2.26 Simulating the opening of line svstem: ( b) transforma t ion to current source; ( c ) res u l t a n t e q u iv a l e n t circu i t .

( l )o Z th

Vl\

_

__

-

,%" ( 1 )

� (l f n

(c)

t m il

1 2.6

O PEN-CONDUCTO R FAULTS

SIS

sequence network of the intact system. Hence, we can use t he bus impedance m atrices Z�Js ' ZblJs ' and Z�Js of the normal configuration of the system to determine the vol tage changes due to the open-conductor faults. But fi rst w e must find expressions for t h e symmetrical components V}O), V} l ), and VP) of the vol tage drops acrOss the fault points p and p i for each type of fault shown in Fig. 1 2.25 . These voltage d rops can be regarded as g iving rise to the following sets of inject ion currents into the sequence networks of the n ormal system configurati on : Positive Sequence At bus At

bus

Va( l )

@) : -

Va(O)

Va(2)

2

@:

Zero Sequence

Negative Sequence

20

2

1 V1I( l )

--

-

21

2 V1I(2)

-

--

2

V (0)

a --

20

2

as shown i n Figs. 1 2.26 and 1 2.27. By multiplying t he bus impedance m atrices ZboJs ) Z blJs , and Zb2Js by current vectors containing only these current i njections, we obtain the fol lowing changes in the symmetrical components of the phase-a voltage of each bus (J) : 6. VI (O) =

Zero sequence :

Positive sequence :

6. VI ( l )

N egative sequence :

6. V( 2) ,

2 1(0) - 2 I�0) n m 2°

2(1) =

=

Va(O)

2 I�n1 )

Va( 1 )

- 2m 2 1(2) III Ir/

Va(2)

-

1m

71

2

2

( 12 .27)

Before developing the equ ations for Va(O), v:,( 1 >, and VP ) for each type of open-con ductor fault, let us de rive express ions for the Thevenin equivalent

' i m p e d a nces o f the s eq u e n ce n etwo rks, as seen from fa u l t points p a n d p .

Looking into the positive- sequence networ k of Fig. 1 2.26( a) between p and p' , we see the imped ance 2��, given by Z ( l ), pp

=

kZ + 1

Zth , m ,, ( - Z I ) z(l) (Il

th . mn

;

- Z

1

+

(1

-

k)Z

1

2

=

_ 21

Z

___ _

Z t( hI ). m "

-

1

( 1 2 .28)

�2)

mm

_

Z(2) @) mn

p

kZ2 ...

) Z(2 th ,

Z(n2n) Z(2 )

=

m n

_

mn

-

Z2

®

Z ( 2)

nm

•

(1

( 2)' Zpp

k )Z2

-

Zn(2)m

p

kZo �

Z (O)

mn

=

Z(O)

Zth ( O) o

Z( O ) - Z (O) ® ll n

I

mm

r

- Zo

...

t

Z (2 )

mn

Z (2)

@)- ®

Zn( 2)n - Z(n2)m

(1

-

k )Zo

Z2

®

�

()

� tTl :;c

.... N

c Z

til

-<

:: ::

�:;c

)r

."

Zm( Om)

-

(0 ) 2 fli T!

Z(O)

Z,<,�!

Zm( On)

E

@)

I(O)

-

)C

Z th . Z;I�:I. ®

t m n

nm

(6) '

)

(II .e'\

()

r

p'

Zth o m n

t

Va( 2

--

nm

) Zp( Op)' y e o

nm

F1GURE 12.27 Simulating the opening of line

I( 2)

a

mn

nm

mn

Va( 2)

( a)

§

(O) Zm(O)m - Zm n

p

Z(2) - Z(2) @)

between points p and p : ( a ) negative-sequence a n d (h) zero - se q u e nce e q u iva l e n t c i rcu i t s .

yae o )

Zo

1 2.6

OPEN-COND UCTOR FAULTS

517

a nd from p to p ' the open-circuit v o lt ag e obtained by voltage d ivision is Open -circuit voltage from p to p' =

2(1)

- zI

th , mn

( Vm

Z1

_

( 1 2 .29)

Before any con d uctor opens, t h e c u rren t Imn in p hase positive sequence a nd is given by

I

II l it

'""c

v"'

-

1""1 i n Eq .

Subst i t ut i ng this expressio n for

a

of the l ine

V

/I

@) - ®

is

( 1 2 .30)

( 1 2.29), we obt a i n

Open-cir cu i t vol tage from p to

p

'

=

Jmn Z��,

( 1 2 .3 1 )

Figure 12.28(a) s hows t he res u l t i ng positive-sequ ence equivalent ci rcu it between points p and p'. Analogous to Eq. ( 1 2.28) we h ave Z(2� = PP

Z t(2) h mn

Z

(O), = zPP

a nd 2

- Z o2

Z (O) th mn

-

20

( 1 2 .32)

which a re the negative- and zero-sequence impedances, respective l y, between p and p i i n Figs. 1 2 .28( b ) and 1 2 .28( c). We can now p roceed to d evelop expres­ si ons for t h e sequence voltage d rops Va( O ) , Va(I ) , and vy ). One open con d uctor Let

us

co n s i d e r

c i rc u i t i n p h a se

(1 ,

o n c o pe n

t h e cu r re n t

c o n d u c t o r as . i n f i g .

1 11 = 0 , a n d

J {.I(O) w h e re

IIi ' '1 nd v o l t a ge (l

Ic

.,

J(I) a

+

1 (2 ) iI

=

0

( 12 .33 )

=0

( 1 2 .34)

a r e t h e s y m m e t r i c a l c o m p o n e n t s of the line c u r ren t s la ' ' from p t o p . S i nce p hases h a nd c arc closed, we a l so have t he

J ( O ) J( I ) a

+

J 2.25 ( a ). Ow i n g to t h e o p e n

so

a

,

and

1 ( 2) a

drops

VP P ' , c

Reso lving the series vol t age d rops across t he fau l t point into their sym metrical

518

CHAPTER 12

U NS YMMETRICAL FAULTS

1a( 1

\.

)

----+-

]

mn

Z( l )'

+

p

O l' Zpp

]a(2)

----+-

r I-

r

Z (2), pp

Va( l)

pp

p'

(a)

-----i

p

r l-

Va( 2 p' I

(6)

)

] (0 )

a -

p

) Zpe Op '

r

yaeO

)

p'

(c) FIGURE 12.28

Look i ng into t h e system between points (c) zero-sequ ence equiva l e n t ci rcu its.

[

I [

p

and

com ponents, we obtain V} D ) V(l)

a Va(2)

1

=

-

3

1

1

1

That is, =

Va( I )

( a ) pos i t ive-sequence, ( b ) negative·se que nce, a n d

[\ I I [ I

a a

1

Va(D)

' p :

=

�)0P ' .

� VpP' . a

a

3

0

Va( 2 )

=

Vpp' . a V,) ,

p .

V , pp , a

3

( 1 2 . 35 )

a

( 1 2 36) .

I n words, this equation states tha t the o pen conductor i n phase a causes equal voltage d rops to appear from p to p' in each of the sequence n etworks. We can satisfy this requirement and that of Eq. ( 1 2.33) by connecting the Theve,nin equ ivalents of the sequence n etworks in parallel at the points p a n d p ' , as shown in Fig . ] 2 .29. From t h is figure the expression for the positive-sequence

1 2.6 OPEN-CONDUCTOR FAULTS

]( 1 ) a

Z(I ),

]m n Z(pI)p '

pp

p

i

p

Vae l ) t+

Z( 2 ),

](2 ) � p

p

pp

i

t

Z(O),

ya( 2)

](0) � p I

t+

yae O)

P .

pp

519

-

Fl G U RE 1 2 .29

Co n ne c t i o n of t h e sequence networks of t h e a nd p' .

c u rr e n t

1� 1 )

i s found to be

1( 1) - 1 1I

7( I �

!! 1 11

PI'

+

m, Z I'(O){' , z IIII

Z ( 2 ), �

" f'

+

Z ( ( I ), 1'1'

( 1 2 .3 7 ) The seq u e nce vol t age d rops V} O ), Va( I ), a n d Va( 2 ) are t he n given by Fig. 1 2.29 as

- V (O) a

=

V (2 )

=

a

=

1

III n

Va( I )

=

1 ( 1 ) Z (2 ), +

zpp (2� Zpp (O),

a

pp

Z (O), pp

Z ( O ), Z ( I � Z ( 2)' pp pp pp O , l + ) ( I ), ) Z( Z Z( , Z m, + Z m, Z( O ), pp

pp

pp

pp

pp

( 1 2 .38 )

pp

The q u antIt Ies on t he right-hand side o f t his equation a re known from t h e i m p e d a nce p aramet e rs of t he sequence n e tworks and t he p refa u l t current i n p h ase a of t h e l i n e ®- @ . Thus, t he curre n ts v'1(O ) / Z U ' VY ) / Z I ' an d VY)/ Z 2 for i nject ion i nto the correspon d i ng seq u e n ce n etworks can be determined from Eq. 0 2 .38). Two open conductors When two con ductors a re ope n , as shown in Fig. 12.25( b), we h ave fau l t cond i t i ons which are t he dua/5 2 o f t hose i n Eqs. ( 1 2 .33) a nd ( 1 2. 34); namely, ( 1 2 .39 )

1

b -

2Dualiry

is

t reated i n m a ny textbooks

0

1

=

c

0

( 1 2 . 40)

520.

CHAPTER 12

UNSYM METRI CAL FAU LTS

Resolving the line currents into their symmetrical components gives ( 12 .4 1 )

Equatio ns ( 1 2.39) and ( I 2.4 1 ) are both sat isfied by co nnect ing the Thcven in equ ival ent of the negative- and zero-seq uence networks in series be tween points p and pi , as shown in Fig. 1 2.30. The seque nce currents a re now expressed by

-I-

Z ( I ), '1'1'

+

( l 2 .42)

Z ( :'1 ") � /'

where I is again the prefault current i n p h ase a of the line ® - ® before the open circuits OCCur in ph ases b and c. The sequence voltage drops are now given by m il

Vel) a

=

(

1 ( 1 ) Z (2), + Z ( O ), a pp

pp

)

=

I

=

I

=

m il

mn

I11 1 /1

(

Z ( I ), pp

pp

pp

+

pp

Z ( 2), + pp

)

Z ( O ),

Z ( l ), z (2 ), + Z (O ),

pp

- Z ( 1 ), Z (2), pp

pp

Z ( I ), + Z (2), + Z ( O ), pp pp pp

Z ( I >, pp

( 1 2.43)

- Z ( I ), Z (O),

-; . '

pp

pp

Z (2), + Z ( O ), pp

pp

In e ach of these equat ions t he right- hand side q u a n t i t i es a re al l known befo re the fault occurs. Therefore, Eq. ( 1 2 .38) can be used to eva l uate t he symmetrical components of the vol tage drops between the fault poi n ts p and pi when an open-conductor fau l t occurs; and Eq . ( 1 2.43) Can be similarly useJ when a fa u l t due to two open co nductors occurs. The net effect of the open conductors on the positive-sequence network is to i ncrease the transfer impedance across the line in which the open -cond uctor fault occurs . For one open co nductor this increase in impedance equ als the parallel combination of the negative- and zero-sequence ne tworks between points p and pi; for two open conductors t he increase in impedance equals the series combination of the negative- a n d ze ro-sequence netwo rks between p.oints p a n d p'.

1 2.6

I

m it

Z( l )

p p'

+

Ia( 1

�

)

p

Z( l ),

Z (2

1a( 2

Vae l

)

p

�

pp'

) 1a( 0

�

p I

j

t+

'

p

�

)

t-

p'

)

521

t+

pp

-

OPEN·CONDU CTOR FAULTS

)

Va( 2

t-

1 (1) a

=

1 (2)

=


[(0) a

t+

1

Z (O), pp

yae O

t

pi

)

-

FlG URE 12.30 Co n n ec t i o n of t h e

po i n t s

p

a nd

' p .

Example

12.8. In the system of Fig. 1 2.5 consider that machine

Solution .

All t

2 is a motor

drawing a load equivalent to 50 MVA at 0.8 power· factor lagging and nominal system voltage of 345 kV at bus G) . Determi ne the change i n voltage at bus G) when the transmission line undergoes ( a ) a one-open-conductor fault and (b) a two-open-conductor fault along its span between buses W and G) . Choose a base of 1 00 M V A , 345 kV i n the transmission l i n e . current in line W- G) choos i n g t h e

12

J

=

as

fo l l ows:

P - jQ V:l*

The sequence networks of F i g . Z\ =

2

2

=

------ = 1 .0 + jO .O

0 .5(0.8 - jO.6) 1 2.6

0 .4

- jO .3 per unit

show that line (1) - G)

jO . l 5 per unit

Zu

has

parameters

= jO .50 per unit

The bus impedance matrices Z�oJs and Z\,IJs = Z�Js are also given in Example 12. l . Designating the open-circuit poi nts of the line as p and pi, we can calculate from

522

CHAPTER 1 2

U NSYMMETRICAL FAU LTS

Eqs.

and 02.32)

0 2. 28)

- ( j0 . 1 5)

2

------- -------

j 0 . 1 696 + jO. 1 696 - 2( j0 . 1 l 04) - j O . 1 5

Z

pp' (U)

-

_

=

_ ------- Z2

--:-:-:-__ u ---:-:::-_

=

jO . 7 1 20

pc r

--::-:-__

(O, + Z(O) - 2 Z (O) - Z Z 22 () 33 23

uni t

"1

- ( f 0 .5 0 ) '

j O . 08 + jO .58 - 2 ( f 0 . 08 ) - jO.50

=

00

Thus, if the line from bus @ to bus G) is opened, then a n infinite impeda nce is seen looking into the zero-sequence network between points p a n d p' of the opening. Figure 1 2.6( b) confirms this fact si nce bus G) would b e i sol a t e d fro m t h e reference by openi ng the connection betwee n bus CV and bus 0.;. . One open conductor In this example Eq. 0 2. 38) becomes =

=

( j0 .7 1 20) ( j0 . 7 1 20) ( 0 . 4 - jO . 3) -. ----- -j 0 . 7 1 2 0 + j O-.7-1 20 0 . 1 068 + j O . 1 424

per unit

and from Eqs. 0 2.27) we now calculate thc sym m e t r i c a l com ponents o f the voltage at bus (1) : Ll V ( 1 ) 3

=

Ll V (2 ) 3

=

Z ( I ) - Z (l) 33 V e l ) 32 a ZI

=(

=(

=

Ll V (0) 3

=

Z (O) - Z ( O ) 32 33 V (O) a Zo

jO . 1 1 04 - jO . 1 696 . j0.15

)

- 0 . 0422 - j O .0562 jO . 08 - j O . 5 8 .

j O .50

)

( 0 . 1 068 + jO. 1 424)

per unit

(\ 0 . 1 068 + j O. 1 424)

- 0 . 1 068 - jO . 1 424

per unit

- 0 . 1 068 - j O . 1424 - 2 ( 0 .0422 + jO .0562) . - - 0 . 1 91 2 - j O . 2548 p e r

unit

1 2.7

Since the prefault voltage at bus G) equals

1 . 0 + jO.O,

IS

Vj

1 2.7

=

V3

=

0 . 8088 - jO .2548 = 0 .848

+

tl V3

=

SUMMARY

523 .

the new voltage at bus G)

( 1 .0 + jO.O) + ( - 0 . 1 9 1 2 - jO.2548)

/

- 1 7 .5°

per unit

Two open conductors Inserting t he infi n ite impedance of the zero-sequence network in series between points p and p' of the positive-sequence network causes an open circuit in the l a t t e r . No power transfer can occur in t he system-confirmation of the fact that power cannot be transferred by on ly one phase conductor of the transmission l ine in t h i s case since t h e zero-sequence network offers no return path for current . S U MMARY

in

shown i n Fig. 1 2.2 a re repl aced by s h o r t c i r c u i t s , t h e i m pe d a nce b e t we e n the fau lt bus ® and the r e fe r e n ce node is t he posi t ive-se que nce i mpeda nce ZJ� in t h e equation devel­ oped for fa u l t s on a power system and is the series imped ance of the Thevenin equivalent of the circu it between bus ® and the reference node. Thus, we can regard Zk� as a single im pedance or the entire positive-sequence network between bus ® and the refe rence with no emfs present. If the voltage Vf is connected in series with this mod ified positive-sequence network, the resu lting circu it, shown in Fig. 1 2. 2( e ), is the Thevenin equivalent of the original posit ive-sequence network. The circu its shown in Fig. 1 2.2 are equ ivalent only in their effect on any external connections made between bus ® and the refer­ e nce node of the original networks. We can easily see that no curren t flows i n t h e branches o f the equivalent circuit i n the absence o f an external connection, b u t current will flow in the branches of the originaL positive-sequence network if any difference exists in the p hase or magni tude of the two emfs in the network. I n Fig. 1 2 .2( b ) t h e cu rrent flowing in the branches in the abse nce of an exte rnal c o n n e c t i o n i s t h e prefa u l t load cu rrent. W h e n t h e o t h e r seq uence n e t w o r k s a r e intercon n ected with the positive­ sequence network of Fig. 1 2 .2( b ) or its equivalent shown in Fig. 1 2.2C e ), the current fl owing out of the network or its equivalent is I};l and the voltage betwee n bus ® and the reference is Vk((� l. With such an external connection, the cu rrent in any branch of the original positive-sequence network of Fig. 1 2 . 2 ( b ) is the positive-sequence cu rrent in phase a of that branch during the fau l t . The prefault component of this cu rrent is included. The current i n any branch of the Thevenin equ ivalent of Fig. 1 2.2( e), however, is only that portion of the actual positive-sequence current fou nd by apportioning J}�) of the fault among the branches represented by zi� according to their impedances and does not i nclude the prefault component. In the p r e cedi n g sect ions w e h ave s e en that the Thcvenin equivalents of the sequ ence networks of a power system can be interconnected so' that solving I I t h e e m rs

a

p o s i t i v e - s e q u e n ce n e two rk l i k e t h a t

+ t1l

I

__

•

Z(

Pos.-seq. n etwork

Vk( a

�

®/ + ] (1) •

fa

Po s . - s e q . n etwork

+t )

V(l

-t

ka

®I Fi�) .

P o s . - s e q . network

I

� vk( a2

-t

)

S H OR T-C I R C U IT F AU LTS

I

G

·

fa

N e g . - s e q . n e twork

I

Zf

+t

VkC oO )

-+

Dou b l e line-to-ground f ault

®l f 1i�) �

I

N e g . - s e q . n e twork

Line-to-line fault

Vh( 2a )

-+

I

Three-phase fau lt

Z{

+t

® I t ] ( 2)

3

+t

) v(l

-t +t) ka

vk( a2

-t + t O) V( _ tk "

•

P o s . -s e c . n e twork

N e g . - s e q . n e tw o r k

®

.

t

1 (0) fa

Ze r o - s o q . n e t w o r k

S i n g l e lin e-to-g round fau l t

® I + ]j�) �

Zero-s e q . n e twork

I

N

+ �

+ -1

� vy ) 1�1) + p

p '.! .

Po s . -seq . n e twork

2) �Va(

1� 2 ) t p

p' l

•

N e g . - s e q . network

I

One open conductor

O P EN-CON DUCTO R FAULTS

+ -1 O) Va(

l�O ) t

�

p

P' l •

Zero - s e q . n e tw o r k

a

�

IeI

)

-=j I'-!- V} I ) IP

I

t ip

I Neg�-Seq.

I

· i

I � Va(2)�

Pos.-seq.

(2)

1a

p' ,--

p' ,.....-

�

1 + Va(O) � -+-

t IP zer�-seq

1(j( 0 )

n e two r k

net ork -I

I

net:ork I I

P r--

Two open conductors

vI�I, VI!), a nd Va(OI, Vyl, a n d Va(2) are t h e

FlGURE 12.31 Summary of the connections of t h e seque nce networks to simula te various t ypes of short-circ u i t fa u l ts t h rough impeda nce Zf.

Vk(;) a re the symmetrical compone nts of p h a se a voltage a t the fault bus ® with respect to referen ce. symmet rical components o f t h e phase a voltage drops a cross t h e open-circ u i t points p a nd p ' .

TABLE 12.1

Summary of equations for s equence voltages and currents at the fault point for various types of faults 1 Line-to-ground fault

�

I:: III I. I. ::l to>

IW

( I ) + Z(2) k k + Z k(Ok) + 3 Zf = Z kk I};)

I::

e:r

III VJ

I---

V(ll Vl III OJ) <0 -

k il

I}�)

=

"0

I}�

- 'h

'k

"

IOl Vk "

Vl�) VPl

=

fa

Ij�)

)

Jj' )(zF) + 710) + ka -

III u I:: III ::l 0OJ VJ

V��), Vl�), V;O), Va(J ),

=

kk

- 1?" )2 k( Ok)

17 r )

-

-

V f Zk( 2k)

+

+

Z

f

1 (2) - - I}!) fa

I}�)

V (2) - - / ( 1 )2 (2)

...

NOle : " II"

=

Iflal ) - ( I ) Zk k

VII) k"

=

Vka(2)

=

V1�)

I}� )

1 (0)

+

7�r )

f"

z i� -

+

Vf Z i�ll ( Z k�

/ (a1 )

f Z (2) kk (l

V ( I ) - V}�) -

0

and

a rc the sym m e t r i c a l co m po n e n t s o f p h os e a r e the sym m e t r i c a l compon e n t s o f

the

a

vol t age at

phase

a

+ -

Zf )

3 Zf ) Z k( Ok) + 3 Z +

(2 ) ZU

Z f.:I.: (O)

+

1a(2)

f

1 (0 )

3 Zf

(l

:' 1}�)Zr

=

Z

=

Vk(O) a

) Z ( II ) - I?(l - " "

V"I

the fa u l t b u s

{/

®

w i t h respect to

)

" IO\ -,- ZpI p). 'I Zpp

)

a

I

=

Z( l \ pp

)

I))

II I I "

=

=

/ ( 2)

pp

pp

'

f'f'

. -;-

a

pp

Z l e ),

1 (0)

z ((l),

(l

pf'

Z ( 2 ), Z (0),

!'�. �

__

7. 1 2 )

, - /1/'- T

7(11)

�J)/"

I 2 ), - 1Ll( 2 )Zpp - I " ) ) L I I l ), iJ

t h e: refe r e n c e .

vo l t a ge d rops a c rOSs t h e oDe n - c i r c l i i t ro i n t , '

Ill

pp

pp

"

I

I mn Z l l )'

- /� ' , Z I � I,

V (2)

=

pp

Two open conductors

Z ( O ), " Z (�), -'- Z tD),

=

v( I )

(l

II)

ll _I

=

- IP )Z(2) kk

i m p l ies p a r a l l e l comb i n a t i o n of i m p e d a n ces, and

+

+

1a( 1 )

V (2) -

k"

a

(ZiT

iF Z (2) ) kk

_ -

ka

JP )Z kk ( 2)

=

=

1(2) _ fa -

= 0

'jll( ,, 7(2) -kk

O n e open conductor

Double l i n e-to-gro u n d fa u l t

Line-to- l i n e fault

Vf

III to>

III ::l

Open-circuit fa ults

Short-circuit faults

:Inu p'.

PI)

I

I

V( I ) = n

( I )' Imn Zpp + =

=

Z (l2fJ),

Va(2)

=

PP

a

1(1) a

PP

+

Z ( O), )

_ / (2)2 (2), (1

f'f'

1 , ( 1 ) ) ,., _ / ( I I )L (O), "

Z(O),

/(1)

I l l )( Z (2 ), (l

+

f

(/

PI'

PP

527

PROBLEMS

the resulting network yields the symmetrical components of current and voltage at the fault . The connections of the sequence networks to simulate various types of short-circuit fau lts, including a symmetrical three-phase fault, are shown in Fi g. 12.3 1 . The sequence networks are indicated schematically by r ec t a n gl es enclosing a h e avy line to represent the reference of the network and a point marked bus ® to represent the point in the network where the fault occurs. The positive-sequ ence network con tains emfs that represent the internal volt­ ages of the machines. Regard less of the prefault voltage p rofile or the particular type of short­ circu i t fau lt occu rring, the only current which causes positive-sequence voltage changes at the buses of the system is the symmetrical component I}�) o f the current 'Ia coming out of phase a of the system at the fault bus ® . These posit ive-sequence vol tage changes can be calcu lated simply by multi plying col umn k of the pos itive-sequence bus impedance matrix Z �\�, by the injected current II,;)' Simila rly, t h e neg at ive- and ze ro-sequence components of the volt age cha nges due to the short-circ u i t fa u l t on the sys t e m a rc caused, respec­ t ively, by the symmetrical components I}�) and IJ�) of the fau lt c ur rent If out of bus ® . These sequ ence voltage changes a re also calculated by mu l tiplying columns ® of Zi,2Js and Z��s by the respect ive current i njections - Ij�) and -

a

. - 1 (0) I" In

a very real sense , t herefore, th ere is only one proced u re for calculating the symmetrical components of the voltage chan ges at the buses of the system when a short-circuit fault occu rs at bus ® -that is, to find I}�), I}�>' and I}�) and to mult iply columns ® of the corresponding bus impedance matrices by the negative values of these currents. For the common types of short-circuit faults the only differences i n the calcu l ations concern t he method of simulating the fau lt a t bus ® and of formu lating the equations for I}�), I}�), and I};). The connections of the Theven in equivalents of t he sequence networks, which provide a ready means of deriving the equ ations for I}�), I}!>, and I};), are summ arized in Fig. 1 2 . 3 1 and the equations t hemselves are set forth in Table 12.1. Fa u lts due to open con ductors i nvolve two inj ections into each s e q u e n ce ne twork a t t h e buses nea rest the ope n i ng in t h e con ductor. Otherw ise, the procedure for caku lating sequence voltage changes in the system is the same as that for short-circuit faults. Equati ons for the sequence volt ages and currents at t h e fa ult are also summarized i n Tab le 1 2. 1 . The reader is rem i nded of the n e e d to a dj u s t the phase a ngles of the sym m e t ri c a l compo n e n t s o f t h e c ur r e n t s and t h e v o l t ages in t h ose p a rts o f the sys tem which are separated from the fau I t bus by �-Y transformers. PRO B LEMS

60-Hz

groundedluandrbogeopencraralotirng at rated voltage load.

12. 1 . A

is

is

r a t e d 5 0 0 MY A , 2 2 kY .

at no

It

i s Y-connected and solidly I t is discon nected from the ,

528

12.2.

CHAPTER 12

U NSYM METRICAL FAULTS

rest of the system. Its reactances are X� = X I = X2 = 0 . 1 5 and Xo = 0 . 05 per unit. Find the ratio of the sub transient line current for a single l ine-to-ground fault to the sub transient line curre n t for a symmetrical t h ree-phase fault. Find the ratio of the sub transient l i ne current for a line-to-l ine faul t to the subtransient current for a symmetrical th ree-phase fa u l t on the genera tor of Prob. 12.1.

i n o h m s t o b e inse rteu i n th e neu tral connec­ tion of the generator of Prob. 1 2 . 1 t o limit the sub transient line cu rrent for a single l ine-to-ground fau l t to t hat for a th ree-phase fau l t . 12.4. With t h e inductive reac tance fou n d i n Prob. 1 2 . 3 i n s e r te d i n t h e n e u t r a l o f t h e gener a tor o f Prob. 12. 1 , find t h e ra t i os o f t h e s u b t r a n s i e n t l i n e c u r r e n t s for t h e fo l lowing fa u lts to t h e s u b t ra n s i e n t l i ne c u r re n t fo r a t h rec-rhasc fa u l t : ( a ) s i n g l e line-to-ground fau l t , ( b ) l i n e - t o - l i n e fa u l t , a n d C C ) d o u b l e l i n e - t o - g ro u n d fa u l t . 12.5. How many ohms o f resistance i n t h e n e u t r a l co n ne c t i o n o f t h e ge nerator o f Prob. 1 2 . 1 wou l d limit the subtransient line current fo r a s i ng l e l i n e- t o -gro u n d fa u l t to that for a three-phase fault? 12.6. A generator r at e d 1 00 M V A , 20 k V h a s X;; = X I X2 = 20% and Xo = 5%. I t s neutr a l is grounded t h rough a reactor of 0.32 n . T h e g e n e r a t o r is op e r a t i n g a t rated vol tage without load and is d iscon nected from th e syste m w h e n a s i n g l e l ine-to-ground fault occurs a t its termina ls. Fin d t h e s u b t ra n s i e n t c u r r e n t i n t h e faulted phase. 12.7. A 1 00-MVA 1 8-kV turbogenerator h aving X;; = Xl = X2 = 20% and X o 5% is about to be connected to a power sy s t e m . The g e n e ra t o r has a c u r r e n t - l i m i t i n g reactor of 0. 1 62 n in the neu tral . Before the generator is connected to t h e system, its voltage is adjusted to 1 6 kV w h e n a d o u b l e l i n e -to-grou n d fa u I t d evelops a t terminals b and c . Fi nd th e i n i t i a l sym m e t ri c a l r oo t m e a n - s q u a re (rms) curren t in t h e g ro u n d a n d in l i n e b . 12.8. The reactances of a generator ra t e d 1 00 M V A , 20 k V arc X;; = X l = X2 = 20% and Xo = 5 %. T h e g e n e r a t o r is c o nn e ct e d t o a 6. - Y t r a n s fo r m e r r a t e el 1 00 M V A . 20Ll-230Y kV, with a reactance of 1 0% . The neu tra l of t h e t ransformer is so l i d l y grounded. The terminal v o l t a g e o f t h e g e n e r a t o r is 2 0 k V w h e n a s i n g l e l i nc - t o ­ grou nd faul t occurs on t h e o p e n - c i r c u i t e d , h i g h -vol tage side of the t r a n s fo r m e r. Find the initial symmetrical rms current in all ph ases of the gen e rator. 12.9. A generator suppl ies a motor t h r o u gh a Y - 6. t r a n s fo r m e r . The generator is connected to th e Y side of th e t ra n s fo rm e r . A fa u l t occu rs b e t w e e n t h e motor terminals and the transformer. The symmetrical components of the subtransient current in t h e motor flowing toward the fau l t are

12.3.

D e t e rmine the inductive

reactance

=

=

-

I� I )

2 1� ) I�O)

=

= =

-

0.8

j

- j2.6 per u n i t

- 2.0

per unit

j3

p e r unit

-

.

0

PROBLEMS

529

From the tra nsformer towa rd the fault

I� I ) = 0.8

I�2)

I�O)

12.10. 12. 1 1 .

12. 1 2.

12.13. 1 2 . 1 4.

= =

-

-

jOA per unit

j 1 .0 per unit

0 per unit

Assu me X;; = X I = X 2 for both the motor and the generator. Describe the type of fault. Find ( a ) thc prcfau l t current, if any, in line a ; (b) the subtra nsient fault current in per unit; and (c) the subtransient current in each phase of the generator in per unit. Using Fig. 12. 1 8, calculate the bus impedance matrices Zhl1s , Zh21s ' and Z\?1s for the network of Example 1 2.6. Solve for the subtransient current in a single l ine-to-ground fau lt fi rst on bus CD and then on bus 0 o f the network of Example 1 2 .6. Usc the bus i mpedance matrices o f Prob. 1 2 . 1 0. A lso, hnd the vol tages to neutral at bus 0 wit h the fault at bus (D . Calculate the subtransient cu rrents i n a l l parts 0 f the system 0 f Example 1 2.6 w it h prcfault current n e g l e c t e d i f t he fa u l t on the low-voltage side of the transformer i s a l i nt-to-l i ne fau lt. Use Z �1s ' Zh21� , and Z ��$ of Prob. 1 2. 1 0. Repeat Prob. 1 2. 1 2 for a double l ine-to-ground fau l t . Each of the machines connected to the two high-vol tage buses s hown i n the single- l ine diagram of Fig. 1 2. 32 is rated 1 00 MYA, 20 kY with reactances of X� = X I = X2 20% a nd Xo 4%. Each th ree-phase transformer is ra ted 100 MYA, 345Y /20ll kY, with leakage reactance of 8%. On a base of 1 00 M YA, 345 kY the reactances of t he transmission line are X I = X2 1 5 % and Xo = 50%. Find the 2 x 2 bus i m pedance m a trix for each of the th ree se quence networks. I f no prefault current is Aowing in the network, hnd the subtransient current to ground for a double l i n e-to-ground fault on lines B and C at bus CD . Repeat for a fau lt at bus 0 . When t he fault is at bus 0, determine the cu rrent in phase b of machine 2 i f the li nes are named so that v:.l l ) leads �,( I ) by 30° . If the p hases a r c named so t h a t I{� I ) leads 1�1 ) by 30° , what letter ( II , h, o r c) wou ld ide ntify t he p h ase of machine 2 w h i c h wou l d carry the current fou nd for phase h above? =

=

=

FI G U R E 1 2 .32 S i n g l e - l i n e d i a g r a m for Proo. 1 2. 1 4 .

12. 15.

Two gene rators G t and G 2 are connected, respectively, through transformers TJ: and T2 to a high -vol tage bus which su ppl ies a t ransmission line. The l ine is open at the far end at which point F a fa u l t occurs. The prefaul t voltage at point F is·

:�

530

CHAPTER 12

5 15

UNSYMMETRICAL F AULTS

kV. Apparatus ratings and reactances are

1000 MVA, 20 kV, Xs = 1 00% X� = X I = X2 = 10% Xo = S% 800 MVA, 22 kV, Xs = 120% X� = X I = X2 = 15% Xo = 8% 1000 MVA, 500Y/20� kV, X = 17.5% 800 MVA, 500Y/22Y kV, X 16.0% XI = 15%, Xo = 40% on a base of 1500 MVA, 500 kV neu tral o f G I i s grounded through a react ance o f 0.04 n . The neutral o f G2 =

The is not grounded . Neutrals of a l l t r a n s fo r m e rs a r c s o l i d l y groun ded. Work o n a base of 1000 M V A , 500 k V in t h e t ra ns m i s s i o n l i n e . N e g l e c t p r cf a u l t cu r r e n t a n d li n d s u b t ra ns i c n t current ( 0 ) i l l p i l ;l s c c o f G I [or ; 1 t i l re e - p h a se fa u l t 'I t P , ( h ) i ll p h a s e B at F for a l i n c - t n - l i n c ["a u l l on l i n c s n a n d C . ( c) i n p i l 'l s e /1 a t P fo r a line-to-ground fault 0 1 1 l i nc /1 , and ( d ) ill p h ase c o f C; 2 f or a l i n e - t o - g ro l i n J fa u l t o n line A . A ss u m e V� ' ) l e a d s �/( ' ) by 30° i n 1', . 12.16. In the network shown in Fig. 1 0 . 1 7 Y - Y - c o n n e c t ed trans formers, each with grounded neutrals, arc at the ends of each tra nsmission line t h J t is not termina t­ ing a t bus G). The transformers connecting the l i nes to bus G) are Y- � , with the neutral of the Y sol idly grounded and the Ll sides connected to bus CD . A l l line reactances shown in Fig. 10. l7 between buses i n c l u d e t h e reactances of the transformers. Zero-seq uence values for t hese lines incl uding transformers are 2.0 times those shown in Fig. 10.17. Both genera tors are Y -connected. Zero-sequence reactances of t h e genera­ tors connected to buses CD and ® arc 0.04 and 0.08 per unit, respectively. The neutral of the generator at bus CD is connected to grou nd through a reactor of 0.02 per unit; the generator at bus G) has a solidly grounded neu tral. Find the bus impedance mat rices Z \}ls ' Zb21s ' and Zbols for t he given network and then compute the subtransient current in per unit ( a ) in a single line-to-ground faul t o n bus (1) a n d ( b ) in the faulted ph ase o f line CD- (1). Assume no prefau l t current is flowing a n d a l l prcfault voltages a t a l l t h e buses is 1 .0& per u n i t . 12.17. T h e network o f Fig. 9 . 2 h a s the l i n e d a t a speci fied i n Tab l e 9.2. Each of the two g e n e r a t o rs cOllnected t o b u se s

CD

am.!

CD

has

X:;

X,

X2

0.25 pel u n i t .

M aking the usual simplifying assu mptions o f Sec. 10.6, determi ne t h e sequence . m atrices Z�ls = Z�ls and usc them to calculate ( a ) The subtransient current in per unit in a l i ne-to- l ine fau l t on bus (1) of the network and ( b ) The fau l t current con tributions from line CD - Q) and line 0)- Q) . Assume that l i nes CD - Q) and G) - Q) arc connected to bus W d i rectly ( no t t hrough transformers) and t h a t a l l positive- a n d negat ive-sequence reactances are identical . 12.18. In the system of Fig. 1 2.9( a ), consider t h a t machine 2 i s a m o t o r d rawing a load equivalent to 80 MVA a t 0.85 power-factor l aggi ng and nom i n a l syste m voltage of 345 kV at bus G) . Determine the change in voltage at bus G) when the transmission line undergoes ( a ) a one-open-conductor fault and ( b ) a two-open ­ conductor fault along its span between buses (1) and G). Choose a base o f 100 MVA, 345 kV in the transm ission line. Consult Examples 1 2 . 1 and 1 2.2 for ZboJs ' Z b�s ' and Zb2Js ' =

=

=

CHAPTER

13

ECON OMIC OPERATI O N O F POWER SYSTEMS

Economic operation is very i mportant for a power system to return a p rofit on the capital invested. Rates fixed by regul atory bod ies and the importance of co nservat ion of fuel place p ressure on power companies to achieve maximum possible efficiency. Maximum efficiency minimizes the cost of a kilowatthour to the consumer and the cost to the company of delivering that k ilowatthour in the face of constantly risi ng p rices for fu e l , l abor, suppl ies, and maintenance . Opera t i o n a l eco n o m i cs i nvo l v i n g power generation and delivery can be subdivided into two pa rts-one d e a l i n g w i t h minimum cost of power production called economic dispa tch and the oth e r dealing with minimum-loss delivery of the generated power to the loads. For any speci fied load condition economi c d ispatch determines the power output of each plant (and each generating unit w i t h i n t h e p l a n t ) w h i c h w i l l m i n i m ize t h e ove r a l l cost of fu e l n e e d e d to serve the

system load. Thus, economic dispatch focuses upon coord inating the production costs at all power pl ants operating on the system and is the major emphasis of t his chapter. The minimum-loss p roblem can assume many forms depending on how control of the power flow in the system is exercised. The economic d ispatch problem and also the minimu m-loss p roblem can be solved by means of the optimal power-flow ( O PF) program. The OPF cal culation can be view e d as a sequence of convent ion al N ewton-R aphson power-flow calculaqons in w hich certai n control lable parameters are automatically adj usted to satisfy the net53 1

532

CHAPTER 1 3

ECONOMIC OPERATION OF POWER SYSTEMS

work constraints while minimizing a specified objective function. In this chapter we consider the classical approach to economic d ispatch and refer the i nter­ ested reader to the citation b elow for details of the OPF approach . ! We study first the most economic d istribution o f the output o f a p lant between the generators, or units, within that plant. The method that we develop also applies to economic scheduling of plant outputs for a given loading of the system without consideration of transmi ssion losses . N ext, we express t ransmis­ sion loss as a function of the outputs of the various plants. Thc n, we d eter m i n c how the output o f each o f the p l a n t s of a syst em i s s c h ed u l e d to achieve t h e minimum cost o f power delivered to the load. Because t he total load of the power system varies t h roughout t h e d ay, coordinated control of the power p l a n t o u t p u ts is n e ce ss a ry to e ns u r e g e n e r a ­ tion-to-Ioad balance so that the system fre q u e ncy w i l l re m a i n as c l o s e as possible to t he nominal operating value, usually 50 or 60 Hz. Accl)rd i n g ly , t h e problem of automatic generation control (AGe) is developed from the ste ady­ state viewpoint. Also, because of the daily load variation, the utility has to decide on the basis of economics which generators t o s t a r t u p , w h i c h to s h u t down, and in what order. T h e computatio nal p r o c e d u re fo r m a k i n g s u c h decisions, c a l l e d unit commitment , i s also developed a t a n i n trod u c t u ry level i n this chapter.

13.1 DISTRIBUTION OF LOAD BETWEEN UNITS WITHIN A PLANT An early attempt at economic d ispatch called for supplying power from only the most efficient plant at light loads. As load increased, power would be supplied by the most efficient plant until the point o f m aximum efficiency of that p lant was reache d . Then, for fu r t h e r i ncrease i n load the n ext m cl',! effici e n t p l a n t would start to feed power to t he system and a t h i rd plant would not be called upon u ntil the point of m ax imum efficiency of the second plant was reached. Even with transmission l osses neglected, t his method fails to m i n i m ize cost. To determine t h e economic distribution of load between t h e various generating u nits (consisting of a turbine, generator, and steam supply), the variable operating costs of the unit m ust be expressed in terms of t he power output. Fuel cost is the principal factor in fossil-fuel plants, and cost of n u clear fue l can also be expressed as a function of output. We b ase our discussion on the economics of fuel cost w ith the realization that other costs which are a function of power output can be i ncluded i n the expression for fue l cost. A typical input-output curve which is a plot of fuel input for a fossil-fuel plant i n British thermal u nits (Btu) per h o u r versus power output of the u n i t i n megawatts i s shown i n Fig. 13.1. The ordinates o f t h e curve are converted to

I H. W. Dom mel and W. F . Tinney, " Opti mal Power- F l ow S o luti ons," IEEE Transactions on Power Apparatus and Systems, vol . PAS-87, Octobe r 1968, p p . 1 866 - 1 8 7 6 .

13.1 1 09

6

X

5

X 1 09

4

-:E.

X

DISTRIBUTION OF LOAD BElWEEN UNITS WITH IN A PLANT

v

1 09

:l

-; 3 Q.

CD

c


2

X

x

1 09

1 09

1 09

533

/

V V

] 00

200

/1

300

Power output.

/

FIGURE 3. 1 400 MW

500

600

I n pu t-out p u t curve for a ge nerating unit showi ng fue l i np u t \'e rsus power output.

dollars per hour by multipl ying t he fu el input by the cost of fuel in dollars per m i l l ion Btu, I f a line i s drawn through the origin to any point o n t h e i nput-output curve , the slope can be expressed in m i l l ions of Btu per hour div ided by t he ou tput in megawatts, or the ratio of fuel input in Btu to energy output in kilowatthours. This ratio is called the heat rate and its reciprocal is the fuel efficiency . Hence, l ower heat rates imply higher fuel efficiency. Maximum fuel efficiency occurs at that point where the slope of the l ine from the origin to a point on the curve is a minimum, that is, at the point where the l in e i s t angent to the curve. For the unit whose input-output curve is shown in Fig. 1 3 . 1 the maximum efficiency is at an output of approximately 280 MW, which requires a n i n p u t of 2.8 X 1 0 9 Btu/h. The heat rate is 10,000 Btu/kWh, and s ince 1 kWh = 3412 Btu, the fuel efficiency is approximately 34%. By comparison, when the output of the u n it is 1 00 MW, the heat rate is 1 1 ,000 Btu/kW h and the fuel efficie ncy is 3 1 % , O f cou rse , t h e fu el requ irement fo r a g i v e n output i s easily converted into dollars per megawatthour. As we shall see, the criterion for d istribution o f the load between any two units is based on whether increas ing the load on o n e unit a s t h e l o a d i s d e c r e a sed o n t h e o t h e r u n i t hy t h e s a m e a mount results i n an i ncrease or decrease in total cost. Thus, we arc concerned with incremental fuel cost , which is determined by the slopes of the input-output curves of the two u n i ts. I f we express the ordinates of the input-output curve i n dollars per hour and let Ii

Pgi

=

input to Unit i , dol lars per hour ( $/h)

=

output of Unit i , megawatts ( MW )

534

CHAPTER 1 3

ECO N O M I C OPERATION OF POWER S YSTEMS

the incremental fuel cost of the unit in dollars per megawat thou r is dfJdPJ.: i ' whereas the average fuel cost i n the same u n i ts i s fJ Pgi. Hence, i f t h e input­ output curve of Unit i is quadratic, we can write $/h

and the u n i t has incremental fuel cost d enoted by

A"

( 13 .1 )

which i s defined by

$ / MW h

( 1 3 .2 )

where a i b" a n d c i are constants. T h e approximate i ncremental fuel cost a t any p articular output is the additional cost in d o l l ars per hour to increase the output by 1 MW. Actually, i ncremental cost is determined by measuring t he slope of the input-output curve and multiplying by cost per Btu in the proper u nits. Since mills ( tenths o f a cent) per k ilowatthour are equal to doll ars per megawatthour, and since a kilowatt is a very small amount of power in comparison with the usual output of a u n i t of a steam plant, incremental fuel cost may be considered as t h e cost of fuel in mi lls per hour to supply an additional kilowatt outpu t . A typical plot of i ncremental fuel cost versus power output is shown i n Fig. 13.2. This figure is obtained by measuring the slope of the input-output curve of Fig. 13.1 for various outputs and by applying a fuel cost of $ 1 .30 per million Btu. However, t he cost of fue l i n terms o f Btu is n o t very predictable, and the reader should not assum e t h a t cost figures throughout this chapter are applicable at any particular time. Figure 1 3.2 shows that i ncremental fuel cost is quite l inear with resp ect t o power output over an appreciable range. In analytical work t h e curve is u s ually approximated by o n e or two straight lines. The dashed l i n e in the figure is a good representation o f the curve. The equation of the line is '

Ai

=

dfi dPgi

-

=

0 .0 1 26 Pgi + 8 . 9

so that when the power output is 300 MW, the incremental cost determined by the l inear approximation is $ 1 2.68/MWh. This value of A i is the approximate additional cost per hour of i ncreasing the output Pg i by 1 MW and the saving in cost per hour of reducing the o utp ut Pg i by 1 MW. The actual incremental cost . at 300 MW is $ 12.50/MWh, but t his power output is near the point of maximum deviation between the actual value and the l inear approximation of incremental cost. For greater accuracy two straight l ines may be drawn to represent this curve in i ts upper and lower range. We now have the background to understand the principle of econe:>mic d ispatch which guides d istribution of l oad among the units with i n one or more .

DISTRIBUTION O F LOAD BETWEEN UNnS WITH I N A PLANT

13.1

16

� u

12

+-'

10

CD

+-'

C Q)

E

Q) "-u c

535

�/

L i nea r

I

� / approximar r-

I

'"

-

-

-

-

-

-/ �

--"

.- - -- - ---- - - - -

100

200

/"

I I

I J

-- Ac t u a l i n cre m e n t a l cost

- - --

----- --- -.. �-

300

400

Pow e r o u t p u t , M W

r=-

---_. _ _ .-

-- -

500

-�

600

FIG U R E 1 3 . 2

I [] c r t: m e n t a l fu e l cos t v e r s u s power o u t p u t for t h e u n i t w h o s e i n p u t - o u t p u t c u rv e i s s h own i n Fig. 13.1.

plants o f the system. For instance, suppose that the total output of a particular plant is suppl ied by two units a n d that the d ivision of load between these units is such that the incremental fuel cost of one is h igher than that of the other. . Now suppose that some of the load is transferred from the unit with t h e higher i ncremental cost to the u n i t with the lower incremental cost. Reduci ng the load on the unit with the h igher i ncremental cost will result in a greater reduction of cost than the increase in cost for adding the same amoun t of load to the unit w i t h the lower incremental cost. The transfer of load from one to the other can be continued with a reduction in total fuel cost until the i ncremental fue l costs of the two u n i ts are equ al . The same reasoning can be extended to a plant with more than two u n i ts. Thus, for econom ical d ivision of load between u ni ts within a pl ant, the criterion is that all units rl'lllst opera te a t the same incremental fuel cost . When the i ncremental fuel cost of each of the units i n a plant i s n early l inear with respect to power output over a range of operation under considera­ tion, equations that rep resent increm ental fuel costs as linear functions of power output will simpl i fy the computations. An economic dispatch sched u le for assigning loads to each unit i n a plant can be prepared by: 1.

3. 2.

Assuming various values of total plant output, Calculating the corresponding i ncremental fuel cost A of the plant, a n d Substituting the value of A for A i i n the equation for the incremental fuel cost of each unit to calculate its output. . ,

536

CHAPTER 13

ECONOMIC OPERATION OF POWER S YSTEMS

curve of A versus plant load establishes the value of A at which each unit should operate for a given total plant load. For a plant with two units operating under economic load distribution t h e A of the plant equals Ai of each unit, and s o

A

( 1 3 .3 )

-

Solving for Pg l and Pg 2 , we obtain Pg l

,\

b,

= --­ a,

Adding these results together

and

Pf{ 2

-

b2

,\ = --­

a2

( 1 3 .4)

a n d t h e n so l v i n g for ,\ g i v e

( 1 3 . 5)

or where

aT =

( 13 6)

( L ­ )-1 . ( E -I ) 1

2

1= 1

bT = aT

.

a I·

2

i= 1

b.

ai

and is the total p l an t output. Equation ( 1 3 .6) is a closed-form sol ution fo r ,\ which applies to a plant with more than two units o n economic d ispatch when the appropriate number of terms is added to the summations o f Eq. ( 1 3.5). For instance, if the plant has K units operating on economic dispatch, then the coefficients of Eq. ( 13.6) ar e given by ( 1 3 .7)

and

.

( 1 3 .8 )

t h e total p lant output Pg T is ( Pg 1 + Pg2 + . . + PgK ). The indivipual output of each of t he K u nits is then calculated from the common value of A

1 3 .1

D ISTRIBUTION O F LOAD BElWEEN UN ITS WITH I N A PLANT

537

given by Eq. ( 13.6). If maximum and minimum loads are specified for e ach unit, some units will be u nable to operate at the same incremental fuel cost as the other u nits and still remain within the l i mits specified for l ight and h e avy loads. Suppose that this occurs for K = 4 and that the calculated value of Pg 4 violates a specified l imit of Unit 4. We then discard the calculated outputs of all four u nits and set the operating value of Pg4 equal to the violated l imit of Unit 4. Returning to Eq. ( 13.6), we recalculate the coefficients a T and b T for the other three units and set the effective economic dispatch value of PgT equal to the total plant load minus the limit value of Pg4 . The resulting value of A then governs the econom ic dispatch of Units 1, 2, and 3 when the actual plant o utput is to be increased or decreased, so long as Unit 4 remains as the only u n i t at a l imit. The equal-incremcntal-fuel-cost criterion which we have developed intu­ i t ively, and now i llustrate numerical ly, will be established mathematical ly in Sec. 1 3 .2. Exa m ple 1 3 . 1 . Incrcmental fucl costs consisting of two units are given by

Al

dfl =

--

dPlt 1

=

0 .0080 Pg 1

+

In

dollars pcr megawatthour for a plant

8 .0

Assume that both units are operating a t all times, that total load varies from 250 to 1 250 MW, and that m aximum and minimum loads on each unit are to be 625 and 1 00 MW, respectively. Find the incremental fuel cost of the plant and the allocation of load between units for the minimum cost of various total loads. Solution. At light loads Unit 1 will have the higher incremental fuel cost and operate at its lower limit of 1 00 MW for which dfl /dPg l is $8.8/MWh. When the output of Unit 2 is also 1 00 MW, df2/dPg 2 is $7.36/MWh. Therefore, a s plant outpu t increases, the addit ional load should come from Unit 2 until df2 /dPg 2 equals $8.8/MWh . Until that point is reached, the incremental fuel cost A of the plant is determined by Unit 2 alone. When the plant load is 250 MW, Unit 2 will supply 1 50 MW with df2 /d P 2 equal to $7.84/MWh. When dfjdP/:2 equals $8.8/MWh, f;

0 .0096 Pg 2 +

P11 2

2.4 -

0 . 0096

6.4

=

= 250

8 .8 MW

and the total p lant o utput Pg T is 350 MW. From t his point on the requi red o u tput of each u n i t for economic load d istribution is found by assuming various values of Pg T ' calculating the corresponding plant A from Eq. ( 1 3 .6), and substituting the value of A i n Eqs. (1 3.4) to compute each unit's outpu t . Resu lts are shown in Table 1 3 . 1 . When Pg T is in the range from 350 to 1 1 75 MW, the plant A is determined by Eq. ( 1 3 .6). At A = 1 2.4 U n i t 2 is operating at its upper limit and additional load

538

CHAPTER 13

ECONOMIC OPERATION OF POWER SYSTEMS

The plant A and outputs of each u n it for variou s values of total output PgT for

TABLE 1 3 . 1

Pg T! MW

Plant

A,

Unit 1

Unit 2

l\f\V

l\f\V

Pg 1 ,

$ / MWh

250

350 500 700 900 1 1 00 1 250 t 1 ndicates t h e o u t p u t of

Pg Z '

7.84 8.80

l OO t

l llO t

1 50 250

9 .45

29 1

1 82

318

1 0.33 1 1 .2 0 1 2 .07 1 2 .40 1 3.00

1 1 75

Example 1 3 . 1

409 500 59 1 625 t 6 25 t

400

509 550 625

t h e u n i t at

its mini mum

(or m Jx i m u m ) l i m i t a n d

the p l a n t A i s t he n e q u a l t o t he i n c r e m e n t a l f u e l cost o f t h e u n i t n o t a t a limit.

must come from Unit 1 , which then determ ines the plant A . Figure 1 3 . 3 shows plant A plotted versus plant output. If we wish to know the d is t ribution of load between the u n i ts for a plant output of 500 MW, we could plot the output of each i nd ividual unit versus plant output, as s hown in Fig. 1 3 .4, from which each u n it's output can be r ead for any plant output. The correct output of each of many u n i ts can be easily computed from Eq. (13.6) by req uiring all u n i t incremental costs to be e qual for any total plant output. For the two un its of the example, given a total output of 500 M W, Pg T

aT

by

( � 1 ) ·- 1 ar(� b2) 1

=

=

+

Q2

+

a1

a2

=

Pl: j

+

( 0 . 008

=

1

= aT

Pg 2 +

(� 0 .008

=

500

1 0.0096 +

MW

). 1

6 .4 0 .0096

)

=

=

4 .363636

X 10- 3

7 .272727

and then for each unit

A

which yields Pg I

=

Pg 2 =

A A

=

a T PgT + b T

- bl

=

=

9 .454545 $/MWh

9 .454545 - 8.0 0 .008

- - 9 .454545 - 6.4 0 . 0096 a1

b2

a2

=

=

1 81 .8182 MW

3 18 . 1 8 18 M W

13.1

D I ST R I B UTION OF LOAD BElWEEN UNITS WITHIN A PLANT

13

539

r--'r--'---T�-.---r--�V�� � � �_� +� 1 2 r--+ --

--

u; o U
.:.

!! co
E

1 0 r----r--�--9 I------!

800

+--'

� 0.

+--'

� 0 ..... c

=:J

-7,Lc-----+----t-------+---

: ____T I o

� 2

��-

--

l l r---�--_+----r---4�--+---�--� /

ill

U c:

I /V

�--

600

200

1) 0 0

F I G C R E 13.3 l ncremc n t a l

600

800

P l a n t o u l p u l . MW

.------.--r- -

�r---

1 00 0

1 200

1 4 00

fu e l

cos t

versus

plant

(l u t p u t w i t h l o t a l p l a n t l o a d e c o n o m i ­ u l ly

u istribuled

fou n t.! i n Exa m p l e

between

u n i ts,

n. 1 .

as

U n it 2

400

200 o

200

400

L-____�__ _

600

800

1 0 00

1 2 00

1 4 00

�______L_____-L____�____�

_ _ __

Plant output, MW FI G U RE 1 3.4 Ou t pu t o f e ach u n i t vers u s p l a n t o l l t r l l t fo r e c o n o m i c a l opera t i on of t h e p l a n t o f Exa m p l e

13. 1 .

Such UCCu r3Cy, howeve r, i s not necessary because o f the uncertainty i n determining exact costs and the use o f an approximate equation in this example to express the incremental costs. The

savings effected by economic d istribution of load rather than some a rhitrary d istribut ion can be fo u n d by in tegra t i n g the exp ression for i ncremen ta l fue l cost a n d by compari ng increases a n d decreases o f cost for the u nits as load i s s h i fted from the most eco n o m i c a l a l location. Example 1 3 .2. Determine t h e saving in fuel cost in doll ars per hour for the economic distribution of a total load of 900 M W between the two units of the plant described in Exa mple 1 3 . 1 compared with equal distribution of the sa'me total load.

540

CHAPTER 13

ECONOMIC OPERATION OF POWER SYSTEMS

Tab le 1 3 . 1 s ho ws that Unit 1 should s upply 400 MW and Unit 2 should supply 500 MW. I f each unit supplies 450 MW, the i ncrease i n cost for Unit 1 I S

Solution.

f

450

�

The

( 0 .008Pg 1

+

8) dPg J

=

(0 .004 Pg2J + 8 Pg 1 +

n e l s when we

cons t a n t (." ] c a c

C1 )

1 45000 4

=

$570 per hour

e va l u a t e a t t h e t w o l i m i t s . S i ll 1 i l a r l y , fo r U n i t

2

-- $548 per hour The negative sign indicates a decrease i n cost, as we expcct for a d ecrease in output. The net increase in cost i s $570 - $548 = $22 per hour. The saving seems small, but this amount saved every hour for a year of conti nuous operation wou l d reduce fuel cost by $192,720 for the year.

The saving effected by e c ono mic distribution of load j ustifies devices for controlling the loading of each unit automatically. We shall consider automatic control of generation later in thi� chapter. But first let us investigate the problem of coordinating transmission losses in the economic distribution of load between plants. 13.2 D ISTRIBUTION OF LOAD BE1WEEN PLANTS

In determining the economic distribution of load between pl ants, we encounter the need to consider losses i n the transmission l ines. Although t he i ncremental fuel cost at one plant bus may be lower t han that of another plant for a given distrib ution of load between the plants, the plant with the lower i ncremental cost at its bus may be m uch farther f rom the load center. The losses in transmission from the plant having the lower i ncremental cost may be so great that economy may dictate lowering the load at the plant with the lower incremental cost and increasing it at the plant w ith the higher incremental cost. Thus, we n eed to coordinate transmission loss into t h e scheduling o f t h e output of e a c h p l a n t for maxim u m economy at a g i v e n level o f system l o a d . For a system with K generating u n its let f = fl + f2 + ' . . + fK

=

K

i

L fi =

( 1 3 .9)

I

where f is the cost function giving the total cost of all the fuel for the ,entire system and is the sum of the fuel costs of the i ndividual units fl ' f2 ' . . . , fK ' The

1 3 .2

DIST R I B UTION OF LOAD BETWEEN PLANTS

541

total megawatt power input to the n e twork from all the u nits is the sum Pg 1 + Pg 2 + . . .

K

+ PgK = L Pg i

( 1 3 . 10 )

i= 1

where Pg I , Pg2 , , Pg K are the i n d ividual outputs of the units injected into t h e ne twork. The total fuel cost f o f t h e system i s a function o f all t h e power plant outputs. The constraining equation on the m inimum value of f is given b y the power bal ance of Eq. (9. 10), which for the p resent purpose we rewrite in the for m •

•

•

PL + PD

K

-

i

L Pg i = 0 =

( 13 . 1 1 )

1

where PD = '[�= 1 Pd i i s the total power received by the loads and PL is the t ransmission loss of the system. O u r objective is to obtain a minimum f for a fixed system load PD subj ect to the power-balance constraint of Eq. ( 1 3. 1 1 ). We now present the procedure for solving such m inimization problems called the method of Lagrange multipliers. The new cost function F is formed by combining the total fuel cost and the equality constra int of Eq. (1 3 . 1 1 ) in the fol lowing manner: ( 1 3 . 1 2)

The augmented cost function F is often called the lagrangian , and we shall see that the parameter A , which we n ow call the Lagrange multiplier , is the effective incremental fuel cost of the system when t ransmission-line losses are taken i nto account. When Ii is give n in dol l a rs per hour and P is in megawatts, F and A a r e expressed i n dol lars per hour ancl dollars per megawa tthour, respect ive ly. The original problem of m i n i mizi ng I s u bj ec t to Eq . ( 1 3 . 1 1 ) is transformed by means of Eq. ( 1 3 . 1 2) into an unconstrained problem in which it is required to m inimize F with respect to A and the generator outputs. Therefore, for minimum cost we requ ire the derivative of F with respect to each Pg i to equal zero, and so

S ince PD I

IS

' fixed and the fue l cost of any one unit varies only if the power

542

CHAPTER 1 3 ECONOMIC OPERATION OF POWER SYSTEMS

output of that unit is varied, Eq . ( 13. 13) yields ( 1 3 . 14)

for each of the generating-u nit outputs Pg i , P;; 2 " ' " P;; K ' Because Ii depends o n only Pgi, the partial deriva tive of I; can be replaced by the fu l l derivative, and Eq. (13. 14) then gives

1

a PI.

O ?h ;

elf; dPh i

( 13.15)

for every value of i . This equation is often written in t h e form ;\ = L

d,! , ' dPg;

-� ,

( 13 . 16)

where L; is called the pena lty fa ctor of plant i and is given by 1

LI =

( 1 3 .17)

The result of Eq. (13 . 16) means that minimum fue l cost is obtained w hen the incremental fuel cost of each unit m ultiplied by its penalty fac t0f is the same for all generating units in the system. The products L;(dfJdPg) are each equal to A, called the system A, which is approximately the cost in dol l a rs per hour to i ncrease the total delivered load by 1 MW. For a system of t hree u nits, not , necessarily in the same power plant, Eq. ( 1 3'. 1 6) yields

\

( 13 . 1 8)

The penalty factor L; depends on 3PL /3Pg ; , w hich is a measure of t h e sensi tivity o f the transmission-system losses t o changes i n Pg ; alone. Generating u nits connected to the sa me bus within a particular power plant have equal access to the transmission system, and so the change in system losses must be the same for a small change in the o u tput of any one o f t hose units. This means that the penalty factors are the same for such u nits located in the same power plant. Therefore, for a plant hav ing, say, three generating units with Ol.JtpU ts Pg l ' Pg 2 , and Pg 3 , the penalty factors L 1 , L 2 , and L3 are equal, and Eq. ( 1 3 . 1 8)

13.3

THE TRANSMISSION-LOSS EQUATI O N

543

then shows that

( 1 3 . 19) Thus, for units connected to a common bus within the same physical power plant we have developed mathematically the same criterion which we reach ed · i n tuitively in Sec. 13. 1 . Equation (13 . 1 6) governs the coordination o f transmission loss into the probJem of economic loading of u nits in plants which are geographi cal ly dispersed throughout the syste m . Accord ingly, the penalty factors of the differ­ e n t p l a n t s n e e d t o be d et e r m i ned which req u i re s t h a t we first express the total t r a n s m i s s i o n l oss of the sys t e m a s a fu n c t i o n o f p l a n t load i ngs. Th is formul ation i s u n d e rt a k e n i n Sec. 1 3 . 3 . 1 3 .3

THE TRANSMISSION-LOSS EQUATION

de rive the transmission-loss equat ion i n terms of powe r output of the pl a nts, we consider a simple system consisting of two generating plants and two loads with the transmission network represented by its bus imped ance matrix. The derivation is undertaken i n two stages. In the first stage we apply a power­ i nvariant transformation to Z bus of the system in order to express the system loss in terms of only generator currents. Thus, the reader may find it useful to review Sec. 8.6. In the second stage we transform the generator currents i nto the power outputs of the plants w hich leads to the desired form of the loss equation for a system with any n u mber K of sources. We begin the formulation us ing, as example, the four-bus system of Fig. 13.5ea) in which nodes CD and W are generator buses, nodes ® and @) a re load buses, and node ® is the system neutral. The case where both generation and load are at the same bus is shown in Fig. 13.5ec), which is explained at the end of this section . The current i njections 13 and 14 at the load buses of Fig. 1 3 . 5 ( a ) are combined toget her to form the composi te system load 1 D given by

To

( 1 3 .20)

Assuming that each load is a constant fraction of the t o t al and

load,

we set ( 1 3 .2 1 )

from which it foll ows that ( 1 3 .22 )

By adding more terms, Eqs. (13 .20) through ( 13.22) can be general ized to systems with more t han two load b uses.

· 544

CHAPTER 13

ECONOMIC OPERATION OF POWER SYSTEMS

Generator 1

CD

Load

II -

@

13 -

14 �

12 -

ZBUS

Jtn (a )

®

I Load

® IO

Ge nerator 2

®

G e nera t or 2

®

+

(b)

(c)

(a) The ex ample four-bus system of Sec. 1 3 . 3 ; (b) the i n terpretation of no-lo a d cu rren t Ino of Eq. ( 1 3 .26); (c) the treatme nt of load c urre n t - 12d a t g e nerator bus (1) . FlGURE 13.5

W e now choose node e qu ations

VI II V2 11

V3 t 1 V4 n

=

@

CD W W @

of Fig. 1 3.S( a ) as reference for t h e nodal

CD 211 22 \ Z] I 24 1

(1) ® 2 \2 2 13 2 23 2 22 Z ]2 Z 3 3 24 2 243

@ 2 14 2 24 2 ]4 2 44

II

12

I]

( 1 3 .23)

14

Double-subscript notation emphasizes the fact that the bus voltages are mea­ sured with respect to reference node @ . Expanding the first row of Eq. ( 1 3 . 23) gives ( 1 3 .2 4 )

Subs tituti ng i n thi s equa tion for 13 resul tant equa tion for 1D yield

=

d3 1D and

14

=

d 4 1D ' then solvi ng the

1 3.3

mE: TRANSM ISSION-LOSS EOUATION

545

i n which the current 1�, called the no-lo a d current , is simply ( 1 3 .26) o

We shall soon see the p hysical m e a ni ng of In , which is a constant current i njecte d into node ® of t he system whenever V' n is constant. By d enoting 'I

=

Zl 1

and ---­

--

d] Z 1 3

+ d4 Z ,4

'2

=

---­ --

d3ZjJ +

d4 Z 1 4

( 13 .27 )

we may simplify the coefficients o f E q . ( 1 3 .25), wh ich then becomes ( 1 3 .28)

Substituting in Eqs. 0 3.2 1 ) fo r ID from Eq . ( 1 3 .28) yields ( 1 3 . 29) ( 13 .30)

We may regard Eqs. ( 13.29) and ( 1 3 .30) as defining the transformation C of "old" currents I, : 12 , 13, and 14 to the set of " n ew" currents I " 12, and I�, just as i n Eq. (8.55); that is,

I,

12

13 14

CD CV G) 0

CD 1

@

®

- d3 t 2

- d3 t 1 - d4 t 1

1

- d 'l, t l - d4 t 1

- d4 t2

:

l [!: 1 [ �: 1 JnO

J

�

( 1 3 .3 1 )

O Jn

c

exp l a i n the row a n d col umn l a b e l s when we provide a physical interpretation of transformation C in Sec. 13 .4. As a result of Eq. ( 1 3 . 3 1 ), the expression for the real power loss of the network takes the form of Eq . ( 8 68) which we now write as

We

shal l

.

,

( 1 3 .3 2 )

w here R b us is the symmetrical real part of Z b us of Eq. (1 3 23) Because of the power-invariant nature of transform ation C, Eq. ( 1 3 32) fully repres � nts the real .

.

.

546

CHAPTER 1 3

ECONOMIC OPERATION OF POW ER SYSTEMS

power loss of the system in terms of the generator currents 11 and 1 2 and the no-load current I�. By fixing upon bus CD as the slack bus in power-flow studies of the system, the current 1 � V1n/ Z 1 1 becomes a constant complex number, which leaves 11 and 12 as the only variables i n the loss expression of Eq. ( 1 3.32). Figure 1 3.5(b ) helps to explain why I/� is called the no-load current. If all load and g e n e r a t i on w e re r e m ove d from the sys t e m a n d t h e vol tage VI II w e r e applied at bus CD , o n ly the current l,�) w ou l d flow through the shunt connec­ tions to node ® . This cu rrent is normally small and relatively constant since it is determined by Thcveni n i m p e d a n ce Z I I ' which i n c l u d e s t he h ig h imped ances of paths associated w i t h l i n e - c h a r g i n g and transformer m a g n e t i z i n g c u r re n t ::, b u t n o t load. At each generator b us w e n o w a ss u m e t h 3 t t h e r ea c t i "e power QK i i s a con stant fraction Si of the re a l pow e r P!: i over t he t i m e p e r i o d o f i n te r e s t . T h i s is equivalent to assuming that c a c h g e n e r a t o r o p e ra t es a t a const a n t p ow er factor over the same period, and so we write =

-

where S 1 Q g l/Pg 1 and S2 = Qg 2/Pg 2 are real numbers. The output currents from the genera tors are then given by =

in w hich a l and a 2 have obvious definitions. From Eqs. (13 .34) 12 , and I� can be expressed i n matrix form by

the

currents 1 1 '

( 1 3 .35)

]CIRbUSC* r�l

and substituting from this equation into Eq. (13.32), we obtain

0: 2

:

1° II

.

