Question

Determine the COP of a heat pump that supplies energyto 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

The rate of energy absorption from the outdoor air is 4400 KJ/h

Explanation:

Coefficient of performance (COP), is an expression of the efficiency of a heat pump. When calculating the COP for a heat pump, the heat output from the condenser (Q) is compared to the power supplied to the compressor (W).

The coefficient of performance, $C O P=\frac{Q_{H}}{W_{i n}}$

We replace the value in the equation with given data,

                     $C O P=\frac{(8000 K J / h)}{1 K W}=8000 K J / h$

By making a conversion, we get

                    $C O P=\frac{(8000 \mathrm{KJ} / \mathrm{h})}{1 \mathrm{KW}} \times \frac{(1 \mathrm{KW})}{3600}=2.22$

The energy balance,  $Q_{L}=Q_{H}-W_{i n}$

Now, replace the values in the equation, we get

                   $Q_{L}=(8000 \mathrm{KJ} / \mathrm{h})-(3600 \mathrm{KJ} / \mathrm{h})=4400 \mathrm{KJ} / \mathrm{h}$