Calculation of fluctuation velocity in this paper

FluidKo · January 4, 2022, 21:19

https://www.researchgate.net/publica..._of_Turbulence

I'm reading this paper.
This paper is claiming there is a problem in k-epsilon model.
Because in k-epsilon model, we assume that eddy viscosity( $\nu_{T}$ ) is isotropy.
But actually in real world, eddy viscosity( $\nu_{T}$ ) is anisotropy in high Reynolds number and is isotropy in low Reynolds number.
It means there can be error when we consider flow with high Reynolds number.

And I have a question in this figure.
How he can calculate fluctuation velocity?

What I know is
fluctuation velocity can be calculated only when we assume that eddy viscosity( $\nu_{T}$ ) is isotropy(like k-epsilon model).
When we use k-epsilon model, we can find k(Turbulence Kinetic Energy) by T.K.E transport eqation and k is same with $\frac{1}{2}$ ( $u^{'2}$ + $v^{'2}$ + $w^{'2}$ ).
Then if k(T.K.E) is found, we can calculate u', v' and w' because u', v' and w' are same each other by assumption of isotropy.

But in this paper, author claims that we should consider flow as anisotropy and he suggests new Eddy viscosity( $\nu_{T}$ ) using [LaTeX Error: Syntax error].
So I think it is impossible to calculate fluctuation velocity cause of anisotropy.
But I think author calculated fluctation velocity using T.K.E because we can find desription in figure that allude he calculated fluctuation velocity by root mean square.
(Figure 8: Profiles of rms velocities perpendicular (v) and parallel (u) to the wall in the impinging jet)

Actually I've thought that assumption of isotropy can be possible in the impinging jet sometimes.
Because impingement occurs nearby wall, and there is a low Reynolds number nearby wall by No slip condition(dominant molecule viscosity).
But, eventhough my deduction is right, I can't understand why there is a difference between v' and u'.
In the case that r/D=2.5, there is a difference between v' and u' that I've marked in uploaded picture.
Difference between v' and u' means this flow is anisotropy and this is contradiction against author's claim.

I'm confused now

Please help me.

Thanks.

LuckyTran · January 5, 2022, 01:04

If you get the tke, and assume that $\overline{u'^2}=\overline{v'^2}=\overline{w'^2}$ , then yes you can get $\overline{u'^2}\approx\frac{2}{3}k$ and from there get the rms-fluctuations of u', v', and w'. A lot of people think this is what isotropy means, but it's not.

In general, let's say you do some RANS with some turbulence model. In your RANS, you will get a solution for the mean velocity field. From this velocity field, you can calculate things like the velocity gradient. You can also calculate the Strain Rate tensor. From the Strain Rate tensor, you can calculate the Reynolds Stress tensor via your eddy viscosity model.

$-\rho \overline{u'_i u'_j} = \mu_t \, \left( \frac{\partial u_i}{\partial x_j} + \frac{\partial u_j}{\partial x_i} - \frac{2}{3} \frac{\partial u_k}{\partial x_k} \delta_{ij} \right) - \frac{2}{3} k \delta_{ij}$

Isotropy means that the turbulent viscosity is a simple scalar coefficient and is not a matrix. Now as long as the velocity field is not trivial, the strain-rate tensor (which is like 3x3 matrix) will not have all 9 entries identical (but of course only 6 are independent), then the Reynolds stresses also will not all be equal. So even for an isotropic turbulence model, you don't necessarily have u', v', and w' all equal and you can indeed can different values for u', v', and w'. This is also the reason why an isotropic model can work for many many flows, it's actually not that restrictive.

And when I say that you CAN calculate the Reynolds stresses, I actually mean MUST. You must be able to calculate all 6 of the Reynold stresses or you will have a closure problem. You personally don't have to be able to calculate this, but your CFD solver and code must.

FluidKo · January 5, 2022, 01:27

Quote:

Originally Posted by LuckyTran

If you get the tke, and assume that $\overline{u'^2}=\overline{v'^2}=\overline{w'^2}$ , then yes you can get $\overline{u'^2}\approx\frac{2}{3}k$ and from there get the rms-fluctuations of u', v', and w'. A lot of people think this is what isotropy means, but it's not.

In general, let's say you do some RANS with some turbulence model. In your RANS, you will get a solution for the mean velocity field. From this velocity field, you can calculate things like the velocity gradient. You can also calculate the Strain Rate tensor. From the Strain Rate tensor, you can calculate the Reynolds Stress tensor via your eddy viscosity model.

