Turbo Machines
.I
A Valan Arasu Department of Mechanical Engineering Thiagarajar Collcgt: of Engineering Madurai
Scanned by: A. Ansari
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111)'
beloved Parents
462
,.
Tl'RDO M"C'I/INES
J
F
Fan losses. 209 Finite stage emciency. 8 First law oflhermodynamics, 1 Flow angle. 42 Flow coemcient. 42, 132 Francis turbine. 395 Free vortex design. 207
Jet pump. 351
Froude's number. 295
Leading edge, 43 Lift. 38.39 Ljungslrom turbine. 263
Friction losses. 209 Fund amental dimensions, 292 G
Gear pump, 347 Geometric similarity, 292 Gibb 's equation. 4 Gland leakage losses. 209 Guide vanes. 392.298 Governing: reaction turbine. 197 Pelion wheel. 388 Kaplnn .urbin •. 400 H
Head. manometric, 331
SIalic. 330 Head developed. 330 Hydrau lic elliciency. 332 Hydraulic pump. 325 Hydraulic turbine. 383
Impeller. 1 J, 64. 325 Impeller efficiency. 333 Impu lse: blading. 190 turbine, 383 Inlet guide vanes. 115 Inward now radial. 255 Ise ntropic efficiency. 130 work. 130 I so~emcie ncy curves, 337
K Kaplan turbine. 400 Kinematic simi larity, 293
L
M
Mach number. 84,85.296 Manometric head. 331,332 Mechanical efficiency. 332 Model analysis. 291 Multi-stage: axia l flow compressor. 141 . 144 cel1lrifugal pump. 337 Muhislaging. 144 Muschel curves, 336 N
Net positive suction head. 334 Newton's second law ormotion. 3 Nozzle loss coefficient, 209
o Overall efficiency. 3,333 p
Partial admission losses, 209 Piston pump. 350 PilCh, 44 Polytropic emciency. 6 Positive displacement pumps. 346 gear. 347 piston. 350 vane, 349 Power tfill1srnitting turbo machines, 441 Pre-heat ractor. 9,205 Pre-whirl, 84 Pressure coemcient, 13 6
Pressu re compound ing, 191 Priming, 335 Propeller tu rbine. 397 Propagation. 137 Pro.orype. 29 1 R
Radi
.heary. 138 Range ofopernlion. 1~] Reaction ratio. 132.202 Re-heat f
Slagnalion pressure. 138 Stagnation temperature rise. 130 Slnll. 136
StanilZ sl ip ractor. 69 SIalic head, 330 Stator blades, . 116 Stator emciency. 135 Stator row. \ 34 Steady now energy equation. J Stodola slip factor equation. 68 Submersible pump, 353 Suction pipe. 327 Super-imposed velocity triangles. 134 Surge. 89. 136 T
Thoma cavitation parameter. 405 Torque converter, 444 Total-to-stalic efficiency. 12 Trailing edge. 43 Turbine: Francis. 395 Kap lan. 391 Pelton. 384
u Un distorted models. 29~ Unit. 287 power. 304 quanrity, 303
v Van e pump. 348 Vaned diffuser, 78 Vandess diffuser. 77 Ve locity compounding. 141 Velocity now angles. 128 Vector angle, 132 Volute. 77 Volumclric efficiency. 333
w Wake, 39 Weber's number, 296 Workdonc factor. 69, 130. 131
REFERENCES
(
INDEX Centrifugal compressor characteristic. A
I . Ag:lfwal . S .K. Fluid Mec hanics and M ac hinery, Tatn Mc;:G raw Hill. New Delhi ,
1997 . ,
Anthony Espos ito , FluiJ Power wilh A ppli c.::ll ions. Prentice Hall. London. 1997 , J. B' ll1 s a1. R.K. A Texthook of Fluid Mec hanics and I-Iydrnulic lvI:!c hincs. Ln xmi
Publi c:ltiulls(P) LId .. New De lhi. I YYH. ~ . Di xllil. S .L .. Fluili Mechanics. Thl!rrlll1l1Yllilnlics ofTurho mllcllinery. Pagaillon
Pn.:s :,> , 1975. 5. Kad :unhi , v.. and Man c har Prasad. Energy Con versi on - Volume III Turbo
Machin ery. New Age Intern ati onal(P ) Ltd .. New Delhi : 1997. o. Khaj uria, P.R., and Duhey, S.P., Gns Turbin es and Propulsive Sys tems , Dhanpa! Rai & So ns, New Delhi , 1992. 7. Sayers, AT., Hydraulic and Compress ible Flow Turbo tvlachincs. McGraw Hill . Lo ndo n. 1992. 8. S hepherd. D.G .. Principles o f Turbo Machinery. Mnc millan Company, New York . 1965. I). Y~lh)'a. S. M .. Turhincs Compressors and Fans. Tal a McGraw Hill. New Delhi, 199 7. III. Yunu s A. Ccn gel .md Michal !.! A. Boles. Thermodynam ics - A n En g ineering Approach. McGraw Hill. New York . 1997 .
Aero foil section. 37 Adiabctic flow, 1:::!8 Air angles. 130 Air lift pump. 352 Angle of incidence, 44, 45
Angular 1110menrum, 3 A'dal compressor, 125 char.::cteristic, 144 multi stage, 144 Axial flow gas turbine rotor. 129 Axial flow rurbine. 383 A'(iai pump. characteristic. 343 velocily triangle, 341 work done. 342
B Best efficiency point, 336 Blade. 37 cascades. 49 chord. 49 Buckingham PI theorem, 290 Buseman slip factor equation, 68
88 Choking. 90 Chord. 44 ConservOltion of mass. 2 Continuity equation, 2 Critical. cavitatiun. 344 NPSH. 334 Cumulative enthalpy drop. 205
D Delivery pipe. 327 Derived dimension. 287 Diagram efficiency. 184 Diffuser. 76 vaned. 78 vandess. 77 Dimensional analysis method. 290 Distorted model. 294 Double regulation, 400 Dl1lft lUbe, 392.398.402 Draft tub I! theory. ~03 Dl1lg.38 Drag coellicient. 39 Dynamic similarily, 293
c Camber angle. 43 Camber line. 43
Cascade. blade angle. 43 curves, 52 dr"g. 38. 39 nomenclarure, 43 pressure loss, 48 secondary losses. 53 Casing, 326 Cavitation. 344 Centrifugal pump, 325 multi stage. 337 vl!locity triHllgll!s, 341
E
Efficiency: fmite stage, 8 hydr.Julic. 332 impeUer. 333 isenlropic. 130 polytropic. 6 static· to-static . 6 total-to-total. 6, II total-to-stalic. 130 volumetric. 333
Euler nwnber.295 pump equation, 4 turbine equation. 4. 183 Exit kinl!tic energy, 83
458
>
ArPENDIX ...: 459
TURBO MACIIINES
AXIAL FLOW TURBINE BLADE DESIGN # include <stdio.h> ff include <math.h> ff include
main ( )
{ doubleCa, Cx1m. Cx2rn, Cx2, i, j. k. I.A1. A2. 81, 82; float aIm. blm, aI, a2. bl, b2. b2m. Om. Dt. Dr, 0, N. Um. Ut, Ur, U. R. elm. X. Cl. Cx1. Cal; i nt x; clrscr ( ); printf ("enter blade tip diameter (m) ="); scanE ("'tf". &Dt); printE ("enter blade root diameter (m) ="); scanE ("'tE". &Dr); printE ("enter mean rotor blade inlet angle (deg.) ="); scanE ("%E". &blm); printE ("enter mean root blade outlet angle (deg.) ="); scanE ("%E", &b2m); printE i"enter mean stator blade outlet angle (deg.) =H); scanf ("%E", &alm); printf ("enter the turbine speed (rpm) ="); scanf ("%E", &N); clrscr ( ) ; /*CONSTANT NOZZLE ANGLE BLADE DESIGN */ Om = (Ot + Dr) / 2; Urn = (3.14*Dm*N) / 60; i alm*3.14/1BO; j blm*3.14/18o; k tan (i) - tan (j); 1 b2m*3.14 / 180; Ca = Urn / k; elm = Ca*cos (i): Cxl~ = Ca*tan (i); for (x=l: x<=2; x++)
{ iE(x==I) {printf ("blade root angles o = Dr;
(deg.): \o\n");
} else {printE ("blade tip angles (deg.): \n\o"); DDt;
} X
pow «Om / Dl.
(sin
(il
*
sin ( i ) ) ;
C1 = X*C1m; Cxl = X*Cx1m; Cal = X*Ca; printE ("Stator blade exit angle =%f \n", aIm); U = 3.I4*D*N / 60; Bl = atan «Cxl - U) ICal); bl = 81.180 / 3.14; print.E ("rotor blade inlet angle =~f \n ", bl); 82 = atan (U/Cal); b2
= 82*lBO/3.14;
printE ("rotor blade exit angle =%E \n b2); R = Cal,,(tan (B2) -tan (81)) I (2.U); printf ("degree of reaction =%f \n\n". R); R
•
} } INPUT enter blnde tip diameter (m);: .75 enter blade root diameter (m) = .45 enter mean rotor blnde inlet angle (deg.);: 0 enter mean root blnde outlet nngle (deg.);: 75 enter mean stator blade outlet angle (deg.);: 75 enter turbine speed (rpm);: 6000
OUTI'UT Blade root
st310r blade e~il angle ;:75.000000 rotor blade inlet angle =57.821709 rotor blade exit angle ;:64.931801 degrec of rC3ction ;:().128160 Blade tip angles (dcg.): stator blade exit angle :;;:75.00Q(X)() rotor blade inlet angle ;:·63.547230 rotor blade e~il angle ;:80.139671 degree of reaction =0.675157
ApPENDIX
-156 ::.- T URno MACIiINES
AXIAL FLOW TURBINE BLADE DESIGN (FOR TURBINE PROBLEMS)
•
Cx! = Cxl rn*Dm/D; Al = atan (Cx1/Ca); a1 = Al.180/3.14; printE ("Stator blade exit angle =t f \n al); U = 3.l4*D*N/60; 81 = atan (tan (AI) ~ U/Ca) i b1 = B1.180/3.14; printi ("rotor blade inlet angle =%f \0-, bI); Cx2 = Cx2m*Dm/D; A2 = atan (Cx2 I Cal ; a2 :: A2.180/3.l4; printE ("stator blade inlet angle =%f \0-. a2) i U = 3.14*D*N/60; 82 = atan (tan (Al) + U/Ca); b2 = B2*180/3.14; printE ("rotor blade exit angle =%f \n-, b2); R = Ca. Itan IB2) -tan IB1I) / 12.UI; printE ("degree of reaction =%f \n\n-, R); R
include <stdio_h> ff include <math.h> n include main ( )
,
doubleCa, Cxlm, Cx2m, Cxl, cx2, i, j. k. I,AI, A2, Bl. B2; Eloiltalm, blm. al, a~, bl, b2. b2m, Dill. Dt, Dr. D. N. Urn. Ut.Ur.U,R; int. x; clrscr ( ); printf ("enter blade tip diameter (m) ="); scanf ("%f", &Dt); printf ("enter blade root diameter (m) ="); scanf ("%f~, &Dr); printE ("enter mean rotor blade inlet angle (deg.) ="); scanE ("%f", &blm): printi ("enter mean root blade outlet angle (deg.) ="); scanE ("%E". &b2mJ; printE ("enter mean stator blade outlet angle (deg.) ="); scanE ("%f" &alm); printE ("enter the turbine speed (rpm) ="); scanf ("%f~, &N); clrscr ( ); Om = (Dt + Dr) I 2; Urn :: (3.14*Dm*N) I 60; alm.3.l4/1BO; j blrn*3.l4/180; k tan Ii) ~ tan (j); 1 b2m*3.14 I 180; Ca :: .Um I k; Cx1m = Ca*tan {il; Cx2m = Ca*t.an (1) ~ Urn; for (x=l; x<=2; x++J
} } INPUT,
I
if I x = =1 I ("blade root angles (deg.): \n\n");
1 print E o = Dr; else lprintf D :: Dt;
("blade tip angles
(deg.):
\ o\n ");
enter blndc tip dinmelcr(m) == 0.75 enter blade root diameter(m) = 0.45 enter mean rotor blade inlet anglc(dcg.):;:: 45 I!nlcr mean rotor blade outlct nngic(deg.):;:: 75 cnler mea n slalDr blade outlet anglc(dcg.} = 76 c:ntcf turbine speed (rpm)=6000.0
Blade root angles: stator blade exit angle =79.418846 rotOr blade inlet angle:;:: 72.061981 stator blade inlet angle =43.9344555 rotor blnde exit angle =72.747986 degree of reuclion =0.028995 Blade lip angles: staIDr blade exit angle ::;72.680054 rotor blade inlet angle =·28.855749 stator blade inl et angle =30.027699 folor blade exit angle =77.028351 degree of reaction =0 .650438
..t, 457
API'ENOIX
if (x::::
=
•
1)
r = rt; U " Ut; printE ("blade tip angles
OUTPUT (deg.): \n\n");
} else
r = r h;
U " Uh: printf ("blade root angles
(deg .. ):
\n \n");
} Cxl = (Cxlm I rm)*r; Cx2 = (Cx2m I rm)*r: KI = (Ca*Ca) + 2* (Cxlm*Cxlm) ; Cal = sqrt (Kl - 2*(Cxl*Cx1): K2 = (Ca*Ca) + 2* (Cx2m*Cx2m) - 2* ((Cx2m I rm) - (Cxlm I rmll *w*nn*rm; Ca2 = sqrt (K2 -2. (Cx2.Cx2) + (2.(Cx2Jr) - (exl / r»
*w*r*r) 1 ; Al " atan IICxl/Cal)): al " 11BO.Al) I 3.14; printf ("inlet air angle = %f \n", all; Bl " atan IIU - Cx l) I Cal); bl " 11BOoBl) I 3.14; printf (-inlet blade angle %f \n·, bl); A2 " atan IICx2 I Ca2)); a2 = 11BOoA2) I 3 .14; printf (Routlet air angle = %f \n R , a2); B2 " atan IIU - Cx2) I Ca2): b2 " 11BOoB2) I 3 . 14; printE (Routlet blade angle = %f \n b2l; R= ((CalIco s (81) ).(Cal/cos (Sl) - (Ca2 /cos (82)* ICa2/cos IB2))) I 120UoICx2-Cxl)); printf ("degree o f reaction = %f \ n\nR,R); R
,
} )
INPUT cnler temperature difference = 20 workdonc filC!Or::::: 0.93 cnrcr menn blnde speed (m/sec) = 200 cntcr nxia l ve locity (m/sec) = 150 ~ntcr
cnll.!r bludc lip speed (m/sec) = 250 cn ler blaclc roOI ~pcc d (m/scc) = 150 rOlational speed (rpm) = 9000
Melln blildc angles (deg.) inlel blnde angle" 45.783056 oul lc l blade angle" 17 .046410 inlct air ang le = J7 .046410 outlet air angh.: = 45.783056 Blnde tip nnglcs (dcg.) inlct nir angle = 22.062051 inlet blndc ang le = 53.646092 out let air angle = 5:'.646093 out lCI blade
....: 455
ApPENDIX ,t5~
.,.
.... 453
TURUO MAClIlNES
INPUT ..:lIh:r h!mp..:rmUfl: dirfcn!ncc = 20 l'nh:r wurkdonc. fnclOr = 0.92 cn t..!r mean blade speed (m/sec) = 210 l'llIcr a.x in l veloci lY lm/sec) = 157.5 l'lw,;r bl'ldl! lip spl!l!d (m/sl!c) = 262.5 l'llIcr blilde root speed (m/SeC) = 157 .5
l:llilstant reaction hlade '.' =n
OUTPUll Mean blade angles (dcg.):
inkl blade angle = 44.935090 outle t bladc angle = 18.601836 inlct air angIe = 18.601836 llUt1ct air ung lc = 44.935090 Blade tip
AXIAL FLOW COMPRESSOR BLADE DESIGN
•
ff
include <stdio.h>
U incl ude <math.h> # include main (
double a1, a2, b1, b2, i, j, k, 1. A1, A2, 81, 82; float td, WF, U, Ca, Cp= 1005.0, R, Urn, Ut, Uh,
W,
Cxlm, Cx2m, r, rh, rt, rm, N, Kl, K2.
exl, Cx2, Cal, Ca2; in t x; printE
("enter temperature difference =-);
Scanf {"%t", &tdl; printE ("enter workdone factor = ~); scanr ("%f", &WF); printf ("enter mean blade speed m/sec) ="); scanf ("%f", &Um); printf ("ente r axial velocity (m /sec) ="); scanf ('''fif", &Ca); printf I "enter the blade tip speed 1m/sec) ="); scanE ("'iif", &Ut); printE ("enter the blade root speed 1m/sec) =-); scanf
("%f",
&Uh)i
printE ("rotational speed (rpm) scanf
("%f~,
=");
&N);
clrscr l ); i (td*Cp) /
/*FORCED VORTEX BLADE DESIGN./ (\oJF*Um*Ca); j = (O.5*2*Um) / Ca; = atan {{i+jl /21; b1 = I1BO.<1 / 3.14;
<
printE ("mean blade angles (deg . ): \n \n"); prin tE ("inlet blade angle %f \n", bl); 1 = atan {j - tan {kll;
=
b2 = 1180.11
/
w =
/60; (2*3.14*N); / {2.3 .1hNI; / {2.3 .1hNI;
3.14;
printE ("outlet blade angle = %f \n", b2) ; princE ("inle t air angle = %E \ n ", al = b2); printf ("outlet air angle = %f \ n \n", a2 = bl); rm
(2*3.1,hN)
(Um*60)
rh = {Uh.601 rt = {Ut.601
/
Cxlm Ca.tan (1); Cx2m Ca*tan (k); for (x = 1; x < = 2; x++)
AI'I'E ~ [}I'\
450 ;;. TURBO MACIIINES
81 = a t a n « U / Ca) b1 = (180 _S1) I 3.14;
else
prin t E (" i n let blade angl e = 'iif \n", ({Um / U). tan (k » );
U = Uri
printE ("blade root angles (deg.)
:\n\n");
a 2 = (I8 0*A 2 ) ,. 3.14; print E (" o u tl et air a n g l e = 'h E \ n", 82 atan «V / Cal - tan (A 2» );
i (td * Cp) / (toW'" U '" Ca); j = (0. 5 >I< 2 ... U) / Ca; k = atan {( i + j) .I 2); b1 = (180 _ kl / 3.14;
} ) else /",FREE VORTEX BLADE DESIGN
*
I
~ (td '" Cp) / (WF '" Urn '" Cal ; j = (0.5 '" 2 '" Urn) I Ca; k = a tan ( (i + j) / 2);
b1 = (180 _ kl / 3.14; printE ("mean blade angles (deg.) :\n \n"); printE ("inlet blade angle = %E \n", bl); 1 = a'tan (j - tan (k) }; b2 = (180 _ 11 13.14; printE ("outlet blade angle = %f \n", b2); printE (ninlet air angle = 'tE \n", a1 = b2); printE ("outlet air angle = %f \n \n", a2 = bl); for (x = 1; x < = 2; x + +)
{ if (x==11 { U = Uti
printE ("blade t1p angles ) else
(deg.)
U = Uri printE ("blade root angles
a 2 );
=
%f \n", b11; printE {~inlet blade angle 1 = atan (i - tan (kll; b2 = (180 _ 11/3.14; printE ("outlet blade angle = %f \n", b2); printE ("inlet air angle = %E \n", a1 = b2); printE ("outlet air angle = %E \n\n" , a2 = bl);
( );
bI );
A2 = atan
}
clrscr
tan (AI»;
(deg.)
\n \n" I;
\n\n" I;
) Al = atan ( (Urn I UI_ tan (111; a1 = (180_A11 I 3.14; printE ("inlet air angle = %f \n", al);
b2 = 1180_8 2 ) I 3.14; printE ( "outlet blade angle = %E \n", b2) ; R = Ca I ( 2* V) * (tan (81) + tan (8 2 )) ; pr i ntE (" d egree of reac t ion = 'i.E \n \ n", RJ;
, )
) INPUT cntcr tempenllure difference::: 20 cnt~r workdonc ractor::: 0.92 cntcr ITlcnn blClde speed (rnlsec) = 210 enter axial velocity (Ill/sec)::: 157 .5 enter bladc lip speed {m/ser.:)::: 261.5 enter bl;]dc rOllt speed (m/sec)::: 157.5 con stant reaction hlade ? ::: y OUTPUT Blade lip angles (deg .): inlel blade 'ngle = 47.687020
outlet blade angle ::: 19.659747 inlet air angle::: 29.659747 outlet air ang lc::: 47.6R7010 Mean blnde anglc .~ (deg.) : inlet blndc nngle::: 44.935090 oUllet blade angle::: [8 .6018:16 inlet air angle::: IR.601 8:16 oUllet air angle::: -I 4 .l}:':'iOl)() Bladc root angles (dcg. ): inlt!l blade anglc::: 43.161696 outlet hlade angle::: :1...11 4 273 inlct nil' angle::: 3.4 I 4273 outlet nir angle ::: 43 .261696
...: ·151
ApPENDIX AXIAL FLOW COMPRESSOR BLADE DESIGN (FOR COMPRESSOR PROBLEMS) fi include <stdio.h> II include <math . h> n include main (
double aI, a2, bl, b2, i. j, k.. 1. Al, A2. 81. 82. q, r , float td. to.'F. U, Ca, Cp=l005 . 0, R. Urn, Ut, U['";
S, t ;
int x; char Q; print f s ca nf
I" enter tempera ture di f ference = ~ I ; ("%f" .
&td);
printf ("enter workdone factor =M); scanf ( "'!f ". &WF); princE ( " enter mean b l ade speed m/sec) (~%f".
scanf
=");
&Um ) ;
printE ("enter axial velocity tm/sec) =n); scanf
("%f",
&Ca);
printE ("enter the blade tip speed tm/sec) scanf
princE scanf
= ");
1"%f", &U t ), ("enLer
("!fiE",
t h e b l ade root speed (m/sec )
= ");
&Ur);
printE ("constant reaction blade? =" ) ; scanf I"%c", &Q), elrser ( I; i f IQ = = 'y') I* CONSTANT REACTION BLADE DESIGN ./
( for Ix=1; x <=3; (:;..:
==
u =
Ut;
if
x+~J
11
{ printf {"blade tip angles (deg.)
\ n\n" I ,
} else if (x
==
2)
{ U
=
Urn;
printf ("mean blade. angles
l
1
(deg.)
:\n\n~);
446
}>
l'uRIlO MACHINES
,,
,,
r.uio. Thus the torque convertor is a more efricient power transmitting device at low speed ral ios. while the fluid coupling is more dridcnl de vice at speed ratio nl!arer to 0.950,.0.98. For automubiles. the speed mlio varies bl!lwccn 0 and 0 .98. TIlL. d are. il is usual IU have: il eomhinal ion of fluid coupling and hydraulic convcnor. so as 10 avoid the inefficie nt rilngl! of operation of each devi..::e: . At low spl!ed rali os. Ihe unit acl as a torque t.:tlllvertor while at high speed rati O!>, Ihl! ~al1lt.: unit works as a lIuid coupling with torque rat io of unity and its t.:fficien C)' increases to about 0 .98 with ,m increasmg spt.:l.!d ratio.
,
". +------'0..,.,------"" ,
,
EXERC'SES 11.5
u~________~__~____~
o
OJ
(In;
In
NOrnA Figure 10.1
Torque COfH'Crler characteristics
IHI51 - - - - - - - - - - ---,,'"
10. 1. 10.2. 10.3. 10.4 . 10.5. 10.6.
10.7. 10.8. 10.9. 10. 10. 10. 11 . 10.12.
10. 13 . 10. 14.
O~________~-------L~ o 0.5 n.'J) I.n
NBINA Figure 10.4
Fluid cOllp'i,,!: characrr-ri.ffics
But the efficiency of the nuid coupling increases continuously with speed rUlio. reaching a ma xim um value of about 95% when the speed ratio is 0 .95. However, the rise in efficiency of Ihe fluid coupling is not as fnst as that of th e torque convertor during initial stages. Infact, lhe torque convertor, running at its design ratio (0.65 - 0.7), has a higher efficiency than that of the fluid coupling at the samc speed
What i.H·t.: power transmitting lurhnm ;'H:hines'! Wbat is hydraulic coupling? E~pl;lin with a neat sketch. lilt.: working of fluid coupling. Derive the efficie ncy ofa fluid coupling. Dl!ri nc : Slip of a fluid coupling. Show that Slip a 1·- '} wheft.:'1 is the Iluid coupling crriciency. In a Iluid coup ling. th e speeds of the driving and drivcn shafts arc HOD rpm and 7 80 rpm res pectively. rind . (a) the e fli cic ncy of the hydraulic coupling and (b) the ~ lip of the coupling. IAns: (a) 97.5% and (b) 2.5<;;-' Whnt is a Torque covertcr? How docs a torque converter differ from il fluid couplin g'! Draw a l1l!a\ .sketch and c~plajn Ihe working , principle of a torque cOllwner'! Dl!ri vl! tile e:<.pn.!ssion for the efficicncy o f a torque converter? What is the role of guide vnnes in a torque convl!rter'! Compare torque convcrtl!r and fluid cOllpling on the basis of speed rati~l. Write a nole on Ihe characteristics of iI nuid coupling. Di sc llss the chamc teristi cs of torque eonvcnnr.
=
Efficiency or IOrq ue convener
TORQUE CONVERTER TIle hydruulic [orqu~ co nverter is n dc.:vicc used to transmit increased or decreased
puwer from one shafl
Power output 'Iu'
another. The torque lmnsmillcd at [he drive n shufl may be mon.: o r less thnn the lorq llc n\'~ilablc ill the driving shan. The lorquc al Ihe driven shaft mny he incrcasc.:d hy
Power input
=
A torque (Ilnvcrtcr (Fig . 10.:!) t:ompriscs (11'(i) pump impclk.r coupled 10 the dri ving slmft ( ii ) turninc runncrcnuplcd 10 Ihl.' driven sh:,f' a nd (iii) Si niaror lixcd guide vane ;I rrangcd bl.'IWCl! rl the pump impdkr and the lurhllll.! runner.
wulT" + Tsl wAT" Wll
w"
(I + TATs)
O bvlDusly. when there arc no guille VU Il!:S. torque converter red uces IU fhlnge cuupling. bl!cuusc. whl!n
Thl!n ,
shari
Figure 1(},2
Tnrqlll' {"o m'crle r
Oi l fl owing from the pump impeller In the turhine nmncr exerts a IOrquc nn Iht.! Thcsc vant:s changl: the direction of now of oil, tht:reby l1l:lIdnl.!
/ T/J=Ts+TAI Whl'r\! T,\ - Torque at driving shafl TI/ - Torque at driven shafl 7~' - Torque at sta ti onary vanes TIle power at any shaft is proportional to the product o r torque and shan speed. T hcll, powcr avai lah le a l shart A is
Further depending upon Ihe design and uril:ntali on or ,guilll! Vunl!S, thl! torque convener may runctio n as a torque mulitiplier or torque diVider. Torque converter acts as a torque muhiplier i.c .. w hen Ts is positive . To achieve th is. Ihe guide vunes arc designed to receivt: u torque rrom the oi l in n direclion opposite to thut t:xertcd on [he driven shun . Ir the guide vanes are designed such that they receive u torque from the oil III thl! samt: (iirt:clion as Ihut orthe driven shoft. then tht: torquccn nvcnc r w ill
CHARACTERISTICS OF FLUID COUPLING & CONVERTER
:'lIld power .. vailab le al)ihaf! B is
PIJ
=
wlJTu
=
wl/(T"
or
+ Tsl
The c haracteristics or nuid coupling Iln d Torque convertor arc shown in Figures 10.3 ifnd I DA. The efficiency and Torqut: ratio (TaIT,,) life runctions ofspccd ratio (N al N,I)' The torque raLio or a torque COnvenor riJlls with increasing s~l!d ralios. while at the same lime the erfici~!1cy increases. At a speed ralio hl!twecn 0.65 and 0.7 (design value), Ill l! transmission effidl!lIcy reac hes ils muximum value of ahout 85%. If the speed ratio exceeds the design value, the efficiency dt:crcasl!S qui le r:Jpid ly.
..... (a) Working Principle In the beginnin g. both the ~harls
ilfC
Lt.:l. N" - Speed of c.Jri ving shaft Nu- Spt.:cd ofdrivcn shnft T" - Torque 01 driving shaft T,, - Torque at driven shaft Then. powe r available at shaft A is
ill rcst. When the driving shan is rotated. th e oil
s tarts moving from the inner r.u..lim 10 the ouler radius o f the pump impeller as shown in th e Fi~. 10.1. The pressure energy and kinetic energy of th e oi l increase allhe o uter
radius of Ihe pump impclh:f.
ill1U powc:r tmnsmitled
10
shaft B is Pfl
a
Nil TIJ
TI1C t.:fficiency cqualion then becomes N IJ x TIJ
l3u\ T,\
Figure J0.1
(b) Efficiency
Tl1 I.e thL: torque lransmiltc:d is the same.
'lFC
Fluid (or) Hydrafdie coupling
The energy-increased oi lthcn enters the runner of the reaction turbine at the OUlcf radius and flows from the ouler radius to Ihe inner radius of the turbine run ner. The oil while nowing through the runner, transfers its energy to the blades of the rullner and makes the runner to ro tate. Thus. linally, the driven shan rotates. The oil then nows back into the pump impeller. Thus, a contin uous ci rcul ation of oil occurs and hence eOnlinuous rotation or both shafts nrc maintained. The power is transmitted hydrau li cally rrom the driv ing shaft 10 the drivt.:n shaft anti the driven shnn is rree rrom engine vibrations. The speed of thl! drivt.:11 shart is nlwoys kss Ihon thl! spl.!cd of the driving shnft by about 2 pcr cenl. The efficiency or the power transmissiun by hydraulic cou pling is about 9R%.
Driven shnft spet.:d Driving shaft spL:ed
=
(c) Slip Slip of fluid coupling is defined as the mtio of the dirferclIl;e of thl.! speeds of tht: dri vi ng nntl driven shafts to IIle spced ur the tlriving sha rt. ~a lh c mati <:ally. I
Slip
I
N J1
\
But power at any shun is givc n by
2rrNT
Power = - --
60
IPower a(Specd x Torque) I
Nn
N" Nil
I
I
-
t7
N,\ or 10 -
The efficiency of hydraulic coupling is derived in the following equation. Power output 'Inuid (oupling Power input Power tran smitted to the driven shaft Powcr avnilnble at l1u: driving shnrt
(o r)
=
1- '1rc
I
The flu id coup ling efrects a smooth transfer of power from th t: enginl.! Iu the transmission. All j~rk in g ilnd roughncss is eliminated by the usc or the nuid coupling. 111b provides smooth (;.Ikc orf and reduces the wear and strain on the drivc trJ in . At high enginL: speeds, the coupling is vl.!ry efficient. Itgivcs ont: 10 one rati o hetween driven ilnd driving members. AI medium speeds. the coup ling is not quite as crfective. At low engine speeds there is little power tran sfer. When the engint: speed is low. there is flO power triln ... rcr. This is same as having a conventional dutch in the disengilgetl position. This allows the coupling to act as a c1ulch. As the engine speed is increased. powcr transfer hccomes morc e rfectivc. The Iluid couplillf! cannOI incrc;'lse the torque above that produced hy the crank shafl.
10_ _ _ _ ___ POWER TRANSMITTING TURBO MACHINES
INTRODUCTION Turbomachincs are crllcgoriscd as (a) Power nbsorbi ng turbomachincs Examp le PUlllpS. fa ns, blowers. and compressors. (b) Power generat ing turhomachin es Example Steam/gas turbines. and hydrauli c turbines. (c) Power transmilling lurbomachincs Example Fluid coupli ngs and torque convertors. The power gcn cr.i.l!in g and power absorbing turbomachines are used respectively 10 product! and absorb power, the fluids fl ow in g throug h them experience a change in enthalpy between the inlet and the outlet. In it power lIunsmiuing turbo mach ine. the Ouid is totally conla in oo in a rigid c<:Is in g and docs not cross the sys lem boundary. The power Iransmi ltin g lurbomnchines essentially serve the saml! purpose as mcchanica ltrnns mi ssion devices such us a gear truin and clutch assembl y. in transmitting power between (wo dirrerent shafts. In a gear lrai n. the d rive is positive i.e. if the driver shaft rotnlcs, th e driVl!1l shaft will necessarily rotate. In a power transmitting turbom achi ne. th ere is no positive drive. since the driver and driven sha fts are not mechanically coup led.
THE HYDRAULIC COUPLING
-
The nuid or hydrauli c coupling is a device used for rrnnsmitting power from driving shon to driven s haft with th e help of fluid (generally oil). Gil is used as the working flu id because or its Slabili lY, non-corrosive nalure and lubricaLing propcnics. There is no mechanic31 con nec tion between the two sh3fts. The hydraulic coupli ng consists of a radial pump impeller mounted o n a driving shaft and a rOldio l now reaclio n turbi ne mounted o n the driven s haft 80th the impeller and runner arc ide ntical in shape and together fonn a casi ng which is completely enclosed and filled with oil.
·US ;.:.
Tt JRDO M "" IIJI\IES
is L!qual to 1.8 m/s. Determin e (a) lii se hnrgc through th e runner and (11 ) width at nutlet if the width at inle t is 20 cm. (MKU- April '97) (Ans: (a) U6 m' / s. (b140 eml 9.18. A reaction turbine works lit 500 rpm under a head of 100 m. 11le dian1l!ler of th e turbine al inlet i:- 100 em and the flow area is 0.35 m 2 . The angles made by absolute and rela li ve velocities at inlet arc 15° and 60° respectively wit h the tnngentinl \'e locity. Detcrmine (a) the volumc fl ow rate. (bl the rowerdcvdnJ1l!d and (l' ) ef(icil!ncy. Assume whirl at ou tl et to be ze ro, IAns: (a) 2.9 m' / s, (b) 2356 kW and (eI82.65% 1 9.1lJ. A Francis turllille wi th an ovcml1 efficiency of70% is required to producc 147 kW. It is working under a hcnd of R m. TIle rl!riphcml veloci ty = o.3....f2i7-i and thl! rad iill vc lClci ty of now at inl et is O.96../IiiH. The wheel nlllS at 200 rpm ilnd the hydraulic lossl.!s in the turbine arc 20% of the av
9.20.
9.27.
9.28.
9.29.
9.30.
9.3 1.
9.32. 9.33. 9 .34 .
9.35. 9.16.
or the turbine
..L1!l
:,..
TUfUJU M ,\OIlNES
H YI)RAIIUc-TI JIUII NI:S """' -Ln
Y. f 7 . Odin!! hydraulic c f!l c iency. l) . f R. J-Iydr;lUlic c nic ieney is 100% when thl! nng lc turned through by the jet in the
horiwnta l plane is - - . Q. 19. Ddille Ihe terms for a Pelton turb ine ,
9.20.
1) •.2 1. 9.22 . 9.23.
9.Z-' . 11.25 . 9.::!fi. 9.27. 9.:!R. 9.21). 9.:\0. 9J I. 903:!. 9.33 . 9.34 . 9035 . 9.36. 9.37. 9.3R. 9 .."\9. 9.-'0.
(a) Nozzle Erfic icncy (bJ Nozzle ve loc ity coe ffic ie nt ll1C n07.Zle efficiency and velocilY coe fficient are relatcd by (u) 11N IC~ = I (h) 'iN = CII (l!) C 1• = '7~ Ddin~ overull efliciency of u Pelton turbine. C la s~ iry radi nl now turbinl.!s. Dr;l\v th~ velocity trian g les for an inwa rd !l ow rodial turbine . What is a draft tub!.!? Why is il used in a radial now turbine '! Ddin!.' tht.' hydrauli l! dficicncy of a rudiaillow turb in c. How t\OI.!S an ill ward now rmJiu lt urbinl! arJjus t uutomutici.llly In thl! load variali nn'! What is an
EXERCISES 9 . 1. Draw a ncat sketch o f a Pelion turbine and descri be the func ti o n of its main componc nl s. 9.1. Obtai n an expressio n for the workdone per seco nd by wa ter on the runner o f a Pelto n whee l. 9. 3. Draw inlet and ou tlct veloci ty tri ang les for a Pcllon turbine and explain them briefiy. 9.4 . Derive an exp ression for maximum efficie ncy of th e Pelion wh ee l giving the re lati o nship between the jet speed and bu cket speeu. 9.5 . De fine the term s: speed rati o. jet rati o. and coeffic ien t of veloc ity.
9.6. Desig n a Pelton wheel for a head of80 m and speed 300 rpm . The PelIOn whc:c:1 devdops 140kW. TakeClI = 0.98. spced ralio =:0.45 and OVl!rallcfficiency=O.H. IAns: (a) D;J.112m. (b) d;8.55em. (e) z;221 9.7. A Pelton wheel is to be desig ned for the following specifications . Power = 1000 kW, head = 200 m, speed = 800 rpm, overaJi efficiency = 0.86 andje:t diame:te:r is not to e:xcecd one-tenth the wheel diamcter. Determine: (a) Wheel diameter. (b) Diameter of the je:L and (c) Number of jeLS required. [AIlS: (a) 0.673 m. (b) 6.73 em. (e) 21 9.8. A Pelton wheel has n mean bucket speed of 10 mls with ajl!l ofwatl!r flowing at the rate of 700 litrcs/sec under u head of 30 m. The buckets denect the jet through an angle of 160°. Calculate the: power and efficiency of the: turbine . eMU-April '98) Assume C u = 0.98. IAns: (a) 187.110 kW. (h) 90.8%1 9.9. A Pelton wheel is having a mean bucket diamcter of I m nnd is running at 1000 rpm . TI1C nct head o n the: Pelton wheel is 700 m. If the side clearance. angle is 15° and disch arge through nozzle is 0. 1 m J Is. find (a) Power available at the nozz le, and (b) hydraulic efficiency of the turbine . IAns: (a) 686.7 kW. (h) 97 .2%1 9. 10. The three jet Pelton wheel is required to generate 10,000 kW under a net head o f 400 m . The blade ang le at o utlet is 15° and the reducti on in relative velocity while passing over the blade is 5%. If the overall eflicie ncy o f the wheel is 80%. Cll = 0 .98 and Speed ratio = 0.46, find (a) the diameter o fuu:jet. (b) total fl ow in m 3 Is nnd (c) the forc e exerted by a sing le jet on the buckcLS in the tangential direction . IAns : (a) 12.5 em (b) 3.1 B m'ls (e) 94.1 kNl 9 . 11 . A Pelton turbine develops 8 MW under 8 net head of 130 m at a speed of 200 rpm . Assuming lIle coefficient of velocity for the nozzle as 0.98, hydmulic efliciency as 87% speed ratio as 0.46 and jel diameter to wheel diameter mtio as 1/9. delennine. (a) the discharge required, (b) th~ diameter of the wheel . (c) the diameter and number of jets required. and (d) the s pecific speed. IAns: (a) 9.6 m'/s. (b) 2.22 m. (e) 0.25 m and 4 and (d) 40.81 9. 12. A Pdton turbine develops 3 MW und~r a head o f 3oo m. TIle overall efficiency of the turbi ne is 83%. If speed ratio = 0.46, e ll 0.98 and s pecific speed is 16.5. then find (a) diameter of the turbine and (b) diameter of lhe jet.. IAns: (a) 1.78 m and (h) 0. 14 ml 9. 13. Draw and Explain the main pans of a radial now reac ti on turbine. 9 . 14 . Draw the inlet and outlet velocity triang les for an in ward now reacton turbine . 9 . 15. De fine the following terms for a radial flow reaction turbine . (a) Hydraul ic efficiency (b) overa ll efficiency. 9. 16. An inward now reactio n turbine has ouler and inner diameters of the wheel as ! m and 0.5 m respec ti ve ly. The vanes are radi al at the inlet and the discharge is radial at th e oULlet.. TIle water enters Lh c vanes al an angle of 10° . Assuming the ve loci ty of now to he constant and is equal to 3 mis, find (a) the speed of wheel and (b) the vane angle at outl et. IAns: (u) 325 rpm. (h) 19.43°1 9. 17 . An inward flow reaction turbine hilS external and internal diameters as 1.1 m and 0.6 m respectively. TIle velocity of now through the runner is constant and
=
434 ;. T UlWO M ,\ CHIN ES
l BO - PI
lun
=
12.9B' 167 .02"
=
PI
-I (
=
8t11.
6.95 ) 33.5 _ 3.34
'1m
p
--(:'-C-'~,U~)-
=
pgQ
30000 x 10' (3.34 x 33.5) 10' x 9.BI x 300 9.B I
- '-g-
x C x 2Y9.5H - --- x 0.91 x 5.97 ~
8.~ X In
U '1m
=
'/0
X
'111 = 0.95
X
0.894
N
=
N
=
Find the operating speed and diameter of lhc run ner of a Kap lan
Y. I . 9.:! . Y.3. l) .4.
Solution H
4.3 m 2
=
Spccd ratio
'10
Flow ratio
= =
=
Speed ratio x JZg H
2 x .)2 x 9.B I x 4.3
=
=
Flow ratio x J2g Ii
=
0.65 x
=
5.97 m/s
J2
lJ.9. 9. 10. 9. 11 . 9. 12. 9. 13.
x 9.BI x 4.3
Delin!.!: Hydrau lic lurbine.
DilTl!rclltia[!.: he[w!.!c ll turhinl.!s :md rulllp .. How ilrc turhinc s da~s incd ':'
pgQH
=
11500 x 10' 0.9 1 x 10) x 9.81 x 4.3
=
299.58 m' /s
hunci ... turhin!.! is n - - spl.!ci!it; specd turbinI.!' . Thl.! Iluid J1 o\\'!'o in the - - direcliun through the PdlOn turhine. \vh .1I is spear in a Pelion turbine'! Describe bridly the huckciS of Pelton wlle!.!1Th e jet uf watl'r is dcnectcd by the !luckels through an angk o f hdwct:n (n) IOOand 12()"
p
Q
.J I. t) rplll
(b) 160 and 105' (0) 130 a"d 140'" 9. 14. \Vh:)! is mc;}nl hy brcakingjcl'! 9.15 , Draw Ihl.! inlet aml outl et vcl o<.: ily tri ang les for a sing ll.! hucket n f Pelh III whl.!cI . 9 . 16 . For t"bximul1l Ene rgy transfer. Ihl.! wht:d velocity is (:1) 2Ct
Nnw. '/0
x I RA
-if x- 8.-3H
(0) Bol h (u ) and (b)
18.4 m/ s
Flow ve loci ty.
C,
nO 00
Energy avui la bh!;I1 the impulsl! turbint: JIlI!.:1 is onIy Kin!.:lie Enl.!rgy. (Tl'udFabe ) Y.S. Wh .. 1 is a rcw.;lion lur!linc '! 9.6. Classify hydraulic turbinl.!s al.:(';llrcling to lh~ cJirl.!ctioll of fl ow through runnt:r. Y. 7. Pellun wheel i ~ <.I high head turbinl.! . lTrut:lful sd 9.R . Low he
0.9 1 0.65
(a) Runner diameter U
;rON
SHORT QUESTIONS
mtio = 2 Flow mtio = 0.65
1150 kW 0.3 D
=
B4 .9%
turbil1!.! h;] ving the foll owing specifications. Rah!d power = 11 500 kW. Average head =4 .3 m. Owrul1 efficiency =9 1%. Diameter of runner boss =0.3 x diameter of runner. speed
Dh
,
(b) Speed
'10
p
-
if
o
(c) Overall efficiency
Example 9.20
,
;(I - O~l! )
=
89.4 %
'I",
If
1
4Q
=
=
"'
:l ( V- - (0.30)') x C
=
(b) Mechanical efficiency
;r
-yIO- - D ;; ) :.: C,
Q
I
( b ) CI
CI / 2 whl.!rc C r is the velocity of jet ul nUL"kcl inkt.
(0 )
,132
Ii Y nllAlJLlC T URD I NES
;.. TURBO MACI-IlNES
E x ample 9 .18 A propeller turbine runnin g al SO rpm has a runner di amell.:r as 6 01 and efreclivc area of flow as :20 01] . The angle of Ihe ru nner bladcs al inlel li nd oull ct urc 1500 and 200 respec ti vel y wil h Ihe tangen t 10 Ihe wheel. Assu ming l:onSlan l ve locilYof now calc ul ate, (a ) discharge. (b) Ihcorc tici.l l power developed, and (c) hydr<.lulic effic iency_
= Cr ! =
= 50 rpm ,
d=6mA rffl'crt'l'1' ='20m 2, /31
= 150
0 •
fJ2 =20
Q ,
Crl
(5.7 1)' 15.7 x 5.8 1 + 2 x 9.8 1 9.8 1 10.96m
uC.•,/g
= C r :!
433
5.7 1 m/s
= =
H
Now.
Solution N
C2
.-
1/11
-,-1- =
1111
84.84%
( 15.7 x 5.8 1)/9.81 10.96
(a) Discharge /I
.
As in K<.Ip lan IU rhi ne, VI
JrDN =- = JTx660x 50 = 15.7 m/, 60
=
= u.
U2
:. U:! = 15.7 m/s
Example 9.19 A Kaplan turbin e has an ou tcrd iumctcrof S m and lnncrdiamr.:tcr as 3 m and deve loping 30,000 kW al 80 rpm under a head uf 12 m. 111e discharge Ihroug h Ihe runn er is 300 m3 /s. If hyd raul ic cflici r.:ncy is 95%, dctcrm inl! (a) inlel and outle l hladc angles. (b) mecha nical efficiency and (cl overall cffi cicnl:Y·
Solution
From outl et veloci ty tri angh: (Refer Fig. 9. 10 1
= u').
tan /J].
Giv!.!n Cr'!. :;:: Discharge
D = 8 m. = 80 rpm .
Dh = 3 m, = 12 m.
N
15.7 15 .7 x tan 100
:;;:
5.7 1 m/s
e = Cr
= 30.000 kW .
p
Q = 300 m) /s
H
(a) Blade angles at inlet and outlet In Kap hm IUrbines V t = U'). as now is ax ial and erl =
'111
= 0.95
e,~
Now,
rl
Jr DN
Q
At/ft.c /jllt X
Cr
'20 x 5.7 1
Q
11 4.2 m'/,
60
U
rr
U
c,' -cx,
15.7 -
pgQ
Pow!.!r developed
(U~x , )
10' x 9.81 x 11 4.2( 15.7 x 5.8 1) 9.8 1 10,1 16.98 kW
H
= 6.95 m/s
4
tan fJ2
fi,
=
6.95 -C"' = - -
=
11.720
U~
33.5
Using Ihe hydraulic efficiency rclatio n C.\·,
(c) Hydraulic efficiency
300
From oullel vt!loci ty tri angle IRefer Fig. 9. 10}
ex,
5.8 1 m/ s
CXI
= 33.5 m/ s
,
- (8' - 3' )
5.7 1
to,,( 180 - 150)
,
rr
From in lel veloci ty Iriangle lRdcr Fig. 9. 101
=
1fx8x80 60
;j(D- - Db) x C,
Discharge, Q
(b) Theoretical power developed
t"n( 180 - fil)
=
111/ x gH
=
= =
U
0.95 x 9.8 1 x 12 33 .5 3.34 mls
Now, from inlet velocity tri angle [Refer Fig. 9. 101 tan (I 80 - fill
=
C,'
l-huRAt !u(' TtJlUII~E5 -i 431
430 r Tunno MACHINES
l 0.88D2
From inlet velocity triangle (Fig. 9.18(a))
34.38
=
:. D
6.22 m
~
=
C'l l
lan 35° 9.92
TIle speed of the turbine is given by
= =
rrDN
U,
60 60 x 21.91
.·. N
Wq
C,rl
0.88 x 20 x 9.81
·V 4
67. }
v'mo
14 . 17 12.1811\/s
(5 .6)'/"
N,
tV.!'1
745.25
Example 9.17 A K.pl.n turbine working undcr • he.d of 20 m develops I J 800 kW. The outer diameter of the runner is 3.5 m and hub diameter 1.75 m. The guide blade angle at the exlrcmc edge of the runner is 35°. TIle hydraulic and overall efficiencies of the turbin e arc 88% and 84% respectively. If the veloci ty of whirl is zero at outlet delcnnine. (i) (ii)
'111 x H x 1-:
U,
N,jp H
Figure 9./8(a)
::;:
'1l1C SpcL'itic speed is given hy
N,
=
Solution P=IIBOOkW. '11/ = 0 .88.
D=3.5m. '111 ::;: 0.B4 and,
Q Q
(i) (a) Runner inlet angle, [.n(180 -
=
fi,)
7R.66
~,
180 -
C"
Wq
iJl
=
101.34°
(i) (b) Runner outlet angle. From out let velocity triangle (Fi.g. 9. 18(h» For a Kap lan turbine VI V1 andCq ::; Cq
tanfh
PRQH
IIROO x 10' 10] x 9.81 x 20 x 0.H4 71.598 m'/s
=
rr,
4
N
71.598
,
,
x (3.5- - 1.75-)
9.92
m/'
Figure Y./8(b)
=
39.2°
(ii) The Speed of turbine
,
4'(D- - Db) x C"
"-
u,
130 ~,
Q is given by
Q
in th!.: negative X -axis direction .
P
= = =
Wq is
=
TIle dischargc through (he runner is
'I"
Dh=1.75m C.Q ::;: 0
12.18-14.17::;:-1.99m/s
The negalive sign indic:ltes that the
Runner inle t and ou tlet vane angle~ at the extreme edge of the runn er and Speed of the turbine.
Givcnlf =20m. Guide blade nngle Q'1 = 35",
14 .17m/s
V I - C.I • Using the rdation for 'III
" x 6.22 67.3 rpm
N
tan3S n
= =
N
U,
X
60
rrD 12.18 x 60 rr x 3.5
= 06.46 rpm
-I~ 8
HVDRAUUc TURBINES "'" 429
;. TUlU10 M AClUNF.S
But w.r! equals Vm s ince
Availab le r ower = pcQ /./ . Hcnc~ flow rate
eX!
is zero . Hence
27 x 106
Q
tan
1000 x 9.81 x 23
=
Out leI blade angle
11 9.7m'/,
fl,
RCIIor spel!J r.l t mean di~lIne(l.!r
V'"
8.21 R, := I'. 26.5
=
Ird... N
60 Jr
= =
x 3.375 x 150
60 26.5 m/s
= 17.2"
Example 9.16 A Kap lan turbine runner is to bl! designed 10 develop 9100 kW. The net availah le. head is 5.6 m . If the speed r.l.lio = 2.09. now ratio = 0 .68 . overall eflicil!l1cy 86 % and the diam!!ler of the hess is If) the diumele r of the runner. Find the diameter o f the runner. its Spel!llllnd thl! spt!ei fi c speed of the turbine.
Solution Given
Power given
10
P = 9 100kW
Power available x '111
runne r
=
27 x 106 x 0.93
=
25.llmW
Flow ralio
=
0.68
(C,r~ = 0)
IV
=
pQVmCx1
25 . 11 , 106
=
1000 x 11 9.7 x 26.5 x C.I I
=
79
I
D-DiamClcr of runner Db-Diameter of boss.
.
Now, speed rallo =
m/+ V;;'] II
.
Q
VI fii"":Tj
v~gH
= =
VI
=
A xial veloc it y
=
C,'
,
=
119.7 x 4 7f x (4 .75 2 - 22) 8.21 m/ s
From Ihe inkt veloci ty triangle rRefer Fig , 9. 1OJ.
eu =
O.6H x (2 x 9.81 x 5.6p
7.13
m/,
The Overall efficie ncy is given by Cr ,
P '10
pgQH
c"
=
:. Q
=
21.91 m/ s
J 2gf/
~(D2_d2)
lan( 180' -illl
, 2.09 x (2 x 9.81 x 5.61'
Flow rat io = - - -
4
=
= 0.86
Db = - D 3
BUllheoretical power given 10 runner is
c" =
Spoou,atio=2.09
H=5 .6m Ove rall EfficiencY'111
=
9 100 x 10' 0.86 x 9.81 x 5.6 x 10' 192.6
26.5 - 7.9
tn ]
/s
The discharge through a Kaplan turbine is given by
Inlet blade angle
#1 = 156.2'
Q
=
At o utlct
192.5
Ia n fh =
ell I \VX~
"
426
>
HYIlll,\ULlCTUItUINES '" 427 TURBO MACHINES
Hence.
Detenninc (i)
(i i) (iii)
fJ I = 24.B2C.
TIle runner vane angles aL the inleL and oULlet.
-
Work <..lone by Ihe water on the runner per kg of water and Hydraulic efficiency
or
Solution
13, =
Given Dl = 0.6 m. N = 200 rpm.
D2
=
1.2 m,
Ii = I qm.
= 15°, 0'2 =90° 0'1
C"
Lan{3, = - '
CrJ
= Cr~
(/,
lan-I (_4_) 12.57
= 4 m/s Hellct.!.lh
=
17 .65
0
(ii) Work done lV / m
=
6.28 x 14.93 93.76 W I (kg /s)
(iii) Hydraulic efficiency VI C., 1
~1/ Figure 9.17
(i) Inlet and outlet vane angles (/,
= =
(/,
= =
W'I"J
=
cx, =
"DIN fiG " x 0.6 x 200 = 6.28 mls 60 rrD2N 60 1T x 1.2 x 200 = 12.57 m/s. 60 C-"l -UI 4 C,' tan 15° = Ian 15°
14.93 mls
=
gH 93.76 9.81 x JO 95.58%
Example 9.15 An axial flow hydraulic turbine has a net head of 23 In <.Icross it . and, whcll running at a spced of 150 rpm, develops 23 mW. The bll.ldc lip and hub diameters are 4.75 and 2.0 m respectively. Ir the hydraulic efficiency is 93 per cenl and the overall efficicncy is 85 per cenl, cakulalc the inlet and oullel bladc angles ilt the mcal1 radius assuming axial now at out lc!.
Solution Mean diameter
D+d
dm
=
2 4.75 + 2 2 3.375 m
Overall efficiency
I
14.93 - 6.28 = 8.65 mls
... Wt l
C"
Ian fJl
'10
=
Power available
= =
\VAl
{3,
=
Lan -
I
(~) 8.65
Power developed Power availablt.:
2) x 10'
0.85 27 mW
.1 :!.I
HYDRAULIC- TuRBINES 4: 425
., T! !;l.IJO M",' IIINF.S
Fn lll1 hyurau ic dficiency relm ion '111 x
C'II
Solution J-I :::: 30 m . D\
gH
0.94 x 9.81 x 70
vanes. Iht! working head.
23.05 28m/s
H
"I
Runn.:r
V: H1C
CX I
f/ = - + --
28
g
2g
-U1=C"1
C"
5.56
ta ncq = Vt V I = Cd / tana l
But C r!
::;
H=
liD D, =
I 2DI
= 0.2935 m
x 0.2935 x 750 60 = 11.53 (
al
= Cd/ Ian 15
c
C
I
PI
WI=Crl
= 3.732C"
D,N
tan _ I
C~
From in ll.:t velocity triangl e, (Re rer Fig. 9. 16)
48.32°
IT
C;
Uf C;,
V,
=
C"l
VIC"'I 2g g = 90". C x ! = VI
11.23"
-=----""---
IT
Crl ::; C'1 '
+~=--+-
Si nce p[
C"
fh
g UIC\! -g
tan _ I (5.56) --
From outle t veloci ty tri angle. I Refer Fig. 9.61 Inn
= 0,
£= U I C r l
C" = C.1I - V I 2B - 23.05
fit
C~!
= 15°.
Since. C.I·~ = 0, C"l = Crz
H=
angks:1\ inl..:! nnd out let
tall P I
fjl :::: 90~. al
= Euler's head + -.-l ::; £ + ---.l 2g 2g
C"
2R
=
= 0.6 m.
C:!
From inlet veloci ty tri angle. Guide blade ang le I Refe r Fig. 9.6]
=
D"!
(a) Speed of the wheel Ir thc.:rc.: is no loss or energy in the guide and runner
V,
"I
= 1.2 m ,
ml'
5.56 -) 1 1.53
V[::::
"
3,732 Crl '
C;I r, + . 2g
(3.732C )1
g 30 = I A2C;,, + 0.051 C'" C" =4.51 6m/s Then
V,
= 3.732 x 4.516 = 16.854 ml' 60 x V,
Speed N = ~ = N
25 ,74"
Cr~. and
60 x 16.854 rr x 1.2 Figuu 9. /6
= 268.24 rpm
(b) Vane angle at exit
WiJt h 4lt ou tle t. D ,b ,
-- =
D, 0 . 1174 m
0.5R7 x 0.0587 0.2935
tan P2
=
V,
4. 516
V; = U2 "D,N IT x 0_6 x 268.24 60= 60
8.43 m/ s
Example 9.13 A Francis turbine working under a head of 30 m has a wheel di:.lmctt:r (If 1.2 In n L th e en trance and 0.6 m at the exi t. The vane angle alth e en tran ces is I)()" and guide blade angle is 15". The woter a[ th e exi t leaves the vanes without any IiIng.cn ti i.!I vc locity and the veloc it y of flow in the runner is cons tant. Neglecting the cffcci of draft tube, and losses in the guide rind runner passngcs determine [he speed IIr thl.! whed in rpm ilnd vane angle at exit. tBV-April '96)
C"
/3,
=
28 .18°
Example 9.14 An ou.\ward flow reacti on turbine has internal and external diam· Clers of the runner as 0.6 m and 1.2 m respectively. The guide blade angle is 15" and ve loc ity or flow through th e runner is constant and equa.1 to 4 mls. If the .spcc~ of the turbine is 200 rpm. Head on the turbine: is 10 m and di scharge at outlet IS mdml.
....., 422
HYDRAULIC TUltiUNES ~ 423
;. T URBO M ACHINES
(v) Width of runner at outlet The discharge through a radial now rcaclion turbine is gi\'cn by (or)
Dlbl
=
b,
=
b,
Solution P = 330 kW
D, 0.9 x 0.2 0.45 0.4 m
=
Design an inward now Francis turb ine whose power output is 330
of lh e runner. Flow velocity is constant and discharge i.Ci radial
D2b1 D]111
=
Example 9 .12
kW under a head 0[70 m running at 750 rpm. 1}1J :;:::: 94%. '10 :;:::: K5%.111C flow ratio ill inlet is 0.15 . The breadth rali o is 0. 1. TIle OUh:r diameter of the runner is twice Ihe inner diameter of runner. The th ickness of the vanes occupy 6% of Ihe: circumtiill area
= 70 m Bre ad lh ratio
/I
Flow ratio = 0. 15 Flow veloci ty
Flow ratio x
(vi) Mass of water flowing through the runner per second
Q
0.1 5 x
pQ
TIl
=
IT
=
=
D,b, C" =
n,
=
1.018 m'/s 10' x 1.018
1;1
=
101 8kg/s
x 0.9 x 0.2 x 1.8
IT
/I _
Q
PI
v?
pg
2g
VIC. ] g
OJ
Q :;: :
Vl
pg H x
9.973 m
2g
1.8 2 ~~
[·:V2
11 ,
= C2 = C..,J
m(VICx ])
1018(9.425 x 10.208) 97.942 kW
(ix) Hydraulic efficiency ~lf
Ac tu al area of now x Velocity of flow
D,
=
0.1 b, =O. ID ,
Q
=
0.94 x rr x 01 x O. ID] x Cr ]
D'I
=
D,
=
Q
0.94 x rr x 0.1 x C r1 0.587 In
= =
V tC.'1
gli 9.425 x 10.208 9.81 x 9.973 98.34%
0.565 0.94 x rr x 0.1 x 5.56
and h,
=
0.565 m'/s
Since, breadth ratio
(viii) Power developed
IV
~o
O.94rr D1iJl x C,.
- - f +-
+ 2x9.8 1 = - -::-=-9.8 1
=
x 9.81 x 70
But.
9.425 x 10.208
lV
.fiiH
P
=
=
[·:c., =
2g
Ii
'10
))0 X 10' 10' x 9.8 1 x 70 x 0.85
v' -1-
=
0.94
Discharge at outlet
- + -and
f{
J2
~lf
oullct.
5.56 mls
(vii) Head at the turbine inlet (H) Ii
750 rpm ;:::: 2D1·
=
N
= 0 . 1 D,
at
= 0. 1 X D,
= 0.0587 m
Tangential speed of the runner at inlet rrDIN
60= 23.05 mls
rr x 0.587 x 750
60
0.85
HVPRAlJl.lcTI!KttlNES ..;{ .;21 -I~O
~
Twwo M ,\CHI NES
From inkt velocity triangle But.
(
PI - P,) +(ZI -
_ __pg
sin lOa
C,=~ sin lOa
l-h!ud loss in runner
=
62+ (33.8' - 7') _ (31.5 x 32.4) 2 x 9.R I 9.8 1
:;:::
Example 9.11 clers as 0.9 III nnd
(b) The velocity of whirl at inlet
1 3 . 6 ~) In
tun IOn
=
C.l:!
= =
An inward now reaction turbine has externa l and internal diam~ OA5 m respec tively. The turbine is running nt 200 rpm nlH.1 the
C'TI
or
WX t
VI VI Wfl
=
0 .9
=
I.S m/s
In
D2
0.45 m
N
at
10°
012
= =
200 rpm
= =
b l = 0 .2 m
C.q -VI rr x 0.9 x 200 rrD,N
--= 60
=
10.208 - 9.425
=
0 .783 m/s
(0.783'
1.963 m/s
(an - I
/31
iJl .
=
/3, =
tan
Fjglln~
9. JS
I
(S) W.XI
-le
8 0.783
--)
66.49 0 tan - I 7f
u--->
+ 1.8')l
(iv) The runner blade angles
90°
(a) The absolute velocity of water at inlet of runner "lc l )
<--
60
9.425 m/s
WI WI
Solution =
fW11+C;I
WI
The abso lute vc loci ty of water at the inlet of runner. The velocity wh irl at the inlet llle rcla!i"e velocity ul th e inlet The rlln ncr blade angles Width of the runner ut the outlet . Mass or water flowing through the runner per second Head at tht.! turbine inlet Powcr developed and hydrauli c cfllc icncy of the turbine
Given D\ C'I Cr !
tan 10" 10.208 ml'
(iii) The relative velocity at inlet
vcJocil Y trian gles and detcnnine Iii) (i ii) l iv) (v) (v i) (v ii) t viii }
1.8 0.1737 10.366 m/s
=
CI
w id th or turbine 111 in.!.:t is 20 em. The ve loc ity of flow through the runner is constnnl anti is equal to 1.8 m /s. TIle guide blndcs mnkc an ang le of 10° 10 thc wngent o f the wheel anti the di~clt .. rgc at the out let of the turhine is radial. Draw the inlet and ou tlet
(i)
= Cq
CI
Z2} = 62 m
(~~)
D1N
V, = - --= 60 V, = 4.712 m/s
Tf
x 0.45 x 200
/32
(an _ ( ( -1.8 -)
/32 =
20.91 0
4.712
60
I
l "I YOnA"uc T u flOiNES "
418 :;:. TUltrJo MACIIINF_"i
(a)(i) Guide vane angle
(or)
125 x IOJ pgQH = 0.74 Hence. now rale Q
= =
oq
125 x UP 0.74 x 1000 x 9.8 1 x 5.5
163 "
3.13m' /s
(ii) Inlet velocity
BUI also
C,
Therefore.
=
=
3. 13
(C 2 +C 1f] );
=
(9.5 2 + 32.4');
,
" x 0 .836 x 0.4(2 x 9.81 x 5.5)!
0.287 m
r.
33. 8 m/s (b) Runner blade entry angle =
Ian f31
Hence, height ofrunner= 0.287 m.
r' _ 1
=
Example 9.10
A Francis turbine has a diameter of 1.4 m and rotates al430 rpm. \Vater enters the runner without shock with a now velocity (C r !) of9.5 m/s und leuves
=
the runner without whirl with an absolute velocity of 7 m/s. The difference between the su m of (he s tatic and potential heads at entrance to the runner and at the exit from the runner is 62 m. If the turbine develops 12250 kW and has a now rate of 12 m J / s o f waler when the net head is 115 m. find
=
(a) the abso lute vel oc ity of the water at entry to the runner and the ang le of the inlet guide vanes. (b) the entry ang le of the runner blades and (c) the head lost in the runner
fl, =
84.6'
= Energy (head) transferred 10 runnt.!r + Head lost in rUl1nr.:r At inlet. PI If, = pg
rrDN
Cr
+ -2g + Z,
At outkt.
60
x 430 x 1.4
p,
H, = .-:.
60
=
9.5
32.4 - 31.5 10.55
(C) Total head across runn er
Runner tip speed
Jr
C" - VI
Hence .
Solution
V,
e. . .
-
31.5 m/s
Power given to runner
pg
C;
+ -= ~J:
-I- Z, -
Now. for l..:ro wllirl at outlet
But C.l1 is zero , s ince Ihen~ is zero whirl at outlet. Hence
C'tl
= =
12250 x 10' x 60 1000 x 12 xrr x 1.4 x 430 32.4 m/s
Hence. loss of hr.:ad in the runner
.
P,_P,) + (CI- C;) V,C-.., = ( ___ - +( Z,-Z,j- ~
~
419
416 ~ TURBO MACHINES
f-1\'uItAuucTUItOlNF_" -l. "'1 7
Now flow veloci ty C rl , is given by
Solution Q/A
Hydraul ic effic iency
0.5/0.113
=
Power givcn to run ncr 'III
4 ...n m/s
Powe r avai lab le rn(V t C'cJ - U2 C x)
f-rom inl et ve loci ty triang le
pgQN C rJ tan PI But s ince lhe How is radial at outle t.
-1.42 x (an(72° )
13.6 m/s
ql/
eXl is zero and 'm' equa ls pQ . Therdon: V[C,C,
=
gfl
Suhst iluting for C.q gives
T
=
-300 x 13 .6
=
-4080 Nm
0.82
,
Q.97(2g flP
=
hence , This is the torque t.:xertcd on the fluid. TI1C to rque exerted by the fluid is + 4080 Nm and is thl! torq ue cxent.:d on the runner. Torque exerted by water on runner = 4080 Nm
cx,
gli
,
c'11 = OAn(2gHP Now. from inlel velocity triangle lRefcr Fig. 9.61
Power ~xc rt ed IV
=
TO)
=
4080 x
=
538 kW
C" ~~
= 260)
From which, in le t guide va ne angle tan .8\
Hydraul ic Efficie ncy, 'If{
=
Power Exerted
538 x 10' 1000 x 9.81 x 0.5 x 124 0.885
[rom wh ich , w ilh V, >
13,
= -36 .20
.~
Example cl.! nt. T he
0.4 0.423 - 0.97 -0.731 as ( ISUO - 36.2" ) = 1-13 .8"
ex,
Runner speed
SK.5%
9.9 An inward now radinl turbine has an overa ll dficiency of 74 per head 1/ across the turhine is 5.5 m li nd the required power output is
nI.!l
x1
'0 give the blildc a ngle PI
=
f-
= Cr,f\V = e c,'- VI = =
pgQN
=
= -l3.4°
XI
Power av'li labll!
538 x 10'
=
0'[
0.4 / 0.423
=
,
"DIN
"
60 60 x 0.97 x (2 x 9.8 1 x 5.5)l IT x 230
1:!5 kW. The runner [angentia l veloc ity is O.97('J..gHP wh ilt.: the flow ve loci ty is
O.4(lg fI)
,
r[ the speed of thl! runner is 230 rpm w ith hydraulic losses accounting [or 18 per cent of thl! energy avai lab le, calcu latt.: the inle t guide vane ex it ang le, the i nk [ anglt.: of the runner vane, the runner diilnlCler ;11 inlet and the height of the ru nner 2.
al inlet. Assume Ihatthc discharge is radial.
Runncr inlet diame te r = 0.836 m . Ovcrilll efficiency Power O ut put
1],. =
Powe r avai lable
,.
"'I
HVDR,\U u c Tu IUl INES ...: 41 5
4 14 :;:. TuneD M ACUINES
:. d
j;;. ~
~
(b) Diameter of the jet (d)
4 x 0. 1999 rr x 33.62
~
d
0.087 m
d
1
IZ
x D
15 + Zd ~ 15 +
1.47 Z x 0.087
Q
IT
0. 133 01
,
'4d-
~
n
23.45
z
4"
~
X (" 1
,
x 0.13 3- > 53.16
Z4
0.739 m·l / s
Q
(d) Size of b uckets
(d) Power develop ed 5d
Width of buckets ~
~
Example 9.7 A Pelton wheel is
~
0.435
5 x 0.087
p
ryO x (pgQH) 0.84 x (1000 x 9.81 x 0.739 x 150)
O. I04m
p
be designed 10 run al 300 rpm under a head o f 150 m. The nozzle diameter is not to exceed one-twelfth lhe wheel .diametcr. The oycrall e ffi cie ncy is 0.84. Determine the diameter of the whee l. dia meter of jet. qua niLily o f water req uired and power developed . Take C II 0.98 and speed ratio = O .4~. 10
=
So lution d 1 - = - 'Iu = 0.84 C l ,
D
IZ
= 0.98
Speed ratio
= 0.46.
91 3.5 kW
1'- ( Exam ple 9.8 An electric ity generating installation uses a Franc is turbine with a rOLOt io nal s p'ced of 1260 rpm. The net head across Ihe turb ine is 124 m and th e vo lume no w rate is 0 .5 mJ/ s. TIle rad ius o f the runner is 0.6 m. the height o f the runner vanes al inlet is 0.03 m and the ang le of thl! inlet g uide vanes is se t at 72:) fro m the radial direc tion. Ass umin g that the abso lu te now veloc it y is radial at the exi t. fin d the torque and pow~r exerted by lhe wale r. Calcu late the hydraul ic efficie ncy.
Solution Fro m the ang ular momentum equ ati on Torqu e.
(a) Diameter of the wheel (D)
=
Veloc ity of jet C I
1"/0 x Power avai lab le at the nonle
III
I.2d ~ 1.2 x 0.087
Deplh of buckets
= 150 m
~
x 1.59
(c) Quantity of water required D
z
N = 300 rpm H
IZ
d
(c) Num b er o f buckets
~
-
CII J2g li 0.98 J"'Z""x'"'9'.8"'I' xC""7l ,," 50
53 .16 m/s
T
Speed ratio x
=
rI C.I I )
-mrtC.q
- pQrl C q
.ffiH
- I OJ )( 0.5 x 0.6Cx,
0.46J2 x 9.8 1 x 150 ~
X1 -
But Cx ] = 0, si nce the now is md ial at ou tlet. a nd therefore T
Velocity o f the wheel (U)
= m (rlC
=
24 .96 m/s
-300Cl " Nm
The inlet area
But, U
rrDN ~
D D
~
60 60 x Z4 .96 rr x 300 1.59 m
where bl is the inl et run ner he ight
= ~
2Jr x 0.6 x 0.03
,
HVDltAUuCTuRBlNe.S .q: 413 412 ;.. TURBO M ,\CIUNP.S
and hence
lllcrcforc. Theoretical Hydraulic Efficiency
d' 0.465 / 0.5 0.93
=
AClual HyJrualic Efficiency is
0.9 x 0.93 0.837
=
U
0.47 x 102 47.94 m/ s
Solution H = 60 m N
= 200 rpm
U x 60
lTD 47.94 x 60 IT x 0.9 1017 rpm
= =
0.8 37
1250x 10'
=
U
= = =
=
=
=
15.44 m/'
=
:. D
= =
143.5 kgj s
Ih~
rr , = -d4
noz.zle diamcler pCITfi2 111=-- -
4
J2
x
9.81 x 60
rrDN 60 60 x 15.44 Jr X 200 1.47 m
(b) Diameter of the jet (d) Overall cfliciency qo
=
Q
=
where A is the nozzle area
whae d is
0.45 x
U
Also rrom co ntinuity equation
II
x 9.81 x 60
speed ralio x J2g H
D
Ht.:l\ce for one noulc, m
J2
33.62 m/s
( t /2mC~)
0.837 x 0.5 x 102' 287 kg/s
= 0.98 5p.:
= =
1250 x 10]
=
C,
C,J2gH
0.98 x
BUI
SUhSLiluling for Ct and so lvin g for the mass now rate 11/
= 100 kW
Velocity or (he buckels
Actual power developed Energy available in jet
AClual Hydraulic Efficiency
P
(a) Diameter of wheel (D) Velocity of jet CI
=
42.3 mm
Example 9.6 Desig n a Pehon wheel ror a head of 60 m and speed 200 rpm . The Pelton wheel develops 100 kW. Take Cu = 0.98, Sp~ed ratio;:: 0.45 and overall efficiency;: 0.85.
Whl!d rOlational spr.!ed
N
= =
d
143.5 x 4 1000 x 102 x rr 1.792 x 10-) m'
P pgQH P pgH " 10
=
100 x 10-' 10] x 9.81 x 60 x 0.85
=
0.1999 m) /s
BUI,
Q
= Area of jel x velocity of jet rr , = '4d- XCi
H YDRAULlCTUlIfIINES ...; 4 11
410 ):> T URno MACIIINf.S
Solution Number of jel' =Z. Shaft power =ZO.OOO HP. I HP =0.736 kW hence. Shaft power = 14.720 kW. D 0. 15 m. H 500 m. C,
=
=
Velocity of each jet
= 1.0.
C u j2gH I. 0
C,
Solution nle Fi g. 9. 14 illustrates the system wi th the veloci ty triangles.
=
J"'z-x--'9".8"'I'--x-"50""0
99.05 m /s
Area of t!
= =
Figure 9.14
0.0177 m'
:. Discharge of each jet
= =
Power ou tput Energy available in jet IV
Hydraulic efficiency A xC ,
0.0177 x 99.05
1.75
m)
(1 / 2mCi)
/5 At entry to nozzle
:. Total dischargeQ
=
2 x 1.75
fI
3.5 m3/s
=
600 - 48
=
552m
Power at the turbine inlct
=
Nozz le vel ocity coe fficient pgQH
C,
10) x 9.81 x 3.5 x 500/10)
Theoretical velocity
17198.75 kW
C, (28 H)It'
Ovt:r311 Efficiency
ShaftPowcr pgQH
14 . 720 17198.75
1l1US,
0.98(2 x 9.81 x 552)' 1'
0.856
102 m/ s
85 .6%
Example 9 .5 The buckets of a Pelton wheel denect the jct through an angle of 170" while the relative velocity oflhe water is reduced by 12 percent due to the bucket friction . Calculate the theoretical hydraulic efficiency from the ve locity triangles for a bucket/jet speed ratio of 0.47 . Under a gross head of 600 m the wheel develops 1250 kW, when the loss of head due to pipe friction between the reservoir and the n07.7.le is 48 m. The bucket circle diameter of the wheel is 900 mm. and Ihere arc Iwo jets. 11le nozzle velocity coefficient is 0.98 . Find the speed of rOlalion of the wheel and the diameter of the nozzles if the actual hydraulic efficiency is 0.9 limes at that calculated ahove .
Now, UIC.'"! - U2C~ ~ V[(V + IVtl - (V - lV,cos(180" - a»]
WI lli
=
V[(C, -V)( I-K co, a ) [
wh ere W2 .:: 1\ Wj. Substituting th e values, IV/m
=
0.47C,(C, - 0.47Ctl(I·-0.88cosI70·'
IV
=
0.465
meT
l
4(}R ,. Tunuo MACHINES
HVORAuuCTuRBINES 4.
In :tctual practice, maximum efficiency takes place when thl! velocity of [he wheel is 0.46 Jimes th~ veloci ty of Ihe jet.
Hence, Power avai lable at the nozzle
= ~2 x 150 x 91.06' = 621.903 kW
i.e.
U
=
0.46.C
U
= =
44.65 m/s
0.46 x 97.06
Power availab le :n lhe nozzle = 621 .903 kW.
(b) Hydraulic efficiency E
The; whee l diameter is where E
rrxN
44.65 x 60 IT x 430 1.98 m
= =
D
discharge through the nozzle Therefore.
mUSl
be equal
Q =
C x
Gd')
d
(~ X;~)'"
C,'
d
0.202
III
IV, IV,
lhe discharge of the turbine .
= =
IV,
C, = 91.06 m/s
C,-U=91.06-41.89 49.17 m/s
= IV, = 49.17 m/s
'I'
W.l!
=
or 202 mm
WJ.· cos 15° 49.17 x (0.97) 47.49 m/s
Then
IV" - U = 47.49 - 41.89 5.61 ·m/s
Wlm
=
U{C X1
+ C.(2)
41.89(91.06 + 5.61)
Solution N
= 1000 rpm .
H
= 400 m.
4049.5 1 W I kg/s
Q = 0.150 m) Is
(a) Power available at the nozzle = ~I1ICi
-
C.l"2)
From the outlet veloci ty Lriangle [Refer Fig. 9.2)
(MKU-No". 1996. April·1998)
111
+
Assuming no loss of relative ve locity
Example 9.3 A PelIon wheel is having n mean bueke! diameter of O.R m and is running al 1000 rpm . The nel head on the Pchon wheel is 400 m. If the side clearance ang!..: is IS" and discharge through the nozzle is 150 lillsec. find (a) Power available at the noi'.l.ic and (h) hydraulic efficiency of the turbine: .
= 0 .8.
U(C X1
From the inlet velocity triangle [Refer Fig. 9.2]
10
(3.12 x ~ ) 97.06 rr
=
=
W
(c) Diameter of the nozzle (d) TIle!
C;/2g Wlmg
U x 60
=
D
Given D
409
= p Q = 10) x 0.150 = 150 kg I s
:. 1JH
-
= =
c 1 =Jet velocity Assumin g U ;:;: 0.46 ela nd U
rr DN
= 60=
rr x 0.8 x 1000
41.89 m/s
:. C,
9 1.06m / s
60
=
I
(4049.5)/9.81) (91.06'/2 x 9.81) 4)2.79 422.63 97.7%
Example 9.4 1\.0 jelS Slrikc a' iIle buckelS of n Pelion wheel. which is having shaft horse power as 20.000. The diameter of each jet is given as 15 COl. If the net head on the turbine is 500 m, find the overall efficiency of lhe turbine. Take CIt = 1.0. (MKU-April '97)
406 ;. T UiUJo MI\CIUNI:"'1
SOLVED PROBLEMS Example 9.1 A gcnenllor is to be driven by a smal l Pelton wheel wi th a head of 91.5 m at the in let to the nOlzle an d di sc harge of 0.04 m) Is. The wheel rotates at 720 rpm and the veloci ty coefficient of th e nozzle is 0.98. Ir the eHki ency or lh e wheel. based on the energy ava il able at cnlIy to the nozzle is 80 per cent and the ralio of bucket speed to the jet speed is 0.46, dctcnnine the wheel-to-jel-di ameter ratio al the ce ntre-li ne or the bucke ts, and the speed of the wheel. What iS,the dimension less power specific speed of th e wheel?
and jct diDmetcr
(~) 'I'
d
Diamclcr ratio D
S"olution
d
=
Power devclopt:d Power avai lable
Overall effir.::iency,)O
p
pgQH~o
=
0.507
=
0.035 14.5
Dimensionle ss power specific speed is given by eq uation I/2 N I'
1000 x 9.8 1 x 0.04 x 9 1.5 x 0 .8
pl/2(gH)5/4
Power developed = 28. 72 kW Velocity c·ocfficicnt
C,
Cv
(2g H) '/2
C,
0.98(2 x 9.81 x 91.5) ' /'
=
41.52 m{s
=
'I' I 720 x (28 .72 x 10)) - x ( 9.8 1 x 60 10)
=
0.0131 rcv
=
0.0 13 1 X 2rr
) l/' ~1.5
,.d
I
Hence. Power spec ific speed = 0.082 rad . A Pellan wheel working under a head of 500 mclc rs. ~roduccs 13 ,000 k W ul 430 rpm. If the efficiency of the wheel is 85% deternllllc (U) dlschargt: of th e tu rb ine (h) diumCler of th e whee l (c) diamelcr or the nozzle. Takt: C!. = 0.9H (~.JU-OC I . '96) Example 9 .2
Therefore, U
0.46C, 0.46 x 41.52
=
19.1m{s
and speed ratio as 0.46.
Solution Given Ii = 500 m .
Also .
wD
U= -
P
=
13. 000kW.
=
l1u
13,000 x 10] 10) x 9.8 1 x SOO x 0.85
720 x 2rr 0.507 m
3. 12m){s
(b) Diameter of the wheel We know the ve loc ity of jct A
= =
Q{C , 0.04 4 1.52 0.963 x IO- J m'
I
fJgH )( '10
2x19. l x60
Jet area
= O.R5
II
p
Q D
N = 430 rpm .
(a) Discharge of the turbine
2
where D is wheel diameter
I
C
= =
Cv j2gH 0 .9 8 x .;r-2-x-9~."'8I"-x----:CSO"'O 97.06 m/s
,
ii .!
I
1
-W4- ;.:.
TURIlO MA CHI NES
EFFICIENCY OF DRAFT TUBE
r---t----.--------,,--(i)
It is defint:d as the ratio the of actual conversion of kinetic energy into pressure I!nc:rgy in the draft tube 10 the kinetic energy available at the draft lube inlet. TId,
=
(V,' - Vl/2g) - lid
(1','/28 )
or
CAVITATION IN TURBINES
x -----L------+------~~~0
Turbine cavitation occurs on the suction surfaces of the blades, at the runner outlet, where the static pre:.sure is a minimum and the absolute velocity high. It should be avoided although it has little effect on the performance of the turbine since it occurs after the runner. Applying the energy equation between the runner outlet and tail-ract! gives
vl- Vl/2g = (PJ Figure 9.13
Atmospheric Pressure Head
=
+X
pg
a
pg
P,
1','
- -\- -2 -\- (H, -\- X) = PI: g
P,
p" = pg
(Pm", -
p,,,,,) /pg
- 2,
( -P"
pg
-\- X
- fls -
(
V,'
[(POlitI - Pullp)/pg - Z2l!H
= )
V,'
-\- ~ -\- II" 2;:
0]
pg
(Vo' - vl) /2g -lid =
Dividing this equrHion by the net helld across the turbine giV!.!s the 'Thoma cavila/ioll parameler ror the turbine
P" --\-X
..·· p, Ihls vnluc of ---=. in equat ion 9.4, we get
~S ub sillullng
+ ZJ + hd
PUlling ZJ = 0, and as the outlet velocity V2 increases, p:! decreases and has its lowest value when the vapour P,. is reached. At this pressure, cavitation occurs (begins) and hence putting PJ equal to Plltm and P:! equal to P uup • the ahove equation becomes
Drafr Tllbe
and pg
P2)/pg - 22
1','
2g - 2.~ - lid
)
The critical value of NPSH at which cavitation occurs is determined from a lest on a model or full size machine in which P:! is decreased until the minimum value al which cavitation begins or the efficiency suddenly decreases is found. Knowing 22 and H it is easy to compute the critical value SigT1lD c , which is the value below a. as given by rhe above equation, for any other similar machine of the same homologous series must not fall. The above equatio n shows that rhe maximum elevation of the turbine above the tail-race is given by
Z2 -nle above equation shows lhat tbe inlet prt;ssun; at the draft lube- inlet is less than atmnsphcrir.: pres sure.
(NPSII)/H
= (Patm
-
PUll p
)/
pg - acH
This equation implies Ihat, as the net head is increased, the turbine elevation above (he tail-race must be decreased. For an excessive head, 22 mighl be negalivc, which implies that excavation would be needed to place the turbine below the level of the tail-race.
H YOR,\ULlC TIJlUIINF.!' ...:. 4()2 ;:.
·W3
TultDo MACHINI:.S
turbine tyre is also selected based on head of water available at the turbine inlet. This method is based on experience and observatio nal fact ors only.
fJead a/ warer iI/metres
Type a/Turbille Kaplan turbine Kaplnn or Francis (preferably Francis) Francis turbine Peli on LUrbine.
0-25 25 - 50 50 - 250 250 and above
The advarllagcs of draft tube are I . The pn:ssurc at the runner ex it is below the atmospheric pressure anLi the turbine operates efficie ntly as if it is pl aced at the tail -race. 2. The kinetic energy (Ci/2g) of waler at the turbine runner outlet is converted into useful pressure energy. 3. The turbine m
TYPES OF DRAFT TUBES DRAFT TU BE Thl! UV:.lllllhk heud is high in an impulse turbine like the pelton wheel and there is nol muc h loss in the overall turbine output nlthough the turbine is placed a metre or two away from the level o f thc tail-race. Whereas, in the case of reaction turbines, the head availi\ble is low and a considerable fraction o f thi s available head would be wasted, if the IUrbines arc placed above the tail-race level
Run!lt!r eXIi
25-
3D Tnjl·rnce
2
~
D
~
Runncr exit
Tail-rDce
2-
2_5D
1llere arc rour types of draft lu bcs, used in rrae\icc . depending upon the nnw co ndition s and Ihe heig ht of the turbine above th e tail-race and so on. Thcy arc namely,
I. 2. 3. 4.
ConicJi draft tu bl! Bell mo uthed or moody spre ading tuhc Si mplc clhow or bent tube Elbow drafttuhc wi th circular inlet and rec tangul ar outlet.
The fina form is the. straight concialtypc stretching frol1l the turbine wthe ta il -race. The seco nd type is also a straight drafttubl! exceptlhat it is bell-sltapl!d. ThIS type of dmfltube has an advantage that il can allow now with whirl component to occur with very smu ll losses at the turbine exit. In any lurbine. the exit absolute velocity usui\lIy has a whirl ct:Jmponenr espccially at part load ope rat io n, the bdl-shaped dmft tube may b~ preferred where the operatio n is at pari load for long pL!rinLis of time. . The third form is the bent draft tub!!, is used when till: lurbine must he located very close to or below the tail- ra ce leve l for so ml! uther c.:onsiLiermions. Howevl!r. the efliciency o f bent draft tube is usually no t as great ,IS th at of the IIrs l IWO ty pe s. 10": fourth form o f drJft tube (Elbow dra fltube) is similar {o the third one except thatlh.: exit shape is squ
DRAFT TUBE THEORY
n
Conside r the draft tube shown in Fig. 9.13. TIle turbine exit is at a hei~ht abovc the tail -race leve l. LeI subscript I denote the conditions at the rotor ex.it .tnd s ubsc.:ripl 2 denote the cond it ions at the out let. Applying Bernoulli's equation to inlct and outlet o f the drafltube taking the draft tube. exit as the datum line we get,
Figure 9. 12
Types oj drafllllbes
It.:vel and leading th e wJter from the turbine outlet to thc tail mce by a tube such Ih;)1 the wilter"reilches almo::;phcric pressure o nly at the tail race. The pipe of grad ually increasing area which is used for discharging water [rom the turbine exil to the t
P
~ p~
V2
+ --.!. + (H, + X) = 2g
V;
p, ~+~+O+"" PK 2N
(9.'1)
whe re x is the dis t
~oo ;..
I
ThRlm M ACH1 NES
and therefore.
E = (U '
+ v eil COI/J, l/K
Ir £ is con s tant ul o ng the hlude: rndius. and C{/ is con stl1nt OVL'r the cro ....~-se:c[i n nal area , then ,I S U 2 inL're3sc.'i rrom hub (0 tip, Ucm fh IllU St decrease [ 0 kl.:ep £ con stanl. Hence , P1 mu st inc rea se rrom hub to lip and Ihl.: hladc musl thcrdorc hI.! Iwist ed .
GOVERNING OF KAPLAN TURBINE
HYORAULlCTURBtNfS " 401
COMPARISON OF HYDRAULIC TURBINE EFFICIENCIES The characteristic curve (Fig.' 9, 11) for the axial now Kaplan turbine is similar to that of the radial How turbine. The maximum efficiency is lower and lhe efficiency curve is much Halter for the impul se turbine. The Francis lurbine pcuks at the highest crtieiency but falls off rapidly at pan load. The Kaplnn turbine has a much nalter curve man the Francis turbine and exhibits a similnr mnximum dficiency of the two important ax.ial now turbines, Dnd hence it has constant average maximum dficiency while the propeller turbine efficiency increases as load increases. Afler reaching a peak value, it fall s off al the maximum load .
Kaplan lurh illl.! ,l;!llVL'rning I.: mploys IwO servo motors a Be: for \lpl.!ruung gllilk vallr.:s (as in f-rallcl s turhin..:) and the other ror operating runner vunl.!S (Figufl.!s lJ. lOlill & Ch» ). Buth the servo Il1 OlOr dis tribution vu lvc:s arc interconnected . There hy, the runn er vanes and guide vanes arc simultaneous ly opermed 5111hal w,lIer pusses Ihrou !! h the hlades with oul s hod ill all load co nJiti ons. This sys te m of g ov"a ning is kno~vn as 'Oo uhle regul at ion' .
100
Frands Kaplan
90 1]
% 80
Gc nC{;Jtor shaft Gencrat or shalt
70
Servo m OlOr IM::?tL Servo mOlor
Rel!ula tioll rod
60 Propeller (fi xed blade)
"" Turbin!! shan 50 O pCrallll !; rutl
Turbinc Sh;lJ\ 40
20
Figure 9, J1 Bl ud c .~
6()
80
100
% of power
Characterislic ClIrl'i!SJo r hydraulic lurbinl!
1
Runner blade
SELECTION OF HYDRAULIC TURBINES The type o f turbine requ ired for a specific purpose can be selected on the basis of me value of {he specific speed. The typl! of turbine for different specific speeds is given
Fignre 9.10(b)
Sel"l'o IIIOfor 1I1t't..:'Ulll islII
" p e rtl/ i ll}; r lfllJlt' r VlJIl/',f
(lJ
below.
Specific Speed
kap/all tII r /Ji lll' , SIrOIl"I/ ;" st'cf i(JII alld Oil redllced .fca/£'
Servo mOlor ror the runner v,mes, con sis ts o f a cylindcr with a pi sto n o p e r~Hed hy high pressure nil wh ic h is s upplied by distrihu[i o n valuL' (not shown in fi!!ure) . The servo motor p is tun is co nnected with nn opcr;lIing rod whi ch moves up an d~dow n and passl.!S through the turhine sha n whic h is made hollow for Ihis purposc. TIle o pcratin g rod trans mit s the mOli on 10 Ih e runner hlades wilh an appropriate! link m e chani ,~ m l'ncl nsed in th e fUnn el' hub.
.....
535 70 450 -
35 70 450 IODD
Type a/Turbine Pelton wheel with single jet. Pelton wheel with two {or} more jcts_ Francis turbine . Kaplan turbine.
Selecti on of turbine is a highly technical job and rt!quires great experience. SeJl!Clion bast!d on specific speed is a scientific method and gives n precise inronnation . The
H VDRAuuc T uKIIINES -i 399
398 ,. T URBO MACHINES
Ir IhcII_vancso·arc fixed 10 th c hub and arc not adJ·uslable, then r" "nCls . ",nownas ,. .tl,e .tu, ~r7e cr lur me on Ihe otb~r hand ir the vancs on the hub arc adjustab lc lhe [ urbjn~ IS now n as· a Kaplan . .an engmeer . K lb . . ruriJmt:. Il is named afler V. Kaplan , a n A usln . ap a.nl tur me IS s uit ab le, whcre a large quan tity of wale r
2. Guide v:mcs J . Huh wilh va nes ~.
Draf! tube
Waler from the pc nslock en tcrs the guide vanes through sc roll casing. TIl e in lel gu id e vancs are fixed and arc situated a[ a plane hig. hcr than thc runner blades such Iha[ the nuid mu st turn throu gh 90° to enter th e runner in the a"jal din.:c lion. Ll)rtd cha ngcs arc effected by ndj ustll1cnt of the runner blade anglc . Thc fun elio n of the guide vane~ is to im pnrt whirl to Ihe nu iu so lh at the radi;:)\ distribution of velocity is the same a.o; in n frec vorl ex. Since this Iype or tu rhinc is useu for low heads and high now rates. the blades must be long i\lld have large chords so that they arc strong enough to transmit th e vcry high torques th at arise. Pitch/Chord rat ios o f 1-1.5 are typ ical for aXIa l !low turbines and thi s resu lt s in fou r, five or six·bladcd ru nner.
