Discussion
A fuel with composition by mass is given as 78 % carbon (C), 16 % hydrogen (H2), 3.2 % sulphur (S), 1.6 % oxygen (O2), and ash. The fuel is fired in a boiler with excess air for the complete combustion and no carbon monoxide (CO) is detected in the flue gases. The dry flue gas contains 0.2 % sulphur dioxide (SO2) by volume. Which one of the following options is closest to the amount of excess air provided (in % by mass)?
Assume molar masses (g mol-1) of C, H2, S, and O2 are 12, 2, 32, and 32, respectively, and the volumetric ratio of nitrogen (N2) and oxygen (O2) in the air as 3.76:1.
Answer: B
The excess air is closest to 7.4 %.
Take 100 g of fuel. In moles: C 6.5, H2 8, S 0.1, fuel O2 0.05. Stoichiometric oxygen:
\[ \begin{aligned} O_{2,st} &= 6.5 + \tfrac{8}{2} + 0.1 - 0.05 \\ &= 10.55\ \text{mol} \end{aligned} \]
With a fraction x of excess air, the dry flue gas holds CO2 6.5, SO2 0.1, excess O2 10.55x and N2 3.76 × 10.55(1 + x). SO2 is 0.2% of it, so the dry gas totals 0.1/0.002 = 50 mol:
\[ \begin{aligned} 6.6 + 39.67 + 50.22x &= 50 \\ x &= 0.074 \end{aligned} \]
So about 7.4% excess air; the percentage is the same by mass, since the excess is a fraction of the same air. Water vapour is excluded because the analysis is on a dry basis; including it, or omitting the fuel oxygen, moves the answer away from the listed 7.4.
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