1 Basic Compressor Principles

7/28/2019 1 Basic Compressor Principles

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Centrifugal Compressors



Adjustable diffuser vanes Adjustable inlet guide vanes





Pressure, temperature and velocity

relationship in a centrifugal compressor



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







• Velocity diagrams




• 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•




• 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



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




• 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.



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




• 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




• 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




• 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




vanes)guidewithout(23.4exampleIn this

1

3 o

o

p

p

vanesguidewith79.3

1

3 o

o

p

p




• 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)




• 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



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



• Pressure versus flow curve for

a real or actual compressor

(ideal curve minus the energy

losses)• The dotted line shows energy

losses.



• 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.



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.



Example

C if l C



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




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.




• (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




• 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.




• 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




• 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




/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




:

/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




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




• 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,





• 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.





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




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




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




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




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





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,




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




• 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





• 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





(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




• 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

1 Basic Compressor Principles

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