We recall that the transpose of a product of matrices equals the reverse-order pro duct of their transposes. For i nstance, if there are three matrices A, B, and C, w e h ave (ABC)T C T BTAT, and taking the complex conjugate o f eacn side =

1 3.3

1Hr: TRANSMISSION-LOSS EOUATION

547

gives (ABCY* = C T* B T *AT* . Thus, we can show that the matrix Ta of Eq. ( 1 3.36) has the convenient p roperty of being equal to the complex conjugate of i ts own transpose. A matrix with this characteristic is called hennitian . 2 Each off-diagonal element m ij of a hermitian matrix is equal to the complex conju­ gate of the corresponding element mji and all the diagonal elements are real n umbers. Consequently, adding T to T: cancels out the imaginary parts of the off-diagonal elements, and we obtain twice the symmetrical real part of Ta , which we denote by a

B 1 0/2 B 20 1 2 B IX)

. --

]

TO' + T: = ----

-

( 1 3 .37)

/

To c o n form to i n d u s t ry practices, w e u s c the sym h o l s h e r e . A d di n g Eq . ( 1 3.36) to i ts complex conjugate and t h e r es u lt g i v e

B 1 0/2,

B201 2 , and Boo

a p p ly i n g

Eq. ( 1 3 . 37) to

( 13 .38)

in which gives

B

12

equals E n . Expanding Eq. ( 13 .38) by row-column multiplication

L L P� i Ei j P�j + L 2

i= w h i c h we

Or

2

1

2

i

j= I

=

1

BiO Pg i

+

Boo

( 1 3 .39)

c a n rearrange into the equivalent form

t h e more general

ve c t o r - m a t ri x fo rm u l a t i o n

( 13 .41)

2An e x a m p l e

o f a h e r m i t i a n m a t r ix is

[

1

�jl

1 +

1

j1

]

.

548

CHAPTER 13

ECONOMIC OPERATION OF POWER SYSTEMS

When the system has K sources rather than j ust two as in our example, the vectors and matrices of Eq. ( 1 3.41) have K rows and/or K columns and the summations of Eq. ( 13 .39) range from 1 to K to yield the general form of the transmission-loss equation. PL

K

K

K

= L L PK i Bij PKi + L B P + iO

i= 1

i= 1 j= l

;.:

i

B oo

( 1 3 .42)

The B terms are called loss coefficients or B-coefficicn ts and the K X K square matrix B, which is always symmetrical , is known simply as the B-malri.:r . The u nits of the loss coefficients are reciprocal megawatts w h e n t h e t h re e - p h a s e power Pg I to Pg K are expressed in megawatts, in w hich case PL will be in megawatts also. The units of Boo match those of PL while BiG is d i mension less. Of course, per-unit coefficients are used in normalized computations. For the system for which they are derived B-coefficients yield the exact loss only for the particular load and operating cond itions used in t h e derivation . The B-coefficients of E q . ( 1 3 .40) are constant as Pg I and Pg 2 vary, only i nsofar as bus voltages at the loads and plants maintain constant magnitude and plant power factors remain constant. Fortunately, the use of constant val u es for the loss coefficients yields reasonably accurate results when the coefficients are calculated for some average operating conditions and if extremely wide sh i fts of l oad between plants or in total load do not occur. In practice, large systems are economically loaded using different sets of loss coefficients calculated for various load conditions. Example

13.3. The fou r-bus system o f Fig. 1 3 . 5 h J S l i ne and

bus data given i n

Ta b l e 1 3 . 2 . The b a s e -case pow e r - f l ow s o l u t i o n i s s h o w n i n Ta b l e U . 3 . Ca l c u l a t e

t he B - coefficients o f t h e system a n d s how t h a t t h e t r a n s m i ssion loss com p u t e d b y the loss for m u l a checks w i t h t he power-flow res u l t s .

Line data and bus data for Example 13.3 t

TABLE 13.2

Bus d a t a

Line data Bus to Bus Lin e Line Li ne

Li ne

CD - @ CD - G) @ - G) 0-@

Series Z

Shunt Y

R

X

B

.00744

.0372

.0775

.01 00S

.0504

. 1 025

.00744

.0372

.0775

.01 272

.0636

. 1 275

p

Q

2 . 20

1 . 3634

2 . S0

1 .7352

Generation Bus

CD 0 G) @

tAl i values in per u n i t on 230-kY, l OO- M YA base .

p

I VI �

LQ".

LDad

1 .0 3.1S

1 .0

1 3.3

THE TRANSMISSION-LOSS EQUATION

549

TABLE 13.3

Power-flow solution for Example 13.3 t Base case Generation Bus

Q

P

CD W ® (1)

Tot a l tAil

Voltage

\

Magnitude (per u nit)

Angle (deg)

1.913 152

1 .872240

1 .0

0.0

3.18

1 . 325439

1 .0

2 .43995

0. 9605 1

- 1 .07932

0.94304

- 2.62658

3 . 1 97679

5 .093 1 52

a l u e s i n per u n i l on 230-kY, I OO - M YA b a s e .

Solution. Each transm ission line is represented by its cqu ival cnt-7T circuit w i t h half of the line-charging susceptance to neutral at both ends. Choosing the neutral node @ as r efere n c e , we c o n s t r u ct the bus i m p e d ance matrix Z bus = R b us + j X bus ' where

R bus

=

X bus =

CD (1) ® @

CD

14

- 0 .795044 - 0.072159 + 2 .9 1 1 963 - 1 .786620

- 0 .072 1 5 9 - 1 .300878 - 1 .786620 + 2. 932995

CD

(1)

G)

@

-

results

-

jQ 4

V4*

- 2 .606321 - - 2 .582784 - 2 .597783 - 2.603�99

- 2 .601379 - 2 . 597783 - 2 .582884 - 2 . 60632 1

0 .9 605 1

/ 1 .07932°

0.94304/ 2.62658° - 2.8

=

X 10-3

- 2 .597783 - · 2.603899 - 2 .60632 1 - 2.5827�4

of Table 1 3 . 3 the load curre nts are calcul a ted, - 2 .2 + j 1 .36340

jQ 3

V* 3 P4

=

@

- 1 .786620 + 2.932995 - 0 . 072 1 5 9 - 1 .300878

From the power-flow P3

®

2 . 9 1 1 963 - 1 .786620 - 0 .795044 - 0 .072 1 5 9

+

CD - - 2 .5R2884 (1) -· 2 .606321 G) - 2 .601 379 @ - 2.597783

13 =

(1)

+

j 1 .73520

=

=

2.694641

3 .493043

L 1 47 . 13 3 e

/ 145 .5863°

550

CHAPTER 13

ECONOMIC OPERATIO N OF POWER S Y STEMS

and we then find that

d3

=

13

d4 =

Thc

quant l t l c s

(\

and

13

+

'4 13 +

14

=

=

14

0 .435473 + jO .006637

0.564527

( 2 of Eq. ( 1 3 . 27)

row - o n c c l e m c n t s of Z ,,"S as fo l l ow s :

CD

-

a rc

j O . OO66 37

C J l c u la t c u from

CD

d�,

d 4 , and t he

0)

Based on the above resul ts, we comp ute the - d, tj terms of Eq. ( 1 3 . 3 1 ) to obtain the current transformation C

c =

and

C T R bus C *

CD® @

wc

=

- 0 . 432705 - jO .007 1 4 3

- 0 . 436644 - jO .0064 1 6

- 0 .432705 - jO .OO7 143

- 0 .560958

- 0 . 5 66037

- 0 . 5 6095 8

+

jO .005884

+

j O .006964

+

jO .005884

then find that

CD r 0 @

4 .282 1 85 - - 0 .030982 0 . 0 .985724

+

+

+

jO

- 0 . 030982 - j O . 0 1 0638 }O

0 .985724 - j O . 005255 1 . 367642

j O . 0 1 0638

5 . 080886

jO .005255

1 .367642 - jO .006039

+

@

0 . 60 1 225

+ +

j O . 006039 jO

l

x 10

The row and column n umbers of the last two e quations are expl ained in Sec. 1 3.4. The p er-unit no-load curren t is calculated to be /0 = n

- v1

__

ZI l

=

1 .0 + jO .O 0 . 0029 1 2 - j2.582884

- 0.000436 - jO.387 1 64

,

THE TRANSMISSION·LOSS EQUATION

1 3 .3

(

)

(

)

and using the base-case power-flow results, we compute from Eq.

uj

1 - jS I =

U2 =

[a.1 l-

VI*

1 - }S2 V2*

1 .872240 1 }. 1 .913152 1 .0� _

=

1 }. 1 .325439 3.180000 1 . 0/ - 2.43995° _

=

T h e h e r m i t i a n m a t r i x T" o f Eq. l'

"

T"

=

� .

[

+

;0 .0

=

1 .0 - jO.97861 5 1 .016538 - jO .373855

l"" j*

- 0 .049448

0 04944R - j0 004538

5 . 963568

0 . 3 75082 - j O . 3 R0069

0 . 1 9497 1

1[

-

+

+

j O . 004538

BIZ

8 10/2 � 20/ 2

B�2 , 1___ B_ _ _ ----1

810/2

820/2

Boo

=

=

=

+

:

) 0 53 95 1 1

) 0 .0

[_! ! 21 : _

Ta , we ob t a i n t he

X

x

10-3

desired

10

-

3

1[ 1

! �� � �

_ _

B 10 / 2

]

]

13 15 2 _ 8 : ____ _�:- +__:�_ __ 820/ 2 Boo 1

This checks w i t h t h e power-flow resu l t of Tabl e

In

+

Jo. 380069

- 0.049�41:; 0 .375082 8 5_.9_6_3)_-6_8- -_0 _. 1_94_9_7_1 _-_0_._04_9_4_4___ 0 .375082 0.194971 + 0.090121

[ l .913152 3 . 1 8 I 1 ] 0.093153 p e r u n it

l' :'

+

8 .3 83 1 83

from which w e calc u l a t e t h e power loss

PI-

.�

0 .3 75082

jO.O 0 q49 7 1 jO.5395 1 1 . 0 . 1. 90 1 2 1

B y extracting the real p a r ts of t h e respective e l e m e nt s B-matrix of per-u n i t loss c o e ffi c ie n t s

BII

(13.34)

( 1 3 . 36) is g i v e n by

a�

8 .383 1 83

=

55!

1 3.3.

Example 1 3 .3 exact agreement between methods of loss calculation is expected since the l oss coefficients were determ ined for the power-flow condi­ tions for which loss was calculated. The amount of error introduced by using the loss coefficients of Example 1 3. 3 for two other operating conditions m ay be seen by examining the results of two converged power-flow solutions shown in Table 1 3.4. The load levels correspond to 90 and 80% of the base-case load of Example 13.3 and Pg 2 is assigned by economic dispat.:: h , as described in S ec.

,

552

CHAPTER 13

ECONOMIC OPERA nON OF POWER SYSTEMS

Comparison of transmission loss calculated by the B-coefficients

TABLE 13.4

of Example 13.3 and by power-flow solutions for different opera ting condition s o f the example system Load level

base:

90% 80%

Pg )

Pg 2

PL

PdJ

Pd4

(slack b u s )

(ccon. d i s p , )

Power fl ow

2.2 1 . 98 1 .76

2.8 2.52 2.24

1 . 9 1 3 1 52 1 .628 1 5 1 1 .35475 1

3.18 2. 947650 2 . 70567 1

0 .09315 2

0 .07580 1

0.09 3 1 5 2 0.076024

0.0604 2 2

0. 060<:>42

1 3 . 5 . I n practice, the loss coefficients a re recalculated a n d

upda ted

on

B - c oeffs.

a pe riod ic

basis using data acquired from the physical power system. So far we have not considered generator b u se s h av i n g local l o a d s . S u p p ose that bus @ of Fig. 1 3 .S( a ) has a load component /2d i n addition to the network injection Since all currents are considered as injections, we can regard the load current as a current entering the network at a d ummy bus, say, number G) , as shown in Fig. 1 3 .S ( c). R b us is then expanded to include a row and column for bus G) with off-diagonal elements identical to t hose of row 2 and column 2, and = We now p roceed to develop the transfor­ m ation exactly as before, mechanically treating bus G) as a load bus with = + 14 + 15 ' This strategy can be injected current where fol lowed i n solving some o f the problems at t h e e n d o f thi s chapter. -

C

12- -12d 12d 255 222Is 12d dsID' ID 13 =

=

13.4 AN INTERPRETATION OF TRANSFORMATION C

Transformation of Eq. ( 1 3 . 3 1 ) has a useful i nterp retation which provides insight and physical meaning to its usc. Let us consider that C is a combi nation of sequential transformations from the " old" currents of Fig. 13.6( a ) to the " n ew" currents, as similarly d iscussed i n Sec. 8.6 . The first transformation fol lows from Eqs. (13.2 1 ) and is written

C

two

C)

I) i2 13 14

CD ill G) @)

CD @ @ 1

1

..r

c(

d3 d4

f:i 1

( 1 3 .43)

By relating individual load currents t o t h e total load current introduces the mathematical concept of system center with current injected Into a hypothetical (that is, nonphysical) node @ of the network, as shown i n

load

iD,iD C)

1 3.4

AN INTERPRETATION OF TRANSFORMATION C

553

Fig. 13.6(b). The current JD is the algebraic sum of all the load currents injected into the network at the buses. The transformer turns ratios dj : 1 shown in Fig. 1 3 .6(b) are current ratios which are the complex-conjugate reciprocals of their corresponding voltage ratios, as explained in connection with Eq. (2.37). The second transformation C 2 defined by

( 1 3 . 44)

o o

Gen erator 2

Generator 1

[� .- I ----. I

II :---- I I

®

@' I

CD

t

Reference

(a)

@)

I

13 14

4-< -=-

In

Loads

®

Generator 2

CD

Gen erator 1

Reference

(6)

(\ � ----=-J 'fY _ 2--r........-- j

Generator 2

I

t2

:

::3

1

)---_- -+-CD_l-+---.... I��

�

G e n e rato r 1

(il\

®

II

l

.1

1 2 12 ::---....

t l In """"::

_=__

rYYY'\

(c)

1

t

1

@� J -

_

Reference

FIGU RE 13.6 Physical i n t e rp re t a t io n t r a n s fo r m a t i o n C

C1C2:

of (a)

gen­ buses; ( 6 ) C 1 e l i m i n a t e s load currents and forms syst e m load cen ter D; ( c ) C2 switches reference from n ode @ to system load center =

t h e original sys t e m o f two

e r a tors and two load

D. All tap-changing transform­

ers are i d eal an d aurren( ratios a r e show n .

554

CHAPTER 13

ECONOMIC OPERATION OF POWER SYSTEMS

removes ID as an independent current and replaces it by the no-load current I�. This is achieved by switching the system reference from node ® to the load center at node @ , as depicted in Fig. 1 3 . 6(c). Here also the transformer turns ratios t; : 1 are current transformation ratios. The combination of and in seque nce is equivalent to the single transformation shown in Eq. ( 1 3 . 3 1 ); that is,

C1

C

C

=

CI

C

2

=

CD W Q) @

-

CD

CD

® ( 1 3 . 45 )

1 .-d

-- d

.1

{

I

{ I

..

- d -,l 2

- d .. { 2

C2

d .. ( I

- d,1 _

.

I

as straightforward mUltiplication shows. The new set of voltages which fol lows each separate transformation can be found using the general formula Vn ew = C* T Vo,�of Eq. (8.63). Applying gives Fig. 1 3 . G ( b ) in which the vol tages at buses W , CD , and @ with respect to reference node C!i) are

CI

CD = @ @

(l) (D Q) @

VI r!

r d� d:] 1

V211

1

V311

�II

This equation shows that V I I I a n d V2 11 are u na l tered a n d t h a t VO n is expressed as a linear combination of the voltages V:l I I and V4 11 with respect to node ® . Tha t i s keeps the r e fe re n c e a t node ® , w h i c h i s t h e same rderence used to const ruct the Z !)US of Eq. ( 1 3 .23). O n t h e o t h e r h a n d , C2 t ra nsforms Vl n , V2 n , a nd VD II of Fig. 1 3 .6( b ) to a new s e t of v o l t ages w i t h n od e @ as refe re nc e , Fig. 1 3.6(c). The new voltages VI D ! V2 0 ! and V,I D are given by ,

CI

lV1 D 1 V2 D

Vn D

CD @ @

CD

II

(1) @ -

1

I

i

I

l j

1 �"

- t '"

2

- t*

C *2 T

]

211

V0 1 1

lV," 1 7 VD,, ] -

V2 n

-

t r VO n

( 1 3 . 47)

O - t I VOIl

Equation ( 1 3.47) reveals buses CD , W, and neutral node @ as the final independent nodes, while the nonphysical load center of node @ is the new

13.5

CLASSICAL ECONOMIC DISPATCH WITH LOSSES

555

reference. Accordingly, the row-column labels of C are those shown in Eq. (13.45), while the rows and columns of the square matrix [ C T R busC* ] of Eq. ( 1 3 .32) can be likewise labeled s e q u e n t i ally with the node numbers CD , G) , and @ since load center D is n ow the reference node. CLASSICAL ECONOMIC DISPATCH WITH LO SSES 13.5

When system loss is neglected , the transmission network is equivalent to a single node to wh ich a l l g e n e r a t ion and load is connected. The penalty factor for each p l a n t t h e n becomes u n i ty a n d the system A is given in the same literal form as Eq. ( 1 3.6). When transmission loss is included, however, the economic d ispatch strategy h as to be iteratively determined by solving the nonlinear coordination equations represented by Eq. ( 1 3 . 1 5 ), which can be written in the form uP' elf; -- - A + A -- = 0 d?g ;

iJP .

( 1 3 .48)

XI

We consider each generat ing unit of the system to have a second-order fuel-cost characteristic in the form of Eq. ( 1 3. 1 ) and a linear incremental fuel cost g ive n by Eq. ( 1 3 .2). The partial derivative term of Eq. ( 1 3 .48), called the incremental loss , is a measure of the sensitivity of the system losses to an incre me n t al change in the output of plant i when al1 other plant outputs are kept fixed. For example, i n a system of two plants the incremental loss of Unit 1 is found from the loss expression of Eq. ( 1 3 .39), which gives ( 13 .49)

Setting i equal to 1 in Eq. (13.48), and then substituting for dfl /dPg l fro m ( 1 3.2) and for 3P[j3Pg ] from Eq. ( 1 3 .49), we obtain

Eq.

( 1 3 .50)

Collecting terms involving Pg I and dividing the resultant equation by A give

( 1 3 .5 1 )

Following exactly the same p rocedu re for 3PL /aPg 2 , we obtain the analogous equation for Unit 2 ( 13 .52 )

556

CHAPTER 13

ECO N O M I C OPERATION OF POWER SYSTEMS

Equations (13.5 1 ) and (13.52) can be rearranged into vector-matrix form

- -b l b2 ( 1 - B 20) - -

( l - B IO) P� 2

A

( 13 .5 3 )

A

and this is the set o f equations for U n its 1 and 2. When t h e system has K sources as in Eq. ( 1 3 .42), the partial d e rivat ive of PI_ with respect to P'r;l gives the genera l formu l a for U n i t i

( :i

-I-

2 EII

)

PI/ I

-I-

K

I:

)= 1 j "" i

2 B1 j Px I =

(1

. _-

B io )

- bAI

Letting i range from 1 t o K, we obtain a system o f l inear sources, which takes on the form of Eq. ( 1 3.53); that is,

( :1

+

2B n

2 B2 1

)

(

2 B ]2 Q2 ).,

+

2 E 22

2 BK 2

2 BK I

)

( QK

A

2BIK

Pg I

2 En

Pg 2

+

2 BKK

)

( 1 3 .54 )

e Cj u a tiolls for a l l K

Pg K

(1 - B

-(1 - B ) - ) 10

20

(1

bl

A b2 A

- BK O ) - -A bK

( 1 3 .55 )

Substituting in Eq. ( 1 3 . 1 1 ) for PL from Eq. ( 13.42), we obtain

which is the power-balance requiremen t for the system as a whole in terms of B-coefficients, the p lant loadings, and the total load. The economic dispatch strategy consists of solving the K equations represented by Eq. (13.55) for those values of power output which also satisfy the power loss and load requirements of Eq. (13.56). There are many different ways to solve Eqs. (13.55) and ( 1 3 .56) for the unknowns Pg 1 > Pg 2 , . . . , Pg K and A . When an initial value of A is chosen in Eq. 0 3 . 5 5) the resulting set of equations becomes l inear. The vaJues o f Pg j , Pg 2 , . . . , Pg K can then be found by any number of solution techniques, such

,

13.5

CLASSICAL ECONOMIC DISPATCH WITH LOSSES

557

as i nverting the coefficient matrix, within the following iterative procedure: Step 1

Specify the system load level PD Step 2

= '£7= 1 Pd j •

For the first iteration, choose initial values for the system A . [One way to make this initializing choice is to assume that losses are zero and calculate i nitial values of A from Eq. (13 .6).] Step 3

Substitute the value of A into Eq. ( 1 3 .55) and solve the resultant system of l inear simultaneous equations for the values of Pg; by some efficient means. S tep 4

Compute the t ransmission loss of

Eq.

Step 5

( 1 3 .42) using values of

Pg ;

from Step 3.

Compare the quant i ty (L� I Pg; - PL ) with PD to check the power balance of Eq. ( 1 3 .56). If power balance is not achieved within a specified tol erance, update the system A by setting

( 1 3 . 57)

k One possible formula for the i ncrement /::d( ) is !1 A(k)

=

)..( k ) _ A( k - l )

" p( � ) - " p(k - l )

--,__ --:-:___----_

K

i..J

;= 1

gl

K

�

i == l

81

[

p(k ) pD + p(Lk ) - '"' L- g i

K 1

i� 1

( 13 .58)

I n Eqs. ( 1 3.57) and ( 1 3 .58) s up e rs c r i p t ( k + 1 ) i n d icates the next iteration being s tarted , s u p e r s c r i p t ( k ) ind icates the i teration just completed, and (k - 1) denotes the immediately preceding iteration.

Step 6

Return to Step 3 and continue the calculations of Steps 3, 4, and 5 u ntil final convergence is achieved.

The final resul ts from the above p rocedure determine both the system A and the economic dispatch outp ut of each generating unit for the specified level of system load. It is i nteresting to note that duri ng each iteration of the overall sol ution Step 3 provides an eco nomic dispatch answer, which is correct · at o ne load l evel even though it may not be the specified load level of the system. ,

558

CHAPTER 1 3

ECONOMIC OPERATION OF POWER SYSTEMS

Example 13.4. The generating uni ts at buses

CD and 0 of the system of Example

13.3

13.1.

have i ncremental fuel costs g iven in Example Calculate the economic load ing o f each unit to meet a total customer load of 500 MW. What is the syste m A a n d w h a t is t h e transmission loss of the system? Determine the penalty factor for

each unit and the incremental fuel cost at each generating bus. The ge n e ra t i n g u n i ts nt t he two d i lTere n t fuel costs i n dol lars per megawat tllOuf given by

Solution.

d!J dPg 1

0 .0080P!: 1

-- =

+

d!2

8.0;

--

dPg 2

=

power p l a n t s h ave i n c re m e n t a l

0 .009 6 P!,' �- +

6

A

B1B20O//22 1 [ 8.383183 5.963568 0.375082 1 10 - 3 0. 1 94971 u.090121 Boo

and P!,' 2 are e x p ressed i n megawa tts. At t h e spec i fi e d load l e v e l o f 500 MW t h e loss c o e fli c i e n t s i n p e r u n i t o n a l aO-MY A base arc g i v e n i n Exa m p l e 1 .3 as

w h ere Pg 1

3

- 0 .049448

- 0 . 049448

=

0 . 1 9497 1

0 . 375082

X

To begin the solution, w e must estimate i n i tial values for A for t h e first i teration. The results of Example 1 3. 1 at the 500-MW load l evel can be u sed for t h is purpose: S tep 1

We are given

PD = 5 .

0

per u n i t on a 100-MVA base.

Step 2

From Example 1 3 . 1 we choose A( I )

=

9.454545.

Step 3

Based on th e estimated value of A( I l , we the e q u a t i o n

0.8

i\( l )

+

2 X 8 .383 1 83 X 1 0 - 3

ou tpu ts

Pg 1 a n d Pg 2 from

3 -2 1 0 3 0.96 2 6 6 10 - 3 ( 1 -0.750164 1 0 3 10- 3)

- 2 X 0 . 049448

X 10-

X

=

c a lcu l a t e t h e

( 1 - . 8 994 2

X

)Jl)

0- 3)

X

+

8.0 A( I ) 6 .4 - (1) A

0 . 049448

X

X 5 .9 35 8

X

Note that per u n i t Q 1 and Q 2 are used in this calculation because al l o t h er

CLASSICAL ECONOMIC DISPATCH WITH LOSSES

1 3.5

559

quantities are in per unit. Owing to the simplicity of the example, this equat ion Pg 1 and Pg can be solved directly to yield results for the first iteration

2

PiP = 1 .512870 per unit;

for

Pi�) = 2 .845238 per unit

Step 4

From the results of Step 3 a n d the given values of the B-coefficients the system power loss is calculated as follows:

=

B I I ( 1 .5 1 2R70) 2

+

2 B 1 2 ( 1 .5 1 2870 ) ( 2 .845238 )

-:- B22 ( 2 . 84 5 238/ =

Step

5

0 .069373 per u n i t

+

Po Pi�» )

Checking the power balance for

Po Pl.} ) +

-

( pil)

+

B I O ( 1 .5 1 2870)

+

B 2 0 ( 2 .84523 8)

+

B oo

5 .00 per unit, we find

=

=

5 .069373 - 4 . 35 8 1 08

=

0 .7 1 1265

w h ich exceeds E = 1 0 - 6 , and so a new value of ,\ must be supplied. The incremental change in ,\ is calculated from Eq. ( 1 3.58) as follows:

p(l») ( P(l) g2 Xl

( .t Pi!)] - (.t Pi?) ]

pD

+ p(LI )

_

+

1= I

S i n c e t h i s is t he fi rs t i t e r a t i o n , ,.\( 0) a n d g ives

A� ( I )

and [he

=

( 9 .454545 - 0)

A(2) A(I)

[

r} p;;» I

1= 1

are

both set equal to zero, which

0 . 7 1 1265 4 .�5H I OH - 0

1

u p d a t e u ,\ t h e n b e c o m e s =

+

� A( I )

Step 6

=

9 .454545

+

1 . 543035

=

=

1 .543035

10 99758

We now return to S tep 3 to repeat the above calculations using ,\( 2 ) during the

second iteration, and so on.

560

CHAPTER 13

ECONOMIC OPERATION OF POWER SYSTEMS

The final converged solution for both system A and the economic loading of the two generating units is found to be A = 9 .839863 $/MWh

Pg 1 =

1 90 .2204 M W

Pg2 = 3 1 9 . 1 0 1 5 M W

For illustrative purposes a convergence criterion of £ example, but such accuracy is not warranted in p ractice.

=

1 0 - 6 w a s u s e d i n this

The transmission loss compu Lcd from the solved va lues of PI: I and PK 2 is 9.32 1 9 1 4 MW at Step 4 of the final iterat ion, and so the tot al generation by the two plants a mounts to 509.32 MW for load a n d losses. The incremental losses of the two

pla nts are

=

2(8.3R3 1 83 X 1 .902204 . . 0 . 04 9448 X 3 . 1 9 1 0 1 5

+

= 0 . 032328

= 2(5 .963568 X 3 . 1 9 1 0 1 5 - 0 .049448 X 1 .902204 = 0.038261

+

0 .375083) X 10

.�

0 . 1 9497 1 ) X 1 0 - 3

and so the penalty factors are given by Lj

=

1

_

1 0 .032328

=

1 .0334 1 ;

L

2

=

1 "1 . 0 .038261

--

.

=

1 .03978

.

The i ncremental fuel costs at the two plant buses are calculated to be

bl

djl dPg 1

=

Q1Pg1

dj2 dPg 2

=

Q2Pg 2 b2 = 0 .96( 3 . 19 1 0 1 5 ) +

+

+

=

0 .80( 1 .902204)

+

8 .0

6 .4

=

=

9 .521763 $/MWh 9 .463374 $/MWh

In this example plant 2 has the lower incremental fue l cost at its bus and carries the b igger share of the 500-MW load. The reader can confirm that the effective incremental cost of supplying the system load (often called the incremental cost

1 3.5

CLASsicAL ECONOMIC DISPATCH WITH LOSSES

of power delivered) checks with the calculations L 1( dfl /dPg l ) 9.839863 $/MWh.

=

561

Lidf2/dPg2) = .

provides

We noted earlier that Step 3 of each iteration of the above procedure valid answers for economic 19ading of the u nits. Those answers are correct at the particular load l evel which gives power bal ance within that iteration. For instance, at Step 3 of the first iteration in Example 1 3.4 the system A is 9.454545 $ /MWh and the generator outputs are computed to be Pi�) 1 5 1 .287 MW and p; �) 284.5238 MW. At Step 4 of the same iteration the corresponding value of pil) is found to be 6.9373 MW. Therefore, a system load PD given by =

=

( p� : ) + Pi i) ) - PLI )

=

wou l d satisfy the power balance in the fol l ow i n g example.

( 4 35 .8 1 08 - 6.9 373 )

=

428 .8735 MW

of the system. We make use of this observation

Calcu late the decrease in production costs o f the two plants of Exa mple 1 3.4 when the system load is red uced from 500 to 429 MW. Example 1 3 . 5 .

Solution. The economic loading of the two plants for a system load level of 500 MW is found in Example 1 3 .4 to be P/: I = 1 90.2 MW and Pg 2 = 3 1 9. 1 MW. In the first i t e r a t i o n of the same exam ple we a lso find that the plant outputs PC I 1 5 1 .3 MW and Pg ? 284.5 MW ensure economic dispatch of the units when the load level is essentially 429 M W . These results are sufficiently accurate for calculating the production-cost red uction between the two load levels as follows: =

=

�fl

=

f 151 . 3 ( 0.0080 Pg I 1 90 . 2

"" ( O .0040P/:-., 1

- f2S4. 5 (

t:.h --

=

:'1 1 '1 . 1

+

+ R . O P!: I

O . O O('6 :J Pg2

8 .0 ) dPg I

1 15 1 .3 ) + CI

1 <)( ) . 2

=

+ 6 .4 ) II)K l

- 364 .34 $ / h

(

) 1284.5 2 ( O . 0048 P!: 2 + 6 .4 Pfi 2 + C2 ."\19.1

=

- 321 .69 $/h

Thus, the total reduction in syst em fu el cost amou nts to $686 per hour.

Procedures have now been developed for coordinating transmission loss of the system into the economic dispatch of those u nits which are already on l ine. In Sec. 13.6 we consider automatic generation control before i nvestigating the unit commitment problem which determines the u nits to be connected on l ine in the first place.

562 13.6

CHAPTER 13

ECONOMIC OPERATIO N OF POWER SYSTEMS

AUTOMATIC GENERATION C ONTROL

Almost all generating companies have tie-line interconnections to n eighboring utilities. Tie lines allow the s haring of generation resources i n emergencies and economies of power production under normal condi:ions o f operation. For purposes of control the entire interconnected system is subdivided i n to con trol areas which usually conform to the boundaries of one or more compa n i es. The net interchange of power over the tie lines of an area is the algebraic d i ffe rence between area generation and area load (plus losses). A schedule is prearranged with n eighboring areas for such tie-line flows, and as long as an area ma�ntains the interchange power on schedule, it is evidently fulfilling i ts primary respons i ­ bility to absorb its own load changes. B u t since each area shares in t h e benefits of interconnected operation, i t is e X D e c t e d al so to share in t he resD o n s i b i l i t y to maintain system frequency. Frequency changes occur because system load varies ran domly throughout the day so that an exact forecast of real power demand can not be assured . The imbalance between real power generation and load demand (plus l osses) throughout the daily load cycle causes kinetic e n ergy of rotation to be either added t o or taken from the on-line generating u n its, and frequency throughout the interconnected system varies as a result. Each cont rol area has a central facility called the energy control center , which m on i tors the system frequency and the actual power flows on i ts tie l i nes to n eighboring areas. The dev iation between desired and actual system frequency is then combined with the devia­ tion from the scheduled net interchange to form a composite measure called the area control error , or simply ACE. To remove a rea control error, the energy control center sends command signals to the generating units at the power plants within its area to control the generator out p u ts so as to restore the net interchange power to scheduled values a n d to assist in restoring the system frequency to its desired value. The monitoring, tel emetering, processing, and control functions are coord i n a t e d w i t h i n t h e individual a r e a b y t h e c o m p u ter­ based automatic generation co ntrol (AGC) sy s t e m at th e energy control center. To help u nderstand the con t rol actions at the power plants, l e t us fi rst consider the boiler-turbine-generator combination of a thermal generating u n it. Most steam turbogenerators (and also hydroturb ines) n o w i n service are equipped with turbine speed governors. The function of the speed governor is to monitor continuously the turbine-generator speed and to control the th rottle valves which adj ust steam flow into the turbine (or the gate position in hydroturbines) in response to changes i n " system speed" or frequency. We use speed and frequency interchangeably since they d escribe proportional quanti­ ties. To permit parallel operation of generating u ni ts, the speed-versus-power output governing characteristic of each u nit has d roop, which means that a decrease i n speed should accompany an increase in load, as depicted by the straight line of Fig. 13.7(a). The per-unit droop or speed regulation Ru of the generating u nit is defined as the magnitude of t h e change in steady-state speed, expressed in per unit of rated speed, w hen the o u tput of the unit is gradually reduced from 1 . 00 per-unit rated power to zero. Thus, per-unit regulation is

1 3.6

AUTOMATIC GENERA TION CONTROL

563

_____ No load

(2 N

I >­ u C III :::J CT �

LL

Rated out p ut

- --- --------- - -- - -- ---- ---------

� o

/

Power out p ut, MW

(a)

- - - - _/ - --

After s u pp lementary control

i

N

IT

I II �

- -- - ---

fo - - ---ij r--

(

LL

--- -

-

--

�(- - ( : -,-I n itial

Final

--� �

----

I

N - W

I I I

Pg o :

FIGURE 13.7 ( a ) Speed-governing character­

•

o

istic of a generating u nit; (b) before/after load increase Il Pg and suppleme ntary control.

Power out p ut, MW

(6)

simply the magnitude of the slope of the speed-versus-power output characteris­ tic when the frequency axis and the power-output axis are each scaled in per u n i t of their respective rated values. From Fig . 1 3 .7( a ) it fol l ows that pe r - u n i t r e g u l a t io n i s given by R"

w h e re

12 It II< S R

=

=

=

=

=

( /2 - I t ) IIR ----- per P" I
l oad fre q u e n cy ( i n H z) a t r a t e d megawat t ra t e d fr e q u e n cy ( i n H z) o f t h e u n i t megawatt base fre q u e n cy ( i n Hz) a t n o

o u t p u t P" R

Multiplying each side of Eq. ( 1 3 .59) by IRISR gives R

=

IR

R li S

R

( 1 3 . 59)

u nit

,.

( 1 3 .60)

564

CHAPTER 1 3

ECON OMIC OPERATION OF POWER SYSTEMS

where R is the magnitude of the slope of the speed-droop characteristic (in Hz/MW). S uppose that t h e u n i t i s supplying output power Pg Q at frequency fa when the load is increased to Pg = Pg O + !1 Pg , as shown i n Fig. 1 3.7(b). As the speed of the unit decreases, the speed governor a llows more steam from the boiler (or water from the gates) t hrough to the turbine to arrest the decline in speed. Equilibrium between input and output power occurs at the new fre­ quency f = ( fo + !1 f ) as shown. According to the slope of , the speed-output characteristic given by Eq. ( 1 3 .60), the frequency change (in Hz) is ( 1 3 .6 1 )

The isolated unit o f Fig. 13.7 would con tinue to ope r a t e a t t h e r e d u c e d frequency f except for the supplementary control a c t i o n o f th e speed changer . The speed control mechanism has a speed changer motor which can parallel-shift the regulation characteristic to the new position shown by the dashed line of Fig. 1 3 .7(b). Effectively, the speed c h a n g e r su p p l e m e n t s t he a c t i o n of the governor b y changing the speed setting to allow more prime-mover energy through to i ncrease the kinetic energy of the generating unit so that it can again o perate at the desired frequency fa while providing the new output Pg . When K generating units are operating i n parallel on the system, their speed-droop characteristics determine how load changes are apportioned among them in the steady state. Consider that the K u nits are synchronously operating at a given frequency when the load changes by !1 P megawatts. Because the units are interconnected by the transmission networks, they are required to operate at speeds corresponding to a common frequency. Accordingly, in the steady-state equilibrium after initial governor action all units will have changed in frequency by the same i ncremental amount !1 f hertz. The corresponding changes in the outputs of the units are given by Eq. ( 13.61) as fol lows: ( 13 .62)

Unit 1 : !1 Pg 1

Unit i :

!1 Pg l.

=

-

- MW R i u fR SR ' !1 f

_ '

( 13 .63)

n 3 .6 4 )

1 3 .6

AUTOMA TIC GENERA TION CONTROL

565

Adding these equations together gives the total change i n output ( 1 3 .65)

from which the system frequency change is

�f

=

(

.

1

--;--::--- --------------:- per uni t R 5R_i + . . + 5 R K 5 L_ + . . . + _ R1u Rill R Ku

( 1 3 .66)

Substituting from Eq. (1 3.66 ) i nto Eq. ( 1 3 . 63), we find the additional output � Pg i of Unit i MW

( 1 3 .67)

which combines with the additional outputs of the other units to satisfy the load change 6. P of the system. The u nits would continue to operate i n synchronism a t the new system frequency except for the supplementary control exercised by the AGC system at the energy control center of the area in w hich the load change occurs. Raise or lower signals are sent to some or all the speed chan gers at the power plants of the particular area. Through such coordinated control of the set points of the speed governors it is possible to bring all the u n i ts of the system back to the desired frequency fa and to achieve any desired load d iv ision within the capabilities of the generating u n its. Therefore, the governors on u nits of the interconnected system tend to maintain load-generation balance rather than a specific speed and the supple­ mentary cont rol of the AGC system within the individual control area functions so as to: •

• • •

Cause the area to absorb its own load changes, Provide the prearranged net interchange with neighbors, Ensure the desired economic dispatch output of each area p lant, a n d Allow the area to do its share t o maintain the desired system frequency_

The ACE is continuously recorded within the energy control center to s how how well the individual area is accomplishing these tasks. The block diagram of Fig. 1 3 .8 indicates the flow of i nformation in a computer controlling a ' particular area. The numbers enclosed by circles adja­ cent to the diagram identify positions on the diagram to simplify our d iscussion of the control operation. The larger circles on the diagram e nclosing the

566

CHAPTER 13

ECO N O M I C OPERATION OF POWER SYSTEMS

B-Coefficients

Tie-line monitoring

Sc.l,eduled net intercha nge Ps

i n te rch a ng e

® -

L

Actual n e t

Pa

+

L

10 B r tJ. r

+

@

-

Area control error

Station

CD

(2)

L

+

Error s i gnal

Tie·line

Tie-line information processi ng

Fr eq u e n c y b i a s setting

® AC E

Non confo r m i n g loads

X

lOB

(

t. r

® Adjust

l o a d i n g d a ta

Fixed · p l a n t g e n e ra tion

@

Compute penalty

-=-

factors

S c h e d u l ed freq u e ncy -

@

L

r.

+

fa

Ac t u a l f r e q u e nc y

Area "

area " -

I nd i v i d u a l

p i an t

ge n e ra t i o n

®

d e s i red

Compute assigned

g e n e ra tion

r----

control error SCE

®

L

-

Total d e s ired p l a n t g e n e ration

+

@

Cont rol sig n a l s to p l a n t s

Tota l actual generation

FIGURE 13.8 Block d iagram il lu s t rat i ng t he co mp ute r co n t ro l l e d opera t ion of a particular area. ·

symbols X or L i ndicate points of m ulti plication or algebraic summation of i ncoming signals. At position 1 processing of informa tion about power flow on tie lines to other control areas is indicated. The actual net i nterchange Pa is positive when net power is out of the area. The scheduled net i nterchange is Ps' At position 2 the scheduled net i nterchange is subtracted from the actual net interchange.3 We shall discuss the con dition where both actual and scheduled net i nterchange are out of the system and therefore posit ive. Position 3 o n the diagram indicates the subtraction of the scheduled frequency fs (for instance, 60 Hz) from the actual freque ncy fa to obtain I1 f,

3

Subtraction of standard or reference value from actual value to obtain the error is the accepted convention of power system engineers and is the n egative of the definition of control error fpund in the literature of control theory.

13.6

AUTOMATIC GENERATION CONTROL

567

the frequency deviation. Position 4 on the diagram i ndicates that the frequency bias setting Bf, a factor with a negative sign and the units MW 10. 1 Hz, is multiplied by 1 0 11/ to obtain a value of megawatts called the frequency bias 00 Bf 11 /). The frequency bias, which is positive when the actual frequency is l ess than the scheduled frequency, is subtracted from ( Pa - P) at position 5 to obtain the ACE, which may be positive or negative. As an equation ( 1 3 .68)

negative ACE means that the area is not generating enough power to send the desired amount out o f the area. There is a deficien cy in net power output. Without frequency bias, the indicated deficiency would be less because t here would be no positive offset 0 0 Bf I1. f) added to P.v (subtracted from P) when actual frequency is less than scheduled frequency and the ACE would be less. The area would prod uce sufficient generation to supply its own load and the prearranged in terchange but would not provide the add itional output to assist neighboring interconnected areas to raise the frequency. Sta tion control error (S CE) is the amount of actual generat ion of all the area p lants minus the d esired generation, as indicated at position 6 of the diagram . This SCE is negative when desired generation is greater than e xisting generation. The key to the whole control operation is the comparison of ACE and SCE. Their difference is an error signal, as indicated at position 7 of the diagram. If both ACE and S CE are negative and equal, the deficiency i n the output from the area equals the excess of the desired generation over the actual generation and no error signal is produced. However, this excess of desired generation will cause a signal, indicated at position 1 1 , to go to the plants to increase their generation and to reduce the magnitude of the SCE; the resulting in crease in output from the area will reduce the magnitude of the ACE at the same time. If ACE i s more negative t han SCE, there will be an error signal to increase the A of the area, and this increase will in turn cause the desired plant generation to increase ( position 9). E a c h plant will receive a signal to increase i t s output as determined by t h e principles of economic d ispatch. This d i s c u s s i o n h a s c on s i d e red spe c i ri ca l l y o n l y the case of scheduled net in terchange out -of the area (positive scheduled net interchange) that is greater t h a n actual net in terchange with A C E equal to or more negative than S e E . The reader should be able to extend the discussion to the other possibilities by referring to Fig. 1 3 .8. Position 1 0 on the diagram indicates the computation of penalty factors for each plant. Here the B-coefficients are stored to calculate aPLlaPg; a n d the penalty factors. The penalty factors are transmitted to the sect ion (position 9), which establishes the individual plant outputs for economic dispatch a n d the total desired plant generation. A

S68

CHAPTER 13

ECONOMIC OPERATION OF POWER SYSTEMS

One other point of importance (not indicated on Fig, 1 3.8) is the offset i n scheduled n e t interchange o f power that varies in proportion to the time error , which is the integral of the per-unit frequency error over time in seconds. The offset is in the direction to help in reducing the integrated difference to zero and thereby to keep electric clocks accurate. Example 1 3.6. Two thermal generating u n its arc operating in paral lel at 60 Hz to supply a total load of 700 M W. Un it 1 , with a rated output of 600 MW and 4% speed-droop characteristic, supplies 400 M W and Unit 2, which has a rated output of 5 00 MW and 5 % speed droop, suppl ies the remaining 300 MW of load. I f the total load increases to 800 M W d e t e r m ine the new loading of each u n i t a n d t h e common freq u e nc y change before a n y supplemelltLiry control a c t i o n occu rs. Neg l ect ,

,

losses.

The initial poin t a of operation on t h e speed regulation characteristic of each unit is shown in Fig. 13.9. For the load i ncrease of 1 00 MW, Eq. ( 1 3 .6fi) gives t he per-unit frequency deviation

Solution.

- 1 00 500 600 + 0 . 04 0 . 05

- 0 .004 per unit

--

Since fR e q u a l s 60 Hz, the frequency change is 0.24 H z and the new frequency of operation i s 59.76 Hz. The l o ad change allocated to each unit is given by Eq. ( 1 3.67) 6. Pg 1 = 6.

P}; 2 =

600/0.04 500 1 00 600 + 0 .05 0 . 04

=

60 MW

--

500/0.05 500 1 00 600 + 0 . 04 0 .05

--

--

=

40 MW

and so Unit 1 supplies 460 MW, whereas Unit 2 supplies 340 MW at the n ew o pe r a t ing points b shown in Fig. 13.9. If supplementary control were appl ied to U ni t 1 alone, the entire 1 00-MW load i ncrease cou ld be absorbed by that u nit by shifting i t s characteristic to t h e final 60-Hz posi t ion at poin t c of Fig. 1 3.9. Unit 2 would then automatically return to i ts original operating point to supply 3 00 MW a t 60 Hz.

The large number of generators and governors within a control area combine to yield an aggregate governing speed-power characteristic for the area as a whole. For relatively small load changes this area characteristic is often assumed linear and then treated l ike that of a single unit of capacity equa l to t hat of the prevailing on-line generation in the area. On this basis, the fo1l0wing

61

N

I >. (,) c Q)

:::J 0Q)

U::

60 59.76 > c:

d s: � ()

59

200

400

460 500

Power output, MW

600

200

300 340

400

500

600

Power o utput. MW

F1GU RE 13.9

Load division between two isolated u nits of different speed-droop c ha r a c t e ristics. Points a show initial distribution of 700-MW load; points b show distribution of 800-MW load at 59.76 Hz, and points c show final operating points of the u nits after supplementary control of Unit 1 .

o tTl Z tTl

�

::l

o z () o z .., ;r:I o r

570

CHAPTER 1 3 ECONOM IC OPERATION OF POWER S YSTEMS

example demonstrates the steady-state operation of AGe for a three-area system in which losses are neglected.

Example 13.7. Three control areas with autonomous AGe systems comprise the interconnected 60-Hz system of Fig. 1 3 . 1 0 ( a). The aggregate speed-d roop characteristics and on-line generating capacities of the areas are Area A : Area

B:

Area

C:

RAu

R llu

R cu

= = =

SU A

0 .0200 per unit ; 0 .0 1 25

per uni t ;

5,Uj

Suc

0 . 0 1 00 pe r unit;

=

=

=

1 6 ,000 M W 1 2 ,000 M W 6,400 MW

Each a r e a has a load level eq\lal to HO'/r of its [ ,I ( e d on- l i n e C i l fl i l c i t y . r:O [ [ c ,\ s o n s of economy area C i s importing 500 M W of i t s load req u i re m e n t s from arca B , a n d 1 00 MW of this i n t e rchange paS!lCS over t h c tic l i nes o f '.\f e a / 1 . which h ,I S a zero scheduled interchange of its own. Determine the s ys t e m frequency devi a tion and the generation changes of each area when a fu lly loaded 400-MW generator is forced out of service in area B . Th e area fre q u e nc y bias se t t ings arc BfA =

BfB =

E!C' =

- 1 200 M W /0 . 1 Hz -

] 5 00 M W /0.1 Hz

-

950 MW /0.1 Hz

Determine the ACE of each area before AGC action begins. The loss of t h e 400- M W u n i t is sensed by t h e other on-line generators as an increase in load, and so the system frequency decreases to a value dete rmined by Eq. ( 13.66) as

Solution.

1 6000 0.0200

+

- 400 1 2000 0 .0 1 25

+

- 10- 3 6

6400 --.0 .0 1 00

per u n i t

Therefore, the frequency decrease i n 0.01 Hz after initial governor action a n d the generators still on-line increase their outputs according to Eq. ( 1 3.63); that is, l::. PgA =

101 6000 X 0 . 0200 6

tJ. PgB =

10 ·3 1 2000 X 0 .0 1 25 6

=

tJ. Pg c =

6400 0 .0100

10-3 6

=

3

--

--

X

--

=

133 MW 160 MW 107 MW

160 MW

10;100 MW 154 MW

1 00 MW

9,600 MW

12,800 MW 60 Hz

Area A

�

1 00 MW

�OO MW

Area B

• •L

Area A

-

59.9 Hz

,

Area B

21 MW ;I> c: -l o � ;I>

(6 )

FIGURE 13. 1 0 ( a ) Normal 60-Hz operation o f three-area sys tem of Exa m p l e 1 3 .9; ( b ) t h loss o f 400-MW u nit in area B before A G e acti

j

n Cl tTl Z tTl : -l

o

z n o

�:
o r

572

CHAPTER

13

ECONOMIC OPERATION

OF

POWER SYSTEMS

Let us suppose that these incremental changes are distributed to the interarea tie lines, as shown in Fig. 1 3 . l O( b). Then, by inspection, the area control error for each area may be written ( A CE ) A

=

( A CE) /I =

(A C E ) (

'

=

( 1 33 - 0) - lO( - 1 200 ) ( - 0 . 0 1 )

=

( 2 60 - 500 ) - I O( -- 1 500) ( - 0 .0 1 ) [ - 3 ()]

-

( - 5 0 ( ) )]

-

I () (

- l) 5 () ) ( - (l . () I )

13

- 390 =

MW MW

1 2 MW

Ideally, the A CE i n areas A and C would be zero. The pred o m i n a t i n g A C E is area B where the 400-M W forced outage occu rred. The A G C system o f area will com m a n d t he o n - l i n e pow e r p l ; l n t s u n d e r i t s c o n t r o l t o i n c r e a s e ge n e r a t i o n o f fs e t t h e l o s s or t h e

A and C

t hen

4()()-MW u n i t

in B to

; 1 1 1 ( \ t o r e s t or c sys t e m fre q u c nc y of h() l I z . A r c a s

re t u rn t o t h e i r origi n a l co n d i t ions.

The frequency error in per unit equals the time error in seconds for every second over which the frequency error persists. Thus, if the per-unit frequency error o f ( - 1 0 - 3/ 6 ) were to last for 1 0 min, then the system time (as given by an electric clock) would be 0 . 1 s slower than independent standard time. 13.7

UNIT COMMITMENT

B ecause the total load of the power system varies throughout the day and reaches a different peak value from one d ay to another, the electric utility has to decide in advance which generators to start up and when to connect them to the network-and t h e sequence in which the operating units should be shut down and for how long. The computational procedure for making such decisions is called unit commitment , and a unit when scheduled for connection to the system is said to be committed. Here we consider the commitment of fossil-fuel units which have d ifferent production costs because of t heir dissimilar efficiencies, d esigns, and fuel types. Although there are many other factors of practical significance which determine when units a re . scheduled on and off to satisfy the operating needs of the system, economics of operation is of major importance. Unlike on-line economic dispatch which economica l ly distributes the actual system load as i t arises to the various units already on-line, unit commitment plans for t h e best set o f units to be available to supply the predicted or forecast load of the system over a future time period. To develop the concept of unit commitment, we consider the p roblem of scheduling fossil-fired thermal units in which the aggregate costs (such as start-up costs, operat ing fuel costs, and shut-down costs) are to be minimized over a daily load cycle. The underlying principles are more easily explained if we d isregard transmissirm loss in the system. Without losses, the transmission n etwork is equivalent to a single plant bus to which all generators and all loads are connected, and the total plant output Pg T then equals the total system load PD' We subdivide the 24-h d ay into discrete intervals or stages and the predicted

13.7 UNIT COMMITMENT

573

2000 1800

� 1400 � 1 600

TI co 0

1200 1000

E

800


en

600

>(J)

400 200 0 0

8

4

1

I

12 Time, h

16

4 3 Stage number

2

20

24

I I

6

5

FI G C RE 1 3 . 1 1 Discrete le v e l s of sys tem load for an exa m p l e d a i ly load cycle.

load of the system will be considered constant over each interval, as exemplified in Fig. 1 3 . 1 1 . The unit commitment procedure then searches for the most economic feasible combination of generating units to serve the forecast load of the system at each stage of the load cycle. The power system with K generating u nits (no two identical) must h ave at l east one u n i t on-line to supply the system load which is never zero ove r the daily load cycle. If each unit can be considered either on (denoted by 1) o r off (denoted by 0), there are 2 K 1 candidate combinations to be examin ed in each stage of the study period. For example, i f K 4, the 1 5 theoretically possibl e combinations for each interval are -

=

C om b i n a t ions

Unit

2

3

4

XI

Xz

I 1

x)

X ot

Xs

X6

X1

Xs

1

0

1

0

0 1 0

0 0

1

0 0

0

1

0

1

X9

X IO

I

0

XII

0 0 0

1

0

I

1

0

XIZ

Xu

X I4

XIS

1

0 1 0 0

0 0

0 0 0 1

0 0 0

1

0

where Xi denotes combination i of the four units. Of course, not all combina­ tions are feasible because of the constraints imposed b y the load level and other practical operating requirements of the system. For example, a combination of u ni ts of total capability less t h a n 1400 MW cannot serve a load of 1400 MW or greater; any such combinatio n is infeasible and can be disregarded over any time ,

574

CHAPTER

13

ECONOMIC OPERATION OF POWER SYSTEMS

in which that level of load occurs. To assist the mathematical formula­ o f the unit commitment problem, let us set

interval tion

xiC k )

=

combination

xi

( 1 3 . 69)

of interval k

and then x /k + 1) represents combi nation Xj of interval (k + 1). If k equals 1 and i equals 9 in the four-unit example, the combination X li } ) means that only Units 1 and 2 are on-l ine during the first time i nte rva l . The prod uction cost incu rred i n supplying powe r over any interval of the daily load cycle depends on wh ich combination of u n i t s i s on-line d u ring that interval . For a given combination X i the minimum production cos t Pi equals the sum of t h e eco n o m i c d is p a t c h cos t s o r t h e i n d i v i d l l
=

min imum production cost

of

combination x , ( k )

( 1 3 .70)

and then P/ k + 1 ) is the minimum production cost of combination x/ k + 1). Besides production cost, in the unit commitment p roblem we milst also consider transition cost , wh ich is t h e cost associ a t e d w i t h ch a n g i n g fro m o n e cornb i n a tion of power-producing un its to another combination. Usual ly, a fixed cost is assigned to shutting down a unit wh ich has been operating on the system since shut-down cost is generally independent of the length of time the u n i t h as been running. However, in practical situations the start-up cost of a unit depends on how l ong the unit has been shut down from previous operation . This is to be expected since the boiler temperature of the unit and the . fuel req u i red to restore operating temperature depend on the duration of cooling. To explain more easily the unit commitment concept, we assu me a fixed start-up cost for each unit and refer the reader to other p ractical considerations treated else­ where. 4 Thus, t ransition cost associated with changing from one combi nation of operating units to another will have fixed sta rt-up and shu t-down com po ne nts denoted by

Tj/k ) = cost of transition from com b i n a ti on x J k ) to c o m b i n a t i on x/ k b etween i n terv a l s

k and

k+1

+

1)

( 13 . 71 )

If each u n i t could be started up or shut down w ithout i ncurring any transition

in a ny o n e hour wou l d b ecome d isj o i n t from and to tally u n r e la t e d to

cost, then from the economic viewpoi nt the p roblem of sched uling un its to operate

t h e sch edul ing problem in any othe r hour of the load cycle. On the other hand, suppose that i t costs $ 1500 to shut down a u n i t and $3000 for each u n i t start-up. Then, to change from combination x /k ) to combination x 3(k + 1) in the above four-unit example, the transition cost T2 3( k ) becomes ($ 1 500 + $ 3 000) = $4500 ,

4 See, for example, C. K. Pang and H. C. Chen, " Optimal Short-Term Thermal U n i t Com m i t m e n t , " IEEE Transactions o n Power Apparatus and Systems, v o l . PAS-95, no. 4 , 1 976, p p . 1 3 36- 1 346:

1 3 .7

UNIT COMMITMENT

575

because Unit 3 is to shut down and Unit 4 is to start up at the beginning of interval (k + 1). Furthermore, the status of units in interval (k + 1 ) affects the cost of transition to interv al (k + 2), and so on. Therefore, transition cost l inks the scheduling decision of any one interval to the scheduling decisions of all the other intervals of the load cycle. Accordingly, the problem of minim izing costs at one stage is tied to the combinations of units chosen for all the other stages, and we say that unit commitment is a multistage or dynamic cost-minimization problem. The dynamic nature of the u n i t commitment problem complicates its solution. Suppose that 10 units are avail able for schedu ling within any one-hour intervat which is not unlikely in practice. Then, theoretically a total of 2 10 1 1023 combinations can be listed. I f it were possible to link each prospective combination of any one hour to each p rospective combination of the next hour of the day, the total n umber of candidate combinations becomes (023) 24 = 1 .726 X 1 072 , which is enormously l arge and u nrealistic to handle. Fortunately, however, the mul tistage decision process of the unit commitment problem can be dimensionally reduced by pract ical constraints of system operations and by a search procedure based on the fol lowing obse rvations: -

=

•

•

•

The daily schedule has N d iscrete t ime intervals or stages, the durations of wh ich are not necessarily equal . Stage 1 precedes stage 2, and so on to the final stage N. A decision m ust be made for each stage k regarding which particular combination of units to operate during that stage. This is the stage k subproblem. To solve for the N decisions, N subproblems are solved sequentially in such a way (called the principle of optimality to be explained l ater) t hat the combi ned best decisions for the N subproblems yield the best overall solution for the original problem.

This strategy greatly reduces the amount of computation t o solve the original u n i t commitment problem, as we shall see. Th e cost Ji / k ) associated w i t h any stage k has two components given by ( 13 .72) w h i c h i s t h e co m b i n e d t ra n s i t i o n a n d p ro d u c t i o n cost i n c u rred by combina t i on

X i during interval k plus the transition cost to combination Xj of the next interval . For ease of explanation i t is assumed that the system load l evels at the beginning and end of the day are the same. Consequently, i t is reasonable to expect that the state of the system is the same at the beginning and at the end of the day. Because of this, when k N, the transition cost TJ N ) becomes zero. We note that the cos t fik ) is t ied by T;/ k) to the decision of the next stage (k + 1). S uppose, for the moment, that we happen to know the best policy =

,

576

CHAPTER

ECONOMIC OPERATION OF POWER SYSTEMS

13

or set of d ecisions (in the sense of m inimum cost ) over the first ( N - 1 ) stages of the d aily load cycle; this is equivalent to assuming that we know how to choose the b est combination of u nits for each of the first ( N - 1) intervals. I f w e agree t h a t combination Xi" i s the best combination for s t a ge ( N - 1 ), then b y searching among all the feasible combi n ations x j of the final stage N , we can find F; . ( N - 1)

=

(x/ N »)

min

{ Pi. ( N

- 1 ) + Ti oj ( N

-

1)

+

F, ( N ) }

( 1 3 .73)

where Fi. ( N - 1) is the mlnimUm cumula tive cost of the finaL two s t a ges starting with combi nation xi.(N - 1 ) and ending with combination x/N ); t h e cumula tive cost Fj( N ) of s t a g e N e q u a l s the p roduction cost P/ N ) s i n c e t h e re is no further transition cost involved. The notation of Eq. ( 1 3 . 73) m e a n s that the search for the minim um-cost d ecision is m ad e over a l l fe asible c o m b i n a t i o n s Xj at stage N. Of course, we do not yet k now which combination i s x i . ( N - 1 ). But for each possible starting combination x /N - 1) i t is straightforward to solve for the best stage-N decision a n d to store the correspond i n g m i n i m u m -cost results in a table for later retrieval when the best combination x i.(N - 1) has been identified. Similarly, starting with the combination x i .( N - 2) at i n t e rv a l ( N - 2), the minimum cumulat ive cost of the final three stages of the study period is given by Fi" ( N - 2)

=

(x/ N - I ))

min

{ Pi" ( N -

2)

+

Ti •j ( N - 2)

+

Pi ( .\' -

where the search is now made among the feasible co m b i n a t io n s ( N - 1). Continuing the above logic, we find the recursive fomlUla Fi. ( k )

=

min (;.) k + I »)

{ Pi o ( k )

+

Ti .j ( k )

+ Fj ( k + I ) }

I)}

( 1 3 .74 )

Xj

of stage

( 1 3 .75)

for the minimum cumu l a t ive cost a t s t age k , w h ere f.:. r a n ges f ro m I t o N . For t h e reasons stated c a rl ier r e g a rd i n g t h e s t a te b e i n g t h e s a m e at t h e beg i n n i n g a n d at t h e end o f t h e d ay w h e n k equals N i n E q . ( 1 3 .75), l�) N ) and F/N + 1) are set t o zero. When k equals 1, the combinat ion xi.(l) i s the known initial-condition i n p ut to the unit comm itment probl e m . The combinations corresponding to subscripts i* and j of Eq. ( 1 3 .75 ) change roks from one stage to the next; the combination x i . ( k ), w hich i nitiates one search among the feasible combinations x/ k + 1), becomes one of the feasible combinations x/k) which enter i nto all searches o f stage (k - 1). This is graphically illus­ trated in Fig. 13.12, which shows a grid constructed for t hree typical consecutive stages (k - 1 ), k, and (k + 1). Suppose that each stage is associated with the load l evel shown and has a choice of o n e of the combinations X l ' X 2 , x3, and x9 prev iously iisted. Then, each node of the grid represents o n e o f the co m b i n a-

1 3.7

18 00

x · (k )

•

X2

'*

X

X3

xg X (k

--

14 00

MW

=

x2

x2

X

Xg 1)

(k ) Stage

(a)

1400 MW

Xl

-

1l 00 MW

x3

1800 M\V

ell

MW

Xl

1)

(k ) Stage

(6)

UNIT COMM ITMENT

(xj (k

+

577

l)}

Xg (k

+

1 1 00

(k

1)

MW

•

Xl

•

Xg

+

1)

FlGURE 13.12

I l l ust rat ion of the minimum cost search proce d u re associa t ed wi t h Eq.

( 1 3 .75) fo r: (a) X j ' ( O set x 2 ; (b) x j .(k 1) set

equal to e q u a l to

-

X I

and th e n to X 2 '

tions and the level o f system load at w hich i t s units are to operate. A n y n ode m a rk e d

X

is

i n fe a s ible

becau s e

i t s combin ation e i t h e r c a n n o t supply the

corresponding level of load or is othe rwise inadmissible. Choosing x i . ( k ) = x 2 i n Eq. ( 1 3 .75) i nitiates a search among the feasible set { X l ' X 2 , X 3 ' x9} for the minimum cumul ative cost Fi. ( k ), as d epicted in Fig. 1 3 . 1 2(a); a similar picture applies when each of the other two feasible combi nations X l and X 3 of stage k is chosen as xi.(k). Repl acing k by ( k 1) i n Eq. (13.75) steps us through to Fig. 1 3 . 1 2( b), which shows that the p revious search-initiating combination has now the role of x/ k ) and enters i nto every search for Fi .(k 1). , -

-

578

CHAPTER 13

ECONOMIC OPERATION OF POWER SYSTEMS

13.8 SOLVING THE UNIT COMMITMENT PROBLEM Equation ( 1 3 .75) is an iterative re lation e mbodying the principle that starting with a given combination X i ' at stage k , the m in imum unit commitment cost is found by minimizing the sum of the current single-stage co'st fJ k ) plus the minimum cumulative cost Fj( k + 1) over the later stages of the study. Th is is one example of the principle of optimality , which states: If the best possible path from A to C passes through interme diate point [J , then the b e s t poss i b le p a t h from B t o C must be t he co r r es p on d i n g p : l r t o r t h e b e s t p a t h r r o m /l t o C . Computationally, we e v a l ua L e o n e d e c i s i o n ,I t a t i m e hegiJ l l l illg w i t h t h e filial stage N and ca r ry t h e m i n i m u m c u m u l a L i v e eos t fu n c t i o n /)ack ward in t i m e t o stage k to fi n d t h e m i n i m u m c u m u l a t i ve cost Fi . ( k ) for the fe a s i b l e co m b i n a t i o n X i " of stage f, Fig. 13. 13. At each stage we build a t a b l e of res u l ts u n t i l we reach stage 1, where the input combination X i ' is d e fi n i t e l y known from t h e i n i t i a l condi tions. The minimum cum ulative cost d ecisions are recove red as w e sweep from stage 1 to stage N searching t h rough the tables already calculated for each stage. The computational p roced ure, known a s dyn a m ic programming , 5 i nvolves two sweeps through each stage k , as depicted i n F i g . 1 3 . 1 3 . I n t h e n r s t sweep, which is computationally intensive, we work hackward computing and record ing for each candidate combination Xi of stage k the minimum F/ k ) and its associated x /k + 1). The second sweep i n the fOlWard direction does not involve a ny processing since with x i ( k ) ident ified we merely en ter the table of results alre ady recorded to retrieve the value FioC k ) and its associated combina­ tion x /k + 1), which becomes X i . ( k + 1) as we move to the next forward stage . . At each stage in the dynamic p rogramming solution of the unit commit­ ment p roblem the economic d ispatch outputs of the available ge nerating units m ust be calculated before we can evaluate the p roduction costs Pio( k ), Fig. 1 3 . 1 3 . For a system with four gene rators t h is means we must establish economic dispatch tables, similar to Table 1 3 . 1 , for each feasible combination of u n i ts at eve ry load level of the daily cycle. If there were no constraints, 1 5 combinations woul d have to be considered at each stage , as we have seen. However, i n practice, constraints always apply. Suppose that t h e four un its have maximum and minimum loading limits and fue l-cost characte rist ics l ike E Q . ( 1 3 . 1 ), with coefficie n ts a i ' bi ' and C i given in Table 1 3. 5 . Then, to supply the load of Fig. 13. 1 1 , at least two of the four generators m ust be on-line at all times of the day. By specifying that Units 1 and 2 always operate (called must-rull un its), the number of feasible comb inations at e a c h s t a g e red u ces from the 1 5 l is t e d earl ier to the 4 combinations x l ' x 2 , X 3 ' and x 9 ' Accordi ngly, in Example 1 3 .8 typical ecpnomic dispatch results are derived for each of the 4 combinations at one of the load levels of Fig. 1 3 . 1 1 . The more complete results of Table 1 3 .6 are then ,

5

See, for example, R . Bellman, Dynamic Programming , P r i nceton U nivers ity Press, Princeto n , NJ,

1957.

'

To stage

(k

579

SOLVING THE UNIT COMMITMENT PROBLEM

13.8

Backward sweep -

1)

From stage

·

C a lculat e t h e transition cost

·

Recall

·

·

N ext

=

x; C k

+

1)

1)

+ 1)

i f any

".

Reta i n t h e m i n i m u m

+

Tj ) ( k )

F; Ck + 1 ) Evalu ate Fj( k ) PiCk + 1 ) + T;; C k ) + Fj(k Record F, (k ) and its associated x ; C k + 1 )

c o m b i n at i o n

(k

Fj( k )

.;

and its associated

x/k

+

1)

�

.'0 . •.

1 ) , recall a n d report Fj• ( k ) and its associated

From stage

(k

-

T o stage ( k

1) Forward

FI G U RE 1 3 . 1 3

x j.

swe e p

S u m m a ry of s t age k calc u l a t ions i n t he dyn a m ic progra m m i n g sol u t ion o f the u ni t p rob l e m of Eq. ( 1 3.75).

