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How to Verify Flow is statistically Stationary? |
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#1 |
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I have a turbulent flow in a pipe in two configurations:
Case A) The flow is Re=5,000. The pipe is not rotating. Case B) The flow is Re=5,000. The pipe is rotating. For each of these cases, how can I verify if the flow is statistically stationary? I was reading about all sorts of tests, is there one which is particularly easy and accurate to apply? I am subtracting out the mean and looking at the fluctuation component in the streamwise direction. |
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#2 | |
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Filippo Maria Denaro
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For example, compute the total kinetic energy in the volume for a long period of time. In general the flow is statistically steady in absence of external time depending forces. |
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#3 |
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A statistically steady flow simply is when the computed statistical mean does not depend on time? This is an oversimplification at best and misleading at worst. To determine if the flow is statistically stationary, you need to:
1. **Perform a time series analysis**: Collect time series data of the flow variable of interest at a fixed point in the flow. 2. **Use appropriate statistical tests**: Apply tests like the Augmented Dickey-Fuller (ADF) test to check for unit roots which indicate non-stationarity. 3. **Analyze the autocorrelation function (ACF)**: Ensure the ACF decreases to zero, which indicates stationarity. 4. **Examine the power spectral density (PSD)**: Check if the PSD remains constant over time. A correct and thorough analysis requires these steps, not just computing the mean over time. |
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#4 | |
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Filippo Maria Denaro
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Your answer says nothing relevanto to the original question. When the NSE are statistically averaged you have the time-derivative of the mean variables. Such derivative must vanish for the flow to be statistically steady. Stop. When you solve DNS/LES, you need to fulfill such condition. URANS is a totally different issue. |
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#5 |
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Your response seems to miss a few critical points regarding statistical stationarity. Firstly, you mentioned that the statistically averaged Navier-Stokes equations (NSE) lead to the time derivative of the mean variables, which is true for statistical steady-state conditions but does not directly address statistical stationarity in the context of turbulence analysis. Statistical stationarity refers to the properties of the fluctuations around the mean, not just the mean flow.
Moreover, when you say, "Such derivative must vanish for the flow to be statistically steady," you are mixing the concept of steady-state (mean flow does not change over time) with statistical stationarity (fluctuations around the mean are time-invariant in their statistical properties). Additionally, regarding your DNS/LES point, it's crucial to remember that when the pipe is rotating, it introduces an external time-dependent force, affecting both the mean flow and the turbulence characteristics. Hence, verifying statistical stationarity in this scenario requires careful time series analysis and appropriate statistical tests as previously mentioned. |
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#6 | |
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Filippo Maria Denaro
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As I wrote, the volume-averaged kinetic energy is indeed what could be simply checked, it is in accord to what Pope states at page 66-68 and Fig.3.21. On the other hand, there is no real need to compute anything, the flow problem addresses by itself if it develops a steady and energy equilibrium flow. If you have a time varying mean inflow, if you have an external driving force, that will break the assumptions since the kinetic energy is not statistically steady. The rotation must be unsteady to introduce a time dependent force. A steady rotation does not change the framework. |
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#7 |
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Your response seems to have some misunderstandings about statistical stationarity. The volume-averaged kinetic energy is useful, but it doesn't fully address whether the flow is statistically stationary. Statistical stationarity involves the statistical properties of the fluctuations around the mean being constant over time, not just the mean flow itself.
