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Lyapunov Alexandr February 22, 2002 07:32

Unsteady flow behind cylinder in k-e model
 
I've use k-e model for calculation of flow around circular cylinder (Re=1e7) and received unsteady flow (like laminar Re=100). Can't it be? Is there any error?

andy February 22, 2002 08:06

Re: Unsteady flow behind cylinder in k-e model
 
If the equations are not fully converged at each time step or if you are using a steady-state relaxation scheme, then the predicted time variation is not valid.

If you have performed a valid time varying simulation then so long as the time scale of the unsteadiness is outside the range of the time scales of the turbulent motion simulated by the k-e model then the solution can be considered valid. This does not mean it will be accurate since it will be subject to both numerical error in time space and the modelling assumptions of the k-e model which are unlikely to hold up particularly well for this class flow.

(We seem to have a theme for the day.)

sylvain February 25, 2002 06:16

Re: Unsteady flow behind cylinder in k-e model
 
Suppose I have made two different computations on the same grid, high Re number, k-eps or anyother RANS equation closure model.

For the first one, I use a steady state numerical scheme.

For the second one, I use the unsteady approach describe by Andy.

Then, I average over the time the solution of the second case. Using the ergodicity properties of this statistically steady flow, I can claim that this time-average solution is solution of the steady RANS equations. This is also the case for the first result. So the the first result and the time average solution of the second compution should be equal.

I have done this for the standart k-eps model, for the Shih, Zu and Lumley k-eps model, for the Spalart Almaras one equation model.

None of them give the correct answer.

I do believe this is due to the fact that RANS is not derived to solve unsteady flows. In fact, the diffusivity effect of the resolve eddies in the second case is also taking into account by the turbulence model. So it is count twice and, as a result, the averaged solution is more diffusive than the first one.

To avoid this, the value of c_nu (in the k-eps model) should be reduce. But who can say how ?

Regards,

Sylvain


andy February 25, 2002 13:12

Re: Unsteady flow behind cylinder in k-e model
 
You seem to be confusing turbulence and flow instability. Vortex shedding is a flow instability it is not turbulence. The scales are all wrong and there is a wacking great hole in the spectra when the two processes are combined. Requiring the lot to be represented by turbulence models which simulate all turbulent transport as a diffusive process is irrational.


sylvain February 25, 2002 13:48

Re: Unsteady flow behind cylinder in k-e model
 
Show me the hole. You will only find it if you make conditionnal average. This is no more RANS.

Sylvain

sylvain February 25, 2002 14:10

Re: Unsteady flow behind cylinder in k-e model
 
Two more points :

The backward facing step is one of the most popular turbulence test case (see the ERCOFAC workshops). The experimental data from Kim, Kline & Johnston are used to compare models and solvers using a steady state scheme. Lesieur for a part, Kim & Moin for an other part have done LES on this test case. Together with the experimental results this computations have showed a vortex shedding behind the step.

Ha-minh and his team, using the semi deterministic approach have showned that a correct unsteady behavior could be obtain by damping the value of c_nu.

andy February 25, 2002 15:20

Re: Unsteady flow behind cylinder in k-e model
 
Not sure what the relevance of all this is and so we may well be at cross purposes.

The motion in low Reynolds number regions that drives turbulence is coherent. So what. Where is the connection with the high Reynolds number flow over an obstacle.

You will have to tell me what term c_nu (presumably a diffusion coefficient) is for if I am to understand the second paragraph.

Originally, I was simply trying to point out that for your steady-state test you are requiring the turbulence model to represent an unsteady motion that is not turbulence. This is not reasonable.

I also fail to understand why you expect your two predictions to give the same answer when you have decomposed and modelled them in two different ways; one definitely unreasonable and one possibly unreasonable depending on the turbulence scales. And why, as a consequence, Reynolds gets a beating about the head for being in some way wrong.


