Page 126 - Handbook of Energy Engineering Calculations

produce 8 molecules of carbon dioxide, 8CO . Similarly, 9 molecules of
2
hydrogen, H , in H combine with 9 atoms of oxygen, O, or 4.5 molecules of
2
l8
oxygen, to form 9 molecules of water, 9H O. Thus, 100 percent, or the
2
stoichiometric, air quantity required for complete combustion of a mole of

fuel, C H , is proportional to 8 + 12.5 moles of oxygen, O .
8 18
2
2. Establish the chemical equation for complete combustion with 100
percent air

With 100 percent air: C H + 12.5O + (3.784 × 12.5)N → 8CO + 9H O +
8 18
2
2
2
2
47.3N , where 3.784 is a derived volumetric ratio of atmospheric nitrogen,
2
(N ), to oxygen, O , in dry air. The (N ) includes small amounts of inert and
2
2
2
inactive gases. See Related Calculations of this procedure.
3. Establish the chemical equation for complete combustion with 400
percent of the stoichiometric air quantity, or 300 percent excess air
With 400 percent air: C H + 50O + (4 × 47.3)N → 8CO + 9H O +
8 18
2
2
2
2
189.2N + (3 × 12.5) O .
2
2
4. Compute the molecular weights of the components in the combustion
process
Molecular weight of C H = [(12 × 8) + (1 × 18)] = 114; O = 16 × 2 = 32;
2
8 18
N = 14 × 2 = 28; CO = [(12 × 1) + (16 × 2)] = 44; H O = [(1 × 2) + (16 ×
2
2
2
1)] = 18.
5. Compute the relative weights of the reactants and products of the
combustion process

Relative weight = moles × molecular weight. Coefficients of the chemical
equation in step 3 represent the number of moles of each component. Hence,
for the reactants, the relative weights are: C H = 1 × 114 = 114; O = 50 ×
2
8 18
32 = 1600; N = 189.2 × 28 = 5298. Total relative weight of the reactants is
2
7012. For the products, the relative weights are: CO = 8 × 44 = 352; H O =
2
2
9 × 18 = 162; N = 189.2 × 28 = 5298; O = 37.5 × 32 = 1200. Total relative
2
2
weight of the products is 7012 also.