Transcript
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. Thecoefficient 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 thisCarnot refrigerator is determined from
5.641K4.257/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/kg147.03kJ/kg173.08K303K257.4
kJ/kg08.17358.9366.26643
HH
LL
L
H
L
H
H
qTT
qTT
hhq
30 C4 3
21160 kPa
QL
QH
T
s
(c) The net work input is determined fromkJ/kg26.0503.14708.173net LH qqw
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 network 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.
8.281R475.4/R532.8
11/
1COP CR,LH TT
(b) Process 4-1 is isentropic, and thus
0.237418589.0
03793.008207.0
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/lbm7.92RBtu/lbm08207.022006.0)37.1578.72(43innet, ssTTw LH
Ideal and Actual Vapor-Compression Cycles
11-4C Yes; the throttling process is an internally irreversible process.
11-5C To make the ideal vapor-compression refrigeration cycle more closely approximate the actualcycle.
11-6C No. Assuming the water is maintained at 10 C in the evaporator, the evaporator pressure will bethe saturation pressure corresponding to this pressure, which is 1.2 kPa. It is not practical to designrefrigeration 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 temperatureof 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 thereversed Carnot cycle, but not so for the ideal vapor-compression refrigeration cycle. This is because the latter cycle involves an irreversible process for which the process path is not known.
11-9C The cycle that involves saturated liquid at 30 C will have a higher COP because, judging from the T-s diagram, it will require a smaller work input for the same refrigeration capacity.
QH
QL
4 3
21
T
s
11-10C The minimum temperature that the refrigerant can be cooled to before throttling is the temperatureof the sink (the cooling medium) since heat is transferred from the refrigerant to the cooling medium.
11-3
11-11 A commercial refrigerator with refrigerant-134a as the working fluid is considered. The quality ofthe refrigerant at the evaporator inlet, the refrigeration load, the COP of the refrigerator, and the theoreticalmaximum refrigeration load for the same power input to the compressor are to be determined.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.Analysis (a) From refrigerant-134a tables (Tables A-11 through A-13)
0.479544
4
34
33
3
22
2
11
1
kJ/kg23.111kPa60
kJ/kg23.111
kJ/kg23.111C42
kPa1200
kJ/kg16.295C65
kPa1200
kJ/kg03.230C34
kPa60
xhP
hh
hTP
hTP
hTP
26 C Water18 C
Win
QL
1.2 MPa 65 C
Expansionvalve
Compressor
Evaporator
Condenser42 C
QH
4
3 2
1 60 kPa -34 C
Using saturated liquid enthalpy at the giventemperature, 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
kW367.8kJ/kg)23.11116kg/s)(295.0455.0()( 32 hhmQ RH
kW513.2kW0.45kJ/kg)03.23016kg/s)(295.0455.0()( in12in QhhmW R
kW5.85513.2367.8inWQQ HL
(c) The COP of the refrigerator is determined from its definition
T
QH
QL
Win·
2
·
·4
3
2
1
2.33513.285.5COP
in
L
WQ
(d) The reversible COP of the refrigerator for thesame temperature limits is
063.51)27330/()27318(
11/
1COPmaxLH TT s
Then, the maximum refrigeration load becomes
kW12.72kW)513.2)(063.5(inmaxmaxL, WCOPQ
11-4
11-12 An ideal vapor-compression refrigeration cycle with refrigerant-134a as the working fluid isconsidered. 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, therefrigerant 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
T
QH
QL4s
·
Win·
·
0.7 MPa
4
32
10.12 MPa
Then the rate of heat removal from the refrigerated space and the power input to the compressor are determined from
s
andkW1.83
kW7.41
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
kW9.2383.141.7inWQQ LH
(c) The COP of the refrigerator is determined from its definition,
4.06kW1.83kW7.41COP
inR W
QL
11-5
11-13 An ideal vapor-compression refrigeration cycle with refrigerant-134a as the working fluid isconsidered. 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, therefrigerant 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
T
QH
QL4s
·
Win·
·
0.9 MPa
4
32
10.12 MPa
Then the rate of heat removal from the refrigerated space and the power input to the compressor are determined from s
andkW2.10
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
kW8.8710.277.6inWQQ LH
(c) The COP of the refrigerator is determined from its definition,
3.23kW2.10kW6.77COP
inR W
QL
11-6
11-14 An ideal vapor-compression refrigeration cycle with refrigerant-134a as the working fluid isconsidered. 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 T
QH
QL4s
·
Win·
·
0.7 MPa
4
32
10.12 MPa
Then, Q kW7.72kJ/kg82.58236.97kg/s0.0541 sL hhm
and 23.4kW1.83kW7.72COP
inR W
QL
Then the percentage increase in and COP becomesQ
4.2%
4.2%
06.406.423.4
COPCOP
COPinIncrease
41.741.772.7inIncrease
R
RR
L
LL Q
QQ s
11-15 [Also solved by EES on enclosed CD] An ideal vapor-compression refrigeration cycle withrefrigerant-134a as the working fluid is considered. The quality of the refrigerant at the end of the throttling process, the COP, and the power input to the compressor are to be determined.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.Analysis (a) In an ideal vapor-compression refrigeration cycle, the compression process is isentropic, therefrigerant 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/kg47.95
kJ/kg47.95liquidsat.
MPa8.0
kJ/kg37.275MPa8.0
KkJ/kg94456.0kJ/kg16.239
vaporsat.kPa140
34
MPa8.0@33
212
2
kPa140@1
kPa140@11
hh
hhP
hss
P
sshhP
f
g
g
T
QH
QL
·
Win·
·
0.8 MPa
4
32
10.14 MPa
The quality of the refrigerant at the end of the throttling process is
0.32208.212
08.2747.95
kPa140@
44
fg
f
hhh
xs
(b) The COP of the refrigerator is determined from its definition,
3.9716.23937.275
47.9516.239COP12
41
inR hh
hhwqL
(c) The power input to the compressor is determined from
kW1.2697.3
kW)60/300(COPR
inLQ
W
11-7
11-16 EES Problem 11-15 is reconsidered. The effect of evaporator pressure on the COP and the power input is to be investigated.Analysis The problem is solved using EES, and the solution is given below.
"Input Data"{P[1]=140 [kPa]}{P[2] = 800 [kPa]Fluid$='R134a'Eta_c=1.0 "Compressor isentropic efficiency" Q_dot_in=300/60 "[kJ/s]"}
"Compressor"x[1]=1 "assume inlet to be saturated vapor"h[1]=enthalpy(Fluid$,P=P[1],x=x[1])T[1]=temperature(Fluid$,h=h[1],P=P[1]) "properties for state 1"s[1]=entropy(Fluid$,T=T[1],x=x[1])h2s=enthalpy(Fluid$,P=P[2],s=s[1]) "Identifies state 2s as isentropic"h[1]+Wcs=h2s "energy balance on isentropic compressor"Wc=Wcs/Eta_c"definition of compressor isentropic efficiency"h[1]+Wc=h[2] "energy balance on real compressor-assumed adiabatic"s[2]=entropy(Fluid$,h=h[2],P=P[2]) "properties for state 2"T[2]=temperature(Fluid$,h=h[2],P=P[2])W_dot_c=m_dot*Wc
"Condenser"P[3]=P[2] "neglect pressure drops across condenser"T[3]=temperature(Fluid$,h=h[3],P=P[3]) "properties for state 3"h[3]=enthalpy(Fluid$,P=P[3],x=0) "properties for state 3"s[3]=entropy(Fluid$,T=T[3],x=0)h[2]=q_out+h[3] "energy balance on condenser"Q_dot_out=m_dot*q_out
"Valve"h[4]=h[3] "energy balance on throttle - isenthalpic"x[4]=quality(Fluid$,h=h[4],P=P[4]) "properties for state 4"s[4]=entropy(Fluid$,h=h[4],P=P[4])T[4]=temperature(Fluid$,h=h[4],P=P[4])
"Evaporator"P[4]=P[1] "neglect pressure drop across evaporator"q_in + h[4]=h[1] "energy balance on evaporator"Q_dot_in=m_dot*q_inCOP=Q_dot_in/W_dot_c "definition of COP"COP_plot = COP W_dot_in = W_dot_c
P1 [kPa] COPplot Win [kW]100 3.216 1.554175 4.656 1.074250 6.315 0.7918325 8.388 0.5961400 11.15 0.4483
11-8
0,0 0,2 0,4 0,6 0,8 1,0 1,2-50
-25
0
25
50
75
100
125
s [kJ/kg-K]
T[C]
800 kPa
140 kPa
R134a
T-s diagram for = 1.0
0 50 100 150 200 250 300101
102
103
104
h [kJ/kg]
P [k
Pa]
31.33 C
-18.8 C
R134a
P-h diagram for = 1.0
0,0 0,2 0,4 0,6 0,8 1,0 1,2-50
-25
0
25
50
75
100
125
s [kJ/kg-K]
T[C]
800 kPa
140 kPa
R134a
T-s diagram for = 0.6
11-9
0.60 0.65 0.70 0.75 0.80 0.85 0.90 0.95 1.002.2
2.4
2.6
2.8
3.0
3.2
3.4
3.6
3.8
4.0
Compressor efficiency
CO
P
COP vs Compressor Efficiency for R134a
100 150 200 250 300 350 4003
4
5
6
7
8
9
10
11
12
P[1] [kPa]
CO
Ppl
ot
100 150 200 250 300 350 4000.4
0.6
0.8
1
1.2
1.4
1.6
P[1] [kPa]
Win
11-10
11-17 A nonideal vapor-compression refrigeration cycle with refrigerant-134a as the working fluid isconsidered. The quality of the refrigerant at the end of the throttling process, the COP, the power input tothe compressor, and the irreversibility rate associated with the compression process are to be determined.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.Analysis (a) The refrigerant enters the compressor as a saturated vapor at the evaporator pressure, and leaves the condenser as saturated liquid at the condenser pressure. From the refrigerant tables (Tables A-12 and A-13),
throttlingkJ/kg47.95
kJ/kg47.95liquidsat.
