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 egge24 April 27, 2012 19:47

Differences between steady and transient simulations

Hi,

I'm simulating a cylinder in a rectangular tank and I would like to know what represents the steady solution.

At high Reynolds numbers in the transient simulation a von Kármán vortex street is formed, so at sucessive time steps vortex tavel and dissipate.

But in the steady simulation, does the solution represents a time average calculation? or what exactly represents?

 LuckyTran April 28, 2012 00:09

Quote:
 Originally Posted by egge24 (Post 357582) Hi, I'm simulating a cylinder in a rectangular tank and I would like to know what represents the steady solution. At high Reynolds numbers in the transient simulation a von Kármán vortex street is formed, so at sucessive time steps vortex tavel and dissipate. But in the steady simulation, does the solution represents a time average calculation? or what exactly represents? Thanks in advance
This discussion is not restricted to just flow variables, so I use the term variables to demonstrate that it is generalized (if difficult to follow just replace it with flow variables).

In a steady simulation, there is no time dependence of any variables. They are annihilated from the start. That is, it is assumed that there is no time-dependence. The governing equations in this case does not include time (transient effects).

In a transient simulation, the time dependence is kept in the variables. The governing equations contain the transient terms and the variables are time-dependent.

A steady simulation will tend towards the steady-state solution (if one exists).

A transient simulation will solve for the time-dependent solution. The time-averaged representation of the transient solution will yield a time-averaged solution.

Only in the case of when the original problem is steady will the steady and transient simulations yield the same results. i.e. if there is no unsteadiness in the problem then the two will yield the same results. The time-averaged solution of the transient simulation will be about the same as the steady solution.

do not confuse unsteadiness with variables changing with time. All flows are inherently variant in time but steady flows will have simple fluctuations about the average whereas unsteady flows will have fluctuations superimposed on top of a time-varying variable.

The von Kármán vortex street is not a steady flow. If you ran a steady simulation (and somehow got it to converge, which would be difficult) it would probably be garbage results and not accurately reflect the true time-averaged solution. To get the proper time-averaged flow field, you need to run an unsteady simulation and then take the average of the variables in time.

 egge24 April 28, 2012 20:52

Thanks LuckyTran! Your explanation help me to clarify my doubts.

Kind Regards

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