Mechanical Engineering - Heat Transfer, Refrigeration and Air Conditioning - Discussion
Discussion Forum : Heat Transfer, Refrigeration and Air Conditioning - Section 7 (Q.No. 18)
18.
A reversible engine has ideal thermal efficiency of 30%. When it is used as a refrigerating machine with all other conditions unchanged, the coefficient of performance will be
Discussion:
2 comments Page 1 of 1.
Bharath said:
6 years ago
COP of heat pump = 1/efficiency of heat engine.
COP of heat pump = 1 + COP of the refrigerator.
So, the answer is 2.33.
COP of heat pump = 1 + COP of the refrigerator.
So, the answer is 2.33.
(1)
Gaurav said:
1 decade ago
C.O.P is reciprocal of efficiency of heat engine.
So 1/0.3 = 3.33 = C.O.P of heat engine.
Now C.O.P of refrigerator = c.o.p of heat engine-1 = 3.33-1 = 2.33.
So 1/0.3 = 3.33 = C.O.P of heat engine.
Now C.O.P of refrigerator = c.o.p of heat engine-1 = 3.33-1 = 2.33.
(1)
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