A HEAT TRANSFER TEXTBOOK - THIRD EDITION Episode 1 Part 10 pdf

214 Transient and multidimensional heat conduction §5.4 Solution. After 1 hr, or 3600 s: Fo = αt r 2 o =  k ρc  20 ◦ C 3600 s (0.05 m) 2 = (0.603 J/m·s·K)(3600 s) (997.6 kg/m 3 )(4180 J/kg·K)(0.0025 m 2 ) = 0.208 Furthermore, Bi −1 = (hr o /k) −1 = [6(0.05)/0.603] −1 = 2.01. There- fore, we read from Fig. 5.9 in the upper left-hand corner: Θ = 0.85 After 1 hr: T center = 0.85(30 −5) ◦ C +5 ◦ C = 26.3 ◦ C To find the time required to bring the center to 10 ◦ C, we first calculate Θ = 10 −5 30 −5 = 0.2 and Bi −1 is still 2.01. Then from Fig. 5.9 we read Fo = 1.29 = αt r 2 o so t = 1.29(997.6)(4180)(0.0025) 0.603 = 22, 300 s = 6hr12min Finally, we look up Φ at Bi = 1/2.01 and Fo = 1.29 in Fig. 5.10, for spheres: Φ = 0.80 =  t 0 Qdt ρc  4 3 πr 3 0  (T i −T ∞ ) so  t 0 Qdt = 997.6(4180)  4 3 π(0.05) 3  (25)(0.80) = 43, 668 J/apple Therefore, for the 12 apples, total energy removal = 12(43.67) = 524 kJ §5.4 Temperature-response charts 215 The temperature-response charts in Fig. 5.7 through Fig. 5.10 are with- out doubt among the most useful available since they can be adapted to a host of physical situations. Nevertheless, hundreds of such charts have been formed for other situations, a number of which have been cataloged by Schneider [5.5]. Analytical solutions are available for hundreds more problems, and any reader who is faced with a complex heat conduction calculation should consult the literature before trying to solve it. An ex- cellent place to begin is Carslaw and Jaeger’s comprehensive treatise on heat conduction [5.6]. Example 5.3 A 1 mm diameter Nichrome (20% Ni, 80% Cr) wire is simultaneously being used as an electric resistance heater and as a resistance ther- mometer in a liquid flow. The laboratory workers who operate it are attempting to measure the boiling heat transfer coefficient, h, by sup- plying an alternating current and measuring the difference between the average temperature of the heater, T av , and the liquid tempera- ture, T ∞ . They get h = 30, 000 W/m 2 K at a wire temperature of 100 ◦ C and are delighted with such a high value. Then a colleague suggests that h is so high because the surface temperature is rapidly oscillating as a result of the alternating current. Is this hypothesis correct? Solution. Heat is being generated in proportion to the product of voltage and current, or as sin 2 ωt, where ω is the frequency of the current in rad/s. If the boiling action removes heat rapidly enough in comparison with the heat capacity of the wire, the surface tempera- ture may well vary significantly. This transient conduction problem was first solved by Jeglic in 1962 [5.7]. It was redone in a different form two years later by Switzer and Lienhard (see, e.g. [5.8]), who gave response curves in the form T max −T av T av −T ∞ = fn ( Bi,ψ ) (5.41) where the left-hand side is the dimensionless range of the tempera- ture oscillation, and ψ = ωδ 2 /α, where δ is a characteristic length [see Problem 5.56]. Because this problem is common and the solu- tion is not widely available, we include the curves for flat plates and cylinders in Fig. 5.11 and Fig. 5.12 respectively. Figure 5.11 Temperature deviation at the surface of a flat plate heated with alternating current. 216 Figure 5.12 Temperature deviation at the surface of a cylinder heated with alternating current. 