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11-1
Chapter 11
REFRIGERATION CYCLES
The Reversed Carnot Cycle
11-1C Because the compression process involves the compression of a liquid-vapor mixture which
requires a compressor that will handle two phases, and the expansion process involves the expansion of
high-moisture content refrigerant.
11-2 A steady-flow Carnot refrigeration cycle with refrigerant-134a 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 TH = 30 C = 303 K and TL = Tsat @ 160 kPa = -15.60 C = 257.4 K, the COP of this
Carnot refrigerator is determined from
5.641K 4.257/K 303
1
1/
1COP CR,
LH TT
(b) From the refrigerant tables (Table A-11),
kJ/kg 58.93
kJ/kg 66.266
C30@4
C30@3
f
g
hh
hh
Thus,
and
kJ/kg 147.03kJ/kg 173.08K 303
K 257.4
kJ/kg 08.17358.9366.26643
HH
LL
L
H
L
H
H
qT
Tq
T
T
q
q
hhq
30 C
4 3
2 1
160 kPa
QL
QH
T
s
(c) The net work input is determined from
kJ/kg 26.0503.14708.173net LH qqw
PROPRIETARY MATERIAL. © 2006 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.
11-2
11-3E A steady-flow Carnot refrigeration cycle with refrigerant-134a as the working fluid is considered.
The coefficient of performance, the quality at the beginning of the heat-absorption 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 TH = Tsat @ 90 psia = 72.78 F = 532.8 R and TL = Tsat @ 30 psia = 15.37 F = 475.4 R.
8.281R 475.4/R 532.8
1
1/
1COP CR,
LH TT
(b) Process 4-1 is isentropic, and thus
0.237418589.0
03793.008207.0
RBtu/lbm 0.08207
14525.005.007481.0
psia 30 @
1
1
psia 90 @ 441
fg
f
fgf
s
ssx
sxsss
(c) Remembering that on a T-s diagram the area enclosed
represents the net work, and s3 = sg @ 90 psia = 0.22006 Btu/lbm·R,
Btu/lbm7.92 RBtu/lbm 08207.022006.0)37.1578.72(43innet, ssTTw LH
Ideal and Actual Vapor-Compression Cycles
11-4C Yes; the throttling process is an internally irreversible process.
11-5C To make the ideal vapor-compression refrigeration cycle more closely approximate the actual
cycle.
11-6C No. Assuming the water is maintained at 10 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 air-conditioning devices that involve such extremely low pressures.
11-7C Allowing a temperature difference of 10 C for effective heat transfer, the condensation temperature
of the refrigerant should be 25 C. The saturation pressure corresponding to 25 C is 0.67 MPa. Therefore,
the recommended pressure would be 0.7 MPa.
11-8C 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 vapor-compression refrigeration cycle. This is because the
latter cycle involves an irreversible process for which the process path is not known.
11-9C The cycle that involves saturated liquid at 30 C will have a higher COP because, judging from the
T-s diagram, it will require a smaller work input for the same refrigeration capacity.
QH
QL
4 3
21
T
s
11-10C The minimum temperature that the refrigerant can be cooled to before throttling is the temperature
of the sink (the cooling medium) since heat is transferred from the refrigerant to the cooling medium.
PROPRIETARY MATERIAL. © 2006 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.
11-3
11-11 A commercial refrigerator with refrigerant-134a as the working fluid is considered. The quality of
the refrigerant at the evaporator inlet, the refrigeration load, the COP of the refrigerator, and the theoretical
maximum refrigeration load for the same power input to the compressor are to be determined.
Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.
