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December 13, 2000, 10:15 
Turbulent lengthscale

#1 
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I am wondering if the lengthscale of turbulence (l=k^(3/2)/eps) is physical variable? So it is possible to estimate size of eddies in turbulent fluid flow by calculating lengthscale with previous equation?


December 14, 2000, 06:06 
Re: Turbulent lengthscale

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From the turbulence course at the university I recall that the length scale you refer to is caracteristic for the larger eddies responsible for the main part of the turbulent transport/diffusion. Other important length scales: the Taylor microscale is frequently said to be representative for the smaller eddy sizes in which most of the dissipation occurs, and the Kolmogorov length scale represents the smallest eddies that can exist in the flow.
Hope this answers your question. Lars Ola 

December 14, 2000, 07:42 
Re: Turbulent lengthscale

#3 
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That was exactly what I was looking for. Thank you Lars Ola.


December 14, 2000, 08:15 
Re: Turbulent lengthscale

#4 
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Near the wall :
l = Kappa * y eps = u_f^{3} / ( Kappa*y ) k = u_f^{2} / sqrt(c_mu) so : l = c_mu^{3/4} * k^{3/2} / eps but this lenght scale is the mixing lenght scale, that's mean it is the lenght scale which correspond to the "eddies" of maximum energy. You musn't forget that the turbulence spectrum is continuus. 

December 14, 2000, 10:22 
Re: Turbulent lengthscale

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The length scale you have mentioned is the largest scale and will be present only in the fully developed turbulence region where "production = diffusion" (equilibrium condition) hypothesis is valid and flow is isotropic. In general, turbulence contains a galaxy of length and time scales. The basic scale is the Kolmogorov, which is the smallest and this is where dissipation of T.K.E. takes place. I am sorry to disagree with Lars, who says that the "most of the dissipation occurs" at the Taylors micro scale.
Yes, you can estimate the size of the eddies only if they are of this size, but not, if the eddies which are smaller then this. GS 

December 14, 2000, 11:59 
Re: Turbulent lengthscale

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Perhaps the correlation length scale is a better estimate of the size of the large eddies. I am not sure but there could be a direct relationship between the length scale computed using velocity correlations and one you have mentioned in fully developed, isotropic turbulence. In general, however, the correlation length scale depends on the size of the coherent structures (if any) but the length scale you have defined depends purely on the turbulent component of the flow. Good example is a turbulent with vonKarman vortex shedding. The shedded vortices are pretty coherent and are not part of the turbulent component of the flow but they can affect the computation of the correlation length scale (since correlations are affected by the presence of these vortices.


December 14, 2000, 15:06 
Re: Turbulent lengthscale

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hi GS,
Lars is correct. The maximum of dissipation occurs near the Taylor Microscale and not at the Kolmogorov microscale. chidu.. 

December 14, 2000, 15:32 
Re: Turbulent lengthscale

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The length scale at which the dissipation is maximum can easily be computed if you have the energy spectrum for the flow. Though it is flow dependent, the assumption of flow universality (in the inertial and dissipation range of scales) can give some idea about the numbers. If you use the Pao's energy spectrum (which gives the combined energy spectrum for the inertial and the dissipation ranges), the wavenumber at which the dissipation spectrum is maximum turns out to be about 56 times less than the Kolmogorov wavenumber. So the length scale of maximum dissipation is about 56 times the Kol. length scale. I am not sure if this is close to the Taylor scale.
PS: For interpretation of Taylor microscale, see the text book by Tennekes and Lumley. The explanation may not be fully clear but I have not come across one which is fully satisfactory. 

December 14, 2000, 15:36 
Re: Turbulent lengthscale

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Kalyan,
that seems correct to me. The Taylor microscale is indeed in that ball park. Hinze's book has a very good part on the various regions of the 3d energy spectrum. We are in agreement here that the maximum dissipation occurs at length scales quite larger then the Kscale. It is easy to see that because, the amount of energy present at the Kscale itself is miniscule. regards, chidu... 

December 14, 2000, 17:45 
Re: Turbulent lengthscale

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I don't know if this helps clarify or adds confusion, but Tennekes and Lumley state on the top of p. 68:
"The Taylor microscale is thus not a characteristic length of the strainrate field and does not represent any group of eddy sizes in which dissipative effects are strong. It is not a dissipation scale, because it is defined with the assistance of a velocity scale which is not relevant for the dissipative eddies." On p. 271 they show that the peak in dissipation occurs in simple "inertial subrange" type turbulence at a length scale equal to 5 times the Kolmogorov scale, a result that is evidently Re independent. On the other hand, the relationship between the Taylor microscale and the Kolmogorov scale is Re dependent (see p. 68 again). All of which seems to indicate that the Kolmogorov, not the Taylor, scale is the appropriate physical length scale associated with dissipation, and the fact that it is 5 (or 6) times the defined scale seems to be unimportant. Keith 

December 14, 2000, 23:07 
Re: Turbulent lengthscale

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Dear Chidu,
I think Keith has given a fitting reply to your argument. I would also recommend you to look at page no. 6768. of the same book(i.e. A first course in turbulence). Taylors microscale has nothing to do with dissipation. Pl. correct me if I am wrong. GS 

December 15, 2000, 04:32 
Re: Turbulent lengthscale

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So do I.
If one looks at the turbulence energy spectrum and at its equation in the case of a homogeneous decaying flow, he will find that the wave length (1/length scale) of maximum energy K_max corresponds to eps/k^(3/2), the Kolmogorov wave length K_kol corresponds to (eps/nu^(3))^(1/4), and that he could define a wave length K_tra which correspond to the last zero of the transfert term and is in between of this two previous wl and defined by : K_tra = sqrt(eps/(nu*k^(2))), I think that this wl corresponds to the Taylor microsacle. And one gets : K_max / K_eta = 1/(Re_t)^(3/4) K_tra / K_eta = 1/(Re_t)^(1/4) where Re_t = (k^2)/(nu*eps) Sylvain 

