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desA December 6, 2006 10:54

Velocity gradient tensor
 
Hi everyone,

Has anyone done research into the velocity gradient tensor? I would appreciate a few links of recent research work in both cfd & experimental.

I have searched reasonably heavily, but so far, I have not located what I'm looking for.

Thanks so much.

desA

agg December 6, 2006 13:59

Re: Velocity gradient tensor
 
Can you be more specific? What is it about the velocity gradient tensor that you are looking for?

desA December 6, 2006 21:21

Re: Velocity gradient tensor
 
Thanks for your reply, agg.

agg wrote:

Can you be more specific? What is it about the velocity gradient tensor that you are looking for?

desA's reply:

Simply-put, has anyone investigated the velocity gradient tensor in relation to the turbulence phenomenon?

desA

agg December 7, 2006 13:55

Re: Velocity gradient tensor
 
I'm not sure if this answers your question, but as an example the velocity gradient tensor has been used in vortex detection (swirling strength) in turbulent flows. Check out this paper for more information:

"Analysis and interpretation of instantaneous turbulent velocity fields" Adrian, Christensen, Liu

Experiments in fluids 29 (2000) 275-290.


Pennysworth December 7, 2006 19:14

Re: Velocity gradient tensor
 
What are you talking about man?!

The velocity gradient IS the main source of turbulence. Look at the production term of Reynolds stresses. Half the contraction of that is the production of turbulent kinetic energy. Gradients in velocity feed the turbulence.


desA (diaw) December 7, 2006 21:41

Re: Velocity gradient tensor
 
Thanks agg & Pennysworth.

Thanks for the link agg. I'd seen a similar reference around 1998 to Liu's work. I'll try & download the reference you've provided. If you perhaps have a copy, I'd be grateful to receive it as my e-mail address. :)

Based on some of my current research, I would say that you may both be correct - vortex motion & the turbulence phenomenon seem to be connected with the velocity gradient tensor, in some way.

Based on earlier work, I am expecting the velocity gradient tensor to be responsible for a jump mechanism. I'm currently calling this the 'flipper' mechanism. I'd love to see what findings other folks have discovered.

desA (diaw)

tom December 8, 2006 12:39

Re: Velocity gradient tensor
 
Hi,

You can have a look at Chong et al. (JFM 357), Ooi et al. (JFM 381) and perhaps some other papers by Cantwell in J. of Fluid Mech. It has been used quite heavily in studies on vortices etc.

agg December 8, 2006 14:17

Re: Velocity gradient tensor
 
What is your e-mail id?

opaque December 8, 2006 17:02

Re: Velocity gradient tensor
 
Dear desA,

From what I understand, the gradient of the velocity vector produces (in 3D) a 3x3 matrix of derivatives. These matrix is not a tensor in its own right because is not "objective" or "frame indifferent"..

In fluid mechanics, we usually refer to the symmetric part of the velocity gradient, Sij = 0.5 [grad (u) + grad^T(u)].. This one is objective and frame indifferent. I imagine you are referring to Sij.

From here, Sij is very much the backbone of the stress behavior in fluid mechanics:

The constitutive equation for stress and velocity is made a function of Sij, and/or its invariants.

1 - Newtonian fluid is an example

2 - Generalized Non-newtonian fluids use the "shear strain rate" as the measure of gradient of velocity.. The "shear strain rate" is just the second invariant of Sij for incompressible fluids.

3 - Pennysworth just mentioned that "velocity gradient" is the main source of turbulence production.. Though correct is just part of the story.. The time derivative of a second order tensor, such as u_iu_j, in an Eulerian frame requires two additional terms to satisfy the "frame indifferent" constraint (regardless of turbulence).. For example, the transport equation for the "elastic part of the stress" in a viscoelastic fluid will be something like

Tensor2_dot + u.grad(Tensor2) +/- (grad(u) Tensor2 + grad(u)^T Tensor2) = sources

the first term is the local time derivative, the second is the advected part and the third is the production of stress due to velocity gradients.. The third exist just to satisfy mathematical constrains, regardless of the meaning of the variable Tensor2 (second order tensor)

I find the following links very helpful:

http://www.grc.nasa.gov/WWW/K-12/Num...2002211716.pdf

In this one, you will find that not all 3x3 matrices are second order tensor, but all second order tensor can be represented by 3x3 matrices..

http://endo.sandia.gov/SEACAS/Docume..._Continuum.pdf

Check for the Frame Indifference section..

