Is Turbulence at Point A Isotropic? Calculation Help

[nightingale]

Homework Statement

Hello everyone, I am having a problem whether or not a turbulence at a specific location (let's say A) is isotropic or not.

I have calculated the two root mean square values of velocity fluctuations measured at the point A in a fully developed turbulent pipe flow.

the first rms= 7.55, which is the velocity component parallel to the pipe axis
the second rms= 0.335, which is the velocity component along the radius
These two components are simultaneously measured.

The question is, is the turbulence isotropic at A?

Do I need a formula to calculate whether the turbulence is isotropic at A?
I am given the density (1.2) and the dynamic viscosity of the fluid (1.8*10^-5).

Any help would be greatly appreciated. Thank you.

 

[Chestermiller]

What does the word "isotropic" mean to you?

Chet

 

[nightingale]

What does the word "isotropic" mean to you?

Chet

That it is uniform in all directions?

I read in wikipedia that "Fluid flow is isotropic if there is no directional preference (e.g. in fully developed 3D turbulence). An example of anisotropy is in flows with a background density as gravity works in only one direction. The apparent surface separating two differing isotropic fluids would be referred to as an isotrope."

Since my flow has two velocity components, does this mean it is not isotropic?

Thank you very much for the help.

 

[Chestermiller]

You are trying to determine whether the turbulence is isotropic, not the overall flow. If the rms velocity fluctuations do not match in two directions, then the turbulence is not isotropic.

Chet

 

[bigfooted]

Since my flow has two velocity components, does this mean it is not isotropic?

You have measurements of two velocity components, but of course, there are three velocity components.

You are trying to determine whether the turbulence is isotropic, not the overall flow. If the rms velocity fluctuations do not match in two directions, then the turbulence is not isotropic.

So, if your measured values were u'u'=0.1 and v'v'=0.1, then you still don't know if the flow is isotropic or not, because the tangential component could be w'w'=0.5 or w'w'=0.0.

Also, turbulent pipe flow is not isotropic, but it becomes more isotropic if you measure further away from the walls.

 

[Chestermiller]

So, if your measured values were u'u'=0.1 and v'v'=0.1, then you still don't know if the flow is isotropic or not, because the tangential component could be w'w'=0.5 or w'w'=0.0.

Well, you would know that it's at least transversely isotropic in the plane that contains u and v. In order for it to be fully isotropic w'w' would have to be equal to 0.1 also.

Also, turbulent pipe flow is not isotropic, but it becomes more isotropic if you measure further away from the walls.

Yes, and...?

 

[bigfooted]

Yes, and...?

He could have answered the question on isotropy without any measurements.

 

[nightingale]

You have measurements of two velocity components, but of course, there are three velocity components.So, if your measured values were u'u'=0.1 and v'v'=0.1, then you still don't know if the flow is isotropic or not, because the tangential component could be w'w'=0.5 or w'w'=0.0.

Also, turbulent pipe flow is not isotropic, but it becomes more isotropic if you measure further away from the walls.

Okay, is it possible to calculate the tangential component from the two rms velocity measurements? Is it by Pythagoras theorem?

The measurements were done at 3 points actually, A, B, and C. But only at A that the two components of velocity were measured. Will it help if I calculate the rms at point B and C?

Thank you very much, as you can see, I'm quite clueless.

 

[nightingale]

You are trying to determine whether the turbulence is isotropic, not the overall flow. If the rms velocity fluctuations do not match in two directions, then the turbulence is not isotropic.

Chet

Thank you, Chet.

I have two different rms values of fluctuations at A (one parallel to the pipe axis and another one along the radius) and thus the turbulence is not isotropic at A.

Is my statement correct? or do I miss something?

Thank you very much.

 

[nightingale]

Another question that has been bothering me is that:
Is the rms fluctuation measured parallel to the pipe is actually the rms fluctuations in the circumferential (tangential)?
Which thus also means the rms fluctuations along the radius is the fluctuations in the radial direction?

Thank you again.

 

[Chestermiller]

Thank you, Chet.

I have two different rms values of fluctuations at A (one parallel to the pipe axis and another one along the radius) and thus the turbulence is not isotropic at A.

Is my statement correct? or do I miss something?

Thank you very much.

In my judgement, your answer is correct.

 

[Chestermiller]

Another question that has been bothering me is that:
Is the rms fluctuation measured parallel to the pipe is actually the rms fluctuations in the circumferential (tangential)?
Which thus also means the rms fluctuations along the radius is the fluctuations in the radial direction?

Thank you again.

The fluctuations in measured parallel to the pipe are in the axial direction, and the fluctuations along the radius are in the radial direction. The fluctuations in the circumferential direction were not measured. That's the best I can judge from the wording.

Chet

 

[nightingale]

In my judgement, your answer is correct.

