[FluidKo]

Hello everybody~!
I'm wondering physical cause and effect between strain and reynolds stress.

In eddy viscosity model, we adopt strain into reynolds stress.

I've heard there is intuitive relation between turbulence and velocity gradient.
Following is what I know.
Let's assume that we consider fluid particle as man.
As we can see from uploaded picture, if man trips over a rock, he will tumble. In this situation we can find that velocity of head and velocity of foot will be different.
It means there is velocity gradient between head and foot.
In the same way, if there is high velocity gradient, there will be turbulence.
Velocity gradient between head and foot can represents velocity gradient between neighboring fluid particles.
Also tumbling can represents turbulence.
We can see it from vortex roll up.

I can understand physical cause and effect between turbulence and velocity gradient.
But in eddy viscosity model, we relate turbulence with strain, not with velocity gradient.

I don't understand what is physical cause and effect between turbulence and strain.
Strain includes velocity gradient over not only one direction but also another direction.
Turbulence shows vortex and vortex rotated with one direction.
It can't rotate to both direction. It is physically impossible.
But why turbulence and strain can be highly related?

Thank you~!

[FMDenaro]

Originally Posted by FluidKo

Hello everybody~!
I'm wondering physical cause and effect between strain and reynolds stress.

In eddy viscosity model, we adopt strain into reynolds stress.

I've heard there is intuitive relation between turbulence and velocity gradient.
Following is what I know.
Let's assume that we consider fluid particle as man.
As we can see from uploaded picture, if man trips over a rock, he will tumble. In this situation we can find that velocity of head and velocity of foot will be different.
It means there is velocity gradient between head and foot.
In the same way, if there is high velocity gradient, there will be turbulence.
Velocity gradient between head and foot can represents velocity gradient between neighboring fluid particles.
Also tumbling can represents turbulence.
We can see it from vortex roll up.

I can understand physical cause and effect between turbulence and velocity gradient.
But in eddy viscosity model, we relate turbulence with strain, not with velocity gradient.

I don't understand what is physical cause and effect between turbulence and strain.
Strain includes velocity gradient over not only one direction but also another direction.
Turbulence shows vortex and vortex rotated with one direction.
It can't rotate to both direction. It is physically impossible.
But why turbulence and strain can be highly related?

Thank you~!

The velocity gradient has almost nothing to do with turbulence, it is a linear term that enters by means of the diffusive flux of momentum at very small lenght scale.
Turbulence is driven by the non linear term, in a quasi inviscid way (at least until the local Re number becomes O(1)).


You are simply considering a closure model. That is the unresolvable fluctuations (a tensor that has its origin from the non linearity) are modelled in a global way, supposing that the effect mimics a supplementary viscosity flux in the momentum equation. Of course, you can substitute such model with a different one where the velocity gradient does not appear at all.

In conclusion, in turbulence the velocity gradient has only the role of dissipate energy at very small lenghts.

[LuckyTran]

If the man simply trips over a rock without falling on his face, the difference in forward speed between his head and his feet is shear motion. A cartwheel is a combination of moving forward and flipping at the same time. Both of these are different parts of the velocity gradient.

You can decompose the velocity gradient into these two symmetric and skew-symmetric parts. The symmetry part is the strain and the skew-symmetric part is the rotation.

By itself, the cartwheel and tumble is not turbulence. You can watch Guiness World Record videos of people doing cartwheels consecutively very fast or for a long time. What they are not doing is not generating turbulence, they are just doing cartwheels. The same is true for fluid flow. Simply tumbling is not turbulence, although it plays a key role in the description of turbulence.

You are correct in that a man moving forward, tripping and tumbling over a rock will generally and intuitively always tumble in the same direction and not the other way. So how does tumbling the other way happen?

