Closure Problem Closure Problem (cont’d)thomas-gomez.net/images/Lectures/Turbulence/turbulence_lectures_part04_tg.pdf1 The turbulence fact : Deﬁnition, observations and universal

1 The turbulence fact : Definition, observations and universal features ofturbulence

2 The governing equations

3 Statistical description of turbulence

4 Turbulence modeling

5 Turbulent wall bounded flows

6 Homogeneous Isotropic Turbulence

7 Homogeneous Shear Flows

8 Results based on the equations of the dynamics in fully developedturbulence

Turbulence modeling// [email protected] 155/374

4 Turbulence modelingClosure problemModels for the closure of the systemFirst order modelsSecond order modelsExercise

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Closure Problem

Momentum equation

Turbulence modeling/Closure problem/ [email protected] 157/374

Closure Problem (cont’d)

Equation for Reynolds stress tensor Rij

@

@tRij +

@

@xk

(ukRij) = �✓

Rjk

@ui

@xk

+ Rik

@uj

@xk

◆� @

@xk

u0iu0

ju0

k

| {z }unknown!!!

�✓

@

@xi

p0u0j+

@

@xj

p0u0i

◆+ 2p0S0

ij

+f 0iu0

j+ f 0

ju0

i+ 2⌫

✓u0

j

@

@xk

S0ik

+ u0i

@

@xk

S0jk

◆

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Models for the closure

Estimate the contribution of the Reynolds stress tensor to the NS equation ?First order =) Eddy viscosity model (EVM) :µt = ⇢⌫t ? ! Rij ! NS equation : u.Classified in terms of number of transport equations solved in addition tothe RANS equations :

Zero equation/algebraic model : Mixing Length, Cebeci-Smith,Baldwin-Lomax, ...One equation : Spalart-Allmaras ⌫t, K, Wolfstein, Baldwin-Barth, ...Two equations : K� ", K� !, K� �, K� L,...Three equations : K� "�A, ...Four equations : v2� f , ...

Second order model =) Solve the Reynolds stress tensor equation : Rij ?ASM : Algebraic Stress ModelRSM : Reynolds Stress Model

Turbulence modeling/Models for the closure of the system/ [email protected] 159/374

First order models

Eddy viscosity modelsTurbulent stresses act similarly to viscousstresses.Turbulent viscosity ⇠ property of the flow.Boussinesq’s Hypothesis 1877

Laminar

⌧ij = µ

✓@ui

@xj

+@uj

@xi

◆� 2

3µ�ij

@uk

@xk

Turbulent

⌧ t

ij = �⇢u0iu0j= µt

✓@ui

@xj

+@uj

@xi

◆� 2

3�ij⇢K

Turbulence modeling/First order models/ [email protected] 160/374

First order modelsEddy viscosity models

@

@tui +

@

@xj

(uiuj) = � @

@xi

(p + ⇢K) +@

@xj

�(µt + µ)Sji

�

µt : turbulent viscosity⇢K : kinetic energy of the fluctuations ⇠ PressureRemarks :

For ⌫ : characteristic spatial scale of molecular motion ⇠ m.f.p. ofmolecules ⌧ scales of macroscopic fluid motionThis clear-cut separation foes not hod between u0

i and ui velocity fields.=) The concept of turbulent viscosity becomes more accurate with theincreasing scale separation.

Problem : How to define the turbulent viscosity µt in terms of the unknowns ofthe dynamics ?

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Zero equation models

Mixing length modelsNo EDP for the tranport of the turbulent stress tensor = nodynamical depedence.A simple algebraic equation is used to close the systemMixing length theory ⇠ characteristic length scale of the eddiesDimensional analysis leads to

⌫t =µt

⇢⇠ ù = `m

✓`m

��du

dy

��

◆

Turbulence modeling/First order models/zero equation [email protected] 162/374

Zero equation models

Example : a very simple model for the boundary layer caseEVM :

⌫t =µt

⇢⇠ ù = `m

✓`m

��du

dy

��

◆

� : boundary layer thickness, : von Kármán constant

(`m = y pour y < �

`m = � pour y � �

Re-injected in the turbulent viscosity expression

⌫t =µt

⇢⇠ ù = `m

✓`m

��du

dy

��

◆

Re-injected in the RANS equations


Zero equation model : The boundary layer case

Reynolds stress tensor

�u0iu0

j= ⌫t

✓@ui

@xj

+@uj

@xi

◆� 2

3�ijK

with

⌫t = `m

✓`m

��du

dy

��

◆

then

�u0v0 = `2m

��du

dy

��2


Zero equation model (cont’d)

Advantages :Simple to implementFast computing timeQuite good predictions for simple flows where experimentalcorrelations for the mixing length exist.Used in higher level models

Drawbacks :No history effect ; purely local.Flows where the turbulent length scale varies : anything withseparation or circulation.Only give mean flow properties and turbulent shear stress.Cannot switch from one type of region to another.Only used for simple external flows.Eddy viscosity is zero if the velocity gradients are zero.Not in commercial CFD code.


