John H. Lienhard IV, John H. Lienhard V. A Heat Transfer Textbook (776116), страница 33
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Therefore, heat must flow from thesuperheated surroundings to the interface, where evaporation occurs.So long as the layer of cooled liquid is thin, we should not suffer toomuch error by using the one-dimensional semi-infinite region solution to predict the heat flow.Transient heat conduction to a semi-infinite region§5.6Thus, we can write the energy balance at the bubble interface: 3dVmW J4π R 2 m2 = ρg hfg 3−q 2mmdt s Q into bubblerate of energy increaseof the bubbleand then substitute eqn. (5.54) for q and 4π R 3 /3 for the volume, V .This givesk(Tsup − Tsat )dR√= ρg hfgdtαπ t(5.57)Integrating eqn. (5.57) from R = 0 at t = 0 up to R at t, we obtainJakob’s prediction:3k∆T2√tR=√π ρg hfg α(5.58)This analysis was done without assuming the curved bubble interfaceto be plane, 24 years after Jakob’s work, by Plesset and Zwick [5.11].
Itwas 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 similarresult:√3√2 3 k∆T√t = 3 RJakob .(5.59)R= √π ρg hfg αBoth predictions are compared with some of the data of Dergarabedian [5.13] in Fig. 5.18. The data and the exact theory match almostperfectly. The simple theory of Jakob et al.
shows the correct dependence on R on√all its variables, but it shows growth rates that are lowby a factor of 3. This is because the expansion of the spherical bubble causes a relative motion of liquid toward the bubble surface, whichhelps to thin the region of thermal influence in the radial direction. Consequently, the temperature gradient and heat transfer rate are higherthan in Jakob’s model, which neglected the liquid motion. Therefore, thetemperature profile flattens out more slowly than Jakob predicts, and thebubble grows more rapidly.Experiment 5.2Touch various objects in the room around you: glass, wood, corkboard, paper, steel, and gold or diamond, if available.
Rank them in231232Transient and multidimensional heat conduction§5.6Figure 5.18 The growth of a vapor bubble—predictions andmeasurements.order of which feels coldest at the first instant of contact (see Problem5.29).The more advanced theory of heat conduction (see, e.g., [5.6]) showsthat if two semi-infinite regions at uniform temperatures T1 and T2 areplaced together suddenly, their interface temperature, Ts , is given by64(kρcp )1Ts − T24=4T1 − T2(kρcp )1 + (kρcp )2If we identify one region with your body (T1 37◦ C) and the other withthe object being touched (T2 20◦ C), we can determine the temperature,Ts , that the surface of your finger will reach upon contact.
Comparethe ranking you obtain experimentally with the ranking given by thisequation.6For semi-infinite regions, initially at uniform temperatures, Ts does not vary withtime. For finite bodies, Ts will eventually change. A constant value of Ts means thateach of the two bodies independently behaves as a semi-infinite body whose surfacetemperature has been changed to Ts at time zero. Consequently, our previous results—eqns. (5.50), (5.51), and (5.54)—apply to each of these bodies while they may be treatedas semi-infinite. We need only replace T∞ by Ts in those equations.Transient heat conduction to a semi-infinite region§5.6Notice that your bloodstream and capillary system provide a heatsource in your finger, so the equation is valid only for a moment.
Thenyou start replacing heat lost to the objects. If you included a diamondamong the objects that you touched, you will notice that it warmed upalmost instantly. Most diamonds are quite small but are possessed of thehighest known value of α. Therefore, they can behave as a semi-infiniteregion only for an instant, and they usually feel warm to the touch.Conduction to a semi-infinite region with a harmonicallyoscillating temperature at the boundarySuppose that we approximate the annual variation of the ambient temperature as sinusoidal and then ask what the influence of this variationwill be beneath the ground.
