LES output

MST · July 13, 2014, 15:13

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.

ghorrocks · July 13, 2014, 20:02

1 - No.
2 - Yes.

MST · July 14, 2014, 02:55

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.

ghorrocks · July 14, 2014, 03:17

You are correct. So I will correct my previous answer to:

1 - Yes.
2 - Yes.

MST · July 16, 2014, 21:37

Glenn, thank you so much for your help.

palmerlee · October 2, 2014, 06:13

Quote:

Originally Posted by MST

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.

ghorrocks · October 5, 2014, 08:27

Yes, that is one of the fundamental assumptions of Reynolds Averaging.

touha · April 3, 2015, 06:54

I would like to ask how to obtain an instantanous velocity with LES in CFX

July 13, 2014, 15:13	LES output	#1
MST New Member Join Date: Dec 2010 Posts: 6 Rep Power: 15	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.

April 3, 2015, 06:54	Velocity in les	#8
touha New Member fathia Join Date: Apr 2015 Posts: 1 Rep Power: 0	I would like to ask how to obtain an instantanous velocity with LES in CFX

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July 13, 2014, 20:02		#2
ghorrocks Super Moderator Glenn Horrocks Join Date: Mar 2009 Location: Sydney, Australia Posts: 17,852 Rep Power: 144	1 - No. 2 - Yes.

July 14, 2014, 02:55		#3
MST New Member Join Date: Dec 2010 Posts: 6 Rep Power: 15	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.

July 14, 2014, 03:17		#4
ghorrocks Super Moderator Glenn Horrocks Join Date: Mar 2009 Location: Sydney, Australia Posts: 17,852 Rep Power: 144	You are correct. So I will correct my previous answer to: 1 - Yes. 2 - Yes.

July 16, 2014, 21:37		#5
MST New Member Join Date: Dec 2010 Posts: 6 Rep Power: 15	Glenn, thank you so much for your help.

October 5, 2014, 08:27		#7
ghorrocks Super Moderator Glenn Horrocks Join Date: Mar 2009 Location: Sydney, Australia Posts: 17,852 Rep Power: 144	Yes, that is one of the fundamental assumptions of Reynolds Averaging.