N
3.036 (-----------) × C[30] (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.