# Fixed Bed Adsorption Column Using Dimensionless Equations

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### Accepted Answer

Torsten
on 1 Dec 2023

Edited: Torsten
on 1 Dec 2023

The usual procedure is to define a grid

0 = X(1) < X(2) < ... < X(n) = 1

approximate the spatial derivative dA/dX in grid point i by

dA(i)/dx ~ (A(i)-A(i-1))/(X(i)-X(i-1)),

keep the time derivatives in their continuous form,

write the PDE system as a system of 2*n ordinary differential equations

A(1) = A_in

dA(i)/dt = - (A(i)-A(i-1))/(X(i)-X(i-1)) - q0/A_in * s(q(i),A(i)) 2<=i<=n

dq(i)/dt = s(q(i),A(i)) 1<=i<=n

and use ode15s to solve the system.

If you need further details, look up "method-of-lines".

##### 7 Comments

Torsten
on 4 Dec 2023

Torsten
on 4 Dec 2023

Here is a short code with the explicit method for the problem:

dy/dt + a*dy/dx = 0

a > 0

0 <= x <= 1

y(x,t=0) = 0 for all x

y(x=0,t) = 1 for all t

It's already very similar to yours.

a = 0.1;

tstart = 0.0;

tend = 2.0;

nt = 1000;

xstart = 0.0;

xend = 1.0;

nx = 1000;

x = linspace(xstart,xend,nx).';

dx = x(2)-x(1);

t = linspace(tstart,tend,nt);

dt = t(2)-t(1);

y = zeros(nx,nt);

y(:,1) = 0.0;

y(1,:) = 1.0;

for it = 2:nt

y(2:nx,it) = y(2:nx,it-1) - dt * a * (y(2:nx,it-1)-y(1:nx-1,it-1))./(x(2:nx)-x(1:nx-1));

end

plot(x,[y(:,1),y(:,500),y(:,1000)])

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