LECTURES 4–11 Geophysical Transfer

1.1 Introduction There are three familiar mechanisms for : conduction, and radia- tion. The distinguishing features of these mechanisms are that involves energy transmission through matter, convection involves the energy that is carried with moving matter and radiation involves energy carried by electromagnetic waves (such as light) 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 material such as Earth’s crust. Let us begin by clarifying what we mean by the term heat flux. A useful approach is to examine the dimensions of several related quantities. We begin by considering some material body, for example a sphere of metal. Imagine that the object is at some high temperate T and that to test the an observer touches the object. In doing so, a certain amount of heat Q is transferred from the object to the observer; clearly the S.I. dimensions of Q are Joules, denoted J. Suppose that P denotes the rate at which heat flows 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 observer felt any discomfort when he or she touched the object? The answer is that neither Q nor P yield information of this sort. What is needed is the rate at which heat flows through some 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 as the 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 distribution and 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 that the lower surface is maintained at a constant temperature T1 and that the upper surface is 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 are maintained at the same temperature. Irrespective of whether this temperature is high or low, there will be no flux of heat through the material. Thus as a provisional observation we write q ∝ T2 − T1. (1.2-1)

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

T − T q ∝ 2 1 . (1.2-2) z2 − z1

1 Additionally, it is clear that the properties of the slab will influence how readily heat can flow through it. If the slab was constructed from metal, the heat flux would be larger than if the slab was constructed of some thermal insulator such as wood or ceramic. We therefore introduce a K characterizing the of the material. The larger the value of K the more conductive is the material; for a thermal insulator K takes small values; thermal conductivity is never negative. Thus we refine our provisional expression to T − T q = K 2 1 . (1.2-3) z2 − z1 A closer examination of (1.2-3) suggests that the expression needs a further improvement. If T2

T − T q = −K 2 1 . (1.2-4) z2 − z1

+z

T2 z = z 2

K

z = z T1 1

Fig. 1.2-1.0 Heat flux through a slab. +x

A more general expression is obtained by substituting ∆z = z2 −z1 and ∆T = T2 −T1 to obtain ∆T q = −K . (1.2-5) ∆z Expressed in the above form (1.2-5) seems like an invitation to let ∆z and ∆T shrink to infinitesimal values to obtain dT q = −K . (1.2-6) dz

2 It is readily confirmed that the dimensions if K are W m−1 deg−1. The above expression is sometimes termed Fourier’s law of conduction and is an example of what physicists call 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 or physics and therefore have lower status than equations such as A = πr2 and E = mc2. Unlike fundamental equations we never question whether constitutive equations are true or 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 only involves temperature gradients in the z direction. Temperature can vary with both space and 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 complication and 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 are unit vectors. Fourier’s law applies to each scalar component of the heat flux vector; thus

∂T q (x,y,z,t)=−K (1.3-2a) x ∂x ∂T q (x,y,z,t)=−K (1.3-2b) y ∂y ∂T q (x,y,z,t)=−K . (1.3-2c) z ∂z

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

∂T ∂T ∂T ∇T = i + j + k. (1.3-3) ∂x ∂y ∂z

∗ 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.

3 Example 1.3.1. Partial derivative of a scalar function Given the scalar function

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

find ∂f/∂x.

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

∂f sin(ay + bz) = exp(−y2) . ∂x (y2 + z2)

Example 1.3.2. Partial derivatives of a scalar function 2 3 Given the scalar function F (x,y,z,t)=x yz sin ω0t, find the partial derivatives with respect to x, y, z and t.

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

∂F =2xyz3 sin ω t ∂x 0 ∂F = x2z3 sin ω t ∂y 0 ∂F =3x2yz2 sin ω t ∂z 0 ∂F = ω x2yz3 cos ω t. ∂t 0 0

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 flux vector q. Answer Separate evaluation of the partial derivatives or T gives ∇T =2Ax i − 2Ay j. Thus from Fourier’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 and having a constant thermal conductivity K. Suppose that the lower boundary of the slab is at z = 0 and the upper boundary at z = h and that the upper and lower boundaries are respectively maintained at constant TS and TB respectively (see Figure 1.4-1).

