L ARGE E DDY S IMULATION (LES) Lars Davidson, - PowerPoint PPT Presentation

N EAR -W ALL R ESOLUTION Biggest problem with LES: near walls, it requires very fine mesh in all directions, not only in the near-wall direction. The reason: violent violent low-speed outward ejections and high-speed in-rushes must be resolved (often called streaks). A resolved these structures in LES requires ∆ x + ≃ 100, ∆ y + min ≃ 1 and ∆ z + ≃ 30 www.tfd.chalmers.se/˜lada Helsinki 4 October 2012 22 / 1

N EAR -W ALL R ESOLUTION Biggest problem with LES: near walls, it requires very fine mesh in all directions, not only in the near-wall direction. The reason: violent violent low-speed outward ejections and high-speed in-rushes must be resolved (often called streaks). A resolved these structures in LES requires ∆ x + ≃ 100, ∆ y + min ≃ 1 and ∆ z + ≃ 30 The object is to develop a near-wall treatment which models the streaks (URANS) ⇒ much larger ∆ x and ∆ z www.tfd.chalmers.se/˜lada Helsinki 4 October 2012 22 / 1

N EAR -W ALL R ESOLUTION Biggest problem with LES: near walls, it requires very fine mesh in all directions, not only in the near-wall direction. The reason: violent violent low-speed outward ejections and high-speed in-rushes must be resolved (often called streaks). A resolved these structures in LES requires ∆ x + ≃ 100, ∆ y + min ≃ 1 and ∆ z + ≃ 30 The object is to develop a near-wall treatment which models the streaks (URANS) ⇒ much larger ∆ x and ∆ z In the presentation we use Hybrid LES-RANS for which the grid requirements are much smaller than for LES www.tfd.chalmers.se/˜lada Helsinki 4 October 2012 22 / 1

N EAR -W ALL R ESOLUTION C ONT ’ D wall y x y + In RANS when using wall-functions, 30 < y + < 100 for the wall-adjacent cells www.tfd.chalmers.se/˜lada Helsinki 4 October 2012 23 / 1

N EAR -W ALL R ESOLUTION C ONT ’ D wall y x y + In RANS when using wall-functions, 30 < y + < 100 for the wall-adjacent cells In LES, ∆ z + ≃ 30 ∆ z + z x www.tfd.chalmers.se/˜lada Helsinki 4 October 2012 23 / 1

N EAR -W ALL R ESOLUTION C ONT ’ D wall y x y + In RANS when using wall-functions, 30 < y + < 100 for the wall-adjacent cells In LES, ∆ z + ≃ 30 ∆ z + EVERYWHERE z x www.tfd.chalmers.se/˜lada Helsinki 4 October 2012 23 / 1

N EAR -W ALL R ESOLUTION C ONT ’ D wall y x y + In RANS when using wall-functions, 30 < y + < 100 for the wall-adjacent cells In LES, ∆ z + ≃ 30 ∆ z + EVERYWHERE AND ∆ x + ≃ 100, z ∆ y + min ≃ 1 x www.tfd.chalmers.se/˜lada Helsinki 4 October 2012 23 / 1

N EAR -W ALL T REATMENT from Hinze (1975) www.tfd.chalmers.se/˜lada Helsinki 4 October 2012 24 / 1

N EAR -W ALL T REATMENT 1.5 1 z 0.5 0 0 1 2 3 4 5 6 x Fluctuating streamwise velocity at y + = 5. DNS of channel flow. We find that the structures in the spanwise direction are very small which requires a very fine mesh in z direction. www.tfd.chalmers.se/˜lada Helsinki 4 October 2012 25 / 1

Z ONAL PANS M ODEL L. Davidson A New Approach of Zonal Hybrid RANS-LES Based on a Two-equation k − ε Model [7] ETMM9, Thessaloniki, 7-9 June 2012 Financed by the EU project ATAAC (Advanced Turbulence Simulation for Aerodynamic Application Challenges) DLR, Airbus UK, Alenia, ANSYS, Beijing Tsinghua University, CFS Engineering, Chalmers, Dassault Aviation, EADS, Eurocopter Deutschland, FOI, Imperial College, IMFT, LFK, NLR, NTS, Numeca, ONERA, Rolls-Royce Deutschland, TU Berlin, TU Darmstadt, UniMAN www.tfd.chalmers.se/˜lada Helsinki 4 October 2012 26 / 1

PANS L OW R EYNOLDS N UMBER M ODEL [17] � ∂ k ∂ k ∂ t + ∂ ( kU j ) = ∂ �� ν + ν t � + ( P k − ε ) ∂ x j ∂ x j σ ku ∂ x j � ∂ε ε 2 ∂ t + ∂ ( ε U j ) �� � ∂ε = ∂ ν + ν t ε k − C ∗ + C ε 1 P k ε 2 ∂ x j ∂ x j σ ε u ∂ x j k k 2 f 2 f 2 ε 2 = C ε 1 + f k k k ν t = C µ f µ ε , C ∗ ( C ε 2 f 2 − C ε 1 ) , σ ku ≡ σ k , σ ε u ≡ σ ε f ε f ε f ε C ε 1 , C ε 2 , σ k , σ ε and C µ same values as [1]. f ε = 1. f 2 and f µ read � R t �� 2 � � − y ∗ � � 2 �� � f 2 = 1 − exp 1 − 0 . 3exp − 3 . 1 6 . 5 � R t �� 2 � � 2 �� � − y ∗ 5 � � f µ = 1 − exp 1 + exp − R 3 / 4 14 200 t Baseline model: f k = 0 . 4. Range of 0 . 2 < f k < 0 . 6 is evaluated www.tfd.chalmers.se/˜lada Helsinki 4 October 2012 27 / 1