VELOCITY TRIANGLES AND WORK DONE Tlte velocit), tri nngles <.IfC usuall y urawn at the me;!n radius . S Ul CI.! conditions change from hub to tip and arc shown in fig urc 9. 10. The now vd ocit y is a"ta l 011 Ihl.! Inlel and outlet. I'-{c nce, C q :; Cq == CfI . The bladc velocity vector Vt is substraclcd from th e abso lutc vc lnci ty vector. CI which is at angle G'I to VI to yield the relative ve loci ty vec tor W t . For shock free cntry ont o the ru nner WI is til the hlildt.: ang le /3t· For max imum dficienc), the whi rl componenl ex! is 7.ero. in wh ich ense th e absolute
Guide blades
--=-0
-
-
velocity nt exit is axial. and then C:! :; Cr~.
U,
,--",-r- Inle t V11T1 C
Boss
ti p
\VA
X
~,
Runner vanes Outlet
vane lip Toil rn
Kaplan turbine
Fig ure 9. 10
Shaft
~
Vdocir.\' ,riellllgeJ for
(In
axial Jlow hytlrouUc ,urbilll'
Euler' s Lurhine eq ual ion gives Draft tube.
E = U(C x , - Cx,l/g and for 7.Cm whirl at ex it
E = UCx,!8 No\\',
Cx , Figure 9.9
Kaplan Tlfrbine
=
U-Cacot( IBO' -/itl
:;
U
+ Co cOL fJ l
I-IVDltAUl.lc TURIUNE..o; ..;,; 397 :\96 ;... Tunoo r...I AC IIIN ES
ring is rOlaled [0 decn!1Isc the flow area between II . ' . . ' " vn ne angles 111U " II" q , r Ie.:: gU ide vanes by chang ing gUide " ' , I ',. c uanllly 0 waler rcael,', ng IIle runner blades 'IS rcd ' d n to nonn"l 'd II u(.:t:, lcrcIorc, , , 11C speed IS bruu!!ht ~" gr.l Uil y and then new'lI ' I dlstrlhution valve •,HI'lil' .' . . A' rell·f I ,' . ' or, .mUlIl, ever and • II,c,'r no rm~ I POSHIOIl stock h) pre vcnt w~llcr hnmmer which m' " , .,. ,I t; va Ve.:: IS provided In Ihe pen .. y ,In:-;c uUC 10 Lhe sudd 'n d " , I . ow passage bel ween Ihe guide hlades. lis fune Lio " . , t; rc ueL lon In (Ie n III pelion turhine. When there is n sudden dce . !~ lSI SI;i1nr to Ihal ~f a (h:fleclor valve npens ilnd diverls lhc water 10 Ih '1 rease In Oil on the turblnc, Ihe relief pressure) is achil'wd in reae;ion lurt.."c ~sal Sr~ce:ITI,1US dhOUblC rcgublion (spcl.!d and . I I ne.: . 1m! ar y \Y en Ih e load I I' lIlcrr.:ases i.e. when [Ill' .srcc:d I,'," ' crease r 'g ] . " . on ,lie tur lJOC , ." "'"e lCIelH:y 10 In [hI.! ol'rnsi1e dirccti'" ll'"O "" 10 "' ".... rl,!.lsc .' l IIC pUSS'ln ' • ~ .• '" b' e.: u aung ring IS moved in allowing more water It) strike the ru~ner hlad~~':'c urea etween till.! gu ide hl;Ldes nnd
Efficiency
Water power input \-Ielld
CHARACTERISTIC CURVES FOR A REACTION TURBINE Curves of waler power input I . hrake power OUlput for conSI 'II1~lgq,U,cC oexc,fI~d by th,e whc~1. I~ow ratc ef(icit.;llcy and . ' _, penlng arc s 10wn III Fig 97 A I full op!.!'[llng, th e flow rnle as shown in F, 97 ' ',. . . , . 11 0 longer independent of the wheel s ced ~~~ '. v.lr.res with the run,nl.!r speed. It is speed is of must int erest sine' Ih c p " bchavl~ur of the turbine al consl<..IOl . ' r.: gcner.lwr runs al a fixed s 'd A,. I ' S0 Ih. ' no,v r',," I load chall"cs C ' . , . . . . 11 ,r.::lsc lanuedbyv·lr' .·I· fl pee .. Si. ll' dectrlcnl frnm lite Fiu 9 R ,1,'" ",c I" . 1 ' e . • 1<1 Ion 0 lIe gil lt! opt:.lllng. It is secn ;" • . r.: au mcreJses slight] . h i d fric tion head loss TI"',, ,'" ro 0 I' I ' "J Y us t e on decreases due 10 Ihe. , .' '-' p p r lona 10 Q- wi ' I . I ' also he nOied Ihatthe cnic""ne . lI C I IS ess at li gh ter 10LlJs, It wi ll . . Y curve at conS lunt spe"d ' nn Impul , lurbine. r.: I's not as nal as III se Torque
FluiLl power input
.j,
Brake power output Figure 9.8
Reaction tIIrbilie characteristics tJl CO/1SitJIII spud
When a reaclion turbinc opcrnting at constanlspced cxp!!rienccs a loaddr.:crense, thc cross scclionnl aren betwc!cn the inlel vanes changes and nngh: 0'1 decreascs. The inlt:1 runn cr urerl is constanl and lhereforc to salisfy the continuity the relative vdoci ty \VI must decreasr.: . The result is thaI the flow onlO the runner is no Innge r shock free and atl!xit C;! may incrcusc. This givcs n higher kinelic energy loss ul the runner exit a!o well as an im;reLise in the whirl component Cx~ down the draft tuhe, TIle now is Ih!.!1l spiral in nature, which decn:ascs the draft lUhe dticiency. Thc dflclency of u n::.u; tiun turbine al lighl loads thererore tends to be le ss than thai nf thl.! pelton wheel. although the design maximum efficiency may be greater. TIIC advanlage or an in\lJard fl ow reaction turbine over
AXI AL FLOW REACTION TURBINE output
N (r,p.m) Figure 9.7
Rt'(/Cl iOIl (IIrbille c/mrrlcreris/I'cs
(If
filII Opt''';'''£:
III axial now reaclion lUrbinc, waler nows parallel to the aX Is of rotalion orlhe shah. In a reaction turbine. Ihe head at thl! inlcL of the lurbine is Iht! sum of pressun: energy and kint!lic energy and .. part of the pressure energy is converted into kinetiC energy ,tS thc wilter fl ows through the runn er. For the axial now reaclion lurhine. thl! !ohaft of the turbine is ven icul. The lower cnd of the shan which is made larger is known as 'hl/h' or ·bm's'. The vanes are fixed on th e hub and hence the hub aC ls as a runnt!f for thl.! axial fl ow reaction turbine, The lWO important axial now reaction turhines arc I. Propeller Turbin!.! and 2. Kaplan Turhine
J94
;
HnlRAUU C T u IUIlt-:ES ...;
TI JRI)O MAClIINt:.S
Power
where Ird is the hend loss in the draft tube . Putting PJ ".... n d Z J cqua I to zero
P2
pg
,
= (\',- -
,
1',-)/2g - Z,
(b) Hydrnulic cfficiencY(I7I1)
+ Ii"
There ~s a limit for the: extent of VJ reduction because of the tube lenot Increases "II also increascs. The draft tube could be vcr I , ,'
Icn~th
~~' ~s the
•
~~~:"~~~~:~:~h~~I~ n?1 e;eeed abouI8~.IO ensure Ihal separalranO;fb~:~~a~I~~!:: elTec t , erc a so an upper 11 mil on the value of Z::! due the cavi ta-tion IS
TIll~
tcrm (P. I
+ Pm ) / pg QH
H~CCIVl!d
by runnl.!r
=
Fluid power availilhle at in lt':l l1angc
=
(P,
+
ws
E P",)/pgQN = - p~QfI
is the theoretical energy transfer pe r uJlit weight of
fluid now. The maximum hydraulic cf(iciency is
10
'III
=
RADIAL FLOW TURBINE LOSSES GOVERNING OF REACTION TURBINES
The losses in teml S of energy balance through the turbine is given by
P
= Pm + Pr + Pc + P, + PI
(9.3)
where P _ P.I, -- shan d power o utput, Pili ;::; mechanical power loss, Pr ;::; runner power loss c - casing an draft I,ube loss, P, ;::; leakage loss and P ;::; water power availabl • Pr + Pc: + PI together IS the hydraulic power loss. c. Runn~r power loss Pr is due 10 friction. shock at the im e ll er entr 'lnd resullS in a head loss Ii, associnled wilh Ihe now
~i~~"10n."
The quantity of water flowin g to th e reaction turbine is controlled by rotnllng the guide blades, These blades are pivoted and connecled by levers and links to the regu laun g ring (Fig, 9 .6(01» , The regulating ring and lever are connected by tWO regulating rods . This regulating lever in turn in connected with regulating shaft. which is operated by the pist on of the servo motor.
..... ·-Guide vanes open
r~le Ihrougl;l~e ru~~;
--Guide vanes closed
Pr = pgQrhr
n~te~~~~ ~~~~~~~o~~ ~~!S r~anuns;: !ri:~snow rale q leaking past the runner and thererorc
t + Q = Q, and with a total head lir
~1C ro ss
Ir--,_I-'
+q
the runner. the lellkage power loss becomes
From Turbinl! Shaft
Disuibulion Valve
t
P, = pgqHr Casing power loss P. is due I II r ' . casi ng and the dr
Pc
= pgQh,
Regulating Ring Figure 9.6(a)
pgQH
= Pm + pg(h,Q, + h, Q + H,q + f,)
Thus. (a) Overall efficiency
ShJft output power Fluid power available at inlet nang c
P,/(pgQH)
I
I
TIle total energy balance of equation 9.3 thus becomes FrtwciJ Tllrbiflt: Col't:rning
When the load on Ihe turbine increases. the speed increa<;es , The lly hJlIs of actuator move a way frollllhc ax.is and the s le eve raises. The dislribution valvc rod is pushed down, Pnrt.2 is open and port-I is c\O!;Cd . TIle high pressurc oil crHers the servo motor Ihrough port-2 nnd pauses the piston to move towards right. The oi l in Ihe right sidl: tbe piston is pushed back inlo the oil sump through port- I and uppcr past the distributing valve. When the piston of the servo motor moves to the right. regulating
or
or
., J; '!
HYDRAlh.JcTURBJNES ... 393 III fa~ili[;lh': \~ :J1l.:r
Ilnw :1I ":(ln s [ant veloc ity Ihrought)utlh~ cin;umrer~ncc oflhe runner. The ..:asing J:-. u:·awlly l11ad~ of Cllncretc, casl steel or platt: stt:c1.
2. Guide vanes The st~lIion;1ry guide van~s arc fix~d on a stiltionary circular wheel which surrounds the runner. The guide vanes allow the water 10 strike the vanes lix~d on the r~nncr without $hock at the inlet. This fixed guide vanes are followed by ilJ.lustabh: gUJde vaneS. The (ross sectional area between the adjuslilble vanl.!s c,m be \'aricd for !lnw control al pari load.
3. Runner It i. . . a circular whed on which a series of radial curved vanes are !ixed.
AI the OUIh!I. (radius r~), the water h.:avcs the blade al an angle tho 10 the langt:nlial velocity vector. The absolute outlet velocit), C2 is the resultant of IV:!, and U~. The now ~clocitjes err and C r2 are directed towards the axis ofroliJtion ant.! an:: given by Q jJ.rrl'II)[ and Q12n ,.~b2 respectively, where b is the height of the runner. Euler's turbine equ;.Iuon gives
E
= W/lllg = (U,c."
- U,Cx,)/g
and E is a max.imum when ;.Ire. equal at the nutlet or a],
eX}
is zero, that is when LIlt: absolute and !low velocitlC:s
= 90"
NET HEAD ACROSS REACTION TURBINE
The waleI' passl's inln lhe ro!l)r where it moves radially through thl!- rOlor V
The net head' H' across the turbine is the difference in the tutal head bl.!tween the inlet nange nnd tht! tail water level. The net heud H is different from the gross head
4. Draft tube The pressure ill the ex il of the rotur of 11 reaction turbine is generally
H[. TilliS
kss II~an Ihe atmospheric pressure. The water at ex.it cannot be directly discharged to the !;.It!. race. A will:. or [lire of gradually increasing mea is used for discharging the water 1mlll tht.: turhIDe exit to Ihe lail race. In other words, the draft tuhe is a tuhe (If ir~l.:rcas~n~ cross sectillnal area which conwrts the kinetic energy of wa[er al the rllrbllle c."'JI rntO pressurl' energy.
Turbine totnl inlet head =
Po pg
+ Vr~ + Z" 28
and turbi nc. loull outil't head =
VELOCITY TRIANGLES AND WORK DONE The inkl :lI1U uutkl velocity trianglc.s for a. runner are shown in tigure 9.6. Water enle rs the runner frolll the inkl guide vanes (at radius TIL with absolute \-clnt lly. al an angll'rtl tn Iho..: uirection ofrnwlion. The tangential vclocitv at the inlet h VI. The relalive velucity vcctor 1V1 oblaind by substfacting VI from CI'at inlet, is al an angk tl[ to Ihe direction of rOiation. /11 is also the inlet blade angle fur shock-free elltry.
Ur
Wx"
where the pn.!ssures arc gauge pr~ssurc. Total head across IUrbine (H)
But ill the tnil-race P-, is atmospheric and 23 is zero. Therefore
I-l
= (P,,/pg + V;/2g + 2") -
V{/2g
Also,
II = II, - hlp - V,'/2g where II j (l is the pi[le friction head loss and the energy given to the runner by the: waler pcr unit weight of now is HI /mu = H - lid - II}: - Ii,
If the water is discharged directly into the wil-race from Ihe mnner outlet, the kinetic energ.y lost would he high. By fitting a draft tube between the runnerout!et and tailrace. a continuous stream of waler is fanned between the two. The tail·race velocity is reduct.:d because of the increase in cross sectional area of thl!- draft tuOc and, because the wi·lr.lce pressure is atmospheric the runner outlet pressure must now be below the atmospheric pressure. Applying the energy cyuation between the runner outlet and
-
tail-race gives
P],
Figllf'(' 9.6
\le/llcir), l/'irlJl~/eJ for a Francis lurbillt'
_ + pg
V; +2, =
~
2g,
•
p]
vi
--'+~ + 2,. + hd pg 2g
390
};:>
TURBO MArlIlN(;S HYllllAI ! UcTI Ill IIINI ;S ....:
The nozzle outJctllrca is redu ced. Hence. the area of jet and the rate of now of water ~ trikin g the huckets is reduced . Consequen tly. the speed which was incre
39 1
1110
." 11( % ) (,(1
20
o
25
50
75
1011
'ii: nl" Full LOIl<j
Figure 9.4
\vrimiflfl o!pe lfOIi _
fllr"ill~ tiJicit'fI( '".
wifh I (IIUI
RADIAL FLOW REACTION TURBINE CHARACTERISTICS OF AN IMPULSE TURBINE The chartlcteristic curves of an impulse turbine for tl constant head are shown In figure 9.3(a) & (b). In Fig. 9.3(a) il is seen thaI the peak values o f efficiency do not v;1ry much ror various gate openi ngs. In Fig. 9.3(b). it is seen that the peak power occurs JI the ~a mc speed irrespect ive of the noz7.1e settings. This is due to the nOZl. le velocity remain ing cons tant in magnilUde direction ns the now rate changes. giving an optimum value of U je , a l a fi xed speed. Windtlge and mechanic:J1 losses and variations in loss coefficienls cau se the s nwll variations.
Reac li o n lurbine means lhal the waler at the inl . . net of lhe: turbine pOssesses kint!L.ic en ergy as we ll ilS pressurc energy A II . s le waler ows through Ih pressure l!nt! rgy gocs on changing' I k ' . t! runner. a part of . 111 0 lnl!lIc energy Thu II rUnner, s under pressure The ' . S lC water thruugh thc .. . runner 15 completely ene/os 'd ' . both the <:
n.
COlIsla nl H ead
MAIN PARTS OF A RADIAL FLOW REACTION TURBINE p 9n
T
~IT
.0
/,//<,- ~ I
511
/
:;
•
'Noi~lc
:
:
,
I
~cll ing
Th e m;un components o r d' I n . a ra la ow reactio n turhine arc (Fig. 9.5) Waler frorn
Srira l casin g
PcnslOd;
\ ----------77"--1--+- RUllfler
Figure 9.3
Efficicl/cy alld po u'a O/lt{1/1i \·er.!f/S speed if! an illipulse turbine
In praclice, one is usua ll y more interested in the fixed speed condition since the gcneralors run at conslnnl speed. The Fig. 9.4 shows that Ihe variation of efficiency wilh load is slight. except al low loads. where the decrease is due 10 changes in Ihe nozzle effic ienc),. and at high loads the increased jet diameter gives ri se to higher buckct losscs.
Figure 9.5
--
7'~'--f--- Gllicle wheel
Radinl flaw fIIrhill~
1, Casing Th e W' I' C I . ,I er rom t IC pen s tocks elller II. , .' .
. . sh
l
r!YDRAUUcTt IHHlNES '"
So, nou.i L' a nd pipL'lmL' troll smission dlicienr.:y
c~,)
(:;)
=
(Z:)
c'
-'-l = -2gH
3H9
of the .nozzle must change. This is accomplished by a spear valve. The position of Ihe spear IS controlled by a servo-mechanism th at senses tht! load changes.
DOUBLE GOVERNING OF PELTON WHEEL All modern pelto.n turbines use Double regul ati on method . nlis mt:thod or governing con tr~ l s Ihe turbllle speed and pressure (i.e . water hammt.:r) in the penstod by the
;md
comb lllt:d spt.:ar and de n ector control. Opt.:rated by the oil prt:ssure governor (Fig. 9.2(a)).
(c) Nozzle Velocity Coefficient
=
Actual jet vcloL:ity Theoreti cal jet velocity
C,
JlgH ThcrL'(ore ..thc n07.7ole cflicicncy bl.!COllll.! S 'IN =
C~
11<"11 I".~n.
I..,'t.
(d) Number of Buckets NUlllhL'r o f bUl:kds
O il
"mOl TUfhin.:
"'1\;0
S,,~U
a runner is given by N
D
= 15+ -
2(/
where 'D' is the pitch diameter of the pellon and d is the diameter of th e jet. The ralio Did is called thejet ratio.
+-li~ar _ ( l' UII1 I1 .
(e) Overall Efficiency '11'
) ")
Power produced
=
Actual energy supplied
=
P, pg QH
\\'hL'n: H is thL' head
GOVERNING OF PELTON WHEEL TURBINE HyJraulic lurhine s a rc usunlly coupled direct ly 10 an elcctricnl gen~ral or il nd s in~c the gCllcr<'lIor must nm at a co nstnnl speed, the speed U of th e turb~ne must .re.mmn constant when th e load changes. II is also desirable to run auhe ma )()nlUm c.fll clc ncy ,lIld thadorc the ralio U I e I, must remain lhe same . 111<1t is the jet velocity must not c han gt:. The only way kfl to adjust to the change in th e turbine load is to c hange the inpu t wata powcr. The input powe r is given by the prnduct pgQlr where fr is L:onstant (a nd hcm;e CI) ane! the o nly v:lriahle is Q. The eh,lJ1ge in now rate is cfrt:eted by n ~l ing that Q ::; Cl A. whe re A is the nna lc area . Si n!.:1.! CI is constan t, th e cross sec tHlOal area
_ N ulolkOpc::n
--- - -
Figuu 9.2(0)
No,,~
Clow:,J
Double HegllimiolJ
The ce ntriruga l governor (or) actuillor is flllachcd to the main shaft of tht: turbine . When the load on Ihe turbine decreases, Ihe speed of the lurhine incrt:a~es and the nyba~ls o f Ihe govc rnor rotate at higher speed and move away fro m the axi s by the cenlTlf~ga l force. TIlcrefore, the sleeve moves upward. TIle motion of tht! fly balls b transmlll~d to the bell crunk le ver and it rotates anuclockwi se. The roller on Ihe cam is raised and the deflector is brought between the nozz le nnd Lhc buckets. Attht: samc limc, t~c double pi.s ton in the distribution value moves down . Now, porr-2 is open and port· I. IS closed. High pressure oil from the sump enters the distributing value through the middle pan and nows down to servo motor striking the len fa ce of the: piston. So the pi ston moves to the righL forcing the spear to more to the right i.e . into the: nozzle.
. tt YIlH ,\II1 .IC T ,
3M6 ... T URnO MA<"" II NES
If the hucket is brought to res t. then suh trhcli ng thc buckct speed V, from the jet ve loci ty C 1 gives th l.! rel ati ve !luid velocity WI on lO the bucket. Thl.! angle turned th roug h hy the je t in th e 110rl7.0lltal plane during its passage over the buc.ke\ surface IS 'ri ' and th e relative t!x.i l vduci ty is WI . If tbe bucket speed vet:lU r U2 IS added to i\/2 in the upproprialc di rec ti on. the absolute veloc it y al exit C2 \~·i.lI hI.! ohtain.cd. should he reali zed tha t the c ompone nt C.r~ o r can be in the p OSlllVC ur negative X di rec lion depending o n the magnitude o f U . Fro m Euler's turbine equation
....;: J H7
Hydraulic Efficiency Tbe hydruu li e cniL'iency is defin ed
'!
c,
ruJl~ I - S
,15
E ne rgy tra nsfern.:u
=
Ene rgy 3"Jilahle in jet
E / (Ci!2g) if 0' = I ~ O -' , the m;u(imum hydraulic c ffi cic ncy is 100 pL'r cenl. In practice. th e dc neclIorJ angle is in the ordcror J 60'·· - 165': to a voi d interrerence with the o nco ming .leI and 1111 i!-. :u': l:nrdin~ly n:Juccu . The rn'l.~ imurn hyuraulk ef'ficiency j:-.
ilnJ sinct: in this cusc C rl is in the negative x . . dircction
W Illi = U I(u
+ IVtl + ( IV,c()J(IHO° -
a) - U)I
' ,y due to rri c Lion across the bucket s urface Assuming no Ioss of r--I' '- u ,,' v".. v"loc, ... (Wt lV2). then
=
Wlm = U(\V! - W!cosa) Therefore. IV
(9. 1)
E = U((C, - U)( I - cosa)J/g N.S The above equation can be optim ized by dirrerentiating with respect to U .
dE
_ =(I
- co,aHC , - 2U) lg
dU
PELTON WHEEL LOSSES AND EFFICIENCIES He1ld losses occur in the pipelines con\'eyi ng the water to the non lt: a nd are ctllllpn.-.et.! o f i'ric ti ll n and hend lnsses. L osses also ()t:cur in the noo.lt: and these ilre e\pn: ssed In lerms of i.I \·t:loc it y I:odricient C 1• (varies fwm D.9S to 0.99). f-ma ll y. tht:re art: windage ilnd friction losses in the whee l it sel f The water suppl y is fr om a reservoir al il ht:ad III J oove the nozzle. As Ih e Iluiu mcwt! s through th e press ure tunnel anu the pt:nstoc k upto the I.!nlry 10 the nou.le . a frict ional head loss 11 / occ urs. A further heJd loss h ill duc to losses in the nozzle takes place so that the head availab le fll r rower ,gcnermio n '- II ex. il fro m the no zzle is /1 '. T ha t is
+ hl/l)
Cr/2g
=
For a maximum, a nd th e n
(a) Pipe-l ine Transmission Efficiency C, = 2U
Energy al
1.!ild
o r pipe lim:
Energy a vailiJble at rese rvoi r
(or) (9.2 )
U = C , /2 BUI
II , - (I, (
H'
=0
in practicc, maximum energy is tran s ferred when the wheel veloci ty is 0.46t imcs
=
ef(\ -
II
H,
H,
(b) Nozzle Efficiency
the velocity of je t. Suhstitu ting e quati o n 9.2 in equ ation 9. 1, we have £01<1:0.
(H , - " 1 )
Energy al nozzle out let liN
E nergy al nozzle inlet
cosa)/4g
In pr;ll:ticc s urface fric ti on o f the bucket is prese nt and Wz
H'
H'
f.
H, - ", H
WI· TIlen equat ion 9. 1
beco mes
=
E = U(C , - U)( I - Kcu.,a)18
C; / 2g H
(or)
where K is lhe rdati ve \'l.!loci ty rati o W2 / \V , .
'IN
C'
= __..L. . 2g H
Table 9.1
Pa ral/li'/l" j-kaJ (111) Max . power (k IV) Best dJici cncy ('Ie)
1. Nozzle and flow contrr,>1 arrangement (Spear) The waleI' from Ihe Pl'lroll wheel
Francis IItrbilll!
100- 1700 55 93
80-500 40
Kaplall UplO
TttJ"/Jilll!
·100
.10 Y4
94
3. According to the head at the inl et of turb in e (a) High head turbine (2S1J-1800 m)
Example : Pelton whed
(b) Medium head turbine (51J-2S0 m)
Example: Francis
(e) Low head turbine (Jess than SO m) Exam ple: Kaplan and propel ler 4. Accor ding to the specific speed of the turbine
(a) Low specific speed turbine
« SO)
Examp le: Pe ll nn wheel
(b) Medium specific speed turbine. (50 < Ns < 2Sp) Exa lllpk: Francis (C) High specific speed turbine. (> 2S0) Exnlllple: Kaplan and propdlcr
reservo ir nows through the penstoc ks al the outl et of \vhll.:h " nou le is liltcd . The nozzle conven s the 10tal he ad at the inlct .Ill thc nOl.7.1 c into ki netic energy. l11e 'Hnllu nl of wa ler striking thl! curved budets of the run ne r is co nt ro lled hy proviJlIlg a !'!peur in Ihe nlll.zlt.:. Thc spl!or is'l co nical needl e which is operated e ither hy a hand wheel or autonl:lIH.:all y in lin axial direction. dependi ng upon the si/.l! of Ihe un it.
2. Runner and buckets Therolating whee l orcirculardis..: iscJl led th..: runner. On Ih e periphery of Ihe runner J number of bucke ls. even ly spact!d . arc fi xed. TIll: shape of the buckets is of a double hemispherical cup or bowl. Each budel is dividt:d into two sy mmetrical pans by a dividing Willi whic h is kn ow n as the spfirtt'r. The Jt:t of waleI' stri kes on Ihe sp litter. The splitter th en divides the jt:t illlo two equal parts and the water comr.:s oul 01 the Outer edge of th e buckct. lllCSC huckcls denet:1 thc jet through all ang le between 160 and 165C. in the sa me plane as Ih t: JCI. Duc 10 thi S dCneclion o f the jet. Ihe mome nlum of the fluid is changed reacl ing on Ihe buckels. A bucket is tht:rcrore pu shed away by the jct.
3. Casing T he casing prevents (he splashing or the waler "nd di sc harges tht: waler tilil r
4. Breakingjet To Slop Iht! run ner with in a short lime. a smull nozz)t: IS provided
PELTON WHEEL
whi ch direct s the jet of water on to Ihe hack of the van es. The jet of W
The pe lt on wheel turbine.: is a pure impulse turbi ne in w h icl~ n jet of fluid leaving the IHlnlc slrikcs Ihe hudclS lixcd to Ihe pcriphay of a rotating whee l. Th~ energy avai lab le allhe ink:1 "f'h..: IUrbine is onl y kinetic e ne rgy. T he pressure at the Inlet and
out kl of the turhine is Jtmospheric. The.: turbine is used for high hC::Jds rang in g from 15010 2000 tn. TIle turbine is named after L. A. Pelton. an 3nlcrican engineer. The fluid fl ows in the tan ge ntia l dir~c ti on.
PARTS OF THE PELTON TURBINE TIle main parts of th e Pdt o n turbine are
3S
!-Ihawn in Fig. 9. 1.
L·.dlcd
VELOCITY TRIANGLES AND WORK DONE FOR PELTON WHEEL The WOller supply is from a consta nt head rescrvoir al an el~l'atioll HI. Dbo\,c the ce ntre-line of the jet. The nozzle at the pen stock end. converts the lota l head JI the in let to the noa h:. into a water jet \. . jth veloc it y Ct. at atmospheric pressure. The veloci ty tri angle for the flo w of nuid onlO and olT a singl e bucket arc shuwn in Fig. 9. 2.
Casing Buckl:t (Vam:s)
L Figllre 9.1
IS
the hrellkillg jf!f. If there is no hrenki ngje t, the runncrdul! to incrt iu goes on rc\'olv mg for a long lime .
Shape: 1)( bu!:kel
Pt'lrvn II/rhifle Figure 9.1
Velocity rriutlglt!J Jo r u rdron rurbjn ~
)82 ~ TURno MACIIlNES
the impeller develops a head of 68 m. Find (a) pump speed, if the im~c\l~r efficiency is 70% (b) the manometric head, assuming 50 percent of. the kmetlc energy at the impeller is wasted and the head 105s in the impeller IS 5 I~. ,and (c) the lowest speed to startlhe pump if VI 0.5 U2 (MU- Apnl 96) IAns: (a) 1500 rpm (b) 42.3 In and (e) 1270 rpm I 8.20. Water is required to be lifted through 110m height from a well. ~umbcr of identical pumps having speed 1000 rpm. specific speed 25 rpm: with a ri~ted discharge of 6 kl/min arc available . Determine how many pumps Will be n:qul.rcd and how they should he connected. [Ans: (a) 4 and ~h) Series I Centrifugal pumps delivering 10 m) Imin of liquid against n hC
=
8.22. A single st age centrifugal pump with impeller diameter as 300 mm rotates at 2000 rpm and lifts 3 m J of water per sec 10 a hci~ht of30 m with:lO e ffici e ncy of 75 per cent. Find the number of stages and dmmcter Of. each Impeller of a similar multistage pump 1O lift 5 m] of water per sec to a height of200 m when rotating a[ 500 rpm. . IAns: (a) 7 and (b) 0.39 m] The impeller of an axjal flow pump is 1.2 m in dlam~ter while the boss 15 0.6 8.23. m in diameter. Find the most suitable speed to proVide a hea~ of 2.5 m. !he velocity of flow through the impeller is 4.5 m/s and the spec~fic ~peed 01 the pump is 335 rpm. Find also the vane angle at inlet at the extenor lipS :nd ncar the boss. Assume no whirl at inlet. {Ans: (a) 340.8 rpm (b) 11.5 and (c)
22.5"J
8.24. Ajet pump is fitted at 3 m above the suction reservoir a~d 19 m belo~v the supp~y reservoir Ii fts water through a total height of 4m. If the Jet pump dellvers.8 ~/mlO of water while using 2.5 lis from the supply reservoir, determine the efhclt.::ncy IAns: 48.9 % ) · o f Jct pump. . . the help of a lme diagram. Describe the following with 8.25 . (a) Je[ pump (b) Air lift pump . . 8.26. Enumerate the advanatages and disadvantages of air lift pump as compared with the centrifugal pump. 8.27. Describe the construction and working of a submersible pump. 8.28. Explain the construction and working of the following pumps with a n~~tsketc h (a) Gear pumps (b) Vane pump (c) Piston pumps
9_ HYDRAUUcTuRBINES
INTRODUCTION Hydraulic turbines convert the hydraulic energy into elcclri"cal energy. The main Iypes of turbines uscd in these days arc the impulse and reaction turbines. Th e predominant type of impul se turbine is the pelton wheel. Reaction tur.bines arc of two lypes I. Radial or Mixell now 2. Axial now. Two types of a,:;j;ll now turhincs exist, one is propeller turbine and the other one is Kaplan turbine . The former has fixed blades whereas the latter has adjustable blades. Francis turbine is an example for radial How turbines. The foll ow ing table su mmarizes the head, power and efficiency values for each type of turbine .
CLASSIFICATION OF HYDRAULIC TURBINES The important classification of hydraulic turbines arc
1. According to the type of energy at the inlet
(a) Impulse turbine (Pelton wheel) Energy available at the turbine inlet is only killeri c energy and the pressure is atmospheric from inlet to the turbine outlet. (b) Reaction turbine Energy available at the turbine inlet is both kinetic energy and pressure energy. Example: Francis, Kaplan and propeller turbincs.
2. According to the direction of flow through the runner
I
1
.j
(a) Tangential flow turbine Water nows along the lungent of the runner. Example Pelton wheel
(b) Radial flow turbine Water nows. in the radial direction through the runner. If the water nows from outwards to inwards radially, the turbine is called the 'In ward radial flo IV turbine' . On the other hand if water nows ratiilllly rrom inwards to outwards, the turbine is known as 'Ollfward radiolflow turbi"e.'
(c) Axial flow turbine Water flows through the runner along the dirccLion parallel to the axis of rotation of the runner. Example Kaplan and propeller. (d) Mixed flow turbine Water flows through the runner in the radial direction and leaves in the direction parallel to the a,:;is of rotation of the runner.
OJ I
,t
I
II
J~n
~.fl.
;...
H \, UH. AU I.IC PUt, u'S "
TIII Ul() l'VI ., C II I:-J FS
-'H I
sc t hack at an angle of 45 <:1 10 Ihe outer rim and the cnlr), of the pump is radl~1. The pump run s at 1000 rpm and thl! ve locity of fl ow through the impeller is conS\il nt al 3m/s. Also, calcu late thc workdone by rh e wall!r pcr kg of wa ler (MV-Ap ril '<;7) and thl! vl!locity and direction of watl!r at out let.
St:llll.' hl.!ad is [hl.! s LIm of [:11 <.:uc tion hl.!acJ :lnu mall lllnL: tri l: hl.!acJ. (h) m:lIwmcuir.: hcnu an d dclivl.!ry hl.!ad. (,-' J suction hl!:Ju and dcli'!l.!ry ht:au.
IAns: (a) IU .8° (b) R92.9 l/kg and \<12K.52 ml, and 6 ' 1 R. I J. A cen!rif u£al pump dd iVl!rs wa la agai nst a net head of 14.5 ml!lers and at a designcJ spccd of 1000 rpm . The vancs arc c urved back aLan angl\! of :ur wit h
S.7. Ot!linc: Ilwnolllclric head . S.M . Dl!fine th e fo llow in g for it centrifuga l pump (al Manolllcl l;C e ffic iency (h\ MecllLlnica l cfficl cm:y tel Overa ll cmc icncy !'i. L) . W hat is l11l.'anl hy ulluimulll slUriing speed lIf i.Il.!c lH ri rugul pumr'.' R. )il. Whal is NPSI/,! X. II , WlllI t is pri ming "! ~ . 12 . Pumrs :Irl! CnOll l.!CICU in para lid 10 (a) Jcvdnp:1 high hc.:ad (h) dl.!vclop a high Jischargl! (c) devdop a hig h power 8. 13. Pumps arc COnlll!CICU in - -- [0 develop a high Ih!.. d, S. l"}. Ddi nc thl.! rhcn omcnon c;}viwl iofl 8. [5. Til e e:.lvilari un panlm Clcr is ddi lH.:d as (al a = lilt/IN P Sf! (hi n = ( NPSI1I > 11." lC ) a::;; NP5 lf j H m R,16. Cnn cnv illl li on he preve nted in thl! ce ntri fuga l pumps'! How'!
the pcriphcry. Tht! impcllt!rdiametl!r is 300 mm and th e outlet width is 50 mm . D":lcrmillc Ihl! dischargc L)f the pump, if manometril: dficil.!ncy is 9)lih . [Ans: U.1 6ij m'\/s l
8. 12. A ccn tri fuga l pump dd ivers 30 litfl!sofwalcr per sccond
B. 13.
8. 14 .
8. 15.
EXERCISES 8. I. Explain tht: worki ng of a sin gll! osla.gl! cen trifugal pump with skl!lchcs. 8.2. Differenliatc hetween thl.! voluw L:a!'> in g and vortC K cosing for the cl!ntrifugal pump ? ~ . J . Ohta in:"ln I.! xprl.'ssion fu r lhe \\'\lrkdollc p..:r kg of wa ter, hy the impdler of a ':":lltnfll!!al pump. SA , Dclin..: thc terms: Suction hcml. Jdivc l) ' hl.!ad , Sialic head and manoml.!tri c hcaJ . S.5. Dcn\'c the ..: xpr..:so; ioll for lhe min im um spcL:d for sturting a cen trifugal pump. S.fl. Draw and disc uss the pe rformnncl! curves of i.l cen trifugal pump. X.7. Whar is a lllultl -s[agt! pum p? Dt!scribe muhi stagt! pump wi th (a) impe ll er.-; in pura Hd ... nd (11) impd lcrs in scries. tl.H. E.,=plain ·lht! phenome non uf caviW lion. What nrc the effects of f.:ilv itation: How G i ll e:lviuHion hl! pr..:\'entcd'! 8.9. A centrifu ga l pu mp has eXlerna l :lIld intern n) impe ller dinmt! te rs :IS 60 e m and 30 em respeclive ly. The vane angl..: at inlet and outlet arc 30<:1 ,lIld 45 respective ly. If th e water enters the impL: lle r at 2.5 meters/sec. Find (a) speed of th e impdlers in rpm. (b) work done per kg of watCf. Q
IAns: (a) 275.66 rpm and (b) 53.34 llkgJ
H.IO. Cu iL' ulatc Ihl! V:.l lll! <.Ing le at inlet of a cl! ll trifugn l pump impeller having 300 mill diameter al illlc! and 600 lIllll diamdl!r at oU Llet. Th~ impdlcr vilnes arc
8. 16 .
a hCIghtof I H mctcrs Ih rough u pipc of 90 m Ions and 100 mm diameicr. If thc ovc rall efficiency of the pump is 75 % find the powc r rl!quired to dri ve the pump. Takl.! f = 0.0 12. IAns : 2U kWI The di ,lIncters llf nn impeller of a ccn tri fugal pump;(1 inkt ;IIIJ outlet arc 30..:m anu 60 c m rl!spcclivel y. Delcnnine the mimmum slD I1ing sp..:..:d, of the pump jf it works ug<.linst u head of 30 rn. [Ans: 8Y 1.8 rplIl) A fou r slagc centrifugal pump has four ide nll cnl impe ller kl!ycd to Ihe same shan . The shafl is running al 400 rpm and the lotill mal1oml!tri~ hl!ad dl!vclopcd hy [h c multi sl
[0
(oj 13K.B m
and (b) 109 kWI
8. 17. A c~ ntrifu ga l pllmp has an im pdlt.:r of I!xtl!rnal diamcter 60 em and intefTl<.IJ diullletl!r)O Clll. The vane angles al inlet and ou tl el an.: ) O und --15 <> respectivel y. The ve loc it y of /l ow is constant at 2.5 mls and thL! veloci ly nt inlet is radial. Find th e pn.:ssurl! rise th rough the impeller, if the pump spel!d IS 276 rpm . IAns: 5...15 m of water] 8. 1R. A cl! ntrifugal pump is working agai nst a hcad of 20 m while rotaling ut thl! nUe of 600 rpm . If the blades arc cu rved back to an angle of 45 ° 10 the tangenl allh c outl Cll ip and Vl!loci ty of now remains constunt at 2 m/s. Calcul atc the Impeller diameter when allihe kinetic e nergy at impe ller uu tlet is W41Sh.:d . IAilS: 0.6)4 JIll 8. 19. A cent rifugal pump handlin g wilte r has backwilrd curved vanes. The impdlc! r tip diamctcr is 500 mm , The rt!lati vc ve loci lY at tip section is --1 51: to the tilngenl 3tl.!x il. If the radial velocity at cx it i!:= 15 mis, the now at tht: in let is rudial and
HY Il RA UI.IC PuMI'S '"
378 }> TURBO MACIIINES
371}
(c) Boss diameter
Also.
Ve loci ty of now
N../Q H J {;I
en
3S x 6J/ 4
N
JTT78
N
134 .22 rpm
V,
= (;I) =
rr x 0.5 x IJ4.22
Q
60
~, =
=
0.j v'r-2~x~9.~ 8~ I x~ 2
=
3. 132 m/s
-V , = -3.51
=
IT,
O. IS O
,
4(0- - D;;)C o IT
w,
= 3.5 1 m/s CII 2
: . lan (3 1 =
0.5 / 2gH
Disc harge through the pump
(b) From inlet velocity diagram rr D" N
=
,
,
4'(0.29- - Dh) x
3. 1 3~
0.0109 ~,
29.67°
0. 11 m
u, Figure 8.JO
Example 8 .21
An ax.ial How pump Iws the fo ll owi ng particul ars: discharge = 180115. head developed = 2m. specific speed = 250, speed ratio = 2.4. Flow ratio = 0.5. Calc ulate (a) speed orthe pump ~ b ) the runner diameter (c) the boss diameter.
Exampl e 8 .22 A jet pu mp fi ned 2.5 m above the suction rcsr.:rvoir and IB m below the supply reservoir lin s wate r through a 100ai heigh t o r "2..7 m. Dete rmine the erfic icncy of the jet pum p when it de live rs 7.5 lis of water while using 2.75 1/ 5 from the supp ly reservoi r. Solution
Solution Q = O.ISO m) /5 Speed rat io
= 2.4
H =2 m Flow ratio
= 0.5
N,
= 250
(a) Pump speed H J{4
N, x fi JI'
250 x (2))/'\
../Q
JO.ISO
N
99 1 rpm
N
= 2.5 m
= 2.75 1/ 5
H" = 2.7 - H, = 2.7 - 2.5 = 0.2 Q.• = 7.5 - Q, = 7.5 - 2. 75
m
H, = 18m Q, + Q, = 7.5 lis Q, = 4.75 1/5
HJ + HJ
= 2.7 m
Therefo re. Je t pump effi ciency,
N../Q
N,
fls
Q,
(b) Runner diameter
Q,(H,
+ HJ)
Q,(N, - H,,) 4.75(2.5 + 0. 2) 2.75( 18 - 0.2) 26.2%
Peripheral vcioci ty
2.4J2gH
V
=
2.4J2 x 9.S I x 2 15.03 m/s
Since.
rr D N
V D D
=
60 60 x 15.03 rr x 99 1 0.29 m
SHORT QUESTIONS 8.1. What is a centri fugal pum p? 8.2. The ce nt rifuga l pump is si mil ar in construction to Lhe francis turbi ne. (TrucIFalsc) 8.3. The efficiency . of the vortex casi ng centri fugal pu mp is [han the efficiency of vo lute cas ing centri fu gal pump. 8.4. Draw the ve loc ity triangl es at inlet and exit of il centrifu gal pu mp. S.5. Wh at is a slip factor? Write the ex press ion for wo rkdone per kg of water of a ce ntrifu gal pum p with fluid sli p.
.r H YnR"L'Lll- PUf>U'S ... 377
NUlllht.:r of s[;Igl.:s
Nnw. the..'
p.lramelcr
ClIvtla\lOI\
140
/I
=
- - = 49
/I
=
5
2SA
=
o
.
=
=
Eq Ualill£! hcacll·o·dficil.:nL of pump· I :.Ind pUlllp·2
H, H,
=
(N'IV,) '( D,D,),
( ~~ )'
=
(~)'(z: )
D,
= =
)2.77 x
=
)2.77 x 0.15 0.25
D"!.
==
pg
P'1'fl/Jo
36
(SU Ill 01
P" pg
I
. head losses)
f
PI .: NPSI I = V,'] pg [ 2X
--I"H I - -
1.8 kP"
P,
=
III
,.
(0.75 x 13.6) - 3.317 - 0 . 1835
6.7
In
Fur case (11 250mm
Cavit'lli on hegan. wher. p. .......!..
P"
D,
Example 8 .19 When a labormory Icst was carried out on a pump, il was found Ihat for iI pump total hca d of36 m at a di sc harge 01'0.05 mJ/s.e:lVitulion began when th e Sum orllle: SIali c pressu re plus the velocity h(!ad at inlet was reduced to ~.5 m. The Jtll10spheri c pressure 750 mm fi g and tht:: V:lpour pressure of water is 1.8 kPa. If the pump is Iu opl.! ral c ill a locution where :nmosphcri c pressure is reduced to 620 mrn IIx ;J.nd tht.: vapour pressure of waler is 1DO Pa. whilt is the va lue of the cnvilat ion par;tnll.:h:r when the pump develops the s;unc Inial hC'ld and disc.::hargl.:'! Is it nel.:l.::isary 11 \ rL'Ju..:e Ihe he ig ht of the pump ahovc the supply and i f so by how mudl'.' Solution
.i
O.09::! l
= -- pg
16
or
.f
From Ihe sll!ady now cnergy equation taking tbe reservoir lev..:! as datum (Z .. = 01 WI.! gel rOf c.::aSl! ( I )
("84)
C500)' 1200 2.77
=
NPSH H 3317
I
v; + -!.- = 3.5 2g
2g
pg
z,
+"
-
Z~I
= 830 Pa
Ji
(0.62 x 13 .6) - 3.317 - 0.0846
h
:::;:;
Since Ihe /low ratt:: is thl.! same," II (ZI -
!l + V/, - PilI/II ] 2g
pg
1.8 x JO)
=
..
=
3.317m==NPSIi
, 5 - ::-::-,----,-,-.;
., Jil "
p,
"
5.03 m
=
II h and the pump must be luwe red hy = l.ll7 III .lI lhe new locatiun .
II
dlstanc.::c.::
Ex ample 8.20 A n axial now pump has:m impeller or out le t diameter 1.0 m . TIle dia uf bns!- IS U.5 I\l. If specific spl.!'ed of pump is 38 and vel ocity of nuw i~ 1. m /s. Suggest a suilllb ic. spet::d of the pump to give a head of 6 m. Also determine vane angle at the en try of Ihe pump, if the lIow is axial at inlet. Solution
Ih e VOlpour pressure. The n
(lg
p~ V.,"!. P"- l i b .--=- + . . . . :. . + Z, = pg .
0, :::;:; I m. D" == 0.5 (aJ Pump speed
In,
Discharge
=
Area of now )( veloc ity of now
=
4(D,- - D h) x ell
9.81 x 10'
=
IT
,
,
IT,
,
-(1- - 0.5-) x 2
4
1.178
m'/s
H YORAULIC Pu ~U's ..... 375 37·1
,. TUllDO MACI1INES
The hC;Jd de "doped hy cnc h pump
C ,l;l
U2 - Wf!
::;
C rl
n'-'l
= ~.21
[1000 X2~r'
m/ s
[.: tan45° = II
:. ex,
-,
[N:,o]'"
From till: oUl lt! t velocity trinngks (Fig 8.2)
~
20.94 - 2.21
~
52 m 18.73 m/ s
Total head dcvclopt.!d (H) = 156 rn.
H
-. No. of pum ps ( n)
From '1m eq unt ion '1m
X
0.9
X
Hm 156 52
(U2 C_(1 )
(g)
~
20.94 x 18.73 9.8 1
35 .99 m
Total bead generated by the pump
" x
H,,, ::; 3 x 35.99
107 .97
PI ::; pgQ(Hm X II )
ryO
'Power output of the pump = pgQ(Hm x 11)
~
P, lllUS.
Shn ft power
Example 8.18
A ce ntri fu gal pump having Lhrce slDgcs in paralld ddivers 360 m
It should be noted that in the first pump. all the three stages arc ill para llel which means that he ad of each Slage will be lhe same i.e 16 m . Where as the: di sc harge of each stage will be 360/ 3 = 120 m) / hL On the contrary. in the seco nd pump the s tages are in series which mean s that each stage would deli ver a di scharge of 450 m] / hr. Where as tol?J head of 140m will be equally co ntribu ted by eae h s tage.
So.
1000 x 9.81 x 0.05 x 107.97 0.8 52959.3 0.8 66 199. 1 W
QI Q, H, 11,
~
120 mJ / hr
~
~ 50 m' / hr
~ ~
16 m the head ge nerated by each stage in the second pum p.
Eq uating the specific speed of pump-I and pump-2. we gCI
~: ~
66. 199 kW
Example 8.17 Find the number of pumps required to take water fr om a deep wel l.under a tota l hend o f 156 m. TI1C pumps ;:Irc ident ical and run at 1000 rpm . The spec ific s peed of each pumr is given as 20 while the rated capac ity o(each pump is 150 lis. (M KU- Nov. '96)
120 450
or
(Z:)' (Z:r
~
(1200)' 1500
(Z:f'
0.64 0 .27
~
1.78
H, N~IOOOrpm.
N,~20.
Q~0.150m'/sec
and
11,
~
1.78
(HIH, )~/'
~
H,
Solution H~15 6 m .
J
of water per hour. agai nst a he
66. 199 kW ~
3
Solution
In
(ii) Shaft power input (P,,)
P,
~
/I
y.
16 ~ 28.4 m
]72
;....
T URlll) M ACH INfS
HYUH :\UI.j(' PU)'II'S
Example B.15 A centrifugal pump hnving external nnd internal diamelers of
2gJ-lm
·.N'2.
n~spl.! c liwly
1.25 m am.I 0.5 m is discharging water al the rate of 2000 litis agninsl a hL"ild of r (1 ill. when running at 300 rpm . The va nes arc curved bnck al an ung!!: of .lOu with thl.! wngl.!nt at the outlet and the ve loci ty of now is constant at 2.5 mIs, Find (a) hyuraulic efficiency nf the pump. (b) power requirl.!d and (c) least SPl!cJ at which aille purnp com mences to worK . (MU-Ocr. '98)
(rr:0'
0.5 m ~no rpm
D[ N
'. N
=
.
U~
X
g
60
en
19.04 - 4.33
295.4 rpm
= 19.64 mls
.,.j,
Solution Given, Number of stage, O'!
11
i f
= :3
= 0.4 m.
b]::::::; 0.02 m.
Ih x 9.81 19.M " 15 .31 52.2%
aI
uutlet = 10%
(i) Head generated by pump Area of now at outlt!t
::::::;
O. 9{rr D'2.b])
xrr x0.4 x0.02
(b) Power required (ps) Nr.:g Jccling 11ll; mechanic.:al losses. the power required 10 run the pump is l!quaJ to Ihe rluid power developed by the impeller
0.023
Ps
HI =
,iIU]Cf~
17111 ::::::;
0.9.
I/O
pQ=1000 )(2 2000 kgl'
:. P.r
r !
Reduc.:tion in an!a
0.9
= 0,8.
N
m'
= 1000 rpm.
Q = 0.05 Ollis Velocity of now at outlt!t
2000 x 19.64 x 15.31
=
60l.3R kW
_Q_=R
,C r2
rr D,!b,!
(c) Minimum starting speed "2gflm
(]f~t)' - (]f~t)' N'
th:= 45 ~
15.31 mjs
IJIII
i.c..
16
(MKU-April '97)
2.5 19.64 - - tan 30°
- - -"tan fi]
l)J~l x
U]CX1 rr x 1.:25 x 300
rr DiN Ul = - - - 60
CI1 =
Hili
l".
Example B.16 A three stage cl!ntrifugal pump has impellers 40 cm in Jiamc:tc:r and 2 em wide at outle\. The vanes nre curved back at the outlet at 451; and they reduce the circumferential area by 10%. The manometric eflicicncy is 90% and the overall dficicncy is 80%. Determine the head gcncralt!d by lhl! pump, when running al 1000 rpm delivering 50 lilres per second. What should bt! tht! shaft power?
(a) HydratJlic (or) Manometric efficiency '1/1i
("6~'
,
(rr X6~2S)' _(rr :0°.5)'
Q ~'2.
373
r- r ,
2
Solution 1.:25 III 16m
""
[(~6~')' - (rr6~') 'J
2gHm
=
0.05 0.0226
=
2.21 m/s
Outlet impeller tangential velocity IT
2gHIII
A,
x 0.4 x 1000 60
=
20.94 m/s
,
,I
370
>-
HYOII ;\ UI.I(' PLiMf'S ..;: 37 1
T Ultoo MA CHINES
(b) Power input
Th e hydrau lic efficiency is
Hili E Ruid Power developed by pum p
'1fI
pgQ H",
=
p
'/
10.1 x 9 . ~1 x 0.0 16 x 42./ 0.76 x I(}-l S.6Y5 kW
Fluid po"..er supplied 10 impe ll er No s lip, thererore,
Example 8 .14 A centriruga l pump with 1.2 III diameter ru ns at 200 rpm amJ pumps 1RSO lis. the average Jirl being 6 m , The angle whic h ,the va nes m,ake at I.!X II wi th the langent til the impeller is 26" a nd the radinl veloCitY or n ow IS 2.5 lTl/~, Dl!tt:rminc th e nlUllometric eflicicm:y and the least spl.!ed' w stan pUmpll1!; agitln ~t a heau 6 111, lhe inkt diameter th e im peller being 0 .6 m ,
ond
C",
gH",
=
or
V'1JlII
=
9.8 1 x 37. 16 98.3D x 0.76
=
4 .871 D mls
or
Solution Give n
D, N
The o Ull e t velocity triangle (\\'ilhout s lip) gives (Refer Fig . 8.2)
filii
=
0.610 200 rpm
= om
D,
1.2 m
Q
1880 lis = 1.88 m.l/s 26<'
fl,
C r2
=
(i) Manometric efficiency (U,C"lg) IT D2N fiO IT x ' 1.2 x 200
D 2(4.87ID - 98 .3 D) Hence,
D' = 0.0495D - 0 .0008
=
.', Impeller diamete r( D)
= 0.214 m C ,I'!
Example 8.13
A sma ll ce ntrirugal pump, when tes ted III N = 2875 rev/min wi lh water, delivered Q 57.2 m J / h and H 42. 1 m at its best effic iency point (11 = 0.76) , Determine the specific speed of the pump allhi s test conditi o n. Compute thc required power input to the pump. (MKU-Nov. '97)
=
=
Wr! C.t~
Solution N = 2875 rpm.
'1m
Q
57.2 J = 3600 = 0.016 m Is.
Hili
= 42.1 m.
= = = =
60 12.57 m / s
Vz - \V.IJ
C" = -2.5- =5. 1_1 m/s ---lan 26
tan fh
12.57 - 5.13 6 x 9.81 12.5 x 7.44
(ii) Least speed to start the pump
= 7.44111 1' = 63.3<;1.
\I is give n by
(a) Specific Speed of the pump N,
=
N.JQ J /4
Hno
= =
. 2875JO.0 16 (42. 1)'/4
22
v x
1.2 x 60
N)' _(If , 0.6 x N)' 60
= 2,96 N! = N=
N :::
2,<).81 x (, 117 .72 I QI),4 ry.;n
zrJu rpm
2,5
"I j S
]68
~
HYI)R AliI.lC PDII'S ....:: 361,1
TUHUO M ACIllNES
ThcrL!fore,
Hili
=
=
th e blalh.:.s occ upy 6 percent of the circumferenlial area and (hI! hydraulic dliclcncy (neglecting s lip) is 76 percenl, whal muSI he the pump impdler diameter?
0.89 x 31.91 28.6 m
Solution
Hene!.!,
28.6 x 10' x 9.81 178.5 kPa Now,
C2 =
(C;~' + C.;l, )'I'
From the continuity equation, the veiocity In the pip!.!!". i!".
v =
Q/A
=
G
Total losses
0.04
=
Pip!.! frictio n losses
=
(
4/1V' - - + 3 (V') 2gd 2g
aud
C.I"}
C:z
=
C" U, - ---~ (:In/h.
= =
0.324 19. 13- - -0 Ian 7 16.49 m/s
=
(0.324' ot- 16.49')' /'
=
16.49 m/s
=
=
H", -
(
NQI/2
1.29')
_ ( 16.49' 28.6 2 x 9.8 1 143.5 kPa
(g 11 )'"
0.075(9.8 I x 37.16)J /4
N =
(0.04) ' I ' ) 1.28 rev/s
=
Flow arc.::l perpcndicu lDr 10 impeller outkt periphery is
pgQHm )0."
28.6
(d) Input power to impeller
=
2.26' 9.81
y
X
2. 16m
N, =
7.43 kW
Ps
2
dClcrmmed.
Ci2g- Ci)
10.1 )( 9 .81 x 0 .0265
+ nlhi..'r losses
4 x 0.005 x 40 +3) V' 0.015 2.~
M.333
= =
(e) Power given to fluid
=
m/'
Tmal n:quircd head = 35 + 2.16 = 37 . 16 m From the defin iti on of the dimensionkss specific speed. the speed of the pump I~
Solving for the static pressure hend
=
= 2.26 2
x 015 )
Power give n to Ouid/1lfJvernll 7.4 3/0.89
Jr D x -
=
0.2950'
10
Now,
C"
8.35 kW
Example 8.12 The basic desig n of a centrifuga l pump has a dimension less specific speed of 0.075 rev. The blades ::Ire fOf"\l/ilrd fut:i ng on the impeller and the out lei angle is 1200 to the tangellt, with an impeller passage width al out let eq ual to onl!-tcnth of th e diam eter. The pump is to be used to pump water from II vertical distanc!.! of)5 m at a now rate of 0.04 m 3 Is. The suction and delivery pipes arc each of 150 mill diameter and have iI combin ed length of 40 m with iI friction faclor of 0.005. Other loss!.!s :It pipe entry, exit, bends ctc are three times the vc.:locity head in the pipes. If
o
=
= = =
Also,
U, =
= =
x 0.94
Q Flow area
0.04 0.2950' 0.136 / -- m 5 0' rrON
rrO x 31.28 98 .30 m/s
I·IYDRAUUC PUMrs ..; 367
366 :;. Tunuo MAC] JINES
35.3 (30 x 14.77/ 9.81) 78.15%
'1m
=
Volume now through the pump is
35.3 45.168
Q
Area x Velocity of now .(2Jrqb) x Cq
2" x 0.051 x 0.064 x I.2Y 0.0265 m1 / s
(c) Pressure rise in the impeller
C;
(b) Stagnation and static pressure rise across the impeller
28
Using the continui ty equat io n r1 C r ]
U2C~2 . - - - Impeller loss - -=-
g 45.168 - 3 - 12.21 29.958
Example 8.11 An impeller with an eye riluiu s of 51 mm and an outsiuc diameter of ~I06 mm rotates at 900 rpm. The inlet and outlet blade angles measured from the
At ou tlet. ltlngcntial impeller velocity is
radi al flow direction arc 75 c and RJ u respectively, while the blade depth i:-: 64 mm. Assuming zero inlet whirl, zero slip and nn hydrauli c efficiency o f 89 percent. Calculate,
(.
900 x 2")
60 . 19.13 m/s.
(a) the volume now rate through the impeller
(b) the stagnation ;lOd static pressure rise across th e impeller (c) the power transferred to lhe nuid and (d) the input power to the impeller
x 0.203
Hydraulic dficiency Total head deve loped '111
Solution Si nce there is no slip,
by pump
Theoretical head dcvdopcd
= Ii,
E
(a) Volume flow rate through the impeller
If thl.: change in potential head across the pump is ignored. the total hl!atl dcvdopcu
TIle ou tlct blade angle measured from the tan gential direction
(P, _PI) (Ci _Ci )
by the pump is given by flm
=
and
fJ, = 90" _75" = 15° At inlet, tangential impeller velocity is
r1. Ct"J
0.051 x 1.29 0.20) 0.324 m/ s
c"
In
fJ;
==
---
pg
+ - - -28
and for an in compressible fluid. the total pressure Ilt!ad ctl· cJTic.:ient is \Y,
pg
COO6~ 2rr) x 0.05 1 =
Figllr~
From the inlet velocity triangle (Refer Fig. 8.29)
C"
(Since zero whi rl Col·] ::;;; 0]
U, CI
4.81 tan 15
1.29 m/s
c
Now,
u,
4.81 m/ s
E = 8.29
U,c" = u, (U, _ IV,,) g g •
C,,)
19 .13 (9 -.1 .1 3 - - 9.B I Ian 7°
19.13 - ( 19. 13 - -00324) 9.81
32.15 m
la07 °
~M
;...
TtJltuo
HYDRAULIC PUMPS '" 365 M AC HINES
Example B.1 0
In a centrifugual pump handli ng liquid waler the head loss in the impell er is 3m. while the pressure gain in !.he casing is 4.7 m ofwaler. which is 38.5% of the absolute kinetic energy at the impeller exit. If the ve loci ty of now al tht: exi t of impeller is 4.64 m/s and tht: impeller tip speed is 30 mls. when the difference in pressure gauge readings installed allhe impelle r inlet and ou tlet is 35.3 m of water find (a) the exit blade anglc and the Euler work input (b) ma nometric efficiency (c) the actual prt:ssure rise in the impeller. Assume slip coefficient of 0.85. (MKU-April '96)
Bu!.
P'J.
PI
-
- -
PC
pg
= Prcssure rise in impc ll cr
.
:hercforc. Press ure risc in impt:lIer
,
= C,r2 UJ,
_
g
Ci + Cf 2g 2g
From inlct velocity triangle. (Fig. 8.27)
Solution Impe ll cr head loss - 3 m
C; 1 = __ )( 4.7 = 12 .21 mls 2g 0.385
IV,
Q
CrJ =
Area of Howat inlet 0.0 15
.
Q
~
-
1\
= 1.492 mls
iT x 0.2 x O.Olfi :. C , = 1.492 mls
nDliJ l
u, Figu re 8.17
u:! = 30 m/s (b..P) impdlu = 35.3 m (a) Exit blad e angle and Euler w ork Input Kinetic energy al impeller ex it
-
=
-
fJ, IV,
C,' = ,,-D-,b-,
0.015 = " )( 0.4 x 0.008 = 1.492 mis
1.492
7f
DJ,N _ 60 -
iT x
0.4 x 1200
60
14.77 mls
C~2
Cx' = -
a .1
The blade angle at exit
= 25 .133 m/s.
tan/h.
= =
22.55 m/ s
C r:!
14.77 = 17.38 mls .85
= -0-
U, - 2.584 = 25.133 - 2.584
.. C"
4.64'
Idl!ul ubsolu tt.! whirl velocity
Figu re 8.28
.'. U! - C.r ! = - 3 0 = 2.584 m/s {an U
U.., = -
= 15.48 OIls
jCi - C;, = /15.48' -
= =
Cr1
Lan
x 9.8 1 x 12.21
From the outlet velocity triangle with fluid slip at impl! lIcr exit (refer Fig. 8.3).
C" U2 - C.t2
U2- C.f:!.= -
Q
= J2
C,
From nudel velocity triangle. (Fig. 8.28) lan fJ .,
C" = 4.64 mis
/1, =
Crl
U2 - C.fl 4.64 30 - 17.38 20.18°
Velocily at impeller exi t Euler work input
Cz
=
JC;z + C;2 /22.55'
=
: . Pressure rise
= =
w1m =
0.85 x 30 x 17.38
+ 1.492'
=
22.6 m/s
25.133 x 22.55 9.81 31.853 m
-
22.6' 2 x 9.81
asU2C.f2
1.492'
+--2 x 9.81
443.19 1/ kg
(b) Ma nometric efficiency IJm
=
-,-,_'.,.1:::",--,-_ = (b.. P )1mI'd Icr (U,C~,lg)
(U,C~,lg)
HYORAULIC PUMPS 4: 363 362
):>
TURDO MACIIINf.5
and
4 x 0.0 12 x 100 x 3. 18' 2 x 9.81 x O. 1 24 .74 m
:. hlo
Wc know thai,
Now, manometric hC
Hm
= =
.. Hm = 20
=
U2 C .t 2
. II . - - - L osscs In Ie'Impe II cr and casIng
Hili
g
U,C~,
--- - - Lo!'is of kinetic energy 8
V,C 2~ (Ci) 28 V,(V, - CV, - 2)' + 4 x' _
(hJ
=
H
=
20 + 24.74 45.26 m
g
2)
wi - 4V, -
792.8
=
p
=0
= =
Solving for U2, wc gCI
v, =
16.94 ml s
but,
V,
=
rrD2 N 60 60 x 16.94 IT x 600 0.5392 m
=
53.92 em
=
D, (or) D,
+ hlo +
v'
-'L 28
3. 18' 9.8 1
+2x
Now using Ihe re lation
4g
g
v, ~ 28
+ lid) + hfo + -
=
P8QHm ry" 1000 x 9.81 x 0.025 x 45.26 0.7 15.86 kW
Example 8.9
Find the risc in pressu re in Ihe impeller of a centrifuga l pump Ihro ugh which water is nowing at the ralcof 15 lis. The intema l·and extemal diameter.; of the impeller arc 20 cm and 40 em respectivel y. 'The widths of impeller at inlel and out lct arc 1.6 em and 0,8 cm. The pump is running at 1200 rpm . TIle water enters tht! impeller radially al inlet and impeller vane angle at outlet is 30°. Neglect losses thro ugh the impe ller.
Solution
Example B.8
A cen trifugal pump hns nn overall efficiency of70pcrcent, supplies 25 Us of water to a height of20 m lhrough a pipe of I Ocm diameter and length of I 00 m. Assume friction co-efficient, f = 0.012 , Estimate the power required to,drivc the pump.
= =
0.015 m1 / s 0.008 m
D,
=
N
0.2 m I 200 rpm
D, = 0.4m C, = C,.
= 0.016 m ::
30"
App lying Bcrnoulli'$ equation at the in let and outlet of the impelle r and neglecting losses from inlctlo outlet. E ncrgy at inlet:;;; E nergy at outlet - workdone by impelle r on water
Solution ~u
D
=
0.7 0. 1 m
= =
Q L
0.025 mJ / s 100 m
H
I
= =
20m 0.012
PI
C~
p,
C;
PS
2g
pg
28
C~2Ul 8
- + - + Z, = -= + ~ + Z, - - -
Loss of head due to friction in pipc
he J o
= 4I L V,'
If inlet and outlet of the impeller arc atlhe same height
"
2gD
where V is the ve locity of water in the pipe.
=
Q -IT- -
- D'
4
= _IT 0.025 = 3.18 m/s x a I' 4
.
(ZI :: Z:!)
pg
~()o
;,..
H YDRAULIC PuMPS "" 361
TURBO M ACHI NES
(c) Absolute velocity of water at exit Cr 1
c, c,
sin 1l'1
(a) Manometric head
4 .24 sin I J.08°
v1 _ V 2
I +(Zo -2,) 2g pgfl, = 13.600 x 9.8 1 x 0.25
(Po-P,)/pg+
Hili
IR.74 m/s
a
33354 N/ m' (Vacuum)
(d) Manometric efficiency
1.5 x 10' N/ m'
Po =
· 'Ilf
Velocity of water in both the suc tion nnd delivery pipes, 1111
(26.57 x 18.25)/9.81 60.7'1,.
.:::;:
Vo
= Vi =
(e) Volumetric efficiency.
=
'I t!
=
'I ..
:;;::;
Q Qt
75 77.25
=
97.09% fI..
Fluid power Shaft power PR
:. Shaft power
:;;::;
(f)
and
'/111
=
flm
=
75 x IO2
J
)
0.55 20.07 kW 20.07 - 1.04 .. Shaft power - mechanical [ . '1m:;;::; Shaft power 20.07
=
18.69 +0 + 0.5
=
19.19 m
1000 x 9.81 x 0.1 x 19.19 22 x 1000 18825.4
25 em of Hg vuccum 0.5 m Di = Do = 15 em
Po P
Q
= =
1.5 bar 22kW 0.1 m' /s
22000
lOSS]
94.8%
=
(
p
Solution fI, 20 - Z, D
1.5 x 10' - (-33354) ) + (5.66' - 5.66') + 0.5 looO x 9.8 1 2g
=
30
Example 8.6 During a test on the cen trifugal pump, lhe roll ow in g reading were obtained. Vaccum gauge reading;. 25 cm of fig, pressure gauge n:ading = 1.5 bar, cffL'.cli ve height between guages ;. 0.5 m, power of electric motor;. 22 kW, discharge of" pump = 1001/5, diameter of suc ti on and ddivcry pipes is eaeh 15 em. Determine m:lOomelrie head and overall dficicncy of the pump.
-
x 0.15'
pgQHm
t10
=
4 4 5.66 m/s
(b) Overall efficiency
fI"
1000 x 9.81 x (
-D-
rr
Thererore,
(f) Mechanical efficiency '10
0.1
Q
~=
=
85 .6%
Example 8.7 A centrifugal pump is working against a head 0(20 m whilc:rotating m the rale of 600 rpm . If the blades are curved back at outlet tip and velocity o~ no~ remains con stant at 2 mls . Calculate the impeller dIameter when 50% of the kmcuc energy at impeller outlet is converted into pressure energy. Solution flm = 20 m, th. = 45°. N = 600 rpm, Cq = CT"l :;;::; 2 mls From ou tlet trinngle for a cerlIifugal pump. (Refer Fig. 8.2)
ex 2 =
=
U, - IV" -
2
=
U, - - - - tan 45°
U2 -
C" -fJtan 2
= U1 -
2
I, I.
i.
I
358
>
HYORAULIC PUM PS .:{ 359
TuRBO MACHIN ES
TOlal quantity of water handled by the pump Q, = Q + QL.uu
From outlet velocity diagram.
=
C.r 2
=
IV.,
C" U, = U, - ---(an Ih 3 37.7 - - --0 = 37 .7 - 5.196 (an 30 32.504 m/s
=
:. Cr l
=
282.7 x 37.7 x 32.504
... \I'
IV
=
77.25 l/ s
= Q/ per side
77.25 - - = 38.625 I/ s 2 38.625 x 10- 3 6.B3 x 10-3 5.66 m/ s
=
346.42 kW Now,
(c) Manometric efficiency
= 75 + 2.25
Cil
= 90"
From inlet veloci lY triangle. (Fi g. B.26 (a»
( U, C.<2ig)
70 , 9.81 ;,7.7 x 32.504 11m
w,
tan Pi
=
Cd U,
~,
=
'. 1.71 °
~
5.66 -9. 16 ~,
u,
60 .04 %
Example 8.5 A radial. single slage, double 5uction, centrifugal pump is manufactured for the following data. D, = 100 mm Q = 751/s N=1750rpm Hm = 30m 0 b2 = 23 mm in total al = 90 Leakage loss = 2.25 I/s Mechanical loss
(b) The absolute water angle Area of now
ilt
~o
=
1.04 kW
outlet = ]'( Dzb2 x Contraction factor
where b-z = 23 x
D, = 290 mm . hi = 25 mm per side
i=
11 .5 mm per side
A,
rr x 0.29 x 0.0115 x 0.B7 9. 12 x lO-'m'
~
= 55%
11, = 27°
Figure 8.26(0)
Velocity of now al outlet
38.625 x 10- ' A, .0.00912 4.24 m/s Q
contraction factor due to vanes thickness :: 0.87. Determine (a) the inlet vane angle (b) the angle at which the waler leaves (he wheel (c) absolute velocity of waler leaving impeller (d) the manometric efficiency and (e) the volumetric and mechanical efficiencies.
~
~
Peripheral speed at outlet
Solution U' =
(a) Inlet vane angle
=
The blade speed at inlet. rrD,N
U, = -
60
=
rr x 0.1 x 1750 =9.16m/s 60
Area of now at inlet
n D, N n x 0.29 x 1750 -w = 60
26.57 m/ s
Now, fh = 27"
where AI
=
rrDlbl x Contraction factor
=
=
rr x 0.1 x 0.025 x 0.B7 6.B3 x 10- ' m'
= =
Velocity of now al inlet
0'2
=
V2 - Wx2
= Vz -
4.24 26.5 7 - - tan 27° I B.25 m/s 4.24 IB.25 13.0Bo
C" tanlh
----
HVORAUUC PuMPS '" 357
356 }- TUiUlO MAClIlNES
a nd from the ou tlet velocity triangle (Refer Fig. 8.2)
TIle slip factor (a) is given by
C.~l
W.r:!
C' 0.77 x 15.92
er :! =
ll
U2 - Cx 2 16.76 - 12.46 = 4.3 m/s WI'2 1an
fh
4.3 x tnn 40° = 3.6 m/ s
;::
12.26 m /s
Now,
= =
lhC.~" - -- =al,·E g 0.77 x 30.B3 23.74 m/ s
if
sin fJ'
2( 1 - (C"IV, )colfJ,)
Then,
2
:.mo
=
D'2b2 x C r 2 0.4 x 0.05 x 3.6 0.2262 m J Is Tr
Jr X
So lutio n 0, = 20,. N = 1200 rpm. D2 = 0.6 m, b:! = 0.05 m
Jr sin I - =-.,.,-:=::--:c=----,=: 2 (J - (I.7B/19) col 300 1
0.77
an:a of now x Velocity
Example 8.4 A cenlfifugal pump havi ng OUler diaml!h.:r equal to two times the inner diameter and running at 1200 rpm, works againsl u total head of 75 m . The ve locilY of flow through the impeller is constant and equal ( 0 3 mls. The vanes arc set back at an angle of 30° at outlet. If the outlet diameter of the impeller is 60 cm and width at outlet is 5 em, determine (a) vane angle at inJct (b) work done per sec by impeller and (c) manometric efficiency. (MKU-Nov. '96)
Thc stodola s lip factor is given by
1-
= = = =
Q
Therefore, the theoretical head with slip is
lim = 75 m,
Crl ;:: C r 2
= 3 m /s.
(a) Vane angle at inlet The inlet velocity triangle for a cemrifugal pump is
8.15
I
shown in Fig. 8.26 Numher of bJudes required = 8.
C"
The outer diameter of an impeller of a centrifugal pump is 40 c m and o utl e t width is 5 em. The pump is running at 800 rpm and working against a head . of 16 nl . The vnnc angle at outlet is 40°. Assuming th e manometric effic iency to be 75%. Determine the discharge. (MKU-Nol'. '89.)
Solution D~ = 0.4 m, /;~ = 0.05 m. N TIll!
i
tan{J1 = -
V,
Example 8.3
= 800 rpm,
Hm
= 16 m ,
{J"1
= 40° , 1]111
==:
0.75.
0,
V,
=
D,
Dj N ~
=
Tr
x 0.3 x 1200 60
w,
3
p, = 9.043
u,
0
Figure 8.26
rr x 0.4 x BOO
=
(b) Workdone p er second
16.76 mls
From Ihc definition of manometric effic iency
'1m Cx:!
= = =
(V,C,·,lg) g X Hili =
U2 11m
12.46 m/s
rrD,N
60 =
u,
flm
9.BI x 16 16.76
X
I
= 18.85m/s
Lan tI, = 18.85
impeller tip speed
60
II
0.6
= -= = -2 = 0.3 m 2
TC
m
=
pQ
10
0.75
=
I
3
rr x 0.6 x 120a
60
= 37 .7 m/s
I
I
I ~
= p(rr 02b, xC,,) X
rr
X
2B2.7 kgls
0.6 x 0.05 x 3
,I
354
)lo
I
TUROO MACHINES
HVDRAUl.IC PUMrs
SOLVED PROBLEMS A centrifugal pump of 1.3 m diameter delivers 3.5 m 3 jmin of water at a tip speed of 10 mjs and a flow velocity of 1.6 m/s. The outlet blade angle is at 30° to the langent at the impeller periphery. Assuming zero whirl at inlet and zero slip, calculate the torque delivered by the impeller.
4211.8 x 0.65
Torque ddivered
10 273.7 Nm
Example 8.1
Solution There is no slip. ThereCore, fh =
fi;
,
Since. there is no inlet whirl component, E
e X]
Impclll!r periphery x Blade depth
FlowareiJ
rr x 0.02 x 0.25
= 0
=
U2 C-'l g
15 .7 x
Flow ve locity CI1
v, = -(U2 g fh.
inclined at 30° to the tangent al impeller outlet. Tbe blades arc 20 mm in depth at the outlet. the impeller is 250 mm in diameter and it rotales at 1450 rpm. The now rate through the pump is 0.028 mJjs and a slip factor of 0.77 may be assumed. Dl!lermine (he theoretical head developed by the impeller and the number of impeller blades. First consider thl! no.slip condition. Assuming hlildl!s of infinitesi nwllhicknes5 the !low area may be calculated as
U1C~1-U]C" 1
Given
Example 8.2 The impell er of a centrifugal pump bas backward.fal.:ing blades
Solution
The Euler head is given by
0.028/( 15.7 x 10- ') 1.78 m/s
From the ou tlet veloci ty triangle (Fig. 8.2)
~
~
wn :mo
tan 30° 1.6
1.781 tan 300 3.08 m/s
tan 30° E
~ 9.81
(10 __ 1._6_) lan 30
Now,
0
V,
DN 160 x 0.25 x 1450/60 . 19 m/s
IT
7.36 m or W/(N/S) Power delivered OV) III
=
Ex (mg) p x Q 1000 x 3.5/60 58.33 kg/s
7.36 x 58.33 x 9.81
.'. Power delivered (W)
IT
Absolule whirl componenl c,-~
Power Angular velocity
1 4211.8 x - (V Ir) U = wr ,
lip speed U
r
The Euler head is
." E=-_V!:,:c~x.c'_--=V~I:C", g
and assuming
CA'I
= 0 (no whirl at inlet) E =
10 m/s
1.3/2 = 0.65 m
Vz - W.tl 19 - 3.08 15.92 m/s
=
4211.8 W Torque delivered
10-' m'
QIA
=
WX2 )
= 30°. From the outlet velocity triangle (Refer Fig. 8.2)
355
=
U2 C .tl K 19 x 15.92
9.81 30.R3 m
;152
:,.. TUltllo MACILINES
HVDRA UI.IC PUMPS "" 353
The no zz le can [ower the pressu re [0 Ihree fourths of almospheric pressure, i.l!. ahout TTl suctiu n head can be obtained. Steam also :>crves (0 prehcat thl! wat!..!!' red 10 the boiler. Wilh waICr at hi gh pressure, a more perfect vaccum ca n be prodUl:eu, sO that a sucion lift of 5.5 (a 6m can be obwined. The jet should be ncar the surface of water. Several jets may he employed' if a large quantity of wl.\tl!r is to be pumped . TIll.! c:lpacilY of a jet pump ran ges upto about 50 lis. If the rale o( tl ow from suppl y resl.!rvoir is Q I and the qU
Q,( H,
+
H,,)
= Q,( H, - Hoi )
III
Air Lift Pump TIle pumps thut utilisl! cu mpressed air [ 0 lift watcr arc know n as .air lift pumps. Thl.! function of comprl!sscd air is to form a mixture of ai r and water and to reduce the mi xture density. Thc dellsit y of a mixturl! of air ilml wJle r is much lower tlwn Ilwl of pure WJler. If such a mixlure is balanced against a water column. th l.! fornH.:r willnst.: much high~r than thl! l;:Iuer. Air tnl: 1
DiKhl1 e
""\ I I
I.
1f='
vicinilY and carie s it upward s to the top of delivery line. The air-water mi xtuft: can risc ab ove tht! water level because of its low density. TIle air is introduced at a considerable depth below the water surface. so that the press ure due [ 0 the column of height L is less than that due to height H. height of water level above the nozzle, thus, causing the flow of walcr to the desirt!d delivery level. The pump disc harges water as long as there is a supply of air. For best results, the lift of th e pump (L - H) should be less than H .
Advantages 1. No moving parts, no valves. No damage due 10 solids in suspension in water. 2. Raises larger quantity of water for a given diameter than any other pump. 3. To drain out water from mines where the compressors arc alre.!ady uvailable.
Disadvantages . I. Vt:ry low dficil!llcy. Th e volume of air in m]( V) ilt atmospheric pn:ssure. required to lift one m·l of watcr through II height (L - H) dt:pt:nds on tbe efficir.:m,'y. 2. Air leakage prohlem.
Submersible Pump As th e nam e implies. the pump and electric motor both arc submerged in wilter. Submersible pumps arc built together as single or multistage centrifugal pumps, directly coupled wilh the electric molor, which is tOlally enclosed. Tht: motor and pump shafts arc supported in water lubricated plain bearings. These.! pumps are vertical centrifugal pumps with radial or mixed flow impcllcn>. Alllhc me.!tallic bearings are wate.!r lubricated and protected against the sand. A non return valve is fiucd to a flange at the top of the pump. The suction line of the pump is situated between the pump and the motor ilnd provided with a perforated strainer. Motor of the submersible pumps arc filled completely with water. The water cools the windings, insulated in a plastic impervious to water. TIle pump shaft is connected to the motor shal'l by a muff coupling. Submersible pu~p s are installedbClow the lowest expectcd water level in the well. These pumps can be usually employed in open wells where frcquent changes of waler leve l make the working of a horizontal pump difficult. Before a submersibl e pump is installed, the motor must be filIed with clean non -acid water free of san d for the purpose of priming. The pump mUSl be started with Ihe valve sli ghlly ope n. If there is a considerable amount of sand in the water, then the pump must be operated with valve partly open, untillhe sand contents arc reduced to acceptable quantity. The valve may then be opened gradually.
Advantages Figure S.25
Air lift pHilip
Fi g. (8. 25) shows the: ;:Iir lift pump. Compressed air, supplied. by a co mpressor il1sWlJed on (or) ahove the ground level, is rassed through one or morc nozzles at the hOllom into the wa leI' at a cons iderab le depth, bl!low the water surf;:lce. The rapid strcam of air al the Il ozzlr.: exit produce a jt!t and entraps the water in the imm ediate
I. High overall efficiency Bnd therefore economical in operation.
2. 3. 4. 5.
No maintenance because of its glandless construction and waler lubrication. No priming is required due tD submerged installations. No suction problem, since fOOl valve is not required, Low noise level.
;
I
! H
H r DllAlJU C P UMPS "" )51
350 }.- TURno M"CII1NliS
During one-half revolution of rotor rOUltion, the volume increases bl.!!\VCl.!n the rotur and cam ring . The resulting volume expansion causes a reduction ofprcss urc. Thi s is the suction process which causes nuid 10 now throug h the inlet port and lillihe void . As the rotor rotiltc s through the second half revolution , the surface o f the cam ring pushes the vanes back into their slots. and the trapped volume is n:dUl.::cd resulting in increase of pressure . This positively ejecL<; the trapped fluid through the discharge pOri .
Center line
-l Cylinder bloc k I - - center line Plllile
COllie
PiSlons
3. Piston Pumps
Cylinder block
A piston pump works on the principle thut II reciprocilting piston C~H1 urilw in fluid when il retr
(a) Axia/ piston pumps Fig . IS .22) shows an axial piston pump. In a xial piston pumps . lhe pistons arc parnllel to the axi s o f the cylinder block. It cons ists o f a cylinder hlock rotating with the drive shaft. A universal link connects lhe block to the drive shaft to provide alignment and pos itive drive. The centre line of the cylinder block is set aI an offset angle relative La the centre line of the dri ve shaft. The cylinder block conlntns it numher of pistons ammgcd along a circle. The piston rods arc.: connected to the drive shaft flange hy ball and 50ckcljoints. The pi stons arc fo rced in and out o f their bores as the distan ce hetween the drive shaft flange und cylinder bloc k L.:hanges.
Oil rurrcuto
Re
Figure 8.23
R{ldial pi.rIOll plllllp
As (he c y linder barrel rolates. Ihe pi slo ns on one side move outward . 'nlis draws in fluid . as eac h cylinder passes th e s uction port s oflh..: pintle . When a pis IOn passes Lht.: point of maximum ec centricity, it is forced inward by Lhe reaction rin g. This force s th e nuid to enter the discharge port o f the pintle .
MISCELLANEOUS TYPES OF PUMP Jet Pump A je l pump is s hown in Fig . (8 .24 ) Jet pumps arc used to lifl Lhe wall.:r in a grcatt!r quantity when a s mall quantity uf wutcr is uvailable deep down the earth. Steam (.or) water at a hi gh pressurc is forced through a fine aperture nozzle, thcreby converttng most of the pressure energy into kineti c energy. It results in lowering of pressure Cilus ing suction to take place . A part o f the suc tion is due probabl y 10 the dr?p .of pressure resulting from conden sati on or the steam . Steam is generall y used an Jet pumps used to fe ed the water to a boiler.
Piston is wilhtlrllwinG rronl bon: pi inlet
From inler Figllre 8,22
Adel' piSlo1I pump
(b) Radial piston pump A radial pump is illustrated in Fig . (S .23) It consists or pintle to direct fluid in and out of the cylinders , a cylinder barrel with pis tons. i:md a rotor containing a reaction ring. Th e pistons remain in constant contact with the reaction ring due to centrirugal force and back pressure on the pistons . Fur pumping action. the reaction ring is moved eccentrically with respect [0 the pintle or shaft a~is.
Figure 8.24
Jelpulllp
....,
;\·HI
HVDRAULIC PUMPS "'"
;... Tl llUlO rVJ..KW NES
TIle ex ternal gea r pump uses spur gems, (which arc noisy al high spceds), helical gears (whil.:h nrc lim itcd 10 low press ure .Ipplication duc to end thrust problems) and herringbone gC~lrs (w hi ch provide greater flow rales s llloolhy).
Inlet
349
Outlet
(b) Internal gear pump Fig. (8 . IR) shows Ih e con figuration orlhl! internal gear pump. llii s rump consists or nn internal gear. a rl!gulill" sp ur ge:tr, a crescent shaped sl!a l and all cx tcrn:lllillusing.
Thrust cage Inl et
Idler rotor
Onlance cup
Crescent seal
Figure 8.20
Inner Figure 8.18
imcrlla/ gear I'Wllf'
As powc r is applied to eilher gear, the motion or Ihe gears draws fl uid from the resa\'oir and rorces it around holh sides or the crescent st!al. which Ole IS as a sea l bl!tween Ihe suction and discharge POrls. Wh.:n Ihe teelh mesh on the sid e opposi le tll thc cresce nt sen l, the fluid is forc ed to enter Ihe disch.ugc pOri of the pump .
(c) Lobe pump Thc lobe pump is illu slflHed in Fi g. (8. 19). 'ntis pump operates in a fas hion similar 10 the external genr pump. But unl ike th e external gear pump. holh lobes arc driven t:xlerna lly so that Ihey do not actu
Rotor housing
Mechanical seal
Serel!' plllllp
There may be on!!, two or Ihree screws. In a 2-scrcw pump, one is rOlor and tbe o~lcr is id ler. In a 3-screw pump ther..: are two idlers on either siLle uf the rotur.Two idlers act as s..:als 10 Ihe power rotor and arc driven by fluid pn.:ssure and not by metOllIic conlOlC I with rolor. The liq uid entcring Ihe inlet passag..: flows iolo the Idl end of the: rOlor whe: rl! it is trapped in pocJ.;...:ts formed by Ihe threads and is carried lowards the di scharge. The movelll..:nl of Ouid is simi lar 10 Ihol o f a nut on Ihe power screw. Thl! advan tages of screw pump over Iht: other IYPI!S of pump are: I. Screw pump operates upto a continuous working pressure of about 150 har. 2. The screw pump is free from pulsation ond vibratio n. So. Ihe pump is silent while: running .
2. Vane Pump Vane pump (Fig. 8.21) cons ists of a rotor disc having a number or radial slots and is splined 10 the drive shaft. The rotor rotatcs inside a cam ring. Eaeh slot has a vane designed to mnte with Ihe surfilce of Ihe cam ring as the rotor turns . There is an ccce ntricity b..:twecn thc rotor and the cam ring. The vanes arc frce 10 slide radi<.illy with Ihe help of springs and thus form the requ ired seal hetwccn thc suction und discharge connections. Centrifugal force keeps Ihe vaneS oul against thc surface of the cam ring . Cam ring surface
Figllre 8.19
Lobe' (ll/llll'
(d) Screw pump The screw pump (Fig. 8.20) is an axial fl ow pos iti ve displaceIlh: nt unit. It consi sts or a rotor or screw to whi ch the source of power i.c. electri c mOlar is direct ly L:Onnccll'd. TIle screw may be single helical or double helical. 11lc ad\'iJ lI1a!;!1! of double helical is Ihal they arc balanced axially. The fluid is carried fnrwarJ In th c dischurgc along Ihe rotor in pockels I'nrmcd hetween tt:elh
!
Figure 8.ll
\ime 1'1111/1'
, I
11l
I jyI)R .... UU C PUMPS '" 347
346 ,. TI)RUO M ,\('IIINES
•
inlet velocity \' i arc meil.<;ured, and u~ is then calculated from the above cljuation. The minimum required NPSfJ or Cfe may then be plolled ror the different degrcl!s of inlet throttling (0 give a curve of (Ie VCSUS now co-efficient. From the steady now energy equation, the energy loss betweelllhc free surface (A) and the inlet side of Ihe pump can be determined . If the datum is placed :ltthe lower reservoir surface, \'1\ 0 and ZA = O. Then the equation inlerms of heads becomes
=
P" / PC = PI / pg where HJu ('lifJrr = Z. equation of 0-. gives
+ "/i + hffJSJ
+ V/"/2g + H:werirm
(a) Ex ternal gea r pumps (h) Internal gear pumps (c) Lobe pllmps (d) Screw pumps 2. Vane pumps 3. Pi sto n pumps The details of the construction and operation of gear. Vimc and piston pumps arc discussed in the following sections.
at in let. Substitution for p./pg in the defining
1. Gear Pumps (a) External gear pump Fig . (8.17) illustrates the operation of an cxt~mal gcar pump which develops now by carrying nuid between the teeth of two mcshmg gears.
If a is above a l · • cavitation will not occur. But in order to achieve this. it may be necessary 10 decrease l is I/j"/illll by decreasing Zi and in some cases the pump may havc to be p laced helow tiie reservoir or pump free su rface. i.c. negative Zj. especially if" li.iS particularly high duc to a long inlet pipe.
Outlet I1rive
METHOD OF PREVENTING CAVITATION TIle following factor s should be taken into account to prevent cavitation:
I. The pressure of Ihe nowing liquid in any part of the hydraulic system should not he all owed to fall below irs vapour pressure. For example. the absolute pressure head should not be below 2.5 m of water, if water is the working medium . 2. TIle specia l materials or coatings such as aluminium . bronze and stai nl ess steel which arc cavitation resistant materials should be used.
POSITIVE DISPLACEMENT PUMPS TIle pumps arc broadly classified into two as (a) Hydrodynamic (or) non- positive displacement pumps; ex: centrifugal and axial pumps and (b) Hydrostatic (or) positive disp lacement pumps: ex : Gear pumps and vane pumps. The former type is used for low-pressure, high-volume now applications.TIlcsc pumps are not self-priming . This is becau se there is 100 much cleareance space between the rotating and stationary elements and to scal ugainstlltlllosphcrie pressure and thus the displacel1l1.:nt bl.!twcen the inletllnd outlet is nOi a positive om:. The laller type of pump ejects a fixed quantity of nuid per revolution of the pump shaft. A pressure relief valve is used to protect the pump against overpressure, because a positive displacement pump continues to ejeci nuid even though the out leI valve is fully closed cllusing an c.xtremcly rapid build up in pressure as the nuid is compressed. Where as. for non·positivc displacement pumps. in such a case. there is no need for safety devices to prevent pump damage. Positi ve displacement pumps can be classified by the lype of motion of internal clements. The motion may be eithl!r rotary or reciprocating. 'nlere nrc essentially of lhree basic types; I . Gear pumps
Inlet Figure 8. / 7 £-af'rllCI/grar pump One of the gears is connected to a drive shaft connect~d . 10 th~ primemover. ~e seco nd gear is driven as it meshes with the driver gear. OlliS camed aroull.d ho~slll.g in ch'lInbers formed between teeth. housing and side \vear plates. Th.c sucllon ~Id~ IS whl!re teeth come out of mesh. and it is the place where the vo lume exp;lI1ds. ?rtnglll.g about a reduction in pressure below atmospheric pressure . Fluid is pu shed mlo thi S voi d by atmospheric pressure hecause the oil supply tan~ i~ ve nted to the atmosphere The discharge side is where · teeth go into mesh, and It IS here where the vo lume
decreases between maling leeth. The theoretical fl ow rale of a gear pump is given by
IQ,,, = V/J
X
N
I
where. Q.h is thl! tlll.: nretical now rate in mJ/min. VD i.s the pump di splaceme~t vn lullll! in m J / rev .a ml N is the pump speed in rpm. The dis placement volume Vo IS iT 1 ' VD = -(D - DelL
4
where, D" - outside diameter of gear teeth. 111 Dj • insidl! diameter o f gear teeth. TIl L - width or gem teeth . 1/1
"
I
~-I·I
:,.. T Ulino t\'I ,H; ltlNES
I-I rDltAUUC' PUp..U'S
JOO HcaJ
\ /", \ ,,( ~ '- ........... '\". PowC'1" -
~ oo
/'
'\".
100
-
--
. . DC'sJgn pOI n!
