ME 470 Vapor and Combined Power Cycles Hw Solutions Inst: Shoeleh Di Julio
Chapter 10, Solution 22.
A steam power plant operates on a simple ideal Rankine cycle between the specified pressure limits. The
thermal efficiency of the cycle, the mass flow rate of the steam, and the temperature rise of the cooling
water are to be determined.
Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.
Analysis (a) From the steam tables (Tables A-4, A-5, and A-6),
h1  h f @ 10 kPa  191.81 kJ/kg
v1  v f @ 10 kPa  0.00101 m 3 /kg
T
w p,in  v1 P2  P1 

3

 1 kJ 

 0.00101 m 3 /kg 7,000  10 kPa 
 1 kPa  m 3 


 7.06 kJ/kg
h2  h1  w p,in  191.81  7.06  198.87 kJ/kg
2
7 MPa
qin
10 kPa
1
P3  7 MPa  h3  3411.4 kJ/kg

T3  500 C  s 3  6.8000 kJ/kg  K
s 4  s f 6.8000  0.6492
P4  10 kPa 

 0.8201
 x4 
s 4  s3
s fg
7.4996

h4  h f  x 4 h fg  191 .81  0.8201 2392 .1  215 3.6 kJ/kg
Thus,
q in  h3  h2  3411 .4  198 .87  3212 .5 kJ/kg
q out  h4  h1  2153 .6  191 .81  1961 .8 kJ/kg
wnet  q in  q out  3212 .5  1961 .8  1250 .7 kJ/kg
and
 th 
(b)
m 
wnet 1250.7 kJ/kg

 38.9%
q in
3212.5 kJ/kg
Wnet
45,000 kJ/s

 36.0 kg/s
wnet 1250.7 kJ/kg
(c) The rate of heat rejection to the cooling water and its temperature rise are
Q out  m q out  35.98 kg/s 1961.8 kJ/kg   70,586 kJ/s
Q out
70,586 kJ/s
Tcoolingwater 

 8.4C
(m c) coolingwater 2000 kg/s 4.18 kJ/kg  C
qout
4
s
Chapter 10, Solution 49.
A steam power plant operates on an ideal reheat-regenerative Rankine cycle with an open feedwater heater.
The mass flow rate of steam through the boiler and the thermal efficiency of the cycle are to be determined.
Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.
Analysis (a) From the steam tables (Tables A-4, A-5, and A-6),
T
h1  h f @ 10 kPa  191.81 kJ/kg
v1  v f @ 10 kPa  0.00101 m 3 /kg

5

 1 kJ 

w pI ,in  v1 P2  P1   0.00101 m 3 /kg 800  10 kPa 
 1 kPa  m 3 


 0.80 kJ/kg
h2  h1  w pI ,in  191.81  0.80  192.61 kJ/kg
P3  0.8 M Pa  h3  h f @ 0.8 MPa  720.87 kJ/kg

3
sat.liquid
 v 3  v f @ 0.8 MPa  0.001115 m /kg

6
0.8 MPa
2
7
y
3
1-y
10 kPa

 1 kJ 

 v 3 P4  P3  0.001115 m 3 /kg 10,000  800 kPa 
 1 kPa  m 3 


 10.26 kJ/kg
w pII ,in
10 MPa
4
8
1
s
h4  h3  w pII ,in  720.87  10.26  731.12 kJ/kg
P5  10 M Pa  h5  3502.0 kJ/kg

T5  550C  s5  6.7585 kJ/kg  K
P6  0.8 M Pa 
 h6  2812.1 kJ/kg

P7  0.8 M Pa  h7  3481.3 kJ/kg

T7  500C  s7  7.8692 kJ/kg  K
5
Boiler
Turbine
6
s 6  s5
1-y
6
4
s8  s f
7.8692  0.6492

 0.9627
P8  10 kPa  x8 
s fg
7.4996
P II

s8  s7
 h  h  x h  191.81  0.9627 2392.1  2494.7 kJ/kg
8
f
8 fg
7
8
y
Open
fwh
Condenser
2
3
1
PI
The fraction of steam extracted is determined from the steady-flow energy balance equation applied to the
feedwater heaters. Noting that Q  W  Δke  Δpe  0 ,
E in  E out  E system0 (steady)  0  E in  E out
 m h   m h
i i
e e

 m 6 h6  m 2 h2  m 3 h3 
 yh6  1  y h2  1h3 
6 / m
 3 ). Solving for y,
where y is the fraction of steam extracted from the turbine (  m
y
Then,
h3  h2 720 .87  192 .61

