question about k-epsilon

lucamirtanini · March 28, 2022, 09:25

Looking here for the theory of the k-epsilon model I have a question:
If k is the mean turbulent kinetic energy, so not dependent on the time, how it is possible to do the derivative in time?
Thanks

FMDenaro · March 28, 2022, 10:57

Quote:

Originally Posted by lucamirtanini

Looking here for the theory of the k-epsilon model I have a question:
If k is the mean turbulent kinetic energy, so not dependent on the time, how it is possible to do the derivative in time?
Thanks

Because in URANS you do not assume a statistical time averaging …
This topic is quite debated also in other forum

LuckyTran · March 28, 2022, 11:03

A triple decomposition allows you to treat temporal variation of statistical variables easily. When you do this you can recognize a coherent part and random part. Most people don't distinguish the time-averaged and unsteady parts of the coherent term because they don't formally use a triple decomposition, but there is also no need to. You can get away with a bit of notational hand-waving and just allow k to be unsteady and it will still be a proper Reynolds-average. Temporal dependence of k is not the issue, you just need to recalibrate your thinking a little bit.

The debate mentioned is what is happening in URANS when you apply an otherwise time-averaged model and try to use it. This topic is indeed contested.

lucamirtanini · March 29, 2022, 04:24

Quote:

Originally Posted by FMDenaro

Because in URANS you do not assume a statistical time averaging …
This topic is quite debated also in other forum

I am reading several textbooks but this passage is not so explained. Which forum? Can you give me the link? I searched but I did not find anything.

lucamirtanini · March 29, 2022, 04:27

Quote:

Originally Posted by LuckyTran

A triple decomposition allows you to treat temporal variation of statistical variables easily. When you do this you can recognize a coherent part and random part. Most people don't distinguish the time-averaged and unsteady parts of the coherent term because they don't formally use a triple decomposition, but there is also no need to. You can get away with a bit of notational hand-waving and just allow k to be unsteady and it will still be a proper Reynolds-average. Temporal dependence of k is not the issue, you just need to recalibrate your thinking a little bit.

The debate mentioned is what is happening in URANS when you apply an otherwise time-averaged model and try to use it. This topic is indeed contested.

Ok. But if I do the triple decomposition which term is used for the turbulent kinetic energy? Just the fluctuating one? If yes, I cannot imagine how it can be time dependent.

FMDenaro · March 29, 2022, 04:54

Quote:

Originally Posted by lucamirtanini

I am reading several textbooks but this passage is not so explained. Which forum? Can you give me the link? I searched but I did not find anything.

For example:

https://www.researchgate.net/post/UR...ompared-to-LES

https://www.linkedin.com/posts/bhara...medium=ios_app

LuckyTran · March 29, 2022, 05:16

The kinetic energy is $\frac{u^2}{2}=\frac{(\overline{u}+u')^2}{2}=\frac{(\overline{u}^2+2 \overline{u} u'+{u'}^2)}{2}$

$\frac{{u'}^2}{2}$ is k in the instantaneous sense. Now take the time-derivative. Nothing says yet that the derivative of k must be 0.

You don't have to go to triple decomposition, just the normal double decomposition is enough. Triple is just for explaining physically what a time-dependent k means.

lucamirtanini · March 29, 2022, 06:07

Quote:

Originally Posted by LuckyTran

The kinetic energy is $\frac{u^2}{2}=\frac{(\overline{u}+u')^2}{2}=\frac{(\overline{u}^2+2 \overline{u} u'+{u'}^2)}{2}$

$\frac{{u'}^2}{2}$ is k in the instantaneous sense. Now take the time-derivative. Nothing says yet that the derivative of k must be 0.

You don't have to go to triple decomposition, just the normal double decomposition is enough. Triple is just for explaining physically what a time-dependent k means.

Ok this what I understood when I heard the explanation in my classroom some years ago. But now I am reading a book where
k= $\frac{{u'}^2}{2}$
have angle bracket which I believe means time averaging and this is what I found also in the link in the OP.

FMDenaro · March 29, 2022, 06:26

Quote:

Originally Posted by lucamirtanini

Ok this what I understood when I heard the explanation in my classroom some years ago. But now I am reading a book where
k= $\frac{{u'}^2}{2}$
have angle bracket which I believe means time averaging and this is what I found also in the link in the OP.

I suppose that <*> stands for the ensemble averaging, the result being an unsteady function. But the devil is in …

lucamirtanini · March 29, 2022, 06:31

Ok Thanks! Now it makes sense. Sometimes in my opinion it is better to specify this details

. Is the ensamble averaging necessary ?