"". /
u=
40
80
120
~45
Wh r.:.n the liquid 1I0ws through a cen tri fuga l pump , the slatic pressure (suction press ure ) at the eye of the impell er is reduced and the velocity increascs. Therefore, there is a danger that cavi tation bubbles rn ayJonn al the inlet to th e impdlcr. When the fluid moves inlo a higher pressure region, these bubbles collapse with tremendous force. givin g ri sc to pressures as high as 3500 atm . Local pining of thl.! impell er can result when th e bubbles collapse on a metallic su rface, anti serious damage can occur by this prolonged cav itati on erosion. Noise is also generated in Ihe form of sharp cracki ng so unds whl!1l cavitation take s place. . Cavitati on is mos tlikdy to occur on the suction side o flhe pump bl!twccn the lower reservoir su rfnce and Ihe pump inlet since it is in thi s region that the lowest prc.ssurc will occ ur. A cavitation parameter is defined as
Efficie:ncy
o
..:
Pump total inl et head above vapour pressure Head developed by pump
AI Ihe inlet nange
Percen tage of design flow rate Figure 8. 15
Now Q is proport ional 10
ell
Characteristic C/ lrt't!S of arial flolI' pump
,md thcrcfnrc
where all pressures arc absolute (or)
Id E/dCncxdE /dQ CX-Ucotlh l For ;IXia l fl ow on inlcL fi2 is relat ively small, and thus for a given pump at i.I g ivt! n speed thl.! h!.!ad - fl ow re lations hip has a Sleep negative slope . Thl! power curve is sim ilarly vcry steep. the power requirement at shut offb cin g perhaps 2- 2.5 times that required nt th e desig n point. This makes for a very expl! nsivc e lec tric motor 10 cover the eventuali ty at th e low fl ow rates ~1nd so lhe fi xed bl.:!.de axial fl ow pump is limited La operation n! the IIxed cksig n point. In variable flow mnchincs. the blade ang le is adjusted so th at Ihc pump runs ,l{ its maximum effic iency at all loads and reduces the shut off power req uiremenl . In Ih e figure 8.15 , the power and head cu rves arc seen to enLera region of instabil ity at aho ut 50 perccn t of lhe design now filte . This is duc to e ll becoming increasing ly small und thereby increasin g the angle of inc ide ncc of How o nto the blade until separation a nd stalling of the blade occurs. Funher head risc at evcn lower fl ow roues. and lhe consequ l.!JlL power ri $c is due 10 th e recirculat io n of the nu id aro und th e blade from [he pressure sid!.! 10 th e s ucti on sidt.! and then up o nto the press ure side o f the next blade. An increased hlad e stagger anglc will o nCe agai n reduce this rec ircul a li o~ , therehy the powl.!r rcqu ir!.!Ol!.!nt.
-
CAVITATION Cavi tati on is dcflned as a phenomenon of formation of vapour bubbles of a nowing liqu id in th e regi o n where tht! locLl I abso lute stati c press ure of the liquid [atls het ow th e vapour press ur!.! of the liquid and th e sudden co llapsin g of these vapo ur hubbies in the high press ure n:gio n.
Every pump has a critical cavi tation number (a,,) , which can only be determined by testing to find the minimum values of NPSH before cavitation occurs. One method is to determine Ihe normal head-flow characteristic of the pump and then to repeat the tes t with th e inlet to the pump progressively throttled so as 10 increase Ihe resistance to now at th e inlet (Fig. 8.16). It will be found thaI for different throttle valve scnings Ih e performance curve will fall away from the normal operating curve al various points 3% head
l\1easure:t.\ Nt'S
NI'SIi
/
--Q, Q, Q, Q-,> Figure 8.16
Q, Q, Q,
Q-,>
Hill Qud NPSH \'usus Q
and one definition of the occurcnce of minimum NPSH is the point at which the: head H drops be low the normal operating charac teri stic by so me arbitrarily selected percentage, usually about 3 percentage. At this condition static inh:t pressure p. and
342 ,. TUlluo MAClIlNES
It should bl.! noted thilt the now area is Ihe i.mnulus formed betwel!n the hub and Ihe bli.ldt! lips. Then we Illay write 111
= pC,,[rr(R,'2 - R~)l
The work dOlle hy the pump is
IW = mU(e" - e,,) I
r
THE ENERGY TRANSFER OR HEAD
Impeller
u
The energy tr4lnsfcr (E) is given by
c,
IV,
c,
W
=
E
mg
=
U(e" - e" J/ g
For maximum energy transfer the absolute Oow velocity must be uxia l ill the inlet i . ~ . CI = Cu' Hence.:, 0'1
= 901>
C ,t l
=0
E
= UC.
11
/1!.
E in lenns of angle fi] is obtained as
Outlet gu ide vnne
e"J/c"
or
CI'! =U- C(lcolfh Figure 8.14
Velocity
trtl;flg/~s
oj axial flolV (lumps
Hencr.: subs tituting for
Since the sU:lIor is fixed, ideally the absolute velocity C I is parallel 10 the trailing edge of the blade (not shown in Fig_ 8.14). The relative velocity WI is obtained by Yeclorially subtracting the impeller tangential velocity U(= UI U2) from CI · WI ideally should be parallel to the rotor leading edge of the rotor. In the exit velocity diagram. the relative velocity vector W2 is parallel to the blade
=
trailing edge.
THE WORK DONE ON THE FLUID From Euler's pump equation. the work done on the Huid
IW = ",(UzC
X} -
UIC xl )
I
Changes in the condition of the fluid take place at a constnnt mean radius, therefore
Also assuming a constant How area from the inlet to outlet, we have
C.I'!
in the above equation.
[E = U(U
- e"cOIIl,)/8 [
TIlis is the maximum energy trnnsfer or head dcveloJ1cd by a n a;(ial now pump. For £ to be consta nt over the whole b lade len gth . the blade angle Ih should hc increased hecause U increases with radius. It is see n from the above equation th .. t lhere mus t be a corresponding increase in the tenn Cn col /3'! . Since Cll is co nstant. /32 shou ld incre'lse. Hence, the blade must be twisted as the radius' increases.
Note: Since thl! density ch,mgc across an axial flow fan is so low, it may be ';011sidercd to he operating w ith an incompressib le nuid and therefore tht: same equations as applied to axial now pumps may be used.
AXIAL PUMP CHARACTERISTICS TypiCi.!1 hcad -now power and efficiency curvcs arc shown in fig 8. J 5. A Sleep neg;J{Jve slope is ev iden t on Ihe head and power curves at low now rail.!s . 111is ca n hI.! c}tplaincd considering the maximum energy trnnsrer equation. For a given blildc design .It fixed speed with :lxinl now at inlet [ E = U(U -
e" col 1;')/8 I
J~O
,·f\·OR.... uuc PuMI'S "' 3-11
,.. T URBO M M·HI NF.5
AxiAL F L OW PUMPS Combined
System resistan ce
I___ ",-l!~in~se ri e s
,.Icurve (1-1 "" H, + KQ 2)
, I
An axial now pump cons isIS of a propeller type impeller running in a casing with fine clearances bel ween Ihe blade lips and the casing inner walls . The Iluid esse nliulJ y passes almost axially thro ugh ulternate rows of the fixed stator b1adcs and movint; ro tor blad es in a muhis tagc axi al now pump . The figun: (8. 13) shows a single Slut;e ax ial now pump.
Ho /
/ /
lI e
/
""-
/'
~
DESCRIPTION
Pump C
Pum D
o
opcrOll ing points 0
Figure 8.11
II
r
C
C+D
Q-C>
7iI'o diJfferelll p UlllpS connected i" series
Flow)
----'::.::. Flow
~~~~~(rre~if.:~~rC~Q2)
Combined in seri es
I
'
I
I I ~ng J e
_
pump
/ /
,, operating point ... Figure ll.11
Two simi/ar pumps cOlmuted ill serieS
Pumps in Series Let 1/ = Number o f identical impellers moullted on the same shaft. Hili := Bend developed by each impell er. Th en the t(llal hl!ad dl!vcl 0pcd x /-fIJI. TIle di sc harge pass ing th ro ugh each impeller is same.
="
Pumps in Parallel Let
II :;:
number o f ident ical pumps arranged in paralle l.
Q:;; Di sc harge from one pum p. TOlal di scharge = /I x Q. E
Smrionary inle! guide vanes
Stationary oUllet guide vanes
Impell er (or) Rotor Figure 8.13
Si/lglf! stage ClxiCJljlolI' pump
The inlet g uide vanes guide the fluid to enter the rotor with 11 purely axial vdocity. The impdl cr blades. however impart a whirl componc:ntto the Iluid . The uUllc:t guide vanes elim inate! Ihe swirl on the o utlet side and turn the fl ow towards the ultis. To ensure a smooth flow without shock at the design condition. Ihe bladc.s of the impeller mus t be twisted . The now area is the same at inlet and outlet and the maltimum head for this type of pump is of the order of 20 m. The usual number of blades lies between two and eight, with a hub diameter to impeller diameter ratio of 0.3 - 0.6. The advanlage of an axial flow pump is its compact conSlruction as well as its ability [0 run at extremely high speeds.
VELOCITY TRIANGLES The in let and outlet velocity triangles nrc drawn for axial now pump (Fig . 8. 14). It will be noticed thaI the impeller blade has an aerefoi! section and that the inlct relative velocity vec tor WI does not impinge tan gentially but rather the blade is inclined 81 an ang le o f incidence j to the relati ve velocity vector Wt. This is similar 10 Ihe angh: of attack of an aerarail in a free strt:am . It is assumed Ihal there is no shock at entry and thot th e nuid leaves the blade tungentially at exit.
33tl
j;>
TUHBO MACHINES
HYDIL\UL!C PUM"S ..:
Q A + Q If but certainly greater than either one. For a very nat (stalic) curve two similar pumps in parallel will deliver nearly twice the flow. (Fig. 8.8)
:rw
Ir pumps A and B are not ide ntical as shown in Fig. 8.9. then the pump B should nllt be run and cannot cven be started up if the opcrating point is above its shutofl' head .
PUMPS COMBINED IN SERIES
1-11
System resistence
eurve , (H = Hs + KQ2) / /
/
To develop a high head, the pumps should he connected in series as shown in (igurc 8.10. The water rrom the suction pipe enters the pump C at inlet and is dischargt.:d iLt the outlet wilh increased pressure. Tilt.: water with increased prcssu n.: from the purnr C is taken to the inlet of the pump D. The pressure of waler
Combined /pumps
(
-- ~' - --, ,
(
\
\
-..... ,
Single pump
\
\
\
\
\
\
A+B
Q~
\
\
From
operating points Figure 8.8
\ \
\
A
pipe
\
\
\
o
To Llc:livcry \
\
\
Two simj/ar pumps connected in parallel
TIle combined brake power is found by adding the brake powers of pumps A and B ilt [hc same head as the operating point. The comhined efficiency equals
/
/
,
I
I pg( Q,I-I-·8)( H m )A+8
Pump B
/
Combined in System pilInllel rcsistcnce curve (1-1 ~ H, + KQl) /
operat ing points
Figure 8.9
B A
A+B
Two differellt p,tmps combilled ill paralfel
Figure B.IO
I
Pump D
I
PllltlpJ in Uri/'S
The discharge through eat.:h pump is the same. The com hi ned head is higher than eilher one. TIlL! pumps in SeriL!5 need not be identical at all, since they merely handle the same disdwrge. The)' may evcn have different specds. although normally both .lre driven hy the same shaft. The combined operating point head for two different pumps in serit.:s will he more than either C or 0 scparately, but not as great as their sum. The combined pumps perfromance curves arc sho\vn in the figures H.IJ .:md H.I:!. . The combined power is the sum of brake power for C and D al the operating pOint now rate.
/
o
/
I
(Ps)A+IJ
Pump A
I
I
Q~
. ' . pg(Qc+o)(H"lc+D The combined effiCiency 15 = . (P.dC+D
336 ,. T UlWO M ACHINES
IiYDltAtJl.IC Pllt.U'S .;( 337
Figure 8.5 shows typical characlCristic curv~s for a ce ntrifuga l pump . The he'ld is <Jpprox im:uely constant at low disc harge and then drops 10 7.ero at Q = QIlItIX ' At thi s spt:ed and impelle r s ize Ihe pump cannot deliver any more nu id than Qmlu , llH: positive slope part o r the head is s how n dashed as mentioned earlier, this regi on c<.In he unswblc Dnd cn use pump surge. The power provided by the pump mo tor rises llIonOfonically wilh d isc harge and then typ ica lly drops orf slightly ncar QI/l/lx. The drop orf reg ion is shuwn dashcd because it is a lso potent ially uns table and can cause mo tor ove rload durin g a trans ient. The efficiency ri ses 10 a maximum at aboul 60 percent of Q",II .I" Th is is Ihe des ign nnw rale (Q f) or Bcs i eniciency po inl (lJEP ), '1 = '}"III.\" The head and po'wcr at th e JJEP will he lerrneJ lin and Pf) re spec lively. NUle th allI is zt!ro al the origin (no now) nllJ at QIII'" (no heLld ). NUl!.! :.Il so thai the ,/ curve is not independcnt but is s im p ly caku lnted rrom Ih e rdaJion
CONSTANT EFFICIENCY OR MUSCHEL CURVES The head versus di scharge curve and efficiency versus discharge curve for various speeds nrc combined 10 ge t the co nSlan l l!fficiency cu rve. The Musche! cu r vcs nrc drawn as fo ll ows .
The v
MULTISTAGE CENTRIFUGAL PUMPS A Ccn lri fu gal pump lital con sists of two or more impdlers mounted o n the samc shaft or on dirferent sha fls is called the I/Ill/listag e cell/rijugal pI/flip. Multistage pumps un! employed to accomp li sh the following two important runctions . I. To produce a high head . 2. To develop a high discharge.
PUMPS COMBINED IN PARALLEL To develop high discharge, Ihe pumps should be connectt:d in p:.lrallcl as shown in the figure 8.7.
O.S 0_6 11.7 08 . 0_8 0 .7
/
,+ , 0.'
Q"
n.s Pump A
N,
Sump
Q--->
Fig'J,.e 8.7
PIIIII/Js;n
puralld
Pump A and pump B lift the water from a common sump and di scharge the water to a common pipe to which th e delivery pipes of each pump is connected. Each of th e pump works Jg
i.e. QA Q---> Figure 8.6
MIIJc he/ curves
+ QB. (Fig. 8.9)
If pump A has ma rc he:Jd than pump 8. pump B cannol be added until the opcrati ng head is below the s hut o frh cad of pump B. Since, the system resi stance curvc ri ses with Q. the combined delivery Q"+H wi ll be less than thl! separate operat ing discharges
334
}.>
H YDR.AULIC PUt.-II'S .;( 335
TURUO MACHINES
where Pj a nd Vi arc the pressure and veloci ty at the pump inlet and P UfJP is the vapo ur pressure o f the liquid . All pressures arc absolute pressures. NPSH is a lso defined as a me:lsure of the e nergy available on th e suc tion s ide of the
For minimum speed
pump . Ifm
in terms of '111 V zC.(!
Hili = '111
X - --
g
Substi tuting the val ue of Hm in the above equation yields
vi - Vr = 2g
NPSH is a commerc i:l l term used by the pump manufaclUrers and indicates the suction head which the pump impeller can produce. In other words , il is the height o f the pump a xis from the water reservoir whic h can be permitted fo r installation.
'111 x
U2 C .\,: g
Now.
PRIMING Priming of l.I ce ntrifu£:ll pump is defined as the operati on in which the suction pipe. casing of the pump a nd a portion o f the delivery pipe upto th e dc;livery val ve is completely filled up wi th the liqu id to be raised by the pump bdore starling the pump. Thus, the air from these parts of the pump is removed. The he;'ld generated by the pump is give n by
and
IE=
TIlcrcforc.
U2~" I
T he ahove e quation is indepe nden t of th e density of th e liquid. This mea ns th at when pump is running in air, the head ge ne rated is in terms o f melre of ai r. If the pump is primed with wate r. the same head is genera ted but is expres sed in metre of water. But as lhe density of air is very low. the gene ra ted hend o f air in te rms o f equivalent metre of wale r head is neg li gib le. It is obvious. th at if impc ller is running in air. il wi ll produce on ly a negligible pressure. which may nOI be sufficient to SUt:k WOller fro m the s ump . To avo id this difficult y, the pump is first primed . i.e. filled up wit h waler.
(~)' -(~)' 28 · ·d· rrN , we get DIV I mg by g60
PERFORMANCE CURVES OF CENTRIFUGAL PUMPS Pc r fOfl11l.1ncc clwrts arc always pl otted for constanl shaft-rota tion speed N. The: basic inde pendent variab le is taken to be the discharge Q. The depe ndent v:lriahles or output arc taken to he head ( Ii) . power(P) and efficiency.
(or)
72~ [Di - DT] = 'III Hence the minimum starling speed is
120 x '711 x C,rl
X
Besl Emcicnc),
D2
rrlDl- Dil
11.l'.H
1
painc !DEI') or DesiGn pninc
H
( un~tablc ;
cause surgc)
un
a nd a centrifugal pump has to run at thi s spee d alleast. 10 di scharge liquid .
NET POSITIVE SUCTION HEAD ~el .Pos itive Su.c ti~n Head ~~PSH} is the head req uired at the pump inlello keep the liqUid from c avllat mg or bOIling. The pump inlet (or) suction s ide is the l ow ~ prcssurc point where cavitaTion w ill first occur. TIle NPSH is defined as
NPSH
p.
V'
pg
28
= ~+ --.!.... -
0 ••,
tJ
P
~ P8
Figf/re 8.5
Typical t'rlZlriJlI galplllllp performan ce curl'es of COIIJIU/Il impd/f!l" ro tari()11
speed
...,
Ii VDllAIJl.l' ?lJMPS ...{ 333
Ir Hi is the (Ot:11 he ad ucross th e impeller, th en j\ Icuknge power loss ca n be defined us
(iii) Overall efficiency
(X.5)
PL = pg Hiq
Fluid power developed by pump Shaft power input
,----=-:-:---, pgQHm
Eq uati o n (8. 4) shows that when the discharge valve of the pump is closc:d. th e leakage fl ow ralc attains its highest value .
Casing Power Loss
=
'1(1
'7!1=---
p,
(iv) Volumetric efficiency Flow rate through pump
If h e is the head luss in th e fluid between the impeller outlet and the pump o utlet flange and the fl nw nlle is Q. then Pc may be de!'inr.:d as
'7 v =
Flow rate th rough impeller
Iry·=~ 1
(8.6) S umming up all the losses gives
(v) Impeller efficiency r--=---~~~~
Fluid power at impeller exit
'Ii = Fluid power supplied to impeller
where' H",' is the manoml.!lric head .
(i) Mechanical efficiency 'lilt
Fluid power supplied to the impeller
=
Power input to the shan P.~ Q,
(hi
+ H;)
Fluid Power supplied to impeller; Auid power developed by Impc llr.:r + Impeller loss The Fluid power supplied to the impeller is also referred 10 a.s the powe:r available al the impeller.
MINIMUM STARTING SPEED
P,
'DIe cen tri fuga l pump will start delivering fluid if and only if the pressure rise in the impell er is more (han or eq ual to manometri c hend (Hm). Otherwise the pump will nOl discharge fluid . although the impeller is rotating. When the impdlr.:r is rOlating the water in contac t with the impeller is al so rotating. This is the case of rorced voncx . In case of forced vortex. the centrifuga l head or head due (0 pressure: the rise in the impeller is 2r; w2r21 = w ______
fJi - Total head across [he impeller h ,.- head loss in the impeller (or)
28
(ii) Manometric (or) hydraulic efficiency AC lUal Head developed hy pump '111
-
Thl!Oft.!licnl head developed by impeller HIII/Ch;
+ Hi)
28
where wr2 - Tange mi al velucity of impe ller at outlet = V2 and wrl- Tan ge ntial velocity or impeller at inlet
= V\
Head du e to pressure ri sc in impeller
Vi Uf
=- - -
(or)
E
=
U,C,., / g = ( Hi
+ h;)
2g
The flow of wa ter wi ll commence, when
Vi - Vr 28
28
HYDRAULIC" PUM!'S
""
3:\ I
Manometric Head (Hm) C;\,
AC;\
It is a head against which a centrifuga l pump has to work. It is denoted hy ffm · It i.-; given by the rollowing expressions. The pump tolil] inlet and outlet heads arc measured at the inkt and oUllet flanges rc'spcctivcly and arc given as Pump total inlet head;:::: Pj I pg + V/12g + Zj Pump total out lct head Pol pg + 12g + Zo Total hcad developed by pump (or) manometric hC<.Id (Hm)
l
Actua l
.[
u, Figure 8.3
VI;
=
=
(Po - P;)/pg
Hs
H; Ddivc:ry pipe I
/
hla +hfo
-
- -
Hm
~p.
--
- -
1
+ I: HlosSt:s + fI/i -t-hfj +"10 + " 1"
The shaft power or energy P.T that is supplied to the pump by the prime mover is not the same as the energy received by the liquid. The difference is mainly dul.! to the following losses: l. Mechanical rriction power loss (Pm) due to the friction be tween the fixed and the rotating parts in the bearing and the stuffing boxcs. 2. Disc friction power loss (PD) due 10 friction between the rotaling faces or the impel1er (or disc) and the liquid . 3. Lealmge nnd recirculation power loss (Pd due to the loss of liquid from the pump (or) recirculation of the liquid in the impeller and 4. Casing power loss (Pl')
H, P;
Z,)
PUMP LOSSES AND EFFICIENCIES
hd
'".~~ .
+ (Z" -
where II f is the fridional head loss III is the other head losses and Hm = Euler head - Losses in the impeller and casing
,.-- -=-= -- --=
tenmrug;!..,-
- V,-}/2g
The type of pump is selected for a specific purpose based on the manomelric head. Hili c
Exit velocity trial/gle wilh slip
HEAD DEVELOPED
I
"
+ (V,;
@-:F h,
Suction pipe
Impeller Power Loss (Disc Friction Power Loss (Pv)) hJj + hrj
------------------- -------
It is caused by an energy or head loss h; in the impeller due to d isc friction. flow separat ion and shock at the impeller entry. The now rate thmugh the impeller is Qr. so the impeller power loss is expressed as
--------- ----
Figure 8.4
I PD = pgQ; h; I
Heads associated with centrifugal pUlllp
Leakage Power Loss
Static Head (H.d It is the vertica l distance between the two levels in the reservoirs or Hs ;:::: hs i.e, Static head is (he sum or the suction head and the delivery head.
+ lid
111e pressure difrerence between the impeller eye and tip can cause a recirculation of a small volume of fluid q, th us reducing the now rate at the oUlletto Q. Thus (8.4)
3:!~
,.
TlIIHKI MACIIlNES
HYDRAULIC PUMPS ""
fJl is Ihe angle suhtelltled by (he blade at the inlet, measured from the tangent to the inlet rJuius, whih! fh IS the blade ungle measured from the tangent at the outlet
r].
and C.1'2 = C] cos
U, WX2
cr2
Thus
ex;!
(8.2)
a,
If,
329
Cr, But by using thl.! cosine rule.
C,
W,
P,
W?
+ Cr)/2 then UICI cos 0'1 = (Uland U.,C, coscr, = (U; - \V~ + C;)/2, Subs~il~ting lh~ ahov; rcJati~n in ~qualion (8.2l. gives E=
Figure 8.2
lIeloeity rriangle:J for a Cl'lIIn/llgal [ll/mp
The fluid I.!ntcrs the blade passage with an absolute velocity C 1 and at ,10 angle. crl
Ihl.! impeller inlet the tangential velocity vector VI = wrl, where w is the 'lI1gular wlocit), of (he impl.!lIcr. The resultant relative velocity of the flow into the blade passage is WI at ~m angle (and not fit due to the shock at the cntry of blade) langentto thc inlet. Similarly, at outlet the relative velocity vector is W2 at an angle land nOI fl2 due tel nuid slip atlhe exit) tangent to the blade. The absolute velocily vcctor C] is oblaineLi by subtrncting the impeller outletlangential velocity vector V'2 from IV] and C] is SCI at an angle Ct'] from the tangent to the blade. As n general case unless othcrwisl.! stated, it will be nssumed that the inlet and outlel blade :mgles arc equal to their corresponding relative flow angles. ( i.e).B~ = f31 (for No-shock cundition) and f3~ = fh (for No-slip at the exit). From Euler's pump equation the work donI.! per second on (he fluid per unil weight o f fluid flowing is 10
,8;,
f1;
IE = WIllig = (U,C,., -
U,C,·,J/g
I
(8.1 )
where, ex is the component of absolute velocity in [he tangenti,ll direction. ' E' is oflen referrcd to as the Euler head and represents the ideal (or) theoretical head developed by the impeller only. . E' ill terms of ansolute velocity is ohtained as follows: From tlie velocity triangles
[(Vi -
V1)
+ (Ci -
Cr)
+ (lVl- Wi)]
2g
(C .,' - C')I'g were, h 1 - represents the increase of kinetic energy of the fluid across the impdl~r, (V~ - V~)/2g represents the energy used in imparting circular motion about Ihe impcl1~r axis [0 the fluid and (Wf - Wl)/2g is the gain of stalic head due 10 reduction of the relative velocity within the impeller. . .. As lhe water enters tbe impeller radially, the absolute velocity of water al IOlct IS III the radial direction and hence al = 90° and C,I'! = 0 because C j = C'L So, equation (8 . 1) reduces to
TIle volume flow rnle of water is
wlH!rc C is 'the radi<.ll component of the absolute velocity and is perpendicular to tht: tangent al i'nlet and outlel while b is Ihe width of lht! blade measured in the:. dirt!ction. When considering lhe slip factor 01. lhe Euler head equalion becomes
Typically, slip factors lie in the region of 0.9, where
al =
Actual C~ Ideal ex
(Refer Fig. 8.3)
r!YDllAUUC FPMI'S "" 3'27 3:!6 y
Tt )1l1l0 MACIUNES
of now and thus increases the pressure of water. nle efticicncy of centrifugal pump having thi s casing is reduced due to the formalion of eddies. Delivery pipe
(b) Vortex casing (Fig 8.1 (b)) If a circular chamber is introduced between the casing and the impeller, then iliat casing is known as \'vrft'X cusing. This l:onsidcr.lbly reduces the loss of energy due to the foonation of eddies. Thus. the dliciem:y uf the pump is more thim lhe efficiency of volute casing ccnlrifugal pump.
(c) Casing with guide blades
Delivery wive
Impeller
(Fig. 8.I(e)) In this 'ypo of casing. the impolkr is shroudcd by a series of guide blades mountcd o n a ring which is known as 'diffuser' . The guide vanes arc designed in such a way that the wilter from the impeller enters the guide vanes without shock. The area of the guide vanes increases, thus. reducing the now velocity through guide vanes and consequently increasing the pressure of wilter. The water rrom the gui de vanes then passes throu gh the surrqunding ca.o;ing, which is in most cases concentric with the impe1\cr. The diffuser is optional and may not be present in a particular design depending upon the size and cost of the pump.
Vone,;
I
ch3mbcr
~casing
/-
~- "-
,/
/
"-
I
\
\
J
\
Suction pipe Sump
\
(
,
~
'
---/" Guide
Figure 8.l(b)
Foot valve nnd strainer
Figure 8./(a)
Centrifugal pump
2. Casing It Is an air-tight passage surrounding the impeller which converts the K.E of water lenvin~ the impeller into pressure energy before the water leaves the casing and enters the delivery pIpe. TIl[~ three commonly used casings are
(a) Volute casing Casing that surrounds the impeller. is of spiml type in which flow area Increpses gradually. TIle increase in area of flow, decreases the velocity
Vortt'x casing
Figurt! 8.l(e)
Y,InC5
ewing wilh gllide van/!.i
3. Suction Pipe and Delivery Pipe A pipe whose o ne end is con nected to ilie inlet of the pump and other end dipped intu the wa ter in a sump is known as the sliction pipt!. A foot valves, fitted atlhc lower ent..! of the suction pipe, opens o nly in thc upward direction. A strainer to remove t..Ichris from water. is
WORK DONE AND VELOCITY TRIANGLES Figure 8.2 shows lhe velocity triangles at the inlet ami outlet lips uf lhe vane fixed to an impeller. The blades arc curved between the inlet radius q and the outlet rnt..!ill s
.....
J::!4 :;..
TURB o M .KIIJNES
run al 1500 rrill. Dl.!terminl.! its power, discharge and head. Find and vl.!rify their SPl.!l'llic spceds. JAns. (a) 0.001 m.l/s (b) 2.:!S m (c) 0.0405 kW and (d) 25.8J 7 . 11. An axi[ll How pump with a rolOr diameter of 30 em hnndlc liquill waler
7.1-1. A model of a Francis turbine of 1/5th of the actual size was lested in a laboratory undl.!r a head of 1.8 m. It develops 3 kW at 360 rpm. Determine the speed' and pm. .·cr dcveloped under 6 m he41d. Also. find the specific speed. [Ans. (a) 131.5 rpm
(b) 456.9 kW and
(e) 299.31
7.15. A model of a Kaplan turbine, 1/12th of the actual turbine size is tested under a head of 3 m. The head for the proloturbine is 7.5 m. The prolOturbine. is designed to produce 6000 kW at a speed of 150 rpm at an efficiency of 83 %. Find spc~d. discharge. power nnd specific speed of the model. [,Ins. (a) 113BA2 rpm. (b)0.4.12m.1 / s. (e) 10.54kWand (d19361 7.16. Whal ,HC unil qU ':lIIti[ic.~ ? Define [he unil quantities for a turbine? Why arc they impnnulH? 7. 17 . Ohtain alll.:xprcssioll for unit spl.!cd, unit discharge and unit powcr for ilturbinc.
pelton lurbine produces [n.ooo kW nfpower while working under u head of 500 Ill . The spel.!d is 300 rpm . Assuming lhe efficiency of tile turbine 10 be 80%, lind the values nf unit quantities. If til(! head on the turbinc falls to 350 m, find the new di scharge , spe!.!d and power for the same efficiency. Verify the specific speed. IAns. (al 0. 114 (b) 13 .4 2 (e) 0.894 (d) 2.13 m'ls (e) 251 rpm . . (f) 5853.8 kW (g) 12.691 7. [9. A FranCIS turbllll! works undera head of 5 m and produces 701W; Ihe discharge through the turbine is 1.5 mJ js and the speed is 180 rpm. Find the unit quantities and new values of speed, discharge and power. when the head increases to 15 7 . 18. t\
m.
[,Ins. (a ) 0.671
(b) 80.5
(e ) 6.261
(d) 2.6 m.1 /s
(e) 311.8 rpm
(f) 363.7 kWJ
-
7.20. A pl.!lton wheel devel ops 5520 kW under a head of 240 m wilh an ovcrnIl !.!~fici c ncy 01'83 % when revol ving at USPl.!l.!U of200rpm. Find the unit quanti!i!.!5. II [he hl.!aJ 011 the same turbin e during off seuson falls to 150 m [hen,· find the di scharge. power and speed for Ihis head. (MU-Ocr. '97) JAns. (a) 13 rpm. 0. 1R2 m.l Is. 1.5 kW (b) 151U rpm, 2.23 m.l /~, 2.73 Il1WJ
8 HYDRAULIC PUMPS CENTRIFUGAL PUMPS
INTRODUCTION The hydraulic machines which conVCrithe ml!chanical energy into hydrllulic energy arc called asplllllps. The hydraulic energy is in the form of pressure energy. Two types of pumps commonly used are centrifugal and axial flow pumps. They arc so named because of the general nature of the fluid flow through the impeller. The combination of centrifugal and axial flow pumps is called 85 mixedjioll' pump wherein part of the liquid flow in the impeller is axial and part is radial. The hydraulic machines that convert mechanical energy into pressure energy, by means of centrifugal force acting on the fluid are called as centrifugal pumps. The centrifugal pump is similar in construction to the Francis lurbine. BUI the difference is tlwtthe fluid flow is in a direction opposite to that in the turbine.
MAIN PARTS OF A CENTRIFUGAL PUMP The parts of a centrifugal pump arc similar La those of a ccnlrifugai compressor. The three important pans of a centrifugal pump arc (1) Impeller (2) Casing and (J) Suction and Delivery pipes (Fig. 8.[0.).
1. Impeller The rotating pan of the centrifugal pump is called the 'Impeller'. The impeller is a rotating solid disc with curve.d blades standing out vertically from the face of the disc. The lips of Ihc blndcs in the impeller arc sometimes covered by another flaL disc to give shrouded blades, otherwise the blade lips are left open and Ihe casing of Ihe pump itself forms tile solid outer wall of the blade passages. The advantage of Ihl! shrouded blade is that flow is prevented from leaking across the blade tips from one passage to another. The impdlcr is mounted on 11 shaft connected to the shaft of an electric motor. As the impeller rotates. the fluid that is drawn into the blade passages at the impel[er inlet is accelerated as it is forced radially outwards. In this way, the Sialic pressure of fluid is raised.
DIMENSIONAL AND MODEL ANALYSIS ~ )2)
322
}>-
TURDO M ACIi I:-JES
7.16. The number or repeating variables are equal to the number or rundamental dimensions of the problem. (Tr.ue/False)
7.17. If a variable is dimensionless. it it"clf is a 1I'-term. (TrucIFalse) 7.18. The product of two 1I'-terms is dimensionless. (True/False) 7.19. What is model analysis? 7.20. \\'hat arc thl! advantages of model analysis? 7.21. Define: Similitude 7.22. Listlhe types elr similarities. 7.23. What is Geometric similarity? 7.24. TI1C rollowing points should be considered in geometric similarity. (a) Flow directions and all angles must be preserved. (b) Orientation should be preserved (e) both (0) and (b) 7.25. What is kinematic similarity? 7.26. What is dynamic similarity? 7.27. What are non-dimensional numbers? 7.28. Detine Reynold's number. 7.29. Mach numher is the square root of the (a) ratio of inertia force to the pressure force (b) ratio of inertia force to the gravity force (c) ratio of inertia force 10 the elastic force 7.30. rroude's number is the square root of the ratio of - - - - - - force to the - - --force. 7.31. Weber's number is the square rool of the ratio o f - - - - - - force to the ----force. 7.37. What arc undistorted models? 7.33. When is a model called a distorted model? 7.34. Define: Specific speed a pump and write the units of each quantity in the ex pressIOn. 7.35. Define: Turbine specific speed and specify the units ofeach variables appearing in the ex.pression. 1.36. The turbine specific speed is given by
7.38. WhH! is model testing ? 7.39. What ;Ire the conditions of camp Ie Ie sim ilarity between a model and it!> pm\Otype ? 7.40. Specific speed of a model is to the speci ric speed of its protoIYp..:· 7.41. What are unit quantities'! 7.42. TIle ullit speed is given by (a) (b) (e)
7.43. What is thL'- usc ofunilljuantities'!
EXERCISES 7.1. Discuss the advantages of dimensional analysis. 7.2. State Buckingham's 1f-thcorem. Describe this theorem for dimensional anal· YSls. 7.J. Show that the discharge of a centrifugal pump is given by
_ND' J [gHm _11_] N'D"ND'p
Q-
where N is the speed of the pump in rpm. D is the diameter of the impeller. g acceleration due to gravity, Hm manometer head . I). viseoc.:iIY of nuid and p the density of the nuid. 7.4. Show by tlimensional analysis. the Power P developed by a hydraulic turhine is g iven by
P= pN'D' J [D;~']
or
(a)
N,fH
7.5.
7.6. 7.7.
(b)
H5/·1
N.Jii Il J /' 7.37. The pump specific speed is given by (e)
(a)
JNQ
N/,fH ./NIH IN/l/
7.8. 7.9.
7.10. 7. 11.
(lim )5/,1
NJQ 7.12.
where p is the density of liquid. N is the speed in rpm, D is the diameter of runner, H is the head and g is the gravitational accelaration. What is model analysis? What arc its advantages? Explain the different types of hydraulic similarities that must ex.ist between a prototype and its model. What do you mt!an by dimensionless numbers. Define and explain four ntH\ dimensional numbers. Deline und derive an t!xprcssion for specific speed of a pump. Derive an expressio n for specific speed of a turbine. How is the model testing of a hydraulic turbomachine mack? A centrifugal pump was tested in a laboratory by a 1:8 model. It consumed 5 kW under a head of 5 m at 450 rpm. If the prototype is to work at 130 m head. determine its power, speed and discharge ratio . IAns. (a) 225 rpm (b) 20480 kW (c)256 1 A ccntrifugi.ll pump discharges water at the rate of 0.167 m" /s at 2000 rpm under a head of 100 m and consumes JOO kW power. A 1:5 scale mudel is to
):::!O ,.. TUiUJO MACIllNES
DIMENSIONAlANDMoDaANAlYSIS ..;,( 321
and
(f) Power
19B6.525(20)'I' (25 )'/ 2
PII(HZ)]/:!
1421.44 kW
1.265 x (160)'"
A Pt!lton turbine produces 5000 kW under a head. of 250.~ <'nd has sr~i.!d or11 n rpm. O\'erall cf(icicnc), or turbine is 85%. Find the unll quantlues: If the hl.!ad falls to 160 III what are th e new values of speed, discharge and power. Fmd also the speci fic speed.
Example 7.13
2560.2 kW
(g) Specific speed
PI = 5000 kW.
N"!..../P"!.
N,
Solution iii = 250 m.
NI = 210 rpm,
'10 = 0.85.
(Ii,)'/"
=
H} = IbO Ill .
I 67.9BJ256li:2 ( 160)'1'
14.94
(a) Unit speed NI
210
= = %J ,.,250
N" = -
=
SHORT QUESTIONS 13.2B
(b) Unit power p _ ~ _ 5000 ,,- H3j"1 - (2S0}-lJ. 2
= 1.265
I
(c) Unit discharge '10
= D.S5 x 10 3 x 9 .81 x 250
2.3 9Y
I );
~ = 2.399 =0.1 52
Q" when
m]
ffi l
fh
,/230
160m
(d) Discharge Q" x
.Jii). v'i6o
0.152 x
1.92 m J /s
(e) Speed
....
Nil
X
.Jii z
13.2B x
=
v'i6o
167.9Brpm
What is Dimensional analysis? What is a Dimension? What is Unit? What are fundamental dimensions and derived dimensions? Give examplC!s. The dimension of power is (a) M L 2T- 2 (b) M L'T- J (c ) M LJT-' 7.6. The dimension of work is - - - 7.7. There is no dimension for efficiency (Truc/FaJsc) 7.8. Match the following (a) Force ·MLT- 1 (b) Dynamic viscosity - none (c) Angle - ML -IT- I (d) Momentum - MLT- 2
7.1. 7 .2. 7.3. 7.4 . 7.5.
7.9. What arc the important advantages of dimensi onal analyis? 7.10. What is dimensional homogeneity? Give an example.
of (he system of units . 7 .11. Dimensionally homogeneous eq uations are 7.12. What is Buckingham IT-lheorem? 7.13. Number of dimensionless terms in Buckingham's Jr-theorem is equal 10 (a) The s um of (Olal no. of variables (n) and the no. of fundamental dimensions (m ) (b) The difference bel ween nand m (c) The product of nand m 7.14. Each IT-term contains (D) m-I variables (b) 1 + m variables (c) m varjub lcs where m is the number of fundamental dimensions. 7.15 . Enumerate, (he points governing lht! choice of repeating variables .
318
).>0
DIM ENSIONAL AN D M OIJEL ANAI ,YS1S
Tunoo MAOIINES
Hm = 5m
H J1 = 8.5 m
D" IO (a) Speed of the model
Pp= 8000kW
P,
P, HJ(!
N J1 =J20rpm
,
N ( H"') '/' (DI') DIU "
=
NI/I
H2"2 6600 x (181'0 (30)'/' 3067 kW
P,
flf! '
5 ) ' / ' (10)120 ( -8.5
= =
Example 7.12 A turbine is to operate under a hCild of 25 m al 200 rpm , Thl,.· discharge is 9 m .1js, If the efficiency is 90% determine the rerfonnan cl! o f the lurhim' under a head of 20 meters. (/) IJU-NfJl'. '(7)
920.4 rpm
(MU-Oct.
(b) Discharge from the model
(~:)' (~:) Qp
Q..
8000 x JO' pgryoHp = JO' x 9.81 x 0.85 x 8.5
=
112.87 m' /s
=
N,../H, ../H, 200 x J20 v'25
N,
0.866 m' /s
(e) Power of model
=
P" ( Dm)' (N.,), Np (920.4)' 8000 ( ~)5 JO 120
=
36.1 kW
=
178.88 rpm
0JI
Using the unit discharge equation
Q,
Example 7.11 A ,urbine devel ops 6600 kW. when running at 100 rpm. The head on the IUrbine is 30 m. If the head on Jhe turbine is reduced to 18 m. determine the speed and power developed by the turbine. Solution Power developed P,
= 6600 kW
Speed N,
H2 = 18 m
= JOO rpm
Speed = N2
Usin g the unit speed equation
= N, N,
= =
~ 77 .45 rpm
8.05
m' /s
Using the unit powcr equation
1}0
./IS
Q,.rfi, ../HI 9 x J20 = v'25
=
Q,
Now,
N, ../H, 100 x
Head If, = 30 m and Power = P2
·~~I
Solution Head on turbine HI = 25 m, Speed N ] = 200 rpm , Discharge QI = 9 lJl ·l/~ Overall efficiency '10 = 90%. Performance of the turbine under the head Hl = 20 m, mcun s 10 find Ihe speed. dischilrge and powt!r developed by lhe lurbinc when working under the head of10 rn . For lhe Head H'!. = 20 m, Speed::: N2 . Discharge:; Q2 and Power::: P2 Using lhe unit speed equation
(920.4) x 112.87 ( ~)' JO 120
For the head
31 '1
Using tht: unit power equation
Solution
Dm=~
'"
:. P,
=
P,
=
0.9 x lOOO x 9.8I x 9 x 25
=
1986525 W
=
1986.525 kW
..,
DIMENSIONAL .... ND MODEl. AN .... LVSIS ~ 317
1[6 :,.. TUJu] o M ,\ C' J-IINES
Exar.lple 7.8 Two geometrica lly simi lar pumps are run ning at the same speed of [000 rpm. One pump has an impeller diam eter of 300 mm and lifls water at Ihl! ratc 3 of 0.02 01 Is against a head of 15 m. Determine the head and impeller diameter of the Olhcr pump 10 deliver a discharge of 0.0 J m1/ 'j.
Solution For pump-l: N, ; IOOOrpm , For pump-2: N'l = 1000 rpm ,
D,;0.3m, Q2
= 0.01
Q,;0.02m.1/"
Solution DIU DJI
;
1
JO
Pm
= 1.84 kW
=5 m
Hili
= 480 rpm
Nrn
H"
= 40 m
(a) Power developed by the prototype Pp
H,;15m
= (~:.)'
(zJ
Pm
(b) Speed of the prototype
mJ /s
(a) Impeller diameter of pump-2
(D~N ),
,
(D~N ),
(~:)'
(~0),C~)480
=
(~:)(~;)
135.76 rpm
(1000) ( ~) 0.02 1000
(10)5 (135.76)' 1.84 480
=
0.5
4163 kW
(c) Specific speed of the prototype
,
;
(0.5)l
x
0.3
(N,)p
=
NpjP"
~ p
0.238 m
( ~), (H') j H,
=
(~), (~:) (~:)
;
em)'
;
-
-
0.3
(0238)' -x 15 9.44 m
A lurbine model of I : 10 develop' 1.84 kW under a head of 5 m of
water at 480 rpm . Find the power deve loped by the prolOlypc under il head of 40 m. Also find the spet!d of the proJOIype. Assume efficiency of both the turbines to be
samc. rind and verify the specific speeds.
(NJ )III
(4163)'/2
87.1
=
(N,)p
Nmffm H;/4
=
480..fi]4 (5)5/' 87.1
=
(NJ ),,,
=
x H,
0.3
Example 7.9
X
(40)5/ 4
Specific spel!d of the model
;
H,
135.76
=
(b) Head developed by the Pump-2
= 81.1
The specific speed of the model is equal to the prototype c!nd thus it is verified.
Example 7.10 A model of a Kaplan turbine, one tenth of the actual size is Ic!sted under a head of 5 m when actual head for proto turbine is 8.5 m. The powl!r to be developed by prototype is 8000 kW. When running at 120 rpm at an overall efficiency of 85%, de,ermine (a) speed (b) di scharge and (0) power of model.
DIMENSIONAl. AND M ODEl. A NALY SIS "' ~
14
};.o
l iS
TURBO M ACIUNf:.s
Head =85 In Flow nue = 20. 000 mJ III 5 . 4 peed- 1490rpm Dlameter= 1200mm Wi.ltcrdensity = 714 kg/ mJ P TI ower:::: rnW (electrical) Ie manufacturer plans to build a model Tes .. . . power to 500 kW and flow to 05 JI . . f t condJllons hm lt th e uvailablc elec tric dficiencies arc
Exam ple 7.7 A geomeuically simi lar model to scale I : 6 of i.I large cen tn fugul pump is tested . Thc prototype paramctcrs nre speed:::: 400 rpm . discharge = 1.7 m) Is. head developed:::: 36.5 m and the power input 720 kW. If the model is tested under a head of 9 m, dctcrmi ne the speed. discharge at which it should be run and the power
Soluti on
Solution
.
Equaling the head. power and flow coefficients for the model and protDlYPI!
Q"
Qm =
PI'
m
=
1111
=
6' 720 kW .
f1m
(Dm)' DI'
(PI')
( 1111)'
(DD". ,)' (DI')' (~) D",
prototype
Q..
Dill -D =03 .
= =
I'
N"INm= 11.1 1 x (0.3)'
Speed ratio
N," x -N ( Dm)' V" p
(6I)' _
x Qr
,, _ _ _ x 1.7 1191.75
400
0.0235 m'/s
(c) Power required by the model Equating the power coeffi cient for the model and prototype
Hm HI'
= =
Head ratio
9 6 x 6 x 4oo 3 .5
1000
Scale ralio
A lso
x Np
(b) Discharge of th e model Equating the nuw coefficients o f the mode l and
( 8 ) ( Dm)4 DI' = (11.I1)'x O.714
Then
D",
1191.75 rpm
Substitute for NI' I N"" thcn
0.5
D I'
"Hp
j
=
p,,!
.,fH.;,
ITT x -
=
Al so
4
Hp;: 36.5 m
prototype
Nil!
;"m = (NI')' (DDm ) ' Nm
= 1.7 mJ Is.
= 9m
(~:)' p
Q'I
(aJ Speed o f th e model Equating thl! head coefficient for the model and
or 20. !lOO ) ( 0.5 x 3600
Nil = 400 rpm .
_Dm = _I
D"
(NI') N (DD,: )'
=
required to dri ve the pump model.
(N
I ')' ( Nm
D )' Dm"
crc)' 3.3
0.3
(D;N')..
=
(D;N')p
Pm
=
(~:)' G:)' " PI'
=
C)' C _
I9175 ___
6
400
2.45 kW
,,720 )'
1
;\ 12 :,.. Twl.OO M ACIlI NES
DIMENSIONAL AND MODEl ANALYSIS ..;, :'\ 1;\
Power of T. 0
- bl - 2
and
DI=2 -
Subs titutin g th e v31uL'S of ill.
bl.
and
CI
=- 2
:. bl
S pecd of model Nm = 1000, Head of modd H", = 8 m. Power of mood P", = 30 kW, Hcnd of prototype Hp = 25 m
4= - 2
(i) Speed of prototype (Np)
in 1C I. we obtai n
Equating th c head coeffic ie nt for the model and prototype (equation 7.20)
][1 = L - 2V - 2p-I.FD
(or)
(~t =(~)/'
Fo
= pL2 V2
][1
,fHp x -Dm
For the second rr -term
- -
.../Hm
fl: 5
M " L oJ T O = LU2(LT-I)I1~("'L -] ) nA'fL -11' - 1 Equaling tht.! powt.!rs of M,
Powe r of M. 0 Power of L, 0 Power of T,O
C2 Q2
+I + b2 -
[;;= ~J
:. c:!.=-l 3C2 -
-b 2 - 1
1
: . b2
=-
.'. a2
= -b2 -
2
=
(ii) Power developed by the prototype Equnting the power coeffic ien t for the model and the prototype (equation 7.19)
1f2=L - 1V - 1p-lp.
(D:N3
rr, = - Ii -
353.5 rpm
I nnda2 =-J
Suhs(ilUling the values of QJ, b2 and c:! in 1C2, we obtain
(or)
t
pL I'
P"
Step-6
=
FD /") _ 0 ( pL'V" pLY -
(or)
C:N3 ),,
(~::)' x
=
p., x
=
30x5' x
=
4143 kW
Subst itutin g thc va lues of Jrl, and 1T2 in .equation (7.34), we get
II
I
- x - xlOOO B 5
L. T on both sides
= = =
xNm
D 1,
(zJ
c~r
(iii) Ratio of the flow rates of two pumps (I.e. model and prototype) Equating the flow coefficie nt orthe model and the prototype (equation 7. 18)
(or) FD = p L'I"g
(~) pLY
Example 7.5 One-fifth scale moue l of a pump w.as tested in a laboratory at 1000
'ne
rpm . hend developed and the powcr input nt thl:! best efficiency point wl:! re found 10 be Rm and 30 kW res pcL·tivclY. If thc prototype pump has to work against a bend of 25 m, dctcnninc it s working speed, the power required to drive iL and the raLio of the now rates handled by the two pumps.
Solution O ne-fifth sca le modc l means that the ratio of Jinear dimensions of a model and its prototype is equal to 1/5.
or
Qp Qm
(Dp)'(NI') D", Nf/I
, C .5) 53 --
=
5 x
=
44. 1875
1000
Example 7.6 Specifications for an ax.ial flow cooJanl pump for one loop or u pressurised waler nuclear rl:!actor arc
DIM!;NSIONAI.,\NnMoOELANAI.YSIS 310
..: 311
j> TURBO MACHINES
Note that.
Equaling Ihe powers of M. L. T o n both sides C::!+I
Powcr of M. 0 Power of L. 0
=
Power of T. 0
=
and Substituting the valucs of az.
bz and Cz
.·. C1=-1
Q2 + b2 - 3cz - I Uz +bz =-2 -b2 -1 : .bz=-1 Q2 =-2-bz =-1
In
7rz. we get
If::! = D- I V-I p-IJL
(or)
~
=
~
and::" is callct.l mllghlle.r.l'!aclor. 'men. Fincliun codlicien[ DVp Re D of a pipe is a function of Reynold's number and roughness faclor.
Example 7.4 The drag force exerted by a flowing nuid on u solid body depenJs upon the length of Ihe body. L. Velocity of now. V. den si ty of nuid p uod viSCOSit y 11. Find an expression for drag force using Buckingham's theorem.
Solution Step-1 The drag force FD depends upon . L. V. P and JL . Therefore. D is a runction of L. V. f1 and IL . Mathematically.
,.
IFo = f(L. V. p. ILl I
JT2::;: - -
DVp
(or)
Third JT-tcrm
f,(Fo.L.II.p.,,)=O
(7 . .13 )
Total no. of variabl es. n = 5.
Step-2
(or)
M O LO T O = L"·l.(LTy)b,(ML -))"L
Equating the powers.
For M. 0 For L. 0 For T. 0 and
=
:. CJ = a + b) - 3c) + l aJ + bJ =-1 c)
Step-3
=
a)
Equation (7.33) can be written as
=
-bJ UJ
Substituting lhe values of UJ. b) and cJ in IfJ
Dimensions of eac h vari able arc Fo=MLT-'. L=M. V=LT-'. p =ML - )andl'=~IL - 'T - ' Hence. number of rundumental dimensions. m = 3. Number of IT-Ierms = n - m:: 5 - 3 = 2.
(7 .34 )
: . b) = 0
=-1
JTJ
Step-4 Each rr -term contains m + I variables, where m is number of repeating variubles and is equal to 3. Choosi ng L. Vand p as repeating variubles. we have
= D-'.\/D .pD E
(or) E
]f)
=
D
1T1
=
L U' Vb, pC, FD
rr2
::;:
L(ll V"l pel J-L
Step-5 Each rr -t~rm is solved by the principle of dimensional humogem:ity. For Ule first If -term
I
Step-5
If)
Substituting the values of 1f!. ]fl, and rr) in equation (7.32). we gt!t
F,(CF':Vp'~)=O
= V" Vb, pC! FD I
substituting dimensions on both sides of If. MOLDTD U".(LT-1)bl(ML - J)CI.M LT- 2 Equating the powers of M. L, T on both sides,
=
(or) Power or M,O
=
Power of L. 0
=
CI
+I
:.
CI =-1
a, + b, - 3c, + I a, = -b, - 4
l
DIMENSLON~1. AND M oncL, Ar.;AI,YSIS ""
30X ;,;.. TULtUO /Vl.o\CIIINF:S
Equating the powl.!rs of M, L. T on both sides, PllWl.!r ('If M, 0 = c:! + I ,', c':! = - I Pl,lWcr of L, 0 ;:;: ( I ::. - Jc:! - 1 ,', Q2 = 3c:! + 1
(12
= -3
+ 1 ::
-2
Ptlw~ror
T,O= - b2-i ,', b).=-i S uhsliluling thl.! val ul.!S of 0:1, b2 :lnd C2 in Jf2,
Solution Step-1 Friction coeffic ient CF is a function of Y, D, J.L. P ilnd :. C,.
or
D - 2 ,w- I ,p-l , jJ-
IT:!.
JOt)
E,
= I(V. D. fJ.. p. E)
II(Cr. V. D. fJ.. P. E) = 0
(7.31 )
Hence, total number of variables n = 6
I'
D2,w,p
Step-2 The dimensions of each variable
Third IT-term
= dimensionless fJ. = M L -I r- I • p =
CF
Y = LT- 1 D = L M L -3. and E= L.
" Number of fundamental dimensions
Substituting the dime nsions on both sides,
m= 3
Equating Ihl.! powcrs on /'1'1 , L afld Ton bOlh sides, Power of M . 0 = CJ :, CJ = 0 Powl.!r of L, 0 = OJ - c) + 3 ,', OJ = 3c) - 3, (J) = - 3
Powcr or T,O
= - b) -
I
,', b) = -1
Suhs tituting the va lues of (/). b) and CJ in]f) D-J,w-I,pO,Q
Then. number of 7f -(enns = n - m = 6 - 3 = 3. Now. equation (7,31) can be writlen as (7.32)
Step-3 Each iT-term contains m + I or 3+1 = 4 variables. where m is equal to the number of rcpeating variable, Choosing D , Y. p as the repeating variables. the threl.! TT ·Icrms are
Q
TTl
DJ ,w
IT:!, ]f)
Step-6 Substituting the vt:llues ofn[, rr::! and
II (
IT)
I'
1],
in equation (7,30)
Q) = 0
= =
D(lIVb1p' ICF Vb:! pC] J.L DQ} Vb) pC] E
D"l
Step-4 Each rr-lenn is solved by thl! principle of dimt!nsional homogeneity, First rr ·tcrm
D 2wp' DJ w
or
Since [he variable C F is dimensionless, it itself is a rr·tcnn, 111l:rcfore, rrl = C,.Scco,?d rr-tccm
Example 7.3 Friction coeffic ieLll (pressure drop per unit length) of a pipe depends on Average speed (V), pipe diameter (D), ViscosilY (Jl), density (p) and inside rouglincs,~ (E), Using Budingham'_ 'i IT -lhcorem, express the friction coefficient of pipe as a function of dimensionless quantities, (MKU-Nol', '98)
Substituting dimensions on either side
)O(i
3.
DIMENSIONAL AND MODEL Af"A.I.VSIS ~
,. TUlmo M ACIIINES
Step-4
Momellwm = Ma~s x Veloc ity
Each rr-Icnn co nt ains til + J vil riahl es. whercm is equal to three and is also a repealing !vi x Length Tim e MLT-'
= = -I. Power
=
Workdone ~:::.:== see
= = =
variab le. Ou t nfs ix variubles 1']. p. JJ. w. D and Q.thrce variables are 10 hcselecled a<; repealing vari ables .. '1' is a dependent variable and should not be selec ted as a repe aling variilble. OUI of the five remai ning variab les. one variabl e should have gcomclrit.: property, the seco nd variable should have fl ow property and Ih e third onc' have nuiJ properly. 1l1ese requirements arc fulfilled by selecting D. w. and p a.<; repea ling variablcs. The rcpeating variables the mselves should not form;1 dimensionlcss lcnn anJ sh ould hnvc themselves fund amental dimensions cquallO m (i.e . .3 here ). Dimensions of D. wand p arc L. M L -J and hcnce the three fundamenlal dimensions exist in D. w, and p and they themselves do not form a dimensionlcs~
Force x Distance Time ( MLT-') x L
r-'.
T ML 2T- J
group.
Example 7.2 The erfic iency of a lurbom achine depends on de nsi ty 'p', dy namic vi~cosity 'tt'
of the nuid , angular velocity 'w', diameter 'D' of th e rotor and the discharge Q. Express' rj' in lerms of the dimensionless parameters. (MU- Apr il '98)
Step-5 Each JT- ICrm is wrillen as rr , rr, rr)
Step-1 The efficiency of a turbomachine (1]) depends on i) p ii) J.L iii) w iv) 0 and v) Q. Hence ',7' is [) function of P, J.L. w. D, Q. Mathematically, ~
.107
= /(p,!",w, D, Q)
. prl . 'I . It Dill . w'IJ . pr~ . Q Dn l
' W
hl
O il! . Wll~ . pr~
=
E
(7.28)
First rr -term
(7.29)
Substituting dimensions on both sides of 1ft
or it can be written as
M O L ° TO ;:: L fI ,
Hence. the 100ai num ber of vari ables (including dependent variable) n=6.
Step-2
.
The value of '1/1' i.e. number of fundamental dimension s for th e problem is obtained by writing dimensions of each vari able. Dimensions of each variable arc 'I P
= =
D
=
Dimensionless ML-),
/'
Q
Land
=
W
=
•
(T - ' )"1 . (M L -3)," . MOL QT U
Equa ling the powers of M.L. r o n both sidl!5 . Power of M. Powers of L. Powers of T.
0 =
c, +0
C! =
a
at = 0 0 = ", + 0 0= -b, +0 : . b, = 0
Subs titu tin g the values of al. b, and Ct in " t . we get
'Hence, number of fund amenlal dimensions. 111
= 3
If a variable is dimensi onless then. it itse lf is a JT-Ierm . Here the variabk 'J]' is a dime nsionl ess and hence 'J]' is a IT-Ierm. As it ex iSIs in first rr-term, '1f! = 'I' . lhere is no need of equ ating the powers. {he value can be ob tained directly.
Step-3 The number of dime nsio nl ess 1i'-tenns is given by
Second rr-term
1=,1-111=6-3=3 Thu s three "-term s ~;ay "I, 1r2 and as
IT)
are formed. Hence equation (7.29) is writte n
/ , (rr" rr" rr))
Substituling the dimensions o n both sides.
=0
(7.30)
MOLor o = Lfl2.(r-I)1I2 (M L -3r 2.ML - I r - I
--.
DIMENSION "L"ND MoUl.:.LAN"LnJ S ...;: ;\05
SlIhSliluling the va lu !.! n r ;..: in t.!qu3Iian 7.22 ~ll1d rearranging the equation. we nhlain
IN,,= ~ I " II
or
= I m, P = Pu PII = K Substituling the va lu e or K in equation 7.29. we gel
When H
2. Unit Discharge Tht.! dis c h ~lrgt: or now th rnugh a turhomac hinc working und..:r :l unit hend 11.": till) is is dcnoh::d by the s)' lllhol (Q" ,. Th..: I.:Xpn.:!>Slllll ror UllIt dI sc harge is give n 'IS: Let, H- I'kmi llf w:tlcr (In th l: lurholnachine Q-Disch:!r);..: passing through th e machill": when ht:ad in II is 'II' II - area of IhH... o r w;u..:r ThL' dist: htlr!;c p:lssi ng thro ugh a !!ivL'JI turhnmil(:hint! unde r 'l h":i.ld . fI' is !!ivc n tH', Q= Area of now,: \'dm;ity • .
P
57f
iJ
giVL'n turhomnch inc ;m:a or !l ow is constant and vl..'lncil)' is rrnrnrlional Iii
Q Cl' vclod ty
crJH
(m)
USE OF UNIT QUANTITIES The behavio ur of a hydraulic turbomnchine working under different heads can be easily known from the va lues of lhe unit quantities. Let HI, Hz be the heads on the turbomachine:. N I. N!. are the corresponding speeds, Q I. Q2 are the discharge. and PI , P,! ure the power developed by the lurbomachinc Using th e defining equations of unit quant ities
Nt
=
QJ Q, - - = - - and .jH, ..,fH]
Q"
Q
Jij
(7.251
PI
P"
3. Unit Power
fll / 2 J
~f11c powt.!.r developed by a IUrbol11achinL' working under 11 unit hC;1J is ca1led Ih e unil po\\'cr. It IS dl!nott.!d hy P/j .
Th e expression for· PII IS oh wined as foll ows, Lcl. l/- Hc;!u of water on the turhomachint! P-Puwcr dc\'clopeJ ny th e tu rhl)/ll:H::hinc under Ihe head ' H' Q- Dlschargc through turhomat:hlnc und er 1Ill! Iwa d . fI ' Till! ove rull c fli cil.!rIl:y is given as
P
P2
= flJ /J 2
Hence, if the speed. discl1W'ge and power developed by a turbine under a panicular head arc known, then by usinp the above three relations. the speed, discharge , and power deve loped by the same turbomachine under a given head can he obtained.
SOLVED PROBLEMS Example 7.1
Determine the dimensions of the following quantitic.:s in M·L-T
Power devdoped Po\V~ r input p
system.
pQII
Solution J. Force -= Mass x Acceleration
'/o(I'QfI)
Nz
.jH,=..,fH]
When 1-/ J 111, Q=Q,., thl.: cons t:lIl1 (K) vuJu e is K = Q ' . .S u hstilUIlllg Iht: v:.dllt: of j( i.lIld reulTilllging t.!qualion 7.27II , WI.: gt.!1
Q"
(7 .27)
= HJ/ Z
P"
lamed ;IS Iilli.1 jloll' VI" I~I/ ;t .diJcll(lI :~e. It
I. Force
2. Torque
3. Momentum
4. Power
Length L-' ., =M T =M x - - -
(or)
Timc-
Pex(Q x II) Since Q ex Jij P a /{V!.
2. Torqlle -= Force x Distance
= =
MLT - ' x L ML 1 T-'!
302
an
d
>
OIMl:NS10NAL AND MODEl. ANALYSIS ""
TtJlwO MAC"lIlNES
Q
N DJ
. known as the flow cocrticiem and denoted as ¢
;2%2 i!' called the Head coefficient or specific head and represented by
IS
= III
(N ~,) .
17 .20 ) (7.18)
fl
(or)
3. Specific power of the model and prototype are same Power of the hydraulic turbomachine is given by
(7 .211
lor)
UNIT QUANTITIES Ipfl'pQHI
Thl! quantities or a hydraulic turbomachinc working under a unit head afe l:alh;J lht.: IIlIil qual/tirie:;. TIlt:! following are the three important unit quantities which mu."1 he
where H is the head of the machine. We know
studied under unit head.
Qa D3 N and .Jii fl' DN
1. Unit Speed
I:. U fl' C, fl'.Jii1
The speed of" turbomachine working under a unit head is termt!d as unit spt!cd. It I~
Pfl'pD'N (D"N') PcrpD 5 N J
denoted by (Nil)· The cxpression for unit speed (N u ) is ontainl!u as follows. Let. N -Specd of a turbomachine under a head H /i - Hea.d under which the turhomachine is working . U-Tangl!ntial vt!locily The tangential velocity. absolute velocity of water and head on -the turbomachim!
or
P pN]D5 = constam
P pN] D5 is known as the power coefficient or specific power and denoted by
P for
model testing
arc related as U a C where C a The tangential velocity (U) is given by
JH
I
U = rrDN
(7. 19)
where Pm
for nHlt.h:1
tl!sling
Applymg tht! nnw coefficient for the model and the prototype
(N~,)
V'
30~
60
I
For a given turhomachinc. the diameter (D) is constant
= P"
:. U cr N
Of
Na U
4. Specific head of the model and prototype are same (or)
Tangential velocity (U) is given by
rrDN
U = - - also
60
Ucr../H (g is dropped since it is a constant) .Jiifl'DN or
H D2 N2
= constant
N=K.Jii
(7.22)
where K is a constant of proportionality. Ifhead on the turbomachine becomes unity. the speed becomes unit speed .i.e. when H=I. N = Nu Substituting these values in cquation 7.22, we get Nu = K
JOO ;... T UlUIo M ,\C! UNES
D IMENS IO NAL AND MODEL ANALYSI S ....;
(or)
MODEL TESTING OF HYDRAULIC TURBO MACHINES
(or)
H5 j'!
= (ConSlanl )-,_ N-
P
(7. 14)
Conslanl is ca lled Ihe Co nSlii nl of proportionality.
Before manufacturing the largl,! sizt!d actual machines, their moods which are in com plele s imilarity with the actua l machines (also called protot ypes) are made. Tests will he con du cted on the mod els and the performance of the prototypes will he pn:dicted. TIll! co mplet!! si milarity between the actual machine and the model will CX,ISI If Ih.: ro llow ing conditions arc sOlli s l1cd .
1. Specific speed of the Model
According to IhL' dcfinitinn o f specific speed. tV
30 I
= N"
whell
jl
= I kW and Ii
I
(N,)",
= I III the ahove cquuliol1 7. 14 reduces tn
constant =
:. P
= N.?
= Specific speed of the prototype
,
= (N,),.
I
or
N,\~
(:(3) = (~J~)
HV!
III
N2
17 . 161 I'
2. Flow coefficient is the same for the model and the prototype
or
The discharge Q for a hydrauli c machine is given by the rdatiun
Q = Area
TIle turbine specific speed N,r is N _ NM j HS/4
(7 . 15 )
N is in rpm . P is in kW and H is in mClres. Table 7.2 S. No. I.
2.
Ranges of specific spet.·ds of fttrbo machilles Turbo machine
Pelton Wheel -Single jCI -Twin jct - Four jet Francis turbine -Rndinl (s low speed)
6.
-Mixed l10w (medium - ex prcss) Kaplan turbine Propeller turbin e (axial) Centrifuga l pumps (slow-high speed) Mixed fl ow pump
7.
Axial fl ow pump
3. 4. 5.
8. 9. 10.
Radial flow co mpresso rs Axial flow s tea m and Gas turbines Axial compressors, blowers
Area x Velocity of now
=
rr DB
where D - diameter of the turbo machine impeller B - width of the impeller and BaD. he nce Thererore .
Q = rr D"1C, Dimensioll/ess specific ...peed 0.02 0 .02 0. 1 0. 14
- 0 ..19 - 0 . 19 - 0.3 - 0.39
0.39 - 2.3 0 .39 0 .65 2.7 1.6 0 .24 1.8
-
0 .65.
3.2
- 5.7
We know that Ihe tangent ial velocity is given by
/T DN
u=-60 or
lIf'DN The tangential veloci ty (U) and now velocity arc reluted tu lhe diameter and speeJ:J!'>
2.3 5.4 .1.6 1.8 4.0
0.4 - 1.4 0.35 - 1.9 1.4 - 20
0 . 17/
UaC,aDN Substituting for C r in equation 7. 17 yit!lds
Q
= /T D'(DN)
(o r)
Q
J5fN
= constant
We O1lso know [hat
UaDN From the ~lhove
1\\'0
~IJ\J
DIM Et-:SION,\L ,\NIl Moo".f A"'''ISSIS ....
298 ... Tunno MACIIINE.<;;
(7.81
The whh.: 7 .2 shows Ihal each Iype o r lUrbomachinc works In a nam:w,; rang!.! (If spct.:ific spc.c.ds. 'nH! spedfic speed is lhe parameter exprc.ssing lbe varintion of all the variahks N. Q and H ur N. P and H which cause similar Oows in turho mac:hint.:$thilt art.: gl!o Oll!lrit.: ally .. imi lar.
equations 'nd 7 • 8 we have < • 7 _7 ..
Derivation of the turbine specific speed
Da JHm
Th e power Jc\'clopcd hy an y tu rbine in ICrrn$ or overall d liciency i:- given hy
N
TIle discharge equation then becomes
Q
Hilt 1.; a - -, 'IvHm) N
or
or
Pa(
Q ::;
H 3/ 2
(constant)~
Nl
Q 1-I)((lS flo. P and g arc conslants)
The ahsolute veloci ty. tangential velocity and head on the turbine arc related as (7 .9)
Ca
From the definition of specific speed N = '!~ ~vhen Hili = 1m and Q = I m'J/ s . Suhstltutlng these values in the abov .
tionality is determined.
JH and
UaC or
(7. 111
UaJH
e equiltlon 7.9. the value of constant or proporBut
Ihl'
tangr.:ntml velocity ' U'
15
given by
rrDN
U = - -
or
60
TIle discharge Q is
17 .121
UaDN From equat ions (7 , 11 ) and (7 12), we have
JH
or
17. iJ)
Da - N
The discharge through the lurhine is given by
Q = =
The pump speciJie speed is
N.,fQ
Ns = - -
11m' 1'
17. 10)
But Area 0' D x 0
Area x Velocity A x C
where B = width si nce Ba D :, ACt D2
2. Specific Speed of Turbine The specific speed or a turbine is defined as the 5 . -unit power when working under'unit I.e d' I ' peed at which the turbine develops a _ t IS expressed as
or rrom equation 7.16 If A a --, and N-
CaJH
N _ N./P
, - (ii)57J
where. N is the speed in rpm. H is the head in m . .. TIle dimensionless spec ific speed (or) the p () a~fid P ]s til: p~wer In kIlowatts. ower specl Ie speed IS gIven by N pill
N,p = P 1/1(8 H)'/4
H ·l / 2
Qa N' Substituting for Q in the power equation, we get Pa
Hl/') (- N 2
.
H
DIMENS LONA L AN Il M ODl:L A NALYS IS ...:
wlll..'r~ K = E las ti c stress a nd A = L Fe
=
C
BUI
p A Lg
The I
4. Weber's Number (We) II is ddincd as Ih!.' squore rOO I orlhl.! rali o of the ine rt ia force of a fl owing lluic.J to the surfm.:e h.:n sion force. II is named after Moritz Weber o f the Poly technic In stitute of Ill-rlin . II is g ivc n ;15
SPECIFIC SPEED Tht! spec ifi c speed is the purameter which docs not explicitly contain the diamch:r of the runner or impeller. It is denoted by the symbol N~. The specific speed is used in c om paring the different lypes of turbo mach ines as every type of lurbomllchinc has different spec ific speed .
1. Pump Specific Speed
Inertia Force (Fj)
The s pecirlc s peed of a centrifugal pump is de ll ned as lhe speed at which the pump delivers one cubi c metre of liquid pcr second against a heaiJ of one metre . It IS
Su rfncc te ns io n r o rcc ( Fs)
j PaALC' wlll::r!.! (J = surfn cc (c nsio n per unit leng th
J K / p = a t vt!locity of sound in the nuid). /vi = -a'
The Mac h number is na med afler Em sl Mach (1838 - 1916).lln Austrian physir.:lsl. When the ro rces duc 10 clastic compression predominatcs in addition (0 the inertial forc e. the dynamic simil:lrilY betwe en the modd a nd prototype isobtamed by equutlng Ihl! lviach nUlIlberofmodcl and the prototype. i.e. (M)mmld = (M)/ru/t>f\'/Jr
~I
=
1
I!lenin Forl:c ( F, J Gravity Force.: (F~) pAC2
w,_
::!t)7
ex pn:sscd ns
illld
A = L:2 whe re N is th e speed in rpm. Q is the discharge in (tn" / sec) and Hm is the manomclnc head in
Wt= __ C_ .fii/ p L H'c'b,:r's
melres. The dimensionless specific speed is givt!n by NQ' /2 (gfi., )'"
l1I ol~e llfll "
is tilt.! law in whic h mode ls are based o n Weber's numbe r. Accordmg to thi S law there is dynamic similurity betwee n the model a nd its prototype .
Expression for pump specific speed
H C IH~ c.
The disciHlrge for a cemrifugal pump is given by the rdation .
IQ = 5. Mach's Number (M) M ~c h
number is de linl"d 'IS the squan: root o f th e ralio o f the incnia fo rce of a fl owin ll nUlL! to the clastic fmce. MlIthc mali cally, jt is define d as C
M
Ine ni:l Force ( Fj)
Elastic Force ( F, )
jPAC' KA
IT
DRC,
I
wht!rc D-Diamcte r o f Ihe pump impeller B-Width o f the impe ller We know Ihal D ct B
Q 01 D'C, The now ve loc ity, tangential velocity and manometric head (Hm) arc rcJah:d us U ct Cra ~
(7 .7 )
DlM ENS IONA I. AN Il M OIJI:L AN:\i.'·SIS ....:
corn:spomJing points iln! I.!qual. Alsu (ile dir!!Clions of the co rrr.:sponding forc!!s at the corrcsponding points s hou ld he the sa Hle (refer Fig . 7.2). Lct (F, )/ = inenia fnrcl.! a t a p(lint in the prototype. (/'~+:) I' = Gravity force at the poinL in the prototype. and (F;)",. (FJl)//I arc corn.:sponding fon:l.!s allhc correspondi ng points in the model.
c
~z
--",
.../
Cr
NON-DIMENSIONAL NUMBERS Non-d imensiona l numbers arc those numbers wh ich arc ohlained by dividin g Ihe inertia force by Ihe viscous force or tilt.: gravity forcc or the prt.:ssurt.: run.:e ur surl';u.:c tension force or tht! elasti c rorce. As this is a ratio of om: for~1.! to another f\ll'\:t· . it is a dimensionless number. These nOll-dimen sional numhers arc also calkd a:cJilllellJiolJfess 1If1l1Jhers .. The foll owing nrt! the important non-dimensional numht.:r:-. whic h arc used as a criteria for the dynamic similarity between a model and ih prolOlypc .
C In2=C r
..- ---.~~
I
1. Reynold's Number (Re) It is defined as thl.! ralio orint.:rtia force of a flowing nuid and the vise(Jus fort:e uf th\.' nuid. It is named after Osborne Reynolds. a British engineer ..... ho tirs{ proposeJ II 111 J 883_ It is e xpressed as
Re Figure 7,1
~y ~
= =
Thl't!(' similari/ies- pmfOf),pe a"d model of a llise
Then , for dynamic s imilarity, we have (Fdfl ;: (Ft;)p ;: F, (Fi)m (F"),,,
where Fr is thc fiHce ratio. M.Hhcmatically, Ncwton's law for any nu id particle requires lhe surn of pressure force (F,,) , gravity force (F~) , and friction force (Ff) eq ual to Ih e uccclerution term . or the inertia force (F;) i.e.
=
IRe
Inert ia force (Fj ) Viscous force (Fu) pAC 2
lL(f)A pCLorCL l' 0
I
where " = ttl P In case of now through pipes , L is taken as diameter 0 , The Jaw in which lhe model s arc based on Reynold 's number is t:alled Reynolu '!'! model law or s imilarity Law. i.e . (Re)/IImJd ;: (Re)lIru(U'YfJ('
2. Euler's Number (Eu) It is defined ilS Ihe square root of Ihe ratio of the inertia force of a nllwinJ; fluid til thc pressure force . Thi s is named arler Leonhard Euler. Mathematic ally. it is c xpre ...... t·t.!
CLASSIFICATION OF HYDRAULIC MODELS hydraulic models, ure classified, bnsed on the scale ratio for the linear dimensions, as follows;
'nlC
as £11
1) Undistorted Models
;:
=
If the sca le ratio for the linear dimensions o f the model and its prototype are same, or if the model is geometrically s im ilar to its prototype, the model is said .to be
Inerlia Force (Fi) Pressure Force (F,I )
jPAC
1
PA C
.,jP/p
fUlJistorted model.
2) Distorted Models
The law in which the models a rc designed on Euler's numhcr is known as Euler's model law i.e (Ett)mfJdtl ;: (£II)p,,"m:vp~
If (he scale ratios for the linear dimensions of the model a nd its prototype arc different or if the model is not geometrically similar to its prototype, then the model is called as distorted model. For distorted models two dirferent scale ratios, one for the horizontal dimensions and the other for the verLical dimensions arc udopted.
It is defined as the square root of the ratio of inertia rorce of allowing nuid to the gravity rorce. It is named after William Froudc , a British naval arc hilel:t. It is cx.pn,:~ s t.:tJ
3. Froude's Number (Fe)
191 ;..
Ttmlio fvIM'III NES DIMENS ION"'. "Nil MOIlEI. ANALYSIS ...: :!lJJ
Thl.' ,·ldl"(l/lta,I.W I" nf Ihe dimr.:nsi onul and model analysis urc I . The pl:,-rormallL"e ur Ihe actualturbomar:hine can be easil y predicted. in advance from ils moud s. The reliubililY depcnd.~ on Ihe degree. or si milarity thaI exists hd W\.'L.' 1I lilL.' IlHluel and the. prolrtlypc. , With rhe hclp of dimensional an:ll ys is, a relationship hClwl!r.:n Ihl: vll ria bll!s inIluencing thl! prnhlr.:m in terms nrlhe dimensionlr.:ss par;Jnletcrs is obtained. This rehttltH1Sh'l' hdps in conduc ting tests on Ihe model. 3. Wi th the help of model tes ting, the mosl economical und safe desi gn may be Sdl!I..' Ir.:U.
4. The homologous points (The points which havl! the samt.! rdative IOCilIIOn) mu sl be rl!ltlll!d hy the same lil1cm··scale ratio. Fig. 7 . 1. shows OJ protOLypr.: and a onr.:-te nth scalr.: modd or a hliJul!. The s..:a l\! ralll1 ilo also applicLlhlr.: 10 thr.: rastcnt!rs used .
The linr.:ar dimensions of Ihe mOlJd arc all onc-tenth of the prototype blade. BUI II!!. angJr.: of ,Iliad: with respect 10 Ihe frt!c slre am is tht! same 10" nOi I ... . All physical dctalis 0 11 the model mUSI be scaled. i.c., the nose radiu s, the surface roughncs... elc. Any departure will lead 10 the vio lation of geomt!tric sim ilarity. Modds which appear similar in shape but violat\! gr.:ometric similarity shu uld nOi be cUOIpart!d.
SIMILITUDE
;-
Si mditll dL' is ucnncu as the simi larity between the model and its prototype in every respect. Thn.:c types of similarities mu st r.:xisl between the modI!! and prototype. They arc: I . [i('onll!lric Silll il.a rilY 2. Killt!malic Similarity 3. Dynamic Similarity
L
,
8 1,
-I= -
=
o~" ~
2m
O}~~ml
)
e,L
".2:'- -
!-
1. Geometric Similarity A mtldd und prototypc an:: geomelri ca lly sim il ar if and only ir all hody dimensions in Ll lllh l: lit!"!:!: l.:lI\lrdinalcs ha\'r.: the SLl Ill!: lin l!nr-scllh: ralio. L!:I Lm - Lenglh ur Illndd /J m • Arr.:adlh of ml)lId Dill . Di:.ullele.r of" IIlCldr.:l \I", - Volullll! of model Am - Arc:t o f model ;md L I1 , ill" D,1o V,I ilml A/I arc the correspondin g vu lu cs of the prototypc. For geomctri c simi larily hctwccn Iht:. model
Homologous points
Fi[:llrt! 7.1
Pr()rOlypt!
t"'"1
{ll/{J
il5 lI!Odd 11';111 CI.'Omt!lric Jimi!rmly
2. Kinematic Similarity A model and prototype art! kint!matically similar ir and only if Ihr.:y have Ihe samc velocity Sl:all! ratio. i.e the motion of IWO systems arc kinematically similar if homologous particles lie at homologous points and ul homologous time (ref!!r Fig. 7.2).
LeI
emj
= Velocity of fluid at point I in model. Cm1 = Velocity of fluid at point 2 in model. and e ,ll , C II ! are corresponding values at the corresponding points of fluid vt!loclly in the prototype. For kinr.:matic similarity, 10 exis l
whl!rl! G r is cu llt!c.J the scale ralio. For arC;1 ratio anti volume ralio, th e relalion should be as given helow. Similarly ror acceleration a/.!
=Llllll
=a,
lIml
(LLm )' = (B)' 8m I'
where C r is the vt!iocity ralio and (I, is Iht! acccianion ralio. The directions of the vdocity in the model and Ihe protol yf>t! shou ld he Iht! same.
I'
I. All angles ;Irc. prr.:scrvcd in geomclril.: similari ty. 2. All now dirct:li ons are preserved. J. The Oriel1l:uiull of Ihe model und prDlDlype, with respcci must hl! id~nlit.:;J1.
3. Dynamic Similarity 10
the surroundin gs
A model and protOlype are dynamically similar if and only if Ihey hav\! Iht! s;lmt! force -scnle (or mass-sealt!) ratio , Thus dynamic similarity is said to exist betwecn the ni odel mid the prototype, if lin: ratios of the corresponding forces acting LIt tht!
290
):>
TURBo MACHINES
Since, Dllllension of L.fI .S = Dimension of R.H.S = LT - 1 equ:Hion C = J2g H is dimensionally homogeneous and can be used in any system of units .
For example. in the problem considered, X2, XJ. and X4 arc n:pealing variahle~ ir Ihe fundamental dimension //I(M . L. T) 3. Then each Jr ternl is wrillen as
=
DIMENSIONAL ANALYSIS METHOD Buckingham PI Theorem The Bud,ingham PI tll!.!orem is one oflhe lIimensional analysis methods ofn:ducing a !lumher ur dimensional variables into a smaller numher of dimensionless groups. The nalll!.! ri CllIlH!S from the mathcrml.lical notation Jr, meaning a producl of variables. If there arc -II' variables in a physical phenomenon and if these variables contain 'm' fundnmental dimen sio ns (M. L. T), then Ihe variables arc arranged inlo J(= n-m) dimension-less terms. Each term is called a Jr-term. The reduced J equals the maximum number of variables which do nOl form a pi among themselves and is always less than or cquillto the number of dimensions describing the variables. Typically, there are six steps invol ved,
I. List and count the 't!' variables involved in the problem. Dimensional analysis will fail, if any imporlant variables arc missing. For example, Let Xl be the dependent variable and X2, X),···. Xn are the independent variables on which X I depends in n physical problem. Then X I is a function of X2, X), ... , Xn and mathematically it is eltprcssed as
(7.4 ) eq uation (7.4) is a dimensionally homogeneous equation. It contains 'n' variables (including dependent variable) equation (7.4) can also be written as
2. List the dimensions of each variable according to M LT9. Detennine the number of fundamental dimensions (say, there are 'm' fundamental dimensions). 3. Find J. according to Buckingham's 'Jf- theorem equation (7.4) can be written in lerms of number of dimensionless groups or tr tenns in which number of ;T- terms is equal 10 (n- m) 1. Hence, equation (7.4) becomes as
(or) (7.5)
4.
variables which don't form a Pi product. Each 7r term contains m + 1 variahle!>, \·.:here . m' is the number of fundamental dimensions and is also called th!.! rcp!.!<1ling variable. Selec~J
Method of Selecting Repeating Variables The number o f repealing variables arc equal to the nurnberof fundamental dimen"loll~ of the problem. The folluwing points govern the choice of rcpe:uing vari ahle" . a. As far as possible. the dcpcmh.:nt variable should not be selected rerl!allll~ variable. b. The repeating varianlc should bc chosen in such a Wily that tlile variahle l~ a ~eomctrie variablc. otilcr variablt:. is a now variable and the third variable I... a Iluid property variable Geometric variables- Length(I ). Diameter ((I). Height (h) ett.:. Flow variables - Vdocily (c), Acceh:ration etc. Fluid property variables Kinematic viscosity (O). Dynamic viscosity (J.d. Density (p). etc. c. Tilc selected variables shou ld nut form a dimensionless group. d. The fl!pcating variables together must have the samc number or funJallH;nlal dimcnsiuns. e. No two rcpealing variables should have the same dimensions. 5. Add one additional variable to the sekcted J variables and form a power proJllcL Each Jr lerm (as given in equation (7.6» is sol\'ed hy Ihe principle of dimensional homogeneity. 6. Write the linal dimensionless function by substituting the \'alues of Jr I. 7T;! . . . ]t'l _ '" in equation (7.5) and check whether all pi groups ilre dimensionless .
MODEL ANALYSIS To predict the performance o f turbo machines such as turbines. compressors. rump!'> etc. Model Ana-lysis is employed. That is, before manufacturing the real (If ilCIU;11 machines, models of the machines arc made ilnd tests ilrc performed on Ihem In !!et the required infonmllion. The model is the small scale rcplicn of the actual mai..'lllnl.!. The actual rnachll1e I!'> called prototype. The study of models of actual mnchinl.!s is callcd as model 01l01r.l'l.l' . The model analysis is actually an c;-.:perimcnlal analysis of finding oul solutioll!,> Ill' complex problems using models of tbe actual system.
::! SS ;.. Tunuo MACHJ NJ:S
DIM ENSIONA l ANDM fJ llIL.A .....\LYSI S ...;
::!Kl)
Tht! dimcn sio ns of mos tl y use d phys ical quantities in turbo mac hin L.:s
Phys ical Quollfirie.r 1I.'ied ill wrbu machill es
I)/i.l'si cal Qual/firy
I. 2.
,.
".5. 6.
7. ~.
9. 10. II. 11. 13. 14. 15. 16. 17. I X. l ~.
Symbo l
Dimensi on
A
L' L' T- '
Area Vo lum e Angular vl!lnci ty Disc harge Accl'krallon dlle II) Ciravity Kin CIl1.Hi l: vi scosil y Forct! \Vcighl Spl!c ilic we ight Dynamic viscos ilY Work. En erg y Power Torque Moml.!l1Lum
M
/\ l1 gk
1/
MLT- ' NO lle ·
TC lTIpcralurl.! Specifi c heLlI Mass Fl ow rale Elliciency
T C,'" C~.
II L~ T - "!I;r l
'"'I
No ne
II
'"
Q g v~
F
,."
mOIll!)' .
M L T -' MLT-'!
w
2 M L- 2 M L- IT - I M L "! T - "! Af L"2 T --·l M L"! T -"!
T
Ihal is th e dim ensionless force coeffi cient F / p C 2 L"! i.Jnd it is the only funt:tion of dimensionless Reynold s numher (pC L /ll ). We shall learn e:o:.(lclly how to make Ihi s rl!dUCli on in thc following section s. The functi on ',1.: ' is different mathemati cully from the origmal function ' {' , hut II ( ontains allthl.! same information. Nothing. IS IOSI In a dimenslOl1
L JT - I LT - :' L '! T - 1
IV
n W, E p
or
2. Th e p redicli(l ll
r-
I . All en urmous S£/I';lI g iI/lim e ami IIWll er.
SUPPOSL . foro.:t: F on a pi.Jrti cul ar bod y immersed in a stream of flUid mov ing with vel oc ity C is a fun cti on ofh ody length L. the nuid dl.!nsiLy p , thl! stream vel oc ity C. i.Jnd thl! tlu id viscos il y 11. it is cx pn: ssL!d as '
.....
f L'.I"/ S
f"mulll("/t'd
Oil CI
sndt' m odt'!.
DIMENSIONAL HOMOGENEITY
Thl.!" impon alll ad v ~lIl1a c e s of uirnt:n sinn;-t l anal ys is besi dl.!s its Illi.lin purpllse In rt:dUl;e variahl es and group !hcm in dirn e nsinnl t!ss form , :Ire
'r
profotype pt!fformall ct' JI"(I III
3. The dt::lermilllllion of ,lit' 1I1(J.'i/ ,'illirable rype of machille nil rhe htlJi.\- OJ maximum eDiciell cyJor (l specified range a/ head, speed olldlltllL' rllft:.
MT - I
ADVANTAGES OF DIM ENSIONAL ANALYSIS
IF ~ f ( L. C . p. /tl l
0/£1
The dilllt! l1si onal annly sis providcs 'scnlin g laws' which can (;OIlVl'r1 data from i.J ( ht! up, small model into design inform ation for an cxpen sive Iilrge pwtOlypt:, Fur exampll.!. One need not bui ld a turhomachinl.! and scc whether il has the mi.Jximulll dcsirt!d dfi ci ency. Onc can measure the dfi(ienq on a smallmodcl and usc LI scaling law to preuictthe effi ciency of a full- scale protolype turhomachint!. For example, we do n' J huild a milli on rupees air plane and sec whethcr it has t!nough lift force . Wt: rneasurt! the lift forc!.! on a small modc! and usc a sl.!aling law to predicllhe lift on tht: full seale protot ype air plant!.
(7. 1J
whe rL mean s ' ;] fun cti on of" a nd is to be Jet ennin ed expl:rimentall y. Gt! nerally. it lak es ahoul IOc xpcrilllental points to dclin e Ll curve. To fin d the dfcc l of body lenglh in eq uatio n 7.1 WI.! shall ha ve 10 run the e," perimenl for 10 Icngths L . For each I. wc sh~1I nee d 10 va lu cs of C, 10 values of p and 10 va lues of IL making a gran d lotal or 10,000 expe rimt:llt s. AI a rale of Rs 100 per c.'l;.pcrimcnl. till! IDlal L!xpcnditun; would be in st!vcrallak.hs. Howeve r. with dimensi on al analy sis. we can immediall::l y reduce th e equati on 7. 1 to an cqui vnlent form
Dimcnsional homogeneity means thilt thl.! dimensi ons or each term 1/\ i.Jn eljuall un on ha th th e siut!!> arc t!quul. l-knc.: e, if th e dimcn sion s of each term 111\ hOlh slue !'. or all l!qui.Jti on arc the same, the equati on i ~ known a!'. dim t'lisiolla//y h{lll/ogt'II(' jJ/l .~ t!fjIHlfirJII . Thl! powcrs of fundamcntal dimensions (L. M, T) on buth Sides or the l.!qu;)ti on wi ll bl.! iucnti cal for a diml.!n si onall y homogeneous I.!quation, Such equutltub arc independent of th e system of unil s (mclric , En£lish or 5.1) Consider the foll ow ing cqunti on
wherc 'C' is vel ocity. H is heighl or any nuid column and g is accelerat ion uut: to gravity. Dilllt!nsion of L.l-1.5 of Ihe above equati on is gi ve n by
Dimension of R.H .S of thc cqui.Jti on is
j (L/T' I:L ~ j~ L '/"r'l Lr '
7 ______
- - -----------
DIMENSIONALAND MODEL ANALYSIS
INTROD UCTION Dimensional ana lysis is a mathematical technique used in rcst.:.m:h work fur de sign and modd testing . II deal s with Ihe dimensions o f thl! physical quanti lies involved in the phenomenon. A dimcll.'iioll is the measure by which a physical variable is expressed quantitatively. Forcxamplc. Leng th is a dimt.:nsion associated with vari;lhlcs such as distance, displacement, width. height and dcnt:ction. A 1111;1 is a particular way of attaching a number (Q the quantitative dimension. Ex: Metl!r and I.:CnlimCI1.:rs afC hath numr.:ri cal units for expressing length. Dimensional analysis is a IlH.:lhod for n.:duL:ing the large number of vanah1c~ 111valved in describing the performance characteristics
or a turbomachinr.: to a nUlllhr.:r
of manageahk dirnensinnlr.:ss groups. For c:
Fu nda mental Dimensions The three basic or primary or rundamental dimensions arc length (L), lime (T) , ;1110 mass, (M). In compressible nuid~, one marl! dirnl!nsio n. apart from the inJere lltleO[ quanlities M,L & T namely lhe temperature () is also consideretl as a rundaml'nta l dimension. II is rderred in short as the MTLO system, when thest! qU
Derived Dim ensions The dimensions which possess more than olle fundamental dimension ;ue called derived (or) .fec(llIdol), dime1lsiolls, For instan ce. ve locity is denoted hy twO baS il' dimensions i.e. distance per unillimc ( LI T), density is denoled by mass per un!! volume (MIL J ) and acCeh!ralion by di sli\ncc per second square (LIT'!). Thl! ex pressions (L I T). (M 1L J) and (LIT2) arc called dimensions or velocity. de nsily :Jnd accclcralion rcspl!ctivciy. They arc called secondary or dl!rivl!d dimensions.
:!S~
;...
TURBO
M . \ ClUNES
6X
(a) De linc spouting vcloc ity (b) Prove that a .rlll " = 0.707 6.9. Show thc skc tch and describle the working prim:iplc of a double rolal ion ward flow ratlia l steam tu rbine stage. 1>.1 O. Draw Ihe entry and exit vcloci ty triangles for a Ljungslrom lurbine. 6. 1 I. Derive the following rclations.
Ihe impeller peripheral spel!d spoUling veloci ty and the Mach number ailiouic ., eXI.
OU I -
(n) ( b)
6. 12 . 1'IH; llL'sign dala Df lin inward !low ex'haust gas turhine arc as rollows: Stagnation pn.;SSLlfl! and tcmpefilluJ'c at nozz le inlet = 700 kPa und ! 075 K.
