In every 8 h shift a plant processes 100 beef carcasses with an average mass of
ID: 1981547 • Letter: I
Question
In every 8 h shift a plant processes 100 beef carcasses with an average mass of 200 kg/carcass. Beef freezes at -2 degree C. The specific heat of fresh beef (i.e. above the freezing temperature) is 3.2 kJ/(kg K) and the specific heat of frozen beef is 1.7 kJ/(kg K). The latent heat of fusion of beef is 233 kJ kg. Determine the power required for a cooler to cool the beef from 30 degree C to 5 degree C in 8h [55.6 kW] From the cooler the carcasses are transferred to a freezer for freezing to -20 degree C in 8 h. Determine the rate of heat removal in the freezer.Explanation / Answer
Mass of beef cooled in 8hrs = 100*200kg Change in temperature = 25 degrees Energy required to cool the beef = 100*200*25*3.2kJ Power =Energy/Time = ( 100*200*25*3.2 ) / ( 8*60*60) = 55.6kW Energy required to change the temperature from 5 degrees to - 20degrees = 100*200*( (5*3.2) + (20*1.7) + (233))kJ Hence the required rate of heat removal = Energy/time = (100*200*283) / (8*60*60) =196.53kW