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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

Respuesta :

As per energy conservation in the reversible engine we can say

[tex]Q_2 + W = Q_1[/tex]

here we know that

[tex]Q_2 = 3300 kJ/h[/tex]

[tex]Q_1 = 4800 kJ/h[/tex]

now from above equation

[tex]3300 + W = 4800[/tex]

[tex]W = 1500 kJ/h[/tex]

now we can convert it into kW

[tex]W = 1500\times \frac{kJ}{3600s}[/tex]

[tex]W = 0.42 kW[/tex]

so above is the power input to the refrigerator

now to find COP we know that

[tex]COP = \frac{Q_2}{W}[/tex]

[tex]COP = \frac{3300}{1500} = 2.2[/tex]

so COP of refrigerator is 2.2

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