Part 29

153 × 1½ = 229.5 cubic feet of air per pound of carbon.

Actual Volume for One Pound Carbon Per Cent Cubic Feet by Volume Carbon Dioxide 32 = 13.91 } Oxygen 16 = 7.00 } = 20.91 per cent Nitrogen 181.5 = 79.09 ----- ------ 229.5 = 100.00

For 100 per cent excess air the volume will be as follows:

153 × 2 = 306 cubic feet of air per pound of carbon.

Actual Volume for One Pound Carbon Per Cent Cubic Feet by Volume Carbon Dioxide 32 = 10.45 } Oxygen 32 = 10.45 } = 20.91 per cent Nitrogen 242 = 79.09 --- ------ 306 = 100.00

In each case the volume of oxygen which combines with the carbon is equal to (cubic feet of air × 20.91 per cent)--32 cubic feet.

It will be seen that no matter what the excess of air supplied, the actual amount of carbon dioxide per pound of carbon remains the same, while the percentage by volume decreases as the excess of air increases. The actual volume of oxygen and the percentage by volume increases with the excess of air, and the percentage of oxygen is, therefore, an indication of the amount of excess air. In each case the sum of the percentages of CO_{2} and O is the same, 20.9. Although the volume of nitrogen increases with the excess of air, its percentage by volume remains the same as it undergoes no change while combustion takes place; its percentage for any amount of air excess, therefore, will be the same after combustion as before, if cooled to the same temperature. It must be borne in mind that the above conditions hold only for the perfect combustion of a pound of pure carbon.

Carbon monoxide (CO) produced by the imperfect combustion of carbon, will occupy twice the volume of the oxygen entering into its composition and will increase the volume of the flue gases over that of the air supplied for combustion in the proportion of

100 + ½ the per cent CO 1 to ----------------------- 100

When pure carbon is the fuel, the sum of the percentages by volume of carbon dioxide, oxygen and one-half of the carbon monoxide, must be in the same ratio to the nitrogen in the flue gases as is the oxygen to the nitrogen in the air supplied, that is, 20.91 to 79.09. When burning coal, however, the percentage of nitrogen is obtained by subtracting the sum of the percentages by volume of the other gases from 100. Thus if an analysis shows 12.5 per cent CO_{2}, 6.5 per cent O, and 0.6 per cent CO, the percentage of nitrogen which ordinarily is the only other constituent of the gas which need be considered, is found as follows:

100 - (12.5 + 6.5 + 0.6) = 80.4 per cent.

The action of the hydrogen in the volatile constituents of the fuel is to increase the apparent percentage of the nitrogen in the flue gases. This is due to the fact that the water vapor formed by the combustion of the hydrogen will condense at a temperature at which the analysis is made, while the nitrogen which accompanied the oxygen with which the hydrogen originally combined maintains its gaseous form and passes into the sampling apparatus with the other gases. For this reason coals containing high percentages of volatile matter will produce a larger quantity of water vapor, and thus increase the apparent percentage of nitrogen.

Air Required and Supplied--When the ultimate analysis of a fuel is known, the air required for complete combustion with no excess can be found as shown in the chapter on combustion, or from the following approximate formula:

Pounds of air required per pound of fuel =

(C O S) 34.56 (- + (H - -) + -) (11) (3 8 8)

where C, H and O equal the percentage by weight of carbon, hydrogen and oxygen in the fuel divided by 100.

When the flue gas analysis is known, the total, amount of air supplied is:

Pounds of air supplied per pound of fuel =

N 3.036 (-----------) × C (12) CO_{2} + CO

where N, CO_{2} and CO are the percentages by volume of nitrogen, carbon dioxide and carbon monoxide in the flue gases, and C the percentage by weight of carbon which is burned from the fuel and passes up the stack as flue gas. This percentage of C which is burned must be distinguished from the percentage of C as found by an ultimate analysis of the fuel. To find the percentage of C which is burned, deduct from the total percentage of carbon as found in the ultimate analysis, the percentage of unconsumed carbon found in the ash. This latter quantity is the difference between the percentage of ash found by an analysis and that as determined by a boiler test. It is usually assumed that the entire combustible element in the ash is carbon, which assumption is practically correct. Thus if the ash in a boiler test were 16 per cent and by an analysis contained 25 per cent of carbon, the percentage of unconsumed carbon would be 16 × .25 = 4 per cent of the total coal burned. If the coal contained by ultimate analysis 80 per cent of carbon the percentage burned, and of which the products of combustion pass up the chimney would be 80 - 4 = 76 per cent, which is the correct figure to use in calculating the total amount of air supplied by formula (12).

The weight of flue gases resulting from the combustion of a pound of dry coal will be the sum of the weights of the air per pound of coal and the combustible per pound of coal, the latter being equal to one minus the percentage of ash as found in the boiler test. The weight of flue gases per pound of dry fuel may, however, be computed directly from the analyses, as shown later, and the direct computation is that ordinarily used.

The ratio of the air actually supplied per pound of fuel to that theoretically required to burn it is:

N 3.036(---------)×C CO_{2}+CO ------------------ (13) C O 34.56(- + H - -) 3 8

in which the letters have the same significance as in formulae (11) and (12).

The ratio of the air supplied per pound of combustible to the amount theoretically required is:

N ------------------ (14) N - 3.782(O - ½CO)

which is derived as follows:

The N in the flue gas is the content of nitrogen in the whole amount of air supplied. The oxygen in the flue gas is that contained in the air supplied and which was not utilized in combustion. This oxygen was accompanied by 3.782 times its volume of nitrogen. The total amount of excess oxygen in the flue gases is (O - ½CO); hence N - 3.782(O - ½CO) represents the nitrogen content in the air actually required for combustion and N ÷ (N - 3.782[O - ½CO]) is the ratio of the air supplied to that required. This ratio minus one will be the proportion of excess air.