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 kilo mole, so the above reaction can be written in terms of kilo mole as

CO + aO2 bCO2

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 bCO2 + cH2O

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 + 5O2 3CO2 + 4H2O

Example:

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

C8H18 + aO2 bCO2 + cH2O

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

C8H18 + 12.5 (O2 + 3.76 N2) 8CO2 + 9H2O + 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; C8H18, n-octane and isooctane. C3H6, cyclopropane; C4H8, cyclobutance; C5H10, cyclopentane. C6H6, benzene; C7H8, toluene; C8H10, xylene.

Excel Program to calculate the (A/F) ratio