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 



71 6 469 

 is, 477,646 X 1.50 = 716,469 B. T. U. or '^^ = 2| tons. But as the 



Zoo, (JUL) 



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 — =^r — = 



450 cubic feet, providing the refrigerating machine is not operated 

 at the time, but as a 7^-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 — — - — =^r '■ 



= 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: 



T= 



(WRm)- (12,000 CHm) 



60 Rb 



Where T= cubic feet of brine in tank. 

 W= weighing of milk in pounds. 



