rayleigh b enard convection and lorenz model rayleigh b
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Rayleigh-B enard Convection and Lorenz Model Rayleigh-B enard - PowerPoint PPT Presentation

Laurette TUCKERMAN laurette@pmmh.espci.fr Rayleigh-B enard Convection and Lorenz Model Rayleigh-B enard Convection Rayleigh-B enard Convection Boussinesq Approximation Calculation and subtraction of the basic state


  1. Laurette TUCKERMAN laurette@pmmh.espci.fr Rayleigh-B´ enard Convection and Lorenz Model

  2. Rayleigh-B´ enard Convection

  3. Rayleigh-B´ enard Convection Boussinesq Approximation Calculation and subtraction of the basic state Non-dimensionalisation Boundary Conditions Linear stability analysis Lorenz Model Inclusion of nonlinear interactions Seek bifurcations

  4. Boussinesq Approximation µ (viscosity ∼ diffusivity of momentum), κ (diffusivity of temperature), ρ (density) constant except in buoyancy force. Valid for T 0 − T 1 not too large. ρ ( T ) = ρ 0 [1 − α ( T − T 0 )] ∇ · U = 0 Governing equations: ρ 0 [ ∂ t + (U · ∇ )] U = µ ∆U − ∇ P − gρ ( T )e z [ ∂ t + (U · ∇ )] T = κ ∆ T ↑ ↑ ↑ advection diffusion buoyancy Boundary conditions: U = 0 z = 0 , d at T = T 0 , 1 z = 0 , d at

  5. Calculation and subtraction of base state ( U ∗ , T ∗ , P ∗ ) Conductive solution: U ∗ = 0 Motionless: T ∗ = T 0 − ( T 0 − T 1 ) z uniform temperature gradient: d 1 + α ( T 0 − T 1 ) z � � ρ ( T ∗ ) = ρ 0 density: d Hydrostatic pressure counterbalances buoyancy force: � P ∗ = − g dz ρ ( T ∗ ) z + α ( T 0 − T 1 ) z 2 � � = P 0 − gρ 0 2 d

  6. Write: T = T ∗ + ˆ P = P ∗ + ˆ T P Buoyancy: ρ ( T ∗ + ˆ T ) = ρ 0 (1 − α ( T ∗ + ˆ T − T 0 )) = ρ 0 (1 − α ( T ∗ − T 0 )) − ρ 0 α ˆ T = ρ ( T ∗ ) − ρ 0 α ˆ T −∇ P − gρ ( T )e z = −∇ P ∗ − gρ ( T ∗ ) − ∇ ˆ P + gρ 0 α ˆ T e z = −∇ ˆ P + gρ 0 α ˆ T e z Advection of temperature: ( U · ∇ ) T = ( U · ∇ ) T ∗ + ( U · ∇ ) ˆ T T 0 − ( T 0 − T 1 ) z � � + ( U · ∇ ) ˆ = ( U · ∇ ) T d = − T 0 − T 1 U · e z + ( U · ∇ ) ˆ T d

  7. Governing equations: ρ 0 [ ∂ t + ( U · ∇ )] U = −∇ ˆ P + gρ 0 α ˆ T e z + µ ∆ U ∇ · U = 0 T = T 0 − T 1 [ ∂ t + ( U · ∇ )] ˆ U · e z + κ ∆ ˆ T d Homogeneous boundary conditions: U = 0 at z = 0 , d ˆ T = 0 at z = 0 , d

