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why RANS velocity profile seems to be developing over time

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Old   July 27, 2022, 09:10
Default why RANS velocity profile seems to be developing over time
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Dear Community

I am doing a CFD-DEM coupled simulation to investigate particle infiltration into created bed. I pour the particles in the cavity area and let them attain equillibrium state (packed bed) and then inject fine particles from inlet along with water at certain mass flow rate. I want my flow to be fully developed and therefore cyclic bc is used, where flow is recirculated from outlet back to inlet and flow is drived with bulk velocity specified in fvOptions. Front and back are also treated as cyclic boundaries. As the case is open channel flow case and thus providing symmetry bc at top boundary to approximate the OCF velocity profile as I am not resolving interphase between air and water explicitly.

As per my understanding the cyclic bc between inlet and oulet should result fully developed OCF velocity profile that should not change over time (as it is RANS solution) but what I observe that my velocity profile seem to develop over time and it takes some time get desired velocity profile. In the attached figure you can see that at time t=1 sec, the velocity is almost constant just after some distance from bottom wall but the next figure at t=14 sec you can see that profile varies from bottom to the top (free surface) as it should. I wonder why my RANS solution is changing over time and flow is not developed at the inlet itself from the beginning of simulation?

Hereb are some more info about the case:
-Turbulece model=kOmegaSST
-yPlus=~140
-Initial turbulence parameters (k, omega) estimated using formula given here in the wiki (changing these values doesnot result anything different as it is an iterative solution and final solution remains the same)

I am not sure if it is my understanding of fully developed flow is correct and also it is trivial that RANS solution also takes some time till they get finally quasi steady state solution. I would be grateful if you could help me understand and may be solve the problem.

Best Regards
Atul
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Old   July 27, 2022, 11:34
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Yes it is rather trivial. The cyclic periodc boundary condition results in the fully developed state after the flow reaches the steady state. Fullydeeloped-ness is a property of the flow (everywhere, especially in the interior) after-the-fact, it is not a condition that you impose all the time and certainly don't achieve it simply by imposing a cyclic boundary condition. A boundary condition is only a constraint applied to boundaries, it says nothing about what happens in the interior.
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Old   July 27, 2022, 12:44
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Hallo Lucky

thanks for your answer, it made things clear.

Just to confirm if I understood it correctly: So fully developed flow is achieved after certain time step till it reaches steady state RANS solution. This development for flow at any time should be same at any cross section between inlet and outlet . here, i mean by that: for example t=0, all of all cross-section should have the same velocity profile as flow is recirculated and this is also true for next time steps e.g t=2,3,...

Am i right?

Best Regards
Atul
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Old   July 27, 2022, 13:32
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No.

Let's say your inlet boundary is at x=0 and your outlet boundary is at x=P and this forms your cyclic pair. Periodic/cyclic boundary condition means that u(x=0)=u(x=P) for any time. It does not imply that u(any x)=u(any other x) for all time nor for any time for that matter. That is, the velocity at any cross section can be whatever it wants except for the cyclic pair at x=0 and x=P. Those are the only special sections of velocity.

Fully developed means u(any)=u(any other x) as t=> infinity. From your geometry however, you never meet this criteria because you have a cavity. So not only does your domain not have a fully developed solution in the classical sense, even if it did you would need to crank it for some time before it arrives at that solution.

What you do have however, is the so-called periodically fully developed flow. Streamwise periodic flow means u(x)=u(x+n*P) for any x and integer n's as t=>infinity, but since your domain consists of only 1 periodic segment, it doesn't mean anything in particular except to redundantly say that you have cyclic boundary conditions and the flow inside can develop however it wants. However, again the flow only arrives at this state when it reaches steady state.
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Old   July 28, 2022, 05:05
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Thanks again for clarifying the things.

Quote:
Fully developed means u(any)=u(any other x) as t=> infinity. From your geometry however, you never meet this criteria because you have a cavity. So not only does your domain not have a fully developed solution in the classical sense, even if it did you would need to crank it for some time before it arrives at that solution.
comment about my geometry: this criteria of fully developed flow, u(any)=u(any other x) as t=> infinity for my geomtery should be possible (atleast in the channel part) because the cavity is fully filled with some particles before the simulation starts and that behaves like a wall made up of particles. the whole geometry and flow is more or less like a channel flow.

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Atul
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Old   July 28, 2022, 09:29
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I'll buy that argument if you initialize it with a fully developed profile and then that is only a statement of the initial condition. But is that how it is being approached? Or do you just impose an initial uniform profile and hit the run button? The flow still evolves temporally afterward as soon as the particles get lifted from the cavity. It's not fully developed for all time and we never really say fully developed for temporally evolving flows.


Still, there is no need to twist terminology. It's already succinct what the constraints are being imposed on the solution at any given time. Just avoid unclear verbiage and you'll confuse yourself a lot less.
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