Turbulence Chapter04

8/16/2019 Turbulence Chapter04

1/13

Prof. Dr.-Ing. Andreas Kempf, Chair of Fluid Dynamics

Original course material by J. Janicka, TU-Darmstadt

TurbulenceLecture 4

Reynolds Averaging


2/13



Reminder: DNS is not practical

"Turbulence in a box"

Re: 40,000Grid: 40963 (70 billion cells)

Computer: Earth SimulatorT. Ishihara, T. Gotoh, Y. Kaneda, Study of High-Reynoldds Number Isotropic Turbulence by Direct Numerical Simulation, Annual Reviewof Fluid Mechanics 2009, 41:165-180.


3/13



•

Do we really need all the details of a DNS? – Time resolved flow field in 3D?

– The entire temporal evolution at every point?

☝ Are time dependent quantities meaningful for other realisations?

– All probability density functions?

– All moments?

– Any other statistical quantity?

• So, what are we really interested in?

– Mean values (first moments)

– Perhaps fluctuation levels (second moments, stresses)

– Perhaps probablity density functions (for combustion)

A less costly approach than DNS is needed!


4/13



•

We are mainly interested in the mean values.• We need equations for mean quantities!

• Reynolds decomposition can provide the mean quantities.

• It is now sufficient to substitute ui and p in the equations to get the

"Reynolds averaged Navier Stokes equations"

u u u ui i i i

= + ! ! =, 0

p p p p= + ! ! =, 0

Reynolds Decomposition

!"

!

!"

! t

u

x

i

i

+ = 0

! "

" !

"

"

"

"

"

" µ

"

"

"

"

"

" # !

u

t u

u

x

p

x x

u

x

u

x

u

x g i j

i

j i j

i

j

j

i

k

k

ij i+ = $ + + $%

&' (

)*+

,--

.

/00

+2

3

✍

u02

i ≥ 0

p02

i ≥ 0

u0

iu0

j ???


5/13



Derivation of Reynolds-Averaged Navier Stokes Eq.


6/13



•

The Conservation equation for mass (constant viscosity and density)

(3.3.1a)

• The Conservation equation for momentum (constant viscosity and

density)

(3.3.2a)

0=

i

i

x

u

!

!

! "

" !

"

"

"

"

"

" µ

"

"

"

"

"

" # !

u

t u

u

x

p

x x

u

x

u

x

u

x g i j

i

j i j

i

j

j

i

k

k

ij i+ = $ + + $%

&'

(

)*

+--

.00

+

2

3

0

i

i j

j

j j

i

i j

i j

i g x x

u

x x

u

x

p

x

uu

t

u !

" "

"

" "

" µ

"

"

"

" !

"

" ! +

##

$

%

&&

'

(++)=+

22 0

i

j j

i

i j

i

j

i g x x

u

x

p

x

uu

t

u++!=+

" "

" #

"

"

$ "

"

"

" 21

Averaged Balance Equations Ia


7/13



Averaged Balance Equations I

The Conservation equation for mass (instantaneous)

(3.3.1)

The Conservation equation for momentum (instantaneous)

(3.3.2)

!"

!

!"

! t

u

x

i

i

+ = 0

! "

" !

"

"

"

"

"

" µ

"

"

"

"

"

" # !

u

t u

u

x

p

x x

u

x

u

x

u

x g i j

i

j i j

i

j

j

i

k

k

ij i+ = $ + + $%

&'

(

)*

+

,--

.

/00

+

2

3


8/13



Applying Reynolds decomposition to the conservation equation formass and momentum, and considering ! and µ to be constant yields:

(3.3.5)

with

(3.3.6)

! = µ / " ( )

!

!

!

!

u

x

u

x

i

i

i

i

+

"= 0

( )!

!

!

!

!

!

u

t

u

t xu u u u u u u ui i

j

i j i j i j i j+

"+ + " + " + " " =

! ! "

+ +

"#

$%

&

'( +

1 1 2 2

)

*

* )

*

* +

*

* *

*

* *

p

x

p

x

u

x x

u

x x g i i

i

j j

i

j j

i

Averaged Balance Equations III


9/13



For constant density, time averaging yields:

After the averaging process, the unknown

Reynolds Stress Tensor arises in the conservation equation

for momentum.

!

Closure problem of turbulence for the Reynolds-Averaged-

Navier Stokes equations.

!

!

u

x

i

i

= 0

!

!

!

! "

!

!

!

! #

!

! !

u

t

u u

x

p

x

u u

x

u

x x g i

i j

j i

i j

j

i

j j

i+ = $ $

% %+ +

1 2

! !u ui j

Reynolds Averaged Navier Stokes Equations

ρu0iu0

j

i =

m2

s21

m3kg = m

2

s21

m3Ns2

m =

N m2


10/13



Interpretation of Reynolds Stresses

•

Similar to viscous stresses

• Example in a shear layer :

!

!

!

! "

!

!

!

! #

!

! !

u

t

u u

x

p

x

u u

x

u

x x g i

i j

j i

i j

j

i

j j

i+ = $ $

% %+ +

1 2

∂ u1

∂ x2> 0

"21,v

"

21,t

x1

x2

"21,v

viscous stress turbulent stress

u1

u1

u01

u2

u0

2u

0

1

u1

u1

u0

1u2

u0

2u

0

1

molecule (discontinuous) fluid particle (continuous)

n e t f l u

x o f h o r i -

z o n t a l m o m e n t u m

"21,t


11/13



The result

The Reynolds Averaged equations for Reynolds averaged velocities

are identical to

the Navier Stokes equations for the instantaneous velocities

with the exception of the Reynolds Stresses.

! Similar CFD codes can be used

!

Turbulent flows behave „similar“ to laminar flows

! A turbulent flow might be thought of as „non-Newtonian“

!

!

u

x

i

i

= 0

!

!

!

! "

!

!

!

! #

!

! !

u

t

u u

x

p

x

u u

x

u

x x g i

i j

j i

i j

j

i

j j

i+ = $ $

% %+ +

1 2


12/13



The sign of Reynolds stresses (example for cosines)

The sign of a Reynolds stress depends on whether u‘ and v‘ arepositively correlated (+), negatively corelated (-), or un-correlated (0).

u0v0

u0 = 0

v0 = 0

u0

, v0

t

v0

u0

u0

v0

u0 v

0

v0


13/13



Problem

How to determine turbulent stresses?• Derive transport equations: Second moment closure

• Exploit similarity to viscous stresses

– Eddy viscosity / Bousinesq approach, with turbulent viscosity

(most common turbulence models)

– How to determine turbulent viscosity?

Documents

Turbulence Chapter04