Question

Average mol Wt of a mixed stream

A mixture of methane and air is capable of being ignited if the mole % of methane is between 5 and 15%. A mixture containing 11 mole% methane in air is flowing at a rate of 700 kg/hr is to be diluted with pure air to reduce methane to the lower flammability limit. Find the flow rate of the dilution air (air entering the system) in (kg-mole/hr).

Answer 1

mass flow rate of air = 700 kg/hr

Mole percent of methane =11% and air =89%

Mole fractions :0.11 CH4 and 0.89 air

Molecular weight of mixture= 0.11* molecular weight of CH4 + 0.89* molecular weight of air

=0.11*16 +0.89*29=27.57

Moles of mixture in 700 kg/ hr= 700/27.57 kgmoles/ hr=25.39 kgmoles/hr

Moles of methane therefore = 25.39*11/100= 2.793 moles/hr

it has to be diluted with moles of air so as to have 5 mole % air

let total moes of mixtture after dilution with air = y

y*5/100= 2.793 moles/hr

y= 2.793*100/5= 55.86 kgmoles/hr

moles of mixture befoe dilution= 25.39 kgmoles/hr and the final moles have become 55.86 kg moles/ hr due to addition of air only

hence moles of air = final moles of mixture- mole of CH4 and air initially= 55.86-25.39=30.47 kgmoles/hr