Centrifugal Compressors
• Classes and comparisons between compressors
Function
Centrifugal
Axial
Engine type
Small engine
Large engine
Mass flow rate
< 15 kg/s
Very large (> 100 kg/s)
Efficiency
Low 86-87 %
High 94 %
# of stages
small
large
Pressure ratio per stage
High (5-7)
Low (<1.5)
Pressure loss
High for more than
one stage
Low, thus allow using
many stages
Fixing and
manufacturing
easy
Not easy
Cost
Cheap, wider
operating range
Very expensive
Centrifugal Compressors
Principle of Operation
Centrifugal compressors consist of stationary casing
containing
a. Rotating impeller (imparts a high velocity of air),
b. Fixed diverging passage (The air is decelerated with rise
in static pressure).
c. Impeller may be single or double-sided
Centrifugal Compressors
•
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Air is sucked into the impeller eye and whirled at high
speed by the vanes of the impeller disc.
The static pressure increases from eye to tip.
Remainder of static pressure rise occurs in diffusers.
Normally half of pressure rise occurs in the impeller
and 50% in diffuser.
Some stagnation pressure loss occurs.
Centrifugal Compressors
Centrifugal Compressors
• Work done and Pressure Rise:
• Absolute velocity of air at impeller tip.
• tangential or whirl component
• radial component.
C2
C w2
Cr 2
•  is the angle given by the direction of
the relative velocity at inlet V1. Also this
is the angle of leading edge of the vane
with tangential direction.
• Slip phenomenon: air trapped between
the impeller vanes does not move with
the impeller, thus air acquire whirl (Cw)
velocity at the tip which is less than u.
• : At ideal conditions, Cr 2  U (impeller tip speed )
Centrifugal Compressors
Centrifugal Compressors
• Velocity diagrams
Centrifugal Compressors
C w2
Slip factor  
;  1
U
0.63 
  1
; ( experiment s : by stanitz);
n
n  number of vanes ( blades)
• Considering unit
mass of air:
• momentum
equation
T  torque  C w 2 r2  C w1 r1 ;
Work  T  C w 2 r2 - 0.0
(for ideal case of no guide vanes)
Utilizing slip factor , thus,
Work  U 2
• Defining a power input factor, 
(due to losses in energy as frictional loss)
, thus,
Work    U 2
Centrifugal Compressors
Energy balance
c p (To 3  To1 )  U 2
Where (To 3  To1 ) : stagnation temperatu re
rise across the compressor
=
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With state 1 as inlet to rotor
“
2 as exit from rotor
“
3 as exit of diffuser
No energy addition in diffuser
Thus
(To3  To1 )  (T0 2  To1 )
Centrifugal Compressors
• Defining c as overall isentropic efficiency, then
overall stagnation pressure ratio is given by :
To3  To1
'
c 
To3  To1
 To
  3'
Po1  To1
Po3
'


  1 To   c (To  To )   1
3
1
  1


T


o1




 c (To3  To1 )   1   cu 2   1
 1 
 1 


To1
c p To1 



 c presents both less ( frictional ) in rotor and diffuser ;  : less (friction) in rotor .
both are limiting work capacity in compressor
 : a factor limiting work capacity of compressor ;  can be increased by
increasing number of vanes, thus
Centrifugal Compressors
• Example 4.1
• The following data are suggested as a basis for
the design of a single-sided centrifugal
compressor:
• Power input factor = =1.04
• Slip factor  = 0.9
• Rotational speed, N= 290 rev/s
• Overall diameter of impeller, D=0.5m
• Eye tip diameter=2re=De=0.3m
• Eye root diameter, D1=2r1=0.15m
• Air mass flow, m=9 kg/s
• Inlet stagnation temperature To1= 295
• Inlet stagnation pressure Po1 = 1.1 bar
• Isentropic efficiency, c=0.78
Centrifugal Compressors
• Requirements are
• (a) to determine the pressure ratio of the
compressor and the power required to drive
it assuming that the velocity of the air at inlet
is axial.
• (b) to calculate the inlet angle of the impeller
vanes at the root and tip of the radii of the
eyes, assuming that the axial inlet velocity is
constant across the eye annulus; and
• (c) to estimate the axial depth of the impeller
channels at the periphery of the impeller.
Centrifugal Compressors
• (a) impeller tip speed
U   r2  2 *  * N * r2  DN
U    0.5  290  455.5m / s
• Temperature equivalent of the work done on unit mass flow of air, is
To3  To1 
U 2
cp
1.04  0.9  455.5 2