Well imagine you have a line of men, a procession of them all moving forward and the same man trips. Now when he tumbles, as his feet roll upward he can kick the person behind him and cause the second man to flip backwards! As his head and arms come down he might touch the person in front of him and cause that person to also flip backwards! So when you can consider groups of people or groups of particles (i.e. a continuum like a fluid) then both can happen. Now these secondary tumbling people can initiate tumbling in their peers. Alternatively there could be no tumbling at all and the man just bumps his neighbors in front and behind which sheers the entire procession. So now you have a model for large scale momentum transport from one person tripping over a rock and imparting that effect onto his peers. The chaotic motions of this entire group of people all stumbling in different ways due to everybody tripping over a bunch of different rocks all at the same time is what leads to turbulence.

[FluidKo]

Originally Posted by FMDenaro

The velocity gradient has almost nothing to do with turbulence, it is a linear term that enters by means of the diffusive flux of momentum at very small lenght scale.
Turbulence is driven by the non linear term, in a quasi inviscid way (at least until the local Re number becomes O(1)).


You are simply considering a closure model. That is the unresolvable fluctuations (a tensor that has its origin from the non linearity) are modelled in a global way, supposing that the effect mimics a supplementary viscosity flux in the momentum equation. Of course, you can substitute such model with a different one where the velocity gradient does not appear at all.

In conclusion, in turbulence the velocity gradient has only the role of dissipate energy at very small lenghts.

I know that relation between turbulence and strain is suggested to close tubulence term.

But what I was wondering is why Boussinesq has chosen turbulence viscosity and strain?
He could choose other physical quantities that follows dimension analysis.
But why did he select turbulent viscosity and strain?
What was physical reason for selection of turbulent viscosity and strain?

I've heard that the reason why we regard reynolds stress term as diffusion term is turbulent stress works similarly to molecule viscous stress.
Similarly to molecule viscous stress, turbulent stress works as loss of flow.

But there can be other ways to model and regard turbulent stress term as loss.
Why did he select turbulent viscosity and strain?
And also turbulence is actively producted in the region with high shear strain.
So what is the physically plausible reason of Boussinesq's hypothesis that can explain relation between turbulence and shear strain?
I was wondering this.

In my intuition, I think explanation of below(line of men with procession) by LuckyTran is plausible.

[FluidKo]

Originally Posted by LuckyTran

If the man simply trips over a rock without falling on his face, the difference in forward speed between his head and his feet is shear motion. A cartwheel is a combination of moving forward and flipping at the same time. Both of these are different parts of the velocity gradient.

You can decompose the velocity gradient into these two symmetric and skew-symmetric parts. The symmetry part is the strain and the skew-symmetric part is the rotation.

By itself, the cartwheel and tumble is not turbulence. You can watch Guiness World Record videos of people doing cartwheels consecutively very fast or for a long time. What they are not doing is not generating turbulence, they are just doing cartwheels. The same is true for fluid flow. Simply tumbling is not turbulence, although it plays a key role in the description of turbulence.

You are correct in that a man moving forward, tripping and tumbling over a rock will generally and intuitively always tumble in the same direction and not the other way. So how does tumbling the other way happen?

Well imagine you have a line of men, a procession of them all moving forward and the same man trips. Now when he tumbles, as his feet roll upward he can kick the person behind him and cause the second man to flip backwards! As his head and arms come down he might touch the person in front of him and cause that person to also flip backwards! So when you can consider groups of people or groups of particles (i.e. a continuum like a fluid) then both can happen. Now these secondary tumbling people can initiate tumbling in their peers. Alternatively there could be no tumbling at all and the man just bumps his neighbors in front and behind which sheers the entire procession. So now you have a model for large scale momentum transport from one person tripping over a rock and imparting that effect onto his peers. The chaotic motions of this entire group of people all stumbling in different ways due to everybody tripping over a bunch of different rocks all at the same time is what leads to turbulence.

I've drawn cartoon of your explanation.
Is this what are you meaning?

[FluidKo]

Originally Posted by LuckyTran

If the man simply trips over a rock without falling on his face, the difference in forward speed between his head and his feet is shear motion. A cartwheel is a combination of moving forward and flipping at the same time. Both of these are different parts of the velocity gradient.

You can decompose the velocity gradient into these two symmetric and skew-symmetric parts. The symmetry part is the strain and the skew-symmetric part is the rotation.

