John H. Lienhard IV, John H. Lienhard V. A Heat Transfer Textbook (776116), страница 34
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The next is to obtain a soft pencil (such as a no. 2 grade) and asoft eraser. We begin with an example that was executed nicely in theinfluential Heat Transfer Notes [5.3] of the mid-twentieth century. Thisexample is shown in Fig. 5.21.The particular example happens to have an axis of symmetry in it.
Weimmediately interpret this as an adiabatic boundary because heat cannotcross it. The problem therefore reduces to the simpler one of sketchinglines in only one half of the area. We illustrate this process in four steps.Notice the following steps and features in this plot:• Begin by dividing the region, by sketching in either a single isothermal or adiabatic line.• Fill in the lines perpendicular to the original line so as to makesquares. Allow the original line to move in such a way as to accommodate squares. This will always require some erasing. Therefore:• Never make the original lines dark and firm.• By successive subdividing of the squares, make the final grid.
Donot make the grid very fine. If you do, you will lose accuracy becausethe lack of perpendicularity and squareness will be less evident tothe eye. Step IV in Fig. 5.21 is as fine a grid as should ever be made.be no component of heat flow normal to them; they must be adiabatic.237Figure 5.21 The evolution of a flux plot.238§5.7Steady multidimensional heat conduction• If you have doubts about whether any large, ill-shaped regions arecorrect, fill them in with an extra isotherm and adiabatic line tobe sure that they resolve into appropriate squares (see the dashedlines in Fig.
5.21).• Fill in the final grid, when you are sure of it, either in hard pencil orpen, and erase any lingering background sketch lines.• Your flow channels need not come out even. Notice that there is anextra 1/7 of a channel in Fig. 5.21. This is simply counted as 1/7 ofa square in eqn. (5.65).• Never allow isotherms or adiabatic lines to intersect themselves.When the sketch is complete, we can return to eqn. (5.65) to computethe heat flux. In this case2(6.14)Nk∆T =k∆T = 3.07 k∆TQ=I4When the authors of [5.3] did this problem, they obtained N/I = 3.00—avalue only 2% below ours.
This kind of agreement is typical when fluxplotting is done with care.Figure 5.22 A flux plot with no axis of symmetry to guideconstruction.239240Transient and multidimensional heat conduction§5.7One must be careful not to grasp at a false axis of symmetry.
Figure5.22 shows a shape similar to the one that we just treated, but with unequal legs. In this case, no lines must enter (or leave) the corners A andB. The reason is that since there is no symmetry, we have no guidanceas to the direction of the lines at these corners. In particular, we knowthat a line leaving A will no longer arrive at B.Example 5.8A structure consists of metal walls, 8 cm apart, with insulating material (k = 0.12 W/m·K) between. Ribs 4 cm long protrude from onewall every 14 cm. They can be assumed to stay at the temperature ofthat wall.
Find the heat flux through the wall if the first wall is at 40◦ Cand the one with ribs is at 0◦ C. Find the temperature in the middle ofthe wall, 2 cm from a rib, as well.Figure 5.23 Heat transfer through a wall with isothermal ribs.Steady multidimensional heat conduction§5.7Solution.
The flux plot for this configuration is shown in Fig. 5.23.For a typical section, there are approximately 5.6 isothermal increments and 6.15 heat flow channels, soQ=2(6.15)Nk∆T =(0.12)(40 − 0) = 10.54 W/mI5.6where the factor of 2 accounts for the fact that there are two halvesin the section. We deduce the temperature for the point of interest,A, by a simple proportionality:Tpoint A =2.1(40 − 0) = 15◦ C5.6The shape factorA heat conduction shape factor S may be defined for steady problemsinvolving two isothermal surfaces as follows:Q ≡ S k∆T .(5.66)Thus far, every steady heat conduction problem we have done has takenthis form. For these situations, the heat flow always equals a function ofthe geometric shape of the body multiplied by k∆T .The shape factor can be obtained analytically, numerically, or throughflux plotting.
For example, let us compare eqn. (5.65) and eqn. (5.66):NWW== (S dimensionless) k∆Tk∆T(5.67)QmmIThis shows S to be dimensionless in a two-dimensional problem, but inthree dimensions S has units of meters:W.(5.68)Q W = (S m) k∆TmIt also follows that the thermal resistance of a two-dimensional body isRt =1kSwhereQ=∆TRt(5.69)For a three-dimensional body, eqn. (5.69) is unchanged except that thedimensions of Q and Rt differ.88Recall that we noted after eqn. (2.22) that the dimensions of Rt changed, dependingon whether or not Q was expressed in a unit-length basis.241242Transient and multidimensional heat conduction§5.7Figure 5.24 The shape factor for two similar bodies of different size.The virtue of the shape factor is that it summarizes a heat conductionsolution in a given configuration. Once S is known, it can be used againand again.
