LECTURES 4–11

Geophysical Heat Transfer

1.1 Introduction

There are three familiar mechanisms for heat transfer: conduction, convection and radia-tion. The distinguishing features of these mechanisms are that thermal conduction involvesenergy transmission through matter, convection involves the energy that is carried withmoving matter and radiation involves energy carried by electromagnetic waves (such aslight) and is most effective when matter is entirely absent.

1.2 Heat Transfer by Conduction

Heat conduction is the usual mechanism for heat transfer in solid material such as Earth’scrust. Let us begin by clarifying what we mean by the term heat flux. A useful approachis to examine the dimensions of several related quantities. We begin by considering somematerial body, for example a sphere of metal. Imagine that the object is at some hightemperate T and that to test the temperature an observer touches the object. In doingso, a certain amount of heat Q is transferred from the object to the observer; clearly theS.I. dimensions of Q are Joules, denoted J. Suppose that P denotes the rate at which heatflows from the object to the observer; it is evident that the S.I. units of P are J s−1 or W.If we knew the exact values of Q and P would we be able to conclude whether the observerfelt any discomfort when he or she touched the object? The answer is that neither Q norP yield information of this sort. What is needed is the rate at which heat flows throughsome area A representing that part of the observer’s body in contact with the hot object.This concept, which describes the rate of energy flow through a given area, is known asthe heat flux and we shall denote it q. The S.I. dimensions of q are W m−2.

We turn to the question of how heat flux q is related to the temperature distributionand thermal properties of a heat-conducting material. Consider a slab of some solid sub-stance that has horizontal boundaries at z = z1 and z = z2 (Figure 1.2-1). Assume thatthe lower surface is maintained at a constant temperature T1 and that the upper surfaceis maintained at a constant temperature T2. Further assume that the system is in steady-state so that no changes occur over time. If T1 = T2 the lower and upper boundaries aremaintained at the same temperature. Irrespective of whether this temperature is high orlow, there will be no flux of heat through the material. Thus as a provisional observationwe write

q ∝ T2 − T1. (1.2-1)

A second feature that seems obvious is that for a given temperature difference T2 −T1 theheat flux through the slab will be smaller if the slab is very thick (z2 � z1) than if theslab is very thin. This observation suggests the refinement

q ∝ T2 − T1

z2 − z1. (1.2-2)

1

Additionally, it is clear that the properties of the slab will influence how readily heat canflow through it. If the slab was constructed from metal, the heat flux would be largerthan if the slab was constructed of some thermal insulator such as wood or ceramic. Wetherefore introduce a physical property K characterizing the thermal conductivity of thematerial. The larger the value of K the more conductive is the material; for a thermalinsulator K takes small values; thermal conductivity is never negative. Thus we refine ourprovisional expression to

q = KT2 − T1

z2 − z1. (1.2-3)

A closer examination of (1.2-3) suggests that the expression needs a further improvement.If T2 < T1 (the lower boundary is hotter than the upper boundary) expression (1.2-3) hasa negative sign, implying that the direction of heat flow is downward from cold to hotrather than from hot to cold. To repair this difficulty we write

q = −KT2 − T1

z2 − z1. (1.2-4)

z = z2

z = z1T1

T2

0

+z

+x

K

Fig. 1.2-1. Heat flux through a slab.

A more general expression is obtained by substituting ∆z = z2−z1 and ∆T = T2−T1

to obtainq = −K

∆T

∆z. (1.2-5)

Expressed in the above form (1.2-5) seems like an invitation to let ∆z and ∆T shrink toinfinitesimal values to obtain

q = −KdT

dz. (1.2-6)

2

It is readily confirmed that the dimensions if K are W m−1 deg−1. The above expressionis sometimes termed Fourier’s law of conduction and is an example of what physicistscall a constitutive relation. Such relations are mathematical expressions that describe,usually in only an approximate way, the behaviour of idealized materials such as the heat-conducting solid. Constitutive equations are not fundamental equations of geometry orphysics and therefore have lower status than equations such as A = πr2 and E = mc2.Unlike fundamental equations we never question whether constitutive equations are trueor false but simply whether they do a good or bad job of approximating reality.

1.3 Vector equation for heat flux

In general q is a vector,∗ not a scalar, so equation (1.2-1) cannot be general because it onlyinvolves temperature gradients in the z direction. Temperature can vary with both spaceand time so the correct mathematical representation of this fact is the function T (x, y, z, t).Clearly there are three possible space derivatives (x, y and z) as well as a time derivative.Ordinary derivatives such as d/dz in (1.2-1) are incapable of expressing this complicationand we must introduce the idea of partial derivatives.

In Cartesian coordinates the heat flux vector field can can be written

q(x, y, z, t) = qx(x, y, z, t) i + qy(x, y, z, t) j + qz(x, y, z, t)k (1.3-1)

where qx, qy and qz are the scalar components of the heat flux vector and i, j and k areunit vectors. Fourier’s law applies to each scalar component of the heat flux vector; thus

qx(x, y, z, t) = −K∂T

∂x(1.3-2a)

qy(x, y, z, t) = −K∂T

∂y(1.3-2b)

qz(x, y, z, t) = −K∂T

∂z. (1.3-2c)

The vector expression of Fourier’s law (1.3-2) can be written using more compact notationby introducing the useful concept of the vector gradient of a scalar field. For the scalarfield T , the gradient of T is written ∇T (pronounced “del T” or “grad T”) and defined as

∇T =∂T

∂xi +

∂T

∂yj +

∂T

∂zk. (1.3-3)

∗ Note that most physics books use bold-face typography, e.g. q, to indicate vectors;in written work it is usual to indicate vectors with arrows, e.g. ~q, or with underlining, e.g.q.

