3. 3
3Global Aspiration …… A World Class University
Brayton Cycle
The Brayton cycle (a.k.a. Joule cycle) is the air-standard ideal
cycle approximation for the gas turbine engine.
4. 4
Process Description Related formula
1-2 Isentropic compression
2-3 Constant pressure heat addition
3-4 Isentropic expansion
4-1 Constant pressure heat rejection
1
2
1
1
2
2
1
−
=
=
k
k
k
T
T
V
V
P
P
1
2
1
1
2
2
1
−
=
=
k
k
k
T
T
V
V
P
P
Brayton Cycle Process Description
Global Aspiration …… A World Class University
13. 13
13Global Aspiration …… A World Class University
ηth Brayton
net
in
out
in
W
Q
Q
Q
, = = −1
where the pressure ratio is rp = P2/P1
ηth Brayton
p
k k
r
, ( )/
= − −
1
1
1
Upon derivation
Thermal efficiency of the Brayton cycle is defined as
15. 15
15Global Aspiration …… A World Class University
For fixed values of Tmin
and Tmax, the net work
of the Brayton cycle
first increases with the
pressure ratio, then
reaches a maximum at
rp = (Tmax/Tmin)k/[2(k - 1)]
,
and finally decreases.
16. 16
16Global Aspiration …… A World Class University
( )
( )43
12
TTc
TTc
W
W
r
p
p
t
c
bw
−
−
==
Back Work Ratio
Part of the work output from turbine is used to drive the
compressor, which in turn requires a work input. Therefore
the back work ratio can be written as
17. 17
17Global Aspiration …… A World Class University
Example GT-1
The ideal air-standard Brayton cycle operates with air
entering the compressor at 95 kPa, 22°C. The pressure
ratio rp is 6:1 and the air leaves the heat addition process at
1100 K. Determine the compressor work and the turbine
work per unit mass flow, the cycle efficiency and the back
work ratio. Assume constant properties.
18. 18
18Global Aspiration …… A World Class University
Cycle Improvement - Regenerative
Therefore, a heat exchanger can be placed between the
hot gases leaving the turbine and the cooler gases leaving
the compressor.
This heat exchanger is called a regenerator or recuperator.
For the Brayton cycle, the turbine exhaust temperature is
greater than the compressor exit temperature.
20. 20
20Global Aspiration …… A World Class University
The regenerator effectiveness εregen is defined as the ratio of
the heat transferred to the compressor gases in the
regenerator to the maximum possible heat transfer to the
compressor gases.
q h h
q h h h h
q
q
h h
h h
regen act
regen
regen
regen act
regen
,
, max '
,
, max
= −
= − = −
= =
−
−
5 2
5 2 4 2
5 2
4 2
ε
21. 21
21Global Aspiration …… A World Class University
For ideal gases using the cold-air-standard assumption
with constant specific heats, the regenerator effectiveness
becomes
5 2
4 2
regen
T T
T T
ε
−
≅
−
Upon derivation the thermal efficiency becomes
23. 23
23Global Aspiration …… A World Class University
Example GT-2
Air enters the compressor of a regenerative gas turbine
engine at 100 kPa and 300 K and is compressed to 800 kPa.
The regenerator has an effectiveness of 65 percent, and the
air enters the turbine at 1200 K. For a compressor efficiency
of 75 percent and a turbine efficiency of 86 percent,
determine
(a) The heat transfer in the regenerator.
(b) The back work ratio.
(c) The cycle thermal efficiency.
Assume air is an ideal gas with constant specific heats.
26. 26
26Global Aspiration …… A World Class University
Example GT-3
An ideal gas turbine with two-stage compression and two-
stage expansion has an overall pressure ratio of 8. Air
enters each stage of the compressor at 300 K and each
stage of turbine at 1300 K. Determine the back work ratio
and the thermal efficiency of the cycle assuming (a) no
regenerators and (b) an ideal regenerator with 100%
effectiveness and (c) a regenerator with 60%
effectiveness. Assume constant specific heats.
33. 33
KJM 281
33Global Aspiration …… A World Class University
Based on the Carnot cycle the heat engine may be composed
of the following components.
