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The difference between steady state and transient 

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November 30, 2018, 01:17 

#21 
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urosgrivc
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Exactly, that is why I mentioned, very rarely, as converging a solution to a point that there is no instabilities is almost impossible, and even if it is possible for some cases, with the change of mesh size or turbulence model you can make it harder to converge again when you include all the details of the flow.
And if it does converge perfectly on a coarse mesh all the details are smudged or dissipated out of the solution, well that is what I meant that it sometimes appear similar when it actually still is not, yes in the numerical world but not in real one. Please give me an example where a steady state and transient solution would be the same: I will just decrease the mesh size and put a more complex turbulent model on it and we will always get to point that these are never the same. Because a transient solution is always a true one. And a steady state one always needs a grain of salt with it. 

November 30, 2018, 04:21 

#22 
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Glenn Horrocks
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I'm sorry, I do not understand your post.
My logic is simple, in the transient Navier Stokes equations if you put the time gradients to zero, (that is you march it out to steady state), then you get the steady state Navier Stokes equations. As they are the same underlying equation then a properly converged solution of them will be the same. Well, sometimes anyway..... the Navier Stokes equations are prone to bifurcations and nonunique solutions
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November 30, 2018, 06:44 

#23 
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Chris Schäfer
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O.K. understood. If Steadystate sets “only” the time dependent terms in the Navier Stokes equations to zero than it is quite obvious to me that Steadystate and transient solutions are the same at least for nonturbulent flow.
But than I would assume that the Statement fully converged Steadystate= Transient, is not valid for turbulent flow? If I have fluctuations in the actual solution I would assume to get different results. Because these cannot be captured by the steady state solution, they depend on time, and you have imposed timeindependence. A nice example can you see on youtube “CFD Transient vs Steadystate" from Ihor Zichenko. He simulates the famous Karman vortex street behind a cylinder. It is obvious for me that you got different results because Steadystate just building up the mean flow. But maybe correct me if I mixed things up or the youtube movie is wrong? @urgosgrivc, we are considering here a general case very all parameters (e.g. mesh ) keeping untouched for both simulations. However turbulence as I explained above is a good point which should be considered by the comparison between SteadyState and transient solution if I’am not wrong? 

November 30, 2018, 07:15 

#24 
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urosgrivc
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I am absolutely sure that you will have fluctuations (rather large ones) in a transient simulation for a case like (flow over a terrain)
And yes steady state will give you a good initial condition or timeaveraged flow values (initialization of transient simulation with the steadystate solution is usually done to save computational time. If you would use transient from the beginning, the flow would need a lot of time to develop through the domain and in many cases, this part (flow development) is not in our interest. Steadystate solver can advance the simulation to its final solution a lot quicker) And transient results will be all over the place, probably some fluctuations repeating at a certain frequency dependant on the size of the vortices, conservation of momentum dictates that small vortices move or change faster than larger ones, this will depend on the terrain and distance from the ground as the further the vortices travel the more they dissipate. If you would be looking at pressure or velocity in any given point in the domain the steadystate solution would give you a fixed number. But transient will give you a timedependent value some kind of a fluctuating curve and if you average out this curve, the steady state number `should be equal` to this timeaveraged one. 

November 30, 2018, 21:23 

#25 
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Glenn Horrocks
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Be careful to separate the physical process, which involves turbulent flows with random fluctuations; from the mathematical modelling of those processes. Turbulent flows are often modelled using turbulence modelling, and most turbulence models start with Reynolds Averaging. Reynolds averaging assumes that you can separate a turbulent flow into a mean flow component and a fluctuation magnitude component, both of which are steady in time. So Reynolds Averaging converts a transient flow field into steady state equations. And then if you do a steady state or transient model (marched out to steady state) of those RANS equations you will get the same answer  again because the underlying equations are the same.
The question of whether RANS is appropriate is another question. urosgrivc: If the flow has transient fluctuations, even small ones, then the steady state simulation cannot fully converge. You cannot compare an unconverged steady state simulation to a converged transient simulation. That is meaningless.
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December 4, 2018, 08:31 

#26 
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Dear Glenn (and also urosgrivis) thanks for your posts.
I getting slowly familiar with that topic and your answer sounds reasonable. But I have 2 further questions in my mind. 1.) If the underlying equations are (except the timedependence) exactly the same, and you can therefore apply also steadystate simulation on turbulent flows why do in some turbulent cases steadystate not converge? Can not be every transient flow somehow statistically steady where you can do the average on? Thats brings me to my second question, can you apply your last post to the kalman vortex shedding case which I mentioned in my previous post (its on youtube). Here you see a converge Steadystate case which represent just the mean flow which is obvious quite different to transient case. Before I read your last post, I was wondering why steadystate not fails here because it is highly transient for me. 

