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Old   February 18, 2009, 07:24
Default Vortex Street
  #1
Blunderbus
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I was wondering if someone could explain what causes a Karman vortex street?

Several years ago I took part in an experiment involving a fluid flow past a cylinder. Powder was sprinkled on the top of the water surface allowing flow patterns to be observed. The experiment allowed us to get a good look at the vortex street pattern downstream of the cylinder.

What I don't understand is the mechanism by which this symmetrical flow suddenly becomes asymmetrical. Is this due to small variations in the cylinders surface/flume wall or some other effect?
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Old   February 18, 2009, 11:57
Default Re: Vortex Street
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Timon
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The asymmetry can be explained from circulation viewpoint. The shedding of a vortex results in a change (with net isolated circulation) must be balanced by the circulation of the around the cylinder surface. This is achieved through a change (ie. asymmetry) in the pressure distribution.
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Old   February 18, 2009, 12:01
Default Re: Vortex Street (revisted)
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Timon
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A reconstruction of last post (because of crappy sentence rearrangements):

The asymmetry can be explained from a circulation viewpoint. The shedding of a vortex (with net circulation) must be balanced by the circulation around the cylinder surface. This is achieved through a change (ie. asymmetry) in the pressure distribution.
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Old   February 18, 2009, 12:49
Default Re: Vortex Street (revisted)
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CFD_nerd
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Isn't circulation an asymmetric phenomena to begin with? it gives rise to a different flow field on either side of the cylinder.

If the flow is initially (percectly) symmetric about a (perfectly) symmetric cylinder then would we see a vortex street? I don't understand how the phenomena begins unless there is some microscopic asymmetry on the flow boundaries, which causes a disturbance in the flow that grows in amplitude - giving rise to asymmetric behaviour.
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Old   February 18, 2009, 13:21
Default Re: Vortex Street (revisted)
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ag
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As the Re number increases the symmetric solution becomes neutrally stable, and any infinitesimal perturbation will cause the symmetric solution to branch into an asymmetric solution. Physically, this could be a slight boundary asymmetry, or an incoming flow irregularity.
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Old   February 19, 2009, 05:55
Default Re: Vortex Street (revisted)
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Timon
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True. The circulation theorem does not provide an explanation for the asymmetric flow as such. It does however give an explanation for the asymmetric pressure distribution (change) when a vortex vortex is shed. As this is a consequence of viscous boundary layer separation, you're obviously right that the asymmetry ultimately is a consequence of some viscous boundary layer disturbance (as is the case for all vortex shedding).

Although the origin is indeed viscous, the process can be understood through inviscid theory. The fact that the vortex street is asymmetric (ie. vortices are shed alternatingly) follows from a small-disturbance stability analysis of the potential vortex street (as done by Karmann). This analysis shows that only the alternate vortex street is conditionally stable for certain geometrical arrangements. See http://eom.springer.de/V/v130110.htm for some additional info.
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Old   February 24, 2009, 19:47
Default Re: Vortex Street
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momentum_waves
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An answer that is rather new, framed from the wave paradigm view of the N-S, is as follows:

Two momentum wave-forms are set up, one from the upper portion of the cylinder, the other from the lower portion. Depending on the flow regime, these waveforms couple & interact. The degree of wave coupling will increase with increasing Reynolds number. With this scenario in mind, turbulence is explained as extremely tight wave coupling, off multiple solution branches.

Take a look at <http://adthermtech.com/documents/RAEFM2_paper_ID024.pdf> and <http://adthermtech.com/documents/RAE...sent_ID024.pdf>

Have fun.

mw...

Momentum waves <www.adthermtech.com/wordpress3>
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