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difference between isotropic and homogeneous in turbulence

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Old   October 28, 2010, 22:13
Default difference between isotropic and homogeneous in turbulence
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Some literature wrote that 'One might argue that no real turbulent flow is isotropic or even homogeneous in the large scales.'
It sounds that homogeneous is stricter than isotropic!
SO who can give me any hints about the difference between isotropic and homogeneous?

Any comments will be appreciated.
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Old   October 29, 2010, 02:10
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Dear Dut_thinker,

Isotropic turbulence demands that there is no mean shear, rotation or buoyancy effects in the flow as this can lead to anisotropy. Homogeneous turbulence is indicative of the fact that there are no mean flow gradients. In a more simpler sense, homogeneity deals with invariance in translation, isotropy deals with invariance in rotation. Take a look at MIT OCW notes on turbulence if you wish for more information.

Regards,

Ganesh
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Old   October 29, 2010, 14:26
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http://books.google.com/books?id=HZs...page&q&f=false
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Old   October 29, 2010, 14:48
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Thanks f-w! This link is very helpful! I'm typing the information below in case the book disappears.

In homogeneous turbulence, the fluctuating velocity field u(x,t) is statistically homogeneous. It is consistent with this definition for the mean velocity gradients d(Ui)dxj to be non-zero but uniform. A good approximation to homogeneous turbulence can be achieved in wind-tunnel experiments and homogeneous turbulence is the simplest class of flows to study using DNS.

A statistically homoegeous field U(x,t) is, by definition, statistically invariant under translations (i.e. shifts in the origin of the coordinate system). If the field is also statistically invariant under rotations and reflections of the coordinate system, then it is (statistically) isotropic.

Reference: Turbulent Flows by S. B. Pope
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Old   October 31, 2010, 00:34
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An isotropic turbulent flow as other members have already replied, is basically a turbulent flow where the time averaged turbulent quantities ( like u r.m.s.) have the same value at each and every location.

A homogeneous flow on the other hand is one where the turbulent quantities at any given location are the same in all the directions. (ex. u r.m.s = v r.m.s.)

example a concrete wall may be isotropic if the concentration of steel and cement is the same at all locations in the wall. But it is not homogeneous because at any point, there may be steel in one direction and cement in the other direction.

Of course it is actually impossible to get either homogeneous or isotropic turbulence in a wind tunnel. But you can get close to it and ppl want to get close to it cause theoretically a homogeneous isotropic turbulence is very attractive.

And ya it is much more impossible to get actual 100% isortopic turbulence cause turbulent flows are by definition dissipative. So the turbulent quantities ( like u r.m.s.) will decay into heat energy with time. Unless you the right amount of energy at the right time it will not be isotropic.
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Old   January 2, 2013, 00:11
Default Mistake maybe?!
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simulationman, Is definition of homogeneous and isotropic in your answer switched maybe?
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Old   January 2, 2013, 12:10
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just to make an example, often the turbulence in channel flow is simulated, the test is composed by two parallel plates that are replicated by periodic boundary conditions. The flow is substained by a forcing pressure gradient.
This flow is simultaneously homogeneous along the two directions of periodicity and inhomogeneous in the direction normal to walls
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Old   November 8, 2015, 00:44
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Homogeneous turbulence: invariance in translation
Isotropic turbulence: invariance in rotation
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Old   January 23, 2021, 07:15
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Quote:
Originally Posted by Jade M View Post
Thanks f-w! This link is very helpful! I'm typing the information below in case the book disappears.

In homogeneous turbulence, the fluctuating velocity field u(x,t) is statistically homogeneous. It is consistent with this definition for the mean velocity gradients d(Ui)dxj to be non-zero but uniform. A good approximation to homogeneous turbulence can be achieved in wind-tunnel experiments and homogeneous turbulence is the simplest class of flows to study using DNS.

A statistically homoegeous field U(x,t) is, by definition, statistically invariant under translations (i.e. shifts in the origin of the coordinate system). If the field is also statistically invariant under rotations and reflections of the coordinate system, then it is (statistically) isotropic.

Reference: Turbulent Flows by S. B. Pope
How can it be possible that for homogeneous turbulence where by definition:

\langle U(x,t)\rangle = \langle U(x+r,t)\rangle

then, the gradient

\frac{\partial{\langle U_i\rangle}}{\partial{x_j}}

be non-zero?!! I mean it MUST be zero!

Quote:
Originally Posted by ganesh View Post
Dear Dut_thinker,
Isotropic turbulence demands that there is no mean shear, rotation or buoyancy effects in the flow as this can lead to anisotropy. Homogeneous turbulence is indicative of the fact that there are no mean flow gradients. In a more simpler sense, homogeneity deals with invariance in translation, isotropy deals with invariance in rotation. Take a look at MIT OCW notes on turbulence if you wish for more information.

Regards,

Ganesh
For isotropic turbulence, we say that the statistical properties of the turbulence are invariant under rotations which means for example at a specific point in the domain we have:
\langle\bar{u^{'2}}\rangle = \langle\bar{v^{'2}}\rangle = \langle\bar{w^{'2}}\rangle
I think this issue has nothing to do with mean shear! Then why mean shear is zero?!
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Old   January 23, 2021, 08:05
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Quote:
Originally Posted by Moreza7 View Post
How can it be possible that for homogeneous turbulence where by definition:

\langle U(x,t)\rangle = \langle U(x+r,t)\rangle

then, the gradient

\frac{\partial{\langle U_i\rangle}}{\partial{x_j}}

be non-zero?!! I mean it MUST be zero!


For isotropic turbulence, we say that the statistical properties of the turbulence are invariant under rotations which means for example at a specific point in the domain we have:
\langle\bar{u^{'2}}\rangle = \langle\bar{v^{'2}}\rangle = \langle\bar{w^{'2}}\rangle
I think this issue has nothing to do with mean shear! Then why mean shear is zero?!



Plane channel flow is an example of flow homogeneous in the streamwise direction while being non-homogeneous in normal direction. It has dU/dy different from zero.
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Old   January 23, 2021, 08:56
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Quote:
Originally Posted by FMDenaro View Post
Plane channel flow is an example of flow homogeneous in the streamwise direction while being non-homogeneous in normal direction. It has dU/dy different from zero.
Thanks for your reply.
Plane channel flow is a partially homogeneous flow (in x and z directions). I think the author is talking about a totally homogeneous flow (in every direction).
Also, if the flow is homogeneous in one direction, it does not mean that velocity must have a constant gradient in other non-homogeneous directions.
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Old   January 23, 2021, 12:11
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I think these statements are the best definitions of isotropic and homogeneous turbulent flow:

A turbulent flow is said to be isotropic if:
  • rotation and buoyancy are not important and can be neglected,
  • there is no mean flow.
Rotation and buoyancy forces tend to suppress vertical motions and create an anisotropy between the vertical and the horizontal directions. The presence of a mean flow with a particular orientation can also introduce anisotropies in the turbulent velocity and pressure fields.

A turbulent flow is said to be homogeneous if:
  • there are no spatial gradients in any averaged quantity.
This is equivalent to assume that the statistics of the turbulent flow are not a function of space.

[MIT OCW]
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