PANS L OW R EYNOLDS N UMBER M ODEL [17] � ∂ k ∂ k ∂ t + ∂ ( kU j ) = ∂ �� ν + ν t � + ( P k − ε ) ∂ x j ∂ x j σ ku ∂ x j � ∂ε ε 2 ∂ t + ∂ ( ε U j ) �� � ∂ε = ∂ ν + ν t ε k − C ∗ + C ε 1 P k ε 2 ∂ x j ∂ x j σ ε u ∂ x j k k 2 f 2 f 2 ε 2 = C ε 1 + f k k k ν t = C µ f µ ε , C ∗ ( C ε 2 f 2 − C ε 1 ) , σ ku ≡ σ k , σ ε u ≡ σ ε f ε f ε f ε C ε 1 , C ε 2 , σ k , σ ε and C µ same values as [1]. f ε = 1. f 2 and f µ read � R t �� 2 � � − y ∗ � � 2 �� � f 2 = 1 − exp 1 − 0 . 3exp − 3 . 1 6 . 5 � R t �� 2 � � 2 �� � − y ∗ 5 � � f µ = 1 − exp 1 + exp − R 3 / 4 14 200 t Baseline model: f k = 0 . 4. Range of 0 . 2 < f k < 0 . 6 is evaluated www.tfd.chalmers.se/˜lada Helsinki 4 October 2012 27 / 1

PANS L OW R EYNOLDS N UMBER M ODEL [17] � ∂ k ∂ k ∂ t + ∂ ( kU j ) = ∂ �� ν + ν t � + ( P k − ε ) ∂ x j ∂ x j σ ku ∂ x j � ∂ε ε 2 ∂ t + ∂ ( ε U j ) �� � ∂ε = ∂ ν + ν t ε k − C ∗ + C ε 1 P k ε 2 ∂ x j ∂ x j σ ε u ∂ x j k k 2 f 2 f 2 ε 2 = 1 . 5 + f k k k ν t = C µ f µ ε , C ∗ ( 1 . 9 − 1 . 5 ) , σ ku ≡ σ k , σ ε u ≡ σ ε f ε f ε f ε C ε 1 , C ε 2 , σ k , σ ε and C µ same values as [1]. f ε = 1. f 2 and f µ read � R t �� 2 � � − y ∗ � � 2 �� � f 2 = 1 − exp 1 − 0 . 3exp − 3 . 1 6 . 5 � R t �� 2 � � 2 �� � − y ∗ 5 � � f µ = 1 − exp 1 + exp − R 3 / 4 14 200 t Baseline model: f k = 0 . 4. Range of 0 . 2 < f k < 0 . 6 is evaluated www.tfd.chalmers.se/˜lada Helsinki 4 October 2012 27 / 1

C HANNEL F LOW : Z ONAL RANS-LES LES, f k < 1 k u , int , ε u , int RANS, f k = 1 . 0 y int y wall x Interface: how to treat k and ε over the interface? They should be reduced from their RANS values to suitable LES values The usual convection and diffusion across the interface is cut off, and new “interface boundary” conditions are prescribed k u , int = f k k RANS Nothing is done for ε x max = 3 . 2 (64 cells), z max = 1 . 6 (64 cells), y dir: 80 − 128 cells CDS in entire region www.tfd.chalmers.se/˜lada Helsinki 4 October 2012 28 / 1

( N x × N z ) = ( 64 × 64 ) . y + int = 500 0 30 -0.2 25 � u ′ v ′ � + 20 -0.4 U + 15 -0.6 10 -0.8 5 0 -1 1 100 1000 30000 0 0.05 0.1 0.15 0.2 y y + Re τ = 4 000 Re τ = 8 000 Re τ = 16 000; Re τ = 32 000. www.tfd.chalmers.se/˜lada Helsinki 4 October 2012 29 / 1

I NTERFACE LOCATION . Re τ = 8 000 . 0 30 25 -0.2 20 � u ′ v ′ � + -0.4 U + 15 -0.6 10 -0.8 5 0 -1 1 100 1000 8000 0 500 1000 1500 2000 y + y + y + = 130 y + = 500 y + = 980 www.tfd.chalmers.se/˜lada Helsinki 4 October 2012 30 / 1

E FFECT OF f k . Re τ = 16 000 . y + int = 500 0 30 25 -0.2 20 � u ′ v ′ � + -0.4 U + 15 -0.6 10 -0.8 5 0 -1 1 100 1000 10000 0 0.05 0.1 0.15 0.2 y y + f k = 0 . 2 f k = 0 . 3 f k = 0 . 5 f k = 0 . 6 www.tfd.chalmers.se/˜lada Helsinki 4 October 2012 31 / 1

