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1
Chapter 6 Fluid Mechanics and Convection 1. Introduction
Any material that flows in response to an applied stress deform
continuously. In solids, stresses are related to rates of strains.
In fluids, stresses are related to rates of strains. Strain rates
in fluids are a result of gradients in the velocities or rates of
displacement of fluid elements. Velocity gradients are equivalent
to strain rate. Newtonian or linear fluid: the rate of strain or
velocity gradient is directly proportional to the applied stress;
the constant proportionality is viscosity. The physics that governs
fluid motions is the principles of mass, momentum, and energy
conservation together with the rheological or constitutive law for
the fluid.
In geodynamics, mantle convection is the manifestation of the
fluid behavior of
the mantle and is responsible for plate tectonics and
continental drift; it plays a dominant role in determining the
thermal structure of the earth. Thermal convection: when a fluid is
heated from within or from below and cooled from above, thermal
convection can occur. The hotter fluid at depth is gravitationally
unstable with respect to the cooler fluid near the upper surface.
Buoyancy forces drive the convective flow. On many scales folding
of crustal rocks can be attributed to the fluid behavior of these
rocks. A fluid instability can also explain the formation of
saltdomes due to diapiric upwelling of a buried layer of salt. 2.
1-D channel flows
The movement of the plates over the surface of the surface of
the earth represents a flow of mantle rock from accretionary plate
boundaries to subduction zones. To maintain mass balance, a
complementary flow of mantle rock from subduction to accretionary
plate boundaries must occur at depth.
A model for this counter flow is asthenospheric flow confined in
a horizontal layer immediately below the lithosphere. Postglacial
rebound data suggest the pressure of a low-viscosity (of the order
of 100 km thick) region beneath the oceanic lithosphere. Seismic
studies show that a region beneath the lithosphere has low seismic
velocity, particularly that shear waves are attenuated. All there
observations favor a low-viscosity region beneath the
lithosphere.
The flow occurs as a result of pressure gradient A10 pp ( A :
the horizontal length of a
section of the channel) or the prescribed motion of one of the
walls 0u .
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2
Shear stress (force per unit area), or gradient in velocity, is
exerted on the horizontal planes in the fluid and at the walls. For
a Newtonain fluid with constant viscosity :
dydu = -----
u : velocity of fluid ( sm ),
: shear stress, tangential stress on the surface normal to y
direction ( 2mN ),
dynamic : viscosity SI unit is Pa s or CGS unit is poise,
kinematic viscosity s
m2 :unit== (SI unit), scm
2 (CGS unit).
The kinematic viscosity is a diffusivity similar to the thermal
diffusivity . hile describes how heat diffuses by molecular
collisions, describes how momentum diffuses.
Prandtl number : Pr (dimensionless quantity)
A fluid with small Prandtl number diffuses heat more rapidly
than it does momentum; the reverse is true for large value of
Pr.
Consider the force balance applied on the element of a fluid as
shown in figure 6-1. (1). The net pressure force on the element in
x direction is ( ) ypp 01 (force per unit length the channel in the
direction normal to the page).
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3
(2). Shear force resulting from velocity gradient is a function
of y. On the upper boundary at y: - shear force in the x direction:
- ( )Ay . On the lower boundary at yy + :
( ) ( ) AA
+=+ y
dydyyy
entranceat pressure :1p
exitat pressure :0p
The net forces from (1) + (2) applied on the element must be
zero if the fluid motion is not accelerated, so
,
( )A
01 ppdyd = , ----
dxdp : horizontal pressure gradient if 0A , where == dx
dpppAA
10
0lim horizontal
pressure gradientdxdp
dyd = . --- momentum equation
When 01 pp > , a pressure difference tends to move the fluid
in the +x direction, the pressure gradient is negative. The
pressure drop in a channel is often expressed in
terms of a hydraulic head H ( )
gppH
01 . The hydraulic head is the height of fluid required to
hydrostatically provide the
( ) ( ) ( ) 001 =
++ AA yydydyypp
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4
applied pressure difference 01 pp . Plug eq
dydu = (constitutive law) into eq ,
dxdp
dyud =2
2
------ Integrating gives
212
21 cycy
dxdpu ++= ------
To solve constants 21 , cc , we must satisfy the boundary
conditions that 0=u at hy = (lower plate is fixed equivalent to
no-slip boundary conditions), and 0uu =
at 0=y . Therefore, eq becomes ( ) 00221 uhyuhyydxdpu +=
------
In Fig. 6-2 (a), if 01 pp = , no pressure gradient, 0=dxdp ,
=
hyuu 10 , the
solution reduces to a linear velocity profile, known as Couette
flow.
In Fig 6-2 (b), if 00 =u , (upper plate is no-slip), ( )hyydxdpu
= 221 . If this is rewritten in terms of y , distance measure from
the center of the channel;
i.e., 22hyyhy == , then
=
421 22 hy
dxdpu .
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5
The velocity profile is parabola that is symmetric about 2hy =
.
3. Asthenospheric counter-flow One model for the flow in the
mantle associated with the movement of the surface plates is a
counter-flow beneath the lithosphere as shown in fig. 6-4.
The lithosphere is assumed to be a rigid plate of thickness Lh ,
moving with velocity
0u . Beneath is the asthenosphere of thickness h and uniform
viscosity . At the base of the asthenosphere, we assume the mantle
is stationary; that is, 0=u at
hy = . At the top of asthenosphere or the base of lithosphere,
0uu = at 0=y . Conservation of mass requires that the flow of
material in the +x direction in the lithosphere must be balanced by
a counter flow in the asthenosphere; that is,
=+ hL udyhu 00 0 ----- Flux of material Flux of material (per
unit distance normal to the page) in the lithosphere in the
asthenosphere By substituting eq of u into eq
02120
3
0 =+ hudxdphhu L , -----
+=2112
20
hh
hu
dxdp L . -----
eq:
( ).hu
dxdphhu
hyuhyy
dxdpudy
,uh
yuhyydxdpu
hhh
21222321
21
03
00
20
0
22
0
002
+=+
=
+=
Substitute eq into eq ,
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6
++=
hy
hy
hh
hyuu L 2
2
0 2161 shown in fig. 6-4. -----
Shear stress LA on the base of the lithosphere at 0=y due to the
counter-flow in the asthenosphere is
+== = hh
hu
dydu L
yLA 322 0
0 -----
The sign in eq indicates that the asthenosphere exerts a drag
force on the base of the lithosphere tending to oppose its
motion.
e.g. ,50,200,100,104 019
yrmmukmhkmhsPa L ==== we get
)22(2.2 barsMPaLA = . The asthenospheric counter-flow requires
that the pressure gradient ( 0>dxdp ), increase with x; i.e., p
must increase in the +x direction i.e. the direction of seafloor
spreading. This increase in pressure with distance from a ridge
could be provided by a hydrostatic head associated with topography.
The ocean floor would have to rise with distance from the
ridge.
The pressure in the asthenosphere at a distance b beneath the
ridge is given by hydrostatic formula:
( )bwwggwp rw ++= ----- w : sea water density,
w : the depth of ocean floor at a distance x from the ridge, :
density of mantle,
rw : ocean floor depth at the ridge.
Differentiate eq with respect to x, ( )dxdwg
dxdp
w = .
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7
Positive dxdp requires a negative
dxdw or an ocean depth that decreases with x.
The slope of the seafloor dxdw is
( ) ( )
+== 2
11212
0
hh
ghu
dxdp
gdxdw L
ww
. -----
e.g. 4233 102.7,10,3300,1000 ==== dxdwsmgmkg
mkg
w . Across the Pacific Ocean, km,x 00010= , this would give a
decrease in depth or
km2.7 . No systematic decrease in ocean depth is observed. The
gravity anomaly predicted by asthenspheric counterflow has also not
been observed in the Pacific. We conclude that the shallow
counterflow model for mantle convection is not correct and that
significant converctive flow occurs beneath the asthensphere. 4.
Pipe flow applications to flows in aquifers and volcanic
conduits.
Consider viscous flow through a circular pipe with a radius R
and a length A , the flow is driven by the pressure difference ( 01
pp ) applied between the sections at a distance A apart. Assume
that the velocity of the fluid along the pipe u depends only on
distance from the center of the pipe r. Find ( )ru by writing a
force balance on a cylindrical control volume of radius r and
length A . (1). The net pressure force on the ends of the
cylindrical control volume is
( ) 201 rpp (a force along the cylinder axis in the direction of
flow). (2). Shear force acting on the cylindrical surface of
control volume, results from the shear stress on the cylindrical
surface ( )r exerting a net frictional force ( )rr A2 on the
control volume. If the flow is steady, not accelerated, the forces
from (1) + (2) must be equal to zero; that is:
( ) Arppr 2012 = , dxdppp = A
01 (pressure gradient along the pipe),
dxdpr
2= . -----
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8
In cylindrical geometry, the shear stress is directly
proportional to the radial gradient of the velocity u :
drdu = . -----
Substitutes into
dxdpr
drdu
2= . ----- Integrate and apply the boundary condition 0=u at Rr
= to give:
( )2241 rR
dxdpu = -----
The velocity profile in the pipe is a parabaloid of revolution,
is known as Poiseuille flow. The maximum velocity in the pipe umax
occurs at 0=r .
dxdpRU 4
2
max = -----
dxdppp =A
10 is negative when 01 pp > , and umax will be positive. The
volumetric flow rate Q through the pipe is the total volume of
fluid passing a
cross-section per unit time. The flow through an annulus of
thickness dr and radius r occurs at the rate ( )rrdru2 ; Q is the
integral of this over a cross section,
drruQR= 0 2 . -----
Substitute into and carry out the integration,
dxdpRQ
8
4= . ----- If we divide Q by the cross-sectional area of the
pipe 2R , we obtain the mean velocity u :
dxdpRu 8
2
= . ----- Compare Umax in and u in :
max21 uu = .
