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
3 Commenti
the cyclist
il 19 Ago 2013
You have written this a bit confusingly. I think it is a bit clearer to write your code as
Re = V*D1 / nu;
% Check for laminar flow.
if Re < 2300
f = 64 / Re;
else
% Here I need to calculate "f" such that it solves the equation
% 0 = 1/sqrt(f)+2*log10(eoverD/3.7 + 2.51/Re/sqrt(f));
end
Notice how I get rid of the confusing use of capital F.
See my suggestion for solving this in my answer below.
harley
il 19 Ago 2013
harley
il 19 Ago 2013
Risposta accettata
Più risposte (1)
Walter Roberson
il 19 Ago 2013
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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