chapter 11 refrigeration cycles · 2019. 7. 12. · 11-1 chapter 11 refrigeration cycles the...
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PROPRIETARY MATERIAL. © 2008 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.
11-1
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
( ) ( ) 5.64=−
=−
=1K4.257/K303
11/
1COP CR,LH TT
(b) From the refrigerant tables (Table A-11),
kJ/kg58.93kJ/kg66.266
C30@4
C30@3====
°
°
f
ghhhh
Thus,
and
( ) kJ/kg 147.03=⎟⎟⎠
⎞⎜⎜⎝
⎛==⎯→⎯=
=−=−=
kJ/kg173.08K303K257.4
kJ/kg08.17358.9366.26643
HH
LL
L
H
L
H
H
qTT
qTT
hhq
(c) The net work input is determined from
kJ/kg 26.05=−=−= 03.14708.173net LH qqw
s
T
QH
QL
160 kPa 1 2
3 430°C
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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.
( ) ( ) 8.28=−
=−
=1R475.4/R532.8
11/
1COP CR,LH TT
(b) Process 4-1 is isentropic, and thus
( ) ( )( )
0.2374=−
=⎟⎟⎠
⎞⎜⎜⎝
⎛ −=
⋅=
+=+==
18589.003793.008207.0
RBtu/lbm0.08207
14525.005.007481.0
psia30@
11
psia90@441
fg
f
fgf
sss
x
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/lbm08207.022006.0)37.1578.72(43innet, =⋅−−=−−= ssTTw LH
s
T
QH
QL1 2
3 4
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11-3
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.
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.
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11-4
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)
0.4795=⎭⎬⎫
==
==
=⎭⎬⎫
°==
=⎭⎬⎫
°==
=⎭⎬⎫
°−==
44
4
34
33
3
22
2
11
1
kJ/kg23.111kPa60
kJ/kg23.111
kJ/kg23.111C42
kPa1200
kJ/kg16.295C65
kPa1200
kJ/kg03.230C34
kPa60
xhP
hh
hTP
hTP
hTP
Using saturated liquid enthalpy at the given temperature, for water we have (Table A-4)
kJ/kg94.108
kJ/kg47.75
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/s0455.0
g75.47)kJ/k94kg/s)(108.(0.25kJ/kg)23.11116.295()()( 1232
=⎯→⎯
−=−−=−
R
R
wwwR
m
mhhmhhm
&
&
&&
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 &&
kW513.2kW0.45kJ/kg)03.23016kg/s)(295.0455.0()( in12in =−−=−−= QhhmW R&&&
kW5.85=−=−= 513.2367.8inWQQ HL&&&
2.33
(c) The COP of the refrigerator is determined from its definition
===513.285.5COP
in
L
WQ&
&
(d) The reversible COP of the refrigerator for the same temperature limits is
063.51)27330/()27318(
11/
1COPmax =−+−+
=−
=LH TT
Then, the maximum refrigeration load becomes
kW12.72=== kW)513.2)(063.5(inmaxmaxL, WCOPQ &&
60 kPa -34°C
1
23
4
QH42°C
Condenser
Evaporator
Compressor
Expansion valve
1.2 MPa 65°C
QL
Win
Water 18°C
26°C
QH
QL 1
2
3
4
s
T
·
·
2
Win ·
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11-5
11-12 An ideal vapor-compression refrigeration cycle with refrigerant-134a as the working fluid is considered. The COP and the power requirement are 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-11, A-12, and A-13),
)throttling( kJ/kg32.107
kJ/kg32.107 liquidsat.
MPa1
kJ/kg29.275 MPa1
KkJ/kg92927.0kJ/kg77.252
vapor sat.
C4
34
MPa1 @33
212
2
C4@1
C4@11
=≅
==⎭⎬⎫=
=⎭⎬⎫
==
⋅====
⎭⎬⎫°=
°
°
hh
hhP
hss
P
sshhT
f
g
g
The mass flow rate of the refrigerant is
kg/s750.2kJ/kg107.32)(252.77
kJ/s400)(41
41 =−
=−
=⎯→⎯−=hh
QmhhmQ L
L
&&&&
The power requirement is
kW61.93=−=−= kJ/kg252.77)29kg/s)(275.750.2()( 12in hhmW &&
6.46
The COP of the refrigerator is determined from its definition,
===kW61.93
kW400COPin
R WQL&
&
QH
QL
4°C 1
2 3
4
1 MPa
s
T
·
Win ·
·
4s
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11-6
11-13 An ideal vapor-compression refrigeration cycle with refrigerant-134a as the working fluid is considered. The mass flow rate of the refrigerant and the power requirement are 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-11, A-12, and A-13),
)throttling( kJ/kg47.95
kJ/kg47.95 liquidsat.kPa800
kJ/kg90.269 kPa800
KkJ/kg92691.0kJ/kg55.255
vapor sat.
kPa400
34
kPa800@33
212
2
kPa400@1
kPa400@11
=≅
==⎭⎬⎫=
=⎭⎬⎫
==
⋅====
⎭⎬⎫=
hh
hhP
hss
P
sshhP
f
g
g
The mass flow rate of the refrigerant is determined from
kg/s0.06247=−
=−
=⎯→⎯−=kJ/kg)47.95(255.55
kJ/s10)(41
41 hhQ
mhhmQ LL
&&&&
The power requirement is
kW0.8964=−=−= kJ/kg255.55)90kg/s)(269.06247.0()( 12in hhmW &&
QH
QL
0.4 MPa 1
2 3
4
0.8 MPa
s
T
·
Win·
·
4s
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11-7
11-14E An ideal vapor-compression refrigeration cycle with refrigerant-134a as the working fluid is considered. The mass flow rate of the refrigerant and the power requirement are 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-11E, A-12E, and A-13E),
)throttling( Btu/lbm869.37
Btu/lbm869.37 liquidsat.psia100
Btu/lbm57.117 psia100
RBtu/lbm22660.0Btu/lbm61.101
vapor sat.
F10
34
psia100@33
212
2
F10@1
F10@11
=≅
==⎭⎬⎫=
=⎭⎬⎫
==
⋅====
⎭⎬⎫°−=
°−
°−
hh
hhP
hss
P
sshhT
f
g
g
The mass flow rate of the refrigerant is
lbm/h376.5=−
=−
=⎯→⎯−=Btu/lbm)869.37(101.61
Btu/h000,24)(41
41 hhQ
mhhmQ LL
&&&&
The power requirement is
kW1.761=⎟⎠⎞
⎜⎝⎛−=−=
Btu/h3412.14kW1Btu/lbm)61.101.57lbm/h)(1175.376()( 12in hhmW &&
QH
QL
-10°F 1
2 3
4
100 psia
s
T
·
Win ·
·
4s
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11-8
11-15E Problem 11-14E is to be repeated if ammonia is used as the refrigerant.
Analysis The problem is solved using EES, and the solution is given below.
"Given" x[1]=1 T[1]=-10 [F] x[3]=0 P[3]=100 [psia] Q_dot_L=24000 [Btu/h]
"Analysis" Fluid$='ammonia' "compressor" h[1]=enthalpy(Fluid$, T=T[1], x=x[1]) s[1]=entropy(Fluid$, T=T[1], x=x[1]) s[2]=s[1] P[2]=P[3] h[2]=enthalpy(Fluid$, P=P[2], s=s[2]) "expansion valve" h[3]=enthalpy(Fluid$, P=P[3], x=x[3]) h[4]=h[3] "cycle" m_dot_R=Q_dot_L/(h[1]-h[4]) W_dot_in=m_dot_R*(h[2]-h[1])*Convert(Btu/h, kW)
Solution for ammonia Fluid$='ammonia' COP_R=5.847 h[1]=615.92 [Btu/lb_m] h[2]=701.99 [Btu/lb_m] h[3]=112.67 [Btu/lb_m] h[4]=112.67 [Btu/lb_m] m_dot_R=47.69 [lbm/h] P[2]=100 [psia] P[3]=100 [psia] Q_dot_L=24000 [Btu/h] s[1]=1.42220 [Btu/lb_m-R] s[2]=1.42220 [Btu/lb_m-R] T[1]=-10 [F] W_dot_in=1.203 [kW] x[1]=1 x[3]=0
Solution for R-134a Fluid$='R134a' COP_R=3.993 h[1]=101.62 [Btu/lb_m] h[2]=117.58 [Btu/lb_m] h[3]=37.87 [Btu/lb_m] h[4]=37.87 [Btu/lb_m] m_dot_R=376.5 [lbm/h] P[2]=100 [psia] P[3]=100 [psia] Q_dot_L=24000 [Btu/h] s[1]=0.22662 [Btu/lb_m-R] s[2]=0.22662 [Btu/lb_m-R] T[1]=-10 [F] W_dot_in=1.761 [kW] x[1]=1 x[3]=0
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11-9
11-16 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/kg82.88
kJ/kg82.88liquidsat.
MPa7.0
C95.34kJ/kg50.273MPa7.0
KkJ/kg94779.0kJ/kg97.236
vaporsat.kPa120
34
MPa7.0 @33
2212
2
kPa120@1
kPa120@11
=≅
==⎭⎬⎫=
°==⎭⎬⎫
==
⋅====
⎭⎬⎫=
hh
hhP
Thss
P
sshhP
f
g
g
Then the rate of heat removal from the refrigerated space and the power input to the compressor are determined from
and ( ) ( )( )
( ) ( )( ) kW1.83
kW 7.41
=−=−=
=−=−=
kJ/kg236.97273.50kg/s0.05
kJ/kg82.8897.236kg/s0.05
12in
41
hhmW
hhmQL
&&
&&
(b) The rate of heat rejection to the environment is determined from
kW9.23=+=+= 83.141.7inWQQ LH&&&
(c) The COP of the refrigerator is determined from its definition,
4.06===kW1.83kW7.41COP
inR W
QL&
&
QH
QL
0.121
2 3
4
0.7 MPa
s
T
·
Win ·
·
4s
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11-10
11-17 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/kg61.101
kJ/kg61.101liquidsat.
MPa9.0
C45.44kJ/kg93.278MPa9.0
KkJ/kg94779.0kJ/kg97.236
vaporsat.kPa120
34
MPa9.0 @33
2212
2
kPa120@1
kPa120@11
=≅
==⎭⎬⎫=
°==⎭⎬⎫
==
⋅====
⎭⎬⎫=
hh
hhP
Thss
P
sshhP
f
g
g
Then the rate of heat removal from the refrigerated space and the power input to the compressor are determined from
and ( ) ( )( )
( ) ( )( ) kW2.10
kW6.77
kJ/kg236.97278.93kg/s0.05
kJ/kg61.10197.236kg/s0.05
12in
41
=−=−=
=−=−=
hhmW
hhmQL
&&
&&
(b) The rate of heat rejection to the environment is determined from
kW8.87=+=+= 10.277.6inWQQ LH&&&
(c) The COP of the refrigerator is determined from its definition,
3.23===kW2.10kW6.77COP
inR W
QL&
&
QH
QL
0.12 MPa1
2 3
4
0.9 MPa
s
T
·
Win ·
·
4s
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11-11
11-18 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/kg58.8248.2142802.049.22
2802.085503.0
09275.033230.0
kPa120@44
kPa120@
34
=+=+=
=−
=⎟⎟⎠
⎞⎜⎜⎝
⎛ −=
fgsfs
fg
fs
hxhh
sss
x
Then,
( ) ( )( ) kW7.72kJ/kg82.58236.97kg/s0.0541 =−=−= sL hhmQ &&
and
23.4kW1.83kW7.72COP
inR ===
WQL&
&
Then the percentage increase in &Q and COP becomes
4.2%
4.2%
=−
=Δ
=
=−
=Δ
=
06.406.423.4
COPCOP
COPin Increase
41.741.772.7in Increase
R
RR
L
LL Q
&
&&
QH
QL
0.12 MPa 1
2 3
4
0.7 MPa
s
T
·
Win ·
·
4s
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11-12
11-19 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/kg98.84
kJ/kg98.84C24MPa65.0
kJ/kg16.281MPa7.0
kJ/kg53.288C50MPa7.0
KkJ/kg97236.0kJ/kg36.246
C10MPa14.0
34
C24@33
3
212
2
22
2
1
1
1
1
=≅
==⎭⎬⎫
°==
=⎭⎬⎫
==
=⎭⎬⎫
°==
⋅==
⎭⎬⎫
°−==
°
hh
hhTP
hss
P
hTP
sh
TP
f
ss
s
Then the rate of heat removal from the refrigerated space and the power input to the compressor are determined from
and ( ) ( )( )
( ) ( )( ) kW5.06
kW19.4
=−=−=
=−=−=
kJ/kg246.36288.53kg/s0.12
kJ/kg84.98246.36kg/s0.12
12in
41
hhmW
hhmQL
&&
&&
(b) The adiabatic efficiency of the compressor is determined from
82.5%=−−
=−−
=36.24653.28836.24616.281
12
12
hhhh s
C
3.83
η
(c) The COP of the refrigerator is determined from its definition,
===kW5.06kW19.4COP
inR W
QL&
&
QH
QL
0.15 MP 1
2
3
4
0.65 MPa
s
T
·
·
2
Win ·
0.14 MPa
0.7 MPa50°C
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11-13
11-20 An air conditioner operating on the ideal vapor-compression refrigeration cycle with refrigerant-134a as the working fluid is considered. The COP of the system 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. The evaporating temperature will be 22-2=20°C. From the refrigerant tables (Tables A-11, A-12, and A-13),
)throttling( kJ/kg32.107
kJ/kg32.107 liquidsat.
MPa1
kJ/kg11.273 MPa1
KkJ/kg92234.0kJ/kg59.261
vapor sat.
C20
34
MPa1 @33
212
2
C20@1
C20@11
=≅
==⎭⎬⎫=
=⎭⎬⎫
==
⋅====
⎭⎬⎫°=
°
°
hh
hhP
hss
P
sshhT
f
g
g
13.39
The COP of the air conditioner is determined from its definition,
=−−
===59.26111.27332.10759.261COP
12
41
inAC -hh
-hhwqL
QH
QL
20°C 1
2 3
4
1 MPa
s
T
·
Win ·
·
4s
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11-14
11-21E A refrigerator operating on the ideal vapor-compression refrigeration cycle with refrigerant-134a as the working fluid is considered. The increase in the COP if the throttling process were replaced by an isentropic expansion 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-11E, A-12E, and A-13E),
expansion)c(isentropi Btu/lbm80.59 F15
)throttling( Btu/lbm339.66
RBtu/lbm12715.0Btu/lbm339.66
liquidsat.psia300
Btu/lbm68.125 psia300
RBtu/lbm22341.0Btu/lbm98.105
vapor sat.
F20
434
4
34
psia300@3
psia300@33
212
2
F20@1
F20@11
=⎭⎬⎫
=°=
=≅
⋅====
⎭⎬⎫=
=⎭⎬⎫
==
⋅====
⎭⎬⎫°=
°
°
s
f
f
g
g
hss
T
hh
sshhP
hss
P
sshhT
2.012
The COP of the refrigerator for the throttling case is
=−−
===98.10568.125
339.6698.105COP12
41
inR -hh
-hhwqL
2.344
The COP of the refrigerator for the isentropic expansion case is
=−−
===98.10568.12580.5998.105COP
12
41
inR -hh
-hhwq sL
The increase in the COP by isentropic expansion is 16.5%.
QH
QL
20°F 1
2 3
4
300 psia
s
T
·
Win ·
·
4s
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PROPRIETARY MATERIAL. © 2008 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
11-22 An ideal vapor-compression refrigeration cycle with refrigerant-134a as the working fluid is considered. The COP of the system and the cooling load are 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-11, A-12, and A-13),
)throttling( kJ/kg51.81
kJ/kg51.81 liquidsat.kPa600
kJ/kg12.267 kPa600
KkJ/kg93766.0kJ/kg51.244
vapor sat.
C10
34
kPa600@33
212
2
C01 @1
C01 @11
=≅
==⎭⎬⎫=
=⎭⎬⎫
==
⋅====
⎭⎬⎫°−=
°−
°−
hh
hhP
hss
P
sshhT
f
g
g
7.21
The COP of the air conditioner is determined from its definition,
=−−
===51.24412.267
51.8151.244COP12
41
inAC -hh
-hhwqL
The cooling load is
kW1141=−=−= kJ/kg)51.8151kg/s)(244.7()( 41 hhmQL &&
QH
QL
-10°C 1
2 3
4
600 kPa
s
T
·
Win ·
·
4s
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11-16
11-23 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/kg33.239MPa14165.0
vaporsat.C5.18
throttlingkJ/kg58.93
kJ/kg58.93C30MPa95.0
kJ/kg20.289MPa0.1
/kgm14605.0KkJ/kg97236.0
kJ/kg36.246
C10kPa140
5
55
34
C30@33
3
212
2
31
1
1
1
1
==
⎭⎬⎫°−=
=≅
=≅⎭⎬⎫
°==
=⎭⎬⎫
==
=⋅=
=
⎭⎬⎫
°−==
°
hPT
hh
hhTP
hss
P
sh
TP
f
ss
v
Then the mass flow rate of the refrigerant and the power input becomes
( ) ( ) ( )[ ] ( ) kW1.88 78.0/kJ/kg246.36289.20kg/s 0.03423/
kg/s0.03423/kgm 0.14605/sm 0.3/60
12in
3
3
1
1
=−=−=
===
Cs hhmW
m
η&&
&&
vV
(b) The rate of heat removal from the refrigerated space is
( ) ( )( ) kW4.99=−=−= kJ/kg93.58239.33kg/s0.0342345 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/kg239.33246.36kg/s0.03423
14065.141
51gain
15
=−=−=
=−=−=Δ
hhmQ
PPP
&&
QH
QL
0.15 MPa1
2s
3
4
0.95 MPa
s
T
·
·
2
Win ·
0.14 MPa-10°C
1 MPa
-18.5°C
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11-17
11-24 EES Problem 11-23 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]"
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11-18
COPplot Win[kW] Qin [kW] ηc 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 10
1
2
3
4
5
6
7
8
9
η c
Win
V1 m3/min1.01.0 0.50.5 0.10.1
0.6 0.65 0.7 0.75 0.8 0.85 0.9 0.95 10
0.5
1
1.5
2
2.5
3
3.5
4
η c
CO
Ppl
ot V1 m3/min1.01.0 0.50.5 0.10.1
0.6 0.65 0.7 0.75 0.8 0.85 0.9 0.95 10
3.6
7.2
10.8
14.4
18
η c
Qin
[kW
] 1.0 1.0 0.50.5 0.10.1
V1 m3/min
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PROPRIETARY MATERIAL. © 2008 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.
11-19
11-25 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)
kJ/kg97.236 vap.)(sat.1
kPa120
kJ/kg87.298C60
kPa8.671
liq.)(sat.0kJ/kg83.86
kJ/kg83.8630.0
kPa120
11
41
22
2
32
33
3
43
44
4
=⎭⎬⎫
===
=⎭⎬⎫
°==
=
=⎭⎬⎫
==
=
=⎭⎬⎫
==
hx
PP
hTP
PP
Pxh
hh
hxP
kPa671.8
The mass flow rate of the refrigerant is determined from
kg/s0.00727=−
=−
=kg236.97)kJ/(298.87
kW45.0
12
in
hhW
m&
&
(c) The refrigeration load and the COP are
kW091.1kJ/kg)83.8697kg/s)(236.0727.0(
)( 41
=−=
−= hhmQL &&
2.43===kW0.45kW091.1COP
in
L
WQ&
&
QH
QL
120 kPa1
2 3
4
s
T
·
Win ·
·
4s
.
QH
60°C Condenser
Evaporator
Compressor
Expansion valve
120 kPa x=0.3 QL
Win
1
23
4
.
.
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11-20
Selecting the Right Refrigerant
11-26C The desirable characteristics of a refrigerant are to have an evaporator pressure which is above the atmospheric pressure, and a condenser pressure which corresponds to a saturation temperature above the temperature of the cooling medium. Other desirable characteristics of a refrigerant include being nontoxic, noncorrosive, nonflammable, chemically stable, having a high enthalpy of vaporization (minimizes the mass flow rate) and, of course, being available at low cost.
11-27C The minimum pressure that the refrigerant needs to be compressed to is the saturation pressure of the refrigerant at 30°C, which is 0.771 MPa. At lower pressures, the refrigerant will have to condense at temperatures lower than the temperature of the surroundings, which cannot happen.
11-28C Allowing a temperature difference of 10°C for effective heat transfer, the evaporation temperature of the refrigerant should be -20°C. The saturation pressure corresponding to -20°C is 0.133 MPa. Therefore, the recommended pressure would be 0.12 MPa.