The mean and maximum velocities in the pipe are directly
proportional to the pressure gradient and inversely proportional to
the viscosity . The result is valid as long as the flow is
laminar.
In fluid mechanics, we often work in terms of dimensionless
variables. Introduce two quantities: a dimensionless pressure
gradient or frictional factor f, and Reynolds number Re to rederive
the relation .
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9
the frictional factor: f dxdp
uR
4 , -----
Reynolds number: Re Du , -----
where D = 2R is the pipes diameter. Use & , we can rewrite
eq as
Re64=f -----
The inverse dependence of f on Reynolds # Re: At sufficiently
higher Re, observed pressure drops become considerably higher than
those given by laminar theory. The flow in the pipe becomes
unsteady with random eddies. This is known as turbulent flow.
The advantage of the dimensionless formulation of the problem is
that the transition to turbulent flow occurs at Re 2200 independent
of the pipe radius, flow velocity, or type of fluid considered (
and ). The mean velocity u corresponding to Re
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10
2200 is 22 smm for water with viscosity sPa = 310 flowing in a
pipe of 0.1 m diameter. This illustrates that most flows of liquids
and gases are in the turbulent regime. No theoretical equivalent to
the Newtonian relation between shear stress and rate of strain as
given in eq . For turbulent flow, it is found by empirically
41
Re31640= .f . -----
5. Artesian aquifer flows
Natural springs are usually due to the flow of groundwater from
a high elevation to a low elevation. The flow takes place through
an aquifer or permeable formation.
In an idealized model for an aquifer in the shape of a
semicircle of R . The entrance of the aquifer lies a distance b
above the exit and its cross section is assumed to be circular with
radius R. The hydrostatic pressure head available to drive flow
through the aquifer is gb , where is the density of water. Since
the overall length of the aquifer is R ( bR >> ), the driving
pressure gradient is
Rgb
dsdp
=
, ----- where s is the distance along the aquifer. The
volumetric flow rate produced by this pressure gradient can be
calculated from eq if the flow is laminar.
Substitute into for dxdp replaced by
dsdp ,
RgbRQ =
8
4
. -----
If the flow is turbulent, we can determine Q by using the
empirical relation between f and Re. Recast eq into dimensional
from:
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11
41
2 23164.04
=
Rudxdp
u
R
. -----
f 1Re The result of rearranging eq so as to determine u is
71
757
474
41
13164.0
24
=
Rdxdpu . -----
Since uRQ 2= , we obtain the volumetric flow rate through the
aquifer for turbulent
flow by multiplying by 2R and substituting for 1 1dp dp gbdx ds
R
= = in eq ,
7197
174
686.7 RRgbQ
= . -----
6. Flow through volcanic pipes.
Another natural example of pipe flow is the flow of magma
through volcanic conduits of nearly circular cross section. The
upward flows of magma is driven by buoyancy of the lighter magma
relative to the denser surrounding rock. At the same depth h, the
lithostatic pressure in the surrounding rock is ghs ( s : density
of rock). The hydrostatic pressure in a stationary column of magma
is ghA ( A : magma density), assuming that the lithostatic and
hydrostatic pressures are equal in the pipe; that is, the walls of
the pipe are free to deform as the magma is driven upward, then the
pressure gradient available to drive the magma up to the
surface
( )gs A . The volumetric flow Q driven by the above pressure
gradient through a volcanic pipe of radius R is from eq in page
8.
( )
44
88gR
dxdpRQ s A== ( ( )g
dxdp
s A = ), -----
if flow is laminar. From eq and uRQ 2= , the volumetric flow for
turbulent
is: ( )[ ]
71
73
74
719
8.14
A
A gRQ s = . -----
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12
7. Conservation of fluid in 2-D Consider a viscous fluid flowing
to 2-D plane x, y the corresponding velocity
components in x & y direction are u and v.
Consider a rectangular element with dimensions x and y as seen
in fig. 6-10. The flow rate per unit area at xx + is
( ) ( ) xxuxuxxu
+=+ . The net flow rate out of the region between x and xx + per
unit area normal to the x direction is:
(1). xxuux
xuu
=+ .
Similarly, the net flow rate per unit area normal to y direction
out of the region between y and yy + is: (2). y
yvvy
yvv
=+
Therefore, the net rate at which fluids flow out of the
rectangular region yx = )1( area across which flow occur( y= ) +
)2( area across which flow occur( x= )
xyyvyx
yu
+= . -----
The total net outward flow rate per unit area of the rectangular
is
yv
xu
yx
xyyvyx
yu
+
=+
.
If the flow is steady (time-independent), and there are no
density variations to consider, there can be no net flow into or
out of the rectangle. The consideration of fluid or continuity
equation is
0=+
yv
xu . -----
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13
In 3-D, 0u v w ux y z
+ + = = GG
, where ( )wvuu ,,= , , ,x y z
= G
This is appropriate for an incompressible fluid. (For
compressible fluid, ( ) 0u =GG .) 8. Force balance in 2-D
Force acting on the control volume; i.e. the rectangular element
yx must be in balance. Include in force balance are the pressure
forces, viscous forces and gravity force. At the moment, we neglect
the inertial force associated with the acceleration of a fluid
element. This is appropriate for the slow motion of very viscous or
high Prandtl number fluids. The earths mantle behaves as a highly
viscous fluid on geological time scales. The viscosity of the
mantle is about 1021 Pas; its density and thermal diffusivity and
about 4000 3m
kg and 1 smm2 . The Prandtl number
of the mantle is about 1023. Therefore, we can ignore the
inertial force.
The balance of pressure, viscous, and gravity forces and the
neglect of inertial force are based on the Newtons second law of
motion with the neglect of its acceleration. Its equivalent to the
conservation of momentum. (1). Pressure Force:
The net pressure force acting on the element yx in the
+x-direction per unit area of the fluid element
( ) ( ) ( ) ( )xp
xxpxxp
yxyxxpyxp
=+=+=
.
using Taylors series: ( ) ( ) xxpxpxxp
+=+
pG
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14
The net pressure force on the element in +y-direction per unit
area of the fluid element
( ) ( ) ( ) ( )yp
yypyyp
yxxyypxyp
=+=+=
. (2). Gravitational Force:
The gravitational body force on a fluid element is its mass
times the acceleration of gravity. The mass for the element yxyx =
. g is the force of gravity per unit area of the element and the
gravity acts in the positive y direction. Thus the net gravitation
force per unit area of the element in the +y direction is
g . (3). Viscous Forces:
The viscous forces acting on the element both parallel and to
the surface of
the element. xx and yy are viscous normal stresses, the viscous
forces per unit area that act to the elements surfaces. The
stresses are positive in the directions shown in fig. 6-12.
If there is no net torque about the center of the element,
yxxy = . The net viscous force in the x-direction per unit
cross-sectional area of the element is
( ) ( ) ( ) ( )yx
yyxyyyx
yxyxx yxyxxxxx
+++ .
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15
Using Taylors series, the above equation simplifies to
yxyxxx
+
.
Similarly, the net viscous force in the y-direction per unit
area is xyxyyy
+
.
For an idealized Newtons fluid, the viscous stresses are
linearly proportional to the velocity gradients. Constitutive
Law:
xu
xx = 2 ,
yv
yy = 2 ,
+
==xv
yu
xyyx . ----- The total normal stress is the sum of the pressure
and the viscous stress; that is:
xupp xxxx
== 2 , -----
yvpp yyyy
== 2 . -----
The minus sign in front of xx and yy are the result of the
opposite sign conventions adopted for (+ for compression) and (+
for extension). The viscous stress is the only contribution to the
shear stress. Use eq to rewrite the viscous forces acting on the
element of the rectangular,
+
+=
+
yxv
yu
xu
yxyxxx
2
2
2
2
2
2 -----
+
+=
+
xyu
xv
yv
xyxyyy
2
2
2
2
2
2 -----
Applying the continuity equation: 00 =
+
=
+
=
+
yv
xu
yyv
xu
xyv
xu
2
22
xu
yxv
=
-----
2
22
yv
yxu
=
----- Use eqs and to replace the mixed derivative terms in eqs
& , we get:
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16
x-direction:
+
2
2
2
2
yu
xu ,
y-direction:
+
2
2
2
2
yv
xv .
the net viscous forces per unit cross-sectional area in x and y
direction, respectively. In 3-D:
+
+
2
2
2
2
2
2
zu
yu
xu x-direction
+
+
2
2
2
2
2
2
zv
yv
xv y-direction 2u G
+
+
2
2
2
2
2
2
zw
yw
xw z-direction
The force balance equations for an incompressible fluid with
very large viscosity undergoing steady flow in 2-D are obtained by
adding the pressure, gravity and viscous forces together and
equating their sum to zero.
For x-direction,
+
+= 2
2
2
2
0yu
xu
xp , -----
y-direction,
+
++= 2
2
2
2
0yv
xvg
yp , -----
and gravity acts only in y-direction. In order to eliminate the
hydrostatic pressure variation in equation , we
introduce gypP = , -----
where the pressure P is the pressure generated by fluid flow.
