John H. Lienhard IV, John H. Lienhard V. A Heat Transfer Textbook (776116), страница 35
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Cylinder of radius R and length L,transferring heat to a parallelisothermal plane; h L2π Lcosh−1 (h/R)6. Same as item 5, but with L → ∞(two-dimensional conduction)2πcosh−1(h/R)7. An isothermal sphere of radius Rtransfers heat to an isothermalplane; R/h < 0.8 (see item 4)4π R1 − R/2h245Table 5.4 Conduction shape factors: Q = S k∆T (con’t).SituationShape factor, S8. An isothermal sphere of radius R,near an insulated plane, transfersheat to a semi-infinite medium atT∞ (see items 4 and 7)4π R1 + R/2hDimensionsmeterSource[5.18]9.
Parallel cylinders exchange heat inan infinite conducting medium−1cosh10. Same as 9, but with cylinderswidely spaced; L R1 and R211. Cylinder of radius Ri surroundedby eccentric cylinder of radiusRo > Ri ; centerlines a distance Lapart (see item 2)12. Isothermal disc of radius R on anotherwise insulated plane conductsheat into a semi-infinite medium atT∞ below it13. Isothermal ellipsoid of semimajoraxis b and semiminor axes aconducts heat into an infinitemedium at T∞ ; b > a (see 4)246cosh−1L2R1cosh−12πL2 − R12 − R222R1 R22πL−1+ cosh2R22πRo2 + Ri2 − L22Ro Ri4R44π b 1 − a2 b24 tanh−1 1 − a2 b2none[5.6]none[5.16]none[5.6]meter[5.6]meter[5.16]§5.8Transient multidimensional heat conduction247Figure 5.26 Resistance vanishes wheretwo isothermal boundaries intersect.The problem of locally vanishing resistanceSuppose that two different temperatures are specified on adjacent sidesof a square, as shown in Fig.
5.26. The shape factor in this case isS=∞N==∞I4(It is futile to try and count channels beyond N 10, but it is clear thatthey multiply without limit in the lower left corner.) The problem is thatwe have violated our rule that isotherms cannot intersect and have created a 1/r singularity. If we actually tried to sustain such a situation,the figure would be correct at some distance from the corner. However,where the isotherms are close to one another, they will necessarily influence and distort one another in such a way as to avoid intersecting. AndS will never really be infinite, as it appears to be in the figure.5.8Transient multidimensional heat conduction—The tactic of superpositionConsider the cooling of a stubby cylinder, such as the one shown inFig.
5.27a. The cylinder is initially at T = Ti , and it is suddenly subjected to a common b.c. on all sides. It has a length 2L and a radius ro .Finding the temperature field in this situation is inherently complicated.248Transient and multidimensional heat conduction§5.8It requires solving the heat conduction equation for T = fn(r , z, t) withb.c.’s of the first, second, or third kind.However, Fig. 5.27a suggests that this can somehow be viewed as acombination of an infinite cylinder and an infinite slab. It turns out thatthe problem can be analyzed from that point of view.If the body is subject to uniform b.c.’s of the first, second, or thirdkind, and if it has a uniform initial temperature, then its temperatureresponse is simply the product of an infinite slab solution and an infinitecylinder solution each having the same boundary and initial conditions.For the case shown in Fig.
5.27a, if the cylinder begins convective cooling into a medium at temperature T∞ at time t = 0, the dimensionaltemperature response is T (r , z, t) − T∞ = Tslab (z, t) − T∞ × Tcyl (r , t) − T∞(5.70a)Observe that the slab has as a characteristic length L, its half thickness,while the cylinder has as its characteristic length R, its radius. In dimensionless form, we may write eqn. (5.70a) asΘ≡T (r , z, t) − T∞= Θinf slab (ξ, Fos , Bis ) Θinf cyl (ρ, Foc , Bic )Ti − T∞(5.70b)For the cylindrical component of the solution,ρ=r,roFoc =αt,ro2andBic =hro,kwhile for the slab component of the solutionξ=z+ 1,LFos =αt,L2andBis =hL.kThe component solutions are none other than those discussed in Sections 5.3–5.5. The proof of the legitimacy of such product solutions isgiven by Carlsaw and Jaeger [5.6, §1.15].Figure 5.27b shows a point inside a one-eighth-infinite region, near thecorner.
This case may be regarded as the product of three semi-infinitebodies. To find the temperature at this point we writeΘ≡T (x1 , x2 , x3 , t) − T∞= [Θsemi (ζ1 , β)] [Θsemi (ζ2 , β)] [Θsemi (ζ3 , β)]Ti − T∞(5.71)Figure 5.27 Various solid bodies whose transient cooling canbe treated as the product of one-dimensional solutions.249250Transient and multidimensional heat conduction§5.8in which Θsemi is either the semi-infinite body solution given by eqn. (5.53)when convection is present at the boundary or the solution given byeqn.