E FFECT OF R ESOLUTION : V ELOCITY ( N x × N z ) = ( 32 × 32 ) ( N x × N z ) = ( 128 × 128 ) 30 30 25 25 20 20 U + U + 15 15 10 10 5 5 0 0 1 100 1000 30000 1 100 1000 30000 y + y + Re τ = 4 000 Re τ = 8 000 Re τ = 16 000; Re τ = 32 000. www.tfd.chalmers.se/˜lada Helsinki 4 October 2012 32 / 1

E FFECT OF R ESOLUTION : R ESOLVED S HEAR S TRESS ( N x × N z ) = ( 32 × 32 ) ( N x × N z ) = ( 128 × 128 ) 0 0 -0.2 -0.2 � u ′ v ′ � + � u ′ v ′ � + -0.4 -0.4 -0.6 -0.6 -0.8 -0.8 -1 -1 0 0.05 0.1 0.15 0.2 0 0.05 0.1 0.15 0.2 y y Re τ = 4 000 Re τ = 8 000 Re τ = 16 000; Re τ = 32 000. www.tfd.chalmers.se/˜lada Helsinki 4 October 2012 33 / 1

E FFECT OF R ESOLUTION : T URBULENT V ISCOSITY Re τ = 4000 Re τ = 8000 0.015 0.015 ν t / ( u τ δ ) 0.01 0.01 0.005 0.005 0 0 0 0.05 0.1 0.15 0.2 0.25 0.3 0 0.05 0.1 0.15 0.2 0.25 0.3 Re τ = 16 000 Re τ = 32 000 0.015 0.015 ν t / ( u τ δ ) 0.01 0.01 0.005 0.005 0 0 0 0.05 0.1 0.15 0.2 0.25 0.3 0 0.05 0.1 0.15 0.2 0.25 0.3 y y ( N x × N z ) = 64 × 64 32 × 32 128 × 128 www.tfd.chalmers.se/˜lada Helsinki 4 October 2012 34 / 1

E FFECT OF R ESOLUTION : T URBULENT V ISCOSITY Re τ = 4000 Re τ = 8000 ! 0.015 0.015 y t i s o c ν t / ( u τ δ ) s 0.01 0.01 i v t n e l 0.005 0.005 u b r u t t 0 0 n 0 0.05 0.1 0.15 0.2 0.25 0.3 0 0.05 0.1 0.15 0.2 0.25 0.3 e d Re τ = 16 000 Re τ = 32 000 n 0.015 0.015 e p e d n ν t / ( u τ δ ) 0.01 0.01 i d i r G 0.005 0.005 0 0 0 0.05 0.1 0.15 0.2 0.25 0.3 0 0.05 0.1 0.15 0.2 0.25 0.3 y y ( N x × N z ) = 64 × 64 32 × 32 128 × 128 www.tfd.chalmers.se/˜lada Helsinki 4 October 2012 34 / 1

SGS M ODELS B ASED ON G RID S IZE When the grid is refined, ν t gets smaller www.tfd.chalmers.se/˜lada Helsinki 4 October 2012 35 / 1

SGS M ODELS B ASED ON G RID S IZE When the grid is refined, ν t gets smaller E P k , res ε sgs κ κ c www.tfd.chalmers.se/˜lada Helsinki 4 October 2012 35 / 1

SGS M ODELS B ASED ON G RID S IZE When the grid is refined, ν t gets smaller E P k , res ε sgs , ∆ ε sgs , 0 . 5 ∆ κ κ c 2 κ c www.tfd.chalmers.se/˜lada Helsinki 4 October 2012 35 / 1

SGS M ODELS B ASED ON G RID S IZE When the grid is refined, ν t gets smaller ε sgs , ∆ = ε sgs , 0 . 5 ∆ s ij � − � τ 12 , t � ∂ � ¯ u � ε sgs = 2 � ν t ¯ s ij ¯ ∂ y E P k , res Grid refinement ⇒ must be accompanied with larger ¯ s ij ¯ s ij ⇒ ¯ s ij ¯ s ij must take place at ε sgs , ∆ higher wavenumbers ε sgs , 0 . 5 ∆ if not ⇒ grid dependent κ κ c 2 κ c www.tfd.chalmers.se/˜lada Helsinki 4 October 2012 35 / 1

∂ ¯ w ′ P OWER D ENSITY S PECTRA OF ν 0 . 5 t ∂ z One-eq k sgs model Zonal PANS 0.02 0.03 w ′ /∂ z ) 2 � 0.025 0.015 0.02 ν t ( ∂ ¯ 0.01 0.015 0.01 0.005 � 0.005 E 0 0 0 20 40 60 80 100 0 20 40 60 80 100 κ z κ z ( N x × N z ) = 64 × 64 32 × 32 128 × 128 www.tfd.chalmers.se/˜lada Helsinki 4 October 2012 36 / 1

SGS DISSIPATION VS . W AVENUMBER Energy spectra of the SGS dissipation show that the peak takes place at surprisingly low wavenumber (length scale corresponding to 10 cells or more). E ( κ ) ε sgs κ κ c www.tfd.chalmers.se/˜lada Helsinki 4 October 2012 37 / 1

