Heat 6-18
Carnot Engines 6-69C No. 6-70C The one that has a source temperature of 600C. This is true because the higher the temperature at which heat is supplied to the working fluid of a heat engine, the higher the thermal efficiency.
6-71 The source and sink temperatures of a Carnot heat engine and the rate of heat supply are given. The thermal efficiency and the power output are to be determined. Assumptions The Carnot heat engine operates steadily. Analysis (a) The thermal efficiency of a Carnot heat engine depends on the source and the sink temperatures only, and is determined from 300 K T 1000 K th,C = 1 - L = 1 - = 0.70 or 70% 1000 K TH 800 kJ/min (b) The power output of this heat engine is determined from the definition of thermal efficiency, & & W = Q = (0.70 )(800 kJ/min ) = 560 kJ/min = 9.33 kW
net, out th H
HE 300 K
6-72 The sink temperature of a Carnot heat engine and the rates of heat supply and heat rejection are given. The source temperature and the thermal efficiency of the engine are to be determined. Assumptions The Carnot heat engine operates steadily.
Q Analysis (a) For reversible cyclic devices we have H Q L T = H rev TL
source 650 kJ HE 250 kJ 24C
Thus the temperature of the source TH must be
Q 650 kJ TH = H TL = 250 kJ (297 K ) = 772.2 K Q L rev
(b) The thermal efficiency of a Carnot heat engine depends on the source and the sink temperatures only, and is determined from 297 K T th,C = 1 - L = 1 - = 0.615 or 61.5% 772.2 K TH
6-73 [Also solved by EES on enclosed CD] The source and sink temperatures of a heat engine and the rate of heat supply are given. The maximum possible power output of this engine is to be determined. Assumptions The heat engine operates steadily. Analysis The highest thermal efficiency a heat engine operating between two specified temperature limits can have is the Carnot efficiency, which is determined from 550C 298 K T th, max = th,C = 1 - L = 1 - = 0.638 or 63.8% 823 K TH 1200 kJ/min Then the maximum power output of this heat engine is determined from the definition of thermal efficiency to be & & W = Q = (0.638)(1200 kJ/min ) = 765.6 kJ/min = 12.8 kW
net, out th H
HE 25C
6-19
6-74 EES Problem 6-73 is reconsidered. The effects of the temperatures of the heat source and the heat sink on the power produced and the cycle thermal efficiency as the source temperature varies from 300C to 1000C and the sink temperature varies from 0C to 50C are to be studied. The power produced and the cycle efficiency against the source temperature for sink temperatures of 0C, 25C, and 50C are to be plotted. Analysis The problem is solved using EES, and the results are tabulated and plotted below. "Input Data from the Diagram Window" {T_H = 550 [C] T_L = 25 [C]} {Q_dot_H = 1200 [kJ/min]} "First Law applied to the heat engine" Q_dot_H - Q_dot_L- W_dot_net = 0 W_dot_net_KW=W_dot_net*convert(kJ/min,kW) "Cycle Thermal Efficiency - Temperatures must be absolute" eta_th = 1 - (T_L + 273)/(T_H + 273) "Definition of cycle efficiency" eta_th=W_dot_net / Q_dot_H th 0.52 0.59 0.65 0.69 0.72 0.75 0.77 0.79 TH [C] 300 400 500 600 700 800 900 1000 WnetkW [kW] 10.47 11.89 12.94 13.75 14.39 14.91 15.35 15.71
0.8 0.75 0.7 0.65
T = 50 C L = 25 C = 0C
th
0.6 0.55 0.5 0.45 0.4 300 400 500 600 700 800 900 1000
Q
H
= 1200 kJ/m in
T
20 18 16 14 12
H
[C]
T = 50 C L = 25 C = 0C
W net,KW [kW ]
10 8 6 4 2 0 300 400 500 600 700 800 900 1000
Q
H
= 1200 kJ/m in
T
H
[C]
6-20
6-75E The sink temperature of a Carnot heat engine, the rate of heat rejection, and the thermal efficiency are given. The power output of the engine and the source temperature are to be determined. Assumptions The Carnot heat engine operates steadily. Analysis (a) The rate of heat input to this heat engine is determined from the definition of thermal efficiency, & Q 800 Btu/min & th = 1 - & L 0.55 = 1 - QH = 1777.8 Btu/min & Q Q
