Chapter 8 Examples
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Course
3140
Subject
Mechanical Engineering
Date
Dec 6, 2023
Type
Pages
13
Uploaded by BrigadierSteelArmadillo15
1
8.25
Superheated steam at 20 MPa, 560
o
C enters the turbine of a vapor power plant.
The
pressure at the exit of the turbine is 0.5 bar, and liquid leaves the condenser at 0.4 bar at 75
o
C.
The pressure is increased to 20.1 MPa across the pump.
The turbine and pump have isentropic
efficiencies of 81 and 85%, respectively.
Cooling water enters the condenser at 20
o
C with a
mass flow rate of 70.7 kg/s and exits the condenser at 38
o
C.
For the cycle, determine
(a) the mass flow rate of steam, in kg/s.
(b) the thermal efficiency.
KNOWN:
Water is the working fluid in a vapor power plant.
Data are given at various states in
the cycle.
FIND:
(a) the mass flow rate of steam, in kg/s and (b) the thermal efficiency.
SCHEMATIC AND GIVEN DATA:
ENGINEERING MODEL:
1.
Each component of the cycle is analyzed as a control volume at steady state.
The control
volumes are shown on the accompanying sketch by dashed lines.
2.
Stray heat transfer in the turbine, condenser, and pump is ignored.
3.
Kinetic and potential energy effects are negligible.
ANALYSIS:
First fix each principal state.
State 1
:
p
1
= 20 MPa (200 bar),
T
1
= 560
o
C →
h
1
= 3423.0 kJ/kg,
s
1
= 6.3705
kJ/kg∙K
State 2s
:
p
2s
=
p
2
= 0.5 bar,
s
2s
=
s
1
= 6.3705
kJ/kg∙K →
x
2s
= 0.8119,
h
2s
= 2212.2 kJ/kg
State 2
:
p
2
= 0.5 bar,
h
2
= 2442.3 kJ/kg (see below)
t
W
±
Turbine
Cooling
water
2
Condenser
Pump
p
W
±
3
4
1
Boiler
in
Q
±
p
3
= 0.4 bar
T
3
= 75
o
C
p
1
= 20 MPa
T
1
= 560
o
C
p
2
= 0.5 bar
p
4
= 20.1 MPa
out
Q
±
h
t
= 81%
h
p
= 85%
6
5
T
5
= 20
o
C
T
6
= 38
o
C
kg/s
7
.
70
cw
m
±
T
4
3
2
1
20 MPa
s
2s
T
1
= 560
o
C
0.5 bar
0.4 bar
20.1 MPa
2
kg
kJ
)
2
.
2212
0
.
3423
)(
81
.
0
(
kg
kJ
0
.
3423
)
(
2
1
t
1
2
2
1
2
1
t
±
±
±
±
o
±
±
s
s
h
h
h
h
h
h
h
h
h
h
= 2442.3 kJ/kg
State 3
:
p
3
= 0.4 bar,
T
3
= 75
o
C
→
From Table A-2
p
3
>
p
sat
@ 75
o
C.
Thus, state 3 is a sub-
cooled liquid state.
Since the pressure is low,
h
3
≈
h
f3
at 75
o
C = 313.93 kJ/kg,
v
3
≈
v
f3
at 75
o
C = 0.0010259 m
3
/kg
State 4
:
p
4
= 20.1 MPa (201 bar),
h
4
= 338.14 kJ/kg (see below)
p
3
4
3
3
4
3
4
3
4
3
p
)
(
)
(
h
h
p
p
h
h
h
h
p
p
±
²
o
±
±
v
v
m
N
1000
kJ
1
kPa
1
m
N
1000
bar
1
kPa
100
85
.
0
bar
)
4
.
0
201
)(
kg
m
0010259
.
0
(
kg
kJ
93
.
313
2
3
4
±
²
h
= 338.14 kJ/kg
State 5
:
T
5
= 20
o
C, liquid →
h
5
≈
h
f5
at 20
o
C = 83.96 kJ/kg
State 6
:
T
6
= 38
o
C, liquid →
h
6
≈
h
f6
at 38
o
C = 159.21 kJ/kg
(a)
The mass flow rate of the steam can be determined by writing an energy balance for the
condenser. With no stray heat transfer with the surroundings and no work, the energy balance
for the condenser reduces to
)
(
)
(
0
6
5
cw
3
2
h
h
m
h
h
m
±
²
±
±
±
where
m
±
is the mass flow rate of the steam and
cw
m
±
is the mass flow rate of the cooling water.
