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The Brayton Cycle
Used in gas turbine Jet engines and power production
The ideal Brayton cycle comprises three reversible processes. It always uses air as the working fluid. An open system Brayton cycle is implemented using the three steady flow devices shown:
The Brayton Cycle Q in
3
Combustor (B)
2
Turbine (T)
Compressor (C)
Air in T
4
1
Exhaust out
3
TH Lines of constant press.
2
4
TL
Wout
1
s
The Brayton Cycle Q
C) An isentropic compression
in
•
2
Q C = 0, •
W C =,
•
m C P (T2 − T1 )
process 1-2.
3
Combustor (B)
Turbine (T)
Compressor (C)
Air in T
4
1
Exhaust out
3
TH Lines of constant press.
2
4
TL
Wout
1
s
The Brayton Cycle Q
B) An isobaric heat addition
in
•
2
W B = 0, •
•
Q B =, m C P (T3 − T2 )
process 2-3.
3
Combustor (B)
Turbine (T)
Compressor (C)
Air in T
4
1
Exhaust out
3
TH Lines of constant press.
2
4
TL
Wout
1
s
The Brayton Cycle Q
B) An isentropic expansion
in
•
2
Q B = 0, •
•
W B =, m (T4 − T3 )
process 3-4.
3
Combustor (B)
Turbine (T)
Compressor (C)
Air in T
4
1
Exhaust out
3
Note that the Turbine and compressor are on the same shaft.
TH Lines of constant press.
2
4
TL
Wout
1
s
This complete cycle is reversible if: 1. the isothermal expansion and compression are reversible, and 2. the expansion and compression are isentropic.
The Brayton Cycle Qin
3
Combustor (B)
2
Turbine (T)
Compressor (C)
Air in T
4
1
Exhaust out
3
TH Lines of constant press.
2
4
TL
Wout
1
s
A fully reversible cycle has not yet been built, but gas turbines running with Brayton cycles are very common in both power production (gas turbines) and transport (jet engines)
The Brayton Cycle Qin
3
Combustor (B)
2
Turbine (T)
Compressor (C)
Air in T
4
1
Exhaust out
3
TH Lines of constant press.
2
4
TL
Wout
1
s
Cutaway P&W JT8D
Example of a Brayton Cycle for Air
For an ideal Brayton cycle running with a flow rate of 4 kg of air per second, find the heat and work values for each of the three processes in the cycle. The cycle will run with a turbine inlet temperature of 1000 °C T and an air inlet 3 temperature of 10 °C. T 1273K 10bar The compressor 2 outlet pressure is a 1bar 10 bar. Air inlet pressure is 1 bar. 4 283K Use n = γ =1.40 and T 1 -1 -1 = 1.005 kJ kg K . H
L
s
Process 1-2 (isentropic compression) Initial temperature = 283 K, initial pressure = 1 bar, n −1 final pressure = 10 bar p2 n T2 = T1 , so 0.4 10 1.4 T2 = 283 =546K 1 =273°C
p1
T TH
1273K
3
10bar 2
1bar
TL
4
283K 1
s
Process 1-2 (isentropic compression) Initial temperature = 283 K, initial pressure = 1 bar, n −1 final pressure = 10 bar p2 n T2 = T1 , so 0.4 10 1.4 T2 = 283 =546K 1 =273°C
p1
T TH
1273K
3
The work put in will be: •
10bar 2
•
− WC = m (T2 − T1 ) =
546K
1bar
4 × 1.005 × (546 − 283)
= 1059 kW
TL
4
283K 1
s
Process 3-4 (adiabatic expansion) Initial temperature = 1273 K, initial pressure = 10 bar, final pressure = 1 bar , To find the final temperature n −1 p2 T2 = T1 1 T2 = 1273 =659K p1 T 10 0.4 1.4
=386°C
The work out will be: •
TH
1273K
3
10bar 2
•
WT = m (T3 − T4 )
546K
1bar
= 4 × 1.005 × (1273 − 659)
= 2467 kW
n
659K TL
4
283K 1
s
Process 2-3 (heat addition) Initial pressure = 10 bar, Initial temperature = 546 K, Final temperature = 1273 K The heat in will be: •
QB = (T3 − T2 )
T
= 4 × 1.005 × (1273 − 546 ) T
1273K
3
H
10bar
= 2921 kW
2
546K
1bar 659K
TL
4
283K 1
s
The total power out is 2467 − 1059 = 1408kW. This gives an efficiency of =0.482 = 48.2%.
