John H. Lienhard IV, John H. Lienhard V. A Heat Transfer Textbook (776116), страница 31
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5.10, forspheres:tQ dt0 Φ = 0.80 =ρc 43 π r03 (Ti − T∞ )sot0Q dt = 997.6(4180)4π (0.05)3 (25)(0.80) = 43, 668 J/apple3Therefore, for the 12 apples,total energy removal = 12(43.67) = 524 kJ§5.4Temperature-response charts215The temperature-response charts in Fig. 5.7 through Fig. 5.10 are without doubt among the most useful available since they can be adapted toa host of physical situations. Nevertheless, hundreds of such charts havebeen formed for other situations, a number of which have been catalogedby Schneider [5.5]. Analytical solutions are available for hundreds moreproblems, and any reader who is faced with a complex heat conductioncalculation should consult the literature before trying to solve it.
An excellent place to begin is Carslaw and Jaeger’s comprehensive treatise onheat conduction [5.6].Example 5.3A 1 mm diameter Nichrome (20% Ni, 80% Cr) wire is simultaneouslybeing used as an electric resistance heater and as a resistance thermometer in a liquid flow. The laboratory workers who operate it areattempting to measure the boiling heat transfer coefficient, h, by supplying an alternating current and measuring the difference betweenthe average temperature of the heater, Tav , and the liquid temperature, T∞ . They get h = 30, 000 W/m2 K at a wire temperature of 100◦ Cand are delighted with such a high value. Then a colleague suggeststhat h is so high because the surface temperature is rapidly oscillatingas a result of the alternating current. Is this hypothesis correct?Solution.
Heat is being generated in proportion to the product ofvoltage and current, or as sin2 ωt, where ω is the frequency of thecurrent in rad/s. If the boiling action removes heat rapidly enough incomparison with the heat capacity of the wire, the surface temperature may well vary significantly. This transient conduction problemwas first solved by Jeglic in 1962 [5.7]. It was redone in a differentform two years later by Switzer and Lienhard (see, e.g. [5.8]), who gaveresponse curves in the formTmax − Tav= fn (Bi, ψ)Tav − T∞(5.41)where the left-hand side is the dimensionless range of the temperature oscillation, and ψ = ωδ2 /α, where δ is a characteristic length[see Problem 5.56]. Because this problem is common and the solution is not widely available, we include the curves for flat plates andcylinders in Fig.
5.11 and Fig. 5.12 respectively.216Figure 5.11 Temperature deviation at the surface of a flat plate heated with alternating current.217Figure 5.12 Temperature deviation at the surface of a cylinder heated with alternating current.218Transient and multidimensional heat conduction§5.5In the present case:30, 000(0.0005)h radius== 1.09k13.8[2π (60)](0.0005)2ωr 2== 27.5α0.00000343Bi =and from the chart for cylinders, Fig. 5.12, we find thatTmax − Tav 0.04Tav − T∞A temperature fluctuation of only 4% is probably not serious.
It therefore appears that the experiment was valid.5.5One-term solutionsAs we have noted previously, when the Fourier number is greater than 0.2or so, the series solutions from eqn. (5.36) may be approximated usingonly their first term:Θ ≈ A1 · f1 · exp −λ̂21 Fo .(5.42)Likewise, the fractional heat loss, Φ, or the mean temperature Θ fromeqn. (5.40), can be approximated using just the first term of eqn. (5.38):Θ = 1 − Φ ≈ D1 exp −λ̂21 Fo .(5.43)Table 5.2 lists the values of λ̂1 , A1 , and D1 for slabs, cylinders, andspheres as a function of the Biot number.
The one-term solution’s error in Θ is less than 0.1% for a sphere with Fo ≥ 0.28 and for a slab withFo ≥ 0.43. These errors are largest for Biot numbers near unity. If highaccuracy is not required, these one-term approximations may generallybe used whenever Fo ≥ 0.2Table 5.2 One-term coefficients for convective cooling [5.1].BiPlateCylinderSphereλ̂1A1D1λ̂1A1D1λ̂1A1D10.010.020.050.099830.140950.221761.00171.00331.00821.00001.00000.99990.141240.199500.314261.00251.00501.01241.00001.00000.99990.173030.244460.385371.00301.00601.01501.00001.00001.00000.100.150.200.300.400.500.600.700.800.900.311050.377880.432840.521790.593240.653270.705070.750560.791030.827401.01611.02371.03111.04501.05801.07011.08141.09181.10161.11070.99980.99950.99920.99830.99710.99560.99400.99220.99030.98820.441680.537610.616970.746460.851580.940771.018441.087251.148971.204841.02461.03651.04831.07121.09311.11431.13451.15391.17241.19020.99980.99950.99920.99830.99700.99540.99360.99160.98930.98690.542280.660860.759310.920791.052791.165561.264401.352521.432031.504421.02981.04451.05921.08801.11641.14411.17131.19781.22361.24880.99980.99960.99930.99850.99740.99600.99440.99250.99040.98801.001.101.201.301.401.501.601.802.002.202.400.860330.890350.917850.943160.966550.988241.008421.044861.076871.105241.130561.11911.12701.13441.14121.14771.15371.15931.16951.17851.18641.19340.98610.98390.98170.97940.97710.97480.97260.96800.96350.95920.95491.255781.302511.345581.385431.422461.456951.489171.547691.599451.645571.686911.20711.22321.23871.25331.26731.28071.29341.31701.33841.35781.37540.98430.98150.97870.97570.97270.96960.96650.96010.95370.94720.94081.570801.631991.688681.741401.790581.836601.879761.958572.028762.091662.148341.27321.29701.32011.34241.36401.38501.40521.44361.47931.51251.54330.98550.98280.98000.97700.97390.97070.96740.96050.95340.94620.93893.004.005.006.008.0010.0020.0050.00100.00∞1.192461.264591.313841.349551.397821.428871.496131.540011.555251.570801.21021.22871.24021.24791.25701.26201.26991.27271.27311.27320.94310.92640.91300.90210.88580.87430.84640.82600.81850.81061.788661.908081.989812.049012.128642.179502.288052.357242.380902.404831.41911.46981.50291.52531.55261.56771.59191.60021.60151.60200.92240.89500.87210.85320.82440.80390.75420.71830.70520.69172.288932.455642.570432.653662.765362.836302.985723.078843.110193.141591.62271.72021.78701.83381.89201.92491.97811.99621.99902.00000.91710.88300.85330.82810.78890.76070.69220.64340.62590.6079219220Transient and multidimensional heat conduction5.6§5.6Transient heat conduction to a semi-infiniteregionIntroductionBronowksi’s classic television series, The Ascent of Man [5.9], includeda brilliant reenactment of the ancient ceremonial procedure by whichthe Japanese forged Samurai swords (see Fig.
