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May 4, 2018, 11:20
#2
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Filippo Maria Denaro
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No, you need to assess the type of flow problem you are solving.
If the problem is laminar, say for example a 2D flow around a cylinder at Re=100 it will not converge to a steady state because the solution is unsteady in its nature due to the vortex shedding. Whatever initial condition you prescribe, after a sufficient numerical transient and if the scheme si sufficiently accurate, the flow develops fully to the same oscillating solution.

But if you are solving turbulent flow problems using a statistical averaging, the steady state represent a specific solution for a time-independent averaged variable (RANS) while solving the unsteady case has a different meaning in the unsteady averaged solution (URANS).

May 4, 2018, 12:15
#3
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Serguei
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 Originally Posted by FMDenaro No, you need to assess the type of flow problem you are solving. If the problem is laminar, say for example a 2D flow around a cylinder at Re=100 it will not converge to a steady state because the solution is unsteady in its nature due to the vortex shedding. Whatever initial condition you prescribe, after a sufficient numerical transient and if the scheme si sufficiently accurate, the flow develops fully to the same oscillating solution. But if you are solving turbulent flow problems using a statistical averaging, the steady state represent a specific solution for a time-independent averaged variable (RANS) while solving the unsteady case has a different meaning in the unsteady averaged solution (URANS).
I am solving turbulent flow problem. As far as I understand, I may use any initial conditions for transient solver. I did use for the initial the steady solution, which, actually, may be not converged. Then I got converged transient solution. See attachment: before 3000 iterations - steady solver, after - transient. So if I want to get the proper solution at any moment of time (doesn't matter what), is it what I am looking for?
Attached Files
 residuals (2).pptx (56.1 KB, 15 views)

May 4, 2018, 12:33
#4
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Filippo Maria Denaro
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 Originally Posted by serguei I am solving turbulent flow problem. As far as I understand, I may use any initial conditions for transient solver. I did use for the initial the steady solution, which, actually, may be not converged. Then I got converged transient solution. See attachment: before 3000 iterations - steady solver, after - transient. So if I want to get the proper solution at any moment of time (doesn't matter what), is it what I am looking for?
The key is to understand if your problem admits or not a physical steady (in statistical meaning) state. For example, a in-cylinder turbulent flow with a cyclic piston movement has no steady state. The same happens for a turbulent flow due to unsteady BC.s where the production of energy is not balanced only by dissipation and there is no energy equilibrium.

Your problem is to understand if a RANS staedy solution is physically suitable for your case or the fact that it does not converge is a signal of its unsteady character. If you assess that RANS is possible, the failing in the convergence is due to numerical and/or modelling issues. Changing to URANS will just add a time derivative that has no physical meaning.

May 4, 2018, 12:53
#5
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 Originally Posted by FMDenaro The key is to understand if your problem admits or not a physical steady (in statistical meaning) state. For example, a in-cylinder turbulent flow with a cyclic piston movement has no steady state. The same happens for a turbulent flow due to unsteady BC.s where the production of energy is not balanced only by dissipation and there is no energy equilibrium. Your problem is to understand if a RANS staedy solution is physically suitable for your case or the fact that it does not converge is a signal of its unsteady character. If you assess that RANS is possible, the failing in the convergence is due to numerical and/or modelling issues. Changing to URANS will just add a time derivative that has no physical meaning.
My flow is type of flow over cylinder with Re about 10000, (but bounded and with asymmetric output). So, it may create the turbulent vortex street or some other type of unsteadiness in the turbulent regime. On another hand, the Re is still looks too low for that.

May 4, 2018, 13:02
#6
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Filippo Maria Denaro
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 Originally Posted by serguei My flow is type of flow over cylinder with Re about 10000, (but bounded and with asymmetric output). So, it may create the turbulent vortex street or some other type of unsteadiness in the turbulent regime. On another hand, the Re is still looks too low for that.

The flow is for sure unsteady in deterministic sense but if you do not add a cylinder oscillations or other unsteady setting in the BC.s, you can solve the statistically steady equations that eliminate any type of unsteady fluctuations from the solution.
Therefore your problem is in the numerical (grid or method) or modelling (turbulence) setting.
You can find in internet many similar problems for the flow over a cylinder

May 4, 2018, 14:31
#7
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 Originally Posted by FMDenaro The flow is for sure unsteady in deterministic sense but if you do not add a cylinder oscillations or other unsteady setting in the BC.s, you can solve the statistically steady equations that eliminate any type of unsteady fluctuations from the solution. Therefore your problem is in the numerical (grid or method) or modelling (turbulence) setting. You can find in internet many similar problems for the flow over a cylinder
The turbulent vortex street behind cylinder or vortex shedding, etc, are unsteady flows by nature, not because external oscillations. As far as I understand, if we try to use steady solver for simulations such flow, we are going to have problems with convergence (as I actually have), and, again, may be because real physics, not because numerical issues.

