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 RodriguezFatz January 17, 2013 11:15

Why y+ < 1

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+ ?

 FMDenaro January 17, 2013 12:03

Quote:
 Originally Posted by RodriguezFatz (Post 402463) 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!

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.

 foolboy007 January 17, 2013 12:37

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

 LuckyTran January 17, 2013 14:47

Quote:
 Originally Posted by RodriguezFatz (Post 402463) 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!
Your logic is pretty much spot on. y+ ~ 1 is a general rule (a very conservative one). You really only need y+<5 to apply the laws in the laminar sublayer. The additional restrictions come from calculation of gradients near the wall (where you now need a few points in the linear region). So if you wanted say 2-3 points in the linear region to resolve the gradients there then you would need correspondingly y+ ~ 1.5 - 2.5. If you are not worried about near wall gradients then you can use y+ ~ 5 and be okay.

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.

 RodriguezFatz January 18, 2013 06:40

Thank you all for your replies!

Quote:
 Originally Posted by FMDenaro (Post 402482) 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.
But all this is also the case for y+ = 2.

Quote:
 Originally Posted by foolboy007 (Post 402488) the fluctuation velocity close to the wall is not linear. it is proportional to y+^2
I should have said that I mean resolved sheaths of RANS turbulence models.

Quote:
 Originally Posted by LuckyTran (Post 402520) Your logic is pretty much spot on. y+ ~ 1 is a general rule (a very conservative one). You really only need y+<5 to apply the laws in the laminar sublayer. The additional restrictions come from calculation of gradients near the wall (where you now need a few points in the linear region). So if you wanted say 2-3 points in the linear region to resolve the gradients there then you would need correspondingly y+ ~ 1.5 - 2.5. If you are not worried about near wall gradients then you can use y+ ~ 5 and be okay.
What are these gradients needed for?

 FMDenaro January 18, 2013 06:45

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 no-slip condition. Apart the wall coefficient, the statistics in the inner part of the flow were acceptable

 RodriguezFatz January 18, 2013 06:56

Quote:
 Originally Posted by FMDenaro (Post 402619) 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 no-slip condition. Apart the wall coefficient, the statistics in the inner part of the flow were acceptable
Ok, I got you. Now I am aware of getting higher accuracy for higher number of gridpoints. My initial question was more like if I am doing complete nonsense with y+ = 3, such as using a log-wall function in the buffer layer that actually rapes the equations.

 FMDenaro January 18, 2013 07:02

Quote:
 Originally Posted by RodriguezFatz (Post 402622) Ok, I got you. Now I am aware of getting higher accuracy for higher number of gridpoints. My initial question was more like if I am doing complete nonsense with y+ = 3, such as using a log-wall function in the buffer layer that actually rapes the equations.
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

 LuckyTran January 18, 2013 08:54

Quote:
 Originally Posted by RodriguezFatz (Post 402617) I should have said that I mean resolved sheaths of RANS turbulence models. What are these gradients needed for?
As you said, the velocity gradients need to be resolved in order to get the proper shear stress. The shear stress is important for both the laminar (molecular diffusion) component of the shear stress. If you are doing RANS (anything that uses Boussinesq hypothesis) then the same shear stress is needed to calculate the turbulent diffusion (for example, k equation in k-epsilon or k-omega).

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"

 RodriguezFatz January 18, 2013 08:57

Quote:
 Originally Posted by LuckyTran (Post 402651) To properly calculate the gradient at solid boundaries, you need sufficient data points there so that the discretization error is not large for gradients (in terms of finite difference approximations to derivatives).
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!

 LuckyTran January 18, 2013 09:14

Quote:
 Originally Posted by RodriguezFatz (Post 402652) 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!
Oh now I see your point. If you know the y+ then you know what the gradient is. But the problem is you don't know the y+.

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.

 FMDenaro January 18, 2013 09:16

Quote:
 Originally Posted by RodriguezFatz (Post 402652) 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!

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...

 RodriguezFatz January 18, 2013 09:30

Quote:
 Originally Posted by FMDenaro (Post 402659) 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...
Yes, that is what I ment.

 RodriguezFatz January 18, 2013 09:40

Quote:
 Originally Posted by LuckyTran (Post 402658) Oh now I see your point. If you know the y+ then you know what the gradient is. But the problem is you don't know the y+. 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.
It seems that we talk at cross-purposes. Maybe it's due to my language problems. :o
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.

 LuckyTran January 18, 2013 09:43

Quote:
 Originally Posted by RodriguezFatz (Post 402662) 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.
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).

 RodriguezFatz January 18, 2013 09:45

I think since we use an iterative solver, we take the gradient of the last iteration and keep doing until convergence.

 flotus1 January 18, 2013 10:22

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.

 RodriguezFatz January 18, 2013 10:33

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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