24 BULLETIN 1335, U. S. DEPARTMENT OF AGRICULTURE 
Total heat of evaporation 
0.32 
For the drier in question this will be. 
=Total heat which must be generated. 
6,952 B. t. u. per minute 
3 32 
21,725 B. t. u. per minute, on the average. One gallon of fuel oil 
yields 148,000 B. t. u. Therefore the burners must consume 
21,725 X60 
148.000 ° °F 9 gallons of fuel oil per hour. 
AMOUNT OF GENERATED HEAT CARRIED BY THE AIR 
Since the thermal efficiency of a heater of the direct radiation 
type, the one selected here, is assumed to be 80,per cent, the amount 
of generated heat carried by the air will be the total heat generated 
multiplied by 0.80, or 21,725 B. t. u. per minute x 0.80, which is 
equivalent to 17,380 B. t. u. per minute, on the average. 
AMOUNT OF HEAT GIVEN UP BY THE AIR 
Heat will be absorbed from the air to meet the requirements 
for total heat of evaporation, which has been estimated in this case 
to be 6,952 B. t. u. per minute. Heat will also be lost through 
radiation, air leakage, and absorption in heating the material, trays, 
and trucks. On this account an allowance of 10 per cent of the 
generated heat being carried by the air will be added—in this case, 
17,380 xX 0.10, or 1,738 B. t. u. per minute. The air in passing 
through the drying chamber will then be required to furnish, on the 
average, 6,952 + 1,738, or 8,690 B. t. u. per minute. 
VOLUME OF AIR REQUIRED 
The volume of air required to give up 8,690 B. t. u. of heat per 
minute is calculated from the specific hows of dry air and water 
vapor. Specific heat is the ratio between the heat required to raise 
(or, conversely, given off by cooling) a given weight of a substance 1 
degree and that required to raise the same weight of water 1 degree, 
the specific heat of water being considered as 1.000. At constant 
pressure the specific heat of dry air is 0.2375, or approximately 0.24, 
while the specific heat of water vapor is 0.475. One cubic foot of a 
mixture of dry air and water vapor under given conditions of 
temperature and humidity will, in dropping 1° F., give up a number of 
B. t.u. represented by the expression: (Pounds of dry air per cubic foot 
x specific heat dry air) + (Pounds of water vapor per cubic foot x 
specific heat of water vapor). In dropping a given number of 
degrees Fahrenheit, the number of B. t. u. given up per cubic foot of 
mixture would be: Drop in degrees Fahrenheit x the foregoing 
expression. epee ae a given number of B. t. u. would be given 
up by the number of cubic feet of mixture represented by the fraction: 
Number of B. t. u. required 
Drop (°F.) x [ (Pounds of dry air per cubic foot x 0.24) + (Pounds 
of water vapor per cubic foot x 0.475) ] 
In the present problem, where an average of 8,690 B. t. u. per 
minute is required from air at 160° F. and 20 per cent relative humid- 
ity, dropping 35° F. in passing through the tunnel, and where the 
weights of dry air and of water vapor in the air at 160° F. and 20: 
