E MBEDDED LES U SING PANS [2] L ARS D AVIDSON 1 AND S HIA -H UI P ENG - - PowerPoint PPT Presentation

e mbedded les u sing pans 2
SMART_READER_LITE
LIVE PREVIEW

E MBEDDED LES U SING PANS [2] L ARS D AVIDSON 1 AND S HIA -H UI P ENG - - PowerPoint PPT Presentation

E MBEDDED LES U SING PANS [2] L ARS D AVIDSON 1 AND S HIA -H UI P ENG 1 , 2 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


slide-1
SLIDE 1

EMBEDDED LES USING PANS [2] LARS DAVIDSON1 AND SHIA-HUI PENG1,2

1Department of Applied Mechanics

Chalmers University of Technology, SE-412 96 Gothenburg, SWEDEN

2FOI, Swedish Defence Research Agency, SE-164 90, Stockholm,

SWEDEN

slide-2
SLIDE 2

PANS LOW REYNOLDS NUMBER MODEL [3]

∂ku ∂t + ∂(kuUj) ∂xj = ∂ ∂xj

  • ν + νu

σku ∂ku ∂xj

  • + (Pu − εu)

∂εu ∂t + ∂(εuUj) ∂xj = ∂ ∂xj

  • ν + νu

σεu ∂εu ∂xj

  • + Cε1Pu

εu ku − C∗

ε2

ε2

u

ku νu = Cµfµ k2

u

εu , C∗

ε2 = Cε1 + fk

fε (Cε2f2 − Cε1), σku ≡ σk f 2

k

fε , σεu ≡ σε f 2

k

fε Cε1, Cε2, σk, σε and Cµ same values as [1]. fε = 1. f2 and fµ read f2 =

  • 1 − exp
  • − y∗

3.1 2 1 − 0.3exp

Rt 6.5 2 fµ =

  • 1 − exp
  • − y∗

14 2 1 + 5 R3/4

t

exp

Rt 200 2 Baseline model: fk = 0.4. Range of 0.2 < fk < 0.6 is evaluated

Davidson& Peng AIAA, Hawaii, 27-30 June 2011 2 / 30

slide-3
SLIDE 3

CHANNEL FLOW: DOMAIN

d

x y δ 2.2δ LES, fk < 1 RANS, fk = 1.0 Interface Interface: Synthetic turbulent fluctuations are introduced as additional convective fluxes in the momentum equations and the continuity equation fk = 0.4 is the baseline value for LES [3]

Davidson& Peng AIAA, Hawaii, 27-30 June 2011 3 / 30

slide-4
SLIDE 4

INLET FLUCTUATIONS

0.5 1 1.5 2 0.5 1 1.5 2

y u′v′, v2

rms, w2 rms, u2 rms/u2 τ

0.1 0.2 0.3 0.4 0.5 0.2 0.4 0.6 0.8 1

w′w′ two-point corr ˆ z 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 − a2)1/2 gives a time integral scale T = 0.015 (∆t = 0.00063)

Davidson& Peng AIAA, Hawaii, 27-30 June 2011 4 / 30

slide-5
SLIDE 5

INTERFACE CONDITIONS FOR ku AND εu

For ku & εu 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?

◮ ku,int = fkkRANS(x = 0.5δ), εu,int = C3/4

µ

k3/2

u,int/ℓsgs, ℓsgs = Cs∆,

∆ = V 1/3

◮ Davidson& Peng AIAA, Hawaii, 27-30 June 2011 5 / 30

slide-6
SLIDE 6

INTERFACE CONDITIONS FOR ku AND εu

For ku & εu 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?

◮ ku,int = fkkRANS(x = 0.5δ), εu,int = C3/4

µ

k3/2

u,int/ℓsgs, ℓsgs = Cs∆,

∆ = V 1/3

◮ Baseline Cs = 0.07; different Cs values are tested Davidson& Peng AIAA, Hawaii, 27-30 June 2011 5 / 30

slide-7
SLIDE 7

CHANNEL FLOW: VELOCITY AND SHEAR STRESSES

10 10

1

10

2

5 10 15 20 25 30

y+ U+

0.5 1 1.5 2 −1 −0.5 0.5 1

y+ u′v′+ x/δ = 0.19 x/δ = 1.25 x/δ = 3

Davidson& Peng AIAA, Hawaii, 27-30 June 2011 6 / 30

slide-8
SLIDE 8

CHANNEL FLOW: STRESSES AND PEAK VALUES VS. x

200 400 600 800 0.5 1 1.5 2 2.5 3 3.5

y/δ resolved stresses x/δ = 3

0.5 1 1.5 2 2.5 3 3.5 2 4 0.5 1 1.5 2 2.5 3 3.5 50 100

x u′u′+

max

νu/νmax u′u′+ u′u′+

max (left)

