Overview of Mixing and Transport. William Young Scripps Institution - - PowerPoint PPT Presentation

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

• f 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

2τ

and x2 = 2Dt

A random walk with 200 steps

x(t) =

N(t)

• k=1

∆k x2 = N(t)∆2 = ∆2 2τ × 2t

5

Einstein’s derivation of the diffusion equation

♦ In each interval τ each particle has a random displacement ∆ with a PDF, φ(∆; τ):

∞

−∞

φ(∆) d∆ = 1, ∆2 ≡

∞

−∞

∆2φ(∆) d∆ < ∞. ♦ 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: ct + Ucx≈Dcxx, U ≡ ∆ τ D ≡ ∆2 2τ , where U and D are independent of τ. (OK provided τ is not too small, or too large.)

6

Einstein’s formula

D = kBT 6πµa

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 |ut|2? ♦ But the nondiffusing example u = U cos ωt confounds us....

8

Diffusion by continuous movements — Taylor (1921)

0.5 1 1.5 2 2.5 3 3.5 4

• 2

2

t u(t), A time series of Lagrangian velocity

♦ For a stationary time-series, u(t), u2 , u2

t ,

u2

tt ,

et cetera are all constants. ♦ This implies uut = 0 and uutt = −u2

t and so on.

9

The correlation function

♦ The Lagrangian velocity correlation function C(t) = u(t0 + t)u(t0) is key in understanding “diffusing power”. ♦ Note C(t) = u2 − t2 2 u2

t + t4

4!u2

tt + · · ·

10

The diffusing power of a velocity field

♦ Taylor’s solution: dx dt = u(t) ⇒ x(t) =

t

0u(t1) dt1.

♦ Multiply by u(t) and ensemble average, dx2 dt = 2

t

0u(t1)u(t) dt,

⇒ dx2 dt = 2

t

0u(t1)u(t)

• ≡C(t−t1)

dt1 . ♦ Our nondiffusing example is u(t) = U cos(ωt + φ), and averag- ing over φ: C(t) = 1

2U2 cos(ωt) .

11

Taylor’s formula: dx2 dt = 2

t

0 C(t1) dt1 .

0.5 1 1.5 2 2.5 3

• 0.2

0.2 0.4 0.6 0.8 1 1.2

t C(t)/U2

D=∫0

∞ C(t) dt

♦ If

∞

0 C(t)dt 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 = ∆2/2τE provided τE ≥ τT . ♦ The Taylor decorrelation time, τT, is defined by D =

∞

0 C(t) dt = U2 rmsτT.

♦ This implies ∆2 = 2U2

rmsτEτT.

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: In = {(n−1)τ < t < nτ} : ψn = Uk−1 cos[k cos θn x+k sin θn y+ϕn] , where θn and ϕn are random phases. (u, v) = (−ψy, ψx) Ukτ is a nondimensional parameter at left θn = π/4

15

Now solve ( ˙ x, ˙ y) = (−ψy, ψx) explicitly in each In

♦ The renovating wave model leads to a random map:

• xn+1

yn+1

• =
• xn + Uτsn sin(kcnxn + ksnyn + ϕn)

yn − Uτcn sin(kcnxn + ksnyn + ϕn)

• ,

where (sn, cn) ≡ (sin θn, cos θn).

t=0 t=20τ

D = τU2/8 independent of k. ♦ Homework: C(t) =??? And evaluate ∂(xn+1,yn+1)

∂(xn,yn)

· · ·

16

The advection-diffusion equation: ct + u·∇c = κ∇2c

♦ 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 {ϕn, θn}.

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=1τ t=2τ t=3τ 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=1τ t=2τ t=3τ t=4τ t=5τ t=6τ t=7τ t=8τ t=9τ t=10τ t=11τ t=12τ

♦ Area ℓ1 × ℓ2 is conserved. But ℓ1 ∝ eγt and ℓ2 ∝ e−γt....

