"76 BULLETIN 98, U. S. DEPARTMENT OF AGRICULTURE. 
refrigeration necessary in 24 hours, when the plant is operated 
under the foregoing conditions, is : 
b. t. u. 
Removing heat coming through walls, floor, and ceiling, 1,316 X 2 (80-32)= 126, 336 
Cooling 1,000 gallons of milk, 1,000 X 8.6 + .95 (75-32)= 351. 310 
477, 646 
Additional refrigeration is also required for cooling the glassware 
and boxes, also a considerable amount is lost due to opening doors 
and the presence of lights and workmen inside the room. As it is 
impossible to calculate the refrigeration lost in opening doors, 
it is customary in practice to allow about 50 per cent additional to 
cover this. Therefore, the total refrigeration required in 24 hours 
716 469 
is, 477,646 X 1.50 = 716,469 B. T. U. or ^qqq = 2\ tons. But as the 
refrigerating machine in a plant of this size is operated only about 
8 hours during the 24, the capacity of the machine will have to be 
three times as large, or 7 \ tons. 
The operation of pasteurizing, cooling the milk to approximately 
45° F., and bottling and storing takes about two hours; conse- 
quently it is necessary to have a large volume of cold brine available 
for this work. The cooling of the brine is accomplished during the 
forenoon, before the milk arrives, and as the temperature of the 
brine rises during the cooling process, it is again cooled down in the 
afternoon and depended upon to hold over temperatures in the 
storage room during the night. 
As a cubic foot of calcium-chlorid brine will absorb about 52 
B. T. U. for each degree rise in temperature, and allowing a 15 
degree rise, 30 to 45 degrees, each cubic foot will take up 52 (30-45) = 
780 B. T. U. 
The volume of brine necessary for cooling the milk will be ' *' — = 
450 cubic feet, providing the refrigerating machine is not operated 
at the time, but as a 7J-ton machine is capable of extracting 12,000 X 
7.5 = 90,000 B. T. U. an hour, or during the two hours taken to cool 
the milk the machine will extiact 90,000x2 = 180,000 B. T. U., 
consequently the actual cubic feet of brine required is L — -^ L 
= 219.6. 
Another method of calculating the amount of brine storage required 
to cool a given amount of milk, based on the capacity of the com- 
pressor used for cooling milk, is as follows : 
r _ (WRm)- (12,000 CHm) 
60 Rb 
Where T= cubic feet of brine in tank. 
"N 7 = weighing of milk in pounds. 
