Question

Problem 027-50E Water enters an ice machine at (60 ID) 'F and leaves as ice at 25'F. Assume latent heat of fusion of ice s 144 Btu/lb) 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 pounds per hour. Give answer in both Btu/hr and hp units

ID = 38

Answer 1

STEP 1: COP of refrigerator COP_R = 2.4 Low Temperature T_L = 26 degree F High Temperature T_H = (60 - ID) degree F Mass of Ice produced m_ice^dot = 28 lbm/h STEP 2: Q_L^dot = m_ice^dot times 144 (Btu/lb) Colling effect Latent heat of fusion = 28 lb times 144 Btu/lb = 4032 Btu Because Total power input w_net, in = Q_L^dot/COP_R = 4032 Btu/2.4 = 1680 Btu/h Conversion factor 1 hp = 2545.468 Btu/h = > w_net, in = 1680/2545.458 np = 0.65999 hp