How can I solve this nested symbolic function ?

Hello, Im doing a metallurgical problem ,to do so I need to make the following code but returns an error when i call the following nested function, by the way I am doing so because I want to export it as an excel add-in .

valor=molienda(10,6,6,6.1,1.02,0.45,0.7,10)

Error using assignin
Attempt to add "z" to a static workspace.

Error in syms (line 262)
assignin('caller', x, xsym);

Error in solution>molienda/mP80 (line 17)
syms Q(x) z

Error in solution>molienda (line 5)
p80=mP80(t);

function a=molienda(F_min,Diametro,Largo,Wi,F80,Vp,fVc,t)

p80=mP80(t);
%Potencia para moler el mineral
W_mineral=Wi*((10/(p80^0.5))-(10/(F80))^(0.5));
P_mineral=W_mineral*F_min*1.341;
%Potencia para mover el medio moledor
Kwb = 4.879*(D^0.3)*(3.2-3*Vp)*fVc*(1-0.1/((2^9)-10*fVc))
W_bolas=80*(Diametro^2)*Largo;
P_bolas=Kwb*W_bolas;
P_total=P_bolas+P_mineral;
a(1)=P_total;
a(2)=P80;
function P_80=mP80(tiempo)
syms Q(x) z
Q(x) = (16.4831*x^(-0.7510)+1.665)*exp(-0.9451*tiempo^0.296);
fun= 80-(100-Q(z))==0;
P_80 = double(solve(fun,z));
end
end

I can´t find the solution ,please if you know how to i would very pleased . Thanks in advance for your help.

Answer 1

[Star Strider]

It would be best to abandon the idea of using the Symbolic Math Toolbox here, since it is not necessary. You can do everything necessary with anonymous functions.

Also, there are discrepancies between the function arguments and the code. I substituted ‘p80’ for ‘F80’ and ‘P80’ in order to make it work. You need to edit ‘molienda’ to resolve all these discrepancies, since I am not certain that the changes I made to it get it to work are correct.

Try this —

valor=molienda(10,6,6,6.1,1.02,0.45,0.7,10)

Equation solved.

fsolve completed because the vector of function values is near zero
as measured by the value of the function tolerance, and
the problem appears regular as measured by the gradient.

Kwb = 10.8133

valor = 1×2

1.0e+05 * 1.8904 0.0000

function a=molienda(F_min,Diametro,Largo,Wi,p80,Vp,fVc,t)

p80=mP80(t);
%Potencia para moler el mineral
W_mineral=Wi*((10/(p80^0.5))-(10/(p80))^(0.5));
P_mineral=W_mineral*F_min*1.341;
%Potencia para mover el medio moledor
Kwb = 4.879*(Diametro^0.3)*(3.2-3*Vp)*fVc*(1-0.1/((2^9)-10*fVc))
W_bolas=80*(Diametro^2)*Largo;
P_bolas=Kwb*W_bolas;
P_total=P_bolas+P_mineral;
a(1)=P_total;
a(2)=p80;
function P_80=mP80(tiempo)
% syms Q(x) z
Q = @(x) (16.4831*x.^(-0.7510)+1.665)*exp(-0.9451*tiempo.^0.296);
fun = @(z) 80-(100-Q(z));
P_80 = fsolve(fun,10);
end
end

.