What is difference between dev2(T(grad(U)) Gauss linear; and dev(T(grad(U))

Mirage · September 29, 2016, 18:42

Hello Foamers

I am running a case in OF3.0.1, already solved in OF2.3.
I just noticed that the syntax for some schemes changed in the fvSchemes.
The fvSchemes has be to be edited to start the simulation.

The old fvSchemes file (OF.2.3) :

Code:

divSchemes
{
...
    div((nuEff*dev(T(grad(U))))) Gauss linear;
...
}

In the new OF version (OF3.0.1) the change must be look like this to run the case:

Code:

divSchemes
{
...
    div((nuEff*dev2(T(grad(U))))) Gauss linear;
...
}

I was wondering, if someone could explain this changes.
Could this change effect the results? What are the difference between the using of dev2 and dev.

Thanks in advance for your help.

Vyssion · September 30, 2016, 07:35

As far as I have found, it is literally just a naming convention to more accurately reflect what is actually happening to the variables/constants when calculating the effective viscous stress.

zarox · February 23, 2017, 08:40

Diffrence between OF2.x and OF3.x and later is the Change of the structure for turbulence model that have been unified between Compressible and incompressible flow. That lead to a Change in the additional "stabilization" term in incompressible flow with dev2 instead of dev.

Would you find here some calculation that explain dev and dev2 difference and their implication in the equation solved.

My simple calculation is for the stress Tensor, with stokes hypothesis:

$\tau = \nu (\nabla u +(\nabla u)^{T})- \nu \frac{2}{3} ( \nabla . u) I$

with $\nabla . (\nabla u) = \Delta u$ and $\nabla . (\nabla u)^{T} = \nabla (\nabla . u)$ and $\nabla . ((\nabla . u) I) = \nabla (\nabla . u)$

$\nabla . \tau = \nu \Delta u +\nu \nabla (\nabla . u) - \nu \frac{2}{3} \nabla (\nabla . u) = \nu \Delta u + \nu \frac{1}{3} \nabla (\nabla . u)$

we retrieve the compressible Navier Stoke equation.

Recalling $tr((\nabla u)^{T})I=tr(\nabla u)I=(\nabla . u) I$

$dev2((\nabla u)^{T})=(\nabla u)^{T} - \frac{2}{3}tr((\nabla u)^{T})I = (\nabla u)^{T} - \frac{2}{3}(\nabla . u) I$

$\nabla . \tau = \nu \Delta u + \nu \nabla . (dev2((\nabla u)^{T})) = \nu \Delta u + \nu \nabla . (\nabla u)^{T} - \nu \frac{2}{3} \nabla . ((\nabla . u) I)$
$\nabla . \tau = \nu \Delta u + \nu \nabla (\nabla . u) - \nu \frac{2}{3} \nabla (\nabla . u)= \nu \Delta u+ \nu \frac{1}{3} \nabla (\nabla . u)$

For incompressible and compressible flow formulation in OF3.x and OF4.x.

$\nabla . \tau = \nu \Delta u+ \nu \frac{1}{3} \nabla (\nabla . u)$

The dev2 use 2/3 factor but it is to lead to 1/3 factor in the stress Tensor for the part with the $\nabla . u$

Before,

$dev((\nabla u)^{T})=(\nabla u)^{T} - \frac{1}{3}tr((\nabla u)^{T})I = (\nabla u)^{T} - \frac{1}{3}(\nabla . u) I$

with

$\nabla . dev((\nabla u)^{T})= \nabla(\nabla . u) - \frac{1}{3}\nabla (\nabla . u) = \frac{2}{3} \nabla (\nabla . u)$

So $\nabla . \tau = \nu \Delta u + \nu \frac{2}{3} \nabla (\nabla . u)$

As I said, the previous implementation for incompressible flow we had

$\nu \Delta u + \nu \frac{2}{3} \nabla (\nabla . u)$

But the $\nabla . u = 0$ in converged flow, so as you said it would be a stabilization term

and for compressible flow as expected for Navier stoke equation for compressible

$\nu \Delta u + \nu \frac{1}{3} \nabla (\nabla . u)$

Now Incompressible and compressible flow have been write to use dev2 in the turbulence model so

$\nu \Delta u + \nu \frac{1}{3} \nabla (\nabla . u)$

But we don't get how is impacted the stability/accurancy by the formulation for incompressible with 1/3 instead of the previous 2/3

By the way, don't be worry if your old case file with the new OpenFOAM3.x and OpenFOAM4.x says

Code:

keyword div((nuEff*dev2(T(grad(U))))) is undefined in dictionary

It is because of the Change of the numerical structure explain with the use of dev2 now instead of dev even for incompressible flow. Mystery remove!

