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dmitriy555 [2]
3 years ago
15

A food department is kept at 2128C by a refrigerator in an environment at 308C. The total heat gain to the food department is es

timated 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.
Physics
1 answer:
Vesna [10]3 years ago
7 0

Answer:

2.2

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=\frac{Desired\ effect}{Work}\\\Rightarrow COP=\frac{Q_L}{W}\\\Rightarrow COP=\frac{3300}{1500}\\\Rightarrow COP=2.2

The COP of the refrigerator is 2.2

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