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

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

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L ARGE E DDY S IMULATION (LES) Lars Davidson, www.tfd.chalmers.se/lada Chalmers University of Technology Gothenburg, Sweden T HREE -D AY CFD C OURSE AT C HALMERS This lecture is a condensed version of the course Unsteady Simulations for

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SLIDE 1

LARGE EDDY SIMULATION (LES)

Lars Davidson, www.tfd.chalmers.se/˜lada Chalmers University of Technology Gothenburg, Sweden

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SLIDE 2

THREE-DAY CFD COURSE AT CHALMERS

◮This lecture is a condensed version of the course Unsteady Simulations for Industrial Flows: LES, DES, hybrid LES-RANS and URANS 5-7 November 2012 at Chalmers, Gothenburg, Sweden Max 16 participants 50% lectures and 50% workshops in front of a PC For info, see http://www.tfd.chalmers.se/˜lada/cfdkurs/cfdkurs.html

www.tfd.chalmers.se/˜lada Helsinki 4 October 2012 2 / 1

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SLIDE 3

LECTURE NOTES

The slides are partly based on the course material at (click here) http://www.tfd.chalmers.se/˜lada/ comp turb model/lecture notes.html This course is part of the MSc programme Applied Mechanics at

  • Chalmers. For Fluid courses, click here

http://www.tfd.chalmers.se/˜lada/ msc/msc-programme.html The MSc programme is presented here http://www.chalmers.se/en/education/programmes/mast

www.tfd.chalmers.se/˜lada Helsinki 4 October 2012 3 / 1

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SLIDE 4

LARGE EDDY SIMULATIONS

GS SGS SGS In LES, large (Grid) Scales (GS) are resolved and the small (Sub-Grid) Scales (SGS) are modelled. LES is suitable for bluff body flows where the flow is governed by large turbulent scales

www.tfd.chalmers.se/˜lada Helsinki 4 October 2012 4 / 1

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SLIDE 5

BLUFF-BODY FLOW: SURFACE-MOUNTED CUBE[14]

Krajnovi´ c & Davidson (AIAA J., 2002)

Snapshots of large turbulent scales illustrated by Q = −∂¯ ui ∂xj ∂¯ uj ∂xi

www.tfd.chalmers.se/˜lada Helsinki 4 October 2012 5 / 1

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SLIDE 6

BLUFF-BODY FLOW: FLOW AROUND A BUS[15]

Krajnovi´ c & Davidson (JFE, 2003)

www.tfd.chalmers.se/˜lada Helsinki 4 October 2012 6 / 1

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SLIDE 7

BLUFF-BODY FLOW: FLOW AROUND A CAR[16]

K r a j n

  • v

i ´ c & D a v i d s

  • n

( J F E , 2 5 )

www.tfd.chalmers.se/˜lada Helsinki 4 October 2012 7 / 1

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SLIDE 8

BLUFF-BODY FLOW: FLOW AROUND A TRAIN[12]

Hemida & Krajnovi´ c, 2006

www.tfd.chalmers.se/˜lada Helsinki 4 October 2012 8 / 1

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SLIDE 9

SEPARATING FLOWS

Wall

TIME-AVERAGED flow and INSTANTANEOUS flow

www.tfd.chalmers.se/˜lada Helsinki 4 October 2012 9 / 1

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SLIDE 10

SEPARATING FLOWS

Wall

TIME-AVERAGED flow and INSTANTANEOUS flow In average there is backflow (negative velocities). Instantaneous, the negative velocities are often positive.

www.tfd.chalmers.se/˜lada Helsinki 4 October 2012 9 / 1

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SLIDE 11

SEPARATING FLOWS

Wall

TIME-AVERAGED flow and INSTANTANEOUS flow In average there is backflow (negative velocities). Instantaneous, the negative velocities are often positive. How easy is it to model fluctuations that are as large as the mean flow?

www.tfd.chalmers.se/˜lada Helsinki 4 October 2012 9 / 1

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SLIDE 12

SEPARATING FLOWS

Wall

TIME-AVERAGED flow and INSTANTANEOUS flow In average there is backflow (negative velocities). Instantaneous, the negative velocities are often positive. How easy is it to model fluctuations that are as large as the mean flow? Is it reasonable to require a turbulence model to fix this?

www.tfd.chalmers.se/˜lada Helsinki 4 October 2012 9 / 1

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SLIDE 13

SEPARATING FLOWS

Wall

TIME-AVERAGED flow and INSTANTANEOUS flow In average there is backflow (negative velocities). Instantaneous, the negative velocities are often positive. How easy is it to model fluctuations that are as large as the mean flow? Is it reasonable to require a turbulence model to fix this? Isn’t it better to RESOLVE the large fluctuations?

www.tfd.chalmers.se/˜lada Helsinki 4 October 2012 9 / 1

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SLIDE 14

TIME AVERAGING AND FILTERING

RANS: time average. This is called Reynolds time averaging: Φ = 1 2T T

−T

Φ(t)dt, Φ = Φ + Φ′ In LES we filter (volume average) the equations. In 1D we get: ¯ Φ(x, t) = 1 ∆x x+0.5∆x

x−0.5∆x

Φ(ξ, t)dξ Φ = ¯ Φ + Φ′′ (1)

no filter

  • ne filter

2 2.5 3 3.5 4 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.5

u , ¯ u x

www.tfd.chalmers.se/˜lada Helsinki 4 October 2012 10 / 1

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SLIDE 15

EQUATIONS

The filtering is defined by the discretization (nothing is done) The filtered Navier-Stokes (N-S) eqns, i.e. the LES eqns, read ∂¯ ui ∂t + ∂ ∂xj ¯ ui ¯ uj

  • = −1

ρ ∂¯ p ∂xi + ν ∂2¯ ui ∂xj∂xj − ∂τij ∂xj , ∂ ¯ ui ∂xi = 0 (2) where the subgrid stresses are given by τij = uiuj − ¯ ui¯ uj Contrary to Reynolds time averaging where u′

i = 0, we have here

u′′

i = 0

¯ ui = ¯ ui

www.tfd.chalmers.se/˜lada Helsinki 4 October 2012 11 / 1

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SLIDE 16

FILTERING: HOW IS EQ. 2 OBTAINED?

