W HY IS THE RANS SOLVER STEADY ? Steady RANS solver = v RANS T v - - PowerPoint PPT Presentation
W HY IS THE RANS SOLVER STEADY ? Steady RANS solver = v RANS T v - - PowerPoint PPT Presentation
S-ZDES: Z ONAL D ETACHED E DDY S IMULATION COUPLED WITH STEADY RANS IN THE WALL REGION Lars Davidson, www.tfd.chalmers.se/lada DES D ETACHED -E DDY S IMULATIONS Problem: the flow in the RANS region is highly unsteady (i.e. URANS)
DES — DETACHED-EDDY SIMULATIONS
Problem:
◮ the flow in the RANS region is highly unsteady (i.e. URANS) ◮ this means that RANS turbulence models (developed for steady
flow) are not accurate
0.05 0.1 0.15 0.2 50 100 150 200 250
Turbulent viscosity y RANS region region LES νt/ν
0.05 0.1 0.15 0.2 1 2 3 4 5
Turbulent kinetic energy y k
: DES; ; 1D steady RANS; : DES resolved k.
www.tfd.chalmers.se/˜lada Workshop, 29 August 2018 2 / 23
DES — DETACHED-EDDY SIMULATIONS
Problem:
◮ the flow in the RANS region is highly unsteady (i.e. URANS) ◮ this means that RANS turbulence models (developed for steady
flow) are not accurate
Solution:
◮ solve the steady equations in the RANS region
0.05 0.1 0.15 0.2 50 100 150 200 250
Turbulent viscosity y RANS region region LES νt/ν
0.05 0.1 0.15 0.2 1 2 3 4 5
Turbulent kinetic energy y k
: DES; ; 1D steady RANS; : DES resolved k.
www.tfd.chalmers.se/˜lada Workshop, 29 August 2018 2 / 23
TWO SOLVERS IN THE ENTIRE DOMAIN
wall Steady RANS solver x y URANS LES wall DES solver δS−RANS
Grey color indicates the solver that drives the flow
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DRIFT TERMS ARE ADDED IN WHITE REGIONS
Steady RANS solver
SRANS
i
= vLES
i
T − vRANS
i
T ∆t
DES solver URANS LES
SLES
i
= vRANS
i
T − ¯ vLES
i
T ∆t ,
Subscript T indicates integration over time T
φ(t)T = 1 T t
−∞
φ(τ) exp(−(t − τ)/T)dτ ⇒ φt
T ≡ φT = aφt−∆t T
+ (1 − a)φt a = exp(−∆t/T).
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WHY IS THE RANS SOLVER STEADY?
Steady RANS solver
SRANS
i
= vRANS
i
T − vRANS
i
T ∆t The RANS solver is called every 10th timestep (can probably be called less often) The solution in the RANS solver stays steady when the drift term, SRANS
i
is steady (constant in time) If the integration time T is too small, there will slightly different steady RANS flow every 10th timestep
◮ Solution: make the steady RANS solver unsteady but use the large
timestep, i.e. 10∆tDES
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PREVIOUS WORK
The present method is similar to those in [1, 2, 3]. The main differences are that
◮ In [1, 3] they use one additional drift terms in the LES momentum
equations to control resolved Reynolds stresses
◮ They include drift terms also in the k and ε equations [1] or the k
equation [3].
◮ In [1, 3] they include five tuning constants in all drift terms. I have
- ne (T).
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TURBULENCE MODELS
Steady RANS solver
EARSM (Explicit Algebraic Stress Model) [4] coupled to Wilcox k − ω model [5]
DES solver URANS LES
DES k − ω model Lengthscale in dissipation term of the k eq.is taken from the IDDES model [6, 7]
RANS and DES turbulence models
www.tfd.chalmers.se/˜lada Workshop, 29 August 2018 7 / 23
NUMERICAL METHOD: CALC-LES & CALC-BFC
CALC-LES [8]: DES solver
◮ Incompressible finite volume method ◮ Pressure-velocity coupling treated with fractional step ◮ Central differencing scheme for momentum eqns ◮ Hybrid 1st order upwind/2nd order central scheme k & ω eqns. ◮ 2nd-order Crank-Nicholson for time discretization
CALC-BFC [9]: RANS solver, called every 10th timestep
◮ Incompressible finite volume method ◮ SIMPLEC ◮ MUSCL: 2nd order bounded upwind scheme for momentum eqns ◮ Hybrid 1st order upwind/2nd order central scheme k & ω eqns. www.tfd.chalmers.se/˜lada Workshop, 29 August 2018 8 / 23
FIRST TEST CASE: CHANNEL FLOW
Reynolds number is Reτ = 5 200. A 32 × 96 × 32 mesh is used xmax = 3.2, zmax = 1.6, 15% stretching in y direction
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CHANNEL FLOW: VELOCITY
1 100 1000 10 20 30 40 50
y+ U+
S-DES S-DES S-DES DES DES
T = 50δ/Ub : DES; : RANS; ◦: DNS. Vertical black lines show DES interface.
