- From fifteen to twenty-two hogs, average weight, 225 pounds.
- From five to six head of cattle, average weight, 700 pounds.
- About forty-five calves, average weight, 80 pounds.
- From fifty to sixty-five sheep, average weight, 60 pounds.
For Cooled Meats.
—For cold storage rooms, where meats which have already been chilled or cooled are stored and held for a greater or lesser period, the refrigeration requirements are not so great as in chill and cooling rooms. The animal heat has been removed and the meats cooled down to a low temperature, consequently but little more refrigeration is required than that necessary to take care of the heat leakage through the insulation, and possibly the recooling of the meat through a range of a few degrees, when the meats may have gained a little in temperature by exposure. Therefore it is estimated that one ton of refrigerating duty will handle 12,000 cubic feet of curing or storage space for temperature of 33° to 35° F.
Freezer Space.
—As will be explained in the chapter relating to refrigerating machine capacity, the capacity of a compressor very rapidly decreases when operating under conditions necessary to produce low freezer temperatures, and owing to the severe conditions imposed it seems necessary to compute that one ton of refrigerating duty will handle about 3,000 feet of freezer space.
Ice Computations.
—Before the application of mechanical refrigeration to packing house purposes, all artificial refrigeration was accomplished by means of ice melting alone, and at that time the packers computed ice melting requirements on a basis of cooling three pounds of meat from 80° F. to as low as it could be cooled by ice melting, for each pound of ice melted. While this rule undoubtedly was the result of practical experience with well constructed coolers, and was in no wise based upon theoretical or heat unit formulas, yet it is interesting to note how close this old rule compares with modern formulas of computing refrigeration. For example, the cooling of 100 head of hogs, averaging 250 pounds dressed, by the packer’s rule, would require—
100 × 2503 = 8,333 pounds, or 4.16 tons ice melting.
And on a heat unit basis, cooling the same number and weight of hogs from 80° to 32° F. would require—
100 × 250 × (80 - 32)288,000 = 4.166 tons refrigeration.