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Example 71
In a vaporcompression refrigeration cycle, R12
evaporates at 40°F and condenses at 110°F. The
compressor has an efficiency of 82%. Air flows over the
evaporator coils at 4 lbm/s. The air enters at 70°F and
exits at 55°F.
a) Find the power input to the compressor in kW
b) Calculate
COP
for the cycle
Solution:
The first law for the evaporator is
()
14
56
Ra
mh
h
h
−=
−
±±
1
Btu
enthalpy(R12,
40,
1)
81.4
lbm
hT
x
==
=
=
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Btu
= enthalpy(R12,
110,
0)
33.5
lbm
hh
T
x
==
=
=
5
Btu
enthalpy(air,
70)
127
lbm
hT
=
6
Btu
enthalpy(air,
55)
123
lbm
=
()
56
14
lbm
Btu
4
127 123
sl
b
m
Btu
81.433.5
lbm
a
R
mh h
m
⎛⎞
−
⎜⎟
−
⎝⎠
−
±
±
lbm
0.301
s
R
m
=
±
To find compressor power,
h
2
will be needed. From the
second law
12
s
s
s
=
1
Btu
entropy(R12,
40,
1)
0.166
lbm R
sT
x
=
=
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 Spring '08
 Kaminski

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