Kolmogorov scale in less than fully developed turbulence

rdemyan · July 31, 2023, 21:25

If the flow is less than fully developed turbulence (and for the sake of argument let's say considerably less), then is the cascade of energy able to "grind down" to the Kolmogorov scale? Or is there perhaps a larger scale at which all of the kinetic energy has been dissipated? Or is the cascade always able to make it to the Kolmogorov scale regardless of the level of turbulence (assuming that there is turbulence). I would think it would be a function of Reynolds number per Lumley's statement in his 1992 paper:

" Note also that, for low Reynolds number, the time shrinks to a very small value, since the
energy is dissipated at essentially the same wave number where it is produced, while for high Reynolds number it goes
to 2L/u." (page 207, J.L. Lumley Phys. Fluids A 4 (2), February 1992.)

Here L is the large scale eddy size and L/u is the turnover time of the largest eddies.

LuckyTran · July 31, 2023, 21:53

What happens at low Reynolds numbers is, there is no longer separation of scales, no energy cascade, and the inertial length scale disappears. But there is always a dissipative scale, though it might not exhibit the -5/3 law.

rdemyan · July 31, 2023, 22:09

Quote:

Originally Posted by LuckyTran

What happens at low Reynolds numbers is, there is no longer separation of scales, no energy cascade, and the inertial length scale disappears. But there is always a dissipative scale, though it might not exhibit the -5/3 law.

When you say there is no longer separation of scales, the inherent assumption is that the large scale eddy size is the same (for this statement). If so, then that means that the smallest scale must be larger and therefore the smallest scale is not the Kolmogorov microscale (even though a Kolmogorov microscale could be calculated??). If what I am saying is true, it probably follows that there will always be a Taylor microscale (even though it's link to physical reality is circumspect). That's because the Taylor microscale would approximately mark the beginning of the dissipation range whereas Kolmogorov scale marks the end.

LuckyTran · July 31, 2023, 22:43

There is no inherent assumption that the large eddy size is the same, they simply vanish.

You can debate if you want to consider the largest scales becoming smaller or the smallest scales becoming larger. However, the only remaining scale is clearly fully dissipative. It would be even weirder to consider that the integral scale or taylor scale suddenly becoming dissipative when it is not behaving like an integral scale nor a taylor scale at all.

Or you can just bring in Occam's razor

Kolmogorov himself studied such cases of transition (including sub-critical transition) where there is only a dissipative scale. Even in the time of Kolmogorov, it was considered that the Taylor microscale is the first to disappear.

You need to be careful what you are calling Kolmogorov scales. If you define Kolmogorov scales as the scales that strictly obey the -5/3 law then Kolmogorov scales never exist.

rdemyan · July 31, 2023, 23:12

Quote:

Originally Posted by LuckyTran

There is no inherent assumption that the large eddy size is the same, they simply vanish.

You can debate if you want to consider the largest scales becoming smaller or the smallest scales becoming larger. However, the only remaining scale is clearly fully dissipative. It would be even weirder to consider that the integral scale or taylor scale suddenly becoming dissipative when it is not behaving like an integral scale nor a taylor scale at all.

Or you can just bring in Occam's razor

Kolmogorov himself studied such cases of transition (including sub-critical transition) where there is only a dissipative scale. Even in the time of Kolmogorov, it was considered that the Taylor microscale is the first to disappear.

You need to be careful what you are calling Kolmogorov scales. If you define Kolmogorov scales as the scales that strictly obey the -5/3 law then Kolmogorov scales never exist.

I have to give some more thought to what you wrote, but I have always been defining the Kolmogorov microscale as:

$\eta=(\frac{\nu^3}{\epsilon})^{1/4}$

FMDenaro · August 1, 2023, 03:43

In an alternative way, the Kolmogorov lenght scale is that one for which the local Reybolds number is 0(1). Therefore you can think it depends only on a characteristic velocity.

