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The Australian Computational Earth Systems Simulator (ACcESS) Multi-Scale Behaviour in the GeoScience I: The Onset of Convection and Interfacial Instabilities by Hans Mühlhaus Overview What is Geodynamics Mantle Convection, Spreading, Folding, Landscape Evolution, Earthquakes, Volcanoes The onset of Natural Convection: Linear instability analysis Numerical Simulations Nusselt Number Remarks on Weakly Non-Linear Analysis Galerkin methods Earth dynamics continental crust convergent plate boundary transform plate boundary divergent plate boundary oceanic spreading ridge trench trench convergent plate boundary strato volcano shield volcano oceanic crust island arc hot spot subducting plate lithosphere asthenosphere TERRA MESH Earth dynamics BENEFITS continental crust convergent plate boundary transform plate boundary divergent plate boundary oceanic spreading ridge trench trench convergent plate boundary strato volcano shield volcano oceanic crust island arc hot spot subducting plate lithosphere asthenosphere 2nd talk: Volcano modelling Montserrat, West Indies 3rd talk: Particle Processes –Spherical particles –Selection of contact physics: –Non-rotational and rotational dynamics –Friction interactions –Linear elastic interactions –Bonded interactions 4th talk: modeling of geological folds Director evolution i Wijn n j n Wijn Wij ( Dki kj Dkj ki ) ij nin j n : the director of the anisotropy W, Wn: spin and director spin D, D’: stretching and its deviatoric part Governing Equations Navier Stokes Equations: v ( v v) g p (v (v)T ) t 0 g g Heat Equation: T c p ( v T ) k 2T t d v dt v(t,x) is the velocity vector, T is the temperature, (4000kg/m2) is the density, (1021Pas) is the viscosity, cp (10-3WK-1s/kg) is the specific heat at constant pressure and k (4Wm-1K-1) is the thermal conductivity Governing Equations, cont. Temperature dependence of density: 0 (1 p (T T0 )) p ( 3 10-5K-1 ) is the thermal expansion coefficient, T0 (288K) is the surface temperature Simplified convection model: T=T0 , v2=0 T,1 , v1=0 x2 H L x1 T=T1 , v2=0 T,1 , v1=0 Consider a square planet…… Governing Equations, cont. Nondimensionalisation: x H~ x, H 2 0c p ~ t t, k v H~ v, [t ] x2 ~ T T0 (T1 T0 )(1 T ) H [t ] Insertion and dropping tildes….. 1 v 1 ( v v) Ra (1 x2 T ) g p (v (v)T ) Pr t g T ( v T ) 2T t 02 gc p (T1 T0 ) H 3 ( 106 108 ) Raleigh Number: Ra k Prandtl Number: Pr / 0 ( 0.25 1018 ) k /( 0c p ) Relevant limit in Geophysics: Pr Governing Equations, cont. Stream function : v1 , 2 x2 v2 ,1 and In this way we satisfy the incompressibility constraint div v=0 identically. Insertion into the velocity equations and the heat equation, assuming infinite Prandtl number gives: 4 RaT,1 0 and T,t (,1T, 2 , 2T,1 ) ,1 2T Dropping nonlinear terms and insert the “Ansatz” (,1 ,T ) sin nx2eimx1 t (m2 n2 2 )2 ,1 m2 RaT , gives: and ( m2 n2 2 )T ,1 m2 Ra (m2 n 2 2 )3 (m 2 n 2 2 ) 2 Thus Marginal instability: 0 (m 2 n 2 2 ) 3 Ra m2 H H m ,n L2 / 2 L1 / 2 For m=1 we obtain: Ra min 27 4 , L1min 2 2 H 4 L1 (( Ra H 4H 2 ) 2 )3 L1 4H 2 ( ) L1 Finite Element Approximations… Ra=104, mesh: 128X128 The Nusselt Number Definition: Nu (Tv V 2 kT, 2 )dV kT, 2 dV total heatflux flux by conduction only V It can be shown that for zero normal velocity b.c.’s and fixed top and bottom Temperature Nu 1 1 1 T T ( v ( v ) ) : ( v ( v ) )dV RaV V 2 Dimensionless MechanicalPower The Nusselt Number Hint for derivation of Nu-Power relationship Ra (1 x2 T )e 2 p (v (v)T ) form (.) v dV V And apply Gauss’s Theorem. For the given b.c.’s it follows that 1 Ra (1 x2 T )v2 dV (v (v)T ) : (v (v)T )dV 2V V Finite Element Approximations… Ra=104, mesh: 128X128 Nusselt Number Finite Element Approximations… Ra=105, mesh: 128X128 Finite Element Approximations… Ra=105, mesh: 128X128 Nusselt Number Finite Element Approximations… Ra=106, mesh: 128X128 Finite Element Approximations… Ra=106, mesh: 128X128 Nusselt Number Finite Element Approximations… Ra=107, mesh: 128X128 