Post on 03-Jun-2018
8/11/2019 HW 9 Solutions -Chapter 7
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HW- Chapter 7- Solutions
(7 th edition)
33,36,44,51,56
7-33 A rigid tank is divided into two equal parts by a partition. One part is filled with compressedliquid water while the other side is evacuated. The partition is removed and water expands into theentire tank. The entropy change of the water during this process is to be determined.
Analysis The properties of the water are (Table A-4)
K kJ/kg0.8313
/kgm0.001017
C60
kPa004
C60@1
3C60@1
1
1
f
f
s sT
P v v
Noting that
/kgm0.002034001017.022 312 v v
K kJ/kg0278.16430.60002524.00261.1
0002524.0001026.0993.3
001026.0002034.0
/kgm340020.0
kPa40
22
22
32
2
fg f
fg
f
s x s s
x P v
v v
v
Then the entropy change of the water becomes
kJ/K0.492K kJ/kg0.83131.0278kg2.512 s smS
7-36 An insulated cylinder is initially filled with saturated liquid water at a specified pressure. The wateris heated electrically at constant pressure. The entropy change of the water during this process is to bedetermined.
Assumptions 1 The kinetic and potential energy changes are negligible. 2 The cylinder is well-insulatedand thus heat transfer is negligible. 3 The thermal energy stored in the cylinder itself is negligible. 4 Thecompression or expansion process is quasi-equilibrium.
Analysis From the steam tables (Tables A-4 through A-6),
K kJ/kg1.4337
kJ/kg467.13
/kgm0.001053
.
kPa150
kPa150@1
kPa150@1
3kPa150@1
1
f
f
f
s s
hhliquid sat
P v v
Also, kg4.75/kgm0.001053
m0.0053
3
1v
V m
We take the contents of the cylinder as the system. This is a closed system since no mass enters orleaves. The energy balance for this stationary closed system can be expressed as
)( 12ine,
out b,ine,
energiesetc. potential, kinetic,internal,inChange
system
massandwork,heat, bynsfer energy tra Net
outin
hhmW
U W W
E E E
2.5 kg
compressed liquid
Vacuum
H2O
150 kPa2200 kJ
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since U + W b = H during a constant pressure quasi-equilibrium process. Solving for h 2,
kJ/kg33.930kg4.75kJ2200
13.467ine,12 mW
hh
Thus,
K kJ/kg6384.27894.52081.04337.1
2081.00.2226
13.46733.930
kJ/kg33.930
kPa150
22
22
2
2
fg f
fg
f
s x s sh
hh xh
P
Then the entropy change of the water becomes
kJ/K 5.72K kJ/kg1.43372.6384kg4.7512 s smS 7-44 A cylinder is initially filled with saturated water vapor at a specified temperature. Heat istransferred to the steam, and it expands in a reversible and isothermal manner until the pressure dropsto a specified value. The heat transfer and the work output for this process are to be determined.
Assumptions 1 The kinetic and potential energy changes are negligible. 2 The cylinder is well-insulated
and thus heat transfer is negligible. 3 The thermal energy stored in the cylinder itself is negligible. 4 Theprocess is stated to be reversible and isothermal.
Analysis From the steam tables (Tables A-4 through A-6),
K kJ/kg6.8177
kJ/kg2631.1kPa800
K kJ/kg6.4302
kJ/kg2594.2
.
C200
2
2
12
2
C200@1
C200@11
s
u
T T
P
s s
uu
vapor sat
T
g
g
The heat transfer for this reversible isothermal process can bedetermined from
kJ 219.9K kJ/kg)6.43026.8177)(kg1.2)(K 473(12 s sTmS T Q We take the contents of the cylinder as the system. This is a closed system since no mass enters or leaves. Theenergy balance for this closed system can be expressed as
)(
)(
12inout b,
12out b,in
energiesetc. potential, kinetic,internal,inChange
system
massandwork,heat, bynsfer energy tra Net
outin
uumQW
uumU W Q
E E E
Substituting, the work done during this process is determined to be
kJ 175.6kJ/kg)2594.22631.1)(kg1.2(kJ9.219out b,W
7-51 Steam is expanded in an isentropic turbine. The work produced is to be determined.
Assumptions 1 This is a steady-flow process since there is no change with time. 2 The process isisentropic (i.e., reversible-adiabatic).
H2O
200 C
sat. vaporQ
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Analysis There is one inlet and two exits. We take the turbine as the system, which is a control volumesince mass crosses the boundary. The energy balance for this steady-flow system can be expressed inthe rate form as
outin
energiesetc. potential, kinetic,internal,inchangeof Rate
(steady) 0system
massandwork,heat, by
nsfer energy tranetof Rate
outin 0
E E
E E E
332211out
out332211
hmhmhmW
W hmhmhm
From a mass balance,
kg/s75.4)kg/s5)(95.0(95.0
kg/s25.0)kg/s5)(05.0(05.0
13
12
mm
mm
Noting that the expansion process is isentropic, theenthalpies at three states are determined as follows:
6)-A(TableK kJ/kg6953.7
kJ/kg4.2682
C100
kPa50
3
3
3
3 s
h
T
P
6)-A(Table kJ/kg3.3979 KkJ/kg6953.7
MPa41
31
1 h s s
P
6)-A(Table kJ/kg1.3309 KkJ/kg6953.7
kPa7002
32
2 h s s
P
Substituting,
kW6328
kJ/kg).4kg/s)(268275.4(kJ/kg).1kg/s)(330925.0(kJ/kg).3kg/s)(39795(332211out
hmhmhmW
7-56 Refrigerant-134a is compressed in an adiabatic compressor reversibly. The process is to besketched on the T-s diagram and the volume flow rate at the inlet is to be determined.
Assumptions The process is steady.
Analysis (b) Noting that the process is isentropic(constant entropy) the inlet and exit states are
obtained from R-134a tables (Tables A-12 and A-13) as follows:
/kgm06360.0
K kJ/kg9301.0kJ/kg88.251
1
kPa320
31
1
1
1
1
v
sh
x
P
kJ/kg35.279 KkJ/kg9301.0
kPa12002
12
2 h s s
P
T
s
3
1
4 MPa
0.7 MPa
2
Steam
4 MPa
50 kPa
700 kPa
T
s
1
21200
320
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We take the compressor as the system, which is a control volume since mass crosses the boundary.Noting that one fluid stream enters and leaves the compressor, the energy balance for this steady-flowsystem can be expressed in the rate form as
)(
0) peke(since
0
12in
2in1
outin
energiesetc. potential, kinetic,internal,inchangeof Rate
(steady) 0system
massandwork,heat, bynsfer energy tranetof Rate
outin
hhmW
QhmW hm
E E
E E E
Solving for the mass flow rate and substituting,
kg/s640.3)kJ/kg88.251(279.35
kW100
12
in
hhW
m
The volume flow rate at the inlet is then,
/sm0.232 3)/kgm06360.0)(kg/s640.3( 311 v V m