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Old   December 2, 2004, 13:09
Default Churchill Friction Factor Equation question
  #1
Nick
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In "A" sub equation - A=(2.4571n(1/(7/Re)^0.9+(0.27e/D)))^16 - what does the "n" stand for?

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Nick
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Old   December 2, 2004, 13:15
Default Re: Churchill Friction Factor Equation question
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Yves
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Is there any chance is says: A = (2.457 ln(1/(7/Re)^0.9 + (0.27e/D)))^16

and ln(...) the natural or Neperian logarithm

ln(...) = log(...)/log(exp(1))
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Old   December 2, 2004, 14:50
Default Re: Churchill Friction Factor Equation question
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Nick
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Yep! That's exactly what it says. I just got an email back from some one, telling me that its not 1n, but its ln. In the font in which I read the equation, the 1 and the l were all but identical.

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Nick
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