 193K
3
1.005  10

p o3
p o1
  c (To3  To1 )   1  0.78  193  3.5
 1 
  4.23
  1 
To1
295 



Centrifugal Compressors
• Power required=
.
m c p (To3  To1 )  9  1.005  193  1746kW
(b) to find the inlet angle it is necessary to determine the
inlet velocity which in this case is axial;
.
i. e. C a1  C1
C a1 1 must satisfy th e continuity equation m  1 A1C a1
where A1 is the flow area at inlet.
Since the density 1 depends upon C1and
both are unknown, a trial and error process is
required.
Centrifugal Compressors
• Flow triangles
• u2=455.5 m/s
u1h   r1h  136.5m / s,
u1t  r1t  273 m / s
• Assume axial flow
• two unknown (,c) in one

2
2

m


C
A


C
d

d
equation but another relation
1 1 1
1 1
1t
1h
4
is given by

2
P1
C1
1 
and To1  T1 
RT1
2c p
Assume 1 and get C1

then
2
c1
then get T1  To1 
2c p
p1  T1   1

and, thus, calculate 1


p o1  To1 

Centrifugal Compressors
• Note this is normal to design for an axial velocity of
about 150 m/s, this providing a suitable compromise
between high flow per unit frontal area and frictional
losses in the intake.
•
Annulus area of impeller eye,
A1 
 (0.32  0.15 2 )
4
 0.053m 2
Based on stagnation conditions:
1   o 
1
po1
RTo1

1.1  100
 1.30kg / m 3
0.287  295
Centrifugal Compressors
m
9
C1  C a1 

 131m /
1 A1 1.30  0.053
Since C1  C a1
, the equivalent dynamic temperature is
2
C1
1312
1.312


 8.5 K
3
2c p
0.201
2  1.005  10
2
T1  To1
p1 
C
 1  295  8.5  286.5 K
2c p
p o1

(To1 / T1 )  1
1 

1.1
295 / 286.5
3.5
 0.992
p1
0.992  100

 1.21kg / m 3
RT1
0.287  286.5
Centrifugal Compressors
checkCa1 :
m
9
C a1 

 140m / s
1 A1 1.21  0.053
final trial :
try C a1  C1 = 145 m/s
equivalent dynamics temperature is
2
C1
145 2
1.45 2


 10.5K
3
2c p 2  1.005  10
0.201
Centrifugal Compressors
2
C
T1  To1  1  295  10.5  284.5 K
2c p
p o1
p1 

(To1 / T1 )
 1

1. 1
295 / 284.5
3 .5
 0.968
p1
0.968  100
1 

 1.185kg / m 3
RT1 0.287  284.5
checkCa1 :
C a1 
m
9

 143m / s
1 A1 1.85  0.053
Centrifugal Compressors
• This is a good agreement and a further trial
using Ca1=143 m/s is unnecessary because a
small change in C has little effect upon .
• For this reason, it is more accurate to use the
final value 143 m/s, rather than the mean of 145
m/s ( the trial value) and 143 m/s.
• The vane angles can now be calculated as
follows:
The peripheral speed, U e , at the impeller eye tip radius 
 re  2  N re   De N    0.3  290  273m / s
and at eye root radius =136.5 m/s,
Centrifugal Compressors
•  at root=tan-1(143/136.5)=46.33,
•  at tip =tan-1143/273=27.65
(c) the shape of the impeller channel between eye and
tip is very much a matter of trial and error.
The aim is to obtain as uniform a change of flow velocity
up the channel as possible, avoiding local decelerations
up the trailing face of the vane.
To estimate the density at the impeller tip, the static
pressure and temperature are found by calculating the
absolute velocity at this and using it in conjunction with
the stagnation pressure which is calculated from the
assumed loss up to this point.
Centrifugal Compressors
Making the choice C r2  Ca1 , thus
Cw2   U  0.9  455.5  410m / s
Cr2  Cw2
2
C2 
2
2c p
2
1.43  4.1

 93.8K
0.201
2
2
m
A
2Cr2
To get  2 , we need to get P2
c  0.78,
loss  0.22,
1/ 2 loss  0.11
the loss in the impeller  0.5(1  c )  0.11
 x , rotor  0.89
Centrifugal Compressors
p o2
p o1
 0.89  193 
 1 

