Production of Enriched Air by Membrane SeparationAuthors: Housam Binous and Ahmed Bellagi
Production of Enriched Air by Membrane SeparationAuthors: Housam Binous and Ahmed Bellagi
Polymer membranes are more permeable to oxygen than nitrogen . This property can be used to enrich the oxygen content of air . Here, permeate mole fraction is plotted versus separation factor for different values of the pressure ratio under the assumption of small stage cut . In addition, the permeate and residue mole fractions are plotted versus stage cut for a separation factor and a pressure ratio equal to 100 and 0.5, respectively .
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.
In[]:=
sol1=Solve[R (α-1) y^2-((α-1) (x+R)+1)y+α x0,y]
Out[]=
y,y
1-R-x+Rα+xα-
-4xα(-R+Rα)+
2
(-1+R+x-Rα-xα)
2(-R+Rα)
1-R-x+Rα+xα+
-4xα(-R+Rα)+
2
(-1+R+x-Rα-xα)
2(-R+Rα)
In[]:=
y[α_]:=y/.sol1[[1]]
In[]:=
x=0.21;
In[]:=
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]}]
Out[]=
Plot of the limit of y when α tends to ∞ verus R for xin=0.2, 0.4 and 0.6.
In[]:=
x=.
In[]:=
lim[a_,r_]:=If[Limit[y[α],αInfinity]<x/R/.xa/.Rr,1,Limit[y[α]/.xa/.Rr,αInfinity]]
In[]:=
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]}]
Out[]=
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:
In[]:=
sol2=Solve[(α-1) (θ+R (1-θ)) y^2-(α θ+(α-1) (xin+R(1-θ))+(1-θ)) y+α xin0,y]
Out[]=
y,y
1-R-xin+Rα+xinα-θ+Rθ+αθ-Rαθ-
-4xinα(-R+Rα-θ+Rθ+αθ-Rαθ)+
2
(-1+R+xin-Rα-xinα+θ-Rθ-αθ+Rαθ)
2(-R+Rα-θ+Rθ+αθ-Rαθ)
1-R-xin+Rα+xinα-θ+Rθ+αθ-Rαθ+
-4xinα(-R+Rα-θ+Rθ+αθ-Rαθ)+
2
(-1+R+xin-Rα-xinα+θ-Rθ-αθ+Rαθ)
2(-R+Rα-θ+Rθ+αθ-Rαθ)
In[]:=
yout=y/.sol2[[1]]
Out[]=
1-R-xin+Rα+xinα-θ+Rθ+αθ-Rαθ-
-4xinα(-R+Rα-θ+Rθ+αθ-Rαθ)+
2
(-1+R+xin-Rα-xinα+θ-Rθ-αθ+Rαθ)
2(-R+Rα-θ+Rθ+αθ-Rαθ)
In[]:=
Off[General::spell1]
In[]:=
xout=(xin-θ yout)/(1-θ);
Feed is air with 21% oxygen and 79% nitrogen
In[]:=
xin=0.21;
Plot of permeate and residue mole fractions versus stage cut for α=100 and R=0.5
In[]:=
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}]
Out[]=
Checking our results
yout when θ tends to 1 should be equal to xin=0.21
In[]:=
yout/.α->100/.R->0.5/.θ->1
Out[]=
0.21
xout when θ tends to 1 should be equal to the following expression:
In[]:=
xout/.α->100/.R->0.5/.xin->0.21/.θ->0.99999
In[]:=
xin=.;x=.;
In[]:=
sol=Solve[R (α-1) xin^2-((α-1) (x+R)+1)xin+α x0,x]//Simplify
In[]:=
sol/.xin->0.21/.α->100/.R->0.5