E LES USING PANS [2, 3] L D - Chalmerslada/slides/slides_embedded_LES.pdfCHANNEL FLOW: DOMAIN d x y...

EMBEDDED LES USING PANS [2, 3]LARS DAVIDSON

Department of Applied MechanicsChalmers University of Technology, SE-412 96 Gothenburg,

SWEDEN

PANS LOW REYNOLDS NUMBER MODEL [4]∂ku

∂t+

∂(kuUj)

∂xj=

∂

∂xj

[(

ν +νu

σku

)

∂ku

∂xj

]

+ (Pu − εu)

∂εu

∂t+

∂(εuUj)

∂xj=

∂

∂xj

[(

ν +νu

σεu

)

∂εu

∂xj

]

+ Cε1Puεu

ku− C∗

ε2ε2

u

ku

νu = Cµfµk2

u

εu, C∗

ε2 = Cε1 +fkfε

(Cε2f2 − Cε1), σku ≡ σkf 2k

fε, σεu ≡ σε

f 2k

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

R3/4t

exp[

−( Rt

200

)2]

}

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

Davidson ( ) www.tfd.chalmers.se/˜lada 2 / 40

CHANNEL FLOW: DOMAIN

d

x

y

δ 2.2δ

LES, fk < 1RANS, fk = 1.0

Interface

Interface: Synthetic turbulent fluctuations are introduced asadditional convective fluxes in the momentum equations and thecontinuity equation

fk = 0.4 is the baseline value for LES [4]


INLET FLUCTUATIONS

0 0.5 1 1.5 20

0.5

1

1.5

2

y

〈u′v ′〉, v2rms, w2

rms, u2rms/u2

τ

0 0.1 0.2 0.3 0.4 0.5

0

0.2

0.4

0.6

0.8

1

w′w

′tw

o-po

intc

orr

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 witha = 0.954, b = (1 − a2)1/2 gives a time integral scale T = 0.015(∆t = 0.00063)


INTERFACE CONDITIONS FOR ku AND εu

For ku & εu we prescribe “inlet” boundary conditions at theinterface.

First, the usual convective and diffusive fluxes at the interface areset to zeroNext, new convective fluxes are added. Which “inlet” valuesshould be used at the interface?

◮ ku,int = fk kRANS(x = 0.5δ), εu,int = C3/4µ k3/2

u,int/ℓsgs, ℓsgs = Cs∆,

∆ = V 1/3

◮


INTERFACE CONDITIONS FOR ku AND εu

For ku & εu we prescribe “inlet” boundary conditions at theinterface.

First, the usual convective and diffusive fluxes at the interface areset to zeroNext, new convective fluxes are added. Which “inlet” valuesshould be used at the interface?

◮ ku,int = fk kRANS(x = 0.5δ), εu,int = C3/4µ k3/2

u,int/ℓsgs, ℓsgs = Cs∆,

∆ = V 1/3

◮ Baseline Cs = 0.07; different Cs values are tested


NUMERICAL METHOD

Incompressible finite volume method

Pressure-velocity coupling treated with fractional step

2nd order central differencing scheme for momentum eqns in LESregion.

2nd order upwind differencing scheme (van Leer) for momentumeqns in RANS region.

Hybrid 1st order upwind/2nd order central scheme k & ε eqns.

2nd -order Crank-Nicholson for time discretization


CHANNEL FLOW: VELOCITY AND SHEAR STRESSES

100

101

102

0

5

10

15

20

25

30

y+

U+

0 0.5 1 1.5 2

−1

−0.5

0

0.5

1

y+〈u

′v′〉+

x/δ = 0.19 x/δ = 1.25 x/δ = 3


CHANNEL FLOW: STRESSES AND PEAK VALUES VS. x

0 200 400 600 8000

0.5

1

1.5

2

2.5

3

3.5

y/δ

reso

lved

stre

sses

x/δ = 3

0 0.5 1 1.5 2 2.5 3 3.50

2

4

0 0.5 1 1.5 2 2.5 3 3.50

50

100

x

〈u′u′〉+ m

ax

〈νu/ν

〉 max

〈u′u′〉+ 〈u′u′〉+max (left)〈v ′v ′〉+ 〈νu〉

+max (right)