11-29 A refrigerator that operates on the ideal vapor-compression cycle with refrigerant-134a is considered. Reasonable pressures for the evaporator and the condenser are to be selected.
Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.
Analysis Allowing a temperature difference of 10°C for effective heat transfer, the evaporation and condensation temperatures of the refrigerant should be -20°C and 35°C, respectively. The saturation pressures corresponding to these temperatures are 0.133 MPa and 0.888 MPa. Therefore, the recommended evaporator and condenser pressures are 0.133 MPa and 0.888 MPa, respectively.
11-30 A heat pump that operates on the ideal vapor-compression cycle with refrigerant-134a is considered. Reasonable pressures for the evaporator and the condenser are to be selected.
Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.
Analysis Allowing a temperature difference of 10°C for effective heat transfer, the evaporation and condensation temperatures of the refrigerant should be 0°C and 32°C, respectively. The saturation pressures corresponding to these temperatures are 0.293 MPa and 0.816 MPa. Therefore, the recommended evaporator and condenser pressures are 0.293 MPa and 0.816 MPa, respectively.
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11-21
Heat Pump Systems
11-31C A heat pump system is more cost effective in Miami because of the low heating loads and high cooling loads at that location.
11-32C A water-source heat pump extracts heat from water instead of air. Water-source heat pumps have higher COPs than the air-source systems because the temperature of water is higher than the temperature of air in winter.
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11-22
11-33 An actual heat pump cycle with R-134a as the refrigerant is considered. The isentropic efficiency of the compressor, the rate of heat supplied to the heated room, the COP of the heat pump, and the COP and the rate of heat supplied to the heated room if this heat pump operated on the ideal vapor-compression cycle between the same pressure limits are to be determined.
Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.
Analysis (a) The properties of refrigerant-134a are (Tables A-11 through A-13)
26.277kPa800
kJ/kg9506.0kJ/kg87.247
C)409.10(kPa200
C09.10kJ/kg91.87
kJ/kg91.87C)306.29(
kPa750
C06.29
kJ/kg76.291C55kPa800
212
2
1
1
1
1
kPasat@200
34
33
3
kPasat@7503
22
2
=⎭⎬⎫
==
==
⎭⎬⎫
°+−==
°−===
=⎭⎬⎫
°−==
°==
=⎭⎬⎫
°==
shss
P
sh
TP
Thh
hTP
TT
hTP
The isentropic efficiency of the compressor is
0.670=−−
=−−
=87.24776.29187.24726.277
12
12
hhhh s
Cη
(b) The rate of heat supplied to the room is
kW3.67=−=−= kJ/kg)91.8776kg/s)(291.018.0()( 32 hhmQH &&
(c) The power input and the COP are
kW790.0kJ/kg)87.24776kg/s)(291.018.0()( 12in =−=−= hhmW &&
4.64===790.067.3COP
inWQH&
&
(d) The ideal vapor-compression cycle analysis of the cycle is as follows:
kJ/kg.K9377.0kJ/kg46.244
kPa200@1
kPa200@1
====
g
g
sshh
kJ/kg25.273MPa800
212
2 =⎭⎬⎫
==
hss
P
34
kPa800@3 kJ/kg47.95hhhh f
===
6.18=−−
=−−
=46.24425.273
47.9525.273COP12
32
hhhh
kW3.20=−=−= kJ/kg)47.9525kg/s)(273.018.0()( 32 hhmQH &&
QH
QL
0.2 MPa1
2 3
4
0.8 MPa
s
T
·
Win ·
·
4s
Q
QL 1
2
3
4
s
T
·
·2
Win·
.
QH750 kPa
Condenser
Evaporator
Compressor
Expansion valve
800 kPa 55°C
QL
Win
1
2 3
4
.
.
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11-23
11-34 A heat pump operating on the ideal vapor-compression refrigeration cycle with refrigerant-134a as the working fluid is considered. The COP and the rate of heat supplied to the evaporator are 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-11, A-12, and A-13),
)throttling( kJ/kg77.117
kJ/kg77.117 liquidsat.kPa1200
kJ/kg00.280 kPa1200
KkJ/kg93210.0kJ/kg72.249
vapor sat.
kPa280
34
kPa1200@33
212
2
kPa280@1
kPa280@11
=≅
==⎭⎬⎫=
=⎭⎬⎫
==
⋅====
⎭⎬⎫=
hh
hhP
hss
P
sshhP
f
g
g
The mass flow rate of the refrigerant is determined from
kg/s0.6605kJ/kg)72.249(280.00
kJ/s20)(12
in12in =
−=
−=⎯→⎯−=
hhW
mhhmW&
&&&
Then the rate of heat supplied to the evaporator is
kW87.15=−=−= kJ/kg)77.11772kg/s)(249.6605.0()( 41 hhmQL &&
The COP of the heat pump is determined from its definition,
5.36=−−
===72.24900.28077.11700.280COP
12
32
inHP -hh
-hhwqH
QH
QL
280 kPa 1
2 3
4
1.2 MPa
s
T
·
Win ·
·
4s
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11-24
11-35 A heat pump operating on a vapor-compression refrigeration cycle with refrigerant-134a as the working fluid is considered. The effect of compressor irreversibilities on the COP of the cycle is to be determined.
Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.
Analysis In this cycle, the refrigerant enters the compressor as a saturated vapor at the evaporator pressure, and leaves the condenser as saturated liquid at the condenser pressure. The compression process is not isentropic. The saturation pressure of refrigerant at −1.25°C is 280 kPa. From the refrigerant tables (Tables A-11, A-12, and A-13),
)throttling( kJ/kg47.95
kJ/kg47.95 liquidsat.kPa800
kJ/kg50.271 kPa800
KkJ/kg93210.0kJ/kg72.249
vapor sat.
kPa280
34
kPa800@33
212
2
kPa280@1
kPa280@11
=≅
==⎭⎬⎫=
=⎭⎬⎫
==
⋅====
⎭⎬⎫=
hh
hhP
hss
P
sshhP
f
s
g
g
The actual enthalpy at the compressor exit is determined by using the compressor efficiency:
kJ/kg34.27585.0
72.24950.27172.249C
1212
12
12C =
−+=
−+=⎯→⎯
−−
=η
ηhh
hhhhhh ss
The COPs of the heat pump for isentropic and irreversible compression cases are
8.082=−−
=−−
==72.24950.271
47.9550.271COP12
32
inidealHP, hh
hhwq
s
sH
7.021=−−
=−−
==72.24934.275
47.9534.275COP12
32
inactualHP, hh
hhwqH
The irreversible compressor decreases the COP by 13.1%.
QH
QL
-1.25°C 1
2s
3
4
0.8 MPa
s
T
·
Win ·
·
4s
2
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11-25
11-36 A heat pump operating on a vapor-compression refrigeration cycle with refrigerant-134a as the working fluid is considered. The effect of superheating at the compressor inlet on the COP of the cycle is to be determined.
Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.
Analysis In this cycle, the compression process is isentropic and leaves the condenser as saturated liquid at the condenser pressure. The refrigerant entering the compressor is superheated by 2°C. The saturation pressure of refrigerant at −1.25°C is 280 kPa. From the refrigerant tables (Tables A-11, A-12, and A-13),
)throttling( kJ/kg47.95
kJ/kg47.95 liquidsat.kPa800
kJ/kg04.274 kPa800
KkJ/kg9403.0kJ/kg96.251
C25.1225.1kPa280
34
kPa800@33
212
2
1
1
1
1
=≅
==⎭⎬⎫=
=⎭⎬⎫
==
⋅==
⎭⎬⎫
°=+−==
hh
hhP
hss
P
sh
TP
f
The states at the inlet and exit of the compressor when the refrigerant enters the compressor as a saturated vapor are
kJ/kg50.271 kPa800
KkJ/kg93210.0kJ/kg72.249
vapor sat.
kPa280
212
2
kPa280@1
kPa280@11
=⎭⎬⎫
==
⋅====
⎭⎬⎫=
s
g
g
hss
P
sshhP
The COPs of the heat pump for the two cases are
8.082=−−
=−−
==72.24950.271
47.9550.271COP12
32
inidealHP, hh
hhwq
s
sH
8.087=−−
=−−
==96.25104.274
47.9504.274COP12
32
inactualHP, hh
hhwqH
The effect of superheating on the COP is negligible.
QH
QL
-1.25°C 1
2
3
4
0.8 MPa
s
T
·
Win ·
·
4s
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11-26
11-37E A heat pump operating on the vapor-compression refrigeration cycle with refrigerant-134a as the working fluid is considered. The effect of subcooling at the exit of the condenser on the power requirement is to be determined.
Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.
Analysis In this cycle, the compression process is isentropic, the refrigerant enters the compressor as a saturated vapor at the evaporator pressure, and leaves the condenser as subcooled liquid. From the refrigerant tables (Tables A-11E, A-12E, and A-13E),
)throttling( Btu/lbm124.45
Btu/lbm124.45 F100psia160
F1005.95.1095.9
Btu/lbm19.119 psia160
KkJ/kg22188.0Btu/lbm81.108
vapor sat.
psia50
34
F100@33
3
psia160@sat3
212
2
psia50@1
psia50@11
=≅
=≅⎭⎬⎫
°==
°=−=−=
=⎭⎬⎫
==
⋅====
⎭⎬⎫=
°
hh
hhTP
TT
hss
P
sshhP
f
g
g
The states at the inlet and exit of the expansion valve when the refrigerant is saturated liquid at the condenser exit are
)throttling( Btu/lbm519.48
Btu/lbm519.48 liquidsat.psia160
34
psia160@33
=≅
==⎭⎬⎫=
hh
hhP
f
The mass flow rate of the refrigerant in the ideal case is
lbm/h0.1415Btu/lbm)519.48(119.19
Btu/h000,100)(32
41 =−
=−
=⎯→⎯−=hh
QmhhmQ H
L
&&&&
The power requirement is
kW4.305=⎟⎠⎞
⎜⎝⎛−=−=
Btu/h3412.14kW1Btu/lbm)81.108.19lbm/h)(1190.1415()( 12in hhmW &&
With subcooling, the mass flow rate and the power input are
lbm/h1.1350Btu/lbm)124.45(119.19
Btu/h000,100)(32
41 =−
=−
=⎯→⎯−=hh
QmhhmQ H
L
&&&&
kW4.107=⎟⎠⎞
⎜⎝⎛−=−=
Btu/h3412.14kW1Btu/lbm)81.108.19lbm/h)(1191.1350()( 12in hhmW &&
Subcooling decreases the power requirement by 4.6%.
QH
QL
50 psia 1
2 3
4
160 psia
s
T
·
Win·
·
4s
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PROPRIETARY MATERIAL. © 2008 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.
11-27
11-38E A heat pump operating on the vapor-compression refrigeration cycle with refrigerant-134a as the working fluid is considered. The effect of superheating at the compressor inlet on the power requirement is to be determined.
Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.
Analysis In this cycle, the compression process is isentropic and leaves the condenser as saturated liquid at the condenser pressure. The refrigerant entering the compressor is superheated by 10°F. The saturation temperature of the refrigerant at 50 psia is 40.23°F. From the refrigerant tables (Tables A-11E, A-12E, and A-13E),
)throttling( Btu/lbm519.48
Btu/lbm519.48 liquidsat.psia160
Btu/lbm71.121 psia160
KkJ/kg2262.0Btu/lbm99.110
F2.501023.40psia50
34
psia160@33
212
2
1
1
1
1
=≅
==⎭⎬⎫=
=⎭⎬⎫
==
⋅==
⎭⎬⎫
°=+==
hh
hhP
hss
P
sh
TP
f
The states at the inlet and exit of the compressor when the refrigerant enters the compressor as a saturated vapor are
Btu/lbm19.119 psia160
KkJ/kg22188.0Btu/lbm81.108
vapor sat.
psia50
212
2
psia50@1
psia50@11
=⎭⎬⎫
==
⋅====
⎭⎬⎫=
hss
P
sshhP
g
g
The mass flow rate of the refrigerant in the ideal case is
lbm/h0.1415Btu/lbm)519.48(119.19
Btu/h000,100)(32
41 =−
=−
=⎯→⎯−=hh
QmhhmQ H
L
&&&&
The power requirement is
kW4.305=⎟⎠⎞
⎜⎝⎛−=−=
Btu/h3412.14kW1Btu/lbm)81.108.19lbm/h)(1190.1415()( 12in hhmW &&
With superheating, the mass flow rate and the power input are
lbm/h3.1366Btu/lbm)519.48(121.71
Btu/h000,100)(32
41 =−
=−
=⎯→⎯−=hh
QmhhmQ H
L
&&&&
kW4.293=⎟⎠⎞
⎜⎝⎛−=−=
Btu/h3412.14kW1Btu/lbm)99.110.71lbm/h)(1213.1366()( 12in hhmW &&
Superheating decreases the power requirement slightly.
QH
QL
50 psia 1
2
3
4
160 psia
s
T
·
Win·
·
4s
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11-28
11-39 A geothermal heat pump is considered. The degrees of subcooling done on the refrigerant in the condenser, the mass flow rate of the refrigerant, the heating load, the COP of the heat pump, the minimum power input are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis (a) From the refrigerant-134a tables (Tables A-11 through A-13)
kJ/kg00.280kPa1400
kJ/kg9223.0kJ/kg59.261
vap.)(sat.1kPa1.572
kJ/kg24.121kPa1.572
23.0C20
212
2
1
1
1
1
43
4
4
4
4
=⎭⎬⎫
==
==
⎭⎬⎫
==
=
==
⎭⎬⎫
=°=
hss
P
sh
xP
hhhP
xT
From the steam tables (Table A-4)
kJ/kg53.167
kJ/kg34.209
C40@2
C50@1
==
==
°
°
fw
fw
hh
hh
The saturation temperature at the condenser pressure of 1400 kPa and the actual temperature at the condenser outlet are
C40.52kPa1400@sat °=T
C59.48kJ24.121
kPa14003
3
3 °=⎭⎬⎫
==
ThP
(from EES)
Then, the degrees of subcooling is C3.81°=−=−=Δ 59.4840.523satsubcool TTT
(b) The rate of heat absorbed from the geothermal water in the evaporator is
kW718.2kJ/kg)53.16734kg/s)(209.065.0()( 21 =−=−= wwwL hhmQ &&
This heat is absorbed by the refrigerant in the evaporator
kg/s0.01936=−
=−
=)kJ/kg24.121(261.59
kW718.2
41 hhQ
m LR
&&
(c) The power input to the compressor, the heating load and the COP are kW6564.0kJ/kg)59.26100kg/s)(280.01936.0()( out12in =−=+−= QhhmW R
&&&
kW3.074=−=−= kJ/kg)24.12100kg/s)(280.01936.0()( 32 hhmQ RH &&
4.68===kW0.6564
kW074.3COPin
H
WQ&
&
(d) The reversible COP of the cycle is
92.12)27350/()27325(1
1/1
1COPrev =++−
=−
=HL TT
The corresponding minimum power input is
kW0.238===92.12kW074.3
COPrevminin,
HQW
&&
QH
QL 1
2
3
4
1.4 MPa
s
T
·
Win ·
·
4s
1.4 MPa s2 = s1
1
2 3
4
QH
20°C x=0.23
Condenser
Evaporator
Compressor Expansion valve
sat. vap.
Win
Water 50°C
40°C
.
.
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11-29
Innovative Refrigeration Systems
11-40C Performing the refrigeration in stages is called cascade refrigeration. In cascade refrigeration, two or more refrigeration cycles operate in series. Cascade refrigerators are more complex and expensive, but they have higher COP's, they can incorporate two or more different refrigerants, and they can achieve much lower temperatures.
11-41C Cascade refrigeration systems have higher COPs than the ordinary refrigeration systems operating between the same pressure limits.
11-42C The saturation pressure of refrigerant-134a at -32°C is 77 kPa, which is below the atmospheric pressure. In reality a pressure below this value should be used. Therefore, a cascade refrigeration system with a different refrigerant at the bottoming cycle is recommended in this case.
11-43C We would favor the two-stage compression refrigeration system with a flash chamber since it is simpler, cheaper, and has better heat transfer characteristics.
11-44C Yes, by expanding the refrigerant in stages in several throttling devices.
11-45C To take advantage of the cooling effect by throttling from high pressures to low pressures.
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11-30
11-46 [Also solved by EES on enclosed CD] A two-stage compression refrigeration system with refrigerant-134a as the working fluid is considered. The fraction of the refrigerant that evaporates as it is throttled to the flash chamber, the rate of heat removed from the refrigerated space, and the COP are to be determined.
Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. 3The flash chamber is adiabatic.
Analysis (a) The enthalpies of the refrigerant at several states are determined from the refrigerant tables (Tables A-11, A-12, and A-13) to be
kJ/kg33.73kJ/kg32.107
kJ/kg31.265
,kJ/kg33.73,kJ/kg32.107,kJ/kg30.259
kJ/kg,16.239
8
6
2
7
5
3
1
==
=
====
hh
h
hhhh
The fraction of the refrigerant that evaporates as it is throttled to the flash chamber is simply the quality at state 6,
0.1828=−
=−
=98.185
33.7332.10766
fg
f
hhh
x
(b) The enthalpy at state 9 is determined from an energy balance on the mixing chamber:
( ) ( )( )( ) ( )( ) kJ/kg21.26431.2651828.0130.2591828.0
11
0
9
26369
outin
(steady)0systemoutin
=−+=−+=
=
=
==−
∑∑h
hxhxh
hmhm
EE
EEE
iiee &&
&&
&&& Δ
also,
kJ/kg97.278KkJ/kg94083.0
MPa14
94
4 =⎭⎬⎫
⋅===
hss
P
Then the rate of heat removed from the refrigerated space and the compressor work input per unit mass of refrigerant flowing through the condenser are
( ) ( )( )
( ) ( )( )
( ) ( )( )( ) ( )( )
kW.039kJ/kg239.16265.31kg/s 0.2043kJ/kg264.21278.97kg/s0.25
kJ/kg73.33239.16kg/s0.2043
kg/s0.2043kg/s0.251828.011
1294incompII,incompI,in
81
6
=−+−=
−+−=+=
=−=−=
=−=−=
hhmhhmWWW
hhmQ
mxm
BA
BL
AB
&&&&&
&&
&&
kW 33.88
(c) The coefficient of performance is determined from
3.75===kW9.03kW33.88COP
innet,R W
QL&
&
QL
0.14 MPa
1
2 5
8
0.5 MPa
s
T
·
41 MPa
6 9A
B 37
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11-31
11-47 EES Problem 11-46 is reconsidered. The effects of the various refrigerants in EES data bank for compressor efficiencies of 80, 90, and 100 percent is to be investigated.
Analysis The problem is solved using EES, and the results are tabulated and plotted below.
