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Calculation of power output and total efficiency for a closed gas turbine

cycle, one axis.

In this cycle heat is transferred to air in a closed cycle by heating from a coal-fired boiler. The

cycle has two compressors with inter-cooling, regenerator and after-cooler (before air entering

the low pressure compressor). Start with drawing a connection scheme for the cycle.

Compressor pressure ratio (π) (both) 2.5

Inlet temperature each compressor 25ºC

Inlet temperature turbine 650ºC

Lowest pressure in cycle 8 bar

Isentropic efficiency compressor (both) 85.0= sk η   

Isentropic efficiency turbine 90.0= sT η   

Pressure loss boiler (air heating) 2 %

Pressure loss, intercooler 5 %

Pressure loss, regenerator (both sides) 5 %Pressure loss, after cooler 5 %

Effectiveness of regenerator* 80 %

Generator efficiency 98.0= g η   

Mechanical efficiency 98.0=mη   

Mass flow in the cycle 185 kg/s

The combustion efficiency is 82% and is defined as the ratio of utilized heat in the boiler tofuel supply

Calculate the power output total electrical efficiency.

*This is the same as temperature efficiency for heat exchangers

Appendix 1: Specific heat ratio as function of gas content and temperature for flue gases

 Nomenclature:

κ = ratio of specific heat (C p/CV)

π = pressure ratiox = gas content in flue gases

t = temperature in ºCT= temperature in K.

Subscripts:s = isentropic

K = compressorT = turbine

REG = regenerator

m = mean value (average)

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1Appendix 1: Specific heat ratio as function of gas content and temperature for flue

gases

1

  “Tabeller and diagram” page 171, version of year 1998 by L. Wester, Sweden, or any other arbitrary handbookwith thermal engineering formulas and tables.

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HINTS:

Hint 1 (-3p): The layout of the gas turbine is shown in the figure.

Boiler 

1

2

3

4

5

6

7

8

1

 

1-2: Compression no 12-3: Inter cooling

3-4: Compression no 24-5: Air heater from GT exhaust

5-6: Heat exchange with coal-fired boiler6-7: Expansion in turbine7-8: Cooling of GT exhaust air- Regeneration to compressed air

8-1: After cooling of GT exhaust before entering the compressor

Hint 2 (-3p): The power output is calculated from

 g  K mT el    P  P  P    η η  *)*(   −=  

where the power output/input is the massflow*the specific heat for air*temperature

difference.

Hint 3 (-4p): The temperature increase in the compressor can be calculated as

][)

1(

112   −=−=∆

−

κ 

κ 

π η 

  K 

SK 

 K 

T T T T   

Remember that κ  is a function of the temperature (appendix 1), thus you have to iterate

 between κ  and ∆T. Start assuming that κ  is 1.40 for the first compressor.

The same procedure has to be done for the turbine when calculating the temperature decrease.

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Solution

5) The figure shows the layout of the closed gas turbine

Boiler 

1

2

3

4

5

6

7

8

1

 

1-2: Compression no 1

2-3: Inter cooling

3-4: Compression no 2

4-5: Air heater from GT exhaust5-6: Heat exchange with coal-fired boiler

6-7: Expansion in turbine

7-8: Cooling of GT exhaust air- Regeneration to compressed air

8-1: After cooling of GT exhaust before entering the compressor

The cycle in a T-s diagram is presented below:

T

s

25 deg C

8 bar 

1

2

3

4

5

6

7

8

 

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The power output from the turbine can be calculated as the turbine power minus the

compressors’ power:

 g  K mT el    P  P  P    η η  *)*(   −=  

Power= mass flow * enthalpy difference

Enthalpy difference =  P c (mean value for temperature interval)* temperature difference

