Fluid mechanics, turbulent flow and turbulencemodeling Lars Davidson Division of Fluid Dynamics Department of Applied Mechanics Chalmers University of Technology SE-412 96 G¨oteborg, Sweden http://www.tfd.chalmers.se/˜lada
[email protected] June 15, 2015 Abstract This course material is used in two courses in the International Master’s programme Applied Mechanics at Chalmers. The two courses are TME225 Mechanics of fluids (Chapters 1-10), and MTF270 Turbulence Modeling (Chapters 11-27). MSc students who follow these courses are supposed to have taken one basic course in fluid mechanics. The lecture notes are also used in the three-day course Unsteady Simulations for Industrial Flows: LES, DES, hybrid LES-RANS and URANS (Chapters 16-27). This document can be downloaded at http://www.tfd.chalmers.se/˜lada/MoF/lecture notes.html and http://www.tfd.chalmers.se/˜lada/comp turb model/lecture notes.html The Fluid courses in the MSc programme are presented at http://www.tfd.chalmers.se/˜lada/msc/msc-programme.html The MSc programme is presented at http://www.chalmers.se/en/education/programmes/masters-info/Pages/Applied-Mechanics.aspx 1 Contents 1 2 3 4 Motion, flow 1.1 Eulerian, Lagrangian, material derivative . . . . . . . . . dv2 ∂v2 1.2 What is the difference between and ? . . . . . . dt ∂t 1.3 Viscous stress, pressure . . . . . . . . . . . . . . . . . . 1.4 Strain rate tensor, vorticity . . . . . . . . . . . . . . . . . 1.5 Product of a symmetric and antisymmetric tensor . . . . 1.6 Deformation, rotation . . . . . . . . . . . . . . . . . . . 1.7 Irrotational and rotational flow . . . . . . . . . . . . . . 1.7.1 Ideal vortex line . . . . . . . . . . . . . . . . . . 1.7.2 Shear flow . . . . . . . . . . . . . . . . . . . . . 1.8 Eigenvalues and and eigenvectors: physical interpretation . . . . . . 11 11 . . . . . . . . . . . . . . . . . . . . . . . . . . . 12 13 14 16 17 19 20 22 22 Governing flow equations 2.1 The Navier-Stokes equation . . . . . . . . . . . . . . . . . . . . . 2.1.1 The continuity equation . . . . . . . . . . . . . . . . . . . 2.1.2 The momentum equation . . . . . . . . . . . . . . . . . . 2.2 The energy equation . . . . . . . . . . . . . . . . . . . . . . . . . 2.3 Transformation of energy . . . . . . . . . . . . . . . . . . . . . . 2.4 Left side of the transport equations . . . . . . . . . . . . . . . . . 2.5 Material particle vs. control volume (Reynolds Transport Theorem) . . . . . . . 24 24 24 24 25 26 27 28 Solutions to the Navier-Stokes equation: three examples 3.1 The Rayleigh problem . . . . . . . . . . . . . . . . 3.2 Flow between two plates . . . . . . . . . . . . . . . 3.2.1 Curved plates . . . . . . . . . . . . . . . . 3.2.2 Flat plates . . . . . . . . . . . . . . . . . . 3.2.3 Force balance . . . . . . . . . . . . . . . . 3.2.4 Balance equation for the kinetic energy . . . 3.3 Two-dimensional boundary layer flow over flat plate . . . . . . . 30 30 33 33 34 36 37 38 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Vorticity equation and potential flow 4.1 Vorticity and rotation . . . . . . . . . . . . . . . . . . . . . . . . . 4.2 The vorticity transport equation in three dimensions . . . . . . . . . 4.3 The vorticity transport equation in two dimensions . . . . . . . . . . 4.3.1 Boundary layer thickness from the Rayleigh problem . . . . 4.4 Potential flow . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.4.1 Complex variables for potential solutions of plane flows . . . 4.4.2 Analytical solutions as f ∝ z n . . . . . . . . . . . . . . . . 4.4.2.1 Parallel flow . . . . . . . . . . . . . . . . . . . . . 4.4.2.2 Stagnation flow . . . . . . . . . . . . . . . . . . . . 4.4.2.3 Flow over a wedge and flow in a concave corner . . 4.4.3 Analytical solutions for a line source . . . . . . . . . . . . . 4.4.4 Analytical solutions for a vortex line . . . . . . . . . . . . . 4.4.5 Analytical solutions for flow around a cylinder . . . . . . . . 4.4.6 Analytical solutions for flow around a cylinder with circulation 4.4.6.1 The Magnus effect . . . . . . . . . . . . . . . . . . 4.4.7 The flow around an airfoil . . . . . . . . . . . . . . . . . . . 2 42 42 44 47 47 50 51 52 53 53 54 55 56 57 60 62 65 3 5 6 Turbulence 5.1 Introduction . . . . . . . . . . . . . . . 5.2 Turbulent scales . . . . . . . . . . . . . 5.3 Energy spectrum . . . . . . . . . . . . . 5.4 The cascade process created by vorticity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 66 66 67 68 72 Turbulent mean flow 6.1 Time averaged Navier-Stokes . . . . . . . . . . . 6.1.1 Boundary-layer approximation . . . . . . 6.2 Wall region in fully developed channel flow . . . 6.3 Reynolds stresses in fully developed channel flow 6.4 Boundary layer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 76 76 77 78 82 84 . . . . . . . . . . . . . . . . 7 Probability density functions 86 8 Transport equations for kinetic energy 8.1 Rules for time averaging . . . . . . . . . . . . . . . . . . . . . . . . 8.1.1 What is the difference between v1′ v2′ and v1′ v2′ ? . . . . . . . 89 89 89 2 8.2 8.3 8.4 8.5 8.6 9 8.1.2 What is the difference between v1′2 and v1′ ? . . . . . . . . . 8.1.3 Show that v¯1 v1′2 = v¯1 v1′2 . . . . . . . . . . . . . . . . . . . 8.1.4 Show that v¯1 = v¯1 . . . . . . . . . . . . . . . . . . . . . . . The Exact k Equation . . . . . . . . . . . . . . . . . . . . . . . . . 8.2.1 Spectral transfer dissipation εκ vs. “true” viscous dissipation, ε The Exact k Equation: 2D Boundary Layers . . . . . . . . . . . . . Spatial vs. spectral energy transfer . . . . . . . . . . . . . . . . . . The overall effect of the transport terms . . . . . . . . . . . . . . . . The transport equation for v¯i v¯i /2 . . . . . . . . . . . . . . . . . . . 90 90 90 90 94 95 96 97 98 Transport equations for Reynolds stresses 9.1 Source terms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.2 Reynolds shear stress vs. the velocity gradient . . . . . . . . . . . . 100 103 105 10 Correlations 10.1 Two-point correlations . . . . . . . . . . . . . . . . . . . . . . . . . 10.2 Auto correlation . . . . . . . . . . . . . . . . . . . . . . . . . . . . 107 107 110 11 Reynolds stress models and two-equation models 11.1 Mean flow equations . . . . . . . . . . . . . 11.1.1 Flow equations . . . . . . . . . . . 11.1.2 Temperature equation . . . . . . . . 11.2 The exact vi′ vj′ equation . . . . . . . . . . . 11.3 The exact vi′ θ′ equation . . . . . . . . . . . 11.4 The k equation . . . . . . . . . . . . . . . . 11.5 The ε equation . . . . . . . . . . . . . . . . 11.6 The Boussinesq assumption . . . . . . . . . 11.7 Modeling assumptions . . . . . . . . . . . . 11.7.1 Production terms . . . . . . . . . . 11.7.2 Diffusion terms . . . . . . . . . . . 11.7.3 Dissipation term, εij . . . . . . . . 11.7.4 Slow pressure-strain term . . . . . . 112 112 112 113 113 115 117 118 119 120 120 121 122 123 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4 11.7.5 Rapid pressure-strain term . . . . . . 11.7.6 Wall model of the pressure-strain term 11.8 The k − ε model . . . . . . . . . . . . . . . . 11.9 The modeled vi′ vj′ equation with IP model . . 11.10 Algebraic Reynolds Stress Model (ASM) . . . 11.11 Explicit ASM (EASM or EARSM) . . . . . . 11.12 Boundary layer flow . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 126 131 133 134 134 135 136 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 138 138 139 142 143 13 Realizability 13.1 Two-component limit . . . . . . . . . . . . . . . . . . . . . . . . . 144 145 14 Non-linear Eddy-viscosity Models 147 15 The V2F Model 15.1 Modified V2F model . . . . . . . . . . . . . . . . . . . . . . . . . . 15.2 Realizable V2F model . . . . . . . . . . . . . . . . . . . . . . . . . 15.3 To ensure that v 2 ≤ 2k/3 . . . . . . . . . . . . . . . . . . . . . . . 150 153 154 154 16 The SST Model 155 17 Overview of RANS models 160 18 Large Eddy Simulations 18.1 Time averaging and filtering . . . . . . . . . . . . . . . . . . . . . . 18.2 Differences between time-averaging (RANS) and space filtering (LES) 18.3 Resolved & SGS scales . . . . . . . . . . . . . . . . . . . . . . . . 18.4 The box-filter and the cut-off filter . . . . . . . . . . . . . . . . . . 18.5 Highest resolved wavenumbers . . . . . . . . . . . . . . . . . . . . 18.6 Subgrid model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18.7 Smagorinsky model vs. mixing-length model . . . . . . . . . . . . . 18.8 Energy path . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18.9 SGS kinetic energy . . . . . . . . . . . . . . . . . . . . . . . . . . 18.10 LES vs. RANS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18.11 The dynamic model . . . . . . . . . . . . . . . . . . . . . . . . . . 18.12 The test filter . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18.13 Stresses on grid, test and intermediate level . . . . . . . . . . . . . . 18.14 Numerical dissipation . . . . . . . . . . . . . . . . . . . . . . . . . 18.15 Scale-similarity Models . . . . . . . . . . . . . . . . . . . . . . . . 18.16 The Bardina Model . . . . . . . . . . . . . . . . . . . . . . . . . . 18.17 Redefined terms in the Bardina Model . . . . . . . . . . . . . . . . 18.18 A dissipative scale-similarity model. . . . . . . . . . . . . . . . . . 18.19 Forcing . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18.20 Numerical method . . . . . . . . . . . . . . . . . . . . . . . . . . . 18.20.1 RANS vs. LES . . . . . . . . . . . . . . . . . . . . . . . . 18.21 One-equation ksgs model . . . . . . . . . . . . . . . . . . . . . . . 161 161 162 163 164 165 166 166 167 167 168 169 169 170 172 173 173 173 174 176 176 177 178 12 Reynolds stress models vs. eddy-viscosity models 12.1 Stable and unstable stratification . . . . . . 12.2 Curvature effects . . . . . . . . . . . . . . . 12.3 Stagnation flow . . . . . . . . . . . . . . . 12.4 RSM/ASM versus k − ε models . . . . . . . . . . 5 18.22 18.23 18.24 18.25 18.26 Smagorinsky model derived from the ksgs equation A dynamic one-equation model . . . . . . . . . . . A Mixed Model Based on a One-Eq. Model . . . . Applied LES . . . . . . . . . . . . . . . . . . . . . Resolution requirements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 178 179 180 180 181 19 Unsteady RANS 19.1 Turbulence Modeling . . . . . . . . . . . . . . . . . . . . . . . . . 19.2 Discretization . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 182 184 186 20 DES 20.1 DES based on two-equation models . . . . . . . . . . . . . . . . . . 20.2 DES based on the k − ω SST model . . . . . . . . . . . . . . . . . 188 188 190 21 Hybrid LES-RANS 21.1 Momentum equations in hybrid LES-RANS . . . . . . . . . . . . . 21.2 The equation for turbulent kinetic energy in hybrid LES-RANS . . . 192 194 194 22 The SAS model 22.1 Resolved motions in unsteady . . . . . . . . . . . . . . . 22.2 The von K´arm´an length scale . . . . . . . . . . . . . . . . 22.3 The second derivative of the velocity . . . . . . . . . . . . 22.4 Evaluation of the von K´arm´an length scale in channel flow 196 196 196 198 198 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23 The PANS Model 201 24 The PITM model 24.1 RANS mode . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24.2 LES mode . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 205 205 205 25 Hybrid LES/RANS for Dummies 25.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . 25.1.1 Reynolds-Averaging Navier-Stokes equations: RANS 25.1.2 Large Eddy Simulations: LES . . . . . . . . . . . . 25.1.3 Zonal LES/RANS . . . . . . . . . . . . . . . . . . . 25.2 The PANS k − ε turbulence model . . . . . . . . . . . . . . 25.3 Zonal LES/RANS: wall modeling . . . . . . . . . . . . . . . 25.3.1 The interface conditions . . . . . . . . . . . . . . . . 25.3.2 Results . . . . . . . . . . . . . . . . . . . . . . . . . 25.4 Zonal LES/RANS: embedded LES . . . . . . . . . . . . . . 25.4.1 The interface conditions . . . . . . . . . . . . . . . . 25.4.2 Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 209 209 209 209 210 210 211 211 212 212 212 214 26 Inlet boundary conditions 26.1 Synthesized turbulence . . . . 26.2 Random angles . . . . . . . . . 26.3 Highest wave number . . . . . 26.4 Smallest wave number . . . . . 26.5 Divide the wave number range 26.6 von K´arm´an spectrum . . . . . 26.7 Computing the fluctuations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 215 215 215 216 216 216 217 217 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6 26.8 26.9 Introducing time correlation . . . . . . . . . . . . . . . . . . . . Anisotropic Synthetic Turbulent Fluctuations . . . . . . . . . . . 26.9.1 Eigenvalues and eigenvectors . . . . . . . . . . . . . . . 26.9.2 Synthetic fluctuations in the principal coordinate system . . . . . . . . . 27 Overview of LES, hybrid LES-RANS and URANS models 28 Best practice guidelines (BPG) 28.1 EU projects . . . . . . . . . . . . . . . . 28.2 Ercoftac workshops . . . . . . . . . . . . 28.3 Ercoftac Classical Database . . . . . . . . 28.4 ERCOFTAC QNET Knowledge Base Wiki . . . . . . . . . . . . . . . . . . . . . . . . 217 220 222 223 224 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . A TME225: ǫ − δ identity 227 227 227 228 228 229 B TME225 Assignment 1 in 2014: laminar flow in a boundary layer B.1 Velocity profiles . . . . . . . . . . . . . . . . . . . . . . . . . B.2 Boundary layer thickness . . . . . . . . . . . . . . . . . . . . B.3 Velocity gradients . . . . . . . . . . . . . . . . . . . . . . . . B.4 Skinfriction . . . . . . . . . . . . . . . . . . . . . . . . . . . B.5 Vorticity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . B.6 Deformation . . . . . . . . . . . . . . . . . . . . . . . . . . . B.7 Dissipation . . . . . . . . . . . . . . . . . . . . . . . . . . . . B.8 Eigenvalues . . . . . . . . . . . . . . . . . . . . . . . . . . . B.9 Eigenvectors . . . . . . . . . . . . . . . . . . . . . . . . . . . B.10 Terms in the v1 equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 230 231 231 232 232 232 233 233 233 234 234 C TME225 Assignment 1 in 2013: laminar flow in a channel C.1 Fully developed region . . . . . . . . . . . . . . . . C.2 Wall shear stress . . . . . . . . . . . . . . . . . . . . C.3 Inlet region . . . . . . . . . . . . . . . . . . . . . . . C.4 Wall-normal velocity in the developing region . . . . C.5 Vorticity . . . . . . . . . . . . . . . . . . . . . . . . C.6 Deformation . . . . . . . . . . . . . . . . . . . . . . C.7 Dissipation . . . . . . . . . . . . . . . . . . . . . . . C.8 Eigenvalues . . . . . . . . . . . . . . . . . . . . . . C.9 Eigenvectors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 235 236 236 236 237 237 237 237 238 238 D TME225: Fourier series D.1 Orthogonal functions . . . . . . . D.2 Trigonometric functions . . . . . . D.2.1 “Length” of ψk . . . . . . D.2.2 Orthogonality of φn and ψk D.2.3 Orthogonality of φn and φk D.2.4 “Length” of φk . . . . . . D.3 Fourier series of a function . . . . D.4 Derivation of Parseval’s formula . D.5 Complex Fourier series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 239 239 240 241 241 241 241 242 242 244 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . E TME225: Why does the energy spectrum, E, have such strange dimensions?245 E.1 Energy spectrum for an ideal vortex . . . . . . . . . . . . . . . . . . 245 . . .9 Pressure-strain terms . . . . . . . . . . . . . M. . . . . . . . . . L. . .11 Do something fun! . . . . . . . . . . . . . . . 276 276 276 277 278 278 280 281 282 282 282 283 283 M MTF270. . . .3 The turbulent kinetic energy equation . . L. . . . . . . . . . . L. . 273 274 L MTF270. . . . . . . . . . .6 Resolved stresses . F. . .3 Part II: StarCCM+ . . . . L. . . . . F. . . . . . . . . . . . . . . . .1 Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .3 Mean flow . .3. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .2 Time averaging . . . L. . . . . .4 Probability density/histogram . . . . . . . . . .4 Brief instruction to begin the asymmetric diffuser case . 287 287 288 288 289 290 290 292 . . . . . . . . . . . . . . . . . . 271 271 272 K MTF270: 1D and 3D energy spectra K. .1 Time history .2 E. . . . . . . . . . . . . . . . . . . F. . . . L. . . . . . . . . An example: Matlab code . . . . . . . . . . . . . L. . . L. . . . . . . L. . . . . . . . . . . . . .1 Time history . . . . . . . . . . . . . . . . . .2 Time averaging . . . . . . . . . . . F. L. . . . . . . . . . . . . . . . . . . . . . . . . . 246 247 251 251 251 252 252 253 253 253 253 253 254 254 G TME225 Learning outcomes 2014 255 H MTF270: Some properties of the pressure-strain term 268 I MTF270: Galilean invariance 269 J MTF270: Computation of wavenumber vector and angles J. . . . . . . . . . . . . . . . . .6 Resolved stresses . . . . . . . . . . . . . . . . . . .1. . . . . .1 Part I: Data of Two-dimensional flow . . . . . .3 An example . . . . .3. .5 Frequency spectrum . . . . . . . . . .1 The wavenumber vector. . . . . . . . . . . . . . . F. . . . . . . . . . . .1 Geometrical quantities .3 Asymmetric diffuser case . . . . .4 The Reynolds stress equations . . . .10 Dissipation . . . . . . .2. . . . . . . . . . . .4 The time-averaged momentum equation F. . . . . . . . . . . . . . . M. . . . . . . . . . . . . . . . . . . . . . . . . . . F. . . . . . . . . . . . . . . . .3. . . . . . . . . . .7 E. . .5 Wall shear stress . . . . . . . .2 Backward Facing Step Tutorial . . . . . . Assignment 1: Reynolds averaged Navier-Stokes L.2 The momentum equations . . L. . . . . . . . .1. . .2 Compute derivatives on a curvi-linear mesh . . . . . . . . . F. . . .7 Fluctuating wall shear stress . . . . . . . . . . . . .8 Production terms . . . . . . . . . . . . . . . . . . . . . . . . . . . . .3 Auto correlation . . . . . . . . . . . . . . . . . M. . . . . . . . . . . . . . κnj . . . . . . . . . . . .1.3. . . . . . . . . . . F TME225 Assignment 2: turbulent flow F. . . . . . . . . . . . . . . . .1 Introduction Tutorial . . . . . . . . . . . . . . . F. . . . . . . . . . . . . . . . . . . . .2 Unit vector σin . . . . M. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . F. . . . . . . . . . . . . . . . . . . . . . . . . M. .1 Energy spectra from two-point correlations . . . . . .1. . . . . . . .7 Resolved production and pressure-strain . . . M. . . . . . Assignment 2: LES M. J. . . . . . N. . O. . . O. . . . . . . Near-wall behavior . . . . . . Energy spectra . . . . . . .1 Introduction . . . . . . . . Something fun . . . . . . . . . . . . . . . . . . . . .7 Turbulent length scales . . . . . . . 292 293 294 294 296 296 296 297 297 N MTF270: Compute energy spectra from LES/DNS data using Matlab N. . . . . . Q. . . . . . . . . . . . .4 Modeled stresses . R. . . . . . . . . . O. DDES and SAS O. . . . . . . . . . . . . . . . DES and DDES Q. . . . . . . . . . . . . . . . . recirculating flow: PANS. . . . . . . . . . . . . Dissipations . . . . . . . . . . . . . . . . . . . . . . . . . . .4 Turbulent kinetic energy . . . . . . . . . . . . . . . . . . . 298 298 298 301 302 O MTF270. . . . . . . . . . . . . Q. . . . . R. N. .2 An example of using FFT . P. . . 315 315 316 316 317 317 317 317 . . . . . .3 Energy spectrum from the two-point correlation . . . . . . . . . . . . P. . . .5 Compute fk . . . . . . . . .1 Rotation from x1∗ − x2∗ to x1 − x2 . .1 Time history . . . . . . . . . . . . . . . .4 Location of interface in DES and DDES . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . R. . . . . . . . . .2 Resolved stresses . . . . . . . . . . . . . . . . . . . . . . . .5 The modelled turbulent shear stress .3 Resolved stresses . . . . . . Two-point correlations . . . . . . . . . .5 Turbulent SGS dissipation . . . . .6 Location of interface in DES and DDES . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . S MTF270: Transformation of tensors S. . . . . . . . . . .4 Energy spectra from the autocorrelation . . . . . . . . . . . . P. . . Assignment 4: Embedded LES with PANS P. .2 Mean velocity profile . . . . . . . . . . . . . . . . . . . . . .10 M. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . O. . . . . . . . . . . . . . . . . . . .8 M.14 M. .9 M. . . . . . . . . . . . . . . . Assignment 6: Hybrid LES-RANS R. . . 304 304 305 305 306 306 306 306 P MTF270. P. . . . . . . . . . . . . . . . . . . . . . . . . .3 Turbulent viscosity . . . . . . .1 Discretization schemes . . . . . 318 318 . . . . . . . . . . . . . . . . . . . . . . . . . .12 M. . . . . . . .8 M. . . . . . . . .11 M. . R. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .15 M. . . . DES. . . . . .5 The modelled turbulent shear stress . . . . . . . . . . . . . . . . . . . . . . . Q. . . . .6 Location of interface in DES and DDES R. . . O. . . . . . . .13 M. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . SGS stresses: Smagorinsky model SGS stresses: WALE model . . . . . . . . . R. . Assignment 3: Zonal PANS. .16 Filtering . . . . . N.7 SAS turbulent length scales . . . . . .4 Turbulent kinetic energy . . Q. . . . . . . . . O. . Test filtering . . . . . . . . . . . . . . . . . . . . . . . . . .3 Resolved stresses . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .2 Mean velocity profile . . . . . . . . . . . . . . . . . . . . . . . . . Assignment 5. . . 311 312 312 313 314 314 R MTF270. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 308 308 309 309 309 310 Q MTF270. .1 Time history . . . .3 The turbulent viscosity . . . . . . . . . . . . . . . . . . . . . . . . . . . . .1 Time history .2 The modeled turbulent shear stress . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 319 320 321 321 321 321 324 U MTF270: Learning outcomes for 2015 325 V References 334 . . . . . . Transformation of a velocity gradient . . . . .2 S. . . . . . . . . . . . . . . . . . . . . . . . . T. . .3 Rotation from x1 − x2 to x1∗ − x2∗ . . . . . . . .4 Analytical solution to Poisson’s equation . . . . . . .3 Green’s third formula . . . .1 Green’s first formula . . . . T. . . . . . . . . T. . . . . .9 S. . . . . . . . . . . . . . . . . . . . T MTF270: Green’s formulas T. . . . . . . . . .2 Green’s second formula . chalmers.se This report can be downloaded at http://www.
[email protected]. Sweden http://www. Davidson Division of Fluid Dynamics.chalmers.se/˜lada.tfd. G¨oteborg. Department of Applied Mechanics Chalmers University of Technology.se/˜lada/MoF/ .10 TME225 Mechanics of fluids L. approach.g. vi = dxi /dt. flow 1. but since we are looking at fixed points in space we need to express the temperature as a function of both time and space. 3. vi = vi (Xi . and then T = T (xi . i. 1. In the Eulerian approach we pick a position. dT /dt. t). or Eulerian. e. Lagrangian. t). This approach is used for fluids.e.1. We can choose to study its motion in two ways: Lagrangian or Eulerian. From classical mechanics. see Fig. We are looking for the temperature gradient. 1 Motion. we know that the velocity of a fluid particle is the time derivative of its space location. t. The chain-rule now gives ∂T ∂T ∂T dxj ∂T dT = = + + vj dt ∂t dt ∂xj ∂t ∂xj (1. T (xi . i. and at time t2 its temperature is T (Xi . t).e. t2 ) T (x2i .2. i. 1. Once we have chosen a particle its starting position is fixed. T .e. In the Eulerian approach it is a little bit more difficult. t1 ) T (Xi . Chapt. see Fig. T (t) and the temperature gradient can be written dT /dt. t2 ).1) Note that we have to use partial derivative on T since it is a function of more than one (independent) variable. at time t1 the temperature of the particle is T (Xi . t1 ). t2 ) Figure 1. In the Lagrangian approach we keep track of its original position (Xi ) and follow its path which is described by xi (Xi . 1. t). T = T (xi ). x1i . In the Lagrangian approach we first pick the particle (this gives its starting position. for example. and watch the particle pass by. This approach is not used for fluids because it is very tricky to define and follow a fluid particle. T (Xi . and temperature varies only with time.1. Motion. The speed of the particle is then expressed as a function of time and its position at time zero. The second term local rate of change . It is however used when simulating movement of particles in fluids (for example soot particles in gasoline-air mixtures in combustion applications). flow 11 T (Xi . material derivative See also [1]. for example. varies in time. Now we want to express how the temperature of a fluid particle varies.1.1: The temperature of a fluid particle described in Lagrangian. The temperature of the fluid. t1 ) Xi T (x1i . Assume a fluid particle is moving along the line in Fig.1 Eulerian.e. i. Xi ). For example. t). The first term on the right side is the local rate of change.1. It may be that the temperature at position xi . is expressed as a function of the position. by this we mean that it describes the variation of T in time at position xi . 1 can be illustrated as follows. the point (x1 . 0. X1 .2 shows a flow path of the fluid particles which can be expressed in time as x1 = exp(t). 1. x2 ) and viL as function of t (and in general also starting location.4) (cf. x2 ) = (1.3 give v1E = x1 . 1.5. 2 = = 1 (the velocity. The flow path is steady in time and it starts at (x1 . The left side in Eq. x2 ) = (2. respectively. X2 ). The velocity at the point dt ∂t ∂v2 ∂v E = 0. for example. Eq. Now we can compute the time derivatives of the v2 velocity as dv2L = exp(−t) dt dv2E ∂v E ∂v E ∂v E = 2 + v1E 2 + v2E 2 = 0 + x1 · 0 − x2 · (−1) = x2 dt ∂t ∂x1 ∂x2 (1.1.3) and Eqs. Equation 1. However. v2L . x2 ) = (0.5.5 and v2 = −0.2 and 1. dt dt dt Consider. increases dv L dv2 by time. rate of change Material derivative . x2 ) = (1. Put your finger out in the blowing wind. Equation 1.2. Here we Students sometimes get confused about the difference between dt ∂t give a simple example. of course. The temperature gradient you’re finger experiences is ∂T /∂t. v2 . The difference betdv2 ∂v2 ween and is now clearly seen by looking at Eq. x2 ) = (1. the velocity.5. 1.1 is called the material derivative and is in this text denoted by dT /dt. dt v2L = dx2 = − exp(−t) dt (1. that Conv. 4. 1). 1.54). term-by-term. gets less negative) .2. 1) in Fig. Actually it increases dt dt dv2 = 2 and v2 = −2 until the end point all the time from the starting point where dt dv2 where = 0. Exercise 1 Write out Eq.1. v2E = −x2 (1. if we (x1 . The superscripts E and L denote Eulerian and Lagrangian. dt We find. x1 . x2 = exp(−t) (1. Note that v1L = v1E and v2L = v2E . which means that it describes the variation of T in space when is passes the point xi .2) and hence x2 = 1/x1 . the only difference is that viE is expressed as function of (t. 1. dT /dt. 1) does not change in time and hence 2 = ∂t ∂t sit on a particle which passes the location (x1 . The temperature gradient you experience is the material derivative. 2) and ends at (x1 .5). Imagine that you’re a fluid particle and that you ride on a bike.5) dv2 dv E dv L = 2 = 2 = x2 = exp(−t).2 gives the velocities v1L = dx1 = exp(t).2 What is the difference between dv2 ∂v2 and ? dt ∂t ∂v2 dv2 and . What is the difference between dv2 ∂v2 and ? dt ∂t 12 on the right side is called the convective rate of change. 1. Figure 1. the first index relates to the surface at which the stress acts and the second index is related to the stress component. 6. respectively. pressure See also [1]. If no tensor notation is used the velocity vector is usually denoted as (u. The diagonal components of σij represent normal stresses and the off-diagonal components of σij represent the shear stresses. v2 . 1.1. w) and the coordinates as (x.j − ρfi = 0 (1. the volume force acts in the middle of the fluid element. u2 . 0) act the three stress components σ11 . v. see Fig.3.3 and 8. 1. 0. v3 ) and the coordinate by x = xi = (x1 .2: Flow path x2 = 1/x1 . you may find other notations of the velocity vector such as ui = (u1 .7) where the first and the second term on the right side represents. z). u3 ). For example.6) Switch notation for the material derivative and derivatives so that ρ dvi ∂σji + ρfi = dt ∂xj (1. As you have learnt earlier. x3 ). ▽: start (t = ln(0. In the present notation we denote the velocity vector by v = vi = (v1 .3a. △: end (t = ln(2)). The first term includes the viscous stress tensor. on a surface whose normal is ni = (1. Viscous stress. x2 . σ12 and σ13 . Chapts. see Fig. In Part I [2] you learnt that the pressure is defined as minus the sum of the normal stress. We have in Part I [2] derived the balance equation for linear momentum which reads ρv˙ i − σji. the net force due to surface and volume forces (σij denotes the stress tensor).e. The filled circle shows the point (x1 .8) . x2 ) = (1. pressure x2 2 13 ▽ 1 △ 0 x1 0 1 2 Figure 1. 1. τij .5)).3 Viscous stress.3b. Stress is force per unit area. 1). i. y. P = −σkk /3 (1. In the literature.1. Chapt. Exercise 3 Write out Eq.3: Stress tensor. P . acting in the middle of the fluid element.e. In that case the thermodynamics pressure. Strain rate tensor. pt . Show how σ21 . however. . σ33 = −P + τ33 . In general. 24. 3. 1. the normal stresses are the sum of the pressure and the viscous stresses. The expression for the viscous stress tensor is found in Eq. 1.4. vorticity 14 (ˆ e1 ) σ12 ti σ11 σ13 x2 fi x2 x1 x1 (a) Stress components and stress vector on a surface. τij . P . pt .5.6. 1.1. see The pressure.8 is nevertheless used. In general. σ11 = −P + τ11 . vorticity See also [1]. σ22 . σkk /3 = −P . σ33 act on a surface with normal vector ni = (0. σ22 = −P + τ22 . (1. acts as a normal stress. is obtained by subtracting the trace. −g. ∂x . i. σ23 act on a surface with normal vector ni = (0. 0).3. 0. and the mechanical pressure. 1. C.3. fi = (0. (b) Volume force. The minus-sign in front of P appears because the pressure acts into the surface.4 Strain rate tensor. Hence we will now discuss the velocity gradients. They will be expressed in the ∂vi velocity gradients.2. 0). 1). pressure is a thermodynamic property. j .9 on matrix form. (ˆ e1 ) Figure 1. which can be obtained – for example – from the ideal gas law. The viscous stress tensor. When there’s no movement. 2. ti Eq. volume (gravitation) force and stress vector. the stress tensor can then be written as σij = −P δij + τij (1. the viscous stresses are zero and then of course the normal stresses are the same as the pressure. may not be the same but Eq.10) Exercise 2 Consider Fig. σ32 . τij . We need an expression for the viscous stresses. 1. 3. Show also how σ31 .9) τij is the deviator of σij .4 at p. from σij . 12 gives ωi = ǫijk (Skj + Ωkj ) = ǫijk Ωkj (1. Inserting Eq. ε331 = 0) ε1jk Skj = ε111 S11 + ε112 S21 + ε113 S31 + ε121 S12 + ε122 S22 + ε123 S32 + ε131 S13 + ε132 S23 + ε133 S33 = 0 · S11 + 0 · S21 + 0 · S31 (1.4.12) where ǫijk is the permuation tensor.17 give Ωmℓ = 1 1 1 εiℓm ωi = εℓmi ωi = − εmℓi ωi 2 2 2 (1. for example. Let us show this for i = 1 by writing out the full equation.11 into Eq. S12 = S21 . . Strain rate tensor.13) The vorticity represents rotation of a fluid particle. . We start by multiplying it with εiℓm so that εiℓm ωi = εiℓm ǫijk Ωkj (1.e. see lecture notes of Toll & Ekh [2]. ε113 = ε221 = .16 and 1. Recall that Sij = Sji (i. i. ω = ∇ × v. see Table A. Now Eqs. 229. .14. 1. or in tensor notation ωi = ǫijk ∂vk ∂xj (1. ε123 = −ε132 = ε231 .16) The ε-δ-identity gives (see Table A.e. .15) + 0 · S12 + 0 · S22 + 1 · S32 + 0 · S13 − 1 · S23 + 0 · S33 = S32 − S23 = 0 Now let us invert Eq. 1. .e.1 at p. i = 3 we get ω3 = ∂v2 /∂x1 − ∂v1 /∂x2 . 229) εiℓm ǫijk Ωkj = (δℓj δmk − δℓk δmj )Ωkj = Ωmℓ − Ωℓm = 2Ωmℓ (1. (1. S13 = S31 . vorticity 15 The velocity gradient tensor can be split into two parts as 1 ∂vi ∂vi ∂vj ∂vj ∂vi + = + − ∂xj 2 ∂xj ∂xj ∂xi ∂xi = 1 2 (1. S23 = S32 ) and ǫijk = −ǫikj = ǫjki etc (i.14) since ǫijk Skj = 0 because the product of a symmetric tensor (Skj ) and an antisymmetric tensor (εijk ) is zero. 1.17) This can easily be proved by writing all the components.1.18) Strain-rate tensor vorticity tensor . 1.1 at p.11) =0 2∂vi /∂xj ∂vi ∂vj + ∂xj ∂xi + 1 2 ∂vi ∂vj − ∂xj ∂xi = Sij + Ωij where Sij is a symmetric tensor called the strain-rate tensor Ωij is a anti-symmetric tensor called the vorticity tensor The vorticity tensor is related to the familiar vorticity vector which is the curl of the velocity vector. If we set. 1. 1. Exercise 6 Write out Eq. 1. This can be shown as follows aij bij = aji bij = −aji bji . 1. 1. (1. Exercise 5 Complete the proof of Eq.5 Product of a symmetric and antisymmetric tensor In this section we show the proof that the product of a symmetric and antisymmetric tensor is zero.19) Ωij = − εijk ωk 2 A much easier way to go from Eq. so that aij bij = −aij bij .14 to Eq. First.20 also for i = 2 and i = 3 and find an expression for Ω12 and Ω13 (cf.12. b11 = −b11 can only be satisfied if b11 = 0.e. 1. Exercise 7 In Eq. It follows that for an antisymmetric tensor all diagonal components must be zero. you should remember how to compute a vector product using Sarrus’ rule. switching indices 1 (1.11? Exercise 8 From you course in linear algebra.21) Exercise 4 Write out the second and third component of the vorticity vector given in Eq. 1. The (inner) product of a symmetric and antisymmetric tensor is always zero. 1. Eq.21 we proved the relation between Ωij and ωi for the off-diagonal components.20) and we get 1 Ω23 = − ω1 2 which indeed is identical to Eq. ω2 and ω3 ). 1.19.14 ω1 = ε123 Ω32 + ε132 Ω23 = Ω32 − Ω23 = −2Ω23 (1. for example. Product of a symmetric and antisymmetric tensor 16 or.19.5. 1. 1. Show that you get the same result as in Eq. Since the indices i and j are both dummy indices we can interchange them in the last expression. • A tensor bij is antisymmetric if bij = −bji .1. we have the definitions: • A tensor aij is symmetric if aij = aji .12 (i. and then that bij = −bji (antisymmetric). where we first used the fact that aij = aji (symmetric). 1. What about the diagonal components of Ωij ? What do you get from Eq.21). Use it to compute the vector product ˆ e1 ˆ e2 ˆ e3 ∂ ∂ ∂ ω = ∇ × v = ∂x ∂x2 ∂x3 1 v1 v2 v3 Verify that this agrees with the expression in tensor notation in Eq.19 is to write out the components of Eq.15 for i = 2 and i = 3. 1. see Eq.5.6. During rotation the fluid particle is not deformed.4. In Fig. 1. Next let us have a look at the deformation caused by Sij . namely shear and elongation (also called extension or dilatation).4. dα/dt = (∂v2 /∂x1 − ∂v1 /∂x2 )/2.e. Chapt. 1 this corresponds to the equation for a straight line y = kx + ℓ where k is the slope which is equal to the derivative of y.11.e. In general. Exercise 10 Consider Fig.e. dy/dx. rotation See also [1]. the rotation is computed as the average. Now we need the velocity at the lower-right corner which is located at x1 + ∆x1 . be divided into two parts: Sij and Ωij . This can of course also be shown be writing out aij bij on component form. 1. 1. 1. 1. 1. 1.1. as shown above. The vorticity is computed as ω3 = ∂v2 /∂x1 − ∂v1 /∂x2 = −2Ω12 = 2dα/dt. The velocity gradient can. The deformation due to elongation is caused by the diagonal terms of Sij . i. The velocity at the lower-left corner is v2 (x1 ). i. We have shown that the latter is connected to rotation of a fluid particle.6. It is computed using the first term in the Taylor series as1 v2 (x1 + ∆x1 ) = v2 (x1 ) + ∆x1 ∂v2 ∂x1 It is assumed that the fluid particle in Fig. Show and formulate the deformation by S23 . and ℓ = v2 (x1 ) rotation . aij bij = a11 b11 + a12 b12 + a13 b13 I II + a21 b21 +a22 b22 + a23 b23 I III + a31 b31 + a32 b32 +a33 b33 = 0 II III The underlined terms are zero. Hence. 1. 1. Deformation. a pure shear deformation by S12 = (∂v1 /∂x2 + ∂v2 /∂x1 )/2 is shown. 1.13 and Exercise 4. i.6. The angle rotation per unit time can be estimated as dα/dt = −∂v1 /∂x2 = ∂v2 /∂x1 .4 is rotated the angle α during the time ∆t. This movement can be illustrated by Fig. Exercise 9 Consider Fig.3. Show and formulate the elongation by S22 .5.4 is estimated as follows. It can be divided into two parts. deformation and elongation as indeed is given by Eq.6 Deformation. terms I cancel each other as do terms II and III. The vertical movement (v2 ) of the lower-right corner (x1 + ∆x1 ) of the particle in Fig. Show and formulate the rotation by ω1 . Exercise 11 Consider Fig. Elongation caused by S11 = ∂v1 /∂x1 is illustrated in Fig. The deformation due to shear is caused by the off-diagonal terms of Sij . rotation 17 This expression says that A = −A which can be only true if A = 0 and hence aij bij = 0. 1. a fluid particle experiences a combination of rotation. the vorticity ω3 can be interpreted as twice the average rotation per unit time of the horizontal edge (∂v2 /∂x1 ) and vertical edge (−∂v1 /∂x2 ). if the fluid element does not rotate as a solid body. 3. rotation 18 ∆x1 − ∂v1 ∆x2 ∆t ∂x2 α ∆x2 α ∂v2 ∆x1 ∆t ∂x1 x2 x1 Figure 1. Here ∂v1 /∂x2 = ∂v2 /∂x1 so that S12 = ∂v1 /∂x2 > 0.4: Rotation of a fluid particle during time ∆t. ∆x1 ∂v1 ∆x2 ∆t ∂x2 α ∆x2 α ∂v2 ∆x1 ∆t ∂x1 x2 x1 Figure 1. Deformation. Here ∂v1 /∂x2 = −∂v2 /∂x1 so that −Ω12 = ω3 /2 = ∂v2 /∂x1 > 0.1.5: Deformation of a fluid particle by shear during time ∆t. .6. Consider a closed line on a surface in the x1 − x2 plane. ti dℓ x2 S x1 Figure 1. When the velocity is integrated along this line and projected onto the line we . The vector.7. see Section 4.7. S. One type of movement was rotation. In this subsection we will give examples of one irrotational and one rotational flow. Irrotational and rotational flow 19 ∆x1 ∂v1 ∆x1 ∆t ∂x1 ∆x2 x2 x1 Figure 1. see Fig. 1. we need to introduce the concept circulation. ti .7: The surface. is enclosing by the line ℓ. We’ll talk more about that later on. Flows are often classified based on rotation: they are rotational (ωi 6= 0) or irrotational (ωi = 0).7 Irrotational and rotational flow In the previous subsection we introduced different types of movement of a fluid particle.1. 1. the latter type is also called inviscid flow or potential flow. there exists a potential. see Fig.4. from which the velocity components can be obtained as ∂Φ (1. 1. denotes the unit tangential vector of the enclosing line.6: Deformation of a fluid particle by elongation during time ∆t. Φ.22) vk = ∂xk Before we talk about the ideal vortex line in the next section. In potential flow. ℓ.4. The lift force is computed as L = ρV Γ (1. take the curl of) and show that the right side becomes zero as it should. 1. is then obtained as vθ = To transform Eq. see Fig. The velocity field in polar coordinates reads vθ = Γ . Multiply Eq. aeronautics and windpower engineering where the lift of a 2D section of an airfoil or a rotorblade is expressed in the circulation for that 2D section. Its potential reads Γθ 2π (1. 1.4.27) where Γ is the circulation.e. 1.7.29) Φ= The velocity. Irrotational and rotational flow 20 obtain the circulation Γ= I vm tm dℓ (1.1 Ideal vortex line The ideal vortex line is an irrotational (potential) flow where the fluid moves along circular paths. Exercise 12 In potential flow ωi = εijk ∂vk /∂xj = 0.30) . 1) is the unit normal vector of the surface S.27 into Cartesian velocity components.23) Using Stokes’s theorem we can relate the circulation to the vorticity as Z Z I Z ∂vk ω3 dS ωi ni dS = ni dS = εijk Γ = vm tm dℓ = ∂xj S S S (1.24 reads in vector notation I Z Z Z Γ = v · tdℓ = (∇ × v) · ndS = ω · ndS = ω3 dS (1.24) where ni = (0.28) 1 ∂Φ Γ = r ∂θ 2πr (1. for example.26) where V is the velocity around the airfoil (for a rotorblade it is the relative velocity. 1.1. since the rotorblade is rotating). The governing equations are derived in Section 4. εijk ∂ 2 Φ/(∂xk ∂xj ) = 0.9.4. Equation 1. 2πr vr = 0 (1.22 by εijk and derivate with respect to xk (i. 0.e. vθ . i. consider Fig. The Cartesian velocity vectors are expressed as x2 x2 = −vθ 2 r (x1 + x22 )1/2 x1 x1 = vθ 2 v2 = vθ cos(θ) = vθ r (x1 + x22 )1/2 v1 = −vθ sin(θ) = −vθ (1. In an ongoing PhD project.8.7. an inviscid simulation method (based on the circulation and vorticity sources) is used to compute the aerodynamic loads for windturbine rotorblades [3].25) S S S The circulation is useful in. 1. 5 0 −0.27 we get v1 = − Γx2 . 5π/4. i.e. 2π(x21 + x22 ) (1.5 −1 −1. see Fig.e.8: Ideal vortex.5 1 1. Thus one must be very careful when using the words “vortex” and ”vorticity”. 1.34) 2 ∂x2 ∂x1 2π (x21 + x22 )2 Hence a fluid particle in an ideal vortex does deform but it does not rotate (i. 1. Since it is a two-dimensional flow (v3 = ∂/∂x3 = 0).27 for which vθ → ∞.31) To verify that this flow is a potential flow. see Fig.8. Irrotational and rotational flow 21 1. its diagonal does not rotate.e. we need to show that the vorticity. The fluid particle (i. Note that generally a vortex has vorticity. its diagonal. That the flow has no vorticity (i. except at the center where the fluid particle does rotate.7. The diagonales are shown as black dashed lines. ωi = εijk ∂vk /∂xj is zero. The locations of the fluid particle is indicated by black. The velocity derivatives are obtained as ∂v1 Γ x21 − x22 . This is true for the whole flow field. π.8). see Section 4. 1.5 0 x1 0.30 into Eq. 2π(x21 + x22 ) v2 = Γx1 .e.e. Note that the deformation is not zero. we only need to compute ω3 = ∂v2 /∂x1 − ∂v1 /∂x2 .2.32) and we get ω3 = Γ 1 (x2 − x21 + x21 − x22 ) = 0 2π (x21 + x22 )2 2 (1. 1.5 Figure 1. ω1 = ω2 = 0. π/4. vorticity .33) which shows that the flow is indeed a potential flow. As a fluid particle moves from position a to b – on its counter-clockwise-rotating path – the particle itself is not rotating. Inserting Eq. i. vortex vs. The fluid particle is shown at θ = 0. This is a singular point as is seen from Eq.4) does not rotate. 3π/2 and −π/6. filled squares. 3π/4.1.5 1 x2 0.5 −1 −0. no rotation) means that a fluid particle moves as illustrated in Fig. The ideal vortex is a very special flow case. x22 1 ∂v1 Γ ∂v2 S12 = = + (1. =− ∂x2 2π (x21 + x22 )2 ∂v2 Γ x22 − x21 = ∂x1 2π (x21 + x22 )2 (1. irrotational (ωi ≡ 0). By vortex we usually mean a recirculation region of the mean flow. It may be little confusing that the flow path forms a vortex but the flow itself has no vorticity. 1. 1. 5. It is rotating in clockwise direction. In the upper half of the channel the vorticity is positive because ∂v1 /∂x2 < 0. h h v2 = 0 where x2 < h/2.1.10.10: A shear flow. indicated by α in Fig. 1.7. ω3 . 3. The vorticity vector for this flow reads 2x2 ∂v1 4 ∂v2 1− − =− ω1 = ω2 = 0. via b to position c it is indeed rotating. 1. Note that the positive rotating direction is defined as the counter-clockwise direction. This is why the vorticity. in the lower half of the channel (x2 < h/2) is negative. .10. see Fig. The fluid particle rotates.2 Shear flow Another example – which is rotational – is the lower half of fully-developed channel flow for which the velocity reads (see Eq. 1. v1 = cx22 . v1 b a v1 c x2 a x1 x3 Figure 1.5.36) When the fluid particle is moving from position a. 1.8 Eigenvalues and and eigenvectors: physical interpretation See also [1]. ω3 = ∂x1 ∂x2 h h (1.max = x2 4x2 1− .35) (1. Eigenvalues and and eigenvectors: physical interpretation vθ 22 θ x2 r θ x1 Figure 1.28) v1 v1. 2.8. Chapt.9: Transformation of vθ into Cartesian components. Furthermore. σij .8. 2. there are no shear stresses on the surfaces of the fluid element. This is the very definition of eigenvectors.11. In the left figure it is oriented along the x1 − x2 coordinate system. 13. σ21 ). The eigenvalues are obtained from the characteristic equation. see right part of Fig. right: rotated to principal coordinate directions. 1. σ22 ) and shear stresses (σ12 . The stresses form a tensor. Since σij is symmetric.PSfrag 1. 1. not imaginary). Chapt. the eigenvectors can be computed.5 at p. i. Note that the with the eigenvectors.11. λ1 and λ2 denote eigenvalues. the eigenvalues are real (i. On the surfaces act normal stresses (σ11 .11). 144. λ1 = σ1′ 1′ and λ2 = σ2′ 2′ . the eigenvalues are the normal stresses in the principal coordinates. v ˆ1 = xˆ1′ and v sign of the eigenvectors is not defined. . the fluid element is rotated α degrees so that its edges are aligned ˆ2 = x ˆ2′ . When the eigenvalues have been obtained.11: A two-dimensional fluid element. v ˆ1 and v ˆ2 denote unit eigenvectors. see Fig. Consider a two-dimensional fluid (or solid) element. There are only normal stresses.e. Left: in original state. In the principal coordinates x1′ − x2′ (right part of Fig.e. Any tensor has eigenvectors and eigenvalues (also called principal vectors and principal values). Given the eigenvectors. 1.5.5 or Eq. which means that the eigenvectors can equally well be chosen as −ˆ v1 and/or −ˆ v2 . Eigenvalues and and eigenvectors: physical interpretation σ23 v ˆ1 v ˆ2 σ21 λ2 λ1 σ12 σ11 x2 23 x2′ x1′ x1 α Figure 1. see [1]. i = 0 Change of notation gives dρ ∂vi =0 +ρ dt ∂xi For incompressible flow (ρ = const) we get ∂vi =0 ∂xi (2. Equation 2.7) In inviscid (potential) flow. 1.2) (2. is constant it can be moved outside the derivative. In this case. This is the Navier-Stokes equations (sometimes the continuity equation is also included in the name “Navier-Stokes”). the Navier-Stokes equation reduces to the Euler equations Euler equations .7. if the flow is incompressible the second term in the parenthesis on the right side is zero because of the continuity equation.1.4 into the balance equations. µ. we get dvi 2 ∂vk ∂τji ∂P ∂ ∂P ρ 2µSij − µ + + ρfi = − + δij + ρfi (2. 5 and 8. 2. If the viscosity.4) Inserting Eq. It is also called the transport equation for momentum.1.5 can now – for constant µ and incompressible flow – be written ρ ∂ 2 vi ∂P dvi +µ + ρfi =− dt ∂xi ∂xj ∂xj (2.1 The continuity equation The first equation is the continuity equation (the balance equation for mass) which reads [2] (2. Chapts.2 The momentum equation The next equation is the momentum equation. Eq.1. Governing flow equations 24 2 Governing flow equations See also [1].5) =− dt ∂xi ∂xj ∂xi ∂xj 3 ∂xk where µ denotes the dynamic viscosity.1 The Navier-Stokes equation 2. If these two requirements are satisfied we can also re-write the first term in the parenthesis as ∂ 2 vi ∂ ∂vj ∂vi ∂ =µ (2µSij ) = µ + (2. We have formulated the constitutive law for Newtonian viscous fluids [2] 2 σij = −P δij + 2µSij − µSkk δij 3 2 τij = 2µSij − µSkk δij 3 (2. 2.2.3) 2. Furthermore.1) ρ˙ + ρvi.6) ∂xj ∂xj ∂xj ∂xi ∂xj ∂xj because of the continuity equation. there are no viscous (friction) forces. 12 is the transport equation for (internal) energy. transformation of kinetic energy into thermal energy).13) where cp is the heat capacity (see Part I) [2] so that Eq. Write out the momentum equation (without using the summation rule) for the x1 direction and show the surface forces and the volume force on a small. The first term on the right side represents reversible heating and cooling due to compression and expansion of the fluid. the velocity should be smaller than approximately 1/3 of the speed of sound) for which du = cp dT (2. Now repeat it for the x2 direction.1.4 and 2. Eqs.i = ρz (2. .7 states that mass times acceleration is equal to the sum of forces forces (per unit volume).j σji + qi.8) Exercise 13 Equation 1. 6. 2. Chapts. 2.11.4 and 8.10 gives du 2 ∂vi ∂ ∂T ρ + 2µSij Sij − µSkk Sii + = −P k (2.11) qi = −k ∂xi Inserting the constitutive laws.14) =Φ+ ρcp dt ∂xi ∂xi 2 High-speed flows relevant for aeronautics will be treated in detail in the course “TME085 Compressible flow” in the MSc programme. u. Sij . We have in Part I [2] derived the energy equation which reads ρu˙ − vi.e. Ωij .10) = σji ρ dt ∂xj ∂xi In Part I [2] we formulated the constitutive law for the heat flux vector (Fourier’s law) ∂T (2. For simplicity. Two of the viscous terms (denoted by Φ) represent irreversible viscous heating (i. 2. these terms are important at high-speed flow2 (for example re-entry from outer space) and for highly viscous flows (lubricants). square fluid element (see lecture notes of Toll & Ekh [2]). qi denotes the conductive heat flux and z the net radiative heat source. Equation 2.2 The energy equation See also [1]. is zero. we neglect the radiation from here on. The energy equation 25 ρ ∂P dvi + ρfi =− dt ∂xi (2.12 gives (cp is assumed to be constant) ∂T ∂ dT k (2. 2.2. into Eq. Now we assume that the flow is incompressible (i. Exercise 14 Formulate the Navier-Stokes equation for incompressible flow but nonconstant viscosity. Change of notation gives ∂vi ∂qi du − (2.9) where u denotes internal energy.2.e. and an anti-symmetric tensor.12) dt ∂xi 3 ∂xi ∂xi Φ where we have used Sij ∂vi /∂xj = Sij (Sij + Ωij ) = Sij Sij because the product of a symmetric tensor. 20 to Eq. k = vi vi /2. the first term on the left side can be re-written 1 d(vi vi ) dk dvi = ρ =ρ (2.15) where α = k/(ρcp ) is the thermal diffusivity. not an lubricant oil). 8. The first term is positive and the second term is negative. the dissipation.21) .15 because one of two assumptions are valid: 1. in Eq.e. not an lubricant oil).10 gives ρ ∂qi ∂σji vi d(u + k) − + ρvi fi = dt ∂xj ∂xi (2.16) where ν = µ/ρ is the kinematic viscosity. 2. Multiply Eq. 2. The fluid is a common liquid (i. Adding Eq. see Eq.18) ρvi dt 2 dt dt (vi vi /2 = k) so that ∂σji dk + ρvi fi (2.12. 2.4). we get dT ∂2T =α dt ∂xi ∂xi (2. The fluid is a gas with low velocity (lower than 1/3 of the speed of sound). Transformation of energy 26 The dissipation term is simplified to Φ = 2µSij Sij because Sii = ∂vi /∂xi = 0.20) This is the transport equation for kinetic energy.e. The Prandtl number is defined as Pr = ν α thermal diffusivity (2. 2. 8.2) ρ ∂vi ∂vi σji dk − σji + ρvi fi = dt ∂xj ∂xj (2. k. The physical meaning of the Prandtl number is the ratio of how well the fluid diffuses momentum to how well it diffuses internal energy (i. see Eq. 1. Φ) is large.e.3 Transformation of energy Now we will derive the equation for the kinetic energy. are you sure that Φ > 0? 2.19) = vi ρ dt ∂xj Re-write the stress-velocity term so that (Trick 1. Exercise 15 Write out and simplify the dissipation term.e.3. One example is the oil flow in a gearbox in a car where the temperature usually is more than 100oC higher when the car is running compared to when it is idle. If we furthermore assume that the heat conductivity coefficient is constant and that the fluid is a gas or a common liquid (i.7 with vi dvi ∂σji ρvi − vi ρfi = 0 (2. temperature). Φ.17) − vi dt ∂xj Using the product rule backwards (Trick 2. this assumption was made when we assumed that the fluid is incompressible 2. is neglected in Eq. The dissipation term.2. Φ. In lubricant oils the viscous heating (i. 20 and 2. It is always positive and represents irreversible heating. • The physical meaning of the a-term in Eq. see Section 5. u + k.).10 and 2. Dissipation is very important in turbulence where transfer of energy takes place at several levels.22 – which includes the viscous stress tensor.21. 2.10. Left side of the transport equations 27 This is an equation for the sum of internal and kinetic energy.e. but this is much smaller than that from the turbulence kinetic energy. Let us take a closer look at Eqs. 2. Eq. σji ∂vi ∂vi ∂vi + τji = −P ∂xj ∂xi ∂xj a (2. 1. d/dt. ψ = vi . and is called viscous dissipation. 2. 2. .10). 2. u (Eq.2. This is a reversible process. The physical process is called production of turbulent kinetic energy. k (Eq. the left sides in transport equations have been formulated using the material derivative.23) This is often called the non-conservative form. thermal energy.22 – which includes the pressure. it can be re-written as ∂ψ ∂ψ ∂vj dρ dψ +ψ =ρ + ρvj +ρ = ρ dt ∂t ∂xj dt ∂xj (2. 2. • The dissipation. • The physical meaning of the b-term in Eq. 2. appears as a sink term in the equation for the kinetic energy.12. Using the continuity equation.4 Left side of the transport equations So far.24) =0 ∂ψ ∂ρ ∂vj ∂ρ ∂ψ + ρvj + vj +ψ +ρ ρ ∂t ∂xj ∂t ∂xj ∂xj nonconservative .e. • Φ does not appear in the equation for the total energy u+k (Eq. 2.22) b=Φ The following things should be noted. First we separate the term σji ∂vi /∂xj in Eqs. which means that kinetic energy is transformed to thermal energy. At the same time we have the usual viscous dissipation from the mean flow to thermal energy. this is turbulence dissipation (or heating).e. temperature etc reads ρ ∂ψ ∂ψ dψ =ρ + ρvj dt ∂t ∂xj (2. 2. see Eq. .20 into work related to the pressure and viscous stresses respectively (see Eq. 2.2. i. the left side of the equation for momentum. T . It is denoted Φ.20) and it appears as a source term in the equation for the internal energy. 2. Then we have transformation of kinetic energy from turbulence kinetic energy to thermal energy. i. no loss of energy but only transformation of energy. Φ. First energy is transferred from the mean flow to the turbulent fluctuations.21).4. u. u + k. P – is heating/cooling by compression/expansion. Let ψ denote a transported quantity (i. u + k. τij – is a dissipation. total energy. this makes sense since Φ represents a energy transfer between u and k and does not affect their sum. This is the transport equation for total energy. For more detail. The transformation of kinetic energy into internal energy takes place through this source term.9). 5 Material particle vs.20) are all given for a fixed control volume.27. for example. the Navier-Stokes equation 2. can be written in three different ways (by use of the chain-rule and the continuity equation) dvi ∂vi ∂ρvi ∂vi ∂ρvj vi = =ρ + ρvj + dt ∂t ∂xj ∂t ∂xj dT ∂T ∂ρvi ∂T ∂ρvj T ρ = =ρ + ρvj + dt ∂t ∂xj ∂t ∂xj ρ (2. 2.26) The continuity equation can also be written in three ways (by use of the chain-rule) ∂ρ ∂ρ ∂vi ∂ρ ∂ρvi ∂vi dρ = +ρ = +ρ + vi + dt ∂xi ∂t ∂xi ∂xi ∂t ∂xi (2. momentum. control volume (Reynolds Transport Theorem) See also lecture notes of Toll & Ekh [2] and [1]. Chapt.28) = dV = dV +Φ + vi + ∂t ∂xi ∂xi ∂t ∂xi V V Z Z ∂Φ = vi ni ΦdS dV + V ∂t S conservative .25) Thus. Vpart . otherwise particles would move across its boundaries. control volume (Reynolds Transport Theorem) 28 The two underlined terms will form a time derivative term. This means that the volume. 2. V .e.27) The forms on the right sides of Eqs. must be moving and it may expand or contract (if the density is non-constant). where Vpart is a volume that includes the same fluid particles all the time. How come? The answer is the Reynolds transport theorem. When solving transport equations (such as the Navier-Stokes) numerically using finite volume methods. d/dt ρdV = 0. the left side of the temperature equation and the Navier-Stokes. energy etc is lost (provided a transport equation for the quantity is solved): “what comes in goes out”. but no mass. ρ ∂ρψ ∂ρvj ψ dψ = + dt ∂t ∂xj (2. The conservation of mass. R Newton’s second law. the energy equations 2. 2.12 and 2. Vpart . The Reynolds transport theorem reads (first line) Z Z dΦ ∂vi d dV ΦdV = +Φ dt Vpart dt ∂xi V Z Z ∂Φ ∂Φ ∂vi Φ ∂Φ ∂vi (2. deformable volume with a collection of particles. The equations we have looked at so far (the continuity equation 2.2.27 are called the conservative form. and the other three terms can be collected into a convective term.26 and 2. The results may be inaccurate due to too coarse a numerical grid. 5. Material particle vs. d/dt ρvi = Fi etc were derived for a collection of particles in the volume Vpart .5.7. to being valid for a fixed volume.2.26 and 2. momentum R and energy) for a collection of material particles. in this way Gauss law can be used to transform the equations from a volume integral to a surface integral and thus ensuring that the transported quantities are conserved. the left sides in the transport equation are always written as the expressions on the right side of Eqs. i.3. In Part I [2] we initially derived all balance equations (mass. which converts the equations from being valid for a moving. the latter is usually fixed (this is not necessary). The divergence theorem was used to obtain the last line and S denotes the bounding surface of volume V . V . on the first line represents the increase or decrease of Vpart during dt. This equation applies to any volume at every instant and the restriction to a collection of a material particles is no longer necessary. ∂vi /∂xi . The divergence of the velocity vector.5. The last term on the last line represents the net flow of Φ across the fixed non-deformable volume. Φ in the equation above can be ρ (mass). ρvi (momentum) or ρu (energy). Material particle vs. 2.10. . in fluid mechanics the transport equations (Eqs.5. . .2. . control volume (Reynolds Transport Theorem) 29 where V denotes a fixed non-deformable volume in space. 2. Hence. ) are valid both for a material collection of particles as well as for a volume. 2.2. 1 is shown in Fig. the moving plate affects the fluid further and further away from the plate.6 0. At the lower boundary (x2 = 0) and at the upper boundary (x2 → ∞) the velocity component v2 = 0. Eq.1: The plate moves to the right with speed V0 for t > 0. So. there is no x1 dependency. 3. Furthermore. v1 (x2 → ∞. this means that the number of independent variables is reduced by one.8 1 v1 /V0 Figure 3. which means that v2 = 0 in the entire domain. The similarity variable. 2. For time greater than zero the plate is moving with the speed V0 . in this case from two (x2 and t) to one (η). Because the plate is infinitely long.1 are v1 (x2 .2) The solution to Eq.7 gives (no body forces. η. µ.1 is a similarity solution. t). f1 = 0) for the v1 velocity component ∂v1 ∂ 2 v1 ρ (3.3) similarity solution .e. 3 Solutions to the Navier-Stokes equation: three examples 3. 3. It turns out that the solution to Eq.3. Hence the flow depends only on x2 and t. i.2. t) = V0 . 3. t = 0) = 0. t) = 0 (3. Solutions to the Navier-Stokes equation: three examples 30 x2 V0 x1 Figure 3. i. ν = µ/ρ. For increasing time (t3 > t2 > t1 ). 3.1 The Rayleigh problem Imagine the sudden motion of an infinitely long flat plate.1.2: The v1 velocity at three different times.2 0. v1 = v1 (x2 .e. 3. rather than the dynamic one. ∂v1 /∂x1 = ∂v3 /∂x3 = 0 so that the continuity equation gives ∂v2 /∂x2 = 0. t3 > t2 > t1 . The boundary conditions for Eq.1) =µ 2 ∂t ∂x2 We will find that the diffusion process depends on the kinematic viscosity. is related to x2 and t as x2 η= √ 2 νt (3.4 0. see Fig. t) and p = p(x2 . v1 (x2 = 0. x2 t3 t2 t1 0 0. we get v1 (x2 .3.7) we conclude that the transformation is suitable.8) Integration a second time gives f = C1 Z η exp(−η ′2 )dη ′ + C2 (3. 3. The Rayleigh problem 31 If the solution of Eq. 3.6) We have now successfully transformed Eq. 3.1) and the boundary conditions (Eq. Eq.1 from ∂/∂t and ∂/∂x2 to d/dη so that it becomes a function of η only. the final solution to √ Eq.5 in Eq.1 gives df d2 f + 2η =0 dη 2 dη (3. it means that the solution for a given fluid will√ be the same (“similar”) for many (infinite) values of x2 and t as long as the ratio x2 / νt is constant. 3.5) Inserting Eqs. 3.10) 0 At the limits. Now let us find out if the boundary conditions.9) 0 The integral above is the error function 2 erf(η) ≡ √ π Z η exp(−η ′2 )dη ′ (3.1 depends only on η. i. t = 0) = 0 ⇒ f (η → ∞) = 0 v1 (x2 = 0.4) We introduce a non-dimensional velocity f= v1 V0 (3. We get ∂v1 dv1 ∂η x2 t−3/2 dv1 1 η dv1 = =− √ =− ∂t dη ∂t 4 ν dη 2 t dη dv1 ∂η 1 dv1 ∂v1 = = √ ∂x2 dη ∂x2 2 νt dη 1 1 d2 v1 ∂ ∂ ∂v1 1 dv1 dv1 ∂ 2 v1 ∂ √ √ = = = = 2 ∂x2 ∂x2 ∂x2 ∂x2 2 νt dη 4νt dη 2 2 νt ∂x2 dη (3. t) = 0 ⇒ f (η → ∞) = 0 Since we managed to transform both the equation (Eq.1 and reduced the number of independent variables from two to one. 3. erf(0) = 0 and erf(η → ∞) = 1.11) . 3. also can be transformed in a physically meaningful way. t) = V0 ⇒ f (η = 0) = 1 (3. Eq. Taking into account the boundary conditions.4 and 3. the error function takes the values 0 and 1. Now let us solve Eq. Integration once gives df = C1 exp(−η 2 ) dη (3.7) v1 (x2 → ∞.7. Now we need to transform the derivatives in Eq. 3. 3.6. 3. 3.2.e.9 is (with C2 = 1 and C1 = −2/ π) f (η) = 1 − erf(η) (3.1. the thickness of the boundary layer.12 at time t = 100 000.e. The solution is presented in Fig.1. v1 .5 0 0 0. 3. numerically using central differences (fj+1 − fj−1 )/(ηj+1 − ηj−1 ). Compare this figure with Fig. ω3 .4 presents the shear stress. 3.12) where we used ν = µ/ρ. 3 2.4. given by Eq. f = v1 /V0 . 3. Often ω3 = − . the magnitude of√the shear stress increases for decreasing η and it is largest at the wall.5 1 0. This is how the graphs in Fig. The Rayleigh problem 32 3 2. 3. is computed from Eq. 30 it is seen that for large times.3.2 were obtained.3: The velocity.5 2 η 1. To compute the velocity. v1 . 3. Then f is obtained from Eq. τw = −ρV0 / πt The vorticity. From the velocity profile we can get the shear stress as τ21 = µ ∂v1 µV0 df µV0 exp −η 2 = √ = −√ ∂x2 2 νt dη πνt (3. 3. 3. we pick a time t and insert x2 and t in Eq. 3.4: The shear stress for water (ν = 10−6 ) obtained from Eq.36) V0 df V0 ∂v1 =− √ =√ exp(−η 2 ) (3. all graphs in that figure collapse into one graph in Fig. As can be seen from Fig.8 1 f Figure 3.2 at p. df /dη.11. 3. The solid line is obtained from Eq.13) ∂x2 dη 2 νt πνt From Fig.2 at p.3. δ.4 0. 3.5 1 0.3.5 0 −6 −5 −4 −3 −2 −1 0 τ21 /(ρV0 ) Figure 3. across the boundary layer is computed from its definition (Eq. 3.5.2 0. increases. 30. τ21 . the moving plate is felt further and further out in the flow.11 and the velocity.5 2 η 1.3. 3. i. Figure 3.12 and circles are obtained by evaluating the derivative. 1.6 0. 3. 3. see Fig.8cm (3. at this point η ≃ 1. say. incompressible flow in a two-dimensional channel. In our case.3 gives δ η = 1. 3.4. The reason is that the flow (with velocity V ) following the curved wall must change its direction. Inserting x2 = δ in Eq.3. Choose suitable values on t1 .3.5. the boundary layer thickness is defined by the position where the local velocity. see Section 4. see Fig. µ = const).2.01 in Fig. t2 and t3 . Exercise 16 Consider the graphs in Fig. ν = µ/ρ rather than by the dynamic one.5: Flow in a horizontal channel. 3. Exercise 17 Consider the graphs in Fig. The inlet part of the channel is shown. the velocity near the walls is larger than in the center. 10 minutes the diffusion length for air and water. The diffusion length can also be used to estimate the thickness of a developing boundary layer.14) It can be seen that the boundary layer thickness increases with t1/2 . 3.3. Find the point f = v1 /V0 = 0.6 νt (3. µ. 3. Exercise 18 Repeat the exercise above for the shear stress.1 Curved plates Provided that the walls at the inlet are well curved.15) As mentioned in the beginning of this section. 3. Note that no scale is used on the x2 axis and that no numbers are given for t1 .1. respectively.5. with constant physical properties (i.2 Flow between two plates Consider steady.2. Create this graph with Matlab for both air and engine oil. v1 (x2 ). Flow between two plates 33 V P1 h P2 P1 V x2 x1 Figure 3.3.11). Equation 3. the diffusion length is determined by the kinematic viscosity. this corresponds to the point where v1 = 0.01V0 . τ21 .2.8 = √ 2 νt √ ⇒ δ = 3.e. 3. 3. After.8cm δwater = 2.8 (we can also use Eq. Create this graph with Matlab. 3. t2 and t3 . reaches 99% of the freestream velocity. are δair = 10.14 can also be used to estimate the diffusion length. see Fig. The physical agent which accomplish this diffusion length . 16) where V denotes the bulk (i. 3. Taking these restrictions into account the continuity equation can be . is higher than the pressure near the wall. ∂v1 /∂x1 = ∂v2 /∂x1 = 0). x1 (i. and Dh = 4A/Sp where Dh .6. it is easier for the flow to sneak along the inner wall where the opposing pressure is smaller than near the outer wall: the result is a higher velocity near the inner wall than near the outer wall. The acceleration and retardation of the flow in the inlet region is “paid for ” by a pressure loss which is rather high in the inlet region. Let us find the governing equations for the fully developed flow region. ∂/∂x3 ).e. x3 (i. and the velocity in this direction is zero.2 Flat plates The flow in the inlet section (Fig. P2 . In a threedimensional duct or in a pipe.5) is two dimensional. in order to force the flow to turn. v3 = 0. if a separation occurs because of sharp corners at the inlet. Hence the pressure in the center of the channel. and the entrance length can. Flow between two plates 34 P2 P1 x2 V x1 Figure 3.2.e V Dh = 0. less opposing pressure) for the fluid to enter the channel near the walls than in the center.e. For large x1 the flow will be fully developed. The same phenomenon occurs in a channel bend. The flow is accelerated in the center because the mass flow at each x1 must be constant because of continuity.016ReDh ≡ 0. Near the inlet the velocity is largest near the wall and further downstream the velocity is retarded near the walls due to the large viscous shear stresses there.6: Flow in a channel bend. the mean) velocity. The flow V approaches the bend and the flow feels that it is approaching a bend through an increased pressure. a swirling motion in the x2 − x3 plane). the region until this occurs is called the entrance region. P1 . is the pressure gradient which forces the flow to follow the wall as closely as possible (if the wall is not sufficiently curved a separation will take place). Since the flow is two-dimensional. This explains the high velocity near the walls.2. the cross-sectional area and the perimeter. Hence. For flow between two plates we get Dh = 2h.e.e. respectively. must be higher than that near the inner wall.016 Dh ν (3. A and Sp denote the hydraulic diameter. for moderately disturbed inflow. i. the pressure difference P2 − P1 creates secondary flow downstream the bend (i.e. in this region the flow does not change with respect to the streamwise coordinate. It is thus easier (i.3. P2 . 3. P1 . be estimated as [4] x1. the pressure loss will be even higher.e. The pressure near the outer wall. see Fig. 3. it does not depend on the third coordinate direction. 21) Integration gives where the integration “constant” C1 may be a function of x1 but not of x2 .5).24) µ 2 = ∂x2 dx1 Since the left side is a function of x2 and the right side is a function of x1 .e. fi = (0. 3. This pressure is zero when the flow is static. 3. when the velocity field is zero. when you want the physical pressure. Integrating Eq.17) Integration gives v2 = C1 and since v2 = 0 at the walls. Since v2 = ∂v1 /∂x1 = 0 the left side is zero so dp ∂ 2 v1 (3. 3. If we denote the pressure at the lower wall (i.5. P .24 is independent of x1 and x2 ) .22.22) Hence the pressure. This is the vertical direction (x2 is positive upwards. Now let us turn our attention to the momentum equation for v2 . at x2 = 0) as p we get P = −ρgx2 + p(x1 ) (3. i. p. i. the ρgx2 as well as the surrounding atmospheric pressure must be added.e. 3. Eq.18) across the entire channel (recall that we are dealing with the part of the channel where the flow is fully developed. The momentum equation can be written (see Eq. see Fig.5). is used in incompressible flow.25 gives h dp x2 v1 = − (3. However.19) ≡ ρv1 dt ∂x1 ∂x2 ∂x2 ∂x2 Since v2 = 0 we get ∂P = −ρg ∂x2 (3. see Fig. in the inlet section v2 6= 0. Flow between two plates 35 simplified as (see Eq. 3. The velocity.25) where h denotes the height of the channel.20) P = −ρgx2 + C1 (x1 ) (3.2. This agrees with our experience that the pressure decreases at high altitudes in the atmosphere and increases the deeper we dive into the sea.e v1 (0) = v1 (h) = 0 (3. 3. The gravity acts in the negative x2 direction. see Fig. Usually the hydrodynamic pressure. 3.24 twice and using Eq.23) where P was replaced by p using Eq.e. −ρg. 24) dv2 ∂v2 ∂v2 ∂P ∂ 2 v2 ρ + ρv2 =− + µ 2 − ρg (3. decreases with vertical height. 0).26) x2 1 − 2µ dx1 h hydrodynamic pressure .3) ∂v2 =0 ∂x2 (3. i. 2. We can now formulate the momentum equation in the streamwise direction ρ ∂v1 ∂v1 dp ∂ 2 v1 dv1 + ρv2 =− +µ 2 ≡ ρv1 dt ∂x1 ∂x2 dx1 ∂x2 (3.e. v1 . is zero at the walls. we conclude that they both are equal to a constant (i. it means that v2 = 0 (3.7 at p. 2.3. 8 0.e. i. The minus sign on the right side appears because the pressure is decreasing for increasing x1 . Flow between two plates 36 1 0.7 which reads for i=1 dv1 ∂σj1 ρ (3.7 Since we know the velocity profile. 3. 3. 3.4 0.31) = dt ∂xj .28.4 0.27) 2µ L 2 2 8µ L We often write Eq.30) τw = µ ∂x2 2 dx1 2 L Actually.7: The velocity profile in fully developed channel flow.mean = h h 3 0 The velocity profile is shown in Fig. 3.2. the pressure is driving the flow.29) 4x2 1 − v1.3. for x2 = h/2. and reads h ∆p h 1 h2 ∆p v1. The negative pressure gradient is constant (see Eq. Eq. this result could have been obtained by simply taking a force balance of a slice of the flow far downstream.24) and can be written as −dp/dx1 = ∆p/L.max Figure 3.6 0. 1.26 on the form v1 v1. Z 2 x2 v1.2 0 0 0. we can compute the wall shear stress. i.28 across the channel.28) The mean velocity (often called the bulk velocity) is obtained by integrating Eq.max (3.e.2 0.max h dx2 = v1.8 1 v1 /v1. we start with Eq. 3. Equation 3.6 x2 /h 0.max = x2 4x2 1− h h (3. This flow is analyzed in Appendix C.max = 1− = (3.2. 3.26 gives h dp h ∆p ∂v1 =− = (3.3 Force balance To formulate a force balance in the x1 direction. The velocity takes its maximum in the center. left and right) so that Z Z Z Z σj1 nj dS (3.2. 0). 0).22 give (the gravitation term on the left and right surface cancels and p and τw are constants and can thus be taken out in front of the integration) 0 = p1 W h − p2 W h − 2τw LW (3. 3. 0.U p1 p2 V walls h x2 τw.L x1 Figure 3. [2] Now use Gauss divergence theorem Z Z ∂σj1 0= σj1 nj dS (3.31 over the volume of a slice (length L).30. Inserting the normal vectors and using Eq.4 Balance equation for the kinetic energy In this subsection we will use the equation for kinetic energy. Hence we integrate Eq. 0).2. 2. Let us integrate this equation in the same way as we did for the force balance.36) where W is the width (in x3 direction) of the two plates (for convenience we set W = 1).8 Z ∂σj1 0= dV (3.33) dV = V ∂xj S The bounding surface consists in our case of four surfaces (lower. The shear stress at the upper and lower surfaces have opposite sign because τw = µ(∂v1 /∂x2 )lower = −µ(∂v1 /∂x2 )upper . see Fig.lower = (0. ni. ni. The left side of Eq. left and right are ni.right = (1. 0.20. Flow between two plates 37 L τw. 1. Forces act on a volume and its bounding surface.34) σj1 nj dS + σj1 nj dS + σj1 nj dS + 0= Slef t Supper Slower Sright The normal vector on the lower. upper. 3.20 .lef t = (−1. 1.3. upper.9 give Z Z Z Z τ21 dS τ21 dS + (−p + τ11 )dS − (−p + τ11 )dS + 0=− Slef t Sright Slower Supper (3. −1. Using this and Eq. Eq. With ∆p = p1 − p2 we get Eq.32) V ∂xj Recall that this is the form on which we originally derived the momentum balance (Newton’s second law) in Part I. The left hand side is zero since the flow is fully developed. 3. 2. 3.8: Force balance of the flow between two plates. ni.35) τ11 = 0 because ∂v1 /∂x1 = 0 (fully developed flow).upper = (0. 0). 3. 8 it is nj = (1. two-dimensional.mean V 3.9 gives 0= ∂vi ∂vi σji − σji + ρvi fi ∂xj ∂xj =0 ∂vj p ∂vi τji ∂vi ∂vi =− + + pδij − τji ∂xj ∂xj ∂xj ∂xj (3. −1. respectively) v1 ∂v1 ∂v1 ∂ 2 v1 + v2 =ν ∂x1 ∂x2 ∂x22 ∂p =0 ∂x2 ∂v1 ∂v2 + =0 ∂x1 ∂x2 (3. Finally we get Z 1 ΦdV (3. The second term is zero on all four surfaces and the first term is zero on the lower and upper surfaces (see Exercises below). v2 = 0 (far from the wall) (3. For example.22 (term b). using Eq.39) V S where S is the surface bounding the volume. 3.∞ . We replace the pressure P with p using Eq.8. The last term corresponds to the viscous dissipation term. Now we apply Eq.mean W h because ρgx2 n1 v1 on the left and right surfaces cancels. loss due to friction).39 to the fluid enclosed by the flat plates in Fig. 2.3. p can be taken out of the integral as it does not depend on x2 .e.3.22 so that Z Z v1 n1 dS (−pv1 + ρgx2 v1 )n1 dS = −(p2 − p1 ) Slef t &Sright Slef t &Sright = ∆pv1.41) where the pressure gradient is omitted in the v1 momentum equation because ∂p/∂x = 0 along a flat plate in infinite surroundings. on the right surface in Fig. 0. The boundary conditions are x2 = 0 : v1 = v2 = 0 (at the wall) x2 → ∞ : v1 → V1. The unit normal vector is denoted by nj which points out from the volume. see Eq. Now we integrate the equation over a volume Z ∂pvj ∂τji vi 0= − + − Φ dV (3. 3.40) ∆p = W hv1. Two-dimensional boundary layer flow over flat plate 38 is zero because we assume that the flow is fully developed.37) Φ On the first line vi fi = v1 f1 + v2 f2 = 0 because v2 = f1 = 0. 3.38) ∂xj ∂xj V Gauss divergence theorem on the two first terms gives Z Z ΦdV 0 = (−pvj + τji vi )nj dS − (3. The third term on the second line pδij ∂vi /∂xj = p∂vi /∂xi = 0 because of continuity.3 Two-dimensional boundary layer flow over flat plate The equations for steady. Φ (i. 0). 3. 0) and on the lower surface it is nj = (0. incompressible boundary layer flow reads (x1 and x2 denote streamwise and wall-normal coordinates.42) . 1. Ψ = (νV1. i.43 into the streamwise momentum equation gives ∂Ψ ∂ 2 Ψ ∂Ψ ∂ 2 Ψ ∂3Ψ − =ν 3 2 ∂x2 ∂x1 ∂x2 ∂x1 ∂x2 ∂x2 (3.∞ = = ∂x2 νx1 x2 (3.1 we replaced x1 and t with the new non-dimensional variable η.44) Inserting Eq.∞ g ′ ∂x2 ∂x2 ∂x2 (3.1 we want to transform the partial differential equation. 3.45 as derivatives of g.∞ x1 )1/2 g + (νV1. g(ξ).43 into the continuity equation ∂v1 ∂v2 ∂2Ψ ∂2Ψ + = − =0 ∂x1 ∂x2 ∂x1 ∂x2 ∂x2 ∂x1 (3.43) With the velocity field expressed in Ψ.45) The boundary conditions for the streamfunction read ∂Ψ = 0 (at the wall) ∂x2 ∂Ψ x2 → ∞ : → V1. It is defined as v1 = ∂Ψ .46) As in Section 3.∞ x1 ) g ′ (νV1.∞ x1 ) g ′ 2 x1 2x1 1/2 1 νV1. 3. Two-dimensional boundary layer flow over flat plate 39 Let’s introduce the streamfunction Ψ.∞ x1 ) = ∂x1 ∂x1 ∂x1 1/2 1 νV1. as 1/2 V1.∞ = (g − ξg ′ ) 2 x1 ∂ξ ′ ∂ ∂Ψ (νV1.e.45.3.47) x2 . Eq. 3. (g ′ denotes dg/dξ) ∂ξ ∂ ∂Ψ 1/2 1/2 g + (νV1.∞ (far from the wall) ∂x2 x2 = 0 : Ψ = (3.∞ 1/2 (3.3. the continuity equations is automatically satisfied which is easily shown by inserting Eq.∞ x1 ) g ξ= νx1 First we need the derivatives ∂ξ/∂x1 and ∂ξ/∂x2 1/2 ∂ξ 1 V1. 3. which is useful when re-writing the twodimensional Navier-Stokes equations. At the same time we define a new dimensionless streamfunction.∞ ξ 1/2 = g − (νV1. into an ordinary differential equation. Now we want to replace x1 and x2 with a new dimensionless variable. In Section 3.49) streamfunction . say ξ.∞ ξ x2 =− =− ∂x1 2 νx1 x1 2x1 1/2 ∂ξ ξ V1.∞ x1 )1/2 = g = V1.48) Now we express the first derivatives of Ψ in Eq. ∂x2 v2 = − ∂Ψ ∂x1 (3. ∞ 1. 3.52) so that 1 ′′ gg + g ′′′ = 0 (3.50) 1.∞ νx1 1/2 g ′′ (3. for the fully developed flow. compute the dissipation.8 (assume fully developed flow).∞ g ′′ =− V1.∞ g ′′ ∂x1 ∂x2 ∂x1 2x1 Inserting Eqs. 3. Exercise 19 For the fully developed flow.∞ ∂x32 νx1 ∂x2 νx1 x2 ∂2Ψ ∂ξ ξ = V1.8. Two-dimensional boundary layer flow over flat plate 40 The second and third derivatives of Ψ read 1/2 V1.50 into Eq. compute the vorticity.∞ g ′′ = V1.28). ωi . Exercise 22 Using the exact solution.∞ ξ ∂2Ψ ′′ ∂ξ = V1. The numerical solution is given in Table 3. .51) 2 Divide by V1. 3. Is it the same as that obtained from the force balance (if not. it should be!). compute the pressure drop.1.39 are zero on the upper and the lower surfaces in Fig.∞ g V1.∞ g = V1. 6].∞ g ′′ − 2x1 ′ 1 2 νV1.∞ x1 1/2 ′ ! (g − ξg ) V1.∞ ′′′ ′′′ ∂ξ = V = V g = V g g ′′′ (3. using the exact solution (Eq.39 is zero also on the left and right surfaces in Fig. 3. 3.∞ g ′′ ∂x22 ∂x2 νx1 x2 2 1/2 ∂3Ψ V1. 3. Φ. Exercise 23 From the dissipation.49 and 3.3.∞ 2 V1.∞ and multiply by x1 gives 1 ξ −g ′ g ′′ − (g − ξg ′ ) g ′′ = g ′′′ 2 2 (3. find the error. Exercise 21 Show that the second term in Eq.45 gives ξ −V1. 3.∞ ξ V1. Exercise 20 Show that the first and second terms in Eq.3.∞ g ′′′ =ν νx1 V1. The flow is analyzed in Appendix B.∞ 1.53) 2 This equation was derived (and solved numerically!) by Blasius in his PhD thesis 1907 [5. 