Question

Bananas are to be cooled from 24°C to 13°C at a rate of 215 kg/h by a refrigeration system. The power input to the refrigerator is 1.5 kW. Determine the rate of cooling, in kJ/min, and the COP of the refrigerator. The specific heat of banana above freezing is 3.35kJ/kg·°C. (Round the final answers to the nearest whole number and two decimal places, respectively.)

Answer 1

Answer:

a) $\dot Q_{L} = 132.046\,\frac{kJ}{min}$, b) $COP_{R} = 1.467$

Explanation:

a) The heat rejected by the bananas is:

$\dot Q_{L} = \dot m \cdot c\cdot (T_{f}-T_{o})$

$\dot Q_{L} = \left(215\,\frac{kg}{h}\right)\cdot \left(\frac{1\,h}{60\,min} \right)\cdot \left(3.35\,\frac{kJ}{kg\cdot ^{\textdegree}C} \right)\cdot (24^{\textdegree}C-13\,^{\textdegree}C)$

$\dot Q_{L} = 132.046\,\frac{kJ}{min}$

b) The coefficient of performance for the refrigeration system is:

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

$COP_{R} = \frac{\left(132.046\,\frac{kJ}{min}\right)\cdot \left(\frac{1\,min}{60\,s} \right)}{1.5\,kW}$

$COP_{R} = 1.467$