That S is nondimensional in two-dimensional configurationsmeans that Q is independent of the size of the body. Thus, in Fig. 5.21, Sis always 3.07—regardless of the size of the figure—and in Example 5.8, Sis 2(6.15)/5.6 = 2.196, whether or not the wall is made larger or smaller.When a body’s breadth is increased so as to increase Q, its thickness inthe direction of heat flow is also increased so as to decrease Q by thesame factor.Example 5.9Calculate the shape factor for a one-quarter section of a thick cylinder.Solution.
We already know Rt for a thick cylinder. It is given byeqn. (2.22). From it we computeScyl =12π=kRtln(ro /ri )so on the case of a quarter-cylinder,S=π2 ln(ro /ri )The quarter-cylinder is pictured in Fig. 5.24 for a radius ratio, ro /ri =3, but for two different sizes.
In both cases S = 1.43. (Note that thesame S is also given by the flux plot shown.)Steady multidimensional heat conduction§5.7Figure 5.25 Heat transfer through athick, hollow sphere.Example 5.10Calculate S for a thick hollow sphere, as shown in Fig. 5.25.Solution. The general solution of the heat diffusion equation inspherical coordinates for purely radial heat flow is:C1+ C2rwhen T = fn(r only). The b.c.’s areT =T (r = ri ) = Tiand T (r = ro ) = Tosubstituting the general solution in the b.c.’s we getC1+ C2 = TiriandC1+ C1 = ToroTherefore,C1 =Ti − Tori roro − riand C2 = Ti −Ti − Tororo − riPutting C1 and C2 in the general solution, and calling Ti − To ≡ ∆T ,we getrori ro−T = Ti + ∆Tr (ro − ri ) ro − riThen4π (ri ro )dT=k∆Tdrro − ri4π (ri ro )mS=ro − riQ = −kAwhere S now has the dimensions of m.243244Transient and multidimensional heat conduction§5.7Table 5.4 includes a number of analytically derived shape factors foruse in calculating the heat flux in different configurations.
Notice thatthese results will not give local temperatures. To obtain that information,one must solve the Laplace equation, ∇2 T = 0, by one of the methodslisted at the beginning of this section. Notice, too, that this table is restricted to bodies with isothermal and insulated boundaries.In the two-dimensional cases, both a hot and a cold surface must bepresent in order to have a steady-state solution; if only a single hot (orcold) body is present, steady state is never reached. For example, a hotisothermal cylinder in a cooler, infinite medium never reaches steadystate with that medium. Likewise, in situations 5, 6, and 7 in the table,the medium far from the isothermal plane must also be at temperatureT2 in order for steady state to occur; otherwise the isothermal plane andthe medium below it would behave as an unsteady, semi-infinite body. Ofcourse, since no real medium is truly infinite, what this means in practiceis that steady state only occurs after the medium “at infinity” comes toa temperature T2 .
Conversely, in three-dimensional situations (such as4, 8, 12, and 13), a body can come to steady state with a surroundinginfinite or semi-infinite medium at a different temperature.Example 5.11A spherical heat source of 6 cm in diameter is buried 30 cm below thesurface of a very large box of soil and kept at 35◦ C. The surface ofthe soil is kept at 21◦ C. If the steady heat transfer rate is 14 W, whatis the thermal conductivity of this sample of soil?Solution.Q = S k∆T =4π Rk∆T1 − R/2hwhere S is that for situation 7 in Table 5.4.
Then1 − (0.06/2) 2(0.3)14 W= 2.545 W/m·Kk=(35 − 21)K4π (0.06/2) mReaders who desire a broader catalogue of shape factors should referto [5.16], [5.18], or [5.19].Table 5.4 Conduction shape factors: Q = S k∆T .SituationShape factor, S1. Conduction through a slabA/LDimensionsmeterSourceExample 2.22. Conduction through wall of a longthick cylinder2πln (ro /ri )noneExample 5.93. Conduction through a thick-walledhollow sphere4π (ro ri )ro − rimeterExample 5.104π RmeterProblems 5.19and 2.15meter[5.16]none[5.16]meter[5.16, 5.17]4. The boundary of a spherical hole ofradius R conducting into an infinitemedium5.
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