(k

+

+

1)

1)

580

,

CHAPTER 13

ECONOMIC OPERATION OF POWER SYSTEMS

TABLE 13.5

Loading limits and coefficients of Equation ( 1 3 . 1 ) for the generating units of E xamples 13.8 and 13.9 Loading limits

Fuel cost parameters

( ��W)' -� 1

Generating unit

1

(MW)

(MW)

1 00 1 00 75 75

625 625 600 500

2 3

4

$

Max,

Min,

number

/

-

bj ,

Ci'

($ / MWh)

($ / h)

.

O . OOHO

H.O

SOO

0.0096 0. 0 1 00 0.0 1 1 0

6.4 7.9 7.5

400 600 400

TABLE 13.6

Economic dispatch outputs and production costs for the unit combinations and system l o a d levels of Examples 1 3 . 8 Comb. Xi

- PD

=

xl

X2 x3

x9

PD Xl

X2

=

( / MWh) $ System ).

1 1 00 MW

1 0.090 1 0.805 1 0. 77 4 1 2.073

1 400 MW 10.804 1 1 .7 1 7 1 1 .7 1 1

x3

x9

PD

1 600 MW

1 1 .280 12.324 12.336

X2

x3 x9

Xl X2

x3 x9

Pg 3 Pg 2 (MW)

Pg 4

261 351 347 509

384 459 456 591

219 290

235

35 1 464 464

459 554 553

290 382

_

=

XI

PD

Pg l

=

1 800

.

_

410 54 1 542 _

_

_

_

298

383 _ _

_

_

_

_

_

_

_

1 3 4 00

469 625t

tDaggers denote loading

_

_

_

_

_

_

_

_

_

558 625t

_

_

_

_

38 7

386 550

12

13

14

286 1 380 1 3758 5 608

3565 4349 43 1 7 5859

2570 33 1 2

2466

3 80 1 5 073 507 3

4349 5419 5407

33 1 2 4 347

3 1 45

_

_

($ / h)

_

_

4452 5 999 60 1 1

440

MW

1 1 .756

_

344

338 442

508 617 618 _

300

11

_

_

_

5803 617 6 6 1 HH _

_ _ _ _ _ _ _ _ _ _

_

_

_

_

_

_

_

_

_

_

_

_

_

_

_

_

_

_

_

_

_

l i m its .

.

_

_

_ _

_

_

_

_

_

_

_

_

_

5 8,428 59,356 58,236

_

_

_

363 1

_

_

_

_

_

_

_

_

_

_

_

_

_

_

_

_

_

_

76,472 79, 184

4 1 26

4394 6458

infeasible infeasible

_

_

70,908 68,976 67,856

4765 _

4 '£ 1;

=

($)

45,848 45 ,848 44 ,792 45,86H

3 1 23

3H41 5069

infeasible 5466 6275

5 1 32 7 063

Fj ( k )

4079

infeasible

and 13.9

_

_

_

_

_

_

_

_

_

_

_

_

-

-

-

_

_

_

_

_

_

_

_

_

_

_

_

-

-

-

13.8

581

SOLVING THE UNIT COMMITMENT PROBLEM

used in Example l 3.9 to solve the unit commitment problem for the same four-unit system and load cycle. We n eglect transmission loss in both of the fol lowing examples to simplify the calculations.

The system load of Fig. 1 3 . 1 1 is to be supplied by combinations of the four generating units of Table 1 3.5. Treating Units 1 and 2 as must-run units, determine the power supplied by the generators of each combination and the corresponding production cost in economically loading the units when the system load level is 1 1 00 MW. Example 13.8.

Solution. Because Units 1 and 2 must operate at all times, thc four combinations

of units to be considcred at each stagc of the daily cyclc are x l > x2, x), and x9, as ind icated abovc . Sincc losses are neglected , economic dispatch of each of these co m b i nat i o n s is determ i ned d i rcctly from Eq . ( 1 3 .6), provided no unit is ope rat ing
m a x i m l l lll or m i n i m u m l o a d i ng l i m i t . We cons i d e r each comb i n a t i o n s e pa ra t ely.

13.1 .

U n i t s 1 a n d 2 h ave i n c r em e n t a l fuel costs I d e n t ic a l to t hose of Accord ingly, w i t h only U n i t s 1 and 2 operating, Table 1 3 . 1 appl ies and the economic dispatch outputs at the 1 1 00-MW load level are Pg I = 509 MW and Pg 2 = 591 MW. The correspon ding hourly production costs of the two un i ts are calculated from Eq . ( 1 3 . l ) using the given parameters as fol lows:

Combilla tioll Exa m p l e

x I) .

II

12

/

0 . 00 4 P I

=

/

O . 00 4 8 P 2

=

8 . 0 Pg 1

+

+

+

6 .4 Pg 2

500

+

1

$5 608 per hour

Pg 1 = 509 =

400 1 Pg 2 = 59 J

=

$5859 per hour

Hence, at stage 1 of Fig. 1 3 . 1 1 the prod uction cost of combination x9 amounts to P9 ( 1 ) = $45,868, which is the accumul ation of ( $5 608 + $5859) for each hour of the 4-h interva l . Combin a tion x J . With Units 1 , 2, and 4 operating, we follow Eq. ( 13.7) t o caleulate the coefficients CI T

bT

=

=

( a ;- I aT

+ a2

(bl

a1

+

1

+ ai

1)

- I

- -) b2 a2

+

b4 a4

=

= ( 0 .008 - 1 + 0 .0096 .- 1 + 0 . 0 1 1 '

aT

( -8 0 .008

+

6 .4 0 . 00 96

+

7.5 ) -0. 1 01

I)- I

=

=

a T Pg T

+

bT

=

3 . 1 243

x

10

- 3 ( 1 1 00 )

+

7 . 3374

=

3 . 1 243

x 10-

MW

t hen

7 . 3374

The i nc r e m ental fuel cost for the three units at the load level of given by Eq . ( 1 3.6) a s A

=

1 100

is

10 .774 MWh

,

and t he corresponding economic dispatch outputs of Units 1, 2, and 4 have the

3

582

CHAPTER 13

ECO N O M I C OPERATION OF POWER SYSTEMS

(rounded-off) values A

Pg I

=

PI(2

=

P11 4 -

A

A

-

bl

al

- b2 lJ2 - b4 lJ 4

1 0 . 774 - 8.0 0.008

=

347 MW

1 0 .774 - 6 .4 () O O 9(j

=

456 MW

1 0 . 774 0 .0 1 1

=

2913 MW

.

7 .S

The hourly prod uction costs o r the t h ree u n its a rc ca lcu lated to be II

=

O .004 Pg21 +

8.0 P,.,. 1

+

500 1 ", , _ .\.:7

=

$3758

per

hour

Therefore, if Units 1 , 2, and 4 of combination x] were to operate during i n terval 1 of Fig. 1 3 . 1 1 , the production cost Pi l ) would equal $44,792, which is fou r times the sum of hourly costs. .

Combination x 2 . The economic dispatch outputs and the production costs of Units 1, 2, and 3 are similarly found by replacing a 4 , b4 , and c 4 i n the last set of calculations by the Unit 3 para meters a J , b], and c ] of Table 1 3.5. The results are aT

=

3 .038

bT

=

7 .4634

A =

10-3

X

1 0 .805 $/MWh

P" I

PK 2

px )

=

35 1 MW

=

459 MW

=

290 MW

Thus, if only U nits 1 , 2, and 3 a r c operating production cost is P 2 (1 ) $45,848. =

at

il

=

i2

=

3801 $/h 43 49 � / h

/3 = 3 3 1 2 $ / h

t h e load level of 1 1 00 M W , the

x / . If all four generating u n its of combination x I were to serve the load of 1 100 MW, the corresponding results are

Combination

aT

=

2. 3805

bT

=

7 .4712

A =

x

10 -3

1 0 .090 $ / MW h

PX I

Pg 2

Pg3 Pg 4

=

261 MW

=

385 MW

=

219 MW

=

235 MW

il = i2

13

14

=

= =

2i:l61 $/h 3565 $/h 2570 $/h 2466 $/h

1 3.8

583

SOLVING TIlE UNIT COMM ITMENT PROBLEM

Since the preceding results apply whenever the system load is 1 100 MW, they are also applicable to stages 1 and 6 of Fig. 1 3 . 1 1 . Table 1 3 . 6 provides the complete results required in the next example. Assuming the start-up cost of each thermal generating unit i s $3000 and the shut-down cost is $ 1 500, determine the optimal unit commitment policy for the four thermal units of Example 1 3 .8 to serve the system load of Fig. 1 3. 1 1. Only the must-run Units 1 and 2 are to operate at the first and final stages o f the load cycle. Example 13.9.

Solution. The dynamic programming solution will be represented graph ically, as shown in Fig. 1 3 . 1 4 . A grid is first constructed for the six stages of the load cycle and thc production cost is entered to the right and below each node from the economic dispatch res u l ts of T a b l e 1 3.6. I n feasible nodes occur at the left a n d right bou ndaries o f th e grid because of specified i n p u t a n d terminating conditions o f t h is examp l c . Each diagonal line of the grid has an associated transition cost be c a u se of t h e change Crom one com b i n a tion of units to another combination. A horizon tal l i nc has zero transition cost beca use it connects two nodes of the same combination, even though at diffe rent load levels. The transition costs of Fig. 1 3 . 1 4 a s s i g n $3000 a n d $ 1 500, respectively, t o each unit stmt-up a n d shut-down a s speci fied. The graphical solu lion of this example is the path of least cumulative cost linking the initial-condition node of stage 1 and the destination node of s tage 6. We begin at stage 6, which has only Units 1 and 2 of combination X9 operating. Setting N equal to 6 and i* equal to 9 in Eq. (1 3.75) gives

.

F9 ( 6)

m i n { P9 ( 6 ) xP)

=

+ 0 + O}

=

$45,868

which is the minimum production cost at stage 6 si nce the set {x/7)} h as no c l e m e n t s . T h e set o f fe a s i b l e combinations at sta g e 6 has only one element, namely, x 9 . We now move to stage 5 to evaluate the minimum cumulative cost of the two final stages. Setting k equal to 5 and {x/6)} equal to {x 9 (6)} in Eq. (13.75), we obtain F, . ( 5 )

=

min { P; . ( 5 )

xlj(6)

+

1';01) ( 5 ) + F<) ( 6 ) }

=

{ P; 0 ( 5 )

+

T;0l)( 5 )

+

$ 45,868 }

stage 5 th e node correspon d i ng to combination xl) is infeasible because Units 1 and 2 alone cannot supply 1 400 M W of system load. Thus, there are three feasible At

n o d e s a t s t a ge 5 corre s p o n d i n g t o the c o m b i n a t i o n s

x l'

x -z ,

and

x],

but

we

do n o t

yet know which of them is on the path of least cumulative cost. Systematica lly, we set x ) , then x 2 , and finally x 3 equal to X i " in the last equation, which gives F ) (5)

=

{ P) ( 5 )

+ T1 , 9 ( 5 ) +

$45 , 868 } = [$58 ,428

+

$3000

+ $45,868]

[$59,356 + $ 1 500 + $45,868] [$58,236

+

$ 1 500

+

$45,86 8 ]

= =

=

$ 1 07,296 $ 1 06,724 $ 1 05 ,604

¥- - - - - - - - 1 4 5,848

183;1768

r-----� lM�OO�----- L--�

254,484 }-------I 258;316

o

o

o

I I I I I I I I I I I

x

- - - - - - -

X:

4 5 ,848

I I I I I I I I I I I

co o

.;i; co

:0 E o

o

)K4 4 1

,792

I I I I I I I I I I

Xg

364,780

363,108

- - - - - - - -

361 ,536

67,856

58,236

x

- - -

------- *-

- -

I I I I I I I I I I I I I

- - - - - - -X

-

- - - - - - - - -x- - - -

I I I I I I I I I I I I I

58,428

- -

)k

4 5 ,848

x

44,792

- f---l 4 5 ,868

4 5 ,848

4 5 , 848

1

FIGURE 13.14

4 5 ,848

2

3

4

Stage number

Dyn amic p rogramming solu tion of the u n i t com m i t m e n t problem of Ex ample D.C).

5

6

1 3.8 SOLVING THE UNIT COMMITMENT PROBLEM

585

The preceding calculations are straightforward because the search for the minimum cumulative cost involves only the single combination x9( 6) of the final stage. We record these results at each of the stage 5 nodes, as shown i n Fig. 1 3 . 1 4, a n d proceed to stage 4 where only combin ations x 1 a n d x2 have sufficient capacity to s erv e the load of 1 800 MW. Hence, the best combination X i . at stage 4 must be either x j or x 2 , but we do not yet know which one. The stage 4 version of Eq. (1 3 .75) is

where x/S) r ange s over the set of fe as i b l e Setting X i ' fl rs t eq u a l to X l ' we obtain

c o mb i n a t i o n s { X j ' x 2 , x :J

The numerical values for this equation are available F I (4)

=

=

m i n { [76, 472

+

from

of stage 5.

Fig. 1 3 . 1 4 , which gives

0 + 1 07 , 296] ; [76 , 472

+

1 500

[ 76, 472

+

1500 + 1 05 , 604] }

m in { [ 1 83 , 768 ] ; [ 1 84 , 696] ; [183, 576] }

=

+

1 06 , 724] ;

$ 1 83 ,576

This resu lt must now be recorded, as shown in Fig. 1 3 . 1 4, and the two values $ 1 83,768 and $ 1 84,696, which could be discarded , are also shown for information only At t h i s p o i n t we have d e term i ned t h e cu m u l t iv e cost from stage 4 to stage 6 fo r t h e path begi n n i n g w i t h c om b in a t i o l l X I a t s t age 4. To eva luate the path sta rting with the feasible combi nat ion X 2 of s t a g e 4, we s e t X i " e q u a l to x2 in the s t a g e 4 v e r s io n of Eq . ( 13 .75) and perform c a l c u l a t i o n s s i m i l a r to those j ust demonstrated. The resu l t i s .

a

F2( 4 )

=

m i n { [ 1 89 , 480] ; [ 1 8 5 , 908 ] ; [ 18 9 , 288 ] }

=

$ 1 8 5 ,908

which is re c o r ded at t h e appropriate node of Fig. 1 3 . J 4. No conclusion can be drawn yet, but when we identify which stage 4 node is on the p a th of l east overall cumulative cost, we select from the n u m e r i ca l resu lts Fj(4) and F2(4). W e now proceed to stage 3 and on to stages 2 and 1 , r e p eat i ng the evaluation of Eq. ( 1 3 .75)

at each feasible node and recording the corresponding m inimum cumulative cost. One of t h e tran sition paths leaving each node is marked with the ( encircled) subscript of the combination from w hi c h t h e l e a s t c u mu l a t ive cost of ,the node is derived.

586

CHAPTER

13

ECONOMIC OPERATION OF POWER SYSTEMS

At stage 1 it is determined that the least overall cumulative cost is $ 3 6 1 ,536 from stage 6 to the initial-condition combination (and vice versa). This total cost is shown in Fig. 1 3 . 1 4 to derive from combination X 3 of stage 2, which in turn derives from combination x3 of stage 3, and so on back to stage 6. The least cost path retraced in Fig. 1 3. 14 shows that the opti mal unit commitment schedule is: Stage

Load leve l,

I

2

3 4 5

6

1 400 1 600 1 800 I 1 00

units

1.2

.r \)

1 . 2. 4

X .I

l . 2. 4

x .1

I . 2, 3, 4

Xl

1 4 ()()

1 . 2. 4

X .1

I I O( )

and the total cost to supply the

$361 ,536 i n this example.

Combin;\ t ion

MW

1,2

X ,)

daily

forecast

load

of

1- i g . 1 3 . 1 1

amo u n b

to

Example 1 3.9 demonstrates the great reduct ion in compu tational effort which is possible by the dynamic programming approach. As show n by the recorded values at the nodes o f Fig. 1 3 . 1 4, the cumulative cost function f",C k ) was evaluated only 27 t imes, which i s the sum (3 + 9 + 6 + 6 + 3 ) o f the i nterstage transitions from stage 1 to stage 6. A brute-force e n umeration a pproach to the same problem would h ave i nvolved a total of 29 1 6 i nterstage t ransitions, which is the product (3 X 9 X 6 X 6 X 3) of l i n e segments shown i n Fig. 1 3 . 14. I ndeed, given the start ing a n d e n d i n g s t a t e s as i n Ex am p l e 1 3 . 9 , and i f there were no infeasible nodes at any of the i n te rm e eJ i a t e s t a g e s , the n u m b e r o f i n t e r stage transitions between t h e 1 5 comb inations x I to X I ) wou l d increase enormously to 15 X 1 5 2 X 1 52 X 1 5 2 X 15 = (225)4 = 2.563 X 1 0 <) . Thus, dy­ namic p r og r amming offers a v i a b l e a pproach to the sol u tion of t h e u n i t com m i t­ ment pr ob l e m w h e n practical cOIl s l r: l i n l s o f system o p e r a t i o l 1 s a rc t a k e n i n to account.

13.9

SUMMARY

The classical solution of the econom i c d i spatch problem is provi ded by the method of Lagrange multipliers. The s ol u t i on s t a t e s that the m i n i m u m fue l cost is obtained when the incremental fue l cost dfJdPgi of each u n i t m u l t i pl i ed by its pen al ty factor Li is the same for all operating u n its of the system. Each of the products L/dfJdPg ) is equal to the system A , w hich is approximately t he co s t i n d o ll a rs per hour to increase the total delivered load by 1 MW; that is,

( 1 3 .76)

PROBLEMS

587

The penalty factor for plant i is defined by 1

( 1 3 .77)

where PL is the total real power transmission loss. The i ncremental loss 8PL /8Pg i is a measure of the sensitivity of the system losses t o an incremental change .in the output of plant i when all other p l a nt outputs are kept fixed . Transmission loss PL c a n be expressed in terms of B -coefficients a n d the power outputs Pgi by PL

K

K

L L, Pgi Bij Pgj

=

i= l j= l

K

+ L B i o Pg i i= \

( 1 3 .78)

Boo

+

The B-coefficie nts, which must have units consistent with those of Pg i' can be determined from the results of a converged power-flow solution by means of a power- invarient transformation based on the real part (R hU) of the system Z bus' An algorithm for solving the classica l economic dispatch probl em is provided i n Sec. 13.5. The fundamentals of automatic ge neration control are explained in Sec. 13.6, where the definitions of area con trol e rror (ACE) and t im e error are introduced. The rudiments of unit commitment are provid ed in S ec 13.7, which explains the principle of optimality u nderlying the dynamic pr og ra m m i n g solu­ tion approach . Figure 1 3 . 1 3 summarizes the overall solution procedure. .

PROBLEMS 1 3 . 1 . F o r a g e n era t i n g u n i t the fuel input i n m i l l io n s o f Btu/h i s expressed as a fu nction of output Pli i n m e gawa tt s by O.032 P; + 5 . 8 PK + 1 20. Determine ( 0 ) The e q u a t i o n for i ncrem e n t a l fu e l cost in d o l l ars per m e g a w a t t hour as a fu nc t io n of Pli i n m egawa t ts b a s ed o n a fue l cost of $2 p e r m i l lion Btu. ( b) Thc avc ragc cost of fu e l p e r megawa t t hour when P � 200 MW. ( c ) Th e approx i m a te a d d i t i o n a l f u e l cost p e r hour t o r a ise the output o f t h e unit from 2 0 0 t o 20 1 MW. A lso, f i n d t h is a d d i t i o n a l cost accu rately a n d compare it =

13.2.

in $ / MWh for four u nits of

w i t h the a pprox i m a t e v a l u e

The i n cre m e n t a l fu e l cost A\

=

dfl

dPg 1

----

=

O . 0 1 2 Pg ]

.

+

9.0

df2

A 2 = -dPg 2

=

a

plant

are

O .0096Pg 2

+

6 .0

588

CHAPTER

13.3.

13.4. 13.5.

13

ECONOMIC OPERATION O F POWER SYSTEMS

Assuming that all four units operate to meet the total plant load of 80 MW, find the incremental fuel cost A of the plant a n d the required output of each unit for economic dispatch. Assu m e that maximum load on each of the fou r units described in Prob. 13.2 is 200, 400, 250 , and 300 MW, respectively, and that minimum load on each unit is 50, 1 00, 80, and 1 10 MW, respectively. W i th t hese m aximum and minimum output limits, find the plant A and MW output of each unit for economic dispatch. Solve Prob. 1 3 .3 when the m i nimum load on U n it 4 is 50 MW rather than 1 10 MW. The incremental fue l costs for two units of a plant arc

where f is in dollars per hour ($/h) and Pf: is in megawa l ls ( M W). I f hoth un its operate a t all times and maximum and minimum loads on each unit are 550 and 1 00 M W , respect ively, pl o t A o f the p lant in $ /MWh versus plant output in MW for economic dispatch as total l o a d v aries from 200 to 1 1 00 \1 W . 13.6. Find the savings in $/h for economic d ispatch of load between the units of Prob. 1 3.5 compared with their sharing the output equally when the total plant output i s 600 MW. 13.7. A power system is supplied by three plants, all of which arc operating on economic d ispatch. At the bus of plan t I the i ncremental cost is S l O.O per MWh, at plant 2 i t is $9.0 per MWh, and at plant 3 it is $ 1 1 .0 per MWh. Which plant has the highest penalty factor and which one h as the lowest penalty factor? Find the penalty factor of plant 1 i f the cost per hour to increase the total delivered load by 1 MW is $ 1 2 .0. 13.8. A power system has two generating plants and B-coe ffi c i e n t s correspond ing to Eq. ( 13.37), which are given in per uni t on a 100-MYA base by

[

5 .0 - 0 .03 0.15 0 .20 - 0 .03 8 .0 -------------+-----0 . 15 0 . 20 0 .06

]

X

10

-3

The i ncremental fuel cost in $/MWh of the generating un its a t the two plants are "- 2

13.9.

=

df2

--

dPI: 2

=

O .0096 f'1: �

�

+

6.0

If plant 1 presently supplies 200 MW and plan t 2 supplies 300 MW, find the penalty factors of each plant. Is the present d ispatch most economical? If not, which plant output should be increased and which one should be decreased? Explain why. Using $ 1 O.0/MWh as the starting value of system A in Example 13.4, perform the necessary calculations during the first itera tion to obtain an updated A . I

PROBLEMS

589

@ of a fou r-bus system is a generator bus and at the same time a load bus. By defining both a generation current and a load current at bus @, as

13.10. Suppose that bus

shown in shown in

13 . 1 1 .

13.12.

13.13.

Fig. 13 .5(c), find the transformation matrix Eq. (1 3.3 1).

C

for this c a s e in the form

The four-bus system 13.5 has bus and line data given in Table 1 3 . 2 . Suppose that t he bus data are slightly modified such that at bus P-generation i s 4.68 u n i t and P-load and -load are 1 .5 per unit a n d 0.9296 per unit, respectively. Using the results of Table 13.3, find t he power-flow solution corresponding to this mod ified bus data. Using the solution t o Prob. 1 3 . 1 0, also find the B-coefficients of this modified problem in which there is load as well as

depicted in Fig.

per

@,

Q

generation at bus @ . Threc genera ting units operating in parallel at 60 Hz have ratings of 300, 500, and 600 MW and h avc spccd-droop characterist ics of 5, 4, and 3 % , respectively. Due to a change in load, an increase in system frequcncy of 0.3 Hz is experienced before an y s u p p l e m e n t a ry co ntrol action occurs. Determine the amount of the changc in sys t e m load and a l s o thc amount of t h e c h a n g e in generation of each unit to absorb the load change. A 60-Hz systcm consisting of thc threc generating units described in Prob. 1 3. 1 2 is connected to a neighboring system via a tie line. Su ppose that a generator in th e neighboring system is forccd out of s e rvice, and that the tie-line flow is observed to increase from the scheduled value of 400 to 63 1 MW. De t e r m ine the amount of the increase in generation of each of the three units and find the A C E of this system whose frequency bias setting is 58 MW /0. 1 Hz. Suppose that it takes 5 min for the AGe of the power system of Prob. 13.13 to command the threc units to increase their generation to restore system frequency to 60 Hz. What is the time error in seconds incurred during this 5-min period? Assume that the initial frequency deviation remains the same throughout this restoration period. Solve Example 13.8 when the system load level is 1 300 MW. If the start-up costs of the four units of Exa mp l e 13.9 are changed to $2500, $3000, $3400, and $2600, and the shut-down costs are changed t o $ 1 500, $ 1200, $ 1 000, and $1400, respectively, find the optimal unit commitment policy. Assume that all other cond itions remain unchanged. Duc to a 400-MW short-term purchase request from the neighboring utility, the demand during the second interval of the day is expected to increase from 1 4 00 to 1 800 MW for the system described in Example 1 3.9. Assuming that other conditions remain unchanged, find the optimal unit commitment policy and the associated total operating cost for the day. Suppose Unit 4 of Example 1 3 . 9 will have to be taken off line for 8 h, beginning at the fifth interval of the day to undergo minor repair work. Determine the optimal un it commitment policy to servc the system load of Fig. 1 3 1 1 and the increase in the operating cost for the day. A diagram similar to Fig. 1 3 . 1 4 is shown in Fig. 1 3 . 1 5 in which directed branches represent transitions from one state, represented by a n od e to a nother state. A ss oc i ated with each directed branch (i, j) is the cost fJk), as defined in Eq. ( 1 3.72). The values o f fJk) a r e given in Table 1 3 .7. Note that index k of f ( k ) does not play any role here, and consequently will now be omitted. If the -

13. 14.

13.1S. 13. 16.

13.17.

13.1 8.

.

13.19.

,

. . IJ

I

590

CHAPTER 1 3

ECONOMIC OPERATION O F POWER SYSTEMS

FI G U R E 1 3 . 1 5 State-transition

diagram

fo r

Prob. 1 3. 1 9.

TABLE 1 3.7

Matrix of costs (or distances) Iij between

states (or nodes) CD and

CD 0 G) @ (]) (i) ® (J) ® ®

CD 20

CD 15

0 of Fig. 1 3 . 1 5

@ 17

CD

(j) @

0J

35

31

38

40

34

39

36

42

®

®

26

22

29

25

41

44

@

33

15

18

value o f lij i s interpreted to be t h e d i s t a nc e between states i and j, t h e n t h e unit commitment problem becomes t h a t o r f i n d i ng the shortest path f r o m t h e origin, represen ted by node CD, to t h e destination, re prese n t e d by node @. The problem of this nature is called the stagecoach problem . Write the backward recurrence eq.u �tion similar to Eq. ( 1 3 .7 5 ) .a nd solve the problem by commencing calculations at the d.e stination and then moving toward the origin . In forward recurrence the process s t a rts w i th the origin and moves toward the destination. Write the forward recurren ce equation, solve the problem, and check the result with that of the bac kw a rd dynamic programming procedure.

CHAPTER

14

Z bU S M ETHO D S

IN CONT INGENCY ANALYSIS

When a line is switched onto or off t h e system through the action of circuit breakers, l ine currents are redistribu ted throughout the network and bus voltages change . The new steady-state bus voltages and line currents can be p redicted by what is called the contingency analysis program. The l arge-scale network models used for contingency evaluation, l ike those used for faul t calculations, d o not have to b e exact because t h e system planners and operators, who must undertake hundreds of studies in a short time period, a re more concerned with knowing if overload levels of current and out-oj-limit vol tages exist than with finding the exact values of those quantities. On this account, a p p r oxima t i o n s are m a d e . O ften resi s t a nce is co n s i d e r e d n egl i g i b l e a n d the n etwork m o d e l becomes purely reactive. Lin e charging and off-nominal t ap c h a n g i n g of t r a n s formers are also frequently omitted. I n many cases l inear models are considered satisfactory and the principle of superposition is then employed. Methods of con tingency ana lysis which use the system Z b us' and a lso Y bus' become attractive . from the computational viewpoint-especially if loads c a n be t r e a t e d as c o n s t a n t c u r r e n t i nj ections i n t o t h e various buses of the system. Removing a line from service can be simulated in the system model by adding the negative of the series impedance of the line between its two end b uses. Thus, methods to examine the steady-state effects of adding lines to an existing system are developed i n Sec. 14. 1 . The concept of compensating currents is mtroduced which allows use of the existing system Z b us without having to modify it. A particular application of this approach to i nterconn�cted power systems is shown in Sec. 1 4.2. 591

592

� R 14

Zbus METHODS IN CONTINGENCY ANALYSIS

In order to quickly evaluate the effects of line outages and generation shifts aimed at relieving overloads, we develop the concept of distribution Jactors i n S ecs. 14.3 and 14.4. It is show n how the distribution factors can be fonnulated from the existing Z bus of the system and how they can be used to study multiple contingencies. The simi l arity of the distribution-factor approach to the dc power-flow method of S ec. 9.7 is explained in Sec. 1 4. 5 . When u ndertaking the contingency analysis of one part of an intercon­ nected power system, we need to represe n t the remaining portion of the overall syste m by a n equivalent network. Section 14.6 discusses the basics of equivalenc­ ing and compares the network equivalents obtainable from both the system Z bu s and the system Y bus ' 14. 1 AD D I N G AN D REMOV I N G M U LTIPLE LIN ES

When considering line additions to or removals from an existing system, i t is not always n ecessary to build a new Z b u s or to calculate new triangular factors of Ybus-especially i f the only i nterest is to establish the impact of the changes on the existin g bus voltages and l i n e flows. An alternative p rocedure is to consider the i njection of compensating currents into the existing system t o account for the effect of the line changes. To illustrate basic concepts, let us consider adding two lines of impedances Za and Zb to an existing system with known Z b u s ' Later we shall consider three or more l in es. Suppose that impedances Za and Zb are to be added between buses @ - @ and ® - ® , respectively, of Fig. 14.1. We assume that the bus voltages VI ' V2 , · . . , VN produced in the original system (without Za and Zb ) by the current i njections 11 , 12 " " , III a re known, and that these injections are fixed in value and therefore are u naffect ed by the addition of Za and Zb' On a per-phase basis, the bus impedance equations for the original system are then given by

1-

VI

Vm

V=

v,.

Vp Vq

VN

CD @) ® ® ® ®

CD

@)

z im

®

zin

®

ZIP

Zlq

ZI N

II

Zm l

Zm m

Zm "

zm p

Zm

q

Zm N

1m

Zn l

Zn m

Zn n

ZIl P

Zn q

Zn N

In

Zpl

Zpm

Zpn

Zpp

Zpq

ZpN

Ip

Zq l

Zqm

Zqn

Z qp

Zqq

ZqN

Iq

ZN l

ZNm

Z Nn

Z Np

Z Nq

Z NN

IN

ZI I

®

®

( 14 . 1 )

ADDING AND REMOVING MULTIPLE LINES

14.1

CD t-V-=1-t-

I1 fT

-

593

I1 _

_ ___

_ 12

o It--I---=' --- --to'I

Z bus

(origi nal system)

I' 2

® I VN

- I_N

__ ____

FlGURE 1 4 . 1

Original system with voltages VI ' V2 , to the current i njec tions I I ' '2 ' . · . , IN · .

.

•

,

VN d u e

and we wish to determine the changes in the b u s vol tages due to adding the two new line impedances. Let V' = [ V; , V; , . . . , V� V denote the vector of bus voltages whieh apply after Z" and Z" have been added. The voltage change at a typical bus ® is then given by ( 1 4 .2 )

The currents 1a and 1b in the added branch impedances to the new bus vol tages by the equations

Za

and

Zb

are rel a ted ( 1 4 .3 )

Figure 1 4.2C a ) shows the new branch currents flowing from bus @ t o bus ® and from bus ® to bus ® . We can rewrite Eqs. (14.3) in the vector-m a trix form ® Z bus (original

system)

-

l

®

V'n

®

Vp'

®

Vq'

�)

?0 (a)

V,;I

- - - - -

Z (I _

_

_

-

f

-

_

_

- - - - - -

Zb _

_

_

V,;

tf v; - V; L-

_

_

_

+

_

Original system o f F i g . 1 4.1 with b u s voltages V; cha nged to Z b o r b y ( b ) equiva lent compe n s a t i n g current i njections. n G U RE 14.2

v,;.

6 1 t;:;;\ I

+

Z bua (original system)

®

. v,,' I -+-_ r

1

V'

__

®I ®

_

P

- Ia

Ia

_ _ _ _ _ _

-

� I q---l-----V'

(b)

_

Ib

vi' b y adding ( a ) i mpeda n ces Z a a n d

594

CHAYTER 14

Zbu, METHODS IN CONTI NGENCY ANALYSIS

form V'1

CD

@)

0 0 0 0

1n 0 [�. [ 0

Zb

a

lb

=

b

1

0

®

®®

-1

®.

�l

-1

V' m V' n

V' - p

V'

=

A cV '

( 1 4 .4)

q

v�-

where Ac IS the branch-to-node i ncidence matrix, which shows the incidence of the two new branches to the nodes of the system. The new branch currents fa and fb have the same effect on the voltages of the original system as two sets of injected currents-{ -1a at bus @)' fa at bus ®} and { fb at bus ®, Ib at bus ® }, as shown in Fig. 14.2(b). These equivalent current injections combine with actual current i njections into the original system to produce the bus voltages V; , V� , . . . , V::,-the same as if the branch impedances Za and Zb had been actually added to the network. In other words, curren ts fa and fb compensate for not modifying Z bu s of the original system to include Za and Zb . On this account, they are called compensating currents. We can express the compensating currents in vector-matrix form as fol lows: -

(D @)

I

® comp = ® ® ®

CD

O

-I fa - fb Ib

@) ®

a

0

-

® ®

®

I

a

b

0

0

-1 1 0 0

0 0

-1 1

0

0

[�: 1 -A�[�: 1

( 1 4 .5)

-

, VN to V{, V; , . . . , V::, can be The changes in the bus voltages from V I ' V2 , calculated by multiplying the original system Z b u s by the vector Icomp of compensating currents. Then, adding Z b us I comp to the vector V of existing bus voltages yields .

.

•

( 1 4 .6)

14.1

595

ADDING AND REMOVING MULTIPLE LINES

This equation shows that the voltage changes at the buses of the original system due to the addition of the branch impedances Za and Zb be twee n the buses @) - @ and ® - @ ' re spect i vely are given by ,

( 1 4 . 7)

where 1 and lb are the compensating currents. The reader will find it usefu l t o check the d imensions of each term in Eq. 04.7) from which the voltage changes .6.V V' - V can be calculated d irectly once values for the currents I a and 10 are determined. We now show how this determination can be made. Premultiplying Eq . ( 14.6) by Ac and then subst ituting for A cV' f rom Eq. ( 14 .4), we obtain a

=

( 1 4 .8 )

Collecting terms which i nvolve la and lb gives ( 1 4.9) Z

where Z is a loop impedance matrix w hich can be fonned directly from the original bus impedance matrix of the system, as we shall soon see. Solving Eq. (14.9) for fa and lb ' we find that ( 1 4 . 1 0)

Note that Vm - Vn and V" - Vq are open -circuit voltage drops between buses @) - ® and ® ® in the origi n a l network, that is, with branch impedances Z a and Zb open in Fig. 14.2(a). These open-circuit voltages are either known or can be easily calcu lated from Eq. ( 1 4 . 1 ) . The definition of matrix Z in Eq. ( 1 4.9) includes the term A c Z bu s A7; , which can be determined as follows: -

A c Z bus A c T

_

-

b

a

@) ® ®

[�

-1

0

0

®

-n

@) ® ® ®

@)

®

®

®

Zmm

Zm

"

Zmp

Zm

Znm

Zn n

Zn p

Zn q

Zpm

Zpn

Zpp

Zpq

Zqm

Zq n

Zqp

Zqq

q

@) ®

® ®

0

1

b

0

-1

a

-1

0

0 1

( 14.11)

596

CHAPTER

14

Zb

... METHODS IN CONTING ENCY ANALYSIS

In this equation we have s hown only t hose elements of Ac and Z b u s which contribute to the calculation. Since all the other elements of A c a re zeros, it is not necessary to d isplay the ful l Z b uS . The indicated multiplications yield

A c Z busA� =

a

b

[

b

a

( Zm m - Z", " ) - ( ZlIm - ZIIII )

( Z"m - Z,," ) - ( Z 'IIII - Z II" )

( 1 4 . 1 2) The diagon al clements i n this equation can be recognized from Fig. 1 4 .3 as the Thevenin impedances Z' I1. ", " and Z'h. !) I/ s e e n when looking into the original system between buses @ - @ and ® - (!j) , respectively. That i s ,

( 1 4 . 13) Substituting from Eq. (14.12) into Eq. (14.9), we obtain b

a

z

( 1 4 . 1 4)

which shows that the compensating currents I a and Ih can be calcul ated by using the known bus voltages V" p V,, , �) ) a n d Vq of the origi nal network and the elements of its Z bu s shown in Eq. 0 4 . 1 4). Thus, Eqs. 0 4 .14) and 0 4.7), i n that order, constitute a two-step p roce­ d u re for the closed-form solution of the voltage changes at the buses of the original system due to the simultaneous addition of branch impedances Z a and Zb. Under the assumption of constant externally injected currents into the original system, we first calculate t he compensating currents using Eq. 0 4. 1 4) and then substitute for these currents i n Eq. 04.7) to find the n ew bus voltages w hich result from adding the new branches. The removal of branch impedances Z a and Zb from the original system can be a nalyzed in a similar manner simply by treating the removals as additions of the negative impedances - Z a and -Zb' as explained in Chap. 8. The elements in the 2 X 2 matrix of Eq. (14 . 1 2) can be calcul ated by using the appropriate elements of columns m, n, p, and q of Z b uS ' or they can be generated from the triangular factors L and U of Yb us , as described in Sec'. 8.5.

14.1

+

'

Zn l!

-

ADDING AND REMOVING MULTIPLE LINES

�

®

Zn m

(a) FI GURE 1 4.3

®

(b)

®.

Theve n i n e quiva l e n t i m p e d a nces l oo k i n g i n t o t h e sys t e m between ( a ) buses buses

and

597

@

and

®

and ( b)

m - n ) and ZC p - q ) are the respective sol utions o f the equaS i nce the vectors z(bus bus tions

m - n) Luz(bus

= -

0

0

1m

1p

1

LUZ(bus p - q)

1/

0

( 14. 15)

=

- 1q 0

we can write ( Zmm - Zmn ) - ( Z nm - Znll )

=

It also follows from Eqs. 0 4. 1 5 ) that

m

( row p

( Zpm - ZpfI ) - ( Z q m - Z q n ) ( Zmp - Zmq ) - ( Znp - Z n q )

( row

-

Z bus

( row

A�

m

-

-

-

m - n) row n ) of z(bus

m -n) row q ) o f z(bus p row n ) o f Z(bus

-

( 14. 16)

q)

i n Eq. 04.7) is an N � 2 matrix in

598

CHAYreR 14

Zbus METHODS IN CONT I N G ENCY ANALYSIS

® Z bUB

Z bus

1

(o riginal system )

FI G U RE 1 4 . 4

( original system )

(a)

A d d i t i on of t h ree b ranches Za . Z;" a n d Z( between ( a ) d i s t i n ct b u s p a irs a n d ( 6 ) n o n d i s t i n c t bus p a i rs i n the ori g i nal sys t e m of k nown Z hll.v

(6)

w hich the columns are equal t o the vectors Zb7:s- n ) and Z bus AT

c

=

[ z(m -n)lz(p-q)] bus

z�;q). T h a t

is, ( 14 . 1 7)

bus

-

When more than two lines are to be added to the original system, the matrix Z is fonned in the manner described above. For i nstance, to add three line impedances Z a ' Zb ' and Z c between the b uses @) - ® , ® ® , a n d 0 - CD , respectively, of Fig. 14.4(a), w e find that b

a

c

Z",p - Zmq - znp + Znq zrp

-

Z ,h . p q

- Zr q

Zsp

+

ZSq

( 1 4 . 1 8) I t is instructive for the reader to determine the elements of Z directly from Z b uS of the system. Other line additions, exemplified i n Fig. 1 4.4(b), can be handled with the same basic procedures si mply by treating each l in e addition as d escribed here. There are many applications for the p rocedures presented in this section, such a s in d ev eloping distribution factors for contingency evalua tion l a t er i n this chapter and i n piecewise methods, which are discussed in the section i mmedi­ ately foll owing.

intfhe

&

14.1. Current injections five-bus system of Fig. 8.9 cause voltages , all i n at selected buses given by V2 = 0.98 0° , V4 = 0.985&, and Vs 1.0 p e r unit. Predict the changes in the b u s voltages when l i ne @ - (i) with 0.04 per-unit reactance and line @ - G) w i th 0.08 per-unit reactance are simultaneously added to the system. Use the triangul a r factors of Y bus specified in Example 8.6.

Example

=

�

.

1 4 .1

ADD ING AND REMOVING MULTIPLE LINES

599

0 - G) and line @ - G) , we need to calculate the corresponding vectors Zb2u�5) and Zb��5) using forward and back substitution, as explained in Sec. 8.5. Continuing the calculations of Example 8.6, the reader can readily show that

Solution. To account for the addition of line

5 Z (4bus- ) -

Z (b2us-S)

-

2 24 234 244 254

225 235 245 255

212 2 22 23 2 242 25 2

-

2 15 2 25 235 245 255

From Eq . ( 1 4. 1 4 ) with m = 2, matrix are determined to be

Z

=

=

=

a

b

[(Z22 [0

a

a

b

[

5,

=

a

- ( - jO. 1 1 25 ) a

j O . 1 525

jO . 1 1 25 ( 1 4. 1 4 ) ,

and

0 0 -j O . 0625 0 jO . 1 1 25

=

-

p =

4, and

q

=

5 the elements of t h e Z

b

- 2 25 ) - ( 25 2 - 255 ) + 2a ( 242 - 245 ) - ( 25 2 - 255 )

U s i n g Eq.

[

n =

jO.1 0 -jO.0625 jO . 1 5 - jO . 1 125

a

0 - ( -j O . 1 125 )

b

2 15

-

Z1 4

+ jO .04

( 2 24 - 22S ) - ( 254 - 255 ) ( 244 - 24S ) - ( 254 - 255 ) + 2b b

]

o - ( - jO . 1 125 )

j 0 . 1 5 - ( -jO . 1 125 )

+

jO .08

b

i O . 1 1 25

jO 3 4 25 .

]

1

we can calcu late fa and fh as the solution of

jO.1125 ] [ fa 1 j O-,1 125 -+jO ,34-2S fb j 0 . 1 525

from which we find fa

=

=

[ V2 - Vs ] V4 V5 -

- jO.158876 and fb

=

=

[

1 .0 - 0 .98 0 .985 - 0.98

]=[ ] 0 .02

0 , 005

' jO. 037587. Accordin� to Eqs. (1 4.7)

·600

CHAPTER 14

and

Zbus METHODS IN CONTINGENCY ANALYSIS

(14.17), the changes in the voltages at all buses are

tN

=

-I Z <2 -5) a bus

-

I

b

Z(4bus- 5)

jO . 1 58876 -jO .0625 o

L ® 13

�

12

-

�'

O .00758 1

I

® Z h 19 __I--=5 __b___--+-__�___r-� I[ Ib

5

® .... !�

-

-

System r

-

I I

:

-

®

6

14 Za

-

Il

o

- jO . 1 1 25

-jO. 1 l25

ft

ft O .005638� O .01 3645 �

0 - jO .037587 -jO .06 25 jO . 1 5

o =

O.003759

jO . l

o

Ia

Tie lines

IS

Is

I I I I I

: - System "

,

-;:-

r

(a)

® 5

_

-

System r

I I I I I I 1 I _ I 1 I

7

-I b

15

Ib

-

...-

1a - 11--

-

I I I I I I

( b)

: r

-

®

17

19

Is

System

II

FIGURE 14.5 ng Two systems with: ( a ) i ntercon n ecting tie-li n e i m p e d a nces Za a n d Zb ; ( b) equiva l e n t compensati current i njections.

1 4.2

PIECEWISE SOLUT ION OF INTERCONNECfED SYSTEMS

601

14.2

PIECEWISE SOLUTION OF' INTERCONNECTED SYSTEMS

The p rocedures developed in Se c. 14. 1 have a particular application in the piecewise solution of a large network. Piecewise methods can be of benefit when individual power companies, joined together by tie l ines to form a large i nterconnected system, are using l inea r system models for p roblem solution such as in fault analysis. We recall tha t a system model is l i near if all currents are directly proportional to the corresponding voltages. As a first step, each utility could form its own Z bus and solve i ts own n e twork as if it were isolated . In a subsequent step results can be exchanged between utilities so that each can mod ify its solutions to account for the effects of the i nterconnections. Figure 1 4 .5( a ) shows two systems with a common reference node. The systems arc independent except for the branches a and b, which are tie lines i n t e r conn e c t i ng t h e two systems. Sy s t e m I conta i ns b uses CD to G) whereas Sys t e m I I con t a i n s b uses ® to ® . Current i nject ions / , to 19 a re assumed known at the buses of the two systems, and it is requ ired to calculate the bus voltages V{ , V� , . . . , V� fo r the overall interconnected system, including the tie-line impedances Z a and Zb ' using the known vol tages available for each separate system. Let Z I denote the 5 X 5 bus imped ance matrix for the stand-alone network of System I conta ining buses CD - G) . Likewise, let Z I I denote the 4 x 4 bus impedance matrix constructed separately for the st and-alone network of System II containing buses ® - ® . Suppose that the current injections I I to Is cause voltages VI ' V2 , . . . , Vs at the buses of System I when the tie l ines are open, and that injections 16 and 19 cause voltages V6 , V7 , . . . , V9 in System I I under the same condition o f open ties. The b u s impedance equations for the overall system without interconnecting tie lines are then given by .

II

VI V2

V =

Vs V6 V7

-

0

Z\ 0

Z II

12

15

( 1 4 . 19)

16 17

--- �

V9

Z bus

19

,

W hen tie-l ine impedances Z a a nd Zb are added between the two systems, tie- l ine currents Ia a nd Ib will flow from b uses @ and G) of System I to b uses ® and (j) of System I I , as shown i n Fig. 14.5(a). We can calcula te the ensuing bus-voltage changes from VI ' Vz, " " V9 to V{ , V� , . . . , V� in the interconnected system by using the compensat;ng currents of Fig. 14.5(b) in the manner described in Sec. 14. 1 . Since I . dd I b are both assumed to flow from System I a

.

602

CHAPTER 14

ZbUl M ETIIODS I N CONTINGENCY ANALYSIS

to System II, we set m = 4, n = 6, p = 5 , and q = 7 in fanning matrix Z of Eq. (14.14) for the present illustrative example, which yields

[ Z(I) Z(Z(22)) Za Z( O Z (2) 1 b

a

a

Z =

b

44

+

66

+

76

( 1) Z 54

( l) + Z 45

+

55

+

( 1 4 .20 )

67

z (2) 77

+

Zh

We have added s u p erscr ip t s ( 1 ) and (2) to identify more read ily the respect ive elements of Z \ and Z u . S i nc e vo ltag es a re known for the separate systems, t h e tie-line currents can b e calculated a s i n Eq. (1 4 . 1 4) a s fol l ows:

[Ia 1 [ Z� ) Z� )

b

a

a

Ib

b

1 Z

+_z_a __Z__4( �_) _+_Z_g_)_ ---l-zW + zW z W + zW + +

-

l

( 1 4 .2 1 )

b

The unknown voltages at the buses of the overall interconnected system, including the tie lines, can now be calcu lated from Eq. ( 1 '+.6), which here assumes the form VI V2

V'1 V'2 v.'5

V5 V6 V7

v.' 6

V'7

lV

�

where i n this case

A c = [AcI I A cId

a

b

0

Zr 0

Z il

A� [ �: 1

( 1 4 .22)

V9

CD (i) G) @ G)

[00 00 .....

0 0

-.....Ad

0

1

0 1

® (]) ® ®

0

- 1

0

-1

A ell

00 �1 ( J.4 .23)

1 4.2

PI ECEWISE SOLUTION OF INTERCON NECTED SYSTEMS

603

®

l z4( 5)

=

®

! z5( 4 )

z"

CD I

( 2) Z 76

'1

FIGURE 14.6

ZI(} )

-- Z67 (2 )

The T h e ve n i n e q u iv a l e n t c i rc u i ts o f Syst e m ] a n d Sys t e m I I o f Fig. 1 4 . 5 a n d t h e pat h s t raced by t h e

ZI(})

l i n e curre n t s la a n d lh ' T h e e l e m e n ts

t h e e l e m e nt s

b e l o n g t o t he b u s i m p e d ance m a t rix Z \ of System I a n d

b e l o n g t o Z l l o f System ] 1 .

The amount of compu tation involved in fi nd ing t h e overal l solution by piecewise methods can be considerab ly reduced, especially since the only matrix inversion is Z - I , which is of smal l d imension depending on the n umber of tie l in es. It is worthwhile to rearran ge m at rix Z of Eq. ( 1 4 . 21 ) in the fol lowing form:

Z

=

@ G)

[

2 44 (1)

2 6(2) 7

2(1) 54

2 77 (2)

t i e l i n es

]

( 1 4 .24)

from Z I I

sInce then it shows how Z is formed by inspection of the bus impedance matrices Z I and Z l l of the separate systems. Evidently, we need only extract from Z I the submatrix of elements with subscripts corresponding to the bound­ ary buses to which the tie l i nes are incident, and l i kewise for Z I I - Then, we add together the two extracted submat rices and the impedances of the tie l ines so t hat the d iagonal entries comhine in Z t o give the impedance of the total path to reference through each t i e l i n e . The Th6ven in eq u ivalent circu its plus the tie l ines of Fig. 1 4.6 graph ically exp lain the in terpretation of Eq. ( 14.21) for our example and also show that the off-diagonal elemen ts of Eq . ( 1 4. 20) represent the total impedances common to t h e p aths through the reference node of the circu lating tie-li ne currents. In the particular system of Fig. 14.5 the t ie l ines h ave d istinct pairs of bou ndary buses- @ - ® for l in e a and (3) - (j) for l i n e b. Figure 1 4.7 shows a more general situ ation w i th the systems now i nterconnected by fou r tie l ines a, b , e, and d between the boundary buses ® ' @ , and ([) of System I, a n d

604

CHAYrER 14

Zbus METIiODS IN CONTINGENCY ANALYSIS

1

FIGURE 1 4.7

Two syste ms interconn ec ted by four t ie l i n e s havin g i m ped a n ces Z . a' e 1 e m e nt o f bus Impe da nce m a trix ZI is and of Z I I is

zHl

zW,

Z Z 0'

( '

an

d Z

d'

T he

typlc ' aI

boundary buses @) and ® of System I I. The two submatrices to be formed separately from the elements of Z I and Z I I are always squ are and symmetrical with row-column dimensions equal to t h e number of tie lines. Thus, for Fig. 1 4 . 7 we have the 4 X 4 sub matrices a a

b C

d

(I)

Z PI'

Z (l

)

PI'

Z(

l)

qp

Z r(

I)

p

Z

b

c

(l)

(I)

) Z(I

I)

(I) Z (lr

pp

Z(1

)

pp

l) Z q( p

(I) Zrp

ZIl q

Zp(

q

(I) Z qq

) Zr( qI

From Z r

d

a

IJ r

(l Z qr

)

(I) Z rr

a

b

c d

(2)

b

c

d

Z fn nl

(2) Z I1lll

(2) Z nJ Il

(2) Z"l n:

) Z n( 2m

2) Z n( n

2) Z n( il

2) Z 11( m

2) Z !(l I n

(2) Zrn nl

2) Z n( n

(2 ) Z In n From

2) Z n( il

Z/�;�

2) Z I(l rn

2) Zn( l lll

Z ll

The rows and columns of both submatrices are l abeled Q, b, c, and d corre­ sponding to the four tie lines, and the entries in each submatrix h ave subscripts relating to the boundary buses of its corresponding system. Note that the diagonal elements in each submatrix are diagonal elements of the bus imp edance matrix for the corresponding system. Each submatrix can be formulated from the corresponding bus i mpedance matrix by using an equation similar to Eq. (14. 1 1) or by applying the following rules sequentially: 1.

For e ach individual system of Fig. 1 4.7, arrange a submatrix with one row and a corresponding column for each tie line.

14.2 PIECEWISE SOLUTION OF INTERCONNECTED SYSTEMS

605

each submatrix, se t the diagonal element for a given tie line equal to the driving-point impedance with subscripts the same as the boundary node to which the tie line is incident within the particular system. For example , in Fig. 1 4.7 tie line c is incident to boundary bus ® of System I and boundary bus ® of System II. Hence in the submatrix for Sys t e m I we set the d iagonal element for tie line c equal to the driving-point impedance Z � � of Z I ' whereas in the submatrix for System II we set the diagonal element equal to Z�� of Z II ' I n each submatrix, set the off-diagonal elements equal to the transfer -impedances with row-column subscripts aligned and consistent with those of t he d iagonal elements. For instance , in the submatrix for System I a l l row c elements have the first su bscript q corresponding to the diagonal element Z��, whereas column b entries all have the second subscript p corresponding to the diagonal element Zl���'

2. In

,

3.

The sub matrices so formed are a dded together and the result is then added to the diagonal matrix of tie-line impedances. For the tie-line connec­ tions of Fig. 14.7 we obtain c d b a 2 z(l) z(l) + z(2) z(1) + z(2) Z + z(l) + z(2) a pr + Z ( ) z(l) + Z( 2 ) z(l) + z(2) + z(2) z(l) + Z(2) Zb + z(l) pr nn nil b pp Z= Z(1) + Z(2) Z( 1 ) + z(2) Z + Z( I ) + Z(2) z(l) + z(2) c z( I ) + Z(2) 1 Z d + z(I) z(rpI ) + Z(2) z(l) + Z( 2) d rr + Z(2) a

pp

mm

pp

TIm

qp

nm

rp

mm

pp

pq

m Tl

mm

mn

nm

pq

qp

nn

c

m fl

qq

nn

rq

mn

qr

nm

mm

( 1 4 .25) an d the tie-line currents are given by vp - v Vp - � Vq - Vn � - v'rr

m

= Z-I

( 14 .26)

open -ci rcu it vol tage d rops

Each diagonal element of Z is a pa th or loop impedance associated with the correspond ing tie-line current which circulates from the reference bus through System I, then through the tie line, and back to the reference bus through System II. The off-diagonal elements of Z are the impedances common to the pa ths of those circulating currents which are the cause of the bus-vol tage

606

OIAn"ER 14

Zbus M ETII ODS I N CONTINGENCY ANALYSIS

changes when the system is interconnected. Thus, Z can be interpreted as a symmetrical loop impedance matrix formed from the bus impedance submatrices

®

® ® 0

®

0

Zpp

Zp q

Zpr

Zqp

Z qq

Z qr

Zrp

Zrq

Z rr

@ ®

Extracted from Z I

@)

®

Zmm

Zmn

Zn m

Znn

Extracted from

ZII

By i nv e r t i n g these submatrices, we o b t a i n b u s admittance s u b m a t r i c c s wh ich represent System I and System I I at their boundary buses on which the tie l ines term inate. Equivalent mesh circuits, called Ward equivalents, can be drawn corresponding to the bus admittance submatrices. Ward equivalents are further discussed in Sec. 1 4.6. The reactance diagram of Fig. 14.8 shows two systems, System I and System II, interconnected by three tie lines a, b, and c. The per-unit reactances o f the elements are a s marked. Note that each system is connected through a reactance to ground, which serves as the common reference for the bus imped ance matrices Z I and Z I J of the separate systems without tie li nes. Table 1 4 . 1 gives the numerical values of Z l a nd Z l l , along with known bus voltages for each individual system without tie li nes. P redict the steady-state voltage changes at the buses of System I when the t hree tie lines are simult aneously closed between the systems. Use the piecewise method based on Z bus '

Example 14.2.

®

) 0 .05 7

5

@

) 0 . 0625 ) 0 . 04

-

j O.05

jO.1

/; lb

jO.OB j5.0 -

Ie

jO.1

j O.05

CD

jO.05

Syste m I

FIGURE 14.8

-

I I I I I I I I

®

j O.OB

- fa

: - Tie lines

®

j O.04

jO.04

•

I I I I I I I I

jO.05

j2.5 -=-

: - System II

Reactance d i agram for Exampl e 1 4.2 show i n g p e r - u n i t values of b ranc h i mpedances. Bounda,ry of buses @) and CD o f System I are connected by t h ree tie lines to b ound ary b uses ® a n g (j) System II.

1 4.2

P IECEWISE SOLUTION OF I NTERCONNEcrED SYSTEMS

607

TABLE 14.1

Bus impedance matrices and voltages for Example 14.2

CD

�

®

@

G)

j5 .000000 j5 .000000 j5 .000000 j5 .000000

j5 .0063 17 j5 .000000 j5 .03585 0 j5 .009476 j5 .01 4529

j5 .042198 j5 .000000 j5 . 009476 j5 .063298 j5 .0 17056

j5 .011 371 j5 .000000 j5 .014529 j5 .017056 j5 .026 1 53

CD - j5 .061466 j5 .000000 Zr =

Z II =

Vr -

W G) @ W

j5 .000000 j5 .0063 1 7 j5 .042 198 j5 .01 137 1

® .

® (J) ® ®

CD @ ®

j2 .549832 j2.5 12121 j2.500000 j2.506734

-

1 .000000 0 .986301 0 .984789 @ 0 .993653 0.998498

�

(j)

j2.512 1 2 1 j2 .527273 j2.500000 j2 .5 1 5 152

®

j2.500000 j2.500000 j 2 .500000 j2.500000

+jO.OOOOOO -jO .083834 -jO.095 1 08 -jO.045583 -jO .05 4 795

VII =

®

j2 .506734 j2 .5 1 5 152 j2.500000 j2.530640

® OJ ®

0 .991467 -jO .039757 0 .999943 + jO .010640 1 .000000 + jO .000000 ® 0 .993820 -jO . 020792

Solution. The main steps in the sol ution i nvolve forming t he matrix Z, finding the

tic-line currents, and using these currents as i nj�ctions i n to System I to determ i ne the bus-vol t age changes . Buses @ and G) are the boundary buses of System I since tie l i nes a a n d b connect to bus @ and tie line c connects to bus G) . Therefore, the subma trix for System I according to our ru l es is

l

a

zll' b z��)

a

c

b

c

Z44( 1 ) Zll' Z44( I) zW Z(54l ) Z54( l ) z W

l

In System V buses ® a n d

a a =

b c

(j)

b

c

[jS .063298 j5 .063298 jS .01 7OS6 1 j5 .063298 j5 .063298 j5 .01 7056 j5 . 01 7 056 j5 .0 17 056 j5.026153

are the boundary buses since tpe tie line

a

608

CHAPTER

14

Z bus METHODS IN CONTINGENCY A N A LYSIS

connects to bus

@ whereas tie lines b and c connect to bus (j) . H ence, we have

66 6 1 [ 6 b

a

[Z (2 ) b Z 7( 2) C Z ( 2)

a

Z ( 27) Z 7(27) Z ( 2)

7

a

b

c

i 2 .549832

j2 . 5 1 2 1 2 1

i2.S 1 2 1 2 1

b j2.5 12121

j 2 .527273

i2 .527273

j2.527273

i2 .527273

c

77

zg) zW

a

=

zw

c

j2 . 5 1 2 1 2 1

1

The diagonal matrix of tic line impedances is

l

a

b c

The

is

b

Z(1) Z = b Z�!) c Z (1 )

.

iO. I O

:

jO.04

j

() Z 44 I (l ) Z44 Z(I)

Z,I ) zw Z ( I)

z ( 2) + b zW c Z (2) zJ 7 a

Zb

+

c

i 7.663 130

j7 .5754 1 9

i 7 . S29 1 77

b j7.5754 1 9 j7 .690571

j 7 .544329

j7 .529 177

j7.544329

[ VV54 -- VV6 ] [

[ 6 67 67 6 1 a

b

c

1

c

a

a

IlIa ] Ie

.

c

t h er efo re c a l c u l a t e d as fo l l ows:

a

=

j o.05

b

[ 4 45 54 54 5 1 [�' l

Z m atrix

a

a

Ie

j7 .593426

J

b

Z ( 2) Z (772 ) (2 ) Z 77

c

2 (2 ) Z 7( 27) Z ( 2) 77

S ince the t ic-line currents la ' lb , and a re chosen with posi tive d irect i on from System I to System II, we have, as i n Eq. ( 14.26) b =

Z- I

V4 - V7

7

=

Z-l

[. IIab ] r Ie

( 0 . 993653 - }0 .045583) - (0.991467 - }0 .039757) (0 .993653 - jO .045583) - (0 .999943 + j O . 0 1 0640) ( 0 .998498 - jO.054795 ) - ( 0 .999943 + } 0 .0 10640)

]

After substituting numerical values for Z and inverting, we then compute

=

0 .361 673 - iO.039494 - 0 . 130972 + jO .0441 1 4 per u ni t - 0.2371 05 - jO.004479 .

1

1 4 .2 P IECEWISE SOLUTION OF INTERCONNEcrED SYSTEMS

609

The branch-to-node incidence matrix for the tie lines has a distinct part for each system,

A,

�

[ Ad I A,II 1

�

�l

CD 0 0

0

@ G) @ W @ (j) ® 0 0 0

0 0 0

1 1 0

0 0 1

- 1 0 0

and from Eqs. ( 14.22) and ( 14.23) the bus-voltage changes given by

0

-1

-1 l>.V1

0 0 0

®

�]

i n System I are

From Z ,

a

CD - j5 .042198

(1) - 0)

@ W CD

=

Q) 0) ® Q)

j5 .000000 j5 .009476 j5 .063298 j5 .017056

b

c

j5 . 042198 j5 .000000 j5 .009476 j5 .063298 j5 .017056

j5 .01 1371 j5 .000000 j5 .014529 j5 .017056 j5 .026153

0 .000849 + jO .024981 0 .000705 + jO .032020 0 . 000684 + jO .033279 - 0.000921 + jO .021461 - 0 . 000667 + jO .034286

l

0 .3 6 1 673 j O . 039494 - 0 . 130972 + j O . 044 1 1 4 - 0 . 237105 - jO . 004479 -

per unit

Exa mple 1 4.3. Dr
Solution. The Theveni n equivalents are shown i n Fig. 14.9 with numerical val ues for the impedances. The open-circuit voltages V4 , Vs , V6 , and V7 have the n umerical values given in Table 1 4. 1 . Replacing the impedances i n the System II equivalent of Fig. 14. 1 0( a ) by their reciprocal values gives the admittance circuit shown i n Fig. 1 4 . 1 0(b). Using the Y-ll transformation formulas of Table 1 .2, we

,

610

CHAPTER 14

Zbus METHODS IN CONTINGENCY ANALYSIS

jO.OO9097

®

(J)

Zc

j O . 0 15 152 +

-:-

Vs

-

Zb

V.

CD

+

j5.01 7056

(6)

za

j O .046242

jO.0377 1 1

j2.512121

FIGURE 14.9 Thevenin equivalent circuits for Exa m p l e 1 4 .3 .

now convert to t h e Ward equivalent m esh circu i t shown in Fig. 14. 1 O ( c). The reader can easily check that the bus adm ittance matrix for the Ward equ ivalent agrees (within round-off e rror) with the i nverse of the bus impedance s ubmat rix extracted from Z II at the boundary buses of System I I . That is,

267 ]-1 2 77

=

[

j2 .549832 j2 .5 1 2 1 2 1

j2.512 12 1 j2 .527273

]

-

1

® (J)

[

@

-j 1 8 . 949386

j 1 8 .835777

j 1 8 .835777

-j 1 9 . 1 1 8533

]

Piecewise methods give exact results for the overall system if currents and voltages a re directly proport ional. For then the system is linear and the principle o f superposition applies, which is t h e basis of the piecewise approach. The assumption of linearity has to be reevaluated in power-flow stu dies, for example, where the active power P and reactive power Q are modeled at the buses rather than the current injections. The load current is therefore a nonlinear function of the applied voltages and the operating point of the ove rall system m ay not be consistent with the individual operating points of the separate systems. In such a case the methods developed in Chap. 9 for solving the power flows in large-scale systems should be used .

1 4.3

ANALYSIS OF SINGLE CONTINGENCIES

jO.015152

611

-j65.997888

®

® jO.03771 1

-j26.517462

j2.512121

(a)

-jO.398070

-

-

(b)

(J) -jO.2828 -j 18.8358 -jO.1 136

®

(c)

Fl G U RE 1 4.10 E q u iva l e n t c i r c u i t s for Sys t e m I I of E x a m p l e 1 4 .3 s h ow i n g :

(a)

T h eve n i n

i mped a nces; ( b ) reciproc a l s

o f T h e ve n i n i mp ed a nces; ( c ) a d m i t t a nce circu i t o f W a r d equiva l e n t .

14.3 ANALYSIS OF SINGLE CO NTI NGENCIES Whenever a transmission lin e or transformer is removed from service, we say

that an outage has occurred. Outages may be planned for purposes of sched­ u led m aintenance or they m ay be forced by weather conditions, faults or other contingencies. A line or transformer is deenergized and isolated from the network by tripping the appropriate circuit breakers. The ensuing current and voltage transients in the network quickly die away and new steady-state operat­ ·ing conditions are established. It is i mportant for both the system operator and

612

CHAPTER 14

Z bus METHODS IN CONTING ENCY ANALYSIS

system planner to be able to evaluate how the line flows and bus voltages will be altered in the new steady state. Overloads due to excessive line c u rr e n ts m ust be avoided and voltages that are too high or too low are not acceptable because they render the system more vulnerable to follow-on (cascading) outages. The large n umb e rs (often, hundreds) of possible outages are analyzed by mea ns of a contingency analysis or contingency evaluation program. Great the

precision is not required in contingency analysis since the primary interest is in knowing whether or not an insecure or vulnerable condition exists in the steady state following any of the outages. Accordingly, to test fo r the effects of line and transformer outages on the bus voltages and line fl.ows in the network, approxi­ mate ac power-flow techniques are generally employed since they can provide a fast solution of the m a ny t e s t cases w h i c h n e e d to be r u n . A n a l t e r n a t i v e l inear method based on t h e syst e m Z hus i s d �s c.: r i betl in t h i s s ec t i on , a n d i n S ec. 1 4. 5 w e show that this Z b u s method is d i re c t l y re l a t e d to t h e d c po w e r flo w a p proxi ­ mation of Chap. 9. Starting with an initial power-flow solution or state estimate of the system, 1 a ll loads and generator inputs to the system are converted to equivalent current injections, which henceforth are regarded as constant. For discussion purposes we assume that network branches are represented by series impedances. We must remember, of course, that if line-charging capacitance and a l l other shunt connections to neutral are omi tted, the neutral becomes isolated and another bus must be chosen as the reference node for the system -

Z b us ·

Before considering contingencies, we first note that the current flow in each line of the system will change if an additional current is injected into any one of the buses. The vol tage changes due to an additional current 11 1m being injected into bus ® of the system are given by

where Z bu s is for the normal configuration of the system. The changes in the voltages of buses CD and (j) can b e w ri t te n ( 1 4 .28)

I See Cha p . 1 5.

14.3 ANALYSIS OF SINGLE CONTINGENCIES

613

buses CD and Q) has impedance Z then the change in the line current from bus CD to bus (j) is

If t h e line connecting

CJ

Il ]'I J.

( 14 .29)

=

This equation leads us to define the current-injection distribution fiactor K '),. . m ' K I), m �

Il [ .

.