  8. Non-dimensionalization Scales: t = d 2 U = κ µκ P = µκ ¯ ˆ ¯ ˆ d 2 ¯ ¯ z = d ¯ z, t, U, T = T , P κ d d 3 gρ 0 α Equations : κ 2 ρ 0 � ¯ = − µκ P + µκ T e z + µκ t + ( ¯ U · ¯ d 3 ¯ ∇ ¯ d 3 ¯ d 3 ¯ ∆ ¯ � ∂ ¯ ∇ ) U U d 3 κ d 2 ¯ ∇ · ¯ U = 0 µκ 2 � ¯ µκ 2 = κ T 0 − T 1 t + ( ¯ U · ¯ ¯ ∆ ¯ ¯ � ∂ ¯ ∇ ) T U · e z + T d 5 gρ 0 α d d d 5 gρ 0 α Dividing through, we obtain: � ¯ µ t + ( ¯ U · ¯ − ¯ ∇ ¯ P + ¯ T e z + ¯ ∆ ¯ � � � ∂ ¯ ∇ ) U = U ρ 0 κ � ¯ = ( T 0 − T 1 ) d 3 gρ 0 α t + ( ¯ U · ¯ U · e z + ¯ ¯ ∆ ¯ � ∂ ¯ ∇ ) T T κµ

  9. Non-dimensional parameters: µ the Prandtl number: P r ≡ ρ 0 κ momentum diffusivity / thermal diffusivity Ra ≡ ( T 0 − T 1 ) d 3 gρ 0 α the Rayleigh number: κµ non-dimensional measure of thermal gradient

  10. Boundary conditions Horizontal direction: periodicity 2 π/q Vertical direction: at z = 0 , 1 T = 0 | z =0 , 1 perfectly conducting plates w = 0 | z =0 , 1 impenetrable plates Rigid boundaries at z = 0 , 1 : u | z =0 , 1 = v | z =0 , 1 = 0 zero tangential velocity Incompressibility ∂ x u + ∂ y v + ∂ z w = 0 = ⇒ ∂ z w = − ( ∂ x u + ∂ y v ) u | z =0 , 1 = v | z =0 , 1 = 0 = ⇒ ∂ x u | z =0 , 1 = ∂ y v | z =0 , 1 = 0 = ⇒ ∂ z w | z =0 , 1 = 0

  11. Free surfaces at z = 0 , 1 to simplify calculations: [ ∂ z u + ∂ x w ] z =0 , 1 = [ ∂ z v + ∂ y w ] z =0 , 1 = 0 zero tangential stress w | z =0 , 1 = 0 = ⇒ ∂ x w | z =0 , 1 = ∂ y w | z =0 , 1 = 0 = ⇒ ∂ z u | z =0 , 1 = ∂ z v | z =0 , 1 = 0 = ⇒ ∂ x ∂ z u | z =0 , 1 = ∂ y ∂ z v | z =0 , 1 = 0 = ⇒ ∂ zz w | z =0 , 1 = − ∂ z ( ∂ x u + ∂ y v ) | z =0 , 1 = 0 Not realistic, but allows trigonometric functions sin( kπz )

  12. Two-dimensional case � u = − ∂ z ψ � U = ∇ × ψ e y = ⇒ = ⇒ ∇ · U = 0 w = ∂ x ψ No-penetration boundary condition: � ψ = ψ 1 at z = 1 0 = w = ∂ x ψ = ⇒ ψ = ψ 0 at z = 0 Horizontal flux: � 1 � 1 dz ∂ z ψ ( x, z ) = − ψ ( x, z )] 1 dz u ( x, z ) = − z =0 = ψ 0 − ψ 1 z =0 z =0 Arbitrary constant = ⇒ ψ 0 = 0 Zero flux = ⇒ ψ 1 = 0 Stress-free: 0 = ∂ z u = − ∂ 2 zz ψ Rigid: 0 = u = ∂ z ψ at z = 0 , 1

  13. Two-dimensional case Temperature equation: ∂ t T + U · ∇ T = RaU · e z + ∆ T U · ∇ T = u ∂ x T + w ∂ z T = − ∂ z ψ ∂ x T + ∂ x ψ ∂ z T ≡ J [ ψ, T ] ∂ t T + J [ ψ, T ] = Ra ∂ x ψ + ∆ T