295 

3.5
 1.582 3.5

p o2
p o1
  imp (To3  To1   1
 1 

To1


To calculate density at exit
Centrifugal Compressors
2
C2

2c p
C r  C 2
2
2
2c p
C 2  u
C r2  C a1 , assume
2
C2
thusT2  To2 
 T2
2c p
togetP2

p o2
po1
'
 To '   1
To2  To1
2 


& c 
 P2
 To 
To2  To1
 1 
thus get 2.
Centrifugal Compressors
p
2
 
/ po2  T2 / To2

3.5
but To2  To3  193  295  488K
2
C2
T2  To2 
 488  93.8  394.2 K , therefore,
2c p
p 2  T2 

p o2  To2 

 1
 394.2 


 488 
 p2
p2
(
)
 po
p o2
 2
3.5
sin ce
 p o2

 p o
 1

, get p 2 as

p o1

p2
p
394.2 

 p 2  p o1 2  1.532 

p o1
p o1
488


3.5
but p o1  1.1, p 2  2.35  1.1  2.58bar
p2
2.58  100
2 

 2.28kg / m 3
RT 2
0.287  394.2
 2.35
Centrifugal Compressors
The required area of cross-section of flow in the
radial direction at the impeller tip is
A
b
m
9

 0.0276 m 2
 2 C r2 2.28  143
A
0.0276

 0.0176m or 1.76 cm
D   0.5
Computational Design of a Centrifugal Compressor
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PROGRAM MAIN
COMMON CP,R,GAMRAT
COMMON VECT(5000,500)
C
C
C
OPEN(30,FILE='D:\Dif\GRIDG.RES
)'
OPEN(5,FILE='C:\CALCULATIONS\Data_PyT10_6.1mps_D50mmFdn.txt )'
OPEN(6,FILE='C:\CALCULATIONS\OUT.txt
)'
OPEN(7,FILE='C:\CALCULATIONS\output data for drawings.txt )'
OPEN(8,FILE='C:\CALCULATIONS\OUT2.txt)'
C
OPEN(30,FILE='C:\Dif\GRIDG.RES
C
OPEN(6,FILE='C:\Dif\Conv 1\GRIDG.OUT
C OPEN(5,FILE='C:\dif\Conv 1\GRIDG.DAT
C OPEN(30,FILE='C:\Dif\Conv
1\GRIDG.RES',FORM='UNFORMATTED
)'
C
OPEN(6,FILE='C:\Dif\GRIDG.OUT
C
OPEN(5,FILE='C:\dif\GRIDG.DAT
C
OPEN(6,FILE='D:\Dif\GRIDG.OUT
C
OPEN(5,FILE='D:\dif\GRIDG.DAT )'
)'
)'
)'
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Computational Design of a Centrifugal Compressor
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PI=22./7.
EPSI=1.05
SIGMA=0.9
RPM=305.
D0=0.6
DIT=0.4
DIR=0.15
FLOW=14
TO1=300
PO1=100.
EFFC=0.8
CP=1005
EFFIMP=0.89
GAMMA=1.4
R=0.287
GAMRAT=GAMMA/(GAMMA-1).
U=PI*D0*RPM
TO13=EPSI*SIGMA*U*U/CP
PO13=(1.+EFFC*TO13/TO1)**GAMRAT
TO3=TO1+TO13
TO2=TO3
PO3=PO1*PO13
POWER=FLOW*CP*TO13/1000.
WRITE(6,11)POWER,TO13,U,PO13
11 FORMAT(2X,'POWER=',E13.4,/2X,'TO13=',E13.5/2X,'U=',E13.5/3X,
'1 Press ratio=',E13.4)//
AI=PI*(DIT**2-DIR**2)/4.
Computational Design of a Centrifugal Compressor
C
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C1=100.
CALL SITER(C1,TO1,PO1,AI,FLOW)
C
WRITE(6,12)C1,EPS,P1,T1,AI
C 12 FORMAT(2X,E13.3/4E13.4)
UE=PI*DIT*RPM
UR=PI*DIR*RPM
ALFAR=ATAN(C1/UR)*180./PI
ALFAT=ATAN(C1/UE)*180./PI
WRITE(6,24)
24 FORMAT(8X,'ALFAT,
ALFAR)/'
WRITE(6,13)ALFAT,ALFAR
13 FORMAT(2X,2E13.3)
C
C
Axial Depth
CR=C1
CW=SIGMA*U
CSQ=CR*CR+CW*CW
PO2=PO1*(1.+EFFIMP*TO13/TO1)**GAMRAT
T2=TO2-CSQ/(2.*CP)
P2=PO2*(T2/TO2)**GAMRAT
RHO2=P2/(R*T2)
A2=FLOW/(RHO2*CR)
AXDEPTH=A2/(PI*D0)
WRITE(6,17)AXDEPTH
17 FORMAT(//10X,'Axial Depth= ', 10X, E13.5)
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Computational Design of a Centrifugal Compressor
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C
CALL PERFORMANCE(POWER,TO1,PO1,EFFC,GAMRAT,CP)
STOP
END
C
SUBROUTINE SITER(C,TO,PO,A1,FLOW)
COMMON CP,R,GAMRAT
C
WRITE(6,102)C,EPS,PO,TO,A1
RHO1=PO/(R*TO)
10 C=FLOW/(RHO1*A1)
T=TO-C*C/(2.*CP)
P=PO*(T/TO)**GAMRAT
C
