Batch distillation of acetone-methanol mixture
​Authors: Housam Binous and Ahmed Bellagi

The kettle of a batch still, operating at 1 atm, is charged with a 50 % mol solution of acetone in methanol. We 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
In[]:=
A1=7.11714;B1=1210.595;C1=229.664;​​A2=8.08097;B2=1582.271;C2=239.726;
In[]:=
PS1=10^(A1-B1/(C1+T));​​PS2=10^(A2-B2/(C2+T));
Activity coefficients using the Van Laar model
In[]:=
G1=Exp[A12(A21(1-x)/(A12x+A21(1-x)))^2]​​G2=Exp[A21(A12x/(A12x+A21(1-x)))^2]
Out[]=
0.221688
2
(1-x)
2
(0.6184(1-x)+0.5797x)

Out[]=
0.207815
2
x
2
(0.6184(1-x)+0.5797x)

In[]:=
A21=0.6184;A12=0.5797;
Total Pressure
In[]:=
P=760;
Integration of Rayleigh's Equation
In[]:=
i=0;
In[]:=
While[i<31,{x=0.2+0.01i,​​sol=FindRoot[xPS1G1+(1-x)PS2G2P,{T,75}],​​T[i]=sol[[1,2]],xa[i]=x,ya[i]=PS1xG1/P/.TT[i]/.xxa[i],i++}]
In[]:=
tbl=Table[{xa[i],1/(ya[i]-xa[i])},{i,0,30}]
Out[]=
{{0.2,7.74431},{0.21,7.6595},{0.22,7.59416},{0.23,7.54643},{0.24,7.51481},{0.25,7.4981},{0.26,7.49534},{0.27,7.50576},{0.28,7.52879},{0.29,7.56399},{0.3,7.61104},{0.31,7.66978},{0.32,7.74013},{0.33,7.8221},{0.34,7.91583},{0.35,8.02153},{0.36,8.13951},{0.37,8.27019},{0.38,8.41407},{0.39,8.57176},{0.4,8.744},{0.41,8.93163},{0.42,9.13564},{0.43,9.35714},{0.44,9.59744},{0.45,9.85803},{0.46,10.1406},{0.47,10.4471},{0.48,10.7797},{0.49,11.141},{0.5,11.534}}
In[]:=
ListPlot[tbl,JoinedTrue,PlotStyleRGBColor[1,0,0]]
Out[]=
0.25
0.30
0.35
0.40
0.45
0.50
8
9
10
11
In[]:=
intg=Interpolation[tbl,InterpolationOrder1]
Out[]=
InterpolatingFunction
Domain: {{0.2,0.5}}
Output: scalar

Fraction remaining in the kettle as a function of mole fraction of acetone in the kettle
In[]:=
L1[x_]:=Exp[-Integrate[intg[t],{t,x,0.5}]]
In[]:=
plt1=Plot[L1[x],{x,0.2,0.5},PlotStyleRGBColor[1,0,0]]
Out[]=
0.25
0.30
0.35
0.40
0.45
0.50
0.2
0.4
0.6
0.8
1.0
Fraction remaining in the kettle when the mole fraction of acetone is equal to 20 % mol
In[]:=
x=.
In[]:=
Exp[-Integrate[intg[x],{x,0.2,0.5}]]
Out[]=
0.0774175
Fraction which must be distilled to get 0.2 mol of acetone in the residue
In[]:=
1-Exp[-Integrate[intg[x],{x,0.2,0.5}]]
Out[]=
0.922583
Second method to get the Fraction remaining in the kettle as a function of the mole fraction of acetone. This method uses NDSolve.
In[]:=
sol=NDSolve[{D[L[x],{x,1}]L[x]/((PS1G1/P-1)x)/.TT[x],​​D[T[x],{x,1}](10^5(-P+(xPS1G1+(1-x)PS2G2))/.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]
Out[]=
LInterpolatingFunction
Domain: {{0.2,0.5}}
Output: scalar
,TInterpolatingFunction
Domain: {{0.2,0.5}}
Output: scalar

In[]:=
plt2=Plot[Evaluate[L[x]]/.sol,{x,0.2,0.5},PlotStyleRGBColor[0,0,1]]
Out[]=
0.25
0.30
0.35
0.40
0.45
0.50
0.2
0.4
0.6
0.8
1.0
Fraction remaining in the kettle when the mole fraction of acetone is equal to 20 % mol
In[]:=
Evaluate[L[x]]/.sol/.x0.2
Out[]=
{0.0774501}
Comparing both methods
In[]:=
Show[plt2,plt1]
Out[]=
0.25
0.30
0.35
0.40
0.45
0.50
0.2
0.4
0.6
0.8
1.0