Rayleigh-B enard convection: a priori estimate on the Boussinesq - - PowerPoint PPT Presentation

Rayleigh-B´ enard convection: a priori estimate on the Boussinesq system that carry physical meaning

• C. Doering & M. Westdickenberg,

& C. Seis, & C. Nobili, & A. Choffrut

Max Planck Institute for Mathematics in the Sciences, Leipzig

Rayleigh-B´ enard convection in the turbulent regime

Rayleigh-B´ enard convection

Temperature advection and diffusion

z = 0 z = 1 T = 0

∂tT + u · ∇T − △T = 0

T = 1

Buoyancy, acceleration, and viscosity

u = 0 1 Pr(∂tu + u · ∇u) + ∇p

= −△u + RaT

1

• ∇ · u = 0

u = 0

Rayleigh number Ra = α g δT h3

ν κ ,

Prandtl number Pr = ν

κ ... in Boussinesq approximation

Beyond the Rayleigh-B´ enard instability Ra < Ra∗: pure diffusion, T = 1 − z, u = 0 Ra > Ra∗: unstable to convection rolls Ra ≫ 1: steady → periodic → “turbulent”

Schlieren picture for Ra ∼ 109

• L. Kadanoff

Efficiency of heat transport: the Nusselt number Diffusion and convection vertical heat flux

heat flux q = Tu − ∇T, vertical heat flux = q ·

1

• Nusselt number = space-time average of q ·

1

• Nu = lim

t↑∞ 1 t 1 Ld−1

t

• (0, L)d−1 × (0, 1) q ·

1

• dx dt′
• L

z = 0 z = 1 y z

Scaling of Nu in Ra and Pr ... Similarity law: Nu = f(Ra, Pr, L)

Grossmann & Lohse ’00 Ahlers et al ’05

... in theory and experiment

First part: Upper bounds on Nu at Pr = ∞ Limitations of the background field method Second part: Upper bounds on Nu at large Pr Re only matters in boundary layer

Malkus’ marginal stability argument...

z = 0 z = 1 T = 0, u = 0

∂tT + u · ∇T − △T = 0 −△u + ∇p = RaT

1

• ∇ · u = 0

T = 1, u = 0

x = Ra−1

3ˆ

x, t = Ra−2

3ˆ

t u = Ra

1 3ˆ

u, Nu = Ra

1 3

Nu

ˆ z = 0 ˆ z = H T = 0, ˆ u = 0

∂ˆ

tT + ˆ

u · ˆ ∇T − ˆ △T = 0 − ˆ △ˆ u + ˆ ∇ˆ p = T

1

• ˆ

∇ · ˆ u = 0

T = 1, ˆ u = 0

.. as a rescaling

Nusselt number Nu independent of height H ≫ 1 Nu ∼ Ra

1 3 for Ra ≫ 1

⇐ ⇒

• Nu ∼ 1 for H ≫ 1

Simulation

(M. Zimmermann):

100 200 300 400 500 600 700 0.05 0.1 0.15 Zeit Nu(t)

NUSSELT−ZAHL

BLAU: ROT: SCHWARZ: H = 250, H = 500, H = 1000, Nu = 0.0648, Varianz = 3.61e−04 Nu = 0.0584, Varianz = 2.15e−04 Nu = 0.0545, Varianz = 1.19e−04

Upper bounds on Nusselt via a priori estimates

• Constantin & Doering (’99):

Nu ln2/3 H, Stokes maximal regularity in L∞

• Doering & O. & Reznikoff (’06):

Nu ln1/3 H, background temperature field method

• O. & Seis (’11):

Nu ln1/15 H, background temperature field method

• Nobili & O. (’16) exponent 1/15 optimal for

background temperature field method

• O. & Seis (’11):

Nu ln2/3 lnH, Stokes maximal regularity in L∞ and background temperature field method

Two insights from rigorous results

• Optimal background temperature profile is non-monotone

stable for H0 log− 1

15 H

yields Nu log

1 15 H

• Optimal background temperature profile

has no physical meaning

Notations Time & horizontal average: · := lim

t↑∞

1 t

t

0 dt′ 1 Ld−1

• (0, L)d−1 dy
• Temperat. fluctuations: θ := T − T

Vertical velocity: w := u ·

1

• L

z = 0 z = H y z

θ ↔ w: Stokes & no-slip b. c. plate & clamped b. c. △2w = − △y θ,

w = ∂zw = 0 for z = 0, H.

