Overview of Mixing and Transport. William Young Scripps Institution of Oceanography June 2004 1
Prehistory of Stirring, Mixing and Transport ♦ Einstein, 1905 — diffusion by discontinuous jumps. ♦ Taylor 1921 — diffusion by continuous movements. ♦ The renovating wave model — Zeldovich 19?? ♦ Eckart, 1948 — stirring versus mixing. ♦ Batchelor, 1952 — exponential stretching of line elements. ♦ Welander, 1955 — visualization of advective distortion. ♦ Molecular diffusivity and eddy diffusivity, correlation functions, enhanced transport, single realizations versus ensemble averages. Turbulent flame propagation, mixing of reactants, the geometry of turbulence. 2
PART I: Eddy Diffusion 3
Diffusion by discontinuous movements ♦ In 1905 Einstein explained Brown’s 1828 observation that sus- pended pollen grains (and also inorganic particles) are in “unin- terrupted and irregular swarming motion”. ♦ Einstein’s assumptions: (i) Particles move independently of one another. (ii) We observe particle positions at time intervals, τ , which are much greater than the interval between molecular collisions. So the motion in one interval is independent of the previous interval. 4
The random walk: D ≡ � ∆ 2 � and � x 2 � = 2 Dt 2 τ A random walk with 200 steps N ( t ) � x ( t ) = ∆ k k =1 � x 2 � = N ( t ) � ∆ 2 � = � ∆ 2 � × 2 t 2 τ 5
Einstein’s derivation of the diffusion equation ♦ In each interval τ each particle has a random displacement ∆ with a PDF, φ (∆; τ ): � ∞ � ∞ � ∆ 2 � ≡ ∆ 2 φ (∆) d∆ < ∞ . φ (∆) d∆ = 1 , −∞ −∞ ♦ If c ( x, t ) is particles per unit length at time t then � ∞ c ( x, t + τ ) = c ( x − ∆ , t ) φ (∆)d∆ . −∞ ♦ Taking τ, ∆ → 0, and Taylor expanding: D ≡ � ∆ 2 � U ≡ � ∆ � c t + Uc x ≈ Dc xx , , 2 τ τ where U and D are independent of τ . (OK provided τ is not too small, or too large.) 6
Einstein’s formula D = k B T 6 πµa 7
Eddy diffusion in a (turbulent) fluid ♦ In a fluid we observe continuous motion of particles. ♦ Example: ocean SOFAR floats. ♦ What is the analog of Einstein’s τ in a moving fluid? ♦ How to quantify the “diffusing power” of a flow? ♦ It is certainly not just KE ∝ �| u | 2 � ! ♦ Certainly changing direction is important e.g., perhaps �| u t | 2 � ? ♦ But the nondiffusing example u = U cos ωt confounds us.... 8
Diffusion by continuous movements — Taylor (1921) A time series of Lagrangian velocity 2 u(t), 0 -2 0 0.5 1 1.5 2 2.5 3 3.5 4 t ♦ For a stationary time-series, u ( t ), � u 2 � , � u 2 � u 2 et cetera t � , tt � , are all constants. ♦ This implies � uu t � = 0 and � uu tt � = −� u 2 t � and so on. 9
The correlation function ♦ The Lagrangian velocity correlation function C ( t ) = � u ( t 0 + t ) u ( t 0 ) � is key in understanding “diffusing power”. ♦ Note C ( t ) = � u 2 � − t 2 t � + t 4 2 � u 2 4! � u 2 tt � + · · · 10
The diffusing power of a velocity field ♦ Taylor’s solution: � t d x d t = u ( t ) ⇒ x ( t ) = 0 u ( t 1 ) d t 1 . ♦ Multiply by u ( t ) and ensemble average, � t � t d x 2 d � x 2 � d t = 2 0 u ( t 1 ) u ( t ) d t, ⇒ = 2 0 � u ( t 1 ) u ( t ) � d t 1 . d t � �� � ≡C ( t − t 1 ) ♦ Our nondiffusing example is u ( t ) = U cos( ωt + φ ), and averag- ing over φ : 2 U 2 cos( ωt ) . C ( t ) = 1 11
d � x 2 � � t Taylor’s formula: = 2 0 C ( t 1 ) d t 1 . d t 1.2 1 ∞ C(t) dt D= ∫ 0 0.8 C(t)/U 2 0.6 0.4 0.2 0 -0.2 0 0.5 1 1.5 2 2.5 3 t � ∞ ♦ If 0 C ( t )d t converges then the eddy diffusivity, D , is a well defined. ♦ But D might be zero (the sea surface) or infinite, (molecular diffusion in d = 2, where C ( t ) ∼ t − 1 ). 12
Reconcile Einstein and Taylor? ♦ In a fluid we can still employ D = � ∆ 2 � / 2 τ E provided τ E ≥ τ T . ♦ The Taylor decorrelation time, τ T , is defined by � ∞ 0 C ( t ) d t = U 2 D = rms τ T . ♦ This implies � ∆ 2 � = 2 U 2 rms τ E τ T . 13
Spatial correlations ♦ Fluid motion is also correlated spatially... ♦ D only provides single-particle information. ♦ To understand spatial correlations we must examine pairs of particles. ♦ To illustrate this, we construct a model that looks a little more like real fluid motion. ♦ We want a smoothly varying velocity field with a well defined D . ♦ We also want to be able to solve the model analytically, also and make efficient simulations. 14
