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

`\dot W_(in) = 0.82363 hp`

Step-by-step 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 * 16 = 4732 Btu/h`

Power is determined as

`\dot W_(in) = \frac{\dot Q_L}COP}`

`\dot W_(in) = (4732)/(2.4) = 1931.42 Btu/h`

`\dot W_(in) =1931.42 Btu * (1 hp)/(2345 Btu/h) = 0.82363 hp`