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Chemical Engineering - Heat Transfer - Discussion

Discussion Forum : Heat Transfer - Section 1 (Q.No. 20)
Heat Transfer
20.
The thermal efficiency of a reversible heat engine operating between two given thermal reservoirs is 0.4. The device is used either as a refrigerator or as a heat pump between the same reservoirs. Then the coefficient of performance as a refrigerator (COP)R and the co-efficient of performance as a heat pump (COP)HP are
(COP)R = (COP)HP = 0.6
(COP)R = 2.5; (COP)HP = 1.5
(COP)R = 1.5; (COP)HP = 2.5
(COP)R = (COP)HP = 2.5
Answer: Option
Explanation:
No answer description is available. Let's discuss.
Discussion:
12 comments Page 2 of 2.
Khushbu chaudhary said:   1 decade ago
Thermal efficiency = 0.4 = 1-(Ql/Qh)
So, Ql/Qh=0.6
Qh/Ql=1/0.6

(COP)R=1/((Qh/Ql)-1) = 0.6/0.4
= 1.5

(COP)HP = (COP)R+1
= 1.5+1 = 2.5
Kshema said:   1 decade ago
Kindly explain the calculation.
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