LECTURES 4–11
Geophysical Heat Transfer
1.1
Introduction
There are three familiar mechanisms for heat transfer: conduction, convection and radiation. The distinguishing features of these mechanisms are that thermal conduction 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 solid 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 temperature 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 substance 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 steadystate 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
(1.2-1)
q ∝ T2 − T1 .
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
q∝
T2 − T1
.
z2 − z1
1
(1.2-2)
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 physical property K characterizing the thermal conductivity 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
T2 − T1
.
(1.2-3)
q=K
z2 − z1
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) has
a negative sign, implying that the direction of heat flow is downward from cold to hot
rather than from hot to cold. To repair this difficulty we write
q = −K
T2 − T1
.
z2 − z1
(1.2-4)
+z
T2
z=z2
K
z=z1
T1
0
Fig. 1.2-1.
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 heatconducting 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
∂x
∂T
qy (x, y, z, t) = −K
∂y
qx (x, y, z, t) = −K
qz (x, y, z, t) = −K
∂T
.
∂z
(1.3-2a)
(1.3-2b)
(1.3-2c)
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.
∂x
∂y
∂z
∗
(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.
3
Example 1.3.1. Partial derivative of a scalar function
Given the scalar function
f (x, y, z) = x exp(−y 2 )
sin(ay + bz)
(y 2 + z 2 )
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(−y 2 )
.
∂x
(y 2 + z 2 )
Example 1.3.2. Partial derivatives of a scalar function
Given the scalar function F (x, y, z, t) = x2 yz 3 sin ω0 t, 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
∂x
∂F
∂y
∂F
∂z
∂F
∂t
= 2xyz 3 sin ω0 t
= x2 z 3 sin ω0 t
= 3x2 yz 2 sin ω0 t
= ω0 x2 yz 3 cos ω0 t.
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 − y 2 ) describes the temperature distribution 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 distribution 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 temperatures TS and TB respectively (see Figure
1.4-1).
+z
TS
z=h
h
0
Fig. 1.4-1.
h.
K
z=0
TB
Steady heat conduction through a homogeneous slab of thickness
∗
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
+x
According to Fourier’s law of conduction q = −K∇T . By inspection of the accompanying 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
= q0 .
(1.4-1)
∂z
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
q0
=− .
dz
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
z + c0
K
∗
(1.4-3)
Differential equations are equations that involve derivatives. In contrast an algebraic
equation involves algebra without derivatives. Examples are
A
d2 Y
dY
+ Y = sin x
+ [BY 2 + C exp(Y ) + Dx]
2
dx
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 undetermined 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) = −(q0 h/K) + c0 which with c0 = TB gives
T (h) = −(q0 h/K) + TB . From (1.4-4b) it follows that TS = −(q0 h/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) = TB − (TB − TS ) .
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 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 2.1 W m−1 deg−1 .
Answer
For steady heat conduction through a slab q0 = K(TB − TS )/h. Thus for the Ross Ice
Shelf q0 = 2.1(−2.0 + 28.0)/400 = 136.5 mW m−2 .
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 temperature 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 illustrates 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 undetermined 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
T (z) = −
q0
z + c0
K
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) = −q0 z/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
q0
=−
dz
K(z)
=−
q0
K0 exp(−αz)
=−
q0 exp(αz)
.
K0
Integration of Fourier’s law gives
T (z) = −
q0
exp(αz) + c0 .
αK0
Applying the boundary conditions T (0) = TB and T (h) = TS to the expression for T (z)
gives
q0
+ c0
T (0) = TB = −
αK0
q0
T (h) = TS = −
exp(αh) + c0 .
αK0
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 )
1.5
exp(αz) − 1
.
exp(αh) − 1
Solution for steady heat flow through a composite slab
Consider a layered medium comprising two homogeneous horizontal slabs (see Figure 1.51). 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
h2
h1
z = h1
z=0
0
TS
2
Fig. 1.5-1. Heat flux through a composite slab.
K2
K1
TB
+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
∂T1
= q0
∂z
∂T2
−K2
= q0 .
∂z
−K1
(1.5-1a)
(1.5-1b)
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
q0
dT1
=−
dz
K1
(1.5-2a)
q0
dT2
=−
.
dz
K2
(1.5-2b)
10
Solving the above differential equations gives
q0
z + c1
K1
q0
T2 (z) = −
z + c2
K2
T1 (z) = −
(1.5-3a)
(1.5-3b)
where c1 and c2 are integration constants. In the above solutions q0 , c1 and c2 are undetermined 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
T1 (h1 ) = T2 (h1 )
T2 (h1 + h2 ) = TS .
(1.5-4a)
(1.5-4b)
(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)
q0 h1
q0 h1
+ TB = T2 (h1 ) = −
+ c2
K1
K2
q0
(h1 + h2 ) + c2 = TS
T2 (h1 + h2 ) = −
K2
T1 (h1 ) = −
(1.5-5b)
(1.5-5c)
yielding (after some algebraic manipulation)
c1 = TB
c2 = TS +
q0 =
(1.5-6a)
K1 (TB − TS )(h1 + h2 )
K2 h1 + K1 h2
K1 K2 (TB − TS )
.