By itself, the cartwheel and tumble is not turbulence. You can watch Guiness World Record videos of people doing cartwheels consecutively very fast or for a long time. What they are not doing is not generating turbulence, they are just doing cartwheels. The same is true for fluid flow. Simply tumbling is not turbulence, although it plays a key role in the description of turbulence.

You are correct in that a man moving forward, tripping and tumbling over a rock will generally and intuitively always tumble in the same direction and not the other way. So how does tumbling the other way happen?

Well imagine you have a line of men, a procession of them all moving forward and the same man trips. Now when he tumbles, as his feet roll upward he can kick the person behind him and cause the second man to flip backwards! As his head and arms come down he might touch the person in front of him and cause that person to also flip backwards! So when you can consider groups of people or groups of particles (i.e. a continuum like a fluid) then both can happen. Now these secondary tumbling people can initiate tumbling in their peers. Alternatively there could be no tumbling at all and the man just bumps his neighbors in front and behind which sheers the entire procession. So now you have a model for large scale momentum transport from one person tripping over a rock and imparting that effect onto his peers. The chaotic motions of this entire group of people all stumbling in different ways due to everybody tripping over a bunch of different rocks all at the same time is what leads to turbulence.

Here are more continuing cartoon.

[LuckyTran]

Boussinesq assumed that turbulence becomes decorrelated very quickly and considered that the mean of the fluctuating momentum flux, should on the average, still obey the averaged Navier-Stokes equations. In other words, Boussinesq assumed that turbulence behaved like random molecular collisions in kinetic theory. When you begin with this presumption, the form of the turbulent fluxes naturally will look exactly like the molecular diffusion terms. A very similar random walk experiment also gives the same Newton's viscosity law. Every transport of a random variable follows this similar pattern (heat conduction, fluid viscosity, binary diffusion, etc).

Where Boussinesq was wrong is: turbulence becomes decorrelated quickly, but not that quickly. It's not random.

[FMDenaro]

Originally Posted by FluidKo

I know that relation between turbulence and strain is suggested to close tubulence term.

But what I was wondering is why Boussinesq has chosen turbulence viscosity and strain?
He could choose other physical quantities that follows dimension analysis.
But why did he select turbulent viscosity and strain?
What was physical reason for selection of turbulent viscosity and strain?

I've heard that the reason why we regard reynolds stress term as diffusion term is turbulent stress works similarly to molecule viscous stress.
Similarly to molecule viscous stress, turbulent stress works as loss of flow.

But there can be other ways to model and regard turbulent stress term as loss.
Why did he select turbulent viscosity and strain?
And also turbulence is actively producted in the region with high shear strain.
So what is the physically plausible reason of Boussinesq's hypothesis that can explain relation between turbulence and shear strain?
I was wondering this.

In my intuition, I think explanation of below(line of men with procession) by LuckyTran is plausible.

LuckyTran already addressed the reason, the observation that turbulence increases the mixing is historically at the basis of an additional diffusive flux.
We are talking of a contribution to the statistically averaged velocity not to the pointwise one.
I just want to address a mathematical issue in this approach. The fluctuations tensor has its nature from the hyperbolic term and is converted in a parabolic term.

[FMDenaro]

Originally Posted by FluidKo

Here are more continuing cartoon.

I don't think you are on a right way with this example. Firs of all, you cannot think about particles if you want to explain the velocity (averaged velocity) gradient that requires a continuous volume of fluid. Then, you have the symmetric gradient with zero trace in the model, only the angle deformation of the volume is involved.

[FluidKo]

Originally Posted by LuckyTran

Boussinesq assumed that turbulence becomes decorrelated very quickly and considered that the mean of the fluctuating momentum flux, should on the average, still obey the averaged Navier-Stokes equations. In other words, Boussinesq assumed that turbulence behaved like random molecular collisions in kinetic theory. When you begin with this presumption, the form of the turbulent fluxes naturally will look exactly like the molecular diffusion terms. A very similar random walk experiment also gives the same Newton's viscosity law. Every transport of a random variable follows this similar pattern (heat conduction, fluid viscosity, binary diffusion, etc).

