Question

A heat pump is to be used for heating a house in winter. The house is to be maintained at 70°F at all times. When the temperature outdoors drops to 40°F, the heat losses from the house are estimated to be 75,000 Btu/h. Determine the minimum power [Btu/h] required to run this heat pump if heat is extracted from the outdoor air at 40°F.

Answer 1

Answer:

$\dot{W_{H} } = 4244.48 Btu/h$

Explanation:

Temperature of the house, $T_{H} = 70^{0} F$

Convert to rankine, $T_{H} = 70^{0}+ 460 = 530 R$

Heat is extracted at 40°F i.e $T_{L} = 40^{0}F = 40 + 460 = 500 R$

Calculate the coefficient of performance of the heat pump, COP

$COP = \frac{T_{H} }{T_{H} - T_{L} } \\COP = \frac{530 }{530 - 500 }\\ COP = \frac{530}{30} \\COP = 17.67$

The minimum power required to run the heat pump is given by the formula:

$\dot{W_{H} } = \frac{\dot{Q_{H} }}{COP}$...............(*)

Where the heat losses from the house, $\dot{Q_{H} } = 75,000 Btu/h$

Substituting these values into * above

$\dot{W_{H} } = \frac{75000}{17.67} \\ \dot{W_{H} } = 4244.48 Btu/h$