December 4, 2018, 17:25 

#27 
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Glenn Horrocks
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These questions lead to some fundamental questions about turbulence.
1) Why do you get turbulence anyway? It is because the flow has sufficient energy that it can overcome viscous dissipation. This means that imperfections are not damped out and grow. In the laminar flow region the flow does not have sufficient energy to overcome viscous dissipation and imperfections are quickly damped out. The intention of a turbulence model is to replace this turbulence energy growth and dissipation with a modelled dissipation equivalent to the turbulence energy cascade. The model is intended to be steady state and adds dissipation to the mean flow to stabilise it to steady state. But no turbulence model is perfect. The dissipation is too much sometimes and not enough in others, so there will be times when there is not enough dissipation, then the flow will be underdamped and instabilities will grow. 2) And a second reason turbulence models fail to obtain a steady state result is when they are used in the wrong application. The von Karman vortex street in a laminar flow is one example. A common error is to think the von Karman vortex street is turbulent  this is not necessarily correct. At low Reynolds number the von Karman vortex street is laminar, but still sheds vorticies. Why isn't this turbulence? Because turbulence is defined as chaotic flow over a wide range of length and time scales. Low Reynolds number von Karman vortex street has just a few length and time scales (ie the shedding frequency and not much else). It does not have large eddies dissipating energy to medium eddies which dissipate energy all the way down to the Kolmogorov length scale (which is microscopic in most cases) and the energy is dissipated through viscosity at the Kolmogorov length scale vorticies. So the low Reynolds number von Karman vortex street is NOT turbulent, and therefore you cannot use a turbulence model to turn it into a steady flow. If you try you will find it does not generate enough dissipation to stabilise the flow and you still get some vortex shedding (but less, as the turbulence model has increased the dissipation). At higher Reynolds numbers where the von Karman vortex street is turbulent this principle still applies. The turbulence model can handle the turbulence, but the vortex street is not turbulent eddies generated using the normal production and dissipation. So existing turbulence models do not handle this distinction, so you are always going to have problems with it.
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December 7, 2018, 03:42 

#28 
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Chris Schäfer
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Dear Glenn, thank you so much for your help.
Ich have just (one last) question regarding this sentence: "But no turbulence model is perfect. The dissipation is too much sometimes and not enough in others, so there will be times when there is not enough dissipation, then the flow will be underdamped and instabilities will grow." But Transient model rely then on the same problematic, therefore I would assume that transient Model fails there too. So where is the difference? 

December 7, 2018, 04:55 

#29 
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A transient model does not need this turbulence dissipation. The flow field can move each time step to follow the way it wants to move and it will converge fine. But if this happens to a steady state simulation it will fail to converge as there is no steady solution to converge towards.
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October 4, 2019, 13:28 

#30  
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George Corner
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Quote:


October 7, 2019, 18:39 

#31 
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Glenn Horrocks
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I do not understand your comment. Your comment appears to have nothing to do with the quote you show. The quote says that a converged steady state simulation, if used as an initial condition for a transient run will result in a transient simulation which does nothing (ie the flow does not evolve, it already has the steady state result). Your comment is about transient simulations converging sometimes when steady state diverges (which I agree with by the way).
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December 5, 2019, 01:31 

#32 
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Dear Glenn ( and also others )
I found this discuss quite useful for me. If I understood correctly, if a flow system has a steady state, the the results for both steadystatesolver and transientsolver will be the same. And the results of a steadystatesolver can be used as a input of the transientsolver. Am I right? So, could you please check a post of mine here? https://www.cfdonline.com/Forums/op...lversame.html I simulates a flow using both the steadystatesolver and transientsolver from OpenFOAM, but got different results. So I want to ask if the steadystatesolver is applied to a system that does not have a steady state, what will happen? Will the results not converge, or it will converge but have no physical meanings? (I found that urosgrivc above said that steadystatesolver results are timeaveraged results) Because what I want to simulate includes slow heat transfer so I want to use steadystatesolver. However the different results make me confused. Would you kindly give me some advises about my simulation? Thank you! 