Rearranging to solve for the mass flow rate of steam gives
)
(
)
(
3
2
5
6
cw
h
h
h
h
m
m
±
±
±
±
Substituting values and solving give
kJ/kg
)
93
.
313
.3
2442
(
kJ/kg
)
96
.
83
.21
159
(
kg/s)
0.7
7
(
±
±
m
±
=
2.50 kg/s
(b)
The thermal efficiency is
3
)
(
)
(
)
(
4
1
3
4
2
1
h
h
h
h
h
h
±
±
±
±
h
Substituting enthalpy values and solving yield
kJ/kg
)
.14
338
3423.0
(
kJ/kg
)
93
.
313
.14
338
(
kJ/kg
)
3
.
2442
3423.0
(
±
±
±
±
h
=
0.3101 (31.01%)
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1
8.33
Steam heated at constant pressure in a steam generator enters the first stage of a
supercritical reheat cycle at 28 MPa, 520
o
C.
Steam exiting the first-stage turbine at 6 MPa is
reheated at constant pressure to 500
o
C.
Each turbine stage has an isentropic efficiency of 78%
while the pump has an isentropic efficiency of 82%.
Saturated liquid exits the condenser that
operates at constant pressure,
p
.
(a) For
p
= 6 kPa, determine the quality of the steam exiting the second stage of the turbine and
the thermal efficiency.
(b) Plot the quantities of part (a) versus
p
ranging from 4 kPa to 70 kPa.
KNOWN:
A supercritical reheat cycle operates with steam as the working fluid.
FIND:
(a) For
p
= 6 kPa, determine the quality of the steam exiting the second stage of the
turbine and the thermal efficiency, and (b) plot the quantities of part (a) versus
p
ranging from 4
kPa to 70 kPa.
SCHEMATIC AND GIVEN DATA:
ENGINEERING MODEL:
1.
Each component of the cycle is analyzed as a control volume at steady state.
The control
volumes are shown on the accompanying sketch by dashed lines.
2.
For all components stray heat transfer is ignored.
3.
Flow through the steam generator, reheater, and condenser is at constant pressure.
4.
Kinetic and potential energy effects are negligible.
ANALYSIS:
First fix each principal state
s
.
State 1:
p
1
= 28 MPa,
T
1
= 520
o
C →
h
1
= 3192.3 kJ/kg,
s
1
= 5.9566 kJ/kg∙K
t
W
±
4
Condenser
out
Q
±
Pump
p
W
±
5
6
1
Steam
Generator
in
Q
±
p
5
=
p
4
x
5
= 0 (saturated liquid)
p
1
= 28 MPa
T
1
= 520
o
C
p
4
=
p
p
6
=
p
1
= 28 MPa
Turbine
2
Turbine
1
2
3
p
2
= 6 MPa
p
3
=
p
2
= 6 MPa
T
3
= 500
o
C
Reheat
Section
h
t1
=
h
t2
= 78%
h
p
= 82%
2
State 2s
:
p
2s
=
p
2
= 6 MPa,
s
2s
=
s
1
= 5.9566 kJ/
kg∙K
→
h
2s
= 2822.2 kJ/kg
State 2
:
p
2
= 6 MPa,
h
2
= 2903.6 kJ/kg (see below)
kg
kJ
)
2
.
2822
3
.
3192
)(
78
.
0
(
kg
kJ
3
.
3192
)
(
2
1
t1
1
2
2
1
2
1
t1
±
±
±
±
o
±
±
s
s
h
h
h
h
h
h
h
h
h
h
= 2903.6 kJ/kg
State 3:
p
3
= 6 MPa,
T
3
= 500
o
C →
h
3
= 3422.2 kJ/kg,
s
3
= 6.8803 kJ/kg∙K
State 4s
:
p
4s
=
p
4
= 6 kPa,
s
4s
=
s
3
= 6.8803 kJ/kg∙K →
x
4s
= 0.8143,
h
4s
= 2118.8 kJ/kg
State 4
:
p
4
= 6 MPa,
h
4
= 2405.5 kJ/kg (see below)
kg
kJ
)
8
.
