charging and discharging processes and discharging processes ... 25.82 kg 1.9367 m /kg ... 0.25 m3...
TRANSCRIPT
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5-77
Charging and Discharging Processes 5-121 A large reservoir supplies steam to a balloon whose initial state is specified. The final temperature in the balloon and the boundary work are to be determined. Analysis Noting that the volume changes linearly with the pressure, the final volume and the initial mass are determined from
(Table A-6) /kgm 9367.1C150kPa 100 3
11
1 =
°==
vTP
Steam 50 m3
100 kPa 150°C
Steam 150 kPa 200°C
331
1
22 m 75)m 50(
kPa 100kPa 150
=== VVPP
kg 82.25/kgm 1.9367
m 503
3
1
11 ===
v
Vm
The final temperature may be determined if we first calculate specific volume at the final state
/kgm 4525.1kg) 25.82(2
m 752
33
1
2
2
22 =
×===
mmVV
v
(Table A-6) C202.5°=
=
=23
2
2
/kgm 4525.1
kPa 150T
P
v
Noting again that the volume changes linearly with the pressure, the boundary work can be determined from
kJ 3125=−+
=−+
= 312
21 m)5075(2
kPa)150100()(2
VVPP
Wb
5-122 Steam in a supply line is allowed to enter an initially evacuated tank. The temperature of the steam in the supply line and the flow work are to be determined. Analysis Flow work of the steam in the supply line is converted to sensible internal energy in the tank. That is,
4 MPa
Initially evacuated
Steam tankline uh =
where (Table A-6) kJ/kg 5.3189C550
MPa 4tank
tank
tank =
°==
uTP
Now, the properties of steam in the line can be calculated
(Table A-6) kJ/kg 5.2901kJ/kg 5.3189
MPa 4
line
line
line
line
=°=
==
uT
hP C389.5
The flow work per unit mass is the difference between enthalpy and internal energy of the steam in the line kJ/kg 288=−=−= 5.29015.3189linelineflow uhw
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
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5-78
5-123 A vertical piston-cylinder device contains air at a specified state. Air is allowed to escape from the cylinder by a valve connected to the cylinder. The final temperature and the boundary work are to be determined. Properties The gas constant of air is R = 0.287 kJ/kg.K (Table A-1).
Air
Air 0.25 m3 600 kPa 300°C
Analysis The initial and final masses in the cylinder are
33
1
11 m 9121.0
K) 27300kJ/kg.K)(3 287.0()m kPa)(0.25 600(
=+
==RTP
mV
kg 2280.0kg) 9121.0(25.025.0 12 === mm
Then the final temperature becomes
K 458.4===kJ/kg.K) 287.0(kg) 2280.0(
)m kPa)(0.05 600( 3
2
22 Rm
PT
V
Noting that pressure remains constant during the process, the boundary work is determined from
kJ 120=−=−= 321 0.05)mkPa)(0.25 (600)( VVPWb
5-124 Helium flows from a supply line to an initially evacuated tank. The flow work of the helium in the supply line and the final temperature of the helium in the tank are to be determined. Properties The properties of helium are R = 2.0769 kJ/kg.K, cp = 5.1926 kJ/kg.K, cv = 3.1156 kJ/kg.K (Table A-2a). Analysis The flow work is determined from its definition but we first determine the specific volume 200 kPa, 120°C
Initially evacuated
Helium
/kgm 0811.4kPa) 200(
K) 27320kJ/kg.K)(1 0769.2( 3line =+
==P
RTv
kJ/kg 816.2=== /kg)m 1kPa)(4.081 200( 3flow vPw
Noting that the flow work in the supply line is converted to sensible internal energy in the tank, the final helium temperature in the tank is determined as follows
K 655.0=→=→=
=+===
tanktanktanktank-
lineline
linetank
kJ/kg.K) 1156.3(kJ/kg 7.2040
kJ/kg 7.2040K) 27320kJ/kg.K)(1 1926.5(
TTTcu
Tchhu
p
v
Alternative Solution: Noting the definition of specific heat ratio, the final temperature in the tank can also be determined from K 655.1=+== K) 273120(667.1linetank kTT
which is practically the same result.
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
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5-79
5-125 An evacuated bottle is surrounded by atmospheric air. A valve is opened, and air is allowed to fill the bottle. The amount of heat transfer through the wall of the bottle when thermal and mechanical equilibrium is established is to be determined. Assumptions 1 This is an unsteady process since the conditions within the device are changing during the process, but it can be analyzed as a uniform-flow process since the state of fluid at the inlet remains constant. 2 Air is an ideal gas with variable specific heats. 3 Kinetic and potential energies are negligible. 4 There are no work interactions involved. 5 The direction of heat transfer is to the air in the bottle (will be verified). Properties The gas constant of air is 0.287 kPa.m3/kg.K (Table A-1). Analysis We take the bottle as the system, which is a control volume since mass crosses the boundary. Noting that the microscopic energies of flowing and nonflowing fluids are represented by enthalpy h and internal energy u, respectively, the mass and energy balances for this uniform-flow system can be expressed as Mass balance: )0 (since initialout2systemoutin ===→∆=− mmmmmmm i
Energy balance:
)0 (since initialout22in
energies etc. potential, kinetic, internal,in Change
system
mass and work,heat,by nsferenergy traNet
outin
≅≅==≅=+
∆=−
pekeEEWumhmQ
EEE
ii
4342143421
100 kPa17°C
8 L Evacuated
Combining the two balances: Q ( )ihum −= 22in where
kJ/kg 206.91kJ/kg 290.16
K 290
kg 0.0096)K 290)(K/kgmkPa 0.287(
)m 0.008)(kPa 100(
2
17-A Table2
3
3
2
22
==
→==
=⋅⋅
==
uh
TT
RTP
m
ii
V
