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November 27, 2003, 09:11 |
Defining time step size
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#1 |
Guest
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OK, this is a simple problem, that I haven't got my head round,
I have a test case to model, which has a characteristic height 0.066m (as used for E in K-E). For running the URANS sims for vortex shedding, in kelvin-helmhotz (K-H) form. I have taken the size of the first K-H instability, and calculated the time the free-stream velocity takes to pass over it, gave it 30 steps over the motion, which results in a timestep of 0.000006 sec. Is that too small? The Cd and Cl traces are extremely unsteady, varying between 10 and 0 for Cl( high as 30 - -15 near start) and 5 and 0 for Cd (30 - -5 near start). Length of 1st K-H approx = 0.005423m (0.000180 sec/periodic motion) Length of test case = 0.371m vel = 30m/s (Re approx 9*10^5) Thanks for any advice. |
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December 1, 2003, 11:07 |
Re: Defining time step size
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#2 |
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is this the right way to go about it? anyone?
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