College of Engineering and Computer Science
Mechanical Engineering Department
Mechanical Engineering 370
Thermodynamics
Fall 2010 Course Number: 14319 Instructor: Larry Caretto
Unit Eleven Homework Solutions, November 30, 2010
1. Consider a 210 MW steam power plant that operates on a simple ideal Rankine cycle.
Steam enters the turbine at 10 MPa and 500oC and is cooled in the condenser to a
pressure of 10 kPa. Show the cycle on a T-s diagram with respect to the saturation lines
and determine (a) the quality of steam at the turbine exit, (b) the thermal efficiency of the
cycle, and (c) the mass flow rate of the steam.
Rankine Cycle Diagram
900
3
800
Temperature (K)
700
600
500
400
2
300
1
4
200
1234-
100
2 Pum p
3 Steam generator
4 Turbine
1 Condenser
0
0
1
2
3
4
5
6
7
8
9
10
Entropy (kJ/kg-K)
Saturation
Condenser
Pump
Steam
Generator
Turbine
The cycle diagram shows the individual steps in the cycle. The increase in temperature in the
pump is typically about 1oC so the isentropic pump step does not really show on the diagram.
The constant pressure heating in the steam generator shows the path of an isobar on a T-s
diagram. In the mixed region, where temperature and pressure are constant, the isobar is a
horizontal line. The condenser, which is completely in the mixed region, has a constant
temperature line to represent the constant pressure process in the condenser.
To compute the quality at the turbine exit, we recognize that this exit state is defined by the
condenser pressure of 10 kPa and an isentropic process such that s out = sin = s(10 MPa, 500oC) =
6.5995 kJ/kg∙K. The outlet quality is thus found from the following equation.
Jacaranda (Engineering) 3519
E-mail: lcaretto@csun.edu
Mail Code
8348
Phone: 818.677.6448
Fax: 818.677.7062
Unit eleven homework solutions
xout 
ME 370, L. S. Caretto, Fall 2010
s out  s f (10 kPa)
s fg (10 kPa)

6.5995 kJ
kg  K
 0.6492 kJ
7.4996 kJ
Page 2
kg  K
xout = 0.793 .
kg  K
In order to compute the efficiency, we need the enthalpy values at all the state points. Following
a conventional Rankine cycle calculation, we find the properties at state one as those of a
saturated liquid at the condenser pressure: h1 = hf(10 kPa) = 191.81 kJ/kg and v1 = 0.001010
m3/kg. The isentropic pump work, |wp| = v1(P2 – P1) where P2 is the same as the inlet pressure to
the turbine, 10 MPa = 10,000 kPa. Thus,
wP1  v1 P2  P1  
0.001010 m 3
10000 kPa  10 kPa 1 kJ 3  10.09 kJ
kg
kPa  m
kg
We then find h2 = h1 + |wP1| = 191.81 kJ/kg + 10.09 kJ/kg = 201.90 kJ/kg.
h3 = h(10 MPa, 500oC) = 3375.1 kJ/kg.
s3 = s(10 MPa, 500oC) = 6.5995 kJ/kg∙K.
As noted above, state 4 is in the mixed region with P4 = 10 kPa and x4 = 0.793. We thus find the
enthalpy from the quality as h4 = hf(P4 = 10 kPa) + x4 hfg(P4 = 10 kPa) = 191.81 + (0.793)(2392.1
kJ/kg) or h4 = 2089.7 kJ/kg.
The heat input to the steam generator, qh = h3 – h2 = 3375.1 kJ/kg – 201.90 kJ/kg = 3173.2 kJ/kg
The condenser heat rejection, ql = |h1 – h4| = |191.81 kJ/kg – 2089.7 kJ/kg| = 1897.9 kJ/kg.
The net work, w = qh - |qL| = 3173.2 kJ/kg| - 1897.9 kJ/kg = 1275.3 kJ/kg.
The efficiency = w / qH = (1275.3 kJ/k ) / (3173.2 kJ/kg) or  = 40.2% .

