Batch distillation of acetone-methanol mixture
Author's Data: Housam BINOUS
Department of Chemical Engineering
National Institute of Applied Sciences and Technology
Tunis, TUNISIA
Email: binoushousam@yahoo.com

The kettle of a batch still, operating at 1 atm, is charged with a 50 % mol solution of acetone in methanol. We would like to compute the fraction of the initial charge that must be distilled to reduce the composition of the solution remaining in the kettle to 20 % mol of acetone. Two methods are proposed, one using FindRoot and the other using NDSolve.

Vapor Pressure using Antoine Equation

A1 = 7.11714; B1 = 1210.595; C1 = 229.664;
A2 = 8.08097; B2 = 1582.271; C2 = 239.726;

PS1 = 10^(A1 - B1/(C1 + T));
PS2 = 10^(A2 - B2/(C2 + T));

Activity coefficients using the Van Laar model

G1 = 
      Exp[A12 (A21 (1 - x)/(A12 x + 
                      A21 (1 - x)))^2]
G2 = 
      Exp[A21 (A12 x/(A12 x + A21 (1 - x)))^2]

A21 = 0.6184; A12 = 0.5797;

Total Pressure

P = 760;

Integration of Rayleigh's Equation

i = 0;

While[
      i < 31, {x = 0.2 + 0.01 i, 
sol = 
          FindRoot[
            x PS1 G1 + (1 - x) PS2 G2 ==  P, {T, 
              75}], 
T[i] = sol[[1, 2]], xa[i] = x, 
        ya[i] = PS1 x G1/P /. T -> T[i] /. 
            x -> xa[i], i++}]

tbl = Table[{xa[i], 1/(ya[i] - xa[i])}, {i, 0, 30}]

ListPlot[tbl, PlotJoined -> True, 
      PlotStyle -> RGBColor[1, 0, 0]]

intg=Interpolation[tbl,InterpolationOrder -> 1]

Fraction remaining in the kettle as a function of mole fraction of acetone in the kettle

L1[x_] := Exp[-Integrate[intg[t], {t, x, 0.5}]]

plt1 = 
      Plot[L1[x], {x, 0.2, 0.5}, 
        PlotStyle -> RGBColor[1, 0, 0]]

Fraction remaining in the kettle when the mole fraction of acetone is equal to 20 % mol

x =.

Exp[-Integrate[intg[x], {x, 0.2, 0.5}]]

Fraction which must be distilled to get 0.2 mol of acetone in the residue

1 - Exp[-Integrate[intg[x], {x, 0.2, 0.5}]]

Second method to get the Fraction remaining in the kettle as a function of the mole fraction of acetone. This method uses NDSolve.

sol=NDSolve[{D[L[x], {x, 1}] ==  
                L[x]/((PS1 G1/P  - 1) x) /. 
              T -> T[x], 
D[
                T[x], {x, 
                  1}] ==  (10^5 (-P + (x PS1 G1 + (1 - 
                                  x) PS2 G2)) /. T -> T[x]), 
            L[0.5] == 1, 
T[0.5] ==  59.7},{L, T},{x, 0.2, 0.5},AccuracyGoal->Infinity,MaxSteps->Infinity,StartingStepSize -> 0.0000001, MaxStepSize ->  0.00001]

plt2 = 
      Plot[Evaluate[L[x]] /. sol, {x, 0.2, 0.5}, 
        PlotStyle -> RGBColor[0, 0, 1]]

Fraction remaining in the kettle when the mole fraction of acetone is equal to 20 % mol

Evaluate[L[x]] /. sol /. x -> 0.2

Comparing both methods

Show[plt2, plt1]