K2 h1 + K1 h2
(1.5-6b)
(1.5-6c)
Thus the final temperature solutions are written
T1 (z) = TB −
K2 (TB − TS )z
K2 h1 + K1 h2
(1.5-7a)
T2 (z) = TS +
K1 (TB − TS )(h1 + h2 − z)
.
K2 h1 + K1 h2
(1.5-7b)
11
1.6
Steady heat flow with cylindrical symmetry
Many interesting problems in engineering heat transfer, for example heat losses from hotwater 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 + y 2 = a2 and x2 + y 2 = 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’
θ
z
(r,θ,z)
r
a
r
b
0
+x
0
K
l
x’
+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 circular 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 measured along the z axis). The corresponding unit vectors are b
r (a unit vector pointing in
b
the direction of increasing r), θ (a unit vector pointing in the direction of increasing θ and
k (a unit vector pointing in the direction of increasing z). Note that like i, j and k these
12
+x
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
r + qθ (r, θ, z, t) θb + qz (r, θ, z, t) k.
q(r, θ, z, t) = qr (r, θ, z, t) b
(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 surface
of the pipe is maintained at a constant temperature Tb . Intuitively, we can infer that for
r, that is the isothermal surfaces are cylindrical and
steady heat flow T = T (r) and q = qr b
the heat flux is purely radial.
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
2πr`qr = P0 .
(1.6-3a)
(1.6-3b)
It can be shown that in cylindrical polar coordinates the gradient can be written
∇T (r, θ, z) =
1 ∂T b ∂T
∂T
b
r+
θ+
k.
∂r
r ∂θ
∂z
(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) it
follows that
P0
dT
=−
.
(1.6-6)
dr
2πr`K
13
The above differential equation (1.6-6) can be integrated to obtain
T (r) = −
P0
ln r + c1.
2π`K
(1.6-7)
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
P0
ln a + c1
2π`K
P0
ln b + c1 .
T (b) = Tb = −
2π`K
T (a) = Ta = −
(1.6-9a)
(1.6-9b)
Solving (1.6-9a) and (1.6-9b) for c1 gives
c1 = Ta +
P0
ln a.
2π`K
(1.6-10)
Substituting this value for c1 into (1.6-9b) gives
P0 =
2π`K(Ta − Tb )
2π`K(Ta − Tb )
=
.
ln b − ln a
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 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)
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 the
amount of heating per unit length of wire. Recalling that P = I 2 R = V I for electrical circuits it evident that the heating strength is P` = I 2 R` where R` is the resistance per unit length of wire. Taking a = 0.0025 m, b = 0.0065 m, Ta = 100 ◦ C,
Tb = 10 ◦ C and K = 1.00 W m−1 deg−1 , we have P` = 2πK(Ta − Tb )/ ln (b/a) =
−1
2π × 1.00 × 90/ ln(0.0065/0.0025) =
p592 W m p. The current flow that will produce this
amount of heating is given by I = P` /R` = 592/0.05 = 108.8 A.
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: b
r (a unit vector pointing in the direction of increasing r, i.e. radially
b
outward from the origin), θb (a unit vector pointing in the direction of increasing θ) and φ
(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) b
r + qθ (r, θ, φ) θb + qφ (r, θ, φ) φb
(1.7-1)
and the gradient of scalar fields such as T can be written
∇T (r, θ, φ) =
1 ∂T b
1 ∂T b
∂T
b
r+
θ+
φ.
∂r
r ∂θ
r sin θ ∂φ
(Proving (1.7-2) is not straightforward.)
15
(1.7-2)
+z
z’
(r, θ,φ)
θ
r
y’
0
+y
x’
φ
+x
Fig. 1.7-1. Spherical polar coordinate system.
+z
b
Tb
r
Ta
a
0
+y
K
Fig. 1.7-2. Steady heat +x
flow through a spherical shell.
Now consider the steady heat flow through a spherical shell having thermal conductivity 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 problem we immediately recognize that the temperature and heat flux respectively have the
r . Denote P (r) as the flow of thermal power
functional forms T = T (r) and q = qr (r) 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)
qr (r) = −K
dT
dr
(1.7-4)
so that (1.7-3) and (1.7-4) combine to give the differential equation
dT
P0 1
=−
.
dr
4πK r2
(1.7-5)
Integrating (1.7-5) gives
P0
+ c1
4πKr
where P0 and c1 are undetermined constants. The boundary conditions are
T (r) =
(1.7-6)
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
+ c1 = Ta
4πKa
P0
+ c1 = Tb .
4πKb
Subtracting (1.7-8b) from (1.7-8a) gives
P0
1 1
−
= Ta − Tb
4πK a
b
from which it follows that
(1.7-8a)
(1.7-8b)
(1.7-9)
P0 = 4πK(Ta − Tb )
ab
.
b−a
(1.7-10)
c1 = Ta − (Ta − Tb )
b
.
b−a
(1.7-11)
Solving (1.7-8a) for c1 gives
Thus from (1.7-6), (1.7-10) and (1.7-11) the complete temperature solution is
a
b 1−
T (r) = Ta − (Ta − Tb )
b−a
r
and the heat flux is
qr (r) = K(Ta − Tb )
17
ab 1
.
b − a r2
(1.7-12)
(1.7-13)
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 . Assuming steady state conditions find the temperature distribution and heat flux in the region
surrounding the cavity.
