About the Reynolds Stress Model input parameter of OpenFOAM in 2D turbulent flow.

watermelon · June 16, 2022, 09:17

Hello everyone, this is watermelon, a new FOAMer.

When I used the LRR model in OpenFOAM v9, I found that the input parameter of R is a symmetric tensor.

I think when i use LRR to calculate the 2D turbulent flow problem, the component Rzz need to be zero.

But when I check my results, i found the component Rzz also has been calculated in 2D turbulent flow, why?

Any suggestion is helpful, thank you!

Best regards,

watermelon

Tobermory · June 18, 2022, 08:18

Turbulence is a 3D phenomenon ... hence even if you zero out one of the normal stresses, the pressure strain terms will repartition the turbulence energy amongst the different Reynolds stresses, including the one that you zero'd out.

Your mean flow may be two dimensional (e.g. a planar shear flow), with zero mean flow in the third direction, but the turbulence will always be three dimensional.

watermelon · June 18, 2022, 08:28

Quote:

Originally Posted by Tobermory

Turbulence is a 3D phenomenon ... hence even if you zero out one of the normal stresses, the pressure strain terms will repartition the turbulence energy amongst the different Reynolds stresses, including the one that you zero'd out.

Your mean flow may be two dimensional (e.g. a planar shear flow), with zero mean flow in the third direction, but the turbulence will always be three dimensional.

Hello Tobermory, thanks for your reply

Yes, you are right, turbulent flow is 3D. What i do not understand is when i calculate a 2D case in LRR, the component of Reynolds Stress in ZZ (Rzz) is not zero.

Because i think in 2D case, the Reynolds Stress should be : Rxx, Rxy, Ryx and Ryy.

But when i use the LRR to calculate the 2D case, i found Rzz also be calculated, but the Rxz, Ryz, Rzx, Rzy are not be calculated. So why the Rzz be calculated in 2D case?

Tobermory · June 18, 2022, 08:32

Umm ... not sure how I can say this differently, my friend. Rzz should NOT be zero, even in a 2D case.

watermelon · June 18, 2022, 08:48

Quote:

Originally Posted by Tobermory

Umm ... not sure how I can say this differently, my friend. Rzz should NOT be zero, even in a 2D case.

Haha, that's OK, thanks again.

Actually, i think when i calculate the 2D case, only the X and Y direction should be calculated. But i found OpenFOAM and Fluent are also have been calculate the Rzz when i use LRR model.

I do not understand how the Rzz was calculated in a 2D case, maybe i should to get more knowledge about turbulence.

Thanks again.

Yours.

Tobermory · June 18, 2022, 08:56

Quote:

Originally Posted by watermelon

Actually, i think when i calculate the 2D case, only the X and Y direction should be calculated. But i found OpenFOAM and Fluent are also have been calculate the Rzz when i use LRR model.

Exactly - this is the key point. In a 2D case, the mean velocity in the direction, W, is not solved for and it is assumed that mean flow variables are homogeneous in the z-direction. However, the LRR closure model still solves for the full Reynolds stress tensor, since even in a 2D case Rzz is not zero. The physical reason is that the pressure strain still redistributes energy to all three normal stresses. Definitely worth reading up on turbulence modelling, esp. second order closure models. Good luck my friend.

watermelon · June 18, 2022, 09:24

Quote:

Originally Posted by Tobermory

Exactly - this is the key point. In a 2D case, the mean velocity in the direction, W, is not solved for and it is assumed that mean flow variables are homogeneous in the z-direction. However, the LRR closure model still solves for the full Reynolds stress tensor, since even in a 2D case Rzz is not zero. The physical reason is that the pressure strain still redistributes energy to all three normal stresses. Definitely worth reading up on turbulence modelling, esp. second order closure models. Good luck my friend.

Thanks, and please allow me to ask a advice for the turbulent kinetic energy.

Actually, turbulent flow is 3D, so the turbulent kinetic energy should be calculated as : $k=\frac{1}{2}\left( \overline{{u}'{u}'}+\overline{{v}'{v}'}+\overline{{w}'{w}'} \right)$ . So, the Boussinesq Hypothesis of eddy viscosity model can write the Reynolds Stress as: $\overline{{{{{u}'}}_{i}}{{{{u}'}}_{j}}}=\frac{2}{\boldsymbol{3}}k{{\delta }_{ij}}-{{\nu }_{t}}\left( \frac{\partial {{{\bar{u}}}_{i}}}{\partial {{x}_{j}}}+\frac{\partial {{{\bar{u}}}_{j}}}{\partial {{x}_{i}}} \right)$

So, can I consider in 2D case, the turbulent kinetic energy can be calculated as : $k=\frac{1}{2}\left( \overline{{u}'{u}'}+\overline{{v}'{v}'} \right)$ , and the Reynolds Stress be written as : $\overline{{{{{u}'}}_{i}}{{{{u}'}}_{j}}}=\frac{2}{\boldsymbol{2}}k{{\delta }_{ij}}-{{\nu }_{t}}\left( \frac{\partial {{{\bar{u}}}_{i}}}{\partial {{x}_{j}}}+\frac{\partial {{{\bar{u}}}_{j}}}{\partial {{x}_{i}}} \right)$ ?

