John H. Lienhard IV, John H. Lienhard V. A Heat Transfer Textbook (776116), страница 32
Текст из файла (страница 32)
(5.50), obtained by a change of variables,is called the error function (erf). Its name arises from its relationship tocertain statistical problems related to the Gaussian distribution, whichdescribes random errors. In Table 5.3, we list values of the error functionand the complementary error function, erfc(x) ≡ 1 − erf(x). Equation(5.50) is also plotted in Fig.
5.15.223Transient and multidimensional heat conduction224§5.6Table 5.3 Error function and complementary error function.ζ 2erf(ζ/2)erfc(ζ/2)0.000.050.100.150.200.300.400.500.600.700.800.901.000.000000.056370.112460.168000.222700.328630.428390.520500.603860.677800.742100.796910.842701.000000.943630.887540.832000.777300.671370.571610.479500.396140.322200.257900.203090.15730ζ 2erf(ζ/2)erfc(ζ/2)1.101.201.301.401.501.601.701.801.82141.902.002.503.000.880210.910310.934010.952290.966110.976350.983790.989090.990000.992790.995320.999590.999980.119800.089690.065990.047710.033890.023650.016210.010910.010000.007210.004680.000410.00002In Fig.
5.15 we see the early-time curves shown in Fig. 5.14 have collapsed into a single curve. This was accomplishedby the similarity trans√formation, as we call it5 : ζ/2 = x/2 αt. From the figure or from Table5.3, we see that Θ ≥ 0.99 whenxζ≥ 1.8214= √22 αtor3x ≥ δ99 ≡ 3.64 αt(5.51)In other words, the local value of (T − T∞ ) is more than 99% of (Ti − T∞ )for positionsin the slab beyond farther from the surface than δ99 =√3.64 αt.Example 5.4For what maximum time can a samurai sword 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 α5The transformation is based upon the “similarity” of spatial an temporal changesin this problem.Transient heat conduction to a semi-infinite region§5.6225Figure 5.15 Temperature distribution ina semi-infinite region.around 1.5 × 10−5 m2 /s). The sword will be semi-infinite until δ99equals the half-thickness. Inverting eqn. (5.51), we findtδ299(0.003 m)2= 0.045 s=23.64 α13.3(1.5)(10)−5 m2 /sThus the quench would be felt at the centerline of the sword withinonly 1/20 s. The thermal diffusivity of clay is smaller than that of steelby a factor of about 30, so the quench time of the coated steel mustcontinue for over 1 s before the temperature of the steel is affectedat all, if the clay and the sword thicknesses are comparable.Equation (5.51) provides an interesting foretaste of the notion of afluid boundary layer.
In the context of Fig. 1.9 and Fig. 1.10, we observe that free stream flow around an object is disturbed in a thick layernear the object because the fluid adheres to it. It turns out that thethickness of this boundary layer of altered flow velocity increases in thedownstream direction.For flow over a flat plate, this thickness is ap√proximately 4.92 νt, where t is the time required for an element of thestream fluid to move from the leading edge of the plate to a point of interest. This is quite similar to eqn. (5.51), except that the thermal diffusivity,α, has been replaced by its counterpart, the kinematic viscosity, ν, andthe constant is a bit larger.
The velocity profile will resemble Fig. 5.15.If we repeated the problem with a boundary condition of the thirdkind, we would expect to get Θ = Θ(Bi, ζ), except that there is no length,L,√ upon which to build a Biot number. Therefore, we must replace L withαt, which has the dimension of length, so√ h αt≡ Θ(ζ, β)(5.52)Θ = Θ ζ,k226Transient and multidimensional heat conduction§5.6√√ The term β ≡ h αt k is like the product: Bi Fo.
The solution of thisproblem (see, e.g., [5.6], §2.7) can be conveniently written in terms of thecomplementary error function, erfc(x) ≡ 1 − erf(x):ζζ2erfc+βΘ = erf + exp βζ + β22(5.53)This result is plotted in Fig. 5.16.Example 5.5Most of us have passed our finger through an 800◦ C candle flame andknow that if we limit exposure to about 1/4 s we will not be burned.Why not?Solution. The short exposure to the flame causes only a very superficial heating, so we consider the finger to be a semi-infinite region and go to eqn. (5.53) to calculate (Tburn − Tflame )/(Ti − Tflame ).
