Question

a refrigerator is to remove heat from the cooled space at a rate of 300 kj/min to maintain its temperature at -8°c. if the air surrounding the refrigerator is at 25°c determine the minimum power input required for this refrigerator.

Answer 1

Surrounding air temperature Ts = 25 degree Celcius

Therefore Ts = 25 + 273 = 298 K

Temperature maintained by refrigerator Tr = -8 degree Celcius

Therefore Ts = -8 + 273 = 265 K

Cooling rate Qr =300 KJ/min

i.e. Qr =5 kW

Coefficient of performance Cp =Tr / (Ts-Tr) = 265/(298-265) = 8.03

Also, Coefficient of performance Cp = Qr/ (Qs-Qr)

8.03 = 5/ (Qs – 5)

Qs = 5.62 kW

Refrigerator’s power input will be:

W = Qs – Qr = 5.62 – 5

Minimum Power input = 0.62 kW

Answer 2

refrigerated space temperature , TL = -8°C
= -8 + 273 = 265K

surrounding air temperature, TH = 25 + 273
= 298K

rate of cooling effect , QL = 300kJ/min
= 300 kJ/60sec
= (300/50) k J/sec
= 5 kW

COPeff = QL/{QH - QL} = TL/{TH - TL}

5kW/(QH - 5) = 265/(298 - 265)

5kW/(QH - 5) = 265/33

QH = 5.6226kW

now, power input required for this refrigerator, Wnet = QH - QL

= 5.6226 - 5 = 0.6226kW