Question

Methanol is combusted with 30% excess air. The product gas contains 3 mol% uncombusted methanol and 4 mol% CO. The remaining carbon exits the furnace as CO₂. Draw and label a flowchart for the process and determine the production rate of CO₂ and water per mole of methanol fed.

Answer 1

Basis : 1 mole of CH3OH

The combustion reaction is CH3OH +3/2 O2---->CO2 + 2H2O

1mole of CH3OH requires   1.5 moles of oxygen

So oxygen required = 1.5 moles Since air contains 80%N2 and 20% O2, moles of air = 1.5/0.2= 7.5 moles

let x= mole of CH3OH unreacted

Mole of CH3OH reacted= 1-x

let y = mole of CH3OH reacted to form CO2 and rest CO through combustion reactions

CH3OH +1.5O2------>CO2 + 2H2O

moles of CO2 formed = y moles of H2O formed= 2y, moles of O2 consumed= 1.5y

1-x-y = moles of CH3OH reacted to form CO

CH3OH + O2----------->CO +2H2O

moles of CO formed= 1-x-y moles of O2 utilized= 1-x-y moles of CO formed =1-x-y moles of H2O=2*(1-x-y)

moles of N2 formed= 9.75*0.8= 7.8 moles unreacted oxygen = 7.5*0.2- (1,5y+1-x-y)

=0.5-0.5y+x  

the product contains 7.8 moles of N2, CO= 1-x-y, H2O= 2y+2-2x-2y= 2-2x, CO2= y, O2= 0,5-0.5y +x, CH3OH= x

total moles =7.8+1-x-y+2-2x +0.5-0.5y+x=11.3-2x-0.5y+x=11.3-x+0.5y

Mole % CO= 100*(1-x-y)/ (11.3-x+0.5y)=4

100-100x-100y= 4*(11.3-2x-0.5y)=45.2-4x-2y

96x+98y =100-45.2= 54.8 (1) x+98/96y= 54.8/96 , x+1.02y= 0.57 , x= 0.57-1.02y (1A)

mole % CH3OH= 100*x/(11.3-x-0.5y)= 3

100x= 3*(11.3- x-0.5y)

100x = 33.9-3x-0.5y

103x+0.5 y= 33.9 (2)

from 1A,   103*(0.57-1.02y)+0.5y= 33.9

58.7- 105.06y +0.5y= 33.9

58.7-33.9 = (105.06-0.5)y, y =0.237 , x = 0.57-1.02*0.237= 0.328

moles of CO2 formed per mole of CO= 0.237, moles of H2O= 2-2*0.328=1.344 moles

CH3OH-1 mole unreacted CH3OH-3 mole % CO= 4 mole %, H2O,CO2 and N2 moles of Air 30% excess