$-\rho \overline{u'_i u'_j} = \mu_t \, \left( \frac{\partial u_i}{\partial x_j} + \frac{\partial u_j}{\partial x_i} - \frac{2}{3} \frac{\partial u_k}{\partial x_k} \delta_{ij} \right) - \frac{2}{3} k \delta_{ij}$

Isotropy means that the turbulent viscosity is a simple scalar coefficient and is not a matrix. Now as long as the velocity field is not trivial, the strain-rate tensor (which is like 3x3 matrix) will not have all 9 entries identical (but of course only 6 are independent), then the Reynolds stresses also will not all be equal. So even for an isotropic turbulence model, you don't necessarily have u', v', and w' all equal and you can indeed can different values for u', v', and w'. This is also the reason why an isotropic model can work for many many flows, it's actually not that restrictive.

And when I say that you CAN calculate the Reynolds stresses, I actually mean MUST. You must be able to calculate all 6 of the Reynold stresses or you will have a closure problem. You personally don't have to be able to calculate this, but your CFD solver and code must.

Actually I've uploaded this question to other website.
And the question of that website has more accurate information about what I know and what I'm confused
So I had to modify this question earlier but I didn't.
Sorry...
Here is the link.
https://www.researchgate.net/post/Ho...in_this_paper2

LuckyTran · January 5, 2022, 02:03

괜찮아

Just to be clear, u', v', and w' in the paper are the root mean squared of the fluctuations and not the instantaneous u'(t), v'(t), w'(t).

Let's suppose you have done CFD and have solved your flow field. So now you have the velocity, and turbulent viscosity, and k. Your eddy viscosity model gives you again:

$-\rho \overline{u'_i u'_j} = \mu_t \, \left( \frac{\partial u_i}{\partial x_j} + \frac{\partial u_j}{\partial x_i} - \frac{2}{3} \frac{\partial u_k}{\partial x_k} \delta_{ij} \right) - \frac{2}{3} k \delta_{ij}$

The diagonal terms on the left hand side are your normal stresses (you need to divide by density of course): $\overline{u'u'}$ , $\overline{v'v'}$ , and $\overline{w'w'}$

From here you get the rms by just taking the square root of the normal stresses.
$u'=\sqrt{\overline{u'u'}}$
$v'=\sqrt{\overline{v'v'}}$
$w'=\sqrt{\overline{w'w'}}$

FluidKo · January 5, 2022, 02:14

Oh I can understand now.
The reason that I was confused is
I have a streotype that I can find the fluctuation velocity at the only case when there is a assumption
' $\overline{u'^2}$ = $\overline{v'^2}$ = $\overline{w'^2}$ '.
And also I've misunderstood ' $\overline{u'^2}$ = $\overline{v'^2}$ = $\overline{w'^2}$ ' is same meaning with isotropy.

Cause of this, I've got a confusion and let you be annoyed.
I've made you describe same things many times.
Sorry.

So you are meaning the assumption of ' $\overline{u'^2}$ = $\overline{v'^2}$ = $\overline{w'^2}$ ' is different with isotropic.
Isotropic means Eddy viscosity is just scalar.
Regardless of the fact that Eddy viscosity is just scalar,
fluctuation velocity of each components in Reynolds Stress can be different.
(u'≠v'≠w')
There is No relation between 'Isotropic' and ' $\overline{u'^2}$ = $\overline{v'^2}$ = $\overline{w'^2}$ '

Then I can calculate fluctation velcoty by
u'= $\sqrt{\overline{u'u'}}$
v'= $\sqrt{\overline{v'v'}}$
w'= $\sqrt{\overline{w'w'}}$
obviously, simply and easily.

So
u'= $\sqrt{\overline{u'u'}}$
v'= $\sqrt{\overline{v'v'}}$
w'= $\sqrt{\overline{w'w'}}$
These look like similar with fluctuation velcotiy and have same dimensions with fluctuation velcotiy.
But actually they are not real fluctuation velcotiy.
Because u'(t), v'(t), w'(t) are different with
u'= $\sqrt{\overline{u'u'}}$
v'= $\sqrt{\overline{v'v'}}$
w'= $\sqrt{\overline{w'w'}}$
these because real fluctuation velocity depends on time.
However
u'= $\sqrt{\overline{u'u'}}$
v'= $\sqrt{\overline{v'v'}}$
w'= $\sqrt{\overline{w'w'}}$
They don't depend on time.
They are just byproduct of calculation that looks similar with real fluctuation velocity and have same dimensions with fluctuation velcotiy.

These doesn't have proper physical meaning that happens in real world.
They are just calculated forcefully because we want to find something that just looks like real fluctuation velcoty[u'(t), v'(t), w'(t)] but are not real fluctuation velocity[u'(t), v'(t), w'(t)].

