Design of a two adiabatic packed bed system
Author's Data: Housam BINOUS
Department of Chemical Engineering
National Institute of Applied Sciences and Technology
Tunis, TUNISIA
Email: binoushousam@yahoo.com

Acknowledgement :
This problem was solved graphically by O. Levenspiel in page 440 of Chemical Reaction Engineering, 3rd Edition, Wiley, 1999.

Off[General::spell1]

Off[InterpolatingFunction::dmval]

R=8.314;

Exothermic Reversible first order reaction

Rate constants

(Subscript[k, 1]) = Exp[7 - 83700/(R T)];

(Subscript[k, 2]) = Exp[18 - 167400/(R T)];

Initial concentration of component A. 
Pressure is 2.494 MPa.

(Subscript[C, 0]) = (2.494 10^6)/(R T);

Heat of reaction

ΔH = -80000;

Rate of reaction

Rate = (Subscript[k, 1]) (Subscript[C, 0]) (1 - X) - (Subscript[k, 2]) (Subscript[C, 0]) X;

Equilibrium conversion versus temperature ( case when reaction rate equals zero ).

sol=Solve[Rate==0,X];

plt=Plot[X/.sol,{T,200,1200},PlotStyle->{Thickness[0.015],RGBColor[0,0,0]},PlotRange->{{200,1200},{0,1}},DisplayFunction ->Identity]

Conversion versus temperature for different values of the reaction rate.

j=0.01;

For[i=1,i<100,i+=1,j+=0.1;Sol[i]=Solve[Log[10,Rate]==-j,X]]

tbl=Table[X/.Sol[i],{i,1,99}];

For[i=1,i<100,i+=5,Plt[i]=Plot[tbl[[i]],{T,200,1200},PlotRange->{0,1},DisplayFunction -> Identity,PlotStyle->Hue[i/100]]]

We get the same result as figure 19.11 page 437 of O. Levenspiel, Chemical Reaction Engineering, 3rd Edition, Wiley, 1999.

pl=Show[plt,Flatten[Table[Plt[i],{i,1,99,5}]],PlotRange->{0,1},DisplayFunction ->$DisplayFunction]

Finding the locus of optimum curve

Opt[i_]:=FindRoot[Flatten[D[X/.Sol[i],{T,1}]]==0,{T,250,1700}]

tbl1=Table[T/.Opt[i],{i,1,99}];

tbl2=Flatten[Table[X/.Sol[i]/.Opt[i],{i,1,99}]];

tbl3=Transpose[{tbl1,tbl2}];

pl1=ListPlot[tbl3,PlotStyle->{Thickness[0.015],RGBColor[0,0,1]},PlotJoined->True,DisplayFunction->Identity]

Show[pl,pl1,PlotRange->{{400,1100},{0,1}},DisplayFunction ->$DisplayFunction]

Interpolation of the optimum conversion versus temperature data

tbl4=Transpose[{tbl2,tbl1}];

rec=Interpolation[tbl4];

Computing optimum temperatures and rates for different conversions

Tbl={rec[0.46],rec[0.58],rec[0.66],rec[0.715],rec[0.785],rec[0.85]};

\(rate[X_, T_] := 
      299975.944190522`\/T\ \ \((E\^\(7 - 10067.356266538369`\/T\)\ \((1 - 
                  X)\) - E\^\(18 - 20134.712533076738`\/T\)\ X)\)\)

rat=Thread[rate[{0.46,0.58,0.66,0.715,0.785,0.85},Tbl]];

flb=Transpose[{{0.46,0.58,0.66,0.715,0.785,0.85},1/rat}];

Tabulation of 1/(-(Subscript[(((Subscript[(r'), A]))), opt])) for different values of (Subscript[X, A])

Join[{{(Subscript[X, A]), 1/(Subscript[((-(Subscript[(r'), A]))), opt])}}, flb] // 
      MatrixForm

Interpolation of 1/(-(Subscript[(((Subscript[(r'), A]))), opt])) versus (Subscript[X, A]) data

InvReactRate=InvReactRate=Interpolation[flb,InterpolationOrder -> 3];

Use of the method of maximization of rectangles to minimize the amount of catalyst needed in the two adiabatic packed bed system with very large recycle ratio (R=∞). We obtain figure E19.2b page 442 of O. Levenspiel, Chemical Reaction Engineering, 3rd Edition, Wiley, 1999.

PLT2=Plot[InvReactRate[x],{x,0.5,1},PlotRange->{{0,1},{0,30}},DisplayFunction->Identity]

sol2=FindRoot[InvReactRate'[ξ]==-(20.59-InvReactRate[ξ])/(0-ξ),{ξ,5.5}]

y=(InvReactRate'[ξ]/.sol2) (x-ξ/.sol2)+InvReactRate[ξ]/.sol2

plt8=Plot[y,{x,0,1},PlotStyle->RGBColor[0,1,1],DisplayFunction->Identity]

<< Graphics`FilledPlot`

plt7=FilledPlot[{InvReactRate[ξ]/.sol2},{x,0,ξ}/.sol2,PlotRange->{{0,1},{0,25}},Fills->RGBColor[0.8,0.2,0.8],
 Curves -> Front,DisplayFunction->Identity]

plt6=FilledPlot[{InvReactRate[0.85]},{x,ξ,0.85}/.sol2,PlotRange->{{0,1},{0,25}},Fills->{GrayLevel[.9]},
 Curves -> Front,DisplayFunction->Identity]

Show[plt6,plt7,plt8,PLT2,Graphics[{RGBColor[0,1,1],Line[{{ξ,20.591}/.sol2,{0,InvReactRate[ξ]/.sol2}}]}],AspectRatio->1,DisplayFunction->$DisplayFunction]

Conversion at exit of first reactor is ξ=0.670871 and Temperature is equal to 808K

ξ/.sol2

rec[ξ/.sol2]

Amount of catalyst needed in the two packed bed reactors are W1 and W2 and they are expressed in Kg.

W1=100 InvReactRate[ξ/.sol2] ξ/.sol2

W2=100 (InvReactRate[ξ/.sol2]-InvReactRate[0.85]) ((ξ/.sol2)-0.85)

Location and duty of heat exchangers. Three heat exchangers will be used to cool the system and their heat duties are equal to Q1, Q2 and Q3. Heat exchangers 1 and 2 will be placed in the recycle loop of the reactors.

Q1=100 ((808-300) 40-80000 0.6708)

Q2=100 ((750-808) 40-80000 (0.85-0.6708))

Q3=100 40. (300-750)

The recommended Design is given in the following sketch