The analysis of the dry exhaust from an internal combustion engine is as follows :

Carbon dioxide $\left( CO _2\right)$ = 15 per cent Carbon monoxide (CO) = 3 per cent

Methane $\left( CH _4\right)$ = 3 per cent Hydrogen $\left(H_2\right)$ = 1 per cent

Oxygen $\left(O_2\right)$ = 2 per cent Nitrogen $\left(N_2\right)$ = 76 per cent

Calculate the proportions by mass of carbon to hydrogen in the fuel, assuming it to be a pure hydrocarbon.

Question

The analysis of the dry exhaust from an internal combustion engine is as follows :

Carbon dioxide [latex]\left( CO _2\right)[/latex] = 15 per cent Carbon monoxide (CO) = 3 per cent

Methane [latex]\left( CH _4\right)[/latex] = 3 per cent Hydrogen [latex]\left(H_2\right)[/latex] = 1 per cent

Oxygen [latex]\left(O_2\right)[/latex] = 2 per cent Nitrogen [latex]\left(N_2\right)[/latex] = 76 per cent

Calculate the proportions by mass of carbon to hydrogen in the fuel, assuming it to be a pure hydrocarbon.

Accepted Answer

Let 1 kg of fuel contain x kg of carbon (C) and y kg hydrogen ([latex]H _2[/latex]). Then considering 1 mole of D.E.G. and introducing X and Y, we can write

[latex]X\left(\frac{x C }{12}+\frac{y}{2} H _2 ight)+Y O _2+\frac{79}{21} Y N _2 \longrightarrow 0.15 CO _2+0.03 CO +0.03 CH _4[/latex]

[latex]+0.01 H _2+0.02 O _2+a H _2 O +0.76 N _2[/latex]

Then,

Nitrogen balance : [latex]\frac{79}{21} Y=0.76[/latex] ∴ Y = 0.202

Oxygen balance : [latex]Y=0.15+\frac{0.03}{2}+0.02+\frac{a}{2}[/latex]

or [latex]0.202=0.15+0.015+0.02+\frac{a}{2}[/latex] ∴ a = 0.034

Carbon balance : [latex]\frac{X x}{12}=0.15+0.03+0.03[/latex] ∴ Xx = 2.52 ...(i)

Hydrogen balance : [latex]\frac{X y}{2}=2 imes 0.03+0.01+a=0.06+0.01+0.034[/latex]

∴ [latex]X y=0.208[/latex] ...(ii)

Dividing equations (i) and (ii), we get

[latex]\frac{X x}{X y}=\frac{2.52}{0.208} \quad ext { or } \quad \frac{x}{y}=12.1[/latex]

i.e., Ratio of C to [latex]H _2 ext { in fuel }=\frac{x}{y}=\frac{12.1}{1}[/latex].

Question 11.30: The analysis of the dry exhaust from an internal combustion ...

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