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subhra March 27, 2004 16:02

steady unsteady
 
Hi, I am a graduate student..Can somebody help me find an answer to following two doubts I have:

1) Do Problems that give rise to physical instabilities will always give rise to numerical instability?

2) What can be the expected consequence of trying to solve numerically the following laminar problem with a steady-state solver?

Flow in a 2-D channel with a cylinder installed at its center(Re=200, say)

An actual experiment with shows a periodic shedding of vortices, the solution never gets time-independent..

Jim Park March 27, 2004 18:25

Re: steady unsteady
 
1. No. Damping (numerical viscosity) can overpower physical instability.

2. Depends on the amount of numerical viscosity. The examples I've seen (hardly all of course) show a steady-state wake (separation, recirculation) behind the obstruction.

subhra March 27, 2004 19:22

Re: steady unsteady
 
Thank you very much. Should I conclude :

(1) With high numerical viscosity what I get from the crazy idea of steady-statesolving is soln. corresponding to much lower reynold's no. than 200 (say 20).

(2) With a low numerical viscosity (say by maintaining second or higher order everywhere ), no steady numerical solution should be possible. (divergence!)

Jim Park March 28, 2004 17:56

Re: steady unsteady
 
1. Yes, that's what I've seen. A high level of damping will force a 'steady state' solution that would have at least some of the features of a steady flow at a much lower Reynolds number.

2. I don't know what would happen in this case. And I sure would be skeptical of any solution obtained.

P. Birken March 29, 2004 05:32

Re: steady unsteady
 
1) What do you mean with physical instability? A blowup? The classical Runge Kutta method can solve the equation x'=x perfectly well. What makes numerical computations very difficult is, if changes in the initial data have a huge impact on the solution. Then you have to be very careful, but it is still possible to compute good solutions.

2) I don't know :)

Tom March 29, 2004 05:47

Re: steady unsteady
 
1. Such a large drop in the Reynolds number would imply that your solution is not converged. In a viscous flow simulation the numerical viscosity should be smaller than the real viscosity.

2. There are two possible outcomes depending on your steady state solver: (i) you reach a steady state - which is a solution to the problem but may not be physically realizable or (ii) the iteration scheme oscillates and so does not converge.

Anton Lyaskin March 29, 2004 06:58

Re: steady unsteady
 
What you get from steady-state solving of such problem - depends on the solver. I've made such an experiment myself once. The result was pretty weird - I've got a "frozen" Karman vortex street which was "switching" half a period per iteration

subhra March 29, 2004 14:12

Re: steady unsteady
 
The 'switching' really interesting and makes some sense too...


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