6.04) Amixture of methane and air is capable of being ignited only if the mole p

Question

6.04) Amixture of methane and air is capable of being ignited only if the mole percent of methane is between 5% (ower and 15% (upper flammability limits. A mixture containing 90 ole methane in air flowing at a rate of 700 kg/h is to be diluted with pure air to reduce the methane concentration to the lower limit. a) Calculate the required flow rate of air in molh. b) Calculate the percent, by mass, of oxygen in the product gas. c) Calculate the flow rate of an 4 methane stream that would have to be added to the original stream to prevent the the mixture from igniting. d Qualitatively explain why there are upper and lower flammability limits.

Explanation / Answer

Mixture contains 9 mole% methane and 91 mole% air

Basis :1 mole of mixture . Molar masses : CH4=16, Air =29

Moles= mass/molar mass

Mass= moles* molar mass

Masses (gm) : CH4= 0.09*16= 1.44, Air =0.91*29 = 26.39

Total mass of 1 mole of mixture = 1.44+26.39 = 27.83 gm

Mass % : CH4= 100*1.44/27.83 = 5.2, Air =100-5.2= 94.8

27.83 kg of mixture is there in 1 kg moles of mixture

700 kg/hr correspond to 700/27.83 kg moles =25.2 kg moles/hr

Let x= moles of air to be used.

Total moles of mixture = 25.2+x

CH4 in this mixture = 25.4*0.09=2.3

Mole % of CH4 required in the resulting mixture = 5%,

Moles of CH4 required = (25.2+x)*5/100 = 2.3

25.2+x = 2.3*100/5= 46

Hence x= 20.8 kgmoles/hr, mass of air = 20.8*29= 603 kg/hr

Mixed stream contains : 750 kg/hr air CH4 mixture and 603kg/hr Air

Total flow rate= 750+603 = 1353 kg/hr

Air ins 750 kg/hr= 750*0.948 =711 kg/hr

Total air = 711+603=1314 kg/hr

Air contains 21% O2, 79% N2 by volume

1 mole of air contains 0.21 mole of O2, 0.79 moles of N2

Masses : O2= 0.21*32= 6.7 and N2= 0.79*32 =25.3

Mass % of O2 in air = 100*6.7/(6.7+25.3)= 20.93%

Mass of O2 in the mixed stream = 0.2093*711= 149 kg/hr

Mass % of O2 = 100*149/1353= 11%

When 47% (assumed to be mole % is to be added, it reaches upper flammability )

Hence let x= moles of CH4 to be added

Hence moles of methane in original mixture = 25.2*0.09 =2.3 moles/hr

Moles of methane to be added= x*0.47 ( x moles of stream containing 47% CH4)

Hence writing methane balance. 2.3+x*0.47= (25.2+x)*15/100 =(25.2+x)*0.15

2.3+0.47x= 3.8+0.15x

Hence 0.32x= 1.5, x= 1.5/0.32= 4.7 moles/hr

Combustible material and air burn only if the fuel concentration lies within well-defined lower and upper limits   which are referred to as flammability limits or explosive limits.

LFL refer to lower limit of fuel that produced flash in presence of ignitable source while UFL refer to Highest concentration (percentage) of a gas or a vapor in air capable of producing a flash of fire in presence of an ignition source

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