217 218 Transient and multidimensional heat conduction §5.5 In the present case: Bi = h radius k = 30, 000(0.0005) 13.8 = 1.09 ωr 2 α = [2π(60)](0.0005) 2 0.00000343 = 27.5 and from the chart for cylinders, Fig. 5.12, we find that T max −T av T av −T ∞  0.04 A temperature fluctuation of only 4% is probably not serious. It there- fore appears that the experiment was valid. 5.5 One-term solutions As we have noted previously, when the Fourier number is greater than 0.2 or so, the series solutions from eqn. (5.36) may be approximated using only their first term: Θ ≈ A 1 ·f 1 ·exp  − ˆ λ 2 1 Fo  . (5.42) Likewise, the fractional heat loss, Φ, or the mean temperature Θ from eqn. (5.40), can be approximated using just the first term of eqn. (5.38): Θ = 1 − Φ ≈ D 1 exp  − ˆ λ 2 1 Fo  . (5.43) Table 5.2 lists the values of ˆ λ 1 , A 1 , and D 1 for slabs, cylinders, and spheres as a function of the Biot number. The one-term solution’s er- ror in Θ is less than 0.1% for a sphere with Fo ≥ 0.28 and for a slab with Fo ≥ 0.43. These errors are largest for Biot numbers near unity. If high accuracy is not required, these one-term approximations may generally be used whenever Fo ≥ 0.2 Table 5.2 One-term coefficients for convective cooling [5.1]. Plate Cylinder Sphere Bi ˆ λ 1 A 1 D 1 ˆ λ 1 A 1 D 1 ˆ λ 1 A 1 D 1 0.01 0.09983 1.0017 1.0000 0.14124 1.0025 1.0000 0.17303 1.0030 1.0000 0.02 0.14095 1.0033 1.0000 0.19950 1.0050 1.0000 0.24446 1.0060 1.0000 0.05 0.22176 1.0082 0.9999 0.31426 1.0124 0.9999 0.38537 1.0150 1.0000 0.10 0.31105 1.0161 0.9998 0.44168 1.0246 0.9998 0.54228 1.0298 0.9998 0.15 0.37788 1.0237 0.9995 0.53761 1.0365 0.9995 0.66086 1.0445 0.9996 0.20 0.43284 1.0311 0.9992 0.61697 1.0483 0.9992 0.75931 1.0592 0.9993 0.30 0.52179 1.0450 0.9983 0.74646 1.0712 0.9983 0.92079 1.0880 0.9985 0.40 0.59324 1.0580 0.9971 0.85158 1.0931 0.9970 1.05279 1.1164 0.9974 0.50 0.65327 1.0701 0.9956 0.94077 1.1143 0.9954 1.16556 1.1441 0.9960 0.60 0.70507 1.0814 0.9940 1.01844 1.1345 0.9936 1.26440 1.1713 0.9944 0.70 0.75056 1.0918 0.9922 1.08725 1.1539 0.9916 1.35252 1.1978 0.9925 0.80 0.79103 1.1016 0.9903 1.14897 1.1724 0.9893 1.43203 1.2236 0.9904 0.90 0.82740 1.1107 0.9882 1.20484 1.1902 0.9869 1.50442 1.2488 0.9880 1.00 0.86033 1.1191 0.9861 1.25578 1.2071 0.9843 1.57080 1.2732 0.9855 1.10 0.89035 1.1270 0.9839 1.30251 1.2232 0.9815 1.63199 1.2970 0.9828 1.20 0.91785 1.1344 0.9817 1.34558 1.2387 0.9787 1.68868 1.3201 0.9800 1.30 0.94316 1.1412 0.9794 1.38543 1.2533 0.9757 1.74140 1.3424 0.9770 1.40 0.96655 1.1477 0.9771 1.42246 1.2673 0.9727 1.79058 1.3640 0.9739 1.50 0.98824 1.1537 0.9748 1.45695 1.2807 0.9696 1.83660 1.3850 0.9707 1.60 1.00842 1.1593 0.9726 1.48917 1.2934 0.9665 1.87976 1.4052 0.9674 1.80 1.04486 1.1695 0.9680 1.54769 1.3170 0.9601 1.95857 1.4436 0.9605 2.00 1.07687 1.1785 0.9635 1.59945 1.3384 0.9537 2.02876 1.4793 0.9534 2.20 1.10524 1.1864 0.9592 1.64557 1.3578 0.9472 2.09166 1.5125 0.9462 2.40 1.13056 1.1934 0.9549 1.68691 1.3754 0.9408 2.14834 1.5433 0.9389 3.00 1.19246 1.2102 0.9431 1.78866 1.4191 0.9224 2.28893 1.6227 0.9171 4.00 1.26459 1.2287 0.9264 1.90808 1.4698 0.8950 2.45564 1.7202 0.8830 5.00 1.31384 1.2402 0.9130 1.98981 1.5029 0.8721 2.57043 1.7870 0.8533 6.00 1.34955 1.2479 0.9021 2.04901 1.5253 0.8532 2.65366 1.8338 0.8281 8.00 1.39782 1.2570 0.8858 2.12864 1.5526 0.8244 2.76536 1.8920 0.7889 10.00 1.42887 1.2620 0.8743 2.17950 1.5677 0.8039 2.83630 1.9249 0.7607 20.00 1.49613 1.2699 0.8464 2.28805 1.5919 0.7542 2.98572 1.9781 0.6922 50.00 1.54001 1.2727 0.8260 2.35724 1.6002 0.7183 3.07884 1.9962 