8.6 A Carnot vapor power cycle operates with water as the working fluid. Saturated liquid
enters the boiler at 1800 lbf/in.2, and saturated vapor enters the turbine (State 1). The
condenser pressure is 1.2 lbf/in.2 The mass flow rate of steam is 1 x 106 lb/h. Data at key
points in the cycle are provided in Fig. P8.6. Determine
(a) the thermal efficiency.
(b) the back work ratio.
(c) the net power developed, in Btu/h.
(d) the rate of heat transfer to the working fluid passing through the boiler, in Btu/h.
State
1
2
3
4
p (lbf/in.2)
1800
1.2
1.2
1800
h (Btu/lb)
1150.4
734.4
840.9
648.3
Fig. P8.6
KNOWN: Water is the working fluid in a Carnot vapor power cycle. Data are given at key
state points in the cycle, and the mass flow rate of steam is known.
FIND: Determine (a) the thermal efficiency, (b) the back work ratio, (c) the net power
developed, and (d) the rate of heat transfer to the working fluid passing through the boiler.
SCHEMATIC AND GIVEN DATA:
State
1
2
3
4
p (lbf/in.2)
1800
1.2
1.2
1800
h (Btu/lb)
1150.4
734.4
840.9
648.3
= 1 x 106 lb/h
p1 = p4 = pboiler = 1800 lbf/in.2
p2 = p3 = pcondenser = 1.2 lbf/in.2
See Cycle 1-2-3’-4’ of Fig. 8.5 for a diagram of the cycle.
ENGINEERING MODEL:
(1) Each component is analyzed as a control volume at steady state.
(2) All processes of the working fluid are internally reversible
(3) The turbine and pump operate adiabatically
(4) Kinetic and potential energy effects are negligible.
ANALYSIS:
The specific enthalpy at each of the principle states is given. The temperatures of the boiler and
condenser are the saturation temperatures corresponding to the respective pressures. From
Table A-3E
T1 = TH = 621.21 oF = 1080.88 oR
T2 = TC = 107.88 oF = 567.55 oR
(a) To find the thermal efficiency, use the boiler and condenser temperatures with the expression
for the Carnot cycle thermal efficiency, as follows:
(b) The back work ratio is
(c) The net power developed is
lb 
Btu

= 238.4×108 Btu/h
Wcycle  1  106 [(1150.4  735.7)  (648.3  472.0)]
h
lb


(d) The rate of heat transfer to the working fluid passing through the boiler is
lb 
Btu

= 502.1×108 Btu/h
Qin  m (h1  h4 )  1  106 (1150.4  648.3)
h
lb

The thermal efficiency can be calculated alternatively as
which is identical to the result of part (a), as expected.
8.12 A nuclear power plant based on the Rankine cycle operates with a boiling-water reactor to
develop net cycle power of 3 MW. Steam exits the reactor core at 100 bar, 520oC and expands
through the turbine to the condenser pressure of 1 bar. Saturated liquid exits the condenser and
is pumped to the reactor pressure of 100 bar. Isentropic efficiencies of the turbine and pump are
81% and 78%, respectively. Cooling water enters the condenser at 15oC with a mass flow rate of
114.79 kg/s. Determine
(a) the thermal efficiency.
(b) the temperature of the cooling water exiting the condenser, in oC.
KNOWN: A nuclear power plant based on the Rankine cycle operates with a boiling-water
reactor between specified pressures of 100 bar and 1 bar to produce 3 MW of net power.
FIND: Determine (a) the thermal efficiency and (b) the temperature of the cooling water exiting
the condenser, in oC.
SCHEMATIC AND GIVEN DATA:
p1 = 100 bar
T1 = 520oC
1
Q in
W t
Turbine
Reactor
Core
2
p2 = 1 bar
Wcycle  3 MW
Condenser
4
p4 = p1 = 100 bar
Q out
6
Cooling water
5
T5 = 15oC
m cooling water  114.79 kg/s
Pump
W p
3
p3 = p2 = 1 bar
x3 = 0 (saturated liquid)
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. Flow through the reactor core and condenser occurs at constant pressure.
3. Stray heat transfer in the turbine, condenser, and pump is ignored.
4. Kinetic and potential energy effects are negligible.
5. Condensate exits the condenser as saturated liquid.
6. For the cooling water, h ≈ hf(T).
1
ANALYSIS: First fix each principal state.
State 1: p1 = 100 bar, T1 = 520oC → h1 = 3425.1 kJ/kg, s1 = 6.6622 kJ/kg∙K
State 2s: p2s = p2 = 1 bar, s2s = s1 = 6.6622 kJ/kg∙K → x2s = 0.8849, h2s = 2415.6 kJ/kg
State 2: p2 = 1 bar, h2 = 2607.4 kJ/kg (see below)
t 
h1  h2
kJ
kJ
 h2  h1  t (h1  h2 s )  3425.1  (0.81)(3425.1  2415.6) = 2607.4 kJ/kg
h1  h2 s
kg
kg
State 3: p3 = 1 bar, saturated liquid → h3 = hf3 = 417.46 kJ/kg, v3 = vf3 = 0.0010432 m3/kg
State 4: p4 = 100 bar, h4 = 430.70 kJ/kg (see below)
p 
v3 ( p4  p3 )
v ( p  p3 )
 h4  h3  3 4
h4  h3
p
m3
N
(0.0010432
)(100  1) bar
1000 2
kJ
100 kPa
1 kJ
kg
m
= 430.70 kJ/kg
h4  417.46 
kg
0.78
1 bar
1 kPa 1000 N  m
State 5: T5 = 15oC, liquid → h5 ≈ hf5 at T5 = 62.99 kJ/kg
(a) The thermal efficiency is

  Wp / m
 (h1  h2 )  (h4  h3 )
Wt / m


Q / m
(h  h )
in
1
4
Substituting enthalpy values and solving yield

(3425.1  2607.4) kJ/kg  (430.70  417.46) kJ/kg
= 0.2687 (26.87%)
(3425.1  430.70) kJ/kg
(b) The temperature of the cooling water exiting the condenser is determined by writing an
energy balance for the condenser, solving for the specific enthalpy of the cooling water exiting
the condenser, and finding the corresponding temperature for the specific enthalpy of the liquid
in Table 2. 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
2
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 specific enthalpy of the cooling water exiting the condenser, h6,
gives
m (h2  h3 )
h6  h5 
m cw
The mass flow rate of steam, m , is determined from the net cycle power

m

m
Wcycle
Wcycle

  Wp / m
 (h1  h2 )  (h4  h3 )
Wt / m
3000 kW
1 kJ/s
= 3.73 kg/s
(3425.1  2607.4) kJ/kg  (430.70  417.46) kJ/kg 1 kW
Solving for the specific enthalpy at the condenser exit gives
h6  62.99 kJ/kg 
(3.73 kg/s) (2607.4  417.46) kJ/kg
= 134.15 kJ/kg
114.79 kg/s
Since the cooling water is liquid, h6 ≈ hf6 at T6. From Table A-2, T6 ≈ 32oC.
3