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June 14, 2020, 23:58 
Grid spacing for turbulent flow

#1 
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I have been reading papers on simulating turbulent flow but they are all vague with regards to grid sizing. I keep seeing 4*pi*delta x 2*delta or 2*pi*h x 4*h. Do I just take these algebras (e.g. delta/h) as 1? None of them assign a value for these algebras and thus, I am not sure what should I used.


June 15, 2020, 06:55 
Delta

#2 
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The itself has a definition. Usually, it is used to define nondimensional grid spacing similar to . In this definition, cell size in a particular direction is used instead of height of the first cell from the wall. A good practice is to maintain about 40 nondimensional spacings in flow direction and half of those in transverse direction.
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June 16, 2020, 11:15 

#3 
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Thanks for the reply but I am still unsure. Can I just use delta=1?\
Or (for brevity, I will write \Delta=D and \delta=d) I was given Dx+=20, Dy+=5, Dz+=10, y+=0.5 and computation domain in 2d units as (pi) x (1) x (0.5*pi) with grid points 68 x 68 x 128. From Pope(2000), delta equals to half height of rectangle channel. Thus, d = 0.5*(2d)*(0.5*pi)? This doesn't makes sense. And why is the domain written in terms of channel's halfwidth? E.g. http://cfd.mace.manchester.ac.uk/cgi...as32_rsol.html Or From this site (https://www.researchgate.net/post/Wh...uirment_of_LES), Half height of channel along x, Lx=0.5*pi*d; Dx+ = dx/Lx = dx/(0.5*pi*d) = 20; dx = 40*pi*d; dx*Nx = (20*pi*d)*68 = 2*pi*d; This doesn't make sense and I think I still don't fully get the dimensionless wall units too. I think my confusion is mainly the notation. How do I interpret this delta? Or does it not matter what delta is as long as my dimensionless value are consistent with what I used for delta? 

June 16, 2020, 11:31 
Notation

#4 
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Forget about the notation. All you need to do is calculate , just like . And then you need to ensure that this value is equal to 2040 when size of the cell is used in the definition.
In terms of channel height, H, and and being number of cells in the streamwise and transverse direction, you need to ensure that grid sizes and follow the relations Across the height, maintain about 120140 cells.
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June 16, 2020, 12:18 

#5 
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However, what physical value should I assign to H in terms of metre?
delta = H / 2; x+ = x / delta; dx+ = dx / delta = x+ / Nx; dx = x / Nx; Subbing in your equation, dx = 16*H*Nx*Nz; Nx and Nz are given while x is given in terms of delta(=0.5*H). Thus, my dx is given in terms of delta or H which means I am still stuck trying to determine value of H or delta? I do have the number of cells given for H but but what value should I give it in terms of metre length? Or does it not matter what physical value I assign to it as long as the above relation holds? Additionally, may I ask where does 16 & 6 come from? 

June 16, 2020, 12:27 
Channel height

#6 
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Channel height is user's choice but is actually the defining variable in Reynolds number. For a fixed Re and fixed fluid, H gets fixed since that is used to define Re.
16 and 6 are not hard and fast values, rather based on best practices developed over the time as well as based on the assumption that the system is periodic. It also depends upon what energy scale do you want to resolve in LES.
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June 16, 2020, 13:27 

#7 
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Thanks so much! I now know how to set up my domain accordingly


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grid, spacing, turbulence 
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