How to iterate xa from 0 to 1 with steps of 0.05 in the function r=F*(k*C*(​1-xa))/(1+​K*C*(1-xa)​). All variables are known except for r.

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I have to iterate the conversion of a reaction (x) in order to get the different values of catalyst mass (m).
The conversion 'x' would go from 0 to 1 with spaces of 0.05.
The function I need to iterate is: m=F*(k_kin*C*(1-x))/(1+k_ads*C*(1-x))
I know the value of all the variables except for m, which is the mass that I'd like to calculate.
F=0.11574; k_kin=1.3*10^-6; C=1.086; k_ads=0.9986
Please, help me. I haven't used matlab in years and I've forgotten how to make a 'for' loop work.
Thank you in advance.

Accepted Answer

Star Strider
Star Strider on 23 May 2019
Everything except ‘x’ are scalars, so you can do this using a vectorization approach without the loop:
F=0.11574; k_kin=1.3*10^-6; C=1.086; k_ads=0.9986;
x = 0:0.05:1;
m=F*(k_kin*C*(1-x))./(1+k_ads*C*(1-x));
figure
plot(x, m)
grid
xlabel('Reaction')
ylabel('Catalyst Mass')
... including the plot! (Note the (./) denoting element-wise division.)

More Answers (2)

Geoff Hayes
Geoff Hayes on 23 May 2019
Edited: Geoff Hayes on 23 May 2019
Marina - try
m = zeros(20,1);
k = 1;
for x = 0:0.05:1
m(k) = F*(k_kin*C*(1-x))/(1+k_ads*C*(1-x))
k = k + 1;
end
edit to above: repalce == with = in assignment to m
  8 Comments
Marina Gonzalez
Marina Gonzalez on 23 May 2019
Thank you! That has really helped. I think I'm doing something wrong because for a conversion of 0, the mass of the catalyst is way bigger than for the total conversion.
Captura.PNG
I'm doing something wrong but I don't know what. I'll think about it.
Thank you once again!

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Marina Gonzalez
Marina Gonzalez on 23 May 2019
It's me again- sorry!
What if the function instead of being the one I said before was m(k) = F*(1+k_ads*C*(1-x))/(k_kin*C*(1-x))
Would anything change? Because when I plot this new equation nothing shows up on the graph
  3 Comments

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