SIatic pressure and temperature at e.'(i t rrom nozzle -= 510 kPa and 995 K. Stalic Pressure and tempcrature al rotor exit = 350 kPa and 918 K. Stagna tion temperature at rotor exit = 920 K. Speed = 26,000 rpm. Mc,ln rotor exi t radius to rotor lip radius = 0.5. n· .~ flow into Ihe rotor is purely radial and 'H exit the flow is axial. Calculale. (a) (olal-(O-lOtal ertlciency (b) outer diamclcroflhe rotor(c) the nozzle and rotor Joss cocfficienlS and (d) The hlade out let angle al the mcan diamcter (measured rrom Ihe radial direction). IAns: (a) 80 % (b) 0.29 m (e) 0. 1625 and 1.15 (d)72 .2°] 0 0. 1:1. A 90 fFR turbine 1135 Ihe following data. Rotordi~mett:r ralio (D t / DII ) = 0.45, rotor speed = 16,000 rpm, noz zle exit air lingle = 20 D , nozzleerfic:iency = 0.95, rotor \vidlh at entry = 5 cm, blade to spouting velocity nltio = 0.66, total -la-static pressure ratio (Pool PJJ = 3.5, ex il pr!.!ssurc = J bar. stagnalion temperature al entry = 650 u C. Assuming constant radial velocity and axiul exi t, ucterminc (a) tht.: rotor diLllHctcr (b) th l! rOior blade exit air angle (e) the mass Ilow rate (d) hub and tip diameter or lip rotor (e) the power dl.!vciopl.!d (r) the IOlal -to-total efficiency (g) no zz le [lml rotor enthalpy loss coeffic ients. IAns: (a) 59 em Ib) 38.9° (e) 14.2 kgls (d) 8.4 em and 44.6 em (e) 3458 kW (f) 92.5 % (g) 0. 126 and 0.338J 6. 14. An IFR tu rbine impulse siage with a llow coe fficient orOA develops 100 kW. The (0Ial-lo-lola1 efficiency is 90 % at 12000 rpm . If the flow rate or air is 2.0 kg/s al an enlry lemperalure or 400 K. del ermine the rotor diameters and air ang les nt the en lry and exit, Ihe nozzle exi t and ang le and the slag nation pressu re ratio across the stage. Assum e zero exil swirl and ConSlant radial velocilY. Take rOlor exi t dinmelcr is 0.8 limes Ih e rOlor inlct diameter. IAns: (a) 35.6 cm and 28.5 cm . (b) PI :;: 21.8 0 • 0'1 = 11.3 0 and flo = 26.6°. (e) 1.68J 6. 15. A small IFR gas turbine, comprising a rin g or nozzle blades. a radial vaned impellcr and axia l diffuser, operales with n total-to-IOfal efficiency of 0.9. At inlet to the stage the stagnation pressure and temperature ore400 kPa and 11 40 K rcspcclivcly. The flow Icaving the lurbine is dirrused to a pressure or 100 kPa and has negligible exit velocity. The nOlzle angle 01 the exit is 16°. DClcrmine
(MV-.-\pril ' ~7) IAns: (a) 580.7 mls (b) 865.7 mls (e) 0 .9731 6. 16. The design data of a Ljungstrom turbine arc specd = 3600 rpm. inncr diaml!ter or the bladl.: ring = 12 cm. blade width = Jcm. blade exit angle - 2~ r . now rOite = 10 kg/so Determine the powcrdeveloped and Lhe enthalpy drop In the blade ring ror ideal flow and optimum conditions. IAns: (a) 12.08 kW and (h) 1.2 k1/kgl 6.17. Determinc Ihe powcr developed by;\ 90" IFR turbin!! whIch has the rollowing. d:lIa. Imj1clll.!rdiamell.!r at cmry = 40 <.:m, impe ller diamete r (mt!an) at t!Xlt = 211 crn. l11a~s nllW ralc = 5 kg/so rj11ll = I H.OOO. iscntropic dficiency =. K51i!:. !low coenicient at entry = O.~. stal ic pressure ralio aL:fOSS the stagc = --I. j1res!> urc at the impeller exit = I hilr. temperntun: at Ihe entry or the stage = 600' C. Assume half the Sialic pressure rali o to occur in the nozzles and thL' volute. The discharge is axial. What is the nozzle angle and width or Ihe impt.:1h:r at cnlf)'? IAns: (al 710.65 kW (h) 16.1' onu 'e) .1 .7 eml
RADI Al. FLOW G AS AN D STEAM T UHUL "' ES .....:
282 ).. TlJrmo M ACl IJN ES
6.10. Draw the Mollier chart for e~pan sion in a 90;' IFR turhinc'?
For rOlar
LH = C,
V,
6.11 . Deline degn:c.: o f reacti on
Cp( T, - T2.<)
j2 C p(To, - T, )
=
j2
=
=
6. J 2. The degree (a) 0.5 (bl 0.25
IVf/2
=
1V:! =
vi 1iOI.78' + 165.88'
jCi + = 194.62 mls
n.ftl
=
:. LR LR
= =
,:<, (P')", P, = ' 1029(385) 527
907.22 K 51X(9 15 - 907.22)
i\n g le. (TruelFalsc ) 6. 14. For it gi ven value of the now coeffic ient. the forward curved vanes give h i.~ltu d egree of react ion compared to the backward curved ,,·anes. (True/Fnlsc) 6.15. As the nnw coefficient incft!ases. the degree of reac ti on of the backward curved vanes and the forward curved vanes. (a) Increases and decreases re specti vely (b) Decrcases and Increases respec tively (c) remains constant turbin e. 6. 16. Lj un g.strom lurhine is a 6. 17. TIl e optimulll r;:ltio of pc.:rip heral veloc ity of bl... dc 10 the relati ve ve loc it y ,II exi t for ma ximum work is (a)
'1,-1
= '1,-/
=
or.lll11
= 2f3
(b) Or'''1'1 ;:: s inJ3 (c) 0".1'1" :;::: cosfi/2
(d) Total-to -to tal efficien cy
6.1 8. TIl e maxi mum work ou tput fro m a Ljungstrom outward now reac tion turbtnt: I},- .s
C' -', 2V'
is tw ice the sq uare of the cxit blade velocity. (TrucIFalse) 6. 19. An outward now radial turbines behaves as a
stage o f the ;Ix i:ll
turbine.
I
101.78' 0.90 1 - 2 x 337.58' 93.95 %
SHORT QUESTIONS a radial fl ow gas IUrbinc? Wh .lI arc the .~pplications or radia l now gas turbines? The fl ow rate In
6. 1. Wh
6.5. 6.6.
6. 1J . T he.: degree n f re
194 .62' /2 0.2128
=
or reac tion of a radial now gas turbine.: with mulal vanes IS
(e) n. 1
x 518 x (925= 9 15) = 10 1.7H IlIls IT D,N = IT x 0.132 x 24. 000 = 165 .88 m l s 60 60
T,
T2.r
6'32. 6.. 6.4.
:!a :\
6.7. The workdonc per unil mass flow in a 90° lFR t . . hlade veloci ty. (TruelFalsc) urbme IS equa l 10 square o f the
6.8. The rOlo r consta nt of an IFR is eq ual to la) "u. re i (b) IIo.rcl-V '/2 (e) 11o 6.9. Draw the veloc ity diagrams for a 90 0 IFR turbine?
EXERCISES 6. 1. 6 .2. 0 .3. 6 .4 .
Draw the sketch· of a 90 1l IFR turbine stage showing its main compo nenh. W hat nrc: Ihe npplications o f IFR turbine? Draw the entry and exit vcloc it y triangles for a 90'" IFR turh ine Prove thilt for a 90 0 IFR lurbine .
(a) \V / I1I -= {j~ (hi '1'1 = I h.5 . Dru w an enth alpy - entropy diagram for now through an in ward -flow radi<J1 lu rbint' slagc fi ttcd with nn cx hnust diffuser. 6.6. Provc tiwi
. , ' It rn'~1 - 1/ 2U 2 ;:: "0]", - 1/ 2U i 6 .7. (a) How is the degree o f reac tion of an IFR turbine stage defilled ? (b) Prove IhJI
R R
I - I I - ,cOII!')
R
1 -V'I -
2
1
180 ;....
R,\DI AL FLOW G .... S ANI) STEAM TuIWJNIiS '" 281
TliltUO M ,\liIlNES
1005(785.99 - 778.78) 524 .76' 12
C/,(Tl - TI.J)
Cr / 2
(a) Total-to-static efficiency Referri ng
10
the Mollier Chart.
0.0526
(h) Rotor enthalpy l oss coefficient
LR IV:!:!
179.48'1 + 222.84 2
=
T, (P')'~' =785.99(_1_)1':1 1.931
'/1-1
=
90.1%
P,
(b) Ro tor diameters
- 651.28) 286.13'/2
Example 6.6 For a radiallurbine stage the pressure and temperature through the stage at design co ndit ions arc: Total pressure and temperature upstream of nozzle arc respec tively 700 kPa and 1145 K, Sialic Pressure and temperature downstream of nou le ilre respect ively 527 kPa nnd 1029 K. At rOior ex it, Sial ic pressure is 385 kPa, static temperature is 915 K and stagnation tempera ture is 925 K. The ratio of rotor exit mean diameter (a rolor inle t diameter is 0.49 nnd speed is 24.000 rpm. Assuming relative flow lit rotor inlet is radial and absolute fl ow at rotor exit axial detcnnine (a) IOla l ~to~slatic efficiency (b) rOiar diameters and (c) enthuipy loss coeffic ient for nonlc and rOIOf. Take r = 1.67 and molecular we ight of gas as 39 .94 (d) IOlal~ t o- lotal efficiency.
S olu t ion
;:;
C,,(TOO - To'!)
U,
=
D,
=
D,
;:;
15 18 x ( 1145 - 925) = 337.58 mls U, x 60 337.58 x 60 ~ = IT x 24,000 0.269 m
=
0.49 x 0.269
=
0.132 m
= 1145 K P, = 527 kP" Tl;:; 915 .K D'!.fluIDI = 0.49 N = 24.000 rpm C, r = 1.67
(c) En tha lpy loss coefficients For nozzle
LN
C/.(T, - T,,) Cr/ 2
C,
J2C/.(TOI - Ttl
Too P2
= U;
\VIm
0.3373
C'I
900.8 K 1145 - 925 1145 - 900.8
:. 11'-J
1005(665.0~
=
(:~) '7'
"" 385)m 1145 ( 700
lVi)2
651.28 K
IV:!
Too - T02 Too - T'!.JJ
Too
286. 13 mls CI'(T~ - T2r }
LR
?na = 700 kPa T, =10~9K To, = 925 K
:::;
11'2 - 1I'2.r
= wi 1'2 = U'1! + Cr'2 =
.. w:!
T"
III -.f
= 385kPa
=
12 x 518(1145 346.66 mls
1029)
, -I
T"
=
R
8.3 14 R = - = - - = 0.208 kJ/kg-K III 39.94 ,. R 1.67 x 0.208 C/, = - - = = 0.518 kl/kg-K r - I 0.67
= =
LN LN
= =
Too (P,), Poo (527) Vel' 1145 700
.1
1021.75 K 518( 1029-1021.75)
346.66'/2 0.0625
278
,. T URBO MAClIINES
(c) Mass now rate for a 90" IFR turbine, R
''1 - II , - - -
R
"0 - "2
=
= 0.5
1.93 I b;) r
0.5C"ITo - T,)
Mass fl ow rule P! A1Crl
In
hut Wl m
= U\2
O.B56 x
0.5Vi T,---
:. T,
vi 493 . 11 2 T-OI- - =923- - C
=
Ci
T02 - 2C
1005
"
681.05 K
p
= 68 1.05 -
=
W
=
T,
V,'
+ 0.5 e
=
Rotor width ;ll
e:~it
'"
179.48' 2 x 1005
Pl(rr D'!.}Cr'!.
= 665 .02 +
0.5
493. 11' 1005
X
I'
Now. hub diameter Lit cx:il
785.99 K
Dz - b'!. = 0 .266 - O.IR I 2
D'!.. II
Too - Tis 'J; 00
T1.~
=
=
_(Too-T') 0.95
923- (
923 - 785.99) 0.95
D!.I
=
D'!.
=
0.4472 m
3.5 (
(g) Nozzle enlhalpy loss coefficient h! -his
,.
778.78) ..
923
x 1005(923 - 645.28)
87. 13%
(:7)' T',) ,=, (-Too
O. I HI 2
IV
--::--:-;;;---:;:----: = 14.24
IIIC,,(7iM) - T2Jj}
,- ,
Poo
+ b2 = 0.266+
(f) Total-la-static efficiency '1, - .
Fronl isen tropic relation.
0.0848 m
and tip diameter at ex:it
778.78 K
=
\
287 x 665.02 = U.524 kg / m'
14.24 0.524 x If x 0.266 x 179.48 0. 181 2 m
.'. b'!.
Too - T, = 0.95
=
I x 105
P1
RT,
N07.l.1c efficiency is given by
.'.
· J.463mW
(e) Hub and tip diameters at exit
Now.
T,
';'U~=14.24 x (493 . 11 21
\V
665.02 K
T,
x 0.59 x OJ)5) x 179.48
(d) Power developed
and
.'. T02
( If
1..t.24 kg / s
111
CI'
1.931 x 10' 1 = 0.H56 kg / m 287 x 785.99
P, RT,
PI
=0.5
C,
=
ci 12 V,I cosex, = 493. 11 1 cos 20" 524.76 m/s
-
r
I
276 ;. TUlwo M ,\CIIINES
Solution 0,
0.09 m
O~ . h
0.025 m
0.062 m 30.000 rpm
PI
1.8 kg/m 3
RADIAL FLOW GAS ANO STEAM TURIlINES "" "2.77
diameters at exit (f) total-to-slatic stage efficiency (g) Nozzle enthulpy loss coeffiCIent and (h) rOlor en lhalpy loss coefficienl.
Solut ion Pun
IT
V, V,
:. C,
=
x 0.09 x 30.000 60
Di 0, b,
14 U7 m/s 14U7/0.707
3.5
P,
I bar
Too
923K
V, C,
0.45
N
16.000 rpm
"I
20'
q"
0.05 m
Cr ,
Cr2
0.447 :< C.I
O.66CJ
C,
jC,,(Too - Tzu)
89.38 m/s
= 1111/ T'
where III
=
(~) 1l
(a) Vo lume flow rate at impeller exit
645.28 K
/2 x
: . C.f
=
Flow arCLl at impeller exit
V, IT,
- 01.,,)
2.5"8 x IO- J x 89.38
r~lIc
ROlor inlet diameter V, x 60
rrN 0.59 m
of mass fl ow is p, Q, = 1.8 x 0.226
0.407 x 141.37'
IV
=
tanfh Cr2
0.45 x 0.59
0.266 m
c,{~~ U,
Cq
= V,
U, "I
Crl = Utlancq
493. I I x Ian 20'
8134 W 8.134kW
=
= (b) Ro tor blade angle at exit
0 .407 kg/s
(b) Power developed is given by
493. I I x 60 rr x 16, 000
Rotor outlet diameter
0.226 m' /s
M
= 493. I I m/s
,
4 2.528 x 10- 3 m 2
The
1005 x (923 - 645.28)
747. 14 m/s 0.66 x 747. 14
':. (0.062' - 0.025')
Q,
0.95
Too ( P, ) ';' = 923 Poo 3.5
P"l Q2
4" (Di,
0.66
Cr
v, =
0.447 x 199.96
Power dc.veloped
=
=
(a) Rotor diameter
199.96 C2
p,
Cr,
C,
179.48m/s
Figure 6.9(a) &: (b)
Example 6 .5 A 90- inward flow radial turbine has the following duta. TOlal-tostatic pressure ratio (POD / P2) = 3.5, exit pressure P2 = I bar, inlet lO1U1 temperature (Too = 913 K, blade to isentropic speed rulio = 0.66. rOiar diameter rmio = 0.45, Speed = 16,000 rpm, noz.zle c:
u,
IT
O,N
60
= IT
x 0.266 x 16.000 60 179.48
tan
fh = 222.84 ~,
=
38.85'
= 222.84 m/s
:!74
~
T URBO MAC III NES
RADIAL flow G AS AND STEM..! T URBINES '" 275
(f) Mach numbers at nozzle and rotor exit M,
c, C, = ;;;= JrRT, = J I.4
M,
=
(a) The impeller tip speed
413.48 x 287 x 471.94
0.9495
.nd
jCj-+Ui vi =
Jr RT,
v,
rrD2m. N
=
rr x 0.21 x 18.0{)()
60 197.92 ml' J I1 8.75'
+ 197.92'
230.8 397.92 0.58
TOI = T] C, =
g) The nozzle and rotor loss coefficients
LN
hi - ill.r
ci/ 2
=
= 532.2
'" Is
u,
C'
+ - '2e "
13.265 x 10' 413.48'/2
jio06 x c,' x (To,
- T,)
LR
0.498
lVi 12
13.265 x
e,
Figure 6.8(b)
v,
cosO' I
=
IU 13)
«,
Fru m th e inl et veloci t)' triangle. Fig. 6.8(b)
uno
"2 - II!\
P
W,
= j 2000 xI.l477(i150 = 560.6 1 mls
0. 1552
LR
C~)'I:: ) ]
(b) Flow angle at nozzle outlet Since Too = TOla nd
Jl.4 x 287 x 394 .08
LN
v,
60
09 [1147 x 1150 x (I -
C,
lO.l
or
230.8'/2
at
=
oos - '
=
18.:1 1"
(~) 560.61
(c) Mach number at nozzle exit Example 6.3 An IFR gas turbine operates with a total-to-tota l efficiency orO.9. At entry 10 the nozzles, the pressure and tempe rature or the gas are 300 kPa and 1150 K respective ly. Atthe oUllet of the nozzle the slatic temperature of gas i~ 740°C and al the outlct of the diffuser the pressure is 100 kPa. TIle gas has negligib le' veloci ty at the diffu ser exit. Find the impeller tip speed, flow angle allhe noulc outlct and Mach number at n07.7. le exit . Assume that the gas enters the impeller radiall y nnd there is 110 whirl at the ex it. Take C p = 11 47 l / kg - K. r = 1.3 3 and R = 284 .55Jkg - K.
C,
~ = Jr RTI =
= Example 6.4
Solution
C,
M,
560.61
JTT3 x
284.5 x 10 13
0.9 1
A small /FR wrbinc run by exhaust gas has the following de slf! n
dal<.t . Rotor inlet lip diameter = 9 em . rO lor ouLlcltip diameter = 6.2 em, rotor OUlIt:1
"" = 0.9 T,
= 74U' C = 1013 K
= 300 k? PO.l = 100 k? Poo
Too
= 1150 K
hu b diameter = 2.5 em. C, I C, 0.447 and V,/C, = 0.707 . Blade speed 30.000 rpm . Delermine (OI l vo lume now rJtc at impeller out let (b) the power developed and lake dl:nsil Y IJf exhau st gas <1t impellers exit LIS 1.8 kg/ m J
=
=
~T2
)0>
RADI .... LFLowG,\S .... NOSTEA""TuRBINES '" '17)
TURBO M ."'C.IIlNES
The width of the rOlor al inlet
we con w rite, h{)(1- hl.'
=
hO I -
=
h lJ
(hoI - hi)
+ (hi
-
b,
Itl,\)
=
C'
6.5 1.499 x 118.75 x n x 0.42 0.02767 m
- ' + 13 .165 = 2C p W
11 8.75
t = -. = sma l
C,
-,---c7.,..,--- = 413.48 m/ s sin 16.69"
Then,
'"
PI WI (rr D,)
2.767 em
(e) The rotor blade height at the exit
flnl - Jib'
=
.113.48' 2000
=
YM.75 kJ / kg
=
98.75 - - = 98.26 K 1.005
+
13.265 ur
Since, C,! -= C,!
Too
1(u
T"
TO! - 9M .26 = 557 - 98.26 458.74 K
Now,
P,
=
POD
= C'I
=
2000 x CII 118.75 m/s
T02
=
401. IK
,', T2
=
40 1.1 -
Pm
(1"',) ;-=J Too
1.4 .:. PI
= =
: , P2
557
und
T he flow un!a
nt
A,
=
P2
=
C'
2C" 413.48'
557 - ~=-:-::::-:; 2 x 1005 471.94 K
PI
-=
R1",
= =
2.03 x 10'
'. A2
=
II"!
287 x 471.94 1.499 kg / m]
( 1"2
=
0.94
To,
f'-' C =
94 08 . )
401.1
= 0.94
x I
= 0.94 bar
//I 6.5 - - = -me,! P2 C 2 (J.94 x 10' P, = 287 x 394.08 RT"!.
0.831 kg/m'
TIli.:n,
P,
=
rotor exil is given by
JUI - _1-
T,
=
11 8.75' 2 x 1005 394 .08 K
Since PO:l = PJ (negligible diffuser exit velocity). Pin = I har
4 (458.74) 0.4
2.03 bur
C;-
+
=
P, r
T,
T02
h"!
6.5 = 0.0658 m' 0.831 x 11 8.75 0.0658 AI n x 0.21 Jr D2.m
=
0.0997 m
=
9.97 c;m
I.;
:no
RAOIALFLowGASANDSTEAM T uRIIINES ""
}o- T URBO MACIIiNES
or
(c) Stage efficiency
11.93 1. 65 x ;r x 0.3 x 120.3 0.0638 m 6.38cm
=
1100 -
lis
Since.
"0)
= 1100 -
Power lV
Example 6.2
=
11.93kg/s
=
mUIC.rr
=
580.15 ·kW
= 11.93 x
1100 - "02 From iscnLrllpic rclnti on
188.5 x 257.98
A si ngle swgc 90° I FR turbine fitted with an exhaust di ffuser
hCUi
th e rollowing dala .
Overall slagc pressure ratio =4 .0. temperature at entry =557 K. diffuse r exi t prl!S5Urc = I ba r. mass flow rate of air = 6.5 kg/s o flow coefficien t ;;: 0.3. rotor tip diameter ;;: 42 cm; mea n diameter at rotor exi t =- 2 1 cm. speed = 18000 rpm . En th alpy losses in Ihe nozzle and the rest orthe s lage arc equa l. Assuming negligible ve locit ies at the nozzle en try and di f[user exit, delcnninc (a) the nozzle ex it air ang le. (h) the r u t or width at en try. (c) the power developed. (d) the st:1ge efficie ncy. (c) the rotor blade heighlllt the.: cxil, (I) Mach numbers al n07.zlc and roto r exits and (g) the nuzzle anJ rolo r loss coefficients,
'. Th.'
0 .673 x 557 = 374 .7 K
\V / m
1000 - /'0' = - -
and
1018.5 6.5 156.69 kJ/kg
Thl!rcfurc.
15fi.69 Cfl{Too - T,lJl)
Solution
1/1
D,
Too 6.5 kg/s .p,
=
0 .42 m
=
4
Pool P-,
=
557 K C" =0.3
U,
OZ.III
PJ
=
Ibm
N
=
18. DUO rpm
U,
D, N
= 60 =
1C
= .p, U,
'h
110) - II J..
60
w,
= 0 .3 x 395.84 = 11 8.75 m /s
'" '" =
&5 .52%
c,
= CI'(7iIJ - T,u) TOJ ::;: T02
~ = err VI
VI
_ , (118.75) 'an 395.84 16.69°
= = =
Figllre 6.8(a)
From th e inlet velocity tri ang le
=
0.8552
Total enthalpy loss
u,
x 0.42 x 18.000
tan 0' 1
=
(d) Rotor width at the entry
= 395.84 lO Is
C"
156.69 1.005(557 - 374.7)
0.2 1 m
(a) The nozzle exit air angle 1C
1I0J
"31$
= hu~
(d) The mass flow rate and power developed 1/1
~71
IV TOI- -
Then , C/,(T02 - T:h .r)
=
1.005(401.1 - 374 .7)
=
26.53 kJ / kg
Given enthalpy losses in the n07.zle an d Ihe res I
(b) Power developed
-
mC11 101 8.5 557 6.5 x 1.005 ~01.1 K
or the stilge ilrc considc rl!d equal.
enthalpy Inss in the noa.le is given by
IV
=
mUf
IV
=
1018.5 kW
= 6.5 x (395.84)'
26.5.1 10, - 10" = - - = 13.265kJ/ kg 2
RADIAl. FLOW G .... SANOSTEAM T URn lNES ::!M'i
;... T UIW O MACHINES
;:;
15.3
=
S tJg~ dficicnc),llrJ'
C,.7U,,[1- (~J '7' ] 1005 x 313 [1 -
W
',' j ]
= = = =
Tot) - TIJ
C, Co,
(e) The air angle and width at the rotor exit C, = C, = 120.3 m/s
: . CI
u., = rr D"N
" Too - hr
th e oUllet velocity tri ang le (C X2 = 0) (Fig . 6.7( b»
C,
liln th
fl,
Rotor
Too - TI ,f
=
,"n-I
=
46.77·
(120.3) 113.1
II I
- II, ---uc--== 0.3 I 58
=
P,/R T, = 2.0 1 x 10'/287 x 332.7
=
2. 105 kg / m'
The muss now rale through thl! rotor is
.(1
0.3 158 x 188.5 x (C,COIC'J)
m
0.3 158 x 188.5 x 120.3(coI 25· ) 15.3 K
TI - T].
From th e workd onelkg
= =
PI AIC'I
=
2.105 x (rr x 0.5 x 0.03) x 120.3
2.105 x (rr x D, x b,) x 120.3 11.93 kg Is
/11
1V1m
litu - To::! :. Jo:!
T,
;:;
7o:!.l ;:; VIC), !
CI!(Too -
AI rotor ex it
p,
= =
48.39 K
P,
=
373 - 48.39 = 324.6 1 K
T,
= = = =
188.5 x 120.3 x
COl
25 °
1005
C;
C;
_e,l
-CI!
T02 - ,=Tc)2 - -,'? 6 1 _ .,_4.
= 0.67
Densi ty of air at ro tor entry PI
I
'-'J=.11.56
[ 141-'56l~ = (0.89)"" Too . 0.67 x 3 = 2.0 1 bar
=
" PI
width at exit R
[
P, " Too 1- (Poo)
P, Poo
u,
x '1N
257.98 mls 257.98/ cos 25' 284.65 m/ s 284.65' 12 1005 x 0.97 41.56 K
= = = =
Figure 6,7(b) Fro m
Cit
C II / casal C, eota t :;;: 120.3 l;ut25'
=
c,
u,
Crf 2 C,,(Too - T,,) Crf 2
=
'IN
71,17 %
60 Tf x 0.3 x 7200 60 = 113.1 m/s
+ 317.4 = 332.7 K
T he nozzle dlicil!ncy is givell hy
1.368( 188.5)'
=
-
317,4 K
120.3' ::! x 1005
/11
: , P2
:. 11.93
P2 A :!C2
P2 1 RT2 Poo / 2 = 3/2 317.4 K
= 1.5 bar
1.5 x 10'/287 x 317.4 1.65 kg/m' 1.65(Tf x Dh
X bh)
x 120.3
...c
269
266 ,.. TURBO MA C'IIIN~
R ADI AL FLowGASANOSTEM,TuRDlNF.S '" :!61
the true o r equivalr!nt blade velocity is 2V2 on account of counter rotation. Therefore. the actual blnde to gas relative velocity ralio should be
CJ'p p,
=
power de vclopcd.
ZV,
W) = cos fJ
Solution
This is sa me as in the fifly percent rc
+ V2 C.'·t
1
,
,
...
(6.Z4)
- "2(Cj - Ci)
From the. velocity trianlge.s at the exits of the first and second rings .
,
e', e',
,
VI + lV I -
2VI WI cosfJ and
eX]
WI cos fJ -
vf + Wl- 2V2 W) cosp and C =
~
Substituting these relations in cqu . 6.24 and assuming VI get .
h, - h, = Z(4V, W, cos
2
2
r, b
Pool P2
IVi
fJ , e"
N "I
'IN
Flow coefficient (¢)
1T
x 0.5 x 7200 60
c, Figure 6.7(aJ
From Ihe inlet ve locity triangle
V,
C.q
-
\VX ]
C,(coterl - cOlfJ;)
V,
Therefore.
:. C,
,
,
., + Wj').
=
"2 W,
h,-h,
=
[(vl-vn -(wi -lVl)]jZ
188 .5
cot Cl' , - cot fJ
120.3 mls
3Uj - U2 - Wi
111 - " 2
7200 rpm 25" 0.97
e, U
:=
+ 4vf - Wi
1( 2+ 4Vi' - Wi., -
= =
2 + IV))
lV~ + (2Ull' - 2lV,(2U,) cos /J
(6.25) Flow coefficient
It should be nOled Ihal in inward flow radial turbines VI > V2 . So, the enthalpy drop (hi - h2) increases by an amount (Ur - Vf)/2 (equ. 6.6a). But, in outward flow radial turbines VI < U2. thereFore enthalpy drop (h, - h2) decreases by a quantity (Vr - Ui)/2 (equ. 6.25). Therefore, For the same size and speed, inward flow radial turbines develop higher power compared to the outward flow Iype.
A cantilever blade type IFR turbine receives air at Poo :; 3 bar, Too ;; 373 K. Other data for this turbine are rotor tip diameter:: 50 cm, rotor exit diameter:; 30 cm. speed:; 7200 rpm, rotor bladc width at entry = 3 em, air angle at rotor entry:: 10°, air angle at nozzle exit = 25 °, nozzle efficiency:: 97%, stage pressure ratio (Pool P2):; 2.0.TIIC radial velocity is constant and the swirl al the rotor exit is zero.
cp, =
Londing coe fficient
V"
=
1
+ t/I,
e,
120.3
V,
= 188.5
0 .638
COI~,
1 + 0.638 co'60' 1.368
SOLVED PROBLEMS Example 6.1
373 K 0.3 m 60", C r ;. = C,
= = = =
Too rio
(a) Flow coefficient and st~ge loading coefficient !rd,N
From the combined velocity triangles, at the exit of Ihe first ring
W~
\
:3 bar 0.5 m 0.03 m 1.0
[JIM)
V1 and WI ~ W), we
.,
fJ - 3V, - V, -
I
V, =~= = 188.5 mls
v,
WJ cos,8 - V2
X2
I
Determint.: (al the flow and loading coefficients. (b) the degree o f reaction and !Io,uge efficiency. (c) Ihe air angle and width at the rotor exit. and (d) the milSS flow rale :lIu.I
(b) Degree of reaction and stage efficiency R
= = =
1- t/I , CO'/J,
2 1 - 0.638 CO' 60' 2
3 1.58%
RADIAL FLOW GAS AND STEAM TURBINES '" 265
VELOCITY TRIANGLES AND STAGE WORK
From thc ve loci ty (riungles.
The vc lOl.:ity triangles ror th e first two s tages arc shown in Fi g, 6,6 , Various veloc iti es at the in le t of th e lirsl stagr.:. are designated by Ih t! suffix o.
\VIm =
UleW I cos
fJ -
Uj}
+ U:!(WJ cosfi
- V2)
Assuming VI :::::: V2 and 'WI ~ lV). th en
= =
W 1111
2U2(\V) cos
fJ -
UJ,)
2(V::'!\V3 cos fJ - Vi)
M ult ip lying a nd dividing by W). we get
,(u,
=
Secolld bloldc rillg
2W ]
;;:;;
u;)
- cosfJ - \V) W2
(6.111
1
2Wf(Grcosp-a})
where G r is the ratio o f the peri pheral veloci ty of blades to the relati ve velocity at !.he and is assumed to be constant i.e,
e~il
err ;;:;; First blade ring
For maximum work.
aw
o
aa,
o.
cos fJ - 2ar (or) Figure 6.(j
\elodty triaflglesJor th e Ljllllgsll"Om turbiill' staGeS
The n: lative vc lnci ly at !he exit of the first stage is WI whic h along wi lh Ihe reripheral n: loc il Y (/ 1 gives the absolute ve locity C I, Th e relative velocity \V2 tt t thl! l!ntry of the sccond siage is ohwi nl!d by subtrac tin g V I rrom Ct . Thus the t.!Xil vd oci lY·lriangle nf lhe li rst s wge anJ lhe cnlry veloc ity lriangle of the seco nd stage arc s upe rim posed nn each ol her. T he same vel oc it y triang le appli es to the roll owi ng stages. Th e rcia(ive veloc i(y ilt the ex it from th e second slagc is W.l and at (he entry lU the third stnge is 11'4 . The comm on ahsnlute velocity atlhis sec tion is C2. The air ang les (13 ) or all stages arc assumed 10 he the sam c. The pcripheral veloci ti es orthe two rows at a given sec ti o n are also olssumed to he equal.
cosfJ 2
=-TIlis is the opt imu m value of a r fo r the maximum work, Then WI cos
p ;;:;; 2V,
lV.l cos
13
1
Stage Work he consit.lcred fu r the purpose of anal ysis . For the second s tag\.!, the work d~cfkg is given by
-
(620)
=
2U2
Tht.! maximum work is obtain ed by substituting the val ue o f a rapI in equation (6,21) Wnllu/m
Thl! firs t h lade row dOl!s nOl form a ge ne ral stage. Therefore. a ny OI her stage. cnn
(6.221
,
,
= 2" Wj cos- fJ
But IV) cos
fJ
=
1U,
(6.13)
Therefore .
lV"wx l nr
2uI
The above Iwo equations show Ihat Ihl! ou tward flow counter rotating radial stages Sill ce
c. , ~
is in Ihe oppos ite dirct.; li l1n.
hl!havl! like an impul se stnge of the axiallype. Bu t th is impression is not true , Because.
R" U'''Lr-1.0WGA~A:-'I)SII;A~ITt : IWI :'\J:S
....:
~lJl
262 ;:. TIJRllO MAC"IIINES
:. R
=
R
=
C'
1_ -_._ "2UICx, _ C.li 2U,
1_
"'2I
(6. 17)
From the inlet veloci ty triangle Fig. 6.3(0)
CJ
1
= VI + Wx,
AN OUTWARD FLOW RADIAL TURBINE (LJUNGSTROM TURBINE)
= UI +C"cot/1 r
. R= .
1_[U'+C"COlfJ'] 2UI
= I _ [ I+
Ljungstrorn turbine is an outward lIow rad ia l turbine (Fig. fi.S(aJ) in whkh the llull! enters the turbine in the axial di rection and Icaves radia ll y a l the cx.it. The areilllf 11m,,' increases towards the exit which accommoJales the expanding ,gas. The mnsl L:or\1 · man ly emp loyed LjungsLro m turbine consists of rin gs of canliiL'ver h lades OIuunh:d on 1\"'0 discs mlaling in oppos itc d irect ions . The cou nle r rlllatlng hladc ring!ot uf a Ljungstrorn lu rbim: is shown in Fig . 6.5(b). Each row of blatl e!> rorm.... a slagc.
(6.18)
"'~col II I ] Figure 6.J(bJ
R=
The h:.M-dcgn:c reaclion (R =O .5) has Ihe ad vnnwge of having" const'm t va lu c of the reac lian at all values of the Claw "'-llcffic icnt. The degree of reaction increases with Iht.: increasing va lues o f Iht.: inlet hlade a n ~ 1e ({fd. Thus for a given va lue of the Ilow coefficient (q,Jl ilc forward curved vane s give a higher dcgrcc of reaction compared to the backward c urved V;.Jncs . The degree of react ion of lh e backward curved vanes decreases with increa!otlng values of Ihe now coeflicicnt. The conve rse is true for the forward curved V;Llle~ .
HI ~"'ICOIII']
When PI = 90 L'(ang le measured fro m the plane or rOlation) i.e .• fo r radial vanes . cot 90°
:. R Consider the expression (fi. 17)
R= I
=
0 I
(6.19)
_.)1C ;'1n
inici
2
_ ex, 2U,
When R = 0 i.e. for an impul se stage C.r , = 2UI and hence til, = 2 a nd R = 1/2 i.c. for the firty pe r ce nt reaction swgc C.I· , = VI and therefore. "', = I Fig. 6 .4 s hows the varialio n of the degree of reaction (at va rious va lues of the flow coefficient) wi th the rOiar blade in le t ang le. 1.0
1..1
Figure
6. ~(a)
All olltwarrlflol\" ratlial lI. d)ill(' tLjllllgS//'OIll Iu/,hllll'}
1.2 1.1
I.n
11'1 lUI
tl .7
IInckwnnl cuned
R 0.(,
L--B"~-~OL.'~:;~~t"~~==---i t "'I " n.' Fnrward culVcd vaoes
D.l n. 0.2 04
0.12!;;0::; ·]0:0-.I!;;O-"?;O~60".-i7"O-.,f.:o-";!;O;--;JIO"O'"\;1O;;1';20;;I;';];;O~,,;;;o7,,"o~r6D ~,. t.!cgtct;'..\ -
Figure 6.4
V(lr;{I(iof/ of deRrce n! n:aclion with rolor blade inlet ungle.,
Figure 6.5(b)
CV/II/tcr I(}Wlillg
billtit
rings o! LjllllgJtnJlI/
/llr/Jillr
RAll l",!. FLOW GAS AND SnA il I TUiUlINt:S ...;
Th l! 10I
Thc energy change may be relnled 10 kinetic energy where the assocmted vduel!) term is known as Ihe Spoflcing velocirydesignulcd as CS. 111U5 rour spoullng veh)cll l e~ . w ith and wi thout a dilTuser and ror 100al-tn-tota l o r tOlal-lU- slalic cunJ illUn:. may he de fined as rollow3
1100 - 11m
\V
1I'.f
" 041
U,'CI
1ClI
110'1,.
+
TIll! appropriate definition would be used depcnding upon the dfil:lenq bcin~ detcrminl!d . Fo r isentropic flow across the (urbine.
16.121
\VIm
= U~ = C}/2
or
, The isc ntro pi l: work bl!twccn th e tow I condi tions at the e ntry and stati c conditi o ns at th e ex it o r th e stage is given by
IV,
1100 - h'1u = CJI{ Too - 72.. >
C"Too
[I -(:':,f']
(6.lJ)
In prac tice. U I / Cslies in the range 0 .68 to 0 .7. Thi s blndl! to gas specd ral io is dcnlJlcd as CIs. '·
DEGREE OF REACTION Degree o r reaction ror a radial flow turbine is defined by
Th !.: IPlal-to-stati c d lic ic ncy is gi vcn hy
'/, -.
IV
" 00 - 11 0'1
\VI-
" 00
SIalic e nthalpy drop in the rotor
R = e.::==::-::::;;::';:::-j;;::-7~"'~= Stagnation en thJlpy drop in tht; singe
"2"
Ui(1 +rJ> , COI /!,) C,JoO[I- ( ~J;l ]
""ui - C,'ToO [1 - (::of] _
II I (6 . (4)
-
112
1100 ho) hI - 112 "01 -
16.151
"0'1
I , , (hoI - h02) - - (Ci - C;)
SPOUTING VELOCITY
2
-
The ise ntropic wl nl t.:nthalpy dro p in the turbine is give n by
I(h oo
h.,,, )
I
when the diffu ser is not till ed . T he e xpressio n becomes (11 00 - IIOJ.. ) when the diffuser is filled .
CO I/(/itin/1.\" with difTu scr
wilh out din'user
Tmlll-to·wwl
C'
-i(
=
(/1{)( 1 -
=
TOfOI-w·stafic
h U.l ,, )
c} 2
C'
C',
2
2
~ .:;:;; ("no - 11 0'1,, )
I
= ( l fN J -
I IJ .. )
(0 . 10)
Ifr th e whirl velocity C X2 is zt:ro at exit and Cz Ass l,lming co nstnnt radia l velocity C r! = C r ] ,
,
"}
,
"}
:. Ci - Ci = Ci - C;1 ;; Cr -
,
= Cf'! I C,.. - _ - [
C'
;]
= C.~1 Figllu 6.3(a)
251\
RADl /\L FI.ow GAS AND STE.;\.'" TUIlBINES
,. T I)IUJQ MACHINES
....;
251)
Total-Io-tota l isentropic efficiency is give n by h
Poo
00
______ 91
PO I
P-
o
1/11
0
=
(Too
To::!,,)
'1" is in the range of 80-90 per cent. TIle losses in::tn IFR tu rbine can be expressed in tl! nns of the noa h: .lntl rntor Ius!> codficicn ts as defined for the .udal stilges. From tile II - s tliagmm, these coefficien ts arc
C;12
II, P,
"I.l
Ci /2
I, and
P,
02, 02$5 OJss
P,
LR =
35 3s
2
2$5
The i.h.:tua l work ou tput of the stage is IV
Mulli,'r clla r' for r'X[WII.firlll ill a 90" ill ward flow radial go.t IIIrbit/l'
m
Eq uatin g the two ex.pressions 1 hOI
"2.1
W£/2
STAGE EFFICIENCIES
2,
FigllTt! 6.J
"2 -
- h01
= UIC.r1 -
(6.6 )
U2 C ,(! I
=
"00 - hU2
=
U,C,fI = U 1 (U,
=
Vi(l+
= "01 - 1101
+ H'q)
From the velocily Iriang le (Fig. 6.1)
It is alrea~y show n that the quan tity 'I' for a cen trifugal comp ressor is given by
sU,
1= "O'd - V2/2 or
where T ;s Therefore.
j]
cons tant
BU I
I(U2 ", - 11.,= I -
V;) - (lV2 - IV;) I -
(Vi, -
2
1V,'- c -I- ';) 2
In the t1i1Tu ser li02 =
fl03.
I
-
C.I !
V,
(fi.6a)
:.
"',
C" cO lfJ l 1+ --
V,
V', nnd -C" V,
=
1>1
1 + ¢ICOlfi,
Then HI/III = 1/IIU~ For a perft.:ct gas (6.7) \\1 / m
T hus.
= CI'(To l -
To~) = U~( I
+ 1/1, cal fJ ,) = ""U I'!
I fi. 10,
Tht.: ise nt ropic work ou tput bc[wt.:en the total co nd iti ons at the entry and c,;i l o f thc stagt.: is (6.8) IV,
If thc losses.in the diffuser
IIfC
ncgl cctcd. then
If>.l I I TOJ ..
=
T02H
;md
~5b
,.
Tl lltliO M ACIIl NI: 5
RADI,\!, FI.OW GASANIl
/~ I\l V~:---
~pucc)
'. (3)
C
r,
r 2 u\terage
C,
~
ROIo, blades
I
Rotor in let
C,~w, ~ u,
f 1
Figure Ii.J(b)
Radial i,,}luw wrbine
ahout 70" , hUlthe vanes can be pivoled to ... lI ow for adju stm enl of the now angle as the load l: hunges. In some turbines, there may be no vanes:.1t all. but a pas::iUgc similar !(l thaI or the vaneless Jiffuscr (di scussed al rcudy ) is fitted . A vane less space ex.isls between the nU 11e1 tip (I f lhe vunes and the rotor. This spact.! is being ut ili sed by the gas ror futher flow aJju stmcnt and hdping in thi: reduction of vibfi.ttory disturban ces
I
Figure 6.1
hncharging , aircrafl and missile auxi li ary dri ves, cryogenics. and gas liquefactinn . Francis (Inwnrd flo w radial) turbine has bee n in use since a long time for hydrormwr gcncr'l1ion. Thl.! Ljungstrom (o u{ward flow radial) turbine is used ill SWam pOWI.!!" generati on. The nUor which is usually manufactured with cost nickel alloy, has hl ades thaI arc curved to change th e now from the radia l (0 thl! axial dirl!Clion. The shrouding for the bludcs is formed hy the casing , and a diffu ser can be filled at the outlet to further fl.-duc c thl' high kin c{i c: energy J{ thut point and thereby incrcJse the I.!l1th.dpy drop across Ihl! nUor.
Velocity friclfIclerJor un inward flow rudinl furb/lll'
and for radial relative velocity at ent ry i.e. for a 90': IFR !Orbi"!.!. fJl = QO" W/m = Vf Isince C.rl = VII
_ t/ll = w_o_rk_d:co,,'-='C,-I-=k~g
From
Ihc
Vl.!locity triangle. CXI = C" :. 1'"
1'01'
thl! 90 0 inward now radinl gas turbines arc shown in Fig.
V,
COl at
= Crl
COl at
V,
=
where ¢t is known as thejloll' coefficiem. For a 90° IF R Iurbinc. V, :::;
(6.2) ell '
Therefore.
V' t/I, = -.l V'I THE THERMODYNAMICS OF FLOW
I ~~
rde rs Ihl.! rotor .and '0 ' subsc ript indi c:llcs [he point of enlry to Ihe nnll.1e V; IIll." ,Ifld secti on 1-3 indicates the diffuSl.!f ou tlet sl.!clion. . \\'l, rk dlllle per unit illass now in the rot or is given hy Euler's turbine equi.ltion Sec tion
C'l l
V'I
VELOCITY DIAGRAMS (1 . 2 .
(6. 1)
The head or siage loading coefficient is defined by
wilhin Ih i: turhine. So me of the ct)mmOn applica{i ons of {be radial turbines nn! in the fields of {ur-
'1IlL.' vei OL.'il y [riangles
'" 257
SC:(;lion UI nu.lius .
I~~
" - - ' ' ' ' Guide "anes ( 2) 'fir==;\<- (or v:me]css
SH .... M Tt IRIHNt;S
If v..'hid vel oc ily is ze ro
The IhCnllOdynamic path followed by the gas is shown on the. Mollier ch:.irI (Fig . 6..3). In the nonie, no work is done. Therefore, hoo = hOI. although Ihc 100ai pressure drops from POo 10 POt because of irrevcrsibilities. Thus.
TIle work done is given by W/111
= hOI -
ho:!
(6.5)
l
6 RADIAL FLOW GAS AND STEAM TURBINES
INTRODUCTION The inward 110w rmkd gas turbines arc usr.:d for applk:llions where the nnw (;lIe IS
vcry low. [or cx.lInplc, (Urbochargcrs for com mere in I (diesel) engines :lnd fire pump ... The radial gas tu rbines nrc vcry compact. the maximum diameter heing nhnut 0 .2 tTl and speeds nrc hi gh. ranging from 40.000 to 1.80.000 rpm. 'nley arc usually of the 901.> type, the bladl!s being J.. r lO the tangent at the periphery of the rotor (lulcr IOlel. The g.IS cmcrs the turbines in the radiol direction and Icaves axially at tht: outlet.
DESCRIPTION A YODinward flow radial turbine is very sim il ar to the centrirugal compressor and thl' only tlilTcrcm:e being Ilwl the gas now is in the oppusite dirc.!t.:liOIi (Fig:-.. 6.I(a) anJ 6.1 (b) . The gas cnlt!rs the sr.:roJl casing, whose cross-sectionJ! :.Ire:! is dct.:re:'lsing ;t~ the gas passes through il. This keeps the veiocilY al the entry 10 the 11I1/.t.le Vllnc!constant :.IS Ihe gas is gradually drawn off on its cin.:um ferellli •.d palh . The nOI.l.h: valles arc cOllverging to incrc;:tsc the kinetic t!ncrgy of lht! gas and tht!Y SL:I the ga:-. anglc fo r entry illlo the rotor. This
Figflre 6.I(a)
IIHl'IIl'dJlolI' radial (IFRJ fIIf'h/l1t'
A XIAL FLOW ST ~AM ANOGAS TuRBlNES .... 153 152 ~ TURBO MACHI NI;S
[he hub . mean and tip sections of the bl ades (a) the blnde angles, (b) degrec of powt!r aUipul of the s tage, (b) the l:ipcl:ific en th alpy drop in th e l:itage in kJ/ kg a nd (c ) the percentage inc rease in relative velocity.
IAns: (a) 11.33 mW, 5.~5 .
(h) 79.98 kJ/kg ond
reac ti o n (c) blade to gas speed ratio. 0 IAns: mean sec tion P I = 60
(e) 50.89%J
An 3.'(inl now gas turbine hn s adcgrecofrc3ction of30% . The hlrldc speed at lhe menn diameter is 300 mrs rind the mass Dow is 2.5 kg/s. The gas ternperature at th l! turbin e: inkt and o utl e t a rc 5000 and :WODC res pec ti ve ly. The fix ed blad e ou tll'l ang k is 101l Ill cus urcd ill lhe same direc tion o f blndl.! spl.!cd . The axia l vdoci ty remains constant :1I 100 mps. Draw veloci ty diagram and determ ine th e relativc veloc ities and power developed. Take C I' 1.005 kJ / kg K. IMKU-Nav. '96)
5.41. An axial flow gas lurbine has rOOI a nd tip diameters o f 600 mm and 75.0 ~m . Rotor speed is 7500 rpm. Rotor bladl! design is based on free vo rtex pnnclpll! ,md absolute ve locity is axial at ex it. 1lle actual change in totalte.mpcra~ure at m(!a n sec ti on is I IODC. Calculate the ai r and blade angles. reaction rall o and blade load in g coefficient al all sectio ns. Take R = 347.2 J/ kg-k and r = 1.3.
(b) 502.5 kWJ
5.36. Two rows o f a velocity compounded impulse turbine hnve a menn blade speed o f 150 m/s and a nm.zle ve locity of 675 mls. The nozzle angle is 200 • The exit a ngles of the firs t movi ng row, fixed a nd second row of moving blades nrc 25° . :!5° and 30° respt!c li vely. T here is a 10% loss of velocity due 10 fric ti on in all blades . The steam flow is 4.5 kg/s. Omw the velocity diagram and delermine
IAns: - 4.9%J 5.39 . What is the percentage of reacti on of a si ngl e stnge turbine opcr'Hing with a tnlal pressure ratio of 3 tlild top temperature of 11 00 K, if the velocity oul of the s tator blade is 550 m/s. and ve loc ity out o f the turb ine is 300 m/s. (MKU- .'Ipril '97)
IAns: 0.64 1
.....
5.40. The blad es of a fret: vortex turbine rotor have in let and out lel an gles of 60 and 65 degrees at a mean diam eter of 100 e m . Tht! correspondin g nozzle a ngle is 7()". The hub 10 tip rati o is 0.6 and the turbine runs al 3600 rpm . Cal c ulate for
=
R
=
(a) th e power outpul and (b) the diagmm efficiency. IA ns: (D) 796.5 kW and (b) 77.7% ) ) .:n. The nozzle ang le in a veloci ty com po unded im pulse turbine is 200 to the mea n b hldc speed wh ic h is 100 m/s fo r eac h movi ng hlades row. The re a rc two rows o f moving bln des wilh c.x it a ng les 26 r• a nd 30°. Between these rows there is n row of fi xed hll.ldes wilh exit nns lc 2W' . The relative ve locity o f stea m drops by 10% during passage through ring o f blades and the final di sc harge is axial . Determ ine the vcJm:ity of steam leavi ng the nozzle s a nd the blade efficiency. If Iht! noa.lc effic ie ncy is 95 % a nd the kinetic e nergy o f s tea m leaving the s tagc is availahle as he al l'nl'rgy. estimate th e state o f steam e nterin g the ne xt stnge. T he condition s o f the steam at nozzle inlet is 8.5 bar and dry sa turated. [Ans: (,) 450 mls (b) 61 % lind (e) 5 bar, X = 0.96J 5.38. In a n impul sc turbi ne desi gned for free vortex flow at th e rotor inlet. the blad e root radius is 25 cm and the blude he ight is 6.3 cm . TIle absolute stenm ve loc ity a t the rotor in let is 450 m/s. the Ouid being directed so us to make an a ngle o f 15° with the whecltangent al the hlade root. If th e speed ratio is 0.4 at Ih e tip. the blade velocity coefficien t is 0.97 and th e difference between th e rotor in let ,md ou tlet ang les is 3" all over the rotor, draw the velocity triangles for Ihe blade ti p and find lhe degree o f reac tion al the sam e pos it ion. IMKU- Mm' ' 97)
=
hu b scc lion f3t R tip section fi t
e
IAns: In) 348 and 472 mis,
fh = 65'" at = 70" 20.4% ond a = 0 .35, 0 71.0 0 • f32 66.3 • a l ;::: 74.8" 41 .4% and a = 0 .2. 0 43.1 " . Ih 65.4". al ;: 65 .6 49.2% and a = 0.54 1
R
lAns: t:rl.m = 72.25° , 1/11 = 2.3 6
R~ 1Il
fhm ;::: 60.93° , fhm = 52.96:. Rm = -17.8 8% "u = 74 . 11° , fll.h = 66,83 , fl'.h = 49.67 , 0
= -49.17% 1/I1.h = 2.98
70.4r, = 4.55%
fi1.1
al,I;:::
Il", = 55,8 2° , R,
= 53.24 "' .
.pl., = 1.911
5.42. A gas turbine slage has an initial absolute pressure of.350 kPa and It:mpc~a.lurc 5650C a nd negli gib le initial ve loc ity. At the mean radlus O. 3~ m , the cond.'tlon: ure nozzle exit absolute static pressure is 207 kPa. nozzle exil now a~glc.:: IS 68 a nd stage reac tion is 0.2. Determine the Dow coefficic!nt. sto.g e loadlOg fact.(Jr~ siage reaction and air and blade ungles at the mean . hub and up. The hub ra~.lu!t and speed arc respectively 0.31 m a nd 8000 rpm . Ass ume that ,Ihe slage IS to ha ... e a free vorte x swirl at thi s speed . Take 1. 148 kJ/kg-K and r:: I.:n . (MKU-Nov. '96) Neglect the losses . IAtU-Oct. '96)
C,. =
I
[Ans:
I
0'1.111
,
I
= 68°, = 0.602,
0'2.111
== -10.3° ,
fJl.m ;: 39.2° .
fi2 .".:::
VtI .III ;: 1.38 Ctt.h Rnr = -0.2 = _ 11.9° , flu. = 55 .3 1", fl, ." = 50 .63" ,p" au = 65 .29°. a:!.1 = - 0,079 ,"'- h = 1.7 6 ,p, = 0.528, R, fl,., = 59.9° .pI.<
55 .!}:\"
= 70.H2 · . = 0.7, = -9. 1' . = 0 .38 = 1.06[
5A3 , The data for an axial turbine SLagc are : hub diameter:: 45 0 mm , tip d iametl.!r :: 750 mm, rotor speed :;: 600 rpm . At the mean section, a, = fh = 75 =. fJl = a, = O. R = 0 .5. Assumin g radial equlibrium a nd constant nou.le angh:. ' d~termine for the hub, mea n and tip sections (a) the absol ulC and rc1a~i\'e a~r anglc.::s (b) blade-Io-gus speed ratio (c) degree of reacti on. Assume ax tal eXit from the stngt: at all sections . IAns: al,l! ;::: 75'" fJu. = 57.87" Q' l,m
au
75° 0 75
Pl.m
th.1
0° - 63.58'
Ih.h = 65
Ih..m = fJ2.1
D ,
75' ,
= 80. 13
0 •
0 .554 , 0.218 0.966, 0.5 1.487, 0 .6751
250
r
TUlmo M ACHI NES
5.8. Prove Ihat " 01, reI
= h02, r eI. across the turbine rotor.
5.9. Define (a) tOlal-(o· IOI:II ~mciency
(b) tota l- to-sialic e fficiency.
5.10. Define (a) Nozzle loss coeffi cient and (b) Rotor loss coeffic ienl
5 . 11 . Define hlnde loading coefficient? What is iI's sig nificance ?
5 . 12. Prove that 1ft = Q'I(tan li1 + Inll {iJ.) 5 . 13. Why arc simple impulse turbiocs no t so common? 5 . 14. Wiull is compounding or slaging? 5 . 15. E.'plain bri e ny a IWO stage pressure compounded impul se lurbine and show the pressure and velocity variati ons across Ihe tu rbine . 5 . 16. What is a velocilY compounded turbine? Draw a Iwo s tage c urti s turbine indicating the pressure and velocity vnriations across it. 5 . 17. Draw the inlet a nd outlet velocity triangles of a two sHlge impul se turbine. 5. J 8. De fine blade velocity coefficienL. How does it vary between an impulse a nd reacti o n turbine. 5 . 19. E xplai n with a neat diagram tllc operations of a reaction turbine. 5 .20. Define: Reaction ratio for a n axial now turbine stage . Compare the d~g ree of reac tion for ax ial now compressor and turbine . 5.21 . De ri ve the following rela tions.
1
R = '2
(a)
(b)
R=
(e) R =
~
2
+
tan (J\)
CII(lan Ih
-
2U
lan O"t>
1+ CII( lana2 - Ianad 2U
5.22. Draw Ihe velocity trinn gles and h-s dii.lgrnm for the following D.xinl now turbine stnges. (a)
R
(e)
R
=0 = 0.5
(b) R < 0 (u)R = 1.0 (e) R > 1.0. 5.23. Prove that for an impUl se turhine
=
(a) a s in 0" 1/2 (b) '7""r.f .din sin:! 5 .24. Prove thaL for it reaction IUrbine (a)
=
a,.
a = sin 0"1 2 sin 1 Q'I
(b) 'Imn.f·J
= I +sin 2 ar
5.25. Prove the following ,Jt 'Is x R.F. 5.26. Expla in the various internilll os~cs in axiul How turbines. 5.27. What is turbine governing? What a rc the different methods of governing of
=
lurbine ? Exphtin hricny.
5.2S. A single whed impulse stea m lurbine has cqui.angular rotor bhldcs thaI den:!. II' 3 .75 kW and produce a torque in the disc o f 1.61 Nm at a menn radiUS IIf 132.5 mill . The rotor receives 0.01 4 kg/s ohtcam fro m nozzles inclined al 711 to the a:l( ial direction and s tea m disch nTgcs fro m th e whed chamber in an aX ial di rection. Find (
[Ans: (a) 60.7
(b) 0.62
(e) 2.Q.l N and (d) IU) NJ
5.29. A 50 percen t rcacli on steam turbine. runningal 450 rpm dcvclops5 MW and h;J ~ a stl!am 'mass flow rale of 6.5 kglkW.hr. At a panicular stage in the expnnsion the 'Ibsulutl! pressure is 85 kPa 011 ;J steam dry ness fr
=
r" by
R=1-(;)2 5.32. Steam with a velocity of 600 mfs e nle rs an impul se turbine row of bladt:s ill an a ngle of 2SO 10 the plane of roLation of the bladcs . Tht: mt:an blade spccd is 255 m/s. The exi t ang le from the bl ades is 30Q , There is a 10% loss in n:1~ti\'c velocity du e 10 friction in the blades. Determine (~) the enlry nnglc oflhe bl~dcs measured from th e ax ial direction (b) the work done per kg ofste.un/sec (e) tht: diagram efficiency (ll) the end thrust per kg o f sleam/sec. [Ans : (a) 4B.5 ° (b) 150.71 kW/(kg/scc) (e) 83.7% nnd (d) 7B N/(kgls)J 5.33. The n07.7.1 e5 of a si m ple impul se IUrbinc arc inclined at an angle of 20° 10 Ihe di rec ti on o f the path o f the moving blades and the steam leave s th c nOlzle~
-
~ ·lK
:;;.. TtJlUlO M "C HI NF.S
n. XI AI.FLOWSTE AMANOGASTuRIIlNES ... 149
"I -
5 .9 . In the rotor of the axial now turbine (iI) II). = co nsta nt. (b)
\V~ "1 - -= /" 2
(e) h,
cl
+ -2
= It.,
-
\V~
-
2
C;
-+----=. 2
5. 10. For il nonnul s ta ge, Ihe siali c Icmpcralure drop across Ihe stage equals the lotal tcmpcralurc drop. ITruclFalseJ 5. 11 . I)(!finc Turhine s tnge 100al-to-Iotal isentropic efficiency. 5. 12 . When is Ihe tnlal·'o-to lal !.!ffil!.!ncy used? S.ll . Odin!.! Blad!.! loading. cneflicicnt. 5. 14. The sil.e of a sta li o n;:lry induslriallurbine is larger than thut of the aircrnft gas IUrhine . [Truc!Fal s!.!}
5. 15 . Ratcau lurhine is an examp le of
5. 10.
5. J7 . 5. Itt 5. 19. 5.20. 5 .21 . 5.12.
5.2l . 5.2-1 . 5. ::!5 . 5.26. 5.27 . 5 .28. 5.21). S.lO. 5.3 I. 5.~ 2 .
-
(a) .Two stage velocity compounded impulse turbine. (b) Simple impul se turbine (c) Two stage pressure compounded impulse turbine. In press ure compounded impul se turbine, the blade height has to be increased IOw
5.33 . Gas turbines Il!nd to be always of the reaction type. Why? 5.34 . Steam turbines are us ually impulse (orJ a mi xture of impulse and rcacuon slag e~ bcc uase of (a ) High pressure ratio. (b) Low overall pressure ratio . (c) Type of working Ouid. 5.35 . The maximum diagram efficiency of an impulse turbine is (a) cot 2 a] (b) sin 2 a, (c) sinO'] 5 .36. The blade to gas speed ratio for maximum diagram efficiency is (a) s in a] (b) 2 sin (e) sin "I /2 5.37. The blade to gas speed ratio for maximum diagrnm efficiency of a reaction turbint: is (a) sin Ci] (b) s in "I /2 (e) 2 s in 0' 1 5 .38. The max.imum stage efficiency of a reacuo n turbine is (n) I + sin 2 Ct'] (b) 1 + sin2a] / 2sin2a, (c) lsin2al / l +sin2al 5.39. The stage dflciency, turbine efficiency and reheat factor arc rdated as (a) 'It = '7$ x R. F (b)'7J =R.Fx'71 (c) R .F=1J, x1J$ 5.40. Annulus flow area interms of hub diameter and blade height is
a,
(0) rr(Dh - ")" (b) rr"(D" +") (e ) rr( Dh + Ii)'
5.41 . What is Govemipg of turbines? How is it done?
EXERCISES 5. 1. Describe (he working principle of an axial flow turbine with a ncat sketch. 5.2. What is an axial How turbine stage? 5 .3. Draw the inlet and DuLlet velocity triangles for an axial flow turbine stage. 5.4. Prove that tbe axial flow turbine siage work \V 1m
= Vel/( Ian PI + Lan th)
5.5. Define Diagram efficiency. What arc the other names of it? 5.6. Draw the h-s diagram for expansion through an axial flow turbine stage.
5.7. The work done factor is not used in axial flow turbine but it is used in llxiul compresso r why?
AXIA.l- FLOW S 'll ; A~1 ANOG"sTUIUJItoil.: S ... 2-17
246 :;:. TORUO MA CIlfNES
(a) Absolute and relative air angles
(a) Absolute and relative air angles Nonie c}l.i t "'ng le, O' Lm
= 0'1./. =
Nozzle exit
75 °.
lun f31.1
tan fiLm IT DmN
Vm
= =
fILm
=
rr
X
(2
X
60 188.49 m/s 2 16. 16 - 188.49 57.92 25.54'
0 .3) 60
X
v,
6000
=
60"
=
#.2.111
=
V" Cul. m
188.49 = 57.92
=
=
78.7 1'
235.62 181.7 1 1.297
188.49 223.78 ·
(c) Degree of reaction
0.842 Rf
= =
Cn I ,'" (Inn (h.Ill - tan f31.m) 2Vm 57.92 x (Inn 72.92' - Ian 25.54) 2 x 188.49 0.427
=
V,
Ct.1
Vm
135.62 47.03
V,
- - = --
Cal,/
(b) Blade-to-gas Speed ratio
(c) Degree of reaction =
=
11,., =
72.92'
Cl.m
rrxO.75x6000 60 = 235.62 m/ s
Since Cx:! = 0
a= -
=
a
=
175 .52 - 235.62 47.03 _ 51.96°
=
(b) Blade-ta-gas speed ratio
=
CII I.,(lanfh, -tan{JI,I)
2V, 47.03 x (Inn78.7 1 -lan(-51.96» 2 x 235.62 0.627
SHORT QUESTIONS
f
5. 1. What is un axia l fl ow steam (or) gas turbine? 5.2. What arc the advantages of an axial now turbine? 5.3. Why is ... ing le or two stage axial flo w compressor preferrc.'CI in ai reral"l jet
(3) Tip Section
Cu
propul sion?
Cl.m
C7 r'''· =
0.3 ( 0.375 el. m
:. CLI
C.[ 1.1
CuI. r
lun f32.l tanfhm
= 75 °
Cx 1.r -V,
rr D,N
=
Par un
R,.,
tl'l.h
=
Cx
Lm
)0.933
x 0.812 x 0.812
CrlL/fI X 0.812
= 0.8 12 223.78 x 0.812
=
181.71 m/s
2 16. 16 x 0.8 12
175.52 m/s
57.92 x 0.812
47.03 m/s
5.4. The blade height is increased towards the low pressure end why? 5.5. Draw the velocity triang les for an axial flow turbine stage? 5.6. The d iagram work per unit mass fl ow is given by (
c.,
(e) U(ln n /I I
+ Inn /I,).
5.7. Define Diagram or blade efficiency (or) Utilisation factor. S.S. Draw (he Mo lli e r diagram for expansion throug h an axial Il ow turbine siage .
:!..J·l
AXIAl-FLOW STEAM ANOGASTulWINES ...{ 245
;.. TI IROO M .-'ICI II NES
(a) Absolute and relative air angles
Gas tcmperature.
C2
T,
TUI -
=
Nozzle exil ang le, aUI
- '-
2C / 74i!
973
2 x 1155.6
= 734.78 K
= 75° =
CI,h'
=
75 .76 m/s
C,r1 . 1J
=
Clh
=
282.74 m/s
(c) At the tip
+"
+ 0.053 =
D, = Do. = 0.53 Gas di schargL.! ang le .
I'm
-
Ia n Ci 1.1
0.583 m
tan f3u
I)
0.583 59.68'
..
u - e-=
C,
cos Cit,l
and
320.27 cos 59.68°
= 634.4
fJ l.l1
CII,Jr-Uh
=
282.74 - 141.37 75.76 61.81'
Cn.h
=
Cn .h
(b) Blade to gas speed ratio
Example 5 .1 B An axia l turbine with conslan t nozzle air ang le (75 °) and zero reaclinn at the hub nlOs at 6000 rpm. It s hub and lip diameters nrc 45 and 75 cm respt:ct ivt!ly. All sections arc designed for maximum utilisation factor. Assuming
Uh Sinal sin 75° --=--=--
=
2
CI.' 0.483
2
(c) Degree of reaction
raJillll!4Uilihriutn co nditions. determine for tbl! hub. mean and tip sections (a) absolute anu relati ve air angle s (b) blade La gas speed ratio (c) degree orreac ti on. Assume axial c:xit from the sl,lgc al all sections.
R"
It) Mean Section
Solution 75~ .
h
--=
Ih.h = /31.11 130." = 61.81'
GlJpt
a, =
WeI.
=
For z.ero reaction sec tion,
mts
Cr 634.4 2 TO! - = 973 - ,--"-,.=cc 2C" 2 X 1155 .6 798.86 K
T,
=
0
or a constant nozzle air angle,
R" = O.
N = 6000.
DII
= 0.45 m.
Df = 0.75
,
Cl.h CI.m
=
C.et .1!
Cal.II
exl.'" = CIII.III
Hub speed.
u"
" x 0.45 x 6000 D"N 60 60 = 141.37 m/s
..
(0.45
IT
= 0 a nd fo r lniJximum utilisation factor GO[II
=
Cl.h
=
U" CUI
=
si ll al
2
141.37 U" = (sin 75" /2) (sina ,/21 292.71 mls
(rm)sm.a
III
I
(1) Hub Section
Si nce R,.
)( cos 75°
·sinal =292.71 x sin 75°
lunal .1II
"0.53
= 292.71
From inlet vt:loci ty triangle
- - Ian62
Ci I . 1
cosal
C (I.h
'm
('" r'''· rm
= =
C I •m
=
C x t.1II
=
Cnl. m
=
+ r, -'II
2
=
(~)0.933 0.225
=
l
rh
+ 0.75)
2
2
2
=03m .
_
- 1.308
292.71 = 223 .78 m/s 1.308 1.308 282.74 C;, I.h 216.16 m/s = 1.308 1.308 75.76 C'IJ./J = 57.92 m/s 1.308 1.308 CI.h
=
242
~
AXIAl. FI.OW ST1:AM AJ ~ D GAsT1 JJtllJr : ES '" 2·1)
T llnno M A("lIINE..";
Assuming thal th ere is no pressure drop in the movi nl:!. hlalk.(T... Mo llie r charI. . -
And
-II
Inn {h." - (an U2.h
U"
;
Inn fI2.h - ta n 54. 1 tan fh."
176.72 87.38 176.72 --+tnn54.1 87.38 3.4
;
;
;
..
To , - T"
C"
973 - 751.49
I
(aJ Degree of reaction
=
{tan Ih.h
r/J"
0.495
=
0. 11
c:!
T,
1(11 - - '-
2Cf, 771.6 K
T,
V'I.J, ~xample 5.17
,iI
pACu
R C"
-
= 1.6 bar
973 K
=
90 - 2H ; 62°
C,
A
;
Dill - II
751.49 K
(16)-
1.333
= 0.5J -
= O.477m
0.053
For rn:e vortex dl:sign C.I I.III · r", = C 1 U · r"
,;, = 20kg/ s
or
=
r,"
=
64.4 30
. tan 0' I.
rr1
Til
0.53 0.477
= - - liln 62'"
Gus veloci ty.
0.333
; 973 ~ 4.5
682.2c0562 ; 320.27 m/s
O.5J m
au ' (-PooP)'
;
; 0.289 kl /kg
pC,
Ii
(aJ At mid section Too
COS'"
In
20 x - 0.718 x 320.27 IOrr 0.053 m and
Om/Ii; 10
,_ ,
kg
; 0.7 18 )
n Dml! = n(lOIr)Jr z ,iI JOlT" = --
D",
Solution
'"
;
=
1)
r ;
105
2R9 x 77 1.6
C/1(r -
A
j(
;
;
1/ 2
H ub dinnH:tcr Dh
-
RT,
3. 15
The roll owing particulars relate to a single stage turbine or frce
1.6
P,
p
vonex. design. TOlal head inlet temperature - 973 K, total head in let pressure - 4.5 bar, staLic head 0 outlet pressure - 1.6 bar. gas now rate - 20 kg/s oNozzle oullet angle - 28 (measured rrom blade velocity). mean blade diam eter to blade height ratio - la, Nozzle loss coe fficient _ 0. 1. Determine the gas velocities. temperatures ilnd discharge angle al the blade mid, root and lip radii. Assume C" = 1.1 55 6 kg/kg·K and r = 1.333.
Too
2 x 1155 .6
Hub diamctcr is QClcrmined as follows
(,an 73.63° - 'an 71.33°) 2
= ,pf, (tan fIu. + tiln Ihh) ; 0.495('an 7 1.33° + 'an 73.63°) ::=
682.21
= 973 -
(bJ At the hub
87.38 ; 0.495
(bJ Blade loading coefficient
l/Iu,
682 .2 m/s
=
Gas tcm perature
!
- (an {JI.I,)
U 176.72
;
2 x 1155 .6 x (221.5 1)
e Lm
2
Co ;
;
- ' +01-' 2C,! . 2C,J
1. I Cr
73.63°
;
fi2.h
(TOI - Til + (T, - T,,) C2 C2
;
I
C,
;
C,
;
C"
320.27
cos Q' I./J
cos 64.43
74 2111/s
=
T I ,) from
A X IAL F1.0W STEAM AND GAS TURBINES "" 241
240 ,. T URBO MACH INI;S
Since
(b) Power developed I\'
=
Um
+ C.r 2)
ri,U(CX I
Cp(tana ,.". - tanfJ, .".)
v.,
+ 0)
lilV(U
tan CLI.n!
75 x (160)'
235.62 tan 75 ° - tan 46° 87.38m / s
=
1. 92 mW
(c) Isentropic Enthalpy drop
87.38
CII
6.11\ 6.11
=
"0 - "2
Flow coe fficient, ¢
110 - h2.f W/ m
= U = 235.62 = 0.37 1/1, = ¢(tunll, + tunp,) =
-= - -
'h 11J I. 92 x JOJ /75
The dilta illmcan radiu s of
tIn
0.37(tan 46°
1/1,
+ tan 75<)
1.76
At foot scc tion, for fre~ vortex flow,
0.85 30. I2kJ / kg
Example 5.16
tan fJl,m
-
axialturbinc s tag~ arc as follows.
ell . lanai ,,,.
RoltJr hl"de angles at cntry ;Jnd t!xit arc 46° nnd 75° rcspl!clivc ly. Nozzle ang le 31 exi l is 75 °. Tht! hub 10 lip ralio is 0.6. Ml!llll rOlor specd is 7,500 rpm . Assum ing frcc vortt!.'( flow cond it iolls dClermine at meiln ;.md rool si!ction (a) dcgrct! of reacti on and hl'hJI,! lo,ldin g coeffic ient Take hub diaml:ll!r ilS450 min .
. r".
ell '
lun au . . rh
'm
: . Ian a,.h
tana, .". -
't,
=
0.6/ 2 tan 75 x 0.45/2 o
4.976
Solution
CLI,
D" = 0.6
v"
D,
N = 7 . 500 rpl11
D" = 0.450m.
78.64'
II
=
IT
D"N = 60
IT
x 0.45' x 7500 = 176.72 m/s
60
A t mean section :
(a) Degree of reaction since 0' 1.111
tan ct1.h
Uh
-
-
C"
= {f2 .m = 75° Rill = 0.5 tan 78.64
(b) Blade loading coefficient
0 -
176.72 B7.JB
2.96
-
#1." =
V.,
lrDmN 60
Dill
011+ Dl 2
= .. VIII
0,45
=
Dli
+ DIi / 0.6 2
+ 0.45 / 0.6
Ian a2 ,h
=
=
1I.6 m IT x 0.6 x 7500
= = 235.62 m/s
0
Similarly,
2
60
71.33
',,,no
tana 2.".-
0.6/2 tan 46" - 0.45/2 54 . 10
= 1.381
23R ,.
AXIAL FLOW STEA M ANOGASTURDINES .... D9
T URBO MACHINES
,;,(8U 2) = 100
:. W
W
X
8
X
1578.08'
(a) Blade angle at inlet
19.74mW
(c) Final state of steam The expansion process is shown on h - s diagram
eL I
(Fig. 5.34).
tan {JI
=
= C 41 tanal
=
C.rl
-
U
C" 180 x 'an 60° = 311.77 mis 311.77 - 280
h
180
o
Po
(b) Blade angle at exit Since R
fh
4
= =
0.5 a l = 60'"
Example 5.15 The conditions of hot gas at the inh:t to a 50% rcact~on turbine with a stalle efficiency of 85% nrc 800 kPa and 900 K. The blade speed IS 160 ~/~ . The mass On ow rate is 75 kg/s and the absolute air ang le at first stage noale eXit IS 700. Assuming maximum utilization factor, Determine the rotor blade anglcs. Power
4S
dcvclo·pcd and isenlropic enthalpy drop across the s tage.
Figure 5.34
Solution ho - 114 ho - II.~l
19.74
X
10'
--=~'-
= 197.4 kJ/kg 100 197.4 - - = 303.69 kJ /kg 0.65 "0 - 197.4 3370 - 197.4 = 3172.6 kJ/kg
=
=
3370 - 303.69 = 3066.31 kJ/kg
= 27 bar,
T4
= 365°C and ~4
rrDCa
" x I x 228.69 0.0146 m
h=--
=
900 K
Po
= 0.85 160 mis
,n
800 kPa 75 kg/s
For maximum utilisation factor condition. bli.lde to gas speed ratio is
U
= -
GUfll
"
~,
WI=C.~
.
= smcl"J
(I
U=Cx
= - U- = -160- = 170.27 m/s
=
U
Since~CI = WZandal :=fh
=
Figllre 5.35
W2sin fh = U = W.rz Oi.c. C2:= ell
O't
R = 0.5
az
=
\/doc:iIY rriallgle.fi
for a 50% reactiofl 5ta,;e with maximum IItilh:.ariOll factor
(a) Rotor blade angles
Solution
w,/1
~C'=C.
{J,=O° . CI sin Ct"1 :. C.r~
L
J4U
= 58.24 m/s
The velocity triangles for a 50% reaction stage at maximum utilization ractor are given in Fig. 5.35. C.L.1 =Cls(nal :=U,:. WI·I =Oi.f. WI C II and
Example 5.14 Air leaves the nozzle of an axial flow turbine stage at an angle of JO"', with an axial velocity of 180 m1s. If the rolor blade speed is 280 mls. Find the rotor blade angle for the reaction of the stage to be 50%. (MU- Oct. '98)
The nozzle ang le (a l) measured from axial direction 0'1 = 90° - 30° = 60° Co = 180 m/s U = 280 m/s
= 70°
Co
(d) Rotor blade height 100 x 0. 105
0'1
~,
U
. sin at sin 70'" C n = Cl cosal = 170.27 x cos 70°
= 0.105 ml / kg
IlJ04
= 0.5
. C,
From Mollier chart
P4
R To
fh = 70° and ::: fil = 0°
-
A XIAl. FLow STEAM ANOGAS TURIHNI!.S ..( 237 236 ;.. T URBO MACIHNES
Since. the rotor blade ungles arc assumed to be equal.
(d) Blade height
,i, x
Ih={J, =64 . 11 '
I')
Assumin g frictionless now over the: blades, we have Allh!.! lirsl !ttage rotor exi t
h,
=
IT
c"
=
CI
h,
Dell COS 0' 1 = 334 .32 x cos 70°
11 4.34 lO is 100 x 0.041 " x I x 114 .34
From the firsl s tage velocity triangle at inlet
IV, sin (J, = 3U Since WI ;::: W2 and
fJ] ;::: fh
= 0.0114 m.
All the second stage rOlor ex it
WI si n f3t
;:::
W2 s in
x
1')4
D C n :::;: 0.0139rn Jr
100 x 0.05 IT X
C.I~
W2 sin fh
1 x 114 .3 4
=
:. e
Xl
Note th~H the blade he ig ht increacs throu g h the turb ine s tages.
For the above proble m, irthe lurhin !.! is two stage curti s turhine wi lh siage efficiency or (15 %. dderm in e (he parameters.
2U
Since C~ ;::: C2 and 0"2 ;::: O'~ (assumed) C ,I sin 0'1,
W~ si n
(a) Rotor blade angles For max imum bl ade dlic icncy. Ihe blade-gas speed rnlio is
Since 1V~ ;:::
U
-=_
C,
ell :::;:
:::;: 'U :::;:
2U ;::: W~ sin
fJ; =
C'xl
/J; + U
U
\Vi llnd ,8; ;::: fJ; wi sin fJ; ;::: Wi sin /32 : :;: U. i.e. W:i ;::: UC.~2
= C .( I = 4U
C, =
u Figure S.JJ
C.~I
or C I .si n 0'1
+ U = 3U = C2sina2
The velocity triangles for the second stnge is shown in
Solution
4
= 3U
Fig. 5.33.
Example 5.13
si nal
fh
From firs l stage velOci lY triang le at exit. Ii,
a"I":::;: -
C; :::;: C2 and \V2, :::;: \V'I
= WI:
W2
x 157 .08 = 668.64 m is s in 70" CI CDsa ] :::::: 668.64 m /sxc Ds 70 a
W;zU
== U - U ;::: D.
4
= 228.69 mis This vc loci ty tri ang les ror the first Slugc Dr a two S(;Jgc cunis impul sc turbine wit h m ax imum blade er. Iic l!.!ncy is shown in Fig . 5.3::! .
Sine!!.
Figure 5.J1 Velocity rric/f/.~' (l!sJo r a cllrtis II/rbine wiTh max imllm
-
UTilizaTion facIO,.
-
C" - U = 4U - U = 3U 11'" 3U 3 x 157.08
:. iJ ,
c:: : ; ~ =
64 . 11 '
228.69
U
U
pi ... f3;
157 .08
c; == CII = 228.69
tan
34,48'
/3; ;::: tJ ~
(b) Power developed IV
=
IV,
= =
\VII
= =
IV, + IVII ,i,U{Cx ' + Cx')
= n,U{4U + 2U)
1;' (6U 2)
,iIU(e~1
,iI(2U
2
)
+ C~2) ;::: ,;,U( 2U + 0)
..., 234
>
TUIOlQ MACHINES AXIAL FLOW STEAM AND GAsTUllfll:-l[:S .;{ 2:\5
For first stage h
1',
o
P,
,
P, 2S 4
-IS
s Figure 5.31
ry"
"0 - "2 Figure S.30
Ve locity (riangle:rjor a roteull tllrbin~ with maximum utilization jacfor
Similarly for the second stnge
p;
= p~
110 - lib
= 53.950.
(bj Power developed
110 - IIJ. 110 - lib
Wi/,;'
= =
WI
liJU(C.~1 m2U
+ C.rl)
2
"0
J1IU(C;1
Work done in second stage
=
+ C;l}
,ilU(2U + 0)
WI/
~ := 0.78 63 .269 kJjkg
WI
+
\VII
1;1(2U'1 41nU 2
+ 2U 2}
4 x 100 x 157.08'
=
To:;:: 500"C
3370 kJ j kg 3370 - 49.35
=
3370 - 63 .269 = 3306.73 kJ j kg
= 3320.65 kJ/ kg
From chart stale of sleam at first slage exit is P2 = 82 bat a, = 0.041 m) j kg
T1_ ::; 470"C
For second stage
Therefore, TOlal work done
49.35 kJ/kg
49.35
=
11, 11,-,
= 2U' =
110 - 112
From Mo1iier chart, Po = 100 bar
Work done in first stage
9.87 mW
(ej Final state of steam The expansion process is shown on " - s diagram (Fig. 5.31).
= = = =
'1I/I
112 - "4 112 - h"J
11, h.~I
"2 -
114
= "z - h4S
= 49.35 kJjkg = 49.35 = 63.269 kJ j kg
=
WI//,iI = 2U'
= =
11, - 11,
ry, 1/ 0.78 3320.65 - 49.35 = 3271.3 kJ/kg
=
3320.64 - 63.269 = 3257.38 kJ/kg
From Mollier chan Pol = 65 bar T.,1 = 445'.C ' .a4 TIlis is the final sta te of the steam at the turbine exit.
-
-
005 •
m) Ik (> ...
AXIAL FLOW STE,o,M AND GAS TURIJI:-.IES '" 233
(d) Power developed
"c.~
IV
~c, ,~~~
)-?U
35 x 394.24 = 3.B mW
IV
Example 5.12
IVA _
The initial pressure and temperature of steam entering a multi-
stage turbine nrc 100 bar and 50Q oC respective ly. The turbine diameter is I m and
speed is 3000 rpm. The exit angle of the first stage nozzle is 70°. The steam flow rate is 100 kg/so Ass uming maximum blade efficiency and equal stagr.: efficiencies of 78%, determine th e rotor blade. angil!s, power developed. final slatl! of steam and lhe blade height if th e turbine is two stage Rateau turbine. Stale the assumptions used.
L J c,-c, u
Solution Figllre5.19
Veloc ity triangles Jor a sing!" impulse slrlge w;lh maximuIII utilization/actor
WI!
(un fit
U
=:
Irln a l
eu
c(/ Ian 7::l C
2C 1 cosO' ]
2
a'i
(a) Rotor blade angles
58.55"
Ii'].
Il,
U
= 58.55°
(b) Flow coefficient = CI casU'1
U
CI sin a l
lanai
2
=
¢(lan
sinal
2
Ian fi t
Ih,)
0.612(Ian SR.55 2
+ Ian 58.55)
be equal. The blnde gas
If
x I x 3000
60
= 157.0B mls
111C veloc ity triangles for a two slage Rateau turbine with max. imum hladc dficiency a nd assu m ing ax.ia l exi t is shown in Fig. 5.30. For the first stage .
~,
p,
= 2U Cx , - U = 2U C ,r ]
=
and C~~
fJ I + tall
10
= 3000 rpm
sin 70
Wq
0.61:2
(c) Blade loading coefficient
V'/
U CI 2U
C] sina l
[an 73"
N
= ~S2 = 0.7B
2 x 157.08 - - - - = 334.32 mls
2
2
= 1m
blade angles nrc assumed
rrDN 60
=
.. C,
ell
~SI
speed ratio ror max.irnum blade efficiency is
G U,"
",,, - 1(1.635)
D
= 70°
= I . 635
- -2.~,
To = 500"C
Po = 100 bar Til = 100 kgls
(a) Rotor blade angles
O. C2 =
= =
U
ell
W-f]
U
lanai
C"
C"
2
Ian
- I
('an 70") - - 2
53.95"
= p,
=U
AXIAL FLOW STEAM .... NOGASTuRU IN ES ..;: 231
.'
230 );;0 TURDO MACIlINES
670 - 249.5
I
= 420.5 m/s
R
=
R
=
28.5
U
I
= 0.285
mUCrr(tanal + lan(2) rr x 0 .5 x 2400 rrDm N
IV
I
Therefore
420.5 - 249.5 2 x 300
(aJ Power output
- --= 60
=
60 62.83 m /s C/t(tanp2 - tan.8l) 2U
=
I
I
R
=
2UR
(c) Blade efficiency
C" ~h
Work done
=
Energy input
=
U(Wq C2
--.!.
+
=
=
IV
110.59 kW
(b) Stage enthalpy drop
R =
2U(lVx ,
=
+ Wx,)R
201 x 10' x 0.285
Example 5.11 enters a simple impulse turbine nozzle and expands adiabatically 10 100 kPa with an efficiency of 90%. The noah: angle is 73 ° 10 the now direction. Assuming optimum conditions. find the rolor bludc
Exam ple 5.10
Steam enters
flow rate of 35 kg/s .
584.76' - 2 - + 57285
Solution
0.8806 or 88.06
= =
q"
angles, flow cO,efficient, blade loading coefficie nt and power dc\'c1opcd for a mass
201 x 10'
=
Po P2
a 50% reaction stage at a pressure of 2.2 bar and
170Cl C of temperature . The rotor run s at 2400 rpm. The rotor mean diameter is 0.5 m and the symmetric rotor and sialor blades have inlet and ex.it angles respec tively of 36" and 19° . Find the actunl sInge power output. If the stage efficiency is 88% find also the enthalpy drop al th e s iage. (MKU-May '97)
For optimum co nditi on.
=
800 kPa
T,
=
973 K
:;:
100 kPa
al
:;:
73 QO
'I"ltr,r .ci ia
Solution
2V
WlsinfJl
=
0.5 2400 rpm
0.5
Po D ..
=
ct, = fh and ct2 =
2.2 bar 0.5 m
PI
To
=
al
=
170'C 36° ct2
=
19'.
= sin2al .and (J1l1'f
U
= -
C,
sin 0'1 = - - . Velocity
2
triangles for a si ngle impulse stage wi th maximum diagram efficiency (o r) ulili1..HiolJ factor is shown in Fig. 5 .29.
WCI
= =
q,
= "" HOI gas at BOO kPa and 700°C
and C, =Ca lcosa, = 2001 cos 70' = 584 .76m/s :. flh
W/III
110.59 0.88 125.67 k1/kg
57285
Since R
2 x 62.83 x 0.5 (an 36° - Ian 19°
I x 62.83 x t64 .38(tal1 36' + tan 19' )
IV
2
Since
R N
=
Power output for J kg of slcnm per sec
Wf~)
+ w;_ _ W2I
2
(t,n /30 - tan Il,) 164.38 m /s
= = =
Clsinal=Csl C.lI
-
U
= 2U -
U :;: U
W.r,:;: U
Since W, :;:::: W2 and PI = {Jz for an ideal impulsc Slag!!.
IV, sin til = W, sin til = U CX!
::::;
0, i.c. C! ::::;
ell
AXIAL FLOW STEAM .... NDG .... sTuRHINES "" ::29
Th~ axial component of the air veloc ity ilt the e:til or the nonle all :tx ial Ol1W h!ilClion stuge is 180 m/s. TIle nozzle inclination [0 [he direction of rnl;J.t ion is 2)0. FillJ thl! rotor blade nnglt!s at the inlet and oUllc l, if Ihl! degree of rL.!';lc ti on shou ld hI! 50% and the hlade speed 180 m/s. Al so for Ihe same blade speed, axial velocity and naule angle, find the degree of r~action. if the absolute ve loci ty at the rOlor o utlet shou ld be axial and equal to the a\lal ve locity ::1\ the inkl. MKU-Apri[ '94
Example 5.8,.
~} r
(b) C, = C" = 180 m/s. C~
is axial. Therefore. the ou tlet velocity triangle will be as shown in Fig. 5 .28( b)
U
IBO
C2
180
tan/h
:::.
-~-=I
: . /12
=
45°
As there is no change in the conditions at Ihe rotor inil!t. f31 is Ihe same .
Solution 180
C"
=
U
m/ ,
= 90 -
" I
= 63°
27
fJ, = 43.9°
R = 0.5 Thu s,
180 m/s
=
R
¢ (Iun 13, - Ian fJll = I x (lan45°-w n43 .9"' )
=
C"
!:In
fil
=
C'i tan at
=
180 x tan 63°
=
353.:!7 m/ s
=
-W'L =
:. PI =
C.r ,
C"
-
.1D.27 - IRO
U
180
C"
43.9<)
I.BB%
Examp le 5 .9 TIle blade speed of an axial now turbine is 300 mls. The mass [Jow ratt! is 2.5 kg/s oThe gas temperature at turbine in let and outlet arc 500° C and 300°C respectively. TIle fixed blade o utlet angle is 70°e. Axial velocity rema ins constant at 200 m/s. Delermine the power developed, degree of reaclio n and blade I!fficil!ncy . [MU - April '96, Oel. '97, Oel. '99 am/ April '991
Solution
w,,~ : cu LJC'
U = 300m/ s T, = 300°C
=
HI
(b)
(al
11I1t't I'e/ority rri(lllglr
Figure S.18(b)
£\'i/ I'docity Iriaflgle
R
tan fJl Y
R
= WJ"2 -
\V.rl
2U
180
U 2R
180
= 70°
To = 500"C Cn = 200 m /s
2.5 kg/s
1;IC 1,(To - T1)
~
2.5 x 1.005 x (500 - 3(0)
=
502.5 kW
. The slage work donclunit ma.ss flow is
2 CI ,
~
(b) Degree of reaction
Dc,gn:l! of reac ti on.
u, q, (tan 1'-
Ii, al
(a) Power developed
U
Figure 5.2S(n)
2
2 0.01 BB (or)
(a) Blade angles
IVf,ilU(IV"
+ IV,,)
C,,(To -
-=- = tan fi '!
~
~+Iil n li l
1 x 0.5 - -1-
+ I. n4 3.9
°
1.962
fJ, Si nce R
= 0.5, fJ'!. =
0'[ :;;;
To)
1.005(500 - 300)
62.99° '" 63°
63 -'l which is give n in th e problem .
:.
W"~
+ \V.q
:;;;
201 x 103
300
:::. 670
201 klfkg
mls
From inlet veloc ity triangle. refer Fig. 5.2
Wrt
CAl -
U :::.
ell Ian al
200 Ian 70° - 300
= =
549.5 - 300
249.:1
m I"
-
U
AX I Al. FLOW STEAM ANI) G AS T lJRH I =" k S .... :2 21
226 ;. TURDO MACI-IINES
From inlet veloc ity triangle.lI the hub. rerer Fig. 5.2 tan
334.83 62.6 1
=
0'1 .1/
From inkl veloci ty triangle atlne tip. refer r-ig. 5.2.
exl., C"
=
72.69"
V,
V"
and tanat.II - tanPI .1I = Cf/
and -C
2,,(0.225)6000 = 60 60 141.37 mls
V,. =
or
2rrfllN
= tan PUr
PUI
200.896 62.61
--=---
tan 0'1.,
= tan at., - Ian PI.I
"
VI
V"
=
tanal.h - -
=
tan79 .4 - - -
=
72"
=
Ian fJI.t
:;;;;
=
C"
"
141.37 62.6 1
Ian f3l.l
V,
tanat.t - -
C"
2,,(0.375)6000 -2"r,N = --,-,--';-;C'.:...c.c..c 60 60 235.62 ml s
=
.
- 29'
f3r.l
2:\5.62
130 0 71.69" - --
62.6 1
Fro m the outlet veloci ty triangle. refer Fig. 5.2. At ou tl et
V" -C
= Ian Ihh -
tan 0'1./1"
CX1,t ·'r
=
C.r 2.1
=
tallthl
=
" BUI tan 0'1.11
AI outlet
CX2.l1
= -C"
CX2.r1l
"1Il
=
45. 16 x 0.3 0.225 60.21 mls
=
th.1
CX1./Il.r",
45 .16 x 0.3 = 36.13 mls 0.375 V, CX2./ -+ - -
ell
eu
235.6 + 36.13 62.61 77"
The degree of reaction at the hub and lip arc Uh
" tan {lZ.h
+ C.t:2 .h
R,.
CII 141.37+60.2 1 62.61 72.75" Cn
11,.. For the tip
C.
=
2Uh (Ian th./! - Ian fJl.h)
=
62.61 (raIl72 .75° - rUII72° ) 2( 141.37) 3. 16%
=
C" 'iij (tan Ih., - tan flu)
and
S imilar eq uati ons as above arc used . At inlet
ex!..