MPa8.0
kJ/kg76.28185.0/16.23937.27516.239
/
kJ/kg37.275MPa8.0
KkJ/kg94456.0kJ/kg16.239
vapor.satkPa140
34
MPa8.0@33
121212
12
212
2
kPa140@1
kPa140@11
hh
hhP
hhhhhhhh
hss
P
sshhP
f
Css
C
ss
g
gT
2
QH· 2s
Win·
3 0.8 MPa
0.14 MPa1
QL·4
s
The quality of the refrigerant at the end of the throttling process is
0.32208.212
08.2747.95
kPa140@
44
fg
f
hhh
x
(b) The COP of the refrigerator is determined from its definition,
3.3716.23976.281
47.9516.239COP12
41
inR hh
hhwqL
(c) The power input to the compressor is determined from
kW1.4837.3kW5
COPinR
LQW
The exergy destruction associated with the compression process is determined from
120
0
0
surr120gen0destroyed ssmT
Tq
ssmTSTX
where
KkJ/kg96483.0kJ/kg76.281
MPa8.0
kg/s0348.0kJ/kg95.47239.16
kJ/s5
22
2
41
shP
hhQ
m L
L
L
Thus,
kW0.210KkJ/kg0.944560.96483kg/s0.0348K298destroyedX
11-11
11-18 A refrigerator with refrigerant-134a as the working fluid is considered. The rate of heat removalfrom 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
T
QH
QL
0.7 MPa 50 C
0.14 MPa-10 C
Win·
2
·
·
s
0.65 MPa24 C
4
3
2s
10.15 MPa
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
(c) The COP of the refrigerator is determined from its definition,
3.83kW5.06kW19.4COP
inR W
QL
11-12
11-19E An ice-making machine operates on the ideal vapor-compression refrigeration cycle, usingrefrigerant-134a as the working fluid. The power input to the ice machine is to be determined.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.Analysis In an ideal vapor-compression refrigeration cycle, the compression process is isentropic, therefrigerant enters the compressor as a saturated vapor at the evaporator pressure, and leaves the condenser as saturated liquid at the condenser pressure. From the refrigerant tables (Tables A-12E and A-13E),
throttlingBtu/lbm39.33
Btu/lbm39.33liquidsat.
psia80
Btu/lbm00.115psia80
RBtu/lbm22567.0Btu/lbm73.102
vaporsat.psia20
34
psia08@33
212
2
psia20@1
psia20@11
hh
hhP
hss
P
sshhP
f
g
g T
QH
QL
·
Win·
·
80 psia
4
32
120 psia
sThe cooling load of this refrigerator is
Btu/s0.7042Btu/lbm169lbm/s15/3600iceice hmQL
Then the mass flow rate of the refrigerant and the power input become
and
hp0.176Btu/s0.7068
hp1Btu/lbm102.73115.00lbm/s0.01016
lbm/s0.01016Btu/lbm33.39102.73
Btu/s0.7042
12in
41
hhmW
hhQm
R
LR
11-13
11-20 A refrigerator with refrigerant-134a as the working fluid is considered. The power input to thecompressor, 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
vT
QH
QL -18.5 C
1 MPa
0.14 MPa-10 C
Win·
2
·
·
s
0.95 MPa30 C
4
3
2s
10.15 MPa
Then the mass flow rate of the refrigerant and the power input becomes
kW1.8878.0/kJ/kg246.36289.20kg/s0.03423/
kg/s0.03423/kgm0.14605/sm0.3/60
12in
3
3
1
1
Cs hhmW
mvV
(b) The rate of heat removal from the refrigerated space is
kW4.99kJ/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
hhmQ
PPP
11-14
11-21 EES Problem 11-20 is reconsidered. The effects of the compressor isentropic efficiency and thecompressor 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=0.78Fluid$='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_inCOP=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]"
11-15
COPplot Win[kW]
Qin[kW]
c[kW]
2.041 0.8149 1.663 0.62.381 0.6985 1.663 0.72.721 0.6112 1.663 0.83.062 0.5433 1.663 0.93.402 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
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
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
c
Qin
[kW
] 1.0 1.0 0.50.5 0.10.1
V1 m3/min
11-16
11-22 A refrigerator uses refrigerant-134a as the working fluid and operates on the ideal vapor-compression refrigeration cycle. The mass flow rate of the refrigerant, the condenser pressure, and the COPof the refrigerator are to be determined.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.Analysis (a) (b) From the refrigerant-134a tables (Tables A-11 through A-13)
4
3 2
1
Win.
120 kPa x=0.3
Expansionvalve
Compressor
Evaporator
Condenser60 C
QH
.
.QL
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 T
QH
QL4s
·
Win·
·4
32
10.12 MPa
kg/s0.00727kg236.97)kJ/(298.87
kW45.0
12
in
hhW
m
(c) The refrigeration load and the COP are
Q kW091.1kJ/kg)83.8697kg/s)(236.0727.0()( 41 hhmL
2.43kW0.45kW091.1COP
in
L
WQ
s
11-17
Selecting the Right Refrigerant
11-23C The desirable characteristics of a refrigerant are to have an evaporator pressure which is above theatmospheric pressure, and a condenser pressure which corresponds to a saturation temperature above thetemperature of the cooling medium. Other desirable characteristics of a refrigerant include being nontoxic,noncorrosive, nonflammable, chemically stable, having a high enthalpy of vaporization (minimizes themass flow rate) and, of course, being available at low cost.
11-24C The minimum pressure that the refrigerant needs to be compressed to is the saturation pressure ofthe 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.
11-25C Allowing a temperature difference of 10 C for effective heat transfer, the evaporation temperatureof 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.
11-26 A refrigerator that operates on the ideal vapor-compression cycle with refrigerant-134a isconsidered. 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 saturationpressures corresponding to these temperatures are 0.133 MPa and 0.888 MPa. Therefore, the recommendedevaporator and condenser pressures are 0.133 MPa and 0.888 MPa, respectively.
11-27 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 saturationpressures corresponding to these temperatures are 0.293 MPa and 0.816 MPa. Therefore, the recommendedevaporator and condenser pressures are 0.293 MPa and 0.816 MPa, respectively.
Heat Pump Systems
11-28C A heat pump system is more cost effective in Miami because of the low heating loads and highcooling loads at that location.
11-29C A water-source heat pump extracts heat from water instead of air. Water-source heat pumps havehigher COPs than the air-source systems because the temperature of water is higher than the temperature of air in winter.
11-18
11-30E A heat pump that operates on the ideal vapor-compression cycle with refrigerant-134a isconsidered. The power input to the heat pump and the electric power saved by using a heat pump instead of a resistance heater 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, therefrigerant enters the compressor as a saturated vapor at the evaporator pressure, and leaves the condenser as saturated liquid at the condenser pressure. From the refrigerant tables (Tables A-12E and A-13E),
throttlingBtu/lbm79.41
Btu/lbm79.41liquidsat.
psia120
Btu/lbm62.116psia120
RBtu/lbm22188.0Btu/lbm81.108
vaporsat.psia50
34
psia120@33
212
2
psia50@1
psia50@11
hh
hhP
hss
P
sshhP
f
g
g
QH
QL
House·
Win·
·
120 psia
4
32
150 psia
T
The mass flow rate of the refrigerant and the power input to the compressor are determined from s
and
Btu/s0.7068=hp1sinceBtu/s.7381Btu/lbm108.81116.62kg/s0.2227
lbm/s0.2227Btu/lbm41.79116.62Btu/s060,000/360
12in
32
hp2.46hhmW
hhQ
m H
H
H
The electrical power required without the heat pump is
Thus,
kW0.7457=hp1sincekW75.1546.258.23
hp.5823Btu/s0.7068
hp1Btu/s060,000/360
insaved
hp21.1WWW
QW
e
He
11-19
11-31 A heat pump that operates on the ideal vapor-compression cycle with refrigerant-134a is considered.The power input to the heat pump is to be determined.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.Analysis In an ideal vapor-compression refrigeration cycle, the compression process is isentropic, therefrigerant 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/kg22.127
kJ/kg22.127liquidsat.