December 15, 2000, 04:33 
Re: Turbulent lengthscale

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I almost expected some controversy on this, among other things because my lecture notes also state that opinions tend to go apart on the interpretation of the Taylor microscale
The statement that this length scale is "representative for the smaller eddy sizes in which most of the dissipation occurs" is followed in my notes by a reference to a publication from 1975 by Hinze. However, later on, the same publication is referred to when stating that "...one can also define a length scale at the maximum of the dissipation spectre". Although I haven't studied the matter thoroughly, I cannot recall having seen any firm statement that the two length scales are actually identical, only that the Taylor scale is "representative"... As such, I believe I was consistent in my posting. So what is the truth? I don't know, but my impression is that the Taylor microscale is a length scale that is used more because it is conventiently defined (and happens to be "representative" of the dissipation spectre maximum) than because it is physically significant. Best regards to you all, Lars Ola 

December 15, 2000, 10:58 
Re: Turbulent lengthscale

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Hi Lars,
I am still disagree with your argument. Why don't you refer to the book by "Uriel Frish" Turbulence. Look at the page nos. 9192 "6.3.2 Effect of a finite viscosity: the dissipation range". On page no. 92 it is clearly mentioned that the "range of scales comparable to or less than "eta"(Kolmogorov dissipation scale = (nu**3/epsilon)**(1/4)) is called the dissipation range". In this range the energy input from nonlinear interactions and the energy drain from viscous dissipation are in exact balance. Which means this the scale (i.e. Kscale) where max. dissipation will occur. GS 

December 15, 2000, 11:13 
Re: Turbulent lengthscale

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Hi GS and others,
This impression that the Taylor MS is representative of the maximum dissipation was something that I vaguely had in my mind. I went back and looked in Hinze. Although not totally clear, I think I was wrong on that. Looks like the only relevant scale in the universal equilibrium range is the Kolmogorov scale. The point I made about the maximum dissipation occuring at a length scale much larger than the Kscale is correct. As pointed out by Keith and Kalyan it is approximately 5 times for the Pao energy spectrum. However, the fact that this ratio is independent of the Reynolds number (alluded to by Keith) is not established beyond doubt. Hinze cites some experimental data where it is seen that this ratio varied from 210. The only association between the Taylor MS (TMS) and dissipation I found is with regard to the final decay period for isotropic turbulence. Here TMS does enter into the picture. In this context Batchelor refers to it as the dissipation length parameter. He adds a footnote that TMS is NOT the length scale of eddies responsible for most of the dissipation. I hope this clears up some of the confusion! regards, chidu... 

December 15, 2000, 11:53 
Re: Turbulent lengthscale

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Hi Chidu,
Still you are partly wrong or right!!! Look at "Turbulence by Uriel Frisch" p.no. 91. We have turbulence energy production eddies, energy containing eddies (inertial range) and energy dissipation eddies. Earlier the inertial range was thought to be of the order of Taylor scale. It has been shown in the K41 framework that, the inertial range actually extends down to the Kdissipation scale. And dissipation takes place in the range of scales comparable to or less than Kscale. GS 

December 15, 2000, 12:20 
Re: Turbulent lengthscale

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Hi GS,
I think there is some misunderstanding here. From my previous email, we can throw away the Taylor MS from the discussion about the maximum dissipation wavenumber. As I said, the only relevant length scale in the dissipation region appears to be the Kscale, the definition of which is standard. All I am saying is that using the same definition, experimental data shows that the wavenumber at which maximum dissipation occurs (let me qualify by assuming isotropic turbulence; for example from ComteBellot and Corrsin's paper) is around 210 times smaller than the Kwavenumber. Translating to lengths, the maxdiss length scale is 210 times larger than Kscale. This does not imply that there is another independent length scale. It just means that there is a multiplicative factor in front of the Kscale. The scaling remains the same. This wavenumber is said to exist in the inertialsubrange. regards, chidu... 

December 18, 2000, 02:18 
Re: Turbulent lengthscale

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So here we are, throwing different literature references at each other. Clearly, not even people who work in this fascinating field of research agree on the issue, or perhaps it is a matter of interpretation or presentation of the research results?
Anyway, it seems established beyond doubt that the length scale that "controls" the dissipation is the Kolmogorov scale. But does anyone know if the Taylor microscale has a clear physical meaning? Regards, Lars Ola 

December 18, 2000, 03:41 
Re: Turbulent lengthscale

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(1). In Wilcox's book, he stated " Lambda*Lambda=mean( u'*u')/ mean( (du'/dx)*(du'/dx) )" as the statistical definition of Lambda, the Taylor microscale. (2). He also stated for high Reynolds number flow, eta << Lambda << l. (3). Then he stated that Lambda is not related to any meaningful range of eddy sizes. (4). So, I think, it is purely a mathematical expression. And people are trying to interpret it under special conditions. It is like saying " July/(11 months)". If you try very hard, you can say that " July/(11 months)" means roughly the summer time. If you are not careful, it can also be interpreted as "711 store". And I don't think that there is a physical meaning for mean(u'*u'), or mean( (du'/dx)*(du'/dx) ). These are statistical definitions only.


December 18, 2000, 03:46 
Re: Turbulent lengthscale

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That's what I thought. Thanks for the answer John. I think I'll be jumping off this string now.
Regs. Lars Ola 

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