TITLE.PM5

FUELS AND COMBUSTION 521

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(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