R, "1
:. Cxu
= =
251. 12xO.3 0.375 200.896 mls
"
,
62.61 2(235.62) (ta n 77" - 10,,( - 29")1
=
64.9%
-
This work done ~qlJill~ the enthalpy drop pcr swgc. Therefore.
1.69 U' or
J 1.59 x 10]
=
=
U
136.7m/s
Solution
n, f3l.m
C"
(c) Blade mean diameter
= = =
0.225 m
r,
45" , co nstant
al.m
= =
0.375 m
+ r,
= 0.225
76° ,
=
f3z,m
75" .
N
6000 rpm
The mean radius.
FJowvdocilY
ell =
Ctcosa l
C,
=
1.43 x U = IA3 x 136.7
Ct
=
195.5 m/s
ell =
J
Y5.5 x cns(90 - :!O" )
rm = rh
2
Urn
60 188.5 m/s
=
From inlet ve locity triangle, at the mean radius, refer Fig . 5.2 rvlass !lnw mil'
. Volume flow rail!
=
1ll1'3n
Vm
Spec ific volum(,!
lanal.1II ":"lan{JI.1II
(rr Dh)C'1
Volume now rale where D - blade
2n(0.3) 6000
=
= 1.618 mJ/ J..:g
Nnw.
2rrrmN 60
=
From SICillll I~blc . ill 1.05 bar, the specific volume or dry stea m is givcn by 1'J.a:
= 0.3 m
2
The melln blade speed
66.9 ml'
=
+ 0.375
188.5 -tan ::-=:==----:-:-:= 62.61 m/s 76" - Ian 45"
lIi;.III1CICr ilnd II - blade heig ht
" 18 .63
=
Cx\. ...
0
=
Cil Ian a 1.111
:=
b2.6J x lan76"
= :!.51.12 m/s
12
1rO(~)X669 =
and from ou tlet velocity triangle at the mean radius, rder Fig. 5.2-
1.618
CX1....
tor)
0=1.31]
= =
WXl .... - VIII == CII tan th.m - Um 62.61 'an 75' - I BB.5
=
45. 16 ml'
For the hub
(d) Rotor speed
For a rrl.!c vone=<. design. U :.:
N
(10
ex . ,
= t,;onsL at all radii. 111creforc.
radius and thl! huh
nO
x 60 x 1.312
13~.7
1T
=
Example 5 .7
1990 rpm
The dala for a free vortex turhine blade arc g ivt:n bdow.
Blade lip dinmeter - 75 un, blade roof diamt.!lcr - 45 em, inici 'Inglc orlhe rolnr hlad!.! at mid hl!ight - 45 " , out h.:t angle of th c nozzle blade at mid hl!ight - 76(" outlt.:t ~lIlg ll! of Ih!.! rotor bladl!. al mid heighl - 75 fl , spt!ed - 6000 rpm. Axial velocity remains COllstant across the rOlOr. Determine for the hub and tip Iii) IH)'1.1.lc exit angle (hl rolor hlade angles (c) Ih~ degre~ ofrl.!actioll.
At the hub inlet "m
=
ex .. /!
=
Cx, .tII
= =
CXI.h . 'h
Cx ..... ·
'm
n, 251.12 x OJ25 0.225 334.83 ml'
b~twcen the mean
A ,\\ ,\L FLow STEAM ,\NO GAS TURBINES .... 223
222 }> TURBO M ACIIINES
By measurement from the velocity dillgram WI Diagram power
= 59 m/s
WX1
+ Wr~ ==
i42 m/s.
mfl(W:fl+Wt~)
=
10.000 3600 x 45.6 x 141
=
17.99 kW
R
P, RF
0.5 0 .75
14 bar 1.04
=
Work done/kg Energy input /kg
=
U(W",
+ IV,,)
and
6.IIs
C(IOO)~ - 59')
3080 - 2270
=
810kJ/kg
x 1.04
6. It
= 78.4%
'I, x 6."s
= =
0.78 x 810 631.8 kl/kg
=
W, 69.5%
=
100 - 59
Diagram power (W) ::;: m D.h the mass flow rate o( steam
=
= Tolal enlhalpy drop per Slage
lV 11.770 /J. h = 631.8 18.63 kg/ s
=
TIl
(e) Enthalpy drop in the moving blades
=
= 2
3.259 kJ/kg
(b) Blade speed Work donclkg
= 2 x 3.259 = 6.5 18 kJ / kg
Example 5.6 Steam enters a 0.5 degree of reaction turbine at 14 bar ,lIld 315 °C and is expanded to a pressure of 0.14 bar. The turbine has a stage efficit!nr.:y of 75% for each stage and the reheat factor is 1.04. The turbine has 20 successive siages and the lot al power output is 11.770 kW. Assuming that all stages develop equal work calculate the steam How rate. Al a certain place in the turbine. the steam has a pressure of 1.05 bar and was dry and saturated. The exit angle of lile blade is 200 and the blnde speed ratio is 0.4. ESlimme the blade speed mean diameter o( th e annulus at this point in the turhine, and the rnlor speed if the blade height is 1/12 of the blade mean diameter.
631.8 N 20 31.59 kJ / kg
=
59
100' - 59'
/J. h
=
(d) Percentage increase in relative velocity WI
0.14 bar 11.770 kW
= 0.78
=
Enthalpy drop per stage
w2 -
=
Thus. actual cnlll.llp)' drop
2C'I - lV', 2 45.6 x 142
'Idirl
=
ryr = 0.75
Ct. then
=
::;:
Wt.: know that Ovcr3 11 efficiency ('II) _ Stage efficiency (t'/S'fll:t') x reheul faclor (R.F)
C'
Since
1'2 IV
(a) Steam flow rate
-1.+ 2
Energy input/kg
315"C 20
TI N
From Mollier chart Ihl.! isentropic enthalpy drop when stt.:am expands from \4 bar and 315 lJ C 10 0.14 bar is
(c) Diagram efficiency
Work done/kg
Solution
sine!.:
0'1
=
U{Wq -I- W.1'l)
= =
U [(C, ' - U)
fh and CI
:=
\V2
+ lV,,]
for a 50% reaction turhin!.:
= U[2C, si n a, - UJ Now U / C, = 0.7
or
C, = 1.43 U
Work done/kg
=
U[2 x (1.43U)sin(90 - 201 - UJ
=
1.69
u'
AXIAI. FI.OWSTEAMANOG .... sTI IRDlNES ...: 221
For un it lIlass nnw nile of Sh:arn. Towl dri vlIl g fnrc!.! = (870 + 295 ) x I AX ial thrus t for first sta ge
= (CIII
= 170 -
C"1)
-
4
= 1165 N
The 135
mCi.ln
blade speed (U)
= 35 I11 /S
= 45.6 m /5
Axiallhru sl ror s~cond singe (C;'l -
C:,:!) =
Axin llhru!<;1 per unit mass fl ow
r~lIc
110 - 100 = 10 m /s
= (J5
+
10) >. I = 45 N
(c) Diagram power = =
1II11( 6 W"
::;;
139.8 kW
+e:. W II ) "1
I x 120( 11 65)/ 10.
A x
=
tubll!. A I J bar and with dry sutufi.ltcd steam
a = 0 .6055:1 m'/ kg 1110 An nulu s arca = -
C"
+ 6. W.I'II )
s in :!Ci I
:::::
=
0.049)8m'
:. D
(e) Maximum diagram efficiency '],Jill
(10.000/3600) x 0.60553 34.2
Now, on nulu s area, A 1r Dh . . where '0' is the mean blade diameter and ' h' is the mean blade hClght
(600)'/2 77.7%
=
=
=
Ci / 2 I :'O( 11 65)
=
ell
Mass now or steam. /II = - - 0 . whe re A is the an nulu s area and 0 is the spec ific volume o r steam lrom the steam
(d) Diagram efficiency 11(6.\\1.,./
= -3 x eu
0.04918 x 0.04
= rr
=
A/rrll
=
0.39 m
Rotor Speed, N
=
si n.:!(90 - 16)
=
92.--1 %
=
Example 5.5
A parson's reaction IUI'hi nc having iden ti cul blad ing dl! li vers dry sa (u nllcd steam ;1. ( J o;lr. The ve locity of stCillll is 100 m/s. The mcan blade he ight is --I C Ol nnd the ex it .lOg lc or the movin g hhliJl! is 20 ;', A I the mean radiu s the axin l now vcl ()cil}, eq ual s 3/·] o f the blilde spcc:d. For a steam fl ow rat e of 10,000 kg/hr, ca lcul a te. (a ) [he rotor speed , in rev/min (b) the power output or s tilge (e) the diagrnm dfic iency (d) the perce ntage in c rease in rclali\'c ve locity in th e mo vin g blades due to expa nsio n III [h ese bl:'ldl.!s (cJ th e cntlw!py drop o rth e s team in tilt! stil ge.
Solution
ux
60
rrD 45 .6 x 60
rr x 0.39 2233 rpm
(b) Diagram p ower S ince the blades arc id entical. the inlet and outh:t velocity triangles will al so hi! identical. From the data 20° CI = 100 m /s the velocilY diagram may he drawn as U :::;: 45.6 m /s a t show n in Fig. 5 .27 . Draw AB = U. Sct off Be alCiI to AS and ~qual Ct· JOIl1 AC . T hen SCI o rr A D ;::: \V:! = Cl (identical blades) al fJz to AB. Jom D wllh B.
=
t?
: 10
c, = 100 m/s 3 ell = - u (a) Rotor speed
" = 0.04 m //I
"
fi2 = 20"
= 1000 kg/ hr
Th e flow ve locity n"
CtCOSa l
C"
101) cos(90 - :'0)
=
34.2 rn /s
E
U.:0.45.6 mls:
Figure 5.27
0
F
AXIAL Fww STEAM A:-InGASTIJRI!INI:.s '" :! 19 218
}:o
TURflo MACIIINr;.'i
\V.q
Example 5.4 The steam in a two row curtis stage leaves the noz.z.les al600 m/s and the blade speed is 120 m/s. Before leaving the stage. it passes through a ring of moving blades. a ring of fixed blades and another ring of moving blades. The nozzle angle is 16", while the discharge angles arc 18 0 for Lhe first moving ring, 21 0, fo r the fixed ring. and 35 0 for the second moving ring. all meOlsurcd relative to the plane of rotation . Assuming 10% drop in ve locity during passage throu gh each ring of blades, draw the ve loci ty triangl es and determine (a) blade inlet angle for each row (h) driving force and axial th rust for each row of moving blade (e) diagram power per kg/s steam now and (d) diagram efficiency. What wou ld be the maximum posHiblc diagram dlicicnt.:y.
Solution
c, =
460 m ls
=
\VI ~
4 10 m/s
C" , CUl
20°
C,
~,
600m /s
11,=
U= 120m / s IVR =0.1
0
1711 m l' 135 m /s 325 ni ls 0
C; = WR . C, = 0.9 x 325 = 292.5 ni ls Thl.! bladc B'C' =
C;
S
k \ ' B' = U . Sct off cl.!d for cach mov ing hlade is same on d ·so La ~ I , ., to ,\'8' . Join A'e' . Measure A'C' represenLJO g WI and caku lalt:
a~ a'l
I 1;::: W
18°
The veloc ity diagrams are drawn as shown in Fig. 5.25 and arc con structed as follows.
= =
Th;}l is the bladl: inkt angle fur firs t row o f moving blade is 20 • . I ' I' .. d I I C' C is the sll:·tm vel oc Ity al I le In ct MCilsurc BD rcpres!.!nling C2 nn Cil cu ale \. I : to the second row o f moving blades.
W2·
a; = lI Dand f3; = 35
=
WR .
tv; I
' D' - W' at a' to A' 8 ' and join 8' with D'. Thi:-. co mpletes the vel oci ty ~Se t a IT I I -:2"'2 . . b· ' F 5 ')6) diagram for th e second stage of two row C'urHS Impulse {ur IIlC ( Ig. ... . Scale I : SO
Stale
J : 5U
c'
E' Fjcur~
5.16
By mC~lsurcment. WI = 190 m/s lV2
Figure 5.25
IV'x,
= IV'x, =
(a) Blade inlet angle for each row Draw a horizontal line AB representing U seLorr BC at 0'1 La AS and equal to Ct. Join AC which represents WI and caluculate W2. lV2 = WR WI . Then set orr AD equa l to W 2 at 112 to AB . Jo in D with B. This completes th e veloci ty diagram for the first stage of the two row curtis impulse turbine (Fig . S.2S).
= 0 .9 x
p; =
190
= 171 mls
155 m ls
C' =II O m / s
140 m js
== 100 m/s
'" C;'l
35°
Th!.! blade inlet angle for the second row o f moving bladcs is 35
0
(b) Driving force and axial thrust
IIV2 = 0.9 IV, I
Driving force for first stage is given by ·l:!. W'1 ==
By measurement from the velocity diagram IV, IV,
= =
W.fl + W'l :::;460+4 10== tO Om /s
Driving forl.:c for sl.:.cond stage is give n by
485 m ls 0.9 x .85 = .36.5 mls
A U
IVXII = IV'.r ,
+ IV,.' ,• =
155 + 140 = 295 m ls
•
STE"~l
A XIAL FLOW
ANoGII. s TuIilllNI:5 "" 217
(a) Steam velocity at exit from the nozzle
(c) The relative velocity ratio
S ilK': the lur~in!.! is impul se, the IOI~ 1 pressure drop wi ll occu r o nly in Ih..: noalt:s.
From the niven J nd calculated d 'Hn. we can d raw the vec tor diag r.nn as show n,ln Fig . 5.24. AfI;r drawing the inlel ve loc ity triangle ABC . locate point ~ by dr~wmg Ih t: pcrpendicu larCE and th en set o ff E F = { W~ , + W"' J} to locate POIOI F . St:t off A D at Ih to AB and leI AD inlerscc l rnc pcrpe ndu.:ular FD al D.
Ass um ing isen trupic l!xpansion. the enthalpy drop can be found rrom [he steam tab ll.!.
=
A I PI =:' hur "1 = h .~ 2747.5 kJ / kg 0.819 kJ / kg - K For iSl!lllrnpic.: l!xpansion, 52 :::: 5[ :\1 P:. ::::; :U~ har. '\"'£:/'2 < Xl: Ihc rdorc. the s lt!J m is wet
-'I
= .,.' =
.\" fl'~
s:! -
'\"""'!
=
"'!f : :
=
6J~
.
Scale I : 50 WI : 180 Ints \\12 " 140 m/s
19 - 1.647
u
5.:\67
O. l)6~
" .r+ x::!I! !..;
+ 0 .964(2170. 1)
=
551.5
=
2643 .5 kJ / kg
Figure 5.14
=
274 7.5 - 2643.5
=
104 kl / kg
By mCi.ls uremCnl from the velocity diagram Rebt ivc Veloc ity al inlet. IVI = 280
Now.
Gain in kinetic energy := Us..:ful he al drop
m/,
Rl.!!:.Hi Vl! Vel oc ity al oull e t. 1V2 :;:::: 240 m/s Relative Velocity ratio =
hOOt> ~ O.lJ .,." 04+ 75i
CI
W, 240 --= :;: : -280 :;: : 0 .857 . WI
(d) Stage efficiency Work done/ kg 'IXWJ.lt = Em:rgy Supplied 10 no:u.lc
·D9.1:! m /s
(b) Diagram efficiency l/./iu
Ene rgy s upplied / kg
=
u
(ali, ) + 2000
=
104
=
106.8 1 kJ /kg
+
751. 2000·
175.65 x 469. J2
imd
'Isttl}:t
=
-
c'0
=
=
IV
=
206 x 10)
IIIU 2. 5 x 175.65 469. 1201 /5 175.65 >< 469. 12
(439. 12)'/2 85.47%
= =
(106.81 x 10) 77.15 %
(or)
= =
' IJill A 'In,,:zll." :;::::
76.92%
0.8547 x 0.90
.
21·'
I> TI IR n o MACII INES
AXIA1.FLOWSTEAM,\NOGAS T u ltliI NES "'" 215
Ve loc ity o f .steam al e.'(il frum the nOZ7.1e
(c) Blade efficiency
C'
---' 2
U(lV X1 'Illin
C,
J2000 x 402
179
896.7 m/s
=
P I .. JO·" ~,
WI .. lll0 '!1 I~
C. , .. J1nrn /~
WJ ,,4J,I,"/,
t":,. lJllmJ. ('..; "' 21(l 'n(~
=
10'
Ma xim um hladc effic iency under optim um condit ion
.,
:=
sm-al
:=
si n:! 70"
al
:=
1)0" :- 20°
:=
70°
= 8H.3%
n :" 1.['
w •• " SSO",' .
. SInge c fflclc ncy ;:;
W.J= "lIlt1m/~
\\'1 "" 0.1W ,
x
896.7' /2 69.4 %
From thi s data . the velocity diagram (Fig. 5.23) ca n be drawn and Ihe following rl!su ll s nre nblained. Draw AB = U. Set o tT Be := CI at oq to AB. Join AC. Measure Pd LCAE). Thcn SCI orr AD 0.7 AC (IV, 0.7 IV,) al /3, p, 10 AB . Join BD.
=
+ W.n)
Cil 2
SCIIlc 1 : IUU
Work done on blade . Total energy supplied 10 blade
279 473
= 58.98% It can also hc found from
0.694 x 0.85
=
58 .99%
Fig"re 5.23
(d) Energy loss in blade friction
a) Axial thrust per kg
=
310-210
= = =
100 N/kg
(b) Power developed per kg of steam/sec. Work donelkg o f steam/so
279 kW /( kg/s) Power developed per kg of Sle
= 279 x 0.9 Steam rale / kw .hr
25 1.1 kW / kg/s 3600 251.1 14.3 kg
2 6::!02 - 434 2
2 98022 J/kg 98.022 kl/kg
Example 5.3 In OJ si ng le row impul se turbine stage . Sleam is su pplied dry and saturated al 5 bar and the exhaus t pres sure is 2.8 bar. There is carryove r velucity or 75 m/s. rrom Ih e previou s stnge nnd the kinetic energy al e.'(it from the nozz le is only 90% uf the theoretica l available energy. TIle nozzle is inclined at 20'" wit h Ih!.' direction or blade rOl<1tlon' a nd blade speed rntio is 0.4 . TIle hladc exil ang le is Jl so 20°. Fur a steam now rate of 2.5 kg/s the o utput or Ih e s iage is 206 kW. ESli rnJh: (a) v~locity of stearn at ex il from tll c nozzle (b) dii:1gram efficiency (c) the relati,,!.: velocity rat io. (d) s iage erric ien cy. .
+ WX2) 300(550 + 380) U(IYX I
= = =
Wf - \Vi
I
Solution
=
(
=
P, 5 bar P, 3 bar m =2.5 kgls W = 206 kW
Co
= 75
mls.
",
= 20°
U /C ,
= 0.4
fh
= 20
2 1:::!
:.-
TI III1Hl M ,\( ' III NI: S
(d) Axial thrust t'l1'IJc (0) ex it ,'doc ity and direction (c) wnrk done per kg of steam (d) axial thrust and Illlw..:r dcvcJoped fo r a stea m now rate of 5 kg/s and (c) dingram or blad..: dficicncy.
Solution
Change in now velocity
c, =
500 Ill is
U = 200 mls
th. = 90 - 25° = 65"0' 1 = 90 - 20'" = 70° Thl.'
diagrum is constructed as fo llows (Fig. 5.::!:!). 1~1I ScI..:t.:t u suiwblc scak (mostl y reduced scn le). (h) Draw AS equal to blatil.' spl!cd U, !~ I Dr~lw Be wit h it s in l; linnti oll Wi lh AS equal [u nozzle angl..: (0'1) ;tIlU S!,!t orr Be l,.'qua l In the steam velocity (C I) on the snme st:ale. IJ) JllIn:\C tu l:OmpJclc Ih..: inlel velocity Iri;tnglc ABC. AC = WI. td SI.!I nIT AD at movi ng bladc cxit ang lc 10 AU . With A as c..:nl rc lind radius cqual hi Ih..: pe.!rccnlagL' 0 1" AC, dmw an arc 10 cut the.! lint! AD at D. AD = W 2 . In this prohl..:m sinL"c th l.! dJ"l.!ct of rriCli on is nl!glcctl.!d, AD=AC. j JIll!! 8D tn cn rnplt:tL' Ihe olLt let VL! lo(., Jty triangll! ABO . BD = C2. Til l! velocity diagriun is show n in Fi g. 5.2:?:.
=
Cui -
=
175 - 135 = 45 Illis
:;;
5 x 45
=
225 N
Axial lhruSl
VclOC lt }'
and power dewlopl.!d W
=
m x (Work done/kg)
=
5 x III 555 kW
(e) Diagram (or) blade efficiency
n
Suk l ; ~O
'7di u
=
U(W.rl
Kinelic energy supplied to the blade.!
W:=W, "' ll~ mI.
=
t:;. ,Mln," "~ '" )/1
I)
+ w.r:!)
Crl1
W, '" 1I~ mO, o,,,j7'
W,'
e'l:!
2
U(WJ1
Work dono / kg
::: ~ 70nu.
C:!
-L:=
w,~ =1 ri ~ mlo
=
. c •• =In m'~ Co.: = IHm..
SOD:!
2
+ \VJ·:!l
III x 10' W I kg III x' IO'
500' 12 I) ~()H "' "
"
Figure 5.22
Th..: foll'l wi ng r..:sults arc ubtained from thl! veloc ity vec tor diagnlm .
(a) Moving blade inlet angle
-
imd dirL'c li nll of sIC am
.11
Example 5.2 TIle bl
on
(b) Exit velocity
c, =
88.8%
160 ml s
Solution U;300 mis ,
cxil
,, = 20' ,
(All),
= 473 kJ/kg,
qN
= 0 .H5,
{J , (blades bl.!ing symmetrical) Usdul heal drop which is converted into kinl!tic cncrgy
(c) Work done per kg of steam lV /III
D. h
= =
2(1)(270 -\- 2R5)
=
IllkW / kg
'I N tl.h J
0,85 x 473
UPV.r 1 + Wr:!)
=
402kJ/kg
\V2
w,
= U.7,
Ih =
210
>
A\"(,\!. FLOW 5T1.;"'" i\NIlGASTuIW1!'.f'_" ...;: :! II
Tunno M i\CI IINI;:'"
Ci 12. where C:! is the absolute velocity of Lhe steam at the blade exit. When this kinclic energy is passed over to the next slage, it is termed "carry oller". In addition 10 these losses, there may be some loss ofheDt energy due 10 'radiation' to Ihe ambient surroundings. In all. the lotal intcrnallosses in a turbine may be 20 to ~O% or so.
restricted 10 Ihe firs t stage of the turbine.lhe n07.7.le areas in till: OIher stages remaining COllstant. II follows that , provided the condition of the SIC<.lnl at the inlcllo the second stage is not materiall y effecled by Ihe changed co nd ition ofl he firs l st;Jge, the absolute pressure o f stea m in front of the second stage nO/.z lc will be directl y propt'rliunalto th e rate o f s tcilm now through the turbine .
(h) Loss due to moisture In the lower stages of the turbine, the steam may become wct os the velocity of water panicles is lowerthan tll.1.( of the steam. So a part of Ihe kinetic energy of sleam is 10SIIO drag Ihe waler partic les along with it.
GOVERNING OF TURBINES In a normal turbine driving an alternatcr, the energy output will vary in accordance with the load . lllc objective of a govcrnor is to m a intain the speed of Ihe turbine conSlanl irrcspective of Ihe load. The perfonnance of the turbine itself depends, 10 a large exlent. on the particular method employed for controlling the suppl y o f steam to the turbine so that the' speed of rotation' will remain constant. The c hief governing methods arc
(a) Throttle governing TIle principle of thi s method b'lsically requires 't"rottUng' of th i.! s team . so as to reduce lhe s team Oow whenever there is a reduction of load on Ihe turbine. In addition to a stop valve, the turbine has a 'double bear ,·ultle' having. seals o f equal or nearly e.qual a reas , so sha ped that the forces on Ihe valve due 10 SIalic pressure and dynamic action arc balanced . This double beat valve actually throttles the steam (Fig. 5.l9). IL is operated by a servo·motor controlled by a centrifugal governor. which is driven by a worm gear at a speed less than that of a turbine.
Figure 5.20
Nuz:.le COlllmJ gOllU1II1I1:
(e) Bypass governing The modern high press ure turhines consist of a numher or stagcs of co mparati vely sma ll mean diamelers of wheel. Ow ing to sma ll heal drop in th e first stage. employing noale control £overn ing is not advisahle . FUriher. In cusc. of higher loads the extra stc om requirell can not be admilled through ;'HlJitiof1al nozz les in the first s tage, due to vilrious reasons. These diflicuhie s of regulation arc overcome by the use or by pass gm'crning (Fig.. 5.21 ). In thi
LJ..~
_
"S
Bypass steam c::::,; ~ :'L.
~To
...
lower slage
,~~~ Fieure 5.21
fJypllSS NovernillJ.:
SOLVED PROBLEMS Figllre 5.19
ThrouJe governing willi double beat valve
(b) Nozzle control governing In this method the nozzles of the turbines ore grouped in two, three (or) more groups and each group of the nozzles is fed with the steam supply conlrolled by valves (Fig. 5.20). Di fferent types of arrangements of valves and groups of nozzles may be employed. But the noz7.le control is necessarily
Example 5.1 The data pertaining to an impul se turhine is as foIlO\,,·s: Stenm velocilY = 500 mIs, blade.: spccd=200 tn/s. Exit angle of nlll"in~ hl
:W.R :,...
T liltBO M ,\('IIJ NES AXIAL FLOW STE.... M ,\NO GAS TURBINES "" 209
Re;lction turbine
Impulse
lurbltle
(b) Nozzle friction losses The friction and eddies cause soml! prt!ssurc drop
Figllre 5.17 (4)
The hl:.ldes of an impluse turbine arc only in action when they arc in front of th e nuzzles, whereas blades of th e rencti on turhine are in action all the time. Jmpul se lu.rbines have the same pressure on the two sides of the rotor blades, \vh e r~as dlff~rc nt pres~urcs exist on the two sides of the moving bl:.ldc of a rcacUon turbine. Becau~e of sln.all pressu re drop in each sLage, the number of stnges required for a reactIOn turhln!: arc much greater than th ose for an impulse turbine ofthl! sa me pllwer.
(5)
(6)
TI~c fluid \'elocit~ and the blade speed for a reacti on turbine arc low as co mpared \\,Ith those of an Impulse turbine. IH) The variatj ~n of dill.gram efficiency wilh hlnde speed rnlio is shown in Fig. 5.1 8. On co mpann~ b~th Imp~lse turbine ~nd reacti on turbine, it is clear from the graph th.at. fo~ a rca~l1on turbine the effiCiency curve is reaso nably flat in the re~ion nl maxlmun dmgram efficiency. This point is of great signifi e3nce as Ihe s ~all
\'ar.int~ons. in th e blade speed rario (~I ), ca n be nc~eptcd without having much "nnallon
III
rhe value of diagram efficiency.
"'-
RI!:lction
turbin e Simpk: impulse
lu rbine
Sin a,
Figflre 5.18
\lariulion of diagram efficicncy wi,h blade .f{lt!ed mlio
INTERNAL LOSSES IN TURBINES An ideal (ur~illc wo.uld develop work equivalent to the isentropi c heal drop of lhe Bul. III practIce, the. actual work oblained from a turbine is mu ch less than the theore tical work (or) isen tropic work. This difference is due to the various losses ellUml!rLlICd as follows, s[cllIn.
(a) Governor valve losses Usually all turbines arc filled with govcrnors for Ihl! purpose of speed regulation. The first loss occurs here In 11ll! form of ·rhrottling· (It thl! main SlOp valve. This los5 may be of the order of 5 \0 IO ?t: in the. nozzle but the most significant loss occurs in thl.! vcJm:ilY of Ihl.! jl!l of the nozzle. This loss mny amount to another 5 to 10%
;.It
the exit
(e) Blade friction losses inlet
(0
Due to this loss, the relative velocity of !ileum allhe the hlade is correspondingly reduced al the exit of the blade
where IVH~.a coe fficient which varies from 0.7 10 0.9, which lakes imo account the loss in velocity dut.! to friction.
(d) Disc friction losses When the disc (or) the turbine wheel fOtatc Sin a Jen se (or) visco us mediulll likc steam, certain fluid resistance is experiem:ed by the.: whed. As a re.:sult, the moving Sleam creates II drag on the stearn whIch SL!ts it in mOlion. Also. a cennin ddinitc 'circa/urio,,' ofsit!um within the wheel openings is developed. thereby in creasing Ihe frictionnl losses. It is difficuh 10 reduce the Insses Jut.! to ~uch drag forces. known as 'disc/riclioll/osst!s', in turbines. This loss is about 10%. (e) Partial admission losses In Ihe first stage Df II high pressun: lurbinc. owing to co mpilra tively small area I"t:quired for the nozzles, the laller extend o\'er the whole pl.!riphcry of the slage: and thus a few blades remain partiall y filled wilh steam, in which the now also gets disturbed considerably. 1l1cre will he certain eddies produced in Ihe channels or the idlt! blades. Since the casing is full of steam. t!vcn these blildes which arc not under the direct inlluence of jets will churn (shake) the steam eddil:s und thus produce 'fall losses' (or) 'windage iosst!J' which arc knuwn as partial admission losses. Attempts may be made to reduce tilese l(l s~ e s hy limn!: slationary shit.:lds around tht: movi ng blades which are not rect.!i ving the SleJIll. (f) Gland leakage losses There is a smull loss of energy in each stage of Ihe turhine , owing LO til(! Icakagl: of steum from one wheel chamber 10 the next through the glands. Here some space hetweeillhe diaphram ami the shaft may hI.! eXls~ing ami lead to leakage. Actually, the functi on of glands at high pressure cnli is t(1 ..:hed. the It.!akage of steam to th e atmosphl:re, while that at the low pressure end j~ to prevent in -leakage of air to the turbine.. However, eve n a best typt: of gland is sus..:..:ptibk ttl leil ks and causes such type of losses. Two lypt.!S of gl
( i)
(g) Residual velocity loss In [he final stage of a IUrhine, the kindie crll:rgy corresponding \ 0 the final absolute velocity of the steam as it leaves the whed is losl wholly (or) partiall y. An·angement ma y he made 10 recover II part of the cn!.!rgy hy reducing the' velocity between the Inst stage and the exhaust hranch . This type of loss is known as the residua l velocity loss, and may he rcducctl also by providing guiu!.! va ne s in the exhaust hood to perform so me diffu ser action. This loss is eq ual 10 Ihe
AXIA l. F LUW STE.'~I ,,~ nGAsTl ; lurl sl ' :-'
::!06 };- TURBO M .... rIllNES
The stage efficiency fOf the given stage is Of
'1/
:!117
'/, R.F
,/.\
Actual enthalpy drop 'I:r=
...
=
1/.\
x R.F
Isentfopic enthalpy drop
FREE VORTEX DESIGN
'1.11
AlB, ;hC,
-=..--=-
Fre.::e.:: vorte.::x pnne.::irk i~ use.::d fur thc dcsig.n oj" lung hlilde.::s.
.
and so on.
. For constant sW-!!nalion enth:..dpy 'll.:rnss Ihc annulu s (~O) = 0 ilnd constant aXial dr
A2B'2
The reheat factor is defined as the ratio of the cumulative l!nthalpy drop Rnnkine enthnlry drop and il is given by
IR .F=
10
the
dC" 1 whirl componellt 0 I' ve ICIC .ilY Cx . . Iy proportliJll;1 . 1 - - = 0 ) . tIc IS.inverSe ( dr to the r:.luius. Rm.lial e.::quilihriulll is achie.::ved in rrcl! VOrtc.x design only when
velocit y
A,B,+A2 B2+"'j A,D
1C \, 1"
= constant I
15 .52)
rl=:;
L AnBrr For 3 stages = c"= ~.':'--:o AtD
(5.491
The.: turbine intcnlal efficiency is defined as the ratio of the total aC(UallUrbine wurk to the Rankine work.
A,C, +A,C2+'" A,D
CONSTANT NOZZLE ANGLE STAGE Constant nozl.k ang le blades lor) straight blades are simpll.! and less I.!xpl.!llSJve. compared to twisicd nozzle blades. For a cons lant absolute air angle lO' I) stag!.! imd r;ldial equilibrillill. the relmi onshir bl.!tween rnl.!an sl.!c tion and any sec ti on a1 radiu !'. f'. i!>.
~ C.I L ", (5.50)
IMPULSE TURBINES VERSUS REACTION TURBINES
rf '/.It = '7.11 = '1.\ is assumed
A,C, '1.\'
= A2B2
AIBI A,CI
AI 81
The salient c.Iiffcn:m:cs between an implusc turbin e and reaction turhinr.: arc sl;J ll.!d he low.
A:;BJ
+ A2C2 + A:;CJ + A2BZ + A:;BJ
(I) In irnpulse turhine.thc !luiu is expande.::d completely mIlle noah: and il remain ..
Il=J
L
AnCn
11 = 1 I/.T
(0.511
11=)
L
AIIBrr
11=1
From equalions (5.50) and (5.5 I),
'1s
=
'1,
x AID
rl_)
L rr = '
AnBn
at constant pn:ssurc during ils passage through tht.: moving hlade!>.. In rl.!ilc tion turhine the Iluid is only partially expanded in thc nol.l.lt.: ilnd tht.: rernainlng. expansions tilke place in the rotor blades. (2) In irnpulsc turhines whl.!l1 the Ouid glides over the moving hladJ.:s. the rcla ti vc veh1city of fluid cither remains constant or reduces slightly due.:: to friction (i.e. H'2 ::: WI )· In reaction turhine. sinn! the Iluic.l is cuntinuously exp;.mding. relative \'e1oci!}' doc s increase ( II'~ > Wd. . (3) Impulse.:: hlades arc of the.:: plat e or profile types and arc sYlllllletrical iiS ~hll\\'1\ 10 Fig. 5. 17 . Re<Jc\ioll turhine blilJe.~ h(lve nerofoil sectilln and ilre.:: il."ymmctrH.:al. The blaue is thickcr at one end (Fig. 5.17) and this provide.::s a ~uiti.lhle shaped passagc for the.:: Iluid to cxpanu .
::!t14
,
AX'AI. r-LOW STI:,\M AN OGASTu ROISr:S "" 205
TI JltUO MACI-II N':S
(e) 100 per cent reaction stage When R = I (equation 5.48) g ives a t = Q2 and C, = C:! . The velocity diagram is in-
(b) Pure impulse stage
In this stngo by deFtlh:n! is no pressure d rop in the rotor. For rcwrs ihle adiab.ltic now, the poi IllS 1.2 and 2s w ill l"lli nddt! ( f\g 5.15 (b» . (c) Negative reaction stage Idea ll y. forro\'l'fslhh: ~dl ~ha ti c fl ow, Ihe poin ts 1.2 and 25 o n the Mo Hie r c h ~rl ~D in ci d c in th e ·zcro rcac lion stuge. Tlh.:rL'fo n::. with isc ntropic flow co nditions prcv;; lLng Ihe ze ro reaclion s t:lge is exactly Ihe same us th e impul se stage. H owcver wh en the now is irreva s ible. Ihey arc no t s;uil e and in fucI nn increase In_.e llllw.lpy occ urs in the rotor of the implusc s iage (hg ..5. 15 (c» . Thi s stage is refern.:d 10 ' IS a negative reli c ti on s tilgC. 'lillian
.Fo ra nc~ati vc reaction s tage. W2 < WI (from cqualion (5.4)) thereby causi ng diffus ion of the the re la~ivc ve loc ity vec tor in the rOlor and a subscq uent rise 10 pressure . Thi s co nditi o n s.hould be avoided. since adverse pre ssure g radients c:lUsing flow sepa rati o n on .he hlade s urfaces results in poor e ffici e ncy.
o
h
clined to the ri g ht. There is nostalic e nthalpy drop in the stator (Fig. 5. 15 (e)).
, Figure S.15(b) Pllre implIlse stage ill a l/ axial gas wrbine
h
o
L---'tc:''-____~ 5
Negatilll' re(IC'rion
(d) 50 percent reaction stage Whcn R = 0.5 from equation (S.47)/h = Q'J (~I ::= (tl ~ IS:l C '. = I~:! and Cl = IV I. Th is results in asy mmetrical vd oci ty diagrum~ 10
enthalpy
111
h
the stat or ilnd rot or arc equal (Fig. 5.15 (d »). Po
11.:,
Figure S.J5(/)
Stage expanslOII walt
reacrio" more thall 100 per Ct'11I ill at. lLriai gas !f"b;ll~
Fi g. 5 . 16 shows the c x. pans ion of steam through a number of turbine slages. A, 8. repres":lIls th e isen lropic cx.pansion in the first stage. TIlt: actual stale: of Sleam with fricti o na l reheatin g is shown hy po int A 1. So Ihe actual heat drop is Ale •. Similarly, Ihe isentropi c "nd ac tual siages of heat drop fo r the succeeding siages arc shown in Fi g . 5 . 16 by A1B2, A]8] and A le2. AJCJ and so on . The drop A I D re prese nts the overall isentropic heat drop (or) Rankine heat drop between the in lel and outleL state of stearn . Th e sum tile isentropic drops in a ll stages of the turbine (A I BI + A2B2 + A,:\B] + ... ) is called the ' CulIlulative elllhalpy drop' . The cumulative enthalpy drop is always greater than Rankine e nthalpy drop (A I D) as the constant pressure lint!s diverge from
or
Po h ho=h t,f-:,'°""-~_-,;-I,/
ho
(f) Reaction more than 100% Increasing the reac ti on ratio to great er than I gives ri se to diffusion in the stator passages or nozzleswithC , < Co(Fig.5.15(f)). This situation should also be avoided because of Ihe likelihood of now separation o n the stalor blade s urfaces .
STAGE EFFICIENCY, TURBINE EFFICIENCY AND REHEAT FACTOR
Figure S.lS(c)
I hi.: drop
I'.
h
IcfliO right on the MolJier c hart.
2s 2
2ss 2s :!
s h
i Figure S.lS(e) Figllre 5. 15(d)
Ficure 5.16 Figure 5.1S(d)
Figure S.lS(e)
50 ,n'l' Ct'1I1 reaclirU/ stage i ll (1II {lxial gas fltrbill e A 100 I't'l' Ccll! reaction stage in (Ill {lxial gus turbine A
£orpullsion proCI!SS in
a
mll/rislage
rurbinl!
202 :> Twmo MACIIINI:S
2-
Cu(lan fJ2 - tan PI)
_-=-2~
=
+sin Zcrl 2 sin z 0'1
IJ/U"I _.1
2U .p(tan til - tan ~I)
2
15.42 )
I + sin 1 0'1
Equatioll (5,46) t::l1I be rearranged into
REACTION RATIO
it
2U
rhl! rl!l.IL'tion ratio U is given hy
=
~h
(:,
01-
Cl) 2
l;lnti'! ;::
(u
+C C" -,)
EquatiOIl (S.47) then becomes
C" (tan a:! (5.44 ) (1.v 1 _ Wl) =("1- /1 1 ) + ' ,
'I - "1
-
2
-
1
+ C ,)J ,-
assumcd to he cons tant through the stage, then
0.5
R
=
= 0
I
(a) Zero reaction stage
+
+
C
tan.a l )
II
2U
If R
= O. from equalio n (5.46), ~, = ~,
and frolll
=
(5.45)
arc shown in the ligurc 5.1 S la).
I',
h
s Figure 5./5((1)
+ IV" + IV" - V) (Wr2 - 'VI·, )(W,I2 + \Vt,) = W X2 - WI', = 2V(IV" + 1\1,,) 2V
~-
[Cu (tall Q:! - tan 0'1 )] 2V '
la)
2V(U
+
Equation (SAS) \\I,! W, . The conditions tlf gas through the stag\!. and Lhc accomp
(2(hOi - hOl)( W,l _ w:!
2V(C,1
+ VIC,,)
= (tana}
1'1 - "2
Remcmht!ring (IHIt hor", = "0'1,.1 then hor -h , and subsliwlin f l ' ,rl O~"I g or In equation (5.44).
=
(5.471
2U
"
"01 - hOl
R
2U
- (;lna l )]
And a third form is give n by subSliluting for
BUI for a normal siage Co = C,_ and smcc, ' I 100 = hOI in the noall!
IS
0.5 + [
2U
C,«(;.Ln f3'!
(5.43)
= (hOi, _ c/)" 2
R
[Cu lan 0'1 - !:!.. - ]
2U
=
", "1
"0 - h'2
(cuIan Ih) _
Then. R
R = SUllie enlhalpy drop across rotor Sialic emh"lpy drop across stage
II
sct:ond foml
\\1'2- W,..
111C reat:lion ralio of an axia l now turbine Vilr' . as the reaclinn rOllin of an axial now co • ~cs \~ Idely from 0 to 100 rt.!r cen l where slagc. mpressor IS usua!!)' .St.!1 al 51) p t:'r Cenl I'or (ill' .
If'C . '
(S ..lt11
(b)
7..uo reuctioll a dal
gClJ
IIIrhl1lt'
The veloci ty tri.mglc is inclined mnre towa rds the left. Idt:all y. for reversible adiabalit: now, the points 1.2 and 2s on the Molliachart slwulLi coincide. and in 1hal case no pressu re drop occurs ;n the rotor.
~no
;;.
AXIALFLOWSTEAMANDGASTlJiUlINES ....
TIJI{ltu MAC IIINES
_. mU Cw'r l + WI·1)
~OI
(or)
=
2WJ2
Wr
15.38)
Suhslituting equations (5.37) & (5.38) in equation (5.36). 2[2UC, sina, - U'[
STAGE EFFICIENCY OF A REACTION TURBINE
15 .09 )
2wi - Wf
Thl! stag e efficiency of a reaction turbine is the rati o of work don e to Ihcc ncrgy input
Using cosine rule. (5.35)
Cf
=
and
+ U"'!.
- 2UC, sinal
C,
Equation (5.39) becomes Rl.!wriling.
=
2Cr - (Cr
(5. 36)
+ U"J.
- 2UC\ sinal)
2 2[2UC\ sino!,! - U }
=
(Cr -
U 2 + 2UCI sinad
2 [2UC I sino,! - U
For a turbin l! of 0.5 n:aciion, W~
and .and
C~ a1
= WI
Considering the nUllll.!rator of equation 5.36 U(H'xl W'I +Wr~
(CX I
-
U)
=
= fi t
+
C ] sinO'] - U
2-
Q
1- ( C]
+ \V2.s in fh + Cl s in a l
U C,
where a = -
2C] sin a l - U
2UCt sin O' ] - U 1
Cons id ering the denominalOr of equation 5.36,
(5.37 )
+2-si na l
Cl 2 2 - --~=----c I - a 2 + 2a sin eq
=
(m )
.'. un\!.1 1 + WI·!) =
2 ---U-),,-=--u--
2-
Ct si n eI') - U
U2
or
+ Wx~),
w.
+ C? - Crl
+ 2UC] sin ad 2Cr ~c--o;-=c'::-::-: (Cf - U2 + lUCI sineq)
(Cf -
fh
2
1540)
d . called thl! blade to gas spec ratio.
Maximum Stage Efficiency 2
Thi s is found from the term 1 - a - 2a sin al· This term should be maximum for the tJs lO be maximum. dry,
= =
do (J
-2a -2sina,
= O. (5.41)
sin at
So,
=
=
2-
2 1 - sin 2 al
1
+ 2 sin- al
AX IAL FLOW Slh\M ·\ NIIGASTl) IUIISI;S ... I 'll)
Unlike the impulse IUrbinc , no noales as such , arc mounted in a reaction (urbinc. The rixcd hlndcs net as noz7.les in which the velocity of the steam is increased nnd they also direct the steam correctly onto tbe moving blades. The diameter of hoth rotor and casing varies (increases) towards the low pressure side. This is mainly due to increase in spcciric vo lume. as the pressu re of Ihe steam decreases. The steam velocity in a reaction turbine is not very high and hence th e speed of Ihe turbine is re/;:ui vely low. A react ion turhine is illustrated di'lgrammatically in the Fig. 5.12. In thi s turhine, the power is obtained mainly by an impul sive force of the incoming SIC,lm and small reactive force of the outgoing steam. F
M
F
M
This. c.xpansion in the. moving blndcs of a reaclion turbine gives un exira rCiJction to Ihe moving blades ovc r that which would be obtaint!lI if the blades wert! impu J.ic. This extra reaction givl!s its name 10 lilt! turbine, the 'reactioll ("rhine' . In a rt! ~II.; ti(}n turbine a stagc is maLic up of a row of ri xt!d bladt!s fnllmveLi hy a row of mnvin); hlades. Stea m cu;celeralion usually occurs ill hath the fhell anti J1II1\' IIl~ hlade rows and helH:e thc slcam passnge hetween the hlad!.!s an: Iluu.!e shapl!d. So. the hladt:~ of reactiolliurhine differ from thai of the impulse turbine . There is an enthalpy drnr in thc steam lIunng its passage through tlie hlJdes whit.::h p rodu~'es the aceder· ;)tinn . Tllc ex Il.:n I ttl whi ch thl.! en· thllipy drop occurs in Ihe moving hladcs is callcd Ih c ·tleRn·(' (!{ rClIt' ·
"
tl llll
If SO l;", Ill' Ill c elllhalpy drop tll;cur.~ in the mo ving hhu.lt.:s. thc slagc will u b~' said Sfl'7r rcac! ion stage . 1\ 5071· Figllre 5./01 Sllfieril/lflVSt'tll·rI(lc/I\·lrltll/~/I·.'i n:OLclilll1 slu~e IS morc commtm in il for (/ rt!cll.·tinll turbine reaction lurhine. BUI in an impulse !urhinc. lhe l:J1tire enthalpy drop oc(;urs in tile (ixl!d hladcs.
IlIlcl pre~~urc
VELOCITY DIAGRAM FOR REACTION TURBINE STAGE
InlCI
Exit pressure
vcloclly
Exit ve locity Figure 5.1 J
Pressure alld l'e/aeiIY varialiam acrass a reaclioll
turbine
The shape of the movi ng blades is so designed to have the reactive force of the leaving steam. 1'r1 accomplish this aren of the outici between the two moving blades will he reduced than Ihal aline in lei Fig . 5.13(a).
Moving blades or impulse IUrbine
.c
The vdm;ily dia£ram for a reaction turbine stage is illustrated in tilc Fig. 5 . 1~ . Thc tliagram illuSlra{ed is symnlet ricnl. showing equal accclcrillions in both li.'(ed ilnJ moving blades Q,nJ hence thi s diagram illustrates the condilion of SOCm reaction . Due to accclc rnti(ln in the Illovi ng hlades W~ > \VI :.lnd thus, other thin); s hcing eqlw1. Ihere is a grealcr cha nge in vd()city of whirl over tht due to thi s rhenomcnon. Howeve r in <.I n:i.lcli on lurbine. there is ,I pressurc drop across cm:h S(;l1;C :.lilt.! tJUl: 10 thi s. together with the hlade annulus area presented, an end thru st will result. As wit h the impulsl! turbine Power = IIIU(dwngc in velocity of whirl)
Mov ing blades or reaclion turbine
Figure 5.J3(a)
11li$ pressure anu velocity variations across a reaction turbine is shown in the Fig. 5.13. TIle !Heam also expnnds in the moving bl
End thrus{ (due III vl.!loL"ity chan);c):;:: m(change in vducity of nUW) However th e sw,ge efficiency must hc reiatcLI to th!.! energy avuilahlc Itlthe slage. fnlbi s C~I ~C energy avnilablc to Ihe stiJge is given hy 6ft:;:: Sp. ent halpy limp in sw,ge. StiJge dlicicnC)' Work done in
:-'Wgl.'
Enthalry drop in "tug!.!'
....,
AX IAL FI.Ow STf.AM .... NO GAsTuRIlIt>:ES "'"
For a two siage turbine.
REACTION BLADING
=
C? / 2
=
8(~}ina' - 16(~}
=
80- sinal - 16a 2
(5 .3 1)
U
where a = - . is the blade speed ratio. C, Fllr maximum cnicicncy, d'/d i"
da - :\2_17
~ s in a l
:.
The pressure reduces through succeed ing stator and rotor rows, thi! velocity is increased at the expense of pressure drop and (his necessitates a blade passage Ihat I~ convergent (owards the ou tlet. For 50 per cenl reaction the stator
B, (small angle)
o (5.m
()
sin 0' \ TIll.' ma ximum
di~gral1l
197
Convcrgem gas PDlhA , >A 2
•
efficie ncy is
strai ght I1t Ir...liling edge __ _~~~tt..~~~~==:-Almost
(5.J3)
'111i.1 . mil,\"
Figure 5.11
The mu .ximum work done is
. IVlm
R~ucfjon
blading
Reaction blad ing is often shroude.d 8tlhe tips, especially if the blades arc long. TIli~ aids in prcvt!nting c,;:cessive vibration by lying the blades together and thu s changing them from can til evers to blades fixed al both ends (Fig. 5. 11 ).
= 2U [2e, (~) - 4U] (5.34)
= 2U14UI = 8U'
From the foreg oing analysi s, we ca n write, in general. for 'n' rows of blndes the uf1timum hlade sf1ccd ratio sin 0'1 :!n
THE REACTION TURBINE Construction of the reac tion turbine is somew hat different from that of thl: impulse. tu rbine.
(5 .340)
I Will'I.. dune 10 Ih e last rnw = - of tolal work. :!" SJllal
Thu s for a three row vd ocity compounded turbine, (he ideal a = - 6- and the Ii.lst rnw of hlaJes would do only 118th .
Ctllllp.~lrIng S ill
at
lhc relatiuns a = _
a = - - lor
4
J
o~
the IOlal work.
Sinai
- 2 - for a si ngle singe impulse turbine and
two row impulse turbine, we find that for the same blade speed
-
and Ihe same no?zl!.! an£le, the slcum vl! locity at the nozzle exi t (C I ) of two rows vdoc it), cnmpoundcd impu ls('" turbine is twice that for a simple impulse turhinc. Since hlade \\'ork. is propnrlional 10 kinelic energy. theoretically Ihe work. of u two row C URTIS st;.Ige is four lime s that of a simpl e stage. for the same hlade speed.
G.ll...ll..l.I.. Steam gland
Filtcd blLide Figure 5.12
Moving blnde
Ileac/iun /tIrbille
Essentially. thc reaction turbine consists of rows of blades moun ted on a drum. TIlese drum blades arc sepuratcd by rows of fixed blades mounted in thL! casing.
AXI,,1. FLOW STl:,\.\I '\r-: I)G,\sTI.' nlltSE~ ..; II}) ]t)4 ;,. Tl IRllD MACHI NES
VELOCITY TRIANGLES OF THE TWO STAGE IMPULSE TURBINE Thl! vclnci l: Iriunglt.:s for a two stage impulse turbine arc shown in Fig. 5. 10. Fig. ) . IOl a) ;tnll (b) show tht.: combined velocity triangles of each stage .
DIAGRAM EFFICIENCY OF A TWO STAGE CURTIS TURBINE Work done pCI' unJlmass !low fro m Ihe lirsl row of moving hlm.lt.::-. IS g iven P)
U(C.1<,
+ C.I-~) + C:! sin cr2)
U(CI sin CYI
1<----- \Vork dooe ---->I
U [( IV, sin fJ ,
+ U)
-1- (II', sin
Assum ing symmetricil l blades wilh no friclioll loss, WI = W :! ilnd fi t =
tt,
2U WI sin
2 UtCI si n CYI - V
I
Work dam: fro m Ihe seco nd row o f mov ing hlade s. Fjg"r~ S./O(a)
SfIfJe";III/1USl!lJ I'e/ocil." l,.j{/1I8/~sJo "Jir:J' JICINt'
There is alwuys a ccrluin loss or vc~ locity Juring the fluw or steam over the hl'IJc ;U1d Ihis loss is lakl!ll into account hy Hllrotillcing n rnclnr cnlled blade "eI(ld,y (,(}f!fficiellt. Ulm.lc velocity cueOicicnt nV,f} is given hy HI H = W:! / Hli . The n:lativc vel oci ty of Sleam in the impulse turbinc blmlc rel1wins canslant as Ihe ste
A.~sUIl1 Jn g thai C;
= C'!. a;
=
a::! and a; = YO = Ute:! sino:!)
Equatio n (5.27 J (;'Ill be wrillen as
= Ujp.v; sin
u Figure 5. 10(b)
SlIperimpmed "floci,.\'
If .tf; =
triangles for .H!COlld stage
r,
f:I; + U)+(W~sinfl~ III
W; sin If; =
I
C; sin a; - U
DUI from equalion 5.2H C; s in a~ ;:: C ~ s in O'~
I
And C~ Sill cr,!
=
or
IV:! Sin fJJ. - U
W, sin fil
2U(lIW" + lIW,,,) . . . B 1a d c (or) dlal!ram effiCiency = .
-
-Ull
fi; (sy mmetric al blades) and tv; = W1(no hlade rricli\IIl). llu:.n (W / b = :!V W; sin fJ;
In rcac liolllUrhinc blnde5lhe steam expands as it nows over the moving blades. This incrc tvl (Rcaclilll' lurbines) tl':! .5. tvl (for Imp ulse turhincs) P(l\\'er fill' Ihe I Wll r()ws of hl mJcs = 111 U (6. 'V. + 6 IV"I ) where 6!V.,/ = change of vc.::lm:i,y of whirl or blade row I (IV:! - WI) 6. HI II I = change vdocity whirl of blndc row 2. (W~ ..... W;)
or
VIC: 1 +C:~)
Cr
- U.
=
{Clsincrl - U)-U
;::
C J si ncr l - 2U
Therefure. Nole Ihallhc rcfcrcn~c here is slillla the kinetic energy of the inpuL sleam (C? /2)
IEnd
Ihrust
= m(6C1I1 + {lCUII ) I
6C", ::: change in ilx.ia l velocity or b ladl! row I . 6Ct~lI::: change in
Nllle thaI cither CUI (or)
Th e tOlal work dnne per unit mass now (W/I11),,,,'11
\vhere
CIlII can be posilive (or) negalive.
=
(lV/mIl
+ (lV / III):!
2UIC, sinal - Uj
+ 2UICI
2U[2CI sinal - 4UI
15 .25 1
Ii, - U II
sinal - 3UI
fl:!
191
>-
TURBO MACHINES
AXIAL FLOW STEAM ANOGASTU RUlNI:.S ...: 19:\
press ure drop of the steam takes place in all the rows of \I,.'ork as nozzles. Nozzle
M
F
fi:~ed
blades s ince they all
to.!
NOt'Lh:
F
M
In S '
l' I-t~r-+
,~
lI ud
a~ ~
{t--"'ijf::=ttll
Specific ,",uhmlc
Specific volume
P' Cundemer pn:»un:
Figure 5.9 Exil \'c lociIY
PreHIlrt!. velocilY and spt!cific I'olulllt! diagrum vJ /11'0 slllgt! l'e/OClIY c()mpounded impulse IUrbi/lt! ,
Condenser
pressure Vdll..: il y of S l ~i\m HI i nl el
Figure 5.8
Pressure. "e/oeity and specific ,'oill/lle diagramJor (/ compoullded illlpulse turbille
(11'(1
stugc preSJl/rt'
Example for veloc ity compounded impulse turbine is 'CURTIS TURBINE'.. , Tht.! specific vo lume of the steam remains co nstant as thr.: steam !lows along the aXI!!. of the turbine . Hence. the blade height is sam!! in all rows. The ve locity co mpou nded turbines are mainly used as drives for centrirugal compressors. pumps. small generators nnd ror driving feed pumps in big power units,
As Ih t! press ure o f th e Sleam g radually decreases. the s pecific volume of the s te am gradua ll y increases. Therefore, the b lade heig hl has to be inc reast.!d towards the low pressure s ide. nle pressure and velocity variations across the turbin e nre show n in Fig. 5.H. Examp les for [hi s type of turbines nrc 'RATEAU TURB INE ' and 'ZOE LLY TURBINE'
TWO STAGE VELOCITY COMPOUNDED IMPULSE TURBINE
-
The arrangemcn t of n two stage velocity com pou nded impulse turbine is 's how n in lh e Fig. 5.9. There is on ly one set of nozzles and two or marc rows of moving bliJdes (in IhlS C.:lse two rows of moving blades). There is a row of fi xed bl"des in hctwc.; n the mov ing hladcs as show n in F ig. 5.9. The fun ctio n or lhl! fixed blades is o nl y t6 d irect Ih!! slr.:a m com in g from Ihe fi rst movin g row to the nexl mov ing row. So. these nrc al so kllown as 'Gu ide blades'. The cil thal py drop takes place o nly in th e no zz le at rhe first ~iI;\ge and il is converted into ki netic energy. Th e kinetic energy of the stenm gained in the nozz les is successive ly absorbr.:d by tllr.: rows or mov ing bl;ldcs "nd finnll y Ihe s t Co\ 1Il is exhaus ted from the lasl row of the blades. TIlr.: var iation o f pressu re and vel oci ty o f Ihe s team along the axis arc also show n in
[he Fig. 5.9. Figllre 5.10
Tw o sla8t! 1't/ociIY COmfl()w/(/~c1 illlJlulsr! IIIrhj,, ~
190 ,. TURDO MAClUNES
A XIAl. FLOW STEA.\1 ANOG "s T"II.HlNIi." ... 191
IMPULSE BLADING Mov ing blade
The entire pressure drop of (he gas occurs in Ihe s tator. In the rotor (he gas velocity changes Ihe direction bUI not magnitude. lmpul~c b lading is employed successfully at Ihe 11Igh·prcssurc end of steam turbines. The ve loci ty o r the steam is increased in the COnVCf!;enl
Non-
1
"
Inlet
P,
ncule row (0 perhaps 800 m/sec
before entering the rolor blades and pnssing ~hrough them at conSlant pressure . A simp le IInr ul sc turbine is shown in Fig. 5.6. Diugram dficicncy is given by equation
5.6. Inlet velocity
1
Exit pressure:
----''----'------..1-._ _
Figure 5.6
Pressure velocity diagram
for a simple impulse turbine r--~=---,,-----, ~
Vny.d
+ W ')
x ---'--'7----':':": I ,
2Ci
iV,1 + Wr1 Ca n be rearranged as lVd + iVx2 iVl si1/ PI + IV1 silt Ih pUll ing fil :::: Ih for impulse or zero reaction Wd + Wr2 WI sill,810 + W R ) Where' tv R' is the relati\le \lelocity ratio
=
=
IV.
= W,/W,
IV", + IVx ' ~ (C , sill '" - V)(I Therefore, 'lDillgrolm
_
-
+ W.)
2U(C, si n '" - V)(I
+ IV.)
CI 2[(U/Cd sin '" - (V/Cd' ] (1
+ IV.)
(5.22)
For maximum diagram efficiency. differentiate the above equation with respect to (U/CI)and eq untc to lcro . Then sin a l - 2UICI =O or . C1
~
V/C ,
~
sin ",/2
= [si n 2 a l /Z] TJ/flnx .diIlK
IIren
F;gllr~
5. 7
pU ~ ;I1:t
/11111111u rtlrb",,. lilt/do
COMPOUNDING (OR) STAGING The vel ocity of stenm entering the turbine b lades is very high. i.e. in lhe order of 1500 mI s, if the en tire pressure drop from boiler pressure to condenser prtssurc is clIrried oul in a single st<1gc nozzle as in simp le impul se turbine (DcI<1v<11 turhinc). As the turbine speed is direc tly proporliona llo the Sleam velocity. the lurbine spced is increased <1S a result of high steam velocity. Speed may he in the order o f 30.nOO rpm. Such high rpm of the turbine ro tors arc not useful for pr<1clicai purposes lJnd a reduction gC<1ring is necessary between the turbine and the ge nerator dri ven by the turbine . There is also a danger of struc tura l failurl.! o f the blades dm: 10 excessive centrirugal stres s. Therefore the veloci ty of the blades is limited to about 400 mls. T he velocit}' of the steam at till! exit of the turbine is nlso high, when <1 sing le s tage of blades :lrC used . Thi s gives ri se to a considernble loss of kinetic energy of
TWO STAGE PRESSURE COMPOUNDED IMPULSE TURBINE x (I
+ W R)
A number of simple impulse IUrbine!' s tages. arranged in series is known a.'" prt!s.'iIlr('. compoul/ding . In this, two stage pressure compounded impulse turbine. two impulse
For an ideal turbine. WR =
I:·
--f--I-__ ConstOln,-
(5 .23)
where a is blade to gas speed ratio. Thus. 'lmf/x.,Ii"N
lllis ~qu'lIio n indicates Ihntlh~ nozzle angle a I sho ul d be as high as possible. th e ideal hdng 90" . But 0'1 is limited by CII ' Large U'I means s maller ' Co ' rc.:suhing in longer blades 10 accommodate the required mass now rate. Typical nOZl.11! a n,glcs arc hctween 65° and 78°. lllc rotor b lade p:,ssages are usually llf constant · area symmetrica l cross sec ti on (Fig. 5 .7) with ink t angle (/31 ) and ou llet angle Uh) of 45 t . Delaval turbilll.! is an e xnmp1c of impulse turhim:.
= s in
2
at
I
(5.24)
turbine (simple) arc placed in series. The turbine is provided with one row of fixed blildcs which work as nozzles ilt the cotry of each row of moving hlades . 'Inc 101<11
-
AXIALFLQWSTEA"-IANOGASTURBINES <{ 189
If
~1_1C . C.x i[ ve l ~c ity j~
~odfl ('· ICJH S I ~
flO(
g iven by
used. Ih e lo [al Lo slatic ef(icicncy in lerms of hlade loss
'Its
=
LI
Category
Prime laeror
c-'_
_
-=
I.
2.
Sialionary industrial turbine: Aircraft gas turbine
Low
Specific fuel consumpLion Minimum weight and a small frontal area
::J
~ "'"i::
1.0
U
(5. /9)
" 0.5~
S.
(5.20) or
/31 + tan thY U
+ tan fJ,) I
slll,tl l ,1Inount of work
IS
1.5
Figllre 5.5
Blade loading coefficil!1It
\lUSUS fl(lw co~Dicic'nt
From Ihe graph of blade loading coefficient versus now coefficient (Fig . 5.5). it is seen that. for a high tOlnl-lo-total efficiency.tht: bladt: loading fact or shou ld be as low as possible. If blude loading factor is low, the How coefficient will be low. n:suhing in a high blade spl.!ed (U = Cf,/t/J), which is eo nsislCnt with allowab le blade stresses. However, the variation of ' 'I,,' with slight variation in the ."',. is very smaiL This is true for wide choice of reaction ratios. In contrast, the lotal-to-stntic efficiency is heavily dependenl upon Ihe re actio n ratio and ' ''IS' can be optimiscd at a given ' ,,,, ' by choosing il suilab lc value of rC<Jction.
BLADE TYPES
The impl i~alin n or: a low now coefficient is Ihal !.he frictional losses are redut~d in ~h~ s~[Jg~, Sin ce, eu IS lo~\'. 1J!, decrcascs with th e decrease in,p . This implies Ihal only U
1.0
~= C"IU
(5.2/ )
wilere ¢ is [he flow coefficient.
done pcr siage. Hen ce . fora required overa ll power output
a large number of stages :lre required .
~ 0.5
The ~vork capac.i ty of Ihl! slage is expressed in terms of il temperature drop coefficient (or) blad~ loadlllg coefficie nt '
~O.93
o~~~~~-L~~~~--
BLADE LOADING COEFFICIENT
I'~,
0 .90
2.0
""E-8 1.5 f-rom equati ons 5.15 . 5. 16, 5. 17 and Ihe above equalion. We gel
Compact and low
High
'1/1
2.5
,.
Large Ilnd
high
Upon rcarr:.mging
Cu(ran
size
and"
normal siage Co = C"! . Hence,
-#1 =
T"rbill~
h OO - h 02
hoo - "2u For
Table 5.1 S.No
'
. In s lation ary industr~aJ power pla~ts where th e specific fuel consumption is of prime Impoflanc~, a la~g~ dlilmcler, rcJalJ vely long IUrbine of low now coefficient and low blad.e loadln~ , gl.Vlng.:I high e~ciency. would probubly be ,lcccpted·. Wherens, gas lu~bme used m a~rcraft propulsIOn have minimum weight and a small (ronlal
Steam turbines a re usually impulse or a mixture of impulse a nd reaclion siages whereas gas [urbine~ lend to be always of the reaction type. The pressure ratio of steam turbines can be of the order of 1000: I but for a gas turbine it is usually within the order of 10: I. So it is obvious that a very long slenm turbine with many reaction stages would be required to reduce the pressure: by a ratip of 1000: I. If the prcssure drop per stage is made large to reduce the numbc:r of stages, the blade leakage loss would increase and lead to reduced efficiency. Therefore rcaction stages are used where the pressure drop per stage is low and also wht!n: the overall pre:ssure ratio of the turbine is relatively low (as in an aero-engine in which three (or) four reaction stages of (or) ncar 50 pcr cent reaction al thl! ml!an radius ar~ employed). So. the shape of the blade varies with the diffcrcnttypcs of stages.
J 86
AXtAL FLow STEAM AN OGAsTtJRU1:-JE..<;
,. TURBO MACIHNES
STATOR (NOZZLE) AND ROTOR LOSSES
Therefore, (h2r - h2.l.f):::: T2{.f!,s -
1wo isentropic efficiencies commonly used in axial flow turbine work arc
And (hI - hl.r )::::: Td·q -
I. Total-Io-Iotal efficiency
""
181
S!u)
51.1)
By e.,(;1Il1ining the Maltier chart. it is clear that
2. TOIal -to-slatic efficiency (S:'-I - s:'-u)
= (SI - Sis)
1. Total-to-total Efficiency (q,,) So, substituting this in the above equation. This is used when the kinetic energy al the outlet of the slage is utilised for producing work . Ex'lmplcs ror such cases arc Ihe propelling nozzle or a turbojet exhau~1. and IIny intcnnediute stage or u multi-sIlIge turbine where the lcaving kinetic en~rgy is used in the rollowing stage. In defining the efficiency, Ihe temperature limits nre taken helwecn tala I temperatures and hence it is rererrcd to as total-to-lotal efficiency.
(5.15)
The dimensionless loss coe frl cients may be defined in two ways. (i) For the nozzle
'Ill
=
'111 =
1100 - 11m "00 -
The enthalpy loss coefficient
holts
110 - 112
Iro - lilts
[for a normal stage Co = C2l
I
LN = (III -1",)/(0.5)Ci
(5.12)
I
(5.16)
(or) the pressure loss coefficient
2. Total-to-static Efficiency ('}")
(Poo -
This is used when the leaving kinetic energy is wasted, and hencc it is not utilised 10 generate work. It is defined as
(5.1)
POI )/( POI - PI)
(Poo - Poil /
npci)
(ii) For the rotor
The enthalpy loss coefficient For a normal stage
= (h2 -
[LR
iI
(5.17)
hlI)/O.5 W
(or) the pressure loss coefficient
Upon rearranging (5.14)
( POI .. , - PO'!",)/(P0'2'I' - P2)
But considering. the slope of a constant pressure line on the h-s diagram given by
Tds
= dll -
~dfl
when p = constant T = (a ll/as)p
(POI", - PO,,,,) /
TIle value of L (or) Yin the stator and rOlor gives the percenta~e drop of energy duc I.~ friclion in the blades, which results in a tola1 pressure and sti,jIlC enthalp~ drop :::Jcr~s.r.; the blades. The losses arc usually in the order of 10-15 per cent and 10 pro~ortlOn . I 1I0W " ·"n' Losses- can be lower ror very low v
For a finite change of enthalpy 6h at conslant pressure tlll '" T tls
npwi)
'I" = (110 -
",)1
[
(110 - II,)
i '( C')] + r; 2
LRWi
+2
T2
LN
(5.1 X)
I X-l
If
1
:,.. T n lUiU M M -IIINES
e,., oF C'I~ tlu:rc w ill be an axiaithrusl in the now direction . Howeve r. il is assu med
[hal [ilt! a xia l velOt'; IY Ca is co nsta nt and th e refore
\V / III
= Vel/ (tana, +
(5.4)
tan cr2)
0>,
I
AXIAL F LOW STEAM
"NOGAsT\lRlltNt:.S '"
as opposed to decclcmting . The cffecl of boundary - layer growth in the lurbine is negligible. For a normal Singe in which Co = CJ., the stalic tl!mpcratu re drop acroSS the stage equa ls the totaite mpc ralUrc drop. To - TJ. Too - T02 [ Co :: C'! I i.e. The turbine s tage to [aI4(O·t o[a l isentropic efficiency is defined as
=
(5 .5)
Equa ti o n (5.5) is u fll..:n referred to as the d iagra rn work per un it mass Aow and a., soc iillt!U wit h this is Ihl.! diagram efficiency. Dia gram efficiency (or)
Actual work dont! by tht! gas
'7'-1
=
'11 -1
= (Too - T02 )/(Too - T02.JJ)
Isentropic work done
(5 .6)
Thl.! diag.ram t.!flicil.!ncy is a lso callt.!d the 'utilisation factor' . The utilisati o n fa c to r , ~ hig.h in most modl.!rll iurb in·cs and hus a va lu e between 90% a nd tJ5 %.
Too - T02
= '7t-1 Too(l - TOluiToo)
Too-T01
= 'IHToo['-(P02IPooJ '; ']
To prove that, across Ihe turbine rotor hOI .rd =
h 02.rri.
WI! proceed as foll ows
The work done per un it mass fl ow rate is
'nle Mollier or h- s diagram ror an axi al OD
Ihm'turbin e is s hoWI} in [he Fig . SA . Tota l prl.!ssllrc [loo :.lIld total t.: n[halp), 1100 re fl!r [0 the stator ink[ conditions. For ad iabatic now through th e sta tor row or nozzle ri ng li oo = hot hU I uwing IU irrc.vcrs ibii i[ies, the 100a i pressure drops 10 PO I at stator o utl ct (or rolor inlet). Expansion 10 P02 and lotal e nthalpy /til:! lakes p lncc in Ihe rOior simi la r In th e axial fl ow compressor. In thi s case al so
IV 1m = h ot - " OJ. (11 00
p.,
h 01
Writ in g
P"
o
"'0
u
u hi -11 2
1 htl!.r'" = hO:!.r ..1 I
The work do nc rer unill11:\ss Aow by the is gi\'cn by
", _"2
g. a ~
\\' /m
= hon - 1102 = hnt - liD'!.
=
e,,)i = ex, +2 e" 12U _ (e,,_-
v = lV.1"!
C'f1
-
C..l 2
+ U = W .11
IV,,: IV"
[ (U _
e,,) + (U + e,,)]
IV,., + IV" IIV _ IV ]
(5.7)
2
or
II' /m = el'(7nl - T01)
= hO!I
!O!
P"
(5.8) Figure 5..1 (lI!
wi - Wf
Mollierc1lQrtJnl' expulI.f iolllhl'OlICli
axial flo II' steam or gas turbille stage
II I
-
Xl
III =
2
Suhslitli ling for W / Ill from equation (5.H) in equation (5.5) Hence,
(5.9) 1 hOl.rd
--
(5 . 10)
Thus
lH aue eflidenc = Dia~m work done cr unit mass fl ow Y \Vor '
h-s DIAGRAM FOR AN AXIAL FLOW T URBINE
I tiS
II sho uld bc noled that lh e work d onc factor().) is not used in eq ua tion (5 .9). This is hccilUse. in a gas or steam ItIrhine . flow Ihrou gh th e blade passage is accelerating
= IIm.rei I
"I
.: ell
= conSlant
(5.11)
I R2 ,.. TURUO MArl liNE.."
A.\!A l. f!.oWSTb\/>I .'\ I\: nG ,\s TIIIUlINES ...;
reduccd in tile moving blades. which nrc /lttached to the turbine hub and recovcred in Ihe fix.et.l stati onary hlades auached to the casing (Fig . 5 . 1).
-4--Sbaft
Stator (or) casmg Figllu 5.l
Steam out A.ialflow
1M 3
'A row of stator hlades rollO\vcd hy u sct of rotor blades is I.:unsilkred to bl! a sta!!e · . The inlello th~stator hlades is designated by the subscript O. Inlet to the rOlOrsec tnln IS referred by suhscript. t and out let from the rotor section is ind1cated with suhscrlpt·::! . All now angles arc measured from the axial dircl.:lion and can! must hl! tilken when th~ flow anglcs are measured from the dire~tion of blade motion I.C. tangcntiill dircl.:llon . The gas leaves the slater blad~s with absolute velocity Cl at angle Ct't ami hy subtracting the hlade velocity vector U. the relative velocity vector at entry to the rotor. WI is determined . In moving across the rotor blade, the !low t1irection is changed and the pressure reduced whi le the ahsolule veloc ity is dccrcased and the rcl:uivc velocily increases. The gas leaves the blm1l: t:.lngcntially at angle fi2 with n:lativc velocity W~ . Vccto rially subtrac ting the blade speed rcsuhs in the abso lu te veloc ity C2 . TIlis !s now the inlet ve locit y 10 the next stator row at angle cr~. which ror iI normill swgc equals Co nt 0'0 . For a normal s[rlge
rurbin~
The density of the gas gradual ly decreases as th e gas moves through the turbine, and to maintain a constant ax.ial flow velocity. the blade height is increased towards the low pressure cnd. The stator row is often termed as the nozzle row and in certain types of steam turbine the nozzle row consists of a set of converging nozzles spaced around Ihe drum . While.examining th e flow through the stage, the following assumptions wi ll be made.
1 C2
= Co and a2 = ao 1
STAGE WORK AND DIAGRAM EFFICIENCY The two vr.:locity triangles are superimposed upon eilch other in Fig . 5 .3. TIle em.. r!! )' tran sfer is gi ven by the Eul er's turhine equation .
1. Row condition s are evaluated a t the mean radius. unless othef\vise stated .
Work done
2. Blade height/mcan radius is small, allowing two·dimcnsional flow theory to be used . 3. Radial veloci ti es arc zero.
VELOCITY TRIANGLES FOR AN AXIAL FLOW TURBINE A sing le turbine stage ,!Od ve locity triangles arc shown in the Fig. 5.2.
u
Figure 5.3
SlIpt'rilllpOJed I·t'lod,." lroillglt!s
IE = U(C"
- C." I/g
I
(5 . 1)
and sinl:c C.{ ~ is in the negalive x direction. the work done per unit muss now is gi ...en
by £g=W j m Figure 5.2
~~/odr)'
trion1:les Jor all a.rial flow gas (ar steam) (!lrbifle slage
=
U{C.r I + C I ~}
(5 .11
=
U(lVJ"I+W'l)
(5 . 3)
5 AxiAL FLOW STEAM AND GAS TURBINES
INTRODUCTION Ax.ial flow steam llm..l gas turbines are turbomachint!s that expand a continuously nowing fluid essenlially in the axial direction . Development of th e axial Row steam and gas turbine was hindered by the need 10 obtain both i1 high flow Ciltc and compression ratio from a compressor to maintain the air requirement for the combustion process and subsequent expansion orthe exhaust gases. Inili:!.lIy. the air was provided by centrifugal comprt!5sors and lah~r the axial now compressors were developed and used in turbojet engines, in which [hI! power developed by Iht! turhine is partly used 10 run the axial flow compressor. Axial Oow turbine has very wide applications ranging from aircraft prupul sion 10 industrial :lnd marine plunts. In this chapler, steam and gas turbines are con sidered together. with the assumption that the same theory applies to both types ofmac.:hines. Thi s is only valid. when the nature of the steam used in steam turbine exists in the superh eated slate which is assumed to behave ns an ideal gas. Nowaduys. the power o utput o f steam turbines varies from few kiloW3US to 660 kW. To have a hi~h power output. su perheated steam using superheater is made to expand in the turbine to below atmospheric pressure in the condenser, tocxlraCtlhe maximum energy frum Ihe steam . Gas turbines an: used as the power unit for large jet aircraft propulsion. bl.!cause they have a very high power 10 weight ratio. In the case of air-craftjct propulsiun, 10 have enough jet thrust, hi gh axial velocities are desirable. Usually one, or at the mosl two row s of nozzles and blades arc required. But, gas turbines which arc used in industrial or mnrine plants require large number of s tages. This is required to reduce the carry ~ over loss, and to rcduce the load on the blades to give tbe turbine a long working lift! . Steam turbines arc used in fossil fuel power stations and for steam dri ven propulsion in ships. although gas turbine propUlsion units are often filled in the s maller class of naval vessel s.
DESCRIPTION The principle of energy extraction from the gas is gradually reducing the high pn:ssurc e nergy by converting it into kinetic energy. This is accomplished oy plissing thc gas alternately through rows of fixed and moving bhldes. The kinetic cnerM-Y of the gas IS
178 );-
A XI A L FLOW COMPRESSORS AND FANS ...:
TURBO MACHINE.';
(a) rotor blade angle inlet. (b) static pressure rise , ec) overall efficiency for a
power input of 1.15 kW, (d) degree of reaclion and (e) pressure coefficient. IAns: (a) 75.7" (b) 59.7 mm WO. (c) 65% (d) 52.6% and (e) 1.89J --1.36 . In an axial now fan the rotor and inlet guide vanes arc symmetri cal and arranged for 50% reaclion . TIle hub and lip diame ters arc 45 em and 75 em res pectively. Speed is 960 rpm . The motor power is 6 kW. The sUltie properties of ai r at inlet are 100 kPa and 305 K. If the fan efficiency is 82% and the mechanical efficie ncy of the: drive is 87%, fi nd . the rotor angles at inlet and exit a nd pressure coclTicient. if the quantity of air handled is 6 mJ I s. IAns: (a) iii = "2 = 52.5". fh = "I = 6.6" and (b) 1.371 ·LH . The first stnge of an axial now compressor develops a pressure ratio of 1.2: I. The inle t pressure nnd temperature arc I bar and 3 1° C respectively. The overall cllicicncy of the compres sor is 83 %. The now coefficie nt is 0.47 . The velocity diagram is symmetrical a nd at mean radius. the ratio of change of whirl velocity to axial veloci ty i5 0.5. Determine the compressor speed if th e mean' diameter is 50 cm . (MKU-April '93)
[Ans: 11056 rpm J 4 .38. TIle condi lion of ai r al the inlet of an axia l air compressor is PI = 768 mm of Hg . TI = 4 1° C. At the mean blade section, the diameter and periphernl velocity arc 500 mm a nd 100 mts respectively. /31 = 5 1°. al = 7°. fh = go . mass now rale is 25 kg/s o Work do ne facto r is 0.95 and mechanical effic ien cy is 92 % s tage efficiency is 88%. De lcrminc (a) a ir ang le a t the stalor entry (b) hladc height at cntry (c) hub to tip ratio (d) stage loading coefficient. (c ) stage pressure rnti(l (f) powcr input a nd (g) relative machine number altht! rotor inlet. IAns: (a) 50.2" (b) 0. 19 In (c) 0.45. (d ) 0.75 (e) 1.08 (I) 204 .7 kW and (g) 0 .33 j 4 .39. An ax iil l cO/llprcs .~or slilgc has meon diame ter 600 mm and runs al 250 rps . TIll': lIl:(ual tcmperalurc rise is 30° C nnd the pressure ratio developed is 1.35 . Inlc[ temperalUre is 35°C and the temperature rise in the rotor is 20°C . Mass now ratc is 50 kg /s and the mechanical efficiency is 85%, determine . (a) power required to drive the compressor (b) degree of react ion (c) loading coefficient (d) slUge efficiency. (MU-Oct. '99) IAns: (a) 1773.5 kW (b) 0.67 (c) 0. 136 and (d) 9 1.9%1 4.40. Thcl'onditions ofnir at the e nt ry of an axial compressor s tage are PI = 768 mm of Hg :Jnd T, J 14 K. The ang les at the me a n blade sections arc 0'1 7D • f1, ::::: 51 0 . fJ1 = 9°. The mean diameter and the peripheral speed arc 500 mm and l OOmIs respectively. Mass flow rate throu gh the stage is 25 kgls, the work done factor is 0.95. mechnnical ellicicncy is 92% and stnge efficiency is 88 %. Assuming free vortex now. determine (a) nir and blade a ngles of rotor. (b) no w co· eHicil!llt. (c) loaciil1l; co· efficie nt at the huh , mean and tip sect ions . IAns: (a) ClI .m r Cl1.m = 50.2" . P I .n! 51 (' . fi1 .11J = ly. 0'1.1 = 5 . 1° 0'1 .1 =40.99 Cl . f11.1 ==60.7° . f3~ . 1 =45.2" , c O'\.,- = 11.2 1°. l:r1.r == 62.68 . Pl.r 32.75° . /h.,- 47.58° (b) 0.74. 0.53.1.18 and (c) 0.75. 0.395.. 1.96J 4.4 I . A n axia l flow cmnpressorcomprises a numhe r of si milar slages with equal work done per sHlge. The :lxinl ve loci ty rema ins conSlant through out lill! c ompres sor.
=
=
=
=
=
=
179
Q vcr
4.42. Find the poly tro pic efficiency of an
4.44 . An nxial f'ln consisting of rotor (no IGV) has the following dn1
=
4.46. An axial c ompressor stage has (he foll owing dalil : pressu rl! and 1l!lllpcrature al enlry arc 1.0 har and 20" C. mcan blad e ring diametcr - 36 f.:1l1 . .. pceJ-1 H.Of)() rpm . blade he ight iltcnlry-6 em, degree of reacti o n-50%.. axia l VcltlC lty-1 XI) m/s. Air an gle s ilt rotor entry il nd stal(lf ex it - 25 °. Assume forced vortex 11 0w. Dc tl.:rmine (a) ro tor blnde air ang lcs (b) degrec of rcal.: lion Ie) ~ pccllic work (d) now codficlenl and (e) luadmg coerrlCLen t
=
=
(cl 0.5 3 1. 0.678. 0.41 8. 0.7929. (d) 0.505. 0.505. 0.505 . (e) 0.5. 0 .5. 0.51
176
;...
TI IRBO M ,\(,'II INI:S
AXIAL F LOW COMPRESSORS AND FANS
th e mass now hy 7 per ce nt while thc blade speed is kept co ns t;)ni and it IS assumed llIlLh.=r II1I S nl.!w cond iti on tlwt the relativt! flow eXit ~l1 g b, for hath the ro tor and .s ta(~~· remain unl:hnngL"d . What is the s til~e loadin g and rC.lctio n il t Ihe nl.!\V condillon? Assume Ihe work done faclnr is 0.9.
.
IAns: 0.435 and 0.64 1
-I.2J . An aXial flow compressor deli vers a IOlni pre ss ure mtio of 6, the tolal heild pressu re an~ tt.!l1lpC~aIUrc a~ e ntry being 0.408 mP.. a nd 300 K re.~pl!c lively. and ~hl.' over:tll IsentropIC effiCiency bdng 82 per ce n!. The deg rec of reaclio n is )0 pe r cent and all stagcs co ntri bu lc a n eq ual amou", of work. A I a panicu lar s lagI.'. th l! hlndL' SPl.!cJ at ~he mean height is 20J m/s. and the axiul veloc it y is 17 1 m/s. If I hI..' ahso lule il lr angh: en te ring the ro tor 111 this slagI.' is 15" and the work don e fac tor is 0.92. detenn ine. (n) th e rolor air in lel angle . (bJ the numbe r of slllgcs required. (c) Ihe s ta lic tcmperaru re o f Ihe air at entry 10 Ihe rotor a nd It!) th~ rutor inlet rcln live mnc hinc nurnbe r.
.
.
lAns: (a)42.6° (b) 12 (e) 284.4 k and (d) 0.6871
--l .2-1 . A nlUltlstage ~xl:l1 compressor is required for compressing air nt 293 K th rou gh :1 pressure milO of 5 to I . Ench stage is 10 be 50% react ion and Ihe mean blade speed 275 mIs, now codlic icn t 0.5. and the slage loadi ng faclor 0.3 arc taken :\... C~ nS L.1.nt for a ll stages. De te rmine the now ang les and the numbe r o f sia ges J:c4u~n:d If til e. siage dlicienq is 88.~%. Take Cp = 1.005 kJ /kg and r -= 1.4 lor 'IIr. A Iso. lind Ih l.! ovcra ll efficien cy of Ihe compressor. . IAns: (a) " I ~ 1', ~ 35" und "2 ~ 1', ~ 52.45° (b) 9 (e) 86.3%1 . .;'25 . An axm l now c(l m~rcsso r has 10 singe s und the follOWing datil apply to each s l l~g: at th e ml.!
' . IAns: (a) 1.14 (b) 5.461 4.26. Each s tage of an a:(Ja l now cumpressor o f 50% reac tion, has th e same m e.m hlad cspeed und
S:1Il1C
flow outlet angle of 30 ° rcl
cod~c.ienl i ~ ~.5. and rcnwins Co ns tant. At emry 10 (irsl stag e th e stagnation L"OIul IlIOn .01 :ur IS I~J3 kPa a nd 278 K and s tatic pressu re is 87.3 kPa and Il ow a n:a IS O.J72 m- . Us ing compress ible now nnalysis , find mass fl ow ra le and Il ow ve loci ty. Pind the shan power when there are 6 such s lages when llIeL"h<1nical effici e ncy is 0.9.
compressor. IAns: (aJ 19,070 rpm (b) 3.07) 4.29, A multi stage axia l flow compressor is 10 have constant axial veloci lY of 150 mls and 50% reactio n. The pressure rauo developed is -1 a nd Iht! inflnitcsimal Singe efficicqcy is 85%. The tcmperntun: al the entry is 20"C. The mean diameter of the bl:ldc ring is 35 c m and speed is 15,000 rpm . The exit angle of !.he blildes in each row is 27°. Calculate Lhe blade angl e at inlet, the number of stage s and pressure ralio of eilch s tage. (MKU-April '96 )
[Ans: (a) 52.92° (b) 6 und (e) 1.381 4.30. A tcn siage axial flow compressor has a pressure ratio of 6.6 and iSl!ntropic effic ie ncy 90%. The compressor hl:ls symmetrica l stages and the compression process is adiabatic . The axial velocity is uni fonn across the stage al 125 mls and the mean blade speed of eac h s lage is 200 mls. If the air at 27°C e nters the compresso r at the ratc of 3 kg/s, detenninc the direclion of air al entry and exit from the rotor and s la tor blades. Also compule the power supplied to air. (MKU-April '92)
[Ans: (a) '"
~
!h =
17.8° und "2
~
1', ~ 51.97° (b)718. 17 kWI
4 .3 1. An axial cOf1)pressor swge has n mean diameter 55 cm and run s at 15.000 rpm. If the actual temperalure rise and pressure ratio developed arc 32:"C and 1.4 respectively, determine (a) the power required to drive the co mpressor while delivering 57 kgls of ai r. Assume mechanical efficiency as 85 per c..:nl a nd ;sn ini tinllempcrature of 35°C (b) the stage efficiency and (c) Ihe degree uf reaction if the temperature at th e rotor exit is 55 °C. (MKU-ApriJ '98)
IAns: (a) 2156.6 kW, (b) 96.8,", (e) 0.6251 4.32. An air compressor has eight stages of eq ual pres sure ralio 1.35. The Il ow ralt! through the compresso r and its overall efficiency an! 50 kg/s and H2% re spcc· lively. Iflhe co nditi on of air at entry ilre 1 bar and 40°C, dctcrmine (a) the statt! of air at the compressor exit . (b) poly[ropic efficiency, (c) efficiency of each stage and (d) the power inpul assuming overall dfic i.:ncy o f the drive as 90%, (MKU-Nol: '961 [Ans: (a) 11.03 bur and 689.8 K, (b) 87.1 % (e) 86.5% und (d) 21.03 mWI 4 .33. A fan takes in 2.5m 3/s o f air at 1.02 bar and 4ZOC and delivers it at 70 em W,G. and 52°C. Determine the mass now nUe through Ihe fan . Ihe power required to drive Ih e fan and the stulic fan efficiency. (MKU-No\'. '96)
IAns:
~ . 27 A n aXliI lllow clHnpressor has conslan l aXia l ve loc ity Ihroug houllhe compressor of 160 m/s . a mean bliJde speed o f244 tn/s and del ive rs a press ure ratio of5: I. E;lc h stagc is lIf 50% re:lL"tio n and the rdativc outl e t air ang les are the samc. 30ll , for cach stage. If n poly tropic efficien cy of 88 per ce nl is assuml.!d, determine the 1ll~lllber o f stagt!s !n Ihe co mpressor. I Ans: 14 )
4.34. An ax.ial ducted fan wit hout any guide vanes has II pressure L"odlicicnt of 0.38 a nd delivers 3kg/s of ai r al 750 rpm. Its hub and lip d iam elCr.i arc 25 em a nd 75 cm re spec tively. If the: L'ond ilions a t the entry arc 1.0 bar a nd 38 ~ C. delcmline (a) air nng lt!s alIhe e ntry and e x. it, (b) prcssurc devd oped in mm W G, (c) fun . efficiency aml (d) power re4uired to drivt! the fan if the oVl!rall cniciency of th e d ri ve is 85%.
-1 .18. A he hc.oP.ler gas turhln c plant consis ts of a four stage ult ial flow compressor. The aX I;}.' co m prL'ssLJr has stage tc mpernture ri se of 30 DC, us ing sy mm e trica l stages wnh
4.35 . An axial fan co nsis ting of rotor only has the foll ow ing data . Huh a nd lip dj . . umetcrs arc 30 and 60 c m respectively. Speed 1000 rpm. ft!lalivc air exit angle 12°. a xial ve loc ity 6 mls. lnld static properties: 10 1 kPa a nd 3 15 K. Deh:rmine
.
-
(MU- A,/rll '97)
Calculate the required rotational speed. Assume a workdone facto r of 0 .86 and a constanl axial velocity of 150 mls. Estimate the total press ure risc ac;:ross the
IA"s: (0)56. 1 kg/s( b) 132. 1 m/s ulId (e) 1102 1 kW I
IAns: (u) 70.85' , 66°. (b) 8.36 mm ofWG.
174
~
T URDO MAClUNE.s
A XI,\ LFLQWCOMI'RESSOHSA:-JIJF"ss "
4.3 1. Deline compressor stage efficiency. ~32 . What is compressor stall ? ·Lt':\ . What is rotilling compressor stall '! ·LJ..J . W hat is compressor s urgi ng':' 435. Why lire lo nger bl ades used in aircraft compressors? -l ..':\o. Deline radial equ ili bri um in compressors. 4 ..17. What arc free vortex blades? 4.38. In free vortex blade dcg ret! of reaclion decreases from lip [0 hub. (Tru clFa lse) 4.39. What are forced vortex blades? 4.40. In muhi slage compression. the slage pressure rise for the same te mpe rature ri st! is equal. (TruelFalse) 4.4 J. Overa ll tolnl pressure ratio o f a multistage axial flow compressors is e4 ual to
~. I
J. Prove
4, J 2. W hat is fn':l: vo rte x blaJ c'? De rive the work t.lone and reacllon rallo for il fr ee
\'l1f1 ex hlad!.! .
la) I1J"'.1:4· :;..(h)
ta) (D. TO.ftll~r,"f) ~" . ,
,
4 . 1h.
4. 17.
~/, · r
(e) (6. TO.III/m lll) ,..:'t
EXERCISES 4. 1. Draw a skc (l.; h of an axi al now co mpressor wit h inlet g ui de vanes and exp lai n th e working princip le of the compressor. 4.2. Draw ve loc ity triang le at th e enLry and ex it for the ax ial compressor stage. o. Draw th e h-s diagram for a complete axial fl ow compressor stage. 4.4. (a) Whal is the work done fact or for an a,.;:ial compressor stage? How does it vary wit h Ihe number of slilges? (b) Define: (a) now coefficient (b) s lilge load ing and (c) pressure c,oefficient for an axial com pressor stage. 4.5. Deflne the degree o f reactio n for an axial now compressor stage. Prove the followi ng relati ons. I (0) R = :1
= II +
R = II + .p( l an iJ, -
Ion adl/2 Ian ", >1/2
4.6. Draw supe rim posed veloci ty triangles forthe fo ll owing axial co mpressor s tages. (a) R 0.5 (hi R < 0.5 (e) R > 0.5
=
4.7. Pro\'e Ihat p] - P,
(a) - U ' l
p
Ih) 11.1
-
= fj)( tana 2 -lanaI)
= (Dof') .•,ugr/pUC,, (t
4.8. What is s urging in ax ial-now compressors? What arc its effects? Describe hriefly.
4.9. What is
R
ON+I
=
ryfllm", ...ullr
(.~) "I,·r/lr - ll ONt-!
