A mixture of methane and air is flammable only if the mole percent of methane is

Question

A mixture of methane and air is flammable only if the mole percent of methane is between 5% and 15%. A mixture containing 9 mole% methane in air flowing at 700 kg/hr is diluted with a stream of pure oxygen to reduce the methane concentration to the lower flammability limit. Calculate the required oxygen flow rate in mol/hr and the resulting mass fractions of methane and oxygen in the product stream. Air may be considered to be 21 mol% O2 and 79 mol% N2 and to have an average molecular weight of 29.0.

Explanation / Answer

Start with a basis. 700 kg mixture. Mol. wt. mixture = (16 x 0.09) + (29 x 0.91) = 27.83 Rate of flow of mix = 700/27.83 = 25.15 moles/hr Methane = 2.264 moles/hr This is 9%. The minimum total flow rate required to reduce this to below 5% = 2.264 x 100/5 = 45.28 moles. Additional moles air needed = 45.28 - 25.15 = 20.13 moles /hr Mass of oxygen = 45.28 x 0.95 x 0.21 x 32 = 289.1 kg/hr

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