1408 = 2921
By comparison, the expected Carnot efficiency is given by, Tsource − Tsink 1273 − 283 η= = = 0.778 Tsource 1273
Also,
= 77.8%.
Net Work 1408 = = 57.1% Total Turbine Work 2467
so about 40% of the turbine power is used to drive the compressor.
The Effect of Isentropic Efficiency on a Brayton Cycle
If the isentropic efficiency of the adiabatic processes is less than 100%, the compression will require more work, and the turbine will produce less work. T This will greatly 3 reduce the output T of the cycle. Lines of 2 H
2s
a
constant press.
4s
TL
4a
1
s
The Effect of Isentropic Efficiency on a Brayton Cycle
On the other hand, the higher outlet temperature from the turbine will reduce slightly the heat input required. T 3
TH 2s
Lines of constant press.
2a
4s
TL
4a
1
s
Example of a Brayton Cycle With an Isentropic Efficiency of <100% For a Brayton cycle running with a flow rate of 4 kg of air per second, find the heat or work values for each of the three processes in the cycle. The cycle will run with a turbine inlet temperature of 1000 °C and an air inlet temperature of 10 °C. The compressor outlet pressure is at 10 bar. Air inlet pressure is 1 bar. The isentropic efficiency of both turbine and compressor is 90%. Use n = γ =1.40 and = 1.005 kJ kg-1 K-1.
Process 1-2 (adiabatic compression) As previous example p2 T2 s = T1 p1
n −1 n
10 , so T2 s = 283 1
0.4 1.4
=546K =273°C
T 3
TH 2s
Lines of constant press.
2a
4s
TL
4a
1
s
Process 1-2 (adiabatic compression) As previous example p2 T2 s = T1 p1
n −1 n
10 , so T2 s = 283 1
With an ηisen of 0.9, the final temperature will be
T2 s − T1 = T2 a = T1 + ηisen 546 − 293 = 283 + 0.9
= 576 K = 303 °C
0.4 1.4
=546K =273°C
T 1273K
TH 2s 546K
TL
3
Lines of constant press.
2a 576K
4s
283K
4a
1
s
Process 1-2 (adiabatic compression) The work in will be: •
•
− WC = m (T1 − T2 a ) = 4 × 1.005 × (576 − 283)
= 1176 kW.
T 1273K
TH 2s 546K
TL
3
Lines of constant press.
2a 576K
4s
283K
4a
1
s
Process 3-4 (adiabatic expansion) As previous example p4 T4 s = T3 p3
n −1 n
1 , so T4 s = 1273 10
0.4 1.4
= 659K =386°C
T 1273K
TH 2s 546K
TL
3
Lines of constant press.
2a 576K
4s
283K
4a
1
s
Process 3-4 (adiabatic expansion) As previous example p4 T4s = T3 p3
n −1 n
1 , so T4s = 1273 10
With an ηisen of 0.9, the final temperature will be
= 659K =386°C
T 1273K
TH
T4 a = T3 + (T4 s − T3 ) × ηisen
2s 546K
= 1273 − (659 − 1273) × 0.9
721 K = 448 °C
0.4 1.4
TL
3
Lines of constant press.
2a 576K
659K 4s
283K
4a
721K
1
s
Process 3-4 (adiabatic expansion) The work out will be: •
•
W T = m (T3 − T4 a ) = 4 × 1.005 × (1273 − 721)
= 2220 kW.
T 1273K
TH 2s 546K
3
Lines of constant press.
2a 576K
659K
TL
4s
283K
4a
721K
1
s
Process 2-3 (isobaric heat addition) Initial pressure = 10 bar, initial temperature = 576 K, final temperature = 1273 K T
The heat in will be: •
1273K
TH
•
3
Lines of constant press.