Finite Element Approximations… Ra=107, mesh: 128X128 Nusselt Number Galerkin Method We consider the ansatz: A(m, t ) sin x2 cos mx1 T B(m, t ) sin x2 cos mx1 C(m, t ) sin 2x2 Insert into: 1 2 ( RaT )dx dx ,1 1 2 1 2 0 and 0 0 2 ( T ( T T ) ,t ,1 ,2 ,2 ,1 ,1 T )T dx1dx2 0 0 0 We obtain: Ram 2 a a 2 ab 2 3 (m ) and 2 2 Ram 3 b 2b 2 a (m 2 ) 3 Rayleigh-Taylor Instabilities x2 x1 Consider two fluids of different densities, the heaviest above the lightest. An horizontal interface separates the two fluids. This situation is unstable because of gravity. Effectively, if the interfaces modified then a pressure want of balance grows. Equilibrium can be found again tanks to surface tension that's why there is a competition between surface tension and gravity. Surface tension is stabilizing instead gravity is destabilizing for this configuration. x1 RT fingers evident in the Crab Nebula Benchmark Problem Rayleigh-Taylor instability benchmark Method mesh γ0 (vrms)max reached at t= van Keken coarse 0.01130 0.003045 212.14 van Keken fine 0.01164 0.003036 209.12 Particle-incell 1024 el 0.01102 0.003098 222 Particle-incell 4096 el 0.01244 0.003090 215 Level set 5175 el 0.01135 0.003116 215.06 Linear Instability Analysis Ground state: vi 0, x2 n h2 p const 1 gx2 , x2 0 and p const 2 gx2 , x2 0 h1 Equilibrium to be satisfied in ground state at time t=t0 and at t0+dt: Stokes equation: h1 p (v (v)T ) g 0 S ( p,v) Continued Equilibrium: d 1/ 2 1/ 2 1/ 2 S S , t S , k vk 0 dt 0 Linear Instability Analysis with ui vi ,t and P p,t x2 n h2 we obtain h1 P,i1/ 2 1/ 2 (ui1,/j2 u1j/,i2 ), j 0 h1 Or, considering the incompressibility constraint: u1j/, 2j 0 P,i1/ 2 1/ 2ui1,/jj2 0 Linear Instability Analysis: Boundary and interface conditions x2 n h2 We assume that the velocities are zero on top and bottom; on the sides we assume that v1=0 and s120, i.e. symmetry boundary. On the interface the velocities as well as the natural boundary termsh have to be continuous. Natural b.c.’s: replace () , j in 1 h1 (P ij (u 1/ 2 1/ 2 1/ 2 i, j u )), j 0 1/ 2 j ,i By n j . The vector ti (P1/ 2 ij 1/ 2 (ui1,/j2 u1j/,i2 ))n j as well as its time derivative have to be continuous on the interface Linear Instability Analysis: Boundary and interface conditions x2 n h2 We assume that the velocities are zero on top and bottom; on the sides we assume that v1=0 and s120, i.e. symmetry boundary. On the interface the velocities as well as the natural boundary termsh have to be continuous. Natural b.c.’s d ((s ij n1j s ij2 n 2j )dA0 ) 0 dt Result: 1 h1 x1 : 1 (u11, 2 u12,1 ) 2 (u12, 2 u22,1 ) 0 Or: x2 : 1u12, 2 2u22, 2 ( P1 P 2 ) ( 1 2 ) gv2 0 Exercises 1. The normal component of the surface velocity of a 2D half plane x2 0 reads v2 V cos kx1 . The half plane is occupied by a Stokes fluid with the viscosity . The normal stress is obtained as s 22 Q cos kx1 . Show that Q 2kV . 2. Consider a Rayleigh-Taylor instability problem involving 2 infinite half-planes; i.e. h1k , h2k 1 . The normal velocity of the Interface reads v2 Vet cos kx1 . Show that ( 1 2 ) g . 2(1 2 )k Hint: Use the relationship Q,gt 0 2kU , U V,t U for a gravity free Half-plane and note that Q,gt 0 Q,t gV . Exercises (folding) w Exercise (folding) cont. Exercise (folding) cont. Excercise 4: Solve convection equations RaT,1 0 4 and T,t (,1T, 2 , 2T,1 ) ,1 2T using the perturbation expansion 1 2 .... T T1 T2 .... t 2 2 2 Ra Ra 0 2 Ra 2 up to terms of order 3 . Hint: Insert into pde’s, collect coefficients of , 2 and 3 Individual coefficients must be equal to zero. Get 1, 2 etc 3. 3 and T3 are a little bit harder to get since the pde’s contain inhomogeneous terms which are proportional to (1, T1) 1. 2. Perturbation solution for the weakly nonlinear problem 4. We require a solubility condition (Fredholm’s alternative) In the present case this just means that the coefficient of the resonant inhomogeneous term must vanish. The coefficient has the form of an ode which can be written as: a 0 m2 Ra 2 1 3 a, 2 a a 2 2 (m ) 2 Racrit Ra