1. Thermodynamic Analyses Gas Turbine Power Plant
Basics in the
Thermodynamic Analyses
of the
Gas Turbine Power Plant
Prof. R. Shanthini
Dept. of C&P Engineering
University of Peradeniya
Sri Lanka

2. fuel hot gases Gen gases to the stack atmospheric air Combustion
chamber
compressed
air
Comp-
ressor
Gas
turbine
Compressor
shaft
Turbine
shaft
Gen
gases
to the
stack
atmospheric
air
Gen stands for Electricity Generator

3. (WGT) fuel hot gases out Gen gases to the stack atmospheric air
W stands for work flow rate and
GT stands for Gas Turbine
fuel
hot
gases
Combustion
chamber
(WGT)
out
compressed
air
Comp-
ressor
Gas
turbine
Gen
gases
to the
stack
atmospheric
air

4. (QCC) (WGT) in hot gases out Gen gases to the stack atmospheric air
Q stands for heat flow rate and
CC stands for Combustion Chamber
(QCC) in
hot
gases
Combustion
chamber
(WGT)
out
compressed
air
Comp-
ressor
Gas
turbine
Gen
gases
to the
stack
atmospheric
air

5. (QCC) (WGT) (WC) in hot gases out Gen in gases to the stack
W stands for work flow rate and
C stands for Compressor
(QCC) in
hot
gases
Combustion
chamber
(WGT)
out
compressed
air 3 2
Comp-
ressor
Gas
turbine
(WC) in
Gen 4 1
gases
to the
stack
atmospheric
air

6. (QCC) (WGT) (WC) in hot gases out Gen in gases to the stack
Combustion
chamber
(WGT)
out
compressed
air 3 2
Comp-
ressor
Gas
turbine
(WC) in
Gen 4 1
gases
to the
stack
atmospheric
air

7. + (WGT) (QGT) = m ( h – h ) + m ( C – C ) / 2 + m g ( Z – Z )
Steady flow energy equation
applied to the flow across turbine:
hot gases 3 +
(WGT)
out
(QGT)
out
Gas
turbine
= m ( h – h ) 3 4 g
+ m ( C – C ) / 2 3 4 2 g
+ m g ( Z – Z ) 3 4 g r
m stands for mass flow rate of gas,
h stands for enthalpy,
C stands for speed of gas flow
g stands for gravitational acceleration, &
Z stands for height above reference level 4 g
gases
to the
stack r

8. = + - (QGT) (WGT) (QGT) = m ( h – h ) + + m ( C – C ) / 2
out
hot gases = +
(WGT)
out
(QGT)
out 3
= m ( h – h ) 3 4 g +
Gas
turbine
+ m ( C – C ) / 2 3 4 2 g
+ m g ( Z – Z ) 3 4 g r
Assumptions:
- Adiabatic condition prevails across the gas turbine 4
gases
to the
stack
- Kinetic energy changes are negligible compared to enthalpy changes
- Potential energy changes are ignored

9. = (WGT) = m ( h – h ) + Assumptions: hot gases out 3 4 g
turbine
Assumptions:
- Adiabatic condition prevails across the gas turbine 4
gases
to the
stack
- Kinetic energy changes are negligible compared to enthalpy changes
- Potential energy changes are ignored

10. = m ( h – h ) (WGT) Assumptions: hot gases out 3 4 g
turbine
Assumptions:
- Adiabatic condition prevails across the gas turbine 4
gases
to the
stack
- Kinetic energy changes are negligible compared to enthalpy changes
- Potential energy changes are ignored

11. = m ( h – h ) (WGT) Assumption: = m C ( T – T ) (WGT) hot gases out 34 g 3
Gas
turbine
Assumption:
- Gases flowing through the turbine behave as ideal gases 4
(WGT)
out
= m C ( T – T ) 3 4 g pg
gases
to the
stack

12. = m ( h – h ) (WGT) T = m C ( T – T ) T out 3 4 g 3 3 4 g pg
Gas
turbine
Temperature at the outlet T 4
Temperature at the inlet 4
Specific heat of gas at
constant pressure
Mass flow rate of gas

13. = m C ( T – T ) (WGT) T =  T = ? out 3 4 g pg 3 changes fixed fixed 4
Gas
turbine
fixed
fixed T 4
= ? 4
free to choose, but we fix it at some value

14. How small should T4 be ? = m C ( T – T ) (WGT) P =  T =  T = ? T P
out
= m C ( T – T ) 3 4 g pg P 3
=  3 T 3
= 
To get maximum work output from the turbine,
Gas
turbine
at the given P 3 4
and
(P stands for pressure) T 4
= ? T 4
should be
as small as possible P 4
= 
How small should T4 be ? 4

15. (WGT) = m C ( T – T ) T P P3 T P P4 3 4 4s out 3 4 g pg 3 3 real flow
Gas
turbine
real
flow T P3
ideal
flow 4 T 4 4s P 4 P4 4
Specific Entropy (s)