The N-S eqns are filtered (=discretized) using Eq. 1 The pressure gradient term, for example, reads ∂p ∂xi = 1 V

  • V

∂p ∂xi dV Now we want to move the derivative out of the integral. It is allowed if V is constant. The filtering volume, V=grid cell which is not constant Fortunately, the error is proportional to V 2, i.e. it is 2nd-order error ∂p ∂xi = ∂ ∂xi 1 V

  • V

pdV

  • + O
  • V 2

= ∂ ∂xi (¯ p) + O

  • V 2

All linear terms are treated in the same way.

www.tfd.chalmers.se/˜lada Helsinki 4 October 2012 12 / 1

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SLIDE 17

NON-LINEAR TERM

First we filter the term and move the derivative out of the integral ∂uiuj ∂xj = ∂ ∂xj 1 V

  • V

uiujdV

  • + O
  • V 2

= ∂ ∂xj (uiuj) + O

  • V 2

We have ∂ ∂xj uiuj; we want ∂ ∂xj ¯ ui ¯ uj Let’s add want we want (on both LHS ans RHS) and subtract want we don’t want This is how we end up with the convective term and the SGS term in Eq. 2, i.e. −∂τij ∂xj = − ∂ ∂xj (uiuj − ¯ ui¯ uj)

www.tfd.chalmers.se/˜lada Helsinki 4 October 2012 13 / 1

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SLIDE 18

LARGE EDDY SIMULATIONS

GS SGS SGS Large scales (GS) are resolved; small scales (SGS) are modelled.

www.tfd.chalmers.se/˜lada Helsinki 4 October 2012 14 / 1

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SLIDE 19

ENERGY SPECTRUM

The limit (cut-off) between GS and SGS is supposed to take place in the inertial subrange (II) I II III κ E(κ) cut-off I: large scales II: inertial subrange, −5/3-range III: dissipation subrange GS SGS

www.tfd.chalmers.se/˜lada Helsinki 4 October 2012 15 / 1

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SLIDE 20

SUBGRID MODEL

We need a subgrid model for the SGS turbulent scales The simplest model is the Smagorinsky model [23]: τij − 1 3δijτkk = −2νsgs¯ sij νsgs = (CS∆)2 2¯ sij¯ sij ≡ (CS∆)2 |¯ s| ¯ sij = 1 2 ∂¯ ui ∂xj + ∂¯ uj ∂xi

  • ,

∆ = (∆VIJK )1/3 (3) A damping function fµ is added to ensure that νsgs ⇒ 0 as y ⇒ 0 fµ = 1 − exp(−y+/26) A more convenient way to dampen the SGS viscosity near the wall is ∆ = min

  • (∆VIJK )1/3 , κy
  • where y is the distance to the nearest wall.

www.tfd.chalmers.se/˜lada Helsinki 4 October 2012 16 / 1

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SLIDE 21

SMAGORINSKY MODEL VS. MIXING-LENGTH MODEL

  • The eddy viscosity in the mixing length model reads in

boundary-layer flow [13, 22] νt = ℓ2

  • ∂U

∂y

  • .
  • Generalized to three dimensions, we have

νt = ℓ2 ∂Ui ∂xj + ∂Uj ∂xi ∂Ui ∂xj 1/2 = ℓ2 2SijSij 1/2 ≡ ℓ2|S|.

  • In the Smagorinsky model the SGS length scale ℓ = CS∆ i.e.

νsgs = (CS∆)2|¯ s| which is the same as Eq. 3

www.tfd.chalmers.se/˜lada Helsinki 4 October 2012 17 / 1

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SLIDE 22

ENERGY PATH

E(κ) E(κ) ∝ κ−5/3 κ inertial range εsgs ≃

  • νsgs

∂u′

i

∂xj ∂u′

i

∂xj

  • ε

dissipating range

www.tfd.chalmers.se/˜lada Helsinki 4 October 2012 18 / 1

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SLIDE 23

LES VS. RANS

LES can handle many flows which RANS (Reynolds Averaged Navier Stokes) cannot; the reason is that in LES large, turbulent scales are

  • resolved. Examples are:
  • Flows with large separation
  • Bluff-body flows (e.g. flow around a car); the wake often includes

large, unsteady, turbulent structures

  • Transition
  • In RANS all turbulent scales are modelled ⇒ inaccurate
  • In LES only small, isotropic turbulent scales are modelled ⇒ accurate

LES is very much more expensive than RANS.

www.tfd.chalmers.se/˜lada Helsinki 4 October 2012 19 / 1

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SLIDE 24

FINITE VOLUME RANS AND LES CODES.

RANS LES Domain 2D or 3D always 3D Time domain steady or unsteady always unsteady Space discretization 2nd order upwind central differencing Time discretization 1st order 2nd order (e.g. C-N) Turbulence model ≥ two-equations zero- or one-eq

www.tfd.chalmers.se/˜lada Helsinki 4 October 2012 20 / 1

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SLIDE 25

TIME AVERAGING IN LES

t1: Start time averaging t2: Stop time averaging t t1: start t2: end ¯ v1

www.tfd.chalmers.se/˜lada Helsinki 4 October 2012 21 / 1

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SLIDE 26

NEAR-WALL RESOLUTION

Biggest problem with LES: near walls, it requires very fine mesh in all directions, not only in the near-wall direction.

www.tfd.chalmers.se/˜lada Helsinki 4 October 2012 22 / 1

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SLIDE 27

NEAR-WALL RESOLUTION

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).

www.tfd.chalmers.se/˜lada Helsinki 4 October 2012 22 / 1

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SLIDE 28

NEAR-WALL RESOLUTION

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

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SLIDE 29

NEAR-WALL RESOLUTION

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

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SLIDE 30

NEAR-WALL RESOLUTION

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

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SLIDE 31

NEAR-WALL RESOLUTION CONT’D

In RANS when using wall-functions, 30 < y+ < 100 for the wall-adjacent cells x y wall y+

www.tfd.chalmers.se/˜lada Helsinki 4 October 2012 23 / 1

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SLIDE 32

NEAR-WALL RESOLUTION CONT’D

In RANS when using wall-functions, 30 < y+ < 100 for the wall-adjacent cells In LES, ∆z+ ≃ 30 x y wall y+ x z ∆z+

www.tfd.chalmers.se/˜lada Helsinki 4 October 2012 23 / 1

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SLIDE 33

NEAR-WALL RESOLUTION CONT’D

In RANS when using wall-functions, 30 < y+ < 100 for the wall-adjacent cells In LES, ∆z+ ≃ 30 EVERYWHERE x y wall y+ x z ∆z+

www.tfd.chalmers.se/˜lada Helsinki 4 October 2012 23 / 1

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SLIDE 34

NEAR-WALL RESOLUTION CONT’D

In RANS when using wall-functions, 30 < y+ < 100 for the wall-adjacent cells In LES, ∆z+ ≃ 30 EVERYWHERE AND ∆x+ ≃ 100, ∆y+

min ≃ 1

x y wall y+ x z ∆z+

www.tfd.chalmers.se/˜lada Helsinki 4 October 2012 23 / 1

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SLIDE 35

NEAR-WALL TREATMENT

from Hinze (1975)

www.tfd.chalmers.se/˜lada Helsinki 4 October 2012 24 / 1

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SLIDE 36

NEAR-WALL TREATMENT

1 2 3 4 5 6 0.5 1 1.5

x z 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

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SLIDE 37

ZONAL PANS MODEL

  • 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

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SLIDE 38

PANS LOW REYNOLDS NUMBER MODEL [17]

∂k ∂t + ∂(kUj) ∂xj = ∂ ∂xj

  • ν + νt

σku ∂k ∂xj

  • + (Pk − ε)