MPa4.1
kJ/kg54.282MPa4.1
KkJ/kg93006.0kJ/kg88.251
vaporsat.kPa320
34
MPa4.1@33
212
2
kPa320@1
kPa320@11
hh
hhP
hss
P
sshhP
f
g
g
QH
QL
House·
Win·
·
1.4 MPa
4
32
10.32 MPa
T
The heating load of this heat pump is determined from
kW.0515C1545CkJ/kg4.18kg/s0.12water12 TTcmQH s
and
Then,
kW2.97kJ/kg251.88282.54kg/s0.09688
kg/s0.09688kJ/kg127.22282.54
kJ/s15.05
12in
32
hhmW
hhQ
qQm
R
H
H
HR
11-20
11-32 A heat pump with refrigerant-134a as the working fluid heats a house by using underground water as the heat source. The power input to the heat pump, the rate of heat absorption from the water, and theincrease in electric power input if an electric resistance heater is used instead of a heat pump are to be determined.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.Analysis (a) From the refrigerant tables (Tables A-12 and A-13),
QH
QL
1 MPa
Water, 8 C
60 CHouse
0 C
Win·
2
·
·
s
4
330 C
10.28 MPa
T
throttlingkJ/kg58.93
kJ/kg58.93C30MPa0.1
kJ/kg38.293C60MPa0.1
kJ/kg83.250C0kPa280
34
C30@33
3
22
2
11
1
hh
hhTP
hTP
hTP
f
The mass flow rate of the refrigerant is
kg/s0.08341kJ/kg93.58293.38
kJ/s0060,000/3,6
32 hhQ
qQm H
H
HR
Then the power input to the compressor becomes
kW3.55kJ/kg250.83293.38kg/s0.0834112in hhmW
(b) The rate of hat absorption from the water is
kW13.12kJ/kg93.58250.83kg/s0.0834141 hhmQL
(c) The electrical power required without the heat pump is
Thus,
kW13.1255.367.16
kW16.67kJ/s3600/000,60
inincrease WWW
QW
e
He
11-21
11-33 EES Problem 11-32 is reconsidered. The effect of the compressor isentropic efficiency on the power input to the compressor and the electric power saved by using a heat pump rather than electric resistanceheating is to be investigated.Analysis The problem is solved using EES, and the solution is given below.
"Input Data""Input Data is supplied in the diagram window""P[1]=280 [kPa] T[1] = 0 [C] P[2] = 1000 [kPa] T[3] = 30 [C] Q_dot_out = 60000 [kJ/h] Eta_c=1.0Fluid$='R134a'""Use ETA_c = 0.623 to obtain T[2] = 60C"
"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])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"{h[2]=enthalpy(Fluid$,P=P[2],T=T[2]) }T[2]=temperature(Fluid$,h=h[2],P=P[2])W_dot_c=m_dot*Wc
"Condenser"P[3] = P[2] 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]=Qout+h[3] "energy balance on condenser"Q_dot_out*convert(kJ/h,kJ/s)=m_dot*Qout
"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"Q_dot_in=m_dot*Q_inCOP=Q_dot_out*convert(kJ/h,kJ/s)/W_dot_c "definition of COP"COP_plot = COP W_dot_in = W_dot_c E_dot_saved = Q_dot_out*convert(kJ/h,kJ/s) - W_dot_c"[kW]"
11-22
Win [kW] c Esaved
3.671 0.6 133.249 0.7 13.422.914 0.8 13.752.641 0.9 14.032.415 1 14.25
0,6 0,65 0,7 0,75 0,8 0,85 0,9 0,95 12,4
2,6
2,8
3
3,2
3,4
3,6
3,8
c
Win
0,6 0,65 0,7 0,75 0,8 0,85 0,9 0,95 112,75
13,05
13,35
13,65
13,95
14,25
c
Esaved
[kW]
11-23
11-34 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 andthe rate of heat supplied to the heated room if this heat pump operated on the ideal vapor-compressioncycle 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
4
3 2
1
Win
.
800 kPa 55 C
Expansionvalve
Compressor
Evaporator
Condenser
750 kPa QH.
.QL
The isentropic efficiency of the compressor is T
QH
QL
Win·
2
·
·4
3
2
1
0.67087.24776.29187.24726.277
12
12
hhhh s
C
(b) The rate of heat supplied to the room is
Q kW3.67kJ/kg)91.8776kg/s)(291.018.0()( 32 hhmH
(c) The power input and the COP are
W kW790.0kJ/kg)87.24776kg/s)(291.018.0()( 12in hhm
s4.64
790.067.3COP
inWQH
(d) The ideal vapor-compression cycle analysis of the cycle is as follows: T
QH
QL4s
·
Win·
·
0.8 MPa
4
32
10.2 MPa
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
s6.18
46.24425.27347.9525.273COP
12
32
hhhh
kW3.20kJ/kg)47.9525kg/s)(273.018.0()( 32 hhmQH
11-24
11-35 A geothermal heat pump is considered. The degrees of subcooling done on the refrigerant in thecondenser, the mass flow rate of the refrigerant, the heating load, the COP of the heat pump, the minimumpower 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)
40 CWater50 C
Win.
sat. vap.
Expansionvalve Compressor
Evaporator
Condenser
20 Cx=0.23
QH.
4
32
1
1.4 MPa s2 = s1
.QL
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 temperatureat the condenser outlet are
T
QH
QL4s
·
Win·
·
1.4MPa
4
3
2
1
C40.52kPa1400@satT
C59.48kJ24.121
kPa14003
3
3 ThP
(from EES)
Then, the degrees of subcooling isC3.8159.4840.523satsubcool TTT
(b) The rate of heat absorbed from the geothermal water in the evaporator is
s
kW718.2kJ/kg)53.16734kg/s)(209.065.0()( 21 wwwL hhmQThis 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
kW3.074kJ/kg)24.12100kg/s)(280.01936.0()( 32 hhmQ RH
4.68kW0.6564
kW074.3COPin
H
WQ
(d) The reversible COP of the cycle is
92.12)27350/()27325(1
1/1
1COPrevHL TT
The corresponding minimum power input is
kW0.23892.12kW074.3
COPrevminin,
HQW
11-25
Innovative Refrigeration Systems
11-36C Performing the refrigeration in stages is called cascade refrigeration. In cascade refrigeration, twoor more refrigeration cycles operate in series. Cascade refrigerators are more complex and expensive, butthey have higher COP's, they can incorporate two or more different refrigerants, and they can achievemuch lower temperatures.
11-37C Cascade refrigeration systems have higher COPs than the ordinary refrigeration systems operatingbetween the same pressure limits.
11-38C The saturation pressure of refrigerant-134a at -32 C is 77 kPa, which is below the atmosphericpressure. In reality a pressure below this value should be used. Therefore, a cascade refrigeration systemwith a different refrigerant at the bottoming cycle is recommended in this case.
11-39C We would favor the two-stage compression refrigeration system with a flash chamber since it issimpler, cheaper, and has better heat transfer characteristics.
11-40C Yes, by expanding the refrigerant in stages in several throttling devices.
11-41C To take advantage of the cooling effect by throttling from high pressures to low pressures.