1: Blasius numerical solution of laminar flow along a flat plate.064546268E-01 1.4 1.998638190E-01 9.550170691E-03 9.6 0.4 6.178762461E-01 9.6 2.230977302E+00 1.423412109E-02 5.2 3.680919063E+00 3.655717258E-01 2.467978611E-06 3.229815738E-01 4.829310173E-01 2.974777682E-0 9.999995242E-01 9.431957680E-03 1 3.648413667E-03 2.6 8.8 6.411179967E-01 9.8 1.2 0.6 5.079259772E+00 5.999754577E-01 9.647091387E-01 3.915419002E-01 9.279238811E+00 5.999997695E-01 41 g′′ 3.989728724E-01 9.669570738E-01 9.6 7.8 2.692360938E+00 2.811933370E-01 9.966634615E-01 2.203207655E-01 5.072505977E+00 1.999962745E-01 9.999990369E-01 9.279213431E+00 6.498039663E+00 2.295180377E-01 6.826835008E-01 9.224092624E-05 6.285698072E-04 7.555182298E-01 9.305746418E+00 2.996117017E-01 9.961553040E-0 9.4 0.3.989372524E-01 2.230071167E-01 3.233296659E-01 9.349939753E-06 1.999216041E-01 9.479212887E+00 6.297800312E-01 3.6 4.085320655E+00 3.700667989E-06 8.590679869E-02 1.2 5.391280556E-01 1.0 8.877895262E-01 9.760814552E-01 9.4 2.396808231E+00 1.481867612E+00 3.115096232E-01 8.297657365E-01 6.314698442E-01 3.165891911E-01 3.283273665E+00 3.4 8.8 7.079881939E+00 4.2 7.2 8.051974749E-02 3.997678702E-01 9.3.222901256E-01 1.279620923E+00 4.129031111E-05 2.0 7.2 2.813103772E-01 7.747581439E-01 6.483509132E-01 2.679220147E+00 5.888247990E+00 3.8 g 0.500243699E-01 7.679356615E+00 4.000000000E+00 6.569094960E+00 1.300791276E-01 3.973463750E-02 1.640779210E-02 1.927659815E-03 1 5.724550211E-01 8.993625417E-01 9.201689553E-04 1.134178897E-02 1 7.402039844E-03 1.017612214E-01 9.948377201E-02 2. .289819351E-01 7.270775140E-05 1.640999715E-03 2.8 8.808627878E-02 8.6 3.758708321E-01 9.937761044E-01 4.080442648E-04 3.320573362E-01 3.562617647E-01 5.079212403E+00 g′ 0.012591814E-02 6.479226847E+00 5.746950094E+00 1.679212609E+00 6.879212471E+00 7.0 3.983754937E-0 9.2 1.460444437E-01 8.929525170E+00 2.327641608E-01 1.4 3.0 1.319838371E-01 3.897261085E-02 2.2 6.999557173E-01 9.8 4.000000000E+00 6.8 3.6 1.061082208E-01 1.116029817E+00 2.879216466E+00 6.079214481E+00 6.187118635E-02 1.4 7.4 4.999866551E-01 9.379487173E-01 3.655988402E-02 5.879295811E+00 5.479457297E+00 4.273892701E-01 3.880290678E+00 4.6 6. Two-dimensional boundary layer flow over flat plate ξ 0 0.0 6.0 2.167567844E-01 5.999980875E-01 9.667515457E-01 2.359298339E-05 4.078653918E-01 2.280917607E-01 2.942455354E-01 9.806151170E-04 6.840065939E-01 1.0 4.695625701E-04 2.999928812E-01 9.8 5.4 5.462841214E-07 Table 3.2 4.0 5.613603195E-01 1. 2. x2 ) P (x1 − 0.4.5∆x1 )n1 (x1 . Note that the v2 momentum equation (see Eqs.5∆x1 ) v1 (x2 ) P (x1 + 0. vorticity is connected to rotation of a fluid particle.6) ∂τji ∂ 2 vi =µ ∂xj ∂xj ∂xj (4. Vorticity equation and potential flow 42 τ21 (x2 + 0. the pressure acts through the center of the fluid particle and is thus not able to affect rotation of the fluid particle. The v1 velocity is given by Eq. 4 Vorticity equation and potential flow 4.1 are in balance.5∆x2 )n2 x1 Figure 4. 18.5∆x2 )n2 τ12 (x1 + 0.1 shows the surface forces acting on a fluid particle in a shear flow. The fluid particle is located in the lower half of fully developed channel flow.4. 15. ωi . 1.5∆x1 ) τ12 (x1 − 0.5 and 2.12 at p. Looking at Fig.1 it is obvious that only the shear stresses are able to rotate the fluid particle. 2. 4. The v1 momentum equation requires that the horizontal viscous stresses balance the pressure difference. Hence τ11 = τ22 = ∂τ12 /∂x1 = 0 and ∂τ21 /∂x2 > 0.4 at p. The v1 velocity field is indicated by dashed vectors.3) . Figure 4.28 and v2 = 0.1) The right side can be re-written using the tensor identity 2 ∂ 2 vj ∂ 2 vk ∂ 2 vj ∂ 2 vi ∂ vj ∂ 2 vi = = − − − εinm εmjk ∂xj ∂xj ∂xj ∂xi ∂xj ∂xi ∂xj ∂xj ∂xj ∂xi ∂xj ∂xn (4.1 Vorticity and rotation Vorticity. In incompressible flow the viscous terms read (see Eqs. 3. 2. As shown in Fig. Let us have a look at the momentum equations in order to show that the viscous terms indeed can be formulated with the vorticity vector.1: Surface forces acting on a fluid particle.2) Let’s verify that ∂ 2 vi ∂ 2 vj − ∂xj ∂xi ∂xj ∂xj = εinm εmjk ∂ 2 vk ∂xj ∂xn (4.4 and 3.5∆x1 )n1 x2 τ21 (x2 − 0. ωi . 1. 4. was introduced in Eq.32) requires that the vertical viscous stresses in Fig. 4. 4. 4. this means that the vorticity can not be created or destroyed in inviscid (friction-less) flow 2. i. Hence.5) At the right side we recognize the vorticity. potential .9 for i = 1 and verify that it is satisfied.1 reads ∂τji ∂ωm = −µεinm ∂xj ∂xn (4.e. ωm = εmjk ∂vk /∂xj .9) Thus. An imbalance in shear stresses (left side of Eq.6) where the first on the right side is zero because of continuity.7.3. The viscous terms are responsible for creating vorticity.1 at p.9). no rotation) or potential flow. unaffected. (The exception is when vorticity is transported into an inviscid region.) Inviscid flow is often called irrotational flow (i.7 using the ε-δ-identity.4) The first term on the right side is zero because of continuity and hence we find that Eq.8) Using Eq. 4. Exercise 24 Prove the first equality of Eq. The viscous terms in the momentum equations can be expressed in ωi . Eq.6 reads ∇2 v = ∇(∇ · v) − ∇ × ∇ × v = −∇ × ω (4.2 can indeed be written as ∂ 2 vi ∂ 2 vk ∂ 2 vj = − εinm εmjk ∂xj ∂xj ∂xj ∂xi ∂xj ∂xn (4. so that ∂ 2 vi ∂ωm = −εinm ∂xj ∂xj ∂xn (4. 229) εinm εmjk ∂ 2 vk ∂ 2 vk ∂ 2 vi ∂ 2 vk = (δij δnk − δik δnj ) = − ∂xj ∂xn ∂xj ∂xn ∂xi ∂xk ∂xj ∂xj (4.e. 4. but also in that case no vorticity is generated or destroyed: it stays constant. The vorticity is always created at boundaries. considering Item 1 this was to be expected.4. 4. 4. see Section 4.1. 4. inviscid flow (i.e.7) In vector notation the identity Eq. so that ∂ωm ∂ 2 vj ∂ 2 vi = − εinm ∂xj ∂xj ∂xj ∂xi ∂xn (4.9) causes a change in vorticity. Exercise 25 Write out Eq.2. The vorticity transport equation in three dimensions 43 Use the ε − δ-identity (see Table A. friction-less flow) has no rotation. there is a one-to-one relation between the viscous term and vorticity: no viscous terms means no vorticity and vice versa. generates vorticity (right side of Eq. The main points that we have learnt in this section are: 1. 14) rotation The volume source is in most engineering flows represented by the gravity which. However. see Eq. 2. v × ω.2 The vorticity transport equation in three dimensions Up to now we have talked quite a lot about vorticity. Inserting Sij = (∂vi /∂xj + ∂vj /∂xi )/2 and multiplying by two gives ∂vi ∂vi ∂vj 2vj − εijk vj ωk (4.e. First.11) = vj + ∂xj ∂xj ∂xi The second term on the right side can be written as (Trick 2. 4. irrotational and potential flow all denote frictionless flow which is equivalent to zero vorticity. Equation 4. The trick we have achieved is to split the convective term into one term without rotation (first term on the right side of Eq.12) where k = vj vj /2. fi = gi .13) rotation The last term on the right side is the vector product of v and ω. The terms inviscid (no friction).2. In vector notation it reads v×ω = 1 ∇(P0 ) ρ (4.14 we get Crocco’s theorem for steady inviscid flow ∂ P P ∂ + k − fi = +k+φ (4. The vorticity transport equation in three dimensions 44 4. we re-write the convective term of the incompressible momentum equation (Eq.13) and one term including rotation (second term on the right side). 16 was used. 8. 1.10) = vj (Sij + Ωij ) = vj Sij − εijk ωk ∂xj 2 where Eq. In this section we will derive the transport equation for vorticity in incompressible flow. 4.4.4) vj ∂vj 1 ∂(vj vj ) ∂k = = ∂xi 2 ∂xi ∂xi (4.24.e. Inserting Eq. but also in this case the vorticity is not affected once it has entered the inviscid flow region.11 can now be written as vj ∂k − εijk vj ωk ∂xi ∂vi = ∂xj no rotation (4.15) εijk vj ωk = ∂xi ρ ∂xi ρ P0 /ρ where ∂φ/∂xi = −fi is the potential of the body force. From Eq. We have learnt that physically it means rotation of a fluid particle and that it is only the viscous terms that can cause rotation of a fluid particle.7) yields ∂vi + ∂t ∂ 2 vi ∂k 1 ∂P +ν + fi − εijk vj ωk = − ∂xi ρ ∂xi ∂xj ∂xj no rotation (4.19 on p.16) frictionless . i.7) as ∂vi 1 vj (4. Eq. i. There is a small difference between the three terms because there may be vorticity in inviscid flow that is convected into the flow at the inlet(s). 4. usually no distinction is made between the three terms. 2.13 into the incompressible momentum equation (Eq. 2. As usual we start with the Navier-Stokes equation.7 at p. 20 and 4.e.18) (4. The vorticity transport equation in three dimensions 45 These equations states that the gradient of stagnation pressure.14 is defined by the cross product εpqi ∂vi /∂xq (∇ × v in vector notation.2. The last term on line 2 is zero because the gravitation vector. we start by applying the operator εpqi ∂/∂xq to the Navier-Stokes equation (Eq. the pressure cannot affect the rotation (i. we have got rid of the pressure gradient term. the vorticity) of a fluid particle since the pressure acts through its center.21) Inserting Eqs. Using the continuity equation (∂vq /∂xq = 0) and Eq. 4.18 can be written εpqi εijk ∂vp ∂ωp ∂vj ωk = ωk − vk ∂xq ∂xk ∂xk The second term on line 2 in Eq.4.17) where the body force fi was replaced by gi . see Exercise 8). the convective term and the diffusive term. Furthermore. p=1 ω1 ∂x1 ∂x2 ∂x3 ∂vp ∂v2 ∂v2 ∂v2 = (4. 4. 4. If we write it term-by-term it reads ∂v1 ∂v1 ∂v1 +ω2 + ω3 . The last term on line 1 is re-written using the ε-δ identity (see Table A. We know that εijk is anti-symmetric in all indices. 4. gi .23) ωk +ω2 + ω3 . because as mentioned in connection to Fig.1 at p.17 can be written as ∂ 3 vi ∂vi ∂ 2 ωp ∂2 νεpqi εpqi =ν =ν ∂xj ∂xj ∂xq ∂xj ∂xj ∂xq ∂xj ∂xj (4. p=2 ω1 ∂xk ∂x ∂x ∂x 1 2 3 ∂v ∂v ∂v ω1 3 +ω2 3 + ω3 3 . is constant. 4. 4. P0 . 4.21 into Eq. p=3 ∂x1 ∂x2 ∂x3 .19. Eq. That makes sense. 4.17 gives finally ∂vp ∂ωp ∂ 2 ωp ∂ωp dωp = ωk +ν ≡ + vk dt ∂t ∂xk ∂xk ∂xj ∂xj (4.1. 229) εpqi εijk ∂vj ωk ∂vj ωk ∂vp ωk ∂vq ωp = (δpj δqk − δpk δqj ) = − ∂xq ∂xq ∂xk ∂xq ∂vp ∂ωp ∂vq ∂ωk + ωk − vq − ωp = vp ∂xk ∂xk ∂xq ∂xq Using the definition of ωi we find that its divergence ∂ωi ∂vk ∂ 2 vk ∂ εijk = εijk = =0 ∂xi ∂xi ∂xj ∂xj ∂xi (4.20) (4. Equation 4. and hence the second term on line 1 and the first term on line 2 are zero (product of a symmetric and an anti-symmetric tensor). Since the vorticity vector in Eq.14) so that ∂2k ∂vj ωk ∂ 2 vi + εpqi − εpqi εijk ∂t∂xq ∂xi ∂xq ∂xq 2 3 1 ∂ P ∂ vi ∂gi = −εpqi + νεpqi + εpqi ρ ∂xi ∂xq ∂xj ∂xj ∂xq ∂xq εpqi (4.22 has a new term on the right-hand side which represents amplification and bending or tilting of the vorticity lines.19) is zero (product of a symmetric and an anti-symmetric tensor).22) We recognize the usual unsteady term. is orthogonal to both the velocity and vorticity vector. 4.22 increasing ω1 and it will stretch the cylinder.23 represent vortex tilting. 4.2) so that ωi = (ω1 . The vorticity transport equation in three dimensions v1 46 ω1 v1 x2 x1 Figure 4. The second term ω2 ∂v1 /∂x2 in line one in Eq. 0. Dashed lines denote fluid element before bending or tilting. but this time with its axis aligned with the x2 axis. We assume that a positive ∂v1 /∂x1 is acting on the fluid cylinder. through the source term ω2 ∂v1 /∂x2 . 4. The volume of the fluid element must stay constant during the stretching (the incompressible continuity equation). it will act as a source in Eq. This means that vorticity in the x2 direction. which means that the radius of the cylinder will decrease. ω2 . take a slender fluid particle. The diagonal terms in this matrix represent vortex stretching.3. 4. Assume is has a vorticity. Hence vortex stretching will either make a fluid element longer and thinner (as in the example above) or shorter and thicker (when ∂v1 /∂x1 < 0). The off-diagonal terms in Eq.3: Vortex tilting. Imagine a slender.2: Vortex stretching. see Fig. Vortex stretching Vortex tilting . cylindrical fluid particle with vorticity ωi and introduce a cylindrical coordinate system with the x1 -axis as the cylinder axis and r2 as the radial coordinate (see Fig.2. Again. 4. Dashed lines denote fluid element before stretching. The velocity gradient ∂v1 /∂x2 will tilt the fluid particle so that it rotates in clock-wise direction.23 gives a contribution to ω1 . 0). and that the velocity surrounding velocity field is v1 = v1 (x2 ). 4. ∂x1 ω2 v1 (x2 ) ∂v1 ∂x1 >0 x2 x1 Figure 4. ∂v1 > 0. 1 we studied the Rayleigh problem (unsteady diffusion). In fully developed channel flow. or the diffusion length. Equation 4. It is only when the wall-normal temperature derivative at the walls are non-zero that a temperature field is created in the domain. with a hot lower wall and a cold upper wall (constant wall temperatures) the heat flux is constant and goes from the lower wall to the upper wall. γ2 = −ν∂ω3 /∂x2 .26b) If the wall-normal temperature derivative ∂T /∂x2 = 0 at both walls (adiabatic walls).24) (Greek indices are used to indicate that they take values 1 or 2). for example. see Section 5.25 is turned into relations for q2 and γ2 by integration γwall = γ2 (4. 4. We have the same situation for the vorticity. This means that vorticity diffuses in the same way as temperature does.3. is constant across the channel (you have plotted this quantity in TME225 Assignment 1). the vorticity flux. v2 . The vorticity transport equation in two dimensions 47 creates vorticity in the x1 direction. This equation is exactly the same as the transport equation for temperature in incompressible flow.15. Hence. in .3. the heat flux at the walls.3 The vorticity transport equation in two dimensions It is obvious that the vortex stretching/tilting has no influence in two dimensions.4.1 Boundary layer thickness from the Rayleigh problem In Section 3. The same is true for ω3 : if ν∂ω3 /∂x2 = −γ2 = 0 at the walls. As shown above. 0) and the vorticity vector reads ωi = (0. vorticity is – in the same way as temperature – generated at the walls. δ. Thus in two dimensions the vorticity equation reads ∂ 2 ω3 dω3 =ν dt ∂xα ∂xα (4. qwall . in this case the vortex stretching/tilting term vanishes because the vorticity vector is orthogonal to the velocity vector (for a 2D flow the velocity vector reads vi = (v1 . The diffusion time. Its gradient. t. 0.25a) (4. Vorticity is useful when explaining why turbulence must be threedimensional.e. will be zero and the temperature will be equal to an arbitrary constant in the entire domain..26a) qwall = q2 (4.4. the flow will not include any vorticity. Vortex stretching and tilting are physical phenomena which act in three dimensions: fluid which initially is two dimensional becomes quickly three dimensional through these phenomena. the vorticity and the temperature equations reduce to ∂ 2 ω3 ∂x22 ∂ 2T 0=k 2 ∂x2 0=ν (4. 4. ω3 ) so that the vector ωk ∂vp /∂xk = 0). the two-dimensional unsteady temperature equation is identical to the two-dimensional unsteady equation for vorticity.25b) For the temperature equation the heat flux is given by q2 = −k∂T /∂x2. see Eq. i. 2. 4.14 can now be used to estimate the thickness of a developing boundary layer (recall that the limit between the boundary layer and the outer free-stream region can be defined by vorticity: inside the vorticity is non-zero and outside it is zero). The boundary layer thickness.3. Eq.4: Boundary layer. This means that . The vorticity transport equation in two dimensions 48 V0 x2 111111111 000000000 1 0 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 1111111 0000000 0 1 00000000000000000000000000000000000000000 11111111111111111111111111111111111111111 00000000000000000000000000000000000000000 11111111111111111111111111111111111111111 x1 δ L Figure 4. In a boundary layer the streamwise pressure gradient is zero. 3. δ. increases for increasing streamwise distance from leading edge (x1 = 0). ∂ 2 v1 . . =0 µ 2. Eqs. for example. the only non-zero terms in the Navier-Stokes equation are the streamwise pressure gradient and the wall-normal diffusion term (see. ∂x2 wall because.7 and 3.23). at the wall. Hence. 2. the flux of vorticity . . ∂ 2 v1 . . ∂ω3 . . = ν =0 (4.27) γ2 = −ν ∂x2 . wall ∂x22 . The proper way to solve the problem is to use Blasius solution.14. ReL = 1/2 L ν Re L Compute what δ(L) you get from Eq. In a boundary layer there is vorticity and outside the boundary layer it is zero (in this flow.28. if we can estimate how far from the wall the vorticity diffuses.3.28) .wall (recall that (∂v2 /∂x1 )wall = 0) along the wall which means that no vorticity is created along the boundary.2cm. B. 4. If we assume that the fluid is air with the speed V0 = 3m/s and that the length of the plate L = 2m we get from Eq.4. but in a developing boundary layer. they are not (v2 6= 0 and ∂v1 /∂x1 6= 0). During this time vorticity will be transported by diffusion in the x2 direction the length δ according to Eq. the vorticity is created at time t = 0+ when the plate instantaneously accelerates from rest to velocity V0 ). The time it takes for a fluid particle to travel from the leading edge of the plate to x = L is L/V0 (in the Rayleigh problem this corresponds to the flow field after time t = L/V0 ). Blasius solution gives (see Eq. Consider the boundary layer in Fig. The vorticity in a developing boundary layer is created at the leading edge of the plate (note that in channel flow. 4. see Section 3. vorticity is indeed created along the walls because in this case the streamwise pressure gradient is not zero). Below we will estimate the boundary layer thickness using the expression derived for the Rayleigh problem. 3. as in Fig. this gives us an estimation of the boundary layer thickness. 3. Exercise 26 Note that the estimate above is not quite accurate because in the Rayleigh problem we assumed that the convective terms are zero. 4. Hence. The vorticity generated at the leading edge is transported along the wall by convection and at the same time it is transported by diffusion away from the wall.1) δ 5 V0 L = (4.14 that δ(L) = 1. . At the end of the plate the boundary thickness is δ(L).4. The vorticity transport equation in two dimensions 49 Exercise 27 Assume that we have a developing flow in a pipe (radius R) or between two flat plates (separation distance h).16. Equation 3. We want to find out how long distance it takes for the the boundary layers to merge.4.14 can be used with δ = R or h. 3.3. . Make a comparison with this and Eq. 1. Using the height. see Eq.22. In steady flow.22 (vi = ∂Φ/∂xi ) into the definition of the vorticity. The gravity ˆ force can be expressed as a force potential. 1. gi = −∂ Φ/∂x i (see Eq. 4. Potential flow 50 4. 0. 4. the work (i.32 gives P ∂Φ v 2 ∂ ˆ =0 + + +Φ (4.31) Since ωi = 0 in potential (irrotational) flow. i. 1. ˆ Inserting gi = −∂ Φ/∂x i in Eq. −g3 ). Eq.32) ∂xi ∂xi 2 ∂xi ρ ∂xi where vi in the unsteady term was replaced by its potential (Eq. We use the Navier-Stokes equation (Eq.33) ∂xi ∂t 2 ρ Bernoulli equation Integration gives the famous Bernoulli equation ∂Φ v 2 P ˆ = C(t) + + +Φ ∂t 2 ρ (4.22). This is easily seen by inserting Eq.e. 4.14) with the viscous term expressed as in Eq.22). Do we have any use of the Navier-Stokes equation? The answer is yes: this equation provides the pressure field.e. gives ∂2Φ ∂ ∂Φ ∂vi = = (4. we get (with fi = gi ) and using k = vi vi /2 = v 2 /2 1 ∂v 2 1 ∂P ∂Φ ∂ + =− + gi (4.7 ∂ωm 1 ∂P ∂k ∂vi − εijk vj ωk = − − εinm + fi + ∂t ∂xi ρ ∂xi ∂xn (4. 1. because it is conservative. The vorticity is then zero by definition since the curl of the divergence is zero.22 into the continuity equation. the integral) depends only on the starting and ending points of the integral: in mathematics this is call an exact differential. 4. Inserting Eq. 2. we get where Φ v2 P + − g3 x3 = C 2 ρ (4.4 Potential flow In potential flow. gh = −g3 x3 . 1.15). Eq.3.30 (together with Eq. This is of great important since many analytical methods exist for the Laplace equation. The velocity field in potential flow is thus given by the continuity equation.e.35) where gi = (0.4.12. the potential satisfies the Laplace equation.30) 0= ∂xi ∂xi ∂xi ∂xi ∂xi i. we get the more familiar form P v2 + + gh = C 2 ρ conservative force (4. 1. The gravity force is conservative because when integrating this force.34) ˆ = −gi xi .4. the velocity vector can be expressed as the gradient of its potential Φ. ωi = ǫijk ∂vk ∂2Φ = ǫijk =0 ∂xj ∂xj ∂xk (4.36) . Eq. 4.29) since ǫijk is anti-symmetric in indices j and k and ∂ 2 Φ/∂xj ∂xk is symmetric in j and k. Another way to derive Eq. 4. x. Φ (Eq.42) Recall that for two-dimensional. A complex derivative of a complex variable is defined as (f (z + z0 ) − f (z))/z0 where z = x+iy and f = u+iv.38 and 4.37) df 1 ∂v ∂u (4. For real functions.1). and in the imaginary coordinate direction. So far the complex plane has been expressed as z = x+iy. f . incompressible flow.4.39 must be the same. is zero. 1.37) 1 ∂u ∂v df = +i dz 2 ∂x ∂x (4.4. the value of a partial derivative.38) and if we approach z0 in the y direction (imaginary direction) we get (see Eq. 3.43) . The streamfunction also satisfies the Laplace equation in potential flow where the vorticity. The derivative df /dz is defined as df 1 ∂ 1 ∂ ∂ ∂ f= (u + iv) = −i −i dz 2 ∂x ∂y 2 ∂x ∂y (4. z¯ = x − iy). Eq. Ψ (Eq. see Eq. We can approach the point z0 both in the real coordinate direction.e. It can also be expressed in polar coordinates (see Fig. ∂f /∂x. 4. ωi . and the velocity potential. Eq. 4.4. The total derivative. 4. ∂x ∂y ∂u ∂v =− ∂y ∂x (4. at x = x0 is defined by making x approach x0 and then evaluating (f (x+x0 )−f (x))/x0 . Potential flow 51 4.40 it to require that f should depend only on z but not on z¯ [7] (¯ z is the complex conjugate of z.43 ∂v2 ∂v1 ∂2Ψ ∂2Ψ + =− + = ω3 = 0 ∂x21 ∂x22 ∂x1 ∂x2 (4.1 Complex variables for potential solutions of plane flows Complex analysis is a suitable tool for studying potential flow. 1. x30 . We start this section by repeating some basics of complex analysis. 3. 4.41) Now we return to fluid mechanics and potential flow.39) = −i dz 2 ∂y ∂y Since Eqs.40) Equations 4.37) 1 ∂u ∂v 1 ∂v ∂u = +i + − 2 ∂x ∂y 2 ∂x ∂y If we approach z0 in the x direction (real direction) we get (see Eq. The complex derivative is defined only if the value of the derivative is independent of how we approach the point z0 . the velocity potential satisfies the Laplace equation.40 are called the Cauchy-Riemann equations. based on the streamfunction. df /dt. y. is defined by approaching the point x10 . i. t as a linear combination of all independent variables (cf.22) f = Φ + iΨ (4.5) z = reiθ = r(cos θ + i sin θ) (4. Let us introduce a complex potential.30. we get ∂u ∂v = . x20 .43). 4. This is easily seen by taking the divergence of the streamfunction. 13.48) . respectively. 1. Real and imaginary axes correspond to the horizontal and vertical axes.45) is one example.47) Taking the first and the second derivatives of Eq.46) The Laplace operator in polar coordinates reads 1 ∂ ∂f 1 ∂2f 2 ∇ f= r + 2 2 r ∂r ∂r r ∂θ (4. see Eq.4. The solution f = C1 z n (4.43 and 1. also satisfies the Laplace equation.e.5: The complex plane in polar coordinates.42 is differentiable. 3.40. 4.46 with respect to r and θ gives ∂f = C1 nrn−1 (cos(nθ) + i sin(nθ)) ∂r 1 ∂ ∂f r = C1 n2 rn−2 (cos(nθ) + i sin(nθ)) r ∂r ∂r ∂f = C1 nrn (− sin(nθ) + i cos(nθ)) ∂θ ∂2f = −C1 n2 rn (cos(nθ) + i sin(nθ)) ∂θ2 (4.44) see Eqs. We write it in polar coordinates n f = C1 reiθ = C1 rn einθ = C1 rn (cos(nθ) + i sin(nθ)) (4. Potential flow 52 y r θ x Figure 4. 4. Eq. since ∂Φ ∂Ψ = = v1 ∂x ∂y and ∂Φ ∂Ψ =− = v2 ∂y ∂x (4.2 Analytical solutions as f ∝ z n Now we will give some examples of f (z) which correspond to useful engineering flows. f .22.4. Hence the complex potential. 4. Let’s first verify that this is a solution of the Laplace equation. i. f also satisfies the Cauchy-Riemann equations. Thus we can conclude that f defined as in Eq.4. Furthermore. df /dz exists. 4. 4.4. see Eq.4. The flow impinges at the wall at x2 = 0. 4. 4.7. corresponds to the imaginary part of f . Eq. is equal to the imaginary part. 4.42 and 4.2.2 Stagnation flow When we set n = 2 in Eqs. 4.2. 4. 4. Ψ. and it is negative to the left and positive to the right.45 and 4.1 Parallel flow When we set n = 1 in Eq.47) is indeed zero.4.e.49) ∇ f= r ∂r ∂r r ∂θ 4. The streamfunction is zero along the symmetry line.46.51) The flow is shown in Fig.42. The polar velocity components are obtained as (see.53) .6.46 we get (inviscid) stagnation flow onto a wall.45 we get (C1 = V∞ ) f = V∞ z = V∞ (x + iy) (4. see Eqs.52) The solution in form of a vector plot and contour plot of the streamfunction is given in Fig. 4. Potential flow 53 y V∞ x Figure 4. 4.44) 1 ∂Ψ = 2r cos(2θ) r ∂θ ∂Ψ = −2r sin(2θ) vθ = − ∂r vr = (4.4. i. so that (C1 = 1) Ψ = r2 sin(2θ) (4.44 gives the velocity components v1 = ∂Ψ = V∞ ∂y and v2 = − ∂Ψ =0 ∂x (4.6: Parallel flow. Ψ. Equation 4. The streamfunction. ∂f 1 ∂2f 1 ∂ 2 r + 2 2 =0 (4.50) The streamfunction. x1 = 0. When we divide the fourth line with r2 and add it to the second line we find that the Laplace equation (Eq. 4.5 1.4.54 on a polar grid.2 except that the velocities are here twice as large because we chose C1 = 1. is given by (Eqs.3 Flow over a wedge and flow in a concave corner Next we set n = 6/5 in Eqs. Figure 4.5 α 0. The streamfunction.5 0 −0.2 −1 −0.5 (a) Vector plot. 4. the imaginary part of f .5 x1 1 (b) Streamlines. 0 0.54) = 2r sin θ(2 cos2 θ − 1) − 4r sin θ cos2 θ = −2r sin θ = −2x2 Recall that since the flow is inviscid (no friction).4.8 0. This gives us (inviscid) flow over a wedge and flow over a concave corner (n should be in the interval 1 < n < 2).5 (b) Streamlines.45 and 4.4 0.5 1 −1 −0.5 1.46) Ψ = r6/5 sin(6θ/5) (4.5 2.46. 4. i.2. and in Cartesian components (see Fig.5 0 x1 0. The lower boundary for x1 < 0 can either be a wall (concave corner) or symmetry line (wedge).7 was generated in Matlab by evaluating Eqs.5 1 1.e.8: Potential flow.42 and 4. a frictionless wall (same as a symmetric boundary).52. (a) Vector plot.8 −0.2 0. 2.5 2 2 1 1 1 0.4 0.2 −0.4 0 0 −0. 1.2 0. 4.55) .5 0 0 −2 1.9) v1 = vr cos θ − vθ sin θ = 2r cos θ cos(2θ) + 2r sin θ sin(2θ) = 2r cos θ(1 − 2 sin2 θ) + 4r sin2 θcosθ = 2r cos θ = 2x1 v2 = vr sin θ + vθ cos θ = 2r sin θ cos(2θ) − 2r cos θ sin(2θ) (4. Stagnation flow. Potential flow 54 0. Figure 4.7: Potential flow.53 and 4.2 x2 x2 0. 4.5 0 x1 0.5 0.5 −1 −0.6 −0.5 x2 x2 2 −1 0 x1 1 −1.6 1 1 0. Note that this flow is the same as we looked at in Section 1.8 1. the boundary condition on the wall is slip.6 0. Figure 4.4. Writing Eq. m ˙ >0 (C1 = 1) and the velocity components read (see Fig. The angle. is one. α.58 on polar form gives f= m ˙ m ˙ m ˙ ln r + ln eiθ = ln reiθ = (ln r + iθ) 2π 2π 2π (4.57) 4. The streamfunction is zero along the lower boundary.4.58) where m ˙ is the strength of the source. x3 . 4.53) 1 ∂Ψ 6 = r cos(6θ/5) r ∂θ 5 6 ∂Ψ = − r sin(6θ/5) vθ = − ∂r 5 vr = (4. the physical meaning of m ˙ is volume flow assuming that the extent of the domain in the third coordinate direction. 4.8. in Fig.8a is given by α= π (n − 1)π = n 6 (4.4.9: Line source. 4.59) .3 Analytical solutions for a line source The complex potential for a line source reads f= m ˙ ln z 2π (4. 4. Potential flow 55 y θ x Figure 4.56) The velocity vector field and the streamfunction are presented in Fig.4. 59) f =− Γ (i ln r − θ) 2π (4.23 and can be expressed as an integral of the vorticity over surface S bounded by line C.61 gives nonphysical flow near origo. 4. When origo is approached. Eq. Hence.7. ˙ dx3 vr rdθ = dx3 2πr 2π 0 0 0 0 0 0 4.63) f = −i 2π which on polar form reads (cf.53) m ˙ 1 ∂Ψ = r ∂θ 2πr ∂Ψ vθ = − =0 ∂r vr = (4. Eq.7.61) over a cylindrical surface as Z 1 Z 2π Z 1 Z 2π Z 1 Z 2π m ˙ m ˙ (4.7. 1.58 and reads Γ ln z (4.4 Analytical solutions for a vortex line A line vortex is another example of a complex potential. This is easily seen by integrating vr (Eq.49 is satisfied. The reason is that the inviscid assumption (zero viscosity) is not valid in this region. 4. see Fig.65) This flow was introduced in Section 1.10. 4. It was introduced in Section 1.60) which shows that Eq.9. it is very similar to Eq. 4. see Eq. Eq.47.61) We find that the physical flow is in the radial direction. 4. Potential flow 56 First. 4. appears in the expression of vθ .4. 1. If m ˙ > 0. 4.62) rdθ = dx3 dθ = m.25 and Fig. The circulation. 1. see Fig. 4. Γ. It was mentioned above that the physical meaning of m ˙ is volume flow. The streamfunction corresponds to the imaginary part of f and we get (see Eq. Eq.1 (where we called it an ideal vortex line) as an example of a flow with no vorticity. vr .64) From the streamfunction (the imaginary part of f ) we get (cf. . the flow is outwards directed and for m ˙ < 0 it is going inwards toward origo. 4. see Eq.4. the velocity. 4.4. we need to make sure that this solution satisfies the Laplace equation.61) 1 ∂Ψ =0 r ∂θ Γ ∂Ψ = vθ = − ∂r 2πr vr = (4. The flow is in the positive θ direction along lines of constant radius. The first and second derivatives read ∂f m ˙ = ∂r 2πr ∂f 1 ∂ r =0 r ∂r ∂r ∂f m ˙ =i ∂θ 2π ∂2f =0 ∂θ2 (4. tends to infinity. It is defined as a closed line integral along line C. 4. Imagine that we move the source and the sink closer to each other and at the same time we increase their strength |m| ˙ so that the product µ = mε ˙ stays constant.50.4.10: Vortex line.66) .5 Analytical solutions for flow around a cylinder The complex potential for the flow around a cylinder can be found by combining a doublet and a parallel flow. see Eq. The resulting complex potential is f= µ πz When adding the complex potential of parallel flow.4. x2 θ V∞ x1 Figure 4.4. we get (4.4. A doublet consists of a line source (strength m) ˙ and sink (strength −m) ˙ separated by a distance ε in the x1 direction (line sources were introduced in Section 4. Potential flow 57 y θ x Figure 4.11: Flow around a cylinder of radius r0 . 4.3). 72) .53) 1 ∂Ψ r02 = V∞ 1 − 2 cos θ vr = r ∂θ r r02 ∂Ψ vθ = − = −V∞ 1 + 2 sin θ ∂r r (4.5 0 Cp 0 −0. 4. The markers show the time-averaged location of separation.5 0 30 60 90 120 150 −1.70) (4.5 0 180 θ (a) CFD of unsteady laminar flow [8].4.75 Figure 4.5 −0. Potential flow 58 rdθ x2 θ x1 Figure 4.12: Flow around a cylinder of radius r0 . r0 .67) Now we define the radius of the cylinder. 4. as r02 = µ/(πV∞ ) so that f= V∞ r02 + V∞ z z (4.5 Cp 1 0. 30 60 90 120 150 180 θ (b) Potential flow.68) (4. Eq. f= µ + V∞ z πz (4.69) On polar form it reads 2 r0 −iθ V∞ r02 iθ iθ + V∞ re = V∞ f= e + re reiθ r 2 r0 = V∞ (cos θ − i sin θ) + r(cos θ + i sin θ) r The streamfunction reads (imaginary part) r02 sin θ Ψ = V∞ r − r Now we can compute the velocity components (see Eq.13: Pressure coefficients.5 −1 −1 −1. Re50 Re100 Re150 Re200 1 0.4.71) (4. Integration of surface pressure. 9 and Fig.4.77) 2π Z 1 Z 2π 4 3 = −r0 dx3 sin θ − sin θ =0 3 0 0 0 . but it is usually much smaller).4. the boundary layers may be thin but the wake is a large part of the domain. 2 CL . 4. Usually the lift force is expressed as a lift coefficient. see Eq. This requirement is not valid neither in the boundary layers nor in the wake. 4.s V∞ − vθ. We assume in Eq. The potential solution agrees rather well with viscous flow up to θ ≃ 20o . 4. The drag coefficient is computed as Z 1 Z 2π FD = − dx (1 − 4 sin2 θ) cos θr0 dθ CD = 3 2 /2 ρV∞ 0 0 (4.11. It should be stressed that although Eqs. 2.7 are solved numerically [8]. To find the lift force. hence these points correspond to the stagnation points. Eqs.76 that the length of the cylinder in the x3 direction is one. FL . Note that the local velocity gets twice as large as the freestream velocity at the top (θ = π/2) and the bottom (θ = −π/2) of the cylinder. 4. Furthermore.68).3 and 2.75) using Eq. we simply integrate the pressure over the surface. we see that the tangential velocity is zero at θ = 0 and π.s P∞ ps V∞ + = + ⇒ ps = P∞ + ρ 2 ρ 2 ρ 2 (4.s = −2V∞ sin θ (4. 4. The lift coefficient is obtained as (see Fig. How do we find the lift and drag force? The only force (per unit area) that acts on the cylinder surface is the pressure (in viscous flow there would also be a viscous stress. We are not interested in the solution inside the cylinder (r < r0 ). which is scaled with the dynamic pressure ρV∞ /2. The velocity field at the cylinder surface.36) 2 2 2 2 vθ. The surface pressure is obtained from Bernoulli equation (see Eq.13 presents the pressure coefficient for potential flow and accurate unsteady CFD of two-dimensional viscous flow [8] (the Reynolds number is sufficiently low for the flow to be laminar).s ps − P∞ = 1 − 4 sin2 θ = 1 − 2 /2 2 ρV∞ V∞ (4.12) Z 1 Z 2π ps FL = − dx sin θr0 dθ CL = 3 2 /2 2 ρV∞ ρV 0 0 ∞ /2 Z 1 Z 2π = −r0 dx3 (1 − 4 sin2 θ) sin θdθ (4. Potential flow 59 We find that vr = 0 for r = r0 as intended (thanks to the definition in Eq.73. see Fig.1.75 are exact they are not realistic because of the strict requirement that the flow should be inviscid. 4. 4. Figure 4.74) where we neglected the gravitation term. reads vr.73) where index s denotes surface.76) 0 0 2π 1 3 = −r0 − cos θ − 4 =0 cos(3θ) − cos θ 12 4 0 The minus sign on the first line appears because pressure acts inwards. 1. r = r0 .73 and 4. 4.s = 0 vθ. The surface pressure is usually expressed as a pressure coefficient Cp ≡ 2 vθ. 69 so that f= V∞ r02 Γ + V∞ z − i ln z z 2π (4.63) to Eq. Here we will introduce the use of additional circulation which alters the locations of the stagnation points and creates lift.4. which is by far the most important one from engineering point of view. The reason is that the pressure is symmetric both with respect to x1 = 0 and x2 = 0. 4.78) . Potential flow 60 a x2 b V∞ θ x1 Γ Figure 4. Same argument for the drag force: the pressure force on the upstream surface cancels that on the downstream surface.14: Flow around a cylinder of radius r0 with additional circulation. Hence. x2 c θ V∞ x1 Γ Figure 4. This approach is used in potential methods for predicting flow around airfoils in aeronautics (mainly helicopters) and windpower engineering. 4.4. inviscid flow around a cylinder creates neither lift nor drag.4. 4.6 Analytical solutions for flow around a cylinder with circulation We will now introduce a second example of potential flow around cylinders. The lift force on the lower surface side cancels the force on the upper side.15: Flow around a cylinder of radius r0 with maximal additional circulation. We add the complex potential of a vortex line (see Eq. 85 give no contribution to the lift.82 gives Γ Γ 2V∞ sin θstag = (4. The pressure is obtained from Bernoulli equation as (see Eq.s Γ = 1 − −2 sin θ + 2 V∞ 2πr0 V∞ 2 4Γ sin θ Γ 2 = 1 − 4 sin θ + − 2πr0 V∞ 2πr0 V∞ Cp = 1 − (4. 4. For a limiting value of the circulation. 4. Γmax = 4πV∞ r0 .84) This corresponds to the maximum value of the circulation for which there is a stagnation point on the cylinder surface.5 that a cylinder without circulation gives neither drag nor lift. Γmax . The larger the circulation. The two positions are indicated with a and b in Fig.70) 2 Γ r0 (cos θ − i sin θ) + r(cos θ + i sin θ) − (i ln r − θ) f = V∞ r 2π The imaginary part gives the streamfunction r2 Γ Ψ = V∞ r − 0 sin θ − ln r r 2π We get the velocity components as (see Eqs.85) We found in Section 4. where vθ. (4.4.4.s = 0.72) r2 1 ∂Ψ = V∞ 1 − 02 cos θ vr = r ∂θ r ∂Ψ r2 Γ vθ = − = −V∞ 1 + 02 sin θ + ∂r r 2πr (4.e.83) ⇒ θstag = arcsin 2πr0 4πr0 V∞ The two angles that satisfy this equation are located in the the first and second quadrants. see Eqs. the larger vθ . denoted with c in Fig. 4. r = r0 .64 and 4. 4. Potential flow 61 On polar form it reads (see Eqs. to increase vθ while leaving vr unaffected.81) The effect of the added vortex line is.79) (4.4. 4.75) 2 2 vθ.65 and 4.82) Now let’s find the location of the stagnation points. reads vr. 4.15.s = 0 vθ.76 that the two first terms in Eq. For circulation larger than Γmax . as expected. the stagnation point will be located above the cylinder.14. Equation 4.76 and 4. i. We found in Eq.s = −2V∞ sin θ + Γ 2πr0 (4.80) (4. The last term cannot give any contribution to the lift because it is constant on the entire . What about the present case? Let’s compute the lift. 4. 4. the two locations s and b will merge at θ = π/2. The velocity at the surface.77. 85 so that Z 1 Z 2π FL ps = − dx sin θr0 dθ 3 2 /2 2 ρV∞ ρV 0 0 ∞ /2 Z 1 Z 2π 2Γ sin θ = −r0 dx3 sin θdθ πr0 V∞ 0 0 2π Γθ Γ 2Γ =− − sin(2θ) =− πV∞ 2πV∞ V∞ 0 CL = (4. The ”lift” force is acting downwards. In the drag integral (see Eq. F . One way to improve the chance that this will happen is to make a loop. Hence. are located at the top of the cylinder and hence the pressure is higher on the top than on the bottom. The drag is. i. which acts downwards so that the ball drops down quickly and (hopefully) hits the table on the other side of the net. 4.77 we found that the first and the second terms in Eq.87) This relation is valid for any body and it is called the Kutta-Joukowski law who – independent of each other – formulated it. where the pressure is largest. Hence you hit it with a .16: Table tennis. Hence. This has interesting application in sports. surface.85 gives no contribution to drag. Hence we only need to include the third term in Eq. for example football. we let the cylinder rotate with speed Ω. The stagnation points.4. however.85 gives rise to a term proportional to sin θ cos θ whose contribution is zero. 4.6. 4. The ball experiences a force. in the negative x2 direction. The reason to the sign of the lift force can easily be seen from Fig. This means that you hit the ball slightly on the top.4. Another example where the Magnus effect is important is golf.e. the additional circulation does not give rise to any drag.4.77). You want the ball to go as far as possible. FL .1 The Magnus effect Circulation around a cylinder is very similar to a rotating cylinder. 4. when you hit it (see Fig.14. The loop uses the Magnus effect. Instead of adding a circulation. table tennis and golf. In Eq. 4. Here the object is often vice versa. Side view. Potential flow 62 Ω x2 F x1 FL Figure 4. Recall that the relative velocity of the air is in the negative x1 direction.86) We find that the lift force on a unit length of the cylinder can be computed from the circulation as FL = −ρV∞ Γ (4.16) and this force makes it rotate with rotation speed Ω (clockwise direction). this term in Eq. 4. A rotating cylinder produces lift. 4. The lift force is downwards because the stagnation points are located on the upper surface. still zero. 4. the ball must hit the table on the side of the opponent. The rotation causes a lift. In table tennis. we only need to consider the third terms. A free-kick uses the Magnus effect.17: Football.4. the ball should turn right towards the goal. Figure 7b in that paper is particularly interesting. The initial velocity of the football is 30 m/s. Potential flow 63 Ω FL F wall goal Figure 4. The reason that the ball turns to the right first after the wall (and not before) is that the forward momentum created by F (the player) is much larger than FL . Imagine there is a free-kick rather close to the opponents’ goal. see Fig. The player hits the ball with his/her left foot on the left side of the ball which creates a force F on the ball. see Fig. The result is a lift force in the positive x2 direction which makes the ball go further. and creates a lift force so that the ball after it has passed the wall turns to the right towards the goal. 4. Top view. If you are interested in football you may be pleased to learn that by use of fluid dynamics it is now scientifically proven that it was much harder to make a good freekick in 2010 worldcup than in 2014 [9]. The player who makes the free-kick wants to make the ball go on the left side of the wall. 4.17. two identical freekicks are made with the football used at the 2013 FIFA Confederations. A final sports example is football. This makes the ball rotate clockwise. As an experiment.4. The relative velocity between the ship and the wind is Vwind + Vship . Top view Vwind Ω Vship FL x2 x1 Figure 4.17. Why? Because the ball was rotated 45 degrees before the second freekick (see . The opponents erects a wall of players between the goal and the location of the free-kick. The freekicks are made 25m from the goal. after the wall of players. slice so that it spins with a positive Ω (counter-clockwise).18: Flettner rotor (in blue) on a ship. The Magnus effect helps to achieve this. Here the lift is used sideways. The ship moves with speed Vship . The result of the two freekicks is that the two footballs reach the goal three meters from each other in the vertical direction. and the wake illustrated in red.e. The boundary layers.19: Airfoil. The diameter of this rotor can be a couple of meter and have a length (i. Potential flow 64 δ(x1 ) V∞ c x2 x1 Figure 4.18.e. at an angle of −π/4. 2c. Figs. x1 = 0 and x1 = c at leading and trailing edge. The Magnus effect creates a force in the orthogonal direction to the relative windspeed. V∞ Figure 4. Rear stagnation point at the upper surface (suction side). see Fig. The Division of Fluid Dynamics recently took part in an EU project where we studied the flow around rotating cylinders in relation to Flettner rotors [10]. Note that if the wind comes from the right instead of from the left.4.4.20: Airfoil. The ship is moving to the right with speed Vship . the rotor should rotate in the counter-clockwise direction. The first Flettner rotors on ships were produced in 1924. δ(x1 ). i. .d) in [9]. The wind comes towards the ship from the left-front (relative wind at an angle of π/4). respectively. 4. height) of 10 − 20 meter. It has recently gained new interest as the cost of fuel is rising. Finally we give an engineering example of the use of the Magnus effect. Streamlines from potential flow. A Flettner rotor is a rotating cylinder (or many) on a ship. The Flettner rotor rotates in the clockwise direction. At the Division of Fluid Dynamics we have an on-going PhD project where we use potential methods for computing the aerodynamic loads for windturbine rotorblades [3]. In aeronautics. The added circulation is negative (clockwise). Streamlines from potential flow with added circulation.4. see Fig.87) FL = ρV∞ Γaeronautic (4. Figure 4. The boundary layer is thinner on the pressure (lower) side than on the suction (upper) side. see Fig.1) shows a two-dimensional airfoil. These methods are still in use in wind engineering and for helicopters.4.20. for example. For low angles of attack (which is the case for. The flow on the pressure (lower) side cannot be expected to make a 180o turn at the trailing edge and then go in the negative x1 direction towards the stagnation point located on the suction side. 16.4. the stagnation point near the trailing edge is located on the suction side which is clearly nonphysical. We want to move the rear stagnation point towards the trailing edge. 4.21. Outside these regions the flow is essentially inviscid.19 (see also Fig. 4.88) .21: Airfoil.4. Rear stagnation point at the trailing edge. When this flow is computed using potential methods. 4. It grows slightly thicker towards the trailing edge (denoted by δ(x1 ) in Fig. an aircraft in cruise conditions) the boundary layers and the wake are thin. The lift of a two-dimensional airfoil (or a two-dimensional section of a three-dimensional airfoil) is then computed as (see Eq. the sign of circulation is usually changed so that Γaeronautic = −Γ. 4. This is achieved by adding a circulation in the clockwise direction.6. The boundary layers and the wake are illustrated in red.7 The flow around an airfoil Flow around airfoils is a good example where potential methods are useful. 4. The magnitude of the circulation is determined by the requirement that the stagnation point should be located at the trailing edge.19). The solution is to move the stagnation points in the same way as we did for the cylinder flow in Section 4. However. The flow around airfoils is a good example where the flow can be treated as inviscid in large part of the flow. This is called the Kutta condition. Potential flow 65 Γ V∞ Figure 4. the location of the front stagnation point is reasonably well captured. Also the flow and combustion in engines. We do not have any exact definition of an turbulent eddy. 2. The increased diffusivity also increases the resistance (wall friction) and heat transfer in internal flows such as in channels and pipes.1 Introduction Almost all fluid flow which we encounter in daily life is turbulent. at least along the walls where wall-jets are formed. Eq. are highly turbulent. the transition to turbulent flow in pipes occurs that ReD ≃ 2300. and in boundary layers at Rex ≃ 500 000. and reduces or delays thereby separation at bluff bodies such as cylinders. VI. Irregularity. The largest eddies extract their energy from the mean flow. Large Reynolds Numbers. jet width. II. airfoils and cars. The turbulence increases the exchange of momentum in e. In turbulent flow the diffusivity increases. The region covered by a large eddy may well enclose also smaller eddies. Hence. Three-Dimensional. Dissipation. The flow consists of a spectrum of different scales (eddy sizes). but they are governed by Navier-Stokes equation. However. Turbulent flow occurs at high Reynolds number. The small eddies receive the kinetic energy from slightly larger eddies. Typical examples are flow around (as well as in) cars. The slightly larger eddies receive their energy from even larger eddies and so on. Continuum. There is no definition on turbulent flow. boundary layer thickness. For example. boundary layers. see below). Even though we have small turbulent scales in the flow they are much larger than the molecular scale and we can treat the flow as a continuum. In turbulent flow we usually divide the velocities in one time-averaged part v¯i .g. and one fluctuating part vi′ so that vi = v¯i + vi′ . Turbulent flow is irregular and chaotic (they may seem random.e.5. The largest eddies are of the order of the flow geometry (i. aeroplanes and buildings are turbulent. we can treat the flow as two-dimensional (if the geometry is two-dimensional). Diffusivity. Air movements in rooms are turbulent. turbulent eddy cascade process . Turbulence 66 5 Turbulence 5. which means that kinetic energy in the small (dissipative) eddies are transformed into thermal energy. It has a characteristic velocity and length (called a velocity and length scale). At the other end of the spectra we have the smallest eddies which are dissipated by viscous forces (stresses) into thermal energy resulting in a temperature increase. The boundary layers and the wakes around and after bluff bodies such as cars. both in piston engines and gas turbines and combustors. when we compute fluid flow it will most likely be turbulent. Turbulent flow is always three-dimensional and unsteady. but we suppose that it exists in a certain region in space for a certain time and that it is subsequently destroyed (by the cascade process or by dissipation. but it has a number of characteristic features (see Pope [11] and Tennekes & Lumley [12]) such as: I. Even though turbulence is chaotic it is deterministic and is described by the Navier-Stokes equations. III. which is independent of time (when the mean flow is steady). aeroplanes and buildings. Turbulent flow is dissipative. This process of transferring energy from the largest turbulent scales (eddies) to the smallest is called the cascade process. etc). when the equations are time averaged. IV.7). V. The energy-containing eddies are denoted by v0 . but they are largest at the smallest eddies. At the smallest scales the frictional forces (viscous stresses) become large and the kinetic energy is transformed (dissipated) into thermal energy. ν.. i.. dissipative scales Figure 5.e. ℓ1 and ℓ2 denotes the size of the eddies in the inertial subrange such that ℓ2 < ℓ1 < ℓ0 . kinetic energy is in this way transferred from the largest scale to the smallest scales. We assume that these scales are determined by viscosity. The smallest scales where dissipation occurs are called the Kolmogorov scales whose velocity scale is denoted by vη . The kinetic energy transferred from eddy-to-eddy (from an eddy to a slightly smaller eddy) is the same per unit time for each eddy size. ℓη is the size of the dissipative eddies.1: Cascade process with a spectrum of eddies. and dissipation. In reality a small fraction is dissipated at all scales. ε. These scales extract kinetic energy from the mean flow which has a time scale comparable to the large scales. times the fluctuating velocity gradient up to the power of two (see Section 8. The dissipation is proportional to the kinematic viscosity. for example). The dissipation is denoted by ε which is energy per unit time and unit mass (ε = [m2 /s3 ]).5. However it is assumed that most of the energy that goes into the large scales per unit time (say 90%) is finally dissipated at the smallest (dissipative) scales.. ∂¯ v1 ∼ t−1 0 ∼ v0 /ℓ0 ∂x2 (5. length scale by ℓη and time scale by τη . Through the cascade process. Turbulent scales 67 flow of kinetic energy ℓη ℓ0 ℓ1 ℓ2 large scales intermediate scales . ν. .1) Part of the kinetic energy of the large scales is lost to slightly smaller scales with which the large scales interact. The argument is as follows.2). The friction forces exist of course at all scales.2 Turbulent scales The largest scales are of the order of the flow geometry (the boundary layer thickness. 5.2. with length scale ℓ0 and velocity scale v0 . The more energy that is to be transformed from kinetic energy to thermal energy. We get two equations.3.7) κn is called the wavenumber. the turbulence fluctuations are composed of a wide range of scales.5) . The Fourier coefficients are given by Z 1 L an = g(x) cos(κn x)dx L −L Z 1 L g(x) sin(κn x)dx bn = L −L Parseval’s formula states that Z L ∞ X L (a2n + b2n ) g 2 (x)dx = a20 + L 2 −L n=1 (5. 5. We can think of them as eddies. g.1. Having assumed that the dissipative scales are determined by viscosity and dissipation.3 Energy spectrum As mentioned above. τη = ε ε 5. It turns out that it is often convenient to use Fourier series to analyze turbulence. We write vη = νa εb 2 2 3 [m/s] = [m /s] [m /s ] (5. ℓη = (5. g(x) = g(x + 2L)). one for meters [m] 1 = 2a + 2b. the larger scales.3) −1 = −a − 3b. In the same way we obtain the expressions for ℓη and τη so that 3 1/4 ν 1/2 ν 1/4 vη = (νε) . ℓη and τη in ν and ε using dimensional analysis. In general.e.4) and one for seconds [s] which give a = b = 1/4.e. see Fig.5. (5.8) . the larger viscosity. the larger the velocity gradients must be. Energy spectrum 68 viscosity: Since the kinetic energy is destroyed by viscous forces it is natural to assume that viscosity plays a part in determining these scales. (5. we can express vη . i. dissipation: The amount of energy per unit time that is to be dissipated is ε.6) where x is a spatial coordinate and κn = nπ L or κ = 2π .2) where below each variable its dimensions are given. with a period of 2L (i. L (5. g(x) = ∞ X 1 a0 + (an cos(κn x) + bn sin(κn x)) 2 n=1 (5. any periodic function. The dimensions of the left and the right side must be the same. can be expressed as a Fourier series. The wavenumber.1.e. For readers not familiar to Fourier series. 5.8 expresses v1′2 as a function of time and the right side expresses v1′2 as a function of frequency. Let now g be a fluctuating velocity component. 5. x) and the right side v1′2 in wavenumber space (vs.6 as a series in time rather than in space. κ ∝ ℓ−1 κ . Z ∞ X E(κ)dκ = L g 2 (κn ) (5. In this case the left side of Eq. corresponds to g 2 (κ) in Eq. . An example of a Fourier series and spectra are given in Appendix E.1.1. see Section 8.8 expresses v1′2 in physical space (vs. 5. wavenumber).2. a brief introduction is given in Appendix D.2: Spectrum for turbulent kinetic energy. thus we can think of wavenumber as proportional to the inverse of an eddy’s diameter. k. see Section 8. The turbulent scales are distributed over a range of scales which extends from the largest scales which interact with the mean flow to the smallest scales where dissipation occurs. i. In that case. i. The total turbulent kinetic energy is obtained by integrating over the whole wavenumber space. It is now convenient to study the kinetic energy of each eddy size in wavenumber space.2.3. energy containing eddies. κn ). 5. see Fig. vi′ vi′ /2.e.e κ ∝ 1/d. I: Range for the large. Intuitively we assume that large eddies have large fluctuating velocities which implies large kinetic energy. E(κ). In wavenumber space the energy of eddies can be expressed as E(κ)dκ (5. The dimension of wavenumber is one over length.8.e. ℓκ . κ. The energy spectrum. then computing the kinetic energy of each eddy size (i. Energy spectrum 69 Pk E(κ) I II εκ ε E ) (κ ∝ − κ 3 5/ III κ Figure 5. isotropic scales. ε. ε.e. i. express g in Eq.9 expresses the contribution from the scales with wavenumber between κ and κ + dκ to the turbulent kinetic energy k. II: the inertial subrange. For a discussion of εκ vs. 5. For a discussion of εκ vs. is probably more familiar to “frequency”. The left side of Eq.9) where Eq. is proportional to the inverse of the length scale of a turbulent eddy. 5. III: Range for small.5.10) k= 0 Think of this equation as a way to compute the kinetic energy by first sorting all eddies by size (i. The reader who is not familiar to the term “wavenumber”. Now let us think about how the kinetic energy of the eddies varies with eddy size. say v1′ . ′ ′ The kinetic energy. is given by the assumption of the cascade process. (5. One can argue that the eddies in this region should be characterized by the spectral transfer of energy per unit time (ε) and the size of the eddies.5. see Eq. This energy transfer takes places via the production term.12) spectral transfer . 5. The energy extracted per unit time by the largest eddies is transferred (per unit time) to slightly smaller scales. Dimensional analysis gives E = [m /s2 ] = 3 κa εb 2 3 [1/m] [m /s ] We get two equations. 1/κ. In this region we have the large eddies which carry most of the energy. Hence the dimension of E is v 2 /κ. The eddies are small and isotropic and it is here that the dissipation occurs. These eddies interact with the mean flow and extract energy from the mean flow.14.e. The eddies in this region are independent of both the large. of an eddy of size (lengthscale). in the transport equation for turbulent kinetic energy. The “transport” in wavenumber space is called spectral transfer. 1/κ. 8. The energy transfer from turbulent kinetic energy to thermal energy (increased temperature) is governed by ε in the transport equation for turbulent kinetic energy. 5. The kinetic energy is the sum of the kinetic energy of the three fluctuating velocity components.10. 5. and finally summing the kinetic energy of all eddy sizes (i. is coming from the large eddies at the lower part of this range and is transferred per unit time to the dissipation range at the higher part. III.i vκ. 5. P k . see Eq. This region is a “transport region” (i. carrying out the integration). for a discussion on the dimension of E.e. Dissipation range.11.5) II. 8. see Appendix E.14.2. Energy spectrum 70 E(κ)dκ). P k = ε. and one for seconds [s] −2 = −3b. Note that the relation P k = {dissipation at small scales}. energy-containing eddies and the eddies in the dissipation range. represents the kinetic energy of all eddies of this size.11) v1 + v2′2 + v3′2 = vi′ vi′ k= 2 2 The spectrum of E is shown in Fig. kκ = vκ. 1 1 ′2 (5.e. in wavenumber space) in the cascade process. that the energy transfer per unit time from eddy-size– to–eddy-size is the same for all eddy sizes. i. The eddies in this region represent the mid-region. The kinetic energy of all eddies (of all size) is computed by Eq. The existence of this region requires that the Reynolds number is high (fully turbulent flow). We find region I.i /2. 5. i. The turbulence is also in this region isotropic. respectively.3. one for meters [m] 3 = −a + 2b. Energy per time unit. see Fig.2. II and III which correspond to: I. see Eq. Note that the physical meaning of E is kinetic energy per unit wavenumber of eddies of size ℓκ ∝ κ−1 . Inertial subrange. The eddies’ velocity and length scales are v0 and ℓ0 . The scales of the eddies are described by the Kolmogorov scales (see Eq.e. 2) εκ ∼ vκ2 v3 ∼ κ ℓκ ℓκ vκ (5. Note that is not true instantaneously. For example if the x1 coordinate direction is rotated 180o the v1′ v2′ should remain the same. · δij . 5. εκ . the energy spectra should exhibit a −5/3decay in the inertial region (region II. Hence.14 for the large energy-containing eddies gives ε0 ∼ v02 v3 ∼ 0 ∼ εκ = ε ℓ0 ℓ0 v0 (5.13) where the Kolmogorov constant CK ≃ 1. Energy spectrum 71 so that b = 2/3 and a = −5/3. rotated 180o). We can now estimate the ratio between the large eddies (with v0 and ℓ0 ) to the Kolmogorov eddies (vη and ℓη ). In the inertial subrange. again.e.2).5. The spectral transfer rate of kinetic energy from eddies of size 1/κ to slightly smaller eddies can be estimated as follows. This is a very important law (Kolmogorov spectrum law or the −5/3 law) which states that. the concept of the cascade process assumes that the energy extracted per unit time by the large turbulent eddies is transferred (per unit time) by non-linear interactions through the inertial range to the dissipative range where the kinetic energy is transformed (per unit time) to thermal energy (increased temperature).15) The dissipation at small scales (large wavenumbers) is determined by how much energy per unit time enters the cascade process at the large scales (small wavenumbers). all shear stresses are zero in isotropic turbulence. Applying Eq. The kinetic energy of the eddy is proportional to vκ2 and the time for one revolution is proportional to ℓκ /vκ .15 give −1/4 v0 v0 = (v0 ℓ0 /ν)1/4 = Re1/4 = (νε)−1/4 v0 = νv03 /ℓ0 vη 3 −1/4 3 −1/4 3 −1/4 ℓ0 ν ν ℓ0 ν = ℓ0 = ℓ0 = = Re3/4 (5.δij which. 5.e.12 we get 2 5 E(κ) = CK ε 3 κ− 3 (5.3. Inserted in Eq. the Reynolds stress tensor for small scales can be written as vi′ vj′ = const. Above we state that the eddies in Region II and III are isotropic. i. isotropic turbulence implies that if a coordinate direction is switched (i.5. shows us that the shear stresses are zero in isotropic turbulence. 67. Furthermore. Fig. i. As discussed on p.e. This is possible only if v1′ v2′ = 0. if the flow is fully turbulent (high Reynolds number).e. i. Equations 5. nothing should change. we know that an isotropic tensor can be written as const. Hence. the fluctuations in all directions are the same so that v1′2 = v2′2 = v3′2 . the energy spectral transfer rate.5 and 5. 5. Hence. Using our knowledge in tensor notation. increased temperature) at the Kolmogorov scales. v1′ v2′ = −v1′ v2′ .e. 5.14) Kinetic energy is transferred per unit time to smaller and smaller eddies until the transfer takes place by dissipation (i. the cascade process assumes that εκ = ε.16) 3 ℓη ε v0 v03 ℓ30 1/2 −1/2 3 1/2 τo ℓ0 νℓ0 v0 ℓ0 v0 = Re1/2 = = τ0 = τη v03 νℓ0 v0 ν isotropic turbulence . This means that – in average – the eddies have no preferred direction. in general v1′ 6= v2′ 6= v3′ . An eddy loses (part of) its kinetic energy during one revolution. for an eddy of length scale 1/κ can be estimated as (see Fig. With a computational grid we must resolve all eddies.e.4. i.3: Family tree of turbulent eddies (see also Table 5.4 The cascade process created by vorticity The interaction between vorticity and velocity gradients is an essential ingredient to create and maintain turbulence. (region II. Five generations. the larger the wavenumber range of the intermediate range where the eddies are independent of both the large scales and the viscosity. The cascade process created by vorticity b 3 b b b 2 3 b 1 2 b 4 72 4 b b b 4 b 4 4 3 b 4 b 4 b 3 4 b b b b b b b b b b b b b b b b b b 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 Figure 5.17) . The large original eddy.22) ∂ωi ∂vi ∂ωi ∂ 2 ωi = ωj +ν + vj ∂t ∂xj ∂xj ∂xj ∂xj ∂vk ωi = ǫijk ∂xj (5. The equation for the instantaneous vorticity (ωi = ω ¯ i + ωi′ ) reads (see Eq. Adapted from [13] where Re = v0 ℓ0 /ν. Two idealized phenomena in this interaction process can be identified: vortex stretching and vortex tilting. with axis aligned in the x1 direction. increases with increasing Reynolds number. the larger the Reynolds number. the disturbances are turned into chaotic.1. This is the very reason why it is so expensive (in terms of computer power) to solve the Navier-Stokes equations. the number of grid cells increases rapidly. see Eq. This means that the eddy range (wavenumber range) of the intermediate region. 5. We find that the ratio of the velocity. is 1st generation. into turbulence. three-dimensional fluctuations. 27.5. the inertial region). 4. Hence. Hence.1). Disturbances are amplified by interaction between the vorticity vector and the velocity gradients. as the Reynolds number increases. length and time scales of the energy-containing eddies to the Kolmogorov eddies increases with increasing Reynolds number. 18) remains constant as the radius of the fluid element decreases. Here a generation is related to a wavenumber in the energy spectrum (Fig. As was mentioned above. see Fig.4. 4. the continuity equation shows that stretching results in a decrease of the radius of a slender fluid element and an increase of the vorticity component (i. x2 or x3 direction.2 this equation is not an ordinary convection-diffusion equation: it has an additional term on the right side which represents amplification and rotation/tilting of the vorticity lines (the first term on the right side). Thus we can assume that there are no viscous stresses acting on the cylindrical fluid element surface which means that the angular momentum r2 ω1′ = const. see Eq. The i = j components of this term represent (see Eq. 5.18 shows that the vorticity increases if the radius decreases (and vice versa). In other words. vortex tilting creates small-scale vorticity in the x2 and x3 direction.1 The large original eddy (1st generation) is aligned in the x1 Vortex stretching . 4. 5.2).5. The creation of multiple eddies by vortex stretching from one original eddies is illustrated in Fig. The increased v2′ velocity component will stretch smaller fluid elements aligned in the x2 direction. 67). 5. 1.3. At the same time. see Fig. As we learnt in Section 4. the length scale of the eddies – whose velocity scale are increased – decreases. For example. They are isotropic.4. A positive ∂v1′ /∂x1 will stretch the cylinder.1: Number of eddies at each generation with their axis aligned in the x1 .23 – is constant. Equation 5. Note that also the circulation.3 illustrates how a large eddy whose axis is oriented in the x1 axis in a few generations creates – through vortex stretching – smaller and smaller eddies with larger and larger velocity gradients. see Fig. The smaller the eddies. 5. 5.5. young generations correspond to high wavenumbers. see Fig. The cascade process created by vorticity generation 1st 2nd 3rd 4th 5th 6th 7th x1 1 0 2 2 6 10 22 73 x2 0 1 1 3 5 11 21 x3 0 1 1 3 5 11 21 Table 5.e. At each stage. the less the original orientation of the large eddy is recalled. The small eddies have no preferred direction. Γ – which is the integral of the tangential velocity round the perimeter.5. 5.2 and from the requirement that the volume must not change (incompressible continuity equation) we find that the radius of the cylinder will decrease. the small eddies “don’t remember” the characteristics of their original ancestor. (5. an extension of a fluid element in one direction (x1 direction) decreases the length scales in the x2 direction and increases ω1′ . Figure 5. the tangential velocity component) aligned with the element. The increased ω1′ means that the velocity fluctuation in the x2 direction is increased.3 and Table 5. ω2′ and ω3′ .23) vortex stretching. This will increase their vorticity ω2′ and decrease their radius. In the same way will the increased ω1′ also stretch a fluid element aligned in the x3 direction and increase ω3′ and decrease its radius. see Fig. We have neglected the viscosity since viscous diffusion at high Reynolds number is much smaller than the turbulent one and since viscous dissipation occurs at small scales (see p. the eddies in the x2 direction create new eddies in the x1 and x3 (3rd generation) and so on. ω3′ v2′ ω2′ ω1′ x2 x3 Figure 5.3. which is equivalent) will tilt the fluid element so that it rotates in the clock-wise direction. ω3′ v1′ 74 ω1′ ω2′ v1′ x2 x1 Figure 5. the second term ω2 ∂v1 /∂x2 in line one in Eq. Once this process has started it continues. The cascade process created by vorticity ∂v1′ ∂x1 >0 ω2′ . The velocity gradient ∂v1 /∂x2 (or ∂v1′ /∂x2 . now with its axis aligned with the x2 axis. v2′ ∂v2′ ∂x2 >0 ∂v1′ > 0. 4. 4. The i 6= j components in the first term on the right side in Eq. The radius of the red fluid element decreases. Its radius decreases (from dashed ∂x1 ω1′ . because vorticity generated by vortex stretching and Vortex tilting .5: The rotation rate of the fluid element (black circles) in Fig.4 increases ∂v ′ and its radius decreases. direction.23 gives a contribution to ω1 (and ω1′ ).4: A fluid element is stretched by line to solid line). Fig.4. As a result. It creates eddies in the x2 and x3 direction (2nd generation). 5. Vortex stretching and vortex tilting qualitatively explain how interaction between vorticity and velocity gradient create vorticity in all three coordinate directions from a disturbance which initially was well defined in one coordinate direction. For each generation the eddies become more and more isotropic as they get smaller. This creates a positive 2 > 0 which stretches the small red ∂x2 fluid element aligned in the x2 direction and increases ω2′ . 4. This shows how vorticity in one direction is transferred to the other two directions through vortex tilting.23 represent vortex tilting. Again.5. take a slender fluid element. 4. The only non-zero component of vorticity vector is ω3 because ω1 = ω2 = Since v3 = 0. From the discussion above we can now understand why turbulence always must be three-dimensional (Item IV on p. we get ωj ∂vi /∂xj = 0. ∂v3 ∂v2 − ≡0 ∂x2 ∂x3 ∂v1 ∂v3 − ≡ 0. If the instantaneous flow is two-dimensional (x1 − x2 plane) we find that the vortex-stretching/tilting term on the right side of Eq. The turbulence is also maintained by these processes. The cascade process created by vorticity 75 vortex tilting interacts with the velocity field and creates further vorticity and so on. ∂x3 ∂x1 . 5. The vorticity and velocity field becomes chaotic and three-dimensional: turbulence has been created.17 vanishes because the vorticity vector and the velocity vector are orthogonal.5. 66). v¯i . 35).3 and v¯i = v¯i .3) One reason why we decompose the variables is that when we measure flow quantities we are usually interested in their mean values rather than their time histories. ¯·. see section 8. i.5) ρ ∂(¯ vj + vj′ ) vi + vi′ )(¯ vi + vi′ ) p + p′ ) vi + vi′ ) ∂ 2 (¯ ∂(¯ ∂(¯ +ρ +µ =− ∂t ∂xj ∂xi ∂xj ∂xj I II III IV Let’s consider the equation term-by-term. vi = v¯i + vi′ p = p¯ + p′ (6. (6.2) v¯i = v¯i + vi′ = v¯i + vi′ where we used the fact that v¯i = v¯i . 6.4) ∂p ∂ 2 vi +µ ∂xi ∂xj ∂xj (6.e. 6.5) The gravitation term. Hence. 6. Eq. 6.1 Time averaged Navier-Stokes When the flow is turbulent it is preferable to decompose the instantaneous variables (for example the velocity components and the pressure) into a mean value and a fluctuating value. The continuity equation and the Navier-Stokes equation for incompressible flow with constant viscosity read ∂vi ∂xi ∂vi vj ∂vi +ρ ρ ∂t ∂xj = 0 = − (6. When we time average Eq.1 we get (6. 6. Turbulent mean flow 76 6 Turbulent mean flow 6. Another reason is that when we want to solve the Navier-Stokes equation numerically it would require a very fine grid to resolve all turbulent scales and it would also require a fine resolution in time (turbulent flow is always unsteady). see Section 8.2 gives vi′ = 0.e. Inserting Eq. p′ = 0 (6. Term I: ∂¯ ∂v ′ ∂¯ ∂¯ vi vi + vi′ ) vi vi ∂(¯ = + i = = ∂t ∂t ∂t ∂t ∂t We assume that the mean flow.1. has been omitted which means that the p is the hydrostatic pressure (i.4. when vi ≡ 0. −ρgi . and hence the term is zero. denotes the time averaged value.6) where we used the fact that vi′ = 0 (see Eq.4) ∂¯ vi + vi′ vi vi ∂¯ ∂v ′ ∂¯ ∂¯ vi = + i = = ∂xi ∂xi ∂xi ∂xi ∂xi (6.7) . Next.4.1) where the bar. see p.6.1 into the continuity equation (6.1. then p ≡ 0. is steady. we use the decomposition in Navier-Stokes equation (Eq. 3 shows that v¯i vj′ = v¯i vj′ = 0 and v¯j vi′ = v¯j vi′ = 0 Hence. jet and wake flow. A new term ρvi′ vj′ appears on the right side of Eq.10) ∂xi 6. The continuity equation applies both for the instantaneous velocity. pressure. The tensor is symmetric (for example v1′ v2′ = v2′ v1′ ). ∂¯ v1 ∂¯ v1 ≪ . Time averaged Navier-Stokes 77 Term II: ∂¯ vj + vj′ ) vi + vi′ )(¯ vi v¯j + v¯i vj′ + vi′ v¯j + vi′ vj′ ∂(¯ = ∂xj ∂xj = vi vj′ ∂¯ ∂vi′ vj′ vi v¯j ∂¯ ∂v ′ v¯j + + i + ∂xj ∂xj ∂xj ∂xj • Section 8. and for the time-averaged velocity.8) = ∂¯ vi ∂ p¯ ∂ µ − + − ρvi′ vj′ ∂xi ∂xj ∂xj (6. channel flow. six stresses) is larger than the number of equations (four: the continuity equation and three components of the Navier-Stokes equations). i. boundary layers along a flat plate. It is an additional stress term due to turbulence (fluctuating velocities) and it is unknown. 6.) the following relations apply v¯2 ≪ v¯1 .1.4 shows that v¯i v¯j = v¯i v¯j .8).11) Reynolds equations closure problem .e. pipe flow.e. vi (Eq. 6. etc.9 which is called the Reynolds stress tensor. ∂x1 ∂x2 (6. Term II reads ∂vi′ vj′ ∂¯ vi v¯j + ∂xj ∂xj Term III: p + p′ ) ∂(¯ ∂ p¯ ∂p′ ∂ p¯ = + = ∂xi ∂xi ∂xi ∂xi Term IV: ∂ 2 v¯i ∂ 2 vi′ ∂ 2 v¯i ∂ 2 (¯ vi + vi′ ) = + = ∂xj ∂xj ∂xj ∂xj ∂xj ∂xj ∂xj ∂xj Now we van finally write the time averaged continuity equation and Navier-Stokes equation ∂¯ vi ∂xi ∂¯ vi v¯j ρ ∂xj = 0 (6.9. 6. ∂vi′ =0 (6.6.1 Boundary-layer approximation For boundary-layer type of flow (i. It represents correlations between fluctuating velocities.1.1. This is called the closure problem: the number of unknowns (ten: three velocity components.9) It is assumed that the mean flow is steady. • Section 8. v¯i (Eq. We need a model for vi′ vj′ to close the equation system in Eq. This equation is the time-averaged Navier-Stokes equation and it is often called the Reynolds equation.1. vi′ . hence it applies also for the fluctuating velocity.4). 6. e. 6.tot x1 and x2 denote the streamwise and wall-normal coordinate. The total shear stress is thus ∂¯ v1 (6.2. Eq. an additional turbulent one – a Reynolds shear stress – appears on the right side of Eq.13.9 using the continuity equation ρ ∂¯ vi ∂¯ vj ∂¯ vi ∂¯ vi v¯j = ρ¯ vj + ρ¯ vi = ρ¯ vj ∂xj ∂xj ∂xj ∂xj (6. In addition to the viscous shear stress. i.12) =0 Using Eq. see Fig. The width (i.6. Wall region in fully developed channel flow 78 x2 v¯1 (x2 ) 2δ xA 2 x1 x3 Figure 6.tot = µ ∂x2 shear stress 6. 6.12. Here the velocity gradient is largest as the velocity drops down to zero at the wall over a very short distance. respectively. length in the x3 direction) of the plates. First we re-write the left side of Eq.9 can be written ∂¯ v1 ∂¯ v1 + ρ¯ v2 ρ¯ v1 ∂x1 ∂x2 ∂¯ v1 ∂ p¯ ∂ ′ ′ µ = − + − ρv1 v2 ∂x1 ∂x2 ∂x2 (6. 6.2 Wall region in fully developed channel flow The region near the wall is very important. Assume steady (∂/∂t = 0).1: Flow between two infinite parallel plates. because they both include the product of one large (¯ v1 or ∂/∂x2 ) and one small (¯ v2 or ∂/∂x1 ) part. 6. Zmax ≫ δ.1. µ∂¯ v1 /∂x2 . Note that the two terms on the left side are of the same order. 6. One important quantity is the wall shear stress which is defined as .14) − ρv1′ v2′ τ12.13) τ12. Zmax . two-dimensional (¯ v3 = ∂/∂x3 = 0) boundary-layer flow. is much larger that the separation between the plates.e. ∂¯ v1 . . τw = µ (6.15) ∂x2 . again. uτ . 6.w From the wall shear stress. consider fully developed channel flow between two infinite plates. we can define a wall friction velocity. as 1/2 τw 2 τw = ρuτ ⇒ uτ = ρ (6.16) In order to take a closer look at the near-wall region. see Fig. let us. In fully developed wall friction velocity .1. 30 (note that h = 2δ) and Eq.tot = τw + x2 = τw 1 − ∂x1 ∂x1 δ (6. i. see Eq. 3. i.7 in [11]) for Reτ = 10 000. Integrating Eq. The buffer region acts as a transition region between these two regions where viscous diffusion of streamwise momentum is .2.18 at p.e. 3.13. δ denotes half width of the channel. 35. The continuity equation gives now v¯2 = 0. 6. is given by Eq. −∂ p¯/∂x1 = τw /δ.2.2: The wall region (adapted from Ch.6. the non-linear convection) term.tot − ρv1′ v2′ = = constant µ ∂x2 ∂x2 (6. 6. the logarithmic region is sometimes called the inertial region. Eq. The streamwise momentum equation.tot . can now be written ∂ p¯ ∂¯ v1 ∂ (6.18) where the total stress. 6. The inner region includes the viscous region (dominated by the viscous diffusion) and the logarithmic region (dominated by turbulent diffusion). Wall region in fully developed channel flow 10−3 10−4 79 10−2 10−1 1 x2 /δ outer region overlap region Wall inner region Centerline log-law region buffer region viscous region + x+ 2 = y 1 5 10 30 100 1000 10000 Figure 6.1 and x+ 2 = x2 uτ /ν denotes the normalized wall distance.e. 6. channel flow. hence they must be constant (see Eq. 6. i.e. ∂¯ v1 /∂x1 = 0.e.13 is zero because we have fully developed flow (∂¯ v1 /∂x1 = 0) and the last term is zero because v¯2 ≡ 0.24 and the text related to this equation).36. because the turbulent stresses stem from the inertial (i. The first term on the left side of Eq. 6. 3.17 from x2 = 0 to x2 gives τ12. see Eq. see Fig.19) At the last step we used the fact that the pressure gradient balances the wall shear stress. the streamwise derivative of the streamwise velocity component is zero (this is the definition of fully developed flow). τ12. − ∂ ∂x2 ∂ p¯ = constant ∂x1 ∂¯ v1 ∂τ12.tot (x2 ) − τw = ∂ p¯ x2 ∂ p¯ x2 ⇒ τ12.14. 6. The wall region can be divided into one outer and one inner region. see Fig.17) 0=− µ + − ρv1′ v2′ ∂x1 ∂x2 ∂x2 We know that the first term is a function only of x1 and the two terms in parenthesis are functions of x2 only. In the logarithmic layer the viscous stress is negligible compared to the Reynolds stress. As we go away from the wall the viscous stress decreases and the turbulent one increases and at x+ 2 ≃ 11 they are approximately equal. Wall region in fully developed channel flow 2000 1 0.8 0 b) 0.16) .19 and Fig.1 100 0. i. further away from the wall it decreases (in fully developed channel flow it decreases linearly with the distance from the wall. : −ρv1′ v2′ /τw . In the inner region. DNS data [14. see Fig. Reτ = 2000.3.4 0.6 0 1 0. b) zoom near the wall.3b).2 0. At the wall. 6. the velocity gradient is directly related to the wall shear stress. : µ(∂¯ v1 /∂x2 )/τw . Note that the total shear stress is constant only close to the wall (Fig. gradually replaced by turbulent diffusion.2 0. the total shear stress is approximately constant and equal to the wall shear stress τw . (see Eq.15 and 6. 6.3a). and the viscous shear stress attains its wall-stress value τw = ρu2τ . see Eq. 15].05 500 x+ 2 0.2. 6.6.3: Reynolds shear stress.2 0. 6. 6. The Reynolds shear stress vanishes at the wall because v1′ = v2′ = 0.8 0 1 Figure 6.6 0.e. a) lower half of the channel.4 0. τw ρ 1 ∂¯ v1 . . = = u2τ = u2τ (6.20) ∂x2 . v¯1 . Often the plus-sign (‘ + ‘) is used to denote inner scaling and equation Eq. we encounter the log-law region.2.4 x2 /δ x+ 2 0.6 1000 0 0 200 0. x2 /δ . it is used in the entire region.8 1500 a) 80 . 0.e. x+ 2 .w µ µ ν Integration gives (recall that both ν and u2τ are constant) v¯1 = 1 2 u x2 + C1 ν τ Since the velocity. 6.1) fairly constant and approximately equal to τw . the friction velocity stemming from the wall shear stress and the viscosity (here we regard viscosity as a quantity related to the wall. see Fig.22) Further away from the wall at 30 . 0. is zero at the wall. is in the region x+ 2 . i.3).003 . Hence the friction velocity. 6. is a suitable velocity scale in the inner logarithmic region.3b. 200 (i. x2 /δ .21 is expressed in inner scaling (or wall scaling) which means that v¯1 and x2 are normalized with quantities related to the wall. 3000 (or 0. ρv1′ v2′ .e. see Fig. x2 /δ 0. uτ . The Reynolds shear stress.21) Equation 6. since the flow is dominated by viscosity). In this region the flow is assumed to be independent of viscosity. the integration constant C1 = 0 so that uτ x2 v¯1 = uτ ν (6. 6.21 can then be written v¯1+ = x+ 2 (6. 6. Reτ = 2000. (6. 2.32) in which a turbulent Reynolds stress is assumed to be equal to the product between the turbulent viscosity and the velocity gradient as −v1′ v2′ = νt ∂¯ v1 ∂x2 (6. i.24 and 6. : DNS What about the length scale? Near the wall.29) Integration gives now v¯1+ = .23) The velocity gradient can now be estimated as ∂¯ v1 uτ = ∂x2 κx2 (6.25 gives (inserting −v1′ v2′ = u2τ ) u2τ = κuτ x2 ∂¯ v1 uτ ∂¯ v1 ⇒ = ∂x2 ∂x2 κx2 (6.2. Another way of deriving the expression in Eq. 6.2. Wall region in fully developed channel flow 2000 81 24 20 v¯1 /uτ 1500 x+ 2 1000 16 12 8 500 4 a) 0 0 5 10 15 20 25 v¯1 /uτ 1 b) 10 100 1000 x+ 2 Figure 6. uτ . data [14.24) based on the velocity scale.27 read ∂¯ v1+ 1 + = ∂x2 κx+ 2 (6. 