Thermodynamic Chapter 6 Thermal Power Plant
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Transcript of Thermodynamic Chapter 6 Thermal Power Plant
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MEC 451THERMODYNAMICS
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CHAPTER 6THERMAL POWER PLANT
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PART 1
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GAS TURBINE POWER PLANT
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Brayton Cycle
The Brayton cycle (a.k.a. Joule cycle) is the air-standard ideal
cycle approximation for the gas turbine engine.
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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
kk
T
T
V
V
P
P
1
2
1
1
2
2
1
k
kk
T
T
V
V
P
P
Brayton Cycle Process Description
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The T-s and P-v diagrams are
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The T-s for actual cycle (including pressure drop and isentropic efficiency)
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outout
outoutoutoutout
inin
ininininin gz
VhmWQgz
VhmWQ
22
2...2...
outoutininin hmhmW...
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.
12
.
hhmW Global Aspiration …… A World Class University
Energy balance: for compressor
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Isentropic efficiency for compressor:
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outout
outoutoutoutout
inin
ininininin gz
VhmWQgz
VhmWQ
22
2...2...
outoutininin hmhmQ...
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.
23
.
hhmQ Global Aspiration …… A World Class University
Energy balance: for boiler
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3
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outout
outoutoutoutout
inin
ininininin gz
VhmWQgz
VhmWQ
22
2...2...
outoutoutinin hmWhm...
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.
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.
hhmW
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4
Energy balance: for turbine
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Isentropic efficiency for turbine:
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outout
outoutoutoutout
inin
ininininin gz
VhmWQgz
VhmWQ
22
2...2...
outoutoutinin Qhmhm...
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.
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.
hhmQ Global Aspiration …… A World Class University
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1
Energy balance: for condenser
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th Brayton net
in
out
in
W
Q
Q
Q, 1
where the pressure ratio is rp = P2/P1
th Braytonpk kr
, ( )/ 11
1
Upon derivation
Thermal efficiency of the Brayton cycle is defined as
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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.
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12
TTc
TTc
W
Wr
p
p
t
cbw
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
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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.
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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.
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The regenerator effectiveness regen 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
regenregen act
regen
,
, max '
,
, max
5 2
5 2 4 2
5 2
4 2
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For ideal gases using the cold-air-standard assumption
with constant specific heats, the regenerator effectiveness
becomes
5 2
4 2regen
T T
T T
Upon derivation the thermal efficiency becomes
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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.
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Cycle Improvement – Intercooling and Reheating
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The T-s diagram for this cycle is shown below.
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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.
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Aircraft Engine
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Alstom GT26 Gas Turbine
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PW 4000 112-INCH FAN ENGINE
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Rolls – Royce Trent
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General Electric (GE) Gas Turbine
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PART 2
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STEAM POWER PLANT
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Based on the Carnot cycle the heat engine may be composed of the following components.
Carnot Cycle
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0.0 1.0 2.0 3.0 4.0 5.0 6.0 7.0 8.0 9.0 10.00
100
200
300
400
500
600
700700
s [kJ/kg-K]
T [
C] 6000 kPa
100 kPa
Carnot Vapor Cycle Using Steam
1
23
4
th Carnot net
in
out
in
L
H
W
Q
Q
Q
T
T
,
1
1
The thermal efficiency of this cycle is given as
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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.
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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:
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The pump work can be determined by:
Cycle Analysis
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2211
PPm
hhmW
hmWhm
pump
pump
The energy balance for boiler:
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3322
hhmQ
hmQhm
in
in
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The energy balance for turbine:
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4433
hhmW
hmWhm
out
out
The energy balance for condenser:
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1144
hhmQ
hmQhm
out
out
Based on the Second Law the thermal efficiency becomes:
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1243
hh
hhhh
q
w
in
netth
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Cycle Improvement
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Actual Cycle
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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.
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Reheat Rankine Cycle
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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
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q
w
in
netth
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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.
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Regenerative Cycle
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Regenerative Cycle with Open FWH
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Regenerative Cycle with Closed FWH
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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.
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Plant Layout
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Plant Monitoring System
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PART 3
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SPECIAL APPLICATION
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Cogeneration
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Cogeneration Stand-Alone System
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Trigeneration Plant
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Combined Cycle
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Karlsruhe Power Station (Germany)
Topping cycle = 2 x Bottoming cycle
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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
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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
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Connaught Bridge Power Station (Klang)
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Binary Vapor Cycle
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