FUELS AND COMBUSTION 531
dharm
\M-therm\Th11-2.pm5
(i)Air-fuel ratio :
The air supplied per 100 moles of dry products is
= 22.65 × 32 +^79
21
F
HG
I
KJ
× 22.65 × 28 = 3110.6 kg
∴ Air-fuel ratio =
3110 6
8 5 12 32 4 1
.
..×+ × = 23.1 kg of air/kg of fuel. (Ans.)
(ii)Per cent theoretical air required for combustion :
Mass fraction of carbon =
12 8 5
12 8 5 32 4 1
×
×+ ×
.
..
= 0.759
Mass fraction of hydrogen =
32 4 1
12 8 5 32 4
.
..
×
×+ = 0.241
Considering 1 kg of fuel, the air required for complete combustion is
= 0 759
8
3
100
23 3
.
.
×FHG IKJ×
L
N
M
O
Q
P + 0 241 8
100
23 3
.
.
L ××
NM
O
QP
= 16.96 kg
∴ Per cent theoretical air required for combustion =^231
16 96
.
.
× 100 = 136.2%. (Ans.)
Example 11.28. The following is the volumetric analysis of the dry exhaust from an inter-
nal combustion engine :
CO 2 = 8.9% ; CO = 8.2% ; H 2 = 4.3% ; CH 4 = 0.5% and N 2 = 78.1%.
If the fuel used is octane (C 8 H 18 ) determine air-fuel ratio on mass basis :
(i)By a carbon balance.(ii)By a hydrogen-oxygen balance.
Solution. (i) As per analysis of dry products, the combustion equation is written as
a C 8 H 18 + 78.1N 2 + 78.1
21
79
F
HG
I
KJ O^2
→ 8.9CO 2 + 8.2CO + 4.3H 2 + 0.5CH 4 + 78.1N 2 + x H 2 O
Carbon balance : 8 a = 8.9 + 8.2 + 0.5 = 17.6 i.e., a = 2.2
∴ Air-fuel (A/F) ratio =
781 28 781^21
79
32
22 8 12 1 18
..
.( )
×+ × ×
×+×
= 2186 8 664 3
250 8
..
.
+ = 28511
250 8
.
.
= 11.37. (Ans.)
(ii) In this case the combustion equation is written as
a C 8 H 18 + b O 2 + b
79
21
F
HG
I
KJ N^2 → 8.9CO^2 + 8.2CO + 4.3H^2 + 0.5CH^4 + b^
79
21
F
HG
I
KJ N^2 + x H^2 O
Carbon balance : 8 a = 8.9 + 8.2 + 0.5 = 17.6 i.e., a = 2.2
Hydrogen balance : 18 a = 4.3 × 2 + 0.5 × 4 + 2x
or 18 × 2.2 = 8.6 + 2 + 2x i.e., x = 14.5
Oxygen balance : 2 b = 8.9 × 2 + 8.2 + x
or 2 b = 17.8 + 8.2 + 14.5 i.e., b = 20.25
∴ Air-fuel (A/F) ratio =
(. )(.)
.( )
20 25 32 20 25^79
21
28
22 8 12 1 18
×+ FHG IKJ×
×+×
=
2781
250 8.
= 11.09. (Ans.)