26 MASS. EXPERIMENT STATION BULLETIN 360 



Estimating the Refrigeration Load 



In estimating the required refrigeration capacity for an apple storage, it will 

 be found that the cooling of the fruit makes up the greater portion of the load. 

 In order to estimate the load, it will be necessary to make certain assumptions. 

 As an illustration, it will be assumed that 10 percent of the capacity of the storage 

 is harvested each day; that apples are brought in at a temperature of 75°; and 

 that average air temperature is 75° and ground temperature, 56°. It is further 

 assumed that loose apples weigh 45 pounds per bushel, including the container, 

 and have a specific heat of .90. 



The refrigeration capacity required to cool 1000 bushels per day is determined 

 as follows: 1000 bushels of apples weigh 45,000 pounds and, having a specific 

 heat of .90, they will give up .90 X45,000, or 40,500 B. t. u., when the temperature 

 is lowered one degree. When cooled from 75° to 32°, the temperature drop is 

 43°, and a total of 43X40,500 or 1,741,500 B. t. u. will be given up. 



In addition to the sensible heat covered by the above calculations, it is also 

 necessary to remove the heat of respiration given off by the apples. The rate at 

 which heat of respiration is generated depends upon the temperature at which 

 apples are stored. Apples stored at 60° give off from 100 to 150 B. t. u. per 

 bushel per day, while at 32° the rate is from 15 to 20 B. t. u. per bushel. As a 

 storage is filled, the heat of respiration to be removed, of course, will increase 

 from day to day, with the greatest amount on the last day of loading. If we 

 assume that the storage is filled in 10 days, the greatest amount of heat will be 

 given up on the 10th day. The daily refrigeration load is about 200 B. t. u. per 

 bushel harvested each day if the storage is carried at 32°, and about 400 B. t. u. 

 if the storage is held at 45°. In the problem we have a.ssumed, cooling 1000 

 bushels a day to 32°, it will be necessary to allow 200,000 B. t. u. for heat of 

 respiration which, when added to 1,741,500 B. t. u. of sensible heat, gives a total 

 of 1,941,500 B. t. u. Inasmuch as a one-ton machine will remove 288,000 B. t. u. 

 in 24 hours, a capacity of 1,941,500-^ 288,000, or 6.7 tons, is required. 



If the apples are to be cooled to 45°, the temperature reduction will be 75° 

 minus 45°, or 30°, making the total sensible heat to be removed 30X40,500, 

 or 1,215,000 B. t. u. The heat of respiration will be 400X1000, or 400,000, 

 and the total heat to be removed per day will be 1,215,000+400,000, or 1,615,000 

 B. t. u. This divided by 288,000 gives a total refrigeration capacity of 5.6 tons 

 needed to cool 1000 bushels per day. 



In order to estimate the heat leakage into a storage, we will assume a one-story 

 building 40 feet wide, 42 feet long, and 10 feet high. The walls are of concrete, 

 6 inches thick and insulated with 3 inches of sheet cork, and half of the wall 

 area is underground. Standard concrete floor construction is used, with 2 inches 

 of sheet cork. The roof is insulated with 4 inches of sheet cork. The heat leakage 

 into this room is shown in Table 6, the data being obtained from Table 4. 



In this example no allowance has been made for air infiltration or the opening 

 of doors. The heat loss due to this cause, however, is relatively small in a fair- 

 sized storage, particularly if apple boxes are moved through ports on roller con- 

 veyors. There is an additional refrigeration load due to the condensation of 

 moisture and subsequent freezing on the cooling coils, but this also represents 

 a small loss and can be neglected in rough calculations when air changes within 

 the storage are kept at a minimum. 



