fsolve with one variabel

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Mohamed Asaad
Mohamed Asaad on 19 Nov 2020
Commented: Mohamed Asaad on 21 Nov 2020
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);

Accepted Answer

Matt J
Matt J on 19 Nov 2020
Edited: Matt J on 19 Nov 2020
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 = 
  3 Comments
Matt J
Matt J on 20 Nov 2020
Presumably, it is because you do not have the Optimization Toolbox. As with Stephan, it works fine when I run it.
Mohamed Asaad
Mohamed Asaad on 21 Nov 2020
Ok thanks for your reply!

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More Answers (1)

Stephan
Stephan on 19 Nov 2020
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)
  3 Comments
Stephan
Stephan on 20 Nov 2020
Edited: Stephan on 20 Nov 2020
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
Mohamed Asaad
Mohamed Asaad on 21 Nov 2020
Ok thanks for your reply!

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