The mass flow, m
210 MW 1000 kJ
W

w 1275.3 kJ
MW  s
kg
m 
165 kg
s
2. Consider a solar-pond power plant that operates on a simple ideal Rankine cycle with
refrigerant-134a as the working fluid. The refrigerant enters the turbine as a saturated
vapor at 1.6 MPa and leaves at 0.7 MPa. The mass flow rate of the refrigerant is 6 kg/s.
Show the cycle on a T-s diagram with respect to the saturation lines and determine (a) the
thermal efficiency and (b) the power output of the plant.
The cycle diagram is shown on the next page. As usual, when the T-s diagram is drawn to scale,
the pump does not appear on the diagram and the constant-pressure heating of the liquid in the
steam generator is very close to the saturation line. This diagram is unusual because there is no
superheating. In addition, the particular inlet and outlet pressures chosen for the turbine are in an
area of the T-s diagram where the slope is nearly vertical. Thus, the isentropic turbine process,
starting at the saturated vapor line lies very close to the saturated vapor line for the entire
process. This is verified by the calculation of the exit quality from the turbine, x4. To compute this
quality we note that the ideal cycle has an isentropic turbine so that s 4 = s3 = sg(1.6 MPa) =
0.90784 kJ/kg∙K. At the condenser pressure of 0.7 MPa and an entropy of 0.90784 kJ/kg∙K, we
find the quality as follows.
x4 
s 4  s f (0.7 MPa )
s fg (0.7 MPa )

0.90784 kJ
kg  K
 0.33230 kJ
0.58763 kJ
kg  K
kg  K
xout = 0.979 .
The value of 98.3% for quality confirms the turbine path in the diagram that is close to the
saturated line for the entire process.
Unit eleven homework solutions
ME 370, L. S. Caretto, Fall 2010
Page 3
Temperature (K)
Next, we do
the usual set
R-134a Rankine Cycle
of calculations
for the
Rankine
400
cycle. In
order to
compute the
350
efficiency, we
need the
300
enthalpy
values at all
the state
250
points.
Following a
conventional
200
Rankine cycle
calculation,
we find the
Saturation
Steam
150
properties at
Generator
state one as
Pump
Turbine
those of a
100
saturated
Condenser
liquid at the
condenser
50
pressure: h1 =
hf(0.7 MPa) =
88.82 kJ/kg
0
and v1 =
0
0.5
1
1.5
2
0.0008331
Entropy (kJ/kg-K)
m3/kg. The
isentropic
pump work, |wp| = v1(P2 – P1) where P2 is the same as the inlet pressure to the turbine, 1.6 MPa =
1600 kPa. Thus,
3
wP1
0.0008331 m
1600 kPa  700 kPa
 v1 P2  P1  
kg
1 kJ
kPa  m
3

0.75 kJ
kg
We then find h2 = h1 + |wP1| = 86.78 kJ/kg + 0.75 kJ/kg = 89.57 kJ/kg.
h3 = hg(1.6 MPa) = 277.863 kJ/kg.
s3 = sg(1.6 MPa) = 0.90784 kJ/kg∙K.
As noted above, state 4 is in the mixed region with P4 = 700 kPa and x4 = 0.979. We thus find the
enthalpy from the quality as h4 = hf(P4 = 700 kPa) + x4 hfg(P4 = 700 kPa) = 88.82 kJ/kg +
(0.979)(176.212 kJ/kg) or h4 = 261.41 kJ/kg.
The heat intput to the steam generator, qh = h3 – h2 = 277.86 kJ/kg – 89.54 kJ/kg = 188.3 kJ/kg
The condenser heat rejection, ql = |h1 – h4| = |88.62 kJ/kg – 261.41 kJ/kg| = 172.6 kJ/kg.
The net work, w = qh - |qL| =188.3 kJ/kg| - 172.6 kJ/kg = 15.70 kJ/kg.
The efficiency = w / qH = (15.70 kJ/k ) / (188.3 kJ/kg) or  = 8.3% .
The power output =
6 kg 15.71 kJ 1 kW
W  m w 
s
kg kJ  s
W  94.2 kW
Unit eleven homework solutions
ME 370, L. S. Caretto, Fall 2010
Page 4
3. Consider a steam power plant that operates on a simple ideal Rankine cycle and has a net
power output of 45 MW. Steam enters the turbine at 7 MPa and 500oC and is cooled in the
condenser to a pressure of 10 kPa by running cooling water from a lake through the
condenser at a rate of 2000 kg/s. Show the cycle on a T-s diagram with respect to the
saturation lines, and determine (a) the thermal efficiency of the cycle, (b) the mass flow
rate of the steam, and (c) the temperature rise of the cooling water.
The diagram for this cycle is similar to the diagram for the cycle in problem 9-16 and is not shown
here.
In order to compute the efficiency, we need the enthalpy values at all the state points. Following
a conventional Rankine cycle calculation, we find the properties at state one as those of a
saturated liquid at the condenser pressure: h1 = hf(10 kPa) = 191.83 kJ/kg and v1 = 0.001010
m3/kg. The isentropic pump work, |wp| = v1(P2 – P1) where P2 is the same as the inlet pressure to
the turbine, 7 MPa = 7,000 kPa. Thus,
wP1  v1 P2  P1  
0.001010 m 3
7000 kPa  10 kPa 1 kJ 3  7.060 kJ
kg
kPa  m
kg
We then find h2 = h1 + |wP1| = 191.81 kJ/kg + 7.06 kJ/kg = 198.87 kJ/kg.
h3 = h(7 MPa, 500oC) =3411.4 kJ/kg.
s3 = s(7 MPa, 500oC) = 6.8000 kJ/kg∙K.
h4 = h(P = Pcond = 10 kPa, s4 = s3). We see that this state is in the mixed region so we have to
compute the quality to determine the enthalpy.
x4 
s 4  s f (10 kPa)
s fg (10 kPa)

