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PROPRIETARY MATERIAL. 2008 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.
7-154
7-203 Refrigerant-134a is expanded adiabatically in a capillary tube. The rate of entropy generation is to be
determined.
Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.
Analysis The rate of entropy generation within the expansion device during this process can be determined
by applying the rate form of the entropy balance on the system. Noting that the system is adiabatic and thus
there is no heat transfer, the entropy balance for this steady-flow system can be expressed as
,
12gen
12gen
gen2211
entropy of change of Rate
(steady) 0system
generation entropy of Rate
gen
mass andheat by ansferentropy trnet of Rate
outin
)(
0
sss
ssmS
Ssmsm
SSSS
'
It may be easily shown with an energy balance that the enthalpy remains constant during the throttling
process. The properties of the refrigerant at the inlet and exit states are (Tables A-11 through A-13)
KkJ/kg 44193.0
KkJ/kg 50.123
0
C50
1
1
1
1 q shxT
KkJ/kg 48038.0)79406.0)(4223.0(14504.0
4223.038.207
92.3550.123
K kJ/kg 50.123
C12
22
2
2
12
2 q
fgf
fg
f
sxss
h
hhx
hh
T
Substituting,
kW/K0.00769 KkJ/kg )44193.0038kg/s)(0.48 2.0()( 12gen ssmS
R-134a
50qC sat. liq.
12qC Capillary tube
PROPRIETARY MATERIAL. 2008 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.
7-166
7-214 The pressure in a hot water tank rises to 2 MPa, and the tank explodes. The explosion energy of the
water is to be determined, and expressed in terms of its TNT equivalence.
Assumptions 1 The expansion process during explosion is isentropic. 2 Kinetic and potential energy
changes are negligible. 3 Heat transfer with the surroundings during explosion is negligible.
Properties The explosion energy of TNT is 3250 kJ/kg. From the steam tables (Tables A-4 through 6)
kJ/kg 811.832.20881889.040.4171889.0
0562.6
3028.14467.2
KkJ/kg 6.0562
kJ/kg 2088.2
,3028.1
,40.417kPa 100
KkJ/kg 2.4467
kJ/kg 906.12
/kgm 0.001177
liquid sat.
MPa 2
22
22
12
2
MPa 2@1
MPa 2@1
3MPa 2@1
1
fgf
fg
f
fg
fg
f
f
f
f
f
uxuu
s
ssx
s
u
s
u
ss
P
ss
uu
vP
v
Analysis We idealize the water tank as a closed system that undergoes a reversible adiabatic process with
negligible changes in kinetic and potential energies. The work done during this idealized process represents
the explosive energy of the tank, and is determined from the closed system energy balance to be
21outb,exp 12outb,energies etc. potential,
kinetic, internal,in Change
system
mass and work,heat,by nsferenergy traNet
outin
)(
uumWE
uumUW
EEE
' '
where
kg 99.67/kgm 0.001177
m 0.0803
3
1
v
Vm
Substituting, kJ 6410kJ/kg811.83906.12kg 67.99exp E which is equivalent to
TNTkg1.972 kJ/kg 3250
kJ 6410TNTm
Water
Tank
2 MPa
PROPRIETARY MATERIAL. 2008 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.
7-174
7-221 Two rigid tanks that contain water at different states are connected by a valve. The valve is opened
and steam flows from tank A to tank B until the pressure in tank A drops to a specified value. Tank B loses
heat to the surroundings. The final temperature in each tank and the entropy generated during this process
are to be determined.
Assumptions 1 Tank A is insulated, and thus heat transfer is negligible. 2 The water that remains in tank A
undergoes a reversible adiabatic process. 3 The thermal energy stored in the tanks themselves is negligible.
4 The system is stationary and thus kinetic and potential energy changes are negligible. 5 There are no
work interactions.
