In a test on a furnace fired with natural gas (composition 95 per cent methane, 5 per cent nitrogen) the following flue gas analysis was obtained: carbon dioxide 9.1 per cent, carbon monoxide 0.2 per cent, oxygen 4.6 per cent, nitrogen 86.1 per cent, all percentages by volume. Calculate the percen

Question

In a test on a furnace fired with natural gas (composition 95 per cent methane, 5 per cent nitrogen) the following flue gas analysis was obtained: carbon dioxide 9.1 per cent, carbon monoxide 0.2 per cent, oxygen 4.6 per cent, nitrogen 86.1 per cent, all percentages by volume. Calculate the percentage excess air flow (percentage above stoichiometric).

Accepted Answer

Reaction: [latex]CH_{4}+ 2O_{2}\longrightarrow CO_{2}+ 2H_{2}O[/latex]

Note: the flue gas analysis is reported on the dry basis, any water formed having been
condensed out.
Nitrogen is the tie component.
Basis: 100 mol, dry flue gas; as the analysis of the flue gas is known, the mols of each
element in the flue gas (flow out) can be easily calculated and related to the flow into the system.
Let the quantity of fuel (natural gas) per 100 mol dry flue gas be X.
Balance on carbon, mols in fuel = mols in flue gas
0.95 X = 9.1 + 0.2, hence X = 9.79 mol
Balance on nitrogen (composition of air [latex]O_{2}[/latex] 21 per cent, [latex]N_{2}[/latex] 79 per cent).
Let Y be the flow of air per 100 mol dry flue gas.
[latex]N_{2}[/latex] in air + [latex]N_{2}[/latex] in fuel = [latex]N_{2}[/latex] in flue gas

[latex]0.79 Y+0.05 imes 9.79=86.1[/latex], hence Y = 108.4 mol

Stoichiometric air; from the reaction equation 1 mol methane requires 2 mol oxygen,

so, stoichiometric air = [latex]9.79 imes 0.95 imes 2 imes \frac{100}{21}=88.6[/latex] mol

Percentage excess air =[latex]\frac{\left(air supplied - stoichiometric air ight) }{ stoichiometric air} imes 100[/latex]
[latex]=\frac{108.4-88.6}{88.6} =22 [/latex] per cent

Question 2.8: In a test on a furnace fired with natural gas (composition 9...

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