In an ideal vapour compression refrigeration cycle, the

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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

Correct Option: A

1– 2 → work done by the compressor
2 – 3 → condenser heat rejected at constant pressure
3 – 4 → throttling
4 – 1 → heat addition in evaporator
Given: h₂ = 283 kJ/ kg
h₃ = 116 kJ/kg = h₄ (from ph curve)
h₁ = 232 kJ/kg
We know

C.O.P =	desired effect	=	Cooling effect
	work input		work done by compressor

C.O.P =	h₁ – h₄	=	232 - 116	= 2.27
	h₂ – h₁		283 - 232