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MEC 451
THERMODYNAMICS
1Global Aspiration …… A World Class University
CHAPTER 6
THERMAL POWER PLANT
2
PART 1
2Global Aspiration …… A World Class University
GAS TURBINE POWER PLANT
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
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
5
5Global Aspiration …… A World Class University
The T-s and P-v diagrams are
6
6Global Aspiration …… A World Class University
The T-s for actual cycle (including pressure drop and
isentropic efficiency)
7
∑∑ 







++++=







++++
out
out
out
outoutoutout
in
in
in
inininin gz
V
hmWQgz
V
hmWQ
22
2...2...
( ) ( )outoutininin hmhmW
...
=+
( )12
.
12
.
hhmW −=
Global Aspiration …… A World Class University
Energy balance: for compressor
8
8Global Aspiration …… A World Class University
Isentropic efficiency for compressor:
9
∑∑ 







++++=







++++
out
out
out
outoutoutout
in
in
in
inininin gz
V
hmWQgz
V
hmWQ
22
2...2...
( ) ( )outoutininin hmhmQ
...
=+
( )23
.
23
.
hhmQ −=
Global Aspiration …… A World Class University
Energy balance: for boiler
2
3
10
∑∑ 







++++=







++++
out
out
out
outoutoutout
in
in
in
inininin gz
V
hmWQgz
V
hmWQ
22
2...2...
( ) ( )outoutoutinin hmWhm
...
+=
( )43
.
34
.
hhmW −=
Global Aspiration …… A World Class University
3
4
Energy balance: for turbine
11
11Global Aspiration …… A World Class University
Isentropic efficiency for turbine:
12
∑∑ 







++++=







++++
out
out
out
outoutoutout
in
in
in
inininin gz
V
hmWQgz
V
hmWQ
22
2...2...
( ) ( ) outoutoutinin Qhmhm
...
+=
( )14
.
41
.
hhmQ −=
Global Aspiration …… A World Class University
4
1
Energy balance: for condenser
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
14
14Global Aspiration …… A World Class University
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
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
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
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.
19
19Global Aspiration …… A World Class University
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
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
22
22Global Aspiration …… A World Class University
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.
24
24Global Aspiration …… A World Class University
Cycle Improvement – Intercooling and Reheating
25
25Global Aspiration …… A World Class University
The T-s diagram for this cycle is shown below.
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.
27
27Global Aspiration …… A World Class University
Aircraft Engine
28
28Global Aspiration …… A World Class University
Alstom GT26 Gas Turbine
29
29Global Aspiration …… A World Class University
PW 4000 112-INCH FAN ENGINE
30
30Global Aspiration …… A World Class University
Rolls – Royce Trent
31
31Global Aspiration …… A World Class University
General Electric (GE) Gas Turbine
32
PART 2
32Global Aspiration …… A World Class University
STEAM POWER PLANT
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
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
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
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:
37
KJM 281
37Global Aspiration …… A World Class University
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
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
40
KJM 281
40Global Aspiration …… A World Class University
Cycle Improvement
41
KJM 281
41Global Aspiration …… A World Class University
42
KJM 281
42Global Aspiration …… A World Class University
Actual Cycle
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.
44
KJM 281
44Global Aspiration …… A World Class University
Reheat Rankine Cycle
45
KJM 281
45Global Aspiration …… A World Class University
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
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.
48
KJM 281
48Global Aspiration …… A World Class University
Regenerative Cycle
49
KJM 281
49Global Aspiration …… A World Class University
Regenerative Cycle with Open FWH
50
KJM 281
50Global Aspiration …… A World Class University
Regenerative Cycle with Closed FWH
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.
52
KJM 281
52Global Aspiration …… A World Class University
Plant Layout
53
KJM 281
53Global Aspiration …… A World Class University
Plant Monitoring System
54
PART 3
54Global Aspiration …… A World Class University
SPECIAL APPLICATION
55
KJM 281
55Global Aspiration …… A World Class University
Cogeneration
56
KJM 281
56Global Aspiration …… A World Class University
Cogeneration Stand-Alone System
57
KJM 281
57Global Aspiration …… A World Class University
Trigeneration Plant
58
KJM 281
58Global Aspiration …… A World Class University
Combined Cycle
59
KJM 281
59Global Aspiration …… A World Class University
Karlsruhe Power Station (Germany)
Topping cycle = 2 x Bottoming cycle
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
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
62
KJM 281
62Global Aspiration …… A World Class University
Connaught Bridge Power Station (Klang)
63
KJM 281
63Global Aspiration …… A World Class University
Binary Vapor Cycle
64
KJM 281
64Global Aspiration …… A World Class University

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Thermodynamic Chapter 6 Thermal Power Plant

