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FUELS AND COMBUSTION 513

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(ii)The percentage composition of dry flue gases is given as below :

Dry Weight Molecular Proportional Percentage volume

products (kg) weight volume

xyz = xy

z

Σz ×^100

CO 0.313 28 0.0112 2.78 per cent (Ans.)

CO 2 2.662 44 0.0605 15.03 per cent (Ans.)

N 2 9.24 28 0.3300 81.97 per cent (Ans.)

SO 2 0.06 64 0.0009 0.22 per cent (Ans.)

Σz = 0.4026

Example 11.9. A fuel (C 10 H 22 ) is burnt using an air-fuel ratio of 13 : 1 by weight. Deter-

mine the complete volumetric analysis of the products of combustion, assuming that the whole

amount of hydrogen burns to form water vapour and there is neither any free oxygen nor any free

carbon. The carbon burns to CO 2 and CO.

Air contains 77% of nitrogen and 23% of oxygen by weight.

Solution. Combustion equation is :

2C 10 H 22 + 31O 2 = 20CO 2 + 22H 2 O

2 × 142 + 31 × 32 = 20 × 44 + 22 × 18

or 284 + 992 = 880 + 396

∴ Air required for complete combustion

=

992 100

284 23

×

× = 15.2 kg/kg of fuel

Air actually supplied = 13 kg/kg of fuel

∴ Deficiency of air = 15.2 – 13 = 2.2 kg/kg of fuel

Also 1 kg of C requires

4

3

100

23

× = 5.8 kg of less air to burn to CO instead of CO 2.

Hence^22

58

.

.

= 0.379 kg C is burnt to CO ;

and

12 10

142

F ×

HG

I

KJ – 0.379 = 0.466 kg of C is burnt to CO^2.

Weight of CO 2 formed = 0.466 ×

11

3

= 1.708 kg

Weight of CO formed = 0.379 ×

7

3

= 0.884 kg

Weight of H 2 O formed =^22

142

F

HG

I

KJ

× 9 = 1.394 kg

Weight of N 2 from air = 13 × 0.77 = 10.01 kg.