Substituting, Qin = (0.0096 kg)(206.91 - 290.16) kJ/kg = - 0.8 kJ → Qout = 0.8 kJ Discussion The negative sign for heat transfer indicates that the assumed direction is wrong. Therefore, we reverse the direction. 5-126 An insulated rigid tank is evacuated. A valve is opened, and air is allowed to fill the tank until mechanical equilibrium is established. The final temperature in the tank is to be determined. Assumptions 1 This is an unsteady process since the conditions within the device are changing during the process, but it can be analyzed as a uniform-flow process since the state of fluid at the inlet remains constant. 2 Air is an ideal gas with constant specific heats. 3 Kinetic and potential energies are negligible. 4 There are no work interactions involved. 5 The device is adiabatic and thus heat transfer is negligible. Properties The specific heat ratio for air at room temperature is k = 1.4 (Table A-2). Analysis We take the tank as the system, which is a control volume since mass crosses the boundary. Noting that the microscopic energies of flowing and nonflowing fluids are represented by enthalpy h and internal energy u, respectively, the mass and energy balances for this uniform-flow system can be expressed as Mass balance: )0 (since initialout2systemoutin ===→∆=− mmmmmmm i
Energy balance:
)0 (since initialout22
energies etc. potential, kinetic, internal,in Change
system
mass and work,heat,by nsferenergy traNet
outin
≅≅==≅≅=
∆=−
pekeEEWQumhm
EEE
ii
4342143421
Air
initially evacuated
Combining the two balances: iipipi kTTccTTcTchu ==→=→= )/(222 vv
Substituting, T2 = × = =1.4 290 K 406 K 133 Co
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
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5-80
5-127 A rigid tank initially contains air at atmospheric conditions. The tank is connected to a supply line, and air is allowed to enter the tank until mechanical equilibrium is established. The mass of air that entered and the amount of heat transfer are to be determined. Assumptions 1 This is an unsteady process since the conditions within the device are changing during the process, but it can be analyzed as a uniform-flow process since the state of fluid at the inlet remains constant. 2 Air is an ideal gas with variable specific heats. 3 Kinetic and potential energies are negligible. 4 There are no work interactions involved. 5 The direction of heat transfer is to the tank (will be verified). Properties The gas constant of air is 0.287 kPa.m3/kg.K (Table A-1). The properties of air are (Table A-17)
kJ/kg 02.250K 350kJ/kg 49.210K 295kJ/kg 17.295K 295
22
11=→==→==→=
uTuThT ii
Analysis (a) We take the tank as the system, which is a control volume since mass crosses the boundary. Noting that the microscopic energies of flowing and nonflowing fluids are represented by enthalpy h and internal energy u, respectively, the mass and energy balances for this uniform-flow system can be expressed as Mass balance: 12systemoutin mmmmmm i −=→∆=−
Energy balance:
)0 (since 1122in
energies etc. potential, kinetic, internal,in Change
system
mass and work,heat,by nsferenergy traNet
outin
≅≅≅−=+
∆=−
pekeWumumhmQ
EEE
ii
4342143421
The initial and the final masses in the tank are
kg 11.946
)K 350)(K/kgmkPa 0.287()m 2)(kPa 600(
kg 2.362)K 295)(K/kgmkPa 0.287(
)m 2)(kPa 100(
3
3
2
22
3
3
1
11
=⋅⋅
==
=⋅⋅
==
RTPm
RTPm
V
V
Q·
V1 = 2 m3 P1 = 100 kPa
T1 = 22°C
Pi = 600 kPa Ti = 22°C
Then from the mass balance, m m mi = − = − =2 1 11946 2 362. . 9.584 kg
(b) The heat transfer during this process is determined from
( )( ) ( )( ) ( )( )kJ 339=→−=
−+−=−+−=
out
1122in
kJ 339kJ/kg 210.49kg 2.362kJ/kg 250.02kg 11.946kJ/kg 295.17kg 9.584
Q
umumhmQ ii
Discussion The negative sign for heat transfer indicates that the assumed direction is wrong. Therefore, we reversed the direction.
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
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5-81
5-128 A rigid tank initially contains saturated R-134a liquid-vapor mixture. The tank is connected to a supply line, and R-134a is allowed to enter the tank. The final temperature in the tank, the mass of R-134a that entered, and the heat transfer are to be determined. Assumptions 1 This is an unsteady process since the conditions within the device are changing during the process, but it can be analyzed as a uniform-flow process since the state of fluid at the inlet remains constant. 2 Kinetic and potential energies are negligible. 3 There are no work interactions involved. 4 The direction of heat transfer is to the tank (will be verified). Properties The properties of refrigerant are (Tables A-11 through A-13)
( )
kJ/kg 335.06100
MPa 1.0
kJ/kg 246.79/kgm 0.02562
vaporsat.kPa 800
kJ/kg 182.9219.1720.762.39/kgm 0.037170.00078870.0527620.70.0007887
7.0C8
kPa 800@2
3kPa 800@22
11
311
1
1
=
°==
====
=
=×+=+==−×+=+=
=°=
ii
i
g
g
fgf
fgf
hCT
P
uuP
uxuux
xT
vv
vvv
R-134a
0.2 m3 R-134a
1 MPa 100°C
Analysis We take the tank as the system, which is a control volume since mass crosses the boundary. Noting that the microscopic energies of flowing and nonflowing fluids are represented by enthalpy h and internal energy u, respectively, the mass and energy balances for this uniform-flow system can be expressed as Mass balance: 12systemoutin mmmmmm i −=→∆=−
Energy balance:
)0 (since 1122in
energies etc. potential, kinetic, internal,in Change
system
mass and work,heat,by nsferenergy traNet
outin
≅≅≅−=+
∆=−
pekeWumumhmQ
EEE
ii
4342143421
(a) The tank contains saturated vapor at the final state at 800 kPa, and thus the final temperature is the saturation temperature at this pressure, 31.31ºC == kPa 800 @sat 2 TT
(b) The initial and the final masses in the tank are
kg 7.81
/kgm 0.02562m 0.2
kg 5.38/kgm 0.03717
m 0.2
3
3
22
3
3
11
===
===
vV
vV
m
m
Then from the mass balance 2.43 kg =−=−= 38.581.712 mmmi
(c) The heat transfer during this process is determined from the energy balance to be
( )( ) ( )( ) ( )(kJ 130=
−+−=−+−=
kJ/kg 182.92kg 5.38kJ/kg 246.79kg 7.81kJ/kg 335.06kg 2.43 1122in umumhmQ ii
