If you compute the spectrum correctly, it should involve only one mode....

This 2D flow is not evolving (apart from viscous decay of the single initial mode) ...

No Kolmogorov, no nothing!!!

I suggest you read U Frisch's book on turbulence

http://www.amazon.com/Turbulence-The-Legacy-A-Kolmogorov/dp/0521457130

Regards

Marc

What do you mean by 2D Taylor-Green (Taylor-Green is 3D...)

Marc. Here is an example of 2D incompressible Ta-Gr vortices. I can't understand why the energy spectrum do not present a decay! I would be happy if you give me any helpful idea regarding this issue! best!! a

Having only one wavenumber : k_x=k_y= \pm 1

your initial data is an exact solution of Euler equation...

So no evolution is expected...

[for Navier-Stokes you should only get viscous decay of the initial data]

The reason is that the stream function \Psi with (u,v)=(\partia_y \psi,-\partia_x \psi) is proportional to the vorticity \omega=curl (u,v)= -(\laplacian \Psi).

so the vorticity is invariant against advection... [this is also true for any axisymmetric stream function]

Beside this point your spectrum look weird: it should be the same as that of the initial data which should involve only 1 wavenumber...

Best

Thank you Marc for your answer. Given a 2D frozen incompressible velocity field, we can compute its 2D fest Fourier transform \hat{u} and \hat{v} and the energy spectrum can easily be determined (please have a look at the picture below).

What does this energy spectrum look like? Is there any decay following the Kolmogorov prediction? And what can we say regarding the flow state: laminar / turbulent? Sorry I can't see things clearly!! I would appreciate it if you could give some help! Thanks! best! a

For a two-dimensional flow, the vortex stretchting mechanism that produces the scale range (and thereby a Kolmogorov -5/3 slope) is absent, so you won't be seeing a "typical" turbulent spectrum for your case. What happens if you initialize a vortical dipole is either a coalesence to a large vortex (if you want: a backscattering effect) or a stable dipole rotation. For the velocity field you showed above, I would could imagine that due to the symmetries, the vortices will stay in place, and just dissipative laminarly.

So in short: a 2D TGV is not turbulent.

If you compute the spectrum correctly, it should involve only one mode....

This 2D flow is not evolving (apart from viscous decay of the single initial mode) ...

No Kolmogorov, no nothing!!!

I suggest you read U Frisch's book on turbulence

http://www.amazon.com/Turbulence-The-Legacy-A-Kolmogorov/dp/0521457130

Regards

Marc

Thank you for you both Marc and Andrea for your answers.

So just to fix the ideas. The laminar kinetic energy spectrum for a 2D incompressible flow undergoes a backscattering effect and large vortices have a tendency to form OR there is just only one mode and it is a nonsense to talk about a distribution of energy with respect to frequency! Please more clarifications!

best!

a

Hello Abdallah,

I found this video demonstrating the upward cascade in 2D flows:

http://www.youtube.com/watch?v=boWeyfYs2EQ

As you can see, from an initial (I am assuming random) distribution of small scale vortices (corresponding to energy in the high wave numbers), the flow develops into large scale structures (with energy at the low wave numbers). This effect is the opposite of the downward cascade in 3D turbulence, where (on average) the energy is transferred from large to small scales. So depending on the initial distribution of the vortices, you will see something like in the video OR (I am assuming that is the case for the symmetrically constrainted TGV) the vortices will just "spin in place".

So far, this behavior is governed by the convection terms in the governing equations. However, if you are taking the viscous effects into account, then the vortices (in 2D) will not only re-merge, but also be dissipated.

I hope this helps!

Regards,

Andrea

thank you Andrea for your powerful answer as well as for the link to the video. I have just leaved two comments there! I can realize now that for 2D incompressible flows, there is a possibility that large vortices have a tendency to form in the detriment of the smaller ones. I have just run an other simulation in a square box with 2049 collocation points but with a random noise and I obtained the same feature.

Thank you again for your answers and clarifications!

best!!

a

Some remarks if you are still interested in this topic: the first paper on the Taylor-Green vortex array was by Taylor G. I., in Philos. Mag., 46 in 1923 and did deal with the two-dimensional version, although nowadays the 3D version is better known. The 3D version was only published in the 1930s by G.I. Taylor and Green.

Any Taylor-Green vortex array is an equilibrium of the 2D Navier-Stokes equation but this does not mean that it is stable. Several studies show that they are only stable for low Reynolds numbers (and low forcing). For moderate or higher Reynolds numbers all Taylor-Green vortex arrays become unstable (only the single vortex array remains probably stable, i.e. u=-cos(k1*x)sin(k2*y) v=sin(k1*x)cos(k2*y) with k1=k2=1 and a rectangular domain with L=2*pi)). When unstable, the other arrays tend to self-organize towards a big vortex with length scale L. If you are simulating an unstable Taylor-Green vortex then it might show some energy at other length scales and, due to the above mentioned self-organization, eventually most energy will be at length scale L (Kraichnan inverse cascade instead of Kolmogorov cascade). However, if you were analyzing a stationary Taylor-Green vortex all the energy (and enstrophy) should be concentrated at the wave number (k1,k2) of the used Taylor-Green vortex.

hope this helps your understanding!

cheers

@Lauret Thank you for your answer! so the energy spectrum for a stationary Ta-Gr vortix with a single vortex array should have a discrete form rather than a continuous one. I may make a mistake in my code. Just I would like to ask you how one can define the Reynolds number in this kind of flows? I mean how one can compute the root-mean-squared velocity in this case?

best

a

given the 2D vector field (u,v) as a function of (x,y), you have to take (u,v)^2=u^2+v^2 and then take the square root sqrt(u^2+v^2). This can be done analytically for the Taylor-Green vortex. If you have a discretized version of the same vortex you can do the same with the vector field only defined on the grid points, but then you have to normalize it such that it is identical to the value computed for the analytical version.

cheers