+z

z = h TS

h K

z = 0 0 TB +x Fig. 1.4-1. Steady heat conduction through a homogeneous slab of thickness h.

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

5 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 some simple predictions about the mathematical form of q and T . In the most general case q = q(x,y,z,t) and T = T (x,y,z,t). These forms allow for arbitrary variation in both and time. The term “steady” rules out the possibility of variations with time, so our first simplification is to appreciate that q = q(x, y, z) and T = T (x, y, z) are the most complex mathematical forms that are consistent with steady heat conduction. Next, we can foresee that because of the slab geometry and the constant-temperature conditions at the slab boundaries, the most complex allowable functional form for the temperature variation is T = T (z). Given that q = −K∇T it immediately follows that the only non-vanishing component of the heat flux vector is the z component. Thus, at worst, q = qz(z) k. There is no heat flux in the x or y directions. Lastly, let us examine the possibility of z variation of the vertical component of heat flux qz(z). If qz varied with z then this would imply that there 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 thermal energy 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 are forced 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 single expression ∂T −K = q . (1.4-1) ∂z 0 Furthermore, because T is only a function of z there is no distinction between the partial derivative ∂/∂z and the ordinary derivative d/dz so (1.4-1) can be written

dT q = − 0 . (1.4-2) dz K 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 q T (z)=− 0 z + c (1.4-3) K 0

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

d2Y dY A +[BY 2 + C exp(Y )+Dx] + Y = sin x dx2 dx (a differential equation) and

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

(an algebraic equation).

6 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. The problem is not completed until these constants have been evaluated. To set values to the two undetermined constants, we require two additional equations. These equations, referred to as boundary conditions, follow from examining the diagram and searching for information that has not yet been used. The two as-yet unused pieces of information are the specified temperature values TB and TS . The equations that apply to these unused pieces 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 boundary of 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 follows that c0 = TB . Also from (1.4-3) we have T (h)=−(q0h/K)+c0 which with c0 = TB gives T (h)=−(q0h/K)+TB . From (1.4-4b) it follows that TS = −(q0h/K)+TB and that q0 = K(TB −TS )/h. Having now evaluated the two undetermined constants we can at last form the complete temperature solution z T (z)=T − (T − T ) . (1.4-5) B B S h

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 but the 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 by climate 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 −28.0 C; the bottom temperature is −2.0 C; the shelf thickness is 400 m. Assuming steady thermal conducting in an infinite slab, calculate the heat flux through the slab. Assume that the thermal conductivity of ice is 2.1Wm−1 deg−1.

Answer

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

7 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 has constant 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 the temperature 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 each boundary; 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 boundary conditions 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 q T (z)=− 0 z + c K 0

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

c0 = TB.

Thus q(z)=qB and T (z)=−qB z/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 applied directly and the result q(z)=qB emerged immediately. It is also possible to apply the boundary condition on q by working entirely with the expression for T (z). Following this tactic, we differentiate T (z)=−q0z/K + c0 to obtain dT (z)/dz = −q0/K and note that the condition q(0) = qB requires that dT (0)/dz = −qB/K, thus q0 = qB as has already been 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 thermal conductivity K(z)=K0 exp(−αz) where K0 and α are constants. The upper boundary of the slab is maintained at constant temperature TS and the lower boundary at constant temperature TB. Find the temperature distribution T (z) and heat flux q(z) in the slab.

8 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 q = − 0 dz K(z) q = − 0 K0 exp(−αz) q exp(αz) = − 0 . K0

Integration of Fourier’s law gives

q0 T (z)=− exp(αz)+c0. αK0

Applying the boundary conditions T (0) = TB and T (h)=TS to the expression for T (z) gives q0 T (0) = TB = − + c0 αK0 q0 T (h)=TS = − exp(αh)+c0. αK0

Solving for c0 and q0 yields αK (T − T ) q = 0 B S 0 exp(αh) − 1 T − T c = T + B S . 0 B exp(αh) − 1 Thus exp(αz) − 1 T (z)=T − (T − T ) . B B S exp(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 has thickness h2 and thermal conductivity K2. The lower boundary of the lower slab is located on the z = 0 plane and is maintained at a constant temperature TB. The upper boundary of the upper slab is located at z = h1 + h2 and is maintained at a constant temperature TS . We seek the heat flux and temperature distribution in each slab.

