BTCs for adsorption with longitudinal diffusion 
under equilibrium assumption
Author's Data: Housam BINOUS
Department of Chemical Engineering
National Institute of Applied Sciences and Technology
Tunis, TUNISIA
Email: binoushousam@yahoo.com

Acknowledgement: The expression for the analytical solution and the governing equations, for adsorption with longitudinal diffusion under equilibrium assumption, were obtained from Harrison at al., Bioseparations Science and Engineering, OUP, 2003 and Leon Lapidus and Neal R. Amundson, "Mathematics of Adsorption in Beds. VI. The Effect of Longitudinal Diffusion in Ion Exchange and Chromatographic Columns", J. Phys. Chem., vol. 56, p. 984, 1952.

Off[NDSolve::"mxsst"]

Off[NDSolve::"bcart"]

Numerical solution for the breakthrough curves for De=1. Results are identical to Harrison at al and Lapidus and Amundson.

Keq = 10; Ka = 1; V = 1; De = 1; α = 0.6;

n[z_, t_] = Keq c[z, t];

\(sol = 
      NDSolve[{De\ D[c[z, t], {z, 2}] == \ 
            V/α\ D[c[z, t], {z, 1}] + 
              D[c[z, t], {t, 1}] + \((1 - α)\)/α\ D[
                  n[z, t], {t, 1}], 
c[0, t] == 1, 
          c[z, 0] == Exp[\(-1000\)\ z]}, {c[z, t]}, {z, 0, 50}, {t, 0, 
          10000}\ , 
        Method -> {"MethodOfLines", \ "SpatialDiscretization" -> {"TensorProductGrid", \ MaxPoints -> 15}}]\)

plt1 = 
      Plot[Evaluate[c[z, t] /. sol /. z -> 50], {t, 0, 1000}, 
        PlotRange -> All, 
        PlotStyle -> RGBColor[0, 0, 1]]

H[t_] := 
      1/2 (1 + 
            Erf[(Sqrt[u^2 t/(4 γ De)]) - 
                Z (Sqrt[ γ/(4 t De)])] + 
            Exp[u Z/De] 
              Erfc[(Sqrt[u^2  t /(4 γ De)]) + 
                  Z (Sqrt[ γ/(4 t De)])])

γ = 1 + Keq (1 - α)/α; Z = 50;

u = V/α;

plt2 = 
      Plot[H[t], {t, 0, 1000}, PlotRange -> All, 
        PlotStyle -> RGBColor[1, 0, 0]]

Show[plt1, plt2, PlotRange -> {{0, 500}, {0, 1}}, 
      Frame -> True]

Numerical solution for the breakthrough curves for De=10. Results are identical to Harrison at al and Lapidus and Amundson.

Keq = 10; Ka = 1; V = 1; De = 10; α = 0.6;

\(sol = 
      NDSolve[{De\ D[c[z, t], {z, 2}] == \ 
            V/α\ D[c[z, t], {z, 1}] + 
              D[c[z, t], {t, 1}] + \((1 - α)\)/α\ D[
                  n[z, t], {t, 1}], 
c[0, t] == 1, 
          c[z, 0] == Exp[\(-1000\)\ z]}, {c[z, t]}, {z, 0, 50}, {t, 0, 
          10000}\ , 
        Method -> {"MethodOfLines", \ "SpatialDiscretization" -> {"TensorProductGrid", MaxPoints -> 15}}]\)

plt3 = 
      Plot[Evaluate[c[z, t] /. sol /. z -> 50], {t, 0, 1000}, 
        PlotRange -> All, 
        PlotStyle -> RGBColor[1, 0, 0]]

H[t_] := 
      1/2 (1 + 
            Erf[(Sqrt[u^2 t/(4 γ De)]) - 
                Z (Sqrt[ γ/(4 t De)])] + 
            Exp[u Z/De] 
              Erfc[(Sqrt[u^2  t /(4 γ De)]) + 
                  Z (Sqrt[ γ/(4 t De)])])

γ = 1 + Keq (1 - α)/α; Z = 50;

u = V/α;

plt4 = 
      Plot[H[t], {t, 0, 1000}, PlotRange -> All, 
        PlotStyle -> RGBColor[0, 0, 1]]

Show[plt3, plt4, PlotRange -> {{0, 1000}, {0, 1}}, 
      Frame -> True]

Effect of longitudinal diffusion on breakthrough curves for De=1 and 10.

Show[plt1, plt3, Frame -> True]