Analysis (a) From refrigerant-134a tables (Tables A-11 through A-13)
0.479544
4
34
33
3
22
2
11
1
kJ/kg 23.111
kPa 60
kJ/kg 23.111
kJ/kg 23.111C42
kPa 1200
kJ/kg 16.295C65
kPa 1200
kJ/kg 03.230C34
kPa 60
xh
P
hh
hT
P
hT
P
hT
P
26 C Water
18 C
Win
QL
1.2 MPa
65 C
Expansion valve
Compressor
Evaporator
Condenser
42 C
QH
4
3 2
1 60 kPa
-34 C Using saturated liquid enthalpy at the given
temperature, for water we have (Table A-4)
kJ/kg 94.108
kJ/kg 47.75
C26 @2
C18 @1
fw
fw
hh
hh
(b) The mass flow rate of the refrigerant may be determined from an energy balance on the compressor
kg/s 0455.0
g75.47)kJ/k94kg/s)(108. (0.25kJ/kg)23.11116.295(
)()( 1232
R
R
wwwR
m
m
hhmhhm
The waste heat transferred from the refrigerant, the compressor power input, and the refrigeration load are
kW 367.8kJ/kg)23.11116kg/s)(295. 0455.0()( 32 hhmQ RH
kW 513.2kW 0.45kJ/kg)03.23016kg/s)(295. 0455.0()( in12in QhhmW R
kW 5.85513.2367.8inWQQ HL
(c) The COP of the refrigerator is determined from its
definition T
QH
QL
Win
·
2
·
·
4
3
2
1
2.33513.2
85.5COP
in
L
W
Q
(d) The reversible COP of the refrigerator for the
same temperature limits is
063.51)27330/()27318(
1
1/
1COPmax
LH TT
s
Then, the maximum refrigeration load becomes
kW 12.72kW) 513.2)(063.5(inmaxmaxL, WCOPQ
PROPRIETARY MATERIAL. © 2006 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.
11-4
11-12 An ideal vapor-compression refrigeration cycle 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
rate of heat rejection to the environment, and the COP are to be determined.
Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.
Analysis (a) 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),
throttlingkJ/kg 82.88
kJ/kg 82.88liquid sat.
MPa 7.0
C95.34kJ/kg 50.273MPa 7.0
KkJ/kg 94779.0
kJ/kg 97.236
vaporsat.
kPa 120
34
MPa 7.0 @ 33
2212
2
kPa 120 @ 1
kPa 120 @ 11
hh
hhP
Thss
P
ss
hhP
f
g
g
T
QH
QL 4s
·
Win
·
·
0.7 MPa
4
3
2
1
0.12 MPa
Then the rate of heat removal from the refrigerated
space and the power input to the compressor are
determined from s
and
kW 1.83
kW 7.41
kJ/kg 236.97273.50kg/s 0.05
kJ/kg 82.8897.236kg/s 0.05
12in
41
hhmW
hhmQL
(b) The rate of heat rejection to the environment is determined from
kW 9.2383.141.7inWQQ LH
(c) The COP of the refrigerator is determined from its definition,
4.06kW 1.83
kW 7.41COP
in
RW
QL
PROPRIETARY MATERIAL. © 2006 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.
11-5
11-13 An ideal vapor-compression refrigeration cycle 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
rate of heat rejection to the environment, and the COP are to be determined.
Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.
Analysis (a) 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),
throttlingkJ/kg 61.101
kJ/kg 61.101liquid sat.
MPa 9.0
C45.44kJ/kg 93.278MPa 9.0
KkJ/kg 94779.0
kJ/kg 97.236
vaporsat.
kPa 120
34
MPa 9.0 @ 33
2212
2
kPa 120 @ 1
kPa 120 @ 11
hh
hhP
Thss
P
ss
hhP
f
g
g
T
QH
QL 4s
·
Win
·
·
0.9 MPa
4
3
2
1
0.12 MPa
Then the rate of heat removal from the
refrigerated space and the power input to the
compressor are determined from s
and
kW2.10
kW 6.77
kJ/kg 236.97278.93kg/s 0.05
kJ/kg 61.10197.236kg/s 0.05
12in
41
hhmW
hhmQL
(b) The rate of heat rejection to the environment is determined from
kW 8.8710.277.6inWQQ LH
(c) The COP of the refrigerator is determined from its definition,
3.23kW 2.10
kW 6.77COP
in
RW
QL
PROPRIETARY MATERIAL. © 2006 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.
11-6
11-14 An ideal vapor-compression refrigeration cycle with refrigerant-134a as the working fluid is
considered. The throttling valve in the cycle is replaced by an isentropic turbine. The percentage increase
in the COP and in the rate of heat removal from the refrigerated space due to this replacement are to be
determined.
Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.