Substituting into yields:
+
+= 2
2
2
2
0yu
xu
xP , -----
+
+= 2
2
2
2
0y
vxv
yP . -----
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9. The stream function The continuity equation in 2-D for an
incompressible fluid can be satisfied if we
introduce a stream function defined such that.
yu
= , x
v = . -----
Substitution of into 0=+
yv
xu (eq ) yields
022
=+
xyyx
. ----- Substitution of into and ,
+
+= 3
3
2
3
0yyxx
P , -----
3 3
3 20Py x y x
= + + . -----
Eliminate the pressure from these equations and obtain a single
differential equation for if we take the partial devirative of eq
w.r.t. y and the partial devirative of w.r.t. x and then add them
together,
4
4
22
4
4
4
04948yyxxxy
++
=>=
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The flows across AB can be calculated from the flows across AP
and PB. Since the mass consideration requires that no net flow is
into or out of the triangle PAB. The volumetric flow rate across AP
into the triangle per unit distance normal to the figure is yu ;
similarly, the flow rate across PB out of the triangle is vdx . The
net flow out of PBA is thus xvyu + ; this must be equal to the
volumetric rate of flow (per unit distance in the 3rd dimension)
into PAB across AB. In terms of the stream function , xvyu + can be
written:
dxx
yy
xvyu =+
=+ . Thus, the small difference d is the volumetric flow rate
between any two points separated by the infinitesimal distance s .
If the points are separated by an arbitrary distance, the integrate
of d between the points,
AB
B
Ad = , -----
gives the volumetric flow rate between the points, i.e. the
difference between the volumes of the stream function at any two
points is the volumetric rate of flow across any line drawn between
the points. 10. Postglacial Rebound
Important information in the fluid behavior of the mantle comes
from studies of the dynamic response of the mantle to loading and
unloading at the surface. Mountains depress the underlying
crust-mantle boundary. The mountain building process is so slow
that dynamic effects can be neglected; i.e. the mantle beneath a
mountain is essentially in hydrostatic equilibrium throughout the
cycle of the mountain. However, the growth and melting of ice
sheets occur sufficiently fast so that dynamic effects are
important in the adjustment of the mantle to the changing surface
load. The thick ice sheet that covers GreenLand has depressed the
surface level several kilometers below the sea level. The load of
ice has forced mantle rock to flow laterally, allowing the earths
surface beneath the ice to subside. When the ice melted about 10000
years ago, the surface rebounded. The rate of rebound can be
determined by dating elevated beaches.
To determine the response of the mantle to the removal of an ice
load, we consider
the flow in a semi-infinite, viscous fluid half-space ( 0>y )
subjected to an initial periodic surface displacement given by
xww mom 2cos= ,
where is the wave length and
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19
The displacement of the surface w loads to a horizontal pressure
gradient due to hydrostatic load. When the surface is displaced
upward (negative w), (w0). The fluid is in driven away from the
region as the displacement w decreases. When the surface is
displaced downward (w>0), the pressure is negative (p
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20
The return of the surface to an undeformed (w = 0) state is
governed by the viscous flow in half-space. The flow can be
determined by solving the biharmonic equation for the stream
function. Since the initial surface displacement is of the form
x2cos , must also vary periodically with x. we can assume the
solution of is
an arbitrary combination of x2sin and apply the method of
separation of variable
and take
( )yYx 2sin= , -----
where ( )yY is to be determined. By substituting into the
biharmonic equation , we obtain
02224
2
2
4
4
=
+
Ydy
Yddy
Yd
. -----
Solution of the constant coefficient differential equation for Y
are of the form ( )myY exp . -----
Substituting into , leads to the solution of m.
0222222
24
22
4 =
=
+
mmm
or 2=m . -----
The two values of m provide two possible solution for Y:
y2exp and
y2exp . ----- The differential equation is 4th order, the
solutions of y should be four; therefore, two additional solutions
are required. It can be verified by direct substitution that
yy 2exp and
yy 2exp . ----- The general solution for Y is the sum of these
four solutions from and ,
+++=
yyyy
DyeCeByeAexsin22222 ,
----- where A, B, C and D are determined by the appropriate
boundary conditions. First, we require the solution to be finite as
y so that C = D = 0. The solution for simplifies to
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21
( )ByAex y +=
22sin . ----
The velocity components u and v can be obtained by
differentiating according to , we find
( )
+==
BByAex
yu
y
22sin
2
, -----
( )ByAexx
vy
+==
22cos2 . -----
Since the part of mantle that behaves like a fluid is overlain
with a rigid lithosphere, we force the horizontal component of the
velocity u to be zero at wy = ; i.e., the no-slip boundary
condition is applied at the upper boundary of the fluid half-space.
Because the vertical displacement w of this boundary is small,
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22
xA
xp 2sin22
3
= on 0=y . -----
Integrating w.r.t. x to give:
xAp 2cos22
2
= on 0=y . -----
yv on 0=y for eq can be found by differentiating eq w.r.t. y
and
then evaluate the result on 0=y . We get
00
=
=yyv . -----
Condition thus simplifies to
xgAw y
2cos222
0
== . ----- The surface displacement w is related to the velocity
fluid by the fact that the time derivative of w is just the
vertical component of the surface velocity; i.e.,
wywy vtw
== =
. -----
Since
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23
displacement, is given by
ggr44 == . -----
The viscosity of the mantle can be established from eq once the
relaxation time for postglacial rebound has been determined.
Quantitative information on the rate of postglacial rebound can
be obtained from elevated beach terraces. Wave action over a period
of time erodes a beach to sea level. If sea level drops or if the
land surface is elevated, a fossil beach can be obtained by
radioactive dating using C14 in shells and driftwoods. The uplift
of the beach terraces is compared with the exponential time
dependence given in eq. Assume that uplift began 10000 years ago.
And mwm 3000 = with m30 of uplift to occur in the future; i.e. mw
30= at 410=t year later (the present).
Previously we have only considered the response to a spatially
periodic surface displacement. Since the problem is linear,
solutions can be superimposed to consider other distribution of
surface displacement. More complete studies of postglacial rebound
include the flexural rigidity of the elastic lithosphere and a
depth dependent viscosity. If the ice sheets continue to melt
during rebound, the sea level increase must be taken into account.
Table 6-2 summarizes the ocean mantle viscosity from rebound
studies. 11. Angle of subduction
As discussed in Chapter 3 (section 17), the oceanic lithosphere
bends as its subducted at a trench. The gravitational body force on
the descending lithosphere is
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24
directed vertically downward. It might be expected that under
this force the lithosphere would bend through 090 and descend
vertically into mantle. However, its observed that oceanic
lithosphere straightens out after subduction and descends at a
finite angle of dip .
One explanation for why the lithosphere descends at an angle
other than 090
(Table 6-3) is that pressure forces due to the induced flows in
the mantle balance the gravitational body forces. The pressure
forces are due to the mantle flow induced by the motion of the
descending lithosphere, they are flow pressures relative to the
hydrostatic pressure. The dip of the lithosphere is thus a
consequence of the balance between the gravitational torque and the
lifting pressure torque.
The pressure forces acting on the lithosphere can be calculated
using 2-D viscous
corner flow model in fig 6-18. The trench is at x = 0. Assume
the surface
0,0 = xy is stationary. The descending lithosphere is the line
extending from 0=x down to the positive x-axis at angle of dip. the
velocity parallel to this line is
U . Distance measured along this line is r . The line divides
the viscous mantle into two corners, the arc and the ocean corner.
The motion of this line divides a flow in the arc corner. The
velocities of the dipping line and the surface induce a flow in he
oceanic corner.
-
25
To solve for the motions in both corners and the flow pressures
on the dipping line, we use the stream functions for the corner
flows that satisfy the biharmonic equation ( 04 = ). For the corner
flow geometry, we write in the form:
( ) ( )
+++= xyDyCxByAx 1tan . -----
where A, B, C, and D are constants determined by boundary
conditions. There are two stream functions with distinct values of
these constants. In the arc and oceanic corner which have different
angles and conditions on their bounding lines. The velocity
corresponding to the stream functions can be obtained from eq
( )1 2 2tan x xu B D Cx Dyy y x y = = + + + , -----
( )1 2 2tan x yv A C Cx Dyx y x y = = + + + + . -----
Recall that:
22
2
21 1
1
1tanyx
xx
xyx
yy +=+
=
, 1 2 2 2 2
2
1tan1
y y yyx x x x yx
= = + +.
The pressure can be found by substituting into and integrating
the
resulting expression for xP
(from
2 2
2 2 0P u ux x y
+ + = ). Alternatively, eq
and can be used to get Py
(from
2 2
2 2 0P v vy x y
+ + = ). Therefore,
( )( )222 yx DyCxP + += . ----- The pressure in is the pressure
relative to the hydrostatic pressure; i.e., the
-
26
pressure associated with flow. General expression for the
constants of integration are complicated, here we
evaluate them for a particular value of the dip angel. For a dip
of 4 , representative
Ryukyu arc, the boundary conditions for the arc corner are:
0== vu on 0,0 >= xy , or 0tan 1 =xy , ----
22Uvu == on xy = , or
4tan 1 =
xy .
Applications of these conditions lead to the expressions for C
and D in the arc corner.