SGS DISSIPATION VS . W AVENUMBER Energy spectra of the SGS dissipation show that the peak takes place at surprisingly low wavenumber (length scale corresponding to 10 cells or more). ε sgs ,κ E ( κ ) ε sgs κ κ c www.tfd.chalmers.se/˜lada Helsinki 4 October 2012 37 / 1

SGS DISSIPATION , Re τ = 8000 s ij � − � τ 12 , t � ∂ � ¯ u � SGS dissipation in the ¯ i ¯ i / 2 eq, ε sgs = 2 � ν t ¯ s ij ¯ u ′ u ′ ∂ y One-eq k sgs model Zonal PANS 15 15 10 10 ε sgs ε sgs 5 5 0 0 0.1 0.15 0.2 0.25 0.3 0.1 0.15 0.2 0.25 0.3 y y ( N x × N z ) = 64 × 64 32 × 32 128 × 128 www.tfd.chalmers.se/˜lada Helsinki 4 October 2012 38 / 1

L OCAL E QUILIBRIUM . Re τ = 4000 , N x × N z = 64 × 64 . k equation ε equation 0.4 4E-3 4E-3 0.03 4E-4 4E-4 0.3 0.02 0.2 2E-3 2E-3 2E-4 2E-4 0.01 0.1 0 0 0 0 0 0 0 0 0 0.1 0.1 0.1 0.2 0.2 0.2 0.3 0.3 0.3 0.4 0.4 0.4 0.5 0.5 0.5 0.6 0.6 0.6 0 0 0 0.1 0.1 0.1 0.2 0.2 0.2 0.3 0.3 0.3 0.4 0.4 0.4 0.5 0.5 0.5 0.6 0.6 0.6 y y � P k � + � C ε 1 P k /ε � + � ε � + � C ε 2 ε 2 / k � + Left vertical axes: URANS region; right vertical axes: LES region. www.tfd.chalmers.se/˜lada Helsinki 4 October 2012 39 / 1

L OCAL E QUILIBRIUM IN ε E QUATION . How can both the k eq. and ε be in local equilibrium?? www.tfd.chalmers.se/˜lada Helsinki 4 October 2012 40 / 1

L OCAL E QUILIBRIUM IN ε E QUATION . How can both the k eq. and ε be in local equilibrium?? If � P k � = � ε � www.tfd.chalmers.se/˜lada Helsinki 4 October 2012 40 / 1

L OCAL E QUILIBRIUM IN ε E QUATION . How can both the k eq. and ε be in local equilibrium?? If � P k � = � ε � then � ε � 2 � ε � � k �� P k �� = C ∗ C 1 � = C ∗ C 1 � k � , because 2 2 www.tfd.chalmers.se/˜lada Helsinki 4 October 2012 40 / 1

L OCAL E QUILIBRIUM IN ε E QUATION . How can both the k eq. and ε be in local equilibrium?? If � P k � = � ε � then � ε � 2 � ε � � k �� P k �� = C ∗ C 1 � = C ∗ C 1 � k � , because 2 2 However, the previous slide shows � ε � ε 2 � � = C ∗ C 1 k P k 2 k www.tfd.chalmers.se/˜lada Helsinki 4 October 2012 40 / 1

L OCAL E QUILIBRIUM IN ε E QUATION . How can both the k eq. and ε be in local equilibrium?? If � P k � = � ε � then � ε � 2 � ε � � k �� P k �� = C ∗ C 1 � = C ∗ C 1 � k � , because 2 2 However, the previous slide shows � ε � ε 2 � � = C ∗ C 1 k P k 2 k Answer: when time-averaging � ab � � = � a �� b � www.tfd.chalmers.se/˜lada Helsinki 4 October 2012 40 / 1

L OCAL E QUILIBRIUM IN ε E QUATION . The answer is because of time averaging ( � ab � < � a �� b � , (see below) 1.2 1.15 1.1 1.05 1 0 0.1 0.2 0.3 0.4 0.5 0.6 y � ε 2 / k � � ε P k / k � � ε 2 � / � k � � ε �� P k � / � k � www.tfd.chalmers.se/˜lada Helsinki 4 October 2012 41 / 1

R ESOLVED A ND M ODELLED T URBULENT K INETIC E NERGY . Modelled: bottom; total: top Resolved 8 8 7 7 � k res + k � 6 6 5 5 � k res � 4 4 3 3 � k � , 2 2 1 1 0 0 0 0.1 0.2 0.3 0.4 0 0.1 0.2 0.3 0.4 y y Re τ = 4 000 Re τ = 8 000 Re τ = 16 000; Re τ = 32 000. www.tfd.chalmers.se/˜lada Helsinki 4 October 2012 42 / 1

C ONCLUDING R EMARKS LRN PANS works well as zonal LES-RANS model for very high Re τ ( > 32 000) The model gives grid independent results The location of the interface is not important (it should not be too close to the wall) Values of 0 . 2 < f k < 0 . 5 have little impact on the results www.tfd.chalmers.se/˜lada Helsinki 4 October 2012 43 / 1