www.tfd.chalmers.se/˜lada Workshop, 29 August 2018 10 / 23
CHANNEL FLOW: TURBULENT QUANTITIES
1000 2000 3000 100 200 300 400 500
y+ νt/ν
0.5 1 1 2 3 4 5
0.005 0.01 1 2 3
y
- : DNS [10];
: DES, resolved turbulence
: DES solver; : RANS solver. Vertical black lines show DES interface.
www.tfd.chalmers.se/˜lada Workshop, 29 August 2018 11 / 23
SECOND TEST CASE: HUMP FLOW
x = −2.1 x = 0 x = 1 y = h y = 0.9 x = 4
The domain of the hump. zmax = 0.2.
The Reynolds number of the hump flow is Rec = 936 000. The mesh has 386 × 120 × 32 cells (x, y, z) Grid from NASA workshop.1 Inlet is located at x = −2.1 and the outlet at x = 4.0,
1https://turbmodels.larc.nasa.gov/nasahump val.html www.tfd.chalmers.se/˜lada Workshop, 29 August 2018 12 / 23
HUMP FLOW: NUMERICAL ISSUES
Steady RANS solver
The drift term in the RANS solver in the LES region (white region) causes unphysical oscillations in the skinfriction The problem was traced to the source term in the pressure correction equation, the continuity error ˙ m Hence, ˙ m was set to zero in the LES region. As a consequence, the RANS velocity field is driven by the drift term, but the RANS pressure is not correct in this region
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HUMP FLOW: Cp & Cf
- 0.5
0.5 1 1.5
- 0.5
0.5 1
x Cp
(A) Pressure coefficient.
- 0.5
0.5 1 1.5
- 5
5 10 10 -3
x Cf
(B) Skinfriction.
T = 20h/Uin. : S-DES, j0 = 33; : S-DES, j0 = 53; : DES
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HUMP FLOW: VELOCITIES
0.5 1 0.12 0.14 0.16 0.18 0.2
0.5 1 0.115 0.12 0.125 0.13
y
(A) x = 0.65
0.5 1 0.05 0.1 0.15 0.2
- 0.3
- 0.2
- 0.1
0.025 0.03 0.035 0.04
(B) x = 0.8
0.5 1 0.05 0.1 0.15 0.2
- 0.3
- 0.2
- 0.1
0.005 0.01 0.015 0.02
(C) x = 0.9
0.5 1 0.05 0.1 0.15 0.2
- 0.2
- 0.1
0.005 0.01 0.015 0.02
y
(D) x = 1.0
0.5 1 0.05 0.1 0.15 0.2
0.1 0.2 0.005 0.01 0.015 0.02
(E) x = 1.1
0.5 1 0.05 0.1 0.15 0.2
0.2 0.4 0.005 0.01 0.015 0.02
(F) x = 1.3
: S-ZDES, j0 = 33; : S-ZDES, j0 = 53; : DES; ◦: exp [11, 12]; +,+: DES interface.
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TURBULENT VISCOSITY (EARSM). S-ZDES, j0 = 53
50 100 0.1 0.2 0.3
50 100 0.002 0.004 0.006 0.008 0.01
y
(A) x = 0.65
200 400 600 0.1 0.2 0.3
200 400 600 0.002 0.004 0.006 0.008 0.01
(B) x = 0.8
200 400 600 800 0.1 0.2 0.3
200 400 600 800 0.002 0.004 0.006 0.008 0.01
(C) x = 0.9
500 1000 0.1 0.2 0.3
500 1000 0.002 0.004 0.006 0.008 0.01
y
(D) x = 1.0
200 400 600 800 0.1 0.2 0.3
200 400 600 800 0.002 0.004 0.006 0.008 0.01
(E) x = 1.1
200 400 600 800 0.1 0.2 0.3
200 400 600 800 0.002 0.004 0.006 0.008 0.01
(F) x = 1.3
: DES solver; : RANS solver; +: DES interface.
