# Darcy friction factor

I have an assingment that says:

How is the friction factor $f$ dependent of the roughness $\epsilon$ and Reynolds number $Re$?

The equaton for the friction factor is $\displaystyle\frac{1}{\sqrt{f}}=1.14-2 \log_{10}\left(\frac{\epsilon}{D_h}+\frac{9.35}{Re\sqrt{f}}\right)$

But how can I say something about this if $f$ is also dependent of it self?

Can anynone help?

It is an implicit function which you have to solve numerically. Typically you would use "fixed point iteration".

You can do this in Excel with some user defined functions where for a given geometry you start with $f=1$ and then use the above $\frac{1}{\sqrt{f}} = L(f)$ a few times until in converges to a value $$f \rightarrow \frac{1}{L^2(f)}$$

In my example with $\epsilon/D_h = 0.005$ and $R_e = 1000$ it converges to $f=0.06561561\ldots$ after 10 iterations.

So after you build a table with various $\epsilon/D_h$ and $R_e$ you can talk about how it behaves.

• So was it decreasing or increasing in your case? Jun 17 '15 at 18:50
• I only did 1 point, so I cannot comment. It is up to you now. Jun 17 '15 at 18:56
• Alternatively you can try $L' = \frac{\partial L}{\partial f}f' + \frac{\partial L}{\partial \epsilon}\epsilon'+\frac{\partial L}{\partial R_e}R_e'$ but it looks to be very messy. Jun 17 '15 at 19:02
• Hmm I think I understand it. So you just pick some values for $\epsilon / D_h$ and $Re$. Then you start with $f=1$ and make iterations until it converges. Then you pick some new values for $\epsilon / D_h$ and $Re$ and make the whole process again. Jun 17 '15 at 19:28
• Yes, this is exactly how you can do it. Jun 18 '15 at 12:10