  14. Velocity equation ∂ t U + ( U · ∇ ) U = P r [ −∇ P + T e z + ∆ U ] Take e y · ∇× : e y · ∇ × ∂ t U = e y · ∇ × ∇ × ∂ t ψ e y = − ∂ t ∆ ψ e y · ∇ × ∇ P = 0 e y · ∇ × T e z = − ∂ x T e y · ∇ × ∆ U = e y · ∇ × ∆ ∇ × ψ e y = − ∆ 2 ψ ∂ t ∆ ψ − e y · ∇ × ( U · ∇ ) U = P r [ ∂ x T + ∆ 2 ψ ] ∇ × ∇ × f = ∇∇ · f − ∆ f

  15. e y · ∇ × ( U · ∇ ) U = ∂ z ( U · ∇ ) u − ∂ x ( U · ∇ ) w = ∂ z ( u∂ x u + w∂ z u ) − ∂ x ( u∂ x w + w∂ z w ) = ∂ z u ∂ x u + ∂ z w ∂ z u − ∂ x u ∂ x w − ∂ x w ∂ z w + u ∂ xz u + w ∂ zz u − u ∂ xx w − w ∂ xz w = ∂ z u ( ∂ x u + ∂ z w ) − ∂ x w ( ∂ x u + ∂ z w ) + u ∂ x ( ∂ z u − ∂ x w ) + w∂ z ( ∂ z u − ∂ x w ) = ( − ∂ z ψ ) ∂ x ( − ∂ zz ψ − ∂ xx ψ ) + ( ∂ x ψ ) ∂ z ( − ∂ zz ψ − ∂ xx ψ ) = ( ∂ z ψ ) ∂ x (∆ ψ ) − ( ∂ x ψ ) ∂ z (∆ ψ ) = − J [ ψ, ∆ ψ ] ∂ t ∆ ψ + J [ ψ, ∆ ψ ] = P r [ ∂ x T + ∆ 2 ψ ]

  16. Linear stability analysis Linearized equations: ∂ t ∆ ψ = P r [ ∂ x T + ∆ 2 ψ ] ∂ t T = Ra ∂ x ψ + ∆ T Solutions: ˆ ψ sin qx sin kπz e λt q ∈ R , k ∈ Z + , λ ∈ C ψ ( x, z, t ) = T ( x, z, t ) = ˆ T cos qx sin kπz e λt ↑ ↑ γ 2 ≡ q 2 + ( kπ ) 2 functions scalars − λγ 2 ˆ T + γ 4 ˆ ψ = P r [ − q ˆ ψ ] ψ − γ 2 ˆ λ ˆ T = Ra q ˆ T

  17. � ˆ � − P r γ 2 P r q/γ 2 � � ˆ � � ψ ψ λ = ˆ ˆ − γ 2 Ra q T T Steady Bifurcation: λ = 0 P r γ 4 − P r Ra q 2 γ 2 = 0 q 2 = ( q 2 + ( kπ ) 2 ) 3 Ra = γ 6 ≡ Ra c ( q, k ) q 2

  18. Convection Threshold Conductive state unstable at ( q, k ) for Ra > Ra c ( q, k )

  19. Conductive state stable if min Ra < q ∈ R Ra c ( q, k ) k ∈ Z + = q 2 3( q 2 + ( kπ ) 2 ) 2 2 q − 2 q ( q 2 + ( kπ ) 2 ) 3 ∂Ra c ( q, k ) 0 = q 4 ∂q 2( q 2 + ( kπ ) 2 ) 2 (3 q 2 − ( q 2 + ( kπ ) 2 ) = q 3 ⇒ q 2 = ( kπ ) 2 = 2 = ( kπ ) 2 / 2 + ( kπ ) 2 ) 3 � q = kπ � = 27 4 ( kπ ) 4 Ra c , k √ ( kπ ) 2 / 2 2 � q = π � = 27 4 ( π ) 4 = 657 . 5 Ra c ≡ Ra c √ , k = 1 2

  20. Rigid Boundaries Calculation follows the same principle, but more complicated. Boundaries damp perturbations = ⇒ higher threshold q c ↓ = ⇒ ℓ c = π/q c ↑ = ⇒ rolls ≈ circular Ra c q c ℓ c 4 π 4 = 657 . 5 27 π stress-free boundaries 1 . 4 √ 2 ≈ 1700 ≈ π ≈ 1 rigid boundaries