11-26
11-42 A two-stage cascade refrigeration system is considered. Each stage operates on the ideal vapor-compression cycle with refrigerant-134a as the working fluid. The mass flow rate of refrigerant through thelower cycle, the rate of heat removal from the refrigerated space, the power input to the compressor, andthe COP of this cascade refrigerator are to be determined.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. 3The heat exchanger is adiabatic. Analysis (a) Each stage of the cascade refrigeration cycle is said to operate on the ideal vapor compressionrefrigeration cycle. Thus the compression process is isentropic, and the refrigerant enters the compressoras a saturated vapor at the evaporator pressure. Also, the refrigerant leaves the condenser as a saturatedliquid at the condenser pressure. The enthalpies of the refrigerant at all 8 states are determined from therefrigerant tables (Tables A-11, A-12, and A-13) to be
kJ/kg47.95kJ/kg91.269
kJ/kg94.63kJ/kg58.260
kJ/kg,47.95,kJ/kg55.255
,kJ/kg94.63,kJ/kg16.239
8
6
4
2
7
5
3
1
hhhh
hhhh
T
The mass flow rate of the refrigerant through thelower cycle is determined from an energy balance on the heat exchanger:
kg/s0.1954kg/s0.2494.6358.26047.9555.255
0
32
85
3285
outin
(steady)0systemoutin
AB
BA
iiee
mhhhh
m
hhmhhm
hmhm
EE
EEE
3
QL
B
A 58
0.8 MPa
6
·
0.4 MPa
4
7 2
1
0.14 MPa
s
(b) The rate of heat removed by a cascade cycle is the rate of heat absorption in the evaporator of the lowest stage. The power input to a cascade cycle is the sum of the power inputs to all of the compressors:
kW7.63
kW34.24
kJ/kg239.16260.58kg/s0.1954kJ/kg255.55269.91kg/s0.24
kJ/kg63.94239.16kg/s0.1954
1256incompII,incompI,in
41
hhmhhmWWW
hhmQ
BA
BL
(c) The COP of this refrigeration system is determined from its definition,
494.kW7.63kW34.24COP
innet,R W
QL
11-27
11-43 A two-stage cascade refrigeration system is considered. Each stage operates on the ideal vapor-compression cycle with refrigerant-134a as the working fluid. The mass flow rate of refrigerant through thelower cycle, the rate of heat removal from the refrigerated space, the power input to the compressor, andthe COP of this cascade refrigerator are to be determined.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. 3The heat exchanger is adiabatic. Analysis (a) Each stage of the cascade refrigeration cycle is said to operate on the ideal vapor compressionrefrigeration cycle. Thus the compression process is isentropic, and the refrigerant enters the compressoras a saturated vapor at the evaporator pressure. Also, the refrigerant leaves the condenser as a saturatedliquid at the condenser pressure. The enthalpies of the refrigerant at all 8 states are determined from therefrigerant tables (Tables A-11, A-12, and A-13) to be
kJ/kg47.95kJ/kg66.268
kJ/kg54.77kJ/kg34.267
kJ/kg,47.95,kJ/kg92.260
,kJ/kg54.77,kJ/kg16.239
8
6
4
2
7
5
3
1
hhhh
hhhh
T
The mass flow rate of the refrigerant through thelower cycle is determined from an energy balance on the heat exchanger:
kg/s0.2092kg/s24.054.7734.26747.9592.260
0
32
85
3285
outin
(steady)0systemoutin
AB
BA
iiee
mhhhh
m
hhmhhmhmhm
EE
EEE
QL
3B
A 58
0.8 MPa
6
·
0.55 MPa
4
7 2
1
0.14 MPa
s
(b) The rate of heat removed by a cascade cycle is the rate of heat absorption in the evaporator of the lowest stage. The power input to a cascade cycle is the sum of the power inputs to all of the compressors:
kW7.75
kW33.81
kJ/kg16.23934.267kg/s2092.0kJ/kg92.26066.268kg/s24.0
kJ/kg54.7716.239kg/s2092.0
1256incompII,compI,inin
41
hhmhhmWWW
hhmQ
BA
BL
(c) The COP of this refrigeration system is determined from its definition,
4.36kW75.7kW81.33COP
innet,R W
QL
11-28
11-44 [Also solved by EES on enclosed CD] A two-stage compression refrigeration system withrefrigerant-134a as the working fluid is considered. The fraction of the refrigerant that evaporates as it isthrottled 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 T
QL
73B
A96
1 MPa 4
·
0.5 MPa
8
5 2
1
0.14 MPaThe fraction of the refrigerant that evaporates as it is throttled to the flash chamber is simply thequality at state 6,
0.182898.185
33.7332.10766
fg
f
hhh
x s
(b) The enthalpy at state 9 is determined from an energy balance on the mixing chamber:
kJ/kg21.26431.2651828.0130.2591828.011
0
9
26369
outin
(steady)0systemoutin
hhxhxh
hmhm
EE
EEE
iiee
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/s0.2043kJ/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
kW33.88
(c) The coefficient of performance is determined from
3.75kW9.03kW33.88COP
innet,R W
QL
11-29
11-45 EES Problem 11-44 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 =1.0Eta_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"h[1]=enthalpy(R134a,P=P[1],x=x1) "properties for state 1"
11-30
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"
compA compB Qout COP0,8 0,8 45,32 2,963
0,8333 0,8333 44,83 3,0940,8667 0,8667 44,39 3,225
0,9 0,9 43,97 3,3570,9333 0,9333 43,59 3,4880,9667 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
11-31
0,8 0,84 0,88 0,92 0,96 142,5
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
comp
Qout
[kW]
COP
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
11-32
11-46 [Also solved by EES on enclosed CD] A two-stage compression refrigeration system withrefrigerant-134a as the working fluid is considered. The fraction of the refrigerant that evaporates as it isthrottled 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
T
QL
73B
A96
1 MPa 4
·
0.32 MPa
8
5 2
1
0.14 MPaThe fraction of the refrigerant that evaporates as it is throttled to the flash chamber is simply the quality at state 6,
0.265171.196
16.5532.10766
fg
f
hhh
x
s(b) The enthalpy at state 9 is determined from an energy balance on the mixing chamber:
kJ/kg84.25490.2552651.0188.2512651.011
0
9
26369
outin
(steady)0systemoutin
hhxhxh
hmhm
EE
EEE
iiee
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
kW33.80
(c) The coefficient of performance is determined from
3.71kW9.10kW33.80COP
innet,R W
QL
11-33
11-47 A two-stage cascade refrigeration cycle is considered. The mass flow rate of the refrigerant throughthe upper cycle, the rate of heat removal from the refrigerated space, and the COP of the refrigerator are tobe 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
Win.Expansion
valveCompressor
Evaporator
Condenser
4
32
1
Win
.Expansionvalve
Compressor
Evaporator
Condenser
QH.
8
76
5
.QL
kJ/kg30.263kPa500
212
2sh
ssP
kJ/kg01.26846.244
46.24430.26380.0 22
12
12
hh
hhhh s
C
kJ/kg33.73kJ/kg33.73
34
kPa500@3
hhhh f
kJ/kg.K9269.0kJ/kg55.255
kPa400@5
kPa004@5
g
g
sshh
kJ/kg33.278kPa1200
656
6sh
ssP
kJ/kg02.28455.255
55.25533.27880.0 66
56
56
hh
hhhh s
C
kJ/kg77.117kJ/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 theheat exchanger
kg/s0.212AA
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
kW25.67kJ/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
2.68566.9
67.25COPin
L
WQ
11-34
11-48 A two-stage cascade refrigeration cycle with a flash chamber is considered. The mass flow rate ofthe refrigerant through the high-pressure compressor, the rate of heat removal from the refrigerated space, the COP of the refrigerator, and the rate of heat removal and the COP if this refrigerator operated on a single-stage 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 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
7
Flashchamber
Expansionvalve Low-press.
Compressor
Evaporator8
9
2
1
Expansionvalve High-press.
Compressor
Condenser
QH
.
6
54
3
.QL
kJ/kg09.261kPa450
212
2sh
ssP
kJ/kg24.26546.244
46.24409.26180.0 22
12
12
hh
hhhh s
C
kJ/kg81.68kJ/kg81.68
kJ/kg77.117kJ/kg77.117
kJ/kg53.257
78
kPa450@7
56
kPa1200@5
kPa450@3
hhhhhhhh
hh
f
f
g
2594.0kPa450
kJ/kg77.1176
6
6 xPh
The mass flow rate of the refrigerant through thehigh pressure compressor is determined from amass balance on the flash chamber
kg/s0.20250.2594-1
kg/s15.01 6
7
xm
m
Also, kg/s05255.015.02025.073 mmm
(b) The enthalpy at state 9 is determined from an energy balance on the mixing chamber:
kJ/kg24.263kJ/kg)53kg/s)(257.(0.05255kJ/kg)24kg/s)(265.(0.15kg/s)2025.0( 99
33279
hh
hmhmhm
Then,
kJ/kg9451.0kJ/kg24.263
kPa4509
9
9 shP
kJ/kg27.284kPa1200
494
4sh
ssP
kJ/kg53.28924.263
24.26327.28480.0 44
94
94
hh
hhhh s
C
11-35
The rate of heat removal from the refrigerated space is
kW26.35kJ/kg)81.6846kg/s)(244.15.0()( 817 hhmQL
(c) The power input and the COP are
kW442.8kJ/kg)24.26353kg/s)(289.2025.0(kJ/kg)46.24424kg/s)(265.15.0()()( 94127in hhmhhmW
3.12442.8
35.26COPin
L
WQ
(d) If this refrigerator operated on a single-stage cycle between the same pressure limits, we would have
kJ/kg.K9377.0kJ/kg46.244
kPa200@1
kPa200@1
g
g
sshh
T
QH
QL
Win·
2
·
·4
3
2
1
kJ/kg84.281kPa1200
212
2sh
ssP
kJ/kg19.29146.244
46.24484.28180.0 22
12
12
hh
hhhh s
C
s
kJ/kg77.117
kJ/kg77.117
34
kPa1200@3
hh
hh f
kW25.66kJ/kg)77.11746kg/s)(244.2025.0()( 41 hhmQL
kW465.9kJ/kg)46.24419kg/s)(291.2025.0()( 12in hhmW
2.71465.9
66.25COPin
L
WQ
11-36
Gas Refrigeration Cycles
11-49C The ideal gas refrigeration cycle is identical to the Brayton cycle, except it operates in the reverseddirection.