May 4, 2018, 14:50
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Filippo Maria Denaro
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 Originally Posted by serguei The turbulent vortex street behind cylinder or vortex shedding, etc, are unsteady flows by nature, not because external oscillations. As far as I understand, if we try to use steady solver for simulations such flow, we are going to have problems with convergence (as I actually have), and, again, may be because real physics, not because numerical issues.
Not at all. You are confusing the physical appearence of the vortex shedding in the pointwise unsteady solution of the NS equation with the statistically averaged solution. In RANS any type of unsteady features is cancelled by the averaging and the solution shows only the averaged field.
If you want to solve the unsteady 3D details of the turbulence behind a cylinder you must use either LES or DNS.

May 4, 2018, 15:09
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 Originally Posted by FMDenaro Not at all. You are confusing the physical appearence of the vortex shedding in the pointwise unsteady solution of the NS equation with the statistically averaged solution. In RANS any type of unsteady features is cancelled by the averaging and the solution shows only the averaged field. If you want to solve the unsteady 3D details of the turbulence behind a cylinder you must use either LES or DNS.
If so, what transient ANSYS FLUENT solver is doing, which available for RANS also?

May 4, 2018, 15:14
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Filippo Maria Denaro
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 Originally Posted by serguei If so, what transient ANSYS FLUENT solver is doing, which available for RANS also?
Unsteady available for performing URANS (aparto from DNS and LES). In such formulation, the unsteady character is due to an external driving force like a periodic motion of a piston or unsteady BC.s.
The URANS variable is theoretically different from the RANS variable.
The RANS formulation must converge to a steady state by definition of the statistical averaging.
I understand that the averaging implied by each formulation can be somehow misleading, you can have a look to the book of Wilcox.

May 4, 2018, 16:06
#11
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 Originally Posted by FMDenaro Unsteady available for performing URANS (aparto from DNS and LES). In such formulation, the unsteady character is due to an external driving force like a periodic motion of a piston or unsteady BC.s. The URANS variable is theoretically different from the RANS variable. The RANS formulation must converge to a steady state by definition of the statistical averaging. I understand that the averaging implied by each formulation can be somehow misleading, you can have a look to the book of Wilcox.
I have some steady solution. The residuals are high, by whatever reasons (may be numerical, as you said). So, I don't trust to this steady solution. But I may use that as initial for transient solution (why not?). Then, I take well converging unsteady (transient) single-time step solution, as you may see from attached plot. Is that OK? Is that correctly obtained solution for unsteady flow?

May 4, 2018, 16:16
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Filippo Maria Denaro
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 Originally Posted by serguei OK, let return to my initial question, please: I have some steady solution. The residuals are high, by whatever reasons (may be numerical, as you said). So, I don't trust to this steady solution. But I may use that as initial for transient solution (why not?). Then, I take well converging unsteady (transient) single-time step solution, as you may see from attached plot. Is that OK? Is that correctly obtained solution for unsteady flow?

Again, you formally can use any type of initial condition for the unsteady solver, that is correct. But that means nothing in terms of the final solution if you are not assessing what are you solving. The turbulent flow around a cylinder is statistically steady and RANS formulation should provide a convergent steady solution. In the literature, for this problem you can find also unsteady solutions obtained using URANS. In my opinion that approach is formally not so clear as the URANS should however converge to a steady state for this problem (the same solution of the RANS). However, the type of solution can be debated in terms of the averaging and the meaning of the solution.
Therefore, in my opinion the unsteady solution you get is likely to be non physical but due to numerical issue. In any case try the URANS, you can check the frequency of the shedding and compare it to other studies.

