fluid dynamics
play

Fluid dynamics Fields Domain R 2 Scalar field f : R Vector - PowerPoint PPT Presentation

Fluid dynamics Fields Domain R 2 Scalar field f : R Vector field f : R 2 Types of derivatives Derivatives measure how something changes due to its parameters f Temporal derivatives t


  1. Fluid dynamics

  2. Fields • Domain Ω ⊆ R 2 • Scalar field f : Ω → R • Vector field f : Ω → R 2

  3. Types of derivatives • Derivatives measure how something changes due to its parameters ∂ f • Temporal derivatives ∂ t • Spatial derivatives � T � ∂ f ∂ x, ∂ f • gradient operator ∇ f = ∂ y ∇ · f = ∂ f x ∂ x + ∂ f y • divergence operator ∂ y ∇ 2 f = ∇ · ( ∇ f ) = ∂ 2 f ∂ x + ∂ 2 f • Laplacian operator ∂ y

  4. Spatial derivatives • Gradient: a vector pointing at the steepest uphill of the function • Divergence: measures the net flux • Laplacian: measures the difference from the average of neighbors

  5. Quiz • What is the laplacian of f(x, y) = 2xy + y 2 ? • (2y, 2x+2y) • 2 • x + 2y • 4

  6. Quiz What’s the gradient at ( π 3 , − π 4 ) ? What about Laplacian?

  7. Representation Density ρ : Ω → [0 , 1] Domain Ω Velocity u : Ω → R 3

  8. Navier-Stoke equations Momentum ∂ t = − ( u · ∇ ) u − 1 ∂ u ρ ∇ p + s ∇ 2 u + f equation Mass equation u : velocity ∇ · u = 0 (Incompressibility p : pressure condition) s : kinematic viscosity f : body force ρ : fluid density

  9. Momentum equation • Each particle represents a little blob of fluid with mass m , a volume V , and a velocity u • The acceleration of the particle a ≡ D u Dt • The Newton’s law mD u Dt = F

  10. Forces acting on fluids • Gravity: m g • Pressure: - ∇ p • Imbalance of higher pressure • Viscosity: µ ∇ · ∇ u • Force that makes particle moves at average speed • dynamic viscosity coefficient: µ

  11. Momentum equation The movement of a blob of fluid mD u Dt = m g − V ∇ p + V µ ∇ · ∇ u Divide by volume ρ D u Dt = ρ g − ∇ p + µ ∇ · ∇ u Rearrange equation giving Navier-Stoke D u Dt + 1 ρ ∇ p = g + s ∇ · ∇ u

  12. Quiz • If we want to add an additional force F to the entire domain of the fluid, how does that change the momentum equation? D u Dt + 1 ρ ∇ p = g + s ∇ · ∇ u

  13. Lagrangian v.s. Eulerian • Lagrangian point of view describes motion as points travel around space over time • Eulerian point of view describes motion as the change of velocity field in a stationary domain • Lagrangian approach is conceptually easier, but Eulerian approach makes the spatial derivatives easier to compute/approximate

  14. Material derivative Dq Dt = ∂ q ∂ t + u ∂ q ∂ x + v ∂ q ∂ y + w ∂ q ∂ z dtq ( t, x ) = ∂ q d ∂ t + ∇ q · d x dt = ∂ q ∂ t + ∇ q · u = Dq Dt

  15. Advection • Advection describe how quantity, q , moves with the velocity field u • Density D ρ Dt = ∂ρ ∂ t + u · ∇ ρ • Velocity Dt = ∂ u D u ∂ t + u · ∇ u

  16. Advection momentum D u Dt + 1 ρ ∇ p = g + s ∇ · ∇ u equation Advection Projection Diffusion Body force ∂ t = − ( u · ∇ ) u − 1 ∂ u ρ ∇ p + s ∇ 2 u + f

  17. Density advection compute derivative of density field: ∂ρ ∂ t = − ( u · ∇ ) ρ integrate density field using ∂ρ ∂ t

  18. Velocity advection compute derivative of velocity field: ∂ u ∂ t = � ( u · r ) u integrate velocity field using ∂ u ∂ t compute derivative of density field: ∂ρ ∂ t = − ( u · ∇ ) ρ integrate density field