 0.2017
h6  h2 2812 .1  192 .61
q in  h5  h4   1  y h7  h6   3502 .0  731 .12   1  0.2017 3481 .3  2812 .1  3305 .1 kJ/kg
q out  1  y h8  h1   1  0.2017 2494 .7  191 .81  1838 .5 kJ/kg
wnet  q in  q out  3305 .1  1838 .5  1466 .6 kJ/kg
W net
80,000 kJ/s

 54.5 kg/s
wnet 1466.1 kJ/kg
and

m
(b)
 th 
wnet 1466.1 kJ/kg

 44.4%
q in
3305.1 kJ/kg
Chapter 10, Solution 50.
A steam power plant operates on an ideal reheat-regenerative Rankine cycle with a closed feedwater heater.
The mass flow rate of steam through the boiler and the thermal efficiency of the cycle are to be determined.
Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.
Analysis
T
5
Turbine
Boiler
4
1-y
Mixing
9
5
10 MPa
6
7
y
8
2
10
4
9
y
3
Closed
fwh
6
0.8 MPa
7
1-y
Condenser
c
h
2
1
10a
3
P II
PI
m
b
e
(a) From the rsteam tables (Tables A-4, A-5, and A-6),
10 kPa
1
8
s
h1  h f @ 10 kPa  191.81 kJ/kg
v1  v f @ 10 kPa  0.00101 m 3 /kg


 1 kJ 

w pI ,in  v1 P2  P1   0.00101 m 3 /kg 10,000  10 kPa 
 1 kPa  m 3 


 10.09 kJ/kg
h2  h1  w pI ,in  191.81  10.09  201.90 kJ/kg
P3  0.8 M Pa  h3  h f @ 0.8 MPa  720.87 kJ/kg

3
sat.liquid
 v 3  v f @ 0.8 MPa  0.001115 m /kg


 1 kJ 

w pII ,in  v 3 P4  P3   0.001115 m 3 /kg 10,000  800 kPa 
 1 kPa  m 3 


 10.26 kJ/kg
h4  h3  w pII ,in  720.87  10.26  731.13 kJ/kg
Also, h4 = h9 = h10 = 731.12 kJ/kg since the two fluid streams that are being mixed have the same enthalpy.
P5  10 MPa  h5  3502 .0 kJ/kg

T5  550 C  s5  6.7585 kJ/kg  K
P6  0.8 MPa 
 h6  2812 .7 kJ/kg
s6  s5

P7  0.8 MPa  h7  3481 .3 kJ/kg

T7  500 C  s7  7.8692 kJ/kg  K
s8  s f
7.8692  0.6492

 0.9627
P8  10 kPa  x8 
s fg
7.4996

s8  s7

h8  h f  x8h fg  191 .81  0.9627 2392 .1  2494 .7 kJ/kg
The fraction of steam extracted is determined from the steady-flow energy balance equation applied to the
feedwater heaters. Noting that Q  W  ke  Δpe  0 ,
E in  E out  E system0 (steady)  0
E in  E out
 m h   m h
i i
e e

 m 2 h9  h2   m 3 h6  h3  
 1  y h9  h2   y h6  h3 
3 / m
 4 ). Solving for y,
where y is the fraction of steam extracted from the turbine (  m
y
h9  h2
h6  h3   h9  h2 

731 .13  201 .90
 0.2019
2812 .7  720 .87  731 .13  201 .90
Then,
q in  h5  h4   1  y h7  h6   3502 .0  731 .13   1  0.2019 3481 .3  2812 .7   3304 .5 kJ/kg
q out  1  y h8  h1   1  0.2019 2494 .7  191 .81  1837 .9 kJ/kg
wnet  q in  q out  3304 .5  1837 .8  1466 .6 kJ/kg
and
m 
W net
80,000 kJ/s

