Two Feed Extraction Column
​Authors: Housam Binous and Ahmed Bellagi

We use two feeds. The first feed which flow rate is 600 Kg/hr consists of acetic acid (35 wt %) in water. The second feed which flow rate is 400 Kg/hr is also a mixture of acetic acid (20 wt %) and water (80 wt %). Optimum feed location for the second feed is computed as well as the number of equilibrium stages needed to achieve an outlet raffinate with 10 % wt acetic acid.
​
Reference:
H. Binous, Equilibrium Staged Separations using Matlab® and Mathematica®, Chemical Engineering Education, 42(2), pp. 69-73, 2008.
​
Liquid Liquid Equilibrium of ternary system Isopropyl ether-water-isopropyl alcohol.
Component 1 = water
Component 2= Acetic Acid
Component 3=Isopropyl ether
In[]:=
tbl1={{98.1,0.69},{97.1,1.41},{95.5,2.89},{91.7,6.42},{84.4,13.30},{71.1,25.50},{58.9,36.70},{45.1,44.30},{37.1,46.40}};
In[]:=
tbl2={{0.5,0.18},{0.7,0.37},{0.8,0.79},{1.0,1.93},{1.9,4.82},{3.9,11.40},{6.9,21.60},{10.8,31.10},{15.1,36.20}};
In[]:=
plt1=ListPlot[tbl1,JoinedTrue,PlotStyleRGBColor[1,0,0],AspectRatio->1]
Out[]=
40
50
60
70
80
90
10
20
30
40
In[]:=
plt2=ListPlot[tbl2,JoinedTrue,PlotStyleRGBColor[0,0,1],AspectRatio->1]
Out[]=
2
4
6
8
10
12
14
5
10
15
20
25
30
35
In[]:=
Show[plt1,plt2,PlotRange{{0,100},{0,100}},EpilogLine[{{100,0},{0,100}}],AspectRatio->1,AxesOrigin->{0,0}]
Out[]=
20
40
60
80
100
20
40
60
80
100
In[]:=
rafnt=Interpolation[tbl1,InterpolationOrder1]
Out[]=
InterpolatingFunction
Domain: {{37.1,98.1}}
Output: scalar

In[]:=
rafnt[90]
Out[]=
8.02219
In[]:=
Extrc=Interpolation[tbl2,InterpolationOrder1]
Out[]=
InterpolatingFunction
Domain: {{0.5,15.1}}
Output: scalar

In[]:=
plt3=Plot[rafnt[x],{x,37.1,98.1},PlotRange{0,100},PlotStyleRGBColor[1,0,0],AspectRatio->1]
Out[]=
40
50
60
70
80
90
20
40
60
80
100
In[]:=
plt4=Plot[Extrc[x],{x,0.5,15.1},PlotRange{0,100},PlotStyleRGBColor[0,0,1],AspectRatio->1]
Out[]=
0
2
4
6
8
10
12
14
20
40
60
80
100
In[]:=
Show[plt3,plt4,PlotRange{{0,100},{0,100}},EpilogLine[{{100,0},{0,100}}],AspectRatio->1,AxesOrigin->{0,0}]
Out[]=
0
20
40
60
80
100
20
40
60
80
100
Using interpolation to get tie lines
Two Feeds and Solvent
Total Feed
Mixing Point
Raffinate Product is such that acetic acid wt % is 10
Extract Product
Two Operating Points
Plotting Feed, Solvent, Extract, Raffinate and Operating Point
Intersection between Line joining Operating Points and Extract Part of Equilibrium curve
Stepping off Stages
Plotting the Hunter and Nash Diagram. 2.89 equilibrium stages are needed to achieve the raffinate product requirement of 10 wt % acetic acid composition.
Checking our results