TOI Com pare
(b) (l::J.lij."tJtrrtll) "rli""=T1
(e)
the co nditi on for rad ia l equilibrium is J d '1 d 2 , - (rC r )-+ -( e," :::: 0 r· dr dr
!!..::11
Ib) R
th ilt
175
s l~dlin g in an axial compressor 51 age? How is il developed? Why is it
ca ll ed ro tating stall?
4. 10. What is radial equilibrium method? E xp lain bri eny.
4. J 8.
4. 19.
=
IAns: (0) 50.24". 35.79" (b) 3 17.5 ml' lei 0.4~ and Id) 260kW I 4.20. An ax ial flow compressor stage is to be des igned for a stugnulio n lemper;J1urt! risc of 20 K. The work done r
=
=
=
=
-
17:! ...
AXIAI.FLOWCOMI'Il.ESSOIlSANUFANS .;,;
T I)IHlD M . \CIII NES
3.5
c"
~
1.11 3 x OA15 ~
e"
7.399 m/ s
~
7.3~9
Ti = 20 ..J2 = 0.36 0 ..,6
=
(d) Rotor inlet and exit angle
[;m(i I
-
C.r I = O. No inle t guide valles 11' / 111 =
Uc. r-:J.=U'!.( I - r/JILln/'1-'!.)
~3 . 395
20.42'0 - 0 .36 tan
fI,
1301
Ii I
Ian
111- +2cp
'"
+ 0.56
tun 65.77
fl t
~
711.13"
(e) Pressure developed oP
=
p:.. ~tLW5
1.11:1 )< MJ.395
=
\)1.H1 N / m'!.
9.46 :x IO-
=
l
rn
9,46 10m of W .G.
SHORT QUESTIONS
-
4. 1. \Vh;'!! is an a:dal now compressor? .J.2. \VII;l! arc [he :Jdvantagcs and disudvanragl!s of an axia l now compressor?
4.3 . Thl.! axial !low comp resso rs arc idl.!al for conSl
4 .5. In thl! compressors . lhe bladl.!s (orm a diverging pussagc and the (luid is dc cc lc rulcJ. (True or Fnlse) 4.b~ The ::uJvanl
(ai "nl", = "02.", (b) hili"'" < flO'!.."d (c) "Ol"e/
65.77'
J7 :l
;>
hO'!. .rd
4 . 16. Draw thc Mollier chart for an axial flow compressor stage. 4.17. The axial flow compressor receives no contribution from the change in tan· gc miai velocity. (True/False) 4.18. Delinc a nd ex press the !OIa!·to·tOla! axiaJ flow compressor cflic.:icncy. 4 . 19. What is work done factor ? 4.20. Work done fnclor - - as the number ofaxial now compressor siage increases. 4 .21. Delin e stage loading . 4.22. Define reaction ratio. 4.23. Rcaclion ratio of n axial now compressor slage is (il) R ¢ tan /3",/2 (b) R :;;: rp tan Pili (c) R :::: 2tfJ Ian /3111 4.24. Ddine now cocfficienl. 4 .'.!5 . Define reaction ratio for incompressible flow machines. 4.26. For a reactio n ratio of 50 per cent (a) the stat ic enthalpy and tcmperalure increase in the stator and rotor arc not equal. (b) Ihe superimposed velocity diagram is nO{ symmt.!lrical. (c) none of the above. 4.2.7. For a reaction r
=
170
~
AXIAL FLOW
TI :nno M"nIlNt:s
Till! dcgrcc
or fI..'Ul:[lun can also be found from C"
2V (,an fl '
R
(a) Overall efficiency
+ 'an Ii,)
'1m X '11 = 0 .9 71.1%
=
~ (lan,81 + tan ,82)
Ideal work.
U2
Example 4.15 An axial blower supp li es air 10 a furnace nt the ralc of 3 kg/s. lllc u[m osphcril: condi tions heing 100 kP" and 3 10 K. The blower cfficlI!ncy is HO~f. <.lOt..Imcchanicul dlicicl1cy is 85%. The power supplicd is ~O kW. Estimu[c thc overi.ll effi~lenl:Y :.tnd pres!'iurc develuped In mm \V.G.
Solution
Or
P, = 100 kPa = 0.85
k~/ s
I/m
T,=310K Power input = 30 kW
(a) Overall efficiency
andU=
=
0,
+
,
/).P
0.68
p
3 x 0.26 = 0.78 III 0.78 + 0.26 = 0 .52 m 2
=
20.42' 04 -- x .
=
83.395 J/kg
/).P
/).H
=
p
Power inpu t
0.68 x 30 x 10' x 1. 124 3 7M3.2 N/m' 779.12mmW .G.
'D"" = '1m
=
rit
0.16 m
D" D,
0.9
'If
3.5 kg/s 1
3 0.79
750 rpm
N
P,
=
98A kPa
Tt
=
)5°C
3.5
=
291.88 W
x
'1m
291.88 0.9 324.32 W
(c) Flow coefficient ttl
c"
pA
Solution OA
=
83.395
Power n:quircd
Example 4.16 An axia l fan without guide vanes has a preSSl1re coefficient of 0.4 and delivers .' .5 k~/s o f nir al 750 rpm . Its hub diameter is 260 mm and hub [0 lip ratio is 113. 111C sliui(.· properties al entry 98.4 kPa ,md 35 uC. delcrmlllc (a) ovcrnll efficil!ncy. if '/111 = 0.9. (b) power required. (e) now c:ocfficicnt (d) rowr inlet and exit angle . (c) 6P in mill ofWG ir 'If = 0.79 .
=
2
/).P -, m
P/RT =100 , 10'/287 x 310 = I. 124 kg/Ill"'
p
rrDIIIN 60
Dh
,it(/).P / pl
=
V'
rr )( 0 .5 2 x. 750 = 20.42 m/s 6U
(b) Pressure developed '10
=
.
x
1
Now
'III )( '1m = 0.8 )( OJ~5
'In
== =
-
U
Ml!an vc\ocity.
Dill
= ;\
0 .79
,it (/).;)
tv
':"P / p
'I" = D.H
X
(b) Power required
0.245 - ([an76.23 '" +lilO 10"' ) 2 0.522 or 52.2%
111
CO~'I'RE5SO ltS "NLlF"'..:~
p
=
2 RTt
= 98.4 X 10' = 1.1 287 x 308
]J kg/m'
':.(D 2 _ Dh') = ~(0.78' - 0.26') 4 ' 4 0.425 m:!
...;
\11
16H
AX I AI.Fl.owCOM I' I( ~SSOlISI\NI)FA:-;~ ...;. 161)
;. T! /fUH) MM' I IiNES
whc.:r~
Dm is [h e Ill l!an rotor dium cter
22.62(22.62 - 5.542 tan 10 )
Dill
.-
U/U
=
0.6 + 0.3 D, + D" = 0,45 = 2 2 Jr x 0.45 x 960 = 22.62 m/s
C"
=
0.245 x 22.62 = 5.542 m/s
=
489.56 J/kg
'" = pQ = 1. 11 25 x 1. 175 = 1.322 kg/s
60
:. Workdonc
The now rate, Q = 0.2 12 x 5.542
m( Wjm)
=
1.322 x 489.56 647.198 W
Q= 1 175m.1/s Ovcm ll cftici ency
(b) Static pressure rise across the stage (l!.P)"
= -2I p(W,-, -
,
Work du ne
=
'1u
IV;-)
Power rc.:qui red
-
647. 198 1000 0.6472
hu,
wf = U~ + C,;
64.72%
~m d
Example 4.14
De termi ne for the fan Slil gC in prob lem no. .:.I. U
(a) rotor bl>1de ang le at the r.:ntry (b) dt!grec of reaction
(l1P) ,u
=
=
1
zP 1
IV'- + C,;, - C,;, , .
,
, ta n-, P2)
C(~
,
Solution
(a) Rotor blade angle at the entry From !hI.: in let vd oc ity tria ngle for fnn stage w ith out any guide vanes (Fig. 4. 11 ), U lan /31 =
ZP IU - - C,; tan- flll
C,~ tan- fll ! = 'iI PU 2II- u1
C"
1
I) " lPU- II- .p- tan-fll !
=
-,C
fl,
=
tan
fl,
_.
76.23<=
2 62
-.- ).
5.542
Figure 4.11
U)2 x loj
p=P/RT
='---:;-:--0 = 1.125 kgjm 287 x 3 16
J
" 1 -0.245 :;1 x 1. 125 )( 22,6:2-[
=
:2R7.27 Njrn ~
=
0.029 m \V.G .
(b) Degree of reaction x tan::! lOll )
R=
(.o.P),OJorl(.o.~o)nat:c:
For an axia l fa n stage co nsis ti ng of only a mtor, (l1P)SLa1: C
(.o.P)rolor
2S7.}? N/ m'
(c) The overafl efficiency =
p( W / 1/1 )
1.1 25 x 489.56 Sincl!
e = 0,
550.76 N/m' 287.27 _ 5'2 550.76 - o. -
,j
\V / m
= uC.r~
U(U - IV,, )
=
U(U - C"tan/h)
:. R :::
52.1%
-
166 ,. Tllltll O MAr· tU NES
(b) Power required to drive the fan (e) The overall tota/-ta-total efficiency
W
mCp ( Tl - Td
Tow l pressure at stator exit PO.'
2.R2 x 1.005(325 - 3151
=
=
Tolal pressure at stator inlet - (6 PO)sllIgc (or at rolor ex it )
=
4.78 - 0.35
=
4.4) cm of waler
28.34.IV
IV
(c) Stati c fan Efficiency
Theoretica l tota l en th alpy change across the stage (6Ptl)st~ll c
P
1000 x 9.81 x (4.43)
Pwg hul 100 --=--~~::::..!... p 1.184
367 J/kg Thl!n,
(ll.hn, h lllGe
'It -,
( A hO)5IilGc
:. 111
(AhOshulgc (6 h oholor
367462.98
Ex ample 4 .13
79.3%
ell PhOlor = (6 Phtngc
An axial fan stage consisting of onl y one rotor has th l! following
3 16 K. Determine the 1low rate. the s\;l lic pressure ri st! .md the ove rall d li clency.
IMKU-Ap,·il ·Y51
357.93 357.93 + 101.58 77.9 %
Solution
Exam~le ~.12 An ax ial flow rim takes in 2.5 m J /s of air at 1.02 bar and 42"C antJ c1ch\'crs H at 75 em W.G. and 52° e. Determ ine the mass now ralC throuoh the fan. Ihl! power rcquiretJ 10 drive the fan and the static fan efficiency. (MU Dc; '97)
D,
= 0 .6 m
D'I
PI
= I.02bar
T I ~:"\16K
=
0 .3
OJ
N == 900 rpm
(a) Flow rate
Solution Q=2.5 m .l /s
325-315 63 _06%
data: rotor bl<Jlk air angle at ex.it: 10°: tip dia: 60 cm: hub dill: 30 cm: speed: 960 rpm: power required: I kW; now coefficient: 0.245; inlct fl ow conditions: 1.0:?' bar and
(f) The degree of reaction for the stage
R
=
P,=1.02bar
T, =3 15K
t:. H= 0.75mW.G.
Q
=
pQ
1/1
=
P, RT,
1.02 x 10'
1.1 28 x 2.5
//I
=
1
= 287 x 3 15 = 1.128 kg/m-
2.82 kg/s
U
AC"
"(' "4 Dr - D;;')
II
(a) Mass flow rate
p
T,=32sK
=
~
=
D.21:! m:!C rr =
=
2 (U.6 - 0 .3')
rr DmN 6U
q,U
Powe r
l' AXI,\I . fl.ow COM I'IIESSOIIS "N il F ,.. .... s
16:'i
(b) The static pressure rise across the rotor
Arca of IInw
A= =
~
(D; - Di) = ~(o2' -
0.01914
0125' )
ell
C, 0.5
III
=
u
=
IT(D"
2 x 60 IT x (!I.1!5 + 0.2) x 3600 120 30.6 m/s
I !':l P lrmnr
= U(C X1
~93.35
=
357.93 N/m'
=
J.tiS em o( watcr
The ilt:llml :.t:Ult: ~Iltha lp)' I.:hange at:ross the stator is
(ci - Ci) j1
eX,)
(26.78' - 22. 1')/2
,Sin!;!.! tlie !lawai inlet is axia l, Cx , = 0 and (rom velocity lrianglcal oUllet (Fig. --1.2) ,
lan fh. x .,n(P r - 20)
U - \\I.\"2 == U - CII
30.6 -
(C!.ho),,,,,,,
22. )
- 135.43
(c) The static pressure rise across the stator
l6.h)SI01I0r
-
114.38 J/kg Th.: thcon:tica l Sialic enthil lpy cha ngj.: across Iht.! stalor is
= =
.10.6 - 22.1 x Ian 35 0
15.l.lm/s
0.75 x
=
30.6 x ( 15.1 3 -0) = 462.98J/kg
85 .79 J/ kg
nl\! ise ntropi c Inial ~nth nl py risi! ,lC ross the rotor (6h o.)rt"or
=
'Ie x (6hO)rOlor
= =
0.9 x r.t62.98J ~
(APU)rOI(lr
=
I1J;1011ur
(l:lh.r)stnIlJ,
(6P)m,lIlr
11~.3 8
p(6 h "' )SI0110r
=
16.7 J/kg
71.1 K4 x HS .79
101.5 8 N/ m' 1.04 em or wakr
p(6.ilo. )rolOr
(d) The change in total pressure across the stator
1. 184 x 416.7
=
493.35 N/m'
==
5.03 em or H20
(/:).Po)mlor
St;lgnntion pressure at [he rotor c:(i l (6.PO)rOlor - Prcssur\! drop a[ intnke
5.03 - 0.25
X (6h)nalor
Till.! stalic pressure n s\! aCross the stator is
Thc tlHaJ pressure rist.! across the ro\Qr is
=
j2
=
The aClu;1I IOI~I I.!lllilalpy risl.! :lcros.~ [he rotor is ( 6 h O)rt',or = U 6Cx
J
493.35 - 1.1 84(26.78' - 22.1' )/2
P hOlur
+ DdN
i:15.
26.78 m/s
= : . (Do
bl;.lllc vdOl.:ily
22.1 mjs
/22~I:!
=
.184 x 0.0 191 4 2::!.1 mjs
=
!c,;+·cI
C, pA
(ei - en /2
(6. Po hOlor - p
(Do P )rOlor
m'
TI1C axia l veloci lY
M~all mill!"
....
+ p (C '
(/:).P)~IDtor
= =
101.58+ 1.184 x (-114 .38)
J
-
-33.85 N/OI'
ThaI is. thL! lotnl pr.:ssurc across the stator drops by
or:: 0.35 em of wa ler.
C')2 /'-
=
, all
amount =
:n.HS Njm-
AXIAl. FLOW C OM1' lt ESSOltS AI'o:ll FAt>1S '"
163
162 ;. TURBO M .... CHINES
=
0'1.1
Vh - C" l , ~
Ian /fl. I.
=
CliVI
150- 11 5.]2 tanfJu
155.89
CIII.I
12.5 e
fl~."
tan
R
2U(Cx ,
53.65"
rl'2.1
- - = --
flU,)
' , C (~ -~
(C(lloh)-
C XLn • - - Xf,
rm
0.212 153.76 x 0.265 = 0.212
= 57.18 m/s
= 192.2 m/,
Axial velocilY at lip inlet
K I - 2C'..... 1.,
26686.125 - 2 x 57.18'
=
cas fiL l
CO!';
2U, (C.(~ .• - C,q,,)
141.9 )' ( 142 )' ( cos 5:'.65° cos 22. 15° 2 x 250 x ( 192.2 - 57. 18) 0.5
V, x 60
'm
142
22.15·
( COl.,)' (C',Ih.,. ),
R
(3) Tip section
- - x,, =
=
(b) Degree of reaction
155.99)' (155.89)' ( cos 36.56° cos 12.Y' 2 x 150 x ( 115.32 - 34.3 1) 0.5
=
192.2
C IIJ ,/
{h, =
2U" (C,fH - C. . . !.h)
250 x 60 = IT X 9000 IT X 9000 0.530 m C'I.. 45 .75 x 0.265
=
tan/h"
-
250 - 57. IM 141.9
Crtl .• 14 2 53.54· U,- eX! " 250-192,:!
=
Cx ,)
-
=
fJl.I
C " ..
(b) Degree of reaction
Tip tJimneler. D,
21.95· V, - C.{".
20147 141.9m/s
Example 4.11 A single-stage axial flaw blow!::, wi th no inlet guide vaneS run~ at 3600 rpm. The rotor tip and hub diameter are 20 and 12 ,5 cm. respectively. ·Illr.: mas!'; flow rate of air is 0.5 kg/s, TIle turning angle of the rotor is 20" towards Iht: axial direction during air flow over the blade. The blade angle at inlet is 5Y If the atmospheric temperature and , pressure arc at I atm and 25°C. respectively. as!';uming constant nxial velocity through the mnchine. find (a) the total pressure orthe air al the exit of the rotor (the rotor total-Io-total efficiency being 90% and the total pressure drop across the intake is 0 .25 cm ofwaterl. (b) the slatic pressure rise across the rotor. (c) the static pressure rise across the stator. if the sta tor efficiency is 75%. (d) thl.! cho.ngc in 10 131 pressure across the stator. (c) the overall total-to-total cfliciency and (I) the degree of reaction for the SIi1ge .
Axial velocity al tip Dullet
,
C.;~ ,
=
K,_
-
"lC' +?(C.,2.< "" .1' , ~ --,
=
"
, 26621.:'1 - 2 x 192.2-
c. f
-
..,)
--
"
+2
(192.2 57.18) , 0.265 - 0.265 x 942.5 x 0.265-
20 1H5.51
:. C.,2 .,
==
wr,-,
Solution (a) Total pressure of air exit of rotor N = 3600 rpm m::;: 0.5
D, == 0.2 m
D,. = 0.125 m
fh-~2=20"
~,
The pressure changes involved arc small so th3lthe flow may he treated as
142 m /s
pressihle . At the inlet. the density of air is
(a) Air and blade angles tan Cil.,
P, = 1.013 bar
= 55"
C'I.' C"l.,
57 . 18 = 141 ,9
Po
1.013 " 1()~ = -Po = .:..c::~~-'= 1.184 kg/m '1 = p
RTo
287 x 29R
IOU'Ill '
160 ;. TlJlOlO MACHI NE-I;
AXIAL flowCOMPRESSORSANOFANS -4: 161
Si milarl y.
For forced . . orh!X design rno
lana:!.,
=
tana2.m· r, Ian 45.71 x 0.8 = 0.82
.'·O':!.I
=
39.4 0
45 .75 x 0.159 :;: 34 .31 m/s 0.2 12 153 .76 x 0.159 x rh= 2 =115.31m /s O. 11
x
•
rh:;:
Axial velocity at hub inlet
and
u,
e-
tan /12 .1
-tuna ... , - .
"50
~50 - tan 39.4
:· th.1
when:
jj
0 :;:
0.85
,
KI
C';l.m
=
40.4 0
:,Co'
L' C
Degree of rC3c li on
IlLl,
R,
=
Ci/(tan .8 1" + lOIn th. ,) 2V, 150(I.n 54.9" + l.n (40.4 "))
=
R,
=
, + 2 C.~, ..,
ISO' + 2 x 45.75' = 26686. 125
=
26686. 125 - 2 x 34.31'
:::
155 .99 mt s
= 24331.77
Axial velocity at hub outlet
2 x 250 0.681
where
Example 4.10 An alternative design proposal to that in example 4.9. is to have forced vorl ex blade design . What then will the air und blade angles and degree of
=
reaction be. Take rota ti onal speed as 9000 rpm .
=
, ' (153.76- 45 .75) 150-+2 x I53 .76--2 - x ( 2rr x 9000) xO'I'-, 0.212 0.212 60 .- -
Solution
=
26621 .32
(1) Mean section The parame ters arc calcu latcd in th e same manner as in examp lc 4.9. The v3 1ucs arc
= =
115.32 34.31) 26621.32 - 2 x 115.32' + 2 ( - - - - - x 942.5 x 0.159' 0.159 0.159 24303 .8
=
155.89 m/s
a \.II1 :;:
th.m =
0'2. 111 :;:
fJl.m
R
16.96
0
= 45 .7 1
0
and
= 0.5
(a) The air and blade angles
(2) Hub section Hub diameter. Dh
Mean diameter. Dill
=
rr x N 0.3 18 m Vm x 60 rr x N 0.424 m
rrx 9000
=
200 x 60 rr x 9000
C.rl.~' :::: C, ' tan Ct l.m :;: 150 x Ian 16.96° = 45.75 m/ s C I ~ .... ;::: CII · lan a2.m :;: 150 x tan45 .7.1 ° = 153 .76 m/s
tanal ./r
=
34.31 C"I.I' - = -155.99 C Hl.h
al, h
=
12.4"
tan fJl.h
=
{Jl.h
=
Ian a2.h r::t2 .h
= =
VII - C IU
CII !.", 36.56'
=
150 - 34.3 1 155.99
C" .• = 115.32
Co,", 36.5"
155.89
AXIAL FI.Ow COMI'ItE.<;SORS .,NO F,,:-;s ...
15S i>
Tormo MACHI NES
or
and rrom lhe reaction ratio
So lvin g for
2RU",
tan fJLIII
+ lan Ih.m
;:::;
c::- ==
Ian fJ I .m
+ Ian
==
I.D
Pl.",
{h.m
150
=
~)
=
1.363
O'~.h
=
53.73"
tiln Ii?.h
=
and {h.m
= =
tan fJl.m
= 1.025
1.33 + 0.72 2 45.71'
and
Since
= 0'2.111
= Pt.m
=
fJl.",
in equation (I).
we
tan Ihm
Ihm Si nce Rm = 0.5, Ct'l.m ;:::; 16.96°
alo",
=
=
45 .71 '
Degree of react ion
Cu (tan {hh
1.025 - 0.72
+ tan /hh)
2V
16.96
(b) Stage air and blade angie at root (or) hub For a free vo rtex design. C-,"j .m . rm
= =
150(tan30.7I ' + t.n(-19.95"» 2 x 150 0.116
(e) Stage air and blade angle at tip
I
(or)
C.fl ./· r/
For free vortex conditi on.
= C,II .m· r",
or CII tan O'l,h . rh
=
Cn13na l.m
"m
rm .r/,
tan O'I.h
=
tan 0'1.",
rm rio
=
Vm = 200 = 1.33 Vh ISO
O'l.h
=
22.1 '
C" . tan au . r,
=
e,l·tanO' I .m · r",
tan 0'1.1
=
Ian Q'l.m . -
Vh Ian
fJl./J PI),
= = =
C,,(lan au,
V"' 200 = O.K V, 250 tan 16.96' x 0.8
lan a I.t
:. al./
+ tan fJl.h)
150 - tan 22. 1 = 0.594 150 30.71'
rm r,
=
rm r,
From velocity triangle [Refer Fig. 4.2].
Simi larl y.
-19.95'
112.1,
fhm
IC.1").h . r h =
- - t,m Ci2.11 C"
- tan 53.73 ' 150 -0.363
get
=
Vh
~
0.5
45. 71 °
Substituting
rm rio ta n 45.71 " x 1.33
tan a2.m .
tan 0'2.11
2 x 0.5 x 200
=
13.71 °
and tan
.B 1.1
V,
=
-
=
250 _ tan 13.71' 150 54.9"
=
Co
-lanai
I
.
= 1.423
159
150
AX I ALFLOWCOMf'ItESSORSANllfMIS ,
:,.. TllItBO M ..... CIIINES
'on 6 1.4 _ ;;-;:-.-:..: 10:;0:;5..;.x.:...:2::0~_ 0.94 x 275. 15 x 150 1.316
=
= =
:.fJl
(c) Stagnation pressure ratio POl
52.77°
~'6ToJ,o, [ 1+ -TOI-
=
Po,
I
(b) Mass flow
[
p,
P,
=
P,
power required
RT, POI
(d) Rotor air angl es
C~~T'
rr DhN
D"
0.6 D, = 0.6(2 x 0.292) = 0 .6(0.584)
~
=
(0584'
tan {JI
=
:· fl,
=
47.71"
=
i (D,1 - Dl) =
x 0.35 x 9000
164 .93 ml s U, 164.93 C, = 150
1.056 kgjm"
A,
=
IT
U" = (;(] =
0.87542 bar
0.35 m
/;\C/1tlTO
27. 197 x 1.005 x 20 546.66 kW
0.87542 x 10' 287 x 208.8 1
p,
;; ;;
T, )'; ' -' ( -To,
"
I ,
=
" 0.9 x 20]'-' 300
+
1.226
P I A , C II
11/
157
60
and tn.n
- 0.05')
tan
tho =
0. 171 7m' :.~,
(or) Flow ure
= =
C1J 6. To
131 - - - AUhel/
1005 x 20 'an 47.71 - ~c:--:.=:;,.:.:,.;=-= 0.94 x 164 .93 x 150 0.235
13.23"
Example 4.9 A
I'm
=
(', +- = -
=
0.2335
"= IiI II<
(0.584 -2-
"I
=
=
A
2rr rlU "
2
0 .35 ) +-2 2
vortex design.
III
r, - rll;;
C.584 -2- -
0.35) T
2;r
=
O. 1717m 2
=
1.056 x 0.17 17 x 150
=
Um = 200 mts C" = 150 mls
=
=
0.2335
Solution U, = 150 mls 6To 20"C
0.117 m X
An axial fl ow compressor stage has blade root, mean and lip vcloc· itics 150.200 and 250 mls respectively. The stage is to he dcsigncd for a stagnation temperature rise of :20°C and an axial velocity of 150 m/s which is constant throughout. The work donI! factor is 0.93. Assum ing 50% reaclion al mean radius. calcul ul!: the stag.e air and bl>1de angles >11 mean, rool and hub and degree of reaction for a rree
X
0. 117
U, = 250 mls I- = 0.93
Rm = 0.5
(a) Stage air and blade angle at mean From 'he work done pcr Slage express ion . we have tan
Ih .m- tan {h.m
=
Ian
13 1.111
=
27 . 197 kg / s -
tan fhlll
C p 6TO )"'U",C(I
0.72
1005 x 20 x 200 x 150
= 0.93
(II
• AX IAL FLOW C OMl'RESSOllS /l. NoFANS "" 155
15·' ;,.. Tl"WO MACIilNES
303 [(2) Vi
=
-
Solution
I]
n.83 79.95 K
= The work done per stage is
-no = CI.(T"
- T,)
u(e.. ] -
(T" - Ttl
= 20°C
C, = 150 m i s
Pu ,
= I bar
r
' u,
q,
). = 0 .94
D'l ;: 0.6 D,
= 300 K
M,.I
= 0 .92
= 0.9
(at tip)
79 .95
=
U
=
be drawn as shown in Fig. 4. 12.
= U(C"
- Cx ,)
T,
C rl
)
U6C.1
ISO:!
C I•
301J - 2 x 1005
U(0 .235 U ) -1005
= '288.81 K
584 .7 m /s U x 60 584.7 x 60
M,., (F ili,)
IV,
--- =
N
"xD
=
C'
TOI - - ' 2C,I
=
= - -
C I•
Speed.
'" To
Assum ing nxial inlet (si nce no inlet gu ide vanes). the ve lOCi ty triangle at in let can
IV
or
N = 90110 rpm
Lj" I
'1
111
0 .92 ( J 1.4 x 287 x 288.8 1)
" x 0.5
=
22334 rpm
313 .39 Oll i s
(b) Absolute velocity of air Sin ce, the ve loc ity diagram is symmetri cal. R = 0.5.
Fig/lr~
R =
and
2U -(C.,., +C,,) 2U - 0.5(2U) + 2U = U
C"
+ C."
=
U
C"2
-
C'rl
=
0.235 U = 137.4 m/s
cos
= 584.7 m/s
fi t =
=
:. iJ,
C,
150
IV,
3\3.39
61.4<:
and
So lving for C'II and C r1 • we get
u,
Cr~
C XI
From the vel ocity triangle
=
W, sin (:II
=
361.05 ml ' 223 .65 m /s
313 .)9 x sin 6 1.4"
275 . 15 ml'
=
Fro m inlet 'Jclocity tri:.IIIg lc I Refer Fig. (4 .2)].
C,
= jCj"+-c,}: =
=
J274~81j
la) Tip radius CII = 0.47 x 584.7
= 274.8 1 m/s
=
+223652
354 .3 2 m/s
r,
Example 4.8
TIle firsl stage of an axial flow compressor is designed ror free 'Jorl e" condition . with no inlet guide Vil nes. The rotational speed is 9000 rpm . and Slag nalion tcmpelTl turc rise is 20c e. The hub·lip ratio is 0.6, the work done raclOr i!'; 0.94 .md isentropic efficiency or the stage is 0.90. Assuming an inl et ve loc ity or 150 mt!'; and ambien t conditions or I bar and 300 K. compute (a ) the tip radius and corresponding rolor ang les. ir the Mach number relative to Ih e tip is limited 10 0.92. (b) mass now en tering th e stage (c) stage stngnatio n pressure ratio
=
U, x 60
2" N 275.15 x 60 ~ 0.292 m 2n x 9000 0.29 2 m
r. Now. (tan
Il, -
tan
flo )
tan th.
=
C1,tJ.To AU,Cn
=
t.1n fi l -
CI'tJ.To AU,C"
4. 12
C,= Co
AXIAL FLOW COMPRESSORS AN D F ANS "'"
:;:.. TlJ ltnO ~'l "CIl I NES
1:'2
=
(b) Stage efficiency
C,,(To,v-t-1 - TOI)
1005(484.6 - 290)
=
N
0 .85;.: 180 )(
e,r x,(tiln-t..t '
T'\s - TI -T] - T\
=
'Is-
-I an 1:\" )
T'[(;:)",
The :Ixia l vd oclt), is U
=
C"
150.4 m/s 195573 16909. 1 11.6 12
N
N Tht::
l1umh~r
308 [(1 .3)'I:l-I]
Ian I1 D+ tan 44 '
=
= '/.1
(c) Reaction ratio R
=
PO""T I
Ttl l
POI
484.6 1.0051n 290 - 0.287 111 5 0.054 kJ / kg - K
Example 4.6
An :Ix ia l compressor has a mean diameter or 60 em and run s at
15.000 rpm . If the actual tcmpcralUre risc and pressure mtio developed arc 30 D e and 1.3 rcspeL:livdy, dctermine (a) p ow~ r required {O dri\'c {Iic com pre sso r whilc lkJivering 57 kg/s of air, assum in g mechanical efficie ncy 86% and initia l tcmpcraturc llf 35"C (b) Ih e slage efficiency and (c) Ih e degn!c of react ion if the temperature
e
= T,
'Jm
= 0.86
R
- T, = ,(l"C
T,
= :'5''C
N = 15.000 rpm . p,
-'P, =11 ..
11/
= 5 7 kgjs
=
Work done
L
57 x 1.005 x 30
=
Solution
PJ
_?
P, - -
= OA7
T,=303K
fiC ... I CI/ = 0.5
D = 0.5 m
(a) Compressor speed D.C,
'" C" 6.C.t
C,'} -
=
q,. 171 H.55/0.R6
=
1998.3 14kW
Ti
ex, = 0.5 ell
= 0.47
0.47 U
=
0_5 (0.47 U)
0.235 U TJs - TI = T,- T,
=
17 1H.55 kW
=
(AT) m gc
P, = 1.01 bar
WI',,,,
Power input
(6T)ro'OI
55 - 35 ---=--30 T..'\-TI 0.67
c.
mC 1,6.T
IV
=
TJ. - T\
7o! = 5Soe
(a) Power required.
......
(tlh)rOlOr (tlh)mgc:
Example 4.7 The first stage or an axial How compressor dc:vclaps a pressurc rUlia of I :2 . The inlcl pressure and temperature arc 1.01 bar and 30"' C n:speclivcly. The overa ll efficiency of compressor is 83%. The How coefficicnl is 0.47. The vclocilY diagram is symmetrical and at Ihe mean radius the ratio or change of whirl vc:iocilY 10 ax ial veloci ty is 0.5. DClcrmine Ihe compressor speed ir the mean diamclcr is 50 cm. Also find Ihe ahsolute velocity oflhe ai r leavi ng the siali onary inlet guide vanes. (MKU-Ma)' '97)
Solution D = 0.6 Tn t!. T
= =
(b) Change of entropy TO N-t-I
30 79.92%
=
of SHigl.!S is 1:2.
C" ln - - - Rln - -
-IJ
T., - T,
180
U
C,,=--'---::'-~ tan Ih + tan Ifl
I S3
TJ - T,
T,[(~~r
=
lie
-IJ
q,
= 0.83
AX IAL FLow COMPRESSORS ANU FANS ..;;
15 1
150 I> TURBO M ACHINES
(c) Overall isentropic Efficiency
The compressor power m C I'(ToN+1 -T01)
IV - - +T0 1
mC"
=
4.5 x IOJ
=
20 x 1.005 223.9 + 288
=
511.9 K
TO,"+,,, _I)/(TON.,
(
+288
TOI
_I)
Tal
hut from isc.:nlropic relation
(a) Pressure at compressor outlet Pressure at compressor ou tlet is detennincd from the following re lati c n.
und
,- ,
TO N + 1 )'II'.rlr-l (
70 . . . +-1
TOI
TOI
Il'hli .4
511.9)"r
Jil' N+ 11o
( 288 6. 12
=
:. Po N + 1
TOI
POl _ POol _ P04 _ PON + 1
~
POl -
-
---p;;;
where _N is the numht:r of stages. PON +!
=
.. POI
P02) N
(
P OI
In or N
51 1.9 = 1.7 8 288
'"
r-J = 1.68 (6. 12) '"
=
_1.68_ - _J =0.872 1.78 - J 87.2%
Example 4 . 5
Si nce pressure ratio for eac h singe is same.
POI -
)
=
'. 'Ie
6. I 2 x 1 = 6.12 bar
(b) Number of stages
(PO N " POI
=
(PoPOI+, )
An axi
Solution a) Number of stages
N
In (PO')
POI
R PO N +- l
/31
5
'it-
=
0.87
180 m l s
I-
=
0.85
POI
U = C"
= 44°
0 .5
02
=
01
TOI
=
290 K
where
Pm
POI
(~:) :~ = G~:) '1W
=
1.24
In (1.24)
= :. Number of stages is 9.
290(5) '" n
In(~)
N
'.:J
+') ,
TOI (PON POI
TOI ....... 1!,
8.42
=:
459.3 K
=
_T.: cO,"'N.!C+!!II,!..---'T0:..c ' + TOI ~,
459.3 - 290 0.87
+ 290 =
= 13"
/32
484.6 K
-
T , .. 8
:,..
AXIALFLOWCOMPRESSORSA:-.;oFt\NS ... 1"9
T U RBO M ACHI NE.S
=
488.9 - 293 3 +'29 0.9 510.67 K
=
N x UC,,(lana:! - lanai)
=
1.005 x 10-1 (510.67 - 293) (8 x 188 x 100) 1.45
= (c) Air velocity in delivery pipe
tan a2 - tan al
3_) "n
438 (
=
3)
419.13K Now,
Since a:!
= fJ l . th c n tan
C;
= T.,- + 2Cr - -J2Cp(To2 - T,)
=
J2x
=
194.75111 /5
1005 x (438 - 419. 13)
Example 4.3
uma l +ta n {J1
An e ight siage axia l fl ow compressor provides an overall pressure ratio of 6: I wi th an overall isentropic efficiency 90%. when the tempe rature air ill inlet is 20°C, The work is divided eq ually nelwcen the slages. A 50 % reaclion is lI sed with a mean blade speed 188 m/s and a conSl ant axial velocity 100 m/s through the compressor. Es timale the power required and blade angles. Assume air to be a
tan fil
(II
50
the veloci ty diagram s
:. al
N
.._,
+> ) -'-
293(6) "n 488.9 K The overall isentropi c efficiency is given hy
'1c =
=
Ian III - 1.45 1.665 - 1.45 = 0.215 12"=1l2
=
Power required per kg of air/so
art! identical. 0'1 fh and a2 = PI· If the compress ion process were isentropic then the temperature of air leaving the compressor stage wou l9 be
(PoPO I
1,45 -+- 1.88 = 1.665 2
=
=
PON
N =8 . POI =6 • ~, = 0 .9. TOI = 293 K. R = 0.5 . U = 188 m/s.
=
(2)
Substituti ng the value of tan PI = 1.665 in (I). we get tanal
C!! = 100 m/s. For 50% reaction turbine, the blades are symmetrical and
= Cu( tanal + tanfJl ) = ~ = 1.88 = !:!... e" 100
=
Solution
:. To 1J +.
= 1.45
From (I) and (1),
or
perfect gus.
L
tan al
From inlet velocity configuration (Fig. 4.2),
=
TOI
PI -
=
III
=
I x 1.005(510.67 - 293)
C,., ( TO N... ,
-
Tud
218.76 kW
Example 4.4 A multistage axial flow compressor absorbs 4.5 mW when de· livering 20 kg/s of air from stagnillion condition of 1 hilT and 288 K. If polytropic efficiency of compression is 0.9 and if the stage stagnation pressure ratio is cansl:lnl. calculate (a) pressure at compressor o utlct . (b) the number of slages . tcl ovt!rall isentrop ic efficiency of compressor. Temperature ri se in the first stage may be taken as 20°e.
TOINot.•" - TO I
TON+ l - TOI
Solution IV = 4.5 mW /1p
= 0.9
POI
tlTo
= I bar
= 20"C
TOl = 28K K T02;; TOl + tlTo = 288+ 20 = 308 K
-
146 ,. TUROO MACHINES
(a) Air flow angle ("'1)
Since R = 0.5.
but. Ih :. 0'1
0'1
=
Ih
and
,f!. - 30\) = 30° _
2 x lOs . :::::::::.....:.:;-:-;: = 2. 11 kg/ m.l 287 x 330
PI
30°
2. 11 x 8.797 x 10- ;\ x 11 4 .27
: . 111
(b) The pressure rise
=
111
WI'"
u 6C.1-
= C,,(T2 - Tt)
Ul;C
2.64 kg/s
(d) Power
- -x +T,
HI
Cp
mC,,(T! - TJ) 2.64 x \.005(356.26 - 330)
Blade mean speed
69.7 kW
IV Tr
U
=
DN
Tr
60= 263.89 m/s
x 0.14 x 36',000 60
263.89 x 100 1005
:. T,
+ 330
356.26 K
Example 4.2 The following data refers to a lest on an axial now compressor. Atmospheric temperature and pressure at inlel arc 18°C and I bar. TOlal head temperatu re in delivery pipe is 165"C. TOIal head pressure in delivery pipe is 3.5 bar. Static pressure in delivery pipe is 3 bar. Calculate (a) lotal head isentropic cniciency. (b) polylropic cficiency. i\nd (c) air Velocity in delivery pipe. (M'!U-ApriJ '9/) Solution
A ssuming thm Lhcre art! no losses in Lhc cam·prcssor, the pressure ratio cn.n be deter-
POI
=
=
(T')"'-' = (356.26)&l T, 330 1.31 x P, = 2.62 bar
Pressure rise
T02, - TOI
rye
To> TO!
=
291
POI
l; P = P, - P, = 2.62 - 2 = 0.62 bar
=
m = PIA ICn
=
:. rk
2U .R
Cn
Ian 13,
+ Ian 13,
In
Ian 30" + tan 60" 114.27 m/s Flow area, A J
C. -
x 0.14 x 0.02
1::
-
POI
_ ' In
= IT Db Tr
)
(Po,)
, - I
=
5
I 4 16.24K 416.24 - 291 438 - 29 1 85.2%
(b) Polytropic efficiency
2 x 263.89 x 0.5
TOI
(Pm) ~ -
=
(c) The amount of air handled
=
8.797 x JO-J m'
=
K
Po"!. == 3.5 bar
(a) Total head isentropic efficiency
T02,
Axial velocity, C a , is given by
=291
TOI
I bar
To, = 273 + 165 = 4)8 K
mined from
(TO') TOI
In 3.5 1.4 I n (4- 38 )
0.4 87.5%
291
Pt. = :\ bar
J -i..J
AXIAl. FLOW COM i'R ESS01LS AND FANS
;... T IIIWU M ACHINES
(5) Frontal Area Ce ntrifu gal compressors have a larger front a l area than that of an axinl now co m~ pressor fo], th e same rating. Thi s makes the a xial now c ompressor mo re s uilnblc for aircraft wo rk .
(6) Working Fluid
," ,I
'I
ThL' pL'rfonnancl! o f cc:n[rifuga l co mpressor is not much afl"l!l: tcd by the acc umul atio n o f de pos its 1m the surface of now passagc whcn workin g with con w minill ing Jluiu s. Unde r s uch co nditi o ns, the performom:e of a .x ial no w c ompressor is gene rally Il11pilircd.
(7) Starting Torque Ce ntri fugal compre ssors need n lo wer s tarting torque than axiat now c ompressors for the sam e capac ity.
o
0.2
0.4
0.6
U.S
(8) Construction Centrifugal compressors IlI.I ve a s impl e, ri gid and re lati ve ly ch e np co nstru ction and al so tess pro ne to ic ing [TOubles at high altitudes.
(9) Multistaging l\'1ultis tag in g is m ore suita ble for a n a xial now compressor wh ere it g ives an increase in pre ss ure w ith neg ligihle losses. Th e number of s tages for the a xial no w compressor vari es from 5 to 1·"
(10) Application C .:: ntri fugal co mprL's sors .huve be e n s uc cess fully used as blowing machin es in sleel mills . compressors for low pressure rcfrigc r.ltio n a nd industrial gases and tur b oc h a rg~ t! rs and supe rc hargers in internal combustion e ng in es . They hnvc also bee n employe d as compressors for s mall gas turbine airc rafts. Axi al nmv compressors are m os tly ust:d in gas turbines and high pressure unit for ind usu ial and large marine gas turbine pl 'lIl[ s.
Figure 4.10
A-riaf 11011' compressor cllLlractuislic Clfrv~.l
2. A small increas..: in mass How will lead to a sharp drop in pressure r:lli~ to point 2. The dens ity also drops sharply so that e l l incr":3ses. l1tis re~ult s III .the .Iarge: dec rease o f the incide nce a ngle in the rear stages , then:by c au smg stalling
III
the
rea r slngcs with negati ve incidence . . 3. The o pe rating poinl m o ves to point 4 if the. spced of the co~prt!.ssor IS. redu.c~~ Both m and e" fall fas ter than the blade spt!cd U resultmg 10 un IIlcft:asc inc idl!nc e a ngle and possible stalling in the first stage . .. 4. Whe n the design speed is increased to poinl 3, the charactensuc I!vcntual1~ be· c omes almost venic;l1 . Thi s will increase bmh the de nsity and pressure rallo as tb e incrc,lst:d s peed allows more air to be passed through the compressor. B~t at the inlet. the mass flow increases faster than the density und ..:hoktng atthc mkt is usuall y the first to occur. All the limiting c o nditions disc ussed above should be avoided tlt ulllimes since they
leud to unstable or inefficient operation.
CHARACTERISTIC CURVE Fig . 4 . 10 s hows [h e c harac te ri sti c c urve of a multi sl<1ge a xial flow c ompressor. C omparin g th is c urve with th a t of th e cCnl rifugnl compressor, it is o bserved [h"l1 the pn.:ss urt: rati o o f the cl: ntrifugal co mpresso r is less se nsitive (0 maSs now vuria[ions at a give n speed than in the a xia l c ompressor. The charilc terist ic curves have [he foll o wing sLllicn[ po inl s.
L
I. Th e desi g n mn ss now and pres sure mli o are at point 1. It is seen Ihut th e desig n po int is vl.! ry c lost! to lhcs urge line (poin15 ) and if the mass now is only s lightl y re du ced , the press ure ralio and d e ns ity in the rcur sttlges will bo th inc rease. Since ell = m /pA . the a xial vel oc ity will dec rease a nd hence the incide nce an glc at will in cre ase sharpl y in the re ar stnges, the reby c ausin g stalling in these stag es.
SOLVED PROBLEMS
Example 4.1 The following data refers to an 8)(inl·fiow compressor: Pt ;;:: 6Qf . turning angle = 300 and 6,e~ ;;:: 100 m i s, degree of reaction ~O% . rp~ ]6:~. ~~an blade diameter 140 mm, inlet pressure and temperature 2 bar
Solution ~, =
60' R = 0 .5 PI
:=
2 bar
= 3~' = 36. 000
~, -~,
N
T, = 273 + 57 =)30 K
D.C, = 100 m/ s D =0.140m b = 0.02 m
AX I,\I . FLOW CO"II'ltF.5S0IlS AN I) r ANS '""' 1·13
1.. 2 ;. TURIIQ MACIlINES
If it is a~s umeulhat we 11<1\"e equal tCllaltcmpcralure risc. in e
~~(N+ l )
stages
i
PO N+I/ PtJI
=
(1"iIN + t / TO\)'j,.r W ':' 11
Also TON+I / Tol = (Jill
P, I
where 6, 7i, is the stage IUtal tcmpcrature ri sc o It is also usua l to assume th ai the poly tropic ant.! tOI:1I . tO· IOI:11 stage isentropic d lit:H:ncies are equal at a va lue (I f abuu l
2
O.H~ .
I
2S
+ N6To) / T,II
I
Using thi s metlwl.i . a very rapid calculation or pressure risc th roug.h the cOll1pre~sm
I I
can he made .
COMPARISON BETWEEN CENTRIFUGAL COMPRESSOR AND AXIAL FLOW COMPRESSOR FiguTt 4. 9
Comprt!ssion pro(."l!.r.J
jn
The comparison bet wee n the main fC;;l.Iurcs of cenlri fugJI and ax.ial How compressors
n multistage compressor
arc given below and the ise ntropic efficiency of tbe co mpression is
(1) Direction of Flow across the Compressor
~, = ("(N+ I), - I")/(/' N+ , -I,,)
If we now. compress from J to 5 in work done IS
Wt/m
number of small fi nite s tages. the isentropic
= (l12.J - "I ) + (IIJs - "2)
+ (11<1.1
- "J)
+ (lIss
- /14)
(2) Pressure Rise per Stage
and for similarly designed stages. the efficiency ''Is' is the samc . Hencc 'Is
= (Ws /m)/(lIN+l
-/q)
where th e n~merator cons ists of a number of isen tropic enthalpy inc reases. But. as lhe entropy Increases through the co mpression, the constant pressure lin es diverge and (lIb -/'1)
+ (IrJ.t
-112)
+ (l14s
- IIJ)
+ (!l5s -1r 4)
> (/I (N+ I),
-
II I)
and thus
I~, >
llel
Th~t is. the ave.ra ll sin gle isc~ tropic comFression effi ciency is less than the stage effiCiency. The d ifferen ce also Increases with pressure ratio and with the number of stages . TI1C overa ll stati c pressure ratio o f a multistage co mpressor can be exprcssed in termS of small stage efficiency as
I
PN + l/ P [ = (TN+ t /TI )'I,..r/(r - 1J
I
In centrifuga l compressors. the flow through the compressor impeller takes place largel y in a pla ne which is perpendicular (0 the axis of the compressor. In axia l !low compressors. lhe now proceeds throughout Ihe compressor in a direction essent iall y parullC\ to the axis of I..h e compresso r.
Centri fugal compressor hJS a hi g.h pressure rise per stage (4 : I l, TIle ax ial now cornpn.:ssnr however gives a pressure ratio of on ly 1.2: 1 per stage. To achieve Ihe desired rn:ssurc ri :;e. the axial now unit has to be provided with a large number or stages which makes the ax ial now compresso r less compact when com pnrcd with the cqu ivalent ccn trifuga l unit.
(3) Isentropic Efficiency The isentropic cfficcnc.:y or centrirugal compressor:> is as high as 80%. Emlier axial now machines had a low isentrop ic efficiency, but with aerofo il blading. a multislilgc axia l now co mprcssor surpasses the maximum centrifuga l compressor efficie ncy by abo u1 4 per ce nt . Tl lc efficiency in bot h cases however declines at high pressure rati o~.
(4) Range of Operation Cc ntri fug ,11 co mprcssors have a wide range of openllion between surging and choking limits. A greater flexibility in operaLion can be achieved by lhe usc o f adju stahlc prcwhirl .alld din-uscr vanes. High efficiency for uxial now compressor is attainet.! only wit hin a n
AX I,'!. FLOW COMl'Rli:SSORS ",,"I) FANS "'"
14\
2;rN where w =- - - and
llll..'lI .
60
(CJ ~",rlll - --
=
. WI
=
COnS(nnl
r
II "' /"111) C - .- -
-
,.
WrlJl(C'~ m -C\· I N,)==Wt. I"::! - \"11
For forced vortCX desi gn, C.I~ = rx~ and C... , = rx\
T hU !
C"
R
'2U ( Ian fil
+ Ian fi~)
In this type of blnde !.he energy transfer increascs from root to tip (Fig . ..UHa) & (b» . There is il limit to hub/tip ratio in order to provide a prnctical minimum value of
[
=
2U (Hl rl
+
:.w = Ur(x1 - xd
IV I !)
axia l veloc ity Cfl ·
[
2U I
lU
-(ell
(II)
+ Cr~)
(bl Allip
At hub (or root)
C,
2U
R=
w,r>(\C,
c.
and for f"n:e voncx co ndition I _
-I- x:! _
XI
2Ur
XI
J _
-
~
+ .\"1
2wr1
· :~l~ re':~I~~n rati ~ varll:s with t!lC r"diu ~. Tht!reforc.
Dt one dcsig ned point o nl y the
Figllre 4.8
\Vtlrk dOlle in aJorced varIa blade
dl.~I1;1I rl: •.lllion rall o can bl.! ohtalOcd. Th iS hladi! dL!sign method has the disacJvanwgc
Ih .1I Ihl.' hl !,! hl.' sl Mach I1I1.Lnbl.'rs occur nr the rotor tip ,llld :11 th c s talur huh.
(c) Constant Reaction Blade
(b) Forced Vortex or Solid Rotation Blades
For true radial cqulibrium, Cli should vary with radius , but conslant reaction blades have been co mmonly used wiLh constant As WI and C:! decrease only slightly frolll tip to root , almost constant Mach number occurs at different radii. So, in constant reaction blade, the altial velocity C II and the reaction ratio' R' are constant at all radii i.e from lip to rool. • R' is equal 10 0.5.
en.
For Ihi s Iype of now Ihe condition is
c,'
I
= C...,
-
= x(cu nslant)
1"1
1"1
Til ~ L .: hic \'e Ih e rndial equilihriulll Ihl.! axial velocity del.Teases I"rorn rom 10 lip . SHI~e C". I~ 1101 constant a ln~ g Ih e hl
W hl... C ,.
= scction radius and K I = C~
:\llllIlh":l:
'11.111
+ 'C~ -
.\ 1.111
C,', = " , _ 2 (C,,)'. ,., + (C" C,') , 1_
-J
_
r
-
-
r
-
r
(Vr-
MULTISTAGE COMPRESSION lolal pressure ratio ncruss a single sUlge is dependent upon the lotal tcmpl.!ratun: ri s\!
111C
I
IV/m = (Ii(N+I), -Ii,)
I
AXIAL Ft.ow COMI'RESSollS ,.,NIl FANS ....:
138 ,. TUHIJU MACIIiNES
RADIAL EQUILIBRIUM THEORY . If the ahsolutc velocity C is res olved into three compont.!l1ts in the tangelltial. axial and radial directinns. [C =
' dr
1 d tlr
,
.
,.2
=
Po
P
c2
22
dP
= -Idr
=0
t + 'c" . dCI + 2rCr r . dC dr elr
. C r df dr
For stcudy nuw cunditions along concentric stream lines. the ve locity. static pressure and density arc constant with timc anddCr = O. LeI " be the radius ora ny cnncenlric strcamlinc. Thcn dilTerentiuting. with respect to r. we get dr
2
dr
Becnus!,!, differcntiating eqn. 4.37 . wt: gel cqn. 4.36a.
+ pC'/2
Po
dPo
d
--(rCr)--I- -(C II )
TIlt.! stagnntion pressure Illay be writtt.!n as
{4 .J6al
r
This I!ljuHlilln is t!quivalclll to
C~ + c} + C; I
2
c2 dC.l dell -..:!.+c , ~-+C(I-d =()
. ,.
1W
dCx pC.r~dr
dC,
+ pCr!~dr
r (or)
de.\"
+2C.l~-
dr
+2CIl
dC" -
dr
(4.341
TYPES OF BLADES
But from fluid Tlwchanics. for plane circulatory flow and for r;]dial equilibrium.
(a) Free Vortex Blade In tilis cast.! the tangential veloc ity distribution tws to hI.! such tl1"l(
C'
,{p
fl2 r
dr
C~ dC r dC" p---..:... +pC.\-· +pCu - r tIr dr
dPo
dr
The energy IransfelTcu to the blades is at the expense of the energy in the Iluid LInd may be c:'(pressed in tcnns of tile stagnation enthalpy change dll". Writing the TcI.'i equation using stagnation properties. we havc
lTd.'; = dho Differentiating w.r.1.
~I
dPo pdr Tds - - -dr tlr
dP" pdr
SuhSlituting equati{ln
4.~n
£III" dr
c~
(4.36)
~-
dr
dell d,. '. CII
=0
=
0 constant
IIV = ute, -
e, 1 - wr(e,
f=rorn Euler's cqualltll1.
C.,)
I
err = constant
---'+ C. r
dC I
I• - -
dr
+
dC" Cu~ dl"
= 0 (Since energy transfer is constant. at all radii)
Td., and - dr
I
For fn.:c vortex bladc. the condition to be satislit.:d is
To simplify this equation the foll owing assumptions nrc made dill!
co nstant
With this condition. the equation 4.37 becomes
in equation 4.35
Tds dr
I' -
Work and reaction ratio in free vortex blade
dllo dr dh o
or ~
I'
. I I ., C ·,s constant 'dong the blade height i.c. from rotor lip 10 st;lIor ,S0. ilXlil ve UCl y I I ' . hub. And reacti on ratio R varies with radius. i.c. it decreases from up to huh.
'I"
Tds dr
IC
(4.35)
i.c. ,h ,'... whirl velocit)' is in versely proponionai to the radius. Applying the condllllHl' bctween \Wo sections. C.I' 1 . r
=
C X 1.,"· rill
= x t constant
C. r~
=
C Xl .,"· rm
= .\"2 constant
where subscript III refers
(0
.r
values al th\! mean radius 'II! '
AXIAL FI.ow COMI'RES.sOrlS AND F A NS .... 137
Thu . .
If C I
= L.l . then IJ..
i..
it
(IlPlp)
'h (.6.11 Lf'/
.'. (.6. p)u r:,!p
11.U . .6.C-r
(4.16)
very c luse :Ipproxirnation of tht! 100al -lo-tmnldTiciency
']11 .
h.lu - hi
'1.1
II] - hi 11 03" - hOI hOJ - hol
n
_ h Ju -hJ
." -
"3 - hi
(427)
Allhaugh Ihe .ahovc expression:; ilrc derived for incomprcssihle now they art.! ncvl'r1heless. ~ va!ld aprn1Xi ma(~on for compressible now, if the slage tempcrillurc (and pressllrt.!) fiSt.! IS sma ll .
this results in [lositive incidence. At starting and low speed operation, lillIe pressure ri se lakes plaee. Where as at the design poinl (i .e. at higher rpm) of the cumprl!ssor. largc pressure rise occ urs from firs! to the last slage. Thus in a 10\\1 spel!d upcra!ion. the abse nce of pres sure in the last stages leads to higher values of elf than thl! designed o ne. and this leads to choking . Choking limits the flow rate and thi s in turn increase!. the posilive incidence in the first stage and bence there is a possihililY of slall. Low speed sta ll occurs in front stages. Al high speeds, the las t siages are more like ly 10 stall because their characteristic is steep and a small reduction in now ratl! produl!es a considerahle reduction of velocity and thu s a large change of incidence. All lhe hladl!s arc not slalled at lhl! same lime. but a stall is propagated from bladl! to blade in a direction opposite to the direction of fOHliion. This propagation spl!l!d is low comparl!d to the rpm (compressor) . TIlUS the stall appeOJrs to he in Ihe dirl!clion of rOintiun. This phenomenon is called ' propagatiolJ' or . rOlatilJg sral/'. Consider a portion ofa blade row shown in Fig. 4 .7. Lei the bladl! A be: sla lll!d. This incn!asl!S the incidellcl! to thl! left of blade A, and' rcducl!s the incidence of the nuw on lO the blades on thc right. The blade B will be the nc)(tto stall.
Inc~cnscd / IIlcldenr
PRESSURECOEFRCENT I'.)
/ /'
It _i~ deJ,i ned a~ [he r;Jtio Ofille actual slagnmion enthalpy change to the kir1l!Lic energy (lJ .l Hurd. whi c h has th(! same .'ipeed as the hladl!s. Thus
/
Decreased [inCidence
(4.2R)
2).,CI/(ti.ln PI
=
-
tun
fh)
U
21.1' ltan fl I
-
ton fl,)
~~----
If Ih.e st.t£e pressure change is so sma ll then thl! nuid can be considered as incom pressible, ' Ililu = III Po) 1 p
(4.30)
So That
Figllre 4.7
CVII/pus.sur .n ull
S unicienlly extensive stall results in a 5unicienl reduction of lolal now and surging .takes plac e. So, sta lling leads 10 surging . .
RADIAL EQUILIBRIUM METHOD 14.3 1)
The press ure l:lJdJicielll usually ranges from 0. 4 to 0.7. This dcfinilioll for
Stall propagation
, (4.29)
v; I' has a llume ri cal value twice Ihe numerical vulueofthe equ:uiull l/J11 = )"rP(tOJnCt:! - wner ,)
COMPRESSOR STALL AND SURGE :-he rn~ s t s i~ni fican! pIH:: no~leno~ affecting the pl!rformance or axial flow compressors I~ the .\ulll. rhl! blade stall IS mi.lIl1ly due 10 Ihe reducl!d now rale (or reduced ell) as
The assumption of two dimensional flow is no longcr valid w hcn Iht! ratio uf huh/lip is less than 0.8. In thl! case of nir craft comprl!ssors. mass flow is high und fronlal area is s mall. To meet this requirement, longl!r bladt!s on a smallcr hub is uscd. So. radiOJI wlocities Sl!t up in the bladl! row changes the muss fiow and Ihl! outll!t vdocily distribution . In such cases, thl! blades should not be dcsignl!d on two diml!nsional principle . RadiOJI now is l!ausl!d by a lempornry imhalance ht!twl!l!n the cl!nlrifugnI and r
1}.j
;..
Ttm uo
MArltl NES
Across the sWim row. Po is cons t
Addi ng tin: equation (4. 19) and (4 .20).the pressure rise in eac h row and co n:' Il.l cnn t! Su. when the ou tlet and inlet ve locity triangl es arc superi mposed, the resulting vdo!.:i t)' diagram is sy mmetrical (Fig. 4.6a).
a normal slagc (C.l =
C,' .giv!.!s ,
(P, - PI> )( :: . p
Case - 2 When R > 0.5. frnm the equation of react ion ratio (4 . 17), il is seen Ih a l th. > a l . and from equa tion 4 . 1N( .1). the stalic enl halpy risc in the rotor is greater Ihan in the stOllar. Si nce f1"!. > 0'1. the superi mposed veloc ity trinngie is skewed to the rig ht (Fig. 4.6b)
Ie; -
=
W;)
(t:J.P).'i,,, ~t = (t:J.P)R," ' I(
+ (lVf - Crl
\4 .2 1 t
+ (6 P )swllIr
From the velOC ity triangles. th e cosi ne rule givt!s
Case - 3 Wlll:n R < 0.5. From tllc equat ion ror reaction ratio 4. 18(01). it is found th at th e s tatic entha lpy rise and pressure rise arc grea tcr in the s ta tor th a n in the rotor anll from elln . 4. 17. we ge t 112 < a t So. th e super im posed ve loc ity trian gle is skewed to the le ft (Fig . 4.6c).
u1 + W 2 - 2UlV COS(~ - fJ)
c' a nd
\V sin
fJ
IV,
or
~ Il.
a,
II,
(4 .221
0,
w~
w.
C~
c,
Subs tituting. t!quati on 4.22 in equatio n 4.2 1
= =
U
U
Ib)R >O.5
1;1 ) R=U.5
(Ul _ 2UlV .'J)_(Ul_2UlV'tl
2U( Wq
W.,~)
-
From the vdocll y diagram . we get
U (c) R < 0.5
Figure 4.6
=
U2
=
WI 2
or WIt - ~1'.1"2
=
·. (P.• - Pil ip
Effect of reClctioll ratio on tile velocity triangles
=U
Vt
+ C.II
:. Wrt
+ C.I :
C .l"l -
U(C.1· :
Cit -
ex,) =
II ]. - 111
14 . 2~
I
Since, fllt" an ise ntropi c prm:css .
1\ reac tion ralio of SO per ce nt is widely used as the ndverse pressure gradie nt over the
slage is s hared equally by the stat or a nd rOlor. TIlis c hoice o f reaction mini mi ses the tendency of Ihe bl ade boundary layers to sepa'rale from th e so lid surfac es, thu}i avoiding lar!,!e stagnatio n pressure losses. A reaction mlio o f 50 per cent is the conditi u n for maximum te mpera ture risc a nd effic ie ncy.
Tds
= 0 = rill -
(dP/PJ an d th t!reforc (t!. h)j"
= t:::.P/p.
T he pressure rise in a renl stnge (i nvolvi ng irreversible prt)f.:essJ pill he dClertmm:d . if the stage dl iciem.:y
IS
know n.
STATIC PRESSURE RISE
STAGE EFFICIENCY
The main fun c ti on of II co mrrcssor is to ra ise the s tatic pressure o f ilir. TIle ideal compressor stage is conside red first. wh ic h has no stagnati on pressure losses . Across the rotor row. Po", is f.:o ns tanl and the equalio n is
Stage eflicil.!ncy is ddined as the rati o of the ist!lI trop u..: enth alpy rise to Ihe al.:tual en thalpy ri se corre sponding to the same Ilnite pressure chn ng..: .
'1.1 =
(4.19 )
61r.1 (6")(1, ./
=
(4 .25)
AX IAL F LOW CUMI'RESSOH.5 A-":I' F.A,t-.'S ...: 1 D
REACTION RATIO The reacIi(lll ratio is a measure of the sialic enthalpy ri se that occurs in the rOior expressed as
C'I Ian fJ2 - CII tan a I 2U
I
CII
-1 +U
:I S
R
R=
St;Jlic I.!n thalpy rise in rotor
=
a nd suhsti tutin g for (II'!. - ""
R
t> and (11 3 -
(CI~
I
=
(4. 14)
+ W}2) ]
- +
1
C.II )
-
=
IV'I -
WI !
IR = (P, = =
(lV"
+ IV" )/2U + Ian /h)/2U
= (ta n /31 + ~an 131)/2 /Jill - mean relati ve ve loc ity vector angle and fl ow co-efficient
where. tan {J",
e'l
U
u- e u - e.q + Wfl l·,
2U
1
-+ 2
_lI...:'.,,,,::--,.,.:C:.;:.<,,, 2U
('un PI -tuna2) ~ tan 0'2>1 / 1
(4. IHI
R (4 . 16)
Pd / (P, -
Pil i
In th e case of compressi ble and irreversible now. a more general dl!li nition of R is in terms of static enlhalpics.
= (Ii, -
/lI )/(Ii, - I,, )
EFFECT OF REACTION RATIO ON THE VELOCITY TRIANGLES Case - 1 Whe n R = 0.5 . The reaction ra ti o R is
axia l vei ocilY = ~:--.,---,.:.. blilde ve loci ty
Similarly, the rc'lction r~Hi o c,m be ex pressed in diffel'l.:nt forms as follows: Substituting for WI 1 in t.!quulion 4. 15
R
(4. 15 )
C,,(ton fJ,
C" tnoll", = U = rp Ian Pm
2
For the case of incompressibl e and revt.!rsibk flnw. tht! t.!xprt!ss ion ust:J for reacti on ratio is of static press ure rise in the rotor 10 the stalic pressure ri se in the swge
I-Icncc.
R
U
= 2: +
R
U - IV \1" Therefore,
!
e,l Ian Ih - e'l Ian 0':0
I
{2U(C·, - Cx,)]
(CJ 2
C ,I !
2 ~U ~ + C, canPI - tana~)
=
(\V.,., + lY"H W" - 11'.,,)
ex, =
-
2U
12U(C" - Cx , )]
BUI C,I: = [j - \VJ'l and
+U
I
III ) in cq uatio-n 4.13.
+ w';)) -
lVI I
I Wq -C. = -2 + 2U
ex,)
(lYf - WiJ/ 12U(C" - C,,)]
(CI~
(.l.I7 )
u -eX!
=
R
= h{)~rd then."'1 -
Also if C 1 = C" lhcn , 113
:2 [I +¢( tan {J2 - tanail l/::!
=
W('2
(4. 13)
2 hI = ( W 1 - Wf)/'1.· - III ;;;; (hOJ - "OJ ) = U(C~, -
(tanti! - lllna,)
Similarl y. subs tilUti ng ror WI2 in equation (..L 15)'
SUllie e nthalpy rist! in stage /I '1 - li l II '}. - /q
Si nce linln'l
I
= 2+
R=
11 ,- -
(I'J - Ii ,)
"I
+ (II,
(4 . 18(a))
- I,,)
when R is 0.5
For a reac ti on rmi o of 50 per cent. the static en thalpy and lempCrDIU re Increase in the stator and rotor ;.tre eq ual. Also from Ihe cquution (4. 17 ),
R
= [I +
13nc,,)[/2
lJO ,.. TUlmo ~ ACJIINES
AX IAl . FLOW CO"'U'ItESSfl n S AN n F ... I'S .c,;: I II
......:----,
the 'H.ia l now compressor receives no contribution from the change in tangential ve loci ty tU). TIle isentropic or overalltOlal-(o-total efficiency is written us Idea l isentropic work input
lip 1.0 r------~"
I
Actual work input
= =
4
Total isentropic enthalpy ri se in th e s lage Actua l enthalpy risc between the same total pressure limits
."n.l.r.1 _-- -hOI 110.1 - "01 c~,
which reduces to Figure 4.4
(4.7)
\'£1,;at;o"
mca n
"I axial
I'cluci,.\' a/llII1: tJ bl{J(iI'
TIle v'lriotion in work done fa c tor ().) with s tage number is show n in the Fi!:! . 4.5. II shows that A decreases as the number o f compressor stage increases.
=
Putting POJ / PIII (ToJ.u/Tot>",-1 the pressure ratio becomes
1.0 (4.8)
~O. 95 9u
The energy input to the fluid will be absorbed in rai sing the pressure and velocity of
'""
th e air and some will be wasted in overcoming various fricti onal losses. However. the who le of th e work input will appear as a stagnation te mperature rise in Ihe air regardless of the isentropi c efficiency. Equating th e work in tenns of tempe rature and air angles.
I(To] -
TO]) = UC"Oan /l , - ,anfh)fC"
I
c
'C 0.85
~ o.8 '=0--4c--:::8--1:'::2:--1:':6:---::2U~
(4.9)
S tage number Figure 4.5
WORK DONE FACTOR In practice 'CII ' is nol cons tan t along the le ngth of the blade and. o work done fa c tor is introd uced. h is de fined as Work do ne factor. A
0.9
o
"0
10
account for this.
Variatiol/ oj wo,k dOl/e fac tor with nUll/her of SllJNt'·(
STAGE LOADING (OR) PRESSURE COEFFICIENT S tage loading (1/1) I' is defined as the ralio o r the power input to the te rm m U ~ .
Actual wo rk absorbing capacity
= ===c;:-==c:-'':-:-':''-'''''':' Ideal work absorbing capaci lY
1/1,,:;::;
Work input 111
U..,
~
He nce.
(4. 10)
(To.1 - Toil = AUC,,('anfj, - 'anfj,) f C" A graph is drawn between thc percentage of blade Icngth and
V'/I
W/mU
=
C (I
(Fig. 4.4).
-
holl/U!
A(C.,,-C.• , )/U
(4.111
).(C,,/U){lana2 - tanad
CII • mean It iJlus trntcs th at it is only at the inlct of the machine th a t the velocity profile. i.e ell . is fairly constant As Ihe air moves through the compressor the change in ell is morc. 11lis is due to the innucnce of the solid boundaries of the rotor and s tator.
2 = (hoJ
VI (I wbcn.:
rp
is the flow coefficicnt.
AI/> (lana1 - lana i)
(4.121
AXIAI.FLOWCOMI'IU:.5SOKSANDF ....NS ..:: 129
diffusion of llil' ab~olul': velocit y IJkcs place in the sli.Hor, where th e Jbsolute ve locit), vector is ilgain lurncd lowards Ihe Jxial direction and a further sUllie prt:ssu re ri sc
h
0)
NOle Ihal all angles arc referred to Ihe a:(ial velocity vector C II . TIle diagra ms arc Jrawn wi!h n brgl.! ~ap (for clarity) hClwecll rOlor and stalor blades. But in prnctice,
(herr.: is o nl y
i.I
0),
0355
sma ll 'Clr.:arance between the lll .
v
)
ENERGY TRANSFER OR STAGE WORK The cnl.!rgy gw..:n tn 'he- air per unit m,I.~S How rate is given by Euler's equation.
= U'].C,.~
~
P,
),
]ss
IW/III
POl
, ....- P",
02
n..:cu rs.
- U,C XI
~
Pul •et Pm ..,
Olrd
~U'
P,
I
nr (4. 1)
E = (V,C.,., - V, C." )/g From Ihl.! ve loc ity triangles. CII is constant through th e stage and VI
= U']. =
P, h ul = hUl
U.
hOI
:lI1d
Figure 4.3
=
C.q
U -
ell Ian /31
ell (Ian PI
C~! - C q
Mollia charI for an w:iol flow compressor S/{Jlft'
Then 1102 - "01 = (/'2 - ht>
- Ian fh)
",,= hlf,!",.
+ (C;~
or (h, - I,,) - (C" - C,' )[2V -
- C;,} /2
= U(C
+ c,' ))/2 =
(C"
XI -
CZ,)
0
Rearranging Thererore E = VC,,(tan ~, - Ian Ih)/g
(112 -lit)
(4.3)
Equlition (4.2) or (4.::q may be used depending upon the information available.
Ilul (W.~~
- wil) =
(\Vf - Wr). sin ce
[I
(":!
-"t)+(W}~ - W;,)/2
o
ell
is constanl. Therefore
\V;
~r
+ (U
The How throu gh Ihe axial flow compressor singe is shown IhermodynamicaHy on the Mollier chart (fig . 4.3) and is simi lar to that of a centri fugal compresso .... Assu min g adiaba tic now through the swge. 110] 11 02, and so eq uation for work s upplied is
=
\V 1m
= "02 -1t0I
(4.4)
-
- w.,.)(W'·2 +
h"+-=h l + - 2 2
MOLLIER CHART
-
o
W-t,)/2
+ (\VAl
The energy transfer may also ~c written in Icrms of the absolute velocity flow angles. £ = UC(j(tana2 - tunadlg
C, »)/2
(h, - I,,) - (C" - C."II(u - C.,,)
(4.2)
(4.51
(or) "O~.rd = hOl.rti where tht! relalive tala I enthalpy is based On the relative velocity. Eq unlion (4.5) shows thaI the to tal enthalpy based on relative velocities in Ihe rOlor is conSlan l across the rolor and thi s result is also valid fo r the' axial 1'ow gar tu,bi"e r010,'. II is already proved that the change in enthalpy for a centrifugal compressor is (1 = "OJd - U~/2J. (4.61
Writing
"0
=
c'
"+.,
A comparison of equation (4.6) with equation (4.5) indicates why tbe t:nlhalpy change in a single stage 3:(ial flow compressor is so low compared to the eentri· fugal compressor. The re lative velocities may be of the snme order of magnitude:, bUI
116 l> TURno
fl.lf1CIlINES
I
G S
It V
VELOCITY TRIANGLES FOR AN AXIAL FLOW COMPRESSOR STAGE In studyi ng the now of the nuid through an axial compressor. it is usua l to consider the changes tak ing place through a compressor stage. The analysis ror now through the slage is assumed to be two dimensional and 10 take place at a mean blade height. The rOlor and stator rows of n stage ilre shown in Fig . 4.2.
)---!O=!'LlbdlJ~liJ:L ~ ~:;~:: Stage - " = - -_ ..
'>___-"~~---------------~---{Discor OJ hub
Figure 4.1
All axial compressor stage'
not fonn part of the compressor stage bUI arc solely to guide Ihe air at the correct angle onlO the fir~ 1 row of moving blades. The blades heighl is decreased as the nuid moves through tbe comprcs~or..This is to maintain a const::mt axial velocity through the compre~sor as tl.H~ ~ensJty m~reascs from Ihe low to hig h pressure regions. A constant a~]al velo.cHY J~ convenient from the poi nt of view of design. but this by no means IS a r:cqUJremcnl. The now through the siage is assumed to take place at a mean blade height. where the blade peripherol ve locities at inlet and ou tlet are the same. There is no now in the radinl direction. Whirl components of velocity exist in the direction of blade mOLion.
WORKING PRINCIPLE The kinetic e nergy is imparted to the air by th e rotating blades. which is then converteu i'l1to a pressure rise. So. the basic principle of working is simi lar to thilt of Ihe cenlrifugnl compressor. Refer:i~g to Fig. 4.1. the air enlers uxially from the right into the inlet guid~ vanes. w~ere II IS denecled by a certain ilngle to impinge on the first row of rOlaling blades ~nh I~C proper angle or l.1uilck. The rOlaling vanes add kinetic energy to the air. There 15 a sllghl pre!'sure risc 10 the air. The air is then discharged al the proper :mgle to the fi~st. row of ~tator blades. where the pressure is further increased by diffusion. The air ]s then directed to Ihe second row of moving blades and the same process is repealed through .the remainin~ compressor slages. In most of the compressors, one 10 Ihrc~ rows or diffuser or stfillgh~ener blildes arc inslalled at Ihe end of the lasl stage 10 !;trmghtcn and slow down the air before it e nters i!Ho the combusti on chamber.
c., Figure 4.2
VelucilY triangles for
CUI
(J.dal flow compressor 5Wf.:t'
A stage consists o f il row of moving blades aUilched 10 the periphery of a rotor hub followed by a row or fixed blades allached to the walls of the outer casi ng. Th e compressor is made up of a number of such stages to give an overall pressure
rati o from the inlet to outlel. Air exits from the previous row or Slator bladers at an angle 0'1 with absolule velocity C I. nle rotor row ha s tilngcntinl veloci ty U. and co mbining the two veloc ity vcclUr~ gives the relntive velocity vector lV I ilt an angle fJl. At rolor row outlet the absolute velocity ve.ctor Cz movcs into Ihe stator row where the now direction is changed from az to O'J with ilbsolule velocity C,l. If the following stage is same as the preceding one. then the s tage is sa id 10 be normal. For a nann ill stage Cl=C3
anda t=O'J
1\'2 is less than WI, showi ng thilt diffusion of relative velocity has taken place wit h some static pressure rise across th e rotor blades. The :lir is turned lOw:lrds the axial direction by the bluc.k ca mber and the effective now area is incrc'lsed from inlet to outlet. thus cil usin g di ffu sion to liike pl:lcc. 5il11il:}r
AxiAL FLOW COMPRESSORS ANOFANs
INTRODUCTION Axial flow compressors and fans arc turbo machines Lhal increase the pressure of lhe gns flowing continuously in lhe axi al direction. Due to lack afknowledge oflhe aero· dynamic behaviour. a reversed rcaclion turbine was used as an axial How compressor in carly days. The efficiency was less than 40 per cent. Then, study of aerodynumic behaviour hdpcd in designing the blades for axial flow compressors. Nowadays. the efficiency of the axial flow compressors surpass the maximum centrifugal compressor efficiency by about 4 per cent. But the efficiency o[the axiaillow compressor is very se nsitive (0 Ihe mass flow rale . Any deviation from the design condition causes Ihe dficit!ncy (0 drop off drastically. Thus the axial flow compressor is ideal for constant load applications such as in aircraft gas turbine engines. Th!!y arc also used in fossil fuel power Simian s, where gas turbines arc used 10 meellhe load exceeding lilt: normal peak load .
Advantages of Axial Flow Compressors I . Axial !law compressor has higher efficiency than radial now compressor. 2. Ax inl now co mpressor gives higher pressure ratio an a single shaft with felalive1y high efficiencies. 3. Press ure ratio of 8: I or even higher can be achieved using muhistagc axial flow compressors. 4. The greatest advantage of the axial flow compressor is its high thrust per unit fronlal area. 5. It can handle large amount of air, inspite of small fronlal area. The main disadvantages are its complexity and cost.
DESCRIPTION OF AN AXIAL FLOW COMPRESSOR An
I;!;!
:.-
T1 11U!\)
tvt ,\I ·IIINI·.S
] .26 . A I.:cnlrirugnl compressor works Wilh no whirl al entry and has radial cxit. n,!: slip ractor is O . ~J. The rolortip speed is 625 m/s. The mass now rale IS 15 kg/s o l'lC ambIent comJilions arc 98 kPa and 23 fJ C. TIle pressure rauo is 7. Cakulate (a) isentropic efficicncy (h) work required per kg of air (c) power supplied for a mcchan ical efficiency of 97%. [Ans: (0) 62% (b) 355.47 kllkg and (e) 5496.93 kWI .~ . 27 . The foIlClwi"f! d
IAns: (a) 82.J'W (b) 385.3 K (e) 2709.3 kJ/kg and (d) U.742 1 3.18. A double -sided cenlrirugal compressor has impeller eye root and tip diamclers I XO mm unu ~OO 111m and is to deliver 16 kg of air per second at 16.0000 rpm . The design ambir.:nt conditions arc 17"C and I bar. Calculale su iHlhle valul! ... for lhe impeller vane ang les al rom and lip of the eye if air is giwn 20" measured from thc radial direction of prewhirl al all radii . 11le ux.ial component of inlet velocity is constant ovcr the eye and is equal 10 150 m/s.
[Ans: ("157.33" and (b) 37.H·1 .3 .29. A cenlrirugal blower takes in 180 mJ jmi n of ai r at PI :::: 1.01 ~ bar and '1 = 43 ~' C .1Od delivers ill'll 750mm \V.G. Taking the effiCIencies of the blower and drive as 80% and 82% respecti vely. delcrmine the power required to drive the blower and the statc of air at cxil. (MV-April '97 & OCI'99) IAns: (aJ 33.65 kW (b) J .0866 bar and 324 .1 KI 330. A ccntrifugnl fan wilh a radial impeller produces., pressure equivalent to 100 cm column of water. The pressure and Icmpcr
IAns: (a) 45.3 kW (bJ 66.4 em (e131.9 elll idl22.6' and (e) 15 .95 em and 7.66cml 3 .3 1. A fan running at 1480 rpm takes in 6 mJ / min of air at inlet co nditions o f Pt = 950111 bar and'l 15°C. If the fan impeller diameter is 40 cm and the h lade tip air angle is 20 a , determine the lOtal pressure developed by the fan and the impeller widlh at exit. The r;Jdial velocity at the exit is 0.2 times the impeller tip speed . Stale the assumptions used.
=
IAns: (aJ 50.67 mm w.G. and (b) J .28 em l 3.32. A eentri fugal compressor runs at 15,000 rpm ::Jnd produces a stagnation pressure ratio of 4 between the impeller inlet and outlet. The stagnation conditiuns of air
at [he exit s uch tital th e exit n.:lativL' velocity is 1.15 m/s and the comprc'isnr t(1tal ~ l o. t owl cf!iciel1 cy is 0.78. Compute the slip, slip l:(u;ITkil!nl and ahsolulC vel (l{:i ty ;11 Ihe compressor ex it if Ihe: rutor diameter ;\I Ihe \lutlet IJot 5KO 111111 . [Ans: (a) 45.9 mls (hi 0.~993 and (e>·n U 10/, 1 3.33 . An airera!'t engine is fillL'd with ;} s ingJc~sidcd ccntrifu£al l:ompres:.or. Thc nirl:raft lIi cs with a speed of 850 km/h at an altitude where [hl! pn::ssure is 0 .2.1 bar and the tempera lUre 217 K. The inlet ducl or the impclkr eye r.:onlam'> li xed V;:1I11.::. whir.:h £i ve the air pre-whirl of 25 r - al all raJii . The inner '--'lit.! lluler lIi:UllI.:lcr o r thc c\'c arc 1HO ilnd 330 mOl rt.:!otpet:l1vcly. The dialllcter " I' Ihc impdler tip is j.H} ;11111 and the rotalional speed Ifl.OOO rpm. ESlim;JIC Ihc s litl!n;Jtiull presJoture allhe cllInpressu r o utlet wllcn Ihc mass /l ow I'> 216 kg/]lIll1 . AS~UIllC lht: ise llirnpi..: dfi cicllq· (0 he O.};. T~ke (hc slip r:II.: (()r it'> (I .Y ~nllIIH: pllwer inrlll fa clor 1IJot 1.0-1. IAns: 1.6:\ harl 3..14 . A centl'lfugLlI fan wilh nn errlcienc), Df 80% runs al 720 rpm . It s imrdlcr JI allleh.:r i ~ I 111 . The imrcllcr lip nngle is backward curved 10 51 (.1 langcnt [1 1 the wheel. Thl.! density or Ihe 'Ii r is 1.25 kg/ m '; nnd lll::Jss now riJlt: is .~ kg/!'. The impcl1cr width at the exit is 10 em. Detl!rminl! the PQwer reljuircd. pressurt: coeffklCI11. stagc reacti on. pressure hcad L1cvdopcd. and flow codlieicnl at c.\ il. Assume I.ert) whirl at inkl ;:md the me(;hanical crflcicncy is "2':''''. 1Ans: (aJ 4 ..15 kW (h I 1.34 Ie) 0.58 Id I 1. 188 kN / m' Hnd (e iCl.~O.1 I 3.35. Air nllWS through a bll1wcr where its lOla I pressure is increilsed l1y 15 el11 411 wa leI' head . The inlcltll[;ll pressure and lemperalUrl.!·arc lOS kPa ;Jlld lS ':C. TilL' IOlill~l o-tu tal efficiency is 75%. Estimate (;) exit to lal pressure and tempe rature and (b) isenlropic and actual change in lOla I head .enthalpy. (MU'''pri( 'C;(J\
[Ans: (H) IOfi.5 kPa. ~S9 5 K and Ihll.152 kJ/kg. 15.16 kJIl.I .1 .30. A cenlriru~al cllmpressor compresses .30 kg or'!If per SCL-. JI run o; al 15 .000 rpm . ' nlC •.111' cntcrs the comprcsso r axially. The rndius at exi t of blade IS ~OO III Ill . The rcialive vcineilY or aIr ell exit lip is I DO m/s. Tht: rclalivr.: air angle at (.;xil is XI) . Find thl! power and idc al head dcvelopeu. tMU-April ' W}I IAns: la16416.5 kW and (hi 213.9 kJ /kgl
12U .... TUIUlO I\I ,\ C III NES
-' . 16. Explain Ihe phenomcna of surging. sla lling and choking in cCnlrifugal compn.!SSOI' sJnge? What is their drcct on [he performance? How 10 minimise or prevent them . . ~ . 17 . A 5RO kW mOlOr drives a cemrifugal compressor of 480 10m OUler diamelCr al a spl!l!d of:20.000 rpm. AI till! impdler o utIet the bludc ang ll! is 26.5 ° measured from the radial direction and th e flow velocity at ex'it from the impeller is 122 m/s. If a mechanical er-ficiency of 95 per ccn! is assumed . what air now i!<. to be expcclcd? Assume thert: is no slip. What arc [he eye lip and hu b diameters if a radiu s ratio of 0.3 is chosen fo r the impcller eye and if the velocity m inlet is 9S Ill/s wilh i'.l!ro whirl? What wi ll he the overalltota l-I O- tOlall scmroplc dficicm:y if an llvcralllOw l pressure fLltio nf 5.5 is rl.!quircd'! Assume thatlhc now inlet is incompressible and ambient :lirconditions are 10 1.3 kPa .:lOd :!8R K. IAns: (a) 2.47 kg/s, (b) 172 mm and 51.6 mm (c ) 81.8 % 1 -' . 1H. A centrifugal compressor impeller hilS 17 radial vanes of!ip diameter 165 mm. It rotates al 46.000 rpm and Ihe air mass flow rale is 0.6 kgls with no whir l at inlel. (,, ) Calcul:ite the theoretical power trans ferred to thl.! olir. AI inlet 10 the impe ller, the mean diameter of the ey e is 63 .5 mm while the annulus height at the eye is 15 mm . The static press ure and temperature at the impeller inlet :Il"e Q.'\ kPa and 293 K respectively. Determine (b) the bludc angle ut the menn diameter at impeller inlet (c) the stagnation temperature :u impdler exit, and (d) the stagnation pressure at impeller e.'( it if the lotal-Io-total efficiency of Iht! impeller is 90 per cenl. IAns: (a) 83.76 kW, (b) 35.4°, (e) 437.8 K, and (d) 338.67 kPaJ 3. 19. A centrifugal compressor is desired to have a total pressure ratio 4 ; I. The inlet eye 10 the compressor impeller is 30 Clll in diameter. The oxial velocity rll in lt.!t IS 1)0 m/s and the mass flow rate is 10 kg/s. The velocity in the de li very duct is I 15 m/s . TIlt! lip speed of the impeller is 450 mls. runs at 16.000 rpm with to[al head iselltropi c efficiency of78 % and pressure cocffleienl is 0.72. The amoient conditions arc I bar and 15'J C. Culculate (a) static pressure and tcmperature at inlet and exit o f compressor (b) slatic pressure rati o (c) work do nelkg o f air (d) theoretical power required to drive th e compressor. (MKU-Apr;1 1993 ,\pr;1 1995) IAns: (n) 279.6 K, 0.9 bar and 3.89 bar, 460.86 K (b) 4. 31 (e) 180.34 kJ/kg and (d) 1803.4 kWI .1 .10. A single -s ided centrifugal compre ssor deli vers 8. I 5 kg/s o f air with a press ure ratio o r 4.-1 [0 I at 18,000 rpm . The t!ntry 10 thc eye for which the internal diameter is 12. 7 em is axial and Ihe mean vcloicty at the eye section is 148 m/s with no pr..:whirl. Stntic conditions nt the eye sectio n arc 15"C nnd I bar. The slip factor is 0.94 i.llld Ihe isclllropi c efficiency of compress ion is 78 %. Nt.!g.lecling losses calculale (a) the risc in temperature during compression if the kinetic en ergy is neg li g ible ( b) the tip speed of impeller eye and tip speed nf impdler outlt'! and (c) ou ter diameter of impeller eye. (MKU-Nol'. '92) IAns: (a) 202.2 K (b) 257 mi, and 464.95 mis, and (c) 49 cm]
CENTRIFUGAL COMl'ltESSOJ{S ..... NO F ..... NS "'" I:! I
3.21. The data of a centrifugal compressor nre given bdow. Outer diameter of the impeller = 50 cm Tip diameter of the eye = 18 em Huh diame ter to the eye = 14 em Specd 16.000 rpm Mass of air handled = 10 kg/s Inlet total pressure = 1.15 bar Inlctlotal wmperaturc = 20(J C Slip factor = 0.91 Total-to-total emciency ::::: 75% for zero whirl at entry. Detaminc Ihe total pressure ratio developed and the power required to drive the compressor. (MU·Ocl. '97) IAns: (a) 3.3 and (b) 1596.7 kWI 3.22. A centrifuga l compressor has inlet guide vanes at the eyc such that fn!e vortcx now is achieved at en try (0 the blades. A( the tip radius oftlte eye the inlet rdative Mach number is not (0 excel!d 0.75 nnd an impeller (Olal-lO·lot<11 c.llicicncy of 0.9 is required. The air leaves the tip of the inlet gU ide vanes wilh a velocity of 90 mIs, the impeller tip diameter is 0.45 m and Ihe outlet diameter is U.76 m . The radial component ofvdocity at exit from the impeller is 50 Ill/s and the impeller rotates at 12.000 rpm. If a slip fnctor of 0.9 is assumed. find Ihe guide vane inlet angle at the tip and the Sialic pressure III impdlcr outlet. Assumt.! TOl = 288 K and POI = 101.3 kPa. IAns: (a) 62.1 ' anJ (b) 247.2 kPal 3.23. A centrifugul compressor compresses ai r at ambient lI!mpcraturc and prt.!ssurc of 288 K and 101.3 kPa respectivdy. The impdlcr runs at a tips peed of 365 m/s. lhe radial velocity at exit from the impeller is 30 mls and the ~ Iip factor is 0.9. Cnlculate the Mach number of the Howat the impeller tip . If thl.! impeller tota]-lO-lotal efficiency is 90 per cent and the now area from the impeller b 0.093 m 1 . Calculatc the mass How rate of air. Assume zero whirl at in let and radial blades. IAns: (a) 0.876 (b) 5. 13 kglsl 3.24 . A compressor operating at a pressure ralio of 3.8 and n speed of 12.000 rpm delivers 8 kg/s of air. The slip factor is assumed to be 0.92. the power factor 1.04 and the overall isentropic efficiency 0.82. Calculate the impeller oullcl diameter. Assume lCro whirl. The Mach numberof lhe air leaving the impeller voncs is to be unity so as 10 ensure that nO shocks occ ur. If the losses in the impeller and the diffuser arc the same. what must he the 3xiOlI depth of the impeller. AI inlet POI = 101.3 kPa and Tot;: 288 K. [AIlS: (a) 0.659 Ol (b) 23.5 mOll 3.25. A centrifugal compressor with an over~1J pressure rati~ of 4 has... an impcll.er . speed of 320 mls and the now area at Impeller oUllet IS 0.12 m-. TIle rad Ial component of the velocity at impeller exit is 30 mls and the slip factor is 0.9. Calculatc the absolute Mach number al the exit and the flow ralC if tht! total \o-,o'al emcieney is 90%. Take TOl = 288 K and POI = I bar. IAns: (a) 0.7M6 and (b) 9.87 kg/sl
J J II
3.23.
,. TURBO MACIIINES 'nlt!
EXERCISES
tilL!0-rclical work done on the air in a centrifugal compressor is
101) OsUj
(bl U;/2 It:) a.\-U,~ .t2
3.1. Draw the skl.:tch of a ce ntrifugal compressor slage mdicatlng the pnnclpal parh. .1 .2. Draw skelches of the three types ofimpclh.:rs and the velOCity triangh.:~"11 theIr entries anJ ex.its. 3 ..1. Draw an enthalpy-entropy diagram for a centrifugal compressor slage showlIlg. s tatic ami stagnalinn values of pn:ssure and enthalpy at variuus sectio ns . 3.4 . Provc th
Ib) J=ho-U'/2 (c) I = "n.rd - \V~ /2 3.27. Dl!line uveralliotal-to-total isentropic efficiency. 3.28. Deline pressure eoerricicnt. :'.2Y. The pressure codficienl and overoJl isentropic efficil!ncy arc related by the
foliowlOg equatJon. (a) !{Jp = 'It . o.l/r,p (h) I{J" = 'h- . OJ (c) '/I- = ri'I,/q.,O.I 3.:m. A simple volute recovers 25~30 per cenl of the available kinetic energy at impdler exit. (Trueif.alse) The diffusion efficiency of a vanelcss diffuser is higher than that of a vaned diffuser. (TruelFalsel J.J2. In a v:lfled diffuser. a more uniform tolal flow occurs when the number of diffuser passage is less than the number of impeller passages. (TruelFalse) lllC divergence angle of the diffusion passage is in the order of la) R - 10" (h) 20-40" (el30-45" .1.34. DeflTlc diffuser efficiency. 3.35. Dl:fine dcgree of reaction. 3.36. The degree of reaction for radial vanes is la' R = (I + r/J~)/2 (hi R = (I - r/J~)/2 Ie) R = (I-
3.5. What is fluid slip? Define slip factor. Give three furmulae
\0
cakulall! the :-.llp
factor. 3.6. Derive Stodola's relation for the slip factor? 3.7. What is pressure coefficient of a cemrifugal compn.!ssor stagc·.' Derive
3.8. Ddine rower input factor and comrrcssor overall isentropIc enit:iency. 3 .9. Prove thaI
where 'P,,-pressure coeffiCient. lit- is compressor eiliL'lenc),. a, ~ :-.llr factllr and f{J ~ power input factor. :U O. Write short notes on (a) Volute or scroll collector (b) Vancless diffuser (c) Vaned diffuser 3.11. How is the degree uf reaction of a centrifugal compressor stage Jdincd .
