TITLE.PM5

FUELS AND COMBUSTION 529

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\M-therm\Th11-2.pm5

i.e., x = p +

q

2

+ r –

0 3125

2

.

= 5 +

10

2 + 0.1562 –

0 3125

2

.

= 9.99

Nitrogen balance : 0.7143 + 2x ×

79

21

= 2s

∴ s = 0 7143

2

. + x × 79
21
= 0 7143
2
. + 9.99 × 79
21
= 37.94
Hence the combustion equation is written as follows :

C 5 H 20 O0.3125N0.7143S0.1562 + 9.99O 2 + 9.99

79

21

F

HG

I

KJ N^2 → 5CO^2 + 10H^2 O + 0.1562SO^2 + 37.94N^2

∴ Stoichiometric air required =

992 32 999^79

21

28

100

..×+ ×F

HG

I
KJ
×
= 13.7 kg/kg of fuel. (Ans.)
(Note. This example can also be solved by tabular method as explained in example 11.20.).
Example 11.26. A sample of fuel has the following percentage composition by weight :
Carbon = 84 per cent Hydrogen = 10 per cent
Oxygen = 3.5 per cent Nitrogen = 1.5 per cent
Ash = 1 per cent
(i)Determine the stoichiometric air-fuel ratio by mass.
(ii)If 20 per cent excess air is supplied, find the percentage composition of dry flue gases by
volume.
Solution. (i) Stoichiometric air fuel ratio :
On the basis of 100 kg of fuel let us assume an equivalent formula of the form :
CaHbOcNd
From the given analysis by weight, we can write
12 a = 84 i.e., a = 7
1 b = 10 i.e., b = 10
16 c = 3.5 i.e., c = 0.218
14 d = 1.5 i.e., d = 0.107
The formula of fuel is
C 7 H 10 O0.218N0.107
The combustion equation is written as

C 7 H 10 O0.218N0.107 + x O 2 + x 79

21

F

HG

I

KJ

N 2 → p CO 2 + q H 2 O + r N 2

Then,
Carbon balance : 7 = p i.e., p = 7
Hydrogen balance : 10 = 2q i.e., q = 5
Oxygen balance : 0.218 + 2x = (2p + q)
or 0.218 + 2x = 2 × 7 + 5 i.e., x = 9.39

Nitrogen balance : 0.107 + 2x

79

21

F

HG

I

KJ= 2r

or 0.107 + 2 × 9.39 ×

79

21

= 2r i.e., r = 35.4