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Is turbulence disripable by chaos?

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Old   March 21, 2001, 15:58
Default Is turbulence disripable by chaos?
  #1
Matthias
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Dear all,

I have two questions concerning turbulence:

(1) From the lectures given in turbulence I know that turbulence is a random process discripable by statistics. A definition of chaos says: "... chaos is a kind of order without periodicity... ". Well, from my point of view turbulence looks like chaos - am I right?

(2) From the derivation of the RANS-Equation I get an additional term rho*u'*v'. Further from the derivation of the k-equation I get an term called dissipation term epsilon:= mu*(du'/dx_j)^2 (averaged). If I consider (theoreticaly) an ideal fluid without viscosity, this term becomes epsilon =0. That means that no dissipation takes place, but the production of turbulence is not effected from the viscosity term. I assume it is possible to produce turbulence in an invicid fluid but what happens with the dissipation of turbulence, is the energy cascade still applicable, or is an invicid flow always laminar?

Thanks for your help

Matthias
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Old   March 21, 2001, 17:16
Default Re: Is turbulence disripable by chaos?
  #2
John C. Chien
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(1). If you take a sequence of pictures, the picture of turbulent flow will be different from time to time, place to place. (2). On the other hand, the picture of chaos has certain clear patterns in it, even though the complete picture might be very complex. (3). So far, I have not seen anything related to "inviscid turbulent flow". Even if you can perform statistical average on inviscid equations, it does not mean that such inviscid turbulent flow can be produced.
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Old   March 21, 2001, 17:48
Default Re: Is turbulence disripable by chaos?
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kalyan
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Interesting questions.

1). Yes, turbulence is indeed chaotic. But the definition "order with out periodicity" is not very precise. For example, you can have quasi periodic motion {e.g. y = sin(t) + sin(sqrt(2)t)}. Chaotic systems do have periodic orbits but they are unstable and have zero "measure".

There are 3 requirements for chaos : i) sensitivity to initial conditions (a formal definition of "sensitivity" can be found in the math text books), ii) topologically transitive (once again see a math text book to find out what it means) iii) I forgot this one.

One of the routes to chaos is through a set of period-doubling bifurcations. Periodic doubling has been known to occur at the onset of turbulence (transition).

ii) This is a difficult question to answer. I am not sure there are satisfactory answers out there. These are answers based on the little knowledge I have.

I have not come across any research that considered three-dimensional inviscid turbulence. Since inviscid flows can not dissipate, energy would probably accumulate at the small scales leading to a potentially destabilizing situation (no one knows what aftermath of the instability would be since, in reality, no fluid has zero viscosity). Mathematically, the phenomenon of finite-time-blow-up in inviscid (Euler) equations has been studied numerically. You can find reference to this work in a book by Lesieur (I hope I got the spelling right).

Inviscid 2-D turbulence has been studied in great detail for decades (mostly using inviscid vortex dynamics). Since the 2-D Euler equations can be put in a Hamiltonian form, many of the basic physics theories have been attempted to study them. Discrete Hamiltonian systems have been constructed using point vortices and statistical mechanics ideas have been applied to them. Several phenomena like atmospheric vortex dynamics, center-guided plasmas and Jupiter's red spot have been explained using these ideas. The easiest theory to understand is the maximum entropy theory that results in the well known "Joyce-Montgomery equation". Others have attempted to derive similar equations using measure theory and ergodic hypothesis (a math concept again).

Getting back to your question, all statistical theories rely more or less on the applicability of ergodicity (and mixing) which would mean that any energy in the fluid dynamic system would be distributed (on an average) equally among all the orthogonal modes. These modes can be wavenumbers in which case the mean kinetic energy spectrum (as a function of the scalar wavenumber magnitude, k) would scale as k^2.
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Old   March 22, 2001, 05:05
Default Re: Is turbulence disripable by chaos?
  #4
sylvain
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The lectures I had in turbulence used to start with a lecture in chaos theory, since my professor believes that turbulence is chaos. A characteristic example is the following :

If one takes a DNS code and has it run on two different cpus (Intel and AMD for example) with exactly the same initial conditions, one won't obtain the same result. The reason is that the way each cpus deal with numerical precision is not the same.

For the second question, introducing the energy spectrum E(K) one gets :

k = integral( from K=0 to K=K_kol ) E(K) dK

epsilon = integral( from K=0 to K=K_kol ) 2 nu K^2 E(K) dK

with E(K) proportional to K^(-5/3) for K large enough but smaller than K_kol, and where K_kol is the wave length associate with the Kolmogorov scale L_kol. As nu tends to zero, L_kol tends to zero and then K_kol tends to infinity. So even if nu tends to zero, it is still possible to have a dissipation rate which is not equal to zero. The physical meaning is :

When nu tends to zero the transformation of turbulent kinetic energy into heat occurs for scales which also tend to zero.

