10-109
10-104 A steam power plant operating on the ideal reheat-regenerative Rankine cycle with three feedwater heaters is
considered. Various items for this system per unit of mass flow rate through the boiler are to be determined.
Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.
High-P
Turbine
Low-P
Turbine
13
Boiler
16
z
14
15
12
Closed
FWH I
10
8
11
9
P IV
Closed
FWH II
7
6
5
n
17
y
x
19
P II
4
Open
FWH
3
m
18
Condenser
PI
2
1
P III
Analysis The compression processes in the pumps and the expansion processes in the turbines are isentropic. Also, the state
of water at the inlet of pumps is saturated liquid. Then, from the steam tables (Tables A-4, A-5, and A-6),
h1
h2
h3
h4
h5
h6
h9
h10
 168.75 kJ/kg
 168.84 kJ/kg
 417.51 kJ/kg
 419.28 kJ/kg
 884.46 kJ/kg
 885.86 kJ/kg
 1008.3 kJ/kg
 1011.8 kJ/kg
h13  3423.1 kJ/kg
h14  3204.5 kJ/kg
h15  3063.6 kJ/kg
h16  2871.0 kJ/kg
h17  3481.3 kJ/kg
h18  2891.5 kJ/kg
h19  2454.7 kJ/kg
For an ideal closed feedwater heater, the feedwater is heated to the exit temperature of the extracted steam, which ideally
leaves the heater as a saturated liquid at the extraction pressure. Then,
P7  1800 kPa

 h7  884.91 kJ/kg
T7  T5  207.1C 
P11  3000 kPa

 h  1008.8 kJ/kg
T11  T9  233.9C  11
Enthalpies at other states and the fractions of steam extracted from the turbines can be determined from mass and energy
balances on cycle components as follows:
Mass Balances:
x  y  z 1
mn  z
Open feedwater heater:
mh18  nh2  zh3
Closed feedwater heater-II:
zh4  yh15  zh7  yh5
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.
10-110
Closed feedwater heater-I:
( y  z )h8  xh14  ( y  z )h11  xh9
Mixing chamber after closed feedwater heater II:
zh7  yh6  ( y  z )h8
Mixing chamber after closed feedwater heater I:
xh10  ( y  z )h11  1h12
Substituting the values and solving the above equations simultaneously using EES, we obtain
h8  885.08 kJ/kg
h12  1009.0 kJ/kg
x  0.05334
y  0.1667
z  0.78000
m  0.07124
n  0.70882
Note that these values may also be obtained by a hand solution by using the equations above with some rearrangements and
substitutions. Other results of the cycle are
wT,out,HP  x(h13  h14 )  y (h13  h15 )  z (h13  h16 )  502.3 kJ/kg
wT,out,LP  m(h17  h18 )  n(h17  h19 )  769.6 kJ/kg
q in  h13  h12  z (h17  h16 )  2890 kJ/kg
q out  n(h19  h1 )  1620 kJ/kg
 th  1 
q out
1620
1
 0.4394  43.9%
q in
2890
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.