error message contain ::: Error using colebrook (line 2) Not enough input arguments. how i can solve it please?

6 vues (au cours des 30 derniers jours)
function F = colebrook(Re,K)
F= (1/(-2*log10((K/3.7)+(2.51/Re*sqrt(F)))))^2;
for i=1:20
Re=i*10^3;
K=0.00002;
colebrook(Re,K)
end
end
  1 commentaire
Walter Roberson
Walter Roberson le 29 Avr 2017
You need to pass arguments when you call the function, just like you do inside the for loop.

Connectez-vous pour commenter.

Réponse acceptée

Walter Roberson
Walter Roberson le 29 Avr 2017
Modifié(e) : Walter Roberson le 29 Avr 2017
colebrook = @(Re,K) (1/(-2*log10((K/3.7)+(2.51/Re*sqrt(F)))))^2;
cb = zeros(1,20);
for i=1:20
Re=i*10^3;
K=0.00002;
cb(i) = colebrook(Re,K);
end
cb
  3 commentaires
Walter Roberson
Walter Roberson le 29 Avr 2017
Wait... you have defined F in terms of F ? Is the idea to find the F such that the two sides balance?
Walter Roberson
Walter Roberson le 29 Avr 2017
Presuming that the idea is to find the F that balances the equation:
colebrook = @(Re,K) (1/(-2*log10((K/3.7)+(2.51/Re*sqrt(F)))))^2;
cf = @(Re,K,F) (1/(-2*log10((K/3.7)+(2.51/Re*sqrt(F)))))^2 - F;
cb = zeros(1,20);
for i=1:20
Re=i*10^3;
K=0.00002;
cb(i) = fzero(@(F) cf(Re,K,F), [0 1]);
end
cb
cb =
Columns 1 through 6
0.0212456669959142 0.0176438993118644 0.0159918090443276 0.0149809406929249 0.0142775420985982 0.0137506154874361
Columns 7 through 12
0.0133364599564298 0.0129997383391223 0.0127190013698845 0.0124803457077015 0.0122742958406683 0.0120941332432202
Columns 13 through 18
0.0119349390546334 0.0117930153254043 0.0116655193666966 0.0115502240640389 0.0114453557666283 0.01134948165294
Columns 19 through 20
0.0112614296189815 0.0111802301156452

Connectez-vous pour commenter.

Plus de réponses (0)

Catégories

En savoir plus sur MATLAB dans Help Center et File Exchange

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by