Two representations of Nusselt number Recall definition Nu =

1 H

H

0 Tw −∂zTdz z = 0 z = H w = ∂zw = 0

△2w = −△yθ

w = ∂zw = 0 z y

Get in addition Nu = θw − d

dzT

for all z, Nu =

H

0 |∇T|2 dz z′ z

More flexible representation of Nusselt number τ(z) with τ(z = 0) = 1 , τ(z = H) = 0

H 1

τ z Decompose T = τ + θ ; keep relation between θ & w Average Nu = −∂zT + w θ w. r. t. −dτ

dz:

Nu =

H

dτ dz ∂zT dz −

H

dτ dz w θ dz

Combine with Nu =

H

0 |∇T|2 dz:

Nu =

H

0 (dτ dz)2 dz −

H

0 |∇θ|2 dz + 2

H

dτ dz w θ dz

Optimal bound through saddle point problem Nu ≤ Nu := min

τ

max

(θ,w)

H

0 (dτ dz)2 dz −

H

0 |∇θ|2 dz + 2

H

dτ dz w θ dz

• min over all τ with τ = 1, 0 for z = 0, H

H 1

τ z max over all (θ, w) with

w = ∂zw = 0

△2w = −△yθ

w = ∂zw = 0 Hopf’43, Nicolaenko&Scheurer&Temam’85, Constantin&Doering’92

Plus I: A marginal stability criterion

• Nu = min

τ

H

0 (dτ dz)2 dz

min over all τ with τ = 1, 0 for z = 0, H with

H

0 |∇θ|2 dz + 2

H

dτ dz w θ dz ≥ 0

for all (θ, w) with △2w = −△y θ

and θ = w = ∂zw = 0 at z = 0, H

... captures transition H∗ to convection rolls

Plus II: Amenable to horizontal Fourier transform ...

• Nu = min

τ

H

0 (dτ dz)2 dz

min over all τ with τ = 1, 0 for z = 0, H with

H

• (k2− d2

dz2)2w

2+ 1

k d dz(k2− d2 dz2)2w

2 dz

+ 2

H

dτ dz w (k2− d2 dz2)2w dz ≥ 0

for all (k, w(z)) with and (k2− d2

dz2)2w = w = dw dz = 0 at z = 0, H

Busse, Howard, Chan ’71, Ierley&Kerswell&Plasting ’05

... monitor bifurcations in optimal k as H ↑ ∞

Stability of logarithmic temperature profile Linear profile: τ = az+b,

dτ dz = a

H

0 aw θ dz ≥ 2

H

0 a(∂zw)2 dz

τ z

light heavy

Log profile: τ = ln

b z−a, dτ dz = 1 z−a

Lemma 1 (Doering & O. & Reznikoff ’06)

H

1 z−aw θ dz ≥ 2

H

1 z−a(∂zw)2 dz

for all (θ, w) with △2w = −△y θ

and w = ∂zw = 0 at z = 0, H

τ z

Non-monotone Ansatz for background temp. profile

1 H

• δ
• δ

τ z

1 2(1 + 1 λ ln z H−z)

λ ≈ ln H δ = boundary layer width Theorem 1 (Doering & O. & Reznikoff ’06, O. & Seis ’11) For δ ≪ ln− 1

15 H have

H

0 |∇θ|2 dz + 2

H

dτ dz w θ dz ≥ 0 for all (θ, w) with △2w = −△y θ and θ = w = ∂zw = 0 at z = 0, H

Hence Nu ≤

• Nu ln

1 15 H

ln− 1

15 is optimal for background field method

Theorem 2 (Nobili & O. ’16) Suppose that

H

dτ dz dz = 1 and

H

0 |∇θ|2 dz + 2

H

dτ dz w θ dz ≥ 0 for all (θ, w) with △2w = −△y θ and θ = w = ∂zw = 0 at z = 0, H.

Then

H

0 (dτ dz)2 dz ln

1 15 H.

In particular

• Nu ∼ ln

1 15 H.

Background field method not optimal Theorem 1 (Doering & O. & Reznikoff ’06, O. & Seis ’11) We have for H ≫ 1 :

• Nu ln

1 15 H

Theorem 2 (Nobili & O. ’16) We have for H ≫ 1 :

• Nu ln

1 15 H.

Theorem 3 (O. & Seis ’11) We have for H ≫ 1 : Nu ln

1 3 ln H

In particular Nu ≪

• Nu.

Theorem 2: Characterization of stable profiles Lemma 2 (Nobili & O. ’16) Suppose that

H

dτ dz w θ dz ≥ 0 for all (θ, w) with △2w = −△y θ and w = ∂zw = 0 at z = 0, H.