v′v′+ νu+

max (right)

w′w′+

Davidson& Peng AIAA, Hawaii, 27-30 June 2011 7 / 30

slide-9
SLIDE 9

CHANNEL FLOW: DIFFERENT Cs VALUE FOR εinterface

ku,int = fkkRANS εu,int = C3/4

µ

k3/2

u,int/ℓsgs, ℓsgs = Cs∆

10 10

1

10

2

5 10 15 20 25 30

y+ U+ x/δ = 3

0.5 1 1.5 2 −1 −0.5 0.5 1

y+ u′v′+ Cs = 0.07 Cs = 0.1 Cs = 0.2

Davidson& Peng AIAA, Hawaii, 27-30 June 2011 8 / 30

slide-10
SLIDE 10

CHANNEL FLOW: DIFFERENT Cs VALUE FOR εinterface

0.2 0.4 0.6 0.8 1 1 2 3 4 5 6

y+ νu/ν x/δ = 3

1 2 3 4 0.85 0.9 0.95 1 1.05

x/δ uτ Cs = 0.07 Cs = 0.1 Cs = 0.2

Davidson& Peng AIAA, Hawaii, 27-30 June 2011 9 / 30

slide-11
SLIDE 11

CHANNEL FLOW: DIFFERENT fk VALUES

10 10

1

10

2

5 10 15 20

y+ U+ x/δ = 3

0.5 1 1.5 2 −1 −0.5 0.5 1

y+ u′v′+ fk = 0.4 fk = 0.2 fk = 0.6

Davidson& Peng AIAA, Hawaii, 27-30 June 2011 10 / 30

slide-12
SLIDE 12

CHANNEL FLOW: DIFFERENT fk VALUES

0.2 0.4 0.6 0.8 1 0.5 1 1.5 2 2.5 3 3.5 4

y+ νu/ν x/δ = 3

1 2 3 4 0.85 0.9 0.95 1 1.05

x/δ uτ fk = 0.4 fk = 0.2 fk = 0.6

Davidson& Peng AIAA, Hawaii, 27-30 June 2011 11 / 30

slide-13
SLIDE 13

HUMP FLOW

xI/c = 0.6 R S NTS 2D RANS PANS Inlet, Separation xS/c = 0.65; reattachment xR/c = 1.1 Rec = 936 000 Uijc

ν

(Uin = c = ρ = 1, ν = 1/Rec H/c = 0.91, h/c = 0.128, x/c = [0.6, 4.2] Mesh: 312 × 120 × 64, Zmax = 0.2c (baseline)

Davidson& Peng AIAA, Hawaii, 27-30 June 2011 12 / 30

slide-14
SLIDE 14

BASELINE INLET FLUCTUATIONS

−1 1 2 3 4 5 6 0.15 0.2 0.25 0.3 0.35 0.4

y/c u′v′, v2

rms, w2 rms, u2 rms/u2 τ

0.02 0.04 0.06 0.08 0.1 0.2 0.4 0.6 0.8 1

w′w′ two-point corr ˆ z 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 − a2)1/2 gives a time integral scale T = 0.038 ∆t = 0.002. 7500 + 7500 time steps (100 hours one core) Fluctuations multiplied by fbl = max {0.5 [1 − tanh(y − ybl − ywall)/b] , 0.02}, ybl = 0.2, b = 0.01.

Davidson& Peng AIAA, Hawaii, 27-30 June 2011 13 / 30

slide-15
SLIDE 15

PRESSURE: AMPLITUDES OF INLET FLUCT

0.5 1 1.5 2 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8

x/c −Cp baseline inlet fluct 1.5× (baseline inlet fluct) 0.5× (baseline inlet fluct)

Davidson& Peng AIAA, Hawaii, 27-30 June 2011 14 / 30

slide-16
SLIDE 16

SKIN FRICTION: AMPLITUDES OF INLET FLUCT

0.5 1 1.5 −2 2 4 6 8 10x 10

−3

x/c Cf

0.6 0.8 1 1.2 1.4 1.6 −2 −1 1 2 3x 10

−3

x/c zoom baseline inlet fluct 1.5× (baseline inlet fluct) 0.5× (baseline inlet fluct)

Davidson& Peng AIAA, Hawaii, 27-30 June 2011 15 / 30

slide-17
SLIDE 17

VELOCITIES: AMPLITUDES OF INLET FLUCT

0.2 0.4 0.6 0.8 1 1.2 0.1 0.15 0.2 0.25

x/c = 65 y

0.5 1 0.05 0.1 0.15 0.2 0.25

x/c = 80

0.5 1 0.05 0.1 0.15 0.2 0.25

x/c = 100 U/Ub y

0.5 1 0.05 0.1 0.15 0.2 0.25

x/c = 110 U/Ub baseline 1.5× (baseline) 0.5× (baseline)