Analysis If the throttling valve in the previous problem is replaced by an isentropic turbine, we would have
s4s = s3 = sf @ 0.7 MPa = 0.33230 kJ/kg·K, and the enthalpy at the turbine exit would be
kJ/kg 58.8248.2142802.049.22
2802.085503.0
09275.033230.0
kPa 120 @44
kPa 120 @
3
4
fgsfs
fg
f
s
hxhh
s
ssx
T
QH
QL 4s
·
Win ·
·
0.7 MPa
4
3
2
1
0.12 MPa
Then, Q kW 7.72kJ/kg 82.58236.97kg/s 0.0541 sL hhm
and 23.4kW 1.83
kW 7.72COP
inR
W
QL
Then the percentage increase in and COP becomes Q
4.2%
4.2%
06.4
06.423.4
COP
COPCOPin Increase
41.7
41.772.7in Increase
R
RR
L
LL
Q
s
11-15 [Also solved by EES on enclosed CD] An ideal vapor-compression refrigeration cycle with
refrigerant-134a as the working fluid is considered. The quality of the refrigerant at the end of the
throttling process, the COP, and the power input to the compressor are to be determined.
Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.
Analysis (a) 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),
throttlingkJ/kg 47.95
kJ/kg 47.95liquid sat.
MPa 8.0
kJ/kg 37.275MPa 8.0
KkJ/kg 94456.0
kJ/kg 16.239
vapor sat.
kPa 140
34
MPa 8.0 @ 33
212
2
kPa 140 @ 1
kPa 140 @ 11
hh
hhP
hss
P
ss
hhP
f
g
g
T
QH
QL
·
Win
·
·
0.8 MPa
4
3
2
1
0.14 MPaThe quality of the refrigerant at the end of the throttling process is
0.32208.212
08.2747.95
kPa 140 @
4
4fg
f
h
hhx
s
(b) The COP of the refrigerator is determined from its definition,
3.9716.23937.275
47.9516.239COP
12
41
inR
hh
hh
w
qL
(c) The power input to the compressor is determined from
kW 1.2697.3
kW)60/300(
COPRin
LQW
PROPRIETARY MATERIAL. © 2006 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.
11-7
11-16 EES Problem 11-15 is reconsidered. The effect of evaporator pressure on the COP and the power
input is to be investigated.
Analysis The problem is solved using EES, and the solution is given below.
"Input Data" {P[1]=140 [kPa]} {P[2] = 800 [kPa] Fluid$='R134a' Eta_c=1.0 "Compressor isentropic efficiency" Q_dot_in=300/60 "[kJ/s]"} "Compressor" x[1]=1 "assume inlet to be saturated vapor" h[1]=enthalpy(Fluid$,P=P[1],x=x[1]) T[1]=temperature(Fluid$,h=h[1],P=P[1]) "properties for state 1" s[1]=entropy(Fluid$,T=T[1],x=x[1]) h2s=enthalpy(Fluid$,P=P[2],s=s[1]) "Identifies state 2s as isentropic" h[1]+Wcs=h2s "energy balance on isentropic compressor" Wc=Wcs/Eta_c"definition of compressor isentropic efficiency" h[1]+Wc=h[2] "energy balance on real compressor-assumed adiabatic" s[2]=entropy(Fluid$,h=h[2],P=P[2]) "properties for state 2" T[2]=temperature(Fluid$,h=h[2],P=P[2]) W_dot_c=m_dot*Wc "Condenser" P[3]=P[2] "neglect pressure drops across condenser" T[3]=temperature(Fluid$,h=h[3],P=P[3]) "properties for state 3" h[3]=enthalpy(Fluid$,P=P[3],x=0) "properties for state 3" s[3]=entropy(Fluid$,T=T[3],x=0) h[2]=q_out+h[3] "energy balance on condenser" Q_dot_out=m_dot*q_out "Valve" h[4]=h[3] "energy balance on throttle - isenthalpic" x[4]=quality(Fluid$,h=h[4],P=P[4]) "properties for state 4" s[4]=entropy(Fluid$,h=h[4],P=P[4]) T[4]=temperature(Fluid$,h=h[4],P=P[4]) "Evaporator" P[4]=P[1] "neglect pressure drop across evaporator" q_in + h[4]=h[1] "energy balance on evaporator" Q_dot_in=m_dot*q_in COP=Q_dot_in/W_dot_c "definition of COP" COP_plot = COP W_dot_in = W_dot_c
P1 [kPa] COPplot Win [kW] 100 3.216 1.554
175 4.656 1.074
250 6.315 0.7918
325 8.388 0.5961
400 11.15 0.4483
PROPRIETARY MATERIAL. © 2006 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.
