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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, te Heat-rejection ratio 10 C 1. 0 C 1. -10 C 1.
Heat-rejection ratio = 1.255 - 0.0095te + 0.0003te^2
Fig. 18-4. At outdoor air temperature = 0 C
Evaporating Temperature, te 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 + 0.475te + 0.0035te
2
For evaporator, ambient = 0 C
2 e
1 e
1 2
t t
t t ln
t t LMTD
qe = UALMTD
At 0 C, cpm = 1.02 kJ/kg.K = 1020 J/kg.K say purely sensible.
qe = wcpmDt = wcpm(t 1 - t 2 ) qe = (2)(1020)(0 - t2)
But,
2 e
e
2 e
t t
0 t ln
0 t q 25 80
Then,
t t
0 t ln 2 e
e (^) =
t t
0 t
2 e
e (^) = −
1.6644te = 2.6644t 2
t 2 = 0.624836te
qe = (2)(1020)(0 - 0.624836te) / 1000 kW qe = -1.274665te kW
qe = 12.9 + 0.475te + 0.0035te^2 = -1.274665te
0.0035te
2
te = -7.485 C
qe = -1.274665(-7.485) qe = 9.541 kW
Heat-rejection ratio = 1.255 - 0.0095(-7.485) + 0.0003(-7.485)^2 Heat-rejection ratio = 1.
qc = (1.343)(9.541)
qc = 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- 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.
kw ( 3600 s/h)( 13 6.8358hr)
= 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. Heating Load = 335 kW
t 4 = mix temperature = (0.15)(-5 C) + (0.85)(21 C) t 4 = 17.1 C
qe = wcp(t 4 - t 3 ) qe = (40 kg/s)(1.0 kJ/kg.K)(17.1 C - 13 C) qe = 164 kW
Condenser qc = qe(1+ COP) / COP qc = 164(1 + 3.2) / 3. qc = 215.25 kW
Supplementary heat = 335 kW - 215.25 kW = 119.75 kW - - - Ans.