  19. Projection Advection Projection Diffusion Body force ∂ t = − ( u · ∇ ) u − 1 ∂ u ρ ∇ p + s ∇ 2 u + f ∇ · u = 0

  20. Divergence free ∇ · u > 0? ∇ · u < 0?

  21. Divergence free ∇ ∙ u = 0 ∇ ∙ u ≠ 0

  22. Projection ∂ t = − ( u · ∇ ) u − 1 ∂ u ρ ∇ p + s ∇ 2 u + f s.t. ∇ · u = 0

  23. Diffusion Advection Projection Diffusion Body force ∂ t = − ( u · ∇ ) u − 1 ∂ u ρ ∇ p + s ∇ 2 u + f ∇ · u = 0

  24. High viscosity fluids

  25. Dropping viscosity • Viscosity plays a minor role in most fluids • Numeric methods introduce errors which can be physically reinterpreted as viscosity • Fluid with no viscosity is called “inviscid”

  26. Body force Advection Projection Diffusion Body force ∂ t = − ( u · ∇ ) u − 1 ∂ u ρ ∇ p + s ∇ 2 u + f ∇ · u = 0

  27. External forces • Gravity • Heat • Surface tension • User-specified forces (stirring coffee)

  28. Boundary conditions • Solid walls • The fluid cannot go in and out of it • Control the normal velocity • Free surfaces • Air can be represented as a region where the pressure is zero • Do not control the velocity at the surface

  29. Solid boundary • Velocity at the boundary u · ˆ n = u solid · ˆ n • No-slip condition: u = u solid • Pressure at the boundary • Make the fluid incompressible AND enforce the solid boundary condition

  30. Physics recap • Physical quantity represented as fields • Navier-Stokes PDE describes the dynamics

  31. Practical challenges • What is the representation for fluids? • use grid structures to represent velocity, pressure, and other quantities of interest • What is the equation of motion for fluids? • approximate Navier-Stokes in a discrete domain • compute Navier-Stokes efficiently

  32. Simple grid structure ( i , j+1 ) p i,j ( i-1 , j ) ( i , j ) ( i+1 , j ) u i,j ( i , j-1 )

  33. Velocity field vector field: u scalar field: u u i,j u i,j scalar field: v v i,j u i,j = ( u i,j , v i,j )

  34. Computing divergence r · u i,j = u i,j +1 � u i,j − 1 + v i +1 ,j � v i − 1 ,j 2 ∆ x cell width: ∆ x v i +1 ,j horizontal vertical scalar field u i,j scalar field u i,j − 1 u i,j +1 v i,j v i − 1 ,j

  35. Derivative evaluation • Evaluating derivatives using scalar fields ∂ t = − ( u · ∇ ) u − 1 ∂ u ρ ∇ p + s ∇ 2 u + f ∂ t = � ( u · r ) u � 1 ∂ u ρ ( r p ) x + s r 2 u + f x ∂ t = � ( u · r ) v � 1 ∂ v ρ ( r p ) y + s r 2 v + f y

  36. MAC grid structure p i,j +1 v i,j +1 / 2 p i +1 ,j p i − 1 ,j p i,j u i − 1 / 2 ,j u i +1 / 2 ,j v i,j − 1 / 2 p i,j − 1

  37. Quiz How do you compute the divergence at u i,j using MAC grid? v i,j +1 / 2 u i − 1 / 2 ,j u i +1 / 2 ,j v i,j − 1 / 2

  38. Explicit integration • Solving for the differential equation explicitly, namely ∂ t = − ( u · ∇ ) u − 1 ∂ u ρ ∇ p + s ∇ 2 u + f u t +1 = u t + h ˙ u ( t ) • You get this...