2118
2
.
3422
)(
78
.
0
(
kg
kJ
2
.
3422
)
(
4
3
t2
3
4
4
3
4
3
t2
±
±
±
±
o
±
±
s
s
h
h
h
h
h
h
h
h
h
h
=
2405.5 kJ/kg
State 5
:
p
5
= 6 kPa
, saturated liquid →
h
5
=
h
f5
= 151.53 kJ/kg,
v
5
=
v
f5
= 0.0010064 m
3
/kg
State 6
:
p
6
=
p
1
= 28 MPa,
h
6
= 185.89 kJ/kg (see below)
p
5
6
5
5
6
5
6
5
6
5
p
)
(
)
(
h
h
p
p
h
h
h
h
p
p
±
²
o
±
±
v
v
m
N
1000
kJ
1
kPa
1
m
N
1000
82
.
0
kPa
)
6
000
,
28
(
kg
m
0010064
.
0
kJ/kg
53
.
151
2
3
6
±
²
h
= 185.89 kJ/kg
(a)
The quality of the steam at the exit of the second stage of the turbine (state 4) is determined
using values from Table
A-
3
,
h
f4
= 151.53 kJ/kg and
h
fg4
= 2415.9 kJ/kg, as follows:
kJ/kg
9
.
2415
kJ/kg
)
53
.
151
5
.
2405
(
4
fg
4
f
4
4
±
±
h
h
h
x
=
0.9330
The cycle thermal efficiency is
m
Q
m
Q
m
W
m
W
m
W
m
Q
m
W
±
±
±
±
±
±
±
±
±
±
±
±
±
±
/
/
/
/
/
/
/
23
61
p
t2
t1
in
cycle
²
±
²
h
)
(
)
(
)
(
)
(
)
(
2
3
6
1
5
6
4
3
2
1
h
h
h
h
h
h
h
h
h
h
±
²
±
±
±
±
²
±
h
3
Substituting enthalpy values and solving yield a thermal efficiency of
kJ/kg
)
6
.
2903
2
.
3422
(
kJ/kg
)
.89
185
.3
3192
(
kJ/kg
)
53
.
151
.89
185
(
kJ/kg
)
5
.
2405
2
.
3422
(
kJ/kg
)
6
.
2903
.3
3192
(
±
²
±
±
±
±
²
±
h
h
=
0.3606 (36.06%)
(b)
For
p
ranging from 4 kPa to 70 kPa, IT gives the following results:
IT Code
// Known Properties
p1 = 28000 // kPa
T1 = 520 // oC
p6 = p1
p3 = 6000 // kPa
T3 = 500 // oC
p2 = p3
p2s = p2
p4 = 6 // kPa
p4s = p4
p5 = p4
x5 = 0
// Known Operating Parameters
eff_t1 = 0.78
eff_t2 = 0.78
eff_p = 0.82
// Calculations for Quality at State 4
h3 = h_PT("Water/Steam", p3, T3)
s3 = s_PT("Water/Steam", p3, T3)
s4s = s3
h4s = h_Ps("Water/Steam", p4s, s4s)
h4 = h3 - eff_t2*(h3 - h4s)
x4 = x_hP("Water/Steam", h4, p4)
// Calculations for Thermal Efficiency
h1 = h_PT("Water/Steam", p1, T1)
s1 = s_PT("Water/Steam", p1, T1)
s2s = s1
h2s = h_Ps("Water/Steam", p2s, s2s)
h2 = h1 - eff_t1*(h1 - h2s)
h5 = hsat_Px("Water/Steam", p5, x5)
v5 = vsat_Px("Water/Steam", p5, x5)
h6 = h5 + (v5*(p6 - p5)/eff_p)
eff_thermal = ((h1 - h2) + (h3 - h4) - (h6 - h5))/((h1 - h6) + (h3 - h2))
IT Output for
p
4
= 6 kPa
eff_thermal
0.3606
h1
3192
h2
2903
h2s
2821
h3
3422
h4
2405
h4s
2118
h5
151
h6
185.4
p2
6000
p2s
6000
p4s
6
p5
6
p6
2.8E4
s1
5.956
s2s
5.956
s3
6.879
s4s
6.879
v5
0.001007
x4
0.933
eff_p
0.82
eff_t1
0.78
eff_t2
0.78
p1
2.8E4
p3
6000
p4
6
T1
520
T3
500
x5
0
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4
Results from IT for p
4
= 6 kPa correspond closely to the results obtained using steam tables
and hand calculations.
