Reynold stress model (RSM) - Lunds tekniska högskolaReynold stress model (RSM) Rixin Yu, 2017 1....

Reynold stress model (RSM)

Rixin Yu, 2017

1

Reynold stress model (RSM)

• Part 1.1 • Characterize the Reynold stress “anisotropy”. • Lumley triangle

• Part 1.2 – Reynold stress equations.

• Derivation • Brief examination

2

What we plan to do now!• Reynold stresses contain information more than k

– k is only the isotropic part of Reynold stress!

• To characterize the Reynold stress “anisotropy”– Using Lumley triangle to represent “anisotropy”.

• Tensor “invariant” concept– It is related to Eigenvalue (Linear algebra) concept.

3

Characterization of Reynold-stress anisotropy

= − 232 = − 23 /2

1: 2nd order tensor 2: symmetrical 2: real valued3: zero trace4: realizable, e.g. ≤ ( + ) +….

Tensor invariant for ?

= ( + + )

( ) ≤ ⋅

Recap: Tensor element values change at a different coordinate

5

#

# #

= 1, 0, 0# = 0, 1, 0

= 0 0 00 0 00 0# = 0 00 0 00 0 0

1st order tensor(vector)

2nd order tensor

Note: such a is NOT realizable, but we just use here for simple demonstration

Characterize of Reynold-stress anisotropy Recap: Tensor elements change values at different coordinate:

6In coordinate

#

# #

= 100

Example: a 2nd order tensor ( ) formed as the product of two 1st order tensor (vector) : and .

In # coordinate

# = 010≡ = 0 0 00 0 00 0

=001 # ≡ # # = 0 00 0 00 0 0

#=100A rotated coordinate #

Original coordinate

Tensor invariants• Tensor elements change values as coordinate changes (kept to be orthogonal)

• Can we find some expressions formed using the tensor element values, those expressions will keep constant under coordinate changes ?

– For 1st order tensor, i.e. = ( , , )• Length of the vector: = + + = # + # + #

– For 2nd order tensor ,• ( ,… , ) = ( # ,… , # )– forexample, ( ,… , ) ≡ + 2 +⋯• ,… , = ( # ,… , # )

• ???7

≡ = 0 0 00 0 00 0 # ≡ # # = 0 00 0 00 0 0#

# #= 100 # = 010

Tensor invariants for 2nd order symmetrical tensor • Yes, we can!

– Treat as matrix, find its 3 eigenvalues

8

= − /2 = ×

= − = 0 | − | = 0 + ⋅ + ⋅ + = 0= + += + += ( − )( − )( − ) = 03 principle

invariants to coordinate change!

Characteristic polynomial

Characterization of Reynold-stress anisotropy-The Lumley triangle

(Real, symmetrical, zero trace tensor)

= + + = = 0= + + = −= = 16 − 12 + 13

33

22

36

26

iib

iib

bIII

bII

==

=−=

ξ

η

Tensor*Tensor Tensorcollapse index low order tensor If reach to first order tensor, scalar (invariant)

= ≠ Compact notation

10

NOT realizable if outside the triangle

Lumley triangle

= + + = 0= + + = −3= = 2

= ⇒ = / − / − /

= / / / b=

= / / b=/ / − /

Two-equations models vs. RSM

12

= −1 ̅ + −= 0

= ( ) + −

= ( ) + −− 23 = − ( + )= 12

Mean continuity:

Mean momentum:

Eddy-viscosity-assumption.

= + Π + −

= ( ∗ ) + −

RSM : 6 transport equations1 transport equations+5 relations

23 +. .

Derive Reynold stress equations

13

Denote = + + − = 0( + ) + ( + ) + 1 − = 0⋅ = ( + ) + ( + ) + 1 − = 0⋅ = ( + ) + ( + ) + 1 − = 0

+ = + + + + ++1 + − ( + ) = 0

Expand first two terms:

Not expand yet

Non-conservative form

Expand then take average: + = 0

14

+ + + + + + 1 ++ + = 0( + )+ + + +

+1 + − 2 + = 0

= 0

+ = −2 +

0

Reynold stress equations

15

+ = − + Π − +

Π = − += + − += −

= − +Production tensor

Dissipation tensor = 2Velocity-pressure-gradient tensor

Viscous diffusionTurbulent

transport

Pressure-rate-of-strain tensor Pressure

transport tensor

Which terms need modelling ?