∂ε ∂t + ∂(εUj) ∂xj = ∂ ∂xj

  • ν + νt

σεu ∂ε ∂xj

  • + Cε1Pk

ε k − C∗

ε2

ε2 k νt = Cµfµ k2 ε , 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

www.tfd.chalmers.se/˜lada Helsinki 4 October 2012 27 / 1

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SLIDE 39

PANS LOW REYNOLDS NUMBER MODEL [17]

∂k ∂t + ∂(kUj) ∂xj = ∂ ∂xj

  • ν + νt

σku ∂k ∂xj

  • + (Pk − ε)

∂ε ∂t + ∂(εUj) ∂xj = ∂ ∂xj

  • ν + νt

σεu ∂ε ∂xj

  • + Cε1Pk

ε k − C∗

ε2

ε2 k νt = Cµfµ k2 ε , 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

www.tfd.chalmers.se/˜lada Helsinki 4 October 2012 27 / 1

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SLIDE 40

PANS LOW REYNOLDS NUMBER MODEL [17]

∂k ∂t + ∂(kUj) ∂xj = ∂ ∂xj

  • ν + νt

σku ∂k ∂xj

  • + (Pk − ε)

∂ε ∂t + ∂(εUj) ∂xj = ∂ ∂xj

  • ν + νt

σεu ∂ε ∂xj

  • + Cε1Pk

ε k − C∗

ε2

ε2 k νt = Cµfµ k2 ε , C∗

ε2 = 1.5 + fk

fε (1.9 − 1.5), 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

www.tfd.chalmers.se/˜lada Helsinki 4 October 2012 27 / 1

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SLIDE 41

CHANNEL FLOW: ZONAL RANS-LES

x y ku,int, εu,int wall yint LES, fk < 1 RANS, fk = 1.0 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 ku,int = fkkRANS Nothing is done for ε xmax = 3.2 (64 cells), zmax = 1.6 (64 cells), y dir: 80 − 128 cells CDS in entire region

www.tfd.chalmers.se/˜lada Helsinki 4 October 2012 28 / 1

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SLIDE 42

(Nx × Nz) = (64 × 64). y+

int = 500

1 100 1000 30000 5 10 15 20 25 30

y+ U+

0.05 0.1 0.15 0.2

  • 1
  • 0.8
  • 0.6
  • 0.4
  • 0.2

y u′v′+ Reτ = 4 000 Reτ = 8 000 Reτ = 16 000; Reτ = 32 000.

www.tfd.chalmers.se/˜lada Helsinki 4 October 2012 29 / 1

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SLIDE 43

INTERFACE LOCATION. Reτ = 8 000.

1 100 1000 8000 5 10 15 20 25 30

y+ U+

500 1000 1500 2000

  • 1
  • 0.8
  • 0.6
  • 0.4
  • 0.2

y+ u′v′+ y+ = 130 y+ = 500 y+ = 980

www.tfd.chalmers.se/˜lada Helsinki 4 October 2012 30 / 1

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SLIDE 44

EFFECT OF fk. Reτ = 16 000. y+

int = 500

1 100 1000 10000 5 10 15 20 25 30

y+ U+

0.05 0.1 0.15 0.2

  • 1
  • 0.8
  • 0.6
  • 0.4
  • 0.2

y u′v′+ fk = 0.2 fk = 0.3 fk = 0.5 fk = 0.6

www.tfd.chalmers.se/˜lada Helsinki 4 October 2012 31 / 1

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SLIDE 45

EFFECT OF RESOLUTION: VELOCITY

1 100 1000 30000 5 10 15 20 25 30

(Nx × Nz) = (32 × 32) y+ U+

1 100 1000 30000 5 10 15 20 25 30

(Nx × Nz) = (128 × 128) y+ U+ Reτ = 4 000 Reτ = 8 000 Reτ = 16 000; Reτ = 32 000.

www.tfd.chalmers.se/˜lada Helsinki 4 October 2012 32 / 1

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SLIDE 46

EFFECT OF RESOLUTION: RESOLVED SHEAR STRESS

0.05 0.1 0.15 0.2

  • 1
  • 0.8
  • 0.6
  • 0.4
  • 0.2

(Nx × Nz) = (32 × 32) y u′v′+

0.05 0.1 0.15 0.2

  • 1
  • 0.8
  • 0.6
  • 0.4
  • 0.2

(Nx × Nz) = (128 × 128) y u′v′+ Reτ = 4 000 Reτ = 8 000 Reτ = 16 000; Reτ = 32 000.

www.tfd.chalmers.se/˜lada Helsinki 4 October 2012 33 / 1

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SLIDE 47

EFFECT OF RESOLUTION: TURBULENT VISCOSITY

0.05 0.1 0.15 0.2 0.25 0.3 0.005 0.01 0.015

Reτ = 4000 νt/(uτδ)

0.05 0.1 0.15 0.2 0.25 0.3 0.005 0.01 0.015

Reτ = 8000

0.05 0.1 0.15 0.2 0.25 0.3 0.005 0.01 0.015

Reτ = 16 000 y νt/(uτδ)

0.05 0.1 0.15 0.2 0.25 0.3 0.005 0.01 0.015

Reτ = 32 000 y (Nx × Nz) = 64 × 64 32 × 32 128 × 128

www.tfd.chalmers.se/˜lada Helsinki 4 October 2012 34 / 1

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SLIDE 48

EFFECT OF RESOLUTION: TURBULENT VISCOSITY

0.05 0.1 0.15 0.2 0.25 0.3 0.005 0.01 0.015

Reτ = 4000 νt/(uτδ)

0.05 0.1 0.15 0.2 0.25 0.3 0.005 0.01 0.015

Reτ = 8000

0.05 0.1 0.15 0.2 0.25 0.3 0.005 0.01 0.015

Reτ = 16 000 y νt/(uτδ)

0.05 0.1 0.15 0.2 0.25 0.3 0.005 0.01 0.015

Reτ = 32 000 y (Nx × Nz) = 64 × 64 32 × 32 128 × 128

G r i d i n d e p e n d e n t t u r b u l e n t v i s c

  • s

i t y !

www.tfd.chalmers.se/˜lada Helsinki 4 October 2012 34 / 1

slide-49
SLIDE 49

SGS MODELS BASED ON GRID SIZE

When the grid is refined, νt gets smaller

www.tfd.chalmers.se/˜lada Helsinki 4 October 2012 35 / 1

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SLIDE 50

SGS MODELS BASED ON GRID SIZE

When the grid is refined, νt gets smaller E εsgs κ Pk,res κc

www.tfd.chalmers.se/˜lada Helsinki 4 October 2012 35 / 1

slide-51
SLIDE 51

SGS MODELS BASED ON GRID SIZE

When the grid is refined, νt gets smaller E εsgs,∆ εsgs,0.5∆ κ Pk,res κc 2κc

www.tfd.chalmers.se/˜lada Helsinki 4 October 2012 35 / 1

slide-52
SLIDE 52

SGS MODELS BASED ON GRID SIZE

When the grid is refined, νt gets smaller E εsgs,∆ εsgs,0.5∆ κ Pk,res κc 2κc εsgs,∆ = εsgs,0.5∆ εsgs = 2νt¯ sij¯ sij − τ12,t∂¯ u ∂y Grid refinement ⇒ must be accompanied with larger ¯ sij¯ sij ⇒ ¯ sij¯ sij must take place at higher wavenumbers if not ⇒ grid dependent

