[smillion]

In layman terms could someone please explain what the steady-state means for simpleFoam solver with turbulent flow?

I mean I always thought turbulent flow was unsteady?

Thanks

[mturcios777]

Though I have never used simpleFoam myself, looking at the solver code it appears that steady state would refer to the mean flow, as the monitored residuals are calculated after the momentum equation is solved (in UEqn.H). At least this makes sense to me with RAS turbulence models, not sure how it would apply to LES.

Obviously it depends on your particular case whether or not a steady mean flow even makes sense or says anything useful. Your choice of timescale is would also be important.

[maysmech]

I have same question.

Any other idea?

[smillion]

Well my case involves modelling the time-averaged solution. If that makes any sense?? In terms of turbulent flow, all I know is that it comprises of mean flow and fluctuating flow.

[mdjames]

I would say the steady-state refers to a mean, fully developed flowfield. While a pipe flow may be turbulent, you can expect it to reach a state where the mean flow profile doesn't change with time.

Am I understanding your question correctly?

[smillion]

Originally Posted by mdjames

I would say the steady-state refers to a mean, fully developed flowfield. While a pipe flow may be turbulent, you can expect it to reach a state where the mean flow profile doesn't change with time.

Am I understanding your question correctly?

Hi mdjames. Thanks for your help. Yeah that seems to make sense to me. But how do you discern whether it has reached fully developed flow? Is it a simple case of observing your post-processing flow or does it fully develop when the solution converges?

[mdjames]

is your flow oscillatory or time-independent as t -> Infinity? Also, is there an analytical solution?

Just a thought, you might get some more help from the General CFD forum as I'm not sure if this is strictly an OpenFOAM question.

[smillion]

Originally Posted by mdjames

is your flow oscillatory or time-independent as t -> Infinity? Also, is there an analytical solution?

Just a thought, you might get some more help from the General CFD forum as I'm not sure if this is strictly an OpenFOAM question.

Hi mdjames.

I am doing this on OpenFOAM. I'm not quite sure what you mean by 'is there an analytical solution?'. But as far as I understand, my flow needs to converge to steady-state within the turbulent flow. Does that make any sense?

Basically I'm just curious as to how turbulent flow can be steady-state when I'm constantly reading in fluid dynamics books that turbulent flow is time dependant? It seems somewhat contradictory to use simpleFoam.

[mdjames]

The flow will be steady-state when the mean flow stops developing. The fluctuations(turbulence) are kind of superimposed on the steady state flow.

Think of a river. If you had to describe the flow to someone, you might say that its velocity peaks in the middle and decays to zero at the banks. However, the river is almost certainly turbulent. Would you call a steadily flowing river time-dependent just because it's turbulent? No, the time dependent part of the flow (eddies) are just along for the ride on the mean flow. It's all a matter of what part of the flow you're interested in.

[maysmech]

thanks.
Is Courant and time step value has important role in simpleFoam?

[smillion]

Originally Posted by mdjames

The flow will be steady-state when the mean flow stops developing. The fluctuations(turbulence) are kind of superimposed on the steady state flow.

Think of a river. If you had to describe the flow to someone, you might say that its velocity peaks in the middle and decays to zero at the banks. However, the river is almost certainly turbulent. Would you call a steadily flowing river time-dependent just because it's turbulent? No, the time dependent part of the flow (eddies) are just along for the ride on the mean flow. It's all a matter of what part of the flow you're interested in.

Ok thanks Matt. I think I understand what you're talking about.