') __

II Im

- Z. un

z-

)m

( 14 .30)

Thus, whenever the current injected i nto bus @ changes by Il Im , the current flow in line CD - CD changes by the amount Il Iij Kij, !:l. 1m ' We are now in a pos ition to consider contingencies. In the normal configu ration of the system if a par ticular line current is excessive, it may be possible to remove the overload by reducing the current inj ected into the system at one bus and correspondingly increasing the current injection at another bus. Suppose that the current injected at bus ® is changed by !:l. Ip whereas the current injected at bus ® is changed by !:l. Iq . By the princi ple of superposition the shift in the current injection from bus ® to bus ® causes the current in line CD - (]) to change by the amount =

m

( 1 4 .31 )

Due t o this application, the current-injection distribution factor is also called the current-shift distribution factor . In Sec. 1 4.5 we shall see that shifting current inj ection from one bus to another is equivalent to shifting real power inputs between the same two buses in the dc power-flow model. On this accou nt, current-shift distribution factors are often called generation-shift distribution factors. Equation 0 4.31 ) shows that precomputed tables of such factors can be used to evaluate how a proposed shift in current injection from bus ® to bus ® cha nges the level of curren t in the line CD - OJ . Let us now consider removing from the network a line or transformer of series impedance Z between buses @ and ® . We know from Sec. 8.3 that the outage can be simulated by adding impedance - Za between the buses in the preoutage Thcvenin equiva lent circu it shown in Fig. 1 4 . 1 1 . Closing switch S connects the impedance - Z a between the buses, and the loop current Ia then flows as shown. With Zmn Zn m ' we see from the figure that a

=

fa

= 2 m ( m

v - V + Znn - 2 Zm n ) - Z m

n

v'n - v"

( 14.32)

tl

Here Vm and Vn are preoutage bus voltages and Z th , mn = Zm m + Znn - 2 Zm n is the preoutage Theveni n impedance between buses @) and ® . The loop ,

614 CHAPTER

14

Zbus

METHODS IN CONTINGENCY A N ALYSIS

s

��I

==�I--�I----� FI G U R E l · t 1 1

Preo u t age Thcvc n i n e q u iv a l e n t

c i rc u i t for sim u l a ti n g ou tage of line

@) - '?!) ,

the same effect on the voltages of the preoutage network as two compensating currents, 11 1m = - Ia into bus @ and 11 In = Ia into bus ® . The expression for the resulting current change 11 Iij from bus CD to bus (]) is given current 1a'

h as

by

/1 /.I),

=

K U. . , m 11 Im

Substituting for

1a

+

K Il. , n 11 In =

[ ( Z,I n

-

z.1 m ) - (Z,J,n - Z·Jm )] Ia Z

( 1 4 .33)

c

from Eq. ( 14.32) gives ( 14 . 34 )

Before the outage of line @ - @ the vol tage drop Vm - v" is due to the current ( 1 4 .3 5 ) From Eqs. 04.34) and (14.35) we find that due to the outage of the line @)- ® is "

the curren t change in the l ine CD - CD

14.3 ANALYS IS OF SINGLE CONTINGENCIES

615

Expressing this change as a fraction of the preoutage current lmn gives ( 14 .36)

where L ij, m n is called the line-outage distribution factor . A similar distribution factor can be defined for the change in current in any other line of the network. For instance, the line-outage distribution factor expressing the change in cur­ rent in line ® - ® of impedance Zb due to the outage of line @ - @ of impedance Z a is ( 1 4 .37)

The distribution factors L 1j, and L pq, m n contain numerator and denominator terms which can be easily generated from the triangle factors of Ybus ' as demonstrated in Secs. 8 .5 and 1 4. 1 . It is also evident from Eqs. ( 1 4. 36) and ( 1 4.37) that tables of distribution factors can be precomputed from the line impedances and the elements of Z bus for the normal system configuration. These tables can be used to evalua te the impact of any single-line outage on the current flows in the remaining l ines. Following the outage of line @ - @ , the new currents in the l ines are given by equations of the form m il

rIJ.

=

l.I).

+

Ll I·I)· = I.I).

+

L I. ). , mn Im n

( 1 4 .38)

Thus, if the actual preoutage current fm n in line § ® is measured or estima ted, Jine overloads due to excessive current changes caused by the loss of that line can be identified from Eq. ( 1 4 .38). The numerical examples which follow illustrate the close agreement often found between the values of the line flows calcula ted by distribution factors and by ac power-flow methods. The same level of agreement is generally not found in correspond ing calculations of bus voltages. -

Figure 1 4. 12 shows a small power system of five buses and six lines, which is the same as System I of Fig. 14.8. When the l ines are represented by their series reactances, the system Z b us is the same as that given for Z I in Table 14.l. The P-Q loads and generation shown i n Fig. 1 4 . 1 2 correspond to a base-case power-flow solution of the system, which yields the bus voltages tabulated under Case a i n Table 1 4.2. Using distribution factors, predict the current i n line G) - ® when line � W is outaged u nder the given operating conditions. Compare the prediction with the exact value obtained from an ac power-flow solution of the same change case. Examp l e 1 4.4.

-

616

CHAPTER 1 4

Zbus

METIIODS IN CONTI NGENCY ANALYSIS

100

®

�

20

®

140

�

� 10

I V5 1

56. 1039

®

) 0 .0 8

)5.0

)0.1

) 0. 05

I VI I

174.9982

Slack

01

= =

0°

CD

�

1 .0

p

---+- Q �

--t-+-

26.3928

FIGURE 14.12

Reactance d i agram for Exa mple 14.4. Vo ltages a n d reacta nces a re i n per u n i t . P a n d Q val u es are i n megawa tts a n d megavars for t he base-case sol u t ion of Table 1 4 .2'. Syst em b a se i s 100 MVA.

The line-outage distribution factor L 53 5 2 is required. Denot i ng t h e series i mpedance of lines cI>- w and cI> - w by ; 5 2 a nd z 5 3 ' respectively, we have fro m Eq. 04 .36)

Solution.

r

Substituting nume rical values, we obta i n L 53 . 5 2

-

-

-

jO.04 ( jS .026153 - jS .OOOOOO) - ( j5 .014529 - jS .OOOOOO) jO.05 ( i5 .026 15 3 + j5 .OOOOOO - 2 X j5 .000000) - jO .04

= 0 . 67 1 5 3 3

�

=

1.0

"----I :� 0

) 0 . 0 625

145

�

)0 .026 1 53

J

,

1

1 4 .3

ANALYSIS OF SINGLE CONTINGENCIES

617

TABLE 14.2

Bus voltages from power-flow solutions of the system of Fig. 14.12 Change cases

Base case

Bus number

b

a

CD

1 . 000000 +jO.OOOOOO

W

1 . 000000

G)

0.98 4 789 -jO.095 1 OR

@

W Case Case

1 .000000 +jO.OOOOOO

0.968853 -j O . l 08 1 08

0.983733 -jO. 1 06285

0.966 186 -jO . 1 24878

0.970492 -jO.095000

0.97 7822 j O OR8536

0 .98 1 780 -jO. 1 2 1 225

0. 975005 - jO. 1 1 6235

0 .97930 1 -jO.066882

0. 9925 8 2

0. 993489 -jO.04 756 1

0 .9908 1 6 -jO.040000

-

.

() V9443() -j O.033446

0 . 998498 - jO ()54 795

0 . 999734 - jO. ()23079

b:

els e :

line

G) - G)

45 - M W

e:

l i nes '

l i ne

G)

- j O ()56858 0 . 996444 - j O DH4 253

o u t of se rvice .

(J)

-

0. 998201 jO. 059904

0.999984 + jO.005738

G) - G) o u t o f s e rv i c e p l u s Case c h a nge G) - G) and w - @ o u t of serv i c e

("

Case d :

Case

+ jO.OOOOOO

D . V<)3(,53 -, 0.0455tlJ

Cha n ge from base

1 . 000000

e

1 .000000 +jO.OOOOOO

+ jO. OOOOOO

0.9863 0 1 -jO. 083834

d

c

s h ift from h u s

t o bus

c

O ther l ine-outage distribution factors calculated in a similar manner are shown i n Table 14.3. From the Case a power- flow results in Table 14.2 ( now indicated by superscript a ) the preoutage cu rren ts in l ines W - @ and W - G) a re calculated as follows: (0 .998498 - jO .054795) - (0 .986301 - jO .083834) jO .04 =

1;.1

=

V:'I 5

-

va J

=

0.725975 - j O .304925 per unit ( 0 .998498 - jO.054795) - (0.984789 - jO .095 1 08) jO .05

'= 0 . 806260 - jO .274 1 80 per unit

The predicted current change in l ine 2

W - G)

therefore is

= 0 .671533 (0.725975 - jO.304925) = 0 .4875 1 6 - jO.204767 per unit

618

CHAPTER 1 4

Z bu. M ETHODS I N CONTI N G ENCY A N A LYSIS

TABLE 14.3

Line-outage and generation-shift d istribution factors for the system of Figure 14. 1 2

8 G I


@) -@ W -W W -W

Line-outage distribution factors -L ii, m n

w -CD @ -0

W -W G) -W @ - G) @ -@

@ -0

1 .000000

1 .000000 - 0, 262295 - o, n7705 - 0.2{J2295 1 .000000

- 0 , 1 48 1 48 1 - 0.328467

i

0. 262295

I

1

0 .328467 I

0. 1 48 1 48

O J)7 1 5 33

0.737705

O)l5 1 8 52

O.2()22�5

- 1 ,000000

- 1 .000000

0 , 1 48 1 48

(J.()7 1 5 33

i

0 . 3 28467

W -W

- 0 1 48 1 4 8

1 . 000000

0 1 48 1 48

- 1 .000000

- \ ,000000

0, 262295

O.XS 1 85 2

0. 737705

.

.

I I

0. 262295 0, 1 48 1 4 8

CD -(j)

Generatio n - s h i ft d i stribution fa ctors-Kj j , p

Cu rre n t c h a n ge in l i n e

I njected

@ -CD

i nto bus

CD W

Outage of line

- 0.61 4656

- 0.38 5 344

0 . 1 0 1 074

0.284270

0.1 0 1 07 4

- 0.385344

- 0, 1 1 3708

0. 1 1 3701:;

0. 2324 70

0.653822

0,232470

0. 1 1 3708

and the p redicted postoutage curre nt from bus

I� 3 = IS3 + tl Is3

= =

(0 .80626 - jO .274 1 80) 1 .293776 - j0 .478947

G) + =

to bus

G)

is

(0.4875 1 6 - jO .204767)

1 .380L - 20 .3 1° per

u nit

W i t h line G) Q) O U I o r se rvice 'l IHI ; i l l operat i n g co n d i t io n s o l h c f\v i sc t he s a m e as in the base case, a new ac power-now sol ution g ives t h e c hange-case results listed u nder Case b in Table 1 4.2. From these results we can calculate the corresponding current in line � - G) as follows: -

vl It3 = -

-

ZS 3

V!

--

(0 .999734 - jO.023079) - (0 .977822 - jO .088536) jO .OS

/ - 1 8 .5 1° -

=

=

-

1 .3091 4 - j0 .438240

=

1 .381

per u n i t

Thus, w e see that the predicted m agnitude o f t h e post outage current com pares very favorably with that calculated by the ac power-flow method. Example 14.5. The only change fro m the input data ' of the base-case power-flow solution of Example 1 4.4 is a shift of 45 MW from the generator a t bus CiJ to the

14.3

619

ANALYSIS OF S INGLE CONTINGENCIES

generator at bus CD i n the system of Fig. 14.12. The bus impedance matrix Z I shown in Table 14.1 therefore appl ies. Using distribution factors, predict the magnitude of the resu l t i n g current i n line ffi- @ and compare t h e predicted value with the exact value calculated from an ac power-flow solut ion of the same

change case.

comparison of results, let us assume t hat the changes in the megawatt i np uts at buses G) and CD a re exactly equal i n per unit to the corresponding changes in the current injections at those two buses. That is, fl. / I = 0.45 + jO.O pe r u nit and fl. Is -0.45 + jO.O per unit. Accord ing to Eq. ( 4.3 ), the current change in l i n e G) - @ is g iv e n by Solution. To facilitate a quick

=

1 1

w he re

and

KS 4 . 5

=

ZS5

- Z45

Z S4

---­

Substitut ing the impedance values from Z l of Table 14 1 , we obtain KS4 1 =

( j5 .01 1371 - j5 .0421 98 ) jO .08

K 54 , s

= - 0.38534

=

=

( j5.026153 - j5 .017056 jO.08

)

0.11371

These and other current-shift distribut ion factors are already tabulated i n Table The current change now is

14. 3 .

fl. /S 4 = =

- 0.38534

( 0.45 + jO .O)

- 0.225 + jO.O per u nit

The predicted value of ff4

x

Is 4

+

0.11371( - 0 .45

+

jO.O)

is found by adding this change to the preou tage curre n t 14.2. Th at is,

calculated from t h e base-case sol u t ion o f Tabl e 1f4

v5a - V4a

(0 .998498 - jO.054795) - (0.993653 - jO.045583) jO.08

= ---4

/�4 = /;4

ZS

+

fl. 1;4

- 0.1 15150 - jO.060563 per u n i t

( - 0. 1 15 150 - jO.060563) + ( - 0.225 + jO.O) ! = - 0.340150 - jO.060563 = 0.345 - 1700 pe r unit

=

With bus CD retai ned as the sl ack bus and the 4S-MW reduction at bus � as the only change from the b ase case, a change-case power-flow calculation gives the ,

620

CHAPTER 1 4

Z'M METH O DS

results l isted u nder Ca se

o f 154 ,

Vsc e I54 . _

ZS4 -

V4c

_ _

=

I N CONT I N G ENCY I\Nl\ i.YSIS

c

in

Table 1 4.2, from which we calculate the exact value

( 0 . 9 96444 - jO .0842 53) - ( 0 . 992582 - jO.056858) jO .08 - 0 .342438 - jO .048275

=

0.346/

-

1 72°

per unit

Thus, w e see that the current-sh i ft dis tribution fa c t o rs g ive a pred i c t e d va l u e of 15 4 , which is acceptably close to the va l li e o b t a i n e d rro m a n ac p o w e r - fl ow s o l u t i o n .

14.4

ANALYSIS O F MULTIPLE CONTIN GENCIES

Contingencies can arise in w hi c h two or more lines a rc trip ped s im u l laneously, or where a line outage has already occurred and a shift in generation is being considered to determine if a line overload caused by the outage can be relieved. If tables of distribution factors for first contingencies a r e available, i t is n o t n ec essary to recompute the tables in order to study the effect of two simultane­ ous contingencies. Although the existing distribution factors assume a nonnal system configuration before the first contingency occurs, they can be combined into formulas for evaluating double contingencies. Examples follow which illustrate how this can be achieved. Suppose that line @- ® is carrying current 1 before it trips out of servic e If an overload occurs on another line CD - 0 due to the outage, it may , be possible to reduce the overload by decreasing the current injected into the system a t some bus ® a n d correspondingly increasing the current injected at another bus ® . Such current-shifting occurs when generation is shifted from a power p lant at bus ® to a power plant at bus ® . The distribution factors of the pr e ce di ng section can be employed to study this possibility of overload relief. Since t h e Z bus model is l inear, the l ine outage and the proposed shift in current will h ave the same combined effect on the overloaded line regardless of the sequence i n which the two events arc considered to occur. For our purposes it is more convenient to consider first the proposed current shift and then the outage of the line @) - ® . According to Eq. ( 14.31), a current shift from bus ® t o bus ® changes the line currents Imn a n d Ijj by t h e amou nts mn

.

11 f. · = K '· l· , P 'J

1:1 1p + K I l. . q

( 1 4 .39 ) 6. lq

The dist ribu t i on factors i n t his equation are those already precomputed for nor mal system configuration and 11 Ip and 11 Iq are t h e proposed changes i !l

the the injected currents at buses ® and @ . As a resul t of shifting the current, the

14.4 ANALYSIS OF MULTIPLE CONTINGENCIES

621

new line currents become F. I)

=

jI. ).

+ fl /,I ).

( 1 4 .40)

Then, if line @) - ® is opened, the additional change i n the current of l ine CD (]) can be calculated using the single l ine-outage distribution factor Lij, as follows: mn

-

( 14 .4 1 )

t h e total change i n the current f;j due to the combination of the l ine-opening and the current-sh ifting. The final value of the steady-state current in l in e CD (]) becomes

This is

-

( 14 .42) Accord ing to Eq . (14.38), the quantity within the brackets represents the current which would flow in line CD - 0 by opening line @ - ® without any genera­ tion shift also occurring. There fore, the net change of current i n the overloaded line due to shifting generation in the network, with line @) - ® already out of service, is given by Eq. ( 14.42) as ( 14 .43)

S ubstituting for

6. (j

and

6. Im n

from Eqs. (14.39) and rearranging, we obtain K;i. q

Thus, K; j , p and K;j, q are new generation-shift d istribution factors which t ake into account the preexjstng outage of line @ - ® . The new distribution factors can be quickly calculated from those already p recomputed for the normal sys t e m con fi g u r a t i o n . Exa mple 14.7 in Sec. 1 4.5 i llustr ates t h e simpli city of the calculations. By evaluating 6. I:�. , the system operators can determine the l evel of overload relief available from the proposed shift o f i nject ions. As another application, let us consider what happens if a second line ® ® is tripped out of service when the line @ - ® is already out of service in the network. If line ® - ® has series impedance Z a and l ine ® - ® has series impedance Zb ' then the simultaneous outage of the two lines can be simul ated by adding Z a and Zb between the respective end b lfses of the -

-

-

622

CHAPTER 1 4

Z bu s

METIiODS IN CONTI NGENCY A NALYSIS

lines. We h ave a similar situation in Sec. 1 4. 1 where equations arc developed for adding new lines to an existing system. Those equations apply directly to the present situation. If we allow for the fact t h at the impedances to be added now have negative signs, Eq. (14.14) becomes

( 14 .4S)

The symbols in this equation have the same mean ing as In Sec. 14. 1 . For example, Vm, � , Vp, and Vq represent bus voltages in the normal system configuration before any line outages have occurred . The voltage drops ( Vm � ) and ( Vp - Vq) are related to the preoutage currents, fmn in line @) - @ and Ipq in line ® - @) ' by -

1

pq

::.;::

Vp - V" Zb

-----

( 1� .46)

New d istribution factors for the two simultaneous outages can be developed by using Eqs. (14.45) and (14.46). It is helpful to present the development in a sequence of steps as follows: Step 1

To find expressions for currents fa and tion factors: •

Ib

in terms of the line-outage distribu­

Divide each row of Eq. ( 14.45) by its d i agonal element to obtain 1 1

( 14 .47) •

Express the new off-diagonal elements i n terms of the line-outage distribution factors Lpq , mn

Lm n , pq

=

( 14.48) =

1 4.4 A NALYSIS O F MULTIPLE CONTINGENCIES

623

The expression obtained is Zb -L pq , mn

1 Za -L mn , pq

� - Vq

1

Zb

•

Zth , mn - Za

Za

( 1 4 .49)

Z t h , pq - Zb

Solve Eq. (14.49) for fa and fb - L p q, m n

- L mn , pq

1

1 =

Zb

1

Za

Za

Zb

( 1 4 .50) •

Substitute for Vm - Vn and � - Vq from Eqs. (14.46) Zb2 Z th , mn - Za

1 =

Z a2

L mn , pq

L pq, mn

Za Z th , p q - Z b Zb

Zb Zth , mn - Za

( 14.5 1 )

I n this equation fm n and fpq represent t h e actual currents i n l ines @) - @ and ® - ® prior to their simul taneous outage. The currents fa and fb are the (nonphysical) loop currents discussed in Sec. 1 4 . 1 . They represent the outage of the two lines and . can be considered as compensating current i njections i nto the nonnal configuration of the system. Step 2

To find the current change •

6. 1ij

due to fa and fb :

Determine the bus-voltage changes due to the current injections ( - fa in to bus @) , fa into bus @) and ( - fb in to bus ®, fb i n t o bus ®). At bu ses CD and CD we obtain ( 14.52)

624 •

CHAPTER 14

Z bul

Zc)

METHODS IN CONTINGENCY A NALYS IS

Determine the current change fl Ii} = (fl V; - fl vj)/ impedance

Z

in line CD (]) of

c

-

( 14 .53) •

Express Eq. ( 1 4.53) in terms _

of

[

the

l i n e - o u t a g e d i s t r i b u t i o n factors

Z u ( 2;/1/ Zc

-

Z;II ) - ( ZjlT/ Zlh , 1Il 1l - Za

-

Zjtl )

1

( 1 4 .54)

L i j, p q =

[ Z lh , ml

The resultant equation is D.. lii

•

- l(

=

Za

-

Za

L 1 ) , 111 11

Substitute for la and 1b from Eq, fl f.I )- -

-

( 14 . 5 1 )

L ij, mn + Lij, p q L p q , mll_ 1 - L p q , mll L m ll , p q

____

J..:.. j . Inn

•

- Zb

b

to obtain

] ( J

Zlh , pZq I{ ':

L..

L -

I j , pq

IJ, m�

1 [ 10 1

+ L.

Lmn ,

L nt l l , p q

p q, mn

( 1 4 .5 5 )

1b

pq

] ][ 1 J

mn Ip q

1

( 14 . 56)

is the effective line-outage distribution factor which expresses the change in the steady-state current of line CD- 0) due to the outage of line @ ® when line ® ® is already removed from the network . A similar statement applies to [;ij,

1}i j ,

mn

-

-

pq'

. Thus, we see that the distribution factors already precom p u t e d fo r sing le­ line outages can be combined to produce distribution factors for two simultane­ ous l ine outages. The above developmen t can be generalized to three or more simultaneous outages, provided the assumption of constant current injections rema ins acceptable-which may not always be the case in practice. ExampJe 14.6. Using distribution factors, predict the change in the current of line

G)- G) when l ines @- @ and @- @ are simultaneously outaged , i n the system of Fig. 14. 1 2. Use the base-case bus data of Example 14.4 and compare results with those obtained by an ac power-flow ca l cu l a t i o n _

1 4.4 AN AL" SIS OF MULTIPLE CONTINGENCIES

Solution.

find that

By setting

5, j

i =

=

3;

m

=

5,

n =

2;

p =

5,

q

=

4

in

625

Eq. 04.56), we

where 152 and 154 are line currents of the base case and

The line-outage distribution factors for single contingencies a re now substituted into these formu las from Table 14.3 to obtain L'S3 , 52 L'S)

,

54

=

=

0 .67 1 533 + 0 .262295 x 0 .328�67 1 - 0 .328467 x 0.737705

=

1

0.262295 + 0 .671533 x 0 .73T'05 1 0 .737705 x 0 .328467

=

1

-

To confinn understanding of these particular numerical results, the reader should ref er to the system diagram of Fig. 14. 12 and consider line W W being taken out of service when line W - (1) is already outaged, and conversely. From Examples 1 4.4 and 1 4.5 we have -

152

153

154

=

0.725975 - jO .304925 per unit

=

0 .806260 - jO.274 1 80 per unit

=

-

0 . 1 1 5 1 50 - jO .060563 per unit

Therefore, /1 /53

=

( 1 .0) (0 .725975 - jO .304925 )

=

0 .6 1 0825 - jO.3654R� p e r u n i t

Adding this cu rrent change to the obtain 153 = 153

+

/1 15 3

=

=

+

( 1 .0)( - 0 .1 1 5 150 - jO .060563)

p r e o u tage

current

(0 .806260 - jO .274 1 80)

+

1 .4 17085 - j O .639668

1 .555

=

IS')

of the base case, we

L - 24 . 3°

(0.610825 - jO.365488) per unit

The ac power-flow results for the system with both lines W- W and 'W - @ out

626

CHAPTER 14

Zbus

M ETHODS IN CONTI NGENCY A NALYSIS

of service are shown in Table

t VSf I53 _

ZS3

V{

=

14.2 under Case

( 0 .999984

+

e,

wh i c h yields

jO .005738) - (0 .979301 - jO .066882) jO .05

= 1 .452400 - j0 .41 3660 = 1 .510/ - 1 5 .9° per

unit

Thus, t h e distribution f a c tor s predict a resu l t for I ISJ ! ' which is comparable to t h a t o f the a c power-flow cal c u l a t i o n . 14.5

C ONTIN GENCY ANALYSIS BY DC

M ODEL

We have already discussed the fact that approximate ac powe r-Row m e t h o d s are also used to analyze contingencies. The simplest method is based on the so-called dc power-flow model described in Sec. 9.7. The de model assumes that: • •

•

The system is lossless and each line is represented by its s e r i e s reactance. Each bus has rated system voltage of 1 .0 per u n i t . Angular differences between voltages at adjacent buses @ and ® are so small that cos om = cos On and sin om sin on = 8m - 8n rad. -

On the basis of these assumptions, the per-unit current flow from bus @ to bus ® in a line of series reactance jXa per unit is a real quantity given by I V,n l ( cos

( cos om

-

cos On ) + j ( sin Om jXa

JXa

8m + j s i n 8", ) -

-

sin 8n)

I V" I ( co s 8/1

+ j s i n 8,, )

( 1 4 .57)

and the per-unit power flow in the same line is ( 1 4.58)

is evident, therefore, that per-unit power and per-unit current are synony­ mous in the dc power-flow model where nominal system voltages are assumed. For convenience, let us agree that equality signs in the equations which fpllow replace the more accurate approximation signs. Then, the current in the line of

It

14.5

per-unit

627

CONTINGENCY ANALYSIS BY DC MODEL

impedance jXc between buses CD and (]) can be written /.I). =

o · - oJ· 1

per unit

( 1 4 .59)

and if the angles of the bus voltages change, the line curren t changes by the amount a /.I).

=

.:l ( O.1 - D}. )

Xc

.:l o · - .:l o · =

__ ' __� )

Xc

per unit

( 14 .60)

If all impedances in the definition of the line-outage distribution factor L ) of Eq. 0 4 . 36) are replaced by their corresponding reactances, we obtain

L , J , tntr . .

,

!liij X"X [ (Xim - XXithI ,mn) - ( XXj", - Xjn ) 1

=

=

_

Jmn

c

,

mn

( 14 .61)

a

correspondence between per-unit power and per-unit current in the l ines then enables us to write

The

Lij,

mn

.:l Pi j

=

Pmn

_

=

Xa [�Xim - Xin ) - (Xjm - Xjn ) 1 Xc Xth ,mn - Xa

( 14.62)

It is apparent that the line-outage distribution factors developed in the p reced­ ing sections by Z bu s methods are essentially the same as those for the dc power-flow method, provided series line reactances are used to represent the l ines. With the lines so represen ted, the current-shift distribution factors of the Z bus method become numerically equal to the generation-shift distribution fa ctors of the dc power-flow model. I f measurements are made of the actual power flows on the physical lines of a system, tables of precomputed distribution factors can be u s e d (IS d e sc r ibed i n the preceding sections. In fact, in the p reced i n g n u m e r i c a l exampl es the magnitudes of the pred icted changes in the current flows are very nearly equal in per unit to the magnitudes of the corresponding megawatt changes obtained by means of distribution factors based on the de power-flow model. The following numerical example illustrates this fact. ,

Usc distribution factors based on the dc power-flow model to predict the change in the power flow PS3 from bus G) to bus G) in the system of Fig. 14. 1 2 due to a generation shift of 45 MW from bus G) to bus CD when line G) (1) has already been taken out of service. All other conditions of load and generation are the same as in Case b of Example 14.4. Exa m p l e 14.7.

-

Solution. Because the l ines in the system of Fig. 14.12 are represented by series reactances, the distribution factors of Table 14.3 apply directly here. ,With per-unit

628

CH APTER 14

Zbul METI-lODS IN CONTING ENCY A N A L YSI$

power replacing per-unit current in Eq. ( 14 .44 ) , and setting i = 5, j = 3; n = 2; p = 5, q = 1 in that equ ation, we obtain

m =

5,

From Table 14.3 we can calculate K;3 . S

K; 3 . 1

LSJ.5 2 KS2, 5

=

KSJ, s +

=

0 .671 5 3 3

=

KS3• 1

+

LS3. S 2 K s 2 . 1

=

0 .232 4 7 0

= 0 . 1 0 1 074

+

+

0.671 533(0 .653822 ) 0 .67 1 533(0.2842 70)

= 0 .291971 and so 6. PS3

= 0.671533

X

( - 0.45)

-I-

0 . 29 1 9 7 1

X

0 . 45

=

- 0 . 1 70803 p e r u n i t

Hence, with line W - @ already out o f service, the power flow i n the line from bus � to bus CD is predicted to d ecrease by 17.1 MW due to the shift i n generation. The power-flow results for Case b show that Pi3 = 131'. 89 MW before the generation spift. Therefore, the predicted value of PS 3 after the 45-MW shift from bus � to bus CD is PS 3 = Pi3

+

� P53

= 1 3 1 .89

+

( - 1 7 . 1 ) = 1 14 .8 MW

Using the full ac power-flow solution for the combined l ine outage and generation shift (Case d of Table 14.2), the reader can show that Ps� 1 1 5. 1 2 MW, and so the predicted value of PS 3 approximates the more accurate ac calculation very closely. =

Previous examples involving distribution factors yield current flows with magni­ tudes very nearly the same in per unit as the real power flows obtainable by the d c power-flow method. 14.6 SYSTEM REDUCTION FOR CONTINGENCY AND FAULT STUDIES

For economic and security reasons individual power companies are connected together by tie lines to form a large interconnected system. If one of the companies wishes to study faults and contingencies on its own internal system, then it must represent the adjoining external system in some manner. Since the individual utility is mostly interested in the line flows and voltages of its own system over which it has some local control, it seeks to represent the external system by a reduced equivalent network.

1 4.6 SYSTEM REDUCfION FOR CONTINGENcY AND FAULT STUDIES

629

Consider the system of Fig. 14.13(a), which shows buses CD to Q) in the internal System I and the remaining buses @ to (j) in the external System II. The systems are connected by two tie lines. The starting point in the contin­ gency analysis of System I is an ac power-flow solution for the entire intercon­ nected system including tie lines and the external system. If the power-flow results are used to convert the loads and generation inputs of all the buses into equivalent current injections, then bus admittance equations can be written as follows:

(j)

® W

G) W

@

CD

V'7

0 0 (J) Yn Y76 Y7 S Y7 4 0 0 0 ® Yfi7 Y66 Y6S Y64 0 W 1;'S 7 YS () Y:�5 YS -l Y5 :1 0 0 @ Y4 7 Y-l (l Y4 � Y.l .l () Y-l 2 0 Y3 5 0 Y3 :1 Y3 2 Y3 1 0 Q) 0 Y24 Y2 1 Yn Y2 1 0 0 (2) 0 Y1 3 Yl 2 Y1 1 0 0 0 CD 0

17 16 15 14

6 V5' V4' 1/:'

( 14 .63)

-

I)

V'3

V2'

12

V'I

II

Ybus

The voltages are shown with primed superscripts to emphasize that they are numerical results in the power-flow solution of the entire interconnected system. The currents I I to 17 are the corresponding equivalent current injections calculated for each bus CD from the equation Ii = ( Pi - jQ)/�/* . The zeros in Y b u s show that the tie lines are the only connections between System I and System II, and elements Y4 2 and YS 3 equal the negatives of the corresponding tie-line admittances. In order to reduce the size of the external system represen­ tation, suppose we eliminate buses (J) and ® using gaussian-elimination formulas l ike those in Eqs. (7.59) and (7. 60), namely, I

yIJ.

=

yI J. -

Y:k Ykj Yk k

r

,

=

/. -

,

The reduced system of equations is then given by

0)

@

W Y;5 Y;4 @ Y;5 Y;4 ® Y3S 0 @ 0 Y24 0 - CD 0

Q) W CD Y5 3 0

Y33 Y23 Y1 3

0

0

Y4 2 0 Y32 Y3 1 Y22 Y2 1 YI 2 Y1 I

Yk

-'-

Yk k

V5' V'4

V3' V'2 V'I

Ik

( 14.64)

I'5

l'4

=

13 12

L II

( 14 .65)

· 630

CHAPTER 1 4

Zb us

M ETHODS I N CONTIN G ENCY ANALYSIS

�----�IL�--

System I- Internal

System II

-+--

-

External

@

12

�----j---j-�--�

I6

®

To bus

To bus

®

®

- Y4

2

---

- Y53

-

1

J

®

I4'

Y,M + Y�5 + Y42 -Y d5

�

YS5

+

Y�

+

Y53

l'5

(b)

(c)

. . . (a) O ne-line d iagram showing System I a nd System II i n t erconnecte d by two tie l ines; ( b ) ' Ward II p l u s equivalent circuit of System I I at buses 0 and G) ; (c) Ward equivalent circu i t of System FIGURE 1 4. 1 3

t i e l ines a t b u s e s

(1)

and

G) .

14.6

SYSTEM REDUCfION FOR CONTINGENCY AND FAULT STUDIES

I� I�,

631

Note that the only new elements are Y� , Y;S ' Y;4 ' and Y;s associated with buses @ and � at which the tie lines terminate. The new current injections into those buses are and which take into account the current injections of the eliminated buses ® and (j) in accordance with Eqs. 04.64). Rows � and @ of Eq. 04.65) enable us to draw the mesh equivalent circuit at buses W and @ of Fig. 14. 13(b). This mesh circuit is usually called a Ward equivalent. 2 If buses W and @ are not of particular interest to System I, then t he node elimination process can be continued un til only buses G) , W , and CD remain. The new

bus admittance equations are given by

Y"32 Y"22

( 1 4 . 66 )

Yl 2

The correspon ding reduced equ ivalent circuit then includes a new Ward equiva­ lent, as shown in Fig. 1 4 . 1 3(c). The currents I� and 1'3 at the boundary buses of System I are equivalent injections which take into account the actual injections of all the eliminated buses of System II, as explained in Sec. 7.6. Multiplying Eq . 0 4.66) by the inverse of its coefficient matrix, we obtain bus impedance equations in the form

G) @ CD The

Z"lZ" Z"Z" Z"Z" [I; ] [Vol Z" Z" Z" 1 G)

@

CD

23

22

21

13

12

11

33

32

31

1"2

II

-

-

V'2 V{

( 1 4 .67)

Z'(, u s

impedance clements have double-primed superscripts to show that they account for the impedances of the entire interconnected system including those of System 1. Distribution factors for analyzing contingencies in Sys tem I can be computed from Z'�u s , as demonstrated in the preceding sections. The above admittance equa tions can be expressed in more g e n e ra l terms. By sequentially numbering the i nternal ( 1 ) buses, the boundary ( B ) bu s e s, and the external ( E) buses, we can write the Y b us equations for the entire intercon2 J.

B . Ward, " Equivalent Circu its for Power-Flow Stud ies,"

pp. 373-382.

Transactions of AlEE, vol. 98, 1 949,

632

CHAPTER 14

nected

Zbus M ETHODS IN CONTINGENCY A N A LYSIS

system in the more general form

i'

( 1 4 .68)

If the

current injections are considered constants, we can eliminate the external nodes as previously described to obtain

[

YIW

----"

YBll - yn/.:y;;i YEn YJB

( 1 4 . 69 )

Example 14.8. The nine-bus interconnected system of Fig. has the bus admittance matrix shown in Table A Newton-Rap hson power-flow solution of the entire system is given in Table Use this solution to convert the P-Q loads and generation into current injections at the buses. By gaussian elimination reduce the representation of System II down to its own boundary buses ® and (J), and then down to the boundary buses (1) and W of System 1 . At each of the two stages of reduction draw the Ward equivalent corresponding to the reduced representation of System II.

14.8

14.14.5.4.

Elements of bus a dmittance matrix for interconnected system of Fig. 14.8t

TABLE 14.4

j4S . 0

-

j2 0 . 0

j25 .0 0 0 0

L...

0 0 0

®

j2 0 . 0

j20 .0

-j52 .9

0 0

j 1 2 .S

0 0 0

tAdmittances are in per unit.

j2 0 . 0

j2S .0

® 0

-j90.0

j I O .O

-j42 .S

j l 0 .0

j2 0 . 0

j2S .0

0 0 0

jl 0.0

j 1 2 .S

0

0 0 0

0 0

j 2S. 0 0

0 0 jlO.0 j 12.S

j2 0 .0

-j62 .S

0

j 2S .0

j2 0 . 0

j20.0

-j82.S j 1 2.S

0 0 0 0

0 0

j2 0 . 0

-

0

j36 .0

j 1 6 .0

0

0 0 0 0

0

j2S.0 j 1 6 .0 -jS 1 .2 jlO.0

CD 0 0 0 0

0

j2 0 . 0

0 j lO . 0 -j30 .0

-

1 4 .6 SYSTEM REDUcrlON FOR CONTINGENCY AND FAULT STUDIES

633

TABLE 14.5

Bus data and power-flow solution for the system of Example 1 4.8t

number

P

CD

Q

1 .75

0.25 1 1

Generation

Bus

Load

p

W

G)

CD

G)

1 .45

in

1 .749950 -jO.25 1 1 1 0

0.986973 -j O.076735

- 1 .402 1 1 4 + jO.21 0339

1 . 00

0.20

0.985577 -jO.086848

- 0.989048 +jO.290096

0.80

0. 1 5

0.993958 - jO.049 1 30

- 0.795430 + jO. 190238

0.998965 -jO.045486

1 .423 2 1 5 -jO. 620698

0.99 1 606 -jO.0(i7204

- 0.993630 + j O . 2 1 8(i23

0. 1 5

0.2759

0.999343 -jO.036254

1 .488995 -jO.330 1 4 5

0.70

0.2058

0. 999 1 4 3 -jO.04 1 392

0 . 674 3 1 3 -jO.634250

0.991 924 -jO.0649 4 1

- 1 . 1 88 1 55 +jO.329855

1 .20

data are

1 .000000 +jO.OOOOOO

1 . 50

® tAl l

Current injection

0. 1 0

1 . 00

®

Voltage

1 .40

0.5553

®

(j)

Q

per

0.25

u nit.

Solution. From the given data we now calculate the equivalent current injections a t

buses

®

19

Is

=

=

and

®

a s follows:

P9 - jQ9 * V9'

=

- 1 . 200000 0 .991924

jO.250000 + jO .064941

+

0 .700000 - jO .605790 0 .999 143 + jO.04 1 392

Pg - jQg V8' *

- 1 . 1 88155 + jO.329855 =

0.6743 1 3 - jO.634250

complete set of current i njections is shown i n Table 14.5. The first stage of reduction does not i nvolve buses of System I, and so we show only the System II (external) portion of the Y nus equations as follows:

A

® ® (j) ®

®

®

(j)

® 0

-j45 .0

. j20 .0

j 25 .0

j20 .0

-j5 2 . 9

j20 . 0

j 12.5

j25 .0

j20 .0

-j90.0

j lO.O

0

j 1 2 .5

j 1 0 .0

-j4 2 .5

V9'

- 1 . 1 88 1 55 + jO.329855

V8'

0.6743 1 3 - jO.634250

V:'6

- 0 .993630 + jO.218623

V'7

1 .488995 - jO.33 0 1 45

634

CHAPTER 14

Zbus METIiODS IN CONTIN G ENCY ANALYSI S

From Eqs. (14.64) the new current injection at bus becomes /8 = /8 -

Y89 /9 = ( 0 .674313 - jO.634250 ) 99

Yo

-

®

upon elimination of bus

®

j20 -.- ( - 1 . 188155 + jO .329855 ) -)4 5

= 0 . 146244 - j0 .487647 per unit

The new per-unit current injection at b u s 0) (0.828909 - jO. 1 4(892), a n d s o gaussian el imination ®

® 0) ® �

-j44 .0 1 1 1 1 1

j31 . 1 1 1 1 1 1

j I 2.)

0) ® � @

()

of bus

-j7 6 . 1 1 1 1 1 1

j l 0.0

j25 . 0

V7'

j12.S

j l O .0

-j42 .S

0

0

j l0.r

0

- j82 . 5

v.' 6

(j)

-jS4 . 1 1 8909

®

j 1 8 .836 1 S 2

j 1 8.8361S2

-j38 .949760

j2S .0

0

j l O .0

j20.0

®

O . 1 4 ()244

V.�'

j31 .l 1 1 1 1 1

Further elimination of bus

r-

G)

®

G)

calculated to b e ® gives

is s i m i l a rly

-

j O.4S7M7

0 . 828909 - jO . 1 46892

- 0 .993630 + j O .2 1 8623

V!)

1 . 423 2 1 5 - j O . 6 1 0698

gives the system of equations

G) 0

j2S .0

0

v'7

j I O .O j20 .0

-j8 2 .S

j 1 2.S

jl2.S

-j62.S

0 . 932288 - j(f.4 9 1 606

v.6'

- 0 .952094 + j O . 080 1 2 1

V'4

- 0 .7 9543 0 + jO . 1 90238

V5'

1 .423215 - j O . 6 2 0698

expected, the only new current i njections are at buses ® and 0) . Con tinuing o n with the second stage of external system elimination, we obtain

As

® @ G) G) CD

G)

-j68 .6 1 4 1 1 1

j23 .426542 j20.0

j25 .0 0

@

j23 .426542

-j43 .632472 0

0

j20 .0

When buses ® and between buses ® and

G)

@

j20.0

j25 . 0

- j 3 6.0

j 1 6.0 -j5 1 .2

0

j 16.0 0

0

j lO.O

CD

0 j20 .0 0

j l O.O

-j30 .0

V'4 v'5

V') V'2

V1'

1 .685300 - 1 . 078084

-

- 1 .402 1 1 4

+

+

-- -�

jO.872232 jO .03345 1

- 0. 989048 + jO .290096 1 . 749950

-

jO . 2 1 03 3 9

jO.25 1 1 1 0

® are eliminated, the Ward equivalent of System I I (j) i s that shown i n Fig. 14. 1 4 ( a). The reader may wish to

14.6 SYSTEM REDUCTION FOR CONTINGENCY AND FAULT STUDIES

17

To bus ®

=

635

0.932288 - }0 .491606 -}0.282757

-} 1 8.836152 -)0. 1 13608

I� System I plus tie li nes

-

=

-

0 .952094

System II

+ ) 0 . 080121

(a)

To bus

@

To b u s

@

To b u s CD

--

1'5

=

1 .685300 - ) 0 .872232

-----1

-) 10.926542

------1

-)20

---

System I

l�

=

- 1.0 78084

+ ) 0 . 03345 1

-- System II plus tie l i nes (b )

Fl G U RE 14.14 (a) Ward e q u i va l e n t b u s e s

®

il n cl

(j)

of S ys t e m I I ; ( h ) W il r d eq u i vil l e n t o f System I I p l u s t i e l i n es.

confirm that the numerical values of the e lements in the figure are consistent with the nu merical entries in the corresponding Y bus equations given above. Similarly, Fig. 1 4 . 1 4( b ) shows the Ward equ ivale nt of the entire System II plus tie lines as s e e n a t t h e b o u n d a ry b u ses @ a n d G) o f Sys tem I. Note that the internal l i n e @ - W of System I is shown separately.

If the current injections can be regarded as constant, then contingen cy analysis studies of System I can proceed with the reduced system of equations. The bus admittance representation and triangular factors of Y b u s can be

636

CHAPTER 14

Zhus METHODS I N CONTING ENCY ANALYSIS

employed or else a Z bus approach . In practice, it is found desirable to maintain a buffer zone of the external system around the internal system when evaluating contingencies on the latter. Within the buffer zone certain generator ( P- V ) buses are retained since they offer a means of voltage and reactive power support for the internal system. In faul t studies the individual operating util ity represents the neighboring systems by a reduced equivalent based on the Z bus of the external system. This reduced equivalent corresponds to a submatrix extracted from the external system Z bus , as demonstrated in Sec. 1 4 .2. I t is worthwhile for the reader to observe that the Ward equivalent obtained in Fig. 1 4 . 1 4( a ) by g a u s s i a n elimi na­ tion coincides with that obtained in Fig. 1 4 . 1 O( c ) by e x t r a c t i ng a s u b m a t ri x from the Z bus of System I I . 14.7

SUMMARY

This chapter explains the application of Z bus methods to contingency analysis. It is shown that these methods are very useful in predicting l i n e- fl ow redistribu­ tions caused by line outages and generation shifts aimed at r e l ieving thermal overloads in the system network. Distribution factors, which are easily gener­ ated from the existing system Z bus ' provide a fast approximate means of assessing the possibility of occurrence of these thermal overloads. As a result, a large n umber of contingencies can be evaluated and a priority listing (or ranking) according to the severity of i mpact can be created. Oth er ac methods of analysis can then be used to examine the most serious contingencies in greater detail. Such ac methods are also preferred when bus-voltage changes are to be p redicted. Basic considerations underlying network reduction and equivalencing are also discussed, and it is shown that the mesh (or Ward) equivalent networks resulting from gaussian elimination of the bus admittance equations yield the same resul ts as the matrix-based equivalents extracted from t h e system Z hu s ' The d iscussion provides a basis for u nd e rs t a n d i n g t h e m a ny o t h e r a p p roaches to equivalencing which are found in industry literature. PROBLEMS 14. 1 . A

four-bus system with

CD Q) ® @

Z bus

given i n per u n i t by

CD

jO.041 jO .03 1 jO.027 jO . O I 8

�

jO .03 1 jO .256 j O . 035 jO .038

®

jO .027 jO .035 j O . 1 58 jO .045

@)

jO .0 1 8 jO .038 jO.045 jO .063

ft

ft

ft

PROBLEMS

637

ft

has bus voltages VI = 1 . O , V3 = a.96 , and V4 = 1 .04 · , V2 0.98 Using the compensation current method, determine the change in voltage at bus W due to the outage of l ine (1)- ® with series impedance jO.3 per u nit. =

14.2. Solve Prob. 1 4. 1 when the ou tage involves ( a ) only line CD- @ of series impedance jO.2 per unit and ( b ) both line and l ine CD - ® of series

(1)- G)

impedance jOJ per u nit.

that line ®- @ of the system of Prob. 1 4 . 1 is actually a double-circuit l ine of combined i m p e d ance jO.2 per unit and that one of the circuits of series impedance jO.4 per unit is to be removed. Using th e compensation current method, determine the change in vol tage at bus (1) due to this outage. 1 4.4. Consider a portion of a large power system, whose Z bus e lements corresponding to the selected buses CD to G) are given in per unit by 14.3. Suppose

CD

(j)

L2:"

® @ G)

CD

0

jO .038 j O.034 jO .036 jO.018 iO.014

jO .034 jO .OS7

jO .044

jO.01 9 iO.Ol3

G)

jO .036 j (L 0 4 4 jO.062 j O . 018 iO.014

0

jO.018 j O.0 1 9 jO.018 iO .028 iO.010

G)

jO.014 jO.013 jO.014 iO .OlD i O .018

L2:"

L2:"

ft

The base-c ;se bus voltages at those selected buses are Vj = 1 .0 V2 = 1 . 1 , V3 = 0.98 V4 = 1 .0 and Vs = 0.99&' all in per unit. Using the com­ pensating current method, determine the change in voltage at bus CD when l ine W - ® of series impedance jO.05 per unit and line @- G) of series impedance jO.08 per unit are both outaged. 14.5. Redo Prob . 1 4 .4 when line (1)- ® and line CD - @ of series imp edan c e jO.lO per unit are outaged. 14.6. Two systems are connected by three branches a , b , and c in th e network of Fig. 1 4 . 1 5 . Using t h e sequential ru les of S e c . 14.2, write the elements of t h e loop impedance matrix Z in symbol ic form. Assume that System I contains b u se s CD through @ and System I I contains buses G) through (j) . Use superscri pts ( I ) and ( 2 ) to denote elements of Z I and Z I I ' respectively. 1 4.7. Form the branch-to-node incidence matrix A c for branches a , b, and c of t h e network of Fig. 14. 15. Then, for the network, determine the matrix Z defined in Eq. (14 .9). Compare your answer with that of Prob. 14.6.

network of Fig. 1 4 . 1 6 where Systc;n �_a_nd System I I are defined . Each of current injections II to 14 is equal to l .OL...=..2.2:, per unit and each of current inj ect i ons Is and 16 is equal to 4.0 - 900 per unit. 14.9. Consider the four-bus system of Fig. 1 4. 1 7 in which line imped ances a r e shown i n p e r unit. Using a generation-shift di st rib ut i on factor based o n t h e dc power-flow method, find the change in power flow in line CD - W when the power genera1 4 . 8 . Use

the piecewise

m e t hod

/

to

determine

the

b u s v o l t ages

for

the

react ance

638

CHAPTER 14

Zbus M ETHODS IN CONTI N G ENCY ANALYSIS

CD

�

- - - - - - - - - -

1

®

®

- - - - - - - - - -

i

G)

(J)

c

1---------_. @


Reference

FIGURE

a, b ,

Reference

14.15

Branches

and c

interconnect two subsystems in Probs. 14.6 and 14.7.

®

jO.25 jO.2

jO.2

j O.2

@ FIGURE

14.16

Network for Prob. 14.8. Parameters are in per

u nit.

Reference

P ROBLEMS

Q) ...,..-,..

639

...... . ,.. ... ® ) 0.5

)8

@ ......

----

®

Fl GURE 14. 1 7

A fou r-bus system for Prob. 1 4 .9.

@ tion at bus is incrementally increased by 0 . 1 per unit. Note : An appropriate bus should be selected as reference to find Z bu $ of the system. 14. 1 0. Consider the power system of Fig. 1 4. 1 2 discussed in Example 1 4.4. Using distribution factors, p redict the current in line CD @ when line @ - (I) is outaged. Assume that the operating condition specified remains the same as in the example. 14.11. The base-case bus voltages at buses @) and @ in a large power system are 1 .02 and 0.99 per u n i t , respectively. Suppose t h a t l ine @) - @ has been removed from service and that the following selected el ements are extracted from the Z bus of the system with l ine @) - @ excluded.

Lg:

-

&

CD (]) @) @

CD

jO.019 jO.015 jO.0l7 jO.0 1 4

(])

jO.015 j O .044 j O . 025 j O .030

@)

jO.017 jO .025 j O . 075 j O .052

®

jO .014 jO .030 jO .052 jO .064

The impedances of lines CD - (]) and @ - @ are jO.OS and jO. 1 per unit, respectively. Using the principle of superposit ion, determine the change in the l ine current of line 0) - CD due to an ou t age of l ine @) - @ . 14.12. Solve Example 1 4 . 6 for the case where only lines CD W and W - G) are simultaneously outaged. @ 14.13. In the five-bus system of Fig. 1 4 . 1 2, consider that line - G) has been taken out of service when an additional i ncremental load of 1 + jO at bus G) has to be met by an additional 1 -M W generation at bus G). Use Eq. ( 1 4.44) to predict the change in cu rrent flow from bus CD to bus G) . 14.14. A five-bus system consisting 0 f buses CD through G) is connected to a partially depicted l arge power system through three tie lines, as shown i n Fig. 1 4 . 1 8 in which line admittances are indicated in per u nit. The equivalent current injections at those five buses are I I 1 .5 + jO , 12 = 0.8 + jO, I) = 0.5 + jO, 14 - 1 .2 + jO, and Is = 0.4 + jO, all i n per unit. By g aussian elimination, reduce the Ybus @ representation of the five-bus system dow n to the boundary buses and ® and draw the corresponding Ward equivalent network.

-

=

-

=

640

CHAPTER 1 4

- - - - -

�

Z hu. METHODS IN CONT I N G ENCY ANALYS I S

-j 2

-) 1

®

-) 4

)1

-

Cl)

-) 4

®

-) 2 -) 1

-) 2

- - - - - +------1---1 ---- ------I

-) 4

®

-) 1

FIGURE 1 4 . 1 8

®

-) 2

@

External system

An i nterconnected system for Prob. 1 4 . 14.

-

14. 15. R edo Prob. 1 4 . 1 4 using matrix inversion as in Eq. ( 14.69). Compare the results with the solution to Prob. 1 4 . 1 4. Now, using gaussian elimination, continue the elimination of all five buses in the external system and then find the resulting W ard equivalent at buses ®, (j) , and ® . Draw an equivalent admittance circuit and m ark the values of all quantities.

CHAPTER

15

STATE ESTI M ATI ON OF P OWER SYSTE M S

Selective monitoring of the generation and transmission system has been provid­ ing the data needed for economic dispatch and load freque ncy control. More recently, however, interconnected power networks have become more complex and the task of securely operating the system has become more d ifficult. To h elp avoid major system failures and regional power blackouts, electric utilities h ave installed more extensive supervisory control and data acquisition (SCADA) throughout the network to support computer-hased systems a t the energy control center. The data bank created is intended for a number of application programs-some to ensure economic system operation and others to assess how secure the system would be if equipment failures and transmission-line outages were to occu r . B e fo re a n y s e c u r i t y a ssessm e n t can he m a d e or co n t rol actions taken, a reliable estimate of t h e existing s t a t e of t h e sys t e m m u s t be determine d . For t his purpose the number of p hysical measurements cannot be restricted to only those quantities required to support conventional power-flow calculations. The inputs to the conventional power-flow program are confined to the P, Q i njections at load buses and P, I V I values at voltage-controlled buses. If even one of these inputs is unavailable, the conventional power-flow solution cannot be obtained. Moreover, gross errors in one or more of the input quantities can cause the power-flow results to become useless. In practice, other yonveniently measured quantities such as P , Q line flows are available, but they cannot be 64 1

·642

CHAPTER 1 5

STATE ESTI MATI ON O F P O W E R SYSTI2MS

used in conventional power-flow calculations. These limitations can be removed by state estimation based on weighted least-squares calculations. The techniques developed in this chapter provide an estimate of t he system state and a quantitative measure of how good the estimate is before it is used for real-time power-flow calculations or on-line system security assessment. Besides the inputs required for conventional power-flow analysis, additional

measurements are made which usually include me g aw a tt a n d m e g a v a r flow in the transmission lines of the system. The unavoidable errors of the measure­ ments are assigned statistical properties and the estimates of the states are subjected to statistica l testing before b e i n g a cce p t e u as s a t isfa c t o ry . Thus, g r oss errors detected in the course of state estimation are a utomatically filtered out. IS.1

TH E M ET H O D OF

LEAST S Q UARES

The electric power transmission system uses wattmeters, varmeters, voltmeters, and current meters to measure real power, reactive power, voltages, and currents, respectively. These continuous or analog q u a n ti ties a r c d1Onitored by current and potential transformers (or other equivalent devices) installed on the l ines a nd on transformers and buses of t he power plants and substations of the system. The analog quantities pass through transdu.cers and analog-to-digital converters, and t he digital outputs are then telemetered to the energy control center over various communication links. The data received at the energy control center is processed by computer to inform the system operators of the present state of the system . The acquired data always contains inaccuracies which are unavoidable since physical measurements (as opposed to numerical calculations) cannot be entirely free of random errors or noise . These errors can be quantified in a statistical sense a nd the estimated values of the quantities being measured are then either accepted as reasonable or rejected if certain measures of accuracy are exceeded. Because of noise, the true values of physical quantities are never known and we have to consider how to calculate the best possible estima tes of the . unknown quantities. The method of least squares is often used to " best fit" measured data relating two or more quantities. Here we apply the method to a simple set of dc measurements which contain errors, and Sec. 1 5 .4 extends the estimation p rocedures to the ac power system. The best estimates are chosen as those w hich minimize the w e ig h t e d s u m o f t h e squares o f the measurement

errors. Consider the simple dc circuit of Fig. 1 5 . 1 with five resistances of 1 n each and two voltage sources VI and V2 of unknown values which are to be

ammeter re adi n gs 2 1 and 2 2 a nd voltmeter readings Z3 and Z 4 ' The symbol z is normally used for measurements regardless of the physical quantity being measured, and likewise, the symbol x applies to qu a n t iti es being estimated. The system model based on elementary estimated. The measurement set consists of

,

15.1

TIlE METI-IOD

OF

LEAST SQUARES

643

circuit analysis expresses the true values of the measured quantities in terms of the network parameters and the true (but unknown) source voltages X l = VI and x 2 = V2 • Then, measurement equations characterizing the meter readings are found by adding error terms to the system model. For Fig. 15.1 we obtain 5 S X I - .!8 X 2

ZI Z2

=

( 15 . 1 )

+

SX2

5

+

e2

( 1 5 . 2)

3

+

"8 X 2

1

+

e3

( 15.3)

� :\ X

+

e4 ----

( 1 5.4)

il x I

Measurements

el

I - SX I

Z3 Z4 -------

+

�X I + I

2

T r u e values from syst e m model

Errors

in which the numerical coefficients are determined by the circuit resistances and the terms el, e 2 , e3, and c4 represent errors in measuring the two currents Z l and Z 2 and the two voltages Z 3 and Z 4 ' Some a uthors use the term residua ls i nstead of errors, so we use both terms interchangeably. If e l , e 2 , e:" and e4 were zero (the ideal case), then any two of the meter readings would give exact and consistent readings from which t he true values Xl and X 2 of VI and V2 could be determined. But in any measurement scheme there are unknown errors which generally follow a statistical pattern, as we shall discuss in Sec. 15.2. Labeling the coefficients of Eqs. (1 5 .1) through ( 15.4) in an obvious way, we obtain ZI

=

hIIXI

+

h 12 X2 + e I = Z I

{rue

( 15 .5)

+ e1

( 1 5 .6) ( 15 .7) ( 15 .8 ) where Z j, ! rue denotes the true value of the measured quantity rearrange Eqs. (15.5) through ( 15.8) into the vector-matrix form e[ e2 e3 e4

=

ZI Z2 Z3 Z4

Z I , t rue Z 2, t rue Z3 , t rue Z4 , true

=

ZI Z2 Z3 Z4

hl1 h 21 h31 h41

h 12 h 22 h 32 h 42

[ :: 1

Zj '

We now

( 1 5 .9)

644

CHAn-ER 15

STATE ESTIMATION OF POWER SYSTEMS

+

FIGURE 1 5 . 1

Simple dc c ircu i t w i t h two a m m eters (A m ) measuring and Z 4 '

Z l

z)

In more compact notation Eq. ( 1 5 .9) can e =

Z - Z ! rue

be

=

and

z2

a n d two v o l t m e t e rs ( Ym ) m e a s u r i n g

written as Z - Hx

( 1 5 . 1 0)

which represents the errors between the actual measurements z and the true (but u nknown) values Z t rue g Hx of the measured quantities. The true values of X l and x 2 cannot be determined, but we can calculate estimates X l and x 2 , as we shall soon see. Substituting these estimates in Eq. ( 1 5 .9) gives estimated values of the errors in the form e1 e2

ZI Z2

"- 3

�

c)

7'

e4

Z4

-

�

Measurements

Estimated errors

h l l h 12 h 2 1 h 22 h ) 1 li n h 4 1 . h 42 Est i mates

[ �: 1

(15.11)

of Lj

Quantities with hats, I such as ej and Xi' are estimates of the corresponding quantities without hats. In Eq. ( 15 . 1 1) the left-hand vector is e, which represents the differences between the actual m easurements Z and their estimated values Z A Hi so that we can write

e=Z

I

-

Z

=

Z

-

Hi = e - H ( i - x)

The symbol is more properly called a circumflex .

( 1 5 . 1 2)

15.1

THE METHOD OF LEAST SQUARES

645

We must now decide upon a criterion for calculating the estimates x 1 and x2 from which e = [ e1 e2 e3 e4Y and z = [ 21 22 23 24V are to be computed. It is not desirable to choose the algebraic sum of the errors to be minimized since positive and negative errors could then offset one another and the estimates would not necessarily be acceptable. It is preferable to minimize the direct sum of the squares of the errors. However, to ensure that measurements from meters of known greater accuracy are treated more favorably than l ess accurate measurements, each term in the sum of squares is multiplied by an appropriate weigh ting factor w to give the objective function ( 1 5. 13 ) We select the best estimates o f the state variables a s those values x I and x 2 which cause the objective function f to take on its minimum value. According t o the usual necessary cond itions for minimizing f, the estimates x I and x 2 are those values of X I and X 2 which satisfy the equations at

ax ,

3f

aX 2

[ Bel

=

x

=

i

2 wj e1 ax }

[ ae,

2 w l el

JX 2

�

Be,

+

+

w2 e Z ax ! w2 e 2

ae2

3x2

-

Be]

+

w3 e 3 ax ,

+ w3e3

ae,

3xz

�

Be, ]

+

w4 e 4 aX l

+ w4 e 4

a e,

aX 2

-

=

]

0

( 1 5 . 1 4)

x

= 0 ( 15 .15) x

The nota tion I x indicates that t h e equations have t o b e evaluated from the state estimates x = [x I x2 V since the true values of t he states are not known. The unknown actual errors ej are then replaced by estimated errors ej, which can be calculated once the state estimates X i are known. Equations ( 1 5 . 14) and ( 1 5 . 1 5 ) i n vector-matrix form become

del dX I Del

de2 dX I dez

d e:, dX I de3

de4 dX I de4

() X

ax ?

cJ X

() X

z

2

2

el

WI

e2

w2 X

A

w] W

e3 e4

w4

A

=

[�]

( 15 . 16)

where W is the diagonal matrix of weighting factors which have special signifi­ cance, as shown in Sec. 15.2. The partial derivatives for substitution in Eq. (15 .16) are found from Eqs. ( 1 5.5) through (15.8) to be constants given by t h e

646

CHAPTER 15

STATE ESTIMATION OF POWER SYSTEMS

elements of H, and so we obtain

e1 c2 c3

=

c4

W..

[�l

( 1 5 . 1 7)

Using the compact notation of Eq. ( 1 5 . 1 2) in Eq. ( 1 5 . 17) yields ( 1 5 . 1 8) Multiplying through x

i n t h i s c<.J u a t i o n a n J s o l v i n g

�

[�: 1

�

C!!_':_��!fr H TWZ G

for :\:

�

G

=-�

[ X l x z l'

g i vc

1 II T WZ

( 15 . 19 )

where i \ and i \ are the weighted least-squares estimates of the state variabl es. Because H is rectangular, the symmetriCal matrix H T WH (often called the gain matrix G) must be inverted as a single entity to yield G - I = ( H T WH) - l , which is also symmetrical. Later in this chapter we discuss the case where G is not invertible due to the lack of sufficient measurements. We expect the weighted least-squares procedure to yield estimates X i ' w hich are close t o t h e true values X i of t h e state variables. An expression for the d ifferences (Xi x) is found by substituting for z = Hx + e in Eq. ( 1 5 . 1 9) to obtain -

( 1 5 .20) G

Canceling G with G- 1 and rearranging the result lead to the equation ( 15 .2 1 )

It i s usefu l to check the dimensions o f each term i n the matrix product of Eq. (1 5 .21), w hich is important for devel oping propert ies of the weighted l east­ squares estimation. For the example circuit of Fig. 1 5 . 1 the matrix G- 1 HT W h as the overall row X column dimensions 2 X 4, w hich means that any one or �ore

1 5.1

T H E METHOD OF LEAST SQUARES

647

of the four errors e l , e 2 , e 3 ' and e 4 can influence the difference between each state estimate and its true value. In other words, the weighted least-squares calculation spreads the effect of the error in any one measurement to some or all the estimates; this characteristic is the basis of t he test for detecting bad data, which is explained in Sec. 1 5 .3. In a quite similar manner, we can compare the calculated values z = Hi of the measured quantities with their actual measurements z by substituting for x x from Eq. (15.21) into Eq. ( 1 5 . 1 2) to obtain -

( 1 5 .22 )

where I is the unit or identity matrix. Immediate use of Eqs. (15.21) and ( 1 5 .22) is not possible without knowing the actu a l errors ei, which is never the practical case. B u t we can use those equations for analyt ical pu rposes after we h ave introduced the statistical prope rties of the errors in Sec. 15 .2. For now let us numerical ly exercise some o f the other formu las developed above in the follow­ ing example.

I n the dc circu it of Fig. 1 5 . 1 the meter readings are Z l 9 . 0 1 A, = 6.98 V, and Z4 5 .01 V. Assuming that the ammeters are more accurate than the voltmeters, let us assign the measurement weights Wl = 1 00, w 2 = 100, w 3 50, and W 4 = 50, respectively. Determine the weighted least­ sq uares estimates of the voltage sources Vl and V2 . Example 1S.1.

Z

2 = 3.02 A,

=

z3

=

=

Solution. The measured currents and voltages can be expressed in terms of the two

voltage sources by elementary circuit analysis using superposition. The results are shown by the system model of Eqs. ( 1 5 . 1 ) through ( 1 5 .4), which have the coefficient matrix H given by H =

l

0.625 - 0 . 1 25 0. 375 0 . 1 25

- 0 . 1 25 0.625 0. 125 0 .375

Calculating the matrix H T W , we obtain HTW

='

=

[

[

0.625 - 0 . 1 25

- 0 . 1 25 0 . 625

0. 375 0. 125

0 . 1 25 0 .375

62.50 - 1 2.50

- 12 .50 62.50

1 8 .75 6 . 25

6 .25 1 8 .75

1

fl] O . .

]

1 00 50

5J

· 648

CHAPTER 15

STATE ESTIMATION OF POWER SYSTEMS

[]

and using this result to evaluate the symmetrical gain matrix G gives

G = H T WH =

=

[

62.50 - 12 .50

- 12 .50 62 .50

[ - 4810 .9375 .4375

- 1 0 .9375 48 .4375

S ubst i t u t i n g n u m erical v a l ues2 in Eq .

=

[ - 48.4375 10 .9375 [ 00.0049 .0218 [ 0.035 1 .2982 1

6 .25 1 8 .75

1 8 .75 6 .25

J

( 1 5. 1 9)

] [

0.035 1 1 .2982

] [ - 6212 .50.50 0.4386 0 .228 1

]

yields

62.50 - 1 0 .9375 - 1 - 1 2 .50 48 .4375

0.0049 0 .021 8

- 0 . 1 25 0 . 625 0 . 1 25 0 .375

0 .625 - 0 . 1 25 0 .375 0 . 1 25

- 1 2.50 62.50

- 12 .50 62.50

0 .2281 0 .4386

1 8 .75 6 .25

1 8 .75 6 .25

1] l 9 .01 A 3 .02 A 6.98 V 5 .0 1 V

6 .25 1 8 .75

=

]

6 .25 1 8 .75

]

z

Z2

ZI Z

J

Z4

[ 1 86 .0261 .0072 V ] V

w hi ch are the estima tes of t h e voltage sources VI and V2 . The est i m ated measurements are calculated from i = H i as follows:

il

£2 zJ

i4

0 .625 - 0 . 1 25 0.375 0 . 1 25

- 0 . 1 25 0 .625 0. 125 0 .375

[

1 6.0072 V 8 .0261 V

]

9 .001 23 A 3 .0 1 544 A

7.00596 V

5 .0 1 070 V

2Numerical r esults in t his chapter were generated by com p u ter. .Answers based on t h e rou n d ed-off values d isplayed here may differ slightly.

15.1

THE METHOD OF LEAST SQUARES

and the estimated errors in the measurements are then given by Eq. ( 1 . 1 1 )

5

e1

e3

9.01 3 .02

e2 e4 �

9 .00123 3 .0 1544 7 .00596 5 .01 070

=

6. 98

5 .0 1

649 as

0 . 00877 A 0.00456 A

-0.02596 V

- 0 .00070 V

The preceding example calculates the state o f the system by the method o f weighted least squares. But knowing t he estimated state i s o f l ittle valu e i f we cannot measure by some means how good tbe estimate is. Consi der, for instance, that the voltmeter readi ng Z 4 of Fig. 1 5 . 1 is 4 . 4 0 rather than t h e 5 . 0 1 V used i n Example 1 5 . 1 . I f the other three meter readings are unchanged, we can calcu late the est imates of the state variables from Eq. ( 1 5 . l 9) as fol lows:

V

[ 1 .2982 0 .035 1 =

[ 15 .86807 7 .75860

[

0 .035 1 1 .2982

VV

0 .4386 0.228 1

1

1

1 9.01 A 0 .228 1 ] 1 3 .02 A 0 .4386 l 6 .98 4 .40

VV

and the estimated measurement e rrors are again given by Eq . (15. 1 1 ) as

9.01 3 .02 6 .98 4 . 40

0 .625 - 0 . 1 25 0 .375 0 . 1 25

- 0 . 1 25 ' 0 .625 [ 1 5 .86807 0 . 1 25 7 .75860 0 .375

VV 1

=

0 .06228 A 0 . 1 5439 A 0 .05965 - 0 .49298

VV

From the practical viewpoi nt i t is not meaningful to compare this second set o f nu merical answers with those o f Example 1 5 . 1 i n order to decide which set i s acceptable. In actu al practice, only o n e set o f measurements i s ava ilable at a ny one time, and so t here is no opport un ity to make comparisons. Given a set of measurements, how t hen can we d ecide i f the corresponding results should be accepted as good estimates of the true values? What criterion for such accep­ tance is reasonable and if a grossly erroneous meter reading is present, can we detect that fact and identify the bad measu rement? These questions can b e answered within a quantifiable level of confidence b y a ttaching statistical mean­ ing to the measurement errors in the least-squares calculations.

