axial compressor theory - stage-by-stage approach - 28th january 2010

Axial Compressor

TheoryTheory

28th January 2010

Prepared by: Cheah CangTo

ctcheah

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stage-by-stage approach

TURBO GROUP – Axial compressor theory

Objective of this document is to introduce the idea of axial compressor Objective of this document is to introduce the idea of axial compressor

sizing based on “stage-by-stage approach”.

This preliminary sizing approach enables axial compressor designer to:

a) calculate blade angle at every single stage (a stage consists of rotor and a) calculate blade angle at every single stage (a stage consists of rotor and

stator)

b) accurately estimate temperature rise (with mean work-done factor) and

pressure rise at each stage

c) to minimize frictional loss (between blade / rotor and air flow) by

2Axial compressor theory – Cheah CangTo

c) to minimize frictional loss (between blade / rotor and air flow) by introducing “de Haller number” at each stage.


Where:

V = velocity relative to the bladeV = velocity relative to the bladeCa = axial velocityC = resultant velocityCw = tangential component of fluid velocity



a

aaC

bCCb 1

11tan ×=×= β

( )tantan αβ +×=+= CabU

a

aaC

aCCa 1

11tan ×=×= α

( )1111

tantan αβ +×=+= aCabU

a

aaC

bCCb 2

22 tan ×=×= β

( )2222


a

aaC

aCCa 2

22tan ×=×= α

( )2222


2211 tantantantan αβαβ +=+∴

−=−=∆ bbaaC

( ) ( )2112

2112

tantantan ββαα −×=−×=∆∴

−=−=∆

aaw

w

CCC

bbaaC



( )CUmCUmworka aw −×××=∆××= tantan.21

ββ

TCmworkb p

aw

∆××=.

21

Equating Euler turbo-machinery equation with energy equation yields:

( )

( )a

pa

CUT

TCmCUm

21

21

tantan

tantan

ββ

ββ

−××=∆

∆××=−×××∴

( )

p

a

C

CUT 21

tantan ββ −××=∆a



de Haller number

With a fixed value of V1, ∆∆∆∆Cw increases With a fixed value of V1, ∆∆∆∆Cw increases as V2 decreases, temperature rise will increase.At the same time, there will be too much of collisions occur when the difference between V1 and V2 is huge, cars will tend between V1 and V2 is huge, cars will tend to disperse / diffuse off the race track.

In other words, high fluid deflection implies a high rate of diffusion, and one implies a high rate of diffusion, and one of the criteria used was the “de Haller” number, defined as V2/V1 > 0.72. Lower values (<0.72) leading to excessive losses.

( )CU tantan ββ −×× ( )

p

a

C

CUT 21

tantan ββ −××=∆

( )21

tan ββ −×=∆ aw CC



rr/rt (at entry) rt (m) N (rev/s)

0.40 0.4772 116.74

0.41 0.4795 116.17

0.42 0.4819 115.59

0.43 0.4844 114.99

Inlet pressure = 101325 Pascal

Inlet temperature = 288.15 K

R (dry air) = 287.0563 J/(kg.K)

Comp. ratio = 20

air flow rate = 100 kg/s p

stagnationstaticC

CTT

×−=

2

2

1

)(1)(1

Ut = 350 m/s

C_a1 = 150 m/s

C_1 = 150 m/s

Cp (atm) = 1003.33 J/(kg.K)

0.43 0.4844 114.99

0.44 0.4870 114.38

0.45 0.4897 113.75

0.46 0.4925 113.09

0.47 0.4955 112.43

0.48 0.4985 111.74

1

)(1

)(1

)(1)(1

−

×=

γ

γ

stagnation

static

stagnationstaticT

TPP

Cp (atm) = 1003.33 J/(kg.K)

gama (atm) = 1.4008 -

T_1 (static) = 276.94 Kelvin

P_1 (static) = 88200.05 Pascal

Density_1 (static) = 1.1095 kg/m3

0.48 0.4985 111.74

0.49 0.5017 111.03

0.50 0.5050 110.31

0.51 0.5084 109.56

0.52 0.5120 108.80

0.53 0.5157 108.01•

m

RT

P

static

static

static×

=)(1

)(1

)(1ρ

Design parameters to start sizing process:

a) tip speed = 350 m/s

b) axial velocity = 150 m/s

0.53 0.5157 108.01

0.54 0.5196 107.20

0.55 0.5237 106.38

0.56 0.5279 105.53

0.57 0.5323 104.65

0.58 0.5369 103.76

−×××

=

•

2

1t

ra

t

r

rC

mr

ρπ

b) axial velocity = 150 m/s

c) hub-tip ratio = 0.4 to 0.6

0.58 0.5369 103.76

0.59 0.5417 102.84

0.60 0.5467 101.90



p

stagnationstaticC

CTT

×−=

2

2

1

)(1)(1

γ

1

)(1

)(1

)(1)(1

−

×=

γ

stagnation

static

stagnationstaticT

TPP

RT

P static

static×

=)(1

)(1ρ

RT static

static×

)(1

)(1

staticexitaxial

exitC

mA

_ρ×

=&



r

UN

××=

π2

72.0_1

2 ≥=V

VHallerde

1V

Mean work-done factor is a function of number is a function of number of stages

( )

p

astage

C

CUT 21

tantan ββλ −=∆


p


[ ] [ ]wwmeanaxialmean CCUCUT

−××=

−×××=∆

λββλ2121

tantan

[ ] [ ]

p

wmeanstage

p

wwmean

p

axialmeanstage

C

CUT

C

CCU

C

CUT

∆××=∆

−××=

−×××=∆

λ

λββλ

a

2121tantan

= −

axial

mean

C

U1

1 tanβ

−

= − wmean CU 21

2 tanβ

=axial

wmean

C

2

2 tanβ

= −

axial

w

C

C 21

2tanα

mean

ww

U

CC

×

+−=Λ

21 21

axialC

cos β

3

2

2

1

cos

cos__

cos

cos__

α

α

β

β

=

=

statorHallerde

rotorHallerde



Stage α1 β1 α2 β2 ∆T Power, kW P.ratio

1 0.00 61.06 34.45 48.30 27.14 2724.33 1.3275

2 29.49 51.18 51.18 29.49 25.96 2610.23 1.2833

3 29.79 51.02 51.02 29.79 24.78 2496.84 1.2479

4 29.48 51.18 51.18 29.48 24.78 2503.19 1.2306

5 29.93 50.95 50.95 29.93 23.60 2390.97 1.2047

6 29.72 51.06 51.06 29.72 23.60 2398.76 1.1931

7 29.53 51.16 51.16 29.53 23.60 2407.46 1.1828

Overall

P.ratio = 20 -

m_dot = 100 kg/s

T_in = 288.15 Kelvin

∆T = 442.50 Kelvin7 29.53 51.16 51.16 29.53 23.60 2407.46 1.1828

8 29.66 51.09 51.09 29.66 23.12 2368.56 1.1700

9 29.51 51.17 51.17 29.51 23.12 2378.45 1.1622

10 29.68 51.08 51.08 29.68 22.65 2340.11 1.1518

11 29.56 51.14 51.14 29.56 22.65 2350.75 1.1457

12 29.75 51.04 51.04 29.75 22.18 2312.53 1.1371

13 29.65 51.10 51.10 29.65 22.18 2323.54 1.1323

14 29.55 51.15 51.15 29.55 22.18 2334.84 1.1278

15 29.84 50.99 51.40 29.06 22.18 2346.38 1.1237

∆T = 442.50 Kelvin

T_out = 730.65 Kelvin

Power = 45797.79 kW

Speed = 6382.52 rpm

15 29.84 50.99 51.40 29.06 22.18 2346.38 1.1237

16 30.91 50.42 52.04 27.76 22.18 2358.10 1.1199

17 31.73 49.96 52.55 26.68 22.18 2369.93 1.1164

18 32.39 49.58 52.98 25.75 22.18 2381.82 1.1131

19 33.36 48.99 53.59 24.34 22.25 2401.00 1.1036



Case study on RB211-24G’s axial compressors:

a. Pressure ratio = 20b. Mass flow rate = 100 kg/sc. LP axial compressor = 7 stagesd. HP axial compressor = 6 stages



RB211-24G LP axial compressor



RB211-24G HP axial compressor



Case study on GE LMS 100’s axial compressors:

a. Pressure ratio = 42a. Pressure ratio = 42b. Mass flow rate = 206.9 kg/sc. LP turbine speed = 5285 rpmd. HP turbine speed = 9362 rpme. Turbine inlet temperature = 1653.15 Kelvinf. Exhaust temperature = 695.05 Kelving. Guaranteed LHV = 8529 kJ/kW.hr (ηηηη = 42.21%)


g. Guaranteed LHV = 8529 kJ/kW.hr (ηηηη = 42.21%)