〈w ′w ′〉+


CHANNEL FLOW: DIFFERENT Cs VALUE FOR εinterface

ku,int = fkkRANS

εu,int = C3/4µ k3/2

u,int/ℓsgs, ℓsgs = Cs∆

100

101

102

0

5

10

15

20

25

30

y+

U+

x/δ = 3

0 0.5 1 1.5 2

−1

−0.5

0

0.5

1

y+

〈u′v′〉+

Cs = 0.07 Cs = 0.1 Cs = 0.2


CHANNEL FLOW: DIFFERENT Cs VALUE FOR εinterface

0 0.2 0.4 0.6 0.8 10

1

2

3

4

5

6

y+

〈νu〉/

ν

x/δ = 3

0 1 2 3 40.85

0.9

0.95

1

1.05

x/δu τ

Cs = 0.07 Cs = 0.1 Cs = 0.2


CHANNEL FLOW: DIFFERENT fk VALUES

100

101

102

0

5

10

15

20

y+

U+

x/δ = 3

0 0.5 1 1.5 2

−1

−0.5

0

0.5

1

y+〈u

′v′〉+

fk = 0.4 fk = 0.2 fk = 0.6


CHANNEL FLOW: DIFFERENT fk VALUES

0 0.2 0.4 0.6 0.8 10

0.5

1

1.5

2

2.5

3

3.5

4

y+

〈νu〉/

ν

x/δ = 3

0 1 2 3 40.85

0.9

0.95

1

1.05

x/δu τ

fk = 0.4 fk = 0.2 fk = 0.6


HUMP FLOWxI/c = 0.6 RS

NTS 2D RANS PANS

Inlet, Separation xS/c = 0.65; reattachment xR/c = 1.1

Rec = 936 000Uijcν (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)


BASELINE INLET FLUCTUATIONS

−1 0 1 2 3 4 5 6

0.15

0.2

0.25

0.3

0.35

0.4

y/c

〈u′v ′〉, v2rms, w2

rms, u2rms/u2

τ

0 0.02 0.04 0.06 0.08 0.1

0

0.2

0.4

0.6

0.8

1

w′w

′tw

o-po

intc

orr

z

Integral length scale L ≃ 0.04 (see 2-p point correlation)Asymmetric time filter (U ′)m = a(U ′)m−1 + b(u′)m witha = 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 byfbl = max {0.5 [1 − tanh(y − ybl − ywall)/b] , 0.02}, ybl = 0.2,b = 0.01.


NUMERICAL METHOD

Incompressible finite volume method

Pressure-velocity coupling treated with fractional step

95% 2nd order central and 5% 2nd order upwind differencingscheme for momentum eqns.

Hybrid 1st order upwind/2nd order central scheme k & ε eqns.

2nd -order Crank-Nicholson for time discretization


PRESSURE: AMPLITUDES OF INLET FLUCT

0.5 1 1.5 20

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

x/c

−C

p

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


SKIN FRICTION: AMPLITUDES OF INLET FLUCT

0 0.5 1 1.5−2

0

2

4

6

8

10x 10

−3

x/c

Cf

0.6 0.8 1 1.2 1.4 1.6

−2

−1

0

1

2

3x 10

−3

x/c

zoom



VELOCITIES: AMPLITUDES OF INLET FLUCT

0 0.2 0.4 0.6 0.8 1 1.20.1

0.15

0.2

0.25

x/c = 65y

0 0.5 10

0.05

0.1

0.15

0.2

0.25

x/c = 80

0 0.5 10

0.05

0.1

0.15

0.2

0.25

x/c = 100

U/Ub

y

0 0.5 10

0.05

0.1

0.15

0.2

0.25

x/c = 110

U/Ubbaseline 1.5× (baseline) 0.5× (baseline)


VELOCITIES: AMPLITUDES OF INLET FLUCT

0 0.2 0.4 0.6 0.8 1 1.20

0.05

0.1

0.15

0.2

0.25

x/c = 120

U/Ub

y

0 0.2 0.4 0.6 0.8 1 1.20

0.05

0.1

0.15

0.2

0.25

x/c = 130

U/Ub

baseline 1.5× (baseline) 0.5× (baseline)


