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The physical meaning of Reynolds stresses? |
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January 28, 2021, 16:15 |
The physical meaning of Reynolds stresses?
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#1 |
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Hello,
I was just thinking about the physical meaning of Reynolds stresses. I know that the existence of Reynolds stresses shows that there is a turbulent flow. Because they are due to the fluctuating components of velocity field. But there are some questions about them: 1- If they decay, does it mean that we have no or weaker turbulence? 2- [For the channel flow] Why do they have their maximum values in the buffer layer and why do they decay by increasing distance from the wall? Does it mean that far from the wall(half of the channel) the turbulence is gone and the regime is laminar? 3- Do they transfer the momentum or caused by the momentum transfer? 4- Are they the reason for more mixing in turbulent flows? |
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January 29, 2021, 05:50 |
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#2 |
Senior Member
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It actually depends from how you define the average in the first place. If it is a Reynolds average (RANS like), the presence of Reynolds stresses alone is not sufficient to call for a turbulent flow. Such flow could be unsteady but laminar, and yet present some stresses. But even if it is an ensemble average (over infinitely many experiments), it is not possible to exclude that some non-linearity would eventually kick in and make some experiments diverge from the others while still being laminar.
However, the magnitude of such stresses, with respect to the mean flow kinetic energy, is typically relevant for turbulent flows only. But, in addition to this, a turbulent flow will also present vorticity and a wide spectrum of scales. So, to answer your questions: 1) The lower the stresses the higher the chances the flow is laminarizing. Yet, this alone is not conclusive about the state of the flow. 2) For wall bounded flows like the channel, the maximum of turbulent kinetic energy production is, indeed, in the buffer layer. For the channel flow it is at the wall (and near it) that interesting stuff happens. For an unconstrained wall bounded flow like an external turbulent boundary layer, at some distance the flow is practically laminar. But, for the channel flow, which is fully developed, turbulence affects the whole section. Yet, as production is near the wall, what you see in the middle of the channel has clearly lower energy, because it is mostly transported/diffused there, not much produced. 3) Not sure I understand what you mean. If you consider the turbulent and resolved kinetic energy equation for the channel, you see that energy is injected in the system by the driving pressure gradient and goes from the averaged flow to the streamwise stresses only trough the (u'*v')_avg * du_avg/dy term. From this, it goes to the other normal ones by a mechanism of pressure redistribution (basically linked to the continuity and the wall presence). Obviously, the term responsible for the transfer from the averaged flow also affects it backward. So you have mean flow energy that goes into the turbulent ones which, in turn, affects the mean flow. It's the non-linearity, you can't really tell apart who causes what. 4) The reynolds stresses are just a quantification of a physical phenomenon, not the phenomenon in itself. So, it is the turbulence that causes more mixing, its structures that erratically move a blob of fluid from one place to another in a way that is not observed in laminar flows. Those movements are associated with temporal/spatial variations of the quantities that, eventually, can be quantified trough the Reynolds stresses. Obviously, from the averaged equations point of view, the stresses are the only thing you observe in them that will signal a turbulent flow; thus, equationwise, they necessarily are the mechanism behind everything turbulent. |
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January 29, 2021, 06:06 |
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#3 | |
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Quote:
I appreciate your help. 1- Is there any [physical] reason that why should this happen in the buffer layer and not in the other layers? (apart from the DNS results that show this phenomenon) 2- So in the channel flow, the maximum production of turbulence is in the buffer layer. Then some small-scales are generated and move to other places [and maybe get bigger and bigger]. So then why we say that energy of the mean flow is given to large scales and they break into small scales and these small scales dissipate the energy [Energy Cascade]? I'm asking this question because, in channel flow, these are the small scales that have the maximum turbulent kinetic energy production! |
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January 29, 2021, 06:14 |
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#4 | |
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Filippo Maria Denaro
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The general idea that the energy cascade has only one direction, that is from large to small scales, is not always true for all flows. For example in 2D turbulence you have inverse cascade. Near to a wall, vortical structures are confined, small scales are anisotropic and energy is redistributed differently. But be aware of the fact that a subdivision in large and small scale implies somehow you are thinking about LES. Reynolds stresses (Paolo addressed this issue) are not as same as the stresses in the Reynolds decomposition. |
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January 29, 2021, 06:24 |
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#5 | ||
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I think the generation of turbulence is quite the oppisite of the energy cascade! These are the small scales that cause the generation of the large scales and the large scales decay downstream! You can see it in the following video (00:50): https://youtu.be/e1TbkLIDWys I didn't get what you mean by: Quote:
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January 29, 2021, 06:25 |
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#6 |
Senior Member
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If you look at, say, the span-wise energy spectrum of the wall-normal velocity component at a distance y+ = 5 from the wall in the channel flow (ok, this seems very specific, but the concept is more general) you will see that most energy is, indeed, at the smallest scales, and from there it flows to both the larger and smaller scales (of course you don't see this dynamic from the spectra alone, but the peak usually is where production happens, and from there it goes everywhere).
So, forget for a moment the classical picture of energy going from large scales to smaller ones. It happens at large, but it is an approximation of a very idealistic situation (yet, most dissipation, in a way or another, typically happens at the small scales). For the channel flow, the main instability that leads to the turbulence is at the wall, with spanwise rolls (whose size is strongly affected by the vicinity to the wall) that break into streamwise ones and grow away from the wall. I'm not an expert on this, but this happens at walls because that's where the main flow is disturbed ,that's where vorticity is produced. Why exactly at a certain location and not another, from the equations it is clear that the responsible mechanism has this peak somewhere there, but I never reasoned on the physical mechanism behind the way the equations for that term appear so. What I previously wrote is that the energy all comes from the average flow, that is, the mechanism that drives the flow. I didn't wrote that this energy gets injected in the large fluctuating scales. As Filippo wrote, that is an LES picture that I didn't mention, I was referring to the Reynolds average, that doesn't have this scale notion in general. |
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January 29, 2021, 06:34 |
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#7 |
Senior Member
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Actually, what happens in the channel flow is that the main mechanism behind turbulence production at the wall is largely indpendent from the mean flow energy, depending from it only trough the mean shear near the wall du_avg/dy.
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January 29, 2021, 06:52 |
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#8 | |
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Filippo Maria Denaro
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You should study the mechanism of wall-generated turbulence, that is what happens in the turbulent BL (viscous sublayer and transitional region). Have a look to this search: https://www.google.com/search?client...rpin+vortex+BL When you see a fluctuating function, say f', it means nothing without indication of the used mean function. There are infinite possible decompositon f=f_mean+f'. Therefore there infinite possible definition of the Reynolds stresses. |
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reynolds stress, turbulence |
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