Hello,

I'd like to solve the equation of Colebrook-White using Smath to find Darcy–Weisbach friction factor (f)

The equation is:



I tried to use solve and roots commands (Smath 0.95.4559), with no sucess.
Is it already possible to find the answer through Smath?

Hi

try using solve(4) instead of solve(2) or changing the "Roots (range)" value in "from 0 to 1" in "Tools" > "Options" > "Calculation"

in the attachment there is an example


regards,

w3b5urf3r

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colebrook_uncommented.sm (13 КиБ) скачан 350 раз(а).

isn't that an implicit equation ? so numerical iteration may be needed, look at this discussion about several methods to solve the equation of Colebrook-White

http://trickofmind.com/?p=1189

Also this post might be of help

http://en.smath.info/forum/yaf_postsm5611_How-to-write-a-recursive-function.aspx#post5611

Hello,

Wrote

Hi

try using solve(4) instead of solve(2) or changing the "Roots (range)" value in "from 0 to 1" in "Tools" > "Options" > "Calculation"

in the attachment there is an example


regards,

w3b5urf3r

It should be noted here that the solve() gives you two solutions. Actually, there are no two solution and I do not know why solve() returned these two. On the other hand, we have to take into account that solve() depends on "Tools" > "Options" > "Calculation" >"Decimal places". If you use, for instance, 4 decimal places these are the solutions

[MATH=eng]el(f.Colebrook,1)=0.011542553663605[/MATH]

[MATH=eng]el(f.Colebrook,2)=0.01216608597662[/MATH]

15 decimal places

[MATH=eng]el(f.Colebrook,1)=0.012166073459707[/MATH]

[MATH=eng]el(f.Colebrook,2)=0.012166080958897[/MATH]

Solve() will return only one solution if your left boundary is not zero, but some small number greater than zero. Have in mind that in the equation you have [MATH=eng]1/sqrt(f)[/MATH].
(I changed displayed decimals with context menu for each result regardless of numbers of decimals chosen from the manu option)

On the other hand, roots() will give you a single solution which does not depend on decimal places chosen from the menu
[MATH=eng]roots(1/sqrt(f#)-(-2)*log({ε/D}/3.7+2.51/{Re*sqrt(f#)},10),f#,0.01)=0.012166080958897[/MATH]

But if you are far from the solution with your guess value, roots() will fail as could be expected when nonlinear equations are involved

[MATH=eng]roots(1/sqrt(f#)-(-2)*log({ε/D}/3.7+2.51/{Re*sqrt(f#)},10),f#,0.03)[/MATH]

That is the "beauty" of strugling with nonlinear equations

Representing 15 decimal places for this equation has no to much sence, I just put them here to point out the problematic results from solve().

Regards,
Radovan

Wrote

Solve() will return only one solution if your left boundary is not zero, but some small number greater than zero. Have in mind that in the equation you have [MATH=eng]1/sqrt(f)[/MATH].

Thank you omorr, I was wondering what generate the 2nd solution

file updated with lower bound little but greater than zero (and Moody Diagram)...


regards,

w3b5urf3r

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You are welcome

Regards,
Radovan

Thank you all, very useful replies