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) ◮ this means that RANS turbulence models (developed for steady flow) are not accurate Turbulent kinetic energy Turbulent viscosity 5 250 RANS 4 200 region 3 150 k ν t /ν LES 2 100 region 1 50 0 0 0 0.05 0.1 0.15 0.2 0 0.05 0.1 0.15 0.2 y y : DES; ; 1D steady RANS; : DES resolved k . www.tfd.chalmers.se/˜lada Workshop, 29 August 2018 2 / 23

DES — D ETACHED -E DDY S IMULATIONS 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 Turbulent kinetic energy Turbulent viscosity 5 250 RANS 4 200 region 3 150 k ν t /ν LES 2 100 region 1 50 0 0 0 0.05 0.1 0.15 0.2 0 0.05 0.1 0.15 0.2 y y : DES; ; 1D steady RANS; : DES resolved k . www.tfd.chalmers.se/˜lada Workshop, 29 August 2018 2 / 23

T WO SOLVERS IN THE ENTIRE DOMAIN Steady RANS solver DES solver δ S − RANS LES y URANS x wall wall Grey color indicates the solver that drives the flow www.tfd.chalmers.se/˜lada Workshop, 29 August 2018 3 / 23

D RIFT TERMS ARE ADDED IN W HITE REGIONS Steady RANS solver DES solver LES URANS = � v LES � T − � v RANS = � v RANS v LES � T � T − � ¯ � T S RANS i i S LES i i , i i ∆ t ∆ t Subscript T indicates integration over time T � t � φ ( t ) � T = 1 φ ( τ ) exp( − ( t − τ ) / T ) d τ ⇒ T −∞ � φ � t T ≡ � φ � T = a � φ � t − ∆ t + ( 1 − a ) φ t T a = exp( − ∆ t / T ) . www.tfd.chalmers.se/˜lada Workshop, 29 August 2018 4 / 23

W HY IS THE RANS SOLVER STEADY ? Steady RANS solver = � v RANS � T − � v RANS � T S RANS i i i ∆ t The RANS solver is called every 10 th timestep (can probably be called less often) The solution in the RANS solver stays steady when the drift term, S RANS is steady (constant in time) i If the integration time T is too small, there will slightly different steady RANS flow every 10 th timestep ◮ Solution: make the steady RANS solver unsteady but use the large timestep, i.e. 10 ∆ t DES www.tfd.chalmers.se/˜lada Workshop, 29 August 2018 5 / 23

P REVIOUS W ORK 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 one ( T ). www.tfd.chalmers.se/˜lada Workshop, 29 August 2018 6 / 23

T URBULENCE M ODELS Steady RANS solver DES solver LES URANS EARSM (Explicit Algebraic DES k − ω model Stress Model) [4] coupled to Lengthscale in dissipation Wilcox k − ω model [5] 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

N UMERICAL M ETHOD : 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 1 st order upwind/2 nd order central scheme k & ω eqns. ◮ 2 nd -order Crank-Nicholson for time discretization CALC-BFC [9]: RANS solver, called every 10 th timestep ◮ Incompressible finite volume method ◮ SIMPLEC ◮ MUSCL: 2nd order bounded upwind scheme for momentum eqns ◮ Hybrid 1 st order upwind/2 nd order central scheme k & ω eqns. www.tfd.chalmers.se/˜lada Workshop, 29 August 2018 8 / 23

F IRST T EST C ASE : C HANNEL FLOW Reynolds number is Re τ = 5 200. A 32 × 96 × 32 mesh is used x max = 3 . 2, z max = 1 . 6, 15 % stretching in y direction www.tfd.chalmers.se/˜lada Workshop, 29 August 2018 9 / 23

C HANNEL FLOW : V ELOCITY 50 40 30 U + 20 DES DES 10 S-DES S-DES 0 S-DES 1 100 1000 y + T = 50 δ/ U b : DES; : RANS; ◦ : DNS. Vertical black lines show DES interface. www.tfd.chalmers.se/˜lada Workshop, 29 August 2018 10 / 23

C HANNEL FLOW : T URBULEN T Q UANTITIES 5 500 3 4 400 2 3 1 300 ν t /ν 0 2 0 0.005 0.01 200 1 100 0 0 0 0.5 1 0 1000 2000 3000 y + 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

S ECOND T EST C ASE : H UMP FLOW x = − 2 . 1 x = 0 x = 1 x = 4 y = 0 . 9 y = h The domain of the hump. z max = 0 . 2. The Reynolds number of the hump flow is Re c = 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, 1 https://turbmodels.larc.nasa.gov/nasahump val.html www.tfd.chalmers.se/˜lada Workshop, 29 August 2018 12 / 23

H UMP FLOW : N UMERICAL I SSUES 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 www.tfd.chalmers.se/˜lada Workshop, 29 August 2018 13 / 23