We want to calculate T − T (where T is thetime-average surface temperature) as a function of: depth, x; thermaldiffusivity, α; frequency of oscillation, ω; amplitude of oscillation, ∆T ;and time, t. There are six variables in K, m, and s, so the problem can berepresented in three dimensionless variables:9ωT −T;Ω ≡ ωt;ξ≡x.Θ≡∆T2αWe pose the problem as follows in these variables. The heat conduction equation is∂Θ1 ∂2Θ=2 ∂ξ 2∂Ωand the b.c.’s areΘξ=0= cos ωtand(5.60)Θξ>0= finite(5.61)No i.c.
is needed because, after the initial transient decays, the remainingsteady oscillation must be periodic.The solution is given by Carslaw and Jaeger (see [5.6, §2.6] or workProblem 5.16). It isΘ (ξ, Ω) = e−ξ cos (Ω − ξ)(5.62)This result is plotted in Fig. 5.19. It shows that the surface temperaturevariation decays exponentially into the region and suffers a phase shiftas it does so.233234Transient and multidimensional heat conduction§5.6Figure 5.19 The temperature variation within a semi-infiniteregion whose temperature varies harmonically at the boundary.Example 5.7How deep in the earth must we dig to find the temperature wave thatwas launched by the coldest part of the last winter if it is now highsummer?Solution.
ω = 2π rad/yr, and Ω = ωt = 0 at the present. First,we must find the depths at which the Ω = 0 curve reaches its local extrema. (We pick the Ω = 0 curve because it gives the highesttemperature at t = 0.)dΘ = −e−ξ cos(0 − ξ) + e−ξ sin(0 − ξ) = 0dξ Ω=0This givestan(0 − ξ) = 1soξ=3π 7π,,...44and the first minimum occurs where ξ = 3π /4 = 2.356, as we can seein Fig. 5.19. Thus,3ξ = x ω/2α = 2.356Steady multidimensional heat conduction§5.7or, if we take α = 0.139×10−6 m2 /s (given in [5.14] for coarse, gravellyearth),:212π x = 2.356= 2.783 m−62 0.139 × 10365(24)(3600)If we dug in the earth, we would find it growing older and colder untilit reached a maximum coldness at a depth of about 2.8 m.
Fartherdown, it would begin to warm up again, but not much. In midwinter(Ω = π ), the reverse would be true.5.7Steady multidimensional heat conductionIntroductionThe general equation for T (r ) during steady conduction in a region ofconstant thermal conductivity, without heat sources, is called Laplace’sequation:∇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.
Wesolved one two-dimensional heat conduction problem in Example 4.1,but this was not difficult because the boundary conditions were made toorder. Depending upon your mathematical background and the specificproblem, the analytical solution of multidimensional problems can beanything from straightforward calculation to a considerable challenge.The reader who wishes to study such analyses in depth should refer to[5.6] or [5.15], where such calculations are discussed in detail.Faced with a steady multidimensional problem, three routes are opento us:• Find out whether or not the analytical solution is already availablein 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.235236Transient and multidimensional heat conduction§5.7Figure 5.20 The two-dimensional flowof heat between two isothermal walls.The flux plotThe method of flux plotting will solve all steady planar problems in whichall boundaries are held at either of two temperatures or are insulated.With a little skill, it will provide accuracies of a few percent.
This accuracyis almost always greater than the accuracy with which the b.c.’s and kcan be specified; and it displays the physical sense of the problem veryclearly.Figure 5.20 shows heat flowing from one isothermal wall to anotherin 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,δTδQ = kδs(5.64)δnNotice that if we arrange things so that δQ, δT , and k are the samefor flow through each rectangle in the flow field, then δs/δn must be thesame for each rectangle. We therefore arbitrarily set the ratio equal tounity, so all the elements appear as distorted squares.The objective then is to sketch the isothermal lines and the adiabatic,77These are lines in the direction of heat flow.
It immediately follows that there can§5.7Steady multidimensional heat conductionor heat flow, lines which run perpendicular to them. This sketch is to bedone subject to two constraints• Isothermal and adiabatic lines must intersect at right angles.• They must subdivide the flow field into elements that are nearlysquare—“nearly” because they have slightly curved sides.Once the grid has been sketched, the temperature anywhere in the fieldcan be read directly from the sketch. And the heat flow per unit depthinto the paper isQ W/m = Nk δTNδs=k∆TδnI(5.65)where N is the number of heat flow channels and I is the number oftemperature increments, ∆T /δT .The first step in constructing a flux plot is to draw the boundaries ofthe region accurately in ink, using either drafting software or a straightedge.
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