A problem regarding the calculation of the turbulence index

tumultous_calmness · October 1, 2024, 07:31

hi all, I have a very fundamental, probably borderline stupid question. First a bit background:

I am trying to come up with a consistent methodology that attempts to locate the transition location on blade surfaces in a turbomachine. After spending so much time on many different methods (which were also not applicable to all cases of simulations), I came across a what seems to be a universal yet robust enough formula, i.e. also in the title, turbulence index, from Spalart&Allmaras - "A One-Equation Turbulence Model for Aerodynamic Flows,", 1994. Which is given as below;

$i_T = \frac{1}{\kappa u_{\tau}}\frac{\partial \hat{\nu}}{\partial n}$ ,

where $\kappa$ is the von karman constant=0.41, $u_{\tau}$ is the friction velocity,

is the wall normal and finally $\hat{\nu}=\kappa y u_{\tau}$ is sort of the modified turbulent viscosity, where

is the wall distance.

My problem is, my normal vector in the software that I am using, is only defined on the solid surface and does not exist for free field. This is a problem since the

is obviously 0 on the wall so subsequently my all

values are also 0. If I'm not wrong

is also not used on the wall but at a small distance.

Is there a smart trick that I can use here? For example not using the wall distance=0 on the surface but simply a very small yet constant $y= \delta$ value? Would such a trick violate the validity of this method, just because my wall normal vector does not extend into the free field?

I know this is a bit futile question as long as one doesn't have the software that I use, but just to gather ideas I wanted to pose this question here. hanks in advance for whatever idea you may suggest me.

sbaffini · October 2, 2024, 12:12

It seems that you need the wall normal derivative of nu_hat.

Depending from where you need it, there are different options.

If you just need it at the first cells near a wall, it is pretty easy (you can either approximate it with a one sided difference or directly retrieve the wall normal from the given face where it is available; you could also use the known near wall behavior of nu_hat to compute the derivative analytically).

If you need it everywhere, you need to compute the gradient of the wall distance scalar. Depending from which software you are using, it might or not be feasible but, once you get that you just do a scalar product with the nu_hat gradient.

sbaffini · October 3, 2024, 17:27

My first answer was before having a chance to actually read again the S&A paper. Also, I couldn't catch your specification of nu_hat from the email (where latex is not converted).

NO, you aren't supposed to use that formula away from walls (worried about n, think about u_tau now...) AND NO, you aren't supposed to use $\hat{\nu}=\kappa y u_{\tau}$ outside the log layer.

So, coming to your question, you can use your formula for

only at walls, where you already have the nu_hat wall normal derivative (i.e., wall diffusion term) or could simply compute it from the cell next to the wall (either one sided at first order or trough cell gradient and wall normal at second order) and you also have u_tau. That's it

tumultous_calmness · October 4, 2024, 06:00

sbaffini, thank you for answer. glad it ignites a discussion here.

I actually understand that with a self-written code for a finite difference formula (there's no wall normal derivative/wall diffusion term inheritely in the program), I can at worst calculate a simple wall normal derivative and get a first idea as to how

behaves over the blade surface.

However what I cannot wrap my head around is that, we know that

, i.e. wall distance, is

on the walls. This already makes $\hat{\nu}=0$ by definition. What am I missing here? Additionally, what about calculating a derivative of a quantity that only exists on wall? My gut tells me there is a subtle way to overcome such a fundamental problem, however my fundamentals knowledge tells me this quantity doesn't make sense.

edit: I must say my background on fluid mechanics and cfd is mostly from atmospheric physics. So I was not formally educated on turbomachinery thus lack a bit of familierty on some basic field specifics.

LuckyTran · October 4, 2024, 09:55

y=0 on wall means nothing. What you are looking for is the gradient. y=0 on walls for example does not mean the wall normal velocity gradient, pressure gradient or wall normal temperature gradient is zero (unless it is the BC). This isn't specific to one industry, it's a fundamental misconception.

sbaffini · October 5, 2024, 06:47

Quote:

Originally Posted by tumultous_calmness

sbaffini, thank you for answer. glad it ignites a discussion here.