Where Boussinesq was wrong is: turbulence becomes decorrelated quickly, but not that quickly. It's not random.

Originally Posted by FMDenaro

I don't think you are on a right way with this example. Firs of all, you cannot think about particles if you want to explain the velocity (averaged velocity) gradient that requires a continuous volume of fluid. Then, you have the symmetric gradient with zero trace in the model, only the angle deformation of the volume is involved.

Sorry, I can't understand...

What I'm wondering is why it is possible to replace turbulent stress() into .

I can understand why Boussinesq adopts mean velocity because we have information about mean field only.
We don't have information about fluctuation that changes every single time.

But why does he suggest model with mean strain rate and turbulent viscosity?
There can be many ways to select other physical quantity.
For example,


If we satisfy dimension, there is no problem to MATHEMATICAL POINT.

But in my opinion, every modeling must have its PHYSICALLY PLAUSIBLE REASONS.
Although there can be many ways to model turbulent stress, why did Boussinesq chosen mean strain rate? not just mean velocity?
What is the physical reason that Boussinesq had no choice but to select mean STRAIN RATE?

The reason I am obsessed with mean strain ratio is because LES also adopts mean strain rate to turbulent stress.
So I've thought there will be specific reason that mean strain rate is adopted often.
Why mean strain is important in physics?

I've heard that Boussinesq has gotten inspiration of modelling with mean strain rate from molecular interaction(collision).
But I have no idea of molecular interaction(collision).
What is relation between molecular collision and mean strain rate?
And how it can be adopted to that represents turbulent and fluctuation?

Thank you~!

[FluidKo]

Originally Posted by LuckyTran

Boussinesq assumed that turbulence becomes decorrelated very quickly and considered that the mean of the fluctuating momentum flux, should on the average, still obey the averaged Navier-Stokes equations. In other words, Boussinesq assumed that turbulence behaved like random molecular collisions in kinetic theory. When you begin with this presumption, the form of the turbulent fluxes naturally will look exactly like the molecular diffusion terms. A very similar random walk experiment also gives the same Newton's viscosity law. Every transport of a random variable follows this similar pattern (heat conduction, fluid viscosity, binary diffusion, etc).

Where Boussinesq was wrong is: turbulence becomes decorrelated quickly, but not that quickly. It's not random.

Originally Posted by FMDenaro

I don't think you are on a right way with this example. Firs of all, you cannot think about particles if you want to explain the velocity (averaged velocity) gradient that requires a continuous volume of fluid. Then, you have the symmetric gradient with zero trace in the model, only the angle deformation of the volume is involved.

Oh I've gotten a new idea.
For example although in laminar flow, there is momentum exchange between two layers by the chaotic motion of molecule.
http://www.homepages.ucl.ac.uk/~uces..._Viscosity.pdf
Although we can find momentum exchange and consequent shear stress, we don't consider motion of molecule because it has microscopic scale that is not continuum.
So we consider only shear stress and that is described by viscosity and strain rate.
(Although we don't consider chaotic motion of molecule, momentum exchange is caused by motion of molecule.)

But in turbulent flow fluid moves with chaotic motion like molecule collision.
Because this motion is macroscopic that is continuum, we have to consider it .
This is expressed by .
In turbulent flow, there is mixing which looks like molecule motion in laminar flow.
And cause of mixing, there is momentum exchange between layers and this is very similar to chaotic motion of molecule.

In laminar flow, momentum exchange(shear stress) that is caused by chaotic motion of molecule is expressed by strain rate and molecule viscosity.
In the same way, for turbulent flow, momentum exchange that is caused by mixing with continuum level can be expressed by strain rate and some viscosity.

If there is strain rate in turbulent flow, there will be velocity difference because strain rate is caused by velocity gradient.
And velocity difference will be compensated by mixing.
Cause of this mixing, momentum will be exchanged between layers.
So important point of mixing and consequent momentum exchange is strain rate.
Because we select strain rate for turbulent stress, we have to select viscosity also for dimensional analysis.

