RANS Boussinesq approximation - local equilibrium

blyatman · September 26, 2021, 00:30

I've read various sources that say that the Boussinesq hypothesis assumes local equilibrium, where the production and dissipation of TKE are balanced. If this is the case, then why do the production and dissipation terms in the TKE equation need to be modeled and/or solved? Don't they simply cancel each other out? This would simplify the k equation to Dk/Dt = turbulent flux or diffusion of k.

LuckyTran · September 26, 2021, 02:06

If you just look at transport equation for k, the production and dissipation terms are in general not equal and do need to be modeled. So just to be clear, canceling those terms and simplifying the tke equation is just plain wrong.

The problem is not really the transport equation for k. The local equilibrium assumption shows up again in another place... Pick your favorite two-equation model that utilizes a transport equation of k (i.e. k-epsilon or k-omega) and just look at the way the turbulent viscosity is determined...

For k-epsilon it is:
$\mu_t = \rho C_{\mu} \frac{k^2}{\epsilon}$

For k-omega it is:
$\mu_t = {\rho k \over \omega }$

Another way to explain how the local equilibrium is buried in these relations is... If production and dissipation were not equal, then you wouldn't be able to get the turbulent viscosity from purely the turbulence scales (k and epsilon) but you would need both turbulence scales and mean flow scales in these relations. This is despite the fact that good two-equations models include production and dissipation terms in the transport equation for k, we can even assume they are modeled correctly! We still run into the local equilibrium hypothesis.

You can of course construct more complex eddy viscosity models that don't make this assumption. But generally all eddy viscosity models that use a simple proportionality constant for the ratio of the Reynolds stresses to mean strain rate will be Boussinesq-like.

blyatman · September 26, 2021, 02:21

Yes the local equilibrium stuff makes sense, but I still don't understand why production and dissipation don't cancel out. Local equilibrium means that production and dissipation balance. So from this definition, the production and dissipation terms are equal, and therefore should cancel out? It seems contradictory to me to assume local equilibrium yet also say that production and dissipation are in general not equal.

LuckyTran · September 26, 2021, 03:27

Models that blatantly cancel out the production and dissipation terms in the tke equation are the zero equation mixing length models. If you think that's the way turbulence should be modeled, feel free to use them...

One equation and two-equation and n-equation models that use transport equations for k do their best to model the transport of k to the best of their ability while still using the eddy viscosity hypothesis.

I can also use the same transport equation for k and epsilon or k and omega and just modify the way the turbulent viscosity is calculated to move away from this local equilibrium issue, but that's not trivial.

blyatman · September 26, 2021, 03:35

I'm not saying that's how I want to use them. I'm just struggling to understand why they don't cancel given my argument above. I don't think my argument is correct since I don't see any sources that use a TKE equation where they cancel out. However, I don't know what's wrong with the reasoning I put forth.

As I said above. Local equilibrium => production = dissipation. Yet production and dissipation are not cancelled out in the k equation. To me, this looks like a simple contradiction that isn't resolved in your original response, and I'm simply just trying to understand how to resolve it.

LuckyTran · September 26, 2021, 05:12

The eddy viscosity model isn't being formally derived mathematically where we start with the "assumption" that production and dissipation are equal and then try to prove a turbulence model out of it.

The terms that need to be modeled are the Reynolds stresses. The Boussinesq approach models the Reynolds stresses a certain way, with a linear relation between the Reynolds stresses and strain rate. This model it turns out is valid only when/where the production and dissipation are equal. We don't actually assume it to be this way ever.

FMDenaro · September 26, 2021, 05:14

Quote:

Originally Posted by blyatman

I've read various sources that say that the Boussinesq hypothesis assumes local equilibrium, where the production and dissipation of TKE are balanced. If this is the case, then why do the production and dissipation terms in the TKE equation need to be modeled and/or solved? Don't they simply cancel each other out? This would simplify the k equation to Dk/Dt = turbulent flux or diffusion of k.

I think that your reasoning is theoreticall acceptable, you can cancel them if the hypothesis of equilibrium is suitable.
On the other hand, now you can use both the simplified TKE and the relation P=eps to develop a turbulence model. What advantage you get in using the former or the latter?

Note that if we consider statistically steady flows, any time-variation is disregarded.

blyatman · September 26, 2021, 05:24

Thanks for your responses. I THINK I sort of get it... So an implication of the Boussinesq is that production and dissipation are equal. However, this is not strictly true, and in general, the production and dissipation terms don't cancel out. However, we still need to keep this implication in mind. If we find that the production and dissipation are approximately in the same ball park, then the Boussinesq hypothesis can still apply. However, if we find that the production and dissipation are disproportionately out of balance, then it means that the Boussinesq hypothesis is technically not valid. Hopefully I'm understanding all of this correctly.