  39. Explicit integration

  40. Stable fluids Invented by Jos Stam Simple, fast, and unconditionally stable

  41. Splitting methods • Suppose we had a system ∂ x ∂ t = f ( x ) = g ( x ) + h ( x ) • We define simulation function S f S f ( x, ∆ t ) : x ( t ) → x ( t ) + ∆ tf ( x ) • Then we could define S f ( x, ∆ t ) : x ( t + ∆ t ) = S g ( x, ∆ t ) ◦ S h ( x, ∆ t )

  42. Splitting methods ∂ t = − ( u · ∇ ) u − 1 ∂ u ρ ∇ p + s ∇ 2 u + f w 0 = u ( x , t ) add force Advect Diffuse Project w 0 ( x ) w 1 ( x ) w 2 ( x ) w 3 ( x ) w 4 ( x ) u ( x , t + Δ t ) = w 4

  43. Splitting methods ∂ t = − ( u · ∇ ) u − 1 ∂ u ρ ∇ p + s ∇ 2 u + f w 0 = u ( x , t ) add force Advect Diffuse Project w 0 ( x ) w 1 ( x ) w 2 ( x ) w 3 ( x ) w 4 ( x ) u ( x , t + Δ t ) = w 4

  44. Body forces

  45. Splitting methods ∂ t = − ( u · ∇ ) u − 1 ∂ u ρ ∇ p + s ∇ 2 u + f w 0 = u ( x , t ) add force Advect Diffuse Project w 0 ( x ) w 1 ( x ) w 2 ( x ) w 3 ( x ) w 4 ( x ) u ( x , t + Δ t ) = w 4

  46. Semi-Lagrangian Advection

  47. Numerical dissipation • Semi-Lagrangian advection tend to smooth out sharp features by averaging the velocity field • The numerical errors result in a different advection equation solved by semi-Largrangian ∂ x = u ∆ x ∂ 2 q ∂ q ∂ t + u ∂ q ∂ x 2 • Smooth out small vortices in inviscid fluids

  48. Splitting methods ∂ t = − ( u · ∇ ) u − 1 ∂ u ρ ∇ p + s ∇ 2 u + f w 0 = u ( x , t ) add force Advect Diffuse Project w 0 ( x ) w 1 ( x ) w 2 ( x ) w 3 ( x ) w 4 ( x ) u ( x , t + Δ t ) = w 4

  49. Diffusion • Solve for the effect of viscosity ∂ w 2 = ν ∇ 2 w 2 ∂ t • Use an implicit method for stable result ( I − ν ∆ t ∇ 2 ) w 3 ( x ) = w 2 ( x )

  50. Splitting methods ∂ t = − ( u · ∇ ) u − 1 ∂ u ρ ∇ p + s ∇ 2 u + f w 0 = u ( x , t ) add force Advect Diffuse Project w 0 ( x ) w 1 ( x ) w 2 ( x ) w 3 ( x ) w 4 ( x ) u ( x , t + Δ t ) = w 4

  51. Projection P( )

  52. Projection • Projection step subtracts off the pressure from the intermediate velocity field w 3 w 4 = w 3 − ∆ t 1 ρ ∇ p • project ( ∆ t, u ) must satisfies two conditions: ∇ · u t +1 = 0 • divergence free: • boundary velocity: u t +1 · ˆ n = u solid · ˆ n

  53. Divergence-free condition u t +1 − u t +1 + v t +1 − v t +1 p i,j +1 2 1 2 1 = 0 ∆ x ∆ x v 2 w 4 = w 3 − ∆ t 1 Replace (from ) u t +1 p i +1 ,j p i − 1 ,j p i,j ρ ∇ p 2 u 1 u 2 u 2 − ∆ t p i +1 ,j − p i,j with v 1 ∆ x ρ Interpolating u 2 = u i +1 ,j + u i,j p i,j − 1 2 − ∆ t = u i +1 ,j + u i,j p i +1 ,j − p i,j resulting in u t +1 2 ∆ x 2 ρ

Download Presentation
Download Policy: The content available on the website is offered to you 'AS IS' for your personal information and use only. It cannot be commercialized, licensed, or distributed on other websites without prior consent from the author. To download a presentation, simply click this link. If you encounter any difficulties during the download process, it's possible that the publisher has removed the file from their server.

Recommend


More recommend