[tomLauriks]

Is it possible that by using steady RANS, convergence for a turbulent flow around a complex bluff body geometry cannot be reached?




This statement was made by the COMSOL (the software that I use) support. However, they only showed me cases with laminar unsteady flow (flow around a cylinder and backward facing step), that did not reach convergence with a stationary solver, while my case is a turbulent flow. And I would expect that a turbulent solver will still find a steady solution, due to the ensemble averaging, which is in practice equal to time averaging, I think. I presume, however, that this depends on the time period over which the flow variables are averaged.
For these reasons, my questions are:
1) Are the steady RANS Navier Stokes equations formulated such that, a specific time over which averaging is performed exists? If yes, how can I determine this time period? If not, why does URANS work? (Thus, why does the time averaging not average over the entire simulation period, and does the transient stadium exist?)
2) Is it indeed possible that for a turbulent flow, a stationary solver will not find a solution, if parts of the flow field are inherently unsteady? Or is this problem probably caused by another problem in my model (bad mesh quality, ....)?

[FMDenaro]

Failing of convergence in solving the RANS equations is quite common. There is no theoretical reason to admit a lack of convergence when the flow has a statistically steady state. Often the problem is due to numerical issue and/or modelling issues

[LuckyTran]

Originally Posted by tomLauriks

1) Are the steady RANS Navier Stokes equations formulated such that, a specific time over which averaging is performed exists? If yes, how can I determine this time period? If not, why does URANS work? (Thus, why does the time averaging not average over the entire simulation period, and does the transient stadium exist?)

In terms of derivation, RANS assumes infinite time average. URANS works because the equations whether you assume infinite-time average or finite-time average is the same. You are using COMSOL so I am not sure the next satement is relevant. But in FVM codes, even supposing that you solve the steady RANS equations, if you use implicit underrelaxatino you will solve even a steady RANS like a URANS (because there is implicit time-step between iterations, there is a relation between under-relaxation factor and local cell Courant number). In other words, numerics are important and can do funny things.

Originally Posted by tomLauriks

2) Is it indeed possible that for a turbulent flow, a stationary solver will not find a solution, if parts of the flow field are inherently unsteady? Or is this problem probably caused by another problem in my model (bad mesh quality, ....)?

Yes it's very possible. Turbulence being defined as random does not really affect the "inherently unsteady" parts (the coherent parts) of the flow. Most vortex shedding examples fall into this category.

But of course bad numerics can also cause problems.

[tomLauriks]

Firstly, both Filippo Maria Denaro and Lucky Tran, thanks for your reply. I'm starting to understand my problem.


Secondly, about the following:

Originally Posted by LuckyTran

In terms of derivation, RANS assumes infinite time average. URANS works because the equations whether you assume infinite-time average or finite-time average is the same.

Could you explain me how the finite-time average in URANS is determined? More specifically, how is the averaging time in URANS determined? (In addition, can the time averaging be different in the different time steps?)

[FMDenaro]

The URANS equations have an implicit assumption about the type of time averaging. As a consequence, the difference in the meaning of the averaging is all contained in the turbulence model...This is somehow similar to the implicit-based LES approach, all the URANS/LES equations seems equal so that is the model that should distinguish the type of averaging. Of course, the numerics introduces a further confusion due to the built-in filter/average in the scheme