Lecture notes on Heat Conduction

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4. Analysis of heat conduction and some steady one-dimensional problems The effects of heat are subject to constant laws which cannot be discovered without the aid of mathematical analysis. The object of the theory which we are about to explain is to demonstrate these laws; it reduces all physical researches on the propagation of heat to problems of the calculus whose elements are given by experiment. The Analytical Theory of Heat, J. Fourier, 1822 4.1 The well-posed problem The heat diffusion equation was derived in Section 2.1 and some atten- tion was given to its solution. Before we go further with heat conduction problems, we must describe how to state such problems so they can re- ally be solved. This is particularly important in approaching the more complicated problems of transient and multidimensional heat conduc- tion that we have avoided up to now. A well-posed heat conduction problem is one in which all the relevant information needed to obtain a unique solution is stated. A well-posed and hence solvable heat conduction problem will always read as follows: FindT(x,y,z,t) such that: 1. ∂T ∇·(k∇T)+q ˙=ρc ∂t for 0tT (whereT can →∞), and for(x,y,z) belonging to 141142 Analysis of heat conduction and some steady one-dimensional problems §4.1 1 some region,R, which might extend to infinity. 2. T =T(x,y,z) at t= 0 i This is called an initial condition, or i.c. (a) Condition 1 above is not imposed att= 0. (b) Only one i.c. is required. However, (c) The i.c. is not needed: i. In the steady-state case:∇·(k∇T)+q ˙= 0. ii. For “periodic” heat transfer, whereq ˙ or the boundary con- ditions vary periodically with time, and where we ignore the starting transient behavior. 3. T must also satisfy two boundary conditions, or b.c.’s, for each co- ordinate. The b.c.’s are very often of three common types. (a) Dirichlet conditions, or b.c.’s of the first kind: T is specified on the boundary ofR fort 0. We saw such b.c.’s in Examples 2.1, 2.2, and 2.5. (b) Neumann conditions, or b.c.’s of the second kind: The derivative ofT normal to the boundary is specified on the boundary ofR fort 0. Such a condition arises when the heat flux,k(∂T/∂x), is specified on a boundary or when , with the 2 help of insulation, we set∂T/∂x equal to zero. (c) b.c.’s of the third kind: A derivative ofT in a direction normal to a boundary is propor- tional to the temperature on that boundary. Such a condition most commonly arises when convection occurs at a boundary, and it is typically expressed as   ∂T  −k =h(T−T )  ∞ bndry ∂x bndry when the body lies to the left of the boundary on thex-coor- dinate. We have already used such a b.c. in Step 4 of Example 2.6, and we have discussed it in Section 1.3 as well. 1 (x,y,z) might be any coordinates describing a positionr:T(x,y,z,t)=T(r, t). 2 Although we write∂T/∂x here, we understand that this might be∂T/∂z,∂T/∂r, or any other derivative in a direction locally normal to the surface on which the b.c. is specified.§4.2 The general solution 143 Figure 4.1 The transient cooling of a body as it might occur, subject to boundary conditions of the first, second, and third kinds. This list of b.c.’s is not complete, by any means, but it includes a great number of important cases. Figure 4.1 shows the transient cooling of body from a constant initial temperature, subject to each of the three b.c.’s described above. Notice that the initial temperature distribution is not subject to the boundary condition, as pointed out previously under 2(a). The eight-point procedure that was outlined in Section 2.2 for solving the heat diffusion equation was contrived in part to assure that a problem will meet the preceding requirements and will be well posed. 