Calculating energy dissipation rate from power spectrum

rdemyan · July 26, 2023, 10:57

The energy dissipation rate is given by:

$\epsilon=\int_0^\infty 2\nu\kappa^2E(\kappa) d\kappa$

But shouldn't the lower value for the integral be the wavelength corresponding to the largest energy containing eddies and not zero. It doesn't make sense to me to have a system where K is equal to zero which corresponds to infinite eddy size.

FMDenaro · July 26, 2023, 11:25

Quote:

Originally Posted by rdemyan

The energy dissipation rate is given by:

$\epsilon=\int_0^\infty 2\nu\kappa^2E(\kappa) d\kappa$

But shouldn't the lower value for the integral be the wavelength corresponding to the largest energy containing eddies and not zero. It doesn't make sense to me to have a system where K is equal to zero which corresponds to infinite eddy size.

k=n*2*pi/L, n=0 corresponds to the constant mean value. However, you have no contribution to the dissipation

rdemyan · July 26, 2023, 12:17

Quote:

Originally Posted by FMDenaro

k=n*2*pi/L, n=0 corresponds to the constant mean value. However, you have no contribution to the dissipation

I'm trying to calculate the cumulative dissipation by numerically integrating this equation using the Pope model for E(K) (Eqn. 6.256 in Pope's book).

$\epsilon=2C\nu\epsilon^{2/3}\eta^{-4/3}\int_0^\infty (\kappa\eta)^{1/3}f_\eta(\kappa\eta)d(\kappa\eta)$

My criterion is that the total cumulative dissipation divided by $\epsilon$ needs to be very close to 100% (this typically happens by the time $\kappa\eta$ reaches around 2). However, the starting value of the integral impacts whether the total cumulative dissipation divided by $\epsilon$ is close to 100%. $c_{\eta}$ is a constant in the dissipation function ( $f_\eta$ ) portion of the Pope model (Eqn. 6.248 in Pope's book), which according to Pope is determined by the criterion that the equation I posted needs to integrate to $\epsilon$ , which I believe means that the cumulative dissipation divided by $\epsilon$ determined by the numerical integration is equal to 100%.

So, for example, if the large eddy size (L) is substituted for 0 in the integral, then if L is equal to 344 microns, $\kappa\eta$ = 0.03785 and $c_{\eta}$ = 0.415, which is very close to Pope's value of 0.4 at high Reynolds number. But if L is equal to 22 microns, then $c_{\eta}$ has to be equal to 0.55, in order for the criteria that the cumulative dissipation divided by $\epsilon$ equals 100% at high values of $\kappa\eta$ .

Ultimately I'm trying to plot the cumulative dissipation divided by $\epsilon$ versus $\kappa\eta$ to determine what the cumulative dissipation value is at a certain value of $\kappa\eta$ . So, for the 2nd example (L = 22 microns) shouldn't a specific value of $\kappa\eta$ (say 0.5) have a higher cumulative dissipation divided by $\epsilon$ than the first example (where L = 344 microns) at $\kappa\eta$ = 0.5?

FMDenaro · July 26, 2023, 12:30

Quote:

Originally Posted by rdemyan

I'm trying to calculate the cumulative dissipation by numerically integrating this equation using the Pope model for E(K) (Eqn. 6.256 in Pope's book).

$\epsilon=2C\nu\epsilon^{2/3}\eta^{-4/3}\int_0^\infty (\kappa\eta)^{1/3}f_\eta(\kappa\eta)d(\kappa\eta)$

My criterion is that the total cumulative dissipation divided by $\epsilon$ needs to be very close to 100% (this typically happens by the time $\kappa\eta$ reaches around 2). However, the starting value of the integral impacts whether the total cumulative dissipation divided by $\epsilon$ is close to 100%. $c_{\eta}$ is a constant in the dissipation function ( $f_\eta$ ) portion of the Pope model (Eqn. 6.248 in Pope's book), which according to Pope is determined by the criterion that the equation I posted needs to integrate to $\epsilon$ , which I believe means that the cumulative dissipation divided by $\epsilon$ determined by the numerical integration is equal to 100%.

