Multiple choice

In an ideal vapour compression refrigeration cycle, the specific enthalpy of refrigerant (in kJ/kg) at the following states is given as: Inlet of condenser: 283 Exit of condenser: 116 Exit of evaporator: 232 The COP of this cycle is

  1. 2.27

  2. 2.75

  3. 3.27

  4. 3.75
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A Correct answer
Explanation

$\text{Substitute the values, we get}\\ COP = \frac{232 - 116}{283 - 232} = \frac{116}{51} = 2.27$

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