Fixed Bed Adsorption Column Using Dimensionless Equations

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tori lansing 2023-11-30

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https://ww2.mathworks.cn/matlabcentral/answers/2054614-fixed-bed-adsorption-column-using-dimensionless-equations

评论： Torsten 2023-12-4

My group and I are trying to write a code for fixed bed adsorption with dimensionless equations and Damkohler's number. I attached an image of the equations we're trying to use and variables. We have a general idea and some code written down of what the variables are meaning but would like some help on where to go with the ODE/PDE code as we're not familiar with them. Any help, guidance, or advice would be greatly appreciated. Thank you very much in advance.

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Answer 1

Torsten 2023-12-1

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https://ww2.mathworks.cn/matlabcentral/answers/2054614-fixed-bed-adsorption-column-using-dimensionless-equations#answer_1363449

编辑：Torsten 2023-12-1

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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".

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tori lansing 2023-12-3

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Torsten, First I want to thank you or all your help and quick responses, it's helping us a lot. Secondly, when we're running what youve given us with our parameters we're running into some isues. Can you take a look and see if maybe we're defining something wrong? We are getting an error that the statement is incomplete. Again, Thank you so much! We appreciate all your help.

tf=3000;
dt=3000/200;
tspan=0:dt:tf; % creates array for length of time
cf=0.015; %initial concentration
nz=200;
np=nz+1;
Cb=zeros(np,1); Cb(1)=cf;
Cp=zeros(np,1);
C0=[Cb;Cp]; % this may need some adjusting, im not sure if this comes out right
% I beleive this establishes the things for the ODE solver
% following the ode solver, the plots are created
% then a function is created for rates that is then recalled in ode solver
% then a for loop to establish the rest of the equations
A_in=input("Input a value for A_in:")
if (A_in>0)&&(A_in<1)
    disp(A_in)
else
    disp('Invalid value for inlet concentration')
end
K=input('Enter a value for the rate constant:')
q_o=input('Enter a value for the total exchange capacity of the system:')
k=0.001
L=10; % meters
K=0.001;
A=0:0.01:A_in;
Ac=A/A_in;
V=0.001;
t=L/V;
delta=tspan./t;
Da=k*A_in*t;
dqdt=zeros(np,1);
s=q_o/A_in;
x=0:0.01:10;
X=x./L;
0 == X(1) < X(2) < ... < X(n) = 1
This statement is incomplete.

Torsten 2023-12-4

To get a foundation in MATLAB programming, I suggest the MATLAB Onramp course to learn about MATLAB basics in an online course free of costs:

https://matlabacademy.mathworks.com/details/matlab-onramp/gettingstarted

Torsten 2023-12-4

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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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Fixed Bed Adsorption Column Using Dimensionless Equations

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