MATHEMATICS O F ADSORPTION I N BEDS. 111
82 1
MATHEMATICS OF ADSORPTIOS I?; BEDS. I11
RADIALFLOW LEOS LAPIDUS
AND
KEAL R. AMUKDSOK
Department of Chemical Enganeerzng, School of Chemistry, I'ntverszty of Mannesota, Mznneapolzs 14,Minnesota Recewed September 16, 1949
In the formation of a radial chromatogram a solution is added to an adsorbent disc of uniform thickness through a central hole whereby the solutes are adsorbed in the form of a symmetrical band. The chromatogram is then developed by adding a sufficient quantity of pure solvent to cause the solutes to move radially through the adsorbent. During development the front and rear portions of the band or bands may exhibit various concentration gradients (sharp or diffuse boundaries). Sumerous papers have appeared in the literature dealing with the application of radial chromatography to laboratory (5, 9) and commercial (2, 4) analysis. However, a theoretical approach to this process has been almost completely neglected. Wilson (10) and Thomas (7) briefly mentioned the procedure necessary to extend their vertical chromatographic equations to the case of a radial disc. Sanborn and h u n d s o n (6) have indicated a few of the basic equations necessary in the analysis. It is the purpose of this paper to present a mathematical solution of single-solute radial chromatography where the relation between the solution concentration and the adsorbate concentration can be expressed in two ways depending upon the mechanism assumed to be controlling: (1) by an adsorption isotherm, and (g) by a kinetic equation. FUNDBMENTAL DIFFERENTIAL EQUATION
Consider an adsorbent bed in the form of an infinite plate of thickness h, with a circular hole of radius R with center at the origin into which the solution is introduced. Also let
r = distance from the origin to any point in the disc c = amount of solute in solution per unit volume solution n = amount of solute adsorbed per unit volume of bed f(c) = n / A = adsorption isotherm A = amount of adsorbent per unit volume of bed t, = volume rate of flow of solution or solvent (assumed constant) V = total volume of solvent poured into disc up to time t T i 0 = total volume of original solution poured into disc D = diffusion constant of solute in solution m = porosity of adsorbent bed.
For a cylindrical layer of adsorbent of thickness Ar, a material balance demands that the amount of solute entering a t r by fluid flow and diffusion minus that leaving at r Ar must be equated to the concentration change which cccurs
+
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LEON LAPIDUS AND NEAL R. AYUNDSON
in the solution plus that change which occurs in the solute adsorbed. Thus, neglecting any vertical concentration gradient one finds -2nrhmD ar 5)
).
= 2nrhAr
- (-2nrhmD
(5);+
ZarhmAr
ar e)
+A,
(2);
r
< i < r 4- Ar
(1)
Divide equation 1 by Ar, assume B , D, and m are constant and let Ar pass to the limit of zero:
ac
8-
ar
an + 2nrh + 2nrhm -atac = 2nhmD at
If diffusion in the radial direction is neglected, the fundamental differential equation results: aC an ac D - 2nrh - 2nrhm - = 0 (3) ar at at
+
+
EQUILIBRIUM THEORY FOR SINGLE SOLUTES
If the assumption is made that instantaneous equilibrium occurs a t every point in the disc between solution and adsorbent, the amount of solute adsorbed can be related to the solution concentration by the adsorption isotherm n = Af(c)
(4)
Equation 3 thus becomes ac
B-
ar
+ 2nrh[m + Af’(c)] atac = 0 -
(5)
wheref’(c) is the derivative off(c). But as V = U t , equation 5 may be alternately written as: ac ac - 2nrh[m Ay(c)] - = 0 (6)
ar
+
+
av
To this equation must be appended the initial condition of the column and the equation rhich describes the condition of the entering solution. These are c ( r , V ) = c O ( V ) , when r = R n(r, V ) = no(r), when V = nhm(r* - R 2 )
(7)
However, the solution of equation 6 is complicated by the fact that one boundary value must be a t r = R. To circumvent this difficulty, the substitution is made that y = r - R and equation 6 becomes
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MATHEM.%TICS O F ADSORPTION IS BEDS. I11
TOW a solution can be obtained a t the much more convenient y that r2 - R* = y*
+2
=
0. Also note
~ y
(9)
We shall now proceed t o derive equations, using the method of the La Place transform, which will completely describe the movements of a solute within the disc for the case of the linear isotherm: n = bc
(10)
The more general isotherm \vi11 be treated in a later communication. While the linear isotherm has often been referred to as impractical, the authors have examined experimental data (8) which indicate that equation 10 holds for a number of systems and in some cases a large range of concentrations. To facilitate the solution we make a change of variable in equation 8 by letting z =
fit
- nhm(r2 - R2)
(11)
and then by further substituting equation 10, there results ac ac -=0 aY + Znhb(y + R ) az
(12)
The boundary conditions necessary using these variables in forming a chromatogram are
Taking the La Place transform of c with respect to z, L[c(y, z)] = M(y, p ) , equations 12 and 13 become: dY
+ 2?rhb(y + R)pM = 0 cu ;%I=-, P
y = o
But equation 14 is a simple linear differential equation whose solution is Jf = C1 exp(-nhb(y2
+ 2Ry)p)
where C, is an arbitrary constant. Applying equation 15,
and thus
(14)
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LEOX LAPIDUS AND KEBL R . AYUSDSON
To take the inverse transform of equation 16, we note that Churchill (3) gives the inverse transform of m exp(-kp) l p as being the stepwise function:
Thus we have :
4Y,
2)
=
iCo ‘0
0
< z < rhb(yz + 2 R y ) z > 7rhb(y2 + 2Ry)
Changing back to the original set of variables and using equation 9, there results:
c(r, V ) =
1:
For developing the band equations 10 and 12 are used again, but in the form:
an an -=0 az/ + 2rhb(y + R) az The appropriate boundary values are c(0,
2)
=
0
n(y, 0) = 0
for y
n(y, 0) = no for y
>a
