A heat pump, with the elements shown in Figure 7.25 (Condenser to valve to evapo

Question

A heat pump, with the elements shown in Figure 7.25 (Condenser to valve to evaporator to compressor to condenser) is to be used to heat a home. On a given day, the evaporator receives heat at 30.0°F and the condenser rejects heat at 70.0°F. The required heat transfer rate from the condenser to the home is 50 x 103 Btu/hour.
a. If the heat pump is operated reversibly, what is the rate of work transfer of energy to the compressor?
b. Express this compressor work rate in kilowatts and calculate the cost per 24.0 hour day if electricity is purchased at 10.0 cents/(kW * hour).

Explanation / Answer

a) For a reversible operating heat pump,

COP = T1 / (T1 - T2)

where, T1 = 70 F = 294.11 K

and T2 = 30 F = 271.89 K

COP = 294.11 / (294.11 - 271.89)

= 13.23

Now heat transferred to the home = Q1 = 50 x 10^3 Btu/hr

COP = Q1 / W in

13.23 = 50 x 10^3 / W in

W in = 3779.29 Btu/hr = rate of work transfer to the compressor.

b) Now 1 Btu = 1.055 KJ

rate of work transfer = 3779.29 x 1.055 /3600 KJ/s = 1.1075 KJ/s

= 1.1075 KW

Now, for a day electricity consumption = 1.1075 x 24 = 26.58 Kw hr

cost of 1 KW hr = 10 cents

Total cost of electicity = 26.58 x 10 = 260.58 cents.

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