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Old   September 27, 2002, 05:43
Default RANS Limitations
  #1
Paul Allen
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Hi, i read somewhere that RANS offers variable sucess in situations of "strong streamline curvature and where non-gradient transport is present".

1) what is meant by stong stream line curvature?

2) what is non-gradient transport. isnt all transport associated with some form of gradient?

3) why does RANS fail in these situations???

thanks Paul.
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Old   September 29, 2002, 06:59
Default Re: RANS Limitations
  #2
Tom
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1) The flow over a turbine blade is an example of a flow with strong stream line curvature: the direction of the mean flow changes rapidly. In fact the flow rotates and this has an influence on the turbulence.

2) When there is shear or rotation the direction of the turbulent flux of, for instance, a scalar is in general not aligned with the direction of the mean gradient of the scalar.

3) Two-equation models like k-eps and k-omega are not capable of taking into account the effect of rotation on turbulence and therefore they give bad results when there is swirl or streamline curvature. Furthermore, this kind of modelling (gradient-diffusion hypothesis) predicts that the direction of the turbulent flux is equal to the mean gradient which is not true in general. Reynolds-stress models, however, can take into account the influence rotation/swirl/stream line curvarture on the flow. So, probably they will do a better job in these cases. Instead of using the gradient-diffusion hypothesis for the turbulent flux of a scalar, you can also try to model the transport equation for the turbulent flux itself and some people have done that. Then you get rid of the assumption that the turbulent flux is aligned with the mean gradient.

Byt the way, these issues are addressed in a number of textbooks on turbulence and modelling. For instance: Pope, Turbulent flows.

Hope this helps,

Tom
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Old   September 29, 2002, 09:58
Default Re: RANS Limitations
  #3
Thomas P. Abraham
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Hello Paul,

1)If you look at the flow in a U-bend, the mean flow goes through a very sharp change in its direction. In this case, we say that the streamline curvature is sharp.

The two-equation models like the k-e and the k-w use Boussinesq assumption which provides a linear constitutive relation between the Reynolds stress and the mean strain. Such an assumption holds good for a simple homogeneous shear flow. When the flow involves more than a simple shear strain (which is true in almost every case), the standard k-e and the k-w models fail.

2)Standard two-equation models do not account for the transport of the Reynolds stresses. If the local effects dominate the transport effects, they perform o.k.

Reynolds Stress Transport Models (RSTM) do not have the above two weaknesses and should perform better in these circumstances.

Thanks,

Thomas
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Old   September 30, 2002, 06:44
Default Re: RANS Limitations
  #4
Dave
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Guys, as I understand it the RNG Ke turbulence model performs better than the Ke model. Why is this ? does the RNG model still not suffer from the underlying faults of the Ke model ?? Dave
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Old   September 30, 2002, 08:54
Default Re: RANS Limitations
  #5
sylvain
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k-eps models are limited by at least three points :

1) they don't correctly take into account the anisotropy of the Reynolds Stress Tensor since the Reynolds Stress Tensor is suppose to be directly proportionnal to Deformation tensor.

2) they don't take into account the "shear sheltering effect".

3) they all make the assumption of spectral equilibrium : the production of dissipation rate is directly proportionnal to the mean velocity gradient.

For point 1, one can take a look at the 2D plane turbulent wall jet (C55 ercoftac database) : the value of y where U is maximum (dU/dy =0) doesn't not match whith the value of y where uv=0, this violate the Boussineq assumption.

The point 2 is exposed in the msg from Tom.

The point 3 is visible behind a backward facing step : the region where k reach its maximum and the region where Eps does the same are predict at the same place with a two equations model. In reality, the production of turbulence is done on the large scale and the dissipation take place on the small scale. During the Kolmogorov cascade, the fluid is convected, so Eps should have reached its maximum later than k.

Reynolds Stress model try to correct the first point.

Non-linear model like RNG or Shih-Zu and Lumley (Fluent's realizable) model try to correct the point 2.

Multi-scale model try to take into account the third one.

Sylvain
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Old   October 1, 2002, 13:20
Default Re: RANS Limitations
  #6
Dave
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Cheers for the explanation sylvain Dave
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Old   October 9, 2002, 04:59
Default Re: RANS Limitations
  #7
hani
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Thomas you said: "Standard two-equation models do not account for the transport of the Reynolds stresses"

what do you mean?
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Old   October 10, 2002, 23:13
Default Re: RANS Limitations
  #8
Thomas P. Abraham
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Hello Hani,

Boussinesq assumption which is used by the standard k-e and k-w models use the local conditions to compute the Reynolds stresses. In reality, the Reynolds stresses could be convected by both the mean and the fluctuating velocities.

RSTMs account for the transport of the Reynolds stresses naturally.

Thanks,

Thomas
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Old   October 14, 2002, 12:03
Default Re: RANS Limitations
  #9
P.Allen
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hi i understand all these points but one made by sylvain.

"they don't correctly take into account the anisotropy of the Reynolds Stress Tensor since the Reynolds Stress Tensor is suppose to be directly proportionnal to Deformation tensor. "

i understand that they are linearly related, but why can it not take acount of anisotropy???

Please correct my mistake from what i say below:

* the deformation tensor is symmetric and often the trace is added to the pressure term. this leaves the deviatoric or anisotropic part - is this correct??? then surely the linear relation is between the stress tensor and a possible anisotropic rate of strain tensor.

then the stress tensor will be anisotropic!?!?!?!

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