Turbo Machines
.I
A Valan Arasu Department of Mechanical Engineering Thiagarajar Collcgt: of Engineering Madurai
Scanned by: A. Ansari
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ISBN 81·259-08-40-1
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"
AU rights ICSIH\lcd. No part 01 IhIS. public 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; xI< 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
..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 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!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
-
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
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.
C ,t l
=0
E
= UC.
11
/1!.
E in lenns of angle fi] is obtained as
Outlet gu ide vnne
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
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
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.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!.!.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 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.
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+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,
=
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 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!. 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.
~ 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 accomp1/2
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(\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 )( :: . 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 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
(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 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
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.'
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 - /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,
. •.•• [ 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 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
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 •..!; 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
(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: -
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
( 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- 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