www.tfd.chalmers.se/˜lada Helsinki 4 October 2012 35 / 1

slide-53
SLIDE 53

POWER DENSITY SPECTRA OF ν0.5

t

∂ ¯ w′ ∂z

20 40 60 80 100 0.005 0.01 0.015 0.02

E

  • νt (∂ ¯

w′/∂z)2 κz One-eq ksgs model

20 40 60 80 100 0.005 0.01 0.015 0.02 0.025 0.03

κz Zonal PANS (Nx × Nz) = 64 × 64 32 × 32 128 × 128

www.tfd.chalmers.se/˜lada Helsinki 4 October 2012 36 / 1

slide-54
SLIDE 54

SGS DISSIPATION VS. WAVENUMBER

Energy spectra of the SGS dissipation show that the peak takes place at surprisingly low wavenumber (length scale corresponding to 10 cells or more). κc κ E(κ) εsgs

www.tfd.chalmers.se/˜lada Helsinki 4 October 2012 37 / 1

slide-55
SLIDE 55

SGS DISSIPATION VS. WAVENUMBER

Energy spectra of the SGS dissipation show that the peak takes place at surprisingly low wavenumber (length scale corresponding to 10 cells or more). κc κ E(κ) εsgs εsgs,κ

www.tfd.chalmers.se/˜lada Helsinki 4 October 2012 37 / 1

slide-56
SLIDE 56

SGS DISSIPATION, Reτ = 8000

SGS dissipation in the ¯ u′

i ¯

u′

i/2 eq, εsgs = 2νt¯

sij¯ sij − τ12,t∂¯ u ∂y

0.1 0.15 0.2 0.25 0.3 5 10 15

y εsgs One-eq ksgs model

0.1 0.15 0.2 0.25 0.3 5 10 15

y εsgs Zonal PANS (Nx × Nz) = 64 × 64 32 × 32 128 × 128

www.tfd.chalmers.se/˜lada Helsinki 4 October 2012 38 / 1

slide-57
SLIDE 57

LOCAL EQUILIBRIUM. Reτ = 4000, Nx × Nz = 64 × 64.

0.1 0.2 0.3 0.4 0.5 0.6 0.1 0.2 0.3 0.4 0.1 0.2 0.3 0.4 0.5 0.6 2E-3 4E-3 0.1 0.2 0.3 0.4 0.5 0.6 2E-3 4E-3

y k equation

0.1 0.2 0.3 0.4 0.5 0.6 0.01 0.02 0.03 0.1 0.2 0.3 0.4 0.5 0.6 2E-4 4E-4 0.1 0.2 0.3 0.4 0.5 0.6 2E-4 4E-4

ε equation y Pk+ ε+ Cε1Pk/ε+ Cε2ε2/k+ Left vertical axes: URANS region; right vertical axes: LES region.

www.tfd.chalmers.se/˜lada Helsinki 4 October 2012 39 / 1

slide-58
SLIDE 58

LOCAL EQUILIBRIUM IN ε EQUATION.

How can both the k eq. and ε be in local equilibrium??

www.tfd.chalmers.se/˜lada Helsinki 4 October 2012 40 / 1

slide-59
SLIDE 59

LOCAL EQUILIBRIUM IN ε EQUATION.

How can both the k eq. and ε be in local equilibrium?? If Pk = ε

www.tfd.chalmers.se/˜lada Helsinki 4 October 2012 40 / 1

slide-60
SLIDE 60

LOCAL EQUILIBRIUM IN ε EQUATION.

How can both the k eq. and ε be in local equilibrium?? If Pk = ε then C1 ε kPk=C∗

2

ε2 k , because C1 = C∗

2

www.tfd.chalmers.se/˜lada Helsinki 4 October 2012 40 / 1

slide-61
SLIDE 61

LOCAL EQUILIBRIUM IN ε EQUATION.

How can both the k eq. and ε be in local equilibrium?? If Pk = ε then C1 ε kPk=C∗

2

ε2 k , because C1 = C∗

2

However, the previous slide shows C1 ε k Pk

  • = C∗

2

ε2 k

  • www.tfd.chalmers.se/˜lada

Helsinki 4 October 2012 40 / 1

slide-62
SLIDE 62

LOCAL EQUILIBRIUM IN ε EQUATION.

How can both the k eq. and ε be in local equilibrium?? If Pk = ε then C1 ε kPk=C∗

2

ε2 k , because C1 = C∗

2

However, the previous slide shows C1 ε k Pk

  • = C∗

2

ε2 k

  • Answer: when time-averaging ab = ab

www.tfd.chalmers.se/˜lada Helsinki 4 October 2012 40 / 1

slide-63
SLIDE 63

LOCAL EQUILIBRIUM IN ε EQUATION.

The answer is because of time averaging (ab < ab, (see below)

0.1 0.2 0.3 0.4 0.5 0.6 1 1.05 1.1 1.15 1.2

y εPk/k εPk/k ε2/k ε2/k

www.tfd.chalmers.se/˜lada Helsinki 4 October 2012 41 / 1

slide-64
SLIDE 64

RESOLVED AND MODELLED TURBULENT KINETIC ENERGY.

0.1 0.2 0.3 0.4 1 2 3 4 5 6 7 8

Resolved y kres

0.1 0.2 0.3 0.4 1 2 3 4 5 6 7 8

Modelled: bottom; total: top y k, kres + k Reτ = 4 000 Reτ = 8 000 Reτ = 16 000; Reτ = 32 000.

www.tfd.chalmers.se/˜lada Helsinki 4 October 2012 42 / 1

slide-65
SLIDE 65

CONCLUDING REMARKS

LRN PANS works well as zonal LES-RANS model for very high Reτ (> 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 < fk < 0.5 have little impact on the results

www.tfd.chalmers.se/˜lada Helsinki 4 October 2012 43 / 1

slide-66
SLIDE 66

HYBRID LES-RANS

Near walls: a RANS one-eq. k or a k − ω model. In core region: a LES one-eq. kSGS model. y x Interface wall wall URANS URANS LES y+

ml

  • Location of interface either pre-defined or automatically computed

www.tfd.chalmers.se/˜lada Helsinki 4 October 2012 44 / 1

slide-67
SLIDE 67

MOMENTUM EQUATIONS

  • The Navier-Stokes, time-averaged in the near-wall regions and

filtered in the core region, reads ∂¯ ui ∂t + ∂ ∂xj ¯ ui ¯ uj

  • = −1

ρ ∂¯ p ∂xi + ∂ ∂xj

  • (ν + νT)∂¯

ui ∂xj

  • νT = νt, y ≤ yml

νT = νsgs, y ≥ yml

  • The equation above: URANS or LES? Same boundary conditions ⇒

same solution!

www.tfd.chalmers.se/˜lada Helsinki 4 October 2012 45 / 1

slide-68
SLIDE 68

TURBULENCE MODEL

  • Use one-equation model in both URANS region and LES region.

