CFD Online Discussion Forums

CFD Online Discussion Forums (
-   Main CFD Forum (
-   -   steady and unsteady flow (

shisheng July 21, 1999 12:51

steady and unsteady flow
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

Re: steady and unsteady flow
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

Re: steady and unsteady flow
(1). It is likely that most real flow problems are unsteady to some extent. (2). You can run a transient code to obtain unsteady results. You can also run a transient code to obtain a steady-state result, if you can bring the solution to steady-state condition. But most likely, you will have some small oscillations and the computed results can not reach the true mathematical steady-state. (3). If you run a steady-state code, there is no way you will be able to get real transient solution. If the iterative process does not converge, then there is no solution at all. (4). So, for steady-state code, only the converged solution is the solution. (5). Unless you have carried out the modelling of the real flow problem properly, the CFD solution alone is just numerical solution to the governing equations and the associated boundary conditions. So, in general, there is no direct relationship between the real flow problem and the CFD solutions, unless the problem is properly modelled.( you can use the steady-state solution to simulate the steady-state condition of a transient process. But if you are interested in the transient behavior of the flow, then the only way to model it is to use the transient code.) (6). So, using a steady-state code will guarantee that you are not going to obtain the transient behavior of the flow. On the other hand, using a transient code will not guarantee that you will obtain a steady-state solution.( most people use transient code to obtain a steady-state solution anyway)

Hongjun Li July 21, 1999 17:06

Re: steady and unsteady flow
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

steady code for unsteady flow, don't understand.
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

Re: steady code for unsteady flow, don't understand.
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

Re: steady and unsteady flow
I think there some methods to solve this kind of problems. not very difficult.

Alexey July 23, 1999 06:00

Re: steady code for unsteady flow, don't understand.
In this case you MUST use unsteady methods only. It is obviously. The relationship between the numerical solution and the real flow physices is the approximation error like this const1*(space_grid_step)**nspace + const2*(time_step)**ntime , where nspace and ntime are orders of accurcy of your numerical scheme in space and time, respectively. Usually, steady methods treat the time step as an iteration parameter which may be very big. Becouse the time approximation error is big too.

All times are GMT -4. The time now is 05:06.