Prove that
where r/J IS now coefficient and {j'J. is hlade outlet <.Ingle. 3.12. Deduce thal the degree of raction R for a centrifugal compressor With raJI,,1 impeller vanes is given hy
R = (I -1>1)/2 3.13. Bndly explain the effect of cach impeller vnne on the perfofl1wnt:c "I ( . :nlrifugaJ compressor. Why is the radial tipped impdlcr 1110:-.t cOllllTlonly u ... ed In centrifugal compressor stages? 3. 14. Explain briefly the purpose of inlet guide vnnes. 3.15. How do Ihe Mach numbers al the entries of the impeller and diffu ... er affcCllhL' no\\' and d 'riciency of i.l centri fugal cumpressor slag.e'! On wh,1I considcratltln\ ,.II"l: thc limiting values of these Mach numbers decided?
CENTltlfUGAI. CO!">II' IH:SSQI«S AN D E\I'~S ""
jlfi ;.. TUIHlO M ,\ CIIINES
SHORT QUESTIONS
(c) Power input · Work
don~/sec II",
II/ I "o~-"1I1 1
=
Work L1llnl.! / Sl.'C
U; - U'1 +II"-wi 1 _ +C;-C'] _ I [ 2 '2 2
=m-
:. POW!!l"
input
=
0.5121.78
=
84.64 W 84.64
= =
+ 55.5 + 921
0.78 108.5 W
Example 3.14 A centrifuga l fan rises the static pressure of air by 14 cm of water, whik run ning at a !! pl.'cd of650 rpm nnd consumi ng 85 metric HP as power. The s tatic press ure ilnd wlllpcram rc of the ai r at the f:ln intake arc respec tivcly 75 cm Hg ilnd ::!jt.c. while th l.: mass How nlle of ai r is 160 kg/ min . Find the exi l static pft..'ssure and Ih..: vn lumc now ralc in m·l /mi n. (MKU - April '9-1) Solution L\ //:;:: 0 . 14
III
1 17
=
or H:!O
Pll\\'l.'r = H5 MHP 1'1 = 27:1 + 15 = ::!9X K Inh.: tril' H.P 0 .735 kW
N 650 rpm H, = 0.75 "' of Hg //I = 260 kg / min
=
(a) Exit static pressure 1', p,+c, p 1', PII ., g H, = 13. 590 x 9.8 1 x 0.75/ I 0' 99.988 kPa
.1. 1. What is centri fu gil l compressor'.' 3.2. Centrifugal compressors and fan s arc employing • dfccI ~ It) ilU.: rca~l· tht.: prcssun.! of the gas. 3.3 . Ccnlrifugal comprl.!ssors ean handl!.! contaminated ga!! . (TrudFabt.:1 3.4. What arc til L.". ad"unga tes and disadvantagcs of centrifugal compre~snnl·.1 3.5. What arc the applic
Fon
3.7. Tht! im pell er vaneS at the eye ure benl why? 3.8. Th e dircclion of now o f gas in the impeller of II ce ntrifugal cn mpn:ssor i!> Axial Radia l (C) TangcllIial 3.9. Impe lh.:r of a cenlrifugal compressor convCriS all the mcdlilnicul energy sup · plied inlo (u)
(b)
(al pn:ss un.: energy o nl y (h) kin ~t ic energy only (e) ki neti c and pressure energies
3. 10. What is the function of a diffuser in n centrifugal compressor'! 3. 11 . Draw th e pressure and velocity variation across a ce ntrifugal cumpn: ssur 3. 12. It is conventional 10 measure blade angles from - - - direction III cCnlflfu!!a l compressors 1. 13 . Thl! air ,now angle at inlet, measured from the radial direc ti on is (a)
45 0
(b) 90" (c) OC
and
61'
=
p",gC,H = 1000 x 9.81 x 0.14/10' 1.3734 kP.
=
99,988
+ 1,3 734
101.36kPil
P'}.
(b) Volume flow rate Q
III /P I
I',
'!.60/p, 1', / RT, = 99,988 x 10' / (287 x 298)
c. Q
=
3. 14. Draw the in let and ou tlet velocity tri.mg lcs for a centrifugal comprc!>snr wi lh mdi allback ward-curved/forwanJ ·curved hlade impeller. 3. 15. What is no-shock condition? 3. 16. What is no-s lip co nditi on? .1.17. Angle fi~ is - - - tilan angle {h due to s lip . 3 . 18. Define n-uid slip in centrifugal compressor. 3.19. How can the nu id s lip be r~ducl!d in cen trifug al compressors? ) ,20. Define s lip faclor. 3.21 . Write the Stodo la slip factor equation. 3.22. The Stn nill. slip f.aclor ..:q uation for radial vanes is
1.169 kg / Ill'
(aJ
OJ
260/1. 169 222.4 III 3/ min
(b)
a,
(cJ
=I-
0.63rr /r.
= I - 0.63,/rr a, = ( I - 0,63rr i/'
;
I
I,
·1
CENTRIFUG .... I. COMPRF_'iSU llS ANIl FAf'.;S -.i.
r j.J ~ TwUJo M .... ClIINES
Solution
From the outlet vel ocity tri angle (Refe r Fig . 3.J(b)) .
I
U:! - C x =
\V q
=
117.8 1 -98.96
=
D'1 = 0.2 m
WI ': ~ U m / s
W2
= 17 m/ !.
N = 1450 rp m
IJIII
= 0.78
p = 1.25 kg/ m"
III
18.85 lOis
c:!
D,= O. 18 m = 0.5 kg/ s
C, =2I m/ s
= ~Sm / s
(a) Stage total pressure rise
;lI1d
Ia n fi2
=
~,
=
flo
=
C,!
6 Po
= p( h., -
ho, )
WI":
I
hUI
' ( -5718.85/ 71.7" t n ll ~
(b) Power input
;,md
'"
[\ 17.8 1 x 98 .96) / 10'
=
(c) Exit stagnation pressure
[(po~ ) '~ - ] Po"
.6Pn
I
U, =
,/.
=
TOI
=
IVl mC" = 11.658/ 1.005
=
11.6 K
6 Pn
=
[ 1 + 'kITo, - TOil ),'-' TO I
t:l Pu
Pm
Po,
=
To:! -
[I
TO I
"
+
=
,
10 1
T02 -
·nercforc . tht! sti.lgc total pressure ri se is
11 .658 kW I (kg/s)
0.75 x 11.6] '" 298
=
1.1 06
Po:!
=
1. 106 x IO U25
P Il:!
=
I J ::!:.07 kPa
A centrifuga l r.m has the fo ll owi ng data: inner diameter of the impeller - 18 em. ouh.: r diameter of Ihe im peller -20 em. the absolute ve loci ty al cn lry is 2 1 m/s and at exi t is 25 mIs, re lative veloci ty at inlet and ex it arc 20 mrs and 17 m/s rcspl:c livc ly. speed - 1450 rpm, now rale - 0.5 kg/s, nnd m olOr efficiency 78 %. Determ ine (a ) stage lotal pressure risc (b) degn:c of reac ti on and (c) powe r required to drive the fan. Assume densilY or ai r as l .25 kg/m J . IMU-Ocl '98 ]
U, =
P
U;_ - U' IV' - IV;_+ C' - C' ] [ 222 1+
I
2
IT
x 0.18,1450
IT
60 x 0.2 x 1450 . 60 = 15. 18 m/s
~:....::.~~:..:.::.::
15. 18' -1 3.67' 2
I
= 1.1.67 m/.:;
+
20'-17' 2
=
[
=
12 1.78 + 55.5 + 921 x 1.25
=
2 11.0 N/ m'
+
25'-2 1' ] 2 x 1.25
(b) Degree of reaction
R =
(6.
P )mrur
(6. PO)mge p(/12 - 'Ii)
=
Ex ample 3.13
U; - U'1
p_
[
=
1.25121.78
=
96.6 N/ m' 96.6 2 11.6 0.457
R
=
R
=
IV' - IV.;' ]
+1
2
_
2
+ 55.5 1
I I~
II:!
>-
CENTRIFUGA l. C OMI'R ESSOKS ANIl FANS
TURno MACIIINES
to the impclkr is 3 kW. Dctenninc th e: fi.ln efficiency, prt!ssurc coe ffi cient, degree: of rcactillil . ["olUtion:!1 speed, and impeller width at c~it.
(c) Degree of reaction
R :;:
Solution
111 = 30e III ::;:
¢~
=
D"!. :;;: 0.466 m
l:> N
--l .29 kg/s
C,
U;
= 0.25
= 0.063 mW .G .
IV = , kW
J
C1·,
1-- -· 2 U"!.
I (19.86) 35.03
=
I -
=
71.7 0/,
'2
(d) Rotational speed
(a) Fan efficiency =
(lIPo) IIlp
=
Ql:>Po = Q(pg l:> N)
2.36 kW
Actua l work done
= = =
.'. Fan effi ciency
=
'If
=
Ideal work d01l1;!
3.82(10' x 9.81 x 0.063)
N
b,
2C.
U,
From nutl!.!t vdo~ilY triangle ( Rdcr Fig . 3.3(h)).
U? x 60 rr D?
=
1435.7 rpm
=
3 78.7%
r1 V'/I :;;: - -
=
35 .03 " 60 rr x 0.466
(e) Impeller w idth at exit
3kW 1.36
(b) Pressure coefficient
Q
3.82
rr D?C,:
rr x 0.466 x 8.711
=
0.298 m
=
29.8 em
Example 3.12 A cen trifugal blower runs ut a spccd or 3000 rpm . Till.! impelh:r ouler diu meier is 75 em and the blades of Ihe impeller arc dcsigncd ror a constant radial veloc it y of 57 m/s. TIlcrc arc no inlet guide Vi1!'l ;":; so Ihallhc absolute vclocil)' at the: inlt!1 is axial. If the degree o f reaction is 0.58. compute th!.! t!x;it hlndt! i.lngle:. A lso delcrminc the power input to the blower (tola l-to- lOlal dficit!ncy of 0 .75) and the exit stagnation pressure. Take Iht! tOlal prcssurt! and temperature at the inlel a!l (MU-April '98)
I aIm illld 25'·c rcspcctiveiy.
Solution N = 3000 rpm C.q = 0 U, [U, - (0 .'25U, 1 Ian 30") J =
:. x 10]
·1.29
057Ui
=
699.93
Ul
=
35.03 m/s
C" C.I !
--
...;: I I J
=
0.25U, = 0.25 x 35.03 = 8.76 ml'
=
U"!. - (C'2 / IUn fh)
:;;:
35.03 - (8 .76/ tan 30°)
= =
19.86 m /s 2 x 19.86 35.03 1.134
=
D, C,
= 0.75 m = ell
c"
= 57 m l'
R = 0.58
'Ie
(a) Exit blade angle 7r
D2N
= 60=
Uz
=
If
x 0.75 )( :\000
60
117.81 m / s
Tile degree o f reaction is
R
=
=
I
Cx ! 2U, 2(1 - R)U, = 2( 1 - 0 .58) 117.81
98.96 m /s
= 0.75
110 j.
Tormo M
CENTIUfUGA LCOMi'RESSU H.S AND F ANS ..{
,\ CIIINES
:. ex,
(b) Volume rate
=
33.55 - (9. 15/ I.n39" ) 22.26 m/s
=
Q
m/p
IV / 111
33.55 x 22.26
3.317/ 1.128
746.823 J / kg
2.94 m' /5
(c) Exit properties of air P, - P,
Fan efficiency '1J
=
7357.5 = 100+ -,ID
=
107.36 kP. Wlm
=
T,
=
82. 1%
(b) Discharge
Cp(Tz - Tl)
T,
Ideal work done / kg Actua l work done / kg 613 . 13 746.823
=
6.P P, + 6.P
III
T, + ~ = 309 + 8256.47 mC II 1005 317.22 K
rrD2b,!C r rr(0.89)(0. 1)(9.15)
Q
=
2.558 m' /s
pQ
= 1.2(2.558) = 3.07 kg / s
Example 3.10 A backward-swcpl centrifugal fan develops a pressu re of75 mm
w.G. It has an impeller diameler of 89 em and runs al 720 rpm . The blade air angle . , lip is 39° and th e width of the impeller is 10 em. Assumi ng a conslan t rad ial velocity of 9. 15 mls and density of 1.2 kg/m J , determ ine the fan efficiency. discharge, power required. stage reacti on. and pressure coefficient.
(c) Power required m :. IV
m(IV / m)
= 3.07(746.823)
2292.7 W
Solution 6. 1i
=
= = 0 .075
In
D, = 0.89m
W.G.
N = 720 rpm
13, = 39"
b, = 0. 1 m
C, = 9. 15m/s
2.2927 kW
(d) Stage reaction
p = 1.2 kg/m'
I _ ~ C.l l
R
(a) Fan efficiency
2 Uz =
Ideal work done/kg
6.P/p
=
=
(p"g6.H) / p
66 .83%
(10' x 9.81 x 0.075) / 1.2 613. 13 J/ kg
I _ ~ (22.26) 2 33.55 0.6683 or
(e) Pressure coefficient
Actual work donclkg IV /m
= U,C",
1/1"
From outlet velocity triangle (Refer Fig. 3.3(b)}, Uz - W.l 1 U, - (C, / lan/3,) rr D,N rr x 0.89 x 720
=
---w- =
31.55 m/s
60
since
=
=
2(22.26) 33.55 1.327
Example 3.11 A backward-swept (fJ 2 = 30°) centri fuga l fan wi th impeller di amctl.!rof 46.6cm is required to deliver 3.82 mJ /s (4.29 kg/s) of air at a total pressure nf6J mill W.G. The flow coeffi cie nt at the impeller exit is 0.25 and the power supplied
CENTRIFUGAtCOMPItESSOR5ANUFANS .... 10K
lOll
;... T URIlO MACHINES
(a) (ii) Impeller eye hub speed
=
•
If
..
jW}, + cj
=
\V,
256.39 m/, 256.39
D"N
60 x 0. 175 X 16.'000 60 146.6 1 m/s
=
Mr. I
JI.4 x 287 x 274.98 0.77
7r
=
Mr. 1
From thi! vt.:inl.:ity triangle at inlel.
c
c,
-.-"- lat
:0
9() - ::!O" =
sin 0'1
7Q"1
\--- U'-----'i
sin 70"
T,
TOI -
2e
= 2MB -
=
.'. p,
)''' - i =
-=-::.c.:.:-,,--
2 )( 1005
U" - C.,
lan fll.ll
:::;
"
288
85 .05 kPa P, 85.05 x 10' 1 RT, = 287 x 274.98 = 1.078 kg/ In '
e" = .,.,.---"--,-
e"
p
6> P
=
=
7357 .5 N/m'
=
= = =
Ideal work done/kg
IS:! __~---'-,--~ = 1.665
146.6 1- ( 152/," n 70")
e" VI - (Cu / lanad
W/m
152 261.79 - (152/ 'an 70") O.7~62
::;
.'. /3,-,
=
Actua l power inpul
36 .36" IV
(b) Maximum Mach number at the eye tI,·la ximum Mach number occurs at the eye lip
WI =
VI
-ex,
261.79- (152/ ,"n70") 206.47 m/ s
7357.5/1.128 6522.6 J l / kg
~B
= =
6522.6 1 0.79 8256.471/ kg
==
Motor power input )( t7m
=
33 x 0.83
27.39 kW
Therefore. the mass now rate
m =
JrRT, lV.,.•
6> P / p
Ideal work done/kg
Actual work done/kg tanfi1.r
= 0.750 m W.G.
P, 100 x 10-' J 1.128 kg/m RT, = 287 x 309 pgAH = IO J x 9.81 x 0.750
=
VII - (CII/tall 0'1)
and
AH
11111:;:: 0.83
(a) Mass rate
100 (274 .9R),n
.
T, = 309 K liB = 0 .79
P, = 100 kPa input power = 33 kW
161.76'
274.98 K "
Po, ( -T, To,
Example 3.9 A ce ntrifugal blower takes in air at 100 kPa and 309 K. It lh! \'c:lops a pressure head o f 750 mm \V.G .. \\.'h ile consuming a power of 33 kW. If the blowe r effic iency ('1IJ) is 79% and mechanical efficiency is 83%, determinl! the mass rale Jnd volume rate and exit properties of air.
Solution
15~
- - = 101.76 m/s
e' - '
:;: ./206.4 7 2-+15i!
=
W Actual work done/kg
27.39 x 10' 8256.47 3.317 kg/'
r
CENTRIFUGAL COMPRESSORS AND FANS ...
106 j;> TURBO MACIIINES
From outlet veloci ty triang le (Refer Fig. 3.3(b» ,
C.I l
C,
From the isentropic relation.
Jeil + e;l
C2
as·
=
U2 = 0.9 x 3jO
107
" 347.95)" 2.806x ( - 397.7
P,
= 315 m/s
1.758 bar
)(3 15)' +(28)' = 316.24 m/s
Since
Therefore.
GsUffe p
=
T02 - T01
p,
0.9 x 350' 1005 109.7 K
=
Mz
PZ
= =
pz A2 Cr1
Pz
=
Poz
So lution 111 double-sided centrifugal compressor impeller. there is an eye on either side of the
irnpeller and the air is taken in on both sides. The double-sided compressor has the advantage that tbe impe ll er is suhjected to approx imately equal slrcsses In the ax.ia l directi on. D" = 0.175 m N = 16. 000 rpm
P, RT2
D,
= 0.3 125 m
TOI = 288 K
rye
c't
rr, , -(D- - D-)
Toz
[ '-'] Po' t (po;)
=
Poz POI
4' h rr , , 4 (0.3125- - 0 175-)
-I
" + 0.9(109.7)J'"
(a) (i) Impeller eye tip speed U,
=
288
=
2.806
Poz
0.0527 m Z
T02 - TOI
[I
=
2.806
X
m = 20 kg/s POI = 100 kPa
Allllu lus arca of flow at the impeller eye
PQ2 is dctcnnined using isentropic efficiency value.
TOI
3.9424 kg/s
=
IlL
c~
--2C p
(b) Mass flow rate m
1.76 x 0.08 x 28
lip diameters of 175 mOl and 312.5 mm and is to de liver 20 kg of air per seconD at 16,000 rpm. Thcdcsign ambient conditions afC 288 K and 100 kPa. Calculate 5uilablt: values for the impeller value angles at the root and tip of eye if the air is given 20 deg. of pre-whirl at all radii. The axial component of inlet velocity is constant over the eye and is about 152 m/s. Also compute the maximum Mach number at the eye .
.j1.4 x 287 x 347.95 0.8458
M,
1.76 kg/m'
Example 3.8 A double· sided centrifugal compressor has impl!llcr cyc:.root ant.]
397.7 _ 3 16.24' 2 x 1005 347.95 K 316.24
=
=
397.7 K Now, T02 -
1.758 x 10' 287 x 347.95
IlL
109.7 + TOi 109.7 + 288
:. T02
=
I
= 2.806 bar
rrD,N
60 IT
x O.3125 x 16,QO[)
60 261.79 m/s
ell
=
152 m/s
10...
C~NTR'FUG A L COMI'RESSOHS A.NO FA.NS "" 105
;.... T URIIO !', ,!,\CIII NES
C' ---.L
7ill -
=
2e"
With IGV
PO .. - . 12 x 1005
= 335 _
Air an gle atlGV exit
327.8 K 1', ) ,-:,
Po, ( -
To,
=
JO~
327.8) H (335
94.53 kPn
=
1',
94 .5 3 x JO' 287 X 327. 8
= 1.005 kg/In'
I
tan
c, = =
5 l. IlOS x (Jf x 0.175 x 0 .075)
p, (rr D",b )
=
-
'" '"
=
'.n
V,
W :c
C"
V,
=
Cu / sin
a,
137.52
mi'
.'. 1',
Since the valucs o f PI and C] are approximately equal to the last iter.llion values, lhe iterati on can nClw be stopped. Thus,
= 120.66 m/,
:=
rrDIIIN=rr x O.175 x I20
=
65.97 m/,
= = =
p, = 1.005 kg/m '
VI
Mr. I
C,
= -V,
Mr. I
This is the air angle ,It entry to Ih e in ducer blade and sin fi, = W,
=
=
C,/SinfJl 120.66/, in 61.33 137.52 01 /'
M r. 1
=
= M r. I
W,
",
=
IV, J rRT, 137.52
J I.4 x 287 x 327.8 0 .379
,
- '-
2C /l
137 .52' 335 - 2 x 1005 325.59 K
111l:n.
= 0,
",120.66
J l.4 x 287 x 325 .59 0.334
Note that the inlet relative Mach number is reduced when IGVs are used .
tan _ , (120.66) - 65.97 6 J.33 "
.'. Il ,
C
C' TUI -
From thc inlet veloc ity tri
",
120.66/ sin 61 .33
=
120.66 m/ s
C,
U,= CIII
_ , (,20.66) -65.97 61.33
=
C~
p, =90
=-
Now.
ond //I
IV,
al
Example 3.7
~ . W,
Detcnnine the absolute Mach number of thc now al the l!xit of a radial vaned impeller of a cemrifugal compressor when the radial component of the velocity at the impdler exit is 28 mls and the slip factor is 0.9. The impeller lip speed is 350 mls. If the impeller area is 0 .08 m:!: and the lolal head isentropic efficiency is 90 %. determine the mass fl ow ratc . Take TO! = 288 K, POI := I bar. (MKV·No\' '95)
Solution Cr~ = 28 m/s 7i,; = 2R8 K
(i.{:=
0 .9
U2 = 350 m/s
(aJ Exit absol ute Mach number
A = 0.08 m:!:
fie = 0.9
IO::!
»
C ENTIUFUGALCOMI'Rf:.SSOKS ANDFANS '"
TURBO MACHINES
rrD,N
v,
" x 0.325 x 16. 500 60
60 280.78 m/'
p,
=
1
13n -
. d " d ' I cloc ily component at lhe inlet of the inducer seclion arc SlIlce the cnsllvan aXla v unknown . their \'~lu es are determined by trial and error method .
Let
(~:)
Po,
p,
P OI
102
'.n-' (~) 280.78 p,
c'
where Dm is impe ller ml!an diame ter and b is the impeller hlm.lc 11I.!lght.
TOl _ _ 1_
2C I ,
0. 1 -1-0.25 = 0 . 175 m 2
To, -I- 185.67 = 298 -I- 185 .67
TOJ
483.67 K
and
1.1
120' 483.67 - 2 x 1005
1.1 =
476.5 K
b
=
.. C,
=
Eltit s tali c pressure
=
Pl
POl Cl)",-' -.-
=
4 ( 476.5 483.67 3.796 bar
=
P.1
IV
= = =
),H
p,
RT, C' T, = To, - -' = 335 2C p P,
mC I' (T03 - TOI) 8.3 x 1.005(185 .67 ) 1.549 mW
Exam p le 3.6 The rollowing data refers to a centrifugill com pressor': tip diam · eler or the eyc·250 mm, hub diameter of the eye·! 00 mm, speed. 120 rps, Mass of air hand lcd·5 kg/s o Inle l stagnation pressure· 102 kPa. inlellolallemperature·335 K. Detemlinc the air ang le al inlet of the inducer blade and inlet relati ve Mach number. If IGV is used thcn determine air angle and rel ative mach number al the exit of IGY. 1M V-Oct '96]
Sol ution
p,
=:: 5 kg/s
0" =0.1 m POI
= 102 kPa
N= 120rps To, = 335 K
=
= = =
328.5 K POI
(T'rS TOI -
= 102
11 4.3' 2 x 1005
C
"
I8 5 335. ),,'
95 .24 kP" 95 .24 x 10' 287 x 328 .5
= 1.01 kg/m'
nnd
c,
5 m p, (" DmiJJ = 1.01 x (n x 0. 175 x 0 .075) 120. 1 m/s
Th~n
D, = 0.25 m
=
=
Now.
(c) Power required IV
0.25 - 0 . 1 0.075 m 2 5 1.061 x In x 0.175 x 0.075) 114 .3 m/ ,
D. -DII 2
10J
=
m
m
c,
.
10.1
X
1.061 kg / m'
(b) Static conditions at exit =
= RTol
287 x 335
. 23.14°
TJ
IO~
CEN1.RI FUGAI.COMI'IU:'SSOKSA~DFANS
100 :,.. T URtlO" t ,\Cl-UN!:S
(i) Eye tip diameter (D,)
and Po~
PI1T
(~'Y~
Po
D2, _ D~
"
.29).1.5 -._> (452 373.28 = 4.:n bar
Pm
=
TO] [(~fj -I]
I)
(~)O."6 _
rye
=
To, - TOl
=
TOl -
0.78 185.67 K
=
= Also
90 .9 %
Example 3 .5 In a rJdial blade cc nlrifugal compressor running at 16,500 rpm, Ihl.! tolu1 press ure ruti o is 4: I when Ihe atmospheri c pressure and temperature arc I atm and 25° C. The dillmclcr of th e hub at impeller eye is 16 cm. The axial veloci ty al inlet and .the absolu tc veloci ty at th e diffuser exit arc both 120 mls . The mass now fate is 8.3 kg/so If the ndiabillic lotal-IO -total efficiency is 78% and pressure coefficient is 0.7, find the main dimensions orthe impe ller. sialic conditions al exit and required power to drive Ihe compressor. (MKU-April' 94.)
TOJ - TOI
=
'-lJosUi lCp
'l'a,
=
'l'1' /q,
=
0.897
=
[
U,
1005(185.67) J 0.897
U, x 60 ~ POI=lbar
POJ=4
011 =0.16m //I = 8,3 kg/s
e" = CI =
POI
120 m/s
","" I
= 0.7
TOl = 298 K C)
q, = 0.78
= 120 m/s
e~ TOl -
_ ,
2e"
= 298 -
rrDhN = 1f x 0. 16 x 16500 60 60 138.23 m/s
120~
From inlet velocity triangle (Refer Fig. 3.3(a)),
290.8'1 K
P,
:. PI
=
=
T, )"' - ' = I x (290.84) - - In ( 'Ii" 298 0 . 91~ bar P 0.9 18 x 10' I _ 1.0998 kg/ m' RT, - 287 x 290.84 POI
-
=
456.1 x 60 x 16. 500
= 1f
(iii) Blade angles at eye hub and eye tip
=---,-,=
2 x 1005
I"
= 0.528 m
(a) Main dimensions of impellers
-
= 0.7/0.78
456.1 m/s
Solution N = 16. 500 rpm
TOl
298 [(4)1'i -I]
164.297
'I,·
4x8.> +0.16' x 1.0998 x 120 0.325 m
=
T02 - TOI
1.013
e'l
(ii) Impeller tip diameter (D2)
(~) ~ -I)
288(
rrpi 1f
D, TOI(
.,
4 4111
D',
Toml head isen tropic efficiency
"J
-(Di- Dk)C"
'"
{Jh
{J"
1
(~:)
=
IaO-
=
Lan
=
40.96°
-I (
120 ) 138.23
""
101
CENTRIfUGAL CO/lll'nESSORS AND FANS
98
,. TURUO MACHINES
[ '-'] P02
o/I P
=
=
U, U,
= =
.". D2
=
=
W"!. +C"!.
IV,
= =
247.34 m /s
T,
=
' To, - Cj 12Cf. = 288 -
T,
= = =
--,-
CpTO' (PO,) - I Ui
1005 x 288 x [(4)11 - I]
VJ
11"I
0,
0.749 433.6 m/s
III
XI
Example 3.4
91.5'1
2 x 1005
283.83 K J rRTI = J I.4 x 287 x 283.83 337.7 m/s
rrD"!.N
U,
Inkl relative M'lch number
"xN
IV,
433.6
= =
+ (85.98)'.
(231.92)'
- --
D
(1 /
a,
M 1,r
" x 200 0.69 m
247.34 337.7 0.732
=
69cm M1.r
=
In a centiirugal compressor with inlct guide vanes, air leaving the
guide vanes has a velocity of91.5 m/s al 70dcg, to tbe tangential direction. Determine the inlet relative Mach number, assuming frictio nlcss now through Ihe guide vanes and impeller tota l head isentropic efficiency. The olher operating conditions are a) Impeller diameter at inlet - 457 mm ' h) Impeller diameter at cxit- 762 mm c) Radial component of veloc ity at impeller cxil- 53.4 mls d) Slip faclO, - 0.9
rr D2N
rr x 0.762 x II. 000
60=
60
438.88 m/s
and
c) Impeller speed - 11,000 rpm <.I) SIalic pressure al impeller c;.;.il - 223 kPa (abs) g) Take TOI 288 K and PIlI 1.0 13 ba,.
=
,
=
Solution ' C, Cx ,
C,I Wx1
U,
91.5 m /s
= =
C, . cos 70"
= = =
C x ! . (an 70Q
= =
W1'I
I,
=
= =
U, 70
31.29 m/s 85 .98 m/s
Ca , W,
Cp (TU2
-
T02 -
Tod TOI
T02 Crl
0.9(438.88)' - (263.21 x 31.29)
=
452.29 K
x 11.000 60
263.21 OIls 263.21 - 31.29 231.92 m /s
164.297 + 288
.. cj C,
= =
.". T2
=
To'- - --2C
=
452.29 -
=
373.28 K
C~l
·C,
60
" x 0.457
164 .297
= = = =
VI - C.r l rrD I N
= = =
53.4 m ls
a, . U, = 0.9 x 438.88
394 .92 'l C .(!
+ C'12
(394.92)'
+ (53.4)'
398 .51 m /s
C;
"(398.51 )'
2 x 1005
" 99
.......
•
96
CE:-lTR1FtJGAL COMI'IU~.s so)l..s AND FANS ....: 97
;.. TlIitRO M "'·IIINF.S
or
:. 'nle hlnde angle at impeller inlct is :W. 9".
Peripheral velocity or impcllt!f lOp ilt outlet
u, Whirl f.:ompOIlt!11I
rrD..,N
=
--=-= (n x 60
=
314. 1601 /s
or absolult! vclocity C.\·l
C,,(7; - T,)
1.1 x 501l01lfiO
is obwincd (rom
b,
.'. The breadth of impellt!r blade ut inlet is 17.7 em.
Now
U:!C t !
P::,! V:!
It- I
T, ( Pp',) .,. = 2BB( 1.5) "'".' = :. C.f!
20 2 x n x 0.3 x 60 0 , 177 m
b,
T,
=
:. \12
329.68 K 1005(329,68 - 2BB )
PI VI
=
T,
133,33 m ls
T,
T,
P,
- - x "'::
I x lOS x 20
= v, =
314 , 16
P,V,
288
x
329,68
!.5 x
la'
15,26 m J Is
Now W.r2
The n
ex 2
U2 -
180,83 mls From the oullet veloc ity vector diagrDnl.
C"' lana, = -
Cr~
or IOn -,
(13~033)
c,
The illade angle i1t inlctto casing is 24.2". C" if Xl
or
IJ, 13,
I B.36" .
=
15,26 2 x rr x 0.6 x 60 0,0675 m
Example 3.3
A single sided ct!nlrirugal compressor is 10 deliver 14 kgls of air when operating at a stagnation pressure ratio or 4: I and n speed or 200 revolUlion/sec , TIle inlet stagnation conditions may be taken as 288 K and 1..0 bar, Assuming a slip faclOr of 0.9, a power input factor or 1.04 and an overall isentropic dficicncy orO.M. cSlimah! the overall diamClcrorlhe impeller. (MKU Nt)l'. /99/)
Solution Given //I
tan - I -60 _)
b,
V:! = 2rrr:!b2Crl
The bn:ildth or impeller blade al oUllet is 6.75 em.
24.2°
tan fi2 =
b,
= =
Q2
31",16-133,33
( IBO ,B3
The blade angle at impeller outlet is 18.36 n
b) Breadth of impeller blade at inlet and outlet IrQ 1 is the dischargt! in nr' /s then
= 14 kgjs =4
Po:!! POI
N
= 200 rps
POI
TO! 0'"
= = =
I bar 288 K 0.9
The pressure cocOicicnl may br.: wriuen as
=
0 .9 x 1..04 x 0,8 0 ,749
rJlI
= 0.8
94
CENTlUFUGALCOMPRESSORSANDFANS "" 95
,. TURBO MACI liNES
iv) Eye external diameter, (D.)
or
Density of air ill enLry. TOb -
TOI
p,
(450.32 - 304 .IY) / 0.8 182.7 K
ii) Impel/er tip speed and impel/er tip diameter
4
'
C p (T02 - Toil
=
1.005 x 182.7
=
183.61 kl/kg
m
=
PIAlel
= =
1.189 x
=
U,
IB3.61 x 10) U,
,
- 0.15-) x 150
0.306 m
:. The eye external diameter D t = 30.6 em
0.95 0.95U, x U,
Solution
Thus the impeller tip speed is 439.62 m/s If 0, is the lip diameter, lhen
D,
4 (D~
law PV 1.5 =con stanL The velocity of flow at inlet and outlet remains constant and is equal to 60mls . Ir the inlet and outlet impeller diameters arc respectively 0.6 m and 1.2 Tn and speed or rotation is SOOO rpm. Find (a) the blade angles al inlet and outlet of the impeller. and the angle al which the air rrom the impeller enters lhe casing: (b) breadth of impeller blade at inlet and outlet. It may be assumed no diffuscr is filted and the whole pressure increase occurs in the impeller and thatlhr.: blades havr.: a negligible thickness.
439.62 m/s
=
rr,
Example 3.2 20 m) of air per second at I bar and IS"C is La be compressed in a centrifugal compressor lhrough a pressure mtio I .S: I . The compression follows the
and C"
,
10
D,
=
J
= 287 x 293 = 1.189 kg/ m
Now,
but C p x actual rise in total temperature
P , l x lO '
RT,
Eye annulu s At = '!..(D 2 _ D2)
From Euler's equation
lV/m
=
Q,
p, -" P, D,
=
JrD,N 60 U, x 60 rrN 439.62 x 60 rr x 20,000
=
0.42 m
:. lip diameter = 0.42 m or 42 em
= 20 mJ Is P, = I bar T, = 288 K = 1.5 C, = C" = 60 m/s D, = 0.6 m
= 1.2 m
N = 5000 rpm
(a) Blade angles and flow angle Pcnpheral velocity or impellcr at inlet U,
=
lTD'IN 60 " x 0.6 x 5000 60 157.1 m/ s
From the inlet velocity vector diagram .
iii) Power required to drive the compressor Power required
til
tan{J1
C, U,
or fil
tan _ I (
x Work done/kg
=
lOx 183.61
=
1836.1 kW
/3,
_ 60_ )
157.1 20.9'
u,
~ c,
CENTIUfUGALCUMPRESSOR5AND FANS'" 93
The areas in equlJtions 0.24). (~.25) lJnd (3.26) rerer to the now areas at the respective locatio ns. Design point
4.0
e::.=
Surge line
~.
0: ~ .1.0
SOLVED PROBLEMS Example 3.1 10 kg of llir per second is 10 be compressed in an uncooll!d centrifugal compressor or the si ngle s ided impeller type . The ambient air conditions arc I bnr and 20 c C, TIle co mpressor runs al 20,000 rev/min. has iSl!n lropi c efficiency o r 80%. and compresses Ihe air from I har slatic pressure to 4.5 bar total pressun: . The. air enters the impeller eye wilh a velocity 150 mJs with no prcwhirl. Assuming thallhe ratio of whirl speed to lip speed is 0.9S. calcu late: i) rise in Iotallcmpernture during compressi o n. if the change in kinetic energy is negligible. ii) the impeller lip speed and lip diamcter, iii) power rt:quired 10 drive the compressor, IV) lhc external diameler or the eye, for which the internal diameter is 15 em.
i§
Solution i) Rise in total temperature of the compressor m = 10 Kgls P, = 1 bar T, = 293 K N = 20,000 rpm
" ~ ~
~
"- 2.0
= 4 .5 bor
=
C, 150 ml' Stagnation Icmpcrilture at inlet
Po,
Cx '
=0
=
T, + C-/2CI'
1.0
0.8 1.0 1.2 mT[II~2/ POI(rclative to design v'lluc) 0.2
Figure 1.10
To,
0.4
0.6
ry,
= 0.95
= 0.8
,
Design muss flnw 0.5) /. Ll..::J.,J...,LJ-,L ...L,J:-Jl...-":...L.L.J
Cxl/U,
150' 293 + 2 x 1005
= =
304.19 K
CelltriJlIgol CDmpressDr characterisTic curves
Stagnation pressure at inlet
CHARACTERISTIC CURVE
Po,
Fig. :\.::!O shows the o\'t.!ra ll pressure rutio lind dticicncy ploued againstm Td(~ / POI at fixc:d speed intervals.or N / 7~1\r.!. 11 is usual to transrcr consta nt e fficienc y points onlO the corrc:sponding con stant s rccd cu rves or [ht! pressure ratio characteristics Hod Ihen join those points togclhl!1' 10 form conSlmll efficil!ncy curves . The foll ow ing salient fea tun.:s can he o bscr\'cd from Ihe graph. I. Al all speed s the r
p
=
C)"'-' ...E.! T,
1(304 , )9) U /O.4 293 1.14 bor
= =
The Icmpernture after isentropic compression from POliO P02 is
PO,)'·'/' ( -POI
=
TOI
=
4.5 304.19 ( 1.14
)0.' /1.4
450.32 K
111e onset of surge occ uring at increas ingly high mass flows, as the speed increases,
-
while the loc us of the lim it of s tabili ty is ca llcd the surge linc. The limit o f ma .. imum flow is usually SCI by chuking in the impeller, while Ihe surge limit of minimulTl mass now is sci by s ialling o r the now into the diffuse r va nes.
Actua l rise in IOta ltemperalure is determined from the definition of isentropic effi· ciency 'Ie =
TOll -
TOI
T02 -
701
90
>
TUllDo MACIIJNf:S
CENTlUFUGAl COMPRESSORS AND F ANS "'\ 91
I'
1:
Increased angle of attack /
/
B
//
A
~
?ne~~e:;red
cho~jl1g o;curs when the relativc veloc ity equals the acoustic velocity r RTI), the above equation becomes (I.e H'I- = OJ
=
attack and dividing by TOI gives
Tr
rRTI
V;l
TOI
2CpToi
2C/,T0I
-+-- - --
u1
.!!.. (I + 2e.0..) Tal
1+ - -'2e TO! "
Since
(
rR
=
ell
r- I
Direction of stall propogation
T'( 1+-r-I)
U'
Flguu 3.19
1+--'-
Air flow direction in rototing stall phenolllenoll
blockage (or) uneven now in the diffuser the blade may stall. Because of this, the mass flow decreases which in turn increases the angle of incidence to the left of blade A. [due to the low mass now rate through the passage, the entering air gets deflected. res ulting in large angle of incidence] whereas angle of incidence decreases to the right of blade A. Thus blade B will be !he nex' stall while blade A will be unstalled and the process is repealed about lhe periphery of the disc . Prolonged cyclic loading and unloading of the rolor blades can lead to fatigue failure or even immediatc ciltastrophic fnilure. The stall propagates in the opposite direction 'to the blade motion at a frequency related to shaft speed. In compressor tes ts. 'rotatin g stall may be audibly recognised as a high frequency ·screech·. At low speed and starting the front stages arc morc likely to stall. But at high speeds, the stall occurs in the last slages. Low speed and slarting slall may be eliminated by variable inlet guide vane rows.
T,
Now
"0' = =
+ (IVf -
__ 2 [ 1+ U'I ] +I 2e To l
(3 .22)
"
IPi/Pol = (T,/Toil ' / r-' I From the conti nuity equation. If!
m j A = pa
=
pAa
=
Po taodTi/ Tod ("+ II /2tr- tJ
(3.23)
[a = aOI x (TI/Toil'/2] Since 110' T 1!2 . Substituting from cquation (3.22) and rearranging.
, = [ rPOIPol[2 ( 1 + Uj'/2C p Toil/r + I]
I
r
+I/2(,.-1)
(3.24)
[aOI = j rRTo; = jrPOI/PoI ] Equation (3.24) imp li es Ihat th..:: chnking mass How rate increases with impeller speed. MJ.'
+ I)](r + I)/ r
- ') 1/2
0.25)
In .thc diffuser passilges, equation (3.25) is also valid with the subscripts changed to the Impeller oUllel conditions.
",+Cr/ 2 I"
2
r
For ise ntropic flow,
'II / .'1
When the mass flow is increased to the ri~ht of point-3 on the negative slope of the characteristic curve (as in Fig. 3.18) a poinl-5 is reached where 110 fUrlhcr increase in mnss flow is possible no matter how wide open the flow control valve is. This indicates thallhe now velocity in the passage reaches the speed of sound at some point within the compressor and the flow chokes. Ch oking means fixed mass now rate regardless of pressure ralio. Choking may take place at the inlet, within the impeller, or in the diffuser section. II will occur in the inlet if stationary guide vanes are filled . In ~uuionary passages like nozzles. the ve locity that is choked is the absolute velocity. In the rolating impeller. it is the relative velocity 'W' that is the choked velocity.
TOI
TOI
'0
Choking
-
2e"Tol
If the air angle of incidence onto blade A is excessive, perhaps due to a partial
'l
V,')/2
III/A = [rP02Pol [2 /( r
+ I)](r + I) / r
-
1]'/2
(3.26)
CENTRIFUGAL COMPRESSORS ANa FANS "" 89
thl! 0plimum \';llu~ of ooth cnn be determined. Tllis maximum value is then \!quilted 1(') the kfl hand sick of equi.ltion 3,21 and the maximum mass !low rate is determined. rig . 3.17 shows thai the hladc angle is almosl co nstant at 60" for maximum mass Jill\\,. Su, oy specifying. the relative Mach number (M, rel), the maximum value ur mass tlnw mav hI.! ca lcu lall'd. Rdati;e Mi..H:: h numhers arc usually restricted to about 0.8 to ensure Ihalthcre is no shock · WOl ve formation iH the impell er inlet.
MACH NUMBER IN THE DIFFUSER Thl! ahso lutl! rvlnc h numht.:rof the Ouid leaving the impeller may exceed unity. T here is nllioss in crJiciL'n cy CiIllSCt! h)' the f"rmafi on of shock waves as long as the radial Oow vdoci tyen is suhsonic. When the consti.ln[ angular momentum with vortex motiun is maintain~d in Ihc v:mdc.:ss space hClwl!l!n impe lle r lip and diffuser, the sup!.!l'so nic di rfusion .,;un takc pl,u:e ill thc \, ..tOeless spm:: e. Thi s redw.:es th e Much numher at inlet lothc dirfusl'r vatle~ til ahnut (Ut High Mm:h numhcrs
CENTRIFUGAL COMPRESSOR CHARACTERISTICS Using the groups of variablt.!s, the characH:ristics of compress ibl e flow machines arc usuall y dcsc rihcd. The characterist ics 3re generall y gi ven as a series of curves of
m.JfOi
.
pUJI PUI plolled against the mass flow parameter - p- - for fixed speed IIltervuls or' 01
N
/.JfOi.
An idealised flxed- spl!l.!d characteristic is shown in Fig. 3. 18. Cons id er a centrifugal compressor deli vering throug.h a flow control valve situated ariel' Iht! diffuscr. Th!.!re is a cerlain pressure ratio POJ / POI, cVl!n 2 if lite valve is full y dosed , and is ) 6 indicall.!d hy point I. ntis pressure ralin is so lely due [("I th e \';.lIles moving the air anoul In 5 Chuk;ng pnin! thl.! impcller. The rrl.!ssurc head sO dl'veinpcd is called "shut of1'" ht.!ad . As the flow cnntrol valve IS opcncd. Ih e air .S\:Irts flowing nnd the diffus l.!rco ntrihute .~ 10 the pressure ratio . Thus. al point 2, the maximum prl!ssurt! ratio is Figure 3.18 Idealised Jued speed clwracleri.rlic fl.! ilChed hut the cfficit.:ncy is jusl oj cenlriJugnl comprt's.rnr below the mu .ximum efficiency.
Further increase in mass now reduces the pressure ratio 10 statc-3. BUI atlhis point, the efficiency is maximum compared with state-2. Thus the value com,:spolllJing 10 poinl-3 is said to be design mass flow rale and the pressure ralio. Further increase in mass flow decreases the pressun! ralio ilnd rCilclH':s zero pressure ratio ill point-4, Corresponding to this point, all the powt.:r absorbt.:d hy the compressor is used 10 overcome the internal friclion and thus the compn:sslOn dlkicncy is zero . Poinl·4 could be reached only theoretica ll y. So, the curv\! just d..:scribcd is not obtainable practically. Butlhc ill:tual curve is differing from the ideal curve dul.! to the fulluwing rcasun~ .
Surging The phenomenon of a momentary increase in the delivcry pressure re~uhing in unsteady. periodic nnd reversal of now through the compressor is called surging. Can· sider a compressor operating at point-:~ on Pm l POI - VS 111 J'lliI / Pol curve, i.e. on th~ negative slope of the curve. A reduction in mass !luw raIl! (due 10 momentary blockage) makes the point to move on to the lefl. Further reduction in mass !low rate incrt.!ases the prt!ssure ratio until it reaches the maximum v;Jiu{!, Ope raling the com· pressor on Ihe nl!gi.ltive slope region (1-2) establishes 'slnhlc operation' . Because, the delivery pressure' PO]' increases. which in lum will controltht! rurtht!r rcduL:tion or flow rale. It is self·correcting. Now, tile compressor is operaling al point 6 on the positive slope (2-4) Oflhc curve. Upon mass now reduction the pressure ratio decreases, until it reaches the POl l POI ax is i.e. z.ero mass flow rate , The maSS flow even becomes negulive through the compressor.When the back-pressure PO) has reduced itself further sufficiently dUt! 10 th e reduced now rate, the positive flow becomes established once again and the compressor picks up until the "restricled mass flow rale" is reached again. wht.!n pressure reduclion takes ploce once again . The compressor operates in un unst4!blt.! rashion . The pressure therefore surges back and forth, if the dowlIStream co"ditjollS are Ill/changed. This phenomenoll is known as ',rurgifl8' or 'pumping', Thu~, when the compressor has 10 ope rate at reduced mass flow rates, the air surges and pulsates throughout Ihe comprl!ssor and Ihe compressor docs not givt! a steady flow of air, Surging, if severe cnough. could lead to failure of the comprt.!ssor pans. Surging occurence can be reduced by making the number of diffuser vanes an odd·numtlcr muhiple of til e impeller vanes. In this way, a pair of diffu .~er passages will be supplied wi lh air from an odd number or vanes and pressure nUl.:tuDlions ure more likdy 10 be evened ou t around the circ umference than ir exact muhiplcs of diffuscr vanes arc employed.
Rotating Stall The phenomenon oj a reduction ill mass flo II' raIl' 1hrollgh Ihe blnrle PCl.I":w,r:r: (Jl hight'r angles of incidellce is kilOit'll as rotating :Hall. It is a separate stall phcnornl!llon, which may lead to surging but can exist on its own in a stahle operating condition. Figure 3. 19 illustrates the air flow directions in a number of blade passages.
-• Ct:.NTIU FUGA I. COM I'It ESSOI1.S ,\ NOFA NS ..;:
Sti
We know
The nClw area is
A,
and w
IT R2k
-
and
[I
DO, = (TO,) l = + ",(,_-_I:...)A_1.ci ] l
U,/R
al
(rrU~k) --,- C, w-
101,
=
is llll! inlellangcnliaJ vdocity orthc im '11' I' 'I . thL' angular vl..' locity. pc cr al t Ie s lroud radius and 'w'
(.\. 19)
2
TI
~h..:rc 'Uj'
C,
", -M,- = 1v11 .n:1 cos fJl
U~C,
=
,
(r RTOI)!
1l1ercforc,
p,(rrR'k)C,
p,
(rRT,l'
no,
ITR\I_r1/R 2)
.. , III
,
a,
IT(R' _ ,')
= = = = =
:. 111
IS
:n
... Tunno MAC'HINtS
IVI(sin' iJ,) . (cos iJ,)
(or)
(3.15)
M,
MI _rd
cos fJl
Substituting foral and M\ from equations (3.19) and (3 .20) respective ly in equation For isentropic relationship
(3.18).
,
M,J , (sin'~, )(eos #,)
."
mw" ITkr P OI(rR T OI)1/2
=
[
;~: [I + ' ~ IM~t [I + ' ~ IM~J
[I +-Z-M, ' -1 'J;;'!n To, =
p,
'"
~
P,/RT,
Suhstituting for PI from equation
.~
;;
(3.16)
=
0
~
u
c
'":x:
'"
Wr(sin PI )(C05 PI ~ 2
. (3.17)
'J~
Ml . rel
0 .2 0 .9 U.MS
0 . 15
O. ID
0.8
0.75
0 .05
o
Figllre 3.17 "
11In., 2RTOI IT k Po,
=
[ 1+
-z-Atl]-L. ,-,
r -I
(,(J
70 80
Oplimisution of the mossflowfllflcriull
It should be remembered that equation 3.21 is applied allhc shroud radius' R' . 'nlc relativc blndc nng!c at radius' R' is {J,. At this radius' R·. the 0l3;(imum vu lul.: or
Mlrd ·al
M' . I.rtlalJ(" sm- fJI )(cos fJd
10 20 3D 40 50
Relative now angle PI at the shroud (deg)
Writing the relati ve Mach number based on the relative velocity IV
=
PUI: 10 I.:\kPa
0.25
.[ I +,--IM 2 '
H'I
]
(3 .21 )
T01: 288 K
~
(3 . 16) In . equation . (3 . 15),
mw2 RTol/ITkPol
C:t 1 ::=O r : 1.4
0.3
(:;~,) [I +' ~ IMIJ.=I
p,
+
] r_ I / ~(r_l)
Mach number
Po,
Nuw
I
r- I , '1 -2- Mi.,,'(cos- #,)
(3.18)
relativc veloci ty onto the blade occurs. So. for a gas of known inlel stagnation cond itions (e.g . the allllosphcrc).lhe right hand sidl! of eq u
,, , CENTRIFUGA L C OMPRESSORS A1"1J F A/I.'S ...:
INLET VELOCITY LIMITATIONS
~CX2~ W~ U:r
rK
u,
Controlling the Mach numbe r allhc eye of a centrifugal compressor a ffc c ts thc Inkt rclntive velocity \VI , Two cases may he e)(umined for the same mass flow rate.:. having uniform absolute velocity Ct, with i'.ero whirl velocity (C,(, = 0) at the entry to a centrifugal compressor,
W2
P,= 96...-.,....-- C,
Forwa rding-facing vanes Figure 3. J.I
H5
,
Rildial vanes
Backward-facing
Valle !;
Centrifugal COlJlpressor ollilet velocit), triangles for I'or-yil/g blelde (1 1111('1 lingle
Case 1: Large eye tip diameter From continuity equation the a~i al velocity C I should be lo w PI A I C I ). Blade speed is high . These arc shown in the velocity Iriangl e. (Fi g 3. 16(a))
em=
Case 2: Small eye tip diameter The a)(ial velocity is large. but the blade speed is sm a ll. It is shown in the vdoc ity triangle diagram (Fig. J . 16(b»
PRE-WHIRL AND INLET GUIDE VANES To restrict the M~ c h numbe r at inlet lo an aCI..'cptohlc value. pre-wh irl should be impafled on Ih ~ air entering the eye. TIli s can be done by plac ing guide va nes at the inlct. Fig . J , 15 clea rl y shows thnithc inle t guide vanes impart a whirl component C,f l (0 the nuid, thus re.ducing WI to an acceptab le value, However, the work capaci ty is rcduccd since C r l is no longer zero.
C,
u,
~
\~u,
(a)
Eye 'ip (or) shroud radius
Ca
,
"
, \ \
- - - -- WilhoUI guide vanes - - - With g uid e \'anes
DE-,f--+-- Hub radius
~
,,
\ WI
, , ,,
Hub
shroud
} E
ye
\'~ Figure j,16
Figure 3. J5
(b)
Effecl of inlet gllidl' L'tllle.)' 011 lhe inl!!1 reililil't! l'I'lnciry
It is nOl ncet!ssary to impa rt pre-wh irl down 10 the hub, as in th is region, th e nuid is nowhere neur so ni c condi ti ons due 10 the lowcr blude spcl'd , The pre-whirl is there fore g rndualJy reduced to 7.e ro by twi sting the inlet guide vanes. Apart from reduci ng the Mach nu mbe r, the pre-whirl has another 'Idvant age of rl!duccd curvature orthe impe ller vanes at inlel. Pre-whirl vanes have the disadvantage o r int roduc ing additional pans and additional weigh ts, which s hould he a n important param e ter to be co ntro lled in jet airp lanes. Also there is 3 danger o f possible icing in the vilnes under unfavourable operating cond ition s i.e, at higher altitudes, Limiting valut!s o f Mach numbe r arc usu a lly kept between 0.7-0.8, for fl ow oVer the blades.
Vo!lociry triul/glt! f or u) large alld b) small inlt!1 ar~u
For bo th of these ex.treme cases, the relative velocity vector WI is high, hUI it must reac h a minimum va lue when movin g from one extreme to another. After dctcnnining thi s relative velociLy by proper design. the Mach number effects ca n be nvoided . Flow into the eye takes place through the annu lu s formed by the shroud mdius 'R' and th e hub r;.Jdiu s 'r ' , For uniform a xia l flow into the eye,
From the vel oc ity triangles (Fig, 3. J 6), we gel
c,
= W, cos il, and U, = IV, sin,8,
" 82
):>
CENTnIFUG " L COMI'R~SSORS "I'll F ,\;..:s "
TURIJO M"CIIINES
The abo . . e equatio n is in the form E = a - bm, where a = Ui Ig and .
>11
p
Forward-f.. cing
b = U,co.p,/pgA. As 'm' increases, E decreases. The characteristic is therefore falling .
(ii) Radial blades
= 90" = 0 IE = a I p,
cot 90°
The energy transferred is constant at all flow rates and hence the charac teristic is neutral.
(iii) Forward-curved blades
Backwardfacing
IE = a + bill I incrcJscs. £ is iflcrcased. The characteristic will then be raising. fh would be lypically 140" for iI multi-bladed centrifugal fan. Thest! equations arc plotted in Fig. 3.12.
When
'III'
"' Figure 3.13
Act/wI cJwraclerislicsfor varying blade ouller Ullg!t'
Forward-bl.!nt blades have higher pressure ratios. But the following disadv;Jn[;Jge:-. arc the hurdles [or its wide range of applications. I. Low efficiency owing to large s lip factor (between I nnd 2).
2. Operating range is closer to th e surge line even under normal running condition~
En~rgy
Figure 3.12
J<E=------ ~2= 96
Theoretical characteristics/or varying ewrlel blade angle
Actual characteristics for various blade ou tlet angles arc shown in Fig. 3.13 : For hath radial and forward facing blades, the power is rising continuously ns the flow rale is increased. In the case of backward-facing vanes the maximum efficiency occurs in the region of maximum power. If 111 increases beyond 'designed 111' (mD), it will resull in a power decrease. and therefore the motor used to drive the compressor may be safely rated at the maximum power. This is said ( a be a '.fe/f limiting characteristic' In case of the radial and forward-facing vanes. if the compressor molar is rated for maxi mum power, then it will be under-utilised most of the time, and extra cost will have to be incurred for the e,;tra rating. Whcreus. if a smaller motor is employed, rated at the design point, then, if 111 increases above I1!D the motor will be overloaded and may fail. So, it is more difficult to decide on a choice of motor for these vanes.
thus narrowing the stable operating range . Bcttercfficiencies can be obtained from backward-bent blades than willi radial \';J lles. but the pressure ratio is lower. So, when a high cor:-.pressor efficiency is desired. machines with backward curved vanes arc u<;::..i. The radial · blade impellers arc usually preferred because I . Ease of manufacturing 2. Lowest unit blade stress for a given diameter and rotational speed (hence lightest weighL.) 3. Equal energy conversion in impeller and diffuser giving higher pressure ratIO'; wi th good efficiency.
Due to these advantages the rndinl blade impeller.; arc used in aircrart centrifugal compressors , Experimental result s show that the slip factor value for raLlial hladc impellers is about 0.9. Hence. where a large pressure rise is required for a machine of small size. radial blades are used. The reason forthe decrease in efficiency in [on"'ardbent blades is that . as the slip fa clor increases (C~~ increases). the energy conversIOn requ)red in the difruser increases as a result of which diffuser inlet VCIOC ilY is higher and the diffusercfllciency rapidly [ails. Therefore it is very rare 10 timJ lI1achin~s wt1h forward curved vanes. /2 increases quitt.: rapidly ;IS f/! Il should be noted that the exit killetic ellergy increases. But ma..:hines with large exit angles (lh) will be less efficient than machine), with small exit angles.
Ci
--.
80
~
TURBO MACHINES
CE:-.ITltln;nAi. CO,\lI'RESSCIitS A:-..I> F ,\I\:S ....: H J
For adiabatic deceleration of rhe fluid rrom absolute velocity Cz to C) with spond ing incrcase orstatic pressure rrom p:!. to p], "02
It !
=
II)
corre-
Substituting cquation (3.13) and equation (3.14) in the equation ror l1egrcc: of rCllCtion. we gC I
(or)
"0]
+ Ci/2 =
11
, t '[ ' '] Vill-!./J2COlP2)!Vi cPi+(I-¢r!col!12)·
R
+ Ci/2
DEGREE OF REACTION
J.( I
The dcgrct! or reacl ion of a centri rugal compressor stagc is given by
=
Change in s tatic enthalpy in the impeller Change in stagna tion enthalpy in the stage
R
1r1 - It I
1102 - hot
1"0] - hot
= hoz -
=
hal as 1t 02 == liOJ]
=
Ir the velocity o r Ihe gas approDching the compressor inlet is negligible (Ct '" 0), then" I ::::: hOI and "2 li 02 -
Cir2
=
h02 -
U;(l -
,
.pi + ( I
Cil2
R
=
(1102 - hOI) -
hOI
= =
VZCx 1 = UI(U1 - Wl"z)
=
-
tP2 COl fil) -,
~:!
cal th}
¢i + I + tfJi
R
I _ rp5cost!c211].
+I-
2 - 21/J2 col .82 - ¢icOSCC 2.82 - 1 + 2rP"! cot /12
-.p, cot ,8,)
2( I - q" cot ,8,)
For ~adial vanes (Il"! = 90"), I , -(I-.p;)
R (3.12)
ci
C;! + (U2 -
ci and
,
ci
+ (U2 - Cq cot 13'].) C;l + Ui(l - C'1!U"!.) cot.82)1
= C;l + uf(l (or)
cj
Vi [(C;I vi) + (l - ¢2 .82)2] vi [.pi + (I - .p2 cot ,8,)'j
=
Cal
Ittll
=
Vi( J -
¢2 Cal .82)
¢2 cot fJ2)
We know rrom the outlet velocity triangle, Cx~ The energy transrer £ U2CX1!g Then,E = V,(U, - C" cot {3,)/g (or) .
1/J"1 cot Ih)2
(3. 13)
From Lhe stage pressure risc ex.pression (eqn. 3.9),
h02 -
-
(i) Backward-curved blades
,
C;!
2
The diITerent blade shapes utilised in impellers of centrirugal compressors l'an h\." classified as (rerer rig . 3.14) 0) Bnckward- racing blades (ji) Radial blades (iii) Farwan.l-facing blildc~
WJ"1)2
C'l cot th.
lV...:!
= (I -
=
Effect of Impeller Blade Shape on Perfonnance
,
CXl + C;l V2- lV.r!
Cx~
and 1/Jp because 1/1 1>
h02 - "01 = U';-(l - ¢2 cOlth) where '4>2' is the flow coeffi cie nt. From the outlet-velocity triangle (Rerer Fig. 3.8(a)J,
and
2CP2 cot fJ2
2( I -
(Ui-U ZC r lcotP2)
,
2
fJ"! - '2¢"! COl fi2 2( I - .p, cot Il,) I _ ¢i{ J + cot 2 13).) + I - 2t/J2 co t Ih 2(1-q"cot,8,) C0l
2(1 r-------::---:--, I - ¢Icosec 2.82
(11m - hOI)
-
.p, cot Il,)-
(3.14)
m where = C,~ pA •
= V2 -
C'1 Cal fi2
78
>-
CENTRIF UGA L C OMrRESSORS AND FANS
TURBO MACIIINES
«
79
St.1tionary diffuser
Diffuser " passage
, ,
.," I
/
\obn~lt!s.r
prbCr
P2'2b,C" P2'2b,C,,/ prb
=
Free vortex flow in diffuser passage
-ru
Figun 3.10
C,
difffUtT
If frictionless flow is assumed, then by conservation of angular momentum (mCxr)
, I "i' ,~
I'
I
and ex = Cx2r2/r But ex Cr (usually) nnd so the absolute velocity or
ex .
»
I.e. Cxr = CX]r2
DifTuscr--~..!j\
,"
b
"
=cons tant (or) C =
'e
is approximately equal to
constant
--r
Our aim is to reduce 'e. To achieve this, 'r' must be large and therefore, for a large
~uction in the outlet kinetic COCTEY: a diffuser with a l~e radius is required.
A vaneless diffuser has wide range of mass now rate. But because or long flow path with'tnis tYpe of diffllser, friction effects arC important and ~.~ : c!fi,cicncy ~s low.
(e) Vaned Diffuser In the vaned diffuser as shown in Fig. 3.11, the vanes are used La diffuse the outlet kinetic energy at a much higher mte, in a shaner length and with a higher efficiency (length of Bow pam and diameter arc reduced) than the vaneless diffuser. A ring of diffuser vanes surrounds the impeller at the outlet, and after Icaving lhe impeUer,;the air moves in logarilhm ic spiral motion across a short vaneless space before entering the diffuser vanes. Once lhe fluid has entered the diffuser passage, the'controlling variable on the rate of diffusion is the divergence angle of the diffuser passage, which is in the order of 8-] 0°, and there should be no separation of boundary layer on lhe passage walls .
,
With diffuser Volute
Figure 3.J J
Vaned diflwer
The points 10 be considered 10 fix up the number of vanes on the diffuser ring arc: 1. Diffusion increases with the increase in the vane number. But increasing the: vane number increases 'the friction Joss. 2. The number of diffuser vanes has no common factor with the number of impeller vanes, BUI. whcn the number of diffuser passages is less than the number of impeller passages, a more uniform total flow occurs. 3. The cross·seclion of lhe diffuser channel shou ld be squared to give a maximum hydrauI!c radius (cross-sectional area/channel perimeter). Change from tbedesign mass flow rate and pressurcratio will change the smooth flow direction into the diffuser passage Bnd will thererore result in lesser efficiency. This may be rectified by utilising variable angle diffuser vanes. The velocity of air leaving the diffuser should be as small as possible as this eases the problem of combustion chamber. The diffuser outlet velocity is usually designed at about 90 m/sec. TIle diffuser efficiency ~D=
~I..:cdc.:.:a::.I-c-c"nd::.',::a:.;Ip,"-y..:d"fO",p,-
-;Actual enthalpy drop
From the II - s diagram (Fig. 3.7), 11)1 - h2 ~D
II) -
=
=
"2
n(T),/T, (T) - T2)
I)
T,[(~t -I] (T) - T,)
76 >
TURBO MACfHNES CENTRIFUGAL COMPRESSORS AND FANS <0{
>/I" = C" (To,, - Tod
(a) Volute or Scroll Collector
Vi Now, isen tropic work == aCluui work x iSl!lHropic effic iency .
= C"(7iI:' - Tod x '1. Thl.!l1.
, VIp
Cr (To' - Tod
=1]"
-
,
U'
BUI
C p {T02 -
Jill)
T1
A simple volute or scroll collector is shown in Fig. 3.9 and consists of a circular passage of increasing cross·secLional area. The feature of the simple volutt: is its low cost. The cross·sectional area increases as the increment of dischargt: increases around (he periphery of the impeller and it is fo und that a conSlant average velocity around the volute results in equal pressures around the compressor casing, and ht:ncc no radial i.hrusl on the shaft. Any deviation in now rate from the design condition will result in a radial thrust. which ultimately results in shaft bt:nding. Of the available kinctic energ y at impeller outlcl.15-30 per cent may be recovered in a simple volute.
= "'a~Vi
TIlus th e pressure coef-ficicnt may he wri Uen
liS
= AnOlhcr definition for 1/1 p is
=
>/I p
Impeller
Work done/ kg
Vin. (Aho) ,
,-
U' / '
:;¥----
where, A~'O is rhe idc.d .s ta~mHion Cnlha lpy change across the stage. If Ihe stage pressure fiSC IS so Sm:l llliJ cn the fluid can be treated as incompressible. (Aho )"
Volule or scroll co/JeclOr
= A Po / p (b) Vaneless Diffuser
so that
>/If =
A Po
(pU}!2J This definition for 1/1 I' has a numerical v
>/I" = (I - q" cotjJ,)
Here diffusion takes plnce in a parallel·sided passage and is governcd by the principle of conservation of angular momentum of the fluid . The radial component of absolute veloci ty is conltolled by the radial cross·secLional area of flow' b' . A vaneless diffuser passage is shown in Fig. 3.10. Mass flow rate' m' al any rad iu s r is given by
1m = pAC, = p(2rrrb)C, I
DIFFUSER
~~~~;Y~I~; i;Pp~i~~:J.n.t rolt! in rhe overill! compression process of a cenlrifugal compr. .
Figure 3.9
Volute of increasing cross-section
c
r Imparts energy 10 the ai r by incrcasin r ' . I ' . kinetic into pressure risco r;r 'II ,I .user oes c~mprcss and Increase Ihc pressure cql!alto 50 percent o[tl1" over' ,1 51,IIIC pressure fi SC. t:
~:en~~;;s .thi~i~parted
e~ergy
l~sr:~i~t~l~d:~~~~~~I:~.r
where 'b' is the width of the diffuser passage perpendicular to the peripheral area of the impeller and is us ually the same as the impeller width. Let the sub.scriplCd variables represent conditions at the impeller outlet and the unsubscripted variables represent conditions at any radius 'r' in the vane less diffuser, then from contiouity equation
74
»
CENTRlFUGALCOMPRESSORSANOFANS '" 75
TURDO MACinNES
Work supplied
hal, -
Cp(T02, CpTOI (
•
u,
hOI
Cr,
Toil
IV,
c,
To,' - I ) TOI
Figure 3.8 (0.)
Vtlocity triangle
From Euler's equation IV
=
(3.8)
III
IV
VI( I
:. >/Ip
= V,(V, - e,2COIIl,) = v2, 1- U2 cot Ih
(e"
=
vI( 1 -
III
where Ro-stagnation pressure ratio From Euler's equation. Work supplied =
VI(I - ¢, col
)
II,)
"" COlli,)
11,)
- ¢2 COl
Vi
>/Ip
(I -
"'2 colli,)
(J .III
Substituting equation (3.1 1) in equation (3.10). we get
R [
,I. V']~ Ro=~= I +~
(3.9)
POI
CpTOI
Equaling (3.8) & (3.9), In tcnns of stalic pressures. following the same procedure. we gc.:1
.-, epTo,(R o'
where
'cP2'
- I)
= vf(l -
¢,COI (3,)
",pVI ] ~
P, [ R=-= 1+-PI
is the now coefficient at the impeller exit. ¢2
= e"
These
twO
CpT,
ratios arc known as slage pressure ratios.
V, Pressure Coefficient
Ro
= [ I + ( I- ¢'COIR')U' ~ 2J ~ epTOI
The loading or pressure coefficient is defined as
'" r
(3.10)
The press ure or loading coefficient is also defined as the ratio of iscnlfopic work to Euler's work. Pressure coefficient. 1/Ip
= Work donc/kg V'2
From the outlet velocity triangle fFig . 3.8(a) J.
For a radial vaned impeller,
..,
Isentropic work
= :.::;:.::;:.::;:.:.::...,-,::.:: Euler'S work
r 72
~ TilnUQ
tv1.... CIIINE."i CENTRJf"UGAlCOMPRESSORSANOFANS "" 73
=
. Since I J h in equllli o n (3.6), the ch ief contribution to the static ent halpy rise is trom the term (Vi - Ur)/2. Usually, CJ.I = 0 is assumed in pre liminary desig n calculati ons. Although , th is is no~ illwilYS ~he case, from the ilc tual energy transfer equation, th e work done on the Jluld per UI1l! mass becomes (W/III)
;
E x g
~ " 01
==
l/IusVf
"0 2
Subs titu tin g
"0 == CI'To and rt.!urmnging the eqn ., we get 10:2 - 7ill ==
v1osui/c"
fl ow velocity ell at inlet must therefore be increased and this increases the loss due to friclion . A compromise is usually made; slip factors of about 0 .9 are being used for a compressor with 19-21 vanes . It may seem that increase in '1/1' increases the energy transfer, but the mte of decrease of isentropic efficien cy with increase in 1/1 negates (nullifies) My appnrent advanLage . So, the ideal condition is to have n power input facto r of unity (l/I = I) . The press ure ratio increases with the impeller lip speed, but material strength shou ld be more as centrifugal stresses are proponional to the square of the lip speed; and for a light alloy impeller, tip speeds are limited to about 460 mfsec . This gives a pressure ratio o r 4 : I. Pressure ratio of 7: J is possible with titanium impell ers. Eq uation ror pressure ratio can be wrinen in terms of fluid properties and How angles. Since I = rRTo t and Cp = rR/(r - I). then POl/POI = I I + TJc l/taA r - I ) , , '
aJ
wh.crt.! , C I' is the .menn specific heat ovc r thi s temperature rangc. S mce:. no work IS done in the d iffu ser, 1101 110] and so
=
Uiloo JJ f-1 (3.7)
The change of pressure ralio with blade Lip speed for various 'TJc ' is shown in Fig. 3.8.
With r~ fcrencc. 1O th e h- s diagram lind eq uation (3.7), a compressor' s ovcrnlltotal. In ' lOtallsentroplc effic iency ''le' is de fin ed as
7
;O~ln~l"i;se~n~l~ro~p~i~c~e:n~lh~":lp~y~ri:sC~bC='~'v~c~e~n~i~nl~c~l~o~nd~o~u~ll~e~1 'h' -_ _ A" Actual enthalpy rise between same lOlal pressure limits liO)" - " 01
where the s~b~cripl 'ss ' represents lhe end st'.l(e on the lotal pressure li nc PO l when . th e process IS Isentropic.
cr=0.92
1jI;1.04
"§ 4 f: ~ ~ ~
=
~
( To~u - To l )
(70.1 .r
'0 1
3
0..
Toll
2 -
«To],,)/ Tal (TO] -
- I) Tal)
380 400 420 440 460 480 500 Blnde tip speed (mls)
BU I.
PO.l / POI
(TOlf)
I To t )r/r- I
;
II -I- Ik(To) - Toil/Tod'N- 1i
=
II + '7c1/!aJUi/(CpTody/r-1
~le s lip fac tor s.hould b~ as hig~ as possi ble, s in ce ilumils (he energy trans fer 10 th e nUl d. eve n und er Ise ntropIc C?ndltions~ a.nd it is seen from Ihe ve loc ity diag ram s lhal CJ."~. appro~ches U2 as (he s lip factor IS Increased . TIl e s lip faclor may be increased h~. Jn ~rea~ lng tl~c nu~her of vn nes but thi s increnses the 'solidify' at the impeller eye, n;suhlOg
340 mls
"'(=1.4
.2
ItO] - hOI
'lr
"01=
In
decrease In the now area at [he inlel. To have a salTle mass fl ow rate, the
Figure 3.8
Overall pressu~ ratio vusus impeller tip spud
STAGE PRESSURE RISE A ND LOADING COEFFIC IENT The stati c pn:ssurc rise in centrifugal stage occurs in the impeller, diffuser and the volu/ e. No changt: in s tagnation enthalpy occ urs in the diffuser and volute. !n th is sec tion, the pressure rise or pressure ratio across the stage for an isenslIopic process is detcnni ncd .
" 70
CENTRJFUGALCOMI~RESSORSANDFI\:-.IS .c{ 71
,. TURDO MACIIINES
Total enthalpy at section-I, i.c. inlet of the impeller. is
'1/1' is also known as 'srage loadillg.coefficient'. Upon leaving the impeller the gas enlers a vanclcss spncc where it moves in a spiral path before entering the diffuser, in which the static pressure is further increased . The clearance between the impeller blades and inner walls of the casing mu st be kept as small as possible to reduce leakage and in some cases the blades themselves are shrouded.
Cl
"0' =11, + 2
and since no shaft work has been done and assuming thaI adiabatic steady now occurs
hoo
MOLLIER CHART Since we arc dea ling with a gas and since the risc in temperature and pressure causes the density to change. il will bcconvenicnllo examine the performance of the machine in terms of Ihe Ihcrmm.lynamic properties of the gas and thi s is done through lhe Mollicr Chart. The II - of diagram for the compression process across the ce ntrifugal compressor is shown in Fig. 3.7.
CJ
110+ -
,
2
C~
+2
=111
(3 ,5)
2, Impeller Work is done on Ihe nuid across the impeller and the static pressure is increased from P lI O P2 . Writing Ihe work done per unit mass on the nuid in terms of enthalpy. we gel
C:
T
wjm
£l 2
P, .c
hOI (from S,F,E,EJ
Thus ,
;
p.,
~
= h02 -
hUI
From Eu ler's pump equation
Po.
~ 0.
-,;
P,
.c
E
Equaling Ihe two equations and afler substituting for Ito
UJ
I = hi
+
C2
hoo =hOI
where 'f' is the impeller constanl.
h02
In genera l
=hOl
I
=
C2
T-
UIC.I"!
= 112 + -f - U2 C.r'2
h+C'/2- UC,
+ (C; + C;/2) - UC~ = h + (IV' - W; + C;l/2 It
Enlr0py, s
Figure J. 7
= ,,+ (\I"
Mollier chart/or a centrifugal compressor
~
The energy equation along a stream line may be written as
"0
= " + ~ = constant Total ent halpy. 1l1crcforc. for the nuid drawn from the atmosphere into the inducer sect ion . the total enthnlpy is
c'
"00 = "0 + ..J!2
- (U -
U'
Cxl' + C;J/2 - UC, C'
+ - 2 - -2 - 2... + VCr' 2 h + W'/2 - U'/2 h
1, Inlet Casing
W'
UC,
-
C' vex + 22
or
I' ~
"0,(,.1-
0'/21
where 1I0 .rel is the total entha lpy based on the relative velocity of lhe fluid. Thus
11,- hi = ((Ui - uil/2l + «IVi - Wil/2)
(3,6)
68
).>
CENTRlFUGALCOM I'RESSOllSANOFANS ..( 69
TUlmo M ACHINES
where A Pond B are functions of th. z andr2/ rt. The SloIJilz slip joctor is given by
2e/sin ~l
[ rI, = I - 0.63rr l td I - (C"I U,)col/'illli Rehllive eddy
is best used in the range 80° < fh. < 90°. If P2 90° . then Us = I - (O.631f Iz). TypicaJly. slip factors li e in tht: region of 0.9. while slip occurs even if the fluid is ideal.
=
Arc AB - 2](('2 /z
ENERGY TRANSFER
,
,
By Euler's pump equation, without slip
1£=l\IllIIg l E Figure 3.6
Th e fe /oril'e eddy be/weell impeller b/ode5
From inlci vclocity triangle (Fig. 3.3(a))
zc~o. Now. the impelle r has an angular veloc ity 'w', so that, relative to the impeller, [he fluid mu st have an angular velocity '-w ' [0 match with (he zero-roralion conditi on. If Ihe rad ius o f a ci rcl e rh al may be inscribed between two s uccessive blades at outlet and .11 a tan gent to the surfaces o f both blades is 'e'. then the slip is given by 6Cx
6Cx
For ideal condition, U2 =
CZl'
from outlet velocity triang le (Fig . 3.3(b»)
E
=
(1fr'lI .. ) sin fh
;:
(U2!zrJ)(rrr2 sin (U'1rr sin fh)/z
= equation (3 . 1)
ex' = 0
= we
The impeller ci rcumference is 2]("r1 and there fore Ihe distance belween the blades is 2JTr~ /z if we ha ve 'z' blades of neg ligible thickness. TIlis may be approximated to 2e!s ;ll p~ and upon rearrungement
e
CU2Cx:! - UtCsd lg
=
U2Cs 2 = Ui
g
(3 .3)
g
and with sl ip, the theoretical work is
,V;
~ E=--g
fhJ
Although equation (3.3) has been modified by the slip factor
(3.4) 10
givr.! equation (3.4).
usU; /g is still th e 'theoretical work' done on the air. si nce slip will be present even
beco~es
as::::
_ U-",,,"-os,,,in::..;::~=-,-::-:1 - -;o;-" z(V, - Cr,cOl~,)
For ~urely radial blades, which are often found in a centrifuga l compressor, be 90 and the Stodola slip factor becomes 1IT,
= I-
"Iz(since cO! 90°
Ih wi ll
if the nuid is fric tion-less (ideal nuid) . In a rea l nuid some of the power supplied by the impeller is used in overcoming losses that have a braking effec t on the wr conveyed by the vanes, and these include windage, disc fr iction and casing friction . The total power per unit weight of flo w is therefore modified by a power input factor.
POWER INPUT FACTOR
= 0) 1
With radial vanes, a very high pressure rise can be obtained, and arc s uitable for high-s peed mnchint:s. The Stodola slip factor equa tion gives best results for the blade ang le in the range C 20 < Ih < 30°. For the range 30° < fJ'1 < 80 0 , Buseman slip factor equation may he employed.
(3.2)
The powe r inpul factor (or) the work factor _
Actuol work supplied
'" - T heoretical work supplied
'y,' ,ypicolly 'ok", volues from 1.03510 1.041. So, the actual energy transfer becomes E = y,rI, Vilg
66
}>
TURBO MACBfNES
CENTRIFUGAL COMPRESSORS AND I; ANS "\ 67
u,
u,
c,'
w.~
c,'
c,
. •.•• [
IV,
--
AeruIl1 wirh 110pl
cr,
!
'"
pz· Ac[ual bladc lingle p;.Angle lit which the nuid ICllves the impdlc:r C . Cin.:ulation around a single billde
Figure 3.4
Velocity triangles/or a badcward curved impeller
When fiz = fJi, there is no fluid slip at the exit where /3' refers to the angle;: of relative velocity vector.
Figure 3.5
Slip and ,'e/oeity distribution in ccnlrijllga/ ptunp impeller btado
Slip factor is defined as
SLIP FACTOR The nuid tcaves the impeller at an angle Pi other than lhe actual blade angle 112. This is due to 'fluid sUp'. Angle Pi is less than angle 112. In centrifugal compressors. the air trapped between the impeller vanes is reluctant to move round with the impeller. and this results in a. higher static pressure on the lending face of the vane than on the tmiling face of the vane. This problem is due to the in ertia of the air. Then the air tends to flow round the edges of the vanes in the clearance space between impeller and casing. One explanation for this is that of the relative eddy hypothesis. Fig. 3.5 shows the pressure distribution built up in the impeller passages due to the motion of the blades. On the leading side of the blade there is a high pressure reg ion while on the trailing side of the blade there is a low pressure region ; the pressure lhus changes across the blade passage. Th~s pressure distribution is similar to that of an aero foil in a free stream and is like· wise associated witll the existence of circulation around the blade, so that on the low pressure side the fluid velocity is increased while on the high pressure side it is decreased, and a non-unifonn velocity distribution results at any radius. Indeed. the flow may separate from the suction surface of the blade. So, the mean direction of the fl ow leavin g the impeller is fJ~ and not 112 as is assumed in the zero-s lip condition. Slip can be reduced by increasing [he number of impeller vanes and reducing the clearance space. Thus C.r2 is reduced to C~2 and the difference .6C.r is defined as the sl ip.
Referring to the above figure, for the no-slip condition. C.r2 = lh - Wz2 and
W.r2
C.r2
= =
/32 ' C, 2 U2 - Cr2cot f32
COl
Sl ip factor
e;, C.r2 =
(ex' - Aex) C.r;2 (3.1 )
Stodola proposed the existcnce of a relative eddy within the blade passages as shown in Fig. (3.6). By definition. a fricrion less fluid which passes t11rough t11C blade pas sages have no rotation. Therefore atlhc outlet of the passage the rotation should bc
64
CENTRIFUGAL COMPRESSORS AND FANS .( 65
j:. TUlWO MACHINES
VELOCITY DIAGRAMS FOR A CENTRIFUGAL COMPRESSOR
ColleCl0r (VOlule)@
Casing )
Diffuser
Figure 3.1
Typical ceTltrifflgal comprl!!Sor
T he principal componenls are Ihe impeller and Ihe diffuser. Wilen the impeller is rotating at high speed, air is drawn in through the eye of the impeller. The absolute velocity of the inflow air is axial. The magnitude and direc tion of the entering relative ve loc ity depends upon the linear velocity of the impeller al the radial posi tion of th e eye as well as the magnitude ilnd direction of the entering absolute velocity. The impelle r vanes at the eye are bent so as to provide shockless entry for the entering fl ow at its re lative enlry angle. TIle air th en flows radially through the impeller passages due 10 centrifuga l fo rce. The lolal mechanical energy driving the compressor is transmitted to the fluid stream in (tIC impeller where it is converted into kinetic energy, pressure il nd heat due to fri ction. The function of the diffus~r is 10 convert th e kinetic energy; of air that leav ing the impeller, into pressure, The air leaving the diffuser is colleCled in a spirnl passage (scroll or volute) from which it is discharged from the compressor. The pressure and velocity vQriatiqn lJcross the compressor is shown in Fig. 3.2.
The gas enters thl! compressor alme eye, in an axial dircction with an absolule veloc ity C I and moves into the inducer section , which cnn be in a separate form or be a pan of Ihe blades. The inducer section transfers the gas o nlo the blades and enables it 10 move smooth ly into the radial direction. Energy is imparted to the gas by the rolating blades, thereby increasing its static pressure; as it moves from radius rl to r2, and the gas moves the blade with absolute ve locity C2. It shou ld be noted Ihat the blades !lIe radial i.e. the blade angle fJ'1 is 90" while the relative velocity vector W'1 is tit angle u fJ; because o r slip. Ideally, the componen l C ,r1 equals UJ,. But it is reduced due 10 slip. The re lative velocity vector WI is obtained by subtracting U I fro m C I . The p, w, nuid enters the blade passages with an u, absolute velocity CI; here C I = Ca. So, the impeller ta ngential velocity vector U, is a L right angle to C I. Where UI wr), 'w' being the angular velocity af the impeller. The resultant relati ve velocity vec(aJ Inll!t \'t/ociry rriang /I! lo r at the inlel is WI as shown in Fig. 3.3(a)
=
- - - -Noslip With slip
Diffuser
_1 ___ _v
Radial di stance
(b)
Clearnncc
Pressure and ve loc ity
Figure 3.2
Prt!ssure and ve/ocifY diagram/or celltrifllgal compressor
Figure J.J
Outid 'tIeloeity Iriangll!
Velocity Irianglt!S for a rtuljal imptllu
(refer Fig. 3.3(b) and Fig. 3.4). For "'TO slip si tuation. III = 90" and so C" = U, and C,~ = W:! where C, is the radial component of the absolute velocity and is perpendicu lar to the tan gen t at inlet and outlet. C"I is the compo nent of the inle! ab'so!u lC veloci ly vector resolved into the tangential direction. W,r and Cor a.n: oflen called as the relative and absolute whirl components, respectively. All angles are measured from lhe tangential direction. When fJl = fJ;, Ihis is referred to QS the '/1 0 shock condirion' at entry. In this case, the fluid moves tangentially onlo the blade.
62
>-
TuRBO MACHINES
EXERCISES 2.1 . ~erive Ihe expression for energy transfer in terms of blade lift and drag c fticlents. . oe
2.2. ExpJa!" (he blade terminologies with a nem skelch. 2.3. Explain the various cascade nome nclalures with a neal and illustrative sketch prove thaI E= 0 + ; - O. 2.4. Wrire nOles on cascade testing. Draw the cascade curves. How is the nom' J value of deflection obtained? Ina Enumeral~ and explain briefly [he different cascade losses. _."6. A 16 m dmOlcter rotor is required to lift and propel a 2500 kg h I' d f J C Icopter al a spec ~ 15 m s. Calc ulate power required by the helicopter assuming a drag co~mclCnl of 0.0056 based on the rOlor area. Ambient condition is J bar and 22 C. (MKU-April '96 & MU-April '99) IAns: 2.24 kW)
;.5.
3________ _ _.__ _ CENTRIFUGAL COMPRESSORS AND FANS
INTRODUCTION Compressors as well as pumps and fans arc the devices used (0 increase the pressure of a fluid. But, Ihey differ in lhe lasks they perform . A fan increases the pressure of n gas sligh ll y and it is mainly used to move a gas around. A compressor is capable of compressing lhe gas 10 very high pressu res. Pumps work very much like compressors except that they handle liquids instead of gases. Centrifugal compressors and fa ns are lurbo machines employi ng centrifugal effec ls to incrl!asc the prl~ssure of the fluid . Single stage ccntrifugaJ compressors have the pressure ratio of 4; I. The besl efficiencies are generally 3 to 4 per cent below lhose obtained from an axial flow compressordesigncd for the same duty. However, at very low mass flow rates, the axial flow compressor efficiency drops down rapidly. The advantages of centrifugal compressor over the axial flow compressor arc (I) smaller length. (2) wide range of mass. flow rale of gas. Although the ccnlrifugal compressor has been superseded by the axial flow com pressor in jel ai rcraft engines. it is usefu l where a short overall eng ine length is required and where il is likely thai deposits will be (ormect in the air passages. Beca.use of the relatively short passage length, loss of performance due 10 bui ld-up deposits will not be as great ilS the axia l compressors. Therefore. the working fluid may even be a contami narcd gas, like exhaust gas.The disadvantages arc-larger frontal area and lower maximum efficiency. If lhe density rat io across the compressor is less than about 1.05, lhe rcrm ,/0'" is used to describe the machine. In that case the fluid is treated as being incompressible; otherwise compressible flow cquations must be used . The Lerm 'blowu' is often used in place of 'fan' . In this chapter, the centrifugal compressor and fan arc considered 10gelher as the theory applied to both machines is the same. The centrifugal compressor is mainly found in turbo chargers.
COMPONENTS AND DESCRIPTION Fig. 3. 1 shows a typical centrifugal compressor.
BLADE THEORY
flO ,. TURBO MACIlIN ES
= = '. CD
33. 13°
2 (0.9 1)
=
0.0823
, x 9.81 x 10'' ) 17.5 cos] ~3 .~3) .."1,,,0-:-:::-:---;-;::;,-_ ( cos- 6:1 1.25 x 100' (
(c) Lift coefficient CI.
lUll') cos alii (Iun 0' )
C,.
+ Ian a:d + CD Inn am
=
2(0.91) cos 33.I3(w.n 40 -\- Irm65)
=
4.60 1
+ 0.OR23 tuJl
~
U J
Example 2 .9
A jet plane which weighs 30,000 N and has a wing ;: rl',l or:2O 1112 nics al a velocilY 01'250 kmlh when the engine delivers 750 kW. 6S r; 1)1 lht! power is uscu to overcome the drag resisl
Solution
=
=
1\1 = .1 0.000 N A 20 m' c 250 kln/h = 69.44 m/ s Power requ ired to overcome drag resistance
= 0.65 x
487.5 - - x 10' = 7020.5 N 69,44
But
Co
= =
CD
=
D O.5p c:!A
7020.5 0.5 x 1.21 x 69.44] 0. 120
X
20
= 30, 000 N : . CL
=
L
Li st the imponam blade tcnninologics. Define (a) base profile. (b) camber line, What is
(it) 0.88 Ellltlx (b) 0.9 Enmr
0.5pc:!t\
30,000 0.5 x 1.21 x 6904.1 2 x 0.5 14
2. 16. 2.17. 2. 18. 2.19. 2.20. 2.2 1. :!.22.
and the
2.23. What is slagger angle? 2.24. ang le represents me angle at which the blade is set in the cas· cade. 2.25. What is c~scade testing? 2.26. Why do (hI! rt:!sults obtained from cascade testing need corrections? 2:27. What is pressure loss coefficient? 2.28. The value or nominal denection angle in terms of maximum defleclion angle is
Tht! lift fmce should he equal 10 the weight of tht:! plant:!. L = 1\1
_ ___ _ drag .
2. 10. The lift is to the drag. 2. 11 . Define (a) lin coefficie nt and (b) drag coefficien t. 2. 12. Lift coeffic ient should be as high as possible for maximum energy trans fer. (True/Fal se) 2. 13. For maximum efficiency, the drag coefficient should as low as poss ible. (Truc/False) 2. 14. Define blade loading factor ilnd flow cocflicient. 2. 15, For a wcll-dt:!s igned hlade, the optimum blade loading factor is
750 = 487.5 kW
D x c
=
2.1. What is the function of bl
(a)
Power required to oVj!rcomc drag rcsist'lOce in terms of drag force is £1\'1,'1) hy
:. D
61
SHORT QUESTIONS
"n - ' [('an 65" -lUn400)/2]
=
(c) 0.8 Emll •r ~()
2.29. What arc the cnscadc losses? 2.30. What arc profile, annulus and tip clearance losses?
58
BLAOETHEORY '" 59
l> TURBO MACHINES
25 30+ 2 42.5°
"'; = , "'I =
107.9 N/m' 107.9 0.5 x 1.25 x 75 2 0.0307
and
(b) Drag coefficien t
a~ - 11
Ct'2
,
am
17.5°
Ct,
==
CD
42.5 - 25
tan-I [(tanat +tnna2)/2]
=
(b) Nominal air angles
=
i,
:.a1.n
,
+0';
(COS'
107.9 ) 38.25) 2( 1/0.9 1) ( 1.25 x 752 . cos248
CD
5 + 42.5 .
0.0365
47.5°
(c) liN coefficient
Nominal cxit~ ang le is dclC.DIlined....fr.ru!1.J,he~ng ~Cmpirical relcuion : anal,n - tana2.n
Ian Q'l.n
Lan Q'l.n
CL
1.55
=
1.0+L5(~
=
tanat,n -
= =
tan 47.5 -
1.55
2( I / 0.91} cos 38 . 25° (~1n 48° - tan 25°} - 0.0365 tan 38.25' ·
=
(~)
0.471
Soluti on Ctl = 40° Ct, = 65' C, = 100 m/s .r/c = 0.91 pO. m = 17.5mmW.G .
25 .22°
Example 2.7 . A compressor casc
2(s/ c)cosQ'm(LanO'I -lan ( 2) - CD lanam
Examp le 2 .8 Air enters the test section ofa turbine blade (0'1 = 40°.0'1 = 65° ) cascade tunnel at lOOmis (p 1.25 kg/m 3). The pitch-chord ririio of the cascade.!' is 0.91. The average loss in the stagnation pressure aerl1ss the cascade is equivalent to 17,5 mm W.G. Determine ror this cascade (a) the pressure loss coefficient, (b-) dll: drug coefficient and (c) the lift coefficient.