(5.50) when the boundary temperature itself is changed at time zero.Several other geometries can also be represented by product solutions. Note that for of these solutions, the value of Θ at t = 0 is one foreach factor in the product.Example 5.12A very long 4 cm square iron rod at Ti = 100◦ C is suddenly immersedin a coolant at T∞ = 20◦ C with h = 800 W/m2 K. What is the temperature on a line 1 cm from one side and 2 cm from the adjoining side,after 10 s?Solution. With reference to Fig. 5.27c, see that the bar may betreated as the product of two slabs, each 4 cm thick.We first evaluateFo1 = Fo2 = αt/L2 =(0.0000226 m2 /s)(10 s) (0.04 m/2)2 = 0.565,and Bi1 = Bi2 = hL k = 800(0.04/2)/76 = 0.2105, and we thenwrite 1xx−1= 0,= , Fo1 , Fo2 , Bi−1,BiΘ12L 1L 22 x= 0, Fo1 = 0.565, Bi−1=4.75= Θ11L 1= 0.93 from upper left-handside of Fig.
5.7× Θ2 1x= , Fo2 = 0.565, Bi−1=4.7522 L 2= 0.91 from interpolationbetween lower lefthand side andupper righthand side of Fig. 5.7Thus, at the axial line of interest,Θ = (0.93)(0.91) = 0.846soT − 20= 0.846100 − 20orT = 87.7◦ CTransient multidimensional heat conduction251Product solutions can also be used to determine the mean temperature, Θ, and the total heat removal, Φ, from a multidimensional object.For example, when two or three solutions (Θ1 , Θ2 , and perhaps Θ3 ) aremultiplied to obtain Θ, the corresponding mean temperature of the multidimensional object is simply the product of the one-dimensional meantemperatures from eqn. (5.40)Θ = Θ1 (Fo1 , Bi1 ) × Θ2 (Fo2 , Bi2 )for two factorsΘ = Θ1 (Fo1 , Bi1 ) × Θ2 (Fo2 , Bi2 ) × Θ3 (Fo3 , Bi3 )(5.72a)for three factors.(5.72b)Since Φ = 1 − Θ, a simple calculation shows that Φ can found from Φ1 ,Φ2 , and Φ3 as follows:Φ = Φ1 + Φ2 (1 − Φ1 )for two factorsΦ = Φ1 + Φ2 (1 − Φ1 ) + Φ3 (1 − Φ2 ) (1 − Φ1 )(5.73a)for three factors.
(5.73b)Example 5.13For the bar described in Example 5.12, what is the mean temperatureafter 10 s and how much heat has been lost at that time?Solution. For the Biot and Fourier numbers given in Example 5.12,we find from Fig. 5.10aΦ1 (Fo1 = 0.565, Bi1 = 0.2105) = 0.10Φ2 (Fo2 = 0.565, Bi2 = 0.2105) = 0.10and, with eqn. (5.73a),Φ = Φ1 + Φ2 (1 − Φ1 ) = 0.19The mean temperature isΘ=T − 20= 1 − Φ = 0.81100 − 20soT = 20 + 80(0.81) = 84.8◦ CChapter 5: Transient and multidimensional heat conduction252Problems5.1Rework Example 5.1, and replot the solution, with one change.This time, insert the thermometer at zero time, at an initialtemperature < (Ti − bT ).5.2A body of known volume and surface area and temperature Tiis suddenly immersed in a bath whose temperature is risingas Tbath = Ti + (T0 − Ti )et/τ .
Let us suppose that h is known,that τ = 10ρcV /hA, and that t is measured from the time ofimmersion. The Biot number of the body is small. Find thetemperature response of the body. Plot the response and thebath temperature as a function of time up to t = 2τ. (Do notuse Laplace transform methods except, perhaps, as a check.)5.3A body of known volume and surface area is immersed ina bath whose temperature is varying sinusoidally with a frequency ω about an average value. The heat transfer coefficientis known and the Biot number is small. Find the temperaturevariation of the body after a long time has passed, and plot italong with the bath temperature.
Comment on any interestingaspects of the solution.A suggested program for solving this problem:• Write the differential equation of response.• To get the particular integral of the complete equation,guess that T − Tmean = C1 cos ωt + C2 sin ωt. Substitutethis in the differential equation and find C1 and C2 valuesthat will make the resulting equation valid.• Write the general solution of the complete equation.
Itwill have one unknown constant in it.• Write any initial condition you wish—the simplest one youcan think of—and use it to get rid of the constant.• Let the time be large and note which terms vanish fromthe solution. Throw them away.• Combine two trigonometric terms in the solution into aterm involving sin(ωt − β), where β = fn(ωT ) is thephase lag of the body temperature.5.4A block of copper floats within a large region of well-stirredmercury. The system is initially at a uniform temperature, Ti .Problems253There is a heat transfer coefficient, hm , on the inside of the thinmetal container of the mercury and another one, hc , betweenthe copper block and the mercury.
The container is then suddenly subjected to a change in ambient temperature from Ti toTs < Ti . Predict the temperature response of the copper block,neglecting the internal resistance of both the copper and themercury. Check your result by seeing that it fits both initialconditions and that it gives the expected behavior at t → ∞.5.5Sketch the electrical circuit that is analogous to the secondorder lumped capacity system treated in the context of Fig. 5.5and explain it fully.5.6A one-inch diameter copper sphere with a thermocouple inits center is mounted as shown in Fig. 5.28 and immersed inwater that is saturated at 211◦ F.
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