?,,,

)***2***(

 K  P T  P 

 g  K  P mT  P el 

T cT c

T cmT cm P 

 K T 

 K T 

∆∆

∆−∆=  ••

η η  

Temperature increase in the compressor  K T ∆  

atratio specifiche

tio pressureracompressor 

efficiencyisentropiccompressor 

 K T 

T T T T 

 K 

SK 

 K 

SK 

 K 

=

==

==

=+=

−=−=∆−

κ 

π 

η 

π η 

κ 

κ 

5.2

85.0

29827325

][

1

)1

(1

12

 

In this equation we have two unknowns; we do not know κ  or the temperature T1. However κ  

is a function of the temperature increase, so we have to start an iteration process (appendix 1)

Assume κ =1.4 (x = 0, air at low temp), then calculate ∆TK  and read the new κ  value, iterating

till the temperatures difference is no bigger than 5

C t 

 K T 

m

 K 

°=+=

=

−=∆

5.772

10525

10515.2815.0

298 )4.1

4.0(

 

With tm = 77,5 °C and x = 0 we find a new κ .

 K T  K  104397.1   =∆⇒=κ   

OK ! Then t2 = 25 + 104 = 129 C 

We can find the specific heat C pk for air with an average temperature of 77.5°C as2 

2 For example “Tabeller and diagram” page 167, version of year 1998 by L. Wester, Sweden, or any other

arbitrary handbook with thermal engineering formulas and tables.

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 K kg kJ  K kg  J c K  P  ,/01.1,/1008   ≈=

 

Temperature increase in turbine T T ∆

:

Here we have to take into regards the pressure losses in the cycle, i.e. the turbine pressure

ratio is not equal to the compressor ratio, as we loose pressure in the heat exchangers.

−=−=∆−

)1

(676

11*

κ 

κ 

π 

η 

T 

ST T    T T T T  

 K T ent  sscoeffici pressurelo

 p

 p

 P 

 P  ssureratioturbinepre

iciencyntropiceff  turbineise

 REG AC 

boiler  REG K  IC  K 

T 

ST 

923273650

99.4

95.0

1*

95.0

1

98.0*95.0*5.2*95.0*5.2

)1(

1*

)1(

1*

)1)(1(**)1(**

90.0

6

1

1

7

6

=+==ℑ

==

ℑ−ℑ−

−−−=

==

==

ξ ξ π ξ π 

π 

η 

 

Again we have one equation and 2 unknowns, so we iterate κ  for expanding flue gases (in

turbine) and temperature decrease, starting with assuming

κ  = 1.35

(remember that the gas content x=0 in this configuration, we are working with a closed air

cycle)

C t 

 K T C  x

C t 

 K T 

m

T 

m

T 

°=

=∆⇒=°==⇒

°=−=

=−=∆

507

286356.1]508,0[

5082

283650

283]99.4

11[923*90.0

35.1/035

κ 

 

OK!

The Specific heat C pt for the expanding flue gases: 

)507;0(   C  x   °=  

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 K kg kJ  K kg  J cT  P  ,/09.1,/1093   ≈=  

Electrical Power:•

m = mass flow = 185 kg/s

W kW  P el 

3.171730098.0*)104*01.1*185*298.0*286*09.1*185(

==−=  

Total efficiency:

?

650

82.0

1*)(*

5

6

56

=

=

=

−

==•

t 

t 

t t cm

 P 

 P 

 P 

comb

comb

 P 

el 

 fuel 

useful 

tot 

o

η 

η 

η 

 

Heat exchanger efficiency 80.0= REGη   

=−

−=

47

45

t t 

t t  REGη  heated air/ inlet temp. difference

C t 

C T t t 

C T t t 

T 

 K 

°=+−=

°=−=∆−=

°=+=∆+=

∆=

317129)129364(80.0

364286650

12910425

5

67

34

22

θ η 

 

21.0

82.0

1*)317650(*08.1*185

17300

,/10844842

317650

=

−

=

=

°=

+=

tot 

m P    K kg  J C t C 

η  

Low efficiency due to high flue gas temperature and low turbine inlet temperature. The

temperature is limited because of the air cooling of boiler tubes.