December 5, 2019, 16:29 

#33 
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Glenn Horrocks
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If the transient simulation is laminar and it settles to steady state then there is nothing which changes with time.
If the transient simulation is turbulent and it settles to a steady state then you are solving the RANS equations so the turbulence is being time averaged, but the large scale flow field has no time derivative (ie does not change with time). Heat transfer + fluid flow often leads to problems due to the different time scales of the physics. Fluid flow is fast and heat transfer is often slow, but you need a small time step to get the fluid flow to converge, but you have long physical times due to the slow heat transfer  meaning run time is very long. A much better way of doing this is to separate the models. This can be done many ways, but a good one is to run a series of fluid flow simulations at different temperatures. From these simulations you extract heat transfer coefficient versus temperature, and then you do a heat transfer only simulation using the HTC versus temperature relation. This approach speeds the simulations as each type of physics can use a time step appropriate for it.
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December 9, 2019, 03:59 

#34 
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Dear Glenn
yes this is the problem I met. the heat transfer is quite a time consuming process so I need to calculate for a long time until it reaches a steady state if I use transient solver. I will try to consider the decoupling way you suggested. And let me make a summary to your reply: a) laminar setting in CFD + long time calculation that the simulation reaches steady state = no variation in the results against time b) RANS setting in CFD + long time calculation that the simulation reaches steady state (no whether the flow in the real world has steady state or not ) = no variation in the results against time ( because in RANS time averaged equations are solved ) c) laminar setting in CFD + short time calculation that the simulation did not reach steady state = variation in the results against time shows the flow developing d) RANS setting in CFD + short time calculation that the simulation did not reach steady state = variation in the results against time am I understanding it correctly? for b), because the RANS model is time averaged, so even for a simulation that simulates a physical flow that won't have steady state can reach a steady state in the CFD showing the time averaged flow field, right? for d), since the RANS model solves time averaged equations, so what will the variation in the results mean? Does it mean the real flow developing during time? Sorry for taking your time and thank you for your reply Best regards. 

December 9, 2019, 16:55 

#35 
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Glenn Horrocks
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I don't understand what you are trying to ask with your 4 variations.
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December 9, 2019, 20:46 

#36 
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Hi Glenn
sorry for my english. but I want to explain what I'm thinking again. for a real flow, despite it is laminar or turbulent, a) if the real flow have a steady state, then if we use CFD to simulate it and use the laminar model in the simulation, and we use the transient solver, after we calculate enough time, the simulation results will reach a state that it won't change anymore. b) despite the real flow have a steady state or not, if we change the laminar model to the RANS model in the simulation, after we calculate enough time from initial, the simulation results will finally reach a state that don't change anymore. because the RANS equations are time averaged equations, so the results are the time averaged fields. c) if the real flow have a steady state, and if we use laminar model but the calculation time is not long enough that the results still have some variations such as velocities, then this results represents the developing process of the flow from the initial to the steady state. d) despite the real flow have a steady state or not, if we use the RANS and calculate form initial, but the calculation time is not long enough, the results are still changing, what is the meaning of this result? is it representing the developing of the flow? of course the laminar or turbulent model should be selected depend on the problem, but here I want to ignore the physical meaning of the calculation results and only want to know if we will get the nochanging results or not. for example, we apply the laminar model to simulate a turbulent flow for a long long time, the results have no meaning because it did not capture the turbulence, but it finally reach a steady state that the flow field don't change. hope you can understand me thank you! Best regards. 

December 9, 2019, 22:50 

#37 
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Glenn Horrocks
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You can use RANS models in transient simulations. This assumes that there is a time scale separation between the turbulent time scales (which are not resolved so their effects are modelled in the RANS approach) to the bulk flow time scales (which are resolved). As long as this separation in time scales is clear then RANS can work. But in many cases the separation of time scales is not clear, which means RANS needs to be used with caution.
But I don't think this is exactly your question, I think your question is whether laminar and RANS approaches can model the transient development of a flow  the answer to that is yes. Be careful about the issues I described in the first paragraph.
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