Computation of Segregation, Hold-back, F-diagrams and E-diagrams for different flow situations
Author's Data: Housam BINOUS
Department of Chemical Engineering
National Institute of Applied Sciences and Technology
Tunis, TUNISIA
Email: binoushousam@yahoo.com

References: 
(1) P. V. Danckwerts, Continuous Flow Systems Distribution of Residence Times, Chemical Engineering Science, vol. 2, n° 1,  pp. 1-13, February 1953.

(2) O. Levenspiel, Chemical Reaction Engineering, 3rd Edition, Wiley, 1999

<< Graphics`FilledPlot`

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Tube legnth

L=140

Tube diameter

Dtu=4.8

Tube volume

V=L Pi (Dtu/2)^2 0.62

Volumetric flow rate

v=4.5

Fractional free volume

ϕ = 0.62;

fluid velocity

u=v/(Pi (Dtu/2)^2 ϕ)

Diffusivity

Di=0.0013 L u

Checking that ((4  D)/Lu) << 1

4 Di/(L u)

We must have (u/L) = (v/V)

u/L

v/V

Perfect mixing case

F1[θ_]:=1-Exp[-v θ/V]

plt1=Plot[F1[θ],{θ,0,10 V/v },PlotStyle->RGBColor[1,0,0],PlotRange->{0,1}]

plt2=FilledPlot[{F1[θ],0},{θ,0,V/v},Fills->RGBColor[0.8,0.2,0.8],PlotRange->{0,1},DisplayFunction->Identity]

plt3=FilledPlot[{F1[θ],1},{θ,V/v,10 V/v},Fills->RGBColor[0.8,0.2,0.8],PlotRange->{0,1},DisplayFunction->Identity]

Show[plt2,plt3,DisplayFunction->$DisplayFunction,PlotRange->{{0,2000},{0,1}}]

Checking our expression

v/V NIntegrate[1-F1[θ],{θ,0,10000}]

Hold-back is equal to 1/e

H1=v/V NIntegrate[F1[θ],{θ,0,V/v}]

1./E

Longitudinal mixing case with approximate expression for the F-function given by P. V. Danckwerts

F2[θ_]:=1/2 (1-Erf[(L-θ u)/(2 Sqrt[θ Di])])

plt4=Plot[F2[θ],{θ,0,10 V/v},PlotStyle->RGBColor[0,0,1],PlotRange->{0,1}]

plt5=FilledPlot[{F2[θ],0},{θ,0,V/v},Fills->RGBColor[0.5,0.8,0.8],PlotRange->{0,1},DisplayFunction->Identity]

plt6=FilledPlot[{F2[θ],1},{θ,V/v,10 V/v},Fills->RGBColor[0.5,0.8,0.8],PlotRange->{0,1},DisplayFunction->Identity]

Show[plt5,plt6,DisplayFunction->$DisplayFunction,PlotRange->{{200,600},{0,1}}]

Checking our expression

NIntegrate[1-F2[θ],{θ,0,3000}]

Hold-back is equal to 0.02

H2=v/V NIntegrate[F2[θ],{θ,0,V/v}]

H2'=v/V NIntegrate[1-F2[θ],{θ,V/v,10 V/v}]

Segregation is equal to 0.348065; a result close to the segregation for the piston-flow case: 1/e=0.367879

Show[plt1,plt4]

sol=FindRoot[F1[θ]==F2[θ],{θ,350}]

NIntegrate[F2[θ]-F1[θ],{θ,θ/.sol,55000}]

NIntegrate[F1[θ]-F2[θ],{θ,0,θ/.sol}]

plt7=FilledPlot[{F2[θ],F1[θ]},{θ,0,θ/.sol},PlotRange->{0,1},DisplayFunction->Identity]

plt8=FilledPlot[{F1[θ],F2[θ]},{θ,θ/.sol,10 V/v},Fills->RGBColor[0.8,0.2,0.8],PlotRange->{0,1},DisplayFunction->Identity]

Show[plt7,plt8,DisplayFunction->$DisplayFunction,PlotRange->{{0,2000},{0,1}}]

Seg=NIntegrate[F2[θ]-F1[θ],{θ,θ/.sol,100000}] v/V

1./E

One can compute the value of the diffusivity Di=0.0729995 from the slope of the F-diagram at θ=V/v .

slope=D[F2[θ],{θ,1}]/.θ->V/v

Solve[2 slope V/v==Sqrt[L u/(Pi x)],x]

E-diagrams for the perfect mixing and Longitudinal mixing case

E1[x_]:=D[F1[θ],{θ,1}]/.θ->x

E2[x_]:=D[F2[θ],{θ,1}]/.θ->x

plt9=Plot[E1[x],{x,0,10 V/v },PlotStyle->RGBColor[1,0,0],PlotRange->All]

plt10=Plot[E2[x],{x,0,10 V/v },PlotStyle->RGBColor[0,0,1],PlotRange->All]

