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 garrison April 7, 2012 03:34

turbulence flat plane boundary

1 Attachment(s)
Hi Foamers!
Recently I am doing a very simple case using S-A-DDES.
The computational domain is 12m long, of which the left side is inlet, the right side is outlet. Inlet velocity is 1m/s, I choose water instead of air so the Reynold number is 1.2e7.
I intend to get a classic turbulent boundary layer velocity profile like this one
http://www.csa.ru/%7Eignatiev/IMAGES/btur_comp.gif

but I failed, I got one like this (pls see my attachment)
It seems I can capture the laminar zone of the BL(Yplus from 0-10), but fail to simulate the log zone, and also, the thickness of the BL should be like 0.17m (from the turbulent BL theory),

but the one I get is far less than 0.17m, it's about 0.01m. You can find in my attachment that the velocity in BL grows to the freestream velocity at about yPlus =80. but theoretically, yPlus=80 should be the log zone, right?

So, I don't know what's wrong with it? My mesh is like this: first point yPlus is below 1, with about 50 points in the BL (the thickness I get from the above formula, 0.17m).
delta t =1e-2, the CFL number is less then 1.

Seems like my boundary layer has not fully developed to a turbulent one. but I got my Reynold number base on the length of the plane (Rex) 1.2e7 which I think is large enough. And I run the case for about 500 seconds, with the inlet velocity 1m/s, the plane length 12m, it should flow through the domain for like 40 times.

Can anyone give me any hint?

Garry

 anishtain4 April 10, 2012 02:01

Dear Garrison

first of all you should scale your y axis too, it's u+ not u/uref, as you see your sublayer shape is not the same either,
second; tell us a little about your solver, your model and if any wall function you are using, and boundary conditions
PS: you mentioned fully developed, there isn't such a thing on external flows

 garrison April 10, 2012 03:09

1 Attachment(s)
Thank you for you reply Anishtain!

Actually I did rescale the y axis,
u+=ux/u*
u*=sqrt(tau/density)
tau=du/dy*mu
du/dy=( u[i+1] - u[i] ) / ( y[i+1] - y[i] )

Am I right?

But I didn't change the x scale to a log one. The new attachment is of a log x axis.

here's my BC:
inlet:
U = 1m/s
nuTilda = 0

outlet:
p = 0

top:

bottomwall:
U = 0
nuTilda = 0

I am confused about the inlet and the outlet conditions. At the beginning of the plate, the BL shout be a laminar one and develop to a turbulent BL with the growth of x (at about Rex= 1e6 ) (please correct me if I'm wrong), according to this I should set the inlet nuTilda to zero.
am I right?

I tried to set the nuTilda to a larger value (like 1e-3) , then the thickness of the BL did grow a lot, but the velocity profile just keep the same shape, when changing the different nuTilda, I even got negtive Ux :confused:

Or should I set the inlet U with a turbulent shape?

And I use pisoFoam, LES(DES) - SADDES turbulent model, without any wall function, the yPlus is less than 1. The LES/DES method should be used to a 3D mesh, but I just have 1 grid in spanwise direction with cyclic BC (not empt), can that be the reason of my problem?

If anyone could tell me how to set the BCs for nuTilda or U or P, it will really help me lot!!

PS. in the attachment below, red points represents x=y ( sublayer ? ), and black points is the u profile I got.

Best Regards

 anishtain4 April 10, 2012 16:05

when you set the inlet boundary condition of turbulent to zero it means you have laminar inlet, which in your case is probable to lead to a laminar boundary layer till the end, since you are not adding any perturbation yourself. there even exist some experiments that flow has not been turbulent for orders larger than 1e5.
There are some methods to determine the k or epsilon in RANS models, but about LES read some papers, I'm not aware of it.
by LES modeling you should have enough grid to capture the kolmogrov scales, read some introductory notes on turbulence about this, 2d turbulent does not exist

 garrison April 10, 2012 21:50

Thank you Anishtain,
Then next I'll try with some finer grids and more grids in spanwise to make it a real 3D case.
Best Regards

 anishtain4 April 11, 2012 08:58

One more hint, when you are using LES you can't see the log-law profile or sublayer as it is unsteady and turbulence, these profiles are averaged ones which would be captured with RANS models when there is no wall-function used. If you want to capture it check them

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