The notebook shows how one can compute the number of equilibrium stages of a distillation column for the separation of a binary mixture using the Ponchon Savarit Graphical Equilibrium Stage Method. The mixture is composed of hexane and octane at a total pressure of 14.696 psia. Feed, distillate and Bottom compositions are 0.5, 0.95 and 0.05, respectively. Feed quality is q = 0.41. We start by computing the enthalpy versus composition diagram. We show how one can plot tie lines and the conjugate line. Then, we compute the number of equilibrium stages at total reflux. The condenser is a partial condenser. 4 stages are required including the partial condenser and reboiler . In addition, we obtain the difference points and L/D minimum. We find L/D minimum = 1.03244 . Difference points when L/D is 1.5 times greater than L/D minimum are obtained. We step off stages in the stripping and rectifying sections to get the following results: the optimum feed stage is stage 3 and the total number of stages required including the partial condenser and reboiler is 7 We conclude by calculating the operating lines if we assume Equimolar Overflow and non Equimolar Overflow.
Suppression of warning messages
In[]:=
Off[FindRoot::lstol];
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Off[General::spell1];
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Off[InterpolatingFunction::dmval]
Vapor Pressures, Normal boiling points and total pressure
Component 1= Hexane
Component 2= Octane
Component 1= Hexane
Component 2= Octane
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A2=6.4141;B2=5947.491;C2=360.26;A1=6.039243;B1=5085.758;C1=382.794;
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PS1=440Exp[A1-B1/(C1+1.8T+32)];PS2=362.1Exp[A2-B2/(C2+1.8T+32)];
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FindRoot[PS114.696,{T,100}]
Out[]=
{T68.9114}
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FindRoot[PS214.696,{T,100}]
Out[]=
{T125.601}
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P=14.696;
Activity coefficients using the Van Larr model
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G1=Exp[A12(A21(1-x)/(A12x+A21(1-x)))^2]/.xx[i]G2=Exp[A21(A12x/(A12x+A21(1-x)))^2]/.xx[i]
Out[]=
A12
2
A21
2
(1-x[i])
2
(A21(1-x[i])+A12x[i])
Out[]=
2
A12
2
x[i]
2
(A21(1-x[i])+A12x[i])
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A21=-0.6731;A12=-0.20100;
Vapor-liquid Equilibrium curve
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h=0.1;i=0;
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Do[{x[i]=hi,T[i]=FindRoot[PPS1G1x[i]+PS2G2(1-x[i]),{T,80}][[1,2]],y[i]=PS1G1x[i]/P/.TT[i]},{i,0,10}]
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tbl1=Table[{x[i],y[i]},{i,0,10}]
Out[]=
{{0.,0.},{0.1,0.303725},{0.2,0.5116},{0.3,0.657371},{0.4,0.761918},{0.5,0.838306},{0.6,0.894805},{0.7,0.936666},{0.8,0.967135},{0.9,0.988008},{1.,1.}}
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plt1=ListPlot[tbl1,JoinedTrue,PlotStyleRGBColor[1,0,0],Epilog{RGBColor[0,1,0],Line[{{0,0},{1,1}}]}]
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Saturated liquid and saturated vapor Enthalpy. h-x and H-y curves.
