Staged separations

One stage

eq1 = xaf F == xa F + ya S

eq2 = ya == m xa

Solve[{eq1,eq2},{ xa, ya}]

Plot[ 1 + Λ,{Λ,.1,10},AxesLabel->{"Λ","separation factor"}]

So we can get the removal we want, we have to pay for it with more S!

Two stages (cross current)

eq1 = xaf F == xa1 F + ya1 S/2

eq2 = ya1 == m xa1

eq3 = ya2 == m xa2

eq4   =F xa1 == F xa2 + ya2 S/2

Solve[{eq1,eq2,eq3,eq4},{ xa1,ya1,ya2,xa2}]

Plot[ {1 + Λ, (1 + Λ/2)^2},{Λ,.1,10},AxesLabel->{"Λ","separation factor"},PlotLegends->{"1 stage","2 stages"}]

Three stages

eq1 = xaf F == xa1 F + ya1 S/3

eq2 = ya1 == m xa1

eq3 = ya2 == m xa2

eq4   =F xa1 == F xa2 + ya2 S/3

eq5   =F xa2 == F xa3+ ya3 S/3

eq6 = ya3 == m xa3

Solve[{eq1,eq2,eq3,eq4,eq5,eq6},{ xa1,ya1,ya2,xa2,ya3, xa3}]

Plot[ {1 + Λ, (1 + Λ/2)^2, (1 + Λ/3)^3},{Λ,.1,10},AxesLabel->{"Λ","separation factor"},PlotLegends->{"1 stage","2 stages","3 stages"}]

Plot[ {1 + Λ, (1 + Λ/2)^2, (1 + Λ/3)^3,(1 + Λ/10)^10, Exp[Λ]},{Λ,.1,2},AxesLabel->{"Λ","separation factor"},PlotLegends->{"1 stage","2 stages","3 stages", "10 stages","infinite stages"}]

For crossflow, how many stages is “infinite”

The formula for ξ for n stages is:

stagen = (1 + Λ/n)^n

Let’s plot the ratio of s for finite stages to s for infinite states.

Plot[{(stagen/Exp[Λ])/.Λ->.5, (stagen/Exp[Λ])/.Λ->1,(stagen/Exp[Λ])/.Λ->2},{n,1,10},PlotStyle->{Green, Orange,Blue},PlotLegends->{"Λ->.5","Λ->1", "Λ->2"},AxesLabel->{"n","\!\(\*FractionBox[\(ξ : \\ n - stages\), \(\(ξ - infinite\)\(\\n\)\)]\)"},PlotRange->{0,1},PlotLabel->"Separation factor ratio for CrossFlow"]

Countercurrent

2 stages

eqcc1 = F xaF+ S ya2 == F xa1 + S ya1

eqcc2 = F xa1+ S yaS == F xa2 + S ya2

Solve[{eqcc1, eqcc2, ya1  == m xa1, ya2 == m xa2},{xa1,xa2,ya2,ya1}]

%/.yaS->0

3 stages

eqcc1 = F xaF+ S ya2 == F xa1 + S ya1

eqcc2 = F xa1+ S ya3 == F xa2 + S ya2

eqcc3 = F xa2+ S yaS == F xa3 + S ya3

Solve[{eqcc1, eqcc2,eqcc3, ya1  == m xa1, ya2 == m xa2, ya3 == m xa3},{xa1,xa2,ya2,ya1,xa3,ya3}]

%/.yaS->0

Simplify[%]

Expand[(F+m S) ((F^2)+(m^2) (S^2))]

3 stages

4 stages counter current

stagencc = Sum[Λ^(i-1),{i,1,n+1}]

stagencc = Sum[Λ^(i),{i,0,n}]//Inactivate

%/.n->3

Activate[%]

stagencc100=stagencc/.n->1000

LogLogPlot[{(stagencc)/.Λ->.5, (stagencc)/.Λ->1.1,(stagencc)/.Λ->2},{n,1,10},PlotStyle->{Green, Orange,Blue},PlotLegends->{"Λ->.5","Λ->1.1", "Λ->2"},AxesLabel->{"n","s: n-stages"},PlotRange->{0,15},PlotLabel->"Separation factor for Countercurrent"]

LogLogPlot[{(stagencc/stagen)/.Λ->.5, (stagencc/stagen)/.Λ->1.1,(stagencc/stagen)/.Λ->2},{n,1,10},PlotStyle->{Green, Orange,Blue},PlotLegends->{"Λ->.5","Λ->1.1", "Λ->2"},AxesLabel->{"n","(countercurrent/crosscurrent)"},PlotLabel->"Separation factor: countercurrent/crosscurrent"]

stagencc/stagen

Limit[%,n->∞]

Limit[%,Λ->∞]

Kremser equation

stagencc = Sum[Λ^(i),{i,0,n}]

Solve[ ξn == %,n]

n/.%[[1]]

FullSimplify[%, Assumptions->{Λ>0}]

Compare cross to counter current on a stage by stage basis

Cross current

stagen

Countercurrent

stagencc = Sum[Λ^(i-1),{i,1,n+1}]

Plot the comparison.  Note that this is valid only at discrete values for n.

Plot[{(stagen/stagencc)/.Λ->.5, (stagen/stagencc)/.Λ->1.001,(stagen/stagencc)/.Λ->2},{n,1,10},PlotStyle->{Green, Orange,Blue},PlotLegends->{"(P/R)->.5","(P/R)->1", "(P/R)->2"},AxesLabel->{"n","\!\(\*FractionBox[\(s : \\ crossflow\), \(\(s : \\ counter\\ flow\)\(\\n\)\)]\)"},PlotRange->{0,1},PlotLabel->"Crossflow v. Counterflow"]

stagencc = Sum[ξ^(i-1),{i,1,n+1}]

(Subscript[s, countercurrent]) =((((((P/R)))^(n+1))-1)/((P/R)-1))

Check the convergence....

stagencc/.{n->100,ξ->.95}

stagencc/.{n->200,ξ->.95}

stagencc/.{n->300,ξ->.95}

Check the convergence....

stagencc/.{n->100,ξ->1.01}

stagencc/.{n->200,ξ->1.01}

stagencc/.{n->300,ξ->1.01}

p m/r == 1

1/4.5

% 1.5