Production of Enriched Air 
by Membrane Separation
Author's Data: Housam BINOUS
Department of Chemical Engineering
National Institute of Applied Sciences and Technology
Tunis, TUNISIA
Email: binoushousam@yahoo.com

Reference:C. J. Geankoplis, Transport Processes and Unit Operations, 3rd edition, Prentice Hall, 1993.

Polymer membranes are more permeable to oxygen than nitrogen. This property can be used to enrich the oxygen content of air.

The separation factor, α, is the ratio of permeabilities and plays a similar role to relative volatility in distillation.

The pressure ration is R.

The fraction of the feed that permeates through the membrane is known as the stage cut, θ, and is given by:

Small stage cut case

If the pressure ration is R=0 (vacuum pump is used in the permeate side) then the permeate mole fraction, y, is given by the following equation:

More generally the permeate mole fraction, y, is the solution of following quadratic equation where x=xin=0.21 (feed is air that contains 21% oxygen and 79% nitrogen).

Plot of the permeate mole fraction versus permeability ratio, α, for R=0, 0.1, 0.2, 0.3, 0.4 and 0.5.

sol1=Solve[R (α-1) y^2-((α-1) (x+R)+1)y+α x==0,y]

y[α_]:=y/.sol1[[1]]

x=0.21;

<< Graphics`Graphics`

LogLinearPlot[{α x/((α-1)x+1)/.R->0,y[α]/.R->0.1,y[α]/.R->0.2,y[α]/.R->0.3,y[α]/.R->0.4,y[α]/.R->0.5},{α,1,10^4},PlotStyle->{RGBColor[1,0,0],RGBColor[1,0,1],RGBColor[0,1,0],RGBColor[0,0,1],RGBColor[0,1,1],RGBColor[1,1,0]}]

Plot of the limit of y when α tends to ∞ verus R for xin=0.2, 0.4 and 0.6.

x=.

lim[a_, r_] := 
      If[Limit[y[α], α -> Infinity] < x/R /. 
            x -> a /. R -> r, 1, 
        Limit[y[α] /. x -> a /. R -> r, α -> 
            Infinity]]

Plot[{lim[0.2,R],lim[0.4,R],lim[0.6,R]},{R,0,1},PlotStyle->{RGBColor[1,0,0],RGBColor[1,0,1],RGBColor[0,1,0]}]

Case where we take stage cut into account. A mass balance gives the following relations:

The permeate mole fraction, yout, is the solution of the following quadratic equation:

sol2=Solve[(α-1) (θ+R (1-θ)) y^2-(α θ+(α-1) (xin+R(1-θ))+(1-θ)) y+α xin==0,y]

yout=y/.sol2[[1]]

Off[General::spell1]

xout=(xin-θ yout)/(1-θ);

Feed is air with 21% oxygen and 79% nitrogen

xin=0.21;

Plot of permeate and residue mole fractions versus stage cut for α=100 and R=0.5

Plot[{yout/.α->100/.R->0.5,xout/.α->100/.R->0.5},{θ,0,1},PlotStyle->{RGBColor[1,0,0],RGBColor[0,0,1]},PlotRange->{0,1}]

Checking our results

yout when θ tends to 1 should be equal to xin=0.21

yout/.α->100/.R->0.5/.θ->1

xout when θ tends to 1 should be equal to the following expression:

xout/.α->100/.R->0.5/.xin->0.21/.θ->0.99999

xin=.;x=.;

sol=Solve[R (α-1) xin^2-((α-1) (x+R)+1)xin+α x==0,x]//Simplify

sol/.xin->0.21/.α->100/.R->0.5