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)
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
.
