8.25 Superheated steam at 20 MPa, 560oC 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 75oC.
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 20oC with a
mass flow rate of 70.7 kg/s and exits the condenser at 38oC. 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:
Q in
p1 = 20 MPa
T1 = 560oC
1
Boiler
T
W t
Turbine
1
ht = 81%
T1 = 560oC
2
p2 = 0.5 bar
20.1 MPa
T6 =
6
p4 = 20.1 MPa
4
Condenser
Pump
W p
3
p3 = 0.4 bar
T3 = 75oC
4
Cooling
water
Q out
5
hp = 85%
20 MPa
38oC
0.5 bar
3
2s
2
0.4 bar
T5 = 20oC
m cw  70.7 kg/s
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: p1 = 20 MPa (200 bar), T1 = 560oC → h1 = 3423.0 kJ/kg, s1 = 6.3705 kJ/kg∙K
State 2s: p2s = p2 = 0.5 bar, s2s = s1 = 6.3705 kJ/kg∙K → x2s = 0.8119, h2s = 2212.2 kJ/kg
State 2: p2 = 0.5 bar, h2 = 2442.3 kJ/kg (see below)
1
s
ht 
h1  h2
kJ
kJ
 h2  h1  ht (h1  h2 s )  3423.0  (0.81)(3423.0  2212.2) = 2442.3 kJ/kg
h1  h2 s
kg
kg
State 3: p3 = 0.4 bar, T3 = 75oC → From Table A-2 p3 > psat @ 75oC. Thus, state 3 is a subcooled liquid state. Since the pressure is low, h3 ≈ hf3 at 75oC = 313.93 kJ/kg,
v3 ≈ vf3 at 75oC = 0.0010259 m3/kg
State 4: p4 = 20.1 MPa (201 bar), h4 = 338.14 kJ/kg (see below)
hp 
h4  313.93
kJ

kg
(0.0010259
v3 ( p4  p3 )
v ( p  p3 )
 h4  h3  3 4
h4  h3
hp
m3
N
)(201  0.4) bar
1000 2
100
kPa
1 kJ
kg
m
= 338.14 kJ/kg
0.85
1 bar
1 kPa 1000 N  m
State 5: T5 = 20oC, liquid → h5 ≈ hf5 at 20oC = 83.96 kJ/kg
State 6: T6 = 38oC, liquid → h6 ≈ hf6 at 38oC = 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
 (h2  h3 )  m
 cw (h5  h6 )
0m
where m is the mass flow rate of the steam and m cw is the mass flow rate of the cooling water.
Rearranging to solve for the mass flow rate of steam gives
m 
m cw (h6  h5 )
(h2  h3 )
Substituting values and solving give

m
(70.7 kg/s) (159.21  83.96) kJ/kg
= 2.50 kg/s
(2442.3  313.93) kJ/kg
(b) The thermal efficiency is
2
h
  Wp / m
 (h1  h2 )  (h4  h3 )
Wt / m


Q / m
(h  h )
in
1
4
Substituting enthalpy values and solving yield
h
(3423.0  2442.3) kJ/kg  (338.14  313.93) kJ/kg
= 0.3101 (31.01%)
(3423.0  338.14) kJ/kg
3
8.35 Steam is the working fluid in the vapor power cycle with reheat shown in Fig. P8.35 with
operational data. The mass flow rate is 2.3 kg/s, and the turbines and pump operate
adiabatically. Steam exits both turbine 1 and turbine 2 as saturated vapor. If the reheat pressure
is 15 bar, determine the power developed by the cycle, in kW, and the cycle thermal efficiency.
KNOWN: A vapor power cycle with reheat operates with steam as the working fluid.
Operational data are provided.
FIND: Determine the power developed by the cycle, in kW, and the cycle thermal efficiency.
SCHEMATIC AND GIVEN DATA:
T
1
State
p
(bar)
h
(kJ/kg)
x
1
160
3353.3
--
2
15
2792.2
1.0
3
15
3169.2
--
4
1.5
2693.6
1.0
5
1.5
467.11
0
6
160
486.74
--
p = 160 bar
3
6
p = 15 bar
2
p = 1.5 bar
5
4
s
Fig. P8.35
Q in
Reheat
Section
Steam
Generator
3
p3 = p2 = 15 bar
2
p2 = 15 bar, x2 = 1.0
1
Turbine
1
p1 = 160 bar
W t
Turbine
2
4
p6 = p1 = 160 bar
p4 = 1.5 bar
x4 = 1.0
6
Condenser
5
Pump
W p
ENGINEERING MODEL:
p5 = p4 = 1.5 bar
x5 = 0 (saturated liquid)
1
Q out
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. The turbines and pump operate adiabatically.
3. Kinetic and potential energy effects are negligible.
ANALYSIS:
The net power developed by the cycle is
Wcycle  Wt1  Wt2  Wp
  2.3 kg/s, mass and energy rate balances for control volumes
For a steam mass flow rate of m
around Turbine 1, Turbine 2, and the pump give, respectively,
kg 
kJ 1 kW

 (h1  h2 )   2.3 (3353.3  2792.2)
Turbine 1: Wt1  m
= 1290.53 kW
s 
kg 1 kJ/s

kg 
kJ 1 kW

 (h3  h4 )   2.3 (3169.2  2693.6)
Turbine 2: Wt2  m
= 1093.88 kW
s 
kg 1 kJ/s

Pump:
kg 
kJ 1 kW

 (h6  h5 )   2.3 (486.74  467.11)
= 45.15 kW
Wp  m
s 
kg 1 kJ/s

Substituting and solving for the net power developed by the cycle give
Wcycle  1290.53 kW  1093.88 kW  45.15 kW = 2339.3 kW
The thermal efficiency is
Wcycle
 
Q
in
The total rate of heat transfer to the working fluid as it passes through the steam generator and
reheater section is determined using mass and energy rate balances as
 (h1  h6 )  (h3  h2 )
Qin  m
Solving for rate of heat transfer gives
kg 
kJ 1 kW

= 7460.2 kW
Qin   2.3 (3353.3  486.74)  (3169.2  2792.2)
s 
kg 1 kJ/s

The thermal efficiency is then
2

Wcycle 2339.3 kW
= 0.314 (31.4%)

Qin
7460.2 kW
3