Page 48 - Handbook of Energy Engineering Calculations

(3.76)(7.942) = 29.861 mol of N . Then 29.861 + 7.942 = 37.803 mol of air,
2
which agrees closely with the 37.823 mol computed in the tabulation. The
difference of 0.02 mol is traceable to roundings.

4. Compute the air required with the stated excess air

With 25 percent excess, the air required for combustion = (125/100)(37.823)
= 47.24 mol.

5. Compute the mols of combustion products
Using data from Table 3, and recalling that the products of combustion of a

sulfur-containing fuel are CO , H O, and SO , and that N and excess O 2
2
2
2
2
pass through the furnace, set up a tabulation thus:

In this calculation, the total moles of CO is obtained from step 2. The moles
2
of H in 100 lb (45 kg) of the fuel, 2.280, is assumed to form H O. In
2
2
addition, the air from step 4, 47.24 mol, contains 0.013 lb of moisture per lb
(0.006 kg/kg) of air. This moisture is converted to moles by dividing the
molecular weight of air, 28.95, by the molecular weight of water, 18, and
multiplying the result by the moisture content of the air, or (28.95/18)(0.013)
= 0.0209, say 0.021 mol of water per mol of air. The product of this and the

moles of air gives the total moles of moisture (water) in the combustion
products per 100 lb (45 kg) of fuel fired. To this is added the moles of O ,
2
0.158, per 100 lb (45 kg) of fuel, because this oxygen is assumed to unite
with hydrogen in the air to form water. The nitrogen in the products of
combustion is that portion of the moles of air required, 47.24 mol from step