Question

Determine the COP of a heat pump that supplies energy to a house at a rate of 8200 kJ/h for each kW of electric power it draws. Also, determine the rate of energy absorption from the outdoor air.

Answer 1

Answer:

$COP_{HP} = 2.278$, $\dot Q_{L} = 1.278\,kW$

Explanation:

The heat supply is:

$Q_{H} = (8200\,\frac{kJ}{h})\cdot (\frac{1\,h}{3600\,s} )$

$Q_{H} = 2.278\,kW$

The Coefficient of Performance of a Heat Pump is:

$COP_{HP} = \frac{\dot Q_{H}}{\dot W}$

$COP_{HP} = \frac{2.278\,kW}{1\,kW}$

$COP_{HP} = 2.278$

The rate of heat absorption from the outdoor air is:

$\dot Q_{L} = \dot Q_{H} - \dot W$

$\dot Q_{L} = 2.278\,kW-1\,kW$

$\dot Q_{L} = 1.278\,kW$