6.8000 kJ
kg  K
 0.6492 kJ
7.4996 kJ
kg  K
xout = 0.8202 .
kg  K
With P4 = 10 kPa and x4 = 0.82027, we find the value of h4 = hf(P4 = 10 kPa) + x4 hfg(P4 = 10 kPa)
= 191.81 + (0.8202)(2392.` kJ/kg) or h4 = 2153.7 kJ/kg.
The heat input to the steam generator, qh = h3 – h2 = 3411.4 kJ/kg – 198.87 kJ/kg = 3212.5 kJ/kg
The condenser heat rejection, ql = |h1 – h4| = |191.81 kJ/kg – 2153.7 kJ/kg| = 1961.8 kJ/kg.
The net work, w = qh - |qL| = 3212.5 kJ/kg| - 1961.8 kJ/kg = 1250.6 kJ/kg.
The efficiency = w / qH = (1250.6 kJ/kg) / (3212.5 kJ/kg) or  = 38.9% .

The mass flow, m
45 MW
W

w 1250.6 kJ
kg
1000 kJ
MW  s
m 
35.98 kg
s
The heat rejection rate from the steam to the cooling water is the product of the mass flow rate
  m q  35.98 kg 1961.8 kJ  70591 kJ
and the value of |qL|: Q
L
L
s
s
s
This heat is added to the cooling water. The cooling water flow is modeled as a steady flow with
negligible kinetic and potential energies. There is no useful work. We model the enthalpy
change of the cooling water by the equation h = cpT since we assume that the effect of
pressure changes on the enthalpy of the relatively incompressible liquid water will be negligible.
Applying the first law to the cooling water then gives the following relationship between the
condenser heat rejection and the cooling water temperature rise, Tcw:
Q cw  Q L  m cw hcw ,out  hcw ,in   m cw c p ,cw Tcw ,out  Tcw ,in   m cw c p ,cw Tcw
Unit eleven homework solutions
ME 370, L. S. Caretto, Fall 2010
Page 5
We can solve this equation for Tcw, and substitute in the given values including the heat capacity
of liquid water to give cp,cw = 4.18 kJ/kg∙K (Table A-3(a) on page 914 of the text for liquid water at
25oC) to obtain the final answer for the temperature rise.
Tcw 
Q L 
m cw c p ,cw
70591 kJ
s

2000 kg 4.18 kJ
s
kg  K
Tcw = 8.44oC .
4. A steam power plant operates on an ideal regenerative Rankine cycle. Steam enters the
turbine at 6 MPa and 450oC and is condensed in the condenser at 20 kPa. Steam is
extracted from the turbine at 0.4 MPa to heat the feedwater in an open feedwater heater.
Water leaves the feedwater heater as a saturated liquid. Show the cycle on a T-s diagram
and determine (a) the net work per kilogram of steam flowing through the boiler and (b) the
thermal efficiency of the cycle.
High Pressure
Turbine (T1)
The diagram of the components
in the cycle is shown on the left.
In terms of the numbered points
on this diagram, the input data
for the problem give P5 = 6 MPa,
T5 = 450oC, Pcond = 20 kPa, and
PFWH = 0.4 MPa.
Low Pressure
Turbine (T2)
5
7
Steam
Generator
8
6
Feedwater
Heater
3
4
2
Pump
(P2)
Condenser
1
Pump
(P1)
For the ideal cycle in which there
are no line losses in pressure or
temperature and no pressure
drops in heat transfer devices,
we have P4 = P5,= 6 MPa, P2 =
P3 = P6 = P7 = PFWH = 0.4 MPa,
and P1 = P8 = Pcond = 20 kPa.
The ideal cycle has isentropic
work devices so s8 = s7 = s6 = s5;
s2 = s1 and s4 = s3. Finally points
1 and 3 are saturated liquid.
As usual, we assume that the individual components are steady-flow devices with negligible
kinetic and potential energies. There is no useful work in the steam generator, feedwater heater,
or condenser. The turbine and pumps are reversible and adiabatic meaning that there is no heat
transfer or entropy change. Thus the first law for each device only one inlet and one outlet is q =
w + hout – hin. We begin by determining the enthalpy at each point in the cycle.
The properties at state one as those of a saturated liquid at the condenser pressure: h 1 = hf(20
kPa) = 251.42 kJ/kg and v1 = 0.001017 m3/kg. The pumps are isentropic and we calculate the
work of the first pump as follows: |wp1| = v1(P2 – P1). Thus,
wP1  v1 P2  P1  
0.001017 m 3
400 kPa  20 kPa 1 kJ 3  0.39 kJ
kg
kg
kPa  m
We then find h2 = h1 + |wP1| = 251.42 kJ/kg + 0.39 kJ/kg = 251.80 kJ/kg.
The properties at state three are also those of a saturated liquid. Here the pressure is the
feedwater heater pressure so that h3 = hf(400 kPa) = 604.66 kJ/kg and v3 = 0.001084 m3/kg. We
use the vP calculation for isentropic pump work for the second pump.
Unit eleven homework solutions
wP2  v3 P4  P3  
ME 370, L. S. Caretto, Fall 2010
Page 6
0.001084 m 3
6000 kPa  400 kPa 1 kJ 3  6.07 kJ
kg
kg
kPa  m
We then find h4 = h3 + |wP2| = 604.66 kJ/kg + 6.07 kJ/kg = 610.73 kJ/kg.
h5 = h(6 MPa, 450oC) = 3302.9 kJ/kg.
s5 = s(6 MPa, 450oC) =6.7219 kJ/kg∙K.
h6 = h(P = PFWH = 400 kPa, s6 = s5). We see that this state is in the mixed region so we have to
compute the quality to determine the enthalpy.
x6 
s 6  s f (400 kPa)
s fg (400 kPa)

