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LECTURES 4–11 Geophysical Heat Transfer 1.1 Introduction There Are

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LECTURES 4–11 Geophysical Heat Transfer 1.1 Introduction There Are 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 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 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 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 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 <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 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 temperatures 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.
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