When you have only a dissipative range you cannot use the -5/3 scalinig for comparison.

rdemyan · August 1, 2023, 11:28

Quote:

Originally Posted by rdemyan

I have to give some more thought to what you wrote, but I have always been defining the Kolmogorov microscale as:

$\eta=(\frac{\nu^3}{\epsilon})^{1/4}$

From Fig. 6.20 b or Fig. 6.14 in Pope, it looks like the steep drop from the end of the inertial range starts at $\kappa\eta$ equal to about 0.3. From Fig. 6.16 in Pope, the cumulative dissipation at $\kappa\eta=0.3$ is about 40%??! I thought an inherent assumption is that dissipation is negligible in the production region and inertial range of the energy spectrum. If so, how could the cumulative dissipation be 40% of the total at the end of the inertial range?

LuckyTran · August 1, 2023, 13:46

Dissipation is negligible as a mechanism of turbulent transport when compared to production. Dissipation is obviously not negligible compared to itself.

dx/x may be a small but dx/dx is O(1)

Local Re>1 is the threshold at which the inertial terms overcome the dissipative effects. It is not a statement that dissipation suddenly disappears.

FMDenaro · August 1, 2023, 13:51

Quote:

Originally Posted by rdemyan

From Fig. 6.20 b or Fig. 6.14 in Pope, it looks like the steep drop from the end of the inertial range starts at $\kappa\eta$ equal to about 0.3. From Fig. 6.16 in Pope, the cumulative dissipation at $\kappa\eta=0.3$ is about 40%??! I thought an inherent assumption is that dissipation is negligible in the production region and inertial range of the energy spectrum. If so, how could the cumulative dissipation be 40% of the total at the end of the inertial range?

The kolmogorov lenght is just the end of the dissipation.
Let us see the dissipation in the physical space, that is mu*S:S. You will see that dissipation depends on the characteristic Re number. Do not take the figure in the textbook to be general, it is at Re_lambda=600, have a look also at Fig. 6.21 and 22 at higher Re.

rdemyan · August 1, 2023, 14:49

Quote:

Originally Posted by LuckyTran

Dissipation is negligible as a mechanism of turbulent transport when compared to production. Dissipation is obviously not negligible compared to itself.

dx/x may be a small but dx/dx is O(1)

Local Re>1 is the threshold at which the inertial terms overcome the dissipative effects. It is not a statement that dissipation suddenly disappears.

I guess I'm not getting it. In Fig. 6.20b in Pope, there is a clear drop in the ordinate value, $\frac{E(\kappa)}{\eta {u_\eta}^2}$ , from the lowest $\kappa\eta$ in the inertial range to the highest $\kappa\eta$ in the inertial range. If it's not dissipation, then what causes this drop in the ordinate? I mean, if there is a continuous input of energy, then why don't all of the wavenumbers in the inertial range have the same value of the ordinate? In the inertial range, isn't the assumption that energy is transferred from a wavenumber to the next highest wavenumber unaffected by any other factors?

PS: I GUESS the answer is that E(K) is defined as the energy per wave number. So if the wave number increases, then the ordinate would have to go down.

FMDenaro · August 1, 2023, 15:26

Quote:

Originally Posted by rdemyan

I guess I'm not getting it. In Fig. 6.20b in Pope, there is a clear drop in the ordinate value, $\frac{E(\kappa)}{\eta {u_\eta}^2}$ , from the lowest $\kappa\eta$ in the inertial range to the highest $\kappa\eta$ in the inertial range. If it's not dissipation, then what causes this drop in the ordinate? I mean, if there is a continuous input of energy, then why don't all of the wavenumbers in the inertial range have the same value of the ordinate? In the inertial range, isn't the assumption that energy is transferred from a wavenumber to the next highest wavenumber unaffected by any other factors?

PS: I GUESS the answer is that E(K) is defined as the energy per wave number. So if the wave number increases, then the ordinate would have to go down.

Energy transfer in real turbulence happens with a degree of correlation. A spectrum with equal density would correspond to a white noise, a totally uncorrelated signal.