Also, saying that steady rotation doesn't change the framework is an oversimplification. Even a steady rotation can affect the turbulence characteristics and, consequently, the statistical properties of the flow. It's essential to consider how these factors might influence your analysis. Regarding the need for computation, it's crucial to note that proper statistical analysis of turbulence often requires detailed computations. Assumptions should be verified with actual data, not just accepted without verification. So, while checking the volume-averaged kinetic energy is a good step, it shouldn't replace a thorough statistical analysis to confirm stationarity. |
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#8 | |
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Lucky
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The test for statistical stationary is exactly as straighforward as described. Calculate the statistic (e.g. the mean) using the definition of the statistic. Do this twice at different times. If they match (within error) then it's stationary. Note that you apply the test for statistical stationarity per statistic. Hence, the recommendation to do it for the mean is an example of such a test, not a claim that it is the only check you need to do. It is readily apparently to someone skilled in the art that there are more statistics than the mean. Turbulence and rotation changes the value of the statistic, not whether or not it is stationary. For example if your sequence is 2, 2, 2, 2, 2, then turning on rotating walls may make your sequence -1.5, -1.5, -1.5 -1.5. You get a completely different result but statistical stationari-ness is statistical stationari-ness. Of course good data requires the data to be good, i.e. higher order statistics all stationary and yielding accurate values. Also you can do a very bad simulation (i.e. a very coarse grid) and still get a statistically stationary result. Statistical stationarity is not a measure of data quality, it is simply a property of data. So, to answer your original question. Calculate your statistics that you care about, the Reynolds stresses or the autocorrelation function or the PSD or whatever, and show that they are time independent (for long time). Volume average kinetic energy is a great and simple way to check (your question explicitly asks for quick and easy) because the kinetic energy is a turbulence statistic, averaging it over the volume is quick and easy, and all cells in the computational domain go into the volume so outliers will cause the average to noticeably. If you want a more specific answer, ask a profound question and don't strike down responses answering exactly your kindergarten question. Filippo explained already that the time-derivative of the statistic needs to be zero. If you want to get even more advanced then you can show that 2nd, 3rd, and higher order derivatives all go to zero... Of course if it fails any of these checks then you know right away that it is not. Like any good statistical test, you always use the null hypothesis. |
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#9 |
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I don't know if there are mathematical proofs of this but, a steady mean flow requires steady Reynolds stresses, and so on going up along the chain of higher order statistics. That is, it is a necessary condition (altough not sufficient). This goes beyond whatever you think you understand about the physics of the flow.
In practice, we do not live in a steady world. Not even your simulation, which somehow had to artificially start somewhere, so there will always be a sufficiently high order statistic showing memory. So, if you're accustomed to science, which comes before NS, you should be aware of these facts: 1) One typically limits the investigation to the things that can or one is willing to measure. If you can't measure a 6th order statistics or don't care about it, then it doesn't really make sense to argue about its state. 2) Linked to the previous and the premise, as nothing is steady and no measure is perfect, you need to set up a threshold for your measure of X to decide on its state. 3) As in any field, there is a community around this topic that, guess what, argued about this and wrote down in papers several possible solutions. Now, you come here and, basically, ask for a TLDR solution for how to determine if a flow is statistically steady. What you got as answer, as already highlighted by others, is directly proportional to the effort you seem to have put in investigating previous works on the topic, which is fair. Not only this, but the volume average of the kinetic energy also has a number of other properties that make it a relevant quantity to monitor, in any case, for unsteady simulations. In contrast, most of the things you mention as actual solution, are largely impractical in a first assessment of the steady state and, in fact, not used at large in the community. Finally, if you think you already know the answer, what are you asking for exactly? |
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#10 |
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Filippo Maria Denaro
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I haven't replied to the sentences of the author of this post since further comments were addressed. I don't know if the autor will reply again, it seems more a post to be noticed. However, as already addressed, if we look at the equations, when <u> does not depend on time, the same must be for <u'u'>. At least in case of incompressible flows where <p> is only an auxiliary variable.
From an energetic point of view, turbulence in equilibrium and statistically steady means that production equates dissipation. That can be computed and verified. This discussion has sense if we are talking about DNS/LES, in RANS you solve directly for the steady state. |
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#11 | |
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Lucky
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I am willing to concede that one could possibly use weird statistical tests like ADF to test for time independence but that is definitely a weird way to do so. Unlike behavioral data of lab rats or humans, it is clear what is population trend. The acceptable range of error that you need to test the null hypothesis is always the next higher order statistic. Since we are dealing with statistical tests on statistics... all the even orders of statistics such as the variance are positive semi-definite, which makes finding their error bounds that much more trivial from a theoretical point of view. Of course, the practical issue is how to actually compute these higher order statistics efficiently on non-uniform grid and then there is the elephant in the room–how many plots is anyone willing to look at |
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#12 |
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Thanks for all your responses. They were insightful to read. I would rather err on questioning something I really don't understand a bit too far, rather than pretend to understand it immediately.
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moderately turbulent flow, pipe flow, stationary |
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