Steve February 25, 2002 18:03

Re: Unsteady flow behind cylinder in k-e model
 
RANS was not derived solely for unsteady flows. The only assumption with regard to unsteady simulations is that the time step you use (and the pysical time scale of the mean unsteadyness) is significantly larger than the time scales that the turbulent eddies operate on. The ensemble averaging process does not add any diffusion to the transient terms, the eddy viscosity acts solely in a spatial sense.

Discrepencies between the steady state and transient solvers is due to the design of the solvers themselves. Steady state solvers typically iterate through psuedo-time towards steady-state, if the problem is unsteady then the residuals will oscillate. However this isn't representitive of the transient solution because the solver will be solving linearised equations. With unsteady solvers there are typically several solves of the linearised equations at each time step, prior to a time step in real time. So I would expect different solutions.

sylvain February 26, 2002 06:36

Re: Unsteady flow behind cylinder in k-e model
 
My point of view is the following :

Experimentally, to obtain RANS values, one has to make ensemble averaging. When looking at flow field behind a cylinder, this may be done the following way :

step 1 : the wind tunnel starts ;

step 2 : at time=t_measure, the velocity at x_measure,y_measure,z_measure is done

step 3 : the wind tunnel stops.

this three steps are repeated until the number of experiments is large enought to get a correct average value.

This procedure give the Reynolds average values at t_measure,x_measure,y_measure,z_measure.

If one wants to see the vortex shedding, he must do the measurements at a time which doesn't depend on the switch on of the wind tunnel, but which is correlated, for exemple, to a pressure value at the skin of the cylinder. This is no more Reynolds average, since the average is done conditionnaly.

The RANS equations are written to give the Reynolds average values of the flow, that's mean, in the case of the wake of a cylinder, a statisticaly steady flow.

k-eps models are written to close the RANS equation, and the first draft by Launder and Spalding didn't show any time derivatives.

What I have tried to show, with my two computations was that applying Reynolds average to unsteady k-eps computation doesn't give the same solution than directly solve steady RANS equations. This was obvious to me, but I just want to check.

The point where we are not agree is just the following : you do believe that in the case of a vortex shedding RANS equations can be used with an unsteady scheme, using the classical k-eps equations. I am just telling that, in this case, k-eps equations must be corrected (damping the turbulent viscosity nu_t by lowering the value of c_nu (nu_t = c_nu * k^2/eps)). The problem is that nobody knows, a priori, how to do this operation.

andy February 26, 2002 07:49

Re: Unsteady flow behind cylinder in k-e model
 
>> The point where we are not agree is just the following : you do believe that in the case of a vortex shedding RANS equations can be used with an unsteady scheme, using the classical k-eps equations. I am just telling that, in this case, k-eps equations must be corrected (damping the turbulent viscosity nu_t by lowering the value of c_nu (nu_t = c_nu * k^2/eps)). The problem is that nobody knows, a priori, how to do this operation.

I was willing to let the earlier stuff go in order to end the thread but here you are incorrectly telling me what I believe. I used the word "reasonable" (I hope everywhere) not correct and indicated under what conditions the assumption would not be reasonable (probably most in practice). Under my conditions of reasonableness you will almost certainly find that your desired correction is small enough to be neglected compared with the errors stemming from modelling turbulence transport as diffusion.

If you want a belief, I would add that such corrections to a poor model for turbulent transport would only be worth persuing for limited specific cases. One such case, for example, would be the wakes from blading in axial turbomachinery where a fair amount of work has been done trying to introduce reasonable Reynolds stress components for the non-turbulent events. Not wholly successful but I would dispute that the reseachers do not know what they are doing.

Jonas Larsson February 26, 2002 07:58

Re: Unsteady flow behind cylinder in k-e model
 
Interesting discussion, thanks guys. I have one small comment concerning your comparison between a steady and an unsteady RANS solution. If you are able to obtain both an unstead and a steady solution using the same model and mesh then something is strange - either your solution should be stable and steady or it should be unsteady - in this case your results are dependent on the numerics and it shouldn't be. I don't see how you can draw any conclusions about k-epsilon from this difference between two numerical methods.