LuckyTran · August 1, 2023, 21:39

E(k) is the energy at the wavenumber k. The energy per wavenumber k would be E(k)/k

Honestly you should plot P(k) versus E(k) versus D(k) to answer your question, then it would be clearer. You are making grandiose statements based on P and D but the entire time you are looking at E.

rdemyan · August 1, 2023, 22:05

Quote:

Originally Posted by LuckyTran

E(k) is the energy at the wavenumber k. The energy per wavenumber k would be E(k)/k

Honestly you should plot P(k) versus E(k) versus D(k) to answer your question, then it would be clearer.

Per Bakker's notes, I think you meant to write that E( $\kappa$ ) = k/ $\kappa$ where k is the turbulent kinetic energy; but I get what you are saying, which is what I meant to state in my P.S. above.

Is P( $\kappa$ ) the production of energy? I haven't seen that term before. But isn't the production at only a very narrow band of low values for k $\eta$ , which is near the integral scale and then I presume it would drop off to zero as the energy starts to get transferred in the inertial regime??

LuckyTran · August 1, 2023, 22:56

I clearly wrote that k is wavenumber. If you decide to use another symbol, that's up to you.

If you have a heat transfer background like I do, k is also thermal conductivity. For physicists, k is boltzmann constant. These arguments are pedantic.

You are putting words in my mouth. I completely disagree that E(k) drops because k is increasing. That is an observation, not an explanation. I encourage your continued reading. Take your time in understanding turbulent transport. There is no rush. You keep flip flopping on whether dissipation is everything, dissipation is nothing, and here again. There is no need to grasp for straws.

rdemyan · August 1, 2023, 23:06

Quote:

Originally Posted by LuckyTran

I clearly wrote that k is wavenumber. If you decide to use another symbol, that's up to you.

No, that's not what I mean. The way you wrote it, it looks like E( $\kappa$ ) is the turbulent kinetic energy, whereas I thought that E( $\kappa$ ) is the energy contained in the wavenumber $\kappa$ . Instead let's call the turbulent kinetic energy TKE.

E( $\kappa$ ) = TKE/ $\kappa$

Source: Slide 26 in http://www.lcad.icmc.usp.br/~buscagl...er_09-kolm.pdf

LuckyTran · August 1, 2023, 23:13

Again, I clearly wrote that k is wavenumber. What really is the difference between E(k) where k is wavenumber versus E(banana) where banana is wavenumber? I am not responsible for your misconceptions.

k is Boltzmann constant, one of the six fundamental constants. If you want to nitpick I can nitpick too. If you want, I can stop using symbols and only write in a wall of text.

rdemyan · August 1, 2023, 23:17

Quote:

Originally Posted by LuckyTran

I clearly wrote that k is wavenumber. If you decide to use another symbol, that's up to you.

If you have a heat transfer background like I do, k is also thermal conductivity. For physicists, k is boltzmann constant. These arguments are pedantic.

You are putting words in my mouth. I completely disagree that E(k) drops because k is increasing. That is an observation, not an explanation. I encourage your continued reading. Take your time in understanding turbulent transport. There is no rush. You keep flip flopping on whether dissipation is everything, dissipation is nothing, and here again. There is no need to grasp for straws.

I did not put words in your mouth and I do not understand your aggressive behaviour. I was very clear that it had nothing to do with your use of the symbol "k".

LuckyTran · August 1, 2023, 23:26

I simply don't agree with your statement that E(k) drops because k is increasing. Please don't attribute that to me.

July 31, 2023, 21:25	Kolmogorov scale in less than fully developed turbulence	#1
rdemyan Member Join Date: Jan 2023 Posts: 64 Rep Power: 3	If the flow is less than fully developed turbulence (and for the sake of argument let's say considerably less), then is the cascade of energy able to "grind down" to the Kolmogorov scale? Or is there perhaps a larger scale at which all of the kinetic energy has been dissipated? Or is the cascade always able to make it to the Kolmogorov scale regardless of the level of turbulence (assuming that there is turbulence). I would think it would be a function of Reynolds number per Lumley's statement in his 1992 paper: " Note also that, for low Reynolds number, the time shrinks to a very small value, since the energy is dissipated at essentially the same wave number where it is produced, while for high Reynolds number it goes to 2L/u." (page 207, J.L. Lumley Phys. Fluids A 4 (2), February 1992.) Here L is the large scale eddy size and L/u is the turnover time of the largest eddies.

