10-41
10-52 A steam power plant operates on an ideal regenerative Rankine cycle with two open feedwater heaters. The net
power output of the power plant 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
7
Turbine
Boiler
7
8
6 10 MPa
10
6
4
9
fwh II
4
5
P III
3
3
0.2 MPa
5 kPa
2
P II
2
Condenser
fwh I
1
8
5 0.6 MPa
1
PI
y
1-y
9
1-y-z
10
s
(a) From the steam tables (Tables A-4, A-5, and A-6),
h1  h f @ 5 kPa  137.75 kJ/kg
v 1  v f @ 5 kPa  0.001005 m 3 /kg


 1 kJ
w pI ,in  v 1 P2  P1   0.001005 m 3 /kg 200  5 kPa 
 1 kPa  m 3

h h w
 137.75  0.20  137.95 kJ/kg
2
1
pI ,in

  0.20 kJ/kg


P3  0.2 MPa  h3  h f @ 0.2 MPa  504.71 kJ/kg

3
sat.liquid
 v 3  v f @ 0.2 MPa  0.001061 m /kg


 1 kJ
w pII ,in  v 3 P4  P3   0.001061 m 3 /kg 600  200 kPa 
 1 kPa  m 3

 0.42 kJ/kg




h4  h3  w pII ,in  504.71  0.42  505.13 kJ/kg
P5  0.6 MPa  h5  h f @ 0.6 MPa  670.38 kJ/kg

3
sat.liquid
 v 5  v f @ 0.6 MPa  0.001101 m /kg


 1 kJ
w pIII ,in  v 5 P6  P5   0.001101 m 3 /kg 10,000  600 kPa 
 1 kPa  m 3

 10.35 kJ/kg




h6  h5  w pIII ,in  670.38  10.35  680.73 kJ/kg
P7  10 MPa  h7  3625.8 kJ/kg

T7  600C  s 7  6.9045 kJ/kg  K
P8  0.6 MPa 
 h8  2821.8 kJ/kg
s8  s 7

s9  s f
6.9045  1.5302

 0.9602
P9  0.2 MPa  x 9 
s fg
5.5968

s9  s7

h9  h f  x 9 h fg  504.71  0.9602 2201.6   2618.7 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.
10-42
s10  s f 6.9045  0.4762

 0.8119
P10  5 kPa  x10 
s fg
7.9176

s10  s 7

h10  h f  x10 h fg  137.75  0.81192423.0  2105.0 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  Δke  Δpe  0 ,
FWH-2:
E in  E out  E system 0 (steady)  0
E in  E out
 m h   m h
i i
e e

 m 8 h8  m 4 h4  m 5 h5 
 yh8  1  y h4  1h5 
8 / m
 5 ). Solving for y,
where y is the fraction of steam extracted from the turbine (  m
y
h5  h4 670.38  505.13

 0.07133
h8  h4 2821.8  505.13
FWH-1:
 m h   m h
i i
e e

 m 9 h9  m 2 h2  m 3 h3 
 zh9  1  y  z h2  1  y h3
9 / m
 5 ) at the second stage. Solving for z,
where z is the fraction of steam extracted from the turbine (  m
z
h3  h2
1  y   504.71  137.95 1  0.07136  0.1373
h9  h2
2618.7  137.95
Then,
qin  h7  h6  3625.8  680.73  2945.0 kJ/kg
qout  1  y  z h10  h1   1  0.07133  0.13732105.0  137.75  1556.8 kJ/kg
wnet  qin  qout  2945.0  1556.8  1388.2 kJ/kg
and
W net  m wnet  22 kg/s 1388.2 kJ/kg   30,540 kW  30.5 MW
(b) The thermal efficiency is
 th  1 
q out
1556.8 kJ/kg
 1
 47.1%
q in
2945.0 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.
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