Example. — How many cubic feet of air are required for the complete combustion of 1 cubic foot of butane, C4H10?
Solution. — Butane contains four atoms of C and ten atoms of H. Hence, substituting in the formula v = ( n + m 4) 4.76, V = (4 + 104)4.76 = 6.5 × 4.76 = 30.94 cu. ft. Ans.
21. When a gas consists of a mixture of various hydrocarbons, the proportion of each hydrocarbon must be ascertained before the required volume of air is completed.
Example. — What volume of air is required for the complete combustion of 1 cubic foot of a gas having the following composition?
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Constituents of Gas Per Cent. Methane, CH4 65 Ethane, C2H6 25 Propane, C3H8 10 100
Solution. — Using the formula of Art. 20, 1 cu. ft. of CH4 requires (1 + 44) 4.76 = 2 X 4.76 = 9.52 cu. ft. of air One cu. ft. of C2H6 requires (2 + 64) 4.76 = 3.5 X 4.76 = 16.66 cu. ft. of air One cu. ft. of C3H8 requires (3 + 84)4.76 = 5 X 4.76 = 23.8 cu. ft. of air
Since there is 65100 cu. ft. of CH4 in 1 cu. ft. of the gas, the volume of air required for the complete combustion of the CH4 will be 9.52 × .65 = 6.19 cu. ft. of air. Similarly, it will require 16.66 × .25 = 4.17 cu. ft. of air to burn the C2H6 and 23.8 × .10 = 2.38 cu. ft. of air to burn the C3H8. The complete combustion of the entire cubic foot of the gas will then require the sum of the amounts necessary to consume the separate portions, or 6.19 + 4.17 + 2.38 = 12.74 cu. ft. Ans.
22. Nearly all gases employed for engine purposes contain, besides hydrocarbons, a greater or less amount of free H, CO, CO2, N, and occasionally small percentages of H2S (sulphureted hydrogen) and O. To find the amount of air required to completely burn a mixture containing these gases, proceed in the same manner as in the example of Art. 21, using the quantities of air required for H, H2S, and CO given in Art. 17, and making, of course, no computation for the CO2 and N, since these two gases are