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);

 采纳的回答

Matt J
Matt J 2020-11-19
编辑:Matt J 2020-11-19
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 个评论

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.
Presumably, it is because you do not have the Optimization Toolbox. As with Stephan, it works fine when I run it.
Ok thanks for your reply!

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更多回答(1 个)

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 个评论

It does not work for me, I get "Undefined function 'fsolve' for input arguments of type 'function_handle'."
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
Ok thanks for your reply!

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