TITLE.PM5

522 ENGINEERING THERMODYNAMICS

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\M-therm\Th11-2.pm5

Hence the analysis of the products is

CO 2 =

0 068

0 467

.

.

× 100 = 14.56%. (Ans.)

H 2 =^005

0 467

.

.

× 100 = 10.7%. (Ans.)

N 2 =

(.. )

.

3 76 0 093

0 467

×

× 100 = 74.88%. (Ans.)

+Example 11.20. Calculate the stoichiometric air-fuel ratio for the combustion of a sam-
ple of dry anthracite of the following composition by mass :
Carbon (C) = 88 per cent Hydrogen (H 2 ) = 4 per cent
Oxygen (O 2 ) = 3.5 per cent Nitrogen (N 2 ) = 1 per cent
Sulphur (S) = 0.5 per cent Ash = 3 per cent
If 30 per cent excess air is supplied determine :
(i)Air-fuel ratio ;
(ii)Wet dry analysis of the products of combustion by volume.
Solution. Stoichiometric air-fuel (A/F) ratio :
In case of a fuel with several constituents a tubular method is advisable, as shown below.
Each constituent is taken separately and the amount of oxygen required for complete combustion
is found from the chemical equation. The oxygen in the fuel is included in the column headed
‘oxygen required’ as a negative quantity.

Mass per Combustion equation Oxygen required per Products per kg of coal

kg coal kg of coal

C 0.88 C + O 2 → CO 2 0.88 ×^3212 = 2.346 kg 0.88 ×^4412 = 3.23 kg CO 2

12 kg + 32 kg → 44 kg

H 2 0.04 2H 2 + O 2 → 2H 2 O 0.04 × 8 = 0.32 kg 0.04 × 9 = 0.36 kg H 2 O

1 kg + 8 kg → 9 kg

O 2 0.035 — — 0.035 kg —

N 2 0.01 — — 0.01 kg N 2

S 0.005 S + O 2 → SO 2 0.005 ×^3232 = 0.005 kg 0.005 ×^6432 = 0.01 kg SO 2

32 kg + 32 kg → 64 kg

Ash 0.03 — — —

Total O 2 = 2.636 kg

From table :

O 2 required per kg of coal = 2.636 kg

∴ Air required per kg of coal = 2 636

0 233

.

.

= 11.31 kg

(where air is assumed to contain 23.3% O 2 by mass)
N 2 associated with this air = 0.767 × 11.31 = 8.67 kg
∴ Total N 2 in products = 8.67 + 0.01 = 8.68 kg