Question

A food department is kept at 2128C by a refrigerator in an environment at 308C. 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

2.2

Step-by-step explanation:

`Q_u` = Heat rejection in the condenser = 3300 kJ/h

`Q_L` = Heat gain to the food department = 4800 kJ/h

Power output is given by

`W=Q_u-Q_L\\\Rightarrow W=4800-3300\\\Rightarrow W=1500\ kJ/h`

COP of a refrigerator is given by

`COP=(Desired\ effect)/(Work)\\\Rightarrow COP=(Q_L)/(W)\\\Rightarrow COP=(3300)/(1500)\\\Rightarrow COP=2.2`

The COP of the refrigerator is 2.2