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3 years ago

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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.

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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