At what critical Reynolds number does vortex shedding begin?

  • #1
40
1
In:

"Fluid Dynamics", Chapter 3 (Turbulence), Section 26,

Landau and Lifchitz analyze the problem of the stability of a steady flow past a body of finite size.

The fluid is assumed to be incompressible and they reach the conclusion that perturbations that deviate from steady flows start to grow when a critical Reynolds number is reached (ASIK, this critical Reynolds number in unrelated to the ##\mathbb{Re}_c## at which the laminar flow becomes turbulent).

They also deduce that the amplitude of the perturbation grows proportional to:

$$A\propto\sqrt{Re-Re_c}$$.

Would this be at the origin of vortex shedding?

What's the name of this critical Reynolds number?

How does this relate to Strouhal's number?

Thank you very much in advance.
 

Answers and Replies

  • #2
40
1
Well this is leading me nowhere again.

This is what I think:

  1. Vortex shedding introduces a NATURAL time scale.
  2. The speed of sound is the maximum velocity at which information travells inside the fluid.
  3. Assume that the incoming fluid velocity increases: the natural time scale decreases and viscidity decreases (eddies live longer) while the speed of sound stays the same.
  4. This situation is untenable. At sufficiently long distances from the obstacle (downstream), eddies become completely aloof of whatever is happening upstream. Sufficiently far away of the obstacle, downstream, you're no longer in the obstacle's future cone.
At some some stage, the general properties of the fluid MUST change, in order of this nonsense (non-causality) not to happen.
 
  • #3
cjl
Science Advisor
1,864
439
Why do you think that objects sufficiently far downstream are unable to be affected? In a subsonic flow, disturbances can propagate from any region in the flow to any other region in the flow, given sufficient time.
 
  • #4
bigfooted
Gold Member
609
122
At ##Re_c## the transition from a steady laminar flow to an unsteady laminar flow occurs. It's just called the critical Reynolds number, which can indeed be confusing because there are many critical Reynolds numbers. For ##Re<Re_c## the flow is steady, so you do not have a shedding frequency. For ##Re>Re_c##, your shedding frequency increases with ##Re##. So ##Re_c## is also the point where your Strouhal curve starts. For flow around a cylinder, ##Re_c=46##
 
  • #5
boneh3ad
Science Advisor
Insights Author
Gold Member
3,204
888
Let's start with a few baseline statements about boundary-layer stability and transition:
(1) In general, there is no critical Reynolds number where laminar flow becomes turbulent. You can apply some simplistic ideas like that to pipe flows and a handful of other special cases, but even then, the underlying physics involve the growth of small disturbances until they become large and break down to turbulence. In other words, even in a pipe flow, passing that critical ##Re## doesn't immediately mean turbulence. It means the flow is unstable and will soon be turbulent.

(2) The comment in Landau & Lifschitz appears to be describing the concept of the minimum critical Reynolds number, ##Re_{cr}##, which is the smallest ##Re## at which small disturbances to the base flow becomes unstable. It should be noted, however, that this does not imply that turbulence is inevitable; it only means that there is at least some degree of instability present, but three may not be enough growth to lead to transition in a given situation. ##Re_{cr}## used to be the state of the art in boundary-layer stability and transition. These days, it is long-deprecated (though still an important piece of the puzzle sometimes).

(3) Consider the generic definition of ##Re##
[tex]Re_{\ell} = \dfrac{U \ell}{\nu}.[/tex]
Also, for illustrative purposes, consider the Blasius boundary layer thickness,
[tex]\delta \propto \sqrt{\dfrac{\nu x}{U}}.[/tex]
So, if we define a ##Re## based on this reference length, we have
[tex]Re_{\delta_r} = R = \dfrac{U}{\nu}\sqrt{\dfrac{\nu x}{U}} = \sqrt{\dfrac{Ux}{\nu}} = \sqrt{Re_x}.[/tex]
What this says, then, is that ##R## is equivalent to the square root of the ##x## Reynolds number. If you look at what Landau & Lifschitz did, their ##A## parameter is related to this concept (though not identical). ##R## is a very common parameter in the study of stability.