11.26) so that Eq. represents the turbulence and has the same dimension as ν.e. Hence νt can be expressed as a product of a turbulent velocity scale and a turbulent length scale. 6. an eddy cannot be larger than the distance to the wall and it is the distance to the wall that sets an upper limit on the eddy-size.6.4: Velocity profiles in fully developed channel flow.41 + 5. and the length scale κx2 . and in the log-law region that gives νt = uτ κx2 (6.24 is to use the Boussinesq assumption (see Eq. Hence it seems reasonable to take the wall distance as the characteristic length scale. [m2 /s]. a constant is added so that ℓ = κx2 . : v¯1 /uτ = (ln x+ : v¯1 /uτ = x+ 2 )/0.27) In non-dimensional form Eqs. νt . 15].25) The turbulent viscosity.28) 1 ln x+ 2 + B or κ v¯1 1 x2 uτ +B = ln uτ κ ν (6. As can be seen in Fig.29 we can define the viscous length scale.2 the log-law applies for x+ 2 .2 – shows that the loglaw fit the DNS up to x+ 2 . The constant. can be expressed using the Boussinesq assumption (see Eq.3. the larger the Reynolds number.5.tot x2 x3 Figure 6. the relevant length scale is the boundary layer thickness. Hence. Reynolds stresses in fully developed channel flow 82 τ32. The resulting velocity law is the defect law x v¯1. 3000 (x2 /δ .2.3). as ν (6. In the outer region of the boundary layer. Figure 6. the upper limit for the validity of the log-law is dependent on Reynolds number. where B is an integration constant.3 Reynolds stresses in fully developed channel flow The flow is two-dimensional (¯ v3 = 0 and ∂/∂x3 = 0). The velocity in the log-region and the outer region (often called the wake region) can be written as πx v¯1 1 2Π 2 (6. is called the von K´arm´an constant.c − v¯1 2 (6. 500 (x2 /δ .30) x+ 2 = x2 /ℓν ⇒ ℓν = uτ Equation 6. 6. κ. ℓν . B = 4. the viscous shear stress ∂¯ v3 ∂¯ v2 τ32 = µ =0 (6. Consider the x2 − x3 plane. 11.38. The constants in the log-law are usually set to κ = 0.34) + ∂x2 ∂x3 log-law . 6.6. 0.32) ∂¯ v3 ∂¯ v2 ′ ′ −ρv2 v3 = µt =0 (6. The turbulent part shear stress.31) = FD uτ δ where c denotes centerline.5: Symmetry plane of channel flow. 0. 6.4 – where the Reynolds number is lower than in Fig.41 and B = 5. the larger the upper limit.1 and Π = 0. 6. From Eq.5 are taken from boundary layer flow [17–19].25).33) + ∂x2 ∂x3 because v¯3 = ∂¯ v2 /∂x3 = 0. v2′ v3′ .32) = ln(y + ) + B + sin2 uτ κ κ 2δ where κ = 0. see Fig. 6.29 is the logarithmic law due to von K´arm´an [16]. Since nothing changes in the x3 direction. 1252 + (−0. 0. in Eq. Consider. 6. −0.17.6a. As can be seen.3. but its gradient. This is the force opposing the movement of the fluid. Forces in the v¯1 equation. The force that balances the pressure gradient is the gradient of the Reynolds shear stress. : −ρ(∂v1′ v2′ /∂x2 )/τw . see Eq.2)2 + 0.25 + 0.2).22 /5 = 0. This opposing force has its origin at the walls due to the viscous wall force (viscous shear stress multiplied by area). and it is also zero since v¯3 = ∂¯ v2 /∂x3 = 0.6.6b which shows the forces in the region away from the wall.e. v¯3 = 0). 6. 6.17 we find that it is not really the shear stress that is interesting.7. Reτ = 2000. : −(∂ p¯/∂x1 )/τw . b) lower half of the channel excluding the near-wall region. Looking at Eq. The pressure must decrease in the streamwise direction so that the pressure gradient term. DNS data [14. Now let’s have a look at the forces in the near-wall region. 6. Reynolds stresses in fully developed channel flow 200 83 2000 1800 1600 150 1400 x+ 2 x+ 2 100 1200 1000 800 50 600 400 0 −150 a) −100 −50 0 50 100 150 b) 200 −2 −1 0 1 2 Figure 6. −∂(ρv1′ v2′ )/∂x2 and µ∂ 2 v¯1 /∂x22 represent. Figure 6. together with the pressure gradient. Another way to describe the behaviour of the pressure is to say that there is a pressure drop. the instantaneous turbulent flow is always threedimensional and unsteady.2.125 + 0. Hence v3 6= 0 and v3′ 6= 0 so that v3′2 6= 0.125. 6. This agrees – fortunately – with our intuition.2 + 0.25.25)2 + 0. the time series v3 = v3′ = (−0. Start by looking at Fig. We can imagine that the fluid (air. for example. −∂ p¯/∂x1 . the Reynolds shear stress is zero at the wall (because the fluctuating velocities are zero at the wall) and increases for increasing wall distance. The pressure gradient is constant and equal to one: this is the force driving the flow. for example) is driven by a fan. The reason is that although the time-averaged flow is two-dimensional (i. see Fig. : µ(∂ 2 v¯1 /∂x22 )/τw . On the other hand.6: Fully developed channel flow. 0. With the same argument. 6. 6.1252 + 0.6 presents the forces. see the red fluid particle in Fig.2)/5 = 0 but v3′2 = v32 = (−0.03475 6= 0. the viscous shear stress is negligible except very near the wall. v1′ v3′ = 0. This gives v¯3 = (−0. the forces acting on the fluid.3 presents the Reynolds and viscous shear stresses for fully developed flow. Here the forces are two orders of magnitude larger than in Fig. a) near the lower wall of the channel.6b but they act over a very thin . The intersection of the two shear stresses takes place at x+ 2 ≃ 11. 15]. ′2 2 However note that v3 = v3 6= 0. Figure 6. It is equal to one near the wall and decreases rapidly for increasing wall distance. 0.17 takes a positive value which pushes the flow in the x1 direction. The gradient of the shear stress. −∂ p¯/∂x1 .125 − 0.125. Boundary layer 84 x2 −ρ ∂v1′ v2′ ∂x2 µ ∂ 2 v¯1 ∂x22 1 − 0 0 1 ∂ p¯ ∂x1 −ρ ∂v1′ v2′ ∂x2 v¯1 Figure 6.8 presents the normal Reynolds stresses. In turbulent flow the velocity profile in the center region is much flatter than in laminar flow (cf.7 at p. the latter is smallest because the turbulent fluctuations are dampened by the wall.36 shows why the pressure drop is larger in the former case compared to the latter.36 at p.4.4 Boundary layer Up to now we have mainly discussed fully developed channel flow. as we did for laminar flow. 3. in a boundary layer flow the . What is the difference between that flow and a boundary layer flow? First. The former is largest because the mean flow is in this direction. 3.7. 36). the streamwise stress is largest and the wall-normal stress is smallest.6). or — in other words – why a larger fan is required to push the flow in turbulent flow than in laminar flow. The red (dashed line) and the blue (solid line) + fluid particle are located at x+ 2 ≃ 400 and x2 ≃ 20.36) As can be seen the pressure drop is directly related to the wall shear stress. In this region the Reynolds shear stress gradient term is driving the flow and the opposing force is the viscous force. ρv2′2 and ρv3′2 . 6. k = vi′ vi′ /2. region (x+ 2 ≤ 40 or x2 /δ < 0. This makes the velocity gradient near the wall (and the wall shear stress. τw ) much larger in turbulent flow than in laminar flow: Eq. see the blue fluid particle in Fig.35) where L is the length of the section. ρv1′2 . Note that this is smaller than v1′2 . 6. 37 and Fig. We get τw ∂ p¯ ∆¯ p = =− L ∂x1 δ (6. see Eq.02). We can of course make a force balance for a section of the channel.7: Forces in a boundary layer.8 at p. 37 which reads 0 = p¯1 Zmax 2δ − p¯2 Zmax 2δ − 2τw LZmax (6. 6. is also included.4 and Fig. respectively (see Fig. As can be seen.6. Fig. 6. The turbulent kinetic energy. 3. 6. Figure 6. the left side of Eq. : ρv1′2 /τw .9: Velocity profiles in a boundary layer along a flat plate. However. 6. : DNS convective terms are not zero (or negligible). Third. ◦: k/u2τ .9: the linear and the log-law regions are very similar for the two flows.e. Reτ = 2000.13 is not zero.41 + 5. : ρv2′2 /τw .6. 2.1) is principally the same as for the fully developed channel flow.1 (compared to approximately x2 /δ ≃ 0. However. Boundary layer 85 2000 100 80 1500 60 x+ 2 x+ 2 1000 40 500 a) 0 0 20 2 4 6 8 b) 0 0 2 4 6 8 Figure 6. 6. 6.13.4. the outer part of the boundary layer is highly intermittent. consisting of turbulent/non-turbulent motion. in boundary layer flow the log-law is valid only up to approximately x2 /δ ≃ 0. i.e. the inner region of a boundary layer (x2 /δ < 0. i.3 in channel flow) . the convective terms must also be taken into account. : ρv3′2 /τw . The flow in a boundary layer is continuously developing.8: Normal Reynolds stresses and turbulent kinetic energy. 2000 24 20 v¯2 /uτ 1500 x+ 2 1000 16 12 8 500 4 a) 0 0 5 10 15 v¯2 /uτ 20 25 b) 1 10 100 1000 x+ 2 Figure 6.2. Second. see Fig. The flow in a boundary layer is described by Eq. increases continuously for increasing x1 . δ. : v¯2 /uτ = (ln x+ : v¯2 /uτ = x+ 2 )/0. in a boundary layer flow the wall shear stress is not determined by the pressure drop. DNS data [14. 15]. data [20]. its thickness. i. Figure 7.5 10 12 14 t 16 18 (b) Point 2. 20 −1. of the v velocity.23.15.5 6 15 1 4 0.5) v ′2 = v ′2 fv′ (v ′ )dv ′ −∞ As in Eq. 6.7. Consider the probability density functions of the fluctuations. probability density function is a useful statistical tool to analyze turbulence. v ′2 = 1 2T Z T −T v ′2 (t)dt root-meansquare RMS standard deviation variance . (see Eq. vrms = 1. With a probability density. v ′2 is usually computed by integrating in time.4) 2T −T where T is “sufficiently” large.1) The RMS is the same as the standard deviation which is equal to the square-root of the variance. From the velocity signals we can compute the probability densities (sometimes called histograms). 76).e. 7. i. The root-mean-square (RMS) can be defined from the second moment as 1/2 vrms = v ′2 (7. i. 7 Probability density functions Some statistical information is obtained by forming the mean and second moments. Horizontal red lines show ±vrms . vrms = 0. Probability density functions 20 1.44. as was done in Section 6. as we do when we define a time average. 20 −6 10 12 14 t 16 18 20 (c) Point 3. see Eq.5 −2 −1 −4 5 0 −5 10 12 14 t 16 18 (a) Point 1. Z T 1 v¯ = vdt (7.e. In order to extract more information. so that Z ∞ fv (v)dv = 1 (7. fv . The second moment corresponds to the variance of the fluctuations (or the square of the RMS.e. for example v¯ and v2′2 .4.5 2 10 v 86 ′ 0 0 −0.2) −∞ Normalize the probability functions. Z ∞ (7.1: Time history of v ′ . the mean velocity is computed as Z ∞ v¯ = vfv (v)dv (7.1). vrms = 2. The mean velocity can of course also be computed by integrating over time. 7.3) −∞ Here we integrate over v.1 at p. 1 presents the time history of the v ′ history at three different points in a flow (note that v ′ = 0). The positive values are often larger than +vrms (the peak is actually close to 8vrms ) but the negative values are seldom smaller than −vrms . 2. Vertical red lines show ±vrms . S = 1.2.15 0. S = −0.2: Probability density functions av time histories in Fig. Let us analyze the data at the three points.2 we can judge whether the PDF is symmetric. 7.1a) shows that there exists large positive values but no large negative values.15 0.2 fv′ 0. Figure 7. this means that the PDF should be symmetric which is confirmed in Fig. This indicates that the distribution of v ′ is skewed towards the positive side. This is confirmed in the PDF distribution.1.77. At this point the time history (Fig. Point 3.2b.2a. so that the skewness.81. Point 1. The extreme values of v ′ are approximately ±1. The time history of the velocity fluctuation (Fig. that the extreme value are small and that positive and negative values are equally frequent (i.1 0. 7. The skewness.7.2 1. The fluctuations at this point are much smaller and the positive values are as large the negative values. A probability density function is symmetric if positive values are as frequent and large as the negative values. The red vertical lines show plus and minus RMS of v ′ .27 and F = (c) Point 3.5 0. but instead of “looking” at the probability density functions. 7.3 2.5vrms .05 0. The PDF function in Fig. Probability density functions 87 3 0.25 2 0. of v ′ is defined as Sv ′ = 1 3 vrms Z ∞ ′3 ′ ′ v fv′ (v )dv = −∞ 1 3 2vrms T Z T −T v ′3 (t)dt skewness . The resulting probability densities functions are shown in Fig.09 and F = 7. 7.25 0.68.1c) shows that the fluctuations are clustered around zero and much values are within ±vrms . which is the skewness.5 0.2c confirms that there are many value around zero. the PDF is symmetric).1b and 7. 7. S. 5. F .1 1 0. Sv′ .5 0. 7. Figure 7.e. In Fig. 7. S = −0. It is defined as Z ∞ v ′3 = v ′3 fv′ (v ′ )dv ′ −∞ 3 and is commonly normalized by vrms . The red horizontal lines indicate the RMS value of v ′ . Point 2. we should use a definition of the degree of symmetry.2b. see Fig. see Figs. 7. The time history shows that the positive and the negative values have the same magnitude. 7.51 and F = (b) Point 2. are given for the three time histories. and the flatness.05 0 −2 0 2 4 6 8 0 −5 10 0 v ′ /vrms 0 −4 5 v ′ /vrms −2 0 2 4 v ′ /vrms (a) Point 1. For a Gaussian distribution (v ′ − vrms )2 1 exp − f (v ′ ) = 2 vrms 2vrms for which F = 3.1a) includes some samples which are very large and hence its flatness is large (see caption in Fig.1c and the caption in Fig. see Fig. Z ∞ 1 F = 4 v ′4 fv′ (v)dv vrms −∞ The fluctuations at Point 1 (see Fig. i. There is yet another statistical quantity which sometimes is used for describing turbulent fluctuations. 7. 7. The flatness gives 4 this information. namely the flatness.7. 7. whereas the fluctuation for Point 3 all mostly clustered within ±2vrms giving a small flatness. but it does not tell us if the time history includes few very large fluctuations or if all are rather close to vrms . flatness .2c.3). 7. 7. The variance (the square of RMS) tells us how large the fluctuations are in average. and it is defined computed from v ′4 and normalized by vrms . Probability density functions 88 Note that f must be normalized (see Eq.e.2a). Transport equations for kinetic energy 89 8 Transport equations for kinetic energy In this section and Section 9 we will derive various transport equations.3·0. There are two tricks which often will be used.04 + 0.2·0. 0.02685 N n=1 1. 0.08·0.06)/4 = 0 N n=1 2. −0. Both tricks simply use the product rule for derivative backwards. −0. Trick 1: Using the product rule we get ∂Ai Bj ∂Bj ∂Ai = Ai + Bj ∂xk ∂xk ∂xk (8.18·0.e. −0.3 + 0.08] v2′ = [0.06)/4 = 0. 0.1 Rules for time averaging 8.3) This trick is usually used backwards.2) and then we call it the “product rule backwards”.15+0. Trick 2: Using the product rule we get ∂Ai ∂Ai ∂Ai 1 1 ∂Ai Ai Ai = Ai = + Ai 2 ∂xj 2 ∂xj ∂xj ∂xj (8.1) This expression can be re-written as Ai ∂Ai ∂Bj ∂Ai Bj = − Bj ∂xk ∂xk ∂xk (8.3.25+0.18 − 0.15. i.n ! =0·0=0 However. the time average of their product is not zero.25.08)/4 = 0 N n=1 1.18.4) 8.e.15 − 0.04−0.n ! N 1 X ′ v N n=1 2.25 + 0.n 2.06] v1′ = v2′ = so that v1′ N 1 X ′ v = (0.n v2′ = N 1 X ′ v N n=1 1.04. v1′ v2′ N 1 X ′ ′ = v v = (0. i.2.n . Ai ∂Ai 1 ∂Ai Ai = ∂xj 2 ∂xj (8.1 What is the difference between v1′ v2′ and v1′ v2′ ? Let the time series of v1′ and v2′ be v1′ = [0.2 − 0.8.n N 1 X ′ v = (0.1. 1. The Exact k Equation 90 2 8.22 + 10 · 0. N 1 X v¯1 v1′2 = N n=1 N 1 X v1.252 + 0.082 )/4 = 0.2.5 and divide by density.2 What is the difference between v1′2 and v1′ ? v1′2 = v2′2 = but 2 v1′ 2 v2′ N 1 X ′2 v = (0.9 from Eq. the left side can be rewritten as vi′ ∂ ∂ ′ vj + vj′ ) − v¯i v¯j = vi′ (¯ vi + vi′ )(¯ v¯i vj + vi′ v¯j + vi′ vj′ .0422 N n=1 1.06)/4] = 0 8.32 + 0.8.04 + 0.2 − 0. multiply by vi′ and time average which gives vi′ ∂ [vi vj − v¯i v¯j ] = ∂xj ∂vi′ vj′ ′ ∂ ∂2 1 [p − p¯] + νvi′ [vi − v¯i ] + v − vi′ ρ ∂xi ∂xj ∂xj ∂xj i (8.15 − 0.n = 10 · (0.152 + 0. k = 21 vi′ vi′ .18 − 0.n N 1 X ′ v N n=1 2.25 + 0. 6.02255 N n=1 2.042 + 0.062 )/4 = 0. is derived from the Navier-Stokes equation.22 + 0.32 + 10 · 0.182 + 0.08)/4] = 0 2 = [(0.n v = N n=1 N n=1 1.n N 1 X ′2 v = (0.082 )/4 = 0.082 )/4 = 0.22 + 0.182 + 0.1. 6. Again.422 8.3 Show that v¯1 v1′2 = v¯1 v1′2 Assume that v¯1 = 10.m N m=1 ! ′2 v1.n = = (10 · 0. ∂xj ∂xj (8.6) .422 ! ! N N X X 1 1 ′2 v¯1 v1′2 = v1. 6. We subtract Eq.n = = N 1 X ′ v N n=1 1.5).5) Using vj = v¯j + vj′ .182 + 10 · 0. we assume incompressible flow (constant density) and constant viscosity (cf.n !2 !2 2 = [(0.4 Show that v¯1 = v¯1 v¯1 = N 1 X = (10 + 10 + 10 + 10)/4 = 10 = v¯1 N n=1 8.2 The Exact k Equation The equation for turbulent kinetic energy.32 + 0.1. Eq.3 + 0. 6 can be written as (replace v¯j by vj′ and use the same technique as in Eq.5 is re-written using Trick 1 1 ∂p′ vi′ 1 ∂v ′ 1 ∂p′ vi′ 1 ∂p′ =− + p′ i = − − vi′ ρ ∂xi ρ ∂xi ρ ∂xi ρ ∂xi (8. 8.2.10) The first term on the right side of Eq. we use Trick 2 v¯j vi′ ∂vi′ ∂xj ! ∂ = v¯j ∂xj 1 ′ ′ vv 2 i i = v¯j ∂ ∂ (k) = (¯ vj k) ∂xj ∂xj (8.14) .10).9) 1 ∂ vj′ vi′ vi′ . The second term on the right side of Eq.e. a turbulent kinetic energy. 8. 8.8. 2 ∂xj (8.12) applying Trick 1 (A = vi′ and B = ∂vi′ /∂xj ). 8. For the first term in Eq. 8. Now we can assemble the transport equation for the fluctuation. 6.13) The last term on the right side of Eq. 8. 8.11.12 we use the same trick as in Eq. The Exact k Equation 91 Using the continuity equation ∂vj′ /∂xj = 0 (see Eq. 8.12 and 8.6 we start using ∂¯ vj /∂xj = 0 vi′ ∂ ∂v ′ (vi′ v¯j ) = v¯j vi′ i ∂xj ∂xj (8. 8. Equations 8. 8.5 is zero because it is time averaging of a ¯ b′ = a ¯b¯′ = 0.7. IV (8. Convection.11) where the continuity equation was used at the last step.13 give ∂ ∂¯ vj k ∂¯ vi = −vi′ vj′ − ∂xj ∂xj ∂xj I II 1 ′ ′ 1 ′ ′ ′ ∂v ′ ∂vi′ ∂k −ν i vj p + vj vi vi − ν ρ 2 ∂xj ∂xj ∂xj III The terms in Eq. the first term is rewritten as vi′ ∂¯ vi ∂ . i.8) Next.9) The third term in Eq.5 can be written νvi′ ∂ 2 vi′ ∂ = νvi′ ∂xj ∂xj ∂x2 ∂vi′ ∂xj ∂ =ν ∂xj ∂v ′ ∂vi′ ∂vi′ ′ vi −ν i ∂xj ∂xj ∂xj (8.9 so that ∂ ∂vi′ ∂vi′ ∂vi′ 1 ′ ′ ′ v vi =ν = + vi ∂xj ∂xj 2 i ∂xj ∂xj ′ ′ 1 ∂ 2 vi′ vi′ ∂ 2k 1 ∂vi vi ∂ =ν =ν ν ∂xj 2 ∂xj 2 ∂xj ∂xj ∂xj ∂xj ∂ ν ∂xj (8.14 have the following meaning.7) For the second term in Eq. v¯i vj′ = vi′ vj′ ∂xj ∂xj (8. 8. 8.9. I. In the fully turbulent region of a boundary layer. 6. i. When the acceleration term and the fluctuating velocity are in opposite directions (i. 8. which is a scalar product between two vectors.6). It can be noted that the production term is an acceleration term. The dissipation term can now be written vi′ Ti′ . The Reynolds stresses (which appear in P k ) are larger for large eddies than for small eddies 2. 5. respectively. How do we know that the majority of the dissipation takes place at the smallest scales? First. vi′ . must be of the same magnitude. when P k > 0). takes places at the smallest scales. it is dissipative). for small wavenumbers. The Exact k Equation 92 II. The last term is viscous diffusion. The large turbulent scales extract energy from the mean flow.e. It is largest for the energy-containing eddies. the product of an inertial force per unit mass (acceleration) and a fluctuating velocity. In order to extract energy from the mean flow. see Fig. The dissipation term stems from the viscous term (see the second term on the last line in Eq. let us investigate how the time scale . vi′ ∂τij′ /∂xj . ∂¯ vi /∂xj . see Eq. There are two arguments for this: 1. When the viscous stress vector is in the opposite direction to the fluctuating velocity. 4. Production.6). this means that the viscose stress vector performs work and transforms kinetic energy into internal energy. The velocity-fluctuation term originates from the convection term (the last term on the right side of Eq. P k .2. The two first terms represent turbulent diffusion by pressure-velocity fluctuations.3) we assume that the viscous dissipation. 8.e. see Fig.15) where C k . 5.6) in the Navier-Stokes equation. The magnitude of the velocity gradient is given by Eq. vj′ ∂¯ vi /∂xj . Dissipation. P k .14.e. and velocity fluctuations.e. Above. III. it is stated that the production takes place at the large energy-containing eddies.1. i. Ti′ = ∂τij′ /∂xj . 8. This term (including the minus sign) is almost always positive. the term is negative (i. A force multiplied with a velocity corresponds to work per unit time. 8. The transport equation for k can also be written in a simplified easy-to-read symbolic form as C k = P k + Dk − ε (8. the time scale of the eddy and the mean velocity gradient. This term originates from the convection term (the first term on the right side of Eq. Dk and ε correspond to terms I-IV in Eq.e. The term (excluding the minus sign) is always positive (it consists of velocity gradients squared). It can be written as . the mean flow performs work on the fluctuating velocity field. This term is responsible for transformation of kinetic energy at small scales to thermal energy. both time scales are proportional to κx2 /uτ .e. ε.1 shows how different velocity profiles create different largest eddies. i. Figure 8.24 and the time scale of a large eddy is also given by ℓ0 /v0 = κx2 /uτ . IV. This requirement is best satisfied by the large scales. The largest eddies created by velocity profile A is much smaller than by velocity profile B. ε.8. i. multiplied by a fluctuating velocity. we assume that the large eddies contribute much more to the production term more than the small eddies. It is largest for large wavenumbers. for example. In the cascade process (see Section 5.2. because the gradient of profile A acts over a much shorter length than the gradient of profile B. The divergence of τij′ is a force vector (per unit mass).2. 2.2. E(κ) ε(κ) εκ εκ+dκ κ dκ κ + dκ κ Figure 8.1: The size of the largest eddies (dashed lines) for different velocity profiles. 5.8. . see Fig.2: Zoom of the energy spectrum for a wavenumber located in Region II or III. The Exact k Equation 93 A B δB δA x2 x1 Figure 8. 16) ∝ ∝ vκ2 ∂x κ ℓκ since ℓκ ∝ κ−1 . We find from ℓ2κ /τκ3 = ℓ20 /τ03 that for decreasing eddy size (decreasing ℓκ ).17. the region may be in the −5/3 region or in the dissipation region. “true” viscous dissipation. ε As a final note to the discussion in the previous section.e. is expressed as the viscosity times the square of the velocity gradient. 5.17) ∝ κ−2/3 ∂x κ Thus the viscous dissipation at wavenumber κ can be estimated as (see the last term in Eq. We assume that no mean flow energy production occurs between κ and κ + dκ. ε.1 and Fig.14. The cascade process assumes that this energy transfer per unit time is the same for each eddy size. see Eq.2.19) where τ0 and ℓ0 are constants (we have chosen the large scales.2)? Recall that the viscous dissipation. see Fig. ε(κ) does indeed increase for increasing wavenumber. ε. i.2. Now we know that the energy spectrum E ∝ vκ2 /κ ∝ κ−5/3 (see Eq. Consider the inertial subrange. (8.1 Spectral transfer dissipation εκ vs. τ0 and ℓ0 ). the −5/3 region). the former is the energy transferred from eddy to eddy per unit time. is equal to the viscous dissipation. If κ and κ + dκ are located in the inertial region (i. τκ = ℓκ ℓ0 2/3 τ0 (8. How large is ε at wavenumber κ (denoted by ε(κ). 8. 8. Now consider Fig.2 which shows a zoom of the energy spectrum. the time scale.14. 8. Turbulent kinetic per unit time energy enters at wavenumber κ at a rate of εκ and leaves at wavenumber κ + dκ a rate of εκ+dκ . see Eq. and the “true” viscous dissipation.e. i. also decreases.18) ε=ν i ∂xj ∂xj ∂x κ i. and the latter is the energy transformed per unit time to internal energy for the entire spectrum (occurring mainly at the small.2. see Section 8. 5. The energy that is transferred in spectral space per unit time.8. If there is viscous dissipation at wavenumber κ (which indeed is the case if the zoomed . i. τκ . then the usual assumption is that εκ ≃ εκ+dκ and that there is no viscous dissipation to internal energy. εκ = ε = vκ3 /ℓκ = ℓ2κ /τκ3 = ℓ20 /τ03 .e.2. 8.20) ∂x κ ℓκ which is the same as Eq.e. dissipative scales). Hence ∂v vκ ∝ ∝ τκ−1 ∝ ℓκ−2/3 ∝ κ2/3 . The energy transferred from eddy-to-eddy per unit time in spectral space can also be used for estimating the velocity gradient of an eddy. i. The velocity gradient for an eddy characterized by velocity vκ and lengthscale ℓκ can be estimated as 1/2 ∂v vκ κ (8. ε. The Exact k Equation 94 varies with eddy size. (8. 5. ε(κ) ≃ 0. it may be useful to look at the difference between the spectral transfer dissipation εκ .13) in the inertial region which gives 1/2 ∂v κ ∝ κ−1/3 κ ∝ κ2/3 (8.e. 8.14) 2 ∂v ∂v ′ ∂vi′ ⇒ ε(κ) ∝ ∝ κ4/3 . 8.e. εκ . The Exact k Equation: 2D Boundary Layers 0.2 a) −0. 8.01 0 0 −0.3 95 0. ε(κ) = εκ+dκ − εκ (8.03 0.04 b) 100 200 300 400 500 x+ 2 Figure 8.3 presents the terms in Eq. region is located in the dissipative region). Note that the production and the dissipation terms close to the wall are two orders of magnitude larger than in the logarithmic region (Fig. ◦: ν∂ k/∂x2 . i. Terms in the k equation scaled by u4τ /ν.3.2 0. 15]. 8.02 0.3a) the other terms do also play a role.8.03 0 10 20 x+ 2 30 40 −0.4b)).02 −0. Closer to the wall (Fig. 8. the former decreases and the magnitude of the latter increases with wall distance and their product takes its maximum at x+ 2 ≃ local equilibrium . The left side is – since the flow is fully developed – zero.3b) all terms are negligible except the production term and the dissipation term which balance each other. 97. The velocity gradient is largest at the wall (see Fig.22 for fully developed channel flow.e. P k = −v1′ v2′ ∂¯ where the shear stress is zero (see Fig. The turbulence kinetic energy is produced by its main source term. isotropic scales for which all derivatives are of the same order and hence the usual boundary-layer approximation ∂/∂x1 ≪ ∂/∂x2 does not apply for these scales. This is called local equilibrium.3 The Exact k Equation: 2D Boundary Layers In 2D boundary-layer flow.01 −0. see p. b) Outer region. Since the region near the wall is dominated by viscosity the turbulent diffusion terms due to pressure and velocity are also small. the production v1 /∂x2 . Reτ = 2000. Figure 8.4a) term.04 0. This is because the dissipation term is at its largest for small.3: Channel flow at Reτ = 2000.1 0.3b). The dissipation term and the viscous diffusion term attain their largest value at the wall and they much be equal to each other since all other terms are zero or negligible. then ε(κ) is simply obtained through an energy balance per unit time. : P k. At the wall the turbulent fluctuations are zero which means that the production term is zero. 8. 8. In the outer region (Fig. the exact k equation reads v2 k ∂¯ v1 k ∂¯ ∂¯ v1 + = −v1′ v2′ ∂x1 ∂x2 ∂x2 ∂k 1 ∂ 1 ′ ′ ∂v ′ ∂vi′ p v2 + v2′ vi′ vi′ − ν − −ν i ∂x2 ρ 2 ∂x2 ∂xj ∂xj (8. 8.1 −0. for which ∂/∂x2 ≫ ∂/∂x1 and v¯2 ≪ v¯1 . DNS data [14. a) Zoom near the wall. +: −∂v2′ vi′ vi′ /2/∂x2 .21) 8. 2 2 : −ε. ▽: −∂v ′ p′ /∂x2 .22) Note that the dissipation includes all derivatives. 8.4. Spatial vs. spectral energy transfer 96 40 40 0.02 20 0.01 35 30 x+ 2 x2 /δ 25 x+ 2 20 15 10 5 0 0 0.2 0.4 0.6 0.8 1 0 ∂¯ v1+ /∂x+ 2 (a) Velocity gradient 0.2 0.4 0.6 0.8 0 1 −v1′ v2′ /u2τ (b) Reynolds shear stress Figure 8.4: Channel flow at Reτ = 2000. DNS data [14, 15]. 11. Since P k is largest here so is also k, see Fig. 6.8. k is transported in the x2 direction by viscous and turbulent diffusion and it is destroyed (i.e. dissipated) by ε. 8.4 Spatial vs. spectral energy transfer In Section 5.3 we discussed spectral transfer of turbulent kinetic energy from large to small eddies (which also applies to the transport of the Reynolds stresses). In Section 8.2 we derived the equation for spatial transport of turbulent kinetic energy. How are the spectral transfer and the spatial transport related? The reason that we in Section 5.3 only talked about spectral transfer was that we assumed homogeneous turbulence in which the spatial derivatives of the time-averaged turbulent quantities are zero, for example ∂v1′2 /∂xi = 0, ∂k/∂xi = 0 etc. (Note that the derivatives of the instantaneous turbulent fluctuations are non-zero even in homogeneous turbulence, i.e. ∂v1′ /∂xi 6= 0; the instantaneous flow field in turbulent flow is – as we mentioned at the very beginning at p. 66 – always three-dimensional and unsteady). In homogeneous turbulence the spatial transport terms (i.e. the convective term, term I, and the diffusion terms, term III in Eq. 8.14) are zero. Hence, in homogeneous turbulence there is no time-averaged spatial transport. However, there is spectral transfer of turbulent kinetic energy which takes place in wavenumber space, from large to small eddies. The production term (term II in Eq. 8.14) corresponds to the process in which large energycontaining eddies extract energy from the mean flow. The dissipation term (term IV in Eq. 8.14) corresponds to transformation of the turbulent kinetic energy at the small eddies to thermal energy. However, real flows are hardly ever homogeneous. Some flows may have one or two homogeneous directions. Consider, for example, fully developed channel turbulent flow. If the channel walls are very long and wide compared to the distance between the walls, 2δ, then the turbulence (and the flow) is homogeneous in the streamwise direction and the spanwise direction, i.e. ∂¯ v1 /∂x1 = 0, ∂vi′2 /∂x1 = 0, ′2 ∂vi /∂x3 = 0 etc. In non-homogeneous turbulence, the cascade process is not valid. Consider a large, turbulent eddy at a position xA 2 (see Fig. 6.1) in fully developed channel flow. The ′ ′ instantaneous turbulent kinetic energy, kκ = vκ,i vκ,i /2, of this eddy may either be transferred in wavenumber space or transported in physical (spatial) space, or both. It may first be transported in physical space towards the center, and there lose its kinetic energy to smaller eddies. This should be kept in mind when thinking in terms of the homogeneous turbulence 8.5. The overall effect of the transport terms 97 cascade process. Large eddies which extract their energy from the mean flow may not give their energy to the slightly smaller eddies as assumed in Figs. 5.2 and 5.1, but kκ may first be transported in physical space and then transferred in spectral space. In the inertial range (Region II), however, the cascade process is still a good approximation even in non-homogeneous turbulence. The reason is that the transfer of turbulent kinetic energy, kκ , from eddy-to-eddy, occurs at a much faster rate than the spatial transport by convection and diffusion. In other words, the time scale of the cascade process is much smaller than that of convection and diffusion which have no time to transport kκ in space before it is passed on to a smaller eddy by the cascade process. We say that the turbulence at these scales is in local equilibrium. The turbulence in the buffer layer and the logarithmic layer of a boundary layer (see Fig. 6.2) is also in local equilibrium. In Townsend [21], this is (approximately) stated as: the local rates of turbulent kinetic energy (i.e. production and dissipation) are so large that aspects of the turbulent motion concerned with these processes are independent of conditions elsewhere in the flow. This statement simply means that production is equal to dissipation, i.e. P k = ε, see Fig. 8.3. In summary, care should be taken in non-homogeneous turbulence, regarding the validity of the cascade process for the large scales (Region I). 8.5 The overall effect of the transport terms The overall effect (i.e. the net effect) of the production term is to increase k, i.e. if we integrate the production term over the entire domain, V , we get Z Pk dV > 0 (8.23) V Similarly, the net effect of the dissipation term is a negative contribution, i.e. Z −εdV < 0 (8.24) V What about the transport terms, i.e. convection and diffusion? Integration of the convection term over the entire volume, V , gives, using Gauss divergence law, Z Z ∂¯ vj k dV = v¯j knj dS (8.25) S V ∂xj where S is the bounding surface of V . This shows that the net effect of the convection term occurs only at the boundaries. Inside the domain, the convection merely transports R k with out adding or subtracting anything to the integral of k, V kdV ; the convection acts as a source term in part of the domain, but in the remaining part of the domain it acts as an equally large sink term. Similarly for the diffusion term, we get Z 1 ′ ′ ′ ∂ 1 ′ ′ ∂k V − v v v + pv −ν 2 j k k ρ j ∂xj V ∂xj (8.26) Z 1 ′ ′ ′ 1 ′ ′ ∂k =− v v v + pv −ν nj dS 2 j k k ρ j ∂xj S The only net contribution occurs at the boundaries. Hence, Eqs. 8.25 and 8.26 show that the transport terms only – as the word implies – transports k without giving any local equilibrium 8.6. The transport equation for v¯i v¯i /2 98 1 0.8 0.6 0.4 0.2 0 0 20 40 60 80 100 x+ 2 Figure 8.5: Channel flow at Reτ = 2000. Comparison of mean and fluctuating dissipation terms, see Eqs. 8.35 and 8.36. Both terms are normalized by u4τ /ν. DNS data [14, 15]. : ν(∂¯ v1 /∂x2 )2 ; : ε. net effect except at the boundaries. Mathematically these terms are called divergence terms, i.e. they can both be written as the divergence of a vector Aj , ∂Aj ∂xj where Aj for the convection and the diffusion term reads v¯j k convection term 1 ′ ′ ′ 1 ′ ′ ∂k Aj = diffusion term v v v + pv −ν − 2 j k k ρ j ∂xj (8.27) (8.28) 8.6 The transport equation for v¯i v¯i /2 The equation for K = v¯i v¯i /2 is derived in the same way as that for vi′ vi′ /2. Multiply the time-averaged Navier-Stokes equations, Eq. 6.9, by v¯i so that v¯i ∂vi′ vj′ ∂¯ vi v¯j ∂ 2 v¯i 1 ∂ p¯ = − v¯i + ν¯ vi − v¯i . ∂xj ρ ∂xi ∂xj ∂xj ∂xj (8.29) Using the continuity equation and Trick 2 the term on the left side can be rewritten as v¯i 1 ∂¯ vi v¯i ∂¯ vi v¯j ∂¯ vi ∂¯ vj K = v¯j v¯i = v¯j 2 = ∂xj ∂xj ∂xj ∂xj (8.30) The viscous term in Eq. 8.29 is rewritten in the same way as the viscous term in Section 8.2, see Eqs. 8.12 and 8.13, i.e. ν¯ vi ∂2K ∂¯ vi ∂¯ vi ∂ 2 v¯i =ν −ν . ∂xj ∂xj ∂xj ∂xj ∂xj ∂xj (8.31) Equations 8.30 and 8.31 inserted in Eq. 8.29 gives ∂vi′ vj′ ∂¯ vj K ∂2K v¯i ∂ p¯ ∂¯ vi ∂¯ vi =ν − −ν − v¯i . ∂xj ∂xj ∂xj ρ ∂xi ∂xj ∂xj ∂xj (8.32) divergence terms 8.6. The transport equation for v¯i v¯i /2 99 The last term is rewritten using Trick 1 as −¯ vi ∂vi′ vj′ ∂¯ vi vi′ vj′ ∂¯ vi =− + vi′ vj′ . ∂xj ∂xj ∂xj (8.33) Note that the first term on the right side differs to the corresponding term in Eq. 8.14 by a factor of two since Trick 2 cannot be used because v¯i 6= vi′ . Inserted in Eq. 8.32 gives (cf. Eq. 8.14) vi ∂¯ vi ∂¯ ∂ ∂¯ vi v¯i ∂ p¯ ∂K ∂¯ vj K ′ ′ ′ ′ = vi vj v¯i vi vj − ν −ν − − ∂xj ∂xj ρ ∂xi ∂xj ∂xj ∂xj ∂xj (8.34) −P k , sink source εmean , sink On the left side we have the usual convective term. On the right side we find: • loss of energy to k due to the production term • work performed by the pressure gradient; in channel flow, for example, this term gives a positive contribution to K since −¯ v1 ∂ p¯/∂x1 > 0 • diffusion by velocity-stress interaction • viscous diffusion • viscous dissipation, εmean Note that the first term in Eq. 8.34 is the same as the first term in Eq. 8.14 but with opposite sign: here we clearly can see that the main source term in the k equation (the production term) appears as a sink term in the K equation. In the K equation the dissipation term and the negative production term (representing loss of kinetic energy to the k field) read −ν ∂¯ vi vi ∂¯ vi ∂¯ + vi′ vj′ , ∂xj ∂xj ∂xj (8.35) and in the k equation the production and the dissipation terms read −vi′ vj′ ∂v ′ ∂vi′ ∂¯ vi −ν i ∂xj ∂xj ∂xj (8.36) The gradient of the time-averaged velocity field, v¯i , is much smoother than the gradient of the fluctuating velocity field, vi′ . Hence, the dissipation by the turbulent fluctuations, ε, in the turbulent region (logarithmic region and further out from walls), is much larger than the dissipation by the mean flow (left side of Eq. 8.35). This is seen in Fig. 8.5 (x+ 2 & 15). The energy flow from the mean flow to internal energy is illustrated in Fig. 8.6. The major part of the energy flow goes from K to k and then to dissipation. In the viscous-dominated wall region, the mean dissipation, ν(∂¯ v1 /∂x2 )2 , is much larger than ε, see Fig. 8.5 for x+ . 10. At the wall, the mean dissipation takes the 2 value ν = 1/2000 (normalized by u4τ /ν). 9. Transport equations for Reynolds stresses 100 ∆T εm n ea ε Pk K k Figure 8.6: Transfer of energy between mean kinetic energy (K), turbulent kinetic energy (k) and internal energy (denoted as an increase in temperature, ∆T ). K = 1 ¯i v¯i and k = 12 vi′ vi′ . 2v 9 Transport equations for Reynolds stresses In Section 8 we derived transport equations for kinetic turbulent energy, k, which is the trace of the Reynolds stress tensor vi′ vj′ divided by two, i.e. k = vi′ vi′ /2. This means that k is equal to twice the sum of the diagonal components of vi′ vj′ , i.e. k = 0.5(v1′2 + v2′2 +v3′2 ). Here we will now derive the transport equation for the Reynolds stress tensor. This is an unknown in the time-averaged Navier-Stokes equations, Eq. 6.9, which must be known before Eq. 6.9 can be solved. The most accurate way to find vi′ vj′ is, of course, to solve a transport equation for it. This is computationally expensive since we then need to solve six additional transport equations (recall that vi′ vj′ is symmetric, i.e. v1′ v2′ = v2′ v1′ etc). Often, some simplifications are introduced, in which vi′ vj′ is modeled by expressing it in a turbulent viscosity and a velocity gradient. Two-equations models are commonly used in these simplified models; no transport equation for vi′ vj′ is solved. Now let’s start to derive the transport equation for vi′ vj′ . This approach is very similar to that we used when deriving the k equation in Section 8.2. Steady, incompressible flow with constant density and viscosity is assumed. Subtract Eq. 6.9 from Eq. 6.5 and divide by density, multiply by vj′ and time average and we obtain vj′ ∂ [vi vk − v¯i v¯k ] = ∂xk ∂v ′ v ′ 1 ∂ ′ ∂ 2 vi′ p + νvj′ + i k vj′ − vj′ ρ ∂xi ∂xk ∂xk ∂xk (9.1) Equation 6.9 is written with the index i as free index, i.e. i = 1, 2 or 3 so that the equation is an equation for v1 , v2 or v3 . Now write Eq. 6.9 as an equation for vj and multiply this equation by vi′ . We get vi′ ∂ [vj vk − v¯j v¯k ] = ∂xk ∂ 2 vj′ ∂vj′ vk′ ′ ∂ ′ 1 p + νvi′ + v − vi′ ρ ∂xj ∂xk ∂xk ∂xk i (9.2) 9. Transport equations for Reynolds stresses 101 It may be noted that Eq. 9.2 is conveniently obtained from Eq. 9.1 by simply switching indices i and j. Adding Eqs. 9.1 and 9.2 together gives vj′ ∂ ∂ [vi vk − v¯i v¯k ] + vi′ [vj vk − v¯j v¯k ] = ∂xk ∂xk 1 ∂p′ 1 ∂p′ − vi′ − vj′ ρ ∂xj ρ ∂xi +νvi′ ∂ 2 vj′ ∂ 2 vi′ + νvj′ ∂xk ∂xk ∂xk ∂xk + (9.3) ∂vj′ vk′ ′ ∂vi′ vk′ ′ v + v ∂xk i ∂xk j Note that each line in the equation is symmetric: if you switch indices i and j in any of the lines nothing changes. This is important since the tensor vi′ vj′ is symmetric. Furthermore, you can check that the equation is correct according to the tensor notation rules. Indices i and j appear once in each term (not more, not less) and index k (the dummy index) appears exactly twice in each term (implying summation). Note that it is correct to use any other index than k in some terms (but you must not use i and j). You could, for example, replace k with m in the first term and with q in the second term; it is permissible, but usually we use the same dummy index in every term. Using vi = v¯i + vi′ , the first line can be rewritten as vj′ ∂ ∂ ′ [¯ vi vk′ + vi′ v¯k + vi′ vk′ ] + vi′ v¯j vk + vj′ v¯k + vj′ vk′ ∂xk ∂xk (9.4) Using the continuity equation the first terms in the two groups are rewritten as vj′ vk′ ∂¯ vi ∂¯ vj + vi′ vk′ ∂xk ∂xk (9.5) We merge the second terms in the two groups in Eq. 9.4. vj′ ∂vj′ v¯k ∂vj′ ∂vi′ v¯k ∂v ′ + vi′ = v¯k vj′ i + v¯k vi′ ∂xk ∂xk ∂xk ∂xk ′ ′ ′ ′ ∂vi vj v¯k ∂vi vj = = v¯k ∂xk ∂xk (9.6) The continuity equation was used twice (to get the right side on the first line and to get the final expression) and the product rule was used backwards to get the second line. Re-writing also the third terms in the two groups in Eq. 9.4 in the same way, the second and the third terms in Eq. 9.4 can be written ∂vi′ vj′ vk′ ∂vi′ vj′ v¯k + ∂xk ∂xk (9.7) The second line in Eq. 9.3 is also re-written using Trick 1 − 1 ∂vj′ 1 ∂ ′ ′ 1 ∂ ′ ′ 1 ′ ∂vi′ + p′ vi p − vj p + p ρ ∂xj ρ ∂xi ρ ∂xj ρ ∂xi (9.8) • Every term – or group of terms – should include the free indices i and j (only once).12) III 1 + p′ ρ ∂vj′ ∂vi′ + ∂xj ∂xi V − 2ν ∂vi′ ∂vj′ ∂xk ∂xk IV Note that the manipulation in Eq.3 is also re-written using Trick 1 ∂vj′ ∂v ′ ∂vj′ ∂ ∂ ∂vi′ ′ ′ ν +ν − 2ν i vi vj ∂xk ∂xk ∂xk ∂xk ∂xk ∂xk The product rule is used backwards to merge the two first terms so that the third line in Eq.10) The third line in Eq.9 allows the diffusion (term III) to be written on a more compact form.5.9) 1 ∂vj′ 1 ∂ ′ ′ 1 ∂ ′ ′ 1 ′ ∂vi′ + p′ vi p − δik vj p + p ρ ∂xk ρ ∂xk ρ ∂xj ρ ∂xi (9.3 reads ! ′ ′ ∂v ∂ ∂v ′ ∂vj′ ∂v j ν vi′ + vj′ i − 2ν i ∂xk ∂xk ∂xk ∂xk ∂xk ! (9. 9. Therefore we re-write the derivative in the two first terms as ∂ ∂ ∂ ∂ = δjk and = δik ∂xj ∂xk ∂xi ∂xk so that −δjk (9.10 and 9. • Every term – or group of terms – should be symmetric in i and j. 9. 9. it is always useful to check that the equation is correct according to the tensor notation rules. Transport equations for Reynolds stresses 102 It will later turn out that it is convenient to express all derivatives as ∂/∂xk . 9.11 so that ∂¯ vi ∂¯ vj ∂ (¯ vk vi′ vj′ ) = −vj′ vk′ − vi′ vk′ ∂xk ∂xk ∂xk I ∂ − ∂xk II ∂vi′ vj′ 1 1 vi′ vj′ vk′ + δjk vi′ p′ + δik vj′ p′ − ν ρ ρ ∂xk ! (9. 9.13) .12 can also be written in a simplified easy-to-read symbolic form as Cij = Pij + Dij + Πij − εij (9.3 are zero because a ¯ b′ = a ¯b¯′ = 0.9. Put the first term in Eq.11) ∂ 2 vi′ vj′ ∂vi′ vj′ ∂vi′ ∂vj′ ∂vi′ ∂vj′ ∂ − 2ν =ν − 2ν =ν ∂xx ∂xk ∂xk ∂xk ∂xk ∂xk ∂xk ∂xk The terms on the fourth line in Eq. • A dummy index (in this case index k) must appear exactly twice (=summation) in every term Equation 9.7 on the left side and the second term on the right side together with Eqs. After a derivation. 9. 9. We can now put everything together. For a normal stress that is larger than v ′2 . Eq. 9. The physical meaning of this term is to redistribute energy between the normal stress components (if we transform Eq.e.1 Source terms In order to analyze the Reynolds stress equation. 9. It is often called the Robin Hood term because it – as Robin Hood – “takes from the rich and gives to the poor”. vi′ vj′ . 9. Since it is an equation for a second-order tensor. Eq.14 is to first derive Eq. A positive source term in a transport equation. Note that the trace of the pressure-strain term is zero. i.12 with the equation for turbulent kinetic energy. It is called the Reynolds stress equations.14.12 to the principal coordinates of vi′ vj′ there are no shear stresses. 9.9. we will now look at the source terms. The average of ′2 = v ′ v ′ /3. Eq. for example Φ. which is called the pressure strain term.15) because of the continuity equation and this is the reason why this term does not appear in the k equation. a production term (II). In both the k and the vi′ vj′ equations there is a convection term (I). 9.17) The second line shows that it is the streamwise and spanwise derivative that operate on time-averaged quantities that are zero.16.16) Now let’s look at this equation for fully developed channel flow for which v¯2 = v¯3 = 0 ∂(·) ∂(·) = =0 ∂x1 ∂x3 (9. see Eq. 8. but since it is symmetric we only need to consider six of them. it consists of nine equations. 1 Πii = p′ ρ ∂v ′ ∂vi′ + i ∂xi ∂xi =0 (9. For 2D boundary layer flow. 9. An alternative – and maybe easier – way to derive Eq. In the vi′ vj′ equation there is a fifth term (V).12 reads ∂ ∂ ∂¯ vi ∂¯ vj (¯ v1 vi′ vj′ ) + (¯ v2 vi′ vj′ ) = −vj′ v2′ − vi′ v2′ ∂x1 ∂x2 ∂x2 ∂x2 ! ′ ′ ∂v v ∂ 1 1 j i − vi′ vj′ v2′ + δj2 vi′ p′ + δi2 vj′ p′ − ν ∂x2 ρ ρ ∂x2 ′ ∂vj′ ∂v ′ ∂vj′ ∂vi 1 + p′ − 2ν i + ρ ∂xj ∂xi ∂xk ∂xk (9. only normal stresses). 8. a diffusion term (III) and a dissipation term (IV).12 and then take the trace (setting i = j) and divide by two.12 is the (exact) transport equation of the Reynolds stress. the the normal stresses is vav av i i pressure-strain term is negative and vice-versa. 9. increases Reynolds stress equations pressure strain Robin Hood . Source terms 103 where Πij denotes the pressure-strain term Πij = p′ ρ ∂vj′ ∂vi′ + ∂xj ∂xi (9.1. Compare Eq.14. not those that operate on instantaneous quantities such as in εij and Πij .14) Equation 9. 20a) P22 (9. a) Zoom near the wall. Now we will look at an important source term in the vi′ vj′ equation. 9.1 0. +: −∂(v2 v1 )/∂x2 . : P11 . : 2 ′2 2 ′ ′2 Π11 . v3′2 (i = j = 3) and v1′ v2′ (i = 1.18) where Q is a heat source. b) Outer region.2. ◦: ν∂ v1 /∂x2 . Q. : −ε11 . 9. T will increase and vice-versa.05 a) −0.15]. DNS data [14.1: Channel flow at Reτ = 2000. namely the production term.1 b) 100 200 300 400 500 x+ 2 Figure 9.16 reads Pij = −vj′ v2′ ∂¯ vi ∂¯ vj − vi′ v2′ ∂x2 ∂x2 (9.1. 2.5 0 10 20 30 40 x+ 2 −0. In the middle there is a heat source. The production term in Eq. Terms in the v1′2 equation scaled by u4τ /ν.5 0.19) For the v1′2 (i = j = 1). the value of Φ. v2′2 (i = j = 2). If Q is positive. see Fig.20d) .2: One-dimensional unsteady heat conduction.05 0 0 −0.9.20b) P33 P12 (9. Source terms 104 0. A simple example is the one-dimensional unsteady heat conduction equation (Eq.15 with vi ≡ 0) ∂T ∂2T =α 2 +Q ∂t ∂x1 (9.20c) (9. j = 2) equations we get ∂¯ v1 ∂x2 ∂¯ v2 = −2v2′ v2′ =0 ∂x2 ∂¯ v3 =0 = −2v3′ v2′ ∂x2 ∂¯ v1 ∂¯ v2 ∂¯ v1 = −v2′ v2′ − v1′ v2′ = −v2′2 ∂x2 ∂x2 ∂x2 P11 = −2v1′ v2′ (9. Q wall wall x1 Figure 9. reaches its maximum at x2 ≃ 11 where also v1′2 takes its maximum (Fig.3 presents the terms in the v3′2 equation. As mentioned previously. The production term – which should be a source term – is here negative. Consider the flow in a boundary layer. note that the dissipation. for example.tfd. For channel flow.2 Reynolds shear stress vs. is zero (as it is in the v1′2 equation). 6. Reynolds shear stress vs. the stresses vi′ vj′ in isotropic turbulence are zero). As for the v2′2 equation.A with (the turbulent fluctuating) velocity v2′ . for the v2′2 equation it goes negative near the wall since the pressurestrain term dampens v2′2 near the wall. This is because dissipation takes place at the smallest scales and they are isotropic. the main source term is the pressure-strain term. the production term. This can also be shown by physical argumentation. rather. v¯1 (x2. acts as the main source term. Figure 9.5 presents the terms in the v1′ v2′ equation. for i 6= j. Furthermore. Another difference is that the pressure diffusion term. ε12 6= 0 because here the wall affects the dissipative scales making them non-isotropic.1 presents the terms in the v1′2 equation (Eq. i. the other way around: v1′ v2′ is negative because its production term. Hence. v2′2 .9.8).2.A ) < v¯1 (x2. also their gradients are zero so that ∂v ′ ∂v ′ ε12 = 2ν 1 2 = 0 (9. Note that in the upper half of the channel ∂¯ v1 /∂x2 < 0 and hence P12 and v1′ v2′ are positive. 9. Π11 . 9.3. Π22 . ε12 . 9.17.se/˜lada/allpaper. P11 . or. That implies there is no correlation between two fluctuating velocity components. Indeed it should be. are shown in Fig.e. If you want to learn more how to derive transport equations of turbulent quantities.B )) comes down to x2. This means that when the particle at x2. see Fig. The terms in the wall-normal stress equation.html 9. Figure 9. At its new location the v1 velocity is in average smaller than at its old. Π22 – the “Robin Hood” term – takes from the “rich” v1′2 equation and gives to the “poor” v2′2 equation energy because v1′2 is large and v2′2 is small. Here we find – as expected – that the pressure-strain term.A (where the .20 shows that P12 is negative (and hence also v1′ v2′ ) in the lower half because ∂¯ v1 /∂x2 > 0 and it is positive in the upper half because ∂¯ v1 /∂x2 < 0. Figure 9. As we saw for the k equation. 9. is negative since ∂¯ v1 /∂x2 > 0.6.16 with i = j = 1). whereas it near the wall gives an important contribution in the v2′2 equation (it balances the negative pressure-strain term).5. is zero. v1′ v2′ = 0 (in general. 2∂v3′ p′ /∂x2 . see [22] which can be downloaded here http://www.1b) the production term balances the pressure-strain term and the dissipation term. Recall that v1′ v2′ is here negative and hence its source must be negative.B to x2. very close to the wall. but it may be noted that here it is positive near the wall. 9.g. Eq. and the dissipation term act as sink terms. 9. 9. the Reynolds shear stress and the velocity gradient ∂¯ v1 /∂x2 have nearly always opposite signs.B ).21) ∂xk ∂xk However. the velocity gradient In boundary-layer type of flow. x+ 2 ≤ 10. the velocity gradient 105 using Eq. e. In the outer region (Fig. see Fig. The pressure-strain term. P12 = −v2′2 ∂¯ v1 /∂x2 .chalmers.B (which has streamwise velocity v1 (x2. A fluid particle is moving downwards (particle drawn with solid line) from x2. ε12 is positive since v1′ v2′ < 0. 015 a) −0. : Π22 . 2 ′2 2 ′ ′ ′2 ′ ▽ : −2∂v2 p /∂x2 .01 −0. b) Outer region.2.3: Channel flow at Reτ = 2000. If we change the sign of the velocity gradient so that ∂¯ v1 /∂x2 < 0 we will find that the sign of v1′ v2′ also changes.05 0. Terms in the v3′2 equation scaled by u4τ /ν. again.015 0.1 −0. a) Zoom near the wall. In cases where the shear stress and the velocity gradient have the same sign (for example.005 0 0 −0.05 0 10 20 30 40 x+ 2 −0. The particle is moving upwards (v2′ > 0). v1′ v2′ < 0.02 b) 100 200 300 400 500 x+ 2 Figure 9. : −ε22 . 15].015 −0.A )) it has an excess of streamwise velocity compared to its new environment at x2.02 0. i. 15]. +: a) Zoom near the wall. Reynolds shear stress vs.02 b) 100 200 300 400 500 x+ 2 Figure 9. : −ε33 .9.005 −0.A .01 0.01 0. b) Outer region. 0. +: −∂(v2 v2 )/∂x2 . DNS data [14. Terms in the v2′2 equation scaled by u4τ /ν.1 0. streamwise velocity is v1 (x2. and it is bringing a deficit in v1 so that v1′ < 0. . −∂(v2′ v3′2 )/∂x2 .05 −0. 9.15 0. the velocity gradient 0. Thus.015 0. ◦: ν∂ 2 v3′2 /∂x22 . v1′ > 0 and the correlation between v1′ and v2′ is in average negative (v1′ v2′ < 0).6) we reach the same conclusion. If we look at the other particle (dashed line in Fig. Thus the streamwise fluctuation is positive.005 0 0 −0. : Π33 . If we study this flow for a long time and average over time we get v1′ v2′ < 0.02 0.e. ◦: ν∂ v2 /∂x2 . DNS data [14.4: Channel flow at Reτ = 2000. in a wall jet) the reason is that the other terms (usually the transport terms) are more important than the production term.15 a) 0 10 20 x+ 2 30 40 −0.05 106 0.005 −0.01 −0. when the two points are moved so close that they merge.5: Channel flow at Reτ = 2000. then B11 = v ′2 (xA 1 ).1 0. x2.05 0.1. We can then form the correlation of v1′ at these two points as C ′ A ′ C B11 (xA (10.05 −0.1) 1 . for example. it is expressed as ′ A ˆ1 ) B11 (xA ˆ1 ) = v1′ (xA 1 . 15]. 10 Correlations 10. x1 ) = v1 (x1 )v1 (x1 ) Often. It is convenient to normalize B11 so that it varies between . on the other hand. If.05 b) 100 200 300 400 500 x+ 2 Figure 9. +: −∂(v1 v2 )/∂x2 . B11 increases.05 0 0 −0.2) A where x ˆ1 = xC 1 − x1 is the separation distance between point A and C. then B11 will go to zero. 2 ′ ′ 2 ′ ′ ′2 ′ ▽ : −∂v2 p /∂x2 .A x2 x1 Figure 9. Terms in the v1′ v2′ equation scaled by u4τ /ν. : Π12 . : P12 . Pick two points along the x1 axis. see Fig. we move point C further and further away from point A.B v2′ > 0 v2′ < 0 v¯1 (x2 ) x2. 10. : −ε12 .1 Two-point correlations Two-point correlations are useful when describing some characteristics of the turbuC lence.x 1 )v1 (x1 + x (10. ◦: ν∂ v1 v2 /∂x2 . It is obvious that if we move point A and C closer to each other.6: Sign of the Reynolds shear stress −ρv1′ v2′ in a boundary layer. say xA 1 and x1 .1 a) 0 10 20 30 x+ 2 40 −0. DNS data [14. a) Zoom near the wall. b) Outer region.15 0.10. Correlations 107 0. the x1 direction. and sample the fluctuating velocity in. is defined as v1. an integral length scale. Two-point correlations 108 B11 Figure 10. 10. Consider a flow where the largest eddies have a length scale of Lint . i.rms (10. Z ∞ B11 (x1 . C A B |xA 1 − x1 | > Lint because for separation distances larger than |x1 − x1 | there is no ′ A ′ C correlation between v1 (x1 ) and v1 (x1 ).e. If the flow is homogeneous (see p. For flows with small eddies.1: Two-point correlation. x ˆ1 . which for v1′ . flows with large eddies will have a two-point correlation function which decreases slowly with separation distance. norm B11 (ˆ x1 ) = 1 v1′ (x1 )v1′ (x1 + xˆ1 ) 2 v1.rms 0 The integral length scale represents the length scale of the large energy-containing eddies. i. Lint . but it is only dependent on the separation of the two points.96) in the x1 direction.5) From the two-point correlation. C |xA 1 − x1 | −1 and +1.1. see Fig. x ˆ1 ) dˆ x1 (10. and the integral length scale is then computed as Z ∞ norm Lint = B11 (ˆ x1 )dˆ x1 (10.7) 0 integral length scale .e. B11 . the two-point correlation.4) RMS is the same as standard deviation (Matlab command std) which is the squareroot of the variance (Matlab command var). decreases rapidly with xˆ1 . The normalized two-point correlation reads norm A B11 (x1 . B11 .rms = v1′2 1/2 (10. can be computed which is defined as the integral of B11 over the separation distance. the two-point correlation does not depend on the location of xA 1 .rms (xA 1 1. If the flow is homogeneous in the x1 direction then Lint does not depend on x1 .3) where subscript rms denotes root-mean-square.rms 1 1 (10. x ˆ1 ) = 1 ′ A v1′ (xA ˆ1 ) 1 )v1 (x1 + x A+x ˆ ) )v (x v1. for example.rms v1. approaches zero for separation distance.2. B11 .10.6) Lint (x1 ) = A C v1. We expect that the two point correlation. Hence. Two-point correlations 109 Lint xA 1 B11 x1 (a) Small integral length scale xA 1 Lint B11 x1 (b) Large integral length scale Figure 10.10. the largest eddies (thick lines) and the integral length scale. .1. Lint .2: Schematic relation between the two-point correlation. is the time separation distance between time A and C. in this case the auto-correlation depends only on time separation.11) 0 The integral time scale represents the time scale of the large energy-containing eddies. B11 (tˆ) = v1′ (t)v1′ (t + tˆ) (10. integral time scale .2.rms 1 + tˆ) (10. the “time direction” is homogeneous and B11 is independent on tA . the correlation of a turbulent fluctuation with a separation in time. tˆ. If the mean flow is steady. the auto correlation reads (10. tˆ) = v1′ (tA )v1′ (tA + tˆ) where tˆ = tC − tA . Tint . The normalized auto-correlation reads norm ˆ B11 (t ) = 1 v ′ (t)v1′ (t 2 v1. Lint .10) In analogy to the integral length scale.2 Auto correlation Auto correlation is a “two-point correlation” in time.e. i.8) B11 (tA . is defined as (assuming steady flow) Z ∞ norm ˆ ˆ Tint = B11 (t)dt (10.10.9) where the right side is time-averaged over t. i.e. Auto correlation 110 10. the integral time scale. If we again choose the v1′ fluctuation. Sweden http://www.10.se This report can be downloaded at http://www. Davidson Division of Fluid Dynamics.chalmers. Department of Applied Mechanics Chalmers University of Technology.se/˜lada.se/˜lada/comp turb model/lecture notes. lada@chalmers. Auto correlation 111 MTF270 Turbulence Modeling L.html .chalmers. G¨oteborg.2.tfd.tfd. 2 is due to buoyancy. p. differentiation gives ∂ρ ∂ρ dθ + dp (11.7) which is the last term in Eq. Now we can re-write the buoyancy source as ρ0 fi = (ρ − ρ0 )gi = −ρ0 β(θ¯ − θ0 )gi (11.1 Flow equations For incompressible turbulent flow. Since the pressure.5 is zero and we introduce the volumetric thermal expansion. If we let density depend on pressure and temperature. 11. For a perfect gas it is simply β = θ−1 (with θ in degrees Kelvin). 11. 6. 11.2. The body force in Eq. is defined as the hydro-dynamic pressure we have re-written the buoyancy source as ρ0 fi = (ρ − ρ0 )gi (11. β. p¯ ≡ 0).11. Then gi = (0. 3.e. v¯i ≡ 0). Time average and we get (see Eq.4) so that p¯ ≡ 0 when v¯i ≡ 0 (note that the true pressure decreases upwards as ρg∆h where ∆h denotes change in height). which agrees well with our intuition.1 Mean flow equations 11.6) dρ = −ρ0 βdθ ⇒ ρ − ρ0 = −βρ0 (θ − θ0 ) where β is a physical property which is tabulated in physical handbooks.2) (note that θ denotes temperature) where ρ0 is a constant reference density. p¯. which means that the density does not depend on pressure. all variables are divided into a mean part (time averaged) and fluctuating part. density differences.1) (11. ∂ρ/∂p = 0. i. 11. the volume force fi = −β(θ¯ − θ0 )gi and the turbulent stress tensor (also called Reynolds stress tensor) is written as: (11. see Eq.5) dρ = ∂θ p ∂p θ Our flow is incompressible.22.8 – has here been re-introduced. Hence the last term in Eq. 77): ∂¯ vi =0 ∂xj ∂ p¯ ∂ 2 v¯i ∂τij ∂ ∂ρ0 v¯i (ρ0 v¯i v¯j ) = − +µ − − βρ0 (θ¯ − θ0 )gi + ∂t ∂xj ∂xi ∂xj ∂xj ∂xj (11. The body force fi – which was omitted for convenience in Eq.7 results in a force vertically upwards. −g) and a large temperature in Eq.e.8 at. it may.3) τij = ρ0 vi′ vj′ The pressure. p¯. then the pressure is zero (i. Reynolds stress tensor . however. so that 1 ∂ρ ⇒ β=− ρ0 ∂θ p (11. The basic form of the buoyancy force is fi = gi where gi denotes gravitational acceleration. 6. Reynolds stress models and two-equation models 112 11 Reynolds stress models and two-equation models 11.1. For the velocity vector this means that vi is divided into a mean part v¯i and a fluctuating part vi′ so that vi = v¯i + vi′ . Consider the case where x3 is vertically upwards. depend on temperature and mixture composition. 0. which means that when the flow is still (i.e.e. denotes the hydrodynamic pressure. i. as exact as the Navier-Stokes equations. (C) with vj′ and Eq. (D) with vi′ .11) qi.1. This is similar to the Reynolds stress tensor on the right side of the time-averaged momentum equation. closure problem . θ.8) where α = k/(ρcp ). An exact equation for the Reynolds stresses can be derived from the Navies-Stokes equation.2). Eq.10) 11. vi′ vj′ . 11.turb ∂ θ¯ = + = −α − vi′ θ′ ρcp ρcp ρcp ∂xi (11. 2. 2. time average and add them together → equation for vi′ vj′ In Section 9 at p. see Eq. It is emphasized that this equation is exact. Eq.9) The last term on the right side is an additional term whose physical meaning is turbulent heat flux vector. 2. Eq. Eq.15 where temperature was denoted by T ) ∂θ ∂vi θ ∂2θ =α + ∂t ∂xi ∂xi ∂xi (11. Introducing θ = θ¯ + θ′ we get ∂ θ¯ ∂¯ ∂ 2 θ¯ ∂v ′ θ′ vi θ¯ =α − i + ∂t ∂xi ∂xi ∂xi ∂xi (11. (B) • Subtract Eq. We must close this equation system.1) and the three momentum equations (Eq.9 reads (cf. (D) • Multiply Eq. (C) • Do the same procedure for vj → equation for vj′ . We need a turbulence model. The exact vi′ vj′ equation 113 11. p¯) plus six turbulent stresses. Eq. We must find some new equations for the turbulent stresses. rather. (A) → equation for vi′ .e. 26. 11. 11. The transport equation for θ reads (see Eq. • Set up the momentum equation for the instantaneous velocity vi = v¯i + vi′ → Eq. it is called the closure problem.15 on p. or. The total heat flux vector – viscous plus turbulent – in Eq. (B) from Eq. (A) • Time average → equation for v¯i . Unfortunately there are ten unknowns. 11.2 The exact vi′ vj′ equation Now we want to solve the time-averaged continuity equation (Eq. is also decomposed into a mean and a fluctuating component as θ = θ¯ + θ′ . 100 these steps are given in some detail. The derivation follows the steps below.11. i. The most comprehensive turbulence model is to derive exact transport equations for the turbulent stresses. More details can also be found in [22] (set the SGS tensor to zero. τija = 0).2 Temperature equation The instantaneous temperature.2.tot qi qi.2. the four usual ones (¯ vi . e.11 can symbolically be written Cij = Pij + Πij + Dij + Gij − εij where Cij Convection Pij Production Πij Pressure-strain Dij Diffusion Gij Buoyancy production εij Dissipation Which terms in Eq. Equation 11.t −gi βvj′ θ′ − gj βvi′ θ′ − 2ν Gij ∂vi′ ∂vj′ ∂xk ∂xk εij where Dij. This is analogous to the momentum equation where we have gradients of viscous and turbulent stresses which correspond to viscous and turbulent diffusion. respectively. Eq. 11. Gij (provided that a transport equation is solved for vi′ θ′ . They are v¯i is obtained from the momentum equation. 11.99 Hence the following terms in Eq.2. The exact vi′ vj′ equation 114 The final vi′ vj′ -equation (Reynolds Stress equation) reads (see Eq.ν . 11. Pij ν • The viscous part of the diffusion term. 11.t + Dij.t and Dij. Eq. Eq.ν D (11.e.ν denote turbulent and viscous diffusion.2 vi′ vj′ is obtained from the modeled RSM equation.11 are known and which are unknown? First. if not.33) . Dij • The buoyancy term. 11. 11.11 are known (i. The total diffusion reads Dij = Dij. i. let’s think about which physical quantities we solve for.11.12) ∂vi′ vj′ ∂vj′ ∂vi′ vj′ p′ ∂vi′ ∂¯ vj ∂¯ vi ′ ′ ′ ′ = −vi vk + − vj vk + + v¯k ∂t ∂xk ∂xk ∂xk ρ ∂xj ∂xi Cij Pij Πij # " 2 p′ vj′ ∂ vi′ vj′ p′ vi′ ∂ vi′ vj′ vk′ + − δik + δjk + ν ∂xk ρ ρ ∂xk ∂xk Dij. Dij .22. Eq. vi′ θ′ is obtained from the Boussinesq assumption. they do not need to be modeled) • The left side • The production term. 9.11) ij. Trick 1) v¯k ∂¯ vk vi′ θ′ ∂vi′ θ′ = ∂xk ∂xk (11.e.2) from the equation for the instantaneous velocity. 11. 11. ∂vk′ /∂xk = 0) vi′ vk′ ∂ θ¯ ∂¯ v + vk′ θ′ ∂xk ∂xk (11.13 multiplied by θ′ and time averaged is zero.13) + ∂t ∂xk ρ ∂xi ∂xk ∂xk ∂xk Multiply Eq.3 The exact vi′ θ′ equation If temperature variations occur we must solve for the mean temperature field. subtract the equation for the mean velocity v¯i (Eq.17 ∂vi′ vk′ θ′ (11.12 with vi′ and multiply Eq. i. The last two terms in Eq. 11.9.15) The second term in the two parenthesis on the first line of Eq.9 from Eq. 11.14 are combined into two production terms (using the continuity equation. 11.14 are re-cast into turbulent diffusion terms using the same procedure as in Eqs.11.8 ∂v ′ θ′ ∂ 2 θ′ ∂ ∂θ′ (vk′ θ¯ + v¯k θ′ + vk′ θ′ ) = α + k + ∂t ∂xk ∂xk ∂xk ∂xk (11. see Eq. add them together and time average ∂ ∂ ∂vi′ θ′ (vk′ θ¯ + v¯k θ′ + vk′ θ′ ) + θ′ (¯ vi vk′ + v¯k vi′ + vi′ vk′ ) + vi′ ∂t ∂xk ∂xk θ′ ∂p′ ∂ 2 θ′ ∂ 2 vi′ =− + αvi′ + νθ′ − gi βθ′ θ′ ρ ∂xi ∂xk ∂xk ∂xk ∂xk (11.13 with θ′ . 11.3. 11. The first term in the two parentheses on line 1 in Eq. vi′ .16 and 11. ∂vi′ vj′ ∂vi′ vj′ θ′ = θ′ = 0 ∂xk ∂xk If you have forgotten the rules for time-averaging.16) +θ = v¯k vi +θ ∂xk ∂xk ∂xk ∂xk Now the two terms can be merged (product rule backwards. 6. The exact vi′ θ′ equation 115 11. subtract Eq.14) The Reynolds stress term in Eq. 11.17) where we used the continuity equation to obtain the right side. see Section 8.1.18) ∂xk .14 are re-written using the continuity equation ! ∂¯ vk θ′ ∂vi′ ∂¯ vk vi′ ∂θ′ ′ ′ ′ ′ vi (11.5) so that ∂v ′ v ′ 1 ∂p′ ∂ 2 vi′ ∂ ∂vi′ (vk′ v¯i + v¯k vi′ + vk′ vi′ ) = − +ν + i k − gi βθ′ (11. vi (Eq. 11. To obtain the equation for the fluctuating temperature.12) To get the equation for the fluctuating velocity. 11. 11. The production term include one term with the mean velocity gradient and one term with the mean temperature gradient.20) Giθ where Piθ . The exact vi′ θ′ equation 116 The viscous diffusion terms on the right side are re-written using the product rule backwards (Trick 1. 11.ν ∂xk ∂xk P r ∂xk ∂xk where Diθ. 11. P r = ν/α ≃ 1. The unknown terms – Πiθ . Hence. respectively. if ν ≃ α (which corresponds to a Prandtl number of unity. 2. It can be noted that there is no usual viscous diffusion term in Eq. 11. Giθ – have to be modeled as usual.20. i. Diθ.14 gives the transport equation for the heat flux vector vi′ θ′ ∂vi′ θ′ ∂ ′ ′ ′ ∂ θ¯ ∂¯ vi θ′ ∂p′ ∂ − − v vθ v¯k vi′ θ′ = −vi′ vk′ − vk′ θ′ + ∂t ∂xk ∂xk ∂xk ρ ∂xi ∂xk k i Piθ Πiθ Diθ.16).t ν ∂ 2 vi′ θ′ ∂v ′ ∂θ′ − (ν + α) i −gi βθ′2 + ν+ P r ∂xk ∂xk ∂xk ∂xk Diθ.22) . Πiθ and Diθ.t ∂ ∂ ∂v ′ ∂θ′ ∂θ′ ∂v ′ +α vi′ +ν θ′ i − (ν + α) i −gi βθ′2 ∂xk ∂xk ∂xk ∂xk ∂xk ∂xk εiθ Diθ. see Eq.21) ! ∂vi′ ∂ 2 vi′ θ′ ∂ ∂ ∂θ′ ′ ′ θ vi ≃2ν −ν −ν ∂xk ∂xk ∂xk ∂xk ∂xk ∂xk ∂ 2 vi′ θ′ ∂ 2 vi′ θ′ ν = 2ν − Diθ.18 and 11.17. Diθ .ν cancels the corresponding term in Eq.15. 11. 11. εiθ . However. if α ≃ ν the transport equation for vi′ θ′ can be simplified as ∂ ∂ ′ ′ ′ ∂vi′ θ′ ∂ θ¯ ∂¯ vi θ′ ∂p′ v¯k vi′ θ′ = −vi′ vk′ − vk′ θ′ + − − v vθ ∂t ∂xk ∂xk ∂xk ρ ∂xi ∂xk k i Piθ Πiθ Diθ. 11. respectively. dissipation and buoyancy term.11.ν .ν εiθ Giθ (11. the diffusion term in Eq.t denote the production. εiθ and Giθ denote viscous diffusion.3. 89) ′ ∂ ∂θ ∂θ′ ∂vk′ ∂ 2 θ′ ∂ ∂θ′ ′ ′ ′ =α −α = αvi αvi vi ∂xk ∂xk ∂xk ∂xk ∂xk ∂xk ∂xk ∂xk ! (11. see p.20 assumes the familiar form ∂θ′ ∂vi′ ∂ ∂ ′ ′ α vi +ν θ ∂xk ∂xk ∂xk ∂xk ! 2 ′ ′ ′ ∂ 2 vi′ θ′ ∂ ∂ ∂ vi θ ∂θ′ ∂v i ′ ′ +ν −α −ν =α θ vi ∂xk ∂xk ∂xk ∂xk ∂xk ∂xk ∂xk ∂xk (11.ν (11.e. The reason is that the viscous diffusion coefficients are different in the vi equation and the θ equation (ν in the former case and α in the latter). On the last line.19 into Eq. this is out of the scope of the present course but the interested reader is referred to [23]. 11.ν = ν + − Diθ. scramble and turbulent diffusion term.19) ′ ′ ′ 2 ∂ vi ∂ ∂vi ∂ ∂vi ∂θ′ ∂vk′ ′ ′ ′ νθ θ =ν −ν = νθ ∂xk ∂xk ∂xk ∂xk ∂xk ∂xk ∂xk ∂xk Inserting Eqs. Often the viscous diffusion is simplified in this way.20 if α = ν. they do not need to be modeled) • The left side • The production term. respectively. The total diffusion reads Dk = Dtk + Dνk . Eq.24) The known quantities in Eq. 11. 11. 11. i. Diθ.95 Hence the following terms in Eq. Piθ • The viscous diffusion term.e. 1 1 ′2 v1 + v2′2 + v3′2 ≡ vi′ vi′ k= 2 2 By taking the trace (setting indices i = j) of the equation for vi′ vj′ and dividing by two we get the equation for the turbulent kinetic energy: ∂v ′ ∂vi′ ∂¯ vi ∂k ∂k + v¯j = − vi′ vj′ −ν i ∂t ∂xj ∂xj ∂xj ∂xj ε Ck Pk − ∂ ∂xj ( vj′ p′ ρ 1 + vi′ vi′ 2 Dtk ) (11.99 vi′ θ′ is obtained from the modeled vi′ θ′ equation Hence the following terms in Eq.4.e. 11. Giθ (provided that a transport equation is solved for θ′2 . 11. Eq. Eq. Eq.ν • The buoyancy term.23 can symbolically be written: C k = P k + Dk + Gk − ε (11. 11. The k equation 117 The same question arises as for the vi′ vj′ equation: which terms need to be modeled? The following quantities are known: v¯i is obtained from the momentum equation.9 vi′ vj′ is obtained from the modeled RSM equation. if not.2 θ¯ is obtained from the temperature equation.23 are known (i.e.22 are known (i. Equation 11.23) 2 +ν ∂ k −gi βvi′ θ′ ∂xj ∂xj k Dνk G where – as in the vi′ vj′ equation – Dtk and Dνk denotes turbulent and viscous diffusion. 11. 11. θ′2 is usually modeled via a relation to k) 11.4 The k equation The turbulent kinetic energy is the sum of all normal Reynolds stresses. Eq.2 k is obtained from the modeled k equation.23 are: v¯i is obtained from the momentum equation. they do not need to be modeled) • The left side .11. Eq. Eq. The turbulent viscosity is then computed as k2 νt = Cµ ε where Cµ = 0. Gij (provided that a transport equation is solved for vi′ θ′ . . a production term due to buoyancy. a production term.24. we must multiply P k and ε by a quantity which has the dimension [1/s].28) Note that this is a modeled equation since we have modeled the production.5 The ε equation Two quantities are usually used in eddy-viscosity model to express the turbulent viscosity. A better choice should be ε/k = [1/s].ν • The buoyancy term. vi′ θ′ is obtained from the Boussinesq assumption. The ε equation 118 k • The viscous diffusion term. destruction and turbulent diffusion terms.e.11. a diffusion term. and to setup a similar equation for ε. ∂ε/∂t).. Dε . we get P ε + Gε − Ψε = ε cε1 P k + cε1 Gk − cε2 ε k (11. P k and ε. ∂k/∂t) whereas the terms in the ε equation have the dimension [m2 /s4 ] (cf.33) 11.27) Dε = ∂xj σε ∂xj The final form of the ε transport equation reads ∂ε ∂ε ε ∂ = (cε1 P k + cε1 Gk − cε2 ε) + + v¯j ∂t ∂xj k ∂xj νt ∂ε ν+ σε ∂xj (11.e. 11. 11. νt ∂ε ∂ ν+ (11. k and ε are used. In the k − ε model. The terms in the k equation have the dimension [m2 /s3 ] (look at the unsteady term. and a destruction term. in the k equation are used to formulate the corresponding terms in the ε equation. Ψε . Di. but it is very complicated and in the end many terms are found negligible. The transport equation should include a convective term. An exact equation for the transport equation for the dissipation ε=ν ∂vi′ ∂vi′ ∂xj ∂xj can be derived (see. 11. 11. One quantity with this dimension is the mean velocity gradient which might be relevant for the production term.38) but with its own turbulent Prandtl number. C ε .09. The turbulent diffusion term is expressed in the same way as that in the k equation (see Eq. σε (see Eq. i.22. 11.26) where we have added new unknown coefficients in front of each term. i.g. It is much easier to look at the k equation. C ε = P ε + Dε + Gε − Ψε (11. P ε . Gε . [24]). e. Hence. Eq. Hence.5.35).25) The production and destruction terms. if not. but not for the destruction. see Eq. but not the left side).11.34) where σθ is the turbulent Prandtl number.6. 2.29 and 11. σθ .2 we replace six turbulent stresses with one new unknown (the turbulent viscosity. Consider the diffusion terms in the incompressible momentum equation in the case of non-constant viscosity (see Eq.33) vi′ θ′ = −αt ∂xi where αt denotes the turbulent thermal diffusivity.32) vi′ vj′ = −νt ∂xj ∂xi 3 3 Now the equation valid also when it is contracted (i. a turbulent) viscosity is introduced to model the unknown Reynolds stresses in Eq.32 is included in Eq. This is of course a drastic simplification.11. vi′ vj′ . i. 2. it defines how efficient the turbulence transports (by diffusion) momentum compared to how efficient it transports thermal energy. we need an equation for the heat flux vector. If the mean temperature equation is solved for. However. it is more common to used an eddy-viscosity model for the heat flux vector. The physical meaning of the turbulent Prandtl number. When Eq.e.6 The Boussinesq assumption In the Boussinesq assumption an eddy (i. 11.2. so that the the diffusion terms can be written ∂ ∂¯ vj ∂¯ vi (ν + νt ) (11. 11. is usually obtained from the turbulent viscosity as αt = νt σθ (11.31) This is identical to the assumption for the Newtonian. by a turbulent viscosity.35) . αt .16.31 is not valid upon contraction 3 (the right side will be zero due to continuity.4. Hence we add the trace of the left side to the right side so that 1 ∂¯ vj ∂¯ vi 2 + δij vk′ vk′ = −2νt s¯ij + δij k + (11. νt . it is an empirical constant which is usually set to 0. 2. The Boussinesq assumption 119 11. νt ). Equation 11. The Boussinesq assumption reads ∂ θ¯ (11. vi′ θ′ .e taking the trace). see Eq. viscous stress for incompressible flow.30) + ∂xj ∂xj ∂xi Identification of Eqs. 2. Eq.7 ≤ σθ ≤ 0. Note that this is the same assumption as Fourier’s law for a Newtonian flux. see Eq. 11.29) ν − vi vj + ∂xj ∂xj ∂xi Now we want to replace the Reynolds stress tensor.30 gives ∂¯ vj ∂¯ vi ′ ′ + −vi vj = νt ∂xj ∂xi (11. is analogous to the physical meaning of the usual Prandtl number.9. The turbulent thermal diffusivity. σθ = 3 Contractation means that i is set to j νt αt (11. 11.e.22. One option is to solve its transport equation. after contraction both left and right side are equal (as they must be) to vi′ vi′ = 2k. 11.5) ∂ ∂¯ vj ∂¯ vi ′ ′ (11. εiθ and ε are dissipation of vi′ vj′ .7. k and ε equations and formulate modeling assumptions for the remaining terms in the Reynolds stress equation. Giθ and Gk are production terms of vi′ vj′ . However. vi′ θ′ and k.g. Diθ. ∂xk ∂xk ∂ θ¯ ∂¯ vi − vk′ θ′ = −vi′ vk′ ∂xk ∂xk Pij = −vi′ vk′ Piθ Pk = ∂¯ vi 1 Pii = −vi′ vj′ 2 ∂xj (11. the Prandtl number (P r). vi′ θ′ and k Πiθ is the pressure-scramble terms of vi′ θ′ Πij is the pressure-strain correlation term. vi′ θ′ and k due to buoyancy Dij. water or air) and its conditions (e. ARSM) Summary of physical meaning: Pij . the thermal diffusivity (α) are physical parameters which depend on the fluid (e.g.36) The k is usually not solved for in RSM but a length-scale equation (i.t .11. In the k − ε model. ε or ω) is always part of an RSM and that equation includes P k . we will introduce a simplified algebraic model. Piθ and P k are production terms of vi′ vj′ .g. 11. the turbulent viscosity (νt ). temperature). Later on.7 Modeling assumptions Now we will compare the modeling assumptions for the unknown terms in the vi′ vj′ . Dtk are the turbulent diffusion terms of vi′ vj′ . mean flow gradients and turbulence). the turbulent thermal diffusivity (αt ) and the turbulent Prandtl number (σθ ) depend on the flow (e.t . which is called the Algebraic Stress Model [ASM] (this model is also called Algebraic Reynolds Stress Model. the Reynolds stresses in the production term are computed .e. Modeling assumptions 120 It is important to recognize that the viscosity (ν). which promotes isotropy of the turbulence εij .7. vi′ θ′ . respectively.1 Production terms In RSM and ASM the production terms are computed exactly ∂¯ vj ∂¯ vi − vj′ vk′ . 11. vi′ θ′ and k Gij . This will give us the Reynolds Stress Model [RSM] (also called the Reynolds Stress underline Transport Model [RSTM]) where a (modeled) transport equation is solved for each stress. The dissipation takes place at the small-scale turbulence. 7.t = 1 ′ ′ ′ νt ∂k v vv =− 2 j i i σk ∂xj (11. 8. + − = s¯ij + Ωij s¯ij = 2 ∂xj ∂xi 2 ∂xj ∂xi ∂xj (11. In other words. Equation 11. dkj.G ∝ v1′ v1′ ∂k ∂k ∂k + v1′ v2′ + v1′ v3′ ∂x1 ∂x2 ∂x3 (11. The turbulent diffusion flux of k is expressed as dkj.t.39 (including the minus sign in Eq. 11. Taking the divergence of Eq. it assumes that the turbulent diffusion term. The incompressibility condition.41) .G .39) Only the triple correlations are included since the pressure diffusion usually is negligible (see Fig. It was derived in [25] from the transport equation of the triple correlation vj′ vi′ vi′ .G ∝ vj′ vk′ ∂k ∂xk (11. 11. A more general gradient hypothesis can be formulated without this limitation.38. vi′ vj′ . was used to obtain the third line.38) ∂ νt ∂ε Dε = ν+ ∂xj σε ∂xj The gradient hypothesis simply assumes that turbulent diffusion acts as to even out all inhomogeneities. e.2 Diffusion terms The diffusion terms in the k and ε-equations in the k − ε model are modeled using the standard gradient hypothesis which reads νt ∂k ∂ ν+ Dk = ∂xj σk ∂xj (11.23) gives the turbulent diffusion term in Eq. 11.40) which is called the general gradient diffusion hypothesis (GGDH). 11.39 assumes that if the gradient is zero in xi direction. then there is no diffusion flux in that direction. transports k from regions where k is large to regions where k is small.3 at p. Modeling assumptions 121 using the Boussinesq assumption.g. In GGDH the turbulent flux dk1. which gives ∂¯ vi 2 ∂¯ vj − δij k + −vi′ vj′ = νt ∂xj ∂xi 3 2 ∂¯ vj ∂¯ vi ∂¯ vi − δij k + P k = νt ∂xj ∂xi 3 ∂xj ∂¯ vi ∂¯ vj ∂¯ vi = νt + = νt 2¯ sij (¯ sij + Ωij ) = 2νt s¯ij s¯ij ∂xj ∂xi ∂xj 1 ∂¯ ∂¯ vi vi vi ∂¯ vj ∂¯ vj 1 ∂¯ . Solving the equations for the Reynolds stresses. ∂¯ vi /∂xi = 0. Ωij = .t. Dtk .7.t.11. opens possibilities for a more advanced model of the turbulent diffusion terms.37) where on the third line we used the fact that s¯ij Ωij = 0 because the product between a symmetric tensor (¯ sij ) and an asymmetric tensor (Ωij ) is zero. 95). for example. is computed as dk1. 44 often causes numerical problems. All shear stresses are zero. in the k equation is obtained by taking the divergence of this equation ∂dkj.26 we take k/ε so that ∂k k (11. 11. 11.t = ∂xm σk ∂xm 11.7. 2.t = ck vk vm ∂xk ε ∂xm Equation 11.e.43) Dtk = = ∂xj ∂xj ε ∂xk This diffusion model may be used when the k equation is solved in an RSM or an ASM. Because of the cascade process and vortex stretching (see Figs. even if ∂k/∂x1 = 0 the diffusion flux dk1.44) ∂vi′ vj′ k ∂ ′ ′ Dij. This means that the velocity fluctuations of the smallscale turbulence have no preferred direction.46) The relations in Eq.38 which for vi′ vj′ reads ! νt ∂vi′ vj′ ∂ (11. Dtk . i.t. 11.11 and 11. vi′ vj′ = 0 if i 6= j because the fluctuations in two different coordinate directions are not correlated. see p. above) must also apply for the gradients of the fluctuations.46 are conveniently expressed in tensor notation as εij = 2 εδij 3 (11. A more stable alternative is to model the diffusion terms as in 11.45) Dij.3) the smallscale turbulence is isotropic.7. i. 11. εij The dissipation term εij (see Eq.t.G k ∂ ∂k ck vj′ vk′ (11. The corresponding diffusion terms for the ε and vi′ vj′ equations read k ∂ε ∂ ε ′ ′ cε vj vk Dt = ∂xj ε ∂xk ! (11. ∂v1′ ∂v1′ ∂v2′ ∂v2′ ∂v3′ ∂v3′ = = ∂xk ∂xk ∂xk ∂xk ∂xk ∂xk ′ ′ ∂vi ∂vj = 0 if i 6= j ∂xk ∂xk (11. 