So, for example, if the large eddy size (L) is substituted for 0 in the integral, then if L is equal to 344 microns, $\kappa\eta$ = 0.03785 and $c_{\eta}$ = 0.415, which is very close to Pope's value of 0.4 at high Reynolds number. But if L is equal to 22 microns, then $c_{\eta}$ has to be equal to 0.55, in order for the criteria that the cumulative dissipation divided by $\epsilon$ equals 100% at high values of $\kappa\eta$ .

Ultimately I'm trying to plot the cumulative dissipation divided by $\epsilon$ versus $\kappa\eta$ to determine what the cumulative dissipation value is at a certain value of $\kappa\eta$ . So, for the 2nd example (L = 22 microns) shouldn't a specific value of $\kappa\eta$ (say 0.5) have a higher cumulative dissipation divided by $\epsilon$ than the first example (where L = 344 microns) at $\kappa\eta$ = 0.5?

The integral lenght of order of microns? Are you sure that is meaningful of a turbulent flow? In microfluidics you have to consider also other effects.
I dont know if the exercise of Pope makes sense at such a low scale.

rdemyan · July 26, 2023, 12:57

Quote:

Originally Posted by FMDenaro

The integral lenght of order of microns? Are you sure that is meaningful of a turbulent flow? In microfluidics you have to consider also other effects.
I dont know if the exercise of Pope makes sense at such a low scale.

I don't see why it wouldn't. The equation given by Pope models the energy spectrum to at least the Kolmogorov microscale. For the two examples I gave, the Kolmogorov microscale is 2 microns and 0.7 microns respectively. Admittedly not fully developed turbulence but it should work I would think.

FMDenaro · July 26, 2023, 13:04

Quote:

Originally Posted by rdemyan

I don't see why it wouldn't. The equation given by Pope models the energy spectrum to at least the Kolmogorov microscale. For the two examples I gave, the Kolmogorov microscale is 2 microns and 0.7 microns respectively. Admittedly not fully developed turbulence but it should work I would think.

Ok, but the Taylor microscale, that defines the onset of the dissipation region, is larger, I doubt you have some fundamental characteristic of turbulence like an inertial range.

rdemyan · July 26, 2023, 13:32

Quote:

Originally Posted by FMDenaro

Ok, but the Taylor microscale, that defines the onset of the dissipation region, is larger, I doubt you have some fundamental characteristic of turbulence like an inertial range.

I think that you are probably right regarding the inertial range specifically for the 22 micron example. However, does that mean that Pope's model can't accurately model the spectrum? Fig. 6.14 in Pope shows experimental energy spectrum data for 17 different systems and Pope claims that his model fits all of this data quite well. Taylor Reynolds numbers for some of the data in Fig. 6.14 are as low as 23. The Taylor Reynolds numbers for the two examples I cited are 63 and 45, respectively. So, it seems to me that Pope's model should work in my case. While the largest scales are small in my system, the energy dissipation rates are very high: 54,000 W/kg for L = 344 microns and 3,700,000 W/kg for L = 22 microns.

One other note: My system is decaying turbulence but some of the data shown in Fig. 6.14 in Pope's book is for grid turbulence (Taylor Reynolds numbers varying from 37 to 540).

So back to my original question: Should the lower bound of the integral be equal to zero or equal to L (examples of 344 microns and 22 microns)?

FMDenaro · July 26, 2023, 13:46

Quote:

Originally Posted by rdemyan

I think that you are probably right regarding the inertial range specifically for the 22 micron example. However, does that mean that Pope's model can't accurately model the spectrum? Fig. 6.14 in Pope shows experimental energy spectrum data for 17 different systems and Pope claims that his model fits all of this data quite well. Taylor Reynolds numbers for some of the data in Fig. 6.14 are as low as 23. The Taylor Reynolds numbers for the two examples I cited are 63 and 45, respectively. So, it seems to me that Pope's model should work in my case. While the largest scales are small in my system, the energy dissipation rates are very high: 54,000 W/kg for L = 344 microns and 3,700,000 W/kg for L = 22 microns.