6.7219 kJ
kg  K
 1.7765 kJ
5.1191 kJ
kg  K
 0.9661
kg  K
h6 = hf(P6 = 400 kPa) + x6 hfg(P6 = 400 kPa) = 604.66 kJ/kg + (0.9661)(2133.4 kJ/kg) or h6 =
2665.7 kJ/kg.
State 7 is the same as state 6 so we have h7 = 2665.7 kJ/kg.
h8 = h(P = Pcond = 20 kPa, s8 = s5). We see that this state is in the mixed region so we have to
compute the quality to determine the enthalpy.
x8 
s8  s f (20 kPa)
s fg (20 kPa)

6.7219 kJ
kg  K
 0.8320 kJ
7.0752 kJ
kg  K
 0.8325
kg  K
h8 = hf(P8 = 20 kPa) + x8 hfg(P8 = 20 kPa) = 251.42 kJ/kg + (0.8325)(2357.5 kJ/kg) or h8 = 2213.97
kJ/kg.
In this cycle there are three distinct mass flow rates at different points in the cycle. These are
 6 a represents the mass flow into the feedwater heater.)
shown in the equations below. (Here, m
m 3  m 4  m 5  m a
m 7  m 8  m 1  m 2  m b
 6a  m
3  m
2  m
 a  m b
m
Taking a mass and energy balance around the feedwater heater gives the following relation for
the mass flow ratio. We can substitute the enthalpy values found above to compute this ratio.
604.66 kJ 2665.67 kJ

m
m 2 h3  h6
kg
kg


 0.8538  b
251.80 kJ 2665.6 kJ
m 3 h2  h6
m a

kg
kg
We can compute the heat input rate for the steam generator, using
m 4  m 5  m a as the mass
flow rate in the steam generator.
Q SG  m a h5  h4 
The power output from the two turbine stages is given by the following equation, which accounts
for the differences in mass flow rate in the two stages.
WT  m a h5  h6   m b h7  h8 
Unit eleven homework solutions
ME 370, L. S. Caretto, Fall 2010
Page 7
Finally, the total power input to the pumps is computed by accounting for the differences in mass
flow rates.
W P  m a wP2  m b wP1
We now have the necessary information to compute the cycle efficiency.

WT  W P W net m a h5  h6   m b h7  h8   m a wP2  m b wP1


m a h5  h4 
Q SG
Q SG
We can divide by the mass flow rate,
 a to get the following equation for the efficiency in terms
m
of the mass flow rate ratio that we found from our analysis of the feedwater heater.


W net
h5  h6   mb h7  h8   wP2  mb wP1
m
m a
m a
 a 
h5  h4 
QSG
m a
In this form, the numerator of the efficiency equation is the net work per unit mass flowing through
the steam generator.
W net
m
m
 h5  h6   b h7  h8   wP2  b wP1
m a
m a
m a
Substituting the values found for the enthalpies in the cycle and the mass flow rate ratio gives the
net work per unit mass flowing through the steam generator as follows:
W net  3302.9 kJ 2665.7 kJ 
 2665.7 kJ 2214.0 kJ 
  (0.8538)

