It is desired to freeze 10,000 loaves of bread each weighing 0.75 kg from an initial room temperature of 18°C to a final temperature of -18°C. The bread-freezing operation is to be carried out in an air-blast freezing tunnel. It is found that the fan motors are rated at a total of 80 horsepower and measurements suggest that they are operating at around 90% of their rating, under which conditions their manufacturer’s data claims a motor efficiency of 86%. If 1 ton of refrigeration is 3.52 kW, estimate the maximum refrigeration load imposed by this freezing installation assuming that fans and motors are all within the freezing tunnel insulation and the heat-loss rate from the tunnel to the ambient air has been found to be 6.3 kW. Extraction rate from freezing bread (maximum) = 104 kW Correct Answer 46 tons of refrigeration Extraction rate from freezing bread (maximum) = 104 kW Fan rated horsepower = 80 Now 0.746 kW = 1 horsepower and the motor is operating at 90% of rating, And so (fan + motor) power = (80 x 0.9) x 0.746 = 53.7 kW With motors + fans in tunnel Heat load from fans + motors = 53.7 kW Heat load from ambient = 6.3 kW Total heat load = (104 + 53.7 + 6.3) kW = 164 kW = 46 tons of refrigeration.