Their Identification is NOT same with real fluctuation velocity[u'(t), v'(t), w'(t)]

Did I understand right?

FluidKo · January 5, 2022, 02:57

Quote:

Originally Posted by LuckyTran

괜찮아

Just to be clear, u', v', and w' in the paper are the root mean squared of the fluctuations and not the instantaneous u'(t), v'(t), w'(t).

Let's suppose you have done CFD and have solved your flow field. So now you have the velocity, and turbulent viscosity, and k. Your eddy viscosity model gives you again:

$-\rho \overline{u'_i u'_j} = \mu_t \, \left( \frac{\partial u_i}{\partial x_j} + \frac{\partial u_j}{\partial x_i} - \frac{2}{3} \frac{\partial u_k}{\partial x_k} \delta_{ij} \right) - \frac{2}{3} k \delta_{ij}$

The diagonal terms on the left hand side are your normal stresses (you need to divide by density of course): $\overline{u'u'}$ , $\overline{v'v'}$ , and $\overline{w'w'}$

From here you get the rms by just taking the square root of the normal stresses.
$u'=\sqrt{\overline{u'u'}}$
$v'=\sqrt{\overline{v'v'}}$
$w'=\sqrt{\overline{w'w'}}$

I've uploaded new message.
Can you check my message?
Thanks

LuckyTran · January 5, 2022, 04:37

All of what you wrote is correct. You have to get from the context or read carefully the definitions in the nomenclature what is the meaning of u'. In this paper u',v', and w' are the rms and not $u'(t)=u(t)-\overline{u}$ . If you read another paper, they might use the other definition.

The rms of the fluctuating velocities anyway are statistics about the turbulence (just like the mean), so they do carry important information. This is especially true for the Reynolds stresses $\overline{u'u'}$ and so on. A lot of people prefer to consider the rms because it has units of m/s just like velocity, and when normalized by the mean it becomes a turbulence intensity. It's easier to discuss using rms rather than working directly in stress units.

FMDenaro · January 5, 2022, 05:16

Reading posts like this, I always wonder what people think about computing fluctuation of velocity from a RANS formulation. Does this operation makes some physical meaning in the results? I really don't believe you can get nothing but a statistical meaning due to the "model".

FluidKo · January 5, 2022, 05:31

Quote:

Originally Posted by LuckyTran

All of what you wrote is correct. You have to get from the context or read carefully the definitions in the nomenclature what is the meaning of u'. In this paper u',v', and w' are the rms and not $u'(t)=u(t)-\overline{u}$ . If you read another paper, they might use the other definition.

The rms of the fluctuating velocities anyway are statistics about the turbulence (just like the mean), so they do carry important information. This is especially true for the Reynolds stresses $\overline{u'u'}$ and so on. A lot of people prefer to consider the rms because it has units of m/s just like velocity, and when normalized by the mean it becomes a turbulence intensity. It's easier to discuss using rms rather than working directly in stress units.

Thanks

Actually I'm not American as you know so I'm not good at English.
Also I'm just senior so I don't have bachelor's degree yet.
But I'm interested in Turbulence so I'm preparing to enter laboratory to get Ph.D.
Eventhough I don't know many thing, you've answered to me sincerely not only today but also yesterday.
I feel gratitude about this.

Thanks

FluidKo · January 5, 2022, 05:52

Quote:

Originally Posted by FMDenaro

Reading posts like this, I always wonder what people think about computing fluctuation of velocity from a RANS formulation. Does this operation makes some physical meaning in the results? I really don't believe you can get nothing but a statistical meaning due to the "model".

I can't understand the meaning of this sentence

' Does this operation makes some physical meaning in the results? I really don't believe you can get nothing but a statistical meaning due to the "model".'

But if I say the role of the RANS that I know, we can get a advantage by taking average to Navier-Stokes equation when we consider turbulent flow.

Turbulence is chaotic and chaotic means it looks like there isn't any pattern.
But if we take average to situation(velocity or pressure) over time, we can find there is a pattern of mean value.

So if we take average to to Navier-Stokes equation when we consider turbulent flow,
we can get information of mean flow.
So if we know mean velocity field that satisfy mass conservation,
(Q=A $\bar{V}$ , Q: flow rate, A: Area that fluid pass through, $\bar{V}$ : Mean velocity)
we can find mean pressure field.