Then

• dτ

dz ≥ 0,

• ˜

H δ dτ dz dz (ln ˜ H δ )

δ

δ 2

dτ dz dz

for all δ ≪ ˜ H. τ z

δ 2 δ

˜ H

Proof of Lemma 2: non-negativity Horizontal Fourier transform y k:

H

dτ dz w(k2 − d2 dz2)2w dz ≥ 0 for all k and all w(z) with w = dw

dz = (k2 − d2 dz2)2w = 0 for z = 0, H.

Limit k ↑ ∞: ∀w

H

dτ dz w2 dz ≥ 0

= ⇒

dτ dz ≥ 0

Limit k ↓ 0: ∀w

H

dτ dz w d4 dz4w dz ≥ 0

Proof of Lemma 2: approximate logarithmic growth Have: ∀w

H

dτ dz w d4 dz4w dz ≥ 0

Change of variables s = ln z, w = z2v: w d4

dz4w = v( d ds + 2)( d ds + 1) d ds( d ds − 1)v

large scales

≈ −2vdv

ds = −dv2 ds , dτ dz dz = dτ ds ds

Get: ∀v

ln H

dτ ds (−dv2 ds ) ds 0

= ⇒

d2τ ds2 = d dszdτ dz 0

Two insights from rigorous results

• Optimal background temperature profile is non-monotone

stable for H0 log− 1

15 H

yields Nu log

1 15 H

• Optimal background temperature profile

has no physical meaning

Second part: Bounds on Nu at large but finite Pr Re only matters in thermal boundary layer

Recall Boussinesq approximation ...

z = 0 z = 1 T = 0

∂tT + u · ∇T − △T = 0

T = 1 u = 0 1 Pr(∂tu + u · ∇u)

−△u + ∇p = RaT

1

• ∇ · u = 0

u = 0

Rayleigh number Ra = α g δT h3

ν κ ,

Prandtl number Pr = ν

κ ... at finite Prandtl number

Theorem 4 (Choffrut&Nobili&0. ’15) Nu

    

(Ra ln Ra)

1 3 for Pr ≥ (Ra ln Ra) 1 3

(Ra ln Ra

Pr

)

1 2

for Pr ≤ (Ra ln Ra)

1 3

    

Constantin & Doering ’96: Nu Ra

1 2 for all Pr

• X. Wang ’07:

Nu Ra

1 3 ln 2 3 Ra for Pr ≫ Ra

Wang’s regime Pr ≫ Ra means Re ≪ 1 in bulk;

• ur regime Pr Ra

1 3 means Re 1

in thermal boundary layer

Follow Constantin&Doering ’99: maximum principle

Starting point: Nu = −∂zT + Tw for all z Average over (0, δ),

use maximum principle 0 ≤ T ≤ 1, use no-slip w = ∂zw = 0 at z = 0: T ∈ [0, 1] w = ∂zw = 0

δ

}

Nu ≤ 1 δ + sup

z∈(0,δ)

w2

1 2 ≤ 1

δ + δ2∂2

z w

for some suitable norm ·

Follow C&D ’99: Maximal regularity for instat. Stokes z = 0 z = 1 u = 0 1 Pr∂tu − △u + ∇p

= − 1

Pru · ∇u + RaT 1

• ∇ · u = 0

u = 0

∇2u 1

Pru · ∇u + RaT

For u·∇u make use of Hardy

1

0 u·∇udz z ≤ 4

1

0 |∇u|2dz and dissipation identity

1

0 |∇u|2dz = Ra(Nu − 1)

For T make use of maximal principle: 0 ≤ T ≤ 1 Almost achieved by f := inff=f1+f0

1

0 |f1|dz z +supz|f0|

A twist on maximal regularity for instationary Stokes

z = 0 z = 1 u = 0

∂tu − △u + ∇p = f ∇ · u = 0

u = 0

f := inff=f1+f0

1

0 |f1|dz z + supz|f0|

• Lemma 3 (Choffrut&Nobili&0. ’15)

Suppose f, u, and p are horizontally banded in the sense of F′f(t, k′, z) = 0 unless 1 ≤ R|k′| ≤ 2 for some R. Then ∂tw + ∇2w + (∂t − ∂2

z )u′ + ∇∇′u f

Summary First part: Upper bounds on Nu at Pr = ∞ background field method is unphysical Second part: Upper bounds on Nu at large Pr For Ra

1 3 to persist Re only matters in boundary layer