Good luck,

Opaque

desA (diaw) December 8, 2006 21:18

Re: Velocity gradient tensor
 
Agg wrote: What is your e-mail id?

desA replies:

e-mail: momentumwaves@gmail.com

Thanks so much agg.

desA (diaw)


desA (diaw) December 8, 2006 22:20

Re: Velocity gradient tensor
 
Thanks very much 'tom' for those references.

desA

desA (diaw) December 8, 2006 23:07

Re: Velocity gradient tensor
 
Hi opaque. Thanks so much for your very detailed input & the references. The NASA reference has a nice approach, especially with the 'action' of a 2nd order tensor on a vector. The questions around what defines a tensor, & the velocity gradient are interesting, in that there do seem to be a few different approaches in the literature.

opaque wrote:

From what I understand, the gradient of the velocity vector produces (in 3D) a 3x3 matrix of derivatives. These matrix is not a tensor in its own right because is not "objective" or "frame indifferent"..

desA replies:

Definition of a tensor:

In terms of the definition of what constitutes a tensor, a useful reference is:

Borisenko A.I., Tarapov I.E., 'Vector & Tensor Analysis with Applications", Dover Publications,1968.

Mase G.T., Mase G.E., "Continuum Mechanics for Engineers", CRC Press, 2ed, 1999.

The dyad product of two vectors provides a tensor (Borisenko, pg 64), based on rotational transformation around the origin of the reference coordinate system. This falls in line with the ?accepted? definition of a tensor. This tensor is rotationally invariant around the origin of the reference frame origin.

Moving onto the dyad product of nabla & velocity, we then have a tensor nabla_v. Mase refers to this as the 'vector gradient' (Mase, pg 35). I've also seen references to nabla_v' being the transpose of the 'velocity gradient matrix'.

So, nabla_v would appear, by definition, at its most primitive level, to be a tensor.

-----------

Invariance & material frame indifference:

Mase (pg 195 - ) has a good look at this concept, in terms of the suitability of constitutive equations. They introduce a number of additional invariance concepts, to which not all tensors will conform. Certain combinations of tensors can be formed to provide this additional invariance requirements. This would fall in with your point 3.

-----------

I'm currently researching nabla_v & its role, if any, in the (momentum-)wave phenomena I've been observing. There is also an interesting apparent link between vortices & waves/oscillations.

Thanks so much for your input.

desA


Tom December 9, 2006 07:44

Re: Velocity gradient tensor
 
A couple of minor points S_ij is usually the deviatoric stress D_ij is the symmetric part of the velocity gradient - in general they are not the same thing.

For frame indifference (see "Continuum mechanics" by Chadwick) all that is required, for a fluid, is that the stress tensor should be a function of D_ij. The Cayley-Hamilton theorem then gives the general result

sigma = -pI + f_0.D + f_1.D^2

where sigma is the stress tensor, p is the pressure and f_0, f_1 are functions of the 3 invariants of D (i.e. tr(D), det(D) and adj(D); tr(D)=0 for an incompressible fluid). This is the general form for the stress tensor for a fluid (usually called a Reiner-Rivlin fluid).

with f_1 = 0 we have

f_0 = const. Newtonian fluid, f_0 proportional to the square root of adj(D) is the Smagorinsky subgrid turbulence model used in LES.

Similar arguments hold for a solid with the velocity gradient replaced by the deformation gradient. The difference between a fluid and a solid is that particle labels in a fluid are interchangeable while in a solid they are not (a consequence of this is Kelvins circulation theorem).

desA (diaw) December 9, 2006 11:36

Re: Velocity gradient tensor
 
Thanks Tom for your excellent insights. Always very much appreciated. (I was hoping you'd look in)

Mase covers some of this, although in not as much detail, in terms of the constitutive equation for the viscous stress tensor (p 286). Again on p288 as the stress tensor.

With Stokes condition & incompressible fluid (isochoric motion) we lose a few more terms, until we finally arrive at the simple form of the incompressible N-S (as we all well know).

dv/dt+ v o nabla_v = n*nabla2_v - (1/d)nabla_p + a

where: dv/dt partial derivative a = body acceleration d = density (constant property fluid) n = kinematic viscosity

Now, the convection term at v.nabla_v is straightforward, but, what of the delta_v (nabla2_v) term?

Can we use:

nabla2_v = nabla o nabla_v ?

ie. v_i,jj = (v_i,j),j

If so, then we seem to have two sets of nabla_v terms - one for convection & the other for the diffusion term (Eulerian form).