23 FORMAT(7X,'C',18x,'EPS',8X,'P',8X,'T',15X,'A1)/'
C
WRITE(6,102)C,EPS,P,T,A1
RHONEW=P/(R*T)
EPS=ABS((RHONEW-RHO1))/RHONEW
IF(EPS.LT.0.001)GO TO 20
RHO1=RHONEW
GO TO 10
20 CONTINUE
WRITE(6,23)
WRITE(6,102)C,EPS,P,T,A1
102 FORMAT(2X,5E13.4)/
Return
End
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Computational Design of a Centrifugal Compressor
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SUBROUTINE PERFORMANCE(POWER,TO1,PO1,EFFC,GAMRAT,CP)
COMMON VECT(5000,500),WMAS(5000,500),BETA(5000,500),PI
FLOW=10.
DFLOW=FLOW/10.
WRITE(6,30)POWER,TO1,PO1,EFFC,GAMRAT,CP
30
FORMAT(6E13.3)
DO 10 I=1,9
TO3=TO1+POWER*1000./FLOW/CP
PO3=PO1*(1.+EFFC*(TO3-TO1)/TO1)**GAMRAT
FLOW=FLOW-DFLOW
C
WRITE(6,20)TO3,PO3
WRITE(6,20)FLOW,PO3/PO1
20
FORMAT(2E13.3)
10 CONTINUE
C
RETURN
END
Centrifugal Compressors
• The Diffuser:
• In the case of gas turbine, the air should exit the
diffuser and enters the combustion chamber at
minimum velocity.
• Thus, design of diffuser requires that only a small
part of strengthening temperature is K.E. normally
u=90m/s at exit of the compressor.
• rapid divergence is not recommended
• optimum angle is 7.0.
• Neglecting losses, thus, angular momentum r
C=constant
• Cr: radial velocity will also decrease.
Centrifugal Compressors
Example 4.2
Consider the design of a diffuser for the compressor
dealt with in the previous example. The following
additional data will be assumed:
Radial width of vaneless space wd =
5 cm
Approximate mean radius of diffuser throat, rm =0.033m
Depth of diffuser passages dd
1.76
Number of diffuser vanes nv
12
Required are (a) the inlet angle of the diffuser vanes and
(b) the throat width of the diffuser passages which are
assumed to be of constant depth
(a)Consider conditions at the radius of the diffuser
vane leading edges, at r2=0.25+0.05=0.3m. Since
in the vaneless space r Cw =constant for constant
angular momentum,
Centrifugal Compressors
C w2
0.25
 410 
 342m / s
0.30
The radial component of velocity can be found by trial
and error. The iteration may be started by assuming
that the temperature equivalent of the resultant
velocity is that corresponding to the whirl velocity, but
only the final trial is given here .
Cw  Cr2
C2
Try Cr2  97 m/s, thus,

2c p
2c p
2
2
2
Centrifugal Compressors
• Ignoring any additional loss between the impeller tip
and diffuser vane leading edges at 0.3m radius, the
stagnation pressure will be that calculated for the
impeller tip, namely it will be that given by
Po 2 / Po1  (1.582)3.5
2
C2
T2  To2 
, T2  488  62.9  425.1K
2c p
3.5
p 2  425.1 
p2 
425.1 

 1.582 
 ,

p o2  488 
p o1 
488 
p 2  3.07  1.1  3.38bar ,  2 
3.5
 3.07
3.38  100
 2.77 kg / m 3
0.287  425.1
Centrifugal Compressors
• Area of cross-section of flow
in radial
• Check on Cr2:
2 *  * 0.3 * 0.0176
 0.0332m 2
• Cr2=Taking Cr as 97.9 m/s,
the angle of the diffuser vane
leading edge for zero
incidence should be
 2  tan 1 (Cr 2 / Cw2 )  tan 1 (97.9 / 342)  16o
Centrifugal Compressors
•
the throat width of the diffuser channels may
be found by a similar calculation for the flow at
the assumed throat radius of 0.33m.
0.25
Cw2  410 
 311m / s
0.33
Try Cr2= 83 m/s
C2
3.112  0.83 2

