[Bodo1993]

Hi, I am wondering what is the main difference between a 2D planner (in x-y plane) and an axisymmetric settings (in r-z plane)?

I have a simple rectangular domain that I simulated in the 2D planner plane, then, I made the same case as axisymmetric. I kept all dimensional values the same for both simulations (e.g. inlet velocity boundary condition, thermophysical properties, etc). However, I got different resulting dynamics. I am wondering what are the settings that should be changed such that the two cases are comparable? Do we have to think about the hydraulic diameter here to equate the Re for both cases?

Looking forward to hearing from you.

[aero_head]

Hello,

I found this thread and thought it may be useful to try to help answer your question: https://forum.ansys.com/discussion/4...utation-domain

Basically, when you are using an axis, the volume of the domain rotating the 2D plane around the axis is being calculated. The means that for an axi-symmetric pipe flow (a cylinder), you take the centre line as middle axis and 1/2 the cross-section as the simulation domain.

The symmetry option is used when you have a symmetry plane (or line in 2D). e.g.: a square. This way, the volume is being calculated with respect to the reference length.

In other words; 2D planar flow is the flow assumed to flow only in a single plane with varying property at different points. Axisymmetric flow is also a 2D flow with a line of symmetry along the plane. Let's consider flow in a pipe (which is obviously 3D). However, if you look at the flow of water from the top view, it is a 2D planar flow (xy plane). If the flow shows a specific property of being similar in nature above and below the center line of the plane, it is said to be axisymmetric flow.

[FMDenaro]

The example I consider useful to explain better is to think about the exact solution for laminar flow in a 2D channel and in a pipe. In the former the solution in the x,y plane repeats itself, infinitely, along the z direction. In the latter, the solution in the r,z plane repeats itself along the azimuthal direction. The physics is different.

[Simbelmynë]

There are a few ways you can approach your question. You can look at the equations, or even better, you can derive the equations using a control volume that fits into a round cylinder vs a control volume that fits into a square cylinder.

[Bodo1993]

Thanks all. That was very helpful!

What I am missing at the moment is how to make the planner 2D and axisymmetric cases comparable? In other words, if we have the results of 2D planner simulation, what should be changed in the problem settings if we want to create a comparable axisymmetric case? What are the conditions for that?
In my case, I am matching the governing dimensionless groups, but that seems not quite enough to achieve comparable dynamics.

Looking forward to hearing from you.

[FMDenaro]

Originally Posted by Bodo1993

Thanks all. That was very helpful!

What I am missing at the moment is how to make the planner 2D and axisymmetric cases comparable? In other words, if we have the results of 2D planner simulation, what should be changed in the problem settings if we want to create a comparable axisymmetric case? What are the conditions for that?
In my case, I am matching the governing dimensionless groups, but that seems not quite enough to achieve comparable dynamics.

Looking forward to hearing from you.

Again, you cannot make really comparable two different physics.

See the Hagen-Poiseulle solution for both planar and axi-symmetric flows and you will see the difference in the equations.

[aero_head]

Originally Posted by FMDenaro

Again, you cannot make really comparable two different physics.

See the Hagen-Poiseulle solution for both planar and axi-symmetric flows and you will see the difference in the equations.

Filippo,

I understand that the Hagen-Poiseulle equation are derived from the Navier-Stokes equations. So, could Bodo (OP) just change the cylindrical coordinates used in the axisymmetric case to cartesian coordinates for the planar case using x=rcosθ,y=rsinθ and z=z compare them that way?

If no, why would this not be acceptable?

[FMDenaro]

Originally Posted by aero_head

Filippo,

The Hagen-Poiseulle equation is derived from the Navier-Stokes equations. So, could Bodo just change the cylindrical coordinates used in the axisymmetric case to cartesian coordinates for the planar case using x=rcosθ,y=rsinθ, and z=z compare them that way?

If no, why would this not be acceptable?

If I understand, the problem is not in starting from the 2D code and changing it in an axi-symmetric version. That would be quite a simple task.

Comparing the two cases, the analytical solutions are quite similar, a quadratic law for the stramwise velocity and a linear law for the pressure.
We should compare the volumetric flow rate, being U_av*H*1 in 2D and U_av*A in the pipe flows. This way one could consider some equivalence.

[aero_head]

Originally Posted by FMDenaro

If I understand, the problem is not in starting from the 2D code and changing it in an axi-symmetric version. That would be quite a simple task.