=
422
22
UC and
=
42
222
2
UD . -----
Thus, the pressure in the arc corner is:
( )[ ]( )222
42
42
yx
yxUPcornerarc
+
+= . -----
On 2
2ryx == , the flow pressure on the top of the descending slab
is:
rU
r
UP 558.8
42
42
=
= . -----
The negative value of the flow pressure on the top of the
descending slab gives the effect of a suction force tending to lift
the slab against the force of the gravity.
The pressure force varies as r1 along the upper surface of the
slab and therefore has
a singularity in the idealized model as 0r . However, the
lifting torque on the slab is the integral of the product rP over
the upper surface of the slab. The lifting torque per unit distance
along the top of the slab is a constant, the torque on the slab is
thus proportional to its length.
The boundary conditions for the oceanic corner are:
Uu = , 0=v on 0,0
-
27
+
+
=4
92
3
231
222
49
2
2
UC , -----
+
+
=2
3122
3222
49 2
UD .
The flow pressure in the oceanic corner is from substituting
into ,
rU
rUP
462.02
49
4232 =
= -----
where the flow pressure P is evaluated on the bottom of the
descending slab. The positive value of P means the induced pressure
on the bottom of the slab also exerts a lifting torque on the slab.
The torque per unit distance along the slab is a constant. The net
lifting torque on the slab is the sum of the torques exerted by
pressures on the top and bottom of the slab. Comparison of and
shows that the torque exerted by the suction pressure in the arc
corner far outweighs the lifting effect of pressure on the bottom
of the slab. 12. Diapirism Diapirism: the buoyant upwelling of
relatively light rocks that rise into the heavier overlying rock;
e.g., salt dome. Initially a layer of salt is deposited at the
surface by evaporation of seawater. Subsequent sedimentation buries
this layer under other heavier sedimentary rocks such as shales and
sandstones. At shallow depth, the strength of the salt layer is
sufficient to prevent gravitational instability from the inducing
flow. As the depth of the salt layer increases, the temperature of
the salt increases because of the thermal gradient. Thermally
activated creep process allow the salt to flow upward to be
replaced by the heavier overlying sedimentary rocks. Eventually the
upward flow of the salt creates salt domes.
The deformation of the rocks above salt domes results in the
formation of impermeable traps for the upward migrating oil and
gas. Other examples of diapirism are in the mountain belt where the
heat flow is high and volcanism heats lower crustal rocks to
sufficiently high temperature so that they can freely flow by
solid-state creep. If the heated rocks at depth are lighter than
the overlying rocks, the deeper rocks will flow upward to form
diapirs.
-
28
We apply the same type of analysis used in postglacial rebound
study to investigate diapirism. Consider the geometry in fig 6-21.
A fluid layer with thickness b and density 1 overlies a second
fluid layer of thickness b but with density 2 . Both fluid layers
have viscosity . The upper boundary of the top layer and the lower
boundary of the bottom layer are rigid surfaces. Take 21 > , the
gravitational instability of heavy fluid overlying light fluid is
known as the Rayleigh-Taylor instability.
Take the undisturbed interface between the superposed fluid
layers to be
byy == ,0 and by = are the upper and lower rigid boundaries. As
a result of
-
29
gravitational instability, the displacement of the disturbed
fluid interface is denoted by
w . Assume that w is given by xww m 2cos0= . The stream function
1 for
the flow in the upper layer has the form of eq . We rewrite eq
using hyperbolic functions instead of exponentials,
+++=
yyDyyCyByAx 2sinh2cosh2sinh2cosh2sin 11111 -----
(2
sinhxx eex
= , 2
coshxx eex
+= ). The stream function 2 for the lower layer is:
+++=
yyDyyCyByAx 2sinh2cosh2sinh2cosh2sin 22222 . ----
The velocity in the layer is found by differentiating 1 and 2
based on eqs ,
+++
++=
yCyDByDyCAxu 2cosh
22sinh
22sin2 1111111 , -
( ) ( )
+++=
yyDByyCAxv 2sinh2cosh2cos2 11111 , -----
+++
++=
yCyDByDyCAxu 2cosh
22sinh
22sin2 2222222 ,
( ) ( )
+++=
yyDByyCAxv 2sinh2cosh2cos2 22222 . -----
We evaluate constants of integration by applying the boundary
conditions of no-slip on by = ; that is,
011 == vu or by = , 022 == vu or by = ,
and the continuity of u and v across the interface. For small
displacements of the interface
-
30
22tanh
22
222
22CbDBbDbCA +=
+ ,
( ) 2222 2tanh bCAbbDB = . -----
Shear stresses must also be continuous across the interface
between the fluid layers. For
-
31
The expression for 2 is obtained by replacing y with y . The
time rate of change of the interface displacement
tw
must be equal to the
vertical of the fluid at the interface. Since
-
32
into
( )
+
==
bbbbAP y 2cosh2sinh
12
22 101
xbbbb
2cos2tanh22cosh2sinh
1
1
2
. -----
By carrying through the same procedure using 2 , we find: ( ) (
) 0102 == = yy PP . -----
becomes ( ) ( ) 0121 2 == yPgw . ----- Eq shows that with the
heavy fluid above light fluid ( )21 > , a downward displacement
of the interface 0>w causes a negative pressure in the upper
fluid layer. This tends to produce a further downward displacement
of the interface leading to instability of the configuration. Upon
substituting eq into , we get
( )
+
=
bbbx
bAgw
2cosh2sinh
12
2cos24 121
1
2 2tanh22cosh2sinh
1
bbbb
. -----
By solving for 1A and substituting the resulting expression into
, we get :
( ) wbb
b
bbb
bgb
tw
+
=
2cosh2sinh
12
2cosh2sinh
12tanh2
4
2
21 . ------
The solution of this equation is
-
33
a
t
eww 0= . ------
with ( )
+
=
bbb
b
bbb
gba
2cosh2sinh
12tanh2
2cosh2sinh
12
4
221
. -----
The quantity a is the growth time ( for 21 > ) of a
disturbance. Fig. 6-23 is a plot of the dimensionless growth time (
) 421 agb as a function of the dimensionless disturbance wave
number
b2 .
If 21 > (heavy fluid overlies on top), the interface is
always unstable; i.e,
0>a . If 21 < (light fluid overlies on top), the interface
is stable because a is negative for all . For large , reduces
to
( )2
1 2
24, 2a gb b
. -----
For every small
-
34
( )1 24 20, a
bgb
. -----
When the heavy fluid lies on the top and the configuration is
unstable, the disturbance with the shortest time constant grows and
dominates the instability. The wavelength that gives the smallest
value for a is:
. b568.2= ----- The rate of growth of this dominant disturbance
is obtained by substituting into with the result.
( )gba 2104.13
= ----- The instability takes longer to grow, the more viscous
the fluid and the smaller the density difference. Though we only
consider the stability problem for small displacements, its
expected that the wavelength of the most rapidly growing small
disturbance closely corresponds to the spacing between fully
developed diapers. A example in N. Germany (fig 6-24). The depth to
the salt large is about 5 Km, and the spacing of the salt dome is
about 10-15 km, in agreement with eq .
-
35
13. Folding Folds are found in both sedimentary and metamorphic
rock on all scales.
Folding occurs under a wide variety of conditions, but is often
associated with compressional tectonics. At relatvely low
temperature, sedimentary rocks flow to produce folds rather than
fracture, pressure solution creep is thought to be a major role.
Sedimentary rocks are often saturated with water. When the
differential stresses are applied to the rock, the minerals
dissolve in regions of high stress and are deposited in regions of
low stress. The result is the deformation of the rock. Pressure
solution Creep of sedimentary rocks can result in a linear
relationship between stress and strain rate, therefore, a Newtonian
fluid behavior. Folds usually form in a pre-existing layered
structure with considerable variation in the material properties of
adjacent layers. If a uniform medium is subjected to compression,
it will be uniformly squeezed as fig 6-25(a). If the medium is
composed of a series of week and strong layers, folding will occur
as fig 6-25 (b). The strong layer is referred to as being
competent; the week layer is referred to being incompetent.
One approach to the quantative study of folding is to consider
an elastic (competent) layer of thickness h embedded between two
semi-infinite Newtonian viscous fluids (incompetent). An end load P
on the elastic layer causes it to buckle.
-
36
The deformation of a thin elastic plate under end loading is
described by the differential equation (3-74). The vertical
component of the normal stress due to flow on the fluids above and
below the plate can be used to determined the force per unit area (
)xq on the plate. The fluids occupy semi-infinite half space.
Assume the deformation of the plate is given by
a
t
m exww
= 2cos . -----
Since the plate forms the boundaries of the fluid half-spaces,
the situation is identical with the post glacial rebound. By
symmetry, the solutions above and below the plate are identical.
Consider the solution below the plate and measure g positive
downward from the base of plate (fig 6-26). The appropriate
solution of the biharmonic equation is ,
+++=
yyyy
DyeCeByeAex22222sin . -----
The velocity should be finite as y and requires C = D = 0. The
rigidity of the plate requires that 0u = on the plate. Assume
-
37
becomes: twPb
= 22 . -----
The pressure TP acting downward on the top of the plate is
relate to the pressure bP acting upward on the base of the plate
by: ( ) ( )xPxP bT = . ----- There is no normal viscous stress on
the plate since
yv
vanishes on 0=y
according to eq 00 =
=yyv . Thus, the net normal stress on the plate is
bbT PPPq 2== . ----- Substitute into , we obtain
( ) ( )t
txwtxq
= ,24,
. ----- With the force per unit area acting on the elastic
plate, the equation for the deflection of the plate is:
4 2
4 2
24 ( )w w wD P q xx x t
+ = = . -----
Substitute eq into , we find:
=
PDa 222
4
. -----
The wavelength corresponding to the smallest value of a is
obtained by setting the derivative of a w.r.t. equal to zero. The
result is:
21
32
=PD . -----
This is the wavelength of the most rapidly growing disturbance.
