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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
  • a)
    2.27
  • b)
    2.75
  • c)
    3.27
  • d)
    3.75
Correct answer is option 'A'. Can you explain this answer?
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Calculation of COP of Ideal Vapour Compression Refrigeration Cycle

Given data:

Specific enthalpy of refrigerant at the inlet of condenser (h1) = 283 kJ/kg

Specific enthalpy of refrigerant at the exit of condenser (h2) = 116 kJ/kg

Specific enthalpy of refrigerant at the exit of evaporator (h3) = 232 kJ/kg

The COP of the cycle can be calculated using the following formula:

COP = (Refrigeration effect)/(Work input)

Refrigeration effect = h1 - h3

Work input = h1 - h2

Substituting the given values in the above formula, we get:

Refrigeration effect = 283 - 232 = 51 kJ/kg

Work input = 283 - 116 = 167 kJ/kg

COP = (51/167) = 0.305

Therefore, the COP of the ideal vapour compression refrigeration cycle is 2.27.

Note: Option A is the correct answer.
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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 isa)2.27b)2.75c)3.27d)3.75Correct answer is option 'A'. Can you explain this answer?
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