Reynold stress equations contain the (= ) equation

16

+ = − + Π − +

+ = − − ′ − 12 +Take half trace (collapse j to i )

Standard k-eq

Π = + − += 0 − 2 ′

= 12 = 22 == 12 = −12 += −

How those RS terms looks like in a turbulent boundary layer?

17

+ = − + Π − + 000 0

+ = − + Π − +

Mean Convection Turb. conv Vis. diff.PressureDissp.Prod.

+ = − + Π − +

+ = − + Π − + + = − + Π − +

+ = − − ′ − 12 +

Distribution of Reynold stres

18

< > < > 0< > < > 00 0 < >000 0

Budget of k equation

19

+ = − − ′ − 12 +Mean Convection Turb. conv Vis. diff.PressureDissp.Prod.

Normalized by∑ = 1

12

′

Put together

20

Π ΠΠ

Π ΠΠ

000 0

Dissipation rate

Simplification: use Kolmogorov’s local isotropy hypothesis:

Accurate dissipation tensor = 2 =

= 23 = 23 0 00 00 0

= = 32 + +3

+ = − + Π − +

The dissipation equation

22

= + −= + −= 23

= 12 = −12 += −

= − 2 ( + )equation

In k − = ⋯

A note for dissipation• Near wall ( y+<20)( = is not valid anymore )

23

Π Π

Π Π

Π ΠΠ Π

Reynold stress models-part 2

24


• Model the pressure-rate-of-strain term:

• Contribution by the slow pressure• Contribution by the rapid pressure

– Rapid Distortion Theory (RDT)• Contribution by the harmonic pressure

25


26

+ = − + Π − +

Π = − += + − += −




transport


transport tensor

Which terms need to be modeled?

The redistribution role of

Π ΠΠ

Π ΠΠ

Reynolds stress at a certain direction( , ) is produced and consumed by a joint contriubtions from three terms: the production term , the disspation term and the Π term (or ) .

The contribution by the last term is signficant, it servers to redistribing the reynolds stresses among different directions.

Indeed summation of over all direction gives zero, i.e = 0.

+. . = − + Π … .000 0

Pressure-rate-of-strain tensor

Responsible for redistribition of fluctuations beween different directions

For pressure fluctuations:

Decompose pressure

28Rapid: Slow: Harmonic:

= +

1 ′ = −2 − −

( ) ( ) ( )

= ⋯ ∉ , = 0 = 0+ = −1 +⋯

Pressure-rate-of-strain tensor

rapid

slow

harmonic

The same can be done for the pressure-rate-of-strain tensor

only important close to walls

29

1 ′ + ′ + ′ = −2 − −1 = −21 = − −

= 0( ) = ( ) + ( ) = ( ) +

∞→ 0

First: turn to the ”slow” pressure

30

1 ′ + ′ + ′ = −2 − −

( ) = ( ) +

Consider homgeneous decaying turbulence ( = 0, but anistropy is still present):

+ = − + Π − + Π = ( ( )+ ( ) + ( )) −

= ( ) −Slow pressure rate-of-strain

A greatly-simplified Reynold stress equation for homogenous turbulence

Slow velocity press-gradient

Understand the slow pressure-rate-of-strain- in decaying homgeneous anistropic turbulence

31

reformulated in terms of anisotropy tensor

= ( ) −= 2 − 13

= 12 ⋅ − 12 ⋅= −2 − + 13 −= − + 2 + 232 Kolmogorov local

isotropy hypothesis

ijij εδε 32=

= −

= + 2

Homogenous flow, no production

Model the slow pressure-rate-of-strain

32

= + + 2 = + ⋅ ⇒ =If term is zero, all 6 elements of anisotropy tensor will increase exponently in time

Rotta(1951) :

A modelling Ideal: assume that turbulence has a natural tendency to become less anisotropic as it decays.

