Pump work consumption is small
compared to worked developed, it is
neglected.
8.33 Steam heated at constant pressure in a steam generator enters the first stage of a
supercritical reheat cycle at 28 MPa, 520oC. Steam exiting the first-stage turbine at 6 MPa is
reheated at constant pressure to 500oC. Each turbine stage has an isentropic efficiency of 78%
while the pump has an isentropic efficiency of 82%. Saturated liquid exits the condenser that
operates at constant pressure, p.
(a) For p = 6 kPa, determine the quality of the steam exiting the second stage of the turbine and
the thermal efficiency.
(b) Plot the quantities of part (a) versus p ranging from 4 kPa to 70 kPa.
KNOWN: A supercritical reheat cycle operates with steam as the working fluid.
FIND: (a) For p = 6 kPa, determine the quality of the steam exiting the second stage of the
turbine and the thermal efficiency, and (b) plot the quantities of part (a) versus p ranging from 4
kPa to 70 kPa.
SCHEMATIC AND GIVEN DATA:
Q in
Reheat
Section
Steam
Generator
p3 = p2 = 6 MPa
T3 = 500oC
3
2
p2 = 6 MPa
1
Turbine
Turbine
1
2
p1 = 28 MPa
o
T1 = 520 C
ht1 = ht2 = 78%
4
p6 = p1 = 28 MPa
W t
p4 = p
6
Condenser
hp = 82%
Q out
5
Pump
W p
p5 = p4
x5 = 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. For all components stray heat transfer is ignored.
3. Flow through the steam generator, reheater, and condenser is at constant pressure.
4. Kinetic and potential energy effects are negligible.
ANALYSIS: First fix each principal state.
State 1: p1 = 28 MPa, T1 = 520oC → h1 = 3192.3 kJ/kg, s1 = 5.9566 kJ/kg∙K
1
State 2s: p2s = p2 = 6 MPa, s2s = s1 = 5.9566 kJ/kg∙K → h2s = 2822.2 kJ/kg
State 2: p2 = 6 MPa, h2 = 2903.6 kJ/kg (see below)
ht1 
h1  h2
kJ
kJ
 h2  h1  ht1(h1  h2 s )  3192.3
 (0.78)(3192.3  2822.2 ) = 2903.6 kJ/kg
h1  h2 s
kg
kg
State 3: p3 = 6 MPa, T3 = 500oC → h3 = 3422.2 kJ/kg, s3 = 6.8803 kJ/kg∙K
State 4s: p4s = p4 = 6 kPa, s4s = s3 = 6.8803 kJ/kg∙K → x4s = 0.8143, h4s = 2118.8 kJ/kg
State 4: p4 = 6 MPa, h4 = 2405.5 kJ/kg (see below)
ht2 
h3  h4
kJ
kJ
 h4  h3  ht2 (h3  h4 s )  3422.2
 (0.78)(3422.2  2118.8)
= 2405.5 kJ/kg
h3  h4 s
kg
kg
State 5: p5 = 6 kPa, saturated liquid → h5 = hf5 = 151.53 kJ/kg, v5 = vf5 = 0.0010064 m3/kg
State 6: p6 = p1 = 28 MPa, h6 = 185.89 kJ/kg (see below)
hp 
v5 ( p6  p5 )
v ( p  p5 )
 h6  h5  5 6
h6  h5
hp
0.0010064
h6  151.53 kJ/kg 
m3
N
(28,000  6)kPa 1000
1 kJ
kg
m2
= 185.89 kJ/kg
0.82
1 kPa 1000 N  m
(a) The quality of the steam at the exit of the second stage of the turbine (state 4) is determined
using values from Table A-4, hf4 = 151.53 kJ/kg and hfg4 = 2415.9 kJ/kg, as follows:
x4 
h4  hf 4 (2405.5  151.53) kJ/kg
= 0.9330