Eddy diffusion and stretching in a moving fluid ♦ Formulate the renovating wave model in d = 2. ♦ Advection with complete loss of memory at intervals of τ : ψ n = Uk − 1 cos[ k cos θ n x + k sin θ n y + ϕ n ] , I n = { ( n − 1) τ < t < nτ } : where θ n and ϕ n are random phases. ( u, v ) = ( − ψ y , ψ x ) Ukτ is a nondimensional parameter at left θ n = π/ 4 15
Now solve ( ˙ x, ˙ y ) = ( − ψ y , ψ x ) explicitly in each I n ♦ The renovating wave model leads to a random map: � � � � x n +1 x n + Uτs n sin( kc n x n + ks n y n + ϕ n ) = , y n − Uτc n sin( kc n x n + ks n y n + ϕ n ) y n +1 where ( s n , c n ) ≡ (sin θ n , cos θ n ). t=20 τ D = τU 2 / 8 independent of k . t=0 ♦ Homework: C ( t ) =??? And evaluate ∂ ( x n +1 ,y n +1 ) · · · ∂ ( x n ,y n ) 16
The advection-diffusion equation: c t + u ·∇ c = κ ∇ 2 c ♦ c ( x , t ) is the “concentration” and the velocity, u ( x , t ), is in- compressible, ∇· u = 0. ♦ Using the renovating wave model for u ( x, t ), we can exhibit single realizations with κ = 0. This is the “method of character- istics”. But really we just iterate the random map... ♦ Compute ensemble averages by integrating over { ϕ n , θ n } . 17
Deformation of a medium sized blob kr ∼ 1 ♦ In a single realization the RW model produces spatially corre- lated deformation (not like molecular diffusion). t=1 τ t=2 τ t=3 τ The renovating wave model with Ukτ = 2 and kr = π . ♦ The IC is a circular blob and kr is a measure of scale separation . ♦ c ( x , t ) = 1 inside the blob and c ( x , t ) = − 1 outside. 18
Stretching of a small blob, kr = π/ 40 ≪ 1 & Ukτ = 1 t=1 τ t=2 τ t=3 τ t=4 τ t=5 τ t=6 τ t=7 τ t=8 τ t=9 τ t=10 τ t=11 τ t=12 τ ♦ Area ℓ 1 × ℓ 2 is conserved. But ℓ 1 ∝ e γt and ℓ 2 ∝ e − γt .... 19
“Eddy Diffusion” of a big blob, kr = 20 π & Ukτ = 1 t=1 τ t=2 τ t=3 τ t=4 τ t=5 τ t=6 τ t=7 τ t=8 τ t=9 τ t=10 τ t=11 τ t=12 τ ♦ The eddy-diffusion limit: small eddies advecting a large-scale tracer distribution. 20
The eddy-diffusion equation — slavishly follow E ♦ In a “decorrelation time”, τ , particles in different realizations have independent random displacements, r . ♦ Assume isotropy, so the displacement, r ≡ | r | , has a pdf g ( r ): � � g ( r ) d 2 r , � r 2 � = r 2 g ( r ) d 2 r , 1 = ( d = 2) . g ( r ) is the same in each time interval of length τ . ♦ If C ( x , t ) is the ensemble-average concentration at time ( x , t ) and � C ( x − r , t ) g ( r )d 2 r . C ( x , t + τ ) = Notice C ( x , t ) = � c ( x , t ) � — tired of all these �� ’s. 21
� C ( x − r , t ) g ( r )d 2 r C ( x , t + τ ) = ♦ Taking τ and r to zero, and freely Taylor expanding: C t = D ∇ 2 C, D = � r 2 � / 4 τ . ♦ Restrictions: � r 2 g ( r ) d 2 r must converge. • The integral � r 2 � = • The decorrelation time, τ , must be finite. • Small r and τ requires scale separation. • Ensemble averages and single-particle descriptors? 22
Details for the RW model? ♦ In the RW model, g ( r ) is the ensemble averaged Green’s func: g ( r ) = � G ( x , τ ) � , G t + u · ∇ G = 0 , G ( x , 0) = δ ( x ) . ♦ I found g ( r ) = 1 H ( τ ∗ − r ) τ 2 − r 2 . � π 2 r (Using nondimensional variables with length scaled by k − 1 and time with ( Uk ) − 1 .) 23
Eddy diffusion of a front c ( x , t ) = ± 1 t=3 τ t=6 τ t=9 τ t=12 τ t=15 τ t=18 τ t=21 τ t=24 τ 24
The ensemble average satisfies C t = D ∇ 2 C . √ ♦ Does the “erf” solution with η = x/ 2 Dt , describe the disper- sion of the front? Not really — in each realization c ( x , t ) = ± 1. ♦ But the ensemble average, C ( x , t ), might be a useful approxi- mation to spatial averages of a single realization e.g., � K ( x − x ′ ) c ( x ′ , t ) d 2 x ′ , ˆ c ( x , t ) ≡ K ( | x | ) is a filter. ♦ In the front problem we can use � L 1 ¯ c ( x, t ) ≡ lim c ( x, y, t )d y , 2 L L →∞ − L and avoid the ˆ ˆ c � = ˆ c issue. ♦ For estimating the rate of chemical reactions “coarse-graining” is definitely not good idea... 25
Asymptotic success of eddy diffusivity, D = U 2 τ/ 8 t=1 t=2 t=3 t=4 150 150 400 100 100 100 200 50 50 50 0 0 0 0 -0.5 0 0.5 -2 0 2 -2 0 2 -2 0 2 t=5 t=10 t=20 t=25 200 150 100 100 150 100 100 50 50 50 50 0 0 0 0 -4 -2 0 2 -5 0 5 -5 0 5 -5 0 5 t=50 t=100 t=400 t=1600 150 150 150 150 100 100 100 100 50 50 50 50 0 0 0 0 -10 0 10 -10 0 10 -20 0 20 40 -50 0 50 The x -coordinates of particles initially at x = 0 in a single realization ( Ukτ = 1). 26
Recommend
More recommend