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Standard X
Physics
Change in State
In an ideal v...
Question

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
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B
2.75
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C
3.27
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D
3.75
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Solution

The correct option is A 2.27
Given data:

h2=283 kJ/kg

h3=116 kJ/kg
=h4

h1=232 kJ/kg



Coefficient of perfomanace

COP=Refrigerating effect(RE)Work input(w)

where RE=h1−h4=232−116=116 kJ/kg

and
w=h2−h1=283−232=51 kJ/kg

∴ COP=11651=2.27

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