16. ) ( T = m C ( T – T ) (WGT) P P = T T P (WGT) = m C ( T – T ) 3 out 34 g pg 3 P 3
For the ideal flow (ideal gas at constant specific entropy):
Gas
turbine T 4s 3 = P ( 4 )
(-1)/ T 4s
where  is the isentropic constant P 4 4
Therefore,
(WGT)
out,ideal
= m C ( T – T ) 3 4s g pg

17. -  (WGT) = m C ( T – T ) = T = Turbine Efficiency out out,ideal T 3 4
Gas
turbine
m C ( T – T ) 3 4 g pg 4s = 4 - = T 3 4 4s

18. ( )    = ( T – ) P = T 4s 4 3 = (WGT) = m C ( T – T )
Governing equations: 3 T 4s 3 = P ( 4 )
(-1)/
Gas
turbine  T 4 = 3
( T 4s – ) 4
(WGT)
out
out,ideal =  T =  T
m C ( T – T ) 3 4s g pg

19. Let‘s do some Excel sheet calculations across the turbine3
Data:
Determine
Gas
turbine
= 350 kg/s m g T 4 C pg
= 1.1 kJ/kg.s T 3
= 1200 K
(WGT)
out 4 P 4
= 1 bar γ
= 1.3
for P3 varying in the range of 2 to 15 bar  T
= 88%

20.  Turbine outlet temperature increases
with decreasing turbine efficiency 
= 88% T

21.  Turbine work output decreases with decreasing turbine efficiency
= 88% T

22. (WGT) fuel hot gases out Gen gases to the stack atmospheric air
Combustion
chamber
(WGT)
out
compressed
air
Comp-
ressor
Gas
turbine
Gen
gases
to the
stack
atmospheric
air

23. (QCC) (WGT) (WC) in hot gases out Gen in gases to the stack
Combustion
chamber
(WGT)
out
compressed
air 3 2
Comp-
ressor
Gas
turbine
(WC) in
Gen 4 1
gases
to the
stack
atmospheric
air

24. (QCC) (WGT) (WC) in hot gases out Gen in gases to the stack
Combustion
chamber
(WGT)
out
compressed
air 3 2
Comp-
ressor
Gas
turbine
(WC) in
Gen 4 1
gases
to the
stack
atmospheric
air

25. (WC) (QC) + m ( h – h ) + m ( C – C ) / 2 + m g ( Z – Z ) =
Steady flow energy equation
applied to the flow across compressor:
compressed
air 2
Subscript a stands for air
Comp-
ressor
(WC) =
(QC)
out
+ m ( h – h ) 2 1 a in 1
+ m ( C – C ) / 2 2 1 a
atmospheric
air
+ m g ( Z – Z ) 3 4 a r

26. Assumptions: (WC) (QC) + m ( h – h ) + m ( C – C ) / 2 + m g ( Z – Z )
- Adiabatic condition prevails across the compressor
compressed
air
- Kinetic energy changes are negligible compared to enthalpy changes 2
- Potential energy changes are ignored
Comp-
ressor
(WC) =
(QC)
out
+ m ( h – h ) 2 1 a in 1
+ m ( C – C ) / 2 2 1 a
atmospheric
air
+ m g ( Z – Z ) 3 4 a r

27. Assumptions: (WC) + m ( h – h ) =
- Adiabatic condition prevails across the compressor
compressed
air
- Kinetic energy changes are negligible compared to enthalpy changes 2
- Potential energy changes are ignored
Comp-
ressor
(WC) =
+ m ( h – h ) 2 1 a in 1
atmospheric
air

28. Assumption: (WC) m ( h – h ) = m C ( T – T ) =
- Air flowing through the compressor behaves as an ideal gas
compressed
air 2
Comp-
ressor
(WC) =
m ( h – h ) 2 1 a in 1
= m C ( T – T ) 2 1 a pa
atmospheric
air

29. T = m C ( T – T ) (WC) T 2 in 2 1 a pa T at the inlet T at the outlet
Comp-
ressor
T at the
inlet 1
T at the outlet T 1
Specific heat of air at
constant pressure
Mass flow rate of air

30. T = ? (WC) = m C ( T – T ) T =  2 in 2 1 a pa fixed changes 1 fixed
= ? 2
(WC) in
= m C ( T – T ) 2 1 a pa
Comp-
ressor
fixed 1
changes T 1
= 
fixed
free to choose, but we fix it at some value

31. How small should T2 be ? T = ? ; P =  (WC) = m C ( T – T ) T =  T P
= ?
; P 2
=  2
(WC) in
= m C ( T – T ) 2 1 a pa
Comp-
ressor
To give minimum work input to the compressor, 1
at the given P 1 2
and T 1
=  T 2
should be
as small as possible P 1
= 
How small should T2 be ?

