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 – 1 – December 2001
The growth question: ρ − ρ 2 , where A 1 2 h Agt = α = ρ + ρ 1 2 But what is α ? � 0.10, … 0.07, 0.06, … 0.03, 0.02 ? H Timescale: T = Ag If δ = h / H , and τ = t / T , δ = α τ 2 then IWPCTM 8 – 2 – December 2001
Experiments Top view 200 mm End view 5 0 0 m m 400 mm IWPCTM 8 – 3 – December 2001
Appropriate modelling (?) IWPCTM 8 – 4 – December 2001
Growth Dimensional analysis/similarity theory h = α Ag t 2 . Single mode Layzer (1955) 2 x π ( ) For x , y a cos ζ = 0 λ dh = if w , dt 2 dw w ( ) ( ) then 2 E Ag 1 E C , + = − − D dt λ where 6 π h � − � E exp . = � � λ � � Experimentally C D ~ 10 � Does this make sense? 2 d h 2 Ag π Early time → linear theory h = 2 dt λ Ag λ Late time → constant velocity w = ∞ C D � h → w ∞ ( t − t 0 ) IWPCTM 8 – 5 – December 2001
Structure Often described as ‘bubbles’… …but more like ‘thermals’ in miscible fluids IWPCTM 8 – 6 – December 2001
Thermals Self-similar r = θ z V = γ r 3 . Buoyancy conserved g ′ V = g ′ γ r 3 = g ′ 0 V 0 . Constant Froude number 2 w 2 F = g r ′ Integrating w = dz / dt 1 2 γ θ 2 z t = ( ) 1 2 2 F g V ′ 0 0 Experimental results → F ≈ 1.2. IWPCTM 8 – 7 – December 2001
Rayleigh-Taylor as thermals Froude number ~ 1.2 (aspect ratio 0.72) � C Thermal ≈ 1.3. Rayleigh-Taylor bubbles a little like thermals → C D ≈ 1.3 But in Rayleigh-Taylor environment • Density field not hydrostatic in ambient � Hydrostatic in mean density � halve buoyancy force → C D ≈ 2.6 • Flow around bubble affected by bubble moving in opposite direction � Drag due to twice rise speed of bubble → C D ≈ 10.4 In agreement with single mode experiments BUT natural R-T has more than one mode IWPCTM 8 – 8 – December 2001
Multi-mode What happens if λ grows with h ? Let λ = ψ h Late times approximation: 1 2 dh Ag � � ( ) 1 2 1 E h = − ψ � � dt C � � D � � Ag ( ) ( ) 2 ( ) 2 � h 1 E t t Ag t t = − ψ − = α − 0 0 C D For C D = 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 0.200 0.180 0.160 0.140 0.120 h/H ∂ h/dt 0.100 0.080 0.060 0.040 0.020 0.000 0.0 10.0 20.0 30.0 40.0 50.0 60.0 70.0 80.0 90.0 100.0 β ψ IWPCTM 8 – 9 – December 2001
Where do the modes begin? How do they interact? � Nonlinear interaction? � Initial perturbation? If modes independent and equal amplitude: 0.500 Key 0.450 δ = α τ 2 with α = 0.06 λ /H = 0.002 0.400 λ /H = 0.004 λ /H = 0.008 λ /H = 0.016 0.350 λ /H = 0.032 λ /H = 0.064 0.300 λ /H = 0.128 δ = h/H λ /H = 0.256 λ /H = 0.512 0.250 λ /H = 1.024 0.200 0.150 0.100 0.050 0.000 0.00 0.50 1.00 1.50 2.00 2.50 3.00 3.50 4.00 τ = ( Ag/H ) 1/2 t Instantaneous nonlinear mode halving interaction when h = λ : 0.500 Key 0.450 δ = α τ 2 with α = 0.033 λ /H = 0.002 0.400 λ /H = 0.004 λ /H = 0.008 λ /H = 0.016 0.350 λ /H = 0.032 λ /H = 0.064 0.300 λ /H = 0.128 δ = h/H λ /H = 0.256 λ /H = 0.512 0.250 0.200 0.150 0.100 0.050 0.000 0.00 0.50 1.00 1.50 2.00 2.50 3.00 3.50 4.00 τ = ( Ag/H ) 1/2 t Which is it? IWPCTM 8 – 10 – December 2001
Mixing See talk by Joanne Holford Energy budget * Can decompose PE into Background PE and Available PE . PE back is the minimum energy state that is achieved by adiabatic rearrangement of fluid parcels. Mixing increases PE back – it cannot decrease it! PE avail is the component of PE that can be converted into KE , heat (through dissipation) and, if mixing occurs, into PE back . = + PE PE Back PE Avail In the absence of external work: D D KE KE KE E avail PE avail PE avail PE avail PE PE back PE back PE back time IWPCTM 8 – 11 – December 2001
Mixing efficiency * E PE PE ∆ ∆ ∆ back Back Back η = − = = Integral ( ) E KE PE PE dt ∆ − ∆ + ∆ ∆ + � ε avail Avail Back 0.500 0.450 0.400 Overall mixing efficiency η 0 0.350 0.300 0.250 0.200 0.150 0.100 Joanne Holford 0.050 0.000 -1.0 0.0 1.0 2.0 3.0 4.0 5.0 6.0 7.0 8.0 9.0 10.0 Angle of tank | α o | PE PE δ δ Back Back η = = instantane ous E KE PE − δ δ + δ Avail Avail 1.00 Joanne Holford 0.80 0.60 η instantaneous 0.40 0.20 0.00 0.0 1.0 2.0 3.0 4.0 5.0 6.0 7.0 8.0 9.0 Time t / τ IWPCTM 8 – 12 – December 2001