H YBRID LES-RANS Near walls: a RANS one-eq. k or a k − ω model. In core region: a LES one-eq. k SGS model. wall URANS Interface LES URANS y y + ml wall x • Location of interface either pre-defined or automatically computed www.tfd.chalmers.se/˜lada Helsinki 4 October 2012 44 / 1

M OMENTUM E QUATIONS • The Navier-Stokes, time-averaged in the near-wall regions and filtered in the core region, reads ∂ ¯ ∂ ¯ ( ν + ν T ) ∂ ¯ ∂ t + ∂ u i = − 1 p + ∂ � u i � � ¯ u i ¯ � u j ∂ x j ρ ∂ x i ∂ x j ∂ x j ν T = ν t , y ≤ y ml ν T = ν sgs , y ≥ y ml • The equation above: URANS or LES? Same boundary conditions ⇒ same solution! www.tfd.chalmers.se/˜lada Helsinki 4 October 2012 45 / 1

T URBULENCE M ODEL • Use one-equation model in both URANS region and LES region. k 3 / 2 ∂ k T ∂ t + ∂ u j k T ) = ∂ � ( ν + ν T ) ∂ k T � T (¯ + P k T − C ε ∂ x j ∂ x j ∂ x j ℓ P k T = 2 ν T ¯ S ij ¯ S ij , ν T = C k ℓ k 1 / 2 T LES-region: k T = k sgs , ν T = ν sgs , ℓ = ∆ = ( δ V ) 1 / 3 URANS-region: k T = k , ν T = ν t , ℓ ≡ ℓ RANS = 2 . 5 n [ 1 − exp ( − Ak 1 / 2 y /ν )] , Chen-Patel model (AIAA J. 1988) Location of interface can be defined by min ( 0 . 65 ∆ , y ) , ∆ = max (∆ x , ∆ y , ∆ z ) www.tfd.chalmers.se/˜lada Helsinki 4 October 2012 46 / 1

D IFFUSER [9] Instantaneous inlet data from channel DNS used. Domain: − 8 ≤ x ≤ 48, 0 ≤ y inlet ≤ 1, 0 ≤ z ≤ 4. x max = 40 gave return flow at the outlet Grid: 258 × 66 × 32. Re = U in H /ν = 18 000, angle 10 o The grid is much too coarse for LES (in the inlet region ∆ z + ≃ 170) Matching plane fixed at y ml at the inlet. In the diffuser it is located along the 2D instantaneous streamline corresponding to y ml . www.tfd.chalmers.se/˜lada Helsinki 4 October 2012 47 / 1

D IFFUSER G EOMETRY . Re = 18 000 , ANGLE 10 o H = 2 δ 7 . 9 H periodic 21 H b.c. no-slip b.c. 4 . 7 H 29 H 4 H convective outlet b.c. www.tfd.chalmers.se/˜lada Helsinki 4 October 2012 48 / 1

D IFFUSER : R ESULTS WITH LES • Velocities. Markers: experiments by Buice & Eaton (1997) x = 3 H 6 14 17 20 24 H x / H = 27 30 34 40 47 H www.tfd.chalmers.se/˜lada Helsinki 4 October 2012 49 / 1

DIFFUSER: R ESULTS W ITH RANS-LES x = 3 H 6 14 17 20 24 H 30 34 40 47 H x / H = 27 forcing; no forcing www.tfd.chalmers.se/˜lada Helsinki 4 October 2012 50 / 1

S HEAR S TRESSES ( × 2 I N L OWER H ALF ) x = 3 H 6 13 19 23 H 33 40 47 H x / H = 26 resolved; modelled www.tfd.chalmers.se/˜lada Helsinki 4 October 2012 51 / 1

RANS-LES: ν t /ν x = 3 H 6 14 17 20 24 H At x = − 7 H ν T , max /ν ≃ 11 At x = 24 H , ν T , max /ν ≃ 450 30 34 40 47 H x / H = 27 forcing; no forcing www.tfd.chalmers.se/˜lada Helsinki 4 October 2012 52 / 1

k − ω MODEL SST-DES DES [24]: Detached Eddy Simulation SST [18, 19]: A combination of the k − ε and the k − ω model � ∂ k ∂ k ∂ t + ∂ u j k ) = ∂ �� ν + ν t � (¯ + P k − β ∗ k ω ∂ x j ∂ x j σ k ∂ x j � ∂ω ∂ω ∂ t + ∂ u j ω ) = ∂ �� ν + ν t � + α P k − βω 2 + . . . (¯ ∂ x j ∂ x j σ ω ∂ x j ν t The dissipation term in the k equation is modified as [19, 25] � � L t β ∗ k ω → β ∗ k ω F DES , F DES = max C DES ∆ , 1 L t = k 1 / 2 ∆ = max { ∆ x 1 , ∆ x 2 , ∆ x 3 } , β ∗ ω ⇒ RANS near walls and LES away from walls www.tfd.chalmers.se/˜lada Helsinki 4 October 2012 53 / 1