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cl2[T_]:=2.8510^2;cl1[T_]:=2.2910^2;
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Do[hl[i]=0.23901(x[i]cl1[T[i]+273.15]+(1-x[i])cl2[T[i]+273.15])T[i],{i,0,10}]
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tbl2=Table[{x[i],hl[i]},{i,0,10}]
Out[]=
{{0.,8555.69},{0.1,7792.94},{0.2,7129.25},{0.3,6546.01},{0.4,6027.61},{0.5,5561.41},{0.6,5137.41},{0.7,4747.97},{0.8,4388.27},{0.9,4058.41},{1.,3771.75}}
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G1=Exp[A12(A21(1-x)/(A12x+A21(1-x)))^2]/.xxhG2=Exp[A21(A12x/(A12x+A21(1-x)))^2]/.xxh
Out[]=
-
0.0910658
2
(1-xh)
2
(-0.6731(1-xh)-0.201xh)
Out[]=
-
0.0271939
2
xh
2
(-0.6731(1-xh)-0.201xh)
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Do[{yh[i]=hi,Th[i]=FindRoot[{PPS1G1xh+PS2G2(1-xh),yh[i]PS1G1xh/P},{T,100},{xh,0.2}][[1,2]]},{i,0,10}]
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tbl3=Table[{yh[i],Th[i]},{i,0,10}]
Out[]=
{{0.,125.601},{0.1,122.905},{0.2,119.993},{0.3,116.821},{0.4,113.327},{0.5,109.426},{0.6,104.986},{0.7,99.7919},{0.8,93.4408},{0.9,84.9737},{1.,68.9114}}
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CG1[T_]:=-4.413+5.28010^-1T-3.11910^-4T^2+6.49410^-8T^3;CG2[T_]:=-6.096+7.71210^-1T-4.19510^-4T^2+8.85510^-8T^3;
In[]:=
dh1=2.8710^4((507.4-273.15)/(507.4-68.9-273.15))^0.38dh2=3.4410^4((568.8-273.15)/(568.8-125.6-273.15))^0.38
Out[]=
32761.7
Out[]=
42445.8
In[]:=
Do[hv[i]=0.23901(yh[i]dh1+(1-yh[i])dh2+(yh[i]CG1[Th[i]+273.15]+(1-yh[i])CG2[Th[i]+273.15])Th[i]),{i,0,10}]
In[]:=
tbl4=Table[{yh[i],hv[i]},{i,0,10}]
Out[]=
{{0.,17359.8},{0.1,16702.7},{0.2,16045.8},{0.3,15388.},{0.4,14727.6},{0.5,14062.6},{0.6,13389.},{0.7,12700.4},{0.8,11984.2},{0.9,11208.3},{1.,10174.1}}
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plt2=ListPlot[tbl2,JoinedTrue,PlotStyleRGBColor[1,0,0]]
Out[]=
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plt3=ListPlot[tbl4,JoinedTrue,PlotStyleRGBColor[0,0,1]]
Out[]=
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Show[plt3,plt2,PlotRange{0,18000},AspectRatio1.5]
Out[]=
h-x and H-y curves with tie lines
Feed, distillate and Bottom compositions are 0.5, 0.95 and 0.05, respectively. Feed quality is q = 0.41. Feed enthalpy is hf.
Plotting Feed Point on Enthalpy composition diagram
Interpolation to get tie lines
Computing the number of equilibrium stages at total reflux. The condenser is a partial condenser. 4 stages are required including the partial condenser and reboiler .
Computing difference points and L/D minimum.
We find L/D minimum = 1.03244.
We find L/D minimum = 1.03244.
Computing difference Points when L/D real is 1.5 times greater than L/D minimum.
Stepping off stages in the rectifying section. We change operating line when the equilibrium tie line crosses the line joining the difference points. The optimum feed stage is stage 3 from the top including the partial condenser. The condenser is a partial condenser.
Stepping off stages in the stripping section. We change operating line when the equilibrium tie line crosses the line joining the difference points. The optimum feed stage is stage 3. The condenser is a partial condenser. A total of 7 stages is required including the partial condenser and reboiler.
Plotting the McCabe and Thiele Diagram. Note that the operating lines are curved since we do not assume Equimolar Overflow. The feed line intersects with the operating lines approximately at x=0.345.
Computing the operating lines if we assume Equimolar Overflow. Comparing the operating lines with (pale blue) and without (dark blue) the Equimolar Overflow assumption.
CITE THIS NOTEBOOK
CITE THIS NOTEBOOK
Ponchon Savarit graphical equilibrium stage method
by Housam Binous and Ahmed Bellagi
Wolfram Community, STAFF PICKS, April 13 2023
https://community.wolfram.com/groups/-/m/t/2897399
by Housam Binous and Ahmed Bellagi
Wolfram Community, STAFF PICKS, April 13 2023
https://community.wolfram.com/groups/-/m/t/2897399