Rationale for having filter for turbulence flow

cfdnewb123 · March 21, 2020, 21:15

"In LES, a low-pass filtering operation is performed so that the resulting filtered velocity field U can be adequately resolved on a relatively coarse grid."

From the above statement, can I interpret that the reason for having a low pass filter especially in turbulent flow is to eliminate oscillations when coarse grid is used? These oscillations are actually eddies smaller than the grid spacing but because coarse grids are used, they come off as oscillations. Thus, am I right to say that the low pass filter used for LES is to have a non-oscillatory flow that can still generate eddies (size larger than grid spacing) for coarser grids?

FMDenaro · March 22, 2020, 05:32

Quote:

Originally Posted by cfdnewb123

"In LES, a low-pass filtering operation is performed so that the resulting filtered velocity field U can be adequately resolved on a relatively coarse grid."

From the above statement, can I interpret that the reason for having a low pass filter especially in turbulent flow is to eliminate oscillations when coarse grid is used? These oscillations are actually eddies smaller than the grid spacing but because coarse grids are used, they come off as oscillations. Thus, am I right to say that the low pass filter used for LES is to have a non-oscillatory flow that can still generate eddies (size larger than grid spacing) for coarser grids?

The idea is that the action of the filter makes "smooth" the variable and, consequently, be better representable on a grid of reasonable size. That makes sense only for turbulent flows. The term "non-oscillatory" is not correct, the filtered field has a wide range of resolve components producing an unsteady field with eddies from the integral to the cut-off lenght scale.

cfdnewb123 · March 28, 2020, 06:11

Thanks for the response. However, I do not have enough cfd experience to understand the significance of representing the effect of unaccounted eddies with turbulence models. Why can't DNS (I assumed is just the full Navier Stokes equation) just be used on a coarser grid? I imagine eddies with lengthscale smaller than grid size will not be capture by the solver and thus, won't this eliminate the small eddies naturally?

Unless we need LES for problems where the turbulent flow started initially from small eddies which need to be resolved? For instance, the Kelvin-Helmholtz instability will not occur if the grids are not fine enough...

FMDenaro · March 28, 2020, 06:23

Quote:

Originally Posted by cfdnewb123

Thanks for the response. However, I do not have enough cfd experience to understand the significance of representing the effect of unaccounted eddies with turbulence models. Why can't DNS (I assumed is just the full Navier Stokes equation) just be used on a coarser grid? I imagine eddies with lengthscale smaller than grid size will not be capture by the solver and thus, won't this eliminate the small eddies naturally?

Unless we need LES for problems where the turbulent flow started initially from small eddies which need to be resolved? For instance, the Kelvin-Helmholtz instability will not occur if the grids are not fine enough...

What you are addressing is the so-called LES no-model in which no sgs model is explicitly supplied but only the numerics acts on.
From a theoretical point of view, owing to the non-linear coupling, the components of the flow that are cutted away by the filtering still affect the resolved filtered components. That justifies the need of an SGS model

LuckyTran · March 29, 2020, 05:17

Just to clarify a little bit more. I would add that non-linear behavior is also the reason we need turbulence models in RANS. In RANS, you apply a temporal filter instead of a spatial one. You still need to account for the influence of the cutted away stuff. You don't need to explicitly know all the details of the things being cutoff, just their influence. The slightly more detailed explanation is:

Statistically, if I have two random variables

and

and let <> brackets denote my linear filtering operator. I can decompose my variable into two parts, $\overline{a}$ and

, the first which survives the filtering operator and the second which is eliminated by it. That is, I define:
$a = \overline{a}+a'$ under that condition that

and $<a>=\overline{a}$

Under operations like addition we are still linear:
$<a+b> = <a>+<b>=\overline{a}+\overline{b}$

Looks pretty promising... Until you try to filter other things like products of variables.
$<ab> = <(\overline{a}+a')(\overline{b}+b')>$ which turns out to be: $\overline{a}\overline{b}+<a'b'>$ or equivalently $\overline{a}\overline{b}+\overline{a'b'}$
That last thing is the stuff that is not cutoff by our filtering operation... Yikes!
Well, we don't need to entire details of

and

, that would require that we know

and

. We only ask that we know the influence of $\overline{a'b'}$ in causing

to deviate from $\overline{a}\overline{b}$
And that's why you need a SGS model in LES and a "turbulence model" in RANS. But in both cases, you are not modeling

and

, you are modeling only their influence $\overline{a'b'}$

And that's why coarse-grid DNS is not DNS but actually a no-SGS LES. In a coarse-grid DNS, you are effectively pretending that $\overline{a'b'}=0$ which you know is incorrect because $u'\neq0$ whenever something has been cutted off (the smaller eddies). True DNS, nothing is being cutted off by the filter and you recover

FMDenaro · March 29, 2020, 05:23

Quote:

Originally Posted by LuckyTran

Just to clarify a little bit more. I would add that non-linear behavior is also the reason we need turbulence models in RANS. In RANS, you apply a temporal filter instead of a spatial one. You still need to account for the influence of the cutted away stuff. You don't need to explicitly know all the details of the things being cutoff, just their influence. The slightly more detailed explanation is:

Statistically, if I have two random variables

and

and let <> brackets denote my linear filtering operator. I can decompose my variable into two parts,

and

, the first which survives the filtering operator and the second which is eliminated by it. That is, I define:
$a = \overline{a}+a'$ under that condition that

and $<a>=\overline{a}$

Under operations like addition we are still linear:
$<a+b> = <a>+<b>=\overline{a}+\overline{b}$

Looks pretty promising... Until you try to filter other things like products of variables.
$<ab> = <(\overline{a}+a')(\overline{b}+b')>$ which turns out to be: $\overline{a}\overline{b}+<a'b'>$ or equivalently $\overline{a}\overline{b}+\overline{a'b'}$
That last thing is the stuff that is not cutoff by our filtering operation... Yikes!
Well, we don't need to entire details of

and

, that would require that we know

and

. We only ask that we know the influence of $\overline{a'b'}$ in causing

to deviate from $\overline{a}\overline{b}$
And that's why you need a SGS model in LES and a "turbulence model" in RANS. But in both cases, you are not modeling

and

, you are modeling only their influence $\overline{a'b'}$

Correct, just a note on the LES decomposition where, differently from RANS, the mixed term (a' b_filt)_filt does not vanish

March 21, 2020, 21:15	Rationale for having filter for turbulence flow	#1
cfdnewb123 Member Join Date: Feb 2019 Posts: 69 Rep Power: 7	"In LES, a low-pass filtering operation is performed so that the resulting filtered velocity field U can be adequately resolved on a relatively coarse grid." From the above statement, can I interpret that the reason for having a low pass filter especially in turbulent flow is to eliminate oscillations when coarse grid is used? These oscillations are actually eddies smaller than the grid spacing but because coarse grids are used, they come off as oscillations. Thus, am I right to say that the low pass filter used for LES is to have a non-oscillatory flow that can still generate eddies (size larger than grid spacing) for coarser grids?

March 29, 2020, 05:17		#5
LuckyTran Senior Member Lucky Join Date: Apr 2011 Location: Orlando, FL USA Posts: 5,754 Rep Power: 66	Just to clarify a little bit more. I would add that non-linear behavior is also the reason we need turbulence models in RANS. In RANS, you apply a temporal filter instead of a spatial one. You still need to account for the influence of the cutted away stuff. You don't need to explicitly know all the details of the things being cutoff, just their influence. The slightly more detailed explanation is: Statistically, if I have two random variables and and let <> brackets denote my linear filtering operator. I can decompose my variable into two parts, $\overline{a}$ and , the first which survives the filtering operator and the second which is eliminated by it. That is, I define: $a = \overline{a}+a'$ under that condition that and $<a>=\overline{a}$ Under operations like addition we are still linear: $<a+b> = <a>+<b>=\overline{a}+\overline{b}$ Looks pretty promising... Until you try to filter other things like products of variables. $<ab> = <(\overline{a}+a')(\overline{b}+b')>$ which turns out to be: $\overline{a}\overline{b}+<a'b'>$ or equivalently $\overline{a}\overline{b}+\overline{a'b'}$ That last thing is the stuff that is not cutoff by our filtering operation... Yikes! Well, we don't need to entire details of and , that would require that we know and . We only ask that we know the influence of $\overline{a'b'}$ in causing to deviate from $\overline{a}\overline{b}$ And that's why you need a SGS model in LES and a "turbulence model" in RANS. But in both cases, you are not modeling and , you are modeling only their influence $\overline{a'b'}$ And that's why coarse-grid DNS is not DNS but actually a no-SGS LES. In a coarse-grid DNS, you are effectively pretending that $\overline{a'b'}=0$ which you know is incorrect because $u'\neq0$ whenever something has been cutted off (the smaller eddies). True DNS, nothing is being cutted off by the filter and you recover sbaffini likes this.

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March 28, 2020, 06:11		#3
cfdnewb123 Member Join Date: Feb 2019 Posts: 69 Rep Power: 7	Thanks for the response. However, I do not have enough cfd experience to understand the significance of representing the effect of unaccounted eddies with turbulence models. Why can't DNS (I assumed is just the full Navier Stokes equation) just be used on a coarser grid? I imagine eddies with lengthscale smaller than grid size will not be capture by the solver and thus, won't this eliminate the small eddies naturally? Unless we need LES for problems where the turbulent flow started initially from small eddies which need to be resolved? For instance, the Kelvin-Helmholtz instability will not occur if the grids are not fine enough...