32. T = m C ( T – T ) (WC) P P3=P2 T P4=P1 P 3 2 4 2s 4s 1 2 in 2 1 a pa 2
Comp-
ressor T
P3=P2 2 4 2s
real flow 4s T 1
P4=P1 P 1 1 1
ideal
flow
Specific Entropy (s)

33. ) ( T = ? (WC) = m C ( T – T ) P =  T = P T =  (WC) = m C ( T – T )2s
= ?
(WC) in
= m C ( T – T ) 2 1 a pa P 2
= 
For the ideal flow (ideal gas at constant specific entropy): 2
Comp-
ressor T 2s 1 = P ( 2 )
(-1)/ 1
Therefore, T 1
= 
(WC)
in,ideal
= m C ( T – T ) 2s 1 a pa P 1
= 

34. -  (WC) = m C ( T – T ) = T = Compressor efficiency in,ideal in C 2s1 a pa 2 =
Comp-
ressor = T 2s 2 1 - 1

35. ( ) / / /    T = ( T + – ) P = T 2 1 2s (WC) = = m C ( T – T )
Governing equations: 2 T 2s 1 = P ( 2 )
(-1)/
Comp-
ressor  C T 2 = 1
( T 2s + – ) / 1 in
(WC)
in,ideal =  C / =  C
m C ( T – T ) 2s 1 a pa /

36. Let‘s do some Excel sheet calculations across the compressor1 2
Comp-
ressor
Data:
= 350 kg/s m a
Determine C pa
= kJ/kg.s T 2 T 1
= 300 K
(WC) in P 1
= 1 bar γ
= 1.4
for P2 varying in the range of 2 to 15 bar  C
= 85%

37.  = 85% Compressor outlet temperature increases
with decreasing compressor efficiency 
= 85% C

38.  = 85% Work input to the compressor increases
with decreasing compressor efficiency 
= 85% C

39. (WGT) W (WC) fuel hot gases out net Gen in gases to the atmospheric
Combustion
chamber
(WGT)
out
compressed
air
Comp-
ressor
Gas
turbine W
net
(WC) in
Gen
gases
to the
stack
atmospheric
air

40. - - = (WC) (WGT) W (WC) = (WGT) W in out net in,ideal out,ideal
Net work output from the turbine is the power available for electricity generation
(WC)
in,ideal =
(WGT)
out,ideal - W
net,ideal
Net work output under ideal conditions is the maximum power available for electricity generation

41. 
= 85% C T
= 88%
and

42. 
= 85% C T
= 88%
and

43. (QCC) W in hot gases net Gen gases to the atmospheric stack air
Combustion
chamber
compressed
air 3 2
Comp-
ressor
Gas
turbine W
net
Gen 4 1
gases
to the
stack
atmospheric
air

44. (QCC) W in hot gases net Gen gases to the atmospheric stack air
Combustion
chamber
compressed
air 3 2
Comp-
ressor
Gas
turbine W
net
Gen 4 1
gases
to the
stack
atmospheric
air

45. (QCC) = m ( h – h ) Assumptions: compressed air hot gases fuel2 3
compressed
air
hot gases
Combustion
chamber
fuel
(QCC)
in,ideal
= m ( h – h ) 3 2 a
Assumptions:
- Kinetic energy changes are negligible
- Potential energy changes are ignored
- Fuel flow rate is negligible compared to the air flow rate

46. (QCC) = m ( h – h ) = m C ( T – T ) Assumption: compressed air2 3
compressed
air
hot gases
Combustion
chamber
fuel
(QCC)
in,ideal
= m ( h – h ) 3 2 a
= m C ( T – T ) 3 2 a pa
Assumption:
- Air flowing through the compressor behaves as an ideal gas

47. / /    = (QCC) (QCC) = m C ( T – T ) compressed air hot gases fuel2 3
compressed
air
hot gases
Combustion
chamber
fuel /  CC
(QCC) in =
(QCC)
in,ideal /  CC
= m C ( T – T ) 3 2 a pa  CC
is the compressor efficiency

48. 2 3
compressed
air
hot gases
Combustion
chamber
fuel
Let‘s do some Excel sheet calculations across the combustion chamber
= 350 kg/s m a
Determine C pa
= kJ/kg.s  CC
= 80%
(QCC) in P 2 3 =

49. 
= 80% CC 
= 80% CC

50.  (QCC) W W (QCC) = in hot net Thermal efficiency Gen net gases to the
atmospheric
air
(QCC) in 1 2
compressed
Comp-
ressor
Combustion
chamber
gases
to the
stack 3 4
Gen
hot W
net
Gas
turbine
Thermal
efficiency W
net  th =
(QCC) in

51. 
= 80% CC

52. C CC 
= 80%
= 85% T
= 88%

53. = - Heat Loss? (QCC) W (QCC) W in hot gases net Gen in net gases
Combustion
chamber
compressed
air 3 2
Comp-
ressor
Gas
turbine W
net
Heat Loss?
Gen 4 =
(QCC) in W
net 1
gases
to the
stack
atmospheric
air

55. Heat Loss (QCC) W in hot gases net Gen gases to the atmosphere
Combustion
chamber
compressed
air 3 2
Comp-
ressor
Gas
turbine W
net
Gen 4
Heat Loss 1
gases
to the atmosphere
through the stack
atmospheric
air

56. Heat is lost with the turbine exhaust gases to the atmosphere through the stack
Should not we make good use of all that heat
that is not only getting wasted
but also pollute the environment in many ways?