John H. Lienhard IV, John H. Lienhard V. A Heat Transfer Textbook (776116), страница 24
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is involved and we identify acharacteristic length, L, in the x-direction. There are seven variablesin three dimensions, or 7 − 3 = 4 pi-groups. Three of these groupsare ones we have dealt with in the past in one form or another:T − T1T2 − T1xΠ2 =LΠ1 =Π3 =hLkdimensionless temperature, which weshall give the name Θdimensionless length, which we call ξwhich we recognize as the Biot number, BiThe fourth group is new to us:Π4 =q̇L2k(T2 − T1 )which compares the heat generation rate tothe rate of heat loss; we call it ΓThus, the solution isΘ = fn (ξ, Bi, Γ )(4.17)In Example 2.1, we undertook such a problem, but it differed in tworespects. There was no convective boundary condition and hence, no h,and only one temperature was specified in the problem.
In this case, thedimensional functional equation was (T − T1 ) = fn x, L, q̇, kso there were only five variables in the same three dimensions. The resulting dimensionless functional equation therefore involved only two§4.4An illustration of dimensional analysis in a complex steady conduction problempi-groups.
One was ξ = x/L and the other is a new one equal to Θ/Γ . Wecall it Φ: xT − T1= fnΦ≡Lq̇L2 /k(4.18)And this is exactly the form of the analytical result, eqn. (2.15).Finally, we must deal with dimensions that convert into one another.For example, kg and N are defined in terms of one another through Newton’s Second Law of Motion. Therefore, they cannot be identified as separate dimensions. The same would appear to be true of J and N·m, sinceboth are dimensions of energy. However, we must discern whether ornot a mechanism exists for interchanging them. If mechanical energyremains distinct from thermal energy in a given problem, then J shouldnot be interpreted as N·m.This issue will prove important when we do the dimensional analysis of several heat transfer problems.
See, for example, the analysesof laminar convection problem at the beginning of Section 6.4, of natural convection in Section 8.3, of film condensation in Section 8.5, and ofpool boiling burnout in Section 9.3. In all of these cases, heat transfernormally occurs without any conversion of heat to work or work to heatand it would be misleading to break J into N·m.Additional examples of dimensional analysis appear throughout thisbook. Dimensional analysis is, indeed, our court of first resort in solvingmost of the new problems that we undertake.4.4An illustration of the use of dimensional analysisin a complex steady conduction problemHeat conduction problems with convective boundary conditions can rapidly grow difficult, even if they start out simple, and so we look for waysto avoid making mistakes.
For one thing, it is wise to take great carethat dimensions are consistent at each stage of the solution. The bestway to do this, and to eliminate a great deal of algebra at the same time,is to nondimensionalize the heat conduction equation before we applythe b.c.’s. This nondimensionalization should be consistent with the pitheorem. We illustrate this idea with a fairly complex example.159160Analysis of heat conduction and some steady one-dimensional problems§4.4Figure 4.5 Heat conduction through a heat-generating slabwith asymmetric boundary conditions.Example 4.7A slab shown in Fig.
4.5 has different temperatures and different heattransfer coefficients on either side and the heat is generated withinit. Calculate the temperature distribution in the slab.Solution. The differential equation isq̇d2 T=−2dxkand the general solution isT =−q̇x 2+ C1 x + C22k(4.19)§4.4An illustration of dimensional analysis in a complex steady conduction problemwith b.c.’sh1 (T1 − T )x=0 = −kdT ,dx x=0h2 (T − T2 )x=L = −kdT .dx x=L(4.20)There are eight variables involved in the problem: (T − T2 ), (T1 − T2 ),x, L, k, h1 , h2 , and q̇; and there are three dimensions: K, W, and m.This results in 8 − 3 = 5 pi-groups. For these we chooseΠ1 ≡ Θ =T − T2,T1 − T2Π4 ≡ Bi2 =h2 L,kΠ2 ≡ ξ =andx,LΠ3 ≡ Bi1 =Π5 ≡ Γ =h1 L,kq̇L2,2k(T1 − T2 )where Γ can be interpreted as a comparison of the heat generated inthe slab to that which could flow through it.Under this nondimensionalization, eqn.
(4.19) becomes5Θ = −Γ ξ 2 + C3 ξ + C4(4.21)and b.c.’s become,Bi1 (1 − Θξ=0 ) = −Θξ=0Bi2 Θξ=1 = −Θξ=1(4.22)where the primes denote differentiation with respect to ξ. Substituting eqn. (4.21) in eqn. (4.22), we obtainBi1 (1 − C4 ) = −C3 ,Bi2 (−Γ + C3 + C4 ) = 2Γ − C3 .(4.23)Substituting the first of eqns. (4.23) in the second we getC4 = 1 +−Bi1 + 2(Bi1 /Bi2 )Γ + Bi1 ΓBi1 + Bi21 Bi2 + Bi21C3 = Bi1 (C4 − 1)Thus, eqn.