71.2 and 5.3 Dissipation term. This gives: 1. A quantity of dimension [s] must be added to get the correct dimension. What applies for the small-scale fluctuations (Items 1 and 2.23.G = ck vj′ vk′ ε ∂xk The diffusion term. v1′2 = v2′2 = v3′2 . see Eqs.42) dkj. . and as in Eq.47) where the factor 2/3 is included so that ε = 12 εii is satisfied. 5. Modeling assumptions 122 Hence.t.11) is active for the small-scale turbulence.e.11.G may be non-zero. 7. Furthermore. and positive for the wall-normal. If this is to happen the kinetic energy in the x1 direction must be larger than that in the x2 and x3 direction. In Section 9 it was shown that for channel flow it is negative for the streamwise equation.48) ∂x2 ∂x3 Indeed. i.11. equations.e. makes a large contribution to the vi′ vj′ equation. i. v3′2 . because it “takes from the rich and gives to the poor”. should be proportional to . the continuity equation gives ∂v2′ /∂x2 + ∂v3′ /∂x3 > 0. moving fluid particles against the pressure gradient.e. the pressure strain re-distributes kinetic energy from both v1′2 and v3′2 to v2′2 . i.1. it was shown that the term acts as to make the turbulence more isotropic. However. If v3′2 ≃ v1′2 . The role of the pressure strain can be described in physical terms as follows.4 Slow pressure-strain term The pressure-strain term. >0 (11. see Fig. v1′2 . Πij . The pressure-strain term is often called the Robin Hood terms. 11.7. v1′2 > v2′2 and v1′2 > v3′2 . v2′2 . decreasing the large normal stresses and the magnitude of the shear stress and increasing the small normal stresses. ∂v2′ ∂v3′ > 0. The kinetic energy lost in the x1 direction is transferred to the x2 and x3 directions and we assume that the collision makes fluid particles move in the other two directions. it acts as a sink term for the shear stress equation. Now let’s assume that v1′2 > v2′2 and v1′2 > v3′2 . 11.e. 11. The amount of kinetic energy transferred from the x1 direction to the x2 and x3 directions. if ∂v1′ /∂x1 < 0. As a result the fluctuating pressure p′ increases at O so that ∂v ′ p′ 1 < 0 ∂x1 The fluid in the x1 direction is performing work.48 we assume that not only their sum is positive but also that they both are positive.1: Physical illustration of the pressure-strain term. In summary. in Eq. Modeling assumptions 123 v1′ v1′ O x2 x1 Figure 11. and spanwise. Assume that two fluid particles with fluctuating velocities v1′ bounce into each other at O so that ∂v1′ /∂x1 < 0. 5 v1∗ 2∗ ′ v′ ′ ′ since v1∗ 2∗ = v2∗ v1∗ . (x1∗ .49) p′ where t denotes a turbulent timescale. the principal axis of vi′ vj′ . respectively. the velocity gradients.53 gives finally ′ ∂v2 3ρ ∂v ′ p′ + 1 ∝ − ρv1′ v2′ ∂x1 ∂x2 4t (11. 126. The expression in Eq.52) so that ′ ∂v1∗ 3ρ ′2 ∂v ′ ′2 (11. We express Eq.49 in the x1∗ and −x2∗ directions (looking at the right side of Eq. S.50 and 11.6b and S. see Appendix S.54 lead as to write Φij.7.11 and 11.e. 11. We use Eqs. Modeling assumptions 124 the difference of their energies.49 in principal coordinates. Thus.6b by v1′ v2′ we get ′2 − v ′2 (11. 11. 11. Replacing u12 in Eq.e. Now we have transformed the right side of Eq. 11. We have introduced the turbulent time scale t = k/ε and a constant c1 .54) This shows that the pressure-strain term acts as a sink term in the shear stress equation.49 (the right side on the first line). x2∗ ). and adding them gives ′ ∂v2′ ∂v ′ ∂v1∗ ∂v ′ + 1 = − 2∗ ∂x1 ∂x2 ∂x1∗ ∂x2∗ the pressure-strain term in Eqs.1.52) Now we apply Eq. 11.55) where Φ denotes the modeled pressure-strain term and subscript 1 means the slow part.6c: replacing u12 and u21 by ∂v1′ /∂x2 and ∂v2′ /∂x1 . i. the concept “slow” and “rapid” is discussed at p. and then transform the equation to (x1 . This pressure-strain model for the slow part was proposed by Rotta in 1951 [26]. .51) (11. Let us show that by transforming the fluctuations to a coordinate system which is rotated an angle α = π/4 then p′ (∂v1′ /∂x2 + ∂v2′ /∂x1 ) ∝ −v1′ v2′ (α = π/4 corresponds to the special case when the normal stresses are equal). 11. i.50) v1′ v2′ = 0.49 can be written ′ ′ ′ ∂v1′ ∂v2∗ ∂v2 ∂v1∗ ′ ′ =p + − p ∂x1 ∂x2 ∂x1∗ ∂x2∗ (11. x2 ) by rotating it angle α = π/4. Next step is to transform the left side. i ∂v1′ ρ 1 h ′2 v1 − v2′2 + v1′2 − v3′2 ∝− ∂x1 t2 ρ ρ 3 ′2 1 ′2 ρ 3 ′2 1 ′2 = − v1′2 − v2 + v3′2 = − v1 − v1 + v2′2 + v3′2 = − v1 − k t 2 t 2 2 t 2 (11. 11.1 ≡ p′ ∂vj′ ∂vi′ + ∂xj ∂xi = −c1 ρ ε k 2 vi′ vj′ − δij k 3 (11. Eqs.52 into Eq. 11. S.49 applies only to the normal stresses.53) p′ v1∗ − v2∗ − 2∗ ∝ − ∂x1∗ ∂x2∗ 2t Inserting Eqs.49 and 11.e. i.11. 11. 59 in Eq.2. 11.57).60) . Thus it should be interesting to re-write Eq. 11.2: Decaying grid turbulence. From this point and further downstream the flow represents homogeneous turbulence which is slowly approaching isotropic turbulence. 11.11. We introduce aij (Eq. Rotta’s model (Eq. see Fig. The grid creates velocity gradients behind the grid which generates turbulence. Modeling assumptions 125 v¯1 x2 a) x2 b) x1 x3 Figure 11. 11.55 behaves for decaying grid turbulence. 11. 11.55) and the model for the dissipation tensor (11.59) Inserting Eq. Let us investigate how Eq. decaying) due to dissipation. Flow from left with velocity v¯1 passes through a grid.7.47) into Eq.58 and dividing by k we obtain v¯1 ε 2 ε ε 2 ε ε daij = −c1 aij − δij + aij + δij = aij (1 − c1 ) dx1 k 3 k k 3 k k (11.56 expressed in the normalized anisotropy Reynolds stress tensor which is defined as vi′ vj′ 2 − δij (11. furthermore the turbulence is slowly dying (i. The exact vi′ vj′ equation for this flow reads (no production or diffusion because of homogeneity) ∂vj′ dvi′ vj′ p′ ∂vi′ − εij (11. then aij = 0. the k equation in decaying grid turbulence reads v¯1 dk = −ε dx1 (11. 11.56.56 so that d(kaij ) 2 ∂k 2 v¯1 = −c1 εaij − δij ε + δij (11. The circles (a) and the thin rectangles (b) illustrates part of the grid which consists of a mesh of circular cylinders.56) = + v¯1 dx1 ρ ∂xj ∂xi Rotta’s pressure-strain model is supposed to reduce anisotropy. Further downstream the velocity gradients are smoothed out and the mean flow becomes constant.e.57) aij = k 3 Note that when the turbulence is isotropic. 11.58) dx1 3 ∂x1 3 Analogously to Eq. 11. 11.55 can be extended by including terms which are non-linear in vi′ vj′ .7. a second-order tensor satisfies its own characteristic equation (see Section 1. In isotropic flow all its components are zero. i. . since according to the Cayley-Hamilton theorem. 11. Φii = 2p′ ∂vi′ /∂xi = 0.e.7. the trace is zero). It is assumed that the effect of the mean gradients is much larger than the effect of the turbulence.e. Modeling assumptions 126 Provided that c1 > 1 Rotta’s model does indeed reduce non-isotropy as it should. 3 ′ v ′ v ′ ) can be this means that terms that are cubic in vi′ vj′ (i.61) ′ ′ vi vj 2 − δij aij = k 3 aij is an anisotropy tensor whose trace is zero.e. see Eq.1 = −c1 ρ εaij + c′1 aik akj − δij akℓ aℓk 3 (11. where a very strong velocity gradient ∂¯ vi /∂xj is imposed so that initially the second term (the slow term) can be neglected.e.63.20 in [27]). This should be so since the exact form of Φij is trace-less. Here we will carry out a mathematical derivation of a model for the rapid part of the pressure-strain term. The model of the slow pressure-strain term in Eq. i. Note that the right side is trace-less (i. 11.5 Rapid pressure-strain term Above a model for the slow part of the pressure-strain term was developed using physical arguments.11. The notation “rapid” comes from a classical problem in turbulence called the rapid distortion problem. The most general form of Φij. vi′ vj′ = vi′ vk′ vk′ vm m j ′ ′ expressed in terms that are linear and quadratic in vi vj . To make it general it is enough to include terms which are quadratic in vi′ vj′ .1 can be formulated as [28] 1 Φij. . . . ∂¯ . vi . 62) . (ε/k) → ∞ (11. ∂xj . e. Subtraction of Equation B from Equation A gives a Poisson equation for the fluctuating pressure p′ 1 ∂ 2 p′ ∂¯ vi ∂vj′ ∂2 ′ ′ vi vj − vi′ vj′ = −2 − (11. ⇒ Equation suming constant viscosity (see Eq. Take the divergence of incompressible equation assuming con Navier-Stokes ∂vi ∂ vj = . 6. Since it includes the fluctuating pressure.9) i. .5) i.63 which gives the most “rapid” response in p′ . ∂xi ∂xj B. . stant viscosity (see Eq. ∂xi ∂xj 2. . ⇒ Equation A. Thus in this case it is the first term in Eq. we start by deriving an exact equation for p′ starting from Navier-Stokes equations. . Take the divergence of incompressible time-averaged equation as Navier-Stokes ∂¯ vi ∂ v¯j = . p′ .63) ρ ∂xj ∂xj ∂xj ∂xi ∂xi ∂xj rapid term slow term . 1. Now we want to derive an exact equation for the pressure-strain term. Πij . The second “slow” term becomes important first at a later stage when turbulence has been generated.e. 6. 11. This means that the last term in square brackets is zero. in volume V . see Fig.64. For a Poisson equation ∂2ϕ =f (11. y2 x1 . Applying Eq.64) ∂xj ∂xj there exists an exact analytical solution given by Green’s formula. the spatial derivative of all time-averaged fluctuating quantities is zero).65) ϕ(x) = − 4π V |y − x| where the integrals at the boundaries vanish because it is assumed that f → 0 at the boundaries. This requirement is not as drastic as it may sound (although very few turbulent flows are homogeneous). ii) the variation of ∂¯ vi /∂xj in space is small. ∂vj′ (y)/∂yi .3. 11. Now make two assumptions in Eq. 11.65 on Eq. see right-side of Eq.e. The factor two in the rapid term appears because when taking the divergence of the convective term there are two identical terms. y1 Figure 11.63 gives ρ p′ (x) = 4π Z ′ dy3 ∂¯ ∂v (y) ∂2 2 vi (y) j vi′ (y)vj′ (y) − vi′ (y)vj′ (y) + |y − x| ∂yj ∂yi ∂yi ∂yj V rapid term (11. see Appendix T (it is derived from Gauss divergence law) Z 1 f (y)dy1 dy2 dy3 (11.3: The exact solution to Eq. y. 11. vi′ (y)vj′ (y). This term is indeed very small compared to the second derivative of the instantaneous fluctuations. The same argument can be used as above: the mean gradient ∂¯ vi /∂xj varies indeed much more slowly than the instantaneous velocity gradient.66) slow term where dy3 = dy1 dy2 dy3 . 11.7. 8.66: i) the turbulence is homogeneous (i. Modeling assumptions 127 y V x x2 .6. The integral is carried out for all points. 11.11. Now we will take a closer look at rapid part (i.67 can be rewritten as ! ′ (x) ∂vj′ (x) ∂vℓ′ (y) ∂v ∂ 2 vj′ (x) ∂ j − vℓ′ (y) = vℓ′ (y) ∂xi ∂yk ∂yk ∂xi ∂yk ∂xi ! ∂vℓ′ (y) ∂2 ′ ∂ ′ ′ (11.68 to Eq. The second term of Mijkℓ in the integral in Eq.71) = v (y)vj (x) − vj (x) ∂yk ∂xi ℓ ∂yk ∂xi ∂2 ′ vℓ (y)vj′ (x) = ∂yk ∂xi . Now multiply Eq.2 (index 2 denotes the rapid part). and second term the rapid term.69 corresponds to moving the mean velocity gradient out of the integral.66 is zero. consider the integral Z L g(ξ)dξ (11.7.66 with ∂vi′ /∂xj + ∂vj′ /∂xi .68) f (x) = |x − ξ| 0 Note that x and ξ are coordinates along the same axis (think of them as two different points along the x axis). the second term) of Mijkℓ . 11. 11. Φij. If the two points.55). 11. are far from each other.70) ρ ∂xj ∂xi ∂xℓ where the first term represents the slow term. Since this term is not a function of y it can be moved in under the integral. i. 11.69) 0 |x − ξ| Going from Eq. Modeling assumptions 128 Assumption i) means that the last term in the integral in Eq. 11. 11. is taken at point x because it has been moved out of the integral.67 can be written on shorter form as ∂vj′ p′ ∂vi′ ∂¯ vk = Aij + Mijkℓ + = Φij. If we assume that g(ξ) varies slowly with ξ. we only need to consider ξ points which are close to x. Φij.2 (11.e. Eq. Equation 11. x and ξ.67) Mijkℓ 1 + 4π Z V ∂vi′ (x) ∂vj′ (x) + ∂xj ∂xi dy3 ∂2 (vk′ (y)vℓ′ (y)) ∂yk ∂yℓ |y − x| Aij Note that the mean velocity gradient. 11.11.e. ∂ 2 vi′ vj′ =0 ∂yi ∂yj Assumption ii) means that the mean velocity gradient can be taken outside the integral. ∂¯ v /∂xℓ . We obtain after time averaging ′ 1 ′ ∂vi (x) ∂vj′ (x) p (x) + ρ ∂xj ∂xi Z ′ ∂¯ vk (x) 1 ∂vi (x) ∂vj′ (x) ∂vℓ′ (y) dy3 = + ∂xℓ 2π V ∂xj ∂xi ∂yk |y − x| (11.68 can be written as Z L dξ f (x) = g(x) (11.1 + Φij. g(ξ) can be moved out of the integral and since x is close to ξ. then the denominator is large and the contribution to the integral is small.1 (see Eq. Hence. In order to understand this better. the two-point correlations are independent of where in space the two points are located. so that aijkℓ is symmetric with respect to i and k.77) Now let us formulate a general expression of aijkℓ which is linear in vi′ vj′ and symmetric in (j. In other words. ∂ ∂ =− ∂xi ∂ri ∂ ∂ = ∂yi ∂ri We can now write Mijkℓ in Eq.e. xi and ri .73) Differentiating Eq. ri ).76) see Appendix H on p.74.78) .71 and 11. yi and ri . 127 gives that ∂/∂xi = 0 (Item I) or ∂/∂yi = 0 (Item II). Modeling assumptions 129 ∂ 2 vj′ (x)/∂yk ∂xi on line 1 is zero because vj′ (x) is not a function of y.e. k).72 gives Equation 11. aijkℓ = aiℓkj (11.11. aijkℓ = akjiℓ (11. or II. Eq. Hence we can replace the spatial derivative by the distance derivative.72 is a coordinate transformation where we replace xi and yi with I. Note that the terms above as well as in Eq.74) (11. i. 11.e.7. For the same reason the last term on line 2 is zero. i.75) = aijkℓ + ajikℓ It can be shown that aijkℓ is symmetric with respect to index j and ℓ (recall that vℓ′ and vj′ are not at the same point but separated by ri ). 268. 11. the two points being x and y.e. Assumption i) at p. 11. using Eqs. We get aijkℓ = c1 δik vj′ vℓ′ + c2 δjℓ vi′ vk′ + c3 (δij vk′ vℓ′ + δkj vi′ vℓ′ + δiℓ vk′ vj′ + δkℓ vi′ vj′ ) + c4 δjℓ δik k + c5 (δij δkℓ + δjk δiℓ )k (11.67 are two-point correlations. 11. 11. ℓ) and (i. Introduce the distance vector between the two points ri = yi − xi (11.67. i.72) ∂ ∂ ∂ = − ∂ri ∂yi ∂xi (11. Furthermore. as Z 1 ∂2 ′ ′ ∂ 2 ′ ′ dr3 Mijkℓ = − vv + vv 2π V ∂rk ∂ri ℓ j ∂rk ∂rj ℓ i |r| (11.75 is independent of in which order the two derivatives are taken. they are only dependent on the distance between the two points (i. c4 + 2c3 + 4c5 = 0 3c1 + 4c3 − 2 = 0. 55 c5 = 20c2 + 6 55 (11. on line 3. k).70 gives ∂¯ vk ∂¯ vk = (aijkℓ + ajikℓ ) φij.82 give four equations c1 + c2 + 5c3 = 0.81) Using Eq. 11.11.78 and 11. Modeling assumptions 130 Each line is symmetric in (j. looking at Eq. Here it is seen that if i = j then Mijkℓ = 0 due to the continuity equation. Let us express all constants in c2 which gives c1 = 4c2 + 10 .85) We find that the c1 term and the second and third part of the c3 term can be merged.81 in Eq. 268) aijiℓ = 2vj′ vℓ′ (11. 11 c4 = − 50c2 + 4 .80 and 11.83) for the five unknown constants. 2c2 + 3c4 + 2c5 = 0 (11.84 into Eq. 11. 11.7. 11. 11.80) + c4 δkℓ k + c5 (3δkℓ + δkℓ )k = vk′ vℓ′ (c1 + c2 + 5c3 ) + kδkℓ (c4 + 2c3 + 4c5 ) Green’s third formula reads (see Appendix H on p.78 gives 0 = c1 δik vi′ vℓ′ + c2 δiℓ vi′ vk′ + c3 (3vk′ vℓ′ + δki vi′ vℓ′ + δiℓ vk′ vi′ + δkℓ vi′ vi′ ) + c4 δiℓ δik k + c5 (3δkℓ + δik δiℓ )k = c1 vk′ vℓ′ + c2 vℓ′ vk′ + c3 (3vk′ vℓ′ + vk′ vℓ′ + vk′ vℓ′ + 2δkℓ k) (11. For example. Furthermore. Consider Eq.75 we get aiikℓ = 0 (11. term 1 & term 3 and term 2 & term 4 are symmetric with respect to j and ℓ and term 1 & term 2 and term 3 & term 4 are symmetric with respect to i and k. ℓ) and (i. 11. the c2 term and the third and fourth part of the c3 term can be merged as .84) Inserting Eq.2 = Mijkℓ ∂xℓ ∂xℓ ∂¯ vi ∂¯ vj ∂¯ vk ∂¯ vk + c2 vi′ vk′ + vi′ vℓ′ + vj′ vk′ = c1 vj′ vℓ′ ∂xℓ ∂xℓ ∂xj ∂xi ∂¯ vj ∂¯ vi ∂¯ vk ∂¯ vk ∂¯ vk +c3 2δij vk′ vℓ′ + vi′ vℓ′ + vj′ vℓ′ + vk′ vj′ + vk′ vi′ ∂xℓ ∂xℓ ∂xℓ ∂xi ∂xj ∂¯ vj ∂¯ vi ∂¯ vi ∂¯ vj +c4 k + c5 k + + ∂xj ∂xi ∂xi ∂xj (11. 11.79) Applying this condition to Eq.78 gives 2vj′ vℓ′ = 3c1 vj′ vℓ′ + c2 δjℓ vi′ vi′ + c3 (δij vi′ vℓ′ + δij vi′ vℓ′ + δiℓ vi′ vj′ + δiℓ vi′ vj′ ) + (3c4 δjℓ + c5 (δij δiℓ + δji δiℓ ))k (11.82) = 3c1 vj′ vℓ′ + 2c2 δjℓ k + 4c3 vj′ vℓ′ + (3c4 + 2c5 )δjℓ )k = vj′ vℓ′ (3c1 + 4c3 ) + δjℓ k(2c2 + 3c4 + 2c5 ) Equations 11.67. 11 c3 = − 3c2 + 2 . 88) Φij. There are two main effects whose underlying physics are entirely different.86) (11.87 satisfy continuity and symmetry conditions. a simpler model has been proposed [29.4. The coefficients for the slow and rapid terms in the LRR and LRR-IP models are summarized in Table 11.11.6 Wall model of the pressure-strain term When we derived the rapid pressure-strain model using Green’s function in Eq.87) where c2 = 0. 11.88: both models represent “return-to-isotropy”. 11.6 (c2 = −1.6 Table 11.88 does satisfy symmetry condition and continuity but it does not satisfy the integral condition in Eq. 11. Modeling assumptions c1 (Eq.2 = − Finally we re-write this equation so that it is expressed in trace-less tensors 2 c2 + 8 Pij − δij P k Φij.88 is commonly called the IP model (IP=Isotropization by Production) . 11.81. In wall-bounded domains it turns out that the effect of the walls must be taken into account. 11. Since the first term is the most important one.88) 131 LRR model 1. a value of γ = 0.5 0. 11. Both the rapid term in the LRR model and the IP model must be modified to include wall modeling. Since the IP model is both simpler and seems to be more accurate than Eq. 11. 11.83 we get 8c2 − 2 6c2 + 4 k 4 − 60c2 c2 + 8 Pij − Dij + P + k¯ sij 11 11 11 55 ∂¯ vk ∂¯ vk Dij = −vi′ vk′ − vj′ vk′ ∂xj ∂xi φij. it is one of the most popular models of the rapid pressure-strain term. using Eq.5 − 0. truncated version of Eq.7. Since two terms are omitted we should expect that the best value of γ should be different than (c2 + 8)/11. 11.7. It might be possible to use a simpler pressure-strain model using one or any two terms.87 it does not satisfy all requirements that Eq. This pressure-strain model is called the LRR model and it was proposed in [29].4) was found to give good agreement with experimental data.88 is a truncated form of Eq.1 11.87. well as the c4 and c5 terms. The effect of the wall is to dampen turbulence.87. 11. Note that Φii = 0 as we required in Eq.55 c2 (Eq. Since Eq. All three terms in Eq.2 = −c2 ρ Pij − δij P k 3 It can be noted that there is a close similarity between the Rotta model and Eq.2 = −ρ 11 3 8c2 − 2 2 60c2 − 4 −ρ Dij − δij P k − ρk¯ sij 11 3 55 (11. The model in Eq. 11. 30] 2 (11. it is often found to give more accurate results [31]. 11.87) c2 (Eq. 11. Although Eq. Equation 11.4 − LRR-IP model 1.1: Constants in the LRR and LRR-IP pressure-strain models. 11. the first expressed in vi′ vj′ and the second in Pij .79. .66 we neglected the influence of any boundaries. 11. 11.88 is a simpler.87 do. 4: Modeling of wall correction in pressure-strain terms. this damping is inviscid (due to pressure) and affects the turbulent fluctuations well into the log-region. The pressure fluctuations dampens the wall-normal fluctuations. k 3/2 /ε.e.1w = c1w v2′2 f k (11. In both cases it is the pressure that informs the fluid particle of the presence of the wall. Moreover. Pressure. It is suitable to include a slow and a fast wall model term. This is true for a fluid particle carried by the wind approaching a building as well as for a fluid particle carried by a fluctuating velocity approaching the wall in a turbulent boundary layer. function f should not exceed one. We need to scale the wall-normal distance with a relevant quantity and the turbulent length scale. Consider a wall.1w = Φ33. In the viscous region the wall model term.1w + Φij.89) where subscript w denotes wall modeling. The IP wall model for the wall-parallel fluctuations reads ε Φ11.w )|.w (xi − xi. the IP wall model reads [32] ε Φ22. 2.4.w ) denotes the distance vector to the wall. the damping function goes to zero since the distance to the wall. 1. see Fig. Viscosity. Φij = Φij. Up to now we have introduced two terms for modeling the pressure-strain term. the slow and the fast term.w (xi − xi. is not relevant and should be zero since it should account only for inviscid damping. As explained above. i.w )|ε (11. Close to the wall the viscous processes (viscous diffusion and dissipation) dominate over the turbulent ones (production and turbulent diffusion).1w .90) where ni.55|ni. the damping effect of the wall should decrease for increasing wall distance. Away from the wall. Φ22. seems to be a good candidate. |ni. in the fully turbulent region. It has nothing to do with viscous damping. When a fluid particle approaches a wall.7.1 + Φij.11.1w = −2c1w v2′2 f k 3 k2 f= 2.2 + Φij. Modeling assumptions 132 x2 x1 Figure 11. the presence of the wall is felt by the fluid particle over a long distance. 11. it is obviously the second of these two processes that we want to include in the wall model. k 3/2 /ε.91) . Since the pressure-strain term includes the fluctuating pressure.w (xi − xi. increases faster than the turbulence length scale.2w (11. Furthermore. For the wall-normal fluctuations. 2w = c2w Φkm.34) we get the modeled equation for k ∂k ∂¯ vi ∂k ∂¯ vj ∂¯ νt ∂ θ¯ vi = νt + + gi β + v¯j ∂t ∂xj ∂xj ∂xi ∂xj σθ ∂xi (11.11.w nm. 11.w f 2 2 (11.09. details on wall-functions and low-Reynolds number models can be found in Sections 3 and 4.8 The k − ε model The exact k equation is given by Eq.w δij − Φki.w nj. the modeled ε equation is obtained from Eq.w − vk vj ni.95) νt ∂k ∂ ν+ −ε + ∂xj σk ∂xj In the same way. Φ12. Φ22.92.e. 1. The k − ε model 133 The requirement that the sum of the pressure strain term should be zero.1w = 0. 11. In that report.38).2 nk.w nm. 1.33 and 11. 11. By inserting the model assumptions for the turbulent diffusion (Eq.98) For details on how to obtain these constants are obtained.1w . the production (Eq. respectively. see Section 3 in Introduction to turbulence models. The general formula for a wall that is not aligned with a Cartesian coordinate axis reads [32] ε 3 ′ ′ 3 ′ ′ ′ n Φij.96) The turbulent viscosity is computed as νt = c µ k2 ε (11.8. 11.1w .w δij − vk vi nk. σk .2 nk.1w = 0.w f k 2 2 (11.1w + Φ22. taking the sum of the eigenvalues).e. 1.1w = − c1w v1′ v2′ f 2 k (11.92) The factor 3/2 is needed to ensure that Φii. σε ) = (0. i.2 ni. Φ21. cε2 .1w .w nj.1w + Φ33. The wall model for the shear stress is set as ε 3 Φ12.w − Φkj.37) and the buoyancy term (Eqs.23.1w = c1w vk′ vm k. You can prove this by rotating the matrix [Φ11. .94) 11.44.1w ] and taking the trace of Φ in the principal coordinates system (i. 1.w nk. 11. cε1 .w nk.93) An analogous wall model is used for the rapid part which reads 3 3 Φij.1w = 0 is satisfied when the coordinate system is rotated. Φii.28 ∂¯ vi ∂ε ε ∂¯ vj ∂¯ vi ∂ε = cε1 νt + + v¯j ∂t ∂xj k ∂xj ∂xi ∂xj νt ε2 ∂ ∂ε ε νt ∂ θ¯ ν+ − cε2 + + cε1 gi k σθ ∂xi k ∂xj σε ∂xj (11. is now satisfied since Φ11.97) The standard values for the coefficients read (cµ .3) (11. 9. pressure-strain and dissipation we get ∂vi′ vj′ = ∂t ∂vi′ vj′ v¯k = ∂xk ∂¯ vj ∂¯ vi − vj′ vk′ −vi′ vk′ ∂xk ∂xk ε 2 −c1 vi′ vj′ − δij k k 3 2 −c2 Pij − δij P k 3 3 ′ ′ ε ′ ′ +c1w ρ [ vk vm nk nm δij − vi vk nk nj k 2 3 ′ ′ − vj vk nk ni ]f 2 3 +c2w [ Φkm.101) . wall.100) In ASM we assume that the transport (convective and diffusive) of vi′ vj′ is related to that of k. rapid part) (viscous diffusion) (turbulent diffusion) −gi βvj′ θ′ − gj βvi′ θ′ (buoyancy production) 2 − εδij (dissipation) 3 (11. slow part) (pressure strain.e.9 The modeled vi′ vj′ equation with IP model With the models for diffusion.2 nk nm δij − Φik. slow part) (pressure strain. wall.100 into the equation above gives Pij + Φij − εij = vi′ vj′ Pk − ε k (11.10 Algebraic Reynolds Stress Model (ASM) The Algebraic Reynolds Stress Model is a simplified Reynolds Stress Model. 11.2 nk ni ]f 2 ∂ 2 vi′ vj′ +ν ∂xk ∂xk " # νt ∂vi′ vj′ ∂ + ∂xm σk ∂xm (unsteady term) (convection) (production) (pressure strain. vi′ vj′ Cij − Dij = C k − Dk k Inserting Eq. The modeled vi′ vj′ equation with IP model 134 11. 114 & 117) as: RSM : Cij − Dij = Pij + Φij − εij k − ε : C k − Dk = P k − ε (11. rapid part) (pressure strain.11.2 nk nj 2 3 − Φjk.99) 11. i. The RSM and k − ε models are written in symbolic form (see p. 1w + Φij. in that expression can be a function of not more than the following five invariants ¯ ji .2w = 0) 2 v1 c1 − 1 + c2 P k /ε k 2 ∂¯ −v1′ v2′ = (1 − c2 ) k 3 (c1 − 1 + P /ε) ε ∂y cµ As can be seen.1w + Φij.e.11. (k /ε )Ωij Ωjk s¯km s¯mi (k 2 /ε2 )¯ sij s¯ji .11.1w = Φij.101 and multiply by k/ε so that 2 k 2 2 k k ′ ′ Pij − δij P − δij k Pij − c1 vi vj − δij k − c2 ε 3 ε 3 3 Collect all vi′ vj′ terms so that " k 2 = Pij −δij P k ε 3 vi′ vj′ k + (Φij. i.102 is an implicit equation for vi′ vj′ . G(n) . Ωij . (11. He assumed that the Reynolds stress tensor can be expressed in the strain-rate tensor. s¯ij .88) and the isotropic model for εij (Eq. Explicit ASM (EASM or EARSM) 135 Thus the transport equation (PDE) for vi′ vj′ has been transformed into an algebraic equation based on the assumption in Eq.1w + Φij. Insert the IP models for Φij.11 Explicit ASM (EASM or EARSM) Equation 11.2w ′ ′ vi vj = δij k + (11. the Reynolds stresses appear both on the left and the right side of the equation. Dividing both sides by P k /ε − 1 + c1 gives finally 2 k (1 − c2 ) Pij − 23 δij P k + Φij. and the vorticity tensor.102 reads (without any wall terms.1w + Φij.104) . 11.103) In two dimension the expression reads vi′ vj′ = k2 k3 2 ¯ kj − Ω ¯ ik s¯kj ) sik Ω kδij + G(1) s¯ij + +G(2) 2 (¯ 3 ε ε (11. 11. (k 3 /ε3 )¯ ¯ ij Ω sij s¯jk s¯ki (k 2 /ε2 )Ω 4 4 ¯ ¯ 3 3 ¯ ¯ (k /ε )Ωij Ωjk s¯ki . It would of course be advantageous to be able to get an explicit expression for the Reynolds stresses. this model can be seen as an extension of an eddy-viscosity model where the cµ constant is made a function of the ratio P k /ε. Now we want to re-write this equation as an equation for vi′ vj′ .47) in Eq.100.2w ) = Pk − ε ε ε k P vi′ vj′ − 1 + c1 = ε 2 2 k Pij − c2 Pij − δij P k + Φij. he showed that the coefficients.2 (Eq. Pope [33] managed to derive an explicit expression for ASM in two dimensions. 11. Furthermore. 11. 11.55) and Φij. 11. Φij. 11.102) 3 ε c1 + P k /ε − 1 In boundary layer flow Eq. i.1 (Eq.2w + δij k(P k /ε − 1 + c1 ) 3 3 where (2/3)δij P k k/ε was added and subtracted at the last line (shown in boxes).2w + δij k(−1 + c1 ) ε 3 3 # 2 2 k − c2 Pij − δij P + Φij.e. Furthermore. it is useful to look at its source terms which to a large degree govern the magnitude of vi′ vj′ . see Section 9. 11.e. Tij8 = s¯im Ω ¯ mk s¯kn Ω ¯ nj − Ω ¯ im Ω ¯ kn Ω ¯ im s¯mk Ω Tij7 = Ω ¯ nj − 2 δij Ω ¯ kn Ω ¯ mk s¯kn s¯nj − s¯im s¯mk Ω ¯ im Ω ¯ mk s¯kn s¯np ¯ pm Ω Tij9 = Ω 3 ¯ pj ¯ mk s¯kn s¯np Ω ¯ pj − Ω ¯ im Ω ¯ np Ω ¯ im s¯mk s¯kn Ω T 10 = Ω ij (11. A large source term in the equation for the v1′2 equation. hence cubic terms or higher can recursively be expressed in linear (¯ sij ) and quadratic tensors (¯ sik s¯kj ).105) where Tijn may depend on the five invariants in Eq. v¯1 = v¯1 (x2 ). for example. Let us study boundary layer flow (Fig. 11.11. 11. vi′ vj′ − 2δij k/3. Eq. In general three-dimensional flow. Equation 11.103. ¯ kj − s¯jk Ω = s¯ik Ω ¯ ki . the Reynolds stress tensor depends on 10 tensors.5) where v¯2 = 0.12. Tij2 10 X G(n) Tijn n=1 1 Tij3 = s¯ik s¯kj − δij s¯ik s¯ki 3 5 ¯ kj ¯ ik s¯km s¯mj − s¯im s¯mk Ω Tij = Ω ¯ ki .5: Boundary layer flow. i. 11. 11. Tijn [33].105 includes only linear and quadratic terms of s¯ij ¯ ij .20 in [27]). Any ASM may be written on the form of Eq.99. ¯ ik Ω ¯ kj − 1 δij Ω ¯ ik Ω Tij4 = Ω 3 ¯ mj − 2 δij Ω ¯ km Ω ¯ mk s¯kp ¯ mk s¯kj + s¯ik Ω ¯ pm Ω ¯ im Ω Tij6 = Ω 3 ¯ kn s¯nj ¯ mk s¯kn s¯nj − s¯im s¯mk Ω ¯ nj .105 is a general form of a non-linear eddy-viscosity model. It may be noted that Eq. Boundary layer flow 136 x2 n v¯1 (x2 ) Xx s x1 Xx Figure 11. The production Pij has the form: Pij = −vi′ vk′ ∂¯ vj ∂¯ vi − vj′ vk′ ∂xk ∂xk .105. will increase v1′2 and vice versa. That is because of Cayley-Hamilton theorem which states that a secondand Ω order tensor satisfies its own characteristic equation (see Section 1. 11.12 Boundary layer flow In order to better understand the Reynolds stress equation. 11.1.105 are symmetric and traceless as required by the left side. note that all terms in Eq. vi′ vj′ − kδij = Tij1 = s¯ij . 2 . v2′2 ). the question arises: what is the main sink term in the v1′ v2′ equation? The answer is. again.88) gives ε 2 Φ22.1 = c1 k − v2′2 > 0 k 3 1 ∂¯ v1 2 Φ22. saves the unfair situation! The IP model for Φij.e.2 (Eq. v1′2 ) and gives to the poor (i. . 11.11.2 = c2 P11 = −c2 v1′ v2′ >0 3 3 ∂x2 Note also that the dissipation term for the v1′ v2′ is zero. which takes from the rich (i.55) and Φij.1 and Φ12.1 and Φij.e. 122). Boundary layer flow 137 In this special case we get: P11 = −2v1′ v2′ P12 = −v2′2 ∂¯ v1 ∂x2 ∂¯ v1 ∂x2 P22 = 0 Is v2′2 zero because its production term P22 is zero? No! The sympathetic term Φij . the pressure strain term Φ12.12. 11. Since the modeled v1′ v2′ does not have any dissipation term. but it takes the value 32 ε for the v1′2 and v2′2 equations (see p.2 (Eq. then the flow is stably stratified. and buoyancy takes it back to its original level 0. Now we will investigate how the Reynolds stress model behaves in stable con¯ ditions.1: Stable stratification due to positive temperature gradient ∂ θ/∂x 3 > 0. This flow situation is called unstable stratification. eddy-viscosity models 138 θ¯2 > θ¯0 ρ2 < ρ0 2 F ¯ ∂ θ/∂x 3 >0 ∂ρ/∂x3 < 0 0 x3 F θ¯1 < θ¯0 ρ1 > ρ0 1 ¯ Figure 12. i. the buoyancy forces the particle back to its original position 0. the situation is reversed. Similarly if a particle originating at level 0. Hence.12. In this way the vertical turbulent fluctuations are dampened.e. Hence.e. i. If a fluid particle at level 0 is displaced upwards to level 2. Here it is lighter than its new environment. whereas in eddy-viscosity models it is modeled. ∂ρ/∂x3 < 0.1) since gi = (0. From the equation for the turbulent heat flux.1 we would then have ρ2 > ρ0 . If it is moved down to level 1 it is heavier than its new environments and it will then continue downwards.e. Consider ∂ θ/∂x 3 > 0. If a fluid particle is displaced from its equilibrium level 0 up to level 2. Eq. v3′ θ′ (i.22 .1 Stable and unstable stratification In flows where buoyancy is dominating. 12 Reynolds stress models vs.e. when ∂ θ/∂x 3 > 0. hence it continues to move upwards. This is illustrated in ¯ Fig. it is heavier then the surrounding at this new level (ρ0 > ρ2 ). turbulent fluctuations are enhanced. 12. The production term due to buoyancy reads (see Eq. 11.1. 12. 12. ∂ θ/∂x 3 > 0). ∂ θ/∂x 3 < 0 and ∂ρ/∂x3 > 0. Cold fluid is ¯ located on top of hot fluid. the temperature has a large effect on the turbulence through the buoyancy term Gij .11) G33 = 2gβv3′ θ′ (12. The reason for the superiority of the former model is in all cases that the production term is treated exactly. 0. see Eq. This means that the density decreases with increasing vertical height. is moved down to level 1.11.e. For the case of unstable stratification. In Fig. If the temperature increases ¯ upwards (i. see Fig. 11. Reynolds stress models vs. eddy-viscosity models In this section we present three fundamental physical processes which Reynolds stress models are able to handle whereas eddy-viscosity models fail. 12. 11. i. −g).1. it is at this location lighter than its new environment. the terms including viscosity are omitted) degenerates to ∂p ρvθ2 − =0 r ∂r (12. In addition they reported significant damping of the turbulence in the shear layer in the outer part of the separation region. 11. we find that the production term. ¯ If the situation in Fig.1 is reversed so that ∂ θ/∂x 3 < 0 the vertical fluctuations are instead augmented. The ratio of boundary layer thickness δ to curvature radius R is a common parameter for quantifying the curvature effects on the turbulence. is that the former reduces k whereas the latter reduces only the vertical fluctuations. An illustrative model case is curved boundary layer flow.3) For gi = (0. the radial inviscid momentum equation (i.11 (take the trace of Gij and divide by two) Gk = 0. it makes v3′ θ′ < 0 so that G33 < 0 (see Eq. 11. Gij .1. The buoyancy term.2.e. Gk .12. Concave curvature destabilizes the turbulence. 12. 12.e. 12. in the k equation reads.2 Curvature effects When the streamlines in boundary layer flow have a convex curvature. is negative. 12. the production term due to temperature gradient ¯ reads P3θ = −v3′2 ∂ θ/∂x 3 < 0 (recall that we assume that buoyancy dominates so that the first term in Eq. especially the shear stress and the Reynolds stress normal to the wall. since G11 = G22 = 0 because the gravity is in the x3 direction (i. When eddy-viscosity models are used.03.2. This is called unstably stratified conditions. 12. 11. Gk < 0 as required. Since the main source term in the v3′ θ′ equation. Thus. the turbulence is stabilized. g1 = g2 = 0). They reported a 50 percent decrease of ρv2′2 (Reynolds stress in the normal direction to the wall) owing to curvature. where they measured δ/R ≃ 0. The reduction of ρv1′2 and −ρv1′ v2′ was also substantial. see Eq. The difference between an eddy-viscosity model and a Reynolds stress model. −g). 0. P3θ .2 is much larger than the second one). see Eq. Note that the horizontal turbulent fluctuations are not affected by the buoyancy term. due to buoyancy yields a damping of the vertical fluctuations as it should.33 gives Gk = −gβ νt ∂ θ¯ σθ ∂x3 (12. A polar coordinate system r − θ with θˆ locally aligned with the streamline is introduced.1. This dampens the turbulence [34. 35]. we find the production term for v3′ θ′ P3θ = −v3′ vk′ ∂ θ¯ ∂¯ v3 − vk′ θ′ ∂xk ∂xk (12. In [36] they carried out an experimental investigation on a configuration simulating the flow near a trailing edge of an airfoil. transport equations are usually not solved for vi′ θ′ . see Fig.33. for the case illustrated in Fig.4) Hence it is seen that in stably stratified conditions.1). Curvature effects 139 with i = 3). The work reviewed by Bradshaw [34] demonstrates that even such small amounts of convex curvature as δ/R = 0. G33 .01 can have a significant effect on the turbulence.5) . Instead the heat flux tensor is modeled with an eddy-viscosity assumption using the Boussinesq assumption. 12.2) In the case illustrated in Fig. As vθ = vθ (r) (with ∂vθ /∂r > 0 and vr = 0).5Gii = −gi βvi′ θ′ (12. it reads Gk = gβv3′ θ′ which with Eq. 12. Here the variables are instantaneous or laminar. since (vθ )A > (vθ )0 . which from Eq. the flow is deflected upwards. At one x1 station. It is discussed below how the Reynolds stress model responds to streamline curvature. it encounters a pressure gradient larger than that to which it was accustomed at r = r0 . when ∂vθ /∂r > 0.3: Streamline curvature occurring when the flow approaches.2.12. which is balanced by the pressure gradient.e. has a stabilizing effect on (turbulent) fluctuations. The centrifugal force exerts a force in the normal direction (outward) on a fluid following the streamline. .3. at least in the radial direction. The streamline is aligned with the θ axis. Eq.2: Flow in a polar coordinate system illustrating streamline curvature. see Fig. a separation region or an obstacle. for example. Instead the centrifugal force drives it back to its original level.5 gives (∂p/∂r)A > (∂p/∂r)0 . Similarly. It is clear from the model problem above that convex curvature. The production terms Pij owing to rotational strains (i. Curvature effects 140 A A 0 0 B r B vθ (r) Figure 12. Since we have assumed that ∂vθ /∂r > 0. if the fluid is displaced inwards to level B. turbulent fluctuation) is displaced outwards to level A. Hence the fluid is forced back to r = r0 . 12. the pressure gradient is smaller here than at r = r0 and cannot keep the fluid at level B. 12.5 shows that the pressure gradient increases with r. 12. The ratio of the normal stresses ρv1′2 to ρv2′2 is typically 5 (or more). If the fluid by some disturbance (e. How will this affect turbulence? Let us study the effect of concave streamline curvature. r v¯1 x2 streamline θ x1 Figure 12.g. Assume that there is a flat-plate boundary layer flow. Curvature effects 141 convex curvature concave curvature ∂Vθ /∂r > 0 ∂Vθ /∂r < 0 stabilizing destabilizing destabilizing stabilizing Table 12. There are two other options. all production is a result of ∂¯ v1 /∂x2 . The turbulence is destabilized owing to concave curvature of the streamlines.2.6b) (12. in the production term are multiplied by the same coefficient (the turbulent viscosity).11): RSM. An increase in the magnitude of ′ ′ P12 will increase −v1 v2 . which continuously increases the Reynolds stresses. ∂¯ v2 /∂x1 ) can be written as (see Eq. : P12 = −v1′2 ∂¯ v1 ∂x2 ∂¯ v2 ∂¯ v1 − v2′2 ∂x1 ∂x2 RSM. Even if ∂¯ v2 /∂x1 is much smaller than ∂¯ v1 /∂x2 it will still contribute non-negligibly to P12 as ρv1′2 is much larger than ρv2′2 .12.6a) ∂¯ v2 ∂x1 ∂¯ v1 ∂¯ v2 + ∂x2 ∂x1 (12. we have assumed concave curvature and positive velocity gradient. It is seen that there is a positive feedback. v1′ v2′ − eq. If the flow (concave curvature) is a wall jet flow where ∂¯ v1 /∂x2 < 0 in the .1: Effect of streamline curvature on turbulence.6d) The terms in boxes appear because of the streamline curvature. ∂¯ v1 /∂x2 and ∂¯ v2 /∂x1 . which in turn will increase P11 and P22 . Above. : P22 = −2v1′ v2′ k−ε k P = νt (12.6c) !2 (12. As long as the streamlines are parallel to the wall.4: The velocity profile for a wall jet. However as soon as the streamlines are deflected. and so on. Note that eddy-viscosity models such as k − ε and k − ω models cannot account for streamline curvature since the two rotational strains. 11. v2′2 − eq. v¯1 (x2 ) x2 x1 Figure 12. ∂¯ v1 /∂x2 . 1. there are more terms resulting from ∂¯ v2 /∂x1 . : P11 = −2v1′ v2′ RSM. v1′2 − eq. Thus the magnitude of P12 will increase (P12 is negative) as ∂¯ v2 /∂x1 > 0. This means that ρv1′2 and ρv2′2 will be larger and the magnitude of P12 will be further increased. It should be noted that concave or convex depends on from which point the streamline is viewed. whereas ASM/RSM do. 12. In RSM. For convenience.5 (P11 + P22 ) = −v1′2 (12. If the streamline (and the wall) is deflected downwards.1. The streamline in Fig. Stagnation flow x2 x2 x1 142 x2 x1 x1 Figure 12. outer part (see Fig. 12. 12.8) where continuity ∂¯ v1 /∂x1 = −∂¯ v2 /∂x2 has been employed. In the k − ε model. and whether there is an increase or decrease in tangential momentum with radial distance from its origin (i. however.3. is concave when viewed from the wall but convex when viewed from the origin of the circle with radius r.7) (12. and destabilizing for ∂¯ v1 /∂x2 < 0. the situation will be as follows: the turbulence is stabilizing when ∂¯ v1 /∂x2 > 0. 12.5) due to ∂¯ v1 /∂x1 and ∂¯ v2 /∂x2 (which are the dominant strains) is: ∂¯ v1 ∂¯ v2 ∂¯ v1 ′2 − v2′2 =− (v − v2′2 ) ∂x1 ∂x2 ∂x1 1 ( 2 2 ) ∂¯ v1 ∂¯ v2 k + k − ε : P = 2νt ∂x1 ∂x2 RSM : 0. for example. . the two terms are added with sign.4) the situation will be reversed: the turbulence will be stabilized. 2. the sign of ∂Vθ /∂r).5: The flow pattern for stagnation flow.3. The production term in the k equations for RSM/ASM and k − ε model in stagnation flow (see Fig.e. The stabilizing or destabilizing effect of streamline curvature depends on the type of curvature (convex or concave).3 Stagnation flow The k − ε model does in this type of flow not model the normal stresses properly. the production will be large because the difference in sign of the two terms is not taken into account because it includes the square of the velocity gradients. these cases are summarized in Table 12.12. v2′2 . v3′2 ) ii) as a consequence of i) it is unable to account for curvature effects iii) as a consequence of i) it is unable to account for irrotational strains (stagnation flow) iv) in boundary layers approaching separation. • Advantages with ASM/RSM: i) the production terms do not need to be modeled ii) thanks to i) it can selectively augment or damp the stresses due to curvature effects (RSM is better than ASM because the convective terms are accounted for) and it is more accurate for boundary layers approaching separation.4 RSM/ASM versus k − ε models • Advantages with k − ε models (or eddy viscosity models): i) simple due to the use of an isotropic eddy (turbulent) viscosity ii) stable via stability-promoting second-order gradients in the mean-flow equations iii) work reasonably well for a large number of engineering flows • Disadvantages: i) isotropic. RSM/ASM versus k − ε models 143 12. the production due to normal stresses is of the same magnitude as that due to shear stresses [37].4.12. • Disadvantages with ASM/RSM: i) RSM is complex and difficult to implement. especially implicit ASM ii) numerically unstable because small stabilizing second-order derivatives in the momentum equations (only viscous diffusion) iii) CPU time consuming . and thus not good in predicting normal stresses (v1′2 . buoyancy etc. s¯11 in Eq.e.2) It is seen that if s¯11 gets too large then v1′2 < 0 which is nonphysical. The equation for finding the eigenvalues of a tensor Cij is (see e.1) These criteria are seldom used in RSMs. e. Assume that the flow is in the x1 direction and that it approaches the wall (see Fig. satisfying the first criteria is actually of importance for eddy-viscosity models in stagnation flow [38]. If a tensor is symmetric. The usual two ones are that all normal stresses should stay positive and that the correlation coefficient for the shear stress should not exceed one.2 is largest.g. Eq. 12.2). [27] or [39]) |Cij − δij λ| = 0 which gives. The Boussinesq assumption for the normal stress v1′2 reads (cf. However. i 6= j vi′2 vj′2 (13. For the strain tensor this means that the off-diagonal components of s¯ij vanish and this is the coordinate system where the diagonal components become largest (e. Here we mean independent of rotation of the coordinate system. Let’s now briefly repeat the concept “invariants”.e. in 2D. [27]). Realizability 144 13 Realizability There are a number of realizability constraints. i..5). vi′2 ≥ 0 for all i |vi′ vj′ | 1/2 ≤ 1 no summation over i and j.7) v1′2 = 2 ∂¯ v1 2 = k − 2νt s¯11 k − 2νt 3 ∂x1 3 (13. 13.13. i. non-realizable. This means something that is independent of the coordinate system. 12.g. This is thus the coordinate system in which the danger of negative v1′2 from Eq.g. then we know that it has real eigenvalues which means that we can rotate the coordinate system so that the off-diagonal components vanish (see. 13. The resulting equation is . . C11 − λ . . C21 . C12 . . =0 C22 − λ . Eq.5) Since Eq.3) (13. 13. it follows that its coefficients I12D and I22D are invariants.3 gives . In 3D.4) λ2 − I12D λ + I22D = 0 I12D = Cii 1 I22D = (Cii Cjj − Cij Cij ) = det(Cij ) 2 (13.5 is the same irrespectively of how the coordinate system is rotated. 13. (13. . . C11 − λ C12 C13 . . . . C21 C22 − λ C23 . . = 0 (13.6) . . C31 C32 C33 − λ . . 5) which is the same in all coordinate systems (hence the name ”invariant”). and Eq.11) The requirement v1′2 ≥ 0 gives now together with Eq.8) It is zero due to the continuity equation.14) .1 Two-component limit Another realizability constraint is to require that when vi′2 approaches zero near walls.13. 13.9) (see Eq.2 and assume incompressible 2D flow.2 is replaced by the largest eigenvalue so that v1′2 = 2 k − 2νt λ1 3 (13.8. 13.11 νt ≤ k k = 3|λ1 | 3 2 s¯ij s¯ij 1/2 (13. The first invariant reads (cf. The solution to Eq.10 is replaced by [38] |λk | = 2¯ sij s¯ij 3 1/2 (13. s¯11 in Eq.1.13) This is a simple modification of an eddy-viscosity model. we apply Eq. 13.2 = ± −I22D 1/2 =± s¯ s¯ 1/2 ij ij 2 (13. Two-component limit 145 which gives λ3 − I13D λ2 + I23D λ − I33D = 0 I13D = Cii 1 I23D = (Cii Cjj − Cij Cij ) 2 1 3D I3 = (2Cij Cjk Cki − 3Cij Cji Ckk + Cii Cjj Ckk ) = det(Cij ) 6 (13. using Eq. 13.7) The invariants are I13D .e.5. it should do so smoothly. 13. (13. 13. 13. As discussed above. One way to ensure this is to require that the derivative of vi′2 should go to zero when vi′2 goes to zero. Let’s go back to Eq. Eq. is λ1. I23D and I23D .12) In 3D. Eq. 13. and it ensures that the normal stresses stay positive.2 in the principal coordinate directions of s¯ij . 13. The second invariant of s¯ij reads I22D = −¯ sij s¯ij /2. Hence.5 is used. 13.10) The eigenvalues of s¯ij correspond to the strains in the principal axis.7 instead of Eq.5) I12D = s¯ii = s¯11 + s¯22 = λ1 + λ2 = 0 (13. 13. 13. i. vi′2 → 0 ⇒ dvi′2 →0 dt (13. 13.1. Two-component limit 146 where d/dt denotes the material derivative (think of Eq. 13.14 in Lagrangian coordinates, i.e. we follow a fluid particle as it approaches the wall). Equation 13.14 requires that when vi′2 approaches zero, the left side (and thus also the right side) of the transport equation of vi′2 should also do so too. Since we are here concerned about the pressure-strain term, we’ll take a look at how it behaves near walls when vi′2 → 0. This is of some relevance in near-wall turbulence where the wall-normal stress goes to zero faster than the wall-parallel ones: this state of turbulence is called the two-component limit [40]. The wall-normal component goes to zero as v2′2 = O(x42 ) whereas v1′2 and v3′2 go to zero as O(x22 ), see Section 4 in Introduction to turbulence models. Neither the form of Φij,2 in Eq. 11.88 nor Eq. 11.87 satisfy the requirement that Φ22,2 = 0 when v2′2 = 0 [28]. In Eq. 11.88, for example, 2 Φ22,2 → γ P k 6= 0 3 (13.15) even if v2′2 = 0. Very complex forms of Φij,2 have been proposed [41] [CL96] which include terms cubic in vi′ vj′ . The CL96 model does satisfy the two-component limit. Another advantage of the CL96 model is that it does not need any wall distances, which is valuable in complex geometries. The models of the slow pressure-strain in Eq. 11.55 (linear model) and Eq. 11.61 (non-linear model) do also not satisfy the two-component limit. The Rotta model (Eq. 11.55), for example, gives Φ22,1 → c1 ρ 2ε 6= 0 3 (13.16) when v2′2 → 0. The only way to ensure that Φ22,1 → 0 is to make c1 → 0 when the wall is approached. A convenient parameter proposed in [40] is A which is an expression of A2 and A3 (the second and third invariant of aij , respectively), i.e. 9 A2 = aij aji , A3 = aij ajk aki , A = 1 − (A2 − A3 ) 8 (13.17) The parameter A = 0 in the two-component limit and A = 1 in isotropic turbulence. Thus A is a suitable parameter to use when damping the constant c1 as the wall is approached. 14. Non-linear Eddy-viscosity Models 147 14 Non-linear Eddy-viscosity Models In traditional eddy-viscosity models the turbulent stress vi′ vj′ is formulated from the Boussinesq assumption, i.e. s¯ij aij = −2νt k vi ∂¯ vj 1 ∂¯ + s¯ij = 2 ∂xj ∂xi (14.1) where the anisotropy tensor is defined as aij ≡ vi′ vj′ 2 − δij k 3 (14.2) The relation between the stress vi′ vj′ and the velocity gradient in Eq. 14.1 is, as can be seen, linear. One way to make eddy-viscosity models more general is to include non-linear terms of the strain-rate (i.e. the velocity gradient) [33]. A subset of the most general form reads [42] aij = −2cµ τ s¯ij 1 2 ¯ kj ¯ ik s¯kj − s¯ik Ω + c1 τ s¯ik s¯kj − s¯ℓk s¯ℓk δij + c2 τ 2 Ω 3 1¯ ¯ 2 ¯ ℓj − Ω ¯ iℓ s¯ℓk s¯kj ¯ ¯ + c3 τ Ωik Ωjk − Ωℓk Ωℓk δij + c4 τ 3 s¯ik s¯kℓ Ω 3 2¯ ¯ 3 ¯ ¯ ¯ ¯ + c5 τ Ωiℓ Ωℓm s¯mj + s¯iℓ Ωℓm Ωmj − Ωmn Ωnℓ s¯ℓm δij 3 3 3¯ ¯ + c6 τ s¯kℓ s¯kℓ s¯ij + c7 τ Ωkℓ Ωkℓ s¯ij vi ∂¯ vj 1 ∂¯ ¯ − Ωij = 2 ∂xj ∂xi (14.3) where τ is a turbulent time scale; for a non-linear k − ε model τ = k/ε, and for a nonlinear k − ω model τ = 1/ω. The tensor groups correspond to a subset of Eq. 11.105: Line 1: Tij1 , Line 2: Tij3 and Tij2 Line 3: Tij4 and Tij5 Line 4: Tij6 ¯ kℓ ¯ kℓ Ω Line 5: Tij1 multiplied by the invariants s¯kℓ s¯kℓ and Ω The expression in Eq. 14.3 is cubic in ∂¯ vi /∂xj . However, note that it is only ¯ ij . This is due to Cayley-Hamilton theorem which states that a quadratic in s¯ij and Ω tensor is only linearly independent up to quadratic terms, see p. 126; this means that, for example, s¯3ij = s¯ik s¯kℓ s¯ℓj can be expressed as a linear combination of s¯2ij = s¯ik s¯kj and s¯ij . aij is symmetric and its trace is zero; it is easily verified that the right side of Eq. 14.3 also has these properties. Examples of non-linear models (sometimes also 14. Non-linear Eddy-viscosity Models 148 called explicit algebraic Reynolds stress models, EARSM) in the literature are the models presented in [42–45]. EARSMs are very popular — especially the model in [45] — in the aeronautical community where explicit time-marching solvers are used. They are computationally cheap, more accurate than linear eddy-viscosity models and they do not give rise to any numerical instabilities as in implicit solvers (like SIMPLE). In implicit solvers a large turbulent viscosity in the diffusion term of the momentum equations is needed to stabilize the solution procedure. v2 = v¯3 = Let’s take a closer look on Eq. 14.3 in fully developed channel flow (¯ ∂/∂x1 = ∂/∂x3 ≡ 0); we obtain a11 1 2 = τ 12 a12 ∂¯ v1 ∂x2 2 (c1 + 6c2 + c3 ) 2 ∂¯ v1 (c1 − 6c2 + c3 ) ∂x2 2 1 2 ∂¯ v1 =− τ (c1 + c3 ) 6 ∂x2 3 1 ∂¯ v1 ∂¯ v1 + τ3 (−c5 + c6 + c7 ) = −cµ τ ∂x2 4 ∂x2 a22 = a33 1 2 τ 12 (14.4) Using values on the constants as in [42], i.e c1 = −0.05, c2 = 0.11, c3 = 0.21, c4 = −0.8 c5 = 0, c6 = −0.5 and c7 = 0.5 we get 2 2 2 ∂¯ v1 v1 0.82 2 ∂¯ kτ ⇒ v1′2 = k + ∂x2 3 12 ∂x2 2 2 2 ¯1 v1 0.5 2 ∂¯ −0.5 2 ∂ u ′2 τ kτ ⇒ v2 = k − = 12 ∂x2 3 12 ∂x2 2 2 2 v1 0.16 2 ∂¯ v1 −0.16 2 ∂¯ τ kτ = ⇒ v3′2 = k − 12 ∂x2 3 12 ∂x2 k ∂¯ v1 = −cµ ε ∂x2 a11 = a22 a33 a12 0.82 2 τ 12 (14.5) We find that indeed the non-linear model gives anisotropic normal Reynolds stresses. In Eqs. 14.4 and 14.5 we have assumed that the only strain is ∂¯ v1 /∂x2 . When we discussed streamline curvature effects at p. 142 we found that it is important to investigate the effect of secondary strains such as ∂¯ v2 /∂x1 . Let’s write down Eq. 14.3 for the strain ∂¯ v2 /∂x1 a11 = a22 a33 a12 1 2 τ 12 ∂¯ v2 ∂x1 2 (c1 − 6c2 + c3 ) 2 ∂¯ v2 (c1 + 6c2 + c3 ) ∂x1 2 ∂¯ v2 1 = − τ2 (c1 + c3 ) 6 ∂x1 3 v2 1 3 ∂¯ (c5 + c6 + c7 ) =− τ 4 ∂x1 1 2 τ = 12 (14.6) 14. Non-linear Eddy-viscosity Models 149 Inserting the constants from [42] (see above) we obtain a11 a22 a33 2 v2 0.5 2 ∂¯ τ =− 12 ∂x1 2 v2 0.82 2 ∂¯ τ = 12 ∂x1 2 v2 0.16 2 ∂¯ τ =− , a12 = 0 12 ∂x1 (14.7) As can be seen the coefficient for a22 is larger than that in Eq. 14.5, and hence the model is slightly more sensitive to the secondary strain, ∂¯ v2 /∂x1 , than to the primary one, ∂¯ v1 /∂x2 . Thus, the non-linear models are able to account for streamline curvature, but due to the choice of constants c5 + c6 + c7 = 0 this effect is weak. 15. The V2F Model 150 15 The V2F Model In the V2F model of [38, 46, 47] two additional equations, apart from the k and εequations, are solved: the wall-normal stress v2′2 and a function f . This is a model which aims at improving modeling of wall effects on the turbulence. Walls affect the fluctuations in the wall-normal direction, v2′2 , in two ways. The wall damping of v2′2 is felt by the turbulence fairly far from the wall (x+ 2 . 200) through the pressure field (i.e. the pressure-strain term) whereas the viscous damping takes place within the viscous and buffer layer (x+ 2 . 10). In usual eddy-viscosity models both these effects are accounted for through damping functions. The damping of v2′2 is in the RSM accounted for through the modeled pressure-strain terms Φ22,1w and Φ22,2w (see Eqs. 11.93 and Eq. 11.94). They go to zero far away from the wall (x+ 2 & 10). In the V2F model the problem of accounting for the wall damping of v2′2 is simply resolved by solving its transport equation. The v2′2 equation in boundary-layer form reads (see Eq. 9.16 at p. 103) " # ∂ρ¯ v1 v2′2 ∂p′ ∂v2′2 ∂ρ¯ v2 v2′2 ∂ (µ + µt ) − 2v2′ + = − ρε22 (15.1) ∂x1 ∂x2 ∂x2 ∂x2 ∂x2 in which the diffusion term has been modeled with an eddy-viscosity assumption, see Eq. 11.45 at p. 122. Note that the production term P22 = 0 because in boundary-layer approximation v¯2 ≪ v¯1 and ∂/∂x1 ≪ ∂/∂x2 . The model for the dissipation term, ε22 , is taken as in RSM (see Eq. 11.47) εmodel = 22 v2′2 ε k Add and subtract εmodel on the right side of Eq. 15.1 yields 22 ∂ρ¯ v1 v2′2 ∂ρ¯ v2 v2′2 + = ∂x1 ∂x2 " # ∂p′ ∂v ′2 ∂ v ′2 v ′2 (µ + µt ) 2 − 2v2′ − ρε22 + ρ 2 ε − ρ 2 ε ∂x2 ∂x2 ∂x2 k k (15.2) In the V2F model P is now defined as 2 ∂p′ v ′2 − ε22 + 2 ε P = − v2′ ρ ∂x2 k (15.3) so that Eq. 15.2 can be written as " # ∂v2′2 ∂ρ¯ v2 v2′2 ∂ ∂ρ¯ v1 v2′2 v ′2 (µ + µt ) + ρP − ρ 2 ε + = ∂x1 ∂x2 ∂x2 ∂x2 k (15.4) P is the source term in the v2′2 -equation above, and it includes the velocity-pressure gradient term and the difference between the exact and the modeled dissipation. Note that this term is commonly split into a diffusion term and the pressure-strain term as v2′ ∂p′ ∂v ′ ∂v ′ p′ = 2 − p′ 2 ∂x2 ∂x2 ∂x2 15. The V2F Model 151 Physically, the main agent for generating wall-normal stress is indeed the pressurestrain term via re-distribution, see example in Section 11.12. A new variable f = P/k is defined and a relaxation equation is formulated for f as ! 2 ′2 Φ 1 2 v ∂ f 22 2 − − L2 2 − f = − ∂x2 ρk T k 3 ν 1/2 k T = max , CT ε ε ! 2 (15.5) Φ22 νt ∂¯ v1 C1 2 v2′2 + C2 = − ρk T 3 k k ∂x2 ( 1/4 ) ν3 k 3/2 , Cη L = CL max ε ε where Φ22 is the IP model of the pressure-strain term, see Eqs. 11.55 and 11.88, the first term being the slow term, and the second the rapid term. The constants are given the following values: cµ = 0.23, CT = 6, cε1 = 1.44, cε2 = 1.9, σk = 0.9, σε = 1.3, C1 = 1.3, C2 = 0.3, CL = 0.2, Cη = 90. The boundary condition for f is obtained from the v2′2 equation. Near the wall, the ′2 v2 equation reads v ′2 ∂ 2 v2′2 + fk − 2 ε (15.6) 0=ν 2 ∂x2 k The first and the last term behave as O(x22 ) as x2 → 0 because Taylor analysis gives v2′2 = O(x42 ), ε = O(x02 ) and k = O(x22 ), see Section 4 Introduction to turbulence Furthermore, ε = 2νk/x22 [48]; using this expression to replace k in Eq. 15.6 gives 0= ∂ 2 v2′2 f εx22 2v2′2 + − 2 ∂x2 2ν 2 x22 (15.7) Assuming that f and ε are constant very close to the wall, this equation turns into an ordinary second-order differential equation with the solution v2′2 = Ax22 + x4 B − εf 2 2 x2 20ν Since v2′2 = O(x42 ) as x2 → 0, both constants must be zero, i.e. A = B = 0, so we get f =− 20ν 2 v2′2 ε x42 (15.8) For more details, see [49]. Above we have derived the v2′2 equation in boundary layer form assuming that x2 is the wall-normal coordinate. In general, three-dimensional flow it reads ∂v 2 v2 ∂ ∂ρ¯ vj v 2 (µ + µt ) + ρf k − ρ ε = ∂xj ∂xj ∂xj k 2 2 ∂ f Φ22 1 v 2 L2 −f =− − − (15.9) ∂xj ∂xj ρk T k 3 Φ22 Pk C1 2 v 2 + C2 = − ρk T 3 k k S=1 L=0.4 is reduced as the wall is approached. then v 2 = v2′2 .S=1 L=0.S=1 L=0. 15. v 2 = v1′2 .12.04. reduced as the wall is approached. this means that the transport equation for v2′2 is turned into an equation for v1′2 .12 In the V2F model a transport equation for the normal stress normal to walls is solved for. the source term in the v2′2 -equation simplifies to Φ22 plus isotropic dissipation (see Eq. This is done automatically since in the general formulation in Eq.6 0. because far from the wall when ∂ 2 f /∂x22 ≃ 0. 15. However.1 where the equation L2 ∂2f −f +S =0 ∂x22 (15. i. 15.5 has the form it has? Far from the wall.5 1 0.2.9. ∂¯ v1 /∂x2 in the expression for Φ22 is replaced by P k . if a wall lies in the x2 − x3 plane. This is how the influence of the source term P in Eq.4 x2 0.4 we get " # ∂ρ¯ v1 v2′2 ∂v2′2 2 ∂ρ¯ v2 v2′2 ∂ (µ + µt ) + ρΦ22 − ρε + = ∂x1 ∂x2 ∂x2 ∂x2 3 (15.11) which is the usual form of the modeled v2′2 -equation with isotropic dissipation. Furthermore.15. the diffusion term ∂ 2 f /∂x22 makes f go from the value of its source term to its (negative) wall value (see Eq.8) over the lengthscale L.5 0 0 0. 15. If the wall lies in the x2 − x3 plane the largest velocity gradient will be ∂¯ v2 /∂x1 or ∂¯ v3 /∂x1 . .2. The influence of the lengthscale L is nicely illustrated: the larger L.e. 15. f approaches the value of the source term as x2 > L.12) has been solved with f = 0 at the wall and with different L and S.1: Illustration of Eq. 15. 15. S=0. for example. as required. 15. Near the wall. The V2F Model 152 2 1. Why does the right side of Eq. 15.08.10) (15. 15. f is. S=2 L=0.S=1 L=0.2 0. The behavior of the equation for f (Eq. so that Eq.1).8 1 Figure 15. This is what happens. As can be seen.5 f L=0.5) is illustrated in Fig. If the wall lies in the x1 − x3 plane. the further away from the wall does f go to its far-field value.2.5 can be written (T = k/ε) kf ≡ P → Φ22 + ε(v2′2 /k − 2/3) When this expression is inserted in Eq. namely f = 0. f equations coupled [49]. Cη L = CL max ε ε (15. In this model they solve for the ratio v2′2 /k instead of for v2′2 which gives a simpler wall boundary condition for f .13) The k and ε-equations are also solved (without damping functions).15. Modified V2F model 153 In the V2F model the turbulent viscosity is computed from νt = Cµ v2′2 T (15.16) Boundary conditions at the walls are k = 0.1. v2′2 = 0 ε = 2νk/x22 (15.1 Modified V2F model In [51] they proposed a modification of the V2F model allowing the simple explicit boundary condition f = 0 at walls. 15.15) (15. Note that the modified model is identical to the original model far from the wall. the boundary conditions are given again k = 0. v 2 = 0 ε = 2νk/x22 f∗ = 0 This modified model is numerically much more stable.6 ε ε ( 3 1/4 ) 3/2 ν k . For convenience. . An alternative is to use the ζ − f model [50] which is more stable. ε and v2′2 . One way to get around that problem is to solve both the k.14) 20ν 2 v2′2 f =− εx42 The boundary condition for f makes the equation system numerically unstable. They introduced a new variable f ∗ = f − 5εv 2 /k 2 and they neglected the term ∂2 −5L ∂xj ∂xj 2 εv 2 k2 The resulting v2′2 and f ∗ -equation read [51] ∂v 2 v2 ∂ ∂¯ vj v 2 (ν + νt ) + kf ∗ − 6 ε = ∂xj ∂xj ∂xj k ∂2f ∗ v2 1 Pk 2 −L2 (C1 − 6) − (C1 − 1) + C2 + f∗ = − ∂xj ∂xj T k 3 k ∂¯ v ∂¯ v ∂¯ v j i i + P k = νt ∂xj ∂xi ∂xj k ν 1/2 T = max . it must for consistency also be used for ε/k in Eq. 15. .√ 1/2 ε 6Cµ v 2 (¯ sij s¯ij ) # " (15.16 reduces to f = right side.2. − (C1 − 6)v − k 3 This modification ensures that v 2 ≤ 2k/3.2 Realizable V2F model The realizable condition for stagnation flow (see p. In the homogeneous region far away from the wall. 15. Then Eq.15) includes the modeled velocitypressure gradient term which is dampened near walls as f goes to zero.15. 15. ∂ 2 f /∂xj ∂xj → 0. and thus v2′2 should be smaller than or equal to 32 k. Since v 2 represents the wall-normal normal stress. i. whereas that for L is not.16 as an upper bound on the source term kf in the v 2 -equation. v 2 denotes the generic wall-normal stress. Furthermore. v2′2 ≤ v1′2 and v2′2 ≤ v3′2 . Thus it should be the smallest one.√ ε sij s¯ij )1/2 6Cµ v 2 (2¯ These realizable conditions have been further investigated by Sveningsson [49. For more details. This is not ensured in the V2F models presented above. if the limit is used in the f equation. A simple modification is to use the right side of Eq. Realizable V2F model 154 15. The source term kf in the v 2 -equation (Eq.15. see [56]. it should be the smallest normal stress.17) k 3/2 k 3/2 L = min . the Laplace term is not negligible. 15. it was found that it is important to impose the limitation on T in a consistent manner.6k T = min .e.52–55].3 To ensure that v 2 ≤ 2k/3 In the V2F model. and they read [51] " # k 0.18) (C1 − 1) + C2 P vsource = min kf. For instance. and as a consequence v 2 gets too large so that v 2 > 23 k. It turns out that in the region far away from the wall. the Laplace term is assumed to be negligible i. and it was found that the limitation on T is indeed important. 2k ε 2 k 2 (15.e. 15. i. Below the simple modification proposed by [56] is presented.e. 144) is used also for the V2F model. Let us express the lefthand side of the ω equation as a combination of the left-hand sides of the ε and the k .15) the pressure reaches its minimum and increases further downstream as the velocity decreases. 60]. Pressure contours. the pressure decreases because the velocity increases. In order to improve both the k − ε and the k − ω model. it is convenient to transform the k − ε model into a k − ω model using the relation ω = ε/(β ∗ k). see Eq. Starting from the leading edge. blue: low pressure 16 The SST Model The SST (Shear Stress Transport) model of [57] is an eddy-viscosity model which includes two main novelties: 1. 16.e. see Fig. At the crest (at x/c ≃ 0. A limitation of the shear stress in adverse pressure gradient regions. it was suggested in [57] to combine the two models. 2. Before doing this. which denotes the material derivative. This region is called the adverse pressure gradient (APG) region. The SST Model 155 Figure 16.1: Flow around an airfoil. Various k−ω models are presented in Section 4 Introduction to turbulence One example of adverse pressure gradient is the flow along the surface of an airfoil. The k − ω model is better than the k − ε model at predicting adverse pressure gradient flow and the standard k − ω model of [58] (see also [48]) does not use any damping functions.16. where β ∗ = cµ . Red: high pressure. dω/dt.23. low-Re number damping functions/terms). 2. It is combination of a k − ω model (in the inner boundary layer) and k − ε model (in the outer region of the boundary layer as well as outside of it).1. assuming steady flow. However. Consider the upper surface (suction side). The lefthand side of the ω equation will consist of the convection term. the disadvantage of the standard k − ω model is that it is dependent on the free-stream value of ω [59. The k − ε model has two main weaknesses: it over-predicts the shear stress in adverse pressure gradient flows because of too large length scale (due to too low dissipation) and it requires near-wall modification (i. one term from the ε equation 1 β∗k Pε (the first term at the right side in Eq.7) . DωT • Production term " ν ∂2ε νω ∂ 2 k + ∗ − 2 β k ∂xj k ∂x2j # (16. Pω 1 β∗k DεT ω − DkT k Destruction.1) Now we have transformed the left side of the ω equation. 16.2) and one from the k equation ω − Pk k (the second term at the right side in Eq.4) Viscous diffusion. 16. Ψω Turbulent diffusion.16. i.e.1) (16. 1 dε dω ε d ε d(1/k) = ∗ = + ∗ ∗ dt dt β k β k dt β dt ε dk 1 dε ω dk 1 dε − ∗ 2 = ∗ − = ∗ β k dt β k dt β k dt k dt (16.6) Pω = • Destruction term Ψω = 1 β∗k Ψε − • Viscous diffusion term νω ∂ 2 k ν ∂ 2 ωk νω ∂ 2 k ν ∂2ε − = − 2 2 ∗ β k ∂xj k ∂xj k ∂x2j k ∂x2j ν ∂ ∂k ω ∂2k ∂ω = ω −ν +k k ∂xj ∂xj ∂xj k ∂x2j " # ν ∂ω ∂k ∂2k ∂k ∂ω ∂2ω ω ∂2k = +ω 2 + +k 2 −ν k ∂xj ∂xj ∂xj ∂xj ∂xj ∂xj k ∂x2j ∂ω ∂ 2ν ∂ω ∂k ν + = k ∂xj ∂xj ∂xj ∂xj Dων = (16. the production of the ω equation will consist of two terms.1) (16. For example. The right side should be transformed in the same manner.3) In the same way we transform the entire right side inserting the modeled equations for k and ε so that 1 Dω ω ω 1 = ∗ Pε − P k − ∗ Ψε − Ψk + Dt β k k β k k Production. Dων 1 ε ω ω Pε − P k = Cε1 ∗ 2 P k − P k β∗k k β k k ω k = (Cε1 − 1) P k (16.5) ε2 ω ω Ψk = Cε2 − ε k k k = (Cε2 − 1)β ∗ ω 2 (16. The SST Model 156 equations by using the chain rule. β = (Cε2 − 1)β ∗ = 0.10) α = Cε1 − 1 = 0. F1 = tanh(ξ 4 ).King model [JK]) which is based on a transport equation of the main shear stress v1′ v2′ .0828.chalmers. we can considerably simplify the turbulence diffusion so that 2νt ∂k ∂ω ∂ νt ∂ω T (16. the viscous part of the cross-diffusion term (second line) is usually neglected (the viscous terms are negligible in the outer region). Still. When introducing this second modification. In the JK model.13) −v1′ v2′ = a1 k . for example. If we assume that σk = σε in the second and third term of the right-hand side.16. At p. the last term in the ω equation (second line in Eq. F1 = 1 in the near-wall region and F1 = 0 in the outer region. Functions of the form ) # " ( √ 4σ k 500ν k ω2 . hence it is multiplied by (1 − F1 ).tfd.3. This brings us to the second modification (see p.44.10) should only be active in the k − ε region. Since the standard k − ω model does not include any cross-diffusion term.075 and βk−ε = 0. 155). the predicted shear stress is too large. predicts adverse pressure gradient flows much better than the k − ω model. 16.8) ω 1 1 ∂νt ∂k ∂ νt ∂ω + − + k σε σk ∂xj ∂xj ∂xj σε ∂xj In the standard k − ε model we have σk = 1 and σε = 1. the v1′ v2′ transport equation is built on Bradshaw’s assumption [62] (16. The SST Model 157 The turbulent diffusion term is obtained as (the derivation is found in [61] which can be downloaded from http://www.12) where βk−ω = 0. (16. the author in [57] noted that a model (the Johnson . is computed as βSST = F1 βk−ω + (1 − F1 )βk−ε (16.se/˜lada) ω νt νt ∂ 2 k 2νt ∂k ∂ω + − + DωT = σε k ∂xj ∂xj k σε σk ∂x2j (16. ξ = min max β ∗ ωy y 2 ω CDω y 2 are used.0828 Since the k − ε model will be used for the outer part of the boundary layer.9) Dω = + σε k ∂xj ∂xj ∂xj σε ∂xj We can now finally write the ε equation formulated as an equation for ω νt ∂ω ω ∂ ∂ ν+ + α P k − βω 2 (¯ vj ω) = ∂xj ∂xj σε ∂xj k 2 νt ∂k ∂ω + ν+ k σε ∂xi ∂xi (16. The βcoefficient. In the SST model the coefficients are smoothly switched from k − ω values in the inner region of the boundary layer to k − ε values in the outer region. 155 it was mentioned that the k − ω model is better than the k − ε model in predicting adverse pressure-gradient flows because it predicts a smaller shear stress.11) . 13. in (a) the Boussinesq assumption together with Eq. 16. and hence (see Eq.15) which explains why the Boussinesq assumption over-predicts the shear stress and works 1/2 poorly in this type of flow (recall that according to experiments −v1′ v2′ ≃ cµ k). 16. 16.16a) (16.k−ω 158 = β ∗1/2 . Then we take the minimum of (a) and (b) so that 1/2 νt = cµ k 1/2 ¯ max(cµ ω.17 (which mimics the Johnson-King model). In boundary layer flow. F2 Ω) (16. because if P k < ε then Ω µ ω since ¯ 2 = ω P k < ωε = cµ ω 2 ¯ 2 = 1 νt Ω Ω νt k k (16. 16.16a. the Boussinesq assumption can be " 2 #1/2 k 1/2 ∂¯ v1 v1 cµ k 2 ∂¯ v1 cµ k 2 ∂¯ P 1/2 1/2 = = cµ k = c k µ ∂x2 ε ∂x2 ε2 ∂x2 ε νt . 16. To summarize the SST modification: 1/2 • the second part.k−ε (16. When the Boussinesq assumption is used in k − ω model for adverse pressure gradient flows. cµ k/Ω in Eq. Eq.13 were used and (b) is taken from the k − ω model.18) Hence.14) It is found from experiments that in boundary layers of adverse pressure gradient flows that the production is much larger than the dissipation (P k ≫ ε) and that −v1′ v2′ ≃ 1/2 cµ k. Eq. Eq. in regions where P k < ε.16b) ¯ is the absolute vorticity (in boundary layer flow Ω ¯ = ∂¯ where Ω v1 /∂x2 ). P k ≫ ε.14 in adverse pressure gradient flow. We have two expressions for the turbulent viscosity 1/2 −v1′ v2′ cµ k = ¯ Johnson-King ¯ Ω Ω 1/2 cµ k k k − ω model νt = = 1/2 ω cµ ω νt = (16. when Ω the Johnson .e. We want (a) to apply only in the boundary layer and hence we multiply it with a function F2 (similar to F1 ) which is 1 near walls and zero elsewhere. To reduce |v1′ v2′ | in Eq.King model. [57] proposed to re-define the turbulent eddy viscosity including the expression in Eq.e. 16. 16. 16. i. It can be seen that νt is ¯ < c1/2 reduced only in regions where P k > ε.17 returns to νt = k/ω as it should.17) ¯ is large).17 reduces νt according to When the production is large (i. The SST Model 1/2 where a1 = cµ written as −v1′ v2′ = k ω νt . It is important to ensure that this limitation is not active in usual boundary layer flows where P k ≃ ε.16.14) − v1′ v2′ 1/2 cµ k ≫1 (16. is active in APG flow . 16. The SST Model 159 • the first part, k/ω in Eq. 16.17 (which corresponds to the usual Boussinesq model), should be active in the remaining part of the flow. Equation 16.18 shows that (it is likely that) the second part is active only in APG regions and not elsewhere. Today, the SST model has been slightly further developed. Two modifications have ¯ in Eq. 16.17 been introduced [63]. The first modification is that the absolute vorticity Ω has been replaced by |¯ s| = (2¯ sij s¯ij )1/2 which comes from the production term using the Boussinesq assumption (see Eq. 11.37), i.e. ∂¯ vi ∂¯ vj ∂¯ vi ¯ ij ) = 2¯ |¯ s |2 = + = 2¯ sij (¯ sij + Ω sij s¯ij ∂xj ∂xi ∂xj (16.19) vi ∂¯ vj ¯ ij = 1 ∂¯ − Ω 2 ∂xj ∂xi ¯ ij = 0 because s¯ij is symmetric and Ω ¯ ij is anti-symmetric. Equation 16.17 where s¯ij Ω with |¯ s| limits νt in stagnation regions similar to Eq. 13.12. The second modification in the SST model is that the production term in the new SST model is limited by 10ε, i.e. Pk,new = min P k , 10ε (16.20) The final form of the SST model is given in Eq. 20.4 at p. 190. 17. Overview of RANS models 160 17 Overview of RANS models This section presents a short overview of the presented RANS models. First the models can be classified as models based on eddy viscosity (i.e. turbulent viscosity) or models in which equations are solved (algebraic or differential) to obtain the Reynolds stress tensor, vi′ vj′ . Eddy-viscosity models, which are based on the Boussinesq assumption, see Eq. 11.32, are • standard k − ε (see Section 11.8) and k − ω models Eddyviscosity models • combination of k − ε and k − ω models such as the SST model, see Section 16. There are more elaborate eddy-viscosity models such as • non-linear models, see Section 14 non-linear models Models not based on the eddy-viscosity are the Reynolds stress models such as • the Reynolds stress transport model (RSM or RSTM), in which transport equations are solved for vi′ vj′ , see Eq. 11.99. RSM • the Algebraic Reynolds stress model, ASM, in which algebraic equations are solved for vi′ vj′ , see Eq. 11.102. ASM explicit ASM • explicit Algebraic Reynolds stress models, in which explicit algebraic equations are solved for vi′ vj′ , see Section 11.11. Finally, there is a class of models which are somewhere in between two-equation eddyviscosity models and Reynolds stress models, and that is • V2F models, see Section 15 The V2F model is an eddy-viscosity model and the model is based on four transport equations. V2F 18. Large Eddy Simulations 161 0.5 0.45 v1 , v¯1 0.4 0.35 0.3 0.25 0.2 no filter one filter 0.15 0.1 2 2.5 3 x1 3.5 4 Figure 18.1: Filtering the velocity. 18 Large Eddy Simulations 18.1 Time averaging and filtering In CFD we time average our equations to get the equations in steady form. This is called Reynolds time averaging: Z T 1 Φ(t)dt, Φ = hΦi + Φ′ (18.1) hΦi = 2T −T (note that we use the notation h.i for time averaging). In LES we filter (volume average) the equations. In 1D we get (see Fig. 18.1) Z x+0.5∆x 1 ¯ Φ(x, t) = Φ(ξ, t)dξ ∆x x−0.5∆x ¯ + Φ′′ Φ=Φ Since in LES we do not average in time, the filtered variables are functions of space and time. The equations for the filtered variables have the same form as Navier-Stokes, i.e. 1 ∂ p¯ ∂ 2 v¯i ∂τij ∂ ∂¯ vi (¯ vi v¯j ) = − +ν − + ∂t ∂xj ρ ∂xi ∂xj ∂xj ∂xj ∂ v¯i =0 ∂xi (18.2) where the subgrid stresses are given by τij = vi vj − v¯i v¯j (18.3) It should be noted that it is formally incorrect to denote τij as a stress since the density is omitted (see, e.g., the text below Eq. 6.9 and Eq. 11.3). Contrary to Reynolds time averaging where hvi′ i = 0, we have here vi′′ 6= 0 v¯i 6= v¯i 18.2. Differences between time-averaging (RANS) and space filtering (LES) 162 This is true for box filters. Note that for the spectral cut-off filter v¯i = v¯i , see p. 164. However, in finite volume methods, box filters are always used. In this course we use box filters, if not otherwise stated. Let’s look at the filtering of Eq. 18.2 in more detail. The pressure gradient term, for example, reads Z 1 ∂p ∂p = dV ∂xi V V ∂xi Now we want to move the derivative out of the integral. When is that allowed? The answer is “if the integration region is not a function of xi ”, i.e. if V is constant. In finite volume methods, the filtering volume, V , is (almost always) identical to the control volume. In general, the size of the control volume varies in space. Fortunately, it can be shown that if V is a function of xi , the error we do when moving the derivative out of the integral is proportional to (∆x)2 [64], i.e. it is an error of second order. Since this is the order of accuracy of our finite volume method anyway, we can accept this error. Now let’s move the derivative out of the integral, i.e. Z ∂ 1 ∂ p¯ ∂p = + O (∆x)2 pdV + O (∆x)2 = ∂xi ∂xi V V ∂xi All linear terms are treated in the same way. Now we take a look at the non-linear term in Eq. 18.2, i.e. the convective term. First we filter the term and move the derivative out of the integral, i.e. Z ∂vi vj ∂ 1 ∂ = (vi vj ) + O (∆x)2 vi vj dV + O (∆x)2 = ∂xj ∂xj V V ∂xj There is still a problem with the formulation of this term: it includes an integral of a product, i.e. vi vj ; we want it to appear like a product of integrals, i.e. v¯i v¯j . To achieve this we simple add the term we want (¯ vi v¯j ) and subtract the one we don’t want ( vi vj ) on both the right and left side. This is how we end up with the convective term and the SGS term in Eq. 18.2. The filtering in Eq. 18.2 is usually achieved through the finite-volume discretization. This means that no additional filtering is done. This is called implicit filtering. Hence, the discretized momentum equations in RANS and LES (Eqs. 6.5 and 18.2) are identical. The only difference is the magnitude of the modeled Reynolds stresses, which are much larger in RANS than in LES. 18.2 Differences between time-averaging (RANS) and space filtering (LES) In RANS, if a variable is time averaged twice (hhvii), it is the same as time averaging once (hvi). This is because hvi is not dependent on time. From Eq. 18.1 we get hhvii = 1 2T Z T −T hvidt = 1 hvi2T = hvi 2T This is obvious if the flow is steady, i.e. ∂hvi/∂t = 0. If the flow is unsteady, we must assume a separation in time scales so that the variation of hvi during the time interval T is negligible, i.e. ∂/∂t ≪ 1/T . In practice this requirement is rarely satisfied. In LES, v¯ 6= v¯ (and since v = v¯ + v ′′ we get v ′′ 6= 0). 18.3. Resolved & SGS scales 163 ∆x x I −1 A x I B I +1 x Figure 18.2: Box filter illustrated for a control volume. Let’s filter v¯I once more (filter size ∆x, see Fig. 18.2. For simplicity we do it in 1D. (Note that subscript I denotes node number.) ! Z 0 Z ∆x/2 Z ∆x/2 1 1 v¯I = v¯(ξ)dξ + v¯(ξ)dξ = v¯(ξ)dξ = ∆x −∆x/2 ∆x −∆x/2 0 ∆x ∆x 1 v¯A + v¯B . = ∆x 2 2 The trapezoidal rule, which is second-order accurate, was used to estimate the integrals. v¯ at locations A and B (see Fig. 18.2) is estimated by linear interpolation, which gives 1 3 3 1 1 v¯I−1 + v¯I + v¯I + v¯I+1 v¯I = 2 4 4 4 4 (18.4) 1 = (¯ vI−1 + 6¯ vI + v¯I+1 ) 6= v¯I 8 18.3 Resolved & SGS scales The basic idea in LES is to resolve (large) grid scales (GS), and to model (small) subgrid-scales (SGS). Even if LES is always made unsteady, we are usually interested in the time-averaged results. This means that we time-average all quantities for post-processing, e.g. h¯ vi i where the brackets denote time-averaging. From the LES equations, Eq. 18.2, we get the resolved (GS) velocities, v¯i . Then we can compute the resolved fluctuations as v¯i′ = v¯i − h¯ vi i. The limit (cut-off) between GS and SGS is supposed to take place in the inertial subrange (II), see Fig. 18.3. I: large, energy-containing scales II: inertial subrange (Kolmogorov −5/3-range) III: dissipation subrange 18.4. The box-filter and the cut-off filter 164 cut-off E(κ) I II resolved scales III SGS κ Figure 18.3: Spectrum of velocity. 18.4 The box-filter and the cut-off filter The filtering is formally defined as (1D) Z ∞ v¯(x) = GB (r)v(x − r)dr −∞ 1/∆, if r ≤ ∆/2 GB (r) = 0, if r > ∆ Z ∞ GB (r)dr = 1 (18.5) −∞ It is often convenient to study the filtering process in the spectral space. The filter in spectral space is particular simple: we simply set the contribution from wavenumbers larger than cut-off to zero. Hence the cut-off filter filters out all scales with wavenumber larger than the cut-off wavenumber κc = π/∆. It is defined as 1/∆ if κ ≤ κc ˆ C (κ) = G (18.6) 0 otherwise The Fourier transform is defined as (see Section D) Z ∞ 1 vˆ(κ) = v(r) exp(−ıκr)dr 2π 0 and its inverse v(r) = Z 0 (18.7) ∞ vˆ(κ) exp(ıκr)dκ (18.8) √ where κ denotes the wavenumber and ı = −1. Note that it is physically meaningful to use Fourier transforms only in a homogeneous coordinate direction; in nonhomogeneous directions the Fourier coefficients – which are not a function of space – have no meaning. Using the convolution theorem (saying that the integrated product of two functions is equal to the product of their Fourier transforms) the filtering in 18.8 sin(κ1 x1 )] (18.2 0 0. what is the cut-off wavenumber? The wave shown in Fig.c L = κ1.4 v1′ 0.25 [1 + 0.3. and let it be a function of x1 .e. see Eq. 18.18.9 and 18.4: Physical and wavenumber space. Sinus curves with different wavenumbers illustrated in physical space.rms . Let’s choose the fluctuating velocity in the x1 direction. its RMS. contrary to the box-filter (see Eq. Furthermore implicit filtering is employed.5 1 1. in other words.4 0.28 and Section E) provided that the coordinate direction is homogeneous. We require it to be homogeneous.4). sin(κ1.c 2∆x1 ) = sin(2π). Eq. 18.5 is conveniently written = = Z Z ∞ 0 ∞ Z0 Z 0 Z ∞ b vˆ(κ) = v¯(κ) = v¯(η) exp(−ıκη)dη 0 Z ∞Z ∞ = exp(−ıκη)GC (ρ)v(η − ρ)dρdη 0 ∞ 0 (18. Highest resolved wavenumbers 165 0.11) and it covers two cells (∆x1 /L = 0. In finite volume methods box filtering is always used.4a reads v1′ = 0.10) using Eqs. v1.5 x1 /L x1 /L (b) One sinus period covering four cells (a) One sinus period covering two cells Figure 18. 