One equation models

EDP for kinetic energy of the fluctuationsEDP for K

K =1

2u0

iu0

i=

1

2Rii

Turbulent viscosity using K

µt = Cµ

pK`m

where Cµ is a free parameter.

Turbulence modeling/First order models/One equation models [email protected] 166/374

One equation models (cont’d)EDP for kinetic energy of the fluctuations

EDP for K from momentum conservation

@

@tK +

@

@xl

(ulK) = �Ril

@ui

@xl

� @

@xl

u0iu0

iu0

l� "

� @

@xl

p0u0l+ ⌫

@2

@xl@xl

K

Model : Transport equation for K

@

@tK +

@

@xl

(ulK) =@

@xl

✓⌫ +

⌫t

�k

◆@K@xl

�+ Pk � "

withPk ⌘ u0

iu0

l

@ui

@xl

' ⌫t

✓@ui

@xj

+@uj

@xi

◆@ui

@xl

" ⌘ ⌫@u0

i

@xl

@u0i

@xl

' K3/2

l=) " = Cd

K3/2

`m

=) 4 free adjustable parameters...Turbulence modeling/First order models/One equation models [email protected] 167/374

One equation models (cont’d)

Spalart-Allmaras model (1994)Modern one-equation models abandoned the K-equationBased on an ad-hoc Transport equation for the eddy viscosity directly

@⌫

@t+ uj

@⌫

@xj

= P⌫ � ✏⌫ +@

@xj

1

⇢

✓µ +

⌫

�⌫

◆@⌫

@xj

�

12 adjustable constants to set ! ! !Boundary/Initial conditions :

Walls : ⌫ = 0Free stream : ideally ⌫ = 0 or ⌫ ⌫

2if problem with the solver


One equation models (cont’d)

Advantages :Inclusion of the history effects.Economical and accurate for : Attached wall-bounded flows, Flowswith mild separation and recirculationDeveloped for use in unstructured codes in the aerospace industryPopular in aeronautics for computing the flow around aero planewings, etc

Drawbacks :Weak predictions for : Massively separated flows, Free shear flows,Decaying turbulence, Jet spreading (⇠ 40% of overprediction on therate of spreading for SP model), Complex internal flows.

Characteristic length scale empirically determinedSA model : ⌫ unaffected by irrotational mean straining


Two equations models

Two unknowns K � "

EDP for K

K =1

2u0

iu0

i=

1

2Rii

EDP equation for the dissipation rate "

"

=) Turbulent viscosity : using K and "

µt ⇠ ul = K1/2

✓K3/2

"

◆=) µt = Cµ

K2

"

Turbulence modeling/First order models/Two equations models [email protected] 170/374

K � " model (cont’d)Transport PDE for the dissipation "

Model : PDE for K

@

@tK +

@

@xl

(ulK) = �Ril

@ui

@xl

� @

@xl

u0iu0

iu0

l� "

� @

@xl

p0u0l+ ⌫

@2

@xl@xl

K

Model for the production term

P = �Ril

@ui

@xl

⇡ 2⌫tSilSil = Cµ

K2

"S2

ij

Model for the diffusion terms (turbulent and pressure)

� @

@xl

⇣u0

iu0

iu0

l+ p0u0

l

⌘⇡ @

@xl

✓⌫t

�K

@K@xl

◆


K � " modelModel : Two transport PDE for K and "

@K@t

+ uj

@K@xj

= Cµ

K2

"

��S��2 � " +

@

@xj

✓✓⌫t

�K+ ⌫

◆@K@xj

◆

@"

@t+ uj

@"

@xj

="

K (C"1P � C"2") +@

@xj

✓✓⌫t

�"

+ ⌫

◆@"

@xj

◆

Two supplementary scalar PDEsTwo unknowns K and " =) Boundary conditions ? Wall functions5 free parameters Cµ, �K, C"1 , C"2 , �" =) Calibration ?Standard values : Launder and Sharma (1974) Cµ = 0.09, �K = 1.0,C"1 = 1.44, C"2 = 1.92, �" = 1.3 (found empirically).Hyp : High Reynolds numbers, isotropyModel for low or transitional Reynolds numbers : K � !, ...


