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-   -   Help with understanding production term in RHS! (https://www.cfd-online.com/Forums/main/247168-help-understanding-production-term-rhs.html)

ramkr January 17, 2023 13:47

Help with understanding production term in RHS!
 
1 Attachment(s)
Hello there!

I am trying to understand the differences between the versions of the 2 eqn Wilcox k-omega model, over the years. I am using the NASA turbulence models site for this.

My doubt here is rather trivial, related to the Production term of the TKE. More specifically, I am trying to understand the mathematical breakdown for how the P_{inc} - Production term for incompressible flow is arrived at, in the below notes page at point 4.

https://turbmodels.larc.nasa.gov/noteonrunning.html

It would be of great help if someone could explain the step-wise breakdown for the P_{inc} clearly. I tried but in vain :o :confused:! Request to see attached image of my write up.

Thanks in advance!

FMDenaro January 17, 2023 14:30

Quote:

Originally Posted by ramkr (Post 843025)
Hello there!

I am trying to understand the differences between the versions of the 2 eqn Wilcox k-omega model, over the years. I am using the NASA turbulence models site for this.

My doubt here is rather trivial, related to the Production term of the TKE. More specifically, I am trying to understand the mathematical breakdown for how the P_{inc} - Production term for incompressible flow is arrived at, in the below notes page at point 4.

https://turbmodels.larc.nasa.gov/noteonrunning.html

It would be of great help if someone could explain the step-wise breakdown for the P_{inc} clearly. I tried but in vain :o :confused:! Request to see attached image of my write up.

Thanks in advance!




Not sure about your doubt, are you asking about the role of the trace of the velocity gradient tensor, that is Div v ? Generally, for incompressible flow we use the symmetric gradient with zero trace. The isotropic term is included in a modified pressure.

LuckyTran January 17, 2023 18:31

There are a few ways to arrive at it but what you need is to derive the transport equation for k. The transport equation for k is a sub-equation that lives inside the total energy equation.

Probably a more straightforward way is to take the dot product of time-accurate velocity with the momentum equation (i.e. take navier-stokes and multiply it by velocity) and then take the Reynolds average. You can then isolate the turbulent kinetic energy from the resulting expression after you subtract the time-averaged bulk energy (the average momentum equation multiplied by the average velocity). A Reynolds stress term will appear and after applying your eddy viscosity model you will arrive at the result you have printed.


It is theoretically very simple to extend the validity of the production term into a compressible formulation (the formula is identical), but you need an eddy viscosity model that holds under these compressible actions.

ramkr January 17, 2023 22:33

Sorry if I confused you, that's not my doubt.
What i wish to understand is how Production term
P=(tau_(ij))*(do u_i)/(do x_j) becomes 2*mu_t*S_(ij)*S_(ij)
for incompressible flow.
As a first step of this process, i understand clearly that (tau_(ij))=2*mu_t*(S_(ij)) for incompressible flow.
Where S_(ij)=0.5((do u_i)/(do x_j)+(do u_j)/(do x_i))
But I'm not able to deduce how they arrived at the final expression for P_inc in step 4. Request to see the attached screenshot for your reference.
https://drive.google.com/file/d/10Pf...w?usp=drivesdk

LuckyTran January 17, 2023 23:13

Expand

\left[ \frac{\partial u_i}{\partial x_j} + \frac{\partial u_j}{\partial x_i} \right]\frac{\partial u_i}{\partial x_j}

And then recognize that, for example:

\frac{\partial u}{\partial y} \frac{\partial v}{\partial x}=\frac{1}{2}\frac{\partial u}{\partial y} \frac{\partial v}{\partial x}+\frac{1}{2}\frac{\partial v}{\partial x} \frac{\partial u}{\partial y}

sbaffini January 18, 2023 04:36

You seem to miss the very fundamental fact that a double contraction between a tensor and a symmetric tensor equals the double contraction between the symmetric part of the first tensor and the originally symmetric tensor.

ramkr January 19, 2023 13:42

Got it!
 
Quote:

Originally Posted by sbaffini (Post 843054)
You seem to miss the very fundamental fact that a double contraction between a tensor and a symmetric tensor equals the double contraction between the symmetric part of the first tensor and the originally symmetric tensor.

Yes, I'm sorry. Just realized this yesterday when expanding and summing up each term of the product between the stress tensor and the velocity derivative:o

Then each term becomes individually representative in that way, as shown.


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