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and pressure. The entropy changes of the helium and the surroundings are to be determined, and it is to be assessed if the
process is reversible, irreversible, or impossible.
Assumptions 1 Helium is an ideal gas with constant specific heats. 2 The cylinder is stationary and thus the kinetic and
potential energy changes are negligible. 3 The thermal energy stored in the cylinder itself is negligible. 4 The compression
or expansion process is quasi-equilibrium.
Analysis (a) The mass of helium is
ft 15psia 25
3
11
P
V
8-125
8-167 Refrigerant-134a is vaporized by air in the evaporator of an air-conditioner. For specified flow rates, the exit
temperature of air and the rate of entropy generation are to be determined for the cases of an insulated and uninsulated
evaporator.
8-126
Substituting,
kW/K 0.00196
KkJ/min 1175.0
K)kJ/kg 0.1551kg/min)( (6.968KkJ/kg 0.3492-0.9479kg/min 2
gen
S
(b) When there is a heat gain from the surroundings at a rate of 30 kJ/min, the steady-flow energy equation reduces to
34air12in TTcmhhmQ pR
Solving for T4,
p
R
cm
hhmQ
TT
air
12in
34
Substituting,
K 4.261
K)kJ/kg 005kg/min)(1. 968.6(
kJ/kg )83.866.99kg/min)(23 (2kJ/min) (30
+C27
4
C11.6T
The entropy generation in this case is determined by applying the entropy balance on an extended system that includes the
evaporator and its immediate surroundings so that the boundary temperature of the extended system is the temperature of
the surrounding air at all times. The entropy balance for the extended system can be expressed as
0
0
gen4air23air1
surr
in
gen44223311
outb,
in
entrop y of
change of Rate
(steady) 0
system
generation
entrop y of Rate
gen
mass andheat by
ansferentrop y trnet of Rate
outin
Ssmsmsmsm
T
Q
Ssmsmsmsm
T
Q
SSSS
RR
or
0
in
34air12gen T
Q
ssmssmS R
where
KkJ/kg 1384.0
K 300
K 261.4
lnK)kJ/kg 005.1(lnln
3
4
3
4
34
0
P
P
R
T
T
css p
8-129
8-170 Helium gas is compressed in an adiabatic closed system with an isentropic efficiency of 80%. The work input and the
final temperature are to be determined.
Assumptions 1 Kinetic and potential energy changes are negligible. 3 The device is adiabatic and thus heat transfer is
negligible. 4 Helium is an ideal gas.
Properties The properties of helium are cv = 3.1156 kJ/kg·K and k = 1.667 (Table A-2b).
)()( 1212in
energies etc. potential,
mass and work,heat,by
TTmcuumUW
v
The isentropic exit temperature is
kPa 900
70.667/1.66
/)1(
2
kk
P
Helium
100 kPa
27°C
8-131
8-172 A gas is adiabatically expanded in a piston-cylinder device with a specified isentropic efficiency. It is to be
determined if air or neon will produce more work.
Assumptions 1 Kinetic and potential energy changes are negligible. 3 The device is adiabatic and thus heat transfer is
negligible. 4 Air and helium are ideal gases.
Properties The properties of air at room temperature are cv = 0.718 kJ/kg·K and k = 1.4. The properties of neon at room
)(
)()(
21out
1212out
energies etc. potential,
kinetic, internal,in Change
system
mass and work,heat,by
nsferenergy traNet
outin
TTmcW
TTmcuumUW
EEE
v
v
kJ/kg 239 203.4)KK)(573kJ/kg 718.0)(90.0()( 21in sv TTcw
Repeating the same calculations for neon,
K 4.134
kPa 3000
kPa 80
)K 273300(
70.667/1.66
/)1(
1
2
12
kk
s
sP
P
TT
kJ/kg 217 134.4)KK)(573kJ/kg 6179.0)(80.0()( 21in sv TTcw
Air will produce more work.
Air
3 MPa
8-134
8-175 Air is compressed in a compressor that is intentionally cooled. The work input, the isothermal efficiency, and the
entropy generation are to be determined.
A-2).
kW 129.0
(b) The power input for a reversible-isothermal process is given by
kPa 1200
2
P
Compressor