4.2 The general solution Once the heat conduction problem has been posed properly, the first step in solving it is to find the general solution of the heat diffusion equation. We have remarked that this is usually the easiest part of the problem. We next consider some examples of general solutions.144 Analysis of heat conduction and some steady one-dimensional problems §4.2 One-dimensional steady heat conduction Problem 4.1 emphasizes the simplicity of finding the general solutions of linear ordinary differential equations, by asking for a table of all general solutions of one-dimensional heat conduction problems. We shall work out some of those results to show what is involved. We begin the heat diffusion equation with constantk andq ˙: q ˙ 1∂T 2 ∇ T+ = (2.11) k α∂t Cartesian coordinates: Steady conduction in the y-direction. Equation (2.11) reduces as follows: 2 2 2 ∂ T ∂ T ∂ T q ˙ 1∂T + + + = 2 2 2 ∂x ∂y ∂z k α∂t          =0 =0 = 0, since steady Therefore, 2 d T q ˙ =− 2 dy k which we integrate twice to get q ˙ 2 T=− y +C y+C 1 2 2k or, ifq ˙= 0, T =C y+C 1 2 Cylindrical coordinates with a heat source: Tangential conduction. This time, we look at the heat flow that results in a ring when two points are held at different temperatures. We now express eqn. (2.11) in cylin- drical coordinates with the help of eqn. (2.13):   2 2 1 ∂ ∂T 1 ∂ T ∂ T q ˙ 1∂T r + + + = 2 2 2 r∂r ∂r r ∂φ ∂z k α∂t             =0 = 0, since steady r=constant =0 Two integrations give 2 r q ˙ 2 T=− φ +C φ+C (4.1) 1 2 2k This would describe, for example, the temperature distribution in the thin ring shown in Fig. 4.2. Here the b.c.’s might consist of temperatures specified at two angular locations, as shown.§4.2 The general solution 145 Figure 4.2 One-dimensional heat conduction in a ring. T = T(t only) IfT is spatially uniform, it can still vary with time. In such cases q ˙ 1∂T 2 ∇ T+ = k α∂t    =0 and∂T/∂t becomes an ordinary derivative. Then, sinceα=k/ρc, dT q ˙ = (4.2) dt ρc This result is consistent with the lumped-capacity solution described in Section 1.3. If the Biot number is low and internal resistance is unimpor- tant, the convective removal of heat from the boundary of a body can be prorated over the volume of the body and interpreted as h(T −T )A body ∞ 3 q ˙ =− W/m (4.3) effective volume and the heat diffusion equation for this case, eqn. (4.2), becomes dT hA =− (T−T ) (4.4) ∞ dt ρcV The general solution in this situation was given in eqn. (1.21). A partic- ular solution was also written in eqn. (1.22).146 Analysis of heat conduction and some steady one-dimensional problems §4.2 Separation of variables: A general solution of multidimensional problems Suppose that the physical situation permits us to throw out all but one of the spatial derivatives in a heat diffusion equation. Suppose, for example, that we wish to predict the transient cooling in a slab as a function of the location within it. If there is no heat generation, the heat diffusion equation is 2 ∂ T 1∂T = (4.5) 2 ∂x α∂t A common trick is to ask: “Can we find a solution in the form of a product of functions of t and x: T =T(t)·X(x)?” To find the answer, we substitute this in eqn. (4.5) and get 1   X T= T X (4.6) α where each prime denotes one differentiation of a function with respect   2 2 to its argument. ThusT = dT/dt andX = d X/dx . Rearranging eqn. (4.6), we get   X 1 T = (4.7a) X α T This is an interesting result in that the left-hand side depends only uponx and the right-hand side depends only upont. Thus, we set both 2 sides equal to the same constant, which we call−λ , instead of, say,λ, for reasons that will be clear in a moment:   X 1T 2 = =−λ a constant (4.7b) X α T It follows that the differential eqn. (4.7a) can be resolved into two ordi- nary differential equations:  2  2 X =−λ X and T =−αλ T (4.8) The general solution of both of these equations are well known and are among the first ones dealt with in any study of differential equations. They are: X(x)=A sinλx+B cosλx for λ ≠ 0 (4.9) X(x)=Ax+B for λ= 0§4.2 The general solution 147 and 2 −αλ t T(t)=Ce for λ ≠ 0 (4.10) T(t)=C for λ= 0 where we use capital letters to denote constants of integration. In ei- ther case, these solutions can be verified by substituting them back into eqn. (4.8). Thus the general solution of eqn. (4.5) can indeed be written in the form of a product, and that product is 2 −αλ t T=XT=e (D sinλx+E cosλx) for λ ≠ 0 (4.11) T=XT=Dx+E for λ= 0 The usefulness of this result depends on whether or not it can be fit to the b.c.’s and the i.c. In this case, we made the functionX(t) take the form of sines and cosines (instead of exponential functions) by placing 2 a minus sign in front ofλ . The sines and cosines make it possible to fit the b.c.’s using Fourier series methods. These