multi-scale behaviour in the geo- science i: the onset of convection and interfacial instabilities...
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Multi-Scale Behaviour in the Geo-Science I: The Onset of Convection
and Interfacial Instabilitiesby
Hans Mühlhaus
The Australian Computational Earth Systems Simulator
(ACcESS)
OverviewWhat is GeodynamicsMantle Convection, Spreading, Folding, Landscape Evolution, Earthquakes, Volcanoes
The onset of Natural Convection:Linear instability analysis
Numerical SimulationsNusselt Number
Remarks on Weakly Non-Linear AnalysisGalerkin methods
hot spot
subducting plate lithosphereasthenosphere
shield volcano
strato volcanotrench
trench
convergentplate boundary
convergentplate boundary
oceanicspreading ridge
divergentplate boundary
transformplate boundary
island arc
continentalcrust
oceaniccrust
Earth dynamics
TERRA MESH
BENEFITS
Earth dynamics
hot spot
subducting plate lithosphereasthenosphere
shield volcano
strato volcanotrench
trench
convergentplate boundary
convergentplate boundary
oceanicspreading ridge
divergentplate boundary
transformplate boundary
island arc
continentalcrust
oceaniccrust
22ndnd talk: Volcano modelling 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
n : the director of the anisotropyW, Wn : spin and director spinD, D’: stretching and its deviatoric part
)( kikjkjkiijn
ij DDWW
ij nin j
jn
iji nWn
Governing Equations
))(()( Tpt
vvgvvv
g
0g
vdt
d
Navier Stokes Equations:
Heat Equation:
TkTt
Tcp
2)(
v
v(t,x) is the velocity vector, T is the temperature, kg/m2) is the density, Pas)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.
))(1( 00 TTp
Temperature dependence of density:
Simplified convection model:
p ( 3 10-5K-1 ) is the thermal expansion
coefficient, T0 (288K) is the surface temperature
x1
x2
T=T1 , v2=0
T=T0 , v2=0
T,1 , v1=0T,1 , v1=0 H
L
Consider a square planet……
Governing Equations, cont.
,~xx H ,~
][
02
tk
cHt
t
p
,~
][vv
t
H
Nondimensionalisation:
)~
1)(( 2010 T
H
xTTTT
))((1
)1()(Pr
12
TpTxRat
vvgg
vvv
TTt
T 2)(
v
Raleigh Number: )1010()( 86
301
20
k
HTTgcRa p
Prandtl Number: )1025.0()/(
/Pr 18
0
0 pck
Relevant limit in
Geophysics:Pr
Insertion and dropping tildes…..
Governing Equations, cont.
2,2
1
x
v 1,2 v
timxexnT 121, sin),(
01,4 RaT TTTT t
21,1,2,2,1,, )(
Stream function
and
In this way we satisfy the incompressibility constraint div v=0 identically.Insertion into the velocity equations and the heat equation, assuming infinitePrandtl number gives:
and
Dropping nonlinear terms and insert the “Ansatz”
, gives:
RaTmnm 21,
2222 )( and 1,222 )( Tnm Thus
2222
32222
)(
)(
nm
nmRam
Marginal instability:
HLRa 22,4
27min1
4
min
For m=1 we obtain:
2/,
2/
)(
0
12
2
3222
L
Hn
L
Hm
m
nmRa
HL1
2
1
322
1
)4
(
))4
((
LH
LH
Ra
Finite Element Approximations…Ra=104, mesh: 128X128
The Nusselt Number
onlyconductionbyflux
heatfluxtotal
dVkT
dVkTTv
Nu
V
V
2,
2,2 )(
Definition:
It can be shown that for zero normal velocity b.c.’s and fixed top and bottom Temperature
dVRaV
NuV
PowerMechanicalessDimensionl
TT
))((:))((2
111 vvvv
The Nusselt NumberHint for derivation of Nu-Power relationship
))(()1( 22TpTxRa vve dV
V v(.)form
And apply Gauss’s Theorem. For the given b.c.’s it follows that
dVdVvTxRaV
TT
V ))((:))((
2
1)1( 22 vvvv
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
212 2sin),(cossin),( xtmCmxxtmBT 12 cossin),( mxxtmA
0))((1
0
21
2
0
21,1,2,2,1,, dxdxTTTTT t
We consider the ansatz:
0)( 211,
1
0
2
0
dxdxTRa
Insert into:
and
We obtain:
abm
Ramaa
322
2
)(
and
3322
22
)(
2a
m
Rambb
Rayleigh-Taylor Instabilities
1x
1x
RT fingers evident in the Crab Nebula
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.