LuckyTran · March 29, 2022, 10:48

Any Reynolds-average will result in the exact same governing equations (i.e URANS) regardless of average you intend or want it to be in your dreams Nothing tells the solver that a time-average or ensemble average was used in this transport equation... And it's not limited to just a time-average or ensemble average either. Since all Reynolds averages will lead to the same outcome, the meaning of rho,u, and phi is therefore open to interpretation at the theoretical level and what spatio-temporal characteristics actually existing in your numerical solution depends on all the details of the numerical methods and underlying closure models used. You can write whatever you want on paper in any document of you think rho, u, and phi are; but solvers work backwards (they solve linear systems to give you rho, u, and phi), they don't go forward (they don't use your interpretation of rho, u, and phi to come up with a transport equation). You can write down time-averages and ensemble-averages and cowabunga-averages all day, numerical solvers don't work that way. If you want to filter DNS data or experimental measurements to recover the appropriate Reynolds-averaged quantity then it matters what averaging you use, but for numerical URANS solvers it does not.

FMDenaro · March 29, 2022, 14:42

Quote:

Originally Posted by LuckyTran

Any Reynolds-average will result in the exact same governing equations (i.e URANS) regardless of average you intend or want it to be in your dreams Nothing tells the solver that a time-average or ensemble average was used in this transport equation... And it's not limited to just a time-average or ensemble average either. Since all Reynolds averages will lead to the same outcome, the meaning of rho,u, and phi is therefore open to interpretation at the theoretical level and what spatio-temporal characteristics actually existing in your numerical solution depends on all the details of the numerical methods and underlying closure models used. You can write whatever you want on paper in any document of you think rho, u, and phi are; but solvers work backwards (they solve linear systems to give you rho, u, and phi), they don't go forward (they don't use your interpretation of rho, u, and phi to come up with a transport equation). You can write down time-averages and ensemble-averages and cowabunga-averages all day, numerical solvers don't work that way. If you want to filter DNS data or experimental measurements to recover the appropriate Reynolds-averaged quantity then it matters what averaging you use, but for numerical URANS solvers it does not.

Yes! And that is one of the issues I highlighted previously!
Actually, the type of mean should be implied by a proper turbulence model, that is a model that acts differently for time or ensemble averaging. But that is one of the lack in URANS formulations.

March 28, 2022, 09:25	question about k-epsilon	#1
lucamirtanini Senior Member luca mirtanini Join Date: Apr 2018 Posts: 165 Rep Power: 8	Looking here for the theory of the k-epsilon model I have a question: If k is the mean turbulent kinetic energy, so not dependent on the time, how it is possible to do the derivative in time? Thanks

March 29, 2022, 06:31		#10
lucamirtanini Senior Member luca mirtanini Join Date: Apr 2018 Posts: 165 Rep Power: 8	Ok Thanks! Now it makes sense. Sometimes in my opinion it is better to specify this details. Is the ensamble averaging necessary ?

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March 28, 2022, 11:03		#3
LuckyTran Senior Member Lucky Join Date: Apr 2011 Location: Orlando, FL USA Posts: 5,761 Rep Power: 66	A triple decomposition allows you to treat temporal variation of statistical variables easily. When you do this you can recognize a coherent part and random part. Most people don't distinguish the time-averaged and unsteady parts of the coherent term because they don't formally use a triple decomposition, but there is also no need to. You can get away with a bit of notational hand-waving and just allow k to be unsteady and it will still be a proper Reynolds-average. Temporal dependence of k is not the issue, you just need to recalibrate your thinking a little bit. The debate mentioned is what is happening in URANS when you apply an otherwise time-averaged model and try to use it. This topic is indeed contested.

March 29, 2022, 05:16		#7
LuckyTran Senior Member Lucky Join Date: Apr 2011 Location: Orlando, FL USA Posts: 5,761 Rep Power: 66	The kinetic energy is $\frac{u^2}{2}=\frac{(\overline{u}+u')^2}{2}=\frac{(\overline{u}^2+2 \overline{u} u'+{u'}^2)}{2}$ $\frac{{u'}^2}{2}$ is k in the instantaneous sense. Now take the time-derivative. Nothing says yet that the derivative of k must be 0. You don't have to go to triple decomposition, just the normal double decomposition is enough. Triple is just for explaining physically what a time-dependent k means.

March 29, 2022, 10:48		#11
LuckyTran Senior Member Lucky Join Date: Apr 2011 Location: Orlando, FL USA Posts: 5,761 Rep Power: 66	Any Reynolds-average will result in the exact same governing equations (i.e URANS) regardless of average you intend or want it to be in your dreams Nothing tells the solver that a time-average or ensemble average was used in this transport equation... And it's not limited to just a time-average or ensemble average either. Since all Reynolds averages will lead to the same outcome, the meaning of rho,u, and phi is therefore open to interpretation at the theoretical level and what spatio-temporal characteristics actually existing in your numerical solution depends on all the details of the numerical methods and underlying closure models used. You can write whatever you want on paper in any document of you think rho, u, and phi are; but solvers work backwards (they solve linear systems to give you rho, u, and phi), they don't go forward (they don't use your interpretation of rho, u, and phi to come up with a transport equation). You can write down time-averages and ensemble-averages and cowabunga-averages all day, numerical solvers don't work that way. If you want to filter DNS data or experimental measurements to recover the appropriate Reynolds-averaged quantity then it matters what averaging you use, but for numerical URANS solvers it does not.