13 rans eqns

RANS equations Equations for k and rij Eddy viscosity

The RANS Equations

Maurizio Quadrio

DIA, Politecnico di Milano

2012

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OUTLINE

1 RANS EQUATIONS

2 EQUATIONS FOR k AND rij

3 EDDY VISCOSITY

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REYNOLDS DECOMPOSITION

Let u(xxx , t) be definite and continuous in Rd ,d = 1,2,3. Letu(xxx) exist in the time-mean sense:

u(xxx) = limT

1T

T0u(xxx , t)dt

REYNOLDS DECOMPOSITION

u(xxx , t) = u(xxx) +u(xxx , t)

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THE FLUCTUATING FIELD

u(xxx , t) is the fluctuating part u(xxx) is independent upon time By definition u = u and u = 0 Coupling between mean and fluctuating field (closure

problem: more unknowns than equations)

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DERIVATION FOR CONTINUITY EQ.NS + REYNOLDSS DECOMPOSITION + TIME AVERAGE

1

uuu = 02

(uuu+uuu)= 03

uuu = uuu

4

uuu = 05

uuu = 0

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MOMENTUM EQUATION

uuu uuu = limT

T0uuu uuudt

= uuu [

limT

T0uuudt

]= uuu

[limT

T0uuudt

]= uuu uuu= 0

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THE FINAL FORM

uuu uuu = uuu2 = uuu2 = uuu2uuu uuu = uuu2 = uuu2

STEADY (RANS)

(uuu uuu) + (uuuuuu

)=1

p+2uuu

UNSTEADY (URANS)

uuu t

+ (uuu uuu) + (uuuuuu

)=1

p+2uuu

Closure is required!7 / 46


THE REYNOLDS STRESSES

uiuj (apparent) Reynolds stresses Momentum diffusion due to turbulent motions (analogy

with viscous stresses) uiuj is a symmetric tensor k = 12uiui uiuj = 23kij +aij aij only determines momentum trasport

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URANSDO THE UNSTEADY RANS MAKE SENSE?

Two separated time scales T1 and T2 T1 must exist in theflowReynoldss decomposition can be redefined as:

uuu(xxx , t) =1T

T/2T/2

uuu(xxx , t + )d T T1

PROBLEMS

Turbulent flows do not admit scale separation Scale T is not clearly defined

Mathematical problem Operative problem

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PROBLEMS WITH RANS1) EFFECTS OF TIME AVERAGING

Aim: to remove unessential information But: details are sometimes important (e.g. combustion) Mapping physics statistics not unique!

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PROBLEMS WITH RANS2) REYNOLDSS DECOMPOSITION

Considering mean values only is limiting Considering mean values only is simple Advantages and disadvantages must be balanced Existence of coherent structures emphasizes limitations

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PROBLEMS WITH RANS3) SMALL-SCALE INTERACTIONS ARE MISSING

Modelling statistics (uiuj ) rather than physics (u

i , uj ) hides

small-scale interactions

Scalar dissipation: RANS can only predict mean fluxes Combustion: RANS miss peak values

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PROBLEMS WITH RANS4) RANS SOLUTIONS AND AVERAGED-NS SOLUTIONS

RANS SOLUTIONS ARE EQUAL TO (TIME-AVERAGED)EXPERIMENTAL DATA?

SSSuuu0?=SSS(t)uuu0

Do averaging- and solution-operators commute?

Theorem: if and only if model for uuuuuu is exact! Difference between averaged solution and RANS solution

is square root of uuuuuu error (modelling error)

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OUTLINE

1 RANS EQUATIONS


3 EDDY VISCOSITY

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EQUATION FOR EE MEAN-FLOW KINETIC ENERGY

E 12uiui

DEDt

+ TTT =P

T i ujuiuj +ui p/2ujsij

P uiujuixj

2sijsij

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EQUATION FOR kk TURBULENT KINETIC ENERGY

k 12uiui

DkDt

+ TTT =P

T i 12uiujuj +u

ip/2ujsij

P uiujuixj

2sijsij

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EQUATION FOR rijrij REYNOLDS STRESS

rij uiujEquation for rij is obtained by:

subtracting eq. for ui from that for ui to get an eq. for ui ; multiplying eq. for ui by that for uj to obtain eq. for uiuj ; time-averaging eq. for uiuj .

( t

+ukxk

)(uiuj

)=

xkuiujuk +Pij + ij ij +2uiuj

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REYNOLDS STRESSES EQUATION

Triple correlation (closure problem):uiujuk

Production tensor:

Pij uiukujxkujuk

uixk

Velocity-pressure gradient tensor:

ij 1

(uip

xj+uj

p

xi

) Dissipation tensor:

ij uixk

ujxk

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TURBULENT KINETIC ENERGY AND REYNOLDSSTRESSES

Trace of uiuj eq. is 2 times the k eq.

