# Supply rate = dissipation rate in Turbulence

1. Aug 16, 2015

### K41

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?

2. Aug 16, 2015

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. Aug 17, 2015

### K41

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. Aug 17, 2015

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. Aug 19, 2015

### K41

I am referring to fully developed flow.

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

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, lets 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, lets 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. Aug 19, 2015

### olivermsun

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. Aug 20, 2015

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/Turbulent-Flows-Stephen-B-Pope/dp/0521598869, 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 $\mathscr{T}(\ell_0) = \varepsilon$.

Last edited by a moderator: May 7, 2017
8. Aug 21, 2015

### K41

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, lets 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

Last edited by a moderator: May 7, 2017
9. Aug 21, 2015

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. Aug 21, 2015

### K41

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. Aug 21, 2015

### K41

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. Aug 22, 2015