Turbulent Kinetic Energy

rietuk · February 19, 2013, 13:25

Hi,
I am conducting a CFD validation against some experimental data.
I am trying to compare turbulent kinetic energy as I am given the RMS of the fluctuating components for the 3 directions in an experimental report.
Am I correct in using the formula k=0.5*(Urms^2+Vrms^2+Wrms^2) to calculate the experimental TKE?
When I have done this the experimental and CFD TKE results are orders of magnitude different but the pressure profile is similar
Thanks for your help

mp121209 · February 19, 2013, 15:16

k' = u'2 + v'2 + w'2

rietuk · February 19, 2013, 15:45

Hi,
So if I have the RMS of the fluctuating components does this equal the u'2 etc. as you mentioned or is it or does it need squaring to equal that?

Thanks for your reply

FMDenaro · February 19, 2013, 16:32

The most important fact in the comparison, is that you must use the same fluctuations that means you must use the same average in the experimental measure and in the numerical simulation!

rietuk · February 19, 2013, 16:39

Ok
From CFD I have Turbulent Kinetic Energy in J/Kg

From experimental I have RMS of fluctuating velocitities u~, v~ and w~, all squared and multiplied the sum by 0.5 to give k.

Is the k comparable to the Turbulent Kinetic Energy from the CFD?

Thanks for your help, much appreciated!

FMDenaro · February 19, 2013, 16:59

Quote:

Originally Posted by rietuk

Ok
From CFD I have Turbulent Kinetic Energy in J/Kg

From experimental I have RMS of fluctuating velocitities u~, v~ and w~, all squared and multiplied the sum by 0.5 to give k.

Is the k comparable to the Turbulent Kinetic Energy from the CFD?

Thanks for your help, much appreciated!

the problem is not in the unit but in the way the measure is done and in the way the CFD is formulated.

When you decompose your velocity in an averaged and a residual component, such a decomposition says you nothing if you don't define which is the averaged velocity... Generally, I suppose that you have in the experiment a Reynolds average, therefore your CFD formulation must reproduce the same average

rietuk · February 19, 2013, 18:32

That makes sense, thank you

The experiment has an average velocity of 1m/s to give the Reynolds number of 40000

To match this Reynolds number the CFD average velocity is 15m/s

How do I scale the CFD TKE or experimental fluctuating velocity components so when I calculate TKE the experimental and CFD TKE are comparable?
Thank you!

February 19, 2013, 13:25	Turbulent Kinetic Energy	#1
rietuk New Member rietuk Join Date: Feb 2013 Posts: 12 Rep Power: 13	Hi, I am conducting a CFD validation against some experimental data. I am trying to compare turbulent kinetic energy as I am given the RMS of the fluctuating components for the 3 directions in an experimental report. Am I correct in using the formula k=0.5*(Urms^2+Vrms^2+Wrms^2) to calculate the experimental TKE? When I have done this the experimental and CFD TKE results are orders of magnitude different but the pressure profile is similar Thanks for your help

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February 19, 2013, 15:16		#2
mp121209 New Member Trần Đình Thắng Join Date: Mar 2012 Location: Moscow Russia, Hanoi Vietnam Posts: 24 Rep Power: 14	k' = u'2 + v'2 + w'2

February 19, 2013, 15:45		#3
rietuk New Member rietuk Join Date: Feb 2013 Posts: 12 Rep Power: 13	Hi, So if I have the RMS of the fluctuating components does this equal the u'2 etc. as you mentioned or is it or does it need squaring to equal that? Thanks for your reply

February 19, 2013, 16:32		#4
FMDenaro Senior Member Filippo Maria Denaro Join Date: Jul 2010 Posts: 6,897 Rep Power: 73	The most important fact in the comparison, is that you must use the same fluctuations that means you must use the same average in the experimental measure and in the numerical simulation!

February 19, 2013, 16:39		#5
rietuk New Member rietuk Join Date: Feb 2013 Posts: 12 Rep Power: 13	Ok From CFD I have Turbulent Kinetic Energy in J/Kg From experimental I have RMS of fluctuating velocitities u~, v~ and w~, all squared and multiplied the sum by 0.5 to give k. Is the k comparable to the Turbulent Kinetic Energy from the CFD? Thanks for your help, much appreciated!

February 19, 2013, 18:32		#7
rietuk New Member rietuk Join Date: Feb 2013 Posts: 12 Rep Power: 13	That makes sense, thank you The experiment has an average velocity of 1m/s to give the Reynolds number of 40000 To match this Reynolds number the CFD average velocity is 15m/s How do I scale the CFD TKE or experimental fluctuating velocity components so when I calculate TKE the experimental and CFD TKE are comparable? Thank you!