)
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
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5-82
5-129E A rigid tank initially contains saturated water vapor. The tank is connected to a supply line, and water vapor is allowed to enter the tank until one-half of the tank is filled with liquid water. The final pressure in the tank, the mass of steam that entered, and the heat transfer are to be determined. Assumptions 1 This is an unsteady process since the conditions within the device are changing during the process, but it can be analyzed as a uniform-flow process since the state of fluid at the inlet remains constant. 2 Kinetic and potential energies are negligible. 3 There are no work interactions involved. 4 The direction of heat transfer is to the tank (will be verified). Properties The properties of water are (Tables A-4E through A-6E)
Q
Steam
Water 3 ft3
300°F Sat. vapor
200 psia 400°F
Btu/lbm 1210.9F400psia 020
Btu/lbm 1099.8,51.269/lbmft 4663.6,01745.0
mixture sat.300
Btu/lbm 1099.8/lbmft 4663.6
vaporsat.F030
32
F030@1
3F030@11
=
°==
====
°=
====
°=
°
°
ii
i
gf
gf
g
g
hTP
uuFT
uuT
vv
vv
Analysis We take the tank as the system, which is a control volume since mass crosses the boundary. Noting that the microscopic energies of flowing and nonflowing fluids are represented by enthalpy h and internal energy u, respectively, the mass and energy balances for this uniform-flow system can be expressed as Mass balance: 12systemoutin mmmmmm i −=→∆=−
Energy balance:
)0 (since 1122in
energies etc. potential, kinetic, internal,in Change
system
mass and work,heat,by nsferenergy traNet
outin
≅≅≅−=+
∆=−
pekeWumumhmQ
EEE
ii
4342143421
(a) The tank contains saturated mixture at the final state at 250°F, and thus the exit pressure is the saturation pressure at this temperature, psia 67.03== °F300 @sat2 PP
(b) The initial and the final masses in the tank are
lbm 20.86232.097.85/lbmft 6.4663.ft 1.5
/lbmft 0.01745ft 1.5
lbm 0.464/lbmft 6.4663
ft 3
3
3
3
3
2
3
3
11
=+=+=+=+=
===
g
g
f
fgf mmm
m
v
V
v
V
vV
Then from the mass balance =−=−= 464.020.8612 mmmi 85.74 lbm
(c) The heat transfer during this process is determined from the energy balance to be
( )( ) ( )(Btu 80,900, =→−=
−+−=−+−=
out
1122in
Btu 90080Btu/lbm 1099.8lbm 0.464Btu 23,425Btu/lbm 1210.9lbm 85.74
Q
umumhmQ ii
)
since Btu 23,4251099.80.232269.5197.85222 =×+×=+== ggff umumumU
Discussion A negative result for heat transfer indicates that the assumed direction is wrong, and should be reversed.
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5-83
5-130 A cylinder initially contains superheated steam. The cylinder is connected to a supply line, and is superheated steam is allowed to enter the cylinder until the volume doubles at constant pressure. The final temperature in the cylinder and the mass of the steam that entered are to be determined. Assumptions 1 This is an unsteady process since the conditions within the device are changing during the process, but it can be analyzed as a uniform-flow process since the state of fluid at the inlet remains constant. 2 The expansion process is quasi-equilibrium. 3 Kinetic and potential energies are negligible. 3 There are no work interactions involved other than boundary work. 4 The device is insulated and thus heat transfer is negligible. Properties The properties of steam are (Tables A-4 through A-6)
kJ/kg 3158.2C350
MPa 1
kJ/kg 2643.3/kgm 0.42503
C200kPa 500
1
31
1
1
=
°==
==
°==
ii
i hTP
uTP v
Analysis (a) We take the cylinder as the system, which is a control volume since mass crosses the boundary. Noting that the microscopic energies of flowing and nonflowing fluids are represented by enthalpy h and internal energy u, respectively, the mass and energy balances for this uniform-flow system can be expressed as
Pi = 1 MPa Ti = 350°C
P = 500 kPa T1 = 200°C
V1 = 0.01 m3
Steam
Mass balance: 12systemoutin mmmmmm i −=→∆=−
Energy balance:
)0 (since 1122outb,
energies etc. potential, kinetic, internal,in Change
system
mass and work,heat,by nsferenergy traNet
outin
≅≅≅−+=
∆=−
pekeQumumWhm
EEE
ii
4342143421
Combining the two relations gives ( ) 112212outb,0 umumhmmW i −+−−=
The boundary work done during this process is
( ) ( )( ) kJ 5mkPa 1
kJ 1m0.010.02kPa 500P
2
1 33
12, =
⋅−=−== ∫ VVVdPW outb
The initial and the final masses in the cylinder are
2
3
2
22
3
3
1
11
m 0.02
kg 0.0235/kgm 0.42503
m 0.01
vvV
vV
==
===
m
m
Substituting, ( ) ( )( )3.26430235.002.02.31580235.002.050 222
−+
−−= u
vv
Then by trial and error (or using EES program), T2 = 261.7°C and v2 = 0.4858 m3/kg (b) The final mass in the cylinder is
kg 0.0412/kgm 0.4858
m 0.023
3
2
22 ===
v
Vm
Then, mi = m2 - m1 = 0.0412 - 0.0235 = 0.0176 kg
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
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5-84
5-131 A cylinder initially contains saturated liquid-vapor mixture of water. The cylinder is connected to a supply line, and the steam is allowed to enter the cylinder until all the liquid is vaporized. The final temperature in the cylinder and the mass of the steam that entered are to be determined. Assumptions 1 This is an unsteady process since the conditions within the device are changing during the process, but it can be analyzed as a uniform-flow process since the state of fluid at the inlet remains constant. 2 The expansion process is quasi-equilibrium. 3 Kinetic and potential energies are negligible. 3 There are no work interactions involved other than boundary work. 4 The device is insulated and thus heat transfer is negligible. Properties The properties of steam are (Tables A-4 through A-6)
kJ/kg 3168.1C350
MPa 0.5
kJ/kg 2706.3 vaporsat.