9 +z

z = h + h 1 2 TS

h K 2 2 z = h 1

h K 1 1 z = 0 T Fig. 1.5-1. 0 Heat flux through a composite slab. B +x

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 logic as that used in section 1.4, we recognize that the unknown temperature distributions have the functional form T1 = T1(z) and T2 = T2(z) and that q1 = q0 k and q2 = q0 k. Note that the heat flux is identical in the two slabs. Were this not the case, there would be a discontinuity in the heat flux as one passed from slab 1 to slab 2. At such a discontinuity thermal energy would be either accumulate or deplete with time so the temperature at the discontinuity would have to change with time—in violation of the assumed condition of steady heat flow. Writing Fourier’s law for the two slabs gives

∂T −K 1 = q (1.5-1a) 1 ∂z 0 ∂T −K 2 = q . (1.5-1b) 2 ∂z 0

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

dT q 1 = − 0 (1.5-2a) dz K1 dT q 2 = − 0 . (1.5-2b) dz K2

10 Solving the above differential equations gives

q0 T1(z)=− z + c1 (1.5-3a) K1 q0 T2(z)=− z + c2 (1.5-3b) K2 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 three additional mathematical conditions to solve for these unknowns. The boundary conditions for 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 the temperature 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) yields three equations in the three unknowns

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

q0h1 q0h1 T1(h1)=− + TB = T2(h1)=− + c2 (1.5-5b) K1 K2 q0 T2(h1 + h2)=− (h1 + h2)+c2 = TS (1.5-5c) K2 yielding (after some algebraic manipulation)

c1 = TB (1.5-6a)

K1(TB − TS )(h1 + h2) c2 = TS + (1.5-6b) K2h1 + K1h2

K1K2(TB − TS ) q0 = . (1.5-6c) K2h1 + K1h2 Thus the final temperature solutions are written

K2(TB − TS )z T1(z)=TB − (1.5-7a) K2h1 + K1h2

K1(TB − TS )(h1 + h2 − z) T2(z)=TS + . (1.5-7b) K2h1 + K1h2

11 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 having inside radius a and outside radius b and length ` and suppose that the pipe is constructed of a substance having constant thermal conductivity K. Figure 1.6-1a shows this pipe and it has been drawn so that the z axis of the Cartesian coordinate system corresponds with the axis of the cylindrical pipe. In such a rectangular coordinate system the inside and outside walls of the pipe are given by the expressions x2 + y2 = a2 and x2 + y2 = b2 which in a two-dimensional (x, y) coordinate system define circles and in a three-dimensional system (x, y, z) define cylinders of radius a and b respectively. It is immediately apparent that 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 a coordinate system that is better suited to the problem under consideration (Fig. 1.6-1b).

(a) (b) +y

+y y’

z θ (r,θ ,z) r a r 0 x’ +x 0 +x b K l +z +z

Fig. 1.6-1. Steady heat conduction through a cylindrical pipe. (a) Pipe geometry. (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 is necessary to introduce three coordinate variables each one of which is uniquely associated with 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 coordinates the three coordinate variables are r (the axial distance measured from the z axis), θ (the anti-clockwise rotation angle measured from the y = 0 plane) and z (the distance mea- sured along the z axis). The corresponding unit vectors are r (a unit vector pointing in the direction of increasing r), θ (a unit vector pointing in the directionb of increasing θ and b k (a unit vector pointing in the direction of increasing z). Note that like i, j and k these

12 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, vector fields such as the heat flux vector q are expressed in terms of the appropriate coordinate variables and unit vectors. Thus the heat flux vector is written

