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iterations process for steady solver and time dependence for unsteady |
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May 4, 2018, 11:16 |
iterations process for steady solver and time dependence for unsteady
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#1 |
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Serguei
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Hi all,
As far as I know, there is some analogy or relation between time dependence of unsteady flow and the iteration process for steady flow simulations. Regarding that, I would like to discuss very common situation. Assume, I don't know, if my flow steady or unsteady. Initially, I do the iterative simulations, using steady solver. The residuals initially drop and then stuck on high level (cycling or random). I stop the steady solver, save the case and data and turn on the unsteady (transient) solver for single-time step simulations. I obtain converged solution. If this solution (for single-time step) makes any sense? Is this is a valid snapshot of the flow in some unidentified moment of time? Then I return to the saved steady and continued that for some more number of iterations. Then I again turn on the unsteady solver. I again obtained the converged unsteady (single-time step) solution. But it is differ from the previous solution. Is this a valid unsteady solution in the some another moment of time? Last edited by serguei; May 7, 2018 at 13:07. |
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May 4, 2018, 12:20 |
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#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). |
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May 4, 2018, 13:15 |
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#3 | |
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Serguei
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May 4, 2018, 13:33 |
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#4 | |
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Filippo Maria Denaro
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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. |
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May 4, 2018, 13:53 |
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#5 | |
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Serguei
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May 4, 2018, 14:02 |
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#6 | |
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Filippo Maria Denaro
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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 |
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May 4, 2018, 15:31 |
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#7 | |
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Serguei
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May 4, 2018, 15:50 |
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#8 | |
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Filippo Maria Denaro
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If you want to solve the unsteady 3D details of the turbulence behind a cylinder you must use either LES or DNS. |
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May 4, 2018, 16:09 |
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#9 | |
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May 4, 2018, 16:14 |
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#10 | |
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Filippo Maria Denaro
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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. |
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May 4, 2018, 17:06 |
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#11 | |
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Serguei
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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? |
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May 4, 2018, 17:16 |
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#12 | |
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Filippo Maria Denaro
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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. |
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May 4, 2018, 17:24 |
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#13 |
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Filippo Maria Denaro
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I suggest a reading of the introduction
https://www.researchgate.net/publica...LIKE_BEHAVIOUR |
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May 4, 2018, 19:53 |
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#14 | |
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Serguei
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May 4, 2018, 19:55 |
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#15 | |
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Serguei
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will read |
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May 5, 2018, 04:25 |
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#16 | |
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Filippo Maria Denaro
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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. |
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May 7, 2018, 10:38 |
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#17 | |
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Last edited by serguei; May 7, 2018 at 13:16. |
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May 7, 2018, 12:53 |
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#18 |
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Alex
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May 7, 2018, 13:08 |
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#19 | |
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Filippo Maria Denaro
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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. |
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May 7, 2018, 13:46 |
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#20 | |
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Serguei
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iterations, steady, unsteady |
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