Water enters an ice machine at 55°F and leaves as ice at 25°F. If the COP of the ice machine is 2.4 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.) The required power input for an ice production rate of 28 lbm/h is

Required power input = 0.77hp

Explanation:

We need to first of all find the cooling load of the ice machine.

The cooling load is expressed as

=(28 lbm/h) * (169 Btu/lbm)

= 4732 Btu/h.

Therefore, to find the power input of the ice machine, we use:

Wm = Ql / COP

Wm = (4732Btu / 2.2)

Wm = 1971.66 Btu/h

Since we are asked to find power, let's convert to hp.

Therefore

Wm = (1971.66 Btu/m) * (1hp/(2545Btu/m)) = 0.7741hp

The required power input for an ice production rate of 28 lbm/h is 0.77hp

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