Thermodynamics of axial compressor and turbine - 3rd December 2009

Gas Turbine Axial

Compressors and Compressors and

Turbines

Thermodynamics calculationsThermodynamics calculations

3rd December 2009Prepared by: Cheah CangTo

Supervised by: James Richard Bryan

TURBO GROUP – Thermodynamics of Axial Compressor and Turbine

Objectives of this discussion:

a) Overview of Brayton cycle.

b) Is to provide the insight of prediction on thermodynamics performances of gas turbine axial compressor and turbine.

c) Note when considering the fundamentals of axial compressor and turbine design, it should be emphasized that successful compressor / turbine design is very much an art, and all the major engine manufacturers have developed a body of knowledge which is kept proprietary for competitive reasons.

2Thermodynamics of axial compressor

a body of knowledge which is kept proprietary for competitive reasons.


Overview of Brayton cycle

The Brayton cycle was first proposed by George Brayton for use in the reciprocating oil-burning engine that he developed around 1870. Today, it is reciprocating oil-burning engine that he developed around 1870. Today, it is used for gas turbines only where both the compression and expansion processes take place in rotating machinery.

Picture taken from Wikipedia

1 to 2: Isentropic air compression1 to 2: Isentropic air compression

2 to 3: Isobaric (constant pressure) heat addition

3 to 4: Isentropic expansion of heated air


4 to 1: Isobaric heat rejection (for closed-cycle)


With input parameters:

a) air mass flow ratea) air mass flow rate

b) compressor compression ratio (note turbine expansion ratio is proportional per Brayton’s P-v diagram discussed earlier)

c) inlet temperaturec) inlet temperature

d) polytropic efficiency

Basically, there are two parameters needed to be evaluated:

a) shaft power requires to perform compression work

b) temperature rises at compressor outlet



Formula for gas power

Power (kJ/s = kW) = mass flow rate (kg/s) x specific heat capacity at constant pressure Power (kJ/s = kW) = mass flow rate (kg/s) x specific heat capacity at constant pressure (kJ/kg.K) x delta temperature (K)

Known / given: mass flow rate

To find: To find:

a) specific heat capacity at constant pressure, Cp

b) delta temperature = T_out – T_in (note: Inlet temperature, T_in is given)b) delta temperature = T_out – T_in (note: Inlet temperature, T_in is given)


Specific heat capacity at constant pressure, Cp is a function of temperature.


Specific heat capacity at constant pressure, Cp is a function of temperature.

What should the temperature be? Inlet temperature or outlet temperature?

The answer is mean temperature, i.e. T_mean = (T_in + T_out)/2 should be used in order to evaluate Cp.

But, how do we calculate T_mean since T_out is an unknown parameter.But, how do we calculate T_mean since T_out is an unknown parameter.

This problem can be solved by iteration method, i.e. take T_mean = T_in (i.e. without T_out) for the first pass of iteration loop until values of T_out converge.

A0 0.992313

A1 0.236688

A2 -1.852148

A3 6.083152

A4 -8.893933

A5 7.097112

A6 -3.234725

A7 0.794571

A8 -0.081873A8 -0.081873

8765432

10008

10007

10006

10005

10004

10003

10002

100010

×+

×+

×+

×+

×+

×+

×+×+= meanmeanmeanmeanmeanmeanmeanmean

p

TA

TA

TA

TA

TA

TA

TA

TAAC



Outlet temperature, T_out is a function of:

a) pressure ratioa) pressure ratio

b) gama = Specific gas constant / (Specific gas constant – specific heat capacity at constant pressure)

c) inlet temperaturec) inlet temperature

d) isentropic efficiency of compressor � it is defined on following page

inTratiopressure

×

−

−

γ

γ

1_

1

in

isentropic

in

out TT +

=

η


ctcheah

Text Box

b) gama = Specific heat capacity at constant pressure / (Specific heat capacity at constant pressure - Specific gas constant)


Isentropic efficiency of compressor is a function of:

a) pressure ratioa) pressure ratio

b) gama

c) polytropic efficiency

1_

1

−

−

ratiopressure γ

γ

1_1

−=

−

ratiopressureisentropic γ

γ

η

1_ −× polytropicratiopressureηγ


1−γ


1_

1_1

1

−

−=

×

−

−

polytropicratiopressure

ratiopressureisentropic

ηγ

γ

γ

γ

η

1_ −ratiopressure

Isentropic efficiency of compressor falls as pressure ratio is increased for the same polytropic efficiency.