∂kT ∂t + ∂ ∂xj (¯ ujkT) = ∂ ∂xj

  • (ν + νT) ∂kT

∂xj

  • + PkT − Cε

k3/2

T

ℓ PkT = 2νT ¯ Sij ¯ Sij, νT = Ckℓk1/2

T

LES-region: kT = ksgs, νT = νsgs, ℓ = ∆ = (δV)1/3 URANS-region: kT = k, νT = νt, ℓ ≡ ℓRANS = 2.5n[1 − exp(−Ak1/2y/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

slide-69
SLIDE 69

DIFFUSER[9]

Instantaneous inlet data from channel DNS used. Domain: −8 ≤ x ≤ 48, 0 ≤ yinlet ≤ 1, 0 ≤ z ≤ 4. xmax = 40 gave return flow at the outlet Grid: 258 × 66 × 32. Re = UinH/ν = 18 000, angle 10o The grid is much too coarse for LES (in the inlet region ∆z+ ≃ 170) Matching plane fixed at yml at the inlet. In the diffuser it is located along the 2D instantaneous streamline corresponding to yml.

www.tfd.chalmers.se/˜lada Helsinki 4 October 2012 47 / 1

slide-70
SLIDE 70

DIFFUSER GEOMETRY. Re = 18 000, ANGLE 10o

H = 2δ 7.9H 21H 29H 4H 4.7H periodic b.c. convective outlet b.c. no-slip b.c.

www.tfd.chalmers.se/˜lada Helsinki 4 October 2012 48 / 1

slide-71
SLIDE 71

DIFFUSER: RESULTS WITH LES

  • Velocities. Markers: experiments by Buice & Eaton (1997)

x = 3H 6 14 17 20 24H x/H = 27 30 34 40 47H

www.tfd.chalmers.se/˜lada Helsinki 4 October 2012 49 / 1

slide-72
SLIDE 72

DIFFUSER: RESULTS WITH RANS-LES

x = 3H 6 14 17 20 24H x/H = 27 30 34 40 47H forcing; no forcing

www.tfd.chalmers.se/˜lada Helsinki 4 October 2012 50 / 1

slide-73
SLIDE 73

SHEAR STRESSES (×2 IN LOWER HALF)

x = 3H 6 13 19 23H x/H = 26 33 40 47H resolved; modelled

www.tfd.chalmers.se/˜lada Helsinki 4 October 2012 51 / 1

slide-74
SLIDE 74

RANS-LES: νt/ν

x = 3H 6 14 17 20 24H x/H = 27 30 34 40 47H forcing; no forcing At x = 24H, νT,max/ν ≃ 450 At x = −7H νT,max/ν ≃ 11

www.tfd.chalmers.se/˜lada Helsinki 4 October 2012 52 / 1

slide-75
SLIDE 75

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 ∂t + ∂ ∂xj (¯ ujk) = ∂ ∂xj

  • ν + νt

σk ∂k ∂xj

  • + Pk − β∗kω

∂ω ∂t + ∂ ∂xj (¯ ujω) = ∂ ∂xj

  • ν + νt

σω ∂ω ∂xj

  • + αPk

νt − βω2+ . . . The dissipation term in the k equation is modified as [19, 25] β∗kω → β∗kωFDES, FDES = max

  • Lt

CDES∆, 1

  • ∆ = max {∆x1, ∆x2, ∆x3} ,

Lt = k1/2 β∗ω ⇒ RANS near walls and LES away from walls

www.tfd.chalmers.se/˜lada Helsinki 4 October 2012 53 / 1

slide-76
SLIDE 76

RESOLUTION

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

slide-77
SLIDE 77

HOW TO ESTIMATE RESOLUTION IN GENERAL? [4, 5]

Energy spectra (both in spanwise direction and time) Two-point correlations Ratio of SGS turbulent kinetic energy ksgs to resolved 0.5u′u′ + v′v′ + w′w′ Ratio of SGS shear stress τsgs,12 to resolved u′v′ Ratio of SGS viscosity, νsgs to molecular, ν Energy spectra of SGS dissipation Comparison of SGS dissipation due to ∂u′

i/∂xj and ∂¯

ui/∂xj

www.tfd.chalmers.se/˜lada Helsinki 4 October 2012 55 / 1

slide-78
SLIDE 78

CHANNEL FLOW, Reτ = 4000, y+ = 440

10 10

1

10

2

10

  • 4

10

  • 3

10

  • 2

10

  • 1

Eww(kz) κz = 2π(kz − 1)/zmax Energy spectra

0.1 0.2 0.3 0.4

  • 0.2

0.2 0.4 0.6 0.8 1

Bww(ˆ z)/w2

rms

ˆ z = z − z0 Two-point correlations (∆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

slide-79
SLIDE 79

CHANNEL FLOW, Reτ = 4000, y+ = 440

10 10

1

10

2

10

  • 4

10

  • 3

10

  • 2

10

  • 1

Eww(kz) κz = 2π(kz − 1)/zmax Energy spectra

0.1 0.2 0.3 0.4

  • 0.2

0.2 0.4 0.6 0.8 1

Bww(ˆ z)/w2

rms

ˆ z = z − z0 Two-point correlations (∆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

slide-80
SLIDE 80

CHANNEL FLOW, Reτ = 4000, y+ = 440

1000 2000 3000 4000 1 2 3 4 5 6

kres y+ kres = (u′2 + v′2 + w′2)/2

1000 2000 3000 4000 0.4 0.5 0.6 0.7 0.8 0.9 1

γ y+ γ =

kres <kT >+kres

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

slide-81
SLIDE 81

CHANNEL FLOW, Reτ = 4000, y+ = 440

1000 2000 3000 4000 1 2 3 4 5 6

kres y+ kres = (u′2 + v′2 + w′2)/2

1000 2000 3000 4000 0.4 0.5 0.6 0.7 0.8 0.9 1

γ y+ γ =

kres <kT >+kres

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

slide-82
SLIDE 82

SGS VS. MOLECULAR VISCOSITY [5]

1 2 3 4 0.2 0.4 0.6 0.8 1

y/H νsgs/ν

2 4 6 8 10 12

  • 4
  • 3
  • 2
  • 1

1

y/H νsgs/ν Nz = 32; Nz = 64; Nz = 128.

x = −H x = 20H

www.tfd.chalmers.se/˜lada Helsinki 4 October 2012 58 / 1

slide-83
SLIDE 83

SGS VS. RESOLVED SHEAR STRESSES

0.05 0.1 0.15 0.2 0.25 0.3 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4

y/H τsgs,12/u′v′

0.01 0.02 0.03 0.04 0.05

  • 3.5
  • 3
  • 2.5
  • 2
  • 1.5
  • 1
  • 0.5

y/H τsgs,12/u′v′ Nz = 32; Nz = 64; Nz = 128.

x = −H x = 20H

www.tfd.chalmers.se/˜lada Helsinki 4 October 2012 59 / 1

slide-84
SLIDE 84

THE PANS MODEL

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

slide-85
SLIDE 85

Embedded LES Using PANS [10, 11] Lars Davidson1 and Shia-Hui Peng1,2 Davidson& Peng

