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September 19, 2012, 02:02 
What's the meaning of the Reynolds number

#1 
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Nick
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Hi,
I've heard a lot, especially in cfd context, that the Reynolds number is the ratio between the inertial to viscous forces. Now what I understand from inertial force is (mass times acceleration) so if this is correct then for pipe flow where the velocity profile doesn't change there must be zero acceleration for fluid particles and yet we define a Reynolds number. What am I missing? Nick 

September 19, 2012, 03:43 

#2  
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Filippo Maria Denaro
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no, "inertial" refers to the lagrangian (convective) term, not to the eulerian (time derivative) term! is the ratio between convective and diffusive fluxes 

September 19, 2012, 04:10 

#3 
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by the convective term you mean (u.del) u where u is the velocity right? well this is also zero for pipe flow for instance where velocity doesn't change on a streamline


September 19, 2012, 04:31 

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September 19, 2012, 05:57 

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Serge A. Suchkov
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rho * u^2 has the dimension of pressure) may be just rho * u ?
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http://openhyperflow2d.googlecode.com 

September 19, 2012, 07:10 

#6 
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September 19, 2012, 07:57 

#7  
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Quote:
where: Dh=Hydraulic diameter of the transversal area V= Velocity of the fluid visc = Cinematic viscosity of the fluid 

September 19, 2012, 08:24 

#8 
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Serge A. Suchkov
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I guess it is term misunderstanding (math vs phys). The Reynolds number it is ratio of forces (inertial vs viscous). Term momentum flux has force (per volume unit) dimention for momentum equation.
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http://openhyperflow2d.googlecode.com Last edited by SergeAS; September 19, 2012 at 08:41. 

September 19, 2012, 12:34 

#9 
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Filippo Maria Denaro
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the meaning of the Re number is even more general ... for example, it can stand for the ratio between the convective to the diffusive time scales...


September 19, 2012, 21:41 

#10 
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Nick
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I guess you don't see my point. Take a control volume in a pipe where velocity profile is not changing with distance. Now if we compute the convective fluxes over the ENTIRE control volume, it is ZERO because the right and left surfaces cancel each other out and there's no flux thru the top and bottom surfaces. On each side (say the right surface of the control volume) the momentum convective flux is Integral of (u ro u.dA )/A , this is nonzero. So at each crosssection the rate of momentum is that Integral (u ro u.dA) which in dimensional analysis is ro u^2 but I don't like the name "the inertial force" it doesn't make much sense to me. Since inertial suggests acceleration. Unless someone has a better physical interpretation for me. Also, for the same example, the lagrangian convective term is actually zero since the velocity of a fluid particle don't change as they convect downstream.


September 20, 2012, 03:11 

#11  
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So I do not understand your doubt 

September 20, 2012, 03:30 

#12 
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Nick
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so again..here's the problem ..saying the reynolds number is the ratio of inertial FORCES to viscous forces is misleading ...here's why I think this is so.... look...from what inertial force means in dynamics : it is the mass times acceleration right? what else could it mean? so when fluid particles DO NOT ACCELERATE (like in a pipe where velocity profile is constance over distance) there is NO inertial forces...but we have a Reynolds number ..so saying that Re is the ratio of inertial to viscous forces is not accurate...hope that makes sense.


September 20, 2012, 03:49 

#13  
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In fluid dynamics the total acceleration is decomposed in a = du/dt + u*grad u du/dt is the Eulerian acceleration u*grad u is the "lagrangian" acceleration u*grad u can be written as div * (u u) using the mass equation. Therefore, we define the Reynolds number as ratio of "partial" acceleration, without involving the eulerian acceleration du/dt. The nondimensional number that takes into account for this acceleration is the Strohual number. 

September 20, 2012, 04:41 

#14 
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Nick
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Yeah u.grad u is zero if.velocity doesn't change with distance.like a pipe situation yet we have Reynolds number


September 20, 2012, 07:05 

#15 
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Filippo Maria Denaro
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but you don't use u *grad u ... you take a characteristic velocity of the problem, for example the averaged velocity U, a characteristic lenght of the problem, that means a lenght in the direction where the velocity changes (in channel where u=u(y) you take L = height).
Then [rho*U^2/H] / [mu* U /L^2] = Re 

September 20, 2012, 08:39 

#16 
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Nick
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I know how.to.calculate the Reynolds number ..my problem was that the term inertial force is not accurate because inertia is zero in that example. I've said this a few times now inertial force is zero if velocity doesn't change with distance.. saying that Reynolds is the ratio between inertial and viscous forces is not accurate


September 20, 2012, 08:48 

#17  
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we simply do not agree on that definition of inertial, no problem Note that, rigorously, in channel flow the motion is driven by the pressure gradient Delta P/L, therefore: P/L = O(rho*U^2/L) that balance the viscous term. Maybe using the "pressure force" is more appealing for you 

September 20, 2012, 08:56 

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Just a quick question for my own edification  if the velocity is not changing with distance, then aren't the viscous forces also zero? Wouldn't the whole flow system become trivial, rendering the Reynolds number moot?


September 20, 2012, 09:07 

#19 
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Filippo Maria Denaro
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In channel flow the velocity changes with distance y alone if the flow is fully devoloped ... however, channel flow can be also the case in which the inviscid velocity inlet (for example a plug profile) develops into a boundary layer... then u = u(x,y). In all cases the Re definition can be suitably introduced


September 20, 2012, 10:08 

#20  
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