John H. Lienhard IV, John H. Lienhard V. A Heat Transfer Textbook (776116), страница 30
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We shall do this somewhatcomplicated task for the simplest case and then look at the results ofsuch calculations in other situations.A simple slab, shown in Fig. 5.6, is initially at a temperature Ti . Thetemperature of the surface of the slab is suddenly changed to Ti , and wewish to calculate the interior temperature profile as a function of time.The heat conduction equation is1 ∂T∂2T=2∂xα ∂t(5.24)with the following b.c.’s and i.c.:T (−L, t > 0) = T (L, t > 0) = T1and T (x, t = 0) = Ti(5.25)In fully dimensionless form, eqn. (5.24) and eqn. (5.25) are∂Θ∂2Θ=2∂ξ∂Fo(5.26)204Transient and multidimensional heat conduction§5.3andΘ(0, Fo) = Θ(2, Fo) = 0and Θ(ξ, 0) = 1(5.27)where we have nondimensionalized the problem in accordance with eqn.(5.4), using Θ ≡ (T − T1 )/(Ti − T1 ) and Fo ≡ αt/L2 ; but for conveniencein solving the equation, we have set ξ equal to (x/L) + 1 instead of x/L.The general solution of eqn.
(5.26) may be found using the separationof variables technique described in Sect. 4.2, leading to the dimensionlessform of eqn. (4.11):Θ = e−λ̂2 Fo!"G sin(λ̂ξ) + E cos(λ̂ξ)(5.28)Direct nondimensionalization of eqn. (4.11) would show that λ̂ ≡ λL,since λ had units of (length)−1 .
The solution therefore appears to haveintroduced a fourth dimensionless group, λ̂. This needs explanation. Thenumber λ, which was introduced in the separation-of-variables process,is called an eigenvalue.2 In the present problem, λ̂ = λL will turn out tobe a number—or rather a sequence of numbers—that is independent ofsystem parameters.Substituting the general solution, eqn. (5.28), in the first b.c. gives0 = e−λ̂2 Fo(0 + E)soE=0and substituting it in the second yields0 = e−λ̂2 Fo!"G sin 2λ̂ so eitherG=0or2λ̂ = 2λ̂n = nπ ,n = 0, 1, 2, . . .In the second case, we are presented with two choices. The first,G = 0, would give Θ ≡ 0 in all situations, so that the initial conditioncould never be accommodated.
(This is what mathematicians call a trivialsolution.) The second choice, λ̂n = nπ /2, actually yields a string ofsolutions, each of the formnπ−n2 π 2 Fo/4ξ(5.29)sinΘ = Gn e22The word eigenvalue is a curious hybrid of the German term eigenwert and itsEnglish translation, characteristic value.Transient conduction in a one-dimensional slab§5.3where Gn is the constant appropriate to the nth one of these solutions.We still face the problem that none of eqns.
(5.29) will fit the initialcondition, Θ(ξ, 0) = 1. To get around this, we remember that the sum ofany number of solutions of a linear differential equation is also a solution.Then we writeΘ=∞$Gn e−n2 π 2 Fo/4n=1πsin n ξ2(5.30)where we drop n = 0 since it gives zero contribution to the series. Andwe arrive, at last, at the problem of choosing the Gn ’s so that eqn. (5.30)will fit the initial condition.πGn sin n ξ = 1Θ (ξ, 0) =2n=1∞$(5.31)The problem of picking the values of Gn that will make this equationtrue is called “making a Fourier series expansion” of the function f (ξ) =1.
We shall not pursue strategies for making Fourier series expansionsin any general way. Instead, we merely show how to accomplish the taskfor the particular problem at hand. We begin with a mathematical trick.We multiply eqn. (5.31) by sin(mπ /2), where m may or may not equaln, and we integrate the result between ξ = 0 and 2.2sin0mπξ2dξ =∞$n=12Gnsin0nπmπξ sinξ dξ22(5.32)(The interchange of summation and integration turns out to be legitimate,although we have not proved, here, that it is.3 ) With the help of a tableof integrals, we find that2050 for n ≠ mnπmπξ sinξ dξ =sin221 for n = mThus, when we complete the integration of eqn.
(5.32), we get5∞2$0 for n ≠ m2mπcosξ =Gn ×−mπ21 for n = m0n=13What is normally required is that the series in eqn. (5.31) be uniformly convergent.205206Transient and multidimensional heat conduction§5.3This reduces to−2 (−1)n − 1 = GnmπsoGn =4nπwhere n is an odd numberSubstituting this result into eqn. (5.30), we finally obtain the solution tothe problem:4Θ (ξ, Fo) =π1 −(nπ /2)2 Fonπeξsinn2n=odd∞$(5.33)Equation (5.33) admits a very nice simplification for large time (or atlarge Fo).