R ESOLUTION For the near-wall region, we know how fine the mesh should be in terms of viscous units (see Slide 22) An appropriate resolution for the fully turbulent part of the boundary layer is δ/ ∆ x ≃ 10 − 20 and δ/ ∆ z ≃ 20 − 40 This may be relevant also for jets and shear layers www.tfd.chalmers.se/˜lada Helsinki 4 October 2012 54 / 1

H OW TO ESTIMATE RESOLUTION IN GENERAL ? [4, 5] Energy spectra (both in spanwise direction and time) Two-point correlations Ratio of SGS turbulent kinetic energy � k sgs � to resolved 0 . 5 � u ′ u ′ + v ′ v ′ + w ′ w ′ � Ratio of SGS shear stress � τ sgs , 12 � to resolved � u ′ v ′ � Ratio of SGS viscosity, � ν sgs � to molecular, ν Energy spectra of SGS dissipation Comparison of SGS dissipation due to ∂ u ′ i /∂ x j and ∂ � ¯ u i � /∂ x j www.tfd.chalmers.se/˜lada Helsinki 4 October 2012 55 / 1

C HANNEL FLOW , Re τ = 4000 , y + = 440 Energy spectra Two-point correlations -1 1 10 rms 0.8 z ) / w 2 E ww ( k z ) -2 0.6 10 B ww (ˆ 0.4 -3 10 0.2 0 -4 10 -0.2 0 1 2 10 10 10 0 0.1 0.2 0.3 0.4 ˆ z = z − z 0 κ z = 2 π ( k z − 1 ) / z max (∆ x , ∆ z ) 0 . 5 ∆ x 0 . 5 ∆ z ◦ 2 ∆ x ; + : 2 ∆ z The (∆ x , ∆ z ) mesh is ( δ/ ∆ x , δ/ ∆ z ) = ( 10 , 20 ) www.tfd.chalmers.se/˜lada Helsinki 4 October 2012 56 / 1

C HANNEL FLOW , Re τ = 4000 , y + = 440 Energy spectra Two-point correlations -1 1 10 rms 0.8 z ) / w 2 E ww ( k z ) -2 0.6 10 B ww (ˆ 0.4 -3 10 0.2 0 -4 10 -0.2 0 1 2 10 10 10 0 0.1 0.2 0.3 0.4 ˆ z = z − z 0 κ z = 2 π ( k z − 1 ) / z max (∆ x , ∆ z ) 0 . 5 ∆ x 0 . 5 ∆ z ◦ 2 ∆ x ; + : 2 ∆ z The (∆ x , ∆ z ) mesh is ( δ/ ∆ x , δ/ ∆ z ) = ( 10 , 20 ) Two-point correlation is better Shows that 2 ∆ z and 2 ∆ x (two-point corr in x ) are too coarse. www.tfd.chalmers.se/˜lada Helsinki 4 October 2012 56 / 1

C HANNEL FLOW , Re τ = 4000 , y + = 440 k res k res = ( u ′ 2 + v ′ 2 + w ′ 2 ) / 2 γ = < k T > + k res 6 1 5 0.9 4 0.8 k res γ 3 0.7 2 0.6 1 0.5 0 0.4 0 1000 2000 3000 4000 0 1000 2000 3000 4000 y + y + Pope [20] suggests γ > 0 . 8 indicates well resolved flow (∆ x , ∆ z ) 0 . 5 ∆ x 0 . 5 ∆ z ◦ 2 ∆ x ; + : 2 ∆ z www.tfd.chalmers.se/˜lada Helsinki 4 October 2012 57 / 1

C HANNEL FLOW , Re τ = 4000 , y + = 440 k res k res = ( u ′ 2 + v ′ 2 + w ′ 2 ) / 2 γ = < k T > + k res 6 1 5 0.9 4 0.8 k res γ 3 0.7 2 0.6 1 0.5 0 0.4 0 1000 2000 3000 4000 0 1000 2000 3000 4000 y + y + Pope [20] suggests γ > 0 . 8 indicates well resolved flow (∆ x , ∆ z ) 0 . 5 ∆ x 0 . 5 ∆ z ◦ 2 ∆ x ; + : 2 ∆ z Pope criterion does not work here www.tfd.chalmers.se/˜lada Helsinki 4 October 2012 57 / 1

SGS VS . M OLECULAR V ISCOSITY [5] x = − H x = 20 H 1 1 0.8 0 0.6 -1 y / H y / H 0.4 -2 0.2 -3 0 -4 0 1 2 3 4 0 2 4 6 8 10 12 � ν sgs � /ν � ν sgs � /ν N z = 32; N z = 64; N z = 128. www.tfd.chalmers.se/˜lada Helsinki 4 October 2012 58 / 1

SGS VS . R ESOLVED S HEAR S TRESSES x = − H x = 20 H -0.5 0.4 0.35 -1 0.3 -1.5 y / H 0.25 y / H -2 0.2 -2.5 0.15 0.1 -3 0.05 -3.5 0 0 0.05 0.1 0.15 0.2 0.25 0.3 0 0.01 0.02 0.03 0.04 0.05 � τ sgs , 12 � / � u ′ v ′ � � τ sgs , 12 � / � u ′ v ′ � N z = 32; N z = 64; N z = 128. www.tfd.chalmers.se/˜lada Helsinki 4 October 2012 59 / 1

