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3.2 Combustion Stoichiometry 65

Most of the time, it is normalized against O 2 , and Eq. (3.14) becomes

1 b
C a H b þ aþ ð O 2 þ3:76N 2 Þ
/ 4
ð3:15Þ

b 3:76 b 1 b
! a CO 2 þ H 2 O þ aþ N 2 þ 1 aþ O 2
2 / 4 / 4
Example 3.4: Fuel lean combustion of octane
Considering a reaction of C 8 H 18 with 10 % excess air by volume, what is the
equivalence ratio of the mixture?
Solution
The stoichiometric combustion of C 8 H 18 with dry air 0:21O 2 þ0:79N 2 Þ is
ð

ð
C 8 H 18 þ59:52 0:21O 2 þ0:79N 2 Þ ! 8CO 2 +9H 2 O + 47N 2
or

ð
C 8 H 18 þ12:5O 2 þ3:76N 2 Þ ! 8CO 2 +9H 2 O + 47N 2
The combustion with 10 % excess is then

ð
C 8 H 18 þ1:10 12:5O 2 þ3:76N 2 Þ ! 8CO 2 +9H 2 O + 47N 2
where 1.10 indicates 10 % excess air.
Oxygen balance is

16 þ 9 þ 2a ¼ 1:1ð12:5Þð2Þ! a ¼ 1:25

Similarly, nitrogen balance gives

b ¼ 1:1ð59:52Þð0:79Þ¼ 51:7

Therefore,

ðA=FÞ s 12:5=1
/ ¼ ¼ ¼ 0:91
ðA=FÞ 1:1 12:5=1
mix
The equivalence ratio can simply be determined by definition

ðA=FÞ s 1
/ ¼ ¼ ¼ 0:91
ðA=FÞ 1:1
mix