Thank you Chet!

Ah I see, is it possible to calculate the fluctuations in the circumferential direction?
The question says, "assuming the rms fluctuations in the circumferential (i.e. tangential) direction to be similar to that of the fluctuations in the radial direction, estimate the turbulent energy at A."

Which is why I was wondering whether the question is referring to the rms values I have already obtained, or I have to calculated another rms value.

Thank you again!

 

[Chestermiller]

Thank you Chet!

Ah I see, is it possible to calculate the fluctuations in the circumferential direction?
The question says, "assuming the rms fluctuations in the circumferential (i.e. tangential) direction to be similar to that of the fluctuations in the radial direction, estimate the turbulent energy at A."

Which is why I was wondering whether the question is referring to the rms values I have already obtained, or I have to calculated another rms value.

Thank you again!

They already said that the fluctuations in the circumferential direction are similar (I interpret this as essentially the same) as the fluctuations in the radial direction.

Chet

 

[nightingale]

They already said that the fluctuations in the circumferential direction are similar (I interpret this as essentially the same) as the fluctuations in the radial direction.

Chet

Thus Chet,
Do you mean that the flow's turbulent kinetic energy at A is thus:

k = 0.5 * (0.335² +0.335²)
k = 0.112225?

*the rms fluctuations along the radius is 0.335.

Thank you very much.

 

[Chestermiller]

Thus Chet,
Do you mean that the flow's turbulent kinetic energy at A is thus:

k = 0.5 * (0.335² +0.335²)
k = 0.112225?

*the rms fluctuations along the radius is 0.335.

Thank you very much.

No. You need to include the contribution from the fluctuation in the circumferential direction.

Chet

 

[nightingale]

No. You need to include the contribution from the fluctuation in the circumferential direction.

Chet

The question says that 'assuming the rms fluctuations in the circumferential (i.e. tangential) direction to be similar to that of the fluctuations in the radial direction, estimate the turbulent energy at A'.

Thus I assume that the rms fluctuations in the circumferential direction = rms fluctuations in the radial direction = 0.335.

I don't understand where I went wrong. Would you please explain?

Thank you very much.

 

[Chestermiller]

k = 0.5 * (0.335² +0.335²+0.335²)

 

[nightingale]

k = 0.5 * (0.335² +0.335²+0.335²)

I'm really sorry, Chet, but I don't quite understand.

If turbulent kinetic energy = 0.5 * (rms u1²+rms u2²+rmsu3²)

where
u₁ is the velocity component parallel to the pipe axis (axial)
u₂ is the velocity component along the radius (radial)
u₃ is the velocity component in the circumferential (tangential)

and u₂=u₃
Then why isn't the kinetic energy equals to
k = 0.5 * (7.55 +0.3352+0.3352)?

Once again, thank you very much for all the help.

 

[Chestermiller]

No. Not the velocity components. The time-dependent fluctuations in the velocity components.

Chet

 

[nightingale]

No. Not the velocity components. The time-dependent fluctuations in the velocity components.

Chet

I'm really sorry Chet, but I couldn't seem to grasp on why yet.

I see, I'm sorry for writing velocity components, I meant the fluctuations in the velocity components. My questions still stands however, why do I have to take 0.335 three times?

"where
u1 is the velocity fluctuations parallel to the pipe axis (axial) = 7.55
u2 is the velocity fluctuations along the radius (radial) = 0.355
u3 is the velocity fluctuations in the circumferential (tangential) = 0.355"

Thank you very much, Chet. I greatly appreciate all your help this far.

 

[nightingale]

No. Not the velocity components. The time-dependent fluctuations in the velocity components.

Chet

Chet, here is the full picture of the problem.

Question a asks me to calculate rms values of fluctuations at B and whether it is isotropic, to which I answer (with your help, thank you very much):
First rms fluctuations= 7.55
Second rms fluctuations= 0.3352
There are two different rms values of fluctuations at B (one parallel to the pipe axis and another one along the radius) and thus the turbulence is not isotropic at B.

Now Question B asks me to calculate the turbulent energy at B, with rms of fluctuations in the circumferential = rms fluctuations in the radial direction.
Which why I thought that the rms values I have calculated could be used in the kinetic energy calculation.

Would you mind explain to me why it is not the rms fluctuations I have calculated (7.55 and 0.3352) before?

Thank you for your patience & continuous guidance, sir!

 

[Chestermiller]

View attachment 90815

Chet, here is the full picture of the problem.

Question a asks me to calculate rms values of fluctuations at B and whether it is isotropic, to which I answer (with your help, thank you very much):
First rms fluctuations= 7.55
Second rms fluctuations= 0.3352
There are two different rms values of fluctuations at B (one parallel to the pipe axis and another one along the radius) and thus the turbulence is not isotropic at B.