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

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 surface of the pipe is maintained at a constant temperature Tb. Intuitively, we can infer that for steady heat flow T = T (r) and q = qrr, that is the isothermal surfaces are cylindrical and the heat flux is purely radial. b Let us now consider a length of pipe ` and a cylindrical surface of radius r where a ≤ r ≤ b. The area of this surface is S(r)=2πr`. The total rate of heat flow through this 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 power flow through the interior surface of the cylinder P (a)=S(a)qr (a) and the exterior surface P (b)=S(b)qr (b). For steady flow, it is necessary for P (a)=P (b). Were this not the case there would be accumulation or depletion of thermal energy within the cylinder and this would lead to an increase or decrease in temperature with time. Thus it is apparent that P (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 1 ∂T ∂T ∇T (r, θ, z)= r + θ + k. (1.6-4) ∂r b r ∂θ b ∂z We shall not prove this result but we shall use it to write

∂T q = −K . (1.6-5) r ∂r

For the case under consideration, T = T (r)so∂/∂r = d/dr. From (1.6-3b) and (1.6-5) it follows that dT P = − 0 . (1.6-6) dr 2πr`K

13 The above differential equation (1.6-6) can be integrated to obtain P T (r)=− 0 ln r + c . (1.6-7) 2π`K 1

Note that the constant power flow P0 and integration constant c1 are both undetermined constants that must be evaluated using boundary conditions. For the given problem the boundary 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 P T (a)=T = − 0 ln a + c (1.6-9a) a 2π`K 1 P T (b)=T = − 0 ln b + c . (1.6-9b) b 2π`K 1

Solving (1.6-9a) and (1.6-9b) for c1 gives P c = T + 0 ln a. (1.6-10) 1 a 2π`K

Substituting this value for c1 into (1.6-9b) gives 2π`K(T − T ) 2π`K(T − T ) P = a b = a b . (1.6-11) 0 ln b − ln a ln (b/a)

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

ln (r/a) T (r)=T − (T − T ) (1.6-12) a a b ln (b/a)

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

K(T − T ) q (r)= a b . (1.6-13) r r ln (b/a)

Example 1.6.1. Melting the insulation of a current-carrying wire Consider an insulated wire carrying an electrical current. The wire has radius 0.25 mm and is surrounded by an insulating vinyl jacket having thickness 0.40 mm. The resistance per unit length of the copper wire is 0.05 ohm m−1 and the vinyl insulation has a thermal conductivity of 1.00 W m−1 deg−1. The outside surface of the wire is maintained at a constant temperature of 10 ◦C. Find the current flow (in Amps) that is required to bring the temperature at the contact between copper and vinyl to 100 ◦C.

14 Answer From (1.6-11) 2π`K(T − T ) P = a b . 0 ln (b/a)

In this problem the heat source is from current flow in the wire and P` = P0/` is the amount of heating per unit length of wire. Recalling that P = I2R = VI for elec- 2 trical circuits it evident that the heating strength is P` = I R` where R` is the re- ◦ sistance per unit length of wire. Taking a =0.0025 m, b =0.0065 m, Ta = 100 C, ◦ −1 −1 Tb =10C and K =1.00 W m deg , we have P` =2πK(Ta − Tb)/ ln (b/a)= 2π × 1.00 × 90/ ln(0.0065/0.0025) = 592 W m−1. The current flow that will produce this amount of heating is given by I = pP`/R` = p592/0.05 = 108.8A.