 


m a
kg
kg
kg
kg




6.07 kJ
0.39 kJ

 (0.8538)
kg
kg
W net 1016.5 kJ

m a
kg
From the equations for the efficiency and the net work, we see that we can use the computed
value of work to simplify the efficiency calculation.
W net
m
m
W net
1016.5 kJ
h5  h6   b h7  h8   wP2  b wP1
m
m a
m a
m a
kg
 a 


h5  h4 
h5  h4  3302.9 kJ  610.73 kJ
Q SG
kg
kg
m a
 = 37.8%
Unit eleven homework solutions
ME 370, L. S. Caretto, Fall 2010
Page 8
5. Repeat problem 4 with the open feedwater heater replaced by a closed feedwater heater.
Assume that the feedwater leaves the heater at the condensation temperature of the
extracted steam and that the extracted steam leaves the heater as a saturated liquid and is
pumped to the line carrying the feedwater.
The diagram of the components
in the cycle is shown on the left.
In the closed feedwater heater,
the feed water flows from point w
to point 4, without mixing with
the extracted steam. (The steam
enters at point 8, transfers heat
to the feed water without mixing,
and leaves at point 3.
7
Steam
Generator
High Pressure
Turbine (T1)
Low Pressure
Turbine (T2)
9
10
8
6
4
Feedwater
Heater
Mixing
Chamber
3
2
5
Condenser
1
Pump
(P1)
Pump
(P2)
In terms of the numbered points
on this diagram, the input data
for the problem give P7 = 6 MPa,
T7 = 450oC, Pcond = 20 kPa, and
P8 = 0.4 MPa. We are also told
that point 3 is a saturated liquid
and T4 has the same
temperature as this saturated
liquid
For the ideal cycle in which there are no line losses in pressure or temperature and no pressure
drops in heat transfer devices, we have P2 = P4 = P5 = P6 = P7 = 6 MPa, P3 = P8 = 0.4 MPa, and
P1 = P10 = Pcond = 20 kPa. The ideal cycle has isentropic work devices so s10 = s9 = s8 = s7; s2 =
s1 and s5 = s3. Finally points 1 and 3 are saturated liquid.
In this cycle there are three distinct mass flow rates at different points in the cycle. These are
 8 a represents the mass flow into the feedwater heater.)
shown in the equations below. (Here, m
m 6  m 7  m a
m 9  m 10  m 1  m 2  m 4  m b
m 8a  m 3  m 5  m c
Taking a mass balance around the mixing chamber gives the following relation among the three
mass flow rates.
m 6  m 4  m 5

m a  m b  m c
As usual, we assume that the individual components are steady-flow devices with negligible
kinetic and potential energies. There is no useful work in the steam generator, feedwater heater,
or condenser. The turbine and pumps are reversible and adiabatic meaning that there is no heat
transfer or entropy change. Thus the first law for each device with only one inlet and one outlet is
q = w + hout – hin. We begin by determining the enthalpy at each point in the cycle.
The properties at state one as those of a saturated liquid at the condenser pressure: h 1 = hf(20
kPa) = 251.42 kJ/kg and v1 = 0.001017 m3/kg. The pumps are isentropic and we calculate the
work of the first pump as follows: |wp1| = v1(P2 – P1). Thus,
wP1  v1 P2  P1  
0.001014 m 3
6000 kPa  20 kPa 1 kJ 3  6.08 kJ
kg
kPa  m
kg
We then find h2 = h1 + |wP1| = 251.42 kJ/kg + 6.08 kJ/kg = 257.50 kJ/kg.
Unit eleven homework solutions
ME 370, L. S. Caretto, Fall 2010
Page 9
The properties at state three are also those of a saturated liquid. Here the pressure is the
feedwater heater pressure so that h3 = hf(400 kPa) = 604.66 kJ/kg and v3 = 0.001084 m3/kg. We
use the vP calculation for isentropic pump work for the second pump.
wP2  v3 P5  P3  
0.001084 m 3
6000 kPa  400 kPa 1 kJ 3  6.07 kJ
kg
kPa  m
kg
We then find h5 = h3 + |wP2| = 604.66 kJ/kg + 6.07 kJ/kg = 610.73 kJ/kg.
According to the problem information T4 has the same temperature as the saturated liquid at point
three. From the saturation tables we find this temperature as 143.61oC. This is a compressed
liquid and we can use the following data in the compressed liquid. We can use a double
interpolation in the compressed liquid tables to find the enthalpy at this point. First we use two
interpolations to find the enthalpy at the desired temperature of 143/61 oC at the two pressures
bounding the given pressure of 6 MPa in the tables.
h(143.61o C ,5 MPa) 
592.18 kJ

kg
h(143.61o C ,10 MPa) 
672.55 kJ 592.18 kJ

kg
kg
160 C  140 C
681.01 kJ 595.45kJ

kg
kg
o
595.45 kJ

kg
o
160 C  140 C
o
o
143.61 C  140 C   606.kg68 kJ
o
o
143.61 C  140 C   610.kg89 kJ
o
o
We can now use these two values to find the desired enthalpy at 6 MPa.
610.89 kJ 606.68 kJ