hfg 4
2415.9 kJ/kg
The cycle thermal efficiency is
h
 Wt1 / m
  Wt2 / m
  Wp / m

Wcycle / m





 Q /m

Q /m
Q /m
in
h
61
23
(h1  h2 )  (h3  h4 )  (h6  h5 )
(h1  h6 )  (h3  h2 )
2
Substituting enthalpy values and solving yield a thermal efficiency of
h
(3192.3  2903.6) kJ/kg  (3422.2  2405.5) kJ/kg  (185.89  151.53) kJ/kg
(3192.3  185.89) kJ/kg  (3422.2  2903.6) kJ/kg
h = 0.3606 (36.06%)
(b) For p ranging from 4 kPa to 70 kPa, IT gives the following results:
IT Code
IT Output for p4 = 6 kPa
// Known Properties
eff_thermal
h1
h2
h2s
h3
h4
h4s
h5
h6
p2
p2s
p4s
p5
p6
s1
s2s
s3
s4s
v5
x4
eff_p
eff_t1
eff_t2
p1
p3
p4
T1
T3
x5
p1 = 28000 // kPa
T1 = 520 // oC
p6 = p1
p3 = 6000 // kPa
T3 = 500 // oC
p2 = p3
p2s = p2
p4 = 6 // kPa
p4s = p4
p5 = p4
x5 = 0
// Known Operating Parameters
eff_t1 = 0.78
eff_t2 = 0.78
eff_p = 0.82
// Calculations for Quality at State 4
h3 = h_PT("Water/Steam", p3, T3)
s3 = s_PT("Water/Steam", p3, T3)
s4s = s3
h4s = h_Ps("Water/Steam", p4s, s4s)
h4 = h3 - eff_t2*(h3 - h4s)
x4 = x_hP("Water/Steam", h4, p4)
// Calculations for Thermal Efficiency
h1 = h_PT("Water/Steam", p1, T1)
s1 = s_PT("Water/Steam", p1, T1)
s2s = s1
h2s = h_Ps("Water/Steam", p2s, s2s)
h2 = h1 - eff_t1*(h1 - h2s)
h5 = hsat_Px("Water/Steam", p5, x5)
v5 = vsat_Px("Water/Steam", p5, x5)
h6 = h5 + (v5*(p6 - p5)/eff_p)
eff_thermal = ((h1 - h2) + (h3 - h4) - (h6 - h5))/((h1 - h6) + (h3 - h2))
3
0.3606
3192
2903
2821
3422
2405
2118
151
185.4
6000
6000
6
6
2.8E4
5.956
5.956
6.879
6.879
0.001007
0.933
0.82
0.78
0.78
2.8E4
6000
6
520
500
0
Results from IT for p4 = 6 kPa correspond closely to the results obtained using steam tables
and hand calculations.
Graphical Results from IT are shown below:
Quality versus Condenser Pressure
Thermal Efficiency versus Condenser Pressure
1
0.4
0.99
0.98
Thermal Efficiency
0.3
Quality
0.97
0.96
0.95
0.2
0.94
0.1
0.93
0.92
0
0
5
10
15
20
25
30 35 40 45
Pressure [kPa]
50
55
60
65
70
0
10
20
30
40
50
60
70
Condenser Pressure (kPa)
In general, as the condenser pressure increases the quality of the steam increases and the
thermal efficiency decreases since the average temperature of heat rejection is higher. As
shown by the Quality versus Condenser Pressure Graph for the conditions of this problem,
when the condenser pressure reaches approximately 70 kPa, the liquid-vapor mixture quality
becomes 1. Above this pressure steam is superheated vapor and the quality is not defined.
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94.46
317.7
94.46
0.1887
94.46
94.46
317.7
2.295×10^6
317.7
2.295×10^6
2.698×10^9
2.698×10^9
37.06%
8.53 Reconsider the cycle of Problem 8.52, but include in the analysis that each turbine stage
and the pump have isentropic efficiencies of 83%.
KNOWN: A regenerative vapor power cycle with one closed feedwater heater operates with
steam as the working fluid. Operational data are provided.
FIND: Determine (a) the cycle thermal efficiency, (b) the mass flow rate into the first turbine
stage, in kg/s, (c) the rate of entropy production in the closed feedwater heater, in kW/K, and (d)
the rate of entropy production in the steam trap, in kW/K.
SCHEMATIC AND GIVEN DATA:
Q in
(1–y)
p1 = 12 MPa
T1 = 560oC
(1)
Steam
Generator
Turbine
1
W t
ht = 83%
Wcycle  330 MW
p2 = 1 MPa
2
(y)
(1–y)
p6 = p1 = 12 MPa
3
p3 = 6 kPa
(1)
(1)
6
p5 = 12 MPa
Q out
(1)
Condenser
5
Closed
Feedwater
Heater
Pump
hp = 83%
Wp
4
p4 = p3 = 6 kPa
x4 = 0 (saturated liquid)
(y)
Trap
7
8
p7 = p2 = 1 MPa
x7 = 0 (saturated liquid)
1
(y)
p8 = 6 kPa
T
1
p = 12 MPa
5
5s
6 7
2s 2
p = 1 MPa
p = 6 kPa
4
3s 3
8
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. All processes of the working fluid are internally reversible except for processes in the
turbines and pumps, heat transfer through a finite temperature difference in the closed feedwater
heater, and throttling in the steam trap.
3. The turbines, pump, closed feedwater heater, and steam trap operate adiabatically.
4. Kinetic and potential energy effects are negligible.
5. Saturated liquid exits the closed feedwater heater, and saturated liquid exits the condenser.
ANALYSIS: First fix each principal state.
State 1 (same as State 1 in problem 8.52): p1 = 12 MPa (120 bar), T1 = 560oC → h1 = 3506.2
kJ/kg, s1 = 6.6840 kJ/kg∙K
State 2s (same as State 2 in problem 8.52) : p2s = p2 = 1 MPa (10 bar), s2s = s1 = 6.6840
kJ/kg∙K → h2s = 2823.3 kJ/kg
State 2: p2 = 1 MPa (10 bar), h2 = 2939.4 kJ/kg (see below) → s2 = 6.9174 kJ/kg∙K
ht 
h1  h2
kJ
kJ
 h2  h1  ht (h1  h2s )  3506.2  (0.83)(3506.2  2823.3) = 2939.4 kJ/kg
h1  h2s
kg
kg
State 3s: p3s = p3 = 6 kPa (0.06 bar), s3s = s2 = 6.9174 kJ/kg∙K → x3s = 0.8191, h3s = 2130.4
kJ/kg
State 3: p3 = 6 kPa (0.06 bar), h3 = 2267.9 kJ/kg (see below) → x3 = 0.8760, s3 = 7.3620
kJ/kg∙K
2
ht 
h2  h3
kJ
kJ
 h3  h2  ht (h2  h3s )  2939.4  (0.83)(2939.4  2130.4) = 2267.9 kJ/kg
h2  h3s
kg
kg
State 4 (same as State 4 in problem 8.52): p4 = 6 kPa (0.06 bar), saturated liquid →
h4 = 151.53 kJ/kg, v4 = 0.0010064 m3/kg, s4 = 0.5210 kJ/kg∙K
State 5: p5 = 1 MPa (10 bar), h5 = 166.07 kJ/kg (see below) → s5 ≈ 0.5677 kJ/kg∙K (assuming
the saturated liquid state corresponding to h5 = hf in Table 2 and interpolating for s5 = sf)
hp 
v4 ( p5  p4 )
v ( p  p4 )
 h5  h4  4 5
h5  h4
hp
m3
N
(0.0010064
)(12000  6) kPa 1000
kJ
1 kJ
kg
m2
h5  151.53 
= 166.07 kJ/kg
kg
0.83
1 kPa 1000 N  m
State 6 (same as State 6 in problem 8.52): p6 = 12 MPa (120 bar), h6 = 606.61 kJ/kg,
s6 = 1.7808 kJ/kg∙K
State 7 (same as State 7 in problem 8.52): p7 = 1 MPa (10 bar), saturated liquid → h7 = 762.81
kJ/kg, s7 = 2.1387 kJ/kg∙K
State 8 (same as State 8 in problem 8.52): p8 = 6 kPa (0.06 bar), h8 =h7 = 762.81 kJ/kg →
s8 = 2.4968 kJ/kg∙K
(a) Applying energy and mass balances to the control volume enclosing the closed feedwater
heater, the fraction of flow, y, extracted at location 2 is
h h
(606.61  166.07) kJ/kg
y 6 5 
= 0.2024
h2  h7 (2939.4  762.81) kJ/kg
For the control volume surrounding the turbine stages
Wt
 (h1  h2 )  (1  y )(h2  h3 )
m1
Wt
kJ
kJ
 (3506.2  2939.4)  (1  0.2024)(2939.4  2267.9) = 1102.4 kJ/kg
m1
kg
kg
For the pump
3
Wp
m1
Wp
m1
 (h5  h4 )
 (166.07  151.53)
kJ
= 14.54 kJ/kg
kg
For the working fluid passing through the steam generator
Qin
kJ
 h1  h6  (3506.2  606.61) = 2899.6 kJ/kg
m1
kg
Thus, the thermal efficiency is
h
1  Wp / m
1 (1102.4  14.54) kJ/kg
Wt / m
= 0.375 (37.5%)