Write P h= , -----
where is the stress in the elastic layer associated with the end
load P , we get:
( )21
21
= Eh ( )
23
112 EhD . -----
Its expected that when folds develop in an elastic layer of rock
surrounded by rock exhibiting fluid behavior, the initial
wavelength of the folds has the dependence on the thickness of the
elastic layer and the applied stress .
The observed dependence of fold wavelength on the thickness of a
fold is given
in Fig 6-27, in good agreement with ( ) 22 101 =E
.
-
38
For GPaE 50= , 25.0= , MPa530= , This is a high stress, but
likely to be about the compressional strength of sedimentary rocks
at a depth of 2-5 km. As the amplitude of a fold increase, its
wavelength decreases somewhat and the bending stress in the elastic
layer exceeds the yield strength of the rock. The elastic layer
either fractures or plastically yield at points of maximum bending
moment that
are at nx21= , ,.......2,1,0=n . If a plastic bending occurs, an
angular or chevron
fold would be expected (fig. 6-28(a)). Folds with nearly
straight limbs of this type are often observed. For a rounded told
(fig 6-28 (b)), its likely that the dominant member has also
deformed in a fluid-like behavior. Assume the competent layer is a
Newtonian fluid with a viscous 1 embedded between two semi-infinite
fluids with viscous ( )010 >> . This approach is often
referred to as the Biot theory of folding. To derive the bending of
a free or isolated plate of viscosity , we follow the theory of
bending of an elastic layer given in section 3-9. The bending
moment is given by (eq 3-61);
= 22
h
h xx ydyM . -----
The longitudinal stress xx in a viscous plate is given by eq
,
-
39
xupxx
= 2 .
For a free plate, yy must vanish its surface, and if the plate
is thin, we can take
0=yy throughout the plate. From eq and 0=yy , we obtain:
yvp
yvpyy
=== 220 . -----
The incompressible continuity eq gives xu
yv
=
, and we can write eq as:
xup
= 2 . ----- By substituting into , we get:
xu
xx = 4 . -----
Eq for the bending moment in viscous plate becomes::
2 2
2 2
4 4 ( 2 )h h
h hxx xxu uM ydy ydyx x
= = = ----- By direct analogy with eq
= 2
2
dxwdyxx , the rate of strain rate x
u is
given by:
txwy
xu
=
2
3
. -----
The sign of this equation is opposite to eq because the rate of
strain xu
and
the strain rate xx have opposite signs. Substitute into and
carry out the integration, we get:
txwhM
= 233
3 . -----
Upon substituting the second derivative w.r.t. x of eq into eq ,
we obtain the general equation for the bending of a thin viscous
plate,
. 3 5 2
4 2'''' '' and '' 3h w wDw Pw q M Dw q P
x t x + = = = ----
2 2
2 2
d M d wq Pdx dx
= + . Solution of this equation give the vertical displacement w
of a viscous plate as a
-
40
function of time. Consider a specific example of a viscous plate
of length L embedded at one end
with a concentrated load aV applied at its other end. Since 0==
qP , eq reduces to
03 4
53
=
txwh -----
Integrate twice w.r.t. x yields:
( ) ( )tfxtfMtx
wh212
33
3+==
----- where ( )tf1 and ( )tf 2 are constants of integration that
can depend on time. Since the overall torque balance given in
eq
( )LXVM a = , must also be applicable to the viscous plate, we
can identify 1f and 2f as
aVf =1 , VaLf =2 . Eq becomes
VaLxVtx
wha +=
2
33
3 . -----
Integrate eq twice more w.r.t. x and satisfy the b.c.s for an
embedded plate,
0==
tww at 0=x , to get
=
323
23 xLVaxtwh . -----
A final integration w.r.t time ( )t and application of the
initial condition 0=w at 0=t , give:
txLhxVw a
=
323
3
2
. -----
Compared eq with eq
=32
2 xLDxVw a , both show the same spatial
-
41
dependence of the deflection, but the deflection of an elastic
plate is time-independent and the deflection of a viscous plate
increases linearly with time. Now return to the viscous folding
problem by considering the buckling of a viscous plate contained
between two semi-infinite viscous fluids. If the approximation
0=yy , we made in the bending moment of a viscous plate, the
plate viscosity 1 must be much larger than the viscosity 0 of the
surrounding half-space. Eq s
solution can take to be the form at
m exww
= 2cos . The responses of the
semi-infinite fluids to the deformation of the viscous plate are
identical to their responses to the bending of an elastic plate.
The force q per unit area on the viscous plate is given by eq .
Substitute into , we obtain:
2
20
4
531 83 x
wPtw
txwh
=
. -----
With w given by eq , we have:
+= 312
2
0 3421 h
Pa
. -----
The wavelength corresponding to the smallest value of a w.r.t.
equal to zero; the result is:
31
0
1
612
=
h . -----
This is the wavelength of the most rapidly growing mode. 14.
Stokes Flow A solid body will rise or fall through a fluid if its
density is different from the density of the fluid. If the body is
less dense, the buoyancy force will cause it to rise; if its more
dense, it will fall. If the fluid is very viscous, the Reynolds
number Re based on the size of the body, the velocity at which the
body moves through the fluid, and the viscosity of the fluid will
be small. In the limit 1
-
42
magma bubbles rise under the buoyancy force. Stokes solution can
be used to estimate the rate of magma ascent as a function of the
size of magma bubbles. We derive an expression for the velocity of
ascent or decent U of a spherical body in a constant-viscosity
fluid with different density. We first calculate the net force or
drag exerted by the fluid on a sphere, and then equate this force
to the buoyancy force responsible for the spheres solution. For
calculating the drag on the sphere due to its steady motion through
the fluid, we can consider the sphere to be fixed and have the
fluid move past the sphere. We dont discuss the transient period
during which the sphere accelerates to its final steady or terminal
velocity.
Consider a sphere radius a centered at the origin of a spherical
coordinate system
( ,,r ) as fig 6-31. The fluid approaches the sphere at =z with
velocity U in the z-direction. The flow is clearly symmetric about
the z-axis. Thus,
neither the velocity nor the pressure p of the fluid depends on
the azimuthal angle . In addition, there is no azimuthal component
of fluid motion; that is, the only non-zero component of fluid
velocity are the radial velocity rU and the meridional velocity u .
The continuity equation and the equations of motion for the slow,
steady, axisymmetric flow of a viscous incompressible fluid one, in
spherical polar
coordinates with 0=u ,
( ) ( ) ururrru r sinsin1100 22 +== , -----
-
43
=+ 02 uVp
( )
+
+
= sinsin22sin
sin110 222
2 urr
uurr
urrrr
p rrr,
-----
+
+
+
=
22222
2 sin2sin
sin1110
ruu
ru
rrur
rrp
r
r
.
----- We need to find a solution subject to the condition that
the fluid velocity approaches the uniform velocity U is the
z-direction as +r . The radial and meridional components of the
uniform velocity are cosU and sinU , respectively. Therefore, we
can write:
cosUur and sinUu as r . ----- We must also satisfy the no-slip
velocity boundary condition or ar = ,
0== uur on ar = . ----- The nature of the boundary conditions
suggests that we try a solution of the form:
( ) cosrfur = and ( ) sinrgu = . ----- Substitute eq into eqs ~
, we obtain:
( )frdrd
rg 2
21= , -----
( )
+
+
= gfdrdfr
drd
rrp 4cos0 22
, -----
( )
+
+
= gfdrdgr
drd
rp 2sin0 2 . -----
We can eliminate the pressure by differentiating eq w.r.t. and
substracting the derivative of eq w.r.t. r to obtain:
( ) ( )
+
++
=r
gfdrdgr
drd
rdrd
rgf
drdfr
drd
r21410 22
22 . -----
The solution of eq and for the function f and g can be found as
simple powers of r . Thus we get:
ncrf = , ------ where c is a constant. Eq gives:
( ) nrncg2
2+= . ----- By substituting eqs and into eq , we find that n
must satisfy:
( )( )( ) .1,2,3,0,0123 ==++ nnnnn
-
44
The function f and g are thus linear combinations of 0r , ,, 23
rr and 1r .