On the other side, if nu equal to zero, then one has to deal with the Euler equation, and I don't think that turbulence mean something in that case.

To conclude, it is possible to deal with turbulence kinetic energy and dissipation rate when nu tends to zero, but not when nu equal zero.

I hope that I am clear enough in my explanation.

Regards,

Sylvain

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Old   March 23, 2001, 07:36
Default Re: Is turbulence disripable by chaos?
  #5
Abhijit Tilak
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well, Matthais

good questions, all of them, sent me to library looking for textbooks. Regarding your first question ,from the limited knowledge i have, u are probably right, bcos as someone pointed out my prof too started to describe turbulence by giving info on chaos. second question I personally think the inviscid turbulence does'nt make much sense to me. A look at k -e equations suggest that e is zero so equation for e makes no sense, A look at k equn suggests that there will be no production (Tau_ij=0) a slight purterbation will not decay. there also wont be any energy cascade bcos there is no way energy can be transferred to lower scales (in fact the question is are the lower scales present in such situation ? Where will the lower scales come from ? since the purturbation does not decay, there will be only one scale in the entire flow domain, The question of energy transfer does not arise! Guys correct me if i am wrong ! ) hope that helps (!)

abhijit tilak aero engg dept iit bombay.
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Old   March 23, 2001, 13:41
Default Re: Is turbulence disripable by chaos?
  #6
kalyan
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K-e model is a model for turbulence derived using the scaling arguments and phenomenology. Such models have been calibrated well to produce good results in certain types of flows but they are NOT the equations governing turbulence.

Small scales vorticity structures (eddies) can be produced by large structures even in inviscid flows (hence the term "inviscid break down"). A good example is the often simulated mixing layer flow. The mean velocity profile has an initial hyperbolic tangent profiles over which some disturbances are imposed. Depending on the type of disturbances, mixing would lead to a cat-eye pattern (primary mode), initial cat-eye patterns which pair up (secondary mode) or the more 3-dimensional sub-harmonic pattern. These organized structures and formed (and break down further into small structures) even without viscosity. Vortex stretching can always lead to creation of thinner and longer vortex tubes from wider vortex tubes. In this way, energy can move between scales.

As I pointed out earlier (if one assumes ergodicity of 2-D Euler flows), energy would tend to equilibriate between "all" possible modes (however small they are) in 2-D inviscid flows. This is clearly not an option since "all modes" include those that have infinitesimal characteristic length scales and the continuum approach breaks down. With some boundary conditions, the equilibriation is constrained and you are left with a large scale mean state of vorticity (2-D Euler flows are hence said to be similar to Bose-Einstein condensates).
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Old   March 24, 2001, 01:26
Default Re: Is turbulence disripable by chaos?
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abhijit tilak
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hi kalyan,

well, by k-e i meant full equations for k-e ( k equn obatined by contracting R_ij equn), Reynolds streess equn & full e equn without any sigmas Pr's. I did'nt get much of ur explanation can u give me some references. I can't digest the idea of energy cascade without viscosity coming into picture. I agree that vortex stretching is possible without viscosity ( i got a equation for vorticity-velocity interaction which tells u what will happen to turbulence whether it dies or is sustained.If in that equation you put viscosity =0 entire rhs collapses & i cant make sense of what remains? )

may be u can throw some light. abhijit
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Old   March 26, 2001, 18:14
Default Re: Is turbulence disripable by chaos?
  #8
kalyan
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Viscosity creates an energy sink at the small dissipative scale. So, in the presence of viscosity, energy transfer is typically from the large scales to the small scales. Most of the energy from eddies of a given size goes to eddies slightly smaller. Thus energy cascades down through a series of scales rather than entering the small dissipative scales directly.

The cascade and inter-scale energy transfer are independent. In fact, energy from small scales (high wavenumbers) does enter the large scales (low wavenumbers), a phenomenon known as "backscatter" (which is non-zero in the inertial range where supposedly one has the cascade). The net transfer of energy from large to small eddies is the difference between the "forward scatter" and "backscatter". Usually forward scatter is larger than backscatter. So, the presence of energy cascade does not in any way rule out both forward and backward tranfer of energy. The expressions for energy transfer from a given wavenumber to another wavenumber can be found in many of the books on turbulence (McComb, Leslie, Leseiur etc.). In fact, most turbulence models (EDQNM, RNG-LES, two-scale DIA) are derived from the specral energy transfer equation.

