We compute the number of equilibrium stages using the Hunter and Nash graphical method of a liquid liquid extraction problem. One feed of acetic acid (35 wt %) in water which flow rate = 1000 kg/hr is extracted using a solvent consisting of pure isopropyl ether which flow rate = 1475 kg/h. The exiting raffinate stream specification is 10 wt % acetic acid.
Liquid Liquid Equilibrium of ternary system Isopropyl Ether-Water-Acetic Acid (data are given in wt. frac.).​​
​​Component 1 = Water Component 2 = Acetic AcidComponent 3 = Isopropyl Ether​ABREVIATIONS:Water = WAcetic Acid = AAIsopropyl Ether = IE
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[]:=
data1=Table[{tbl1[[i]][[1]],tbl1[[i]][[2]],100-tbl1[[i]][[1]]-tbl1[[i]][[2]]},{i,1,Length[tbl1]}]
Out[]=
{{98.1,0.69,1.21},{97.1,1.41,1.49},{95.5,2.89,1.61},{91.7,6.42,1.88},{84.4,13.3,2.3},{71.1,25.5,3.4},{58.9,36.7,4.4},{45.1,44.3,10.6},{37.1,46.4,16.5}}
In[]:=
data2=Table[{tbl2[[i]][[1]],tbl2[[i]][[2]],100-tbl2[[i]][[1]]-tbl2[[i]][[2]]},{i,1,Length[tbl2]}]
Out[]=
{{0.5,0.18,99.32},{0.7,0.37,98.93},{0.8,0.79,98.41},{1.,1.93,97.07},{1.9,4.82,93.28},{3.9,11.4,84.7},{6.9,21.6,71.5},{10.8,31.1,58.1},{15.1,36.2,48.7}}
In[]:=
Pt={{2,1}/2.,{0,Sqrt[3.]}/2};(*transformationfromrectangulartoequilateral*)
In[]:=
plt1=Show[TernaryListPlot[{data1,data2},JoinedTrue,PlotStyle{{Thick,Darker@Red},{Thick,Darker@Blue}},AspectRatio1,GridLinesAutomatic,Frame->True],Graphics[{Text[Style["W",14,Darker@Green],Pt.{0.5,-0.1}],Text[Style["IE",14,Darker@Green],Pt.{-0.1,0.5}],Text[Style["AA",14,Darker@Green],Pt.{0.6,0.5}]}]]
Out[]=
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}}/100
Out[]=
{{0.981,0.0069},{0.971,0.0141},{0.955,0.0289},{0.917,0.0642},{0.844,0.133},{0.711,0.255},{0.589,0.367},{0.451,0.443},{0.371,0.464}}
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}}/100
Out[]=
{{0.005,0.0018},{0.007,0.0037},{0.008,0.0079},{0.01,0.0193},{0.019,0.0482},{0.039,0.114},{0.069,0.216},{0.108,0.311},{0.151,0.362}}
In[]:=
rafnt=Evaluate[Interpolation[tbl1,InterpolationOrder1]]
Out[]=
InterpolatingFunction
Domain: {{0.371,0.981}}
Output: scalar

In[]:=
rafnt[0.90]
Out[]=
0.0802219
In[]:=
Extrc=Evaluate[Interpolation[tbl2,InterpolationOrder1]]
Out[]=
InterpolatingFunction
Domain: {{0.005,0.151}}
Output: scalar

Using interpolation to get tie lines
In[]:=
tbl3={{0.5,98.1},{0.7,97.1},{0.8,95.5},{1.0,91.7},{1.9,84.4},{3.9,71.1},{6.9,58.9},{10.8,45.1},{15.1,37.1}}/100;
In[]:=
tie1=Interpolation[tbl3,InterpolationOrder1]
Out[]=
InterpolatingFunction
Domain: {{0.005,0.151}}
Output: scalar

In[]:=
tbl4={{98.1,0.5},{97.1,0.7},{95.5,0.8},{91.7,1.0},{84.4,1.9},{71.1,3.9},{58.9,6.9},{45.1,10.8},{37.1,15.1}}/100;
In[]:=
tie2=Interpolation[tbl4,InterpolationOrder1]
Out[]=
InterpolatingFunction
Domain: {{0.371,0.981}}
Output: scalar

