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Effacer les filtres

numerical root finding procedures

10 vues (au cours des 30 derniers jours)
harley
harley le 19 Août 2013
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 commentaires
harley
harley le 19 Août 2013
thanks for that, when i run i get in the command window;
Exiting fzero: aborting search for an interval containing a sign change because complex function value encountered during search. (Function value at -0.28 is -9.229-3.1086i.) Check function or try again with a different starting value.
code is
D = 0.1;
e = 0.0015e-3;
nu = 1.01e-6;
eoverD = e/D;
%
V = 2;
%
Re = V*D / nu;
%
if Re < 2300
f = 64 / Re;
else
darbyFormula = @(x) 1/sqrt(x)+2*log10(eoverD/3.7 + 2.51/Re/sqrt(x));
f = fzero(darbyFormula,1);
end
harley
harley le 19 Août 2013
got it working, thanks

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the cyclist
the cyclist le 19 Août 2013
Modifié(e) : the cyclist le 19 Août 2013
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)

Plus de réponses (1)

Walter Roberson
Walter Roberson le 19 Août 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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