K � " model (cont’d)

Advantages :Massively used, implemented in numerous CFD codes.Spatial variation of the turbulent kinetic energy.Simple to implement.Quite good predictions of the simple sheared flows.Stable calculations

Drawbacks :Not quite efficient for complex flows : recirculations, strong anisotropy,

swirling and rotating flows, flows with strong separation, axis symmetric jets,...

Ad hoc equation for ".Valid only in the fully developed turbulence zone.Wall functions implementation needed.Over-prediction of K in the strong shear regions.Over dissipating at all scales of the flows (stabilizing effect).


K � " model (cont’d)Simulations : mean velocity field



Simulations : K and "


K � " model (cont’d)ComparisonsPredicted turbulent viscosity around a transonic airfoil=) Spalart and Chien models for shear layer...

from http ://www.innovative-cfd.com



ComparisonsPredicted surface pressure coefficient and shock location2.3 degrees angle of attack and a Mach number of 0.729

The only real differences for this case lie in the predicted shock location onthe upper surface. The more sophisticated models are not always the bestones to use.


K � ! model (Wilcox 1993, Menter 1994, ...)Model : Two transport PDE for K and !

@K@t

+ uj

@K@xj

= Cµ

K!

��S��2 � " +

@

@xj

✓✓⌫t

�K+ ⌫

◆@K@xj

◆

@!

@t+ uj

@!

@xj

= CµC!1 |S|2 � C!2!2 +

@

@xj

✓✓⌫t

�!

+ ⌫

◆@!

@xj

◆

with

! = "/K and ⌫t = Cµ

K!

.

5 free parameters Cµ, �K, C!1 , C!2 , �! =) Calibration ?Developed for Boundary layer flows.Possibly with streamwise pressure gradients.


Generic formulation for two equations models

K � � models

Model : K � � with � = Kl"m

Dimensional analysis : ⌫t = CµK2+l/m��1/m

Standard formulation for �

@�

@t+ uj

@�

@xj

=�

K (C�1P � C�2") +@

@xj

✓✓⌫t

��

+ ⌫

◆@�

@xj

◆

5 free-parameters.

Turbulence modeling/First order models/Generic form [email protected] 179/374

Generic formulation for two equations modelsK � � models

� = Kl"m

Table: Examples of two-equations turbulence models for incompressible flows.� = Kl"m

Model l mChou (1945), Launder, ... K � " 0 1

Kolmogorov (1942) , Saffman, Wilcox, Menter ... K � ! -1 1Cousteix (1997), Aupoix ... K � ' -1/2 1

Rotta (1951), Smith ... K � l 3/2 -1Speziale (1990) K � ⌧ 1 -1Zeierman (1986) K � K⌧ 2 -1

Saffman (1970), Launder, Spalding, Wilcox ... K � !2 -2 2Rotta (1968), Rodi, Spalding ... K � Kl 5/2 -1

Glushko (1971) ... K � l2 3 -2

Turbulence modeling/First order models/Generic form [email protected] 180/374

Second order models

PrincipleUse the governing equations of the dynamics to directly determine thecomponents of the 2nd-order Reynolds stress tensor Rij , instead ofusing the Boussinesq’s hypothesis analogy.Efficient for anisotropic flows

ExamplesASM : Algebrabic Stress ModelRSM : Reynolds Stress Model

Turbulence modeling/Second order models/Principle [email protected] 181/374

Reynolds stress tensor equation

Rij models

@

@tRij +

@

@xk

(ukRij)| {z }

I

= �✓

Rjk

@ui

@xk

+ Rik

@uj

@xk

◆

| {z }II

� @

@xk

u0iu0

ju0

k

| {z }III

�✓

@

@xi

p0u0j+

@

@xj

p0u0i

◆

| {z }IV

+ 2p0S0ij| {z }

V

+ f 0iu0

j+ f 0

ju0

i| {z }V I

+ 2⌫

✓u0

j

@

@xk

S0ik

+ u0i

@

@xk

S0jk

◆

| {z }V II

I, II : exact termsIII, IV , V , V II =) Model

Turbulence modeling/Second order models/Reynolds stress model [email protected] 182/374