general methods are not developed in this book; however, a complete Fourier series solution is presented for one problem in Section 5.3. The preceding simple methods for obtaining general solutions of lin- ear partial d.e.’s is called the method of separation of variables. It can be applied to all kinds of linear d.e.’s. Consider, for example, two-dimen- sional steady heat conduction without heat sources: 2 2 ∂ T ∂ T + = 0 (4.12) 2 2 ∂x ∂y SetT=XY and get   X Y 2 =− =−λ X Y whereλ can be an imaginary number. Then ⎫ ⎬ X=A sinλx+B cosλx forλ ≠ 0 λy −λy ⎭ Y=Ce +De 1 X=Ax+B forλ= 0 Y=Cy+D The general solution is −λy λy T =(E sinλx+F cosλx)(e +Ge ) forλ ≠ 0 (4.13) T =(Ex+F)(y+G) forλ= 0148 Analysis of heat conduction and some steady one-dimensional problems §4.2 Figure 4.3 A two-dimensional slab maintained at a constant temperature on the sides and subjected to a sinusoidal varia- tion of temperature on one face. Example 4.1 ◦ A long slab is cooled to 0 C on both sides and a blowtorch is turned on the top edge, giving an approximately sinusoidal temperature dis- tribution along the top, as shown in Fig. 4.3. Find the temperature distribution within the slab. Solution. The general solution is given by eqn. (4.13). We must therefore identify the appropriate b.c.’s and then fit the general solu- tion to it. Those b.c.’s are: x on the top surface : T(x, 0)=A sinπ L on the sides : T(0orL,y)= 0 asy →∞ : T(x,y→∞)= 0 Substitute eqn. (4.13) in the third b.c.: (E sinλx+F cosλx)(0+G·∞)= 0 The only way that this can be true for allx is ifG = 0. Substitute eqn. (4.13), withG= 0, into the second b.c.: −λy (O+F)e = 0§4.2 The general solution 149 soF also equals 0. Substitute eqn. (4.13) withG=F= 0, into the first b.c.: x E(sinλx)=A sinπ L It follows that A = E and λ = π/L. Then eqn. (4.13) becomes the particular solution that satisfies the b.c.’s:   x −πy/L T =A sinπ e L Thus, the sinusoidal variation of temperature at the top of the slab is attenuated exponentially at lower positions in the slab. At a position ofy = 2L below the top,T will be 0.0019A sinπx/L. The tempera- ture distribution in thex-direction will still be sinusoidal, but it will have less than 1/500 of the amplitude aty= 0. Consider some important features of this and other solutions: • The b.c. at y = 0 is a special one that works very well with this particular general solution. If we had tried to fit the equation to a general temperature distribution,T(x,y = 0)= fn(x), it would not have been obvious how to proceed. Actually, this is the kind of problem that Fourier solved with the help of his Fourier series method. We discuss this matter in more detail in Chapter 5. • Not all forms of general solutions lend themselves to a particular set of boundary and/or initial conditions. In this example, we made the process look simple, but more often than not, it is in fitting a general solution to a set of boundary conditions that we get stuck. • Normally, on formulating a problem, we must approximate real be- havior in stating the b.c.’s. It is advisable to consider what kind of assumption will put the b.c.’s in a form compatible with the gen- eral solution. The temperature distribution imposed on the slab by the blowtorch in Example 4.1 might just as well have been ap- proximated as a parabola. But as small as the difference between a parabola and a sine function might be, the latter b.c. was far easier to accommodate. • The twin issues of existence and uniqueness of solutions require a comment here: It has been established that solutions to all well- posed heat diffusion problems are unique. Furthermore, we know150 Analysis of heat conduction and some steady one-dimensional problems §4.3 from our experience that if we describe a physical process correctly, a unique outcome exists. Therefore, we are normally safe to leave these issues to a mathematician—at least in the sort of problems we discuss here. • Given that a unique solution exists, we accept any solution as cor- rect since we have carved it to fit the boundary conditions. In this sense, the solution of differential equations is often more of an in- centive than a formal operation. The person who does it best is often the person who has done it before and so has a large assort- ment of tricks up his or her sleeve. 