5.13). The metal is heated,folded, beaten, and formed, over and over, to create a blade of remarkabletoughness and flexibility. When the blade is formed to its final configuration, a tapered sheath of clay is baked on the outside of it, so the crosssection is as shown in Fig. 5.13. The red-hot blade with the clay sheath isthen subjected to a rapid quenching, which cools the uninsulated cuttingedge quickly and the back part of the blade very slowly. The result is alayer of case-hardening that is hardest at the edge and less hard at pointsfarther from the edge.Figure 5.13 The ceremonial case-hardening of a Samurai sword.§5.6Transient heat conduction to a semi-infinite region221Figure 5.14 The initial cooling of a thinsword blade.
Prior to t = t4 , the blademight as well be infinitely thick insofar ascooling is concerned.The blade is then tough and ductile, so it will not break, but has a finehard outer shell that can be honed to sharpness. We need only look alittle way up the side of the clay sheath to find a cross section that wasthick enough to prevent the blade from experiencing the sudden effectsof the cooling quench. The success of the process actually relies on thefailure 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 influenceof 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 viewthe depth of a body as infinite with respect to the thickness of the regionthat has felt the heat transfer process?”Consider next the cooling process within the blade in the absence ofthe clay retardant and when h is very large.
Actually, our considerationswill apply initially to any finite body whose boundary suddenly changestemperature. The temperature distribution, in this case, is sketched inFig. 5.14 for four sequential times. Only the fourth curve—that for whicht = t4 —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 infinitely 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 analysisThe calculation of the temperature distribution in a semi-infinite regionposes 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 anice way with the help of dimensional analysis.222Transient and multidimensional heat conduction§5.6When the one boundary of a semi-infinite region, initially at T = Ti ,is suddenly cooled (or heated) to a new temperature, T∞ , as in Fig. 5.14,the dimensional function equation isT − T∞ = fn [t, x, α, (Ti − T∞ )]where there is no characteristic length or time. Since there are five variables in ◦ C, s, and m, we should look for two dimensional groups.T − T∞x= fn √(5.44)T −Tαt i ∞ΘζThe very important thing that we learn from this exercise in dimensional analysis is that position and time collapse into one independentvariable.
This means that the heat conduction equation and its b.c.s musttransform from a partial differential equation into√ a simpler ordinary differential equation in the single variable, ζ = x αt. Thus, we transformeach side of∂2T1 ∂T=2∂xα ∂tas follows, where we call Ti − T∞ ≡ ∆T :∂T∂Θ∂Θ ∂ζx∂Θ= (Ti − T∞ )= ∆T= ∆T − √;∂t∂ζ ∂t∂t2t αt ∂ζ∂Θ ∂ζ∆T ∂Θ∂T= ∆T=√;∂x∂ζ ∂xαt ∂ζand∂2T∆T ∂ 2 Θ∆T ∂ 2 Θ ∂ζ√==.∂x 2αt ∂ζ 2αt ∂ζ 2 ∂xSubstituting the first and last of these derivatives in the heat conductionequation, we getd2 Θζ dΘ=−2dζ2 dζ(5.45)Notice that we changed from partial to total derivative notation, sinceΘ now depends solely on ζ.
The i.c. for eqn. (5.45) isT (t = 0) = TiorΘ (ζ → ∞) = 1(5.46)Transient heat conduction to a semi-infinite region§5.6and the one known b.c. isT (x = 0) = T∞orΘ (ζ = 0) = 0(5.47)If we call dΘ/dζ ≡ χ, then eqn. (5.45) becomes the first-order equationζdχ=− χdζ2which can be integrated once to getχ≡dΘ2= C1 e−ζ /4dζ(5.48)and we integrate this a second time to getζΘ = C1e−ζ2 /40dζ +Θ(0)(5.49)= 0 accordingto the b.c.The b.c. is now satisfied, and we need only substitute eqn. (5.49) in thei.c., eqn. (5.46), to solve for C1 :∞2e−ζ /4 dζ1 = C10The definite integral is given by integral tables as√π , so1C1 = √πThus the solution to the problem of conduction in a semi-infinite region,subject to a b.c. of the first kind is1Θ= √πζ0e−ζ2 /42dζ = √π ζ/202e−s ds ≡ erf(ζ/2)(5.50)The second integral in eqn.
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