These are not the rms fluctuations. There are the rms velocities. To get the rms fluctuations, you first calculate the time-averaged velocity, then you subtract that from the instantaneous velocities to get the velocity fluctuations, and then you evaluate the rms value of the velocity fluctuations. Didn't you learn this from your textbook or in class? Now please, go back and report back to me with the time-averaged velocities and the correct rms values of the fluctuations for part B.

Chet

 

[nightingale]

These are not the rms fluctuations. There are the rms velocities. To get the rms fluctuations, you first calculate the time-averaged velocity, then you subtract that from the instantaneous velocities to get the velocity fluctuations, and then you evaluate the rms value of the velocity fluctuations. Didn't you learn this from your textbook or in class? Now please, go back and report back to me with the time-averaged velocities and the correct rms values of the fluctuations for part B.

Chet

Thank you for clearing that up Chet! Unfortunately I don't have a textbook or lecture slides on this, but I have borrowed some books from the library.

I learned that:
U(x,t) =U (x,t) + u(x,t)

Where U is the velocity, U is the time averaged velocity, u is the fluctuations.
I read that:

U (x,t) = ½T ∫ U(x,t +t') dt'
But I'm not quite sure on how much T I should take (although I read that it has to be significantly large), and I don't really know how to solve that (with t' and dt'), so I decided on taking another method.

Here I calculate the average of the velocity u (axial) and v (radial), and obtained 7.5375 and -0.0125.
I take this as the time-averaged velocity.
And thus I calculate the fluctuations.

The first fluctuations:
u(x,t) = 7.6 - 7.5375
u(x,t) = 0.0625
Did the same to the rest.

Finally I calculate that the rms fluctuation values for axial direction and radial direction and obtained 0.43857 and 0.33307. The flow is thus not isotropic at B.I also read that the turbulence kinetic energy equals to
k = ½ (u'²+v'²+w'²)
I have obtained the fluctuations u and v (Assuming that my rms fluctuations are correct) and w=v= 0.33307 then the turbulence kinetic energy=
k = 0.5 * (0.43857²+0.33307²+0.33307²)

Am I heading in the right direction? Thank you, Sir.

 

[Chestermiller]

Thank you for clearing that up Chet! Unfortunately I don't have a textbook or lecture slides on this, but I have borrowed some books from the library.

I learned that:
U(x,t) =U (x,t) + u(x,t)

Where U is the velocity, U is the time averaged velocity, u is the fluctuations.
I read that:

U (x,t) = ½T ∫ U(x,t +t') dt'
But I'm not quite sure on how much T I should take (although I read that it has to be significantly large), and I don't really know how to solve that (with t' and dt'), so I decided on taking another method.

Here I calculate the average of the velocity u (axial) and v (radial), and obtained 7.5375 and -0.0125.
I take this as the time-averaged velocity.
And thus I calculate the fluctuations.

The first fluctuations:
u(x,t) = 7.6 - 7.5375
u(x,t) = 0.0625
Did the same to the rest.

Finally I calculate that the rms fluctuation values for axial direction and radial direction and obtained 0.43857 and 0.33307. The flow is thus not isotropic at B.I also read that the turbulence kinetic energy equals to
k = ½ (u'²+v'²+w'²)
I have obtained the fluctuations u and v (Assuming that my rms fluctuations are correct) and w=v= 0.33307 then the turbulence kinetic energy=
k = 0.5 * (0.43857²+0.33307²+0.33307²)

Am I heading in the right direction? Thank you, Sir.

Yes.

 

[nightingale]

Yes.

Chet, would it be alright if I ask you about the subsequent questions? I'm worried that I might be making mistakes, like the ones I did with the fluctuations.
The next question asks me to calculate the magnitude of shear stress at B, and I read from a book that the total stress for turbulent flow is:
τ = ρv(dU/dx) - ρu'v' (I did not consider the isotropic stress)

Where v is the kinematic viscosity.
I can calculate the turbulent shear stress:
τ = -ρu'v'
where u' and v' are the fluctuating components.

Thus the shear stress is:
τ = -ρu'v'
τ = -1.2 * 0.333 * 0.43857
Do I have to consider the fluctuations in the tangential direction?

What bothers me is the the forth question, the viscous shear stress at B. The viscous shear stress formula is:
τ = ρv(dU₁/dx₂)
I think that the U here is the time average velocity in the axial direction... but I don't understand what is (dU₁/dx₂). (
τ = 1.8 * 10^-5* (d7.5375/dx)
τ = 1.8 * 10^-5*1
τ = 1.8 * 10^-5

I can only think of doing it this way, but I doubt that is actually correct. Could you please let me know what is dU₁/dx₂?
Another book (Pope, 2000) actually writes that:
τ = ρv(dU₁/dx₂+dU₂/dx₁)

Which formula do I use? and again, do I have to consider the fluctuations in the tangential direction?
Thank you, Sir.

