error message contain ::: Error using colebrook (line 2) Not enough input arguments. how i can solve it please?
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Moath Maqableh
el 29 de Abr. de 2017
Comentada: Walter Roberson
el 29 de Abr. de 2017

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 comentario
Walter Roberson
el 29 de Abr. de 2017
You need to pass arguments when you call the function, just like you do inside the for loop.
Respuesta aceptada
Walter Roberson
el 29 de Abr. de 2017
Editada: Walter Roberson
el 29 de Abr. de 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 comentarios
Walter Roberson
el 29 de Abr. de 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
el 29 de Abr. de 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
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