Carnot Cycle
34. 34
KJM 281
34Global Aspiration …… A World Class University
0.0 1.0 2.0 3.0 4.0 5.0 6.0 7.0 8.0 9.0 10.0
0
100
200
300
400
500
600
700700
s [kJ/kg-K]
T[C]
6000 kPa
100 kPa
Carnot Vapor Cycle Using Steam
1
2
3
4
ηth Carnot
net
in
out
in
L
H
W
Q
Q
Q
T
T
, = = −
= −
1
1
The thermal efficiency of this cycle is given as
35. 35
KJM 281
35Global Aspiration …… A World Class University
Reasons why the Carnot cycle is not used:
Pumping process 1-2 requires the pumping of a mixture
of saturated liquid and vapor.
Low quality steam at the turbine exit.
The impracticalities of the Carnot cycle can be eliminated by:
Condensing the steam completely in the condenser.
Superheating the steam to take advantage of a higher
temperature.
36. 36
KJM 281
36Global Aspiration …… A World Class University
Process Description
1-2 Isentropic compression in pump
2-3 Constant pressure heat addition in boiler
3-4 Isentropic expansion in turbine
4-1 Constant pressure heat rejection in condenser
Rankine Cycle
The simple Rankine cycle has the same component layout
as the Carnot cycle. The processes in a simple Rankine
cycle are:
38. 38
KJM 281
38Global Aspiration …… A World Class University
The pump work can be determined by:
Cycle Analysis
( )
( )121
12
2211
PPm
hhmW
hmWhm
pump
pump
−=
−=
=+
•
••
•••
υ
The energy balance for boiler:
( )23
3322
hhmQ
hmQhm
in
in
−=
=+
••
•••
39. 39
KJM 281
39Global Aspiration …… A World Class University
The energy balance for turbine:
( )43
4433
hhmW
hmWhm
out
out
−=
+=
••
•••
The energy balance for condenser:
( )14
1144
hhmQ
hmQhm
out
out
−=
+=
••
•••
Based on the Second Law the thermal efficiency becomes:
( ) ( )
( )23
1243
hh
hhhh
q
w
in
net
th
−
−−−
==η
43. 43
KJM 281
43Global Aspiration …… A World Class University
Example SPP-1
A power plant operates on a simple Rankine cycle. Steam
enters the turbine at 3 MPa and 350°C and is condensed
in the condenser at a pressure of 75 kPa. Determine (a)
the work net, (b) the heat input in the boiler, and (c) the
thermal efficiency of the cycle.
46. 46
KJM 281
46Global Aspiration …… A World Class University
Component Process First Law Analysis
Boiler Constant Pressure qin
= (h3
- h2
) + (h5
- h4
)
Turbine Isentropic wout
= (h3
- h4
) + (h5
- h6
)
Condenser Constant Pressure qout
= (h6
- h1
)
Pump Isentropic win
= (h2
- h1
) = v1
(P2
- P1
)
( ) ( ) ( )
( ) ( )4523
126543
hhhh
hhhhhh
q
w
in
net
th
−+−
−−−+−
==η
47. 47
KJM 281
47Global Aspiration …… A World Class University
Example SPP-2
Consider a steam power plant operating on the ideal reheat
Rankine cycle. Steam enters the high-pressure turbine at
15 MPa and 600°C and is condensed in the condenser at a
pressure of 10 kPa. If the moisture content of the steam at
the exit of the low-pressure turbine is not to exceed 10.4
percent, determine (a) the pressure at which the steam
should be reheated, and (b) the thermal efficiency of the
cycle. Assume the steam is reheated to the inlet
temperature of the high-pressure turbine.
51. 51
KJM 281
51Global Aspiration …… A World Class University
Example SPP-3
Consider a steam power plant operating on the ideal
regenerative Rankine with one open feedwater heater. Steam
enters the turbine at 150 bar and 600°C and is condensed in
the condenser at a pressure of 0.1 bar. Some steam leaves
the turbine at a pressure of 12 bar and enters the open
feedwater heater. Determine the fraction of steam extracted
from the turbine and the thermal efficiency of the cycle.
59. 59
KJM 281
59Global Aspiration …… A World Class University
Karlsruhe Power Station (Germany)
Topping cycle = 2 x Bottoming cycle
60. 60
KJM 281
60Global Aspiration …… A World Class University
Sultan Ismail Power Station (Paka)
Combined Cycle (808 MW)
Gas Turbine – 137 MW x 4 Units
Steam Turbine – 130 MW x 2 Unit
Topping cycle = 2 x Bottoming cycle
61. 61
KJM 281
61Global Aspiration …… A World Class University
Pasir Gudang Power Station (Johor)
Combined Cycle (404 MW)
Gas Turbine – 137 MW x 2 Units
Steam Turbine – 130 MW x 1 Unit
Topping cycle = 2 x Bottoming cycle