Chapter18 Refrigeração e Ar Condicionado - Stoecker and Jones solution

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CHAPTER 18 - HEAT PUMPS 
 Page 1 of 4 
18-1. An air-source heat pump uses a compressor with the performance characteristics shown in Fig. 18-4. The 
evaporator has an air-side area of 80 m2 and a U-value of 25 W/m
2
.K. The airflow rate through the 
evaporator is 2 kg/s, and the condensing temperature is 40 C. Using the heat-rejection ratios of a hermetic 
compressor from Fig. 12-12, determine the heating capacity of the heat pump when the outdoor-air 
temperature is 0C. 
 
Solution: 
 use Fig. 18-4 and Fig. 12-2. 
 
 Fig. 12-2 at 40 C Condensing temperature. 
 Evaporating Temperature, t
e
 Heat-rejection ratio 
 10 C 1.19 
 0 C 1.255 
 -10 C 1.38 
 
 Heat-rejection ratio = 1.255 - 0.0095t
e
 + 0.0003t
e
2
 
 
 Fig. 18-4. At outdoor air temperature = 0 C 
 
 Evaporating Temperature, t
e
 Rate of evaporator heat transfer 
 -10 C 8.5 kw 
 0 C 12.9 kw 
 10 C 18.0 kw 
 
 Rate of evaporator heat transfer = 12.9 + 0.475t
e
 + 0.0035t
e
2
 
 
 For evaporator, ambient = 0 C 
 








−
−
−
=
e2
e1
21
tt
tt
ln
tt
LMTD
 
 
 q
e
 = UALMTD 
 
 At 0 C, c
pm
 = 1.02 kJ/kg.K = 1020 J/kg.K say purely sensible. 
 
 q
e
 = wc
pm
Dt = wc
pm
(t
1
 - t
2
) 
 q
e
 = (2)(1020)(0 - t2) 
 
 But, 
 
 
( )( ) ( )








−
−
−
=
e2
e
2
e
tt
t0
ln
t0
8025q
 
 Then, 
 
1.02
1
tt
t0
ln
e2
e
=







−
−
 
 
2.6655
tt
t0
e2
e
=
−
−
 
 1.6644t
e
 = 2.6644t
2
 
CHAPTER 18 - HEAT PUMPS 
 Page 2 of 4 
 t
2
 = 0.624836t
e
 
 
 q
e
 = (2)(1020)(0 - 0.624836te) / 1000 kW 
 q
e
 = -1.274665t
e
 kW 
 
 q
e
 = 12.9 + 0.475t
e
 + 0.0035t
e
2
 = -1.274665te 
 0.0035t
e
2
 + 1.749665t
e
 + 12.9 = 0 
 
 t
e
 = -7.485 C 
 
 q
e
 = -1.274665(-7.485) 
 q
e
 = 9.541 kW 
 
 Heat-rejection ratio = 1.255 - 0.0095(-7.485) + 0.0003(-7.485)
2
 
 Heat-rejection ratio = 1.343 
 
 q
c
 = (1.343)(9.541) 
 
 q
c
 = 12.8 kW - - - Ans. 
 
18-2. The heat pump and structure whose characteristics are shown in Fig. 18-6 are in a region where the deisgn 
outdoor temperature is -15 C. The compressor of the heat pump uses two cylinders to carry the base load 
and brings a third into service when needed. The third cylinder has a capacity equal to either of the other 
cylinders. How much supplementary resistance heat must be available at an outdoor temperature of -15 C? 
 
Solution: 
 Use Fig. 18-6. 
 At -15 C 
 Heat loss of structure = 17.8 kW 
 Heating capacity = 8.0 kW 
 
 For two-cylinder = 8.0 kW 
 
 For three-cylinder = (3/2)(8.0 kW) = 12.0 kW 
 
 Supplementary resistance heat = 17.8 kW - 12.0 kW 
 = 5.8 kW - - - Ans. 
 
18-3. The air-source heat pump referred to in Figs. 18-4 and 18-5 operates 2500 h during the heating season, in 
which the average outdoor temperature is 5 C. The efficiency of the compressor motor is 80 percent, the 
motor for the outdoor air fan draws 0.7 kW, and the cost of electricity is 6 cents per kilowatt-hour. What is the 
heating cost for this season. 
 
