RANS modeling The turbulent viscosity assumption Modelling of turbulent flows: RANS and LES Turbulenzmodelle in der Str¨ omungsmechanik: RANS und LES Markus Uhlmann Institut f¨ ur Hydromechanik Karlsruher Institut f¨ ur Technologie www.ifh.kit.edu SS 2012 Lecture 3 1 / 24
RANS modeling The turbulent viscosity assumption LECTURE 3 Introduction to RANS modelling 2 / 24
RANS modeling The turbulent viscosity assumption Questions to be answered in the present lecture How can the Reynolds-averaged equations be closed? What are the different types of models commonly used? Do simple eddy viscosity models allow for acceptable predictions? 3 / 24
RANS modeling The turbulent viscosity assumption The challenge of turbulence Recap of the salient features of turbulent flows ◮ 3D, time-dependent, random flow field ◮ largest scales are comparable to characteristic flow size → geometry-dependent, not universal ◮ wide range of scales: τ η /T ∼ Re − 1 / 2 , η/L ∼ Re − 3 / 4 ◮ wall flows: energetic motions scale with viscous units δ ν /h ∼ Re − 0 . 88 ◮ non-linear & non-local dynamics 4 / 24
RANS modeling The turbulent viscosity assumption General criteria for assessing turbulence models Level of description ◮ how much information can be extracted from the results? Computational requirements & development time ◮ how much effort needs to be invested in the solution? Accuracy ◮ how precise and trustworthy are the results? Range of applicability ◮ how general is the model? 5 / 24
RANS modeling The turbulent viscosity assumption Possible discrepancies between computation & experiment (adapted from Pope “Turbulent flows”) 6 / 24
RANS modeling The turbulent viscosity assumption Reynolds averaging procedure – need for modeling ◮ decompose velocity field into mean and fluctuation: u ( x , t ) = � u ( x , t ) � + u ′ ( x , t ) ◮ average continuity & momentum equations: � u i � ,i = 0 ∂ t � u i � + ( � u i �� u j � ) ,j + 1 ν � u i � ,jj − � u ′ i u ′ ρ � p � ,i = j � ,j ◮ task of RANS models: → supply the unclosed Reynolds stresses � u ′ i u ′ j � 7 / 24
RANS modeling The turbulent viscosity assumption Reynolds averaging – the closure problem Averaging always introduces more unknowns than equations ◮ transport equation for the n th moment → contains ( n + 1) th moment . . . and so on ⇒ requires closure at some level ◮ the higher the level, the more terms need modeling Most successful closures: ◮ n = 1 : turbulent viscosity models ◮ n = 2 : Reynolds stress models 8 / 24
RANS modeling The turbulent viscosity assumption Common types of RANS models Models based on the turbulent viscosity hypothesis j � = − ν T ( � u i � ,j + � u j � ,i ) + 2 � u ′ i u ′ 3 δ ij k ◮ turbulent viscosity ν T needs to be specified (modeled) Reynolds-stress transport models ¯ D � u ′ i u ′ j � = . . . ¯ D t ◮ various unknown terms (cf. lecture 5) Non-linear turbulent viscosity models � u ′ i u ′ j � = non-linear-function ( � u i � ,j , k, ε, . . . ) (cf. lecture 7) 9 / 24
Generalities RANS modeling Algebraic TVMs The turbulent viscosity assumption One-equation models Assumptions behind Boussinesq’s hypothesis j � − 2 3 k δ ij = − 2 ν T ¯ � u ′ i u ′ S ij Reynolds stress assumed proportional to local mean strain rate 1. mechanisms generating Reynolds stress are assumed local → transport effects neglected 2. turbulent stress and mean strain are assumed aligned → this stems from the linearity of the relation � assumptions in general not true! 10 / 24
Generalities RANS modeling Algebraic TVMs The turbulent viscosity assumption One-equation models The locality assumption: example of failure � Straight section Axisymmetric Experiments demonstrate: Turbulence contraction generating Straight section grid x 1 − ◮ importance of history − S ij = 0 S 11 k − − 1 − S 22 = S 33 = − k effects 2 S ij = 0 ◮ contraction with ¯ 0.20 Contraction Straight Section S ij = cst 0.10 b 22 b ij but: increasing anisotropy 0.00 ◮ ¯ S ij =0 in straight section -0.10 but: non-zero stress b 11 -0.20 -0.30 0.0 0.5 1.0 0.0 0.2 0.4 0.6 t ε /k Turbulent viscosity models S λ t will not work in this case! � exp. Tucker (1970) • exp. Warhaft (1980) (from Pope “Turbulent flows”) 11 / 24