"Input Data" "P[1]=140 [kPa] P[4] = 1000 [kPa] P[6]=500 [kPa] Eta_compB =1.0 Eta_compA =1.0" m_dot_A=0.25 [kg/s]
"High Pressure Compressor A" P[9]=P[6] h4s=enthalpy(R134a,P=P[4],s=s[9]) "State 4s is the isentropic value of state 4" h[9]+w_compAs=h4s "energy balance on isentropic compressor" w_compA=w_compAs/Eta_compA"definition of compressor isentropic efficiency" h[9]+w_compA=h[4] "energy balance on real compressor-assumed adiabatic" s[4]=entropy(R134a,h=h[4],P=P[4]) "properties for state 4" T[4]=temperature(R134a,h=h[4],P=P[4]) W_dot_compA=m_dot_A*w_compA
"Condenser" P[5]=P[4] "neglect pressure drops across condenser" T[5]=temperature(R134a,P=P[5],x=0) "properties for state 5, assumes sat. liq. at cond. exit" h[5]=enthalpy(R134a,T=T[5],x=0) "properties for state 5" s[5]=entropy(R134a,T=T[5],x=0) h[4]=q_out+h[5] "energy balance on condenser" Q_dot_out = m_dot_A*q_out
"Throttle Valve A" h[6]=h[5] "energy balance on throttle - isenthalpic" x6=quality(R134a,h=h[6],P=P[6]) "properties for state 6" s[6]=entropy(R134a,h=h[6],P=P[6]) T[6]=temperature(R134a,h=h[6],P=P[6])
"Flash Chamber" m_dot_B = (1-x6) * m_dot_A P[7] = P[6] h[7]=enthalpy(R134a, P=P[7], x=0) s[7]=entropy(R134a,h=h[7],P=P[7]) T[7]=temperature(R134a,h=h[7],P=P[7])
"Mixing Chamber" x6*m_dot_A*h[3] + m_dot_B*h[2] =(x6* m_dot_A + m_dot_B)*h[9] P[3] = P[6] h[3]=enthalpy(R134a, P=P[3], x=1) "properties for state 3" s[3]=entropy(R134a,P=P[3],x=1) T[3]=temperature(R134a,P=P[3],x=x1) s[9]=entropy(R134a,h=h[9],P=P[9]) "properties for state 9" T[9]=temperature(R134a,h=h[9],P=P[9])
"Low Pressure Compressor B" x1=1 "assume flow to compressor inlet to be saturated vapor"
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11-32
h[1]=enthalpy(R134a,P=P[1],x=x1) "properties for state 1" T[1]=temperature(R134a,P=P[1], x=x1) s[1]=entropy(R134a,P=P[1],x=x1) P[2]=P[6] h2s=enthalpy(R134a,P=P[2],s=s[1]) " state 2s is isentropic state at comp. exit" h[1]+w_compBs=h2s "energy balance on isentropic compressor" w_compB=w_compBs/Eta_compB"definition of compressor isentropic efficiency" h[1]+w_compB=h[2] "energy balance on real compressor-assumed adiabatic" s[2]=entropy(R134a,h=h[2],P=P[2]) "properties for state 2" T[2]=temperature(R134a,h=h[2],P=P[2]) W_dot_compB=m_dot_B*w_compB
"Throttle Valve B" h[8]=h[7] "energy balance on throttle - isenthalpic" x8=quality(R134a,h=h[8],P=P[8]) "properties for state 8" s[8]=entropy(R134a,h=h[8],P=P[8]) T[8]=temperature(R134a,h=h[8],P=P[8])
"Evaporator" P[8]=P[1] "neglect pressure drop across evaporator" q_in + h[8]=h[1] "energy balance on evaporator" Q_dot_in=m_dot_B*q_in
"Cycle Statistics" W_dot_in_total = W_dot_compA + W_dot_compB COP=Q_dot_in/W_dot_in_total "definition of COP"
ηcompA ηcompB Qout COP 0,8 0,8 45.32 2.963
0,8333 0,8333 44.83 3.094 0,8667 0,8667 44.39 3.225
0,9 0,9 43.97 3.357 0,9333 0,9333 43.59 3.488 0,9667 0,9667 43.24 3.619
1 1 42.91 3.751
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
]
1000 kPa
500 kPa
140 kPa
R134a
1
2
39
45
67
8
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11-33
0,8 0,84 0,88 0,92 0,96 1
42,5
43
43,5
44
44,5
45
45,5
2,9
3
3,1
3,2
3,3
3,4
3,5
3,6
3,7
3,8
ηcomp
Qou
t[k
W]
CO
P
200 300 400 500 600 700 800 900 1000 11003.45
3.50
3.55
3.60
3.65
3.70
3.75
3.80
P[6] [kPa]
CO
P
COP vs Flash Chamber Pressure, P6
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11-34
11-48 [Also solved by EES on enclosed CD] A two-stage compression refrigeration system with refrigerant-134a as the working fluid is considered. The fraction of the refrigerant that evaporates as it is throttled to the flash chamber, the rate of heat removed from the refrigerated space, and the COP are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. 3The flash chamber is adiabatic. Analysis (a) The enthalpies of the refrigerant at several states are determined from the refrigerant tables (Tables A-11, A-12, and A-13) to be
kJ/kg16.55kJ/kg32.107
kJ/kg90.255
,kJ/kg16.55,kJ/kg32.107,kJ/kg88.251
kJ/kg,16.239
8
6
2
7
5
3
1
==
=
====
hh
h
hhhh
The fraction of the refrigerant that evaporates as it is throttled to the flash chamber is simply the quality at state 6,
0.2651=−
=−
=71.196
16.5532.10766
fg
f
hhh
x
(b) The enthalpy at state 9 is determined from an energy balance on the mixing chamber:
( ) ( )( )( ) ( )( ) kJ/kg84.25490.2552651.0188.2512651.0
11
0
9
26369
outin
(steady)0systemoutin
=−+=−+=
=
=
==−
∑∑h
hxhxh
hmhm
EE
EEE
iiee &&
&&
&&& Δ
and
KkJ/kg94074.0kJ/kg84.254
MPa32.09
9
9 ⋅=⎭⎬⎫
==
shP
also, kJ/kg94.278KkJ/kg94074.0
MPa14
94
4 =⎭⎬⎫
⋅===
hss
P
Then the rate of heat removed from the refrigerated space and the compressor work input per unit mass of refrigerant flowing through the condenser are
( ) ( )( )
( ) ( )( )
( ) ( )( )( ) ( )( )
kW9.10kJ/kg239.16255.90kg/s0.1837kJ/kg254.84278.94kg/s0.25
kJ/kg55.16239.16kg/s0.1837
kg/s0.1837kg/s0.250.265111
1294incompII,incompI,in
81
6
=−+−=
−+−=+=
=−=−=
=−=−=
hhmhhmWWW
hhmQ
mxm
BA
BL
AB
&&&&&
&&
&&
kW 33.80
(c) The coefficient of performance is determined from
3.71===kW9.10kW33.80COP
innet,R W
QL&
&
QL
0.14 MPa
1
2 5
8
0.32 MPa
s
T
·
41 MPa
6 9
A
B3
7
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11-35
11-49 A two-stage cascade refrigeration cycle is considered. The mass flow rate of the refrigerant through the upper cycle, the rate of heat removal from the refrigerated space, 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) The properties are to be obtained from the refrigerant tables (Tables A-11 through A-13):
kJ/kg.K9377.0kJ/kg46.244
kPa200@1
kPa200@1
====
g
g
sshh
kJ/kg30.263kPa500
212
2 =⎭⎬⎫
==
shss
P
kJ/kg01.268
46.24446.24430.26380.0 2
2
12
12
=⎯→⎯−
−=
−−
=
hh
hhhh s
Cη
kJ/kg33.73
kJ/kg33.73
34
kPa500@3
====
hhhh f
kJ/kg.K9269.0kJ/kg55.255
kPa400@5
kPa004@5
====
g
g
sshh
kJ/kg33.278kPa1200
656
6 =⎭⎬⎫
==
shss
P
kJ/kg02.284
55.25555.25533.27880.0 6
6
56
56
=⎯→⎯−
−=
−−
=
hh
hhhh s
Cη
kJ/kg77.117
kJ/kg77.117
78
kPa1200@7
====
hhhh f
The mass flow rate of the refrigerant through the upper cycle is determined from an energy balance on the heat exchanger
kg/s0.212=⎯→⎯−=−
−=−
AA
BA
mm
hhmhhm
&&
&&
kJ/kg)33.7301kg/s)(268.15.0(kJ/kg)77.117.55255(
)()( 3285
(b) The rate of heat removal from the refrigerated space is
kW25.67=−=−= kJ/kg)33.7346kg/s)(244.15.0()( 41 hhmQ BL &&
(c) The power input and the COP are
kW566.9kJ/kg)46.24401kg/s)(268.212.0(kJ/kg)55.25502kg/s)(284.15.0()()( 1256in
=−+−=−+−= hhmhhmW BA &&&
2.68===566.9
67.25COPin
L
WQ&
&
.
5
6 7
8
QH
Condenser
Evaporator
Compressor
Expansion valve
Win
1
2 3
4
Condenser
Evaporator
Compressor
Expansion valve
QL
Win
.
.
.
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11-36
11-50 A two-evaporator compression refrigeration cycle with refrigerant-134a as the working fluid is considered. The cooling rate of the high-temperature evaporator, the power required by the compressor, and the COP of the system are to be determined.
Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.
Analysis From the refrigerant tables (Tables A-11, A-12, and A-13),
kJ/kg44.234 vapor sat.
C4.26
kJ/kg45.250 vapor sat.
C0
)throttling( kJ/kg47.95
kJ/kg47.95 liquidsat.kPa800
C4.26@77
C0@55
364
kPa800@33
==⎭⎬⎫°−=
==⎭⎬⎫°=
=≅=
==⎭⎬⎫=
°−
°
g
g
f
hhT
hhT
hhh
hhP
The mass flow rate through the low-temperature evaporator is found by
kg/s05757.0kJ/kg)47.95(234.44
kJ/s8)(67
2672 =−
=−
=⎯→⎯−=hh
QmhhmQ L
L
&&&&
The mass flow rate through the warmer evaporator is then
kg/s04243.005757.01.021 =−=−= mmm &&&
Applying an energy balance to the point in the system where the two evaporator streams are recombined gives
kJ/kg23.2411.0
)44.234)(05757.0()45.250)(04243.0(7251117251 =
+=
+=⎯→⎯=+
mhmhm
hhmhmhm&
&&&&&
Then,
3
45
6
Compressor
Expansion valve1
2
Evaporator 27
Evaporator 1
Condenser
Expansion valve
Expansion valve
QH
QL
-26.4°C 1
2
3
6
800 kPa
s
T
·
Win·
·
4 5
7
0°C
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11-37
kJ/kg26.286 kPa800
KkJ/kg9789.0 kJ/kg23.241
kPa100
212
2
11
C26.4@sat 1
=⎭⎬⎫
==
⋅=⎭⎬⎫
=≅= °−
hss
P
sh
PP
The cooling rate of the high-temperature evaporator is
kW6.58=−=−= kJ/kg)47.95(250.45)kg/s04243.0()( 451 hhmQL &&
The power input to the compressor is
kW4.50=−=−= kJ/kg)23.24126kg/s)(286.1.0()( 12in hhmW &&
The COP of this refrigeration system is determined from its definition,
3.24=+
==kW4.50
kW6.58)(8COP
inR W
QL&
&
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11-38
11-51E A two-evaporator compression refrigeration cycle with refrigerant-134a as the working fluid is considered. The power required by the compressor and the COP of the system are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.
Analysis From the refrigerant tables (Tables A-11E, A-12E, and A-13E),
Btu/lbm99.100 vapor sat.
psia15
Btu/lbm32.105 vapor sat.
psia30
)throttling( Btu/lbm304.45
Btu/lbm304.45 liquidsat.psia140
psia15@77
psia03 @55
364
psia140@33
==⎭⎬⎫=
==⎭⎬⎫=
=≅=
==⎭⎬⎫=
g
g
f
hhP
hhP
hhh
hhP
The mass flow rates through the high-temperature and low-temperature evaporators are found by
lbm/h99.49Btu/lbm)304.45(105.32
Btu/h3000)(45
1,14511, =
−=
−=⎯→⎯−=
hhQ
mhhmQ LL
&&&&
lbm/h58.179Btu/lbm)304.45(100.99
Btu/h000,10)(67
2,26722, =
−=
−=⎯→⎯−=
hhQ
mhhmQ LL
&&&&
Applying an energy balance to the point in the system where the two evaporator streams are recombined gives
Btu/lbm93.10158.17999.49
)99.100)(58.179()32.105)(99.49()(21
725111217251 =
++
=++
=⎯→⎯+=+mm
hmhmhhmmhmhm&&
&&&&&&
Then,
Btu/lbm24.122 psia140
RBtu/lbm2293.0 Btu/lbm93.101
psia15
212
2
11
1
=⎭⎬⎫
==
⋅=⎭⎬⎫
==
hss
P
shP
The power input to the compressor is
kW1.366=⎟⎠⎞
⎜⎝⎛−+=−+=
Btu/h3412.14kW1Btu/lbm)93.10124lbm/h(122.)58.17999.49())(( 1221in hhmmW &&&
2.79
The COP of this refrigeration system is determined from its definition,
=⎟⎠⎞
⎜⎝⎛+
==Btu/h3412.14
kW1kW1.366
Btu/h3000)(10,000COP
inR W
QL&
&
3
45
6
Compressor
Expansion valve1
2
Evaporator 27
Evaporator 1
Condenser
Expansion valve
Expansion valve
QH
QL
15 psia 1
2
3
6
140 psia
s
T
·
Win·
·
45
7
30 psia
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PROPRIETARY MATERIAL. © 2008 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.
11-39
11-52E A two-evaporator compression refrigeration cycle with refrigerant-134a as the working fluid is considered. The power required by the compressor and the COP of the system are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.
Analysis From the refrigerant tables (Tables A-11E, A-12E, and A-13E),
Btu/lbm99.100 vapor sat.
psia15
Btu/lbm11.110 vapor sat.
psia60
)throttling( Btu/lbm304.45
Btu/lbm304.45 liquidsat.psia140
psia15@77
psia06 @55
364
psia140@33
==⎭⎬⎫=
==⎭⎬⎫=
=≅=
==⎭⎬⎫=
g
g
f
hhP
hhP
hhh
hhP
The mass flow rates through the high-temperature and low-temperature evaporators are found by
lbm/h7.740Btu/lbm)304.45(110.11
Btu/h000,48)(45
1,14511, =
−=
−=⎯→⎯−=
hhQ
mhhmQ LL
&&&&
lbm/h6.179Btu/lbm)304.45(100.99
Btu/h000,10)(67
2,26722, =
−=
−=⎯→⎯−=
hhQ
mhhmQ LL
&&&&
Applying an energy balance to the point in the system where the two evaporator streams are recombined gives
Btu/lbm33.10858.1797.740
)99.100)(58.179()11.110)(7.740()(21
725111217251 =
++
=++
=⎯→⎯+=+mm
hmhmhhmmhmhm&&
&&&&&&
Then,
Btu/lbm45.130 psia140
RBtu/lbm2430.0 Btu/lbm33.108
psia15
212
2
11
1
=⎭⎬⎫
==
⋅=⎭⎬⎫
==
hss
P
shP
The power input to the compressor is
kW5.966=⎟⎠⎞
⎜⎝⎛−+=−+=
Btu/h3412.14kW1Btu/lbm)33.10845lbm/h(130.)6.1797.740())(( 1221in hhmmW &&&
2.85
The COP of this refrigeration system is determined from its definition,
=⎟⎠⎞
⎜⎝⎛+
==Btu/h3412.14
kW1kW5.966
Btu/h000),10(48,000COP
inR W
QL&
&
3
45
6
Compressor
Expansion valve1
2
Evaporator I 7
Evaporator II
Condenser
Expansion valve
Expansion valve
QH
QL
15 psia 1
2
3
6
140 psia
s
T
·
Win·
·
45
7
60 psia
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PROPRIETARY MATERIAL. © 2008 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.
11-40
11-53 A two-stage compression refrigeration system with a separation unit is considered. The mass flow rate through the two compressors, the power used by the compressors, and the system’s COP are to be determined.
Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.
Analysis From the refrigerant tables (Tables A-11, A-12, and A-13),
kJ/kg74.252 kPa200
KkJ/kg96866.0kJ/kg86.225
vapor sat.
C40
)throttling( kJ/kg43.38
kJ/kg43.38 liquidsat.
C1.10
)throttling( kJ/kg47.95
kJ/kg47.95 liquidsat.kPa800
kJ/kg24.273 kPa800
KkJ/kg93773.0kJ/kg46.244
vapor sat.
C1.10
878
C10.1@sat 8
C40@7
C40@77
56
C1.10@55
34
kPa800@33
212
2
C1.10@1
C1.10@11
=⎭⎬⎫
===
⋅====
⎭⎬⎫°−=
=≅
==⎭⎬⎫°−=
=≅
==⎭⎬⎫=
=⎭⎬⎫
==
⋅====
⎭⎬⎫°−=
°−
°−
°−
°−
°−
°−
hssPP
sshhT
hh
hhT
hh
hhP
hss
P
sshhT
g
g
f
f
g
g
The mass flow rate through the evaporator is determined from
kg/s0.1601=−
=−
=⎯→⎯−=kJ/kg)43.38(225.86
kJ/s30)(67
6676 hhQ
mhhmQ LL
&&&&
An energy balance on the separator gives
kg/s0.2303=−−
=−−
=⎯→⎯−=−47.9546.24443.3874.252)1601.0()()(
41
5862412586 hh
hhmmhhmhhm &&&&
The total power input to the compressors is
kW10.93=−+−=
−+−=kJ/kg)46.24424kg/s)(273.2303.0(kJ/kg)86.22574kg/s)(252.1601.0(
)()( 122786in hhmhhmW &&&
The COP of this refrigeration system is determined from its definition,
2.74===kW10.93
kW30COPin
R WQL&
&
3
4 5
6
Compressor
Expansion valve
1
2
Evaporator7
Separator
Condenser
Expansion valve
Compressor
8
QL
-40°C 7
8 3
6
s
T
·
20.8 MPa
4 15
-10.1°C
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PROPRIETARY MATERIAL. © 2008 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.
11-41
11-54E A two-stage compression refrigeration system with a separation unit is considered. The cooling load and the system’s COP are to be determined.
Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.
Analysis From the refrigerant tables (Tables A-11E, A-12E, and A-13E),
Btu/lbm28.116 psia120
RBtu/lbm22127.0Btu/lbm11.110
vapor sat.
psia60
)throttling( kJ/kg787.41
Btu/lbm787.41 liquidsat.psia120
)throttling( Btu/lbm339.66
Btu/lbm339.66 liquidsat.psia300
Btu/lbm06.123 psia300
RBtu/lbm21924.0Btu/lbm16.115
vapor sat.
psia120
878
8
psia60@7
psia60@77
56
psia120@55
34
psia300@33
212
2
psia120@1
psia120@11
=⎭⎬⎫
==
⋅====
⎭⎬⎫=
=≅
==⎭⎬⎫=
=≅
==⎭⎬⎫=
=⎭⎬⎫
==
⋅====
⎭⎬⎫=
hss
P
sshhP
hh
hhP
hh
hhP
hss
P
sshhP
g
g
f
f
g
g
An energy balance on the separator gives
6641
5862412586 5258.1
339.6616.115787.4128.116)()( mm
hhhh
mmhhmhhm &&&&&& =−−
=−−
=⎯→⎯−=−
The total power input to the compressors is given by
)(5258.1)()()(
126786
122786in
hhmhhmhhmhhmW
−+−=−+−=&&
&&&
Solving for 6m& ,
lbm/h4681Btu/lbm)16.11506.123(5258.1)11.110(116.28
Btu/h)14.341225()(5258.1)( 1278
in6 =
−+−×
=−+−
=hhhh
Wm
&&
The cooling effect produced by this system is then
Btu/h319,800=−=−= Btu/lbm)787.41.11lbm/h)(1104681()( 676 hhmQL &&
The COP of this refrigeration system is determined from its definition,
3.75===kW25
kW412.14)(319,800/3COP
inR W
QL&
&
3
4 5
6
Compressor
Expansion valve
1
2
Evaporator7
Separator
Condenser
Expansion valve
Compressor
8
QL
60 psia7
8 3
6
s
T
·
2300 psia
4 15120 psia
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11-42
11-55 A two-stage cascade refrigeration system is considered. Each stage operates on the ideal vapor-compression cycle with upper cycle using water and lower cycle using refrigerant-134a as the working fluids. The mass flow rate of R-134a and water in their respective cycles and the overall COP of this system are to be determined.
Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. 3The heat exchanger is adiabatic.
Analysis From the water and refrigerant tables (Tables A-4, A-5, A-6, A-11, A-12, and A-13),
)throttling( kJ/kg94.63
kJ/kg94.63 liquidsat.kPa400
kJ/kg59.267 kPa400
KkJ/kg96866.0kJ/kg86.225
vapor sat.
C40
)throttling( kJ/kg44.858
kJ/kg44.858 liquidsat.
MPa6.1
kJ/kg4.5083 MPa6.1
KkJ/kg0249.9kJ/kg1.2510
vapor sat.
C5
78
kPa040@77
656
6
C40@5
C40@55
34
MPa6.1 @33
212
2
C5@1
C5@11
=≅
==⎭⎬⎫=
=⎭⎬⎫
==
⋅====
⎭⎬⎫°−=
=≅
==⎭⎬⎫=
=⎭⎬⎫
==
⋅====
⎭⎬⎫°=
°−
°−
°
°
hh
hhP
hss
P
sshhT
hh
hhP
hss
P
sshhT
f
g
g
f
g
g
The mass flow rate of R-134a is determined from
kg/s0.1235=−
=−
=⎯→⎯−=kJ/kg)94.63(225.86
kJ/s20)(85
85 hhQ
mhhmQ LRRL
&&&&
An energy balance on the heat exchanger gives the mass flow rate of water
kg/s0.01523=−−
=−−
=⎯→⎯
−=−
44.8581.251094.6359.267kg/s)1235.0(
)()(
41
76
4176
hhhh
mm
hhmhhm
Rw
wR
&&
&&
The total power input to the compressors is
kJ/s35.44kJ/kg)1.2510.4kg/s)(508301523.0(kJ/kg)86.22559kg/s)(267.1235.0(
)()( 1256in
=−+−=
−+−= hhmhhmW wR &&&
The COP of this refrigeration system is determined from its definition,
0.451===kJ/s44.35
kJ/s20COPin
R WQL&
&
QL -40°C 5
6 3
8
s
T
·
2 1.6 MPa
4 1 7
5°C 400 kPa
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11-43
11-56 A two-stage vapor-compression refrigeration system with refrigerant-134a as the working fluid is considered. The process with the greatest exergy destruction is to be determined.
Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.
Analysis From Prob. 11-55 and the water and refrigerant tables (Tables A-4, A-5, A-6, A-11, A-12, and A-13),
K303C30K303C30K243C30
kJ/kg0.4225kJ/kg92.161
kg/s01751.0kg/s1235.0
KkJ/kg27423.0KkJ/kg24757.0
KkJ/kg96866.0KkJ/kg0869.3KkJ/kg3435.2
KkJ/kg0249.9
0
32
85
8
7
65
4
3
21
=°==°==°−=
=−==−=
==
⋅=⋅=
⋅==⋅=⋅=
⋅==
TTT
hhqhhq
mmss
ssss
ss
H
L
H
L
w
R&
&
⎟
The exergy destruction during a process of a stream from an inlet state to exit state is given by
⎟⎠
⎞⎜⎜⎝
⎛+−−==
sink
out
source
in0gen0dest T
qT
qssTsTx ie
Application of this equation for each process of the cycle gives
[ ][ ] kJ/s52.4)96866.024757.0)(1235.0()0869.30249.9)(01751.0()303(
)()(
kJ/s05.1243
92.16127423.096866.0)303)(1235.0(
kJ/s00.1)24757.027423.0)(303)(1235.0()(
kJ/s94.3)3435.20869.3)(303)(01751.0()(
kJ/s53.38303
0.42250249.93435.2)K303)(01751.0(
67R41w0exchheat destroyed,
850R85destroyed,
780R78destroyed,
340w34destroyed,
23023destroyed,
=−+−=
−+−=
=⎟⎠⎞
⎜⎝⎛ −−=
⎟⎟⎠
⎞⎜⎜⎝
⎛−−=
=−=−=
=−=−=
=⎟⎠⎞
⎜⎝⎛ +−=
⎟⎟⎠
⎞⎜⎜⎝
⎛+−=
ssmssmTX
Tq
ssTmX
ssTmX
ssTmX
Tq
ssTmX
L
L
H
Hw
&&&
&&
&&
&&
&&
For isentropic processes, the exergy destruction is zero:
0
0
56destroyed,
12destroyed,
=
=
X
X&
&
Note that heat is absorbed from a reservoir at -30°C (243 K) and rejected to a reservoir at 30°C (303 K), which is also taken as the dead state temperature. Alternatively, one may use the standard 25°C (298 K) as the dead state temperature, and perform the calculations accordingly. The greatest exergy destruction occurs in the condenser.
QL-40°C 5
6 3
8
s
T
·
21.6 MPa
4 17
5°C 400 kPa
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11-44
Gas Refrigeration Cycles
11-57C The ideal gas refrigeration cycle is identical to the Brayton cycle, except it operates in the reversed direction.
11-58C The reversed Stirling cycle is identical to the Stirling cycle, except it operates in the reversed direction. Remembering that the Stirling cycle is a totally reversible cycle, the reversed Stirling cycle is also totally reversible, and thus its COP is
COPR,Stirling =−
11T TH L/
11-59C In the ideal gas refrigeration cycle, the heat absorption and the heat rejection processes occur at constant pressure instead of at constant temperature.
11-60C In aircraft cooling, the atmospheric air is compressed by a compressor, cooled by the surrounding air, and expanded in a turbine. The cool air leaving the turbine is then directly routed to the cabin.
11-61C No; because h = h(T) for ideal gases, and the temperature of air will not drop during a throttling (h1 = h2) process.
11-62C By regeneration.
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11-45
11-63 [Also solved by EES on enclosed CD] An ideal-gas refrigeration cycle with air as the working fluid is considered. The rate of refrigeration, the net power input, and the COP are to be determined.
Assumptions 1 Steady operating conditions exist. 2 Air is an ideal gas with variable specific heats. 3Kinetic 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),
T hP
T hP
r
r
1 1
1 3
285 2851411584
320 320 291 7375
1
3
= ⎯ →⎯ ==
= ⎯ →⎯ ==
K kJ / kg
K kJ / kg
..
..
Thus,
( )
( )kJ/kg76201
K8201347507375125050
kJ/kg17452 K4450792515841
50250
4
43
4
2
21
2
34
12
.h
.T..PPPP
.h
.T..PPPP
rr
rr
==⎯→⎯=⎟
⎠⎞
⎜⎝⎛==
==⎯→⎯=⎟
⎠⎞
⎜⎝⎛==
Then the rate of refrigeration is
( ) ( ) ( )( ) kW 6.67=−=−== kJ/kg201.76285.14kg/s0.0841refrig hhmqmQ L &&&
(b) The net power input is determined from
& & &W W Wnet, in comp, in turb, out= −
where
( ) ( )( )( ) ( )( ) kW9.48kJ/kg201.76320.29kg/s0.08
kW13.36kJ/kg285.14452.17kg/s0.08
43outturb,
12incomp,
=−=−=
=−=−=
hhmWhhmW
&&
&&
Thus,
& . .Wnet, in = − =13 36 9 48 3.88 kW
(c) The COP of this ideal gas refrigeration cycle is determined from
COP 6.67 kW3.88 kWR
net, in= = =
&
&Q
WL 1.72
s
T
1
2 QH
47°C 12°C
3
4
·
QRefrig ·
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11-46
11-64 EES Problem 11-63 is reconsidered. The effects of compressor and turbine isentropic efficiencies on the rate of refrigeration, the net power input, and the COP are to be investigated. Analysis The problem is solved using EES, and the solution is given below.
"Input data" T[1] = 12 [C] P[1]= 50 [kPa] T[3] = 47 [C] P[3]=250 [kPa] m_dot=0.08 [kg/s] Eta_comp = 1.00 Eta_turb = 1.0 "Compressor anaysis" s[1]=ENTROPY(Air,T=T[1],P=P[1]) s2s=s[1] "For the ideal case the entropies are constant across the compressor" P[2] = P[3] s2s=ENTROPY(Air,T=Ts2,P=P[2])"Ts2 is the isentropic value of T[2] at compressor exit" Eta_comp = W_dot_comp_isen/W_dot_comp "compressor adiabatic efficiency, W_dot_comp > W_dot_comp_isen" m_dot*h[1] + W_dot_comp_isen = m_dot*hs2"SSSF First Law for the isentropic compressor, assuming: adiabatic, ke=pe=0, m_dot is the mass flow rate in kg/s" h[1]=ENTHALPY(Air,T=T[1]) hs2=ENTHALPY(Air,T=Ts2) m_dot*h[1] + W_dot_comp = m_dot*h[2]"SSSF First Law for the actual compressor, assuming: adiabatic, ke=pe=0" h[2]=ENTHALPY(Air,T=T[2]) s[2]=ENTROPY(Air,h=h[2],P=P[2]) "Heat Rejection Process 2-3, assumed SSSF constant pressure process" m_dot*h[2] + Q_dot_out = m_dot*h[3]"SSSF First Law for the heat exchanger, assuming W=0, ke=pe=0" h[3]=ENTHALPY(Air,T=T[3]) "Turbine analysis" s[3]=ENTROPY(Air,T=T[3],P=P[3]) s4s=s[3] "For the ideal case the entropies are constant across the turbine" P[4] = P[1] s4s=ENTROPY(Air,T=Ts4,P=P[4])"Ts4 is the isentropic value of T[4] at turbine exit" Eta_turb = W_dot_turb /W_dot_turb_isen "turbine adiabatic efficiency, W_dot_turb_isen > W_dot_turb" m_dot*h[3] = W_dot_turb_isen + m_dot*hs4"SSSF First Law for the isentropic turbine, assuming: adiabatic, ke=pe=0" hs4=ENTHALPY(Air,T=Ts4) m_dot*h[3] = W_dot_turb + m_dot*h[4]"SSSF First Law for the actual compressor, assuming: adiabatic, ke=pe=0" h[4]=ENTHALPY(Air,T=T[4]) s[4]=ENTROPY(Air,h=h[4],P=P[4]) "Refrigeration effect:" m_dot*h[4] + Q_dot_Refrig = m_dot*h[1] "Cycle analysis" W_dot_in_net=W_dot_comp-W_dot_turb"External work supplied to compressor" COP= Q_dot_Refrig/W_dot_in_net "The following is for plotting data only:" Ts[1]=Ts2 ss[1]=s2s Ts[2]=Ts4 ss[2]=s4s
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11-47
COP ηcomp ηturb QRefrig [kW]
Winnet [kW]
0.6937 0.7 1 6.667 9.612 0.9229 0.8 1 6.667 7.224 1.242 0.9 1 6.667 5.368 1.717 1 1 6.667 3.882
0,7 0,75 0,8 0,85 0,9 0,95 10
1
2
3
4
5
6
7
ηcomp
QR
efrig
[kW
]
η turb
0.70.70.850.851.01.0
0,7 0,75 0,8 0,85 0,9 0,95 10
2
4
6
8
10
12
ηcomp
Win
;net
[kW
]
η turb
0.70.70.850.851.01.0
0,7 0,75 0,8 0,85 0,9 0,95 10
0,5
1
1,5
2
ηcomp
CO
P
η turb
0.70.70.850.851.01.0
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11-48
11-65 [Also solved by EES on enclosed CD] An ideal-gas refrigeration cycle with air as the working fluid is considered. The rate of refrigeration, the net power input, and the COP are to be determined.
Assumptions 1 Steady operating conditions exist. 2 Air is an ideal gas with variable specific heats. 3Kinetic and potential energy changes are negligible.
Analysis (a) We assume 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),
T hP
T hP
r
r
1 1
1 3
285 2851411584
320 320 291 7375
1
3
= ⎯ →⎯ ==
= ⎯ →⎯ ==
K kJ / kg
K kJ / kg
..
..
Thus,
( )
( )kJ/kg76201
K8201347507375125050
kJ/kg17452 K4450792515841
50250
4
43
4
2
21
2
34
12
.h
.T..PPPP
.h
.T..PPPP
s
srr
s
srr
==⎯→⎯=⎟
⎠⎞
⎜⎝⎛==
==⎯→⎯=⎟
⎠⎞
⎜⎝⎛==
Also,
( )( )( )
kJ/kg54219762012932085029320
433443
43
.....
hhhhhhhh
sTs
T
=−−=
−η−=⎯→⎯−−
=η
Then the rate of refrigeration is
( ) ( ) ( )( ) kW 5.25=−=−== kJ/kg219.54285.14kg/s0.0841refrig hhmqmQ L &&&
(b) The net power input is determined from
& & &W W Wnet, in comp, in turb, out= −
where
( ) ( )( ) ( )[ ] ( )
( ) ( )( ) kW8.06kJ/kg219.54320.29kg/s0.08
kW16.700.80/kJ/kg285.14452.17kg/s0.08
43outturb,
1212incomp,
=−=−=
=−=η−=−=
hhmW
/hhmhhmW Cs
&&
&&&
Thus,
& . .Wnet, in = − =16 70 8 06 8.64 kW
(c) The COP of this ideal gas refrigeration cycle is determined from
COP 5.25 kW8.64 kWR
net, in= = =
&
&Q
WL 0.61
s
T
1
2QH
47°C12°C
3
4s
·
QRefrig ·
2
1 4
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PROPRIETARY MATERIAL. © 2008 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.
11-49
11-66 A gas refrigeration cycle with helium as the working fluid is considered. The minimum temperature in the cycle, the COP, and the mass flow rate of the helium are to be determined.
Assumptions 1 Steady operating conditions exist. 2 Helium is an ideal gas with constant specific heats. 3Kinetic and potential energy changes are negligible.
Properties The properties of helium are cp = 5.1926 kJ/kg·K and k = 1.667 (Table A-2).
Analysis (a) From the isentropic relations, ( )
( )( )
( )( ) K1.208
31K323
K2.4083K263
667.1/667.0k/1k
3
434
667.1/667.0k/1k
1
212
=⎟⎠⎞
⎜⎝⎛=⎟⎟
⎠⎜
⎞⎜⎝
⎛=
==⎟⎟⎠
⎜⎞
⎜⎝
⎛=
−
−
PP
TT
PP
TT
s
s
and
( ) ( )( )
( ) ( ) ( )K5.444
80.0/2632.408263/
1.20832380.0323
121212
12
12
12
min
433443
43
43
43
=−+=−+=⎯→⎯
−−
=−−
=
==−−=−−=⎯→⎯
−−
=−−
=
Csss
C
sTss
T
TTTTTTTT
hhhh
TTTTT
TTTT
hhhh
ηη
ηηK231.1
(b) The COP of this gas refrigeration cycle is determined from
( ) ( )
( ) ( )
( ) ( ) 0.356=−−−
−=
−−−−
=
−−−−
=
−==
1.2313232635.4441.231263
COP
4312
41
4312
41
outturb,incomp,innet,R
TTTTTT
hhhhhh
wwq
wq LL
(c) The mass flow rate of helium is determined from
( ) ( )( ) kg/s 0.109=−⋅
=−
=−
==K231.1263KkJ/kg5.1926
kJ/s18
41
refrig
41
refrigrefrig
TTcQ
hhQ
mpL
&&&&
s
T
1
2QH
50°C-10°C
3
4s
·
QRefrig ·
2
1 4
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11-50
11-67 A regenerative gas refrigeration cycle using air as the working fluid is considered. The effectiveness of the regenerator, the rate of heat removal from the refrigerated space, the COP of the cycle, and the refrigeration load and the COP if this system operated on the simple gas refrigeration cycle are to be determined.
Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. 3Air is an ideal gas with constant specific heats.
Properties The properties of air at room temperature are cp = 1.005 kJ/kg·K and k = 1.4 (Table A-2).
Analysis (a) From the isentropic relations,
( )
( )( ) K4.4325K2.273 4.1/4.0k/1k
1
212 ==⎟⎟
⎠⎜
⎞⎜⎝
⎛=
−
PP
TT s
K5.472
2.2732.2734.43280.0 2
2
12
12
12
12
=⎯→⎯−
−=
−−
=−−
=
TT
TTTT
hhhh ss
Cη
The temperature at state 4 can be determined by solving the following two equations simultaneously:
( ) 4.1/4.0
4
k/1k
4
545 5
1⎟⎠⎞
⎜⎝⎛=⎟⎟
⎠⎜
⎞⎜⎝
⎛=
−
TPP
TT s
ssT TT
Thhhh
54
4
54
54 2.19385.0
−−
=→−−
=η
Using EES, we obtain T4 = 281.3 K.
An energy balance on the regenerator may be written as
or,
( ) ( )
K3.2463.2812.3082.2734316
61436143
=+−=+−=
−=−⎯→⎯−=−
TTTT
TTTTTTcmTTcm pp &&
The effectiveness of the regenerator is
0.434=−−
=−−
=−−
=3.2462.3083.2812.308
63
43
63
43regen TT
TThhhh
ε
(b) The refrigeration load is
kW21.36=−=−= K)2.19346.3kJ/kg.K)(25kg/s)(1.004.0()( 56 TTcmQ pL &&
(c) The turbine and compressor powers and the COP of the cycle are
kW13.80K)2.27372.5kJ/kg.K)(45kg/s)(1.004.0()( 12inC, =−=−= TTcmW p&&
kW43.35kJ/kg)2.19381.3kJ/kg.K)(25kg/s)(1.004.0()( 54outT, =−=−= TTcmW p&&
0.478=−
=−
==43.3513.80
36.21COPoutT,inC,innet, WW
QW
Q LL&&
&
&
&
s
T
1
3
0°C
56
4
2sQH ·
QRefrig ·
35°C
Qrege
2
5-80°C
.
3
45
6
QH
Compressor
12
QL
Heat Exch.
Heat Exch.
Regenerator
Turbine
.
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11-51
(d) The simple gas refrigeration cycle analysis is as follows:
( )
( ) K6.19451K2.3081 4.1/4.0k/1k
34 =⎟⎠⎞
⎜⎝⎛=⎟
⎠⎞
⎜⎝⎛=
−
rTT s
K6.2116.1942.308
2.30885.0 4
4
43
43 =⎯→⎯−
−=⎯→⎯
−−
= TT
TTTT
sTη
kW24.74=
−=
−=
kJ/kg)6.21173.2kJ/kg.K)(25kg/s)(1.004.0(
)( 41 TTcmQ pL &&
[ ]kW32.41
kJ/kg)6.211(308.2)2.273(472.5kJ/kg.K)5kg/s)(1.004.0(
)()( 4312innet,
=−−−=
−−−= TTcmTTcmW pp &&&
0.599===32.4174.24COP
innet,WQL&
&
s
T
1
2QH
35°C 0°C
3
4s
·
QRefrig ·
2
1 4
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11-52
11-68E An ideal gas refrigeration cycle with air as the working fluid has a compression ratio of 4. The COP of the cycle is to be determined.
Assumptions 1 Steady operating conditions exist. 2 Air is an ideal gas with constant specific heats. 3Kinetic and potential energy changes are negligible.
Properties The properties of air at room temperature are cp = 0.240 Btu/lbm·R and k = 1.4 (Table A-2Ea).
Analysis From the isentropic relations,
R8.37641R)560(
R7.668R)(4)450(
4.1/4.0/)1(
3
434
4.1/4.0/)1(
1
212
=⎟⎠⎞
⎜⎝⎛=⎟⎟
⎠⎜
⎞⎜⎝
⎛=
==⎟⎟⎠
⎜⎞
⎜⎝
⎛=
−
−
kk
kk
PP
TT
PP
TT
The COP of this ideal gas refrigeration cycle is determined from
2.06=−−−
−=
−−−−
=
−−−−
=
−==
)8.376560()4507.668(8.376450)()(
)()(
COP
4312
41
4312
41
outturb,incomp,innet,R
TTTTTT
hhhhhh
wwq
wq LL
s
T
1
2 QH
100°F-10°F
3
4
·
QRefrig ·
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11-53
11-69E An gas refrigeration cycle with air as the working fluid has a compression ratio of 4. The COP of the cycle is to be determined.
Assumptions 1 Steady operating conditions exist. 2 Air is an ideal gas with constant specific heats. 3Kinetic and potential energy changes are negligible.
Properties The properties of air at room temperature are cp = 0.240 Btu/lbm·R and k = 1.4 (Table A-2Ea).
Analysis From the isentropic relations,
R9.402psia19psia6R)560(
R7.668R)(4)450(
4.1/4.0/)1(
3
434
4.1/4.0/)1(
1
212
=⎟⎟⎠
⎞⎜⎜⎝
⎛=⎟⎟
⎠⎜
⎞⎜⎝
⎛=
==⎟⎟⎠
⎜⎞
⎜⎝
⎛=
−
−
kk
s
kk
s
PP
TT
PP
TT
and
R1.470)87.0/()4507.668(450/)(
R3.412)9.402560)(94.0(560)(
121212
12
12
12
433443
43
43
43
=−+=−+=⎯→⎯
−−
=−−
=
=−−=−−=⎯→⎯
−−
=−−
=
Csss
C
sTss
T
TTTTTTTT
hhhh
TTTTTTTT
hhhh
ηη
ηη
The COP of this gas refrigeration cycle is determined from
0.364=−−−
−=
−−−−
=
−−−−
=
−==
)3.412560()4504.701(3.412450)()(
)()(
COP
4312
41
4312
41
outturb,incomp,innet,R
TTTTTT
hhhhhh
wwq
wq LL
s
T
1
2s QH
100°F -10°F
3
4s
·
QRefrig ·
2
1 4
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11-54
11-70 An ideal gas refrigeration cycle with air as the working fluid provides 15 kW of cooling. The mass flow rate of air and the rates of heat addition and rejection are to be determined.
Assumptions 1 Steady operating conditions exist. 2 Air is an ideal gas with constant specific heats. 3Kinetic and potential energy changes are negligible.
Properties The properties of air at room temperature are cp = 1.005 kJ/kg·K and k = 1.4 (Table A-2a).
Analysis From the isentropic relations,
K3.191kPa500kPa100K)303(
K1.464kPa100kPa500K)293(
4.1/4.0/)1(
3
434
4.1/4.0/)1(
1
212
=⎟⎠⎞
⎜⎝⎛=⎟⎟
⎠⎜
⎞⎜⎝
⎛=
=⎟⎠⎞
⎜⎝⎛=⎟⎟
⎠⎜
⎞⎜⎝
⎛=
−
−
kk
kk
PP
TT
PP
TT
The mass flow rate of the air is determined from
kg/s0.1468=−⋅
=−
=⎯→⎯−=K)3.191K)(293kJ/kg(1.005
kJ/s15)(
)(41
Refrig41Refrig TTc
QmTTcmQ
pp
&&&&
The rate of heat addition to the cycle is the same as the rate of cooling,
kW15== Refrigin QQ &&
The rate of heat rejection from the cycle is
kW23.8=−⋅=−= K)303K)(464.1kJ/kg5kg/s)(1.001468.0()( 32 TTcmQ pH &&
s
T
1
2 QH
30°C20°C
3
4
·
QRefrig ·
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11-55
11-71 An ideal gas refrigeration cycle with air as the working fluid is considered. The minimum pressure ratio for this system to operate properly is to be determined.
