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numerical root finding procedures

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harley
harley am 19 Aug. 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 Kommentare
harley
harley am 19 Aug. 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 am 19 Aug. 2013
got it working, thanks

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Akzeptierte Antwort

the cyclist
the cyclist am 19 Aug. 2013
Bearbeitet: the cyclist am 19 Aug. 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)

Weitere Antworten (1)

Walter Roberson
Walter Roberson am 19 Aug. 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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