1.55 1.0+ 1.5(1)
=
= =
1.083
1.0 + 1.5
Cl'2,n
lan-' [(Ian 48' +1.n25°)/2]
Q'l,n - 0'1
in
=
2 2(S/C)(POm/pCfHcosJa",/cos ad
=
=
=
PO.m ---, = 0.5pC,
(Of x 9.81 x I p
=
= 1.25 kg/ m'
(a) Pressure l oss coefficient Po,,,,
0.5pcl PO,m
=
= 1.25 kg/m J
(a) Pressure loss coefficient
17 .5
at = 48° 0'2 = 25° pO. rn 11 mm W.G.
p
11 10' x 9.81 x - , 10
(b) Drag coefficien t CD
=
0'
0.5 x 1.25 x 1002 0.0275
56 :.- TUHliO M ,\C III NES
BLADETUffiRY ..;: 57
(d) Number of blades
Since there is no incidence on the blade Cl'J = Q' I Circ umfere nce at mean radius
=
=
9
Pitch ilt mean radi us
a t - Cl'1
2rr r NI
39. 11 0
9
S 2IT x 0.212 Alternatively
0.0296
45 Since
j
9- 0
E
9
9
=
0
Note Ihat cascildc air a ngle is equa l to Iht: compressor relative air angle. That is
= (JI
nnd [X2
lllcrcfore. deri vation angle is
= f31
(a) Nominal deflection angle
=
0', -
=
45 - 15
=
30 1l
[X2
==
/31 -
/32
o =
0.233(39.11)
" =
9. 11
0
Ii. The blade stagger ascade the blade stagger is given by
(b) Deviation angle 8
q, \ "
mO(sjc) '/1
0.2:l(211Ic)'
III
For a drcular afC c.:ambcr 2a jc
,-
0
023(1)+0.1 x
o
0.26 x 9(0.8)
o
0.2339
G~)
'I'
=
39. I I 45 - · - 2 25.45 0
Example 2 .6 A compressor cascade is constructed from ci rc ular arc acre-foil bl:ldes (camber nngle = 25°) SCI al a stagger ang le of 30° with a pitch-chord ratio of 1.0. TIle momentum thickness chord ralio is 0 .031 . The nomin al value of incidence is 5° . Determine the cascade blade angles and the nomina1air ang les. (MKU-April'95)
Solution
(c) Blade camber angle
fl
= 25°
¢x = 30
1l
sIc = 1
in
=5
1l
(a) Cascade blade angles
9
a;
:. 9
2
Cl'l -
1
+ O. I (a,/50)
0.26
but
'"
=I
:. m
0.2339
I - 0.233 30 0.767 39.1 10
=
(0 ) the blildc stagger
al
=0 -
E
tc.:) Ihe devi:.IIi on ungJc
= 45° p, = 15
- "
= 0
(b) Ihl.! blade camber angle
En
+i
E= 9
Example 2.5 DCh.!fllline fo r!.l co mpressor hl .IlJe with u c ircular nrc.: ci.llnbcr line anJ Ihl! fLlII l)wing JOlla. Pitch-c hord ratio - O.H. rd:.Hive uir angll! a t inlet - 45°, relat ive air illlgie ,II out kt -15". Assume 7.cro inc idence. (il) nominal ddll!C li nn an gle
Solution sl,· = 0 .8 P ,
+ 0,2339
45 - 15 + 0.2338
=
0'2 -
0
[X~
-
o'~
-
+0 0'2 + 0.2330
e
0'2
=
,
[XI -
0
'2 = 30
0
54
BLADE THEORY
}:. TURBO MACHINES
Solution IV =980N
..::
55
Determine (a) the mean radius, (b) the blade height. (c) the pitch nnd chord and (d)
C=5m/s Co = 1.3
the number of blades.
p= 1.22
Solution
Drag force
=
D
111
C'
'. A =
980/
p
= 1.1
ell
kg/m]
U = 200 mls
= 157 m/s Ilc=3 slc=0.8
(a) Mean radius
II' = 980 N
but D
= 25 kg/s
N = 150 revls
CO,p·Z·A
(13 x 1.22 x ~)
U
=
2rrN 200 2rr x 150
49.4) m 2
=
Projected area of the hemispherical parachute
=
A
=
:. D
(b) Blade height 49.43 x 4
J
rr
The blade height is found from the annulus area of now.
= 7.93 m
A
Example 2.3 A wing ofa small airplane is rectangular in plan (10 m x 1.2 m).
Solution 240 x 101 3600 = 66.67m/s
C=
=
1.1 x 157
0.145
LID= 10
(C=
2:r-:;:; 0.145 2n x O.212 0.11 m
J(O.IL)'+L' 1.01 L 2
(c) The chord and pitch
The weight lhal the plane can carry is the lifl force.
20 x 10-' :. L = (1.01)'/' = 19.9 kN
. Blade aspect rn\10
span (I)
= chord ( C ) :. C
and L pAC'/2
m'
:. Blade height
F=20kN
JD'+L2
F'
pea
=
TIle total force.
F
111
=
25
The total aerodynamic force acting on the wing. moving al240 kmlh. is 20 kN. If the lin-drag ralio is 10. calculale the coefficient of lin and the total weight the plane can carry. Take p of air = 1.2 kg/m].
A = lOx 1.2= 12m'
0.212 m
19.9 x 10 3
=
66.67'
1.2 x 12 x - 2 -
0.622
Example 2.4 An axial now compressor has the following design dala : mass now rate of air - 25 kg/so density - 1.1 kg/m]. axial velocity - 157 mls. rotational speed - J50 rev/so Mean blade speed - 200 m/s. rotor blade aspect ratio - 3, pilch chord ratio - 0.8.
1 3 0.11 3
0.037 m Brade pitch 5
=
Pilch - chord ratio x chord
=
0.8 x 0.037
0.0296 m
-
BLADETUEORY
Turbine rolor blades are designed based on Ihe cascade dal<\ whic h is. simi lar 10 th e compressor rolor h lades. The lifl and drag coefficients are obtained from the cascade data curves. The drag coefficient is expressed as CD = 2(s/c)(Pom/pCrHcos3a",/cos2a:d
4. Tip clearance loss Loss due 10 tip clearance is causcd due 10 the Jeaknge of the now from the pressure side to the suctio n side of the blades through lht! lip c1cllrancc. This loss is sometimes considered a.s a secondary loss. Losses in a blade cascade of compressor and turbine arc principally the same. The: magnitude and mechanism of these losses differ from compressor 10 turbine .
amI the lift coefficient
whcrc
0'/11
= l
SOLVED PROBLEMS
= 2(sjc) cos 0'/11 {Ian at(f}an a:d(9<:v tan Om
-rl (tan a:Qtan 0'1)
2]
_r..n!ssurC"IO'S'S~mntre1T" elermined and an estimation of the efficiency be made. TIle drag coefficient must again be modified due to the blades actu::IlIy being in annular form. R.eal~xis l &lIthe hub and tip while the ideal How pauern is diswrbcd h~! the preceding and succeeding blades. We crag coefficient IS modified
by~ance @SI C ~c
s
~
,n"
a
ow
given by
~ CDC = flCl(kC' ISll _ ~I
f[CDS~
;-~
7 5.,' 1cJ. :., ' )
= >
~C(O'l\cll.o\(y \ .\~ T
~t ':-.>-t; :p 1.,> l'
Example 2.1 An aerofoil having a chord length of 2.25 m and a span of 13.5 m moves aI a velocity of 125 mls through standard atmosphere at an elevation of 2500 m. The anglc or auack being 5°25'. Calculate the weight which the wing carries and the power required to drive the aerofoil. Take corresponding to i = 5°25'. CL 0.465 ,nd CD = 0.022. Density of air = 1.25 kg/m'-
=
Solution
c = 2.25 m c = 125 m js CD = 0.022
1= 13.5 m
CL =0.465 = 1.25 kg/m'
P
weight carried by the aeroroil shou ld be equiJlto the lift force.
.
,
w~.w Ceil[ . C ween lhec
cL=CL · P·-·A 2 125 0.465 x 1.25 x "'2' x (2.25 x 13.5)
... IV
now)' / (1+.Hub radiUS) .
Blade out Ie! area normal to ). = [ . [( Bll.ldcs tlli el area nomw llo flow
TIp
137.93 kN
radIUS
Drag force
Hcnc:c for a turhine,
ICDT =CD+CDC+CDSI
Losses in a Cascade 11H! following losses occur in a cascade.
1. Profile loss This occurs due to the boundary layer growlh on the blade surfiJce. 111is loss increases when the boundary layer separates from the blade sur fa ce and ·is governcd by hlade profile fat given Row conditions.
2. Annulus loss This occurs due to Ihe boundary layer growth on the Hoors and ccilings of the blade passages.
C
' CD ' P ' -·A 2
D
wbere CD is obtained from cascade dala. TI1US the linear cascade data may be effectIvely used 10 determine the lift and drag cocfticienls for lhc cascade and [hen be modi lied by Ihc addition of annu lus drag and sccondary losses in order to apHfoximatc the drag coemcie nt for an annular cascade.
-
53
3. Secondary loss This occurs at the hub and tip du e 10 the three dimensional nalure! of thc now and the blade curvalure.
CASCADE LIFT AND DRAG COEFFICIENTS FOR TURBINE BLADES
CI.
'"
1252 0.022 x 1.25 x -2- x (2 .25 x 13.5)
=
6525.88 N D x c
Power required
6525.88 x 125
=
8 15 .735 kW
Example 2.2 Calcula!!! the diameter of a parachute 10 be used for dropping an object weighing 980 N so that the maximum lenninal velocity of dropping is S mls. The drag codficicnt for the parachute, which may be Ireal\!d as hemispherical, is 1.3. The density of air is 1.22 kg/m)
50
}>
TwU]o MACIIIN~
BLADE THEORY .:{ 51
,
,---------======-]1.5 1.0
o..'i G'
y
2
~ t~-; ., <:.
t \:,
c, Figure 2.13
Velocities and/or-us in a cascade
- -- --
' 0< .
_I
~
O< I~
o LL====::::;:=:=:::l~
- 20
'V 'f
----..
VI(
Figur~
~ ._; '?
2.14
-10 - 5 0 5 Incidence angle:, I (deg)
- IS
Lilt and drag
coeJJici~/lrs
vasw inddl.'nu angle
'.0,------ - - - - - - ,
.,
=
The lift C 2scosa ta - t~lL£ps!.txm.l..=..{sPOm..Eosam) tana~ llH!n the l ift ~fficlcnt CL = L/O.5pe!c
= 2(s/c} cos Ctm(tan a l -
1.0
- - - - j o.,
0.'
lana2) - CD tan am
Th air inlet velocit ell the incidence angle j and the air inlet an Ie known..Ihe.de,yjatiwUlllgle E. is read from e grag etw e a d i for the angle of incidence and
n Air O\IIJelllngl~ . nl (ikg)
Figure 2.15
Cascade lift co~ffici~nf vt!rsus air outlet OIlg/~
b
To the rofile dril a
owing (s/c), values of POm/O.SpC2 can be read from the same ra h (E. VS i) for-vatimrs·'mel ence.angles nnd substitution of these variab es mto e ualions for CD and C curves of C nd C a fie oHe against the incidence an Ie as shown in Fi .2.14. Finall ainsl the'air Ie Q'2 for the nominal value o(,£n-!or-a-whele-sefics-o( different geomclry cascades. to give the varHifion or C ( with air outlet angle for a particular (s /e) rauo (Fig. 2.15). The drag coefficient is very small in comparison with CL nnd is therefore oflen ignored. SO-lharcquauon ror CL becomes -
ttE~sJcosam(tanQ'1 -tnna~
10
e
two further dra s must be added to lake the
whcr@ s lhq'bI,dc heigh\. 'n~:::=:::::::==:.J ~dary IOsi'cs by
...
ITCos =
O.OI8c!.! j
TIle Iota I drag coefficient is given by
]...S -(! (Oroll.--~ O .) ~
(le ,"
ICOT=CO+COA+Cos l
+-0,00 7..7-1- Q,018C L2
48 ;.:. TURAO M AC lJlNES
BLAOETIlEORY
incidence upto a maximum deflection angle (e IlUJ •f ). This is [he s[all p'o int where scp· a:,arion occurs on the suction sl!I.fat:e-of-the..Qlade . Bu t thi s '-Ingle may not be well
=::::::: \ .
I
~endi"nt mainJ~gilchkhord ratio an'!€]. A plot between En and a l for different values of (sic) is shown in Fig. 2.12. These curves are particularly useful to the designer when any two the three variables an: fixed.
((/J;
or
---
1
.-"~
"'" "9
40 35
"
30
E u
~
J'
25&
'0
Ii:u
20
6 U
ff»
"-
"
u
Trniling edge positions
0.3
~N_
-u
,~
Ji'.
~
~
0.2
~~
c.o;' c
.e ;; ~
Mean
0.1
c
'" "
0;
0
Figure 2,/0
2 4 3 5 6 Distance along trailing edge of cuscade
7
______- --A·ir-ou
Figure 2.11
Cascade deflection Qnd preJSure loJS cun'es at om: ancle oj incidence "\I'
r - - - - - - - - - ---,40
,.-----,
YJe.C()
----.-,<~-". ,fin ........ J -
0.100
, "U
-
r -
)II
:
20 .,
0 .050
d..
0 .025
Minimum pressure
<;-f:!
- 20
-1 5
-10
los .~
-5
(I'-, -
oull~ t
angle'
PJl
= p(C; -
,,,tot NI DIX'
C~)/2 - (POI - Po,)..
7?
5t"""
where the difference (POI - Pol) is obtained from the cascade tl!sl. It should be noted thai POI > POl. because no work is in the cnscadc and the flow is proceeded irreversible. Hence , the above equation will be written as
l'.P = p(Cr - Ch/2 - Po...
10
rmean
=
=
(P2 - P,) nn Porn (POI - Po,) .. The summation of all forces ac mg r JO me control volume x and)' directions must equal to me rate o~ change of momentum of the air towards these directions . Consideri ng the forces and the changes in velocity in the directions x and )', the following relations are obtained for drag and lift. where 6P
deflection and pressilre Joss Cll,:!!
defined in some dcsi os and i[ is [aken as [h e angle of incidence, where lhe - pressorclass.i wic/! 100mu . epols ows al, orawidcrange ofincidence, me pressure loss IS air y constant, and it is poss ible to select an angle of dcOection En (ca lled /lominal dejlecIiOlI allgle) which is al so compatible with low pressure loss as representative by Ihe panicular design , The I!~S nerally O.Be m . It has been taken as 0.8 times IDt!.JllllX.imum defl ec ti on angle (i.e. /I de termined from large number of cascade tests that lhe namm a
~----
Cascade nominal defleclion angle versus air
.
~ :;
Incidence, I (dell ) Figurr1;-l.l-C~afl
(deg)
CASCADE LIFT AND DRAG COEFFICIENTS
:!i
Twice minimum [
z
FiS. 2.13 shows two blades of a cascade having chord. C and pitch. s. At sections I and 2. the 100ai pressures are POI and Po2 respectively with corresponding . . e1ocities of C] and C2 . ThefleilSi~ across the cascade is assumed to be negligible. The stalic pressure change across me. cascade IS g iven by S L.... --
...G
c 10
0.075
'"
.g
- -
tangle, a
[ Drag D =
oS Pam' cos ani
j
Dividing the drag by 0.5pC;IC gives the drag coefficient (CD) CD
=2((sjc) (Po.../ pC,;) cos am
... 1.3
46
>
TWlnO MACIII Nf:.s
BLAIJETIlEOKY
-.l ·17
De.viation angle nle diffcrence between the air angle and blade angle at exit is referred to as deviation.
,
' (~ = ai -all
•
It is different from the deviation angle for compressor cascade and the difference is due 10 the dirrerenl conven ti on useJ to define the stagger angle and the exit angles in turbine cascades.
"
Air deflection angle The denection 'lngle for the turbine cil5cade is defined as
Wind
Fluid ou t
~
the sum of Ihe ::li r ang les al Lhc entry and exil.
cf~aJ+;P ntis is again differe nl [rpm IhjH of Lhe comoressor cascade. It is also expressed in terms of othe r angles as
t...Blade cascade
<=0+;-6
Olord
line Camlmr line
tip , the blade row.
Figure 2.B (0)
ascade Ifomef/cfatllre
CASCADE TESTING AND CURVES The flow around blades in compressors and turbines is different from now around isolated aero·foils. because of Lhe dfect of adjacent blades, i.e. Lhe gas flow around a blade is affected by the flow adjacent blade. 1l1is increases as the solidity (cf.t) increases. and I of a blade section. groups of blades of conSlont profile are mounted in parallel fashion at the end of a wind tunnel as s~own in the Fig. 2.9. The number of blades comp.~ ..1l:u;..cas. ~de has to be suffi clcn@mrtly-8"OrtOwTth an ~~) to eliminatc
may necessary to carry 10 picture of the now in the blades. Fig. 2.9 is known ilS a lincilr cascilde and can be imagined a<; II row of compressor . o r turbine blades unwound from the rotor to form the cascade. The air deflection
I
4"- :0- TUROO MACIIINES
~f tl 1
-=/
(/\ t
The deviallon
hi BL.... OETIlEORY 0( 45
a~rc~ IS caused by the air not remainmg aunched to the blade over
its total curvature. 6
isgiven by the empirical rcJruionship
ct3(S/C)I/i]) where
-
~£€l(a,~n
and
N1.::::=, O. 2..-
~ . is the distance alon
nrc cambcr
Ill!.!, \
2a c)
Ihe chord to the point of maximum camber. For a circular I , nnd this bin e ann is often eh_oscn.
oUl let air
\1£ =a,-a'r\ From the definit;ollS-Of.dirreren t..angles , 'l cID1.bJ?£c:n_tbaUhsy, are related by the rollowing expression. Figure 2.8
'-----'
cascade flolllellt.'lalllres .
b) Blade camber angle The differcnce between the inlct nnd outlet camber er all Ie cnote ·by O. Mmhcmmltnny
anglcs is called t~
crGG21P
i) Stagger angle The angle between the aIial..difcet·ion-ilnd.Lh~Qfd is known as stagger angle. denoted by® and it represents lheanglc al which the bI~j_S!!t in the cascade.
c) Chord The c hord is 'Ihe maximum width of the blade profile in a direclion parallel 10 Ihe chord linc, i.c. the distance between [h e blade leading and trailing edges, and is dl!notcd by c. 1 ·\ .r_ \ot"I~ r,.\~:r 't '" a. d) Pitch Th pilch dr blade spitcing, desi nated s is [he distance between the corresponding palMt IS expressed Cit er y Ihe pitch to chord rallO (s c or a lernaUve y lhe so lidi ty (c Is). When Ihe hlades nrc evenly spaced around a rOior. the pilch is Ihe circumrerence at any radius dividl!d by Ihe number or blades.
e) Air angle Thl.! angle between the direction of veloci ty rdati ve c..bladc and tht.! axis or the j] c row IS I C lr allg e an IS enOle The air angles at the inlet and oU ll el arc at an a2 respecll vely. The cascade air ung le is equal 10 the compressor or turbine relative air a n g l e . ' ". , f) Angle of incidence camber
the inlel air angle and the inlet
~R~~ CASCADE NOMENCLATURE TIle setting of blades in a turbine cascade is invariably at a slagger angle (¢,), i.c. the chord lines of the trubine blades arc tilted lowards the bind..: curvature as showp in Fig. 2.8(0).
Camber angle The tangents 10 the camber line althe entry and exit make: the camber angh!s and with the aXIal olrectiOn. - In contrast to the compressor cascade. the blade camber Bngle for l_he turbine cascade: is defined a.i.!.~ sum ofmemlct and outlet camber angles.
a;
a;
(le=a;+jJ) The air ang lcs at and a2 nrc diffctent from the blade angles
0'; and 0'1
Angle of incidence The difrerence between the nir nnd the blade angle at the entry is known as th~ angle or incidence.
li=al-a~1 g) Deviation angle
-
As in Ihe compressor cascade, thc incidence angle can bt: positive or negative. Aow at n large positive incidence is nssocialed with positive sl1lll, i.e. flow separation on the suction side of the blade; conversely, a large negative incidence is associated with negative stall, i.e. now separation on the pressure side orthe blnde.
42
BLADETIJEORY
~ TURBO M ACHINES
wt},f,;.Q;JheJla.Kthr.u"g!1.Qne blade passage of width loading factor is given by
' 5'
r
has been considered. The blade
..... 43
Max. imum thickness Equally spaced .. lations
Lending edge ""'"7 f---+----....:=--7\-E- Tr3iling edge
Power delivered
mU 2 (11110) / U '
=
Position of
x
...K = maximum C
Fx /(pC"isU)
thickness
I<----c Base profile
",(c/s) sec {l",ICL -1- CD tan tI", I
where f/J is call ed the now coefficient and is defined as.
= Axial velocity (ell ) Blade speed (U)
(~?r
r::-: 1~~UY.Jl · ~ . um e fr.Jlclcncy rna~ lI~ Ihe mean !low angJ~1 bou ~a nd. substi-
tutlnE tl1l5 Into lie Blade loading faclor egua .on, the expression for optimum blade loading fac tor 1/1/1111' is obtained. .,.
rjJ(C/S)[CL
'Plopt
../2
=
+ Col
If CD iSJTluch.smallillhan CL . .which usually occ urs in ,lhe case of a well-designed blade. then
-
.,.
IOf'tr'/'l
=
Pammf!fers u.red ill deJcribillg bladi' shnpes alld ro njif,:llrtlliutl.s oj blCJdt'J
2. Maximum thickness 11 is a use ful parameter forde:;cribing an aero-foil and it is expresscd as a pcn:cntage of the blade length.
?
cfJ
Figllre J.7
rjJ(C/s)CL
../2
3. Position of maximum thickness
It is an another usdul parameter which is spec ified as a perce ntage of the blade length.
4. Leading edge (Nose)
It is usually a circular arc blended into the main
profile and spec ified by its radius as
II
percentage of the maximum thickness .
5. Trailing edge It is ideally sharp. i.e. of zero radius. bUI as this is impossible from strength considerations. it is al so a circu lar arc specified as a percentage of the ma ximum lh ickness.
6. Camber line If the • • is of the iineOlr pro fil e is g iven some predetermined curvature. then it is called cam· her lill e. TIle base profile is fitted o n \0 the curved camber line. This camber line is formed ei ther by one or more circulOlr arcs or o ne or lwo parabolic: arcs . A sing le circular o r parabolic arc is quite common as it is geometrically simple
Profile statiolls
Camber line
[Fig. 2.7\a» ) .
BLADE TERMINOLOGY
CASCADE NOMENCLATURE
Blade profiles arc usually of aero-foil shape for optimum perform.lnt.:c. But. si mple geometrical shapes composed of circular arcs and straight lines are used when cost is more important than the efficiency. Many blade profiles are fomled by bending a symmetrical aero-foil sec lion on a curved mean line. The parameters used in describing blade shapes and configurations of blades (Fig . 2.7) are as follows
A cascnde is a row of geomelricOllly similar blades arranged al equal d isLances from each o th er nnd aligned to the now direction . The various impurtant no menclatures (Fig. 2.8) of a compressor cascade arc
1. Base profile It is defined by dividing the major a;.:;is into equally spaced stations designated as a percentage of the blade len gth and specifying the hcighi from axis to profile at each statio n.
.W :,..
TURBO MACHINES
BLAOETIIEORV .;( 41
dfClpS nbruptly hecnuse the separmioo pain! moves bad 10 give a smaller wake on the down strcnm side Qf blade . In compressors, thl! blades form a diverging passage i. e the arCil at inlet is less than at out lc t and therefore the nuid is decelerOled in the pussage . In turbines. the blades form a converging passage. i . c. the area at inlet is grenter than that aI outlet.. Thus. the fluiJ i~ accelerated in the pa ssage. A nuid c nn be acce lerated over a wide range with high efficiency. but the process of diffusion ca nnot be carried out so rapidly due 10 the onsel of separation on the suction sid !.! of Lhe blad es, antJ consequent s lnlling . This is si.mililr to the includell angle of a diffuser being too great nnd separation taking place nlang the diffuser walls. The maximum ratl.! of efficient diffusion within the blade rows is cquivp.lcnt to a ~ angle of ahouCl c or 'I he curvature of compressor blades is less when compared to the curvature o f turhin !.! hlades. Becau se, if the rate or change of compressor b lade: profile is high. now separation wi ll occurdlle to udwrst! pressure gradient. Whereas in II turbine, the pressur..: gradient is favourable and with a vcry lorge curvature. i.e. 90° or even more, con he employed without severe losses . As a result . the angular turning of [he relative veloci ty vector is much gre.lIcr in the turbine Ihan in th e compressor. Typical blade seclions are shown in Figs. 2.4(a) a nd 2.4(b).
Force acting in the direction of the blade rotation (.t direction) is given by
= :=
L cos L cos
Pm + D sin 13m ,om (1 + (CD/CLl
tan
Pm)
~W,
8)
I Figure 2.5
Lift and drag forces on a compressor rotor blade
where W,
.
CD =
'"
(a)
CUlI/preHOI' blude passage
(b)
" .r----~-~F. .
, (O.5plV;;A)
and CL =
w~v
p.
D
2
Tllrbine blade pauuge
L , (O.5pWmA)
where the blade area A is th:::roduct of the chord 'c and th@ W lhl e,de height) and putting
-
-------
t
F, Figure 2.6
lVm .= -
.....
Consider a rotor hlnde shown in Fig . 2.5. wi th relative velocity vectors 'WI and W2 at angles PI and f11. This system is similar [0 now ovcr an acro-foil. so that lift and drag forces will be set up on the blade. TI~ drag force is acting in tbe line of the mean velocity vector Will al angle /3111 to the axial direction and (hI! lilt lorce acts perpendicular to Ihis . The forces on the air' wi ll act in the opposite direction as shown in lhe Fig. 2.6. The resultant force experienced by the air is therefore given by the vec tor R in the figure.
Ruolving blade forres
into tlrt dirt!Clion of rotation
Figure 2..1
ENERGY TRANSFER IN TERMS OF LIFT AND DRAG COEFFICIENTS
0
Co -
cos fJm
----
we have
F.r
=
pC,;(cI)CLsecjJm[l
+ (CD/CLltan/lml
L
~e power delivered to the air is given by
F,U
= m(llilo)
where ~ho is the.change iQJolal enthalpy across the rotor and
In
'=
EEllh .
/
3R
)>
BI.AI)ETItEOllY
TIJIluo M . . . C\lIN!iS
:w
and the pressure has a much lower value. wi th the nuid ncar the surface leaving in a ~o~ encrgy region. called th~Fig . 2.4). Measuremenl5 of pressure and velocily In the wnke can give the loss 0 energy due (0 the prc:-ence of the body and the drag can be ca lculated . Alternatively. the drag can be measured as an actual fo rce by means of n bahll1ce from the wind (unneltest.
aem-foil. Fig . 2.2 shows a non-symmetrical aero-foil placed in a s tream o f gas. Unlike the symmetrical area-foil. in this case, there is .1 pronounce Istur ane hich results - 111 greater oca de cellon a ow. 0 introduce suc h a high dellectlon over lhe gas stream . the ae ro-foil must exert a force on it, and hen ce an equal and opposite force of reaction is exerted by the gns on the aerofoil. The components of the resultant force arc disl:ussed in the following section.
~
LIFT AND DRAG COEFFICIENTS Figure 2.2 Flow pClff/'m around a nOIlsymmetrical (Jem-foil
Lift coefficient is a meas ure of the ability of il given section to support a weight when caused to move through a Ouid. <15 in the case of aeroplane wing. or. alternatively. In trnnsfcr energy to a fiuid. as in a pump or compressor. or to transfcr energy to a rotor when a nuid is caused to now over it. as in turbines. It is defined as
L C L = --:-,...::=-.,-
DRAG AND LIFT
(O.5plV~AI
TIle res ultant force normal to the aero·foil consists o f two components. namely lin and drag. The forces of lift and drag on a blade section nre shown in Fig. 2.3 with lift nonnal to the direction of the approach veloc ity and the drag parallel to it. The lift is due to an unbalanced force (pressure di s tribulion) over me aero-foil surface and is denoted as L. The drag deuno ted as D. i!'i due 10 the shearing stress at the surfnce and the consequent boundary layer. TIIC drag force is made up of a friction drag. duc to the pure skin friction ..;. Drug ~ effects, and a pressure drag. due to an unIbl ('I balonced pressure distribution around the blade . Figure 2.3 The forces of lift (lIId drag on TIle boundary layeris usually laminar for aJ Compressor blade bJ TIlrbine bladl!
~7 7 D~r ~ "
e'l
/l,el
a short distance downstream oflhe lending edge. then it becomes turbulent. The drag due to a laminor boundary is less than it turbulent layer. TIlUS for low pure friction drag. it is imporlant to maintain alaminar boundary layer over as much of the surface itS possible. If the pressure gradicnt is !'icvere, lhal is if the rate of change of aero-foil profile is too rapid. then the fluid in the boundary layer is brou hI to rest and leaves the surface in confu sed eddies. n is phenomenon is called eparollo ~ak-alV'd", or('flo~v re\'er.ra~ Fig. 2.4 and manifests lise In several olhcr ways as well as in the simBoundary layer Wake ple case of diffusion in a straight duct. Flow separalion due to ad--> verse pressure gradient decreases the lift "nd increases the drag. In practice. the ndverse pressure gradient I ncar the tail causes a thick boundFigure 2.4 Separation alld Wake ary layer and possibly separation,
(2.1 I
Drag coefficient is a measure of the loss of energy associated wi th the uscfullask (If produc ing li ft. It is defined as Co
D
= (O.5plV,;AI ,
(2.2J
where lYm is the mean relative velocity. ;\ is the area of the body .md the fac.:lm of 0.5 i!'i in serted for convenience as O.5p \V~ is defined as dY'lOmic pfuJttre. Some care is nceded in the evaluation of area associated with a given value of C/). por bodies of revolution which arc symme trica l ,Ibout an il;(is and parallclto the now. A is taken ilS the projected area normal to the direction of the now (e.g. spheres. cylinders. elc .), For olher bodies (e.g. blades, aeroplane wings. etc .) which arc nonnally dlher unsymmetrical or not aligned parallel to the now. or both, the area is evaluated in terms specifically defined as required. It may be noted tlla! CD as given in equation (2.2) is the ralio of the aClUal dr~g force 10 the force which would be exertc.:d if the reprCSCnlJlive area of the body were acted upon by the dynamic pressure. It is ilpparent that the maximum energy tran sfer implies (he Ia.rgcst possible fluid deflection or lift codficicnl. while maximum efficiency requires the lowcst poss ihlt:: loss of pressure or drag coefficienl. TIle condition s for il blade scction should allcmpt 10 approac h those for laminar flow over a flat plate. as this gives the lowcst possible drag coefficienl. But it is difficult to ach ie ve this in prac tice . because (I) blades must have the curvatufC to ch an ge the di rect ion of the fluid . introducinll
a pressure gradient and a tendency for flow separation, (2) hlndes mu st have a finite thickness from considerations of strcngth •.
The best co nditions arc a) th e~ ul ve a shape such that Ihe b) for the flow ta.bc..abolo!C-!· re-rr.::';;;':~;:J:!::::':;::'::';;';;:';;;;;;;;':;':;;:;;::~'40l1!1
-
36
;... TlilUiO M .-\CIIINl:.S
1.25. Th~ initial S \:lIC of air nowing through a compressor is p\ :; 1.02 bar, T1 ::;: 300 K. The c.x it pressure is "2.5 bar ilnd Ihe compressor crticiency is 75%. Determine the infinill.!simal efficiency o f the compressor. Comment on the deviation in the efficiency. [Ans: 78%] 1.26. The ove rall pressure ratio across 0 three stage gas turbine is 11 and its efficiency is 88% . If th e pressure raLio of each slage is the same and the inlet temperature is 1500 K delcrminc (a) pre ssure ratio of ench singe, (b) polytropic efficiency and (c) stage erficiency. IAns: (a) 2.22 (b) 83 .7% and (e) 85.2%J 1.3 and CIl/C u 1.4,cnlculate the polytropic 1.27. Fortm index of expansio n ofll effi c iency. IMV,OcI.'97] IAns.80.77 %J 1.2M. Air ente rs a compressor at {1 stalic condition of 150 kPa and 15°C ilnd a ve locity of 50 m/s. At theexitlhc static conditions arc 0.3 MPa and 100°C and u velocity of 100 m/s. E\'a luate (a) isen tropi c and nctual changes in e nth alpy and (b) total efficien cy. IN/V, Oct. '96] [A ns. (a) 63.39 kJlkg and 85.43 kJlkg (b) 75.29 %1 1.29. In a cen trifugal compressor the air is compressed \0 double the pressure. The inlet tempera ture is 27 °C and th e final temperature is J 07°C. Calc ulate the cfliciency of the compressor ami the power required 10 drive il if 30 kg/min of air is compressed. IMU, Apr. '98] [Ans. (a) 82 . 13% and (b) 40.2 kW) 1.30. A cent rifuga l compressor takes in air at 101 kPa and 25 °C Bnd compresses it through a pressure ratio of 3.5 : I. llH! index of compression" I .65 because of frictional healing. The mass flow rate of ai r handled by the compressor is 29 kg/so Find (a) overall efficie ncy of compressor and (b) power supplied by the motor with mechanical efficiency of 95% IAns. (a) 67.4% and (b) 5809.2 kW) 0 1.31 . Hot air enters a 3 slage turbine with tala I head properties of750 kPa and 900 e at the rale of 25 kg/so The final exit pressure is 105 kPa . The pressure at exil of I and II stages are 500 kPa nnd 250 kPa respective ly. TIle individual singe efficiencies for Ihe 3 s tages are each 75 % . Find (a) reheal faclor (b) overall efficien cy and (c) power developed . IAns. (a) 1.05 (b) 78.75 % and (c) 9980 kWI
=
2 BLADE THEORY
=
=
The energy transfer in lurbomachines is effected by changing the anguli.lr momentum of the fluid . T he cha nge in angular momentum is cause!d by the dynamic acti on of one! or more rotaling blade rows . llle dynamic actio n of Ihe rutating blade rows, sets up forcl.!s between the blade row and the fluid, while Iht! components of these forces in the di reclion of blade moti on g ive rise \0 the energy transfer be tween the blades nnd fluid . The th eory of compressor and turbine blades is discussed in this chapler.
AERO-FOIL SECTION An aero- foi l may be defined as u streamli ned form, hou nded principillly by two fl;.IIIened curves whose length and widlh are very large. relative to its · thickness. Acro-foi l is classified as symmetrical aero-foil and non -symmet rical aero-foil.
Symmetrical Aero-foil The acro-foil· whose axis of symmetry is parallel 10 the direction of undisturbed velocity of approach is called symmerrical aero-foil. The fl ow pattern aro und a symmetrical aero-foil placcd)n a Slream of gos is show n in Fig. 2.1. The now divides around lhe aero- foil al Ihe leading edge and then rejoins at the trailing edge. Though Ihere is some local disturbance. th ere is no permanent denection of the main stream. The forces excrled in this case arc only due to friction and the local disturbance .
Figure 2. J
Flow pauern around a symmrtricul uao-Joil
Non-symmetrical Aero-foil
-
If the nero-foil axis is inclined at an angle 'i', calh:d as Ihe ongle of attack, 10 the direction of the undisturbed approaching flow, then it is called;>.!; lIo/l-symmetrical
34 i> T URBO MACIIINES I . ~.
Tak ing log on either side. ij I'
(
0.4 ) TA
In (1/2)
In 0.865
:. rip
73.23%
Then
1'IT
=
1)3('·1.'1''''·') ( -2 1-
(DJ(1;l)
78.76%
=
(b) Power developed IV
= =
whcre I
8
25 x 1.005 x
873 [I -
m
07Jf~;"']
7738 kW
(c) Reheat factor ijT
R.F
= 0.7876
ij,
0.75
1.05
EXERCISES 1. 1. Define Turbo mnch inc. 1.2. I .J. 1.4. 1.5_
How are devit:cs pumping gases classified? What are sh ro uded nn d unshrouded IUrbo machines? Class ify turbo machines on the basi s of work transfer. Define the types of turbo machines based o n fluid movement th rough the ma chine. 1.6. Derivc the general Eulcr's c)(prcssion for a turbo mac hine. 1.7. Define the followin g efficiencies of power absorbing tu rbo machi nes. (a) Total -to-Tolal Efficiency (b) Static-to-StD.tic Efficiency
Define thl.! fnllnw ing i!fficicncil!.S o f pCl\vcr genl.!r..IIm g tu rbo mOlchincs. (a) Tota l-to -TOIal Efficic ncy tbl To tal -to-S lillic Efficiency 1.9. Draw th e h-s diagram with static ilnd stag nation slate s for the compressIOn and expansion processes for a gas. 1. 10. What is prehe::J.l factor in a multistage compressor? Pruve Ihat prch eat fa ctor IS less than unity. 1. 11 . Prove tllatthe cnmpressnr stage efficie ncy i~ greilterthan Ihe co mpressor overall efficiency. 1. 12. What is reheat ractor in a multistage turbine'! Prove th aI R.F is greatcr than unity. 1. 13. Prove thaI the turbine overall dfic iency is greater than the turbine stage efficie ncy. 1.14. De fine polytropic crficiency of n compressor. 1.15 . Derive the polytropic compressio n effi cie ncy through an infinitesimal compression stage . 1. 16. Definc polytropic cxpl.lnsion efTic iency. 1.17. Derive the polytropic expan sion efficiency through un infinite simal turblll!.: stage. 1. 1H. Derive the reheat fiJt: tor in term s of thl! stage and ovcrall pressure ratios. 1. 19. Air nows through a blower wherein its total pressure is increased hy 15 elll ur liqu id water. TIle inl et tOlnl pressure and the temperature o rlile air an: 1.05 har and J5 e C respect ively. TIle IOtal·to-total d ficicn cy is 70'70. Evaluatc (a) the c)( it total pressurc. (b) the exi l isentropic total temperature. and (c) the: isentro) it.: and actual clwngcs in tota l en th alpy. IMU-Apri/ '961 IAns. (al 1.065 bar. (b) 2R9.2 K. and (e) 1.206 kJlk g. 1.723 kJlkg l 1.20. A com prcssor has a lo tal-to-total cfTi cie ncy of 80% and an ovcralltotal pressure rLlli o of 5: I . CaJc: ulate tb c small s tage emciency of ::Ie co mprcsso r. IAns. 83.9%J 1. 2 1. Ai r fl ows through an air turbine where its stagnatio n pressure is decreased in the ratio 5: I. T he tolal · lO-tolnl effic ie ncy is 0.8 and the air now rate is 5 kg/s. If the total power output is 405 kW. Find (a) thc inlct to tal temperature, (b) tlte actual exi t totnl temperature. (c) the netu .. 1 exit staliC tcmper.1Iure if th e exi t now veloc ity is 100 m/s. and (a) the tota!·to·static cfficil.!ncy of the turbinc . IAns. (a) 273 K. (b) 192.4 K. (e) 187.4 K. and (d) 767< 1 1.22. A lu rbi ne hus a small stagc efficiency of 84% a n~ an ovc rallto ta! pressure r"till o f 4.5: I. Cilkulall.! the tCltai-to ·{Olal efficil.! ncy o f the turhine . I An.... 86.74% ) 1.23. A gas turbine is required to dcvelop 7360 kW wi th an air now rale uf 50 kg/s. Ir the turl:l inc inlet temperature and press ure arc I OOO"C and 8 bar respectively. Calculate the exi t temperaturc and pressure ir the isentropic dfi cie ncy o r the turbino is 90 %. IAns. (a) 1127 K. ilnd (h) 5 harl 1.24. A low pressure air compressor increases the air pressure by Ison mm W.G . If the initial and finiil comJ ilions or nir arc PI :: J .02 bar. TI = JOO K and Tl = 3 15 K. determine and compare the compressor and the infinitesimal stage cmeioneics. [Ans: (,,) 78% and (il) 78.8 " I
32 :;:. TUROO M AC' HINES
Solution P, _ -
P,
-= :'1
8 ASJCCONCEI'TSOI: TuKBOM.-.CiIiNES '"
33
(d) The rotal-ro-sraric efficiency '11/ = 0.8
til
= 5 kg/s
II' = 500 kW
TOI - To:! Ta l - T2J
=
(a) Inlet total temperature
c~ 2C p 0.63 1701 = 0.63 1 x 337 T02J' -
TU!r
2 12.65 K
T
TOI -
-
lOO:!
500 x IOJ 5 x 1005
IV
TO]
212.65 - 2 x 1105 20768 K 99.5 337 - 207.68 76.94%
mC p 99.5 K
fJ,-S
111C turbine tOlal-w-total efficiency is TOI -
To:!
TUI -
Ttl:!,
TOI
(PU:!) 7!
Example 1.14 In II thr~e s tage IUrbi ne thc prl!ssure rntio of e
POI
U) TOI -
:. TOI - T02, TO! -
0.631 TOI
'!:j TO!
= 0.63 1Tol
TO]
Solution N = 3 P, / P,
=2
q,
= 0.75
T,
= 600 + 273 = 873 K
(a) Overall efficiency
q" 99.5 0.8 337 K
=
TOI
(b) The actual exit total temperature TOI -
= =
To:!
.'. T02
Tm
=
99.5 TOI - 99.5 337 - 99.5 237.5 K
'11' is dctcnnined using 'Is expressio n.
.(i'})"'(';') -
1q,
=
(c) The actual exit static temperature
1-(i'}J~ P, 1-
c~ To"l- - -
•
0.75
=
2C"
100' 237.5 - 2 x 1005 232.5 K
P,
(!)"P
1-
G)"'(1cl)
=
0.865
H)
cr 2
"
m
= 25 kg /'
n AS I(' CONCf:1'TSOFTuRlJoM .... CHINES "" 31
30 }.> TURBO M .... CHIN ES
Solution N = 4
Example 1.12 Air nows through a blower where in its tala I pressure is increased
p,
(a) Overall pressure ratio
= ( ;; =
= 0.65
~,
--==1.2 P,
m = 45kg/s
2. 1
r
= (12)'
by 20 cm W.G . The inlet tolal pressure and tala I temperature of air are 1.04 har and I SoC respec ti \'ely. The total-I o-Iotal efficincy is 72% . Evaluale (a) the exit inial prcs!'ure and IOlallemperalure (b) iscntropic and actual changes in lotal enthalpy.
Solution
"'Po
70,
= 0 .2 m W.G .
= 273 +
Po,
= 1.04 bar
18 = 291 K
'Iu = 0.72
(a) Exit total pressure and temperature
(b) Overall efficiency
P2)N(' ~ "
( = P, -
~,.
-I
--'-''-'-':.....,~~-
"'Po P,
IO J x 9.81 x 0.2
=
= 0.01962+ 1.04
"'Po + P, 1.0596 bar
P2)N( if. ) -I ( -P, '1/1 is
= 1962 N/m'
obtained from the following equation:
(i,( -, P2) ;~J ( -P,
-,
0.'
0.65
'1tr
=
1.0596) - \'-l - 1] 29 1 [ ( 1.04
To! -
TOI
0.72
(1.2)'" - I
:::;:
293.16K
0.'
(1.2) ;;;;= - I
0.
(1.2) ;;;;=
1.0823
Taking log on both sides, we get
(b) Isentropic and actual changes in total enthalpy
"'''0
~p
0.6589 ( 1.2)'( Vi I _ I
:. '1c
=
ry,
=
"'' 0
C,,("'Tol
= Cp(T02 -
Tod
1.005(293. 16 - 291) 2. 171 kJ/kg
4.11,4
( 1.2) IIlIllr.TJ - I 62.28%
and isenlropic change in lolaI cnthalpy is 6110.1
(c) Power required
(6/rO) X 'III
2. 171 x 0.72
mCp(TN+ ' - T,l
IV
TN +I
,-,
=
T, (PN+') - - '" P, 293(2.1 l
0.' =""'"
404 K
W IV
=
45 x 1.005(404 - 293) 5019.98 kW
"'''0..
1.563 kJ / kg
Example 1.13 Air flows through nn air turbine where its stagnati on pressure is decreased in the ralio 5: I. The lotal-to-tolal efficiency is 0.8 and the air How rate is 5 kg/so Ir the total power output is 500 kW, find (a) inlet total temperature (h) the actual exit tala I temperature (e) the actual exi t SL1lic temperature if the flow velocily is 100 mls and (d) lhe (alal-'a-Slalic efficiency a fthe device. (MSU, Nov. '96(
2R ;;-
T UIHlO MACIIINES
BASICCONCEPTSOFTURuoMhCllINI:.s ,
TI - TN + 1 =
r}r(TJ
(A T)"p" r,,1f
(~) PN+ I
"rT,
=
0.88 x 1500(1-
=
654 .68 K
=
655 K
= =
'Ip
(
I',
=
7N + 1 = ~p
(I -T.V--..,, )
I' 'hI
=
=
(bj For the second stage
TN+!.r)
-
1-
T2 - U~ T)sroge
TJ
'.') --;-
(A>1i)
=
",UIl1 (T'73 ) ,,';.11 = (1281.67) 1063.34 =
2.183
1500 - (,55 = H45 K
--In -T,/(P,) - In - ( In (1500/845) In(ll) r r - I
TtH I
Stage efficiency
PN + I
~)
1-
0.4 0.837
=
655 3 = 218.33 T, -2 18.33= 1500 -2 18.33 1281.67 K
T4
=
T3 - (AT)s/ugl!
= =
1063.34 - 218.33 845K
Pressure ratio
,
Pressure ratio.
p,
P,
~
-
=
( r,
=
1.93
1- (2.:83) 11 85.2%
(cj For the third stage
""3
~) ~ptr-iJ
I)~ (2.183
=
'h.I
'1s.2
(AT)uperull
pi
1281.67 - 218.33 1063.34 K
Pressure ratio
(aj For the first stage
P,
I
~
= (~) /rJ'i'n"S" 1281.67
=
(i)"",·11 =
=
2.61
. ('~;34)-
Slage' efficiency
Slag\: effic iency
...".....,
I
1- ( 2.61
qel r -I)
,
l]s.J
(~:r-
Th.3
I-(~:) 'l.r.1
,
= 11-
1/1.1. 1
=
1.93
(-1.93I )11
84.95%
)--.:<
=
,
(I- ) ""Ys''''
=
--
29
=
1- ( - I 2.61 85.51%
)1:l
Example. 1.11 Each stage of a 4 stage air compressor delivering 45 kg of air per second operates at a pressure ratio of 1.2, with a stage efficiency of 65%. Calculate overall efficiency and pressure rJ.Lio. Calculate powerrcquircd to drive the compressor if air temperature at inlel is 20°C. [MU, April '96 & Oct. '97]
....., BASIC CONCEPTS OF TURBO MACHINES . . '27
26 ,.. TURBO MACHINES
Actual exit temperature is determined from the overall compressor efficiency
Solution
(: r e
T, = 1500 K
'IT = 0.88
2.2
=
1 2)
TN + ls - T,
q,.
1cJ
TN+I - TI
= 0.798
560.56 - 308
---0--:'---
0.798 x 1500 = 1197.45 K
I sex)
-
T, - T" =
O.SS( 1500 - 1197.45)
7~
1233.76 K
q"
and P,v+1 =8.16 x PI =8.16bar
(b) Polytropic efficiency
r ) In (T,/T,) ( r-I In(P,/P,)
~) ( 0.4
In( 15001 1233.76)
'1/i
(
=
0.4) _ 1_n.:...(8_.1_6.:...)_ ( 1.4 In(623.7/308)
Note that the P?lytropic efficiency is less than the turbine efficiency This is on . account of rehealJng.
Example 1 .9 An air . compressor has eight . stages of equal pressure ratio I 3 TI n le • o~v rate through th.e. compressor and ils overall efficiency arc 45 kg/s ilnd 80'o/~ respccuvcly. If the conditIOns of air at entry arc I bar and 35°C d . '1 I f ' ' etermlne (a) sa e 0 mr al compressor exit, (b) polytropic efficiency. and (c) efficiency of r.!rlch stage. IM KU - NOI' '98/
=
85'"
(c) Stage efficiency ,-
'1.~
=
p,
m = 45kg/s
'Ie
+ 35 =
(:;)'
-I
1'-1
(1.3)'11 _ I
= 0.8
< I~
308 K 'I>
(a) State of air at compressor exit Overall pressure ratio.
,
(~~)~-
Solution
PI = I bar TI = 273
~) In(PN + IIPtJ r In(TN+I ITtJ
In(2.2)
86.7 %
P; = I.3
+ 30S
623.7 K
7"" +1
-;-;15:;;0-;;'0':::-=-:1-;-19~7:"'.4~5
T,
=
0.8
1500- T,
T, - T2
=
t!X·
pression.
=
(1.3) "'"" - I 84.43%
Since the pressure ratio in each swge is the same, the stage efficiencies arc also the P2 ) ' _ R - ( 1.3) = 8. 16 ( PI
" P +,)';' ---p;=(8.16)" N
(
1.82 Ideal exit temperature
same.
Example 1.10 The overall pn!ssure ratio through a three stage gas turbine is 11 .0 and dficiency is 88 %. TIle temperature at inlet is 1500 K. Ir the temperature ri.r.c in .each stagt: is the sa me. determine ror cach stnge (a) pressure ratio alld (b) stage efficiency
Solution = =
1.82 x TI = 1.82 x 308 560.56 K
P ,vt l
P,
=
II
11r = 0.88
T, = 150n K
:!-I :.> TURBO MACHIN r:S
B ASICCONCEPTSOF TURHO MACItINES 0{
Example 1.6
A low pressu re air compressor develops a ressure of 1400 mm = 1.0 1 bar, TI = . , 2 = _0 K. detcfn ine comprlo! ssor LInd th e infinitesimal slage t!fficiencies. [MKU -April' 99 ]
~f (he initial and final slale s of air arc PI
Alternative method: <=.I
Solutio ·
( ::) , -I rye
1400mm W.G.
t:.P
1.0lbar, TI = 30SK,
=
IAxIOl ,9.B I
T,
2S
=
= )20K
PI
13734 Pa
ry"
0. 13734 bar
=
t' ,
( -P,
=
I
-I
(, . 14734) 11 1.01
_I
D.'
(,.14734) rr.;; _ I 1.01
75.67%
Example 1.7 A high pressure compressor changl!S the slate. of a~r from P,
= PI
PI = 1.01 bar, T\ ;::: 305 K to P2 = 3 bar. The compressor efficiency Determine the infinitesimal efficiency of the compressor.
= 1.01 +0.13734 = 1.14734 bur
+ t:.P
From isentropic rel at ion T" TI
T"
Solution P,
(::f
=
1.037 , TI
TI
= 305 K
P, = 3 bur
?,/ PI
3/1.01
[J
T"
= 1.037 ,
TI (;:) ';oJ = 305(2.97)
=
416.27 K T" - TI
T2 - T,
=
=
TI
=
3 16.2B - 305 320 - 305
75.23%
ryp
r- I
In (P,/ PJl
r
III (T,/TJl 1.4
=
In( 320/30S)
0.7S8B
= 4S3.36K
(' - I) In(?,/ PJl r In(T,/TJl
(OA) T:4
111(3/1.01)
111(453.36/ 305)
78.5%
N~tc thatlhe infinitesimal stage efficiency is greater than the compressor r.:fficit::ncy. This difference is due to preheating.
75.88 %
Since the pressure rise in the compre ssor is low, the two eac h ot her.
= =
!..c'....:...!.) In(1.1 4734/ 1.01)
416.27 - 305 0.75 14B.36
14B.36 + 305
T,
(b) Infinitesimal stage efficiency
(
'K
305
(a) Compr essor efficiency
Tl -
= 2.97 bar
=
316.2BSK
T1s - T]
75% .
JI . ~
(1.147 34) 1.01 1.037
= = = =
= 1.0 I bur
IS
cfficicncie.~
arc close to
Example 1.8 The pressure ratio across a gas turbine is 2.~ and. c~ficiency is 88%. The temperature of gas at inlet is 1500 K delcnnine polytropIc efhclcncy.
::!2
>-
BAS1CCONCEPTSOFTuRBOMACHINE.S ..:: 1;\
TURBO MACHINES
The pressure ratio is
Solution Pm =46 POI
Example 1.4 A power generating turbo machine dcvelops 100 kW output when thc now through the device is 0.1 m3 /s of oil having a density of 800 kg/m J . The total-to-total efficiency is 75 %. Evaluate (a) the change in tolal pressure of the oil, and (blthc change in static pressure of the oil if the inlet and exit flow velocities arc ~ and 10 mts respectively.
Solution
= 100 kW
Q = 0. 1 m'ls Mass now rate of oil m = pQ W
p = BOO kg/m)
~H = 0.75
N = 4 Pm/POI = 0.4 For the firsl two stages, rh = 0.86 For the last two stages. '7.1 = 0.84 (Refer Reheat Factor Section.)
'"
Po,)';' ( -" POI
1-
I'
=I-(OA)C'
0.23 Th!! actual work output rrom the first two slages is
= =
1V1.2
= BOO x 0.1 = BO kgls
CPf.LTol [I
+ (I
C p TOl (0.23) [I
- f.L~,)) x~,
+ (I
- (0.23 x 0 .B6) 1 x 0.86
0.356 Cp TOl
TIle change in total enlhalpy
-WI'" = lOO/BO -1.25 kJ/kg
The actual work output rrom the last two stagcs is
The iscntropic change in total enthalpy
(6."0,)
=
-1.25 0.75 -1.67kJ/kg 6."0 '1,_,
T03 -
6.P
=
=
0.644 TOl
=
Cp TOl (0.644)(0.23) [I + (I - (0.23)(0.84»J 0.B4
=
0.225Cp Tol
Total actual work output rrom the turbine
IV = IV1.2
(C' - C')
~OOOI
BOO( 10' - 3') -13.4 - --:::=-=:-' 2000 x 100 -13.4 - 0.364
=
lV,,,
800 x (-1.67) 100 - 13.4 bar
(b) The change in static pressure 6. Po-P
TOI(I - f.L~.)' = TOI(I - (0.23)(0.B6))'
To)
(a) The change in total pressure of the oil 6. Po = p(6."o,)
the temperature arter the end or the first two stages is
111t!
+ IV).,
= 0.5BI CpTOI
total isentropic work due to a single stage compression is CpTol [1-(1-Il)NJ
IV,
=
-13.Bbar
CpTOI [I - (I - 0.23)' J
0.649 CI' TOl
The negative sign implies that the pressure decreases during un expansion process.
Example 1.5 In a four stage turbine hnndling air, the stagnation preSsure rntio between the el(it and the inlet of ench Singe is 0.4. The stage efficiencies of the first two stages are 86% each. while those or the last two stages are 84% each. Find the overall efficiency of the turbine.
Overall turbine crficicncy
IV,IW 89.5%
= 0.581/0.649
...,
BASICCONCEPTSOFTURBOMACI llNES 4. 21 20 ,. TURllO M ACHINES
Solution
Solution Po, ; 7 bar To,; 830 K
= 1.5 bar r; 1.3
70, ; 1100 K C, ; 250m/s
PO'!
(a) Tota l-to-Tota l efficiency
6.110
"u:! -
To,
303 K
hOI
Po,; I bar
(a) Finding the type of turbo machine TO! - T02 TOl - T02,
Since the change in enthalpy is positive (6 kJlkg) Ihis turbo machine would bl! a work absorbing machine.
(b) Exit total temperature
Using isentropic re lation
For air as
11
perfect gas,
~ho =
C I' ~ To · 6.110
:. T02 - TOI
;
Therefore,
= =
1100-830 =0.B21 1100 - 771.1
Cp
+
To,
;
6 303 + 1.005
;
308.97 K or 35 .97°C
T"
;
C"
;
:. Cp
T"
;
;
'h-J
;
;
"01 - "02 hal - Ills T01 -
C;
C
p
(i) Ir the fluid is air
82.1% 1102, - hOI
;
11110
;
(c) To tal p ress ure ratio
tJ/-/
(b) Total-to-Static efficiency TJ,- .r
= 6 kJ/kg
;
"02 -
CI'TO,[(~r -I]
Tal - To:! Tal - T2,r
---, 2Cp rR ii 8.314 - - and R; - ; - - ; 0.2897kJ/kg - K r- I M 28.7 1.3 x 0.2897 ; 1.255 kJ /kg - K 0.3 250' 771.12 x 1255 746.19K
;
= ;
;
I
Example 1.3 Suppose a IUrbo machine is operated such that the change in total enthalpy is 6 kl/kg or fluid when the inlet total temperature is 30 D e and the inlet total pressure is I bar. (a) What general type of turbo machine would this be? (b) What is the exit tOlaltempcrature if the fluid is air? (c) What is the tolal pressure ratio across the machine ir [he adiahatic [olal·lo-lOlnl efficiency is 75% (i) if [he fluid is nir and (ii) if tile fluid is liquid wnler,
Cp(TOl - Tod + ry,_,(llll0l]'I'-'
[
CpTol
[I
" 0.75X6]'" 1.005 x 303
+
1.053
Oi) If lht! nuid is liquid water
1100 - 830 0 6 1100 - 746.19; .73 76.3 %
hOI
Cp(T02, -Tod Cp(To:! - Tal)
11 Po
; pllllo,
where 11110,
;
ry.-,(Illlo); 0.75 x 6
4.5 kJ/kg 11 Po
Po,
4.5 x 1000 45 bar 45
+ I = 46 bar
18 ;;:. TUilBO MACI~INES
BA SIC CONCEI'TSOF Tultoo MACI11 "' ES ...
The
= =
It}
For an isen tropic com pressio n,
+ To, + ... + TON+Jl Cp'hI'TOI [I + (I-I'q,) + (I -I'q,)' + .. . + ( I _I'~,)N-I]
OO/::"Po
Cpry.,I' ITOI
The (enns within the brackets are of th e fonn 1 + r +,2
+ ... + , 11 -
1;
=
1000(10-1) x 10' 0.9 kJ / kg
hich is a By definition
geometric series with common rali o,. TIle solution is I - ,11 . 1- r In this case, , ( I -ILTJJ)' Then, the equation reduces to
=
=
CpTOI
[I -
TJJ
L
6.110. 6.110
(I -llq,)N]
611o,
:. 6110
N
= ~(/::"Po)
I
0.9
% ;: 0.75
( 1.25) lVJ/
=
/-)
1.2 kJ/kg
nlis is the work of compression. The isentropic work output from a singJF e eltpansion from POI to PO(N+[ ) may - - be obta iiiCdTrom eqn. (1.25) by setting ,= I. IV,
(b) If the fluid is air as a perfect gas /::,.110,
CpTOI [I - ( I _I' )N ] CpTol
[I - R';]
( 1.26)
r-I TOl,
N
i=1
r
CpTol - 1 - (I -I'q,) N].
[I - ( I (
R.F=
;;: Tot ( Po')-rPOI 1.4 - I
=
L IV,,/W,
_li~,)N]
'h[ I-R~- I /']
- Toil
From isentropic relation
Wh~:: PO!,o.I+ ,)i~p-resents the overall pressure ratio. Since R.F
= C /'CTOl,
L [ CpTol
)
-
<=.1 J 1 - Ro'
283(10)
1.4
546. 4 K T herefore.
/::,.110,
1.005( 546.4 - 283)
=
264.70 kJ /kg
and
6.11 0 ,
SOLVED PROBLEMS
q"
Example 1.1 TIle inilial and
finaltolal pressures o f a fluid afe I bar and 10 bar respec ti ve ly. The initial total temperature is lODe. What is th e work of compression for adiabatic steady now wi lh a total-to-total efficiency of75 % if (a) Lhe fl uid is liquid water and (b) Ihe fluid is air as a perfect gas .
= =
264.70
0.75 352.94 kJ / kg
Hence, the adiabatic wo rk of com pression per kg of nuid is 352.94 kJ .
Solution POI
= 1 bar
POl
= 10 bar
TOI = 283 K.
q"
= 0.75
(a) If the fluid is liquid water S ince the fluid is in compressible
"0 = constant =
~,p =: density of water. p
Example 1.2 Gases from a combustion chamber enter a gas turbine al a 10lill pressure of7 bar and a (olai temperature o f 11 00 K. 1l1C tOlal pressure and lotallCIll' perature at the turbine ex-it arc 1.5 bar and 830 K. Take r =: 1.3 and molecular weight of gases = 28.7. Evaluate total-to- total efficiency and Ihe total-Io-static efficiency if lhe ex it ve locity is 250 mls. Assume adiabatic steady now. ,
16 }> TURBO M ACHINES
BAslrCoNCEt'TSOf"TuItBOMAl"·IIINF.5 ..:
For u turbine with N-slages, we have N
L
11'"
1:::: 1
- --
>
( 1.16)
Ws N
The ratio of
L
WSI
10
Wr
11
The reheat ractor for the expansion of a perfect gas in an N-slagc: turbine. assuming thut the stage efficienc ies '1.r and the pressure ratios Po, / Poi-t-I' where; = 1.2. · ··. n fo r all Ihe stages arc equal, is expressed in terms of stage pressure ratio as follows . For the first slagc,
is called the reheat faclOr (R.F).
i_I
so that ( 1.22)
F
L
IV"
; :: 1
( 1.17)
. = -I-V- >
-
Let
----
. of The milgnitude e cat filclor in mulLista e tll' . ~ ( 1.03 or 1.04. 1ft le stage efficiencies were the same, the totaillctual work outpu;'-rom ffie various individ ual stages would be ( 1.18)
Then
and
( 1.19) wbere '11 is the overall turbine efficiency. Combining equations (1.18) and (1.19), we get N
'It
L
IV"
i=1
- =--
TOI - T02
The actual work output lhat would be obtained from a single stage expansion is
or
,-
1JJTolll
( 1.23)
For th c..sccond stnge
r
T02 - TO), = T021 - (POJ/ Pm)
""] r
( 1.20)
but
PO] PQ2 -=-
From eqn . (1.16), we find that
P02
'11> '/.1 That is. the QvcraJlturbine e fficiency '11 is greater than ule turbine siage efficiencies
POI
TIlcrcforc
lis .
Combining eqns. (I.I?) and (1 .20), we have the following relation .
c(Rz;::;J
and (1.21)
Consider again /. Fig.( 1.7) for a first stage cxp.lIlsion. It is seen thai the final stale -02 may be obtained after lin ideal expansion from 01 - 02s followed by a 'reheating' of the fluid from state 02.r to 02 at constant pressure (TCh. > T02,) Hence.lhe fluid at 02 has a grcateravailahi lity than the fluid at 02.r(hol > 1102,) . An cxpnnsion from state 02 to a lower pressure must necessarily resuh in a larger output IIHIO that obtainable from stale 02.r. 111is effect thai makes RF> I is called the reheat
t:JJect .
....
T02 - T03
= '1.r T021l
or (1.24 )
For the Nth stage
14
»
BASIC"CONC£I'TSOFTURDoMACHINES 4: 15 TUIlDQ MACHINES
4. Finite Stage Efficiency The stage efficiency, considering static vuJue of temperature and pressure (Fig . 1.6.), is defined as
to a rinal state'· 02 which has higher entropy than th at or state Ob. The corresponding exit conditions ror the second stage are 03s and 03 respectively. . . The isentropic work done by the two stage turbine is the sum oflhe stage IsentrOpI C worb . Pm
h
01 \
Pm
\
The ,stage efficiency Clln now be expressed in terms of polytropic efficiency
,
\ 0'
02,
\
"- \
POJ
"03
and OJ, 03 ss
Therefore.
1-
c:) "';-"
Figure 1.7
,
q, = --'-'--'-''-,-..,.-
l-(~f
The same equation can be used to determine the overall efficiency of a multistage turbine. ex.cept that the stage pressure ratio is replaced by the overall pressure ratio. The overall efficiency, for n N·stage turbine with a constant stage pressure ratio, can ~e expressed as .
PN+I = (P2)N P,
P,
Rtlltar tffecr in a muirisragt rurbint:
I: W"
+ W.I!
=
Hlsi
:;::
(hOI - holt)
; :=1
(1.14)
+ (1102
- hOJs)
Consider now a turbine in which the fluid expands rrom POI to PO) in one smgc. TI,e iscnLropic work is (1.15)
In both the cases, the actual work done is W = hOI - 110J
'1,
=
The constant pressure line diverges in the II - s diagram as the entropy increases. Therefore, the isentropic enlha.lpy drop across a stage incrca.c;cs for a. constant stag· nation pressure drop fi Po across each stage. Consequently, [he s~m o~ the stage isentropic enthalpy drops is greater than the isentropic enthalpy drop to a stog ie stage expansion. TIlal is
REHEAT FACTOR IN TURBINES Consider a turbine with two stages where the nuid (perfect gas) expands from POI to POJ as shown on the II - s diagram. Fig. 1.7. Slate-OJ is the initial condition 01 the enlly of the first stage and 02s is the condition that would be reached at the first stage exit if the expansion process had been isentropic, The actual expansion leads
or
BASICCONCEPTS OfTURDOMACllINES "" 13 I ~ ;... T URDO MACHINES
2. Total-to.statlc Efficiency
But dTs
II is an efficiency in which til e ideal work is based on stagnation property at inlet and static property at outlet.
dT := 1'/ 1'
T
(~)
and co nstan t;:
P
r
lhen
dT
r-I
T=1I p ' r
3. Polytropic Efficiency stages. Then
(0
(1.11 )
Integratin g between lht: limits of the overall expansion between PI and f2
A turbine stage can be cons idered as made up of aTL infinite nu mber of s mnJI or infin ites-
imal
dP
P
in an ' infinitesimal turbine
accoun t for expn nsiu rl
In(i,) ;
q"
r~ I
ln (;;)
Rearrangi ng. P,
h
T2 ; Tt
\
p
\
(I. 12)
Pt
Assuming the irreversible adiabatic expansion (1-2) as equivalent process with index n, the temperature and pressure are related by
P-dP
If!'S
(P,)"'(~) (0
a poJyLTopic
\
"'
(1. 13) \
P, 2
Equating eqns. (1.12) nnd (1.13) ,
Fjgur~
1.6
E:rpllllsiun process in infin itesimal and fillite fur-bint
sfflg~s
Comparing the powers,
siage . it small singe o r infinites imal stage or po lytropi c e ffi ciency is defined . Consider a small stage (Fig. 1.6) between press ures P and P -dp. The effici ency of th is turbi ne stage is d efin ed as
qp
_~ -r- n (r-I)
or r
(I. 9)
For 'In isentropi c process
qp
n-i I) . -n-
Alternati vely. the index of ex pans}on in the actual process is expressed as T - (.-, - ) ~ Con"ant
r
(1.10)
11
P '
Differentiatin g eq n. ( J. I 0). we get
dT,
= (r -
~ Const ant [p-
When 1'/ p
= -....,----",--r - (r - 1)qp
. . 'de£ ilh = I . 11 == r. The actual ex pansion of process curve (1- 2) comel W
th e isentropic expansion line (I - 2s).
10
)0
BASlcCoNCErrSOFTuRBoMACHINES
TuRBO MACHINES
where 'Ie is the overall compressor efficiency and Ws is the isentropic work.
= "°3
W.r
11
-
1t0l
TIle actual work input is the same for both si ngle stage and multistage compression processes.
IV = "OJ - hOI
«
11
Consider again Fig. (1.4) for a first stage compression, state 02 may be obtained after nn ideal compression from 01 to 02s followed by "prcheatin" of the fluid from state 021 to 02 al constant pressure (To; > To2.s). Thi s inh erent thermodynamic elTect that reduces the efficiency ofa multistage compressor is caJled the preheat effect.
EFFICI ENCIES OF TURB INES
Then from equ's 1.7 and J .7a, The enthalpy-entropy diagram for now both reversible and irreversib le th rough N
L
'Is
IV"
1=1
- = -- 'Ie
a turbine is shown in Fig. 1.5 . TIle static condition of the fluid at inlet is dcwnnincd ~y stale I. with s tate 01. as the corresponding stagnation state. ( I.H)
Ws
The final slatic properties arc dctcnnincd by the s tate 2, with 02. as the
Since the constant pressure lines diverge in the direction of increasing entropy on h-s diagram, lhe isentropic enthalpy risc across each stage increases even [or a conslnnl stagnation pressure rise 6. Po across each stage. TIlen, the sum of the stage isentropic enthalpy rises is greatcr than the isentropic enthalpy rise in asing le siage compression. For a two slage compressor (h02, - hOI)
+ (ho),
'0' 01
"
- h02) > (h03" - hOI)
POl
i.c. E;=I Wfi > Ws
02
For N stages, P, N
LIV"
1=1
IV,
2
> I
2S
Equation (1.8) can be wrillen as
IV, = =:-;;;--'--= P.F
---==--=:...- 2:[:1 Wrj
N
That is, the ratio of Ws to
L
( 1.8")
W Si is known as the Preheat factor (P.F)
"';""~ ' ~~~ (
P.F. =
IV,
N < I LI=I IV"
The preneat facto r is less than unity. Then, equation (1.8a) becomes
'1e < I ~,
Figure 1.5
Reversible and irreversible I!xpan.1ion proCt!ssn
corresponding stagnation s tate . If the proccss were reversible, the final fl uid stat ic Slate would be 2f and the s tagnation state wou ld be 02s . Process I - 2 is lhe actual expa nsion process and process 1 - 2.t' is the isentropl(; or idcnl ex pa ns ion process. In turbin es, the efficiencies may be defined using either the static or the stagnation properties of the fluid or even a co mbination of both . The commonly used turbine efficiencies are
1. Total-to-Total Efficiency It is an efficiency based on stagnation properties at inlet and outle!.
or
i.e., the overall compressor efficiency '1c is less than the compressor stage erficiencies
'h ·
'7t - r =
hOI - "02 hOI -
hOb
--
8
;... TURBO MACHINES
BASIC CONCEPTS Of TURBO MACHINES ....: 9
4. Finite Stage Efficiency A slagt! with a tinitlo! prl!ssure drop is a finite stage. Taking stalic v
=
PREHEAT FACTOR IN COMPRESSORS Consider a two stage compressor working between POI and PO] an shown in Fig. 1.4 . Tn isentropic now, the outlet conditions of the gas for the first stage and second stage arc at 02s and 03 s respectivdy, where as the actual oullct
The finite stage or sInge efficiency can be expressed in terms of the smLll1 stage or polytropic efficiency 03 s S
i.Jnd
Therefore,
01
Figure 1.4
For a multistage compressor with a constant stage pressure ratio, (he overall pressure ral io is given by
where N is tbe number of stages and Therefore, the overall efficiency is
PN -H
,
is the pressure al the end of the N,h stage.
( 1.6)
W = l/Tls I:Ws;
,
L
;=1
Wsi is the isentropic work input to the two stage compressor and is the sum
i=1
'h
P,)N(:.i;!) ( -PI
of the stage isentropic works. For n compressor with
=
PN+I) ';' ( -PI,
- I
stages.
w=
1
N
-L:WSi
( 1.7)
TIs i=]
Consider now, a single stage compressor raising the fluid pressure from POI The actual work input [hal would be supplied is
[0 PO].
I
PN+I),," - I ( -PI
'/I'
-I
The nveTliIl efficiency in lerms of overall prCSSUfl! ratio is
-
cOlllpr~ssor
conditions are at 02 and 03. Corresponding to first stage and second stage respectively. The subscript's' refers to constant entropy and '0' refers to total conditions of the fluid. If the stage efficiencies were the same, the total actual work input to the different individual stages would be
where
rIc
Preheal effecl in a nJullislage
(1.78)
6
;...
B,.\SICCONCEI'TSOFTURBOMACHINES "
TUReo M,.\CHINES
7
Differentiating equation (1.2) and substituting eq uation (1.1). we get
1. Total-to-Total Efficiency II is an efficiency based on stagnation properties at cnuy and exit. Widtnl
hOlJ - 1101
WnClunl
1102 - hOI
'1'-1 = - -- =
dT· 'I" = [,
~
1 x P-(i)]d P x constant
Constant va lue is obtained from equation (1.2). Therefore.
2. Static-to-Static Efficiency ( 1.3)
It is an efficiency based on stalic properties at entry and exit Integrati ng hetween the limit s of {he full compression from In
3. Polytropic Efficiency A compressor slage can be viewed as made up of an infinite number of sma ll stages. To accounl for a compression in an infinitesimal stage, polytropic efficiency is defined for an elemental compression process, Consider a small compressor stage as shown in Fig. 1.3 between pressures P and P+dP. The polytropic effic iency of a compressor stage is defined as
dT,
(I.I )
ryp = dT
we get
(T,Tl) = -,-, -I. 1·In (Pl) PI
TJ"
Rearranging. ( 1.4)
If the Irreversible adIabatic compresslO proccs is-assumed 10 be equivalent 10 a ~roJ;ess wJt po Ylr0P'C Indcx, n, lhe following rclaliansh~p between tempera urc and pressure wilJCXist.
--
( 1.5)
P,
h
PliO P"!
Equating eqns (104) and (1.5). !.=..I
(~:t = C;)
P+dP I
I
'"
.
Comparing the power,
P
r-I
I
dT,
~
'7{1 . r
P,
=
II-I II
or r -
1
II
'1P = - , - . ;;-=I s Figur~
J.J
COlllpr~ssion
prouss in
infinit~simal
and finilt! compressor stages
For an isentropic process, the relationsh ip between pressure and temperature is given by
~ =CQnstant
( 1.2)
{TIlc poJyLTapic efficiency is also ca ed as small stage or infinitesimal stage efficiency 1 ". A I ital val ue of palYlro ic efficiency for a compressor i{O.8 and in the initial T]J were 'll IS the SInge e IClen!;),. design calculallon It is often nssumed lha '1 AlternatIve y, po ytropic index of compressIOn 10 lhe actual process is
4
,.
BAsrCCONCEPTSOFTURlloMACHIN1.;5
~ 5
TUlWO MACHINES
Thus Si nce wr
Tds
=U
where U is the impeller tangentiul velocity
:. 1IV = m(U,Cx2 - VIC.,,) I This equation is known as the general form of Euler's equation. Euler's turhine equation is
Iw ::;: m(U,C.r l -
U2C.r2) >
01
Euler's pump equation is
= dh -
udP
The second Tds equation is extensively used in the study of compressible machines. In terms of stagnation propenies
1I0w
Tds = dho - uodPo
For un incompressible fluid unde!Soing an .isentropic process (Le. ds ::;: 0) as in funs, the ideal change in stagnation enthalpy IS
(I(t.hol, = "ot.Po =
t.~I'I
Since v'" = l/ po and Po = P
rl\~v-=--m~(U~,-C~x-'--~U~I~C~x~d->~Ol
EFFICIENCIES OF COMPRESSORS 4. The Second Law of Thermodynamics The second law of thermodynamics leads to the definition of Entropy. and is defined as 1 8 Q",=TdS
I
nlro chan is caused b ealtransfcr. ~~ and rrevcrsibl li~The entropy change during a process IS pasl Ive or an Irreversible process or zero for a rcvl!rsible process. TIlUS, work producing devices such as turbines, deliver marc work, and work cons uming devices such as pumps and compressors consume less work when they operate reversibly. The differential form of the conservation of energy equation for a closed stationary sys,em (.1 fixed mass) can be expressed for a reversible process as 8Q,tlJ - 8W,tll
Fig.I .2 shows the reversible and irreversible adiablltie compression processes on the e nthnlpy-enLr0py diagram. The initial condition of lhe Huid is represented by state-I . The stagnation poin~ corresponding to this stale is 01. The final condition of the Huid is denoted by state-2 and the corresponding stagnation point is 02. If the process were reversible, the final fluid static and stagnation conditions would be 2S and 025 respectively.
= dU
BtH
6Q,t!u OW'l'lI
= TdS = PdV
Thus TdS
=dU + PdV
(or) on unil mass ba sis
Figure 1.2
ITds = all + Pdu I Th is e'lu.ation is known as lhe First Tds equation or Gibb's equation. The second TrIoi' equation is ohtained by eliminating rill from the first Tds equation by using the definition of enthalpy (II = Jl + Pv)
.-
(lh =dll + Pdu+udP
Rel'ersible and irrel'l~rsible compression procesus
Process 1-2 is the actual compression process and is accompanied by an increase in entropy. Process 1-2S is the ideal compression process. The efficiencies of compressors may be defined in terms of either stagnation or stalic properties of the fluid or even a combination of barh . The following afe Ihe commonly used compressor efficiencies:
2
}>
TURBO MACHINF.5 BAS!CCONCEI'TSOFTURllOMACIUNES '"
The turbo machines can also be classified by the energy transfer from or to thc rotuting blades, which arc fixed onlo a shaft. In Ihc work absorbing machines the fluid pressure (or) head,., {in the case of hydraulic machines) (or) the enthalpy (for compressible flow machines) increases from inlelto outlet. But in work delivering mnchinc.o; the fluid pressure or enthalpy, decreases from the inlet to the outlct. TIle product change in head or enthalpy, nnd the mass now rale of the fluid through the machine, represents the energy absorbed by (or) extracted from the rolating blades. In turbo machines, the energy transfer is accomplished by changing the angula"r momentum of the fluid and so the shapes of the blades and rotating members differ from one type to another. Turbo machines can also be c1assifed based on the direction of flow of fluid across the rotating member. If the flow is axial, the machine is called an axial flow machine. If the flow is only radial, it is known as radial flow or celltrifugal machille. If the flow is partly axial and partly radinl, the machine is known as mixedflow machine.
or
IQ -
W -111[6.11
+ 6ke + 6peJ I
~bis equation is known as steady flow energy equation(SFEE}. A turbo machine, bClng o.pcrated essentially under the sam!.! conditions for long periods of time, can be convcnlcnlly analysed us a steady now device. This equation, when applied to a turho machine, may be simplified pertaining to !.he type of turbo machine, bcc;mse many of the terms arc zero (or) get cancelled with others.
3. The Newton's Second Law of Motion Accord~ng to this law, the sum of all the forces acting on a controlled volume in a particular direction is equal to the rate of change of momentum of the fluid across the controlled volume in the same direction.
BASIC LAWS AND GOVERNING EQUATIONS The basic laws of thermodynamics and fluid mechanics are used in turbo machines. The important laws and governing ~quations used in turbo machi.nes are as follows:
1. The Principle of Conservation of Mass The conservation of mass is one of the most fundamental principles in nature. Mass, like energy is a conserved property, and it cannot be created or destroyed. The conservation of mass principle for a controlled volume undergoing a steady flow process. requires that the ma.,c;s flow rate(m) across the controlled volume remains constant. Mathematically,
-'"''''*~fF ~
Figure 1,1
where . subscripts I and 2 denote the inlet and outlet conditions respectively. The conservation of mass equation is onen referred to as the continuity equntion in fluid _mechanics. In compressible flow machines, the mass flow rate (kg/s) is e,r;cIusively used while in hydraulic machines the volume flow rate (m] Is) is preferred. .
2. The First Law of Thermodynamics The first law of thermodynamics which is also known as the conservation of energy principle s.lates that energy can neither be created nor destroyed; it can only change from one fonn to another. TIle conservation of energy equation for a general steady now system can be expressed verbally ns (HealtransferredJ - [Shaft work]:::; (Mass now rote) [(Change in enthalpy per unit mass) + (Change in kinetic energy per unit mass) + (Change in potential energy per unit mass)]
J
--
- >-- - - f
MOllellunt ojfluid particle across Q controlled \'o/umt'!
In turbo machines, thc impellers arc rotating and the power output IS expressed a~ the p.roduct of torque and angUlar velocity and so angular momentum !S Iht! pnmc parameter. Consider a fluid particle moving across a controlle~ vol.urne as shown in Fig 1.1. The Fluid particle travels from point A to point B whde sImultaneously moving from a radius r! to radius r2. If C... I and C... 2 arc components of abliolute velocities in the tangential direction, then the sum of alllhc torques acting on the system is equal to the rate of change of angular momentum Mathematically, .
If the machine revolves with angular velocity w, then the power (W) is
xvi ,
T UIWO
MACIIINF..s.
Efficiency of Draft Tube Cavitation in Turbines
Solved Problems 406 Short Questions 435 Exercises 436
10. Power Transmitting Turbo Machines Inrroduction 11,e Hydraulic Coupling Working Principle Efficiency Slip Torque Converter Characteristics of Fluid Coupling & Convener
£\'ercises
4-1 7
Appendix 449 References -160 Index .f61
405 405
441 441 441 442 442 443 444
BASIC CONCEPTS OF TURBO MACHINES
~45
DEFINITION 'ur a machine is definc ClS cner y or imparts cne y. to a i one or more rotalin ntinuou win stream of fluid b the d nam'c blade rows. The prefix '(uebo'lis a latin word meaning' pin or '~hir ) implying Lhat turbo machines rotate in some way. Ir.lhe machine adds energy to the fluid. it is commonly called a pump. If it eXlracts energy. Ihen it is called a turbine . A device which pumps Jiguids is simply called a urn if it u s ases' then three different terms rna be used de odin u 0 the ressure . e chiev d. nearl .<1?:,.b!,f u e ris (e device is called a (13£1., betwet;n.. . 7 and bsolu il is called a 1) owe and J' ve 3 bar absolute pressure it is called a '(Ompressor) The difference between the turbo machine and the positive displncement machine is that in the fonner, (he fluid is moving continuously across the machine unlike in the latter. where the fluid enlers a.closed chamber, which is isolated from the inlet and outlct sections of the mnchine for a very short period of time within which work is done on or by Ihe fluid .
CLASSIFICATION OF TURBO MACHINES
-
P' mem er IS cnc as rou In sue a way t at I e working fluid cpnnP! be djvened 10 flow around the edges of the impeller, it is caJ cd a shrouded turbo machine. Examples of this are turbines, pumps, elc. If the fluid flows around the edges of ihe impeller which is nOI shrouded, then it is called an unshrouded turbo machine. Examples of this ar~r e:ro-generator and ( arrC:rart propej!elj) - -- Turbomachines may rail into anyone of the two classes depending on whether work is done by the fluid on the rOlating member (examples: hydraulic turbine. gas turbine. elc.) or work is done by the rolming member on the fluid (examples: pump. compressor. etc.).
xiv r
TUllDo
M..\CIIINE.'i
Main Parts of a Centrifugal POOlP . Impeller Casing Suction Pipe and Delivery Pipe Work done and Velocity Triangles Head Developed . SIalic Head (H,) Manometric Head (H ) Pwnp Losses and Effici;ncies Impeller Power Loss (Disc Friclion Power Loss (P oJ) Leakage Powcr Loss Casing Power Loss Mechanical Efficiency Manometric (or) Hydraulic Efficiency ,. Overall Efficiency Volumetric Efficiency Impeller Efficiency MinimOOl Starting Speed Nel Positive Suction Head Priming Performance Curves of Centrifugal Pumps Constant Efficiency or Muschel Curves Mullistage Centrifugal Pumps Purnps Combined in Parallel Pumps Combined in Series Pumps in Series Pumps in Parallel . Axial Flow Pumps Description Velocity Triangles The Work done on tile Fluid The Energy Transfer (or) Head Axial Pump Characteristics Cavitation Method of Pre venting Cavitation Positive Displacement Pumps Gear Pumps Vane Pump Piston Pump Miscellaneous Types of Pumps Jel Pump
315 315 326 327 327 330 330 331 331 331 331 332 332 332 333 333 333 333 334 335 335 336 337 337 , 339 340 340 341 341 341 342 343 343 344 346 346 347 349 350 351 351
•I
Air Lift Pump Submersible Pump Solved Problems 354 SharI QllesTions 379 Exercises 380 9. Hydraulic Turbines lntroduction . Classification ofHydrauJic Turbines According 10 Ihe Type of Energy al tile Inlel According to the Direction of Flow through the Runner According 10 the Head al tile Inlet of Turbine According 10 the Specific Speed of tile Turbine Pelton Wheel . Parts of the Pelton Turbine Velocity Triangles and Work Done for Pelton Wheel Hydraulic Efficiency Pelton Wheel Losses and Efficiencies Pipe. Line Transmission Efficienc), Nozzle Efficiency Nozzle Velocity Coefficient Number of Buckets Overall Efficiency Governing of Pelton Wheel Turbine Double Governing of Pelton Wheel Characteristics of an Impulse Turbine Radial Flow Reaction Turbine Main Pans of a Radial Flow Reaction Turbine Velocity Triangles and Work done Net Head Across Reaction Turbine Radial Flow Turbine Losses Governing ofRcaction Turbines Characteristic Curves for a Reaction Turbine Axial Flow Reaction Turbine Velocity Triangles and Work Done Governing of Kaplan Turbine Comparison of Hydraulic Turbine Efficiencies Selction of Hydraulic Turbines Draft Tube Types of Draft Tubes Draft Tube Theory
352 353
3B3
383 383 383 383 384 384 384 384 385 387 387 387 387 388 388 388 388 389 390 391 391 392 393 394 395 396 397 399 400
401 401 402 403 403
CUNIl:NI)
xii
,
, xiii
TUltBO MAC'lIlNES
7, Dimension a l and Model Analysis
Stator (Nozzle) and Rotar Losses Total-to-TOlal Efticiency (~,,)
186 186
Total-ta-Static Efticiency (~u) Blade Loading Coefficient Blade Types Impulse Blading Compounding (or) Staging
186 188 189 190 191 191 192 194 195 197 197 199 200 201 202 205 207 207 207 208 210
Two Stage Pressure Compounded Impulse Turbine Two Stage Velocity Compounded Impulse Turbine
Velocity Triangles of the Two Slage Impulse Turbine Diagram Efficiency of a Two Stage Curtis Turbine
Reaction Blading The Reaction Turbine
Velocity Triangle Diagram for Reac tion Turbine Stage Swge Efficiency ora Reaclion Turbine
Maximwn Stage Efficiency Reaction Ratio
Stagc Efficiency, Turbine Efficiency and Reheat Factor Free Vortex Design Constant Nozzle Angle Stage Impulse Turbines Versus Reaction Turbines Internal Losses in Turbines
Governing of Turbines Solved Problems 211 Short Quest;ons 147 Exercises 2-19
6, Radial Flow Gas and Steam Turbines Introduction Description
Velocity Diagrams The Thennodynamics of Flow Stage Efficiencies Spouting Velocity Degree of Reaction
An Ourward Flow Radial Turbine (LJungstrom Turbine) Velocity Triangles and Stage Work Stage Work Solved Problems 266 Short Questions 282 Exercises 183
•
Introduction Fundamental Dimensions Derived Dimensions
Advantages of Dimensional Analysis Dimensional Homogeneity
Dimensional Analysis Method Buckingham PI Theorem Method of Selecting Repealing Variables Model Analysis Similirude Geometric Similarity Kinematic Similarity Dynamic Similarity Classification of Hydraulic Models Undistarted Models Distorted Models Non-dimensional Nwnbers Reynold's Nwnber (Re) Euler's Nwnber (Eu) Froude's Nwnber (F,) Weber's Number (W,l Mach's Number (M) Specific Speed Pump Specific Speed Expression for Pwnp Specific Speed Specific Speed of Turbine Deri vation of the Turbine Specific Speed Model Testing of Hydraulic Turbo Machines
255 255 · 255 256 257 259 ',60 261 263 264 264
Unit Quantities
Unit Speed Unit Discharge Unit Power
Use of Unit Quantities Solved Problems 305 Short Questions 321 £urcises 323
'J
287
287 287 287 288 289 290 290 291 291 292
292 293 293 294 294 294 295 295 295 295 296 296 297 297 297 298 299 301 303 303 304 304 305
8, Hydraulic P umps
325
Centrifugal Pwnps Introduction
325 325
-, x ,.
TURDO MAClnNE."
Cmm::NT~
Cascaded Testing and Curves
46
Cascade Lift and Drag Coe!ficients , Cascade Lift and Drag Coefficients Turbine Blades Losses in a Cascade Solved Problems 53 Short Questions 61 Exercises 62
49 52 52
fo',
3. Centrifugal Compressors and Fans Introduction Components and Description Velocity Diagrams for a Centrifugal Compressor Slip Factor Energy Transfer Power Input Factor-
Mollier Chart Inlet Casing Impeller Stage Pressure Rise and Loading Coefficient Pressure Coefficient
Diffuser Volute or Scroll Collector Vaneless Diffuser Vaned Diffuser Degree of Reaction EfTect of Impeller Blade Shape on Performance . Pre-whirl and Inlet Guide Vanes Inlet Velocity Limitations Mach Number in TIle Diffuser Centrifugal Compressor Characteristic Surging Rotating Stall Choking Characteristic Curve Solved Problems 93 SharI Queslions Il7 Exercises Il9 4. Axial Flow Compressors and Fans Introduction Advantages of Axial Flow Compressors Description of an Axial Flow Compressor
•
Working Principle Velocity Triangles for an Axial Flow Compressor Stage Energy Transfer or Stage Work
Mollier Chart Work Done Factor Stage Loading (or) Pressure Coefficient Reaction Ratio Effect of Reaction Ratio on the Velocity Triangles Static Pressure Rise Stage Efficiency
63 63 63 65
Pressure Coefficient
Compressor Stall and Surge Radial Equilibrium Method Radial Equilibriwn TI1eory Types of Blades
66
69 69 70 70 71 73 75 76 77 . 77 78 80 81 84 85 88 88 89 89
Free Vortex Blade Forced Vortex or Solid Rotation Blades Constant Reaction Blade
Multistage Compression Comparison Between Centrifugal Compressor and Axial Flow Compressor
Direction of Flow Across the Compressor Pressure Rise per Stage Isotropic Efficiency Range of Operation Frontal Area Working Fluid Starting Torque
Construction Multistaging Applications Characteristic Curve Solved Problems 145 Shorf Questions 172 Exercises 174
90
92
5. Axial Flow Steam and Gas Turbines 125
125 125 125
"
Introduction Description Velocity Triangles for an Axial FlowTurbines Siage Work and Diagram Efficiency h-s Diagram for and Axial Flow Turbine
.... xi
126 127 128 128 130 131 132
133 134 135 136 136 137 138 139 139
140 141 141 143 143 143 143 143 144 144 144 144 144 144 144
181 181 181 182 183 184
-
"iii ,..
TUROO tvl Ac HINE."i
This book will be useful to tCtlchers and studen ts of Mechanical Enginel!ring. candidates of AMTE. competiti ve examinations like UPSC, -TNPSC and GATE and prac ticing engineers ..
•
CONTENTS
I \\'elcomc comments and suggesti ons from readers.
The materi al in this book is based on the concepts already developed over the years by various authors. which can be read ily gathered fTom the list of references given at the end of the book. lowe my gratitude to all of them.
Pre/ace
J acknowledge with appreciation the encouragement and suggestions given
vii
1. Basic Concepts ofThrbo Machines
hy my colleagues as we ll as my students.
Definition Classification of Turbo Machines Basic Laws and Governing Equations The Principle of Conservation of Mass The First Law of Thermodynamics The Newton ' 5 Second Law of Motion The Second Law ofThermodynumics Efficiencies of Compressors Total-to-Total Efficiency Static-to-Static Efficiency Polytropic Efuciency Finite Stage Efficiency Preheat Factor in Compressors Efficiencies ofTurbines Tolal-to-Total Efficiency Total-to-Static Efficiency Polytropic Efficiency Finite Stage Efficiency Reheat Factor in Turbines
I am grateful to my family members for their moral support and sustained encourage ment throughout the preparation of this book. I place on record my sincere gratitude to my co ll ege Chairman Dr. Tm!. Radha Thiagarajan, th e Co rrespondent Thiru. Karumuthu T. Kannan and the Ptincipal Dr. Y. A bhaikumar for their support. I sincere ly thank Mr. P.K. Madhava n, Vikas Publishing House Pl't Ltd, fo r publishing uli s book. [ am indebted to my studen ts Mr. H. Prnsa nna and Mr. A. Pradeep Chakkaravarthy for their help in proof reading.
A Valan Arasu
Solved Problems Exercises 34
2 2 2 3 4
5 6 6 6 8 9 II II 12 12 14 14
18
2. Blade Theory
"
1
Aero-foil Section Symmetrical Aero-foil Non Symmetrical Aero-foil Drag and Lift Lift and Drag Coefficients Energy Transfer in Tenns of Lift and Drag Coefficients Blade Terminology Cascade Nomenclature Turbine Cascade Nomenclature
37 37
37 37 38 39 40 42 43
45
,I.
PREFACE
The objective of this book is to cover comprehensively the basics of turbo
machines. Turbo machines has been an essential subject in the engineering curriculum and is often perceived as a tough subject mainly due to the unavailability of standard textbooks. I have made an attempt here to provide adequate material for the study of both compressible and incompressible flow turbo machines. The book adopts the 51 system of units throughout . It include s nu-
merous solved and unsolved problems in each chapter to assist students in understanding the subject. Chapter I introduces the basics of turbo machines. The Euler's equation has been derived in this chapter. Chapter 2 deals with the geometries ofbladcs and blnde rows of different types of turbo machines. The blowers and fans arc
low-pressure compressible flow machines and are principally the same as the compressors. Therefore, these machines are combined with compressors. Centrifugal compressors, blowers and fans are discussed in chaptcr J . Axial flow compressors and fans are described in chapter 4. In the salllc chapter. multi-stage compressors are also covered . I\:
6 deals with inward and outward flow radial turbines. Velocity triangles and enthalpy - entropy diagrams have been frequently used to explain the thermo-
dynamic aspects of these compressible flow machines.
"
Chapter 7 includes dimensional and model analysis of turbo machincs and types of similarities. A brief discussion on non-dimensional numbers is prcsented here. Centrifugal and axial flow pumps are discussed in detail in chapter 8, It also includes the study of positive displacement pumps. Hydraulic turbines such as Pelton, Francis and Kaplan turbines are discussed in chapter 9. Governing of
hydraulic turbines is also included here. Power transmiaing turbo machines and their characteristics are discussed in detail in chapter 10.
The reference direction in velocity triangles for centrifugal machines is thc tangential direction and that ofaxial machines is the axial direction. This is the convention adopted here. Computer software has been developed for somc
selected problems in turbo machines and is given in the Appendix.
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Turbo Machines
,I
A Valan Arasu Department of Mechanical Engineering Thiugurajar Collcgt: of Engineering Madurai
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