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I was wondering about the symmetry axis boundary condition in commercial CFD solvers such as ANSYS Fluent.

If the problem is the flow through a round pipe or out of a round nozzle, it is natural to set up a solver in axisymmetric space. That means a 3D flow domain $(x, r, θ)$ can be sliced along the $θ$ direction to provide a 2D slice in $(x, r)$. A symmetry axis boundary condition is used to close the problem along $x$ for $r=0$. Physically, axisymmetry is $\frac{\partial f}{\partial \theta}=0$, where $f$ represents velocity or pressure. Practically, a symmetry axis boundary condition is implemented as $\frac{\partial f}{\partial r}=0$ at $r=0$ in solvers such as Fluent. Correct me if I am wrong!

Does a symmetric boundary condition guarantee that the resultant flowfield will be axisymmetric everywhere?

For example, turbulent flow in a pipe can be non-axisymmetric.

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  • $\begingroup$ I don't see why the physical condition is $\partial f/\partial\theta=0$. $\endgroup$
    – Pu Zhang
    Nov 23, 2023 at 4:07
  • $\begingroup$ The geometry domain is cylindrical and r=0 is the axis on which a symmetry condition is applied. If the 'symmetry condition' is axisymmetry, then physically $\frac{\partial f }{\partial \theta}=0$. Is there another interpretation? $\endgroup$ Nov 23, 2023 at 4:57
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    $\begingroup$ Probably it is a good idea for you to insert a figure to illustrate the coordinate system in your mind. I'm afraid we have some misunderstanding on this. $\endgroup$
    – Pu Zhang
    Nov 25, 2023 at 3:07

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Since your problem is axisymmetric, your resultant flowfield does not depend on the angle. However, it depends on the radial and longitudinal coordinates if you are solving a stationary flow problem. That means it can be characterized by the function $f(r,z)$. If your problem is dynamic, the flowfield depends also on time: $f(r,z,t)$.

As for the boundary conditions of axisymmetric systems, you can generally have more than one type of condition at the axis, depending on the physical problem you are solving. You may have:

$${\partial f(r,z) \over \partial r} \Bigg |_{r=0} = 0 \tag 1$$

, but you may also have for example the following boundary condition:

$$f(0,z)=0 \tag 2$$

As I said, it depends on the problem you are solving.

To answer your question, just because you have one of the above-mentioned conditions at the axis, does not mean your flowfield must be axisymmetric. The fact that you restrained your flowfield function to not depend on the angle means that your flowfield must be axisymmetric.

In fact, you can have a non-axisymmetric flowfield and still have the condition $(1)$ be part of your solution, as long as at the axis there is a local optimum of $f$ in the radial direction.

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