CFD Online Discussion Forums (http://www.cfd-online.com/Forums/)
-   Main CFD Forum (http://www.cfd-online.com/Forums/main/)
-   -   3D axisymmetric flow in cylindrical coordinate = 2D cartesian flow? (http://www.cfd-online.com/Forums/main/117070-3d-axisymmetric-flow-cylindrical-coordinate-2d-cartesian-flow.html)

 shubiaohewan May 1, 2013 11:33

3D axisymmetric flow in cylindrical coordinate = 2D cartesian flow?

Hi All,

I'm simulating a axisymmetric flow in 3D cylindrical coordinate. Since it's axisymmetric, I'm thinking if I can re-write the code in 2D cartesian, just by looking at one plane with constant azimuthal angle.

The governing equations in 3D cylindrical coordinate has additionally terms like , f being either velocity or pressure. If I want to write the code in 2D, then terms like these should be discarded (because now it's cartesian).

My question is, now it seems to me that the governing equations are different for this same problem. I know this may sound absurd, could any one explain it? Thanks.

Shu

 Far May 1, 2013 14:51

You can solve 3d cylindrical problem in cartesian coordinates as 2d problem in cylindrical coordinates.

 FMDenaro May 1, 2013 15:00

but the axi-symmetric geometry does not imply the same simmetry in the flow... that is realized for laminar steady flows as well as for statistically averaged turbulent flows but, in general, the flow can be fully three-dimensional also for symmetric geometries ...

 shubiaohewan May 1, 2013 16:10

Quote:
 Originally Posted by Far (Post 424483) You can solve 3d cylindrical problem in cartesian coordinates as 2d problem in cylindrical coordinates.
Hi Far,

Could you be more specific? What will you deal with those f/r terms? In 3D cylindrical coordinate, they are singularities, while in 2D, no such problem.

Thanks.
Shu

 shubiaohewan May 1, 2013 16:12

Quote:
 Originally Posted by FMDenaro (Post 424485) but the axi-symmetric geometry does not imply the same simmetry in the flow... that is realized for laminar steady flows as well as for statistically averaged turbulent flows but, in general, the flow can be fully three-dimensional also for symmetric geometries ...
Hi FMDenaro,

I have some problem following what you said... Do you think I can do 2D cartesian for 3D axisymmetric flow? What physics will I lost when I do this?

Thanks,
Shu

 FMDenaro May 1, 2013 16:15

Quote:
 Originally Posted by shubiaohewan (Post 424500) Hi FMDenaro, I have some problem following what you said... Do you think I can do 2D cartesian for 3D axisymmetric flow? What physics will I lost when I do this? Thanks, Shu
As I wrote, this approximation is correct only for laminar steady flows or for statistical averaged (RANS) turbulent flows.
The flow solution in DNS/LES/URANS can not be modelled correctly by simmetry conditions

 RodriguezFatz May 2, 2013 04:13

Quote:
 Originally Posted by shubiaohewan (Post 424500) Hi FMDenaro, I have some problem following what you said... Do you think I can do 2D cartesian for 3D axisymmetric flow? What physics will I lost when I do this? Thanks, Shu
If you have 3d axis-symmetry it means that you have a rotational body and also the flow is rotationally symmetrical. You can not model this by 2d cartesian coordinats, because this a complete different symmetry. If you use 2d axisymmetric then you will indeed get exactly the same result.

 FMDenaro May 2, 2013 04:39

Quote:
 Originally Posted by RodriguezFatz (Post 424577) If you have 3d axis-symmetry it means that you have a rotational body and also the flow is rotationally symmetrical. You can not model this by 2d cartesian coordinats, because this a complete different symmetry. If you use 2d axisymmetric then you will indeed get exactly the same result.

onset of instability can destroy the rotational flow symmetry even for a geometrical symmetry (just think of a flow around a cylinder at Re>40 - 50)...this happens, for example, in round jets, in pipes, etc.
In turbulent statistically steady flows, you can use RANS formulations and recover the symmetry but you solve for a statistical field.

 rmh26 May 2, 2013 12:43

He isn't asking about whether or not have can assume axisymmetric flow, he is asking how to write down the equations. If the flow is axisymmetric you can indeed reduce the problem to two dimension, however you are not reducing it to Cartesian coordinates. You will still have the 1/r terms in the equations to deal with. Basically take the full 3d cylindrical equations and set all of the terms with either a theta velocity or a theta derivative to zero and you will get you new set of equations. The 1/r terms don't cause singularities if your boundary conditions are correct.

 RodriguezFatz May 2, 2013 13:36

Alright, now it makes sense. Shubiaohewan, all the terms f/r are terms, where f is the derivative of something. Now, for r->0, also f->0 (due to the symmetry). If you rummage in your memmory and use de L'Hospitals rule, you will see, that the (dy/dr)/r for r->0 actually is the second derivative, dy^2/dr^2, in axisymmetric cases. So for the inner points, where r=0, you have a different set of equations, because you need to discretize second instead of first derivatives. All fluxes are zero there.

 FMDenaro May 2, 2013 14:08

Quote:
 Originally Posted by rmh26 (Post 424739) He isn't asking about whether or not have can assume axisymmetric flow, he is asking how to write down the equations. If the flow is axisymmetric you can indeed reduce the problem to two dimension, however you are not reducing it to Cartesian coordinates. You will still have the 1/r terms in the equations to deal with. Basically take the full 3d cylindrical equations and set all of the terms with either a theta velocity or a theta derivative to zero and you will get you new set of equations. The 1/r terms don't cause singularities if your boundary conditions are correct.

I agree but I would remark that using symmetric conditions requires some knowledge about the type of flow to solve for. Depending on the physics, you can realize a simulation both in the (r,z) and (r,theta) planes. Or you need to solve the 3D flows ...

 All times are GMT -4. The time now is 07:14.