Question

g A food department is kept at -12oC by a refrigerator in an environment at 30oC. The total heat gain to the food department is estimated to be 3300 kJ/h, which should be transferred out of the food department by the refrigerator. The heat rejection from the refrigerator to the environment is 4800 kJ/h. Determine the power input required by the refrigerator, in kW and the COP of the refrigerator. Is the refrigeration cycle reversible, irreversible, or impossible

Answer 1

Answer:

a) $\dot W = 0.417\,kW$, b) $COP_{R} = 2.198$, c) Irreversible.

Explanation:

a) The power input required by the refrigerator is:

$\dot W = \dot Q_{H} - \dot Q_{L}$

$\dot W = \left(4800\,\frac{kJ}{h} - 3300\,\frac{kJ}{h}\right)\cdot \left(\frac{1}{3600} \,\frac{h}{s} \right)$

$\dot W = 0.417\,kW$

b) The Coefficient of Performance of the refrigerator is:

$COP_{R} = \frac{\dot Q_{L}}{\dot W}$

$COP_{R} = \frac{3300\,\frac{kJ}{h} }{(0.417\,kW)\cdot \left(3600\,\frac{s}{h} \right)}$

$COP_{R} = 2.198$

c) The maximum ideal Coefficient of Performance of the refrigeration is given by the inverse Carnot's Cycle:

$COP_{R,ideal} = \frac{T_{L}}{T_{H}-T_{L}}$

$COP_{R,ideal} = \frac{261.15\,K}{303.15\,K - 261.15\,K}$

$COP_{R,ideal} = 6.218$

The refrigeration cycle is irreversible, as $COP_{R} < COP_{R,ideal}$.