650

CHAPTER 15

15.2

STATE ESTIMATION OF POWER SYSTEMS

STATISTICS, ERRORS, AND ESTIMATES

Computer simulations such as power-flow stud ies give exact answers, but i n reality we n ever know t h e absol utely true state of a p hysically operating system. Even when great care is taken to ensure accuracy, unavoidable random noise enters i n to the measurement p rocess to distort more or less the p hysical results. However, repeated measurements of the same quantity u nder carefu lly con­ trol led conditions reveal certain statistical p roperties from which the true value can be estimated. If the measured va lues are plotted as a fu nction of their relative frequency of occurrence, a h istogram is obtained to which a continuous curve can be fitted as the nu mber of measurements increases (theoretically, to an i nfi n i t e n u mber). The continuous curve most commonly encou ntered is the bell-shaped fu nction p( z ) shown in Fig. 15.2. The function p( z ), cal l ed the gaussian or normal probability density JUllelioll , has the fo r m u l a = p( z )

1 2 2 7T if 1

1 ( :: -1-1- )"

[ - '2

( 1 5 .23)

-u-

and the variable representing the val ues of the measured quantity along the horizontal axis is known a s the gaussian or normal random variable z . Areas

p(z)

=

1

(T .fi; t:

1 - 2

(Z-JL - )2

p(z)

p(y)

a

S h ad ed area =

=

-4

J.L - 2 (]"

J.L

a

-2

o

y

'

b

y

"

J.L

+

2

Pr e a < z < b ) Pr ( y' < .Y < y" )

z-scale

2 (]" 4

y-s cale

The gaussian probability density fu nction p( z ) a n d standard gaussian probab i l i ty dens i ty fu nction p( y ) 0 btained by setting y (z - J-L) /u . FIGURE 1 5.2

=

STATISTICS, ERRORS, AND ESTIMATES

15.2

651

u nder the curve give the probabilities associated with the corresponding inter­ vals of the horizontal axis. For examp le, the probability that z takes on values between the po int s a and b of Fig. 15 .2 is the shaded area given by Pr ( a

< z < b)

=

fbp ( z ) dz =

I

,fj;; (j 2 7T

a

fb - "2 a

£

1

( Z )2 --;-

/L

dz

( 1 5 . 24)

where the symbol Pr( ' ) means the probability of ( . ) occurring . The total area u nder the curve p( z) between and + equals 1 because the value of z is certain (with probability equal to 1 or 100%) to lie between arbitrarily l arge extreme values. The gaussian probability density function of Eq. ( 1 5.23) is completely determined once the parameters J.L a nd (J are known. Figure 1 5 .2 shows that p ( z ) has i ts maximum value when z equals J-t , which is the expected value of z denoted by E[ z ] and defined by -

J-L

00

00

= E[ z ]

=

f

+ 00

-- - 00

z p ( z ) dz

( 1 5 .25 )

The expected value J-t i s often called t h e mean since t h e values of z are sym metrically clustered about J-L . The degree to which the curve of p(z) spreads out about f-L (that is, the width) depends on the variance (J2 of z defined by ( 1 5 .26) Thus, the variance of z is the expected value of the square of th e deviations of z from its mean ; smaller values of (j 2 provide n arrower and higher curves with each having unity area centered about J-t. Th e positive square root of the variance is the standard devia tion (j of z. The mean and variance are important parameters of many commonly encountered probability density functions. Each such function has a curve with underlying areas which quantify the probabilities of the associated random variable. Th e gaussian distribution plays a very important role in measurement statistics. Because the density p( z ) cannot be directly integrated, the areas u nder the curve, such as that shown in Fig. 1 5 .2, have been t abulated for the standard gaussian density function pC y ), which has the parameters J-t y 0 and (Jy = 1 . Separate tables of the gaussian d istribution for other values of J-L and 0' a re unnecessary since we can rescale the horizontal axis of Fig. 1 5.2, and thereby transform from z to y , using the change-of-variable formula =

y =

z - J.L

---

u

( 1 5 .27)

652

CHAPTER 15

STATE ESTIMATION OF POWER SYSTEMS

TABLE 15.1

The standard gaussian distributiont a

Pr(a)

a

Pr(a)

a

Pr(a)

a

Pr(a)

0.48928 0. 49 0 6 1 0.49 1 SO 0.49286 0.49379

. 05

0.01 994

.80

0.288 1 4

1 .55

0.43943

2.30

.10

0.03983

.85

0 .30234

1 .60

0.44520

2.35

.15

0.05962

.90

0.3 1 594

1 .65

0.45053

.20

0.07926 0.09871

.95

0.32894

1 .70

0.45543

1 . 00

0.34 1 34

1 .75

0.45994

2.40 2.45 2.50

.30

0 . 1 1791

1 .05

0.35 3 1 4

1 .8 0

0.46407

2.55

.35

0 . 1 3683

1 . 10

0 . 36433

1 .85

0.46784

.40

0 . 1 5 542

1.15

0.37493

1 .90

0.47 1 28

.45

0. 1 7364

0.38493

l .95

0.4744 1

.50

0 . 1 9 146

1 .20

0.39435

2.00

0. 47726

2.60 2 .65 2.70 2.75

0.47982

.25

1 .25

.55

0.20884

1 .3 0

0.40320

2.05

.60

0.22575

1 .35

0.4 1 149

2. 1 0

.65

0.24 2 1 5

1 .40

0.4 1 924

2. 1 5

0.48422

.70 .75

0.25804

1 .45

0.42647

2.20

0.48 6 1 0

0.27337

1 .50

0 .433 1 9

22

. 5

0.48778

tPr( a) is the value of f2; 1

J:

2 rr 0

E

2 .RO

2.85 2 . 90 2.95 3.00

0.482 1 4

0.49461 0.4953·� 0.49597 0.49653 0. 4 9 70 2 O.497·;� 0.497::; 1

0.498 1 3 0.4 98 � 1 0.49865

- IY (Jy. I

2

It is evident from Eq. (15.27) that dz = O'dy , and if z takes on values between a and b, then y must lie between t he corresponding numbers y ' = (a - J.1J/O' and y " = (b - 1-'-)/0'. Hence, substituting for y in Eq. (15 .24), we obtain

Pr ( a

<

z

<

b) =

1 � y 2 7T'

f

y

y '

" £

.

- -2 y 2 dy = Pr ( y , < y < y ) I

"

( 15 .28)

which shows that y (with J.L y = 0 and O'y = 1 ) is the random variable with standard gaussian probability density. Thus, changing scale according to Eq. ( 1 5.27) allows us to use the tabulated values of y, given in Table 1 5 . 1 , for any random variable z with normal probability density function. I n t he new scale y tells us how many standard d eviations fro m the mean the corresponding value of z lies, Fig. 15.2. The area under the gaussian density curve is d istributed as follows: 68% within the z-lim its I-'- + 0', 95 % within the z-lirnits I-'- ± 20', and 99% within the z-limits J.L ± 30'. Accordingly, the p robability is 99% that the value o f the gaussian random variable z lies within 3(7' of its mean /-L . Henceforth, w e assume that t h e noise terms e l , ez , e3, and e4 are independent gaussian random variables with zero means and the respective variances 0' 12 , (Ti, (Tt, and (Tt Two random variables ei and ej are indepe n dent

1 5.2

when E(ejej)

=

STATISTICS, ERRORS, AND

E$TIMATES

653 ·

0 for i =1= j. The zero-mean assumption implies that the error in

each measurement has equal probability of taking on a positive or negative value of a given magnitude. In Sec. 1 5.3 we encounter th e product of th e v ect or e and its transpose T e = [ e l e 2 e3 e4 ], which is given by ee T =

e 2I

el e2 e3 e4

[ el

e2

e3

e4 ]

=

e2 e l e3 e l e4 e l

e l e2 e 22 e) e2

e4 e2

e l e3 e2 e3 e 32 e4e 3

e 1 e4 e 2 e4 e 3 e4 e 42

( 1 5 .29 )

The expected value of ee T is found by calculating the expected value of each entry in the matrix of Eq. ( 1 5 .29). The expected values of all the off-diagonal elements are zero because the errors are assumed to be independent; the expected values of the diagonal elements are nonzero and correspond to t h e variance E[e n = u/ for i from 1 to 4. The resultant d iagonal matrix is u s ual ly assigned the symbol R, and so the expected value of Eq. (15 .29) becomes u I2

E [ d] ( 1 5 . 30)

With this background, we can now consider some statistical properties of the weighted least-squares estimation procedure. In Eq. (1 5.5) the measurement z 1 is the sum of the gau ss i an random variable e l and the constant term (h I l x 1 + h I2 X 2 ), which represents the true value Z I . true of The addition of the constant term to e 1 shifts the curve o f e l to t he right by the amount of the true value (or to the left by the same amount if the true value is negative). But the shift in no way alters the shape and spread of the probability density function of e I ' Hence, Z 1 also has a gaussian probability density function with a mean value J.L I equal to the true value Z I , true and a v a r i a nce o} e q u a l to t h a t of e l . S i m i l a r remarks apply t o Z 2 ' z 3 ' and Z 4 ' which also have means equal to their true values, and variances ui, ui, and al corresponding to the variances of e 2 , e3, and e4 , respectively. Thus, meters with smaller error variances have narrower curves and provide more accurate mea­ surements. In formulating the objective function f of Eq. ( 1 5 . 13), preferential weighti ng is given to the more accurate measurements by choosing the weigh t Wj as the reciprocal of the corresponding variance u/. Errors of smaller variance thus have greater weight. Henceforth, we specify the we i g hti ng matrix W of , Z I '

654

CHAPTER 15

Eqs. 05.21)

STATE ESTIMATION OF POWER SYSTEMS

and 0 5 .22) as 1

1 ( 1 5 .3 1 )

1 1

and the gain rna trix G then becomes ( 1 5 .32)

We can regard the elements in the weighting matrix of Example I S . 1 as the reciprocals of the error variances a} = a-{ = 1 / 1 00 for meters 1 and 2 and 2 2 U3 = 0'4 = 1 / 50 for meters 3 and 4. The corresponding standard deviations of the meter errors are 0' 1 = 0'2 = 0. 1 A and 0'3 = 0'4 (Ii / 1 0) V. Thus, the probability is 99% that ammeters 1 and 2 of Fig. 15. 1 , when functioning properly , will give readings within 30' or 0.3 A of the true values of their measured currents. Comparable statements apply to the two voltmeters of Fig. 1 5 . 1 if their errors are also gaussian random variables. From Eq. (15 .25), which defines the expected value of the random variable Z, we can show that E[az + b] = aE[ z ] + b if a and b are constants. There­ fore, taking the expected value of each side of Eq. ( 15.21) gives =

E[ e ] ] £ [ e2 ] £ [ e3 ] £ [ e4 ]

[�1

( 15 .33 )

This equation uses the fact that x 1 a n d x 2 are definite numbers equal to the true (but unknown) values of the state variables while e l , e 2 , e3, and e4 are gaussian random variables with zero expected values. It follows from Eq. (15 .33) th at ( 15 .34)

which

implies that the weighted least-squares estimate of each state variabl y has an expected value equal to the true value. In a similar ma-nner, we can use

TEST FOR BAD DATA

15.3

Eq. (15 .22) E

655

to show that

e1

e2

e3 e4

ZI - Z1 Z 2 Z2 "

=

E

and since Zj equals Zj ,

Z3 - Z3 Z4 - Z 4

e1

-

A

=

[I

-

HG - 1 H TR - l ] E

"

tru e

+

ej ,

e2

e3 e4

-

0 0 0 0

( 15 . 3 5 )

it follows from Eq (15 .35) that .

( 1 5 .36)

Equations ( 1 5 .34) and (15.36) state that the weighted least-squares estimates of the state variables and the measured quantities, on the average, are equal to their true values-an obviously desirable property, and the estimates are said to be un biased . Accordingly, in Example 1 5 . 1 the sol u t i o n s of 1 6 .0072 V for VI and 8.026 1 V for V2 c a n be acce p t e d a s u n b iased e s t i ma tes of the true values, provided no bad measurement data are present. The presence of bad data can be detected as described in the next section. 15.3

TEST FOR BAD DATA

When the system model is correct and the measurements are accurate, there is good reason to accept the state estimates calculated by the weighted least­ squares estimator. But if a measurement is grossly erroneous or bad, it should be detected and then identified so that it can be removed from the estimator calculations. The statistical properties of the measurement errors facilitate such detection and identification.3 Each estimated measurement error ej = ( Zj - z) is a gaussian random variable with zero mean, as shown by E q . ( 1 5 .35). A formula for the variance of ej can be determined from Eq. (15.22) in two steps. First, with W replaced by R \ we postmultiply e by its transpose eT to obtain --

The left-hand side of this equation represents a square matrix l ike that of Eq. ( 1 5 .29); the two matrices in brackets on the right-hand side contain only T cons t a n t e l e m e n ts and are t ra nsposes of one a nother. The t e r m e e is actu ally the same matrix given in Eq. ( 1 5 .29) and its expected value R is the diagonal matrix of Eq. (15.30). In the second step we take the expected value of each side 3

T h i s section i s based o n 1. F . D op azo,

O . A . KJ i t i n , a n d A . M. Sasson, G ross Measurement E r r o r s

Sys t e m s : D e t e c t i o n a n d I d en t i fic a t i on of

PlCA Conference, 1 973.

,"

"

St ate

IEEE

Estimation of Power

in Proceedings of the

656

CHAPTER 15

STATE ESTIMATION OF POWER SYSTEMS

of Eq. (1S.37) to find that ( 1 5 .38)

-R R

Multiplying R by the bracketed matrix [I o ut R to the right from the result give

-I

HG - I H T ] and then factoring

- HG - 1 H T R - 1 ] [ R - H G - 1 H T ] = [ I - HG - 1 H 'J' R - 1 ] [ I - HG 1 H F R I ] R ( 1 5 . 39 ) The re ader should show that the m at rix [I - H G I H T R - ] multi p l ied by i tself E[ eeT ] = [ I

idempotent

remains unaltered (it is called a n simplifies a s follows:

I

-

matrix),

and

s o Eq. ( 1 5.39 )

- i) i n the diagonal r;; rr t ries,

( 1 5 .40)

The square matrix ee T takes the form of Eq. ( 1 S . 2 9) with typical entries given by

elj ' Therefore, substituting for

ej

( zj

=

w e.:

obtain

( 15 .4 1 )

where R�"j is the symbol for the j th diagonal element of the matrix R' = R - HG - 1 H T. Since ej has zero mean value, Eq. (15.41) is actually the formula for the variance of ej, which makes ,;R�"j the standard deviation. Dividing each ,.� i d e of Eq. (1S.41) by the number Rj j gives

"

i '

E

eJ

- i

[ ( � )21 [( y'RJ 21 . z·

= E

)

)

= 1

( 1 5 .42 )

('j

Thus, in the manner of Eq. (15.27) the deviation of the estimated error (e) from i ts mean (0), divided by the standard deviation Rjj ), yields the standard gaussian random variable

e, )

-0

( 15 .43 )

IR'� y H j)

with zero mean and variance equal to 1, as shown by Eq. (15.42). Matrices R of Eq. (1S.30) and R' (R - HG - 1 H T ) of Eq. (15.40) are called It can be also shown using Eq. (15.21) that the covariance matrix of x - x given by E[(x - x)(x - X) T ] is G - 1 = (H T R - 1 1-I ) - 1 . =

covariance matrices.

15.3 TEST FOR BAD DATA

657

TABLE 1 5.2

Summary of residue terms and covariant matricest Term

Statement

Expression

Covariance matrix:j:

e = z - Hx

(measured - true\

z - Hx

R

ex = x - x

(true - estimated),.

- G - 1 H TR - 1 e

(measured - esti mated)z

(I - H G - I H T R - I ) e

e

=

z

- Z

i

and

=

G- 1

=

R'

(R - H G - 1 H T )

=

( H T R - 1 H) - 1

Hx

tA l l resi d u es h ave zero means. :j:D i a go n a l t e rms a re the e rror v a r i a n c e s .

A summary of some of the terms encountered above and their statistical p roperties is presented in Table 1 5 .2. We have al ready discussed the fact that t he true measurement error e j is n ever known in engineering applications. The best that can be done is to calculate t h e estimated e r ror ej, wh ich then replaces ej in t h e objective function. Accordingly, we substitute from Eq. (15 .43) into Eq. ( 1 5 . 1 3) to obt a i n ( 15 . 44 )

where Nm is the number of measurements a !: d the weighting factor Wj is set equal to l /a/. The weigh ted sum of squares f is i tself a random variable which . has a well-known probability d istribut ion with values (areas) already tabulated i n m any textbooks on statistics. I n order to use t hose tables, we need to know the mean value of f, which is simple to determine, as we now demonstrate. Multiplying the nu merator and the denominator of the jth term i n Eq. ( 1 5 .44) by the calculated varia nce Rjj of the measurement error and taking t he expected value of the resu lt, g ives

£ [ /]

=

E [ � R�; e�Z 1 . )" ' I

(T. J

=

R,11

N,,, "" L....

j '"-

1

R.'. , --.!.!.. E

�.

2

�

[

(

z

_

J

_

,

i , )2

Rjj

,

]

( 1 5 .45)

The expected value in the right-hand summat ion of Eq. ( 1 5 .45) equals 1 according to Eq. (15.42), and so we obtain for the mean of f E

[ f]

=

E

[� e: 1 � Rj; j = 1 O'j

=

j=

1

<Jj

( 15 .46)

658

CHAPTER 15

STATE ESTI MATION OF POWER SYSTEMS

It is shown i n Example 15.2 that the right-hand side of this equation has the numerical value 2 = (4 - 2) for the four measurements and two state variables of the circuit of Fig. 15.1. More generally, it can be demonstrated (Prob. 15.4) that the expected value of f is always numerically equal to the number of degrees of freedom ( Nm - N); that is,

E[f]

=

�

i...J

Eli:! iJ�] -

J. = 1

a· J

2

=

NIII

-

NS

( 1 5 .47)

when t here is a se t of Nm measurements and Ns i ndependent state variables to be est imated. Thus, the mean value of f is an integer which is found simply by subtracting the number of state variables from the number of measurements. The number ( Nm - N) is also cal led the redundancy o f the meas u reme nt scheme, and i t is obviously large when t here are many more measurements than state

variables.

Just as ej has the standard flaussian distribution, statistical theory shows that the weighted sum of squares f of Eq. (15.44) has the chi-square distribution xl where X i s the Greek letter chi, k = ( Nm - N) is t he number of degrees at' freedom, an d a relates to the area under the X 2, curve. The chi-square a distribution very closely matches the standard gaussian distribution when k is l a r g e (k > 30), whi ch is often the case in power system applications. Figure 1 5.3 shows the probability density function of Xl, a for a representative small value of k . As usual, the total area under the curve equals 1, but here i t is not a'

P7

degrees o f freedom

Area Area (1 a)

Q

-

o FlGURE 15.3

2

Xk. a

Probability d ensity function p{X 2 ) of the chi-square distribution degrees of fre edom.

xi

a

for a (small) value of k

1 5.3 TEST FOR BAD DATA TABLE 15.3

Values of area

a

to the right of

X

2

X !,

=

a

in Fig. 15.3 a

a

k

2 3 4 5 6 7

8

9 10

659

0.05

0.025

0.01

0.005

k

0.05

0.025

0.01

3.84 5.99 7.82 9.49 1 1 .07

5.02 7.38 9 .35 1 1.14 12.83

6.64 9.21 1 1.35 13 .28 15.09

7.88 10.60 1 2.84 14 .86 1 6.75

11 12 13 14 15

19.68 2 1 .03 22.36 23.69 25.00

21.92 23.34 24.74 26. 1 2 27.49

24.73 26.22 27.69 29.14 30.58

3 1 .32 32.80

12.59 14.07 15.51 1 6.92 1 8 .3 1

14.45 1 6.01 17.54 19.02 20.48

1 6.81 18.48 20.09 2 1 .67 23.21

1 8.55 20.28 21 .96 23.59 25. 1 9

16

26.30 27.59 28.87 30.14

28.85 30.19 31 .53 32.85

3 1 .4 1

34 . 1 7

32.00 33.4 1 34.81 36. 19 37.57

34.27 35.72 37. 1 6 38.58 40.00

17

18 19 20

0.005

26 .7 6

28.30 29.82

symmetrically distributed. The area u nd er the curve to the right of xl a i n Fig. 1 5 . 3 equals a, which is the probability that f exceeds X r a ' The remai � ing area under the curve is the probability (1 - a ) that the calculated value of the weighted sum of squares f� with k degrees of freedom, will take on a value less than X l, a ; that is, Pr ( f <

x l, a )

= (1 - a)

( 15 .48)

Based on this equation, the critical value of the statistic f can be determined using the tabulated values o f x t a given in Table 15.3. For example, choosing a = 0.01 and k = ( Nm - N) = 2, we can condjde that the calculated valu e of f is less than the critical valu e of 9.21 with probability (1 - 0.01) or 99% confidence since xi 0 '0 1 Q.21 in Table 1 5 . 3 . Thus, the chi-square distribution of f p rovides a test for detection of bad measurements. The procedure is as fol lows: =

•

•

•

•

Use the raw measurements Z j from the system to determine the weighted least-squares estimates X i of the system states from Eq. (15 . 1 9 ). Substitute the estimates Xi in the equation z = Hx to calculate the estimated values ij of the measurements and hence the est imated errors ej = Zj - ij • Evaluate the weighted sum of squares I = L.f;\ ef /a/For the appropriate number of degrees of freedom k = ( Nm - Ns ) and a specified probability a , determine whether or not the value of f is less than the critical value corresponding to a . In practice, this means we check that the inequality

f < Xk2 , 1r ,..

( 1 5 .49)

660

•

CHAPTER 15

STATE ESTIMATION OF POWER SYSTEMS

is satisfied. If it is, then the ineasured raw data and the state estimates are accepted as being accurate . . When the requirement of inequality (15.49) is not met, there is reason to suspect the presence of at least one bad measurement. Upon such detection, omit the measurement corresponding to the largest standardized error, namely, ( z j - i)j ...jRjj of EG,.' ( 1 5 .43), and reevaluat � the state estimates along with the sum of squares f. If the new value of f satisfies the chi-square test of inequality ( 1 5 .49), then the omitted measurement has been successfully identi­ fied as the bad data point.

The sum of squares of the estimated errors will be large when bad measure­ ments are present. Thus, detection of bad data points is readily accomplished by the chi-square test. The identification of the particular bad data is not so easily undertaken. In practical power system applications the number of degrees of freedom is large, which allows discarding a group of measurements correspond­ ing to the largest standardized residuals. Although there is still no guarantee that the largest standardized errors always indicate the bad measurements, the problem of identifying the gross errors is thereby eased. In each of the following examples we assume that all the measurement errors are gaussian random variables. In Example 1 5.2 the results of Example 1 5 . 1 are used to determine f when the meter accuracies are known. Then, in Examples 15.3 and 15.4 we check for the presence of bad data in the measure­ ment set. Suppose that the weighting factors W I to W4 in Example are the corresponding reciprocals of t he e rror variances for the four meters of Fig. Eva l uate the expected value of the sum of squares f o f t h e measurement residuals.

15.1

Example 15.2.

15.1.

First, l e t us evaluate the d iagonal clements of R ' R - H G . I H T before forming the sum of squa res for Eq. To t ake advantage of previous c alculations in Example we write the matrix R ' in the form R ' = (1 - H G - 1 H T R - I )R. Then, from Example the m atrix H G - 1 H T R - 1 t akes o n the value

Solution.

=

15.1,

(15.46). 15.1

]

-0.125 [ 1.2982 0. 0351 0.4 386 0.2281 ] 0.625 0.2281 0.4386 1. 2 982 0.125 0. 0 351 0.375 Only t h e diagonal elements resulting from this equation are required for Eq. (15.46), a n d so multiplying the first row of H by the first column of G - ) H TR � I , the

15.3 TEST FOR BAD

DATA

661

second row by the second column, and so on, we obtain x

X

X

0.1930

0.8070

X

X

X

where X denotes the nonzero e lements w hich do not have to be eval uated. The d iagonal elements of R' are then found to be R 'I I R �2 R '3 3 R �4

W J 1

[ 0 .S07

0 . 807

u2]

OlJ}

0 . 1 93

u22

u32

ul

where u/ i s the error variance of the jth meter. The expected value of corresponding to Eq. ( 1 5 .46) is calculated for N = 4, as follows:

i Rj;

j= 1

m

=

aj

= =

(1

(1

4

-

0 .807) u? a]2

+

+ 1 + 1 4- 1 ) -

2 = 2

-

(1

-

0 .807)u{ a}

(0 .807

+

0 .807

+

+

(1

-

0.1

9

0 . 193)ul u3

3

4-

+

(1

0 .1 3)

-

f

0.193 ) ai al

9

which is the number of degrees of freedom when the system of Example 1 5 . 1 has four measuremen ts from which to estimate the two state variables. It is appa rent from the way in which the solution is set forth in this example that the sum of the d i agonal elements of the matrix H G I H T R - 1 numerically equ als the number of state variables, Ns ' -

Using the chi-square test of i nequality (15.49), check for the presence of bad data in the raw measurements of Example 1 5 . 1 . Choose a = Exam ple 1 5.3.

0.01.

Solution. Since there are fou r measurements and two state variables in Exam ple Based on the 1 5 . 1 , we choose k = and find from Table 1 5 .3 t hat xl = results for cj in Example 1 5 . 1 , the estimated sum of sq � ares f is calculated as

2

a

9.2 1.

662

CHAPTER 1 5

follows:

f=

STATE ESTIMATION OF POWER SYSTEMS

4 e� 1: � = lOOe? +

100 ei 50 e� 50 et 2 2 2 2 = 100(0.00877) + 100(0. 00456) + 50(0.02596) + 50(0.00070) = 0.043507 w hich is obviously less than 9. 2 1. Therefore, we conclude (with 99% confidence) that the raw measurement set of Example 15.1 has no bad m easurements. Example

given by

j= 1

+

+

Ui

15.4. Suppose that the raw m eas u re m en t s e t [ ZI

for

the

system o f

Fig. 1 5 . 1 is

Z 4 f = [ 9.0 1 A 3.02 A 6.98 V 4 .40 V f and the meters are the same as in Example 15. 2 . Check for the presence of bad data using the chi-square test for 0. 0 1. Eliminate .any bad data detected and Z2

z)

a =

calculate the resultant state estimates from the

reduced

data set.

The raw measurement set is the same as at the end of Sec. estimate the errors to be

15.1 where we f [ [ el e2 e3 e4 ] T = 0.0 6228 0.15439 0. 0 5965 -0. 4 9298 Substituting these numerical values in Eq. (15. 4 4), we find that

Solution.

f= L 4

j=

e�

�

I (Tj

=

100 e �

+

1 00 e �

+

50ej

+

50e�

= 100(0.06228 2 + 0.15439 2 ) + 50(0. 05965 2 + 0. 49298 2 ) 15.1009 This value of f exceeds Xi. O.OI = 9.21, and so we conclude that there is at least one =

bad measurement. The standardized error estimates are next calculated using the diagonal elements Rjj f r om Example 1 5 . 2 as fol lows:

0.06228 2 v'QT0.06228 eI = -1. 4 178 93j100 yR'1 l ...;( 1 - 0.807)aI 0.15439 i = V 0.15439 = 3.5144 e2 = °.l93/100 JR�2 V(1 - 0.807)a =

153 TEST F O R BAD DATA

e3

"fR'33

=

e4

yR'44

=

0.05965 y( 1 - 0.193 ) aJ -0.49298 /( 1 - 0.193 ) al

=

0.05965

�0.807 /50

-0.49298

';0 .807/50

663

0.4695 -

3.

8804

The magnitude of the l argest standardized e rror corresponds to measurement Z 4 in this case. Therefore, we identify Z 4 as the bad measurement and omit it . from the state-estimation calculations. For the th ree remaining measurements the revised matrices H and H T R- I are given by

H

=

[ -0.125 0.625

-0.125 0.625 0. 375 0.125

and the gain G

=

]

HTR - I =

a l on g w i t h its i nverse become

[ -13.47.268125 5625 - 41.13.420625 8125 J m a t r ix

12. 5 0 18.6.7255 ] [ - 6212..5500 - 62.50

G- I

=

0.007391 ] [0.0. 000739 230431 0.026522

The state estimates based on the three retained measurements are

[ ]

0.17391 0.47826 1 9.01 1 .5 6522 . 0.30435 3.02 6.98

=

[ 16.0074 1 8.0265 V

V

1 [ 3.9.00013157 1 [il l = [ -0.1250.625 -0.1250.625 1 [ 16.0074 8.0265

and the revised estimates of the same three measured quantities are

�2

Z3

=

0.375 0.125

7. 0061

A A V

The estimated errors are calculated to be

from which the sum of squares

f � '[7=

I

ef/

a/ is found to equal 0.0435. Since

664

CHAPTER 15

STATE ESTIMATION OF POWER SYSTEMS

there are now only three measurements from which to estimate the two state variables, we compare f = for one wit h the chi-square value xr o.o l = degree of freedom and conclude that no more bad data exist. The reader is encouraged to check that the matrix HG - I H T R - 1 of this example has diagonal elements which numerically add up to 2, the n umber of state v�riables (see Prob. The new state estimates based on the three good m easurements essentially match those of Example 1 5 . 1 .

0.0435

6.64

15.3).

15.4

POWER SYSTEM STATE ESTIMATI ON

The state of the ac power system is expressed by the voltage magni tudes and phase angles at the buses. Although relative phase angles of bus voltages cannot be measured, they can be calculated using real-time data acquired from the system. These data are processed by the state estimator , a computer program which calculates voltage magnitudes a n d relative phase angles of the system buses. While the state estimator gives results similar to those available from a conventional power-flow program, the input d ata and calculation procedure are quite different . The state estimator operates on real-time inputs which are of two kinds-data and status information. The on/off status of switching devices (such as circuit breakers, disconnect switches, and transformer taps) determines the network configuration, which changes whenever those devices operate. Remote terminal units located on the system to report such changes also monitor ( 1) analog data in the form of megawatt and megavar flows on all major lines, (2) P and Q loading of generators a n d transformers, and (3) voltage magnitudes at most of the buses of the system. Every few seconds the remote units are scanned and the complete set of m easurements is telemetered to the energy control center. Thus, the state estimator draws on a larger data base than that previously described for the conventional power flow. In practical state estimation the number of actual measurements is far greater than the number of data inputs required by the planning-type power flow of Table 9. 1 . As a result, there a re many more equations to solve than there are unknown state variables. This redundancy is necessary, as we have seen, because measurements are sometimes grossly i n error or unavailable owing to malfunctions in the data-gathering systems. Direct use of the raw measurements is not advisable, and some form of data filtering is needed before the raw data are used in compu tations. S ta t e estimation p e r form s this filtering function by processing the set of measurements as a whole so as to obtain a mean or average-value estimate of all the system state variables. I n this way the best fit of the entire set of input data is obtained rather than a separate processing of the individual measurements. The estimated state of the simple dc circuit of Sec. 1 5 . 1 is given by the closed-fonn solution of Eq. 05. 19) bec ause the circuit equations are linear. For the ac power system the measurement equations are nonlinear and itefSitive solutions are required as in the Newton-Raphson power-flow procedure.

15.4

POWER SYSTEM STATE ESTIMATI O N

66S

Suppose, for instance, that the measurement equations are of the form ( 1 5 .50) ( 1 5 .5 1 ) ( 1 5 .52) ( 15 .53)

where, in general, h i ' h 2 ' h 3 ' and h 4 are nonlinear functions which express the measured quantities in terms of t he state variables, and e 1 , e2 , e 3 , and e 4 are gaussian random variable noise terms. The true values of x I and x 2 are not known and have to be estimated from the measurements Z I ' Z 2 ' Z 3 ' and Z 4 · We begin by form ing the weighted sum of squares of the errors with weights Wj chosen equal to the reciprocal of the corresponding error variances a/ , wh ich gives

( 1 5 .54)

before, the estimates x 1 and x 2 which minimize f must satisfy Eq. ( 1 5 . 1 6) and that equation now takes on the form

As

1

2 a!

ah l ax ! ah I aX 2

ah 2 ax ! ah 2 aX 2

Dh ) aX I ah aX 2 �

aJz4 ax ! ah 4 iJ x

2

1

X2) I2 - h 2 ( X l ' X2 ) z) h )( x l ' X z ) Z4 h4(XI ' xz) II

2 cr 2

-

-

0- 2 :\

i

R- 1

-

1

crl

h l( X l '

=

[�l

( 15 .55)

666

. CHAITER 15

Noting

STATE ESTIMATION OF POWER SYSTEMS

t hat the partial derivative terms depend on x 1 and x 2 • let us define

Hx

=

ah 1

ah 1 ax ] ah 2 ax ( ah3 ax (

aX 2 ah2 aX 2 ah 3 aX 2 Ah 4 AX2

A iz 4

Ax ]

and

( 1 5 .56)

then we can write Eq. ( 1 5 .55) in the m o r e compact form

( 15 .57)

To solve this equation for t h e state estimates x 1 and X 2 , we follow t h e same procedure as for the Newton-Raphson power flow. For example, using Eq. (9.31) to linearize h / x l , x2) about the initial point ( x �O), x �O» gives

h ]( x I ' x 2 )

=

h I ( x (IO) ' x (2O» )

+

ah (0) � X (1O) _1 aX l

+

ah (0) � X 2(O) _I uX2 .:l

( 1 5 .58 )

where, as b efore, L1 x)O) = x�l) - x;O) represents the typical state-variable correc­ tion and X) I ) is the first calculated value of Xi' Similarly, expanding h l x I ' x 2 ), h3( x ! , x 2 ) , and h / x l > X 2 ) ; subst ituting the expansions in Eq. (15.57); and rearranging yield

2 1 - h l ( x i°>' x �O » ) 22 - h 2 ( X �O\ x�O» ) H ( O) T R - I 23 - h 3 ( X �O), x&O») x

( 15 .59)

24 - h4( x�O), x �O» ) w hich all q uantities with the superscript (0) are computed at t h e initial values x�O) a n d x�O). We require the correction s �x 1 and L1X 2 to be (approximately) zero in order to satisfy Eq. ( 1 5 .57), and so the calculations of Eq. (15.59) are continued b y using L1 X�k) X�k + I ) - X � k ) to form the more general iterative in

=

15.4

POWER SYSTEM STATE ESTIMATION

667

equation Z1

-

h 1 ( X(k) l ' X(k») 2 ( 1 5 .60)

For convenience the iteration counter k has been omitted from H x in E q. 05.60). However, at each iteration the elements of the jacobian Hx and the quantities Z j h / x ik ), X�k ») are evalua ted from the latest available values of the state variables until two success ive solutions have converged to within a specified precision i n de x E ; that i s u n t i l Ix�k + l ) - x � k ) 1 < E for every i. When there are N, state variables a n d a larger n umber Nm o f measure­ m e n t s , the r e ct a n g u l a r form o f H fo r i t e ra t i o n k i s g i v e n by -

,

x

H = x

ah ] ( k ) ax ,

all , ( k ) aX 2

) 3 h z (k aX 1

ah 2 aX2

ah N ( k )

ah N ( k )

m

ah 1 (k ) Bh2 (k)

(k)

( 15 . 6 1 )

In

and Eq. 05 .60) assumes the general iterative form z1 Z2

-

-

(

) X(k» )

h 1 X(l k ) ' X 2( k ) ' . . . ' x (Nsk »

h 2 ( X (l k ) ' X (2k ) ' . . . '

Ns

( 15 .62) At convergence the solution X(k + 1 ) corresponds to the weighted least-squares estimates of the state variables, which we denote by x N• �

]

T

( 15 .63)

668 To •

CHAnER 1 5

STATE ESTIMATION OF POWER SYSTEMS

check the estimates using the same statistical tests as before, wc

Compute the corresponding measurement errors using ( 15 . 64 )

• •

Evaluate the sum of squares f according to Eq. ( 1 5.44), and Apply the chi-square test of Sec. 15.3 to check for the presence of b ad measuremen ts.

To illustrate the procedures for linearizing and iteratively solving the a c power system equations to obtain estimates of the states, let us consider Examples 1 5 .5 and 1 5 .6, w hich now follow. 15.5. Two voltmeters, two varmeters, and one wattmeter are installed on the system of Fig. 9.3 to measure the following five quantities:

Example

bus

(1) voltage magnitude:

bus CD vol tage m agnitude:

bus CD reactive-power injection :

l ine P-flow from bus CD to bus

(1) :

line Q-flow from bus @ to bus CD :

z1

Z2

= 1 V2 ! = I VI I

z3 = Q1

Z 4 = Pl 2

Z 5 = Q2 '

as shown i n Fig . 15.4 on page 670. Formulate the l i nearized equat ions calculating the weighted least-squares estimates of the system states.

for

In this two-bus example we choose 0 , = 0° as the reference angle and the three u nknown state variables as X I = O2, X2 = I V2 1 , and x 3 = I VI I . The first two state variables are the same as for the power-flow study of Example 9.3. The third state variable x 3 = I VI I is necessary here (unlike i n the power-flow problems) becau s e the magnitude of the voltage at bus CD is u ncertain and has to be Solution .

estimated.

Expressions for the line flows in terms of the state variables are given by Eqs. (9.36) and (9.37) as follows:

For t h e reactive power Q 1 i njected into the network at bus CD of Fi g. 15.4,

1 5.4

POWER SYSTEM STATE ESTIMATION

669

Eq. (9.39) leads to

Note that Q r includes the reactive power in the reactance of j6.0 per unit from bus CD to neutral. The above equations allow us to write the following expressions for the measurement errors in the k t h iteration as

e( k ) I

e( k ) 2

=

=

z z

I

-

hI

X(k)) ( X( k ) X(k) 2 J

'

,

=

3

2 ' X(k)) 2 - h 2 ( X(k1 ) ' X(k) 3

=

z z

1

2

) - X(k 2 -

X(k) 3

In the same i teration the partial derivatives of according to Eq. ( 1 5 . 61), which now becomes ah 1

(k)

ax ) ah 2

H (k) x

=

(k)

ah 1

aX 2

ah 2

aX l

aX 2

ah3 (k ) aX

ah3

l

ah 4

ax ) (! h

s

ax )

(k)

(k )

aX 2

ah 4

aX 2 ah s

aX 2

(k ) (k)

(k ) (k)

(k )

0 .. 0 4X (3k )X (k) 2 sin X (1 k ) - 4 X (3k )X (k) 2 cos X (I k) 4 X(k 3 )X(k) 1 2 sin X (k)

ah l

a X3

h ), h2,

and

h3

are evaluated

(k)

ah 2 ( k ) aX 3 ah (k ) 3

aX 3 ah4

aX 3 ah ",

(k ) (k )

aX 3

1 0 ) X( k) - 4 X (k 3 cos 1 k) - 4 X (k) 3 sin X (I

8 x 2 - 4 X (3k) cos X (I k )

0 1 Z5 X ( k) - 4 X(k) cos X(k) 1 2 3 3 ) (k - 4 X (2k ) sin X 1 - 4 X (2k ) cos X (k) 1

670

CHAPTER 15

PI

Ql ( Z 3 ) I

STATE ESTI MATION OF POWER SYSTEMS

1V1 1 & CD

I V2 1&

P1 2(Z4)

P2 1

)0.25

-

Q12

-+--

Pz

2

-

-

---+-

--+-

Q 2 1(Z S)

j6.0

Q2

I

tJ

0 .3

06

FI G URE 15.4

T h e one-line d i a gram of Example 1 5 .5 showing t h e q u a n t it ies 1 V2 1 , I V) I . Q ) . P1 2 ' a n d Q21 cons tituting t h e m easurement s e t { Z l ' z 2 ' z 3 ' Z 4 , z s } ·

Based o n H �k ) and eJ k ) , the counterpart of Eq. 05.62) for calculating the state estimates is of the form

[ 1 X �k + J ) X (2k 1 ) +

x (3k

+

e lk ) e (2k )

(k)

e3

_

1)

e

�k )

e 5( k )

with the weighting matrix of reciprocal variances given by

1 1 1 1 1

1 5 .4 POWER SYSTEM STATE ESTIMATION

671

The equations developed in the preceding example are important illustrations of the general procedure and fonn the basis of the numerical calculations in Examples 15.6 and 15.7 for estimating the system state. Tclemetcrcd measurements on the physical system corresponding 1 5 .4 show that the p e r - un i t values of the five measured quantities are

Example 1 5.6.

to Fig.

z\

=

I V2 1

0 . 92

=

Z 5 = Q 2 1 = 0 .305

The variances of the measurement errors are specified 2

2

(T I = (T 2 =

( 0 .01 ) 2

a} = a}

=

( 0 .02) 2

in

per unit as

t h e equations o f Example 1 5 .5, compute t h e weigh ted least-squares estim ates of the state variables X I = ° 2 , X2 = I V2 1 , and X 3 = I VI I of Fig. 1 5 .4. Check the results using the statistical tests of Sec. 15.3.

Using

Solution. We begin the i terative calculations by using flat-start values x \O) = 0° and x �O) = 1 .0 per unit for the three state variables which yield sin x \O) = 0 and cos x �O) = 1. Substituting these n umerical values in the expressions of Example 15.5 gives the measurement errors of the first iteration:

x �O)

=

e �O)

e �O)

Simil a rly,

=

=;=

the

z) z2

- x �O)

- x �O)

= =

0 .92 -

1 . 00 =

1 .02 - 1 .00

numerical form of

H (x�

=

- 0 .08 0 .02

the j acobian

=

0 0 0 -4 0

becomes

1 0 -4 0 4

0

1

13 '3

0 -4

.

672

CHAPTER 15 STATE ESTIMATION OF POWER SYSTEMS

The gain m atrix G �O)

G ;"

�

[!

=

H�O)TR 1 H �O) is now eva l ua ted as follows: -

-:J

0 -40 -4 0 0 1 0 1 2 (0.01 ) 1/ (0.01 13 J

1

x

(0.02)

2 1 2 (0.015)

1

(0.02/ � 10 4 X

[r 7

�

[�

-!l

00 - 01 - 1.7778 0 1 1.0833 0

l

00 0 01 0 -4 -40 40 - 04 1

\3 J

09.0000 � X 10 4 -8.3333 9.6944 -

8 333 .

The n umerical form of Eq. ( 1 5 . 62) for the

first

iteration becomes

) e (IO ) e�O ) e�O ) e�O ) e�O

o o

o -1

1 1.0833

- 1.o7778 o

Jl

-0.0.0200 0800 383 0.0.0.543980 05'0

133 ] -8.3333 ]- [ --0.1.02631

1 5 .4 POWER SYSTEM STATE ESTIMATION o

[ ]

1

o

-8. 3333 9.6944

= -0.1495 00..99796127

673

0.1898

These values of the state variables act as input data to the second iteration and computations continue until convergence is reached. The converged values of the x/s are chosen as the state-variable estimates Xi w ith X I = 82 X2

=

x)

=

=

I V2 1

-

0.1762 rad

= - 10.0955°

= 0.9578 per unit I VI I = 0.9843 per unit

= - x 2 = 0.92 - 0.9578 = -0.0378 e2 z2 = 1. 02 - 0. 9843 = 0.0357

The corresponding estimates of the measurement errors i n per unit are el

ZI

2)

=

and the weighted

sum

of squares of these errors becomes

/= .t ( :. ] 2 J

=

I

=

)

( - 0 . 0378) 2 ( )2

0.0 1

+

+

( - 0 .0630) (0.015) 2

2

-----,-, 48 -: = -----: ( 0 .0357) 2 (0.0 1 ) 2

+

+

(0 .28 1 0) 2 (0.02) 2

(0.3 0) 2 (0 .02)2

545

If f is chi-sq uare distribu ted, the number of degrees of freedom k equals since there are five measuremen ts and three state variables to be estimated. For a 99% confidence interval (a 0.0 1) the requirement f � Xi.. a must be satisfied. In Table . 2 1, and it is now clear for k 2 and a 0.01 , we find that xi 0 '0 1 � that the calculated value f 545 is excessive, indic t ing the presence of bad data. Consequently, . we cannot accept the calculated values of the state variables as being accurate. If there is only one bad measurement, we can follow the procedu re of Sec. 15.3 i n an attempt to identify it. But if there are two or more grQSS errors in

15.3

=

=

==

=9

2

'674

CHAPTER J5

STATE ESTIMATION OF POWER SYSTEMS

the measurement set, our procedures break down since Nm Ns = = and ' redundancy is lost when two or more measurements are d iscarded, The calculations u nderlying the identification procedure are provided in Example

5-3 2 15.7.

-

Example 15.7.

I dentify the bad data detected in the measurement set of Example

15.calculate 6 . Discard the erroneous data, and if sufficient good m easurements rema I n, the new system state. In an e ffort to identify the bad measurements let us compute t h e standardized residuals in the given measurement s e t . To d o so, w e first determine the di agonal c lements o f t h e cova r i a n ce m (\ t r ix R' R H xG ; I H '� ' = ( I H x G : I I"' �'R - I )R using t h e es t i m a tes Xi of Exa m p l e in the following sequence of calculations:

Solution.

15.6

-

=

Hx

1

00

=

0

- 4 x 3 X 2 cos X I

Based on R - 1 matrix are

8

x2

-

A

)x)

- 4 x 3 sin X l

4 x 3 x 2 sin x I

=

25

- 4 x 3 cos X I

4 x 3 x 2 sin X I

4 x 3 cos X I

01

- 4.i 2 cos

- 4 x 2 sin x I

XI

- 4 x 2 cos X I

1 0 0 -0.06610 -3.08761 0.16303 -3. 125 0.3.67863 901 -3, 7716 719 -0,76610 of Example 15.6, the correspondin g numerical elements of the gain 4 .4

4 1. 2 169 ] - 8.1.1239 -7.6575 10 5 518 . 66 -7.6575 41

9

x

and the reader can now determine the diagonal elements of the matrix H x G x- 1 HTR -1 x

which sum to Nm

=

X

=

X X

X X X

0.5 618 0.4976 0.5307 0.9656 X X X X

X X X

X X

X

X X X X .4443 0

3. Diagonal entries In the covariant matrix R' are readily ,

15.4 POWER SYSTEM STATE ESTIMATIO N

computed

as

675

follows: 0 . 4382 =

The calculated

0 .5024

0 .4693

ej of Example 15. 6 correspond to standardized errors e 1 V -0. 0378 4 5.7 106 0.4382 10p::; e2 0.0357 4 5.0419 {R'22 /0 .5024 100.2810 4 20. 5 079 10..jR:l) ./1 e4 -0. 0630 4 -22.6559 ..jR'44 VO.0773 10jRes' V 0.3 480 4 23.3403 .. S5 2. 2229 10--

-

=

X

X

A

eJ

=

--

.8772 X

=

=

X

--

=

=

X

We note that large standardized errors are associated with measurements z 3 ' Z 4 , and z s ' In order to preserve redundancy, only one measurement can be d iscarded, which we choose to be z5 corresponding to the l argest standardized e rror. From

676

CHAPTER 15

xiO) ZI'

STATE ESTIMATION O F POWER SYSTEMS

z3' and Z4 in the original measurement set and the fiat-start values 0, x�O) x�O) 1 .0 per unit we repeat the state-estimation calculations,

ZZ '

=

=

=

which yield

i1

= f> 2

x2

= I V2 1

-0.1600 rad 9.1673° 0.9223 u n =

=

per

-

it

The reader can verify t he following n u merical values of the corresponding jacobian H ll a n d gain matrix G x :

".

�

Gx

=

r -0-3.�7978053

[ -0.6_19134670 - 1. 6904

1 0 -4.0.6482 0 175 -0. 4670 -4.5.2217 6877

]

4_0.X5876S8 4 --4.1. 06904 877 X 106. 9998

]

The new estimates of the measu rement errors are found to be

=

0.92 -0.9223 -0. 0023 1.02 -1. 0 174 0.0026

=

0.509 - 0. 5978

=

=

0. 0002

The weighted sum of squares (with one degree of freedom) amounts to

f

=

( - .0023 ) 2 ( 0.0026 ) 2 ( - .0022 ) 2 ( 0.0002 ) 2 0.1355 0. 02 0.0 15 0.01 0.0 1 xt 6.64 15.3. X3 �� 99% nfi n +

+

+

=

which i s certainly less than 0 01 = of Table Consequently, the estimates X l ' X 2 , and of t his exampl an be accepted with co de c e as reasonable values of the state variables. We con clude, therefore, that Z s is the only b ad d ata point in the original measurement set.

15.5

THE

STRUcrURE AND FORMATION O F H x

677

Bad data can originate from m any causes. One such cause arises when a met er is improperly connected with the leads reversed so that it reads upscale

(positive) rather than downscale (negative). For instance, in the original data set of Example 15.6 Z 5 is incorrectly reported to be equal to 0.305 per unit. If we change the sign of this measurement by setting Z 5 0.305 per unit in the original data and repeat the state-estimation calculations using all five measure­ ments, we obtain =

XI

= 02

X2

=

- 0 . 1 599 rad

I V2 1

-

=

-

- 9 . 1616°

0 .9222 per unit

which are acceptable results with the sum of squares f = 0 . 1 42 1 . Thus, state estimation can sometimes revea l when properly fu nctioning meters ha v e been installed backward on the physical system.

THE STRUCTURE OF H x 15.5

AND

FORMATION

The phase angles of the voltages a t different substations of the system cannot be economically measured, but voltage magnitudes are routinely mo n i t or e d I n order to estimate ' both voltage angles and magnitudes, w e m us t choose the a ngle at one of the N buses of the system as refere�ce for all other angles, which leaves N - 1 angles and N ma g nit u d es to be calculated by Eq. (15.62). Hence, the state-estimation jacobian H x ' unl ike the square jacob ian J of conventional Newton-Raphson power flow, always has (2 N - 1) col u m ns and a larger num­ ber Nm of rows, as shown in Eq. (15.61). Each row of H x corresponds uniquely to one of t he measured quant ities ind icated in the tra nsmission-l ine equivalent c i r c u i t of F ig . 15.5, namely, .

•

•

•

•

•

The voltage magnitude I V; I at a typical bus (j) , The active power Pi injected into t h e network at bus (j) , The reactive power Q; injected into the network at bus (j) , T h e active power fl o w Pi} a t b u s CD or Pj; a t bus 0 in t h e line connect ing b uses (j) and (]) , and Th e react ive power flow Qij at bus (j) or Qji at bus (]) in the line connecting buses CD a n d (]) .

Although line-charging capacitance is shown in Fig. 15.5, line-charging megavars

678

CHAPTER 15

STATE ESTIMATION OF POWER SYSTEMS

CD

- y.IJ.

=

· . + j· BI). ) - ( G I)

_L

FIGURE 15.5 The

TI _- EQUival ent

L..._ ... _

-

-

Ci r � ----.J

per-phase represe n t a tion of t h e transmission l i ne from bus CD to bus (f) s howing the bus i nj e c tions , vol t age m ag n i t u des, a n d line flows w h ich can be measu r e d . Li n e c h a r g i n g is repre s e n t e d b y susceptance B;j a n d 1'; j is t h e e l e m e n t of Y bus e q u a l to the negative of the s e r i e s a d m i t t a nce of the line.

cannot be separately measured since they are distrib u ted al ong the l en g t h of the line and form part of the Qij and Qji measurements. For simplicity, other shunt-connected reactors or capacitors are not shown, but it should be under- . stood that P and Q in any lumped external connections from the typical bus CD to neutral can be measured and included as part of the net injections, Pi and Q;. This is demonstrated in Example 1 5 .5 w here the injection Q \ includes the reactive power flow in the reactance from bus CD t o neutral. If I V; I , Pi ' and Qi are measured at every bus and the flows Pi j , �'i ' Q ij ' and Qji are measured i n every line, the measurement set is full and Hx has a total of Nm (3N + 4 B ) rows, where B is the number of network branches or lines. The ratio of rows to columns in H x (which is equivalent to the ratio of . measurements to state variables) is called the redundancy factor , and it equals (3N + 4 B )/(2 N 1 ), or approximately 4.5 in a large fully monitored power system with an average of 1 .5 branches per bus; that is, with B /N = 1 .5 . In practice, the redundancy factor can be somewhat s m a l l e r s i nce the measure­ ment set may be less than full while the number of state variables remains at 2N 1 . When the number of measurements equals the number of state variables, there is no redundancy, H is square, and the coefficient matrix (H� R - I H ) - l H�R- l of Eq. (15.62) reduces to H � I if H x is also invertible. In that case the weighting matrix R - } h as no effect; all the raw measurements m ust be used directly (even if erroneous); and if any one measurement is missing, the system cannot be solved. Thus, an adequate number of redundant measurements from strategic points of the system is absolutely essential to ensure a good state estimate, and this makes the jacobian rectangular in the state estimator. =

-

-

x

15.5 THE STRUcrURE AND FORMATION OF

H.

679

When the full set of measurements is numbered so that those of the same type are grouped together, the measurement vector takes on the fonn

Zl

I V1 1

ZN

I VN I

ZN+ l

Z2N

PN

Z2N+ l

Q1

Z 3N

Z =

PI

Z 3N+ 1 Z 3N + B Z3 N + B + I Z3N+2B Z 3N + 2B + I z 3 N + 3B Z 3 N + 3B + l

Z 3N + 4B

QN pI.}.

N voltage magnitudes I V; I

N

bus injections Pi

N

bus injenions Qi

B line flows

Pjj

B l ine flows

Pji

B line flows

Qij

B line flows

Qji

( 15 .65)

--

p}I. . --

Q jj --

Qj;

and the corresponding jacobian H x has the symbolic block form shown in Table 1 5 .4 on page 681 .

680

CHAPTER 15

STATE ESTI MATION OF POWER SYSTEMS

Note that H x has a row for P I and for Q l and a column corresponding to 0 1 ' This is because in state-estimation calculations a reference angle must be specified at one of the buses which we have chosen to be bus CD , as in Example 15.5 . The voltage magnitude I VI I does not have to be specified, which makes bus CD different from the slack bus of the power-flow problem. The zero elements in block 1 of Table 1 5 .4 result from the fact that the measurement I �I does not explicitly depend on O J so that a I V; I / a o} equals zero for all values of i and j. Also, for a full measurement set the diagonal elements of block 2 are unity because a l V; I /a l Vj l = 1 when i equals j and otherwise is zero. The real power injected into the system at bus CD i s given by Eq. (9.38), which is repeated here as I VI ] but not for

Pi = 1 V: 1 2 G ii

+

L I V; Vn V; l cos ( O + N

n = n

in

n

I

*i

On - OJ

( 1 5 . 66)

and likewise, Eq. (9.39) expresses the reactive power injected into bus CiJ Qj =

-

{ 1 V: 1 2Bii

+

t 1 V; V,r Y;n l sin ( Oin

n=l n

*i

-

+

on

- ) 0; )

( 1 5 .67)

The injection Pi includes all the active power flows from bus CD to other buses of the system so that for the particular line CD (j) the active power flow Pij from bus CD to bus 0 is given by

. - . -

pIJ.

I VI 1 2 GIJ + 1 V VJ YI J I cos (0" IJ

=

I

+

0J - 0 I. )

-

( 15 . 68)

-

The first term - I V; 1 2 Gij is necessary because line CD (]) contributes the series conductance Gij to the calculation of the self-conductance Gii of Eq. (15 .66). Likewise, the reactive power flow from bus CD to bus 0 on line CD (J) can be written from Eq. (15 .67) as

.

-

-

( 1 5 . 69 )

where B;j/2 is the line-charging susceptance and B ij is the series susceptance contributed by line CD (J) to the self-susceptanc e Bii· The expressions for �'i and QJ I. are obtained from Eqs. ( 1 5 .68) and ( 1 5 .69) simply by interchanging

15.5 THE STRUcnJRE AND. FORMA TION OF H . TABLE 15.4

Typical elements in the block form of state-estimation jacobian H x

r-

® ® ...

-

0

1 0

1

.

0

aPI

-

.

aP I

..

-

aPN

.

-

aPN

aQl

.

-

.

-

.

.

-

aQ l

.

CiQtv t30 2

=

"

.

aPij t30i

.

.

.

aPji

-

.

.

.

.

.

.

.

aQ ij aoi

dO) i

-

isr6l

--

.

.

.

aPi, ao)

.

...

aPji ao)

aQ ij

.

.

. .

.

.

. . .

iJ<Sj

'

aQ) i

_

"

a8)

t30i -

N

-

1

.

,

.

aQ l

ol Vl i

al VI I

rM �R

.

[UM E�

.

.

.

.

.

--

,

.

--

a WN I aPN a Wtv l

aQ l

aPij

--

al Vi I

a!j i al Vi I

.

.

..

.

.

\f

--

al vN I

...

oQtv

.

aP I

.

. . .

--

t30tv

a8i

. .

CiQtv

.

.

(II VI I

--

dOtv

--

aPN

--

aotv

ao 2

'

a i vi l

J3rl

.

t30 2

Hx

aP I

--

t30N

ao 2

I

11r2l

1

aQN

..

. .

.

aQ ij a l v, l

aQji al ViI

.

.

.

.

.

.

alVN I

aPi,

.

,

al "}l

aPji al Vi I

.

.

.

)

.

a l Vii aQ 'j

aQj i al"}l

' "

.

,

.

-

N

.

--

,

.

N b us I nj' e ctIOns .

B line flows Pi)

Qj

681

682

CHAPTER 15

STATE ESTIMATION OF POWER SYSTEMS

Formulas ror elements or the state-estimation jacobian TABLE 15.5

a l Vi i

--

aSj

aPj -

aSj

aPj -

aSj

=

a Qj aSj

a Qj -

aSj

I - ViVjY;j l sin(Oij + OJ

N

1 Vi Vn Y;n I s i n(fJin +

n =1 n "' i

=

aPjj -

aSj

aPjj -

aSj

aPjj -

aSj

aPjj

-

as;

O J

aSj O J -

aOj

CJQji

--

aSj

a Qjj

--

- 1 VJj Y;j 1 c o s( Oj}

N

=

I VYnY;n I c os(O;n

L

=

n=1 n "' i

-

=

=

=

=

=

- 0)

,

+

on

- 8)

. . 8 - 8·) J

I

· l sin(OIJ + oJ· I VI VY J IJ

8·)

0, -

OJ )

..

+ +

I V, Vi Y;j l cos(o, j +

0;

OJ

I

- OJ ) - OJ)

. \ co s(O' I + 0J - 0 '' ) I V, VY J IJ

l Vi Vj Y;j l co s (eij

+

OJ

- I VjVjY;j l c o s ( oij + OJ

2

1

1

O Pi

0 1 JIj 1 oP;

- 0)

- 0)

4

a l V, I

I

I

7

8

a Q;

a 1 JIj 1

a -a I Vi 1 OP'j

O P;j

a l Jlj I 0 1 10 1

I

aPj ;

a 1 JIj 1 aPj ,

--

l9

In

1

a l V, I

rJj -,) I Vi I

a I Vi i Vj

--

1 11

12

I

a Qji

a Qj;

0 1 Jlj l

l13 1 4

--

0 1 Vj l

I

a l Vi I

a i I/, I

j );

--

2 1 V, I Gij +

=

--

15 f6l

'1=

=

1

=

--

I :,

oU

=

--

- 0)

OJ - 8)

I VI V) YIJ· l s in(O I) +

- 1 V, Jlj Y;i 1 sin( fJij

-

On

+

1 V, Jlj Y;j l s i n(O"j

=

=

a l Vi i -a 1 Vj 1

all i and j

= =L

-

ao;

0

"x

=

L

,, = 1 " "" i

- 2 1 Vi l E" -

=

N

1 V" Y, n l si n(8'n +

j 11 * 1 �-

On

I v, Y' ) 1 cos( 8'· ) + o)·

- 8)

- 0·) I

- 8·) ,

- OJ)

I Vj Y;) I cos( 8ij + 0 ; - 0)

=

(

=

=

0)

2 1 Vj I G;i + I Vi Y;j l c o s(8;j + 0,

=

=

- 8,)

-t

l cos(eIJ + 0J 2 1 V' I G 'J + I VY ) ')

=

=

I s i nC O , j

- : V, Y"

L

n

- 0)

I Vn Y,n I cas (O /n + 0" - 0)

N

=

I v. Y,j 1 c os( Ojj + O J

- 2 1 VI 1 -" + B' I2 - 2! VI I

B�

,

(-

E ;j 2

+

B ')

)

)

-

I V; Y;, i sin(O'J

- I Vj Y; , l sin(O;j

-

+ +

I V; Y;j I sin( e;) +

OJ

- 0)

OJ - OJ)

8,

-

- I Vj Y;j l sin(Oij + OJ -

0)

0)

! 5.5

683

THE STRUCTURE AND FORMAT!ON OF H.

su bscript s i and j on all voltage m agnitudes and angles. Differentiating the eq ua tio ns for Pi' Qi' Pij , and Qij with respect �o the state variables 82, ' 8N and I VI I , ' . " I VN I , we obtain expressions for the remaining elements of H x , as shown in Table 15.5. •

•

,

The equations of Table 15.5 d e ter m in e the pattern of zero and nonzero elements in the jacobian H x ' which appears in Ta bl e 15.4. We have already remarked that in H " for a ful l measurement set: • •

Each

element of block 1 is zero. Each off-diagonal elemen t of block 2 is zero and each diagonal element is 1 .

Le t

now examine the pattern o f zero a n d nonzero elements i n blocks 3, 4, 5 , and 6 , which i n volve d e r i va t ives of the r cal and r ea ct iv e power injections into t h e b u s es . When t h e measu re m e n t set is full, blocks 4 and 6 are square (N X N) and h ave off-d iagon a l clements with I Yij l as a m u l ti plying factor. Therefore, if all Pi a n d Qi measurements are n umbered according to their numbers, we find us

thdL:

•

Blocks 4 and 6 follow exactly the same zero and nonzero pattern as the system Ybus ' This is because the (row, column) locations (i, j) and (j, i) in those blocks have nonzero entries if, and only if, I:j is nonzero.

Allowing for the fact that there is no column for •

01

in

HX)

we note:

Blocks 3 and 5 have the same zero and nonzero pattern as Y bus with i t s first column omitted.

Again, assuming a full measurement set, let us examine blocks 7 through 14, which invo l ve P and Q line flows. I f t he B measurements in each of the subsets (PI)' { Pjl}, (Qi)' and (Qj) are numbered in exactly the same sequence as the network branches, then: •

•

Blocks 8, 1 0, 12, and 14 have exactly the same zero and nonzero structure as th e network branch-to-bus incidence matrix A. This is because each line flow is i n c i d e n t to i t s two t e rm i n a l b u se s . Blocks 7, 9, 1 1 , and 13 have the same zero and nonzero structure as A with its first column omitted.

When the measurement set is not full, those rows corresponding to the missing measurements are simply deleted from the full jacobian H " . These observations yiel d t h e structure o f the complete jacobifln H " i n a straightforward manner. Since H x has many zeros, the gain matrix G" =

684

CHAITER

15

STATE ESTI MATION OF POWER SYSTEMS

H�R- 1Hx also has many zeros and further rules can be developed for identify­ ing the locations tional efficiency .

of the nonzero elements of

Gx

in order to enhance computa­

In the example which follows we demonstrate the formation of Hx for the particular case where only line flows and one bus-voltage magnitude are measured. At least one voltage measurement is always required in order to establish the voltage level of the system. Example I S . 8 . I n the system of Fig. 1 5 .6 ( o n page 686), Pij a n d Q ij m e asu re m e n ts are made at the ends of eac h l i ne and vol t age magn itude I V3 1 is also measured. No bus i njected powers a n d no oth e r bus vo l t a ges are m easu r e d . D e t e r m i n e t h e s t r uc t u r e of I I x , write t h e p a r l i a l de rivat ive f o r m u f i ts n o n z e ro e l e m e nt s , a n d e x p r e ss any fo u r o f t h e m i n l i t e r ;.} ] fo r m .

Figure 1 5 .6 has N 4 buses and B. 4 branches. Choosing 01 0° as the reference angle leaves O 2 , 0), 0 .\ , l VI I , I V2 1 , I V3 1 , a n d 1 V4 1 as t h e s e v e n ( Ns = 2 N 1 ) state variables to be estimated from t h e 1 7 (4 B + 1 ) m e a s u r e m e n t s . Therefore, the redundancy factor is ( 4 B + l )/Ns 17/7 = 2 . 4 . Since I V3 1 is the only voltage m e a s u r e m e n t a n d n e it h e r Pi n o r Qi a r e b ei n g m e a s ur e d , H h a s o n ly one row corresponding to a I V3 1 / a I V3 1 = 1 in blocks 1 through 6 of the general state-estim ation jacobian. Let us mark the branches b l to b4 , as shown in Fig. 15 .6, and order the line-flow measurements accordingly as follows:

Solution.

=

=

=

-

=

x

Branch bI :

Measurements

--

Then, the

P14

P4 1

Q 14

Q41

b2 :

P13

P3 1

Q 13

Q3 1

b3:

P23

P32

Q 23

Q3 2

b4 :

P24

P4 2

Q 24

Q 42

bran c h - t o-bus i n cid e n c e

A =

bl

b2 b3 b4

m a t rix fo r t h e n e t w o rk i s

CD W 1

0

1 0 0

0 1 1

G)

@

0

- 1

-1 -1 0

0 0 -1

which shows the zero and nonzero p attern of blocks 7 through 1 4 of H for this particular example. With a column for each s tate variable and a row for each x

15.5

measurement, H x assumes the form 52

53

54

0

0

0

0

0

--

0

--

a P23

--

aD 2 a P24

--

aD 2

--

aD3 0 0

0

--

a FJ !

I VI I .

I V3 1

0

1

0

--

0

0

--

0

�-

0

--

0

0

0

--

�-

0

0

--

a l v2 1

0

--

0

0

--

a P l4 aD 4

0

a P l4

a l v1 1 aP 1 3

a l v1 1

a P24

--

aD 4

D I V2 1 a P2 4

a l v3 1 a P23

a l V3 1

-

a P4 1

a P4 1

aD 4

a l Vj l oP] 1

-

aP23

aP l 3

a p3 ]

--

0

--

--

0

0

--

a P3 2

a P32

a l v2 1

a l V3 1

-

0

-

0

--

0

0

0

--

0

0

0

--

0

---

0

0

--

a Q24 a0 4

0

aD 2

a P4 2 aD 2

0

a Q 23

aD 3

a P3 2 aD 3

aQJ3 aD 3 a Q 23

aD 2 a Q 24 aD2

aD �

0

0

0

a Q 32

aQ3 1 aD ] aQ 32

aD 2

aD 3

0

a Q42 aD 2

0

I�I

I V2 1

0

a P32

I

a03 a P23

0

--

Hx =

a p l ,)

THE STRUcrURE AND FORMATIO N OF H.

aP4 2 aD4

aQ 14 aD 4

a l V1 1

a Q l4

a l v1 1 aQ13

a l v1 1

a P42

a l V2 1

a Q 23

a l V2 1 a Q Z4

a l v3 1

aQ13

a l v3 1 a Q 23 a l V3 1

aPl 4

a l V4 1

a P2 4

a l V4 1 a P4 1

a l v4 1 0 0

a P4 2 a l V4 1 aQ l4

--

a l v4 1 0

--

0

a l v2 1

0

--

--

0

0

--

0

--

iJ Q 3 1 a l VI I

0

0

0

--

0

--

a Q4 1 aD 4

a Q4 2 aD 4

a Q4 1

a l v\ 1

a Q 32

a l v2 1 a Q42

a l v2 1

aQ3 1

a Q 24 a l V4 1 a Q4 1

a l v4 1

--

0

--

0

0

--

a l V3 1 a Q 32 a l v3 1

a Q4 2

a l v4 1

685

686

CHAPTER 15

STATE ESTIMATION OF POWER SYSTEMS

....�I_ . ® P3 1

!t

Q 3 1 --+- P32 Q32

FIGURE

15.6

One-line d i agram for Exampl e

to

b4•

15.8

--

I

showing t he ?ij a n d Qij measurements on l i ne s marked b 1

The expressions for a Pn/a8 3 ' ap23 /a l V2 1 , a Q 24 /ao 2 ' and a Q24/a l V4 1 are given by Table 15.5 as follows:

a \ V2 1 J P23

The expressions for other elements can be similarly obtained.

