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EART162: PLANETARY INTERIORS Francis Nimmo F.Nimmo EART162 Spring 10 Last week - Seismology • Seismic velocities tell us about interior properties Vs G Vp K 43 G • Adams-Williamson equation allows us to relate density directly to seismic velocities d (r ) (r ) g (r ) 2 4 2 dr V p 3 Vs • Travel-time curves can be used to infer seismic velocities as a function of depth • Midterm F.Nimmo EART162 Spring 10 This Week – Fluid Flow & Convection • • • • • • • Fluid flow and Navier-Stokes Simple examples and scaling arguments Post-glacial rebound Rayleigh-Taylor instabilities What is convection? Rayleigh number and boundary layer thickness See Turcotte and Schubert ch. 6 F.Nimmo EART162 Spring 10 Viscosity • Young’s modulus gives the stress required to cause a given deformation (strain) – applies to a solid • Viscosity is the stress required to cause a given strain rate – applies to a fluid • Viscosity is basically the fluid’s resistance to flow elastic E Young’s modulus viscous viscosity • Kinematic viscosity measured in Pa s • [Dynamic viscosity n/ measured in m2s-1] • Typical values for viscosity: water 10-3 Pa s, basaltic lava 104 Pa s, ice near melting 1014 Pa s, mantle 1021 Pa s • Viscosity often temperature-dependent (see Week 3) F.Nimmo EART162 Spring 10 Defining Viscosity Recall • • Viscosity is the stress generated for a given strain rate • Example – moving plate: u d (Shear) stress required to generate u velocity gradient u / d (= y ) Viscosity = d / u • Example – moving lava flow: Driving shear stress = gd sinq a d Surface velocity = gd2 sinq / q e.g. Hawaiian flow =104 Pa s q=5o d=3m gives u=2 ms-1 (walking pace) F.Nimmo EART162 Spring 10 Adding in pressure • In 1D, shear stress (now using t) is t y • Let’s assume u only varies in the y direction u t y Fluid velocity u t y y y x t P P x P x x Viscous force (x direction, per unit volume): t Fv y Pressure force (x direction, per unit volume): P Fp x Why the minus sign? F.Nimmo EART162 Spring 10 Putting it together • Total force/volume = viscous + pressure effects t P 2u P Ft Fv Fp 2 y x y x • We can use F=ma to derive the response to this force Du ma V What does this mean? Dt • So the 1D equation of motion in the x direction is 2u Du P 2 Dt x y What does each term represent? • In the y-direction, we would also have to add in buoyancy forces (due to gravity) Fb gF.Nimmo EART162 Spring 10 Navier-Stokes • We can write the general (3D) formula in a more compact form given below – the Navier-Stokes equation • The formula is really a mnemonic – it contains all the physics you’re likely to need in a single equation • The vector form given here is general (not just Cartesian) u 2 u u u P g yˆ t Yuk! Inertial term. Pressure Source of turbulence. gradient See next slide. Buoyancy force (e.g. Diffusion-like viscosity Zero for steady2 term. Warning: u is thermal or electromagnetic) state flows ŷ is a unit vector complicated, especially in non-Cartesian geom. F.Nimmo EART162 Spring 10 Reynolds number 2u u u u 2 P F y t • Is the inertial or viscous term more important? • We can use a scaling argument to get the ratio Re: uL Here L is a characteristic a Re lengthscale of the problem • Re is the Reynolds number and tells us whether a flow is turbulent (inertial forces dominate) or not • Fortunately, many geological situations allow us to neglect inertial forces (Re<<1) • E.g. what is Re for the convecting mantle? F.Nimmo EART162 Spring 10 Example 1 – Channel Flow u P u u u 2 F y x t 2 L y x 0 u P2 P1 2d (Here u doesn’t vary in x-direction) • 2D channel, steady state, u=0 at y=+d and y=-d a 1 P1 P0 2 2 u d y 2 L • Max. velocity (at centreline) = (P/L) d2/2 • Does this result make sense? • We could have derived a similar answer from a scaling argument – how? F.Nimmo EART162 Spring 10 Example 2 – falling sphere u u u u 2 P F y t 2 Steady-state. What are the important terms? a r u An order of magnitude argument gives drag force ~ ur Is this dimensionally correct? The full answer is 6pur, first derived by George Stokes in 1851 (apparently under exam conditions) By balancing the drag force against the excess weight of the sphere (4pr3g/3 ) we can obtain the terminal velocity (here is the density contrast between sphere and fluid) F.Nimmo EART162 Spring 10 Example 3 – spreading flow u u u u 2 P F y t 2 y h(x,t) u d x Low Re, roughly steady-state. What are the important terms? 