Comparing the two cases, the analytical solutions are quite similar, a quadratic law for the stramwise velocity and a linear law for the pressure.
We should compare the volumetric flow rate, being U_av*H*1 in 2D and U_av*A in the pipe flows. This way one could consider some equivalence.

Ah yes, that makes more sense as to what Bodo is asking to do. You are of course correct in that comparing the two cases would not be so simple. It is interesting that a viable solution would be to compare the volumetric flow rates of the two cases, that is a good idea.

[Bodo1993]

Originally Posted by FMDenaro

If I understand, the problem is not in starting from the 2D code and changing it in an axi-symmetric version. That would be quite a simple task.

Comparing the two cases, the analytical solutions are quite similar, a quadratic law for the stramwise velocity and a linear law for the pressure.
We should compare the volumetric flow rate, being U_av*H*1 in 2D and U_av*A in the pipe flows. This way one could consider some equivalence.

Thanks.
If I understood correctly, in the case when we input the average velocity as a boundary condition, this will make:
U_av_axisymmetric = U_av_2D * (4/(pi*D)). Assuming H in 2D = D in axisymmetric.
That also means the ratio of Re numbers (Re_axisymmetric/Re_2D) = (4/(pi*D). Depending on the dimensional value of D, this may result in comparing two different flow states (laminar/turbulent) for both cases. For example, if D<<1, then Re_axisymmetric >> Re_2D.

[FMDenaro]

Originally Posted by Bodo1993

Thanks.
If I understood correctly, in the case when we input the average velocity as a boundary condition, this will make:
U_av_axisymmetric = U_av_2D * (4/(pi*D)). Assuming H in 2D = D in axisymmetric.
That also means the ratio of Re numbers (Re_axisymmetric/Re_2D) = (4/(pi*D). Depending on the dimensional value of D, this may result in comparing two different flow states (laminar/turbulent) for both cases. For example, if D<<1, then Re_axisymmetric >> Re_2D.

There is no sense in comparing 2D channel and 3D pipe in turbulence

[Bodo1993]

Originally Posted by FMDenaro

There is no sense in comparing 2D channel and 3D pipe in turbulence

Right. A hypothetical example here if we were to match the dimensional volumetric flow rate: let H = D = 0.001 m and Re_2D = 200, then, Re_axisymmetric = 2.5 e5. That is: laminar flow for 2D and turbulent flow for axisymmetric, which are obviously not comparable.

If matching the volume flow rate would not make the cases comparable, what would it be that has to be matched/changed to make them comparable?
Can they be comparable from the first place?

I would appreciate a clarification.

[LuckyTran]

Originally Posted by Bodo1993

what would it be that has to be matched/changed to make them comparable?
Can they be comparable from the first place?

Exactly what is being compared though? You need state such things. If you want to match volumetric flow rates and Reynolds numbers, then you make the circular diameter the same as the square channel width and you get flow through a square or circular pipe with the same volumetric flowrate for the same inlet velocity. But what else are you comparing? It's too vaguely defined (actually not at all).


The general solutions are not comparable in the sense that a circle is not a square. If you're asking whether I can make a circle fit inside a square (but not make a circle into a square), yes that can be done.

[FMDenaro]

Originally Posted by LuckyTran

Exactly what is being compared though? You need state such things. If you want to match volumetric flow rates and Reynolds numbers, then you make the circular diameter the same as the square channel width and you get flow through a square or circular pipe with the same volumetric flowrate for the same inlet velocity. But what else are you comparing? It's too vaguely defined (actually not at all).


The general solutions are not comparable in the sense that a circle is not a square. If you're asking whether I can make a circle fit inside a square (but not make a circle into a square), yes that can be done.

Actually, the original question is about 2D case, a solution that repeats itself infinitely along the third dimension, differently from the square channel. Thus the volumetric flow rate is just a finite value when considered for unit of lenght.

I agree, I don't see a sense in the question without more details.

[picpic]

Originally Posted by Bodo1993

Right. A hypothetical example here if we were to match the dimensional volumetric flow rate: let H = D = 0.001 m and Re_2D = 200, then, Re_axisymmetric = 2.5 e5. That is: laminar flow for 2D and turbulent flow for axisymmetric, which are obviously not comparable.

If matching the volume flow rate would not make the cases comparable, what would it be that has to be matched/changed to make them comparable?
Can they be comparable from the first place?

I would appreciate a clarification.

Might be a bit late. But why not set the same inlet velocity instead of Re when comparing?