15.6 SUM MA R Y

687

A state-estimation algorithm based on measurements of only line flows and the required bus-voltage magnitude is described in the literature to which the reader is referred for further details.4 Because Hx has many zeros, the gain matrix G x = H � R - \ H also has many zeros, and so Gx is never explicitly inverted as implied by Eq. (15.60). Instead LDU triangular factorization (described in Chap. 7), along w it h P - 8 and Q - I VI decoupling (described in Sec. 9.7), and optimal ordering (de­ scribed in Sec. B.1 of the Appendix) can be applied for efficient computations and enhanced solution times of state estimation. x

15.6

SUMMARY

State estimation determines the existing operating conditions of the system which are required for real-time control. The network model normally used covers the transmission portion of the operating area. Hence, the parameter values for all transmission Jines, transformers, capacitor banks, and i ntercon nec­ tions are required as in power-flow stud ies. I n th is chapter it is assumed that accurate parameters are known and on this basis state-estimation calculations employing the basic weighted least-squares approach are described. The state estimates are the bus-voltage angles and magnitudes which are computed by Eq. (15.62) from a set of redundant measurements. When bad measurements are detected, the state estimates are no longer reliable, which means that grossly e rroneous data have to be identified and filtered out by statistical tests. The diagonal elements of the covariance matrix R' R - H x G; 1 H� are used to calculate the largest standardized residuals which help in identifying the bad measurements. In industry studies most of the computation time is spent in evaluating the gain matrix G x = H�R - ) H x , which is very large, is symmetric, and contains many zeros (very sparse). Because of its sparse nature, G x is not explicitly inverted. Instead, optimal ordering and LDU factorization are employed. Con­ vergence upon the solution of Eq. (15.62) can be influenced by where the meters are placed on the system and the types of measurements (whether injections, voltage magnitudes, or line flows). Redundancy is important but a full set of measurements is not necessary and may not be desirable from the computa­ tional viewpoint. =

41 . F. Do pazo, O. A. Klitin, G. W. S t agg, a n d L. S. Van Slyck, " State Calculation of Power Systems from Line Flow Measurements," IEEE Transactions on Power Apparatus and Systems, vol. PAS·89, no. 7, September/October 1 970.

688

CHAnER 1 5

STATE ESTIMATION O F P OWER SYSTEMS

PROBLEMS 15.1.

The circuit of Fig. 15.1 is redrawn i n Fig. 1 5 .7, in which three loop current variables are identified as x I > x 2 ' and X 3 ' Although not shown, ammeters and voltmeters with the same accuracy a re assumed to be installed a s in Fig. 1 5 . 1 , and the meter readings are also a ss u med to be the same as those in Example 1 5 . 1 . Determine the weighted least-squares estimates o f the three loop currents. Using the estimated loop currents, determine the sou rce voltages VI and V2 and compare the results with those o f Example 1 5 . 1 .

1 n

CD

1 n

1

f::\

n" x3 )

FIGURE IS.7

Circuit of Fig. 1 5 . 1 with t hree loop currents X I ' x 2 , a n d x) a n d two n o d e s l a b e l e d

CD

and

Q) .

15.2. Show that E[(x - x.)(x X)T ] G I , where G is the gain m atrix. 15.3. Show that the sum of the diagonal elements in the matrix HG - I H T R - 1 In E q ( 1 5.40) is numerically equal to t h e number of state variables. 15.4. Prove Eq. ( 1 5.47). 15.5. Consider the voltages at the two nodes l ab e le d CD and (1) in the circuit of Fig. 1 5 .7 to be state variables. Using the ammeters and voltmeters connected as shown in Fig. 15.1 and their readings given in Example 1 5 . 1 , determine the weighted least-squares estimates of these node voltages. Using the result, determine the source voltages VI and V2 and compare the results with those of Example 1 5 . 1 . A lso, calculate the e x pe cte d value of the s u m of squares of the m easu reme nt residuals using E q ( 1 5 .46) and check your answer using Eq. ( 1 5 .47). 15.6. Five ammeters numbered A 1 to A s are used in the dc circuit of Fig. 1 5 .8 to d etermine the two unknown source currents II and 12 , The standard deviations of the meter errors are 0.2 A for meters A 2 and A s and 0.1 A for the other thre e meters. The readings o f t h e five meters are 0.12, 1 . 18, 3.7, 0.81, and 7 . 1 A, respectively. (a) Determine the weighted least-squ ares estimates of the source currents II and 12 , -

=

-

.

.

PROBLEMS

689

FIGURE 15.8 D C c i rc u i t with five a m m e t ers.

(b)

1 5 .7.

I S .g.

15.9.

1 5 . 10.

Using the chi-squ are test of Eq. ( 1 5 .49) for a = 0.01 , check for the presence of bad data in the measurements. (c) Eliminate any bad data detected in part ( b ) and find the weighted least-squares estimates of the source currents using the red uced data set. C d) Apply the chi-square test for a = 0.0 1 to the resu lts of part (c) to check if the result is statistically acceptable. Redo Prob. 1 5 . 6 when the u nknowns to be determined are not the sou rce currents but, rather, the voltages at the three nodes labeled CD , W , and ® in Fig. 1 5 .8. Consider the circuit of Fig. 1 5.8 for which accu racy of the ammeters and t heir readings are the same as those specified in Prob. 1 5 .6. As in Prob. 1 5 .7, the voltages at the three nodes labeled CD , @, and ® arc to be estimated without first finding the source currents. ( a ) Suppose that meters A 4 and A s are found to be out of order, and therefore only three measu rements z , = 0. 1 2, z 2 = 1 . 18, and z) = 3.7 are avai l able. Determine the weighted least-squ ares estimates of the nodal voltages and the estimated errors e" e2 , and e3 ' ( b ) This time suppose meters A 2 and A 5 are out of order and the remaining three meters are working. Using three measurements z , = 0. 1 2, z ) = 3.7, and Z 4 = 0.8 1 , can the nodal voltages be estimated without finding the source currents first? Explain why by examining the matrix G . Suppose that the two voltage sources i n Exar.1ple 1 5 . 1 have been replaced with new ones, and the meter readings now show z , = 2.9 A, Z 2 = 1 0 . 2 A, z ) = 5. 1 V, and Z4 = 7. 2 V. ( a ) Determine t h e w e i g h t ed l e as t- s q u a res est imates of t h e new s o u rce vo l t ages. ( b ) Using the chi-squarc test for a = 0.005, detcct bad data. (c) Eliminate the bad data and determine again the weighted least-squares estimates of the source voltages. ( d) Check your result in part (c) again using the chi-square test. Five wattm�ters are installed on the four-bus system of Fig. 15.9 to measure line real power flows, where per-unit reactances of the l ines are X1 2 = 0.05, X13 = 0.1,

690

CHAPTER 15

STATE

ESTIMATION OF POW E R SYSTEMS

60 MW

CD

40 MW

........ .,.. """" ... ""' ®

.......

@ -.......

--.....,...-- @

-

70 MW

50 MW

FIGURE 15.9 Four-bus n e twork.

X23 = 0.04, X24 = 0.0625, and X34 = 0.08. Suppose that the meter readings show

that

Z l = P12 =

0 .34 per unit

Z2 = P 1 3 =

0 .26 per unit

ZJ

0 . 17

=

P2.1 =

Z 4 = P24 Z5

=

=

P34 =

per u n it

-

0 .24 p e r unit

-

0 .22 per unit

where the variances of t he measurement e rrors in per unit are given by (a)

(b)

Apply the dc power-flow method of Sec. to this network with bus CD as reference and determine the corresponding H m atrix. Then, compute the weighted least-squares estimates of t he phase angles of the bus voltages i n radians. Using the chi-square test for a = 0.0 1 , identify two bad measurements. Between the two bad measurements one is not worse than the other as far as

9.7

PROBLEMS

15.11.

691

accuracy is concerned. Explain why. If both bad measurements are eliminated simultaneously, would it be possible to estimate the states of the system? (c) For the two bad measurements identified in part (b), determine t he relation­ ship between the two error estimates in terms of the reactances of the corresponding two l ines. ( d ) Eliminate one of the bad measurements :dentified in part (b) and determine the weighted least-squares estimates of the phase angles of the bus voltages using the reduced data set. Do the sam e for the other bad measurement. By comparing the two resul ts, i dentify the buses at which the estimated phase angles are equal i n the two cases. In the four-bus system of Prob. suppose that the variance of the measure­ ment error for Z s is and that all the other data remai n the same. Qualitatively describe how the newly estimated values 24 and is of the measure­ ments will differ from those obtained in Prob. Verify your answer by recal c u la t i n g t h e w e i g h t e d least-squa res estimates of t h e ph ase angles ( in radians) of t h e bus vol tages a n d t h e correspon d i n g 2. Suppose that a l i n e of i m p e d a n c e per unit is added between buses CD and @ in th e ne twork of Fig. and lhat a wattmeter is installed on this line at bus CD . The variance of the measurement error for this added wattmeter is assu med to be t h e same as that of t he others. The meter readings now show

(0.OS)2 IS.10

IS.10.

1 5 . 1 2.

jO. 0 2S IS. 9 ,

Z1

Z2

=

P1 2

= P13

= =

Z3 = P23 = Z4

=

P24

=

Z 5 = P34 = z6 = P 1 4 =

0.32 per 0. 24 per 0.16 per unit -0.29 per unit -0.27 per unit O.OS per unit u nit

unit

Find the H matrix that describes the dc power flow with bus CD as reference and compute the weighted least-squares estimates of the phase angles o f the bus voltages in radians. (b) Using the chi-square test for a = el iminate any bad data and recompute the weighted least-squares estimates of the phase angles of the bus voltages. Check your result again using the chi-sq..l are test for a = In the four-bus system described i n Prob. suppose that the wattmeter o n line CD - 0 is out o f order and t h a t the readings o f the remaining five wattmeters are the same as those specified in P r o b ( a ) Apply t h e de power-flow analysis with bLS CD as reference and determin e the H matrix. Then, compute the weighted least-squares estimates of the phase · angles of the bus voltages in radians. (b) Using the chi-square test for a = identify two bad measurements. Eliminate one of them and compute the weighted least-squares estimates of the bus-voltage ph ase angles. Restore the elimin ated bad measurement and remove the second one before recompt:ting the estimates of the bus-voltage ,

(a)

0.01,

15.13.

IS.12 . IS.12. 0.0 1,

0.0 1.

692

15.14.

CHAPTER 15

STATE ESTIMATION OF POWER SYSTEMS

phase angles. Compare the two sets of results and identify the buses at which the estimated angles are equal in the two cases. Does the presence of line CD -@ (but with no line measurement) affect the identification of those b uses? Compare the identified buses with those identified in Prob. 1 5 . 1 0( d ). Three voltmeters and four wattmeters are installed on the three-bus system of Fig. 15. 10, where per-unit reactances of the l ines are X I 2 0. 1 , X13 0.08, and X21 = 0 05 . The per-unit values of the three voltm e t e r measurements a r e Z 1 I VI I = 1 .01, Z z 1.02, and z ) = I V) I 0.98. The read ings o f t h e two I V2 1 wattmeters measuring MW generation at buses CD a nd G) are Z4 0.48 per unit and 0.33 per unit, respectively. The measurement of the wattmeter on line CD - 3 at bus CD shows Z6 0 4 1 per unit and that of the wattmeter on line @ - 3 at bus G) is Z 7 0.38 per unit. The variances of t he measurement errors are given in p e r unit as .

z�=

(a)

=

=

=

= =

=

.

=

=

Use bus CD as reference to find expressions for the elements of t h e

W: Wattmeter Vm : Voltmeter

F1GURE 15.10

Three-bus network.

=

m a tr ix

PROBLEMS

H�k ) and those of the measurement errors e�k) in terms of state variables, as done in Example 1 5.5. (b) Using the initial value of 1.0 per unit for all bus voltages, find the values of the state variables that will be obtained at the end of the first iteration of the weighted least-squares state-estimation process. Application of the weighted least-squares state estimation to the three-bus system with all the measurements described in Prob. 1 5. 1 4 yields the fol lowing estimates of the states:

&

15.15.

6?3

I VI I

L

1 .0 1 09

per unit

I V2 1 = 1 .0 1 87

per unit

02

I V3 1 = 0 .9804

per unti

83

=

==-

0 .0 1 0 1 radians 0 .0308 radians

The sequence of di-a gonal elements in the covariance matrix R' is 0.8637

X 1 0 - 6, 3 3 0 . 1 882 X 1 0 - 5, 0.21 89 X 1 0 - 6, 0.7591 X 10 - , 0.8786 X 1 0 - , 0. 1 8 1 2 X l O - z , and 0 . 1 532 X 1 0 - 2 . Find the estimates of the measurement errors e i and the corre­

1 5 . 16.

15.17.

sponding standardized errors. Solve Prob. 1 5 . 1 4 when the two wattmeters installed on l ines CD - ® and @ - ® are replaced with two varmeters and their readings are 0.08 and 0.24 per u nit, respectively. Suppose that real and reactive power flows are measured at both ends of each of t he five lines in the four-bus system of Fig. 1 5.9 using ten wattmeters and ten varmeters. The voltage magnitude is measured at bus @ only, and bus inj ected powers are not measured at all. ( a ) Determine the structure of H x by writing the partial derivative form o f i ts nonzero elements, as shown in Example 15.8. Assume that line-flow measure­ ments are ordered in the following sequence: CD-G), CD - G) , @ - G), @ - 0 , and G) - @ (and the same sequence also in reverse directions). ( b ) Suppose that the clements of the Y bus of the network are given by YIf

(c)

15. 1 8.

= G- - + }B = I Y · I If

.

If

If

I e. . L3

and that the total charging susceptance of line CD (j) is B;j' Write out nonlinear fu nctions which express the measured quantities PZI and QZI in terms o f s t a t e

vari ables.

-

In terms of state variables write out the expressions, similar to those given in Example 15.8, for the nonzero elements in the rows of the m atrix Hx corresponding to measurements P2 1 and Q 2 1 ' The method o f Example 1 5 . 8 based on measurements of only line flows (plus a voltage measurement at one bus) is applied to the three-bus system of Fig. 15.10 using three wattmeters and three varmeters. The per-unit values of the '

694

CHAPTER 15

STATE ESTIMATION OF POWER SYSTEMS

measurements are

Z I = I VI I = 1 .0 Z2

Z3 Z4

P 12 = 0 .097

== = =

P 1 3 = 0 .383

P23

Z5 = z6

Z7

Ql 2

== == = Q13

Q23

- 0 .101

0.048 0.276·

0 .427

where the varia nces of a l l the measu r e m e n ts are (0.02)2. The pe r - u n i t r e a c t a n ces of t h e l i nes a re as specified in P rob. 1 5 . 1 4. Using b u s CD as r e fe re n ce and fiat-start va l ues, find t h e val ues o f t he s ta t e variables t h a t w i l l he obtained a t t h e e n d o f t h e fir s t i t e r a t i o n o f t h e we i g h t e d l e a s t -sq u a res s t a t e est i m a t io n .

CHAPTER

16

POWER SYSTEM STABILITY

When ac generators were driven by reciprocating steam engines, one of the major problems in the operation of machinery was caused by sustained oscilla­ tions in speed or hunting due to the periodic variations in the torque applied to' the generators. The resulting periodic variations in voltage and frequency were transmitted to the motors connected to the system. Oscillations of the motors caused by the variations in voltage and frequency sometimes caused the motors to Jose synchronism entirely if their natural frequency of oscillation coincided with the frequency of oscillation caused by the engines driving the generators. Damper windings were first used to minimize hunting by the damping action of the losses resulting from the currents induced in the damper windings by any relative motion between the rotor and the rotating field set up by the armature current. The use of turbines has reduced the problem of hunting although it is still present where the prime mover is a diesel engine. Maintaining synchronism between the various parts of a power system becomes increasingly difficult, grow.

h o w ever,

16.1

as

t h e sys tems a n d i n te rco n n ec t i o n s between systems co n t i n u e to

THE STABILITY PROBLEM

Stability studies which evaluate the impact of disturbances on the electrome­ chanical dynamic behavior of the power system are of two types-transient and steady state . Transient stability studies are very commonly undertqken by elec­ tric utility planning departments responsible for ensuring proper dynamic per695

696

CHAPTER

16

POWER SYSTEM STABILITY

formance of the system. The system models used in such studies are extensive because present-day power systems are vast, heavi ly interconnected systems with h undreds of machines which can interact through the medium of their extra-high-voltage and ultra-high-voltage networks. These machines have associ­ ated excitation systems and turbine-governing control systems which in some but not all cases are modeled in order to reflect properly correct dynamic perfor­ mance of the system. If the resultant nonlinear differential and algebraic equations of the overall system are to be solved, then either direct methods or iterative step-by-step procedures must be used. In this chapter we emphasize transient stability considerations and introduce basic iterative procedures used in transient stability studies. Before doing so, however, let us first discuss certain terms commonly encountered in stability analysis. l A power system is in a steady-state operating condition if all the measured (or calculated) physical quantities describing the operating condition of the system can be considered constant for purposes of analysis. When operating in a steady-state condition if a sudden change or sequence of changes occurs in one or more of the parameters of the system, or in one or more of its operating quantities, we say that the system has undergone a disturbance from its steady-state operating condition. Disturbances can be large or- small depending on their origin. A large disturbance is one for which the nonlinear equations describing the dynamics of the power system cannot be validly linearized for purp oses of analysis. Transmission system faults, sudden load changes, loss of generating units, and line switching are examples of large disturbances. If the p ower system is operating in a steady-state condition and i t undergoes change which can be properly analyzed by linearized versions of its dynamic and algebraic equations, we say that a small disturbance h as occurred. A change in the gain of the automatic voltage regula tor in the excitation system of a large generating unit could be an example of a small disturbance. The power system is steady-state stable for a particular steady-state operating condition if, follow­ ing a small disturbance, it returns to essentially the same steady-state condition of operation. However, if following a l a rge d isturbance, a s i g n i ft c a n t l y d i ffe re n t but acceptable steady-state operating condition is attained, we say that t h e system i s transiently stable. Steady-state stability studies are usually less extensive in scope than transient stability studies and often involve a single machine operating into an infinite bus or just a few machines undergoing one or more small disturbances. Thus, steady-state stability studies examine the stability of the system under small incremental variations in parameters or operating conditions about a steady-state equilibrium point. The nonlinear differential and algebraic equa­ tions of the system are replaced by a set of linear equations which are then I For further discussion, see " Proposed Terms a n d D efi nitions for Power System Stability," A Task Force Report of the System Dynamic Performance Subcom mittee, IEEE Transactions on Power Apparatus and Systems, vol. PAS 1 0 1 , July 1 982, pp. 1 894- 1 898. •

I

16.1 THE STABILITY PROBLEM

697

solved by methods of linear analysis to determine if the system is steady-state stable. S ince transient stability studies involve l arge disturbances, linearization of the system equations is not permitted. Transient stability is sometimes studied on a first-swing rather than a multiswing basis. First-swing transient stability studies use a reasonably simple generator model consisting of the transient internal voltage E; behind transient reactance X�; in such studies the excitation systems and turbine-governing control systems of the generating units are not represented. Usually, the time period under study is the first second following a sys tem fault or other large disturbance. If the machines of the system are found to remain essentially in synchronism within the first second, the system is regarded as being transiently stable. Multiswing stability studies extend over a longer study period, and therefore the effects of the generating units ' control systems must be considered since they can affect the dynamic performa nce of the units during the extended period . Machine models of greater sophistication are then needed to properly reflect the behavior of t he system. Thus, excitation systems and turbine-governing cont rol systems may 01' may not be represented in steady-state and transient stability studies d epending on the objectives. In all stability studies the obj ective is to determine whet her or not the rotors of the machines being perturbed return to constant speed operation. Obviously, this means that the rotor speeds have departed at l east temporarily from synchronous speed. To facilitate computation, three funda­ mental assumptions therefore are made in all stability studies: 1.

Only synchronous frequency cu rrents and vol tages are considered in the stator windings and the power system. Consequently, dc offset currents and harmonic components are n eglected. 2. Symmetrical components are used in the representation of unbal anced faults. 3. Generated voltage is considered unaffected by machine speed variations. These assumpt ions permit the use of phasor a lgebra for the transmISS I on network and solution by power-flow techniques using 60-Hz parameters. Also, negative- and zero-sequence networks can be incorporated into the positive­ sequence network at the faul t point. As we shall see, three-phase balanced fau lts are generally considered. However, in some special studies circuit-breaker clearing operation may be such that consideration of unbal anced conditions is unavoidable. 2

2 For information beyond the scope of t his book, see P. M. Anderson and A. A. Fouad, System Control and Sta bility, The Iowa S t a t e University Press, Ames, lA, 1 977.

Power

698

CHAPTER 16

POWER SYSTEM STABI LITY

16.2 ROTOR D YNAMICS AND THE SWING EQUATION

The equation governing rotor motion of a synchronous machine is based on the elementary p rinciple in dynamics which states that accelerating torque is the product of t h e moment of inertia of the rotor times its angular acceleration. In the MKS (meter-kilogram-second) system of units this equation can be written for the synchronous gene rator in the form ( 16.1) w here the symbols have the following mea n i n gs : J

Om

t Tm Te

Ta

the total moment of inertia of t h e rotor m asses, i n kg-m 2 the angular d isplacement of the rotor w i th respect to a stationary axis, i n mechanical rad ians (rad) time, i n seconds (s) the mechanical or shaft torqu e supplied by the prime mover less retarding torque due to rotational losses, i n N-m the net electrical or electromagnetic torque, in N-m the net accelerating torque, in N-m

The mechanical torque Tm and the electrical torque Te are considered positive for the synchronous generator. This means that Tm is the resultant shaft torque - which tends to accelerate the rotor in the positive 8m direction of rotation, as shown in Fig. 1 6.l(a). Under steady-state operation of the generator Till and Te are equal and the accelerating torque Ta is zero. In this case there is n o acceleration or deceleration of the rotor masses and t h e resultant constant speed is the synchronous speed. The rotating m asses, which include the rotor of the generator and the prime mover, are said to be ill synchronism with the other machines operating at synchronous speed in the power sys tem. The prime

-

,

T",

(a) FIGURE 16.1

(a)

(b)

(6)

Representation of a machi n e rotor comparing d i rec t i o n o f rota tion a n d mechanical a n d elec trical torques for:

a generator;

a motor.

16.2

ROTOR DYNAMICS AND THE SWING EQUATIO N

699

mover may be a hydroturbine or a steam turbine, for which models of different levels of complexity exist to represent their effect on Tm ' In this text Tm i s considered constant at any given operating cond ition. This assumption i s a fair one for generators even though input from the prime mover is controlled by governors. Governors do not act u ntil after a change in speed is sensed, and so they are not considered effective d uring the time period i n which rotor dyna mics are of interest in our stability studies here. The electrical torque Te corresponds to the net air-gap power in the m achine, and thus accounts for the total output power of the generator p lu s I J I 2 R losses in the armatu re winding. 1n the synchronous motor the direction of power flow is opposite to that in the generator. Accordingly, for a motor both T,1I and Te in Eq. ( 1 6 . 1 ) are reversed in sign, as shown in Fig. 1 6 . 1 ( b). Te then corresponds to the air-gap power supplied by the electrical system to d rive the rotor, whereas Tm represents the counter torque of the load and rotational losses tending to retard the rotor. Since em is measured w i t h respect to a stationary reference axis on the stator, it is a n abso l u te measure o f rotor angle. Consequently, i t continuously increases with time even at constant synchronous speed. Since the rotor speed relative to synchronous speed is of in terest, it is more convenient to measure the

rotor angular position with respect to a reference axis which rotates at syn­ chronous s pe e d . Therefore, we define

( 1 6 .2) where Wsm is the synchronous speed of the machine in mechan ical radians per s eco n d and o m i s the angular displacement of the rotor, i n mechanical radians, from the synchronously rotating reference axis. The derivatives of Eq. (1 6.2) with respect to time are dem --

dt

a nd ( I (L l ) s h uws

= Wsm +

d 2 em

d 2 0m

dt 2

dt 2

d Om dt

( 1 6 .3) ( 1 6 .4 )

dOm/dl is constant and equals the synchronous speed only when dOl/jdl is zero. Therefore, dom /dt eq u a t i o n

t l1 < l t t h e ro t o r a l l g u l a r v e l oc i ty

re p rese n t s t h e d e v i a t i o n or t h e rotor speed from synchronism and t h e u n i t s o f

measure are mechan ica l radians per second. Equation ( 1 6.4) represents the rotor acce l e r a t i o n measured in mechanical radians per second-squa red. Substituting Eq. ( 16.4) i n Eq. ( 16 . 1 ), we obtain

( 1 6 .5)

700

CHAPTER 16

POWER SYSTEM STABI LITY

It is convenient for notational purposes to introduce

Wm =

( 1 6 .6)

for the angular velocity of the rotor. We recal l from elementary dynamics that power equals torque times angular velocity , and so multiplying Eq. (16.5) by W m ' w e obtain

( 16 .7) where Pm Pe Pa

=

=

=

shaft power input to the machine less rotationa l loss�s electrical power crossing i ts a i r gap accelerating power which accounts for any unbalance between those two quantities

Usually, we neglect rotational losses and armature I I I R losses and think of Pm as power suppl ied by the prime mover and Pe as the electrical power output. The coefficient JWm is the angular momentum of the rotor; at synchronous speed W sm it is denoted by M and cal l ed the inertia constant of the machine. Obviously, the units in which M is expressed must correspond to those of J and w rn • A careful check of the un its in each term of Eq. (1 6.7) shows that M is expressed in joule-seconds per mechan ical rad ian, and we write 2

( 16 .8) While we have used M in this equation , the coefficient is not a constant in the strictest sense because Wm does not equal synchronous speed under all condi­ tions of operation . However, in practice, Wm does not d iffer si.gnificantly from synchronous speed when the machine is stable, and since power is more convenient in calculations than torque, Eq. ( 1 6 .8) is preferred . In machine data supplied for stability studies another constant related to inertia is often encoun­ tered. This is the so-called H constant , which is defined by H=

and

stored k inetic energy in m egajoules at synchronous speed machine rating in MY A

H=

1J 2 2'

Wsm Smach

---

� M wsm Smach

--

MJ/MVA

( 1 6 .9)

1 6.2

ROTOR D YNAMICS AND THE SWING EOUATION

701

where Sm a ch is the three-ph ase r ating of the machine i n megavoltamperes. Solving for M in Eq. (16.9), we obtain

M= and substituting f9r

M

2H --

W sm

S m ach MJ / mech rad

( 16 .1 0 )

in Eq. (16.8), we find Pa

S m ach

( 16 . 1 1 )

This equation leads to a very simple resul t . Note that Dm is expressed in mechan ical radians in the numera tor of Eq. (16. 1 1 ) , whereas W s m is exp ressed in mechanical radians per second in the denom i n ator. Therefore, we can write the equation in the form

( 1 6 . 1 2) provided both D and Ws have consistent u n its, which m ay be mechanical o r electrical degrees o r radi ans. H a n d t have consistent units since megaj ou les per megavoltampere is i n u n i ts of time in seconds and Pa , Pm ' and Pe m us t be in per unit on the same base as H . When the subscript m is associated w i t h w , ws ' and D , it means mechanical u ni ts are being used ; otherwise, electrical u ni ts are implied. Accordingly, Ws is t h e synchronous speed in electrical u ni ts . For a system with an electrical frequency o f f hertz Eq. 06. 12) becomes H d 28

7T

f dt 2 = Pa

-

Pm

Pe per u n i t

( 16.13)

when D i s in electrical radians, w h i l e

( 1 6 .14) appl ies when D is in electrical de grees. Equation 06.12), called the swing equation of the machine, is the funda­ mental equation which governs t h e rotational dynamics of the synchronous machine in stability studies. We n ote t hat it is a second-order differential

702

CHAPTER 16

POWER SYSTEM STABILITY

equation, w hich can be written as the two first-order differenti al equations 2H d w -

Ws

-

dt

d8 dl

=

pm

-

P per unit e

= w - ws

( 1 6 . 15) ( 16 . 1 6)

in w hich w, W S ' and 8 involve electrica l rad i a ns or electrical degrees. We use the various equivalent fo rms of the swi ng equat ion t h roughout this ch apter to determi ne t he stability of a m achine within a p o w e r system. When the swing equat ion is solved, we obta in th e e x p rc s s i on fo r 8 as a fu nct i on of time . A graph of the solution is called the swing curve of the machine and inspection o f the swing curves of all the m achines of the system wil l show whether the machines remain in synchronism after a disturbance. 16.3 FURTHER CONSIDERATIONS OF THE SWING EQUATION

The megavoltampere (MY A) base used i n Eq . ( 1 6 . 1 1 ) is the machine rating Smach which is introduced by the definition o f H. In a stabili ty study of a power system with many synchronous machines only one MYA base common to all parts of the system can be chosen. Since the right-hand side of the swing equation for each machine must b e expressed in per u nit on this common system base, it is clear that H on t h e l eft-hand side of each swing equation must also be consistent with the system base. This is accomplished by converting H for e ach machine based on its own individual rating to a value determined by the system base, Ssystem ' Equation ( 1 6 . 1 1 ), m U ltiplied on each side by the ratio (Smach/Ssystc m ), leads to the conversion formula . S mach Hsy�,cm = Hmach S system

( 1 6 . 17)

i n w hich the subscript for each term indicates the corresponding base being used. In industry studies the system base that is usually chosen is 1 00 M YA. The inertia constant M is rarely used in practice and the forms of the swing equation involving H are more often encoun tered. This i s because the value o f M varies widely with the s iz e and type of the machine, whereas H assumes a much narrower range o f values, as shown in Table 16. 1 . Machine manufacturers also use the symbol WR 2 to specify for the rotating parts of a generating unit (i ncluding the prime mover) the weight in pounds multiplied by the square of the radius of gyration in feet. Hence, WR 2 /32.2 is the moment of ' inertia of the machine in slug-feet squared.

1 6.3

FURTHER CONSIDERATIONS OF THE SWI NG EQUATION

703

TABLE 16.1

Typical inertia constants of synchronous machinest

MJ / MVA

Inertia constant,

Type of machine

Tu rb i n e generator: Con densing, 1 800 r /min 3600 r /min N o n con d e n s i n g , 3600 r /min

Ht

9-6 7-4 4-3

Wa t e rwheel ge n erator: Slow-speed, <

200 r /min

2 -3 2 -4

r

H i g h -spe e d , > 200 / m i n

Synch ronous con d e nser;§ Large

1 .25 l .00 2 . 00

Small Synch ron o u s motor w i t h load v a r i e s from 1 .0 to 5 .0 u n d

h i g h e r for h e avy flywhee l s

t R c n r i n t c d oy p e r m i s s i o n of t h e A B B Pow e r T & D Co m p a -I Y , I n c.

f ro m Electrical Transmission and Distrifm t iofl Reference Book .

:j: W h e r e r a n ge is g iv e n , t h e fi r s t fig u re a p p l I e s to m a c h i nes of s m a l l e r m egavol t a mpere rat i ng. § H y d roge n -coo l e d , 25% l ess.

Develop a formula to calculate the H constant from U'R 2 and then evaluate H for a nuclear generating unit rated at 1 333 MVA, 1 800 r /min with WR 2 5,820,000 Ib-ft 2 .

Example 1 6 . 1 .

=

Solution.

The kinetic energy (KE) of rotation in foot-pounds at synchronous speed KE

=

2 [ 2 7T ( r /min) ] 2 -21 WR ft-lb 32.2 60 --

Since 550 ft-Ib/s equals 746 W, i t fol lows that 1 ft-lb equals 746/ 550 1. Hence, converting foot-pounds to megajou lcs and d ividing by the machine rating i n megavoltampcres, we obtain H=

(

746 550

x

10-

('

) � WR2 [21T(r/min) ]2 2 3 2 .2

S mach

60

which yields upon simplification H =

2 .3 1

X

1O - IO WR 2 ( r/min) 2

--

--

Smach

704

CHA PTER ·1 6

POWER SYSTEM STABILITY

Inserting the given machine data in t his equation, we obt ain

6 2 1° 2.3 1 10- (5.82 10 )(1800) 3. 27 MJ/MYA 1333 to a 100-MYA system base, we obtain 3.27 1333 100 43.56 MJ/MYA

H =

Converting

H

X

H

X

X

=

--

=

=

In a stabi l i ty study for a large system with m a ny m a ch i n e s ge o g r a p h i c a l l y d ispersed over a w i d e a rea i t is des i ra b l e to m i n i m i ze t h e n u m be r o f s w i n g equations t o b e so l ve d . T h is c a n b e done i f t h e t ra ns m i ss io n - l i n e fa u l t , o r o t h e r disturbance on the system, affects the machines within a power plant so that their rotors swing together. I n such cases the machines within the plant can be combi ned into a single equivalent machine just as if their rotors were mechani­ cally coupled and only one swing equ ation must be written for them. Consider a power plant with two generators con nected to the same bus which is electrically remote from the network disturbances. The swing equ ations on the common system base are

( 1 6 . 1 8)

( 1 6 . 1 9) Add ing the equ ations together and denoting 0 1 and 0 2 by swing together, we obtain -- -2 w dt s

=

(j

smce the rot ors

P,n - Pe per u n i t

( 16 .20)

w here H = ( H I + H2 ), Pm = ( Pm l + Pm 2 ), and Pe = ( Pe l + Pe 2 ). This single equation, which is in the form of Eq. ( 1 6. 1 2 ), can be solved to represent the p lant dynamics.

Two genera ting u nits operate i n parallel within the same power plant and h ave the following ratings: Example 16.2.

U n it

60-Hz

1 : 500 MY A, 0.85 power factor, 20 kY, 3600 r/min 4. 8 MJ/MYA HI

=

16.3 FURTHER CONSIDERATIONS OF THE SWING EOUATION

Unit 2 :

=

H2

Calculate the equivalent Solution.

power factor, 22 kV ,

1 333 MVA , 0 . 9

H

705

1800 r/min

3 .27 MJ/MVA

constant for the two units on a

100-MVA

base.

The total kinetic energy (KE) of rotation of the two mach ines is KE � (4 .8

Therefore, the

H

5 00 )

x

+

(3 .27

x

1333)

=

6759 MJ

constant for the equivalent machine on l OO-MVA base is H

=

67.59 MJ/MVA

and this value can be used in a single swing equation, provided the machines swing together so that their rotor angles are in step at each instant of time. Machines which swing together are called coherent machines. It is noted t ha t w h e n both Ws a n d 0 are expressed i n electrical degrees or radi ans, the swing equations for coherent machines can be combined together even though, as in the example, the rated speeds are d ifferent. This fact is often used i n stability studies involving many m achines in order to reduce the number of swin g equations which need t o b e solved . For any pair of noncoherent m achines in a system swing equations similar to Eqs. ( 16. 1 8) and ( 1 6. 1 9) can be written . Dividing each equation by its left-hand -side coefficient and subtracting the resultan t equations, we obtain ( 1 6 .21 )

Mul tiplying each side by H 1 H2/( H )

� WS

(

1

d2 ( 0 ] - O2 ) H] H 2 H I + H2 dt 2

=

+

H2 ) and rearranging, we find t h at

Pm l H2 - Pm 2 H1 HI + H2

_

Pe ] H 2 - Pe 2 H 1 H I + H2

( 1 6 .22)

which also may be written more Simply in the form of the basic swing equation, Eq. ( 1 6. 1 2), as fol lows: ( 1 6 .23)

Here the relat ive angle 0 ] 2 equ als 0 ] - 82 , and an equivalent inertia and

706

CHAPTER 16

POWER SYSTEM STAI3ILlTY

weighted input and output powers arc defined by H1 2

=

H I H2

HI

+

H2

---­

( 16.2 4) ( 1 6 .25) ( 1 6 .26)

A noteworthy application of these equations conce rns

a two- mach ine syste m h aving only one genera tor (machine one) and a synchronous motor (machi ne two) connected b y a network of pure re actances. Wh atever change occurs i n the generator output is thus absorbed by the motor, and we can write

( 16 .27) Under these cond itions, Pm l 2

=

Pm ' Pe l 2

=

Pe , and Eq.

(1 6.22) reduces to

which is also the format of Eq. (16. 12) for a single mach ine. Equation (1 6.22) demonstra tes that stability of a machine within a system is a relative p roperty associated with its dynamic behavior with respect to the other machines of the syst c m . T h e rotor angle of onc mach i ne, say, 6 \ , c a n be ch ose n for co m par iso n w i t h l h � ro t o r ; \ \ l g l c o r e a c h o ( h � r m ;\ c h i n c , s y m bo l i z e d by 6 2 , In order to be stable the a n g u l a r d i f fe re n ces be twee n a l l mach ines must d ecrease after the fina l switching ope ration-such as t h e open ing of a c i r c u i t b reaker to clear a fault. Al though we may choose to plot the angle between a m achine's rotor and a synchronously rot ating reference axis, i t is the relative a n gl es between machines which are important. Our discussion above empha­ sizes t h e r e l a tive n a t u re of t h e sys t e m s t a b i l i ty p r o perty a n d shows that the essential fea tures of stability study are revealed by consideration of two­ m achine problems. Such prob lems are of two types: those having one machine of fin ite i nertia swinging with respect to an infin ite bus and those having two fini te inerti a machines swinging with respect to each other. An infinite bus m ay be considered for stability purposes as a bus a t which there is located a machine of constant internal voltage, h aving zero impedance and infinite inertia. The point of connection of a generator to a large power system may be regarded as , such a bus. I n a l l cases the swing equation assumes the form of Eq. (16. 1 2), each

16.4

THE POWER-ANGLE EQUATION

707

term of which must be exp licit ly d escribed b efore it can be solved. The e qu ation for Pe is essential to this description, and we now p roceed to its characterization for a general two-machine syste m. 1 6.4

THE POWER-ANGLE EQUATION

In the swing equation for the generator the input mechanical power from the prime mover, Pm ' will be considered constant. As we have mentioned previ­ ously, this is a reasonable assumption because con d i tions in the electrical network can be expected to change before the control governor can cause the turbine to react . Since Pm in Eq. (1 6. 1 2) is constant, the electrical power output Pe will determine whether the ro tor accele rates, decelerates, or remains at synchronous speed. When Pe equals Pm ' the machine operates at steady­ state synchronous speed; when PI' cha nges from t h is val ue, t he rotor deviates from synchronous speed . Changes in Pe a re determined by condi tions on the transmission and distribu tion networks and the loads on the system to w hich the genera tor supplies power. Electrical network d isturbances resulting from severe load changes, network fau l ts, or circu i t-breaker operations may cause the generator output Pe to change rapidly, in which case e lectromechanical tran­ sients exist. Our fundamental assump tion is that the effect of machine speed variations upon the generated voltage is negligible so that the manner in w hich Pe changes is determined by the powe r-flow equations applicab le to the state of the electrical network and by the model chosen to represen t the elect rical behavior of the machine. Each synchronous machine is represented for tran­ sient stability studies by its transient internal voltage E; in series with the transient reactance X� , as shown i n Fig. 1 6.2( a ) in which � is the terminal vol tage. This corresponds to the steady-state representation in which syn-

I

I

I

v,

(0)

(6)

.... ... ... ../

- - - -

Reference

FIGURE 1 6.2 P h a sor d i agram of a synchronous mach i n e for transient s t a b i l ity s t u d i es .

708

CHAPTER 16

POWER SYSTEM STABI LITY

FIG U RE 16.3

Schematic d iagram for stab i l i ty stu d ies. Tra n s i e n t reacta nces associated w ith £; a n d £2 a r e incl u de d i n the t ra n sm ission network.

Tra nsmission network

chronous reactance Xd is in series w i t h the synchronous i n ternal or no-load voltage Ej• Armature resistance is n egligible in m ost cases so that the phasor diagram of Fig. 16.2( b ) applies. Since each m achine must be considered relative to the system of which it is part, the p h asor angles of the m ac hine quantities a re m easured with respect to the common system reference. Figure 16.3 schematica l ly represe n ts a generato r s u p plying power t hrough a transmission system to a receiving-end system a t bus CD . The recta ngle represents the transmission system of l i nea r passive compone nts, s u c h a s tra ns­ formers, transmission lines, and capacitors, and incl udes the t ransient reactance of the generator. Therefore, the vol tage E� represents the tra nsient in te r n al voltage o f the generator at bus CD . The voltage E; at the receiving end is regarded here as that of an i nfinite bus or as the transient internal voltage of a synchronous motor whose transient reactance is inc l uded i n the n etwork. Later we shall consider the case of two generators supplying constant-impedance loads within the network. The bus admittance matrix for the network reduced to two nodes in addition to the reference n o d e is

( 1 6 .28)

From Eq. (9.4) we have Pk

+ jQk

=

N

Vk L ( Y rr V:J * k rr = 1

Letting k and N equal 1 and 2, respectively, and substituting obtain

( 1 6 . 29 ) £'

for V, we ( 1 6 . 3 0)

If w e define

16.4 THE POWER·ANGLE EQU ATION

709

Eq. ( 1 6.30) yields

( 1 6 .3 1 ) ( 1 6 . 3 2)

Similar equa tions apply at bus CD by substituting subscripts 1 for 2 and 2 for 1 i n the two p receding equations. If we let

and define a new angle y such t hat

we obtain from Eqs. (16.3 1 ) a n d ( 1 6.32)

( 1 6 . 33) ( 1 6 .3 4 ) Equation (1 6.33) may be written more simply as

( 1 6 .35) where

( 1 6 .36)

S ince P I represents the electric power output of the generator (armature loss neglected), we have replaced it by Pe in Eq. (16.35), which is often called the power-a ngle equa tion ; its graph as a function of 8 is called the power-angle curv e . The p ar a m e t e r s PC ' Pm ax , and y a rc c o ns tan t s fo r a given netwo rk configuration and constant voltage magnitudes I E � I and I E; I . When the net­ work is considered without resistance, all the elements of Y bus are susceptances, and then G J J and y are both zero. The power-angle equation which then applies for the pure reactance network is simply the familiar equation ( 1 6 .37) where Pmax

=

I E� I I E� I /X and X is the transfer reactance between E; a n d E� .

Exa mple 1 6.3. The single-line diagram of Fig. 16.4 shows a generator connected through parallel transmission lines to a larg e metropolitan system considered as an infinit e bus. The machine is delivering 1 .0 per-u:1it power and both the terminal voltage and the i nfinite-bus voltage are 1 .0 per unit. Numbers on the ., diagram

710

CHAPTER 1 6 POWER SYSTEM STABILITY

j O. 4

)0 . 1

Xd = �

F I G U RE 1 6.4

O n e- l i ne d iagra m for Exam p l es 16.3 and 1 6. 4 . P o i n t P is at the c e n ter o f t h e line.

p

j O. 4

indicate the val ues of the reactances on a common system base. The t ransient reactance of the generator i s 0 .20 per unit as i ndicated. Determine the power-angle equation for the g iven sy s t e m ope rat ing conditions. The reactance d iagram for the system is shown in Fig. reacta nce between t he term inal vol tage and the i n finite bus is

Solution.

X

0. 10

=

+

0.4 2

=

I Vc l l VI X

L

- j3 . 33 3

- j3 . 3 3 3

(a)

Q)

- j 5.0 (b)

�

1 . 05& "

FIGURE 16.5

( 1 . 0 ) ( l .0 ) 0.3

SJr1 Ci =

determ ined by

1 .0

l . 0/0°

jO . 40

VI

(--�

=

is

jO . 40

I

E�,

Sin a

The series

0.3 per unit

and therefore the 1 .0 per-unit power output of the generator ---

1 6 .5( a ).

- j2 . 5 P

1 . 0LQ:

-

- j2 . 5

- j5. 0

- j 5.0 (e)

- j5 . 0

(?)

1 . 0LQ:

-=

_ Reactance d i ag r a m for: (a) the prefau l t n etwork for Exa m p l e 1 6.3 w i th i m pedan ces in p,er u n i t ; (b) and ( c); the fau l ted n etwork for Exa m p l e 1 6.4 with t h e s a m e i mpedan ces con verted t o a d m i t t ances a nd m arked i n pe r u n it.

i6.4 THE POWER-ANGLE EOUA TION

711

where V is the voltage of the infinite bus a n d a i s the angle of t he terminal voltage relative to the infinite bus. Solving for a, we obtain a

=

sin - 1 0 . 3

17.458°

=

so that the terminal voltage is �

=

1 .0/ 1 7 .4580

0 . 954

=

l .O� l .O�

+

jO .300 per unit

The output current from the generator is now calculated as I

=

=

= =

l .0

+

------'==--

jO.3

jO . 1 535

1 .0 1 2

=

/ 8 .729°

per unit

and the t ransient internal vol tage is t hen found to be E;

=

=

( 0 .954 + iO .30) 0 . 923 - jO .5

=

j( 0 .2) ( 1 .0

+

+

jO . 1 535)

1 .050/ 28 .440 per unit

The power-angle equation relating the transie�l t internal voltage E; and t he infini te-bus voltage V is determ ined by the total series reactance X

=

0 .2

+

0.1

+

0 .4

- =

2

0.5 per unit

Hence, the desired equation is Pe

where

0 is

=

( 1 .050 ) ( 1 .0) sin 0 0 .5

=

2 . 1 0 sin 0 per unit

the mach ine rotor angle with respect to the infinite bus.

The powe r-angle equation of the p receding example is plotted in Fig. 1 6. 6 . Note that the mechanical input power Pili is constant and intersects the p 2. 1

l .0

o

- - - - - - - - - - -� -_.__

' Pe

.....- P,

=

2 . 1 0 si n o

1 . 50 s i n O

��------�--�'��'- - P m =

- Pe

=

�------ - �------��

2 8 . 44°

90"

1 5 1 . 56°

0 .808 sin b

1 80"

FIGURE 16.6 P l ot of powe r-angle curv.es found in Examples 1 6 . 3 to 1 6. 5 .

712

CHAPTER 16 POWER SYSTEM STABI LITY

sinusoidal power-angle curve at t h e operating angle 00 24.44°. This i s t h e initial angular position of t h e generator rotor co r r espond i n g to the given operating conditions. The swing e quation for t h e m achine may be writt e n =

-- --

H

d20

1 .0 - 2 . 1 0 sin 0 per u n i t

180! df 2

( 1 6 . 38 )

where H i s i n megajouIcs per megavoltampere, f i s the electrical fre q u e n cy o f t h e system, and 0 i s i n electrical degrees. We can e asily check t he results o f Example 16 . 3 since under the given opera ting co n d itio ns , Pe = 2 . l 0 s i n 28. 44 ° = 1 .0 per u n it , which corresponds exact ly to the mecha nical p ow e r i n p u t P'" a nd

zcro. I n the next example wc determ inc t he powe r-angle cqU<.l l IQI1 fo r t h e s a m e system with a t h r ee - p h a s e fault at P, the m idpoint of one of the transmission l ines. Positive acceleration is sh own to exist due to the fault. t h e acce l e r a t i o n is

The system of Example 1 6.3 is operating u n d e r t h e i ndicated conditions when a three-phase fault occurs at point P of Fig. 16.4. Determ ine the power-angle equation for the system with the fau l t 0 n and the corresponding swing equation. Take H = 5 MJ / MV A.

Example 1 6.4.

The reactance diagram is shown in Fig. 16.5 ( b ) with t h e f a u l t on the system at point P. Values shown are admittances in per unit. The effect of the short circuit caused by the fault is clearly shown by redrawing the reactance d iagram, as in Fig. 1 6 . S(c). As calculated in Example 1 6 . 3 : transient internal based on the assumption of voltage of the generator remains at E; = 1 .05 constant flux linkage in the machine. The net transfer admittance connecting the voltage sources remains to be determined. The buses are n u mbe r ed as shown and the Y hus is formed by inspection of Fig. 1 6. S ( c ) as fol lows:

Solution.

/28.440

Y b us

=

CD

@

®

[

CD - j 3 .333 0 .000 j 3 .333

CD 0.00 - j 7 . 50 j 2.50

G) j3 .333 j2.500 j - 1 0.833

j

B us W has no external source con nection, and it may be removed by the node elimination procedure of Sec. 7.4 to yield the reduced bus admittance matrix CD CD [ - j 2.308

@

jO .769

CD

j O .769 -j6. 923

]

16.4 THE POWER·ANGLE EOUATI O N

713

The magnitude 9f the transfer admittance is 0.769, and thus

The power-angle equation with the fault on the system is therefore Pe

=

0 . 808 sin (j per unit

and the corresponding swing equation is 5 d 2 (j 1 80! dt 2

--

--

1 .0 - 0.808 sin 0 per unit

( 1 6 .39)

Due to its inertia, the rotor cannot change position instantly upon occurrence of the fault. Therefore, the rotor angle 0 is i nitially 28 .44°, t h e same as in Examp l e 1 6. 3 , and the electrical power output i s Pe O.80R sin 28.440 0.385. The i nitial accelerating power is =

Pa

=

1 .0 - 0.385

=

=

0 .61 5 per unit

and the initial acceleration is positive with the value given by 2 = dt

d 2 (j

-

1 801 ( 0 . 6 1 5 ) = 22. 14/ elec deg/s2 5

--

where 1 is the system frequency. Relaying schemes sensing the fault on the line will act to clear t he fau l t by simultaneous opening of the line-end breakers. When this occurs, another power­ angle equation applies because of the network change. The fault on the system of Example 1 6.4 is cleared by simul taneous opening of the circuit breakers at each end of the affected line. Determine the power-angle equation and the swing equation for the postfault period .

Example 1 6.5.

Inspection of Fig. 1 6.5(a) shows that u pon removal of the fau lted line, net transfer admittance across the system is

Solutio n .

the

1 0.1

-------

j(0 .2

+

+

0 .4)

=

-j 1 .429 per unit

so that in the bus admittance matrix Y1 2

=

j 1 .429

Therefore, the postfault powe r-angle equation is Pe

=

( 1 .05) ( 1 .0) ( 1 .429)sin o

=

1 .500 sin 0

714

CHAPTER 1 6 POWER SYSTEM STABILITY

and the correspond ing swing equat ion is 5 d28 -1 80! dt 2

--

The

=

1 .0

-

1 .500 sin 8

a ccele r a t i on at the i nstant of clea ring the fau l t

position of the rotor at that t i m e . The through 1 6. 5 arc compared i n Fig. 1 6.6.

power- a n g l e

a n gu l a r cu rves for Examples 1 6.3 d e p e n d s o n t he

SYNCHRONIZING POWER COEFFI CIENTS

16.5

In Example 1 6.3 t h e o p e r a t i n g po i n t on t h e s i n u so i d a l Pl' c u rv e of F i g . 1 6 . 6 is fou nd to be at Do = 20.44°, when� t h e m e c h a ni c a l pow e r i n p u t J�l l e q u a l s t he e lectrical power output Pc ' In the same figure it is also seen that Pc equals Pill at D = 15 1.56°, and t his might appear to be an eq ually acceptable operating point. However, this is not the case a s now shown. A commonsense requiremen t for an acceptable opera t i ng po i n t i s that the generator should not lose synchronism w h e n smal l tempora ry ch anges occu r i n t h e electrical power output from t h e mach ine. To examine t h i s re quirement for fixed mechanical input power P"" cons ider small incremental changes in the operating point parameters; that is, consider

( 1 6 .40) where the subscript zero denotes the steady-state operating po i nt val ues and the subscript delta ( � ) identi fies the incremental va ri ations from those va l ues. S ubst ituting Eqs. ( 1 6 .40) in Eq . ( 1 6.37 ), we obtain the power-angle eq uation for the gener al two-mach i ne system in the form

S ince Dt:. is a small incremental displacement from Do , and

( 16 .41 )

a n d when strict equali ty is assumed, the previous equation becomes

( 1 6 .42) At the initial operating point Do

( 1 6 .43)

16.5

SYNCH RON IZING POWER COEFFICIENTS

715

and it then follows from Eq. (16.42) that

( 1 6 .44 ) Substituting the incremental variables of Eq. ( 16.40) in the basic swing equation, Eq. (16.12), we obtain

( 1 6 .45 ) Replacing the right-hand side o f this equation b y Eq. (16.44) and transposing terms, we obtain 1 ( 6 .46 )

since Do is a constant value. Noting that P cos Do is the slope of t h e at t h e a n g l e Do , w e denote this slope as Sp and define i t a s max

p u w e r - a n gl e c u rv e

( 1 6 .47 ) where Sp is called the synchronizing power coefficient . When Sp is used in Eq. ( 1 6.46), the swing equation governing the incremental rotor-angle variations m ay be rewritten in the form

( 1 6 .48) This is a l inear, second -order d ifferential equation, the solut ion to which depends on t h e a l gebraic sign of Sp . When Sp is posit ive, the solution ot:;.C t ) corresponds to that of simple harmonic motior. ; such motion is represented by the oscillations o f an undamped swinging pendulum.] When Sp is negative, the solut ion Dt:,( t ) in c r e a s es exponent ially wi thout limit . Th e refore, in Fig. 1 6. 6 the operating point Do = 28 .44° is a point of stable equi l ibrium, in the sense t h a t t h e ro t o r- a ng l e swing i s bounded fol lowing a small perturbation. I n the physica l situa tion damping will restore t h e rotor angle t o D o following t h e temporary electrical perturbation. On the other hand, the point D 1 5 1 .5 6° is a point of =

0, which h a s t h e g eneral solution The e q u ation o f sim ple harmonic m otion i s d 2 x/dt 2 + w�x A .cos w n t + B sin w n t w i th constants . A and B d eterm ined by the initial conditi ons . The s o l u tion w h e n plotted is an u ndamped sinusoid o f ang u l a r frequ ency wn . 3

=

716

CHAPTER 1 6

POWER S YSTEM STA B I LITY

b

( a ) Pe n d u l u m

a

I I

I

,-ll .. "'...... - .

<

I'

I I

I

\

b

\ ,

\

\ \

FI G U R E 1 6 . 7

\ -.)0 �� (' - � ....-...,..,.. ",

Pe n d u l u m a n d r() t �t t i ll g d i s k t o i l ­ l u s t r a t e a ru t m sw i n g i n g w i t h r e ­ spect t o

( Ii ) P e n d u l u m on d i s k

;In

i n fi n i t e b u s .

u nstable eq uil ibrium si nce Sf) i s neg:t l ive t he re . So, this po int i s not a v�t l i d operating point. The c hanging position of the generator rotor swi nging with resp ect to t he infinite bus may be visualized by an a nalogy. Consider a pend u l u m swi nging from a p ivot on a stationary frame, as shown in Fig. 1 6 . 7( a ). Poin ts a a nd c a re the m axim u m poi nts of the osc il lation of t he pendulum abo u t the eq u i l ib ri u m point b. Damping will eventually bring the pen dulum to rest at b. Now imagine a d isk rotating in a clockwise d irection about the pivot of t he pendulum, as shown in Fig. 1 6.7(b), and superimpose the motion of the pe ndul um on the motion of the disk. When the pen dulum is moving from a to c , the combined angular velocity is slower than that o f the d isk. When the pe ndulum is moving from c to a , the combined angular velocity is faster than that o f t h e disk. At points a and c the angular velocity of t he pendulum alone i s zero an d t he combined angular velocity equals that of the disk. I f the angular veloci ty of the d isk corresponds to the synchronous speed of the rotor, and if the motion of t h e pendulum alone represents the swinging of the rotor with respect to an i nfinite bus, the superimposed motion of the p e n d u l u m on t hat of the d isk represents the actual angular motion of the rotOL From t he above d i scu s s i o n we cone! ude that the s o l ut i o n o f Eq . ( ] 6 . 4 8 ) represents sinusoid a l osci llations, provided the synchron izing powe r coe fficien t S p is positive. The angular frequency of t h e undamped oscillations i s give n by

( 1 6 .49) which corresponds to a frequency of oscillation given by

( 1 6 .50) The machine of Example is operating at 0 = 28. 44° when it is subjected to a slight temporary electrical system disturba nce. Determ ine the Example 16.6.

16.3

16.6 EQUAL-AREA CRITERION OF STABIUTY

717

frequency and period of oscillation of the machine rotor if the disturbance is removed before the prime mover responds. Take H 5 MJ /MVA. =

Solution. The applicabl e swing equation is Eq. (1 6.48) and the synchronizing power

coefficient at the opera ting poin t is

Sp The angular frequency w

n

�

/

2 . l 0 cos 28 .44°

=

=

1 .8466

of oscillation is t herefore w,Sp

) 377 X 1 .8466 V 2X5

�

2H

=

8 .343 e1ec ra d / s

The correspondin g frequency of oscillation is f

n

=

8 .343

-2 77"

and the period of oscill ation is T

=

1

-

f,1

=

=

1 . 33 Hz

0 .753 S

The above example is an important one from the practical viewpoint since it indicates the order of magnitude of t h e frequencies which can be superimposed upon the nominal 60-Hz frequency in a l arge system having many intercon­ nected machines. As load on t h e system c hanges randomly throughout the d ay, i ntermachine oscil lations i nvolving frequencies of the order of 1 Hz tend to arise, but these are qu ickly damped out by the various d amping influences caused by the prime mover, the system loads, and the machine itself. It is worthwhile to note that even i f the t ransmission system in our example contains res istance, nonetheless, the swinging of t he rotor is harmon ic and undamped . Prob lem 16.8 examines t h e effect of resistance on the synchronizing power coefficient and the frequency of oscill ations. In a l ater section we again d iscuss the concept of synchronizing coefficients. In the next section we examine a method of determining stability u nder transient conditions caused by large dis turbances. 16.6

EQUAL-AREA CRITERI O N O F STABILITY In Sec. 16.4 we developed swing equations which are nonlinear in nature. Formal solutions of such equations cannot be explicitly found. Even i n the case of a s ingle machine swinging with respect to an infinite bus, it is very difficult to obtain l iteral-form solutions, and computer methods are therefor� nonnally

718

CH APTER 16

POWER SYSTEM STABI LITY

FIGURE 1 6.8 O n e - l i n e d i a g r a m of t h e sys l e m of F ig. 1 6 .4

with the add i t ion of a short t ra nsm ission l i n e .

use d . To exam ine the stability of a two-mach ine syst em without sol v i ng t he swing equation, a direct app roach is possible as now discussed . The system shown in Fig. 1 6.8 i s t h e same a s that co nsid ered previously excep t for t he add ition of a short transmission line. I n i t i a l l y , c i rc u i t breaker A is con s i d ered to b e c l o s e u w h i l e c i rc u i t b re ; l K e r fJ a t t h e oppos i t e e nd o f t he s h o r t line is open. Therefore, the initial opera ting cond it ions of Example 1 6.3 may be considered unaltered . At point P close to the bus a th ree -phase fault occurs and is cleared by circuit breaker A after a short period of time. Thus, the effective t ransmission system is unal tered except whi le the fa ult i s on . The short circuit caused by the fault is effect ive ly at the bus, and so t h e e lect rical power outpu t from the generator is zero until the fault is cleared . The physical con ditions before, during, and after the fau lt can be understood by analyz ing t h e power-angle curves of Fig. 1 6.9. The generator is operating initially at synchronous speed w ith a rotor angle of 00 ' and the input mechanical power Pm equals the output electri cal power Pe , as shown at point a in Fig. 1 6. 9( a ). W h en the fault occurs at t 0, the electrical power output is suddenly zero while the input mechanical power is u naltered, as shown in Fig. 1 6.9( 6 ). The d i fference in power must be accou nted for by a rate of change of stored kinetic energy in the rotor masses. This can be accomplished only by an increase in speed wh ich results from the cons tant accelerating power Pm . I f we denote the time t o clear the fau lt by tc' then the acceleration is constant for time ' less than Ic and is given by =

WS

-P 2H

In

( 1 6 .5 1 )

While the fau l t is on, the velocity increase above synchronous speed is fou nd by integrating th is equation to obtain

do dt

-

=

fl - P 0

Ws

dt 2H m

( 1 6 .52 )

A further i ntegration with respect to t ime yields for the rotor angle ( 1,6 .53 )

16.6 EQUAL-AREA CRITERION OF STABILITY

719

Pm r-��----4r-

(a) e

o

60

e

FIGURE 16.9 Power-angle cu rves for the genera tor s hown in Fig. 1 6.8. Areas A I a n d A2 are e q u a l as are a reas A 3 and A 4 .

Equations 0 6.52) and 06.53) show that the velocity of the rotor relative to synchronous speed increases linearly with t ime while the rotor angle advances from Do to the angle D c at clearing; that is, the angle (5 goes from b to c i n Fig. 16.9( .0 ). At t h e i n st a nt of fa u l t clearing t he i n c re a se in roto r s p e e d and the angle s e p a r a t i o n between the ge nerator and the infi n ite b us are given, respectively, by

( 16.54) and

( 16.55)

When the fault is cleared at the angle D c , the electrical power output abruptly increases to a value corresponding to point d on the power-angle curve . At d the electrical power output exceeds the mechanical power input, and thus the

720

CHAPTER 16

POWER SYSTEM STABILITY

accelerating power is neg ative. As a consequence, the rotor slows down as Pe goes from d to e in Fig. 1 6.9(c). At e the rotor speed is again synchronous although t h e rotor angle has advanced to ox ' The angle Ox is determined b y the fact that areas A 1 a nd A 2 must be equal, as expl ained later. The accelerating power a t e is still neg ative (retarding), and so the rotor cannot remain at synchronous speed but must continue to slow down. The rel at ive velocity is negative and the rotor angle moves back from Ox a t e along the power-angle curve of Fig. 1 6.9(c) to poi n t a a t wh ich the rotor speed i s l e ss th a n syn­ chronous. From a to f the mechanical pow e r exc e e d s t h e e l e c t r i c a l powe r and the rotor increases speed aga i n until i t reaches sy nchronism at J. Po i n t J is located so that areas A 3 and A 4 are equ al. In the absence of d amping the rotor would continue to oscillate in the sequence f-a-e, e-a -J, and so on, with synchronous speed occurring a t e a n d f. We shall soon show that the shaded areas A l and A 2 i n F i g . 16.9(b) m u s t be equal, and similarly, areas A 3 and A 4 in Fig. 16.9( c ) must be equal. I n a system w here one m achine is swinging with respect to an infinite bus we may use this principle of equality of areas, called t h e equal-area criterion , to d etermine the stability of the system under transient conditions without solvin g the swing equ ation. Although t h e criterion is not appl icable to multimach ine systems, it h elps in understanding how certain factors influence the transient st ability of any system. The d erivation of the equ al- are a criterion is made for one m ach ine and an infinite bus although the considera tions in Sec. 16.3 show th a t the method can be readily ad apted to general two-machine syst ems. The swing equ ation for t h e m achine connected to t h e bus is ( 1 6 .56 ) Define the angular veloc ity

of

the

r o t or r e l a t ive

Wr = dt = W

do

Differenti ating

obtain

Eq.

( 16.57) w i t h

respect t o

l

-

to synch ronous speed

W,

( 16 .57)

a n d s u b s t i t u ting

i n Eq.

2H d W r -

Ws

- =p -p dt m

0 6.5 6), w e

( 1 6 . 5 8)

e

When the rotor speed is synchronous, it is clear th at

by

w

equ als

Ws

and

,

wr

IS

1 6.6

EQUAL-AREA CRITERION OF STABILITY

721

zero. Multiplying both sides of Eq. (16.58) by Wr = dojdt , we have H

- 2 wr WS

dWr -

dt

=

do ( Pm - P ) e dt

( 1 6 . 5 9)

The left-hand side of this equat ion can be rewritten to give ( 1 6 .60) Multiplying by dt and integrating, we obtain ( 1 6 .6 1 ) The subscripts for the W r terms correspond to th ose for the ° l imits. That is, the rotor speed W r l corresponds to that at the angle 0 , and Wr2 corresponds to ° 2 , S i nce W r represe nts t h e departure of the rotor speed from synch ronous speed, we readily see that if the rotor speed is synchronous at °1 and ° 2 , then, correspondingly, W r l W r 2 O. U nder this condition, E q . (16.61) becomes =

=

( 1 6 .62) This equation applies to any two points ° 1 and °2 on the power-angle d i agram, provided they are points at w h ich the rotor speed is synchronous. In Fig. 1 6.9(b) two such points are a and e corresponding to °0 and ox ' respectively. If w e perform the in tegration of Eq. (16.62) i n two steps, w e can write ( 1 6 .63) or

( 1 6 .64)

The left in tegral applies to the fault period, whereas the right integral corre­ sponds to the immediate postfa u l t period u p to t he point of maximum swing ox ' I n Fig. 1 6.9( b ) PI' is zero d u ring the fault. The shaded area A 1 is given by t he left-hand side of Eq. ( 1 6.64) and the shaded area A 2 is given by the right-h and side. So, the two areas A 1 and A 2 are equal. S ince the rotor speed is also synchronous at ° x and at Oy in Fig. 1 6 .9(c), the same reasoning as above shows that A 3 equals A 4 . The areas A 1 and A 4 are directly proportional t o the increase i n k inetic energy o f the rotor w hile i t is

722

CHAPTER 16

POWER SYSTEM STA UI LITY

p

Powe r-an g le c u rve s howing the c r i t i c a l clearing angle ocr ' A r e a s A 1 a n d A 2 a re equ a l .

FIGUR E 1 6 . 1 0

"

accelerating, whereas areas A 2 and A :; a r e proport ional to the decrease in k inetic energy of t h e rotor w h i l e i t i s d e c e l e ra t i n g . This c a n he s e e n h y i nspection of both sides of E q . ( 1 6.6 1 ). Th e rd'ure, t h e e q u a l - a r e a c r i t e r i o n s l a t e s that the kinetic energy added to the rotor following a fault must be remove d a fter the faul t in order to restore the rotor to synchronous spee d . The shaded area A l is dependent on t h e t i m e taken to clear the fau l t . I f there is delay in clearing, the angle 0c is increased; l ikewise, the area A I i ncreases and the equal-area criterion requi res that area A 2 also increase to restore the rotor to synchronous speed at a l arger angle of m aximum swi ng OX " If the delay i n clearing is prolonged so that the rotor angle ° swings beyond the angle 0max i n Fig. 16.9, then the rotor speed at t h at point on the power- angle curve is above synchronous speed when positive accelera ting power is again encountered. U nder the influence of this positive accelerating powe r, the angle o will increase without limit and instability results. Therefore, there is a critical angle for clearing the fault in order to satisfy the requirements of the equal-area criterion for stability. This angle, called the critical clearing angle ocn is shown in Fig. 16. 10. The corresponding critical time for removing the fault is called t h e critical clearing time t c r ' Thus, t h e critical clearing time i s t h e maximum elapsed time from the initiation of the fault until its isolation such that the power system is transiently stable. In the particular case of Fig. 1 6 . 10 both the critical clearing angle a nd the critical clearing time can be calcul ated as fol lows. The rectangular area A I is

( 1 6 .65 ) w hile the area

A2

is

( 1 6 .66) =

Pmax( cos o cr - cos 0max )

-

Pm ( 0m a x - ocr)

1 6.6

EQUAL-AREA CRITERION OF STABILITY

723

Equating the expressions for A l and A 2 and transposing terms, yield ( 1 6 .67) We see from the sinusoidal power-angle curve that 0max

= 7r

and

-

00 elec rad

( 1 6 .68) ( 1 6 .69)

Substituting for 0 max and Pm in Eq. ( 1 6.67), simplifying the result, and solving for t he critical clearing angle ocr ' we obtain ( 1 6 .70) Substituting this value of ocr in the left-hand side of Eq. ( 1 6.55) yields ( 1 6 .7 1 ) from which we find the critical clearing time

I CT

( 1 6 .72)

=

Example 1 6.7. Calculate the critical clearing angle and the critical clearing time for the system of Fig. 16.8 when the system is subjected to a three-phase fault at point P on the short transmission l ine. The initial conditions are the same as those in Exa mple 1 6 .3, and H = 5 MJ /MVA

Solution .

