Membrane Separation 
of a Ternary Mixture
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.

Feed is a ternary mixture containing components A, B and C.  Feed flow rate is (Subscript[q, f]) = 
        1.0 (10^4) (cm^3) (STP)/s. Feed composition is (Subscript[x, fA]) = 0.25,  (Subscript[x, fB]) = 0.55,  
      (Subscript[x, fC]) = 0.20.

Membrane thickness is t=2.54 (10^(-3))cm.

The fraction of the feed that permeates through the membrane is known as the stage cut, θ=0.25, and is given by:

Permeabilities are expressed in \(TraditionalForm\`cm\^3\ \((STP)\)\ cm/\(\((s\ cm\^2\ cmHg)\)\(.\)\)\) Permeabiilties of components A, B and C are equal to (Subscript[(P'), A]) = 
        200 (10^(-10)),  (Subscript[(P'), B]) = 50 (10^(-10)) and (Subscript[(P'), C]) = 
          25 (10^(-10)) .

The pressure of the feed side is (Subscript[p, h]) = 300 cmHg while (Subscript[p, l])=30 cmHg is the pressure on the permeate side.

The three rate of permeation equations are called rate1, rate2 and rate3 where (Subscript[A, m])is the membrane area.

rate1 = 
      (Subscript[q, P]) (Subscript[y, PA]) == ((Subscript[(P'), A])/t) 
          (Subscript[A, m]) ((Subscript[p, h]) (Subscript[x, OA]) - (Subscript[p, l]) (Subscript[y, PA]))

rate2 = 
      (Subscript[q, P]) (Subscript[y, PB]) == ((Subscript[(P'), B])/t) 
          (Subscript[A, m]) ((Subscript[p, h]) (Subscript[x, OB]) - (Subscript[p, l]) (Subscript[y, PB]))

rate3 = 
      (Subscript[q, P]) (Subscript[y, PC]) == ((Subscript[(P'), C])/t) 
          (Subscript[A, m]) ((Subscript[p, h]) (Subscript[x, OC]) - (Subscript[p, l]) (Subscript[y, PC]))

The three material balance equations are called matbalance1, matbalance2 and matbalance3.

matbalance1 = 
      (Subscript[x, OA]) == (((Subscript[x, fA]) - θ (Subscript[y, PA]))/(1 - θ))

matbalance2 = 
      (Subscript[x, OB]) == (((Subscript[x, fB]) - θ (Subscript[y, PB]))/(1 - θ))

matbalance3 = 
      (Subscript[x, OC]) == (((Subscript[x, fC]) - θ (Subscript[y, PC]))/(1 - θ))

The two summation rules are called summation1 and  summation2.

summation1 = (Subscript[y, PA]) + (Subscript[y, PB]) + (Subscript[y, PC]) == 1

summation2 = (Subscript[x, OA]) + (Subscript[x, OB]) + (Subscript[x, OC]) == 1

(Subscript[x, fA]) = 0.25; (Subscript[x, fB]) = 0.55; (Subscript[x, fC]) = 0.20;

θ = 0.25;

(Subscript[(P'), A]) = 200 10^-10; (Subscript[(P'), B]) = 50 10^-10; (Subscript[(P'), C]) = 
      25 10^-10;

t=2.54 10^-3;

(Subscript[q, P]) = θ (Subscript[q, f]);

(Subscript[q, f]) = 1.0 10^4;

(Subscript[p, h]) = 300; (Subscript[p, l]) = 30;

Using FindRoot, we solve the seven simultaneous equations where the seven unknowns are (Subscript[A, m]),  (Subscript[y, PA]),  (Subscript[y, PB]),  (Subscript[y, PC]),  (Subscript[x, OA]),  
      (Subscript[x, OB]) and
(Subscript[x, OC]). Results in agreement with those obtained using the iteration method presented by C. J. Geankoplis, Transport Processes and Unit Operations, 3rd edition, Prentice Hall, 1993. Permeate mole fraction of components A, B and C are 0.4552, 0.4502 and 0.0944. Residue mole fraction of components A, B and C are 0.1815, 0.5832 and 0.2351. The membrane area is 3.5417 (10^6)(cm^2).

FindRoot[{rate1, rate2, rate3, matbalance1, matbalance2, matbalance3, 
        summation1}, {(Subscript[y, PA]), 0.2}, {(Subscript[y, PB]), 0.2}, {(Subscript[y, PC]), 0.2}, {(Subscript[A, m]), 
        10^6}, {(Subscript[x, OA]), 0.2}, {(Subscript[x, OB]), 0.2}, {(Subscript[x, OC]), 0.2}]