John H. Lienhard IV, John H. Lienhard V. A Heat Transfer Textbook (776116), страница 22
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If there is no heat generation, the heat diffusionequation is1 ∂T∂2T=2∂xα ∂t(4.5)A common trick is to ask: “Can we find a solution in the form of a productof functions of t and x: T = T (t) · X(x)?” To find the answer, wesubstitute this in eqn. (4.5) and getX T =1 T Xα(4.6)where each prime denotes one differentiation of a function with respectto its argument. Thus T = dT/dt and X = d2 X/dx 2 . Rearrangingeqn. (4.6), we get1 TX =Xα T(4.7a)This is an interesting result in that the left-hand side depends onlyupon x and the right-hand side depends only upon t. Thus, we set bothsides equal to the same constant, which we call −λ2 , instead of, say, λ,for reasons that will be clear in a moment:1 TX == −λ2Xα Ta constant(4.7b)It follows that the differential eqn.
(4.7a) can be resolved into two ordinary differential equations:X = −λ2 Xand T = −α λ2 T(4.8)The general solution of both of these equations are well known andare among the first ones dealt with in any study of differential equations.They are:X(x) = A sin λx + B cos λxX(x) = Ax + Bfor λ ≠ 0for λ = 0(4.9)The general solution§4.2147andT (t) = Ce−αλT (t) = C2tfor λ ≠ 0for λ = 0(4.10)where we use capital letters to denote constants of integration. [In either case, these solutions can be verified by substituting them back intoeqn. (4.8).] Thus the general solution of eqn. (4.5) can indeed be writtenin the form of a product, and that product is2T = XT = e−αλ t (D sin λx + E cos λx) for λ ≠ 0T = XT = Dx + Efor λ = 0(4.11)The usefulness of this result depends on whether or not it can be fitto the b.c.’s and the i.c. In this case, we made the function X(t) take theform of sines and cosines (instead of exponential functions) by placinga minus sign in front of λ2 .
The sines and cosines make it possible to fitthe b.c.’s using Fourier series methods. These general methods are notdeveloped in this book; however, a complete Fourier series solution ispresented for one problem in Section 5.3.The preceding simple methods for obtaining general solutions of linear partial d.e.’s is called the method of separation of variables.
It can beapplied to all kinds of linear d.e.’s. Consider, for example, two-dimensional steady heat conduction without heat sources:∂2T∂2T+=0∂x 2∂y 2(4.12)Set T = XY and getY X =−= −λ2XYwhere λ can be an imaginary number. Then⎫X = A sin λx + B cos λx ⎬Y = Ceλy + De−λyX = Ax + BY = Cy + D⎭for λ ≠ 01for λ = 0The general solution isT = (E sin λx + F cos λx)(e−λy + Geλy ) for λ ≠ 0T = (Ex + F )(y + G)for λ = 0(4.13)148Analysis of heat conduction and some steady one-dimensional problems§4.2Figure 4.3 A two-dimensional slab maintained at a constanttemperature on the sides and subjected to a sinusoidal variation of temperature on one face.Example 4.1A long slab is cooled to 0◦ C on both sides and a blowtorch is turnedon the top edge, giving an approximately sinusoidal temperature distribution along the top, as shown in Fig. 4.3.
Find the temperaturedistribution within the slab.Solution. The general solution is given by eqn. (4.13). We musttherefore identify the appropriate b.c.’s and then fit the general solution to it. Those b.c.’s are:on the top surface :on the sides :as y → ∞ :T (x, 0) = A sin πxLT (0 or L, y) = 0T (x, y → ∞) = 0Substitute eqn. (4.13) in the third b.c.:(E sin λx + F cos λx)(0 + G · ∞) = 0The only way that this can be true for all x is if G = 0. Substituteeqn. (4.13), with G = 0, into the second b.c.:(O + F )e−λy = 0§4.2The general solutionso F also equals 0. Substitute eqn.
