```10-57
10-60 A steam power plant that operates on an ideal regenerative Rankine cycle with a closed feedwater heater is
considered. The temperature of the steam at the inlet of the closed feedwater heater, the mass flow rate of the steam
extracted from the turbine for the closed feedwater heater, the net power output, and the thermal efficiency 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 @ 20 kPa  251.42 kJ/kg
High-P
turbine
5
v 1  v f @ 20 kPa  0.001017 m 3 /kg
w pI ,in  v 1 P2  P1  /  p
Low-P
turbine
Boiler
1
 (0.001017 m /kg )(8000  20 kPa )
0.88
 9.22 kJ/kg
3
6
1-y
h2  h1  w pI ,in
 251.42  9.223
 260.65 kJ/kg
P3  1 MPa  h3  h f @ 1 MPa  762.51 kJ/kg

sat. liquid  v 3  v f @ 1 MPa  0.001127 m 3 /kg
7
4
Mixing
Cham.
10
9
8
Closed
fwh
y
Cond.
2
11
PII
3
1
PI
w pII ,in  v 3 P11  P3  /  p
 (0.001127 m 3 /kg )(8000  1000 kPa ) / 0.88
 8.97 kJ/kg
h11  h3  w pII ,in  762.51  8.97  771.48 kJ/kg
Also, h4 = h10 = h11 = 771.48 kJ/kg since the two fluid streams which are being mixed have the same enthalpy.
P5  8 MPa  h5  3399.5 kJ/kg

T5  500C  s 5  6.7266 kJ/kg  K
P6  3 MPa 
h6 s  3104.7 kJ/kg
s 6  s5

T 
h5  h6

 h6  h5   T h5  h6 s 
h5  h6 s
 3399.5  0.883399.5  3104.7   3140.1 kJ/kg
P7  3 MPa  h7  3457.2 kJ/kg

T7  500C  s 7  7.2359 kJ/kg  K
P8  1 MPa 
h8 s  3117.1 kJ/kg
s8  s 7

T 
h7  h8

 h8  h7   T h7  h8s 
h7  h8 s
 3457.2  0.883457.2  3117.1  3157.9 kJ/kg
P8  1 MPa

T8  349.9C
h8  3157.9 kJ/kg 
PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
preparation. If you are a student using this Manual, you are using it without permission.
10-58
P9  20 kPa 
h9 s  2385.2 kJ/kg
s9  s 7

T 
h7  h9

 h9  h7   T h7  h9 s 
h7  h9 s
 3457.2  0.883457.2  2385.2  2513.9 kJ/kg
The fraction of steam extracted from the low pressure turbine for closed feedwater heater is determined from the steadyflow energy balance equation applied to the feedwater heater. Noting that Q  W  ke  pe  0 ,
1  y h10  h2   y h8  h3 
(1  y )(771.48  260.65)  y (3157.9  762.51) 
 y  0.1758
The corresponding mass flow rate is
m 8  ym 5  (0.1758)(15 kg/s)  2.637 kg/s
(c) Then,
q in  h5  h4  h7  h6  3399.5  771.48  3457.2  3140.1  2945.2 kJ/kg
q out  1  y h9  h1   1  0.17582513.9  251.42  1864.8 kJ/kg
and
W net  m (q in  q out )  (15 kg/s)(2945.8  1864.8)kJ/kg  16,206 kW
(b) The thermal efficiency is determined from
 th  1 
q out
1864.8 kJ/kg
1
 0.3668  36.7%
2945.8 kJ/kg
q in
PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
preparation. If you are a student using this Manual, you are using it without permission.
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