In Example 1 ().3

t h e pow er- a ng l e e q u a t i o n Pe

=

00

=

P

max

sin {j

28 .44°

=

=

and i n i t ial rotor

2. l O s i n {j

anglc

0.496 clec rad

Mechan ical input power Pm is 1 .0 per u nit, and Eq. (1 6.70) then gives

=

81 .697°

=

1 .426 elec rad

are

724

CHAPTER 16

POWER SYSTEM STABI LITY

Using this value

with the other k nown quantities in Eq. ( 1 6.72) yields t cr =

which is

60 Hz.

4 X 5( 1 .426 - 0.496) 377 X 1

=

0.222 s

equivalent to a critical clearing time of 13.3 cycles on a frequency base of

This example serves to establ ish the conce pt of crit ical clearing t i m e which is essential to the design of proper relaying schemes for fault clea ring. In more general cases the critical clearing time cannot be exp licitly found wi t hout solving the swi ng equations by computer s i m u l a t i o n .

FURTHER APPLICATIO N S OF THE EQUAL-AREA CRITERIO N 1 6.7

T h e equa l-area criterion is a very use ful means for analyzing stability o f a system of two m achines, or of a single machine supplied from an infinite bus. However, the computer is the only p ractical way to detenn i n e the stability o f a large system. Because the equal-area criterion is so helpfu l in understanding transient stability, we continue to examine it b riefly be fore discussing the d etermination of swing curves by the computer approach. When a generator is supplyin g power to an infinite bus over parallel transmission l ines, opening one of the l ines may cause the gene rator to lose synchronism even though the load cou l d be supp lied over the remai ning l ine u nder steady-state conditions. If a three-phase short circuit occurs on the bus to whieh two parallel lines are connected, no power can be transmitted over either line. This is essentially the case in Example 1 6.7. However, if the fa ult is at the end o f one of the l ines, ope ning b re a k e rs at both ends of the l i n e w i l l isolate the fault from the system and allow power to flow through the other parallel l ine. When a three-phase fault occurs a t some point on a double-ci rcu it line other than on the paralleling buses or at the extreme ends of the line, there is some i mpedance between the paralleling buses and the fault. Therefore, some power is transmitted while the faul t is sti l l on the system. The power-angle equation in Example 1 6.4 demo n s t r a tes t h is fact. When power is transmitted d uring a fault, the equal-area criterion is applied, as shown in Fig. 1 6. 1 1 , which is similar to the power-angle diagram o f Fig. 1 6.6. Before the fault P sin 0 is the power which can be transmitted; during the faul t r 1 Pm ax sin 0 is the power which can be transmitted; and r2 Pmax sin 0 is the power which can be transmi tted after the fault is cleared by switching at t h e instant when 0 = Dcr ' Examination of Fig. 1 6. 1 1 shows that ocr is the ' critical cle aring angle i n t h is case. By evaluating the areas A ) an9 A 2 max

16.7 FURTHER APPLI CATIONS OF THE EQUAL-AREA CRITERION

725

p

- rl

Pmax s i n tl

FIGURE 1 6 . 1 1 Equal-area criterion applied t o fault clearing when power is transmitted during the fault. Areas A l and A 2 are equal.

using the procedural steps of the p revious section, we find that ( 1 6 .73)

l iteral-form solution for the critical clearing time f e r is not possible in t h i s case. For the particular system and fault l ocation shown in Fig. 1 6.8 t h e applicable values are T I = 0, T 2 = 1 , a n d Eq. ( 1 6.73) then reduces t o Eq. (16.67). S hort-circuit faults which do not involve a l l three phases allow some power to be transmitted over the unaffected phases. S uch fau lts are represented by connecting an impedance (rat h e r than a short circuit) between the fault point and the reference node in the positive-sequence impedance diagram. The l arger the i mpedance shunted across the positive-sequence network to represent the fault, the larger the power transmitted during the fault. The amount of power transmitted during the fault affects the value of A I for any given clearing angle. Thus, smaller values of T l result i n greater distu rbances to the system, as smaller a mounts of p o we r are transmitted during the fault. Consequently, the area Al of acceleration is larger. In order of increasing severity (that is, decreasing T I Pmax) the variou s faults a re: A

1.

A

2. A 3. A 4. A

single l ine- to-ground faul t line-to-line fault double l ine-to-ground fau l t t hree- phase fault

The single line-to-ground fault occurs most frequently, and t h e three-phase fault is the least frequent. For complete reliability a system should be d esigned for transient stability for three-phase faults at the worst locations, and this is virtually the universal practice.

726

CHAPTER 16

POWER SYSTEM STABILITY

Determine the critical clearing angle for the three-phase fau lt described i n Examples 1 6.4 and 1 6.5 when the initial system configuration and prefault operating conditions are as described in Example 1 6.3.

Example 16.8.

Solution.

The power-angle equations obtained in the previous examples are Pmax sin o = 2 . 1 00 sin o

Bcfore t h e fau l t :

n

r ) Pmax sin

Dur i g t h e fau l t :

r2 Pmax sin 0

t

Aft er t h e fau l : Hence,

r)

=

0.808 -2 . 100

=

(j

0 . 385

r2

=

=

=

0 .808 sin

8

1 .500 sin 0

1 .500 -= 0 .7 1 4 2 . 1 00

From Example 16.3 we have 00

=

and from Fig. 1 6. 1 1 we calcu late O max

=

1800 - sin - 1

28 . 44°

=

[ -- 1 1 .000 1 .500

0 .496 rad

=

1 38 . 1 90°

=

2 .4 12

rad

Therefore, inserting numerical values in Eq. 0 6.73), we obtain

cos O cr

=

=

1 .0 () (2.4 1 2 - 0 .4 96) 0 .71 4 cos ( 1 38 . 1--9°) -----0 .385 cos ( 28 .44°) 2 . 1 0 ------��

+

--.-8 -4-.7-O0 3 5 1

-

----

-

0.127

Hence,

To determine the critical clearin g time for this example, we must obtain the swing curve of 0 versus t. In Sec. 1 6 . 9 we discuss one method o f computing such swing curves.

16.8 MULTIMACHI N E STA B ILITY STUDIES: CLASSICAL REPRESENTATION

16.8 MULTIMACHINE STABILIlY STUDIES: CLASSICAL REPRESENTATION

727

The equal-area criterion cannot be used directly in systems where three or more machines are represented. Although the physical phenomena observed in the two-machine problems are basically the same as in the multimachine case, nonetheless, the compl exity of the numerical computations increases with the n umber of machines considered in a transient stability study. When a multi­ machine system operates under e lectromechanical transient conditions, in­ termachine oscillations occur t h rough the med ium of the transmission syste m connecting the mac hines. If any one machine coul d be consid ered to act alone as the single osci llating source, it would send into the interconnected system an electromechanical osci llation d eterm ined by its inertia and synchronizing power. A typical frequency of such an osc illa tion is of the order of 1 -2 Hz, and this is superimposed upon the nominal 60-Hz frequency of the system. When many m ach ine rotors arc simultaneously u nd ergo ing transient oscillation, the swing curves reflect the combined presence of many such oscillations. Therefore, the t ransmission system frequency is not u nd u ly perturbed from nominal frequency, and the assumption is made that the 60-Hz network parameters are still appl icable. To ease the complexity of system modeling, a nd thereby the compu ­ tational burden, the fol lowing a dditional assumptions are commonly made I n transient stab ility studies:

1. The mechanical power input to each machine remains constant d uring the entire period of the swing curve computation. 2. Damping power is negligible. 3. Each machine may be represe nted by a constant transient reactance in series with a constant transient intern a l voltage. 4. The mechanical rotor angle of each machine coincides with 0, the electrical phase angle of the transient internal voltage. 5. Al l loads may be considered as shunt imped ances to ground with values determined by conditions prevailing immediately prior to the transient condi­ t ions.

The system stabil ity model based on these assumptions is cal led the classical stability mo del, and studies which use this model are called classical stability s t u dies. These assumptions, which we shall adopt, are i n addition to the fundamental assumptions set forth in Sec. 16.1 for all stability studies. Of cou rse, detailed computer p rograms with more soph isticated machine and load models are available to modify one or more of assumptions 1 to 5. Throughout t h is chapter, however, the classical model is used to study system disturbances originating from three-phase faults. The system condi tions b efore the faul t occurs, and the network configura­ t ion both during and after its occurrence, must be known III any transient

'728

CHAYIER 16

POWER SYSTEM STABILITY

stability study, as we preliminary steps are

1. 2.

have seen. Consequently, required:

m

the multimachine case two

The steady-state prefault conditions for the system are calculated usmg a production-type power-flow program. The prefault network rep resentation is determined and then modified to account for the fault and for the postfault conditions.

From the first preliminary step we know the values of power, reactive power, and voltage at each generator terminal and load bus, with all angles measured with respect to the slack bus. The transient internal vol tage of each generator is then calculated using the equation E

=

v,

+

( 1 6.74)

jX� I

where V, is the corresponding terminal voltage and I is the output current. Each load is converted into a constant admittance to ground at its bus using the e qu ation .

( 1 6 .75 ) where PL + jQL is the load and I VL I is the magnitude of the corresponding bus voltage. The bus admittance matrix which is used for the p refault power-flow calcul ation is now augmented to include the transient reactance of each genera­ tor and the shunt admittance of each load, as suggested in Fig. 16.1 2. Note that the injected current is zero at all buses except the internal buses o f the generators. In the second preliminary step the bus admittance matrix is modified to correspond to t he faulted and postfa u l t conditions. S i n c e o n l y the g e n e ra t o r internal buses have injections, all other buses can be el iminated b y Kron

I

B o u n d a ry o f a u g m e n t e d n etwork

r-- - - - - - - - - - -- - - - - - - - - - l

CD: I

�-----r�

Xd1

E2

I I I I I

Tra n s m i s s i o n n etwork

+

Ei -

: I

YLs l

I I I I I I

I

-

E3

+ II I

-

L

_ _ _ _ _ _

_ _ _ _

_

_ _ _ _ _ _ _ _ _ _

I -1

FIGURE 1 6 . 1 2 Augm ented network o f

a

power syste m .

16.8

MULTIMACHINE STABILITY STU DIES: CLASSICAL REPRESENTATION

729

reduction. The dimensions of the modified matrices then correspond to the n umber of generators. During and after the fault the power flow i nto the network from each generator is calcu lated by the correspond ing power-angle equation. For example, in Fig. 1 6. 12 the power out of generator 1 is given by

Pe l

-

I E;

1 2C

II

+

IE; l i E; " Yl 2 l cos ( 0 1 2

-

e 12)

+

I E; l i E; I I Y1 3 l cos ( 0 1 3

-

(13)

( 1 6 .76)

where ° 1 2 equals ° 1 ° 2 , S imilar equat ions are written for Pe 2 and Pe3 using the I:j elements of the 3 X 3 bus admi t t ance ma trices appropriate to the fau lt or postfault condition . The Pei expressions form part of the swing equ ations -

2 N-I

--

d28 _ df

ws

I

--

2

_ . - ..

r

mi

-

p<-;

1 = 1 , 2, 3

( 1 6 . 77)

represent t he mot ion of e a c h rotor d uring t he fau l t anQ postfa u l t perio�s. The solutions depend on t he loca tio n and d ura t ion of the fault, and the Ybus resulting when the fa ulted l in e is removed. The basic p rocedures used i n computer programs for classical stability studies a r e revealed in the fol lowing examples. which

Example 16.9. 60-Hz, 230-kV transmission system shown in Fig. 16.13 h a s two generators of finite inertia and an infinite bus. The transformer and line data are given in Table 16. 2. three-phase fault occurs on line @ - @ near bus @ . Using the prefault power-flow solution given i n Table 16.3, determine the swing equation for each machine during the fault period. The generators have reactances

A

A

L4 L5 FIGURE 16.13 One-line d iagram for Examp l e 1 6.9.

T® OGenerator

2

730

CHAPTER 16

POWER S YSTEM STABI LITY

�ne and transformer data for Example

TABLE 16.2

Bus

to

R

bus

CD � Transformer CD -G) Line G) � Line G) -W ( 1 ) Line G) -G)<2) Line @ -G)

Series Z

Transformer

tAil

16.9t Shunt Y

X

B

0.022 0.040 0.007

0.040

0.082

0.008

0.047

0.098

0.008

0.047

0.098

0.0 1 8

0.1 10

0 .226

val ues i n p e r u n i t On n O · k V , ! O()- M V A \l;ISC.

TABLE 1 6.3

Bus data and prefault load-How valuest Q

Generation Voltage

P

1.030/8.88° 1.020/6.38° 1.000LQ: 1.018/4.68° 1.01 1/2.27°

3 . 500

Bus

CD @ ® @ �

tValues

are

and

1 .850

Load P

Q

1 .00

0.44

0.50

0. 1 6

0.712 0.298

i n per unit o n 230-kV, lOO-MVA base.

H

values expressed on a 1 00-MVA base as follows:

Generator 1 : 400 MVA, 20 kV , X:J

Generator 2 : 250 MVA, 1 8 k V , Xd

= =

0.067 per u n i t ,

H

0 . 1 0 per unit,

H

=

=

1 1 .2

MJ /MVA

8.0 MJ/MVA

In order to formulate the swing equations, we must first determine the transient internal voltages. Based on the data of Table 1 6 .3, t h e c u r r e n t into the network at bus CD is

Solution.

II =

( P I + jQ I ) * --v -* I-­

/ - 8.88°

3 .50 - jO.712 1 .030

Similarly, the current into the network at bus 12

=

( P2 + jQ 2 ) *

--v -*-2

@

1 .850 - jO.29B

/ - 6.38°

=

is

------ =

1 .020

3 .468

/ - 2 .6 19° L

1 .837 � 2.77e _

16.8

From

MULTIMACHINE STABI LITY STUDI ES: CLASSICAL REPRESENTATION

731

Eq. ( 16.74) we then calculate 0 E; 1. 030/ 8. 880 jO. 067 3. 4 68/ -2. 6 19° 1.100/ 20.82 jO.lO 1.837/ -2.77e 1.065/ 16.190 E; 1 020/ 6 . 3 8 =

=

0

.

At the infinite bus

+

x

+

x

=

=

have

we

and so

The P-Q loads at buses @ and W arc converted into equivalent shunt admittances using Eq. 0 6.75 ), which yields YL 4

Yo

=

=

1 .00 - j0.44 ., ( .0 l r

1 8

0 .50 - jO . 1 6 ( 1 1 )2

0 .9649 - j0.4246 p e r u n it

=

0 .4 92 - jO.1565 per u it 8

=

1 .0

n

The prefault bus admittance matrix is now modified to include the load admittances and the t r a n s i e n t r e a c t a n c e s of the machines. Buses CD and @ designate the fictitious internal nodes behind the transient reactances of the machines. So, i n the prcfault bus admittance matrix, for example, YJ I Y34

The sum

=

1

jO .067

+

jO .022

- j l 1 .236

1 =

-

0 .007

jO .040

+

=

per u n i t

-4.2450 j24.2571 per u n it

of the a d m i t t a n ces c o n n e c t e d to

+

huscs

G), @, a n d W

s h u n t e a p ;l c i [ (I nces o r [ h e [ r; l I 1 s rn i s s i oll l i n e s . S o , ( I [ b u s

Y44

=

- j l 1 .236 +

=

0. 0 1 8

+

+

jO .082 2

jO. 1 1 0

+

jO .226 2

--

+

@

4 .2450

must include the

we h ave

- j 24

.2571

+ 0 . %49 - j0 .4246

6 .6587 - j44 .6175 per unit

The new prefault bus admittance matrix is d isplayed as Table 16.4 on page Bus @ must be short-circuited to the reference to represent the fault. Row 4 and column 4 of Table 1 6. 4 thereby disappear because node @ is n ow m erged with the reference node. Next, the row and column representing bus G) are eliminated by Kron reduction, and we obtain the bus admittance matrix shown i n

733.

732

.:

CHAPTER 16

POWER SYSTEM STABILITY

the upper half of Table 1 6. 5 . The faulted-system Y bus shows that bus CD decouples from the other buses during the fault and that bus CD is connected directly to bus G) . This reflects the physical fact that the short circuit at bus @ reduces to zero the power injected into the system from generator 1 and causes generator 2 to deliver its power rad ially to bus (1) . Under fau lt conditions, t h e power-angle equations based on values from Tab le 1 6. 5 a re

=

=

( 1 .06S f ( O . 1J62) �

+

( 1 .()()S ) ( J .(l ) ( 5 . 1 () ()5 ) cos( 8 2 -

0 . 1 5 4 5 + 5 .5023 s i n ( 8 2 · · 0 . 755°)

90 .755°)

pcr u n i t

Thereforc, while the fault is o n the system, the desired swing equations (values of Pm ! and Pm 2 from Table 1 6.3) are

1 801 "

=

--

1 1 .2

=

{

e\ec deg/s 2

- l 0. 1 545 5 . 5023 S i n ( 8 , - 0 .755" ) I� � � 1 801 [ 1 .6955 - 5 . 5 023 s i n ( 0 2 - 0 . 755° ) ] e c dcg/s 2 8.0

s:o

] 80/

�

0.5)

� 1 .85

+

--

---

------

----

el

1}

y

The three-phase fault in Example 1 6 . 9 is cleared by simultaneously the circuit breakers at the ends of the faulted line. Determine the swing equations for the postfault period. Exa m ple 16.10.

ope n i n g

'.

Solution. Since the fault is cleared by removing l i ne @ - G)' the prefault Ybus of Table 1 6.4 must be modified again. This is accomplished by substituting zero for Y45 and YS4 and by subtracting the series admittance of line @ - w and the capacit ive susceptance of one-half the l ine from e lements Y44 and Y55 of Table 16.4. The reduced bus admittance m atrix applicabl e to the postfault network is shown in the lower half of Table 1 6 . 5 . The zero elements in the first and second rows reflect the fact that generators 1 and 2 are not in terconnected when line

,

16.8 MULTIMACHINE STABILITY STUDIES: CLASSICAL REPRESENTATI ON

733

TABLE 16.4

Elements of prefault bus admittance matrix for Example 16.9t

CD

CD

®

@

®

-jl1.2360 0.0 0.0

0.0 -j7. 1 429 0.0

j l 1 .2360

0.0

0.0 0.0 1 1 .284 1 -j65.4731 - 4.2450 +j24.2571 - 7.0392 + j4 1 .3550

j l 1 .2360 0.0 - 4.2450 +j24.2571 6.6588 -j44.6175 - 1 .4488 + j8.8538

0.0 j7. 1 429 - 7.0392 + j4 1 .3550 - 1 .4488 + j8.8538 8.9772 -j57.2972

Bus

CD W ®

@

m

j7.1429

0.0

t A d m i t t ances i n per

u n it.

TABLE 1 6.5

Elements of faulted and postfault bus adm itta nce matrices for Exam ple 1 6.9t

CD

Bus

CD

0.0000 - j 1 1.2360 (1 1 .2360L - 90° ) 0.0 + jO.O

W

® CD

W

CD

Faulted network

0.0 + jO.O

0

®

0.0 + jO.O

0.0 + jO.O

0.1362 - j6.2737 (6.2752 / - 88.7563° ) - 0.681 + j5.1661 (5.1 665/ 90.7552° )

- 0.0681 + j5. l661 (5.1665 /90.7552° ) 5.7986 - j35.6299 (36.0987/ - 80.7564° )

PostfauIt network

0.5005 - j7.7897 (7.8058L - 86.3237° ) 0.0 + jO.O 0.22 16 + j7.6291 (7.6323/91 .6638° ) -

tAdmittances

0.0 + jO.O 0.1591 - j6. 1 168 (6. 1 189/ - 88.5101° ) - 0.090 1 + j6.0975 (6.0982 / 90.8466° )

- 0.2216 + j7.6291 (7.6323/9 1 .6638° ) - 0.0901 + j6.0975 (6.0982 / 90.8466° ) 1 .392 7 - j 13.8728 ( 1 3.9426L - 84.2672° )

in per u n i t .

@ - � is removed. Accordingly, each generator is connected radially to the infinite bus, and we can write power-angle equations for the postfault conditions as fol lows: ==

=

( 1 .100) 2 (0 .5005) 0.6056

+

( 1 .100) ( 1 .0) (7 .6323)cos ( 8 1 - 91 .664°)

+ 8 .3955 sin e 8 )

- 1 .664°) per unit

734

CHAPTER 16 POWER SYSTEM STABI LITY

= (1 .065)\0.1591) + ( 1 .065 ) ( 1 .0)(6 .0982)cos( 8 2 - 90 .847°) = 0 . 1 804 6 .4934 sin( 0 2 - 0 .84r) per unit +

For the postfault period the applicable swing equations a re given b y d 20 1 -2 dl

{ 3 .5 - [ O . GO SG -1 1 .2 1 80!

=

=

and

d202 dt 2

=

I HOf -- [ 2 . 894 4 1 1 .2

+

8 . J 955 s i n ( o l

- 8 .3955 s i n ( o l -

1 80 ! - { l . S5 - [ 0 . lS04 8.0

+

- 1 .6(i4°) ] }

l .G64°) ]

e!ec dcg/s 2

6 .4934 sin ( 0 2 - O .S4r) ] }

= -- [ 1 .6696 - 6 .4934 sin ( 8 2 - 0. S 4 r ) ]

I SO! S .O

c l ec deg/s 2

The power-angle equations obtained in Examples 16.9 and 16.10 are of the form of Eq. (16.35), and the corresponding swing equations assume the form 2 d 8 dt 2

180! -=[ H

pm - pc

-P

max:

sin ( S - y)]

( 16 .78)

The bracketed right-hand term represents the accelerating power of the rotor. Accordingly, we may write Eq. (16.78) as d2S

dt 2

180!

=

-- Pa clcc deg /s2 H

where

( 1 6 .79) ( 1 6 . 80 )

In the next section we discuss how to solve equations of the form of Eq. (16.79) in order to obtain S as a function of t ime for specified clearing times. STEP-BY-STEP SOLUTION O F THE SWING CURVE 16.9

For large systems we depend on the compu ter, which determines 8 versus t for all machines i n which we are interested; and S may be plotted versus t for a machine to obtain the swing curve of that m achine. The angle 8 is calculated as a function of t ime over a period long e nough to determine whether S will increase without limit or reach a maximum and start to decrease. Although the I

16.9 STEP-BY-STEP SOLUTION OF THE S WING CURVE

735

latter result usually indicates stability, on a n actual system where a n u mber of variables are taken into account it m ay be necessary to plot 0 versus ove r a long enough interval to be sure 0 will not increase again without returning to a low value. By determining swing curves for various clearing times, we can find the length of time permitted before clearing a fault. Standard interrupting times for circuit breakers and their associated relays are commonly 8, 5, 3, or 2 cycles after a fault occurs, and thus breaker speeds may be specified. Calculations should be made for a fault i n the position which will allow the least transfe r of

t

power from the machine a n d for the most severe type of faul t for which

protection against loss of stability is justified. A number of different methods is available for the numerical evaluation of second-order differential equations in step-by-step computations for small incre­ ments of the independent variable. The more elaborate methods are p ractical only when the computations are performed on a computer. The step-by-step method used for hand calculations is necessarily simpler than some of the methods recommended for computers. In the methods for hand calculation the change in the angular position of the rotor during a short interval of time is computed by making the followi ng assumptions: 1.

2.

The accelerating power Pa computed at the beginning of an interval is constant from the middle of the p receding interval to the middle of the interval considered. Throughout any interval the angular velocity is constant at the value com­ puted for the middle of the interval.

Of course, neither of the above assumptions is accurate since 0 is changing continuously and both Pa and w are functions of O. As the time interval is decreased, the computed swing curve becomes more accurate. Figure 1 6 . 1 4C a ) will help in visualizing the assumptions. The accelerating p ower is computed for the points enclosed in circles at the ends of the n 2, n 1 , and n intervals, which are the beginnings of t h e n 1 , n , and n + 1 i ntervals. The step curve of Pa in Fig. 1 6. 14( a ) results from the assumptions that Pa is constant between the midpoints of the intervals. Similarly, wr ' the excess of the a ngular velocity w over the synchronous angular velocity w is shown in Fig. 16. 1 4(6) a s a step curvc t h a t i s const a n t throughout the i nterval at the value computed for the midpoint. Between the ordinates n �. and n 'i there is a change of speed caused by the constant accelerating power. The change In speed is the product of the acceleration and the time interval, and so -

-

-

5'

-

-

d 20 wr, n - I /2 - wr , n - 3/2 = dt 2

/). t

1 80!

=

P , -1 H an

( 1 6 .8 1 )

�t ,

.

·736

CHAPTER 16

POWER SYSTEM STABI LiTY

Pa l" - 2) Pa l,, - I I

c---..--

, , , I , , I ,

- - -

- -

---

, ,

, ,

n

n - 2 , n - l ,

, I

, , I

Assumed Act ual

I

,

n - l. 2

Assu med Actual

I n 2

FlGURE 1 6. 1 4 n - 2

n- l

n

Actual a n d assumed v alues of Pa , (; a s fu n c tions o f time.

Wr ,

and

The change in 8 over an interval is t he p roduct of W for the in terval and t he time of the interval. Thus, the change in 8 d u ring t he n 1 interval is r

-

( 16 .82) and during

the n t h in terval

( 16 .83) Subtracting Eq. (16.82) from Eq. ( 16.8 3 ) and substituting Eq. (16.81) In the resulting equation eliminate all values o f W r ' and we find that

( 16.8 4 )

1 6.9

STEP-BY-STEP SOLUTION OF THE SWING CURVE

where

737

( 16 .85 )

Equation (16 . 84) is importan t for the step-by-step solu tion of the swing equation, for it shows how to c a l c u late the change in 0 during an interval based on the accelerating power for that interval and the change in 0 for the p receding interval. The accelerating power is calculated at the beginn ing of each new interval and the solution progresses until enough points are obtained for plotting the swing curve. Greater accuracy is obtained when !1t is small. A value of !1 t = 0.05 s is u sually satisfactory. The occurrence of a fault causes a discontinuity in the accelerating power Pa , w hich has a zero value before the fault and a nonzero value immediately following the fault. The d iscontinuity occurs at the beginn ing of the interval when t = O. Reference to Fig. 1 6 . 1 4 shows that our method of calculation assumes that the accelerat ing power computed at the beginning of an interval is constant from the midd le of the preceding interval to the middle of the interval being considered. When the fau lt occurs, we have two values of Pa at the beginning of an i n terval, and we must take the average of these two va) ues as the constant accelerating power. The p rocedure is illustrated in the following example. Prepare a table showing the steps taken to plot the swing curve of m achine 2 for the fault on the 60-Hz system of Examples 16.9 and 16.10. The fault is cleared at 0.225 s by simultaneously opening the circuit breakers at the ends of the faulted line.

Example 1 6 . 1 1 .

Without loss o f generality, we consider the detailed computations for machine 2. Computations to plot the swing curve for machine 1 are left to the student. Accord ingly, we drop the subscript 2 as the indication of the machine nu mber from all symbols i n what fol lows. All calculations are made in per unit on a 1 00-M VA base. For the time interv al t:.t 0 05 s the parameter k applicable to machine 2 is

Solution.

=

k = W h e n t h e fau lt i n s t antly. H e n c e ,

I RO!

-

H

') (�l r =

I SO x 60

---

8 .0

occurs at t 0, t h e from Example 1 6.9 =

X

.

( 0 05 ) 2 .

=

3 .375

elcc

rotor a n g l e of mach i n e

deg

2 cannot change

and during the fault Pe

=

0 . 1 545

+

5 .5023 sin ( o - 0.755° ) per unit

.,

738

CHAPTER 16

POWER SYSTEM STABILITY

Therefore, as already seen in Example 1 6.9,

Pa = Pm - Pe = 1 .6955 - 5 .5 023 si n e 0 - 0 .755°) per unit

At the beginning of the first interval there is a discontinuity in the accelerating power of each machine. Immediately before the fault occurs Pa = O. I mmediately after the fault occurs Pa = 1 .6955 - 5 .5023 sin( 1 6 . 1 9° - 0.755°) The average value of Pa at that

t

k Pn

=0

is

= 3 .375

i

x

X

0.23 1 0

=

=

0.23 1

0. l 155

per u ni t

p e r u n i t . W e t h e n fi nd

0 . 1 1 55 = 0 . 3898°

Identifying the time intervals by nume ri cal subscripts, we find that t h e change i n the rotor angle o f machine 2 as time advances over the first i n terva l from 0 to D t is given by �Ol = 0

+

0 .3898

=

0.3898°

At the end of the first time interval we then have 0,

and

-

y

=

1 6 .5798° - 0 .755° = 1 5 .8248°

At t = � t = 0.05 s we find that [ kPa , I = 3 .375 ( Pm - Pc )

- Pmax

si n ( 8 1 - y ) ]

= 3 .375 [ 1 .6955 - 5 .5023 sin( 1 5 .8248°) ] = 0.6583°

and

it

follows that

the i ncrease in t h e rotor a n g l e ovc r the second t i m e

interval is

At the end of the second time i nterval The subsequent steps in the computations are shown i n Table 1 6.6. Note that the postfaul t equation found i n Example 1 6. 1 0 is needed. In Table 1 6.6 the terms Pmax sin ( o - y ), Pa , and 0" have values computed at the time t shown in the first column but �on is the change in the rotor angle during the interval that begins at t he time i ndicated. For example, in the row for t = 0.10 S the angle 1 7.6279 ° is the first value calculated and is found by adding the change in angle during the preceding time interval (0.05 to 0.10 s) to the angle at t = 0.05 s. Next, Pmax sin ( o y ) is calculated for 0 = 1 7.6279°. Then, Pa = ( Pm P) - P sinCcS y) and kPa are calcul ated . The v alue of kPa is 0.3323°, which is added. to the angular change of 1 .048 1 ° d uring the preceding i n terval to find the -

max

-

-

1 6.9 STEP-BY-STEP SOLUTION OFTH E SWING CURVE

' 739

Computation of swing curve for m achine 2 of Example 16 .11 for clearing at 0.225 s k ( 1 80f/H X 6. t )2 3 .375 elec deg. Before clearing Pm - Pc 1 .6955 p . U . , Pmax = 5 .5023 p.u., and

=

=

TABLE 16.6

'Y

=

I, s

=

0.75 5°. After clearing these values become 1 .6696, 6.4934, and 0.847, respectively.

O> n

-y ) , elec deg -

Pm ax s i n ( & n 'Y ) , per unit -

0 0 + o av

15 .435

0.05

1 5 .8248

0.10

1 6 .8729

Pa ,

per un i t

kPa,n - 1 ' elec deg

0.00 0.23 10 0. 1 1 55

0.3 898

1 .5 005

0 . 1 950

0.6583

1 .5 970

0.0985

0.3323

1 . 4 644

a&n'

elec deg

&n'

eJec deg 1 6. 1 9 1 6 19 16.19 .

0.3898 1 6.5 798

1 .0 48 1 17 .6279 1 . 3804

0.15

1 8 .2533

1 .7234

- 0.0279

1 9.0083

- 0.0942 1 .2862

0.20

1 9. 5 395

1 . 8403

- 0 . 1 448

- 0. 48 86

0.25

20.24 5 1

2.2470

- 0.5774

- 1 .9487

20.2945 0.7976

2 1 .0921 - 1 .15 1 1

0.30

1 9.0940

2 . 1 24 1

- 0.4545

1 9.94 1 0

- 1 .534 - 2.6852

0.35

1 6 .4088

1 .8343

- 0. 1 647

- 0.5559

0.40

1 3 . 1678

1 . 4792

0 . 1 904

0. 6425

1 7.2558 -

3 .24 1 0 1 4 . 0 1 48

- 2.5985 0.45

1 0.5693

1 .1911

0.4785

1 1 . 4 1 63

1 .6 1 5 1 - 0. 9833

0.50

9.5860

1 .0 8 1 3

0 .5883

1 0.433 0

1 .9854 1 . 0020

0.55

1 0.5 880

1 . 1 93 1

0.4765

1 1 . 43 50

1 .6081 2.6101

0.60

1 3 . 1 98 1

1 .4 826

0 . 1 870

0.63 1 2

0.65

1 6 .4395

1 .8376

- 0 . 1 680

- 0.5672

1 4 . 04 5 1 3.24 1 4

1 7.2865 2 .6742

0.70 0.75

1 9. 1 1 3 7 20.2468

2. 1 262 2.2471

- 0.4566 - 0.5775

- 1 . 54 1 1

1 9 . 9607 1 .1331

2 1 .0938

- 1 .9492 - 0. 8 1 6 1

0.80

1 9.4307

2. 1 60 1

- 0.4905

20.2777

- 1 .655 6 - 2.4716

0.85

1 7 .8061

change of l .3804° during the interval beginning at t = 0 . 1 0 s. This value added to 17.6279° gives the value 0 = 1 9.0083° at l = 0.15 s. Note that the value of Pm - Pc changes at 0.25 s because the fault is cleared at 0.225 s. The angle 'Y h asalso changed from 0.755° to 0.847°.

Whenever a fault is cleared, a discontinuity occurs in the acceleratin g power Pa . When clearing i s a t 0.225 s , a s i n Table 16.6, n o special a p proach is

' 740

CHAITER 16

POWER SYSTEM STABI LITY

since our procedure assumes a d iscontinuity at the middle of an interval. At the beginning of the interval fol lowing clearing the assumed constant value of Pa is that determined for 8 at the beginning of the interval following clearing. When clearing is at the beginning of an interval such as at 3 cycles (0.05 s ) , two values of accelerating power resul t from t he two expressions for the power output of the generator. One a pplies during and one after cleari n g the fault. For t h e system of Example 1 6. 1 1 if the d iscontinuity occ u rs at 0.05 s, t he average of the two values is assumed as the constant val ue of p{/ from 0.025 to 0.075 s. The procedure is the same as tha t fol lowed upon o cc u rr en c e of the faul t at t = 0 , a s demonstrated in Table 1 6. 6 . Following t h e same pro c e u u r e s a s i n Tab l e 1 6.6, w e c a n d e t e r m i n e 0 versus I for m ac h i ne 1 for c l e a r i n g a t 0.025 s a n d for bo t h m �l � h i I 1 C S for c l e a r i n g at 0.05 s . In t h e next sect i o n w e see com pu ter p r i n t o u t s o f ;j v e rs u s ( tor b m h machines calcula ted for clearing a t 0.05 a n d 0.225 s . Swing curves p lotted for the two machi nes i n Fig. 16. 1 5 show that m achine 1 is u nstable for clearing at 0.225 s. For clearing at 0.20 s, however, it can b e shown t hat the system is stable. The equal-area criterion confirms that t h e actual critical clearing time i s between 0.20 and 0.225 s (see Prob. 1 6 . 16).

required

. -- - - - ,--- _ .

�.- .

-

- --- - - ---_ .--_...

--- _.

-

_

. . . .. _. _-

.

.

. . r- -

. . . . ..

--..

Mach ine 1

/ J

30°


00


25°

ro

u

... u
Q)

-c

20°

1 5°

o

/

I I I

V

+ -----L

./'

�

T

- --- -- -- ----

0. 1

I

0.3

0 .4

� -+ -....

/

/

�/

t, seco n d s 1

0.52 s

I

\

0.2

FIGURE 1 6 . 15

Swing curves for machi nes

1\ 1

I

0.5

Machine 2

0 .6

I

0.7

and 2 of Examples 1 6 . 9 to 1 6. 1 1 for c l e a r i n g at 0.225 s .

0.8

COMPUTER PROGRAMS FOR TRANSIENT STABILITY STUDIES

16.10

741

Figure 16.1 5 shows that the change in the rotor angle of machine 2 is quite small, even though the fault is not cleared until 13.5 cycles after it occurs. It is interesting, therefore, to calculate the approximate frequency of oscillation of the rotor using the linearization procedure of Sec. 1 6.5. The synchronizing power 'coefficient calculated from the postfault power-angle equation for ma­ chine 2 is given by Sp

= =

dPe dB

d

=

dB

[ 0 . 1 804 + 6 .4934 sin ( B - 0.847°) ]

6.4934 cos ( B - 0 . 84r )

We note from Table 1 6.6 that the angle of machine 2 varies between 10.43° and 21 .09°. Using either angle makes little difference i n the value found for Sp . I f we use the average value of 1 5.76°, we find that Sp

=

6 .274 p e r-unit power/elec rad

and by Eq. ( 1 6.50) the frequency of oscillation is In

=

377 X 6 .274 2X8

1 2 7T"

1 .935 Hz

from which the period of oscillation is calculated to be T

=

1 n I

1 =

1 .935

=

0.517 s

Figure 1 6 . 1 5 and Table 1 6.6 confirm this value of T for machine 2. When faults are of shorter duration than 0.225 s, even more accurate results can be expected since the swing of the rotor is correspondingly smaller. In the above examples it is possible to calculate the swing curves for each machine separately because of the fault location considered. When other fault locations arc chosen, intermachine oscillations occur because the two generato rs do not decouple. The swing-curve computations are then more unwieldy. For such cases manual calcul ations are time-consum ing and should bc avoided. Computer programs of great versatility are generally available and should be used . 16.10

COMPUTER PROGRAMS FOR TRANSIENT STABILITY STUDIES

Present-day computer programs for transient stability studies have evolved from two basic needs 0) the requirement to study very large interconnected systems with numerous machines and (2) the need to represent machines and their associated control systems by more detailed models. The classical mach i ne

742

CHAPTER 1 6 POWER SYSTEM STABILITY

representation is suitable for many studies. However, more elaborate models may be required to represent modern turboalternators with dynamic character­ istics determined by the many technological advances in the design of machine and control systems. The simplest possible synchronous machine model is that used in classical stability studies. The more complicated two-axis machine models of Chap. 3 provide for direct- and quadrature-axis nux cond itions during the subtransie n t a n d transient periods following a system d isturbance. By providing for varying flux linkages of the field winding in the d i rect axis, these models al low represe n­ tation of the action of the continuously acting automatic voltage regulator and excitation system with which all . modern machines are equipped. Turbine control systems, w hich automatically gov ern the mechanical power input to the generating u nit, also have dynamic response characteristics which can influence rotor dynamics. If these control sche mes a re to be represented, the model of the generating unit must be further extended. The more complex generator models give rise to a larger number of differen tial and algebraic equations for each machi ne. In large system studies many generators supply widely dispersed load centers through extensive trans­ mission systems whose performance also must be represented by a very large number of algebraic equations. Therefore, two sets of equations need to be solved simultaneously for each interval of time fol lowing the occurrence of a system disturbance. One set consists of t h e algebraic equations for the steady­ state behavior of the network and its loads, and the algebraic equa tions rela ting � ·and £' of the synchronous machines. The other set consists of the differential equations which describe the dynamic electromechanical performance of the machines and their associated control systems. The Newton-Raphson power-flow p rocedure described in Chap. 9 is the most commonly used solution technique for the network equations . Any one of several well-known step-by-step p rocedures may be chosen for numerical i n te­ gration of the differential equations. The fourth-order Runge-Kutta method is very often used in product ion-type transient stability programs. Other methods known as the Euler method, the mod ified Euler method, the trapezoidal method, a n d p redictor-corrector m e t llOds similar t o the step-by-step method developed in Sec. 1 6 .9 are al ternatives. Each of these methods has advan tages and disadvantages associated with numerical stability, time-step size, computa­ tional effort per integration step, and accuracy of solutions obtained. 4 Table 16 . 7 sh ows the computer printo u t for swing curves of mach ines 1 and 2 of Example 1 6. 1 1 for clearing a t 0 .225 s and at 0.05 s. These results were obtained by use of a production-type stability program, which couples a Newton-Raphson power-flow program w i t h a fourth-order Runge-Kutta proce-

Analysis, Chaps. 9 a n d 10, McGraw-Hill, Inc. N ew York, 1 968. 4

For further informa t ion, see G. W . S t agg and A . H. EI-Ab iad, Computer Methods

in

Power System

16. 1 1

F ACTORS AFFECTING TRANSIENT STABILITY

Computer printout of swing curves for machines 1 and 2 of Examples 16.9 to 1 6 . 1 1 for clearing at 0.225 and 0.05 S

743

TABLE 16.7

Clearing at 0.225 Time

Mach. 1 angle

s

Clearing at 0.05 Mach. 2 angle

0.00 0.05 0.10

20.8 25 . 1 37.7

1 6.2 1 6.6 1 7.6

0.15 0.20 0.25 0.30 0.35 0.40 0 . 45 0.50 0.55 0.60 0.65 0.70 0.75

58.7 88. 1 123.1 151.1 175 .5 205 . 1 249.9 3 1 9.3 40 1.0 489.9 566.0 656.4 767.7

1 9.0 20.3 20.9 1 9.9 1 7. 4 1 4. 3 1 1 .8 1 0. 7 1 1 .4 1 3.7 1 6.8 1 9.4 20.8

Time 0 .00

0.05 0. 1 0 0. 1 5 0.20 0.25 0.30 0.35 0.40 0.45 '

0.50 0.55 0.60 0.65 0.70 0 . 75

Mach. 1 angle

s

Mach. 2 angle

20.8 25 . 1 32.9

1 6.2 1 6.6 1 7.2

37.3 36.8 31.7 23.4 1 4.6 8.6 6.5

1 7.2 1 6.7 1 5 .9 1 5 .0 1 4.4 1 4.3 1 4. 7 1 5 .6 1 6.4 17.1 1 7.2 16.8 1 6.0

1 0. 1 1 7. 7 26.6 34.0 37.6 36.2

dure. It is interesting to compare the closeness of the hand-calculated values of Table 16.6 with those for machine 2 i n Table 16 . 7 for the case where the fault is cleared at 0.225 s. The assumption of constant admittances for the loads allows us to absorb these admittances into Y bu s and thereby to avoid power-flow calcul ations, which are required when more accurate solutions using Runge-Kutta calculations are desired. The l atter, being of the fou rth order, require four iterative power-flow computations per time step. 1 6 . 1 1 FACTORS AFFECTI N G T RAN SIENT STABILITY

Two factors which indicate the relative stability of a generating unit are ( 1 ) the angular swing of the machine during and following fault conditions and (2) the critical clearing t ime. It i s a p p a r e n t from t his chapter t h a t b o t h t h e H con s t a n t and the transient reactance X� of the generating unit have a direct effect on both of these factors. Equations (16.84) and (16.85) show that the smaller the H constant, the larger the angular swing during a ny time interval. On the other hand, Eq. (1 6.36) shows that Pmax. decreases as the transient reactance of the machine increases. This is so because the transient reactance forms part of the overall series reactance which is the reciprocal of the transfer admittance of ! he system.

744

CHAPTER 16 POWER SYSTEM STABI LITY

Examination of Fig. 16. 1 1 shows that all three power curves are lowered when Pmax is decreased. Accordingly, for a given shaft power Pm ' the initial rotor angle D o is i ncreased, D max is decreased, and a smaller difference between D o and Dcr exists for a smaller Pmax ' The net result is that a decreased P constrains a machine to swing through a smaller angle from its original position before it reaches the critical clearing angle. Thus, any developmen ts which lower the H constant and increase transient reactance X� of the machine cause the critical clearing time to decrease and lessen the probability of maintaining stability under transient conditions. As power systems continually increase in size, there may be a corresponding necd for higher-rated generating u ni ts. These larger un its have advanccd cooling systems which allow higher-rated capacities without comparable increase in rotor size. As a result, H con s ta n ts continue to decrease with potential a d v e rs e i m p a c t on generating u n i t stability. At the same time this uprating process tends to result in higher transient and synchronous reactances, which makes the task of designing a reliable and stable system even more challenging. Fortunately, stability con t ro l techniques and transmission system designs h ave also been evolving to increase overall system stability. The con trol schemes include: rn a)(.

• • • •

Excitation systems Turbine valve control Single-pole operation of circuit breakers Faster fault clearing times

System design strategies aimed at lowering system reactance include: • • •

Minimum transformer reactance S eries capacitor compensation of l ines Additional transmission lines

When a fault occurs, the voltages at all buses of the system are reduced. At generator terminals the reduced voltages are sensed by the automatic voltage regulators which act within the excitation system to restore generator term i n al voltages. The general effect of the excitation system is to reduce the initial rotor angle swin g fol lowing the fau l t . This is accomplished by boosting the voltage applied to the field winding of the generator through action of the amplifiers i n the forward path of the voltage regulators. The increased air-gap fl ux exerts a restraining torque on the rotor, which tends to slow down its motion. Moder n excitation systems employing thyristor controls can respond rapidly to bus-volt­ age reduction and can effect from 0.5 to 1 .5 cycles gain in critical clearing times for three-phase faults on the high-side bus of the generator step-up transformer. Modern e lectrohydraulic turb i n e-governing systems have the ability to close turbine valves to reduce unit acceleration during severe system fault's near

16. 1 2 SUMMARY

745

the unit. Immediately upon detecting differences between mechanical input and electrical output, control action initiates the valve closing, which reduces the power input. A gain of 1 to 2 cycles in critical clearing time can be ach i eved. Reducing the reactance of the system during fault conditions i n cre ases rIP and decreases the acceleration area of Fig. 16. 1 1 . The possibility of maintaining stability is thereby enhanced. Since single-phase faults occur more often than three-phase faults, relaying schemes which allow independent or selective circuit-breaker pole operation can be used to clear the faulted p h ase while keeping the unfaulted phases intact. Separate relay systems, trip coils, and operating mechanisms can be provided for each pole so as to mitigate s tuck­ breaker contingencies following three-phase faults. Independent-pole operation of critical circuit breakers can extend the critical clearing time by 2 to 5 cycles depending on whether one or two poles fail to open under fault conditions. Such gain in critical clearing time can be important especially if backup clearing times are a problem for system stability. Reducing the reacta nce of a transmission l ine is another way of ra i sing P Compensating for line reactance by series capacitors is often an economi­ cal means of increasing stability . Increasing the number of parallel lines be­ tween two points i s a common means of reducing reactance. W hen parallel transmission l ines are used instead of a single line, some power is transferred over the remaining l i ne even during a three-phase fault on one of the lines-un­ less the fault occurs at a paralleling bus. For other types of faults o n one l ine more power is transferred during the fault if there are two l ines in parallel t h an is transferred over a single fau lted line. For more than two lines i n p arallel the power transferred during the faul t is even greater. Power transferred into the system is subtracted from power input to the generator to obtain accelerating power. Thus, the more power is transferred into the system during a fault, the lower the acceleration of the machine rotor and the greater the degree of stability. m ax

max .

16.12

S U lVI MAR Y

with e l e m e n t ar y p r i n c i p l e s o f r o t a t i o n a l m o t i o n , t h e swi n g e q u a t i o n gove rn i n g the electromechanical dynamic behavior of e a c h generating unit is d evel oped . The swing equ ation is shown to be nonlinear because the electrical power o utput from the generating unit is a nonl inear function of the rotor angle. B ecause of this nonlinearity, i terative step-by-step methods of solution of the swing e qua­ tion are generally required. In the special case of two finite machines (or one machi ne operating into an infinite bus) the equal-area criterion of stability can be used to calculate the critical clearing angle. It is shown, however, that fi nding the critical clearing time (which is the maximum elapsed time from the initiation of a faul t until its isolation such that the system is transiently stable) generally requires a numerical solu tion of the swing equation.

Th i s c h a p t e r p res e n t s t h e b , I S i cs o r r)Ower sys t e m s t a b i l i t y a n a l ys i s . S t a r t i n g

746

CHAPTER 16 POWER SYSTEM STABI LITY

Classical stability studies and their u nderlying assumptions are explained for the multimachine case and a simple step-by-step procedure for solving the swing equations of the system is illustrated numerically. A basis is thereby provided for further study of the more powerful numerical techniques employed in industry-based production-type computer p rograms. Transient stability of the power system is affected by many other factors related to the design of the system network, its protection system, and the control schemes associated with each of the gene rating units. These factors are discussed in summary form. PROBLEM S 16. 1 .

1 6.2.

1 6.3.

1 6.4. 16.5.

A 60-Hz fou r-pole t urboge n e r a t or r a t e d SOO M V t-\ , 2 2 kV h a s an inert ia consta n t o f H = 7.5 MJ jMVA. Find ( a ) the kinetic e nergy stored i n th e rotor a t syn­ chronous speed and (b) th e angular accel eration if the e lectrical power developed is 400 MW when the input l ess the rota tiona l losses is 740,000 hp. If the accelerat ion computed for the ge ne rator described in Prob. 1 6 . 1 is constant for a period of 1 5 cycles, find the ch a nge in (5 in electrical d eg r e e s i n that period and the speed in revolutions per mi nute at t he c n d of 1 5 cycles. Ass u m e that the generator is synchronized with a large system a n d has n o acce lerating torq u e before the 1S-cycle period begi ns. T h e generator of Prob. 16.1 is d e l iveri ng rated megavol tamperes at 0.8 power-fac­ tor l ag when a fault reduces the e lectric power output by 40%. Determi ne the accelerating torque in n ewton-meters a t the time the fault occurs. Neglect l osses and assume constant power inpu t to the shaft. Determine the WR 2 of the generator of Prob. 16. 1 . A generator having H 6 MJ j MVA is connected to a synch ronous motor h av i ng H 4 MJ jMVA through a network of reactan ces. The generator is de l ivering power of 1 .0 per un it to the motor when a fault occurs which reduces the del ivered power. At the time whe n the reduced power del ivered is 0.6 per u n i t, d e t e r m i n e the a n g u l a r acce l e r a t i o n o f t h e generator w i t h respect t o t h e motor. =

=

16.6. A

pow er sys t e m i s i dc n t ica l t o t h a t o j E x a m p l e I () . :\ , e xce p t t i 1 < 1 l l i l e i m p e d a n c e or

e ach o f the paral lel t ra n s m i ss i o n l i n e s is ) 0 . 5 , l ml t h e d e l ive r e d po e r is 0.0 per u n i t when both the termi nal voltage of th e mach ine and the vol tage of the infinite b us are 1 .0 per unit.' De termin e the power-a ngle equation for t he system dur ing the specified operating conditions . I f a three-phase fau It occurs on th e power syste m of Prob. 1 6.6 at a point on one of the transmission l i nes at a d i s t a nce of 30% of the l i n e length away from the sending-end terminal of t h e l i ne, dete rmine ( a ) the power-an gle equation du ring the fault and (b) t he swi ng equation. Assume that the system is operating under the .conditions specified i n P rob. 1 6.6 when the fault occu rs. Let H = 5 . 0 MJ jMVA, as in Example 1 6.4. Series resistance in the transmission network resu l ts in positive val ues for Pc and y in Eq. ( 16.80). For a given e lectrical power output, show the effects of resistance on the synchronizing coefficient Sp ' the frequ ency of rotor oscillations, and the da mping of these oscillations. w

. 16.7.

1 6.8.

PROBLEMS 1 6.9.

16.10.

1 6. 1 1 .

1 6 . 12 .

16.13.

1 6. 1 4.

16.15.

A generator having H = 6.0 MJ /MVA is delivering power of 1 .0 per unit to an infinite bus through a purely reactive network when the occurrence of a fault reduces the generator output power to zero. The maximum power that could be delivered is 2.5 per unit. When the fault is cleared, the original network condi­ tions again exist. Determine the critical clearing angle and critical clearing time. 60-Hz generator is supplying 60% of P to an infinite bus through a reactive network. A fault occurs which increases the reactance of the network between the generator internal voltage and the infinite bus by 400%. When the fault is cleared, the maximum power that can be delivered is 80% of the original maximum value. Determine the critical clearing angle for the condition described. If t h e generator of Prob. 1 6. 1 0 has an inertia constant of H = 6 MJ/MVA and Pm ( equ al to 0.6 Pma) is 1 .0 per-unit power, fi n d the critical clearing time for the condit ion of Prob. 1 6. 1 0. Use 6.t = 0.05 to plot the necessary swing curve. For the system and fault conditions described in Probs. 1 6.6 and 1 6.7, determine t h e power-angle equation if the fault is cleared by the simultaneous opening of breakers at both ends of t he faulted line at 4.5 cycles after t he fault occurs. Then,

A

max

plot the swing curve of the generator t hrough t = 0.25 s. Extend Table 1 6.6 to fin d 0 a t t = 1 .00 s. Calcu late the swing cu rve for machine 2 of Examples 1 6.9 th rough 1 6. 1 1 for fault clearing at 0.05 s by the method described in Sec. 1 6.9. Compare the resul ts with t he values obtained by the production-type program and l isted in Tabl e 1 6.7. If the three-phase fault on the system of Example 1 6.9 occurs on line G) - � at bus G) and is cleared by the simultaneous opening of breakers at both ends of the line at 4.5 cycles after the fault occurs, prepare a table like that of Table 1 6.6 to plot the swing curve of machine 2 t hrough t 0.30 s. By applying the equal-area criterion to the swing curves obtained in Examples 1 6.9 and 1 6. 1 0 for mach ine 1, (a) derive an e quation for t he critical c1ea�ing angle, ( b ) solve t he equation by trial and error to evaluate ocr and (c) use Eq. ( 1 6.72) to find the critical clearing time. =

1 6. 1 6 .

747

APPENDIX

A

TABLE A.l

Typical range of transformer reactancest Power transformers 25,000 kVA and l a rger

Nominal system voltage, kV 34 . 5 69 1 15 1 38 161 230 345 500 700

t

Forced-air-cooled, % 5-8

Forced-oll-cooled, % 9-14

6- 1 0

1 0- 1 6

6- 1 1

1 0-20

6-13

1 0-22

6- 1 4

1 1-25

7- 1 6

1 2-27

8-1 7

1 3-28

1 0-20

1 6-34

1 1-2 1

1 9-35

P erce nt on rated kilovoltampere base. Typical transformers Ilre now designed for the minimum

1 %.

reactance value shown. Dis tribution transformers have considerably lower reac tance. Resistances of transformers are usually lower than

748

APPENDIX A

749

TABLE A.2

Typical reactances of three-phase synchronous machinest Values are per unit. For each reactance a range of values is l isted below the typical value*

Turbine-genera tors 2-pole Conventional cooled 1 .76

Xd

1 .7- 1 .82

Xq

1 .66 . 1 .63- 1 .69

X�

0.2 1

X" d

Xl

0. 1 8-0.23

Salient-pole generators

4-pole

Con ductor cooled

1.95

Conven tiona I cooled

Conductor cooled

1 . 38

1 . 87

l .7 2-2. 1 7

1 . 2 1 - 1 . 55

1 .6-2. 1 3

1 . 93

1 .3 5

\ .7 1 -2. 1 4

1 . l 7 - 1 . 52

0.33 0.264 0 . 3 8 7

0.26 0.25-0.27

With dampers

Wi t h o u t

da mpers

1

0.6- 1 .5

1

0.6- 1 . 5

1.82

0.6

0.6

1 . 56-2.07

0.4-0.8

0.4-0.8

0.4 1

0.32

0.32

0.3 5-0.467

0.25-0.5

0. 25-0. 5

0. 1 3

0.28

0. 1 9

0.29

0.2

0.30

0. 1 1 -0. 1 4

0.23-0. 3 2 3

0. 1 84-0. 1 97

0.269-0.32

0. 1 3-0.32

0.2-0. 5

= X�

0.2

0.40

0. 1 3-0.3 2

0.30-0.45

= X�

= X�

= X�

Xo � t

O a t a furnished by ABB Power T & 0 Company, Inc.

§Xo

i

Reactances of older machines will generally be close to minimum values. varies

X:;.

so critically with

from 0 . 1 to 0.7 of

armature winding pitch that a n average value can hardly be given .

Variation

is

TABLE A.3

Electrical characteristics of bare aluminum conductors steel-reinforced

(ACS R)t

Resistance Reactance per conductor I -it spacing, 60 H z A c , 60 H z

Code word

Aluminum

Stranding

Layers of

area, cmil

AI/St

a l um i nu m

I

,

�

Capacitive

GMR D., (t

I n ductive

X ,

Xa , n/mi

M n · mi

Qutsid.e

Dc, 20°C,

20°C,

50 °C

diame�r, in

n/ 1 ,OOO ft

fl/mi

n/mi

0 . 609

0 . 0646

0 . 34 8 8

0 . 38 3 1

0 . 0 1 98

0 . 476

0 . 1 09 0

0 . 64 2

0 . 0640

0 . 34 5 2

0 . 3792

0 . 02 1 7

0 . 4 65

0 . 1 07 4

Waxwi ng

266 , 800

1 8 /1

2

Partridge

266 , 800

26/7

2

Os t ric h

3 00 , 000

26/7

2

0 . 68 0

0 . 0569

0 . 3070

0 . 33 7 2

0 . 02 2 9

0 . 4 58

0 . 1 05 7

336 , 400

18/1

2

0 . 684

0 . 05 1 2

0 . 2767

0 . 3037

0 . 02 2 2

0 . 462

0 . 1055

336 , 400

26/7

2

0 . 72 1

0 . 0507

0 . 27 3 7

0 . 3006

0 . 0243

0 . 451

0 . 1 04 0

Oriole

336 , 400

30/7

2

0 . 74 1

0 . 0504

0 . 27 1 9

0 . 2987

0 . 02 5 5

0 . 44 5

0 . 1 03 2

Merlin

Li nnet

Chickadee

397 , 500

18/1

2

0 . 743

0 . 04 3 3

0 . 2342

0 . 2572

0 . 024 1

0 . 45 2

0 . 1031

Ibi�

397 , 500

26/7

2

0 . 783

0 . 04 3 0

0 . 2323

0 . 2551

0 . 0264

0 . 44 1

0 . 1015

Pelican

4 77 , 000

1 8/ 1

2

0. 814

0 . 03 6 1

0 . 1 957

0 . 2148

0 . 0264

0 . 441

0 . 1 004

Flicker

477 , 000

24/7

2

0 . 84 6

0 . 0359

0 . 1943

0 . 2 1 34

0 . 0284

0 . 432

0 . 0992

Hawk

477 , 000

26/7

2

0 . 858

0 . 03 5 7

0. 193 !

0 . 2 1 20

0 . 0289

0 . 4 30

0 . 0988

Hen

477 , 000

30/7

2

0 . 883

0 . 0355

0 . 1919

0 . 2 1 07

0 . 0304

0 . 424

0 . 0980

Osprey

556 , 500

1 8/ 1

Parakeet

556 , 500

24/7

)

2

0 . 879

0 . 0309

0 . 1 679

0 . 1 84 3

0 . 0284

0 . 432

0 . 09 8 1

2

0 . 9 14

0 . 0308

0 . 1 669

0 . 1 83 2

0 . 03 0 6

0 . 4 23

0 . 09 6 9 0 . 09 6 5

Dove

556 , 500

26/7

2

0 . 927

0 . 0307

0 . 1 663

0 . 1 826

0 . 03 1 4

0 . 420

Rook

636 , 000

24/7

2

0 . 97 7

0 . 0269

0 . 146!

0 . 1 603

0 . 03 2 7

0. 415

0 . 09 5 0

G ros benk

636 , 000

26/7

2

0 . 990

0 . 0268

0 . 1 4 5-1

0 . 1 596

0 . 0335

0. 412

0 . 09 4 6

Drake

795 , 000

26/7

2

1 . 108

0 . 02 1 5

0 . 1 1 72

0 . 1 284

0 . 0373

0 . 39 9

0 . 09 1 2

Tern

795 . 000

45/7

3

1 . 063

0 .0217

0 . 1 1 88

0 . 1 302

0 . 03 5 2

0 . 406

0 . 09 2 5

Rail

954 , 000

45/7

Card i n a l

9 5 4 , 000

54/7

1 , 033 , 500

:.I

Ortolan Bl uej a y

1 , 1 1 3 , 000

Finch

1 , 1 1 3 , 000

Bobolink

1 , 4 3 1 , 000

Pl o v er

1 , 4 3 1 . 000

Falcon

1 , 590 , 000

Bluebird

2 , 1 56 , 000

Bittern

1 , 27 2 , 000

PbeB.Ba n t

1 . 27 2 . 000

Lap","ing

1 , 590 , 000

� Data,

,

t Most used multilayer sizes. by perm i ss io n

I ,

from Aluminum

1 . 1 65

0.0181

0 . 09 9 7

0 . 1 092

0 . 0386

0 . 39 5

0 . 0897

3

1 . 1 96

0 . 0 1 80

0 . 0988

0 . 1 082

0 . 0402

0 . 39 0

0 . 08 9 0

45/7

3

1 . 213

0 . 0 1 67

0 . 092·1

0. 1011

0 . 0� 0 2

0 . 390

0 . 0885

45/7

3

1 . 259

0 . 0 1 55

0 . 086 1

0 . 094 1

0 . 04 1 5

0 . 386

0 . 0874

54/ 1 9

3

1 . 29 3

0 . 01 5 5

0 . 0856

0 . 0937

0 . 0436

0 . 380

0 . 0866

45/ 7

3

1 . 345

0 . 0136

0 . 0762

0 . 0832

0 . 04 4 4

0 . 378

0 . 08 5 5

54/ 1 9

3

1 . 382

0 . 0135

0 . 07 5 1

0 . 0821

0 . 0466

0 . 372

0 . 08 4 7

45/7

3

1 . 427

0. 0121

0 . 068-1

0 . 0746

0 . 04 7 0

0 . 37 1

0 . 08 3 7

0 . 0673

0 . 0735

0 . 04 9 4

0 . 365

0 . 08 2 9

O . 062:l

0 . 0678

0 . 04 9 8

0 . 364

0 . 08 2 2

54/ 1 9

3

1 . 465

0 . 0 1 20

45/7

3

1 . 502

0 . 0109

54/ 1 9

3

1 . 54 5

0 . 0 1 08

84/ 1 9

4

1 . 762

0 . 0080

A ssociation,

1

0 . 06 1 2

0 . 0667

0 . 0523

0 . 358

0 . 08 1 4

0 . 0476

0 . 05 1 5

0 . 0586

0 . 344

0 . 0776

A luminum Electrical ConductQr Handbook, 2nd ed., Washington, D . C . , 1 98 2 .

TABLE A.4

Inductive reactance spacing factor

Xd

a t 60 Hzt (oh m s per mile per conductor) Sepnrution

Inches Feet

0 1 2 3 4 5 6 7 8 9 10 11 12 13

14

15 16 17 18 19 20 21 22 23 24 '25 26

27

28 29

30 31

32 33 34 35 36 37 38 39 40

41 42 43

44 45

46 47 48 49

0

. . . . . . 0 0 . 01',4 1 0 . 1 333 0 . 1682 0 . 1 953 0 . 2 1 74 0 . 2361 0 . 2523 0 . 2666 0 . 2794 0 . 29 1 0 0 . 30 1 5 0 . 3112 0 . 3202 0 . 3286 0 . 3364 0 . 3438 0 . 3 507 0 . 3573 0 . 3635 0 . 3694 0 . 37 5 1 0 . 3805 0 . 3856 0 . 3906 0 . 3953 0 . 3999 0 . 4043 0 . 4086 0 . 4 1 27 0 . 4 1 67 0 . 4205 0 . 4243 0 . 4279 0 . 43 1 4 0 . 4 34 8 0 . 4382 0 . 44 14 0 . 4445 0 . 4476 0 . 4 506 0 . 4535 0 . 4&64 0 . 4592 0 . 4619 0 . 4646 0 . 4672

1

- 0 . 30 1 .'; 0 . 0097 0 . 08 9 1 0 . 1366 0 . 1707 0 . 1 973 0 . 2191 0 . 2376

2

- 0 . 2 1 74 0 . 0 1 87 0 . 0938 0 . 1 399 0 . 1732 0 . 1993 0 . 2207 0 . 2390

3

- 0 . 1 682 0 . 027 1 0 . 0984 0 . 1 4 30 0 . 1 756 0 . 20 1 2 0 . 2224 0 . 2404

4

1)

6

7

8

9

10

11

- 0 . 1 333 0 . 03 4 9 0 . 1 028 0 . 1461 0 . 1 77 9 0 . 20 3 1 0 :2240 0 . 24 1 8

- 0 . 1 062 0 . 04 2 3 O . 107 1 0. 1491 0 . 1 802 0 . 2050 0 . 2256 0 . 24 3 1

- 0 . 084 1 0 . 04 9 2 0. 1 1 12 0 . 1 520 0 . 1 825 0 . 2069 0 . 2 27 1 0 . 24 4 5

- 0 . 0654 0 . 0558 0 . 1 1 52 0 . 1 54 9 0 . 1847 0 . 2087 0 . 2287 0 . 2458

- 0 . 0492 0 . 0620 0 . 1 1 90 0 . 1 577 0 . 1869 0 . 2 1 05 0 . 2302 0 . 24 7 2

- 0 . 03 4 9 0 . 0679 0 . 1 22 7 0 . 1 604 0 . 1891 0 . 2 1 23 0 . 23 1 7 0 . 24 85

- 0 . 0221 0 . 07 3 5 0 . 1 26 4 0 . 1 63 1 0. 1912 0 . 2 14 0 0 . 23 3 2 0 . 24gS

- 0 . 0 1 06 0 . 0789 0 . 1 29 9 0 . 1 65 7 0 . 1 93 3 0 . 2 1 57 0 . 2347 0 . 25 1 1

At 60 Hz, in !l/mi per cond uctor X d = 0 . 2794 log d d = separati on, ft For three-phase li nes d = D.q

0 . 4697 0 . 4722

tFrom Electrical TraltSmission and Distribution Reference Book, by permission

of the ABB Power T & D Comp any, I nc .

TABLE A.S

Shunt capacitance-reactance spacing factor Xd at 10 Hz

(megaohm-miles per conductor)

Sepsration Feet

0

Inches

1

2

3

4

5

6

7

8

9

10

11

- 0 . 0 1 20 0 . 0 1 52 0 . 0291 0 . 0385 0 . 04 57 0 . 05 1 5 0 . 0563 0 . 0604

- 0 . 0085 0 . 0 1 66 0 . 0300 0 . 0392 0 . 0462 0 . 05 1 9 0 . 0567 0 . 0608

- 0 . 0054 0 . 0180 0 . 0309 0 . 0399 0 . 0467 0 . 0523 0 . 0570 0 . 06 1 1

- 0 . 0026 0 . 01 9 3 0 . 0318 0 . 0405 0 . 0473 0 . 0527 0 . 0574 0 . 06 1 4

,

0

1 2 3 4

5 6 7 8

9 10 11 12 13 14 15 16 17 18 19

20 21 22

23 24

25 26 27 28 29 30 31 32 33 34

35

36 37 38 39 40 41 42 43 44 45 46 47 48 49

. . . . . . 0 0 . 0206 0 . 0326 0 . 04 1 1 0 . 0478 0 . 0532 0 . 0577 0 . 06 1 7 0 . 0652 0 . 0683 0 . 07 1 1 0 . 0737 0 . 0761 0 . 0783 0 . 0803 0 . 0823 0 . 084 1 0 . 0858 0 . 0874 0 . 0889 0 . 0903 0 . 09 1 7 0 . 0930 0 . 0943 0 . 0955 0 . 0967 0 . 0978 0 . 0989 0 . 0999 0 . 1009 0 . 1019 0 . 1028 0 . 1037 0 . 1046 0 . 1055 0 . 1063 0. 1071 0 . 1 079 0 . 1087 0 . 1094 0 . 1 102 0 . 1 109 0 . 1 1 16 0 . 1 123 0 . 1 129 0 . 1 1 36 0 . 1 142 0 . 1 149 0 . 1 155

From Electrical

- 0 . 0737 0 . 0024 0 . 02 1 8 0 . 0334 0 . 04 1 7 0 . 0482 0 . 0536 0 . 0581

Transmission

- 0 . 0.S32 0 . 00!6 0 . 0229 0 . 0342 0 . 04 23 0 . 0487 0 . 0540 0 :0584

- 0 . 04 1 1 0 . 0066 0 . 0241 0 . 0350 0 . 0429 0 . 0492 0 . 0544 0 . 0588

- 0 . 0326 0 . 008 5 0 . 02 5 1 0 . 0357 0 . 0435 0 . 0497 0 . 0548 0 . 05 9 1

- 0 . 02 60 0 . 0 1 03 0 . 0262 0 . 0365 0 . 04 4 1 0 . 050 1 0 . 0552 0 . 0594

- 0 . 0206 0 . 0 1 20 0 . 0272 0 . 0372 0 . 0446 0 . 0506 0 . 0555 0 . 0598

- 0 . 0 1 60 0 . 0 1 36 0 . 0282 0 . 0379 0 . 0452 0 . 05 1 0 0 . 0559 0 . 060 1

At 60 Hz:. in M O ' mi per conductor X d' = 0.06831 log d d = separati on, ft For three-phlUle l i nes d .. D e q

.