 May 4, 2018, 16:24 #13 Senior Member   Filippo Maria Denaro Join Date: Jul 2010 Posts: 6,777 Rep Power: 71 I suggest a reading of the introduction https://www.researchgate.net/publica...LIKE_BEHAVIOUR

May 4, 2018, 18:53
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 Originally Posted by FMDenaro Again, you formally can use any type of initial condition for the unsteady solver, that is correct. But that means nothing in terms of the final solution if you are not assessing what are you solving. The turbulent flow around a cylinder is statistically steady and RANS formulation should provide a convergent steady solution. In the literature, for this problem you can find also unsteady solutions obtained using URANS. In my opinion that approach is formally not so clear as the URANS should however converge to a steady state for this problem (the same solution of the RANS). However, the type of solution can be debated in terms of the averaging and the meaning of the solution. Therefore, in my opinion the unsteady solution you get is likely to be non physical but due to numerical issue. In any case try the URANS, you can check the frequency of the shedding and compare it to other studies.
You said: "The turbulent flow around a cylinder is statistically steady". This is debatable. I meat in the literature some mentions about turbulent vortex street. And physically it does make a sense. But, any way, if I got the converged transient solution, it should be doesn't matter, from what initial I got it. For sure, it may be converged, but still is not physical. But if that is physical, may be you are not right?

May 4, 2018, 18:55
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 Originally Posted by FMDenaro I suggest a reading of the introduction https://www.researchgate.net/publica...LIKE_BEHAVIOUR
thank you,

May 5, 2018, 03:25
#16
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Filippo Maria Denaro
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 Originally Posted by serguei You said: "The turbulent flow around a cylinder is statistically steady". This is debatable. I meat in the literature some mentions about turbulent vortex street. And physically it does make a sense. But, any way, if I got the converged transient solution, it should be doesn't matter, from what initial I got it. For sure, it may be converged, but still is not physical. But if that is physical, may be you are not right?

You are largely confusing the physical instantaneous aspect of the vortex shedding (that you can simulate using DNS or LES) from the appearence of the statistically averaged solution (that you approach via RANS/URANS). I strongly suggest to study better such issues. Having a commercial code where you can click on some GUI option and having some numerical solution to plot has nothing to do with the understanding of fluid dynamics.

May 7, 2018, 09:38
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 Originally Posted by FMDenaro You are largely confusing the physical instantaneous aspect of the vortex shedding (that you can simulate using DNS or LES) from the appearence of the statistically averaged solution (that you approach via RANS/URANS). I strongly suggest to study better such issues. Having a commercial code where you can click on some GUI option and having some numerical solution to plot has nothing to do with the understanding of fluid dynamics.
thank you for teaching. You are absolutely right theoretically, from general point of view. But in the real CFD life, some particular issues is not so easy to explain from the general concepts in the couple of sentences. One of those questions is my question from the beginning of this discussion.

Last edited by serguei; May 7, 2018 at 12:16.

 May 7, 2018, 11:53 #18 Super Moderator     Alex Join Date: Jun 2012 Location: Germany Posts: 3,400 Rep Power: 47 The results of your unsteady flow at any given moment t< depend on the initial conditions. In your case the initial conditions are prescribed by the instant you stopped using the steady solver. So yes, the converged unsteady solution is physically valid. But it depends on the initial conditions. serguei likes this.

May 7, 2018, 12:08
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Filippo Maria Denaro
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 Originally Posted by serguei thank you for teaching. You are absolutely right theoretically, from general point of view. But in the real CFD life, some particular issues is not so easy to explain from the general concepts in the couple of sentences. One of those question is my question from the beginning of this discussion.

Generally, if you have un unsteady solution in deterministic sense, it is fundamental to prescribe the correct and physical initial and (unsteady) Bc.s. For example that is fundamental if you are using an exact unsteady solution of the NS equations to assess your code.

However, you are simulating a statistical solution in a turbulent flow, therefore you can start from any initial condition (also from scratch) but you have to wait until the run gives a physically correlated solution. That means you have your solution running a numerical transient that means physically nothing for a certain number of time units. For example immagine the flow in a cylinder subject to a periodical motion of a piston. You start from some initial condition but you have to assess how many cycles are needed before the numerical transient is ended. This is particularly critical for DNS/LES and somehow less critical for URANS.

In your specific case, I suggest starting from your initial approximatively-steady solution, then let run the code for several time units controlling the evolution of total kinetic energy in time. You should see a numerical transient until this quantity oscillates around a well defined averaged level. From now on, the numerical solution has a meaningful unsteady feature that you can analyse while discarding the previous solutions.

That apart my criticism about the real meaning of URANS in your problem.

May 7, 2018, 12:46
#20
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Serguei
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when you say:"but you have to wait until the run gives a physically correlated solution", do you mean when numerical transient is over, and some my control value, as you say:" oscillates around a well defined averaged level"? As far, as I understand, it not necessarily should be integral value, like total kinetic energy, but, for example,velocity magnitude in some control point? Isn't?