Chemical Engineering - Heat Transfer - Discussion
Discussion Forum : Heat Transfer - Section 1 (Q.No. 20)
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
Discussion:
12 comments Page 2 of 2.
Nakachew Munwyelet said:
6 years ago
(COP)R = Qc/(Qh-Qc)=(Qc/Qh)/(1-Qc/Qh)
Thermal efficiency = (Qh-Qc)/Qh = 1-Qc/Qh.
Qc/Qh = 1 - Thermal efficiency.
(COP)R = (1-Thermal efficiency)/Thermal efficiency,
(COP)hp = Qh/(Qh-Qc)=1/(1-Qc/Qh) = 1/Thermal efficiency,
(COP)R = (1-Thermal efficiency)/Thermal efficiency,
= (1/Thermal efficiency)-1 = (COP)hp-1.
(COP)R = (COP)hp-1.
(COP)hp = (COP)R+1.
Thermal efficiency = (Qh-Qc)/Qh = 1-Qc/Qh.
Qc/Qh = 1 - Thermal efficiency.
(COP)R = (1-Thermal efficiency)/Thermal efficiency,
(COP)hp = Qh/(Qh-Qc)=1/(1-Qc/Qh) = 1/Thermal efficiency,
(COP)R = (1-Thermal efficiency)/Thermal efficiency,
= (1/Thermal efficiency)-1 = (COP)hp-1.
(COP)R = (COP)hp-1.
(COP)hp = (COP)R+1.
Harika said:
3 years ago
Nice explanation. Thankyou.
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