.

and

Distribution Reference Book. by

permission of the ABB Power T & 0 Company. Inc.

APPENDIX A

753

TABLE A.6

ABeD constants for various networks VV'v

Is

+t �

IR

Z

-

t 1-

-

Vs

u

c -

VR Series i m pedance

Is

A B c D

+

= = = =

1 Z o

1

IR

---to-

---

+

A = 1 B = O c = y D 1

+

I

y

Vs

VR

=

S h u n t a d m i tt a n ce

IR

Is +

-�

+

i

Vs I

U n symmetrical

U n sy m metrica l

1 + Y2Z Z Y1 + Yz + Z Y1 Y2 1 + YIZ

7r

A \ A 2 + RIC, A IB2 + BID! C = A 2CI + CzDI

A

+ +---+ +

B

+ �---.

A j Bj CjDj

+

+

N etworks i n para l l e l

=

D = BzCI + DID1

N etworks in cascade

--

= = = =

-+

.---� +

A B C D

IR

+

+

= 1 + YZI = Zl + Z2 + YZ1Z2 = Y = 1 + YZ2

T

Z

Is

A B C D

=

+

+

A = (A IBz A1BI)/(B1 B 2) B = BIBd(BI + B1) C = CI + C2 + (A I - A 2) (D2 - DI) /(B1 + B,) D == (BzD1 + BID1)/ (B1 + B2)

754

APPENDIX A

A.l DISTRIBUTED WINDINGS OF THE SYNCHRONOUS MACHINE

The field and armature windings of the synchronous machine described in Sec. 3 . 1 are d istributed in slots around the periphery of the air gap. In the round-rotor case we now show that these windings can be treated as concen­ trated coils in the idealized mode l of the machine. Figure A. l(a) represents a two-pole machine of length I with a uniform air gap of width g , which is very much smaller than the radius r of the round rotor. Compared to the air gap, the stator and rotor iron parts have very high m agnetic p ermeability and therefore have negligibly small magnetic field i nten­ s i ty A d istributed a phase winding of the type used in the armatu re of the three-phase synchronous machine has a total of No turns set in to, say, four slots per pole as shown. All three phases a , b, and c of the armature therefore occupy a total of 24 slots that are symmetrically d istributed around the periph­ ery of the air gap. Cyli ndrical coordina tes z , x , and 8d are used to measure: .

1. 2.

3.

Distance z along the length 1 of the machine, Radial d istance x perpendicular to the length, and Angular displacement 8d counterclockwise around the air gap.

Because the air-gap width g is so small relative to the radius r of the rotor, for g iven values of z and 8d we can indicate a position in the air gap by setting x = r. The net axial current in each p h ase is zero, as i ndicated by the equal n umber of dots and crosses. The dot (and the cross) i ndicates positive d irection of current flow out of (and into) the plane of the paper in Fig. A. I . When positive current ia flows i n the winding, the flux pattern of Fig. A . 1 ( b ) is produced with north (N) and south (S) poles as shown. The flux crossin g the air gap passes radially through the elemental area r X ded X dz shown in Fig. A. 1(a). The flux l ines are consistent with the right-hand rule of Sec. 2 . 1 , and such flux l inking the winding is considered positive: In this way self-inductance is positive. Let Ha«()d ) denote the air-gap d istribution of the magnetic field intensity due to ia. Ha«()d) may be determined i n a straightforward manner by applying Ampere's and Gauss's l aws as now explained: I •

Unrolling the air gap as depicted i n Fig. A.2( a ) yields the developed d iagram of F ig. A.2(b ), which shows the conductors located around the periphery.

Electromechanics, The R o nald Press Co., New York, NY, 1 965 . iT-his section follows t h e development i n Chap.

9 o f N . L. Schmitz a n d D. W . Novot ny,

IntrodJ1ctory

A.I

DISTRIBUTED WINDINGS OF THE SYNCH RONOUS MACHINE

755

Air gap

(a)

N

Air gap

s

(b) FI G U RE A . I

Rep rese ntation of a two-po le-nons a l i e n t rotor mach ine showing: ( a ) elemental port ion o f a n i m aginary air-gap su rface su rrou nd ing the rotor; ( 6 ) t h e pattern o f fl u x d u e t o positive current i a into (cross) and out of (dot) plane of the p aper.

756 '

APPENDIX A

/��'<�':"'<'=':" I' ()d

=

0

'�"y ''''

: ....'

FIG U RE A.2 ( a ) U n rolling the a i r gap of Fig. A . l ; ( b ) developed d i agram showing p a t h mllfJq ; ( c ) conductor-cou n t i n g fu nc­ tion n }OJ); (d) w i n d i n g fu n c t io n Na(Od ); ( c ) m a gnetic field i n t e n s i ty Ha(8d) due to curre n t i Q •

(a)

•

Ampere's law applied along closed paths such as that marked mnpq In Fig. A.2( b) gives

¢

m npq

H · ds

=

net current enclosed

(A.1)

Those portions of the paths i n the h ighly permeable parts o f the rotor and stator make zero contribution to the line integral because H is considered negligible in iron compared to air. Let us choose the positive direction of H to be inward from the stator to the rotor. In the air gap at angular displacement 8d t h e line i ntegral of the radial field intensity Ha(8d) along the path m-n yields Ra(8d) X g . Likewise, the contribution is Ha(O) X g along the air-gap portion of path p�q at 0° angular d isplacement. Equation (A. I) then gives -

( A .2) •

For each choice of 8d in Fig. A.2(b) the net current enclosed is obtained simply by counting the number of dots and crosses of the conductors inside the path mnpq. Counting co n d u ctors i n this manner, w e obtain ( A .3 )

where n a(8d) d enotes the number of conductors with dots minus the number with crosses inside the path of integration. Figure A.2(e) shows a sketch of ni6d) assuming that the current changes from zero to ia (dot) o r i a (cn)ss) at the center of the conductor. -

A.I

..

�p

(b)

g

T Na

(c)

4 Na

q

i

-

I

I

--

1 1 1 1

o

8d

- - - - - - - - - - - --r-----r--� l

-:-I I

- - -----

-

2g

0

(c) Na i a

- -

-----

-:I

---

-------

--- -

-

--

1 -r- - -r-----I

--� 1 1

27T ----

27T

7T

I I

I

I I

I 1 1 I I I

Ra(8d)

I I I I

I I I 1

1 I

I 1

1- - - - -

- - - - - - -

1 I 1 1 1 I

.

8d

( ConI inued )

1 1 I I I

---.....,. ----- - 4 I I I I

1 1 I I

0

- - -- - - I

1

2g

FIGURE A...2

na(8d)

- ---------- �----- ------

o

Na i a

� - - --- - - - - i

- - l - --l---l- - - t - - - T - - - - -------

- --

r

4

m

1- _ _ - - -- - - -r----' - - - .L - _ _ ...J...._..,.. _ _ _ _ _ _ _ _ _ _ -I

4 I: 2 Na

-

Stator

r

o

(d )

DISTRIBUTED WINDI NGS OF THE SYNCHRONOUS MACHINE

+I I I I I I I I

7r

- - - - - -

- - - - - -

-1

I I I I 1 I

27T

757

758 •

APPEN D I X A

Rearranging Eq. ( A 3) we obtain .

,

( A A)

Gauss's law for magn etic fields states that the LOtal magnetic fl ux cross i n g back and forth in the air gap must be zero. Therefore, integrating H) 8d ) over a cylindrical surface surrounding the rotor just inside the air gap must yield zero, which means that the average value of H(J ( 8r1 ) m u s t be zero. Conse­ quently, thc avcrage v a l u e o f t h e r i g h t - h a n d s i d e o f E q . C A . 4 ) i s zero, and we obtain ( A.S)

I II - - fl a . av g

where n a ' av is the average value of n a( 8d ) around for HaCO) i n Eq. CAA) enables us to write

t he

periphery.

S u b st i t u t ing

( A .6)

The winding junction Na(8d) n aC8 d) 11 a , has zero average value a n d d iffers from n aC 8d ) by the constant value n a .av ' Hence, it can be graphically constructed by shifting the sketch of n a( 8d) vertically down by the amount n a, av so that it yields a zero-average value, as demonstrated in Fig. A . 2( d ). The magnetic field intensity Ha(8d) due to the current i a in the distributed winding is then found by mul tiplying Na( 8d ) by lalg , as shown in Fig. A.2 e e ) . =

-

av

The simple graphical construction of Fig. A . 2 g ives a s t a i rcase fu n c t i o n for and Ha(8d), but it is not difficult to envision that in an a ct u a l m ach ine the slot spacing and conductor layers of the distributed windings could be arranged so that the fundamental component of Na( 8d ) predominates and conforms to the s i n usoid a l d istributions shown in Fig. A.3(a). We note that the peak value of N/8d) is Na l 2 the number of turns per pole. Figure A.3(b) depicts the fundamental component of a flux-density distri­ bution labeled B/8d), which is considered to be caused by current i k in a n arbitrary d istributed winding k around t h e air gap. This winding may b e any one of the p h ases (including phase a i tself) or the field winding. Due to the flux density Bk(8d ), the elemental area (r X d 8d X dz ) on the rotor surface of Fig. A.ICa) has flux Bk(8d )( r X d8d X dz ) p assing normally through it in a radial direction. This flux links the stator winding a according to the spatial distfibu­ t ion Na(8d ), as n ow demonstrated.

Na(8d)

,

A.I

DISTRIBUTED WINDINGS O F THE SYNCHRONOUS MACHINE

759

. ... .

foo-....· ) Rotor

Stator

r---�---+----r--+---'

o

7T

0d l

(a)

(6)

0d 2

+ 1T

+

7T

0d3

+

0d 4 +

1T

FIGURE A.3 Fu n d a mental sinusoidal compone n t s of: ( a ) w i n d i n g fu n c t io n current ik in distributed winding k .

1T

0d

N)Od); (6) flux density BkCOd) due to

Integrating the flux density B/ 8d ) over t h e entire length I o f the rotor a n d around the circumference of t h e a i r gap, w e obtain from Fig. A.3

( A .7) where A ak represents the t otal fl ux linkage with the a winding due to the current ik in wind.i ng k . Each cross section of the air gap along the axial length o f the machine has the same winding distribution and the same flux density Bk (8d ) . Hence, the bracketed term i n Eq. C A .7) is independent of z so that integration with respect to z yields

( A.8)

760 Let

APPENDIX A US

now consider the intervals of integration subdivided as follows:

( A .9)

Symbolically su mm ing the four preceding equations yields

( A . 10)

The rig h t h and side of Eq. (A. I0) can be written more compactly in terms of the function n a( 8d ) shown in Fig. A.2( e). Therefore, we have -

(A.11)

Substi tuting for

n ( 8d ) a

=

Na (8 d) + n a,

av '

we obtain from Eqs. (A.8) and (A. l } ) ( A . 12)

The seco nd integral term on the right-hand s ide of this equation equals zero since the average value of Bk(8d ) around the peri phery of the air gap is zero, and we obtain (A . 1 3)

A. I

1.

DISTRIBUTED WINDINGS OF THE SYNCHRONOUS MACHINE

Suppose that the current ik is actually in the a winding itself; that is, k Then, substituting in Eq. (A.13) for Ha(8d) from Eq. (A.6), we obtain

761 =

a.

(A. 14)

Considering only the fundamental component that

-

Na l cos 8 d of Na(8d), we find

(A. I S )

2.

where Lua is the self-inductance of a winding and Na l equals the effective turns per pole of the fundamental component. Note that Lao is a positive constan t value. Suppose that the current i k is actually in a distributed b winding identical to, but shifted 1200 from, the a winding. Setting k b in Eq. (A.1 3 ) and again considering only fundamental components, we obtain for the mutual induc­ tance L ab (or L b ) between windings =

This equation can be Simplified by using the identity cos a cos (3

=

- Hcos( a

-

(3 )

+

cos( a

+

( A . 1 7)

(3 ) ]

to yield

( A . 1 8) When Bk(8d) arises from current i in a n identical c winding displaced 2400 from the axis of the a phase, similar calculation shows that the mutual inductance L ac (or Lc) between the a and c windings is c

( A . 1 9) 3.

We note that Lab and L ac are equal negative constants. If B k ( fJd) is due to current if in the field winding a t the i nstant when its axis is at an arbitrary counterclockwise angle 8 d wi t h respect to the a wind ing, the mutual inductance Laf (or L f) due to the fun damental com po n e n t s of ,

762

APPENDIX A

the

spatial d istributions of the two windings is g iven by

Nfl is the turns per pole of t he fundamental component of the f winding

and the dummy symbol (J;, is used for the variable of integration because (Jd has an arbitrary but definite value in Eq. (A.20). Applying the trigonometric identity of Eq. (A. 1 7) yields

Integration over the 2rr interval gives 2rr co s (Jd for integrand and zero for the second term so that

th e fi rst

term of t he

( A .22) We note that L a! varies cosinusoidally with the position (Jd of the rotor. When the field winding rotates cou nterclockwise, it causes flux to link the stator windings in the sequence a-b-c, and the l inkage with phases b and c are exactly the same as with the a phase except that they occur later. This means that the field has exactly the same magnetizing effect o n t h e b p hase along the (J d = 1 20° axis and on the c phase along the (Jd = 240° axis as it has on the a phase at (]d 0°. Hence, the mutual i nductance L b! between the b phase and the field, and L e! between the c phase and the field, m ust be of the same form as Eq. (A.22); that is, =

( A .23)

-

'

( A .24)

Also, results similar to Eqs. (A.IS) and (A. 18) c a n be found for the b and c windings of the round-rotor machine. Summarizing, we have the fol lowing: •

Th e self-inductances o f the distributed armature windings are positive con­ sta nts, which we now call L s so that L =L s

aa

= Lbb = L

ec

=

fL o-rr rl

· - N 21

--

g .

a

( A .2S)

A2

P-TRANSFORMATION OF STATOR QUANTITIES

763

T h e mutual inductances between the armature windings are negative con­

•

stants, which we now call -M s so that

J-L o 7T rl

---

2g

•

Nll l 2

(A.26)

The mutual inductance between the field winding and each of the stator w indi ngs varies with the rotor pos i tio n ()d as a cosinusoidal function with maximum val u e Mf J.L o 7T rINa l NI l /g so t h a t =

( A-27)

In the round-rotor machine (and, indeed, in the salient-pole ma c h i ne also) the field winding has a constant self-inductance L ff. This is because the field winding on the d-axis produces flux through a s imilar magnetic path in the stator for all positions of the rotor (neglecting t h e small effect of armature slots). A.2 P-TRANSFORMATION OF STATOR QUANTITIES

To transform n-b-c s t a tor flux linkages to d-q-O quantities by means of matrix P of Eq. (3.42), rearrange the flux-linkage expressions of Eq . (3.41) as follows:

( A.28)

Now substitute for the a -b-c fl ux linkages and currents from Eqs. (3.43) to obtain

p- I

Ad Aq

Ao

Laa =

L ba L ea

Lab L bb

L e i>

L ac

L bc

L ec

p- l

'd lq

'0

+

La! Lb

J If

L eI

( A . 29)

764

APPEN D I X A

Multiplying across Eq. (A.29) by P gives A

d

Aq

Ao

According to Table

L aa

= p Lba

L bc p -

L bb

L ca

3. 1 ,

Lac

L ab

L ee

L e i>

lq lo

.

L af

+ p L bf I f Le

( A 30 )

i

we have

0

- L m

l

ld

o

0

0

1

1

2( ed : ) 2 - cos 2 ( ed + ; ] cos 2 ( ed + � )

ed ( ed + : ) COS2(ed 5:)

- cos 2

cos

cos 2

+

and i t can be shown that

ed

cos 8" sin 1

Ii

+

ed -

cos( 8" sine

1

Ii

1 20°) 1 20°)

1

1

1

(

cos 2 8j

cos2 ( ed

+ +

-

�)

3 2 Ti

)

cos( Btl 240°) s i n e Bd - 240°) 1.

Ii

( A .31 )

-T

( A . 32)

Substitut ing from E q s . (A.3 l) a n d ( A . 3 2) i n to Eq . (A. 3 D) and simplifying, we obtain Eq. (3 .44). To transfo rm a-b-c stator voltages to d-q-O quantities by means of matrix P, rearrange Eqs. (3.47) as follows: ( A .33)

A.2 P-TRANSFORMATION OF STATOR OUANTITIES

765

Multiplying across Eq. (A.33) by P gives, according to Eqs. (3. 43),

(A.34) Preserving order while using the derivative-of-a-product rule in Eq. (A.34), we obtain

(A .35)

Substituting for P from Eq. (A.32) and simplifying yield Eqs. (3.48).

APPENDIX

B

B.l

SPARSITY

AND NEAR-OPTIMAL ORDERING

Section 7.6 shows that the equivalent n etwork corresponding to the coefficient matrix at each step of the gaussian-elimination procedure has one node l ess than that for the previous step. However, the number of branches in t h e equivalent network a t each step depends on t h e order in which t h e nodal equations are processed. To see the effect of ordering, consider the graph of the four-bus system shown in Fig. B.1 ( a). In this example the off-diagonal elements are nonzero in row 1 and column 1 of Y b U because of the branches connected from bus CD to buses @ , G) , and 4 . All the remaining off-diagonal elements of Yb u s are zero since buses 2 ,' G) and @ are not directly con nected to one another. If bus CD is p rocessed first in g a u s s i a n e l i m i n a t i o n of the nodal equat ions, the effect on Yb us is similar to the usual Y -6. transforma­ tion symbolically indicated by

�

CD @ G) @ CD - X X X X @ X X X ® X X @ X Initial

Y bu s

=

Q)

Q)

@

W G) @

l: :J 0 X

0

...

Y bus after Step 1

,

in which the symbol X indicates an element which is initially nonzero and 0 represents a fill-in for a previously zero element. The fill-ins are symmetrical

Bol

®

(a)

SPARSITY AND NEAR-OPTIMAL ORDERING

767

®

/

®

@

/

/

/

/

/

/

/

/

/ I

/ \ \

\

___

\

rp

\

\Y

__

\

\

_ _ _ _ _

\

\

\

\

(b)

\

®

®

original removed fill-in

®

(c)

FIG URE B . l The effects of ordering on node e l i m ination: (a) the original graph and node n u mbers; ( 6 ) the three new branches a fter e l i m inating node CD ; (c) the reordered nodes. and

result from the Step Y:)(new)

=

1 c a l c u l a t io n s

1ji(neW) =

�j

� l Ylj -

i

Y1 1

and

j

=

2, 3, 4

(B.l)

because the subtractive term o n the right-hand side of this equation is nonzero. S ince fill-ins and branch additions are equivalent, the new branches joining node-pairs 0- G) , 0- @ , and ®- @ of Fig. B.1(b) represent the fill-ins of the new 3 X 3 Y b u s ' If we now renumber the nodes as shown in Fig. B.1(c) and solve the equations in the sequence of the new node numbers, we find after Step 1 that

CD @ G) @

CD 0 ® @ x

x

x x

x

x

x

I nitial Y h u s

x

x

x

=>

@ Q)

@

0 G) @

[: �l x

x

Y bus a fter Step 1

Therefore , by nod e renumbering fill-ins m ay be avoided and new branches need

not be added to the network graph. The above example demonstrates two general rules for altering the system graph when the voltage variable at node ® is eliminated in the course of gaussian elimination: 1.

Remove node ® and all its incident branches from the system graph. In particular, remove branch km i f node ® is radially con n eFted to the network, that is, if node ® is connected to only one other node @.

768

APPENDIX B

a new branch between each pair of nodes CD and CD i f, and only i f, they are both directly connected to node ® bu t not to each other before n o d e ® i s eliminated.

2. Add

Using these rules, we find that the network graph has one node less at each step of gaussian elimination and the number o f branches incident to the remaining no d e s may be made to vary. Our aim is to avoid adding branches if at all possible, which is equivalent to avoiding symmetrical fill-ins in the coefficient matrix. Hence, eliminating radially connected nodes and nodes with the fewest number of connected branches is evidently a desirable graphical strategy, which translates into the following scheme of ordering the system equations for gaussian elimination: •

•

At Step 1 of the forward elimination process, select the variable to be eliminated corresponding to the diagonal element of the row with the most zero e ntries. If two or more variables meet this condition, select one which causes the fewest fill-ins for the next step. At e ach subsequent step, select the variable to be eliminated by applying the same rule to the reduced coefficient m atrix.

This ordering scheme is said to be near-optimal because in practical use it gives excellent reduction in the accumulation of fill-ins even though it does not guarantee t he absolute minimum (optimum) number. To illustrate the ordering scheme graphically we: • •

•

Draw a graph corresponding to Y (thus omitting the reference node). At Step 1, select the first node for elimination from the graph which h as t h e fewest i ncident branches and which creates the fewest new branches a s determined b y rules ( 1 ) and (2). bus

At e ach subsequent step, update the count of branches at the remaining buses using rules (1) and (2) and then apply the Step 1 selection criterion to the u p d ated graph.

More than one ordered sequence of nodes may be found since two or more nodes with exactly the same number of i ncident branches may qualify for elimin a tion at some steps. In any event, limiting the branch count at a bus limits the fill-ins in the upper and lower (LU) factors of the corresponding coefficient matrix. Th ese graphical features o f gaussian e limination, which apply to any squ are m atrix with symmetrical nonzero pattern, are illustrated in the following num e rical examples. B.lo The grap h shown in Fig. B.2(a) is for a 5 5 Y bu s system (the reference node not e xp l i c i t l y represented). Graphically determine a sequen.ce i n which buses @ , Cfi) , 0 , @ , a n d 0 should b e numbered so as to minimize the n umber of fi ll-ins i n the L U factors of Y bu s . Example

X

B.l

SPARSITY AND NEAR·OPTIMAL O RDERING

769

Node @ of Fig. B.2(a ) with only one incident branch is chosen for Step 1 elimination accord ing to rule (1). With node @ elimin ate d and branch ca effectively removed ( indicated by thin line) as far as further calculations are concerned, we u pdate the cou nt of active branches (that is, heavy and fill-in lines) a t each remaining node. We then have nodes @ and 0 with three branches stilI active and nodes @) and ® with only two, Fig. B.2 ( b). The elimi na ti o n of either node @) or ® in Step 2 is equally acceptable for two reasons. First, each has the fewest active branches, and second, the elimination of either one will not create a new branch because they connect to two nodes @ and 0 , which are already connected to each other-rule (2). Let us agree to el iminate node @) i n S tep 2, in w h ich c a s e nodes @ , (£) , and 0 become equal candidates for Step 3 with two active branches remai n ing at each, Fig. B . 2( c ). In fact, nodes @ ' ® , and 0 can be e limin ated in any ord e r without creat ing fil l-ins, and so we label t hem nodes W , @ , and (3) , respectively, as shown in Figs. B . 2 ( d ) and B.2 ( e). W hen the nodal equations of the system are ordered according to the node numbers of Fig. B.2(c), the rows and col umns of t he corresponding Y nus assume the nonzero pattern Solution.

CD @ G) @ W

CD G) ® 0 Ci) x x

x

X X

x

x X

X

X

X

X

X

x X

X X

It is readily shown that successive elimination of the node variables in the order Ve , Vd , Va ' Vh , and Ve creates no fill-ins in the triangular factors L an d U. Thus, the original sparsi ty of Y b u s is p reserved in its factors. On the other hand, elimina ting i10des in the strict a l phabetical order of Fig. B.2( a) yields after node @ is e l iminated the 4 X 4 Y hlls given by @ ® CD @ (0

@ ® X

x

x

X

X

X

x

X

CD X

x

(if) (0 X

X

X

X

x X X

=

® CD @ 0

® CD @ 0 x

0

0

0

0 X

0

x

x

0

0

0

X

X

X

X

corresponding to Fig. B.2( f ) in 'w hich there are fou r fill-in branches. Thus, the alphabetical order of sol u tion would not be desirable in this example since the original sparsity woul d then be completely l ost and the triangul ar matrices L and U would require more storage and computer time for calculations. Number the nodes of the graph of Fig. B.3(a ) in aI1 optimal order for the triangular factorization of the correspond ing Ybll S ' Exa m ple

B.2.

-

770

APPENDIX B

® CD

CD

Cb)

G') J

I \

®

/ (" - - - - - .- - - - - \ - -

\

(d)

\

d

Fill-in

branch

®

@

®

"

\

cn

(e)

A graphical i l l ustra t i on of optimal ordering in Exa m p l e 8. 1 ; ( a ) t h e original gra p h ; ( 6 ) a ft e r S t e p I elimin ation o f node 0 ; (c) after Step 2 e l i m i n a tion o f node 0 ; C d ) after S t e p 3 e l i m i n a t io n o f node @; ( e ) o p t i m a l order of t h e node e l i m i n a t ions. Also, ( n t h e fi l l - i n branches fo l lowi ng S t e p I elimination of node @ . FI G U RE B.2

®

®

®

@

�------�.-------�------�--------- .

(a)

®

(j)

®

,/

,/

,/

,/

,/

,/

(6 )

,/

,/ '"

®

®

FIGURE B.3 The graph of Example B.2: ( a ) 10 nodes to be o p t i m a l l y o rde red ; ( 6 ) fi l l - i n (dashed) b r a nc h a fter

Step

6.

In t hi s exa m p l e m ore t h a n o n e o p t i m a l l y o r d e r ed seq u e n c e c a n b e expected since at some steps of e l i m i n a t i o n m ore than one node has t he s a m e number of incident active branches. For i nstance, nodes (£) and ® of Fig. B . 3 ( a ) have one radial branch each, and s o we m ay eliminate node (£) in S t e p 1 and node ® in Step 2, or vice versa. After Step 2 branches ed and hg are made in active and are effectively removed from the graph. Updating the count of active b ranches at the remaining nodes, we recognize nodes @' (f) , (J) , and Q) ( each with two branches still active) as candidates for Step 3 elimination. To eliminate either node @ or node ([) at this stage would create a fill-in branch in accordance w ith rule (2), and so we focus u p o n nodes CD and Q) as t h e o n l y acceptablt: choices i n S t e p Solution.

B.2 SPARSITY OF THE JACOBIAN

771

3. Selecting node (j) removes branches ji and jd from the l ist of active branches and no new branch needs to be added since nodes (J) and @ are a l ready connected . In Step 4 we eliminate node (J) aiong with its single active br a nc h Uf. Continuing this process of node e l imination and active branch designation in Fig. B.3(a), we arrive at the solution

Step number Nod e elim inated No. of act ive bran ch e s

R e su l t i n g fi l l

-

ins

1

2

1 0

1 0

3

4

5

® 0 (J) CD @) 2 0

1

1 0

0

6

CD 2 0

7

@ 2 2

8

9

® (j) 2 0

1 0

10

® 0 0

The above s o l u t i o n s h ows t h a t two fi l l i n s arc unavoidable. This is be c au s e i n e l i m i n a t i n g n o d e ® i n S t e r 7 t h e n e w h r; t n c h hI shown h y t h e dashed l i n e i n Fig. B3( h ) is c re a t e d h e t we c n n o d e s ® ; l l l d w h i c h were not r re v io l l s l y co n n e cted. I ns p e c t i o n or Fig. B . : H f) ) s h ows t h a t onc f i l l - i l l b r a n c h m u st o c c u r regard less of which one of the nodes ® , ® , o r ® is ch o s e n for e l i m i n at i o n in Step 7. -

Besides, the 6 is

n o n ze ro

(j) ,

([) ,

p a ttern of the reduced coefficient matrix resulting from Step

@

® CD ®

@

x x

x

® CD x

x x

x

x

x

® x

x

x

regard less of the order in which we ch oose the row and corresponding col umn of each o f t h e r e m a inin g nodes @' ® , CD, or @ . Thus, the reader can confirm by gaussian e l i m i nation that two symmetric fi l l - i n s w i l l o c c u r even i n the optimal o r d e r i ng o f t h e nodes.

When bus numbers are assigned to Fig. B.3( a ) in accordance with the step numbers of the above solution, the rows and columns of Ybus will be optimally ordered for gaussian el im ination, and as a result, the triangul a r factors L and U will requ i re min imum storage and com puting time for solving the nodal equa­ tions.

B.2

SPARSI TY OF

THE JACOBIAN

The sparsity of bus admittance matrices of large-scale power systems i s dis­ cussed in Sees. 7.9 and B . 1 . Th e pa ttern in which the zero elements occur in Y bus is almost identical to that found in the fou r submatrices J 1 1 , J 1 2 , J 2 1 , and J 22 o f the jacobian. We ca n easily obse rve this sparsity by w riting �he m ismatch equa tions for the four-bus system of Example 9.5 with symbols, rather t h a n with

772

APPENDIX B

numerical

v al u e s , i n the following form:

G) 0

N 33 + 2 1 VJ I " C J) -N 4 .,

o

()

o

o

-M

33

- '- j V.; :1 c8 .13

6°2

6 P2

uo ,

� P,

60�

� P4

u l V2 1 ! Ve l

:J. Q c

! V3 1

:J. Q .i

---

6 1 V, 1

( B .2)

are zeros i n the jacobian because the partial derivatives with Y23 or Y3 2 as a multiplier are zero since bus W is not di rectly connected to bus G) in the single-line d iagram of Fig. 9.2. Of course, the slack bus has no rows and no columns in Eq. (B.2), and there is no row for Q4 and no column for � ! V4 ! / I V4 1 since bus @ is voltage controlled. These missing rows and columns constitute t h e only sparsity differences between the four submatrices of the jacobian and t h e system Y b u s of this example. If the five equations represented by Eq. (B.2) were solved by gaussian elimination in the same order in which they are shown, unnecessary flU-in elements woul d be generated because the equations a re not yet ordered in accordance w it h the rules of Sec. B. 1 . By noting that the zero elements of Eq. (B.2) occur symmetrically in each row and column associated with the same bus, w e are led to rearrange the equations in to the form

There

M>2

6 !V2 1

---

6. Q z

6. 0 3

6. PJ

1 V2 1

o

o

o

o

6. P2

t::. I V3 1 I VJ I 6. 0 4

t::. QJ 6. P4

(B.3)

B.2 SPARSITY OF THE JACOBIAN

773

In this equation we have rearranged the jacobian matri� so that the two state-variable corrections of each load bus and the associated power mis­ matches, f1 P and f1 Q, are grouped together in pairs. This pairwise grouping of the P and Q equations of the load buses clearly shows the correspondence between the sparsity of the jacobian and that of Ybus ' Comparing the jacobian of Eq. (B.3) with the system Y b u s given in Table 9.4, we see that the sparsity of Ybus is replicated in the jacobian on a 2 X 2 block basis for each of the buses-ex­ cept for the slack bus and the voltage-controlled buses which have rows and columns excluded, as di s cu s sed previously. Therefore, the rules of Sec. B.l for orde ring the network nodes apply directly to the jacobian of the mismatch equations written in the form of Eq. (B.3). This sparsity correspondence between Y bus and the jacobian is to be expected since those elements directly proportional to I �j I will be zero in both matrices if there is no branch admittance connecting buses CD and 0) in the network. The fol lowing numerical example illustrates a solution of the mismatch equations by optimal ordering and triangular factorization of the jacobian. Example B.3. Order the rows o f the numerical jacobi an given in Sec. 9.4 following Example 9.5 so as t o minimize fill-in elements when the lower- and upper-triangular factors are calculated. Numerically evaluate the triangular factors and from the mismatch equations calculate the voltage corrections in the first iteration of the Newton-Raphson procedure. Solution. In this example the calculated results are displayed to three decimal figures although the calculations have been performed to much greater precision on the computer. Combining in p airs the P and Q m ismatch equations f or each of the load buses 0 and ® in Example 9.5 yields

@)

-

@

45 .443

8 .882

o

o

- 26 .365

- 1 .597

- 9 .089

44 .229

o

o

5 . 273

- 0.447

o

o

o

o

26 .365

- 5 .273

4 1 .269

8 . 1 33

- 1 5 .4 2 1

- 1 .940

8 .2 5 4

40.459

3 . 084

- 0 .835

- 1 5 .4 2 1

- 3 .084

4 1 .786

-

--------��---- -----�- �

J acobian

Corrections

2.213

Mismatches

The numerical values in the reordered jacobian are not symmetrically equal, but the zero elements do fol low a sym metrical pattern. Consequently, the coefficient matrix has an equivalent linear graph to which the ordering rules of Sec. B.l apply directly. We may the reby confirm that the rows and colu mns of the above equation

774

APPENDIX B

are already optimally ordered for gaussia n e l imination. The c lements in the first column of the j acobian constitute the first c o lumn of the lower-triangular factor L; a n d the elements of the first row, when d ivided by the in itial pivo t 45.443, constitute the first row of the upper-triangu l ar factor U. Having recorded these elements, we follow the usual procedure of eliminating the firs t row and column from the jacobian a n d thereby obtain the second column of L and the sec ond row of U , a nd so on. The ca lcul a t ions y i e l d the m a t rix L g i v e n by

45 .443 - 9 .089

46 .005

o

o

o

o

- 26 .365

- 1 .597 - 0 .447 . - 1 .940 - 0 . 835

Xl x2

4 1 .269 - 8 .254 - 1 5 .421

- 0 . 1 20

42 .086 - 0 .045

X3

20 . 727

X4

Xs

2.213

-

Here the x 's denote the intermed iate resu l ts t o be calculated b y forward substitution . The solved values of the x 's are shown along with the upper-triangu l ar m atrix U in t h e system of equations L

0. 1 95

1

x

0

0

- 0 .580

0

0

0 . 000

1

0 . 1 97

- 0 . 374

1

602

- 0 .035

6 1, V2 !

- 0 .01 7

6 03

- 0 .047

! V2 !

0 . 000

6 i VJ I

1

�O4

I V) I

- 0 .029 0 .027 �

U

Back substitution yields the volt age corrections of t he first

11 1 VI I I ( per unit) V

cv

G)

x

i t e ra t i o n :

@

118 (rad.)

- 0.016

- 0.03 1

0.027

--

- 0.017

- 0.029

-

In accordance w ith Eqs. (9.49) and (9.50), we add these corrections to the original (fiat-start) values specified in Example 9.5 to obtain the updated voltages:

(deg.) I V I (per unit)

8

@

G)

- 0.931

- 1 .788

0.983

0.97 1

@

1 .544 -

B.2

SPARSITY OF THE JACOBIAN

775

from which the new jacobian elements and the power mismatches of ' the second iteration are calculated . The final converged solution is shown in Fig. 9.4.

Because of the order in which we have solved the mismatch equations of this ex amp le nO fill-ins have occurred and the zero elements of the original j acobian are repeated in the triangular matrices L and U. The elements 0.000 appearing in U are actually the result of truncating the numerical results to d isplay only three significant decimal figures. It is straightforward to confirm that the numerical product of L and U e q u a l s the original jacobian matrix within the rou nd-off accu racy d isplayed. ,

INDEX

ABCD constants, 201 i n circle-di agram equations, 216, 2 1 7 for networks (table), 753 in power-flow equa tions, 2 1 5 Acceleration factors, 339 A d mittance: branch, 240 drivi n g point, 32 equival ent admi ttance, ne twork,

25 1 -255

mutual

of nodes, 32 in pow e r Il o w s tu d ies, 329,

admi ttance,

;IS d a t
337

-

measurement of, 285 primitive, 240, 246 s e l f-admi ttance, of nodes, 32 as d a ta i n powe r-flow stud ies, 329,

337

measurement of, 285 transfer, 32 A d m i ttance matrix ( see Bus adm ittance mat rix) Alternating-cu rrent (ac) circuits: ba l a nced three-phase, 1 4-25 double-subscript not ation for, 4-5

Alternating-curre nt ( Can t . ) powe r in balanced t h ree-phase,

24-25

power in single-phase, 5 - 1 0 single-subscript notation for, 3-4 Al ternato r ( see Synchronous machine) American T\ational Standards Institute (ANSI): graph ical symbols for d iagrams,

34-36

g u i d e for circu it-breaker

404-405 A r e a c o n t ro l 566-572 er ror (ACE), 562 frequency change, 564-566, 5 70 time e rror, 568, 572 Armature reacti o n , 98, 1 00, 1 07 Attenuat ion constant, 205 ;l p p l i c ;l t i o n s , ,

Autorratic generation control (AGC),

532, 562-568 Autotransformers, 7 1 -72

B coefficients, 547-548 calcu la tion of (example), 548-551 I

777

778

INDEX

B coefficients ( Con t.)

comparison with power flow, 5 5 1 -552 matrix of, 547-548 Back substitution, 264, 266-267 Branch admittances, 240, 246, 250 Branch equations, 240 Branch impedances, 240, 245, 250 Building block m a trix, fo r Y b!I" 33 , 242 Bundled conductors, 1 64 capaci t iv e r e a c t a nce o f, 1 K () - I KK inductive rea c t a n c e o f, I h 4 - 1 h) Bus a d m ittance e q u a t i o n s , s o l u t i o n o r, 263- 27 1 Bus a d mittance matrix: building-block formation of, 33, 242-25 1 d e fin i tion o f, 3 2 formation rules, 32 formation using A , 261 -263 measurement of element val u e s , 285 m odification of, 255-257 m utual coupling in, 245-25 1 Bus impedance m atrix: calculating clements o f, 300 - 3 1 0 c hanging reference o f, 3 1 3-3 1 6 definition o f, 284 Jrivi ng-r>oi n l i l1 l p c u ( l I 1 C C S u r, 2 K 4 m e a s u re m e n t o f, 2K6 equiv a l e n t n e t w o r k I'm t h r c e - p h a s e fau l ts, 395 -402 e q u iv a l e n t n e tw o r k fo r u nsym m e t rical fa u l t s, 524-525 forma tion o f, 30 1 - 305 m utual coupling i n , 3 1 6-322 m od ification of, 294-2<)<) negative-sequence, 47 1 -472 positive-sequence, 471 -472 power-invariant t ra nsformations with, 3 1 0-3 1 2 transfer i mpedances of, 284 measure m e n t o f, 287 in t h ree-phase faul t analysis, 390-395

Bus i m p e u a n c e m a t r i x ( CUll f . ) i n u nsymmetrical fau l t a nalysis, 470-5 27 z e ro-seq u e n c e , 47 1 -472 Buses, d e fi n it i o n of. 259 C a p a ci t a n ce :

c a l c u l a t i o n by

m od i fi e d g e o m e t r i c

m e �l n d i s t a c e , I K o - 1 0 0

n

d e li n i t io n . 1 4 2, 1 7()

I l C ll I L l i . 1 75

�l nd 10

e a r t h . 1 � 3 - 1 �()

1 75

a n d n o n u ll d o r m c h a rge u i s t r i bu t io n .

of

1 88 - 1 90

par

sum

allcl

m a ry

ci rc u i t t h re e

- p h a s e l i ne s .

a n d s t ra n d e d c o n d u c t o r s , 1 7 6

of

of calcu l a t i

on s ,

t h ree - p hase l i n e s w i t h

spacing, 1 77 - I SO

of t h re e - p h as e

191

equiva l e n t

lines with

u n sym m e tric a l

1 S 0 - 1 83

s p a c i ng,

t wo-w i r e l i n e s. 1 7 3 - 1 7 6 ( See also Reactance)

t r a n s pos i t i o n t o h a l a n c c . 1 80 - 1 8 2

of

I - ft s p a c i n g , 1 7 6

C : l [J : l c i t i\'c r C :k t a n c c : at

!\ CS R

( ; Ihlc,

7 .') ( )

o r, 7 5 2

s p a c i n g rac t or , 1 7 ()

l ;l h l c

Ca p a c i t o rs :

(IS

, 292-294

g e n e ra t o r s of r e ac t ive <)- ! O

to c

on t r o l

o

b u s v l t age

p ow e r ,

218-220 i mp e d an c e , 204-206,

fo r s e r i e s com p e n s a t i o n , C h a ra c t e ri s tic

226 termin ation of line by , 206, 227 C h a r gi n g c u r r e n t , 1 70, 1 79- 1 80 Ch i-square p ro b a b i l i ty d i str i bu t i o n , 658 v a l u e s of ( t a b l e ) , 659

IN DEX

Circle d iagrams of transmission l i n e s P a n d Q, 2 1 6-21 7 Circuit breakers, 380, 402-406 Compensating currents, calculation by Z 'o u s ' 594-5 9 8 example of, 5 9 8-600

Compensation factor: series, 2 1 8 shunt, 220 Compl ex power, 1 0 Con ductance, 1 4 1 , 1 70 Cond uctors: characteristics of ACS R (t able), 750 types, 1 42- 1 4 3 Connection m a t r i x ( see I n c i d e nc e

mat rix A)

C o n t i n g e nc y

an a l y s i s : by dc m odel, 626 - 628 by d istri b u t ion factors: m u ltiple outages, 620-626 single outage, 6 1 1 - 620 Con t ingency eval u a t ion ( see Contingency ana lysis) Con t ro l : of P a n d Q over transmission l i n e s b y generator adjustm e n t , 1 05 - 1 1 6

o f P a n d Q over transmission l i n e s b y transformers, 76-80,

779

Current (Cont.) reflected wave of, 205, 227, 229 for short transmission lines (equations), 1 96

133 - 1 36, 383-387 transient i n R- L circuits, 3 8 1 -383

subtransient,

transient in synchronous m achines, 1 32- 1 33

6- Y transformers: pe r-phase imped ance of, 62 p h ase shift i n , 64- 7 1 , 454-459 Dependent v a r i a b les ( see State variables) D e t c c t ion

of

bad

data

by c h i - s q u a re

65�-660 example of, 6t1 1 - 664 tcst,

D i rect current (dc) transm ission, 23 1 -233

D istributed winding, synchronous machine, 754-762 D i stribution factors: c u rrent -injection, 6 1 3 cu rre nt-shift, 6 1 3 generation-shift, 6 1 3 l i ne -ou tage, 6 1 5 Double l i ne-to-ground fau lts: t h rough impedance, 478-479, 494-496

3 6 1 - 3 67

of plant generat ion, 5 62 - 5 68 of power exch (� n ge b e t w e e n a r e a s ,

562-568 Corona, 1 4 1 , 1 70

worked examples, 496-499 D o u b l e s u bsc r i p t n o t a t i o n , 4-5 D r i v i n g po i nt impedances of nod es, 284

Cava r i a nce m a t r i x , ({i () - () 5 7

m e a s u re m e n t o f,

286

C r i t i c a l c l e a r i n g a n gl e , 722-724

C r i t i c a l c \ c ;l r i n g l i m c ,

722 - 727>

Cu rre n t :

d i re c t i o n

o f, 3

i ncident wave of, 205 , 225 , 227, 229 fo r l o n g t r a n s m i s s i on l i nes (equations), 204 -205 , 208 fo r med ium-length transm ission l ines (equations), 20 1 m o m e n t a ry i n circu i t b r e a k e r s , 403 ,

Econom i c d i s p a t c h ( see Economic

E a r t h , c l rc c t on c a p a c i t a nc e , 1 83 - 1 85

load

between p l a n ts ; Econom i c load dist ribu t ion betwee n u n its in a plant) Economic load d istribution be tween plants: d i s t ri b u t i o n

exam p l e , 5 5 8-5 6 1

780

I N D EX

Economic load between p lants (Con t .) mathematical d evelopment, 540-543, 555-557 s avings effected by, 5 6 1 Eco nomic load distribution betv,' een units in a plant, 532-537 example of, 537 5 3 () i nt u itive concept, 535 mathematical devclopmc n t . 53 5 - 5 37 savings c ffcctcd by, 5 3 9 - 5 4 ( ) Economic o p e r a t i o n o r powe r :-, y s t e m s , -

5 3 1 - 5 H ()

Eddy-cu rrc n t loss, 5 2 Equal-area criterion, 720-726 Equivalence of sources, 239-240 Equivalent admittance network, 25 1 -255 Equivalent circuit: for bus impedance matrix, 395 -402 of long transmission l i ne, 21 2-2 1 4 o f m ed ium-length t ransmission line, 200-201 of short transmission l i ne, 1 96 o f synchronous machines, 1 00- 1 02, 1 17-1 3 1 steady-state, 1 00- 1 02 subtransient, 1 29- 1 3 1 transient, 1 17- 1 29 of transformers, 5 1 -53 Equivalent equilateral s p a ci n g Do,:q , 1 63, 1 8 1 Equ ivalent network between buses, 290-291 Estimation methods: least-squares, 642-644 weighted least-squares, 645-647 . example of, 647-649 Euler's identity, 2 External equivalents: by gaussian elimination, 628-632 by Z hus method, 603- 606

Fault calculations: using Y hus , 4 1 0-4 1 1 using Z hu" 390-395 by equivalent circuits, 395 -402 Faults: c u r r e n t s and voltages of, 5 2 6 d e fI n i t io n 3S() d o u b l e l i n e - t o - g r o u n d , 494 -499 l i ne-e n d , 39t3-402 l i ne-to- l i n e , 488-494 o p e n co n u u c t o r 5 1 2 - 5 2 3 ,

-

,

s e q u e n c e n e t w or k co n n e c t i o n s fo r ,

5 24 - 5 2 5

Fi e l d i n tensity: e l e c t ric, 1 7 1

magnetic, 1 47- 14t3 F l u x l inkages: of coi ls, 43, 48-49 i nte rnal , 1 4 6 - 1 4 9 of one conductor in a group , 1 53- 1 55 partial, 1 4 6 - 1 4 9 Fortescue, C. L., 4 16 Forward elimination , 263-266 Fuel cost, 532-543 Fuel efficiency, 533 Frequency bias, 567 Frequency changes, 5 64 5 66 , 5 7 0 t i me

-

e r ro r s

o f, 5 6 8 - 5 7 2

r que n cy of osci llation, 7 1 6 -7 1 7 , 740· 74 1

F e

Gain m a t r i x , 6 4 6 Gauss-Seidel method of power-flow stud ies, 3 3 5 - 3 42 G a u ss i a n e l i m i n a t i o n , 26 3 - 27 1

back subst itution, 264, 266-267 example of, 267-27 1 forward elim ination, 263-266 General ized circui t constants ( see A B eD constants) Geometric mean distance (GMD): capacitive calculations, 186- 1 9 1 conductor to conductor, 1 5 7

,

I N DEX

Geometric mean distance ( Cont .) i nductance calculations, 1 57- 1 66 mutual, 157 self (see Geometric mean radius) Geometric mean radius (GM R): ACSR (table), 750 of bundled conductors, 1 86 - 1 88 definition of, 157 calculations to compute inductance, 157- 1 66 calculations to compute capacitance, 1 86- 1 91

Graph, di rected, 258 Ground wi res, 193, 221

781

Impedance(s) ( Cont.) o f transformers, 53 -57, 62 table, 748 unsymmetrical sequence circui ts, 457-461

zero-sequence ( see Zero-sequence impedance) ( See also Reactance) Impedance m atrix ( see Bus i mped ance matrix) I ncidence matrix A: branch-to-node, 257-263 defin ition of, 259 Incidence voltage and c urre n t, 205, 225-230

H constants (table), 703 calculation of, 701 -706 defini tion of, 700 Heat rate, 533 Hyperbolic functions, 207-21 0 Hysteresis loss, 52

Imped ance(s): branch, 240 ch aracteristic ( see Characte ristic imped ance) 6- and Y-connected sequence circuits, 429-435 di agrams, 36-37 driving point of nodes, 284 measurement of, 286 measured through transformers, 45 , 62

negative-sequence (see Negative­ sequence imped ance) positive-sequence ( see Posi t ivesequence impedance) primitive, 240, 245 surge, 206 Thevenin, 289, 29 1 transfer of nodes, 284 measurement of, 287

I ncremental fuel cost, 533-537 Independ ent buses, 259 Independent nodes, 259 Inductance: calculation by method of geometric mean distance, 1 5 7 - 1 66 damper winding, 1 1 9, 1 29 definition, 146- 1 47 direct-axis, 1 20 direct-axis sub transient, 1 3 0 direct-axis transient, 1 28 and internal fl ux, 1 46- 149 mutual, 1 46 of parallel-circuit three-phase l ine, 1 88 - 1 89

quad ra tu re-axis, 1 20 round -rotor, 9 1 -93 salient-pole rotor, 1 1 9- 1 20 of single-phase two-wire line, 1 5 1 - 153

summary of calculations, 1 66 of three-phase line with equivalent spacing, 1 6 1 o f th ree-p hase l i n e with unsymmetrical spacing, 1 6 1 - 1 63

transposition to balance, 1 62 zero-sequence, 1 20 (See also Reactance)

782

IN DEX

( .'ICC

Inductive reactance: at 1-ft spacing, 160 ACSR table, 750 spacing factor, 1 60 ACSR table, 75 1 Inertia constant, 700 (See also H constants) Infinite bus, 1 03 in stability stud ies, 706 Infinite l ine, 205, 227 Input-output cu rve, 533 I nterchange of p o w er , 562-567 Interconnected power systems: automatic con t rol of, 562-567 external equivalents, 628-632 p iecewise solution of, 601 -61 0

Load d ispatch i n g

Jacobian matrix: elements of, 352 formulas for, 375 power flow, 344 J acobian matrix state estimation: elements of, 682 example o f, 685 form of, 666-667, 68 1 structure of, 677-683

Magnetically coupled coi ls, 46-5 1 coi l admittances/impedances, 49 equivalent circui ts, 50-5 1 flu x li nka ge equations, 48 l eakage and magnetizing inductances, 50-5 1 M a t rix ( see Bus admittance m a t rix; Bus impedance matrix; G a i n matrix; I ncidence matrix A; J acob i a n matrix; Weigh t i n g

Kinetic e ne rgy, 700, 703 Kron reduction, 27 1 -274 formula for, 273 Lagrangian multiplier, 54 1 Lightning: arresters, 23 1 ground wires for protection against, 1 93, 221 Line additions/removals, 5 92-600 and compensating currents, 594-598 Line charging (megavars), 356,

359-360

Line-to-line faults (see Faul ts)

Economic l o a d d istribution between p l a n t s ; Economic load d istribution between u n its in a plant) Load-flow con t rol by transform ers,

76-80, 3 6 1 -367 Load-flow studies ( see Powe r-flow s t u d i es)

Long- l i n e equations: " hyperbolic form o f, 207-2 1 0 i n terpretation of, 205 -207 sol u t ion of, 202-204 Loss coe ffi c i e n ts ( sec B coe ffici e n t s ) Losses a s fu nction o f generati on,

543-548 Losses of system, formu l a for, 3 1 2-3 1 3

-

m a t ri x)

Measurement eq uations, 643 , 665 Medium-length l ine, 200-202 M i smatches: equations for, 344 power, 347 .. in power-flow stud i es, 33 1 Mutual admitt ance of nodes ( see Adm ittance) Mutually coupled branches, 245 -255 add i t i on to system, 3 1 6-3 1 9 ad m ittan ces , 246, 250 equivalent network, 25 1 -25 5

impedances, 245, 250 removal from syst em 320-321 ,

I N D EX

783

Per-unit q uantities (Cont.) selection of base, 56-59 for three-phase quantities, 6 1 - 63 Per-unit reactance: of synchronous machines (table), 749 of three-winding transformers, 72-75 of transformers (table), 748 ( See also Per-unit quantities) Phase constant, 205 Phase sequence, 1 5 - 1 9 Phase shift in three-phase transformers, 59-69, 458-459 Piecewise solution of systems, 601 - 606 example of, 606-609 Polarity m a r k s : for tra nsformers, 43, 66-67 for vol tages, 3 - 5 Positive-sequence components, 4 1 7-421 Posit ive-sequence impedance, 432 of circu it el ements, 429-458 Posit ive-sequence networks ( see Reactance diagrams; Sequence networks) Powe r: average value, 8 complex, 1 0 control o f flow by transformers, O ne-line di agrams, 34-3 6 76-80, 361 -367 Opt imal power flow (O PF), 5 3 1 -532 direction of flow, 1 1 - 1 3 Ordering, near optimal, 279-280 instantaneous, 5-6 react ive, 9, 1 2- 1 3 in single-phase circu its, 5- 1 0 Park's transformat ion, 1 1 9- 1 25, by symmet rica l components, 763-765 427-429 Penalty factor, 542 in three-phase circu its, 24-25 Per-phase equivalent (see Single-phase transmitted over transmission line, equivalent) 215-217 Per-unit quantities: ( See also React ive power) advantages of, 80-81 Power angle, 1 07- 108 ( See a lso Power-angle curves; change of base, 29-30 Power-angle equations) definition of, 25

Negative-sequence circu its, 429-458 Negative-sequence components, 417-422 Negative-sequence impedance, 432 of circuit elements, 429-458 of synchronous machines, 442-449 table of, 749 Negative-sequence networks (see Sequence networks) Network, equivalent between nod es, 290-29 1 Network (or system) red uction, 27 1 -274, 628-632 Networks: A B eD constants of (tahle), 753 ( See also Sequence networks) Newton-Raphson met hod for power­ flow stud ies, 342-349 Nodal admittance matrix, 238, 241 formation of, 248-25 1 Node: definition of, 30 elimination of, 27 1 -274 formula for, 273 equ ations, 30-32 for power-flow problems, 330

784

I N DEX

Power-angle curves, 709 in equal-area analysis, 7 17-725 ( See also Power-angle equations) Power-angle equ ations, 707-714, 729 definition of, 709 examples of, 709-714, 730-734 Power factor, 8, 9 Power-flow equations, 330, 347 summary of, 374-375 Power-now p r o ble m :

Re actance (Cont . ) i nductive ( see Inductive reactance) sequence (positive, negative , and zero), 431-432 of circui t elemen ts, 429-458 of synchronous machines (tabl e )

Power- now s t u d i e s: d a t a ro r , 334 by dc method, 373 by decoupled method, 368-373 by G a u s s Se i d e l method, 335-342 by Newton-Raphson method,

Reactance o f b u n d l ed capacitive, 1 � 3 - 1 � � inductive, 1 64- 1 65

fo r m u l a t i o n

sU ll l m a ry o f,

329-335 33 4

of,

-

342-353

numerical resul ts, 356-36 1 voltage-controlled buses, 340-341 ,

353

Power-invariant transformations,

3 1 0- 3 1 3 o f c urrents, 5 45 -546

load center i nterpretation of,

552-555

Power-loss formula, using Z huS '

3 10-3 1 2

Power tri;mgle, 1 0- 1 1 Primi tive : admittances, 240, 246 impedances, 240, 245 Probability: ch i-square d istribution, 658-659 definitions of, 65 1 Probability density function: gaussian (norma!), 650 standard gaussian, 65 1 table of values of, 652 Propagation constant, 204 Reactance: .

capacitive (see Capacitive reactance) diagrams, 36-37

,

749

subtransient, 130 synch ronous, 102

transformer leakage , 5 1 tr a n s i e n t 1 2R

( Sec a /so C a p a c i t a n c e , I n d u c t a n ce ) ,

co n d u c t o r s :

Reactive compensation of transmission lines: series, 2 1 8- 2 1 9

s h u n t , 2 1 8 , 220

Reactive power: definition, 8 direction of flow, 1 1 - 1 3 generated by c a p a c i t o r s, 9- 1 0 sign of, 9 , 1 3 of synchronous mach ines with overexci tation, 1 05- 108 of synch ronous m achines with underexcitation, 105- 108 Reactive powe r flow: by mac h i n e excitation, 1 05 - 1 08 by regu l a t i n g transformer, 78-80 Redundancy i n state estimation, 664 factor of, 678 Reflected vol tage and current, 98,

1 1 7- 1 20

Reflection coefficient, 227 Regulating t r a ns fo rm e r s , admittance model of, 362 Regulation ( see Voltage regulation) Resistance, 1 43 - 1 46 ACS R table, 750 Self-admittance of nodes (see Admittance) Sequence circuits: of impedances ( see Impedances)

I N DEX

Sequence circu its (Cont .) sy n c hr o n o u s m a c h i n e , 442-447 t r a n s forme rs, 449- 45 7 6.- Y,

6-6 ,

453 -454

454-457

Y-Y, 449-452 t r a n s m ission l i ne, 4 3 5 - 4 4 1 S e q u e nce compon e n t s ( see Symme t r i c al compo n e n t s)

S e q u e nc e i m p e d a nces, 432

S e q u e n ce ne tworks, 46 1 - 4 6 7 i n t e rcon n e c t i o n fo r fa u l t c a l c u l a t i o n s , 484- 495 , 5 1 8 - 5 25 S hort-c i rc u i t m e g a vo l t a m p c r e s , 402-403

S h o r t t r a n s m ission l i n e , 1 96 - 1 9 7 S i n g l e - l i n e d i agrams ( see O n e - l i n e d i ag n l m s) S i n g l e- l i n e -to-gro u n d fa u l t s ( sec Fa u l t s ) S i n g l e - p h a se e q u i v a l e n t , 2 1 S i n gl e - p h a s e l oads, 46 1 S i n gl e - s ubscript not a t i o n , 3 -4 S k i n e ffe c t , 1 4 5 S l ack b us, 3 3 2-335

S o u rces, e q uivalence of, 2 3 9 - 240 S p a r s i ty : n e a r-op t i m a l ord e r i n g , 766-7 7 1 of j a cob i a n , 7 7 1 -775 of m a t r i ces, 279-280 S p e e d regu l a t i o n, 562-504 S t a b i l i ty : d e fi n i t i on of, 095-697 fa c t o rs a ffec t i ng , 74 3 - 745 rotor dynam ics fo r, 698- 702 types of, 695 -697 ( See also S t a b i l i ty s t u d i es) S ta b i l ity s t u d i e s : ass u m p t ions for a l l , 0 9 7 ass u m pt i o n s for c l ass i c a l m ac h i n e mod e l , 727 rotor d y n a m i cs, 698- 702 sw i n g e q u a t ion d e rived , 698- 702 types of, 695 -697 S t a te e s t i m a t i o n of power sys t e m : e x a m p l e s of, 668-67 6

785

State estima tion of (Cont.) i terative solu t ions, 667-668 mathematical development, 664-668 m e as u r e m e n t vector form, 679 State variables, 3 3 3 , 3 64 S t a t i o n co n t rol error (seE), 5 67 S t a t is t i cs :

e xpected val u e , 65 1 mea n , 65 1 resid u a l s (errors), 643 s t a n d a rd d e v i a t i o n , 65 1 v a r i a n ce , 65 1 S u rge a rresters, 23 1 S u rge imped a nce l o a d i ng (S I L), 206 S w i n g cu rY e s , 702 co m p u t e r so l u t i o n o f, 74 1 -743 s t e p - by-step s o l u t ion o f, 734-740 Sw i n g e q u a t io n s, 70 1 -702 ( See a lso S w i n g c u rves) S ym met r i c a l co m p o n ents: deAn i t ion of, 4 1 7- 4 2 1 p h ase s h i ft i n Y a n d 6 c i r c u i ts, 422-427 p h ase s h i ft in Y -6 t r a n sfor m er s , 64- 7 1 , 454-459 powe r i n terms of, 427-429 of u n sym m e t rical p h asors, 4 1 8 - 420 S y n c h ronizing power coe ffi c i e n t s , 7 1 4-7 1 7 S y nc h ronous mach i n e : a rm a t u re re aet i u n , 9 8 - 1 00 c o n s t r u c t i o n , 88 - 9 1 d i s t r i b u ted w i n d i n gs o f, 754-762 eq u iva l e n t c i rcu i ts ( see Eq u iva l e n t circu i t ) exc i t a t i o n , R 8 , 1 05 - 1 1 7 system c o n tro l , 9 8 , 1 0 5 - 1 1 7 vo l t age v a r i a t i o n , 94, 1 05 - 1 1 7 fre q u e n cy, 9 1 i n d u c t a nce (see I nductance) l o a d i n g capab i l i ty d i a g ra m , 1 1 O - 1 l7 const ruct i o n of, 1 1 4 - 1 1 5 use of, I 1 6 - 1 1 7 Pa rk's t ra n s fo r m a t i o n , \ 1 9 - 1 27 , 763- 765

786

I N D EX .....

Synchronous machine (Cont .) of stator c u rr e n ts, 1 20- 1 23 of stator flux lin k a ge s, 1 20, 763-764 of stator vol t a g e s , 1 23 - 1 24, 764-765 ph asor dia g r a m for , l O J , I Do power angle, 1 07- 1 \ 0

,

e xc i t a t i o n

,

-

s t e ad y - s t a t e s t a b i l i t y l i m i t , 1 1 5

t i m e const a n ts, 1 ] 0 - 1 ] 2 -

( see V o l t a ges)

m o d e l , 1 1 7 - 1 27

Theve n i n i m p e d a n ce between nodes,

291 Thcven i n 's theorem : in calculating t h ree p h a s e fa u l t curren ts, 386-387, 392-402 network equ ivalent, 290-29 1 i n representing power systems, 288-29 1 , 395 -40 1 , 473 -477, 484-526 i n Z ous met hods, 287-29 4 -

a

f, 28G

i m p e d a n c e , � - Y , 22 powe r i n v a r i a n t , 3 J 0 - 3 1 3 i n c a l c u l a tio n of l o ss e s , 545-546 T r a n s fo r m c rs : a u to t ra n s form e rs, 7 1 -72 d o t c o n vc n t i o n for. 4 3

T r a n s forIllll t io n s :

1 05- 1 1 0 r e a c t iv e p owe r o pe r a t i n g c h a r t , 1 1 3-1 17 rou nd -rotor fl u x- l i n kage equat ions, 93-97 s a l i e n t - p o le rotor fl u x l i n k a ge e q ua t io n s , 1 1 8- 1 2 1 seq u en c e ( p os i t i v e , negat ive, zero ) reactance, 444-446 subtransient, 1 30 sy n c h ron o u s , 1 02 table of, 749 transi ent, 1 28 s h or t c i rc u i t c u r r e n lS, 1 ] 2 - 1 ] () s h o r t - c i rcu i t s i m u la t i o n , 1 2 5 , 1 27- 1 29 speed, 90-9 1 two a x i s v o l tages

T� �e e - p h a s e c i rc u i t s : power i n , 24-25 volt age a n d current i n , 1 4 -23 Th ree-phase fa u l ts, 380-402 To r q u e a n g l e ( see Powe r a n g l e )

- ' -

m e a s ur e m e n t

1O� by

�, . l�

T r a n s fe r i m p e d a n ces, 284

r e a c t i v e power d e l iv e r e d t o syste m ,

reactive power c o n t r o l

-;.

: �:

e d d y - c u r r e n t l o ss i n . 52

e qui v a l e n t circuits of, S 1 -53 for ze r o - s e q u c n ce c ur re n t , 449-456

h y st e r es i s loss i n ,

52

i m p e d a n c e s e e n by. 4 5 , 6 2 ideal, 4 1 -46

l e a kage r e a c t a n c e of, SO-5 1 l o a d - t a p c h a n g i n g ( LTC), 76 m a g ne t i z i n g c u r r e n t i n , 5 2

3 () i - ] 6 7 o p e n -c i rc u i t t e s t , S 5 p h a se s h i ft in Y-� , 6 4 6 9 po i a r i Ly m a r k i ngs to r , 4 .3 , 64 - 6 7 o ff- n o m i n a l t u r n s

r a t i o , 76-79, -

3 6 1 -3 6 7

- reactances (tabid, 748 r e g u l a t i n g , H) - 80 ,

o f vo l t a g e m a g n i t u d e ,

78-79,

362-364 o f vo l t a g e p h a s e- a n g l e , 78-7CJ, 365, 367 s ho r t -c i rcu i t t e s t , 54 t a p - c h a n g i n g u n d e r l o a d (TC UL), 76 t h ree-phase, 59-63 t h r e e - w i n d i n g , 72-74 T r a n s i e n t a n a l ys is of t r a n s m i ss i o n l i nes:

228 -229 reflection coefficient, 227-229 reflections, 226-230

l a t t i ce d i agrams,

t ra v e l i n g waves, 222-226 i n c i d e n t , 225- 229

I NDEX

Transient analysis ( Cont.) reflected, 226-230 velocity of, 223-225 Transient stability, 696-697 factors affecting, 7 43 - 74 5 ( See also Stability stud i es) Transients on transmission l in es, 22 1 -23 1 Transmission l ine s : c l assification by lengt h , 1 95 long, 202-215 m e dium-length, 20 0- 2 0 2 short, 1 96- 1 97

t ra nsients on, 22 1 - 2 3 I ero s e q u enc e r e a c t a n c c o r. 4 3 � -440 Transposition of I ransmissioll l i n e s : a n d capacitance, 1 t)0- 1 82 and ind uctance, 1 6 1 1 62 Trit.lngu lar factorization, 2 7 4 - 279 lind Z bus e l e m e n t s , 307-- 3 1 0 i n faul t studies, 4 1 0 - 4 ] 1 z

-

-

Unit

commitment : mathematical developm ent, 572-578 m ethod of sol u tion, 578-58 1 dyn amic programming examples, 581 5 8 6 transi tion (start -up / sh u t-down) costs, 5 74 - 575 U nsym m et r i c a l fa u l ts ( see Fau lts) U n sym m e t rical ser i es impedaJl ces, -

459-46 1

Velocity of propagat ion, 207, 225 Voltage-co n t ro l l e d buses i n rmweT­ flow s t u d i e s , l 4 ( ) - - l 4 1 , l S l

Vo l t a g e r e g u l ;l l i o l l :

d e fi n iti o n of, 53 equation using constant A, 202 reduction by reactive com pensation, 2 1 9-22 1 -

787 ·

Vol tages:

and capacitors, 2 18-219, 292-294

d-q-O, 123 - 1 27

exc itation, 94, 1 05 - 1 1 7 generated, 94- 1 00, 1 26 - 1 27 incident wave of, 205, 225, 226, 229 no-load, 9 4 open-circuit, 94 pol arity m arks for, 3 - 4 refl ected wave of, 205 , 226-229 and specificat ion a t b u s 1 06 - 1 08 s u b t ra n s i e n t i n tern a l 9 4 , 3 84 - 3 85 sy n c h r o n o u s i n t e r n a l 9 4 i n t h re e - p h a s e ci rc u i t s , 1 4 - 1 9 t r;l nsie n l inte rnal, 3 R 4 - 3 R5 ,

,

,

Ward equivalents, 600, 629-63 1 examples of, 609-6 1 1 , 632-635 Wavele ngth. 20fi-207 Waves, i ncid e n t and refl ected, 205, 225-229 We ighting facto r , 645 Weig hting m atrix, 645, 653 6 5 4 -

( see Bus admittance m at r i x ) y -tl transforme rs ( see 6- Y

y 1m

Z blls

transformers)

( see Bus i mpedance m a t r ix)

Zero-se quence c ircuits: o f tl a n d Y i m p e d a n c e s

,

4 2 9 - 435

o r s y m m e t r i c a l t ransmission I i n c , tl - Y o f

I

r ; l I 1 s !o r m C fs, 4 4 {) - 45 6

435 -439

c ne

o r sy n c h ro n o u s m a h i

,

442-447

Ze ro- s e q u c n c e c o m p o n e n t s, 4 1 7 - 420

i m p e cJ a nc c 43 1 of circuit elemen ts, 429- 4 58 of synchronous machines, 442-447 table of, 7 4 9 Zero-sequence networks, 4 6 1 - 4 6 6

Ze ro - se q u e n c e

,