Phase Envelope for Ethane-Benzene Mixture ​
​Authors: Housam Binous and Ahmed Bellagi

We compute phase envelopes for two binary mixtures containing ethane and benzene (20 % ethane and 80 % benzene; 70 % ethane and 30 % benzene) . This binary system exhibits a peculiar behavior called retrograde condensation (e . g . We get condensation as we increase the temperature or decrease the pressure) . We use the SRK equation of state to compute equilibrium constants . The critical points are indicated and were obtained from HYSYS 3.2 . The computed phase envelopes are similar to what is obtained using HYSYS 3.2 .
Reference: Nasri and H . Binous, Applications of the Soave - Redlich - Kwong Equation of State using Mathematica®, Journal of Chemical Engineering of Japan, 40 (6), pp . 534 - 538, 2007.
Acknowledgement :
This problem was presented in Thermodynamique, Application au Génie Chimique et à l'Industrie Pétrolière by Jean Vidal, Technip, 1997.
In[]:=
Off[General::"spell1"]
In[]:=
Off[FindRoot::"lstol"]
In[]:=
Off[Unset::"norep"]
Reduced Pressure and Temperature
In[]:=
Tre[i_]:=T/Tc[i]
In[]:=
Pre[i_]:=P/Pc[i]
Equilibrium constants using the ϕ-ϕ method and the Soave-Redlich-Kwong Equation of State
In[]:=
Ap[i_]:=0.42747a[i]Pre[i]/Tre[i]^2
In[]:=
Bp[i_]:=0.08664Pre[i]/Tre[i]
In[]:=
ϕ
v
[i_]:=Exp[(
Z
v
-1)Bp[i]/
B
v
-Log[
Z
v
-
B
v
]-
A
v
/
B
v
(2Ap[i]^0.5/
A
v
^0.5-Bp[i]/
B
v
)Log[(
Z
v
+
B
v
)/
Z
v
]];
In[]:=
ϕ
l
[i_]:=Exp[(
Z
l
-1)Bp[i]/
B
l
-Log[
Z
l
-
B
l
]-
A
l
/
B
l
(2Ap[i]^0.5/
A
l
^0.5-Bp[i]/
B
l
)Log[(
Z
l
+
B
l
)/
Z
l
]];
In[]:=
a[i_]:=(1+m[i](1-Tre[i]^0.5))^2
In[]:=
m[i_]:=0.480+1.574ω[i]-0.176ω[i]^2
In[]:=
A
v
=
2
∑
i=1
2
∑
j=1
y[i]y[j]Ab[i,j];
In[]:=
Ab[i_,j_]=(Ap[i]Ap[j])^0.5;
In[]:=
B
v
=
2
∑
i=1
y[i]Bp[i];
In[]:=
A
l
=
2
∑
i=1
2
∑
j=1
x[i]x[j]Ab[i,j];
In[]:=
B
l
=
2
∑
i=1
x[i]Bp[i];
In[]:=
K[i_]:=
ϕ
l
[i]/
ϕ
v
[i];
Critical Pressure and Temperature and acentric factor for Ethane and Benzene
In[]:=
Pc[1]=709.8;Tc[1]=550.0;ω[1]=0.1064;
In[]:=
Pc[2]=714.2;Tc[2]=1012.7;ω[2]=0.2116;
Phase Envelope for Mixture 1: 20% Ethane and 80% Benzene
In[]:=
P=5×14.69;
In[]:=
x[1]=0.2;x[2]=1-x[1];
In[]:=
y[1]=.;y[2]=.;
In[]:=
T=.
In[]:=
sol[1]=FindRoot[{y[1]K[1]x[1],y[2]K[2]x[2],y[1]+y[2]1,
Z
l
^3-
Z
l
^2+
Z
l
(
A
l
-
B
l
-
B
l
^2)-
A
l
B
l
0,
Z
v
^3-
Z
v
^2+
Z
v
(
A
v
-
B
v
-
B
v
^2)-
A
v
B
v
0},{y[1],0.005},{y[2],0.995},{T,400},{
Z
l
,0.05},{
Z
v
,0.9},MaxIterations1000000]//Chop
Out[]=
{y[1]0.994927,y[2]0.00507289,T480.637,
Z
l
0.0201611,
Z
v
0.944335}
In[]:=
Te[1]=T/.sol[1];Pe[1]=5×14.69;
In[]:=
i=.
