how is TKE defined?

dkoletsos · May 22, 2020, 17:17

I understand that the turbulent kinetic energy is defined as k=1/2(<u_i*u_i>)
where does the factor of 1/2 come from? Is this pure convention? Has an integral been taken? Is there a more general equation, an averaging equation that brings in the factor 1/2?

stud-many · May 23, 2020, 07:01

As a kinetic Energy is defined as 1/2 m c**2 , it makes sense to define the tke as 1/2sum(ui'ui'). i guess it would also make sense to multiply this with the fluid-density. wouldn't it? I was only working on incompressible turbulent flows so far. for me, it was pretty handy not to have the density in there

sbaffini · May 23, 2020, 08:00

There are several possible answers to this, but for all of them it really is a matter of convention. When you look at it from the fluid mechanics point of view, you really just need to look at it from the following perspective:

$\phi \frac{\partial \phi}{\partial t} = \frac{1}{2}\frac{\partial \phi^2}{\partial t}$

as the factor really comes out from differentiation. This is for the general kinetic energy. When you apply any average or filter to that, you end up with its turbulent counterpart that, obviously, has the same factor.

From a more general perspective in mechanics, having the factor and the mass in the definition makes it an exact differential which is cool for some manipulations.

May 22, 2020, 17:17	how is TKE defined?	#1
dkoletsos New Member dimitri koletsos Join Date: May 2020 Posts: 3 Rep Power: 5	I understand that the turbulent kinetic energy is defined as k=1/2(<u_i*u_i>) where does the factor of 1/2 come from? Is this pure convention? Has an integral been taken? Is there a more general equation, an averaging equation that brings in the factor 1/2?

May 23, 2020, 07:01		#2
stud-many New Member Malte Join Date: Nov 2017 Posts: 20 Rep Power: 8	As a kinetic Energy is defined as 1/2 m c**2 , it makes sense to define the tke as 1/2sum(ui'ui'). i guess it would also make sense to multiply this with the fluid-density. wouldn't it? I was only working on incompressible turbulent flows so far. for me, it was pretty handy not to have the density in there dkoletsos likes this.

May 23, 2020, 08:00		#3
sbaffini Senior Member Paolo Lampitella Join Date: Mar 2009 Location: Italy Posts: 2,151 Blog Entries: 29 Rep Power: 39	There are several possible answers to this, but for all of them it really is a matter of convention. When you look at it from the fluid mechanics point of view, you really just need to look at it from the following perspective: $\phi \frac{\partial \phi}{\partial t} = \frac{1}{2}\frac{\partial \phi^2}{\partial t}$ as the factor really comes out from differentiation. This is for the general kinetic energy. When you apply any average or filter to that, you end up with its turbulent counterpart that, obviously, has the same factor. From a more general perspective in mechanics, having the factor and the mass in the definition makes it an exact differential which is cool for some manipulations. dkoletsos likes this.

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