I actually understand that with a self-written code for a finite difference formula (there's no wall normal derivative/wall diffusion term inheritely in the program), I can at worst calculate a simple wall normal derivative and get a first idea as to how

behaves over the blade surface.

However what I cannot wrap my head around is that, we know that

, i.e. wall distance, is

on the walls. This already makes $\hat{\nu}=0$ by definition. What am I missing here? Additionally, what about calculating a derivative of a quantity that only exists on wall? My gut tells me there is a subtle way to overcome such a fundamental problem, however my fundamentals knowledge tells me this quantity doesn't make sense.

edit: I must say my background on fluid mechanics and cfd is mostly from atmospheric physics. So I was not formally educated on turbomachinery thus lack a bit of familierty on some basic field specifics.

As mentioned by LuckyTran, a function can have any finite value at n=0 and this still says nothing about its d/dn derivative at n=0 (using here, for correctness, the wall normal coordinate n, which is the one actually involved, while y being a superflous and confusing simplification in your case).

Imagine a function f like:

f = an + b

then you have df/dn = a. As you can probably better see now, b doesn't affect df/dn.

Now, your nu_hat is indeed 0 at walls by definition, but certainly not its derivative (unless it turns out so from the computation).

P.S. For equilibrium boundary layers nu_hat is meant to be linear from the wall up to the end of the log-layer (my bad, in my previous post I wrongly worded this concept, and it seemed that it holded ONLY in the log-layer, which is not). Still, this is not valid in general so, again, don't rely on this in general.

tumultous_calmness · October 9, 2024, 10:50

Hi again, thank you once more for comments. I think I have to explain myself more in detail. Essentially the problem of evaluating this term at the wall apeears when we look at the expansion of the derivative:

$\frac{1}{\kappa u_{\tau}}\frac{\partial (y\kappa u_{\tau})}{\partial n} =\frac{\partial y}{\partial n} + \frac{y}{\kappa}\frac{\partial \kappa}{\partial n} + \frac{y}{ u_{\tau}}\frac{\partial u_{\tau}}{\partial n}$

In above equation 2nd term is 0, regardless of where you evaluate it since $\kappa$ is a constant. 3rd term is not 0 in the free field, but 0 at the wall, since

at the wall. Then all we are left is $\frac{\partial y}{\partial n}$ , which in no way measures any turbulent condition from a given system.

Am I making a mistake here with the expansion?

sbaffini · October 9, 2024, 11:18

What you need to understand is that your expression for nu_hat is not general, but only valid for equilibrium boundary layers

LuckyTran · October 9, 2024, 16:29

Quote:

Originally Posted by tumultous_calmness

$\frac{1}{\kappa u_{\tau}}\frac{\partial (y\kappa u_{\tau})}{\partial n} =\frac{\partial y}{\partial n} + \frac{y}{\kappa}\frac{\partial \kappa}{\partial n} + \frac{y}{ u_{\tau}}\frac{\partial u_{\tau}}{\partial n}$

I'm not sure where you are even going with this expansion.

You have committed the grave sin of dividing by 0. 0/0 is not 1. Hence, you are in fact, not left with only dy/dn

Additionaly, y is n and you have arrived dn/dn is ???? which it should be the identity relation. And of course, that clearly tells you nothing.

Put another way, if I say df/dy = df/dy and I put in y=0, that still doesn't tell me anything because I never knew what was df/dy to begin with.

tumultous_calmness · October 10, 2024, 04:38

Quote:

Originally Posted by LuckyTran

I'm not sure where you are even going with this expansion.

You have committed the grave sin of dividing by 0. 0/0 is not 1. Hence, you are in fact, not left with only dy/dn

Additionaly, y is n and you have arrived dn/dn is ???? which it should be the identity relation. And of course, that clearly tells you nothing.