18. 18.5 0 0.2 0 node 1 2 0 3 −0.e. If we define this as the cut-off wavenumber we get κ1.14).9) ˆv = G ˆ C vˆ = vˆ ˆ C Gˆ vˆ = G (18.5.2 v1′ 0. i.5 Highest resolved wavenumbers Any function can be expressed as a Fourier series such as Eq.5 1 1.c 2∆x1 = 2π (i. Thus.12) . for the cut-off filter it is vice versa.e. does not vary with x1 . i. what is the highest wavenumber that is resolved? Or.2 1 2 3 4 5 −0. Now we ask the question: on a given grid. nothing happens when we filter twice in spectral space. 18.c = 2π/(2∆x1 ) = π/∆x1 (18. This means that the filtering is the same as the discretization (=integration over the control volume which is equal to the filter volume. 18. D.9) exp(−ıκρ) exp(−ıκ(η − ρ))GC (ρ)v(η − ρ)dρdη ∞ ˆ C (κ)ˆ exp(−ıκρ) exp(−ıκξ)GC (ρ)v(ξ)dξdρ = G v (κ) 0 If we filter twice with the cut-off filter we get (see Eq. v1′ .6. 18.8 (see Section 5. Eq. The box filter is sharp in physical space but not in wavenumber space. recall that 2π is one period) so that κ1.5). The simplest model is the Smagorinsky model [65]: 1 ∂¯ vi ∂¯ vj τij − δij τkk = −νsgs = −2νsgs s¯ij + 3 ∂xj ∂xi (18. see Eq. If we require that the highest resolved wavenumber should be covered by four cells (∆x1 /L = 0.e.7 Smagorinsky model vs.c = 2π/(4∆x1 ) = π/(2∆x1 ). Near the wall.065 [66] to CS = 0. Disadvantage of Smagorinsky model: the “constant” CS is not constant. 18.4b.6. A damping function fµ is added to ensure this fµ = 1 − exp(−x+ 2 /26) (18.14) The scalar |¯ s| is the norm (i. but it is flow-dependent. 69] . It is found to vary in the range from CS = 0.25 [67].15) A more convenient way to dampen the SGS viscosity near the wall is simply to use the RANS length scale as an upper limit.13) 2 2p s| νsgs = (CS ∆) 2¯ sij s¯ij ≡ (CS ∆) |¯ and the filter-width is taken as the local grid size 1/3 ∆ = (∆VIJK ) (18.19. then the cut-off wavenumber is given by κ1. the SGS viscosity becomes quite large since the velocity gradient is very large at the wall. Subgrid model 166 It is of course questionable if v1′ in Fig.18. n o 1/3 ∆ = min (∆VIJK ) . mixing-length model The eddy viscosity according to the mixing length theory reads in boundary-layer flow [68.e. However this is the usual definition of the cut-off wavenumber. because the SGS turbulent fluctuations near a wall go to zero.16) where n is the distance to the nearest wall. as in Fig. the “length”) of ∂¯ vi /∂xj + ∂¯ vj /∂xi in the Boussinesq assumption. However.6 Subgrid model We need a subgrid model to model the turbulent scales which cannot be resolved by the grid and the discretization scheme. 16. 18.25). 18. 18. i. so must the SGS viscosity. κn (18.4a really is resolved since the sinus wave covers only two cells. . . v1 . 2 . ∂¯ . νt = ℓ .. ∂x2 . Generalized to three dimensions. νt = ℓ ∂xj ∂xi ∂xj In the Smagorinsky model the SGS turbulent length scale corresponds to ℓ = CS ∆ so that νsgs = (CS ∆)2 |¯ s| which is the same as Eq.13 . we have 1/2 ∂¯ vj ∂¯ vi ∂¯ vi 2 + = ℓ2 (2¯ sij s¯ij )1/2 ≡ ℓ2 |¯ s|. 18. see Eq.9 SGS kinetic energy The SGS kinetic energy ksgs can be estimated from the Kolmogorov −5/3 law.5. transfer of energy from kinetic energy to internal energy which corresponds to an increase in temperature) takes place at all wave numbers.e. 18.8 Energy path The path of kinetic energy is illustrated in Fig. It should be mentioned that this process is an idealized one. κc ) gives the SGS kinetic energy Z ∞ E(κ)dκ ksgs = κc The Kolmogorov −5/3 law now gives Z ∞ CK κ−5/3 ε2/3 dκ ksgs = κc (18. 18.17) from the resolved turbulence. The SGS kinetic energy is then transferred to higher wave-numbers via the cascade effect and the kinetic energy is finally dissipated (ε=physical dissipation) in the dissipation range. SGS kinetic energy is dissipated εsgs = −τij s¯ij = 2νsgs s¯ij s¯ij (18.5: Energy spectrum. and the dissipation increases for increasing wave number. The total turbulent kinetic energy is obtained from the energy spectrum as Z ∞ E(κ)dκ k= 0 Changing the lower integration limit to wavenumbers larger than cut-off (i.18.18. but in reality dissipation (i.e. 18. At cut-off. Energy path E(κ) 167 PK in e rti al εs gs E(κ) ∝ κ ra = ng ε P ks −5/3 e gs dissipating range κc κ Figure 18. This is a good approximation.8. We assume that ALL dissipation takes place in the dissipation range. 8.18) . This energy is transferred to the SGS scales and act as production term (Pksgs ) in the ksgs equation. 5 (Note that for these high wavenumbers.g.10 LES vs. turbulent structures o Transition • In RANS all turbulent scales are modeled ⇒ inaccurate • In LES only small. the Kolmogorov spectrum ought to be replaced by the Kolmogorov-Pau spectrum in which an exponential decaying function is added for high wavenumbers [68. the wake often includes large.19) In the same way as ksgs can be computed from Eq. is obtained from Z κc E(κ)dκ kres = 0 18.10. Carrying out the integration and replacing κc with π/∆ we get ksgs 3 = CK 2 ∆ε π 2/3 (18. where CK = 1. Chapter 3]). unsteady. . the reason is that in LES large.6: Energy spectrum with grid and test filter. LES vs.18. flow around a car). isotropic turbulent scales are modeled ⇒ accurate • LES is very much more expensive than RANS. Examples are: o Flows with large separation o Bluff-body flows (e. RANS 168 cut-off. kres . turbulent scales are resolved. the resolved turbulent kinetic energy. RANS LES can handle many flows which RANS (Reynolds Averaged Navier Stokes) cannot.18. 18. grid filter test filter E(κ) SGS grid scales I SGS test filter scales II resolved on test filter III resolved on grid filter κ z{ κ = π/ ∆ κc = π/∆ Figure 18. If we apply two filters to Navier-Stokes [grid filter and a second. For exz { z{ ample. 18. )] where ∆ = 2∆ we get z{ z{ z{ z{ z{ 1 ∂ p¯ ∂ v¯ i ∂ 2 v¯ i ∂Tij ∂ +ν − + v¯ i v¯ j = − ∂t ∂xj ρ ∂xi ∂xj ∂xj ∂xj where the subgrid stresses on the test level now are given by z { z{ z{ Tij = vi vj − v¯ i v¯ j z{ z{ z{ z{ z{ z{ 1 ∂ p¯ ∂ v¯ i ∂ 2 v¯ i ∂ τ ij ∂ +ν − + v¯ i v¯ j = − ∂t ∂xj ρ ∂xi ∂xj ∂xj ∂xj z { z{ z{ ∂ − v¯i v¯j − v¯ i v¯ j ∂xj Identification of Eqs. the 1D example in Fig. 18. see Fig. 18. ∆ = 2∆.e.20 and 18.20) (18.24) The trace of this relation reads z{ Lii ≡ Tii − τ ii With this expression we can re-formulate Eq.7 v¯ is computed as (∆x = 2∆x) ! Z P Z E Z E z{ 1 1 v¯ = v¯dx + v¯dx v¯dx = 2∆x W 2∆x W P 1 1 v¯W + v¯P v¯P + v¯E (18. The test-filtered variables are computed by integration over the test filter.21) (18. The dynamic model 169 18.23) (18.11 The dynamic model In this model of [70] the constant C is not arbitrarily chosen (or optimized). but it is computed. 18.11. coarser filter (test z{ filter.12 The test filter z{ The test filter is twice the size of the grid filter.22 gives z { z{ z{ z{ v¯i v¯j − v¯ i v¯ j + τ ij = Tij The dynamic Leonard stresses are now defined as z { z{ z{ z{ Lij ≡ v¯i v¯j − v¯ i v¯ j = Tij − τ ij (18. 18. i.26) = (¯ vw ∆x + v¯e ∆x) = + 2∆x 2 2 2 1 vW + 2¯ vP + v¯E ) = (¯ 4 . denoted by z{ .22) (18.24 as z{ 1 1 1 z{ Lij − δij Lkk = Tij − δij Tkk − τ ij − δij τ kk 3 3 3 (18.18.6.25) In the energy spectrum. the test filter is located at lower wave number than the grid filter. J−1/2. J.K−1/2 (18.K+1/2 +¯ vI−1/2. J. filtering at the test level is carried out in the same way by integrating over the test cell assuming linear variation of the variables [71].8) z{ 1 v¯ I. I. K Figure 18.J+1/2.27) +¯ vI−1/2.J+1/2.K+1/2 + v¯I+1/2.K−1/2 + v¯I+1/2. J + 1. K I − 1.8: A 2D test filter control volume. in the z{ range between ∆ and ∆. test and intermediate level The stresses on the grid level. J − 1.K−1/2 + v¯I+1/2. test level and intermediate level (dynamic Leonard stresses) have the form τij = vi vj − v¯i v¯j stresses with ℓ < ∆ z{ z { z{ z{ Tij = vi vj − v¯ i v¯ j stresses with ℓ < ∆ z{ z{ Lij = Tij − τ ij stresses with ∆ < ℓ < ∆ Thus the dynamic Leonard stresses represent the stresses with lengthscale. K I + 1.K = (¯ vI−1/2. ℓ.K+1/2 + v¯I+1/2. Stresses on grid.13. .e.J−1/2.K+1/2 ) 18. K I.J.7: Control volume for grid and test filter. (see Fig.K−1/2 8 +¯ vI−1/2. J.J−1/2. K x2 x1 I.J+1/2. For 3D.18. i.J−1/2.13 Stresses on grid. test and intermediate level 170 test filter grid filter E P W ∆x ∆x Figure 18.J+1/2. 18. 18. In real 3D flows. Usually local averaging and clipping (i. and it has been found necessary to average C in homogeneous direction(s).28) τij − δij τkk = −2C∆2 |¯ 3 where z{2 z{ z{ 1 Tij − δij Tkk = −2C ∆ | s¯ | s¯ ij 3 (18.28 is not squared (cf. C = C(xi . 18.e.31a gives 1 ∂Q (18.31a) (18. | s¯ | = 2 s¯ ij s¯ ij 2 ∂xj ∂xi Note that C in Eq. Let us define the error as the difference between the left-hand side and the right-hand side of Eq.18. Eq. Lilly [72] suggested to satisfy Eq. test and intermediate level 171 Assume now that the same functional form for the subgrid stresses that is used at the grid level (τij ) also can be used at the test filter level (Tij ).31 is re-written so that Lij Mij (18. 179). Carrying out the derivation of 18. Equation 18. 18.e.32) = 4Mij Lij − δij Lkk + 2CMij = 0 ∂C 3 Since ∂ 2 Q/∂C 2 = 8Mij Mij > 0 it is a minimum.30) Lij − δij Lkk = −2C ∆ | s¯ | s¯ ij − ∆ |¯ s|¯ sij 3 Note that the “constant” C really is a function of both space and time.166). substituting this equation and Eq.25 gives ! z { z{2 z{ z{ 1 2 (18. i.30 raised to the power of two. . C must be clipped to ensure that the total viscosity stays positive (ν + νsgs ≥ 0). has a minimum (or maximum) when ∂Q/∂C = 0. there is no homogeneous direction.13 at p. 18. Furthermore.29 into Eq.30 in a least-square sense. Use of one-equation models solve these numerical problems (see p. and we have five (¯ sij is symmetric and traceless) equations for C. Applying the test filter to Eq. requiring that C stays within pre-defined limits) of the dynamic coefficient is used.29) z{ z{ z{ z{ z{ 1/2 1 ∂ v¯ i ∂ v¯ j z{ s¯ ij = + . 18. 18. C should be compared with CS2 .e.28 (assuming that C varies slowly). This causes numerical problems. the Smagorinsky model. Hence.13. t). If we use the Smagorinsky model we get 1 s|¯ sij (18.30 is a tensor equation.33) C=− 2Mij Mij It turns out that the dynamic coefficient C fluctuates wildly both in space and time. 2 1 Lij − δij Lkk + 2CMij 3 ! z { z{2 z{ z{ 2 Mij = ∆ | s¯ | s¯ ij − ∆ |¯ s|¯ sij Q= (18. i. Q. Stresses on grid. 18. Equation 18.31b) The error. 18. The total diffusion term will have the form ∂ ∂¯ v diffusion term = (ν + νsgs + νnum ) (18.tfd. Indeed. It can be shown using Neumann stability analysis that all upwind schemes are dissipative (see Further reading at http://www. The first-derivative in the convective term is estimated by first-order upwind differencing as (finite difference.chalmers. νnum = 0. see Eq.35.14.9) v¯I − v¯I−1 ∂¯ v = v¯I + O (∆x) (18.35) O(∆x) where the second term on the right side corresponds to the error term in Eq.e. 18. 18. Eq. MILES [73]). because this class of schemes does not give any numerical dissipation. All upwind schemes give numerical dissipation in addition to the modeled SGS dissipation. to dampen) resolved turbulent fluctuations. This means that the total dissipation due to SGS viscosity and numerical viscosity is (cf. This is the reason why in LES we should use a central differencing scheme. the second term on the right-hand side will act as an additional (numerical) diffusion term. Numerical dissipation 172 v¯I I −1 I +1 I Figure 18. 18. 18. here we focus on ensuring proper dissipation through an SGS model rather than via upwind differencing. Taylor expansion gives v¯I−1 = v¯I − ∆x so that ∂¯ v 1 ∂ 2 v¯ + (∆x)2 2 + O (∆x)3 ∂x 2 ∂x v¯I − v¯I−1 ∂¯ v 1 ∂ 2 v¯ = − ∆x 2 + O (∆x)2 ∆x ∂x 2 ∂x Insert this into Eq.9: Numerical dissipation.34. 18. When this expression is inserted into the LES momentum equations.17) εsgs+num = 2(νsgs + νnum )¯ sij s¯ij . there are LES-methods in which upwind schemes are used to create dissipation and where no SGS model is used at all (e.5|¯ v|∆x.36) ∂x ∂x where the additional numerical viscosity. The SGS model is – hopefully – designed to give a proper amount of dissipation.14 Numerical dissipation The main function of an SGS model is to dissipate (i. see Fig.34) v¯ ∂x exact ∆x where we have assumed v¯I > 0.se/˜lada/comp turb model/).g.34 ∂¯ v ∂¯ v 1 ∂ 2 v¯ v¯ vO (∆x)2 = v¯ − ∆x¯ v 2 +¯ ∂x exact ∂x 2 ∂x (18. However.18. Below it is shown that first-order upwind schemes are dissipative. see [22]. if the cross term is neglected. i. Scale-similarity Models 173 For more details on derivation of equations transport equations of turbulent kinetic energies.18. and the sum of the cross term Cij and the Reynolds term is modeled as [74. It was found that this model was not sufficiently dissipative. but only the sum Lij + Cij is (see Appendix I). and the last term is denoted the Reynolds SGS stress.38) M and Rij = 0 (superscript M denotes Modeled). Note that the Leonard stresses.37 it seems natural to assume that the cross term is responsible for the interaction between resolved scales (¯ vi ) and modeled scales (vi′′ ).e. Lij . κc . .15 Scale-similarity Models In the models presented in the previous sections (the Smagorinsky and the dynamic models) the total SGS stress τij = vi vj − v¯i v¯j was modeled with an eddy-viscosity hypothesis. 18. 18. the term in square brackets is denoted cross terms.17 Redefined terms in the Bardina Model In [75] it was found that the Leonard term Lij and the cross term Cij are not Galilean invariant by themselves.39) 18. In scale-similarity models the total stress is split up as τij = vi vj − v¯i v¯j = (¯ vj + vj′′ ) − v¯i v¯j vi + vi′′ )(¯ = v¯i v¯j + v¯i vj′′ + v¯j vi′′ + vi′′ vj′′ − v¯i v¯j h i vi v¯j − v¯i v¯j ) + v¯i vj′′ + v¯j vi′′ + vi′′ vj′′ = (¯ where the term in brackets is denoted the Leonard stresses. are computable. (smaller than ∆) are similar to the ones just below κc (larger than ∆). Thus τij = Lij + Cij + Rij Lij = v¯i v¯j − v¯i v¯j Cij = v¯i vj′′ + v¯j vi′′ (18.37) Rij = vi′′ vj′′ . As a consequence. because then the modeled momentum equations do not satisfy Galilean invariance. since Cij includes both scales. 18. 75] M Cij = cr (¯ vi v¯j − v¯i v¯j ) (18.15. Looking at Eq. they are exact and don’t need to be modeled. hence the word ”scale-similar”. and thus a Smagorinsky model was added M Cij = cr (¯ vi v¯j − v¯i v¯j ) M Rij = −2CS2 ∆2 |¯ s|¯ sij (18. the Leonard stresses must not be computed explicitly. In scale-similarity models the main idea is that the turbulent scales just above cutoff wavenumber.16 The Bardina Model In the Bardina model the Leonard stresses Lij are computed explicitly. these models are called mixed models. [77] presents and evaluates a dissipative scale-similarity model. An eddy-viscosity model has to be added to make the model sufficiently dissipative. The stresses in the Bardina model can be redefined to make them Galilean invariant for any value cr A modified Leonard stress tensor Lm ij is defined as [76] m m τijm = τij = Cij + Lm ij + Rij Lm vi v¯j − v¯i v¯j ) ij = cr (¯ m Cij =0 (18. and the total SGS stress τij is modeled as τij = v¯i v¯j − v¯i v¯j − 2(CS ∆)2 |¯ s|¯ sij (18. 174 E(κ) 10 0 ε+ SGS −10 ε− SGS −20 −30 0 50 100 150 x+ 2 κc κ Figure 18. M Lm ij = Lij + Cij In order to make the model sufficiently dissipative a Smagorinsky model is added. +: −εSGS .40) m Rij = Rij = vi′′ vj′′ Note that the modified Leonard stresses is the same as the “unmodified” one plus the modeled cross term Cij in the Bardina model with cr = 1 (right-hand side of Eq. Reτ = 500. Above it was mentioned that when the first scale-similarity model was proposed it was found that it is not sufficiently dissipative [74].10: Dissipation terms and production term from DNS data. 18.e. 1 m∗ ∗ ∗ vi + Vi )(¯ vj + Vj ) − (¯ vi + Vi ) (¯ vj + Vj ) L = v¯i∗ v¯j∗ − v¯i v¯j = (¯ cr ij = v¯i v¯j + v¯i Vj + v¯j Vi − v¯i v¯j − v¯i Vj − Vi v¯j 1 = v¯i v¯j − v¯i v¯j = Lm cr ij (18. : −ε+ : −ε− SGS .18. The filtered Navier-Stokes read d¯ vi ∂ 2 v¯i ∂τik 1 ∂ p¯ =ν − + dt ρ ∂xi ∂xk ∂xk ∂xk (18. Ref.41) Below we verify that the modified Leonard stress is Galilean invariant. i.38). 963 mesh data filtered onto a 483 mesh.42) 18.18. A dissipative scale-similarity model. SGS .43) .18 A dissipative scale-similarity model. The SGS stress tensor is given by τik = vi vk − v¯i v¯k . the aim is to dampen turbulent resolved fluctuations.45. SGS dissipation are defined as the sum of all instants when εSGS is positive and negative. turbulent kinetic energy. Let us take a closer look at the equation for the resolved. max(εSGS . it is not sufficiently dissipative. It is the diffusion term in this equation which appears in the first term on the right side (first line) in Eq.e. This could be achieved by setting ∂τik /∂xk = 0 when its sign is different from that of v¯i′ (see the last term in Eq. Eq. the scale-similarity model is slightly dissipative (i. 18. In stability analysis the concern is to dampen numerical oscillations. 18. To ensure that εSGS > 0.45). εSGS > 0) . The second derivative on the right side is usually called a diffusion term because it acts like a diffusion transport term.45. that we want to dissipate. It is shown in Neumann analysis that the diffusion term in the Navier-Stokes equations is dissipative. but the forward and back scatter dissipation are both much larger than εSGS .e. respectively. The last term. v¯′ . i. in connection with SGS models. A dissipative scale-similarity model. it acts as a dissipation term). it dampens numerical oscillations. for details. see [77]. when it is negative.18. εSGS .18. back scattering occurs. and back scatter. 18.45. 18. Figure 18. . i. (18. respectively. Usually we do not. i. This would work if we were solving for k. the two last terms on the second line in Eq. we set −∂τik /∂xk to zero when its sign is different from that of the viscous diffusion term (cf. for example.e. εSGS in Eq. is the viscous dissipation term which is always positive. v¯i′ . εSGS . When analyzing the stability properties of discretized equations to an imposed disturbance. The forward scatter. Chapter 8 in [78]).e. forward scattering takes place (i.44) When it is modeled with the standard scale-similarity model. and the equations that we do solve (the filtered Navier-Stokes equations) are not directly affected by the dissipation term. 18. which reads 2 ′ dk vk′ v¯i′ v¯i′ i p′ v¯i′ i 1 ∂h¯ ∂h¯ vi i ∂h¯ ∂ v¯i ′ + + =ν v¯i − + h¯ vk′ v¯i′ i dt ∂xk ∂xi 2 ∂xk ∂xk ∂xk 2 ′ ∂τik ∂ v¯i ′ ∂τik ′ ∂τik ′ v¯i = ν − v¯ − v¯ = (18. Instead we have to modify the SGS stress tensor as it appears in the filtered NavierStokes equations. using Neumann analysis (see.10 presents SGS dissipation. As can be seen. This is achieved by defining a sign function. 175 where d/dt and τik denote the material derivative and the SGS stress tensor. ε.e. 18. computed from filtered − DNS data.45) ∂xk ∂xk ∂xk ∂xk i ∂xk i 2 ′ ′ ∂ k ∂τik ′ vi ∂¯ vi ∂¯ ν − −ν v¯ ∂xk ∂xk ∂xk ∂xk ∂xk i ε εSGS The first term on the last line is the viscous diffusion term and the second term. k = hvi′ vi′ i/2. is a source term arising from the SGS stress tensor. When it is positive. One way to make the SGS stress tensor strictly dissipative is to set the back scatter to zero. 0).43. ε+ SGS . since it is the resolved turbulent fluctuations. this term is referred to as a dissipation term. we must consider the filtered Navier-Stokes equations for the fluctuating velocity. However. εSGS .45). k in Eq. which can be positive or negative. j−1 ) (note that j denotes node number and i is a tensor v¯f.i = 0. v¯i ∂xi n+1/2 ∂ p¯ n+1/2 + α∆t ⇒ v¯i∗ = v¯i ∂xi n+1/2 Take the divergence of Eq.47b. central differencing in space. Forcing 176 18.e. In hybrid LES-RANS. .i = v¯f. viscous and SGS terms. 18. Solve the discretized filtered Navier-Stokes equation.i − α∆t ∂ p¯n+1 ∂xi (18. The face velocities are corrected as n+1 ∗ v¯f.47a) (18.46) ∂ p¯n+1/2 ∂ p¯n −α∆t − (1 − α)∆t ∂xi ∂xi where H includes convective. This new approach is the focus of [79]. Create an intermediate velocity field v¯i∗ from Eq. v¯2 and v¯3 . In the 3D MG we use a plane-by-plane 2D MG. the stresses can then be used as forcing at the interface between URANS and LES.18.47a. 18. ∂ p¯n n+1/2 − (1 − α)∆t v¯i∗ = v¯in + ∆tH v¯n .47b and require that ∂¯ vf.19 Forcing An alternative way to modify the scale-similarity model is to omit the forward scatter. More details can be found [81]. Eq.47b) /∂xi = 0 so that ∗ vf.i 1 ∂¯ ∂ 2 p¯n+1 = ∂xi ∂xi ∆tα ∂xi (18. The discretized momentum equations read n+1/2 n+1/2 = v¯in + ∆tH v¯n . and Crank-Nicolson (α = 0. for v¯1 .19.46. The face velocities n+1/2 n+1/2 n+1/2 + v¯i. 18.e.5(¯ vi.48) The Poisson equation for p¯n+1 is solved with an efficient multigrid method [80]. 2.j index) do not satisfy continuity. i. to include instants when the subgrid stresses act as counter-gradient diffusion. In SIMPLE notation this equation reads X ∂ p¯n+1/2 n+1/2 = aP v¯i ∆V anb v¯n+1/2 + SU − α∆t ∂xi nb where SU includes all source terms except the implicit pressure. 18. i.49) A few iterations (typically two) solving the momentum equations and the Poisson pressure equation are required each time step to obtain convergence.i (18. finite volume method with collocated grid arrangement.5) in time is briefly described below. v¯i v¯i (18. Create an intermediate velocity field by subtracting the implicit pressure gradient from Eq. 1.20 Numerical method A numerical method based on an implicit. 18. 83] are often used for LES. see Table 18. Use linear interpolation to obtain the intermediate velocity field. 18. 7. 3. 8.49 6. Hence. v¯1 t t1 : start t2 : end Figure 18.1: Differences between a finite volume RANS and LES code.20.48) is solved with an efficient multigrid method [80]. Compute the face velocities (which satisfy continuity) from the pressure and the intermediate face velocity from Eq. 18.1. Since the Poisson solver in [80] is a nested MG solver. What are the specific requirements to carry out LES with a finite volume code? If you have a RANS finite volume code. LES Above a numerical procedure suitable for LES was described. . in general. An LES code is actually simpler than a RANS code. However. for example pressurecorrection methods such as SIMPLE [82. The Poisson equation (Eq. Next time step. C-N) zero. 18. v¯f. at the face 4. it is very simple to transform that into an LES code. The turbulent viscosity is computed.1 RANS vs.20. 5. any numerical procedure for RANS can also be used for LES.or one-equation Table 18. Both the discretization scheme and and the turbulence model are simpler in LES than in RANS. Numerical method 177 RANS 2D or 3D steady or unsteady 2nd order upwind 1st order more than two-equations Domain Time domain Space discretization Time discretization Turbulence model LES always 3D always unsteady central differencing 2nd order (e. Step 1 to 4 is performed till convergence (normally two or three iterations) is reached.i .18.11: Time averaging in LES. it is difficult to parallelize with MPI (Message Passing Interface) on large Linux clusters. when we do large simulations (> 20M cells) we use a traditional SIMPLE method.g. 2 on p. The equation can be written on the same form as the RANS k-equation. 18. production and dissipation in Eq. length of a recirculation region). Time averaging can start at time t1 when the flow seems to have reached fully developed conditions. As mentioned above. 100L/V . is equivalent to the SGS dissipation in the equation for the resolved turbulent kinetic energy (look at the flow of kinetic energy discussed at the end of [84]). The theoretical statistical error varies with number of independent samples. There are two reasons: first. see Fig. as N −1/2 . 100 time units.14. turbulence models in LES are simple. 18. V (free-stream or bulk velocity). a value of 10 may be sufficient if L is the length of a recirculation region. 18. M. see Section 18. The question then arises. Pksgs . Pksgs = 2νsgs s¯ij s¯ij .51) . 18. We start the windtunnel: when has the flow (and the turbulence) reached fully developed conditions so that we can start to measure the flow? Next question: for how long should we carry out the measurements? Both in LES and the windtunnel. hence a second-order (or higher) central differencing scheme should be employed. N . and a length scale. ∂ksgs ∂ ∂ ∂ksgs (ν + νsgs ) + Pksgs − ε (¯ vj ksgs ) = + ∂t ∂xj ∂xj ∂xj (18. no equation for the turbulent length scale is required since the turbulent length scale can be taken as the filter width.21 One-equation ksgs model A one-equation model can be used to model the SGS turbulent kinetic energy.18.13. ε = Cε ∆ Note that the production term.22 Smagorinsky model derived from the ksgs equation We can use the one-equation model to derive the Smagorinsky model.50 are in balance so that Pksgs = 2νsgs s¯ij s¯ij = ε (18. L (width of a wake or a body.e. In LES we are doing unsteady simulations.21. ∆ ∝ κII . 288. The cut-off takes place in the inertial subrange where diffusion and convection in the ksgs equation are negligible (their time scales are too large so they have no time to adapt to rapid changes in the velocity gradients.11. ∆. see Eq. The time discretization should also be non-dissipative.12. Hence. 18. It is difficult to judge for how long one should carry out time averaging.e. The length scale in the Smagorinsky model is the filter width. and use this to estimate the required averaging time. One-equation ksgs model 178 It is important to use a non-dissipative discretization scheme which does not introduce any additional numerical dissipation.50) 3/2 ksgs 1/2 νsgs = ck ∆ksgs . i. 18. second. the recorded time history of the v¯1 velocity at a point may look like in Fig. Eq. Usually it is a good idea to form a characteristic time scale from a velocity. The Crank-Nicolson scheme is suitable. s¯ij ). only the small-scale turbulence is modeled and. i. for example. a windtunnel. may be a suitable averaging time for the flow around a bluff body. when can we start to time average and for how long? This is exactly the same question we must ask ourselves whenever doing an experiment in. it has the great advantage that the coefficients are computed rather than being prescribed. 18. isotropic.j . III: Range for small. I: Range for the large. independent of the large scales (ℓ) and the dissipative scales (ν).53) Eq.52) Dimensional analysis yields a = 1/3. The equation for the subgrid kinetic energy reads [85. We can write the SGS viscosity as νsgs = εa (CS ∆)b (18.53 gives 3 νsgs = (CS ∆)4 ε = (CS ∆)4 νsgs (2¯ sij s¯ij ) ⇒ νsgs = (CS ∆)2 |¯ s| (18.12: Spectrum for k.18. In real applications ad-hoc local smoothing and clipping is used. 169) is the numerical instability associated with the negative values and large variation of the C coefficient. To ensure numerical stability. 86] (see also [87. a constant value (in space) of C (Chom ) is used in the diffusion term in Eq. energy containing eddies.55) 1 2 = −2C∆ksgs s¯ij 1 2 with νef f = ν + 2Chom ∆ksgs . 18.23.55 and in the momentum equations. (18. dissipative scales. being a dynamic model. The C in the production term Pksgs is computed dynamically (cf. Eq.33). 18. Let us replace ε by SGS viscosity and ∆. Furthermore. Usually this problem is fixed by averaging the coefficient in some homogeneous flow direction.23 A dynamic one-equation model One of the drawbacks of the dynamic model of [70] (see p. 18. The main object when developing this model was that it should be applicable to real industrial flows. 88]) ∂ ∂ ∂ksgs (¯ vj ksgs ) = Pksgs + + ∂t ∂xj ∂xj Pksgs = −τija v¯i.54) 1/2 |¯ s| = (2¯ sij s¯ij ) which is the Smagorinsky model. II: the inertial subrange for isotropic scales. A dynamic one-equation model E(κ) 179 I II III κ Figure 18. . 18. b = 4/3 so that νsgs = (CS ∆)4/3 ε1/3 . Below a dynamic one-equation model is presented.51 substituted into Eq. τija 3/2 ∂ksgs ksgs νef f − C∗ ∂xj ∆ (18. detailed SUV [100]. the flow around a simplified bus [91. 18. i. P ksgs − C∗ ksgs ∆ ∆ so that z {! z{ z{ 1 n 3/2 ∆ n+1 K (18.57) Since the turbulence on both the grid level and the test level should be in local equilibrium (in the inertial −5/3 region).e. The equations for ksgs and K read in symbolic form 3/2 ksgs ∆ K 3/2 − C∗ z{ ∆ T (ksgs ) ≡ Cksgs − Dksgs = Pksgs − C∗ T (K) ≡ CK − DK = P K (18. ∆ K2 The idea is to put the local dynamic coefficients in the source terms. 91]. the flow and heat transfer in a square rotating duct [92. In this model a dynamic one-equation SGS model is solved. {3/2 z{ K 3/2 1z = P K − C∗ z{ . 18. We have also done some work on buoyancy-affected flows [104–110]. This is a big advantage compared to the standard dynamic model of Germano (see discussion on p. 1 1 2 2 s¯ij s¯ij ixyz = 2Chom h∆ksgs s¯ij s¯ij ixyz h2C∆ksgs The dissipation term εksgs is estimated as: 3/2 εksgs ≡ νTf (vi. 18. . vi. An even better approximation should be to assume T (ksgs ) = T (K).25 Applied LES At the Department we used LES for applied flows such as flow around a cube [90.24 A Mixed Model Based on a One-Eq. i. the left-hand side of the two equations in Eq. i. 94]. 18.55). Model 180 Chom is computed by requiring that Chom should yield the same total production of ksgs as C. trains and buses subjected to sidewinds and wind gusts [101–103].j ) = C∗ ksgs . and the scale-similarity part is estimated in a similar way as in Eq. in the production and the dissipation terms of the ksgs equation (Eq.18.24.e.57 should be close to zero. ∆ (18. 171).93]. Model Recently a new dynamic scale-similarity model was presented by [89]. In this way the dynamic coefficients C and C∗ don’t need to be clipped or averaged in any way. 18.j .41. 99]. a simplified car [95–97] and the flow around an airfoil [98.56) Now we want to find a dynamic equation for C∗ .58) C∗ = P − P ksgs + C∗ ksgs 3 . A Mixed Model Based on a One-Eq.e. These structures must be resolved in an LES in order to achieve accurate results. 40).13: Energy spectra in fully developed channel flow [112].18. we can afford δ/∆x1 and δ/∆x3 in the region of 10 − 20 and 20 − 40. the flow in recirculation regions and shear layer can. this kind of resolution can hardly ever be afforded. the computational resource requirement of accurate LES is prohibitively large. For wall-bounded flows at high Reynolds numbers of engineering interest. Indeed. 116]. This is similar to the requirement in RANS (Reynolds-Averaged Navier-Stokes) using low-Re number models. + ∆x+ 1 = 1300 and ∆x3 = 1800. The resolution requirements in wall-parallel planes for a wellresolved LES in the near-wall region expressed in wall units are approximately 100 (streamwise) and 30 (spanwise).33. The geometry is shown in Fig. Then the spectra of the resolved turbulence will exhibit −5/3 range. the turbulence in the recirculation region and in the shear layer downstream the bump turned out to be well resolved.26. 18. 20). 10). 18. 69. 18. 18. ◦: (5. δ). δ denotes half : (10. The suggestion in these works is that two-point correlations is the best way to estimate if an LES is sufficiently resolved or not. the spectra in the boundary layer will look something like that shown in Fig. often called the streak process.max Figure 18. . the requirement of near-wall grid resolution is the main reason why LES is too expensive for engineering flows. δ/∆x3 ). In outer scaling (i. 20). +: (10. In this case.e. : channel width. which was one of the lessons learned in the LESFOIL project [115. At low to medium Reynolds numbers the streak process is responsible for the major part of the turbulence production. In [114] the flow (Re ≃ 106 ) over a bump was computed.13 [112]. 20).113] different ways to estimate the resolution of an LES were investigated. comparing the resolution to the boundary layer thickness. The turbulence in the boundary layer on the bump was very poorly resolved: ∆x1 /δin = 0.26 Resolution requirements The near-wall grid spacing should be about one wall unit in the wall-normal direction. This enables resolution of the near-wall turbulent structures in the viscous sub-layer and the buffer layer consisting of high-speed inrushes and low-speed ejections [111]. Energy spectra are actually not very reliable to judge if a LES simulation is well resolved or not. (20. Nevertheless. : (10. Resolution requirements 181 Thick dashed line −5/3 slope −1 E33 (k3 ) 10 −2 10 −3 10 −4 10 0 1 10 10 2 10 κ3 = 2πk3 /x3.44. In applied LES.14. respectively. In [112. see figure on p. ∆x3 /δin = 0. Number of cells expressed as (δ/∆x1 . see Fig. Even if the turbulence in boundary layer seldom can be resolved.15. unless fine grids are used. In URANS the usual Reynolds decomposition is employed.0035 (near the wall).13.14: Onera bump. 19 Unsteady RANS To perform an accurate LES. This causes problems. Unsteady RANS 182 L1 V1. 1 v¯(t) = 2T Z t+T t−T v(t)dt.in H y h x L2 L Figure 18. near walls.max Figure 18. However. LES will probably give poor predictions in these regions. which can be captured by a fairly coarse mesh.1) . v = v¯ + v ′′ (19. where the flow is governed by large.19. LES is very good for wake flow. turbulent structures. : x2 /H = 0. a very fine mesh must be used. Thick dashed line shows −5/3 slope. Computational domain (not to scale). : x2 /H = 0. x2 /H = 0.34 (in the shear layer). often denoted URANS (Unsteady RANS). for example. i.e.15: Energy spectra E33 (κ3 ) in the recirculation region and the shear layer : downstream the bump (x1 /H = 1. An alternative to LES for industrial flows can be unsteady RANS (ReynoldsAveraged Navier-Stokes). −1 E33 (κ3 ) 10 −2 10 −3 10 10 100 200 κ3 = 2πk3 /x3.2). if attached boundary layers are important. but with the transient (unsteady) term retained. the URANS equations read µt = c µ ρ (19. The modeled turbulent fluctuation. and the modeled. x3 .1. Unsteady RANS 183 The URANS equations are the usual RANS equations. t). t) and vi′′ vj′′ = vi′′ vj′′ (x1 . but also function of time. v ′′ . h¯ v i. : v.e. If the flow has strong vortex shedding. v¯. on incompressible form they read ∂vi′′ vj′′ ∂¯ vi 1 ∂ p¯ ∂ 2 v¯i ∂ (¯ vi v¯j ) = − +ν − + ∂t ∂xj ρ ∂xi ∂xj ∂xj ∂xj ∂ v¯i =0 ∂xi (19. h¯ vi h¯ vi t Figure 19. which means that we can decompose the results from an URANS as a time-averaged part. x3 . p¯ = p¯(x1 . v¯′ .1: Decomposition of velocities in URANS. v = v¯ + v ′′ = h¯ v i + v¯′ + v ′′ (19. h¯ v i. i. one is often interested only in the time-averaged flow. i. What type of turbulence model should be used in URANS? That depends on type of flow.19. x2 . x2 .5) (19. x2 . the standard high-Re number k − ε model can be used.6) ∂¯ vi ∂ p¯ ∂ ∂ρ¯ vi v¯k ∂ρ¯ vi (µ + µt ) =− + + ∂t ∂xk ∂xi ∂xk ∂xk (19. 19.e. turbulent fluctuation.7) v. when this is added to h¯ v i + v¯′ we obtain v. Even if the results from URANS are unsteady. .4) = + ∂t ∂xj ∂xj σk ∂xj ∂ vj ε ∂ρε ∂ρ¯ = + ∂t ∂xj ∂xj µt ∂ε ε c1ε P k − cε2 ρε µ+ + σε ∂xj k k2 ε With an eddy-viscosity. v ′′ . a resolved fluctuation. is not shown in the figure.3) see Fig. x3 . v¯i = v¯i (x1 . : v¯. i. We denote here the time-averaged velocity as h¯ v i.2) Note that the dependent variables are now not only function of the space coordinates. µt ∂k ∂ vj k ∂ρk ∂ρ¯ µ+ + P k − ρε (19. t).e. The V2F and non-linear eddy-viscosity models are better. the question arises how the results should be time averaged. Turbulence Modeling 184 Figure 19.4 and 19. νt . So we are doing unsteady simulations. The common definition of URANS is that the turbulent length scale is not determined by the grid.5) was used in [117] for two-dimensional URANS simulations computing the flow around a triangular flame-holder in a channel. In practice this requirement is often not satisfied [81]. Eq. i. i. and this was discussed in connection to Fig. The URANS momentum equation and the LES momentum equation are exactly the same. 19. 19. part of the turbulence is modeled (v ′′ ) and part of the turbulence is resolved (¯ v ′ ). 19. In Fig.3 the v¯2 velocity in a point above the flame-holder is shown and it can be seen that the velocity varies with time in a sinusoidal manner. i. know how they were time averaged? Or if they were volume filtered? The answer is that they don’t. see Fig. The standard k − ε model (Eq. which dampens the resolved fluctuations too much. we . This issue is the same when doing LES. In URANS. is much larger than the SGS viscosity.1 Turbulence Modeling In URANS.e. except that we denote the turbulent viscosity in the former case by νt and in the latter case by νsgs .19. When we’re doing URANS. in Eq. If we want to compare computed turbulence with experimental turbulence. when should we start to average and for how long. v¯′ . 19. should have a much smaller time scale than the resolved ones.2.7.2: Configuration of the flow past a triangular flameholder. What we care about is that the turbulence model and the discretization scheme should not be too dissipative. In URANS we do usually not care about scale separation. This flow has a very regular vortex shedding. Flow from left to right Usually the standard k − ε model is not a good URANS model because it gives too much modeled dissipation (i. hence.e. how do the momentum equation.1 should be much smaller than the resolved time scale. too large turbulent viscosity).11. they should not kill the resolved fluctuations. T . νsgs . This is called scale separation. much more of the turbulence is modeled than in LES. On the other hand. whereas in LES it is.e.e. the turbulent viscosity. 19.1. the modeled turbulent fluctuations. 18. 19. v¯′ . v ′′ . and. How is this possible? The theoretical answer is that the time. and the flow actually has a scale separation. but still we time average the equations. One cycle of the v¯2 velocity in a cell near the upper-right corner of the flameholder. 119] when . right figure: x = 1. It can be seen that here the resolved and the modeled turbulence are of the same magnitude. Figure 19. must add these two parts together. 19. the reason for this is that the turbulence model is too dissipative. If the turbulence model in URANS generates ”too much” eddy viscosity.1.43H.4: 2D URANS k − ε simulations compared with experiment [117]. Solid lines: total turbulent kinetic energy.1H (x = 0 at the downstream vertical plane of the flame-holder). dashed lines: resolved turbulent kinetic energy: ∗: experimental data. the flow may not become unsteady at all. Left figure: x = 0.3: 2D URANS k − ε simulations [117].19. It was found in [118. Turbulence Modeling 185 Figure 19. Profiles downstream the flameholder are shown in Fig. because the unsteadiness is dampened out.4. How dissipative a scheme needs to be in order to be stable is flow dependent. The reason why the unsteadiness in these computations was not dampened out is that the vortex shedding in this flow is very strong. Discretization 186 using URANS for the flow around a surface-mounted cube and around a car. In general. It is seen that with LES and central differencing nonphysical oscillations are present (this was also found in [90]). However.e. The reason is that we don’t want to dampen out resolved. even if a central differencing scheme is used. this gives first-order accuracy in both space and time. LES with the Mars scheme (in which some numerical dissipation is present) and URANS with the central scheme (where the modeling dissipation is larger than in LES) no such nonphysical oscillations are present. 19. both the discretization and the turbulence model have high dissipation. Two different discretization schemes were used: the central scheme and the Mars scheme (a blend between central differencing and a bounded upwind scheme of second-order accuracy). In [118] LES and URANS simulations were carried out of the flow around a surfacemounted cube (Fig. 19. In Fig. and was successfully applied in URANS-simulations for these two flows. it may work with no dissipation at all (i. that the standard k − ε model was too dissipative.6H) using URANS and LES with central differencing are shown together with URANS and the Mars scheme.5) with a coarse mesh using wall-functions.2 Discretization In LES it is well-known that non-dissipative discretization schemes should be used. whereas for industrially complex flows maybe a bounded second-order scheme must be used. central differencing).2.6 the time-averaged velocity profile upstream of the cube (x1 = −0. This is to some extent true also for URANS. If turbulent unsteady inlet fluctuations are used. The turbulence model that was used was the standard k − ε model. the hybrid discretization scheme for the convective terms was used together with fully implicit first-order discretization in time. The main reason to the nonphysical oscillations is that the predicted flow in this region does not have any resolved fluctuations. resolved fluctuations dominate over any numerical oscillations. for some simple flows. For time discretization. the second-order accurate Crank-Nicolson works in most cases. Non-linear models like that of [120] was found to be less dissipative. 19.19. the nonphysical oscillations do usually not appear. a discretization scheme which has little numerical dissipation should be used. . In this case the turbulent. turbulent fluctuations. In the predictions on the flame-holder presented above. Thus. Figure 19.5: URANS simulations of the flow around a surface-mounted cube. . Discretization 187 Figure 19.6: URANS simulations of the flow around a surface-mounted cube.19.2. Velocity profiles upstream the cube [118]. ∆xζ ). This term is sometimes neglected [123]. Ψ = Cw1 ρfw d (20.1) νt = ν˜t f1 The production term P and the destruction term Ψ have the form ν˜t P = Cb1 ρ s¯ + 2 2 f2 ν˜t κ d 2 ν˜t 1/2 s¯ = (2¯ sij s¯ij ) .1 DES based on two-equation models The model described above is a one-equation model. The modeled length scale is reduced and the consequence is that the destruction term Ψ increases. i. k 3/2 /ε or k 1/2 /ω) or the filter width ∆. In these models the turbulent length scale is either obtained from the two turbulent quantities (e.20. The constant Cdes is usually set to 0. 4. The aim is to treat the boundary layer with RANS and capture the outer detached eddies with LES. νt = k 1/2 ℓt . ∆xη . In [122] the DES model was proposed in which d is taken as the minimum of the RANS turbulent length scale d and the cell length ∆ = max(∆xξ . Outside the turbulent boundary layer d > Cdes ∆ so that the model operates in LES mode. A model based on the k − ε model can read νt ∂ ∂ ∂k ∂k ν+ + P k − εT (¯ vj k) = + ∂t ∂xj ∂xj σk ∂xj νt ε ∂ ∂ ∂ε ∂ε ν+ + (C1 P k − C2 ε) (¯ vj ε) = + ∂t ∂xj ∂xj σε ∂xj k P k = 2νt s¯ij s¯ij . DES 188 20 DES DES (Detached Eddy Simulation) is a mix of LES and URANS. However.e. which in many situations is not a relevant turbulent length scale. 121. Cdes ∆) (20. this would give rise to an increased production term P through the second term (see Eq. The model was originally developed for wings at very high angles of attack.g. Sect. DES models based on two-equation models were proposed [124–126]. ∆xη and ∆xζ denote the cell length in the three grid directions ξ. 20. which gives a reduction in the turbulent viscosity ν˜t . In the boundary layer d < Cdes ∆ and thus the model operates in RANS mode.3) ∆xξ . Recently. In RANS mode it takes its length scale from the wall distance.2) d in the RANS SA model is equal to the distance to the nearest wall. this second term is a viscous term and is active only close to the wall. It can be written [115.6] ∂ρ˜ νt Cb2 ρ ∂ ν˜t ∂ ν˜t ∂ ∂ρ¯ vj ν˜t µ + µt ∂ ν˜t + = +P −Ψ + ∂t ∂xj ∂xj σν˜t ∂xj σν˜t ∂xj ∂xj (20.65.2). 20. d˜ = min(d. η and ζ. The RANS model that was originally used was the one-equation model proposed in [121]. A reduced ν˜t gives a smaller production term P so that the turbulent viscosity is further reduced. At first sight it may seem that as the model switches from RANS mode to LES mode thus reducing d. 132–135] at the location where the model switches from RANS mode to LES mode. ε is needed as soon as the model switches to RANS model again. Cε ∆ In other models [63. When the model switches to LES mode. However.20. A third alternative is to modify only the turbulent length scale appearing in the turbulent viscosity [127]. see Section 23) and PITM [130] (Partially Integrated Transport Modeling. The result is that the dissipation in the k equation increases so that k decreases which gives a reduced νt . One way to get around this is to impose turbulence fluctuations as forcing conditions [84. and the turbulent dissipation. 124] only the dissipation term. DES based on two-equation models 189 The turbulent length scale. ℓt . In these models ε increases because of its reduced destruction term which decreases both k and νt . Ck ∆ ε k 3/2 εT = max ε. When the grid is sufficiently fine. say. it will take some distance L before the momentum equations start to resolve any turbulence. The forcing is added in the form of a source term (per unit volume) in the momentum equations. see Section 24). εT . 127] k 3/2 ℓt = min Cµ . This poses a major problem with this type of models. the turbulence is supposed to be represented by resolved turbulence. In the RANS mode the major part of the turbulence is modeled. 129] (Partially Averaged Navier-Stokes. the length scale is taken as ∆. are computed as [126. A rather new approach is to reduce the destruction term in the ε equation as in PANS [128. . but ε is not used. A low-Reynolds number PANS was recently proposed [129] in which the near-wall modifications were taken from the AKN model [131]. If the switch occurs at location x1 .1. εT is modified. This is exactly what happens at an inlet in an LES simulation if no real turbulence is given as inlet boundary conditions. In regions where the turbulent length scales are taken from ∆ (LES mode) the ε-equation is still solved. All coefficients are blended between the k − ω and the k − ε model using the function F1 . In DES the dissipation term in the k equation is modified as [63] Lt ∗ ∗ .k−ω = 0. the DES modification is meant to switch the turbulent length scale from a RANS length scale (∝ k 1/2 /ω) to a LES length scale (∝ ∆) when the grid is sufficiently fine.09. η = max .0828. σk.44.20. When FDES is larger than one. α = F1 αk−ω + (1 − F1 )αk−ε (20. βk−ε = 0. that a large part of the turbulence is resolved rather than being modeled. Again. Hence it is crucial to generate an appropriate grid.1 β kω → β kωFDES .3 αk−ω = 5/9. FDES = max CDES ∆ ∆ = max {∆x1 . 63] νt ∂ ∂k ν+ + Pk − β ∗ kω ∂xj σk ∂xj νt Pk ∂ ∂ω ν+ +α − βω 2 ∂xj σω ∂xj νt 1 ∂k ∂ω + 2(1 − F1 )σω2 ω ∂xi ∂xi " ( √ ) # 4σ k k 500ν ω k−ε F1 = tanh(ξ 4 ). σω.61.k−ε = 0. ∆x3 } . for example.k−ε = 1.2 DES based on the k − ω SST model The standard k − ω model SST reads [57.k−ω = 0. β ∗ ωd d2 ω ∂ ∂k (¯ vj k) = + ∂t ∂xj ∂ω ∂ (¯ vj ω) = + ∂t ∂xj where d is the distance to the closest wall node.856. ξ = min max . for α.5 σk. The result is.7 shows that it is the grid that determines the location where the model switches between RANS and LES.4) β ∗ ωd d2 ω CDω d2 1 ∂k ∂ω . ∆x2 . k 1/2 Lt = ∗ β ω (20.2. a1 = 0. DES based on the k − ω SST model 190 20. With a smaller turbulent viscosity in the momentum equations. |¯ s|F2 ) 1/2 2k 500ν F2 = tanh(η 2 ). σω. (20. 10−10 CDω = max 2σωk−ε ω ∂xi ∂xi a1 k νt = max(a1 ω. hopefully. βk−ω = 3/40.5) The constants take the following values: β ∗ = 0. Equation 20.6) (20. αk−ε = 0.7) where CDES = 0.85. the dissipation term in the k equation increases which in turn decreases k and thereby also the turbulent viscosity. the modeled dissipation (i. . The SST model behaves as a k − ω model near the wall where F1 = 1 and a k − ε model far from walls (F1 = 0).e the damping) is reduced and the flow is induced to go unsteady. 20. This results in a poorly resolved LES and hence inaccurate predictions.20. δ). Different proposals have been made [136.4) of the SST model. .8) CDES ∆ where FS is taken as F1 or F2 (see Eq. This means that the flow in the boundary layer is treated in LES mode with too a coarse mesh. the further out from the wall does the switch take place. This is called DDES (Delayed DES). 1 (20. In [63] F = F2 . DES based on the k − ω SST model 191 The larger the maximum cell size (usually ∆x) is made.2.137] to protect the boundary layer from the LES mode Lt FDES = max (1 − FS ). In some flows it may occur that the FDES term switches to DES in the boundary layer because ∆z is too small (smaller than the boundary layer thickness. . An event of a high-speed in-rush is illustrated in Fig. see Fig. Indeed. and the momentum deficit in the inner layer increases for increasing x1 . the computational resource requirement of accurate LES is prohibitively large. There are also events which occurs in the other direction. v1′ in the viscous wall region. The resolution requirements in wall-parallel planes for a well-resolved LES in the near-wall region expressed in wall units are approximately 100 (streamwise) and 30 (spanwise). As a result the mid-part is lifted up even more and a tip of a hairpin vortex is formed. These events are called bursts or ejections.1). The object of hybrid LES-RANS (and of DES) is to eliminate the requirement of high near-wall resolution in wall-parallel planes. It can be seen that an inflection point is created in the instantaneous velocity profile. These structures must be resolved in an LES in order to achieve accurate results. 21. which was one of the lessons learned in the LESFOIL project [115. Figure 21. The mid-part of the vortex line experiences now a higher v¯1 velocity (denoted by U in the figure) than the remaining part of the vortex line. The vorticity of the legs of the hairpin lift each other through self-induction which helps lifting the tip even more. see Fig. 21. The process by which the events are formed is similar to the later stage in the transition process from laminar to turbulent flow. for wall-bounded flows at high Reynolds numbers of engineering interest. In the near-wall region (the URANS region). The near-wall grid spacing should be about one wall unit in the wall-normal direction. This enables resolution of the near-wall turbulent structures in the viscous sub-layer and the buffer layer consisting of high-speed in-rushes and low-speed ejections [111]. a low-Re number RANS turbulence model (usually an eddy-viscosity model) is used.21. The streamwise distance between the events is related to the boundary layer thickness (4δ.e. This is the main reason why the grid must be very fine in the spanwise direction.1). a turbulent fluctuation – the mid-part of the vortex line is somewhat lifted up away from the wall. In the lower part of the figure the spanwise vortex line is shown. In the LES region. the requirement of near-wall grid resolution is the main reason why LES is too expensive for engineering flows. the turbulent structures very elongated in the streamwise direction. As can be seen.g. Eventually the momentum deficit becomes too large and the highspeed fluid rushes in compensating for the momentum deficit. i. Thus. coarser . Hybrid LES-RANS 192 21 Hybrid LES-RANS When simulating bluff body flows. The spanwise separation between sweeps and bursts is very small (approximately 100 viscous units. In the x1 −x2 plane (upper part of Fig. 116]. This is similar to the requirement in RANS using low-Re number models. Initially it is a straight line. respectively) are shown and mean velocity profiles (denoted by U and U as the hairpin vortex is created. The in-rush event is also called a sweep.3. 21. often called the streak process. LES (Large Eddy Simulation) is the ideal method.1) the instantaneous ¯ in the figure.2 presents the instantaneous field of the streamwise velocity fluctuation. the usual LES is used. lowspeed fluid is ejected away from the wall. see Fig. U .1. it is a challenging task to make accurate predictions of wallbounded flows with LES. 21. The idea is that the effect of the near-wall turbulent structures should be modeled by the RANS turbulence model rather than being resolved. On the other hand. In the outer region (the LES region). but due to a disturbance – e. 21. Bluff body flows are dominated by large turbulent scales that can be resolved by LES without too fine a resolution and accurate results can thus be obtained at an affordable cost. At low to medium Reynolds numbers the streak process is responsible for the major part of the turbulence production. 3: The LES and URANS region. DNS of channel flow [84]. Hybrid LES-RANS 193 Figure 21. wall URANS region LES region URANS region x2 wall x1 Figure 21.2: Fluctuating streamwise velocity in a wall-parallel plane at x+ 2 = 5. x+ 2.21. x3 x1 Figure 21.1: Illustration of near-wall turbulence (taken from [68]).ml . 2 The equation for turbulent kinetic energy in hybrid LES-RANS A one-equation model is employed in both the URANS region and the LES region. for x2 > x2. in this form.014k 1/2n/ν)] 1. the boundary layer thickness) rather than the near-wall turbulent processes. 21. No special treatment is used in the equations at the matching plane except that the form of the turbulent viscosity and the turbulent length scale are different in the two regions. see Eq. and the turbulent SGS viscosity is used in the LES region.07k 1/2 ℓ 1. The turbulent RANS viscosity is used in the URANS region. τij = −2νT s¯ij In the inner region (x2 ≤ x2.1. Momentum equations in hybrid LES-RANS URANS region ℓ νT Cε −3/4 κcµ n[1 − exp(−0.g. in the outer region (x2 > x2. whereas hybrid LES-RANS aims at covering only the inner part of the boundary layer with URANS.1 Momentum equations in hybrid LES-RANS The incompressible Navier-Stokes equations with an added turbulent/SGS viscosity read ∂¯ vi ∂¯ vi 1 ∂ p¯ ∂ ∂ (ν + νT ) (21.1.21.1) (¯ vi v¯j ) = − + + ∂t ∂xj ρ ∂xi ∂xj ∂xj where νT = νt (νt denotes the turbulent RANS viscosity) for x2 ≤ x2. νT . respectively. kT = 0.138]. The main difference is that the original DES aims at covering the whole attached boundary layer with URANS. When prescibing the location of the RANS-LES interface.ml ) kT corresponds to the RANS turbulent kinetic energy. see Table 21. 21. k.41).ml .137.1) and kT .3) and. grid spacing in wall-parallel planes can be used. DES is similar hybrid LES-RANS.2k 1/2 n/ν)] 1/4 κcµ k 1/2 n[1 − exp(−0. the velocity profile may show unphysical behaviour near the interface because of the rapid variation of the tur- .0 194 LES region ℓ=∆ 0. 139].2.2) ∂t ∂xj ∂xj ∂xj ℓ PkT = −τij s¯ij . At the walls. and. is computed from an algebraic turbulent length scale (see Table 21. In later work DES has been used as a wall model [133. The unsteady momentum equations are solved throughout the computational domain. the latter is obtained by solving its transport equation. Hybrid LES-RANS is similar to DES (Detached Eddy Simulations) [122. n and κ denote the distance to the nearest wall and von K´arm´an constant (= 0.05 Table 21. ∆ = (δV )1/3 . The turbulent viscosity. νT = νsgs . 21. The grid resolution in this region is presumably dictated by the requirement of resolving the largest turbulent scales in the flow (which are related to the outer length scales.ml (see Fig. which reads 3/2 k ∂kT ∂ ∂ ∂kT (ν + νT ) + PkT − Cε T (¯ vj kT ) = + (21.1: Turbulent viscosity and turbulent length scales in the URANS and LES regions. e.ml ) it corresponds to the subgrid-scale kinetic turbulent energy (ksgs ). 21. 133.21. 134]. 112.2. Much work on forcing have been presented in order to allieviate this problem [84. . The equation for turbulent kinetic energy in hybrid LES-RANS 195 bulence viscosity. One way to improve a RANS model’s ability to resolve large-scale motions is to use the SAS (Scale. Hence. This leads to a more accurate prediction of the flow.Adaptive Simulation) model 22.1 Resolved motions in unsteady When doing URANS or DES.1D .e. Another reason is that if the numerical solution wants to go unsteady. high turbulent viscosity often dampens out these instabilities. One reason is that there may not be any steady solution. because if the flow wants to go unsteady. part of the turbulent spectrum — will be resolved instead of being modeled. the equations will not converge.2 The von K´arm´an length scale The von K´arm´an length scale LvK. In many cases this is an undesired feature. it is usually a bad idea to force the equations to stay steady. The SAS model 196 22 The SAS model 22. the large turbulent scales — i.22. In URANS or in DES operating in RANS mode. the momentum equations are triggered through instabilities to go unsteady in regions where the grid is fine enough. . . ∂h¯ v i/∂x2 . . . = κ. 2 ∂ h¯ v i/∂x2 . applicable in three dimensions. Menter & Egorov [140] derived a new k − kL model using the von K´arm´an length scale. O. .2) ∂xj ∂xj ∂xk ∂xk There are other options how to computed this second derivative.3D = κ |¯ s| |U ′′ | (22. as noted in [140]. The first and second derivatives in Eq. the Smagorinsky model. Finally. they called this model the SST-SAS model. The von K´arm´an length scale is smaller for an instantaneous velocity profile than for a time averaged velocity. The second derivative can be generalized in a number of ways. see Fig. We want to extend this expression to a general one. This model is described in more detail below.1 are given in boundary layer form. In the SAS model it is taken as 2 0. see Eq. the general three-dimensional expression for the von K´arm´an length scale reads LvK.1. based on the k − k 1/2 L model of Rotta [142].1) 2 which includes the second velocity gradient is a suitable length scale for detecting unsteadiness.3) In [141] they derived a one-equation νt turbulence model where the von K´arm´an length scale was used. the von K´arm´an length scale decreases when the momentum equations resolve (part of) the turbulence spectrum. This is interesting because. The model was called the SAS model.5 ∂ v¯i ∂ 2 v¯i U ′′ = (22. for example. In the same way as in. 22. in [143] they modified the k − ω-SST model to include the SAS features. Later. 22. Hence. (22.5. we take the first derivative as |¯ s| = (2¯ sij s¯ij )1/2 . 22. Menter & Egorov [143] introduced a SAS-term in the ω equation.4.1 .1: Velocity profiles from a DNS of channel flow.3D where Lt denotes a RANS length scale.e. Eq. see Fig. 22. 22.1 • This results in a decrease in the von K´arm´an length scale. LvK. this is an undesirable feature. The SST-SAS model The k − ω SST model is given in Eq. The von K´arm´an length scale 197 1 0. 190 (see also the section starting at p.22. The additional term reads k 1/2 Lt .2. perhaps.1D . The last item in the list above is the main object of the SAS model. grow.3D ωcµ When unsteadiness occurs — i. The reaction to local unsteadiness in a eddy-viscosity model without the SAS feature is as follows: the increased local velocity gradients will create additional production of turbulent kinetic energy and give an increased turbulent viscosity which will dampen/kill the local unsteadiness. i. Lt = 1/4 (22.3. Eq.4 at p.4) ζ˜2 κ|¯ s |2 LvK. 20.3D LvK.4 0. 22. Thus it seems reasonable to add a new production term proportional to Pω Lt /LvK. Solid line: time-averaged velocity with length scale Lx.6 LvK.3D • As a consequence the additional source. As mentioned in the introduction to this chapter. dashed line: instantaneous velocity with length scale LvK.e. when the momentum equations attempt to resolve part of the turbulence spectrum — this term reacts as follows: • Local unsteadiness will create velocity gradients which decrease the turbulent length scale.1D 0. Eq.3D . in the ω equation increases • This gives an increase in ω and hence a decrease in νt • The decreased turbulent viscosity will allow the unsteadiness to stay alive and. The production term in the ω equation in the k−ω-SST model reads Pω = αP k /νt ∝ |¯ s|2 . when the von K´arm´an length scale becomes small.2 0 0 5 10 v¯1 15 20 Figure 22. To decrease the turbulent viscosity we should increase ω. 155) Now. .8 x2 0. The object of this term is to decrease the turbulent viscosity when unsteadiness is detected. In finite volume methods there are two main options for computing second derivatives.2 the turbulent length scale. Near the wall LvK. 22.3.2. we need to compute the second velocity gradients. 22. The von K´arm´an length scale based on instantaneous velocities.3D i. 22. is evaluated using DNS data of fully developed channel flow.1D increases because the time-average second derivative. 22.2. Option I: compute the first derivatives at the faces ∂v vj+1 − vj vj − vj−1 ∂v . hLvK. is presented in Fig. Option I was used unless otherwise stated. its magnitude is close to ∆x2 which confirms that the von K´arm´an length scale is related to the smallest resolvable scales.3D 2k 1 ∂ω ∂ω 1 ∂k ∂k T2 = max .3D i increases slightly whereas ∆x2 continues to decrease. = = ∂x2 j+1/2 ∆x2 ∂x2 j−1/2 ∆x2 and then ⇒ ∂2v ∂x22 = j (∆x2 )2 ∂ 4 v vj+1 − 2vj + vj−1 + (∆x2 )2 12 ∂x42 Option II: compute the first derivatives at the center ∂v vj+2 − vj vj − vj−2 ∂v . σΦ ω 2 ∂xj ∂xj k 2 ∂xj ∂xj (22. 22. T2 is included to make sure that the model in steady flow works as a k − ω SST model.4 Evaluation of the von K´arm´an length scale in channel flow In Fig.2. hLvK. 22. For x2 > 0.5) k 1/2 L= 1/4 ωcµ Note that the term T1 is the “real” additional SAS term. This implies that the smallest scales that can be resolved are related to the grid scale. hLvK. 22. The second derivative of the velocity 198 When incorporating the additional production term (Eq. based on the averaged velocity profile h¯ v1 i = h¯ v1 i(x2 ) is also included in Fig.22.1D .2. . 0) L T1 = ζ˜2 κS 2 LvK. The von K´arm´an length scale.4) in the k − ω-SST model. LvK. When using DNS data only viscous dissipation of resolved turbulence affects the equations. see [143]) PSAS = FSAS max (T1 − T2 . Closer to the wall. and as can be seen it is much larger than hLvK. the last term in the ω equation is replaced by (for further details.3D i.3 The second derivative of the velocity To compute U ′′ in Eq. = = ∂x2 j+1 2∆x2 ∂x2 j−1 2∆x2 and then ⇒ ∂2v ∂x22 j = vj+2 − 2vj + vj−2 (∆x2 )2 ∂ 4 v + 4(∆x2 )2 3 ∂x42 In [144].3D i. 963 mesh. : hLvK.04 0.4 0. : (∆x1 ∆x2 ∆x3 )1/3 .04 0.02 0.25 0 0. Left: global view.2 0.1D . : LvK.8 1 0 0 x2 /δ x2 /δ Figure 22. Reτ = 2000. : hLvK. ◦: x2 -stretching of 9%.15 0. x2 -stretching of 17%.1 20 40 0.03 0.01 0. ∆x2 . 32 × 64 × 32 mesh.065.016.4 0. and as a results hLvK.39. In Fig.01 0.3D i.08 0. hLvK. from a 1D RANS simulation at Reτ = . ◦: ∆x2 .1D .031δ). ∆x1 /δ = 0.6 x2 /δ 0. When using hybrid LES-RANS.3D i follows closely LvK.05 0 0 0.02 15 20 25 0.5 (∆x1 ∆x2 ∆x3 ) . No such behavior is seen for the three-dimensional formulation.08 0. ∆x3 /δ = 0.1 0.1 0. ℓk−ω . Hybrid LESRANS. The resolved turbulence is dissipated by a modeled dissipation.3D i. ∆x3 /δ = 0.22.2 0. ∆x1 /δ = 0.25 0. As a result.02 0.8 1 0 0 x2 /δ Figure 22.3 0. ∂ 2 h¯ v1 i/∂x22 .2 0.04 0.1D .03 0. −2hνT s¯ij s¯ij i (νT denotes SGS or RANS turbulent viscosity). part of the turbulence is resolved and part of the turbulence is modeled. Close to the wall in the URANS region (x2 < 0. : LvK.2: Turbulent length scales in fully developed channel flow. DNS. Left: global view. The RANS turbulent length scale. data from hybrid LES-RANS are used (taken from [84]). : 1/4 1/3 0.3.05 0 0 0. the length scale of the smallest resolved turbulence is larger in hybrid LES-RANS than in DNS. right: zoom.04 0. and νT ≫ ν.19.06 0.05 0.3D i. Reτ = 500. Evaluation of the von K´arm´an length scale in channel flow 0 100 200 x+ 2 300 400 500 0 0. the resolved turbulence is dampened by the high turbulent viscosity. goes to zero as the wall is approached.06 0.02 60 80 100 0. right: zoom. x+ x+ 2 2 0 500 1500 2000 0.4. 22.15 0.05 0.3: Turbulent length scales in fully developed channel flow.2 0. +: ℓk−ω = k /(cµ ω).6 0.1 199 x+ 2 5 10 0. ∆x2 < (∆x1 ∆x2 ∆x3 )1/3 near the wall. In the hybrid simulations. LvK. (∆x1 ∆x3 ∆x3 )1/3 > 2hLvK. However.1D .3.4.1D . it can be noted that the three-dimensional filter width is more that twice as large as the three-dimensional formulation of the von K´arm´an length scale. goes to zero because the velocity derivative goes to zero.e.5δ). 22. the von K´arm´an length scale. LvK. In the inner region (x2 < 0. its behavior is close to that of the von K´arm´an length scale. In the center region the RANS turbulent length scale continues to increase which is physically correct. flow in an asymmetric diffusor and flow over an axi-symmetric hill. whereas far from the wall ∆x2 > (∆x1 ∆x2 ∆x3 )1/3 because of the stretching in the x2 direction and because of small ∆x1 and ∆x3 .3D i. i. the SST-SAS model has been evaluated in channel flow. Two filter scales are included in Figs. . Evaluation of the von K´arm´an length scale in channel flow 200 2000 with the k − ω SST model is also included in Fig.22.2 and 22. 22. In [144].3. In the DNS-simulations. The PANS Model 201 23 The PANS Model The PANS method uses the so-called “partial averaging” concept. The left side can be re-written ∂k ∂ku ∂k ∂ku ∂ku ∂ku ∂ku fk = = + (V¯j − v¯j ) + V¯j + V¯j + v¯j ∂t ∂xj ∂t ∂xj ∂t ∂xj ∂xj (23. (for simplicity we omit the buoyancy term) ∂ νt ∂k ∂k ∂k k ¯ = fk P − ε + ν+ + Vj fk (23. namely f¯ = P(F ). where vi indicates instantaneous velocity components. i.2) where τij is the central second moment resulting from the partial averaging for the nonlinear terms.e. fk and fε . In order to formulate the PANS eddy viscosity. where P denotes the partial-averaging operator. The last term on the right side in Eq. We consider incompressible flows.5) The convective term must be expressed in v¯j (the PANS averaged velocity) rather than in V¯j (the RANS averaged velocity).5 is usually neglected. we use f¯ to denote the partially-averaged part.2.23. where s¯ij is the strain-rate tensor of the computed flow and νu is the PANS eddy viscosity. 23. that is τij = (P(vi vj ) − v¯i v¯j ). 11.3) The extent of the resolved part is now determined by fk and fε .95) in the k − ε model by fk . Usually fε = 1. relating the unresolved to the resolved fluctuating scales. because it is v¯j that transports ku because v¯j represents the PANS resolved part of vj . 146] they employed the standard k − ε model as the base model. and fε is the ratio between the unresolved (εu ) and the total (ε) dissipation rates. which corresponds to a filtering operation for a portion of the fluctuating scales [145].4) ∂t ∂xj ∂xj σk ∂xj where Vi denotes the RANS velocity. This term is similar to the Reynolds stress tensor resulting from the Reynolds averaging in RANS or to the subgrid-scale (SGS) stress tensor after the spatial filtering in LES. have been introduced. To close the system of the partially-averaged Navier-Stokes equations. the partially-averaged turbulent kinetic energy. as in RANS and LES. Applying the partial averaging to the governing equations gives ∂¯ vi =0 ∂xi ∂¯ vi 1 ∂ p¯ ∂ ∂(¯ vi v¯j ) =− + + ∂t ∂xj ρ ∂xi ∂xj (23. For an instantaneous flow variable. F .1) ν ∂¯ vi − τij ∂xj (23. In [145. so that νu = Cµ ku2 /εu . These give k= εu ku and ε = fk fε (23. 23. a model is needed for τij . fε < 1 implies that (large) dissipative scales are resolved. Parameter fk defines the ratio of unresolved (partially-averaged) turbulent kinetic energy (ku ) to the total kinetic energy (k). ku and its dissipation rate εu . . two parameters. they defined in [145] another two quantities. The ku equation is derived by multiplying the RANS k equation (Eq. For simplicity. we also use the terminology of Reynolds stresses for the term τij in Eq. In the derivation of the transport equations for ku and εu . In [145] they proposed using the conventional eddy viscosity concept so that τij = −2νu s¯ij . and the strain rate of PANS-resolved flow field.10) ∂t ∂xj ∂xj σku ∂xj where the production term.e.e. 23.3 νt ∂ νt ∂εu ∂ε ∂ ν+ = ν+ fε ∂xj σε ∂xj ∂xj σε ∂xj ∂ νu ∂εu = ν+ ∂xj σεu ∂xj where σεku = σε k fk2 fε (23. ∂εu ∂ε ∂(εV¯j ) ∂(εu v¯j ) = fε + + ∂t ∂xj ∂t ∂xj ε νt ε2 ∂ε ∂ ν+ + Cε1 Pk − Cε2 = fε ∂xj σε ∂xj k k νu = c µ The diffusion term is re-written using Eq.8) fk P k − ε = Pu − εu This relation implies Pk = 1 εu (Pu − εu ) + fk fε (23. 23. is expressed in terms of the PANS eddy viscosity.6) (23.e. 23.23.14) (23.9) Using Eqs.3 νt ∂ νt ∂ku ∂k ∂ ν+ = ν+ fk ∂xj σk ∂xj ∂xj σk ∂xj ∂ νu ∂ku = ν+ ∂xj σku ∂xj where σku = σk fk2 fε (23. ∂¯ vj ∂¯ vi ∂¯ vi + Pu = νu (23.7) The sum of the source terms in Eq. i. i. i. (23.6 and 23.5.12) (23. The PANS Model 202 The diffusion term is re-written using Eq.4 must be equal to the sum of the source terms of the ku equation.15) .11) ∂xj ∂xi ∂xj where ku2 ε The εu equation is derived by multiplying the RANS ε equation by fε . νu .8 the final transport equation for ku can now be written as ∂ku νu ∂ku ∂ ∂(ku v¯j ) ν+ + Pu − εu = + (23. 23. Pu .13) (23. 23. ku . The PANS Model 203 In the same way.7. see Eqs. 23.4.10.9) εu fk 1 ε2 fk ε ε2 εu = Cε1 − Cε2 u (Pu − εu ) + fε Cε1 Pk − Cε2 k k ku fk fε fε ku 2 2 2 εu ε ε fk ε fk = Cε1 Pu − Cε1 u + Cε1 u − Cε2 u (23. is actually formulated in terms of the RANS turbulent viscosity from the base model. By referring to Eqs.20.17) (23. In other models. 23. 23.16) ku ku ku fε fε ku 2 εu ∗ εu = Cε1 Pu − Cε2 ku ku where ∗ Cε2 = Cε1 + fk (Cε2 − Cε1 ) fε The εu equation in the PANS model now takes the following form 2 ∂εu εu νu ∂εu ∂ ∂(εu v¯j ) ∗ εu ν+ + Cε1 Pu = − Cε2 + ∂t ∂xj ∂xj σεu ∂xj ku ku (23. but not the diffusion term. 23. is a unique property of the PANS model. the RANS) turbulent viscosity has been used for ku .19(a) suggests that the turbulent transport for the PANS-modeled turbulent kinetic energy. the the additional term (V¯j −¯ vj )∂εu /∂xj has been neglected.23. whereas on the right-hand side of Eq.19(a) suggests that. was derived by multiplying the RANS equation for k by fk which was assumed to be constant in space and in time. for improved modeling of near-wall turbulence. ε or ω equation is modified. The modification of the diffusion coefficient. the modeled turbulent diffusion in the PANS formulation is a factor of σk /σku = fε /fk2 larger than in Eq. which reads νsgs ∂ksgs ∂ (23.e.19(a) the total (i. A Low Reynolds number PANS model was presented in [148]. for example.19b) = ∂xj σku ∂xj The expression on the right-hand side of Eq. Eq. such as DES [147].20) ∂xj σk ∂xj In Eq.19(b). 23. the turbulent diffusion term was obtained as ∂ ∂ νt ∂k νt ∂ku fk = (23. when used as an SGS model.12 and 23. The PANS equation for ku . 23. 23.20 the SGS turbulent viscosity is invoked for the transport of ksgs . With fε = 1 and fk = 0. This is different from the turbulent diffusion in subgrid scale (SGS) modeling of LES with a one-equation ksgs model. this factor is larger than six. The only difference between PANS and PITM is that in the former model the diffusion coefficients in the k and ε are modified. the production and destruction terms are re-formulated as (using Eqs.10 and 23.3 and 23.6. X-LES [126] and PITM [130]. the sink term in the k. Equation 23. 23. . σku .19a) ∂xj σk ∂xj ∂xj σk ∂xj νu ∂ku ∂ (23. A recently developed LRN PANS model is employed. 23.18) As in the ku equation. σk = 1.09 (23. i.e.5.9.21) fk2 f2 .4. Cε2 = Cε1 + εu fε (23. σε = 1. Cε2 = 1.22) . σεu ≡ σε k fε fε The modifications introduced by the PANS modeling as compared to its parent RANS model are highlighted by boxes. The model constants take the same values as in the LRN model [131].23. The PANS Model 204 which reads [129] νu ∂ku ∂ ∂(ku v¯j ) ∂ku ν+ + (Pu − εu ) = + ∂t ∂xj ∂xj σku ∂xj 2 ∂εu εu νu ∂εu ∂ ∂(εu v¯j ) ∗ εu ν+ + Cε1 Pu = − Cε2 + ∂t ∂xj ∂xj σεu ∂xj ku ku νu = Cµ fµ σku ≡ σk ku2 ∗ fk (Cε2 f2 − Cε1 ) . Cµ = 0. Cε1 = 1.4. 24.24.sgs dt ksgs ksgs 3 Tsgs dLt 3 T dLt ksgs Cε2.3) Differentiation of Eq.6) 24. Eq. 24.4 k 3/2 dLt 3 k 1/2 k dε (P − ε) − c 2 =c dt 2 Lt Lt dt and using Eq. represents scales with wavelength larger that the cut-off.2) (24. 24. The production P k is replaced by Pksgs = εsgs (see Fig. The dissipation can be estimated as ε=c k 3/2 Lt (24.4) Inserting Eq.7) .3 gives 3ε k ε dLt 3 Pk − ε T dLt ε2 dε = (P − ε) − = − dt 2k Lt dt 2 T Lt dt k (24. 24. κc .1 RANS mode Consider homogeneous turbulence. e. The PITM model 205 24 The PITM model PITM is an acronym for Partially Integrated Transport Model [130. ksgs .5. 18. T . the modeled turbulent kinetic energy.sgs = Cε1.3 gives 3 k 1/2 dk k 3/2 dLt dε =c −c 2 dt 2 Lt dt Lt dt (24. Now.2. Comparing Eqs. In an SGS model. see Section 3.5) where T = k/ε. decay of grid generated turbulence.g.5 can be written as ε2 εεsgs dε − Cε2.2 we find that Cε1 = 1.5 and 24. The time scale.4 in [48] and Fig.sgs = + = + 2 Lt dt 2 Lt dt ktot (24.5 and Cε2 = Cε1 + T dLt Lt dt (24. k represents all turbulence.1 into Eq.5). 18. 24. see Fig. is replaced by Tsgs = ksgs /ε since the time scale of the small scales is different (it is smaller) than that of the large scales. 11. 24. The viscous dissipation is the same in RANS and LES. 24. The k and ε equations can for this flow be simplified as dk = Pk − ε dt (24. 149].1) dε ε = (cε1 P k − cε2 ε) dt k where d/dt = v¯1 d/dx. In the RANS model.2 LES mode Now let’s transform the RANS k − ε model above to an SGS k − ε model. Homogeneous turbulence. LES mode E 206 Fc Ec ε ε κ−5/3 dκ κc κd κ Figure 24.11.e. When the cut-off does not move in time (which implies that dksgs /dt).12 we get dksgs dκc = Fc − Ec −ε (24.e. Let’s introduce a new dissipative length scale of wavenumber κd which is much larger than κc ε κd − κc = ξ 3/2 (24. ∞]. resolved and SGS). at the cut-off.8) ksgs The constant is very large since κc ≪ κd .10 and 24. 24. note that κ ≃ ∞. When the cut-off does not move we get simply Fc = ε (24. is dκc (24. εsgs .1.24. Taking the time derivative of Eq. in decaying homogeneous grid turbulence.1: Spectral energy balance in PITM. see Eq. 24. because Eq.11) When κc is time dependent as. 24. 24.18. for example. Recall that k is the area below E in Fig. κ− c ] into [κ+ .10) where Fc = F (κc ) denotes the spectral transfer from wavenumber range [0. 24. i. thus we can write κd ≃ ξ ε 3/2 ksgs (24.3 is still a valid estimate ε.13) dt dt Comparing Eq.1. 18. where ktot denotes the total turbulence (i. see d sgs c Eq. we can set-up the balance of ksgs when κc moves dκc during the time interval dt.12) where Ec dκ denotes the spectral energy in the slice dκc . 24. 24.14) εsgs = Fc − Ec dt .2. Note that L is not replaced by Lsgs .1.13 with Eq. then Fc = εsgs = Pksgs (24. κc . we find the spectral energy transfer.9) Consider spectral balance of ksgs in homogeneous turbulence. see Fig. dksgs = Fc dt − Ec dκc − εdt (24. 17) − ksgs ε ξ 3/2 2 dt ksgs dt With Eq. Now derivate Eq.2.9 with respect to time and insert Eq.20) This equation can now be written dε ε2 3 3 ksgs Fd ε εsgs − = − −1 dt 2 ksgs ksgs 2 κd Ed ε |{z} {z } | Cε1.sgs = − Cε2 − −1 = + 2 ktot κd Ed ε 2 ktot 2 (24. i. 24.15) where ε is the energy transfer rate across κd .sgs → Cε1 Cε2.24. 24.16) dt Ed ksgs It may be noted that both the nominator and denominator of the right side go to zero as κd → ∞. 24.23 in the expression for Cε2.21 gives 3 ksgs ktot 3 3 ksgs Fd Cε2. 24.22) Fd −1 ε (24. then ksgs → ktot so that Cε1. 24. 24.sgs (24.24) ksgs = Cε1 + (Cε2 − Cε1 ) ktot . ε = Fd − Ed dκd dt (24.e.15 ! d ε Fd − ε ξ 3/2 = (24.16 with Eq.23) It may be noted that both the term in the parenthesis and Ed go to zero as κd → ∞. 24.21) Cε2.14 at wavenumber κd .14 and then Eq. The left side is re-written as # " 1 dε 3 −5/2 dksgs (24.13 we get " # 1 dε 3 −5/2 dκc ξ 3/2 − ksgs ε Fc − Ec −ε 2 dt ksgs dt Replacing the left side of Eq.sgs When κc → 0 (RANS). 24. 24. Inserting Eq.9) dε 3 ε Fd − ε 3 ε 3/2 Fd − ε + (εsgs − ε) = ε + (εsgs − ε) = ksgs dt ξEd 2 ksgs κd Ed 2 ksgs (24. 