HW #5
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University of Illinois, Chicago *
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325
Subject
Mechanical Engineering
Date
Apr 3, 2024
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1 8.27
Steam is the working fluid in the ideal reheat cycle shown in Fig. P8.27 together with operational data. If the mass flow rate is 1.3 kg/s, determine the power developed by the cycle, in kW, and the cycle thermal efficiency. KNOWN:
An ideal reheat cycle 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: State p
(bar) T
(
o
C) h
(kJ/kg) 1 140 520.0 3377.8 2 15 201.2 2800.0 3 15 428.9 3318.5 4 1 99.63 2675.5 5 1 99.63 417.46 6 140 431.96 Fig. P8.27 6
s
T
5
2
1
p
= 1 bar
p
= 140 bar
3
4
p
= 15 bar
t
W
±
4
Condenser
out
Q
±
Pump
p
W
±
5
6
1
Steam
Generator
in
Q
±
p
5
= p
4
= 1 bar
x
5
= 0 (saturated liquid)
p
1
= 140 bar
p
4
= 1 bar
p
6
= p
1
= 140 bar
Turbine
2
Turbine
1
2
3
p
2
= 15 bar
p
3
= p
2
= 15 bar
Reheat
Section
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2 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. 3.
The turbine and pump operate adiabatically. 4.
Kinetic and potential energy effects are negligible. 5.
Condensate exits the condenser as saturated liquid. ANALYSIS: The net power developed by the cycle is p
t2
t1
cycle
W
W
W
W
±
±
±
±
±
²
Mass and energy rate balances for control volumes around the two turbine stages and the pump give, respectively, Turbine 1: )
(
2
1
t1
h
h
m
W
±
±
±
Turbine 2: )
(
4
3
t2
h
h
m
W
±
±
±
Pump: )
(
5
6
p
h
h
m
W
±
±
±
where m
±
is the mass flow rate of the steam. Solving for the net power developed by the cycle yields )]
(
)
(
)
[(
5
6
4
3
2
1
cycle
h
h
h
h
h
h
m
W
±
±
±
²
±
±
±
kJ/s
1
kW
1
kg
kJ
)
46
.
417
96
.
431
(
kg
kJ
)
5
.
2675
5
.
3318
(
kg
kJ
)
0
.
2800
8
.
3377
(
s
kg
3
.
1
cycle
»
¼
º
«
¬
ª
±
±
±
²
±
¸
¹
·
¨
©
§
W
±
cycle
W
±
= 1568.2 kW
The thermal efficiency is in
cycle
Q
W
±
±
K
The total rate of heat transfer to the working fluid as it passes through the steam generator and reheater is determined using mass and energy rate balances as >
@
)
(
)
(
2
3
6
1
in
h
h
h
h
m
Q
±
²
±
±
±
3 Solving for rate of heat transfer gives >
@
kJ/s
1
kW
1
kg
kJ
)
0
.
2800
5
.
3318
(
)
96
.
431
8
.
3377
(
s
kg
3
.
1
in
±
²
±
¸
¹
·
¨
©
§
Q
±
= 4503.6 kW The thermal efficiency is then kW
503.6
4
kW
568.2
1
in
cycle
Q
W
±
±
K
= 0.3482 (34.82%)
1 8.33
Steam heated at constant pressure in a steam generator enters the first stage of a supercritical reheat cycle at 28 MPa, 520
o
C. Steam exiting the first-stage turbine at 6 MPa is reheated at constant pressure to 500
o
C. 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: 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:
p
1
= 28 MPa, T
1
= 520
o
C → h
1
= 3192.3 kJ/kg, s
1
= 5.9566 kJ/kg∙K
t
W
±
4
Condenser
out
Q
±
Pump
p
W
±
5
6
1
Steam
Generator
in
Q
±
p
5
= p
4
x
5
= 0 (saturated liquid)
p
1
= 28 MPa
T
1
= 520
o
C
p
4
= p
p
6
= p
1
= 28 MPa
Turbine
2
Turbine
1
2
3
p
2
= 6 MPa
p
3
= p
2
= 6 MPa
T
3
= 500
o
C
Reheat
Section
h
t1
= h
t2
= 78%
h
p
= 82%
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2 State 2s
: p
2s
= p
2
= 6 MPa, s
2s
= s
1
= 5.9566 kJ/
kg∙K
→ h
2s
= 2822.2 kJ/kg State 2
: p
2
= 6 MPa, h
2
= 2903.6 kJ/kg (see below) kg
kJ
)
2
.