Have I missed something glaringly obvious in my approach? I'd welcome input here.

desA (diaw)


Tom December 9, 2006 12:53

Re: Velocity gradient tensor
 
"Can we use:

nabla2_v = nabla o nabla_v ?

ie. v_i,jj = (v_i,j),j"

Strictly speaking the viscous term is

div( gradv + (gradv)^T )

which is, in component form

( v_i.j + v_j,i ),j = v_i,jj + v_j,ij

For an incompressible fluid v_j,j = 0 so the last term vanishes yielding your formula. Basically, as pointed out in an earlier post, v_i,j is not a tensor since it is not invariant under reflections (i.e. moving from a left handed system to a right handed one). The reason why it's not invariant is best seen by writing

v_i,j = (v_i,j + v_j,i)/2 + ( v_i,j - v_j,i )/2

The first term on the right is the symmetric part of the velocity gradient which is invariant under the orthogonal group O(3) while the second is the anti(or skew)-symmetric part which is not invariant with respect to reflections. Basically this last term is related to the rotation of the fluid (vorticity) and so when viewed in a mirror changes sign. Since the stresses must be observer independent they cannot depend upon the anti-symmetric part of the velocity gradient.

Strictly speaking div is only defined for Tensors and it is defined via, for some tensor T and any choice of constant vector a,

[div(T)].a = div(Ta),

which is not the same as your formula! Try appying this formula to T_ij = u_i,j and T_ij = u_i,j + u_j,i - only one of them gives the correct answer.

"Continuum Mechanics: Consise theory and problems" by Peter Chadwick is worth looking at.

Hope my ramblings aren't too confusing, Tom.

desA (diaw) December 10, 2006 05:48

Re: Velocity gradient tensor
 
Thanks for your input, Tom. As always, lots of food for thought. While you are on-thread, I'd like to probe a little deeper, if I may.

(Tom): Basically, as pointed out in an earlier post, v_i,j is not a tensor since it is not invariant under reflections (i.e. moving from a left handed system to a right handed one).

(desA): Fair-enough. So, in order to strictly be a tensor, it seems that it has to be invariant under *both* rotation & reflection. (I had been considering the rotation definition.)

A few thoughts: Would v_i,j be classed as a pseudo-tensor - if it is invariant under basis rotation (I'll re-check the basis rotation)? Although this system is not a tensor, can the principal values & principal directions still be computed (per usual approach)?

If so, would there be any significance in the eigenalues becoming complex.

What part does nabla_v play, if any, in the creation of jump/shock phenomena in the Euler equations? In other words, what drives the shock - pressure, velocity, or both together?

------------

(Tom): Strictly speaking div is only defined for Tensors and it is defined via, for some tensor T and any choice of constant vector a,

[div(T)].a = div(Ta),

which is not the same as your formula! Try appying this formula to T_ij = u_i,j and T_ij = u_i,j + u_j,i - only one of them gives the correct answer.

(desA): This, I had not come across, or certainly hadn't thought about. I'll work through the implications. Excellent.

So, basically, it is not possible to decompose nabla2_v => nabla o nabla_v, since nabla_v is not a tensor. Fair enough.

-----------

(Tom): "Continuum Mechanics: Concise theory and problems" by Peter Chadwick is worth looking at.

(desA): I see it's a Dover publication. I'll get one on order. Thanks so much for the reference.

----------------

(Tom): Hope my ramblings aren't too confusing, Tom.

(desA): I'm always very grateful for your wise input. These e-discussions have been invaluable in shaping a large number of my academic sorties along this research path. I will always be extremely grateful for your kindness. (If anyone rambles on, it's me :)

desA

Tom December 10, 2006 10:14

Re: Velocity gradient tensor
 
"(desA): Fair-enough. So, in order to strictly be a tensor, it seems that it has to be invariant under *both* rotation & reflection. (I had been considering the rotation definition.) "

Basically Tensors fall into 2 categories (1) those invariant under the rotation group SO(3) and (2) those invariant under reflections and rotations O(3). The decomposition of the velocity gradient into

L_ij = u_i,j = (u_i,j+u_j,i)/2 + (u_i,j-u_j,i)/2 = D_ij + W_ij

creates a symmetric and skew-symmetric tensor. D_ij is the deformation tensor which is invariant under O(3) while W_ij is the spin tensor which is invariant only under SO(3). Since the physics must be invariant under the larger O(3) group the stress tensor must only depend upon D_ij. As pointed out above by the other "tom" the W_ij term is important in vortex dynamics (as should be obvious from its form).

It was incorrect of me to say that L_ij was not a tensor - I meant that physically it's only the symmetric part that is important in the stresses because the Navier-Stokes equations are in covariant form so that invariance under reflections is important. I think your problem is that your contracting the wrong index in your div; e.g. div(T) = T_ij,i not T_ij,j !

Hope this clears things a bit (it's a long time since I studied tensors - close to 20 years).

"What part does nabla_v play, if any, in the creation of jump/shock phenomena in the Euler equations? In other words, what drives the shock - pressure, velocity, or both together?"

