fsolve with one variabel

Question

Mohamed Asaad 2020-11-19

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https://ww2.mathworks.cn/matlabcentral/answers/653048-fsolve-with-one-variabel

评论： Mohamed Asaad 2020-11-21

Hi!

Could any one help me with solvin this problem usin "fsolve"? When i run this i get: "Undefined function 'fsolve' for input arguments of type 'function_handle'" The versison of Matlab on my computer is: Mataab R2020b- academic use.

syms x
Tinc   =   5+273;
Tinv   =   45+273;
mv     =   1000;
Cpv    =   4180;
A      =   0.032*(21-2); 
h      =   20;
U      =   2*h;
Qc=@(x)mv*Cpv(Tinv-x);
Q=@(x)U*A*(x-Tinc);
r=@(x)Q(x)-Qc(x);
Tutc=fsolve(r,30);

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Answer 1

Matt J 2020-11-19

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编辑：Matt J 2020-11-19

在 MATLAB Online 中打开

I don't understand why you would be using Symbolic Math Toolbox variables,

syms x

unless you were planning to use solve,

Tinc = 5+273;

Tinv = 45+273;

mv = 1000;

Cpv = 4180;

A = 0.032*(21-2);

h = 20;

U = 2*h;

Qc=mv*Cpv*(Tinv-x);

Q=U*A*(x-Tinc);

Tutc=solve(Q==Qc)

Tutc =

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Matt J 2020-11-20

Presumably, it is because you do not have the Optimization Toolbox. As with Stephan, it works fine when I run it.

Mohamed Asaad 2020-11-21

Ok thanks for your reply!

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Answer 2

Stephan 2020-11-19

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https://ww2.mathworks.cn/matlabcentral/answers/653048-fsolve-with-one-variabel#answer_549433

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Tinc   =   5+273;
Tinv   =   45+273;
mv     =   1000;
Cpv    =   4180;
A      =   0.032*(21-2); 
h      =   20;
U      =   2*h;
Qc=@(x)mv*Cpv*(Tinv-x);
Q=@(x)U*A*(x-Tinc);
r=@(x)Q(x)-Qc(x);
Tutc=fsolve(r,30)

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Stephan 2020-11-20

编辑：Stephan 2020-11-20

在 MATLAB Online 中打开

In R2020a it works for me:

Equation solved, solver stalled.
fsolve stopped because the relative size of the current step is less than the
value of the step size tolerance squared and the vector of function values
is near zero as measured by the value of the function tolerance.
<stopping criteria details>
Tutc =
  317.9998