(4.21) becomes2(Bi1 Bi2 ) + Bi12(Bi1 Bi2 ) + Bi12ξ−ξ +Θ=1+Γ1 + Bi1 Bi2 + Bi1Bi1 + Bi21 Bi2 + Bi21Bi1Bi1ξ−−21 + Bi1 Bi2 + Bi1Bi1 + Bi1 Bi2 + Bi215(4.24)The rearrangement of the dimensional equations into dimensionless form isstraightforward algebra. If the results shown here are not immediately obvious toyou, sketch the calculation on a piece of paper.161162Analysis of heat conduction and some steady one-dimensional problems§4.4This is a complicated result and one that would have required enormouspatience and accuracy to obtain without first simplifying the problemstatement as we did. If the heat transfer coefficients were the same oneither side of the wall, then Bi1 = Bi2 ≡ Bi, and eqn.
(4.24) would reduceto ξ + 1/Bi(4.25)Θ = 1 + Γ ξ − ξ 2 + 1/Bi −1 + 2/Biwhich is a very great simplification.Equation (4.25) is plotted on the left-hand side of Fig. 4.5 for Bi equalto 0, 1, and ∞ and for Γ equal to 0, 0.1, and 1. The following featuresshould be noted:• When Γ 0.1, the heat generation can be ignored.• When Γ 1, Θ → Γ /Bi + Γ (ξ − ξ 2 ). This is a simple parabolic temperature distribution displaced upward an amount that depends onthe relative external resistance, as reflected in the Biot number.• If both Γ and 1/Bi become large, Θ → Γ /Bi. This means that wheninternal resistance is low and the heat generation is great, the slabtemperature is constant and quite high.If T2 were equal to T1 in this problem, Γ would go to infinity.
In sucha situation, we should redo the dimensional analysis of the problem. Thedimensional functional equation now shows (T − T1 ) to be a function ofx, L, k, h, and q̇. There are six variables in three dimensions, so thereare three pi-groupsT − T1= fn (ξ, Bi)q̇L/hwhere the dependent variable is like Φ [recall eqn. (4.18)] multiplied byBi. We can put eqn. (4.25) in this form by multiplying both sides of it byh(T1 − T2 )/q̇δ. The result is 1h(T − T1 )1 = Bi ξ − ξ 2 +22q̇L(4.26)The result is plotted on the right-hand side of Fig.
4.5. The followingfeatures of the graph are of interest:• Heat generation is the only “force” giving rise to temperature nonuniformity. Since it is symmetric, the graph is also symmetric.Fin design§4.5• When Bi 1, the slab temperature approaches a uniform valueequal to T1 + q̇L/2h. (In this case, we would have solved the problem with far greater ease by using a simple lumped-capacity heatbalance, since it is no longer a heat conduction problem.)• When Bi > 100, the temperature distribution is a very large parabolawith ½ added to it. In this case, the problem could have been solvedusing boundary conditions of the first kind because the surfacetemperature stays very close to T∞ (recall Fig. 1.11).4.5Fin designThe purpose of finsThe convective removal of heat from a surface can be substantially improved if we put extensions on that surface to increase its area.
Theseextensions can take a variety of forms. Figure 4.6, for example, showsmany different ways in which the surface of commercial heat exchangertubing can be extended with protrusions of a kind we call fins.Figure 4.7 shows another very interesting application of fins in a heatexchanger design. This picture is taken from an issue of Science magazine [4.5], which presents an intriguing argument by Farlow, Thompson,and Rosner. They offered evidence suggesting that the strange rows offins on the back of the Stegosaurus were used to shed excess body heatafter strenuous activity, which is consistent with recent suspicions thatStegosaurus was warm-blooded.These examples involve some rather complicated fins.
But the analysis of a straight fin protruding from a wall displays the essential featuresof all fin behavior. This analysis has direct application to a host of problems.Analysis of a one-dimensional finThe equations. Figure 4.8 shows a one-dimensional fin protruding froma wall. The wall—and the roots of the fin—are at a temperature T0 , whichis either greater or less than the ambient temperature, T∞ . The lengthof the fin is cooled or heated through a heat transfer coefficient, h, bythe ambient fluid.
The heat transfer coefficient will be assumed uniform,although (as we see in Part III) that can introduce serious error in boil-163a.Eight examples of externally finned tubing:1) and 2) typical commercial circular fins of constantthickness; 3) and 4) serrated circular fins and dimpledspirally-wound circular fins, both intended to improveconvection; 5) spirally-wound copper coils outside andinside; 6) and 8) bristle fins, spirally wound and machined from base metal; 7) a spirally indented tube toimprove convection and increase surface area.b.
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