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June 17, 2011, 12:15 
Comparing measured values with LES results

#1 
Member
N. A.
Join Date: May 2010
Posts: 64
Rep Power: 7 
Dear LES Foamers,
I have a question regarding compainrg LES results with that of measured data, for ex. Temperature measurements. Usually measured data for T is time average data, where T_avg_measured = Integral (T,dt)/Integral(dt). (Eq. 1) where the integral over time ranges from 0 to t. The T is the instantaneous value and dt depends upon sampling frequency. In OpenFOAM, we can solve Mean values and extract prime squared values as: T { mean on; prime2Mean on; base time; } (Eq. 2) where T is the resolved part and the subgrid scale is modeled. The instataneous value will be actually: T_instantaneous = T + T', (Eq. 3) where T is the resolved part and for which we are solving the LES. What I understand is that if the mean value calculation of LES solution in openFOAM is the timeaveraged value of T (resolved part). T_mean_resolved = Integral (T, dt)/Integral(dt), (Eq. 4) where dt ranges from lets say 0 to t and T is the resolved part. But an apple to apple comparison would be Mean value calculated from the instantaneous value and not the resolved value. Such as: T_mean_instant = Integral (T_instantaneous, dt)/Integral(dt) (Eq. 5) What I understand is that the mean value calcualted from OpenFOAM is using Eq. 4 and its not very appropraite to comapre with Eq 1 above with measured data. I would like to comapre mean value as per Eq. 5 to comapre with Eq. 1 data, but openFOAM does not calculate mean value of instantaneous values. Or does it? I would like to open up a discussion on this point and would like to know inputs on how to comapred experimental measured time averaged values with LES results. Thanks, Nir 

June 22, 2011, 04:23 

#2  
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Gregor Olenik
Join Date: Jun 2009
Location: http://greole.github.io/
Posts: 78
Rep Power: 8 
Hi,
I have the feeling that you are confused by the difference between filtering and averaging. Quote:
Your Eq. 3 should look like T (t) = T_filtered(t) + T_subgrid(t). (Eq1) Using the time average (<>) now will leed to <T> = <T_filtered(t)> + <T_subgrid(t)>. Where <T_subgrid(t)> ≈ 0 (Eq.2) (because to subgrid scales are assumed to be homogeneous and isotropic). Therefore <T> ≈ <T_filtered(t)> and a comparisson between time averaged experimental and time averaged and filtered LES values can be made. gregor btw.: you can proof my eq.2 by inserting reynolds assumption into my eq.1 and doing a time average <T> = <T_filtered> + <T_filtered(t)'> + <T_subgrid> + <T_subgrid(t)'> where everything except <T_filtered> is zero. ( <T_subgrid> = 0 because of the isotropic and homogeneous nature of subgrid scale fluctuations) Obviously these assumptions don't hold if your filter size is too large and your subgrid scales contain non isotropic turbulence. 

June 22, 2011, 10:21 

#3 
Member
N. A.
Join Date: May 2010
Posts: 64
Rep Power: 7 
Greg,
Thanks for your response and initiating discussion. My understanding is that for LES by definition <T_subgrid(t)> is not equal to 0 (or to be precise its not necessary that it will be zero always). Hence I am not quite convinced that <T_filtered(t)>=<T_instantaneous_experiments(t)> Can you proove it or give a reference where it is proved? Thanks, Nir 

June 27, 2011, 07:05 

#4  
Member
Gregor Olenik
Join Date: Jun 2009
Location: http://greole.github.io/
Posts: 78
Rep Power: 8 
Hi,
Quote:
http://dx.doi.org/10.1007/s1049400791289 and there are is a chapter in a german text book by J.Fröhlich "Large Eddy Simulation turbulenter Strömungen" (chapter 5.9.1) And of course reading the Pope might (or might not) enlighten you The point is, the smaller the scales are, the more homogeneous and isotropic they become. Therefore an average over smallest scales of the first moments can often (not always) be neglected (look at fig 13.2 p.564 in Pope's book, this might give you an idea). gregor 

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