# What's the meaning of the Reynolds number

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 September 19, 2012, 02:02 What's the meaning of the Reynolds number #1 Senior Member   Nick Join Date: Nov 2010 Posts: 125 Rep Power: 7 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
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Filippo Maria Denaro
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Quote:
 Originally Posted by Nick R 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

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 Senior Member   Nick Join Date: Nov 2010 Posts: 125 Rep Power: 7 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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Filippo Maria Denaro
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 Originally Posted by Nick R 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
I mean the ratio of the fluxes, the convective flux is rho * u^2

September 19, 2012, 05:57
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Serge A. Suchkov
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 Originally Posted by FMDenaro I mean the ratio of the fluxes, the convective flux is rho * u^2
rho * u^2 has the dimension of pressure) may be just rho * u ?
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September 19, 2012, 07:10
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Filippo Maria Denaro
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 Originally Posted by SergeAS rho * u^2 has the dimension of pressure) may be just rho * u ?
no, not at all, the ratio is
[convective flux]/[diffusive flux] = rho*u^2/ (mu * u/L) = rho*u*L/mu = Re

September 19, 2012, 07:57
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Marco
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Quote:
 Originally Posted by Nick R 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
You can define it as Re=Dh*V/visc

where:

Dh=Hydraulic diameter of the transversal area
V= Velocity of the fluid
visc = Cinematic viscosity of the fluid

September 19, 2012, 08:24
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Serge A. Suchkov
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 Originally Posted by FMDenaro no, not at all, the ratio is [convective flux]/[diffusive flux] = rho*u^2/ (mu * u/L) = rho*u*L/mu = Re
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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Last edited by SergeAS; September 19, 2012 at 08:41.

 September 19, 2012, 12:34 #9 Senior Member   Filippo Maria Denaro Join Date: Jul 2010 Posts: 2,500 Rep Power: 31 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 Senior Member   Nick Join Date: Nov 2010 Posts: 125 Rep Power: 7 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 non-zero. So at each cross-section 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
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Filippo Maria Denaro
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Quote:
 Originally Posted by Nick R 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 non-zero. So at each cross-section 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.
u^2/L is actually an acceleration .... as well as mu*u/rho/L^2 ... the ratio is again the Re number... Note that in channel the surface integral of the mass flux (rho*u) is zero over the volume, not the mass flux alon ... the same comment is valid for the integral of the convective flux, not for the single flux ....
So I do not understand your doubt

 September 20, 2012, 03:30 #12 Senior Member   Nick Join Date: Nov 2010 Posts: 125 Rep Power: 7 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
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Filippo Maria Denaro
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 Originally Posted by Nick R 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.
The classical physics of a rigid body expresses the Newton law as m*a = F, this is a Lagrangian framework wherin a is the total acceleration. Here you see the inertial term...
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 non-dimensional number that takes into account for this acceleration is the Strohual number.

 September 20, 2012, 04:41 #14 Senior Member   Nick Join Date: Nov 2010 Posts: 125 Rep Power: 7 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 Senior Member   Filippo Maria Denaro Join Date: Jul 2010 Posts: 2,500 Rep Power: 31 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 Senior Member   Nick Join Date: Nov 2010 Posts: 125 Rep Power: 7 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
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Filippo Maria Denaro
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Quote:
 Originally Posted by Nick R 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

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 #18 Senior Member   Join Date: Jul 2009 Posts: 230 Rep Power: 11 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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 Originally Posted by agd 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?
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
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Nick
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Quote:
 Originally Posted by FMDenaro 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
How do you define inertial force? You.previously said that it is the convective term in the navier stokes equations..that term is zero for.the.described channel flow.

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