kPa 002
kJ/kg .618252201.60.6504.716.0
kPa 002
kPa 200@22
111
1
=
°==
===
=×+=+=
==
ii
i
g
fgf
hTP
hhP
hxhhxP
Pi = 0.5 MPa Ti = 350°C
(P = 200 kPa) m1 = 10 kg
H2O
Analysis (a) The cylinder contains saturated vapor at the final state at a pressure of 200 kPa, thus the final temperature in the cylinder must be T2 = Tsat @ 200 kPa = 120.2°C (b) We take the cylinder as the system, which is a control volume since mass crosses the boundary. Noting that the microscopic energies of flowing and nonflowing fluids are represented by enthalpy h and internal energy u, respectively, the mass and energy balances for this uniform-flow system can be expressed as Mass balance: 12systemoutin mmmmmm i −=→∆=−
Energy balance:
)0 (since 1122outb,
energies etc. potential, kinetic, internal,in Change
system
mass and work,heat,by nsferenergy traNet
outin
≅≅≅−+=
∆=−
pekeQumumWhm
EEE
ii
4342143421
Combining the two relations gives ( ) 112212outb,0 umumhmmW i −+−−=
or, ( ) 112212 hmhmhmm i0 −+−−=
since the boundary work and ∆U combine into ∆H for constant pressure expansion and compression processes. Solving for m2 and substituting,
( )( ) ( ) kg 9.072kg 10
kJ/kg 2706.33168.1kJ/kg .618253168.1
12
12 =
−−
=−−
= mhhhhm
i
i
Thus, mi = m2 - m1 = 29.07 - 10 = 19.07 kg
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
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5-85
5-132 A rigid tank initially contains saturated R-134a vapor. The tank is connected to a supply line, and R-134a is allowed to enter the tank. The mass of the R-134a that entered and the heat transfer are to be determined. Assumptions 1 This is an unsteady process since the conditions within the device are changing during the process, but it can be analyzed as a uniform-flow process since the state of fluid at the inlet remains constant. 2 Kinetic and potential energies are negligible. 3 There are no work interactions involved. 4 The direction of heat transfer is to the tank (will be verified). Properties The properties of refrigerant are (Tables A-11 through A-13)
Q
R-134a
R-134a 0.12 m3 1 MPa
Sat. vapor
1.2 MPa 36°C
kJ/kg .30102C36MPa 1.2
kJ/kg 116.70/kgm 0.0008934
liquid sat.MPa 1.2
kJ/kg 250.68/kgm 0.02031
vaporsat.MPa 1
C36@
MPa 1.2@2
3MPa 1.2@22
MPa 1@1
3MPa 1@11
==
°==
====
=
====
=
°fii
i
f
f
g
g
hhTP
uuP
uuP
vv
vv
Analysis We take the tank as the system, which is a control volume since mass crosses the boundary. Noting that the microscopic energies of flowing and nonflowing fluids are represented by enthalpy h and internal energy u, respectively, the mass and energy balances for this uniform-flow system can be expressed as Mass balance: 12systemoutin mmmmmm i −=→∆=−
Energy balance:
)0 (since 1122in
energies etc. potential, kinetic, internal,in Change
system
mass and work,heat,by nsferenergy traNet
outin
≅≅≅−=+
∆=−
pekeWumumhmQ
EEE
ii
4342143421
(a) The initial and the final masses in the tank are
kg 134.31/kgm 0.0008934
m 0.12
kg 91.5/kgm 0.02031
m 0.12
3
3
2
22
3
3
1
11
===
===
v
V
v
V
m
m
Then from the mass balance kg 128.4=−=−= 91.531.13412 mmmi
(c) The heat transfer during this process is determined from the energy balance to be
( )( ) ( )( ) ( )( kJ 1057=
−+−=−+−=
kJ/kg 250.68kg 5.91kJ/kg 116.70kg 134.31kJ/kg 102.30kg 128.4 1122in umumhmQ ii
)
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
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5-86
5-133 A rigid tank initially contains saturated liquid water. A valve at the bottom of the tank is opened, and half of the mass in liquid form is withdrawn from the tank. The temperature in the tank is maintained constant. The amount of heat transfer is to be determined. Assumptions 1 This is an unsteady process since the conditions within the device are changing during the process, but it can be analyzed as a uniform-flow process since the state of fluid leaving the device remains constant. 2 Kinetic and potential energies are negligible. 3 There are no work interactions involved. 4 The direction of heat transfer is to the tank (will be verified). Properties The properties of water are (Tables A-4 through A-6)
Q
H2O Sat. liquid T = 200°C V = 0.3 m3
kJ/kg 852.26liquid sat.
C200
kJ/kg 850.46/kgm 0.001157
liquid sat.C200
C200@
C200@1
3C200@11
==
=
====
°=
o
o
o
o
fee
f
f
hhT
uuT vv
Analysis We take the tank as the system, which is a control volume since mass crosses the boundary. Noting that the microscopic energies of flowing and nonflowing fluids are represented by enthalpy h and internal energy u, respectively, the mass and energy balances for this uniform-flow system can be expressed as Mass balance: 21systemoutin mmmmmm e −=→∆=−
Energy balance:
)0 (since 1122in
energies etc. potential, kinetic, internal,in Change
system
mass and work,heat,by nsferenergy traNet
outin
≅≅≅−+=
∆=−
pekeWumumhmQ
EEE
ee
4342143421
The initial and the final masses in the tank are
( ) kg 129.7kg 259.4
kg 259.4/kgm0.001157
m 0.3
21
121
2
3
3
1
11
===
===
mm
mv
V
Then from the mass balance, kg 129.77.1294.25921 =−=−= mmme
Now we determine the final internal energy,
( )( ) kJ/kg 866.467.1743009171.046.850009171.0
C200
009171.0001157.012721.0001157.0002313.0
/kgm 0.002313kg 129.7
m 0.3
222
2
22
33
22
=+=+=
=°=
=−−
=−
=
===
fgf
fg
f
uxuuxT
x
m
v
vv
Vv
Then the heat transfer during this process is determined from the energy balance by substitution to be
( )( ) ( )( ) ( )( )
kJ 2308=−+= kJ/kg 850.46kg 259.4kJ/kg 866.46kg 129.7kJ/kg 852.26kg 129.7Q
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5-87
5-134 A rigid tank initially contains saturated liquid-vapor mixture of refrigerant-134a. A valve at the bottom of the tank is opened, and liquid is withdrawn from the tank at constant pressure until no liquid remains inside. The amount of heat transfer is to be determined. Assumptions 1 This is an unsteady process since the conditions within the device are changing during the process, but it can be analyzed as a uniform-flow process since the state of fluid leaving the device remains constant. 2 Kinetic and potential energies are negligible. 3 There are no work interactions involved. 4 The direction of heat transfer is to the tank (will be verified). Properties The properties of R-134a are (Tables A-11 through A-13)
R-134a Sat. vapor
P = 800 kPa V = 0.12 m3 Q
kJ/kg 95.47liquid sat.
