Use fzero() to calculate x required to achieve V/F for every combination of V and F.

Hi,

I have a formula ((x)/((k*c0^n)(1-x)^n) that predicts the conversion of a reaction (x) based on different reactor volumes (V) and flow rates (F). I would like to use fzero() to find the conversion (x) required to achieve V/F for every combination of V and F.

I then want to plot x against V and F using surf().

The code I have so far is below, I'm struggling to use fzero() as each time I run it I get errors and Im not sure if that's because my initial guess does not contain a root or If I have to re-arrange my V/F equation to make x the product before using fzero().

Thanks

k = 0.9;
n = 2;
c0 = 2; %initial concentration (mol/m^3)
V = 100:10:200;
F = 10:1:100;
%initial guess = 0.5 (can be anything)
V/F = @(x) (x)/((k*c0.^n)*((1-x)^n));
fun = @(x) V/F;
x = fzero(fun,0.5)

%To plot
%surf()
%figure(5) %figure 5

[Rik]

http://web.archive.org/web/20201031075934/https://webcache.googleusercontent.com/search?q=cache%3AT7Yo6Qlfz_UJ%3Ahttps%3A%2F%2Fwww.mathworks.com%2Fmatlabcentral%2Fanswers%2F628538-use-fzero-to-calculate-x-required-to-achieve-v-f-for-every-combination-of-v-and-f+&cd=1&hl=en&ct=clnk&gl=us

[Rena Berman]

(Answers Dev) Restored edit

Answer 1

[Star Strider]

The fzero function is a root-finder, so the function it is given to solve must equate to 0 for any delection of independent variables. I implemented that approach in the revised coding of ‘fun’ here:

k = 0.9;
n = 2;
c0 = 2; %initial concentration (mol/m^3)
V = 100:10:200;
F = 10:1:100;
V = linspace(100, 200, 25);
F = linspace(10, 100, 50);
%initial guess = 0.5 (can be anything)
VF = @(x) (x)./((k*c0.^n)*((1-x).^n));                      % Not Directly Used
fun = @(x,V,F) V./F - (x)./((k*c0.^n)*((1-x).^n));
for k1 = 1:numel(V)
    for k2 = 1:numel(F)
        x(k2,k1) = fzero(@(x)fun(x,V(k1),F(k2)),0.5);
    end
end

figure
surfc(V, F, x)
grid on
xlabel('V')
ylabel('F')
zlabel('Conversion')
view(130,10)

produces:

.

[Cate may]

Thank you! Exactly what I expected

[Star Strider]

As always, my pleasure!