TITLE.PM5

FUELS AND COMBUSTION 525

dharm

\M-therm\Th11-2.pm5

Example 11.22. Find the stoichiometric air-fuel ratio for the combustion of ethyl alcohol
(C 2 H 6 O), in a petrol engine. Calculate the air-fuel ratios for the extreme mixture strengths of
80 per cent and 130 per cent. Determine also the wet and dry analyses by volume of the exhaust
gas for each mixture strength.
Solution. The equation for combustion of ethyl alcohol is as follows :
C 2 H 6 O + 3O 2 → 2CO 2 + 3H 2 O
— Since there are two atoms of carbon in each mole of C 2 H 6 O then there must be two
moles of CO 2 in the products, giving two atoms of carbon on each side of the equation.
— Similarly, since there are six atoms of hydrogen in each mole of ethyl alcohol then there
must be three moles of H 2 O in the products, giving six atoms of hydrogen on each side
of the equation.
— Then balancing the atoms of oxygen, it is seen that there are (2 × 2 + 3) = 7 atoms on
the right hand side of the equation, hence seven atoms must appear on the left hand
side of the equation. There is one atom of oxygen in ethyl alcohol, therefore a further six
atoms of oxygen must be supplied, and hence three moles of oxygen are required as
shown.
Since the O 2 is supplied as air, the associated N 2 must appear in the equation,

i.e., C 2 H 6 O + 3O 2 + 3 ×^79
21
N 2 → 2CO 2 + 3H 2 O + 3 ×^79
21
N 2
One mole of fuel has a mass of (2 × 12 + 1 × 6 + 16) = 46 kg. Three moles of oxygen have a
mass of (3 × 32) = 96 kg.

∴ O 2 required per kg of fuel =^96

46

= 2.09 kg

∴ Stoichiometric A/F ratio =

209

0233

.

.

= 8.96/1. (Ans.)

Considering a mixture strength of 80% :

Now, mixture strength =

Stoichiometric A/F ratio

Actual A/F ratio

i.e., 0.8 =

8.96 / 1

Actual A/F ratio

∴ Actual A/F ratio =

896

08

.

.

= 11.2/1. (Ans.)

This means that 1/0.8 or 1.25 times as much air is supplied as is necessary for complete

combustion. The exhaust will therefore contain 0.25 stoichiometric oxygen.

i.e., C 2 H 6 O + 1.25 3379

(^2221)
F ON+×
HG
I
KJ
→ 2CO 2 + 3H 2 O + 0.25 × 3O 2 + 1.25 × 3 ×^79
21
N 2
i.e., The products are :
2 moles CO 2 + 3 moles H 2 O + 0.75 moles O 2 + 14.1 moles N 2
The total moles = 2 + 3 + 0.75 + 14.1 = 19.85
Hence wet analysis is :
CO 2 =
2
19 85.
× 100 = 10.08%. (Ans.)
H 2 O =^3
19 85.
× 100 = 15.11%. (Ans.)