T HE PANS M ODEL The PANS model is a modified k − ε model It can operate both in RANS mode and LES mode In the present work a low-Reynolds turbulence version of the PANS is used A method how to implement embedded LES is proposed It is evaluated for channel flow and hump flow www.tfd.chalmers.se/˜lada Helsinki 4 October 2012 60 / 1

Embedded LES Using PANS [10, 11] Lars Davidson 1 and Shia-Hui Peng 1 , 2 Davidson& Peng 1 Department of Applied Mechanics Chalmers University of Technology, SE-412 96 Gothenburg, SWEDEN 2 FOI, Swedish Defence Research Agency, SE-164 90, Stockholm, SWEDEN

PANS L OW R EYNOLDS N UMBER M ODEL [17] � ∂ k u ∂ k u ∂ t + ∂ ( k u U j ) = ∂ �� ν + ν u � + ( P u − ε u ) ∂ x j ∂ x j σ ku ∂ x j � ∂ε u ε 2 ∂ t + ∂ ( ε u U j ) �� � ∂ε u = ∂ ν + ν u ε u u − C ∗ + C ε 1 P u ε 2 ∂ x j ∂ x j σ ε u ∂ x j k u k u k 2 f 2 f 2 ε 2 = C ε 1 + f k u k k ν u = C µ f µ , C ∗ ( C ε 2 f 2 − C ε 1 ) , σ ku ≡ σ k , σ ε u ≡ σ ε ε u f ε f ε f ε C ε 1 , C ε 2 , σ k , σ ε and C µ same values as [1]. f ε = 1. f 2 and f µ read � R t �� 2 � � − y ∗ � � 2 �� � f 2 = 1 − exp 1 − 0 . 3exp − 3 . 1 6 . 5 � R t �� 2 � � 2 �� � − y ∗ 5 � � f µ = 1 − exp 1 + − exp R 3 / 4 14 200 t Baseline model: f k = 0 . 4. Range of 0 . 2 < f k < 0 . 6 is evaluated www.tfd.chalmers.se/˜lada Helsinki 4 October 2012 62 / 1

C HANNEL FLOW : D OMAIN Interface LES, f k < 1 RANS, f k = 1 . 0 d y x δ 2 . 2 δ Interface: Synthetic turbulent fluctuations are introduced as additional convective fluxes in the momentum equations and the continuity equation f k = 0 . 4 is the baseline value for LES [17] www.tfd.chalmers.se/˜lada Helsinki 4 October 2012 63 / 1

I NLET FLUCTUATIONS 2 1 w ′ w ′ two-point corr 0.8 1.5 0.6 y 1 0.4 0.2 0.5 0 0 0 0.5 1 1.5 2 0 0.1 0.2 0.3 0.4 0.5 � u ′ v ′ � , v 2 rms , w 2 rms , u 2 rms / u 2 ˆ z τ Anisotropic synthetic fluctuations, u ′ , v ′ , w ′ , Integral length scale L ≃ 0 . 13 (see 2-p point correlation) Asymmetric time filter ( U ′ ) m = a ( U ′ ) m − 1 + b ( u ′ ) m with a = 0 . 954 , b = ( 1 − a 2 ) 1 / 2 gives a time integral scale T = 0 . 015 ( ∆ t = 0 . 00063) www.tfd.chalmers.se/˜lada Helsinki 4 October 2012 64 / 1

I NTERFACE C ONDITIONS FOR k u AND ε u For k u & ε u we prescribe “inlet” boundary conditions at the interface. First, the usual convective and diffusive fluxes at the interface are set to zero Next, new convective fluxes are added. Which “inlet” values should be used at the interface? ◮ k u , int = f k k RANS ( x = 0 . 5 δ ) , ε u , int = C 3 / 4 k 3 / 2 u , int /ℓ sgs , ℓ sgs = C s ∆ , µ ∆ = V 1 / 3 ◮ www.tfd.chalmers.se/˜lada Helsinki 4 October 2012 65 / 1

I NTERFACE C ONDITIONS FOR k u AND ε u For k u & ε u we prescribe “inlet” boundary conditions at the interface. First, the usual convective and diffusive fluxes at the interface are set to zero Next, new convective fluxes are added. Which “inlet” values should be used at the interface? ◮ k u , int = f k k RANS ( x = 0 . 5 δ ) , ε u , int = C 3 / 4 k 3 / 2 u , int /ℓ sgs , ℓ sgs = C s ∆ , µ ∆ = V 1 / 3 ◮ Baseline C s = 0 . 07; different C s values are tested www.tfd.chalmers.se/˜lada Helsinki 4 October 2012 65 / 1

C HANNEL F LOW : V ELOCITY AND S HEAR S TRESSES 30 1 25 0.5 20 � u ′ v ′ � + U + 0 15 10 -0.5 5 -1 0 0 1 2 0 0.5 1 1.5 2 10 10 10 y + y + x /δ = 0 . 19 x /δ = 1 . 25 x /δ = 3 www.tfd.chalmers.se/˜lada Helsinki 4 October 2012 66 / 1