Actually it is imposssible to capture fluctuation velocities of each components on every time.
In real world, fluctuation depends on time, but we've taken a average Navier-Stokes equation(so there isn't time dependent term) then give this to computer so that computer can't print output.
Irregular fluctuation can be captured in real world.
If we want to capture information of irregular fluctuation, we should do experiment.

This is just my opinion. I'm just interested in turbulence but I'm not good at this.
So notice that I can be false.
Thanks

FMDenaro · January 5, 2022, 06:03

Quote:

Originally Posted by FluidKo

I can't understand the meaning of this sentence

' Does this operation makes some physical meaning in the results? I really don't believe you can get nothing but a statistical meaning due to the "model".'

But if I say the role of the RANS that I know, we can get a advantage by taking average to Navier-Stokes equation when we consider turbulent flow.

Turbulence is chaotic and chaotic means it looks like there isn't any pattern.
But if we take average to situation(velocity or pressure) over time, we can find there is a pattern of mean value.

So if we take average to to Navier-Stokes equation when we consider turbulent flow,
we can get information of mean flow.
So if we know mean velocity field that satisfy mass conservation,
(Q=A $\bar{V}$ , Q: flow rate, A: Area that fluid pass through, $\bar{V}$ : Mean velocity)
we can find mean pressure field.

Actually it is imposssible to capture fluctuation velocities of each components on every time.
In real world, fluctuation depends on time, but we've taken a average Navier-Stokes equation(so there isn't time dependent term) then give this to computer so that computer can't print output.
Irregular fluctuation can be captured in real world.
If we want to capture information of irregular fluctuation, we should do experiment.

This is just my opinion. I'm just interested in turbulence but I'm not good at this.
So notice that I can be false.
Thanks

Exactly ! Why people think to compute any type of fluctuation from a RANS solution? No meaning at all.

FluidKo · January 5, 2022, 06:40

Quote:

Originally Posted by FMDenaro

Exactly ! Why people think to compute any type of fluctuation from a RANS solution? No meaning at all.

Aha you wanted to confirm me whether I know right.
I couldn't catch the intention what are you meaning
because you know English is not my mother language.

I was confused there is a u' in paper.
So suddenly I thought "what I've known is wrong? fluctuation can be captured in CFD?"
Because I'm not good at turbulence and English so it is hard to understand paper.
That is why I was confused.

Thanks

January 5, 2022, 01:04		#2
LuckyTran Senior Member Lucky Join Date: Apr 2011 Location: Orlando, FL USA Posts: 5,843 Rep Power: 68	If you get the tke, and assume that $\overline{u'^2}=\overline{v'^2}=\overline{w'^2}$ , then yes you can get $\overline{u'^2}\approx\frac{2}{3}k$ and from there get the rms-fluctuations of u', v', and w'. A lot of people think this is what isotropy means, but it's not. In general, let's say you do some RANS with some turbulence model. In your RANS, you will get a solution for the mean velocity field. From this velocity field, you can calculate things like the velocity gradient. You can also calculate the Strain Rate tensor. From the Strain Rate tensor, you can calculate the Reynolds Stress tensor via your eddy viscosity model. $-\rho \overline{u'_i u'_j} = \mu_t \, \left( \frac{\partial u_i}{\partial x_j} + \frac{\partial u_j}{\partial x_i} - \frac{2}{3} \frac{\partial u_k}{\partial x_k} \delta_{ij} \right) - \frac{2}{3} k \delta_{ij}$ Isotropy means that the turbulent viscosity is a simple scalar coefficient and is not a matrix. Now as long as the velocity field is not trivial, the strain-rate tensor (which is like 3x3 matrix) will not have all 9 entries identical (but of course only 6 are independent), then the Reynolds stresses also will not all be equal. So even for an isotropic turbulence model, you don't necessarily have u', v', and w' all equal and you can indeed can different values for u', v', and w'. This is also the reason why an isotropic model can work for many many flows, it's actually not that restrictive. And when I say that you CAN calculate the Reynolds stresses, I actually mean MUST. You must be able to calculate all 6 of the Reynold stresses or you will have a closure problem. You personally don't have to be able to calculate this, but your CFD solver and code must. FluidKo likes this.

