Department of Mechanical Engineering
ME 322 – Mechanical Engineering
Thermodynamics
Lecture 26
Use of Regeneration in Vapor Power Cycles
What is Regeneration?
• Goal of regeneration
– Reduce the fuel input requirements (Qin)
– 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 to
heat the boiler feedwater
2
Regeneration – Open FWH
Increased temperature
into the boiler due to
regenerative heating
3
Keeping Track of Mass Flow Splits
Define a mass flow fraction,
y1  1
yn 
y2
y3
y7
y6
y5
y4
mn
m1

mass flow rate at any state n
mass flow rate entering the HPT
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!
4
Regeneration – Closed FWH
There are two types of closed feedwater heaters
Closed FWH with
Drain Pumped
Forward
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Closed FWH with
Drain Cascaded
Backward
Regeneration – Closed FWH
Example – Closed FWH with Drain Cascaded Backward
y1  1
y3
y2
y6
y5
y4
y8
y7
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Regeneration – Multiple FWH
7
Regeneration Example
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
P1  1500 psia
T1  900F
P2  100 psia
P3  1 psia
P6  100 psia
x6  0
P5  100 psia
P7  P1
P4  1 psia
x4  0
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Known Properties
P1  1500 psia
T1  900F
P2  100 psia
P3  1 psia
P6  100 psia
x6  0
P5  100 psia
P7  P1
P4  1 psia
x4  0
The next step is to build the property table
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Unknown Properties
P1  1500 psia
T1  900F
P2  100 psia
P3  1 psia
P6  100 psia
x6  0
P5  100 psia
P7  P1
P4  1 psia
x4  0
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P1  1500 psia
T1  900F
Array Table
P2  100 psia
The resulting property table ...
P3  1 psia
P6  100 psia
x6  0
P5  100 psia
P7  P1
P4  1 psia
x4  0
Now, we can proceed with the thermodynamics!
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P1  1500 psia
T1  900F
Boiler Modeling
P2  100 psia
The heat transfer rate at the boiler
can be found by applying the First
Law,
Qin  m1  h1  h 7 
P3  1 psia
P6  100 psia
x6  0
P5  100 psia
P7  P1
No flow rate information is given. However, we can find the
heat transferred per lbm of steam entering the HPT,
Qin
qin 
 h1  h7
m1
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P4  1 psia
x4  0
P1  1500 psia
T1  900F
Turbine Modeling
P2  100 psia
The thermal efficiency of the cycle
is given by,
P3  1 psia
Wnet Wt  Wp Wt / m1  Wp / m1
th 


Qin
Qin
Qin / m1
P6  100 psia
x6  0
P5  100 psia
P7  P1
P4  1 psia
x4  0
The turbine power delivered is,
Wt  m1h1  m 2 h2  m3h3
m2
m3
Wt
 h1 
h2 
h3
m1
m1
m1
wt 
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Wt
 h1  y2 h 2  y3h3
m1
The flow fractions need to
be determined!
P1  1500 psia
T1  900F
Pump Modeling
P2  100 psia
P3  1 psia
There are two pumps in the cycle.
Therefore,
P6  100 psia
x6  0
Wp  Wp1  Wp 2
P7  P1
P4  1 psia
x4  0
Wp  m 4  h5  h 4   m 6  h 7  h 6 

m4
wp 
Wp
Wp
m1
m1
m1
h
5
 h4  
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m6
m1
h
7
 h6 
 y4  h5  h 4   y6  h 7  h 6 
Then ... th 
P5  100 psia
Wt / m1  Wp / m1
Qin / m1

wt  wp
qin
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).
P1  1500 psia
T1  900F
Mass Conservation
P2  100 psia
P3  1 psia
The flow fractions must be found.
The easy flow fractions are ...
y1  y 6  y 7  1
y3  y4  y5
P6  100 psia
x6  0
P7  P1
Conservation of mass applied to the FWH gives,
m 2  m5  m6
m2
m1

m5
m1

m6
m1
y2  y5  y6
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P5  100 psia
P4  1 psia
x4  0
P1  1500 psia
T1  900F
Closing the System
P2  100 psia
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,
P3  1 psia
P6  100 psia
x6  0
P5  100 psia
P7  P1
P4  1 psia
x4  0
m 2 h 2  m 5 h5  m 6 h 6
m2
m1
h2 
m5
m1
h5 
m6
m1
h6
y 2 h 2  y 5 h5  y 6 h 6
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
P1  1500 psia
T1  900F
The updated property table ...
P2  100 psia
P3  1 psia
P6  100 psia
x6  0
P5  100 psia
P7  P1
P4  1 psia
x4  0
From previous analysis,
th 
wt  wp
qin
wt  h1  y2 h 2  y3h3
wp  y4  h5  h 4   y6  h 7  h 6 
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Cycle Performance Parameters
P1  1500 psia
T1  900F
The heat rate of the cycle is,
HR 
Q in
W net
Q in / m1
P2  100 psia
qin


W t / m1  W p / m1 wt  wp
P3  1 psia
P6  100 psia
x6  0
P5  100 psia
P7  P1
P4  1 psia
x4  0
EES Solution (Key Variables):
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