fsolve with one variabel
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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);
0 件のコメント
採用された回答
Matt J
2020 年 11 月 19 日
編集済み: Matt J
2020 年 11 月 19 日
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)
3 件のコメント
Matt J
2020 年 11 月 20 日
Presumably, it is because you do not have the Optimization Toolbox. As with Stephan, it works fine when I run it.
その他の回答 (1 件)
Stephan
2020 年 11 月 19 日
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 件のコメント
Stephan
2020 年 11 月 20 日
編集済み: Stephan
2020 年 11 月 20 日
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
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