11-8
0,0 0,2 0,4 0,6 0,8 1,0 1,2-50
-25
0
25
50
75
100
125
s [kJ/kg-K]
T[C]
800 kPa
140 kPa
R134a
T-s diagram for = 1.0
0 50 100 150 200 250 300
101
102
103
104
h [kJ/kg]
P [
kP
a]
31.33 C
-18.8 C
R134a
P-h diagram for = 1.0
0,0 0,2 0,4 0,6 0,8 1,0 1,2-50
-25
0
25
50
75
100
125
s [kJ/kg-K]
T[C]
800 kPa
140 kPa
R134a
T-s diagram for = 0.6
PROPRIETARY MATERIAL. © 2006 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.
11-9
0.60 0.65 0.70 0.75 0.80 0.85 0.90 0.95 1.00
2.2
2.4
2.6
2.8
3.0
3.2
3.4
3.6
3.8
4.0
Compressor efficiency
CO
P
COP vs Compressor Efficiency for R134a
100 150 200 250 300 350 400
3
4
5
6
7
8
9
10
11
12
P[1] [kPa]
CO
Pp
lot
100 150 200 250 300 350 400
0.4
0.6
0.8
1
1.2
1.4
1.6
P[1] [kPa]
Win
PROPRIETARY MATERIAL. © 2006 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.
11-10
11-17 A nonideal vapor-compression refrigeration cycle with refrigerant-134a as the working fluid is
considered. The quality of the refrigerant at the end of the throttling process, the COP, the power input to
the compressor, and the irreversibility rate associated with the compression process are to be determined.
Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.
Analysis (a) 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),
throttlingkJ/kg 47.95
kJ/kg 47.95liquid sat.
MPa 8.0
kJ/kg 76.281
85.0/16.23937.27516.239
/
kJ/kg 37.275MPa 8.0
KkJ/kg 94456.0
kJ/kg 16.239
vapor.sat
kPa 140
34
MPa 8.0 @ 33
121212
12
212
2
kPa 140 @1
kPa 140 @ 11
hh
hhP
hhhhhh
hh
hss
P
ss
hhP
f
Css
C
ss
g
g
T2
QH · 2s
Win
·
3 0.8 MPa
0.14 MPa1
QL ·4
s
The quality of the refrigerant at the end of the throttling process is
0.32208.212
08.2747.95
kPa 140 @
4
4fg
f
h
hhx
(b) The COP of the refrigerator is determined from its definition,
3.3716.23976.281
47.9516.239COP
12
41
inR
hh
hh
w
qL
(c) The power input to the compressor is determined from
kW 1.4837.3
kW 5
COPin
R
LQW
The exergy destruction associated with the compression process is determined from
120
0
0
surr120gen0destroyed ssmT
T
qssmTSTX
where
KkJ/kg 96483.0kJ/kg 76.281
MPa 8.0
kg/s 0348.0kJ/kg 95.47239.16
kJ/s 5
22
2
41
sh
P
hh
Q
q
Qm L
L
L
Thus,
kW 0.210KkJ/kg 0.944560.96483kg/s 0.0348K 298destroyedX
PROPRIETARY MATERIAL. © 2006 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.
11-11
11-18 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),
throttlingkJ/kg 98.84
kJ/kg 98.84C24
MPa 65.0
kJ/kg 16.281MPa 7.0
kJ/kg 53.288C50
MPa 7.0
KkJ/kg 97236.0
kJ/kg 36.246
C10
MPa 14.0
34
C24 @ 33
3
212
2
22
2
1
1
1
1
hh
hhT
P
hss
P
hT
P
s
h
T
P
f
ss
s
T
QH
QL
0.7 MPa
50 C
0.14 MPa
-10 C
Win
·
2
·
·
s
0.65 MPa
24 C
4
3
2s
1 0.15 MPa
Then the rate of heat removal from the refrigerated space and
the power input to the compressor are determined from
and
kW 5.06
kW19.4
kJ/kg 246.36288.53kg/s 0.12
kJ/kg 84.98246.36kg/s 0.12
12in
41
hhmW
hhmQL
(b) The adiabatic efficiency of the compressor is determined from
82.5%36.24653.288
36.24616.281
12
12
hh
hh sC
(c) The COP of the refrigerator is determined from its definition,
3.83kW 5.06
kW 19.4COP
in
RW
QL
PROPRIETARY MATERIAL. © 2006 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.
