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An ideal refrigerator working on a reversed Rankine cycle has a capacity of 3 tons. The COP of the unit is found to be 6. The capacity of the motor required to run the unit is (take 1 T = 210 kJ/min)
1. 1.85 kW
2. 1.75 kW
3. 1.65 kW
4. 1.50 kW

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Correct Answer - Option 2 : 1.75 kW

Concept:

\(COP = \frac{{Referigeration\;Capacity}}{{Work\;input}}\)

1 Ton of Refrigerant = 210 kJ/min = 3.5 kW

Calculation:

Given, Refrigeration capacity = 3 tons = 3 × 3.5 kW = 10.5 kW, COP = 6

\(6 = \frac{{10.5\;kW}}{{{W_{In}}}}\)

Win = 1.75 kW

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