# vortex shedding frequency

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 August 22, 2013, 14:57 vortex shedding frequency #1 New Member   Ali Join Date: Nov 2012 Posts: 21 Rep Power: 5 Hi all Will somebody help me that what is the most simple way to calculate the vortex shedding frequency or the Strouhal number for flow past a circular cylinder?(2D)

 August 22, 2013, 15:12 #2 Senior Member   Join Date: Jul 2009 Posts: 232 Rep Power: 11 Compute the lift, store the time history of the lift, and then do an FFT of the lift history. Or, simply measure the period of the cycle from the lift history and invert.

 August 22, 2013, 15:17 #3 Super Moderator   Sijal Join Date: Mar 2009 Location: Islamabad Posts: 4,301 Blog Entries: 6 Rep Power: 43 CFD is the most comperhensie and simple way to solve these problems.

August 22, 2013, 15:35
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Quote:
 Originally Posted by agd Compute the lift, store the time history of the lift, and then do an FFT of the lift history. Or, simply measure the period of the cycle from the lift history and invert.

That's absolutely correct, just make sure that you take enough cycles into account before doing FFT, excluding any initial transients (e.g. when starting from potential or constant flow)

August 30, 2013, 20:37
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Quote:
 Originally Posted by agd Compute the lift, store the time history of the lift, and then do an FFT of the lift history. Or, simply measure the period of the cycle from the lift history and invert.
Yes but first computing the lift is not straightforward,and secondly theoretically the lift for a cylinder should be null because of the symmetry of the cylinder.

Just record the time history of the vertical velocity on a fix point in the wake, it's enoughand perform an FFT as Agd said.

 August 30, 2013, 22:30 #6 Senior Member   Join Date: Jul 2009 Posts: 232 Rep Power: 11 While the time-averaged lift will be zero, the instantaneous lift force should oscillate at the shedding frequency. Computing the lift is extra effort, but it can provide additional information on the quality of the solver - for example by checking that the time-averaged lift actually does equal zero.

August 31, 2013, 07:15
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Quote:
 Originally Posted by agd While the time-averaged lift will be zero, the instantaneous lift force should oscillate at the shedding frequency. Computing the lift is extra effort, but it can provide additional information on the quality of the solver - for example by checking that the time-averaged lift actually does equal zero.

The lift of a symmetric body is null. You can not lift a sphere or cylinder with a circular base.
For the drag force your are right

 August 31, 2013, 09:15 #8 Senior Member   Join Date: Jul 2009 Posts: 232 Rep Power: 11 Um, no. See for example http://www.dtic.mil/dtic/tr/fulltext/u2/a273243.pdf The instantaneous lift force on the cylinder is not zero. Th average lift force is zero. The average drag will not be zero, and the instantaneous drag will oscillate at twice the Strouhal frequency.

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