Analysis (a) The steam in tank A undergoes a reversible, adiabatic process, and thus s2 = s1. From the
steam tables (Tables A-4 through A-6),
Tank A:
kJ/kg 2125.9kJ/kg 1982.17895.011.561 /kgm 0.47850001073.060582.07895.0001073.07895.0
3200.5
6717.18717.5
mixture sat.
kPa 300
KkJ/kg 5.87171191.58.07765.1
kJ/kg 2163.39.19488.022.604
/kgm 0.37015001084.046242.08.0001084.0
8.0
kPa 400
,2,2
3,2,2
,2
,2
kPa 300@sat,2
12
1
1,1
1,1
31,1
1
1
q
fgAfA
fgAfA
fg
fA
A
A
fgfA
fgfA
fgfA
uxuu
x
s
ssx
TT
ss
P
sxss
uxuu
x
x
P
vvv
vvv
C133.52
Tank B:
KkJ/kg 7.7100
kJ/kg 2731.4
/kgm 1.1989
C250
kPa 200
,1
,1
3,1
1
1
q B
B
B
s
uT
Pv
The initial and the final masses in tank A are
and
kg 0.4180/kgm 0.47850
m 0.2
kg 0.5403/kgm 0.37015
m 0.2
3
3
,2,2
3
3
,1,1
A
AA
A
AA
m
m
v
V
v
V
Thus, 0.5403 - 0.4180 = 0.1223 kg of mass flows into tank B. Then,
kg 3.12231223.031223.0,1,2 BB mm The final specific volume of steam in tank B is determined from
/kgm 1.1519kg 3.1223
/kgm 1.1989kg 3 33
,2
11
,2,2
B
B
B
BB
m
m
m
vVv
We take the entire contents of both tanks as the system, which is a closed system. The energy balance for
this stationary closed system can be expressed as
A
steam
V = 0.2 m3
P = 400 kPa
x = 0.8
B
steam
m = 3 kg
T = 250qC P = 200 kPa
u600 kJ
PROPRIETARY MATERIAL. 2008 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.
7-175
BA
BA
umumumumQ
WUUUQ
EEE
11221122out
out
energies etc. potential, kinetic, internal,in Change
system
mass and work,heat,by nsferenergy traNet
outin
0)=PE=KE (since )()( '' ' '
Substituting,
^ ` ^ `
kJ/kg 2522.0
4.273131223.33.21635403.09.2125418.0600
,2
,2 B Bu u Thus,
KkJ/kg 7.2274kJ/kg 2522.0
/kgm 1.1519
,2
,2
,2
3,2 q
B BBB s Tuv C113.2 (b) The total entropy generation during this process is determined by applying the entropy balance on an
extended system that includes both tanks and their immediate surroundings so that the boundary
temperature of the extended system is the temperature of the surroundings at all times. It gives
,
BAgensurrb,
out
entropyin Change
system
generationEntropy
gen
mass andheat by ansferentropy trNet
outin
SSST
Q
SSSS
'' '
Rearranging and substituting, the total entropy generated during this process is determined to be ^ ` ^ `
kJ/K0.916 ''
K 273
kJ 6007100.732274.71223.38717.55403.08717.5418.0
surrb,
out11221122
surrb,
outgen
T
Qsmsmsmsm
T
QSSS
BABA
PROPRIETARY MATERIAL. 2008 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.
7-196
7-235 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 and the amount of entropy generated 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. 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 o' mmmmmmm i Energy balance:
)0peke (since initialout22in
energies etc. potential, kinetic, internal,in Change
system
mass and work,heat,by nsferenergy traNet
outin
## # '
EEWumhmQ
EEE
ii
Combining the two balances: ihumQ 22in where
kJ/kg 206.91
kJ/kg 290.16K 290
kg 0.0060K 290K/kgmkPa 0.287
m 0.005kPa 100
2
17-A Table2
3
3
2
22
o
u
hTT
RT
Pm
ii
V
Substituting,
Qin = (0.0060 kg)(206.91 - 290.16) kJ/kg = - 0.5 kJ o Qout = 0.5 kJ Note that the negative sign for heat transfer indicates that the assumed direction is wrong. Therefore, we
reverse the direction.
