fsolve with one variabel

조회 수: 1 (최근 30일)
Mohamed Asaad
Mohamed Asaad 2020년 11월 19일
댓글: Mohamed Asaad 2020년 11월 21일
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);

채택된 답변

Matt J
Matt J 2020년 11월 19일
편집: Matt J 2020년 11월 19일
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 = 
  댓글 수: 3
Matt J
Matt J 2020년 11월 20일
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 2020년 11월 21일
Ok thanks for your reply!

댓글을 달려면 로그인하십시오.

추가 답변 (1개)

Stephan
Stephan 2020년 11월 19일
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)
  댓글 수: 3
Stephan
Stephan 2020년 11월 20일
편집: Stephan 2020년 11월 20일
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 2020년 11월 21일
Ok thanks for your reply!

댓글을 달려면 로그인하십시오.

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