Please, help me fix the first code. why is 1st code not working but the 2nd code is?

2021 11 7
1 个回答
更新时间：2021 11 7
5 次查看（30 天）

0 个投票

Ca0 = 8.24;
Cb0 = 0.824;
K = 6.98*10^17;
C0 = Ca0 + Cb0;
K1 = 0.00083287;
K2 = 0.2172;
K4 = 1.341;
K5 = 1.60;
r1 = 0.00522;
Ca = (Cb0*((Ca0/Cb0) - X));
F = @(X)((K1*C0*(Ca - ((3*X*Cb0*X*Cb0)/(K*Cb0*(1 - X)))))/(((K1*3*X*Cb0*X*Cb0)/(K*Cb0*(1 - X))) + (K2*Cb0*(1 - X)) + (K5*X*Cb0) + (K4*K5*3*X*Cb0*X*Cb0) + 1)) - r1
roots = fsolve(F,0)
----------------------------------------------------------------------------------------------------------------------------
The code above returns error message: Undefined function or variable 'X'.
But the code below runs successfully after replacing value of Ca in F.
-----------------------------------------------------------------------------------------------------------------------------
Ca0 = 8.24;
Cb0 = 0.824;
K = 6.98*10^17;
C0 = Ca0 + Cb0;
K1 = 0.00083287;
K2 = 0.2172;
K4 = 1.341;
K5 = 1.60;
r1 = 0.00522;
F = @(X)((K1*C0*((Cb0*((Ca0/Cb0) - X)) - ((3*X*Cb0*X*Cb0)/(K*Cb0*(1 - X)))))/(((K1*3*X*Cb0*X*Cb0)/(K*Cb0*(1 - X))) + (K2*Cb0*(1 - X)) + (K5*X*Cb0) + (K4*K5*3*X*Cb0*X*Cb0) + 1)) - r1
roots = fsolve(F,0)
In addition, how can I put the 1st format (that is, where Ca is defined separately and then Ca placed in F) in a function and solve

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回答（1 个）

Star Strider
Star Strider 2021-11-7
Ran in:

0 个投票

Ran in:
Since ‘Ca’ is actually a function of ‘X’ it has to be coded as such —
Ca = @(X) (Cb0*((Ca0/Cb0) - X));
Ca0 = 8.24;
Cb0 = 0.824;
K = 6.98*10^17;
C0 = Ca0 + Cb0;
K1 = 0.00083287;
K2 = 0.2172;
K4 = 1.341;
K5 = 1.60;
r1 = 0.00522;
Ca = @(X) (Cb0*((Ca0/Cb0) - X));
F = @(X)((K1*C0*(Ca(X) - ((3*X*Cb0*X*Cb0)/(K*Cb0*(1 - X)))))/(((K1*3*X*Cb0*X*Cb0)/(K*Cb0*(1 - X))) + (K2*Cb0*(1 - X)) + (K5*X*Cb0) + (K4*K5*3*X*Cb0*X*Cb0) + 1)) - r1
F = function_handle with value:
@(X)((K1*C0*(Ca(X)-((3*X*Cb0*X*Cb0)/(K*Cb0*(1-X)))))/(((K1*3*X*Cb0*X*Cb0)/(K*Cb0*(1-X)))+(K2*Cb0*(1-X))+(K5*X*Cb0)+(K4*K5*3*X*Cb0*X*Cb0)+1))-r1
roots = fsolve(F,0)
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.
roots = 1.3226
And with that change, it works!
.

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