The entropy generated during this process is determined by applying the entropy balance on an
extended system that includes the bottle and its immediate surroundings so that the boundary temperature
of the extended system is the temperature of the surroundings at all times. The entropy balance for it can be
expressed as
,
220
1122tankgeninb,
out
entropyin Change
system
generationEntropy
gen
mass andheat by ansferentropy trNet
outin
smsmsmSST
Qsm
SSSS
ii ' '
Therefore, the total entropy generated during this process is
kJ/K0.0017 K 290
kJ 0.5
surr
out
outb,
out022
outb,
out22gen
T
Q
T
Qssm
T
QsmsmS iii
5 L
Evacuated
10 kPa
17qC
PROPRIETARY MATERIAL. 2008 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.
7-200
7-238 The turbocharger of an internal combustion engine consisting of a turbine driven by hot exhaust
gases and a compressor driven by the turbine is considered. The air temperature at the compressor exit and
the isentropic efficiency of the compressor are to be determined.
Assumptions 1 Steady operating conditions exist. 2
Kinetic and potential energy changes are negligible. 3
Exhaust gases have air properties and air is an ideal gas
with constant specific heats.
Properties The specific heat of exhaust gases at the
average temperature of 425C is cp = 1.075 kJ/kg.K and
properties of air at an anticipated average temperature of
100C are cp = 1.011 kJ/kg.K and k =1.397 (Table A-2).
Analysis (a) The turbine power output is determined from
kW 075.1C400)-C)(450kJ/kg. 5kg/s)(1.07 02.0(
)( 21exhT qq TTcmW p For a mechanical efficiency of 95% between the turbine and the compressor,
kW 021.1kW) 075.1)(95.0(TC WW m K Then, the air temperature at the compressor exit becomes
C126.1q qq
2
2
12airC
C70)-C)(kJ/kg. 1kg/s)(1.01 018.0(kW 021.1
)(
T
T
TTcmW p
(b) The air temperature at the compressor exit for the case of isentropic process is
C106K 379kPa 95
kPa 135K) 27370(
1)/1.397-(1.397/)1(
1
212 q
kks
P
PTT
The isentropic efficiency of the compressor is determined to be
0.642 701.126 7010612 12C TT TT sK
Compressor Turbine
Exh. gas
450qC 0.02 kg/s
Air, 70qC 95 kPa
0.018 kg/s 400qC
135 kPa
PROPRIETARY MATERIAL. 2008 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.
7-202
7-240 A cryogenic turbine in a natural gas liquefaction plant produces 350 kW of power. The efficiency of
the turbine is to be determined.
Assumptions 1 The turbine operates steadily. 2 The properties of methane is used for natural gas.
Properties The density of natural gas is given to be 423.8 kg/m3.
Analysis The maximum possible power that can be obtained from this
turbine for the given inlet and exit pressures can be determined from
kW 2.480kPa)3004000(kg/m 8.423
kg/s) 55()(
3outinmax PPmW U
Given the actual power, the efficiency of this cryogenic turbine becomes
72.9%0.729 kW 2.480
kW 350
maxW
W
K
This efficiency is also known as hydraulic efficiency since the cryogenic
turbine handles natural gas in liquid state as the hydraulic turbine handles
liquid water.
Cryogenic
turbine
LNG, 40 bar
-160qC, 55 kg/s
3 bar
7-203.pdf (p.1)7-203.pdf (p.2)7-214.pdf (p.3)7-214.pdf (p.4)7-221.pdf (p.5)7-221.pdf (p.6-7)7-235.pdf (p.8)7-235.pdf (p.9)7-238.pdf (p.10)7-238.pdf (p.11)7-240.pdf (p.12)7-240.pdf (p.13)