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 August 13, 2013, 08:31 Re=5 Mio. Still steady? #1 Member   Shenren Xu Join Date: Jan 2011 Location: London, U.K. Posts: 64 Rep Power: 9 In this validation example provided by NASA, that runs SA turbulence model RANS on zero pressure gradient flat plate, http://turbmodels.larc.nasa.gov/flatplate_sa.html how come the RANS solver can still converge to steady state solution when Re is around 5 million? If even (much) finer mesh is used, i.e., DNS, I suppose one shall pick up the unsteadiness. The question is, how fine the mesh needs to be in order to NOT converge to steady state? Is it possible to predict this critical mesh density a priori? Cheers, Shenren

 August 13, 2013, 10:20 #2 Senior Member   Filippo Maria Denaro Join Date: Jul 2010 Posts: 3,752 Rep Power: 41 From your question I suppose you are not familiar with RANS/URANS/LES/DNS formulations... The RANS equation are obtained from the DNS ones by applying the Reynolds average, which implies a statistical steady state: (x) = (1/T) Int [t0, t0+T] f(x,t) dt for T going to infinite

August 13, 2013, 10:54
#3
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I think if you run your RANS code with very small time step and
very fine mesh, you are doing DNS. Isn't that the case?

Quote:
 Originally Posted by FMDenaro From your question I suppose you are not familiar with RANS/URANS/LES/DNS formulations... The RANS equation are obtained from the DNS ones by applying the Reynolds average, which implies a statistical steady state: (x) = (1/T) Int [t0, t0+T] f(x,t) dt for T going to infinite

August 13, 2013, 10:58
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Quote:
 Originally Posted by Shenren_CN I think if you run your RANS code with very small time step and very fine mesh, you are doing DNS. Isn't that the case?
No, never... first, in RANS you do not solve a time marching formulation but only for a steady state. Then, even if h->0 the RANS model will never give a vanishing contribution so to tend towards the DNS solution. This happens only for LES.

August 13, 2013, 11:18
#5
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1) Why can't I run RANS using a small time step to get an unsteady solution? If your time step is small enough, it shall presumably pick up fluctuation with very small temporal scale.

2) Why not? Could you please point me to any reference?

3) What equation is discretized in a DNS solver then?

Quote:
 Originally Posted by FMDenaro No, never... first, in RANS you do not solve a time marching formulation but only for a steady state. Then, even if h->0 the RANS model will never give a vanishing contribution so to tend towards the DNS solution. This happens only for LES.

August 13, 2013, 11:25
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Quote:
 Originally Posted by Shenren_CN 1) Why can't I run RANS using a small time step to get an unsteady solution? If your time step is small enough, it shall presumably pick up fluctuation with very small temporal scale. 2) Why not? Could you please point me to any reference? 3) What equation is discretized in a DNS solver then?

The RANS equations are steady by definition, no time-derivative exists, there is no time-step.
The URANS formulation is a special version of the RANS wherein the time derivative exists but the resolved time frequency is quite low compared to the turbulence ones (see for example the flow in a cylinder with a moving piston). In any case the URANS solution would not converge to the DNS one for small time step.
The DNS formulation requires to solve the original Navier-Stokes system of equations over all the characteristic scales of turbulence.

See the book of Pope for general topic on turbulence and the book of Ferziger and Peric for CFD related issues.

August 13, 2013, 11:35
#7
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1) Okay, so you call that URANS... Then, I alter my question to be: why does not URANS with REALLY REALLY REALLY small time step converge to DNS result? Shouldn't the 'resolved time frequency' determined by your time step (which is prescribed by the user), and not an intrinsic feature of this particular method URANS?

2) Again, how is the equation you are working with in DNS from the RANS equation, or URANS equation, except using a small time and length scale? I agree with you that it WANTS to resolve flow features of all scales, but how is it done?

3) Anyways, thanks for the patience and I really appreciate your reply. Cheers

Quote:
 Originally Posted by FMDenaro The RANS equations are steady by definition, no time-derivative exists, there is no time-step. The URANS formulation is a special version of the RANS wherein the time derivative exists but the resolved time frequency is quite low compared to the turbulence ones (see for example the flow in a cylinder with a moving piston). In any case the URANS solution would not converge to the DNS one for small time step. The DNS formulation requires to solve the original Navier-Stokes system of equations over all the characteristic scales of turbulence. See the book of Pope for general topic on turbulence and the book of Ferziger and Peric for CFD related issues.

 August 13, 2013, 11:41 #8 Senior Member   Filippo Maria Denaro Join Date: Jul 2010 Posts: 3,752 Rep Power: 41 in URANS you can also try using a very small time step but the variable you solve is always a statistical variable . That means that even for dt->0 the contribution of the turbulence model will not vanish in general . Only if your URANS turbulence modelling is built in such a way that vanishes for dt->0, h->0 you would tend towards DNS. DNS is very simple, use the Navier-Stokes equation without any average/filtering/modelling and solve over a computational grid having h of the order of the Kolmogorov scale.

 August 13, 2013, 11:44 #9 Senior Member   Filippo Maria Denaro Join Date: Jul 2010 Posts: 3,752 Rep Power: 41

August 13, 2013, 11:57
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What if I run my URANS solver (without turbulence modeling, as if running for laminar flow) over a grid having h of the order of the Kolmogorov scale? Does it produce your DNS result? (And at the same time, using the same time step you are using for your DNS solver, of course.)

