1 basic compressor principles
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Centrifugal Compressors
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Adjustable diffuser vanes Adjustable inlet guide vanes
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Pressure, temperature and velocity
relationship in a centrifugal compressor
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Centrifugal Compressors• Work done and Pressure Rise:
• Absolute velocity of air at impeller tip.• tangential or whirl component• radial component.
• is the angle given by the direction of therelative velocity at inlet V1. Also this is the
angle of leading edge of the vane withtangential direction.
• Slip phenomenon: air trapped between theimpeller vanes does not move with theimpeller, thus air acquire whirl (Cx) velocityat the tip which is less than u.
• :
2C
2 xC
2r C
) ( , 2 speed tipimpeller U C conditionsideal At x
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Centrifugal Compressors
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Centrifugal Compressors
• Velocity diagrams
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Centrifugal Compressors
• Considering unitmass of air:
• momentumequation
blades)(vanesof number n
stanitz); by:sexperiment(;n
0.631
1 ;U
Cfactor Slip x2
2
22
1122
UWork
thus,,factor slipUtilizing
vanes)guidenoof caseideal(for
0.0-TWork
;
r C
r C r C torqueT
x
x x
2UWork
thus,,
loss)frictionalasenergyinlossesto(due
factor,input power aDefining•
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Centrifugal Compressors
• With state 1 as inlet to rotor
• “ 2 as exit from rotor
• “ 3 as exit of diffuser
•
No energy addition in diffuser• Thus
)(13 oo T T
=
)(120 oT T
compressor theacrossrise
retemperatustagnation :)( Where
)(
balanceEnergy
13
2
13
oo
oo p
T T
U T T c
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Centrifugal Compressors• Defining c as overall isentropic efficiency, then overall
stagnation pressure ratio is given by :
13
13
'
oo
oo
cT T
T T
11
'
'
1
131
1
3
1
3)(
o
ooco
o
o
o
o
T
T T T
T
T
P
P
121
11
13 1)(
1
o p
c
o
ooc
T c
u
T
T T
compressor of capacitywork limitingfactor a:
compressor incapacitywork limitingare both
.rotor in(friction)less:;diffuser androtor in)frictional(less both presents
c
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Centrifugal Compressors
• The Diffuser:
• In the case of gas turbine, the air should exit the diffuser
and enters the combustion chamber at minimumvelocity.
• 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 rC=constant
• Cr: radial velocity will also decrease.
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Centrifugal CompressorsCompressibility Effects• At the impeller inlet,( eye of the impeller), the relative
velocity is high and could be very close to sound values.
0.91.308/338/ 111 RT V M t
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
supersonic1.01.06/11 RT V M t
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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 asaltitude.
• By trial and error, the value of Ca can be determined from Ca
and , C1 and C1 can be determined. Then value V1 can be
determined which is smaller =239 m/s.
82.0239
RT
M
For this design, the flow is subsonic at altitude.
Trying smC a /1501
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Centrifugal Compressors
• For 30 pre whirl
• C1=150/cos30=173.2
mkg bar p
c
C T T
p
/14.1,918.0 ,1.280
2
1
2
101
239
56273149
/8630tan149.
149053.0*148.1
9 on,check
22
1
1
1
t
x
a
v
smC vel
C
7.01020*280*287.04.1
239
M
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Centrifugal Compressors
• In spite of the advantage, it has a disadvantage of reducing the pressure ratio of compressor.
2/)(
/
,/1
11
2
013
1113
1
1
3
t haveragec
pc x
oc
o
o
uuuu
cuC uT
whereT T P P
ratio. pressureinreductionhenceand
of reductiontoleadlwhich wilvaluehasC130a1 T
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Centrifugal Compressors
vanes)guidewithout(23.4exampleIn this
1
3 o
o
p
p
vanesguidewith79.3
1
3 o
o
p
p
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Centrifugal Compressors
• Vaneless diffusers:
• For vaneless diffuser, no problem, it can handle supersonicflow 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 lessthan impeller blades.
• N (vanes)<N (impeller)
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Centrifugal Compressors
• Non-dimensional quantities for compressorcharacteristics:
• 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.