1.7 Steady heat flow with spherical symmetry The spherical polar coordinate system is a second example of what are referred to as curvilinear coordinate systems. In this system the fundamental coordinates are r (radial distance from the origin), θ (colatitude angle measured from the +z axis) and φ (azimuth angle). Figure 1.7-1 shows a spherical polar coordinate system and its relationship to a conventional Cartesian coordinate system. Three unit vectors are associated with the (r, θ, φ) coordinates: r (a unit vector pointing in the direction of increasing r, i.e. radially outward from the origin),b θ (a unit vector pointing in the direction of increasing θ) and φ b b (a unit vector pointing in the direction of increasing φ). Vector fields such as the heat flux vector q can be expressed in spherical polar coordinates as follows:

q(r, θ, φ)=qr(r, θ, φ) r + qθ(r, θ, φ) θ + qφ(r, θ, φ) φ (1.7-1) b b b and the gradient of scalar fields such as T can be written

∂T 1 ∂T 1 ∂T ∇T (r, θ, φ)= r + θ + φ. (1.7-2) ∂r b r ∂θ b r sin θ ∂φ b

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

15 +z z’

(r, θ , φ ) θ

r y’ +y 0 x’ φ +x

Fig. 1.7-1. Spherical polar coordinate system.

+z

b Tb r Ta a +y 0

K

Fig. 1.7-2. Steady heat+ flowx 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 of the shell is maintained at constant temperature Ta and the exterior boundary at constant temperature 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 the functional forms T = T (r) and q = qr(r) r. Denote P (r) as the flow of thermal power b

16 across the spherical surface S(r)=4πr2 where a ≤ r ≤ b. For steady-state conditions it is apparent 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) dT q (r)=−K (1.7-4) r dr so that (1.7-3) and (1.7-4) combine to give the differential equation dT P 1 = − 0 . (1.7-5) dr 4πK r2 Integrating (1.7-5) gives P T (r)= 0 + c (1.7-6) 4πKr 1 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 P 0 + c = T (1.7-8a) 4πKa 1 a P 0 + c = T . (1.7-8b) 4πKb 1 b Subtracting (1.7-8b) from (1.7-8a) gives P 1 1 0  −  = T − T (1.7-9) 4πK a b a b from which it follows that ab P =4πK(T − T ) . (1.7-10) 0 a b b − a

Solving (1.7-8a) for c1 gives b c = T − (T − T ) . (1.7-11) 1 a a b b − a Thus from (1.7-6), (1.7-10) and (1.7-11) the complete temperature solution is b a T (r)=T − (T − T ) 1 −  (1.7-12) a a b b − a r and the heat flux is ab 1 q (r)=K(T − T ) . (1.7-13) r a b b − a r2

17 Example 1.7.1. Steady heat flux from a spherical cavity Suppose that a spherical cavity of radius a exists in an infinite homogeneous medium having thermal conductivity K (Fig. 1.7-3). The cavity wall is maintained at constant temperature 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 region surrounding the cavity.

a Ta K

➛ T0 T

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 from the corresponding expressions for a spherical shell. Note that a spherical cavity is simply a shell having an infinite outside radius. Taking Tb = T0 and let a →∞ a q (r)=K(T − T ) r a 0 r2 a T (r)=T − (T − T ) 1 −  a a 0 r

1.8 Measuring thermal conductivity and geothermal flux This section remains to be written.

18 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 a substance results in an increase of temperature. When substances are at their melting temperature (referred to as the solidus), the addition of heat leads to a phase change from the 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 resting on a platform of bedrock. Suppose that the temperature at the ice–rock contact is 0◦C and that a constant upward heat flux q1 flows from the rock (Figure 1.9-1). Consider the heat flowing through a cross-sectional area A of bed for a time interval ∆t. The amount of 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 cannot result in an elevation of ice temperature. Thus a volume of ice V = A∆z will be melted where ∆z represents the thickness of the melted layer. The amount of thermal energy required 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 Jkg−1 is the latent heat of melting for ice. Equating QIN and QV gives

q1A∆t = ρLA∆z (1.9-3) from which we obtain ∆z q = 1 . (1.9-4) ∆t ρL In the limit for small ∆z and ∆t (1.9-4) gives

dz q = 1 . (1.9-5) dt ρL

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

19 (a) (b)

q

K ➛ 2 2 q A

Ice ➛ 2

Rock q ∆ ➛ 1 z K1 q

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

If the ice mass is not isothermal at 0◦C but has a non-vanishing temperature gradient dT2/dz at the ice–rock boundary then there will be a heat flux q2.IfK2 is the thermal conductivity of ice then the magnitude of this heat flux is q2 = −K2dT2/dz. Clearly this heat flux represents energy flow away from the boundary and is not therefore an energy flow that can promote melting at the boundary. The heat flux available for melting, denoted qm, is the difference between the in-flowing and out-flowing fluxes qm = q1 − q2 and the melting rate is thus q q − q v = m = 1 2 . (1.9-6) m ρL ρL