2822
3
.
3192
)(
78
.
0
(
kg
kJ
3
.
3192
)
(
2
1
t1
1
2
2
1
2
1
t1
±
±
±
±
o
±
±
s
s
h
h
h
h
h
h
h
h
h
h
= 2903.6 kJ/kg State 3:
p
3
= 6 MPa, T
3
= 500
o
C → h
3
= 3422.2 kJ/kg, s
3
= 6.8803 kJ/kg∙K
State 4s
: p
4s
= p
4
= 6 kPa, s
4s
= s
3
= 6.8803 kJ/kg∙K → x
4s
= 0.8143, h
4s
= 2118.8 kJ/kg State 4
: p
4
= 6 MPa, h
4
= 2405.5 kJ/kg (see below) kg
kJ
)
8
.
2118
2
.
3422
)(
78
.
0
(
kg
kJ
2
.
3422
)
(
4
3
t2
3
4
4
3
4
3
t2
±
±
±
±
o
±
±
s
s
h
h
h
h
h
h
h
h
h
h
= 2405.5 kJ/kg
State 5
: p
5
= 6 kPa
, saturated liquid → h
5
= h
f5
= 151.53 kJ/kg, v
5
= v
f5
= 0.0010064 m
3
/kg State 6
: p
6
= p
1
= 28 MPa, h
6
= 185.89 kJ/kg (see below) p
5
6
5
5
6
5
6
5
6
5
p
)
(
)
(
h
h
p
p
h
h
h
h
p
p
±
²
o
±
±
v
v
m
N
1000
kJ
1
kPa
1
m
N
1000
82
.
0
kPa
)
6
000
,
28
(
kg
m
0010064
.
0
kJ/kg
53
.
151
2
3
6
±
²
h
= 185.89 kJ/kg (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, h
f4
= 151.53 kJ/kg and h
fg4
= 2415.9 kJ/kg, as follows: kJ/kg
9
.
2415
kJ/kg
)
53
.
151
5
.
2405
(
4
fg
4
f
4
4
±
±
h
h
h
x
= 0.9330
The cycle thermal efficiency is m
Q
m
Q
m
W
m
W
m
W
m
Q
m
W
±
±
±
±
±
±
±
±
±
±
±
±
±
±
/
/
/
/
/
/
/
23
61
p
t2
t1
in
cycle
²
±
²
h
)
(
)
(
)
(
)
(
)
(
2
3
6
1
5
6
4
3
2
1
h
h
h
h
h
h
h
h
h
h
±
²
±
±
±
±
²
±
h
3 Substituting enthalpy values and solving yield a thermal efficiency of kJ/kg
)
6
.
2903
2
.
3422
(
kJ/kg
)
.89
185
.3
3192
(
kJ/kg
)
53
.
151
.89
185
(
kJ/kg
)
5
.
2405
2
.
3422
(
kJ/kg
)
6
.
