Supply rate = dissipation rate in Turbulence

In summary: L of the large eddies. So how can the dissipation rate be proportional to a supply rate which is determined by large scale eddies?The supply rate and the dissipation rate are generally considered proportional (or equal, typically) due to the conservation of energy. If you the large-scale motions are dumping energy into the smaller scales at a certain rate all the way down to the dissipative scales, then the dissipation rate must be the same as the supply rate to avoid a build up of energy. The different in order of magnitude between the scales can be reconciled by the fact that there are a whole lot more eddies in the dissipative scale than at the larger scales.
  • #1
K41
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1
So in many books, authors state that the supply rate, u^3/l is proportional to the dissipation rate ɛ, in a turbulent flow i.e:

u^3/l ~ ɛ

where u and l are characteristic velocity and length scales of the large eddies.

Many will also state that the small scale eddies have a time period or "eddy turn over time" which is considerable smaller than the large scale eddies.

So my question is, if the small scale eddies have orders of magnitude different turn over times, how can the dissipation rate be proportional to a supply rate which is determined by large scale eddies?
 
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  • #2
I am not sure if I follow your line of reasoning in the question. The supply rate and the dissipation rate are generally considered proportional (or equal, typically) due to the conservation of energy. If you the large-scale motions are dumping energy into the smaller scales at a certain rate all the way down to the dissipative scales, then the dissipation rate must be the same as the supply rate to avoid a build up of energy. The different in order of magnitude between the scales can be reconciled by the fact that there are a whole lot more eddies in the dissipative scale than at the larger scales.
 
  • #3
So if you have say, 10J on energy in a large scale eddy. Say for instance, the supply rate is 1J/S. So the dissipation rate must also be ~1J/S. So in ten seconds, all our energy will have passed down to a small scale where it is dissipated in, ten seconds.

So if the large scale eddies have a time scale of 10s and there was only one large eddy, could you say from this one large eddy that to dissipate the energy, required 100 small-scale eddies which had a time scale of 0.1s? If this is the case, does the large scale eddy immediately "create" these small eddies simultaneously or are they created one after another. My problem with the "lots more eddies at small scale" answer is that to me, for the rates to be equal, it would mean that we would need a succession of small eddies created, that is, one after another so that each one adds up to dissipate the energy that was there. If you had all the eddies created at once, the time that is passed remains the same, still 0.1 seconds to dissipate all that energy. :s
 
  • #4
You are absolutely correct, you would need a succession of smaller eddies to be created, and that is exactly what happens. While that is occurring, the traditional theories that most students study in turbulence (e.g. Kolmogorov's theories) are not valid, as it is not a fully-developed turbulent flow. It is still transitional. Once the turbulence is fully developed, there are a range of scales in the flow ranging from the integral scales, ##L##, all the way down to the Kolmogorov scales, ##\eta##, where the ##O(L)## scales contain most of the energy, energy is dissipated at scales of ##O(\eta)##, and the scales satisfying ##\eta \ll \ell \ll L##, called the inertial subrange, really just exist to pass energy down to increasingly smaller scales until they reach ##O(\eta)##, where ##Re = O(1)## and viscous dissipation occurs.
 
  • #5
boneh3ad said:
You are absolutely correct, you would need a succession of smaller eddies to be created, and that is exactly what happens. While that is occurring, the traditional theories that most students study in turbulence (e.g. Kolmogorov's theories) are not valid, as it is not a fully-developed turbulent flow. It is still transitional. Once the turbulence is fully developed, there are a range of scales in the flow ranging from the integral scales, ##L##, all the way down to the Kolmogorov scales, ##\eta##, where the ##O(L)## scales contain most of the energy, energy is dissipated at scales of ##O(\eta)##, and the scales satisfying ##\eta \ll \ell \ll L##, called the inertial subrange, really just exist to pass energy down to increasingly smaller scales until they reach ##O(\eta)##, where ##Re = O(1)## and viscous dissipation occurs.

I am referring to fully developed flow.

Yes, so smaller and smaller eddies are created. But my point originally is this:

turbulence.jpg

The sources say that supply rate is proportional to dissipation rate. Yet, if you look at this diagram, a typical diagram you see energy cascades, the large eddies, in red, have a turn over time of L/u. So they will take, let's say, ten seconds to transfer their energy to the next scale which is in orange. Another ten seconds will then pass til the energy reaches the smaller scale denoted by yellow. Now eventually, we reach the smallest scale. There are 15 small eddies, denoted green. Assume they are the same size such that their turn over times are all the same, let's say l_small/u_small. So they have a life-time which is considerably less than the large eddies. 1/ Time = Rate. So if lifetime goes down, rate must go up.Now, if just those 15 eddies are sufficient to dissipate the energy that the red eddies transferred to them, the dissipation rate cannot equal the supply rate since those small eddies will have dissipated all the energy by the time their life time is up which, by definition (and I believe through experiment), be shorter than the time period the large eddies. The only way the dissipation rate can be equivalent is if:

a) The yellow eddies, upon a second turn over time, create another stage of green eddies which are further required to dissipate the original energy from the large scale etc.
b) An even smaller scale is created which seems unlikely because viscosity will immediately prevent this (assume the red scale is the smallest possible scale in this flow).

