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ME470 Refrigeration Cycles HW Solutions
Inst: Shoeleh Di Julio
Chapter 11, Solution 18.
A refrigerator with refrigerant-134a as the working fluid is considered. The rate of heat removal from the
refrigerated space, the power input to the compressor, the isentropic efficiency of the compressor, and the
COP of the refrigerator are to be determined.
Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.
Analysis (a) From the refrigerant tables (Tables A-12 and A-13),
P1  0.14 MPa  h1  246 .36 kJ/kg
 s  0.97236 kJ/kg  K
T1  10 C
 1
T
P2  0.7 MPa 
 h2  288 .53 kJ/kg
T2  50 C

P2 s  0.7 MPa 
 h2 s  281 .16 kJ/kg
s 2 s  s1

P3  0.65 MPa 
 h3  h f
T3  24 C

@ 24C
0.65
MPa
24C
2s
·
QH
2 0.7 MPa
50C
·
Win
3
 84 .98 kJ/kg
h4  h3  84 .98 kJ/kg throttling 
0.15 MPa
4
·
QL
1
0.14 MPa
-10C
s
Then the rate of heat removal from the refrigerated space and
the power input to the compressor are determined from
Q L  m h1  h4   0.12 kg/s 246.36  84.98  kJ/kg  19.4 kW
and
W in  m h2  h1   0.12 kg/s 288.53  246.36  kJ/kg  5.06 kW
(b) The adiabatic efficiency of the compressor is determined from
C 
h2 s  h1 281 .16  246 .36

 82.5%
h2  h1
288 .53  246 .36
(c) The COP of the refrigerator is determined from its definition,
Q
19.4 kW
COP R  L 
 3.83
W in 5.06 kW
Chapter 11, Solution 31.
A heat pump that operates on the ideal vapor-compression cycle with refrigerant-134a is considered. The
power input to the heat pump is to be determined.
Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.
Analysis In an ideal vapor-compression refrigeration cycle, the compression process is isentropic, the
refrigerant enters the compressor as a saturated vapor at the evaporator pressure, and leaves the condenser
as saturated liquid at the condenser pressure. From the refrigerant tables (Tables A-12 and A-13),
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P1  320 kPa  h1  h g @320 kPa  251 .88 kJ/kg
s  s
sat. vapor
g @ 320 kPa  0.93006 kJ/kg  K
 1
P2  1.4 MPa
s 2  s1
T

 h2  282 .54 kJ/kg

P3  1.4 MPa 
 h3  h f
sat. liquid

h4  h3  127 .22 kJ/kg
@ 1.4 MPa
House
·
QH
2
3 1.4 MPa
 127 .22 kJ/kg
throttling 
·
Win
0.32 MPa
The heating load of this heat pump is determined from
1
·
QL
4
Q H  m cT2  T1 water
s
 0.12 kg/s 4.18 kJ/kg  C45  15 C  15 .05 kW
and
m R 
Q H
Q H
15.05 kJ/s


 0.09688 kg/s
qH
h2  h3 282.54  127.22  kJ/kg
Then,
W in  m R h2  h1   0.09688 kg/s 282.54  251.88  kJ/kg  2.97 kW
Chapter 11, Solution 32.
A heat pump with refrigerant-134a as the working fluid heats a house by using underground water as the
heat source. The power input to the heat pump, the rate of heat absorption from the water, and the increase
in electric power input if an electric resistance heater is used instead of a heat pump are to be determined.
Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.
Analysis (a) From the refrigerant tables
(Tables A-12 and A-13),
P1  280 kPa 
 h1  250 .83 kJ/kg
T1  0C

P2  1.0 MPa 
 h2  293 .38 kJ/kg
T2  60 C

P3  1.0 MPa 
 h3  h f
T3  30 C

@ 30C
T
60C
·
QH
·
Win
30C 1 MPa
3
 93 .58 kJ/kg
h4  h3  93 .58 kJ/kg throttling 
0.28 MPa
4
The mass flow rate of the refrigerant is
Q
Q H
60,000/3,6 00 kJ/s
m R  H 

 0.08341 kg/s
qH
h2  h3 293.38  93.58  kJ/kg
Then the power input to the compressor becomes
 h2  h1   0.08341kg/s293.38 250.83 kJ/kg  3.55 kW
Win  m
(b) The rate of hat absorption from the water is
2
House
·
QL
1
0C
Water, 8C
s
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 h1  h4   0.08341kg/s250.83 93.58 kJ/kg  13.12 kW
Q L  m
(c) The electrical power required without the heat pump is
W e  Q H  60 ,000 / 3600 kJ/s  16.67 kW
Thus,
W increase  W e  W in  16 .67  3.55  13.12 kW
Chapter 11, Solution 42.
A two-stage cascade refrigeration system is considered. Each stage operates on the ideal vapor-compression
cycle with refrigerant-134a as the working fluid. The mass flow rate of refrigerant through the lower cycle,
the rate of heat removal from the refrigerated space, the power input to the compressor, and the COP of this
cascade refrigerator are to be determined.
Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. 3
The heat exchanger is adiabatic.
Analysis (a) Each stage of the cascade refrigeration cycle is said to operate on the ideal vapor compression
refrigeration cycle. Thus the compression process is isentropic, and the refrigerant enters the compressor as
a saturated vapor at the evaporator pressure. Also, the refrigerant leaves the condenser as a saturated liquid
at the condenser pressure. The enthalpies of the refrigerant at all 8 states are determined from the
refrigerant tables (Tables A-11, A-12, and A-13) to be
h1  239 .16 kJ/kg ,
h2  260 .58 kJ/kg
h3  63 .94 kJ/kg ,
h4  63 .94 kJ/kg
h5  255 .55 kJ/kg , h6  269 .91 kJ/kg
h7  95 .47 kJ/kg,
T
0.8 MPa
h8  95 .47 kJ/kg
The mass flow rate of the refrigerant through the
lower cycle is determined from an energy balance
on the heat exchanger:
E in  E out  E system0 (steady)  0
E in  E out
6
0.4 MPa
7
3
8
2
A
5
B
4
·
QL
 m h   m h
e e
0.14 MPa
1
s
i i
m A h5  h8   m B h2  h3 
m B 
h5  h8
255 .55  95 .47
0.24 kg/s   0.1954 kg/s
m A 
h2  h3
260 .58  63 .94
(b) The rate of heat removed by a cascade cycle is the rate of heat absorption in the evaporator of the
lowest stage. The power input to a cascade cycle is the sum of the power inputs to all of the compressors:
Q L  m B h1  h4   0.1954 kg/s 239.16  63.94  kJ/kg  34.24 kW
W in  W compI, in  W compII, in  m A h6  h5   m B h2  h1 
 0.24 kg/s 269.91  255.55  kJ/kg  0.1954 kg/s 260.58  239.16  kJ/kg
 7.63 kW
(c) The COP of this refrigeration system is determined from its definition,
Q L
34.24 kW
COP R 