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Example 1.3.1. Partial derivative of a scalar function

Given the scalar function

f(x, y, z) = x exp(−y2)sin(ay + bz)(y2 + z2)

find ∂f/∂x.

Answer

To calculate the partial derivative with respect to x, we follow the simple prescription oftreating y and z as constants and differentiating with respect to x:

∂f

∂x= exp(−y2)

sin(ay + bz)(y2 + z2)

.

Example 1.3.2. Partial derivatives of a scalar function

Given the scalar function F (x, y, z, t) = x2yz3 sinω0t, find the partial derivatives withrespect to x, y, z and t.

Answer

Following the same procedure as in the previous example we have:

∂F

∂x= 2xyz3 sinω0t

∂F

∂y= x2z3 sinω0t

∂F

∂z= 3x2yz2 sinω0t

∂F

∂t= ω0x

2yz3 cosω0t.

Example 1.3.2. Gradient of a scalar field

Assume that the scalar function f(x, y, z) describes some scalar field and that

f(x, y, z) = ax + by + cyz

where a, b and c are constants. Evaluate ∇f

4

Answer

Separate evaluation of the partial derivatives gives ∂f/∂x = a, ∂f/∂y = b + cz and∂f/∂z = cy. Thus

∇f = a i + (b + cz) j + cy k.

Example 1.3.3. Evaluation of the heat flux

Assume that the scalar function T (x, y, z) = A(x2 − y2) describes the temperature distri-bution within a medium having constant thermal conductivity K. Evaluate the heat fluxvector q.

Answer

Separate evaluation of the partial derivatives or T gives ∇T = 2Ax i − 2Ay j. Thus fromFourier’s law q = −2AK(x i − y j).

1.4 Solution for steady heat flow through a homogeneous slab

Here we show how Fourier’s law can be integrated to obtain the steady temperature dis-tribution in a horizontal homogeneous∗ slab. Consider a horizontal slab of thickness h andhaving a constant thermal conductivity K. Suppose that the lower boundary of the slabis at z = 0 and the upper boundary at z = h and that the upper and lower boundariesare respectively maintained at constant temperatures TS and TB respectively (see Figure1.4-1).

z = h

z = 0 TB

TS

0

+z

Kh

+xFig. 1.4-1. Steady heat conduction through a homogeneous slab of thicknessh.

∗ In physics the term homogeneous indicates that there is no spatial variation of physicalproperties. Thus, for a homogeneous slab, every part of the slab has the same propertiesand these properties have no spatial dependence. In a homogeneous slab the thermalconductivity K is therefore a constant.

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According to Fourier’s law of conduction q = −K∇T . By inspection of the accom-panying diagram and noting the significance of the term “steady” we can make somesimple predictions about the mathematical form of q and T . In the most general caseq = q(x, y, z, t) and T = T (x, y, z, t). These forms allow for arbitrary variation in bothand time. The term “steady” rules out the possibility of variations with time, so our firstsimplification is to appreciate that q = q(x, y, z) and T = T (x, y, z) are the most complexmathematical forms that are consistent with steady heat conduction. Next, we can foreseethat because of the slab geometry and the constant-temperature conditions at the slabboundaries, the most complex allowable functional form for the temperature variation isT = T (z). Given that q = −K∇T it immediately follows that the only non-vanishingcomponent of the heat flux vector is the z component. Thus, at worst, q = qz(z)k. Thereis no heat flux in the x or y directions. Lastly, let us examine the possibility of z variationof the vertical component of heat flux qz(z). If qz varied with z then this would imply thatthere was an imbalance between out-flow and in-flow of heat in various regions of the slab.If such an imbalance existed then this would lead to accumulation or depletion of thermalenergy in that region which, in turn, would result in an increase or decrease of temperature.Such a change would violate the assumption of steady thermal conditions. Thus we areforced to conclude that qz = q0 where q0 is a constant. Given these preliminary insights,we now proceed to calculate the temperature distribution and heat flux.

Because, as we have seen, T = T (z) and q = q0 k, Fourier’s law simplifies to the singleexpression

−K∂T

∂z= q0. (1.4-1)

Furthermore, because T is only a function of z there is no distinction between the partialderivative ∂/∂z and the ordinary derivative d/dz so (1.4-1) can be written

dT

dz= − q0

K. (1.4-2)

Equation (1.4-2) can be regarded as a first-order ordinary differential equation.∗ We solve(1.4-2) by integrating both sides of the equation to obtain

T (z) = − q0

Kz + c0 (1.4-3)

∗ Differential equations are equations that involve derivatives. In contrast an algebraicequation involves algebra without derivatives. Examples are

Ad2Y

dx2+ [BY 2 + C exp(Y ) + Dx]

dY

dx+ Y = sin x

(a differential equation) and

AY + x2[BY 2 + C exp(Y ) + Dx]Y = sin x

(an algebraic equation).

