Toy models for Rayleigh- Taylor instability: Stuart Dalziel - - PDF document

IWPCTM 8 – 1 – December 2001

Toy models for Rayleigh- Taylor instability:

Stuart Dalziel

Department of Applied Mathematics and Theoretical Physics University of Cambridge

International Workshop on the Physics of Compressible Turbulent Mixing 14 December 2001 (9:10 – 9:30) with thanks to Joanne Holford (DAMTP) David Youngs (AWE)

IWPCTM 8 – 2 – December 2001

The growth question:

2

Agt h α = , where

2 1 2 1

ρ ρ ρ ρ + − = A

But what is α ?

0.10, … 0.07, 0.06, … 0.03, 0.02 ? Timescale: Ag H T = If δ = h/H, and τ = t/T, then δ = α τ2

IWPCTM 8 – 3 – December 2001

Experiments

400 mm 5 m m 200 mm End view Top view

IWPCTM 8 – 4 – December 2001

Appropriate modelling (?)

IWPCTM 8 – 5 – December 2001

Growth

Dimensional analysis/similarity theory

h = α Ag t2.

Single mode

Layzer (1955)

For

( )

λ π ζ x a y x 2 cos , = if w dt dh = , then

( ) ( )

λ

2

1 2 w C E Ag dt dw E

D

− − = + , where

• −

= λ πh E 6 exp . Experimentally CD ~ 10 Does this make sense? Early time → linear theory h Ag dt h d λ π 2

2 2

= Late time → constant velocity

D

C Ag w λ =

∞

h → w∞(t − t0)

IWPCTM 8 – 6 – December 2001

Structure

Often described as ‘bubbles’… …but more like ‘thermals’ in miscible fluids

IWPCTM 8 – 7 – December 2001

Thermals

Self-similar r = θ z V = γ r3. Buoyancy conserved g′ V = g′ γ r3 = g′0V0. Constant Froude number r g w F ′ =

2 2

Integrating w = dz/dt

( )

t z V g F = ′

2 2 1 2 1

2 θ γ Experimental results → F ≈ 1.2.

IWPCTM 8 – 8 – December 2001

Rayleigh-Taylor as thermals

Froude number ~ 1.2 (aspect ratio 0.72) CThermal ≈ 1.3. Rayleigh-Taylor bubbles a little like thermals → CD ≈ 1.3 But in Rayleigh-Taylor environment

• Density field not hydrostatic in ambient

Hydrostatic in mean density halve buoyancy force → CD ≈ 2.6

• Flow around bubble affected by bubble moving in opposite

direction Drag due to twice rise speed of bubble → CD ≈ 10.4 In agreement with single mode experiments BUT natural R-T has more than one mode

IWPCTM 8 – 9 – December 2001

Multi-mode

What happens if λ grows with h? Let λ = ψ h Late times approximation:

( )

2 1 2 1

1 h E C Ag dt dh

D

• −

= ψ

• (

) ( ) ( )

2 2

1 t t Ag t t E C Ag h

D

− = − − = α ψ For CD = 10 and ψ = 1, α = 0.025. [Full Layzer growth with ψ = 1 gives α = 0.023.] Growth rate maximised with ψ ~ 10 giving α ~ 0.103

0.0 10.0 20.0 30.0 40.0 50.0 60.0 70.0 80.0 90.0 100.0 0.000 0.020 0.040 0.060 0.080 0.100 0.120 0.140 0.160 0.180 0.200

β h/H

ψ ∂h/dt

IWPCTM 8 – 10 – December 2001

Where do the modes begin? How do they interact? Nonlinear interaction? Initial perturbation? If modes independent and equal amplitude:

0.00 0.50 1.00 1.50 2.00 2.50 3.00 3.50 4.00 0.000 0.050 0.100 0.150 0.200 0.250 0.300 0.350 0.400 0.450 0.500

τ = (Ag/H)1/2t δ = h/H

Key δ = α τ2 with α = 0.06 λ/H = 0.002 λ/H = 0.004 λ/H = 0.008 λ/H = 0.016 λ/H = 0.032 λ/H = 0.064 λ/H = 0.128 λ/H = 0.256 λ/H = 0.512 λ/H = 1.024

Instantaneous nonlinear mode halving interaction when h = λ:

0.00 0.50 1.00 1.50 2.00 2.50 3.00 3.50 4.00 0.000 0.050 0.100 0.150 0.200 0.250 0.300 0.350 0.400 0.450 0.500

τ = (Ag/H)1/2t δ = h/H

Key δ = α τ2 with α = 0.033 λ/H = 0.002 λ/H = 0.004 λ/H = 0.008 λ/H = 0.016 λ/H = 0.032 λ/H = 0.064 λ/H = 0.128 λ/H = 0.256 λ/H = 0.512

Which is it?