kPa 800
kJ/kg 246.79/kgm 0.025621
vaporsat.kPa 800
kJ/kg 79.246= kJ/kg, 79.94
/kgm 0.025621= /kg,m 0.0008458 kPa 800
kPa 800@
kPa 800@2
3kPa 800@22
331
===
====
=
=
=→=
fee
g
g
gf
gf
hhP
uuP
uu
P
vv
vv
Analysis We take the tank as the system, which is a control volume since mass crosses the boundary. Noting that the microscopic energies of flowing and nonflowing fluids are represented by enthalpy h and internal energy u, respectively, the mass and energy balances for this uniform-flow system can be expressed as Mass balance: 21systemoutin mmmmmm e −=→∆=−
Energy balance:
)0 (since 1122in
energies etc. potential, kinetic, internal,in Change
system
mass and work,heat,by nsferenergy traNet
outin
≅≅≅−+=
∆=−
pekeWumumhmQ
EEE
ee
4342143421
The initial mass, initial internal energy, and final mass in the tank are
( )( ) ( )( )
kg 684.4/kgm 0.025621
m 20.1
kJ 2.4229246.793.51394.7935.47
kg .9838.5133.4735/kgm 0.025621m 0.750.12
/kgm 0.0008458m 0.250.12
3
3
22
111
3
3
3
3
1
===
=+=+==
=+=×
+×
=+=+=
vV
v
V
v
V
m
umumumU
mmm
ggff
g
g
f
fgf
Then from the mass and energy balances, kg .3034684.498.3821 =−=−= mmme
( )( ) ( )( ) kJ 201.2=−+= kJ 4229kJ/kg 246.79kg 4.684kJ/kg 95.47kg 34.30inQ
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5-88
5-135E A rigid tank initially contains saturated liquid-vapor mixture of R-134a. A valve at the top of the tank is opened, and vapor is allowed to escape at constant pressure until all the liquid in the tank disappears. The amount of heat transfer is to be determined. Assumptions 1 This is an unsteady process since the conditions within the device are changing during the process, but it can be analyzed as a uniform-flow process since the state of fluid leaving the device remains constant. 2 Kinetic and potential energies are negligible. 3 There are no work interactions involved. Properties The properties of R-134a are (Tables A-11E through A-13E)
QR-134a
Sat. vapor P = 100 psia V = 4 ft3
Btu/lbm 113.83 vaporsat.
psia 100
Btu/lbm 104.99/lbmft 0.4776
vaporsat.psia 100
Btu/lbm 99.104 Btu/lbm, 623.37
/lbmft 0.4776 /lbm,ft 0.01332 psia 100
psia 100@
psia 100@2
3psia 100@22
331
===
====
=
==
==→=
gee
g
g
gf
gf
hhP
uuvP
uu
P
v
vv
Analysis We take the tank as the system, which is a control volume since mass crosses the boundary. Noting that the microscopic energies of flowing and nonflowing fluids are represented by enthalpy h and internal energy u, respectively, the mass and energy balances for this uniform-flow system can be expressed as Mass balance: 21systemoutin mmmmmm e −=→∆=−
Energy balance:
)0 (since 1122in
energies etc. potential, kinetic, internal,in Change
system
mass and work,heat,by nsferenergy traNet
outin
≅≅≅−=−
∆=−
pekeWumumhmQ
EEE
ee
4342143421
The initial mass, initial internal energy, and final mass in the tank are
( )( ) ( )( )
lbm 8.375/lbmft 0.4776
ft 4
Btu 2962104.996.7037.62360.04
lbm 66.746.7060.04/lbmft 0.4776
ft 0.84/lbmft 0.01332
ft 0.24
3
3
22
111
3
3
3
3
1
===
=+=+==
=+=×
+×
=+=+=
vV
v
V
v
V
m
umumumU
mmm
ggff
g
g
f
fgf
Then from the mass and energy balances, lbm 58.37375.874.6621 =−=−= mmme
( )( ) ( )( )Btu 4561=
−+=−+=
Btu 2962Btu/lbm 104.99lbm 8.375Btu/lbm 113.83lbm 58.371122in umumhmQ ee
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
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5-89
5-136 A rigid tank initially contains superheated steam. A valve at the top of the tank is opened, and vapor is allowed to escape at constant pressure until the temperature rises to 500°C. The amount of heat transfer is to be determined. Assumptions 1 This is an unsteady process since the conditions within the device are changing during the process, but it can be analyzed as a uniform-flow process by using constant average properties for the steam leaving the tank. 2 Kinetic and potential energies are negligible. 3 There are no work interactions involved. 4 The direction of heat transfer is to the tank (will be verified).