Checking our functions

NIntegrate[E1[x],{x,0,20 V/v}]

NIntegrate[E2[x],{x,0,20 V/v}]

Functions given by O. Levenspiel give same F-diagram

E6[θ_]:=v/V Sqrt[1/(4 Pi Di/(u L))] Exp[-(1-θ v/V)^2/(4 θ v/V Di/(u L))]

F6[x_]:=NIntegrate[E6[θ],{θ,0,x}]

plt19=Plot[F6[x],{x,0,1.5 V/v},PlotStyle->RGBColor[1,0,1],DisplayFunction->Identity]

E7[θ_]:=v/V Sqrt[1/(4 Pi Di/(u L))] Exp[-(1-θ v/V)^2/(4 Di/(u L))]

F7[x_]:=NIntegrate[E7[θ],{θ,0,x}]

plt20=Plot[F7[x],{x,0,1.5 V/v},PlotStyle->RGBColor[0,1,1],DisplayFunction->Identity]

Show[plt20,plt4,plt19,DisplayFunction->$DisplayFunction]

Laminar flow in pipes with u=0.401096

F3[θ_]:=If[v θ/V<0.5,0,1-V^2/(4 v^2 θ^2)]

plt11=Plot[F3[θ],{θ,0,8 V/v },PlotStyle->RGBColor[0,0,1],PlotRange->All]

Show[plt1,plt11]

sol1=FindRoot[F1[θ]==F3[θ],{θ,250}]

sol2=FindRoot[F1[θ]==F3[θ],{θ,550}]

a2=NIntegrate[F3[θ]-F1[θ],{θ,θ/.sol1,θ/.sol2}]

a1=NIntegrate[F1[θ]-F3[θ],{θ,0,θ/.sol1}]

a3=NIntegrate[F1[θ]-F3[θ],{θ,θ/.sol2,10 v/V}]

a1+a3-a2

Segregation is equal to -0.184113; a result between the segregation for the piston-flow case: 1/e=0.367879 and the segregation when most the space is dead water: -1.

seg2=-a2 v/V

plt11=FilledPlot[{F3[θ],F1[θ]},{θ,0,θ/.sol1},PlotRange->{0,1}]

plt12=FilledPlot[{F3[θ],F1[θ]},{θ,θ/.sol1,θ/.sol2},Fills->RGBColor[0.8,0.8,0.1],PlotRange->{0,1}]

plt13=FilledPlot[{F1[θ],F3[θ]},{θ,θ/.sol2,10 V/v},PlotRange->{0.98,1}]

Show[plt12,plt11,plt13,DisplayFunction->$DisplayFunction,PlotRange->{{0,3500},{0,1}}]

Hold-back is exactly equal to (1/4)=0.25

NIntegrate[1-F3[θ],{θ,0,100000}]

v/V NIntegrate[1-F3[θ],{θ,V/v,100000}]

v/V NIntegrate[F3[θ],{θ,0,V/v}]

plt14=FilledPlot[{F3[θ],0},{θ,0,V/v},PlotRange->{0,1},DisplayFunction->Identity]

plt15=FilledPlot[{F3[θ],1},{θ,V/v,10 V/v},Fills->RGBColor[0.8,0.2,0.8],PlotRange->{0,1},DisplayFunction->Identity]

Show[plt14,plt15,DisplayFunction->$DisplayFunction,PlotRange->{{0,2000},{0,1}}]

E-diagrams for the laminar pipe flow case

E3[x_]:=D[F3[θ],{θ,1}]/.θ->x

plt21=Plot[E3[x],{x,0,3 V/v },PlotStyle->RGBColor[1,0,0],PlotRange->All]

Piston-flow case

F4[θ_]:=UnitStep[θ-V/v]

plt15=Plot[F4[θ],{θ,0,8 V/v },PlotStyle->RGBColor[0,0,1],PlotRange->All]

Show[plt1,plt15]

plt16=FilledPlot[{F4[θ],F1[θ]},{θ,0,V/v},PlotRange->{0,1},DisplayFunction->Identity]

plt17=FilledPlot[{F1[θ],F4[θ]},{θ,V/v,10 V/v},Fills->RGBColor[0.8,0.2,0.8],PlotRange->{0,1},DisplayFunction->Identity]

Show[plt17,plt16,DisplayFunction->$DisplayFunction,PlotRange->{{0,2000},{0,1}}]

Segregation in the case of piston-flow is equal to 1/e

NIntegrate[F4[θ]-F1[θ],{θ,V/v,10 V/v}]

NIntegrate[F1[θ]-F4[θ],{θ,0,V/v}]

NIntegrate[F1[θ]-F4[θ],{θ,0,V/v}] v/V

1./E

E-function for the piston-flow case is δ( x - (( V)/v))

E4[x_]:=D[F4[θ],{θ,1}]/.θ->x

E4[x]