606.68 kJ
kg
kg
6 MPa  5 MPa   607.53 kJ
h4 

kg
10 MPa  5 MPa
kg
The mixing chamber has no heat or work, but is has three different mass flow rates. Thus the
first law and mass conservation equations for this device can be written as shown below and
manipulated to get an equation for h6 in terms of mass flow rate ratios.
m 6  m 4  m 5

m 6 h6  m 4 h4  m 5h5
m a  m b  m c

h6 

m b m c

1
m a m a
 m 
m
m
m 4
h4  5 h5  b h4  1  b h5
m 6
m 6
m a
 m a 
Thus, we can compute h6 if we know the mass flow rate ratio in the above equation. We can find
this mass flow rate ratio from an analysis of the closed feedwater heater. Application of the first
law for no heat and work (and recognizing that the two streams in this device do not mix) gives
the following result.
m 2 h2  m 8a h8  m 3 h3  m 4 h4
 m b h2  m c h8  m c h3  m b h4

m b h3  h8

m c h2  h4
We have already seen how to compute h1, h3, and h4, and we will determine h8 below. Thus we
will be able to compute the mass flow rate ratio
compute the ratio
m b
m b
shown above from enthalpy values. To
m c
required to compute h6, we have to make the following computations.
m a
Unit eleven homework solutions
ME 370, L. S. Caretto, Fall 2010
Page 10
m b
m b
m b
h3  h8
m b m c
m c
m c
h2  h4




h3  h8
m a m a m b  m c m b
1
1
m c
m c
m c
h2  h4
We continue to find enthalpy values, using the conventional methods for the isentropic turbine
work.
h7 = h(6 MPa, 450oC) = 3302.9 kJ/kg.
s7 = s(6 MPa, 450oC) =6.7219 kJ/kg∙K.
h8 = h(P = P8 = 400 kPa, s8 = s7). We see that this state is in the mixed region so we have to
compute the quality to determine the enthalpy.
x8 
s8  s f (400 kPa)
s fg (400 kPa)

6.7219 kJ
kg  K
 1.7765 kJ
5.1191 kJ
kg  K
 0.9661
kg  K
h8 = hf(P8 = 400 kPa) + x8 hfg(P8 = 400 kPa) = 604.66 kJ/kg + (0.9661)(2133.4 kJ/kg) or h8 =
2665.67 kJ/kg.
State 9 is the same as state 9 so we have h9 = 2665.67 kJ/kg.
h10 = h(P = Pcond = 20 kPa, s10 = s7). We see that this state is in the mixed region so we have to
compute the quality to determine the enthalpy.
x10 
s10  s f (20 kPa)
s fg (20 kPa)

6.7219 kJ
kg  K
 0.8320 kJ
7.0752 kJ
kg  K
 0.8325
kg  K
h10 = hf(P10 = 20 kPa) + x10 hfg(P10 = 20 kPa) = 251.42 kJ/kg + (0.8325)(2357.5 kJ/kg) or h10 =
2213.97 kJ/kg.
We now have all the enthalpy values required to compute the mass flow rate ratios
604.66 kJ 2665.67 kJ

m b h3  h8
kg
kg


 5.888
257.50 607.53 kJ
m c h2  h4

kg
kg

mb
m b
m b
m c
m c m a 0.8548
5.888


 0.8548


 0.14158
m a m b  m c 5.888  1
m a m b
5.888
m c
m c
With the value just found for
h6 
 m
m b
h4  1  b
m a
 m a
m b
, we can compute h6:
m a

 607.52 kJ 
 610.73 kJ  607.99 kJ
h5  (0.8548)
  1  0.8548
 
kg
kg
kg





Unit eleven homework solutions
ME 370, L. S. Caretto, Fall 2010
We can compute the heat input rate for the steam generator, using
Page 11
m 6  m 7  m a as the mass
flow rate in the steam generator.
Q SG  m a h7  h6 
The power output from the two turbine stages is given by the following equation, which accounts
for the differences in mass flow rate in the two stages.
WT  m a h7  h8   m b h9  h10 
Finally, the total power input to the pumps is computed by accounting for the differences in mass
flow rates.
W P  m c wP2  m b wP1
We now have the necessary information to compute the cycle efficiency.
WT  W P W net m a h7  h8   m b h9  h10   m c wP2  m b wP1



m a h7  h6 
Q SG
Q SG
We can divide by the mass flow rate,
 a to get the following equation for the efficiency in terms
m
of the mass flow rate ratio that we found from our analysis of the feedwater heater.


