Question

5. 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 14.0 mole% methane in air flowing at a rate of 600. 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 kmol/h and the percent by mass of oxygen in the product gas.

Answer 1

Block diagram

Average molecular weight of inlet gas

Mavg = (molecular weight x mol fraction) of air + (molecular weight x mol fraction) of CH4

= 29 x 0.86 + 16 x 0.14

= 27.18 g/mol

Molar flow rate of inlet gas = mass flow rate / Mavg

= (600 x 1000 g/h) / (27.18 g/mol)

= 22075 mol/h

Molar flow rate of inlet CH4 = 22075 mol/h x 0.14

= 3090.5 mol/h

Molar flow rate of inlet air = 22075 mol/h x 0.86

= 18984.5 mol/h

CH4 balance

3090.5 = n1 x 0.05

n1 = 61810 mol/h

Moles of air in outlet stream = 0.95 x 61810 = 58719.5 mol/h

Overall balance

Moles of inlet gas + moles of pure air = moles of outlet gas

22075 + pure air = 61810

Pure air = 39735 mol/hr x 1kmol/1000mol

= 39.735 kmol/hr

Moles of oxygen in outlet gas = 0.21 x moles of air in outlet gas

= 0.21 x 58719.5

= 12331.095 mol/h

Mass flow rate of oxygen in outlet stream

= moles of O2 in outlet gas x molecular weight of O2

= 12331.095 mol/h x 32 g/mol

= 394595.04 g/h

Average molecular weight of outlet gas

M = (molecular weight x mol fraction) of air + (molecular weight x mol fraction) of CH4

= 29 x 0.95 + 16 x 0.05

= 28.35 g/mol

Mass flow rate of outlet gas = moles of outlet gas x M

= 61810 mol/h x 28.35 g/mol

= 1752313.5 g/h

Mass % of O2 in outlet gas

= mass of O2 in outlet gas x 100 / Mass flow rate of outlet gas

= 394595.04 x 100 / 1752313.5

= 22.52 %