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Figure 8.2: Sketch of the cooling model of the oceanic lithosphere.

We will consider the situation that the temperature is stationary. This does not imply

that the ow vanishes it means that the partial time-derivatives vanish: @T=@t = 0,

@v=@t = 0.

Problem l: Show that in the absence of heat sources (S = 0) the conservation equation

(8.26) reduces to:

2!

2

U @T = @ T +@ T : (8.39)

@x2 @z 2

@x

In general the thickness of the oceanic lithosphere is less than 100 km, whereas the width

of ocean basins is several thousand kilometers.

Problem m: Use this fact to explain that the following expression is a reasonable ap-

proximation to (8.39):

@T = @ 2 T :

U @x (8.40)

@z 2

Problem n: Show that with the replacement = x=U this expression is identical to the

heat equation (8.28).

Note that is the time it has taken the oceanic plate to move from its point of creation

(x = 0) to the point of consideration (x), hence the time simply is the age of the oceanic

lithosphere. This implies that solutions of the one-dimensional heat equation can be used

to describe the cooling of oceanic lithosphere with the age of the lithosphere taken as the

time-variable. Accounting for cooling with such a model leads to a prediction of the depth

of the ocean that increases as t with the age of the lithosphere. For ages less than about

p

100 Myear this is in very good agreement with the observed ocean depth 45].

8.5 The explosion of a nuclear bomb

As an example of the use of conservation equations we will now study the condition under

which a ball of Uranium of Plutonium can explode through a nuclear chain reaction. The

starting point is once again the general conservation law (8.3), where Q is the concentration

N(t) of neutrons per unit volume. We will assume that the material is solid and assume

8.5. THE EXPLOSION OF A NUCLEAR BOMB 83

there is no ow: v = 0. The neutron concentration is a ected by two processes. First, the

neutrons experience normal di usion. For simplicity we assume that the neutron current

is given by expression (8.25): J = . Second, neutrons are produced in the nuclear

; rN

chain reaction. For example, when an atom of U235 absorbs one neutron, it may ssion and

emit three free neutrons. This e ectively constitutes a source of neutrons. The intensity

of this source depends on the neutrons that are around to produce ssion of atoms. This

implies that the source term is proportional to the neutron concentration: S = N, where

is a positive constant that depends on the details of the nuclear reaction.

Problem a: Show that the neutron concentration satis es:

@N = 2N + N: (8.41)

@t r

This equation needs to be supplemented with boundary conditions. We will assume that

the material that ssions is a sphere with radius R. At the edge of the sphere the neutron

concentration vanishes while at the center of the sphere the neutron concentration must

remain nite for nite times:

N(r = R t) = 0 and N(r = 0 t) is finite : (8.42)

We restrict our attention to solutions that are spherically symmetric: N = N(r t).

Problem b: Apply separation of variables by writing the neutron concentration as N(r t) =

F(r)H(t) and show that F(r) and H(t) satisfy the following equations:

@H = H (8.43)

@t

2F + ( )F = 0 (8.44)

;

r

where is a separation constant that is not yet known.

Problem c: Show that for positive there is an exponential growth of the neutron con-

centration with characteristic growth time = 1= .

Problem d: Use the expression of the Laplacian in spherical coordinates to rewrite (8.44).

Make the substitution F(r) = f(r)=r and show that f(r) satis es:

@2f + ( )f = 0 : (8.45)

;

@r2

Problem e: Derive the boundary conditions at r = 0 and r = R for f(r).

Problem f: Show that equation (8.45) with the boundary condition derived in problem

e can only be satis ed when

2

n for integer n :

= (8.46)

R

;

CHAPTER 8. CONSERVATION LAWS

84

Problem g: Show that for n = 0 the neutron concentration vanishes so that we only

need to consider values n 1.

Equation (8.46) gives the growth rate of the neutron concentration. It can be seen that

the e ects of unstable nuclear reactions and of neutron di usion oppose each other. The

-term accounts for the growth of the neutron concentration through ssion reactions, this

term makes the inverse growth rate more positive. Conversely, the -term accounts for

di usion, this term gives a negative contribution to .

Problem h: What value of n gives the largest growth rate? Show that exponential growth

of the neutron concentration (i.e. a nuclear explosion) can only occur when

r

R> : (8.47)

This implies that a nuclear explosion can only occur when the ball of ssionable material is

larger than a certain critical size. If the size is smaller than the critical size, more neutrons

di use out of the ball than are created by ssion, hence the nuclear reaction stops. In

some of the earliest nuclear devices an explosion was created by bringing two halve spheres

that each were a stable together to form one whole sphere that was unstable.

Problem g: Suppose you had a ball of ssionable material that is just unstable and that

you shape this material in a cube rather than a ball. Do you expect this cube to be

stable or unstable? Don't use any equations!

8.6 Viscosity and the Navier-Stokes equation

Many uids exhibit a certain degree of viscosity. In this section it will be shown that

viscosity can be seen as an ad-hoc description of the momentum current in a uid by

small-scale movements in the uid. Starting point of the analysis is the equation of

momentum conservation in a uid:

@( v) + ( vv) = F (8:13) again:

@t r

In a real uid, motion takes places at a large range of length scales from microscopic eddies

to organized motions with a size comparable to the size of the uid body. Whenever we

describe a uid, it is impossible to account for the motions at the very small length scales.

This not only so in analytical descriptions, but it is in particular the case in numerical

simulations of uid ow. For example, in current weather prediction schemes the motion

of the air is computed on a grid with a distance of about 100 km between the gridpoints.

When you look at the weather it is obvious that there is considerable motion at smaller

length scales (e.g. cumulus clouds indicating convection, fronts, etc.). In general one

cannot simply ignore the motion at these short length scales because these small-scale

uid motions transport signi cant amounts of momentum, heat and other quantities such

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