QSTEAM 2 MPa
Properties The properties of water are (Tables A-4 through A-6)
kJ/kg 3468.3,kJ/kg 3116.9/kgm 0.17568
C500MPa 2
kJ/kg 3024.2,kJ/kg 2773.2/kgm 0.12551
C300MPa 2
22
32
2
2
11
31
1
1
===
°==
===
°==
huTP
huTP
v
v
Analysis We take the tank as the system, which is a control volume since mass crosses the boundary. Noting that the microscopic energies of flowing and nonflowing fluids are represented by enthalpy h and internal energy u, respectively, the mass and energy balances for this uniform-flow system can be expressed as Mass balance: 21systemoutin mmmmmm e −=→∆=−
Energy balance:
)0 (since 1122in
energies etc. potential, kinetic, internal,in Change
system
mass and work,heat,by nsferenergy traNet
outin
≅≅≅−=−
∆=−
pekeWumumhmQ
EEE
ee
4342143421
The state and thus the enthalpy of the steam leaving the tank is changing during this process. But for simplicity, we assume constant properties for the exiting steam at the average values. Thus,
kJ/kg 3246.22
kJ/kg 3468.33024.22
21 =+
=+
≅hh
he
The initial and the final masses in the tank are
kg1.138/kgm 0.17568
m 0.2
kg 1.594/kgm 0.12551
m 0.2
3
3
2
22
3
3
1
11
===
===
v
V
v
V
m
m
Then from the mass and energy balance relations, kg 0.456138.1594.121 =−=−= mmme
( )( ) ( )( ) ( )(kJ 606.8=
−+=−+=
kJ/kg 2773.2kg 1.594kJ/kg 3116.9kg 1.138kJ/kg 3246.2kg 0.4561122 umumhmQ eein
)
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
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5-90
5-137 A pressure cooker is initially half-filled with liquid water. If the pressure cooker is not to run out of liquid water for 1 h, the highest rate of heat transfer allowed is to be determined. Assumptions 1 This is an unsteady process since the conditions within the device are changing during the process, but it can be analyzed as a uniform-flow process since the state of fluid leaving the device remains constant. 2 Kinetic and potential energies are negligible. 3 There are no work interactions involved. Properties The properties of water are (Tables A-4 through A-6)
Q&
kJ/kg 2700.2 vaporsat.
kPa 175
kJ/kg 2524.5/kgm 1.0036
vaporsat.kPa 175
kJ/kg 5.2524 kJ/kg, 82.486
/kgm 1.0037 /kg,m 0.001057 kPa 175
kPa 175@
kPa 175@2
3kPa 175@22
331
===
====
=
==
==→=
gee
g
g
gf
gf
hhP
uuP
uu
P
vv
vv
PressureCooker
4 L 175 kPa
Analysis We take the cooker as the system, which is a control volume since mass crosses the boundary. Noting that the microscopic energies of flowing and nonflowing fluids are represented by enthalpy h and internal energy u, respectively, the mass and energy balances for this uniform-flow system can be expressed as Mass balance: 21systemoutin mmmmmm e −=→∆=−
Energy balance:
)0 (since 1122in
energies etc. potential, kinetic, internal,in Change
system
mass and work,heat,by nsferenergy traNet
outin
≅≅≅−=−
∆=−
pekeWumumhmQ
EEE
ee
4342143421
The initial mass, initial internal energy, and final mass in the tank are
( )( ) ( )( )
kg 0.004/kgm 1.0037
m 0.004
kJ 926.62524.50.002486.821.893
kg 1.8950.0021.893/kgm 1.0036
m 0.002/kgm 0.001057
m 0.002
3
3
22
111
3
3
3
3
1
===
=+=+==
=+=+=+=+=
vV
v
V
v
V
m
umumumU
mmm
ggff
g
g
f
fgf
Then from the mass and energy balances, kg 1.891004.0895.121 =−=−= mmme
( )( ) ( )( ) kJ 4188kJ 926.6kJ/kg 2524.5kg 0.004kJ/kg 2700.2kg 1.8911122in
=−+=−+= umumhmQ ee
Thus,
kW 1.163===s 3600kJ 4188
tQQ∆
&
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
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5-91
5-138 An insulated rigid tank initially contains helium gas at high pressure. A valve is opened, and half of the mass of helium is allowed to escape. The final temperature and pressure in the tank are to be determined. Assumptions 1 This is an unsteady process since the conditions within the device are changing during the process, but it can be analyzed as a uniform-flow process by using constant average properties for the helium leaving the tank. 2 Kinetic and potential energies are negligible. 3 There are no work interactions involved. 4 The tank is insulated and thus heat transfer is negligible. 5 Helium is an ideal gas with constant specific heats. Properties The specific heat ratio of helium is k =1.667 (Table A-2). Analysis We take the tank as the system, which is a control volume since mass crosses the boundary. Noting that the microscopic energies of flowing and nonflowing fluids are represented by enthalpy h and internal energy u, respectively, the mass and energy balances for this uniform-flow system can be expressed as Mass balance: 21systemoutin mmmmmm e −=→∆=−
m m m me212 1 2
12 1= → = =(given) m
He 0.08 m3 2 MPa 80°C
Energy balance:
)0 (since 1122
energies etc. potential, kinetic, internal,in Change
system
mass and work,heat,by nsferenergy traNet
outin
≅≅≅≅−=−
∆=−
pekeQWumumhm
EEE
ee
4342143421
Note that the state and thus the enthalpy of helium leaving the tank is changing during this process. But for simplicity, we assume constant properties for the exiting steam at the average values.
