A mixture of methane and air is capable of being ignited only if the mole pecent

Question

A mixture of methane and air is capable of being ignited only if the mole pecent of methane is between 5% and 15%. A mixture containing 12mole% methane in air flowing at a rate of 900.00kg/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 by mass of oxygen in the product gas. (note Air may be taken to consist of 21 mole % o2 and 79% N2 and to have an average molecular weight of 29.0)

Explanation / Answer

We have 900kg of mixture

Molecular weight of mixture:

Molecular weight of CH4 : (16 x 0.12 )= 1.92

Molecular weight of Air : (29 x 0.88 mol) = 25.52

Molecular weight of the mixture = 27.44

Rate of flow of mixture = 900 / 27.44 = 32.7 moles/hour

Rate of flow of Methane for 12 mol % = 3.924 moles/hr.

The minimum flow required to reduce below 5% is 3.924 moles/hr * 100/5 = 78.48 moles/hr.

Flow rate of air = 78.48 - 32.7 = 45.78 moles/ hr.

Percentage Mass of oxygen = 78.48 * 0.21 * 0.95*32 =501.016 kg/hr.

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