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HUMP FLOW: SHEAR STRESSES. S-ZDES, j0 = 53
- 4
- 3
- 2
- 1
10 -3 0.1 0.2 0.3
- 4
- 3
- 2
- 1
10 -3 0.005 0.01
y
(A) x = 0.65
- 0.03
- 0.02
- 0.01
0.1 0.2 0.3
- 2
- 1.5
- 1
- 0.5
10 -3 0.005 0.01
(B) x = 0.8
- 0.03
- 0.02
- 0.01
0.1 0.2 0.3
- 2
- 1
1 10 -3 0.005 0.01
(C) x = 0.9
- 0.03
- 0.02
- 0.01
0.1 0.2 0.3
- 2
- 1
1 10 -3 0.005 0.01
y
(D) x = 1.0
- 0.03
- 0.02
- 0.01
0.1 0.2 0.3
- 2
- 1
1 2 10 -3 0.005 0.01
(E) x = 1.1
- 0.03
- 0.02
- 0.01
0.1 0.2 0.3
- 6
- 4
- 2
10 -3 0.005 0.01
(F) x = 1.3
: DES solver, resolved; : RANS solver; : DES solver, modeled.
www.tfd.chalmers.se/˜lada Workshop, 29 August 2018 17 / 23
HUMP FLOW: WRONG p IN RANS REGION
1 2 3
- 0.5
0.5 1
x Cp
Pressure coefficient. : LES; : RANS, j0 = 53; : RANS, j0 = 33
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CONCLUSIONS
A new steady RANS coupled to DES (S-ZDES) is proposed. Very good results . . . but the hump results are maybe/probablby contaminated by a numerical fix Drawback: it is dependent on the lower limit of integration time, T for the hump flow
◮ T = 10h/Uin too small (h is hump height) ◮ T = 20 and 50 give indentical results ◮ For T = 100 we must more than double developing+sampling time
to 345 + 345 (7.3 + 7.3 throughflow times)
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REFERENCES I
[1]
- H. Xiao and P
. Jenna. A consistent dual-mesh framework for hybrid LES/RANS modeling. Journal of Computational Physics, 231:1848–1865, 2012. [2]
- B. de Laage de Meux, B. Audebert, R. Manceau, and R. Perrin.
Anisotropic linear forcing for synthetic turbulence generation in large eddy simulation and hybrid RANS/LES modeling. Physics of Fluids A, 27(035115), 2015. [3]
- R. Tunstall, D. Laurence, R. Prosser, and A. Skillen.
Towards a generalised dual-mesh hybrid les/rans framework with improved consistency. Computers & Fluids, 157:73–83, 2017.
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REFERENCES II
[4]
- S. Wallin and A. V. Johansson.
A new explicit algebraic Reynolds stress model for incompressible and compressible turbulent flows. Journal of Fluid Mechanics, 403:89–132, 2000. [5]
- D. C. Wilcox.
Reassessment of the scale-determining equation. AIAA Journal, 26(11):1299–1310, 1988. [6]
- M. L. Shur, P
. R. Spalart, M. Kh. Strelets, and A. K. Travin. A hybrid RANS-LES approach with delayed-DES and wall-modelled LES capabilities. International Journal of Heat and Fluid Flow, 29:1638–1649, 2008.
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REFERENCES III
[7]
- S. Arvidson, L. Davidson, and S.-H. Peng.
Interface methods for grey-area mitigation in turbulence-resolving hybrid RANS-LES. International Journal of Heat and Fluid Flow, 73:236–257, 2018. [8]
- L. Davidson.
CALC-LES: A Fortran code for LES and hybrid LES-RANS. Technical report, Division of Fluid Dynamics, Dept. of Mechanics and Maritime Sciences, Chalmers University of Technology, Gothenburg, 2018. [9]
- L. Davidson and B. Farhanieh.
CALC-BFC: A finite-volume code employing collocated variable arrangement and cartesian velocity components for computation
- f fluid flow and heat transfer in complex three-dimensional
geometries.
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REFERENCES IV
- Rept. 95/11, Dept. of Thermo and Fluid Dynamics, Chalmers
University of Technology, Gothenburg, 1995. [10] M. Lee and R. D. Moser. Direct numerical simulation of turbulent channel flow up to Reτ ≈ 5200. Journal of Fluid Mechanics, 774:395–415, 2015. [11] D. Greenblatt, K. B. Paschal, C.-S. Yao, J. Harris, N. W. Schaeffler, and A. E. Washburn. A separation control CFD validation test case. Part 1: Baseline & steady suction. AIAA-2004-2220, 2004. [12] D. Greenblatt, K. B. Paschal, C.-S. Yao, and J. Harris. A separation control CFD validation test case Part 1: Zero efflux
- scillatory blowing.
AIAA-2005-0485, 2005.
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