1Department of Applied Mechanics

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

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

SWEDEN

slide-86
SLIDE 86

PANS LOW REYNOLDS NUMBER MODEL [17]

∂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

www.tfd.chalmers.se/˜lada Helsinki 4 October 2012 62 / 1

slide-87
SLIDE 87

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 [17]

www.tfd.chalmers.se/˜lada Helsinki 4 October 2012 63 / 1

slide-88
SLIDE 88

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)

www.tfd.chalmers.se/˜lada Helsinki 4 October 2012 64 / 1

slide-89
SLIDE 89

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

◮ www.tfd.chalmers.se/˜lada Helsinki 4 October 2012 65 / 1

slide-90
SLIDE 90

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 www.tfd.chalmers.se/˜lada Helsinki 4 October 2012 65 / 1

slide-91
SLIDE 91

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

www.tfd.chalmers.se/˜lada Helsinki 4 October 2012 66 / 1

slide-92
SLIDE 92

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′+

www.tfd.chalmers.se/˜lada Helsinki 4 October 2012 67 / 1

slide-93
SLIDE 93

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

www.tfd.chalmers.se/˜lada Helsinki 4 October 2012 68 / 1

slide-94
SLIDE 94

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

www.tfd.chalmers.se/˜lada Helsinki 4 October 2012 69 / 1

slide-95
SLIDE 95

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

www.tfd.chalmers.se/˜lada Helsinki 4 October 2012 70 / 1

slide-96
SLIDE 96

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

www.tfd.chalmers.se/˜lada Helsinki 4 October 2012 71 / 1

slide-97
SLIDE 97

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)

www.tfd.chalmers.se/˜lada Helsinki 4 October 2012 72 / 1

slide-98
SLIDE 98

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.

www.tfd.chalmers.se/˜lada Helsinki 4 October 2012 73 / 1

slide-99
SLIDE 99

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)

www.tfd.chalmers.se/˜lada Helsinki 4 October 2012 74 / 1

slide-100
SLIDE 100

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)

www.tfd.chalmers.se/˜lada Helsinki 4 October 2012 75 / 1

slide-101
SLIDE 101

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)

www.tfd.chalmers.se/˜lada Helsinki 4 October 2012 76 / 1

slide-102
SLIDE 102

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)

www.tfd.chalmers.se/˜lada Helsinki 4 October 2012 77 / 1

slide-103
SLIDE 103

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)

www.tfd.chalmers.se/˜lada Helsinki 4 October 2012 78 / 1

slide-104
SLIDE 104

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)

www.tfd.chalmers.se/˜lada Helsinki 4 October 2012 79 / 1

slide-105
SLIDE 105

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)

www.tfd.chalmers.se/˜lada Helsinki 4 October 2012 80 / 1

slide-106
SLIDE 106

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)

www.tfd.chalmers.se/˜lada Helsinki 4 October 2012 81 / 1

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SLIDE 107

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

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SLIDE 108

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

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SLIDE 109

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

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SLIDE 110

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

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SLIDE 111

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

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SLIDE 112

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

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SLIDE 113

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

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SLIDE 114

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

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SLIDE 115

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

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SLIDE 116

PANS: CONCLUDING REMARKS

Embedded LES with k − ε PANS and Synthetic b.c.

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SLIDE 117

PANS: CONCLUDING REMARKS

Embedded LES with k − ε PANS and Synthetic b.c. Channel flow

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SLIDE 118

PANS: CONCLUDING REMARKS

Embedded LES with k − ε PANS and Synthetic b.c. Channel flow

◮ Isotropic fluctuations work well for channel flow www.tfd.chalmers.se/˜lada Helsinki 4 October 2012 91 / 1

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SLIDE 119

PANS: CONCLUDING REMARKS

Embedded LES with k − ε PANS and Synthetic b.c. Channel flow

◮ Isotropic fluctuations work well for channel flow ◮ Strong dependence on interface ku & εu values www.tfd.chalmers.se/˜lada Helsinki 4 October 2012 91 / 1

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SLIDE 120

PANS: CONCLUDING REMARKS

Embedded LES with k − ε PANS and Synthetic b.c. Channel flow

◮ Isotropic fluctuations work well for channel flow ◮ Strong dependence on interface ku & εu values ◮ No strong dependence on amplitude, L or T of fluctuations www.tfd.chalmers.se/˜lada Helsinki 4 October 2012 91 / 1

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SLIDE 121

PANS: CONCLUDING REMARKS CONT’D

Hump flow

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SLIDE 122

PANS: CONCLUDING REMARKS CONT’D

Hump flow

◮ PANS & synthetic inlet b.c. with fk everywhere gives good results

except Cf (error > 50%)

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SLIDE 123

PANS: CONCLUDING REMARKS CONT’D

Hump flow

◮ PANS & synthetic inlet b.c. with fk everywhere gives good results

except Cf (error > 50%)

◮ With embedded isotropic fluctuations, interface must be located far

upstream

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SLIDE 124

PANS: CONCLUDING REMARKS CONT’D

Hump flow

◮ PANS & synthetic inlet b.c. with fk everywhere gives good results

except Cf (error > 50%)

◮ With embedded isotropic fluctuations, interface must be located far

upstream

◮ With embedded anisotropic fluctuations, good results are obtained,

still poor Cf

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SLIDE 125

PANS: CONCLUDING REMARKS CONT’D

Hump flow

◮ PANS & synthetic inlet b.c. with fk everywhere gives good results

except Cf (error > 50%)

◮ With embedded isotropic fluctuations, interface must be located far

upstream

◮ With embedded anisotropic fluctuations, good results are obtained,

still poor Cf

◮ On-going work . . . www.tfd.chalmers.se/˜lada Helsinki 4 October 2012 92 / 1

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SLIDE 126

Large Eddy Simulation of Heat Transfer in Boundary layer and Backstep Flow Using PANS [6] Lars Davidson THMT-12, Palermo, Sept 2012

slide-127
SLIDE 127

PANS LOW REYNOLDS NUMBER MODEL [17]

∂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.