Pii = 2P

ii =2

xi

uip

ii = 2

Difference between dissipation and pseudo-dissipation:

2sijsij uixk

uixk

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OUTLINE

1 RANS EQUATIONS


3 EDDY VISCOSITY

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CLASSIFICATION OF TURBULENCE MODELS

MODELS BASED ONEDDY-VISCOSITY

rij is given through aneddy-viscosity

more developed less recent

MODELS FOR THEREYNOLDS-STRESS TENSOR

a model for rij is givendirectly

less developed more recent

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THE GRADIENT-DIFFUSION HYPOTHESISAN EXAMPLE FOR THE PASSIVE SCALAR

(uuu ) + uuu = 2

Scalar flux vector uuu : direction and magnitude of the(turbulent) transport of

Hypothesis: this vector is aligned to the mean scalargradient

Turbulent diffusivity t(xxx) (positive scalar quantity):

uuu =t(xxx)

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THE TURBULENT DIFFUSIVITY

Define the effective diffusivity e(xxx):

e(xxx) = + t(xxx)

The mean scalar equation is closed as:

(uuu ) = (e(xxx))

Problem: t(xxx) must be known

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THE TURBULENT-VISCOSITY HYPOTHESIS

(uuu uuu) + (uuuuuu

)=1

p+2uuu

Reynolds stresses tensor uuuuuu: effect of turbulentfluctuations on the mean motion

Hypothesis: this tensor is aligned to the meanrate-of-strain tensor

Eddy viscosity t(xxx) (positive scalar quantity):

aij uiuj +23kij = t(xxx)

(uixj

+ujxi

)

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THE EDDY VISCOSITY

Define the effective viscosity e(xxx):

e(xxx) = +t(xxx)

The mean momentum equation is closed as:

(uuu uuu) =p+ (e(xxx)uuu)

Problem: t(xxx) must be known.

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APPRAISAL OF THE BOUSINNESQS HYPOTHESIS

1 Intrinsic hypothesis: aij depends upon mean velocitygradients only

2 Specific hypothesis:

aij =2tsijmodelled after laminar flows

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1) THE INTRINSIC HYPOTHESISTURBULENCE DOES HAVE MEMORY!

Experiment: sudden axisymmetric contraction after a grid(Uberoi 1956, Sk/ = 2.1)

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EVOLUTION OF ANISOTROPYNORMALIZED ANISOTROPIES bij = aij/2k

Open symbols: larger strain Sk/ = 55

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INTRINSIC HYPOTHESIS IS INCORRECT

Contraction: normalized anisotropies bij = aij/2k are notconstant

For large strain bij depend on the total amount of meanstrain

Straight section: anisotropy is not zero and decreases onthe turbulence timescale k/

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THE MOLECULAR ANALOGY

VISCOUS FLUIDKinetic theory of gases:

velocity scale: meanmolecular speed c

length scale: mean freepath

kinematic viscosity 12c

TURBULENT FLOWEmpirical argument:

velocity scale: turbulencevelocity scale u

length scale: turbulencelength scale `

turbulent viscosity t u`

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EXAMPLESIMPLE SHEAR FLOW WITH S = u1/x2 U/L

VISCOUS FLUID

Fluid = ensemble ofmolecules

Molecular timescalem = /c

Shear timescale S =S 1mS

cS =

LUc

= KnM 1

TURBULENT FLOW

Turbulent flow = ensembleof eddies

Turbulent timescalet = k/

Shear timescale S =S 1tS S k

= O(1)

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IS THE INTRINSIC HYPOTHESIS WORTH SOMETHING?TURBULENT VISCOSITY IS OFTEN USEFUL

In simple shear flow the mean velocity gradients change slowly:

Local mean velocity gradients characterize the history ofmean distortion

Reynolds stress balance is dominated by local processes WhenP/ 1 the turbulent-viscosity hypothesis is correct

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2) THE SPECIFIC HYPOTHESISTURBULENCE IS NOT A NEWTONIAN FLUID

INCONSISTENT!

aij =2tsij

In turbulent shear flows sii = 0 but normal Reynoldsstresses are not!

aij is not aligned with sij

NOT INVARIANT

aij =2tsijTensorial relation is not general

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SPECIFIC HYPOTHESIS

INVARIANT BUT NOT OBJECTIVE

aij = t ,ijklskl

Not rotation-invariant

OBJECTIVE RELATION

aij = 1t ij +2t sij +

3t sikskj

Unable to reproduce simple experimental situations...A costitutive eq. for turbulence does not need to be objective(Coriolis, etc)

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SPECIFIC HYPOTHESIS: WHY A LINEAR LAW?

NEWTONIAN FLUID

Straining small comparedto molecular scales:

S /c 1

Small departure fromequilibrium

Linear dependence ofstress tensor on velocitygradient tensor

TURBULENT FLOW

Straining large comparedto turbulence scales:

S k/ > 1

Large departure fromequilibrium

More general dependencyof Reynolds stresses onvelocity gradient tensor

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BEYOND A LINEAR LAW

Non-linear laws: e.g. memory effect

aij = A23kij

0

M()[uixj

(t ) + ujxi

(t )]d

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LIMITATIONS OF EDDY VISCOSITY CONCEPT

Eddy viscosity not fully adequate for: Flows with abrupt change of shear rate Flows over curved surfaces Flows in ducts with secondary motions and/or separations Rotating or stratified flows Three-dimensional flows Many others...

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RANS RESULTS HAVE LIMITED RELIABILITY...FREITAS, SELETED BENCHMARKS FROM COMMERCIAL CFD CODES, J.FLUIDS ENG. 1995

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... BUT CAN BE USED NONETHELESS!TOROROSSO (1)

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... BUT CAN BE USED NONETHELESS!UPPER RESPIRATORY AIRWAYS (1)

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RANS equationsEquations for k and rijEddy viscosity

13 rans eqns

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