2a
QB = m (T3 − T2 a )
576K
= 4 × 1.005 × (1273 − 576 ) 4a
= 2803 kW
TL
721K
283K 1
s
The total power out is 2220 − 1176 = 1044 kW. This compares with 1408 kW for a Brayton cycle with an isentropic compressor and turbine. As
Net Work 1044 = = 47.0% Total Turbine Work 2220
Over the half the power produced from the turbine goes into driving the compressor.
The total power out is 2220 − 1176 = 1044 kW. This gives an overall cycle efficiency of 1044 = 0.372 = 37.2%. 2803
This compares with 48.2% for a Brayton cycle with an isentropic compressor and turbine. This corresponds to a 23% reduction in cycle efficiency. If the isentropic efficiency of the compressor and turbine is reduced to 80%, the overall cycle efficiency will drop to 25%!
The effect of Combustor Pressure Drop on a Brayton Cycle The pressure drop across the combustor means that the expansion starts from a reduced pressure, compared to the ideal case. This in turn will raise the outlet T temperature from 3 the turbine and thus T reduce the power 2 Lines of from the turbine. constant i
H
a
press. 4i
TL
1
s
The effect of Combustor Pressure Drop on a Brayton Cycle The pressure drop across the combustor means that the expansion starts from a reduced pressure, compared to the ideal case. This in turn will raise the outlet Pressure T temperature from drop through 3 the turbine and thus T combustor 3 reduce the power 2 Lines of from the turbine. constant i
H
a
a
4a
press. 4i
TL
Temperature difference due to pressure drop
1
s
Example of a Brayton Cycle With a combustor pressure drop A Brayton cycle runs with a flow rate of 4 kg s-1of air. Find the heat and work values for the three processes in the cycle. The cycle uses a turbine inlet temperature of 1000 °C and an air inlet temperature of 10 °C and pressure of 1 bar. The compressor outlet pressure is 10 bar. The pressure drop across the combustor is 5% of the pressure rise in the compressor. The isentropic efficiency of both turbine and compressor is 90%. Use n = γ =1.40 and = 1.005 kJ kg-1 K-1.
Process 1-2 (adiabatic compression) As previous example p2 T2 s = T1 p1
n −1 n
10 , so T2 s = 283 1
With an ηisen of 0.9, the final temperature will be
=546K =273°C Pressure drop through combustor
T 3i
TH
T2 s − T1 = T2 a = T1 + Lines of ηisen constant press. 546 − 293 = 283 + 0.9 T
= 576 K = 303 °C
0.4 1.4
L
3a 576K
2a 4a 4i
Temperature difference due to pressure drop
283K 1
s
Process 1-2 (adiabatic compression) The work in will be: •
•
− WC = m (T1 − T2 a ) = 4 × 1.005 × (576 − 283)
= 1176 kW. Pressure drop through combustor
T 3i
TH
Lines of constant press.
3a 576K
2a 4a 4i
TL
Temperature difference due to pressure drop
283K 1
s
Process 2-3 (heat addition) Pressure drop = 0.05×(10−1) = 0.45 bar Initial pressure = 10 bar, final pressure = 9.55 bar, initial temperature = 576 k, final temperature = 1273 K Pressure The heat in will be: T drop •
•
QB = m (T3 − T2 a )
TH
= 4 × 1.005 × (1273 − 576 )
= 2803 kW.
Lines of constant press.
through combustor
3i 3a 576K
2a 4a 4i
TL
Temperature difference due to pressure drop
283K 1
s
Process 2-3 (heat addition) Note that the pressure drop does not effect the heat input term. However, it will increase the entropy of the gas, reducing its potential for doing work. This will manifest itself in the next section •
•
QB = m (T3 − T2 a )
3i
TH
= 4 × 1.005 × (1273 − 576 )
= 2803 kW.
Pressure drop through combustor
T
Lines of constant press.
3a 576K
2a 4a 4i
TL
Temperature difference due to pressure drop
283K 1
s
Process 3-4 (adiabatic expansion) Initial temperature = 1273K, initial pressure = 9.55 bar, final pressure = 1 bar. To find the final temperature, initially assume 100% isentropic efficiency p4 a T4i = T3 p3i
n −1 n
1273K
TH
, so 1 T4i = 1273 9.55
Pressure drop through combustor
T
0.4 1.4
= 668K = 395°C
Lines of constant press.