H H
Then the power output of this heat engine can be determined from
& & Wnet,out = th QH = (0.55)(1777.8 Btu/min ) = 977.8 Btu/min = 23.1 hp
TH HE 800 Btu/min 60F
(b) For reversible cyclic devices we have
& QH Q & L
T = H rev TL
Thus the temperature of the source TH must be
& Q 1777.8 Btu/min TH = H TL = 800 Btu/min (520 R ) = 1155.6 R Q & L rev
6-76 The source and sink temperatures of a OTEC (Ocean Thermal Energy Conversion) power plant are given. The maximum thermal efficiency is to be determined. Assumptions The power plant operates steadily. Analysis The highest thermal efficiency a heat engine operating between two specified temperature limits can have is the Carnot efficiency, which is determined from
24C HE 3C
W
th, max = th,C = 1 -
TL 276 K = 1- = 0.071 or 7.1% TH 297 K
6-77 The source and sink temperatures of a geothermal power plant are given. The maximum thermal efficiency is to be determined. Assumptions The power plant operates steadily. Analysis The highest thermal efficiency a heat engine operating between two specified temperature limits can have is the Carnot efficiency, which is determined from
140C HE 20C
W
th, max = th,C = 1 -
TL 20 + 273 K = 1- = 0.291 or 29.1% TH 140 + 273 K
6-21
6-78 An inventor claims to have developed a heat engine. The inventor reports temperature, heat transfer, and work output measurements. The claim is to be evaluated. Analysis The highest thermal efficiency a heat engine operating between two specified temperature limits can have is the Carnot efficiency, which is determined from
th, max = th,C = 1 -
TL 290 K = 1- = 0.42 or 42% TH 500 K
500 K 700 kJ HE 290 K 300 kJ
The actual thermal efficiency of the heat engine in question is
th =
Wnet 300 kJ = = 0.429 or 42.9% QH 700 kJ
which is greater than the maximum possible thermal efficiency. Therefore, this heat engine is a PMM2 and the claim is false.
6-79E An inventor claims to have developed a heat engine. The inventor reports temperature, heat transfer, and work output measurements. The claim is to be evaluated. Analysis The highest thermal efficiency a heat engine operating between two specified temperature limits can have is the Carnot efficiency, which is determined from
th, max = th,C = 1 -
TL 540 R = 1- = 0.40 or 40% TH 900 R
900 R 300 Btu HE 540 R 160 Btu
The actual thermal efficiency of the heat engine in question is
th =
Wnet 160 Btu = = 0.533 or 53.3% QH 300 Btu
which is greater than the maximum possible thermal efficiency. Therefore, this heat engine is a PMM2 and the claim is false.
6-22
6-80 A geothermal power plant uses geothermal liquid water at 160C at a specified rate as the heat source. The actual and maximum possible thermal efficiencies and the rate of heat rejected from this power plant are to be determined. Assumptions 1 The power plant operates steadily. 2 The kinetic and potential energy changes are zero. 3 Steam properties are used for geothermal water. Properties Using saturated liquid properties, the source and the sink state enthalpies of geothermal water are (Table A-4)
Tsource = 160C hsource = 675.47 kJ/kg xsource = 0 Tsink = 25C hsink = 104.83 kJ/kg xsink = 0
Analysis (a) The rate of heat input to the plant may be taken as the enthalpy difference between the source and the sink for the power plant
& & Qin = mgeo (hsource - hsink ) = (440 kg/s)(675.47 - 104.83) kJ/kg = 251,083 kW
The actual thermal efficiency is
th =
& W net,out 22 MW = = 0.0876 = 8.8% & 251.083 MW Qin
(b) The maximum thermal efficiency is the thermal efficiency of a reversible heat engine operating between the source and sink temperatures
th, max = 1 -
TL (25 + 273) K = 1- = 0.312 = 31.2% TH (160 + 273) K
(c) Finally, the rate of heat rejection is
& & & Qout = Qin - Wnet,out = 251.1 - 22 = 229.1 MW
6-23
Carnot Refrigerators and Heat Pumps 6-81C By increasing TL or by decreasing TH.
