Question

A refrigeration system is required to freeze 0.487 kg of ice in 1.37 minutes. The freezer is at 24.4 deg F and the room air is at 65.5 deg F.

What is the minimum horsepower of motor required to operate this refrigeration system in h.p.?

Answer 1

Solution-;

Given data,  

m=0.487 kg

Then we know, 1 hp = 745.7 N

Temperature is ,
T1 = 24.4 F = 268.93 K
T2 = 65.5 F = 291.48 K

Time, t = 1.37 minutes

To convert seconds so,

t= 1.37*60sec

Time , t = 82.2 second

Energy required per sec = mLf / t

Then put values..

= (0.487kg*334*1000) / 82.2 second

= 1978.81 W

Heat / work = T1/T2-T1

So, put values
Heat/work is

= 268.93 / (291.48 - 268.93)
Heat/work = 11.926

W = 1978.81 / 11.926
= 165.924 Watts

Power = W / 1hp

Power = 165.924 / 745.7
= 0.2225 hp (Answer)

I hope help you !!