24
332
1 rcrc
rccf +++= ,
24
33
21 222
rcr
cr
ccg += , where ,, 32,1 ccc and 4c are constants. The velocity
components ru and u are
given by
cos243321
+++= rcrc
rccur . -----
sin2222
43
32
1
+= rcr
cr
ccu . -----
Since ru and u must satisfy conditions as r , Uc =1 , and 04 =c
. -----
The no-slip condition on ar = (eq) requires that:
2
3
2Uac = and
23
3aUc = . -----
The final expressions for ru and u are:
cos23
21 3
3
+=
ra
raUur , -----
sin43
41 3
3
=
ra
raUu . -----
The pressure associated with the flow can be found by
substituting eqs and into eq and integrating with respect to ,
cos2
32r
aUp = . ----- Both pressure forces and viscous forces act on the
surface of the sphere. By symmetry, the net force on the sphere
must be in the negative z-direction. This net force is the drag D
on the sphere. We first calculate the contribution of the pressure
forces to the drag. The pressure force on the sphere acts in the
negative radial direction. The component of this force is, per unit
area of the surface,
2cos2
3cosaUp = . -----
The pressure drag pD is obtained by integrating the product of
the force per unit
area with the surface area element da sin2 2 over the entire
surface of the sphere, == 0 2 2cossin3 aUdaUDp . -----
The viscous stresses acting on an area element of the spheres
surface are the radial
-
45
viscous stress rr . ( )
ar
rarrr r
u
==
= 2 , -----
and the tangential stress r
( )ar
rarr
urr
ur
r=
=
+
=
1 . -----
By substituting eqs & into & , we find: ( ) 0==arrr ,
----- ( )
aU
arr 2sin3 == . -----
The nonzero tangential stress r is a force per unit area in the
direction. The component of this force per unit area in the
negative z-direction is:
( )a
Ur 2
sin3sin2 = . -----
The viscous drag vD is found by integrating the product of this
quantity with
the surface area element da sin2 2 over the entire surface of
the sphere, == 0 3 4sin3 aUdaUDv . -----
The total drag on the sphere is the sum of the pressure drag and
the viscous drag:
aUDDD vp 6=+= . ----- This is the well-known Stokes formula for
the drag on a sphere moving with a small constant velocity through
a viscous incompressible fluid. Stokes resistance law is often
written in dimensionless form by normalizing the drag with the
product of the
pressure 221 Uf ( f is the density of the fluid) and the
cross-sectional area of the
sphere 2a . The dimensionless drag coefficient DC is thus:
( )2 212 24
12
Df
f
DCReUaU a
= = , -----
where the Reynold number is given by ( )2fU aRe = . -----
The Stokes drag formula can be used to determine the velocity of
a sphere rising buoyantly through a fluid by equating the drag to
the gravitational driving force. If
the density of the sphere s is less than the density of the
fluid f . The net
-
46
upward buoyang force according to Archimedes principal is
( ) = 334 agF sf ----- Set this equal to the drag on the sphere
aU6 and solve for the upward velocity U to obtain
( )
92 2ga
U sf= .
This result is only valid for the Reynold number is of order 1
or smaller. For larger eR #, the flow of a fluid about a sphere
becomes quite complex. Vortices are generated and the flow becomes
unsteady. The dependence of the drag coefficient DC for a sphere on
eR # is given in fig. 6-23. The Stokes flow from eq
is a valid approximation for eR < 1. The sharp drop in DC at
5103=eR
is associated with the transition to turbulent flow. The
dependence of DC on eR for a sphere is similar to the dependence of
f on eR for pipe flow given in fig 6-7.
In terms of the drag coefficient, the upward velocity of a
sphere from eqs , is given by:
( ) 1283
f s
D f
agU
C
=
. ------
The drag coefficient DC can be obtained from eR # and fig 6-23.
We can estimate the velocity of magma ascent through the
lithosphere. Refractory peridotite xenoliths with a maximum
dimension of about 0.3 m have been found in the basaltic lava
erupted in 1801 at Hawaii. An upper limit on the size of xenoliths
that can be entrained is obtained by setting the relative velocity
U equal to the flow velocity of the magma. The viscosity of the
basaltic magma is estimated to be 10 Pas. Assume
-
47
3600 mkg
ms = and ma 15.0= , we find from eq that
smU 3= ( hrkm8.10 ). The corresponding value of eR # from eq
with
32700 mkg
f = is 243. Therefore, the Stokes formula is only approximately
valid. Using eq and the empirical correlation given in Fig. 6-32,
we find
smU 87.0= and 70=eR . On the other hand, we take a typical
mantle velocity
yrmm10 , 3100 m
kg= , Pas2110= , and 210 smg = . Form eq , we find that the
spherical bodies with radii less 38 km will be entrained in mantle
flows. The conclusion is that sizable inhomogeneities can be
carried with the mantle rocks during mantle convection. One model
for magma migration is that magma bodies move through the
mantle
because of the differential buoyancy of the liquid. The velocity
of a bubble of
low-viscosity fluid moving through a high-viscosity fluid
because of buoyancy is
given by
( )f
bfgaU
3
2 = , -----
where b is the density of the bubble and f is the density of the
surrounding fluid, and f is the viscosity of the ambient fluid. Eq
differs from eq in that the boundary conditions for is free-slip;
that is, 0,0 == rru at ar = . Taking Pasm
kgkma bf21
3 10,600,5.0 === , we find
yrmmU 016.0= . Even for a relatively large magma body, the
migration velocity is
about 13 orders of magnitude smaller than that deduced from
xenoliths.
The calculated velocity of hrmm016.0 is unreasonably slow,
because it would take
about Gyr10 to migrate 100 km. If the magma doesnt penetrate the
lithosphere by
diaspirism, an alternative mechanism is hydrofracturing. Liquid
under pressure can
fracture rock, it has been suggested that the pressure caused by
the differential
buoyancy of magma can result in the propagation of a fracture
through the lithosphere
along with the magma migrates.
-
48
15. Plume Heads and Tails A simple steady-state model for the
ascent of a plume head through the mantle is given in fig 6-33. The
plume head is modeled as a spherical diapir whose velocity is given
by the stokes flow solution. Using the solution of prob. 6-23, we
can write the terminal velocity U of the ascending plume head from
eq ,
( )m
pmgaU
3
2 = , ------
where a is the radius of the plume head, p is the density of the
hot plume rock, m
is the density of the surrounding mantle rock. We take pT to be
the mean
temperature of the plume rock and 1T to be the temperature of
the surrounding rock.
From eq , we get
( )1TTpvmmp = . ------ Substitute eq into to get the ascent
velocity of the plume head,
( )m
pvm TTgaU 3
12 = . ------
The plume tail is modeled as a cylindrical pipe and the buoyancy
driven volume flux
pQ of the plume rock is given by eq ,
( )p
pmp
gRQ
48
= , ------
-
49
where R is the radius of the plume tail and p is the viscosity
of the plume rock. A measure of the strength of a plume is the
buoyancy flux b, which is defined by
( )pmpQB = . ----- A combination of eqs , , gives
( )p
vpm TTgRB 22124
8= . -----
The total heat flux in a plume HQ is related to the volume flux
by
( ) pppmH QTTcQ 1= , ----- where pc is the specific heat at
constant pressure. A combination of eqs ,
, gives
v
pH
BcQ = , -----
In the steady-state model, the plume head neither gains nor
loses fluid; this requires that the mean flow velocity in the plume
tail equals the ascent velocity of the plume head U ,
therefore,
URQp2= . -----
Once the plume flux B has been specified along with the other
parameters, the radius of the plume tail R can be determined by eq
, the heat flux in the plume HQ by , the ascent velocity of the
plume head U by , and the radius of the plume head a by . As
pointed out in section 1-6, the ascent of mantle plumes forms the
topographic swells, referred as hotspots. The buoyancy flux
associated with a mantle plume can be determined from the rate of
hotspot swell formation. We hypothesize that the excess mass
associated with the swell is compensated by the mass deficit of the
hot (light) plume rock impinging on the base of the lithosphere.
Thus the buoyancy flux B is given by
( ) PSwm uAB = , ----- where m is the mantle density, w is the
water density, SA is the cross-sectional area of the swell in a
vertical cross section perpendicular to the plume track, and Pu is
the plate speed relative to a fixed hotspot reference frame.
-
50
Ex. The Hawaii hotspot.
Take yrmmuP 90= , 2331 km.AS = (taken from fig. 1-19), 32300
m
kgww = ,
and we find skgB 3104.7 = . Taking KkgkJ.c op 251= and
15103 = Kov , the plume heat flux from eq is WQH
11103= , slightly less than 1% total surface heat flux. The
radius of the Hawaii plume R from is km84 . (assume
skgB 3104.7 = , Pasp 1910= , KTT op 2001 = , 15103 = Kov ,
33300 mkg
m = , and 28.9 smg = ). The radius 84 kmR= is relatively small
and explains why plumes are very different to observe seismically.
From eqs and
, the volume flux in Hawaii plume yrkmQp
312= . However, the volume
flux of basalt vQ required to create the Hawaii Islands and
seamount chains is
estimated to be yrkm31.0 . Thus, it only needs to melt ~1% of
the plume flux to
generate hotspot volcanics at Hawaii. The buoyancy fluxes for 43
plumes are given in Table 6-4. The total buoyancy flux
for three plumes skgB 3105.58 = . Taking KkgkJ.c op 251= and
15103 = Kov , the total plume heat flux from W.QH 13102440 =
(~5.5% of the total globe heat flow W.Q 1310434 = ). In section
4-23, the basal heating of the oceanic and continental lithosphere
mQ is W.
1310581 . Therefore, the derived plume heat flux is only 15% of
the total heat flux associated with the basal heating of the
lithosphere. The difference can be attributed either to plumes
that
-
51
impinge on the base of the lithosphere but are too small to have
a surface expression or to secondary mantle convection involving in
the lower part of the lithosphere. Ex. 2. Runion hotspot and the
flood basalt province of the Deccan Traps in fig. 1-22.
From table 6-4, the present buoyancy flux of the Runion plume
skgB 3104.1 = .
