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Old   July 13, 2014, 14:13
Default LES output
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MST
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Hi Everyone,

I have some fundamental question about LES techniques and its output variables in CFX.
the velocity in LES is decomposed into filtered velocity (resolved part) and modeled (unresolved part).
U=[u]+u'', say [u] is the resolved velocity and u'' is unresolved component.
my questions are:
1- Is it correct to say that the resolved part [u] is the instantaneous velocity of resolved part?
2- The CFX outputs velocity and velocity.trnave. is the velocity represents the instantaneous velocity [u] and velocity.trnave represents the mean flow of the of the resolved part?

thank you so much in advance.
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Old   July 13, 2014, 19:02
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1 - No.
2 - Yes.
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Old   July 14, 2014, 01:55
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MST
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Glenn,
Thank you so much. I may did not cast my question properly. For me, if 2 is correct, 1 is correct as well.
When we say instantaneous velocity U, that means U=mean velocity + fluctuating, this is clear for me when we talk about RANS, however, in LES, U= filtered velocity (resolved)+ modeled (unresolved)
My question is:
Does the filtered velocity in LES contain both (mean velocity + fluctuating) of resolved field? i.e. instantaneous velocity of the resolved field.
Similarly for the modeled field.
If this is correct, then, in LES, U= filtered velocity+ modeled= (mean velocity + fluctuating) of resolved + (mean velocity + fluctuating) of unresolved.
So, the full velocity of the mean flow= (mean velocity) of resolved + (mean velocity) of unresolved.
Rms velocity= rms velocity of resolved + rms of unresolved.
Thus, the TKE=TKE resolved + TKE unresolved.
TKE resolved is calculated using rms velocity of resolved flied.
TKE unresolved is calculated using rms velocity of unresolved flied.
Is it correct explanation? If not, could you please comment on them?
I really appreciate you help.
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Old   July 14, 2014, 02:17
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Glenn Horrocks
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You are correct. So I will correct my previous answer to:

1 - Yes.
2 - Yes.
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Old   July 16, 2014, 20:37
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MST
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Glenn, thank you so much for your help.
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Old   October 2, 2014, 05:13
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Peter
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Quote:
Originally Posted by MST View Post
Glenn,
Thank you so much. I may did not cast my question properly. For me, if 2 is correct, 1 is correct as well.
When we say instantaneous velocity U, that means U=mean velocity + fluctuating, this is clear for me when we talk about RANS, however, in LES, U= filtered velocity (resolved)+ modeled (unresolved)
My question is:
Does the filtered velocity in LES contain both (mean velocity + fluctuating) of resolved field? i.e. instantaneous velocity of the resolved field.
Similarly for the modeled field.
If this is correct, then, in LES, U= filtered velocity+ modeled= (mean velocity + fluctuating) of resolved + (mean velocity + fluctuating) of unresolved.
So, the full velocity of the mean flow= (mean velocity) of resolved + (mean velocity) of unresolved.
Rms velocity= rms velocity of resolved + rms of unresolved.
Thus, the TKE=TKE resolved + TKE unresolved.
TKE resolved is calculated using rms velocity of resolved flied.
TKE unresolved is calculated using rms velocity of unresolved flied.
Is it correct explanation? If not, could you please comment on them?
I really appreciate you help.
I wonder if "(mean velocity) of unresolved" is equal to zero.
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Old   October 5, 2014, 07:27
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Yes, that is one of the fundamental assumptions of Reynolds Averaging.
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Old   April 3, 2015, 05:54
Smile Velocity in les
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I would like to ask how to obtain an instantanous velocity with LES in CFX
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