One general comment:
(4) Subsonic flows are governed by elliptical equations, and as such, what occurs at one location in the flow field affects the rest of the flow field. The influence of a body does die away at a distance, but it is never truly zero.

And some comments on vortex shedding in light of the above:
(5) While vortex shedding does not have a whole lot of shape-independent rules of thumb, it is generally a result of boundary-layer separation. This also means that it is inextricably linked to the idea of laminar-turbulent transition.

(6) For example, consider a cylinder. As ##Re## increases, the boundary layer separates and leads to large separation vortices. As ##Re## continues to increase, these trailing vortices begin to shed, creating the familiar vortex street with laminar vortices. At higher ##Re##, the boundary layers transition, causing the separation point to move rearward, the wake to shrink, and the vortices to stop shedding. At a higher ##Re##, the now-turbulent vortices begin shedding again. So, as you can see, there is no basic principle for a vortex shedding ##Re##.

(7) The above example was for circular cylinders (and spheres). There are infinitely many possible shapes that will experience vortex shedding, and they will all have different characteristics. The reason that ##R=\sqrt{Re_x}## is relevant is that it is a good parameter for correlations with laminar-turbulent transition, but it is not a hard and fast rule. Even on a subsonic flat plate with zero pressure gradient, we can't tell you what ##Re_{tr}## is in general, let alone more complicated shapes.
 
  • Like
Likes Anashim
  • #6
40
1
Thank you very much for your post.
 
  • #7
40
1
(4) Subsonic flows are governed by elliptical equations, and as such, what occurs at one location in the flow field affects the rest of the flow field. The influence of a body does die away at a distance, but it is never truly zero.
However, in fully developed turbulent flows, you have the "two-thirds law". How is this long-range correlation consistent with an elliptical PDE?
 
  • #8
boneh3ad
Science Advisor
Insights Author
Gold Member
3,204
888
Do you mind explaining why you think said correlation is not consistent with an elliptical PDE?

I'll preface the rest of this by noting that my field is not classical turbulence so I don't recall all of the details of Kolmogorov's work, and all of my texts on the subject are at work.

That being said, there is at least one assumption in Kolmogorov's work, that of isotropic and homogeneous turbulence, which is rarely a great description of a real flow.

However, assuming that turbulence can be captured exactly by the Navier-Stokes equations (this is generally agreed upon but not mathematically proven), then any such flow must still obey the rules of elliptical equations.
 
  • #9
40
1
Do you mind explaining why you think said correlation is not consistent with an elliptical PDE?
I am not a mathematician, but in Field Theory long-range correlations are always associated with massless propagators. Propagators is something I would immediately associate with a wave equation (i.e. an hyperbolic PDE, not an elliptic one). I'm sorry, but I cannot be more specific, as I said, I'm not a mathematician.
 
  • #10
boneh3ad
Science Advisor
Insights Author
Gold Member
3,204
888
The Kolmogorov 2/3 law is not a long-range hypothesis, though. It requires that the distance between the two points, ##r##, fall within the inertial subrange, which generally encompasses length scales that are quite small. It's a statistical treatment of the flow assuming homogeneous and isotropic turbulence as well, so it's only approximate in reality and only if the flow locally approximates those conditions. Certainly, applying it to a wake at two points that are very different in character would break those underlying assumptions.
 

Related Threads on At what critical Reynolds number does vortex shedding begin?

  • Last Post
Replies
1
Views
2K
Replies
3
Views
3K
  • Last Post
Replies
4
Views
3K
Replies
3
Views
5K
Replies
10
Views
886
  • Last Post
Replies
2
Views
2K
  • Last Post
Replies
10
Views
3K
  • Last Post
Replies
14
Views
10K
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
3
Views
7K
Top