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 .

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).

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

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 .

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 ,

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 = .

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= . -----

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 .

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

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:

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 = . -----

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 . -----

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

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

+++ .

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:

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

++

=>=

18

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

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

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

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,

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

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

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.