Question

A vegetable freezing plant requires 35 tons of refrigeration. The freezing temperature is – 38°C while the ambient temperature is 25°C. If the performance of the plant is 30% of the theoretical reversed Carnot cycle working within the same temperature limits, calculate the power required (given: 1 ton of refrigeration = 210 kJ/min).

Answer 1

`\dot W = 109.375\,kW`

Step-by-step explanation:

The theoretical Coefficient of Performance of a refrigerator is:

`COP_(R, ideal) = (T_(L))/(T_(H)-T_(L))`

`COP_(R,ideal) = (235.15\,K)/(298.15\,K-235.15\,K)`

`COP_(R,ideal) = 3.733`

The real Coefficient of Performance is:

`COP_(R) = 0.3\cdot COP_(R,ideal)`

`COP_(R) = 1.120`

The power required to freeze the vegetables is:

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

`\dot W = (\left(35\,tonr \right)\cdot \left((210\,(kJ)/(min) )/(1\,tonr) \right)\cdot \left((1\,min)/(60\,s) \right))/(1.120)`

`\dot W = 109.375\,kW`