I think this is why Boussinesq has selected strain rate and turbulent viscosity for modeling of turbulent stress.
What do you think about this?
Is this right?

Thank you~!

[FMDenaro]

Yes, this is the correct way to think about. On a side the macroscopic model that drives to write the molecular diffusive flux of momentum, on the other side the modelling of the unresolved fluctuation in terms of a supplementary eddy viscosity-based momentum diffusion (a contribute that enters in evaluation of the averaged velocity).

[FluidKo]

Originally Posted by FMDenaro

Yes, this is the correct way to think about. On a side the macroscopic model that drives to write the molecular diffusive flux of momentum, on the other side the modelling of the unresolved fluctuation in terms of a supplementary eddy viscosity-based momentum diffusion (a contribute that enters in evaluation of the averaged velocity).

Originally Posted by LuckyTran

Boussinesq assumed that turbulence becomes decorrelated very quickly and considered that the mean of the fluctuating momentum flux, should on the average, still obey the averaged Navier-Stokes equations. In other words, Boussinesq assumed that turbulence behaved like random molecular collisions in kinetic theory. When you begin with this presumption, the form of the turbulent fluxes naturally will look exactly like the molecular diffusion terms. A very similar random walk experiment also gives the same Newton's viscosity law. Every transport of a random variable follows this similar pattern (heat conduction, fluid viscosity, binary diffusion, etc).

Where Boussinesq was wrong is: turbulence becomes decorrelated quickly, but not that quickly. It's not random.

Thank you so much to you guys.
I couldn't understand when I've heard that inspiration of modeling for turbulent stress is based on molecule interaction.
Because we don't consider molecule interaction in fluid dynamics that considers continuum.
Also I couldn't understand when I've heard that we consider turbulent stress as molecule stress in laminar flow because I've thought turbulent flow and laminar flow are totally different.
But in terms of
momentum exchange by chaotic motion(turbulent flow: mixing, laminar flow: molecule interaction) and strain rate
and fact that momentum exchange works as stress,
they are similar.
Thanks for you guys and writer of this lecture note.
http://www.homepages.ucl.ac.uk/~uces..._Viscosity.pdf
Without you guys, I won't understand it.
Thank you~!

[LuckyTran]

A fluid element can do two things. It can move (translate and rotate). It can undergo a shape change or deformation. The bulk motion is described by velocity and angular velocity. The shape change is described by the velocity gradient tensor (if we strictly talking about 1st order deformations). The gradient tensor can be decomposed into a strain rate tensor and a spin tensor. Since Navier-Stokes is a conservation of linear momentum, the two closure parameters in your closure model will be either the velocity and/or the strain-rate tensor (the spin tensor will be in the angular momentum equation, i.e. the vorticity equation). Strain-rate tensor and velocity gradient mean pretty much the same thing in terms of phenomenon.

So why the strain-rate and not the mean velocity?

In kinetic theory you have a collection of particles each with their own energies that collide with and exchange energy with each other. Yet there is a mean bulk or total volume average energy. For example, pressure is the bulk average of linear momentum of each individual particle. Temperature is the bulk average of the kinetic energy of each particle. The ratio of mean to molecular kinetic energy is Boltzmann's constant.

When pairs of particles interact, high energy particles give their energy to low energy particles and low energy particles receive energy from high energy particles (and not the other way around because that would violate equipartition). During a transport process, the net exchange of fluctuating energy in kinetic theory is therefore driven by differences in the bulk averages. Take Fourier's law of heat conduction for example. The internal energy exchanged is proportional to the temperature difference between two points with the Boltzmann constant being replaced by the thermal conductivity. Similarly, Ohm's law replaced electrical permitivity by electrical conductance. And so on. Note that these are all archetypes of mixing length models.

Fluctuations of velocity therefore should exchange momentum proportional to the difference (i.e. gradient) of velocity with the molecular viscosity being replaced by the turbulent viscosity. If two particles come from locations that have the same bulk energy, they won't exchange anything. So it makes sense that velocity by itself isn't the cause just like two objects having the same temperature won't exchange heat. Again, a mixing model is implied for the bulk properties.


