Question

Water enters an ice machine at 55°F and leaves as ice at 25°F. If the COP of the ice machine is 2.45 during this operation, determine the required power input for an ice production rate of 28 lbm/h. (169 Btu of energy needs to be removed from each lbm of water at 55°F to turn it into ice at 25°F.)

Answer 1

Answer:

$\dot W_{in} = 0.82363 hp$

Explanation:

Given data:

T_1 = 55 degree F

t_2 - 25 degree F

COP = 2.45

Production rate = 28 lbm/h

we knwo that cooling load is given

$\dot Q_L = \dot m q_L$

$\dot Q_L = 28 lbm/h \times 16 = 4732 Btu/h$

Power is determined as

$\dot W_{in} = \frac{\dot Q_L}COP}$

$\dot W_{in} = \frac{4732}{2.4} = 1931.42 Btu/h$

$\dot W_{in} =1931.42 Btu \times \frac{1 hp}{2345 Btu/h} = 0.82363 hp$