Question

A food department is kept at â12°c by a refrigerator in an environment at 30°c. the total heat gain to the food department is estimated to be 3300 kj/h and the heat rejection in the condenser is 4800 kj/h. determine the power input to the compressor, in kw and the cop of the refrigerator

Answer 1

As per energy conservation in the reversible engine we can say

$Q_2 + W = Q_1$

here we know that

$Q_2 = 3300 kJ/h$

$Q_1 = 4800 kJ/h$

now from above equation

$3300 + W = 4800$

$W = 1500 kJ/h$

now we can convert it into kW

$W = 1500\times \frac{kJ}{3600s}$

$W = 0.42 kW$

so above is the power input to the refrigerator

now to find COP we know that

$COP = \frac{Q_2}{W}$

$COP = \frac{3300}{1500} = 2.2$

so COP of refrigerator is 2.2