Flash Distillation of a Mixture of Hydrocarbons
Author's Data: Housam BINOUS
Department of Chemical Engineering
National Institute of Applied Sciences and Technology
Tunis, TUNISIA
Email: binoushousam@yahoo.com

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)).

p = 3.5 14.696

θ = 1.8 300

K1[1] = Exp[-970688.5625 1/θ^2 + 7.15059 - 
              0.76984 Log[p] + 6.90224/p] /. 
          p -> 51.436 /. θ -> 540.

K1[2] = Exp[-1280557/θ^2 + 7.94986 - 0.96455 Log[p]] /. 
          p -> 51.436 /. θ -> 540.

K1[3] = Exp[-1166846/θ^2 + 7.72668 - 0.92213 Log[p]] /. 
          p -> 51.436 /. θ -> 540.

K1[4] = Exp[-1524891/θ^2 + 7.33129 - 0.89143 Log[p]] /. 
          p -> 51.436 /. θ -> 540.

K1[5] = 
Exp[-1481583/θ^2 + 7.58071 - 
              0.93159 Log[p]] /. p -> 51.436 /. θ -> 
          540.

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.

second virial coefficients from Pitzer-Curl equations

T = 300;P = 3.5 760;  (* 
      pressure unit is mmHg and temperature unit is Kelvin *)

B0[i_] := 
      0.083 - 0.422 Tre[i]^(-1.6)
B1[i_] := 
      0.139 - 0.172 Tre[i]^(-4.2)
B[i_] := 
      R Tc[i]/Pc[i] (B0[i] + w[i] B1[i])
Tre[i_] := T/Tc[i]
R = 
        62.36 1000;
  
Tc[1] =  369.8;
Pc[1] =  42.5 760;
w[1] =  0.153;
Vc[1] =  203;
Zc[1] = 0.274;
  
Tc[2] = 425.2;
Pc[2] = 38 760;
w[2] =  0.199;
Vc[2] =  255;
Zc[2] = 
        0.283;
  
Tc[3] = 408.2;
Pc[3] = 36.5 760;
w[3] =  0.183;
Vc[3] =  263;
Zc[3] = 0.2;
  
Tc[4] = 469.7;
Pc[4] = 33.7 760;
w[4] =  0.251;
Vc[4] =  304;
Zc[4] = 
        0.262;
  
Tc[5] = 460.4;
Pc[5] = 33.9 760;
w[5] =  0.227;
Vc[5] =  306;
Zc[5] = 0.271;

Table[B[i], {i, 1, 5}]

Liquid Molar Volumes using Rackett equation

v[i_] := Vc[i] Zc[i]^((1 - Tre[i])^0.2857)

Table[v[i], {i, 1, 5}]

Vapor Pressures at 300K

Psat[i_] := 
      Exp[C1[i] + C2[i]/T + C3[i] Log[T] + 
            C4[i] T^C5[i]] 760/(1.01325 10^5)

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] = 1.0919 10^-5; C4[2] = 9.4509 10^-6; 
    C4[3] = 2.0063 10^-2; C4[4] = 9.6171 10^-6; 
    C4[5] = 8.9795 10^-6;
C5[1] = 2; C5[2] = 2; C5[3] = 1; C5[4] = 2; C5[5] = 2;

Table[Psat[i], {i, 1, 5}]

K2[i_] := 
      Psat[i]/P Exp[(v[i] - 
                B[i]) (P - Psat[i])/(R T)]

Comparing the equilibrium constants computed with both methods

Table[K2[i], {i, 1, 5}]

Table[K1[i], {i, 1, 5}]

Feed composition

z[1] = 0.2; z[2] = 0.3; z[3] = 0.4; z[4] = 0.05; 
    z[5] = 0.05;

Sloving the Rachford-Rice Equation

sol1=FindRoot[(∑_(i = 1)^5) (((K1[i] - 1) z[
                      i])/(1 + (K1[i] - 1) FV))==0,{FV, 0.6}]

FV = sol1[[1, 2]]

Computing Liquid and Vapor phase compositions

Eq[i_] := 
      z[i] ==  x[i] (1 - FV) + K1[i] x[i] FV

sol2 = 
      Solve[Table[Eq[i], {i, 1, 5}], Table[x[i], {i, 1, 5}]]

Do[x[i] = sol2[[1, i, 2]], {i, 1, 5}]

Do[y[i] = K1[i] x[i], {i, 1, 5}]

Table[y[i], {i, 1, 5}]

Checking our solution

(∑_(i = 1)^5) x[i]

(∑_(i = 1)^5) y[i]

(∑_(i = 1)^5) x[i]-(∑_(i = 1)^5) y[i]