A company owns a refrigeration system whose refrigeration capacity is 200 tons (1 ton of refrigeration =
Question:
A company owns a refrigeration system whose refrigeration capacity is 200 tons (1 ton of refrigeration = 211 kJ/min), and you are to design a forced-air cooling system for fruits whose diameters do not exceed 7 cm under the following conditions:
The fruits are to be cooled from 28°C to an average temperature of 8°C. The air temperature is to remain above - 2°C and below 10°C at all times, and the velocity of air approaching the fruits must remain under 2 m/s. The cooling section can be as wide as 3.5 m and as high as 2 m.
Assuming reasonable values for the average fruit density, specific heat, and porosity (the fraction of air volume in a box), recommend reasonable values for the quantities related to the thermal aspects of the forced-air cooling, including (a) how long the fruits need to remain in the cooling section, (b) the length of the cooling section, (c) the air velocity approaching the cooling section, (d) the product cooling capacity of the system, in kg fruit/h, (e) the volume flow rate of air, and (f) the type of heat exchanger for the evaporator and the surface area on the air side.
Step by Step Answer:
Fundamentals of Thermal-Fluid Sciences
ISBN: 978-0078027680
5th edition
Authors: Yunus A. Cengel, Robert H. Turner, John M. Cimbala