transient problem run in steady state
when I run a timedependent problem using the steady state solver, what are my solutions representing? Is it a snapshot in time, or is it timeaveraged. 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 timeaveraged, if I ran it long enough, would it show symmetric flow? Just wondering, thanks

Re: transient problem run in steady state
final results

Re: transient problem run in steady state
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.

Re: transient problem run in steady state
a final result implying it is a snapshot in time because the flow field changes with time? thanks

Re: transient problem run in steady state
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 1e03, 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.

All times are GMT 4. The time now is 06:56. 