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where c0 is an integration constant. Note that (1.4-3) contains two undetermined constants:q0 (the constant but as yet unknown heat flux) and c0 the integration constant. Theproblem is not completed until these constants have been evaluated. To set values tothe two undetermined constants, we require two additional equations. These equations,referred to as boundary conditions, follow from examining the diagram and searching forinformation that has not yet been used. The two as-yet unused pieces of information arethe specified temperature values TB and TS . The equations that apply to these unusedpieces of information are

T (0) = TB (1.4-4a)

T (h) = TS (1.4-4b)

which are termed boundary conditions because they describe conditions at the boundaryof the slab. By applying these boundary conditions to (1.4-3) we can evaluate the unde-termined constants. First note that (1.4-3) gives T (0) = c0; thus from (1.4-4a) it followsthat c0 = TB . Also from (1.4-3) we have T (h) = −(q0h/K) + c0 which with c0 = TB givesT (h) = −(q0h/K) + TB . From (1.4-4b) it follows that TS = −(q0h/K) + TB and thatq0 = K(TB −TS)/h. Having now evaluated the two undetermined constants we can at lastform the complete temperature solution

T (z) = TB − (TB − TS)z

h. (1.4-5)

Note that (1.4-5) predicts a linear variation of temperature with depth in the slab.Note also that the heat flux q0 = K(TB − TS)/h depends on the thermal conductivity butthe temperature profile is independent of thermal conductivity.

Example 1.4.1. Heat flux through the Ross Ice Shelf

Ice shelves are the floating fringes of ice sheets. Their surface temperature is controlled byclimate and their bottom temperature is close to the freezing temperature of sea water.The surface temperature of the Ross Ice Shelf in West Antarctica is −28.0C; the bottomtemperature is −2.0C; the shelf thickness is 400m. Assuming steady thermal conductingin an infinite slab, calculate the heat flux through the slab. Assume that the thermalconductivity of ice is 2.1W m−1 deg−1.

Answer

For steady heat conduction through a slab q0 = K(TB − TS)/h. Thus for the Ross IceShelf q0 = 2.1(−2.0 + 28.0)/400 = 136.5mW m−2.

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Example 1.4.2. Heat flux through a slab having specified basal temperature and heat flux

Consider the steady heat flux through a homogeneous slab of thickness h. The slab hasconstant thermal conductivity K and the lower boundary is maintained at constant tem-perature TB. There is a constant heat flux qB across the basal boundary. Find thetemperature distribution T (z) and heat flux q(z) in the slab. NOTE: This example illus-trates that it is not necessary that there be a boundary condition associated with eachboundary; it is sufficient that there are as many boundary conditions as there are undeter-mined constants. In this example there are two undetermined constants and two boundaryconditions are associated with a single boundary.

Answer

As in the previous cases, the heat flux is constant through the slab so that

q(z) = q0

and integration of Fourier’s law yields the result

T (z) = − q0

Kz + c0

where q0 and c0 are undetermined constants. The boundary conditions are T (0) = TB andq(0) = qB which gives the results

q0 = qB

c0 = TB.

Thus q(z) = qB and T (z) = −qBz/K + TB.An alternative approach to applying the boundary conditions is less direct but equally

correct. Note that in the foregoing example the boundary condition on q(z) was applieddirectly and the result q(z) = qB emerged immediately. It is also possible to apply theboundary condition on q by working entirely with the expression for T (z). Following thistactic, we differentiate T (z) = −q0z/K + c0 to obtain dT (z)/dz = −q0/K and note thatthe condition q(0) = qB requires that dT (0)/dz = −qB/K, thus q0 = qB as has alreadybeen demonstrated.

Example 1.4.3. Heat flux through a slab having spatially-varying thermal conductivity

Consider the steady heat flux through a slab having thickness h and depth-varying thermalconductivity K(z) = K0 exp(−αz) where K0 and α are constants. The upper boundaryof the slab is maintained at constant temperature TS and the lower boundary at constanttemperature TB. Find the temperature distribution T (z) and heat flux q(z) in the slab.

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Answer

Vertical variation of the thermal conductivity has no affect on the fact that

q(z) = q0

(i.e. there is no vertical variation in heat flux). From Fourier’s law

dT

dz= − q0

K(z)

= − q0

K0 exp(−αz)

= −q0 exp(αz)K0

.

Integration of Fourier’s law gives

T (z) = − q0

αK0exp(αz) + c0.

Applying the boundary conditions T (0) = TB and T (h) = TS to the expression for T (z)gives

T (0) = TB = − q0

αK0+ c0

T (h) = TS = − q0

αK0exp(αh) + c0.

Solving for c0 and q0 yields

q0 =αK0(TB − TS)exp(αh) − 1

c0 = TB +TB − TS

exp(αh) − 1.

Thus

T (z) = TB − (TB − TS)exp(αz) − 1exp(αh) − 1

.

1.5 Solution for steady heat flow through a composite slab

Consider a layered medium comprising two homogeneous horizontal slabs (see Figure 1.5-1). The lower slab has thickness h1 and thermal conductivity K1 and the upper slab hasthickness h2 and thermal conductivity K2. The lower boundary of the lower slab is locatedon the z = 0 plane and is maintained at a constant temperature TB. The upper boundaryof the upper slab is located at z = h1 + h2 and is maintained at a constant temperatureTS . We seek the heat flux and temperature distribution in each slab.

9

z = 0

TB

TS

0

+z

z = h1

z = h + h21

h2

h1

K2

K1

+xFig. 1.5-1. Heat flux through a composite slab.