RESOLVED AND MODELLED (< 0) SHEAR STRESSES

−0.005 0.005 0.010.1

0.15

0.2

0.25

x/c = 0.65y

0 0.01 0.02 0.03 0.040

0.05

0.1

0.15

0.2

0.25

x/c = 0.80

0 0.01 0.02 0.03 0.040

0.05

0.1

0.15

0.2x/c = 1.00

〈τ12,u〉, 〈−u′v ′〉/U2b

y

0 0.01 0.02 0.030

0.05

0.1

0.15

0.2x/c = 1.10

〈τ12,u〉, 〈−u′v ′〉/U2b



SHEAR STRESSES: AMPLITUDES OF INLET FLUCT

Resolved and Modelled (< 0) Shear stresses

0 0.01 0.02 0.030

0.05

0.1

0.15

0.2x/c = 1.20

〈τ12,u〉, 〈−u′v ′〉/U2b

y

0 0.01 0.02 0.030

0.05

0.1

0.15

0.2x/c = 1.30

〈τ12,u〉, 〈−u′v ′〉/U2b



TURB VISCOSITY: AMPLITUDES OF INLET FLUCT

0 2 4 6 8 100.1

0.12

0.14

0.16

0.18

0.2x/c = 0.65

y

0 10 20 30 40 50 60 700

0.05

0.1

0.15

0.2x/c = 0.80

0 10 20 30 40 50 60 700

0.05

0.1

0.15

0.2x/c = 1.00

〈νt〉/ν

y

0 10 20 30 40 50 60 700

0.05

0.1

0.15

0.2x/c = 1.10

〈νt〉/νbaseline 1.5× (baseline) 0.5× (baseline)