 54.5 kg/s
wnet 1467.1 kJ/kg
q out
1837.8 kJ/kg
 1
 44.4%
q in
3304.5 kJ/kg
 th  1 
(b)
Chapter 10, Solution 67.
A cogeneration plant is to generate power and process heat. Part of the steam extracted from the turbine at a relatively
high pressure is used for process heating. The net power produced and the utilization factor of the plant are to be
determined.
Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.
Analysis From the steam tables (Tables A-4, A-5, and A-6),
h1  h f @ 10 kPa  191.81 kJ/kg
v1  v f
@ 10 kPa
 0.00101 m 3 /kg
6
wpI,in  v1 P2  P1 


 1 kJ 

 0.00101 m 3 /kg 600  10 kPa 
 1 kPa  m 3 


 0.60 kJ/kg
Turbine
Boiler
7
8
h2  h1  wpI,in  191.81  0.60  192.41 kJ/kg
Process
heater
5
h3  h f
@ 0.6 MPa
 670.38 kJ/kg
Mixing chamber:
E  E
 E
in
out
0 (steady)
system
 m h   m h
i i
e e
 0  E in  E out
 m 4 h4  m 2 h2  m 3 h3
Condenser
3
1
P II
PI
4
2
or,
m 2 h2  m 3 h3 22.50 192.41  7.50 670.38

 311.90 kJ/kg
m 4
30
v 4  v f @ h f  311.90 kJ/kg  0.001026 m 3 /kg
h4 
w pII,in  v 4 P5  P4 

T

 1 kJ 

 0.001026 m 3 /kg 7000  600 kPa 
 1 kPa  m 3 


 6.57 kJ/kg
6
P6  7 M Pa
T6  500C
7 MPa
·
Qin
0.6 MPa
4 3 ·
Qproces
5
h5  h4  w pII,in  311.90  6.57  318.47 kJ/kg
 h6  3411.4 kJ/kg

 s 6  6.8000 kJ/kg  K
2
P7  0.6 MPa 
 h7  2774.6 kJ/kg
s7  s6

s
10 kPa
·
Qout
1
6.8000  0.6492

 0.8201
P8  10 kPa  x8 
s fg
7.4996

s8  s 6
 h  h  x h  191.81  0.82012392.1  2153.6 kJ/kg
8
f
8 fg
s8  s f
Then,
7
8
s
W T, out  m 6 h6  h7   m 8 h7  h8 
 30 kg/s 3411.4  2774.6kJ/kg  22.5 kg/s 2774.6  2153.6kJ/kg  33,077 kW
W p,in  m 1 wpI,in  m 4 wpII,in  22.5 kg/s 0.60 kJ/kg   30 kg/s 6.57 kJ/kg   210.6 kW
W net  W T, out  W p,in  33,077  210.6  32,866 kW
Also,
 7 h7  h3   7.5 kg/s 2774.6  670.38 kJ/kg  15,782 kW
Q process  m
Q in  m 5 h6  h5   30 kg/s 3411.4  318.47   92,788 kW
and
u 
W net  Q process 32,866  15,782

 52.4%
92,788
Q in
Chapter 10, Solution 70.
A cogeneration plant modified with regeneration is to generate power and process heat. The mass flow rate
of steam through the boiler for a net power output of 15 MW is to be determined.
Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.
Analysis
From the steam tables (Tables A-4, A-5, and A-6),
6
Turbine
Boiler
7
8
Process
heater
5
Condenser
9
4
P II
3
1
PI
fwh
2
h1  h f
@ 10 kPa
 191.81 kJ/kg
v1  v f
@ 10 kPa
 0.00101 m 3 /kg
wpI,in  v1 P2  P1 


 1 kJ 

 0.00101 m 3 /kg 400  10 kPa 
 1 kPa  m 3 


 0.39 kJ/kg
h2  h1  wpI,in  191.81  0.39  192.20 kJ/kg
h3  h4  h9  h f
v4  v f
@ 0.4 MPa
wpII,in  v 4 P5  P4 

@ 0.4 MPa
 604.66 kJ/kg
 0.001084 m 3 /kg

 1 kJ 

 0.001084 m 3 /kg 6000  400 kPa 
 1 kPa  m 3 


 6.07 kJ/kg
T
6
h5  h4  wpII,in  604.66  6.07  610.73 kJ/kg
P6  6 MPa  h6  3302 .9 kJ/kg

T6  450 C  s 6  6.7219 kJ/kg  K
6 MPa
5
3,4,9
2
0.4 MPa
s 7  s f 6.7219  1.7765
10 kPa

