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January 17, 2013, 11:15 
Why y+ < 1

#1 
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Philipp
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Hi all,
I allways wondered about why  for resolving the viscous sublayer  the threshold for y+ is supposed to be "1". Why not anywhere in the linear region, thus "<5" ? As far as I understood it is needed to get correct boundary conditions for turbulent quantities (such as k) and for the velocity gradient at the wall. Now, since the viscous layer is linear, the velocity gradient would be the same for y+ = 2. Which of the other values would be incorrect for higher y+ ? Thanks in advance!
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January 17, 2013, 12:03 

#2  
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Filippo Maria Denaro
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The key, is to see y+ as a local Reynold number based on the computational grid length. y+ = O(1) means that you are able to solve in a region where the diffusive (and dissipative) effetcs are of the same order of the convective ones. That means also that a fluid particle the is advected in the computational cell has the same characteristic diffusive time. 

January 17, 2013, 12:37 

#3 
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Guan
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the fluctuation velocity close to the wall is not linear. it is proportional to y+^2


January 17, 2013, 14:47 

#4  
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Actually, it is much more important to have sufficient cells across the entire inner layer to resolve the variables changes within the inner layer. You can actually achieve solutions superior to y+<0.5 but not enough cells in inner layer, using y+ ~ 5 and with enough cells to completely resolve the inner layer. This is actually the mesh criteria for LES and DNS style grids where y+ ~ 3 is considered sufficient as long as the most important regions are resolved. 

January 18, 2013, 06:40 

#5  
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Thank you all for your replies!
Quote:
Quote:
Quote:
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January 18, 2013, 06:45 

#6 
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Filippo Maria Denaro
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Actually, I wrote O(1) ... resolving with some grid point below y+=1 is a sort of guarantee that you are able to solve the boundary layer profile taking into account for an accurate evaluation of the wall stress.
Howevewr, in our group LESinItaly, we did several comparison for LES using an unresolved grid and prescribing the noslip condition. Apart the wall coefficient, the statistics in the inner part of the flow were acceptable 

January 18, 2013, 06:56 

#7  
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Quote:
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January 18, 2013, 07:02 

#8 
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Filippo Maria Denaro
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I think that the key is to forget for a while the theoretical results for turbulence over flat plate/channel/pipe .... in practical applications is so rare to have some boundary that can be "assumed" as a theoretical wall ... From my experience, linear law, log law, etc. are useful to understand some features of the flow but I doubt that can be used to prescribe general boundary condition when your first grid point close to wall exceeds y+=1, expecially in LES. Maybe in RANS this approach can be better justified


January 18, 2013, 08:54 

#9  
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To properly calculate the gradient at solid boundaries, you need sufficient data points there so that the discretization error is not large for gradients (basically finite difference approximations to derivatives). If on top of fluid flow you are also trying to simulate convective heat transfer (or mass transfer). You need to have finer grids because the thermal & concentration profile can only be properly resolved only after the hydrodynamic profile has already been resolved. i.e. the thermal & concentration gradients are guaranteed to be always less accurate than the velocity gradient. I'm glad we were able to have this discussion as I find people nowadays that blindly accept the y+~ 1 rule, and then apply a huge stretch factor to their grids and then claiming that they have "resolved the viscous sublayer" 

January 18, 2013, 08:57 

#10 
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What I was trying to say: When the profile is linear, it does not matter if I have y+=1 or y+=4, because due to linearity it's always the same gradient!
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January 18, 2013, 09:14 

#11  
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It is linear yes but it is an unknown linear gradient. That is because of the unknown friction velocity (recall the formulas for u+ and y+). Hence although you know the gradients to be linear, you do not know the value of the gradient. Put another way, you don't know the value of y+ until you know the friction velocity, which you do not know until you have solved the problem. Until then, you must solve for y+ (solve for friction velocity). To solve for friction velocity, you must somehow resolve the shear stresses in that region. Hence your grid requirements are motivated by the need to solve for these gradients. 

January 18, 2013, 09:16 

#12  
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Filippo Maria Denaro
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again, what you addressed is a result of a statistical theory... then, if you solve for the statistical variable, as in RANS, what you say can be somehow right. But for other formulation you do not solve for the statistically averaged velocity... 

January 18, 2013, 09:30 

#13 
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Yes, that is what I ment.
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January 18, 2013, 09:40 

#14  
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Imagine a standard universal wall velocity profile. Now you solve the equations numerically on a grid. Somewhere in your discretized equations the friction velocity appears. For the velocity gradient you will use the first cell's velocity divided by the distance to the wall. This is a pretty good assumption, since you know from the books, that the velocity profile is linear. Now, what I was trying to say is, that this velocity gradient will allways have  exaclty  the same value, whether it is calculated at y+=1, 2,3,4,5. Since the velocity is a linear function there, the "sample rate" of your numerical grid is not important.
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January 18, 2013, 09:43 

#15 
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But you do not know this velocity in the first place. And that is why you cannot compute the y+ and velocity profile. You can only arrive at this velocity by properly accounting for the fluid stress (by solving the momentum balance).


January 18, 2013, 09:45 

#16 
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I think since we use an iterative solver, we take the gradient of the last iteration and keep doing until convergence.
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January 18, 2013, 10:22 

#17 
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Alex
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Let me join the discussion with a new thought:
If the y+ of the first cell is 5, the extent of the next cells can only be equal or larger than 5 for a mesh with a reasonable expansion ratio. While the velocity is a linear function in the region of y+<5, it becomes nonlinear beyond this point and may not be resolved adequately with cells larger than y+=5. Thus a y+ of 1 is a conservative estimate to ensure that also the nonlinear region y+>5 is resolved reasonably. 

January 18, 2013, 10:33 

#18 
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Philipp
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Flotus, I totally agree. But this is "just" an accuracy problem. You don't abuse any model equation with complete nonsense by ignoring it, right?
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