TURB VISCOSITY: AMPLITUDES OF INLET FLUCT

0 10 20 30 40 50 60 700

0.05

0.1

0.15

0.2x/c = 1.20

〈νt〉/ν

y

0 10 20 30 40 50 60 700

0.05

0.1

0.15

0.2x/c = 1.30

〈νt〉/ν



PRESSURE: fk = 0.3; NO INLET FLUCT; fk = 0.5

0.5 1 1.5 20

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

x/c

−C

p

fk = 0.3 no inlet fluct fk = 0.5


SKIN FRICTION: fk = 0.3; NO INLET FLUCT; fk = 0.5

0 0.5 1 1.5−2

0

2

4

6

8

10x 10

−3

x/c

Cf

0.6 0.8 1 1.2 1.4 1.6

−2

−1

0

1

2

3x 10

−3

x/c

zoom



VELOCITIES: fk = 0.3; NO INLET FLUCT; fk = 0.5

0 0.2 0.4 0.6 0.8 1 1.20.1

0.15

0.2

0.25

x/c = 65y

0 0.5 10

0.05

0.1

0.15

0.2

0.25

x/c = 80

0 0.5 10

0.05

0.1

0.15

0.2

0.25

x/c = 100

U/Ub

y

0 0.5 10

0.05

0.1

0.15

0.2

0.25

x/c = 110

U/Ubfk = 0.3 no inlet fluct fk = 0.5


VELOCITIES: fk = 0.3; NO INLET FLUCT; fk = 0.5

0 0.2 0.4 0.6 0.8 1 1.20

0.05

0.1

0.15

0.2

0.25

x/c = 120

U/Ub

y

0 0.2 0.4 0.6 0.8 1 1.20

0.05

0.1

0.15

0.2

0.25

x/c = 130

U/Ub




−0.0025 0.0025 0.0050.1

0.12

0.14

0.16

0.18

0.2x/c = 0.65

y

0 0.01 0.02 0.03 0.040

0.05

0.1

0.15

0.2

0.25

x/c = 0.80

0 0.01 0.02 0.03 0.040

0.05

0.1

0.15

0.2x/c = 1.00

〈τ12,u〉, 〈−u′v ′〉/U2b

y

0 0.01 0.02 0.030

0.05

0.1

0.15

0.2x/c = 1.10

〈τ12,u〉, 〈−u′v ′〉/U2b

fk = 0.3 no inlet fluct fk = 0.5Davidson ( ) www.tfd.chalmers.se/˜lada 28 / 40

SHEAR STRESSES: fk = 0.3; NO INLET FLUCT; fk = 0.5


0 0.01 0.02 0.030

0.05

0.1

0.15

0.2x/c = 1.20

〈τ12,u〉, 〈−u′v ′〉/U2b

y

0 0.01 0.02 0.030

0.05

0.1

0.15

0.2x/c = 1.30

〈τ12,u〉, 〈−u′v ′〉/U2b



TURB VISCOSITY: fk = 0.3; NO INLET FLUCT; fk = 0.5

0 2 4 6 8 100.1

0.12

0.14

0.16

0.18

0.2x/c = 0.65

y

0 50 100 1500

0.05

0.1

0.15

0.2x/c = 0.80

0 50 100 1500

0.05

0.1

0.15

0.2x/c = 1.00

〈νt〉/ν

y

0 50 100 1500

0.05

0.1

0.15

0.2x/c = 1.10

〈νt〉/νfk = 0.3 no inlet fluct fk = 0.5


TURB VISCOSITY: fk = 0.3; NO INLET FLUCT; fk = 0.5

0 50 100 1500

0.05

0.1

0.15

0.2x/c = 1.20

〈νt〉/ν

y

0 50 100 1500

0.05

0.1

0.15

0.2x/c = 1.30

〈νt〉/ν



PRESSURE: WALE, Nk = 128, Zmax = 0.4, CDS

0.5 1 1.50

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

x/c

−C

p



SKIN FRICTION: WALE, Nk = 128, Zmax = 0.4, CDS

0 0.5 1 1.5−2

0

2

4

6

8

10x 10

−3

x/c

Cf

0.6 0.8 1 1.2 1.4 1.6

−2

−1

0

1

2

3x 10

−3

x/c

zoom



VELOCITIES: WALE, Nk = 128, Zmax = 0.4, CDS

0 0.2 0.4 0.6 0.8 1 1.20.1

0.15

0.2

0.25

x/c = 65y

0 0.5 10

0.05

0.1

0.15

0.2

0.25

x/c = 80

0 0.5 10

0.05

0.1

0.15

0.2

0.25

x/c = 100

U/Ub

y

0 0.5 10

0.05

0.1

0.15

0.2

0.25

x/c = 110

U/UbWALE Nk = 128, Zmax = 0.4 CDS


VELOCITIES: WALE, Nk = 128, Zmax = 0.4, CDS

0 0.2 0.4 0.6 0.8 1 1.20

0.05

0.1

0.15

0.2

0.25

x/c = 120

U/Ub

y

0 0.2 0.4 0.6 0.8 1 1.20

0.05

0.1

0.15

0.2

0.25

x/c = 130

U/Ub

WALE Nk = 128, Zmax = 0.4 CDS



−0.02 −0.01 0 0.01 0.020.1

0.15

0.2

0.25

x/c = 0.65y

0 0.01 0.02 0.03 0.040

0.05

0.1

0.15

0.2

0.25

x/c = 0.80

0 0.01 0.02 0.03 0.040

0.05

0.1

0.15

0.2x/c = 1.00

〈τ12,u〉, 〈−u′v ′〉/U2b

y

0 0.01 0.02 0.030

0.05

0.1

0.15

0.2x/c = 1.10

〈τ12,u〉, 〈−u′v ′〉/U2b



SHEAR STRESSES: WALE, Nk = 128, Zmax = 0.4, CDS


0 0.01 0.02 0.030

0.05

0.1

0.15

0.2x/c = 1.20

〈τ12,u〉, 〈−u′v ′〉/U2b

y

0 0.01 0.02 0.030

0.05

0.1

0.15

0.2x/c = 1.30

〈τ12,u〉, 〈−u′v ′〉/U2b



TURB VISCOSITY: WALE, Nk = 128, Zmax = 0.4, CDS

0 2 4 6 8 100.1

0.12

0.14

0.16

0.18

0.2x/c = 0.65

y

0 10 20 30 40 50 60 700

0.05

0.1

0.15

0.2x/c = 0.80

0 20 40 60 800

0.05

0.1

0.15

0.2x/c = 1.00

〈νt〉/ν

y

0 10 20 30 40 50 60 700

0.05

0.1

0.15

0.2x/c = 1.10

〈νt〉/νWALE Nk = 128, Zmax = 0.4 CDS


TURB VISCOSITY: WALE, Nk = 128, Zmax = 0.4, CDS

0 10 20 30 40 50 60 700

0.05

0.1

0.15

0.2x/c = 1.20

〈νt〉/ν

y

0 10 20 30 40 50 60 700

0.05

0.1

0.15

0.2x/c = 1.30

〈νt〉/ν



CONCLUDING REMARKS

LRN PANS has been shown to work well as an embedded LESmethod

Channel flow: At two δ downstream the interface, the resolvedturbulence in good agreement with DNS data and the wall frictionvelocity has reached 99% of its fully developed value.

Channel flow: The treatment of the modelled ku and εu across theinterface is important.

LRN PANS predicts the hump flow well. LRN PANS better thanWALE.

Hump flow: CDS gives unphysical fluctuations near the inlet. Thebecause larger the smaller the inlet fluctuations


[1] ABE, K., KONDOH, T., AND NAGANO, Y.A new turbulence model for predicting fluid flow and heat transfer inseparating 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 paper2011-3108 (27-30 June, Honolulu, Hawaii, 2011).

[3] DAVIDSON, L., AND PENG, S.-H.Embedded LES using PANS applied to flow in a channel and overa hump (submitted).AIAA Journal (2012).

[4] MA, J., PENG, S.-H., DAVIDSON, L., AND WANG, F.A low Reynolds number variant of Partially-AveragedNavier-Stokes model for turbulence.International Journal of Heat and Fluid Flow 32 (2011), 652–669.10.1016/j.ijheatfluidflow.2011.02.001.


E LES USING PANS [2, 3] L D - Chalmerslada/slides/slides_embedded_LES.pdfCHANNEL FLOW: DOMAIN d x y...

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