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

a)
`\dot Q_(L) = 132.046\,(kJ)/(min)`, b)
`COP_(R) = 1.467`

Step-by-step 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\,(kg)/(h)\right)\cdot \left((1\,h)/(60\,min) \right)\cdot \left(3.35\,(kJ)/(kg\cdot ^(\textdegree)C) \right)\cdot (24^(\textdegree)C-13\,^(\textdegree)C)`

`\dot Q_(L) = 132.046\,(kJ)/(min)`

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

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

`COP_(R) = (\left(132.046\,(kJ)/(min)\right)\cdot \left((1\,min)/(60\,s) \right))/(1.5\,kW)`

`COP_(R) = 1.467`