TITLE.PM5

FUELS AND COMBUSTION 509

dharm

\M-therm\Th11-1.pm5

(ii) To get volumetric composition of the product of combustion let us use tabular method.

Name of gas Weight (x) Molecular Proportional Percentage volume

weight (y)

volume (z) =

(x)

(y)

=

(z)

(z)

100

Σ

×

CO 2 3.109 44 0.0707 12.51 per cent. (Ans.)

O 2 0.521 32 0.0163 2.89 per cent. (Ans.)

N 2 13.38 28 0.4780 84.60 per cent. (Ans.)

Σz = 0.5650

+Example 11.5. Percentage volumetric analysis of a sample of flue gases of a coal fired
boiler gave CO 2 = 10.4 ; CO = 0.2 ; O 2 = 7.8 and N 2 = 81.6 (by difference). Gravemetric percentage
analysis of coal was C = 78, H 2 = 6, O 2 = 3 and incombustible = 13. Estimate :
(i)Weight of dry flue gases per kg of fuel.
(ii)Weight of excess air per kg of fuel.
Solution.
Element, wt. (kg) O 2 reqd. (kg) Dry products (kg)

C = 0.78 0.78 ×^83 = 2.08 0.78 ×^113 = 2.86 (CO 2 )

H 2 = 0.06 0.06 × 8 = 0.48

O 2 = 0.03

Total O 2 = 2.56

Minimum wt. of air needed for combustion = (2.56 – .03) ×
100
23
= 11 kg.
(i)Weight of dry flue gases per kg of fuel :
To determine the wt. of flue gases per kg of fuel let us use tabular method to convert
volumetric analysis to analysis by weight.

Name of gas Volume per Molecular Relative volume Weight per kg of

m^3 of flue gas weight z = x × y flue gas

(x) (y) = Σzz

CO 2 0.104 44 4.576 0.1525

CO 0.002 28 0.056 0.0019

N 2 0.816 28 22.848 0.7616

O 2 0.078 32 2.496 0.0832

Σz = 29.976 (say 30)

Amount of carbon present per kg of gases

= Amount of carbon in 0.1525 kg of CO 2 + Amount of carbon present in 0.0019 kg of CO

=

3

11 × 0.1525 +

3

7 × 0.0019 = 0.0416 + 0.0008 = 0.0424 kg.