One other note: My system is decaying turbulence but some of the data shown in Fig. 6.14 in Pope's book is for grid turbulence (Taylor Reynolds numbers varying from 37 to 540).

So back to my original question: Should the lower bound of the integral be equal to zero or equal to L (examples of 344 microns and 22 microns)?

If you see the figure 6.14, the spectra at Re_lambda=36 (BL flow) shows practically no inertial range. The dissipation range starts after few symbols. Pope wrote often "for high Reynolds flow" to highlight that fact that in turbulence you have a range of inertial transfer energy (that is almost inviscid). Your flow problems has an integral scale that is practically also the Taylor microscale. And, being your problem in decaying condition, the range will be totally dissipating.

FMDenaro · July 26, 2023, 13:55

Integration is performed from k=0, fig. 6.16. Try to plot your dissipation spectra.

rdemyan · July 26, 2023, 13:58

Quote:

Originally Posted by FMDenaro

If you see the figure 6.14, the spectra at Re_lambda=36 (BL flow) shows practically no inertial range. The dissipation range starts after few symbols. Pope wrote often "for high Reynolds flow" to highlight that fact that in turbulence you have a range of inertial transfer energy (that is almost inviscid). Your flow problems has an integral scale that is practically also the Taylor microscale. And, being your problem in decaying condition, the range will be totally dissipating.

I see a BL 23 but not a BL 36. I agree with your comments, but on Fig 6.14 for BL 23, Pope is still able to fit the data using his spectrum model. So while your comments are certainly correct, his model still fits that BL 23 data - so I should be able use his model for my system. For the BL data that you cite, do you think he used a lower value for the integral of zero?

Sorry, I didn't see your post just above.

FMDenaro · July 26, 2023, 14:06

Quote:

Originally Posted by rdemyan

I see a BL 23 but not a BL 36. I agree with your comments, but on Fig 6.14 for BL 23, Pope is still able to fit the data using his spectrum model. So while your comments are certainly correct, his model still fits that BL 23 data - so I should be able use his model for my system. For the BL data that you cite, do you think he used a lower value for the integral of zero?

Sorry, I didn't see your post just above.

I suppose the fit cna somehow work but is not really meaningful at such a low Re number.

Try to replicate the computation of the energy spectra and dissipative spectra. I am curious to see if you can get a relevant difference between the integral lenght and the Taylor microscale.
The integral starts always from k=0.

rdemyan · July 26, 2023, 14:10

Quote:

Originally Posted by FMDenaro

Integration is performed from k=0, fig. 6.16. Try to plot your dissipation spectra.

Yes, Fig. 6.16, the dissipation spectrum, is what I am using as a guide. The problem is that if I start from k = 0, there is virtually no difference between my two examples. As I think about it some more, that probably makes sense. Fig. 6.14 shows that all of the 17 models converge once the dissipation range is reached. It's only at lower values of $\kappa\eta$ that there is a difference between the 17 systems. So, that would seem to suggest that Fig. 6.16, the dissipation spectra, is pretty accurate for all of the 17 systems shown in Fig. 6.14. Would you agree with that?

But, why does Pope say that the coefficient, $c_\eta$ , needs to be changed from the high Re value of 0.4 based on 100% cumulative dissipation equaling the energy dissipation rate?

FMDenaro · July 26, 2023, 14:32

Quote:

Originally Posted by rdemyan

Yes, Fig. 6.16, the dissipation spectrum, is what I am using as a guide. The problem is that if I start from k = 0, there is virtually no difference between my two examples. As I think about it some more, that probably makes sense. Fig. 6.14 shows that all of the 17 models converge once the dissipation range is reached. It's only at lower values of $\kappa\eta$ that there is a difference between the 17 systems. So, that would seem to suggest that Fig. 6.16, the dissipation spectra, is pretty accurate for all of the 17 systems shown in Fig. 6.14. Would you agree with that?