1 . TH / TL - 1
6-82C It is the COP that a Carnot refrigerator would have, COPR =
6-83C No. At best (when everything is reversible), the increase in the work produced will be equal to the work consumed by the refrigerator. In reality, the work consumed by the refrigerator will always be greater than the additional work produced, resulting in a decrease in the thermal efficiency of the power plant. 6-84C No. At best (when everything is reversible), the increase in the work produced will be equal to the work consumed by the refrigerator. In reality, the work consumed by the refrigerator will always be greater than the additional work produced, resulting in a decrease in the thermal efficiency of the power plant. 6-85C Bad idea. At best (when everything is reversible), the increase in the work produced will be equal to the work consumed by the heat pump. In reality, the work consumed by the heat pump will always be greater than the additional work produced, resulting in a decrease in the thermal efficiency of the power plant.
6-86 The refrigerated space and the environment temperatures of a Carnot refrigerator and the power consumption are given. The rate of heat removal from the refrigerated space is to be determined. Assumptions The Carnot refrigerator operates steadily. Analysis The coefficient of performance of a Carnot refrigerator depends on the temperature limits in the cycle only, and is determined from
COPR, C = 1 1 = = 14.5 (TH / TL ) - 1 (22 + 273K )/(3 + 273K ) - 1
22C R 2 kW 3C
The rate of heat removal from the refrigerated space is determined from the definition of the coefficient of performance of a refrigerator,
& & QL = COPRWnet,in = (14.5)(2 kW ) = 29.0 kW = 1740 kJ/min
6-24
6-87 The refrigerated space and the environment temperatures for a refrigerator and the rate of heat removal from the refrigerated space are given. The minimum power input required is to be determined. Assumptions The refrigerator operates steadily. Analysis The power input to a refrigerator will be a minimum when the refrigerator operates in a reversible manner. The coefficient of performance of a reversible refrigerator depends on the temperature limits in the cycle only, and is determined from
COPR, rev = 1 1 = = 8.03 (TH / TL ) - 1 (25 + 273 K )/ (- 8 + 273 K ) - 1
25C R 300 kJ/min -8C
The power input to this refrigerator is determined from the definition of the coefficient of performance of a refrigerator, & QL 300 kJ/min & Wnet,in, min = = = 37.36 kJ/min = 0.623 kW COPR, max 8.03
6-88 The cooled space and the outdoors temperatures for a Carnot air-conditioner and the rate of heat removal from the air-conditioned room are given. The power input required is to be determined. Assumptions The air-conditioner operates steadily. Analysis The COP of a Carnot air conditioner (or Carnot refrigerator) depends on the temperature limits in the cycle only, and is determined from
COPR, C = 1 1 = = 27.0 (TH / TL ) - 1 (35 + 273 K )/ (24 + 273 K ) - 1
35C A/C House 24C
The power input to this refrigerator is determined from the definition of the coefficient of performance of a refrigerator, & QL 750 kJ/min & Wnet,in = = = 27.8 kJ/min = 0.463 kW COPR, max 27.0
6-89E The cooled space and the outdoors temperatures for an air-conditioner and the power consumption are given. The maximum rate of heat removal from the air-conditioned space is to be determined. Assumptions The air-conditioner operates steadily. Analysis The rate of heat removal from a house will be a maximum when the air-conditioning system operates in a reversible manner. The coefficient of performance of a reversible air-conditioner (or refrigerator) depends on the temperature limits in the cycle only, and is determined from
COPR, rev = 1 1 = = 29.6 (TH / TL ) - 1 (90 + 460 R )/ (72 + 460 R ) - 1
90F
The rate of heat removal from the house is determined from the definition of the coefficient of performance of a refrigerator,
42.41 Btu/min & & = 6277 Btu/min QL = COPRWnet,in = (29.6 )(5 hp ) 1 hp
A/C House 72F
5 hp
6-25
6-90 The refrigerated space temperature, the COP, and the power input of a Carnot refrigerator are given. The rate of heat removal from the refrigerated space and its temperature are to be determined. Assumptions The refrigerator operates steadily. Analysis (a) The rate of heat removal from the refrigerated space is determined from the definition of the COP of a refrigerator,
& & QL = COPRWnet,in = (4.5)(0.5 kW ) = 2.25 kW = 135 kJ/min
25C R TL 500 W COP = 4.5
(b) The temperature of the refrigerated space TL is determined from the coefficient of performance relation for a Carnot refrigerator,
COPR, rev 1 1 = 4.5 = (TH / TL ) - 1 (25 + 273 K )/TL - 1
It yields TL = 243.8 K = -29.2C
6-91 An inventor claims to have developed a refrigerator. The inventor reports temperature and COP measurements. The claim is to be evaluated. Analysis The highest coefficient of performance a refrigerator can have when removing heat from a cool medium at -12C to a warmer medium at 25C is
COPR, max = COPR, rev =
(TH / TL ) - 1 (25 + 273 K )/ (- 12 + 273 K ) - 1
1
=
1
= 7.1
25C
The COP claimed by the inventor is 6.5, which is below this maximum value, thus the claim is reasonable. However, it is not probable.