IWPCTM 8 – 11 – December 2001

Mixing

See talk by Joanne Holford

Energy budget *

Can decompose PE into Background PE and Available PE. PEback is the minimum energy state that is achieved by adiabatic rearrangement of fluid parcels. Mixing increases PEback – it cannot decrease it! PEavail is the component of PE that can be converted into KE, heat (through dissipation) and, if mixing occurs, into PEback. = + PE PEBack PEAvail In the absence of external work:

PE PEavail PEback KE PEavail PEback KE D PEavail PEback KE D Eavail time

IWPCTM 8 – 12 – December 2001

Mixing efficiency *

( )

Avail Back Back Back avail back Integral

PE KE PE dt PE PE E E ∆ + ∆ − ∆ = + ∆ ∆ = ∆ ∆ − =

ε

η

• 1.0

0.0 1.0 2.0 3.0 4.0 5.0 6.0 7.0 8.0 9.0 10.0 0.000 0.050 0.100 0.150 0.200 0.250 0.300 0.350 0.400 0.450 0.500

Angle of tank |αo| Overall mixing efficiency η0

Avail Back Avail Back

• us

instantane

PE KE PE E PE δ δ δ δ δ η + = − =

0.0 1.0 2.0 3.0 4.0 5.0 6.0 7.0 8.0 9.0 0.00 0.20 0.40 0.60 0.80 1.00

Time t/τ ηinstantaneous

Joanne Holford Joanne Holford

IWPCTM 8 – 13 – December 2001

Thermal

Entrainment into a thermal

wA dt dV β = … β = 0.18.

Energetics of a thermal

Mixing efficiency not well defined: depends on size of domain!

Rayleigh-Taylor

2h ρ1 ρ2 H 2λ

IWPCTM 8 – 14 – December 2001

h = αAgt2 δ = ατ2 w = 2αAgt ω = 2ατ V = 2L2h

Total potential energy

4 2 *

4 1 τ α − = = PE PE PE

Total Total

Background potential energy

2h ρ1 ρ2 H ρb ρa

Changes due to entrainment between counter-flowing streams. Invoke entrainment hypothesis: ue = βw Area of entrainment independent of h ⇔ depth of entrainment comparable with λ entraining area = ϕ × plan area.

( )

4 2 *

1 τ ϕβα − − =

Back

PE

IWPCTM 8 – 15 – December 2001

Available potential energy

( )

4 2 * * *

4 2 τ α ϕβ + − = − =

Back Tot Avail

PE PE PE

0.0 1.0 2.0 3.0 4.0 5.0

• 1.00
• 0.50

0.00 0.50 1.00

Time t/τ Potential energy PE(t)/PE(0)

Key Experiment: PETot Experiment: PEBack Model: PEBack Model: PETot

( )

4 1 2 4

2

• +

= ϕβ α τ

Joanne Holford’s experiments

IWPCTM 8 – 16 – December 2001

Kinetic energy

4 3 *

16 τ σα = KE

0.0 1.0 2.0 3.0 4.0 5.0 0.000 0.050 0.100 0.150 0.200 0.250 0.300 0.350 0.400

Time t/τ Kinetic energy KE(t)/PE(0)

Key Experiments Model: σ = 1 Model: σ = 2

Available energy changing

( )

3 2 * * *

16 4 4 τ α σα ϕβ τ τ τ − + − = + = d dPE d dKE d dE

Avail Avail

Hence, energy is lost whenever α < ¼ (for β = 0, σ = 1).

Joanne Holford’s experiments

IWPCTM 8 – 17 – December 2001

Instantaneous mixing efficiency

σα ϕβ ϕβ τ τ τ η 16 4

* * *

− + = + − = d dKE d dPE d dPE

Avail Back Inst

So for ϕ = 16, β = 0.18, σ = 1, and α = 0.06, then ηInst = 0.49.