H UMP FLOW : C p & C f 10 -3 10 1 5 0.5 C p C f 0 0 -0.5 -5 -0.5 0 0.5 1 1.5 -0.5 0 0.5 1 1.5 x x ( A ) Pressure coefficient. ( B ) Skinfriction. T = 20 h / U in . : S-DES, j 0 = 33; : S-DES, j 0 = 53; : DES www.tfd.chalmers.se/˜lada Workshop, 29 August 2018 14 / 23

H UMP FLOW : V ELOCITIES 0.2 0.2 0.2 0.13 0.04 0.02 0.18 0.125 0.035 0.015 0.15 0.15 0.03 0.01 0.12 0.16 y 0.1 0.025 0.115 0.1 0.005 0 0.5 1 -0.3 -0.2 -0.1 0 -0.3 -0.2 -0.1 0 0.14 0.05 0.05 0.12 0 0.5 1 0 0.5 1 0 0.5 1 ( A ) x = 0 . 65 ( B ) x = 0 . 8 ( C ) x = 0 . 9 0.2 0.2 0.2 0.02 0.02 0.02 0.015 0.015 0.015 0.15 0.15 0.15 0.01 0.01 0.01 0.005 0.005 0.005 y 0.1 0.1 0.1 0 0 0 -0.2 -0.1 0 0 0.1 0.2 0 0.2 0.4 0.05 0.05 0.05 0 0 0 0 0.5 1 0 0.5 1 0 0.5 1 ( D ) x = 1 . 0 ( E ) x = 1 . 1 ( F ) x = 1 . 3 : S-ZDES, j 0 = 33; : S-ZDES, j 0 = 53; : DES; ◦ : exp [11, 12]; + , + : DES interface. www.tfd.chalmers.se/˜lada Workshop, 29 August 2018 15 / 23

T URBULENT VISCOSITY (EARSM). S-ZDES, j 0 = 53 0.3 0.3 0.3 0.01 0.01 0.01 0.008 0.008 0.008 0.006 0.006 0.006 0.2 0.2 0.2 0.004 0.004 0.004 0.002 0.002 0.002 y 0 0 0 0 200 400 600 800 0 50 100 0 200 400 600 0.1 0.1 0.1 0 0 0 0 50 100 0 200 400 600 800 0 200 400 600 ( A ) x = 0 . 65 ( B ) x = 0 . 8 ( C ) x = 0 . 9 0.3 0.3 0.3 0.01 0.01 0.01 0.008 0.008 0.008 0.006 0.006 0.006 0.2 0.2 0.2 0.004 0.004 0.004 0.002 0.002 y 0.002 0 0 0 0 500 1000 0 200 400 600 800 0 200 400 600 800 0.1 0.1 0.1 0 0 0 0 500 1000 0 200 400 600 800 0 200 400 600 800 ( D ) x = 1 . 0 ( E ) x = 1 . 1 ( F ) x = 1 . 3 : DES solver; : RANS solver; + : DES interface. www.tfd.chalmers.se/˜lada Workshop, 29 August 2018 16 / 23

H UMP FLOW : SHEAR STRESSES . S-ZDES, j 0 = 53 0.3 0.3 0.3 0.01 0.01 0.01 0.005 0.005 0.005 0.2 0.2 0.2 y 0 0 0 -4 -3 -2 -1 0 -2 -1.5 -1 -0.5 0 -2 -1 0 1 10 -3 10 -3 10 -3 0.1 0.1 0.1 0 0 0 -4 -3 -2 -1 0 -0.03 -0.02 -0.01 0 -0.03 -0.02 -0.01 0 10 -3 ( A ) x = 0 . 65 ( B ) x = 0 . 8 ( C ) x = 0 . 9 0.3 0.3 0.3 0.01 0.01 0.01 0.005 0.005 0.005 0.2 0.2 0.2 y 0 0 0 -2 -1 0 1 -2 -1 0 1 2 -6 -4 -2 0 10 -3 10 -3 10 -3 0.1 0.1 0.1 0 0 0 -0.03 -0.02 -0.01 0 -0.03 -0.02 -0.01 0 -0.03 -0.02 -0.01 0 ( D ) x = 1 . 0 ( E ) x = 1 . 1 ( F ) x = 1 . 3 : DES solver, resolved; : RANS solver; : DES solver, modeled. www.tfd.chalmers.se/˜lada Workshop, 29 August 2018 17 / 23

H UMP FLOW : WRONG p IN RANS REGION 1 0.5 C p 0 -0.5 0 1 2 3 x Pressure coefficient. : LES; : RANS, j 0 = 53; : RANS, j 0 = 33 www.tfd.chalmers.se/˜lada Workshop, 29 August 2018 18 / 23

C ONCLUSIONS 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 = 10 h / U in 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) www.tfd.chalmers.se/˜lada Workshop, 29 August 2018 19 / 23