FUELS AND COMBUSTION 531dharm
\M-therm\Th11-2.pm5(i)Air-fuel ratio :
The air supplied per 100 moles of dry products is
= 22.65 × 32 +^79
21F
HGI
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×
×+ ×.
..
= 0.759Mass 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 7598
3100
23 3
.
.
×FHG IKJ×
L
N
MO
Q
P + 0 241 8100
23 3
.
.L ××
NMO
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
79F
HGI
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 asa C 8 H 18 + b O 2 + b79
21F
HGI
KJ N^2 → 8.9CO^2 + 8.2CO + 4.3H^2 + 0.5CH^4 + b^79
21F
HGI
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.)