January 5, 2022, 02:03		#4
LuckyTran Senior Member Lucky Join Date: Apr 2011 Location: Orlando, FL USA Posts: 5,843 Rep Power: 68	괜찮아 Just to be clear, u', v', and w' in the paper are the root mean squared of the fluctuations and not the instantaneous u'(t), v'(t), w'(t). Let's suppose you have done CFD and have solved your flow field. So now you have the velocity, and turbulent viscosity, and k. Your eddy viscosity model gives you again: $-\rho \overline{u'_i u'_j} = \mu_t \, \left( \frac{\partial u_i}{\partial x_j} + \frac{\partial u_j}{\partial x_i} - \frac{2}{3} \frac{\partial u_k}{\partial x_k} \delta_{ij} \right) - \frac{2}{3} k \delta_{ij}$ The diagonal terms on the left hand side are your normal stresses (you need to divide by density of course): $\overline{u'u'}$ , $\overline{v'v'}$ , and $\overline{w'w'}$ From here you get the rms by just taking the square root of the normal stresses. $u'=\sqrt{\overline{u'u'}}$ $v'=\sqrt{\overline{v'v'}}$ $w'=\sqrt{\overline{w'w'}}$ FluidKo likes this.

January 5, 2022, 02:14		#5
FluidKo Senior Member - Join Date: Oct 2021 Location: - Posts: 142 Rep Power: 6	Oh I can understand now. The reason that I was confused is I have a streotype that I can find the fluctuation velocity at the only case when there is a assumption ' $\overline{u'^2}$ = $\overline{v'^2}$ = $\overline{w'^2}$ '. And also I've misunderstood ' $\overline{u'^2}$ = $\overline{v'^2}$ = $\overline{w'^2}$ ' is same meaning with isotropy. Cause of this, I've got a confusion and let you be annoyed. I've made you describe same things many times. Sorry. So you are meaning the assumption of ' $\overline{u'^2}$ = $\overline{v'^2}$ = $\overline{w'^2}$ ' is different with isotropic. Isotropic means Eddy viscosity is just scalar. Regardless of the fact that Eddy viscosity is just scalar, fluctuation velocity of each components in Reynolds Stress can be different. (u'≠v'≠w') There is No relation between 'Isotropic' and ' $\overline{u'^2}$ = $\overline{v'^2}$ = $\overline{w'^2}$ ' Then I can calculate fluctation velcoty by u'= $\sqrt{\overline{u'u'}}$ v'= $\sqrt{\overline{v'v'}}$ w'= $\sqrt{\overline{w'w'}}$ obviously, simply and easily. So u'= $\sqrt{\overline{u'u'}}$ v'= $\sqrt{\overline{v'v'}}$ w'= $\sqrt{\overline{w'w'}}$ These look like similar with fluctuation velcotiy and have same dimensions with fluctuation velcotiy. But actually they are not real fluctuation velcotiy. Because u'(t), v'(t), w'(t) are different with u'= $\sqrt{\overline{u'u'}}$ v'= $\sqrt{\overline{v'v'}}$ w'= $\sqrt{\overline{w'w'}}$ these because real fluctuation velocity depends on time. However u'= $\sqrt{\overline{u'u'}}$ v'= $\sqrt{\overline{v'v'}}$ w'= $\sqrt{\overline{w'w'}}$ They don't depend on time. They are just byproduct of calculation that looks similar with real fluctuation velocity and have same dimensions with fluctuation velcotiy. These doesn't have proper physical meaning that happens in real world. They are just calculated forcefully because we want to find something that just looks like real fluctuation velcoty[u'(t), v'(t), w'(t)] but are not real fluctuation velocity[u'(t), v'(t), w'(t)]. Their Identification is NOT same with real fluctuation velocity[u'(t), v'(t), w'(t)] Did I understand right? Last edited by FluidKo; January 5, 2022 at 04:07.

January 5, 2022, 04:37		#7
LuckyTran Senior Member Lucky Join Date: Apr 2011 Location: Orlando, FL USA Posts: 5,843 Rep Power: 68	All of what you wrote is correct. You have to get from the context or read carefully the definitions in the nomenclature what is the meaning of u'. In this paper u',v', and w' are the rms and not $u'(t)=u(t)-\overline{u}$ . If you read another paper, they might use the other definition. The rms of the fluctuating velocities anyway are statistics about the turbulence (just like the mean), so they do carry important information. This is especially true for the Reynolds stresses $\overline{u'u'}$ and so on. A lot of people prefer to consider the rms because it has units of m/s just like velocity, and when normalized by the mean it becomes a turbulence intensity. It's easier to discuss using rms rather than working directly in stress units. FluidKo likes this.

January 5, 2022, 05:16		#8
FMDenaro Senior Member Filippo Maria Denaro Join Date: Jul 2010 Posts: 7,053 Rep Power: 75	Reading posts like this, I always wonder what people think about computing fluctuation of velocity from a RANS formulation. Does this operation makes some physical meaning in the results? I really don't believe you can get nothing but a statistical meaning due to the "model". LuckyTran and FluidKo like this.

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