R COP= 6.5 -12C
6-92 An experimentalist claims to have developed a refrigerator. The experimentalist reports temperature, heat transfer, and work input measurements. The claim is to be evaluated. Analysis The highest coefficient of performance a refrigerator can have when removing heat from a cool medium at -30C to a warmer medium at 25C is
COPR, max = COPR, rev =
(TH / TL ) - 1 (25 + 273 K )/ (- 30 + 273 K ) - 1
1
=
1
= 4.42
25C
The work consumed the by actual refrigerator during this experiment is
& Wnet,in = Wnet,in t = (2 kJ/s )(20 60 s ) = 2400 kJ
R 2 kW 30,000 kJ -30C
Then the coefficient of performance of this refrigerator becomes
COPR = QL 30,000kJ = = 12.5 Wnet,in 2400kJ
which is above the maximum value. Therefore, these measurements are not reasonable.
6-26
6-93E An air-conditioning system maintains a house at a specified temperature. The rate of heat gain of the house and the rate of internal heat generation are given. The maximum power input required is to be determined. Assumptions The air-conditioner operates steadily. Analysis The power input to an air-conditioning system will be a minimum when the air-conditioner operates in a reversible manner. The coefficient of performance of a reversible air-conditioner (or refrigerator) depends on the temperature limits in the cycle only, and is determined from
COPR, rev = 1 1 = = 26.75 (TH / TL ) - 1 (95 + 460 R )/ (75 + 460 R ) - 1
95F A/C 800 kJ/min House 75F
The cooling load of this air-conditioning system is the sum of the heat gain from the outside and the heat generated within the house,
& Q L = 800 + 100 = 900 Btu/min
The power input to this refrigerator is determined from the definition of the coefficient of performance of a refrigerator, & 900 Btu/min QL & = = 33.6 Btu/min = 0.79 hp Wnet,in, min = COPR, max 26.75
6-94 A heat pump maintains a house at a specified temperature. The rate of heat loss of the house is given. The minimum power input required is to be determined. Assumptions The heat pump operates steadily. Analysis The power input to a heat pump will be a minimum when the heat pump operates in a reversible manner. The COP of a reversible heat pump depends on the temperature limits in the cycle only, and is determined from
COPHP, rev = 1 1 = = 10.2 1 - (TL / TH ) 1 - (- 5 + 273 K )/ (24 + 273 K )
80,000 kJ/h House 24C
The required power input to this reversible heat pump is determined from the definition of the coefficient of performance to be
& Wnet,in, min = & 80,000 kJ/h 1 h QH = 3600 s = 2.18 kW COPHP 10.2
HP -5C
which is the minimum power input required.
6-27
6-95 A heat pump maintains a house at a specified temperature. The rate of heat loss of the house and the power consumption of the heat pump are given. It is to be determined if this heat pump can do the job. Assumptions The heat pump operates steadily. Analysis The power input to a heat pump will be a minimum when the heat pump operates in a reversible manner. The coefficient of performance of a reversible heat pump depends on the temperature limits in the cycle only, and is determined from
COPHP, rev = 1 1 = = 14.75 1 - (TL / TH ) 1 - (2 + 273 K )/ (22 + 273 K )
110,000 kJ/h
The required power input to this reversible heat pump is determined from the definition of the coefficient of performance to be
& Wnet,in, min = & 110,000 kJ/h 1 h QH = 3600 s = 2.07 kW COPHP 14.75
House 22C
HP
5 kW
This heat pump is powerful enough since 5 kW > 2.07 kW.