Now you still have the option of gradient, gradient squared, gradient cubed, and transcendental functions of the gradient. However, Boussinesq did what everyone likes to do, take Taylor series of everything. And so you end up with the linear strain rate relation of the reynolds stresses being proportional to the strain rate (or gradient of velocity).

[FluidKo]

Originally Posted by LuckyTran

A fluid element can do two things. It can move (translate and rotate). It can undergo a shape change or deformation. The bulk motion is described by velocity and angular velocity. The shape change is described by the velocity gradient tensor (if we strictly talking about 1st order deformations). The gradient tensor can be decomposed into a strain rate tensor and a spin tensor. Since Navier-Stokes is a conservation of linear momentum, the two closure parameters in your closure model will be either the velocity and/or the strain-rate tensor (the spin tensor will be in the angular momentum equation, i.e. the vorticity equation). Strain-rate tensor and velocity gradient mean pretty much the same thing in terms of phenomenon.

So why the strain-rate and not the mean velocity?

In kinetic theory you have a collection of particles each with their own energies that collide with and exchange energy with each other. Yet there is a mean bulk or total volume average energy. For example, pressure is the bulk average of linear momentum of each individual particle. Temperature is the bulk average of the kinetic energy of each particle. The ratio of mean to molecular kinetic energy is Boltzmann's constant.

When pairs of particles interact, high energy particles give their energy to low energy particles and low energy particles receive energy from high energy particles (and not the other way around because that would violate equipartition). During a transport process, the net exchange of fluctuating energy in kinetic theory is therefore driven by differences in the bulk averages. Take Fourier's law of heat conduction for example. The internal energy exchanged is proportional to the temperature difference between two points with the Boltzmann constant being replaced by the thermal conductivity. Similarly, Ohm's law replaced electrical permitivity by electrical conductance. And so on. Note that these are all archetypes of mixing length models.

Fluctuations of velocity therefore should exchange momentum proportional to the difference (i.e. gradient) of velocity with the molecular viscosity being replaced by the turbulent viscosity. If two particles come from locations that have the same bulk energy, they won't exchange anything. So it makes sense that velocity by itself isn't the cause just like two objects having the same temperature won't exchange heat. Again, a mixing model is implied for the bulk properties.


Now you still have the option of gradient, gradient squared, gradient cubed, and transcendental functions of the gradient. However, Boussinesq did what everyone likes to do, take Taylor series of everything. And so you end up with the linear strain rate relation of the reynolds stresses being proportional to the strain rate (or gradient of velocity).

By the way I have one more question.
I understand momenntum exchange works as shear stress in laminar flow.
But as uploaded cartoon, I can't understand why momentum exchange in turbulent flow can be loss?

What I've guess is following
Let's assume there is mixing between fluid elements in outer layer and viscous subalyer.
Although faster fluid element in outer layer interacts with fluid element in viscous sualyer, velocity of fluid element in viscous sublayer won't be changed because in viscous sublayer, there isn't active velocity change by strong molecule(laminar) viscosity.

But I'm not sure whether I'm right or not.
Because upper my guess can be applied to situation of mixing between fluid element in outer layer and fluid element in viscous sublayer.
But there can be mixing between fluid elements that all of those are in same outer layer.
In this case I think I can't apply my guess.

What is right thing?
Why momentum exchange can work as shear stress(loss) eventhough there is no change of momentum?

P.S_1
I've guessed following also.
In turbulent flow, there is always ACTIVE mixing.
Cause of active mixing, flow elements are always mixed to nearby viscous sublayer.
So strictly there is no actual momentum loss by mixing between fluid elements in outer layer.
But event that fluid elements are mixed to viscous sublayer always exist so there is momentum loss.
I'm not sure. What is right?

P.S_2
I think the reason why I'm confused is following.
Shape of velocity profile of turbulent flow and laminar flow are diffrent.
In laminar flow, although there is mixing by molecule interation, alwyas there is velocity gradient.
Eventhough there is mixing and consequent momentum exchange, velocity depends on location.
Near wall region is always solver that the region far from the wall.
So I can understand that momentum of fluid that is far from the wall is corroded by the fluid that is nearby wall.