Assumptions 1 Steady operating conditions exist. 2 Air is an ideal gas with constant specific heats. 3Kinetic and potential energy changes are negligible.
Properties The properties of air at room temperature are cp = 1.005 kJ/kg·K and k = 1.4 (Table A-2a).
Analysis An energy balance on process 4-1 gives
K1.248
KkJ/kg1.005kJ/kg20
K268
)(
Refrig14
41Refrig
=⋅
−=−=
−=
p
p
cq
TT
TTcq
The minimum temperature at the turbine inlet would be the same as that to which the heat is rejected. That is,
K2933 =T
Then the minimum pressure ratio is determined from the isentropic relation to be
1.79=⎟⎠⎞
⎜⎝⎛=⎟⎟
⎠⎜
⎞⎜⎝
⎛=
− 4.0/4.1)1/(
4
3
4
3
K248.1K293
kk
TT
PP
s
T
1
2 QH
20°C -5°C
3
4
·
QRefrig ·
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11-56
11-72 An ideal gas refrigeration cycle with with two stages of compression with intercooling using air as the working fluid is considered. The COP of this system 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. 3Kinetic and potential energy changes are negligible.
Properties The properties of air at room temperature are cp = 1.005 kJ/kg·K and k = 1.4 (Table A-2a).
Analysis From the isentropic relations,
K2.128161K)283(
K5.420K)(4)283(
K9.378K)(4)255(
4.1/4.0/)1(
5
656
4.1/4.0/)1(
3
434
4.1/4.0/)1(
1
212
=⎟⎠⎞
⎜⎝⎛=⎟⎟
⎠⎜
⎞⎜⎝
⎛=
==⎟⎟⎠
⎜⎞
⎜⎝
⎛=
==⎟⎟⎠
⎜⎞
⎜⎝
⎛=
−
−
−
kk
kk
kk
PP
TT
PP
TT
PP
TT
The COP of this ideal gas refrigeration cycle is determined from
1.19=−−−+−
−=
−−−+−−
=
−−−+−−
=
−==
)2.128283()2835.420()2559.378(2.128255
)()()(
)()()(
COP
653412
61
653412
61
outturb,incomp,innet,R
TTTTTTTT
hhhhhhhh
wwq
wq LL
The mass flow rate of the air is determined from
kg/s0.163=−⋅
=−
=⎯→⎯−=K)2.128K)(255kJ/kg(1.005
kJ/s)3600/000,75()(
)(61
Refrig61Refrig TTc
QmTTcmQ
pp
&&&&
10°C
s
T
1
2
3
4
5
6
-18°C
3
4 5
6
. Q
1
2
Turbine
Q .
Heat exch.
Heat exch.
. Q
Heat exch.
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11-57
11-73 A gas refrigeration cycle with with two stages of compression with intercooling using air as the working fluid is considered. The COP of this system 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. Properties The properties of air at room temperature are cp = 1.005 kJ/kg·K and k = 1.4 (Table A-2a).
Analysis From the isentropic relations,
K2.128161K)283(
K5.420K)(4)283(
K9.378K)(4)255(
4.1/4.0/)1(
5
656
4.1/4.0/)1(
3
434
4.1/4.0/)1(
1
212
=⎟⎠⎞
⎜⎝⎛=⎟⎟
⎠⎜
⎞⎜⎝
⎛=
==⎟⎟⎠
⎜⎞
⎜⎝
⎛=
==⎟⎟⎠
⎜⎞
⎜⎝
⎛=
−
−
−
kk
s
kk
s
kk
s
PP
TT
PP
TT
PP
TT
and
K9.135)2.128283)(95.0(283)(
K8.44485.0/)2835.420(283/)(
K8.40085.0/)2559.378(255/)(
655665
65
65
65
343434
34
34
34
121212
12
12
12
=−−=−−=⎯→⎯−−
=−−
=
=−+=−+=⎯→⎯−−
=−−
=
=−+=−+=⎯→⎯−−
=−−
=
sTss
T
Csss
C
Csss
C
TTTTTTTT
hhhh
TTTTTTTT
hhhh
TTTTTTTT
hhhh
ηη
ηη
ηη
The COP of this ideal gas refrigeration cycle is determined from
0.742=−−−+−
−=
−−−+−−
=
−−−+−−
=
−==
)9.135283()2838.444()2558.400(9.135255
)()()(
)()()(
COP
653412
61
653412
61
outturb,incomp,innet,R
TTTTTTTT
hhhhhhhh
wwq
wq LL
The mass flow rate of the air is determined from
kg/s0.174=−⋅
=−
=⎯→⎯−=K)9.135K)(255kJ/kg(1.005
kJ/s)3600/000,75()(
)(61
Refrig61Refrig TTc
QmTTcmQ
pp
&&&&
10°C
s
T
1
2
3
4
5
6
-18°C
2s4s
6s
3
4 5
6
. Q
1
2
Turbine
Q .
Heat exch.
Heat exch.
. Q
Heat exch.
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11-58
Absorption Refrigeration Systems
11-74C Absorption refrigeration is the kind of refrigeration that involves the absorption of the refrigerant during part of the cycle. In absorption refrigeration cycles, the refrigerant is compressed in the liquid phase instead of in the vapor form.
11-75C The main advantage of absorption refrigeration is its being economical in the presence of an inexpensive heat source. Its disadvantages include being expensive, complex, and requiring an external heat source.
11-76C In absorption refrigeration, water can be used as the refrigerant in air conditioning applications since the temperature of water never needs to fall below the freezing point.
11-77C The fluid in the absorber is cooled to maximize the refrigerant content of the liquid; the fluid in the generator is heated to maximize the refrigerant content of the vapor.
11-78C The coefficient of performance of absorption refrigeration systems is defined as
geninpump,genR inputrequired
outputdesiredCOPQQ
WQQ LL ≅
+==
11-79C The rectifier separates the water from NH3 and returns it to the generator. The regenerator transfers some heat from the water-rich solution leaving the generator to the NH3-rich solution leaving the pump.
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11-59
11-80 The COP of an absorption refrigeration system that operates at specified conditions is given. It is to be determined whether the given COP value is possible.
Analysis The maximum COP that this refrigeration system can have is
142268300
268 K403 K30011COP
0
0maxR, .
TTT
TT
L
L
s=⎟
⎠⎞
⎜⎝⎛
−⎟⎠⎞
⎜⎝⎛ −⎟ =⎟
⎠
⎞⎜⎜⎝
⎟⎛
−⎟⎠
⎞⎜⎜⎝
⎛−=
which is slightly greater than 2. Thus the claim is possible, but not probable.
11-81 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
2.64=⎟⎠⎞
⎜⎝⎛
−⎟⎠⎞
⎜⎝⎛ −⎟ =⎟
⎠
⎞⎜⎜⎝
⎟⎛
−⎟⎠
⎞⎜⎜⎝
⎛−=
273298273
K393K29811COP
0
0maxR,
L
L
s TTT
TT
11-82 The conditions at which an absorption refrigeration system operates are specified. The maximum rate at which this system can remove heat from the refrigerated space is to be determined.
Analysis The maximum COP that this refrigeration system can have is
151243298
243K403K29811COP
0
0maxR, .
TTT
TT
L
L
s=⎟
⎠⎞
⎜⎝⎛
−⎟⎟⎠
⎞⎜⎜⎝
⎛−⎟ =⎟
⎠
⎞⎜⎜⎝
⎟⎛
−⎟⎠
⎞⎜⎜⎝
⎛−=
Thus,
( )( ) kJ/h105.75 5×=×== kJ/h10515.1COP 5genmaxR,maxL, QQ &&
11-83E The conditions at which an absorption refrigeration system operates are specified. The COP is also given. The maximum rate at which this system can remove heat from the refrigerated space is to be determined.
Analysis For a COP = 0.55, the rate at which this system can remove heat from the refrigerated space is
( )( ) Btu/h100.55 5×=== Btu/h10550COP 5genR .QQL&&
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11-60
11-84 A reversible absorption refrigerator consists of a reversible heat engine and a reversible refrigerator. The rate at which the steam condenses, the power input to the reversible refrigerator, and the second law efficiency of an actual chiller are to be determined.
Properties The enthalpy of vaporization of water at 200°C is hfg = 1939.8 kJ/kg (Table A-4).
Analysis (a) The thermal efficiency of the reversible heat engine is
370.0K)15.273200(
K)15.27325(11 0revth, =
++
−=−=sT
Tη
The COP of the reversible refrigerator is
52.7K)15.27310()15.27325(
K)15.27310(COP0
revR, =+−−+
+−=
−=
L
L
TTT
The COP of the reversible absorption refrigerator is
78.2)52.7)(370.0(COPCOP revR,revth,revabs, ===η
The heat input to the reversible heat engine is
kW911.72.78
kW22COP revabs,
in === LQQ
&&
Then, the rate at which the steam condenses becomes
kg/s0.00408===kJ/kg1939.8kJ/s911.7in
fgs h
Qm
&&
(b) The power input to the refrigerator is equal to the power output from the heat engine
kW2.93==== )kW191.7)(370.0(inrevth,HEout,Rin, QWW &&& η
(c) The second-law efficiency of an actual absorption chiller with a COP of 0.7 is
0.252===2.78
7.0COPCOP
revabs,
actualIIη
Rev. HE
T0
Ts
Rev.Ref.
TL
T0
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11-61
Special Topic: Thermoelectric Power Generation and Refrigeration Systems
11-85C The circuit that incorporates both thermal and electrical effects is called a thermoelectric circuit.
11-86C When two wires made from different metals joined at both ends (junctions) forming a closed circuit and one of the joints is heated, a current flows continuously in the circuit. This is called the Seebeck effect. When a small current is passed through the junction of two dissimilar wires, the junction is cooled. This is called the Peltier effect.
11-87C No.
11-88C No.
11-89C Yes.
11-90C When a thermoelectric circuit is broken, the current will cease to flow, and we can measure the voltage generated in the circuit by a voltmeter. The voltage generated is a function of the temperature difference, and the temperature can be measured by simply measuring voltages.
11-91C The performance of thermoelectric refrigerators improves considerably when semiconductors are used instead of metals.
11-92C The efficiency of a thermoelectric generator is limited by the Carnot efficiency because a thermoelectric generator fits into the definition of a heat engine with electrons serving as the working fluid.
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11-62
11-93E A thermoelectric generator that operates at specified conditions is considered. The maximum thermal efficiency this thermoelectric generator can have is to be determined.
Analysis The maximum thermal efficiency of this thermoelectric generator is the Carnot efficiency,
%3.31=−=−==R800R550
11Carnotth,maxth,H
L
TT
ηη
11-94 A thermoelectric refrigerator that operates at specified conditions is considered. The maximum COP this thermoelectric refrigerator can have and the minimum required power input are to be determined.
Analysis The maximum COP of this thermoelectric refrigerator is the COP of a Carnot refrigerator operating between the same temperature limits,
Thus, ( ) ( ) ( )
W12.1
10.72
===
=−
=−
==
72.10 W130
COP
1K268/K2931
1/1COPCOP
maxminin,
CarnotR,max
L
LH
QW
TT
&&
11-95 A thermoelectric cooler that operates at specified conditions with a given COP is considered. The required power input to the thermoelectric cooler is to be determined.
Analysis The required power input is determined from the definition of COPR,
COPCOP
180 W0.15R
inin
R= ⎯ →⎯ = = =
&
&&
&QW
W QL L 1200 W
11-96E A thermoelectric cooler that operates at specified conditions with a given COP is considered. The required power input to the thermoelectric cooler is to be determined.
Analysis The required power input is determined from the definition of COPR,
hp3.14=Btu/min33.310.15Btu/min20
COPCOP
Rin
inR ===⎯→⎯= LL Q
WWQ &
&&
&
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11-63
11-97 A thermoelectric refrigerator powered by a car battery cools 9 canned drinks in 12 h. The average COP of this refrigerator is to be determined.
Assumptions Heat transfer through the walls of the refrigerator is negligible.
Properties The properties of canned drinks are the same as those of water at room temperature, ρ = 1 kg/L and cp = 4.18 kJ/kg·°C (Table A-3).
Analysis The cooling rate of the refrigerator is simply the rate of decrease of the energy of the canned drinks,
W6.71=kW00671.0s 360012
kJ290
kJ290=C3)-C)(25kJ/kgkg)(4.1815.3(kg3.15=L)0kg/L)(0.351(9
coolingcooling
cooling
=×
=Δ
=
°°⋅=Δ=×==
tQ
Q
TmcQm
&
Vρ
The electric power consumed by the refrigerator is
W36=A)V)(312(in == IW V&
Then the COP of the refrigerator becomes
COP W36 W
cooling
in= = = ≈&
&. .
Q
W
6 71 0 200.186
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11-64
11-98E A thermoelectric cooler is said to cool a 12-oz drink or to heat a cup of coffee in about 15 min. The average rate of heat removal from the drink, the average rate of heat supply to the coffee, and the electric power drawn from the battery of the car are to be determined.
Assumptions Heat transfer through the walls of the refrigerator is negligible.
Properties The properties of canned drinks are the same as those of water at room temperature, cp = 1.0 Btu/lbm.°F (Table A-3E).
Analysis (a) The average cooling rate of the refrigerator is simply the rate of decrease of the energy content of the canned drinks,
W36.2=⎟⎠⎞
⎜⎝⎛
×=
Δ=
°°⋅=Δ=
Btu1J 1055
s 6015Btu84.30
Btu30.84=F38)-F)(78Btu/lbm lbm)(1.0 771.0(
coolingcooling
cooling
tQ
Q
TmcQ p
&
(b) The average heating rate of the refrigerator is simply the rate of increase of the energy content of the canned drinks,
W49.7=⎟⎠⎞
⎜⎝⎛
×=
Δ=
°°⋅=Δ=
Btu1J 1055
s 6015Btu4.42
Btu42.4=F75)-F)(130Btu/lbm lbm)(1.0771.0(
heatingheating
heating
tQ
Q
TmcQ p
&
(c) The electric power drawn from the car battery during cooling and heating is
W441
W181
.1.2
W7.49COP
2.112.01COPCOP
0.2 W2.36
COP
heating
heatingheatingin,
coolingheating
cooling
coolingcoolingin,
===
=+=+=
===
QW
QW
&&
&&
11-99 The maximum power a thermoelectric generator can produce is to be determined.
Analysis The maximum thermal efficiency this thermoelectric generator can have is
1420K353K30311maxth, .
TT
H
L =−=−=η
Thus,
kW39.4==== kJ/h 142,000kJ/h)(0.142)(106inmaxth,maxout, QW && η
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11-65
Review Problems
11-100 A steady-flow Carnot refrigeration cycle with refrigerant-134a as the working fluid is considered. The COP, the condenser and evaporator pressures, 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) The COP of this refrigeration cycle is determined from
( ) ( ) ( ) 5.06=−
=−
=1K253/K303
11/
1COP CR,LH TT
(b) The condenser and evaporative pressures are (Table A-11)
kPa770.64kPa132.82
====
°
°−
C03@satcond
C20@satevap
PPPP
(c) The net work input is determined from
( ) ( )( )( ) ( )( ) kJ/kg82.19591.21280.049.25
kJ/kg43.5791.21215.049.25
C20@22
C20@11
=+=+=
=+=+=
°−
°−
fgf
fgf
hxhhhxhh
kJ/kg27.35===
=−=−=
06.5kJ/kg138.4
COP
kJ/kg4.13843.5782.195
Rinnet,
12
L
Lq
w
hhq
s
T
qL1 2
3 430°C
-20°C
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11-66
11-101 A heat pump that operates on the ideal vapor-compression cycle with refrigerant-134a as the working fluid is used to heat a house. The rate of heat supply to the house, the volume flow rate of the refrigerant at the compressor inlet, and the COP of this heat pump 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),
( )throttling kJ/kg61.101
kJ/kg61.101liquidsat.
MPa9.0
kJ/kg75.275MPa9.0
/kgm099867.0KkJ/kg93773.0
kJ/kg46.244
vaporsat.kPa200
34
MPa0.9@33
212
2
3kPa200@1
kPa200@1
kPa200@11
=≅
==⎭⎬⎫=
=⎭⎬⎫
==
==
⋅====
⎭⎬⎫=
hh
hhP
hss
P
sshh
P
f
g
g
g
vv
The rate of heat supply to the house is determined from
( ) ( )( ) kW55.73=−=−= kJ/kg.61101275.75kg/s0.3232 hhmQH &&
(b) The volume flow rate of the refrigerant at the compressor inlet is
( )( ) /sm 0.0320 3=== /kgm0.099867kg/s0.32 311 vV m&&
(c) The COP of t his heat pump is determined from
5.57=−−
=−−
==46.24475.27561.10175.275COP
12
32
inR hh
hhwqL
11-102 A relation for the COP of the two-stage refrigeration system with a flash chamber shown in Fig. 11-12 is to be derived.
Analysis The coefficient of performance is determined from
COPRin
=q
wL
where
( )( )
( )( ) ( )( )94126incompII,incompI,in
66816
11
with1
hhhhxwww
hhh
xhhxqfg
fL
−+−−=+=
−=−−=
QH
QL
200 kPa 1
2 3
4
0.9 MPa
s
T
·
Win ·
·
House
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11-67
11-103 A two-stage compression refrigeration system using refrigerant-134a as the working fluid is considered. The fraction of the refrigerant that evaporates as it is throttled to the flash chamber, the amount of heat removed from the refrigerated space, the compressor work, and the COP are to be determined.
Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. 3The flashing chamber is adiabatic.
Analysis (a) The enthalpies of the refrigerant at several states are determined from the refrigerant tables to be (Tables A-11, A-12, and A-13)
kJ/kg94.63kJ/kg47.95
kJ/kg58.260
,kJ/kg94.63,kJ/kg47.95
,kJ/kg55.255,kJ/kg16.239
8
6
2
7
5
3
1
=
=
=
=
===
hh
h
hhhh
The fraction of the refrigerant that evaporates as it is throttled to the flash chamber is simply the quality at state 6,
0.1646=−
=−
=62.191
94.6347.9566
fg
f
hhh
x
(b) The enthalpy at state 9 is determined from an energy balance on the mixing chamber:
( ) ( )( )( ) ( )( ) kJ/kg75.25958.2601646.0155.2551646.0
11
0
9
26369
outin(steady)0
systemoutin
=−+=−+=
=
=→==−
∑∑h
hxhxh
hmhm
EEEEE
iiee &&
&&&&& Δ
KkJ/kg94168.0kJ/kg75.259
MPa4.09
9
9 ⋅=⎭⎬⎫
==
shP
Also,
kJ/kg47.274KkJ/kg94168.0
MPa8.04
94
4 =⎭⎬⎫
⋅===
hss
P
Then the amount of heat removed from the refrigerated space and the compressor work input per unit mass of refrigerant flowing through the condenser are
( )( ) ( )( )
( )( ) ( )( )( )( ) ( ) kJ/kg32.6
kJ/kg146.4
=−+−−=−+−−=+=
=−−=−−=
kJ/kg259.75274.47kJ/kg239.16260.580.1646111
kJ/kg63.94239.161646.011
94126incompII,incompI,in
816
hhhhxwww
hhxqL
(c) The coefficient of performance is determined from
4.49===kJ/kg32.6kJ/kg146.4COP
inR w
qL
qL
0.14 MPa
1
2 5
8
0.4 MPa
s
T
40.8 MPa
6 9A
B 37
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11-68
11-104 A refrigerator operating on a vapor-compression refrigeration cycle with refrigerant-134a as the working fluid is considered. The process with the greatest exergy loss is to be determined.
Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.
Analysis In this cycle, the refrigerant enters the compressor as a saturated vapor at the evaporator pressure, and leaves the condenser as saturated liquid at the condenser pressure. The compression process is not isentropic. From the refrigerant tables (Tables A-11, A-12, and A-13),
KkJ/kg34605.0 kJ/kg82.88
C10
)throttling( kJ/kg82.88
KkJ/kg33230.0kJ/kg82.88
liquidsat.kPa700
kJ/kg38.270 kPa700
KkJ/kg93766.0kJ/kg51.244
vapor sat.