  21. Lorenz Model: including nonlinear interactions J [ ψ, ∆ ψ ] = J [ ψ, − γ 2 ψ ] = ∂ x ψ ∂ z ( − γ 2 ψ ) − ∂ x ( − γ 2 ψ ) ∂ z ψ = 0 ψ ˆ ˆ J [ ψ, T ] = T [ ∂ x (sin qx sin πz ) ∂ z (cos qx sin πz ) − ∂ x (cos qx sin πz ) ∂ z (sin qx sin πz )] ψ ˆ ˆ = T qπ [cos qx sin πz cos qx cos πz + sin qx sin πz sin qx cos πz ] T qπ (cos 2 qx + sin 2 qx ) sin πz cos πz + ˆ ψ ˆ T qπ ψ ˆ ˆ = sin 2 πz 2 ↑ ↑ ↑ ↑ functions scalars

  22. ˆ ψ ( x, z, t ) = ψ ( t ) sin qx sin πz T ( x, z, t ) = ˆ T 1 ( t ) cos qx sin πz + ˆ T 2 ( t ) sin 2 πz ˆ ψ ˆ J [ ψ, T 2 ] = T 2 [ ∂ x (sin qx sin πz ) ∂ z (sin 2 πz ) − ∂ x (sin 2 πz ) ∂ z (sin qx sin πz )] ψ ˆ ˆ = T 2 q 2 π cos qx sin πz cos 2 πz ψ ˆ ˆ = T 2 q π cos qx (sin πz + sin 3 πz ) Including ˆ T 3 ( t ) cos qx sin 3 πz = ⇒ new terms = ⇒ Closure problem for nonlinear equations Lorenz (1963) proposed stopping at T 2 .

  23. Lorenz Model T 1 /γ 2 − γ 2 ˆ ∂ t ˆ ψ = P r ( q ˆ ψ ) sin qx sin πz ψ − γ 2 ˆ ∂ t ˆ T 1 + qπ ˆ ψ ˆ Ra q ˆ T 2 = T 1 cos qx sin πz T 2 + qπ T 1 = − (2 π ) 2 ˆ ∂ t ˆ ψ ˆ ˆ T 2 sin 2 πz 2 Define: πq 2 πq 2 πq 2 γ 2 ˆ 2 γ 6 ˆ 2 γ 6 ˆ X ≡ √ ψ, Y ≡ √ T 1 , Z ≡ √ T 2 , r ≡ q 2 b ≡ 4 π 2 γ 2 = 8 τ ≡ γ 2 t, γ 6 Ra, 3 , σ ≡ P r Famous Lorenz Model: ˙ X = σ ( Y − X ) ˙ Y = − XZ + rX − Y ˙ Z = XY − bZ

  24. σ = P r (often set to 10, its value for water) r = Ra/Ra c Damping = ⇒ − σX , − Y , − bZ Advection = ⇒ XZ , XY Symmetry between ( X, Y, Z ) and ( − X, − Y, Z )

  25. Lorenz Model Pitchfork Bifurcation Steady states: 0 = σ ( Y − X ) = ⇒ X = Y 0 = − XZ + rX − Y = ⇒ X = 0 or Z = r − 1 � 0 = XY − bZ = ⇒ Z = 0 or X = Y = ± b ( r − 1)     � �   0 b ( r − 1) − b ( r − 1)  , � �  , 0 b ( r − 1) − b ( r − 1)         0 r − 1 r − 1

  26. Jacobian:   − σ σ 0 Df = r − Z − 1 − X   Y X − b For ( X, Y, Z ) = (0 , 0 , 0) :   − σ σ 0 Df (0 , 0 , 0) = r − 1 0   0 0 − b Eigenvalues: λ 1 + λ 2 = T r = − σ − 1 < 0 λ 1 λ 2 = Det = σ (1 − r ) λ 3 = − b < 0

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