Eddy Viscosity Models and Reynolds Stress Models

JuPa · August 20, 2013, 18:14

Hi,

Depending on how far the solver thinks the flow is away from the wall, the k - omega SST model will blend between the BSL k - omega and k - epsilon models, with the added terms to account for the transport of turbulent shear stress.

However far from a wall the k - omega SST model would have inherited the flaws of the standard k - epsilon model (since it's used far from the wall). Hence is there a particular reason why there isn't a model which blends between the BSL k - omega and RNG or Realiazable k - epsilon models? To me it just seems to be an obvious step in the right direction toward an accurate, robust and computationally cheap turbulence model.

The same question applies to epsilon and omega based Reynolds stress models.

Furthermore as computers have become relatively cheap and have increased power (compared to 5 to 10 years ago) is there a reason why the Reynolds stress models aren't being used as much as they should? I read many papers where the authors have just used 2 equation models. Why aren't Reynold stress models being used? They should be more accurate shouldn't they, since they compute the Reynolds stresses directly. Or is this just a CFD myth that the Reynolds stress models are more accurate than 2 equation eddy viscosity models?

ghorrocks · August 20, 2013, 19:56

I do not think the BSL k-w or RNG models are much better than plain old k-e away from the wall. There might be classes of flows where they are better but when you look at the whole spectrum of flows the gains by changing models is thin (or non-existent). Also the mathematical complexities of implementing a transition may make these alternate models difficult - I am no turbulence expert so I cannot comment on this.

Why is RSM models not more widely used? Again, because they are not more accurate in many classes of flows, and they are much more difficult to implement. A 2-eqn model with proper convergence and mesh refinement will be better than a RSM model where convergence was not achieved and mesh refinement is impossible.

RSM is more accurate for flows where turbulence anisotropy is important. For many flows this is not important and therefore there is little extra accuracy from RSM, but a lot of extra pain with more difficult convergence. So yes, I would say that over the whole spectrum of flows, it is a myth that RSM is more accurate than a good 2-eqn model. There are classes of flows where RSM is more accurate, but I would not generalise that.

And finally - keep in mind CFD has several purposes. Scientists and mathematicians use CFD to push the limits of known physics and mathematics. Engineers use CFD to model the design to lead to design improvements. CFX was written for engineers, with the principle of "accurate enough". There is no point getting an extra 1% accurate if it makes the simulation 10 times longer to run and the extra accuracy is not required anyway.

August 20, 2013, 18:14	Eddy Viscosity Models and Reynolds Stress Models	#1
JuPa Senior Member Mr CFD Join Date: Jun 2012 Location: Britain Posts: 361 Rep Power: 15	Hi, Depending on how far the solver thinks the flow is away from the wall, the k - omega SST model will blend between the BSL k - omega and k - epsilon models, with the added terms to account for the transport of turbulent shear stress. However far from a wall the k - omega SST model would have inherited the flaws of the standard k - epsilon model (since it's used far from the wall). Hence is there a particular reason why there isn't a model which blends between the BSL k - omega and RNG or Realiazable k - epsilon models? To me it just seems to be an obvious step in the right direction toward an accurate, robust and computationally cheap turbulence model. The same question applies to epsilon and omega based Reynolds stress models. Furthermore as computers have become relatively cheap and have increased power (compared to 5 to 10 years ago) is there a reason why the Reynolds stress models aren't being used as much as they should? I read many papers where the authors have just used 2 equation models. Why aren't Reynold stress models being used? They should be more accurate shouldn't they, since they compute the Reynolds stresses directly. Or is this just a CFD myth that the Reynolds stress models are more accurate than 2 equation eddy viscosity models?

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August 20, 2013, 19:56		#2
ghorrocks Super Moderator Glenn Horrocks Join Date: Mar 2009 Location: Sydney, Australia Posts: 17,871 Rep Power: 144	I do not think the BSL k-w or RNG models are much better than plain old k-e away from the wall. There might be classes of flows where they are better but when you look at the whole spectrum of flows the gains by changing models is thin (or non-existent). Also the mathematical complexities of implementing a transition may make these alternate models difficult - I am no turbulence expert so I cannot comment on this. Why is RSM models not more widely used? Again, because they are not more accurate in many classes of flows, and they are much more difficult to implement. A 2-eqn model with proper convergence and mesh refinement will be better than a RSM model where convergence was not achieved and mesh refinement is impossible. RSM is more accurate for flows where turbulence anisotropy is important. For many flows this is not important and therefore there is little extra accuracy from RSM, but a lot of extra pain with more difficult convergence. So yes, I would say that over the whole spectrum of flows, it is a myth that RSM is more accurate than a good 2-eqn model. There are classes of flows where RSM is more accurate, but I would not generalise that. And finally - keep in mind CFD has several purposes. Scientists and mathematicians use CFD to push the limits of known physics and mathematics. Engineers use CFD to model the design to lead to design improvements. CFX was written for engineers, with the principle of "accurate enough". There is no point getting an extra 1% accurate if it makes the simulation 10 times longer to run and the extra accuracy is not required anyway.