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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
[2009]
  • a)
    2.27
  • b)
    2.75
  • c)
    3.27
  • d)
    3.75
Correct answer is option 'A'. Can you explain this answer?
Verified Answer
In an ideal vapour compression refrigeration cycle, the specific entha...

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

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Most Upvoted Answer
In an ideal vapour compression refrigeration cycle, the specific entha...

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

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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 : 283Exit of condenser : 116Exit of evaporator : 232The COP of this cycle is[2009]a)2.27b)2.75c)3.27d)3.75Correct answer is option 'A'. Can you explain this answer?
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