September 26, 2021, 00:30	RANS Boussinesq approximation - local equilibrium	#1
blyatman Member BM Join Date: Sep 2021 Posts: 35 Rep Power: 4	I've read various sources that say that the Boussinesq hypothesis assumes local equilibrium, where the production and dissipation of TKE are balanced. If this is the case, then why do the production and dissipation terms in the TKE equation need to be modeled and/or solved? Don't they simply cancel each other out? This would simplify the k equation to Dk/Dt = turbulent flux or diffusion of k.

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September 26, 2021, 02:06		#2
LuckyTran Senior Member Lucky Join Date: Apr 2011 Location: Orlando, FL USA Posts: 5,674 Rep Power: 65	If you just look at transport equation for k, the production and dissipation terms are in general not equal and do need to be modeled. So just to be clear, canceling those terms and simplifying the tke equation is just plain wrong. The problem is not really the transport equation for k. The local equilibrium assumption shows up again in another place... Pick your favorite two-equation model that utilizes a transport equation of k (i.e. k-epsilon or k-omega) and just look at the way the turbulent viscosity is determined... For k-epsilon it is: $\mu_t = \rho C_{\mu} \frac{k^2}{\epsilon}$ For k-omega it is: $\mu_t = {\rho k \over \omega }$ Another way to explain how the local equilibrium is buried in these relations is... If production and dissipation were not equal, then you wouldn't be able to get the turbulent viscosity from purely the turbulence scales (k and epsilon) but you would need both turbulence scales and mean flow scales in these relations. This is despite the fact that good two-equations models include production and dissipation terms in the transport equation for k, we can even assume they are modeled correctly! We still run into the local equilibrium hypothesis. You can of course construct more complex eddy viscosity models that don't make this assumption. But generally all eddy viscosity models that use a simple proportionality constant for the ratio of the Reynolds stresses to mean strain rate will be Boussinesq-like.

September 26, 2021, 02:21		#3
blyatman Member BM Join Date: Sep 2021 Posts: 35 Rep Power: 4	Yes the local equilibrium stuff makes sense, but I still don't understand why production and dissipation don't cancel out. Local equilibrium means that production and dissipation balance. So from this definition, the production and dissipation terms are equal, and therefore should cancel out? It seems contradictory to me to assume local equilibrium yet also say that production and dissipation are in general not equal.

September 26, 2021, 03:27		#4
LuckyTran Senior Member Lucky Join Date: Apr 2011 Location: Orlando, FL USA Posts: 5,674 Rep Power: 65	Models that blatantly cancel out the production and dissipation terms in the tke equation are the zero equation mixing length models. If you think that's the way turbulence should be modeled, feel free to use them... One equation and two-equation and n-equation models that use transport equations for k do their best to model the transport of k to the best of their ability while still using the eddy viscosity hypothesis. I can also use the same transport equation for k and epsilon or k and omega and just modify the way the turbulent viscosity is calculated to move away from this local equilibrium issue, but that's not trivial.

September 26, 2021, 03:35		#5
blyatman Member BM Join Date: Sep 2021 Posts: 35 Rep Power: 4	I'm not saying that's how I want to use them. I'm just struggling to understand why they don't cancel given my argument above. I don't think my argument is correct since I don't see any sources that use a TKE equation where they cancel out. However, I don't know what's wrong with the reasoning I put forth. As I said above. Local equilibrium => production = dissipation. Yet production and dissipation are not cancelled out in the k equation. To me, this looks like a simple contradiction that isn't resolved in your original response, and I'm simply just trying to understand how to resolve it.

September 26, 2021, 05:12		#6
LuckyTran Senior Member Lucky Join Date: Apr 2011 Location: Orlando, FL USA Posts: 5,674 Rep Power: 65	The eddy viscosity model isn't being formally derived mathematically where we start with the "assumption" that production and dissipation are equal and then try to prove a turbulence model out of it. The terms that need to be modeled are the Reynolds stresses. The Boussinesq approach models the Reynolds stresses a certain way, with a linear relation between the Reynolds stresses and strain rate. This model it turns out is valid only when/where the production and dissipation are equal. We don't actually assume it to be this way ever.

September 26, 2021, 05:24		#8
blyatman Member BM Join Date: Sep 2021 Posts: 35 Rep Power: 4	Thanks for your responses. I THINK I sort of get it... So an implication of the Boussinesq is that production and dissipation are equal. However, this is not strictly true, and in general, the production and dissipation terms don't cancel out. However, we still need to keep this implication in mind. If we find that the production and dissipation are approximately in the same ball park, then the Boussinesq hypothesis can still apply. However, if we find that the production and dissipation are disproportionately out of balance, then it means that the Boussinesq hypothesis is technically not valid. Hopefully I'm understanding all of this correctly.