Suppose that we wish to evaluate Θ at the outer center of theslab—at x = 0 or ξ = 1. Then4×Θ (0, Fo) =π⎫⎧ ⎪⎪2 2 ⎬⎨211π3π5πexp −Fo − exp −Fo + exp −Fo + · · ·⎪⎪23252⎭⎩ = 0.085 at Fo = 1= 0.781 at Fo = 0.1= 0.976 at Fo = 0.01 10−10 at Fo = 1= 0.036 at Fo = 0.1= 0.267 at Fo = 0.01 10−27 at Fo = 1= 0.0004 at Fo = 0.1= 0.108 at Fo = 0.01Thus for values of Fo somewhat greater than 0.1, only the first term inthe series need be used in the solution (except at points very close to theboundaries).
We discuss these one-term solutions in Sect. 5.5. Before wemove to this matter, let us see what happens to the preceding problemif the slab is subjected to b.c.’s of the third kind.Suppose that the walls of the slab had been cooled by symmetricalconvection such that the b.c.’s were∂T ∂T h(T∞ − T )x=−L = −kand h(T − T∞ )x=L = −k∂x x=−L∂x x=Lor in dimensionless form, using Θ ≡ (T −T∞ )/(Ti −T∞ ) and ξ = (x/L)+1,∂Θ1∂Θ=−and=0−ΘBi∂ξ∂ξξ=0ξ=0ξ=1Transient conduction in a one-dimensional slab§5.3207Table 5.1 Terms of series solutions for slabs, cylinders, andspheres. J0 and J1 are Bessel functions of the first kind.SlabCylinderSphereAnfn2 sin λ̂nxcos λ̂nLλ̂n + sin λ̂n cos λ̂n2 J1(λ̂n )2λ̂n J0(λ̂n ) + J12(λ̂n )2Equation for λ̂nJ0sin λ̂n − λ̂n cos λ̂nλ̂n − sin λ̂n cos λ̂nroλ̂n rrλ̂nrosincot λ̂n =λ̂n J1(λ̂n ) = Biro J0(λ̂n )λ̂n rroλ̂n cot λ̂n = 1 − BiroThe solution is somewhat harder to find than eqn. (5.33) was, but theresult is4∞ 2 sin λ̂ cos[λ̂ (ξ − 1)]$nnΘ=exp −λ̂2n Fo(5.34)λ̂n + sin λ̂n cos λ̂nn=1where the values of λ̂n are given as a function of n and Bi = hL/k by thetranscendental equationcot λ̂n =λ̂nBi(5.35)The successive positive roots of this equation, which are λ̂n = λ̂1 , λ̂2 ,λ̂3 , .
. . , depend upon Bi. Thus, Θ = fn(ξ, Fo, Bi), as we would expect. Thisresult, although more complicated than the result for b.c.’s of the firstkind, still reduces to a single term for Fo 0.2.Similar series solutions can be constructed for cylinders and spheresthat are convectively cooled at their outer surface, r = ro .
The solutionsfor slab, cylinders, and spheres all have the formΘ=∞$T − T∞=An exp −λ̂2n Fo fnTi − T∞n=1(5.36)where the coefficients An , the functions fn , and the equations for thedimensionless eigenvalues λ̂n are given in Table 5.1.4λ̂nBiLSee, for example, [5.1, §2.3.4] or [5.2, §3.4.3] for details of this calculation.208Transient and multidimensional heat conduction5.4§5.4Temperature-response chartsFigure 5.7 is a graphical presentation of eqn.
(5.34) for 0 Fo 1.5 andfor six x-planes in the slab. (Remember that the x-coordinate goes fromzero in the center to L on the boundary, while ξ goes from 0 up to 2 inthe preceding solution.)Notice that, with the exception of points for which 1/Bi < 0.25 onthe outside boundary, the curves are all straight lines when Fo 0.2.Since the coordinates are semilogarithmic, this portion of the graph corresponds to the lead term—the only term that retains any importance—in eqn.
