Integral Length Scale - 2D

WhiteShadow · May 15, 2018, 08:05

Hello Everyone,

I want to calculate the integral length scale directly from the energy spectrum. According to Wiki (https://en.wikipedia.org/wiki/Integral_length_scale)
this can be done via

$l=\frac{\int_0^\infty E(k)k^{-1}\,\text{d}k}{\int_0^\infty E(k)\,\text{d}k}.$

But I am working on 2D-flows and I have reason to suspect, that this formula should include a factor of 2 in this case, though I couldn't find anything about this.
Does anyone know whether there is indeed a factor of 2 or if I am mistaken?

Thanks a lot and greetings

FMDenaro · May 16, 2018, 04:09

What is your specific reasoning for the factor 2 in 2D?

Actually, my opinion is that any evalution of the integral lenght is for the order of magnitude, so that a factor of 2 would not change so much.
See https://www.researchgate.net/publica...Numerical_Data

WhiteShadow · May 16, 2018, 04:48

I used a gaussian Spectrum with

$E(k)=\frac 18 \sigma^2 k^3 L^4 \text{exp}(-\frac{k^2L^2}{4})$ .

I calculated the longitudinal correlation function via

$R_{11}(r \vec{e}_1)=\int_{R^2}e^{i \vec{k} \cdot \vec{e}_1 r}\Phi_{11}(\vec{k}) \, \text{d}\vec{k}$ .

In this case I obtained

$R_{11}(r \vec{e}_1)/v_{\text{rms}}^2= f(r) = e^{-\frac{r^2}{L^2}}$ .

The integral length scale

is the integral over the correlation function

$l=\int_0^\infty f(r) \,\text{d}r = \sqrt{\pi}/2 L$ .

But if I calculate the same quantity via the method in my first post,
I get

$l = \sqrt{\pi}/4 L$ .

LuckyTran · May 17, 2018, 23:52

These formulae look quite arbitrary. What is the result for 3D?

WhiteShadow · May 18, 2018, 11:46

I'm sorry, but I do not understand. What do you think is arbitrary?

FMDenaro · May 19, 2018, 17:56

I think you could start by the general definition in terms of the integral of the autocorrelation, considering that spectra and correlation are related each other by direct/inverse Fourier transformation.
But in general the determination of the integral scale is somehow still debated, as you can see for example here
http://www.turbulence-online.com/Pub...apers/WG02.pdf

May 15, 2018, 08:05	Integral Length Scale - 2D	#1
WhiteShadow New Member Join Date: May 2017 Posts: 10 Rep Power: 9	Hello Everyone, I want to calculate the integral length scale directly from the energy spectrum. According to Wiki (https://en.wikipedia.org/wiki/Integral_length_scale) this can be done via $l=\frac{\int_0^\infty E(k)k^{-1}\,\text{d}k}{\int_0^\infty E(k)\,\text{d}k}.$ But I am working on 2D-flows and I have reason to suspect, that this formula should include a factor of 2 in this case, though I couldn't find anything about this. Does anyone know whether there is indeed a factor of 2 or if I am mistaken? Thanks a lot and greetings

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May 16, 2018, 04:09		#2
FMDenaro Senior Member Filippo Maria Denaro Join Date: Jul 2010 Posts: 6,877 Rep Power: 73	What is your specific reasoning for the factor 2 in 2D? Actually, my opinion is that any evalution of the integral lenght is for the order of magnitude, so that a factor of 2 would not change so much. See https://www.researchgate.net/publica...Numerical_Data

May 16, 2018, 04:48		#3
WhiteShadow New Member Join Date: May 2017 Posts: 10 Rep Power: 9	I used a gaussian Spectrum with $E(k)=\frac 18 \sigma^2 k^3 L^4 \text{exp}(-\frac{k^2L^2}{4})$ . I calculated the longitudinal correlation function via $R_{11}(r \vec{e}_1)=\int_{R^2}e^{i \vec{k} \cdot \vec{e}_1 r}\Phi_{11}(\vec{k}) \, \text{d}\vec{k}$ . In this case I obtained $R_{11}(r \vec{e}_1)/v_{\text{rms}}^2= f(r) = e^{-\frac{r^2}{L^2}}$ . The integral length scale is the integral over the correlation function $l=\int_0^\infty f(r) \,\text{d}r = \sqrt{\pi}/2 L$ . But if I calculate the same quantity via the method in my first post, I get $l = \sqrt{\pi}/4 L$ .

May 17, 2018, 23:52		#4
LuckyTran Senior Member Lucky Join Date: Apr 2011 Location: Orlando, FL USA Posts: 5,752 Rep Power: 66	These formulae look quite arbitrary. What is the result for 3D?

May 18, 2018, 11:46		#5
WhiteShadow New Member Join Date: May 2017 Posts: 10 Rep Power: 9	I'm sorry, but I do not understand. What do you think is arbitrary?

May 19, 2018, 17:56		#6
FMDenaro Senior Member Filippo Maria Denaro Join Date: Jul 2010 Posts: 6,877 Rep Power: 73	I think you could start by the general definition in terms of the integral of the autocorrelation, considering that spectra and correlation are related each other by direct/inverse Fourier transformation. But in general the determination of the integral scale is somehow still debated, as you can see for example here http://www.turbulence-online.com/Pub...apers/WG02.pdf