Transcript Eart162 - UCSC Earth & Planetary Sciences

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
•
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•
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•
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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
velocity gradient u / d (= uy )
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   uy
• Let’s assume u only varies in the y direction
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):
Fp  
P
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
yˆ 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 6pur, first derived by George Stokes in
1851 (apparently under exam conditions)
By balancing the drag force against the excess weight of the
sphere (4pr3g/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
u2  u1
l
b
b
u 2 u1l
2   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
 2T H
u
 2 
t
z
z
Cp
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 Advective term
Evolution t
z 
 x
• 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 (T12T0 )
(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 
4/3
•Does this equation
 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

2
   u  u     u  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
gaTd 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)
V1  xy
v(y)
u(x+x)
v(y+y)
V2  xt[v( y  y)  v( y)]
 yt[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  jz
  3j
P
 3j 

   2  3   0
x
 x z z 
  2v  2v 
P

   2  2   0
z
z 
 x
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 kx1  kz
j
v


• We have dw/dt =
x :
dw
dt
 kz
 kAe
coskx1  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  2Ak 2 coskx 2
• Finally, eliminating A from 1 and 2 we get (at last!):
1 dw g
This ought to look

familiar . . .
w dt 2k
F.Nimmo EART162 Spring 10
Postglacial rebound (concluded)
 g 
1 dw g

 w  w0 exp
t
w dt 2k
 2k 
• 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
 x y 
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
 x y 
  2v  2v 
Dv
P


   2  2   g
Dt
y
 x y 
F.Nimmo EART162 Spring 10
Putting it together
  2u  2u 
Du
P
• x-direction 

   2  2 
Dt
x
 x y 
  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