It's the density equation (coupled to the pressure via the equation of state) that causes the shock - this is why incompressible flows do not have shocks. The pressure-density coupling is why the speed of sound, and hence Mach number, is so important in shock formation. The different types of jump condition are discussed in Chadwicks book but only the shock-wave is possible in a compressible fluid. The best place to find out more about shock waves is the book by Courant & Friedrichs or, the cheaper, "High speed flow" by C.J. Chapman.

desA (diaw) December 10, 2006 10:50

Re: Velocity gradient tensor
 
Thanks again Tom for your input. Things are now a lot more clear.

(Tom): It was incorrect of me to say that L_ij was not a tensor - I meant that physically it's only the symmetric part that is important in the stresses because the Navier-Stokes equations are in covariant form so that invariance under reflections is important.

(desA): This makes sense.

(Tom): I think your problem is that your contracting the wrong index in your div; e.g. div(T) = T_ij,i not T_ij,j !

Hope this clears things a bit (it's a long time since I studied tensors - close to 20 years).

(desA): From direct tensor notation development, I'm getting - substituting for d_ij onto either nabla, or T_ij, it comes out the same way (developed in cartesian tensor form):

div(T) = T_ji,j *e_i (vector)

I'll get down to settling this one in my head during the next few days. Thanks so much for the pointers. (I'm pretty much hooked on tensors at the moment - they are so useful).

(desA): "What part does nabla_v play, if any, in the creation of jump/shock phenomena in the Euler equations? In other words, what drives the shock - pressure, velocity, or both together?"

(Tom): It's the density equation (coupled to the pressure via the equation of state) that causes the shock - this is why incompressible flows do not have shocks. The pressure-density coupling is why the speed of sound, and hence Mach number, is so important in shock formation. The different types of jump condition are discussed in Chadwicks book but only the shock-wave is possible in a compressible fluid. The best place to find out more about shock waves is the book by Courant & Friedrichs or, the cheaper, "High speed flow" by C.J. Chapman.

(desA): I've not worked in the compressible region, but would guess that the variable density will also enter into the Continuity equation, basically turning it into a pressure form. If so, then viola - a pressure convection-type wave equation, driven along by nabla_v! If so, this makes perfect sense now & I can see in my mind's eye how this all strings together.

Thanks for the Courant & Friedrichs, Chapman links. I've got access to the first one, if I remember our library correctly.

---------

Oscillation-type waveforms:

Now, that we have firmly established that incompressible flows cannot evidence pressure-type waves, due to fixed density, would it, in your opinion, be possible for these flows to show oscillation-type solutions? If so, then I may just happen to be in some luck, & my previous year's research has all been worthwhile :)

If oscillation solutions are possible, could these be considered as time-decaying wave-forms? These should display resonance-type & vibration-type effects. If it does seem plausible for these wave-forms to be evidenced, then I'm going to clean up my current paper I'm working on & submit it promptly. :)

desA


desA (diaw) December 10, 2006 12:29

Re: Velocity gradient tensor (errata)
 
Correction:

(desA):... If so, then viola - a pressure convection-type wave equation, driven along by -div(v)!


Tom December 10, 2006 16:31

Re: Velocity gradient tensor
 
"Now, that we have firmly established that incompressible flows cannot evidence pressure-type waves, due to fixed density, would it, in your opinion, be possible for these flows to show oscillation-type solutions?"

Yes this is well known. A simple example is Taylor-Couette flow when instability is lost through a Hopf bifurcation. Another example is ABC flow which can also loose stability via a Hopf bifurcation. There are lots of examples of this in the literature on weakly nonlinear stability theory.

"If oscillation solutions are possible, could these be considered as time-decaying wave-forms? These should display resonance-type & vibration-type effects..."

Strictly speaking if they are decaying they won't exhibit resonance. Resonance requires mode interaction at a bifurcation point; i.e. a point where stability is lost to two or more modes whose wavenumber and/or frequency are integral multiples. You can get transient growth via this mechanism but it does not result in finite amplitude equilibrium solutions.

"(desA): I've not worked in the compressible region, but would guess that the variable density will also enter into the Continuity equation, basically turning it into a pressure form. If so, then viola - a pressure convection-type wave equation, driven along by nabla_v! If so, this makes perfect sense now & I can see in my mind's eye how this all strings together."

I think you're referring here to the well-known acoustic wave equation. Lighthill did something like this in the 50's or 60's. For a barotropic fluid P=f(rho) and so rho can be eliminated from the continuity equation to get an evolution equation for P (alternatively you can write rho=g(P) and eliminate P from the momentum equations).


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