Swamee & Jain Friction Factor Equation
3

Question

Swamee & Jain Friction Factor Equation

Mech85 (Mechanical)

(OP)

9 Feb 04 00:26

Hello,

The original Swamee & Jain explicit equation for friction factor calculation for turbulent flow in pipes has f = 0.25/[(log.....)]^2.

I have also seen on the web f = 1.325/[(log.....)]^2. The difference is NOT the factor of four beween the Fanning friction factor and the Darcy friction factor. Can anyone explain the discrepancy?
Thanks

smckennz (Mechanical) · Answer 1 · 2004-02-09 03:12:30

Hi Perth
never seen the 1.325, but the 0.25 is the one I memorised after checking it against Shacham and Colebrook White with good correlation.
I expect it may be a mistake.

If you are from Perth Australia, you probably got Swamee Jain 0.25 from the Hardie Iplex PE manual. Where did the 1.325 come from?
Cheers

Steve McKenzie

25362 (Chemical) · Answer 2 · 2004-02-09 07:19:06

To Perth1, may be I'm wrong but it appears that

(ln x)²/(log x)² = 1.325/0.25= 5.3

Id est,

(0.25)(ln x)²= (1.325)(log x)²

Thus, one formula is based on log x, and the other on ln x.

Would I be right ?

Mech85 (Mechanical) · Answer 3 · 2004-02-09 20:24:34

25362,

Yes, you're correct - I have gone back and checked the two articles using 1.325 in the formula and they DO use ln rather than log. Thanks for spotting the difference.

quark (Mechanical) · Answer 4 · 2004-02-10 12:19:25

That's a brilliant... (Ok, I resisted myself saying it a guess

). One star for you.

Regards,

Eng-Tips.com : Solving your problems before you get them.

25362 (Chemical) · Answer 5 · 2004-02-12 07:03:14

Quark. Thanks for the undeserved star.
"Things should be made as simple as possible, but not any simpler." Albert Einstein

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