A gaseous fuel contains the following analysis by volume: 5% CO2: 40% H2; 40% CH

Question

A gaseous fuel contains the following analysis by volume: 5% CO2: 40% H2; 40% CH4; 15% N2. Determine the stoichiometric air and the % content of each DRY product. (4.76m3, 89.7% N2, 10.3% CO2)

Explanation / Answer

Let the total weight of the fuel be 100 grams. 5 grams CO2 40 grams H2 40 grams CH4 15 grams N2 Combustion equations of H2 and CH4Let the total weight of the fuel be 100 grams. Combustion equations of H2 and CH4 H2 + 0.5 ( O2+3.76N2) --------> H20 + 1.88 N2 (2g + 68.64g ---> 18g + 52.64 g) ( mass of each reactant and products, this implies 2 grams of h2 requires the respective mass of air and will give the mass of the products) CH4 + 2 ( O2+3.76N2) --------> CO2 + 2 H20 + 7.56 N2 (16g + 278.68 g ---> 44g + 36g + 211.68 g) ( mass of each reactant and products, this implies 16 grams of ch4 requires the respective mass of air and will give the mass of the products) Reaction 1: combustion of H2 2 grams of h2 will give 18 grams of h2o we have 40 grams of h2, which will give 40 *18/2 = 360 grams of h20. Reaction 2: Combustion of CH4 16 grams of ch4 will give 44 grams of co2 we have 40 grams of ch4, which will give 40 *44/16 = 110 grams of co2. 16 grams of ch4 will give 36 grams of h2o we have 40 grams of ch4, which will give 40 *36/16 = 90 grams of h2o. our final mixture will contain 5 grams + 110 grams (from combustion of CH4) = 115 grams of co2 15 grams + 52.64 grams (from combustion of H2) + 211.68 grams (combustion of CH4) = 279.32 of N2 360 grams (from combustion of H2) + 90 grams (from combustion of CH4) = 450 grams of h2o Total mass of mixture = 115 + 279.32 + 450 = 844.32 grams 115 / 844.32 = 13.62 % by mass of CO2 279.32 / 844.32 = 33.08 % by mass of N2 450 / 844.32 = 53.30 % by mass of H2O 29.16% CO2 and 70.84% N2

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