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SLIDE 128

NUMERICAL METHOD

Incompressible finite volume method Pressure-velocity coupling treated with fractional step Differencing scheme for momentum eqns:

◮ 95% 2nd order central and 5% 2nd order upwind differencing

scheme (baseline) OR

◮ 100% 2nd order central differencing

Hybrid 1st order upwind/2nd order central scheme k & ε eqns. 2nd-order Crank-Nicholson for time discretization

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SLIDE 129

BOUNDARY LAYER FLOW: DOMAIN

x y L H δinlet Inlet: δinlet = 1 (covered by 45 cells), Reθ = 3 600, Uin = ρ = 1. Stretching 1.12 up to y/δ ≃ 1. Domain: L/δin = 3.2, H/δin = 15.6, Zmax = 1.5δin Resolution: ∆z+

in ≃ 27, ∆x+ in ≃ 54

Grid: 66 × 96 × 64 (x, y, z)

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SLIDE 130

ANISOTROPIC SYNTHETIC FLUCTUATIONS: I [3, 2, 8]

Prescribe an homogeneous Reynolds tensor, uiuj (here from DNS)

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SLIDE 131

ANISOTROPIC SYNTHETIC FLUCTUATIONS: I [3, 2, 8]

( u

′ 1

u

′ 1

)

λ

x1,λ ( u

′ 2

u

′ 2

)

λ

x2,λ u′

1,λu′ 2,λ = 0

Prescribe an homogeneous Reynolds tensor, uiuj (here from DNS) isotropic fluctuations in principal directions, (u′

1u′ 1)λ = (u′ 2u′ 2)λ,

u′

1,λu′ 2,λ = 0

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SLIDE 132

ANISOTROPIC SYNTHETIC FLUCTUATIONS: I [3, 2, 8]

( u

′ 1

u

′ 1

  • λ

x1,λ ( u

′ 2

u

′ 2

  • λ

x2,λ u′

1,λu′ 2,λ = 0

Prescribe an homogeneous Reynolds tensor, uiuj (here from DNS) isotropic fluctuations in principal directions, (u′

1u′ 1)λ = (u′ 2u′ 2)λ,

u′

1,λu′ 2,λ = 0

re-scale the normal components, (u′

1u′ 1)λ > (u′ 2u′ 2)λ,

u′

1,λu′ 2,λ = 0

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SLIDE 133

ANISOTROPIC SYNTHETIC FLUCTUATIONS: II

u′

1u′ 1 > u′ 2u′ 2

x1 u′

2u′ 2

x2 u′

1u′ 2 = 0

Transform from (x1,λ, x2,λ) to (x1, x2) u′2

1

u′2

2

= 23, u′2

1

u′2

3

= 5 from (u′

1u′ 1)peak in DNS channel flow, Reτ = 500

www.tfd.chalmers.se/˜lada Helsinki 4 October 2012 98 / 1

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SLIDE 134

INLET CONDITIONS FOR ku AND εu AS IN [10]

A pre-cursor RANS simulation using the AKN model (i.e. PANS with fk = 1) is carried out. At Reθ = 3 600, URANS, VRANS, kRANS are taken. ¯ uin = URANS + u′

synt, ¯

vin = VRANS + v′

synt, ¯

win = w′

synt

Anisotropic synthetic fluctuations are used. The fluctuations are scaled with ku/ku,max. ku,in = fkkRANS, εu,in = C3/4

µ

k3/2

u,in/ℓsgs, ℓsgs = Cs∆, ∆ = V 1/3,

Cs = 0.05

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SLIDE 135

INLET TURB. FLUCTUATION, TWO-POINT CORRELATIONS

  • 2

2 4 6 200 400 600 800 1000

stresses y/H

0.1 0.2 0.3 0.4 0.5 0.2 0.4 0.6 0.8 1

ˆ z/δ, ˆ z/H Bww(ˆ z) Two-point correlation u+2

rms,

v+2

rms,

w+2

rms

u′v′+

  • : inlet;

x = 3δin

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SLIDE 136

BOUNDARY LAYER: VELOCITY AND SKIN FRICTION

1 10 50 1000 5 10 15 20 25

y+ U+ 100%CDS

0.5 1 1.5 2 2.5 3 2.6 2.8 3 3.2 3.4 3.6 x 10

  • 3

x Cf x = δin; x = 2δin; x = 3δin; : DNS [21] 100% CDS; 100% CDS, Uin from AKN; 25% larger inlet fluct.; 25% larger inlet fluct., Cs = 0.07; markers: 0.37 (log10Rex)−2.584 (+: AKN; ◦: DNS)

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SLIDE 137

REYNOLDS STRESSES

  • 1

1 2 3 500 1000 1500

y+ u′v′ u′u′

  • 1

1 2 3 500 1000 1500

y+ u′v′ v′v′, w′w′, u′u′ x = δin; x = 2δin; x = 3δin. x = 3δin; Markers: DNS [21]

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SLIDE 138

BACKWARD FACING STEP: DOMAIN

4.05H 21H 4H H x y qw ReH = 28 000 Experiments by Vogel & Eaton [26] Mean inlet profiles from RANS (same as in boundary layer) Grid: 336 × 120 in x × y plane. Zmax = 1.6H, Nk = 64, ∆z+

in = 31.

Anisotropic synthetic fluctuations, u′, v′, w′ (same as for boundary layer flow); no fluctuations for t′ Constant heat flux, qw, on lower wall.

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SLIDE 139

BACKSTEP FLOW. SKIN FRICTION AND STANTON

NUMBER

  • 5

5 10 15 20

  • 2
  • 1

1 2 3 4 x 10

  • 3

x/H Cf

5 10 15 1 1.5 2 2.5 3 3.5 4x 10

  • 3

x/H St PANS; PANS, 50% smaller inlet fluctuations; WALE; •: PANS, no inlet fluctuations; : 2D RANS; ◦,•: experiments [26].

www.tfd.chalmers.se/˜lada Helsinki 4 October 2012 104 / 1

slide-140
SLIDE 140

BACKSTEP FLOW: VELOCITIES.

0.2 0.4 0.6 0.8 1 1 1.2 1.4 1.6 1.8 2 2.2 2.4 2.6

¯ u/Uin x = −1.13H

  • 0.2

0.2 0.4 0.6 0.8 1 0.5 1 1.5 2 2.5

¯ u/Uin x = 3.2H

0.2 0.4 0.6 0.8 1 0.5 1 1.5 2 2.5

¯ u/Uin x = 14.86H PANS; PANS, 50% smaller inlet fluctuations; WALE;

  • : PANS, no inlet fluctuations;

: 2D RANS; ◦: experiments [26].

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SLIDE 141

BACKSTEP FLOW: RESOLVED STREAMWISE STRESS.

0.05 0.1 0.15 1 1.2 1.4 1.6 1.8 2 2.2 2.4 2.6

urms/Uin y/H x = −1.13H

0.05 0.1 0.15 0.2 0.5 1 1.5 2 2.5

urms/Uin x = 3.2H

0.05 0.1 0.15 0.5 1 1.5 2 2.5

urms/Uin x = 14.86H PANS; PANS, 50% smaller inlet fluctuations; WALE;

  • : PANS, no inlet fluctuations;

: 2D RANS; ◦: experiments [26].

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SLIDE 142

BACKSTEP FLOW: TURBULENT VISCOSITIES.

2 4 6 8 1 1.2 1.4 1.6 1.8 2 2.2 2.4 2.6

νu/ν y/H x = −1.13H

5 10 15 20 0.5 1 1.5 2 2.5

νu/ν x = 3.2H

5 10 15 0.5 1 1.5 2 2.5

νu/ν x = 14.86H PANS; PANS, 50% smaller inlet fluctuations; WALE;

  • : PANS, no inlet fluctuations;

: 2D RANS/10;

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SLIDE 143

FORWARD/BACKWARD FLOW

Fraction of time, γ, when the flow along the bottom wall is in the downstream direction.

