Question

A heat pump maintains a dwelling at 68 F. When operating steadily, the power input to the heat pump is 5 hp, and the heat pump receives energy by heat transfer from 55 F well water at a rate of 500 Btu/min. determine the coefficient of performace. Evaluating electricity at $ 0.10 per kW.h determine the cost in a a month when the heat pump operates for 300 hours.

Answer 1

Given:

Work done by the pump, \(W_{P}=5 \mathrm{hp}\).

Heat from cold reservoir, \(Q_{C}=500 \mathrm{Btu} / \mathrm{min}\).

Higher temperature, \(T_{H}=68^{\circ} \mathrm{F}\).

Colder temperature, \(T_{C}=55^{\circ} \mathrm{C}\).

Calculate the COP of the heat pump.

$$ \begin{aligned} \mathrm{COP} &=\frac{Q_{H}}{W_{P}} \\ &=\frac{W+Q_{C}}{W_{P}} \\ &=\frac{5 \mathrm{hp}+500 \frac{\mathrm{Btu}}{\min } \frac{1 \mathrm{~min}}{60 \mathrm{sec}} \frac{1055.05 \mathrm{~J}}{1 \mathrm{Btu}} \frac{1 \mathrm{hp}}{746 \mathrm{Watt}}}{5 \mathrm{hp}} \\ &=\frac{5+11.786}{5} \\ &=3.357 \end{aligned} $$

Calculate the cost of electricity.

$$ \begin{aligned} \text { Cost } &=(\text { Cost } / \mathrm{kW} \cdot \mathrm{h}) \times W_{p} \times t \\ &=\frac{0.10 \$}{1 \mathrm{~kW} \cdot \mathrm{h}} \times 5 \mathrm{hp} \times \frac{0.746 \mathrm{~kW}}{1 \mathrm{hp}} \times 300 \mathrm{hr} \\ &=111.9 \$ \end{aligned} $$