So on one hand they say that the turn over time of these eddies are small and they have diagrams showing that a few large eddies create many many small eddies, yet they still suggest that the RATE of supply remains proportional to RATE of dissipation, even though the TIME (or DURATION) of dissipation will go down because we are summing the dissipation of all the eddies in that stage that were originally created from the large red eddies, hence the dissipation RATE must go up, hence it cannot equal the supply rate?

I'm not suggesting that the theory is wrong, I just don't understand what it is I'm missing here.
 

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  • #6
Why is it assumed that ##T \sim \dfrac{L}{U} \gg \dfrac{l_\mathrm{small}}{u_\mathrm{small}} \sim t_\mathrm{small}##? (This doesn't necessarily follow from ##{L}/{l_\mathrm{small}} \gg 1##.)
 
  • #7
The energy transfer rate at a given scale, ##\ell##, is ##\mathscr{T}(\ell) = u(\ell)^2/\tau(\ell)##. If you take a look in a source like https://www.amazon.com/dp/0521598869/?tag=pfamazon01-20, you can see that ##u(\ell) = (\varepsilon\ell)^{1/3} \sim u_0(\ell/\ell_0)^{1/3}## and ##\tau(\ell) = (\ell^2/\varepsilon)^{1/3} \sim \tau_0(\ell/\ell_0)^{2/3}##, where ##_0## quantities are taken at the very large, energy containing scales. From there you can show that ##\varepsilon = \mathscr{T}(\ell) = u(\ell)^2/\tau(\ell)##, and using those relations in terms of ##\ell/\ell_0##, you can see that the ##\ell##-dependence drops out and you are just left with ##\mathscr{T}(\ell_0) = \varepsilon##.
 
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  • #8
boneh3ad said:
The energy transfer rate at a given scale, ##\ell##, is ##\mathscr{T}(\ell) = u(\ell)^2/\tau(\ell)##. If you take a look in a source like https://www.amazon.com/dp/0521598869/?tag=pfamazon01-20, you can see that ##u(\ell) = (\varepsilon\ell)^{1/3} \sim u_0(\ell/\ell_0)^{1/3}## and ##\tau(\ell) = (\ell^2/\varepsilon)^{1/3} \sim \tau_0(\ell/\ell_0)^{2/3}##, where ##_0## quantities are taken at the very large, energy containing scales. From there you can show that ##\varepsilon = \mathscr{T}(\ell) = u(\ell)^2/\tau(\ell)##, and using those relations in terms of ##\ell/\ell_0##, you can see that the ##\ell##-dependence drops out and you are just left with ##\mathscr{T}(\ell_0) = \varepsilon##.

But even if you use the Kolmogorov relationships:

Lets say we have Re = 50,000, kinematic viscosity of air is 1.343*10^-5 and our system scale is 0.5m

This gives us a velocity of 1.343m/s. So for simplicity, let's assume our large eddy scale eddy velocity and lengths are 1.3m/s and 0.5m respectively. Now this gives a large eddy time scale of T ~0.38secs and a supply rate of ~4.4 J/s per unit mass.

We've explicitly used the time scale of 0.38secs. So in 0.38 secs, 4.4J of energy per unit mass is transferred to the lower scale.

Now we use Kolmogorov's relationships, specifically, t_n / T ~ Re^-0.5. So t_n is the small eddy time and T is the aforementioned large eddy time scale. Following the numbers above, t_n = 0.0017 secs.

So by the time the small scale eddy dies, it can only have dissipated ~0.02J of energy. This arises from the logic that the dissipate RATE must be the same as the SUPPLY RATE, and the small eddies die after 0.0017 secs so we use the supply rate to calculate how much energy per unit time they can dissipate during the span of their lifetime, i.e.:

0.38 ::: 4.4
0.0017 ::: X

x = 4.4 / (0.38/0.0017) = 0.02J.

It can't dissipate 4.4J of energy in its life time because then it would clearly have dissipated energy at a rate of 2588J/s per unit mass which would break the theory.

So what is happening? It would take over 200 further small eddies to be generated, one after the another, to dissipate all the energy that was passed down at the same rate. These would need to be the same size, the same length, the same time and still part of the same stage of the cascade. Now the diagrams I attached earlier suggest that these eddies (of the same size and still in the same stage of the cascade) may actually exist simultaneously, rather than are generated one after the other. But again, if they are all generated simultaneously, we increase the energy we dissipate per second.

