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May 19, 2003, 20:25 
help on isotropic turbulence initial condition

#1 
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Hello All:
I am doing LES on decaying of isotropic turbulence. I get a very strange problem on how to set the initial condition. I am using Rogallo's code (1981). He defined two random components of the velocity field in the plane orthogonal to k. Q1 = sqrt( E(k)/4pi k^2 ).exp(i.theta1).cos(alpha) Q2 = sqrt( E(k)/4pi k^2 ).exp(i.theta1).sin(alpha) Two unit vectors are defined as e1 = ( k * omega )/ ( k*omega ) e2 = ( k * e1 ) / ( k*e1 ) then u = Q1.e1 + Q2.e2 it is supposed to get an isotropic, homogeneous velocity field. But I can't get it, the initial velocity field I get is not isotropic. The integral length scale L11, L22 and L33 are not equal. Also u_rms = v_rms ~ 0.6, but w_rms is very small ~ 1.e6. Can't understand why that happens ...... Anyone can give me some suggestions? thank you very much! 

May 20, 2003, 14:22 
Re: help on isotropic turbulence initial condition

#2 
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Do you mean Rogallo's method or did you are actually using the code written by Rogallo?


May 24, 2003, 02:37 
Re: help on isotropic turbulence initial condition

#3 
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Thank you! both actually. Made some modification of his original code but the method for the isotropical initial condition isnot changed I just found the reason. It becauses of the random function.
But I got another question. It's a free decaying case, u, v, w should have the same behaviour. But I find enrgy of u ( u^2 ) decease fast than v and w. So at the end of run, it becomes anisotropic ...... Anyone has idea why that happens? I run the same code with another initial condition which got from DNS result. It runs quite well. But we want to try some new initial condition ( pulse initial condition ) for our study. Thanks a lot! 

May 24, 2003, 04:42 
Re: help on isotropic turbulence initial condition

#4 
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Hi,
That is not so surprising. In general you will always find some anisotropy at the end of the run, but hopefully it is small. I suppose you have some 'seed' in your random number generator. Changing seed results in statistically independent initial velocity fields but with the same mean energy spectrum etc. If you do a couple of decay runs with different seeds, thus different initial velocity fields, and take the average of the runs then you should find approximately isotropic results.Thus isotropic results are only to be expected in the mean. The anisotropy is probably worse when the integral length scale is large in comparison with the box size. Tom 

May 24, 2003, 16:22 
Re: help on isotropic turbulence initial condition

#5 
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Thank you very much! Tom.
You solved my big problem! Yes, it is true. In my case, the integral length scale is large, about 1/3 of the box size. The reason is my energy peak is at first mode ( k = 1 ), while the integral scale is like uiuj(1)/q^2 in spectral space. Now I understand why other people set the energy peak at high mode like k > 8. I spent a lot of time on this problem because I only change the initial conditon why I can't get the right behavior. Another consideration is for k = 1, there are only 8 points fall in this shell, it's really hard to get a real random phases. But for k > 8, there are thousands points I think. I have an another question. If we want to watch the behavior of integral length sacles, I think I have 2 ways to do it now: 1. Generate several initial conditon by choosing different seeds, and average them to get a new isotropic initial condition. Then run with that initial condition once to get the result. 2. Run with diffferent initial conditions, but average the final results instead Which way is better? Thanks a lot! Your help will be highly appreciated! 

May 26, 2003, 07:57 
Re: help on isotropic turbulence initial condition

#6 
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Hi,
The second method is the best. The first method doesn't help you. You don't get a better initial field by averaging. But you can ofcourse average the integral length scales in the three directions. By the way, if you want to study the integral length scales it is perhaps a good idea to look at the the twopoint correlations. The twopoint correlation should approach zero for long correlation lengths. If it doesn't (in other words the box is too small) then you can't say much about the behaviour of the integral length scales. Tom 

June 7, 2003, 21:17 
Re: help on isotropic turbulence initial condition

#7 
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thank you! Tom! spent 2 weeks work on that well, still not quite successful ...


March 12, 2010, 19:03 

#8 
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cfdnewbie
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Hello dear colleagues,
I was just wondering whether this forum / topic is still active and if there's someone around who could help me out with Rogallo's procedure? Thank you all in advance! A CFDnewbie 

April 28, 2011, 16:55 
turbulence generator modeling in wind tunnel (orifice perforated plate OPP))

#9 
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mehdi farahnak
Join Date: Apr 2011
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hello every one.
i want to model a isotropic turbulent flow with openfoam. my case is a turbulence generator in a wind tunnel. i just know the velocity of air at entrance of the tunnel i want to model it with simplefoam. what boundary condition must i choose for P and U and ....? please help me 

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