11-50C 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,Stirling1
1T TH L/
11-51C In the ideal gas refrigeration cycle, the heat absorption and the heat rejection processes occur at constant pressure instead of at constant temperature.
11-52C In aircraft cooling, the atmospheric air is compressed by a compressor, cooled by the surroundingair, and expanded in a turbine. The cool air leaving the turbine is then directly routed to the cabin.
11-53C No; because h = h(T) for ideal gases, and the temperature of air will not drop during a throttling(h1 = h2) process.
11-54C By regeneration.
11-37
11-55 An ideal-gas refrigeration cycle with air as the working fluid is considered. The maximum and minimum temperatures in the cycle, the COP, and the rate of refrigeration are to be determined.Assumptions 1 Steady operating conditions exist. 2 Air is an ideal gas with variable specific heats. 3Kinetic and potential energy changes are negligible.Analysis (a) We assume both the turbine and the compressor to be isentropic, the turbine inlet temperatureto 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
250 050 7329
300191 386
1
3
250 K kJ / kg
300 K kJ / kg
..
..
T
4
3
QH· 2
1QRefrig·
-23 C27 C
Thus,
kJ/kg9721846203861
31
kJ/kg6034219872732903
4
min43
4
2
max21
2
34
12
.hTT..P
PPP
.hTT..P
PPP
rr
rr
K219.0
K342.2s
(b) The COP of this ideal gas refrigeration cycle is determined from
COPRnet, in comp, in turb, out
qw
qw w
L L
whereq h h
w h hw h h
L 1 4
2 1
3 4
250 05 218 97 3108342 60 250 05 92 5530019 218 97 8122
. . .
. . .. . .
kJ / kgkJ / kgkJ / kg
comp, in
turb, out
Thus, COPR31 08
92 55 8122.
. .2.74
(c) The rate of refrigeration is determined to be
kJ/s2.49kJ/kg31.08kg/s0.08refrig LqmQ
11-38
11-56 [Also solved by EES on enclosed CD] An ideal-gas refrigeration cycle with air as the working fluidis 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 temperatureto 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 k
K k
..
..
J / kg
J / kg
T
4
3
QH· 2
1QRefrig·
47 C12 C
Thus,
kJ/kg76201 K82013475073751
25050
kJ/kg17452 K4450792515841
50250
4
43
4
2
21
2
34
12
.h
.T..PPPP
.h
.T..PPPP
rr
rr s
Then the rate of refrigeration is
kW6.67kJ/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/s0.08kW13.36kJ/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
QW
L 1.72
11-39
11-57 EES Problem 11-56 is reconsidered. The effects of compressor and turbine isentropic efficiencies onthe 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]=Ts2ss[1]=s2sTs[2]=Ts4ss[2]=s4s
11-40
COP comp turb QRefrig[kW]
Winnet[kW]
0.6937 0.7 1 6.667 9.6120.9229 0.8 1 6.667 7.2241.242 0.9 1 6.667 5.3681.717 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
comp
QRefrig
[kW]
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
comp
Win;net
[kW]
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
comp
COP
turb
0.70.70.850.851.01.0
11-41
11-58E 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 temperatureto 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-17E),
T hP
T hP
r
r
1 1
1 3
500 119 4810590
580 138 661 7800
1
3
R B
R B
..
..
tu / lbm
tu / lbm
T
4
3
QH· 2
1QRefrig·
120 F40 F
Thus,
Btu/lbm14101R44235933078001
3010
Btu/lbm68163R9683177305901
1030
4
43
4
2
21
2
34
12
.h.T..P
PPP
.h.T..P
PPP
rr
rr s
Then the rate of refrigeration is
Btu/s9.17 Btu/lbm101.14119.48lbm/s0.541refrig hhmqmQ L
(b) The net power input is determined from
W W Wnet, in comp, in turb, out
where
Btu/s18.79 Btu/lbm101.14138.66lbm/s0.5 Btu/s22.10Btu/lbm119.48163.68lbm/s0.5
43outturb,
12incomp,
hhmWhhmW
Thus, . . .Wnet, in Btu / s2210 18 76 3 34 4.73 hp
(c) The COP of this ideal gas refrigeration cycle is determined from
COP 9.17 Btu / s3.34 Btu / sR
net, in
QW
L 2.75
11-42
11-59 [Also solved by EES on enclosed CD] An ideal-gas refrigeration cycle with air as the working fluidis 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 thecompressor 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 k
K k
..
..
J / kg
J / kg
4
2
4s
3
QH·
2
11QRefrig·
T
47 C12 CThus,
kJ/kg76201 K82013475073751
25050
kJ/kg17452 K4450792515841
50250
4
43
4
2
21
2
34
12
.h
.T..PPPP
.h
.T..PPPP
s
srr
s
srrs
Also,
kJ/kg54219762012932085029320
433443
43
.....
hhhhhhhh
sTs
T
Then the rate of refrigeration is
kW5.25kJ/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,
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
QW
L 0.61
11-43
11-60 A gas refrigeration cycle with helium as the working fluid is considered. The minimum temperaturein 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·Kand k = 1.667 (Table A-2).
4
2
4s
3
QH·
2s
11QRefrig·
TAnalysis (a) From the isentropic relations,
K1.20831K323
K2.4083K263
667.1/667.0k/1k
3
434
667.1/667.0k/1k
1
212
PP
TT
PP
TT
s
s 50 C-10 C
sand
K5.44480.0/2632.408263/
1.20832380.0323
121212
12
12
12
min
433443
43
43
43
Csss
C
sTss
T
TTTTTTTT
hhhh
TTTTT
TTTT
hhhh
K231.1
(b) The COP of this ideal gas refrigeration cycle is determined from
0.3561.2313232635.444
1.231263
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/s0.109K231.1263KkJ/kg5.1926
kJ/s18
41
refrig
41
refrigrefrig
TTcQ
hhQ
mpL
11-44
11-61 An ideal-gas refrigeration cycle with air as the working fluid is considered. The lowest temperaturethat can be obtained by this cycle, the COP, and the mass flow rate of air are to be determined.Assumptions 1 Steady operating conditions exist. 2 Air is an ideal gas with constant specific heats. 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-2). Analysis (a) The lowest temperature in the cycle occurs at the turbine exit. From the isentropic relations,
min
4140k1k
4
545
4140k1k
1
212
K617341 K258
C122.3 K33954K266
T.PPTT
.PPTT
././
././
C99.4
Qrege
QH· 2
4
6QRefrig·
5
-15 C
-7 C
3
1
T
(b) From an energy balance on the regenerator, 27 C
6143
outin
(steady)0systemoutin 0
hhmhhmhmhm
EE
EEE
iiee
sor,
or,C49C15C27C74316
61436143
TTTT
TTTTTTcmTTcm pp
Then the COP of this ideal gas refrigeration cycle is determined from
1.12C4.9915C73.122
C4.99C49
COP
5412
56
5412
56
outturb,incomp,innet,R
TTTTTT
hhhhhh
wwq
wq LL
(c) The mass flow rate is determined from
kg/s0.237C99.449CkJ/kg1.005
kJ/s12
56
refrig
56
refrigrefrig
TTcQ
hhQ
mpL
11-45
11-62 An ideal-gas refrigeration cycle with air as the working fluid is considered. The lowest temperaturethat can be obtained by this cycle, the COP, and the mass flow rate of air are to be determined.Assumptions 1 Steady operating conditions exist. 2 Air is an ideal gas with constant specific heats. 3Kinetic and potential energy changes are negligible.Properties The properties of air at room temperatureare cp = 1.005 kJ/kg·K and k = 1.4 (Table A-2).