2903
.3
3192
(
±
²
±
±
±
±
²
±
h
h
= 0.3606 (36.06%) (b)
For p
ranging from 4 kPa to 70 kPa, IT gives the following results: IT Code // Known Properties 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))
IT Output for p
4
= 6 kPa
eff_thermal 0.3606 h1 3192 h2 2903 h2s 2821 h3 3422 h4 2405 h4s 2118 h5 151 h6 185.4 p2 6000 p2s 6000 p4s 6 p5 6 p6 2.8E4 s1 5.956 s2s 5.956 s3 6.879 s4s 6.879 v5 0.001007 x4 0.933 eff_p 0.82 eff_t1 0.78 eff_t2 0.78 p1 2.8E4 p3 6000 p4 6 T1 520 T3 500 x5 0
4 Results from IT for p
4
= 6 kPa correspond closely to the results obtained using steam tables and hand calculations. Graphical Results from IT are shown below: 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. Quality versus Condenser Pressure
Pressure [kPa]
70
65
60
55
50
45
40
35
30
25
20
15
10
5
0
1
0.99
0.98
0.97
0.96
0.95
0.94
0.93
0.92
Thermal Efficiency versus Condenser Pressure
Condenser Pressure (kPa)
70
60
50
40
30
20
10
0
0.4
0.3
0.2
0.1
0
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1 8.40
A power plant operates on a regenerative vapor power cycle with one open feedwater heater. Steam enters the first turbine stage at 12 MPa, 560
o
C and expands to 1 MPa, where some of the steam is extracted and diverted to the open feedwater heater operating at 1 MPa. The remaining steam expands through the second turbine stage to the condenser pressure of 6 kPa. Saturated liquid exits the open feedwater heater at 1 MPa. The net power output for the cycle is 330 MW. For isentropic processes in the turbines and pumps, 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 open feedwater heater, in kW/K. KNOWN:
A regenerative vapor power cycle with one open 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, and (c) the rate of entropy production in the open feedwater heater, in kW/K. SCHEMATIC AND GIVEN DATA: t
W
±
3
Condenser
out
Q
±
Pump
2
p2
W
±
4
7
1
Steam
Generator
in
Q
±
p
4
= p
3
= 6 kPa
x
4
= 0 (saturated liquid)
p
1
= 12 MPa
T
1
= 560
o
C
p
3
= 6 kPa
p
7
= p
1
= 12 MPa
MW
330
cycle
W
±
Open
Feedwater
Heater
Pump
1
p1
W
±
5
6
2
(1)
(1)
(1
–
y
)
(1
–
y
)
(1
–
y
)
(
y
)
p
6
= p
5
= p
2
= 1 MPa
x
6
= 0 (saturated liquid)
Turbine
p
2
= 1 MPa
2 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 mixing in the open feedwater heater. 3.
The turbines, pumps, and open feedwater heater operate adiabatically. 4.
Kinetic and potential energy effects are negligible. 5.
Saturated liquid exits the open feedwater heater, and saturated liquid exits the condenser. ANALYSIS: First fix each principal state. State 1
: p
1
= 12 MPa (120 bar), T
1
= 560
o
C → h
1
= 3506.2 kJ/kg, s
1
= 6.6840 kJ/kg∙K
State 2
: p
2
= 1 MPa (10 bar), s
2
= s
1
= 6.6840 kJ/kg∙K
→ h
2
= 2823.3 kJ/kg State 3
: p
3
= 6 kPa (0.06 bar), s
3
= s
1
= 6.6840 kJ/kg∙K → x
3
= 0.7892, h
3
= 2058.2 kJ/kg State 4
: p
4
= 6 kPa (0.06 bar), saturated liquid → h
4
= 151.53 kJ/kg, s
4
= 0.521
0 kJ/kg∙K
, v
4
= 0.0010064 m
3
/kg State 5
: p
5
= p
2
= 1 MPa (10 bar), s
5
= s
4
= 0.521
0 kJ/kg∙K → )
(
4
5
4
4
5
p
p
h
h
±
²
|
v
m
N
1000
kJ
1
kPa
1
N/m
1000
kPa
)
6
1000
(
kg
m
0010064
.
0
kg
kJ
53
.
151
2
3
5
±
¸
¸
¹
·
¨
¨
©
§
²
h
= 152.53 kJ/kg State 6
: p
6
= 1 MPa (10 bar), saturated liquid → h
6
= 762.81 kJ/kg, s
6
= 2.1387 kJ/kg∙K
, v
6
= 0.0011273 m
3
/kg s
T
6
5
2
1
p
= 6 kPa
p
= 12 MPa
3
4
p
= 1 MPa
7
3 State 7
: p
7
= p
1
= 12 MPa (120 bar), s
7
= s
6
= 2.1387 kJ/kg∙K → )
(
6
7
6
6
7
p
p
h
h
±
²
|
v
m
N
1000
kJ
1
kPa
1
N/m
1000
kPa
)
1000
12000
(
kg
m
0011273
.
0
kg
kJ
81
.
762
2
3
7
±
¸
¸
¹
·
¨
¨
©
§
²
h
= 775.21 kJ/kg (a)
Applying energy and mass balances to the control volume enclosing the open feedwater heater, the fraction of flow, y
, extracted at location 2 is kJ/kg
)
53
.
152
3
.
2823
(
kJ/kg
)
53
.
152
81
.
762
(
5
2
5
6
±
±
±
±
h
h
h
h
y
= 0.2285 For the control volume surrounding the turbine stages )
)(
1
(
)
(
3
2
2
1
1
t
h
h
y
h
h
m
W
±
±
²
±
±
±
kg
kJ
)
2
.