But in turbulent flow, I think there is no velocity difference between 2 mixed points after mixing as we can see from my uploaded picture(momentum exchange in outer layer).
What is wrong of my conjecture?

Thank you~!

[FMDenaro]

Originally Posted by FluidKo

By the way I have one more question.
I understand momenntum exchange works as shear stress in laminar flow.
But as uploaded cartoon, I can't understand why momentum exchange in turbulent flow can be loss?

What I've guess is following
Let's assume there is mixing between fluid elements in outer layer and viscous subalyer.
Although faster fluid element in outer layer interacts with fluid element in viscous sualyer, velocity of fluid element in viscous sublayer won't be changed because in viscous sublayer, there isn't active velocity change by strong molecule(laminar) viscosity.

But I'm not sure whether I'm right or not.
Because upper my guess can be applied to situation of mixing between fluid element in outer layer and fluid element in viscous sublayer.
But there can be mixing between fluid elements that all of those are in same outer layer.
In this case I think I can't apply my guess.

What is right thing?
Why momentum exchange can work as shear stress(loss) eventhough there is no change of momentum?

P.S_1
I've guessed following also.
In turbulent flow, there is always ACTIVE mixing.
Cause of active mixing, flow elements are always mixed to nearby viscous sublayer.
So strictly there is no actual momentum loss by mixing between fluid elements in outer layer.
But event that fluid elements are mixed to viscous sublayer always exist so there is momentum loss.
I'm not sure. What is right?

P.S_2
I think the reason why I'm confused is following.
Shape of velocity profile of turbulent flow and laminar flow are diffrent.
In laminar flow, although there is mixing by molecule interation, alwyas there is velocity gradient.
Eventhough there is mixing and consequent momentum exchange, velocity depends on location.
Near wall region is always solver that the region far from the wall.
So I can understand that momentum of fluid that is far from the wall is corroded by the fluid that is nearby wall.

But in turbulent flow, I think there is no velocity difference between 2 mixed points after mixing as we can see from my uploaded picture(momentum exchange in outer layer).
What is wrong of my conjecture?

Thank you~!

I suggest you to start forgetting any wall and confined turbulence. Also forget the shape of the velocity profile in a BL since it represents only the statistical velocity.
In turbulence you have 3D vortical unsteady structures, even without walls. These structures are subject to the stretching of the vorticity and are stirred, elongated and so on. Define a characteristic lenght and velocity for these structures. Then evaluate the Re number associated, this is a local and field dependent number. Large structures have large Re number and the diffusive momentum flux can be disregarded when compared to the inertial one. But the stretching acts to generate via via smaller structures and the local Re diminuishes. Thus, at a certain scale the momentum flux (that is the molecular viscosity x velocity gradient) has the same importance of the inertial and produces kinetic energy to dissipate. At Re=O(1) you reached the lenght scale of the vortical structures denoted Kolmogorov lenght scale. At this level the diffusive momentum flux is so relevant that no smaller vortical structures are produces and all the energy is dissipated.

This framekwork is represented by a classical energy spectra in the homogenous isotropic turbulence (no walls at all).
In conclusion, the velocity gradient need to be very high to allow its product with the molecular viscosity to be relevant.


In case of confined walls, the local Re number is somehow substituted by the local y+ law.

[LuckyTran]

In the first cartoon, you have mixing between two fluid elements and this is the same for a laminar or turbulent flow.

The second second you have mixing between a layer of fluid next to a wall with the wall boundary itself (a momentum sink). And this interaction is also the same for laminar and a turbulent flow.

Note that N-S equations is a conservation of momentum. There is no momentum loss in laminar or turbulent flows unless there is a sink (i.e. a no slip wall or some black hole). The loss of momentum is caused by the wall, not because the flow is turbulent or laminar. Your confusion likely stems from the misuse of the word "laminar/viscous sublayer" to describe how the momentum exchange takes place near wall boundaries.

