Hi, all

I am doing some simulations about a steady circular jet in cross flow that flushing over. What I use is a domain consisting of a channel and a circular pipe, with hybrid mesh, K-e standard or RNG model. In my steady simulation, I found that the flow field is not fully symmetrical on the two sides, though all bounary conditions are strictly set symmetrical. I first wondered this might be due to the false diffusion from the mesh, for the unstructured mesh is not 100% symmetrical. However, I have read some experiment paper stating that this asymmetry does exist to some degree, even though the experiment set-up being symmetrical to very high accuracy.

I want to have some research into this problem. But I don't know how to eliminate those error from computation. Does anyone have good ideas?

Thanks in advance.

At high Reynolds numbers, jets in cross flows are asymmetrical when fed from a plenum/annulus through a hole. Essentially the flow through the hole contains a vortex like that flowing out of the hole in a bath. It can spin either way. The holes in gas turbine combustion chambers will often contain splitters and/or non-circular holes in order to control this asymmetry which creates unevenness in the temperature profile through the turbines.

Asymmetrical solutions with symetrical boundary conditions are not uncommon in fluid dynamics and this is only of many such examples.

My point of view is different:

If you use the steady solver, you have to find the steady solution, nummerically a steady state solution can be only, and only symmetric. This is particularly true if you use a RANS turbulence model, all fluctuations are considered as turbulence, the mean flow should be steady and symmetric. If you do not find such kind of solution with your steady solver, you can be sure that you have a problem with the grid(unsymmetric) or the discretization scheme(could be unstable).

But we should define what asymetric means.

It is acceptable: Few percents difference between both sides, with small oscillations arround the symmetric solution.

Not acceptable: Strong oscillations, big flow structure created...

But of course this does not mean that the real flow is symmetric. It is possible to have a real steady and unsymmetric flow, in that case you should first simulate it with the steady solver, so you will find the symmetric solution.

Then switch to the unsteady solver, perturbate unsymmetrically the solution, and see if the solution converges to a steady and unsymmetrical solution. As example the flow could choose a side (left or right) to develop a recirculation zone.

Then try the opposit perturbation on the symmetrical solution, and see if the recirculation zone is developping in the opposit side than previous.

If not....there is a problem, the solution can not choose preferentially one side.

> If you use the steady solver, you have to find the steady solution, nummerically a steady state solution can be only, and only symmetric.

This is incorrect. I suspect you are confusing situations like vortex shedding where meaningful steady solutions may not exist with multiple/asymmetric solutions due to the nonlinearity of the governing equations. Setting the time derivative terms to zero still leaves nonlinear terms and so changes nothing.

Perhaps the classical example is diffusing a symmetrically confined flow a little too much so it separates. The separated flow will preferentially attach to give an asymmetrical solution which may be steady or unsteady depending on the geometry. Increasing the rate of diffusion further will usually restore a symmetrical solution.

Let's be more precise...in the case of multiple steady states.

An usual CFD mistake is to introduce asymmetry within symmetric systems (through the grid). Then when they perform a simulation they can find an unsymmetric solution.

Typically these kind of unsymmetricity are found for non-linear flows(moderate or high Re). So we should be highly carefull in those conditions.

I agree that nonlinearity does not need time derivative to act on the flow behavior. But if you do not introduce any assymetricity you will never get an assymetric solution, even in highly non linear systems..

Let us consider a symmetrical geometry, with a symmetrical grid. Set the time derivative to zero, and introduce an initial symmetric condition....Your steady state solution should stay symmetric...

Now if you introduce a perturbation to the symmetricity, you can then obtain another steady state (more stable)...

I am very carefull when I deal with nonlinearity, the dependance on the initial conditions is very important.

So I advise to find first the steady symmetric solution so that I can be sure that there is no numerical asymmetricity introduced. Then I introduce BY MYSELF the asymmetry and observe the new steady state if it exists...

> Let us consider a symmetrical geometry, with a symmetrical grid. Set the time derivative to zero, and introduce an initial symmetric condition....Your steady state solution should stay symmetric...

No. If the symmetric solution is unstable in the manner of, say, the symmetric solution for a fluidic switch or my previous diffuser example you will always obtain a stable (and physical realizable) asymmetric solution unless the solution procedure introduces fully symmetrical roundoff errors at all stages. The latter is only going to happen in very exceptional circumstances.

In practice, the only way to obtain an unstable, unrealizable, symmetric solution is to force it by halving the grid and imposing a symmetry boundary condition down the middle. You can then use this solution and its reflection on the full grid to show it satisfies the governing equations. However, if you continue with the solution procedure, numerical roundoff will ensure this unstable solution is unstable and it will be replaced by a stable asymmetric one.

How can you have unsymmetrical roundoff errors if the grid is symmetric ??

> How can you have unsymmetrical roundoff errors if the grid is symmetric ??

By sweeping through the grid preferentially. If symmetry is about the mid I plane and you perform forward/backward sweeps from I=1 to I=numi your truncation errors will not in general be symmetrically distributed.

If the grid is symmetric about something other than zero your metrics will not be symmetric. For example, if the symmetry is about x=2.0 then dx=2.4-2.0 will differ from dx=2.0-1.6 by the size of the truncation error.

Etc...

It is difficult to maintain numerical symmetry in a real codes even though it is one of the first checks one uses to check coding.

I never faced such kind of problem, Within Fluent I have always achieved symmetry in the solution...

But a slight disturbance leads to unstability or unsymetricity.

This is maybe because I use no preferential sweeping....

It was an interesting discussion, thanks to you Andy!!