Ta
a
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 → ∞
qr (r) = K(Ta − T0 )
a
r2
a
T (r) = Ta − (Ta − T0 ) 1 −
r
1.8
Measuring thermal conductivity and geothermal flux
This section remains to be written.
18
1.9
1.9.1
Special interface conditions
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 = q1 A∆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 J kg−1 is the latent heat of
melting for ice. Equating QIN and QV gives
from which we obtain
q1 A∆t = ρLA∆z
(1.9-3)
q1
∆z
=
.
∆t
ρL
(1.9-4)
In the limit for small ∆z and ∆t (1.9-4) gives
dz
q1
=
.
dt
ρL
(1.9-5)
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
Rock
q
1
A
2
q
K1
2
q
➛ ➛
Ice
K2
➛
(b)
➛
(a)
∆z
1
q
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 . If K2 is the thermal
conductivity of ice then the magnitude of this heat flux is q2 = −K2 dT2 /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
qm
q1 − q2
=
.
(1.9-6)
vm =
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 pressureadjusted 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 J kg−1 .
Answer
The heat flux toward the base of the glacier is q1 = 0.07 W m−2 and that flowing into
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−2 . The melting rate is
therefore vm = qM /ρL = 0.039/(900 × 3.335 × 105 ) = 1.30 × 10−10 m s−1 = 4.1 mm a−1 .
20
1.9.2
Heat generation by sliding
A second interface condition that has special relevance to geology and geophysics 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
(1.9-7)
q2 = q1 + qf .
➛ ➛
In other words, the outgoing heat flux q2 exceeds the incoming flux q1 by an amount qf .
q
(Medium 2)
2
v
➟
q
(Medium 1)
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
(1.9-9)
qf = τ v.
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 s a−1 . Thus the sliding velocity is v = 0.01/3.15576 ×
107 m s−1 and the frictional contribution to heat flux is qf = 0.01(5 × 106 )/3.15576 × 107 =
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
(1.9-10)
q2 − q1 = qf − qm
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
−K2
dz
dT2
dT2
+ K1
= vτ − ρL .
dz
dz
dt
22
(1.9-11)
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 = 10 m a−1 ; (b) if v = 100 m a−1 .
Answer
Noting that q2 = 0 for an isothermal ice mass, (1.9-10) gives
1
dz
=
(vτ + q1 ) .
dt
ρL
For case (a) the melting rate is 2.72 × 10−10 m s−1 = 8.6 mm a−1. For case (b) the melting
rate is 3.46 × 10−2 m s−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] = W m−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.
dz
(1.10-2)
Because q = −K dT /dz, (1.10-2) yields the expression
d
dz
dT
K
= −a(z).
dz
23
(1.10-3)
➛
∆z
q+∆q
➛
A
q
a
Fig. 1.10-1. Slab of radioactive material.
+z
0
q0
➛
T0
+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
= a0
dz
yielding
q(z) = a0 z + c1
where c1 is an integration constant. From the boundary condition q(0) = q0 it follows that
q0 = c1 and thus q(z) = q0 + a0 z. 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 − a0 d = 0 yielding d = q0 /a0 as the depth at which
heat flux vanishes. The temperature solution is found from
−K
dT
= q(z)
dz
which yields the differential equation
dT
1
= − (q0 + a0 z).
dz
K
Integrating the above gives
T (z) = −
a0 2
q0
z−
z + c2
K
2K
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
T (z) = T0 −
1.10.2
a0 2
q0
z−
z .
K
2K
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
(1.10-4)
P (r + ∆r) = P (r) + 4πr2 a(r) ∆r
which for small ∆r yields the differential equation
dP
= 4πr2 a.
dr
25
(1.10-5)
Noting that P (r) = 4πr2 q(r) it follows that
1 d 2
r
q(r)
= a(r).
r2 dr
Further noting that q = −K dT /dr Equation (1.10-6) can be expressed
dT
1 d
2
r K
= −a.
r2 dr
dr
(1.10-6)
(1.10-7)
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
r q(r) = a0 r2
dr
we find that
c1
a0
r+ 2
3
r
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
a0
r.
q(r) =
3
Now turning attention to the temperature distribution, from q(r) = −K dT /dr we have
q(r) =
a0
dT
=−
r.
dr
3K
Integrating with respect to r gives
T (r) = −
a0 2
r + c2 .
6K
Applying the boundary condition T (r0 ) = T0 gives c2 = T0 + a0 r02 /6K, from which it
follows that
a0 2
(r − r2 ).
T (r) = T0 +
6K 0
Incidentally, the temperature at the centre of the sphere is T (0) = T0 + a0 r02 /6K.
26