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 Lake Vostok, near the Russian research station Vostok, in East Antarctica. Take the surface temperature at Vostock as −58.0 ◦C and the ice–bed contact as −3.2 C (the pressure- adjusted melting temperature of ice). The ice thickness is 3700 m. Assuming that the geothermal flux flowing into the base of the ice sheet is 0.07 W m−2, find the bottom melting rate. The density of ice is 900 kg m−3 and the latent heat of melting is 3.335 × 105 Jkg−1.

Answer −2 The heat flux toward the base of the glacier is q1 =0.07 W m and that flowing into −2 the base of the glacier is q2 = K(TB − TS )/h =2.1(−3.2+58.0)/3700 = 0.031 W m . −2 The heat flux available for melting is therefore qm =0.039 W m . The melting rate is 5 −10 −1 −1 therefore vm = qM /ρL =0.039/(900 × 3.335 × 10 )=1.30 × 10 ms =4.1mma .

20 1.9.2 Heat generation by sliding A second interface condition that has special relevance to geology and is the case of frictional heat generation as one medium slides relative to another. Examples include the frictional heating that accompanies motion on faults (leading to the geological feature known as slickensides), the frictional interaction of tectonic plates (resulting in an additional contribution to geothermal heat flow) and the sliding of a glacier or ice sheet over its bed. Consider the plane contact between two semi-infinite media (Figure 1.9-2). If there is no relative motion between the two media and no melting at the contact then the heat flux across the boundary is continuous, i.e. q1 = q2. Now suppose that the upper medium is sliding over the lower medium at some constant velocity v and that there is frictional heat generation produced by this sliding. This friction will introduce an additional contribution qf 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 .

q v (Medium 2) ➛ 2 ➟ (Medium 1) q

➛ 1

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 seems reasonable to postulate that qf is proportional to the sliding rate v and to some measure of 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 the media. Dimensional analysis of (1.9-8) leads to the conclusion that [A] = Pa (i.e. A has the dimensions of pressure). Let us first consider the possibility that A is in fact the contact pressure 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 = mg where g =9.80 m s−2 and the pressure (force per unit area) on the ski surface p = mg/A.If the skier waxes the skis, they run faster because friction is reduced even though the contact pressure remains the same. This simple thought experiment demonstrates that the contact pressure p is not the quantity that we require. In fact what is needed is the shear stress τ acting at the interface between the two media. Like pressure this is a measure of force per unit area, but for shear stress the force is tangential to the surface. The correct expression for 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 cm a−1 and that the shear stress at the interface between the lithosphere and asthenosphere is 5 MPa. What is the 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 sa−1. Thus the sliding velocity is v =0.01/3.15576 × 7 −1 6 7 10 ms and the frictional contribution to heat flux is qf =0.01(5×10 )/3.15576×10 = 0.001585 W m−2.

1.9.3 Sliding with melting The case of combined sliding and melting is relevant to glaciers, ice sheets and certain tectonic situations. When a glacier is sliding over its bed there is a possibility that frictional melting can make an important contribution. The faster the sliding, the faster the rate of melting. Because sliding is lubricated by water there is a positive feedback that might result in a runaway increase in flow rate. Some glaciers, in fact, display a flow instability called “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 flux escaping from the interface into overlying ice, qf is the frictionally-generated heat flux and qm is the heat flux that is extracted during the melting process. An alternative expression that conveys the same physics as (1.9-10) is

dT dT dz −K 2 + K 2 = vτ − ρL . (1.9-11) 2 dz 1 dz dt

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 the geothermal flux is 0.05 W m−2 and basal shear stress is τ =105 Pa. What is the rate of basal melting (a) if v =10ma−1; (b) if v = 100 m a−1.