But, why does Pope say that the coefficient, $c_\eta$ , needs to be changed from the high Re value of 0.4 based on 100% cumulative dissipation equaling the energy dissipation rate?

The fact you see no differences in your two cases should confirm you are working at a very low Re number.

I don't have now in mind that part of the Pope discussion, however he showed the case at high Re number 600.

LuckyTran · July 26, 2023, 19:45

6.256 uses a model valid only in the dissipative range (i.e. infinite reynolds number). Why are you are not at least using 6.246 if you want to see what happens at lower ranges?

Any model function you use for the dissipation spectra must satisfy

Quote:

Originally Posted by rdemyan

$\epsilon=\int_0^\infty 2\nu\kappa^2E(\kappa) d\kappa$

integrated from 0 to infinity. This comes from the definition of turbulent dissipation rate which also results in the contribution at exactly 0 frequency to be 0. So technically you can start integrated at some infinitesimal start at 0+dk. Infinitesimally small is not equivalent to arbitrarily large. They are quite opposite.

rdemyan · July 26, 2023, 20:34

Quote:

Originally Posted by LuckyTran

6.256 uses a model valid only in the dissipative range (i.e. infinite reynolds number). Why are you are not at least using 6.246 if you want to see what happens at lower ranges?

Any model function you use for the dissipation spectra must satisfy

integrated from 0 to infinity. This comes from the definition of turbulent dissipation rate which also results in the contribution at exactly 0 frequency to be 0. So technically you can start integrated at some infinitesimal start at 0+dk. Infinitesimally small is not equivalent to arbitrarily large. They are quite opposite.

Thanks for your comment. Based on my discussions with FMDenaro, I am now looking at 6.246, but my main interest is the dissipation range (but that could change). It's been instructive to run the calcs for 6.246. Things are pretty much as expected based on Fig. 6.14 except that the cumulative value of k from the spectrum is not always equaling k calculated from my energy dissipation expression. I'm still checking for errors and trying to figure out why there is a difference.

LuckyTran · July 27, 2023, 08:09

If dissipative scales are your fetish then feel free to indulge

But then you should not expect that integrating a subset of the spectra matches the mean in a universal way especially considering that the largest scales are the least universal amongst scales. In other words, you not only know that integrating from L to infinity cannot add up to 100%, it must add up to less by a function that is Reynolds number dependent. You need a new criteria

rdemyan · July 27, 2023, 20:37

Quote:

Originally Posted by FMDenaro

If you see the figure 6.14, the spectra at Re_lambda=36 (BL flow) shows practically no inertial range. The dissipation range starts after few symbols. Pope wrote often "for high Reynolds flow" to highlight that fact that in turbulence you have a range of inertial transfer energy (that is almost inviscid). Your flow problems has an integral scale that is practically also the Taylor microscale. And, being your problem in decaying condition, the range will be totally dissipating.

@FMDenaro: Attached are charts similar to the form shown in Fig. 6.13 in Pope. The orange line represents the -5/3 slope for the inertial range. They can be roughly compared with Fig. 6.14 in terms of the ordinate values based on Re; even though Fig. 6.14 is for $E_{11}(\kappa)$ and the attached are for $E(\kappa)$ . As you expected the inertial range is relatively short at about a decade on the $\kappa\eta$ scale for Re = 63 and about a factor of 3 to 4 for Re = 45.

LuckyTran · July 28, 2023, 00:35

Yes you have discovered the concept of separation of scales in turbulent flows, which increases with increasing Reynolds number. You can imagine that at low enough Re when you become wholly laminar and you have only dissipative scales.

Btw your E(k) is assuming local isotropy and isn't actually more general than E11.

FMDenaro · July 28, 2023, 04:12

Quote:

Originally Posted by rdemyan

@FMDenaro: Attached are charts similar to the form shown in Fig. 6.13 in Pope. The orange line represents the -5/3 slope for the inertial range. They can be roughly compared with Fig. 6.14 in terms of the ordinate values based on Re; even though Fig. 6.14 is for $E_{11}(\kappa)$ and the attached are for $E(\kappa)$ . As you expected the inertial range is relatively short at about a decade on the $\kappa\eta$ scale for Re = 63 and about a factor of 3 to 4 for Re = 45.