19

“Eddy Diffusion” of a big blob, kr = 20π & Ukτ = 1

t=1τ t=2τ t=3τ t=4τ t=5τ t=6τ t=7τ t=8τ t=9τ t=10τ t=11τ t=12τ

♦ 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): 1 =

• g(r) d2r,

r2 =

• r2g(r) d2r,

(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, t + τ) =

• C(x − r, t)g(r)d2r .

Notice C(x, t) = c(x, t) — tired of all these ’s.

21

C(x, t + τ) =

C(x − r, t)g(r)d2r

♦ Taking τ and r to zero, and freely Taylor expanding: Ct = D∇2C, D = r2/4τ . ♦ Restrictions:

• The integral r2 =

r2g(r) d2r must converge.

• 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), Gt + u·∇G = 0, G(x, 0) = δ(x). ♦ I found g(r) = 1 π2 H(τ∗ − r) r

• τ2 − r2 .

(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=3τ t=6τ t=9τ t=12τ t=15τ t=18τ t=21τ t=24τ 24

The ensemble average satisfies Ct = D∇2C.

♦ 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., ˆ c(x, t) ≡

• K(x − x′)c(x′, t) d2x′,

K(|x|) is a filter. ♦ In the front problem we can use ¯ c(x, t) ≡ lim

L→∞

1 2L

L

−L

c(x, y, t)dy , 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 = U2τ/8

• 0.5

0.5 200 400

t=1

• 2

2 50 100 150

t=2

• 2

2 50 100

t=3

• 2

2 50 100 150

t=4

• 4
• 2

2 50 100

t=5

• 5

5 50 100 150 200

t=10

• 5

5 50 100

t=20

• 5

5 50 100 150

t=25

• 10

10 50 100 150

t=50

• 10

10 50 100 150

t=100

• 20

20 40 50 100 150

t=400

• 50

50 50 100 150

t=1600

The x-coordinates of particles initially at x = 0 in a single realization (Ukτ = 1).

26

Diffusing Front Homework

♦ For the front problem show that C(x, t) = erf(η) , η = x 2 √ Dt . ♦ If c′ ≡ c − C, then the “variance” is υ ≡ c′2. Show that the variance satisfies: υt = D∇2υ + 2DC2

x source

. ♦ Show that the solution is υ(x, t) = 1 − erf2(η) , the easy way (pure thought). Check this the hard way by solving the υ equation.

27

PART II: Stirring and Mixing

28

Cream and coffee — Eckart 1948

Initial: there are distinct interfaces separating globules of cream and coffee. Within each globule, the concentration of cream is nearly constant and the concentration gradient is close to zero. There is a very large con- centration gradient at the interfaces between globs of coffee and cream. But the interfaces are small in number and not of great area, so the average gradient in the coffee mug is small. Stirring: the cream is mechanically swirled and folded, and molecular diffu- sion is unimportant. During this second stage the area of the interface and concentration gradients increase. Mixing: the gradients suddenly disappear and the fluid becomes homoge- neous; molecular diffusion is responsible for the sudden mixing. ♦ Note the importance of the mixing stage to chemical reactions.

29

A Welander 1955 scrapbook

30

Welander: distortion of passive scalar

31

More distortion and expansion

32

Expansion of a blob

♦ If the blob is small at t = 0 then the center goes on a random walk while the principal axes of the ellipse behave exponentially: ℓ1(t) ∝ eγt and ℓ2(t) ∼ e−γt . This is B52’s argument that in the Lagrangian frame the flow provides a time-scale — the strain γ — but no length scale. ♦ There is an important “microlength”: ℓκ ≡

• κ/γ.

♦ At very large times, there is a filamentary mess and the growing separation of distant filaments is relative diffusion, with diffusivity L ≈ 2 × 2D × t. ♦ Between these two extreme limits, things are interesting e.g., Richardson L ≈ √ εt3.

33

Stirring versus mixing in a steady laminar flow

t=1 t=2 t=4 t=8 t=16 t=32

Solution of ct + (1 − r2)cθ = (8 × 10−4)∇2c. The transit time round a streamline at r is T = 2π/(1 − r2).

♦ Differential advection increases |∇c| (stirring). Then κ wipes

• ut both |∇c| and c2 (mixing).