0.6434 100.00 1.55525 1.2731 0.8185 2.38090 1.6015 0.7052 3.11019 1.9990 0.6259 ∞ 1.57080 1.2732 0.8106 2.40483 1.6020 0.6917 3.14159 2.0000 0.6079 219 220 Transient and multidimensional heat conduction §5.6 5.6 Transient heat conduction to a semi-infinite region Introduction Bronowksi’s classic television series, The Ascent of Man [5.9], included a brilliant reenactment of the ancient ceremonial procedure by which the Japanese forged Samurai swords (see Fig. 5.13). The metal is heated, folded, beaten, and formed, over and over, to create a blade of remarkable toughness and flexibility. When the blade is formed to its final configu- ration, a tapered sheath of clay is baked on the outside of it, so the cross section is as shown in Fig. 5.13. The red-hot blade with the clay sheath is then subjected to a rapid quenching, which cools the uninsulated cutting edge quickly and the back part of the blade very slowly. The result is a layer of case-hardening that is hardest at the edge and less hard at points farther from the edge. Figure 5.13 The ceremonial case-hardening of a Samurai sword. §5.6 Transient heat conduction to a semi-infinite region 221 Figure 5.14 The initial cooling of a thin sword blade. Prior to t = t 4 , the blade might as well be infinitely thick insofar as cooling is concerned. The blade is then tough and ductile, so it will not break, but has a fine hard outer shell that can be honed to sharpness. We need only look a little way up the side of the clay sheath to find a cross section that was thick enough to prevent the blade from experiencing the sudden effects of the cooling quench. The success of the process actually relies on the failure of the cooling to penetrate the clay very deeply in a short time. Now we wish to ask: “How can we say whether or not the influence of a heating or cooling process is restricted to the surface of a body?” Or if we turn the question around: “Under what conditions can we view the depth of a body as infinite with respect to the thickness of the region that has felt the heat transfer process?” Consider next the cooling process within the blade in the absence of the clay retardant and when h is very large. Actually, our considerations will apply initially to any finite body whose boundary suddenly changes temperature. The temperature distribution, in this case, is sketched in Fig. 5.14 for four sequential times. Only the fourth curve—that for which t = t 4 —is noticeably influenced by the opposite wall. Up to that time, the wall might as well have infinite depth. Since any body subjected to a sudden change of temperature is in- finitely large in comparison with the initial region of temperature change, we must learn how to treat heat transfer in this period. Solution aided by dimensional analysis The calculation of the temperature distribution in a semi-infinite region poses a difficulty in that we can impose a definite b.c. at only one position— the exposed boundary. We shall be able to get around that difficulty in a nice way with the help of dimensional analysis. 