24.sgs in Eq.sgs → Cε2 = 3 ktot − 2 κd Ed (24.18 gives " # 1 dε 3 −5/2 dκc Fd − ε ξ 3/2 − ksgs ε Fc − Ec −ε = 2 dt Ed ksgs dt (24.19) We get (using first Eq.18) (24. LES mode 207 We can write a similar equation as Eq. 24 is identical to the expression for Cε2 pative scales are resolved so that fε = 1).24. Cε2. see Eq.sgs = Cε1 + fk (Cε2 − Cε1 ) (24.21. see Eq. LES mode 208 When ksgs /ktot → 0 (only a small part of the turbulent kinetic energy is modeled).17 (provided that no dissiEq. Indeed. Then Cε2. 23.2. when ksgs /ktot → 1 (everything is modeled).sgs . Cε2. the PITM model is in LES mode.e.sgs is reduced which increases ε and reduces the modeled turbulent kinetic energy and the turbulent viscosity. It should be noted that this model is very similar to the PANS model. 0.sgs → Cε1 .25) In both models the Cε2 coefficients are modified. i. a usual RANS k − ε model). The only difference between the PANS and PITM models is that the diffusion coefficients in the ku and εu are modified in the former model.42 1 + γ(Lt /∆)2/3 . 23. Instead they compute it using the ratio of the integral and the modeled lengthscales as Cε2 = Cε1 + 3/2 where Lt = ktot /ε and γ = 0. i. ∗ .42. In [130] they do not use ksgs /ktot to compute Cε2. then the PITM model is in RANS mode (i. 24.e.e. On the other hand. 1 Introduction Fluid flow problems are governed by the Navier-Stokes equations 1 ∂p ∂ 2 vi ∂vi vj ∂vi =− +ν + ∂t ∂xj ρ ∂xi ∂xj ∂xj (25. the velocity vector and the pressure are split into a time-averaged part (hvi i and hpi) and a fluctuating part (vi′ and p′ ). νt . i. in case of a boundary layer). 25. In turbulent flow. All turbulent fluctuation are modeled with a turbulence model and the results when solving Eq. CFD (Computationally Fluid Dynamics) based on finite volume methods is used extensively to solve the RANS equations. The resulting equation is called the RANS (Reynolds-Averaging Navier-Stokes) equations ∂hvi′ vj′ i 1 ∂hpi ∂ 2 hvi i ∂hvi ihvj i =− +ν − ∂xj ρ ∂xi ∂xj ∂xj ∂xj 1 ∂hpi ∂hvi i ∂ =− (ν + νt ) + ρ ∂xi ∂xj ∂xj (25. often called eddies.2 the unknown Reynolds stresses are expressed by a turbulence model in which a new unknown variable is introduced which is called the turbulent viscosity. the velocity and pressure are unsteady and vi and p include all turbulent motions.1. On the right side of Eq.1 Reynolds-Averaging Navier-Stokes equations: RANS In order to be able to solve the Navier-Stokes equations with a reasonable computational cost. respectively.2 are highly dependent on the accuracy of the turbulence model. This has the unfortunate consequence – unless one is a fan of huge computer centers – that it is computationally extremely expensive to solve the Navier-Stokes equations for large Reynolds numbers. 25. The smallest scales are related to the eddies where dissipation takes place.1.2. The ratio of the largest to the smallest eddies increases with Reynolds number. 25.e. Re = |vi |δ/ν. Hybrid LES/RANS for Dummies 209 25 Hybrid LES/RANS for Dummies 25.2) The last term on the first line is called the Reynolds stress and it is unknown and must be modeled. i.e. The LES equations read 1 ∂ p¯ ∂ 2 v¯i ∂τij ∂¯ vi v¯j ∂¯ vi =− +ν − + ∂t ∂xj ρ ∂xi ∂xj ∂xj ∂xj 1 ∂ p¯ ∂¯ vi ∂ =− (ν + νsgs ) + ρ ∂xi ∂xj ∂xj (25. 25. The spatial scale of these eddies vary widely in magnitude where the largest eddies are proportional to the size of the largest physical length (for example the boundary layer thickness. The ratio of νt to ν may be of the order of 1000 or larger.3) . p = hpi + p′ . In industry today. p is the pressure and ν and ρ are the viscosity and density of the fluid. Eq. 25. δ. vi = hvi i + vi′ .2 Large Eddy Simulations: LES A method more accurate than RANS is LES (Large Eddy Simulations) in which only the small eddies (fluctuations whose eddies are smaller than the computational cell) are modeled with a turbulence model. where the kinetic energy of the eddies is transformed into internal energy causing increased temperature.25.1) where vi denotes the velocity vector. the spatial scales of the “large” turbulent eddies which should be resolved by LES are in reality rather small.e. we have describe how to use the zonal LES/RANS method for flows near walls. One example is prediction of aeroacoustic noise created by the turbulence around an external mirror on a vehicle [100].25. 25. Hence. The reason is entirely due to physics: near the walls. i. The difference of νsgs compared to νt in Eq. This term must also – as in Eq.2 and 25.2 is that it includes only the effect of the small eddies.2 The PANS k − ε turbulence model In the present work. see Fig. The ratio of νsgs to ν is of the order of 1 to 100. When the flow near walls is of importance.2 – be modeled.3 Zonal LES/RANS Equations 25. see Fig. 25.e. The last term on the first line includes the Reynolds stresses of the small eddies.e. 25.4. νsgs . These methods are called Detached Eddy Simulation (DES). 25.4 we will present simulations using embedded LES in a simplified configuration represented by the flow in a channel in which RANS is used upstream of the interface and LES is used downstream of it. and at the second line it has been modeled with a SGS turbulent viscosity.e. a RANS turbulence model is used for the turbulent viscosity. Furthermore. the PANS k − ε model is used to simulate wall-bounded flow at high Reynolds number as well as embedded LES. which are called SGS (sub-grid stresses). |¯ vi′ v¯j′ |/|τij | (see Eqs. but in order to predict the noise in the region of the external mirror we must predict the large turbulence fluctuations and hence LES must be used in this region. The turbulence model reads [128.4) Near the walls.2.3 can be written in a same form as ∂¯ vi ∂¯ vi 1 ∂ p¯ ∂ ∂¯ vi v¯j (ν + νT ) =− + + ∂t ∂xj ρ ∂xi ∂xj ∂xj (25. Another form of zonal LES/RANS is embedded LES. because the large. νT = νsgs . Above. is turns out that LES is prohibitively expensive because very fine cells must be used there. the ratio of the resolved to the modeled turbulence. νT = νt and away from the walls an LES turbulence model is employed.2 and 25. i. hybrid LES/RANS or zonal LES/RANS. in which an LES region is embedded in a RANS region. The disadvantage of LES is that it is much more expensive than RANS because a finer mesh must be used and because the equations are solved in four dimensions (time and three spatial directions) whereas RANS can be solved in steady state (no time dependence). The PANS k − ε turbulence model 210 Note that the time dependence term (the first term on the left side of the first line) has been retained. .3) is much larger than one. 25. Note that the time dependence term is now retained also in the RANS region: near the wall we are using an unsteady RANS. All proposed methods combines RANS and LES where RANS is used near walls and LES is used some distance away from the walls. URANS.1. However. their spatial scales get smaller for increasing Reynolds number. 25. the resolved) fluctuations are part of v¯i and p¯ and are not modeled with the turbulence model. In Section 25. i. time dependent turbulent (i. The focus here is zonal LES/RANS. Much research has the last ten years been carried out to circumvent this problem. LES is much more accurate than RANS because only a small part of the turbulence is modeled with the turbulence SGS model whereas in RANS all turbulence is modeled. 25. The flow around the vehicle can be computed with RANS.1. 23. the turbulence model in Eqs. Cε2 = 1. fk = 1.09 (25. When fk is ∗ decreased (to 0. 25.8 decreases because k decreases and ε increases.8 acts as a RANS turbulence model (large turbulent viscosity) when fk = 1 and it acts as an LES SGS turbulence model (small turbulent viscosity) when fk = 0.5–25. 25. see Eq.5.4. the turbulent viscosity νT should be an SGS viscosity and in the latter region it should be an RANS viscosity.1 The interface conditions The interface plane (see Fig. 25.7 decreases.1) separates the URANS regions near the walls and the LES region in the core region.4 in the present study). In the former region. fk = 0. In the LES region fk = 0.2 Figure 25.3.7) k2 .3.4 and in the URANS region fk = 1. Cε1 = 1. 25.5) (25. This is achieved by setting the usual convection and diffusion fluxes of k at the interface to zero. and • νT in Eq.1: The LES and URANS regions. As a result νT = Cµ • ε increases because the destruction term (last term in Eq.5) is the main sink term in the k equation increases. Fully developed channel flow. Periodic boundary conditions are applied at the left and right boundaries. the model acts as a standard k − ε RANS model giving a large turbulent viscosity. 129]. The key elements in the present use of the PANS k − ε model are highlighted in red. New fluxes are introduced using smaller SGS values [150].0 x1 3.6 which is the main sink term) in the ε equation decreases.4 x2 interface URANS. When fk in Eq.21 (here in a slightly simplified form to enhance readability) ∂k ∂k¯ νT ∂ vj ∂k ν+ + Pk − ε = + ∂t ∂xj ∂xj σk ∂xj ∂ε ∂ε¯ ∂ vj = + ∂t ∂xj ∂xj 2 ε νT ∂ε ∗ ε ν+ + Cε1 Pk − Cε2 σε ∂xj k k ∗ Cε2 = Cε1 + fk (Cε2 − Cε1 ). fk = 1.25.0 interface 2 LES. 25. Zonal LES/RANS: wall modeling 211 URANS. 25.6) (25. 25.3 Zonal LES/RANS: wall modeling 25. Hence νT must decrease rapidly when going from the URANS region to the LES region. 25.8) ε Note that k and ε are always positive. Hence. Cµ = 0. Cε2 in Eq. • k decreases because ε (last term in Eq.9 (25. .7 is equal to one. 6 (δ = uτ = 1).2 (b) Resolved shear stresses Figure 25.1 The interface conditions The interface plane is now vertical. : Reτ = 8 000. More detailed results can be found in [150]. Zonal LES/RANS: embedded LES 212 0 30 −0. Hence.2. 25.1. 25. 25.3. 25. Anisotropic synthetic turbulent fluctuations are used [151.05 0.3 for three different resolutions in the x1 − x3 plane. The grid in the x2 direction varies between 80 and 128 cells depending on Reynolds number.2b) increase. (Nx × Nz ) = (64 × 64) Reτ = 4 000. As can be seen. the predicted velocity profiles are in good agreement with the log-law which represents experiments.max = 1. The size of the domain is x1.2.max = 3. The turbulent viscosity profiles are shown in Fig. The interface is set to x+ 2 ≃ 500 for all grids. 16 000 and 32 000. It is interesting to note that the turbulent viscosity is not affected by the grid resolution. 25. ////: Reτ = 32 000. 8 000. 25.4 15 −0. The interface conditions for k and ε are treated in the same way as in Section 25.2b presents the resolved shear stresses. : 25.3) to trigger the flow into turbulence-resolving mode.max = 2 and x3. The baseline mesh has 64 × 64 cells in the streamwise (x1 ) and spanwise (x3 ) directions. The interface is shown by thick dashed lines and it moves towards the wall for increasing Reynolds number since it is located at x+ 2 ≃ 500 for all Reynolds numbers.4.4 Zonal LES/RANS: embedded LES 25. .8 5 0 1 100 1000 30000 −1 0 x+ 2 (a) Velocities 0. the model yields grid independent results contrary to other LES/RANS models.15 0.1 x2 0. see Fig.3. The turbulent viscosity (Fig.2 Results Fully developed channel flow is computed for Reynolds numbers Reτ = uτ δ/ν = 4 000. : Reτ = 16 000.6 10 −0.2 hv1′ v2′ i+ 25 U+ 20 −0.4.3) is sharply reduced when going across the interface from the URANS region to the LES region and the resolved fluctuations (the Reynolds shear stress in Fig. The difference is now that “inlet” turbulent fluctuations must be added to the LES v¯i equations (Eq. This shows that the model is switching from RANS mode to LES mode as it should. Figure 25.2: Velocities and resolved shear stresses. 152]. The velocity profiles and the resolved shear stresses are presented in Fig.25. x2.4. 25. 2 0.5: Channel flow with inlet and outlet.25. fk = 1. Interface 2 URANS. fk = 0.015 0.3 (a) Reτ = 4000. hνT /νimax (maxx2 hu′1 u′1 i) 25 0 100 1/2 30 U+ νT /(uτ δ) 25.8 (b) (Nx × Nz ) = (64 × 64).01 νT /(uτ δ) 0. : (Nx × Nz ) = (32 × 32).25 0.2 0.1 0. Zonal LES/RANS: embedded LES . ////: Reτ = 32 000. : (Nx × Nz ) = (64 × 64).15 x2 0. 1 : Reτ = : Reτ = Figure 25.19.4: The LES and URANS regions. (b) maximum resolved streamwise turbulent fluctuations and turbulent viscosity versus x1 .4.5 : Turbulent viscosity (right x2 axis).4 1 2.005 0 0 0. The log-law is plotted in 1. (a) Velocities. x1 = x1 = 3. The left boundary is an inlet and the right boundary is an outlet. symbols.213 0.05 0.3: Turbulent viscosity. : Reτ = 8 000. 4 000.5 3 0 3.6 0. x1 = 0. (b) 50 0.2 x2 x1 Figure 25. 4 20 2 15 10 5 0 10 1 0 0 2 10 10 x+ 2 (a) Velocities.5 1 1. : (Nx × Nz ) = (128 × 128) 0.5 x1 2 2.015 0.0 LES. : maximum streamwise fluctuations (left x2 axis) Figure 25.005 0 0 0. 16 000.4 x2 0.01 0. 2 × 2 × 1. With a 3.4.e.25.2 Results The Reynolds number for the channel flow is Reτ = 950. At x1 = 3. The resolved streamwise velocity fluctuations are zero in the RANS region.5b). Inlet conditions at x = 0 are created by computing fully developed channel flow with the PANS k − ε model in RANS mode (i. Figure 25.4. and the maximum resolved values increase sharply over the interface thanks to the imposed synthetic turbulent “inlet” fluctuations. 25. a mesh with 64 × 80 × 64 cells is used in the streamwise (x1 ). Zonal LES/RANS: embedded LES 214 25.4. 1. the wall-normal (x2 ) and the spanwise (x3 ) direction. x1 = 0. with fk = 1). 25. as they should (Fig. Hence. The results are presented in more detail in [150]. the predicted velocity agrees very well with the experimental log-law profile.5a presents the mean velocity and the resolved shear stresses at three streamwise locations. The turbulent viscosity is reduced at the interface from its peak RANS value of approximately 80 to a small LES value of approximately one (these values are both fairly low because of the low Reynolds number).25 and 3 (recall that the interface is located at x1 = 1). it is seen that the present model successfully switches from RANS to LES across the interface. see Fig.6 domain.19. . The new coefficient. 155.2 Random angles The angles ϕn and θn determine the direction of the wavenumber vector κ. The synthesized turbulence at one time step is generated as follows. D. are related to an and bn as 1/2 bn αn = arctan cn = a2n + b2n (26.g. However. and the phase angle. 157. k and ε). 26.17. see Section 5. 156] for creating turbulence for generating noise. Inlet boundary conditions 215 26 Inlet boundary conditions In RANS it is sufficient to supply profiles of the mean quantities such as velocity and temperature plus the turbulent quantities (e. see Appendix J. cn . 26. resolved fluctuations. in unsteady simulations (LES.2) an A general form for a turbulent velocity field can thus be written as v′ (x) = 2 N X n=1 u ˆn cos(κn · x + ψ n )σ n (26. URANS. the time history corresponds to turbulent. 26. phase and direction of Fourier mode n. ψ n and σin are the amplitude. The eddies are generated randomly in the inflow plane and then convected through it.1. Another method based partly on synthesized fluctuations is the vortex method [154]. . . The method is able to reproduce first and second-order statistics as well as two-point correlations. A third method is to take resolved fluctuations at a plane downstream of the inlet plane. αn denotes the direction of the velocity vector. bn = cn sin(αn ). It is based on a superposition of coherent eddies where each eddy is described by a shape function that is localized in space. see Eq.3 and Eq.1 Synthesized turbulence The method described below was developed in [84. One way is to use a pre-cursor DNS or well resolved LES of channel flow. ) the time history of the velocity and temperature need to be prescribed. 158]. For more details. In some flows it is critical to prescribe reasonable turbulent fluctuations. αn . 153]. re-scale them and use them as inlet fluctuations.3 and Eq. . This method is limited to fairly low Reynolds numbers and it is difficult (or impossible) to re-scale the DNS fluctuations to higher Reynolds numbers. but in many flows it seems to be sufficient to prescribe constant (in time) profiles [114. Below we present a method of generating synthesized inlet fluctuations.26. A turbulent fluctuating velocity fluctuating field (whose average is zero) can be expressed using a Fourier series. There are different ways to create turbulent inlet boundary conditions. It was later further developed for inlet boundary conditions [151. DES .3) where u ˆn . 26. Let us re-write this formula as an cos(nx) + bn sin(nx) = cn cos(αn ) cos(nx) + cn sin(αn ) sin(nx) = cn cos(nx − αn ) (26.1) where an = cn cos(α) . v′ . 3 Highest wave number Define the highest wave number based on mesh resolution κmax = 2π/(2∆) (see Section 18.26. ∆η = x2.max /N3 .e. κni .3. α = 1. Factor p should be larger than one to make the largest scales larger than those corresponding to κe .1δin is suitable. A value p = 2 is suitable. κmax − κ1 . In [151. The turbulent length scale.1: The wave-number vector.453. of size ∆κ. respectively. 158] it was found that Lt ≃ 0. The fluctuations are set to zero at the wall and are then interpolated to the inlet plane of the CFD grid (the x2 − x3 plane). Highest wave number 216 ξ2n σin αn x3 ξ3n ξ3n ξ1n ξ1n θ n κn i 0 x1 x2 ϕn Figure 26.5 Divide the wave number range Divide the wavenumber space. and the velocity unit vector. where ∆ is the grid spacing. 157. Often the smallest grid spacing near the wall is too small. and then a slightly larger values may be chosen. where η denotes the wall-normal direction and N2 and N3 denote the number of cells in the x2 and x3 direction. where κe = α9π/(55Lt). equally large. Lt . 26. into N modes. ∆x3 = x3. σin . may be estimated in the same way as in RANS simulations.max /N2 .5).4 Smallest wave number Define the smallest wave number from κ1 = κe /p. The fluctuations are generated on a grid with equidistant spacing (or on a weakly stretched mesh). i. are orthogonal (in physical space) for each wave number n. 26. Lt ∝ δ where δ denotes the inlet boundary layer thickness. . 26. Z ∞ k= E(κ)dκ (26. The code for generating the isotropic fluctuations can be downloaded here http://www. 26. They are independent of each other and their time correlation will thus be zero.7) 0 26.3 to be computed.chalmers. This is nonphysical.e.453 cE = √ π Γ(1/3) where Γ(z) = Z (26.4 and Fig. V3′ .26. v1′ =2 N X u ˆn cos(β n )σ1 n=1 v2′ = 2 N X u ˆn cos(β n )σ2 n=1 v3′ = 2 N X (26.5) 0 which gives [68] 4 Γ(17/6) ≃ 1. In this way inlet fluctuating velocity fields (v1′ .3 is then obtained from u ˆn = (E(κ)∆κ)1/2 E(κ) = cE 2 u2rms (κ/κe )4 e[−2(κ/κη ) ] κe [1 + (κ/κe )2 ]17/6 κ = (κi κi )1/2 . σin and ψ n .8) u ˆn cos(β n )σ3 n=1 β n = k1n x1 + k2n x2 + k3n x3 + ψ n where u ˆn is computed from Eq. see Eq.4) κη = ε1/4 ν −3/4 The coefficient cE is obtained by integrating the energy spectrum over all wavenumbers to get the turbulent kinetic energy. κnj .ht 26.8 Introducing time correlation A fluctuating velocity field is generated each time step as described above.4. i. 26.tfd. are . 26. V2′ . new fluctuating velocity fields.se/˜lada/projects/inlet-boundary-conditions/proright. von K´arm´an spectrum 217 26. 26. v2′ .e. i. allows the expression in Eq. (26.2. The amplitude u ˆn of each mode in Eq.6. V1′ .6 von K´arm´an spectrum A modified von K´arm´an spectrum is chosen. 26.7 Computing the fluctuations Having u ˆn . To create correlation in time. v3′ ) are created at the inlet x2 −x3 plane.6) ∞ ′ e−z xz−1 dz ′ (26. The inlet boundary conditions are prescribed as (we assume that the inlet is located at x1 = 0 and that the mean velocity is constant in the spanwise direction.in as [159] + x+ x2 2 ≤ 5 + + V1.5 which ensures that hV1′2 i = hv1′2 i (h·i denotes averaging). V2. The mean inlet profiles. t) ′ ′ ′ where v1. x3 . a RANS solution or from the law of the wall.in .in (x2 ) + v2. x2 .26. For more detail. t) = V2.in (x2 ) + u′1. are either taken from experimental data.in . see Section 26.8. V1.9 is a convenient way to prescribe the turbulent time scale of the fluctuations. The time correlation of will be equal to exp(−tˆ/Tint ) (26. x3 . V3.in (x2 .in = (V3′ )m (see Eq. x3 . if V2.12) ′ v3.8. t) v¯3 (0.in (x2 . 26.in (x2 .05 + 5 ln(x2 ) 5 < x+ (26.in .2: Modified von K´arm´an spectrum computed based on an asymmetric time filter (V1′ )m = a(V1′ )m−1 + b(v1′ )m (V2′ )m = a(V2′ )m−1 + b(v2′ )m (V3′ )m = a(V3′ )m−1 + (26. Introducing time correlation E(κ) κ1 218 κe E(κ) ∝ κ−5/3 E(κn ) ∆κn log(κ) κn Figure 26.10) where ∆t and Tint denote the computational time step and the integral time scale. for example. x3 .9).13) 2 < 30 1 + + ln(x ) + B x ≥ 30 2 2 κ .in = 0 we can estimate V1.in = (V2′ )m and v3. t) = V1. 26. v2.in = V3. x3 . The second coefficient is taken as b = (1 − a2 )0.11) where tˆ is the time separation and thus Eq. x2 . x3 . t) = V3. respectively. x3 ) v¯1 (0.9) b(v3′ )m where m denotes the time step number and a = exp(−∆t/Tint ) (26.in = (V1′ )m . x2 .in = −3. t) ′ v¯2 (0.in (x2 ) + (26. 26.8. Introducing time correlation 219 1 0.8 0.6 B(τ ) 0.4 0.2 0 0 0.2 0.4 0.6 τ 0.8 1 Figure 26.3: Auto correlation, B(τ ) = hv1′ (t)v1′ (t − τ )t (averaged over time, t). Eq. 26.11; : computed from synthetic data, (V1′ )m , see Eq. 26.9. : where κ = 0.4 and B = 5.2. The method to prescribed fluctuating inlet boundary conditions have been used for channel flow [151], for diffusor flow [153] as well as for the flow over a bump and an axisymmetric hill [160]. Equation 26.9 introduces a time correlation with an integral time scale Tint . In order to understand Eq. 26.10, Equation 26.9 is written for m = N . . . 1 where N denotes number of time steps ′ UN ′ UN −1 ′ UN −2 U3′ U2′ U1′ ′ ′ = aUN −1 + buN ′ ′ = aUN −2 + buN −1 ′ ′ = aUN −3 + buN −2 .. . (26.14) = = = aU2′ aU1′ aU0′ + + + bu′3 bu′2 bu′1 With U0′ = 0 we get U1′ = bu′1 U2′ = abu′1 + bu′2 U3′ = a2 bu′1 + abu′2 + bu′3 U4′ U5′ = = 3 a bu′1 a4 bu′1 + + 2 a bu′2 a3 bu′2 which can be written as ′ Um =b + + m X k=1 abu′3 + bu′4 a2 bu′3 + abu′4 am−k u′k (26.15) + bu′5 (26.16) 26.9. Anisotropic Synthetic Turbulent Fluctuations 220 using Eq. 26.10 with tˆ = t/Tint . Let’s show that hU ′2 i = hu′2 i. Equation 26.16 gives hU ′2 i = m m m 1 X ′ ′ b2 X m−k ′ X m−i ′ a ui Uk Uk = 2 a uk m m i=1 k=1 (26.17) k=1 Recall that the synthetic fields u′k and u′i are independent which means that hu′k u′i i = 0 for k 6= i. Hence hU ′2 i = m m k=1 k=1 X b2 X 2(m−k) ′ ′ a uk uk = hu′2 i(1 − a2 ) a2k m (26.18) The requirement is that (1 − a2 ) m X k=1 a2k = (1 − a2 ) m X ck = 1 (26.19) k=1 Pm The sum k=1 ck has for large m the value Sm = 1/(1 − c) = 1/(1 − a2 ) which shows that the requirement in Eq. 26.19 is satisfied. The time correlation between many time steps with time step separation m − n = r ≥ 0 reads hU ′ U ′ ir = n m X X b2 an−p u′p am−k u′k (m − r)n p=1 b2 = m−r k=1 m−r X k=1 a2(m−k) a−r u′2 k ′ ′ = hu u ia (26.20) −r using Eqs. 26.18 and 26.19 (assuming m ≫ r to obtain the second line and the last expression; many samples are needed to get good statistics). Inserting Eq. 26.10 into Eq. 26.20 we get the normalized autocorrelation hU ′ U ′ ir = exp(−tˆ/Tint ) B11 (tˆ)norm = hu′2 i (26.21) where tˆ = r∆t. The integral time scale can be computed from the autocorrelation as (see Eq. 10.11) Z ∞ Z ∞ norm ˆ ˆ exp(−tˆ/Tint )dtˆ B11 (t)dt = Tint = (26.22) 0 0 ∞ ˆ = −Tint exp(−t/Tint ) 0 = −Tint (0 − 1) = Tint This shows that the one-sided filter in Eq. 26.9 introduces an integral time scale Tint as prescribed by the definition of a in Eq. 26.10. The integral time scale can be taken as Tint = Lint /Ub using Taylor’s hypothesis, where Ub and Lint denote the inlet bulk velocity and the integral length scale, respectively. 26.9 Anisotropic Synthetic Turbulent Fluctuations Isotropic fluctuations are generated above. However, turbulence is generally anisotropic. The method for generating anisotropic turbulence is presented in [84,152,155,161] and below it is described in detail. The method can be summarized by the following steps. 26.9. Anisotropic Synthetic Turbulent Fluctuations 221 1. A Reynolds stress tensor, hvi′ vj′ i, is taken from DNS data for turbulent channel flow. Since the generated turbulence is homogeneous, it is sufficient to choose one location of the DNS data. The Reynolds stresses at y + ≃ 16 of the DNS channel data at Reτ = 590 [162] are used. Here hv1′ v1′ i – and hence the degree of anisotropy – is largest. The stress tensor reads 7.67 −0.662 0 hvi′ vj′ i = −0.662 0.32 0 (26.23) 0 0 1.50 e∗i , are computed for 2. The principal directions (the eigenvectors, see Fig. 26.4), ˆ ′ ′ the hvi vj i tensor, see Section 26.9.1. The eigenvalues are normalized so that their sum is equal to three. This ensures that the kinetic energy of the synthetic fluctuations does not change during transformation. 3. Isotropic synthetic fluctuations, u′i,iso , are generated in the principal directions of hu′i u′j i. 4. Now make the isotropic fluctuations non-isotropic according to hvi′ vj′ i. This is done by multiplying the isotropic synthetic fluctuations in the ˆ e∗i directions by √ (i) the square-root of the normalized eigenvalues, λ √ norm . In Section 26.9.2 this ∗n is done by multiplying the unit vector σi by λ(i) norm , see Eq. 26.27 and Fig. 26.5. This gives a new field of fluctuations q (1) ′ (u1 )aniso = λnorm (u′1 )iso q (2) (26.24) (u′2 )aniso = λnorm (u′2 )iso q (3) (u′3 )aniso = λnorm (u′3 )iso √ The wavenumber vector, κj , is divided by λ(i) to ensure that the (u′i )aniso velocity field is divergence free, i.e. s q 1 (j) ∗ ∗ σj κj = λnorm σj κj = σj κj = 0, (26.25) (j) λnorm see Eq. J.3. Note that there is still no correlation between the (u′i )aniso fluctuations, which means that the shear stresses are zero (for example, h(u′1 u′2 )aniso i = 0). 5. The (u′i )aniso fluctuations are transformed to the computational coordinate system, xi ; these anisotropic fluctuations are denoted (u′i )synt . The transformation reads (u′i )synt = Rij (u′j )aniso (26.26) where Rij is the transformation matrix which is defined be the eigenvectors, see Section 26.9.1. The (u′i )synt fluctuations are now used in Eq. 26.9 replacing ′ vi.in . The Reynolds stress tensor of the synthetic anisotropic fluctuations is now identical to the DNS Reynolds stress tensor, i.e. h(u′i u′j )synt i = hvi′ vj′ i 26.9. Anisotropic Synthetic Turbulent Fluctuations 222 6. Since the u′i,synt are homogeneous, the Reynolds stresses, hu′i,synt u′j,synt i, have constant values in the inlet plane. One can choose to scale the inlet fluctuations by a k profile taken from DNS, experiments or a RANS simulation. The Matlab code for generating isotropic or anisotropic fluctuations can be downloaded [163]. 26.9.1 Eigenvalues and eigenvectors The normalized eigenvalues and eigenvectors may conveniently be computed with Matlab (or the freeware Octave) as stress=[7.6684 -6.6206e-01 0; ... -6.6206e-01 3.1974e-01 0; ... 0 0 1.4997]; diag_sum=trace(stress)/3 stress=stress/diag_sum % ensures that the sum % of the eigenvalues=3 [R,lambda] = eig(stress) v_1_temp=[R(1,1);R(2,1); R(3,1)]; v_2_temp=[R(1,2);R(2,2); R(3,2)]; v_3_temp=[R(1,3);R(2,3); R(3,3)]; lambda_1_temp=lambda(1,1); lambda_2_temp=lambda(2,2); lambda_3_temp=lambda(3,3); where stress is taken from Eq. 26.23. Matlab defines the smallest eigenvalue as the first one and the largest as the last. Here we define the first eigenvalue (streamwise direction) as the largest and the second (wall-normal direction) as the smallest, i.e. v1=v3_temp; v2=v1_temp; v3=v2_temp; lambda_1=lambda_3_temp; lambda_2=lambda_1_temp; lambda_3=lambda_2_temp; Make sure that Matlab has defined the first eigenvector in the first or the third quadrant, and the second eigenvector in the second or the fourth quadrant, see Fig. 26.4. If not, change sign on some of the eigenvector components. % switch sign on 12 and 21 to fix the above requirements v1_new(2)=-v1_new(2); v2_new(1)=-v2_new(1); 26.9. Anisotropic Synthetic Turbulent Fluctuations ˆ e∗2 223 ˆ e∗1 x2 x1 Figure 26.4: Eigenvectors ˆ e∗1 and ˆ e∗2 . x1 and x2 denote streamwise and wall-normal direction, respectively. √ λ(1) ˆ e∗1 √ λ(2) ˆ e∗2 x2 x1 Figure 26.5: Eigenvectors multiplied by the √ x1 and x2 denote streamwise √ eigenvalues. and wall-normal direction, respectively. λ(1) and λ(2) define RMS of (u′1 )aniso and (u′2 )aniso , respectively. 26.9.2 Synthetic fluctuations in the principal coordinate system The equation for generating the synthetic fluctuations in the principal coordinate system, (u′i )aniso , is similar to Eq. 26.3. The difference is that we now do it in the transformed coordinate system, and hence we have to involve the eigenvector matrix, Rij , and the eigenvalues, λ(i) . The equation reads [155, 161] ′ ∗ u aniso (x ) = 2 N X n=1 κ∗ = κ∗i = u ˆn cos(κ∗n · x∗ + ψ n )σ ∗n r 1 Rji κj , λ(i) σ ∗n = σi∗n The superscript ∗ denotes the principal coordinates. p = λ(i) Rji σjn (26.27) 27. Overview of LES, hybrid LES-RANS and URANS models 224 27 Overview of LES, hybrid LES-RANS and URANS models DNS (Direct Numerical Simulation) is the most accurate method available. In DNS we solve the unsteady, 3D Navier-Stokes equations (see Eq. 2.7) numerically without any turbulence model. This gives the exact solution of the flow field in time and space. We can afford to do DNS only for low Reynolds number. The higher the Reynolds number gets, the finer the grid must be because the smallest turbulent length and time scales, which we must resolve in DNS, decrease with Re−3/4 and Re−1/2 , respectively, see Eq. 5.16. Hence the cell size in each coordinate direction of our CFD grid must decrease with Re−3/4 and the time step must decrease with Re−1/2 . Let’s take an example. If the Reynolds number increases by a factor of two, the number of cells increases by ∆Re3/4 ∆Re3/4 ∆Re3/4 ∆Re1/2 = ∆Re11/4 = 211/4 = 6.7 x-direction y-direction z-direction DNS (27.1) time Above we assume that the lengthscales are reduced when the Reynolds numbers is increased. This implies that we assume that the Reynolds number is increased due to an increase in velocity or a decrease in viscosity. We can, of course, also consider the change of Reynolds number by changing the size of the object. For example, it is affordable to compute the flow around a small car such as those we played with as kids (for this car of, say, length of 5cm, the Reynolds number is very small). As we increase the size of the car we must increase the number of cells (the smallest cells cannot be enlarged, because the smallest turbulent scales will not increase). Also the time step cannot be increased, but we must compute longer time (i.e. increase the number of timesteps) in order to capture the largest time scales (assuming that the velocity of the small and the large car is the same). Having realized that DNS is not feasible, we turn to LES, see Section 18. Here, the smallest scales are modeled, and only the eddies that are larger than the grid are resolved by the (filtered) Navier-Stokes. With LES, we can make the smallest grid cells somewhat larger (the cell side, say, 2 − 3 times larger). However, it is found that LES needs very fine resolution near walls, see Section 21. To find an approximate solution to this problem we use RANS near the walls and LES away from the walls. The models which we have looked at are DES (Section 20), hybrid LES-RANS (Section 21), SAS (Section 22) and PANS (Section 23); see also Section 25 where PANS and Zonal PANS are discussed. As stated above, the LES must at high Reynolds number be combined with a URANS treatment of the near-wall flow region. In the literature, there are different methods for bridging this problem such as Detached Eddy Simulation (DES) [122, 138, 147] hybrid LES/RANS [164] and Scale-Adapted Simulations (SAS) [165, 166] (for a review, see [167]). The two first classes of models take the SGS length scale from the cell size whereas the last (SAS) involves the von K´arm´an lengthscale. The DES, hybrid LES/RANS and the SAS models have one thing in common: in the LES region, the turbulent viscosity is reduced. This is achieved in different ways. In some models, the turbulent viscosity is reduced indirectly by reducing the dissipation term in the k equation, see Eq. 20.7, as in two-equation DES [168]. In other models, such as the two-equation XLES model [126] and in the one-equation hybrid LESRANS [84], it is accomplished by reducing the length scale in both the expression for LES The source term is activated by resolved turbulence. when the grid is refined in streamwise and spanwise directions. directly. When the momentum equations are in turbulence-resolving mode. ε. In the SAS model based on the k − ω model. PSAS . they can be defined as a second-generation URANS model (a 2G-URANS model [167]). Hybrid LES-RANS. hence it can be classified as an URANS model (first generation). SAS. 23. and. 23. second. When a large part of the boundary layer is covered by LES. PANS. the interface moves closer to the wall. A RANS model is used in unsteady mode. PANS and PITM models have in [167] been classified as second-generation URANS models. and can hence be classified as URANS models. in the ω equation. the turbulent viscosity is reduced by an additional source term. Hence they are usually defined as an URANS model (defined below). This is also a model that can operate both in LES and RANS mode. it is called WM-LES (Wall-Modeled LES). 22. second. DES. In the partially averaged Navier-Stokes (PANS) model [128] and the Partially Integrated Transport Model (PITM) [130. A RANS models is used near the walls and LES is used away from the walls. in steady flow it is inactive. and. see Eq. the turbulent viscosity is reduced by decreasing the destruction term in the dissipation (ε) equation which increases ε. A short description of the models are given here.1. The LES lengthscale is the filterwidth. Overview of LES. the turbulent viscosity is reduced because of the enhancement of ε (νt = cµ k 2 /ε. the turbulent kinetic energy (k) decreases because of the increased dissipation term.5. Unless the flow is prone to go unsteady. This decreases the turbulent viscosity in two ways: first. see Eq. On the other hand. Since the models usually aim at resolving a substantial part of the turbulence spectrum. However. These models are able to operate both in LES and RANS mode. a large part of the turbulence spectrum is usually resolved which is in contrast to standard URANS models. These two models do not use the filter width. Hybrid LES-RANS and WM-LES can be considered to be the same thing. The interface is usually defined automatic. νt . PITM. The only difference in the models is that in the PANS model the turbulent diffusion coefficients in the k and ε equations are modified. see Eq. URANS. The difference between DES and hybrid LES-RANS is that the original DES covers the entire boundary layer by RANS whereas hybrid LESRANS treats most of the boundary layer in LES mode.17.2 and Table 21. or 2G-URANS models. In LES mode the models do not use the filterwidth as a lengthscale. because ω appears in the denominator in the expression for the turbulent viscosity. In the original DES the entire boundary layer is covered by RANS.27. In unsteady mode this model usually resolved less turbulence than the other models mentioned above. see Eq. because k is reduced due to the increased dissipation term β ∗ kω. In unsteady mode the model does not use the filterwidth as a lengthscale. WM-LES . PSAS increases which increases ω. This decreases the turbulent viscosity in two ways: first. In unsteady mode the model does not use the filterwidth as a lengthscale. 21. 149]. The PANS model and the PITM models are very similar to each other although their derivations are completely different. The LES lengthscale is the filterwidth. hybrid LES-RANS and URANS models 225 the turbulent viscosity as well as for the dissipation term in the k equation.12). PANS. Usually a RANS model developed for steady flow is used. Lowest accuracy and CPU cost .27. Overview of LES. WM-LES. The models listed above can be ranked in terms of accuracy and CPU cost: 1. SAS.e. same as RANS). DES. PITM. Hybrid LES-RANS. Only a small part of the turbulence is resolved. 4. 3. hybrid LES-RANS and URANS models 226 the results will be steady (i. Highest accuracy and CPU cost 2. URANS. manchester. Two different CFD codes usually give the same results on the same grid.manchester.mace. or that the discretization scheme in one code was more diffusive than in the other.ercoftac.org/publications/ercoftac best practice guidelines/single-phase flows spf/ . Best practice guidelines (BPG) 227 28 Best practice guidelines (BPG) In the early days of CFD. The main reason for this improved situation is because of workshops and EU projects where academics.manchester.2 Ercoftac workshops Workshops are organized by Ercoftac (European Research Community On Flow.se/˜lada/projects/lesfoil/proright. Then they compare and discuss their results.ac.mace.uk/ATAAC/WebHome 28.uk/desider ATAAC: Advanced Turbulence Simulation for Aerodynamic Application Challenges http://cfd.ercoftac. there could be substantial differences between the results. Even if the same grid and the same turbulence model were used. turbulence models etc are defined and the participants in the workshops and EU projects carry out CFD simulations for these test cases.org/fileadmin/user upload/bigfiles/sig15/database/index.1 EU projects Four EU projects in which the author has taken part can be mentioned LESFOIL: Large Eddy Simulation of Flow Around Airfoils http://www.tfd.uk/flomania/ DESIDER: Detached Eddy Simulation for Industrial Aerodynamics http://cfd. Today the situation is much improved.mace. The Special Interest Group Sig15 is focused on evaluating turbulence models. The outcome from all workshop are presented here http://www. Test cases with mandatory grids.html Ercoftac also organizes workshops and courses on Best Practice Guidelines. 28. Turbulence And Combustion). There could be small differences in the implementation of the boundary conditions in the two codes.chalmers. different CFD codes used to give different results. boundary conditions.ac. engineers from industry and CFD software vendors regularly meet and discuss different aspects of CFD. The reasons to these differences could be that the turbulence model was not implemented in exactly the same way in the two codes. The publication Industrial Computational Fluid Dynamics of Single-Phase Flows can be ordered on Ercoftac www page http://www.28.html FLOMANIA: Flow Physics Modelling: An Integrated Approach http://cfd.ac. ercoftac. Ercoftac Classical Database 228 28. Each Application Area is comprised of Application Challenges. which includes some 100 experimental investigations.ercoftac. These are generic. These are realistic industrial test cases which can be used to judge the competency and limitations of CFD for a given Application Area. Underlying Flow Regimes. Turbomachinery. visit ERCOFTAC QNET Knowledge Base Wiki http://www. External Aerodynamics.3.4 ERCOFTAC QNET Knowledge Base Wiki The QNET is also the responsibility of Ercoftac. Chemical and Process Engineering.3 Ercoftac Classical Database A Classical Database. For more information. Combustion and Heat Transfer etc.org/products and services/classic collection database 28.org/products and services/wiki/ . can be found at Ercoftac’s www page http://www. Here you find descriptions of how CFD simulations of more than 60 different flows were carried out. well-studied test cases capturing important elements of the key flow physics encountered across the Application Areas. The flows are divided into Application Areas.28. These are sector disciplines such as Built Environment. . i 1 2 1 n 2 1 2 j 1 1 2 k 2 2 1 ǫinm ǫmjk ǫ12m ǫm12 = ǫ123 ǫ312 = 1 · 1 = 1 ǫ21m ǫm12 = ǫ213 ǫ312 = −1 · 1 = −1 ǫ12m ǫm21 = ǫ123 ǫ321 = 1 · −1 = −1 δij δnk − δik δnj 1−0=1 0 − 1 = −1 0 − 1 = −1 1 3 1 3 1 3 1 1 3 3 3 1 ǫ13m ǫm13 = ǫ132 ǫ213 = −1 · −1 = 1 ǫ31m ǫm13 = ǫ312 ǫ213 = 1 · −1 = −1 ǫ13m ǫm31 = ǫ132 ǫ231 = −1 · 1 = −1 1−0=1 0 − 1 = −1 0 − 1 = −1 2 3 2 3 2 3 2 2 3 3 3 2 ǫ23m ǫm23 = ǫ231 ǫ123 = 1 · 1 = 1 ǫ32m ǫm23 = ǫ321 ǫ123 = −1 · 1 = −1 ǫ23m ǫm32 = ǫ231 ǫ132 = 1 · −1 = −1 1−0=1 0 − 1 = −1 0 − 1 = −1 Table A. TME225: ǫ − δ identity A 229 TME225: ǫ − δ identity The ǫ − δ identity reads ǫinm ǫmjk = ǫmin ǫmjk = ǫnmi ǫmjk = δij δnk − δik δnj In Table A.A.1: The components of the ǫ − δ identity which are non-zero.1 the components of the ǫ − δ identity are given. We denote the first index as i and the second index as j. i.nj) are the v1 values at the upper boundary. .j). download the data files boundary layer data. x1 = L (i=ni) Neumann boundary conditions are used.B. Open Matlab and execute boundary layer.se/˜lada/MoF/. v2 = 0.dat and blasius. Download also the file with the Blasius solution. i. ni = 252 points in the x1 direction and nj = 200 points in the x2 direction. Lower boundary. the corresponding Matlab arrays are u2d.289m. There are three field variables. TME225 Assignment 1 in 2014: laminar flow in a boundary layer 230 B TME225 Assignment 1 in 2014: laminar flow in a boundary layer You will get results of a developing two-dimensional boundary-layer flow (i. Inlet boundary. The height of the domain is H = 1.∞ = 1.1) are the v1 values at the lower boundary – i≤ i=i lower sym: symmetric boundary condition – i> i=i lower sym: wall • u2d(:.chalmers. B. u2d(i. xc. v2d and p2d.m Open boundary layer.461m and x0 = 0.e. i≤ i=i lower sym: v2 = ∂v1 /∂x2 = 0.e.1).19m.e.dat. i. flow along a flat plate. i. Upper boundary. ∂v1 /∂x1 = ∂v2 /∂x1 = 0. yc.e.m in an editor and make sure that you understand it. You can also use Octave on Linux/Ubuntu. x2 = 0 (j=1) x1 ≤ 0.yp. The field variables are stored at the cell centers denoted xp. find out and write down the governing equations From the course www page http://www. x2 = H (j=nj) v2 = ∂v1 /∂x2 = 0.e. The simulations have been done with CALC-BFC [169] using the QUICK discretizatin scheme [170].e. see Fig. Octave is a Matlab clone which can be downloaded for free.e. blasius.tfd. • First. The flow is steady and incompressible. i> i=i lower sym: v1 = v2 = 0. The mesh has 252 × 200 grid points. the fluid is air of 20o C and V1.dat and the m-file boundary layer. v2 and p. v1 .dat. i.dat.m which reads the data and plot some results. a wall Outlet.∞ = 1m/s . L = 2. x1 = −x0 (i=1) v1 = V1. a symmetry boundary condition You will use Matlab to analyze the data. a symmetry boundary condition x1 > 0. i. Hence • u2d(:. 1 – B. • u2d(1.1 (the Blasius solution is loaded in boundary layer. At the end of this Assignment the group should write and submit a report (in English). Compute and plot δ99 vs.∞ 1/2 (B. this boundary layer thickness is denoted δ99 .∞ is taken from the data file and v2. but we don’t recommend it). What happens with the streamwise velocity. B. Divide the report into sections corresponding to the sections B. • u2d(ni.Blasius /V1. δ.1. right? The boundary layer thickness. 3. near the wall as you go downstream from x1 = 0? Why does v1 behave as it does? What happens with v2 ? Does the behavior of v2 has anything to do with v1 ? Plot profiles of v1 vs. ξ.1) The disadvantage of δ99 is that this definition is entirely dependent on the behavior of the velocity profile near the edge of the boundary layer.1. see Table 3. these should clearly be described and presented.:) are the v1 values at the outlet. The results should also be discussed and – as far as you can – explained.Blasius /V1.2 Boundary layer thickness Look at the v1 profiles vs.∞ is obtained from Eqs.10. The work should be carried out in groups of two (you may also do it on your own. Plot also v2 /V1. Now plot v1 /V1.∞ with Blasius solution.m). Note that when you compute ξ you must use x1 .1 Velocity profiles The flat plate starts at x1 = 0. v1 .B. Compare v1 /V1. x2 and ν from the Navier-Stokes solution.∞ vs. can be defined in many ways. x1 and compare it with the Blasius solution δ99. Velocity profiles 231 V∞ H x2 V∞ x1 boundary layer δ(x1 ) x0 L Figure B. x2 at a couple of x1 locations. There are two other. The boundary layer gets thicker and thicker as you go downstream. v1. In some sections you need to make derivations. more . One way is the boundary layer thickness at the x2 value where the local velocity is 99% of the freestream velocity. Present the results in each section with a figure (or a numerical value).Blasius = 5 νx1 V1.1: Flow along a flat plate (not to scale).43 and 3.49.∞ .47 and compare with Blasius solution. see Eq. x2 you plotted at different x1 locatations in Section B. B. 3.:) are the v1 values at the inlet.∞ and v2 /V1. 4 Skinfriction The skinfriction. 2.∞ V1. we assume that v2 ≪ v1 and ∂v1 /∂x2 ≫ ∂v1 /∂x1 : are these assumptions satisfied in this flow? B. x2 at a couple of x1 locations. x1 and δ increases vs. B. 2. When using Eq.4? how large is it at the wall? Plot Cf vs.664 V1. x1 and compare it with the Blasius solution 1. is the same as τ12 = τ21 . Velocity gradients 232 stable. is a dimensionless wall shear stress defined as Cf = τw 1 2 2 V1.j). We don’t want to carry out the integration in Eq. Plot δ ∗ and θ vs.∞ 0 The upper limit of the integrals is infinity. recall that the flow is incompressible. ωi . you find that v1 has a maximum outside the boundary layer.4.4 (note that only one component of ωi is non-zero).∞ x1 /ν Cf decreases vs. δ ∗ .4. see Section 1.j+1) < u2d(i.1.∞ B. x1 and compare with the Blasius solution 1/2 νx1 ∗ δBlasius = 1.∞ (B. x1 : do you see any connection? B.328 Cf. Furthermore. they are defined as Z ∞ v1 ∗ dx2 δ = 1− V1.5 Vorticity Do you expect the flow to be irrotational anywhere? Let’s find out by computing the vorticity vector.∞ 0 (B. the displacement thickness. at the wall. Where is it largest? Plot the vorticity also upstream of the plate. x2 at a couple of x1 locations.2 outside of this maximum. definitions of boundary layer thickness.Blasius = p V1.∞ 1/2 νx1 θBlasius = 0. Cf . Hence. τw . terminate the integration when u2d(i. ∗ because this will give an incorrect estimate of δBlasius and θBlasius . How does this velocity derivative change as you move downstream along the plate? Do you see any connection with the change of the boundary layer thickness with x1 and the change of ∂v1 /∂x2 with x1 ? In boundary layer approximations. Plot the vorticity above the plate vs.3.B. 2.2) Z ∞ v1 v1 1− dx2 θ= V1. what about ∂v2 /∂x1 in Eq.3 Velocity gradients Compute and plot ∂v1 /∂x2 vs. and the momentum thickness. Now. see Eq. If you look carefully at the velocity profiles which you plotted in Section B. θ. is the flow rotational anywhere? Why? Why not? .721 V1.3) The wall shear stress. Sij . Tb .8)? Pick an x1 location above the plate. versus x2 . Now you can compute the increase in bulk temperature. This is discussed in some detail at p.e. where n and lambda denote eigenvalues and eigenvectors. 2. see Eq. Use the Matlab command eig.4. (see Eq.5. 27.8 Eigenvalues Compute and plot the eigenvalues of the viscous stress tensor. we have only two off-diagonal terms (which ones?). we divided the velocity gradient into a strain-rate tensor. re-write the left side on conservative form (see Section 2. the temperature. tau_21 tau_22]. respectively? Compare Ωij with the vorticity. 2.4) 0 v1 dx2 When you compute the convective flux in Eq. τij . What is the physical meaning of the eigenvalues (see Chapter 1. Assume that the lower (x2 = 0) and the upper (x2 = H) boundaries are adiabatic.14 to compute the temperature increase that is created by the flow (i. ωi . furthermore we can neglect the heat flux by conduction. Are they similar? Any comment? B. see Eq. Ωij . Deformation B. which you plotted in Section B. Plot and compare one of the off-diagonal term of Sij and Ωij . tau 21.6. for example. Note that tau 11. Plot also the four stress components. q1 .6 233 Deformation In Section 1. Is (Are) anyone(s) negligible? How does the largest component of τij compare with the largest eigenvalue? Any thoughts? And again: what is the physical meaning of the eigenvalues? .11) at the inlet and outlet.4) and then apply the Gauss divergence theorem.6. Use Eq. If you have computed the four elements of the τij matrix you can use the following commands: tau=[tau_11 tau_12. 2.e. The bulk temperature is defined as Rh v1 T dx2 Tb = R0 h (B.9 at the outlet.12. from inlet to outlet. This means that dissipation increases the internal energy. Where are they largest? Why? What is the physical meaning of Sij and Ωij . Since the flow is two-dimensional. and a vorticity tensor. by dissipation).12. tau 22 are scalars and hence the coding above must be inserted in for loops. What is the physical meaning of the dissipation? Where do you expect it to be largest? Where is it largest? How large is is upstream the plate? The dissipation appears as a source term in the equation for internal energy. [n. you get Z h v1 T dx2 0 which indeed is very similar to the bulk temperature in Eq. B. Next. respectively.7 Dissipation Compute and plot the dissipation. Start by integrating the dissipation over the entire computational domain. τij .B. 2. Plot one eigenvalue as a x − y graph (eigenvalue versus x2 ).lambda]=eig(tau). B. Φ = τji ∂vi /∂xj . 2. i. tau 12. x2 at a couple of x1 locations. of course. How large is the gradient of vorticity. 3. Plot the three terms vs. what could be the reason? Recall that in Eq. Use the Matlab command quiver to plot the field of the eigenvectors. perpendicular to each other. ∂ω/∂x2 . Let’s verify that. Maybe it is not zero? Plot it and check! If it’s not zero.41 are equal: are they? If not. i. check if the left side and the right side of Eq. 3.9 234 Eigenvectors Compute and plot the eigenvectors of τij . Try to analyze why the eigenvectors behave as they do.41 we assume that ∂p/∂x1 = 0.B. It is enough to plot one of them. what could be the reason? How large is ∂ 2 v1 /∂x22 at the wall? How large should it be? (Hint: which other terms in the v1 equation are non-zero?). see Eq. An eigenvector is. Recall that at each point you will get two eigenvectors. 1.e. 3. Plot also the sum of the terms. The left side and the right side of this equation must be equal. both v ˆ1 and −ˆ v1 in Fig.41. B. a vector. Eigenvectors B. Recall that the sign of the eigenvector is not defined (for example.10 Terms in the v1 equation The Navier-Stokes equation for v1 includes only three terms. at the wall? Look at the discussion in connection to Eq. .11 at p.9.27. 4. 23 are eigenvectors). The results should also be discussed and – as far as you can – explained. The grid is 199 × 28.se/˜lada/MoF/. v2 2d and p 2d. We denote the first index as i and the second index as j. Open Matlab and execute channel flow.C. see Fig. You will use Matlab to analyze the data. Open channel flow. the corresponding Matlab arrays are v1 2d. . Present the results in each section with a figure (or a numerical value). v1 2d(1.m in an editor and make sure that you understand it. i.:) are the v1 values at the outlet.m which reads the data and plot some results. ni = 199 grid points in the x1 direction and nj = 28 grid points in the x2 direction. v2 and p.9. The height of the channel is h = 0. C TME225 Assignment 1 in 2013: laminar flow in a channel You will get results of a developing two-dimensional channel flow (i.01m and L = 0. Octave is a Matlab clone which can be downloaded for free. You can also use Octave on Linux/Ubuntu.1 – C. but we don’t recommend it).e. download the data file channel flow data. In some sections you need to make derivations. v1 . Hence in v1 2d(:.B:. Divide the report into sections corresponding to the sections C.nj) are the v1 values at the upper wall.1) are the v1 values at the lower wall. you cannot assume that the flow is fully developed). • First. C.:) are the v1 values at the inlet. i.9. The field variables are stored at these grid points. At the end of this Assignment the group should write and submit a report (in English). The work should be carried out in groups of two (you may also do it on your own.j). find out and write down the governing equations (N. the fluid is air of 20o C. The simulations have been done with Calc-BFC [169]. these should clearly be described and presented. v1 2d(:. flow between two parallel plates). From the course www page http://www. v1 2d(i.6385m.1. TME225 Assignment 1 in 2013: laminar flow in a channel 235 V h x2 x1 L Figure C. There are three field variables. The flow is steady and incompressible.1: Flow between two plates (not to scale).tfd.e. v1 2d(ni.dat and the m-file channel flow.e. The inlet boundary condition (left boundary) is v1 = Vin = 0.chalmers. 2.L (index L denotes Lower). How well does this agree with your values? In the fully developed region. What x1 value do you get? In Section 3.01. τw. ∂v1 /∂x1 = 0.2).1 Fully developed region Fully developed conditions mean that the flow does not change in the streamwise direction. Fully developed region 236 C.C. for example.1. What value should it take in the fully developed region (see Section 3. 99% of its final value.2. a distance taken from the literature is given. If we define “fully developed” as the location where the velocity gradient in the center becomes smaller than 0. Look at the vertical velocity component.2 Wall shear stress On the lower wall. how long distance from the inlet does the flow become fully developed? Another way to define fully developed conditions can be the x1 position where the centerline velocity has reached. |∂v1 /∂x1 | < 0. compare the velocity profile with the analytical profile (see Section 3. for example)? C.01. is computed as . the wall shear stress.e.2.2. i.2)? What value does it take (at x2 = h/4.e. v2 . i. ∂v1 . . 1) ∂x2 .L = µ (C. τw.L ≡ τ21.w. 1. −1. i. If you. C. 2. τw.9) to the fully developed profile somewhere downstream.1. Chapt.9) but at the wall ∂v2 /∂x1 = 0.4.2 to compute the wall shear stress at the upper wall. Plot also ∂v1 /∂x1 . Plot the two wall shear stresses in the same figure. and hence the v1 velocity in the center must increase due to continuity. Z h ξ(x1 ) = v1 (x1 .2) (ˆ n) (C. Plot v1 in the center and near the wall as a function of x1 .9 and ing) normal vector is n 2.L Recall that τ12 = µ(∂v1 /∂x2 + ∂v2 /∂x1 ) (see Eqs.U . If you use Eq. Why does it behave as it does? Now we will compute the wall shear stress at the upper wall. Plot τw. x2 )dx2 0 what do you get? How does ξ(x1 ) vary in the x1 direction? How should it vary? . nj = (0.4 and 1. the normal vector points out from the surface (i. The v1 velocity is decelerated in the near-wall regions.e.2) ti = τji nj which is a general way to compute the stress vector on a surface whose (outward pointˆ = nj . The expression for τij can be found in Eqs. you get the incorrect sign.3. use Cauchy’s formula (see Fig.30).3 Inlet region In the inlet region the flow is developing from its inlet profile (v1 = V = 0. 0)). 2. Skk = 0 because of the continuity equation. Use Eq. integrate v1 . Instead.3 and [1]. On the top wall. Eq. How do they compare? In the fully developed region. 1. 3. recall that the flow in incompressible. compare with the analytical value (see Eq. C.L versus x1 .e. C. for a fixed x1 . 4. 3. q1 . Plot it in the fully developed region as ω3 vs. What about v2 ? How do you explain the behaviour of v2 ? C. 2. Plot and compare one of the off-diagonal term of Sij and Ωij .6. What is the physical meaning of the dissipation? Where do you expect it to be largest? Where is it largest? Any difference it its behaviour in the inlet region compared to in the fully developed region? The dissipation appears as a source term in the equation for internal energy.C. Ωij . Sij .9 at the outlet. in the developing region. v1 near the walls decreases for increasing x1 . respectively? Compare Ωij with the vorticity. ωi . 2. re-write the left side on conservative form (see Section 2. the temperature.e.5 Vorticity Do you expect the flow to be irrotational anywhere? Let’s find out by computing the vorticity vector ωi .3 we found that.5. furthermore we can neglect the heat flux by conduction.e. you plotted in Section C.12. The bulk temperature is defined as Rh v1 T dx2 Tb = R0 h v dx2 0 1 (C. This means that dissipation increases the internal energy.11) at the inlet and outlet. 2.4 Wall-normal velocity in the developing region In Section C. Where is it largest? Plot the vorticity also in the inlet and developing regions. is the flow rotational anywhere? Why? Why not? C. from inlet to outlet. x2 . see Section 1.4. what happens with the vorticity in the inlet region? Now. 27. Wall-normal velocity in the developing region 237 C. we have only two off-diagonal terms (which ones?).7 Dissipation Compute and plot the dissipation. see Eq.14 to compute the temperature increase that is created by the flow (i. see Eq. Use Eq. 2. Assume that the upper and the lower wall are adiabatic.6 Deformation In Section 1. C. Since the flow is two-dimensional. we divided the velocity gradient into a strain-rate tensor. i.4) and then apply the Gauss divergence theorem. 2. . and a vorticity tensor. Where are they largest? Why? What is the physical meaning of Sij and Ωij . (see Eq. Now you can compute the increase in bulk temperature.4 (note that only one component of ωi is non-zero).12. you get Z h v1 T dx2 0 which indeed is very similar to the bulk temperature in Eq. Are they similar? Any comment? C. Φ = τji ∂vi /∂xj . Tb . This is discussed in some detail at p. Next. by dissipation). Start by integrating the dissipation over the entire computational domain. for example.3) When you compute the convective flux in Eq. τij . Eigenvalues 238 C. Recall that the sign of the eigenvector is not defined (for example. What is the physical meaning of the eigenvalues (see Chapter 1. respectively. 23 are eigenvectors). tau 12.8)? Pick an x1 location where the flow is fully developed. Use the Matlab command eig.11 at p.8. .C. An eigenvector is.9 Eigenvectors Compute and plot the eigenvectors of τij . a vector.8 Eigenvalues Compute and plot the eigenvalues of the viscous stress tensor. tau 22 are scalars and hence the coding above must be inserted in for loops. It is enough to plot one of them. 1.lambda]=eig(tau). perpendicular to each other. Recall that at each point you will get two eigenvectors. tau_21 tau_22]. Note that tau 11. Try to analyze why the eigenvectors behave as they do. both v ˆ1 and −ˆ v1 in Fig. of course. Is (Are) anyone(s) negligible? How does the largest component of τij compare with the largest eigenvalue? Any thoughts? And again: what is the physical meaning of the eigenvalues? C. Plot also the four stress components. versus x2 . τij . If you have computed the four elements of the τij matrix you can use the following commands: tau=[tau_11 tau_12. Use the Matlab command quiver to plot the field of the eigenvectors. Plot one eigenvalue as a x− y graph (eigenvalue versus x2 ). [n. where n and lambda denote eigenvalues and eigenvectors. tau 21. D. ci . ci . D.1.1) see Fig. Hence the coordinates. in physical space which form an orthogonal base in R3 (i. Let us call them basis functions.e. Any vector. Now define the scalar product of two vectors.e. T = c1 V1 + c2 V2 + c3 V3 (D. D TME225: Fourier series Here a brief introduction to Fourier series extracted from [171] is given. D. V3 . the second line follows because of the orthogonality of Vi . TME225: Fourier series 239 V2 T (V2 |T) V1 (V1 |T) Figure D.1 Orthogonal functions Consider three vectors. are determined by ci = (T|Vi )/|Vi |2 (D. The coordinates.1: Scalar product. V2 . as a · b = (a|b). D. their scalar products are zero).2) 2 where |Vi | denotes the length of Vi . i. in R3 can now be expressed in these three vectors.1 and Vi which gives (T|Vi ) = (c1 V1 |Vi ) + (c2 V2 |Vi ) + (c3 V3 |Vi ) = (c1 V1 |V1 ) + (c2 V2 |V2 ) + (c3 V3 |V3 ) 2 2 2 = c1 |V1 | + c2 |V2 | + c3 |V3 | = ci |Vi | (D.3) . can be determined by making a scalar product of Eq. T. a and b. V1 . the coordinates. D. . . as in Eq. i. .3. we usually start by doing an intelligent “guess”. over the interval [a. Trigonometric functions 240 Now let us define an infinite (∞-dimensional) functional space. is defined as (f |gn ) = Z b f (x)gn (x)dx (D. . gi .e. and then we prove that it is correct. = = ci (gi |gi ) = ci ||gi ||2 (D. Eq. ||gn || = (gn |gn )1/2 = Z a b !1/2 gn (x)gn (x)dx (D. is ||gn || 3.2.1).4) a Then. is |Vi | 3.1. gn .6 is non-zero only if i = n. π]. ci .5) n=1 As above. f and gn .7) Similar to Eq. |Vi | which is the length of Vi in Eq. . the “coordinates” can be found from (switch from index i to n) cn = (f |gn )/||gn ||2 (D. B. In mathematics. sin x.e.8) The “coordinates”. The length of the basis vector. sin(nx). .9) Let us now summarize and compare the basis functions in physical space and the basis functions in functional space. (f |gi ) = ∞ X n=1 cn (gn |gi ) (D. be expressed as ∞ X f= cn g n (D. ci+1 (gi+1 |gi ) . in a similar way to Eq. cn (cf. Vi .e.2 Trigonometric functions Here we choose gn as trigonometric functions which are periodic in [−π. f with the basis functions. are called the Fourier coefficients to f in system {g}∞ 1 and ||gn || is the “length” of gn (cf. 1. Any function in [a. D. π]. ci (gi |gi ) .D. sin(2x). Any vector in R3 can be expressed in the orthogonal basis vectors Vi 2. D. D. D. in Eq. The coordinates of Vi are computed as ci = (T|Vi )/|Vi |2 1.6) Since we know that all gn are orthogonal.10) . D. . The length of the basis function.3).2. (f |gi ) = (c1 g1 |gi ) + (c2 g2 |gi ) . . i. . .] (D. cn . So let us “guess” that the trigonometric series [1. with orthogonal basis functions {g}∞ 1 . b]. Multiply. . i. we must now find the “coordinates”. cos(nx). The coordinates of gn are computed as cn = (f |gn )/||gn ||2 D. The “scalar product” of two functions. . . The question is now how to choose the orthogonal function system {g}∞ 1 on the interval [−π. any function can. cos x. b] can be expressed in the orthogonal basis functions gn 2. Orthogonality of ψn and ψk Z Z π 1 π [cos((n + k)x) + cos((n − k)x)] dx (ψn |ψk ) = cos(nx) cos(kx)dx = 2 −π −π π 1 1 1 = sin((n + k)x) + sin((n − k)x) = 0 for k 6= n 2 n+k n−k −π (D. .13) D.).12) D. .14) because cos((n + k)π) = cos(−(n + k)π) and cos((n − k)π) = cos(−(n − k)π). . i. .e.2. . 2. that the integral of the product of any two functions φk and ψk is zero on B[−π.16) 2 Z π 2 .2.D. . .) and ψk (x) = cos(kx) (k = 0. D. (D. . . 3.2.e.4 “Length” of φk π 1 x − sin(2kx) = π for k ≥ 1 (φk |φk ) = ||φk || = sin (kx)dx = 2 4k −π −π (D. Trigonometric functions 241 is an orthogonal system. . for n = 2k = 2. π] and we need to compute their “length” (i.11) where φk (x) = sin(kx) (k = 1.15) D. . gn (x) = ψk (x). D. their norm).10 can be defined as φk (x). The function system in Eq. 4. Now we need to show that they are orthogonal.1 “Length” of ψk Z 2 (ψk |ψk ) = ||ψk || = 2 (ψ0 |ψ0 ) = ||ψ0 || = Z π 1 x + sin(2kx) = π for k > 0 cos (kx)dx = 2 4k −π −π π 2 π −π 1 · dx = 2π (D. 1.2 Orthogonality of φn and ψk Z Z π 1 π [sin((n + k)x) + sin((n − k)x)] dx (φn |ψk ) = sin(nx) cos(kx)dx = 2 −π −π π 1 1 1 cos((n + k)x) + cos((n − k)x) =− =0 2 n+k n−k −π (D.2. .3 Orthogonality of φn and φk Z π Z 1 π (φn |φk ) = sin(nx) sin(kx)dx = [cos((n − k)x) − cos((n + k)x)] dx 2 −π −π π 1 1 1 sin((n − k)x) − sin((n + k)x) = = 0 for k 6= n 2 n−k n+k −π (D.2. for n = 2k + 1 = 1. Taking the average of f (i. i.22) n=1 Now we want to prove that the Fourier coefficients are the best choice to minimize the difference N X an gn || (D.e.D.18a) (D.11 forms an orthogonal system of functions. Fourier series of a function 242 D. gives (see Eq. 2π.4 Derivation of Parseval’s formula Parseval’s formula reads Z π −π (f (x))2 dx = ∞ X π 2 a0 + π (a2n + b2n ) 2 n=1 (D. If we set c = a0 /2. Assume that we want to approximate the function f as well as possible with an orthogonal series ∞ X an g n (D. if f¯ = 0 then a0 = 0.19c) Note that a0 /2 corresponds to the average of f .18c) where n > 0. D.20) f (x)dx = 2π −π 2π 2 2 Hence.e.3 Fourier series of a function Now that we have proved that {g}∞ 1 in Eq.19a) Z π 1 a0 1 a0 f¯ = · 2π = (D.19b) (D.21) We will try to prove this formula. we know that we can express any periodic function.19a) (D.17) n=1 where x is a spatial coordinate.18b) (D.18b. integrating f from −π to π) and dividing with the integration length. The Fourier coeffients are given by Z 1 π bn = (f |φn )/||φn ||2 = f (x) sin(nx)dx π −π Z π 1 f (x) cos(nx)dx an = (f |ψn )/||ψn ||2 = π −π Z π 1 c = (f |ψ0 )/||ψ0 ||2 = f (x)dx 2π −π (D. D. then a0 is obtained from Eq. D.23) ||f − n=1 . ∞ a0 X + (an cos(nx) + bn sin(nx)) 2 n=1 Z 1 π bn = (f |φn )/||φn ||2 = f (x) sin(nx)dx π −π Z 1 π f (x) cos(nx)dx an = (f |ψn )/||ψn ||2 = π −π f (x) = (D.3. f (with a period of 2π) in {g}∞ 1 as ∞ X f (x) = c + (an cos(nx) + bn sin(nx)) (D. D. 4. Using the definition of the norm and the laws of scalar product we can write .D. Derivation of Parseval’s formula 243 Later we will let N → ∞. ! N N N . X X X . 2 ||f − an gn || = f − an gn . f − ak g k . and its magnitude gets closer and closer to that of the right side. Appendix N describes in detail how to create energy spectra from two-point correlations. which gives Parseval’s formula.e. cn so that ||f − N X n=1 an gn ||2 = ||f ||2 − N X n=1 ||gn ||2 c2n (D.2. This means that the series on the left side is convergent. ||ψn || and ||φn || (see Sections D.5 and D. D. . the length of ||ψ0 ||. D.26) The left side must always be positive and hence N X n=1 ||gn ||2 c2n ≤ ||f ||2 = Z π (f (x))2 dx for all N (D. {g}N 1 .24 is thus minimized if the coefficients an are chosen as the Fourier coefficients. Using the Fourier coefficients in Eq. i. but it will always stay smaller than ||f ||2 .2.1 and D. 2 ||f || ≡ Z π 2 (f (x)) dx = −π = 2π a20 2 N X n=1 +π N X n=1 ||gn ||2 c2n a2n + b2n = = ||ψ0 || a20 2 ∞ X + N X n=1 ||ψn ||a2n + ||φn ||b2n π 2 a +π (a2n + b2n ) 2 0 n=1 Note that 2π and π on the second line are the “length” of ||gn ||. n=1 = (f |f ) − N X n=1 = (f |f ) − 2 N X n=1 n=1 an (f |gn ) − an (f |gn ) + N X k=1 N X n=1 k=1 N X N X ak (f |gk ) + n=1 k=1 an ak (gn |gk ) = (D. Expressing f in the second term using the Fourier coefficients cn (see Eqs.24) a2n (gn |gn ) because of the orthogonality of the function system.8) gives N X (f |f ) − 2 n=1 an cn (gn |gn ) + = ||f ||2 + = ||f ||2 + N X n=1 N X n=1 N X n=1 a2n (gn |gn ) ||gn ||2 a2n − 2an cn 2 ||gn ||2 (an − cn ) − N X n=1 (D.4). D.19 and letting N → ∞ it can be shown that we get equality of the left and right side. the magnitude of the left side increases.25) ||gn ||2 c2n The left side of Eq.27) −π As N is made larger. D.19 gives the Fourier series of a real function. D.28 we get Z π 1 1 ∗ c−n = cn = f (x)(cos(nx) + ı sin(nx))dx = (an + ıbn ).19.31) see Eq. Complex Fourier series 244 D.29) f (x)(cos(nx) − ı sin(nx))dx = (an − ıbn ). For negative n in Eq.28 is equivalent to Eq. It is more convenient to express a Fourier series in complex variables even if the function f itself is real.28b) −π where the Fourier coefficients. read – assuming that f is real – according to Eq. Below we verify that if f is real. The Fourier coefficients. .5.31 into Eq.D. a0 . cn .32) which verifies that the complex Fourier series for a real function f is indeed identical to the usual formulation in Eq. D. D.28 reads Z π 1 1 c0 = f (x)dx = a0 2π −π 2 (D. D. a0 . then Eq. D.28a) (D. D.28 Z π 1 1 cn = (D. D. Eq. This means that the real coefficients are multiplied by a factor two except the first coefficient. cn . are complex. n > 0 2π −π 2 where an and bn are given by Eq.30 and D. which makes up for the factor 21 in front of a0 in Eq.30) 2π −π 2 where c∗n denotes the complex conjugate.29. cn .19. D.28 is that the imaginary coefficients for negative and positive n cancel whereas the real coefficients add.19 although the Fourier coefficients. D. n > 0 (D.28 over the formulation in Eq.19. D. On complex form it reads f (x) = cn = ∞ X cn exp(ınx)) n=−∞ Z π 1 2π f (x) exp(−ınx)dx (D. The trick in the formulation in Eq.19 is that we don’t need any special definition for the first Fourier coefficient.5 Complex Fourier series Equation D. One advantage of Eq. D. For n = 0. D. D.28 gives f (x) = = = ∞ 1 1X a0 + (an − ıbn ) exp(ınx) + (an + ıbn ) exp(−ınx) 2 2 n=1 ∞ 1X 1 a0 + (an − ıbn )(cos(nx) + ı sin(nx)) + (an + ıbn )(cos(nx) − ı sin(nx)) 2 2 n=1 ∞ ∞ X X 1 1 a0 + an cos(nx) − ı2 bn sin(nx) = a0 + an cos(nx) + bn sin(nx) 2 2 n=1 n=1 (D. D.19. are complex. Inserting Eqs. Since we have chosen to express the energy spectrum as a function of the wavenumber.1) κ1 We could also express E as a function of the turbulent length scale of the eddies. E. see Eq. 1. vr = 0. 5. i. see Eqs. Equation 7. The energy spectrum is a spectral density function in a similar way as fv (v) in Eq. A similar dimension analysis for the kinetic energy of v ′ gives vκ′2 ∝ κ−2/3 .2 is the first moment. 7.2. k. TME225: Why does the energy spectrum. 5. dimension analysis gives E ∝ κ−5/3 .5 is one over velocity. k = vi′ vi′ . E . The reason is that it is a spectral density so that the kinetic energy. The dimension of fv (v) in Eq. i. The kinetic energy.5 defines the second moment. 5. However.3) k= 8π 2 (R2 − R1 ) R1 r2 From this we can define a kinetic energy spectrum as E ideal (r) = so that k= Z Γ 8π 2 (R2 − R1 )r2 R2 E ideal (r)dr (E. has the strange dimension v 2 /ℓ.12 and 5. have such strange dimensions? The energy spectrum. k = vθ2 /2 ∝ r−2 . integrating vκ′2 over all wavenumbers does not give any useful integral quantity.4 shows that E ideal ∝ r−3 . 1. 7.1 Energy spectrum for an ideal vortex The velocity field for an ideal vortex is given by vθ = Γ/(2πr). As in the previous section.4) (E. have such strange dimensions? 245 E TME225: Why does the energy spectrum.8. see Eq. 7.10 and 5. where Γ denotes the circulation. as mentioned above. Z R2 Γdr 1 (E. 7. 5.2 and fv′ (v ′ ) in Eq.13.10.5) R1 Equation E. .e. ℓκ (κ = 2π/ℓκ . E. However. is computed by integrating over all wavenumbers. see Eq.7).2) [m ] ℓκ As can be seen.E.27 and Fig. we find that the dependence of k on the radius is one degree higher. The integral of the energy spectrum in the inertial region can be estimated as (see Eqs.e. the energy spectrum E ℓ (ℓκ ) obeys the −1/3 law and E(κ) obeys the 2/3 −5/3 law.13) Z κ2 2 5 κ− 3 dκ k = CK ε 3 (E. vκ′2 varies as κ−2/3 (or ℓκ ). E. the difference is that fv (v) in Eq. for the vortex between r = R1 and r = R2 is obtained by integrating vθ2 /2 from R1 to R2 . The energy spectrum is then integrated as Z κ2 Z κ2 5 2 3 E(κ)dκ = CK ε k= κ− 3 dκ κ1 κ1 [m2 ] κ−2/3 Z ℓ2 2 5 dκ dℓκ = −CK 2πε 3 κ− 3 ℓ−2 κ dℓκ dℓ κ κ1 ℓ1 Z ℓ1 ε 23 Z ℓ1 1 − E ℓ (ℓκ )dκ ℓκ 3 dℓκ = = CK 2π ℓ ℓ2 2 2/3 2 2 = CK ε 3 Z κ2 5 κ− 3 (E. see Eq.e. The period of the four terms is L.19 corresponds to 2π/(L/m) with m = (1. 2.8 shown in Fig. i. L/3.5 v (x) = a1 cos x + a2 cos x L/1 L/2 (E.8 to have an energy spectrum of −5/3. For simplicity we make it symmetric so that only the cosine part needs to be used.8) 2π(n − 1)) 2π(n − 1)) 0. Eq.6 for N = 16 discrete points with ∆x = 1/(N − 1) as 2π(n − 1)) 2π(n − 1)) 0.5 + a3 cos + a4 cos N/3 N/4 where (n − 1)/N = x/L and n = [1. 1].6 is continuous. 3.5 ′ 0. the integral in Eq. . 3 · 3π/L and 4 · 2π/L.7). N ]. and L/4 corresponding to wavenumber 2π/L.e.5 0. E. Am = N 1 X ′ 2πm(n − 1) v (x) cos N n=1 N (E. E.1 shows how v ′ varies over x/L. L/3 and L/4. D. a1 = 1.8 decrease as m−5/3 .3). i. E. Now let’s make a DFT of v ′ to get the energy spectrum (see Matlab code in Section E. E.18 Z Z 1 π ′ 2πmx 1 π ′ dx (E. a computational grid.5 0.5 0. D.2. The four terms in E. In unsteady CFD simulations we are always dealing with discrete points. let’s express Eq. The first coefficient. i. Now we want v ′ in Eq. it is given for any x/L = [0. E. D. We make it with four terms. which is zero.e. i. 4).e. tells us that the kinetic energy of an eddy of wavenumber κ is simply the square of its Fourier coefficient. D. It reads then (see Eq. a2 = 2−5/3 . Hence we let the ratio of the ak coefficients in Eq. a3 = 3−5/3 . E.2 246 An example Let’s generate a fluctuating velocity. Parseval’s formula. v ′ .E. in Eq. a4 = 4−5/3 (E.7) 0 Equation Eq. Z L v ′ dx = 0 (E. An example E. D.10) v (x) cos(κx)dx = v (x) cos am = π −π π −π L is replaced by a summation over discrete points. using a Fourier series. E.6) 2π 2π 0.1 can be regarded as the velocity fluctuations at one time instant of four eddies of lengthscale L. i.9) Figure E.e.19 because the mean of the fluctuation v ′ is zero.5 + a3 cos x + a4 cos x L/3 L/4 where the wavenumber n in Eq. Hence.19) 2π 2π ′ 0.21. a0 = 0.5 + a2 cos v (x) = a1 cos N/1 N/2 (E. 2 · 2π/L. L/2.11) where (n − 1)/N = x/L (note that m = 0 corresponds to the mean. In DFT. L/2. 3 An example: Matlab code close all clear all % number of cells N=16. E. Figure E.2. E ℓ (ℓκ ). L/4 m = 3. see −1/3 Matlab code in Section E. E.3. 0 0 10 ℓ Evv Evv 10 −1 10 −2 10 1 10 κ= −1 0 10 2π(m−1) L (a) Dashed thick line shows κ−5/3 . ℓκ 10 (b) Dashed thick line shows κ−1/3 . L/3 m = 2. E. : term 2 (m = 2). Figure E. L/2 m = 1. Matlab code is given in Section E. An example: Matlab code 247 m = 4.8 vs.E. L 2 1 0 0 0. ◦: Evv = A2m . term 3 (m = 3).2: Energy spectrum of v ′ . E. versus eddy size.2b shows the computed energy spectrum.2a.3.2 0. to achieve this).3.1: v ′ in Eq.6 0. am . The total energy is now computed as hv ′2 ix = N X A2m = m=1 N X Evv (m) (E. ◦: term 4 (m = 4). It decays as ℓκ as it should.4 0. Matlab code is given in Section E. . see Fig. see Eq. Now plot the energy spectrum. : Figure E. Evv = A2m versus wavenumber.8 1 x/L : term 1 (m = 1).3. x/L.12) m=1 where h·ix denotes averaging over x. It can be seen that it decays as κ−5/3 as expected (recall that we chose the Fourier coefficients. thick line: v ′ . W_cos(m)=W_cos(m)+a*cos(arg1)/N.’k-.2) hold plot(x_over_L.’linew’.5*cos(3*arg2).2) plot(x_over_L.E. x_over_L=(n-1)/N.N).5*cos(2*arg2(i))+a3ˆ0.5*cos(4*arg2). arg2(i)=2*pi*x_over_L(i). for m=1:N for i=1:N a=v(i). v(i)=a1ˆ0. a3=3ˆ(-5/3).5*cos(2*arg2).’k-’. % E_vv=mˆ(-5/3) a1=1.’linew’.’linew’.N).f2. f3=a3ˆ0.w.’o’.’.f1.f4.’linew’.f3.4) h=gca set(h.5*c end % take DFT W_cos=zeros(1.5*cos(3*arg2(i))+a4ˆ0.’linew’.5*cos(arg2(i))+a2ˆ0.’r--’. f2=a2ˆ0.’fontsi’. plot(x_over_L. end end % Note that all elements of W_sin are zero since v(i) is symmetric %******************************************************************* figure(1) f1=a1ˆ0. W_sin(m)=W_sin(m)+a*sin(arg1)/N. a2=2ˆ(-5/3). n=1:1:N. An example: Matlab code 248 L=1. f4=a4ˆ0. for i=1:N arg2(i)=2*pi*(i-1)/N.2) plot(x_over_L. W_sin=zeros(1.[20]) xlabel(’x’) ylabel(’y’) . a4=4ˆ(-5/3).2) plot(x_over_L.3. arg1=2*pi*(m-1)*(i-1)/N.5*cos(arg2). ’xscale’.ˆ2+W_sin. yynoll=1. yyy(2)=yyy(1)*(xxx(2)/xxx(1))ˆ(-5/3). yyy(1)=yynoll.’log’) set(h. end set(h.ps -depsc2 % %******************************************************************* . An example: Matlab code 249 axis([0 1 -1 3]) print vprim_vs_L.ˆ2.2) hold h=gca set(h. for i=1:N int_wave=int_wave+PW(i).’yscale’.’bo’. plot only one side of the symmetric spectrum and % multiply by two so that all energy is accounted for plot(mx(1:N/2).E. for i=1:N int_phys=int_phys+v(i). end % compute the average of energy in wavenumber space int_wave=0.ˆ2/N.01 1]) % plot -5/3 line xxx=[5 50].’linew’.ps -depsc2 % %******************************************************************* % figure(2) % % the power spectrum is equal to W*conj(W) = W_cosˆ2+W_sinˆ2 PW=W_cos.’fontsi’. % plot power spectrum.4) % compute the average of energy in physical space int_phys=0. plot(xxx.[20]) xlabel(’x’) ylabel(’y’) print spectra_vs_kappa.’log’) axis([5 50 0.2*PW(1:N/2).yyy. mx=2*pi*(n-1)/L.’r--’.’linew’.3. /mx.1 1.yyy. yyy(2)=yyy(1)*(xxx(2)/xxx(1))ˆ(-1/3).3.*lx.E.[20]) xlabel(’x’) ylabel(’y’) print spectra_vs_L.’yscale’. set(h.9. An example: Matlab code 250 % figure(3) % compute the length corresponding the wavenumber lx=2*pi.’log’) set(h.’xscale’. % multiply PW by lxˆ(-2) to get the energy spectrum E(lx) PW_L=PW. plot(xxx.1 1] yynoll=0.’fontsi’.’r--’.2 0.’bo’.4 1]) % plot -1/3 line xxx=[0.’linew’. yyy(1)=yynoll.’log’) axis([0. % plot power spectrum plot(lx(1:N/2).2*PW_L(1:N/2).2) hold h=gca set(h.’linew’.4).ˆ(-2).ps -depsc2 . 5 and x2 = 235.min = 3 and v1. What is the differences between v1 and v2 ? F. x2 = 8. The file has eight columns of v1 and v2 at four nodes: x2 /δ = 0. Start the program pl time. Start Matlab and run the program pl time. Study the time history of the blue line (x2 /δ = 0.1–F.m which loads and plots the time history of v1 . the buffer layer and in the logarithmic layer.8. change the plot command to plot(t. The time history of v1 at x2 /δ = 0. x2 = 53.min = 21. MikTex (www.chalmers. You should write a report where you analyze the results following the heading F. Add the following code (before the plotting section) .0654 and ∆z = 0.95. F. for example.107 are shown.’bo’) Use this technique to zoom.0039. to look at the details of the time history. x2 /δ = 0.dat with the time history of v1 and v2 . Use the Matlab.1 Time history At the course home page http://www. It is available on Linux. Octave is a Matlab clone which can be downloaded for free. The four points are located in the viscous sublayer. 