Question 6.4: Heating a House by a Heat Pump A heat pump is used to meet t...

The COP of a heat pump is given. The power consumption and the rate of heat absorption are to be determined.
Assumptions Steady operating conditions exist.
Analysis (a) The power consumed by this heat pump, shown in Fig. 6–25, is determined from the definition of the coefficient of performance to be

\dot{W}_{\text {net,in }}=\frac{\dot{Q}_{H}}{ COP _{ HP }}=\frac{80,000 kJ / h }{2.5}= 3 2 , 0 0 0 kJ / h \text { (or } 8.9 kW \text { ) }

(b) The house is losing heat at a rate of 80,000 kJ/h. If the house is to be maintained at a constant temperature of 20°C, the heat pump must deliver heat to the house at the same rate, that is, at a rate of 80,000 kJ/h. Then the rate of heat transfer from the outdoor becomes

\dot{Q}_{L}=\dot{Q}_{H}-\dot{W}_{\text {net,in }}=(80,000-32,000) kJ / h =48,000 kJ / h

Discussion Note that 48,000 of the 80,000 kJ/h heat delivered to the house is actually extracted from the cold outdoor air. Therefore, we are paying only for the 32,000-kJ/h energy that is supplied as electrical work to the heat pump. If we were to use an electric resistance heater instead, we would have to supply the entire 80,000 kJ/h to the resistance heater as electric energy. This would mean a heating bill that is 2.5 times higher. This explains the popularity of heat pumps as heating systems and why they are preferred to simple electric resistance heaters despite their considerably higher initial cost.