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-   -   Why does unsteady problem tend to be steady? (http://www.cfd-online.com/Forums/main/4415-why-does-unsteady-problem-tend-steady.html)

wang February 14, 2002 06:10

Why does unsteady problem tend to be steady?
 
Hi, all

I have just began to calculate the flow around a circular cylinder. At the beginning, the flow seems to be quasi-periodic. But at the end, it goes to a steady stage with two vortices behind the cylinder. What would be an explanation because it is unphysically? Can anyone give an explanation about why or how this would happen?

Thank you very much.

Fred Uckfield February 14, 2002 07:02

Re: Why does unsteady problem tend to be steady?
 
Too much diffusion (probably numerical) can not sustain transient behaviour.

wang February 14, 2002 09:07

Re: Why does unsteady problem tend to be steady?
 
thank you Fred Uckfield,

but how can I avoid such a problem. Or how could I evaluate the result. For example, If I have a airfoil at high angle of attack, how I can get the "physical" values of lift or drag, or vortex-shed frequency?

thanks

Smith February 14, 2002 09:32

Re: Why does unsteady problem tend to be steady?
 
What is the Reynolds number? Remember you only get periodc vortex shedding in the range 60 < Re < 5000

wang February 14, 2002 10:31

Re: Why does unsteady problem tend to be steady?
 
Thank you Smith,

but the situation is, even in such a range of Re number, the simulation (using RANS) goes to a steady state. What could I do to evaluate the result ?

Thanks

sylvain February 14, 2002 12:25

Re: Why does unsteady problem tend to be steady?
 
RA in RANS means Reynolds Average, the average is made over realization. That's mean, you make N experiments (realizations) and at the same time and at the same position for each realization you measure the velocity and the pressure. Then you average the N values for each quantities and you get the Reynolds average value of this quantities.

For the flow past a cylinder, I doubt that for each realization at the same time and at the same position you will have the same eddy. So the average over the realisation is steady. That doesn't mean that the flow is steady, that's mean the flow is statictically steady.

Anyway, that's doesn't prevent you from finding unsteady RANS solution, but I wonder what the turbulent model mean in such a case.

Cheers,

Sylvain

Steve February 14, 2002 16:58

Re: Why does unsteady problem tend to be steady?
 
The ensemble averaging used to derive the RANS equations is based on a turbulent time scale that is several orders of magnitude smaller than the period of vortex shedding for this flow. Thus ensemble averaging removes the small scale turbulent fluctuations, while leaving the transient behaviour of the mean flow. Therefore the simulations should still be transient. The flow isnt statistically steady.

The problem with the mentioned simulations is that the boundary layer is not turbulent in the Reynolds number range 60 < Re < 5000. The boundary layer is laminar and the wake (possibly) tubulent. RANS models will unphysically introduce turbulent diffusion in the boundary layer that will damp the boundary layer instabilities that cause this flow to be unsteady.

Therefore you should use the standard NS equations up to a Reynolds number of around 10000. When doing this you will not resolve any turbulent entrainment into the wake however with regards to vortex shedding and force predictions it is accurate modelling of the laminar boundary layer that is important.

Above Re = 10e5 transition becomes a problem, upsteam of the separation point the boundary layer is partially tubulent and consequently drag is reduced. For this situation you need a good transition model (but does one exist). Above Re=10e6 the flow is fully turbulent and RANS simualtions should predict the correct unsteady behaviour (although many models will underpredict the separation region due to stagnation effects and problems in the defect layer).

Hope this all helps, Steve


sylvain February 15, 2002 04:24

Re: Why does unsteady problem tend to be steady?
 
I do believe that turbulent spectrum is continuous. Therefore there is no gap between the largest scales which correspond to the eddies of the wake and the kolmogorov scale, and this either for the spatial or the temporal spectrum. So how could a RANS model do the difference?

When using a RANS model in CFD, the cut-off between resolve and unresolve time/space scales is driving by the time step and the mesh size.

In this case of unsteady flows, LES or DES should be used to correctly predict the unsteadyness of the flow, since both of them introduce this cut-off in there derivation.

Just one more remark, this does NOT mean that transient computation can't be done with RANS model. I am just talking about unsteady computation. For example, if you are looking to the intake of an ICE, the used of RANS models is correct since the time dependence is coming form the movement of the geometry.

Cheers,

Sylvain

P.Fonteijn February 17, 2002 17:13

Re: Why does unsteady problem tend to be steady?
 
- Which tubulence model do you use? k-omega is suitable for these simulation but you need a fine grid.

- Which differencing scheme do you use? Upwind can damp out everything...

Pascale

wang February 19, 2002 06:08

Re: Why does unsteady problem tend to be steady?
 
Thanks Pascale,

I use low-Rey k-eps model with SMART spatial discretization.

P.Fonteijn February 19, 2002 16:56

Re: Why does unsteady problem tend to be steady?
 
We have used low Reynolds k-omega successfully using a second order differencing scheme. We used a fine grid, with y+ going down to 1. Using Upwind no oscillations were observed.

I don't know SMART. What is it?

Pascale

wang February 20, 2002 06:16

Re: Why does unsteady problem tend to be steady?
 
thanks Steve for your help,

but I'm not sure about several points of your comments.

1. "Therefore you should use the standard NS equations up to a Reynolds number of around 10000." What do you mean about the "standard NS" equation? DNS or RANS with standard wall function, or something else?

2. what do you mean about "Above Re = 10e5 transition becomes a problem, upsteam of the separation point the boundary layer is partially tubulent " of "PARTIALLY TURBULENT"?

3. At Re>10^6, do you mean that there will be a perfect periodic solution predicted by RANS and the result will not be damped?

thanks a lot

alain February 20, 2002 08:00

Re: Why does unsteady problem tend to be steady?
 
Steve is perfectly right : Try a laminar simulation and you will have the vortex shedding in the correct reynolds range.

If you really want to use a turbulence model for any reason you should use a very fine grid with small time step (10-4; 10-5 s). in this approach NS will be averaged only for small scale of turbulence thank to turbulence model which act in a similar way of the subscale filter in LES.

These approach is often call VLES

regards alain

Steve February 20, 2002 15:11

Re: Why does unsteady problem tend to be steady?
 
1. By Standard NS I mean the Navier Stokes equations. Since the boundary layer is laminar there is no need for DNS type grid resolution, a coarse grid will be fine.

2. By partially turbulent I mean the boundary layer will initially be laminar (near the stagnation point), and at some stage before the separation point turbulent transition will occur.

3. At 10^6 and above I think that the flow is quite chaotic and there is unlikely to be a clear period. RANS will damp out some of this behaviour, however RANS should at least be able to predict the drag.

wang February 21, 2002 07:09

Re: Why does unsteady problem tend to be steady?
 
Hi Pascale,

thanks for your tip. SMART method is a higher-order polynomial-based discretization scheme for the convection term. Detailes can be found in "Curvature-Compensated Convective Transport: SMART, a New Boundedness-preserving Transport Algorithm" by Gaskell, P.H. and Lau, A.K.C.,in "International Journal for Numerical Methods in Fluids". Vol.8., 617-641 (1988).

wang February 21, 2002 07:11

Re: Why does unsteady problem tend to be steady?
 
thanks a lot, Steve.


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