 

[Chestermiller]

Chet, would it be alright if I ask you about the subsequent questions? I'm worried that I might be making mistakes, like the ones I did with the fluctuations.
The next question asks me to calculate the magnitude of shear stress at B, and I read from a book that the total stress for turbulent flow is:
τ = ρv(dU/dx) - ρu'v' (I did not consider the isotropic stress)

Where v is the kinematic viscosity.
I can calculate the turbulent shear stress:
τ = -ρu'v'
where u' and v' are the fluctuating components.

Thus the shear stress is:
τ = -ρu'v'
τ = -1.2 * 0.333 * 0.43857
Do I have to consider the fluctuations in the tangential direction?

This is not correct. You don't use the rms values of the fluctuations in the two directions. You use the average of the product of the individual fluctuations in the two directions.

What bothers me is the the forth question, the viscous shear stress at B. The viscous shear stress formula is:
τ = ρv(dU₁/dx₂)
I think that the U here is the time average velocity in the axial direction... but I don't understand what is (dU₁/dx₂). (
τ = 1.8 * 10^-5* (d7.5375/dx)
τ = 1.8 * 10^-5*1
τ = 1.8 * 10^-5

I can only think of doing it this way, but I doubt that is actually correct. Could you please let me know what is dU₁/dx₂?
Another book (Pope, 2000) actually writes that:
τ = ρv(dU₁/dx₂+dU₂/dx₁)

Which formula do I use?

You need to use the average velocities at the points A, B, and C to get the derivative of the average velocity.

and again, do I have to consider the fluctuations in the tangential direction?

No. Not for this.

 

[nightingale]

This is not correct. You don't use the rms values of the fluctuations in the two directions. You use the average of the product of the individual fluctuations in the two directions.

Thank you, Sir. I found that the average fluctuations u' and v' are both zero. Thus the shear stress is zero.
τ = -ρu'v'
τ = -1.2 * 0 * 0
τ = 0
Will this be acceptable?

You need to use the average velocities at the points A, B, and C to get the derivative of the average velocity.

I found the average velocities at A, B and C respectively are:
A= 7.525
B= 7.5375
C =7.8625

I was thinking that the derivative of the average velocity may be obtained if I plot them on a chart, taking the gradient from the equation?

Thus the derivative of the average velocity is 0.4219?

Thank you very much for your help.

 

[Chestermiller]

Thank you, Sir. I found that the average fluctuations u' and v' are both zero. Thus the shear stress is zero.
τ = -ρu'v'
τ = -1.2 * 0 * 0
τ = 0
Will this be acceptable?

No. I said the "average of the products," not the "product of the averages."

I found the average velocities at A, B and C respectively are:
A= 7.525
B= 7.5375
C =7.8625

I was thinking that the derivative of the average velocity may be obtained if I plot them on a chart, taking the gradient from the equation?

Thus the derivative of the average velocity is 0.4219?

I don't know how you got that average velocity plot, but it doesn't look right. How did you end up with the funny shape?
For the velocity gradient, I get: (7.8625 - 7.525)/0.8

Chet

 

[nightingale]

No. I said the "average of the products," not the "product of the averages."

I tried searching about the difference between the 'average of products' and the 'product of averages'. But I still don't quite get it yet. Do you mean I should reduce the each velocity components with each other and find the fluctuations?

At B:
u' =u1 - u2 = 7.6 - 8 = -0.4
u' =u2 - u3 = 8 - 8.1 = -0.1
u' =u3 - u4 = 8.1 - 8 = 0.1
u' =u4 - u5 = 8 - 7.5 = 0.5
u' =u5 - u6 = 7.5 - 7.1 = 0.4
u' =u6 - u7 = 7.1 - 6.9 = 0.2
u' =u7 - u8 = 6.9 - 7.1 = -0.2
Average u' = (-0.4-0.1+0.1+0.5+0.4+0.2-0.2)/7 or 7.6 - 7.1/7
Average u' = 0.07143

v' =v1 - v2 = 0.3 - 0.1 = 0.2
v' =v2 - v3 = 0.1 + 0.3 = 0.4
v' =v3 - v4 = -0.3 + 0.6 = 0.3
v' =v4 - v5 = -0.6 + 0.3 =-0.3
v' =v5 - v6 = -0.3 - 0 = -0.3
v' =v6 - v7 = 0.0 - 0.4 = -0.4
v' =v7 - v8 = 0.4 - 0.3 = 0.1
Average v' = (0.2+0.4+0.3-0.3-0.3-0.4+0.1)/7 or 0.3 - 0.3/7
Average v' = 0

And thus
τ = -ρu'v'
τ = -ρ* 0.07143 *0
Is this correct?
Thank you kind Sir.