Put another way, if I say df/dy = df/dy and I put in y=0, that still doesn't tell me anything because I never knew what was df/dy to begin with.

excuse me but you are not being polite, why are you spending time with these words instead of showing what you see wrong? I highly suspect that you are fighting with an imaginary person in your head.

Coming back to the discussion, where is the division by 0?? y is not n btw, y is the closest distance to the wall, where n is normal to the blade surface. towards hub and shroud surfaces there is significant difference between the two.

from this point on I will only reply to people who are taking my point seriously and not lose time try to repeat the same things over and over again.

sbaffini · October 11, 2024, 10:25

Quote:

Originally Posted by tumultous_calmness

excuse me but you are not being polite, why are you spending time with these words instead of showing what you see wrong? I highly suspect that you are fighting with an imaginary person in your head.

Coming back to the discussion, where is the division by 0?? y is not n btw, y is the closest distance to the wall, where n is normal to the blade surface. towards hub and shroud surfaces there is significant difference between the two.

from this point on I will only reply to people who are taking my point seriously and not lose time try to repeat the same things over and over again.

While I don't get either where the 0/0 should come from, it is certainly true that y=n in our context. Indeed, the segment connecting any fluid point with the closest point on a wall surface is always perpendicular to that surface. The only exception to this is when at the closest point the normal is not well defined, like cusps. In this case any theory is certainly far from its applicability ranges but, in any case, the two points are well defined; also, the normal doesn't exist itself there.

In addition to this, note that u_tau is only defined at walls. It only exists there because its definition involves a derivative taken at the wall. You could, kind of arbitrarily, associate an u_tau to a fluid point away from walls by locating the nearest point on a wall and its associated u_tau, but that's not how u_tau should be used in general (also, equilibrium flows are, again, special).

So, again, forget your expression for u_tau and just compute the derivative as in the original expression which, let me further clarify it, just involves two wall derivatives and k, so it is only valid at walls and/or their adjacent cells.

tumultous_calmness · October 15, 2024, 07:07

Quote:

Originally Posted by sbaffini

While I don't get either where the 0/0 should come from, it is certainly true that y=n in our context. Indeed, the segment connecting any fluid point with the closest point on a wall surface is always perpendicular to that surface. The only exception to this is when at the closest point the normal is not well defined, like cusps. In this case any theory is certainly far from its applicability ranges but, in any case, the two points are well defined; also, the normal doesn't exist itself there.

In addition to this, note that u_tau is only defined at walls. It only exists there because its definition involves a derivative taken at the wall. You could, kind of arbitrarily, associate an u_tau to a fluid point away from walls by locating the nearest point on a wall and its associated u_tau, but that's not how u_tau should be used in general (also, equilibrium flows are, again, special).

So, again, forget your expression for u_tau and just compute the derivative as in the original expression which, let me further clarify it, just involves two wall derivatives and k, so it is only valid at walls and/or their adjacent cells.

Thanks for your patient explanations @sbaffini. I followed your idea and simply calculated a rough derivative on the wall normal direction and divided it by the $u_{\tau} = \sqrt{\nu\Omega}$ , for $\Omega = \lvert \vec{\nabla}\times\vec{W} \lvert$ , which is suggested to be used by Spalart&Allmaras('92). It definitely shows me interesting results, however they are not perfectly correct.

Not to prolonge the discussion unnecessarily but for anyone who might see this and want to know what I did, I simply took the derivative as:

$i_T \approx \frac{1}{\kappa u_{\tau}}\frac{\hat{\nu}^{(1st Cell)} - \hat{\nu}^{(Wall)}}{\Delta d}$ ,

where $\Delta d$ is the distance between the blade wall and the first cell. A typical

distribution should show a sharp change from 0 to 1, right around the location of transition from laminar to turbulence. Values I get seem to find very well the approximate transition location however there's no sharp change from 0 to 1 whatsoever. It hovers from blade tip until the transition between 0.7-0.9 and then after the transition goes above 1. These results could be very well specific to my simulated case (for example, I am not sure how much this index is valid for internal flows, almost all examples I found of its usage was applied on external flows...), however for my initial attempt it seems to be not entirely but somewhat wrong.