Thanks a lot!

Tobermory · June 18, 2022, 09:27

Nope. Last time of saying this. Even in a "2D case", as I explained above, the turbulence remains 3D. So the turbulence energy is the sum of ALL 3 normal stresses, and Rzz = w'w' is not zero.

watermelon · June 18, 2022, 09:30

Ok, thanks a lot~~

Wish you good health and success in your career.

Best Regards.

watermelon · June 19, 2022, 08:44

OK, thank you guys.

I think I have known this problem. In Turbulent Flows, which is written by Pope, in section 4.2, prof. Pope has interpret the 2D case in turbulent flow, and the Reynolds Stress is $\left[ \begin{matrix} {{R}_{11}} & {{R}_{12}} & 0 \\ {{R}_{21}} & {{R}_{22}} & 0 \\ 0 & 0 & {{R}_{33}} \\ \end{matrix} \right]$ , this is the 2D-3C (two-dimensional, 3 components) case.

And what i have suppose the 2D case, the Reynolds Stress is : $\left[ \begin{matrix} {{R}_{11}} & {{R}_{12}} & 0 \\ {{R}_{21}} & {{R}_{22}} & 0 \\ 0 & 0 & 0 \\ \end{matrix} \right]$ , this is the 2D-2C (two-dimensional, 2 components) case, and the 2D-2C is the "upper boundary" of the Lumley triangle.

So, the 2D-2C is a special case of 2D-3C, and 2D-2C is usually in the near-wall turbulence condition (Pope Chapter 9).

And I also derived the second-order equation in 2D case, the $\overline{{w}'{w}'}$ can be calculated.

Thanks!

Best Regards,

watermelon

June 16, 2022, 09:17	About the Reynolds Stress Model input parameter of OpenFOAM in 2D turbulent flow.	#1
watermelon New Member Xingguang Zhou Join Date: May 2022 Location: CHINA Posts: 14 Rep Power: 4	Hello everyone, this is watermelon, a new FOAMer. When I used the LRR model in OpenFOAM v9, I found that the input parameter of R is a symmetric tensor. I think when i use LRR to calculate the 2D turbulent flow problem, the component Rzz need to be zero. But when I check my results, i found the component Rzz also has been calculated in 2D turbulent flow, why? Any suggestion is helpful, thank you! Best regards, watermelon

June 18, 2022, 08:18		#2
Tobermory Senior Member Join Date: Apr 2020 Location: UK Posts: 736 Rep Power: 14	Turbulence is a 3D phenomenon ... hence even if you zero out one of the normal stresses, the pressure strain terms will repartition the turbulence energy amongst the different Reynolds stresses, including the one that you zero'd out. Your mean flow may be two dimensional (e.g. a planar shear flow), with zero mean flow in the third direction, but the turbulence will always be three dimensional. watermelon likes this.

June 18, 2022, 09:27		#8
Tobermory Senior Member Join Date: Apr 2020 Location: UK Posts: 736 Rep Power: 14	Nope. Last time of saying this. Even in a "2D case", as I explained above, the turbulence remains 3D. So the turbulence energy is the sum of ALL 3 normal stresses, and Rzz = w'w' is not zero. watermelon likes this.

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June 18, 2022, 08:32		#4
Tobermory Senior Member Join Date: Apr 2020 Location: UK Posts: 736 Rep Power: 14	Umm ... not sure how I can say this differently, my friend. Rzz should NOT be zero, even in a 2D case.

June 18, 2022, 09:30		#9
watermelon New Member Xingguang Zhou Join Date: May 2022 Location: CHINA Posts: 14 Rep Power: 4	Ok, thanks a lot~~ Wish you good health and success in your career. Best Regards.

June 19, 2022, 08:44		#10
watermelon New Member Xingguang Zhou Join Date: May 2022 Location: CHINA Posts: 14 Rep Power: 4	OK, thank you guys. I think I have known this problem. In Turbulent Flows, which is written by Pope, in section 4.2, prof. Pope has interpret the 2D case in turbulent flow, and the Reynolds Stress is $\left[ \begin{matrix} {{R}_{11}} & {{R}_{12}} & 0 \\ {{R}_{21}} & {{R}_{22}} & 0 \\ 0 & 0 & {{R}_{33}} \\ \end{matrix} \right]$ , this is the 2D-3C (two-dimensional, 3 components) case. And what i have suppose the 2D case, the Reynolds Stress is : $\left[ \begin{matrix} {{R}_{11}} & {{R}_{12}} & 0 \\ {{R}_{21}} & {{R}_{22}} & 0 \\ 0 & 0 & 0 \\ \end{matrix} \right]$ , this is the 2D-2C (two-dimensional, 2 components) case, and the 2D-2C is the "upper boundary" of the Lumley triangle. So, the 2D-2C is a special case of 2D-3C, and 2D-2C is usually in the near-wall turbulence condition (Pope Chapter 9). And I also derived the second-order equation in 2D case, the $\overline{{w}'{w}'}$ can be calculated. Thanks! Best Regards, watermelon