In the previous slide, we learned that isentropic efficiency of compressor falls as

compression ratio increases.

But why GT manufacturers are building high pressure ratio machines (e.gBut why GT manufacturers are building high pressure ratio machines (e.g

LMS100, compression ratio of 40:1)?

Reasons are:Reasons are:

a) Isentropic efficiency of turbine increases as expansion ratio increases.

b) Higher OVERALL (combination of compressors, combustors, diffusers, turbine b) Higher OVERALL (combination of compressors, combustors, diffusers, turbine

expanders, etc) efficiency. Overall efficiency means net shaft output power

divided by fuel input power. Compression work is taking energy from the fuel, in

contrast turbine is extracting work from it. Turbine work is defined by:contrast turbine is extracting work from it. Turbine work is defined by:

∆××=•

TCpmPower turbineturbine

−××=∆−

γ

γ 1_

_

11

ratioExpansion

EffTT turbineisentropicinturbine


_ ratioExpansion


In order to maintain maximum temperature drop across turbine, it is necessary to

have higher expansion ratio, this means the pressure shall be kept high at

turbine inlet and low at turbine outlet. This leads to the reason why high power turbine inlet and low at turbine outlet. This leads to the reason why high power

rating GT requires high compression ratio. Although it is true that a higher T_in

(combustion outlet temperature) will improve efficiency (refer Brayton cycle), for

the purposes of this discussion T_in is assumed constant since all turbine the purposes of this discussion T_in is assumed constant since all turbine

manufacturers have similar combustor outlet temperatures to maximize

efficiency.

c) Mismatch (in term of air density) between compressor and turbine affecting

overall GT efficiency occurs on high compression ratio machine at high ambient overall GT efficiency occurs on high compression ratio machine at high ambient

temperature. This effect can be clearly seen when comparing temperature de-

rate curves between low pressure ratio industrial gas turbines versus high

pressure ratio aero-derivative gas turbines. However, the topic is out of scope of pressure ratio aero-derivative gas turbines. However, the topic is out of scope of

this discussion.



−××=∆−

γ

γ 1_

_

11

ratioExpansion


Delta T across turbine vs expansion ratio

800

γ_ ratioExpansion

700

750

800

Delt

a T

em

pera

ture

(K

)

600

650

Delt

a T

em

pera

ture

(K

)

500

550

Delt

a T

em

pera

ture

(K

)

400

450

0 5 10 15 20 25 30 35 40

Expansion ratio


Expansion ratio

•


turbineturbine TCpmPower ∆××=•

Turbine power output vs expansion ratio

90000

95000

80000

85000

Tu

rbin

e s

haft

po

wer

(kW

)

70000

75000

Tu

rbin

e s

haft

po

wer

(kW

)

55000

60000

65000

Tu

rbin

e s

haft

po

wer

(kW

)

50000

55000

0 5 10 15 20 25 30 35 40

Expansion ratio



2.00

Specific work output vs pressure ratio T3/T1 = 2

T3/T1 = 3

T3/T1 = 4

T3/T1 = 5

1.50

T_in

)]

T3/T1 = 5

1.00

Sp

ecif

ic w

ork

ou

tpu

t [W

/(C

p*T

_in

0.50

Sp

ecif

ic w

ork

ou

tpu

t [W

/(

0.00

0 5 10 15 20 25 30

-0.50

Pressure ratio



2.00

Specific work output vs pressure ratio T3/T1 = 2

T3/T1 = 3

T3/T1 = 4

1.50

T_in

)]

T3/T1 = 5

PR=17

1.00

Sp

ecif

ic w

ork

ou

tpu

t [W

/(C

p*T

_in PR=17

Optimum pressure ratio for a

0.50

Sp

ecif

ic w

ork

ou

tpu

t [W

/(

PR=11Optimum pressure ratio for a given temperature ratio, T3/T1.

0.00

0 5 10 15 20 25 30

Sp

ecif

ic w

ork

ou

tpu

t [W

/(

PR=7

PR=3

-0.50

Pressure ratio


Pressure ratio

OVERALL (combination of compressors, combustors, diffusers, turbine expanders, etc) efficiency

TURBO GROUP – Thermodynamics of Axial Compressor and TurbineOVERALL (combination of compressors, combustors, diffusers, turbine expanders, etc) efficiency

increases as compression ratio increases.