[FluidKo]

Originally Posted by FMDenaro

I suggest you to start forgetting any wall and confined turbulence. Also forget the shape of the velocity profile in a BL since it represents only the statistical velocity.
In turbulence you have 3D vortical unsteady structures, even without walls. These structures are subject to the stretching of the vorticity and are stirred, elongated and so on. Define a characteristic lenght and velocity for these structures. Then evaluate the Re number associated, this is a local and field dependent number. Large structures have large Re number and the diffusive momentum flux can be disregarded when compared to the inertial one. But the stretching acts to generate via via smaller structures and the local Re diminuishes. Thus, at a certain scale the momentum flux (that is the molecular viscosity x velocity gradient) has the same importance of the inertial and produces kinetic energy to dissipate. At Re=O(1) you reached the lenght scale of the vortical structures denoted Kolmogorov lenght scale. At this level the diffusive momentum flux is so relevant that no smaller vortical structures are produces and all the energy is dissipated.

This framekwork is represented by a classical energy spectra in the homogenous isotropic turbulence (no walls at all).
In conclusion, the velocity gradient need to be very high to allow its product with the molecular viscosity to be relevant.


In case of confined walls, the local Re number is somehow substituted by the local y+ law.

Actually I was known the process of turbulence dissipation.
But I was wondering why turbulence dissipation can be expressed by strain rate.
So I just have thought about it.
Following is my conjecture.

If there is a velocity gradient, we can see there is velocity difference.
By the property of turbulence, fluid elements are mixed so that there is momentum exchange between fluid elements.
If velocity difference(velocity gradient) is high, then it means quantity of momentum exchange is high.
And high quantity of momentum exchange means that size of eddy is big as uploaded picture.
Eddy looks like motion but it will become loss.
It will be dissipated by heat or noise.(energy cascade)
So it means as size of eddy becomes bigger, quantity of dissipated energy becomes bigger.

In conclusion, dissipated energy can be expressed by eddy(energy cascade).
And the size of eddy is determined by momentum exchange(velocity gradient).
That is why we consider turbulent stress using strain rate(velocity gradient)
What do you think about it?

Thank you~!

[FluidKo]

Originally Posted by LuckyTran

In the first cartoon, you have mixing between two fluid elements and this is the same for a laminar or turbulent flow.

The second second you have mixing between a layer of fluid next to a wall with the wall boundary itself (a momentum sink). And this interaction is also the same for laminar and a turbulent flow.

Note that N-S equations is a conservation of momentum. There is no momentum loss in laminar or turbulent flows unless there is a sink (i.e. a no slip wall or some black hole). The loss of momentum is caused by the wall, not because the flow is turbulent or laminar. Your confusion likely stems from the misuse of the word "laminar/viscous sublayer" to describe how the momentum exchange takes place near wall boundaries.

I understand loss of momentum is caused by only wall.
But in RANS turbulent stress is loss also.
I've thought about why turbulent stress can be expressed by strain rate below.
What do you think about it?

Thank you~!

[LuckyTran]

Turbulent dissipation rate is the rate at which turbulent kinetic energy is converted into heat. The work done by viscous forces is u*tau. We know exactly what tau is, it comes from Newton's viscosity law. The turbulent dissipation rate is simply the fluctuating component of this term.

All you need to do is apply the Reynolds decomposition (put u= ubar+u') and apply it to the viscous term. Then you take the average and and then you can separate the viscous term into a term containing only the mean velocity and another term which contains only fluctuating velocities which is the turbulent dissipation rate. Hint: the inner product of any vector/tensor with itself is positive semi-definite. Unlike the Reynolds stresses, which fundamentally come from the advection term, the turbulent dissipation rate comes from the viscous term. What we don't know is what are the instantaneous velocities and how to express the mean dissipation in terms of only mean strain rates but we know that the instantaneous dissipation rate is exactly related to the instantaneous strain via Newton's viscosity law. There is no doubt that the turbulent dissipation rate is related to the strain, the question is what exactly does this closure model look like.