2058
3
.
2823
)(
2285
.
0
1
(
kg
kJ
)
3
.
2823
2
.
3506
(
1
t
±
±
²
±
m
W
±
±
= 1273.2 kJ/kg For the pumps )
)(
1
(
)
(
4
5
6
7
1
p
h
h
y
h
h
m
W
±
±
²
±
±
±
kg
kJ
)
53
.
151
53
.
152
)(
2285
.
0
1
(
kg
kJ
)
81
.
762
21
.
775
(
1
p
±
±
²
±
m
W
±
±
= 13.17 kJ/kg For the working fluid passing through the steam generator kg
kJ
)
21
.
775
2
.
3506
(
7
1
1
in
±
±
h
h
m
Q
±
±
= 2731.0 kJ/kg Thus, the thermal efficiency is kJ/kg
0
.
2731
kJ/kg
)
17
.
13
2
.
1273
(
/
/
/
1
in
1
p
1
t
±
±
m
Q
m
W
m
W
±
±
±
±
±
±
K
= 0.461 (46.1%)
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4 (b)
The net
power developed is )
/
/
(
1
p
1
t
1
cycle
m
W
m
W
m
W
±
±
±
±
±
±
±
Thus, )
/
/
(
1
p
1
t
cycle
1
m
W
m
W
W
m
±
±
±
±
±
±
±
MW
1
s
kJ
1000
kg
kJ
)
17
.
13
2
.
1273
(
MW
330
1
±
m
±
= 261.9 kg/s
(c)
The rate of entropy production in the open feedwater heater is determined using the steady-
state form of the entropy rate balance: cv
0
V
±
±
±
±
²
±
²
¦
¦
¦
e
e
e
i
i
i
j
j
j
s
m
s
m
T
Q
Since the feedwater heater is adiabatic, the heat transfer term drops. Thus, 5
5
2
2
6
6
cv
s
m
s
m
s
m
s
m
s
m
i
i
i
e
e
e
±
±
±
±
±
±
±
±
±
¦
¦
V
]
)
1
(
[
5
2
6
1
cv
s
y
ys
s
m
±
±
±
±
±
V
kJ/s
1
kW
1
K
kg
kJ
)]
5210
.
0
)(
2285
.
0
1
(
)
6840
.
6
)(
2285
.
0
(
1387
.
2
[
s
kg
9
.
261
cv
±
±
±
V
±
= 54.86 kW/K
Mixing of streams within the open feedwater heater is a source of irreversibility that produces entropy.
1 8.52
As indicated in Fig. P8.52, a power plant similar to that in Fig. 8.11 operates on a regenerative vapor power cycle with one closed feedwater heater. Steam enters the first turbine stage at state 1 where pressure is 12 MPa and temperature is 560
o
C. Steam expands to state 2 where pressure is 1 MPa and some of the steam is extracted and diverted to the closed feedwater heater. Condensate exits the feedwater heater at state 7 as saturated liquid at a pressure of 1 MPa, undergoes a throttling process through a trap to a pressure of 6 kPa at state 8, and then enters the condenser. The remaining steam expands through the second turbine stage to a pressure of 6 kPa at state 3 and then enters the condenser. Saturated liquid feedwater exiting the condenser at state 4 at a pressure of 6 kPa enters a pump and exits the pump at a pressure of 12 MPa. The feedwater then flows through the closed feedwater heater, exiting at state 6 with a pressure of 12 MPa. The net power output for the cycle is 330 MW. For isentropic processes in each turbine stage and the pump, 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. (d) the rate of entropy production in the steam trap, in kW/K. 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: State p (kPa) T (
o
C) h (kJ/kg) s (kJ/kg∙K)
x 1 12,000 560 3506.2 6.6840 2 1,000 2823.3 6.6840 3 6 2058.2 6.6840 0.7892 4 6 151.53 0.5210 0 5 12,000 163.60 0.5210 6 12,000 606.61 1.7808 7 1,000 762.81 2.1387 0 8 6 762.81 2.4968 0.2530 P8.52 s
T
8
5
2
1
p
= 6 kPa
p
= 12 MPa
3
4
p
= 1 MPa
7
6
2 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 heat transfer through a finite temperature difference in the closed feedwater heater and throttling through the 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: (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 kJ/kg
)
81
.