Let us denote the temperature distribution and heat flux in the lower slab as T1(z)and q1(z) respectively and for the upper slab T2(z) and q2(z). Following the same logicas that used in section 1.4, we recognize that the unknown temperature distributions havethe functional form T1 = T1(z) and T2 = T2(z) and that q1 = q0 k and q2 = q0 k. Notethat the heat flux is identical in the two slabs. Were this not the case, there would be adiscontinuity in the heat flux as one passed from slab 1 to slab 2. At such a discontinuitythermal energy would be either accumulate or deplete with time so the temperature at thediscontinuity would have to change with time—in violation of the assumed condition ofsteady heat flow.

Writing Fourier’s law for the two slabs gives

−K1∂T1

∂z= q0 (1.5-1a)

−K2∂T2

∂z= q0. (1.5-1b)

As in section (1.4), because T1 = T1(z) and T2 = T2(z), there is no distinction betweenpartial z derivates and ordinary z derivatives. Thus (1.5-1) can be written

dT1

dz= − q0

K1(1.5-2a)

dT2

dz= − q0

K2. (1.5-2b)

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Solving the above differential equations gives

T1(z) = − q0

K1z + c1 (1.5-3a)

T2(z) = − q0

K2z + c2 (1.5-3b)

where c1 and c2 are integration constants. In the above solutions q0, c1 and c2 are un-determined constants. Because there are three undetermined constants, we require threeadditional mathematical conditions to solve for these unknowns. The boundary conditionsfor this two-layered slab are

T1(0) = TB (1.5-4a)

T1(h1) = T2(h1) (1.5-4b)

T2(h1 + h2) = TS . (1.5-4c)

Conditions (1.5-4a) and (1.5-4c) are reminiscent of those found in section (1.4). Condition(1.5-4b) is a statement that temperature varies continuously across boundaries; thus thetemperature at the upper boundary of slab 1 and the lower boundary of slab 2 are identical.

Applying the boundary conditions (1.5-4) to the solution expression (1.5-3) yieldsthree equations in the three unknowns

T1(0) = c1 = TB (1.5-5a)

T1(h1) = −q0h1

K1+ TB = T2(h1) = −q0h1

K2+ c2 (1.5-5b)

T2(h1 + h2) = − q0

K2(h1 + h2) + c2 = TS (1.5-5c)

yielding (after some algebraic manipulation)

c1 = TB (1.5-6a)

c2 = TS +K1(TB − TS)(h1 + h2)

K2h1 + K1h2(1.5-6b)

q0 =K1K2(TB − TS)K2h1 + K1h2

. (1.5-6c)

Thus the final temperature solutions are written

T1(z) = TB − K2(TB − TS)zK2h1 + K1h2

(1.5-7a)

T2(z) = TS +K1(TB − TS)(h1 + h2 − z)

K2h1 + K1h2. (1.5-7b)

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1.6 Steady heat flow with cylindrical symmetry

Many interesting problems in engineering heat transfer, for example heat losses from hot-water pipes, involve cylindrical geometry. To begin, consider a cylindrical pipe havinginside radius a and outside radius b and length ` and suppose that the pipe is constructedof a substance having constant thermal conductivity K. Figure 1.6-1a shows this pipe andit has been drawn so that the z axis of the Cartesian coordinate system corresponds withthe axis of the cylindrical pipe. In such a rectangular coordinate system the inside andoutside walls of the pipe are given by the expressions x2 + y2 = a2 and x2 + y2 = b2 whichin a two-dimensional (x, y) coordinate system define circles and in a three-dimensionalsystem (x, y, z) define cylinders of radius a and b respectively. It is immediately apparentthat cylindrical objects do not fit comfortably in a rectangular coordinate system. Thus,as a first step to solving the problem of heat conduction in circular pipes, we introduce acoordinate system that is better suited to the problem under consideration (Fig. 1.6-1b).

+x

+y

+zl

a

bK

0

r

(a)

+x

+y

+z

0 x’

y’

rθz

(b)

(r, ,z)θ

Fig. 1.6-1. Steady heat conduction through a cylindrical pipe. (a) Pipegeometry. (b) Cylindrical polar coordinate system.

The cylindrical polar coordinate system is a three-dimensional outgrowth of the cir-cular polar coordinate system. To describe the location of points in three dimensions it isnecessary to introduce three coordinate variables each one of which is uniquely associatedwith spatial direction. In a rectangular (Cartesian) system these coordinates are (x, y, z)and the three directions are the unit vectors i, j and k. In cylindrical polar coordinatesthe three coordinate variables are r (the axial distance measured from the z axis), θ (theanti-clockwise rotation angle measured from the y = 0 plane) and z (the distance mea-sured along the z axis). The corresponding unit vectors are r̂ (a unit vector pointing inthe direction of increasing r), θ̂ (a unit vector pointing in the direction of increasing θ andk (a unit vector pointing in the direction of increasing z). Note that like i, j and k these

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unit vectors are mutually orthogonal but unlike i, j and k their alignment relative to x,y and z axis depends on spatial position (r, θ, z). In cylindrical polar coordinates, vectorfields such as the heat flux vector q are expressed in terms of the appropriate coordinatevariables and unit vectors. Thus the heat flux vector is written

q(r, θ, z, t) = qr(r, θ, z, t) r̂ + qθ(r, θ, z, t) θ̂ + qz(r, θ, z, t)k. (1.6-1)

Note that the heat flux vector q has three scalar components (qr, qθ, qz).Continuing with our analysis of heat transfer in pipes, let us assume that the inside

surface of the pipe is maintained at a constant temperature Ta and the outside surfaceof the pipe is maintained at a constant temperature Tb. Intuitively, we can infer that forsteady heat flow T = T (r) and q = qr r̂, that is the isothermal surfaces are cylindrical andthe heat flux is purely radial.

