Liquid chromatography with 
linear adsorption equilibrium and plate model 
Author's Data: Housam BINOUS
Department of Chemical Engineering
National Institute of Applied Sciences and Technology
Tunis, TUNISIA
Email: binoushousam@yahoo.com

Injection of components A and B

(Subscript[CA, 0])[
        t_] := (Subscript[CA, init]) (UnitStep[t] - UnitStep[t - 10])

(Subscript[CA, init]) = 0.23;

(Subscript[CB, 0])[
        t_] := (Subscript[CB, init]) (UnitStep[t] - UnitStep[t - 10])

(Subscript[CB, init]) = 0.23;

Plot[(Subscript[CA, 0])[t], {t, 0, 200}, 
      PlotStyle -> RGBColor[1, 0, 0]]

Column diameter=Dc
Void fraction of the bed=ϵ
Volumetric flow rate=L

L=0.4;

ϵ = 0.4;

Dc=4;

Column length=Lbed

Lbed=25;

Column volume=V\(TraditionalForm\`\_tot\)

(Subscript[V, tot])=Lbed*Pi*Dc*Dc/4.

Number of stages and Volume of a stage

NS=200;

(Subscript[V, stage]) = (Subscript[V, tot])/NS

Equilibrium constants of components A and B

(Subscript[K, A]) = 3;

(Subscript[K, B]) = 1.8;

Components balances for liquid phase combined with the linear equilibrium relationships.

eq[i_] := 
      D[(Subscript[CA, i])[t], {t, 1}] == 
        L ((Subscript[CA, i - 1])[t] - 
                (Subscript[CA, i])[t])/(ϵ (Subscript[V, stage]) + (1 - ϵ) (Subscript[V, stage]) (Subscript[K, A]))

eq2[i_] := 
      D[(Subscript[CB, i])[t], {t, 1}] == 
        L ((Subscript[CB, i - 1])[t] - 
                (Subscript[CB, i])[t])/(ϵ (Subscript[V, stage]) + (1 - ϵ) (Subscript[V, stage]) (Subscript[K, B]))

Solving the set of differential equations

eqn=Table[eq[i],{i,1,200}];

initialCond = Table[(Subscript[CA, i])[0] == 0, {i, 1, 200}];

sys=Flatten[Join[eqn,initialCond]];

Unknown = Table[(Subscript[CA, i]), {i, 1, 200}];

sol=NDSolve[sys,Unknown,{t,0,10000}];

eqn2=Table[eq2[i],{i,1,200}];

initialCond2 = Table[(Subscript[CB, i])[0] == 0, {i, 1, 200}];

sys2=Flatten[Join[eqn2,initialCond2]];

Unknown2 = Table[(Subscript[CB, i]), {i, 1, 200}];

sol2=NDSolve[sys2,Unknown2,{t,0,3000}];

Plot of the concentration profile of species A versus time at different positions of the column. As peak moves down the column, it gets widers and shorter.

Plot of the concentration profile of components A and B at the exit of the column versus time. Peaks are quite symetric because of the linear equilibrium assumption.

plt0 = 
      Plot[Evaluate[(Subscript[CA, 50])[t]] /. sol, {t, 0, 3000}, PlotRange -> All, 
        PlotStyle -> RGBColor[0, 0, 1], 
        DisplayFunction -> Identity]

plt1 = 
      Plot[Evaluate[(Subscript[CA, 100])[t]] /. sol, {t, 0, 3000}, PlotRange -> All, 
        PlotStyle -> RGBColor[1, 0, 0], 
        DisplayFunction -> Identity]

plt2 = 
      Plot[Evaluate[(Subscript[CA, 200])[t]] /. sol, {t, 0, 3000}, PlotRange -> All, 
        PlotStyle -> RGBColor[0, 1, 0], 
        DisplayFunction -> Identity]

Show[plt0,plt1,plt2,DisplayFunction->$DisplayFunction]

plt3 = 
      Plot[Evaluate[(Subscript[CB, 200])[t]] /. sol2, {t, 0, 3000}, PlotRange -> All, 
        PlotStyle -> RGBColor[1, 0, 1], 
        DisplayFunction -> Identity]

Show[plt2,plt3,DisplayFunction->$DisplayFunction]