11-12
11-19E An ice-making machine operates on the ideal vapor-compression refrigeration cycle, using
refrigerant-134a as the working fluid. The power input to the ice machine 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-12E and A-13E),
throttlingBtu/lbm 39.33
Btu/lbm 39.33liquid sat.
psia 80
Btu/lbm 00.115psia 80
RBtu/lbm 22567.0
Btu/lbm 73.102
vapor sat.
psia 20
34
psia 08 @ 33
212
2
psia 20 @ 1
psia 20 @ 11
hh
hhP
hss
P
ss
hhP
f
g
g
T
QH
QL
·
Win
·
·
80 psia
4
3
2
120 psia
sThe cooling load of this refrigerator is
Btu/s 0.7042Btu/lbm 169lbm/s 15/3600iceice hmQL
Then the mass flow rate of the refrigerant and the power input become
and
hp 0.176Btu/s 0.7068
hp 1Btu/lbm 102.73115.00lbm/s 0.01016
lbm/s 0.01016Btu/lbm 33.39102.73
Btu/s 0.7042
12in
41
hhmW
hh
Qm
R
LR
PROPRIETARY MATERIAL. © 2006 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.
11-13
11-20 A refrigerator with refrigerant-134a as the working fluid is considered. The power input to the
compressor, the rate of heat removal from the refrigerated space, and the pressure drop and the rate of heat
gain in the line between the evaporator and the compressor 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),
kJ/kg 33.239
MPa 14165.0
vapor sat.
C5.18
throttlingkJ/kg 58.93
kJ/kg 58.93C30
MPa 95.0
kJ/kg 20.289MPa 0.1
/kgm 14605.0
KkJ/kg 97236.0
kJ/kg 36.246
C10
kPa 140
5
55
34
C 30 @ 33
3
212
2
31
1
1
1
1
h
PT
hh
hhT
P
hss
P
s
h
T
P
f
ss
v
T
QH
QL -18.5 C
1 MPa
0.14 MPa
-10 C
Win
·
2
·
·
s
0.95 MPa
30 C
4
3
2s
1 0.15 MPa
Then the mass flow rate of the refrigerant and the power input becomes
kW1.88 78.0/kJ/kg 246.36289.20kg/s 0.03423/
kg/s 0.03423/kgm 0.14605
/sm 0.3/60
12in
3
3
1
1
Cs hhmW
mv
V
(b) The rate of heat removal from the refrigerated space is
kW 4.99kJ/kg 93.58239.33kg/s 0.0342345 hhmQL
(c) The pressure drop and the heat gain in the line between the evaporator and the compressor are
and
kW0.241
1.65
kJ/kg 239.33246.36kg/s 0.03423
14065.141
51gain
15
hhmQ
PPP
PROPRIETARY MATERIAL. © 2006 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.
11-14
11-21 EES Problem 11-20 is reconsidered. The effects of the compressor isentropic efficiency and the
compressor inlet volume flow rate on the power input and the rate of refrigeration are to be investigated.
Analysis The problem is solved using EES, and the solution is given below.