4.3 Dimensional analysis Introduction Most universities place the first course in heat transfer after an introduc- tion to fluid mechanics: and most fluid mechanics courses include some dimensional analysis. This is normally treated using the familiar method of indices, which is seemingly straightforward to teach but is cumbersome and sometimes misleading to use. It is rather well presented in 4.1. The method we develop here is far simpler to use than the method of indices, and it does much to protect us from the common errors we might fall into. We refer to it as the method of functional replacement. The importance of dimensional analysis to heat transfer can be made clearer by recalling Example 2.6, which (like most problems in Part I) in- volved several variables. Theses variables included the dependent vari- 3 able of temperature,(T −T); the major independent variable, which ∞ i was the radius,r; and five system parameters,r,r ,h,k, and(T −T). i o ∞ i By reorganizing the solution into dimensionless groups eqn. (2.24), we reduced the total number of variables to only four: ⎡ ⎤ T−T i ⎢ ⎥ = fn⎣ r r, r r, Bi ⎦ (2.24a) i o i T −T ∞ i          two system parameters dependent variable indep. var. This solution offered a number of advantages over the dimensional solution. For one thing, it permitted us to plot all conceivable solutions 3 Notice that we do not call T a variable. It is simply the reference temperature i ◦ against which the problem is worked. If it happened to be 0 C, we would not notice its subtraction from the other temperatures.§4.3 Dimensional analysis 151 for a particular shape of cylinder, (r /r), in a single figure, Fig. 2.13. o i For another, it allowed us to study the simultaneous roles ofh,k andr o in defining the character of the solution. By combining them as a Biot number, we were able to say—even before we had solved the problem— whether or not external convection really had to be considered. The nondimensionalization made it possible for us to consider, simul- taneously, the behavior of all similar systems of heat conduction through cylinders. Thus a large, highly conducting cylinder might be similar in its behavior to a small cylinder with a lower thermal conductivity. Finally, we shall discover that, by nondimensionalizing a problem be- fore we solve it, we can often greatly simplify the process of solving it. Our next aim is to map out a method for nondimensionalization prob- lems before we have solved then, or, indeed, before we have even written the equations that must be solved. The key to the method is a result called the Buckingham pi-theorem. The Buckingham pi-theorem The attention of scientific workers was drawn very strongly toward the question of similarity at about the beginning of World War I. Buckingham first organized previous thinking and developed his famous theorem in 1914 in the Physical Review 4.2, and he expanded upon the idea in the Transactions of the ASME one year later 4.3. Lord Rayleigh almost si- multaneously discussed the problem with great clarity in 1915 4.4. To understand Buckingham’s theorem, we must first overcome one concep- tual hurdle, which, if it is clear to the student, will make everything that follows extremely simple. Let us explain that hurdle first. Suppose thaty depends onr,x,z and so on: y=y(r,x,z,...) We can take any one variable—say, x—and arbitrarily multiply it (or it raised to a power) by any other variables in the equation, without altering the truth of the functional equation, like this: y y 2 = x r,x,xz x x Many people find such a rearrangement disturbing when they first see it. That is because these are not algebraic equations — they are functional equations. We have said only that ify depends uponr,x, andz that it 2 will likewise depend uponx r,x, andxz. Suppose, for example, that we gave the functional equation the following algebraic form: −z y=y(r,x,z)=r(sinx)e152 Analysis of heat conduction and some steady one-dimensional problems §4.3 This need only be rearranged to put it in terms of the desired modified 2 variables andx itself(y/x,x r,x, andxz): 2 y x r xz = (sinx) exp − 3 x x x We can do any such multiplying or dividing of powers of any variable we wish without invalidating any functional equation that we choose to write. This simple fact