From , the radius of the plume conduit kmR 55= ; from and , the
volume flux yr
kmQp3
2.2= ; from , the mean ascent velocity of the plume
yrmU 23.0= . Assume the strength of the Runion plume has
remained constant
for the last Myr60 , taking Pasm2110= , we find from that the
radius of the
plume head kma 336= . The corresponding volume of the plume head
38102.1 kmVPH = . The volume of the basalts in the Deccan Traps
36105.1 kmVB . Thus, it needs to melt ~1% of the plume head to
form the flood basalts of the Deccan Traps. Assuming the volume
flux of the Runion plume
yrkmQP
32.2= and has remained constant over the Myr60 life time of the
plume,
the total volume flux through the plume tail has been 38103.1 km
. For the ascent velocity of the plume head U equal to yr
m23.0 , it could take about Myr12 for
the plume head to ascend from the CMB to the surface.
-
52
16. Pipe flow with heat addition We now consider the problems
involving both fluid and heat transfer. Consider
an example of the heating of water in an aquifer as the heat
balance on a thin, cylindrical shell of fluid in a pipe. The
thickness of the shell is r , and its length is
x (fig. 6-33). The heat conducted out of the cylindrical surface
at rr + per unit time is:
( ) ( )rrxqrr r ++2 ,
where ( )rrqr + is the radial heat flux at rr + . The heat
conducted into the shell across the inner cylindrical surface
is:
( )rxqr r2 . Since r is small, we can expand:
( ) ( ) ........rr
qrqrrq rrr ++=+
By neglecting higher power of r , we can write the net rate at
which heat is conducted into the cylindrical shell through. Its
inner and outer surface as: Its inner and outer surface as:
( ) ( ) ( )[ ] rqdrdqrxrrqrrrrqx rrrr
+=++ 22 . -----
In cylindrical coordinates, the radial heat flux rq is related
to the radial temperature
gradient rT
by Fouriers law of heat conduction; i.e.,
rTkqr
= , ----- where k is the thermal conductivity of the fluid. Eq
for the net effect of
radial heat conduction can thus be rewritten as:
+
rT
rTrrkx 2
2
2 . The amount of heat converted out of the shell at xx + by the
velocity ( )ru per unit time is given by:
( )xxcTrur +2 .
-
53
The amount of heat converted into the shell at x per unit time
is given by ( )xcTrur 2 .
By expanding ( ) ( ) ........++=+ x
xTxTxxT , we find the net rate at which fluids
carry heat out of the shell is
( ) ( )[ ] xxTcrurxTxxTcrur
=+ 22 . ----- If the flow is steady so that the temperature of
the fluid doesnt change with time and if axial heat conduction is
unimportant compared with advection of heat by flow, the net
effects of radial heat conduction and axial heat advection must
balance. Therefore, we can equate with ,
+
=
rT
rrTk
xTuc 12
2
. -----
By equating axial heat advection to radial heat conduction, we
also assumed that viscous dissipation or frictional heating in the
fluid is negligible. We can determine the temperature distribution
in the pipe using eq for the laminar flow in section 6-4. The
velocity as a function of radius can be expressed in terms of the
mean velocity u by combining eqs & to give
=
2
12Rruu . -----
If the wall temperature of the pipe is wT , changing linearly
along its length; i.e.,
21 cxcTw += , ----- where 1c and 2c are constants. Accordingly,
we assume that the temperature of the fluid is given by
( ) ( )rTrcxcT w +=++= 21 . ----- (In this situation, the net
contribution of axial heat conduction to the heat balance of a
small cylindrical shell vanishes). Thus is the difference between
the fluid temperature and the wall temperature. Substitute and into
,
+=
drd
rdrdkc
Rruc 112 2
2
1
2
. -----
The boundary conditions are wTT = at Rr = , ----- 0=rq at 0=r .
-----
The latter is required because there is no line source or sink
of heat along the axis of the pipe. Condition is satisfied if
-
54
0==Rr , ----- and condition with the Fouriers law yields
00
=
=rdr
d . ----- The solution of eq that satisfies these boundary
conditions is
+= 4
4
2
221 43
8 Rr
Rr
kRcuc . -----
The heat flux to the wall wq can be found by substituting eq
into Fourier law and evaluate the result at Rr = ,
121 Rcucqw = . (c: specific heat) -----
The heat flux is a constant, independent of x . If 01 >c , wT
increases in the direction of flow, the heat flows through the wall
of the pipe into the fluid. If 01
-
55
conductivity k , the conductive heat flux cq would be: ( )Dh
kqD
TTkq wwc == . -----
Thus the Nu # is:
c
w
qq
Nu = ----- and heat transfer with fluid flow through the pipe is
4.36 times more efficient than conductive heat transport through an
equivalent stationary fluid layer across which the same temperature
difference is applied. 17. Aquifer model for hot springs
The result derived from the precious section can be used to
study the heating water flowing through an aquifer surrounded by
hot rocks. Consider the same semicircular aquifer with circular
cross section in fig. 6-9, the heat convected along the aquifer is
balanced by the heat lost or gained by conduction to the walls. We
can write:
( )TTRhdsTducR w = 22 , -----
where s is the distance measured along the aquifer from the
entrance, u is the mean velocity in the aquifer, T is the
flow-averaged temperature of the aquifer fluid and wT is the
temperature of the aquifer wall rock. Assume a laminar flow so that
the heat transfer coefficient h in eq can be used. The coordinate s
can be related to the angle by
Rs = . ----- Assume the wall temperature wT is related to local
thermal gradient by
0sin TRTw += , ----- where 0T is the surface temperature.
Substitution of eqs , , into eq , yields
( )TTRkdTd
RucR += 0
2
sin1148
. ------ The equation can be simplified through the introduction
of the Pclet number Pe defined by
kRucPe . -----
The Pe number is a dimensionless measure of the mean velocity of
the flow through the aquifer. Its related to Re and Pr already
introduces. Since (thermal diffusivity ) =
ck , Pe can be written as
-
56
Pe uR= . -----
Using the definitions of DuRe ( D = 2R ) and
Pr , we can rewrite 1 2 1Pe Re Pr2 2
u R = = . -----
The simplification of eq is done by the introduction of a
dimensionless temperature defined by:
RTT
= 0 . -----
With eqs & , we can put eq into the form
sin
4811 =+ d
dPeRR . -----
This is a linear 1st-order differential equation that can be
integrated using an integration factor. With the boundary condition
that the water entering the aquifer is
at the surface temperature, 0TT = or 0= at 0= , the solution can
be written 12
11481
1148
1148expcossin
1148
+
+=
RPeR
RPeR
RPeR
RPeR . -----
The dimensionless temperature e at the exit of the aquifer, = ,
is
2
11481
11481
1148exp
+
+
=
RPeR
RPeR
RPeR
e
. -----
e (the exit temperature) is plotted as a function of RRPe
in fig 6-35. e is
-
57
maximum when 5=RRPe . Thus, for given values of all parameters
other than u ,
there is a particular flow rate through the aquifer that
maximizes e . The maximum e is about one-half of the maximum wall
temperature at the base of the aquifer,
since 21=e , corresponds to RTTe += 2
10 , and wT at 2
= is
( )RTRT
-
58
18. Thermal Convection Plate tectonics is a consequence of
thermal convection in the mantle driven
largely by radiogenic heat sources and the cooling of the earth.
When a fluid is heated from below or from within and cooled from
above, it will has dense cool fluid near the upper boundary and hot
light fluid at depth. The situation is gravitationally unstable,
and the cool fluids tend to sink and the hot fluid tends to
rise.
Density variation caused by thermal expansion lead to buoyancy
forces that drive thermal convection. Thus its essential to account
for density variations in the gravitational body force term of the
conservation of momentum equation. However, the density variations
are sufficiently small so that they can be neglected in the
continuity equation. This is known as the Boussinesq Approximation.
It allows us to use the incompressible conservation of mass
equation . The force balance equations & are also applicable by
adding the vertical buoyancy forces due to small density
variations, let
+= 0 , ----- where 0 is a reference density and 0
-
59
Density variations due to temperature changes are given by: (
)00 TTv = , ------
where v is the volumetric coefficient of thermal expansion and
0T is the reference temperature corresponding to 0 . Substitution
of into gives
( )0022
2
2
0 TTgy
vxv
yP
v
+
+= -----
The last term in , the buoyancy force per unit volume depends on
the temperature. Thus the velocity field cant be determined without
simultaneously solving for the temperature field. Therefore we
require the heat equation. The energy balance must account for heat
transport by both conduction and convection.