TL
3i 3a
576K
2a 4a 4as
283K 1
668K 4i Temperature difference due to pressure drop
s
Process 3-4 (adiabatic expansion) T4i= 668K = 395°C With an isentropic efficiency of 0.9, the final temperature will be:
T4 a = T3 + (T4 − T3 ) ×η isen = 1273 − (668 − 1273) × 0.9 Pressure drop through combustor
T
= 729K = 456°C
1273K
TH
Lines of constant press.
TL
3i 3a
576K
2a 4a 4as
283K 1
729K
721K 4i Temperature difference due to pressure drop
s
Process 3-4 (adiabatic expansion) The work out will be •
•
W = m (T4 − T3 ) = 4 × 1.005 × (1273 − 729 )
= 2189 kW Pressure drop through combustor
T 1273K
TH
Lines of constant press.
TL
3a 576K
2a 4a
283K 1
s
729K
The total power out is 2189 − 1176 = 1012kW. This compares with 1044 kW for a Brayton cycle with no pressure drop across the combustor As
Net Work 1012 = = 46.2% Total Turbine Work 2189
Again, over the half the power produced from the turbine goes into driving the compressor.
The total power out is 2189 − 1176 = 1012kW. The overall cycle efficiency is 1012 = 2803
0.36.1 = 36.1%
This compares with 37.2% for a Brayton cycle with the same isentropic efficiencies, but no pressure loss through the combustor. A five percent pressure drop in the combustor thus leads to a 3% drop in cycle efficiency.
To Summarise These Results; Turbine inlet temperature = 1000 °C Air inlet temperature = 10 °C Air inlet pressure = 1 bar Compressor outlet pressure = 10 bar. The effect of isentropic efficiency and pressure drop through the combustor are:
Condition
1
2
3
Isentropic efficiencies (%)
100
90
90
Combustor pressure drop (b)
0
0
0.45
Heat input rate (kW)
2921
2803
2803
Net power output (kW)
1408
1044
1012
Net work/Turbine work (%)
57.1
47.0
46.2
Cycle efficiency (%)
48.2
37.2
36.1
Drop from previous (%)
N/A
22.8
2.96
Thus endeth the course
Used in gas turbine Jet engines and power production
The ideal Brayton cycle comprises three reversible processes. It always uses air as the working fluid. An open system Brayton cycle is implemented using the three steady flow devices shown:
The Brayton Cycle Q in
3
Combustor (B)
2
Turbine (T)
Compressor (C)
Air in T
4
1
Exhaust out
3
TH Lines of constant press.
2
4
TL
Wout
1
s
The Brayton Cycle Q
C) An isentropic compression
in
•
2
Q C = 0, •
W C =,
•
m C P (T2 − T1 )
process 1-2.
3
Combustor (B)
Turbine (T)
Compressor (C)
Air in T
4
1
Exhaust out
3
TH Lines of constant press.
2
4
TL
Wout
1
s
The Brayton Cycle Q
B) An isobaric heat addition
in
•
2
W B = 0, •
•
Q B =, m C P (T3 − T2 )
process 2-3.
3
Combustor (B)
Turbine (T)
Compressor (C)
Air in T
4
1
Exhaust out
3
TH Lines of constant press.
2
4
TL
Wout
1
s
The Brayton Cycle Q
B) An isentropic expansion
in
•
2
Q B = 0, •
•
W B =, m (T4 − T3 )
process 3-4.
3
Combustor (B)
Turbine (T)
Compressor (C)
Air in T
4
1
Exhaust out
3
Note that the Turbine and compressor are on the same shaft.
TH Lines of constant press.
2
4
TL
Wout
1
s
This complete cycle is reversible if: 1. the isothermal expansion and compression are reversible, and 2. the expansion and compression are isentropic.
The Brayton Cycle Qin
3
Combustor (B)
2
Turbine (T)
Compressor (C)
Air in T
4
1
Exhaust out
3
TH Lines of constant press.
2
4
TL
Wout
1
s
A fully reversible cycle has not yet been built, but gas turbines running with Brayton cycles are very common in both power production (gas turbines) and transport (jet engines)
The Brayton Cycle Qin
3
Combustor (B)
2
Turbine (T)
Compressor (C)
Air in T
4
1
Exhaust out
3
TH Lines of constant press.
