lecture 26
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Lecture 26. Use of Regeneration in Vapor Power Cycles. What is Regeneration?. Goal of regeneration Reduce the fuel input requirements (Q in ) Increase the temperature of the feedwater entering the boiler (increases average Th in the cycle Result of regeneration - PowerPoint PPT PresentationTRANSCRIPT
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Department of Mechanical EngineeringME 322 – Mechanical Engineering
Thermodynamics
Lecture 26
Use of Regeneration in Vapor Power Cycles
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What is Regeneration?• Goal of regeneration
– Reduce the fuel input requirements (Q in)– Increase the temperature of the feedwater entering the boiler
(increases average Th in the cycle• Result of regeneration
– Increased thermal efficiency• Energy source for regeneration
– High pressure steam from the turbines• Regeneration equipment
– Feedwater heater (FWH)– This is a heat exchanger that utilizes the high pressure steam
extracted from the turbine toheat the boiler feedwater
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Regeneration – Open FWH
Increased temperature into the boiler due to regenerative heating
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Keeping Track of Mass Flow Splits
1 1y
2y
3y
4y
5y6y7y
Define a mass flow fraction,
1
mass flow rate at any state mass flow rate entering the HPT
nn
m nym
Determination of the flow fractions requires application of the conservation of mass throughout the cycle and the conservation of energy around the feedwater heater(s).
Note: If a mass flow rate is known or can be calculated, then the flow fraction approach is not necessary!
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Regeneration – Closed FWHThere are two types of closed feedwater heaters
Closed FWH with Drain Pumped
Forward
Closed FWH with Drain Cascaded
Backward
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Regeneration – Closed FWHExample – Closed FWH with Drain Cascaded Backward
1 1y
2y3y
4y
5y6y
7y8y
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Regeneration – Multiple FWH
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Regeneration Example
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Given: A Rankine cycle is operating with one open feedwater heater. Steam enters the high pressure turbine at 1500 psia, 900°F. The steam expands through the high pressure turbine to 100 psia where some of the steam is extracted and diverted to an open feedwater heater. The remaining steam expands through the low pressure turbine to the condenser pressure of 1 psia. Saturated liquid exits the feedwater heater and the condenser.
Find:(a) the boiler heat transfer per lbm of steam entering the
high pressure turbine(b) the thermal efficiency of the cycle(c) the heat rate of the cycle
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Regeneration Cycle
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1
1
1500 psia900 F
PT
2 100 psiaP
3 1 psiaP
4
4
1 psia0
Px
5 100 psiaP 6
6
100 psia0
Px
7 1P P
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Known Properties
10
1
1
1500 psia900 F
PT
2 100 psiaP
3 1 psiaP
4
4
1 psia0
Px
5 100 psiaP 6
6
100 psia0
Px
7 1P P
The next step is to build the property table
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Unknown Properties
11
1
1
1500 psia900 F
PT
2 100 psiaP
3 1 psiaP
4
4
1 psia0
Px
5 100 psiaP 6
6
100 psia0
Px
7 1P P
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Array Table
12
1
1
1500 psia900 F
PT
2 100 psiaP
3 1 psiaP
4
4
1 psia0
Px
5 100 psiaP 6
6
100 psia0
Px
7 1P P
The resulting property table ...
Now, we can proceed with the thermodynamics!
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Boiler Modeling
13
1
1
1500 psia900 F
PT
2 100 psiaP
3 1 psiaP
4
4
1 psia0
Px
5 100 psiaP 6
6
100 psia0
Px
7 1P P
The heat transfer rate at the boiler can be found by applying the First Law,
1 1 7inQ m h h
No flow rate information is given. However, we can find the heat transferred per lbm of steam entering the HPT,
1 71
inin
Qq h h
m
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Turbine Modeling
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1
1
1500 psia900 F
PT
2 100 psiaP
3 1 psiaP
4
4
1 psia0
Px
5 100 psiaP 6
6
100 psia0
Px
7 1P P
The thermal efficiency of the cycle is given by,
t pnetth
in in
W WWQ Q
The turbine power delivered is,
1 1 2 2 3 3tW m h m h m h
2 31 2 3
1 1 1
tm mW
h h hm m m
1 2 2 3 31
ttWw h y h y hm
The flow fractions need to be determined!
1 1
1
/ //
t p
in
W m W mQ m
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Pump Modeling
15
1
1
1500 psia900 F
PT
2 100 psiaP
3 1 psiaP
4
4
1 psia0
Px
5 100 psiaP 6
6
100 psia0
Px
7 1P P
There are two pumps in the cycle. Therefore,
1 2p p pW W W
4 5 4 6 7 6pW m h h m h h
4 65 4 7 6
1 1 1
p m mWh h h h
m m m
4 5 4 6 7 61
pp
Ww y h h y h h
m
1 1
1
/ //
t p t pth
in in
W m W m w wQ m q
Then ...
This is an important step in the analysis. All specific energy transfers need to be based on the same flow rate. The common value is chosen to be the inlet to the high pressure turbine (HPT).
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Mass Conservation
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1
1
1500 psia900 F
PT
2 100 psiaP
3 1 psiaP
4
4
1 psia0
Px
5 100 psiaP 6
6
100 psia0
Px
7 1P P
The flow fractions must be found. The easy flow fractions are ...
2 5 6m m m
2 5 6
1 1 1
m m mm m m
2 5 6y y y
1 6 7 1y y y
3 4 5y y y
Conservation of mass applied to the FWH gives,
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Closing the System
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1
1
1500 psia900 F
PT
2 100 psiaP
3 1 psiaP
4
4
1 psia0
Px
5 100 psiaP 6
6
100 psia0
Px
7 1P P
Where is the missing equation? Mass is conserved in the FWH, but so is energy. Therefore, we need to apply the First Law to the FWH,
2 2 5 5 6 6m h m h m h
2 5 62 5 6
1 1 1
m m mh h h
m m m
2 2 5 5 6 6y h y h y h
The equations can be solved! The result is a new property table with a column for the mass flow fractions.
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Augmented Array
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1
1
1500 psia900 F
PT
2 100 psiaP
3 1 psiaP
4
4
1 psia0
Px
5 100 psiaP 6
6
100 psia0
Px
7 1P P
The updated property table ...
From previous analysis,t p
thin
w wq
1 2 2 3 3tw h y h y h
4 5 4 6 7 6pw y h h y h h
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Cycle Performance Parameters
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1
1
1500 psia900 F
PT
2 100 psiaP
3 1 psiaP
4
4
1 psia0
Px
5 100 psiaP 6
6
100 psia0
Px
7 1P P
The heat rate of the cycle is,
1
1 1
/HR
/ /in in in
net t p t p
Q Q m qW W m W m w w
EES Solution (Key Variables):