= −2 = − − 23= − − 1 Linear, return-to-isotropy

33

22

6

6

ii

ii

b

b

=

=

ξ

η

We need this term in Reynolds stress eq.

(Nonlinear) Model of slow pressure-rate-of-strain

33

Linear: c ⋅

Rotta(1951) :

= − − 1Nonlinear: C ⋅

= 1 + 12 ( ) + 12 − 13…can be (linearly) expressed by

and ; due to Cayley-Hamilton theoremfor 2nd order tensor .Characteristically polynomial: + ⋅ + ⋅ + = 0Plug into the characteristic polynomail+ ⋅ + ⋅ + = 0

(Sakar & Speziale, 1990)( ) = −3.4; = 4.2

Although is zero trace, is not nessary zero trace.

?

Now turn to rapid pressure : Rapid distortion theroy(RDT)What is Rapid distortion?

Turbulence-to-Mean strean time scale ratio is large! 1/ = / ≫ 1Rapid mean-flow distortion:→ −∞ ; → +∞satisfy continuity,+ + = 0

turbulent velocity fluctuations ( )

Rapid distortion theroy(RDT) for homogenous turbulenceThe governing equations for fluctuating velocity and pressure reduce to a set of linear P.D.E.s.

1 ′ + ′ = −2 − −+ + + + 1 + ′ − ′ = 0

Eq. for pressure

There is no nonlinear term anymore (powerful assumption), analytical results can be found. How? Firstly lets start from a single Fourier wave.

Eq.s for the fluctuating velocity

A 1D example“Rapid distortion” of a almostly frozen “scalar fluctuation”

+

-

+

-

++- -

⋅

⋅

The wave number| ( )|increases with time for this particular “distortion”

A kind of elastic process

After a while

Initially

A glimpse of rapid distortion theroy in 2D

37

“rapid” deformation of 2D “frozen” fluctuating scalar field

+

+

+

-

-

-

-

++

+--

--

Mean strain( ⇐ / =0 Same area after distortion)Through , ( ) rotates, changes magnitude | |

+

+

+

-

-

-

-

+

+

+

--

-

-

evolve as ( )⋅( )

“Initially” a single flutuating wave ⋅

Rapid distortion theroy: single-wave result

38

′ = − ′ ( − 2 ̂ ̂ )̂ = − ̂ ( − ̂ ̂ )

Following this “single” time-evolving wave (t), the fluctuation amplitude ( )evolves independent of length of wave-vector | |, only on its orientation .

=Rotate direction

Actual turbulence fluctuations should be summed for all possible wave vectors. Since RDT equations is linear, allowing each wave to evolve independently(then to be summed)

′ ⋅

Rapid distortion theroy: Linearly-summed-all-waves

39

is a fourth order tensor, depends on direction of Fourier mode (partially contained in the velocity-spectrum) , which is not contained in the Reynolds-stress ( ).

′ ⋅= + ( )

Rapid-straining turbulence in a periodic domain :

( ) = 2 ( + )= ∗ ̂ ̂

Two turbulence states may start with thesame but different , their temporal evolutions are not uniquely determined by . This can be a limitation for modelling ( ), same applies to ( ). Two routes return-

to-isotropy for ( )

An example: shear rapid distortion

40Initial, isotropic turbulence

= + ( )Evolve in time due to combined effect of ( )

= − 232 Later, anisotropy ++

t=0

direction 1

direction 2

t>0

RDT model for rapid pressure ( )Initial, isotropic turbulence under rapid, mean distortion :

( ) = 45 = −35= − −= −23 + Ω −⋯= −( ) = −35 ( − 23 ) = ,zero trace

RDT

Anisotropy production

= + ( )

Counteracts 60% of production

Only valid for

Part 3• Reynolds stress models

– Combined models for the pressure-rate-of-strain terms• Slow pressure model + rapid pressure model

– Model the turbulence and pressure transport terms• Gradient diffusion

– Model dissipation• Isotropic model • Non-isotropic model (Near wall consideration)