 51.5K , T2  488  51.5  436.5K
2c p
0.201
2
p2 
436.5 
 1.582 

p o1 
488 
3.5
 3.37, p 2  3.37  1.1  3.71bar
3.71  100
2 
 2.96kg / m 3
0.287  436.5
Centrifugal Compressors
• Area in radial direction=A (radial) = 2Db =0.0365
Get
C r2
m9
(check ) C r2 
 C r2  83.3
Aradi  2
 ( direction of flow)  tan
-1
(
Ath  Ar sin   0.0945 m 2
C r2
C 2
Ath  n * b ( width of throat )
 width  4.4cm
)  15
0
Centrifugal Compressors
Compressibility Effects
• At the impeller inlet,( eye of the impeller), the relative
velocity is high and could be very close to sound values.
M 1  V1t / RT1  308/338  0.91.
No problem at sea level conditions, however at
high altitude ( aircraft engine), speed of sound
decreases and we might have supersonic flow.
For example at 11000 m, T=217 K
M 1  V1t / RT  1.06  1.0 supersonic
Centrifugal Compressors
• we try to avoid this by having guide vanes and it is better
to be variable in the case of change of conditions, such
as altitude.
• By trial and error, the value of Ca can be determined
from Ca and , C1t 9and C1t can be determined. Then
value V1t9can be determined which is smaller.=239 m/s.
M
239
RT
 0.82
For this design, the flow is subsonic at
altitude.
C a  150m / s
Trying
1
Centrifugal Compressors
• For 30 pre whirl
• C1=150/cos30=173.2
T1  T0 
C1
2
2c p
 280.1, p1  0.918bar ,   1.14kg / m
9
check on, C a1 
 149
1.148 * 0.053
vel.C1  149 tan 30  86m / s
v1t 
149 2  273  56 
M 
2
 239
239
1.4  0.287 * 280 *1020
 0.7
Centrifugal Compressors
• In spite of the advantage, it has a disadvantage of
reducing the pressure ratio of compressor.
Po3
Po1

 1   c To13 / T1  1 , where


T013   u 2  C1 u c / c p
u c  u average  (u1h  u1t ) / 2
C a1 has value which wil l lead to reduction of T013
and hence reduction in pressure ratio.
Centrifugal Compressors
In this example
po3
po3
po1
po1
 4.23( without guide vanes)
 3.79with guide vanes
for details see
text book
Centrifugal Compressors
• Vaneless diffusers:
• For vaneless diffuser, no problem, it can handle
supersonic flow while vaned diffuser can’t.
• At the exit of the vaneless diffuser, C3=355,
M2=0.56<1.0, which is subsonic and is ok for vaned
diffuser.
• Advantages of vane less diffuser:
– Mach number M2 could be supersonic without
– Vaneless space will eliminate any non-uniformity of
the flow coming out of the impeller ( jets and wakes).
– This is good to avoid any problem in exciting the
vanes.
– As a normal practice, no. of vanes in the diffuser is
less than impeller blades.
• N (vanes)<N (impeller)
Centrifugal Compressors
• Non-dimensional quantities for compressor
characteristics:
• D=diameter,
N=rpm,
m=mass flow rate
• po1=inlet pressure,
po2=exit pressure
• T01=inlet temperature, To2=exit temperature
• N=no. of variables
• M=basic dimensions
• there are 7 variables, 3basic dimensions (M,L,T)
• and  terms 7-3=4.
m RTo1 ND
Po 2 / Po1 , To 2 / To1 ,
,
2
D Po1
RTo1
For same compressor
m To1 N
,
Po1
To1
Centrifugal Compressors
 Stall
• Defined as the (aerodynamic stall) or the breakaway of the flow from the suction side of the blades.
• A multi-staged compressor may operate safely with
one or more stages stalled and the rest of the
stages unstalled . but performance is not optimum.
Due to higher losses when the stall is formed.
 Surge
• Is a special fluctuation of mass flow rate in and out
of the engine. No running under this condition.
• Surge is associated with a sudden drop in delivery
pressure and with violent aerodynamic pulsation
which is transmitted throughout the whole machine.
Centrifugal Compressors
Centrifugal Compressors