That k-epsilon often is too diffusive and can damp-out unsteadyness is quite well known - the litterature is full of modifications of k-eps to fix this problem. One example that comes to my mind is the Kato-Launder fix, which, if I remember correctly was first developed to predict unsteady vortex-shedding behind cylinders (square shaped I think). By modifying c_my you can probably get a similar effect.

I don't see how these problems with k-epsilon can mean that all unsteady solutions obtained with k-epsilon are wrong.

You bring up an interesting point about the "hole in the spectra". Are you saying that in all unsteady k-eps simulations you will get a continious spectra down to the "grid-size filter"? I would expect that it should be possible to obtain, for example, certain vortex-shedding frequencies that are not part of a continious spectra.

sylvain February 26, 2002 10:15

Re: Unsteady flow behind cylinder in k-e model
 
>> I was willing to let the earlier stuff go in order to end the thread but here you are incorrectly telling me what I believe.

Sorry if I had made an offence, I am not american nor english, but I try to explain myself in our language and it's sometime difficult for me to put some "nuances" in my threat.

About k-eps models, I'm just using this one as an exemple because it is the most popular one. But I have the same grief for all other RANS closure models, this includes Reynolds Average Stress models.

Anyway, what I only want to do is to warn people about the use of RANS modeling for unsteady case, but doing 3DURANS is better than nothing.

sylvain February 26, 2002 11:09

Re: Unsteady flow behind cylinder in k-e model
 
>> If you are able to obtain both an unstead and a steady solution using the same model and mesh then something is strange - either your solution should be stable and steady or it should be unsteady - in this case your results are dependent on the numerics and it shouldn't be.

Yes, to obtain the steady solution I use a implicit scheme with "local time step". If it is not sufficient to avoid ascillation, I sometime use half a geometrie. For the unsteady solution I use a fully explicit scheme with a small enough time step.

>> I don't see how these problems with k-epsilon can mean that all unsteady solutions obtained with k-epsilon are wrong.

I am just saying that RANS equations are not derived to modelized unsteady flows. The only time dependance they can handle is moving boundaries conditions.

>> Are you saying that in all unsteady k-eps simulations you will get a continious spectra down to the "grid-size filter"?

I wouldn't say so for a k-eps simulations, but I think this could be true with an LES. To confirm that, I will be happy to see the E11(K) spectrum in the wake of a cylinder. Does anybody have this ?


Jonas Larsson February 26, 2002 11:48

Re: Unsteady flow behind cylinder in k-e model
 
Then you are just comparing errors produced by two different numerical methods - has nothing to two with whether or not k-epsilon can handle unsteady flows.

If you have a clear scale separation you can easily derive an unsteady RANS version by averging out only turbulent scales. I see no principal problem with this. It is another debate whether or not a certain flow has this scale separation - many engineering flows do not.

I agree though that k-epsilon is often too diffusive and tends to kill unsteadyness and that you often have to use tuned k-epsilon models to have a chance to capture, for example, correct vortex shedding frequencies behind bluff bodies.

Jonas Larsson February 26, 2002 12:16

Re: Unsteady flow behind cylinder in k-e model
 
Can be okay - I think that vortex shedding re-appears at Re numbers above 5 million or so. The Strouhal number should be about twice as big as for the laminar case though. Schlichting has a nice plot of different experimental data on this - page 32 in my edition 7.

sylvain February 26, 2002 12:50

Re: Unsteady flow behind cylinder in k-e model
 
>> Then you are just comparing errors produced by two different numerical methods - has nothing to two with whether or not k-epsilon can handle unsteady flows.

I would like not to, numerical methods are tools, not models. But I am not so sure that the behavior I was talking about is just a consequence of numerical method.

>> If you have a clear scale separation you can easily derive an unsteady RANS version by averging out only turbulent scales. I see no principal problem with this. It is another debate whether or not a certain flow has this scale separation - many engineering flows do not.

I still don't understand how this can be possible, since the average method used to derived RANS equations is ensemble average. Since you don't know on which size of the cylinder the first eddy will come, the two sides are equiprobables and the ensemble average is symetric. So the Reynolds average description of the flow has to be symetric if you want to compare it with experimental data.



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