is at the heart of the important example that follows. Example 4.2 Consider the heat exchanger problem described in Fig. 3.15. The “un- known,” or dependent variable, in the problem is either of the exit temperatures. Without any knowledge of heat exchanger analysis, we can write the functional equation on the basis of our physical under- standing of the problem: ⎡ ⎤ ⎢ ⎥  ⎢ ⎥ T −T = fn C ,C , T −T ,U ,A (4.14) c c max min h c out in ⎣ in in ⎦                 2 2 K W/K W/K K W/m K m where the dimensions of each term are noted under the quotation. We want to know how many dimensionless groups the variables in eqn. (4.14) should reduce to. To determine this number, we use the idea explained above—that is, that we can arbitrarily pick one vari- able from the equation and divide or multiply it into other variables. Then—one at a time—we select a variable that has one of the dimen- sions. We divide or multiply it by the other variables in the equation that have that dimension in such a way as to eliminate the dimension from them. We do this first with the variable(T −T ), which has the di- h c in in mension of K. ⎡ T −T c c ⎢ out in = fn⎣ C (T −T ),C (T −T ), max c min c h h in in in in T −T       h c in in    W W ⎤ dimensionless ⎥ ⎥ (T −T ),U(T −T ),A h c h c in in in in ⎦        2 2 K W/m m§4.3 Dimensional analysis 153 The interesting thing about the equation in this form is that the only remaining term in it with the units of K is (T −T ). No such h c in in term can exist in the equation because it is impossible to achieve dimensional homogeneity without another term in K to balance it. Therefore, we must remove it. ⎡ ⎤ ⎢ ⎥ T −T c c out in ⎢ ⎥ = fn C (T −T ),C (T −T ),U(T −T ),A max h c min h c h c in in in in in in ⎣ ⎦           T −T c h in in    2 2 W W W/m m dimensionless 2 Now the equation has only two dimensions in it—W and m . Next, we 2 multiplyU(T −T ) byA to get rid of m in the second-to-last term. h c in in 2 Accordingly, the termA (m ) can no longer stay in the equation, and we have ⎡ ⎤ T −T c c ⎢ ⎥ out in = fn⎣ C (T −T ),C (T −T ),UA(T −T ),⎦ max h c min h c h c in in in in in in          T −T h c in in    W W W dimensionless Next, we divide the first and third terms on the right by the second. This leaves onlyC (T −T ), with the dimensions of W. That term min h c in in must then be removed, and we are left with the completely dimension- less result:   T −T C UA c c max out in = fn , (4.15) T −T C C h c min min in in Equation (4.15) has exactly the same functional form as eqn. (3.21), which we obtained by direct analysis. Notice that we removed one variable from eqn. (4.14) for each di- mension in which the variables are expressed. If there aren variables— including the dependent variable—expressed inm dimensions, we then expect to be able to express the equation in (n−m) dimensionless groups, or pi-groups, as Buckingham called them. This fact is expressed by the Buckingham pi-theorem, which we state formally in the following way:154 Analysis of heat conduction and some steady one-dimensional problems §4.3 A physical relationship among n variables, which can be ex- pressed in a minimum ofm dimensions, can be rearranged into a relationship among(n−m) independent dimensionless groups of the original variables. Two important qualifications have been italicized. They will be explained in detail in subsequent examples. Buckingham called the dimensionless groups pi-groups and identified them as Π ,Π ,...,Π . Normally we call Π the dependent variable 1 2 n−m 1 and retain Π as independent variables. Thus, the dimensional 2→(n−m) functional equation reduces to a dimensionless functional equation of the form Π = fn(Π ,Π ,...,Π ) (4.16) 1 2 3 n−m Applications of the pi-theorem Example 4.3 Is eqn. (2.24) consistent with the pi-theorem? Solution. To find out, we first write the dimensional functional equation for Example 2.6: T−T = fn r,r,r , h ,k ,(T −T) i i o ∞ i                2 K m m m W/m K W/m·K K There are seven variables(n= 7) in three dimensions, K, m, and W (m= 3). Therefore, we look for 7− 3= 4 pi-groups. There are four pi-groups in eqn. (2.24): T−T r r hr i o o Π = , Π = , Π = , Π = ≡ Bi. 1 2 3 4 T −T r r k ∞ i i i Consider two features of this result. First, the minimum number of dimensions was three. If we