Overall GT efficiency versus compression ratioO

vera

ll g

as t

urb

ine e

ffic

ien

cy

Overa

ll g

as t

urb

ine e

ffic

ien

cy

Overa

ll g

as t

urb

ine e

ffic

ien

cy

( )rationcompressiooverall _ln09979.007641.0 ×+=η ( )rationcompressiooverall _ln09979.007641.0 ×+=η

Note: Overall GT efficiency is derived from machine manufacturers’ published heat rate..

Compression ratio

Gas turbine heat rate data courtesy of James Bryan [GSGnet.net (2009)]


Compression ratio


GT thermal efficiency versus pressure ratio:

comparison between Brayton and actual cycle

0.6

0.7

1

0.5

0.6

Brayton

Actualγ

γη

1

1

2

11

−

−=

P

P

Brayton

0.4

0.3

×+= 2ln09979.007641.0

P

Pactualη

0.1

0.2

1

P

0.0

0.1


0 5 10 15 20 25 30 35 40


GT thermal efficiency versus pressure ratio:

comparison between Brayton and actual cycle

0.6

0.7

0.5

0.6

Brayton

Actual

Reduction of thermal efficiency due

to irreversible losses.

0.4

to irreversible losses.

0.3

0.1

0.2

0.0

0.1


0 5 10 15 20 25 30 35 40


But, hang on a second…previous slide tells us overall GT efficiency goes up as

compression ratio increases (or temperature drop across turbine increases as compression ratio increases (or temperature drop across turbine increases as

expansion ratio increases), then why GT manufacturer don’t produce high

efficiency machine, let say more than 50% at the expense of high compression

ratio? ratio?

The reason behind this, at least what I have in mind is explained on the following

pages.



Let us revisit “Delta temperature versus expansion ratio” curve presented earlier.

Delta T across turbine vs expansion ratioDelta T across turbine vs expansion ratio

750

800

650

700

Delt

a T

em

pera

ture

(K

)

3

25

Temperature drop is more sensitive on expansion

550

600

Delt

a T

em

pera

ture

(K

)

65

Temperature drop is more sensitive on expansion

ratio of low pressure range compared to high

pressure range. It means rate of change of

efficiency is decreasing with increasing expansion

ratio.

450

500

3

From this point onwards, please be absolute clear

that (don’t confuse):

a) Overall GT efficiency increases as expansion

ratio increases.

b) Rate of change of overall GT efficiency

decreases as expansion ratio increases.400

0 5 10 15 20 25 30 35 40

Expansion ratio

decreases as expansion ratio increases.



In order to quantify sensitivity of efficiency as a function of expansion ratio, the following

expression is derived: by differentiating the curve function described previously to obtain

slope gradient.

Rate of change of efficiency = delta temperature turbine / delta expansion ratioRate of change of efficiency = delta temperature turbine / delta expansion ratio

−××=∆1

−××=∆−

γ

γ 1_

_

11

ratioExpansion


These are nearly constant

( )24812.0_1

−−=∆ ratioExpansionT

Gama = 1.33 (for turbine)

24812.1_24812.0

−×= ratioExpansiondT

( )24812.0_1

−−=∆ ratioExpansionT

( )24812.1

_24812.0_

−×= ratioExpansionratioExpansiond

dT

It is proposed that this new function is used as the basis for gas turbine

performance prediction and comparison within a database environment.


performance prediction and comparison within a database environment.


Rate of change of overall GT efficiency as a function of expansion ratio is now defined:

( )24812.1

_24812.0_

−×= ratioExpansionratioExpansiond

dT

( )ratioExpansiond

dT

_ ( )_ ratioExpansiond

0.12

This curve (dT/d[expansion ratio]) enables us to visualize how sensitive is

overall GT efficiency as a function of expansion ratio.

( )ratioExpansiond _

0.08

0.1

overall GT efficiency as a function of expansion ratio.