Statistical or not, it's just how you look at the RANS equation. And if
the Reynolds stress is not taken into account, should that be exactly the same as your NS equation discretized for DNS solver?

Quote:
 Originally Posted by FMDenaro in URANS you can also try using a very small time step but the variable you solve is always a statistical variable . That means that even for dt->0 the contribution of the turbulence model will not vanish in general . Only if your URANS turbulence modelling is built in such a way that vanishes for dt->0, h->0 you would tend towards DNS. DNS is very simple, use the Navier-Stokes equation without any average/filtering/modelling and solve over a computational grid having h of the order of the Kolmogorov scale.

August 13, 2013, 12:04
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Quote:
 Originally Posted by Shenren_CN What if I run my URANS solver (without turbulence modeling, as if running for laminar flow) over a grid having h of the order of the Kolmogorov scale? Does it produce your DNS result? (And at the same time, using the same time step you are using for your DNS solver, of course.) Statistical or not, it's just how you look at the RANS equation. And if the Reynolds stress is not taken into account, should that be exactly the same as your NS equation discretized for DNS solver?

yes, if you eliminate from your URANS code all the turbulence models and use dt and h to resolve all scales, you will get a DNS solution.

August 13, 2013, 12:09
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I apologize for not having stating clearly that I meant running RANS/URANS without modeling the turbulence, otherwise I would have saved a lot of your time explaining to me. Thanks a lot for the help!

Cheers,
Shenren

Quote:
 Originally Posted by FMDenaro yes, if you eliminate from your URANS code all the turbulence models and use dt and h to resolve all scales, you will get a DNS solution.

August 13, 2013, 12:13
#13
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Having confirmed that RANS/URANS without modeling turbulence, if run with small dt, and fine mesh, is identical to DNS.

Then back to my original question, would the flow starts to get unsteady if finer mesh is used? If yes, then, would we ever get mesh convergence?

Quote:
 Originally Posted by Shenren_CN I apologize for not having stating clearly that I meant running RANS/URANS without modeling the turbulence, otherwise I would have saved a lot of your time explaining to me. Thanks a lot for the help! Cheers, Shenren

August 13, 2013, 12:14
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Quote:
 Originally Posted by Shenren_CN I apologize for not having stating clearly that I meant running RANS/URANS without modeling the turbulence, otherwise I would have saved a lot of your time explaining to me. Thanks a lot for the help! Cheers, Shenren
that's ok now, a advantage of using LES codes (with implicit filtering) is that the solution automatically tends towards the DNS for vanishing computational steps, without entering into code to cancel the call to subroutines for turbulence modelling.

August 13, 2013, 12:28
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Quote:
 Originally Posted by Shenren_CN Having confirmed that RANS/URANS without modeling turbulence, if run with small dt, and fine mesh, is identical to DNS. Then back to my original question, would the flow starts to get unsteady if finer mesh is used? If yes, then, would we ever get mesh convergence?
if you solve in DNS the flow over a flat plate you will see a small steady 2D laminar region, then the transition region wherein the flow develops its unsteady nature until a full developed turbulence region.
Of course the computational is very expensive...

see this recent paper http://journals.cambridge.org/action...ne&aid=5876700

August 13, 2013, 12:30
#16
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This is exactly what I have been looking for. Thanks!

Quote:
 Originally Posted by FMDenaro if you solve in DNS the flow over a flat plate you will see a small steady 2D laminar region, then the transition region wherein the flow develops its unsteady nature until a full developed turbulence region. Of course the computational is very expensive... see this recent paper http://journals.cambridge.org/action...ne&aid=5876700

August 14, 2013, 06:49
#17
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This is interesting to know. So the difference is that in RANS/URANS with turbulence modelling, it assumes fluctuation of all scales are modelled, while in LES, the code models only the part of fluctuation that's sub-grid scale, and it will automatically vanish when sub-grid scale tends towards zero?

Quote:
 Originally Posted by FMDenaro that's ok now, a advantage of using LES codes (with implicit filtering) is that the solution automatically tends towards the DNS for vanishing computational steps, without entering into code to cancel the call to subroutines for turbulence modelling.

August 14, 2013, 06:56
#18
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Quote:
 Originally Posted by Shenren_CN This is interesting to know. So the difference is that in RANS/URANS with turbulence modelling, it assumes fluctuation of all scales are modelled, while in LES, the code models only the part of fluctuation that's sub-grid scale, and it will automatically vanish when sub-grid scale tends towards zero?
that's correct, but only for LES with implicit filtering.
Conversely, using explicit filtering the filter width is taken fixed, independently from the computational step which can also tends to zero producing a grid independent LES solution, not a DNS one.

August 14, 2013, 09:16
#19
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I have very little knowledge about LES. This is a good introduction for me. I learnt something useful, thanks!

Quote:
 Originally Posted by FMDenaro that's correct, but only for LES with implicit filtering. Conversely, using explicit filtering the filter width is taken fixed, independently from the computational step which can also tends to zero producing a grid independent LES solution, not a DNS one.

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