# boundery conditions for Karman vortex street

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 November 3, 2003, 08:15 boundery conditions for Karman vortex street #1 Rainer Guest   Posts: n/a Hello guys, i´ve got a big problem. I´ve tried the karman vortex street wich Star CD. But it doesn´t work. Is there a trick in the setup of the boundaries? The cylinder has a wall, no-slip condition. The other walls are wall, slip conditions. My inlet is 1,5 m/s and the outlet is pressure out. The Grid ist completely symmetrical. I´ve tried the K-E low reynolds in all forms, including square and cubic. I´ve tried a laminar solution but this also didn´t work. Hope someone can help me Regards Rainer

 November 3, 2003, 10:03 Re: boundery conditions for Karman vortex street #2 Brian Guest   Posts: n/a Umm, sorry to ask the obvious, but you ARE doing this as a transient solution, right? Make sure that your Reynolds Number is in the correct range for vortex shedding (above 40 I think). Also, your timestep must be quite small to capture the onset of instability, and I have found that using double-precision helps. If you still don't get vortices, try putting in a small trigger by deleting a tiny cell on one side of your obstacle, introducing a slight asymmetry. A Google search for "von Karman vortex Reynolds Number" will yield lots of advice. GL.

 November 3, 2003, 10:39 Re: boundery conditions for Karman vortex street #3 Rainer Guest   Posts: n/a Thank you brian. My solution ist transient, I forgot to mention. Reynolds Number ist about 400 I think. The timesteps are 0,001 seconds. So I will try use double-precision. I´ve also tried to set the cylinder out of symmetry, but this doesn´t work. So I will try the trick with the cell. Regards Rainer

 November 4, 2003, 04:49 Re: boundery conditions for Karman vortex street #4 Anton Lyaskin Guest   Posts: n/a How far is your outlet from the cylinder? And what do you get instead of vortex street? A symmetrical pair of vortices? May be you just have to wait longer for the onset of instability... You can also estimate the frequency of vortex shedding from consideration Sh=0.21 - your time step must be small enough to have at least 20 time steps on a period. And may be you have to use Cranck-Nicolson, which is more accurate in time.

 November 4, 2003, 05:26 Re: boundery conditions for Karman vortex street #5 Rainer Guest   Posts: n/a Hi Anton, the outlet is about 1500mm behind the cylinder. The cylinder-diameter is 100mm, the grid is 300mm wide, the offset off the inlet to the cylinder is 250mm. I´ve got no vortices at all. The result looks like it is calculated with steady state. That was the first result I got. It was calculated transiend but it looks like steady state. Now I will try the trick of Brian, I´ve deleted one cell of the cylinder and the first result looks pretty good I think. Hope it would get better. Regards Rainer

 November 5, 2003, 07:50 Re: boundery conditions for Karman vortex street #6 John L Guest   Posts: n/a What is your inlet condition and the distance from the cylinder. It's essential to set you inlet away from the cylinder and have proper inlet condition (velocity and k-e profiles) - to do this, you can simulate a simple channel flow with same mass flow rate and use the results as inlet b.c.

 November 6, 2003, 07:06 Re: boundery conditions for Karman vortex street #7 Anton Lyaskin Guest   Posts: n/a With Re=400 and D=100mm you should have velocity of 6e-2 m/s only. So you'll have to wait for more than 4 seconds until flow from the inlet will reach your cylinder! And the onset of vortex shedding will be after some time of the same order. So probably you really have to wait longer. Besides the width of the channel is too small and the blockage effect can damp the oscillations (I was making such simulations before). You can try making your grid about 5 times wider.