C HANNEL F LOW : STRESSES AND PEAK VALUES VS . x x /δ = 3 4 100 3.5 resolved stresses 3 � ν u /ν � max max 2.5 � u ′ u ′ � + 2 2 50 1.5 1 0.5 0 0 0 0 0.5 0.5 1 1 1.5 1.5 2 2 2.5 2.5 3 3 3.5 3.5 0 x 0 200 400 600 800 y /δ � u ′ u ′ � + � u ′ u ′ � + max (left) � v ′ v ′ � + � ν u � + max (right) � w ′ w ′ � + www.tfd.chalmers.se/˜lada Helsinki 4 October 2012 67 / 1

C HANNEL F LOW : DIFFERENT C s VALUE FOR ε interface k u , int = f k k RANS ε u , int = C 3 / 4 k 3 / 2 u , int /ℓ sgs , ℓ sgs = C s ∆ µ x /δ = 3 30 1 25 0.5 20 � u ′ v ′ � + U + 0 15 10 -0.5 5 -1 0 0 1 2 0 0.5 1 1.5 2 10 10 10 y + y + C s = 0 . 07 C s = 0 . 1 C s = 0 . 2 www.tfd.chalmers.se/˜lada Helsinki 4 October 2012 68 / 1

C HANNEL F LOW : DIFFERENT C s VALUE FOR ε interface x /δ = 3 6 1.05 5 1 4 � ν u � /ν u τ 3 0.95 2 0.9 1 0 0.85 0 0.2 0.4 0.6 0.8 1 0 1 2 3 4 y + x /δ C s = 0 . 07 C s = 0 . 1 C s = 0 . 2 www.tfd.chalmers.se/˜lada Helsinki 4 October 2012 69 / 1

C HANNEL F LOW : DIFFERENT f k VALUES x /δ = 3 1 20 0.5 � u ′ v ′ � + 15 U + 0 10 -0.5 5 -1 0 0 1 2 0 0.5 1 1.5 2 10 10 10 y + y + f k = 0 . 4 f k = 0 . 2 f k = 0 . 6 www.tfd.chalmers.se/˜lada Helsinki 4 October 2012 70 / 1

C HANNEL F LOW : DIFFERENT f k VALUES x /δ = 3 4 1.05 3.5 3 1 � ν u � /ν 2.5 u τ 2 0.95 1.5 1 0.9 0.5 0 0.85 0 0.2 0.4 0.6 0.8 1 0 1 2 3 4 y + x /δ f k = 0 . 4 f k = 0 . 2 f k = 0 . 6 www.tfd.chalmers.se/˜lada Helsinki 4 October 2012 71 / 1

H UMP F LOW x I / c = 0 . 6 S R NTS 2D RANS PANS Inlet, Separation x S / c = 0 . 65; reattachment x R / c = 1 . 1 Re c = 936 000 U ij c ( U in = c = ρ = 1, ν = 1 / Re c ν H / c = 0 . 91, h / c = 0 . 128, x / c = [ 0 . 6 , 4 . 2 ] Mesh: 312 × 120 × 64, Z max = 0 . 2 c (baseline) www.tfd.chalmers.se/˜lada Helsinki 4 October 2012 72 / 1

B ASELINE I NLET FLUCTUATIONS 0.4 1 w ′ w ′ two-point corr 0.35 0.8 0.3 0.6 y / c 0.25 0.4 0.2 0.2 0.15 0 -1 0 1 2 3 4 5 6 0 0.02 0.04 0.06 0.08 0.1 � u ′ v ′ � , v 2 rms , w 2 rms , u 2 rms / u 2 ˆ z τ Integral length scale L ≃ 0 . 04 (see 2-p point correlation) Asymmetric time filter ( U ′ ) m = a ( U ′ ) m − 1 + b ( u ′ ) m with a = 0 . 954 , b = ( 1 − a 2 ) 1 / 2 gives a time integral scale T = 0 . 038 ∆ t = 0 . 002. 7500 + 7500 time steps (100 hours one core) Fluctuations multiplied by f bl = max { 0 . 5 [ 1 − tanh ( y − y bl − y wall ) / b ] , 0 . 02 } , y bl = 0 . 2, b = 0 . 01. www.tfd.chalmers.se/˜lada Helsinki 4 October 2012 73 / 1

P RESSURE : A MPLITUDES O F I NLET F LUCT 0.8 0.7 0.6 0.5 − C p 0.4 0.3 0.2 0.1 0 0.5 1 1.5 2 x / c 1 . 5 × (baseline inlet fluct) baseline inlet fluct 0 . 5 × (baseline inlet fluct) www.tfd.chalmers.se/˜lada Helsinki 4 October 2012 74 / 1

S KIN F RICTION : A MPLITUDES O F I NLET F LUCT zoom -3 3x 10 -3 10x 10 2 8 6 1 4 C f 0 2 -1 0 -2 -2 0 0.5 1 1.5 0.6 0.8 1 1.2 1.4 1.6 x / c x / c baseline inlet fluct 1 . 5 × (baseline inlet fluct) 0 . 5 × (baseline inlet fluct) www.tfd.chalmers.se/˜lada Helsinki 4 October 2012 75 / 1