762
3
.
2823
(
kJ/kg
)
60
.
163
61
.
606
(
7
2
5
6
±
±
±
±
h
h
h
h
y
= 0.2150 For the control volume surrounding the turbine stages )
)(
1
(
)
(
3
2
2
1
1
t
h
h
y
h
h
m
W
±
±
²
±
±
±
kg
kJ
)
2
.
2058
3
.
2823
)(
2150
.
0
1
(
kg
kJ
)
3
.
2823
2
.
3506
(
1
t
±
±
²
±
m
W
±
±
= 1283.5 kJ/kg t
W
±
3
Condenser
out
Q
±
4
6
1
Steam
Generator
in
Q
±
p
4
= p
3
= 6 kPa
x
4
= 0 (saturated liquid)
p
1
= 12 MPa
T
1
= 560
o
C
p
3
= 6 kPa
p
6
= p
1
= 12 MPa
MW
330
cycle
W
±
Closed
Feedwater
Heater
Pump
7
2
(1)
(1)
(1
–
y
)
(1
–
y
)
(1)
(
y
)
p
7
= p
2
= 1 MPa
x
7
= 0 (saturated liquid)
Turbine
5
Trap
p
8
= 6 kPa
p
2
= 1 MPa
p
5
= 12 MPa
(
y
)
(
y
)
(1)
p
W
±
8
Your preview ends here
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3 For the pump )
(
4
5
1
p
h
h
m
W
±
±
±
kg
kJ
)
53
.
151
60
.
163
(
1
p
±
m
W
±
±
= 12.07 kJ/kg For the working fluid passing through the steam generator kg
kJ
)
61
.
606
2
.
3506
(
6
1
1
in
±
±
h
h
m
Q
±
±
= 2899.6 kJ/kg Thus, the thermal efficiency is kJ/kg
6
.
2899
kJ/kg
)
07
.
12
5
.
1283
(
/
/
/
1
in
1
p
1
t
±
±
m
Q
m
W
m
W
±
±
±
±
±
±
K
= 0.438 (43.8%)
(b)
The net
power developed is )
/
/
(
1
p
1
t
1
cycle
m
W
m
W
m
W
±
±
±
±
±
±
±
Thus, )
/
/
(
1
p
1
t
cycle
1
m
W
m
W
W
m
±
±
±
±
±
±
±
MW
1
s
kJ
1000
kg
kJ
)
07
.
12
5
.
1283
(
MW
330
1
±
m
±
= 259.6 kg/s
(c)
The rate of entropy production in the closed feedwater heater is determined using the steady-
state form of the entropy rate balance: cv
0
V
±
±
±
±
²
±
²
¦
¦
¦
e
e
e
i
i
i
j
j
j
s
m
s
m
T
Q
Since the feedwater heater is adiabatic, the heat transfer term drops. Thus,
4 2
2
5
5
7
7
6
6
cv
s
m
s
m
s
m
s
m
s
m
s
m
i
i
i
e
e
e
±
±
±
±
±
±
±
±
±
²
±
¦
¦
V
)]
(
[
2
7
5
6
1
cv
s
s
y
s
s
m
±
²
±
±
±
V
kJ/s
1
kW
1
K
kg
kJ
)]
6840
.
6
1387
.
2
)(
2150
.
0
(
5210
.
0
7808
.
1
[
s
kg
6
.
259
cv
±
²
±
V
±
= 73.35 kW/K
Heat transfer between a finite temperature difference within the closed feedwater heater is a source of irreversibility that produces entropy. (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: cv
)
(
0
V
±
±
±
²
±
²
¦
e
i
j
j
j
s
s
m
T
Q
where m
±
is the mass flow rate through the steam trap. Since the steam trap is adiabatic, the heat transfer term drops. Thus, )
(
)
(
)
(
7
8
1
7
8
7
cv
s
s
m
y
s
s
m
s
s
m
i
e
±
±
±
±
±
±
±
V
kJ/s
1
kW
1
K
kg
kJ
)]
1387
.
2
4968
.
2
(
s
kg
6
.
259
)
2150
.
0
(
cv
±
¸
¹
·
¨
©
§
V
±
= 19.99 kW/K
The throttling process in the steam trap is a source of irreversibility that produces entropy.
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