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111
Chapter 11
REFRIGERATION CYCLES
The Reversed Carnot Cycle
111C
Because the compression process involves the compression of a liquidvapor mixture which
requires a compressor that will handle two phases, and the expansion process involves the expansion of
highmoisture content refrigerant.
112
A steadyflow Carnot refrigeration cycle with refrigerant134a as the working fluid is considered. The
coefficient of performance, the amount of heat absorbed from the refrigerated space, and the net work input
are to be determined.
Assumptions
1
Steady operating conditions exist.
2
Kinetic and potential energy changes are negligible.
Analysis
(
a
)
Noting that
T
H
= 30
q
C = 303 K and
T
L
=
T
sat @ 160 kPa
= 15.60
q
C = 257.4 K, the COP of this
Carnot refrigerator is determined from
+,
+
,
5.64
0
0
1
K
4
.
257
/
K
303
1
1
/
1
COP
C
R,
L
H
T
T
(
b
)
From the refrigerant tables (Table A11),
kJ/kg
58
.
93
kJ/kg
66
.
266
C
30
@
4
C
30
@
3
q
q
f
g
h
h
h
h
Thus,
and
kJ/kg
147.03
¸
¸
¹
·
¨
¨
©
§
¡o
¡
0
0
kJ/kg
173.08
K
303
K
257.4
kJ/kg
08
.
173
58
.
93
66
.
266
4
3
H
H
L
L
L
H
L
H
H
q
T
T
q
T
T
q
q
h
h
q
30
q
C
43
2
1
160 kPa
Q
L
Q
H
T
s
(
c
)
The net work input is determined from
kJ/kg
26.05
0
0
03
.
147
08
.
173
net
L
H
q
q
w
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View Full Document 112
113E
A steadyflow Carnot refrigeration cycle with refrigerant134a as the working fluid is considered.
The coefficient of performance, the quality at the beginning of the heatabsorption process, and the net
work input are to be determined.
Assumptions
1
Steady operating conditions exist.
2
Kinetic and potential energy changes are negligible.
Analysis
(
a
) Noting that
T
H
=
T
sat @ 90 psia
= 72.78
q
F = 532.8 R and
T
L
=
T
sat @ 30 psia
= 15.37
q
F = 475.4 R.
+,
8.28
0
0
1
R
475.4
/
R
532.8
1
1
/
1
COP
C
R,
L
H
T
T
(
b
)
Process 41 is isentropic, and thus
+ ,
+
,
0.2374
0
¸
¸
¹
·
¨
¨
©
§
0
¢
.
.
18589
.
0
03793
.
0
08207
.
0
R
Btu/lbm
0.08207
14525
.
0
05
.
0
07481
.
0
psia
30
@
1
1
psia
90
@
4
4
1
fg
f
fg
f
s
s
s
x
s
x
s
s
s
(
c
)
Remembering that on a
Ts
diagram the area enclosed
represents the net work, and
s
3
=
s
g @ 90 psia
= 0.22006 Btu/lbm·R,
+
,
+
,
Btu/lbm
7.92
R
Btu/lbm
08207
.
0
22006
.
0
)
37
.
15
78
.
72
(
4
3
in
net,
¢
0
0
0
0
s
s
T
T
w
L
H
Ideal and Actual VaporCompression Cycles
114C
Yes; the throttling process is an internally irreversible process.
115C
To make the ideal vaporcompression refrigeration cycle more closely approximate the actual
cycle.
116C
No.
Assuming the water is maintained at 10
q
C in the evaporator, the evaporator pressure will be
the saturation pressure corresponding to this pressure, which is 1.2 kPa.
It is not practical to design
refrigeration or airconditioning devices that involve such extremely low pressures.
117C
Allowing a temperature difference of 10
q
C for effective heat transfer, the condensation temperature
of the refrigerant should be 25
q
C. The saturation pressure corresponding to 25
q
C is 0.67 MPa.
Therefore,
the recommended pressure would be 0.7 MPa.
118C
The area enclosed by the cyclic curve on a
T

s
diagram represents the net work input for the
reversed Carnot cycle, but not so for the ideal vaporcompression refrigeration cycle.
This is because the
latter cycle involves an irreversible process for which the process path is not known.
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This note was uploaded on 09/21/2009 for the course ME 311 taught by Professor Ferrenberg during the Fall '09 term at Nevada.
 Fall '09
 ferrenberg

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