[srv537]

The Boussinesq assumption is used to calculate Reynolds turbulent stress in most of the models(ke, komega etc) except DNS.

all these model predict the same (isotropic) all normal component of Reynolds stress (averaged u'u' =v'v' =w'w') but shear stress prediction is good.

i tried ke, komega, SST and 2D LES and in all these model normal stress distribution was same, isotropic in nature and did not match with the experiments but shear stress (u'v') distribution was quite good predicted.

the reason as per my knowledge is eddy viscosity term is 3-4 order smaller than the kinetic energy term (2/3k) thats why i am getting same value of all normal stress(u'u' =v'v'=w'w').

now my question is if i want to predict normal turbulent stress then i dont have any option except DNS or i am doing something wrong.

Thank you.

Regards,
srv

[FMDenaro]

Just to give a clarification: the residuals in RANS and LES are two different things like onion and apple.

Furthermore, DNS is not a model and 2D LES has no physical meaning apart from few cases in geophysical flows.

[srv537]

Originally Posted by FMDenaro

Just to give a clarification: the residuals in RANS and LES are two different things like onion and apple.

Furthermore, DNS is not a model and 2D LES has no physical meaning apart from few cases in geophysical flows.

so sir if i will use 3D LES then it will predict normal stress accurately?

[FMDenaro]

your question is not clear...

First, in LES the SGS model is not only based on the Bousinnesq hypothesis, there are many other models that are not based on the eddy viscosity assumption.
Furthermore, when you model the unresolved stress you have a deviatoric and isotropic part. In case of incompressible flow, the isotropic part is not modelled at all but it is included in a modified pressure function.
Finally, 3D LES does not mean you "predict" better the unresolved terms. You simply allows for the physics of the large flow scale to evolve correctly.

[davidwilcox]

In 2D flows, from the continuity equation, du/dx = -dv/dy.
If you look at the normal stress formulation for eddy viscosity models, say uu for example, [uu=-2*Eddy_Viscosity(du/dx)+ (2/3)*Turbulence kinetic energy].
What does this tell you when you're calculating vv? You can probably get away with it (IF YOU'RE VERY LUCKY) in 3D flows.

Now, with regards to accuracy, it's hard to replicate the normal stresses for complex cases even with the RSM. But just because you could not match the experimental data, do not be disheartened. It is the model's inherent nature of predicting such stresses. You could create your own model some day that might solve turbulence.

[davidwilcox]

Just to add on, Peter Bradshaw once wrote that it is not accurate to relate the stresses with strain rate as in the eddy viscosity formulation . It is more logical to relate it to k. In his model uv=a*k where a roughly equals 0.3( the Bradshaw's or Taylor's constant).


Sent from my iPhone using CFD Online Forum mobile app