Thermal Entrainment into a thermal dV wA = β dt … β = 0.18. Energetics of a thermal Mixing efficiency not well defined: depends on size of domain! Rayleigh-Taylor ρ 2 2 λ 2 h H ρ 1 IWPCTM 8 – 13 – December 2001
h = α Agt 2 δ = ατ 2 w = 2 α Agt ω = 2 ατ V = 2 L 2 h Total potential energy PE * 2 4 Total PE 1 4 = = − α τ Total PE 0 Background potential energy ρ 1 ρ a 2 h H ρ b ρ 2 Changes due to entrainment between counter-flowing streams. Invoke entrainment hypothesis: u e = β w Area of entrainment independent of h ⇔ depth of entrainment comparable with λ � entraining area = ϕ × plan area . ( ) * 2 4 PE 1 = − − ϕβα τ Back IWPCTM 8 – 14 – December 2001
Available potential energy ( ) * * * 2 4 PE PE PE 2 4 = − = − + ϕβ α τ Avail Tot Back 1.00 1 4 2 � � τ = 2 4 � � ( ) � � α + ϕβ � � 0.50 Potential energy PE( t )/PE(0) Key Experiment: PE Tot Experiment: PE Back Model: PE Back Model: PE Tot 0.00 -0.50 Joanne Holford’s experiments -1.00 0.0 1.0 2.0 3.0 4.0 5.0 Time t / τ IWPCTM 8 – 15 – December 2001
Kinetic energy * 3 4 KE 16 = σα τ 0.400 0.350 0.300 Key Kinetic energy KE( t )/PE(0) Experiments Model: σ = 1 0.250 Model: σ = 2 0.200 0.150 Joanne Holford’s 0.100 experiments 0.050 0.000 0.0 1.0 2.0 3.0 4.0 5.0 Time t / τ Available energy changing * * * dE dKE dPE Avail Avail = + d d d τ τ τ ( ) 2 3 4 4 16 = − + ϕβ − σα α τ Hence, energy is lost whenever α < ¼ (for β = 0, σ = 1). IWPCTM 8 – 16 – December 2001
Instantaneous mixing efficiency * dPE Back d τ η = − Inst * * dPE dKE Avail + d d τ τ ϕβ = 4 16 + ϕβ − σα So for ϕ = 16, β = 0.18, σ = 1, and α = 0.06, then η Inst = 0.49. 1.00 0.80 Thermal prediction Joanne Holford’s 0.60 η instantaneous experiments 0.40 0.20 0.00 0.0 1.0 2.0 3.0 4.0 5.0 6.0 7.0 8.0 9.0 Time t / τ IWPCTM 8 – 17 – December 2001
Integral mixing efficiency If there no mixing after reaching the bottom… ( ) ( ) bot 0 PE PE − Back Back η = Integral ( ) 0 PE Avail 1 η = ϕβ Integral 8 For ϕ = 16 and β = 0.18, then η Integral = 0.36. IWPCTM 8 – 18 – December 2001
If there is mixing after reaching the bottom… 1 ( ) � � * bot E 1 4 = + σα − ϕβ � � Avail 4 � � ( ) ( ) After bot bot If E E , then ∆ = η Back stab Avail 1 1 1 � � 1 4 η = ϕβ + η + σα − ϕβ � � Integral stab 8 2 4 � � For η stab = 0.2, then η Integral = 0.41. 0.500 0.450 0.400 Overall mixing efficiency η 0 0.350 0.300 Joanne Holford’s 0.250 experiments 0.200 0.150 0.100 0.050 0.000 -1.0 0.0 1.0 2.0 3.0 4.0 5.0 6.0 7.0 8.0 9.0 10.0 Angle of tank | α o | IWPCTM 8 – 19 – December 2001
Extensions * ρ 2 2 λ 2 h ∆ c H ρ 1 Let ∆ c be the fractional displacement of the centroid of the bubble from z = 0. → * PE δ Back η = − Inst * * PE KE δ + δ Avail ϕβ = ( ) 4 4 4 16 + ϕβ − − ϕβ ∆ − σα c Pyramid ( ∆ c = 1/4): η Inst = 0.6. Parabolic ( ∆ c = 1/6): η Inst = 0.56. (gives linear mean concentration) IWPCTM 8 – 20 – December 2001
How can we avoid having to specify C D ? Shell model GOY model (Gledzer–Ohkitani–Yamada): dU ( ) * * * * * * n ak U U bk U U ck U U = + + n n 1 n 2 n 1 n 1 n 1 n 2 n 1 n 2 + + − − + − − − dt 2 k U F − ν + n n n with k n = β n k 0 , a = 1, b = − ε and c = − 1 + ε . In Rayleigh-Taylor instability, energy input at all scales. dU ( ( ) ) n k U U k U U 1 k U U = − ε − − ε n n 1 n 2 n 1 n 1 n 1 n 2 n 1 n 2 + + − − + − − − dt 2 k U F − ν + n n n 2 dw w Recall Layzer model: ( ) ( ) 2 E Ag 1 E C + = − − D dt λ 1 E − n Hence F A g , where = n n 2 E + n 6 h π � − � n E exp and A n = A h n / h . = � � n � λ � n � � The mode penetrations h n and total penetration h are obtained from dh = n U and h max h . = n n dt n IWPCTM 8 – 21 – December 2001
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