"Input Data" "T[5]=-18.5 [C] P[1]=140 [kPa] T[1] = -10 [C]} V_dot[1]=0.1 [m^3/min] P[2] = 1000 [kPa] P[3]=950 [kPa] T[3] = 30 [C] Eta_c=0.78 Fluid$='R134a'" "Compressor" h[1]=enthalpy(Fluid$,P=P[1],T=T[1]) "properties for state 1" s[1]=entropy(Fluid$,P=P[1],T=T[1]) v[1]=volume(Fluid$,P=P[1],T=T[1])"[m^3/kg]" m_dot=V_dot[1]/v[1]*convert(m^3/min,m^3/s)"[kg/s]" h2s=enthalpy(Fluid$,P=P[2],s=s[1]) "Identifies state 2s as isentropic" h[1]+Wcs=h2s "energy balance on isentropic compressor" Wc=Wcs/Eta_c"definition of compressor isentropic efficiency" h[1]+Wc=h[2] "energy balance on real compressor-assumed adiabatic" s[2]=entropy(Fluid$,h=h[2],P=P[2]) "properties for state 2" T[2]=temperature(Fluid$,h=h[2],P=P[2]) W_dot_c=m_dot*Wc "Condenser" h[3]=enthalpy(Fluid$,P=P[3],T=T[3]) "properties for state 3" s[3]=entropy(Fluid$,P=P[3],T=T[3]) h[2]=q_out+h[3] "energy balance on condenser" Q_dot_out=m_dot*q_out "Throttle Valve" h[4]=h[3] "energy balance on throttle - isenthalpic" x[4]=quality(Fluid$,h=h[4],P=P[4]) "properties for state 4" s[4]=entropy(Fluid$,h=h[4],P=P[4]) T[4]=temperature(Fluid$,h=h[4],P=P[4]) "Evaporator" P[4]=pressure(Fluid$,T=T[5],x=0)"pressure=Psat at evaporator exit temp." P[5] = P[4] h[5]=enthalpy(Fluid$,T=T[5],x=1) "properties for state 5" q_in + h[4]=h[5] "energy balance on evaporator" Q_dot_in=m_dot*q_in COP=Q_dot_in/W_dot_c "definition of COP" COP_plot = COP W_dot_in = W_dot_c Q_dot_line5to1=m_dot*(h[1]-h[5])"[kW]"
PROPRIETARY MATERIAL. © 2006 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.
11-15
COPplot Win [kW]
Qin [kW]
c [kW]
2.041 0.8149 1.663 0.6
2.381 0.6985 1.663 0.7
2.721 0.6112 1.663 0.8
3.062 0.5433 1.663 0.9
3.402 0.4889 1.663 1
0.6 0.65 0.7 0.75 0.8 0.85 0.9 0.95 1
0
1
2
3
4
5
6
7
8
9
c
Win
V1 m3/min
1.01.0 0.50.5 0.10.1
0.6 0.65 0.7 0.75 0.8 0.85 0.9 0.95 1
0
0.5
1
1.5
2
2.5
3
3.5
4
c
CO
Pp
lot
V1 m3/min
1.01.0 0.50.5 0.10.1
0.6 0.65 0.7 0.75 0.8 0.85 0.9 0.95 1
0
3.6
7.2
10.8
14.4
18
c
Qin
[k
W] 1.0 1.0 0.50.5 0.10.1
V1 m3/min
PROPRIETARY MATERIAL. © 2006 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.
11-16
11-22 A refrigerator uses refrigerant-134a as the working fluid and operates on the ideal vapor-
compression refrigeration cycle. The mass flow rate of the refrigerant, the condenser pressure, 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) (b) From the refrigerant-134a tables (Tables A-11 through A-13)
4
32
1
Win .
120 kPa x=0.3
Expansion valve
Compressor
Evaporator
Condenser
60 C
QH
.
.QL
kJ/kg 97.236 vap.)(sat. 1
kPa 120
kJ/kg 87.298C60
kPa 8.671
liq.) (sat. 0
kJ/kg 83.86
kJ/kg 83.8630.0
kPa 120
11
41
22
2
32
33
3
43
44
4
hx
PP
hT
P
PP
Px
h
hh
hx
P
kPa 671.8
The mass flow rate of the refrigerant is
determined from T
QH
QL 4s
·
Win ·
·4
3
2
1
0.12 MPa
kg/s 0.00727kg236.97)kJ/(298.87
kW 45.0
12
in
hh
Wm
(c) The refrigeration load and the COP are
Q kW 091.1kJ/kg)83.8697kg/s)(236. 0727.0()( 41 hhmL
2.43kW 0.45
kW 091.1COP
in
L
W
Q
s
PROPRIETARY MATERIAL. © 2006 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.