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Heat Transfer - Engineering

Q1:

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

A (COP)R = (COP)HP = 0.6

B (COP)R = 2.5; (COP)HP = 1.5

C (COP)R = 1.5; (COP)HP = 2.5

D (COP)R = (COP)HP = 2.5

ANS:C - (COP)R = 1.5; (COP)HP = 2.5

The coefficient of performance (COP) for a refrigerator and a heat pump can be related to the thermal efficiency (η) of a reversible heat engine by the following equations: COPR​=Win​QL​​=QH​−QL​QL​​ COPHP​=Win​QH​​=QH​−QL​QH​​ Where:

  • QL​ is the heat removed from the low-temperature reservoir (for refrigerator).
  • QH​ is the heat added to the high-temperature reservoir (for heat pump).
  • inWin​ is the work input to the system.
Given that the thermal efficiency (η) of the reversible heat engine is 0.4, we have: η=0.4=QH​Wout​​ 0.4QH​QH​−QL​​=0.4 =0.41−QH​QL​​=0.4 0.6QH​QL​​=0.6 QL​=0.6×QH​ Substituting this into the equations for COP, we get: QH​−0.6QH​0.6QH​​=0.40.6​=1.5 COPHP​=QH​−0.6QH​QH​​=0.41​=2.5 Therefore, the correct answer is: (COP)�=1.5;(COP)HP=2.5(COP)R​=1.5;(COP)HP​=2.5