numerical root finding procedures
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trying to solve the part under else. eover and Re are known, but still leaves me with (f) on both side of the original colebrook equation. 1/sqrt(f) = -2*log10(eoverD/3.7 + 2.51/Re/sqrt(f)). Please help, a bit stuck.
Re = V*D1 / nu;
% Check for laminar flow.
if Re < 2300
F = 64 / Re;
else
F(f)=1/sqrt(f)+2*log10(eoverD/3.7 + 2.51/Re/sqrt(f));
end
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the cyclist
2013년 8월 19일
편집: the cyclist
2013년 8월 19일
You should be able to use the function fzero() to solve for f in your implicit equation.
>> doc fzero
for details.
I think this will do it, but definitely check:
darbyFormula = @(x) 1/sqrt(x)+2*log10(eoverD/3.7 + 2.51/Re/sqrt(x));
f = fzero(darbyFormula,1)
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Walter Roberson
2013년 8월 19일
If you do some algebraic manipulation, you get
x = 0.3340248829e22 / (-0.5020000000e11 * lambertw(.4586822894 * Re * exp(.1239681863 * eoverD * Re)) + 6223202955 * eoverD * Re)^2
with no searching (provided that eoverD already has a value)
lambertw is in the Symbolic Toolbox. If you do not have that, then see http://www.mathworks.com/matlabcentral/newsreader/view_thread/32527
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