had written watts as J/s, we would have had four dimensions instead. But Joules never appear in that particular problem independently of seconds. They always appear as a ratio and should not be separated. (If we had worked in English units, this would have seemed more confusing, since there is no name for Btu/sec unless§4.3 Dimensional analysis 155 we first convert it to horsepower.) The failure to identify dimensions that are consistently grouped together is one of the major errors that the beginner makes in using the pi-theorem. The second feature is the independence of the groups. This means that we may pick any four dimensionless arrangements of variables, so long as no group or groups can be made into any other group by math- ematical manipulation. For example, suppose that someone suggested that there was a fifth pi-group in Example 4.3: 2 hr Π = 5 k It is easy to see thatΠ can be written as 5 2 2 2 2 hr r r Π o 2 i Π = = Bi 5 k r r Π i o 3 ThereforeΠ is not independent of the existing groups, nor will we ever 5 find a fifth grouping that is. Another matter that is frequently made much of is that of identifying the pi-groups once the variables are identified for a given problem. (The method of indices 4.1 is a cumbersome arithmetic strategy for doing this but it is perfectly correct.) We shall find the groups by using either of two methods: 1. The groups can always be obtained formally by repeating the simple elimination-of-dimensions procedure that was used to derive the pi-theorem in Example 4.2. 2. One may simply arrange the variables into the required number of independent dimensionless groups by inspection. In any method, one must make judgments in the process of combining variables and these decisions can lead to different arrangements of the pi-groups. Therefore, if the problem can be solved by inspection, there is no advantage to be gained by the use of a more formal procedure. The methods of dimensional analysis can be used to help find the solution of many physical problems. We offer the following example, not entirely with tongue in cheek: Example 4.4 Einstein might well have noted that the energy equivalent,e, of a rest156 Analysis of heat conduction and some steady one-dimensional problems §4.3 mass,m , depended on the velocity of light,c , before he developed o o the special relativity theory. He would then have had the following dimensional functional equation:   2 kg· m e N·mor e = fn(c m/s,m kg) o o 2 s The minimum number of dimensions is only two: kg and m/s, so we look for 3− 2 = 1 pi-group. To find it formally, we eliminated the dimension of mass frome by dividing it bym (kg). Thus, o 2   e m = fn c m/s,m kg o o 2    m s o this must be removed because it is the only term with mass in it Then we eliminate the dimension of velocity (m/s) by dividinge/m o 2 byc : o e = fn(c m/s) o 2 m c o o This timec must be removed from the function on the right, since it o is the only term with the dimensions m/s. This gives the result (which could have been written by inspection once it was known that there could only be one pi-group): e Π = = fn(no other groups)= constant 1 2 m c o o or 2 e= constant· m c o o Of course, it required Einstein’s relativity theory to tell us that the constant is unity. Example 4.5 What is the velocity of efflux of liquid from the tank shown in Fig. 4.4? Solution. In this case we can guess that the velocity,V, might de- pend on gravity,g, and the headH. We might be tempted to include§4.3 Dimensional analysis 157 Figure 4.4 Efflux of liquid from a tank. the density as well until we realize thatg is already a force per unit mass. To understand this, we can use English units and divideg by the 4 2 2 conversion factor, g . Thus (g ft/s )/(g lb ·ft/lb s )=g lb /lb . c c m f f m Then V = fn H,g    2 m/s m/s m so there are three variables in two dimensions, and we look for 3−2= 1 pi-groups. It would have to be V 3 Π = = fn(no other pi-groups)= constant 1 gH or 4 V = constant· gH The analytical study of fluid mechanics tells us that this form is √ 2 correct and that the constant is 2. The groupV /gh, by the way, is called a Froude number, Fr (pronounced “Frood”). It compares inertial forces to gravitational forces. Fr is about 1000 for a pitched baseball, and it is between 1 and 10 for