2x
Benchmark Problem
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-in-cell 1024 el 0.01102 0.003098 222
Particle-in-cell 4096 el 0.01244 0.003090 215
Level set 5175 el 0.01135 0.003116 215.06
Rayleigh-Taylor instability
benchmark
Linear Instability Analysis
1h
1h
00
2/1,
2/1,
2/1
kkt vSSSdt
d
Equilibrium to be satisfied in ground state at time t=t0 and at t0+dt:
Ground state:
0,
0,,0
222
221
xgxconstp
andxgxconstpvi
2h2x n
Continued Equilibrium:
0))((),(
v
gvvpS
Tp
Stokes equation:
Linear Instability Analysis
1h
0)( ,2/1
,2/1
,2/12/1
, jijjii uuP
02/1,
2/12/1, jjii uP
obtainwe
andwith ,, ttii pPvu
2h2x n
Or, considering the incompressibility constraint: 02/1
, jju1h
Linear Instability Analysis: Boundary and interface conditions
j,)(
1h
1h
We assume that the velocities are zero on top and bottom; on the sides we assume that v1=0 and i.e. symmetry boundary.On the interface the velocities as well as the natural boundary terms have to be continuous.Natural b.c.’s: replace
2h2x n
as well as its time derivative have to be continuous on the interface
By . The vector
in
0))(( ,2/1
,2/1
,2/12/1 jijjiij uuP
jn
jijjiiji nuuPt ))(( 2/1,
2/1,
2/12/1
Linear Instability Analysis: Boundary and interface conditions
0))(( 0221 dAnn
dt
djijjij
1h
1h
We assume that the velocities are zero on top and bottom; on the sides we assume that v1=0 and i.e. symmetry boundary.On the interface the velocities as well as the natural boundary terms have to be continuous.Natural b.c.’s
2h2x n
Or:
Result:
0)()(:
0)()(:
221212
2,221
2,21
2
21,2
22,1
211,2
12,1
11
gvPPuux
uuuux
Exercises
1. The normal component of the surface velocity of a 2D half plane reads . The half plane is occupied by a Stokes fluid with the viscosity The normal stress is obtained as . Show that .
2. Consider a Rayleigh-Taylor instability problem involving 2 infinite half-planes; i.e. . The normal velocity of the
Interface reads . Show that .
Hint: Use the relationship for a gravity free Half-plane and note that .
02 x12 cos kxVv
122 cos kxQ kVQ 2
12 cos kxVev tk
g
)(2
)(
21
21
1, 21 khkh
UVUUkQ tgt
,0
, ,2gVQQ t
gt
,0
,
Exercises (folding)
w
Exercise (folding) cont.
Exercise (folding) cont.
Excercise 4:
01,4 RaT TTTT t
21,1,2,2,1,, )(
Solve convection equations
and
using the perturbation expansion
up to terms of order .
Hint:
....22
1 ....22
1 TTT t2 22
0 RaRaRa
3
1. Insert into pde’s, collect coefficients of and 2. Individual coefficients must be equal to zero. Get
etc and T3 are a little bit harder to get since the pde’s contain inhomogeneous terms which areproportional to (
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:
3222
22
, 2
1
)(aa
m
Rama
a
RaRacrit
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