Davidson& Peng AIAA, Hawaii, 27-30 June 2011 16 / 30

slide-18
SLIDE 18

VELOCITIES: AMPLITUDES OF INLET FLUCT

0.2 0.4 0.6 0.8 1 1.2 0.05 0.1 0.15 0.2 0.25

x/c = 120 U/Ub y

0.2 0.4 0.6 0.8 1 1.2 0.05 0.1 0.15 0.2 0.25

x/c = 130 U/Ub baseline 1.5× (baseline) 0.5× (baseline)

Davidson& Peng AIAA, Hawaii, 27-30 June 2011 17 / 30

slide-19
SLIDE 19

RESOLVED AND MODELLED (< 0) SHEAR STRESSES

−5 5 10 15 x 10

−3

0.1 0.15 0.2 0.25

x/c = 0.65 y

0.01 0.02 0.03 0.05 0.1 0.15 0.2 0.25

x/c = 0.80

0.01 0.02 0.03 0.04 0.05 0.1 0.15 0.2

x/c = 1.00 τ12,u, −u′v′/U2

b

y

0.01 0.02 0.03 0.05 0.1 0.15 0.2

x/c = 1.10 τ12,u, −u′v′/U2

b

baseline 1.5× (baseline) 0.5× (baseline)

Davidson& Peng AIAA, Hawaii, 27-30 June 2011 18 / 30

slide-20
SLIDE 20

SHEAR STRESSES: AMPLITUDES OF INLET FLUCT

Resolved and Modelled (< 0) Shear stresses

0.01 0.02 0.03 0.05 0.1 0.15 0.2

x/c = 1.20 τ12,u, −u′v′/U2

b

y

0.01 0.02 0.03 0.05 0.1 0.15 0.2

x/c = 1.30 τ12,u, −u′v′/U2

b

baseline inlet fluct 1.5× (baseline inlet fluct) 0.5× (baseline inlet fluct)

Davidson& Peng AIAA, Hawaii, 27-30 June 2011 19 / 30

slide-21
SLIDE 21

TURB VISCOSITY: AMPLITUDES OF INLET FLUCT

5 10 15 20 0.1 0.12 0.14 0.16 0.18 0.2

x/c = 0.65 y

20 40 60 80 100 120 0.05 0.1 0.15 0.2

x/c = 0.80

20 40 60 80 100 120 0.05 0.1 0.15 0.2

x/c = 1.00 νt/ν y

20 40 60 80 100 120 0.05 0.1 0.15 0.2

x/c = 1.10 νt/ν baseline 1.5× (baseline) 0.5× (baseline)

Davidson& Peng AIAA, Hawaii, 27-30 June 2011 20 / 30

slide-22
SLIDE 22

TURB VISCOSITY: AMPLITUDES OF INLET FLUCT

20 40 60 80 100 120 0.05 0.1 0.15 0.2

x/c = 1.20 νt/ν y

20 40 60 80 100 120 0.05 0.1 0.15 0.2

x/c = 1.30 νt/ν baseline 1.5× (baseline) 0.5× (baseline)