Let us now consider a length of pipe ` and a cylindrical surface of radius r wherea ≤ r ≤ b. The area of this surface is S(r) = 2πr`. The total rate of heat flow throughthis surface (i.e. the power measured in W) is

P (r) = S(r)qr(r), (1.6-2)

simply the product of surface area and heat flux. Special cases of (1.6-2) are the powerflow through the interior surface of the cylinder P (a) = S(a)qr(a) and the exterior surfaceP (b) = S(b)qr(b). For steady flow, it is necessary for P (a) = P (b). Were this not the casethere would be accumulation or depletion of thermal energy within the cylinder and thiswould lead to an increase or decrease in temperature with time. Thus it is apparent thatP (r) is a constant, say P0, and from (1.6-2) it follows that

S(r)qr (r) = P0 (1.6-3a)

2πr`qr = P0. (1.6-3b)

It can be shown that in cylindrical polar coordinates the gradient can be written

∇T (r, θ, z) =∂T

∂rr̂ +

1r

∂T

∂θθ̂ +

∂T

∂zk. (1.6-4)

We shall not prove this result but we shall use it to write

qr = −K∂T

∂r. (1.6-5)

For the case under consideration, T = T (r) so ∂/∂r = d/dr. From (1.6-3b) and (1.6-5) itfollows that

dT

dr= − P0

2πr`K. (1.6-6)

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The above differential equation (1.6-6) can be integrated to obtain

T (r) = − P0

2π`Kln r + c1. (1.6-7)

Note that the constant power flow P0 and integration constant c1 are both undeterminedconstants that must be evaluated using boundary conditions. For the given problem theboundary conditions are

T (a) = Ta (1.6-8a)

T (b) = Tb. (1.6-8b)

Applying (1.6-8a) and (1.6-8b) to (1.6-7) respectively gives

T (a) = Ta = − P0

2π`Kln a + c1 (1.6-9a)

T (b) = Tb = − P0

2π`Kln b + c1. (1.6-9b)

Solving (1.6-9a) and (1.6-9b) for c1 gives

c1 = Ta +P0

2π`Kln a. (1.6-10)

Substituting this value for c1 into (1.6-9b) gives

P0 =2π`K(Ta − Tb)

ln b − ln a=

2π`K(Ta − Tb)ln (b/a)

. (1.6-11)

Applying these values (1.6-7) gives the final temperature solution

T (r) = Ta − (Ta − Tb)ln (r/a)ln (b/a)

(1.6-12)

and substituting (1.6-11) into (1.6-3b) gives the final heat flux solution

qr(r) =K(Ta − Tb)r ln (b/a)

. (1.6-13)

Example 1.6.1. Melting the insulation of a current-carrying wire

Consider an insulated copper wire carrying an electrical current. The wire has radius0.25mm and is surrounded by an insulating vinyl jacket having thickness 0.40mm. Theresistance per unit length of the copper wire is 0.05 ohmm−1 and the vinyl insulation hasa thermal conductivity of 1.00W m−1 deg−1. The outside surface of the wire is maintainedat a constant temperature of 10 ◦C. Find the current flow (in Amps) that is required tobring the temperature at the contact between copper and vinyl to 100 ◦C.

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Answer

From (1.6-11)

P0 =2π`K(Ta − Tb)

ln (b/a).

In this problem the heat source is from current flow in the wire and P` = P0/` is theamount of heating per unit length of wire. Recalling that P = I2R = V I for elec-trical circuits it evident that the heating strength is P` = I2R` where R` is the re-sistance per unit length of wire. Taking a = 0.0025m, b = 0.0065m, Ta = 100 ◦C,Tb = 10 ◦C and K = 1.00W m−1 deg−1, we have P` = 2πK(Ta − Tb)/ ln (b/a) =2π × 1.00 × 90/ ln(0.0065/0.0025) = 592W m−1. The current flow that will produce thisamount of heating is given by I =

√P`/R` =

√592/0.05 = 108.8A.

1.7 Steady heat flow with spherical symmetry

The spherical polar coordinate system is a second example of what are referred to ascurvilinear coordinate systems. In this system the fundamental coordinates are r (radialdistance from the origin), θ (colatitude angle measured from the +z axis) and φ (azimuthangle). Figure 1.7-1 shows a spherical polar coordinate system and its relationship toa conventional Cartesian coordinate system. Three unit vectors are associated with the(r, θ, φ) coordinates: r̂ (a unit vector pointing in the direction of increasing r, i.e. radiallyoutward from the origin), θ̂ (a unit vector pointing in the direction of increasing θ) and φ̂(a unit vector pointing in the direction of increasing φ). Vector fields such as the heat fluxvector q can be expressed in spherical polar coordinates as follows:

q(r, θ, φ) = qr(r, θ, φ) r̂ + qθ(r, θ, φ) θ̂ + qφ(r, θ, φ) φ̂ (1.7-1)

and the gradient of scalar fields such as T can be written

∇T (r, θ, φ) =∂T

∂rr̂ +

1r

∂T

∂θθ̂ +

1r sin θ

∂T

∂φφ̂. (1.7-2)

(Proving (1.7-2) is not straightforward.)

