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I am attempting to try to find out if there is any effect of Von Karman vortices on a group of wind speed readings where it is presumed that due to a mountain nearby the data collection spot Von Karman vortices were formed which would register in the data. Based on what I have read the vortex shedding frequency which should be present in the data is simply calculated from the formula $F= \mathbb{St}\cdot \frac{V}{L}$ with $\mathbb{St}$ as the Strouhal number, $V$ as the velocity and $L$ as the characteristic length. From what I understand the Strouhal number is essentially constant for a given flow, but when I looked up how to find the Strouhal number it was defined as $\mathbb{St} = F\cdot \frac{L}V$, which is the exact same relationship as given in the first equation for the computation of the shedding frequency. My question then is in two parts:

1) How do I calculate the Strouhal number for a flow so that I can calculate the shedding frequency?

2) What should I use as my characteristic length for a landscape feature such as a mountain or cliff etc as everything online tells only in the case of uniform cylinders and using their diameter instead of something like a cone which can be used as an approximation of a mountain or other landmass.

thank you for any help.

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1 Answer 1

up vote 6 down vote accepted

So how this is done is a bit of a black art, much like how you choose what to use for other non-dimensional numbers in fluids (like Reynolds number). But you sort of have it backwards. You don't want to compute $St$ to find the shedding frequency; $St$ is good if you want to compare flows over different conditions but want to show the physics is the same, or it's good if you were given the number, velocity, and length scale from somebody and need to compute the frequency.

The best way, if you can, to compute the shedding frequency is to take a time signal of your velocity and compute the FFT of it and look for spikes in low-frequency ranges. Depending on the frequencies of the vortices, you may not have the appropriate temporal resolution. For instance, if the vortex is shedding at 1 Hz (1 vortex per second), then you need to take at least 2 samples per second due to frequency aliasing.

Assuming you have a frequency computed, your next choice is the velocity. Again, there is some art in choosing the correct one. The best choice here is the wind-speed ahead of the mountain/cliff if you have that data (or outside of the shed vortices behind the mountain so you are getting a "free-stream" velocity). Any velocity you choose from inside the vortex shedding would have to be averaged to remove the vortex effects. But it's unlikely you'll have steady data and so your average will be a moving one, not very helpful.

Lastly is the choice of length. Cylinders choose the diameter typically; bluff bodies or backward facing steps choose the step height. So for a cliff, the height of the cliff is a good number. Same could be said for a mountain.

Ultimately though, what you choose doesn't matter so long as you choose it explicitly and consistently! So if you want the length scale to be the height of the cliff, always use that one when comparing $St$ from various conditions. Same goes for the other terms; if you are publishing, please be very clear about what you are choosing. It's very frustrating to try and compare results with incomplete information and makes it very suspicious when authors don't list their reference scales and only present the resulting non-dimensional number!

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Thank you so much. This really helps out a lot. –  Patrick Jul 8 '13 at 6:47
@Patrick No problem, happy to help. If this answered your question, don't forget to upvote it and/or select it as the right answer so the question doesn't stay open and "unanswered" forever! –  tpg2114 Jul 8 '13 at 23:45
This did answer my question. Thanks for the answer. Unfortunately I cannot up vote your answer as it requires 15 reputation and I only have 13. your answer deserves it though. –  Patrick Jul 9 '13 at 18:36
@Patrick Eventually you'll get there :) –  tpg2114 Jul 9 '13 at 18:37
Thanks. and now I do, literally 1 minute later. that is a +1 on the answer –  Patrick Jul 9 '13 at 18:40

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