Answer

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

dz 1 = (vτ + q ) . dt ρL 1

For case (a) the melting rate is 2.72 × 10−10 ms−1 =8.6mma−1. For case (b) the melting rate is 3.46 × 10−2 ms−1 =3.46 cm a−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 also important sources of heat. The radioactivity of crust and mantle rocks, though slight, is nevertheless an important contribution to the geothermal flux that escapes through Earth’s surface.

1.10.1 Slab geometry Thus far we have assumed that no internal heat sources are present. If the layers contain radioactive materials this cannot be true. Here we examine the effect of radioactive heat production on geothermal flux. Suppose the rate of internal heat generation per unit volume is a (which can be either constant or spatially variable). Consider the rate of production in a block of material having base area A and thickness ∆z (Figure 1.10-1). Noting that [a]= Wm−3, the rate of heat flow into the volume A∆z in time ∆t is qA and the outflow is (q +∆q)A. Because the 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 = a. (1.10-2) dz

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

d dT K  = −a(z). (1.10-3) dz dz

23 q+∆q

A ➛

∆z a

q Fig. 1.10-1. Slab of radioactive material. ➛

+z

q T0 ➛ 0 0 +x a

K

-d q=0

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 flux is constant at q0. Note that the heat flux will decrease with depth until at some depth d the heat flux will vanish. Assume a positive-upward coordinate system (Figure 1.10-2) so that z = −d is the level at which heat flux vanishes. Find the variation of temperature and heat flux with z and find d.

24 Answer

With a = a0 dq = a dz 0 yielding q(z)=a0z + c1

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

dT −K = q(z) dz which yields the differential equation

dT 1 = − (q + a z). dz K 0 0 Integrating the above gives q a T (z)=− 0 z − 0 z2 + c K 2K 2

where c2 is an integration constant. From the boundary condition T (0) = T0, it follows that c2 = T0 and the temperature distribution is given by q a T (z)=T − 0 z − 0 z2. 0 K 2K

1.10.2 Spherical geometry The case of spherical geometry is especially relevant to discussions of the thermal structure of moons and planets. Suppose that the distribution of radioactive heat sources has radial geometry so that a(r) is the rate of heat production per unit volume and r is radial distance from the centre of the sphere. The radius of the sphere is assumed to be r0. Inside the sphere r ≤ r0, the rate of heat flow through the spherical surface having radius r is denoted P (r) and the heat flow at radial distance r +∆r is denoted P (r +∆r). It is immediately apparent that P (r +∆r)=P (r)+4πr2a(r)∆r (1.10-4) which for small ∆r yields the differential equation

dP =4πr2a. (1.10-5) dr

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

1 d r2q(r) = a(r). (1.10-6) r2 dr Further noting that q = −K dT/dr Equation (1.10-6) can be expressed

1 d dT r2K  = −a. (1.10-7) r2 dr dr

Example 1.10.2. Radioactively heated sphere

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

Answer Starting from the equation d 2 2 r q(r) = a0r dr we find that a c q(r)= 0 r + 1 3 r2

where c1 is an integration constant. Note that at r = 0 the above expression leads to an infinite heat flux unless c1 vanishes. Infinite heat flux at the centre of the sphere is non-physical (in fact the heat flux would be expected to vanish there since there can be no heat source contained within a vanishingly small sphere centred at r = 0). Thus it follows that a q(r)= 0 r. 3 Now turning attention to the temperature distribution, from q(r)=−K dT/dr we have

dT a = − 0 r. dr 3K Integrating with respect to r gives a T (r)=− 0 r2 + c . 6K 2

2 Applying the boundary condition T (r0)=T0 gives c2 = T0 + a0r0/6K, from which it follows that a T (r)=T + 0 (r2 − r2). 0 6K 0 2 Incidentally, the temperature at the centre of the sphere is T (0) = T0 + a0r0/6K.

26