Generalities RANS modeling Algebraic TVMs The turbulent viscosity assumption One-equation models Assumption of stress/strain alignment b ij = − ν T ¯ Boussinesq: S ij k DNS data for channel flow 0 400 800 1200 1600 2000 But, data shows: 0.4 b 11 0.2 ◮ even in simple equilibrium flows 0 b 33 b ij → anisotropy NOT aligned with b 22 −0.2 b 12 mean strain rate −0.4 ◮ example: plane channel flow 0 0.2 0.4 0.6 0.8 1 y/h ◮ problem worse in more complex (Jimenez et al., Re τ = 2000 ) flows 12 / 24
Generalities RANS modeling Algebraic TVMs The turbulent viscosity assumption One-equation models The analogy: Newtonian stress/turbulent viscosity Kinetic theory for ideal gases → Newtonian stress law 2 ¯ with: ν ≈ 1 − σ ij /ρ − p/ρδ ij = − 2 νS ij Cλ ◮ ¯ C mean molecular speed, λ mean free path C S = O (10 − 10 ) λ ◮ time scale ratio in shear flow: ¯ Eddy viscosity hypothesis for turbulent flow j � − 2 3 k δ ij = − 2 ν T ¯ � u ′ i u ′ S ij ◮ typical time scale ratio: k ε S = O (1) ◮ local equilibrium assumption in general NOT valid! 13 / 24
Generalities RANS modeling Algebraic TVMs The turbulent viscosity assumption One-equation models Linear turbulent viscosity models How can the turbulent viscosity ν T be determined? ◮ uniform turbulent viscosity (cf. lecture on jet flow) ◮ algebraic expressions (mixing-length etc.) ◮ one-equation models ( k -model, Spalart-Allmaras) ◮ two-equation models ( k - ε , k - ω ) (cf. lecture 4) 14 / 24
Generalities RANS modeling Algebraic TVMs The turbulent viscosity assumption One-equation models Mixing-length model (Prandtl 1925) Consider two-dimensional shear flow (channel or BL) ◮ dimensionally: ν T = u ∗ · ℓ m ◮ fluid “lump” travels δy = ℓ m ◮ maintains original u ( y ) ◮ for constant shear S : u’=u(y)−u(y+lm) u ′ = −S · ℓ m ������ ������ ������ ������ ������ ������ ������ ������ ������ ������ ������ ������ ������ ������ ������ ������ ������ ������ ������ ������ ◮ Prandtl’s approximation: lm � � y d � u � u ∗ ≈ ℓ m � � ������ ������ � � ������ ������ ������ ������ u(y),v’(y) ������ ������ d y � � ������ ������ ������ ������ ������ ������ ������ ������ ������ ������ x � � d � u � � � ν T = ℓ 2 ⇒ � � m d y � � 15 / 24
Generalities RANS modeling Algebraic TVMs The turbulent viscosity assumption One-equation models Mixing-length coefficients for different flows Self-similar free shear flows α plane wake 0.180 ◮ mixing length: ℓ m = α · r 1 / 2 mixing layer 0.071 plane jet 0.098 round jet 0.080 (from Wilcox 2006) Fully-developed wall-bounded shear flows ◮ van Driest function for buffer and log-region: A + = 26 ℓ m = κy (1 − exp( − y + /A + )) ◮ simple cut-off for the outer region: max( ℓ m ) = 0 . 09 δ ◮ more elaborate models for boundary layers: Cebeci & Smith (1967), Baldwin & Lomax (1978) 16 / 24
Generalities RANS modeling Algebraic TVMs The turbulent viscosity assumption One-equation models Assessment of mixing-length models Advantage ◮ numerically efficient: only solve averaged Navier-Stokes + algebraic expressions Drawbacks ◮ turbulent velocity scale entirely determined by mean flow ◮ incompleteness: flow-dependent mixing length 17 / 24
Generalities RANS modeling Algebraic TVMs The turbulent viscosity assumption One-equation models Turbulent kinetic energy model j � − 2 ν T = u ∗ · ℓ ∗ 3 k δ ij = − 2 ν T ¯ � u ′ i u ′ S ij Determine characteristic velocity u ∗ from TKE ◮ u ∗ often not given by mean flow e.g. decaying grid turbulence ◮ Kolmogorov (1942), Prandtl (1945): √ u ∗ = c and: ℓ ∗ = ℓ m k with: c = 0 . 55 , ⇒ determine k from transport equation � ℓ m still needs to be provided flow by flow 18 / 24
Generalities RANS modeling Algebraic TVMs The turbulent viscosity assumption One-equation models Turbulent kinetic energy model: closure The TKE transport equation ¯ D k 1 2 � u ′ i u ′ i u ′ j � + � u ′ j p ′ � /ρ − νk ,j D t − P = − − ˜ ε ¯ � �� � ˜ T ′ ,j ◮ production term closed through Boussinesq hypothesis ◮ model for dissipation from high-Re assumption: with: C D = c 3 (from log-law) ε = C D k 3 / 2 /ℓ m ˜ ◮ model for flux term from gradient-transport hypothesis: � � ν + ν T T ′ = − ˜ ∇ k with: σ k = 1 σ k 19 / 24
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