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