Physics Help Forum loss coefficient

 Jan 22nd 2016, 07:11 AM #1 Senior Member   Join Date: Jun 2014 Posts: 306 loss coefficient i cant understand why the author wrote that the loss coeeicient (including the friction factor) tend to be independent of the Reynold number at the large Reynold number , why ? Attached Thumbnails
 Feb 3rd 2016, 05:14 AM #2 Senior Member   Join Date: Jun 2014 Posts: 306 can someone try to reply please? sorry , i didn't mean to bump this
 Feb 3rd 2016, 05:24 AM #3 Junior Member   Join Date: Feb 2016 Posts: 4 Generally these coefficients are empirical data. when you calculate the losses using this coefficient they are also multiplied by the squared of the velocity. So in some way is like Re it is all already inside in the coefficient. Is something less accurate but good enough for engineering purposes.
Feb 3rd 2016, 07:35 AM   #4
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 Originally Posted by fambros85 Generally these coefficients are empirical data. when you calculate the losses using this coefficient they are also multiplied by the squared of the velocity. So in some way is like Re it is all already inside in the coefficient. Is something less accurate but good enough for engineering purposes.
can you explain further?

 Feb 3rd 2016, 01:48 PM #5 Senior Member     Join Date: Apr 2008 Location: Bedford, England Posts: 668 At high Reynolds numbers the flow will be fully turbulent. In general flows vary very little with changes in Reynolds number until they reach a critical value when they change quite abruptly from laminar to turbulent flows. Below this critical point they change little and above this critical point they change little. Reynolds number gives a measure of where this critical point will be for the flow you are considering. Provided all your flow is above this critical value (which is almost a certainty for flow in a pipe), then the loss coefficient will not vary significantly with Reynolds number. __________________ You have GOT to Laugh !
Feb 3rd 2016, 10:11 PM   #6
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 Originally Posted by MBW At high Reynolds numbers the flow will be fully turbulent. In general flows vary very little with changes in Reynolds number until they reach a critical value when they change quite abruptly from laminar to turbulent flows. Below this critical point they change little and above this critical point they change little. Reynolds number gives a measure of where this critical point will be for the flow you are considering. Provided all your flow is above this critical value (which is almost a certainty for flow in a pipe), then the loss coefficient will not vary significantly with Reynolds number.
can you prove this ?

 Feb 5th 2016, 01:51 PM #7 Senior Member     Join Date: Apr 2008 Location: Bedford, England Posts: 668 Have a look at this: Moody Diagram Note how the on the right hand side of the diagram (high Reynolds number) the variation of friction coefficient with Reynolds number is almost flat. I came across this which seems to address exactly the issues you are looking at: Laminar_turbulent.pdf Note particularly the Moody diagram on page 14. Note in laminar flow there is a direct linear relationship between friction factor and Reynolds number. In turbulent flow the relationship is much less direct, and in "fully" turbulent flow there is hardly any connection at all. __________________ You have GOT to Laugh !

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