Question:-
In a vapour compression refrigeration plant, the enthalpy values at different points are :
1. Enthalpy at exit of the evaporator = 350 kJ/kg
2. Enthalpy at exit of the compressor 375 kJ/kg
3. Enthalpy at exit of the condenser = 225 kJ/kg
The refrigerating efficiency of the plant is 0.8. What is the power required per kW of cooling to be produced?
Option (A)
0.25 kW
Option (B)
4.0 kW
Option(C)
12.5 kW
Option(D)
11 kW
Correct Option:
(A)
Question Solution:
h1 = 350 KJ/kg
h2 = 375 KJ/kg
h3 = h4 = 225 KJ/kg
(COP)th = {(h1-h4) / (h2-h1)} = {(350-225)/(375-350)} = 125/25 = 5
Refrigeration efficiency = (COP)actual / (COP)Theoretical
Þ COPactual = 0.8 x 5 = 4
Þ R.E / Work done (Power) = 4
Þ Work done (power) / refrigerating effect = 1/4 = 0.25

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