  • 1. 1 MEC 451 THERMODYNAMICS 1Global Aspiration …… A World Class University CHAPTER 6 THERMAL POWER PLANT
  • 2. 2 PART 1 2Global Aspiration …… A World Class University GAS TURBINE POWER PLANT
  • 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
  • 5. 5 5Global Aspiration …… A World Class University The T-s and P-v diagrams are
  • 6. 6 6Global Aspiration …… A World Class University The T-s for actual cycle (including pressure drop and isentropic efficiency)
  • 7. 7 ∑∑         ++++=        ++++ out out out outoutoutout in in in inininin gz V hmWQgz V hmWQ 22 2...2... ( ) ( )outoutininin hmhmW ... =+ ( )12 . 12 . hhmW −= Global Aspiration …… A World Class University Energy balance: for compressor
  • 8. 8 8Global Aspiration …… A World Class University Isentropic efficiency for compressor:
  • 9. 9 ∑∑         ++++=        ++++ out out out outoutoutout in in in inininin gz V hmWQgz V hmWQ 22 2...2... ( ) ( )outoutininin hmhmQ ... =+ ( )23 . 23 . hhmQ −= Global Aspiration …… A World Class University Energy balance: for boiler 2 3
  • 10. 10 ∑∑         ++++=        ++++ out out out outoutoutout in in in inininin gz V hmWQgz V hmWQ 22 2...2... ( ) ( )outoutoutinin hmWhm ... += ( )43 . 34 . hhmW −= Global Aspiration …… A World Class University 3 4 Energy balance: for turbine
  • 11. 11 11Global Aspiration …… A World Class University Isentropic efficiency for turbine:
  • 12. 12 ∑∑         ++++=        ++++ out out out outoutoutout in in in inininin gz V hmWQgz V hmWQ 22 2...2... ( ) ( ) outoutoutinin Qhmhm ... += ( )14 . 41 . hhmQ −= Global Aspiration …… A World Class University 4 1 Energy balance: for condenser
  • 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
  • 14. 14 14Global Aspiration …… A World Class University
  • 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.
  • 19. 19 19Global Aspiration …… A World Class University
  • 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
  • 22. 22 22Global Aspiration …… A World Class University
  • 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.
  • 24. 24 24Global Aspiration …… A World Class University Cycle Improvement – Intercooling and Reheating
  • 25. 25 25Global Aspiration …… A World Class University The T-s diagram for this cycle is shown below.
  • 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.
  • 27. 27 27Global Aspiration …… A World Class University Aircraft Engine
  • 28. 28 28Global Aspiration …… A World Class University Alstom GT26 Gas Turbine
  • 29. 29 29Global Aspiration …… A World Class University PW 4000 112-INCH FAN ENGINE
  • 30. 30 30Global Aspiration …… A World Class University Rolls – Royce Trent
  • 31. 31 31Global Aspiration …… A World Class University General Electric (GE) Gas Turbine
  • 32. 32 PART 2 32Global Aspiration …… A World Class University STEAM POWER PLANT
  • 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:
  • 37. 37 KJM 281 37Global Aspiration …… A World Class University
  • 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
  • 40. 40 KJM 281 40Global Aspiration …… A World Class University Cycle Improvement
  • 41. 41 KJM 281 41Global Aspiration …… A World Class University
  • 42. 42 KJM 281 42Global Aspiration …… A World Class University Actual Cycle
  • 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.
  • 44. 44 KJM 281 44Global Aspiration …… A World Class University Reheat Rankine Cycle
  • 45. 45 KJM 281 45Global Aspiration …… A World Class University
  • 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.
  • 48. 48 KJM 281 48Global Aspiration …… A World Class University Regenerative Cycle
  • 49. 49 KJM 281 49Global Aspiration …… A World Class University Regenerative Cycle with Open FWH
  • 50. 50 KJM 281 50Global Aspiration …… A World Class University Regenerative Cycle with Closed FWH
  • 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.
  • 52. 52 KJM 281 52Global Aspiration …… A World Class University Plant Layout
  • 53. 53 KJM 281 53Global Aspiration …… A World Class University Plant Monitoring System
  • 54. 54 PART 3 54Global Aspiration …… A World Class University SPECIAL APPLICATION
  • 55. 55 KJM 281 55Global Aspiration …… A World Class University Cogeneration
  • 56. 56 KJM 281 56Global Aspiration …… A World Class University Cogeneration Stand-Alone System
  • 57. 57 KJM 281 57Global Aspiration …… A World Class University Trigeneration Plant
  • 58. 58 KJM 281 58Global Aspiration …… A World Class University Combined 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
  • 62. 62 KJM 281 62Global Aspiration …… A World Class University Connaught Bridge Power Station (Klang)
  • 63. 63 KJM 281 63Global Aspiration …… A World Class University Binary Vapor Cycle
  • 64. 64 KJM 281 64Global Aspiration …… A World Class University