6-96 A heat pump that consumes 5-kW of power when operating maintains a house at a specified temperature. The house is losing heat in proportion to the temperature difference between the indoors and the outdoors. The lowest outdoor temperature for which this heat pump can do the job is to be determined. Assumptions The heat pump operates steadily. Analysis Denoting the outdoor temperature by TL, the heating load of this house can be expressed as
& QH = (5400 kJ/h K )(294 - TL ) = (1.5 kW/K )(294 - TL )K
The coefficient of performance of a Carnot heat pump depends on the temperature limits in the cycle only, and can be expressed as
COPHP = 1 1 = 1 - (TL / TH ) 1 - TL /(294 K) & QH & W
5400 kJ/h.K House 21C
or, as
COPHP = =
(1.5 kW/K )(294 - TL )K
6 kW
HP TL 6 kW
net,in
Equating the two relations above and solving for TL, we obtain TL = 259.7 K = -13.3C
6-28
6-97 A heat pump maintains a house at a specified temperature in winter. The maximum COPs of the heat pump for different outdoor temperatures are to be determined. Analysis The coefficient of performance of a heat pump will be a maximum when the heat pump operates in a reversible manner. The coefficient of performance of a reversible heat pump depends on the temperature limits in the cycle only, and is determined for all three cases above to be
COPHP ,rev = COPHP ,rev = COPHP ,rev = 1 - (T L / T H ) 1 = 1 = 29.3 1 - (10 + 273K )/ (20 + 273K )
20C HP TL
1 1 = = 11.7 1 - (T L / T H ) 1 - (- 5 + 273K )/ (20 + 273K ) 1 - (T L / T H ) 1 = 1 = 5.86 1 - (- 30 + 273K )/ (20 + 273K )
6-98E A heat pump maintains a house at a specified temperature. The rate of heat loss of the house is given. The minimum power inputs required for different source temperatures are to be determined. Assumptions The heat pump operates steadily. Analysis (a) The power input to a heat pump will be a minimum when the heat pump operates in a reversible manner. If the outdoor air at 25F is used as the heat source, the COP of the heat pump and the required power input are determined to be
COPHP, max = COPHP, rev = 1 1 = = 10.15 1 - (TL / TH ) 1 - (25 + 460 R )/ (78 + 460 R )
55,000 Btu/h
and
& Wnet,in, min = & 55,000 Btu/h 1 hp QH = 2545 Btu/h = 2.13 hp COPHP, max 10.15
House 78F
(b) If the well-water at 50F is used as the heat source, the COP of the heat pump and the required power input are determined to be
COPHP, max = COPHP, rev = 1 1 = = 19.2 1 - (TL / TH ) 1 - (50 + 460 R )/ (78 + 460 R )
HP
25F or 50F
and
& Wnet,in, min = & 55,000 Btu/h 1 hp QH = 2545 Btu/h = 1.13 hp COPHP, max 19.2
6-29
6-99 A Carnot heat pump consumes 8-kW of power when operating, and maintains a house at a specified temperature. The average rate of heat loss of the house in a particular day is given. The actual running time of the heat pump that day, the heating cost, and the cost if resistance heating is used instead are to be determined. Analysis (a) The coefficient of performance of this Carnot heat pump depends on the temperature limits in the cycle only, and is determined from
COPHP, rev = 1 1 = = 16.3 1 - (TL / TH ) 1 - (2 + 273 K )/ (20 + 273 K )