the water flowing over the spillway of a dam. 4 One can always divide any variable by a conversion factor without changing it.158 Analysis of heat conduction and some steady one-dimensional problems §4.3 Example 4.6 Obtain the dimensionless functional equation for the temperature distribution during steady conduction in a slab with a heat source,q ˙. Solution. In such a case, there might be one or two specified tem- peratures in the problem: T orT . Thus the dimensional functional 1 2 equation is ⎡ ⎤ ⎢ ⎥ ⎢ ⎥ T−T = fn (T −T ),x,L, q ˙,k , h 1 2 1 ⎣ ⎦               3 2 K K m W/m W/m·K W/m K where we presume that a convective b.c. is involved and we identify a characteristic length,L, in thex-direction. There are seven variables in three dimensions, or 7− 3= 4 pi-groups. Three of these groups are ones we have dealt with in the past in one form or another: T−T 1 dimensionless temperature, which we Π = 1 shall give the nameΘ T −T 2 1 x Π = dimensionless length, which we callξ 2 L hL which we recognize as the Biot number, Bi Π = 3 k The fourth group is new to us: 2 qL ˙ which compares the heat generation rate to Π = 4 the rate of heat loss; we call itΓ k(T −T ) 2 1 Thus, the solution is Θ= fn(ξ, Bi,Γ) (4.17) In Example 2.1, we undertook such a problem, but it differed in two respects. There was no convective boundary condition and hence, noh, and only one temperature was specified in the problem. In this case, the dimensional functional equation was  ˙ (T−T )= fn x,L,q,k 1 so there were only five variables in the same three dimensions. The re- sulting dimensionless functional equation therefore involved only two§4.4 An illustration of dimensional analysis in a complex steady conduction problem 159 pi-groups. One wasξ=x/L and the other is a new one equal toΘ/Γ.We call itΦ:   T−T x 1 Φ≡ = fn (4.18) 2 qL ˙ /k L And this is exactly the form of the analytical result, eqn. (2.15). Finally, we must deal with dimensions that convert into one another. For example, kg and N are defined in terms of one another through New- ton’s Second Law of Motion. Therefore, they cannot be identified as sep- arate dimensions. The same would appear to be true of J and N·m, since both are dimensions of energy. However, we must discern whether or not a mechanism exists for interchanging them. If mechanical energy remains distinct from thermal energy in a given problem, then J should not be interpreted as N·m. This issue will prove important when we do the dimensional anal- ysis of several heat transfer problems. See, for example, the analyses of laminar convection problem at the beginning of Section 6.4, of natu- ral convection in Section 8.3, of film condensation in Section 8.5, and of pool boiling burnout in Section 9.3. In all of these cases, heat transfer normally occurs without any conversion of heat to work or work to heat and it would be misleading to break J into N·m. Additional examples of dimensional analysis appear throughout this book. Dimensional analysis is, indeed, our court of first resort in solving most of the new problems that we undertake. 4.4 An illustration of the use of dimensional analysis in a complex steady conduction problem Heat conduction problems with convective boundary conditions can rap- idly grow difficult, even if they start out simple, and so we look for ways to avoid making mistakes. For one thing, it is wise to take great care that dimensions are consistent at each stage of the solution. The best way to do this, and to eliminate a great deal of algebra at the same time, is to nondimensionalize the heat conduction equation before we apply the b.c.’s. This nondimensionalization should be consistent with the pi- theorem. We illustrate this idea with a fairly complex example.160 Analysis of heat conduction and some steady one-dimensional problems §4.4 Figure 4.5 Heat conduction through a heat-generating slab with asymmetric boundary conditions. Example 4.7 A slab shown in Fig. 4.5 has different temperatures and different heat transfer coefficients on either side and the heat is generated within it. Calculate the temperature distribution in the slab. Solution. The differential equation is 2 ˙ d T q =− 2 dx k and the general solution is 2 qx ˙ T=− +C x+C (4.19) 1 2 2k

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