FUELS AND COMBUSTION 521
dharm
\M-therm\Th11-2.pm5
(ii)Air-fuel ratio A/F :
The air-fuel ratio on a mole basis is
2.2 + 8.27 = 10.47 moles air/mole fuel. (Ans.)
The air-fuel ratio on a mass basis is found by introducing the molecular weights
A/F = 10 47 28 97
12 1 4
..
()
×
+×
= 18.96 kg air/kg fuel. (Ans.)
The theoretical air-fuel ratio is found by writing the combustion equation for theoretical air,
CH 4 + 2O 2 + 2
79
21
F
HG
I
KJ N^2 → CO^2 + 2H^2 O + (2)
79
21
F
HG
I
KJ N^2
A/Ftheor.=
2279
21
28 97
12 1 4
+ F
HG
I
KJ
L
N
M
O
Q
P
+×
().
()
= 17.24 kg air/kg fuel. (Ans.)
(iii)Percent theoretical air :
Per cent theoretical air = 18 96
17 24
.
.
× 100 = 110%. (Ans.)
Example 11.19. The gravimetric analysis of a sample of coal is given as 82% C, 10% H 2
and 8% ash. Calculate :
(i)The stoichiometric A/F ratio ; (ii)The analysis of the products by volume ;
Solution. (i) The stoichiometric A/F ratio :
1 kg of coal contains 0.82 kg C and 0.10 kg H 2.
∴ 1 kg of coal contains^082
12
. moles C and^010
2
.
moles H 2
Let the oxygen required for complete combustion = x moles
Then the nitrogen supplied with the oxygen = x×^79
21
= 3.76x moles
For 1 kg of coal the combustion equation is therefore as follows :
082
12
. C + 010
2
. H
2 + x CO 2 + 3.76x N 2 →^ a CO 2 + b H 2 O + 3.76 x N 2
Then, Carbon balance : 082
12
. = a ∴ a = 0.068 moles
Hydrogen balance : 2 ×^010
2
. = 2b ∴ b = 0.05 moles
Oxygen balance : 2 x = 2a + b ∴ x =
2 0 068 0 05
2
F ×+
HG
I
KJ
..
= 0.093 moles
The mass of 1 mole of oxygen is 32 kg, therefore, the mass of O 2 supplied per kg of coal
= 32 × 0.093 = 2.976 kg
i.e., Stoichiometric A/F ratio = 2 976
0 233
.
.
= 12.77 (Ans.)
(where air is assumed to contain 23.3% O 2 and 76.7% N 2 by mass)
Total moles of products = a + b + 3.76x = 0.068 + 0.05 + 3.76 × 0.093 = 0.467 moles