2
4
TL
Wout
1
s
Cutaway P&W JT8D
Example of a Brayton Cycle for Air
For an ideal Brayton cycle running with a flow rate of 4 kg of air per second, find the heat and work values for each of the three processes in the cycle. The cycle will run with a turbine inlet temperature of 1000 °C T and an air inlet 3 temperature of 10 °C. T 1273K 10bar The compressor 2 outlet pressure is a 1bar 10 bar. Air inlet pressure is 1 bar. 4 283K Use n = γ =1.40 and T 1 -1 -1 = 1.005 kJ kg K . H
L
s
Process 1-2 (isentropic compression) Initial temperature = 283 K, initial pressure = 1 bar, n −1 final pressure = 10 bar p2 n T2 = T1 , so 0.4 10 1.4 T2 = 283 =546K 1 =273°C
p1
T TH
1273K
3
10bar 2
1bar
TL
4
283K 1
s
Process 1-2 (isentropic compression) Initial temperature = 283 K, initial pressure = 1 bar, n −1 final pressure = 10 bar p2 n T2 = T1 , so 0.4 10 1.4 T2 = 283 =546K 1 =273°C
p1
T TH
1273K
3
The work put in will be: •
10bar 2
•
− WC = m (T2 − T1 ) =
546K
1bar
4 × 1.005 × (546 − 283)
= 1059 kW
TL
4
283K 1
s
Process 3-4 (adiabatic expansion) Initial temperature = 1273 K, initial pressure = 10 bar, final pressure = 1 bar , To find the final temperature n −1 p2 T2 = T1 1 T2 = 1273 =659K p1 T 10 0.4 1.4
=386°C
The work out will be: •
TH
1273K
3
10bar 2
•
WT = m (T3 − T4 )
546K
1bar
= 4 × 1.005 × (1273 − 659)
= 2467 kW
n
659K TL
4
283K 1
s
Process 2-3 (heat addition) Initial pressure = 10 bar, Initial temperature = 546 K, Final temperature = 1273 K The heat in will be: •
QB = (T3 − T2 )
T
= 4 × 1.005 × (1273 − 546 ) T
1273K
3
H
10bar
= 2921 kW
2
546K
1bar 659K
TL
4
283K 1
s
The total power out is 2467 − 1059 = 1408kW. This gives an efficiency of =0.482 = 48.2%.
1408 = 2921
By comparison, the expected Carnot efficiency is given by, Tsource − Tsink 1273 − 283 η= = = 0.778 Tsource 1273
Also,
= 77.8%.
Net Work 1408 = = 57.1% Total Turbine Work 2467
so about 40% of the turbine power is used to drive the compressor.
The Effect of Isentropic Efficiency on a Brayton Cycle
If the isentropic efficiency of the adiabatic processes is less than 100%, the compression will require more work, and the turbine will produce less work. T This will greatly 3 reduce the output T of the cycle. Lines of 2 H
2s
a
constant press.
4s
TL
4a
1
s
The Effect of Isentropic Efficiency on a Brayton Cycle
On the other hand, the higher outlet temperature from the turbine will reduce slightly the heat input required. T 3
TH 2s
Lines of constant press.
2a
4s
TL
4a
1
s
Example of a Brayton Cycle With an Isentropic Efficiency of <100% For a Brayton cycle running with a flow rate of 4 kg of air per second, find the heat or work values for each of the three processes in the cycle. The cycle will run with a turbine inlet temperature of 1000 °C and an air inlet temperature of 10 °C. The compressor outlet pressure is at 10 bar. Air inlet pressure is 1 bar. The isentropic efficiency of both turbine and compressor is 90%. Use n = γ =1.40 and = 1.005 kJ kg-1 K-1.
Process 1-2 (adiabatic compression) As previous example p2 T2 s = T1 p1
n −1 n
10 , so T2 s = 283 1
0.4 1.4
=546K =273°C
T 3
TH 2s
Lines of constant press.
2a
4s
TL
4a
1
s
Process 1-2 (adiabatic compression) As previous example p2 T2 s = T1 p1
n −1 n
10 , so T2 s = 283 1
With an ηisen of 0.9, the final temperature will be
T2 s − T1 = T2 a = T1 + ηisen 546 − 293 = 283 + 0.9
= 576 K = 303 °C
0.4 1.4
=546K =273°C
T 1273K
TH 2s 546K
TL
3
Lines of constant press.