C10
44
4
34
kPa700@3
kPa700@33
212
2
C10@1
C10@11
⋅=⎭⎬⎫
=°−=
=≅
⋅====
⎭⎬⎫=
=⎭⎬⎫
==
⋅====
⎭⎬⎫°−=
°−
°−
shT
hh
sshhP
hss
P
sshhT
f
f
s
g
g
The actual enthalpy at the compressor exit is determined by using the compressor efficiency:
kJ/kg95.27485.0
51.24438.27051.244C
1212
12
12C =
−+=
−+=⎯→⎯
−−
=η
ηhh
hhhhhh ss
and kJ/kg95252.0 kJ/kg95.274
kPa7002
2
2 =⎭⎬⎫
==
shP
The heat added in the evaporator and that rejected in the condenser are
kJ/kg186.13kJ/kg)82.8895.274(kJ/kg155.69kJ/kg)82.88244.51(
32
41
=−=−==−=−=
hhqhhq
H
L
⎟
The exergy destruction during a process of a stream from an inlet state to exit state is given by
⎟⎠
⎞⎜⎜⎝
⎛+−−==
sink
out
source
in0gen0dest T
qT
qssTsTx ie
Application of this equation for each process of the cycle gives
kJ/kg6.29=⎟⎠
⎞⎜⎝
⎛ −−⎟ =⎟⎠
⎞⎜⎜⎝
⎛−−=
=⋅−=−=
=⎟⎠
⎞⎜⎝
⎛ +−⎟ =⎟⎠
⎞⎜⎜⎝
⎛+−=
=⋅−=−=
K273kJ/kg69.155
34605.093766.0)K295(
kJ/kg06.4KkJ/kg)33230.034605.0)(K295()(
kJ/kg17.3K295kJ/kg13.186
95252.033230.0)K295(
kJ/kg38.4KkJ/kg)93766.095252.0)(K295()(
41041destroyed,
34034destroyed,
23023destroyed,
12012destroyed,
L
L
H
H
Tq
ssTx
ssTxTq
ssTx
ssTx
The greatest exergy destruction occurs in the evaporator. Note that heat is absorbed from freezing water at 0°C (273 K) and rejected to the ambient air at 22°C (295 K), which is also taken as the dead state temperature. Alternatively, one may use the standard 25°C (298 K) as the dead state temperature, and perform the calculations accordingly.
QH
QL
-10°C 1
2s
3
4
0.7 MPa
s
T
·
Win ·
·
4s
2
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11-69
11-105 A refrigerator operating on a vapor-compression refrigeration cycle with refrigerant-134a as the working fluid is considered. The process with the greatest exergy loss is to be determined.
Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.
Analysis From the refrigerant tables (Tables A-11, A-12, and A-13),
KkJ/kg33146.0 kJ/kg98.84
C10
)throttling( kJ/kg98.84
KkJ/kg31958.0kJ/kg98.84
C247.27.26
7.2kPa700
kJ/kg38.270 kPa700
KkJ/kg93766.0kJ/kg51.244
vapor sat.
C10
44
4
34
C24@3
C24@3kPa700@sat 3
3
212
2
C10@1
C10@11
⋅=⎭⎬⎫
=°−=
=≅
⋅=≅=≅
⎪⎭
⎪⎬
⎫
°=−=−=
=
=⎭⎬⎫
==
⋅====
⎭⎬⎫°−=
°
°
°−
°−
shT
hh
sshh
TTP
hss
P
sshhT
f
f
s
g
g
The actual enthalpy at the compressor exit is determined by using the compressor efficiency:
kJ/kg95.27485.0
51.24438.27051.244C
1212
12
12C =
−+=
−+=⎯→⎯
−−
=η
ηhh
hhhhhh ss
and kJ/kg95252.0 kJ/kg95.274
kPa7002
2
2 =⎭⎬⎫
==
shP
The heat added in the evaporator and that rejected in the condenser are
kJ/kg189.97kJ/kg)98.8495.274(kJ/kg159.53kJ/kg)98.84244.51(
32
41
=−=−==−=−=
hhqhhq
H
L
⎟
The exergy destruction during a process of a stream from an inlet state to exit state is given by
⎟⎠
⎞⎜⎜⎝
⎛+−−==
sink
out
source
in0gen0dest T
qT
qssTsTx ie
Application of this equation for each process of the cycle gives
kJ/kg6.44=⎟⎠
⎞⎜⎝
⎛ −−⎟ =⎟⎠
⎞⎜⎜⎝
⎛−−=
=⋅−=−=
=⎟⎠
⎞⎜⎝
⎛ +−⎟ =⎟⎠
⎞⎜⎜⎝
⎛+−=
=⋅−=−=
K273kJ/kg53.159
33146.093766.0)K295(
kJ/kg50.3KkJ/kg)31958.033146.0)(K295()(
kJ/kg25.3K295kJ/kg97.189
95252.031958.0)K295(
kJ/kg38.4KkJ/kg)93766.095252.0)(K295()(
41041destroyed,
34034destroyed,
23023destroyed,
12012destroyed,
L
L
H
H
Tq
ssTx
ssTxTq
ssTx
ssTx
The greatest exergy destruction occurs in the evaporator. Note that heat is absorbed from freezing water at 0°C (273 K) and rejected to the ambient air at 22°C (295 K), which is also taken as the dead state temperature. Alternatively, one may use the standard 25°C (298 K) as the dead state temperature, and perform the calculations accordingly.
QH
QL
-10°C 1
2s
3
4
0.7 MPa
s
T
·
Win ·
· 2
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11-70
11-106 A refrigerator operating on a vapor-compression refrigeration cycle with refrigerant-134a as the working fluid is considered. The process with the greatest exergy loss is to be determined.
Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.
Analysis In this cycle, the refrigerant leaves the condenser as saturated liquid at the condenser pressure. The compression process is not isentropic. From the refrigerant tables (Tables A-11, A-12, and A-13),
KkJ/kg4988.05089.0
kJ/kg77.117C37
)throttling( kJ/kg77.117
KkJ/kg42441.0kJ/kg77.117
liquidsat.
MPa2.1
kJ/kg11.298 MPa2.1
KkJ/kg9865.0kJ/kg09.233
C30737kPa60
4
4
4
4
34
MPa1.2@3
MPa1.2@33
212
2
1
1
1
C37@sat 1
⋅==
⎭⎬⎫
=°−=
=≅
⋅====
⎭⎬⎫=
=⎭⎬⎫
==
⋅==
⎭⎬⎫
°−=+−=== °−
sx
hT
hh
sshhP
hss
P
sh
TPP
f
f
s
The actual enthalpy at the compressor exit is determined by using the compressor efficiency:
kJ/kg33.30590.0
09.23311.29809.233C
1212
12
12C =
−+=
−+=⎯→⎯
−−
=η
ηhh
hhhhhh ss
and KkJ/kg0075.1 Btu/lbm33.305
MPa2.12
2
2 ⋅=⎭⎬⎫
==
shP
The heat added in the evaporator and that rejected in the condenser are
kJ/kg56.187kJ/kg)77.11733.305(kJ/kg32.115kJ/kg)77.11709.233(
32
41
=−=−==−=−=
hhqhhq
H
L
⎟
The exergy destruction during a process of a stream from an inlet state to exit state is given by
⎟⎠
⎞⎜⎜⎝
⎛+−−==
sink
out
source
in0gen0dest T
qT
qssTsTx ie
Application of this equation for each process of the cycle gives
kJ/kg57.1K273)34(
kJ/kg32.1154988.09865.0)K303(
KkJ/kg)42441.04988.0)(K303()(
kJ/kg88.10K303kJ/kg56.187
0075.142441.0)K303(
kJ/kg36.6KkJ/kg)9865.00075.1)(K303()(
41041destroyed,
34034destroyed,
23023destroyed,
12012destroyed,
⎟ =⎟⎠
⎞⎜⎜⎝
⎛+−
−−⎟ =⎟⎠
⎞⎜⎜⎝
⎛−−=
=⋅−=−=
=⎟⎠
⎞⎜⎝
⎛ +−⎟ =⎟⎠
⎞⎜⎜⎝
⎛+−=
=⋅−=−=
L
L
H
H
Tq
ssTx
ssTxTq
ssTx
ssTx
kJ/kg22.54
The greatest exergy destruction occurs in the expansion valve. Note that heat is absorbed from fruits at -34°C (239 K) and rejected to the ambient air at 30°C (303 K), which is also taken as the dead state temperature. Alternatively, one may use the standard 25°C (298 K) as the dead state temperature, and perform the calculations accordingly.
QH
QL
-37°C 1
2s
3
4
1.2 MPa
s
T
·
Win ·
·
4s
2
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PROPRIETARY MATERIAL. © 2008 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.
11-71
11-107 A refrigerator operating on a vapor-compression refrigeration cycle with refrigerant-134a as the working fluid is considered. The process with the greatest exergy loss is to be determined.
Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.
Analysis From the refrigerant tables (Tables A-11, A-12, and A-13),
KkJ/kg4585.04665.0
kJ/kg26.108C37
)throttling( kJ/kg26.108
KkJ/kg39486.0kJ/kg26.108
C403.63.46
3.6MPa2.1
kJ/kg11.298 MPa2.1
KkJ/kg9865.0kJ/kg09.233
C30737kPa60
4
4
4
4
34
C40@3
[email protected]@sat 3
3
212
2
1
1
1
C37@sat 1
⋅==
⎭⎬⎫
=°−=
=≅
⋅=≅=≅
⎪⎭
⎪⎬
⎫
°=−=−=
=
=⎭⎬⎫
==
⋅==
⎭⎬⎫
°−=+−===
°
°
°−
sx
hT
hh
sshh
TTP
hss
P
sh
TPP
f
f
s
The actual enthalpy at the compressor exit is determined by using the compressor efficiency:
kJ/kg33.30590.0
09.23311.29809.233C
1212
12
12C =
−+=
−+=⎯→⎯
−−
=η
ηhh
hhhhhh ss
and KkJ/kg0075.1 Btu/lbm33.305
MPa2.12
2
2 ⋅=⎭⎬⎫
==
shP
The heat added in the evaporator and that rejected in the condenser are
kJ/kg07.197kJ/kg)26.10833.305(kJ/kg83.124kJ/kg)26.10809.233(
32
41
=−=−==−=−=
hhqhhq
H
L
⎟
The exergy destruction during a process of a stream from an inlet state to exit state is given by
⎟⎠
⎞⎜⎜⎝
⎛+−−==
sink
out
source
in0gen0dest T
qT
qssTsTx ie
Application of this equation for each process of the cycle gives
kJ/kg73.1K273)34(
kJ/kg83.1244585.09865.0)K303(
KkJ/kg)39486.04585.0)(K303()(
kJ/kg44.11K303kJ/kg07.197
0075.139486.0)K303(
kJ/kg36.6KkJ/kg)9865.00075.1)(K303()(
41041destroyed,
34034destroyed,
23023destroyed,
12012destroyed,
⎟ =⎟⎠
⎞⎜⎜⎝
⎛+−
−−⎟ =⎟⎠
⎞⎜⎜⎝
⎛−−=
=⋅−=−=
=⎟⎠
⎞⎜⎝
⎛ +−⎟ =⎟⎠
⎞⎜⎜⎝
⎛+−=
=⋅−=−=
L
L
H
H
Tq
ssTx
ssTxTq
ssTx
ssTx
kJ/kg19.28
The greatest exergy destruction occurs in the expansion valve. Note that heat is absorbed from fruits at -34°C (239 K) and rejected to the ambient air at 30°C (303 K), which is also taken as the dead state temperature. Alternatively, one may use the standard 25°C (298 K) as the dead state temperature, and perform the calculations accordingly.
QH
QL
-37°C 1
2s
3
4
1.2 MPa
s
T
·
Win ·
· 2
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PROPRIETARY MATERIAL. © 2008 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.
11-72
11-108E A two-evaporator compression refrigeration cycle with refrigerant-134a as the working fluid is considered. The cooling load of both evaporators per unit of flow through the compressor and the COP of the system are to be determined.
Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.
Analysis From the refrigerant tables (Tables A-11E, A-12E, and A-13E),
Btu/lbm68.98 vapor sat.
F5.29
Btu/lbm40.107 vapor sat.
F30
)throttling( Btu/lbm519.48
Btu/lbm519.48 liquidsat.psia160
F5.29@77
F30@55
364
psia160@33
==⎭⎬⎫°−=
==⎭⎬⎫°=
=≅=
==⎭⎬⎫=
°−
°
g
g
f
hhT
hhT
hhh
hhP
For a unit mass flowing through the compressor, the fraction of mass flowing through Evaporator II is denoted by x and that through Evaporator I is y (y = 1-x). From the cooling loads specification,
)(2)(
2
6745
2 vape,1 vape,
hhyhhx
QQ LL
−=−
= &&
where
yx −= 1
Combining these results and solving for y gives
3698.0)519.4840.107()519.4868.98(2
519.4840.107)()(2 4567
45 =−+−
−=
−+−−
=hhhh
hhy
Then,
6302.03698.011 =−=−= yx
Applying an energy balance to the point in the system where the two evaporator streams are recombined gives
3
45
6
Compressor
Expansion valve1
2
Evaporator 27
Evaporator 1
Condenser
Expansion valve
Expansion valve
QH
QL
-29.5°F 1
2
3
6
160 psia
s
T
·
Win·
·
45
7
30°F
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11-73
Btu/lbm18.1041
)68.98)(3698.0()40.107)(6302.0(1
751175 =
+=
+=⎯→⎯=+
yhxhhhyhxh
Then,
Btu/lbm14.131 psia160
RBtu/lbm2418.0 Btu/lbm18.104
psia10
212
2
11
F29.5@sat 1
=⎭⎬⎫
==
⋅=⎭⎬⎫
=≅= °−
hss
P
sh
PP
The cooling load of both evaporators per unit mass through the compressor is
Btu/lbm55.66=−+−=
−+−=Btu/lbm)519.48(98.68)3698.0(Btu/lbm)519.48(107.40)6302.0(
)()( 6745 hhyhhxqL
The work input to the compressor is
Btu/lbm26.96Btu/lbm)18.104(131.1412in =−=−= hhw
2.06
The COP of this refrigeration system is determined from its definition,
===Btu/lbm26.96Btu/lbm55.66COP
inR w
qL
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11-74
11-109E A two-evaporator compression refrigeration cycle with refrigerant-134a as the working fluid is considered. The process with the greatest exergy destruction is to be determined.
Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.
Analysis From Prob. 11-107E and the refrigerant tables (Tables A-11E, A-12E, and A-13E),
Btu/lbm621.82Btu/lbm161.50Btu/lbm881.58
3698.016302.0
RBtu/lbm22948.0RBtu/lbm11286.0RBtu/lbm22260.0RBtu/lbm10238.0RBtu/lbm09774.0
RBtu/lbm2418.0
6767,
4545,
7
6
5
4
3
21
==−==−=
=−==
⋅=⋅=⋅=⋅=⋅=
⋅==
H
L
L
qhhqhhq
xyx
sssss
ss
⎟
The exergy destruction during a process of a stream from an inlet state to exit state is given by
⎟⎠
⎞⎜⎜⎝
⎛+−−==
sink
out
source
in0gen0dest T
qT
qssTsTx ie
Application of this equation for each process of the cycle gives the exergy destructions per unit mass flowing through the compressor:
[ ] Btu/lbm3.18=−−=
−−=
=⎟⎠⎞
⎜⎝⎛ −−=
⎟⎟⎠
⎞⎜⎜⎝
⎛−−=
=⎟⎠⎞
⎜⎝⎛ −−=
⎟⎟⎠
⎞⎜⎜⎝
⎛−−=
=⋅−×+×=
−+=
=⎟⎠⎞
⎜⎝⎛ +−⎟ =⎟
⎠
⎞⎜⎜⎝
⎛+−=
)11286.0)(3698.0()22260.0)(6302.0(2418.0)R555(
)(
Btu/lbm54.0R440
Btu/lbm161.5011286.022948.0)R555)(3698.0(
Btu/lbm44.0R495
Btu/lbm881.5810238.022260.0)R555)(6302.0(
Btu/lbm73.4RBtu/lbm)09774.011286.03698.010238.06302.0)(R555(
)(
Btu/lbm2.67R555
Btu/lbm621.822418.009774.0)R555(
6510mixingdestroyed,
67,67067destroyed,
45,45045destroyed,
3640346destroyed,
23023destroyed,
ysxssTX
Tq
ssyTx
Tq
ssxTx
sysxsTxTq
ssTx
L
L
L
L
H
H
&
For isentropic processes, the exergy destruction is zero:
012destroyed, =X&
The greatest exergy destruction occurs during the mixing process. Note that heat is absorbed in evaporator 2 from a reservoir at -20°F (440 R), in evaporator 1 from a reservoir at 35°F (495 R), and rejected to a reservoir at 95°F (555 R), which is also taken as the dead state temperature.
QH
QL
-29.5°F 1
2
3
6
160 psia
s
T
·
Win ·
·
45
7
30°F
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PROPRIETARY MATERIAL. © 2008 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.
11-75
11-110 A two-stage compression refrigeration system with a separation unit is considered. The rate of cooling and the power requirement are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.
Analysis From the refrigerant tables (Tables A-11, A-12, and A-13),
kJ/kg51.264 kPa400
KkJ/kg95813.0kJ/kg91.230
vapor sat.
C32
)throttling( kJ/kg94.63
kJ/kg94.63 liquidsat.C9.8
)throttling( kJ/kg22.127
kJ/kg22.127 liquidsat.kPa1400
kJ/kg49.281 kPa1400
KkJ/kg92691.0kJ/kg55.255
vapor sat.
C9.8
878
C8.9@sat 8
C32@7
C32@77
56
C9.8@55
34
kPa0014@33
212
2
C9.8@1
C9.8@11
=⎭⎬⎫
===
⋅====
⎭⎬⎫°−=
=≅
==⎭⎬⎫°=
=≅
==⎭⎬⎫=
=⎭⎬⎫
==
⋅====
⎭⎬⎫°=
°
°−
°−
°
°
°
hssPP
sshhT
hh
hhT
hh
hhP
hss
P
sshhT
g
g
f
f
g
g
An energy balance on the separator gives
kg/s280.194.6351.26422.12755.255kg/s)2()()(
58
4126412586 =
−−
=−−
=⎯→⎯−=−hhhh
mmhhmhhm &&&&
The rate of cooling produced by this system is then
kJ/s213.7=−=−= kJ/kg)94.6391kg/s)(230.280.1()( 676 hhmQL &&
The total power input to the compressors is
kW94.89=−+−=−+−=
kJ/kg)55.25549kg/s)(281.2(kJ/kg)91.23051kg/s)(264.280.1()()( 122786in hhmhhmW &&&
3
45
6
Compressor
Expansion valve
1
2
Evaporator 7
Separator
Condenser
Expansion valve
Compressor
8
QL
-32°C 7
8 3
6
s
T
·
2 1.4 MPa
4 1 5
8.9°C
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11-76
11-111 A two-stage vapor-compression refrigeration system with refrigerant-134a as the working fluid is considered. The process with the greatest exergy destruction is to be determined.
Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.
Analysis From Prob. 11-109 and the refrigerant tables (Tables A-11, A-12, and A-13),
K29827325K25527318kJ/kg27.154kJ/kg97.166
kg/s280.1kg/s2
KkJ/kg95813.0KkJ/kg2658.0KkJ/kg24761.0
KkJ/kg4720.0KkJ/kg45315.0
KkJ/kg92691.0
0
32
67
lower
upper
87
6
5
4
3
21
=+===+−=
=−==−=
==
⋅==⋅=⋅=
⋅=⋅=
⋅==
TTT
hhqhhq
mm
ssssss
ss
H
L
H
L
&
&
⎟
The exergy destruction during a process of a stream from an inlet state to exit state is given by
⎟⎠
⎞⎜⎜⎝
⎛+−−==
sink
out
source
in0gen0dest T
qT
qssTsTx ie
Application of this equation for each process of the cycle gives
[ ][ ] kW11.0)4720.092691.0)(kg/s2()95813.024761.0)(kg/s280.1()K298(
)()(
kW32.14K255kJ/kg97.1662658.095813.0)K298)(kg/s280.1(
kW94.6KkJ/kg)24761.02658.0)(K298)(kg/s280.1()(
kW23.11KkJ/kg)45315.04720.0)(K298)(kg/s2()(
K298kJ/kg27.15492691.045315.0)K298)(kg/s2(
41upper85lower0separatordestroyed,
670lower67destroyed,
560lower56destroyed,
340upper34destroyed,
230upper23destroyed,
=−+−=
−−−=
=⎟⎠
⎞⎜⎝
⎛ −−=
⎟⎟⎠
⎞⎜⎜⎝
⎛−−=
=⋅−=−=
=⋅−=−=
=⎟⎠
⎞⎜⎝
⎛ +−=
⎟⎟⎠
⎞⎜⎜⎝
⎛+−=
ssmssmTX
Tq
ssTmX
ssTmX
ssTmX
Tq
ssTmX
L
L
H
H
&&&
&&
&&
&&
&&
kW26.18
For isentropic processes, the exergy destruction is zero:
0
0
78destroyed,
12destroyed,
=
=
X
X&
&
Note that heat is absorbed from a reservoir at 0°F (460 R) and rejected to the standard ambient air at 77°F (537 R), which is also taken as the dead state temperature. The greatest exergy destruction occurs during the condensation process.