222 Transient and multidimensional heat conduction §5.6 When the one boundary of a semi-infinite region, initially at T = T i , is suddenly cooled (or heated) to a new temperature, T ∞ , as in Fig. 5.14, the dimensional function equation is T − T ∞ = fn [ t, x, α, (T i −T ∞ ) ] where there is no characteristic length or time. Since there are five vari- ables in ◦ C, s, and m, we should look for two dimensional groups. T − T ∞ T i −T ∞    Θ = fn  x √ αt    ζ  (5.44) The very important thing that we learn from this exercise in dimen- sional analysis is that position and time collapse into one independent variable. This means that the heat conduction equation and its b.c.s must transform from a partial differential equation into a simpler ordinary dif- ferential equation in the single variable, ζ = x  √ αt. Thus, we transform each side of ∂ 2 T ∂x 2 = 1 α ∂T ∂t as follows, where we call T i −T ∞ ≡ ∆T : ∂T ∂t = (T i −T ∞ ) ∂Θ ∂t = ∆T ∂Θ ∂ζ ∂ζ ∂t = ∆T  − x 2t √ αt  ∂Θ ∂ζ ; ∂T ∂x = ∆T ∂Θ ∂ζ ∂ζ ∂x = ∆T √ αt ∂Θ ∂ζ ; and ∂ 2 T ∂x 2 = ∆T √ αt ∂ 2 Θ ∂ζ 2 ∂ζ ∂x = ∆T αt ∂ 2 Θ ∂ζ 2 . Substituting the first and last of these derivatives in the heat conduction equation, we get d 2 Θ dζ 2 =− ζ 2 dΘ dζ (5.45) Notice that we changed from partial to total derivative notation, since Θ now depends solely on ζ. The i.c. for eqn. (5.45)is T(t = 0) = T i or Θ ( ζ →∞ ) = 1 (5.46) §5.6 Transient heat conduction to a semi-infinite region 223 and the one known b.c. is T(x = 0) = T ∞ or Θ ( ζ = 0 ) = 0 (5.47) If we call dΘ/dζ ≡ χ, then eqn. (5.45) becomes the first-order equa- tion dχ dζ =− ζ 2 χ which can be integrated once to get χ ≡ dΘ dζ = C 1 e −ζ 2 /4 (5.48) and we integrate this a second time to get Θ = C 1  ζ 0 e −ζ 2 /4 dζ + Θ(0)    = 0 according to the b.c. (5.49) The b.c. is now satisfied, and we need only substitute eqn. (5.49)inthe i.c., eqn. (5.46), to solve for C 1 : 1 = C 1  ∞ 0 e −ζ 2 /4 dζ The definite integral is given by integral tables as √ π,so C 1 = 1 √ π Thus the solution to the problem of conduction in a semi-infinite region, subject to a b.c. of the first kind is Θ = 1 √ π  ζ 0 e −ζ 2 /4 dζ = 2 √ π  ζ/2 0 e −s 2 ds ≡ erf(ζ/2) (5.50) The second integral in eqn. (5.50), obtained by a change of variables, is called the error function (erf). Its name arises from its relationship to certain statistical problems related to the Gaussian distribution, which describes random errors. In Table 5.3, we list values of the error function and the complementary error function, erfc(x) ≡ 1 − erf(x). Equation (5.50) is also plotted in Fig. 5.15. [...]... Superheated Liquid This prediction is relevant to a large variety of processes, ranging from nuclear thermodynamics to the direct-contact heat exchange It was originally presented by Max Jakob and others in the early 19 30s (see, e.g., [5 .10 , Chap I]) Jakob (pronounced Yah -kob) was an important figure in heat transfer during the 19 20s and 19 30s He left Nazi Germany in 19 36 to come to the United States... 0.20309 0 .15 730 ζ 2 erf(ζ/2) erfc(ζ/2) 1. 10 1. 20 1. 30 1. 40 1. 50 1. 60 1. 70 1. 80 1. 8 214 1. 90 2.00 2.50 3.00 0.880 21 0. 910 31 0.934 01 0.95229 0.96 611 0.97635 0.98379 0.98909 0.99000 0.99279 0.99532 0.99959 0.99998 0 .11 980 0.08969 0.06599 0.047 71 0.03389 0.02365 0. 016 21 0. 010 91 0. 010 00 0.007 21 0.00468 0.000 41 0.00002 In Fig 5 .15 we see the early-time curves shown in Fig 5 .14 have collapsed into a single... to a semi-infinite region with a harmonically oscillating temperature at the boundary Suppose that we approximate the annual variation of the ambient temperature as sinusoidal and then ask what the influence of this variation will be beneath the ground We want to calculate T − T (where T is the time-average surface temperature) as a function of: depth, x; thermal diffusivity, α; frequency of oscillation,... Plesset and Zwick [5 .11 ] It was verified in a more exact way after another 