House 20C 82,000 kJ/h
The amount of heat the house lost that day is
& QH = QH (1 day ) = (82,000 kJ/h )(24 h ) = 1,968,000 kJ
Then the required work input to this Carnot heat pump is determined from the definition of the coefficient of performance to be
Wnet,in = 1,968,000 kJ QH = = 120,736 kJ COPHP 16.3
HP
8 kW
Thus the length of time the heat pump ran that day is
t = Wnet,in 120,736 kJ = = 15,092 s = 4.19 h & W 8 kJ/s
net,in
2C
(b) The total heating cost that day is
& Cost = W price = Wnet,in t (price ) = (8 kW )(4.19 h )(0.085 $/kWh ) = $2.85
(
)
(c) If resistance heating were used, the entire heating load for that day would have to be met by electrical energy. Therefore, the heating system would consume 1,968,000 kJ of electricity that would cost
1 kWh New Cost = QH price = (1,968,000kJ ) 3600 kJ (0.085 $/kWh ) = $46.47
6-30
6-100 A Carnot heat engine is used to drive a Carnot refrigerator. The maximum rate of heat removal from the refrigerated space and the total rate of heat rejection to the ambient air are to be determined. Assumptions The heat engine and the refrigerator operate steadily. Analysis (a) The highest thermal efficiency a heat engine operating between two specified temperature limits can have is the Carnot efficiency, which is determined from
th, max = th,C = 1 -
TL 300 K = 1- = 0.744 TH 1173 K
900C 800 kJ/min
-5C
Then the maximum power output of this heat engine is determined from the definition of thermal efficiency to be
& & Wnet,out = th QH = (0.744 )(800 kJ/min ) = 595.2 kJ/min & which is also the power input to the refrigerator, Wnet,in .
HE
R
27C
The rate of heat removal from the refrigerated space will be a maximum if a Carnot refrigerator is used. The COP of the Carnot refrigerator is
COPR, rev = 1 1 = = 8.37 (TH / TL ) - 1 (27 + 273 K )/ (- 5 + 273 K ) - 1
Then the rate of heat removal from the refrigerated space becomes
& & QL , R = COPR, rev Wnet,in = (8.37 )(595.2 kJ/min ) = 4982 kJ/min
(
)(
)
(b) The total rate of heat rejection to the ambient air is the sum of the heat rejected by the heat engine & & ( QL , HE ) and the heat discarded by the refrigerator ( QH , R ),
& & & QL , HE = QH , HE - Wnet,out = 800 - 595.2 = 204.8 kJ/min & & & QH , R = QL , R + Wnet,in = 4982 + 595.2 = 5577.2 kJ/min
and
& & & Qambient = QL, HE + QH , R = 204.8 + 5577.2 = 5782 kJ/min
6-31
6-101E A Carnot heat engine is used to drive a Carnot refrigerator. The maximum rate of heat removal from the refrigerated space and the total rate of heat rejection to the ambient air are to be determined. Assumptions The heat engine and the refrigerator operate steadily. Analysis (a) The highest thermal efficiency a heat engine operating between two specified temperature limits can have is the Carnot efficiency, which is determined from
th, max = th,C = 1 -
TL 540 R = 1- = 0.75 TH 2160 R
1700F 700 Btu/min HE 80F
20F
Then the maximum power output of this heat engine is determined from the definition of thermal efficiency to be
& & Wnet,out = th QH = (0.75)(700 Btu/min ) = 525 Btu/min & which is also the power input to the refrigerator, Wnet,in .
