Question

A mixture of methane and air is capable of being ignited only if the mole percent of methane is between 5% and 15%. A mixture containing 9.0 mole% methane in air flowing at a rate of 7.00 x 10^2 kg/h is to be diluted with pure air to reduce the methane concentration to the lower flammability limit. Calculate the required flow rate of air in mol/h and the percent mass by mass of oxygen in the product gas. (Note: Air may be taken to consist of 21 mole% 02 and 79% N2 and to have an average molecular weight of 29.0)

Answer 1

We have 700 Kg/h of the mixture.

First we calculate the moles of mixture = (16*0.09) + (29*0.91) = 27.83 mol

q(mixture) = 700/27.83 = 25.15 moles/hr

The 9% of Methane = 2.264 moles/hr

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The minimum total flow rate required to reduce this to below 5% = 2.264*100/5 = 45.28 moles.

Additional moles air needed = 45.28 - 25.15 = 20.13 moles/hr

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O2 Mass = (qt(min) * 95 %mol) * (21 %mol*32 amuO2) = (45.28*0.95) * (0.21*32) = 289.1 kg/hr O2

%O2 Mass = O2mass / (Total mass + O2mass) *100 = 289.1/ (700+289.1) * 100 = 29.22 % O2

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