11
1
11
2
1
1212
,
,,/,/
oo
o
oo
o
oooo
T
N
P
T mcompressor same For
RT
ND
P D
RT m
T T P P
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Characteristic Pressure
versus Flow Plot
• Ideal characteristic pressure
versus flow plot for the
theoretical compressor is a
straight line that slopesdownward to the right
• Ideal characteristic plot is
affected by various energy
losses in a real compressor
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• Pressure versus flow curve for
a real or actual compressor
(ideal curve minus the energy
losses)• The dotted line shows energy
losses.
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• Actual curve is not usableover the entire range of zero-tomaximum flow.
• Useful part of the actual
pressure curve is between theleft and right limit areas.(pressure in this middle areawill decrease as flowincreases in a predictable and
stable fashion)• Unusable area on the left is
where surge occurs and theunusable area on the right iswhere choke occurs.
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Centrifugal Compressors Stall
• Defined as the (aerodynamic stall) or the break-away of the flow from the suction side of the blades.
• A multi-staged compressor may operate safely with oneor more stages stalled and the rest of the stagesunstalled . 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 theengine. No running under this condition.
• Surge is associated with a sudden drop in deliverypressure and with violent aerodynamic pulsation which istransmitted throughout the whole machine.
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Example
C if l C
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Centrifugal Compressors• Example 4.1
• The following data are suggested as a basis for thedesign 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
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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.
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Centrifugal Compressors
• (a) impeller tip speed
smU /5.4552905.0 • Temperature equivalent of the work done on unit mass flow of air, is
K c
U
T T p
oo 19310005.1
5.4559.004.13
22
13
23.4295
19378.01
)(1
5.31
1
13
1
3
o
ooc
o
o
T
T T
p
p
DNr * N**2 22 r U
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Centrifugal Compressors
• Power required=kW T T cm oo p 1746193005.19)( 13
.
(b) to find the inlet angle it is necessary to determine the inlet
velocity which in this case is axial;
11 e.. C C i a
.
inlet.atareaflowtheisA1where
mequationcontinuityesatisfy thmust 11111 aa C AC
Since the density 1 depends upon C1and both
are unknown, a trial and error process is
required.
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Centrifugal Compressors
• Flow triangles•
u2=455.5 m/s
• Assume axial flow• two unknown (,c) in one
equation but another relation isgiven by 2
1
2
111111
4
ht d d C AC m
p
oc
C T T and
RT
P
2
2
11
1
11 1
1
111
2
1 111
calculate thus,and,
2 getthenCgetand
11
1
oo
p
o
T
T
p
pthen
ccT T Assume
smr u
smr u
t t
hh
/273
,/5.136
11
11
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Centrifugal Compressors
• Note this is normal to design for an axial velocity of about 150m/s, this providing a suitable compromise between high flowper unit frontal area and frictional losses in the intake.
• Annulus area of impeller eye,
222
1 053.04
)15.03.0( m A
Based on stagnation conditions:
3
1 /30.1295287.0
1001.1
1
1
1mkg
RT
p
o
o
o
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Centrifugal Compressors
/131053.030.1
911
1 1m
AmC C a
11 aC C Since , the equivalent dynamic temperature is
3
1
11
5.3
1
1
1
2
11
2
3
22
1
/21.15.286287.0
100992.0
992.05.286/295
1.1
)/(
5.2865.82952
5.8201.0
31.1
10005.12
131
2
1
1
1
mkg RT
p
T T
p p
K c
C T T
K c
C
o
o
p
o
p
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Centrifugal Compressors
:
/140053.021.1
9
:
111
1
trial f inal
sm A
mC
checkC
a
a
m/s145= try 11C C a
equivalent dynamics temperature is
K c
C
p
5.10201.0
45.1
10005.12
145
2
2
3
22
1
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Centrifugal Compressors
sm A
mC
checkC
mkg RT
p
T T
p p
K c
C T T
a
a
o
o
p
o
/143053.085.1
9
:
/185.15.284287.0
100968.0
968.0
5.284/295
1.1
)/(
5.2845.102952
11
3
1
11
5.3
11
1
2
11
1
1
1
1
1
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Centrifugal Compressors
• This is a good agreement and a further trial usingCa1=143 m/s is unnecessary because a small changein C has little effect upon .
