Download presentation

Published byYessenia Stedman Modified over 3 years ago

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 3**

4 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 3**

4 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 turbine**

3 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 2s**

1 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 compressor**

1 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 fuel**

2 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 air**

2 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 fuel**

2 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?

Similar presentations

Presentation is loading. Please wait....

OK

CHAPTER 5: Mass and Energy Analysis of Control Volumes

CHAPTER 5: Mass and Energy Analysis of Control Volumes

© 2018 SlidePlayer.com Inc.

All rights reserved.

Ads by Google

Ppt on review of related literature on research Ppt on hydroelectric plants in india Ppt on percentage for class 6 Ppt on computer malwares for mac Convert word doc to ppt online shopping Ppt on non biodegradable wastewater Ppt on himalaya in hindi Ppt on personality development in hindi Download ppt on national festivals of india Download ppt on area of parallelogram