R
The rate of heat removal from the refrigerated space will be a maximum if a Carnot refrigerator is used. The COP of the Carnot refrigerator is
COPR, rev = 1 1 = = 8.0 (TH / TL ) - 1 (80 + 460 R )/ (20 + 460 R ) - 1
Then the rate of heat removal from the refrigerated space becomes
& & QL , R = COPR, rev Wnet,in = (8.0 )(525 Btu/min ) = 4200 Btu/min
(
)(
)
(b) The total rate of heat rejection to the ambient air is the sum of the heat rejected by the heat engine & & ( QL , HE ) and the heat discarded by the refrigerator ( QH , R ),
& & & QL , HE = QH , HE - Wnet,out = 700 - 525 = 175 Btu/min & & & QH , R = QL , R + Wnet,in = 4200 + 525 = 4725 Btu/min
and
& & & Qambient = QL , HE + QH , R = 175 + 4725 = 4900 Btu/min
6-32
6-102 A commercial refrigerator with R-134a as the working fluid is considered. The condenser inlet and exit states are specified. The mass flow rate of the refrigerant, the refrigeration load, the COP, and the minimum power input to the compressor are to be determined. Assumptions 1 The refrigerator operates steadily. 2 The kinetic and potential energy changes are zero. Properties The properties of R-134a and water are (Steam and R-134a tables)
P = 1.2 MPa 1 h1 = 278.27 kJ/kg T1 = 50C T2 = Tsat@1.2 MPa + Tsubcool = 46.3 - 5 = 41.3C P2 = 1.2 MPa h2 = 110.17 kJ/kg T2 = 41.3C Tw,1 = 18C hw,1 = 75.54 kJ/kg x w,1 = 0 Tw, 2 = 26C hw, 2 = 109.01 kJ/kg x w, 2 = 0
26C 1.2 MPa 5C subcool QH Condenser
Water 18C 1.2 MPa 50C
Expansion valve
Win Compressor
Evaporator QL
Analysis (a) The rate of heat transferred to the water is the energy change of the water from inlet to exit
& & Q H = m w (hw, 2 - hw,1 ) = (0.25 kg/s)(109.01 - 75.54) kJ/kg = 8.367 kW
The energy decrease of the refrigerant is equal to the energy increase of the water in the condenser. That is,
& & Q H = m R (h1 - h2 ) m R = & & QH 8.367 kW = = 0.0498 kg/s h1 - h2 (278.27 - 110.17) kJ/kg
(b) The refrigeration load is
& & & Q L = Q H - Win = 8.37 - 3.30 = 5.07 kW
(c) The COP of the refrigerator is determined from its definition,
COP =
& Q L 5.07 kW = = 1.54 & 3.3 kW Win
1 1 = = 4.49 TH / TL - 1 (18 + 273) /(-35 + 273) - 1
(d) The COP of a reversible refrigerator operating between the same temperature limits is
COPmax =
Then, the minimum power input to the compressor for the same refrigeration load would be
& Win, min = & QL 5.07 kW = = 1.13 kW COPmax 4.49
6-33
6-103 An air-conditioner with R-134a as the working fluid is considered. The compressor inlet and exit states are specified. The actual and maximum COPs and the minimum volume flow rate of the refrigerant at the compressor inlet are to be determined. Assumptions 1 The air-conditioner operates steadily. 2 The kinetic and potential energy changes are zero. Properties The properties of R-134a at the compressor inlet and exit states are (Tables A-11 through A-13)
P1 = 500 kPa h1 = 259.30 kJ/kg 3 x1 = 1 v1 = 0.04112 m /kg P2 = 1.2 MPa h2 = 278.27 kJ/kg T2 = 50C
QH Condenser Expansion valve 1.2 MPa 50C Win Compressor 500 kPa sat. vap.
Analysis (a) The mass flow rate of the refrigerant and the power consumption of the compressor are
1 m 3 1 min 100 L/min 1000 L 60 s & V1 & mR = = = 0.04053 kg/s 3 v1 0.04112 m /kg
Evaporator QL
& & Win = m R (h2 - h1 ) = (0.04053 kg/s)(278.27 - 259.30) kJ/kg = 0.7686 kW
The heat gains to the room must be rejected by the air-conditioner. That is,
1 min & & & Q L = Q heat + Qequipment = (250 kJ/min) + 0.9 kW = 5.067 kW 60 s
Then, the actual COP becomes
COP = & QL 5.067 kW = = 6.59 & Win 0.7686 kW 1 1 = = 37.4 T H / T L - 1 (34 + 273) /(26 + 273) - 1 & QL 5.067 kW = = 0.1356 kW COPmax 37.38
(b) The COP of a reversible refrigerator operating between the same temperature limits is
COPmax
(c) The minimum power input to the compressor for the same refrigeration load would be
& Win, min =
The minimum mass flow rate is & Win, min 0.1356 kW & mR, min = = = 0.007149 kg/s h2 - h1 (278.27 - 259.30) kJ/kg Finally, the minimum volume flow rate at the compressor inlet is
& V&min,1 = mR, minv 1 = (0.007149 kg/s)(0.04112 m 3 /kg) = 0.000294 m 3 /s = 17.64 L/min