Multicomponent Distillation
Case of Pentane-Hexane-Heptane System
Author's Data: Housam BINOUS
Department of Chemical Engineering
National Institute of Applied Sciences and Technology
Tunis, TUNISIA
Email: binoushousam@yahoo.com

Pentane = component 1
Hexane = component 2
Heptane = component 3 

Relative volatilities: (Subscript[α, 13])=6.35, (Subscript[α, 23])=2.47

Feed composition: 30% Pentane, 30% Hexane, 40% Heptane

Feed is saturated liquid

Distillate composition: 95% Pentane, 4.9% Hexane, 0.1% Heptane

Bottom composition: 5% Pentane

Reference: Doherty and Malone, Conceptual Design of Distillation Systems, McGraw-Hill, 2001

Overall Material Balance gives Bottom composition:
5% Pentane, 39.7% Hexane, 55.3% Heptane

Solve[{F ==  D + B,  
        F 0.3 ==  
          D 0.95 + B 0.05, 
F 0.3 ==  
          D 0.049 +  B Hexane_Bottom,  
        D/F ==  (0.3 - 0.05)/(0.95 - 0.05)}, {F, B, D, 
        Hexane_Bottom}]

Calculation of the liquid composition profile in the Rectifying Section with 10 Stages. Reflux ratio is equal to 2.5.

r = 2.5;

d[1] = 0.95; d[2] = 0.049; d[3] = 0.001; Ne = 10;

xr[1, Ne] = 0.95; xr[2, Ne] = 0.049; xr[3, Ne] = 0.001;

R[i_, n_] :=  
      yr[i, n - 1] ==  
        r/(r + 1) xr[i, n] + d[i]/(r + 1)

a[1] = 6.35; a[2] = 2.47; a[3] = 1;

sol = 
        Solve[{xr[1, Ne] ==  
              a[1] xr[1, Ne - 1]/
                  Sum[a[j] xr[j, Ne - 1], {j, 1, 3}], 
xr[
                2, Ne] ==  
              a[2] xr[2, Ne - 1]/
                  Sum[a[j] xr[j, Ne - 1], {j, 1, 
                      3}], 
1 ==  
              Sum[xr[j, Ne - 1], {j, 1, 3}]}, {xr[1, Ne - 1], xr[2, Ne - 1], 
            xr[3, Ne - 1]}];

xr[1, Ne - 1] = sol[[1, 1, 2]];

xr[2, Ne - 1] = sol[[1, 2, 2]];

xr[3, Ne - 1] = sol[[1, 3, 2]];

Eq1[i_, n_] :=  
      xr[i, n] ==   yr[i, n]/a[i] 1/
            Sum[ yr[j, n]/a[j], {j, 1, 3}]

Sys1 = Table[R[i, n], {i, 1, 3}, {n, 1, Ne - 1}];

Sys2 = Table[Eq1[i, n], {i, 1, 3}, {n, 0, Ne - 2}];

System1 = Flatten[Join[Sys1, Sys2]];

unk1 = Table[xr[i, n], {i, 1, 3}, {n, 0, Ne - 2}];

unk2 = Table[yr[i, n], {i, 1, 3}, {n, 0, Ne - 2}];

Unknown1 = Flatten[Join[unk1, unk2]];

sol1 = NSolve[System1, Unknown1];

xr[1, 0] = sol1[[1, 1, 2]];

xr[2, 0] = sol1[[1, 2, 2]];

Do[xr[1, i] = sol1[[1, 3 +  i, 2]], {i, 1, 8}]

Do[xr[2, i] = sol1[[1, 11 +  i, 2]], {i, 1, 8}]

tbl = Table[{xr[1, i], xr[2, i]}, {i, 0, Ne - 1}];

plt1 = 
      ListPlot[tbl, PlotJoined -> True, 
        PlotStyle -> RGBColor[1, 0, 0], 
        PlotRange -> {{0, 1}, {0, 1}}, AspectRatio -> 1, 
        Epilog -> {RGBColor[0, 0, 0], 
            Line[{{1, 0}, {0, 1}}]}]

plt2 = 
      ListPlot[tbl, 
        PlotStyle -> {RGBColor[1, 0, 0], PointSize[0.02]}]

Show[plt1, plt2]

Calculation of the liquid composition profile in the Stripping Section with 10 Stages. Reboil ratio is eaqual to 1.35.

S[i_, n_] :=  
      ys[i, n] ==  (s + 1)/s xs[i, n + 1] - b[i]/s

b[1] = 0.05; b[2] = 0.397; b[3] = 0.553;

s = 1.35;

Do[{xs[i, 0] = b[i]}, {i, 1, 3}];

Sys3 = Table[S[i, n], {i, 1, 3}, {n, 0, Ne - 1}];

Eq2[i_, n_] :=  
      ys[i, n] ==   xs[i, n] a[i] 1/
            Sum[ xs[j, n] a[j], {j, 1, 3}]

Sys4 = Table[Eq2[i, n], {i, 1, 3}, {n, 0, Ne - 1}];

System2 = Flatten[Join[Sys3, Sys4]];

unk3 = Table[xs[i, n], {i, 1, 3}, {n, 1, Ne}];

unk4 = Table[ys[i, n], {i, 1, 3}, {n, 1, Ne}];

Unknown2 = Flatten[Join[unk3, unk4]];

sol2 = NSolve[System2, Unknown2];

xs[1, 10] = sol2[[1, 1, 2]];

xs[2, 10] = sol2[[1, 2, 2]];

xs[1, 1] = sol2[[1, 56, 2]];

xs[2, 1] = sol2[[1, 57, 2]];

Do[xs[1, -i + 9] = sol2[[1, 31 + 3 i, 2]], {i, 0, 
        7}]

Do[xs[2, -i + 9] = sol2[[1, 32 + 3 i, 2]], {i, 0, 
        7}]

tb2 = Table[{xs[1, i], xs[2, i]}, {i, 0, Ne}];

plt3 = 
      ListPlot[tb2, PlotJoined -> True, 
        PlotStyle -> RGBColor[0, 0, 1], 
        PlotRange -> {{0, 1}, {0, 1}}, AspectRatio -> 1, 
        Epilog -> {RGBColor[0, 0, 0], 
            Line[{{1, 0}, {0, 1}}]}]

plt4 = 
      ListPlot[tb2, 
        PlotStyle -> {RGBColor[0, 0, 1], PointSize[0.02]}]

Show[plt3, plt4]

Final composition profile of the ternary ideal mixture of pentane, hexane and heptane.

The number of theoretical stages required is obtained by looking at the intersection of the stripping and rectifying composition profiles. This number is equal to 8.3 stages (Stripping section: 5 stages and rectifying section 4.3 stages). Feed is introduced five satges from the bottom of the column.

Overall Material Balance is shown as green line joining bottom, distillate and feed.

Purple Line shows the total condenser (it is a tie line).

Show[plt1, plt2, plt3, plt4, 
      Graphics[{{PointSize[ .03], RGBColor[0, 1, 0], Point[{0.3, 0.3}]}, 
          PointSize[ .02], RGBColor[1, 0, 1], 
          Point[{xr[1, Ne], xr[2, Ne]}]}], 
      Epilog -> {{RGBColor[0, 1, 0], 
            Line[{{0.05, 0.397}, {0.95, 0.049}}]}, {RGBColor[1, 0, 1], 
            Line[{{xr[1, Ne - 1], xr[2, Ne - 1]}, {xr[1, Ne], 
                  xr[2, Ne]}}]}, 
{RGBColor[0, 0, 0], 
            Line[{{1, 0}, {0, 1}}]}}]