Colebrook friction factor
Colebrook friction factor
(OP)
What are the origins for different Colebrook friction factor equations shown in a few texts?
This one I have used in the past is 1/f^(-0.5) = -2Log(ε/(3.7Re)+2.51/(f^0.5*Re))
Not sure why John McKetta and Ron Darby use these coefficients: 1/f^(-0.5) = -4Log(ε/(3.7Re)+1.26/(f^0.5*Re))
This one I have used in the past is 1/f^(-0.5) = -2Log(ε/(3.7Re)+2.51/(f^0.5*Re))
Not sure why John McKetta and Ron Darby use these coefficients: 1/f^(-0.5) = -4Log(ε/(3.7Re)+1.26/(f^0.5*Re))





RE: Colebrook friction factor
Have a look at the FAQs at the top of this forum - some references are given in the FAQ written by Quark.
Katmar Software
Engineering & Risk Analysis Software
http://katmarsoftware.com
RE: Colebrook friction factor
I know its not the question you asked, but you can save yourself a lot of grief if you switch to the Churchill equation. http://www.eng-tips.com/faqs.cfm?fid=1236
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RE: Colebrook friction factor
In my earlier post I said that there are two main forms of the Colebrook friction factor - the Fanning and Moody (or Darcy) factors. There is in fact a third form, which I call the Stanton friction factor. This friction factor is 1/2 of the Fanning factor and 1/8 of the Moody factor. It is very important to know which friction factor you are using because the coefficient in the Darcy-Weisbach equation for frictional pressure drop must change accordingly.
The Stanton form is very rarely used, but one of the authors who does use it is Churchill. The original publication by Churchill (Chem Eng, Nov 7, 1977) and the form given by Quark in the FAQ above both use the Stanton friction factor.
Many young (and not so young) engineers have discovered that their pressure drop calculations are out by a factor of 2, 4 or 8 depending on which friction factor they have used with which form of the D-W equation.
Katmar Software
Engineering & Risk Analysis Software
http://katmarsoftware.com
RE: Colebrook friction factor