Reynolds stress tensor equationRij models

@

@tRij +

@

@xk

(ukRij) = Pij+⇧ij + Dij + "ij

Exact :Pij = �

✓Rjk

@ui

@xk

+ Rik

@uj

@xk

◆

Approximation needed = model :

⇧ij = 2p0S0ij

where S0ij

=1

2

✓@u0

i

@xj

+@u0

j

@xi

◆

Dij = � @

@xk

hu0

iu0

ju0

k+ p0u0

i�jk + p0u0

j�ik

i

"ij = 2⌫@u0

i

@xk

@u0j

@xk



Rij models : Reference Linear Model@

@tRij +

@

@xk

(ukRij) = Pij + ⇧ij + Dij + "ij

Exact expression :⇧ij = 2p0S0

ij

Model of Rotta :⇧ij = ⇧(1)

ij+ ⇧(2)

ij

with⇧(1)

ij= �C1

✓Pij � Pkk

3�ij

◆

⇧(2)ij

= �2C2

✓Rij

Rnn

� �ij

3

◆


Reynolds stress tensor equationRij models

@

@tRij +

@

@xk


Exact :

"ij = 2⌫@u0

i

@xk

@u0j

@xk

Model of Hanjalic & Launder : Local isotropy

"ij =2

3"�ij

with

@"

@t+

@

@xk

(uk") = C"

@

@xj

✓K"

Rij

@"

@xi

◆+ C"1

"

KPkk

2� "2

K



Rij models@

@tRij +

@

@xk


Exact expression : Diffusion term

Dij = � @

@xk

hu0

iu0

ju0

k+ p0u0

i�jk + p0u0

j�ik

i

Extended gradient model :

Dij = CD

@

@xn

✓k

"Rnm

@Rij

@xm

◆


RLM model

AdvantagesBetter efficiency compared to the K � " modelBetter approximation of the mean velocity fieldBetter tendencies for the second order quantities : K, ", ...

DrawbacksIncompressible 3D : 3 (ui) + 6 (Rij) + 1 (") = 10 unknowns=) 10 scalar equations

More free parameters to calibrate...Still far from universality...


Exercise

Mixing length model for shear layer problem

Let’s consider a 2D shear layer turbulent flow with a mean velocity field asu = (u(y, t), 0, 0).

The boundary conditions for the velocity reads u(y = ±1) = ± 12Us.

�(t) is the mixing layer width defined such as u(y = ± �

2 ) = ± 25Us.

We use the following expression for the Reynolds stress components :

u0iu0j=

2

3k�ij � ⌫T

✓@ui

@xj

+@uj

@xi

◆, (22)

where the turbulent viscosity reads

⌫T = `2m

��@u

@y

�� (Smagorinsky 1963). (23)

Turbulence modeling/Exercise/ [email protected] 188/374

ExerciseMixing length model for shear layer problem (cont’d)

1 Write the equation for u(y, t). Is that equation closed ?

2 The mixing length hypothesis for the eddy viscosity model is used considering a uniform

mixing length across the flow and proportional to its width, i.e. `m = ↵�(t), where ↵ is a

given constant. Determine the governing equation for u(y, t).

3 Show that we can obtain a self similar solution defined as u(y, t) = Usf(⇠) where

⇠(y, t) = y/�(t) and f(⇠) satisfies

� S⇠f0 = 2↵

2f0f00

, (24)

where S is a parameter to express in terms of Us and �(t).

4 Show that the equation (24) admits two solutions, denoted f1 and f2 including three

constants.

5 Write this solution in the three parts of the flow : firstly for |⇠| > ⇠?, then for |⇠| < ⇠

?, show

that

f =3

4

⇠

⇠?�

1

4

✓⇠

⇠?

◆3

,

where ⇠?

is defined by f0(±⇠

?) = 0.Hint : Show that the increasing rate S of the mixing layer width can be expressed in terms of

the mixing layer length constant ↵ and of ⇠?.

6 Give an approximation for ⇠?

by considering that, by definition of �(t), one has f( 12 ) = 2

5 .

7 Plot ⌫T as function of y.

Turbulence modeling/Exercise/ [email protected] 189/374

Documents

Closure Problem Closure Problem (cont’d)thomas-gomez.net/images/Lectures/Turbulence/turbulence_lectures_part04_tg.pdf1 The turbulence fact : Deﬁnition, observations and universal