• For this reason, it is more accurate to use the finalvalue 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:
sm N Dr N The
ee /2732903.02 r radiustipeyeimpeller at the ,Uspeed, peripheral
e
e
and at eye root radius =136.5 m/s,
Centrifugal Compressors
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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 absolutevelocity at this and using it in conjunction with the stagnation
pressure which is calculated from the assumed loss up to this
point.
Centrifugal Compressors
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Centrifugal Compressors
thusC a ,CchoicetheMaking 1r2
2
2
2 2 2 222
2
2 2
0.9 455.5 410 /
1.43 4.193.8
2 0.201
w
r w
p
C U m sC C
C K c
m ACr
2 2
,
get , we need to get P
0.78, 0.22, 1/ 2 0.11
loss in the impeller 0.5(1 ) 0.11
0.89
c
c
x rotor
To
loss loss
the
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Centrifugal Compressors
1
5.3
5.3
1
13
1
2
1
2
(1
582.1295
19389.01
o
ooimp
o
o
o
o
T
T T
p
p
p
p
To calculate density at exit
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Centrifugal Compressors
2
'1'
2
2
2
22
222
2
12
12
1
2
1
2
2
12
2
2
&
2
,
22
P T T
T T
T
T
p
p
togetP
T c
C T thusT
assumeC C
uC
c
C C
c
C
oo
oo
c
o
o
o
o
p
o
ar
p
xr
p
thus get 2.
C t if l C
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Centrifugal Compressors
K T T but T T p p oooo 488295193 //3222
5.3
22
ceT
T
p
p
therefore K c
C T T
oo
p
o
sin488
2.394
,,2.3948.934882
5.31
22
22
2
22
2
3
2
22
2
5.3
22
2
222
/28.2
2.394287.0
10058.2
58.21.135.2 ,1.1
35.2488
2.394532.1
,)(
1
1
1
1
11
2
22
mkg
RT
p
bar p pbut
p p p p
p p
as p
p get
p
p
p
p
p
p
o
o
o
o
oo
o
oo
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Centrifugal Compressors
The required area of cross-section of flow in the radialdirection at the impeller tip is
2
2
m0.027614328.2
9
2
r C
m A
cmor m D
Ab 76.1 0176.0
5.0
0276.0
Centrifugal Compressors
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Centrifugal Compressors
Example 4.2
Consider the design of a diffuser for the compressor dealt
with in the previous example. The following additional datawill 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,
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Centrifugal Compressors
smC x /34230.0
25.0410
2
The radial component of velocity can be found by trial and
error. The iteration may be started by assuming that thetemperature equivalent of the resultant velocity is that
corresponding to the whirl velocity, but only the final trial
is given here .
p
x
p c
Cr C
c
C
22 thus,m/s,97Cr2Try
2
2
22
2
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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
3
22
5.3
2
5.3
2
2
2
22
/77.21.425287.0
10038.3
,38.31.107.3
07.3488
1.425582.1 ,
488
1.425
1.4259.62488 ,
2
12
2
mkg bar p
p
p
p
p
K T
c
C T T
oo
p
o
5.3
12 )582.1(/ oo P P
Centrifugal Compressors
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Centrifugal Compressors
• Area of cross-section of flow inradial
• Check on Cr2:
• Cr2=Taking Cr as 97.9 m/s, theangle of the diffuser vaneleading edge for zero incidenceshould be
20332.0
0176.0*3.0**2
m
o
xr C C 16)342/9.97(tan)/(tan 1
22
1
2
Centrifugal Compressors
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Centrifugal Compressors
(b) 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.
sm /31133.0
25.0410Cx2
Try Cr2= 83 m/s
3
2
2
5.3
2
2
222
2
/96.25.436287.0
10071.3
71.31.137.3 ,37.3488
5.436582.1
5.4365.51488 ,5.51201.0
83.011.3
2
1
mkg
bar p p
p
K T K c
C
o
p
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Centrifugal Compressors
• Area in radial direction=A (radial) = 2Db =0.0365
3.839 )(
22
2
2
r
radi
r
r
C A
mC check
C Get
0
2
1-
15)( tanflow)of direction(2
Cx
C r
throat of widthbn A
m A A
th
r th
) ( *
0945.0sin2