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+x

+y

+z

0

r

θ

φ

(r, , )θ φ

z’

x’

y’

Fig. 1.7-1. Spherical polar coordinate system.

+x

+y

+z

0

a

r

b

Ta

Tb

K

Fig. 1.7-2. Steady heat flow through a spherical shell.

Now consider the steady heat flow through a spherical shell having thermal conduc-tivity K, inside radius a and outside radius b (Figure 1.7-2). The interior boundary ofthe shell is maintained at constant temperature Ta and the exterior boundary at constanttemperature Tb. From the assumptions of steady heat flow and the geometry of the prob-lem we immediately recognize that the temperature and heat flux respectively have thefunctional forms T = T (r) and q = qr(r) r̂. Denote P (r) as the flow of thermal power

16

across the spherical surface S(r) = 4πr2 where a ≤ r ≤ b. For steady-state conditions it isapparent that P (a) = P (b) and P (r) = P0. Thus

qr(r)S(r) = P0. (1.7-3)

From Fourier’s law of conduction and (1.7-2)

qr(r) = −KdT

dr(1.7-4)

so that (1.7-3) and (1.7-4) combine to give the differential equation

dT

dr= − P0

4πK

1r2

. (1.7-5)

Integrating (1.7-5) gives

T (r) =P0

4πKr+ c1 (1.7-6)

where P0 and c1 are undetermined constants. The boundary conditions are

T (a) = Ta (1.7-7a)

T (b) = Tb. (1.7-7b)

Applying (1.7-7a) and (1.7-7b) to (1.7-6) gives

P0

4πKa+ c1 = Ta (1.7-8a)

P0

4πKb+ c1 = Tb. (1.7-8b)

Subtracting (1.7-8b) from (1.7-8a) gives

P0

4πK

(1a− 1

b

)= Ta − Tb (1.7-9)

from which it follows thatP0 = 4πK(Ta − Tb)

ab

b − a. (1.7-10)

Solving (1.7-8a) for c1 gives

c1 = Ta − (Ta − Tb)b

b − a. (1.7-11)

Thus from (1.7-6), (1.7-10) and (1.7-11) the complete temperature solution is

T (r) = Ta − (Ta − Tb)b

b − a

(1 − a

r

)(1.7-12)

and the heat flux isqr(r) = K(Ta − Tb)

ab

b − a

1r2

. (1.7-13)

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Example 1.7.1. Steady heat flux from a spherical cavity

Suppose that a spherical cavity of radius a exists in an infinite homogeneous mediumhaving thermal conductivity K (Fig. 1.7-3). The cavity wall is maintained at constanttemperature Ta and at a very great distance from the cavity the temperature is T0. Assum-ing steady state conditions find the temperature distribution and heat flux in the regionsurrounding the cavity.

aTa

T ➛ T0

K

Fig. 1.7-3. Spherical cavity in a conducting medium.

Answer

The heat flux and temperature relations for a similar cavity can be readily obtained fromthe corresponding expressions for a spherical shell. Note that a spherical cavity is simplya shell having an infinite outside radius. Taking Tb = T0 and let a → ∞

qr(r) = K(Ta − T0)a

r2

T (r) = Ta − (Ta − T0)(1 − a

r

)

1.8 Measuring thermal conductivity and geothermal flux

This section remains to be written.

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1.9 Special interface conditions

1.9.1 Melting at the interface

Thus far we have restricted our discussion to situations where the addition of heat to asubstance results in an increase of temperature. When substances are at their meltingtemperature (referred to as the solidus), the addition of heat leads to a phase change fromthe solid to liquid state. This situation is interesting both geologically and geophysically.

Let us examine the situation when a slab of ice at its solidus temperature 0◦C is restingon a platform of bedrock. Suppose that the temperature at the ice–rock contact is 0◦Cand that a constant upward heat flux q1 flows from the rock (Figure 1.9-1). Consider theheat flowing through a cross-sectional area A of bed for a time interval ∆t. The amountof heat flowing across this area is

QIN = q1A∆t. (1.9-1)

Because the ice is assumed to be at its melting temperature, the addition of heat cannotresult in an elevation of ice temperature. Thus a volume of ice V = A∆z will be meltedwhere ∆z represents the thickness of the melted layer. The amount of thermal energyrequired to melt this volume of ice is

QV = ρLA∆z (1.9-2)

where ρ = 900 kg m−3 is the density of ice and L = 3.35 × 105 J kg−1 is the latent heat ofmelting for ice. Equating QIN and QV gives

q1A∆t = ρLA∆z (1.9-3)

from which we obtain∆z

∆t=

q1

ρL. (1.9-4)

In the limit for small ∆z and ∆t (1.9-4) gives

dz

dt=

q1

ρL. (1.9-5)

We can recognize that dz/dt has dimensions of velocity and it is in fact the rate of meltingof ice which we shall denote vm.

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Ice

RockK1

K2

➛q1

➛q2

➛q1

➛q2A

∆z

(a) (b)

Fig. 1.9-1. Melting at the boundary between two materials. (a) Contactbetween ice and rock. (b) Block of melted ice.