Some observations:
1) the inertial range in the energy spectra is also more extended than I expected. Note that the slope -5/3 in the dissipation range has no meaning as term of comparison.

2) The model dissipation spectra in Pope shows a clear exponential decay, your figure seems different, that should be the effect of the too low Re number.

3) You wrote you are working on a decadying turbulence, that means you do not have an energy equilibrium state. Energy and dissipation spectra will change during the decaying.

rdemyan · July 28, 2023, 10:28

Quote:

Originally Posted by FMDenaro

Some observations:
1) the inertial range in the energy spectra is also more extended than I expected. Note that the slope -5/3 in the dissipation range has no meaning as term of comparison.

2) The model dissipation spectra in Pope shows a clear exponential decay, your figure seems different, that should be the effect of the too low Re number.

3) You wrote you are working on a decaying turbulence, that means you do not have an energy equilibrium state. Energy and dissipation spectra will change during the decaying.

@FMDenaro: I don't really notice a difference in the dissipation range decay between the Pope model and the charts (the charts were generated using the Pope model, so.....). The turbulence dies out after essentially one turnover time of the largest eddies. In terms of calculating the energy dissipation rate, I do assume an equilibrium between production and dissipation; however, there is a time lag which per Lumley is a function of Re with a maximum value of twice the large eddy turnover time (he calculated this assuming that energy is transferred between adjacent scales only). In DNS simulations, Pearson's group comes up with a maximum of one turnover time which they attribute to energy being directly passed to much smaller scales as opposed to Lumley's assumption.

@LuckyTran: The reason I am interested in the dissipation spectra is because the size of the micromixing scale appears to be a function of this time lag. The longer the time lag the more time available for the scales to be "ground down". The micromixing scale appears to be a function of the cumulative dissipation, which cumulative dissipation I am calculating by using Pope's model. Also, I was comparing the values of E(K) with E11(K) and according to Pope's Fig. 6.11, there is a difference. I just wanted to see if the values shown on my charts were in the right ball park for the Reynolds numbers (which they appear to be).

July 26, 2023, 10:57	Calculating energy dissipation rate from power spectrum	#1
rdemyan Member Join Date: Jan 2023 Posts: 64 Rep Power: 3	The energy dissipation rate is given by: $\epsilon=\int_0^\infty 2\nu\kappa^2E(\kappa) d\kappa$ But shouldn't the lower value for the integral be the wavelength corresponding to the largest energy containing eddies and not zero. It doesn't make sense to me to have a system where K is equal to zero which corresponds to infinite eddy size.

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July 26, 2023, 13:55		#9
FMDenaro Senior Member Filippo Maria Denaro Join Date: Jul 2010 Posts: 6,773 Rep Power: 71	Integration is performed from k=0, fig. 6.16. Try to plot your dissipation spectra.

July 27, 2023, 08:09		#16
LuckyTran Senior Member Lucky Join Date: Apr 2011 Location: Orlando, FL USA Posts: 5,675 Rep Power: 66	If dissipative scales are your fetish then feel free to indulge But then you should not expect that integrating a subset of the spectra matches the mean in a universal way especially considering that the largest scales are the least universal amongst scales. In other words, you not only know that integrating from L to infinity cannot add up to 100%, it must add up to less by a function that is Reynolds number dependent. You need a new criteria

July 28, 2023, 00:35		#18
LuckyTran Senior Member Lucky Join Date: Apr 2011 Location: Orlando, FL USA Posts: 5,675 Rep Power: 66	Yes you have discovered the concept of separation of scales in turbulent flows, which increases with increasing Reynolds number. You can imagine that at low enough Re when you become wholly laminar and you have only dissipative scales. Btw your E(k) is assuming local isotropy and isn't actually more general than E11.