QL
-32°C7
8 3
6
s
T
·
2 1.4 MPa
4 1 5
8.9°C
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11-77
11-112 A regenerative gas refrigeration cycle with helium as the working fluid is considered. The temperature of the helium at the turbine inlet, the COP of the cycle, and the net power input required are to be determined.
Assumptions 1 Steady operating conditions exist. 2 Helium is an ideal gas with constant specific heats at room temperature. 3 Kinetic and potential energy changes are negligible.
Properties The properties of helium are cp = 5.1926 kJ/kg·K and k = 1.667 (Table A-2).
Analysis (a) The temperature of the helium at the turbine inlet is determined from an energy balance on the regenerator,
( ) ( )6143
outin
(steady)0systemoutin 0
hhmhhmhmhm
EE
EEE
iiee −=−⎯→⎯=
=
=Δ=−
∑∑ &&&&
&&
&&&
or,
Thus,
( ) ( )
( ) ( ) K278C25C10C206134
61436143
=°=°−+°−−°=+−=
−=−⎯→⎯−=−
C5TTTT
TTTTTTcmTTcm pp &&
(b) From the isentropic relations, ( )
( )( )
( )( ) C93.9K1179
31 K278
C135.2 K24083 K263
66716670k1k
4
545
66716670k1k
1
212
°−==⎟⎠⎞
⎜⎝⎛=⎟⎟
⎠⎜
⎞⎜⎝
⎛=
°===⎟⎟⎠
⎜⎞
⎜⎝
⎛=
−
−
.PPTT
.PPTT
././
././
Then the COP of this ideal gas refrigeration cycle is determined from
( ) ( )
( ) ( )( )
( )[ ] ( )[ ] 1.49=°−−−°−−
°−−°−=
−−−−
=
−−−−
=−
==
C93.95C10135.2C93.9C25
COP
5412
56
5412
56
outturb,incomp,innet,R
TTTTTT
hhhhhh
wwq
wq LL
(c) The net power input is determined from
( ) ( )[ ]( ) ( )[ ]
( )( ) ( )[ ] ( )[ ]( )kW108.2=
−−−−−°⋅=
−−−=
−−−=−=
93.9510135.2CkJ/kg 5.1926kg/s0.455412
5412outturb,incomp,innet,
TTTTcm
hhhhmWWW
p&
&&&&
s
T
1
3
-10°C
-25°C 56
4
2QH ·
QRefrig ·
20°C
Qregen ·
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11-78
11-113 An absorption refrigeration system operating at specified conditions is considered. The minimum rate of heat supply required is to be determined.
Analysis The maximum COP that this refrigeration system can have is
259.1263298
263K358K298
11COP0
0maxR, =⎟
⎠⎞
⎜⎝⎛
−⎟⎟⎠
⎞⎜⎜⎝
⎛−⎟ =⎟
⎠
⎞⎜⎜⎝
⎟⎛
−⎟⎠
⎞⎜⎜⎝
⎛−=
L
L
s TTT
TT
Thus,
kW53.9 1.259
kW12COP maxR,
mingen, === LQQ
&&
11-114 EES Problem 11-113 is reconsidered. The effect of the source temperature on the minimum rate of heat supply is to be investigated.
Analysis The problem is solved using EES, and the solution is given below.
"Input Data:" T_L = -10 [C] T_0 = 25 [C] T_s = 85 [C] Q_dot_L = 8 [kW]
"The maximum COP that this refrigeration system can have is:" COP_R_max = (1-(T_0+273)/(T_s+273))*((T_L+273)/(T_0 - T_L))
"The minimum rate of heat supply is:" Q_dot_gen_min = Q_dot_L/COP_R_max
Qgenmin [kW] Ts [C] 13.76 50 8.996 65 6.833 80 5.295 100 4.237 125 3.603 150 2.878 200 2.475 250
50 90 130 170 210 2502
4
6
8
10
12
14
Ts [C]
Qge
n;m
in[k
W]
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11-79
11-115 A room is cooled adequately by a 5000 Btu/h window air-conditioning unit. The rate of heat gain of the room when the air-conditioner is running continuously is to be determined.
Assumptions 1 The heat gain includes heat transfer through the walls and the roof, infiltration heat gain, solar heat gain, internal heat gain, etc. 2 Steady operating conditions exist.
Analysis The rate of heat gain of the room in steady operation is simply equal to the cooling rate of the air-conditioning system,
& &Q Qheat gain cooling= = 5,000 Btu / h
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11-80
11-116 A regenerative gas refrigeration cycle using air as the working fluid is considered. The effectiveness of the regenerator, the rate of heat removal from the refrigerated space, the COP of the cycle, and the refrigeration load and the COP if this system operated on the simple gas refrigeration cycle are to be determined.
Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. 3Air is an ideal gas with variable specific heats.
Analysis (a) For this problem, we use the properties of air from EES:
kJ/kg50.433kPa500
kJ/kg.K6110.5C0
kPa100kJ/kg40.273C0
212
2
11
1
11
=⎭⎬⎫
==
=⎭⎬⎫
°==
=⎯→⎯°=
shss
P
sTP
hT
kJ/kg63.308C35
kJ/kg52.47340.273
40.27350.43380.0
33
2
2
12
12
=⎯→⎯°=
=−
−=
−−
=
hT
hh
hhhh s
Cη
For the turbine inlet and exit we have
kJ/kg45.193C80 55 =⎯→⎯°−= hT
sT hh
hhhT
54
54
44 ?
−−
=
=⎯→⎯=
η
=⎭⎬⎫
==
=⎭⎬⎫
==
=⎭⎬⎫
°==
shss
P
sTP
sTP
545
5
44
4
11
1
kPa500
?kPa500
kJ/kg.K6110.5C0
kPa100
We can determine the temperature at the turbine inlet from EES using the above relations. A hand solution would require a trial-error approach.
T4 = 281.8 K, h4 = 282.08 kJ/kg
An energy balance on the regenerator gives
kJ/kg85.24608.28263.30840.2734316 =+−=+−= hhhh
The effectiveness of the regenerator is determined from
0.430=−−
=−−
=85.24663.30808.28263.308
63
43regen hh
hhε
s
T
1
3
0°C
5s6
4
2s QH·
QRefrig ·
35°C
Qregen
2
5-80°C
.
3
45
6
QH
Compressor
12
QL
Heat Exch.
Heat Exch.
Regenerator
Turbine
.
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11-81
(b) The refrigeration load is
kW21.36=−=−= kJ/kg)45.19385kg/s)(246.4.0()( 56 hhmQL &&
(c) The turbine and compressor powers and the COP of the cycle are
kW05.80kJ/kg)40.27352kg/s)(473.4.0()( 12inC, =−=−= hhmW &&
kW45.35kJ/kg)45.19308kg/s)(282.4.0()( 54outT, =−=−= hhmW &&
0.479=−
=−
==45.3505.80
36.21COPoutT,inC,innet, WW
QW
Q LL&&
&
&
&
(d) The simple gas refrigeration cycle analysis is as follows:
kJ/kg63.308kJ/kg52.473kJ/kg40.273
3
2
1
===
hhh
kJ/kg2704.5C35kPa500
33
3 =⎭⎬⎫
°==
sTP
kJ/kg.K52.194kPa100
434
1 =⎭⎬⎫
==
shss
P
kJ/kg64.21152.19463.308
63.30885.0 4
4
43
43 =⎯→⎯−
−=⎯→⎯
−−
= hh
hhhh
sTη
kW24.70=−=−= kJ/kg)64.21140kg/s)(273.4.0()( 41 hhmQL &&
[ ] kW25.41kJ/kg)64.211(308.63)40.273(473.52kg/s)4.0()()( 4312innet, =−−−=−−−= hhmhhmW &&&
0.599===25.4170.24COP
innet,WQL&
&
s
T
1
2QH
35°C 0°C
3
4s
·
QRefrig ·
2
14
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11-82
11-117 A heat pump water heater has a COP of 2.2 and consumes 2 kW when running. It is to be determined if this heat pump can be used to meet the cooling needs of a room by absorbing heat from it.
Assumptions The COP of the heat pump remains constant whether heat is absorbed from the outdoor air or room air.
Analysis The COP of the heat pump is given to be 2.2. Then the COP of the air-conditioning system becomes
COP COPair-cond heat pump= − = − =1 2 2 1 12. .
Then the rate of cooling (heat absorption from the air) becomes
& & ( . )( .Q Wincooling air-cond= COP kW) kW = 8640 kJ / h= =12 2 2 4
since 1 kW = 3600 kJ/h. We conclude that this heat pump can meet the cooling needs of the room since its cooling rate is greater than the rate of heat gain of the room.
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11-83
11-118 An innovative vapor-compression refrigeration system with a heat exchanger is considered. The system’s COP is to be determined.
Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.
Analysis From the refrigerant tables (Tables A-11, A-12, and A-13),
kPa200kJ/kg46.244
vapor sat.
C1.10
)throttling( kJ/kg32.79
kJ/kg32.79 kPa800
C203.113.31
3.11
kJ/kg47.95 liquidsat.kPa800
C10.1@sat 6
C1.10@66
45
C20@4
4
kPa800@sat 4
kPa800@33
====
⎭⎬⎫°−=
=≅
=≅⎪⎭
⎪⎬
⎫
=°=−=
−=
==⎭⎬⎫=
°−
°−
°
PPhhT
hh
hhP
TT
hhP
g
f
f
An energy balance on the heat exchanger gives
kJ/kg61.26046.24432.7947.95)()( 64314361 =+−=+−=⎯→⎯−=− hhhhhhmhhm &&
Then,
kJ/kg17.292 kPa800
KkJ/kg9970.0 kJ/kg61.260
kPa200
212
2
11
1
=⎭⎬⎫
==
⋅=⎭⎬⎫
==
hss
P
shP
The COP of this refrigeration system is determined from its definition,
5.23=−−
=−−
==61.26017.292
32.7946.244COP12
56
inR hh
hhwqL
3
4
5
6
Compressor
Throttle valve
1
2
Heat exchanger
Condenser
Evaporator
QH
QL
-10.1°C 1
2
3
5
800 kPa
s
T
·
Win ·
·
6
4
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PROPRIETARY MATERIAL. © 2008 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.
11-84
11-119 An innovative vapor-compression refrigeration system with a heat exchanger is considered. The system’s COP is to be determined.
Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.
Analysis From the refrigerant tables (Tables A-11, A-12, and A-13),
kPa200kJ/kg46.244
vapor sat.
C1.10
)throttling( kJ/kg43.65
kJ/kg43.65 kPa800
C103.213.31
3.21
kJ/kg47.95 liquidsat.kPa800
C10.1@sat 6
C1.10@66
45
C10@4
4
kPa800@sat 4
kPa800@33
====
⎭⎬⎫°−=
=≅
=≅⎪⎭
⎪⎬
⎫
=°=−=
−=
==⎭⎬⎫=
°−
°−
°
PPhhT
hh
hhP
TT
hhP
g
f
f
An energy balance on the heat exchanger gives
kJ/kg50.27446.24443.6547.95)()( 64314361 =+−=+−=⎯→⎯−=− hhhhhhmhhm &&
Then,
kJ/kg28.308 kPa800
KkJ/kg0449.1 kJ/kg50.274
kPa200
212
2
11
1
=⎭⎬⎫
==
⋅=⎭⎬⎫
==
hss
P
shP
The COP of this refrigeration system is determined from its definition,
5.30=−−
=−−
==50.27428.30843.6546.244COP
12
56
inR hh
hhwqL
3
4
5
6
Compressor
Throttle valve
1
2
Heat exchanger
Condenser
Evaporator
QH
QL
200 kPa 1
2
3
5
800 kPa
s
T
·
Win ·
·
6
4
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PROPRIETARY MATERIAL. © 2008 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.
11-85
11-120 An ideal gas refrigeration cycle with with three stages of compression with intercooling using air as the working fluid is considered. The COP of this system is to be determined.
Assumptions 1 Steady operating conditions exist. 2 Air is an ideal gas with constant specific heats. 3Kinetic and potential energy changes are negligible.
Properties The properties of air at room temperature are cp = 1.005 kJ/kg·K and k = 1.4 (Table A-2a).
Analysis From the isentropic relations,
K5.72555
1K)288(
K1.456K)(5)288(
K7.400K)(5)253(
4.1/4.0/)1(
7
878
4.1/4.0/)1(
3
4364
4.1/4.0/)1(
1
212
=⎟⎠⎞
⎜⎝⎛
××=⎟⎟
⎠⎜
⎞⎜⎝
⎛=
==⎟⎟⎠
⎜⎞
⎜⎝
⎛==
==⎟⎟⎠
⎜⎞
⎜⎝
⎛=
−
−
−
kk
kk
kk
PP
TT
PP
TTT
PP
TT
The COP of this ideal gas refrigeration cycle is determined from
0.673=−−−+−
−=
−−−+−−
=
−−−+−+−−
=
−==
)5.72288()2881.456(2)2537.400(5.72253
)()(2)(
)()()()(
COP
873412
81
87563412
81
outturb,incomp,innet,R
TTTTTTTT
hhhhhhhhhh
wwq
wq LL
15°C
s
T
1
2
3
4
5
6
-20°C
7
8
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11-86
11-121 An ideal vapor-compression refrigeration cycle with refrigerant-22 as the working fluid is considered. The evaporator is located inside the air handler of building. The hardware and the T-s diagram for this heat pump application are to be sketched. The COP of the unit and the ratio of volume flow rate of air entering the air handler to mass flow rate of R-22 through the air handler 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-22 data from the problem statement,
)throttling( kJ/kg101
kJ/kg101 liquidsat.kPa1728
kJ/kg7.283 kPa1728
KkJ/kg9344.0kJ/kg1.248
vapor sat.
C5
34
kPa1728@33
212
2
C5@1
C5@11
=≅
==⎭⎬⎫=
=⎭⎬⎫
==
⋅====
⎭⎬⎫°−=
°−
°−
hh
hhP
hss
P
sshhT
f
g
g
(b) The COP of the heat pump is determined from its definition,
5.13=−−
=−−
==1.2487.283
1017.283COP12
32
inHP hh
hhwqH
(c) An energy balance on the condenser gives
TcTcmhhmQ pa
apaRH Δ=Δ=−=
vV&
&&& )( 32
Rearranging, we obtain the ratio of volume flow rate of air entering the air handler to mass flow rate of R-22 through the air handler
R22/s)kgair/min)/((m462 3=
=
⋅
−=
Δ−
=
R22/s)air/s)/(kg(m699.7
)K20)(K)kJ/kg005.1)(/kgm847.0/1(kJ/kg)1017.283(
)/1(3
332
Tchh
m pR
a
v
V&
&
Note that the specific volume of air is obtained from ideal gas equation taking the pressure of air to be 101 kPa and using the room temperature of air (25°C = 298 K) to be 0.847 m3/kg.
QH
QL
-5°C 1
2 3
4
45°C
s
T
·
Win ·
·
4s1
2 3
4
-5°C
Condenser
Evaporator
Compressor Expansion
valve
sat. vap.QL
Win
Air
45°C
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11-87
11-122 An ideal vapor-compression refrigeration cycle with refrigerant-134a as the working fluid is considered. Cooling water flows through the water jacket surrounding the condenser. To produce ice, potable water is supplied to the chiller section of the refrigeration cycle. The hardware and the T-s diagram for this refrigerant-ice making system are to be sketched. The mass flow rates of the refrigerant and the potable water 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-134a data from the problem statement,
)throttling( kJ/kg77.117
kJ/kg77.117 liquidsat.kPa1200
kJ/kg07.284 kPa1200
KkJ/kg94456.0kJ/kg16.239
vapor sat.
kPa140
34
kPa1200@33
212
2
kPa140@1
kPa140@11
=≅
==⎭⎬⎫=
=⎭⎬⎫
==
⋅====
⎭⎬⎫=
hh
hhP
hss
P
sshhT
f
g
g
(b) An energy balance on the condenser gives
TcmhhmQ pwRH Δ=−= &&& )( 32
Solving for the mass flow rate of the refrigerant
kg/s50.3=−
⋅=
−
Δ=
kJ/kg)77.11707.284()K10)(K)kJ/kg18.4)(kg/s200(
32 hhTcm
m pwR
&&
(c) An energy balance on the evaporator gives
ifwRL hmhhmQ &&& =−= )( 41
Solving for the mass flow rate of the potable water
kg/s18.3=−
=−
=kJ/kg333
kJ/kg)77.11716.239)(kg/s3.50()( 41
if
Rw h
hhmm
&&
QH
QL
140 kPa 1
2 3
4
1.2 MPa
s
T
·
Win ·
·
4s1
2 3
4
140 kPa
Condenser
Evaporator
Compressor Expansion
valve
sat. vap.
Win
Potable water
1.2 MPa
Water 200 kg/s
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11-88
11-123 A vortex tube receives compressed air at 500 kPa and 300 K, and supplies 25 percent of it as cold air and the rest as hot air. The COP of the vortex tube is to be compared to that of a reversed Brayton cycle for the same pressure ratio; the exit temperature of the hot fluid stream and the COP are to be determined; and it is to be shown if this process violates the second law.
Assumptions 1 The vortex tube is adiabatic. 2 Air is an ideal gas with constant specific heats at room temperature. 3 Steady operating conditions exist.
Properties The gas constant of air is 0.287 kJ/kg.K (Table A-1). The specific heat of air at room temperature is cp = 1.005 kJ/kg.K (Table A-2). The enthalpy of air at absolute temperature T can be expressed in terms of specific heats as h = cpT.
Analysis (a) The COP of the vortex tube is much lower than the COP of a reversed Brayton cycle of the same pressure ratio since the vortex tube involves vortices, which are highly irreversible. Owing to this irreversibility, the minimum temperature that can be obtained by the vortex tube is not as low as the one that can be obtained by the revered Brayton cycle.
(b) We take the vortex tube as the system. This is a steady flow system with one inlet and two exits, and it involves no heat or work interactions. Then the steady-flow energy balance equation for this system & &E Ein out= for a unit mass flow rate at the inlet ( &m1 kg / s)= 1 can be expressed as
321
332211
332211
75.025.01 TcTcTc
TcmTcmTcmhmhmhm
ppp
ppp
+=
+=+=
&&&
&&&
Canceling cp and solving for T3 gives
K307.3=
×−=
−=
75.027825.0300
75.025.0 21
3TT
T
Therefore, the hot air stream will leave the vortex tube at an average temperature of 307.3 K.
(c) The entropy balance for this steady flow system & & &S S Sin out gen− + = 0 can be expressed as with one inlet and two exits, and it involves no heat or work interactions. Then the steady-flow entropy balance equation for this system for a unit mass flow rate at the inlet ( &m1 kg / s)= 1 can be expressed
⎟⎟⎠
⎞⎜⎜⎝
⎛−⎟ +⎟
⎠
⎞⎜⎜⎝
⎛−=
−+−=−+−=
+−+=−+=
−=
1
3
1
3
1
2
1
2
1312
133122
1323322113322
inoutgen
lnln75.0lnln25.0
)(75.0)(25.0)()(
)(
PP
RTT
cPP
RTT
c
ssssssmssm
smmsmsmsmsmsm
SSS
pp
&&
&&&&&&&
&&&
Substituting the known quantities, the rate of entropy generation is determined to be
0>kW/K 461.0kPa500kPa100lnkJ/kg.K)287.0(
K300K3.307lnkJ/kg.K)005.1(75.0
kPa500kPa100lnkJ/kg.K)287.0(
K300K278lnkJ/kg.K)005.1(25.0gen
=
⎟⎠⎞
⎜⎝⎛ −+
⎟⎠⎞
⎜⎝⎛ −=S&
which is a positive quantity. Therefore, this process satisfies the 2nd law of thermodynamics.