Itturns out that the burn threshold of human skin, Tburn , is about 65◦ C.(That is why 140◦ F or 60◦ C tap water is considered to be “scalding.”)Therefore, we shall calculate how long it will take for the surface temperature of the finger to rise from body temperature (37◦ C) to 65◦ C,when it is protected by an assumed h 100 W/m2 K.
We shall assumethat the thermal conductivity of human flesh equals that of its majorcomponent—water—and that the thermal diffusivity is equal to theknown value for beef. ThenΘ=βζ =hx=0k65 − 800= 0.96337 − 800since x = 0 at the surface21002 (0.135 × 10−6 )th αtβ === 0.0034(t s)k20.6322The situation is quite far into the corner of Fig. 5.16. We read β2 0.001, which corresponds with t 0.3 s.
For greater accuracy, wemust go to eqn. (5.53):30.0034t+e0.0034t0.963 = erf0erfc0+ =0Figure 5.16 The cooling of a semi-infinite region by an environment at T∞ , through a heat transfer coefficient, h.227228Transient and multidimensional heat conduction§5.6By trial and error, we get t 0.33 s. In fact, it can be shown that2βΘ(ζ = 0, β) 1 − √πfor β 1√which can be solved directly for β = (1 − 0.963) π /2 = 0.03279,leading to the same answer.Thus, it would require about 1/3 s to bring the skin to the burnpoint.Experiment 5.1Immerse your hand in the subfreezing air in the freezer compartmentof your refrigerator.
Next immerse your finger in a mixture of ice cubesand water, but do not move it. Then, immerse your finger in a mixture ofice cubes and water , swirling it around as you do so. Describe your initialsensation in each case, and explain the differences in terms of Fig. 5.16.What variable has changed from one case to another?Heat transferHeat will be removed from the exposed surface of a semi-infinite region,with a b.c. of either the first or the third kind, in accordance with Fourier’slaw:dΘk(T−T)∂T ∞i√=q = −k∂x x=0dζαtζ=0Differentiating Θ as given by eqn. (5.50), we obtain, for the b.c.
of thefirst kind,k(T∞ − Ti )√q=αt12√ e−ζ /4πζ=0=k(T∞ − Ti )√π αt(5.54)Thus, q decreases with increasing time, as t −1/2 . When the temperatureof the surface is first changed, the heat removal rate is enormous. Thenit drops off rapidly.It often occurs that we suddenly apply a specified input heat flux,qw , at the boundary of a semi-infinite region. In such a case, we canTransient heat conduction to a semi-infinite region§5.6differentiate the heat diffusion equation with respect to x, soα∂2T∂3T=∂x 3∂t∂xWhen we substitute q = −k ∂T /∂x in this, we obtainα∂q∂2q=2∂x∂twith the b.c.’s:q(x = 0, t > 0) = qwq(x 0, t = 0) = 0ororqw − q =0qw x=0qw − q =1qw t=0What we have done here is quite elegant.
We have made the problemof predicting the local heat flux q into exactly the same form as that ofpredicting the local temperature in a semi-infinite region subjected to astep change of wall temperature. Therefore, the solution must be thesame:xqw − q√.(5.55)= erfqw2 αtThe temperature distribution is obtained by integrating Fourier’s law. Atthe wall, for example:0 TwqdxdT = −Ti∞ kwhere Ti = T (x → ∞) and Tw = T (x = 0). Then3qw ∞Tw = Ti +erfc(x/2 αt) dxk 0This becomesTw∞qw 3= Ti +αterfc(ζ/2) dζk0√=2/ πsoqwTw (t) = Ti + 2k2αtπ(5.56)229230Transient and multidimensional heat conduction§5.6Figure 5.17 A bubble growing in asuperheated liquid.Example 5.6Predicting the Growth Rate of a Vapor Bubblein an Infinite Superheated LiquidThis prediction is relevant to a large variety of processes, rangingfrom nuclear thermodynamics to the direct-contact heat exchange.
Itwas originally presented by Max Jakob and others in the early 1930s(see, e.g., [5.10, Chap. I]). Jakob (pronounced Yah -kob) was an important figure in heat transfer during the 1920s and 1930s. He leftNazi Germany in 1936 to come to the United States. We encounterhis name again later.Figure 5.17 shows how growth occurs. When a liquid is superheated to a temperature somewhat above its boiling point, a smallgas or vapor cavity in that liquid will grow. (That is what happens inthe 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-equilibriumcoexistence of liquid and vapor.
![](https://s.studizba.com/z.php?f=/templates/si/i/noavatar.png&w=100&h=100&t=1"){kind=avatar}
