Question

Determine the COP of a heat pump that supplies energy to a house at a rate of 8000 kJ/h for each kW of electric power it draws. Also, determine the rate of energy absorption from the outdoor air.

Answer 1

Solution;

Given;

$\dot Q =8000kJ/h$

$W=1kW$

$COP_{HP}=\frac{Q}{W}=(\frac{8000kJ/h}{1kW})(\frac{1kW}{3600kJ/h})=2.22$

$\dot Q_L=\dot Q_{HP}-\dot W$

$\dot Q_L=8000-1(3600)=4400kJ/h$