 4.49

7.63 kW
Wnet,in
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COP R 
Q L
34.24 kW

 4.49

7.63 kW
Wnet,in
Chapter 11, Solution 55.
An ideal-gas refrigeration cycle with air as the working fluid is considered. The maximum and minimum
temperatures in the cycle, the COP, and the rate of refrigeration are to be determined.
Assumptions 1 Steady operating conditions exist. 2 Air is an ideal gas with variable specific heats. 3
Kinetic and potential energy changes are negligible.
Analysis (a) We assume both the turbine and the compressor to be isentropic, the turbine inlet temperature
to be the temperature of the surroundings, and the compressor inlet temperature to be the temperature of the
refrigerated space. From the air table (Table A-17),
T1  250 K


T1  300 K


h1  250.05 kJ / kg
Pr 1  0.7329
h3  30019
. kJ / kg
Pr 3  1.386
T
·
QH
27C
-23C
3
Thus,
Pr2 
P2
Pr  30.7329   2.1987 
 T2  Tmax  342.2 K
P1 1
h2  342 .60 kJ/kg
Pr4 
P4
1
Pr   1.386   0.462 
 T4  Tmin  219.0 K
P3 3  3 
h4  218 .97 kJ/kg
2
4
1
·
QRefrig
s
(b) The COP of this ideal gas refrigeration cycle is determined from
COPR 
qL
wnet, in

qL
wcomp, in  w turb, out
where
q L  h1  h4  250.05  218.97  31.08 kJ / kg
wcomp, in  h2  h1  342.60  250.05  92.55 kJ / kg
wturb, out  h3  h4  30019
.  218.97  81.22 kJ / kg
Thus,
COPR 
31.08
 2.74
92.55  81.22
(c) The rate of refrigeration is determined to be
Q refrig  m qL   0.08 kg/s31.08 kJ/kg   2.49 kJ/s
Chapter 11, Solution 61.
An ideal-gas refrigeration cycle with air as the working fluid is considered. The lowest temperature that can
be obtained by this cycle, the COP, and the mass flow rate of air are to be determined.
Assumptions 1 Steady operating conditions exist. 2 Air is an ideal gas with constant specific heats. 3
Kinetic and potential energy changes are negligible.
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Properties The properties of air at room temperature are cp = 1.005 kJ/kg·K and k = 1.4 (Table A-2).
Analysis (a) The lowest temperature in the cycle occurs at the turbine exit. From the isentropic relations,
P 
T2  T1  2 
 P1 
k 1 / k
P 
T5  T4  5 
 P4 
 266 K 4 0.4 / 1.4  395 .3 K  122.3 C
k 1 / k
1
 258 K  
4
0.4 / 1.4
 173 .6 K  99.4C  Tmin
(b) From an energy balance on the regenerator,
E in  E out  E system0 (steady)  0

E in  E out
m e he 

m i hi 
 m h3  h4   m h1  h6 
T
·
QH
3
27C
-7C
-15C
2
Qrege
4
5
1
n
6
·
QRefrig
or,
s
m c p T3  T4   m c p T1  T6  
 T3  T4  T1  T6
or,
T6  T1  T3  T4   7C   27 C   15 C   49 C
Then the COP of this ideal gas refrigeration cycle is determined from
COP R 
qL
qL

wnet,in wcomp, in  w turb,out

h 6  h5
h2  h1   h4  h5 

T6  T5
T2  T1   T4  T5 

 49 C   99 .4C 
 1.12
122 .3   7 C   15   99 .4C
(c) The mass flow rate is determined from
Q refrig
Q refrig
Q refrig
12 kJ/s
m 



 0.237 kg/s
qL
h6  h5 c p T6  T5  1.005 kJ/kg  C 49   99.4 C
Chapter 11, Solution 71.
The conditions at which an absorption refrigeration system operates are specified. The maximum COP this
absorption refrigeration system can have is to be determined.
Analysis The maximum COP that this refrigeration system can have is
 T
COP R, max  1  0
 Ts
 TL

 T  T
L
 0
 
298 K  273 
  1 
  393 K  298  273   2.64

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