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![DNS Order
Required Fourier Modes (per direction):
N ≈1.6
L
η
⎛
⎝
⎜
⎞
⎠
⎟=1.6⋅ ReL
3
4 ≈ 0.4Rλ
3
4
Required Fourier Modes (3-space):
N3
≈ 0.06⋅ Reλ
9
4
Required Floating Point Operations
Ops ≈ O N3
[ ]
3
( )](https://cdn.statically.io/img/image.slidesharecdn.com/presentation1-181211162700/85/LES-from-first-principles-15-320.jpg)
Uploaded byMichael Munroe
LES from first principles
The document discusses various fluid dynamics concepts including large eddy simulation, Reynolds number, and momentum equations. It outlines the distinctions between direct numerical simulation (DNS), Reynolds averaged Navier-Stokes (RANS), and the employed conservation laws within fluid dynamics. Additionally, it presents computational implications for various scenarios such as soccer balls, aircraft, and turbulence modeling methods.
LES from first principles
- 1.
- 2. Reynolds Number General Equation: U= fluid speed L = Characteristic Length ν = Kinematic Viscosity Where: Re = U ⋅ L ν Examples Soccer Ball ≈ 1e5 Falcon ≈ 1e6 (at 200mph) Airplane ≈ 2e7 Boeing 747 ≈ 6e8
- 3. Pre-Smagorinski DNS -Direct Numerical Simulation Very accurate, but very costly RANS - Reynolds Averaged NS Very fast, but loses transient features Experimentally driven approximation (Empirical Methods)
- 4. DNS - DirectNumerical Simulation
- 5. Acceleration of FluidParticle dU = ∂U ∂xi dxi + ∂U ∂x j dx j + ∂U ∂xk dxk + ∂U ∂t dt An infinitesimal change of spatial velocity of a fluid particle is: The acceleration is therefore dU dt = ∂U ∂xi dxi dt + ∂U ∂x j dx j dt + ∂U ∂xk dxk dt + ∂U ∂t dt dt = ∂U ∂x1 U1 + ∂U ∂x2 U2 + ∂U ∂x3 U3 + ∂U ∂t =∂tU +Ui∂xi U j =∂tU +∂xi U jUi( )
- 6. Substantial Derivative Spatial Changeof Velocity: DU Dt = ∂U ∂x1 U1 + ∂U ∂x2 U2 + ∂U ∂x3 U3 + ∂U ∂t Spatial Change of Vector Property A D Dt A( ) =∂t A +∂xi AjUi( ) D Dt A( ) = ∂A ∂t + U ⋅ ∇( )A or
- 7. Conservation of Mass Conservationof Mass d dt M( ) system = 0 ⇒ η =1 ∂ ∂t ρ⋅ dV CV ∫ + ρ ˜U ⋅ d ˜A CS ∫ = 0 ∂ ∂t dV CV ∫ + ˜U ⋅ d ˜A CS ∫ = 0 ˜U ⋅ d ˜A CS ∫ = 0 ˜U ⋅ d ˜A CS ∫ = ∇ • ˜U( )dV CV ∫ = 0 CV form (via RTT) Incompressibility (constant density) Harmonic BC (const shape & vol) Divergence Theorem
- 8. Continuity (continued) ∇ • ˜U()dV CV ∫ = 0 ∇ • ˜U( ) = ˜A A = 1 V ˜A⋅ dV parcel ∫ ∇ • ˜U( )dV CV ∫ = ∇ • U( ) CV = 0 lim CV →δ ∇ • U( ) CV = ∇ • ˜U( ) Change of Variable Mean Value Thm Substitute and Simplify Take the Limit Continuity Equation ∇ • ˜U = 0
- 9. Force Tensor Note: This samederivation informs Solid Deformation, Convective and Conductive Heat Transfer and Mass Transfer.
