how to code equation

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Cesar Cardenas
Cesar Cardenas 2023년 2월 28일
편집: Voss 2023년 2월 28일
Nu = (0.032.*(Pr).^(1/3)).*((rho.*Uinf.*x/mu).^(1/2)).*(1 -(x0/x).^(3/4).^(-1/3);
Hello, is this code right? not sure, any help will be appreciated thanks.
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Les Beckham
Les Beckham 2023년 2월 28일
It looks like it might be right, but that depends on the size and class of all of the variables; are they scalars, vectors, 2d arrays, something else, integers, complex, double. The details matter.
Did you try to run the code? If so, what happened? If you got an error message, cut and paste all of the text of the error message into an edit of your question, or a new comment.
Since we can't see your Matlab workspace, we can't answer your question until you share more information.
Cesar Cardenas
Cesar Cardenas 2023년 2월 28일
I have not run the code, but I see this on the workspace "parse error at EOL"
Nu = (0.032.*(Pr).^(1/3)).*((rho.*Uinf.*x/mu).^(1/2)).*(1 -(x0/x).^(3/4).^(-1/3);

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Voss
Voss 2023년 2월 28일
편집: Voss 2023년 2월 28일
Nu = (0.032.*(Pr).^(1/3)).*((rho.*Uinf.*x/mu).^(1/2)).*(1 -(x0/x).^(3/4).^(-1/3);
% ^^ you need another closing parenthesis here
% ^ ^ ^ ^ ^ ^ extraneous parentheses (that don't affect the operation)
% ^^^^^ should be 0.332
After correcting those things:
Nu = 0.332.*Pr.^(1/3).*(rho.*Uinf.*x/mu).^(1/2).*(1 -(x0/x).^(3/4)).^(-1/3);
And some of those / may have to be ./, depending on whether x and/or mu are scalars. Doesn't hurt to just use ./

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