Solution: 
 Use Fig. 18-4 and Fig. 18-5. 
 Outdoor air temperature = 5 C. 
 Fig. 18-5 
 Heating capacity = 15.4 kW 
 Evaporator heat-transfer rate = 12 kW 
 Compressor Power = 3.4 kW 
 
 Power to compressor motor = (3.4)(2500)($0.06) / (0.80) 
 = $ 637.50 
CHAPTER 18 - HEAT PUMPS 
 Page 3 of 4 
 Power to outdoor air fan motor = $ 150.00 
 
 Heating Cost = $ 637.50 + $ 150.00 = $ 787.50 - - - Ans. 
 
18-4. A decentralized heat pump serves a building whose air-distribution system is divided into one interior and 
one perimeter zone. The system uses a heat rejector, water heater, and storage tank (with a water capacity 
of 60 m
3
) but no solar collector. The heat rejector comes into service when the temperature of the return-loop 
water reaches 32 C, and the boiler supplies supplementary heat when the return-loop water temperature 
drops to 15 C. Neither component operates when the loop water temperature is between 15 and 32 C. The 
heating and cooling loads of the different zones for two periods of a certain day as shown in Table 18-1. The 
loop water temperature is 15 C at the start of the day (7 A.M.). The decentralized heat pumps operate with 
COP of 3.0. Determine the magnitude of (a) the total heat rejection at the heat rejector from 7 A.M. yo 6 P.M. 
and (b) the supplementary heat provided from 6 P.M. to 7 A.M. 
 
Table 18-1 Heating and Cooling loads in Prob. 18-4. 
 Interior zone Perimeter zone 
 
 Heating, kW Cooling, kW Heating, kW Cooling, kW 
7 A.M. to 6 P.M. -------------- 260 ------------- 40 
6 P.M. to 7 A.M. -------------- 50 320 ------------ 
 
Solution: 
 
 Weight of water in storage tank. 
 V = 60 m
3
 
 at 24 C, ρ = 997.4 kg/m3 
 w = (997.4)(60) = 59,884 kg 
 
 Storage tank heat 
 = (59,884 kg)(4,190 kJ/kg.K)(32 - 15 K) 
 = 4.266 GJ 
 
(a) Heating time = 
 
( ) ( )s/h3600
3
31
kW40260
kJ4,265,537





 +
+
=
 
 = 2.962 hra 
 From 7 A.M. to 6 P.M. = 11 hrs 
 
 Total heat rejection 
 = (260 + 40 kW)[(1+ 3) / 3](3600 s/h)(11 - 2.962 hr) 
 = 11,574,720 kJ 
 = 11.6 GJ - - - Ans. 
 
 (b) Supplementary heat 
 
 Storage tank heat = 4,265,537 kJ 
 
 
( ) kJ4,265,537s/h3600kw
3
31
50
31
3
320(Time) =










 +
−





+
 
 Time = 6.8358 hrs 
 
 Supplementary heat 
CHAPTER 18 - HEAT PUMPS 
 Page 4 of 4 
 
( )( )hr6.835813s/h3600kw
3
31
50
31
3
320 −










 +
−





+
=
 
 = 3,846,461 kJ 
 = 3.85 GJ - - - Ans. 
 
18-5. The internal-source heat pump using the double-bundle heat pump shown in Fig. 18-9 is to satisfy a heating 
load of 335 kW when the outdoor temperature is -5 C, the return air temperature is 21 C, and the 
temperature of the cool supply air is 13 C. The minimum percentage of outdoor air specified for ventilation is 
15 percent, and the flow rate of cool supply air is 40 kg/s. If the COP of the heat pump at this condition is 3.2, 
how much power must be provided by the supplementary heater? 
 
Solution: 
 Outdoor air = -5 C, 15 % flow rate 
 Return air = 21 C 
 t
3
 = Cool supply air = 13 C, w = 40 kg/s 
 COP = 3.2 
 Heating Load = 335 kW 
 
 t
4
 = mix temperature = (0.15)(-5 C) + (0.85)(21 C) 
 t
4
 = 17.1 C 
 
 q
e
 = wc
p
(t
4
 - t
3
) 
 q
e
 = (40 kg/s)(1.0 kJ/kg.K)(17.1 C - 13 C) 
 q
e
 = 164 kW 
 
 Condenser 
 q
c
 = q
e
(1+ COP) / COP 
 q
c
 = 164(1 + 3.2) / 3.2 
 q
c
 = 215.25 kW 
 
 Supplementary heat = 335 kW - 215.25 kW 
 = 119.75 kW - - - Ans. 
 
 
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