- 10. Newtonian Shear Viscosity Assumptions: •Linear • Isotropic • Homogenous • Constant Value
- 11. Momentum d ˜F =dm D ˜U Dt ⇒ d ˜F dV = ρ D ˜U Dt Definition of Momentum Momentum along 1 axis ∂σ11 ∂x1 + ∂τ12 ∂x1 + ∂τ13 ∂x1 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟= ρ ∂U1 ∂x1 U1 + ∂U1 ∂x2 U2 + ∂U1 ∂x3 U3 + ∂U1 ∂t ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ Generalized Momentum ∂xi ˜Fij = ρ Ui∂xi U j +∂tUi( )
- 12. Momentum (cont) Momentum ρ Ui∂xi Uj +∂tUi( ) =∂xi ˜Fij Shear ρ ∂ ˜U ∂t + ρ ˜U ⋅ ∇( ) ˜U = − ∇p ρ + μ∇ 2 ˜U ρ = constant μ = constant ⇒ ν = μ ρ = constant
- 13. Navier Stokes Equations MomentumEquation Continuity Equation ∂ ˜U ∂t + ˜U • ∇( ) ˜U −ν∇ 2 ˜U + ∇p ρ = 0 ∇ ⋅ ˜U = 0 Assumptions: • Continuum Hypothesis • Ideal Gas, Perfect Caloric Gas • Steady, Homogenous, Isotropic, nonzero, finite: • density, viscosity, thermal conductivity, specific heats Energy Equation ρ ∂e ∂t + ˜U • ∇( )e = ∂Q ∂t + k∇ 2 T + Φ Calorically perfect ideal gas p = γ −1( )ρe T = γ −1( )e R
- 14. Navier Stokes Closure Numberof Equations: • Continuity = 1 • Momentum = 3 • Energy =1 • Cal. Perf. Ideal Gas =2 Total is 7 Number of Unknowns: • density •Pressure •Energy •Temperature • velocity in x-direction • velocity in y-direction • velocity in z - direction Total is 7 Number of Equations = Number of Unknowns Therefore this is a closed system
- 15. DNS Order Required FourierModes (per direction): N ≈1.6 L η ⎛ ⎝ ⎜ ⎞ ⎠ ⎟=1.6⋅ ReL 3 4 ≈ 0.4Rλ 3 4 Required Fourier Modes (3-space): N3 ≈ 0.06⋅ Reλ 9 4 Required Floating Point Operations Ops ≈ O N3 [ ] 3 ( )
- 16. DNS Computability Soccer Ball≈ 1.5e36 Falcon ≈ 1.4e42 (at 200mph) Airplane ≈ 3.8e48 Boeing 747 ≈ 3.7e59 Implications (operations*): Implications (years**): Soccer Ball ≈ 4.9e16 Falcon ≈ 4.4e22 (at 200mph) Airplane ≈ 1.2e29 Boeing 747 ≈ 1.1e40 *Assuming a Teraflop computer ** Age of the universe ~ 15e9 years DNS applied to engineering problems: Totally Accurate Very Computationally Intensive
- 17. RANS - AveragedNavier Stokes
- 18. Reynolds Decomposition U v x,t( )= U v x,t( ) + v u v x,t( ) Velocity = Mean + Deviatoric where U = lim T → ∞ 1 T U ⋅ dt 0 T ∫ This is familiar.