Graphical Results from IT are shown below:
In general, as the condenser pressure increases the quality of the steam increases and the
thermal efficiency decreases since the average temperature of heat rejection is higher.
As
shown by the Quality versus Condenser Pressure Graph for the conditions of this problem,
when the condenser pressure reaches approximately 70 kPa, the liquid-vapor mixture quality
becomes 1.
Above this pressure steam is superheated vapor and the quality is not defined.
Quality versus Condenser Pressure
Pressure [kPa]
70
65
60
55
50
45
40
35
30
25
20
15
10
5
0
1
0.99
0.98
0.97
0.96
0.95
0.94
0.93
0.92
Thermal Efficiency versus Condenser Pressure
Condenser Pressure (kPa)
70
60
50
40
30
20
10
0
0.4
0.3
0.2
0.1
0
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1
8.40
A power plant operates on a regenerative vapor power cycle with one open feedwater
heater.
Steam enters the first turbine stage at 12 MPa, 560
o
C and expands to 1 MPa, where some
of the steam is extracted and diverted to the open feedwater heater operating at 1 MPa.
The
remaining steam expands through the second turbine stage to the condenser pressure of 6 kPa.
Saturated liquid exits the open feedwater heater at 1 MPa.
The net power output for the cycle is
330 MW.
For isentropic processes in the turbines and pumps, determine
(a) the cycle thermal efficiency.
(b) the mass flow rate into the first turbine stage, in kg/s.
(c) the rate of entropy production in the open feedwater heater, in kW/K.
KNOWN:
A regenerative vapor power cycle with one open feedwater heater operates with
steam as the working fluid.
Operational data are provided.
FIND:
Determine (a) the cycle thermal efficiency, (b) the mass flow rate into the first turbine
stage, in kg/s, and (c) the rate of entropy production in the open feedwater heater, in kW/K.
SCHEMATIC AND GIVEN DATA:
t
W
±
3
Condenser
out
Q
±
Pump
2
p2
W
±
4
7
1
Steam
Generator
in
Q
±
p
4
=
p
3
= 6 kPa
x
4
= 0 (saturated liquid)
p
1
= 12 MPa
T
1
= 560
o
C
p
3
= 6 kPa
p
7
=
p
1
= 12
MPa
MW
330
cycle
W
±
Open
Feedwater
Heater
Pump
1
p1
W
±
5
6
2
(1)
(1)
(1
–
y
)
(1
–
y
)
(1
–
y
)
(
y
)
p
6
=
p
5
=
p
2
= 1 MPa
x
6
= 0 (saturated liquid)
Turbine
p
2
= 1 MPa
2
ENGINEERING MODEL:
1.
Each component of the cycle is analyzed as a control volume at steady state.
The control
volumes are shown on the accompanying sketch by dashed lines.
2.
All processes of the working fluid are internally reversible except for mixing in the open
feedwater heater.
3.
The turbines, pumps, and open feedwater heater operate adiabatically.
4.
Kinetic and potential energy effects are negligible.
5.
Saturated liquid exits the open feedwater heater, and saturated liquid exits the condenser.
ANALYSIS:
First fix each principal state.
State 1
:
p
1
= 12 MPa (120 bar),
T
1
= 560
o
C →
h
1
= 3506.2 kJ/kg,
s
1
= 6.6840 kJ/kg∙K
State 2
:
p
2
= 1 MPa (10 bar),
s
2
=
s
1
= 6.6840 kJ/kg∙K
→
h
2
= 2823.3 kJ/kg
State 3
:
p
3
= 6 kPa (0.06 bar),
s
3
=
s
1
= 6.6840 kJ/kg∙K →
x
3
= 0.7892,
h
3
= 2058.2 kJ/kg
State 4
:
p
4
= 6 kPa (0.06 bar), saturated liquid →
h
4
= 151.53 kJ/kg,
s
4
= 0.521
0 kJ/kg∙K
,
v
4
= 0.0010064 m
3
/kg
State 5
:
p
5
=
p
2
= 1 MPa (10 bar),
s
5
=
s
4
= 0.521
0 kJ/kg∙K →
)
(
4
5
4
4
5
p
p
h
h
±
²
|
v
m
N
1000
kJ
1
kPa
1
N/m
1000
kPa
)
6
1000
(
kg
m
0010064
.
