fsolve with one variabel

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);

 Risposta accettata

Matt J
Matt J il 19 Nov 2020
Modificato: Matt J il 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 Commenti

I don't know either why I am doing that, I am just trying to solve this problem by using fsolve. I tried with different ways to do that even without syms x as follows:
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)
"Undefined function 'fsolve' for input arguments of type 'function_handle'."
but I did not work either.
Matt J
Matt J il 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 il 21 Nov 2020
Ok thanks for your reply!

Accedi per commentare.

Più risposte (1)

Stephan
Stephan il 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 Commenti

Mohamed Asaad
Mohamed Asaad il 20 Nov 2020
It does not work for me, I get "Undefined function 'fsolve' for input arguments of type 'function_handle'."
Stephan
Stephan il 20 Nov 2020
Modificato: Stephan il 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 il 21 Nov 2020
Ok thanks for your reply!

Accedi per commentare.

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