If the ice mass is not isothermal at 0◦C but has a non-vanishing temperature gradientdT2/dz at the ice–rock boundary then there will be a heat flux q2. If K2 is the thermalconductivity of ice then the magnitude of this heat flux is q2 = −K2dT2/dz. Clearly thisheat flux represents energy flow away from the boundary and is not therefore an energy flowthat can promote melting at the boundary. The heat flux available for melting, denotedqm, is the difference between the in-flowing and out-flowing fluxes qm = q1 − q2 and themelting rate is thus

vm =qm

ρL=

q1 − q2

ρL. (1.9-6)

Example 1.9.1. Rate of melting at the base of the Antarctic Ice Sheet

Lakes are known to exist beneath the Antarctic Ice Sheet. The best-studied site is LakeVostok, near the Russian research station Vostok, in East Antarctica. Take the surfacetemperature at Vostock as −58.0 ◦C and the ice–bed contact as −3.2C (the pressure-adjusted melting temperature of ice). The ice thickness is 3700m. Assuming that thegeothermal flux flowing into the base of the ice sheet is 0.07W m−2, find the bottom meltingrate. The density of ice is 900 kg m−3 and the latent heat of melting is 3.335× 105 J kg−1.

Answer

The heat flux toward the base of the glacier is q1 = 0.07W m−2 and that flowing intothe base of the glacier is q2 = K(TB − TS)/h = 2.1(−3.2 + 58.0)/3700 = 0.031W m−2.The heat flux available for melting is therefore qm = 0.039W m−2. The melting rate istherefore vm = qM/ρL = 0.039/(900 × 3.335 × 105) = 1.30× 10−10 ms−1 = 4.1mma−1.

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1.9.2 Heat generation by sliding

A second interface condition that has special relevance to geology and geophysics is thecase of frictional heat generation as one medium slides relative to another. Examplesinclude the frictional heating that accompanies motion on faults (leading to the geologicalfeature known as slickensides), the frictional interaction of tectonic plates (resulting in anadditional contribution to geothermal heat flow) and the sliding of a glacier or ice sheetover its bed.

Consider the plane contact between two semi-infinite media (Figure 1.9-2). If there isno relative motion between the two media and no melting at the contact then the heat fluxacross the boundary is continuous, i.e. q1 = q2. Now suppose that the upper medium issliding over the lower medium at some constant velocity v and that there is frictional heatgeneration produced by this sliding. This friction will introduce an additional contributionqf to the heat flux so that

q2 = q1 + qf . (1.9-7)

In other words, the outgoing heat flux q2 exceeds the incoming flux q1 by an amount qf .

(Medium 2)

(Medium 1) ➛q1

➛q2 ➟v

Fig. 1.9-2. Frictional heat generated by one slab sliding over another.

We aim to establish the mathematical form of the frictional heat flux. It seemsreasonable to postulate that qf is proportional to the sliding rate v and to some measureof the frictional interaction of the two media. Let us propose that

qf = Av (1.9-8)

where A is a constant of proportionality that indicates the degree of contact between themedia. Dimensional analysis of (1.9-8) leads to the conclusion that [A] = Pa (i.e. A has thedimensions of pressure). Let us first consider the possibility that A is in fact the contactpressure p between the two media. Does this make sense? Consider a skier of known mass

21

m skiing on skis of known surface area A. The downward gravitational force is F = mgwhere g = 9.80ms−2 and the pressure (force per unit area) on the ski surface p = mg/A. Ifthe skier waxes the skis, they run faster because friction is reduced even though the contactpressure remains the same. This simple thought experiment demonstrates that the contactpressure p is not the quantity that we require. In fact what is needed is the shear stress τacting at the interface between the two media. Like pressure this is a measure of force perunit area, but for shear stress the force is tangential to the surface. The correct expressionfor frictional heat flux is

qf = τv. (1.9-9)

Note that the foregoing discussion is not a proof, merely a plausibility argument.

Example 1.9.3. Frictional heat generated by the lithospheric sliding

Suppose a lithospheric plate is moving over the asthenosphere at 1 cma−1 and that theshear stress at the interface between the lithosphere and asthenosphere is 5MPa. What isthe contribution of this interaction to the geothermal heat flux?

Answer

The first step is to convert the velocity to standard S.I. units using the fact that there are(60)2(24)(365.25) = 3.15576 × 107 s a−1. Thus the sliding velocity is v = 0.01/3.15576 ×107 ms−1 and the frictional contribution to heat flux is qf = 0.01(5×106)/3.15576×107 =0.001585W m−2.

1.9.3 Sliding with melting

The case of combined sliding and melting is relevant to glaciers, ice sheets and certaintectonic situations. When a glacier is sliding over its bed there is a possibility that frictionalmelting can make an important contribution. The faster the sliding, the faster the rateof melting. Because sliding is lubricated by water there is a positive feedback that mightresult in a runaway increase in flow rate. Some glaciers, in fact, display a flow instabilitycalled “surging” so the possible role of frictional melting deserves a close examination.Balancing the heat fluxes gives

q2 − q1 = qf − qm (1.9-10)

where q1 is the geothermal heat flux flowing upward to the interface, q2 is the heat fluxescaping from the interface into overlying ice, qf is the frictionally-generated heat flux andqm is the heat flux that is extracted during the melting process. An alternative expressionthat conveys the same physics as (1.9-10) is

−K2dT2

dz+ K1

dT2

dz= vτ − ρL

dz

dt. (1.9-11)

22

Example 1.9.4. Frictional melting of an isothermal glacier

Consider an ice mass of density 900 kg m−3 which is isothermal at 0◦C. Suppose thegeothermal flux is 0.05W m−2 and basal shear stress is τ = 105 Pa. What is the rate ofbasal melting (a) if v = 10ma−1; (b) if v = 100ma−1.

