Water enters an ice machine at 55 degree farenheit and leave

Water enters an ice machine at 55 degree farenheit and leaves as ice at 25 degree farenheit. The COP of the machine is 3. 100lb/hr of ice produced. Heat is rejected to the surroundings at 100 degree farenheit.

1.) What power input is required?

2.) What is the COP of a perfect ice machine?

Solution

Given: m =100 lb/hr,COP = 3

1) The cooling load can be calculated by,

Q_L = m*(169 BTU/lb) = 16900 BTU/hr

So, Power input required = W_in = Q_L/COP = 5633.333 BTU/hr = 5633.33/2545 hp = 2.2135 hp.

2) COP of perfect ice machine = 100/(55-25) = 3.333