Flash Distillation of a Mixture of Hydrocarbons
Authors: Housam Binous and Ahmed Bellagi
Flash Distillation of a Mixture of Hydrocarbons
Authors: Housam Binous and Ahmed Bellagi
Authors: Housam Binous and Ahmed Bellagi
We use two methods to compute the equilibrium constants at T = 300 K and P = 3.5 bars for a mixture of propane, n - butane, iso - butane, iso - pentane and n - pentane. Then, the Rachford - Rice Equation is solved to compute the fraction vaporized when the feed is 20 mole % propane, 30 mole % n - butane, 40 mole % iso - butane, 5 mole % n - pentane and 5 mole % iso - pentane . Finally, both liquid and vapor phase compositions are calculated .
Computing Equilibrium Constants for propane, n-butane, iso-butane, n-pentane and iso-pentane at P=3.5 bars and T=300 K.
Component 1: propane
Component 2: n-butane
Component 3: iso-butane
Component 4: n-pentane
Component 5: iso-pentane
Computing Equilibrium Constants for propane, n-butane, iso-butane, n-pentane and iso-pentane at P=3.5 bars and T=300 K.
Component 1: propane
Component 2: n-butane
Component 3: iso-butane
Component 4: n-pentane
Component 5: iso-pentane
Equilibrium constants using the equations that fit the DePriester Charts (McWilliams, Chem. Eng., 80 (25), 138 (1973)).
In[]:=
p=3.5×14.696(*pressureunitispsi*)
Out[]=
51.436
In[]:=
θ=1.8×300(*temperatureunitisRankinedegree*)
Out[]=
540.
In[]:=
K1[1]=Exp[-970688.56251/θ^2+7.15059-0.76984Log[p]+6.90224/p]/.p51.436/.θ540.
Out[]=
2.51561
In[]:=
K1[2]=Exp[-1280557/θ^2+7.94986-0.96455Log[p]]/.p51.436/.θ540.
Out[]=
0.784838
In[]:=
K1[3]=Exp[-1166846/θ^2+7.72668-0.92213Log[p]]/.p51.436/.θ540.
Out[]=
1.09598
In[]:=
K1[4]=Exp[-1524891/θ^2+7.33129-0.89143Log[p]]/.p51.436/.θ540.
Out[]=
0.243987
In[]:=
K1[5]=Exp[-1481583/θ^2+7.58071-0.93159Log[p]]/.p51.436/.θ540.
Out[]=
0.310074
Equilibrium constants computation with pressure effects included by
using the Poynting correction factor and the fugacity coefficient in the vapor phase. The fugacity coefficient in the vapor phase is computed with the second virial coefficient.
using the Poynting correction factor and the fugacity coefficient in the vapor phase. The fugacity coefficient in the vapor phase is computed with the second virial coefficient.
second virial coefficients from Pitzer-Curl equations
In[]:=
T=300;P=3.5×760;(*pressureunitismmHgandtemperatureunitisKelvin*)
In[]:=
B0[i_]:=0.083-0.422Tre[i]^(-1.6)B1[i_]:=0.139-0.172Tre[i]^(-4.2)B[i_]:=RTc[i]/Pc[i](B0[i]+w[i]B1[i])Tre[i_]:=T/Tc[i]R=62.36×1000;(*propane*)Tc[1]=369.8;Pc[1]=42.5×760;w[1]=0.153;Vc[1]=203;Zc[1]=0.274;(*n-butane*)Tc[2]=425.2;Pc[2]=38×760;w[2]=0.199;Vc[2]=255;Zc[2]=0.283;(*isobutane*)Tc[3]=408.2;Pc[3]=36.5×760;w[3]=0.183;Vc[3]=263;Zc[3]=0.2;(*n-pentane*)Tc[4]=469.7;Pc[4]=33.7×760;w[4]=0.251;Vc[4]=304;Zc[4]=0.262;(*isopentane*)Tc[5]=460.4;Pc[5]=33.9×760;w[5]=0.227;Vc[5]=306;Zc[5]=0.271;
In[]:=
Table[B[i],{i,1,5}]
Out[]=
{-391.842,-711.347,-639.641,-1178.5,-1068.45}
Liquid Molar Volumes using Rackett equation
In[]:=
v[i_]:=Vc[i]Zc[i]^((1-Tre[i])^0.2857)
In[]:=
Table[v[i],{i,1,5}]
Out[]=
{90.848,104.702,87.4267,111.682,116.461}
Vapor Pressures at 300K
In[]:=
Psat[i_]:=Exp[C1[i]+C2[i]/T+C3[i]Log[T]+C4[i]T^C5[i]]760/(1.0132510^5)
In[]:=
C1[1]=59.078;C1[2]=66.343;C1[3]=100.18;C1[4]=78.741;C1[5]=72.35;C2[1]=-3492.6;C2[2]=-4363.2;C2[3]=-4841.9;C2[4]=-5420.3;C2[5]=-5010.9;C3[1]=-6.0669;C3[2]=-7.046;C3[3]=-13.541;C3[4]=-8.8253;C3[5]=-7.883;C4[1]=1.091910^-5;C4[2]=9.450910^-6;C4[3]=2.006310^-2;C4[4]=9.617110^-6;C4[5]=8.979510^-6;C5[1]=2;C5[2]=2;C5[3]=1;C5[4]=2;C5[5]=2;
In[]:=
Table[Psat[i],{i,1,5}]
Out[]=
{7492.53,1935.19,2783.72,549.226,734.704}
In[]:=
K2[i_]:=Psat[i]/PExp[(v[i]-B[i])(P-Psat[i])/(RT)]
Comparing the equilibrium constants computed with both methods
In[]:=
Table[K2[i],{i,1,5}]
Out[]=
{2.48655,0.750883,1.04149,0.23883,0.312025}
In[]:=
Table[K1[i],{i,1,5}]
Out[]=
{2.51561,0.784838,1.09598,0.243987,0.310074}
Feed composition
In[]:=
z[1]=0.2;z[2]=0.3;z[3]=0.4;z[4]=0.05;z[5]=0.05;
Solving the Rachford-Rice Equation
In[]:=
sol1=FindRoot0,{FV,0.6}
5
∑
i=1
(K1[i]-1)z[i]
1+(K1[i]-1)FV
Out[]=
{FV0.577085}
In[]:=
FV=sol1[[1,2]]
Out[]=
0.577085
Computing Liquid and Vapor phase compositions
In[]:=
Eq[i_]:=z[i]x[i](1-FV)+K1[i]x[i]FV
In[]:=
sol2=Solve[Table[Eq[i],{i,1,5}],Table[x[i],{i,1,5}]]
Out[]=
{{x[1]0.106687,x[2]0.342531,x[3]0.379008,x[4]0.0886971,x[5]0.0830766}}
In[]:=
Do[x[i]=sol2[[1,i,2]],{i,1,5}]
In[]:=
Do[y[i]=K1[i]x[i],{i,1,5}]
In[]:=
Table[y[i],{i,1,5}]
Out[]=
{0.268384,0.268831,0.415384,0.021641,0.0257599}
Checking our solution
In[]:=
5
∑
i=1
Out[]=
1.
In[]:=
5
∑
i=1
Out[]=
1.
In[]:=
5
∑
i=1
5
∑
i=1
Out[]=
0.