Combustion and Fuel Chemistry

 

 

Consider the reaction between two molecules of carbon monoxide (CO) and one molecule of oxygen (O₂) to produce two molecules of carbon dioxide (CO₂):

2 CO + O₂ → 2 CO₂

There is a conservation of mass, and conservation of the number of atoms.

It is often convenient to consider a unite quantity of fuel, for instance a kilomole, so the above reaction can be written in terms of kilo mole as

CO + a O₂ → b CO₂

 

CO + ½ O₂ → CO₂

CO + ½ (O₂ + 79/21 N₂) → CO₂ +  ½ (79/21) N₂

Relative mass

(12 + 16) + ½ (16 x 2 + 79/21 x 14 x 2) → (12 + 16 x 2) + ½ (3.76) 14 x 2

28 + ½ (32 + 105.33) → (12 + 32) + 52.64

28 + 68.667 → 44 + 52.64

1+ 68.667/28 → 96.64/28

1 + 2.45 → 3.45

The stoichiometric ratio 2.45.

Fuels are often mixtures of hydrocarbons, with bonds between carbon atoms, and between hydrogen and carbon atoms. During combustion these bonds are broken, and new bonds are formed with oxygen atoms, accompanied by a release of chemical energy. The products are carbon dioxide and water.

 

If sufficient oxygen is available, a hydrocarbon fuel can be completely oxidized. The carbon in the fuel is then converted to carbon dioxide CO₂ and the hydrogen to water H₂O. For example, consider the overall chemical equation for the complete combustion of one mole of propane C₃H₈:

C₃H₈ + a O₂ → b CO₂ + c H₂O

A carbon balance between the reactants and products gives b = 3. A hydrogen balance gives 2c = 8, or c = 4. An oxygen balance gives 2b + c = 10 =  2a, or a = 5. Thus the above equation becomes:

C₃H₈ + 5 O₂ → 3 CO₂ + 4 H₂O

Example:

Find the A/F ratio of stoichiometric combustion (complete combustion) of C₈H₁₈.

C₈H₁₈ + a O₂ → b CO₂ + c H₂O

A carbon balance gives b = 8,

A hydrogen balance gives 2c = 18, or c = 9,

An oxygen balance gives 2a = 2b + c = 2 x 8 + 9 = 16 + 9 = 25, or a = 12.5

Where oxygen 20.95 ≈ 21 % and nitrogen 79.05 ≈ 79

The combustion equation becomes:

C₈H₁₈ + 12.5 (O₂ + 79/21 N₂) → 8 CO₂ + 9 H₂O + 12.5 (79/21) N₂

C₈H₁₈ + 12.5 (O₂ + 3.76 N₂) → 8 CO₂ + 9 H₂O + 12.5 (3.76) N₂

Where the molecular weight:

C = 12.011 ≈ 12

H = 1.008 ≈ 1

O = 16

N = 14.08 ≈ 14

Relative mass:

(12 x 8 + 1x18) +12.5 (16x2 + 3.76 x 14x2) → 8 (12 +16x2) + 9 (1x2 + 16) + 47 (14x2)

(96+18) + 12.5 (32+105.28) → 8 (12+32) + 9 (2+16) + 47 (28)

114 + 12.5 (137.28) → 8 (44) + 9 (18) + 47 (28)

114 + 1716 → 352 + 162 +  1316

114 + 1716 → 1830

1 + 1716/114 → 1830/114

1 + 15.05 → 16.05

The stoichiometric (A/F) is 15.05.

Examples of fuel compositions are: CH₄, methane; C₂H₆, ethane; C₂H₈, propane; C₈H_18, n-octane and isooctane. C₃H₆, cyclopropane; C₄H₈, cyclobutance; C₅H₁₀, cyclopentane. C₆H₆, benzene; C₇H₈, toluene; C₈H₁₀, xylene.

 

Excel Program to calculate the (A/F) ratio