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7nadin3 [17]
2 years ago
12

A refrigerator has a cooling load of 50 kW. It has a COP of 2. It is run by a heat engine which consumes 50 kW of heat to supply

the necessary power to the refrigerator. The efficiency of the heat engine is:________________.
Engineering
1 answer:
Sunny_sXe [5.5K]2 years ago
8 0

Answer:

50%

Explanation:

<u>Given information</u>

Cooling load=50 kW

COP=2

Consumption=50 kW

<u>Calculations</u>

Revised input is given by cooling load/COP=50/2=25 kW

Efficiency= Work output/ Revised input=25/50=0.5

Efficiency=0.5*100=50%

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