• A hierarchy of other turbulence models– Algebraic RS models

• Simple but effective approaches inspired from RS-models

– Nonlinear eddy-viscosity models– Elliptical relaxation model

42

Put things together A basic combined model for pressure-rate-of-strain term

= ( ) + ( ) = − − 23 − ( − 23 )

= −2 = − − 23( ) = −35 ( − 23 )

(Launder, Reece, Rodi , 1975)

Simply, widely used, resonable-performence

Linear Rotta model for slow pressure :

Isotropization of Production for rapid pressure

LRR-IP

Other “advanced” pressure-rate-of-strain modelsAll these ”advanced” models can be described in a general form:

=== − 13== + − 23= Ω − Ω= + − 23= Ω − Ω= − 13

Key modelling ingredients:

1: Input tensor: ,

Tensor requirement1.1 Symmetry (antisymmetric)Ω1.2 deviatric (zero trace)−231.3 Normalize (non-dimentional)= ⋅

2: Linear / nonlinearLinear:

slow: c ⋅rapid: c ⋅ or c ⋅ Ω

Nonlinear: slow: C ⋅rapid: none ( Mijkl is linear)

…….

Other “advanced” pressure-rate-of-strain modelssome examples with given coeffcients

=LRR-IPLaunder et al 1975

LRR-QILaunder et al 1975

SSGSpeziale et al 1991−2 −2 − − ∗

0 043 45 − 6 ∗2 611 2 + 32 211 10 − 7

0 0 0

0 0 0

0 0 0

Note that only SSG has non-linear return to isotropy.








46

Modelling Reynold stress equations

47

+ = − + Π − +

Π = − += + − += −




transport


transport tensor

Inhomogenous tubulence(Model turbulent-transport term)

ikj

jki

kjit

ijkpupuuuuT δρ

δρ

′′+

′′+′′′=)(

Normally the pressure transport is neglected (or implicitly included in the modelling of the turbulent transport) on the basis of the energy budget.

Turbulent transport

Pressure transport

Gradient diffusion models:

k

jis

tijk x

uukCT∂

′′∂−=

ε

2)( Shir (1971)

l

jilks

tijk x

uuuukCT

∂

′′∂′′−=

ε)( Daly & Harlow (1970)

⎟⎟⎠

⎞⎜⎜⎝

⎛

∂

′′∂+

∂′′∂

+∂

′′∂−=′′′

k

ji

j

ki

i

kjskji x

uuxuu

xuukCuuu

ε

2

⎟⎟⎠

⎞⎜⎜⎝

⎛∂

′′∂′′+

∂′′∂′′+

∂

′′∂′′−=′′′

l

jilk

l

kilj

l

kjliskji x

uuuu

xuuuu

xuu

uukCuuuε

Mellor & Herring (1973)

Hanjalic & Launder (1972) 48








49

Dissipation rate

Simplification: use Kolmogorov’s local isotropy hypothesis:

Accurate dissipation tensor = 2 =

= 23 = 23 0 00 00 0

= = 32 + +3

+ = − + Π − +

Dissipation rate

Hence, dissipation can be anisotropic close to walls, but also in moderate Reynolds number flow and for large mean rate of strain.

Near wall effects:• Low Reynolds number• High shear rate• Two-component turbulence ( fluctuaiton in direction is grealty inhibited)• Wall blocking (pressure reflections)

A demonstration of anisotropic dissipation near wall using DNS

• Near wall ( y+<20), = does not hold.

52

Π Π

Π Π

Π ΠΠ Π

(To be isotropic, it should be ≈ ≈ and ≈ 0)

Dissipation rate

Near walls region(anisotopic)

53

Two component behavior when approching wall → 0(Exact analysis: series expansion)

= ′ ′ ≠ 2 ≠ 2= 2 ′ ′ ≠ 2= 4 ′ ′

∗ = ′ ′ Using a blending function

= 23a simple model by Rotta (1951)

= ∗ + (1 − ) 23

Far from wall (isotropic)

(Not very accurate!, but not very far off)

= 1aty = 0and → 0 → ∞

Note: the isotropic part of disspation is not zero at the wall = 0!