Study on LMS 100 axial compressors (T_in = 15 degree C)

With intercooler, turbine rpm Without intercooler, turbine rpm Without intercooler, free rpm

LP compressor HP compressor LP compressor HP compressor LP compressor HP compressor

Pressure ratio 4 10.5 4 10.5 4 10.5

T_in (K) 288.15 310.95 288.15 446.81 288.15 446.85T_in (K) 288.15 310.95 288.15 446.81 288.15 446.85

T_out (K) 446.9 647.68 446.81 905.84 446.85 905.44

Shaft power (MW) 33.17 71.62 33.15 101.56 33.16 101.46

RPM 5283.99 9331.32 5285 9362 4359.26 7654.32

Mass flow rate (kg/s) 206.9 206.9 206.9 206.9 206.9 206.9

Turbine inlet temperature (K) 1653.15 1653.15 1653.15Turbine inlet temperature (K) 1653.15 1653.15 1653.15

Cp (J/kg.K) 1025.26 1049.06 1049.06

Fuel required for combustion to achieve TIT (MW) 213.29 162.20 162.29

Total turbine output (MW) 203.24 207.96 207.96

Net turbine output (MW) 98.44 73.24 73.33

Efficiency = Net power output / fuel input 46.16% 45.15% 45.19%

With a mere increase of 1 percent in GT efficiency from non-intercooled to intercooled configuration, what is the real benefit from intercooled GT with added complexity (air-to-water or air-to-air intercooler) on the total package?

Reason # 1As we can see from table above, total turbine power output almost constant i.e. from 203.24 MW (with intercooler) to 207.96 As we can see from table above, total turbine power output almost constant i.e. from 203.24 MW (with intercooler) to 207.96 MW (without intercooler), because temperature drop across HP turbine, LP turbine and Power turbine is constant (from 1653.15 Kelvin to 695.05 K).

However net turbine power output is relatively high on intercooled GT because it takes more fuel input and less shaft power is required for HP compressor (intercooler removes approximately 28 MW of heat from compressed air before it enters HP is required for HP compressor (intercooler removes approximately 28 MW of heat from compressed air before it enters HP compressor, i.e. 28 MW of fuel is wasted).

In other words, with the same GT footprint (excluding auxiliaries, in this case intercool system) it is possible to increase turbine’s work output i.e. from 73.24 MW to 98.44 MW.



With intercooler, turbine rpm

LP compressor HP compressor

Pressure ratio 4 10.5

T_in (K) 288.15 310.95T_in (K) 288.15 310.95

T_out (K) 446.9 647.68

Shaft power (MW) 33.17 71.62

RPM 5283.99 9331.32

Mass flow rate (kg/s) 206.9 206.9

Turbine inlet temperature (K) 1653.15Turbine inlet temperature (K) 1653.15

Cp (J/kg.K) 1025.26

Fuel required for combustion to achieve TIT (MW) 213.29

Total turbine output (MW) 203.24

Net turbine output (MW) 98.44

Efficiency = Net power output / fuel input 46.16%

Reason # 2

Power loss versus ambient temperature

Due to constant temperature (310.95 Kelvin) input on HP compressor, this leads to:

a. Constant power output on inlet temperature range from -5 deg. C to 23 deg. C, means ISO output is valid from -5 deg. C to 23 deg. C.b. Relatively less power loss at 23 deg. C and above (compared to other GT).Hence GE’s aim is to market LMS 100 in high ambient temperature regions.

Power loss occurred at 23 deg. C and above is due to:

a. LP compressor’s performance, it takes more shaft power when ambient temperature rises. b. Air-to-air intercooler, because cooling medium is ambient air.


b. Air-to-air intercooler, because cooling medium is ambient air.


ReferencesReferences

1. Gas Turbine Theory: Cohen, Rogers and Saravanamuttoo, 4th edition (1996)

2. Gas Turbine Performance: P.P. Walsh, P. Fletcher, 2nd edition (2004)2. Gas Turbine Performance: P.P. Walsh, P. Fletcher, 2nd edition (2004)

3. The Jet Engine: Rolls-Royce plc, 5th edition (1996)


End of note

axial compressor theory - stage-by-stage approach - 28th january 2010

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