Page 87 - Air pollution and greenhouse gases from basic concepts to engineering applications for air emission control

3.1 Air–Fuel Ratio 61

Solution
Note that one mole of air is simplified as a mixture of 0.21 mol of O 2 and 0.79 mol
of N 2 ; the molar weight of air is 28.82 g/mole.

(a) Assuming the mixture is in the same container and reached steady state
without chemical reactions. The A/F based on volume is simply

n a
ðA=FÞ ¼ ¼ 15
mix
n f
(b) The equivalence ratio is determined as

ðA=FÞ s 16:66
/ ¼ ¼ ¼ 1:11
ðA=FÞ 15
mix
It is a fuel rich mixture.

3.2 Combustion Stoichiometry

The combustion is stoichiometric when all the atoms in the fuel are burned into
their corresponding oxides, for instance, all the carbon to CO 2 and all hydrogen to
H 2 O. Consider a hydrocarbon fuel with a carbon and b hydrogen atoms in each
molecule, the stoichiometric combustion process of a hydrocarbon fuel with a
molecular formula C a H b can be described as

C a H b þ aO 2 ! bCO 2 þ cH 2 O ð3:4Þ

where the values of a, b and c depend on α and β.
A simple mass balance leads to the answers of a, b and c in terms of α and β (see
Example 3.2), and Eq. (3.4) becomes

b b

C a H b þ a þ O 2 ! aCO 2 þ H 2 O ð3:5Þ
4 2
It indicates that in order to achieve the above stoichiometric combustion, for
each hydrocarbon molecule, (a þ b=4) oxygen molecules are required to convert
the carbon and hydrogen into carbon dioxide and water, respectively.
In general, all liquid and gaseous fossil fuels are mixtures of multiple compo-
nents. For ease of engineering analysis, however, they can be simplified as an
average formula C a H b , where α and β stand for the numbers of carbon and
hydrogen atoms in the fuel, respectively. For a natural gas, its formula is simplified
as CH 4, and for a typical liquid fuel, e.g., gasoline, a ¼ 8 and b ¼ 18.