In[]:=
For[i=2,i<64,P=(4+i)14.69;x[1]=0.2;x[2]=1-x[1];sol[i]=FindRoot[{y[1]K[1]x[1],y[2]K[2]x[2],y[1]+y[2]1,
Z
l
^3-
Z
l
^2+
Z
l
(
A
l
-
B
l
-
B
l
^2)-
A
l
B
l
0,
Z
v
^3-
Z
v
^2+
Z
v
(
A
v
-
B
v
-
B
v
^2)-
A
v
B
v
0},{y[1],y[1]/.sol[i-1]},{y[2],y[2]/.sol[i-1]},{T,800},{
Z
l
,
Z
l
/.sol[i-1]},{
Z
v
,
Z
v
/.sol[i-1]},MaxIterations1000000]//Chop;Te[i]=T/.sol[i];Pe[i]=P;Print[i," ",Te[i]," ",Pe[i]," ",
Z
l
/.sol[i]," ",
Z
v
/.sol[i]];i++]
2 495.717 88.14 0.0236746 0.938399
3 509.431 102.83 0.0271119 0.933073
4 522.114 117.52 0.0304855 0.928234
5 533.992 132.21 0.0338051 0.923789
6 545.223 146.9 0.0370784 0.919669
7 555.921 161.59 0.0403118 0.915814
8 566.174 176.28 0.0435109 0.912177
9 576.05 190.97 0.0466803 0.908717
10 585.6 205.66 0.0498245 0.905399
11 594.868 220.35 0.0529472 0.902194
12 603.888 235.04 0.0560522 0.899075
13 612.689 249.73 0.0591426 0.896016
14 621.293 264.42 0.0622219 0.892999
15 629.722 279.11 0.065293 0.890002
16 637.991 293.8 0.0683588 0.887008
17 646.115 308.49 0.0714223 0.884001
18 654.106 323.18 0.0744864 0.880967
19 661.973 337.87 0.0775537 0.877891
20 669.727 352.56 0.0806272 0.87476
21 677.375 367.25 0.0837095 0.871564
22 684.923 381.94 0.0868036 0.86829
23 692.378 396.63 0.0899123 0.864929
24 699.744 411.32 0.0930383 0.86147
25 707.026 426.01 0.0961848 0.857904
26 714.228 440.7 0.0993546 0.854222
27 721.355 455.39 0.102551 0.850416
28 728.408 470.08 0.105777 0.846477
29 735.392 484.77 0.109036 0.842398
30 742.308 499.46 0.112331 0.83817
31 749.159 514.15 0.115666 0.833787
32 755.949 528.84 0.119045 0.82924
33 762.678 543.53 0.122471 0.824522
34 769.35 558.22 0.12595 0.819626
35 775.966 572.91 0.129484 0.814544
36 782.529 587.6 0.13308 0.809269
37 789.041 602.29 0.136742 0.803791
38 795.504 616.98 0.140475 0.798103
39 801.92 631.67 0.144286 0.792196
40 808.292 646.36 0.148182 0.786059
41 814.623 661.05 0.152169 0.779683
42 820.915 675.74 0.156255 0.773056
43 827.171 690.43 0.160449 0.766166
44 833.395 705.12 0.164762 0.758998
45 839.589 719.81 0.169204 0.751536
46 845.759 734.5 0.173788 0.743764
47 851.909 749.19 0.178529 0.735658
48 858.044 763.88 0.183444 0.727197
49 864.169 778.57 0.188553 0.718351
50 870.292 793.26 0.193878 0.709087
51 876.421 807.95 0.19945 0.699363
52 882.567 822.64 0.205302 0.689131
53 888.74 837.33 0.211478 0.678327
54 894.957 852.02 0.218032 0.666872
55 901.238 866.71 0.225037 0.654662
56 907.608 881.4 0.232589 0.641556
57 914.107 896.09 0.240825 0.627357
58 920.788 910.78 0.249945 0.61177
59 927.741 925.47 0.260274 0.594324
60 935.124 940.16 0.272394 0.574173
61 943.287 954.85 0.287585 0.549459
62 953.448 969.54 0.310282 0.513758
63 966.368 984.23 0.347619 0.484124
In[]:=
tbl1=Append[Table[{((Te[i]-459.67)-32)/1.8+273.15,Pe[i]/14.69},{i,1,64}],{((513)-32)/1.8+273.15,968/14.69}];
In[]:=
plt1=ListPlot[tbl1,JoinedTrue,PlotStyleRGBColor[1,0,0],PlotRange{{200,600},{0,100}}]
Out[]=
200
300
400
500
600
0
20
40
60
80
100
In[]:=
P=5×14.69;
In[]:=
y[1]=0.2;y[2]=1-x[1];
Phase Envelope for Mixture 2: 70% Ethane and 30% Benzene
Plotting both envelopes together. This mixture present a peculiar phenomenon called retrograde condensation.