Page 328 - Wastewater Solids Incineration Systems

Appendix A Combustion Fundamentals 289

Reaction products (RHS)

(0.87 lb C/12 (lb/lb-mole C) 1.0 (lb-mole CO /lb-mole C)
2
44 (lb/lb-mole O ) 3.19 lb CO /lb oil from carbon combustion
2 2
(0.13 lb H /2 (lb/lb-mole H ) 1.0 (lb-mole H O/lb-mole H )
2 2 2 2
18 (lb/lb-mole H O) 1.17 lb O /lb oil from hydrogen combustion
2 2
4.36 lb (CO H O)/lb oil total
2 2
Overall equation based on oxygen
1 lb oil 3.36 lb O 3.19 lb CO 1.17 lb H O
2 2 2
Calculation of nitrogen and stoichiometric air:
N 3.36 lb O 0.7685 lb N /0.2315 lb O 11.15 lb N /lb oil
2 2 2 2 2
Air 3.36 lb O 11.15 lb N 14.51 lb air/lb oil
2 2
Overall equation based on air

1 lb oil 14.51 lb air 3.19 lb CO 1.17 lb H O 11.15 lb N
2 2 2
Adding 20% excess air:

XS air 0.2 14.51 lb air/lb oil 2.90 lb XS air/lb oil
0.2 3.36 lb O /lb oil 0.67 lb O /lb oil
2 2
N in XS air 0.2 11.15 lb N /lb oil 2.23 lb N /lb oil
2 2 2
Overall equation based on 20% XS air:

1 lb oil 17.41 lb air 3.19 lb CO 1.17 lb H O 0.67 lb O 13.38 lb N
2 2 2 2
18.41 lb reactants 18.41 lb products (17.24 lb dry products)

Calculating the number of moles of each component of the flue gas is done by
dividing the calculated mass by the respective molecular weight of each component,
as follows:

3.19 lb CO /44 (lb/lb-mole CO ) 0.0725 lb-mol CO 11.4%
2 2 2
1.17 lb H O/18 (lb/lb-mole H O) 0.0650 lb-mol H O 10.2%
2 2 2
0.67 lb O /32 (lb/lb-mole O ) 0.0209 lb-mol O 3.3%
2 2 2
13.38 lb N /28 (lb/lb-mole N ) 0.4779 lb-mol N 75.1%
2 2 2
Total 0.6363 lb-mol 100.0%