| Pounds of air required per pound of fuel | = | 34.56 | ( | | + | ( | H | - | | ) | + | | ) | [29] ( 11 ) |
where C, H and O equal the percentage by weight of carbon, hydrogen and oxygen in the fuel divided by 100.
[Pg 157]
When the flue gas analysis is known, the total, amount of air supplied is:
| Pounds of air supplied per pound of fuel | = | 3.036 | ( | | N | | ––––––––––––––––– | | CO 2 + CO |
| ) | × | C | [30] | ( 12 ) |
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:
| 3.036 | ( | | N | | ––––––––––––––––– | | CO 2 + CO |
| ) | × | C |
| ( 13 ) |
| ––––––––––––––––––––––––––––––––––––––––––––– |
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