Isobaric Vapor-Liquid Equilibrium Diagram of Ethanol-Ethyl Acetate using UNIQUAC Model​
​Authors: Housam Binous and Ahmed Bellagi

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Off[General::"spell1"]
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Clear[x,T,xe,Te,ye]
Vapor Pressure and Normal Boiling Points for Ethyl Acetate (component 1) and Ethanol (component 2)
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A1=88.376;B1=-7147.86;C1=0.0;D1=-10.9917;E1=8.5461310^-006;​​F1=2.0;
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Psat1=Exp[A1+B1/(T+C1)+D1Log[T]+E1T^F1]
Out[]=
88.376+8.54613×
-6
10
2.
T
-
7147.86
0.+T

10.9917
T
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FindRoot[Psat1==101.325,{T,250}]
Out[]=
{T350.199}
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A2=86.486;B2=-7931.1;C2=0.0;D2=-10.2498;E2=6.3894910^-006;​​F2=2.0;
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Psat2=Exp[A2+B2/(T+C2)+D2Log[T]+E2T^F2]
Out[]=
86.486+6.38949×
-6
10
2.
T
-
7931.1
0.+T

10.2498
T
In[]:=
FindRoot[Psat2==101.325,{T,250}]
Out[]=
{T351.322}
Activity coefficients using UNIQUAC model
Component 1 = Ethyl Acetate
Component 2 = Ethanol
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A[1,2]=-0.2733;A[2,1]=0.6541;B[1,2]=-159.1655;B[2,1]=-155.2346;
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A[1,1]=0;A[2,2]=0;B[1,1]=0;B[2,2]=0;
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tau[j_,i_]:=Exp[A[j,i]+B[j,i]/T]
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Gam1=Exp[Log[Gam1C]+Log[Gam1R]];
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Gam2=Exp[Log[Gam2C]+Log[Gam2R]];
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Gam1C=ExpLog[phi[1]/x[1]]+z2q[1]Log[v[1]/phi[1]]+l[1]-phi[1]x[1]
2
∑
j=1
x[j]l[j];
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Gam2C=ExpLog[phi[2]/x[2]]+z2q[2]Log[v[2]/phi[2]]+l[2]-phi[2]x[2]
2
∑
j=1
x[j]l[j];
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r[1]=3.47858;r[2]=2.10547;
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q[1]=3.11590;q[2]=1.97200;
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l[i_]:=z/2(r[i]-q[i])-(r[i]-1)
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z=10;
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Gam1R=Expq[1]1-Log
2
∑
j=1
v[j]tau[j,1]-
2
∑
j=1
v[j]tau[1,j]
2
∑
k=1
v[k]tau[k,j]
;
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Gam2R=Expq[2]1-Log
2
∑
j=1
v[j]tau[j,2]-
2
∑
j=1
v[j]tau[2,j]
2
∑
k=1
v[k]tau[k,j]
;
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phi[i_]:=r[i]x[i]
2
∑
j=1
r[j]x[j]
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v[i_]:=q[i]x[i]
2
∑
j=1
q[j]x[j]
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GAM1=Gam1/.x[1]x/.x[2]1-x;
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GAM2=Gam2/.x[1]x/.x[2]1-x;
VLE data computation using UNIQUAC Model
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i=0;
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P=101.325;
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While[i<101,{x=i0.01,​​sol=FindRoot[Psat1xGAM1+Psat2(1-x)GAM2P,{T,357,358}];xe[i]=x,ye[i]=Psat1xGAM1/P/.sol,​​Te[i]=T/.sol,Print[i],i++}]
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Comparing data obtained from UNIQUAC Model and data from Aspen-PLUS