Axial compressor theory - stage-wise isentropic efficiency - 18th March 2010

Axial compressor

theory

Stage-wise isentropic efficiencyStage-wise isentropic efficiency18th March 2010Prepared by: Cheah CangTo

TURBO GROUP – Axial compressor theory - Stage-wise isentropic efficiency

From previous works, air deflection angle and air outlet angle are calculated. Moving on, in this chapter we will find:

a. Spacing (pitch) between blades

b. Chord

2Axial compressor theory - Stage-wise isentropic efficiency – Cheah CangTo


Design deflection curves

40

50

S/C = 1.5 from CRS

S/C = 1.0 from CRS

S/C = 0.5 from CRS

S/C = 1.5 (curve fit)



to find

Analysis of the values of nominal deflection determined from a large number of tests covering different forms of cascade, has shown that its value is MAINLY dependent on the pitch/chord ratio and air outlet angle.

s/c = 0.5

s/c = 1.0

s/c = 1.5


0

10

20

30

-10 0 10 20 30 40 50 60 70

Air outlet angle, degrees

Air

de

fle

ctio

n,

de

gre

es

to find

prediction based on S/C

s/c = 1.5


Determination of chord length will now depend on the pitch, which itself is dependent on the number of blades in the row. When making a choice for this number, the aspect ratio of the blade, “h/c” has to be considered because of its effect on secondary losses.

number of blades, n = 2 x pi x mean radius / pitch

note: h/c = 3 for initial guess, iterative calculations require. (h = blade height)


Blade heightChord


Referring to the diagram of forces acting on the cascade, the static pressure rise across the blades is given by:

( )

( )

_22

_2

2

2

1

_

0201

0102

2

2

2

1

2

101

2

20212

coscos2

2

:

2

2

1

2

1

wVV

p

wVVp

wppdefine

ppVVp

VpVpppp

aa −

−

=∆

−−=∆

=−

−+−=∆

−−

−=−=∆

αα

ρ

ρ

ρ

ρρ

a

a

a


( )_

2

2

1

2

2

2

2

1

2

2

2

1

2

_

2

2

1

2

2

21

tantan2

tantancos

1

cos

1:

cos

1

cos

1

2

coscos2

wV

p

note

wV

p

a

a

−−=∆∴

−=−

−

−=∆

ααρ

αααα

αα

ρ

αα

a

( ) ( )( )

21

21

121

1221

1

tantantan2:

tantan2

1tan

tantan2

1

tan

tantantan2

1

tan

ααα

αα

ααααα

α

+=

+=

+

=

+−

= −−−

m

a

aa

a

aaa

m

or

V

VV

V

VVV


Force acting along the cascade (from consideration of momentum changes) per unit length is given by:

( ) ( )21

2

21tantantantan ααρααρ −×=−×= aaaa VsVVVsF



Drag force,DPm cCVD

2

2

1ρ=

( ) ( )[ ]

( ) ( )

m

mmama

mmmamamm

swD

swVsVsD

swVsVspsFD

α

ααααραααρ

αααααραααραα

cos

cossintantansintantan

coscostantantansintantancossin

_

_

21

2

21

2

_

21

2

21

2

=∴

+−−−=

+−−−=∆−=

a

3____

_2

coscoscoscos

cos2

1

αααα

αρ mDPm

wswswssw

swcCV =

EquatingDPm cCVD

2

2

1ρ= with

mswD αcos_

= yields:

( by definition of CDP )


1

2

3

2

1

_

1

22

1

3_

2

3_

2

2

_

2

_

2

_

cos

cos

2

1

cos

cos

2

1

cos

2

1

cos

cos

2

1

cos

2

1

2

1

cos

α

α

ρ

α

α

ρ

α

ρα

α

ρ

α

ρρ

α

mDP

mDP

a

m

m

a

m

m

m

m

mDP

V

w

c

sC

V

w

c

sC

V

w

c

s

V

w

c

s

V

w

c

s

cV

swC

××=∴

××=

××=

××=××==

a

a


Lift force,

Equating with yields:

( by definition of CL )Lm cCVL2

2

1ρ=

( ) ( )[ ]

( )[ ] mmmma

mmmamamm

swVsL

swVsVspsFL

ααααααρ

αααααραααραα

sinsintancostantan

sinsintantantancostantansincos

_

21

2

_

21

2

21

2

−+−=

−−+−=∆+=

a

Lm cCVL2

2

1ρ= ( )[ ] mmmma swVsL ααααααρ sinsintancostantan

_

21

2 −+−=

( )[ ]

( )[ ]

( )[ ]

−

+−=

−+−=

cV

sw

cV

VsC

swVscCV

m

m

m

mmmaL

mmmmaLm

_

2

_

2

21

2

_

21

22

2

1

sin

2

1

sintancostantan

sinsintancostantan2

1

ρ

α

ρ

αααααρ

ααααααρρ

a


( )[ ]( )

( )[ ]

( )[ ]( )

( )[ ]

−

+−=

−

+−=

−

+−=

−

+−=

cV

sw

c

sC

cV

sw

cV

sVC

cV

sw

cV

sVC

cV

sw

cV

sVC

a

mmmmmmL

a

mm

a

mmmmaL

m

a

m

m

a

mmmaL

m

m

m

mmmaL

2

2_

23

21

2

2_

2

23

21

2

2

2

_

2

2

21

2

2

_

2

21

2

2

1

cossincossintancostantan2

2

1


cos2

1

sin

cos

sintancostantan2

2

1

sinsintancostantan2

ρ

αααααααα

ρ

αααααααα

αρ

α

α

ααααα

ρ

αααααα

a

a

a

a

continue on next page...