(4.13) with G = F = 0, into the firstb.c.:E(sin λx) = A sin πxLIt follows that A = E and λ = π /L. Then eqn. (4.13) becomes theparticular solution that satisfies the b.c.’s:xe−π y/LT = A sin πLThus, the sinusoidal variation of temperature at the top of the slab isattenuated exponentially at lower positions in the slab. At a positionof y = 2L below the top, T will be 0.0019 A sin π x/L. The temperature distribution in the x-direction will still be sinusoidal, but it willhave less than 1/500 of the amplitude at y = 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 thisparticular general solution. If we had tried to fit the equation toa general temperature distribution, T (x, y = 0) = fn(x), it wouldnot have been obvious how to proceed.
Actually, this is the kindof problem that Fourier solved with the help of his Fourier seriesmethod. We discuss this matter in more detail in Chapter 5.• Not all forms of general solutions lend themselves to a particularset of boundary and/or initial conditions. In this example, we madethe process look simple, but more often than not, it is in fitting ageneral solution to a set of boundary conditions that we get stuck.• Normally, on formulating a problem, we must approximate real behavior in stating the b.c.’s. It is advisable to consider what kind ofassumption will put the b.c.’s in a form compatible with the general solution. The temperature distribution imposed on the slabby the blowtorch in Example 4.1 might just as well have been approximated as a parabola.
But as small as the difference between aparabola and a sine function might be, the latter b.c. was far easierto accommodate.• The twin issues of existence and uniqueness of solutions requirea comment here: It has been established that solutions to all wellposed heat diffusion problems are unique. Furthermore, we know149150Analysis of heat conduction and some steady one-dimensional problems§4.3from our experience that if we describe a physical process correctly,a unique outcome exists.
Therefore, we are normally safe to leavethese issues to a mathematician—at least in the sort of problemswe discuss here.• Given that a unique solution exists, we accept any solution as correct since we have carved it to fit the boundary conditions. In thissense, the solution of differential equations is often more of an incentive than a formal operation. The person who does it best isoften the person who has done it before and so has a large assortment of tricks up his or her sleeve.4.3Dimensional analysisIntroductionMost universities place the first course in heat transfer after an introduction to fluid mechanics: and most fluid mechanics courses include somedimensional analysis.
This is normally treated using the familiar methodof indices, which is seemingly straightforward to teach but is cumbersomeand sometimes misleading to use. It is rather well presented in [4.1].The method we develop here is far simpler to use than the methodof indices, and it does much to protect us from the common errors wemight fall into.
We refer to it as the method of functional replacement.The importance of dimensional analysis to heat transfer can be madeclearer by recalling Example 2.6, which (like most problems in Part I) involved several variables. Theses variables included the dependent variable of temperature, (T∞ − Ti );3 the major independent variable, whichwas the radius, r ; and five system parameters, ri , ro , h, k, and (T∞ − Ti ).By reorganizing the solution into dimensionless groups [eqn. (2.24)], wereduced the total number of variables to only four:⎤⎡T − Ti⎥⎢= fn⎣ r ri ,ro ri ,Bi(2.24a)⎦T∞ − Ti dependent variableindep. var. two system parametersThis solution offered a number of advantages over the dimensionalsolution. For one thing, it permitted us to plot all conceivable solutions3Notice that we do not call Ti a variable.
It is simply the reference temperatureagainst which the problem is worked. If it happened to be 0◦ C, we would not notice itssubtraction from the other temperatures.Dimensional analysis§4.3for a particular shape of cylinder, (ro /ri ), in a single figure, Fig. 2.13.For another, it allowed us to study the simultaneous roles of h, k and roin defining the character of the solution. By combining them as a Biotnumber, 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, simultaneously, the behavior of all similar systems of heat conduction throughcylinders.
Thus a large, highly conducting cylinder might be similar inits behavior to a small cylinder with a lower thermal conductivity.Finally, we shall discover that, by nondimensionalizing a problem before we solve it, we can often greatly simplify the process of solving it.Our next aim is to map out a method for nondimensionalization problems before we have solved then, or, indeed, before we have even writtenthe equations that must be solved. The key to the method is a resultcalled the Buckingham pi-theorem.The Buckingham pi-theoremThe attention of scientific workers was drawn very strongly toward thequestion of similarity at about the beginning of World War I. Buckinghamfirst organized previous thinking and developed his famous theorem in1914 in the Physical Review [4.2], and he expanded upon the idea in theTransactions of the ASME one year later [4.3]. Lord Rayleigh almost simultaneously discussed the problem with great clarity in 1915 [4.4].
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