Combining the mass and energy balances: 0 12 1
12 1 2 1 1= + −m h m u m ue
Dividing by m1/2 1221
12 22
020 TcTcTT
coruuh pe vv −++
=−+=
Dividing by cv: ( ) vcckTTTTk p / since420 1221 =−++=
Solving for T2: ( )( )
( )( ) ( ) K 225=+−
=+−
= K 3531.66721.6674
24
12 TkkT
The final pressure in the tank is
( ) kPa 637===→= kPa 2000353225
21
121
222
22
11
2
1 PTmTm
PRTmRTm
PPV
V
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5-92
5-139E An insulated rigid tank equipped with an electric heater initially contains pressurized air. A valve is opened, and air is allowed to escape at constant temperature until the pressure inside drops to 30 psia. The amount of electrical work transferred is to be determined. Assumptions 1 This is an unsteady process since the conditions within the device are changing during the process, but it can be analyzed as a uniform-flow process since the exit temperature (and enthalpy) of air remains constant. 2 Kinetic and potential energies are negligible. 3 The tank is insulated and thus heat transfer is negligible. 4 Air is an ideal gas with variable specific heats. Properties The gas constant of air is R =0.3704 psia.ft3/lbm.R (Table A-1E). The properties of air are (Table A-17E)
T hT uT u
i i= → == → == → =
580 R 138.66 Btu / lbm580 R 98.90 Btu / lbm580 R 98.90 Btu / lbm
1 1
2 2
Analysis We take the tank as the system, which is a control volume since mass crosses the boundary. Noting that the microscopic energies of flowing and nonflowing fluids are represented by enthalpy h and internal energy u, respectively, the mass and energy balances for this uniform-flow system can be expressed as Mass balance: 21systemoutin mmmmmm e −=→∆=−
Energy balance:
)0 (since 1122ine,
energies etc. potential, kinetic, internal,in Change
system
mass and work,heat,by nsferenergy traNet
outin
≅≅≅−=−
∆=−
pekeQumumhmW
EEE
ee
4342143421
The initial and the final masses of air in the tank are
( )( )( )( )
( )( )( )( )
lbm 8.38R 580R/lbmftpsia 0.3704
ft 60psia 30
lbm 20.95R 580R/lbmftpsia 0.3704
ft 60psia 75
3
3
2
22
3
3
1
11
=⋅⋅
==
=⋅⋅
==
RTP
m
RTP
m
V
V
We
AIR60 ft3
75 psia 120°F
Then from the mass and energy balances, m m me = − = − =1 2 20 95 8 38. . 12.57 lbm
( )( ) ( )( ) ( )( )Btu 500=
−+=
−+=
Btu/lbm 98.90lbm 20.95Btu/lbm 98.90lbm 8.38Btu/lbm 138.66lbm 12.571122ine, umumhmW ee
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5-93
5-140 A vertical cylinder initially contains air at room temperature. Now a valve is opened, and air is allowed to escape at constant pressure and temperature until the volume of the cylinder goes down by half. The amount air that left the cylinder and the amount of heat transfer are to be determined. Assumptions 1 This is an unsteady process since the conditions within the device are changing during the process, but it can be analyzed as a uniform-flow process since the exit temperature (and enthalpy) of air remains constant. 2 Kinetic and potential energies are negligible. 3 There are no work interactions. 4 Air is an ideal gas with constant specific heats. 5 The direction of heat transfer is to the cylinder (will be verified). Properties The gas constant of air is R = 0.287 kPa.m3/kg.K (Table A-1). Analysis (a) We take the cylinder as the system, which is a control volume since mass crosses the boundary. Noting that the microscopic energies of flowing and nonflowing fluids are represented by enthalpy h and internal energy u, respectively, the mass and energy balances for this uniform-flow system can be expressed as Mass balance: 21systemoutin mmmmmm e −=→∆=−
Energy balance:
)0 (since 1122inb,in
energies etc. potential, kinetic, internal,in Change
system
mass and work,heat,by nsferenergy traNet
outin
≅≅−=−+
∆=−
pekeumumhmWQ
EEE
ee
4342143421
AIR
300 kPa 0.2 m3 20°C
The initial and the final masses of air in the cylinder are
( )( )( )( )
( )( )( )( ) 12
13
3
2
222
3
3
1
111
kg 0.357K 293K/kgmkPa 0.287
m 0.1kPa 300
kg 0.714K 293K/kgmkPa 0.287
m 0.2kPa 300
mRTP
m
RTP
m
==⋅⋅
==
=⋅⋅
==
V
V
Then from the mass balance, m m me = − = − =1 2 0 714 0 357. . 0.357 kg
(b) This is a constant pressure process, and thus the Wb and the ∆U terms can be combined into ∆H to yield Q m h m h m he e= + −2 2 1 1
Noting that the temperature of the air remains constant during this process, we have hi = h1 = h2 = h.
Also, 121
2 mmme == . Thus,
( ) 0=−+= hmmmQ 1121
121
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5-94
5-141 A balloon is initially filled with helium gas at atmospheric conditions. The tank is connected to a supply line, and helium is allowed to enter the balloon until the pressure rises from 100 to 150 kPa. The final temperature in the balloon is to be determined. Assumptions 1 This is an unsteady process since the conditions within the device are changing during the process, but it can be analyzed as a uniform-flow process since the state of fluid at the inlet remains constant. 2 Helium is an ideal gas with constant specific heats. 3 The expansion process is quasi-equilibrium. 4 Kinetic and potential energies are negligible. 5 There are no work interactions involved other than boundary work. 6 Heat transfer is negligible. Properties The gas constant of helium is R = 2.0769 kJ/kg·K (Table A-1). The specific heats of helium are cp = 5.1926 and cv = 3.1156 kJ/kg·K (Table A-2a). Analysis We take the cylinder as the system, which is a control volume since mass crosses the boundary. Noting that the microscopic energies of flowing and nonflowing fluids are represented by enthalpy h and internal energy u, respectively, the mass and energy balances for this uniform-flow system can be expressed as Mass balance: 12systemoutin mmmmmm i −=→∆=−
Energy balance:
)0 (since 1122outb,
energies etc. potential, kinetic, internal,in Change
system
mass and work,heat,by nsferenergy traNet
outin
≅≅≅−+=
∆=−
pekeQumumWhm
EEE
ii
4342143421
He 25°C
150 kPa
( )( )( )( )
( )( )( )
( )( )kg 7041.74
KK/kgmkPa 2.0769m 97.5kPa 150
m 97.5m 65kPa 100kPa 150
kg 61.10K 295K/kgmkPa 2.0769
m 65kPa 100
223
3
2
222
331
1
22
2
1
2
1
3
3
1
111
TTRTPm
PP
PP
RTPm
=⋅⋅
==
===→=
=⋅⋅
==
V
VVVV
V
He
22°C 100 kPa
Then from the mass balance,
m m mTi = − = −2 1
2
7041 74 10 61. . kg
Noting that P varies linearly with V, the boundary work done during this process is
( ) ( ) ( ) kJ 4062.5m6597.52
kPa1501002
