Question

Determine the COP of a refrigerator that removes heat from the food compartment at a rate of 5060 kJ/h for each kW of power it consumes. Also, determine the rate of heat rejection to the outside air.

Answer 1

Answer:

$COP_{R} = 1.406$, $\dot Q_{H} = 2.406 W$

Explanation:

The COP of a refrigerator is modelled after this expression:

$COP_{R} = \frac{\dot Q_{L}}{\dot W} \\COP_{R} = \frac{(5060 \frac{kJ}{h})(\frac{1 h}{3600 s})}{1 kW}\\COP_{R} = 1.406$

The rate of heat rejection to the outside air is:

$\dot Q_{H} = \dot Q_{L} + \dot W\\\dot Q_{H} = (5060 \frac{kJ}{h} )\cdot (\frac{1}{3600 s} ) + 1 kW\\\dot Q_{H} = 2.406 kW$