Complete Mixing Model for Gas Separation by Membranes 
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.

The separation factor for this binary mixture (components A and B) that we want to separate by membrane permeation is α=((Subscript[(P '), A])/(Subscript[(P '), B]))=10.

α = 10;

The pressure of the feed side is (Subscript[p, h]) = 190 cmHg while (Subscript[p, l])=19 cmHg is the pressure on the permeate side. The pressure ratio is R=((Subscript[p, l])/(Subscript[p, h])) = 0.1.

(Subscript[p, l]) = 19;

(Subscript[p, h]) = 190;

R=0.1;

Stage cut is equal to θ=0.2.

θ = 0.2;

t= 2.54 (10^(-3))cm is membrane thickness. \(TraditionalForm\`q\_O = \(\((1 - θ)\)\(\ \)\(q\_f\)\(\ \)\)\)is reject out flow rate. (Subscript[q, f]) = 1 (10^6) (cm^3) (STP)/s is feed flow rate. Feed mole fraction is xin=0.209. (Subscript[(P'), A])=500 \(TraditionalForm\`10\^\(-10\)\ cm\^3\ \((STP)\)\ \(\(cm\)\(/\)\((s\ cm\^2\ cmHg)\)\(\ \)\)\)is the permeability of component A. The mole fraction y in the permeate side is given by:

t=2.54 10^-3;

(Subscript[q, f]) = 10^6;

(Subscript[(P'), A]) = 500 10^-10;

xin=0.209;

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=0.506809;

A material balance gives the reject out mole fraction

Off[General::spell1]

xout=(xin-θ yout)/(1-θ)

Membrane area is given by the following relation:

(Subscript[A, m]) = ((θ (Subscript[q, f]) yout)/(((Subscript[(P'), A])/t) ((Subscript[p, h]) xout - (Subscript[p, l]) yout)))