5
2
Qrege
27 C
QH· 2s
4
6QRefrig·
5
-15 C
-7 C
3
1
T
s
Analysis (a) The lowest temperature in the cycle occurs at the turbine exit. From the isentropicrelations,
C99.4K617341 K258
C122.3K33954 K266
4140k1k
4
545
4140k1k
1
212
.PPTT
.PPTT
././
s
././
s
and
C4165750731227
4991580015
121212
12
12
12
min
544554
54
54
54
.././TTTT
TTTT
hhhh
T..TTTT
TTTT
hhhh
Csss
C
sTss
TC82.5
(b) From an energy balance on the regenerator,
6143
outin(steady)0
systemoutin 0
hhmhhmhmhm
EEEEE
iiee
or,
or,C49C15C27C74316
61436143
TTTT
TTTTTTcmTTcm pp
Then the COP of this ideal gas refrigeration cycle is determined from
0.32C5.8215C74.165
C5.82C49
COP
5412
56
5412
56
outturb,incomp,innet,R
TTTTTT
hhhhhh
wwq
wq LL
(c) The mass flow rate is determined from
kg/s0.356C82.549CkJ/kg1.005
kJ/s12
56
refrig
56
refrigrefrig
TTcQ
hhQ
mpL
11-46
11-63 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 P
PTsT
Turbine
RegeneratorHeatExch.
HeatExch.
QL.
21
Compressor
6
54
3
.QH
K5.4722.273
2.2734.43280.0 22
12
12
12
12
TT
TTTT
hhhh ss
C
The temperature at state 4 can be determined by solving the following twoequations simultaneously:
4.1/4.0
4
k/1k
4
545 5
1TPP
TT s
ssT TT
Thhhh
54
4
54
54 2.19385.0
Using EES, we obtain T4 = 281.3 K.An energy balance on the regenerator may be written as
5
2
Qregen
QH· 2s
4
6QRefrig·
5s
3
1
s
T
or,K3.2463.2812.3082.2734316
61436143
TTTT
TTTTTTcmTTcm pp
35 C
0 CThe effectiveness of the regenerator is
0.4343.2462.3083.2812.308
63
43
63
43regen TT
TThhhh
-80 C
(b) The refrigeration load is
kW21.36K)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
0.47843.3513.80
36.21COPoutT,inC,innet, WW
QW
Q LL
11-47
(d) The simple gas refrigeration cycle analysis is as follows:
4
2
4s
3
QH·
2
11QRefrig·
K6.19451K2.3081 4.1/4.0k/1k
34 rTsT
T
K6.2116.1942.308
2.30885.0 4
4
43
43 TT
TTTT
sT
35 C 0 C
kW24.74kJ/kg)6.21173.2kJ/kg.K)(25kg/s)(1.004.0(
)( 41 TTcmQ pL
s
kW32.41kJ/kg)6.211(308.2)2.273(472.5kJ/kg.K)5kg/s)(1.004.0(
)()( 4312innet, TTcmTTcmW pp
0.59932.4174.24COP
innet,WQL
11-48
Absorption Refrigeration Systems
11-64C Absorption refrigeration is the kind of refrigeration that involves the absorption of the refrigerantduring part of the cycle. In absorption refrigeration cycles, the refrigerant is compressed in the liquidphase instead of in the vapor form.
11-65C 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 externalheat source.
11-66C In absorption refrigeration, water can be used as the refrigerant in air conditioning applicationssince the temperature of water never needs to fall below the freezing point.
11-67C The fluid in the absorber is cooled to maximize the refrigerant content of the liquid; the fluid inthe generator is heated to maximize the refrigerant content of the vapor.
11-68C The coefficient of performance of absorption refrigeration systems is defined as
geninpump,genR inputrequired
outputdesiredCOPQQ
WQQ LL
11-69C 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 thepump.
11-70 The COP of an absorption refrigeration system that operates at specified conditions is given. It is tobe 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.
11-71 The conditions at which an absorption refrigeration system operates are specified. The maximumCOP this absorption refrigeration system can have is to be determined.Analysis The maximum COP that this refrigeration system can have is
2.64273298
273K393K29811COP
0
0maxR,
L
L
s TTT
TT
11-49
11-72 The conditions at which an absorption refrigeration system operates are specified. The maximumrate 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 5kJ/h10515.1COP 5genmaxR,maxL, QQ
11-73E 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
11-74 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 lawefficiency of an actual chiller are to be determined.Properties The enthalpy of vaporization of water at 200C 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,
sTT
The COP of the reversible refrigerator is
52.7K)15.27310()15.27325(
K)15.27310(COP0
revR,L
L
TTT
T0
TL
Rev.Ref.
Ts
T0
Rev.HE
The COP of the reversible absorption refrigerator is 78.2)52.7)(370.0(COPCOP revR,revth,revabs,
The heat input to the reversible heat engine is
kW911.72.78
kW22COP revabs,
inLQ
Q
Then, the rate at which the steam condenses becomes
kg/s0.00408kJ/kg1939.8kJ/s911.7in
fgs h
Qm
(b) The power input to the refrigerator is equal to the power output from the heat enginekW2.93)kW191.7)(370.0(inrevth,HEout,Rin, QWW
(c) The second-law efficiency of an actual absorption chiller with a COP of 0.7 is
0.2522.78
7.0COPCOP
revabs,
actualII
11-50
Special Topic: Thermoelectric Power Generation and Refrigeration Systems
11-75C The circuit that incorporates both thermal and electrical effects is called a thermoelectric circuit.
11-76C When two wires made from different metals joined at both ends (junctions) forming a closedcircuit and one of the joints is heated, a current flows continuously in the circuit. This is called theSeebeck effect. When a small current is passed through the junction of two dissimilar wires, the junction iscooled. This is called the Peltier effect.
11-77C No.
11-78C No.
11-79C Yes.
11-80C 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 temperaturedifference, and the temperature can be measured by simply measuring voltages.
11-81C The performance of thermoelectric refrigerators improves considerably when semiconductors are used instead of metals.
11-82C 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.
11-83E A thermoelectric generator that operates at specified conditions is considered. The maximumthermal efficiency this thermoelectric generator can have is to be determined.Analysis The maximum thermal efficiency of this thermoelectric generator is the Carnot efficiency,
%3.31R800R550
11Carnotth,maxth,H
L
TT
11-84 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
11-51
11-85 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
11-86E A thermoelectric cooler that operates at specified conditions with a given COP is considered. Therequired 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 QW
WQ
11-87 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, = 1 kg/Land 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 canneddrinks,
W6.71=kW00671.0s360012
kJ290
kJ290=C3)-C)(25kJ/kgkg)(4.1815.3(kg3.15=L)0kg/L)(0.351(9
coolingcooling
cooling
tQ
Q
TmcQm V
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
W6 71 0 200.186
11-52
11-88E 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 theelectric 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,
W36.2Btu1
J1055s6015
Btu84.30
Btu30.84=F38)-F)(78Btu/lbmlbm)(1.0771.0(
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,
W49.7Btu1
J1055s6015
Btu4.42
Btu42.4=F75)-F)(130Btu/lbmlbm)(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
11-89 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
Thus,
kW39.4kJ/h142,000kJ/h)(0.142)(106inmaxth,maxout, QW
11-53
Review Problems
11-90 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
5.061K253/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/kg82.19591.21280.049.25kJ/kg43.5791.21215.049.25
C20@22
C20@11
fgf
fgf
hxhhhxhh -20 C
30 C
qL
4 3
21
T
s
kJ/kg27.3506.5kJ/kg138.4
COP
kJ/kg4.13843.5782.195
Rinnet,
12
L
Lq
w
hhq
11-91 A large refrigeration plant that operates on the ideal vapor-compression cycle with refrigerant-134aas the working fluid is considered. The mass flow rate of the refrigerant, the power input to thecompressor, and the mass flow rate of the cooling 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, therefrigerant 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
MPa0.7@33
2212
2
kPa120@1
kPa120@11
hh
hhP
Thss
P
sshhP
f
g
g
T
QH
QL
·
Win·
·
0.7 MPa
4
32
10.12 MPa
The mass flow rate of the refrigerant is determined from
kg/s0.675kJ/kg88.82236.97
kJ/s100
41 hhQ
m Ls
(b) The power input to the compressor iskW24.7kJ/kg236.97273.50kg/s0.67512in hhmW
(c) The mass flow rate of the cooling water is determined from
kg/s3.73C8CkJ/kg4.18
kJ/s124.7)(
kW124.7kJ/kg88.82273.50kg/s0.675
watercooling
32
TcQ
m
hhmQ
p
H
H
11-54
11-92 EES Problem 11-91 is reconsidered. The effect of evaporator pressure on the COP and the power input is to be investigated.Analysis The problem is solved using EES, and the solution is given below.