In[]:=
tbl5=Table[Line[{Pt.{i,Extrc[i]},Pt.{tie1[i],rafnt[tie1[i]]}}],{i,1/100,15/100,2/100}]
Out[]=
{Line[{{0.01965,0.0167143},{0.9491,0.0555988}}],Line[{{0.072195,0.0730839},{0.8709,0.173292}}],Line[{{0.1257,0.131116},{0.8143,0.256401}}],Line[{{0.179218,0.189171},{0.769936,0.319519}}],Line[{{0.223577,0.231362},{0.718654,0.353272}}],Line[{{0.266686,0.271388},{0.669267,0.384495}}],Line[{{0.298547,0.291931},{0.636942,0.392954}}],Line[{{0.330407,0.312474},{0.604616,0.401413}}]}
In[]:=
Show[plt1,Graphics[{Darker@Cyan,tbl5}]]
Out[]=
Feed and Solvent
In[]:=
F=1000.;S=1475.;xF[1]=0.65;xF[2]=0.35;xF[3]=0;​​xS[1]=0;xS[2]=0;xS[3]=1.00;
Mixing Point
In[]:=
sol1=Solve[{MF+S,MxM[1]FxF[1]+SxS[1],​​MxM[2]FxF[2]+SxS[2],MxM[3]FxF[3]+SxS[3]},{M,xM[1],xM[2],xM[3]}]
Solve
:Solve was unable to solve the system with inexact coefficients. The answer was obtained by solving a corresponding exact system and numericizing the result.
Out[]=
{{M2475.,xM[1]0.262626,xM[2]0.141414,xM[3]0.59596}}
In[]:=
xM[1]=sol1[[1,2,2]]
Out[]=
0.262626
In[]:=
xM[2]=sol1[[1,3,2]]
Out[]=
0.141414
In[]:=
xM[3]=sol1[[1,4,2]]
Out[]=
0.59596
In[]:=
M=sol1[[1,1,2]]
Out[]=
2475.
Raffinate Product is such that acetic acid wt % is 10
In[]:=
xR[2]=10/100.
Out[]=
0.1
In[]:=
xR[1]=FindRoot[rafnt[x]xR[2],{x,0.80}][[1,2]]
Out[]=
0.879015
In[]:=
xR[3]=1-xR[1]-xR[2]
Out[]=
0.0209855
Extract Product
In[]:=
sys={R+EPM,xR[1]R+xE[1]EPMxM[1],xR[2]R+xE[2]EPMxM[2],xR[3]R+xE[3]EPMxM[3],Extrc[xE[1]]xE[2]}
Out[]=
EP+R2475.,0.879015R+EPxE[1]650.,0.1R+EPxE[2]350.,0.0209855R+EPxE[3]1475.,InterpolatingFunction
Domain: {{0.005,0.151}}
Output: scalar
[xE[1]]xE[2]
In[]:=
sol2=Quiet@Quiet@FindRoot[sys,{EP,1800},{R,150},{xE[1],5/100},{xE[2],15/100},{xE[3],80/100}]​​
Out[]=
{EP1842.97,R632.033,xE[1]0.0512402,xE[2]0.155617,xE[3]0.793143}
In[]:=
EP=sol2[[1,2]]
Operating Point
Plotting Feed, Solvent, Extract, Raffinate and Operating Point
Stepping off Stages
Plotting the Hunter and Nash Diagram. A little less than 6 stages (5.35 equilibrium stages) are needed to achieve the raffinate product requirement of 10 wt % acetic acid composition. Tie lines are shown in dashed green. Operating lines are plotted in magenta. The difference point is indicated in orange.
Checking our results

References

◼
  • [1] H. Binous, Equilibrium Staged Separations using Matlab® and Mathematica®, Chemical Engineering Education, 42(2), pp. 69-73, 2008.
    • ◼
  • [2] H Binous, Liquid-liquid equilibrium and extraction using Mathematica, Computers in Education Journal, 16(1), pp. 78-81, 2006.
    • ◼
  • [3] P.C. Wankat, Equilibrium Staged Separations, Prentice Hall, Englewood Cliffs, N.J. (1988). (example 18-2, pp. 595).
    • ◼
  • [4] T. G. Hunter and A. W. Nash, J. Soc. Chem.Ind., (53) 95T-102T, 1934

    • CITE THIS NOTEBOOK

    Hunter and Nash Graphical Equilibrium Stage Method using Equilateral Diagrams​
    by Housam Binous and Ahmed Bellagi
    Wolfram Community, STAFF PICKS, February 5, 2024
    ​https://community.wolfram.com/groups/-/m/t/3116346