79.tfd.’b-’. you can use the zoom buttons above the figure. A 96 × 96 × 96 mesh has been used. t = 10 and t = 11 and v1. y (x2 ) and z (x3 ) respectively. The streamwise.miktex.0176.0033 (every second time step). In DNS the unsteady. You can also use Octave on Linux/Ubuntu. All data have been made non-dimensional by uτ and ρ. With uτ = 1 and ν = 1/Reτ = 1/500 this cor+ + + respond to x+ 2 = 1.F. for example. The Re number based on the friction velocity and the half channel width is Reτ = uτ h/ν = 500 (h = ρ = uτ = 1 so that ν = 1/Reτ ). Periodic boundary conditions were applied in the x and z direction (homogeneous directions). three-dimensional Navier-Stokes equations are solved numerically. How does the time variation of v1 vary for different positions? Plot also v2 at the four different positions. Make a zoom between. Plot v1 for all four nodes. Alternatively.0176) more in detail. This is conveniently done with the command axis([10 11 3 21]) In order to see the value at each sampling time step. wall-normal and spanwise directions are denoted by x (x1 ).org) which is free to download.11.se/˜lada/MoF/ you find a file u v time 4nodes. see Fig. and thus what you see is really v1 /uτ . It is recommended (but the not required) that you use LATEX(an example of how to write in LATEXis available on the course www page). Recall that the velocities have been scaled with the friction velocity uτ . 6. On Windows you can use.0164.u2. TME225 Assignment 2: turbulent flow F 251 TME225 Assignment 2: turbulent flow In this exercise you will use data from a Direct Numerical Simulation (DNS) for fully developed channel flow.u2.47.0176 and x2 /δ = 0. You can do the assignment on your own or in a group of two.2 at p.t. The sampling time step is ∆t = 0.2 Time averaging Compute the average of the v1 velocity at node 2. The cell size in x and z directions are ∆x = 0. x2 /δ = 0.107 and x2 /δ = 0. b = v¯1 dx2 (F. Let us investigate how these forces (the pressure gradient. Start by trying with 100 samples as umean_100=mean(u2(1:100)) What is the maximum and minimum value of v1 ? Compare those to the mean.2). Compute and plot also the instantaneous fluctuations. see Eq.2) This equation is very similar to fully developed laminar flow which you studied in Assignment 1.) . u1_fluct=u1-u1_mean.mat does not include p¯. Do the same exercise for the other three nodes. Plot it both as linearlinear plot and a log-linear plot (cf. 6. the difference is that we now have an additional term which is the derivative of the Reynolds shear stress. respectively? F. use x+ 2 for the wall distance. v1 = x2 . y. node 1 and nj are located at the lower and upper wall. 3.m which reads the data files.F.3 Mean flow All data in the data files below have been stored every 10th time step. The data file includes v1 . v2 and v3 from the same DNS as above. Recall that all terms in the equation above represent forces (per unit mass). v1 = κ ln x2 + B (κ = 0. (the file uvw inst small. Download the file uvw inst small. V1. + + In the log-linear plot. Compute the mean velocity. Include the linear law. F. the viscous term and the Reynolds stress term) affect fluid particles located at different x2 locations. set ∂ p¯/∂x = −1.b and centerline velocity. Find out how many samples must be used to get a correct mean value. How far out from the wall does the velocity profile follow the linear law? At what x+ 2 does it start to follow the log-law? Compute the bulk velocity Z 2h 1 V1. v1′ at node 1. + + −1 and the log law. for example. Mean flow 252 umean=mean(u2) Here the number of samples is n = 5000 (the entire u2 array).mat. but now you are given the time history of all x2 nodes at one chosen x1 and x3 node. Fig. Compute and plot the three terms.c .24. There are nj = 98 nodes in the x2 direction. Since the flow is fully developed and two dimensional we get 0=− ∂ 2 v¯1 ∂v ′ v ′ 1 ∂ p¯ +ν − 1 2 2 ρ ∂x1 ∂x2 ∂x2 (F.dat and the Matlab file pl vel.4 The time-averaged momentum equation Let us time average the streamwise momentum equation. Your data are instantaneous.1) 2h 0 (recall that h denote half the channel width) What is the Reynolds number based on V1. is computed as u1_mean=mean(u1). respectively.4).41 is the von K´arm´an constant and B = 5.3. They should be equal. Plot the normal stresses in one figure and the shear stresses in one figure (plot the stresses over the entire channel. 9. This is a measure of the fluctuating tangential force on the wall due to turbulence. Pij . from x2 = 0 to x2 = 2h).5) .5 Wall shear stress Compute the wall shear stress at both walls. The root-mean-square (RMS) is then defined as 1/2 φrms = φ′2 (F.6? F. Which shear stresses are zero? F. φ. any fluctuating variable. you can compute the fluctuations as vi′ = vi − v¯i (F.3 you computed the mean velocities. are largest). What about the viscous term? Is it always negative? Where is it largest? At that point? which term balances it? How large is the third term? The pressure term should be a driving force. This is probably the most common form of fluid-solid interaction.8 Production terms In order to understand why a stress is large.5. Usually.3) Now you can easily compute all stresses vi′ vj′ . is large (there may be exceptions when other terms. a stress is large when its production term. see Eq. 9.7 Fluctuating wall shear stress In the same way as the velocity.9 Pressure-strain terms The pressure-strain term reads (see Eq. it is useful to look at its transport equation. F.F. can be decomposed into a mean and fluctuation as φ = φ¯ + φ′ . Are they? F. what about the sign of the corresponding shear stress plotted in Section F. Which ones are zero (or close to)? Does any production term change sign at the centerline? If so. Plot the production terms for all non-zero stresses across the entire channel. In general. From the instantaneous and the mean velocity.e.14) p′ Πij = ρ ∂vj′ ∂vi′ + ∂xj ∂xi (F. such as the diffusion term. If heat transfer is involved. Where is the Reynolds shear stress positive and where is it negative? F. i.6 Resolved stresses In Section F.4) Compute the RMS of the wall shear stress. the fluctuating temperature at the wall inducing fluctuating heat transfer may be damaging to the material of the walls causing material fatigue. Wall shear stress 253 If a term is positive it means that it pushes the fluid particle in the positive x1 direction.12. the wall shear stress can be decomposed into a mean value and a fluctuation. There are many interesting things yet to be analyzed! . Compute and plot ∂v ′ ∂vi′ (F. For which stresses is it negative and positive? Why? Which term Πpij is the largest source and sink term. 9. Now think of some other way to analyze the data. the difference between Πpij and Πij is small. k and ∆T is illustrated in Fig.8) εmean = ν ∂xk ∂xk The flow of kinetic energy between K. Using the velocities stored in uvw inst small. F. The pressure diffusion term in the v2′2 equation – which is the difference between Eqs. Where is it largest? In which equation does this quantity appear? Let us now consider the equations for the mean kinetic energy. The dissipation in the K equation reads vi ∂¯ vi ∂¯ (F. respectively? F.6 The dissipations. are defined in Eqs. x3 ). x2 . ε and εmean . for the three normal stresses and the shear stress across the channel. Plot the pressure strain.mat (see Section F. 8. (x1 ±∆x1 . K = v¯i v¯i /2 (Eq.11 Do something fun! You have been provided with a lot of data which you have analyzed in many ways.mat and the Matlab file pl diss. x2 . The data file includes the time history of the velocities along five x2 lines [(x1 . Which is large and which is small? How is the major part of the kinetic energy transformed from K to ∆T ? Is it transformed via k or directly from K to ∆T ? F. Πpij . Download the data file p inst small. Dissipation 254 Our data are obtained from incompressible simulations. This allows you to compute all the three spatial derivatives of p′ .F. 9. Hence.14. x2 .e. we prefer to compute the velocity-pressure gradient term Πpij = − vj′ ∂p′ vi′ ∂p′ − .3.6.3).mat and the Matlab file pl press strain.34) and turbulent kinetic energy.m which reads it.10 Dissipation The physical meaning of dissipation.dat. k = vi′ vi′ /2 (Eq. ε. Compute and plot also εmean and P k . diss inst. i. x2 .7) ε=ν i ∂xk ∂xk see Eq.8. increased temperature.14).m which reads the data file. Download the files y half. x3 ). is transformation of turbulent kinetic energy into internal energy. The data cover only the lower half of the channel. 9.3 and 9.6 (the two first terms in Eq. The time histories of the pressure along five x2 lines [(x1 . in which the pressure may vary unphysically in time (∂p/∂t does not appear in the equations). x2 . x2 .7 and F. Hence. respectively. F. x3 ) and (x1 . 8.6) see the second line in Eq. F. x3 ± ∆x3 )] are stored in this file.10.5 and F. x3 ±∆x3 )] so that you can compute all spatial derivatives. (x1 ± ∆x1 .8) – is small except very close to the wall (see Figs. x3 ) and (x1 . 8.5). ρ ∂xj ρ ∂xi (F. you can compute all the terms in Eq. 8. se/˜lada/flow viz. 1. Watch the on-line lecture Vorticity. Explain the physical meaning of diagonal and off-diagonal components of Sij 8. that act on a Cartesian surface whose normal vector is ni = (1. 1. 0).4 in the lecture notes) iii.3 and the Lecture notes of Toll & Ekh [2]) 5. ∂T ∂T ii. 2. The flow in the tank moves like a solid body. What is the definition of irrotational flow? 10.se/˜lada/flow viz. σij . Show also the stress vector. tniˆ . part 1 at http://www.G. How does the vorticity meter move? This is a curved flow with vorticity. What is a vortex line? ii. Start from Eq. 0. 3.2. 1. Show the relation between the stress tensor. Explain the physical meaning of the eigenvectors and the eigenvalues of the stress tensor (see Section 1. (see the Lecture notes of Toll & Ekh [2]) 6. Show that the product of a symmetric and an antisymmetric tensor is zero. Consider the flow in Section 1. (see Eq.html i. The teacher talks of ωA and ωB . He puts the vorticity meter into the tank. 1. What is the physical meaning of irrotational flow? 11. what does it denote? (cf. Derive the relation between the vorticity vector and the vorticity tensor 12.8 and the Lecture notes of Toll & Ekh [2]) 14.tfd. 1. iv.18) 13.chalmers. where ω = 0. Show that ∂v1 /∂t is different from dv1 /dt. TME225 Learning outcomes 2014 G 255 TME225 Learning outcomes 2014 TME225 Learning outcomes 2014: week 1 1. 7. What happens with the vorticity meter? (cf. Explain the difference between Lagrangian and Eulerian description of the motion of a fluid particle.html i. tniˆ . σij . The teacher puts the vorticity meter into a flow in a straight channel (near a wall). C.10) . and the stress vector. The formula is nicely explained in Part 2 + vi ∂t ∂xi 4.2 and Fig.chalmers. Watch the on-line lecture Eulerian and Lagrangian Description. Fig.16 and express the vorticity tensor as a function of the vorticity vector (Eq.tfd. The teacher shows the rotating tank (after 3 minutes). Show which stress components. Part 1 describes the difference between Lagrangian and Eulerian points and velocities. Explain the physical meaning of Ωij 9. part 1 – 3 at http://www.5(ωA + ωB ). Fig. 1. x. respectively? The Bernoulli equation gives then the velocity. Now repeat it for the x2 direction. the teacher introduces a spiral vortex tank (in the lecture notes this is called an ideal vortex).7) . 1. vii.10 and 4.5 (use the formulas in the Formula sheet (it can be found on the course www page)) 17.25). The teacher presents a two-dimensional wing. 1. Γ. Eqs. What happens with the vorticity when we get very close to the center? Does it still remain zero? What happens with the tangential velocity? (see Eq. How large is the vorticity and the circulation in the spiral vortex tank? Does is matter if you include the center? xiii. initially there is only vorticity near the outer wall.29) ix. Eq. square fluid element (see lecture notes of Toll & Ekh [2]). How does the vorticity meter at the different locations? Where is the vorticity smallest/largest? Explain why. and its relation to vorticity (cf. r. vi. After 10:35. After 15:25 minutes. 1. the vortex meter is shown in the spiral vortex tank. 1.29 in the lecture notes. 16.7 states that mass times acceleration is equal to the sum of forces (per unit volume). Derive the Navier-Stokes equation. Fig. 2. what does he say about ωB ? What conclusion does is draw about ω? This is a straight flow with vorticity. He says that one of the “vorticity legs” (ωA in Item 14ii above) is parallel to the wall.G. the explanation uses the fact that the tangential velocity. As time increases. the teacher looks at the rotating tank again. xiv. Write out the momentum equation (without using the summation rule) for the x1 direction and show the surface forces and the volume force on a small. He starts to rotate the tank. the teacher shows the figure of a boundary layer.23 and 1. 2. After 8:40 minutes. respectively? The velocity difference creates a circulation. a bit further out and finally at the edge of the boundary layer. This illustrates that as long as there is an imbalance in the shear stresses. where is it high and low. The teachers explains the concept of circulation. the flow in the entire tank has vorticity (and circulation). vorticity (and circulation) spread toward the center. the teacher puts the vorticity meter at different locations in the boundary layer. TME225 Learning outcomes 2014 256 v. After 4:20 minutes. How large is the vorticity and the circulation in the rotating tank? xii. is inversely proportional to the radius. Γ. see Eq.8 in the lecture notes).1. see Figs. Simplify the Navier-Stokes equation for incompressible flow and constant viscosity (Eq. viii. 15. Finally. How does the teacher explain that the vorticity is zero (look at the figure he talks about after 6 minutes). he puts it near the solid wall. vorticity (and circulation) is changed (usually created). How does the vortex meter behave? (cf. Equation 1. What is a vortex core? xi. vθ . Where is the pressure low and high. After 4:30 minutes. 20.12 (again.15). Show how the left side of the transport equations can be written on conservative and non-conservative form . Explain the energy transfer between kinetic energy. u 5. Eq. k. 3. What is the physical meaning of the different terms? 2. Eq. use the Formula sheet). Derive the transport equation for kinetic energy. TME225 Learning outcomes 2014 257 TME225 Learning outcomes 2014: week 3 1. u. 2. Derive the transport equation for the internal energy. and internal energy. 2. Simplify the the transport equation for internal energy to the case when the flow is incompressible (Eq.G. What is the physical meaning of the different terms? 4. vi vi /2. 2. 18. 3.26) 6. 3. Explain the flow physics in a channel bend (Fig. Derive the flow equations for fully developed flow between two parallel plates. derive the flow equation governing the Rayleigh problem expressed in f and η. Ψ 7. 3. Show that the continuity equation is automatically satisfied in 2D when the velocity is expressed in the streamfunction. Derive the Blasius equation.5). 3.45). 3. 3. how are they expressed in the similarity variable η? 2. . a gradient) of shear stresses. Explain the flow physics at the entrance (smooth curved walls) to a plane channel (Fig. derive the equation for two-dimensional boundary-layer flow expressed in the streamfunction (Eq. 4. 5. what are the boundary conditions in time (t) and space (x2 ).e.41. Explain why pressure cannot create vorticity. Start from Eq. Start from Eq. Eq. 3.e. 3.G.6). Show how the boundary layer thickness can be estimated from the Rayleigh problem using f and η (Fig.22 and 3. Starting from Eq.53. 3. 3. i. fully developed channel flow (Eqs.45 9. Explain (using words and a figure) why vorticity can be created only by an imbalance (i.41 8. TME225 Learning outcomes 2014 258 TME225 Learning outcomes 2014: week 4 1. Starting from the Navier-Stokes equations (see Formula sheet).3) 3. Eq. The Navier-Stokes equation can be re-written on the form ∂vi + ∂t ∂ 2 vi ∂k 1 ∂p +ν + fi − εijk vj ωk = − ∂x ρ ∂xi ∂xj ∂xj | {z } |{z}i rotation no rotation Derive the transport equation (3D) for the vorticity vector. is zero 3. Γ. Show that the divergence of the vorticity vector. ωi .chalmers. What value does the fluid velocity take at the surface? What is this boundary conditions called: slip or no-slip? How do they define the boundary layer thickness? iii. the vorticity) created? Where is the vorticity created in “your” channel flow (TME225 Assignment 1)? The vorticity is created at different locations in the flat-plate boundary layer and in the channel flow: can you explain why? (hint: in the former case . 1. How is the circulation. 4.22 2. TME225 Learning outcomes 2014 TME225 Learning outcomes 2014: week 5 259 June 15. defined? (cf.html i.23) How is it related to vorticity? How do they compute Γ for a unit length (> δ) of the boundary layer? How large is it? How does it change when we move downstream on the plate? v. Eq. with Eq. 3. Show that the vortex stretching/tilting term is zero in two-dimensional flow 5.tfd. Watch the on-line lecture Boundary layers parts 1 at http://www. How is the wall shear stress defined? How does it change when we move downstream? (how does this compare with the channel flow in TME225 Assignment 1?) iv. How does the boundary layer thickness change when we move downstream? ii. Show the similarities between the vorticity and temperature transport equations in fully developed flow between two parallel plates r r ν 1 δ 7. 8. Explain vortex stretching and vortex tilting 4.G.e. Use the diffusion of vorticity to show that ∝ = (see also ℓ Uℓ Re Eq. Where is the circulation (i. 2015 1.14).24 6. 4. Consider the flow over the flat plate. Derive the 2D equation transport equation for the vorticity vector from the 3D transport equation.se/˜lada/flow viz. ∂p ∂ 2 v1 . . =µ 2. = 0. 1) .3]) vi.wall [see Section 4. Section. ∂x1 ∂x2 wall but not in the latter. 4.3. How do they estimate the boundary layer thickness? (cf. this has an implication for γ2. tfd. stall occurs at a higher angle of incidence compared when the boundary layer is laminar. Consider the airfoil: when the boundary layer on the upper (suction) side is turbulent. ii.se/˜lada/flow viz. Watch the on-line lecture Boundary layers part 2 at http://www.html i. How large are the largest scales? What is dissipation? What dimensions does it have? Which eddies extract energy from the mean flow? Why are these these eddies “best” at extracting energy from the mean flow? 13. iii.chalmers. Vortex generator are place on the suction side in order prevent or delay separation. The upper one is turbulent and the lower one is laminar.se/˜lada/flow viz. What characterizes turbulence? Explain the characteristics. TME225 Learning outcomes 2014 260 9. Why is the vorticity level higher after the contraction? iv. How? ii. τη . The second part of the movie deals with turbulent flow: we’ll talk about that in the next lecture (and the remaining ones). Try to explain why separation is delayed. How does the boundary layer thickness change at a given x when we increase the velocity? Explain why. Try to explain why. Why? vi. Why is the turbulent wall shear stress larger for the turbulent boundary layer? What about the amount of circulation (and vorticity) in the laminar and turbulent boundary layer? How are they distributed? v.tfd. the separation region is suppressed. the length scale. my figures in the ’summary of lectures’) Explain the differences. What happens when the angle of the diffusor increases? vii. What is the difference in the two velocity profiles? (cf. Watch the on-line lecture Boundary layers parts 2 (second half) & 3 at http://www. ℓη and the time scale. Two boundary layers – one on each side of the plate – are shown. When the flow along the lower wall of the diffusor is tripped into turbulent flow.html i. What is a turbulent eddy? 12. Explain the cascade process. vη . What are the Kolmogorov scales? Use dimensional analysis to derive the expression for the velocity scale. . Consider the flow in a divergent channel (a diffusor): what happens with the boundary layer thickness and the wall shear stress? vi. What do we mean by a “separated boundary layer”? How large is the wall shear stress at the separation point? viii. iv. 10. 11.G. Consider the flow in a contraction: what happens with the boundary layer thickness after the contraction? iii. Is the wall shear stress lower or higher after the contraction? Why? v.chalmers. The flow is “tripped” into turbulence. How is the energy transfer from eddy-to-eddy. 15. try to explain how turbulence creates a Reynolds shear stress. v2′ ) fluctuations. What is its physical meaning? Describe the physical process of vortex stretching which creates smaller and smaller eddies. The energy spectrum consists of thee subregions: which? describe their characteristics. The movie laminar shows flow in a pipe. iv. computed? Use dimensional analysis to derive the −5/3 Kolmogorov law. the loss) decreases. Show and discuss the family tree of turbulence eddies and their vorticity. main difference? In which flow is the wall shear stress τw = µ ∂x2 laminar or turbulent? ii. Make a figure of the energy spectrum. and the pressure drop increases. The center part of the pipe is colored with blue dye and the wall region is colored with red dye: by looking at this flow.se/˜lada/flow viz.tfd. what happens with the pressure drop? 19.html i. and the pressure drop (i. the dye does not mix with the water. Draw a laminar and turbulent velocity profile for pipe flow. Dye is introduced into the pipe. estimated? Show how the ratio of the large eddies to the dissipative eddies depends on the Reynolds number. the resistance. εκ . Watch the on-line lecture Turbulence part 1 at http://www. . TME225 Learning outcomes 2014 261 14. Write the vortex stretching/tilting term in tensor notation. k. Discuss the connection between mixing and the cross-stream (i. Why does the irregular motion of wave on the sea not qualify as turbulence? ii. 18. In the movie they show that the flow can remain laminar up to 8 000. the velocity near the wall is larger than in laminar flow.chalmers.G.se/˜lada/flow viz.e. the drag. Why? How does the characteristics of the water flow coming out of the pipe change due to the second decrease of viscosity? iv. It is usually said that the flow in a pipe gets turbulent at a Reynolds number of 2300. What is the ∂¯ v1 largest. v. For laminar flow. What does isotropic turbulence mean? 16. in turbulent flow it does.html i.tfd. Why? iii. E(κ). How is the turbulence syndrome defined? iii. How do they achieve that? v. Why? The viscosity is further decreased. Describe the cascade process created by vorticity. In turbulent flow. The viscosity is decreased. Show the flow of turbulent kinetic energy in the energy spectrum. When the mixing occurs. 17.e. Show that in 2D flow the vortex stretching/tilting term vanishes. how is the turbulent kinetic energy. Given the energy spectrum. Watch the on-line lecture Turbulence part 2 at http://www. Try to explain the increased pressure drop in turbulent flow with the increased mixing.chalmers. 6. the water passes down the cascade) and the turbulent cascade process. Explain the analogy of a water fall (cascade of water. Make a figure and show where these regions are valid (Fig. The wall region is divided into an inner and outer region.G. The two turbulent jet flows have the same energy input and hence the same dissipation. TME225 Learning outcomes 2014 262 vi. ix. buffer layer and log-layer. Use this fact to explain why the smallest scales in the high Reynolds number jet must be smaller that those in the low Reynolds number jet. The inner region is furthermore divided into a viscous sublayer. one at low Reynolds number and one at high Reynolds number. 20. uτ . Use the decomposition vi = v¯i + vi′ to derive the time-averaged Navier-Stokes equation. They look very similar in one way and very different in another way. they explain the cascade process. 2 and v 22. At the end of the presentation of the jet flow. defined? Define x+ ¯1+ . A new terms appears: what is it called? Simplify the time-averaged Navier-Stokes equation for boundary layer flow. viii. Which scales are similar and which are different? vii. How is the friction velocity.2) . Two turbulent jet flows are shown. What is the total shear stress? 21. light fluid (yellow). TME225 Learning outcomes 2014 TME225 Learning outcomes 2014: week 6 263 June 15. k. Make a figure and show how the velocity and Reynolds shear stress vary across the channel. see Eq. 5. Study how the two fluids mix downstream of the partition.G. For which eddies is the production largest? Why? 6.se/˜lada/flow viz. Which two terms balance each other in the outer region? Which term drives (“pushes”) the flow in the x1 direction? Which two terms are large in the inner region? Which term drives the flow? 4. In the cascade process. Show this. How is the mixing affected? This flow situation is called unstable . which scales die first? The scenes of the clouds show this in a nice way. 2. In fully developed turbulent channel flow. the situation is reversed: cold. In the next example.16) iii. negative source) term? 5. we assume that the dissipation is largest at the smallest scales. Derive the exact transport equation for turbulent kinetic energy. 6. iv. 8. there are a couple of physical phenomena which may inhibit turbulence and keep the flow laminar: mention three. Watch the on-line lecture Turbulence part 3 at http://www. better or worse than in the previous example? This flow situation is called stable stratification.21). show how the three terms vary across the channel. v. What are the suitable velocity and length scales in the inertial region? Derive the log-law. Consider fully developed turbulent channel flow.e.19). the time-averaged Navier-Stokes consists only of three terms. ε(κ) ∝ κ4/3 . What do they mean? ii. Which terms do only transport k? Which is the main source term? Main sink (i. What are the relevant velocity and length scales in the viscous-dominated region? Derive the linear velocity law in this region (Eq.e. What happens with the small scales when the Reynolds number is increased? What happens with the large scales? Hence.18 at p. 94. The two fluids are identical. how does the ratio of the large scales to the small scales change when the Reynolds number increases (see Eq. heavy fluid (dark blue) is moving on top of hot. Discuss the physical meaning of the different terms in the k equation.html i. 3. After that.tfd. the fluid on the top is hot (yellow) and light. The movie says that there is a similarity of the small scales in a channel flow and in a jet flow. i. In which region (viscous sublayer. 2015 1. Consider the flow in the channel where the fluid on the top (red) and the bottom (yellow) are separated by a horizontal partition. In the last example. In decaying turbulence. 6. how do the fluids mix downstream of the partition. Even though the Reynolds number may be large. and the one at the bottom (dark blue) is cold (heavy).chalmers. buffer layer or log-layer) does the viscous stress dominate? In which region is the turbulent shear stress large? Integrate the boundary layer equations and show that the total shear stress varies as 1 − x2 /δ (Eq. for the shear stress: how large is it? 16. Derive the exact transport equation for turbulent Reynolds stress. 9. 14. The mean normal stress ′2 = v ′ v ′ /3. what sign will the pressure-strain term have for can be defined as vav i i ′2 ? What the normal stresses which are. 12. Consider the dissipation term. One term appears in both the k and the K equations: which one? Consider the dissipation terms in the k and the K equations: which is largest near the wall and away from the wall.G. Discuss the physical meaning of the different terms in the vi′ vj′ equation. TME225 Learning outcomes 2014 264 stratification. Which ones? Which terms are non-zero at the wall? 8. respectively? Why is there no pressure-strain term in the k equation? 17. You can read about stable/unstable stratification in Section 12. v2′2 . P k = −v1′ v2′ ∂¯ v1 /∂x2 . respectively? Consider the production term for v1′ v2′ : what sign does it have in the lower and upper part of the channel. the k equation is dominated by two terms. Consider the pressure-strain term in the vi′ vj′ equation. Show that the role of the convection and diffusion terms is purely to transport the quantity (k for example) and that they give no net effect except at the boundaries (use the Gauss divergence theorem). Is this consistent with the sign of P12 ? .26. vi′ vj′ . 11. Consider fully developed channel flow: how are the expressions for the production terms in the vi′ vj′ equations simplified? Which production terms are zero and non-zero.22). Discuss the physical meaning of the different terms. How is homogeneous turbulence defined? 10. 8. see Eqs.1 at p. Given the exact k equation. Derive the exact transport equation for mean kinetic energy. respectively? Show where they appear in the energy spectrum. Take the trace of the vi′ vj′ equation to obtain the k equation. respectively. ε12 . 7. Consider channel flow and use physical reasoning to show that v1′ v2′ must be negative and positive in the lower and upper half of the channel. Give an example of how they are combined in non-homogeneous turbulence. largest? In order to explain this. give the equation for boundary-layer flow (Eq. K. 138. 13. Where is the production term. larger and smaller than vav role does Π12 has? What sign? Why do we call the pressure-strain term the Robin Hood term? 15. respectively. Compare in meteorology where heating of the ground may cause unstable stratification or when inversion causes stable stratification.25 and 8. All spatial derivatives are kept in the dissipation term: why? In the turbulent region of the boundary layer. Discuss the difference of spatial transport of k and spectral transfer of k. v3′2 and v1′ v2′ equations? Which are the largest terms at the wall? Which terms are zero at the wall? 18. show how −v1′ v2′ and ∂¯ v1 /∂x2 vary near the wall. Consider the fully turbulent region in fully developed channel flow: which are the main source and sink terms in the v1′2 . 8. 2015 1.G. . Define the two-point correlation. TME225 Learning outcomes 2014 TME225 Learning outcomes 2014: week 7 265 June 15. How is it normalized? What physical meaning does it have? The integral time scale is defined in analogy to the integral length scale: show how it is defined. How is it normalized? What is the physical meaning of the two-point correlation? How is it related to the largest eddies? How is the integral length scale defined? 2. Define the auto correlation. A venturi meter is a pipe that consists of a contraction and an expansion (i. What happens with the velocity and pressure.2. Vs ds dp . In the movie they show that the acceleration along s.html i. They increase the speed in the venturi meter. The water flow goes through the contraction.html i. What is the static pressure? How can it be measured? What is the difference between the stagnation and the static pressures? v. The pressure decreases slowly downstream. Why? vi. Fluid particles become thinner and elongated in the contraction.e.tfd.chalmers.G. Eq. Explain why. The pressure now increases in the downstream direction. The bulk velocities at the inlet and outlet are equal.se/˜lada/flow viz.chalmers. this means that the pressure in the contraction region must decrease as we increase the velocity of the water.tfd. How large is the pressure at the surface of the pipe relative to the surrounding pressure? iii. Explain the relation between streamline curvature and pressure (cf. viii.e. TME225 Learning outcomes 2014 266 TME225 Learning outcomes 2014: just for fun! 1. There is a pressure drop. The pressure difference in the contraction region and the outlet increases. dVs . Since there is atmospheric pressure at the outlet.html i. but still the pressure at the outlet is lower than that at the inlet. i.tfd. The water from the tap which impinges on the horizontal pipe attaches to the surface of the pipe because of the Coanda effect. What happens with the pressure drop when there is a separation in the diffusor? vii. Watch the on-line lecture Pressure field and acceleration part 2 at http://www. . Why? iii. is related to iii. Try to explain.se/˜lada/flow viz. Section 3. Watch the on-line lecture Pressure field and acceleration part 1 at http://www. Why? ii. 2. Compare this relation with the three-dimensional the pressure gradient ds form of Navier-Stokes equations for incompressible flow. This is called cavitation (this causes large damages in water turbines). What is the stagnation pressure? How large is the velocity at a stagnation point? iv. ii. Finally the water starts to boil. What is the Coanda effect? ii. although the water temperature may be around 10o C. Watch the on-line lecture Pressure field and acceleration part 3 at http://www. Explain how suction can be created by blowing in a pipe. a diffusor).7 2.1).chalmers. 3. The bleeders (outlets) are opened.se/˜lada/flow viz. Water flow in a manifold (a pipe with many outlets) is presented. At the end of the contraction. Explain why. .G. TME225 Learning outcomes 2014 267 iv. there is an adverse pressure gradient (∂p/∂x > 0). H. 11.3) where the boundary integrals have been omitted.75 at p. 11. MTF270: Some properties of the pressure-strain term H 268 MTF270: Some properties of the pressure-strain term In this Appendix we will investigate some properties of aijkℓ in Eq. aijkℓ = aiℓkj (H.H. (H.3 gives vj′ vℓ′ = − 1 4π Z V ∂ 2 vℓ′ vj′ dy3 1 = aijiℓ ∂xi ∂xi |y − x| 2 where the last equality is given by Equation 11.4) . H. i.e.1) Since Eq. 129.1 in the definition of aijkℓ in Eq. Setting ϕ = vℓ′ vj′ in Eq. Introduce the two-point correlation function Bjℓ (r) = vj′ (x)vℓ′ (x + r) Define the point x′ = x + r so that Bjℓ (r) = vj′ (x′ − r)vℓ′ (x′ ) = vℓ′ (x′ )vj′ (x′ − r) = Bℓj (−r) We get ∂Bℓj (−r) ∂ 2 Bjℓ (r) ∂ 2 Bℓj (−r) ∂Bjℓ (r) =− ⇒ = ∂ri ∂ri ∂rk ∂ri ∂rk ∂ri (H.2) Green’s third formula (it is derived from Gauss divergence law) reads Z ∇2 ϕ 1 dy3 ϕ(x) = − 4π V |y − x| (H.75 is integrated over r3 covering both r and −r (recall that vℓ′ and vj′ are separated by r). aijkℓ is symmetric with respect to index j and ℓ.75. This is shown by transforming both the exact M (i. it is Galilean invariant. 172]. and thus the terms are not Galilean invariant. Now. t∗ = t.3. cr = 1 (see Eq.5) . ∗ L∗ij + Cij = Lij + Cij .e. xi )coordinate system to the (t∗ . For the Leonard and the cross term we get (note that since Vi is constant Vi = V¯i = V¯i ) vi + Vi )(¯ vj + Vj ) − (¯ vi + Vi )(¯ vj + Vj ) L∗ij = v¯i∗ v¯j∗ − v¯i∗ v¯j∗ = (¯ = v¯i v¯j + v¯i Vj + v¯j Vi − v¯i v¯j − v¯i Vj − Vi v¯j = v¯i v¯j − v¯i v¯j + Vj (¯ vi − v¯i ) + Vi (¯ vj − v¯j ) = Lij − Vj vi′′ − Vi vj′′ (I. I.e. Cij transforms as in Eq. The exact form of Cij Cij (Eq. Galilean invariance means that the equations do not change if the coordinate system is moving with a constant speed Vk . 18. (I.1) we get ∂x∗j ∂φ ∂t∗ ∂φ ∂φ ∂φ(xi . x∗k = xk + Vk t. x∗i ) (I.39). However. v¯k∗ = v¯k + Vk By differentiating a variable φ = φ(t ∗ .I. MTF270: Galilean invariance 269 I MTF270: Galilean invariance Below we repeat some of the details of the derivation given in [75].3 we find that the Leonard term and the cross term are different in the two coordinate systems.3) ∗ vi + Vi )vj′′ + (¯ vj + Vj )vi′′ = Cij = v¯i∗ vj′′∗ + v¯j∗ vi′′∗ = (¯ = v¯i vj′′ + vj′′ Vi + v¯j vi′′ + vi′′ Vj = Cij + vj′′ Vi + vi′′ Vj From Eq.4) The requirement for the Bardina model to be Galilean invariant is that the constant must be one. i. The Bardina term transforms as ∗ ∗ ∗M Cij = cr (¯ vi∗ v¯j∗ − v¯i v¯j ) i h vi + Vi )(¯ vj + Vj ) vi + Vi )(¯ vj + Vj ) − (¯ = cr (¯ = cr [¯ vi v¯j − v¯i v¯j − (¯ v i − v¯i )Vj − (¯ v j − v¯j )Vi ] M ′′ ′′ = Cij + cr v i Vj + v j Vi .e. note that the sum is. I. Transforming the material derivative from the (t. I. t) = + = ∗ ∂xk ∂xk ∂xj ∂xk ∂t∗ ∂x∗k ∂φ(xi . Let’s denote the moving coordinate system by ∗. Eq. let’s look at the Leonard term and the cross term. i. ∂t ∂xk It shows that the left hand side does not depend on whether the coordinate system moves or not.37) and the modelled one. Since the filtering operation is Galilean invariant [75].2 is it easy to show that the Navier-Stokes (both with and without filter) is Galilean invariant [75.e. 18. (I. 18.38). i. t) ∂φ ∂x∗k ∂φ ∂t∗ ∂φ ∂φ = Vk ∗ + ∗ . = + ∗ ∂t ∂t ∂xk ∂t ∂t∗ ∂xk ∂t (I.2) From Eq. we have v¯k∗ = v¯k + Vk and consequently also vk′′∗ = vk′′ . x∗i )-coordinate system gives ∂φ ∂φ ∂φ ∂φ ∂φ = ∗ + Vk ∗ + (vk∗ − Vk ) ∗ + vk ∂t ∂xk ∂t ∂xk ∂xk ∂φ ∂φ = ∗ + vk∗ ∗ . we have Cij to make the Bardina model Galilean invariant the Leonard stress must be computed explicitly. Note that in order the exact stress.I. but here this does not matter. Cij 6= Cij .4). Cij (see Eq. because provided cr = 1 the M modelled stress. I. Cij . as for ∗M M + L∗ij = Cij + Lij . transforms in the same way as the exact one. Cij . Thus. . MTF270: Galilean invariance 270 ∗M M As is seen. J. Compute the wavenumber vector. κnj x3 dAi θn κni x2 n ϕ x1 Figure J. κn1 = sin(θn ) cos(ϕn ) κn2 = sin(θn ) sin(ϕn ) κn3 = cos(θn ) p(ϕn ) = 1/(2π) p(ψ n ) = 1/(2π) p(θn ) = 1/2 sin(θ) p(αn ) = 1/(2π) 0 ≤ ϕn ≤ 2π 0 ≤ ψ n ≤ 2π 0 ≤ θn ≤ π 0 ≤ αn ≤ 2π Table J. MTF270: Computation of wavenumber vector and angles 271 J MTF270: Computation of wavenumber vector and angles For each mode n. i. κnj .1) .J. αn and θn (see Figs.1: The probability of a randomly selected direction of a wave in wave-space is the same for all dAi on the shell of a sphere. J. using the angles in Section J according to Fig. The probability distributions are given in Table J. see Fig. They are chosen so as to give a uniform distribution over a spherical shell of the direction of the wavenumber vector.1 and 26.e. J.1: Probability distributions of the random variables.1. J. (J. create random angles ϕn .1.1) and random phase ψ n .1 The wavenumber vector.1. 0. 0) (0. Hence. σin κni = 0 (superscript n denotes Fourier mode n). 1.2 Unit vector σin Continuity requires that the unit vector. vector κni . J. 0) (0.1 and J. Table J. 0) αn 0 90 (0.3) σin will lie in a plane normal to the σ1n = cos(ϕn ) cos(θn ) cos(αn ) − sin(ϕn ) sin(αn ) σ2n = sin(ϕn ) cos(θn ) cos(αn ) + cos(ϕn ) sin(αn ) σ3n = − sin(θn ) cos(αn ) (J.4) The direction of σin in this plane (the ξ1n − ξ2n plane) is randomly chosen through αn . 0. see Fig. 0) (−1. Unit vector σin 272 κn i (1. 1) (0.4. This gives (J.e. 1. 1) (0.2) i.2: Examples of value of κni .J. 0) 0 90 Table J. J. 1. 0. 0) 0 90 (0. 0.2 gives the direction of the two vectors in the case that κi is along one coordinate direction and α = 0 and α = 90o . −1) (−1. 0) σin (0.2. 0. 1. 0) (1. This can be seen by taking the divergence of Eq. σin and αn from Eqs.1. 26. −1) (0. and κnj are orthogonal. 0. 0. 0. 26. σin . .3 which gives ∇ · v′ = 2 N X n=1 u ˆn cos(κn · x + ψ n )σ n · κn (J. κ2 or κ3 ). −∞ < κ2 < ∞ and −∞ < κ3 < ∞. K. D. form a Fourier-transform pair (see Eq.e.2) can be expressed by the energy spectrum tensor as [68. The Reynolds stress ρv1′2 . The diagonal components of the two-point correlation tensor are real and symmetric and hence the antisymmetric part of exp(−ıκ1 x ˆ1 ) – i.6) for i = j.3) Eij (κ1 ) = 2 −∞ A factor of two is included because E ∝ Ψii /2 is used to define a energy spectrum for the turbulent kinetic energy k = vi′ vi′ /2. The v1′2 can be computed both from the .5 are simplified as Z Z ∞ 1 +∞ Eij (κ1 ) cos(κ1 x ˆ1 )dκ1 = Eij (κ1 ) cos(κ1 xˆ1 )dκ1 Bij (ˆ x1 ) = 2 −∞ 0 Z Z 2 +∞ 1 +∞ Bij (ˆ x1 ) cos(κ1 x ˆ1 )dˆ x1 = Bij (ˆ x1 ) cos(κ1 xˆ1 )dˆ x1 Eij (κ1 ) = π −∞ π 0 (K. and the one-dimensional spectrum. D. Both in experiments and in LES it is usually sufficient to study the twopoint correlation and the energy spectra along a line.e.28b) Z +∞ 1 Ψij (κ) = Bij (ˆ x) exp(−ıκm rm )dˆ x1 dˆ x2 dˆ x3 (K.10. x2 . one-dimensional energy spectra. the sinus part – is zero and Eqs. respectively. Chapter 3] (see Eq.1) −∞ where xˆm and κm are the separation vector the two points and the wavenumber vector. κ (κ1 . form a Fourier-transform pair. i. Note that the maximum magnitude of the wavenumber vector contributing to Eij (κ1 ) is very large since it includes all κ2 and κ3 . is equal to the two-point correlation tensor ρBij with with zero separation distance. are often used. MTF270: 1D and 3D energy spectra K 273 MTF270: 1D and 3D energy spectra The general two-point correlation Bij of vi′ and vj′ (see Eq. x ˆm .4 and K.8). They are formed by integrating over a wavenumber plane.28a) Z +∞ Bij (x1 .3. Hence. The two-point correlation. Eij (κ1 ). Bij . K. Ψij .8 and K. 10. Bij (ˆ x1 ). 18.e. Z 1 +∞ Bij (ˆ x1 ) = Eij (κ1 ) exp(ıκ1 x ˆ1 )dκ1 (K.4) 2 −∞ Z +∞ 2 Bij (ˆ x1 ) exp(−ıκ1 x ˆ1 )dˆ x1 (K. see Eqs.5) Eij (κ1 ) = 2π −∞ where Eij is twice the Fourier transform of Bij because of the factor two in Eq. reads Z 1 +∞ Ψij (κ)dκ2 dκ3 (K. The one-dimensional two-point correlation. for example. i. which are a function of scalar wavenumber. Eij (κ). x3 )) = Ψij (κ) exp(ıκm x ˆm )dκ1 dκ2 dκ3 (K. for example. the energy spectrum for the wavenumber κ1 . and the energy spectrum tensor.K. The complex Fourier transform exp(ıκm x ˆm ) is defined in Appendix D (see Eq. K. for example.2) (2π)3 −∞ The separation between the two points is described by a general three-dimensional vector. 4.9) The integral in Eq. • The energy spectrum.e. K.4).10) k= 2 0 where 4πκ2 is the surface area of the shell. 0. K. Eq. k.4. Ψij (κ).12) Below the properties of the three energy spectra are summarized. K. Z 1 ∞ 4πκ2 Ψii dκ (K.6 we stated that the one-dimensional energy spectra and the two-point correlations form Fourier-transform pairs.8) (K. Eij (κ1 ). D. The proof is given in this section. an be written as k= k= 1 2 Z 0 1 2 ∞ E11 (κ1 )κ1 + Z 1 2 +∞ −∞ Z ∞ 0 Ψii (κ)dκ1 dκ2 dκ3 Z 1 ∞ E22 (κ2 )κ2 + E33 (κ3 )dκ3 2 0 (K. see ˆ be the Fourier coefficient of the velocity fluctuaParseval’s formula. E(κ) = 2πκ2 Ψii so that Z κ E(κ)dκ. corresponds to the square of the Fourier coefficient of the velocity fluctuation (see Parseval’s formula.8 has no directional dependence: it results in a scalar. The energy spectrum is given by the square of the Fourier coefficients. k = vi′ vi′ /2. Take the covariance of the .1. 0) = Ψ11 (κ)dκ1 dκ2 dκ3 −∞ Z ∞ v1′2 = B11 (0) = E11 (κ1 )dκ1 (K. • The one-dimensional spectrum.1 ) and one-dimensional spectrum (Eq. K. for example. We now define an energy spectrum. is a tensor which is a function of a scalar (one component of κm ). |κ| ≡ κ. D. E(κ).6) Z v1′2 +∞ = B11 (x1 . Energy spectra from two-point correlations 274 three-dimensional spectrum (Eq. is a scalar which is a function of the length of the wavenumber vector. Let u tion u′ in the x direction which is periodic with period 2L. is a tensor which is a function of the wavenumber vector. k= (K. Instead of integrating over dκ1 dκ2 dκ3 we can integrate over a shell with radius κ = (κi κi )1/2 and letting the radius go from zero to infinity.7) 0 Hence the turbulent kinetic energy. E11 (κ1 ) = vˆ12 (κ1 ) (K. • The three-dimensional spectrum tensor. K. i.1 Energy spectra from two-point correlations In connection to Eqs.e.K.5 and K.11) 0 The energy spectrum E11 (κ1 ). Eq. K. i. 28b) of the two-point correlation B11 . it is seen that the Fourier transform of a two-point correlation (in this example hB11 (x1 )i) indeed gives the corresponding one-dimensional energy spectrum (in this example E11 (κ1 ) = h(ˆ u(κ))2 i).4 except the factor of two.13) where h·i denotes averaging over time.15 for κ = κ′ is evaluated as (see “length of of ψk ” in Appendix D. Since we are performing a Fourier transform in x we must assume that this direction is homogeneous. see Appendix D. K. K.1. The integral in Eq. D.17) 2L Hence. and the integral of these functions is zero unless κ = κ′ .15 can now be written 1 ˆ11 (x)i = hB ˆ11 (x)i hˆ u(κ)ˆ u(κ)i = (L + L)hB (K. all averaged turbulence quantities are independent of x and the two-point correlation is not dependent of x (or x′ ) but only on the separation distance x − x′ . K. D. B and symmetric since B11 is real and symmetric.14) The second integral (in parenthesis) is the Fourier transform (which corresponds to the Fourier coefficient in discrete space. Hence we replace x′ by x + x′′ so that * ! + Z L Z L−x 1 ′ ′′ ′ ′′ ′′ hˆ u(κ)ˆ u(κ )i = u(x)u(x + x ) exp(−ı(κx + κ (x + x ))dx dx 4L2 −L −L−x ! + * Z L Z L−x 1 1 ′ ′′ ′ ′′ ′′ exp(−ı(κ + κ )x) B11 (x ) exp(−ıκ x ))dx dx = 2L −L 2L −L−x (K. see discussion in connection to Eq.13. Eq. Energy spectra from two-point correlations 275 Fourier coefficients. Furthermore.16) L x L 4πx = + sin =L 2 8π L −L In the same way. . the “length of of φk ” in Eq. 108. + * Z L ′ ′ ′′ 1 ˆ exp(−ı(κ + κ )x))dx (K. The remaining integral in Eq. this equation corresponds to Eq.e. see Eq.16 is also L. i. They are orthogonal functions. u ˆ(κ′ ) and uˆ(κ) where κ and κ′ denote two different wavenumbers and x and x′ denote two points separated in the x directions so that + * Z L Z L 1 1 ′ ′ ′ ′ ′ hˆ u(κ)ˆ u(κ )i = u(x) exp(−ıκx)dx u(x ) exp(−ıκ x )dx 2L −L 2L −L * + Z LZ L 1 ′ ′ ′ ′ = u(x)u(x ) exp(−ı(κx + κ x )dxdx 4L2 −L −L (K.5) and since it does not depend where B ˆ11 is real on the spatial coordinate it has been moved out of the integral. i. Equation K. D.15 includes trigonometric function with wavelengths κ and κ′ . and use ψ1 = cos(2πx/L)) Z L 2πx (ψ1 |ψ1 ) = ||ψ1 ||2 = cos2 dx L −L (K.e.15) hˆ u(κ)ˆ u(κ )i = B11 (x ) 2L −L ˆ11 denotes the Fourier transform of B11 (cf. 10.K.6 on p. K. (v1′2 .1 Part I: Data of Two-dimensional flow You can do the assignment on your own or in a group of two.m. v2′2 . Assignment 1: Reynolds averaged Navier-Stokes 276 L MTF270.m) which reads the data and plots the vector field and the pressure contours. You will analyze one of the following flows: Case 1: Flow over a wavy wall (small wave) [173. It is available on Linux.lyx. Download the data from http://www.miktex. L. the deadline can be found at the Student Portal. At the wwwpage you can download a M-file (pl vect. ρ = 1) based on the bulk velocity in the channel at x = 0 and the channel height. You will use Matlab. MTF270. You must also download the function dphidx dy. ρ = 1) based on the bulk velocity in the channel at x = 0 and the channel height. p¯) and turbulent quantities. should be submitted electronically at the Student Portal www.org) or MikTex (www. Lyx (www. v3′2 . Contact the teacher to get a Case No. ρ = 1) based on the bulk velocity in the channel at x = 0 and the channel height. i. It is recommended (but the not required) that you use LATEX(an example of how to write in LATEXis available on the course www page). The two-dimensional time-averaged Navier-Stokes for the x1 momentum reads (the density is set to one.1. You can also use Octave on Linux/Ubuntu. v1′ v2′ . along with the Matlab files(s).e. Use Matlab or Octave to read data files of the mean flow (¯ v1 . and ε).org) which are both free to download. Make sure you put this function in the directory where you execute pl vect. see Section L. Although some of the data are probably not fully accurate.chalmers.m which computes the gradients. Re = 10 000 (ν = 1 · 10−4 .2.student. The report.1) . 173]. Octave is a Matlab clone which can be downloaded for free. ρ = 1) based on the bulk velocity in the channel at x = 0 and the channel height. On Windows you can use. In which regions do you expect the turbulence to be important? Now let’s find out. in this exercise we consider the data to be exact. Case 3: Flow over a hill Re = 10 595 (ν = 1/10595.se/˜lada/comp turb model.1 Analysis Study the flow.tfd. Case 4: Flow over two hills Re = 10 595 (ν = 1/10595. Re = 10 000 (ν = 1 · 10−4 . Case 2: Flow over a wavy wall (large wave) [173. ρ = 1) ∂¯ v1 v¯2 ∂ p¯ ∂ 2 v¯1 ∂v ′2 ∂ 2 v¯1 ∂v ′ v ′ ∂¯ v1 v¯1 + =− +ν − 1 +ν − 1 2 2 2 ∂x1 ∂x2 ∂x1 ∂x1 ∂x1 ∂x2 ∂x2 ∂¯ v1 v¯2 ∂¯ v2 v¯2 ∂ p¯ ∂ 2 v¯2 ∂v1′ v2′ ∂ 2 v¯2 ∂v2′2 + =− +ν − + ν − ∂x1 ∂x2 ∂x2 ∂x21 ∂x1 ∂x22 ∂x2 (L. Periodic boundary conditions are imposed in streamwise (x1 ) and spanwise (x3 ) directions in all flows.portal. The work should be carried out in groups of two (if you want to work on you own that is also possible) .L. for example.se. Assignment 1: Reynolds averaged NavierStokes L. You’ll use data from a coarse DNS. 173]. v¯2 . The database corresponds to a two-dimensional flow.’fontsize’.y. In v1 /∂x1 and ∂¯ v1 /∂x2 . Make also a zoom near the walls. with ’surf’ and ’quiver’.1.2) .4].2) % linewidth=2 you may want to zoom in on y=[0 0. pl vect.m loads the data file and plots the profiles of v1′2 at some x stations. v¯1 and p¯.m the function dphidx dy.1 0.01] and u=[-0.L. To enhance readability you may omit the small terms or use two plots per vertical grid line.m is used to compute ∂¯ Use this function to compute all derivatives that you need. Plot the stresses along vertical grid lines at these two locations using the Matlab command plot(x. One way to find these locations is to use the Matlab surf command.− ∂x1 ∂x2 .[20]) title(’velocity’. You will need to compute the derivatives of e. Find two (or more) x1 locations (vertical grid lines) where the v1′2 stress is large and small.’fontsize’.2. For example.[20]) ylabel(’y/H’.1. for a x − y plot plot(u. you can use the command h1=gca.1. Now let’s think of the forces as vectors. L.4 0 0. Please make sure that in your report the numbering on the axis and the text in the legend is large enough. this is achieved by axis([-0. e. Plot also all terms in Eq. the velocity vector field and a contour plot of velocity gradient ∂¯ v1 /∂x2 .y).[20]) %the number ’20’ gives the fontsize The size of the labels and the title is similarly controlled by xlabel(’x/H’. Part I: Data of Two-dimensional flow 277 Recall that all the terms on the right-hand side represent x components of forces per unit volume.1.g. L.g. respectively.2 The momentum equations The file pl vect. Which terms are negligible? Can you explain why they are negligible? What about the viscous terms: where do they play an important role? Which terms are non-zero at the wall? (you can show that on paper).’fontsize’. So far we have looked at the v¯1 -momentum equation. The gradient of the normal stresses in the x1 − x2 plane represent the force vector (see Eq.1. L. Compute all terms in Eq.1 0.[20]) Assignment 1. N.’fontsize’.’linew’.2) ! ∂v1′2 ∂v2′2 FN = − (L.01]) The ’axis’ command can be used together with any plot. set(h1. Assignment 1. 1. involves k an ε. Plot also vector fields of the shear stress. Plot the vector field FN to find out some features. |ni. Zoom up in interesting regions.4) Assignment 1.n | = (x(I.3 The turbulent kinetic energy equation The exact transport equation for the turbulent kinetic energy. The production term consists of the sum of four terms.3. Φij . what happens with the vector field? Zoom-in on interesting regions. N. These should be taken from the 2D RANS simulation (they are loaded in pl vect.4 The Reynolds stress equations The modeled transport equation for the Reynolds stresses can be written as ∂ 2 vi′ vj′ ∂ ′ ′ v¯k vi vj = ν + Pij + Φij + Dij − εij ∂xk ∂xk ∂xk ∂¯ vj ∂¯ vi − vj′ vk′ Pij = −vi′ vk′ ∂xk ∂xk (L. Explain why it is large at some locations and small at others.5) The pressure-strain term.3. Plot the production term along the two grid lines. the lower wall is the closest wall to node (I.e.3).6) The damping function.2) ! ∂v1′ v2′ ∂v1′ v2′ FS = − (L. FS (see Eq.L.6. Do you have local equilibrium (i.1. Anything interesting? L.− ∂x2 ∂x1 Identify the first term in Eqs. In the damping function.93 and 11. 11. J) − x(I. reads ∂ ∂2k (¯ vj k) = ν + Pk + Dk − ε ∂xj ∂xj ∂xj ∂¯ vi Pk = −vi′ vj′ ∂xj (L. Plot the dissipation and compare it with the production. 11.w (xi −xi. L.1. . f (see Eq. L.55.88. Note that FN and FS are forces per unit volume ([N/m3 ]). Assignment 1.90).2 and L. the pressure gradient and the viscous terms. L. 1))2 + (y(I.1. Dij .5. 11.3) .2 and L.w )| denotes the distance to the nearest wall. 11. need to be modeled. J). f . L. If. and the diffusion term. J) − y(I. Part I: Data of Two-dimensional flow 278 and the corresponding force vector due to the shear stresses reads (see Eq.94. for example. Assignment 1.4. k.89. 1))2 (L. two of which involve the shear stress while the other include the normal stresses. Eq. Compare the contributions due the shear stress and the normal stresses.m).3 in the momentum equation for v1 . Write out the momentum equation also for v¯2 and identify the other two terms in Eqs. P k ≃ ε) anywhere? L. 11. Here we use the models in Eqs. then 1/2 |xi − xi. 1. When v2′2 reaches a maximum or a minimum along a grid line normal to the wall. Assignment 1. Part I: Data of Two-dimensional flow 279 2. and compute then ni.4) is usually positive. νt = Cµ k 2 /ε (L. When using the Boussinesq assumption the production of turbulent kinetic energy (use k and ε from the 2D RANS simulation) P k = 2νt s¯ij s¯ij (L. The exact production of turbulent kinetic energy (see Eq. explain why. 0. 1).5. Compare the eddy-viscosity stresses with two of the Reynolds stresses from the database. c2 . c2w . To compute ni. and the GGDH is then replaced by a simple eddy viscosity model ! ∂ νt ∂vi′ vj′ Dij = .w from si.L.8.11) 3. 1. Make also a zoom-in near walls. use two plots per vertical grid line or simply omit terms that are negligible. To find out to what degree the exact Reynolds stress and the strain rate are parallel. Use k and ε from the 2D RANS simulation.w (see Eq.9) is always positive. Try to explain why some terms are large and vice versa. Is that the case for you? Assignment 1. Use the simple eddy viscosity model for the turbulent diffusion term.9. that the lower wall is the closest wall to cell (I.1. 0. L. c1w .94). L. σk ) = (0. The reason why the eddy-viscosity production in Eq.w .e vi′ vj′ = −2νt s¯ij + (2k/3)δij where νt = cµ k 2 /ε.09. If so. vi′ vj′ . 1) Assignment 1. 11. It can however become negative.6. Another way to explain this fact is that the modeled Reynolds stress. i. one can compute the eigenvectors. si. The diffusion terms Dij and Dε can be modeled using the Generalized Gradient Diffusion Hypothesis GGDH of [174] ! k ∂vi′ vj′ ∂ Dij = cuk um (L.8) ∂xm σk ∂xm The following constants should be used: (cµ . Use k and ε from the 2D RANS simulation. Usually.7) ∂xm ε ∂xk This diffusion model can cause numerical problems. compute first the vector which is parallel to the wall.3. L. .7. c1 . s¯ij are parallel.5. then ni. and the strain rate tensor. a stress is large in locations where its production (or pressurestrain) term is large. If the figure becomes too crowdy. Choose two stresses.9 must be positive is of course that neither νt nor s¯ij s¯ij can go negative. again.w = (0.w for the general case (see Eqs. If we assume.4 in the entire domain to investigate if the production is negative anywhere. Assignment 1.93 and 11. L. Compute the stresses using the Boussinesq assumption. 0. Plot the different terms in the equations for one vertical grid line fairly close to the inlet (not too close!). Compute the exact production in Eq. J) and that the lower wall is horizontal. The eigenvalues correspond to the normal stresses in the direction of the eigenvectors.nw .ne . On the www page you download RANS and DNS data.11. The eigenvalues correspond to the normal strain rate in the direction of the eigenvectors (see Section 13). φ. i. x2.1). compute it as v1. In order to do that we use Gauss’ law over a control volume centered at face e (dashed control volume in Fig.e. you will get eigenvectors in the direction (±1.se ). x2. On a Cartesian grid it is more convenient to use the built-in Matlab function gradient. including Cartesian ones. Zoom in on interesting regions.10) Let’s compute ∂v1 /∂x1 . In which regions differ the eigenvectors of the Reynolds stress tensor and those of the strain rate tensor most? This should indicate regions in which an eddy-viscosity model would perform poorly.sw .P = 1 (v1.L. reads Z Z ∂φ φni dA dV = V ∂xi A To compute ∂v1 /∂x1 we set φ = v1 and i = 1 which gives Z Z ∂v1 v1 n1 dA dV = A V ∂x1 Assuming that ∂v1 /∂x1 is constant in the volume V we obtain Z 1 ∂v1 = v1 n1 dA ∂x1 V A . Since the two eigenvectors are perpendicular to each other it is sufficient to plot one of them (for example. the off-diagonal components) dominate.2.2 Compute derivatives on a curvi-linear mesh In this section we describe how the derivatives on a curvi-linear grid are computed in the provided Matlab function dphidx dy. L.nw ) and (x1. (x1. Compute the eigenvalues and eigenvectors of the strain rate tensor. 1). but the approach used below works for all meshes. The divergence theorem for a scalar. the diagonal components) dominate the direction of the eigenvectors will be along the x1 and x2 axes (explain why!). Zoom in on interesting regions. If the shear strain rates (i. s¯ij .m.10. x2. (x1.e. When you need a variable. Assignment 1.e. thus we get two eigenvectors and two eigenvalues. Make also a comparison of eigenvectors and eigenvalues in fully-developed channel flow. L.se + v1. Compute derivatives on a curvi-linear mesh 280 Assignment 1. The data you have been given.ne ). Plot the eigenvectors as a vector field.sw ). ±1) and if the normal strain rates (i. all represent the same principal coordinate system).sw + v1. at (x1. say v1 . at the center of the cell. Compute the eigenvalues and eigenvectors of the Reynolds stresses. vi′ vj′ . x1 and x2 and all variables are stored at the grid points. Zoom in on interesting regions. −1). (−1. the eigenvectors (1.nw + v1. −1) and (1. Our flow is 2D. x2.se . 1). (−1.ne ) 4 (L. s2e ) P ne = (n1e . .2. n2e ) E W sw se S Figure L.se de s2e = x2. s and the unit vector in the z-direction s·n =0 s × zˆ = n. For the east face. ).L. nw.11) .w 1 {(v1 An1 )e + (v1 An1 )n + (v1 An1 )w + (v1 An1 )s } V L.1) ∂v1 1 = ∂x1 V X (v1 n1 A)i = i=e. nw.se de de = q (x1.ne − x2.2. for example. .1: Control volume.ne − x1. (L.ne − x2. Coordinates x1 . The relation between the normal vector n.ne − x1. gives us the normal vector for the east face as n1e = s2e n2e = −s1e .). Compute derivatives on a curvi-linear mesh 281 N ne nw se = (s1e . x2 are given at the corners (ne.w.se )2 (note that the area of the east face Ae is equal to de since ∆z = 1). L. .se )2 + (x2. The velocity v1 is stored at the corners (ne. .n. In discrete form we can write (see Fig. we get s1e = x1...1 Geometrical quantities It is useful to first compute the unit vectors s along the control volume. 3 282 Part II: StarCCM+ In this task. To access the user guide. 1.011) will be used. 4. 3. a macro file and experimental data have been provided. Tick the Power-On-Demand box and fill the license box • Ask the assistant for the POD license. • Ask the assistant for the POD license 4. They can be downloaded from the course homepage. 3. Tick the Power-On-Demand box and fill the license box. type StarCMM+ 2. The tutorial data can be downloaded from the course homepage L. . Open a terminal window. click Help → Tutorial Guide • Introduction tutorial can be found directly on the left hand side.3.L. L. a commercial CFD software (StarCCM+ 9. To start a new simulation.06. For simplicity. 1. This tutorial is a good bridge to the asymmetric diffuser case. click File → new simulation • Other method: click the new simulation icon on the top left corner of the screen. Open a terminal window.1 Introduction Tutorial This tutorial is intended to first time user of the sTARCCM+. a mesh. 5. Several turbulence models will be used and the results will be compared with experimental data. The tutorials can be found in the StarCCM+ user guide.3. To access the user guide.2 Backward Facing Step Tutorial This tutorial is a good bridge before doing the asymmetric diffuser case. The task is to do simulation of an asymmetric diffuser. Part II: StarCCM+ L. The introduction tutorial (not mandatory) is a good starting point to get used to workflow of the sTARCCM+. To start a new simulation. In the terminal window. click File → new simulation • Other method is by clicking the new simulation icon on the top left corner of the screen.3. click Help → Tutorial Guide • The steady flow: backward facing step tutorial can be found inside Incompressible Flow folder. Here are some steps to access the tutorial. In the terminal window. several inlet velocity profiles. The next tutorial is the steady backward facing step tutorial. Here are some steps to access the tutorial. Before doing the task. type sTARCCM+ 2. it is recommended to first do some of the tutorials of the StarCCM+. Make other interesting comparisons! L. Start by importing the mesh • Download the mesh from the course homepage • On StarCCM+ window → Click File → Import → Import Volume Mesh • Load your mesh → AsymmetricDiffuser 2D.3. Plot the turbulence production. 1. The tutorial data can be downloaded from the course homepage. Now you can choose a new turbulence model. How much larger is the turbulence in the shear layer compared to the flow in the boundary layer upstream of the step? How large is the turbulence intensities in the two regions. Assignment 1. segregated flow and tick Temporary Storage Retained Run one iteration to create the date and then open a Scalar Scene and plot the turbulence production. Do this by going to the solvers folder. El-Behery & Hamed [175] 3. Choose a new turbulence model (right click on Continua/Fluid and select Select models . It is highly recommended to read these references before doing the case.3.ccm 4.4 Brief instruction to begin the asymmetric diffuser case 1. Untick All y+ Wall treatment and then untick the turbulence model.. Select turbulence model • (If Physics 1 is not present) Right click on Continua → New → Physics continuum . L. Buice & Eaton [176] 4. Part II: StarCCM+ 283 5. Run a couple different turbulence model and compare the results.12.13. The Reynolds number (Re = 20 000) is based on the bulk velocity (Ub = 20m/s) and the inlet channel height (H = 15mm).L. Compare the turbulence levels in the shear layer for different turbulence models.3 Asymmetric diffuser case In this case 2D Asymmetric diffuser will be studied. 2. P k . Davidson [144]..2: Flow through an asymmetric plane diffuser Ercoftac database.3. A short recirculation region is usually connected to high turbulence in the shear layer bounding the recirculation region. The case is built based on these references. Ercoftac case 8.. Start the StarCCM+ • Open a terminal window → type StarCCM+ 2. Start a new simulation • Click the new simulation icon → tick the power on demand box • Fill the license box with the POD license 3. Assignment 1. Continue to work on the backward facing step flow. Set the Inlet boundary condition • Expand Tools folder → right click on Tables → choose New tables → File Table • Load inlet velocity profile which match the chosen turbulence model. Set the stopping criteria • Right click on the Report folder → go to New Report → click Pressure Drop • In the Pressure Drop 1 properties window (a) High Pressure = Inlet (b) Low Pressure = Outlet • Right click on the Pressure Drop 1 → click on create monitor and plot from report • Right click on the Stopping criteria folder →choose create new criterion → create from monitor → click on Pressure Drop 1 monitor • In the Pressure Drop 1 Monitor criterion properties window . • Expand Region folder → Boundaries → inlet (a) Physics Conditions → Velocity Specification → Method → Components (b) Physics Values → Velocity → Method → Table (x. Part II: StarCCM+ 284 • (If Physics 1 is present) Right click on Physics 1 → Select model • In the new window select: (a) (b) (c) (d) (e) (f) Steady Gas Segregated flow Constant Density Turbulent Choose turbulence model that you want to use. Set the Gas properties • Expand Physics 1 folder → Gas → Air → Material Properties → Dynamic Viscosity → Constant → Value = 6E − 5P a − s 6. since the provided mesh has a low Y+ value.y. iv. iii. Table = Inlet velocity according to the turbulence model Table: X-Data = velocity[i] Table: Y-Data = velocity[j] Table: Z-Data = velocity[k] 7.z) i. Extract more data from the simulation (for example: Turbulent kinetic energy production) • Expand Solvers folder → Segregated Flow → tick temporary storage retained 8. ii.3.y. 5.L. Choose only turbulence model that have All y + wall treatment.z) Table (x. 56m X17 = 16. Go to Parts → select all the derived parts • Expand the X-Y Plot 1 (a) X-type → Type = Scalar → Scalar Function → Field Function = Velocity[i] (b) Y-types → Y Type 1 → Type = Direction → Vector Quantity → Value = [0. z = 0 • Display → New Geometry Displayer • Specify the x-coordinate according to the experimental data (see below) • Click Apply for each of the x-coordinate (a) (b) (c) (d) (e) (f) (g) (h) (i) (j) X − 6 = −5.09m X30 = 30. • Right click on Derived Parts folder → New Part → Section → Plane • Input Parts → Select → All the region • Set normal → x = 1.L.32m X24 = 23.59m X06 = 5.71m X27 = 27. Create planes to extract the results at the same location as the experimental data. Part II: StarCCM+ 285 (a) Criterion option = Asymptotic (b) Click on Asymptotic limit • set [max-min] value = 1E − 4 → Number of samples = 100 • If the simulation stop before convergence is reached. y = 0.98m X14 = 13.48m X34 = 33.87m 10. This will disabled the maximum iteration criteria. 9.3. set the [max-min] value = 1E − 5 • Expand the Stopping Criteria folder → Go to Maximum Steps → in the Maximum Steps properties windows untick the enabled.93m X20 = 20.87m X03 = 2. Insert experimental data • Download the experimental data from the course homepage • Expand Tools folder • Right click on Tables folder → choose Create a New Table → click File Table → Load all the experimental data .0] – Go to each Plane Section and change the X Offset accordingly 11. Create the X-Y plot to monitor the result • Right click on Plots folder → choose New Plot → click on X-Y Plot • On the new X-Y Plot properties window (a) i.1. how much does the pressure increase from inlet to outlet).e. How large is the pressure recovery (i. compare the shear stresses v1′ v2′ .L. If the recirculation region is different. In the backstep flow. the reason is maybe connected to the turbulence level in the shear layer above the recirculation region. Are the results different? Where do the difference appear? In the plane channel upstream of the diffuser? Compare turbulent quantities such a k and νt (or if you chose to use a Reynolds stress model. Run the simulation • Go to Solution → Run • Other method: click icon which show a running man. Assignment 1. Or maybe the reason is that the incoming boundary layers are different. Part II: StarCCM+ 286 • Go back to X-Y plot monitor → right click on Data Series folder → Add Data → Select all the experimental data • Expand Data series folder → select all the experimental data by click the first data hold shift button and click the last data → right click on the highlighted area → click swap column 12.14.15. you found that one turbulence model was better than the other one(s). Assignment 1. Is the same turbulence model best for this flow? Make other interesting comparisons! .3. Choose a new turbulence model Run the same turbulence models as for the backstep case and compare the results. Look at the results. Run the program pl time. h. y.se/˜lada/comp turb model/ you find a file u v time 4nodes. The cell size in x and z directions are ∆x = 0. t = 10 and t = 11 and umin = 3 and umin = 21. change the plot command to plot(t. see Fig.dat with the time history of u and v.tfd.0033 (every second time step). The size of the domain is (L. M. Assignment 2: LES 287 M MTF270. Assignment 2: LES You can do the assignment on your own or in a group of two.m which loads and plots the time history of u. y/δ = 0.’bo’) Use this technique to zoom and to look at the details of the time history.1) The Re number based on the friction velocity and the half channel width is Reτ = uτ h/ν = 500 (h = ρ = uτ = 1 so that ν = 1/Reτ ). for example. The file has eight columns of u and v at four nodes: y/δ = 0.0176.M. It is recommended (but the not required) that you use LATEX(an example of how to write in LATEXis available on the course www page). Use Matlab or Octave.miktex. Alternatively.47.m. you can use the zoom buttons above the figure.8. The streamwise. and thus what you see is really u/uτ .0654 and ∆z = 0.0176 and y/δ = 0. y/δ = 0. y (x2 ) and z (x3 ) respectively.chalmers. A 96 × 96 × 96 mesh has been used. y + = 8. This is conveniently done with the command axis([10 11 3 21]) In order to see the value at each sampling time step.1. How does the time variation of u vary for different positions? Why? Plot also v at the four different positions.0164. With uτ = 1 and ν = 1/Reτ = 1/500 this correspond to y + = 1. The equations that have been solved are ∂vi =0 ∂xi ∂vi ∂p 1 ∂ 2 vi ∂ (vi vj ) = δi1 − + + ∂t ∂xj ∂xi Reτ ∂xj ∂xj (M.95. Download the Matlab/Octave program pl time.u2. It is available on Linux. The sampling time step is 0. Study the time history of the blue line (y/δ = 0. Periodic boundary conditions were applied in the x and z direction (homogeneous directions).’b-’.107 and y/δ = 0.org) which are both free to download.org) or MikTex (www. y + = 53. Plot u for all four nodes.0176) in greater detail.1 Time history At the www-page http://www. z). for example. Octave is a Matlab clone which can be downloaded for free. wall-normal and spanwise directions are denoted by x (x1 ). Make a zoom between.5 and y + = 235. The time history of u at y/δ = 0.M.t. Lyx (www. What is the differences between u and v? .lyx.0039. On Windows you can use. Zmax) in (x. Recall that the velocities have been scaled with the friction velocity uτ .u2. You will receive data from a DNS of fully developed flow in a channel. MTF270.107 are shown. 2 Compute the autocorrelation for v1′ using the Matlab command imax=500.2.2) We see that the statistical error decreases as N −1/2 . In Section M.11 not to infinity.2 but use samples every Tint second. Plot the autocorrelation as plot(t(1:imax). In reality it is not only the number of samples that is relevant.10. int_T_1=trapz(two_uu_1_mat)*dt. as 1 v1. Add the following code (before the plotting section) umean=mean(u2) Here the number of samples is n = 5000 (the entire u2 array).2) xlabel(’t’) ylabel(’u’) handle=gca set(handle. You investigated how few samples you could use.two_uu_1_mat(1:imax).’fontsi’.e.rms error = √ N hv1 i (M. If the samples are separated by Tint seconds they are independent. Hence. Note that the autocorr norm command returns the normalized autocorrelation. To find out if two samples are independent.M.imax). Tint . Time averaging 288 M. The theoretical estimate of the statistical error varies with number of independent samples. .e. provided that the samples are independent.3 Auto correlation Auto correlation is defined in Section 10. two_uu_1_mat=autocorr(u1. Start by trying with 100 samples as umean_100=mean(u2(1:100)) Do the same exercise for the other three nodes. N .1 you time averaged the velocities to get the mean value. B11 . where we set the maximum separation in time to imax = 500 (i. re-do the averaging you did in Section M. Plot the normalized autocorrelation and compute the integral time scales also for the other three points. i. Find out how many samples must be used to get a correct mean value.’linew’. 10.2 Time averaging Compute the average of the u velocity at node 2. but also that they are independent. M. but to imax · ∆t). see Eq.[20]) Compute the integral time scale Tint as dt=t(1). we carry out the integration in Eq. it is convenient to use the integral time scale. 10. S. Compute and plot the PDFs of the points close to the wall and you will find that they are more skewed. This hypothesis assumes that – if the turbulence level is not too strong – the velocity fluctuation at point x and time t is the same as that at time (t − τ ) at point (x−ξ) where τ = (x−ξ)/hui (it takes time τ for the particle to travel from point (x−ξ) to point x with a velocity hui).’fontsi’.M.3.4 Probability density/histogram Histogram (also called probability density function. . The variable pdf 3 is a vector of length 20 whose elements gives the number of samples in each bin. When we want to find the velocity fluctuation at point (x − ξ) at time (t − τ ) we can instead take it at point x at time t.2) xlabel(’u3’) ylabel(’PDF’) handle=gca set(handle. see Section 7. fv . i. S=mean(u3_fluct. is a variable that quantify the skewness and it is defined as Z ∞ 1 Sv ′ = 3 v ′3 fv′ (v ′ )dv ′ vrms −∞ Compute it as urms3=std(u3_fluct). Probability density/histogram 289 Let us use Taylor’s frozen turbulence hypothesis to compute the integral length scale. Compute the PDF as u3_fluct=u3-mean(u3).3) Lint = hui 0 Compute the integral lengthscale. as we do when we define a time average. max(u3 )]. [min(u3 ). the mean velocity is computed as Z ∞ hvi = vfv (v)dv (M.ˆ3)/urms3ˆ3. The hypothesis assumes that the turbulence is frozen between point (x − ξ) and x. plot(u3_pdf. Verify that the magnitude of S is large for the walls close to the wall.e. of the v velocity. M. PDF) can give additional useful information. where urms3 = hv3′2 i1/2 .’linew’. Plot the histogram as norm3=sum(pdf3)*(u3_pdf(2)-u3_pdf(1)). The mean velocity can of course also be computed by integrating over time. [pdf3 u3_pdf]= hist(u3_fluct.5) −∞ Here we integrate over v.20) Here we have divided u3 into 20 bins which span the variation of u3 . With a probability density. 7. You find that the PDF is rather symmetric.pdf3/norm3.[20]) where norm3 is the integral in Eq.4.4) −∞ Normalize the probability density function so that Z ∞ fv (v)dv = 1 (M. Skewness. The Taylor hypothesis makes it possible to compute the integral lengthscale from the integral timescale as Z ∞ norm ˆ ˆ B11 (t)dt = huiTint (M. 2) Has the region in which the energy spectrum decays fast vanished? If not. you find that for even higher frequencies the kinetic energy decays even faster. You do a FFT of that signal to get the Fourier coefficients ai and then plot a2i . The Matlab file pl spectrum. It uses the Matlab function pwelch.’linew’. Actually. and do a FFT of that signal.6 Resolved stresses At the www-page (http://www.f_L]=pwelch(u4(i).5 Frequency spectrum One way to verify that the LES you have performed resolves the turbulence properly. In this case you must average over many instants to get a reasonably smooth spectrum.se/˜lada/comp turb model) you find data files with three instantaneous flow fields (statistically independent). As you can see from the plot.5. It should be mentioned that spectra may not be a reliable measure of resolution [112. Since the data files are rather large. v (v2 ). and the coarser the sampling.’r--’.e. The reason is that we are using nmax=256 samples.chalmers. The data files are Matlab binary files.[]. use even coarser sampling (set m=8. 113]. You may note that the low region of the frequency spectrum is also modified. w (v3 )and p (made nondimensional by uτ and ρ). how much energy resides in each frequency. i=1:m:n. the inverse of the wave length.e. [px_L. This is usually the case when the maximum CFL number (Courant-Friedrichs-Lewy condition) is set to one. Look at the spectra for the three other points (make changes in pl spectrum. i. the higher the lowest frequencies. every 4th of the samples. Then you get the frequency spectrum. is to look at the spectra of the resolved turbulence. since CFL in many points is much smaller than one (CFL=1 in one cell means that a fluid particle is transported across that cell during one time step). The same data file as in the previous exercise is used (u v time 4nodes. (u = hui + u′ ) and plots a2i . in a well-resolved LES we want to have the cut-off in the inertial subrange where the kinetic energy decays as the wave number (or frequency) up to the power of −5/3. The coarser sampling that is used. plot(f_L. Run the program by typing pl spectrum As we mentioned in the Lecture Notes. note that you must then set nmax=256. M. dt=m*t(1).px_L. u′ . say. Frequency spectrum 290 M. this is the case.tfd.