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July 31, 2023, 21:53		#2
LuckyTran Senior Member Lucky Join Date: Apr 2011 Location: Orlando, FL USA Posts: 5,675 Rep Power: 66	What happens at low Reynolds numbers is, there is no longer separation of scales, no energy cascade, and the inertial length scale disappears. But there is always a dissipative scale, though it might not exhibit the -5/3 law.

July 31, 2023, 22:43		#4
LuckyTran Senior Member Lucky Join Date: Apr 2011 Location: Orlando, FL USA Posts: 5,675 Rep Power: 66	There is no inherent assumption that the large eddy size is the same, they simply vanish. You can debate if you want to consider the largest scales becoming smaller or the smallest scales becoming larger. However, the only remaining scale is clearly fully dissipative. It would be even weirder to consider that the integral scale or taylor scale suddenly becoming dissipative when it is not behaving like an integral scale nor a taylor scale at all. Or you can just bring in Occam's razor Kolmogorov himself studied such cases of transition (including sub-critical transition) where there is only a dissipative scale. Even in the time of Kolmogorov, it was considered that the Taylor microscale is the first to disappear. You need to be careful what you are calling Kolmogorov scales. If you define Kolmogorov scales as the scales that strictly obey the -5/3 law then Kolmogorov scales never exist.

August 1, 2023, 03:43		#6
FMDenaro Senior Member Filippo Maria Denaro Join Date: Jul 2010 Posts: 6,773 Rep Power: 71	In an alternative way, the Kolmogorov lenght scale is that one for which the local Reybolds number is 0(1). Therefore you can think it depends only on a characteristic velocity. When you have only a dissipative range you cannot use the -5/3 scalinig for comparison.

August 1, 2023, 13:46		#8
LuckyTran Senior Member Lucky Join Date: Apr 2011 Location: Orlando, FL USA Posts: 5,675 Rep Power: 66	Dissipation is negligible as a mechanism of turbulent transport when compared to production. Dissipation is obviously not negligible compared to itself. dx/x may be a small but dx/dx is O(1) Local Re>1 is the threshold at which the inertial terms overcome the dissipative effects. It is not a statement that dissipation suddenly disappears.

August 1, 2023, 21:39		#12
LuckyTran Senior Member Lucky Join Date: Apr 2011 Location: Orlando, FL USA Posts: 5,675 Rep Power: 66	E(k) is the energy at the wavenumber k. The energy per wavenumber k would be E(k)/k Honestly you should plot P(k) versus E(k) versus D(k) to answer your question, then it would be clearer. You are making grandiose statements based on P and D but the entire time you are looking at E.

August 1, 2023, 22:56		#14
LuckyTran Senior Member Lucky Join Date: Apr 2011 Location: Orlando, FL USA Posts: 5,675 Rep Power: 66	I clearly wrote that k is wavenumber. If you decide to use another symbol, that's up to you. If you have a heat transfer background like I do, k is also thermal conductivity. For physicists, k is boltzmann constant. These arguments are pedantic. You are putting words in my mouth. I completely disagree that E(k) drops because k is increasing. That is an observation, not an explanation. I encourage your continued reading. Take your time in understanding turbulent transport. There is no rush. You keep flip flopping on whether dissipation is everything, dissipation is nothing, and here again. There is no need to grasp for straws.

August 1, 2023, 23:13		#16
LuckyTran Senior Member Lucky Join Date: Apr 2011 Location: Orlando, FL USA Posts: 5,675 Rep Power: 66	Again, I clearly wrote that k is wavenumber. What really is the difference between E(k) where k is wavenumber versus E(banana) where banana is wavenumber? I am not responsible for your misconceptions. k is Boltzmann constant, one of the six fundamental constants. If you want to nitpick I can nitpick too. If you want, I can stop using symbols and only write in a wall of text.

August 1, 2023, 23:26		#18
LuckyTran Senior Member Lucky Join Date: Apr 2011 Location: Orlando, FL USA Posts: 5,675 Rep Power: 66	I simply don't agree with your statement that E(k) drops because k is increasing. Please don't attribute that to me.