0
kg
kJ
53
.
151
2
3
5
±
¸
¸
¹
·
¨
¨
©
§
²
h
= 152.53 kJ/kg
State 6
:
p
6
= 1 MPa (10 bar), saturated liquid →
h
6
= 762.81 kJ/kg,
s
6
= 2.1387 kJ/kg∙K
,
v
6
= 0.0011273 m
3
/kg
s
T
6
5
2
1
p
= 6 kPa
p
= 12 MPa
3
4
p
= 1 MPa
7
3
State 7
:
p
7
=
p
1
= 12 MPa (120 bar),
s
7
=
s
6
= 2.1387 kJ/kg∙K →
)
(
6
7
6
6
7
p
p
h
h
±
²
|
v
m
N
1000
kJ
1
kPa
1
N/m
1000
kPa
)
1000
12000
(
kg
m
0011273
.
0
kg
kJ
81
.
762
2
3
7
±
¸
¸
¹
·
¨
¨
©
§
²
h
= 775.21 kJ/kg
(a)
Applying energy and mass balances to the control volume enclosing the open feedwater
heater, the fraction of flow,
y
, extracted at location 2 is
kJ/kg
)
53
.
152
3
.
2823
(
kJ/kg
)
53
.
152
81
.
762
(
5
2
5
6
±
±
±
±
h
h
h
h
y
= 0.2285
For the control volume surrounding the turbine stages
)
)(
1
(
)
(
3
2
2
1
1
t
h
h
y
h
h
m
W
±
±
²
±
±
±
kg
kJ
)
2
.
2058
3
.
2823
)(
2285
.
0
1
(
kg
kJ
)
3
.
2823
2
.
3506
(
1
t
±
±
²
±
m
W
±
±
= 1273.2 kJ/kg
For the pumps
)
)(
1
(
)
(
4
5
6
7
1
p
h
h
y
h
h
m
W
±
±
²
±
±
±
kg
kJ
)
53
.
151
53
.
152
)(
2285
.
0
1
(
kg
kJ
)
81
.
762
21
.
775
(
1
p
±
±
²
±
m
W
±
±
= 13.17 kJ/kg
For the working fluid passing through the steam generator
kg
kJ
)
21
.
775
2
.
3506
(
7
1
1
in
±
±
h
h
m
Q
±
±
= 2731.0 kJ/kg
Thus, the thermal efficiency is
kJ/kg
0
.
2731
kJ/kg
)
17
.
13
2
.
1273
(
/
/
/
1
in
1
p
1
t
±
±
m
Q
m
W
m
W
±
±
±
±
±
±
K
=
0.461 (46.1%)
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4
(b)
The
net
power developed is
)
/
/
(
1
p
1
t
1
cycle
m
W
m
W
m
W
±
±
±
±
±
±
±
Thus,
)
/
/
(
1
p
1
t
cycle
1
m
W
m
W
W
m
±
±
±
±
±
±
±
MW
1
s
kJ
1000
kg
kJ
)
17
.
13
2
.
1273
(
MW
330
1
±
m
±
=
261.9 kg/s
(c)
The rate of entropy production in the open feedwater heater is determined using the steady-
state form of the entropy rate balance:
cv
0
V
±
±
±
±
²
±
²
¦
¦
¦
e
e
e
i
i
i
j
j
j
s
m
s
m
T
Q
Since the feedwater heater is adiabatic, the heat transfer term drops.
Thus,
5
5
2
2
6
6
cv
s
m
s
m
s
m
s
m
s
m
i
i
i
e
e
e
±
±
±
±
±
±
±
±
±
¦
¦
V
]
)
1
(
[
5
2
6
1
cv
s
y
ys
s
m
±
±
±
±
±
V
kJ/s
1
kW
1
K
kg
kJ
)]
5210
.
0
)(
2285
.
0
1
(
)
6840
.
6
)(
2285
.
0
(
1387
.
2
[
s
kg
9
.
261
cv
±
±
±
V
±
=
54.86 kW/K
Mixing of streams within the open feedwater heater is a source of irreversibility that produces
entropy.
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