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Old   August 22, 2013, 14:57
Default vortex shedding frequency
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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)
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Old   August 22, 2013, 15:12
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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.
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Old   August 22, 2013, 15:17
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Old   August 22, 2013, 15:35
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Quote:
Originally Posted by agd View Post
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)
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Old   August 30, 2013, 20:37
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Quote:
Originally Posted by agd View Post
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.
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Old   August 30, 2013, 22:30
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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.
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Old   August 31, 2013, 07:15
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Quote:
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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
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Old   August 31, 2013, 09:15
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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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