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The COP of an ideal refrigerator of capacity 2.5 T is 5. The power of the motor required to run the plant is : (a) 1.15 kW (b) 1.35 kW (c) 1.55 kW (d) 1.75 kW |
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Answer» (d) 1.75 kW The COP of a refrigerator, COP = \(\cfrac{Refrigeration \,effect}{Compression \,work}\) = \(\cfrac RW\) 5.0 = \(\cfrac{2.5\times3.5}W\) W = \(\cfrac{2.5\times3.5}5\) = 1.75 kW |
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