Part II Analysis of Heat Conduction 139 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: Find T(x,y,z,t) such that: 1. ∇·(k∇T)+ ˙ q = ρc ∂T ∂t for 0 <t<T (where T can →∞), and for (x,y,z) belonging to 141 142 Analysis of heat conduction and some steady one-dimensional problems §4.1 some region, R, which might extend to infinity. 1 2. T = T i (x,y,z) at t = 0 This is called an initial condition, or i.c. (a) Condition 1 above is not imposed at t = 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, where ˙ q 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 of R for t>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 of T normal to the boundary is specified on the boundary of R for t>0. Such a condition arises when the heat flux, k(∂T /∂x), is specified on a boundary or when , with the help of insulation, we set ∂T/∂x equal to zero. 2 (c) b.c.’s of the third kind: A derivative of T 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 −k ∂T ∂x     bndry = h(T − T ∞ ) bndry when the body lies to the left of the boundary on the x-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 position  r : 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 constant k and ˙ q: ∇ 2 T + ˙ q k = 1 α ∂T ∂t (2.11) Cartesian coordinates: Steady conduction in the y-direction. Equation (2.11) reduces as follows: ∂ 2 T ∂x 2  =0 + ∂ 2 T ∂y 2 + ∂ 2 T ∂z 2  =0 + ˙ q k = 1 α ∂T ∂t    = 0, since steady Therefore, d 2 T dy 2 =− ˙ q k which we integrate twice to get T =− ˙ q 2k y 2 +C 1 y + C 2 or, if ˙ q = 0, T = C 1 y + C 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): 1 r ∂ ∂r  r ∂T ∂r     =0 + 1 r 2 ∂ 2 T ∂φ 2    r =constant + ∂ 2 T ∂z 2  =0 + ˙ q k = 1 α ∂T ∂t    = 0, since steady Two integrations give T =− r 2 ˙ q 2k φ 2 +C 1 φ + C 2 (4.1) 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) If T is spatially uniform, it can still vary with time. In such cases ∇ 2 T   =0 + ˙ q k = 1 α ∂T ∂t and ∂T/∂t becomes an ordinary derivative. Then, since α = k/ρc, dT dt = ˙ q ρc (4.2) 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 ˙ q effective =− h(T body −T ∞ )A volume W/m 3 (4.3) and the heat diffusion equation for this case, eqn. (4.2), becomes dT dt =− hA ρcV (T −T ∞ ) (4.4) 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 ∂x 2 = 1 α ∂T ∂t (4.5) 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 X  T= 1 α T  X (4.6) where each prime denotes one differentiation of a function with respect to its argument. Thus T  = dT/dt and X  = d 2 X/dx 2 . Rearranging eqn. (4.6), we get X  X = 1 α T  T (4.7a) This is an interesting result in that the left-hand side depends only upon x and the right-hand side depends only upon t. Thus, we set both sides equal to the same constant, which we call −λ 2 , instead of, say, λ, for reasons that will be clear in a moment: X  X = 1 α T  T =−λ 2 a constant (4.7b) It follows that the differential eqn. (4.7a) can be resolved into two ordi- nary differential equations: X  =−λ 2 X and T  =−αλ 2 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 cos λx for λ ≠ 0 X(x) = Ax + B for λ = 0 (4.9) §4.2 The general solution 147 and T (t) = Ce −αλ 2 t for λ ≠ 0 T (t) = C for λ = 0 (4.10) 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 T =XT =e −αλ 2 t (D sin λx +E cos λx) for λ ≠ 0 T =XT =Dx +E for λ = 0 (4.11) 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 function X(t) take the form of sines and cosines (instead of exponential functions) by placing a minus sign in front of λ 2 . 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 T ∂x 2 + ∂ 2 T ∂y 2 = 0 (4.12) Set T =XYand get X  X =− Y  Y =−λ 2 where λ can be an imaginary number. Then X=A sin λx + B cos λx Y=Ce λy +De −λy    for λ ≠ 0 X=Ax +B Y=Cy + D  for λ = 0 The general solution is T = (E sinλx + F cos λx)(e −λy +Ge λy ) for λ ≠ 0 T = (Ex +F)(y +G) for λ = 0 (4.13) 148 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: on the top surface : T(x,0) = A sin π x L on the sides : T(0orL, y) = 0 as y →∞: T(x,y →∞) = 0 Substitute eqn. (4.13) in the third b.c.: (E sin λx +F cos λx)(0 +G ·∞) = 0 The only way that this can be true for all x is if G = 0. Substitute eqn. (4.13), with G = 0, into the second b.c.: (O + F)e −λy = 0 [...]... complex example 15 9 16 0 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... Bi1 Bi2 + Bi1 Bi1 + Bi2 Bi2 + Bi2 1 1 Bi1 Bi1 ξ− − 1 + Bi1 Bi2 + Bi1 Bi1 + Bi2 Bi2 + Bi2 1 1 (4.24) The rearrangement of the dimensional equations into dimensionless form is straightforward algebra If the results shown here are not immediately obvious to you, sketch the calculation on a piece of paper 16 1 16 2 Analysis of heat conduction and some steady one-dimensional problems §4.4 This is a complicated... 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 15 8 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... problems Analysis of a one-dimensional fin The equations Figure 4.8 shows a one-dimensional fin protruding from a wall The wall—and the roots of the fin—are at a temperature T0 , which is either greater or less than the ambient temperature, T∞ The length of the fin is cooled or heated through a heat transfer coefficient, h, by the ambient fluid The heat transfer coefficient will be assumed uniform, although (as... 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 wellposed heat diffusion problems are unique Furthermore, we know 14 9 15 0 Analysis of heat conduction and some steady one-dimensional problems... equal to 0, 1, and ∞ and for Γ equal to 0, 0 .1, and 1 The following features should be noted: • When Γ 0 .1, the heat generation can be ignored • When Γ 1, Θ → Γ /Bi + Γ (ξ − ξ 2 ) This is a simple parabolic temperature distribution displaced upward an amount that depends on the relative external resistance, as reflected in the Biot number • If both Γ and 1/ Bi become large, Θ → Γ /Bi This means that... 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 pitheorem We illustrate this idea with a fairly... several heat transfer problems See, for example, the analyses of laminar convection problem at the beginning of Section 6.4, of natural 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... temperature nonuniformity Since it is symmetric, the graph is also symmetric Fin design §4.5 • When Bi 1, the slab temperature approaches a uniform value ˙ equal to T1 + qL/2h (In this case, we would have solved the problem with far greater ease by using a simple lumped-capacity heat balance, since it is no longer a heat conduction problem.) • When Bi > 10 0, the temperature distribution is a very large... large parabola with ½ added to it In this case, the problem could have been solved using boundary conditions of the first kind because the surface temperature stays very close to T∞ (recall Fig 1. 11) 4.5 Fin design The purpose of fins The convective removal of heat from a surface can be substantially improved if we put extensions on that surface to increase its area These extensions can take a variety . 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. 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- able. gravitational forces. Fr is about 10 00 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