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transient problem run in steady state

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Old   March 11, 2008, 23:38
Default transient problem run in steady state
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luigi
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when I run a time-dependent problem using the steady state solver, what are my solutions representing? Is it a snapshot in time, or is it time-averaged. For example, if I am modeling vortex shedding, but running the problem in steady state, what do my velocity contour plots represent? If it is time-averaged, if I ran it long enough, would it show symmetric flow? Just wondering, thanks
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Old   March 12, 2008, 02:04
Default Re: transient problem run in steady state
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Syed Shabbar Raza
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final results
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Old   March 12, 2008, 14:36
Default Re: transient problem run in steady state
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Paolo Lampitella
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I'm not sure, but maybe you should not be able to obtain a final result because the solver will diverge or something like that. Maybe it's problem dependent... i never tryed it.
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Old   March 12, 2008, 14:46
Default Re: transient problem run in steady state
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luigi
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a final result implying it is a snapshot in time because the flow field changes with time? thanks
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Old   March 13, 2008, 07:54
Default Re: transient problem run in steady state
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Isaac
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If the problem is truly time dependent, the solution you get solving it as steady state will not be a correct representation. I remember working on a problem of votex shedding where the residuals appeared as converged (setady state) but on closer observation, it was evident that the mass was not converged between in and out. The residuals showed oscillations. So though they were below 1e-03, the solution was not really converged. You can probably look at it as "one of the solutions" but not the correct representation. That is what I think. Just an educated opinion.
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