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September 27, 2002, 04:43 
RANS Limitations

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1) what is meant by stong stream line curvature? 2) what is nongradient transport. isnt all transport associated with some form of gradient? 3) why does RANS fail in these situations??? thanks Paul. 

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September 29, 2002, 05:59 
Re: RANS Limitations

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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) Twoequation models like keps and komega 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 (gradientdiffusion hypothesis) predicts that the direction of the turbulent flux is equal to the mean gradient which is not true in general. Reynoldsstress 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 gradientdiffusion 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 

September 29, 2002, 08:58 
Re: RANS Limitations

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Hello Paul,
1)If you look at the flow in a Ubend, the mean flow goes through a very sharp change in its direction. In this case, we say that the streamline curvature is sharp. The twoequation models like the ke and the kw 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 ke and the kw models fail. 2)Standard twoequation 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 

September 30, 2002, 05:44 
Re: RANS Limitations

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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


September 30, 2002, 07:54 
Re: RANS Limitations

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keps 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. Nonlinear model like RNG or ShihZu and Lumley (Fluent's realizable) model try to correct the point 2. Multiscale model try to take into account the third one. Sylvain 

October 1, 2002, 12:20 
Re: RANS Limitations

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Cheers for the explanation sylvain Dave


October 9, 2002, 03:59 
Re: RANS Limitations

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Thomas you said: "Standard twoequation models do not account for the transport of the Reynolds stresses"
what do you mean? 

October 10, 2002, 22:13 
Re: RANS Limitations

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Hello Hani,
Boussinesq assumption which is used by the standard ke and kw 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 

October 14, 2002, 11:03 
Re: RANS Limitations

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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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