"Input Data""P[1]=120 [kPa]"P[2] = 700 [kPa] Q_dot_in= 100 [kW] DELTAT_cw = 8 [C] C_P_cw = 4.18 [kJ/kg-K]Fluid$='R134a'Eta_c=1.0 "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"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$,h=h[2],P=P[2]) "properties for state 2"{h[2]=enthalpy(Fluid$,P=P[2],T=T[2]) }T[2]=temperature(Fluid$,h=h[2],P=P[2])W_dot_c=m_dot*Wc
"Condenser"P[3] = P[2] h[3]=enthalpy(Fluid$,P=P[3],x=0) "properties for state 3"s[3]=entropy(Fluid$,P=P[3],x=0)h[2]=Qout+h[3] "energy balance on condenser"Q_dot_out=m_dot*Qout
"Throttle Valve"h[4]=h[3] "energy balance on throttle - isenthalpic"x[4]=quality(Fluid$,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"Q_dot_in=m_dot*Q_inCOP=Q_dot_in/W_dot_c "definition of COP"COP_plot = COP W_dot_in = W_dot_c m_dot_cw*C_P_cw*DELTAT_cw = Q_dot_out
11-55
P1 [kPa] COP WC [kW]120 4,056 24,66150 4,743 21,09180 5,475 18,27210 6,27 15,95240 7,146 13,99270 8,126 12,31300 9,235 10,83330 10,51 9,517360 11,99 8,34390 13,75 7,274
100 150 200 250 300 350 400
4
8
12
16
5
9
13
17
21
25
P[1] [kPa]
COP
Wc
[kW]
11-56
11-93 A large refrigeration plant operates on the vapor-compression cycle with refrigerant-134a as theworking fluid. The mass flow rate of the refrigerant, the power input to the compressor, the mass flow rateof the cooling water, and the rate of exergy destruction associated with the compression process are to bedetermined.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.Analysis (a) The refrigerant enters the compressor as a saturated vapor at the evaporator pressure, and leaves the condenser as saturated liquid at the condenser pressure. From the refrigerant tables (Tables A-12and A-13),
throttlingkJ/kg82.88
kJ/kg82.88liquidsat.
MPa7.0
C95.34kJ/kg50.273MPa7.0
KkJ/kg94779.0kJ/kg97.236
vaporsat.kPa120
34
MPa0.7@33
2212
2
kPa120@1
kPa120@11
hh
hhP
Thss
P
sshhP
f
sss
g
g
T
QH
QL
·
Win·
·
0.7 MPa
4
32
10.12 MPa
The mass flow rate of the refrigerant is determined froms
kg/s0.675kJ/kg88.82236.97
kJ/s100
41 hhQ
m L
(b) The actual enthalpy at the compressor exit is
kJ/kg285.6775.0/97.23650.27397.236/1212
12
12Cs
sC hhhh
hhhh
Thus, kW32.9kJ/kg236.97285.67kg/s0.67512in hhmW
(c) The mass flow rate of the cooling water is determined from
and
kg/s3.97C8CkJ/kg4.18
kJ/s132.9)(
kW132.9kJ/kg88.82285.67kg/s0.675
watercooling
32
TcQ
m
hhmQ
p
H
H
The exergy destruction associated with this adiabatic compression process is determined from
( )X T S T m sdestroyed gen0 0 2 s1
where
KkJ/kg98655.0kJ/kg67.285
MPa7.02
2
2 shP
Thus,
kW7.80KkJ/kg0.947790.98655kg/s0.675K298destroyedX
11-57
11-94 A heat pump that operates on the ideal vapor-compression cycle with refrigerant-134a as theworking fluid is used to heat a house. The rate of heat supply to the house, the volume flow rate of therefrigerant 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, therefrigerant 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
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
vvQH
QL
House·
Win·
·
0.9 MPa
4
32
1200 kPa
T
sThe rate of heat supply to the house is determined from
kW55.73kJ/kg.61101275.75kg/s0.3232 hhmQH
(b) The volume flow rate of the refrigerant at the compressor inlet is
/sm0.0320 3/kgm0.099867kg/s0.32 311 vV m
(c) The COP of t his heat pump is determined from
5.5746.24475.27561.10175.275COP
12
32
inR hh
hhwqL
11-95 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
qw
L
where
94126incompII,incompI,in
66816
11
with1
hhhhxwww
hhh
xhhxqfg
fL
11-58
11-96 A two-stage compression refrigeration system using refrigerant-134a as the working fluid isconsidered. The fraction of the refrigerant that evaporates as it is throttled to the flash chamber, the amountof 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
T
qL
73B
A96
0.8 MPa 4
0.4 MPa
8
5 2
1
0.14 MPaThe fraction of the refrigerant that evaporates as it isthrottled to the flash chamber is simply the quality at state 6,
0.164662.191
94.6347.9566
fg
f
hhh
x
s(b) The enthalpy at state 9 is determined from an energy balance on the mixing chamber:
kJ/kg75.25958.2601646.0155.2551646.011
0
9
26369
outin(steady)0
systemoutin
hhxhxh
hmhm
EEEEE
iiee
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 massof 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
4.49kJ/kg32.6kJ/kg146.4COP
inR w
qL
11-97 An aircraft on the ground is to be cooled by a gas refrigeration cycle operating with air on an open cycle. The temperature of the air leaving the turbine is to be determined.Assumptions 1 Steady operating conditions exist. 2 Air is an ideal gas with constant specific heats at roomtemperature. 3 Kinetic and potential energy changes are negligible.Properties The specific heat ratio of air at room temperature is k = 1.4 (Table A-2).Analysis Assuming the turbine to be isentropic, the air temperature at the turbine exit is determined from
11-59
C9.0 K264kPa250kPa100 K343
4140k1k
3
434
././
PPTT
11-60
11-98 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 tobe 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 theregenerator,
Qregen·
QH·
2
4
6QRefrig·
5-25 C
-10 C
3
1
T
6143
outin
(steady)0systemoutin 0
hhmhhmhmhm
EE
EEE
iiee
20 C
or,
Thus,K278C25C10C206134
61436143
C5TTTT
TTTTTTcmTTcm pp
s
(b) From the isentropic relations,
C93.9K117931 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
1.49C93.95C10135.2
C93.9C25
COP
5412
56
5412
56
outturb,incomp,innet,R
TTTTTT
hhhhhh
wwq
wq LL
(c) The net power input is determined from
kW108.293.9510135.2CkJ/kg5.1926kg/s0.4554125412outturb,incomp,innet, TTTTcmhhhhmWWW p
11-99 An absorption refrigeration system operating at specified conditions is considered. The minimumrate 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, kW53.91.259
kW12COP maxR,
mingen,LQ
Q
11-61
11-100 EES Problem 11-99 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] 13.76 508.996 656.833 805.295 1004.237 1253.603 1502.878 2002.475 250
50 90 130 170 210 2502
4
6
8
10
12
14
Ts [C]
Qgen;min
[kW]
11-62
11-101 A house is cooled adequately by a 3.5 ton air-conditioning unit. The rate of heat gain of the house 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 Noting that 1 ton of refrigeration is equivalent to a cooling rate of 211 kJ/min, the rate of heatgain of the house in steady operation is simply equal to the cooling rate of the air-conditioning system,
( .Q Qheat gain cooling ton)(211 kJ / min) = 738.5 kJ / min =3 5 44,310 kJ / h
11-102 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
11-63
11-103 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 tobe 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
Turbine
RegeneratorHeatExch.
HeatExch.
QL.
21
Compressor
6
54
3
.QH
kJ/kg63.308C35
kJ/kg52.47340.273
40.27350.43380.0
33
2
2
12
12
hT
hh
hhhh s
C
For the turbine inlet and exit we have
kJ/kg45.193C80 55 hT
sT hh
hhhT
54
54
44 ?
5
2
Qregen
QH· 2s
4
6QRefrig·
5s
3
1
s
T
35 C
shss
P
sTP
sTP
545
2
44
4
11
5
kPa500
?kPa500
kJ/kg.K6110.5C0
kPa100 0 C
-80 C
We can determine the temperature at the turbine inlet from EES using the above relations. A hand solutionwould require a trial-error approach.
T4 = 281.8 K, h4 = 282.08 kJ/kgAn energy balance on the regenerator gives
kJ/kg85.24608.28263.30840.2734316 hhhh
The effectiveness of the regenerator is determined from
0.43085.24663.30808.28263.308
63
43regen hh
hh
(b) The refrigeration load is
kW21.36kJ/kg)45.19385kg/s)(246.4.0()( 56 hhmQL
11-64
(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
0.47945.3505.80
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
4
2
4s
3
QH·
2
11QRefrig·
T
35 CkJ/kg2704.5
C35kPa500
33
3 sTP
0 C
kJ/kg.K52.194kPa100
434
1sh
ssP
s
kJ/kg64.21152.19463.308
63.30885.0 4
4
43
43 hh
hhhh
sT
kW24.70kJ/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
0.59925.4170.24COP
innet,WQL
11-65
11-104 An air-conditioner with refrigerant-134a as the refrigerant is considered. The temperature of therefrigerant at the compressor exit, the rate of heat generated by the people in the room, the COP of the air-conditioner, and the minimum volume flow rate of the refrigerant at the compressor inlet 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/kg77.117
kJ/kg77.117
39.277kPa1200
kJ/kg9240.0kJ/kg04112.0
kJ/kg30.259
1kPa500
34
kPa@12003
212
2
1
1
1
1
1
hh
hh
hss
Ps
h
xP
f
s
v
.