Davidson& Peng AIAA, Hawaii, 27-30 June 2011 21 / 30

slide-23
SLIDE 23

PRESSURE: fk = 0.5; NO INLET FLUCT; Nk = 128

0.5 1 1.5 2 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8

x/c −Cp Nk = 128 no inlet fluct fk = 0.5

Davidson& Peng AIAA, Hawaii, 27-30 June 2011 22 / 30

slide-24
SLIDE 24

SKIN FRICTION: fk = 0.5; NO INLET FLUCT; Nk = 128

0.5 1 1.5 −2 2 4 6 8 10x 10

−3

x/c Cf

0.6 0.8 1 1.2 1.4 1.6 −2 −1 1 2 3x 10

−3

x/c zoom Nk = 128 no inlet fluct fk = 0.5

Davidson& Peng AIAA, Hawaii, 27-30 June 2011 23 / 30

slide-25
SLIDE 25

VELOCITIES: fk = 0.5; NO INLET FLUCT; Nk = 128

0.2 0.4 0.6 0.8 1 1.2 0.1 0.15 0.2 0.25

x/c = 65 y

0.5 1 0.05 0.1 0.15 0.2 0.25

x/c = 80

0.5 1 0.05 0.1 0.15 0.2 0.25

x/c = 100 U/Ub y

0.5 1 0.05 0.1 0.15 0.2 0.25

x/c = 110 U/Ub Nk = 128 no inlet fluct fk = 0.5

Davidson& Peng AIAA, Hawaii, 27-30 June 2011 24 / 30

slide-26
SLIDE 26

VELOCITIES: fk = 0.5; NO INLET FLUCT; Nk = 128

0.2 0.4 0.6 0.8 1 1.2 0.05 0.1 0.15 0.2 0.25

x/c = 120 U/Ub y

0.2 0.4 0.6 0.8 1 1.2 0.05 0.1 0.15 0.2 0.25

x/c = 130 U/Ub Nk = 128 no inlet fluct fk = 0.5

Davidson& Peng AIAA, Hawaii, 27-30 June 2011 25 / 30

slide-27
SLIDE 27

RESOLVED AND MODELLED (< 0) SHEAR STRESSES

−5 5 x 10

−3

0.1 0.15 0.2 0.25

x/c = 0.65 y

0.01 0.02 0.03 0.04 0.05 0.1 0.15 0.2 0.25

x/c = 0.80

0.01 0.02 0.03 0.04 0.05 0.1 0.15 0.2

x/c = 1.00 τ12,u, −u′v′/U2

b

y

0.01 0.02 0.03 0.05 0.1 0.15 0.2

x/c = 1.10 τ12,u, −u′v′/U2

b

Nk = 128 no inlet fluct fk = 0.5

Davidson& Peng AIAA, Hawaii, 27-30 June 2011 26 / 30

slide-28
SLIDE 28

SHEAR STRESSES: fk = 0.5; NO INLET FLUCT; Nk = 128

Resolved and Modelled (< 0) Shear stresses

0.01 0.02 0.03 0.05 0.1 0.15 0.2

x/c = 1.20 τ12,u, −u′v′/U2

b

y

0.01 0.02 0.03 0.05 0.1 0.15 0.2

x/c = 1.30 τ12,u, −u′v′/U2

b

Nk = 128 no inlet fluct fk = 0.5

Davidson& Peng AIAA, Hawaii, 27-30 June 2011 27 / 30

slide-29
SLIDE 29

TURB VISCOSITY: fk = 0.5; NO INLET FLUCT; Nk = 128

5 10 15 20 0.1 0.12 0.14 0.16 0.18 0.2

x/c = 0.65 y

50 100 150 200 250 0.05 0.1 0.15 0.2

x/c = 0.80

50 100 150 200 250 0.05 0.1 0.15 0.2

x/c = 1.00 νt/ν y

50 100 150 200 250 0.05 0.1 0.15 0.2

x/c = 1.10 νt/ν Nk = 128 no inlet fluct fk = 0.5

Davidson& Peng AIAA, Hawaii, 27-30 June 2011 28 / 30

slide-30
SLIDE 30

TURB VISCOSITY: fk = 0.5; NO INLET FLUCT; Nk = 128

50 100 150 200 250 0.05 0.1 0.15 0.2

x/c = 1.20 νt/ν y

50 100 150 200 250 0.05 0.1 0.15 0.2

x/c = 1.30 νt/ν Nk = 128 no inlet fluct fk = 0.5

Davidson& Peng AIAA, Hawaii, 27-30 June 2011 29 / 30

slide-31
SLIDE 31

CONCLUDING REMARKS

LRN PANS has been shown to work well as an embedded LES method Channel flow: At two δ downstream the interface, the resolved turbulence in good agreement with DNS data and the wall friction velocity has reached 99% of its fully developed value. Channel flow: The treatment of the modelled ku and εu across the interface is important. LRN PANS predicts the hump flow well but the recover rate sligtly too slow Hump flow: large (small) inlet fluctuations gives a smaller (larger) recirculation

Davidson& Peng AIAA, Hawaii, 27-30 June 2011 30 / 30

slide-32
SLIDE 32

[1] ABE, K., KONDOH, T., AND NAGANO, Y. A new turbulence model for predicting fluid flow and heat transfer in separating and reattaching flows - 1. Flow field calculations.

  • Int. J. Heat Mass Transfer 37 (1994), 139–151.

[2] DAVIDSON, L., AND PENG, S.-H. Emdedded LES with PANS. In 6th AIAA Theoretical Fluid Mechanics Conference, AIAA paper 2011-3108 (27-30 June, Honolulu, Hawaii, 2011). [3] MA, J., PENG, S.-H., DAVIDSON, L., AND WANG, F. A low Reynolds number variant of Partially-Averaged Navier-Stokes model for turbulence. International Journal of Heat and Fluid Flow 32 (2011), 652–669. 10.1016/j.ijheatfluidflow.2011.02.001.

Davidson& Peng AIAA, Hawaii, 27-30 June 2011 30 / 30