W net
h7  h8   mb h9  h10   wP2  mb wP1
m
m a
m a
 a 
h7  h6 
QSG
m a
In this form, the numerator of the efficiency equation is the net work per unit mass flowing through
the steam generator.
W net
m
m
m
 h7  h8   b h9  h10   c wP2  b wP1
m a
m a
m a
m a
Substituting the values found for the enthalpies in the cycle and the mass flow rate ratio gives the
net work per unit mass flowing through the steam generator as follows:
W net  3302.9 kJ 2665.7 kJ 
 2665.7 kJ 2214.0 kJ 
  (0.8548)

 


m a
kg
kg
kg
kg




6.07 kJ
6.08 kJ
 0.14518
 (0.8548)
kg
kg
W net 1017.3 kJ

m a
kg
Unit eleven homework solutions
ME 370, L. S. Caretto, Fall 2010
Page 12
From the equations for the efficiency and the net work, we see that we can use the computed
value of work to simplify the efficiency calculation.



W net
W net
1017.3 kJ
h7  h8   mb h9  h10   mc wP2  mb wP1
m
m a
m a
m a
m a
kg
 a 


h7  h6 
h5  h4  3302.9 kJ  607.99 kJ
QSG
kg
kg
m a
 = 37.7%
6
A steam power plant operates on an ideal reheat-regenerative Rankine cycle and has a net
power output of 80 MW. Steam enters the high-pressure turbine at 10 MPa and 550oC and
leaves at 0.8 MPa. Some of the steam is extracted at this pressure to heat the feedwater in
an open feedwater heater. The rest of the steam is reheated to 500 oC and is expanded in
the low pressure turbine to the condenser pressure of 10 kPa. Show the cycle on a T-s
diagram and determine (a) the mass flow rate of steam flowing through the boiler and (b)
the thermal efficiency of the cycle.
High Pressure
Turbine (T1)
Low Pressure
Turbine (T2)
5
6
7
Steam
Generator
8
Feedwater
Heater
3
4
The diagram of the components
in the cycle is shown on the left.
In terms of the numbered points
on this diagram, the input data
for the problem give P5 = 10
MPa, T5 = 550oC, Pcond = 10
kPa, and PFWH = 0.8 MPa, and
T7 = 500oC.
2
Pump
(P2)
Condenser
1
Pump
(P1)
For the ideal cycle in which there
are no line losses in pressure or
temperature and no pressure
drops in heat transfer devices,
we have P4 = P5,= 10 MPa, P2 =
P3 = P6 = P7 = PFWH = 0.8 MPa,
and P1 = P8 = Pcond = 20 kPa.
The ideal cycle has isentropic
work devices so s8 = s7, s6 = s5;
s2 = s1 and s4 = s3. Finally points
1 and 3 are saturated liquid.
As usual, we assume that the individual components are steady-flow devices with negligible
kinetic and potential energies. There is no useful work in the steam generator, feedwater heater,
or condenser. The turbine and pumps are reversible and adiabatic meaning that there is no heat
transfer or entropy change. Thus the first law for each device only one inlet and one outlet is q =
w + hout – hin. We begin by determining the enthalpy at each point in the cycle.
The properties at state one as those of a saturated liquid at the condenser pressure: h 1 = hf(10
kPa) = 191.81 kJ/kg and v1 = 0.001010 m3/kg. The pumps are isentropic and we calculate the
work of the first pump as follows: |wp1| = v1(P2 – P1). Thus,
wP1  v1 P2  P1  
0.001010 m 3
800 kPa  10 kPa 1 kJ 3  0.80 kJ
kg
kPa  m
kg
We then find h2 = h1 + |wP1| = 191.81 kJ/kg + 0.80 kJ/kg = 192.61 kJ/kg.
Unit eleven homework solutions
ME 370, L. S. Caretto, Fall 2010
Page 13
The properties at state three are also those of a saturated liquid. Here the pressure is the
feedwater heater pressure so that h3 = hf(800 kPa) = 720.87 kJ/kg and v3 = 0.001115 m3/kg. We
use the vP calculation for isentropic pump work for the second pump.
wP2  v3 P4  P3  
0.001115 m 3
10000 kPa  800 kPa 1 kJ 3  10.26 kJ
kg
kPa  m
kg
We then find h4 = h3 + |wP2| = 720.87 kJ/kg + 10.26 kJ/kg = 731.13 kJ/kg.
h5 = h(10 MPa, 550oC) = 3502.0 kJ/kg.
s5 = s(1 MPa, 550oC) =6.7585 kJ/kg∙K.
h6 = h(P = PFWH = 800 kPa, s6 = s5). This state is in the gas region so we have to find h6 by
interpolation between the first two rows at 800 kPa. This gives h6 = 2812,8 kJ/kg.
h7 = h(0.8 MPa, 500oC) = 3481.3 kJ/kg.
s7 = s(0.8 MPa, 500oC) =7.8692 kJ/kg∙K.
h8 = h(P = Pcond = 10 kPa, s8 = s7). We see that this state is in the mixed region so we have to
compute the quality to determine the enthalpy.
x8 
s8  s f (10 kPa)
s fg (10 kPa)