5 years by Scriven [5 .12 ] These calculations are more complicated, but they lead to a very similar result: √ √ 2 3 k∆T √ t = 3 RJakob (5.59) R= √ π ρg hfg α Both predictions are compared with some of the data of Dergarabedian [5 .13 ] in Fig 5 .18 The data and the exact theory match almost perfectly The simple theory of Jakob et al... encounter his name again later Figure 5 .17 shows how growth occurs When a liquid is superheated to a temperature somewhat above its boiling point, a small gas or vapor cavity in that liquid will grow (That is what happens in the superheated water at the bottom of a teakettle.) This bubble grows into the surrounding liquid because its boundary is kept at the saturation temperature, Tsat , by the near-equilibrium... be analyzed as a semiinfinite region after it is quenched, if it has no clay coating and hexternal ∞? Solution First, we must guess the half-thickness of the sword (say, 3 mm) and its material (probably wrought iron with an average α 5 The transformation is based upon the “similarity” of spatial an temporal changes in this problem Transient heat conduction to a semi-infinite region §5.6 225 Figure 5 .15 ... heat flowing from one isothermal wall to another in a regime that does not conform to any convenient coordinate scheme We identify a series of channels, each which carries the same heat flow, δQ W/m We also include a set of equally spaced isotherms, δT apart, between the walls Since the heat fluxes in all channels are the same, δQ = k δT δs δn (5.64) Notice that if we arrange things so that δQ, δT , and... with an example that was executed nicely in the influential Heat Transfer Notes [5.3] of the mid-twentieth century This example is shown in Fig 5. 21 The particular example happens to have an axis of symmetry in it We immediately interpret this as an adiabatic boundary because heat cannot cross it The problem therefore reduces to the simpler one of sketching lines in only one half of the area We illustrate... not the analytical solution is already available in a heat conduction text or in other published literature • Solve the problem (a) Analytically (b) Numerically • Obtain the solution graphically if the problem is two-dimensional It is to the last of these options that we give our attention next 235 236 Transient and multidimensional heat conduction §5.7 Figure 5.20 The two-dimensional flow of heat between... region of constant thermal conductivity, without heat sources, is called Laplace’s equation: ∇2 T = 0 (5.63) It looks easier to solve than it is, since [recall eqn (2 .12 ) and eqn (2 .14 )] the Laplacian, ∇2 T , is a sum of several second partial derivatives We solved one two-dimensional heat conduction problem in Example 4 .1, but this was not difficult because the boundary conditions were made to order Depending . 1. 14 41 0.9960 0.60 0.70507 1. 0 814 0.9940 1. 018 44 1. 1345 0.9936 1. 26440 1. 1 713 0.9944 0.70 0.75056 1. 0 918 0.9922 1. 08725 1. 1539 0.9 916 1. 35252 1. 1978 0.9925 0.80 0.7 910 3 1. 1 016 0.9903 1. 14897 1. 1724. 0.89035 1. 1270 0.9839 1. 302 51 1.2232 0.9 815 1. 6 319 9 1. 2970 0.9828 1. 20 0. 917 85 1. 1344 0.9 817 1. 34558 1. 2387 0.9787 1. 68868 1. 32 01 0.9800 1. 30 0.94 316 1. 1 412 0.9794 1. 38543 1. 2533 0.9757 1. 7 414 0 1. 3424. 1. 1724 0.9893 1. 43203 1. 2236 0.9904 0.90 0.82740 1. 110 7 0.9882 1. 20484 1. 1902 0.9869 1. 50442 1. 2488 0.9880 1. 00 0.86033 1. 119 1 0.98 61 1.25578 1. 20 71 0.9843 1. 57080 1. 2732 0.9855 1. 10 0.89035 1. 1270