2 4 6 8 10 12 14 0.2 0.4 0.6 0.8 1

x/H γ PANS; PANS, 50% smaller inlet fluctuations; WALE;

  • : PANS, no inlet fluctuations; ◦: experiments [26].

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SLIDE 144

SHEAR STRESSES. x = 3.2H

  • 5

5 10 15 x 10

  • 4

0.02 0.04 0.06 0.08 0.1

y/H PANS

  • 5

5 10 15 x 10

  • 4

0.02 0.04 0.06 0.08 0.1

RANS 2νt¯ s12; ν ∂¯ u ∂y ; −u′v′; ◦: 2νt¯ s12−u′v′.

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SLIDE 145

SHEAR STRESSES. x = 14.86

0.2 0.4 0.6 0.8 1 x 10

  • 3

0.02 0.04 0.06 0.08 0.1

y/H PANS

0.2 0.4 0.6 0.8 1 x 10

  • 3

0.02 0.04 0.06 0.08 0.1

RANS 2νt¯ s12; ν ∂¯ u ∂y ; −u′v′; ◦: 2νt¯ s12−u′v′.

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slide-146
SLIDE 146

TERMS IN THE ¯ u EQUATION. x = 3.2H

  • 0.06
  • 0.04
  • 0.02

0.02 0.04 0.06 0.02 0.04 0.06 0.08 0.1

y/H PANS

  • 0.06
  • 0.04
  • 0.02

0.02 0.04 0.06 0.02 0.04 0.06 0.08 0.1

RANS ∂ ∂y (2νt¯ s12); ν ∂2¯ u ∂y2 ; −∂¯ u¯ u ∂x ; +: −∂¯ u¯ v ∂y ; ⋆: −∂¯ p ∂x , △: −∂u′v′ ∂y .

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slide-147
SLIDE 147

TERMS IN THE ¯ u EQUATION. x = 14.86H

  • 0.06
  • 0.04
  • 0.02

0.02 0.04 0.06 0.02 0.04 0.06 0.08 0.1

y/H PANS

  • 0.06
  • 0.04
  • 0.02

0.02 0.04 0.06 0.02 0.04 0.06 0.08 0.1

RANS ∂ ∂y (2νt¯ s12); ν ∂2¯ u ∂y2 ; −∂¯ u¯ u ∂x ; +: −∂¯ u¯ v ∂y ; ⋆: −∂¯ p ∂x , △: −∂u′v′ ∂y .

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SLIDE 148

HEAT FLUXES. x = 3.2H

  • 1
  • 0.8
  • 0.6
  • 0.4
  • 0.2

0.02 0.04 0.06 0.08 0.1

y/H PANS

  • 1
  • 0.8
  • 0.6
  • 0.4
  • 0.2

0.02 0.04 0.06 0.08 0.1

RANS νt σt ∂ ¯ T ∂y

  • ;

ν σℓ ∂¯ T ∂y ; −vθ. ◦: νt σt ∂ ¯ T ∂y

  • − vθ.

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SLIDE 149

HEAT FLUXES. x = 14.86H

  • 1
  • 0.8
  • 0.6
  • 0.4
  • 0.2

0.02 0.04 0.06 0.08 0.1

y/H PANS

  • 1
  • 0.8
  • 0.6
  • 0.4
  • 0.2

0.02 0.04 0.06 0.08 0.1

RANS νt σt ∂ ¯ T ∂y

  • ;

ν σℓ ∂¯ T ∂y ; −vθ. ◦: νt σt ∂ ¯ T ∂y

  • − vθ.

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SLIDE 150

TERMS IN THE ¯ T EQUATION. x = 3.2H

  • 100
  • 50

50 100 0.02 0.04 0.06 0.08 0.1

y

PANS

  • 100
  • 50

50 100 0.02 0.04 0.06 0.08 0.1

RANS ∂ ∂y νt σt ∂¯ T ∂y

  • ;

ν σℓ ∂2¯ T ∂y2 ; −∂¯ u¯ T ∂x ; +: −∂¯ v¯ T ∂y ; △: −∂vθ ∂y .

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SLIDE 151

TERMS IN THE ¯ T EQUATION. x = 14.86H

  • 50

50 0.02 0.04 0.06 0.08 0.1

y/H PANS

  • 50

50 0.02 0.04 0.06 0.08 0.1

RANS ∂ ∂y νt σt ∂¯ T ∂y

  • ;

ν σℓ ∂2¯ T ∂y2 ; −∂¯ u¯ T ∂x ; +: −∂¯ v¯ T ∂y ; △: −∂vθ ∂y .

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SLIDE 152

CONCLUDING REMARKS

Developing boundary layer

◮ Synthetic fluctuations give fully developed conditions after a couple

  • f boundary layer thicknesses

◮ 5% upwinding dampens resolved fluctuations; can be compensated

by 25% larger inlet fluctuations

Backstep flow

◮ Very good agreement with experiments ◮ 2D RANS predicts turbulent diffusion surprisingly well ◮ Synthetic inlet fluctuations give an improved Stanton number;

  • therwise small effect in the reciculation region

◮ LRN PANS and WALE equally good ◮ 5% upwinding has a negligble effect in the recirculation region www.tfd.chalmers.se/˜lada Helsinki 4 October 2012 117 / 1

slide-153
SLIDE 153

REFERENCES I

[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] BILLSON, M. Computational Techniques for Turbulence Generated Noise. PhD thesis, Dept. of Thermo and Fluid Dynamics, Chalmers University of Technology, G¨

  • teborg, Sweden, 2004.

[3] BILLSON, M., ERIKSSON, L.-E., AND DAVIDSON, L. Modeling of synthetic anisotropic turbulence and its sound emission. The 10th AIAA/CEAS Aeroacoustics Conference, AIAA 2004-2857, Manchester, United Kindom, 2004.

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SLIDE 154

REFERENCES II

[4] DAVIDSON, L. Large eddy simulations: how to evaluate resolution. International Journal of Heat and Fluid Flow 30, 5 (2009), 1016–1025. [5] DAVIDSON, L. How to estimate the resolution of an LES of recirculating flow. In ERCOFTAC (2010), M. V. Salvetti, B. Geurts, J. Meyers, and P . Sagaut, Eds., vol. 16 of Quality and Reliability of Large-Eddy Simulations II, Springer, pp. 269–286. [6] DAVIDSON, L. Large eddy simulation of heat transfer in boundary layer and backstep flow using pans (to be presented). In Turbulence, Heat and Mass Transfer, THMT-12 (Palermo, Sicily/Italy, 2012).

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SLIDE 155

REFERENCES III

[7] DAVIDSON, L. A new approach of zonal hybrid RANS-LES based on a two-equation k − ε model. In ETMM9: International ERCOFTAC Symposium on Turbulence Modelling and Measurements (Thessaloniki, Greece, 2012). [8] DAVIDSON, L., AND BILLSON, M. Hybrid LES/RANS using synthesized turbulence for forcing at the interface. International Journal of Heat and Fluid Flow 27, 6 (2006), 1028–1042. [9] DAVIDSON, L., AND DAHLSTR ¨

OM, S.

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