 0.9661
P7  0.4 MPa  x 7 
1
s fg
5.1191

s7  s6
 h  h  x h  604 .66  0.9661 2133 .4  2665 .7 kJ/kg
7
f
7 fg
s8  s f 6.7219  0.6492

 0.8097
P8  10 kPa  x8 
s fg
7.4996

s8  s 6
 h  h  x h  191 .81  0.8097 2392 .1  2128 .7 kJ/kg
8
f
8 fg
Then, per kg of steam flowing through the boiler, we have
wT, out  h6  h7   0.4h7  h8 
 3302 .9  2665 .7 kJ/kg  0.42665 .7  2128 .7 kJ/kg
 852.0 kJ/kg
wp,in  0.4 wpI,in  wpII,in
 0.4 0.39 kJ/kg   6.07 kJ/kg 
 6.23 kJ/kg
wnet  wT, out  wp,in  852 .0  6.23  845 .8 kJ/kg
Thus,
m 
Wnet 15,000 kJ/s

 17.73 kg/s
wnet 845.8 kJ/kg
7
8
s
Chapter 10, Solution 76.
A combined gas-steam power cycle is considered. The topping cycle is a gas-turbine cycle and the bottoming cycle is a
simple ideal Rankine cycle. The mass flow rate of the steam, the net power output, and the thermal efficiency of the
combined cycle are to be determined.
Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. 3 Air is an
ideal gas with constant specific heats.
Properties The properties of air at room temperature are cp = 1.005 kJ/kg·K and k = 1.4 (Table A-2).
Analysis (a) The analysis of gas cycle yields
P
T6  T5  6
 P5
Q  m h
in




k 1 / k
 300 K 16 
T
0. 4 / 1. 4
 662.5 K
1500 K
7
 air c p T7  T6 
air 7  h6   m
 14 kg/s 1.005 kJ/kg  K 1500  662.5 K  11,784 kW
·
Qin
W C ,gas  m air h6  h5   m air c p T6  T5 
 14 kg/s 1.005 kJ/kg  K 662.5  300  K  5100 kW
P
T8  T7  8
 P7




k 1 / k
 1
 1500K  
 16 
8
6
 679.3 K
300 K
5
2
W net,gas  W T ,gas  W C ,gas  11,547  5,100  6447 kW
1
9 420 K
STEAM
CYCLE 15 kPa
·
4
Qout
From the steam tables (Tables A-4, A-5, and A-6),
h1  h f @ 15 kPa  225.94 kJ/kg
v1  v f @ 15 kPa  0.001014 m 3 /kg


 1 kJ 
  10.12 kJ/kg
wpI,in  v1 P2  P1   0.001014 m 3 /kg 10,000  15 kPa 
 1 kPa  m 3 


h2  h1  wpI,in  225.94  10.13  236.06 kJ/kg
P3  10 M Pa  h3  3097.0 kJ/kg
T3  400C  s 3  6.2141 kJ/kg  K
s4  s f
6.2141  0.7549
P4  15 kPa  x 4 

 0.7528
s fg
7.2522

s 4  s3
 h  h  x h  225.94  0.75282372.3  2011.8 kJ/kg
4
f
4 fg
Noting that Q  W  Δke  Δpe  0 for the heat exchanger, the steady-flow energy balance equation yields
E in  E out  E system0 (steady)  0  E in  E out
 m h   m h
i i
m s 
(b)
e e
 m s h3  h2   m air h8  h9 
c p T8  T9 
h8  h9
1.005 kJ/kg  K 679.3  420 K 14 kg/s   1.275 kg/s
m air 
m air 
3097.0  236.06 kJ/kg
h3  h2
h3  h2
W T, steam  m s h3  h4   1.275 kg/s 3097.0  2011.5 kJ/kg  1384 kW
W p,steam  m s w p  1.275 kg/s 10.12 kJ/kg   12.9 kW
W net,steam  W T, steam  W p,steam  1384  12.9  1371 kW
and
W net  W net,steam  W net,gas  1371  6448  7819 kW
(c)
 th 
W net
7819 kW

 66.4%
11,784 kW
Q in
3
400C
10 MPa
0. 4 / 1. 4
W T ,gas  m air h7  h8   m air c p T7  T8 
 14 kg/s 1.005 kJ/kg  K 1500  679.3 K  11,547 kW
GAS
CYCLE
s