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June 18, 2002, 22:18 |
Karman vortex street
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#1 |
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Hi, everybody.
I have a trouble with the calculation of the Karman vortex street. It is a problem of flowing over a cylinder. The parameters that I used in Fluent(2D Segregated) are listed below: ++++++++++++++++++++++++++++++++++++++++++++++++++ +++++ The size of domain :length:40m width:15m The diameter of cylinder :2m The location of cylinder 15,7.5) Boundary conditions : left boundary -- Velocity Inlet right boundary -- Pressure Outlet top and bottom boundary -- Wall cylinder -- Wall The properties of air :density:1.225kg/m3 viscosity:1.7894e-05kg/m-s X-Velocity in Velocity Inlet :7.3037e-04m/s (constant),so Re=100. Viscous Model :Laminar Time step size :0.01s Number of Time steps :20000 ++++++++++++++++++++++++++++++++++++++++++++++++++ +++++ But I did not see the Karman vortex street. What shall I do? Please give me some advice, thanks a lot. |
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June 19, 2002, 05:17 |
Re: Karman vortex street
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#2 |
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try wather instead of air
use realizable k-eps-turbulence-model instead of laminar flow. good luck, Laika, still orbiting |
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June 19, 2002, 21:46 |
Re: Karman vortex street
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#3 |
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I have done the calculation with your method, but it doesn't work. Any suggestions?
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June 19, 2002, 22:52 |
Re: Karman vortex street
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#4 |
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At that low reynolds number the flow may be steady. Either perturb the flow with asymmetric initial conditions. Or try a slightly higher Reynolds number. From memory I think it is more like Re=300 when the vortex street develops.
Don't use a turbulence model because the flow isn't turbulent and turbulence with just damp unsteadyness. Steve |
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June 20, 2002, 01:04 |
Re: Karman vortex street
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#5 |
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Anyway, at first, I think that you should try to give pertubation to the steady solution to see the unsteady flow. Otherwise, even though the flow is unsteady, you can see the unsteady flow after (maybe) couple of millions time step, at which time, numerical truncation error can be 'artificial pertubation.
Sincerely, Jinwook |
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June 24, 2002, 12:00 |
Re: Karman vortex street
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#6 |
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Besides the other comments on making sure you start your time-accurate simulation from some "perturbed" state, you may also want to increase your time step to something larger that 0.01 s, such as 16 sec.
I say increase your time step based on the following: Let time setp dt = (period of lift)/1000 (actually even a few hundred time steps per period is probably enough, but 1000 should be plenty). period of lift = 1/freq = D/(U * St) D = cylinder diameter = 2 m U = uniform inlet speed = 7.3037e-04 m/s St = Strouhal number = 0.165 for Re=100 cylinder period of lift = 1.66e+04 sec dt = 1.55e+04/1000 = 16.6 sec |
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