Answer:
The required power is [tex]P_i = 0.2692 \ hp[/tex]
Explanation:
From the question we are told that
The temperature of the water entering ice machine is [tex]T_1 = 55 ^o F[/tex]
The temperature of the water leaving is [tex]T_2 = 25 ^oF[/tex]
The COP of the ice machine is [tex]COP= 3.7[/tex]
The production rate of an ice [tex]\r m = 15.0 \ lbm /hr[/tex]
The energy that needs to removed from each lbm of water at 55 F is [tex]E = 169 Btu[/tex]
Generally the cooling load of the ice machine is mathematically represented as
[tex]\r Q_L = \r m * E[/tex]
[tex]\r Q_L =15 * 169[/tex]
=> [tex]\r Q_L = 2535 \ Btu/h[/tex]
Generally the COP of the ice machine is mathematically represented as
[tex]COP = \frac{\r Q_L}{P_i}[/tex]
Here [tex]P_i[/tex] is the net power input needed to successfully run the ice machine
So
[tex]3.7 = \frac{2535}{P_i}[/tex]
=> [tex]P_i = 685 \ Btu/h[/tex]
Converting to horsepower
[tex]P_i = \frac{685}{2545}[/tex]
=> [tex]P_i = 0.2692 \ hp[/tex]