I appreciate your help btw, I know I'm probably just being dumb here lol
 
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  • #9
Nothing says all of those small eddies are generated simultaneously or sequentially. In reality, it would be a constant process of small eddies being formed and dying such that the dissipation rate is consistent with the rate at which energy is being supplied.

Those simple figures are great for getting a visual, but it isn't as if each eddy simultaneously breaks up into two more smaller eddies as it would appear in the image. The energy cascades a constant, poorly-understood process. However, predictions made using these hypotheses have been tested experimentally and found to be accurate.
 
  • #10
boneh3ad said:
Nothing says all of those small eddies are generated simultaneously or sequentially. In reality, it would be a constant process of small eddies being formed and dying such that the dissipation rate is consistent with the rate at which energy is being supplied.

Those simple figures are great for getting a visual, but it isn't as if each eddy simultaneously breaks up into two more smaller eddies as it would appear in the image. The energy cascades a constant, poorly-understood process. However, predictions made using these hypotheses have been tested experimentally and found to be accurate.

So if someone were to ask me, is it fair to respond and say we still don't really know exactly how, at each stage, the forming and dying process of these smaller eddies occur to allow for the dissipation and supply rate to be consistent? And secondly, to sum up, can we also just say that what we have is the Richardson cascade, a phenomenological description which describes a general overview of energy dissipation in turbulence, and Kolmogorov's theory quantifying the small scales at which dissipation occurs, but the details at each stage are still, as you put "poorly understood"?
 
  • #11
olivermsun said:
Why is it assumed that ##T \sim \dfrac{L}{U} \gg \dfrac{l_\mathrm{small}}{u_\mathrm{small}} \sim t_\mathrm{small}##? (This doesn't necessarily follow from ##{L}/{l_\mathrm{small}} \gg 1##.)

Thanks for replying. I just wanted to respond to your post separately. Your right, we can't from just l/L, assume that. We have to make other assumptions (Kolmogorov "K41" theory). I just left out the details because at the time I felt they weren't immediately relevant to the point I was trying to make.
 
  • #12
djpailo said:
So if someone were to ask me, is it fair to respond and say we still don't really know exactly how, at each stage, the forming and dying process of these smaller eddies occur to allow for the dissipation and supply rate to be consistent? And secondly, to sum up, can we also just say that what we have is the Richardson cascade, a phenomenological description which describes a general overview of energy dissipation in turbulence, and Kolmogorov's theory quantifying the small scales at which dissipation occurs, but the details at each stage are still, as you put "poorly understood"?

That's probably mostly fair. I seem to recall Pope saying pretty much exactly what I said in his book: basically that the precise mechanism/morphology of the energy cascade is fairly poorly understood but that the predictions based on that theory have been borne out experimentally with regularity. That was written in 2005-ish, I believe, but I've not come across anything contradicting it yet.
 

1. What is the significance of the supply rate being equal to the dissipation rate in turbulence?

The supply rate being equal to the dissipation rate in turbulence is a fundamental relationship that helps us understand how energy is transferred and dissipated in turbulent flows. It indicates that the rate at which energy is supplied to the flow is balanced by the rate at which it is dissipated, resulting in a state of equilibrium.

2. How is the supply rate related to the dissipation rate in turbulence?

The supply rate and the dissipation rate are directly proportional to each other. This means that as the supply rate increases, the dissipation rate also increases, and vice versa. This relationship is described by the Kolmogorov energy spectrum, which is a fundamental equation in the study of turbulence.

3. What factors affect the supply rate and dissipation rate in turbulence?

There are several factors that can influence the supply rate and dissipation rate in turbulence. These include the velocity and length scales of the flow, the viscosity of the fluid, and the characteristics of the turbulence-producing mechanism. Additionally, external factors such as surface roughness and flow geometry can also impact these rates.

4. How is the supply rate and dissipation rate measured in turbulence?

The supply rate and dissipation rate in turbulence can be measured using various experimental techniques, such as hot-wire anemometry and laser Doppler velocimetry. These methods involve directly measuring the velocity fluctuations and energy dissipation within the flow, which can then be used to calculate the supply and dissipation rates.

5. What are the practical applications of understanding the supply rate and dissipation rate in turbulence?

Understanding the supply rate and dissipation rate in turbulence is crucial in various fields, including aerodynamics, atmospheric science, and oceanography. It helps us predict and control the behavior of turbulent flows, which is essential for designing efficient and safe structures and vehicles. Additionally, this knowledge can also aid in the development of improved turbulence models for numerical simulations.

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