Q / m
2899.6 kJ/kg
in
1
(b) The net power developed is
1(Wt / m
1  Wp / m
 1)
Wcycle  m
Thus,
m1 
Wcycle
(Wt / m1  Wp / m1)
kJ
s = 303.3 kg/s
m1 
kJ 1 M W
(1102.4  14.54)
kg
1000
330 M W
(c) The rate of entropy production in the closed feedwater heater is determined using the steadystate form of the entropy rate balance:
0
j
Q j
Tj
  m i si   m e se   cv
i
e
Since the feedwater heater is adiabatic, the heat transfer term drops. Thus,
 cv   m e se   m i si  m 6 s6  m 7 s7  m 5s5  m 2 s2
e
i
4
 cv  m1[s6  s5  y(s7  s2 )]
 cv  303.3
kg
kJ 1 kW
= 74.58 kW/K
[1.7808  0.5677  (0.2024)(2.1387  6.9174)]
s
kg  K 1 kJ/s
(d) The rate of entropy production in the steam trap is determined using the one-inlet, one-exit,
steady-state form of the entropy rate balance:
0
Q j
Tj
j
 m ( si  se )   cv
where m is the mass flow rate through the steam trap.
Since the steam trap is adiabatic, the heat transfer term drops. Thus,
 cv  m (se  si )  m 7 (s8  s7 )  ym1(s8  s7 )


 cv  (0.2024) 303.3
kg 
kJ 1 kW
= 21.98 kW/K
(2.4968  2.1387)]
s 
kg  K 1 kJ/s
Compared to the ideal cycle in problem 8.52, the presence of internal irreversibilities in the
turbine stages and the pump results in lower cycle thermal efficiency, higher required mass
flow rate of steam entering the first-stage turbine, and greater rate of entropy production in
the closed feedwater heater. Although the inlet and exit states for the steam trap are the same
as those in Problem 8.52, the rate of entropy production during the throttling process is
greater since the mass flow rate is higher.
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h1 - h2
3833.9 - 2949.1
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