(5.34). When we take the logarithm of the one-term version ofeqn. (5.34), the result is2 sin λ̂1 cos[λ̂1 (ξ − 1)]−λ̂21 Foln Θ lnλ̂1 + sin λ̂1 cos λ̂1Θ-intercept at Fo = 0 ofthe straight portion ofthe curveslope of thestraight portionof the curveIf Fo is greater than 1.5, the following options are then available to us forsolving the problem:• Extrapolate the given curves using a straightedge.• Evaluate Θ using the first term of eqn. (5.34), as discussed in Sect. 5.5.• If Bi is small, use a lumped-capacity result.Figure 5.8 and Fig. 5.9 are similar graphs for cylinders and spheres.Everything that we have said in general about Fig. 5.7 is also true forthese graphs. They were simply calculated from different solutions, andthe numerical values on them are somewhat different. These charts arefrom [5.3, Chap.
5], although such charts are often called Heisler charts,after a collection of related charts subsequently published by Heisler[5.4].Another useful kind of chart derivable from eqn. (5.34) is one thatgives heat removal from a body up to a time of interest:⌠tt∂T ⎮⌡Q dt = − kAdt∂x surface00⌠ Fo Ti − T∞ ∂Θ L2⎮dFo= −⌡ kAL∂ξ surface α0209Figure 5.7 The transient temperature distribution in a slab at six positions: x/L = 0 is the center,x/L = 1 is one outside boundary.210Figure 5.8 The transient temperature distribution in a long cylinder of radius ro at six positions:r /ro = 0 is the centerline; r /ro = 1 is the outside boundary.211Figure 5.9 The transient temperature distribution in a sphere of radius ro at six positions: r /ro = 0is the center; r /ro = 1 is the outside boundary.212Transient and multidimensional heat conduction§5.4Dividing this by the total energy of the body above T∞ , we get a quantity, Φ, which approaches unity as t → ∞ and the energy is all transferredto the surroundings:t⌠ Fo∂Θ ⎮dFo= −⌡Φ≡ρcV (Ti − T∞ )∂ξsurface0Q dt0(5.37)where the volume, V = AL.
Substituting the appropriate temperaturedistribution [e.g., eqn. (5.34) for a slab] in eqn. (5.37), we obtain Φ(Fo, Bi)in the form of an infinite seriesΦ (Fo, Bi) = 1 −∞$Dn exp −λ̂2n Fo(5.38)n=1The coefficients Dn are different functions of λ̂n — and thusof Bi — forslabs, cylinders, and spheres (e.g., for a slab Dn = An sin λ̂n λ̂n ). Thesefunctions can be used to plot Φ(Fo, Bi) once and for all. Such curves aregiven in Fig. 5.10.The quantity Φ has a close relationship to the mean temperature ofa body at any time, T (t). Specifically, the energy lost as heat by time tdetermines the difference between the initial temperature and the meantemperature at time tt0!"!"Q dt = U (0) − U (t) = ρcV Ti − T (t) .(5.39)Thus, if we define Θ as follows, we find the relationship of T (t) to ΦtQ(t) dtT (t) − T∞0Θ≡= 1 − Φ.=1−Ti − T∞ρcV (Ti − T∞ )(5.40)Example 5.2A dozen approximately spherical apples, 10 cm in diameter are takenfrom a 30◦ C environment and laid out on a rack in a refrigerator at5◦ C.
They have approximately the same physical properties as water,and h is approximately 6 W/m2 K as the result of natural convection.What will be the temperature of the centers of the apples after 1 hr?How long will it take to bring the centers to 10◦ C? How much heatwill the refrigerator have to carry away to get the centers to 10◦ C?Figure 5.10 The heat removal from suddenly-cooled bodies asa function of h and time.213214Transient and multidimensional heat conduction§5.4Solution.
After 1 hr, or 3600 s:3600 skαtFo = 2 =ρc(0.05m)2ro20◦ C(0.603 J/m·s·K)(3600 s)= 0.208=(997.6 kg/m3 )(4180 J/kg·K)(0.0025 m2 )Furthermore, Bi−1 = (hro /k)−1 = [6(0.05)/0.603]−1 = 2.01. Therefore, we read from Fig. 5.9 in the upper left-hand corner:Θ = 0.85After 1 hr:Tcenter = 0.85(30 − 5)◦ C + 5◦ C = 26.3◦ CTo find the time required to bring the center to 10◦ C, we firstcalculateΘ=10 − 5= 0.230 − 5and Bi−1 is still 2.01. Then from Fig. 5.9 we readFo = 1.29 =αtro2sot=1.29(997.6)(4180)(0.0025)= 22, 300 s = 6 hr 12 min0.603Finally, we look up Φ at Bi = 1/2.01 and Fo = 1.29 in Fig.
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