# Homework Help: Isotropic Turbulence Flow

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1. Oct 23, 2015

### nightingale

1. The problem statement, all variables and given/known data
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.

2. Oct 23, 2015

### Staff: Mentor

What does the word "isotropic" mean to you?

Chet

3. Oct 24, 2015

### nightingale

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.

4. Oct 24, 2015

### Staff: Mentor

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

5. Oct 24, 2015

### bigfooted

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.

6. Oct 24, 2015

### Staff: Mentor

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.
Yes, and....?

7. Oct 25, 2015

### bigfooted

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

8. Oct 25, 2015

### nightingale

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.

9. Oct 25, 2015

### nightingale

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.

10. Oct 25, 2015

### 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.

11. Oct 25, 2015

### Staff: Mentor

12. Oct 25, 2015

### Staff: Mentor

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

13. Oct 25, 2015

### nightingale

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!

14. Oct 25, 2015

### Staff: Mentor

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

15. Oct 26, 2015

### nightingale

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.

16. Oct 26, 2015

### Staff: Mentor

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

Chet

17. Oct 26, 2015

### nightingale

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.

18. Oct 26, 2015

### Staff: Mentor

k = 0.5 * (0.3352 +0.3352+0.3352)

19. Oct 26, 2015

### nightingale

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)
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.

20. Oct 26, 2015

### Staff: Mentor

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

Chet

21. Oct 26, 2015

### nightingale

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
u3 is the velocity fluctuations in the circumferential (tangential) = 0.355"

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

22. Oct 26, 2015

### nightingale

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!

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23. Oct 26, 2015

### Staff: Mentor

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

24. Oct 27, 2015

### nightingale

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.

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.

Last edited: Oct 27, 2015
25. Oct 27, 2015

Yes.