34

Summary

♦ Differential advection produces spectacular distortion of pas- sive scalars. ♦ Advected tracer fields rapidly develop much smaller length scales than those in the IC and in the velocity field. ♦ The distinction between reversible stirring and and irreversible mixing is important e.g., chemical reaction requires molecular “contact”. ♦ Eddy diffusivity misses all this fun because it only provides information about single particle statistics.

35

A prototypical straining flow

♦ A prototypical example: 2D straining flow, ψ = −αxy: ct + αxcx − αycy = κ∇2c . Note the important length ℓ =

• κ/α.

36

More strain: ct + αxcx − αycy = κ∇2c

♦ There is a steady solution, −αycy = κcyy, or : cy = A exp

• − y2

2ℓ2

• ,

c(x, ±∞, t) = ±

π

2Aℓ.

• 2
• 1.5
• 1
• 0.5

0.5 1 1.5 2

• 1
• 0.5

0.5 1

y The erf solution

♦ ℓ≡

• κ/α is the scale on which advection can balance diffusion.

37

Unsteady solution of ct + αxcx − αycy = κ∇2c

♦ Define Lagrangian coordinates, (a, b, τ) by considering ( ˙ x, ˙ y) = α(x, −y), ⇒ (x, y) =

• eαta, e−αtb
• ,

and t = τ. ♦ (a, b) = (xe−αt, ye+αt) = initial position of the particle which is at (x, y) at time t. ♦ The punchline is ∂t + αx∂x − αy∂y = ∂τ, and cτ = κe−2ατcaa + κe2ατcbb . ♦ There is an industry devoted to this method e.g., J-LT’s lects.

38

Straining and a plane wave eikx+ily

♦ The solution is c(x, y, t) = A(τ) exp [ika + ilb] , = A(t) exp

• ike−αtx + ile+αty
• ,

(1) with dA dt = −κ

• k2e−2αt + l2e+2αt

A . ♦ The decay is ultimately superexponential, A ∼ exp

• −(κl2/α)eαt

.

39

Illustration of superexponential decay

0.5 1 1.5 2 2.5 3 3.5 2 4 6 8 10 12

α t 〈 ∇ c ⋅ ∇ c 〉 using kl=1, 1/2, 1/4, 1/8

♦ An illustrative solution is c(x, y, t) = exp

• −ℓ2k2 sinh 2αt
• cos[ke−αtx] cos[ke+αty]

and ∇c·∇c peaks at αt∗ ∼ − ln(kℓ), where ℓ ≡

• κ/α.

40

Straining homework: ct + αxcx − αycy = κ∇2c

♦ With c(x, y, 0) = δ(x)δ(y), obtain the Gaussian solution c(x, y, t) = 1 2πpq exp

• − x2

2p2 − y2 2q2

• ,

where p(t) and q(t) are p2 ≡ κ α(e2αt − 1) , q2 ≡ κ α

• 1 − e−2αt

. ♦ Note that cmax(t) = c(0, 0, t) ∝ e−αt is independent of κ! ♦ Explain why the decay isn’t superexponential? ♦ Do this problem before J-LT spills the beans.

41

THE END

42

The industry

♦ The Lagrangian coordinate trick works for: ct + σx·∇c = κ∇2c , where

σx =

  

ux uy uz vx vy vz wx wy wz

     

x y z

  

♦ The small scale (ℓc ≪ ℓu) structure of c can be analyzed in a lot of detail...

43

Significance of κ|∇c|2

♦ Start with ct + u·∇c = κ∇2c + s and ∇·u = 0. ♦ Then multiply by c and integrate d dt

• c2 dV =
• sc dV
• production

− κ

• |∇c|2dV
• dissipation

♦ Some standard manipulations

• c × κ∇2c dV =
• c × κcn dA
• =0

−κ

• |∇c|2dV

and

• c × u·∇c dV =
• ∇·(uc2/2) dV =
• (c2/2)u·ˆ

ndA = 0

44

d dt

c2 dV = sc dV − κ |∇c|2dV

♦ κ|∇c|2 is the rate of dissipation of c2-stuff.

45