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Old Jan 22nd 2016, 07:11 AM   #1
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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 ?
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Old Feb 3rd 2016, 05:14 AM   #2
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can someone try to reply please? sorry , i didn't mean to bump this
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Old Feb 3rd 2016, 05:24 AM   #3
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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.
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Old Feb 3rd 2016, 07:35 AM   #4
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Originally Posted by fambros85 View Post
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?
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Old Feb 3rd 2016, 01:48 PM   #5
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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.
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Old Feb 3rd 2016, 10:11 PM   #6
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Originally Posted by MBW View Post
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 ?
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Old Feb 5th 2016, 01:51 PM   #7
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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.
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