The following is the analysis (by weight) of a chemical fuel :

Carbon = 60 per cent ; Hydrogen = 20 per cent ; Oxygen = 5 per cent ; Sulphur = 5 per cent and Nitrogen = 10 per cent.

Find the stoichiometric amount of air required for complete combustion of this fuel.

Question

The following is the analysis (by weight) of a chemical fuel :

Carbon = 60 per cent ; Hydrogen = 20 per cent ; Oxygen = 5 per cent ; Sulphur = 5 per cent and Nitrogen = 10 per cent.

Find the stoichiometric amount of air required for complete combustion of this fuel.

Accepted Answer

On the basis of 100 kg fuel let us assume an equivalent formula of the form :

[latex]C _a H _b O _c N _d S _e[/latex]

From the given analysis by weight, we can write

12a = 60 or a = 5

1b = 20 or b = 20

16c = 5 or c = 0.3125

14d = 10 or d = 0.7143

32e = 5 or e = 0.1562

Then the formula of the fuel can be written as

[latex]C _5 H _{20} O _{0.3125} N _{0.7143} S _{0.1562}[/latex]

The combustion equation is

[latex]C _5 H _{20} O _{0.3125} N _{0.7143} S _{0.1562}+x O _2+x\left(\frac{79}{21} ight) N _2 \longrightarrow p CO _2+q H _2 O +r SO _2+s N _2[/latex]

Then,

Carbon balance : 5 = p ∴ p = 5

Hydrogen balance : 20 = 2q ∴ q = 10

Sulphur balance : 0.1562 = r ∴ r = 0.1562

Oxygen balance : 0.3125 + 2x = (2p + q + 2r)

i.e., [latex]x=p+\frac{q}{2}+r-\frac{0.3125}{2}=5+\frac{10}{2}+0.1562-\frac{0.3125}{2}=9.99[/latex]

Nitrogen balance : [latex]0.7143+2 x imes \frac{79}{21}=2 s[/latex]

∴ [latex]s=\frac{0.7143}{2}+x imes \frac{79}{21}=\frac{0.7143}{2}+9.99 imes \frac{79}{21}=37.94[/latex]

Hence the combustion equation is written as follows :

[latex]C _5 H _{20} O _{0.3125} N _{0.7143} S _{0.1562}+9.99 O _2+9.99\left(\frac{79}{21} ight) N _2 \longrightarrow 5 CO _2+10 H _2 O +0.1562 SO _2+37.94 N _2[/latex]

∴ Stoichiometric air required [latex]=\frac{9.92 imes 32+9.99 imes\left(\frac{79}{21} ight) imes 28}{100}= 1 3 . 7[/latex] kg/kg of fuel.

(Note. This example can also be solved by tabular method as explained in example 11.20.).

Question 11.25: The following is the analysis (by weight) of a chemical fuel...

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