( )mmmmm

mmmm

c

a

c

ac

c

baa

c

aba

ac

ab

c

a

ac

ab

c

a

c

b

a

b

c

a

ααααα

αααα

coscossintancos

cossintan,cos

3

2

3

22

3

23

3

22

3

3

23

3

22

2

2

2

3

3

3

===+

=+

=+=+∴

=××==

m

m

mm

mmm

mmm

a

b

a

c

c

ba

c

a

c

ba

c

a

c

b

αα

αα

ααα

ααα

tancos

cossin?

cos,cossin

?coscossin

3

3

3

2

3

2

3

3

3

3

2

2

2

2

32

==×==

==×=⇒

×=

a

( )[ ]

−+−

=

cV

sw

c

sC mmmmmm

L2

2_

23

21

1


ρ

αααααααα


( )[ ]

( )[ ] ( )

( )mDP

mL

mmmmmmL

a

mmmL

Cc

sC

V

w

c

s

c

s

cV

sw

c

sC

cV

sw

c

sC

αααα

α

αα

ρ

ααα

αρ

ααααα

ρ

ααααα

tancostantan2

cos

tancos

2

1

costantan2

cos2

1

costancostantan2

2

1

costancostantan2

21

1

2

3

2

1

_

21

1

22

1

3_

21

2

3_

21

−−

=∴

××−−

=

−

−=

−

−=

a

a

cV

ca

2

2

1ρ


CDP and CL can be calculated by using data from the following two curves.

note: These two curves are plotted based on extensive cascade tests



Mean deflection

25

30

35

40


0

5

10

15

20

-20 -15 -10 -5 0 5 10

Incidence


Mean stagnation pressure loss

0.05

0.06

0.07

0.08 2

1

_

2

1V

w

ρ


0.00

0.01

0.02

0.03

0.04

-20 -15 -10 -5 0 5 10

Incidence


DSDADPD

DA

LDS

CCCC

h

sC

CC

++=

=

=

02.0

018.02

=3

2

_

cos1 αρ m

D

s

C

V

wCoefficient of loss:

Let ,theoretical static pressure rise:

( )

( )

( )22

2

2

2

1

2

2

2

2

1

2

2

_

2

2

1

2

2

secsec

cos

1

cos

1

2

tantan2

tantan2

ααρ

αα

ρ

ααρ

ααρ

−=∆

−=∆

−=∆

−−=∆

Vp

Vp

Vp

wV

p

a

altheoretica

altheoretica

a

a

a

a

0_

=w


1

2

2

1

cos

cos

2 α

αρ m

c

sV ( )

2

2

1

2

2

1

2

2

1

22

1

1

2

2

22

1

1

2

2

2

1

22

2

2

1

2

cos

cos1

2

cos

cos1

2sec

sec1

2

sec

sec1

2

sec

secsec2

α

α

ρ

α

αρ

α

αρ

α

ααρ

ααρ

−=∆

∴

−=

−=∆

−=∆

−=∆

V

p

VVp

Vp

Vp

ltheoretica

ltheoretica

altheoretica

altheoretica

a

a

a

∆

−=

2

1

2

1

_

2

1

2

1

1

V

p

V

w

ltheoretica

isentropic

ρ

ρ

η


Example: Solar Mars 90 Calculated isentropic efficiency for every single stage.



Example: LM2500Calculated isentropic efficiency for every single stage.



Example: LM2500+Calculated isentropic efficiency for every single stage.



Example: LM2500+G4Calculated isentropic efficiency for every single stage.



Parameter LM2500 LM2500+ LM2500+G4 Unit

Overall performance

ISO power 23262 31076 33679 kW

heat rate 9611 8782 8782 kJ/kW.hr

Eff_thermal 37.46 40.99 40.99 %

gas power 29.93 40.06 44.03 MW

Comparison between LM2500, LM2500+ and LM2500+G4


Compressor

(ISO conditions)

gas power 29.93 40.06 44.03 MW

mass flow rate 68.50 83.80 89.50 kg/s

pressure ratio 17.90 21.50 23.00 -

number of stages 16 17 17 -

rotational speed 6885 6278 6124 rpm

t_out 711.13 749.12 761.90 Kelvin

delta T 422.98 460.97 473.75 Kelvin

Eff_isentropic * 84.26 84.19 84.44 %

End of note

* based on stage-wise isentropic efficiency

Documents

Axial compressor theory - stage-wise isentropic efficiency - 18th March 2010