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

a)
`\dot W = 0.417\,kW`, b)
`COP_(R) = 2.198`, c) Irreversible.

Step-by-step explanation:

a) The power input required by the refrigerator is:

`\dot W = \dot Q_(H) - \dot Q_(L)`

`\dot W = \left(4800\,(kJ)/(h) - 3300\,(kJ)/(h)\right)\cdot \left((1)/(3600) \,(h)/(s) \right)`

`\dot W = 0.417\,kW`

b) The Coefficient of Performance of the refrigerator is:

`COP_(R) = (\dot Q_(L))/(\dot W)`

`COP_(R) = (3300\,(kJ)/(h) )/((0.417\,kW)\cdot \left(3600\,(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) = (T_(L))/(T_(H)-T_(L))`

`COP_(R,ideal) = (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)`.