312
21 =−+
=−+
= VVPP
Wb
Using specific heats, the energy balance relation reduces to W 1122outb, TcmTcmTcm ipi vv +−=
Substituting,
( )( ) ( ) ( )( )(2951156.361.101156.374.70412981926.561.1074.70415.4062 222
+−
−= T
TT)
It yields T2 = 333.6 K
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5-95
5-142 An insulated piston-cylinder device with a linear spring is applying force to the piston. A valve at the bottom of the cylinder is opened, and refrigerant is allowed to escape. The amount of refrigerant that escapes and the final temperature of the refrigerant are to be determined. Assumptions 1 This is an unsteady process since the conditions within the device are changing during the process, but it can be analyzed as a uniform-flow process assuming that the state of fluid leaving the device remains constant. 2 Kinetic and potential energies are negligible. Properties The initial properties of R-134a are (Tables A-11 through A-13)
kJ/kg 11.354kJ/kg 03.325
/kgm 02423.0
C120MPa 2.1
1
1
31
1
1
===
°==
hu
TP v
Analysis We take the tank as the system, which is a control volume since mass crosses the boundary. Noting that the microscopic energies of flowing and nonflowing fluids are represented by enthalpy h and internal energy u, respectively, the mass and energy balances for this uniform-flow system can be expressed as Mass balance: 21systemoutin mmmmmm e −=→∆=−
Energy balance:
)0 (since 1122inb,
energies etc. potential, kinetic, internal,in Change
system
mass and work,heat,by nsferenergy traNet
outin
≅≅≅−=−
∆=−
pekeQumumhmW
EEE
ee
4342143421
The initial mass and the relations for the final and exiting masses are
2
3
21
2
3
2
22
3
3
1
11
m 0.502.33
m 0.5
kg 02.33/kgm 0.02423
m 0.8
v
vV
vV
−=−=
==
===
mmm
vm
m
e
Noting that the spring is linear, the boundary work can be determined from
R-134a 0.8 m3
1.2 MPa 120°C
kJ 270m0.5)-0.8(2
kPa 600)(1200)(2
321
21inb, =
+=−
+= VV
PPW
Substituting the energy balance,
kJ/kg) kg)(325.03 02.33(m 5.0m 5.002.33270 22
3
2
3−
=
−− uhe vv
(Eq. 1)
where the enthalpy of exiting fluid is assumed to be the average of initial and final enthalpies of the refrigerant in the cylinder. That is,
2kJ/kg) 11.354(
2221 hhh
he+
=+
=
Final state properties of the refrigerant (h2, u2, and v2) are all functions of final pressure (known) and temperature (unknown). The solution may be obtained by a trial-error approach by trying different final state temperatures until Eq. (1) is satisfied. Or solving the above equations simultaneously using an equation solver with built-in thermodynamic functions such as EES, we obtain T2 = 96.8°C, me = 22.47 kg, h2 = 336.20 kJ/kg, u2 = 307.77 kJ/kg, v2 = 0.04739 m3/kg, m2 = 10.55 kg
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5-96
5-143 Steam flowing in a supply line is allowed to enter into an insulated tank until a specified state is achieved in the tank. The mass of the steam that has entered and the pressure of the steam in the supply line are to be determined. Assumptions 1 This is an unsteady process since the conditions within the device are changing during the process, but it can be analyzed as a uniform-flow process since the state of fluid entering the tank remains constant. 2 Kinetic and potential energies are negligible. Properties The initial and final properties of steam in the tank are (Tables A-5 and A-6)
400°C
Sat. vapor 2 m3
1 MPa
Steam
kJ/kg 8.2582/kgm 19436.0
vap.)(sat. 1MPa 1
1
31
1
1
==
==
uxP v
kJ/kg 2.2773
/kgm 12551.0C300
MPa 2
2
32
1
2
==
°==
uTP v
Analysis We take the tank as the system, which is a control volume since mass crosses the boundary. Noting that the microscopic energies of flowing and nonflowing fluids are represented by enthalpy h and internal energy u, respectively, the mass and energy balances for this uniform-flow system can be expressed as Mass balance: 12systemoutin mmmmmm i −=→∆=−
Energy balance:
)0 (since 1122
energies etc. potential, kinetic, internal,in Change
system
mass and work,heat,by nsferenergy traNet
outin
≅≅≅−=
∆=−
pekeQumumhm
EEE
ii
4342143421
The initial and final masses and the mass that has entered are
kg 5.645=−=−=
===
===
29.1094.15
kg 94.15/kgm 0.12551
m 2
kg 29.10/kgm 0.19436
m 2
12
3
3
22
3
3
11
mmm
m
m
i
vV
vV
Substituting,
kJ/kg 3.3120kJ/kg) kg)(2582.8 29.10(kJ/kg) kg)(2773.2 94.15(kg) 645.5( =→−= ii hh
The pressure in the supply line is
(determined from EES) kPa 8931=
°==
ii
i PTh
C400kJ/kg 3.3120
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5-97
5-144 Steam at a specified state is allowed to enter a piston-cylinder device in which steam undergoes a constant pressure expansion process. The amount of mass that enters and the amount of heat transfer are to be determined. Assumptions 1 This is an unsteady process since the conditions within the device are changing during the process, but it can be analyzed as a uniform-flow process since the state of fluid entering the device remains constant. 2 Kinetic and potential energies are negligible. Properties The properties of steam at various states are (Tables A-4 through A-6)
12
33
1
11 /kgm 16667.0
kg 0.6m 0.1
PPm
=
===V
v
Q
Steam5 MPa500°C
Steam 0.6 kg 0.1 m3
800 kPa
kJ/kg 4.2004/kgm 16667.0
kPa 80013
1
1 =
=
=u
v
P
kJ/kg 9.2715
/kgm 29321.0C250kPa 800
2
32
2
2
==
°==
uTP v
kJ/kg 7.3434C500
MPa 5=
°==
ii
i hTP
Analysis (a) We take the tank as the system, which is a control volume since mass crosses the boundary. Noting that the microscopic energies of flowing and nonflowing fluids are represented by enthalpy h and internal energy u, respectively, the mass and energy balances for this uniform-flow system can be expressed as Mass balance: 12systemoutin mmmmmm i −=→∆=−
Energy balance:
)0 (since 1122outb,in
energies etc. potential, kinetic, internal,in Change
system
mass and work,heat,by nsferenergy traNet
outin
≅≅−=+−
∆=−
pekeumumhmWQ
EEE
ii
4342143421
Noting that the pressure remains constant, the boundary work is determined from
kJ 80m0.1)0.12(kPa) (800)( 312outb, =−×=−= VVPW
The final mass and the mass that has entered are
kg 0.082=−=−=
===
6.0682.0
kg 682.0/kgm 0.29321
m 0.2
12
3
3
2
22
mmm
m
i
v
V
(b) Finally, substituting into energy balance equation
kJ 447.9=−=+−
in
in
kJ/kg) kg)(2004.4 6.0(kJ/kg) kg)(2715.9 682.0(kJ/kg) kg)(3434.7 082.0(kJ 80
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.