Question

A refrigeration system is required to freeze 0.336 kg of ice in 1.80 minutes. The freezer is at 24.8 deg F and the room air is at 68.7 deg F.

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

Answer 1

Solution-;

Given data,  

m=0.336 kg

1 hp = 745.7 N
T1 = 24.8 F = 269.04 K
T2 = 68.7 F = 293.54 K

Time, t = 1.80 minutes = 1.80*60

Time , t = 108 second

Energy required per sec = mLf / t

Then put values..

= (0.336*334*1000) / 108

= 1039.11 W

Heat / work = T1/T2-T1

Here, T1= 269.04 K

T2 = 293.54 K

So, put values
Heat/work = 269.04 / (293.54 - 269.04)
Heat/work = 10.98

W = 1039.11 / 10.98
= 94.64 Watts

Power = W / 1hp

Power = 94.64 / 745.7
= 0.127 hp (Answer)

I hope help you !!