Note: All 6 Reynold stresses(containing ) will always be zero at wall.

For all regions:

Dissipation rate

A more accurate model accounting for the two-component behaviour closeto the wall:

∗ = ′ ′ + ′ ′ + ′ ′ + ′ ′ ⁄1 + 52 ′ ′

= ∗ + 1 − 23

: Wall normal direction vector

Simiarly blend this near-wall model with the isotropic model

For all regions:

55

The harmonic pressure

At y=0 the pressure fluctuations are:′ = ′Now consider the Poisson equation for pressure fluctuations:

( )jijijii

j

j

i

kkuuuu

xxxu

xu

xxp ′′−′′

∂∂∂

−∂

′∂∂∂

−=∂∂′∂ 22

21ρ

This is independent of viscous effects. Hence for the rapid and slow parts one uses the inviscid boundary condition and for the harmonic part the viscous boundary condition, i.e.:

= ′= 0 = 0Note that p(h) does not affect beyond y+=15








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57

Can Reynold-stress equations be reduced?

In previous slides, ”modeling” of all nessary terms in 6-Reynold Stress equations and 1 disspation equaiton are presented.

However solving 7 RS model equations are more expensive than standard 2-equation models , is there other simplficaton to avoid solving 5 extra equations while retaining some benefits of RS models in terms of better describing RS anistropy. (It should be better than those models based on Boussineq eddy-viscosity assmption).

Yes, that is algerbric stress model.

Compare Reynold stress and k-equations

+ + = + − 23Six eqs.

+ + = − j=i , take halfone k-eq.

Let’s call it , then this must be called

contains time/spatial-derivative for the unknown ,can we simplify 6-elemement- by represented it using the single-element- without derivative ?

+ = − + Π − + Regroup R-S equation

Algebraic stress model

+ + = + − 23Six eqs.

+ + = − × ×Gained Benefits:(Simple Algebraic relations to compute , no need to solve P.D.E.s)

Rodi’s Ideal 1972

Whatever pressure-models:LRR-IP ; LRR-QI ;SSG , etc…

Price to pay: (Weak equilibrium assumption)

Approximation ≈ × = ,may not be accurate.

Isotropic dissipation

Six Reynolds stress multiply one k-eq

Algebraic stress modelsWeak-equilibrium assumption (Rodi, 1972)

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛ ′′

∂∂

+⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛ ′′

∂∂

+⎥⎥⎦

⎤

⎢⎢⎣

⎡

∂∂

+∂∂′′

=∂

′′∂+

∂

′′∂

kjuiu

kxkuk

juiu

tk

kxk

kutk

kjuiu

kxjuiu

kut

juiu

=≈ × 12= 2 + 23

Algebraic stress model

+ + = + − 23

+ + = − ×

Whatever pressure-models:LRR-IP ; LRR-QI ;SSG , etc…

Isotropic dissipation

1: Solve one expensive transport equation for k

− = + − 233: Solve five light weight algebraic relations to compute the remained

−

The Algebraic stress model replaces boussineqassumption. − = − + = −2 to be used in the (A) Three averaged momentum eq. (B) Production terms ( )

2: Solve another expensive transport equation for

Algebraic stress models vs. k-e model

62

Non-linear eddy viscosity model

63

Match the unit dimension

Linear: c ⋅ , or c ⋅ ΩNonlinear: C ⋅ Ω , or C ⋅− 23 = −2

Nonlinear model with possible combinations of tensor product of 2nd order,

+ + + Ω + Ω + Ω Ω + Ω Ω

and Ω are the symmetrical and antisymmertical part of mean velocity strain tensor

Elliptical relaxation model

64

Motivation: “Non-local” model of the anisotropic redistribution term ( ( ) ≡ Π − Π ) for inhomogeneous turbulence (Wall bounded flow)

Previous "local" model (note: neglect the difference − = − )∗, = ∗, , ∗, , ∗, , | ∗Desired model ( ∗, ) = (.., ∗,..) + remote effects