2a 576K
4s
283K
4a
1
s
Process 1-2 (adiabatic compression) The work in will be: •
•
− WC = m (T1 − T2 a ) = 4 × 1.005 × (576 − 283)
= 1176 kW.
T 1273K
TH 2s 546K
TL
3
Lines of constant press.
2a 576K
4s
283K
4a
1
s
Process 3-4 (adiabatic expansion) As previous example p4 T4 s = T3 p3
n −1 n
1 , so T4 s = 1273 10
0.4 1.4
= 659K =386°C
T 1273K
TH 2s 546K
TL
3
Lines of constant press.
2a 576K
4s
283K
4a
1
s
Process 3-4 (adiabatic expansion) As previous example p4 T4s = T3 p3
n −1 n
1 , so T4s = 1273 10
With an ηisen of 0.9, the final temperature will be
= 659K =386°C
T 1273K
TH
T4 a = T3 + (T4 s − T3 ) × ηisen
2s 546K
= 1273 − (659 − 1273) × 0.9
721 K = 448 °C
0.4 1.4
TL
3
Lines of constant press.
2a 576K
659K 4s
283K
4a
721K
1
s
Process 3-4 (adiabatic expansion) The work out will be: •
•
W T = m (T3 − T4 a ) = 4 × 1.005 × (1273 − 721)
= 2220 kW.
T 1273K
TH 2s 546K
3
Lines of constant press.
2a 576K
659K
TL
4s
283K
4a
721K
1
s
Process 2-3 (isobaric heat addition) Initial pressure = 10 bar, initial temperature = 576 K, final temperature = 1273 K T
The heat in will be: •
1273K
TH
•
3
Lines of constant press.
2a
QB = m (T3 − T2 a )
576K
= 4 × 1.005 × (1273 − 576 ) 4a
= 2803 kW
TL
721K
283K 1
s
The total power out is 2220 − 1176 = 1044 kW. This compares with 1408 kW for a Brayton cycle with an isentropic compressor and turbine. As
Net Work 1044 = = 47.0% Total Turbine Work 2220
Over the half the power produced from the turbine goes into driving the compressor.
The total power out is 2220 − 1176 = 1044 kW. This gives an overall cycle efficiency of 1044 = 0.372 = 37.2%. 2803
This compares with 48.2% for a Brayton cycle with an isentropic compressor and turbine. This corresponds to a 23% reduction in cycle efficiency. If the isentropic efficiency of the compressor and turbine is reduced to 80%, the overall cycle efficiency will drop to 25%!
The effect of Combustor Pressure Drop on a Brayton Cycle The pressure drop across the combustor means that the expansion starts from a reduced pressure, compared to the ideal case. This in turn will raise the outlet T temperature from 3 the turbine and thus T reduce the power 2 Lines of from the turbine. constant i
H
a
press. 4i
TL
1
s
The effect of Combustor Pressure Drop on a Brayton Cycle The pressure drop across the combustor means that the expansion starts from a reduced pressure, compared to the ideal case. This in turn will raise the outlet Pressure T temperature from drop through 3 the turbine and thus T combustor 3 reduce the power 2 Lines of from the turbine. constant i
H
a
a
4a
press. 4i
TL
Temperature difference due to pressure drop
1
s
Example of a Brayton Cycle With a combustor pressure drop A Brayton cycle runs with a flow rate of 4 kg s-1of air. Find the heat and work values for the three processes in the cycle. The cycle uses a turbine inlet temperature of 1000 °C and an air inlet temperature of 10 °C and pressure of 1 bar. The compressor outlet pressure is 10 bar. The pressure drop across the combustor is 5% of the pressure rise in the compressor. The isentropic efficiency of both turbine and compressor is 90%. Use n = γ =1.40 and = 1.005 kJ kg-1 K-1.