34°C
26°C
4
3 2
1
Win
1200 kPa
Expansionvalve
Compressor
Condenser
500 kPa
.QH
.QL
EvaporatorkJ/kg42.283
30.25930.25939.27775.0 2
2
12
12
hh
hhhh s
C
C54.522
2
kJ/kg42.283kPa1200
ThP
(b) The mass flow rate of the refrigerant is T
QH
QL
Win·
2
·
·4
3
2s
1
kg/s04053.0/kgm0.04112
s60min1
L1000m1L/min)100(
3
3
1
1
v
Vm
The refrigeration load is
kW737.5kJ/kg)77.11730kg/s)(259.04053.0()( 41 hhmQL
which is the total heat removed from the room. Then, the rate of heat generated by the people in the room is determined from
s
kW0.67kW)9.060/250737.5(equipheatpeople QQQQ L
(c) The power input and the COP are kW9774.0kJ/kg)30.25942kg/s)(283.04053.0()( 12in hhmW
5.879774.0737.5COP
in
L
WQ
(d) The reversible COP of the cycle is
38.371)27326/()27334(
11/
1COPrevLH TT
The corresponding minimum power input is
kW1535.038.37kW737.5
COPrev
Lminin,
QW
The minimum mass and volume flow rates are
kg/s006366.0kJ/kg)30.259(283.42
kW1535.0
12
minin,min hh
Wm
L/min15.7/s)m0002617.0(/kg)m112kg/s)(0.04006366.0( 331minmin1, vV m
11-66
11-105 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 airor room air.Analysis The COP of the heat pump is given to be 2.2. Then the COP of the air-conditioning systembecomes
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 / h12 2 2 4
since 1 kW = 3600 kJ/h. We conclude that this heat pump can meet the cooling needs of the room since itscooling rate is greater than the rate of heat gain of the room.
11-67
11-106 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 roomtemperature. 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 roomtemperature 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 thisirreversibility, the minimum temperature that can be obtained by the vortex tube is not as low as the onethat 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 itinvolves no heat or work interactions. Then the steady-flow energy balance equation for this system
for a unit mass flow rate at the inletE Ein out (m1 kg / s)1 can be expressed as
321
332211
332211
75.025.01 TcTcTc
TcmTcmTcmhmhmhm
ppp
ppp
Warm air Cold air
Compressedair
2 3
1
Canceling cp and solving for T3 gives
K307.375.0
27825.030075.025.0 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 can be expressed as with one inletand two exits, and it involves no heat or work interactions. Then the steady-flow entropy balance equationfor this system for a unit mass flow rate at the inlet
S S Sin out gen 0
(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/K461.0kPa500kPa100lnkJ/kg.K)287.0(
K300K3.307lnkJ/kg.K)005.1(75.0
kPa500kPa100lnkJ/kg.K)287.0(
K300K278lnkJ/kg.K)005.1(25.0genS
which is a positive quantity. Therefore, this process satisfies the 2nd law of thermodynamics.
11-68
(d) For a unit mass flow rate at the inlet (m1 kg / s)1 , the cooling rate and the power input to thecompressor 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
W
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
TT T
L
H L ( )300 27812.6
Discussion Note that the COP of the vortex refrigerator is a small fraction of the COP of a Carnotrefrigerator operating between the same temperature limits.
11-69
11-107 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 roomtemperature. 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 roomtemperature 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 thisirreversibility, the minimum temperature that can be obtained by the vortex tube is not as low as the onethat 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 itinvolves no heat or work interactions. Then the steady-flow entropy balance equation for this system
for a unit mass flow rate at the inletE Ein out (m1 kg / s)1 can be expressed as
321
332211
332211
75.025.01 TcTcTc
TcmTcmTcmhmhmhm
ppp
ppp
Warm air Cold air
Compressedair
2 3
1
Canceling cp and solving for T3 gives
K307.375.0
27825.030075.025.0 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 can be expressed as with one inletand two exits, and it involves no heat or work interactions. Then the steady-flow energy balance equationfor this system for a unit mass flow rate at the inlet
S S Sin out gen 0
(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/K5130kPa600kPa100kJ/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.
11-70
(d) For a unit mass flow rate at the inlet (m1 kg / s)1 , the cooling rate and the power input to thecompressor 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
W
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
12.6K)278300(
K278COPCarnotLH
L
TTT
Discussion Note that the COP of the vortex refrigerator is a small fraction of the COP of a Carnotrefrigerator operating between the same temperature limits.
11-71
11-108 EES The effect of the evaporator pressure on the COP of an ideal vapor-compression refrigerationcycle 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 c P1 [kPa]1.851 0.7 1002.863 0.7 2004.014 0.7 3005.462 0.7 4007.424 0.7 500
100 150 200 250 300 350 400 450 5000
2
4
6
8
10
P[1] [kPa]
COP
comp
1.01.00.70.7
11-72
11-109 EES The effect of the condenser pressure on the COP of an ideal vapor-compression refrigerationcycle 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"
400 600 800 1000 1200 14000
1
2
3
4
5
6
7
8
P[2] [kPa]
COP
comp
1.01.00.70.7
COP c P2 [kPa]4.935 0.7 4003.04 0.7 650
2.258 0.7 9001.803 0.7 11501.492 0.7 1400
11-73
Fundamentals of Engineering (FE) Exam Problems
11-110 Consider a heat pump that operates on the reversed Carnot cycle with R-134a as the working fluidexecuted 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 cycleis(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 numericalvalues).
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)+273TL=TEMPERATURE(R134a,x=0,P=P2)+273q_H=h_fgCOP=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"
11-111 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 numericalvalues).
TH=20+273TL=-5+273Q_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"
11-74
11-112 A refrigerator operates on the ideal vapor compression refrigeration cycle with R-134a as theworking fluid between the pressure limits of 120 kPa and 800 kPa. If the rate of heat removal from therefrigerated 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 numericalvalues).
P1=120 "kPa"P2=800 "kPa"P3=P2P4=P1s2=s1Q_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"
11-113 A heat pump operates on the ideal vapor compression refrigeration cycle with R-134a as theworking fluid between the pressure limits of 0.32 MPa and 1.2 MPa. If the mass flow rate of the refrigerantis 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 numericalvalues).
P1=320 "kPa"P2=1200 "kPa"P3=P2P4=P1s2=s1m=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)
11-75
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) "Usingh_fg at P1"
11-114 An ideal vapor compression refrigeration cycle with R-134a as the working fluid operates betweenthe pressure limits of 120 kPa and 1000 kPa. The mass fraction of the refrigerant that is in the liquid phaseat 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 numericalvalues).
P1=120 "kPa"P2=1000 "kPa"P3=P2P4=P1h1=ENTHALPY(R134a,x=1,P=P1)h3=ENTHALPY(R134a,x=0,P=P3)h4=h3x4=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)
11-115 Consider a heat pump that operates on the ideal vapor compression refrigeration cycle with R-134aas the working fluid between the pressure limits of 0.32 MPa and 1.2 MPa. The coefficient of performanceof 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 numericalvalues).
P1=320 "kPa"P2=1200 "kPa"P3=P2
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P4=P1s2=s1h1=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=h3COP_HP=qH/WinWin=h2-h1qH=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)"
11-116 An ideal gas refrigeration cycle using air as the working fluid operates between the pressure limitsof 80 kPa and 280 kPa. Air is cooled to 35 C before entering the turbine. The lowest temperature of thiscycle 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 numericalvalues).
k=1.4P1= 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"
11-117 Consider an ideal gas refrigeration cycle using helium as the working fluid. Helium enters thecompressor at 100 kPa and –10°C and is compressed to 250 kPa. Helium is then cooled to 20°C before itenters 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 numericalvalues).
11-77
k=1.667Cp=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"
11-118 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 geothermalsource 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 numericalvalues).
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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11-119 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 kPaand 50 C, and leaves the condenser as saturated liquid at 800 kPa. The coefficient of performance of thisrefrigerator 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 numericalvalues).
P1=160 "kPa"P2=800 "kPa"T2=50 "C"P3=P2P4=P1h1=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=h3COP_R=qL/WinWin=h2-h1qL=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"
11-120 ··· 11-129 Design and Essay Problems
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