sbaffini · October 15, 2024, 15:23

Quote:

Originally Posted by tumultous_calmness

Thanks for your patient explanations @sbaffini. I followed your idea and simply calculated a rough derivative on the wall normal direction and divided it by the $u_{\tau} = \sqrt{\nu\Omega}$ , for $\Omega = \lvert \vec{\nabla}\times\vec{W} \lvert$ , which is suggested to be used by Spalart&Allmaras('92). It definitely shows me interesting results, however they are not perfectly correct.

Not to prolonge the discussion unnecessarily but for anyone who might see this and want to know what I did, I simply took the derivative as:

$i_T \approx \frac{1}{\kappa u_{\tau}}\frac{\hat{\nu}^{(1st Cell)} - \hat{\nu}^{(Wall)}}{\Delta d}$ ,

where $\Delta d$ is the distance between the blade wall and the first cell. A typical

distribution should show a sharp change from 0 to 1, right around the location of transition from laminar to turbulence. Values I get seem to find very well the approximate transition location however there's no sharp change from 0 to 1 whatsoever. It hovers from blade tip until the transition between 0.7-0.9 and then after the transition goes above 1. These results could be very well specific to my simulated case (for example, I am not sure how much this index is valid for internal flows, almost all examples I found of its usage was applied on external flows...), however for my initial attempt it seems to be not entirely but somewhat wrong.

Few things that might affect your results:

1) u_tau is not random, or sort of, it has to be computed in the exact same way it enters the computation (tau_wall = rho u_tau^2)

2) tau_wall involves the wall normal derivative of the wall parallel velocity at the center of the cell face on the wall, not some quantity evaluated at the near wall cell center

3) the nu_hat wall normal derivative must be taken in the same way of the one involved in tau_wall, which implies taking care of the correct wall normal and distance for non regular grids

Now, there might be reasons for which you can't do all of this, but I would consider them mandatory before any futher inquire on the method and/or implementation

October 1, 2024, 07:31	A problem regarding the calculation of the turbulence index	#1
tumultous_calmness New Member Tumul Calm Join Date: Aug 2023 Location: Europe Posts: 5 Rep Power: 3	hi all, I have a very fundamental, probably borderline stupid question. First a bit background: I am trying to come up with a consistent methodology that attempts to locate the transition location on blade surfaces in a turbomachine. After spending so much time on many different methods (which were also not applicable to all cases of simulations), I came across a what seems to be a universal yet robust enough formula, i.e. also in the title, turbulence index, from Spalart&Allmaras - "A One-Equation Turbulence Model for Aerodynamic Flows,", 1994. Which is given as below; $i_T = \frac{1}{\kappa u_{\tau}}\frac{\partial \hat{\nu}}{\partial n}$ , where $\kappa$ is the von karman constant=0.41, $u_{\tau}$ is the friction velocity, is the wall normal and finally $\hat{\nu}=\kappa y u_{\tau}$ is sort of the modified turbulent viscosity, where is the wall distance. My problem is, my normal vector in the software that I am using, is only defined on the solid surface and does not exist for free field. This is a problem since the is obviously 0 on the wall so subsequently my all values are also 0. If I'm not wrong is also not used on the wall but at a small distance. Is there a smart trick that I can use here? For example not using the wall distance=0 on the surface but simply a very small yet constant $y= \delta$ value? Would such a trick violate the validity of this method, just because my wall normal vector does not extend into the free field? I know this is a bit futile question as long as one doesn't have the software that I use, but just to gather ideas I wanted to pose this question here. hanks in advance for whatever idea you may suggest me.