When there is no viscosity, there is still energy transfer from eddies of a given size to eddies of all other sizes. One cannot say whether the overall (net=forward-backward) transfer is from the large scale to the small scales or vice-versa. If I had to guess based on what I have read, I would still say overall transfer of energy is from the large scales to the small scales. If the small scale dissipation is non-existent, the energy would keep entering smaller and smaller scales and eventually, you would breach the continuum limit.

In a numerical code, the energy accumulates at the smallest resolved scale and the code would eventually blow up. However, it is hard to trust numerical results for a perfectly inviscid fluid. There is always some numerical error (dissipative, dispersive, de-aliasing) and the closure problem (the ability to represent a continuum system with finite degrees of freedom). I suspect that the transfer of energy to the small scales (before large accumulation) MAY be due to the fact that the numerical truncation creates a numerical bias (e.g. a virtual small scale dissipation range). After a while, if the numerical dissipation at the small scales can not handle the incoming energy and the accumulation begins. I have also read some where in a statistical physics book that it is dissipation that enables us to use discrete representations (like in CFD, FEM etc.) to describe non-linear continuum systems (like turbulence, non-linear structural dynamics) and obtain good results. So, what this means is that a numerical simulation of inviscid DNS is not possible. And I do not have any idea about the other alternative, i.e., to actual experiment with inviscid fluid.

For theoretical studies on 2-D inviscid turbulence, you can search the literature by following people : Joyce, Montgomery, Onsager, Berdichevsky, Miller, Aref etc. You can also use terms like "ergodic" and "turbulence" and "point vortex dynamics" together. If you can not find anything, e-mail me and I can send you some papers.
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Old   March 26, 2001, 18:40
Default Re: Is turbulence disripable by chaos?
  #9
John C. Chien
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(1). Inviscid means no viscous loss. (2). "Inviscid turbulence" means turbulent flow which does not create viscous loss. (3). Is it something useful for aircraft designer? to study "inviscid turbulent flow" to save the fuel cost?
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Old   March 27, 2001, 13:04
Default Re: Is turbulence disripable by chaos?
  #10
kalyan
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I am not sure you can save fuel cost by studying fuel cost since you can not make air inviscid. There, however, have been attempts to study turbulent flow (lift and drag) around airfoils using inviscid type equations. The idea is puzzling initially, but it is based on the asymptotic high Reynolds number similarity of turbulent flows (turbulent flows attain an asymptotic structure beyond a certain value of Re).

I have mostly seen 2D vorticity-stream function method used in this context. In this method, if the flow is viscous, a two-step (operator-splitting) scheme is used to integrate the equations. The first step is a inviscid step where the vorticity field is advection using the velocity field and the second step consists of viscous evolution of vorticity (Methods like the Random-Vortex method, Vortex-in-cell, Vortex-Cloud are some examples of these methods). Invisid flows only require the normal velocity to be zero at the walls but viscous flows need an additional no-slip conditions. So, you need a viscous step and additional boundary conditions (on stream-function at the walls) if the fluid is viscous. In the method mentioned above, the viscous step is dropped (since the viscosity is very small) but the no-slip condition is retained. So, one ends up solving the inviscid equations with viscous boundary conditions. The no-slip condition is usually hard to satisfy (given the lack of a diffusive operator). So, some people have chosen a few discrete elements on the wall and obtained the required solution by minimizing the mean-square of the tangential velocities at these points. I have not seen a rigorous experimental validation of this approach. I am convinced that it will give you the right lift coefficient but I am not sure whether the drag would come out right. If it does, it may be laminar drag or turbulent drag. It will perhaps be turbulent drag if the vorticity field attains a state of chaos (known to happen in point-vortex systems).

This apporach does not seem mathematically sound. But, so is the conventional method used to predict lift around airfoils using inviscid flow equations and the Kutta condition. Constant density inviscid flows are incapable of producing vorticity by themselves. If the initial vorticity is zero in a domain, non-zero vorticity can never be produced inside it. So, you would end up with zero force on any body (zero lift and drag). For airfoils, one imposes a condition that the trailing edge is a separating (stagnation) point (which is the Kutta condition). This condition is justifiable only because any real fluid (i.e., viscous fluid) can not bend around a sharp (infinite curvature) corner. When the Kutta condition is imposed, the airfoil generates vorticity and as a result, lift. The fact remains that inviscid equations are not fully sufficient to predict lift and have to be supplemented with a condition that is a viscous consequence. The approach described above is similar.

NOTE : For flows around real aircraft, there is some skin friction drag. But for most aircraft (high subsonic/transonic), the lift related inviscid drag (due to the wing-tip vortices) is larger. This is more the case with flows with supersonic pockets since the wave drag adds to the inviscid drag. So, it doesn't help a great deal to minimize skin friction. That's why Boeing did the easiest thing to reduce drag. They added the winglets on 747 to reduce lift-induced drag.
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