Answer

Noting that q2 = 0 for an isothermal ice mass, (1.9-10) gives

dz

dt=

1ρL

(vτ + q1) .

For case (a) the melting rate is 2.72× 10−10 ms−1 = 8.6mma−1. For case (b) the meltingrate is 3.46× 10−2 ms−1 = 3.46 cma−1.

1.10 Distributed heat sources and heat flux

Earth’s crust and mantle are not only media through which heat can flow but are alsoimportant sources of heat. The radioactivity of crust and mantle rocks, though slight, isnevertheless an important contribution to the geothermal flux that escapes through Earth’ssurface.

1.10.1 Slab geometry

Thus far we have assumed that no internal heat sources are present. If the layers containradioactive materials this cannot be true. Here we examine the effect of radioactive heatproduction on geothermal flux.

Suppose the rate of internal heat generation per unit volume is a (which can be eitherconstant or spatially variable). Consider the rate of production in a block of materialhaving base area A and thickness ∆z (Figure 1.10-1). Noting that [a] = W m−3, the rateof heat flow into the volume A∆z in time ∆t is qA and the outflow is (q + ∆q)A. Becausethe internal heat production rate is aA∆z it follows that

(q + ∆q)A = qA + aA∆z (1.10-1)

which simplifies to ∆q/∆z = a or, in the limit of small ∆z,

dq

dz= a. (1.10-2)

Because q = −K dT/dz, (1.10-2) yields the expression

d

dz

(K

dT

dz

)= −a(z). (1.10-3)

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∆z

A

a

q➛

q+∆q➛

Fig. 1.10-1. Slab of radioactive material.

a

K

+z

+x

T0

q=0-d

0

q0➛

Fig. 1.10-2. Heat flux and temperature in a slab of radioactive material.

Example 1.10.1. Heat flux and temperature distribution in a radioactively-heated slab.

Suppose that the radioactive heat production is constant with depth, i.e. a(z) = a0.Assume that the surface temperature (at z = 0) is held at T0 and the surface heat fluxis constant at q0. Note that the heat flux will decrease with depth until at some depth dthe heat flux will vanish. Assume a positive-upward coordinate system (Figure 1.10-2) sothat z = −d is the level at which heat flux vanishes. Find the variation of temperatureand heat flux with z and find d.

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Answer

With a = a0dq

dz= a0

yieldingq(z) = a0z + c1

where c1 is an integration constant. From the boundary condition q(0) = q0 it follows thatq0 = c1 and thus q(z) = q0 + a0z. Note that z is negative in the lower half space thusq(z) is maximum at z = 0 and decreases as depth increases and z decreases. The depthat which q(z) vanishes is q(−d) = q0 − a0d = 0 yielding d = q0/a0 as the depth at whichheat flux vanishes. The temperature solution is found from

−KdT

dz= q(z)

which yields the differential equation

dT

dz= − 1

K(q0 + a0z).

Integrating the above gives

T (z) = − q0

Kz − a0

2Kz2 + c2

where c2 is an integration constant. From the boundary condition T (0) = T0, it followsthat c2 = T0 and the temperature distribution is given by

T (z) = T0 −q0

Kz − a0

2Kz2.

1.10.2 Spherical geometry

The case of spherical geometry is especially relevant to discussions of the thermal structureof moons and planets. Suppose that the distribution of radioactive heat sources has radialgeometry so that a(r) is the rate of heat production per unit volume and r is radial distancefrom the centre of the sphere. The radius of the sphere is assumed to be r0. Inside thesphere r ≤ r0, the rate of heat flow through the spherical surface having radius r is denotedP (r) and the heat flow at radial distance r + ∆r is denoted P (r + ∆r). It is immediatelyapparent that

P (r + ∆r) = P (r) + 4πr2a(r)∆r (1.10-4)

which for small ∆r yields the differential equation

dP

dr= 4πr2a. (1.10-5)

25

Noting that P (r) = 4πr2q(r) it follows that

1r2

d

dr

[r2q(r)

]= a(r). (1.10-6)

Further noting that q = −K dT/dr Equation (1.10-6) can be expressed

1r2

d

dr

[r2K

dT

dr

]= −a. (1.10-7)

Example 1.10.2. Radioactively heated sphere

Consider a sphere of radius r0 having constant thermal conductivity K and a spatiallyuniform distribution of radioactive heat sources a(r) = a0. Assume that the surfacetemperature of the sphere is held constant at temperature T0. Find the distribution ofheat flux and temperature with radial distance inside the sphere.

Answer

Starting from the equationd

dr

[r2q(r)

]= a0r

2

we find thatq(r) =

a0

3r +

c1

r2

where c1 is an integration constant. Note that at r = 0 the above expression leads toan infinite heat flux unless c1 vanishes. Infinite heat flux at the centre of the sphere isnon-physical (in fact the heat flux would be expected to vanish there since there can be noheat source contained within a vanishingly small sphere centred at r = 0). Thus it followsthat

q(r) =a0

3r.

Now turning attention to the temperature distribution, from q(r) = −K dT/dr we have

dT

dr= − a0

3Kr.

Integrating with respect to r gives

T (r) = − a0

6Kr2 + c2.

Applying the boundary condition T (r0) = T0 gives c2 = T0 + a0r20/6K, from which it

follows thatT (r) = T0 +

a0

6K(r2

0 − r2).

Incidentally, the temperature at the centre of the sphere is T (0) = T0 + a0r20/6K.

26