+ + # = + ( ) −+ + ∗ = + ( ) +

( ) ≡ ( ) − ( − )Use this particular form of eq. is to (i) avoid extra model for , (ii) ( ) = 0 at wall, easy to handle

(x, y, z, t) − L ( , , , ) + + = ( , , , )( , , , )

Elliptical relaxation model

65

6 simple relations: , , , = , , , x, y, z, t6EllipticalrelaxationPoissonequationstosolve .+ + ∗ = + ( ) +

, , , = − − 1 − 23 − ( − 23 )

− =

= max( , )Time scale: = max( , 6 )

If this 1 is removed, it is standard LRR-IP model for the term (different than )

∗ = Now ( ) is able to receive non-local, remote contributions, it also recovers the original model if L = 0 and ?

66

Π = (Π − Π ) +( Π )= ( ) − ( )= ( ) − ( )(13Π ) = − 13 +

− 13 2 −− 23= − 23( ) = 23

Eq. 11.140, page 429

Eq. 11.6, p.389

Eq. 11.3, page 388Π = −1 +

compare eq. (11.138) in page 429 with eq. (7.187) in page 315, pope’s book.

Π = −1 + 1 + − 1 += −

Decomposition 1:

Decomposition 2:

This slide summarizes the relations between and ( ); and

Eq. 11.3, p. 388

( ) = (1 + )

Eq. 11.5, p. 389

Eq. 7.187, p.317

Eq. 7.186, p.317Eq. 7.192, p.319

Eq. 7.193, p. 319

Additional slides

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68

A) ( , )B) , ≡ , , , , ,C) , ≡ , ∗ ∗, , ∗, , ,D) , ≡ = ( , )

Transport equation of in general P.D.E. form : (Be aware of boundary conditions)

Select ( , ) from this set

, + , + ( , ∗ ) = , ∗ − , ∗

Reynold stress models

When Reynolds stress model equations are solved, we will get all 6 elements of Reynold stress at our disposal, our interests goes to

characterize the Reynold stress anisotropy.

69

, =23 0 00 23 00 0 23

Perfect isotropy: , ≡ ∗ ∗∗

23 − 0 00 23 − 00 0 23 + ( + )23 023 00 0 23

some tentative examples of deviation from perfect isotropy base on “tensor structure”

23 − 023 − 00 0 23 + ( + )

a) unequal diagonal terms” b): nonzero off-diagonal term” c): combined a & b

To characterize the Reynold stress anisotropy, it naturally goes with (i) first remove from the all diagonal terms and(ii) then normalize by 2 , that gives us the anisotropic tensor :

70

23 − 0 00 23 − 00 0 23 + ( + )23 023 00 0 23

23 − 023 − 00 0 23 + ( + )

= − 232 = − 23 /2

− ∗ 0 00 − ∗ 00 0 ∗ + ∗ 0 ∗ 0∗ 0 00 0 0 − ∗ ∗ 0∗ − ∗ 00 0 ∗ + ∗Superscript * represented division by 2

a) unequal diagonal terms” b): nonzero off-diagonal term” c): combined

Using tensor invariants (i.e. expression formed by eigenvalues of ) is a better way to characterize the anisotropic tensor than a simple tensor-structure based classification of anisotropy

71

− ∗ 0 00 − ∗ 00 0 ∗ + ∗ 0 ∗ 0∗ 00 0 − ∗ ∗ 0∗ − ∗ 00 0 ∗ + ∗a) unequal diagonal terms” b): nonzero off-diagonal term” c): combined

− = − ∗ 0∗ −0 − = − ∗ = 0 = ∗, = − ∗, = 0+ ∗ 0 00 − ∗ 00 0 0

Choosing another coordinate, the (b) “non-zero off-diagonal” problem is equivalent to (a) “unequal diagonal terms”

Tensor invariants gives a coordinate-independent description of anisotropy

Plot the resolved 5 anisotropic Reynolds stresses into the Lumley triangles of 2-dimentions

72

= + + = 0= + + = −3= = 2

= − 23 /2

Reynold stress model (RSM) - Lunds tekniska högskolaReynold stress model (RSM) Rixin Yu, 2017 1....

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