error message contain ::: Error using colebrook (line 2) Not enough input arguments. how i can solve it please?
3 次查看(过去 30 天)
显示 更早的评论

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 个评论
Walter Roberson
2017-4-29
You need to pass arguments when you call the function, just like you do inside the for loop.
采纳的回答
Walter Roberson
2017-4-29
编辑:Walter Roberson
2017-4-29
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 个评论
Walter Roberson
2017-4-29
Wait... you have defined F in terms of F ? Is the idea to find the F such that the two sides balance?
Walter Roberson
2017-4-29
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
更多回答(0 个)
另请参阅
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!