0.0 1.0 2.0 3.0 4.0 5.0 6.0 7.0 8.0 9.0 0.00 0.20 0.40 0.60 0.80 1.00

Time t/τ ηinstantaneous

Joanne Holford’s experiments Thermal prediction

IWPCTM 8 – 18 – December 2001

Integral mixing efficiency

If there no mixing after reaching the bottom…

( ) ( ) ( )

Avail Back bot Back Integral

PE PE PE − = η ϕβ η 8 1 =

Integral

For ϕ = 16 and β = 0.18, then ηIntegral = 0.36.

IWPCTM 8 – 19 – December 2001

If there is mixing after reaching the bottom…

( )

• −

+ = ϕβ σα 4 1 4 1

* bot Avail

E If

( ) ( )

bot Avail stab bot After Back

E E η = ∆ , then

• −

+ + = ϕβ σα η ϕβ η 4 1 4 1 2 1 8 1

stab Integral

For ηstab = 0.2, then ηIntegral = 0.41.

• 1.0

0.0 1.0 2.0 3.0 4.0 5.0 6.0 7.0 8.0 9.0 10.0 0.000 0.050 0.100 0.150 0.200 0.250 0.300 0.350 0.400 0.450 0.500

Angle of tank |αo| Overall mixing efficiency η0

Joanne Holford’s experiments

IWPCTM 8 – 20 – December 2001

Extensions *

2h ρ1 ρ2 H 2λ ∆c

Let ∆c be the fractional displacement of the centroid of the bubble from z = 0. →

( )

σα ϕβ ϕβ ϕβ δ δ δ η 16 4 4 4

* * *

− ∆ − − + = + − =

c Avail Back Inst

KE PE PE Pyramid (∆c = 1/4): ηInst = 0.6. Parabolic (∆c = 1/6): ηInst = 0.56. (gives linear mean concentration)

IWPCTM 8 – 21 – December 2001

How can we avoid having to specify CD?

Shell model

GOY model (Gledzer–Ohkitani–Yamada):

( )

n n n n n n n n n n n n n

F U k U U ck U U bk U U ak dt dU + − + + =

− − − + − − + + 2 * 2 * 1 2 * 1 * 1 1 * 2 * 1

ν with kn = β nk0, a = 1, b = −ε and c = −1 + ε. In Rayleigh-Taylor instability, energy input at all scales. ( ) ( )

n n n n n n n n n n n n n

F U k U U k U U k U U k dt dU + − − − − =

− − − + − − + + 2 2 1 2 1 1 1 2 1

1 ν ε ε

Recall Layzer model: (

) ( )

λ

2

1 2 w C E Ag dt dw E

D

− − = + Hence

n n n n

E E g A F + − = 2 1 , where

• −

=

n n n

h E λ π 6 exp and An = A hn/h. The mode penetrations hn and total penetration h are obtained from

n n

U dt dh = and

n n

h h max = .

IWPCTM 8 – 22 – December 2001

0.0 2.0 4.0 6.0 8.0 10.0 12.0 14.0 16.0 18.0 20.0 0.000 0.050 0.100 0.150 0.200 0.250 0.300 0.350 0.400 0.450 0.500

Time (t) Penetration (h)

Y=F; F=.7830*1.62000E-03*X*X; Erms=1.4199E-02 Y=F; F=-8.2640E-03+6.3999E-03*X+.5568*1.62000E-03*X*X; Erms=2.7973E-03

Approximate quadratic growth Coefficient depends on initial spectrum Possible to replicate α ~ 0.06

IWPCTM 8 – 23 – December 2001

Conclusions

General

• Initial conditions are important for gross features
• Internal details relatively insensitive to initial conditions
• Appropriate modelling of initial conditions gives close

agreement

Thermals model

• Single-mode growth rate consistent with isolated thermal
• Simple model for transfer between modes replicates t2 growth
• Mixing efficiency consistent with thermal entrainment

Shell model

• Baroclinic input at all scales
• Very simple model replicates t2 growth
• Growth rate sensitive to initial spectrum

An explanation?

No, but it helps.