2 g h 2 P h u u( y) y ay b g 2 x x x y h u Conservation of mass gives (why?) dy t x h g d 3 2 h a As long as d >>h, we get: 2 t 3 x What kind of equation is this? Does it make physical sense? Where might we apply it on Earth? F.Nimmo EART162 Spring 10 Postglacial Rebound ice L mantle mantle w • Postglacial rebound problem: How long does it take for the mantle to rebound? • Two approaches: – Scaling argument – Stream function j – see T&S u u • Scaling argument: 2 P t x 2 • Assume u is constant (steady flow) and that u ~ dw/dt • We end up with 1 dw gL a ~ decay constant w dt • What does this equation mean? F.Nimmo EART162 Spring 10 Prediction and Observations • Scaling argument gives: w w0 exp( t / t ) How does this time constant compare with that for spreading flow? t~ gL Hudson’s Bay deglaciation: L~1000 km, t=2.6 ka So ~2x1021 Pa s So we can infer the viscosity of the mantle http://www.geo.ucalgary.ca/~wu/TUDelft/Introduction.pdf A longer wavelength load would sample the mantle to greater depths – higher viscosity F.Nimmo EART162 Spring 10 Rayleigh-Taylor Instability b 1 m b 2 m • This situation is gravitationally unstable if 2 < 1 : any infinitesimal perturbation will grow • What wavelength perturbation grows most rapidly? • The full solution is v. complicated (see T&S 6-12) – so let’s try and think about it physically . . . L u1 L l l u1 u2 1 2u1 l u 2 u1 l b b 2 u 2 u1l 2 b b F.Nimmo EART162 Spring 10 R-T Instability (cont’d) • Recall from Week 5: dissipation per unit volume m 2 • We have two contributions to total dissipation (1 , 2) • By adding the two contributions, we get a 2 l 4 2 Ptot mu1 L 3 b l 2 term 1 term • What wavelength minimizes the dissipation? • We end up with dissipation minimized at lmin=1.26 b • This compares pretty well with the full answer (2.57b) and saves us about six pages of maths F.Nimmo EART162 Spring 10 R-T instability (cont’d) • The layer thickness determines which wavelength minimizes viscous dissipation • This wavelength is the one that will grow fastest • So surface features (wavelength) tell us something about the interior structure (layer thickness) Salt domes in S Iran. Dome spacing of ~15 km suggests salt layer thickness of ~5 km, in agreement with seismic observations ~50km F.Nimmo EART162 Spring 10 Convection Cold - dense • Convection arises because fluids expand and decrease in density when heated Fluid • The situation on the right is gravitationally unstable – hot fluid will tend to rise • But viscous forces oppose fluid motion, so Hot - less dense there is a competition between viscous and (thermal) buoyancy forces • So convection will only initiate if the buoyancy forces are big enough • What is the expression for thermal buoyancy forces? F.Nimmo EART162 Spring 10 Conductive heat transfer • Diffusion equation (1D, Cartesian) T T T H u 2 t z z Cp 2 Advected Conductive component component Heat production • Thermal diffusivity =k/Cp (m2s-1) • Diffusion timescale: 2 t~ d F.Nimmo EART162 Spring 10 Convection equations • There are two: one controlling the evolution of temperature, the other the evolution of velocity • They are coupled because temperature affects flow (via buoyancy force) and flow affects temperature (via the advective term) 2 2 v v v Navier- 2 2 P gaT Buoyancy force Stokes t x z Note that here the N-S equation is neglecting the inertial term 2 2 T T T Thermal K 2 2 v T Evolution t z x Advective term • It is this coupling that makes solving convection problems hard F.Nimmo EART162 Spring 10 Initiation of Convection • Recall buoyancy forces favour motion, viscous forces oppose it d • Another way of looking at the problem