September 29, 2016, 18:42	What is difference between dev2(T(grad(U)) Gauss linear; and dev(T(grad(U))	#1
Mirage Member Mirage Join Date: Jul 2012 Posts: 43 Rep Power: 13	Hello Foamers I am running a case in OF3.0.1, already solved in OF2.3. I just noticed that the syntax for some schemes changed in the fvSchemes. The fvSchemes has be to be edited to start the simulation. The old fvSchemes file (OF.2.3) : Code: divSchemes { ... div((nuEff*dev(T(grad(U))))) Gauss linear; ... } In the new OF version (OF3.0.1) the change must be look like this to run the case: Code: divSchemes { ... div((nuEff*dev2(T(grad(U))))) Gauss linear; ... } I was wondering, if someone could explain this changes. Could this change effect the results? What are the difference between the using of dev2 and dev. Thanks in advance for your help. Dipsomaniac likes this.

September 30, 2016, 07:35		#2
Vyssion New Member Vyssion Join Date: Mar 2016 Posts: 12 Rep Power: 10	As far as I have found, it is literally just a naming convention to more accurately reflect what is actually happening to the variables/constants when calculating the effective viscous stress. __________________ *Optimism, pessimism, fck that: We're going make it happen. As God is my bloody witness, I'm hell-bent on making it work.** - Elon Musk in 2008, after three unsuccessful Falcon launches.

February 23, 2017, 08:40	dev and dev2 difference and implication in equation solved	#3
zarox Member Emeline Noel Join Date: Dec 2013 Location: Paris Posts: 31 Rep Power: 12	Diffrence between OF2.x and OF3.x and later is the Change of the structure for turbulence model that have been unified between Compressible and incompressible flow. That lead to a Change in the additional "stabilization" term in incompressible flow with dev2 instead of dev. Would you find here some calculation that explain dev and dev2 difference and their implication in the equation solved. My simple calculation is for the stress Tensor, with stokes hypothesis: $\tau = \nu (\nabla u +(\nabla u)^{T})- \nu \frac{2}{3} ( \nabla . u) I$ with $\nabla . (\nabla u) = \Delta u$ and $\nabla . (\nabla u)^{T} = \nabla (\nabla . u)$ and $\nabla . ((\nabla . u) I) = \nabla (\nabla . u)$ $\nabla . \tau = \nu \Delta u +\nu \nabla (\nabla . u) - \nu \frac{2}{3} \nabla (\nabla . u) = \nu \Delta u + \nu \frac{1}{3} \nabla (\nabla . u)$ we retrieve the compressible Navier Stoke equation. Recalling $tr((\nabla u)^{T})I=tr(\nabla u)I=(\nabla . u) I$ $dev2((\nabla u)^{T})=(\nabla u)^{T} - \frac{2}{3}tr((\nabla u)^{T})I = (\nabla u)^{T} - \frac{2}{3}(\nabla . u) I$ $\nabla . \tau = \nu \Delta u + \nu \nabla . (dev2((\nabla u)^{T})) = \nu \Delta u + \nu \nabla . (\nabla u)^{T} - \nu \frac{2}{3} \nabla . ((\nabla . u) I)$ $\nabla . \tau = \nu \Delta u + \nu \nabla (\nabla . u) - \nu \frac{2}{3} \nabla (\nabla . u)= \nu \Delta u+ \nu \frac{1}{3} \nabla (\nabla . u)$ For incompressible and compressible flow formulation in OF3.x and OF4.x. $\nabla . \tau = \nu \Delta u+ \nu \frac{1}{3} \nabla (\nabla . u)$ The dev2 use 2/3 factor but it is to lead to 1/3 factor in the stress Tensor for the part with the $\nabla . u$ Before, $dev((\nabla u)^{T})=(\nabla u)^{T} - \frac{1}{3}tr((\nabla u)^{T})I = (\nabla u)^{T} - \frac{1}{3}(\nabla . u) I$ with $\nabla . dev((\nabla u)^{T})= \nabla(\nabla . u) - \frac{1}{3}\nabla (\nabla . u) = \frac{2}{3} \nabla (\nabla . u)$ So $\nabla . \tau = \nu \Delta u + \nu \frac{2}{3} \nabla (\nabla . u)$ As I said, the previous implementation for incompressible flow we had $\nu \Delta u + \nu \frac{2}{3} \nabla (\nabla . u)$ But the $\nabla . u = 0$ in converged flow, so as you said it would be a stabilization term and for compressible flow as expected for Navier stoke equation for compressible $\nu \Delta u + \nu \frac{1}{3} \nabla (\nabla . u)$ Now Incompressible and compressible flow have been write to use dev2 in the turbulence model so $\nu \Delta u + \nu \frac{1}{3} \nabla (\nabla . u)$ But we don't get how is impacted the stability/accurancy by the formulation for incompressible with 1/3 instead of the previous 2/3 By the way, don't be worry if your old case file with the new OpenFOAM3.x and OpenFOAM4.x says Code: keyword div((nuEff*dev2(T(grad(U))))) is undefined in dictionary It is because of the Change of the numerical structure explain with the use of dev2 now instead of dev even for incompressible flow. Mystery remove! fu-ki-pa, Mirage, acgnipper and 27 others like this.