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Old   March 28, 2022, 09:25
Default question about k-epsilon
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luca mirtanini
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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
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Old   March 28, 2022, 10:57
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Quote:
Originally Posted by lucamirtanini View Post
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
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Old   March 28, 2022, 11:03
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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.
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Old   March 29, 2022, 04:24
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Originally Posted by FMDenaro View Post
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.
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Old   March 29, 2022, 04:27
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Quote:
Originally Posted by LuckyTran View Post
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.
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Old   March 29, 2022, 04:54
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Quote:
Originally Posted by lucamirtanini View Post
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
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Old   March 29, 2022, 05:16
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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.
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Old   March 29, 2022, 06:07
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Originally Posted by LuckyTran View Post
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.
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Old   March 29, 2022, 06:26
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Quote:
Originally Posted by lucamirtanini View Post
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 …
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Old   March 29, 2022, 06:31
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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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Old   March 29, 2022, 10:48
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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.
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Old   March 29, 2022, 14:42
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Quote:
Originally Posted by LuckyTran View Post
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.
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