COMBUSTION

COMBUSTION EQUATIONS

Let us begin our review of this particular variety of chemical-

reaction equations by considering the combustion of propane in
a pure oxygen environment. The chemical reaction is
represented by:

C3H8 + 5 O2 → 3 CO2 + 4 H2O

Note that the number of moles of the elements on the left-hand

side may not equal the number of moles on the right-hand side.
However, the number of atoms of an element must remain the
same before, after, and during a chemical reaction; this
demands that the mass of each element be conserved during
combustion.
In writing the equation we have demonstrated some knowledge
of the products of the reaction. Unless otherwise stated we will
assume complete combustion: the products of the
combustion of a hydrocarbon fuel will be H2O and CO2.
Incomplete combustion results in products that contain H2,
CO, C, and/or OH.
For a simple chemical reaction, such as the combustion of
propane, we can immediately write down a balanced chemical
equation. For more complex reactions the following systematic
method proves useful:
1. Set the number of moles of fuel equal to 1.
2. Balance CO2 with number of C from the fuel.
3. Balance H2O with H from the fuel.
4. Balance 02 from CO2 and H2O.
For the combustion of propane we assumed that the process
occurred in a pure oxygen environment. Actually, such a
combustion process would normally occur in air. For our
purposes we assume that air consists of 21% 0 2 and 79% N2 by
volume so that for each mole of 02 in a reaction we will have:
79 mol N 2
=3. 76
21 mol O 2

Thus, on the (simplistic) assumption that N2 will not undergo

any chemical reaction, the chemical reaction of the combustion
of propane is replaced by:

C3H8 + 5(O2 + 3.76N2) → 3CO2 + 4H2O + 18.8N2

The minimum amount of air that supplies sufficient 0 2, for the
complete combustion of the fuel is called theoretical air or
stoichiometric air. When complete combustion is achieved
with theoretical air, the products contain no O2, as in the our
reaction. In practice, it is found that if complete combustion is to
occur, air must be supplied in an amount greater than
theoretical air. This is due to the chemical kinetics and
molecular activity of the reactants and products. Thus we often
speak in terms of percent theoretical air or percent excess
air, where:
% theoretical air = 100% + % excess air
Slightly insufficient air results in CO being formed; some
hydrocarbons may result from larger deficiencies.
The parameter that relates the amount of air used in a
combustion process is the air-fuel ratio (AF),which is the ratio
of the mass of air to the mass of fuel. The reciprocal is the fuel-
air ratio (FA). Thus:
m m
AF= air FA= fuel
mfuel mair

Again, considering propane combustion with theoretical air, the

air-fuel ratio is:
m 5⋅4 . 76⋅29 kg air
AF= air = =15 . 69
mfuel 1⋅44 kg fuel
where we have used the molecular weight of air as 29 kg/kmol
and that of propane as 44 kg/kmol. If, for the combustion of
propane, AF > 15.69, a lean mixture occurs; if AF < 15.69, a
rich mixture results.
The combustion of hydrocarbon fuels involves H2O in the
products of combustion. The calculation of the dew point of the
products is often of interest; it is the saturation temperature at
the partial pressure of the water vapour. If the temperature
drops below the dew point, the water vapour begins to
condense. The condensate usually contains corrosive
elements, and thus it is often important to ensure that the
temperature of the products does not fall below the dew point.

EXAMPLE 01
Butane is burned with dry air at an air-fuel ratio of 20. Calculate
( a ) the percent excess air, ( b )the volume percentage of CO2
in the products, and ( c ) the dew-point temperature of the
products.
The reaction equation for theoretical air is:
C4H10+ 6.5(O2 + 3.76N2) → 4CO2+ 5H2O + 24.44N2
( a ) The air-fuel ratio for theoretical air is:
mair 6. 5⋅4 . 76⋅29 kg air
AF th = = =15 . 47
mfuel 1⋅58 kg fuel

This represents 100% theoretical air. The actual air-fuel ratio is

20. The excess air is then:
AF act −AF th 20−15 . 47
% excess air =
( AFth )
⋅( 100 % ) =(15 . 47 )
⋅( 100 % )=29 .28 %

( b ) The reaction equation with 129.28% theoretical air is:

C4H10 + (6.5)(1.2928)(O2 + 3.76N2) → 4CO2 + 5H20 + 1.903O2 + 31.6N2

The volume percentage is obtained using the total moles in the

products of combustion. For CO2, we have:
 4
% CO2 = ( )
42 .5
⋅(100 % )=9. 41 %

( c ) To find the dew-point temperature of the products we need

the partial pressure of the water vapour. It is found using the
mole fraction to be:
5
pv =z H O⋅patm =
2 ( )
42 .5
⋅(100 )=11.76 kPa

where we have assumed an atmospheric pressure of 100 kPa.

Using Table C-2 (Properties of Saturated H 2O – Pressure
Table), we find the dew-point temperature to be Td.p.= 49°C.

EXAMPLE 02
Butane is burned with 90% theoretical air. Calculate the volume
percentage of CO in the products and the air-fuel ratio. Assume
no hydrocarbons in the products.
For incomplete combustion we add CO to the products of
combustion. Using the reaction equation from Example 01, we
have:

C4H10 + (0.9)(6.5)(O2 + 3.76N2) → aCO2 + 5H2O + 22N2 + bCO

With atomic balances on the carbon and oxygen we find:
C: 4=a+b
O: 11.7 = 2a + 5 + b
a = 2.7 and b = 1.3
The volume percentage of CO is then:
1 .3
% CO =
31 ( )
⋅( 100 % ) =4 . 19 %

The air-fuel ratio is:

mair 0. 9⋅6 .5⋅4 . 76⋅29 kg air
AF= = =13 . 92
mfuel 1⋅58 kgfuel
EXAMPLE 03
Butane is burned with dry air, and volumetric analysis of the
products on a dry basis (the water vapour is not measured)
gives: 11.0% CO2,1.0% CO, 3.5% O2, and 84.5% N2. Determine
the percent theoretical air.
The problem is solved assuming that there is 100 moles of dry
products. The chemical equation is:
aC4H10 + b(O2 + 3.76N2) → 11CO2 + 1CO + 3.5O2 + 84.5N2 + cH2O

We perform the following balances:

C: 4a = 11 + 1 :. a = 3
H: 10a = 2c :. c = 15
0: 2b = 22 + 1 + 7 + c :. b = 22.5
A balance on the nitrogen allows a check: 3.76b = 84.5, or b =
22.47. This is quite close, so the above values are acceptable.
Dividing through the chemical equation by the value of a so that
we have 1 mol fuel:
C4H10 + 7.5(O2 + 3.76N2) → 3.67CO2 + 0.33CO + 1.17O2 + 28.17N2 +
5H2O

Comparing this with the combustion equation of Example 01

using theoretical air, we find:
7.5
( )
% theoretical air =
6.5
⋅( 100 % )=107 . 7 %
EXAMPLE 04

Volumetric analysis of the products of combustion of an

unknown hydrocarbon, measured on a dry basis, gives 10.4%
CO2, 1.2% CO, 2.8% O2, and 85.6% N2. Determine the
composition of the hydrocarbon and the percent theoretical air.
The chemical equation for 100 mol dry products is:
CaHb + c(O2 + 3.76N2) → 10.4CO2 + 1.2CO + 2.8O2 + 85.6N2 + dH2O

Balancing each element:

C: a = 10.4 + 1.2 :. a = 11.6
N: 3.76c= 85.6 :. c = 22.8
0: 2c = 20.8 + 1.2 + 5.6 + d :. d = 18.9
H: b = 2d :. b = 37.9

The chemical formula for the fuel is C11.6H37.9. This could

represent a mixture of hydrocarbons, but it is not any species
listed in Appendix B, since the ratio of hydrogen atoms to
carbon atoms is 3.27 = 13/4.
To find the percent theoretical air we must have the chemical
equation using 100% theoretical air:
C11.6H37.9+ 21.08(O2 + 3.76N2) → 11.6CO2 + 18.95H2O + 79.26N2

Using the number of moles of air from the actual chemical

equation, we find:
22. 8
(
% theoretical air = )
21. 08
⋅( 100 % )=108 %