A liquid fuel containing 88 wt% C and 12% H is burned with air, comprising 21 vo

Question

A liquid fuel containing 88 wt% C and 12% H is burned with air, comprising 21 vol% O2 and 79% N2, in a combustor. The combustion product is called “flue gas”, and the flue gas leaving the combustor is then passed through a desiccant column, to completely removed water from the flue gas. The resulting dry flue gas is then passed through a gas analyser, to analyse for the composition of the dry flue gas, and the analytical results are as follows: CO2 13.4% by vol., O2 3.6%, and N2 83.0%.

If the amount of fuel fed into the combustor is 100 g, calculate the amount of the dry flue gas, in g, and determine the percentage excess of O2.

Explanation / Answer

Mass of N in Flue Gas = (100)83.8/ 79.0 = 106.08

106.08/0.838 = 126.58 g of dry flue gas

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