[]. The high frequencies exist because we over-resolve in time compared to the resolution in space. Thus we want the resolved turbulence to have a behaviour like this for high frequencies. Investigate this by using only.m does a FFT of the time history of fluctuating velocity.M. One can analyze the time history of a variable at a point. % 4*dt m=4.dat). i. The other way to do it is to look at the instantaneous velocity along a grid line.e. it is recommended that you do all tasks using only . i. You find a Matlab/Octave program at the www-page which reads the data and computes the mean velocity. The data files include the instantaneous variables u (v1 ).1/dt). the lower frequencies appear in the spectrum. Two-point correlations are usually better.m).nmax. Then you get the energy spectrum as a function of the wave number. p = hpi + p′ The symbol h. data files ’1’. It is only in this special academic test case where we have three homogeneous directions (x. z.c. Use the definition to compute the stresses. Wait with analysis of the results till you have done next part. for example hv1′ v2′ i = h(v1 − hv1 i) (v2 − hv2 i)i = hv1 v2 i − hv1 hv2 ii − hv2 hv1 ii + hhv1 ihv2 ii = hv1 v2 i − 2hv1 ihv2 i + hv1 ihv2 i = hv1 v2 i − hv1 ihv2 i.i denotes averaging in the homogeneous directions x and z. (M. Compute the six stresses of the stress tensor.6) .M. hvi′ vj′ i. periodic b.6. Note that in reality h. When everything works. x2 j 2h x3 k X3. We decompose the instantaneous variables in time-averaged and fluctuating quantities as vi = hvi i + vi′ . then use also data files ’2’ and ’3’ averaging by use of the three files. Figure M. x2 j h¯ v1 i 2h x1 i L (a) x1 − x2 plane. t) where we can – in addition to time averaging – also can use x and z averaging.max (b) x2 − x3 plane.c.1: Channel domain. Resolved stresses 291 periodic b.i always denote time averaging. max /2. Above. if we let T → ∞ when time averaging. 2. in Item 1 and 2 you have just carried out explicit filtering.M. Compare v¯1 and v¯2 with v1 and v2 .7 Resolved production and pressure-strain Compute the production term and the pressure-strain terms ∂hv1 i ∂y ∂hv 1i = −2hv1′ v2′ i ∂y ∂hv1 i = −hv2′ v2′ i ∂y ′ ∂v = 2 p′ 1 ∂x ′ ∂v ∂v ′ = p′ 1 + p′ 2 ∂y ∂x ′ ∂v = 2 p′ 2 ∂y Pk = −hv1′ v2′ i P11 P12 Φ11 Φ12 Φ22 Do the production terms and the pressure-strain term appear as you had expected? (see the previous course TME225) Now analyze the fluctuations in the previous subsection. we use only three time steps (three files). If we would use many more time steps – or. but now for a 2D filter (x1 and x3 direction). e.e. Which stresses do you think are symmetric with respect to the centerline? or anti-symmetric? What’s the reason? When averaging. 1 T →∞ 2T hφi = lim Z +T φdt −T then some of the stresses would be zero: which ones? Why? M. Discuss the differences between no filter. Filter v1 and v2 to get v¯1 and v¯2 using a 1D box-filter (in the x1 direction) with filter width ∆ = 2∆x1 (this corresponds to a test filter. 18. we use an implicit filter). In LES we almost always assume that the filter width is equal to the control volume (i. Repeat Item 1.7. Compare the results along the same lines as in Item 1 and 2. 1. in general.g.27. 18.8 Filtering Plot v1 and v2 along x1 at two different x2 values at x3 = x3. see Eq. the formula for a 3D filter is given in Eq.26. Resolved production and pressure-strain 292 M. . ∆ = 2∆x1 and ∆ = 4∆x1 . Do the same thing again but with a filter width of ∆ = 4∆x1 (now you must derive the expression on your own!). we should really have used a 3D filter. ∆ = 2∆x1 and ∆ = 4∆x1 . which reads p τij = −2νsgs s¯ij .9 SGS stresses: Smagorinsky model Plot v1 and v2 along x1 at two different x2 values at x3 = x3. Compute the SGS stress τ12 from the Smagorinsky model. 18.1.27.7) s¯ij = + 2 ∂xj ∂xi fµ = 1 − exp(−x+ 2 /26) The filtered velocities. v¯i . Discuss the differences between no filter. Before doing the 2D filter. the formula for a 3D filter is given in Eq.M. Compare v¯1 and v¯2 with v1 and v2 . are taken from Section M. νsgs = (Cs fµ ∆)2 2¯ sij s¯ij 1 ∂¯ vi ∂¯ vj (M. but in order to keep it simple. Repeat Item 1. look in the Lecture Notes how a 3D filter is done. 2. Compare the results along the same lines as in Item 1 and 2. we use an implicit filter). In LES we almost always assume that the filter width is equal to the control volume (i. we use the 2D filter. Any thoughts? .8 using the 2D filter (in x1 and x3 ). SGS stresses: Smagorinsky model 293 cut-off E(κ) I II resolved scales III SGS κ Figure M. in Item 1 and 2 you have just carried out explicit filtering. see Eq. Filter v1 and v2 to get v¯1 and v¯2 using a 1D box-filter (in the x1 direction) with filter width ∆ = 2∆x1 (this corresponds to a test filter.e. but now for a 2D filter (x1 and x3 direction). Above. Plot them across the channel.9.2: Spectrum with cut-off. 1. The constant Cs = 0. M.max /2. Do the same thing again but with a filter width of ∆ = 4∆x1 (now you must derive the expression on your own!). 18. Compare the SGS stress hτ12 i with the resolved stress hu′ v ′ i and compare the SGS viscosity with the physical one.26. vi = v¯i + vi′′ = h¯ vi i + v¯i′ + vi′′ (M. vi . Compute the SGS stresses from the definition τij = vi vj − v¯i v¯j (M.9 and filter all velocities to obtain v¯1 .9. v¯2 and v¯3 .1. 8. fµ . M. At cut-off. compute the filter length as ∆ = min{κn.12) . h¯ vi i. but now we have a decomposition into time-averaged filtered velocity. v¯i′ . Since the cut-off is assumed to be located in the inertial sub-range (II).9) with Cm = 0.8 and M.10 SGS stresses: WALE model Repeat the Task in Section M. see Fig. what is the relation between εsgs and ε? Considering the cascade process.2.4 (von K`arm`an constant). ∆} (M.10) ∂¯ vi i ∂xj (M.e. resolved fluctuation. the SGS dissipation is at high Re numbers equal to the dissipation. i.11 Dissipations Compute the dissipation ε=ν ∂vi′ ∂vi′ ∂xj ∂xj and plot ε across the channel. Introduce a 2D filter (2∆x1 and 2∆x3 ) as in Sections M. SGS stresses: WALE model 294 As an alternative to the damping function. In this case you should set fµ = 1. but now for the WALE model by [177].M.10. which reads ∂¯ vi 2 . g = gik gkj ∂xj ij 1 1 2 2 2 − δij gkk gij + gji s¯dij = 2 3 3/2 s¯dij s¯dij 2 νsgs = (Cm ∆) 5/4 5/2 (¯ sij s¯ij ) + s¯dij s¯dij gij = (M. This is the case also in LES. M.8) where n is the distance to the nearest wall and κ = 0. see Eq. the production term in the equation for turbulent kinetic energy appears as a sink term in the equation for the mean kinetic energy.34.11) and compute the SGS dissipation εsgs = −hτij Now. In LES we introduce a filter which is assumed to cut off the spectrum at κc in the inertial region. and ′′ SGS fluctuation. kinetic energy is extracted from the resolved flow by the SGS dissipation εsgs . what did you expect? Recall that when we do traditional Reynolds decomposition.325 which corresponds to Cs = 0. M. 4 can be downloaded from course home page .6) ∂¯ vi′ ∂¯ vi ∂h¯ vi i ε′sgs = −τij′ = − τij + hτij i (M. we compute it with the same formula as in Eq.se/˜lada).chalmers.M.3: Transfer of kinetic turbulent energy. Hence the SGS dissipation in Eq. M.11.e. M. k¯ = 21 h¯ vi′ v¯i′ i and 2 vi ih¯ 1 ′′ ′′ ksgs = 2 hvi vi i. v¯i′ .13) ∂xj ∂xj ∂xj h¯ v1′ v¯2′ i ε′sgs Note that ε′sgs is defined using the fluctuating velocity gradient (see [22]4 ).tfd. The exact SGS dissipation is computed as (since ε′sgs is a product of two fluctuating quantities. M.3) ∂h¯ v1 i : dissipation (which is equal to production with minus sign) by resolved ∂x2 ¯ equation turbulence in the K * ′ + ∂¯ vi ∂¯ ∂¯ v ′ ∂¯ vi vi′ = νsgs ≃ hνsgs i i: SGS dissipation term in the k¯ equa∂xj ∂xj ∂xj ∂xj tion. M. ∆T denotes increase in internal energy. 86] (can be downloaded from www.11. no decomposition into time-averaged. Dissipations 295 hksgs i ≃ j ihs¯i i εs′ gs ∂h¯ vi i v¯i′ v¯j′ ∂xj k¯ ε ¯ K 2 hν sg hs¯ij si ′ v¯ i ∂ ∂ v¯ i ∂xj ν ∂xj ν ∂ hv¯ ∂ x i i ∂ hv¯ j ∂x i i j ′ ∆T ¯ = 1 h¯ vi i and k¯ = 21 h¯ vi′ v¯i′ i Figure M. i. 20 in [84]) The transport equation for h 12 v¯i′ v¯i′ i is derived in [22]. and resolved fluctuations. dissipation.3 (cf.chalmers. The flow of kinetic energy can be illustrated as in Fig. is made. h¯ vi i. Fig. K 2 vi ih¯ denote time-averaged kinetic and resolved turbulent kinetic energy. (can be downloaded from www.11 appears as an instantaneous production term in the equation for ksgs [85. ¯ = 1 h¯ Now we have three equations for kinetic energy: K vi i. contrary to εsgs = Pksgs in Eq. compare and discuss the four dissipations (see Fig. M. respectively. When deriving the ksgs equation.se/˜lada). This is the modelled SGS dissipation. Plot (along x2 ).tfd. M.1 and Fig. the two-point correlations in Section 10. (M. z{ Use this definition (1D filter.17) . but be careful when integrating the correlation above in the x3 direction.16) L1 (x2 ) = 2 v1. which is defined as L3 (x2 ) = 1 2 v3. We have 96 cells in the x3 direction.14) Lij = v¯i v¯j − v¯ i v¯ j and compare it (across the channel) with the resolved stress hv1′ v2′ i and the SGS stress hτ12 i defined in Eq.e.15) I=1 K=1 where xK 3 and ζm are the spanwise locations of the two points.e. In z{ dynamic models.10. In order to estimate m. ζ3 )dζ3 . and one of the points (x13 ) is at K = 1 then the other (x13 − 2∆) is at K = 95. it is no point showing both symmetric parts. Do the quantities exhibit the near-wall behaviour that you expected? M. x3 − ζm ) 96 × 96 (M.13 Near-wall behavior What is the near-wall behavior of hv1 i. for v1 . for example. ζm ) = 96 X 96 X 1 ′ I K v1′ (xI1 . L1 . ζm = 2∆x3 . ∆ = 2∆. Plot the two-point correlation at a couple of x2 positions.e.rms Z ∞ B33 (x2 . plot the quantities in log-log coordinates. we often define the test filter as twice the usual filter. show only half of it (cf.. When plotting two-point correlations. Compute and plot the integral length scale. x2 .rms 0 Compute also L3 . i. If. Take advantage of the fact that the flow is periodic. N. hv1′2 i and hv2′2 i (i. ∆ = 4∆x1 ) to compute the dynamic Leonard stress hL12 i from the definition z { z{ z{ (M. which is defined by Z ∞ 1 B11 (x2 . Test filtering 296 vi′ ∂¯ vi′ ∂¯ : viscous dissipation term in the k¯ equation ν ∂xj ∂xj 2 v1 i ∂h¯ v1 i ∂¯ v1 ∂h¯ ¯ equation. what is m in hv1 i = O(xm 2 )). x2 .12 Test filtering Above the filtered velocities were computed using the filter width ∆ = 2∆x1 .12.1). M. xK 3 )v1 (x1 . 0 What’s the difference between L1 and L3 ? Did you expect this difference? (M. ζ1 )dζ1 . Do you expect the magnitude of stresses to be similar? M. ≃ hνsgs i : SGS dissipation term in the K νsgs ∂x2 ∂x2 ∂x2 M.14 Two-point correlations The two-point correlation for u′ B11 (x2 . i. 18 is satisfied.rms = N X ˆ11 /N B (M. M. You can find some details on how to use Matlab’s FFT in Appendix N.18) 1 see Appendix N. M. The reason is that the Fourier coefficients correspond to the energy spectrum. the sum of all coefficients should give the energy. When plotting two energy spectrum. Confirm that Eq. Energy spectra 297 M.15. If you have computed the Fourier coefficients properly. the energy spectrum in Fig.16 Something fun Think of an interesting turbulent quantity and plot it and analyze it! . and if we integrate the energy spectrum over all wave numbers we get the total energy.15 Energy spectra The energy spectrum of any second moment can be obtained by taking the FFT of the corresponding two-point correlation. for example. M. N. we get ˆ11 (κz ) = F F T (B11 ) B and summation gives 2 v1. Plot the energy spectra at a couple of x2 locations.5 b). it is no point showing both symmetric parts. When we take the FFT of Eq. The energy spectrum of any second moment can be obtained by taking the FFT of the corresponding two-point correlation.M. show only half of it (cf.15. in the the continuous FFT Z 1 L c u(x) exp(−ıκx)dx.e. Consider the function u = 1 + cos(2πx/L) = 1 + cos(2π(n − 1)/N ) (N.2) where m denotes summation in homogeneous directions (i. An alternative way is to Fourier transform of a (time-averaged) two-point correlation.2) B(x3 .2. we are interested to look at the energy spectra. Then we show how to derive the energy spectrum from a spatial two-point correlation. so we expect the Fourier coefficients to be real.N.1 Introduction When analyzing DNS or LES data.e. In the following section we give a simple example how to use Matlab to Fourier transform a signal where we know the answer. 10. If we want to have the energy spectrum versus wavenumber. from a two-point correlation in time). The ratio (n − 1)/N corresponds to the physical coordinate. N.5) U (κ) = L −L .4) n=1 1≤k≤N √ where k is the non-dimensional wavenumber and ı = −1.2 An example of using FFT Here we will present a simple example.3) where L = 3 is the length of the domain and N = 16 is the number of discrete points. which is defined as (see Eq. Let’s use this function as input vector for the discrete Fourier transform (DFT) using Matlab. In Matlab the DFT of u is defined as (type help fft at the Matlab prompt) N X −ı2π(k − 1)(n − 1) U (k) = un exp N (N.e. N.e. x ˆ3 ) = hv3′ (x3 − x ˆ3 )v3′ (x3 )i (N. By taking the Fourier transform of the time signal (a fluctuating turbulent velocity) and then taking the square of the Fourier coefficients we obtain the energy spectrum versus frequency. N. we Fourier transform N instantaneous signals in space and then time average the N Fourier transforms. Finally. at which wave numbers) the fluctuating kinetic turbulent energy reside. MTF270: Compute energy spectra from LES/DNS data using Matlab N 298 MTF270: Compute energy spectra from LES/DNS data using Matlab N. Here we assume that x3 is an homogeneous direction so that B33 is independent of x3 . B33 = B33 (ˆ x3 ). κ = 2π/L (N.1) where x ˆ3 is the separation between the two points. From these we can find out in which turbulence scales (i. see Fig. some comments are given on how to create an energy spectrum versus frequency from an autocorrelation (i. The two-point correlation for an infinite channel flow is shown in Fig. The function u is symmetric. On discrete form the expression for B33 reads B33 (k∆z) = M 1 X ′ v (x3 − k∆z)v3′ (x3 ) M m=1 3 (N. x.1. time plus spatial homogeneous directions). B33 (ˆ x3 ). i. 2 0 −0.2: The u function. i. 16 nodes are used.6 0.N.5 u 1 0.1: Two-point correlation.4 the period [0. n=1:1:N.5). B(ˆ x3 ) = hv3′ (x3 − x ˆ3 )v3′ (x3 )i.4 0.2 0 0. 1. An example of using FFT 299 Note that the discrete Fourier U (k) coefficients in Eq.3 0. it can be noted that in Eq.5 3 . 2π] is used whereas the formulation in Eq.5 x 2 Figure N.5 0.4 must be divided by N .6 x ˆ3 Figure N.2 1 B33 (ˆ x3 ) 0. N. The u function is shown in Fig. u=1+cos(2*pi*(n-1)/N). π]. In Matlab. U (k)/N . N.1 0.8 0.5 is based on the interval [−π. node 1 is located at x = 0 and node 16 is located at 15L/16.2. 2.4 0.5 1 1.2.4 1.2 0. N. N. of DNS data in channel flow taken from [84]. N. N. 2 2 16 1.5 9 0 0 0.e. in order to correspond to the Fourier coefficients U c (N corresponds to L in Eq. we generate the function u in Eq. Furthermore.3 using the commands N=16. N. end end Note that since u is symmetric. This is easily verified from Eq. The energy.5.5. U (16)/N = 0.7) .5 N n=1 n (N. and versus wavenumber κ = 2π(n − 1)/L. for k=1:N for n=1:N arg1=2*pi*(k-1)*(n-1)/N. N.6) n=1 In wavenumber space the energy is – according to Parseval’s formula. In Fig.N.4 directly in Matlab as U=zeros(1. Instead of using the built-in fft command in Matlab we can program Eq.3.3. i. N. Type U=fft(u). The first Fourier coefficient corresponds – as always – to the mean of u.3. we have only used cos(−x) = cos(x) (the symmetric part of exp(−ıx)).2. see Eq. hu2 i. correspond to the cosine functions (there must be two since U is symmetric). N.e. U (1)/N = hui.4 by inserting k = 1. i. N. N = 9.e. For k = 2 we have cos(2π(n − 1)/N ) and for k = N cos((N − 1)2π(n − 1)/N ) = cos(−2π(n − 1)/N ) = cos(2π(n − 1)/N ) which corresponds to cos(2πx/L) in Eq. 16] corresponds to the negative. D. N. The remaining coefficients are zero. Two of them. k. U (2)/N = 0. Eqs.5). Since the function u includes only one cosine function and a mean (which is equal to one) only three Fourier coefficient are non-zero. symmetric part of the wavenumbers in the physical formulation (cf. of the signal in Fig. U(k)=U(k)+u(n)*cos(-arg1). The resulting Fourier coefficients are shown in Fig.4 and N.4 – equal to the integral of the square of the Fourier coefficients. U/N is plotted versus non-dimensional wavenumber. N. It can be noted that the interval [k = N/2 + 1.3.N). N.2 can be computed as hu2 i = 1 L Z L u2 (x)dx = 0 N X u2n /N = 1.5 (N. Z 0 ∞ U 2 (κ)dκ = N 1 X 2 U /N = 1. An example of using FFT 300 Now we take the discrete Fourier transform of u. 1 hu i = L 2 see Fig. 3 Energy spectrum from the two-point correlation Now that we have learnt how to use the FFT command in Matlab.8 0.max ≃ 1.2 0. Thus.4 0. which is the . k.5b shows the same energy spectra in log-log scale (only half of the spectrum is included). N. κ. B33 (ˆ x3 ) = hv3′ (x3 )v3′ (x3 + xˆ3 )i.55.8) The simulations in [84] have been carried out with periodic boundary conditions in x3 direction (and x1 ).4 reads ˆ33 (k) = B N X B33 (n) exp n=1 −ı2π(k − 1)(n − 1) N (N.N. Figure N.8.e. and hence B33 (ˆ x3 ) is symmetric.5 1 1. ˆ33 (k) = B N X n=1 B33 (n) cos 2π(k − 1)(n − 1) N (N.2 1 B33 (ˆ x3 ) 0. i.4.8 0. where x3.3.6 0. N. N.9) ˆ33 κ3 are presented versus wavenumber κ3 = In Fig.1 and Fig. 20 25 30 (b) Versus wavenumber. N. of DNS data in channel flow taken from [84]. see Fig.6 U/N U/N 0.5 x ˆ3 Figure N.max .4: Periodic two-point correlation.1. N.2 0 −0.5a the Fourier coefficients B 2π(n − 1)/x3. Energy spectrum from the two-point correlation 1 1 0.3: The U/N Fourier coefficients. let’s use it on our two-point correlation in Eq. it is sufficient to use the cosine part of Eq.2 0 0. Figure N. N.6 301 0. Equation N.8 0.4 0. see Fig.4.2 0 0 5 10 0 0 15 5 10 15 κ k (a) Versus non-dimensional wavenumber. N. 1.4 0.4 96 1 1. 15 −2 10 0.05 0 0 −4 100 200 300 κ3 400 10 1 10 2 κ3 10 (b) Log-log scale (a) Linear-linear scale Figure N. N.e. v3′2 = Z ∞ 0 N X ˆ33 (κ3 )dκ3 = 1 ˆ33 (n) = 1. versus frequency. N. x ˆ3 )dˆ x3 = (N. common way to present energy spectra. Dashed line in b) show −5/3 slope. κ3 .12) The two-point correlation for zero separation is equal to v3′2 .e. f . Taken from [84]. say.4 Energy spectra from the autocorrelation When computing the energy spectra of the v3′ velocity. the Fourier coefficient for the first non-dimensional wavenumber. the time series of v3′ (t) is commonly Fourier transformed and the energy spectrum is obtained by plotting the square of the Fourier coefficients versus frequency. i. Another way to obtain v3′2 is to integrate the energy spectrum in Fig.4. ˆ33 (1)/N is equal to the mean of B33 . Fourier transform each subset and then average. i. B33 (0) = v3′2 = 1.51.1 −3 10 0. The dashed line shows the −5/3 slope which indicates that the energy spectra from the DNS follows the Kolmogorov −5/3 decay.5.13) N.10) compare with Eq.6) Z ∞ 1 hB33 i Lint (x3 ) = B33 (x3 .25 −1 10 B33 (k)/N 0. the command pwelch is a convenient command which does all this. Energy spectra from the autocorrelation 302 0.N. .e. i.9. i. We can also split the time signal into a number subsets.2 0. B hB33 i = N 1 ˆ 1 X B33 (n) ≡ B 33 (1) N n=1 N (N. As usual.5: The energy spectrum of v3′2 versus wavenumber. Note that this is almost the same expression as that for the integral length scale which reads (see Eq. In Matlab.e.51 B B N n=1 (N. 10.11) ′2 v3 0 v3′2 Hence the integral length scale is related to the first Fourier mode as Lint = ˆ33 (1) B N v3′2 (N. 3. First we create the autocorrelation B33 (τ ) = hv3′ (t)v3′ (t + τ )i (this can be seen as a ˆ33 (f ) in two-point correlation in time). Energy spectra from the autocorrelation 303 In the previous section we computed the energy spectrum versus wavenumber by Fourier transforming the two-point correlation. Then B33 (τ ) is Fourier transformed to get B the same way as in Section N. We can use the same approach in time. .N.4. w (v3 ) and kT (made non-dimensional by uτ and ρ). this correspond to x+ 2 = 22.06. Assignment 3: Zonal PANS. You find a Matlab/Octave program at the www page which reads the data and computes the mean velocity. The time history of v¯1 at x2 /δ = 0. ’3’ and ’4’. x2 /δ = 0. How do you expect that the difference in νt affects the time history of v¯1 . MTF270.0203. Since the data files are rather large.se/˜lada/allpaper. Recall that the velocities have been scaled with the friction velocity uτ .0028. The file has nine columns of v¯1 and v¯3 at five nodes (and time): x2 /δ = 0. x2 = 516 and x2 = 1600. Hence. How does the time variation of v¯1 differ for different positions? Recall that the two points closest the wall are located in the URANS region and the other two are located in the LES region. With uτ = 1 and ν = 1/Reτ = 1/8000. All data have been made non-dimensional by uτ and ρ.05 and ∆x3 = 0. Does the time history of v¯1 behave as you expect? What about v¯3 ? Compute the autocorrelation of the five points imax=500.dat with the time history of v¯1 and v¯2 .20. The data files are Matlab binary files. 28 cells (29 nodes including the boundary) are located in the URANS region at each wall.0364. Periodic boundary conditions were applied in the x1 and x3 direction (homogeneous directions). Octave is a Matlab clone which can be downloaded for free.1 Time history At the www page you find a file u v time 5nodes zonal pans. Assignment 3: Zonal PANS. O. DES. and thus what you see is really v¯1 /uτ . Plot v¯1 for all five nodes. At the course www page you find data files with instantaneous flow fields (statistically independent).tfd. three nodes are located in the URANS region and two nodes in the LES region. + + + x+ 2 = 162. DES. x2 = 291. and average by use of all four files. In the URANS region the turbulent viscosity is much larger than in the LES region. x2 /δ = 0. Use the Matlab program pl time zonal pans to load and plot the time history of v¯1 . Use Matlab or Octave on Linux/Ubuntu.html . The matching line is located at gridline number 29 at which x+ 2 ≃ 500.0645 and x2 /δ = 0. it is recommended that you do all tasks using only data files ’1’.250E −4 (every time step). To study the profiles in closer detail. The sampling time step is 6.chalmers. x2 /δ = 0. The data are taken from [150] 5 . Use one of these programs to analyze the data. use the axis-command in the same way as when you studied the DNS data. You will also find a file with time history of u. v (v2 ). DDES and SAS In this exercise you will use data from a Zonal PANS for fully developed channel flow. The Re number based on the friction velocity and the half channel width is Reτ = uτ h/ν = 8000.imax). DDES and SAS O 304 MTF270. The data files include the instantaneous variables u (v1 ).O. When everything works. 5 this paper can be downloaded from http://www. then use also data files ’2’. The cell size in x1 and x3 directions are ∆x1 = 0. respectively.025. x2 /δ = 0. two_uu_1_mat=autocorr(u1. A 64 × 96 × 64 mesh has been used.0203 and x2 /δ = 0.0645 are shown. plot(t(1:imax). Compute the integral timescale dt=t(1). Plot it in a new figure (a figure is created by the command figure(2)). int_T_1=trapz(two_uu_1_mat)*dt. we want to look at the time-averaged results.5 hv1′2 i + hv2′2 i + hv3′2 i and plot it in a new figure. vfluct=v3d(i.k)-umean(j). .j.’fontsi’.m to look at the mean velocity profiles. O.two_uu_1_mat(1:imax).O. but if you use only one file this may not be the case due to too few samples). Use more than one file to perform a better averaging.k)-vmean(j). uv(j)=uv(j)+ufluct*vfluct. Plot the autocorrelation. The time averaged velocity profile is compared with the log profile (markers).[20]) How does it compare to the integral timescale? Compute the autocorrelation and integral timescale also for the other three points.2. Now compute hv1′ v2′ i. Use the file pl uvw zonal pans. pl uvw zonal pans.’linew’.1). There are four files with instantaneous values of v¯1 .j. Mean velocity profile 305 Above we set the maximum separation in time to 500 samples. end end end uv=uv/ni/nk.2 Mean velocity profile After having performed a Zonal PANS.2) xlabel(’t’) ylabel(’B_{uu}’) handle=gca set(handle. Compute also the resolved turbulent kinetic energy kres = 0. Here’s an example how to do: uv=zeros(nj. Do you see any difference between the points in the URANS region and the LES region? O.m reads the instantaneous v¯1 field and performs an averaging in the homogeneous directions x1 and x3 . Compute first vmean (this quantity should be very small. for k=1:nk for j=1:nj for i=1:ni ufluct=u3d(i.3 Resolved stresses We want to find out how much of the turbulence that has been resolved and how much that has been modelled. it’s given in the paper). In SST-DES. Compute the turbulent viscosity as 0. ∆y . the location of the interface is computed using k and ω. ) will be used.4). F1 or F2 (see Eq.3).7 SAS turbulent length scales Compute the 1D von K´arm´an length scale defined as .O. Recall that ν = 1/8000. ε (file ed1 zonal pans. . ). te2 zonal pans.m. Let’s use F2 as the shielding function as in [63].mat. the boundary layer is shielded with a damping function. 115 in relation to Fig. In SA-DES. Let’s compare that with DES and DDES. if you’re ambitious you may include it.7.4. . i. the shielding function (see Eq. How does the location compare with gridline 29 and SA-DES? In DDES. kT (file te1 zonal pans. ∆z }. can be downloaded from the www page and loaded at the beginning of pl uvw zonal pans. .8) may be one of the blending functions. the interface is defined as the location where the wall distance is equal to CDES ∆ where ∆ = max{∆x . . respectively? What about the sum? The magnitude of resolved and modelled turbulent kinetic energies is discussed on p. Turbulent kinetic energy 306 O.09k 2 /ε (we neglect the damping function which is active only in the viscous sublayer. Does DDES work.5 The modelled turbulent shear stress We have computed the resolved shear stress. 20. does it make the model to be in RANS mode in the entire boundary layer? O. the interface in the present simulations are prescribed along a fixed grid line (No 29). Are they smooth across the interface? Is the resolved shear stress large in the URANS region? Should it be large? Why/why not? O. Which is largest? Which is largest in the URANS region and in the LES region. 20. see Eq. Compute ω from ε/(β ∗ k) and compute the location using Eq. 20. . Compute the modelled shear stress from the Boussinesq assumption ∂¯ v2 ∂¯ v1 + τ12 = −2νT s¯12 = −νT ∂x2 ∂x1 Plot it and compare with the resolved shear stress (see Section O. . kT (te1 zonal pans.mat.mat.mat.4 Turbulent kinetic energy Now plot and compare the resolved and modelled turbulent kinetic energies.3. 20.6 Location of interface in DES and DDES As mentioned above. Note that the modelled turbulent kinetic energy. 6 in [150]. and the dissipation. ). . How does this compare with gridline number 29? In SST-DES. . . Let’s find the modelled shear stress. The modelled turbulent kinetic energy. O.e. . . ∂h¯ v1 i/∂x2 . . . 1D = κ . LvK. 2 ∂ h¯ v1 i/∂x22 . Zoom up near the wall.e. How does it behave (i. what is n in O(xn2 )? What should n be? . (O. of h¯ v1 i.1) Note that you should take the derivatives of the averaged v¯1 velocity. i.e. e.O. . SAS turbulent length scales 307 Compare that with the von K´arm´an length scale defined from instantaneous v¯1 . i.7. . . ∂¯ v1 /∂x2 . . (O.1D.2) LvK.inst = κ . . 2 ∂ v¯1 /∂x22 . the first and second derivative must be defined in 3D. 144] .1D ? When we’re doing real 3D simulations. How does it compare with LvK. One way of defining the von K´arm´an length scale in 3D is [143. . . S . LvK.3D.inst = κ . . ′′ . . U 0.4.5) (O.e. O. . T1 in Eq.6.3) The second derivative is then computed as U ′′2 = ∂ 2 v¯1 ∂x21 ∂ 2 v¯2 ∂x21 ∂ 2 v¯3 ∂x21 2 2 2 + + + ∂ 2 v¯1 ∂x22 ∂ 2 v¯2 ∂x22 ∂ 2 v¯3 ∂x22 2 2 2 + + + ∂ 2 v¯1 ∂x23 ∂ 2 v¯2 ∂x23 ∂ 2 v¯3 ∂x23 2 2 (O. Compare them with Eq.5 S = (2νt s¯ij s¯ij ) 0. 22.5 2 ∂ 2 v¯i ∂ v¯i U ′′ = ∂xj ∂xj ∂xj ∂xj (O.4 and O.5) and what effect does it give to ω? Another way to compute the second derivative is [178] U ′′ = U ′′2 ∂ 2 v¯i ∂ 2 v¯i ∂xj ∂xk ∂xj ∂xk ∂ 2 v¯1 ∂x1 ∂x2 2 0. O.3 and O.4) 2 Plot the von K´arm´an length scale using Eqs. O.5 2 ∂ 2 v¯1 = +2 +2 ∂x1 ∂x3 2 2 2 2 2 2 ∂ v¯1 ∂ v¯1 ∂ v¯1 +2 + ∂x22 ∂x2 ∂x3 ∂x23 2 2 2 2 2 2 ∂ v¯2 ∂ v¯2 ∂ v¯2 +2 +2 ∂x21 ∂x1 ∂x2 ∂x1 ∂x3 2 2 2 2 2 2 ∂ v¯2 ∂ v¯2 ∂ v¯2 +2 + ∂x22 ∂x2 ∂x3 ∂x23 2 2 2 2 2 2 ∂ v¯3 ∂ v¯3 ∂ v¯3 + 2 + 2 ∂x21 ∂x1 ∂x2 ∂x1 ∂x3 2 2 2 2 2 2 ∂ v¯3 ∂ v¯3 ∂ v¯3 +2 + ∂x22 ∂x2 ∂x3 ∂x23 ∂ 2 v¯1 ∂x21 2 (O. What’s the difference? What effect do the different length scales give for PSAS (i.6) Plot and compare the von K´arm´an length scales using the second derivatives defined in Eqs.1. 23.4). In the present simulations.10) and the εu (Eq.6 domain.18) equations are solved. The interface condition for εu is computed with the baseline value Cs = 0. In this exercise you will use data from an embedded PANS of channel flow.2) with fkLES = 0.0 LES. 23. we have set ρ = 1.95 2.1.12. is set from kinter and an SGS length scale. δ = 1 and uτ ≃ 1.6 − 103) × 24 in viscous units.1) and the interface condition for ku is computed as kinter = fkLES kRAN S (P.dat with the time history of v¯2 . the wall-normal (y) and the spanwise (z) direction. P.P. the streamwise (x). ℓsgs . δ.25 x2 x1 Figure P. P. uτ . Inlet conditions at x = 0 are created by computing fully developed channel flow with the LRN PANS model in RANS mode (i. see Fig.2 × 2 × 1. Assignment 4: Embedded LES with PANS P 308 MTF270. Anisotropic synthetic fluctuations are added at the interface. fk = fkLES < 1 0.1: Channel flow configuration. The RANS part extends up to x1 = 0.1. respectively. The resolution is approximately (the wall shear stress varies slightly along the wall) 48 × (0. fk = 1. and half the channel width. The ku (Eq. The interface conditions on ku and εu will make the turbulent viscosity steeply decrease from its large values in the RANS region to much smaller values appropriate for the LES region. P. which is estimated from the Smagorinsky model as ℓsgs = Cs ∆ (P.07. 23. The Reynolds number for the channel flow is Reτ = 950 based on the friction velocity. a mesh with 64 × 80 × 64 cells is used in. with fk = 1).95. The modeled dissipation. The PANS model is a modified k − ε model which can operate both in RANS mode and LES mode.4.1. The interface separates the RANS and the LES regions.e. where kRAN S is taken at x = 0. εinter . downstream the equations operate in LES mode ((i.0139 (x+ 2 ≃ 13) and .5. The turbulent viscosity is computed from Eq. With a 3.1 Time history At the www-page you find a file u time interior.e. Assignment 4: Embedded LES with PANS Interface 2 RANS. see Fig. see Fig. fk = 0. The data are taken from [152]. P. MTF270. The file has eight columns of v¯2 along two lines: x2 = 0. 24 (x+ 2 ≃ 230). pl uuvvww 2d.2.000625 (every time step).e. resolved and modeled stresses. The hv1′2 i profiles are very different in the RANS region (x1 < 0.0139 at x1 = 0.95) and in the LES (x1 > 0. In Matlab figure 2. Try to understand the coding. x2 in both regions.4) 0 where B norm = B(τ )/B(0) so that B norm (0) = 1. x1 (figure 2).2. In the Matlab file the integral timescale is computed from the autocorrelation. The autocorrelation is defined as Z ∞ B(τ ) = v(t)v(t − τ )dt (P. Run pl uuvvww 2d. You can also use Octave on Linux/Ubuntu. don’t you? P.4 Modeled stresses In Section P.925. Where is it large and where is it small? Why? Now plot it also vs. Compute the modeled Reynolds stresses from the Boussinesq assumption . right? P.2 you looked at the resolved Reynolds stresses.P. P. skip the rest of this section and proceed to Section P. Use the Matlab/Octave program pl time pans to load and plot the time history of v¯2 . Now let’s look at the modeled stresses.e. To study the profiles in closer detail. The time history of v¯2 at x2 = 0. 2. 14 in [179] we can then compute the prescribed integral timescale. Something drastically happens at x1 = 0. aren’t they? Why? This can also be seen in figure 2 where hv1′2 i is plotted vs. Octave is a Matlab clone which can be downloaded for free. When prescribing the time correlation of the synthetic fluctuations. they are located at x1 = 0. x1 = 1. Normalize it with hνi. both in the RANS region and in the LES region. Use the file pl uuvvww 2d.m to look at the mean quantities such as velocity.95. 1.175 and x1 = 2. Plot v¯2 for the other nodes and study the differences. turbulent viscosities etc.3 Turbulent viscosity Plot the turbulent viscosity vs.675 are shown. hv1′ v2′ i. 1.3) 0 Study the coding and try to understand it. Use Matlab. the autocorrelation is plotted.175. Resolved stresses 309 x2 = 0. for x1 < 0. i.95. You find the same difference between RANS and LES region as for hv1′2 i. Why is there such a big difference in the fluctuations? If you’re not interested in integral time scales. i. The resolved stresses hv1′2 i are plotted vs x2 (figure 1) and vs. x1 Now plot the resolved shear stresses. uu 2d. Compute the autocorrelations and the integral timescales. The integral time scale is defined as Z ∞ T = B norm (τ )dτ (P. x1 .675. Plot the resolved stress also in the RANS region.2 Resolved stresses Now we will look at the time-averaged results.m reads the fields and transforms them into 2D arrays such as u 2d.775.175.775 and x1 = 1. The sampling time step is 0. the integral timescale T is used. Two x1 stations are shown in figure 1. plot hνu i/ν. use the axis-command in the same way as when you studied the DNS data.95). The constant a is in [152] set to 0.954 and from Eq. P. 8. P. plot also this quantity.5.2. hvi′ vj′ mod i = −hνu i ∂h¯ vi i ∂h¯ vj i + ∂xj ∂xi 2 + δij hku i 3 (P. the production term in the ku equation includes both mean and fluctuating strain rates since ∂¯ vi ∂¯ vj ∂¯ vi Pu = εsgs = νu + ∂xj ∂xi ∂xj which in the Matlab file is stored as pksgs 2d. goes directly from K In the LES region. Turbulent SGS dissipation E(κ) 310 Pk in e rti al εs gs E(κ) ∝ κ ra = ng e ε P u −5/3 dissipating range κ κc Figure P. The energy flow is visualized in Fig.5) Compare the resolved and the modeled shear stress as well as the streamwise and the normal stresses in the RANS region and in the LES region.2: Energy spectrum. The resolved turbulence extracts kinetic energy via the production term. In both the RANS and the LES region the process of viscous dissipation takes place via εu . however. there is no resolved turbulence. Hence the kinetic energy ¯ to the modeled turbulence. P. which represents a source term in the k equation (Eq. K finally to internal energy via dissipation. ksgs (or ku ) and ¯ mostly goes to resolved turbulence. In RANS mode. does the relation Pu = εu hold? .e. Is the turbulence in local equilibrium.3 where the energy in K ¯ then to modeled turbulence. ku . Investigate the relation between Pu = εsgs and the production due to the resolved turbulence P k = −hvi′ vj′ i ∂h¯ vi i ∂xj Compare also P k in the LES region and in the RANS region. P k . Hence.5 Turbulent SGS dissipation In an LES the resolved turbulent fluctuations can be represented by a energy spectrum as in Fig.34). M. Now consider the LES region. k.14) and a sink term in the ¯ equation (Eq. 8. εu . i. Strelets in St. This ratio was used to scale the local height of the channel and thus modifying the contour shape of the upper wall.6) conditions are taken from 2D RANS SST k − ω simulations carried out by the group of Prof. The results are presented in detail in [150. they are also described in detail at http://cfdval2004.038.6 from the RANS simulation are used together with V3 = 0 as mean inlet velocities to which the fluctuating velocity V1′ .c . in order to compensate for the blockage effect of the side plates in the computation. The Reynolds number of the hump flow is Rec = 936 000. 179] 6 1 0. respectively. The computed integral length scale for the synthetic inlet fluctuations is L ≃ 0. Experiments were conducted by Greenblatt [180. the surface shape of the upper wall (above the hump) is modified and the upper wall is moved slightly downwards. h.4 0.Q.6 0.128. recirculating flow: PANS.1. DES and DDES Q 311 MTF270. and the inlet mean velocity at the centerline. Grid. recirculating flow: PANS.6 1 x1 2 Figure Q. are given by H/c = 0. 152.2. based on the hump length. Q. Neumann conditions are used at the outflow section located at x1 = 4. The distributions of V1 and V2 at x1 = 0.2 0 0. H.se/˜lada/allpaper.91 and h/c = 0.html .040 and the integral time scale T ≃ 0.181].nasa.8 x2 0. Uin. Inflow boundary (at x1 = 0.tfd. There are side-wall effects (3D flow) near the side plates in the experiment. Petersburg.larc. Assignment 5. Q. The maximum height of the hump. Every 4th grid line is shown. see Fig. The baseline mesh has 312 × 120 × 64 cells with Zmax = 0.gov/case3.chalmers. The ratio of the local cross-sectional area of the side plates (facing the flow) to the cross sectional area of the tunnel enclosed by the side plates was computed.2.html. It is noted that the prescribed inlet integral length scale is rather large [152] (approximately equal to the 6 these papers can be downloaded from http://www. Slip conditions are used at the upper wall and symmetric boundary conditions are used on the spanwise boundaries. and the channel height. c. V2′ and V2′ are superimposed.1. The grid is shown in Fig. Hence. DES and DDES In this exercise you will use data from PANS and Zonal PANS for for the flow over a hump. MTF270. Assignment 5.1: Hump flow. 17. Discretization schemes 312 inflow boundary layer thickness). 179]. the square of the velocities and a number of other quantities are read. 12 in [179]. do the nonphysical fluctuations disappear (they do. Compare with hv1′ v2′ i in Fig.65 (gridline 10 is chosen which corresponds to x 2d(10.1. Furthermore. Pure central differences sometimes give problems is flow regions like this.652).65. Consider the plot of h¯ v1 i and hv2′ v2′ i at x = 0. Q. Octave is a Matlab clone which can be downloaded for free.65 vanish quickly further downstream. x . Use one of these programs to analyze the data. At the course www page you find data files with time and spanwise averaged flow fields. Q. the m file plots contours of hv1′ v1′ i.2.dat. vectz aiaa paper. xy hump. don’t they?). Use Matlab or Octave on Linux/Ubuntu.65.046 (x1 = 0. The peak of hv2′ v2′ i. Large nonphysical fluctuations were found in regions far from the airfoil.1)= 0.8 . The resolved stresses are computed as in Eq. Consider the resolved fluctuations: note how strongly they increase when going from the attached boundary layer (x1 = 0. in these studies the problems were solved by using 100% van Leer scheme in the far-field regions. It is tempting to draw the conclusion that these nonphysical fluctuations are caused by the inlet synthetic fluctuations.m. gridline i=10) to 0. This gives an indication how well the turbulence has been resolved.2 The modeled turbulent shear stress As usual.1 Discretization schemes You will analyze results from two publications [152.dat and x065 off. 1. In [152] the same simulations were carried out but now using a blend of central differencing (95%) and van Leer scheme (5%). see Fig. This is the main reason why the nonphysical fluctuations at x1 = 0. gridline i=45). Similar (much worse) problems with central differences were encountered in LES of the flow around an airfoil [99. Analyze these results by loading file vectz aiaa journal.65) to the recirculating region. The m-file reads the data files. The reason is that synthetic fluctuations with a large integral length scale are efficient in generating resolved turbulent fluctuations [160]. 183].0017 (x1 = 0. The velocity profile shows that the boundary layer extends up to x2 ≃ 0. PANS etc.24 is almost 5 times larger than in the boundary layer. the peak value at x2 ≃ 0. when analysing results of DES. M. However. The main difference is that in [179] pure central differencing was used whereas in [152] a blend of 95% central differencing and 5% bounded second-order upwind scheme (van Leer scheme [182]) was employed. The experimental values are also included in the plots at x = 0. a vector plot as well as h¯ v1 i and hv2′ v2′ i at x = 0.65. if much turbulence is resolved it may be either good . Start by downloading pl vect hump. the hv2′ v2′ i exhibits large values outside the boundary layer.dat in pl vect hump. The velocities.Q.800. 0.dat. increases from 0. These fluctuations are nonphysical and stem from the use of central differencing. the grid. when increasing the magnitude of the inlet synthetic fluctuations the magnitude of the nonphysical fluctuation diminishes. However. The nonphysical fluctuations appear in the outer region where no physical resolved turbulence is present. for example. Now plot hv2′ v2′ i at locations further downstream.m. we want to find out how much of the turbulence is modeled and how much is resolved. 13a in [179].6. right? That might lead you to conclude that the turbulence is better resolved in the boundary layer than in the recirculating region. Recall that the turbulent diffusion in. but that is not the reason. You find that the turbulent viscosity is large in the recirculation region.4. A suitable quantity to look at is the shear stress which is much more relevant than the turbulent kinetic energy. much larger than the normal stresses (at least in boundary layer flows). gridline i=10) and the recirculating region (e.g. It is true that the discretization schemes used in the two papers are slightly different. it means that the turbulent diffusion in the k and ε equations are 1/0.e.4. gridline i=45).4 [179] (Q. ∂hvi′ vj′ i ∂¯ vi ∂ νt and − (Q. Plot the turbulent viscosity.2) 1. 113].8). Plot the viscosity. the k equation reads ∂ νt ∂k (Q. the shear stress is the largest diffusion term in the momentum equations. i.e. . fk2 σε = 1.g. Q.3 The turbulent viscosity It may be interesting to look at the turbulent viscosity.3 corresponds to the original PANS model. You find that it is much smaller than for the data of [152].2. It would actually be even more relevant to look at the divergence of these terms. When estimating the influence of the modeled turbulence.42 ≃ 6 times larger in [152] than in [179]. preferably the ratio of the modeled to the resolved shear stress [112. The difference stems from the fact that the turbulent Prandtl numbers in the k and ε equations were in the two papers set as σk = 1.4) ∂xj σk ∂xj Since fk = 0. But that would be an incorrect conclusion. see Eq. 20. in the recirculation region. Go back and load the results of [179] (vectz aiaa paper. In recirculating regions it is usually good but it may be disastrous in boundary layer regions close to the wall if the mesh is not fine enough (this is the very reason why DDES was proposed. Where is the modeled shear stress large and where is it small? Look at other x1 locations.4 fk2 [152] (Q.4 . as mentioned in Section Q. νt /ν. The turbulent viscosity 313 or bad. This explains why the peaks of k are much larger in [179] compared to [152].1) ∂xj ∂xj ∂xj because these are the terms that indeed appear in the momentum equations.dat).3) σε = Equation Q. Now plot the resolved (uv 2d) and the modelled shear stress (uv model 2d) in the boundary layer region (e. The consequence is that peaks in k and ε (and also νt ) are reduced in the former case compared to the latter (this is the physical role played by diffusion: it transports from regions of high k to regions of low k).3.Q. we should look at the ratio of modeled stresses to the resolved one. You have noted that the turbulent viscosities are rather large. νt /ν is larger than 100. and. we should not look at the turbulent viscosity because it does not appear as a term in the momentum equations. i. Usually it is plotted scaled with ν. σk = 1. for example. νt /ν (vis 2d/viscos) in the boundary layer and in the recirculating region. 7.6) fk = cµ−2/3 .4. in the RANS region near the lower wall fk = 1 (interface at grid line No 32). Q. fk = 0. How does the location compare with gridline 32 and SA-DES? In DDES.4 in the entire domain.4 and 1)? Compare it also with the definition of fk = k/ktotal (cf.4 in the LES region or it is computed from Eq. Location of interface in DES and DDES 314 Hence.4).5) 2 ∂ cµ kRAN νt. F1 or F2 (see Eq.2/64) In SA-DES. ktotal = 1 hv1′2 i + hv2′2 i + hv3′2 i + hki 2 (Q.3. 179].4 ∂xj (Q. It is bigger or smaller than the prescribed values (0. 20. the interface is defined as the location where the wall distance is equal to CDES ∆ where ∆ = max{∆x . does it make the model to be in RANS mode in the entire boundary layer? What about the separation region? Q. 179] ∆z = 0. How is k/ktotal related to Eq. the RANS turbulent viscosity appears in the turbulent diffusion of k (and ε).dat in pl vect hump.m.8) may be one of the blending functions.6 and plot it at a couple of x1 locations (cf.4 Location of interface in DES and DDES The results analyzed above were from LES simulations [152.7) Compute fk from Eq. 26 in [150]). fk = 0. In the Zonal PANS simulations [150]. 20. the shielding function (see Eq.e. see Eq. because the turbulent diffusion term reads (recall that fk = k/ktotal = k/kRAN S where kRAN S denotes the turbulent kinetic energy in a RANS simulation) ∂ ∂ νt fk2 ∂k cµ k 2 ∂k = ∂xj 1.5 Compute fk In [152.3). Does DDES work.2/32 (note that in [152. RANS) viscosity is responsible for the transport of the modeled turbulent kinetic energy. 14) . The expression for computing fk reads [184] 3/2 k ∆ (Q. in the original PANS model (Eq Q.Q.4 ∂xj ∂xj εfk2 1. Q. Load the file vectz zonal pans. ∆y . Recall that ∆z = 0. Fig. i. Lt = total Lt hεi where ∆ = (∆V )1/3 where ∆V denotes the cell volume and ktotal denotes resolved plus modeled turbulent kinetic energy. 20. Thus the total (i. Q. the location of the interface is computed using k and ω.6? (see the discussion in [150] leading to Eq.4 ∂xj cf. In SST-DES. Fig. 26 in [150]).4 ∂xj ∂xj 1. 18 and 19 in [129]. the boundary layer is shielded with a damping function.e.e. i.179] (i. How does this compare with gridline number 32? In SST-DES. Let’s use F2 as the shielding function as in [63]. 20. Now we will analyze results from Zonal PANS [150] where the interface is prescribed along a fixed grid line (No 32).e. Let’s compare that with DES and DDES. Q.6. Compute ω from ε/(β ∗ k) and compute the location using Eq. Eqs. the PANS model was used in LES mode). ∆z }.RAN S ∂k ∂ S ∂k = = ∂xj ε1. Since the data files are rather large.015 and x2 /δ = 0. The data files are Matlab binary files. two_uu_1_mat=autocorr(u1. The Re number based on the friction velocity and the half channel width is Reτ = uτ h/ν = 8000. Does the time history of v¯1 behave as you expect? What about v¯2 ? Compute the autocorrelation of the four points imax=500. A 64 × 96 × 64 mesh has been used.imax). All data have been made non-dimensional by uτ and ρ. but the domain and Reynolds number is taken from [185]. two nodes are located in the URANS region and two nodes in the LES region. In the URANS region the turbulent viscosity is much larger than in the LES region. At the course www page you find data files with instantaneous flow fields (statistically independent).R. Assignment 6: Hybrid LES-RANS R 315 MTF270. Assignment 6: Hybrid LES-RANS In this exercise you will use data from a Hybrid LES-RANS for fully developed channel flow.06. The file has nine columns of v¯1 and v¯3 at four nodes (and time): x2 /δ = 0. use the axis-command in the same way as when you studied the DNS data. R. How does the time variation of v¯1 differ for different positions? Recall that the two points closest the wall are located in the URANS region and the other two are located in the LES region. it is recommended that you do all tasks using only data files ’1’. Compute the integral timescale .025. x2 = 120. Above we set the maximum separation in time to 500 samples. The turbulence model is the same as in [84] (no forcing). The data files include the instantaneous variables u (v1 ). The matching line is located at gridline number 29 at which x+ 2 ≃ 500. and thus what you see is really v¯1 /uτ . To study the profiles in closer detail. this correspond to x+ 2 = 22.1 Time history At the www page you find a file u v time 4nodes hybrid. x2 /δ = 0.015. Plot v¯1 for all four nodes. and average by use of all four files. 28 cells (29 nodes including the boundary) are located in the URANS region at each wall. w (v3 ) and kT (made non-dimensional by uτ and ρ). You find a Matlab/Octave program at the www page which reads the data and computes the mean velocity. Use Matlab or Octave on Linux/Ubuntu. How do you expect that the difference in νt affects the time history of v¯1 . The cell size in x1 and x3 directions are ∆x1 = 0.dat with the time history of v¯1 and v¯3 .35. x2 /δ = 0. v (v2 ). Hence.05 and ∆x3 = 0.099 and x2 /δ = 0. Use one of these programs to analyze the data. Recall that the velocities have been scaled with the friction velocity uτ . Use the Matlab program pl time hybrid to load and plot the time history of v¯1 . Periodic boundary conditions were applied in the x1 and x3 direction (homogeneous directions). Octave is a Matlab clone which can be downloaded for free. MTF270.0028.250E − 4 (every time step). x2 = 792 and + x2 = 2800. respectively. ’3’ and ’4’. With uτ = 1 + + and ν = 1/Reτ = 1/8000. then use also data files ’2’. You will also find a file with time history of u. x2 /δ = 0. When everything works. The sampling time step is 6. The time history of v¯1 at x2 /δ = 0.35 are shown. Use more than one file to perform a better averaging. Compute first vmean (this quantity should be very small.[20]) How does it compare to the integral timescale? Compute the autocorrelation and integral timescale also for the other three points. end end end uv=uv/ni/nk.3 Resolved stresses We want to find out how much of the turbulence that has been resolved and how much that has been modelled. Plot the autocorrelation. but if you use only one file this may not be the case due to too few samples).m to look at the mean velocity profiles. The time averaged velocity profile is compared with the log profile (markers). Mean velocity profile 316 dt=t(1). int_T_1=trapz(two_uu_1_mat)*dt.2) xlabel(’t’) ylabel(’B_{uu}’) handle=gca set(handle. R.2 Mean velocity profile After having performed a hybrid LES-RANS. Now compute hv1′ v2′ i. Plot it in a new figure (a figure is created by the command figure(2)). plot(t(1:imax).k)-umean(j). Do you see any difference between the points in the URANS region and the LES region? R. .’linew’. Use the file pl uvw hybrid.5 hv1′2 i + hv2′2 i + hv3′2 i and plot it in a new figure. vfluct=v3d(i. uv(j)=uv(j)+ufluct*vfluct.1). There are four files with instantaneous values of v¯1 .j.two_uu_1_mat(1:imax).R.j. Here’s an example how to do: uv=zeros(nj.2.m reads the instantaneous v¯1 field and performs an averaging in the homogeneous directions x1 and x3 .k)-vmean(j).’fontsi’. we want to look at the time-averaged results. Compute also the resolved turbulent kinetic energy kres = 0. pl uvw hybrid. for k=1:nk for j=1:nj for i=1:ni ufluct=u3d(i. In SA-DES.R. ∆z .mat. Which is largest? Which is largest in the URANS region and in the LES region. .07kT ℓ Table R.6 Location of interface in DES and DDES As mentioned above. It is computed as δV = (∆x1 ∆x2 ∆x3 ). Recall that ν = 1/8000.4 Turbulent kinetic energy Now plot and compare the resolved and modelled turbulent kinetic energies.3.5kT n[1 − exp(−0. the location of the interface is computed using k and ω.2kT n/ν)] 1/2 1/2 0.1 and do the usual averaging. kT (file te1 hybrid. . R. 20.1: Expressions for ℓ and νT in the LES and URANS regions.5 The modelled turbulent shear stress We have computed the resolved shear stress. te2 hybrid. When computing ∆. will be used. Let’s find the modelled shear stress. n denotes the distance from the wall.4. you need the volume. . see Eq. look at the beginning of the m-file. respectively? What about the sum? The magnitude of resolved and modelled turbulent kinetic energies is discussed in the last subsection in [84]. ∆x1 and ∆x3 are constant and ∆x2 is stored in the array dy(j). R. 20.mat.m. ). One would expect that (∆x1 ∆x2 ∆x3 )1/3 < ℓURAN S everywhere.3). ∆y . Let’s compare that with DES and DDES. Which is largest? Any surprises? Compare them with ∆x2 and (∆x1 ∆x2 ∆x3 )1/3 . the interface is defined as the location where the wall distance is equal to CDES ∆ where ∆ = max ∆x . of the cells. Where is it large and where is it small? (Recall that the URANS region is located in the first 28 cells). ).mat.09 · 2.1.7 Turbulent length scales Compute and plot the turbulent length scales given in Table R. The modelled turbulent kinetic energy. Turbulent kinetic energy ℓ νT URANS region 1/2 2. R. Compute ω from νT and k as ω = νT /k) and compute the location using Eq.7. Is this the case? . . use more files. Are they smooth across the interface? Is the resolved shear stress large in the URANS region? Should it be large? Why/why not? R. Note that the modelled turbulent kinetic energy. δV . . Plot the ℓSGS and ℓURAN S length scales in both regions. Compute the modelled shear stress from the Boussinesq assumption ∂¯ v2 ∂¯ v1 + τ12 = −2νT s¯12 = −νT ∂x2 ∂x1 Plot it and compare with the resolved shear stress (see Section R. . How does this compare with gridline number 29? In SST-DES. Plot hνT i/ν. can be downloaded from the www page and loaded at the beginning of pl uvw hybrid. Is it smooth? Do you need more samples? If so.5n[1 − exp(−0.014kT n/ν)] 317 LES region ℓ = ∆ = (δV )1/3 1/2 0. kT (te1 hybrid. the interface in the present simulations is prescribed along a fixed grid line (No 29). Compute the turbulent viscosity according to Table R. the bij is given by b11 = cos α.1) where bij denotes the cosine between the axis bij = cos (xi .6a) (S. xj∗ ) (S. (S.5) Inserting Eq.6d) . S.g. Equation S. the transformation reads uij = bik bjm uk∗m∗ As an example.6c) (S.2) In Fig. x2 ).1.S. MTF270: Transformation of tensors 318 x2 x∗2 x∗1 β α x1 Figure S.4) √ b22 = 1/ 2 (S. e. x2∗ ) and (x1 . b21 = 1/ 2.3) The relations bik bjk = bki bkj = δij are fulfilled as they should. b12 = cos β = − cos(π/2 − α) = − sin(α) b21 = cos(π/2 − α) = sin α. b12 = −1/ 2.3 gives √ √ √ b11 = 1/ 2. set α = π/4.4 gives u11 = b11 b11 u1∗1∗ + b12 b11 u2∗1∗ + b11 b12 u1∗2∗ + b12 b12 u2∗2∗ 1 = (u1∗1∗ − u2∗1∗ − u1∗2∗ + u2∗2∗ ) 2 u12 = b11 b21 u1∗1∗ + b12 b21 u2∗1∗ + b11 b22 u1∗2∗ + b12 b22 u2∗2∗ 1 = (u1∗1∗ − u2∗1∗ + u1∗2∗ − u2∗2∗ ) 2 u21 = b21 b11 u1∗1∗ + b22 b11 u2∗1∗ + b21 b12 u1∗2∗ + b22 b12 u2∗2∗ 1 = (u1∗1∗ + u2∗1∗ − u1∗2∗ − u2∗2∗ ) 2 u22 = b21 b21 u1∗1∗ + b22 b21 u2∗1∗ + b21 b22 u1∗2∗ + b22 b22 u2∗2∗ 1 = (u1∗1∗ + u2∗1∗ + u1∗2∗ + u2∗2∗ ) 2 (S. b22 = cos α (S. S.1 Rotation from x1∗ − x2∗ to x1 − x2 The rotation of a vector from the xi∗ coordinate system to xi reads (see. Chapter 1 in [27]) ui = bij uj∗ (S. S.5 into Eq.1: Transformation between the coordinate systems (x1∗ . S MTF270: Transformation of tensors S. For a second-order tensor.6b) (S.. The eigenvalues. Rotation from x1 − x2 to x1∗ − x2∗ S. s22 = 0 (S. s2∗1∗ = 0.e. S.2. Let us choose π/4 and 3π/4 for which the transformation in Eq. The eigenvectors for Eq.11b) (S.2 319 Rotation from x1 − x2 to x1∗ − x2∗ Now we do the same transformation as in Section S.8 s1∗1∗ = s12 . S. S. λ(k) . Consider fully developed flow in a channel. respectively.1. 2 ∂x2 λ(2) ≡ s2∗2∗ = −s12 = − 1 ∂v1 2 ∂x2 (S.11d) The fully developed channel flow is obtained by inserting Eq.11a) (S. from x1 − x2 to x1∗ − x2∗ . s12 = 1 ∂v1 .9) where c11 = cos α. see Appendix C. i.e.10) see Fig.1 but backwards. c12 = cos(π/2 − α) = sin α c21 = cos β = − sin α.7) so that the strain-rate tensor.12) Since the off-diagonal elements are zero it confirms that the coordinate system x1∗ − x2∗ with α = π/4 is indeed a principal coordinate system.1 correspond now to the streamwise and wallnormal directions in the channel. i. Let the x1∗ − x2∗ coordinate system denote the eigenvectors.S. S. xj ) (S. s2∗2∗ = −s21 (S.11c) (S. c22 = cos α (S.8) The x1 and x2 coordinates in Fig. 2 ∂x2 s21 = s12 . sij .12.9 reads (α = π/4) s1∗1∗ = c11 c11 s11 + c12 c11 s21 + c11 c12 s12 + c12 c12 s22 1 = (s11 + s21 + s12 + s22 ) 2 s1∗2∗ = c11 c21 s11 + c12 c21 s21 + c11 c22 s12 + c12 c22 s22 1 = (−s11 − s21 + s12 + s22 ) 2 s2∗1∗ = c21 c11 s11 + c22 c11 s21 + c21 c12 s12 + c22 c12 s22 1 = (−s11 + s21 − s12 + s22 ) 2 s2∗2∗ = c21 c21 s11 + c22 c21 s21 + c21 c22 s12 + c22 c22 s22 1 = (s11 − s21 − s12 + s22 ) 2 (S. for which ∂v1 = v2 = 0 ∂x1 (S.13) . s1∗2∗ = 0. λ(1) ≡ s1∗1∗ = s12 = 1 ∂v1 . of sij correspond to the diagonal elements in Eq. ±3π/4. reads s11 = 0. The transformation from x1 − x2 to x1∗ − x2∗ reads si∗j∗ = cik cjm skm . It can be seen that the relation cji = bij is satisfied as it should.8 are any two orthogonal vectors with angles ±π/4. S. S. cij = cos (xi∗ . 13 and S.15) It can be seen that ∂v1∗ /∂x1∗ = s1∗1∗ and ∂v1∗ /∂x2∗ + ∂v2∗ /∂x1∗ = 2s1∗2∗ = 0 (see Eqs. ∂x1∗ ∂x2∗ 2 ∂x2 ∂v2∗ ∂v2∗ 1 ∂v1 = =− . S. S. . Transformation of a velocity gradient S.11a-S.e.S. fully-developed channel flow) ∂v1∗ 1 ∂v1 ∂v1∗ = = .11d with α = π/4 1 (A11 + A21 + A12 + A22 ) 2 1 = (−A11 − A21 + A12 + A22 ) 2 1 = (−A11 + A21 − A12 + A22 ) 2 1 = (A11 − A21 − A12 + A22 ) 2 A1∗1∗ = A1∗2∗ A2∗1∗ A2∗2∗ (S.14) Insert Eq. S. S.7 (i. ∂x1∗ ∂x2∗ 2 ∂x2 (S.3 320 Transformation of a velocity gradient Consider the velocity gradient Aij = ∂vi /∂xj . Apply the transformation from the x1 − x2 system to the principal coordinate system x1∗ − x2∗ in Eqs.10 with α = π/4 and replace Aij by the velocity gradient use Eq.15) as it should.3. see .1 Green’s first formula Gauss divergence law reads Z V ∂Fi dV = ∂xi Z Fi ni dS (T.7) As usual we are considering a volume V with bounding surface S and normal vector ni .2 Green’s second formula Switching ϕ and ψ in Eq.6) This is Green’s second formula.2) ∂x ∂x ∂x i i i S V The left side is re-written as ∂ ∂xi ∂ψ ∂2ψ ∂ψ ∂ϕ ϕ =ϕ + ∂xi ∂xi ∂xi ∂xi ∂xi which inserted in Eq. T.3) (T. V .6.T.2 gives Z Z Z ∂ψ ∂ψ ∂ϕ ∂2ψ ϕ dV + dV = ni dS ϕ ∂x ∂x ∂x ∂x ∂x i i i i i V S V (T.3 Green’s third formula In Green’s second formula. In the last section we will derive the analytical solution to the Poisson equation. T. Replacing Fi by ϕ ∂xi Z Z ∂ψ ∂ψ ∂ ϕ ϕ dV = ni dS (T.5 from T. T.5) (T.4 gives Z Z Z ∂ϕ ∂ϕ ∂ψ ∂2ϕ dV + dV = ψ ni dS ψ ∂xi ∂xi ∂xi V ∂xi ∂xi S V Subtract Eq. consider a small sphere in V . T. Since function ψ(r) is singular for r = rP .4) This is Green’s first formula. set ψ(r) = 1 |r − rP | (T. T. and ni is the normal vector of S ∂ψ gives pointing out of V .1) S where S is the bounding surface of the volume. The derivations below are partly taken from [186]. Eq. T. T.4 gives Z Z ∂2ψ ∂2ϕ ∂ϕ ∂ψ ϕ −ψ −ψ dV = ϕ ni dS ∂xi ∂xi ∂xi ∂xi ∂xi ∂xi V S (T. MTF270: Green’s formulas 321 T MTF270: Green’s formulas In this appendix we will derive Green’s three formulas from Gauss divergence law. i.6 we need the first and the second derivative of ψ.9) ri 3ri 3 1 r2 3 = −3 3 + = − = 0 + r r r4 r3 r r4 To get the right side on the second line we used the fact that ri ri = r2 .11) ri − riP ∂ϕ 1 −ϕ + (−nεi )dS − 3 |r − r | |r − r | ∂x ε P P i S IS∗ . T. We get Z Z ∂2ϕ ∂ϕ ri − riP 1 1 ni dS dV = −ϕ − − |r − rP |3 |r − rP | ∂xi S V −S ε |r − rP | ∂xi ∂xi Z (T.1. i.i 1 ∂ =− ∂xi |r − rP | |r − rP |3 (T. T. respectively. T. Apply Green’s second formula for this volume which has the bounding surfaces S and S ε with normal vectors ni (outwards) and nεi (inwards). Green’s third formula 322 ni S V nεi r Sε rP x2 x1 Figure T.8 and T.10) 2 ∂ 1 =0 ∂xi ∂xi |r − rP | for ri 6= riP .8) ∂xi r r2 r since the derivative of a distance X is a vector along the increment of the distance. In V there is a small sphere S ε located at rP with radius ε and normal vector nεi . ∂X/∂xi = Xi /X where X = |Xi |. Now we replace ri = r by r − rP = ri − rP. for V excluding the sphere S ε .3.i in Eqs. T.e. Fig.1: Green’s third formula.9 which gives ri − rP. The first derivative of 1/ri is computed as ∂r/∂xi ri 1 ∂ =− =− 3 (T. A volume V with bounding surface S with normal vector ni .1.T. see Fig. The second derivative is obtained as ∂ ri ∂ri 1 ∂r 3ri 1 ∂2 =− = − + ∂xi ∂xi r ∂xi r3 ∂xi r3 ∂xi r4 (T.e. In Eq. 19 is not a problem. nεi . b]. In one dimension this theorem simply states that for the integral of a function.15) + i P.e. |r − rP | = ε (T. Note the minus sign in front of the normal vector in the S ε integral. Consider a small sphere with radius r = |ra − rP | centered at point P . T.3.i ∂ϕ 2 IS ε = ϕ + (r − r ) ε dΩ = dΩ (T. this is because the normal vector must point out of the volume V − S ε . In the sphere the normal vector. T.12.e.i ) IS ε = ϕ(rQ ) (rQ ) dΩ (T. Applying this theorem to the integral in Eq. V − S ε .e. i. correspond to the direction from point rP . ε. In spherical coordinates the volume element can then be expressed as dV = r2 sin θdrdθdα = r2 drdΩ (T.14 in the last integral in Eq.13 and T.i | (T. T.15 gives Z Z ∂ϕ dΩ + rQ.i denotes a point on S ε .e.16) a holds. there exists (at least) on point for which the the relation Z b g(x)dx = (a − b)g(xQ ) (T. i. S ε . nεi = ri − rP. T. b].i is the radius of sphere S ε . for sphere S ε can be expressed in spherical coordinates as dS = ε2 Ω = ε2 sin θdθdα (T. Inserted in Eq.11 gives Z Z ϕ ∂ϕ ri − rP. As we let Q → P . i.13) The surface area.17 reads lim IS ε = 4πϕ(rQ ) since R ε→0 Sε (T. where xQ denotes a point on [a.i − rP.19) This is Green’s third formula.11 gives Z ∂2ϕ 1 dV V |r − rP | ∂xi ∂xi Z Z 1 ∂ϕ 1 ri − riP 1 + n dS + ni dS ϕ i 4π S |r − rP |3 4π S |r − rP | ∂xi ϕ(rP ) = − 1 4π (T. i. Green’s third formula 323 where the volume integral is taken over the volume V but excluding the sphere S ε .i r − rP = |r − rP | |ri − rP. The singularity 1/|r − rP | in the volume integral in Eq.T. over an interval [a. g(x).i in order to make its length equal to one. The length of the vector ri − rP. dS.18) dΩ = 4π. . T. Inserting Eqs.12) where we have normalized the vector ri − rP. into the sphere.14) where Ω is the solid angle. the radius. T. T.20) Hence it is seen that volume element dV goes to zero faster than the singularity 1/|r − rP |.17) ∂xi Sε Sε where rQ ≡ rQ.i ε2 ε2 ∂xi ∂xi Sε Sε To re-write this integral we will use the mean value theorem for integrals. of sphere S ε goes to zero so that the integral in Eq. 25) ϕ(rP ) = − 4π V |r − rP | This is the analytical solution to Poisson’s equation.13 and T.22 reads Z 1 f (r) dV (T.23) n dS = ϕ ϕdΩ ϕn n dS = i i i 4π S |r − rP |3 4πR2 S 4π S using ni ni = 1.13. T. T. Analytical solution to Poisson’s equation T. T. gives Z 1 f (r) dV ϕ(rP ) = − 4π V |r − rP | (T. Using Eqs.21.4.24) Z Z 1 1 ∂2ϕ dV = = f dV 4πR S ∂xi ∂xi 4πR V This integral also goes to zero for large R since we have assumed that f is limited.14. .12. T.22) Z Z 1 ∂ϕ 1 ri − riP 1 + n dS + n dS ϕ i i 4π S |r − rP |3 4π S |r − rP | ∂xi We choose the volume as a large sphere with radius R. the first surface integral can be written as Z Z Z 1 1 ri − riP 1 (T. This integral goes to zero for large R since ϕ → 0 as R → ∞. T.21 as Z Z 1 ∂ϕ 1 1 ∂ϕ ni dS = ni dS 4π S |r − rP | ∂xi 4πR S ∂xi (T. T. Eq.4 324 Analytical solution to Poisson’s equation Poisson’s equation reads ∂2ϕ =f ∂xj ∂xj (T. Hence the final form of Eq.T.19. T. Gauss divergence law and Eq.21) where we assume that ϕ goes to zero at infinity and that the right side is limited.22 can be re-written using Eq. T. Green’s third formula. Eq. The second surface integral in Eq. 6. 11. How is the production term modeled in the k − ε model (Boussinesq)? Show how it can be expressed in s¯ij 9. the term which includes the triple correlation) in the k equation is modeled. 11. θ. modeled in the Boussinesq approach? .22. Show how the turbulent diffusion (i.2 on p.U. 7.e. vi′ vj′ .11 (see Section 11. 10. 113) how to derive the transport equation for vi′ θ′ . How are the Reynolds stress. and the transport equation ¯ show the principles (in the same way as is done for the v ′ v ′ equation on for θ. Given the modeled k equation.2 on p. What is the expression for the total heat flux that appears in the θ¯ equation? 4. ρgi . Discuss the physical meaning of the different terms. Eq. Given the transport equation for the temperature. re-written in incompressible flow? 2. How is the buoyancy term. Eq. i j Section 11. vi′ θ′ . Which terms in the vi′ vj′ equation need to be modeled? Explain the physical meaning of the different terms in the vi′ vj′ equation. Which terms need to be modeled? 3. and the turbulent heat flux. 8. 113) 5. Show the principles how to derive the transport equation for vi′ vj′ . MTF270: Learning outcomes for 2015 U 325 MTF270: Learning outcomes for 2015 Week 1 1. Derive the Boussinesq assumption. derive the modeled ε equation. modeled pressure-strain terms and modeled dissipation terms for a simple shear flow (e.63 and U. Show that for decaying grid anisotropy tensor is defined as aij = 2k turbulence. Eq. respectively. Derive the algebraic stress model (ASM). In some stress equations there is no production terms nor any dissipation term. Derive the exact Poisson equation for the pressure fluctuation. 3. What sign does it have? Give also the expressions for Φ11. 11. The slow pressure-strain model reads Φij. Which are the “slow” and “rapid” terms? Why are they called “slow” and “rapid”? 7. Give the expression for the production terms. 2.1w . Discuss and show how the dissipation term. 11.66) for the pressure-strain term. MTF270: Learning outcomes for 2015 326 Week 2 1.U.1) Use Eqs. then a model for the shear stress is also obtained. channel flow. .2 = −c2 Pij − 23 δij P k . The v′ v′ i j − 32 δij . 6.1 to derive the exact analytical solution (Eq. Describe the physical effect of stable stratification and unstable stratification on turbulence. the model for the slow pressure-strain model indeed acts as to make the turbulence more isotropic if c1 > 1. 4.63.1 = −c1 ρ kε vi′ vj′ − 32 δij k and Φij. jet flow . Using physical reasoning. 11. What main assumption is made? 8.g.1w 9. How come? Which is the main source term (or sink term) in these equations? 10.1w and Φ33. Derive the slow pressure strain term. for wall effects. The modeled slow and rapid pressure strain term read Φij. Show that if the model is expressed in the principal directins. ). .1 = −c1 ρ kε vi′ vj′ − 32 δij k . For a Poisson equation ∂2ϕ =f ∂xj ∂xj there exists an exact analytical solution Z 1 f (y)dy1 dy2 dy3 ϕ(x) = − 4π V |y − x| (U. the model for the pressure-strain term above is formulated only for the normal streses. 5. . εij . is modeled. boundary layer. Φ22. Use physical reasoning to derive a model for the diagonal components of the pressure-strain term (slow part). 7. Streamline curvature: now consider a boundary layer where the streamlines are curved away from the wall (concave curvature). Show that in the Reynolds stress model. Show that the Reynolds stress model dampens and increases the vertical fluctuation in stable and unstable stratification. What is a realizability constraint? There are two main realizability constraints on the normal and the shear stresses: which ones? 6. are obtained from |Sij − δij λ| = 0. there is only a small production of turbulence whereas eddy-viscosity models (such as the k − ε model) give a large production of turbulence. 3. 4. Consider buoyancy-dominated flow with x3 vertically upwards. The three terms read 1 2 c1 τ s¯ik s¯kj − s¯ℓk s¯ℓk δij 3 ¯ kj ¯ ik s¯kj − s¯ik Ω +c2 τ 2 Ω 2¯ ¯ 3 ¯ ¯ ¯ ¯ +c5 τ Ωiℓ Ωℓm s¯mj + s¯iℓ Ωℓm Ωmj − Ωmn Ωnℓ s¯ℓm δij 3 Show that each term has the same properties as aij .12).U. Which equations are solved in the V2F model? . Hint: the eigenvalues. traceless and symmetric. Show that the Reynolds stress model gives an enhanced turbulence production (as it should) because of positive feedback between the production terms. λ2 . What is a non-linear eddy-viscosity model? When formulating a non-linear model. Show that the turbulence is dampened if ∂vθ /∂r > 0 and that it is enhanced if the sign of ∂vθ /∂r is negative. as it should. 5. Show that the effect of streamline curvature in the k − ε model is much smaller. λ1 . respectively. In which coordinate system is the risk largest for negative normal stresses? Derive an expression (2D) how to avoid negative normal stresses by reducing the turbulent viscosity (Eq. 13. Show also that k in the k − ε model is affected in the same way. 2. The production term for the vi′ vj′ and the vi′ θ′ equations read Gij = −gi βvj′ θ′ − gj βvi′ θ′ . non-dimensional. i.e. Show that the Boussinesq assumption may give negative normal stresses. MTF270: Learning outcomes for 2015 327 Week 3 1. Piθ = −vi′ vk′ ∂ θ¯ ∂xk respectively (we assume that the velocity gradient is negligible). Consider stagnation flow. Consider streamline curvature for a streamline formed as a circular arc (convex curvature). the anisotropy tensor aij = −2νt s¯ij /k is often used. 8. In the V2F model. The f equation in the V2F model reads Φ22 1 ∂2f − L2 2 − f = − ∂x2 k T 2 v2′2 − k 3 ! . The transport equation for v2′2 reads (the turbulent diffusion term is modeled) " # ∂ρ¯ v1 v2′2 ∂v2′2 ∂ρ¯ v v2′2 ∂ (µ + µt ) −2v ′ ∂p′ /∂x2 −ρε22 + = ∂x1 ∂x2 ∂x2 ∂x2 | 2 {z } ρΦ22 Show how this equation is re-written in the V2F model.9)? 11. How does f enter into the v2′2 equation? Show that far from the walls.1). MTF270: Learning outcomes for 2015 328 9. 15. In which region is each model used and why? How is ω expressed in k and ε? . 10. T ∝ k . the v 2 equation is solved: what is the difference between v2′2 and v 2 (see the discussion in connection to Eq. What does the acronym SST mean? The SST model is a combination of the k − ε and the k − ω model. ε L∝ k 3/2 ε Explain how the magnitude of the right side and L affect f (Fig 15. the f and the v2′2 equation) returns to the v2′2 equation in the Reynolds stress model.e.U. the V2F model (i. Show how a sinus wave sin(κc x) corresponding to cut-off is represented on a grid with two and four cells. 5. Show the three different regions (the large energycontaining scales. 4. M.U. grid (i. κc are located in the spectrum. see Fig. the −5/3 range and the dissipating scales).e resolved) scales and the cut-off. ¯ K ¯ 12. What is the purpose of the shear stress limiter in the SST model? Show that the eddy-viscosity assumption gives too high shear stress in APG since P k /ε ≫ 1 (Eq. Consider a 1D finite volume grid. a blending function F1 is used. Derive a transport equation for ω from the k and ε transport equations. 16. what does this function do? 3. 7. Consider the energy spectrum and discuss the physical meaning of Pksgs and εsgs . Show the difference between volume averaging (filtering) in LES and timeaveraging in RANS. In the SST model. How is κc related to the grid size ∆x for these cases? 10. the destruction and the viscous diffusion terms. 6. ¯ and the ksgs equations.3. 8. respectively. Derive the one-equation ksgs equation 11. How are k and ksgs computed from the energy spectrum? . Carry out a second filtering of v¯ at node I and show that v¯I 6= v¯I . Consider the energy spectrum. Where should the cut-off be located? Show where the SGS scales. you only need to do the production. MTF270: Learning outcomes for 2015 329 Week 4 1. 2. Consider the spatial derivative of the pressure in the filtered Navier-Stokes: show that the derivative can be moved outside the filtering integral (it gives an additional second-order term). The filtered non-linear term has the form ∂vi vj ∂xj Show that it can be re-written as ∂¯ vi v¯j ∂xj giving an additional term − ∂τij ∂ ∂ (vi vj ) + (¯ vi v¯j ) = − ∂xj ∂xj ∂xj on the right side. 9. Discuss the energy path in connection to the source and sink terms in the k.14). τij . Eq. .test Draw an energy spectrum and show which wavenumber range k. Use this relation and derive the final expression for the dynamic coefficient. MTF270: Learning outcomes for 2015 330 1. The equation you dervived above is a tensor equation for C. Use Eq. ksgs. 18. 3.2) v¯i v¯j − v¯ i v¯ j + τ ij = Lij + τ ij = Tij ¯ ksgs . cover.6 and 18.22) 2. Tij . C. and the test filter SGS stress.U. U. Derive the Smagorinsky model in two different ways (Sections 18.2 and derive the relation ! z { z{2 z{ z{ 1 2 s|¯ sij Lij − δij Lkk = −2C ∆ | s¯ | s¯ ij − ∆ |¯ 3 4.33. What is a test filter? Grid and test filter Navier-Stokes equation and derive the relation z { z{ z{ z{ z{ (U. Formulate the Smagorinsky model for the grid filter SGS stress. What is DDES? Why was it invented? . What are the five main differences between a RANS finite volume CFD code and a LES finite volume CFD code? What do you need to consider in LES when you want to compute time-averaged quantities? (see Fig. Which equations and which term? What is the effect on the modeled. an additional diffusion term and dissipation terms appear because of a numerical SGS viscosity 2. What are they called? What does the word “scale-similar” mean? Two of the terms are not Galilean invariant: which ones? What is Galilean invariance? 4. how fine does the mesh need to be in the wall region be? Why does it need to be that fine? 2 ν˜t . Tint . τij is written as three different terms. be used?(see Section M. The modified (reduced) length scale in two-equation DES models can be introduced in different equations. What is DES? The length scale in the RANS S-A model reads d computed in the corresponding DES model? 6. We usually define the SGS stress tensor as τij = vi′ vj′ − v¯i v¯j .3) 3. When doing LES. Derive these three terms. turbulent quantities? 8. How can the integral time scale. How is the length scale computed in a k − ε two-equation DES model? Where in a boundary layer does the DES model switch from RANS to LES? 7. how is it 5. MTF270: Learning outcomes for 2015 331 Week 5 1. Show that when a first-order upwind schemes is used for the convection term.11). 18. In scale-similarity models.U. 23. Assume that fε = 1. Describe hybrid LES-RANS based on a one-equation model. How is the von K´arm´an length scale defined? An additional source term is introduced in the ω equation: what is the form of this term? Describe how this source reduces the turbulent viscosity. respectively? . When is the term large and small. How is the instantaneous velocity decomposed? What turbulence models are used? What is scale separation? 3. 2.21. What is the main modification compared to the standard k − ε model? What is the physical meaning of fk ? 5.1 and 19. Describe URANS. Describe the SAS model. Consider the ∗ destruction term in the ε equation and the coefficient Cε2 . Explain what happens if fk is reduced from 1 (RANS mode) to fk = 0.U.4 (LES mode). Discuss the choice of discretization scheme and turbulence model in URANS (see Sections 19. MTF270: Learning outcomes for 2015 332 Week 6 1. Describe the PANS model. The PANS equations are given in Eqs.2) 4. 6. for the energy-containing eddies. determined? With this method. the generated shear stress is zero: why? How is the correlation in time achieved? 2. κmax . MTF270: Learning outcomes for 2015 333 Week 7 1. see http://www. κe .tfd.se/˜lada/slides/slides qles davidson.U. determined? How are the maximum and minimum wavelengths.se/˜lada/slides/slides inlet.pdf What form on the spectrum is assumed? How is the wavenumber.pdf Which method is good/bad? Which is best? .chalmers.chalmers. κmin . Mention four different ways to estimate the resolution of an LES that you have made. see Chapter 26 and first 13 slides in http://www. Give a short description of the method to generate synthetic turbulent inlet fluctuations.tfd. [12] H. Norberg. New York. R. Zhang.es/ftp/channels/data/statistics/. [13] P. Jim´enez. 2003. G¨ottingen. 2007. [5] H. Dept. [4] J. Wojtkowiak and C. [15] S. 2006. 1972. of Applied Mechanics. References V 334 References [1] J. [9] S. Springer-Verlag. Cambridge University Press. Abedi. [6] W.V. 39:347–370. Quantitative numerical analysis of flow past a circular cylinder at Reynolds number between 50 and 200. Sweden. Germany. Hager. [3] H. G¨oteborg. Blasius: A life in research and education. Golubev. Physics of Fluids A. Sweden. Jim´enez. Hong and T. Ekh. University of G¨ottingen. Development of vortex filament method for aerodynamic loads on rotor blades. G¨oteborg. 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