- 19. Reynolds Continuity Standard Conservationof Mass (Continuity) In terms of Reynold Decomposition ∇ •U = 0 ∇ • u + U( ) = 0 ⇒ ∇ • u = 0 ⇒ ∇ • U = 0
- 20. Mean OF SubstantialDerivative but UiU j = Ui U j + uiuj therefore ∂ ∂xi UiU j =∂xi Ui U j + uiuj( ) = Ui ∂xi U j + U j ∂xi Ui +∂xi uiuj = Ui ∂xi U j +∂xi uiuj D Dt U j = ∂ U j ∂t + ∂ ∂xi UiU j The mean is Finally D Dt U j = ∂ U j ∂t + Ui ∂ U j ∂xi + ∂ uiuj ∂xi Remembering ∇ • U =∂xi Ui = 0
- 21. Mean of NonlinearTerm UiU j = Ui + ui( )⋅ U j + uj( ) = Ui U j + U j ui + Ui uj + uiuj = Ui U j + uiuj The mean is Because U j ui = U j ⋅ lim T → ∞ 1 T ui ⋅ dt 0 T ∫ = U j ⋅ 0 = 0 Ui uj = Ui ⋅ lim T → ∞ 1 T uj ⋅ dt 0 T ∫ = Ui ⋅ 0 = 0
- 22. Mean-Substantial Derivative The MeanOF the Substantial Derivitave D Dt U j = ∂ U j ∂t + Ui ∂ U j ∂xi + ∂ uiuj ∂xi Therefore the Mean-Substantial Derivative is D Dt = ∂ ∂t + Ui ∂ ∂xi or D Dt = ∂ ∂t + Ui • ∇
- 23. Mean Momentum DU j Dt = DUj Dt +∂xi uiuj ⇒ ν∇ 2 U j − ∂xi p ρ = DU j Dt +∂xi uiuj ⇒ DU j Dt =ν∇ 2 U j − ∂xi p ρ −∂xi uiuj
- 24. Alternate Mean Momentum Considerthe following DU j Dt =ν∇ 2 U j − ∂xi p ρ −∂xi uiuj ⇒ ρ DU j Dt =∂xi μ ∂x j Ui +∂xi U j( ) + p δij − ρ uiuj( ) Where δij = 0, i ≠ j 1, i = j ⎧ ⎨ ⎩ ⎫ ⎬ ⎭ uiuj = u1u1 u1u2 u1u3 u2u1 u2u2 u2u3 u3u1 u3u2 u3u3 ⎡ ⎣ ⎢ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎥
- 25. Boussinesq Idea Boussinesq (1877)suggested that Deviatoric Reynolds Stress is proportional to Mean Rate of Strain uiuj = 2 3 kδij −2νT Sij Sij = ∂x j Ui +∂xi U j 2 Where Strain is: While is called:νT Eddy Viscosity Or Turbulent Viscosity Stress = constant x Strain And k = 1 2 uiui Or The only term capable of transporting momentum is the AN-isotropic stress component, therefore the isotropic terms are unscaled.
- 26. Time-Averaged Navier Stokes TimeAveraged Reynolds Stress • Very accurate for mean behavior • Apply to High Reynolds Number flows • Inapplicable to transient flows time-dependent solution is destroyed by averaging DU j Dt =∂xi νeff ∂x j Ui +∂xi U j( ) + ∂x j ρ p − 3ρk( ) ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ νeff =ν +νT ˜x,t( )
- 27. LES - LargeEddy Simulation
- 28. Smagorinski Idea Smagorinski suggesteda multi-model approach: 1. Numeric Solution for Large Scales • Grid Scale (GS) = Grid Size or Larger 2. Statistical Parametrization for Small Scales • Sub-Grid Scale is Smaller than Grid Size Range: Inertial Dissipative 80%Energy: k 20% Model: Simulate Parametrize Resolve 80% + Guess 20% Less CPU use Show transient phenomenology
- 29. Balance of Momentum DU j Dt =ν∇ 2 U j − ∂xi p ρ −∂xi uiuj ∂xi uiuj → ∂xi τ ij R D U j Dt =ν∇ 2 U j − ∂xi p ρ −∂xi τ ij R Averaged Momentum Notation Change Useful Expression Large Scale Modelling Looks like Time Averaged.
- 30. Viscosity Parametrization τij r = −2νTSij νT = l s 2 S l s 2 = Cs ⋅ Δ( ) νT = CsΔ( )S Boussinesq Idea (eddy viscosity) Define Eddy viscosity as a function of characteristic filtered rate of strain. Smagorinski Length Smagorinski Coefficient Filter WidthEddy Viscosity
- 31. (continued) λs = Δ Ck ⋅af ⋅ S3 S2 3 2 ⎛ ⎝ ⎜ ⎜ ⎜ ⎞ ⎠ ⎟ ⎟ ⎟ −1 2 af ≡ 2 kΔ( ) 1 3 ˆG κ( ) 2 Δ⋅ dκ∫ S2 = af Ckε 2 3 Δ −4 3 Smagorinski Length Filter Dependent Parameter Approximate Characteristic Filtered Rate of Strain
- 32. Example (Lilly 1967) Assumed: •Filtered Strain Rate can be determined from the Kolmogorov Spectrum using Equation 1. • A Sharp Spectral Filter is used as the filtering function. (1) S3 ≈ S2 3 2 (2) CS = l S Δ = 1 π 2 3Ck ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 3 4 ≈ 0.17 (3) νT = 0.17⋅ Δ( )⋅ S
- 33. Energy Implication P ≡νTS2 Rate of Transfer of Energy The mean strain is always positive. For all positive viscosity values there is unidirectional transfer. There is no backscatter.
