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 shisheng July 21, 1999 12:51

Hi, experts if the flow physices is unsteady, can I use the steady method (code) to simulate this flow problem? if not, what is the wrong? or the simulation is not convegent or the simulation results are false?

Best wishes!

 clifford bradford July 21, 1999 16:31

you can simulate anything you want, the computer doesn't care (and you might have convergence and all that) but as for you results - they'll be as trustworthy as a politician. you can modify steady codes to make them unsteady and some codes can do both but if the flow is unsteady the code must be as well. be sensible and don't cut corners

 John C. Chien July 21, 1999 16:54

 Hongjun Li July 21, 1999 17:06

It depends on the numerical schemes. Some schemes are good for both steady and unsteady solutions but most of them are not. Many schemes were developed to get final steady solutions and do not care HOW to get there, so there are many 'acceleration methods' to speed up convergence rate. (for example, local time step, pre-conditioning, relaxation, enthalpy damping, etc.). The intermidiate solutions (not converged solutions) have no physical meanings at all. Only the final converged solutions are physically meaningful. In another word, the destination is right but the path is a short-cut. Those schemes can not be used for unsteady flow calculations.

 shisheng July 22, 1999 06:10

thnak you, but don't understand

Hi, all,

thank you for your contributions. I mean if the flow physices is unsteady, for example, flow around a cylinde, Reynolds number is about in the transient regime, can we us the code for steady state (without time derivative item) to get the solution ? (naturally steady solution), but the flow phsices is unsteady, if we can get the numerical solution? is this solution phsically meaningful?

Best regards

 John C. Chien July 22, 1999 10:06

Re: thnak you, but don't understand

(1). As I said, there is no direct relationship between the numerical solution and the real flow, unless the flow is properly modelled. (2). You really have some difficulties in describing your problem. For flow over a cylinder, as the Reynolds number is increased to a certain range, the flow in the wake region will become unsteady. That is the physics, the real flow problem. (3). On the other hand, if you obtain a solution to the steady-state equation at the same Reynolds number, it is still all right because you are solving a mathematical problem. This has nothing to do with the real problem at all. The same is true, if you obtain a solution to the steady-state Euler equation, the solution will be inviscid solution and steady. In the real world, there is no steady-state, inviscid problem or solutions. (4). So, the key issue here is the "proper modeling" and the " proper interpretation" of the "CFD model" and the "CFD solutions". That is one has to make a "proper connection before and after performing CFD analysis". (5). In other word, steady and symmetric solution to the Navier-Stokes equations at a high Reynolds number for flow over a cylinder is just a mathematical solution. Without the "modeling" and "interpretation" stages, the solution has no physical meaning at all.

 Patrick Godon July 22, 1999 11:49

Why don't you just use a time-dependent code for your unsteady problem !? There are plenty for free in the Academia. Why would you use a steady modeling for an unsteady problem? I don't understand. If you know that the flow has no steady state solution, then again your model gonna be wrong and the flow you are modeling will give you a wrong answer, even if you'll a get a solution.

 shisheng July 22, 1999 13:01

thnak you John and Patrick,

thank you both for the suggestion, because I want to know the relationship between the numerical solution and the real flow physices. if my geometries are complex, at low Reynolds number the flow is perhaps steady, but when the Reynolds number is increased step by step, the flow changs from steady to unsteady (or periodic), I don't know at which Reynolds number this change will happen, so I want to know if at a Reynoldy number I can get not only steady solution, but also unsteady solution. But I don't know which solution is real solution. steaty or unsteady? which is the real solution?

Beat regards

 clifford bradford July 22, 1999 18:17

Re: thnak you, but don't understand

as john say's, if you use a steady code you will get a steady result but this is only a solution to the "equations of steady flow" which are only an approximation to the real thing.(even the full unsteady reynolds averaged blah blah blah equations are only an approximation to the solution) if you want to see unsteady phenomena like vortex shedding you must use an unsteady code there's no way around it. even if you're modelling a situation (say at higher Re) where where the wake is "steady" the steady solution will only be a kind of "smearing" of the real thing. to see the fine features in the wake which are by definition unsteady you need an unsteady code.

 Jason Wang July 22, 1999 20:17