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October 2, 2024, 12:12		#2
sbaffini Senior Member Paolo Lampitella Join Date: Mar 2009 Location: Italy Posts: 2,195 Blog Entries: 29 Rep Power: 39	It seems that you need the wall normal derivative of nu_hat. Depending from where you need it, there are different options. If you just need it at the first cells near a wall, it is pretty easy (you can either approximate it with a one sided difference or directly retrieve the wall normal from the given face where it is available; you could also use the known near wall behavior of nu_hat to compute the derivative analytically). If you need it everywhere, you need to compute the gradient of the wall distance scalar. Depending from which software you are using, it might or not be feasible but, once you get that you just do a scalar product with the nu_hat gradient.

October 3, 2024, 17:27		#3
sbaffini Senior Member Paolo Lampitella Join Date: Mar 2009 Location: Italy Posts: 2,195 Blog Entries: 29 Rep Power: 39	My first answer was before having a chance to actually read again the S&A paper. Also, I couldn't catch your specification of nu_hat from the email (where latex is not converted). NO, you aren't supposed to use that formula away from walls (worried about n, think about u_tau now...) AND NO, you aren't supposed to use $\hat{\nu}=\kappa y u_{\tau}$ outside the log layer. So, coming to your question, you can use your formula for only at walls, where you already have the nu_hat wall normal derivative (i.e., wall diffusion term) or could simply compute it from the cell next to the wall (either one sided at first order or trough cell gradient and wall normal at second order) and you also have u_tau. That's it

October 4, 2024, 06:00		#4
tumultous_calmness New Member Tumul Calm Join Date: Aug 2023 Location: Europe Posts: 5 Rep Power: 3	sbaffini, thank you for answer. glad it ignites a discussion here. I actually understand that with a self-written code for a finite difference formula (there's no wall normal derivative/wall diffusion term inheritely in the program), I can at worst calculate a simple wall normal derivative and get a first idea as to how behaves over the blade surface. However what I cannot wrap my head around is that, we know that , i.e. wall distance, is on the walls. This already makes $\hat{\nu}=0$ by definition. What am I missing here? Additionally, what about calculating a derivative of a quantity that only exists on wall? My gut tells me there is a subtle way to overcome such a fundamental problem, however my fundamentals knowledge tells me this quantity doesn't make sense. edit: I must say my background on fluid mechanics and cfd is mostly from atmospheric physics. So I was not formally educated on turbomachinery thus lack a bit of familierty on some basic field specifics.

October 4, 2024, 09:55		#5
LuckyTran Senior Member Lucky Join Date: Apr 2011 Location: Orlando, FL USA Posts: 5,761 Rep Power: 66	y=0 on wall means nothing. What you are looking for is the gradient. y=0 on walls for example does not mean the wall normal velocity gradient, pressure gradient or wall normal temperature gradient is zero (unless it is the BC). This isn't specific to one industry, it's a fundamental misconception.

October 9, 2024, 10:50		#7
tumultous_calmness New Member Tumul Calm Join Date: Aug 2023 Location: Europe Posts: 5 Rep Power: 3	Hi again, thank you once more for comments. I think I have to explain myself more in detail. Essentially the problem of evaluating this term at the wall apeears when we look at the expansion of the derivative: $\frac{1}{\kappa u_{\tau}}\frac{\partial (y\kappa u_{\tau})}{\partial n} =\frac{\partial y}{\partial n} + \frac{y}{\kappa}\frac{\partial \kappa}{\partial n} + \frac{y}{ u_{\tau}}\frac{\partial u_{\tau}}{\partial n}$ In above equation 2nd term is 0, regardless of where you evaluate it since $\kappa$ is a constant. 3rd term is not 0 in the free field, but 0 at the wall, since at the wall. Then all we are left is $\frac{\partial y}{\partial n}$ , which in no way measures any turbulent condition from a given system. Am I making a mistake here with the expansion?

October 9, 2024, 11:18		#8
sbaffini Senior Member Paolo Lampitella Join Date: Mar 2009 Location: Italy Posts: 2,195 Blog Entries: 29 Rep Power: 39	What you need to understand is that your expression for nu_hat is not general, but only valid for equilibrium boundary layers