- 34. Computational Issues Strengths • LessCPU Overhead than DNS • Performs at High Reynolds Numbers • Retains Transient Phenomena Weaknesses • No Backscatter • Poor performance at boundaries (walls) • Over-predicts losses for Inhomogenous turbulent flows
- 35. LES Computability Soccer Ball≈ 9.0e28 Falcon ≈ 5.2e33 (at 200mph) Airplane ≈ 7.4e38 Boeing 747 ≈ 4.5e47 Implications (operations*): Implications (years**): Soccer Ball ≈ 2.8e9 Falcon ≈ 1.6e14 (at 200mph) Airplane ≈ 2.3e19 Boeing 747 ≈ 1.4e28 *Assuming a Teraflop computer ** Age of the universe ~ 15e9 years NOW: the soccer ball can be mostly characterized in UNDER the age of the universe! That’s a change of 7 orders of magnitude!
- 36. Bibliography Lumley. J. L.,Yaglom. A. M., (March 2001) A Century of Turbulence. Flow, Turbulence, and Combustion, 66, 241-286 Fox, McDonald, and Pritchard (2004) Introduction to Fluid Mechanics. John Wiley & Sons Pope. S. B. (2000). Turbulent Flows. Cambridge: University Press Tannehill. J. C., Anderson. D. A., Pletcher. R. H. (1997) Computational Fluid Mechanics and Heat Transfer (2nd ed). Washington: Taylor & Francis Publishers Yoshizawa. A. (1998) Hydrodynamic and Magnetohydrodynamic Turbulent Flows: Modeling and Statistical Theory. Fluid Mechanics and Its Applications. (Vol. 48). Boston: Kluwer Academic Publishers
- 37.
- 39. Observation The dissipative regionundergoes a transformation in phenomenology when k is ~ kc/10.
- 40. Heisenberg ∆t ⋅ ΔE≥ h 2 Δt = L2 ν ΔE = 1 2 Iω2 I = 1 2 mr2 = 1 2 ρL2 L 2 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 2 Heisenberg Frisch (pp103) Ang. Mom. Mom. Of Inertia Substitution ω2 = 2πr U U = L ν r = 1 2 L ω2 = 2π L 2 ν L ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 2 Ang. Vel. Velocity Radius Ang. Vel. Finally ρνπ 2 L6 8 ≥ h 2
- 41. Heisenberg vs Kolmogorov ρνπ2 LH 6 4 ≥ h ⇒ LH ≥ 4h ρνπ2 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 1 6 Re≈1 @ Kc Conclusion: Heisenberg can be within one decade from Kolmogorov name rho nu Lh (µµ) Λκ ( µµ) Λη/Λκ ηγ−273κ 13595 0.000000124 7.2 6.61 1.10 ν2−100κ 3.4388 0.000002 18.1 53.18 0.34 αιρ−100κ 3.5562 0.000002 18.0 53.18 0.34 χο−200κ 1.6888 0.00000752 16.4 143.60 0.11 χο2−280κ 1.9022 0.00000736 16.1 141.31 0.11 η2−100κ 0.24255 0.0000174 19.7 269.41 0.07 νη3−300κ 0.6894 0.0000147 17.0 237.40 0.07 αιρ−300κ 1.1614 0.00001589 15.4 251.68 0.06 ηε−100κ 0.4871 0.0000198 17.1 296.82 0.06 η20−380κ 0.5863 0.00002168 16.4 317.72 0.05
- 42. Relevant Phenomena Tunneling Matter Energy Condensates Implosion Explosions (Bosenova) Radical Change of model
Editor's Notes
- #35 Because the assumption of homogenous turbulence, the inhomogenous turbulence at walls is poorly described by this method.