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

Answer:

$\dot W = 109.375\,kW$

Explanation:

The theoretical Coefficient of Performance of a refrigerator is:

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

$COP_{R,ideal} = \frac{235.15\,K}{298.15\,K-235.15\,K}$

$COP_{R,ideal} = 3.733$

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:

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

$\dot W = \frac{\left(35\,tonr \right)\cdot \left(\frac{210\,\frac{kJ}{min} }{1\,tonr} \right)\cdot \left(\frac{1\,min}{60\,s} \right)}{1.120}$

$\dot W = 109.375\,kW$