TKE budget equation in LES

doctorWho · March 11, 2014, 23:58

Hi,

I am calculating TKE budget equation by using solutions from LES simulation.
From filtered momentum equation of incompressible flow,(Please note than variables are filtered/resolved value and filter symbol is omitted)
$\frac{\partial u_i }{\partial t} + \frac{\partial}{\partial x_j} (u_i u_j) = -\frac{1}{\rho} \frac{\partial p}{\partial x_i} + \nu \frac{\partial}{\partial x_j}(\frac{\partial u_i}{\partial x_j} + \frac{\partial u_j}{\partial x_i}) - \frac{\partial \tau_{ij}}{\partial x_j}$
By using linear eddy-viscosity model, $\tau_{ij} = -2 \nu_t S_{ij}$ , the above equation reduces to
$\frac{\partial u_i }{\partial t} + \frac{\partial}{\partial x_j} (u_i u_j) = -\frac{1}{\rho} \frac{\partial p}{\partial x_i} + (\nu + \nu_t)\frac{\partial}{\partial x_j}(\frac{\partial u_i}{\partial x_j} + \frac{\partial u_j}{\partial x_i})$

I could get tke budget equation using the above equation with time averaging like RANS
$\frac{\partial k }{\partial t} + \bar{u_j}\frac{\partial k}{\partial x_j} = -\bar{u_i' u_j'} \frac{\partial \bar{u_i}}{\partial x_j} - \frac{1}{\rho} \frac{\partial \bar{u_i' p'}}{\partial x_j} - \frac{1}{2} \frac{\partial \bar{u_i' u_i' u_j'}}{\partial x_j} + (\nu+\nu_t) \frac{\partial^2 k}{\partial x_j^2} - (\nu+\nu_t) \bar{\frac{\partial u_i'}{\partial x_j}\frac{\partial u_i'}{\partial x_j}}$
where bar means time mean and tke

and resolved fluctuation $u_i'=u_i-\bar{u}_i.$

this equation is basically identical to the counterpart from RANS except for $\nu+\nu_t$ instead of $\nu$

When I applied this equation to the results of flow around a circular cylinder, I found there is imbalance around shear layer otherwise balance is made.

I have checked my discretizing each term and it looks good. When I compared resolved turbulent stress (u'v', u'w' and v'w') to subgrid stress ( $\tau_{ij}$ ), resolved stress is more than 99%, that means I have resolved most of turbulent field. How can I explain this imbalance around shear layer?

Thanks in advance.

FMDenaro · March 12, 2014, 03:49

According to the book of Sagaut, the equations 3.30 to 3.33 are in the form you should use. Now, is your solution statistically steady? What about production and dissipation budget?

doctorWho · March 12, 2014, 10:02

FMDenaro

Yes, the solution is statistically steady. As I understand, equation 3.30 of Sagaut is equation for instantaneous resolved kinetic energy not resolved turbulent kinetic energy. Also equation 3.33 is equation for unresolved kinetic energy but mine is equation for resolved turbulent kinetic energy which is following the time averaging of RANS.
Do you mean that I cannot use tke equation in LES?

As I check production and dissipation around shear layer, production is about 10 times larger than dissipation.

FMDenaro · March 12, 2014, 11:27

To be more clear, when you compute a LES solution your velocity field is

ULES = G*u = u -u'LES

then you can compute the satistical average

<ULES> = <u> -<u'LES>

Therefore, being <u> = URANS = u -u'RANS

<ULES> = URANS -<u'LES>

You can see the difference ...

March 11, 2014, 23:58	TKE budget equation in LES	#1
doctorWho New Member SSSSS Join Date: Jun 2011 Posts: 29 Rep Power: 16	Hi, I am calculating TKE budget equation by using solutions from LES simulation. From filtered momentum equation of incompressible flow,(Please note than variables are filtered/resolved value and filter symbol is omitted) $\frac{\partial u_i }{\partial t} + \frac{\partial}{\partial x_j} (u_i u_j) = -\frac{1}{\rho} \frac{\partial p}{\partial x_i} + \nu \frac{\partial}{\partial x_j}(\frac{\partial u_i}{\partial x_j} + \frac{\partial u_j}{\partial x_i}) - \frac{\partial \tau_{ij}}{\partial x_j}$ By using linear eddy-viscosity model, $\tau_{ij} = -2 \nu_t S_{ij}$ , the above equation reduces to $\frac{\partial u_i }{\partial t} + \frac{\partial}{\partial x_j} (u_i u_j) = -\frac{1}{\rho} \frac{\partial p}{\partial x_i} + (\nu + \nu_t)\frac{\partial}{\partial x_j}(\frac{\partial u_i}{\partial x_j} + \frac{\partial u_j}{\partial x_i})$ I could get tke budget equation using the above equation with time averaging like RANS $\frac{\partial k }{\partial t} + \bar{u_j}\frac{\partial k}{\partial x_j} = -\bar{u_i' u_j'} \frac{\partial \bar{u_i}}{\partial x_j} - \frac{1}{\rho} \frac{\partial \bar{u_i' p'}}{\partial x_j} - \frac{1}{2} \frac{\partial \bar{u_i' u_i' u_j'}}{\partial x_j} + (\nu+\nu_t) \frac{\partial^2 k}{\partial x_j^2} - (\nu+\nu_t) \bar{\frac{\partial u_i'}{\partial x_j}\frac{\partial u_i'}{\partial x_j}}$ where bar means time mean and tke and resolved fluctuation $u_i'=u_i-\bar{u}_i.$ this equation is basically identical to the counterpart from RANS except for $\nu+\nu_t$ instead of $\nu$ When I applied this equation to the results of flow around a circular cylinder, I found there is imbalance around shear layer otherwise balance is made. I have checked my discretizing each term and it looks good. When I compared resolved turbulent stress (u'v', u'w' and v'w') to subgrid stress ( $\tau_{ij}$ ), resolved stress is more than 99%, that means I have resolved most of turbulent field. How can I explain this imbalance around shear layer? Thanks in advance. Last edited by doctorWho; March 12, 2014 at 10:04.

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March 12, 2014, 03:49		#2
FMDenaro Senior Member Filippo Maria Denaro Join Date: Jul 2010 Posts: 6,983 Rep Power: 73	According to the book of Sagaut, the equations 3.30 to 3.33 are in the form you should use. Now, is your solution statistically steady? What about production and dissipation budget?

March 12, 2014, 10:02		#3
doctorWho New Member SSSSS Join Date: Jun 2011 Posts: 29 Rep Power: 16	FMDenaro Yes, the solution is statistically steady. As I understand, equation 3.30 of Sagaut is equation for instantaneous resolved kinetic energy not resolved turbulent kinetic energy. Also equation 3.33 is equation for unresolved kinetic energy but mine is equation for resolved turbulent kinetic energy which is following the time averaging of RANS. Do you mean that I cannot use tke equation in LES? As I check production and dissipation around shear layer, production is about 10 times larger than dissipation.

March 12, 2014, 11:27		#4
FMDenaro Senior Member Filippo Maria Denaro Join Date: Jul 2010 Posts: 6,983 Rep Power: 73	To be more clear, when you compute a LES solution your velocity field is ULES = G*u = u -u'LES then you can compute the satistical average <ULES> = <u> -<u'LES> Therefore, being <u> = URANS = u -u'RANS <ULES> = URANS -<u'LES> You can see the difference ...