(a) Let us consider CH4 combustion in air with an equivalence ratio of 0.9, where we...

4.1 Ratings (787 Votes)

(1)flow rate of CH₄ =0.9 x 400Kg/s

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â =360Kg/s

mols of air in a second= 400x1000g/29gmol^-1(considering oxygen and nitrigen present in air)

^{Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â} =13793.1034mol

mols of CH₄ in a second= 360x1000g/16gmol^-1

^{Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â} =22500.0000mol

mol ratio= 13793.1034mol:22500.0000mol

number of nitrogen moles=79% x13793.1034mol=10853.89mol

number of oxygen moles=21% x22500.0000mol=4725mol

CH4+2O2----------> CO2 + 2H2O accordingly 11250 mols of methane reacts in a second

OUT PUT STEAM

number of water mols =4725 mol

number of carbondioxide mols=2362.5 mol

number of nitrogen molesl=10853.89mol

remaining methane mols=22500- 2362.5 mol=20137.5mol

toal number of mols in output=38078.89mol

MOL FRACTIONS

WATER=11250/38078.89=0.29

carbondioxide=2362.5/38078.89=0.06

nitrogen=10853.89/38078.89=0.28

methane=20137.5/38078.89=0.52

BY MULTIPLYING EACH MOLS BY MOLECULAR WEIGHT OU CAN CALCULATE THE MASS FRACTIONS

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