Process 1-2 (adiabatic compression) As previous example p2 T2 s = T1 p1
n −1 n
10 , so T2 s = 283 1
With an ηisen of 0.9, the final temperature will be
=546K =273°C Pressure drop through combustor
T 3i
TH
T2 s − T1 = T2 a = T1 + Lines of ηisen constant press. 546 − 293 = 283 + 0.9 T
= 576 K = 303 °C
0.4 1.4
L
3a 576K
2a 4a 4i
Temperature difference due to pressure drop
283K 1
s
Process 1-2 (adiabatic compression) The work in will be: •
•
− WC = m (T1 − T2 a ) = 4 × 1.005 × (576 − 283)
= 1176 kW. Pressure drop through combustor
T 3i
TH
Lines of constant press.
3a 576K
2a 4a 4i
TL
Temperature difference due to pressure drop
283K 1
s
Process 2-3 (heat addition) Pressure drop = 0.05×(10−1) = 0.45 bar Initial pressure = 10 bar, final pressure = 9.55 bar, initial temperature = 576 k, final temperature = 1273 K Pressure The heat in will be: T drop •
•
QB = m (T3 − T2 a )
TH
= 4 × 1.005 × (1273 − 576 )
= 2803 kW.
Lines of constant press.
through combustor
3i 3a 576K
2a 4a 4i
TL
Temperature difference due to pressure drop
283K 1
s
Process 2-3 (heat addition) Note that the pressure drop does not effect the heat input term. However, it will increase the entropy of the gas, reducing its potential for doing work. This will manifest itself in the next section •
•
QB = m (T3 − T2 a )
3i
TH
= 4 × 1.005 × (1273 − 576 )
= 2803 kW.
Pressure drop through combustor
T
Lines of constant press.
3a 576K
2a 4a 4i
TL
Temperature difference due to pressure drop
283K 1
s
Process 3-4 (adiabatic expansion) Initial temperature = 1273K, initial pressure = 9.55 bar, final pressure = 1 bar. To find the final temperature, initially assume 100% isentropic efficiency p4 a T4i = T3 p3i
n −1 n
1273K
TH
, so 1 T4i = 1273 9.55
Pressure drop through combustor
T
0.4 1.4
= 668K = 395°C
Lines of constant press.
TL
3i 3a
576K
2a 4a 4as
283K 1
668K 4i Temperature difference due to pressure drop
s
Process 3-4 (adiabatic expansion) T4i= 668K = 395°C With an isentropic efficiency of 0.9, the final temperature will be:
T4 a = T3 + (T4 − T3 ) ×η isen = 1273 − (668 − 1273) × 0.9 Pressure drop through combustor
T
= 729K = 456°C
1273K
TH
Lines of constant press.
TL
3i 3a
576K
2a 4a 4as
283K 1
729K
721K 4i Temperature difference due to pressure drop
s
Process 3-4 (adiabatic expansion) The work out will be •
•
W = m (T4 − T3 ) = 4 × 1.005 × (1273 − 729 )
= 2189 kW Pressure drop through combustor
T 1273K
TH
Lines of constant press.
TL
3a 576K
2a 4a
283K 1
s
729K
The total power out is 2189 − 1176 = 1012kW. This compares with 1044 kW for a Brayton cycle with no pressure drop across the combustor As
Net Work 1012 = = 46.2% Total Turbine Work 2189
Again, over the half the power produced from the turbine goes into driving the compressor.
The total power out is 2189 − 1176 = 1012kW. The overall cycle efficiency is 1012 = 2803
0.36.1 = 36.1%
This compares with 37.2% for a Brayton cycle with the same isentropic efficiencies, but no pressure loss through the combustor. A five percent pressure drop in the combustor thus leads to a 3% drop in cycle efficiency.
To Summarise These Results; Turbine inlet temperature = 1000 °C Air inlet temperature = 10 °C Air inlet pressure = 1 bar Compressor outlet pressure = 10 bar. The effect of isentropic efficiency and pressure drop through the combustor are:
Condition
1
2
3
Isentropic efficiencies (%)
100
90
90
Combustor pressure drop (b)
0
0
0.45
Heat input rate (kW)
2921
2803
2803
Net power output (kW)
1408
1044
1012
Net work/Turbine work (%)
57.1
47.0
46.2
Cycle efficiency (%)
48.2
37.2
36.1
Drop from previous (%)
N/A
22.8
2.96
Thus endeth the course
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