Consider a small 2-D element shown in fig. 6-37. Since the
thermal energy content of the fluid is cT per unit volume, an
amount of heat ycTu is transported across the right side of the
element by velocity u in x-direction, this is the energy flow per
unit time and per unit length in the direction to the figure. Then
the energy flow rate per unit area at xx + is ( ) xcTu
xcTu
+ . The net energy advected out of the element per unit time and
per unit length due to flow in the x-direction is thus
( ) ( ) yxcTux
ycTuxcTux
cTu =
+ . -----
The same analysis in the y-direction gives convection
( ) ( ) yxcTvy
xcTvycTvy
cTv =
+ . -----
The net rate at which the heat is conducted out of the element
unit length has been derived in ,
-
60
+
22
2
2
yT
xTk -----
Conservation of energy requires that the combined transport of
energy out of the element by conduction and convection must be
balanced by the change in the energy content of the element. The
thermal energy of the fluid is cT per unit volume. Thus, this
quantity changes at the rate ( ) yxcT
t
per unit length of the fluid. By combining the effects of
conduction, convection and thermal inertia, we obtain
( ) ( ) ( ) 022
2
2
=+
+
+
cvT
ycuT
xyT
xTkcT
t . -----
By treating and c as constants and noting that ( ) ( )
yTv
xTu
yv
xuT
yTv
xTuvT
yuT
x +
=
+
++
=+
, = 0 (continuity eq.)
and kc
= , we get the heat equation for 2-D flows
+
=+
+
2
2
2
2
yT
xT
yTv
xTu
tT , -----
TTutT 2=+
K . In , we have neglected frictional heating in the fluid
associated with the resistance to flow and compressional heating
associated with the work done by pressure force in moving the
fluid. Note: Material derivative In fluid mechanics, ones concern
is normally with the fluid velocity as a function of space and
time, rather than trying to follow any particle displacements or
paths in solid mechanics as called the Lagrangian description. The
description of the flow at every fixed point as a function of time
is called the Eulerian formulation. For any quantity ( , , , )Q x y
z t that represents a given property of the fluid, the total
differential change of the quantity Q o associated with the changes
dx, dy, dz, dt is given by
. Q Q Q QdQ dx dy dz dtx y z t
= + + + The spatial increments by following a paricular particle
are , , .Therefore, the time derivative of of a particular particle
is,
dx udt dy vdy dz wdzQ
= = =
-
61
( ) , where ( , , ).dQ DQ Q Q Q Q Qu v w u Q u u v wdt Dt t x y
z t = = + + + = + = GG G The quantity /DQ Dt is termed the
substantial derivative, particle derivative or material derivative
and /Q t is called the local derivative. 19. Linear stability
analysis for the onset of thermal convection in
a layer of fluid heated from below.
At 2by = , cold boundary 0TT =
At 2by = , hot boundary 1TT =
( )01 TT > Assume no heat source in the fluid, buoyancy
forces tends to drive convection when fluids near the heated lower
boundary becomes hotter and lighter than the overlying fluid and
tend to rise. Upper boundary is denser than the fluid below and
tends to sink. However, the motion doesnt take place for small
temperature differences across the layer because the viscous
resistance of the medium to flow must be overcome. We now
determined the minimum ( )01 TT required for convection to occur.
For sufficiently small ( )01 TT , the fluid is stationary ( 0== vu
), and we can assume that a steady
=
0t
conductive state with 0=x
exits. The energy
equation then simplifies to
022
=dy
Td c , -----
where subscript c indicates that this is a conduction solution.
The eq satisfies
the b.cs 0TT = at 2by = and 1TT = at 2
by += , so the temperature profile is
yb
TTTTTc 0101 2
++= . ----- The stationary conductive state will persist until
01 TT reaches a critical value at which even the slightest further
increase in temperature difference will cause the layer to become
unstable and convection to occur. Thus, at the onset of convection,
the
-
62
fluid temperature is nearly conduction temperature profile and
the temperature difference T ,
( )y
bTTTT
TTTT c 0101 2+= , -----
is arbitrary small. The convective velocities u , v are also
infinitesimal when motion first takes place. The energy eq that
pertains to the onset of convection can be written in terms of T by
solving for T and substituting into . We get
( )
+
=++
+
2
2
2
201
yT
xT
bTTv
yTv
xTu
tT -----
vuT ,, are small quantities, the nonlinear terms xTu
and
yTv
on the left
side of are much smaller than the remaining linear terms. Thus,
they can be neglected and can be written as:
( )
+
=+
2
2
2
2
01 yT
xTTT
bv
tT . -----
The neglect of the nonlinear terms, the terms involving products
of the small
quantities ,, vu and T , is a standard mathematical approach to
problems of stability. Its known as a linearized stability
analysis. Its valid for the study of the onset of convection when
the motion and the thermal disturbance are infinitesimal. To
summaries, the equations for the small perturbations of temperature
T , velocities
vu , and pressure p when the fluid layer becomes unstable
are
0=+
yv
xu , -----
+
+= 2
2
2
2
0yu
xu
xP , -----
+
+= 2
2
2
2
00 yv
xvTg
yP
v , -----
( ) 2 21 0 2 2T v T TT Tt b x y + = +
. -----
We have takes the buoyancy force at any point in the layer to
depend only on the
-
63
departure of the fluid temperature from the basic conduction
temperature at the point ( )Tgv 0 . Eqs - are solved subject to the
following b.c.s: the surface
2by = are
isothermal and no flow occurs across them; i.e.,
0== vT on 2by = . -----
If the boundaries of the layer are solid surface,
0=u on 2by = . (no-slip conduction) -----
If the surfaces 2by = are free surfaces; i.e., if there is
nothing at
2by = to exert
a shear stress on the fluid, u needs not vanish on the
boundaries. Instead, the shear stress yx must be zero on 2
by = ,
0=yx on 02 =+
=yu
xvby on
2by = -----
Because 0=v on 2by = , and 0
xv on
2by = , the free surface bcs are
0=
yu on
2by = . -----
A simple analytic solution can be found if the free surface
conditions are adopted. Introduce the stream function defined in
eqs that automatically satisfies the continuity equation . Eqs -
can be written:
+
= 33
2
3
0yyxx
P , -----
+
=
xyxTg
yP
v 2
3
3
3
00 , -----
( )
+
=+
2
2
2
2
011
yT
xT
xTT
btT . -----
Eliminate the pressure from eqs & yields:
xTg
yyxx v
+
+= 04
4
22
4
4
4
20 . -----
Eqs and can be solved for and T by the method of separation
of
-
64
variables. The boundary conditions & are automatically
satisfied by the solutions of the form:
texby
= 2sincos0 , -----
texbyTT
= 2coscos0 . ----- For 0> , The disturbance will grow with
time, the heated layer is convectively unstable. For 0
-
65
32 22
2
2 2
2
4
Ra4
b
b
+ > . -----
The growth rate is negative and there is stability if Ra is less
than right side of eq . Convection just sets in when 0= , which
occurs when
32 22
2
cr 2 2
2
4
Ra Ra4
b
b
+ = . -----
The critical value of Rayleigh number crRa makes the onset of
convection. If
crRaRa < , disturbance decays with time; if crRa Ra> ,
perturbation will grow exponentially with time. Eq shows that crRa
is a function of the wavelength of the disturbance (see fig. 6-39).
If Ra number and disturbance wavelength are such that the point
lies above the curve, the perturbation of wavelength is unstable;
if the point lies below the curve, convection cant occur with
disturbance of wavelength .
Eg. If Ra= 2000, disturbance with 8.0 2 b 5.4 are convectively
unstable.
For b2
0.8 and b2
5.4, convection cant occur. Fig 6-39 shows that there is
a minimum value of crRa . If Ra lies below the minimum, all
disturbances decay,
-
66
the layer is stable, and convection cant occur.
The value of b2 at which Ra is a minimum can be obtained by
setting the
derivative of the right side of w.r.t. b2 equal to zero,
2 3 22 2 2 2 2 2 2 22 2cr
2 2 2 2
Ra 4 4 2 4 2 43 2 2 02
b b b b b bb
= + + =
or 2
2 =b . -----
The value of the wavelength corresponding to the smallest value
of the critical Rayleigh number is
b22= . ----- Substitution of this value back to eq gives the
minimum critical Rayleigh number
( ) 4cr 27min Ra 657.54= = . -----
When Ra exceeds crRa for convection to occur, one can think of
the temperature difference 01 TT across the layer as having to
exceed a certain minimum value or the viscosity of the fluid having
to lie below a critical value before convection sets in. For
examples, by increasing 01 TT with other quantities fixed, when Ra
reaches 657.5 (for heating from below with stress-free boundaries),
convection sets in and the
aspect ratio of each convection cell is 2 . The minimum crRa and
the
corresponding disturbance wavelength for no-slip velocity
boundary conditions can be solved numerically. For that case, min
crRa 1707.8= and b016.2= .
The linear stability analysis for the onset of convection can be
carried out for a fluid layer heated uniformly from within and
cooled from above. The lower boundary is assumed to be insulating;
i.e. no heat flows across the boundary. The fluid near the top is
cooled and denser; therefore buoyancy forces can drive fluid motion
if they are strong enough to overcome the viscous resistance. Thin
type of instability is directly applicable to the earths mantle.
Since the earths interior is heated by decay of the radioactive
elements and the near-surface rocks are cooled by heat conduction.
The appropriate Ra for a fluid layer heated from within is
2 50Ra vH
gHbk
= , -----
-
67
where H is the rate of internal heat generation per unit mass.
For no-slip bcs, the
minimum crRa is 2722 and the associated value of b2 is 2.63; for
free-slip bcs,
min crRa 867.8= and the associated value of b2 is 1.79.
We can estimate the value of Ra for the mantle. Based on the
postglacial
rebound studies, we take Pas2110= . The rock properties take
KmWk o4= ,
smm21= , and 15103 = Kov , 210 smg = and an average density
30 4000 mkg= . Based on Chapter 4s discussion, the heat
source
kgWH 12109 = . If convection is restricted to the upper mantle,
take kmb 700= ,
we find 6Ra 2 10= . For the entire mantle convection, take kmb
2880= , we find 9Ra 2 10= . In either case, Ra is much greater than
the minimum crRa . It was
essentially this calculation that led Arthur Holmes to propose
in 1931 that thermal convection in the mantle was responsible for
driving continental drift.
20. A transient boundary layer theory for finite-amplitude
thermal
convection. The linear stability analysis determines whether or
not thermal convection occurs, but its not useful to determine the
structure of convection when Ra exceeds crRa . Because its linear,
the linear stability analysis cant predict the magnitude of
finite-amplitude convection flows.