Question

7–91 A refrigerator is to remove heat from the cooled space at a rate of 300 kJ/min to maintain its temperature at 28°C. If the air surrounding the refrigerator is at 25°C, determine the minimum power input required for this refrigerator. Answer: 0.623 kW

Answer 1

minimum power input required for this refrigerator is 0.623 kW

Step-by-step explanation:

given data

rate = 300 kJ/min

maintain temperature T1 = -8°C = 265 K

air surrounding the refrigerator T2 = 25°C = 298 K

solution

we get here first COP that is

COP =
`(T1)/(T2-T1)`

COP =
`(265)/(298-265)` = 8.03

and we know here

COP =
`(Q2)/(W)`

8.03 =
`(300)/(60W)`

W = 0.623 kW