is there are two competing timescales – what are they? a Top temperature T0 d Incipient upwelling Hot layer Bottom temp. T1 • Whether or not convection occurs is governed by the dimensionless (Rayleigh) number Ra: ga (T1 T0 )d Ra 3 • Convection only occurs if Ra is greater than the critical Rayleigh number, ~ 1000 (depends a bit onF.Nimmo geometry) EART162 Spring 10 Constant viscosity convection • Convection results in hot and cold boundary layers and an isothermal interior • In constant-viscosity convection, top and bottom b.l. have same thickness • • • • cold T0 T0 (T0+T1)/2 Isothermal interior hot T1 d T1 Heat is conducted across boundary layers Fconv k (T12T0 ) (T1 T0 ) In the absence of convection, heat flux Fcond k d So convection gives higher heat fluxes than conduction The Nusselt number defines the convective efficiency: F conv d Nu Fcond 2 F.Nimmo EART162 Spring 10 Boundary layer thickness d • We can balance the timescale for conductive thickening of the cold boundary layer against the timescale for the cold blob to descend to obtain an expression for the b.l. thickness : a ~ d Ra 1 / 3 • So the boundary layer gets thinner as convection becomes more vigorous • Also note that is independent of d. Why? • We can therefore calculate the convective heat flux: 1/ 3 (T1 T0 ) (T1 T0 ) 1/ 3 k ga 4/3 T1 T0 Fconv k ~k Ra 2 2d 2 F.Nimmo EART162 Spring 10 Example - Earth 1/ 3 k ga •Does this equation 4/3 T1 T0 Fconv make sense? 2 • Plug in some parameters for the terrestrial mantle: =3000 kg m-3, g=10 ms-2, a=3x10-5 K-1, =10-6 m2s-1, =3x1021 Pa s, k=3 W m-1K-1, (T1-T0) =1500 K • We get a convective heat flux of 170 mWm-2 • This is about a factor of 2 larger than the actual terrestrial heat flux (~80 mWm-2) – not bad! • NB for other planets (lacking plate tectonics), tends to be bigger than these simple calculations would predict, and the convective heat flux smaller • Given the heat flux, we can calculate thermal evolution F.Nimmo EART162 Spring 10 Summary • Fluid dynamics can be applied to a wide variety of geophysical problems • Navier-Stokes equation describes fluid flow: u u u 2u P g yˆ t • Post-glacial rebound timescale: t ~ gL • Behaviour of fluid during convection is determined by a single dimensionless number, the Rayleigh number Ra gaTd 3 Ra F.Nimmo EART162 Spring 10 End of lecture • Supplementary material follows F.Nimmo EART162 Spring 10 Thermodynamics & Adiabat • A packet of convecting material is often moving fast enough that it exchanges no energy with its surroundings • What factors control whether this is true? • As the convecting material rises, it will expand (due to reduced pressure) and thus do work (W = P dV) • This work must come from the internal energy of the material, so it cools • The resulting change in temperature as a function of pressure (dT/dP) is called an adiabat • Adiabats explain e.g. why mountains are cooler than valleys F.Nimmo EART162 Spring 10 Adiabatic Gradient (1) • If no energy is added or taken away, the entropy of the system stays constant dQ • Entropy S is defined by dS T Here dQ is the amount of energy added to the system (so if dQ=0, then dS=0 also and the system is adiabatic) • What we want is T at constant S. How do we get it? P • We need some definitions: Q T P mC p Specific heat capacity (at constant P) a V1 VT P Thermal expansivity S P T VT P Maxwell’s identity F.Nimmo EART162 Spring 10 T Adiabatic Gradient (2) • We can assemble these pieces to get the z adiabatic temperature gradient: adiabat dT aT dP C p • NB You’re not going to be expected to reproduce the derivation, but you do need to learn the final result • An often more useful expression can be obtained by converting pressure to depth (how?) a dT agT dz Cp • What are typical values for terrestrial planets? F.Nimmo EART162 Spring 10 Incompressibility & Stream Function • In many fluids the total volume doesn’t change x y u(x) v(y) u(x+x) v(y+y) V1 xy V2 xt[v( y y ) v( y )] yt[u ( x x) u ( x)] a u v u 0 Incompressibility If V1=V2 then condition x y • We can set up