1
3 2
Compressed air
Cold air Warm air
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11-89
(d) For a unit mass flow rate at the inlet ( &m1 kg / s)= 1 , the cooling rate and the power input to the compressor are determined to
kW5.53=278)K-00kJ/kg.K)(35kg/s)(1.0025.0(
)()( c1c1cooling
=
−=−= TTcmhhmQ pcc &&&
kW1.1571kPa100kPa500
80.0)14.1(K)00kJ/kg.K)(37kg/s)(0.281(
1)1(
4.1/)14.1(
/)1(
0
1
comp
00incomp,
=⎥⎥⎦
⎤
⎢⎢⎣
⎡−⎟
⎠⎞
⎜⎝⎛
−=
⎥⎥⎦
⎤⎢⎢⎣
⎡−⎟⎟
⎠⎜
⎞⎜⎝
⎛−
=
−
− kk
PP
kRTm
Wη
&&
Then the COP of the vortex refrigerator becomes
0350.kW1.157kW5.53COP
incomp,
cooling ===W
Q&
&
The COP of a Carnot refrigerator operating between the same temperature limits of 300 K and 278 K is
COP 278 K KCarnot =
−=
−=
T
T TL
H L ( )300 27812.6
Discussion Note that the COP of the vortex refrigerator is a small fraction of the COP of a Carnot refrigerator operating between the same temperature limits.
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11-90
11-124 A vortex tube receives compressed air at 600 kPa and 300 K, and supplies 25 percent of it as cold air and the rest as hot air. The COP of the vortex tube is to be compared to that of a reversed Brayton cycle for the same pressure ratio; the exit temperature of the hot fluid stream and the COP are to be determined; and it is to be shown if this process violates the second law.
Assumptions 1 The vortex tube is adiabatic. 2 Air is an ideal gas with constant specific heats at room temperature. 3 Steady operating conditions exist.
Properties The gas constant of air is 0.287 kJ/kg.K (Table A-1). The specific heat of air at room temperature is cp = 1.005 kJ/kg.K (Table A-2). The enthalpy of air at absolute temperature T can be expressed in terms of specific heats as h = cp T.
Analysis (a) The COP of the vortex tube is much lower than the COP of a reversed Brayton cycle of the same pressure ratio since the vortex tube involves vortices, which are highly irreversible. Owing to this irreversibility, the minimum temperature that can be obtained by the vortex tube is not as low as the one that can be obtained by the revered Brayton cycle.
(b) We take the vortex tube as the system. This is a steady flow system with one inlet and two exits, and it involves no heat or work interactions. Then the steady-flow entropy balance equation for this system & &E Ein out= for a unit mass flow rate at the inlet ( &m1 kg / s)= 1 can be expressed as
321
332211
332211
75.025.01 TcTcTc
TcmTcmTcmhmhmhm
ppp
ppp
+=
+=+=
&&&
&&&
Canceling cp and solving for T3 gives
K307.3=×−
=
−=
75.027825.0300
75.025.0 21
3TT
T
Therefore, the hot air stream will leave the vortex tube at an average temperature of 307.3 K.
(c) The entropy balance for this steady flow system & & &S S Sin out gen− + = 0 can be expressed as with one inlet and two exits, and it involves no heat or work interactions. Then the steady-flow energy balance equation for this system for a unit mass flow rate at the inlet ( &m1 kg / s)= 1 can be expressed
⎟⎟⎠
⎞⎜⎜⎝
⎛−⎟ +⎟
⎠
⎞⎜⎜⎝
⎛−=
−+−=−+−=
+−+=−+=
−=
1
3
1
3
1
2
1
2
1312
133122
1323322113322
inoutgen
lnln75.0lnln25.0
)(75.0)(25.0)()(
)(
PP
RTT
cPP
RTT
c
ssssssmssm
smmsmsmsmsmsm
SSS
pp
&&
&&&&&&&
&&&
Substituting the known quantities, the rate of entropy generation is determined to be
0>kW/K 5130kPa600kPa100kJ/kg.K)ln2870(
K300 K3307kJ/kg.K)ln0051(750
kPa600kPa100kJ/kg.K)ln2870(
K300 K278kJ/kg.K)ln0051(250gen
.
....
...S
=
⎟⎠⎞
⎜⎝⎛ −+
⎟⎠⎞
⎜⎝⎛ −=&
which is a positive quantity. Therefore, this process satisfies the 2nd law of thermodynamics.
1
3 2
Compressed air
Cold air Warm air
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11-91
(d) For a unit mass flow rate at the inlet ( &m1 kg / s)= 1 , the cooling rate and the power input to the compressor are determined to
kW5.53=278)K-00kJ/kg.K)(35kg/s)(1.0025.0(
)()( c1c1cooling
=
−=−= TTcmhhmQ pcc &&&
kW9.1791kPa100kPa600
80.0)14.1(K)00kJ/kg.K)(37kg/s)(0.281(
1)1(
4.1/)14.1(
/)1(
0
1
comp
00incomp,
=⎥⎥⎦
⎤
⎢⎢⎣
⎡−⎟
⎠⎞
⎜⎝⎛
−=
⎥⎥⎦
⎤⎢⎢⎣
⎡−⎟⎟
⎠⎜
⎞⎜⎝
⎛−
=
−
− kk
PP
kRTm
Wη
&&
Then the COP of the vortex refrigerator becomes
0310.kW9.179kW5.53COP
incomp,
cooling ===W
Q&
&
The COP of a Carnot refrigerator operating between the same temperature limits of 300 K and 278 K is
12.6=−
=−
=K)278300(
K278COPCarnotLH
L
TTT
Discussion Note that the COP of the vortex refrigerator is a small fraction of the COP of a Carnot refrigerator operating between the same temperature limits.
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11-92
11-125 EES The effect of the evaporator pressure on the COP of an ideal vapor-compression refrigeration cycle with R-134a as the working fluid is to be investigated.
Analysis The problem is solved using EES, and the solution is given below.
"Input Data" P[1]=100 [kPa] P[2] = 1000 [kPa] Fluid$='R134a' Eta_c=0.7 "Compressor isentropic efficiency" "Compressor" h[1]=enthalpy(Fluid$,P=P[1],x=1) "properties for state 1" s[1]=entropy(Fluid$,P=P[1],x=1) T[1]=temperature(Fluid$,h=h[1],P=P[1]) h2s=enthalpy(Fluid$,P=P[2],s=s[1]) "Identifies state 2s as isentropic" h[1]+Wcs=h2s "energy balance on isentropic compressor" W_c=Wcs/Eta_c"definition of compressor isentropic efficiency" h[1]+W_c=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]) "Condenser" P[3] = P[2] h[3]=enthalpy(Fluid$,P=P[3],x=0) "properties for state 3" s[3]=entropy(Fluid$,P=P[3],x=0) h[2]=Qout+h[3] "energy balance on condenser" "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$,h=h[4],P=P[4]) "Evaporator" P[4]= P[1] Q_in + h[4]=h[1] "energy balance on evaporator" "Coefficient of Performance:" COP=Q_in/W_c "definition of COP"
COP ηc P1 [kPa] 1.851 0.7 100 2.863 0.7 200 4.014 0.7 300 5.462 0.7 400 7.424 0.7 500
100 150 200 250 300 350 400 450 5000
2
4
6
8
10
P[1] [kPa]
CO
P
η comp
1.01.00.70.7
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11-93
11-126 EES The effect of the condenser pressure on the COP of an ideal vapor-compression refrigeration cycle with R-134a as the working fluid is to be investigated.
Analysis The problem is solved using EES, and the solution is given below.
"Input Data" P[1]=120 [kPa] P[2] = 400 [kPa] Fluid$='R134a' Eta_c=0.7 "Compressor isentropic efficiency" "Compressor" h[1]=enthalpy(Fluid$,P=P[1],x=1) "properties for state 1" s[1]=entropy(Fluid$,P=P[1],x=1) T[1]=temperature(Fluid$,h=h[1],P=P[1]) h2s=enthalpy(Fluid$,P=P[2],s=s[1]) "Identifies state 2s as isentropic" h[1]+Wcs=h2s "energy balance on isentropic compressor" W_c=Wcs/Eta_c"definition of compressor isentropic efficiency" h[1]+W_c=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]) "Condenser" P[3] = P[2] h[3]=enthalpy(Fluid$,P=P[3],x=0) "properties for state 3" s[3]=entropy(Fluid$,P=P[3],x=0) h[2]=Qout+h[3] "energy balance on condenser" "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]= P[1] Q_in + h[4]=h[1] "energy balance on evaporator" "Coefficient of Performance:" COP=Q_in/W_c "definition of COP"
COP ηc P2 [kPa] 4.935 0.7 400 3.04 0.7 650
2.258 0.7 900 1.803 0.7 1150 1.492 0.7 1400
400 600 800 1000 1200 14000
1
2
3
4
5
6
7
8
P[2] [kPa]
CO
P
η comp
1.01.00.70.7
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PROPRIETARY MATERIAL. © 2008 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.
11-94
Fundamentals of Engineering (FE) Exam Problems
11-127 Consider a heat pump that operates on the reversed Carnot cycle with R-134a as the working fluid executed under the saturation dome between the pressure limits of 140 kPa and 800 kPa. R-134a changes from saturated vapor to saturated liquid during the heat rejection process. The net work input for this cycle is
(a) 28 kJ/kg (b) 34 kJ/kg (c) 49 kJ/kg (d) 144 kJ/kg (e) 275 kJ/kg
Answer (a) 28 kJ/kg
Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values).
P1=800 "kPa" P2=140 "kPa" h_fg=ENTHALPY(R134a,x=1,P=P1)-ENTHALPY(R134a,x=0,P=P1) TH=TEMPERATURE(R134a,x=0,P=P1)+273 TL=TEMPERATURE(R134a,x=0,P=P2)+273 q_H=h_fg COP=TH/(TH-TL) w_net=q_H/COP
"Some Wrong Solutions with Common Mistakes:" W1_work = q_H/COP1; COP1=TL/(TH-TL) "Using COP of regrigerator" W2_work = q_H/COP2; COP2=(TH-273)/(TH-TL) "Using C instead of K" W3_work = h_fg3/COP; h_fg3= ENTHALPY(R134a,x=1,P=P2)-ENTHALPY(R134a,x=0,P=P2) "Using h_fg at P2" W4_work = q_H*TL/TH "Using the wrong relation"
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11-95
11-128 A refrigerator removes heat from a refrigerated space at –5°C at a rate of 0.35 kJ/s and rejects it to an environment at 20°C. The minimum required power input is
(a) 30 W (b) 33 W (c) 56 W (d) 124 W (e) 350 W
Answer (b) 33 W
Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values).
TH=20+273 TL=-5+273 Q_L=0.35 "kJ/s" COP_max=TL/(TH-TL) w_min=Q_L/COP_max
"Some Wrong Solutions with Common Mistakes:" W1_work = Q_L/COP1; COP1=TH/(TH-TL) "Using COP of heat pump" W2_work = Q_L/COP2; COP2=(TH-273)/(TH-TL) "Using C instead of K" W3_work = Q_L*TL/TH "Using the wrong relation" W4_work = Q_L "Taking the rate of refrigeration as power input"
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11-96
11-129 A refrigerator operates on the ideal vapor compression refrigeration cycle with R-134a as the working fluid between the pressure limits of 120 kPa and 800 kPa. If the rate of heat removal from the refrigerated space is 32 kJ/s, the mass flow rate of the refrigerant is
(a) 0.19 kg/s (b) 0.15 kg/s (c) 0.23 kg/s (d) 0.28 kg/s (e) 0.81 kg/s
Answer (c) 0.23 kg/s
Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values).
P1=120 "kPa" P2=800 "kPa" P3=P2 P4=P1 s2=s1 Q_refrig=32 "kJ/s" m=Q_refrig/(h1-h4) h1=ENTHALPY(R134a,x=1,P=P1) s1=ENTROPY(R134a,x=1,P=P1) h2=ENTHALPY(R134a,s=s2,P=P2) h3=ENTHALPY(R134a,x=0,P=P3) h4=h3
"Some Wrong Solutions with Common Mistakes:" W1_mass = Q_refrig/(h2-h1) "Using wrong enthalpies, for W_in" W2_mass = Q_refrig/(h2-h3) "Using wrong enthalpies, for Q_H" W3_mass = Q_refrig/(h1-h44); h44=ENTHALPY(R134a,x=0,P=P4) "Using wrong enthalpy h4 (at P4)" W4_mass = Q_refrig/h_fg; h_fg=ENTHALPY(R134a,x=1,P=P2) - ENTHALPY(R134a,x=0,P=P2) "Using h_fg at P2"
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11-97
11-130 A heat pump operates on the ideal vapor compression refrigeration cycle with R-134a as the working fluid between the pressure limits of 0.32 MPa and 1.2 MPa. If the mass flow rate of the refrigerant is 0.193 kg/s, the rate of heat supply by the heat pump to the heated space is
(a) 3.3 kW (b) 23 kW (c) 26 kW (d) 31 kW (e) 45 kW
Answer (d) 31 kW
Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values).
P1=320 "kPa" P2=1200 "kPa" P3=P2 P4=P1 s2=s1 m=0.193 "kg/s" Q_supply=m*(h2-h3) "kJ/s" h1=ENTHALPY(R134a,x=1,P=P1) s1=ENTROPY(R134a,x=1,P=P1) h2=ENTHALPY(R134a,s=s2,P=P2) h3=ENTHALPY(R134a,x=0,P=P3) h4=h3
"Some Wrong Solutions with Common Mistakes:" W1_Qh = m*(h2-h1) "Using wrong enthalpies, for W_in" W2_Qh = m*(h1-h4) "Using wrong enthalpies, for Q_L" W3_Qh = m*(h22-h4); h22=ENTHALPY(R134a,x=1,P=P2) "Using wrong enthalpy h2 (hg at P2)" W4_Qh = m*h_fg; h_fg=ENTHALPY(R134a,x=1,P=P1) - ENTHALPY(R134a,x=0,P=P1) "Using h_fg at P1"
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11-98
11-131 An ideal vapor compression refrigeration cycle with R-134a as the working fluid operates between the pressure limits of 120 kPa and 1000 kPa. The mass fraction of the refrigerant that is in the liquid phase at the inlet of the evaporator is
(a) 0.65 (b) 0.60 (c) 0.40 (d) 0.55 (e) 0.35
Answer (b) 0.60
Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values).
P1=120 "kPa" P2=1000 "kPa" P3=P2 P4=P1 h1=ENTHALPY(R134a,x=1,P=P1) h3=ENTHALPY(R134a,x=0,P=P3) h4=h3 x4=QUALITY(R134a,h=h4,P=P4) liquid=1-x4
"Some Wrong Solutions with Common Mistakes:" W1_liquid = x4 "Taking quality as liquid content" W2_liquid = 0 "Assuming superheated vapor" W3_liquid = 1-x4s; x4s=QUALITY(R134a,s=s3,P=P4) "Assuming isentropic expansion" s3=ENTROPY(R134a,x=0,P=P3)
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11-99
11-132 Consider a heat pump that operates on the ideal vapor compression refrigeration cycle with R-134a as the working fluid between the pressure limits of 0.32 MPa and 1.2 MPa. The coefficient of performance of this heat pump is
(a) 0.17 (b) 1.2 (c) 3.1 (d) 4.9 (e) 5.9
Answer (e) 5.9
Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values).
P1=320 "kPa" P2=1200 "kPa" P3=P2 P4=P1 s2=s1 h1=ENTHALPY(R134a,x=1,P=P1) s1=ENTROPY(R134a,x=1,P=P1) h2=ENTHALPY(R134a,s=s2,P=P2) h3=ENTHALPY(R134a,x=0,P=P3) h4=h3 COP_HP=qH/Win Win=h2-h1 qH=h2-h3
"Some Wrong Solutions with Common Mistakes:" W1_COP = (h1-h4)/(h2-h1) "COP of refrigerator" W2_COP = (h1-h4)/(h2-h3) "Using wrong enthalpies, QL/QH" W3_COP = (h22-h3)/(h22-h1); h22=ENTHALPY(R134a,x=1,P=P2) "Using wrong enthalpy h2 (hg at P2)"
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11-100
11-133 An ideal gas refrigeration cycle using air as the working fluid operates between the pressure limits of 80 kPa and 280 kPa. Air is cooled to 35°C before entering the turbine. The lowest temperature of this cycle is
(a) –58°C (b) -26°C (c) 0°C (d) 11°C (e) 24°C
Answer (a) –58°C
Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values).
k=1.4 P1= 80 "kPa" P2=280 "kPa" T3=35+273 "K" "Mimimum temperature is the turbine exit temperature" T4=T3*(P1/P2)^((k-1)/k) - 273
"Some Wrong Solutions with Common Mistakes:" W1_Tmin = (T3-273)*(P1/P2)^((k-1)/k) "Using C instead of K" W2_Tmin = T3*(P1/P2)^((k-1)) - 273 "Using wrong exponent" W3_Tmin = T3*(P1/P2)^k - 273 "Using wrong exponent"
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11-101
11-134 Consider an ideal gas refrigeration cycle using helium as the working fluid. Helium enters the compressor at 100 kPa and –10°C and is compressed to 250 kPa. Helium is then cooled to 20°C before it enters the turbine. For a mass flow rate of 0.2 kg/s, the net power input required is
(a) 9.3 kW (b) 27.6 kW (c) 48.8 kW (d) 93.5 kW (e) 119 kW
Answer (b) 27.6 kW
Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values).
k=1.667 Cp=5.1926 "kJ/kg.K" P1= 100 "kPa" T1=-10+273 "K" P2=250 "kPa" T3=20+273 "K" m=0.2 "kg/s" "Mimimum temperature is the turbine exit temperature" T2=T1*(P2/P1)^((k-1)/k) T4=T3*(P1/P2)^((k-1)/k) W_netin=m*Cp*((T2-T1)-(T3-T4))
"Some Wrong Solutions with Common Mistakes:" W1_Win = m*Cp*((T22-T1)-(T3-T44)); T22=T1*P2/P1; T44=T3*P1/P2 "Using wrong relations for temps" W2_Win = m*Cp*(T2-T1) "Ignoring turbine work" W3_Win=m*1.005*((T2B-T1)-(T3-T4B)); T2B=T1*(P2/P1)^((kB-1)/kB); T4B=T3*(P1/P2)^((kB-1)/kB); kB=1.4 "Using air properties" W4_Win=m*Cp*((T2A-(T1-273))-(T3-273-T4A)); T2A=(T1-273)*(P2/P1)^((k-1)/k); T4A=(T3-273)*(P1/P2)^((k-1)/k) "Using C instead of K"
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11-102
11-135 An absorption air-conditioning system is to remove heat from the conditioned space at 20°C at a rate of 150 kJ/s while operating in an environment at 35°C. Heat is to be supplied from a geothermal source at 140°C. The minimum rate of heat supply required is
(a) 86 kJ/s (b) 21 kJ/s (c) 30 kJ/s (d) 61 kJ/s (e) 150 kJ/s
Answer (c) 30 kJ/s
Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values).
TL=20+273 "K" Q_refrig=150 "kJ/s" To=35+273 "K" Ts=140+273 "K" COP_max=(1-To/Ts)*(TL/(To-TL)) Q_in=Q_refrig/COP_max
"Some Wrong Solutions with Common Mistakes:" W1_Qin = Q_refrig "Taking COP = 1" W2_Qin = Q_refrig/COP2; COP2=TL/(Ts-TL) "Wrong COP expression" W3_Qin = Q_refrig/COP3; COP3=(1-To/Ts)*(Ts/(To-TL)) "Wrong COP expression, COP_HP" W4_Qin = Q_refrig*COP_max "Multiplying by COP instead of dividing"
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11-103
11-136 Consider a refrigerator that operates on the vapor compression refrigeration cycle with R-134a as the working fluid. The refrigerant enters the compressor as saturated vapor at 160 kPa, and exits at 800 kPa and 50°C, and leaves the condenser as saturated liquid at 800 kPa. The coefficient of performance of this refrigerator is
(a) 2.6 (b) 1.0 (c) 4.2 (d) 3.2 (e) 4.4
Answer (d) 3.2
Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values).
P1=160 "kPa" P2=800 "kPa" T2=50 "C" P3=P2 P4=P1 h1=ENTHALPY(R134a,x=1,P=P1) s1=ENTROPY(R134a,x=1,P=P1) h2=ENTHALPY(R134a,T=T2,P=P2) h3=ENTHALPY(R134a,x=0,P=P3) h4=h3 COP_R=qL/Win Win=h2-h1 qL=h1-h4
"Some Wrong Solutions with Common Mistakes:" W1_COP = (h2-h3)/(h2-h1) "COP of heat pump" W2_COP = (h1-h4)/(h2-h3) "Using wrong enthalpies, QL/QH" W3_COP = (h1-h4)/(h2s-h1); h2s=ENTHALPY(R134a,s=s1,P=P2) "Assuming isentropic compression"
11-137 ··· 11-145 Design and Essay Problems
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