7.8692 kJ
kg  K
 0.6492 kJ
7.4996 kJ
kg  K
 0.9627
kg  K
h8 = hf(P8 = 10 kPa) + x8 hfg(P8 = 10 kPa) = 191.81 kJ/kg + (0.9627)(2392.1 kJ/kg) or h8 = 2494.7
kJ/kg.
In this cycle there are three distinct mass flow rates at different points in the cycle. These are
 6 a represents the mass flow into the feedwater heater.)
shown in the equations below. (Here, m
m 3  m 4  m 5  m a
m 7  m 8  m 1  m 2  m b
 6a  m
3  m
2  m
 a  m b
m
Taking a mass and energy balance around the feedwater heater gives the following relation for
the mass flow ratio. We can substitute the enthalpy values found above to compute this ratio.
720.87 kJ 2812,8 kJ

m
m 2 h3  h6
kg
kg


 0.7984  b
m 3 h2  h6 192.61 kJ 2812.8 kJ
m a

kg
kg
We can compute the heat input rate for the steam generator, using
m 4  m 5  m a as the mass
flow rate for the initial the steam generator flow and mb for the reheat flow.
Q SG  m a h5  h4   m b h7  h6 
The power output from the two turbine stages is given by the following equation, which accounts
for the differences in mass flow rate in the two stages.
WT  m a h5  h6   m b h7  h8 
Finally, the total power input to the pumps is computed by accounting for the differences in mass
flow rates.
Unit eleven homework solutions
ME 370, L. S. Caretto, Fall 2010
Page 14
W P  m a wP2  m b wP1
We now have the necessary information to compute the cycle efficiency.

WT  W P W net m a h5  h6   m b h7  h8   m a wP2  m b wP1


m a h5  h4   m b h7  h6 
Q SG
Q SG
We can divide by the mass flow rate,
 a to get the following equation for the efficiency in terms
m
of the mass flow rate ratio that we found from our analysis of the feedwater heater.


W net
h5  h6   mb h7  h8   wP2  mb wP1
m
m a
m a
 a 

QSG
h5  h4   mb h7  h6 
m a
m a
In this form, the numerator of the efficiency equation is the net work per unit mass flowing through
the steam generator.
W net
m
m
 h5  h6   b h7  h8   wP2  b wP1
m a
m a
m a
Substituting the values found for the enthalpies in the cycle and the mass flow rate ratio gives the
net work per unit mass flowing through the steam generator as follows:
W net  3502.0 kJ 2812.8 kJ 
 3481.3 kJ 2494.7 kJ 
  (0.7984)

 


m a
kg
kg
kg
kg





10.26 kJ
0.80 kJ 1465.9 kJ
 (0.7984)

kg
kg
kg
From this specific work, we can find the mass flow rate required for a power output of 80 MW.
 1000 kJ 
80 MW 


Wnet
MW  s 

m a 

1465.9 kJ
W net
m a
kg
m a 
54.6 kg
s
From the equations for the efficiency and the net work, we see that we can use the computed
value of work to simplify the efficiency calculation.


W net
W net
h5  h6   mb h7  h8   wP2  mb wP1
m
m a
m a
m a
 a 



QSG
h5  h4   mb h7  h8 
h5  h4  mb h7  h8 
m a
m a
m a
Unit eleven homework solutions
ME 370, L. S. Caretto, Fall 2010
1465.9 kJ
kg

 3481.3 kJ 2812.8 kJ 
3502.0 kJ 731.13 kJ


 (0.7984)

kg
kg
kg
kg


Page 15
 = 44.4%
An alternative approach for finding the efficiency is to determine the heat loss in the condenser.
Q cond  m b h1  h8  . Since this is the rejected heat, we can use the following approach for
computing the efficiency.
m b
h1  h8 



QSG  Qcond
Wnet
m b h1  h8 
m a


 1
 1
m
m a h5  h4   m b h7  h6 
Q SG
Q SG
h5  h4   b h7  h6 
m a
Applying the results previously found to this equation gives...
 2494.7 kJ 191.81 kJ 

(0.7984)

kg
kg


  1
 44.4%
 3481.3 kJ 2812.8 kJ 
3502.0 kJ 731.13 kJ


 (0.7984)

kg
kg
kg
kg