a stream function j which automatically satisfies incompressibility and describes both the horizontal and the vertical velocities: j j Note that these satisfy incompressibility u v y x F.Nimmo EART162 Spring 10 Stream Function j • Only works in 2 dimensions • Its usefulness is we replace u,v with one variable j 2u 2u P 2 2 0 x z x u j z 3j 3j P 2 3 0 x x z z 2v 2v P 2 2 0 z x z v jx Check signs here! 3j 3j P 3 2 0 z z x x Differentiate LH eqn. w.r.t. z and RH w.r.t x a 4j 4j 4j 2 2 2 4 0 4 x x z z j 0 4 The velocity field of any 2D viscous flow satisfies this equation F.Nimmo EART162 Spring 10 Postglacial rebound and j (1) • Biharmonic equation for viscous fluid flow 4j 4j 4j 2 2 2 4 0 4 x x z z • Assume (why?) j is a periodic function j=sin kx Y(y) Here k is the wavenumber = 2p/l • After a bit of algebra, we get ky ky ky ky a j sin kx Ae Bye Ce Dye • All that is left (!) is to determine the constants which are set by the boundary conditions – in real problems, this is often the hardest bit • What are the boundary conditions? • u=0 at z=0, v=dw/dt at z=0, u=v=0 at large z F.Nimmo EART162 Spring 10 Postglacial rebound and j (2) • Applying the boundary conditions we get a j Ae kz sin kx1 kz j v • We have dw/dt = x : dw dt kAe kz cos kx1 kz 1 • Vert. viscous stress at surface (z=0) balances deformation: gw zz P 2 vz Why can we ignore this term? • For steady flow, we can derive P from Navier-Stokes 2u 2u P 2 2 0 x z x a P 2Ak 2 cos kx 2 • Finally, eliminating A from 1 and 2 we get (at last!): 1 dw g This ought to look familiar . . . w dt 2k F.Nimmo EART162 Spring 10 Postglacial rebound (concluded) g 1 dw g w w0 exp t w dt 2k 2k • So we get exponential decay of topography, with a time constant depending on wavenumber (k) and viscosity () • Same result as we got with the scaling argument! • Relaxation time depends on wavelength of load • Relaxation time depends on viscosity of fluid F.Nimmo EART162 Spring 10 Convection Cold - dense • Convection arises because fluids expand and decrease in density when heated Fluid • The situation on the right is gravitationally unstable – hot fluid will tend to rise • But viscous forces oppose fluid motion, so Hot - less dense there is a competition between viscous and (thermal) buoyancy forces • So convection will only initiate if the buoyancy forces are big enough • Note that this is different to the Rayleigh-Taylor case: thermal buoyancy forces decay with time (diffusion), compositional ones don’t • What is the expression for thermal buoyancy F.Nimmo forces? EART162 Spring 10 F.Nimmo EART162 Spring 10 Two Dimensions . . . • In 1D, shear stress (now using t) is x txy v y • In 2D, there are three different stresses: txy t xy uy vx t xx 2 ux t yy 2 yv txx u t uy tyy Shear stress Normal stresses • Where do the factors of 2 come from? p(y)x p(x)y y x p(y+y)x • Force due to pressure (x direction, per unit cross-sectional area): p(x+x)y P Fp x a 1 F.Nimmo EART162 Spring 10 Viscous forces on an element (1) x y txy txy v u y x tyy txx • Viscous force (x direction, per unit cross-sectional area): a t xx t xy Fv x y 2 • Total force balance given by viscous + pressure forces 1 + 2 • After some algebra, we get total force in x-direction: 2u 2u P Ft ( x) 2 2 x y x a Note that force in x-direction only depends on velocity in x-direction and the x-gradient of pressure F.Nimmo EART162 Spring 10 Viscous forces on an element (2) • In the y-direction, body forces can also be important Fb g • Otherwise, the analysis is the same as before 2 2 P v v Ft ( y) 2 2 g y x y • We can use F=ma to derive the response to this force Du ma V What does this mean? Dt • So the equations of motion in x and y directions are 2u 2u Du P 2 2 Dt x y x 2v 2v Dv P 2 2 g Dt y x y F.Nimmo EART162 Spring 10 Putting it together 2u 2u Du P 2 2 • x-direction Dt x y x 2v 2v Dv P • y-direction 2 2 g Dt y x y Pressure gradient • Special cases: Viscous terms Body force – Steady-state – Du/Dt=0 – One-dimension (e.g. v=0, u only varies in y direction) 2u Du P 2 Dt x y F.Nimmo EART162 Spring 10