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3.2 Combustion Stoichiometry 67

1 b
C a H b þ aþ ð O 2 þ3:76N 2 Þ
/ 4

3:76 b
! xCO 2 þ 1 xð½ ÞCOŠþ yH 2 O+ 1 yÞH 2 Šþ aþ N 2 ð3:18Þ
ð
½
/ 4

1 b 1 þ x þ y
þ a þ O 2 þ
/ 4 2
3.2.4 Complex Fossil Fuel Combustion Stoichiometry

As mentioned above, it is unusual to have pure hydrocarbon fuel. There are always
atoms other than hydrogen and carbon in fossil fuels. For example, sulfur and
oxygen are typical elements in fossil fuels, coal, oil or gas. There are also other
additives in oil and gas. As a result, the combustion stoichiometry of these fuels
becomes complicated. For a general fossil fuel the corresponding stoichiometric
combustion can be described as, only for example,

C a H b O o Cl q S s N n þ a 0:21O 2 þ0:79N 2 Þ
ð
ð3:19Þ
! bCO 2 þ cH 2 O þ dHCl þ eN 2 þ f SO 2
Then the atom balances give

Carbon balance : a ¼ b ð1Þ

Hydrogen balance : b ¼ 2c þ d ð2Þ
Chloride ClðÞ balance : q ¼ d ð3Þ

Sulfur balance : s ¼ f ð4Þ

Oxygen balance : o þ 20:21Þa ¼ 2b þ c þ 2f ð5Þ
ð
Nitrogen balance : n þ 20:79Þa ¼ 2e ð6Þ
ð
Solving the above six equations, the unknown coefficients, a through f, can be
determined. The chemical reaction formula for the stoichiometric combustion of the
complex fuel becomes,

b q o
C a H b O o Cl q S s N n þ 4:76 a þ þ s ð 0:21O 2 þ 0:79N 2 Þ
4 4 2

b q n b q o
! aCO 2 þ H 2 O þ qHCl þ þ 3:76 a þ þ s N 2 þ sSO 2
2 2 4 4 2
ð3:20Þ