LM 6000

Compression ratio = 28.1

LMS 100

Compression ratio = 40

0.06

0.08

0.04

0

0.02

0 10 20 30 40 50 60


0 10 20 30 40 50 60

Expansion

ratio


Overall GT efficiency merely increased by 3.43%increased by 3.43%

Ov

era

ll G

T e

ffic

ien

cy

Ov

era

ll G

T e

ffic

ien

cy

Ov

era

ll G

T e

ffic

ien

cy


Compression ratio


It’s time to put all theories discussed in the previous section into practical work. Attempt the following example:

Calculate the isentropic efficiency, outlet temperature and power input for a compressor (of gas turbine) of 20:1 pressure ratio, polytropic efficiency of 91.07%, with a mass flow of 100 kg/s and an inlet temperature of 35 deg. C.

Also insert a parametric table to investigate the effects (on the outlet temperature and power input) of changing inlet temperature to 30 and 38 deg. C respectively.respectively.


Input parameters


Input parametersa) Compression ratio 20:1b) Polytropic efficiency of 91.07%c) Mass flow of 100 kg/sc) Mass flow of 100 kg/sd) Inlet temperature of 35 deg. C.

But, something is missing...inlet pressure?Then make assumption, machine is located at mean sea level, i.e. atmospheric pressure = 101325 pascal

e) Inlet pressure = 101325 pascale) Inlet pressure = 101325 pascal

Parameters pass 1 pass 2 pass 3 pass 4 pass 5 pass 6 pass 7 pass 8 pass 9 Unit

Cp 1004.33 1039.11 1036.61 1036.78 1036.77 1036.77 1036.77 1036.77 1036.77 J/(kg.K)

gama 1.4002 1.3817 1.3830 1.3829 1.3829 1.3829 1.3829 1.3829 1.3829 -

T_out 788.16 764.59 766.21 766.10 766.10 766.10 766.10 766.10 766.10 K

Outlet temperature is converging from 4th

T_out 788.16 764.59 766.21 766.10 766.10 766.10 766.10 766.10 766.10 K

T_mean 308.15 548.16 536.37 537.18 537.12 537.13 537.13 537.13 537.13 K

Isentropic efficiency 0.8694 0.8694 0.8694 0.8694 0.8694 0.8694 0.8694 0.8694 0.8694 -

Outlet temperature is converging from 4th

iteration onwards.



Parameters pass 1 pass 2 pass 3 pass 4 pass 5 pass 6 pass 7 pass 8 pass 9 Unit

Cp 1004.33 1039.11 1036.61 1036.78 1036.77 1036.77 1036.77 1036.77 1036.77 J/(kg.K)

gama 1.4002 1.3817 1.3830 1.3829 1.3829 1.3829 1.3829 1.3829 1.3829 -gama 1.4002 1.3817 1.3830 1.3829 1.3829 1.3829 1.3829 1.3829 1.3829 -

T_out 788.16 764.59 766.21 766.10 766.10 766.10 766.10 766.10 766.10 K

T_mean 308.15 548.16 536.37 537.18 537.12 537.13 537.13 537.13 537.13 K

Isentropic efficiency 0.8694 0.8694 0.8694 0.8694 0.8694 0.8694 0.8694 0.8694 0.8694 -

Outlet temperature, T_out = 766.10 K or 492.95 deg. C

Shaft power = 100 (kg/s) x 1036.77 (J/kg.K) x [766.10 - 308.15] (K)

= 100 x 1036.77 x 457.95= 100 x 1036.77 x 457.95

= 47478882.15 J/s

= 47479 kW



Parametric table

Effects on shaft power and outlet temperature as inlet temperature changes.

T_in (C) Power (kW) T_out (deg.C) Isentropic efficiency

30 46749 481.65 0.8693

35 47479 492.95 0.8694

changes.

35 47479 492.95 0.8694

38 47917 499.71 0.8694

Shaft power vs inlet temperature Outlet temperature vs inlet temperature

47600

47800

48000500.0

Ou

tle

t te

mp

era

ture

(d

eg

. C

)

47000

47200

47400

Sh

aft

po

wer

(kW

)

490.0

Ou

tle

t te

mp

era

ture

(d

eg

. C

)

46600

46800

30 31 32 33 34 35 36 37 38 39

Inlet temperature (deg. C)

480.0

30 31 32 33 34 35 36 37 38 39

Inlet temperature (deg. C)

Ou

tle

t te

mp

era

ture

(d

eg

. C

)


Inlet temperature (deg. C) Inlet temperature (deg. C)

Thermodynamics of axial compressor and turbine - 3rd December 2009

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Transcript of Thermodynamics of axial compressor and turbine - 3rd December 2009