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

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Presentation on theme: "Basics in the Thermodynamic Analyses of the Gas Turbine Power Plant Prof. R. Shanthini Dept. of C & P Engineering University of Peradeniya Sri Lanka."— Presentation transcript:

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

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

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

5 atmospheric air (W GT ) out Comp- ressor (W C ) in (Q CC ) in Gen compressed air hot gases Combustion chamber Gas turbine gases to the stack W stands for work flow rate and C stands for Compressor

6 atmospheric air (W GT ) out Comp- ressor (Q CC ) in Gen hot gases compressed air 3 Combustion chamber Gas turbine gases to the stack (W C ) in

7 (W GT ) out (Q GT ) out = m ( h – h ) 34 g + m ( C – C ) / g hot gases gases to the stack Gas turbine 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 Steady flow energy equation applied to the flow across turbine: g + m g ( Z – Z ) 3 4 g r r

8 + (Q GT ) out = + m ( C – C ) / g Assumptions: - Adiabatic condition prevails across the gas turbine - Kinetic energy changes are negligible compared to enthalpy changes = m ( h – h ) 34 g + - (Q GT ) out 3 4 hot gases gases to the stack Gas turbine + m g ( Z – Z ) 3 4 g r - Potential energy changes are ignored (W GT ) out

9 (W GT ) out = 3 4 hot gases gases to the stack Gas turbine Assumptions: - Adiabatic condition prevails across the gas turbine - Kinetic energy changes are negligible compared to enthalpy changes - Potential energy changes are ignored = m ( h – h ) 34 g +

10 (W GT ) out = m ( h – h ) 34 g 3 4 hot gases gases to the stack Gas turbine Assumptions: - Adiabatic condition prevails across the gas turbine - Kinetic energy changes are negligible compared to enthalpy changes - Potential energy changes are ignored

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

12 (W GT ) out 3 = m C ( T – T ) 34 gpg T 4 T 3 Specific heat of gas at constant pressure Mass flow rate of gas 4 Temperature at the outlet Temperature at the inlet Gas turbine m ( h – h ) 34 g =

13 3 T 4 T 3 = 4 fixed changes fixed free to choose, but we fix it at some value = ? (W GT ) out = m C ( T – T ) 34 gpg Gas turbine

14 3 T 4 T 3 = 4 = ? P 4 = P 3 = should be as small as possible T 4 How small should T 4 be ? (W GT ) out = m C ( T – T ) 34 gpg Gas turbine To get maximum work output from the turbine, at the given P 3 P 4 and ( P stands for pressure)

15 3 T 4 T 3 4 P 4 P 3 (W GT ) out = m C ( T – T ) 34 gpg T Specific Entropy (s) 3 4s P3P3 P4P4 4 real flow ideal flow Gas turbine

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

17 3 4 T = (W GT ) out (W GT ) out,ideal m C ( T – T ) 34 gpg m C ( T – T ) 34s gpg = Turbine Efficiency - = T 3 T 3 T 4 T 4s - Gas turbine

18 3 4 T T 4 = T 3 ( T 3 T 4s –– ) T 3 T = P ( 4 P 3 ) ( -1)/ (W GT ) out (W GT ) out,ideal = T = T m C ( T – T ) 34s gpg Gas turbine Governing equations:

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

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

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

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

23 atmospheric air gases to the stack (W GT ) out Comp- ressor (W C ) in (Q CC ) in Gen compressed air hot gases Combustion chamber Gas turbine

24 atmospheric air gases to the stack (W GT ) out Comp- ressor (W C ) in (Q CC ) in Gen 3 compressed air hot gases Combustion chamber Gas turbine

25 atmospheric air Comp- ressor (W C ) in 1 = (Q C ) out + m ( h – h ) 21 a + m ( C – C ) / a 2 compressed air Subscript a stands for air + m g ( Z – Z ) 3 4 a r Steady flow energy equation applied to the flow across compressor:

26 atmospheric air Comp- ressor (W C ) in 1 = (Q C ) out + m ( h – h ) 21 a + m ( C – C ) / a 2 compressed air + m g ( Z – Z ) 3 4 a r Assumptions: - Adiabatic condition prevails across the compressor - Kinetic energy changes are negligible compared to enthalpy changes - Potential energy changes are ignored

27 atmospheric air Comp- ressor (W C ) in 1 = + m ( h – h ) 21 a 2 compressed air Assumptions: - Adiabatic condition prevails across the compressor - Kinetic energy changes are negligible compared to enthalpy changes - Potential energy changes are ignored

28 atmospheric air Comp- ressor (W C ) in 1 = m ( h – h ) 21 a = m C ( T – T ) 21 apa 2 compressed air Assumption: - Air flowing through the compressor behaves as an ideal gas

29 1 2 T 1 T 2 (W C ) in = m C ( T – T ) 21 apa T at the inlet T at the outlet Specific heat of air at constant pressure Mass flow rate of air Comp- ressor

30 1 2 T 1 T 2 (W C ) in = m C ( T – T ) 21 apa fixed changes fixed free to choose, but we fix it at some value = = ? Comp- ressor

31 1 T 1 = P 1 = T 2 = ? (W C ) in = m C ( T – T ) 21 apa ; P 2 = 2 should be as small as possible T 2 How small should T 2 be ? at the given P 1 P 2 and Comp- ressor To give minimum work input to the compressor,

32 T 3 4s P 3 =P 2 4 real flow ideal flow (W C ) in = m C ( T – T ) 21 apa 1 T 1 P 1 T 2 P s 2 P 4 =P 1 Comp- ressor Specific Entropy (s)

33 1 T 2s 1 T = P ( 2 P 1 ) ( -1)/ T 1 = P 1 = T 2s = ? (W C ) in,ideal = m C ( T – T ) 2s1 apa P 2 = 2 (W C ) in = m C ( T – T ) 21 apa Comp- ressor For the ideal flow (ideal gas at constant specific entropy): Therefore,

34 1 C = (W C ) in,ideal (W C ) in = T 2s T 2 T 1 T m C ( T – T ) 2s1 apa m C ( T – T ) 21 apa = Compressor efficiency Comp- ressor

35 T 2s 1 T = P ( 2 P 1 ) ( -1)/ 1 2 C T 2 = T 1 ( T 2s T 1 + –) / = C m C ( T – T ) 2s1 apa / Comp- ressor in (W C ) in,ideal = C / Governing equations:

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

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

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

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

40 = (W C ) in (W GT ) out W net - (W C ) in,ideal = (W GT ) out,ideal - W net,ideal Net work output from the turbine is the power available for electricity generation 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 atmospheric air gases to the stack (Q CC ) in Gen compressed air hot gases W net Comp- ressor Combustion chamber Gas turbine

44 atmospheric air gases to the stack (Q CC ) in Gen compressed air hot gases W net Comp- ressor Combustion chamber Gas turbine

45 (Q CC ) in,ideal = m ( h – h ) 32 a Assumptions: - Kinetic energy changes are negligible - Potential energy changes are ignored - Fuel flow rate is negligible compared to the air flow rate 2 3 compressed air hot gases Combustion chamber fuel

46 2 3 compressed air hot gases Combustion chamber fuel = m C ( T – T ) 32 apa Assumption: - Air flowing through the compressor behaves as an ideal gas (Q CC ) in,ideal = m ( h – h ) 32 a

47 (Q CC ) in 2 3 compressed air hot gases Combustion chamber fuel = m C ( T – T ) 32 apa is the compressor efficiency (Q CC ) in,ideal = CC / CC CC /

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

49 = 80% CC = 80% CC

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

51 = 80% CC

52 C = 80% = 85% T = 88%

53 atmospheric air gases to the stack (Q CC ) in Gen compressed air hot gases W net Heat Loss? Comp- ressor Combustion chamber Gas turbine = - (Q CC ) in W net

54

55 atmospheric air gases to the atmosphere through the stack (Q CC ) in Gen compressed air hot gases W net Heat Loss Comp- ressor Combustion chamber Gas turbine

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?


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