John H. Lienhard IV, John H. Lienhard V. A Heat Transfer Textbook (776116), страница 25
Текст из файла (страница 25)
An array of commercial internally finned tubing(photo courtesy of Noranda Metal Industries, Inc.)Figure 4.6 Some of the many varieties of finned tubes.164Fin design§4.5165Figure 4.7 The Stegosaurus with whatmight have been cooling fins (etching byDaniel Rosner).ing, condensing, or other natural convection situations, and will not bestrictly accurate even in forced convection.The tip may or may not exchange heat with the surroundings througha heat transfer coefficient, hL , which would generally differ from h. Thelength of the fin is L, its uniform cross-sectional area is A, and its circumferential perimeter is P .The characteristic dimension of the fin in the transverse direction(normal to the x-axis) is taken to be A/P . Thus, for a circular cylindricalfin, A/P = π (radius)2 /(2π radius) = (radius/2).
We define a Biot number for conduction in the transverse direction, based on this dimension,and require that it be small:Bifin =h(A/P )1k(4.27)166Analysis of heat conduction and some steady one-dimensional problems§4.5Figure 4.8 The analysis of a one-dimensional fin.This condition means that the transverse variation of T at any axial position, x, is much less than (Tsurface − T∞ ). Thus, T T (x only) and theheat flow can be treated as one-dimensional.An energy balance on the thin slice of the fin shown in Fig. 4.8 gives−kAdT dT + h(P δx)(T − T∞ )x = 0+kAdx x+δxdx x(4.28)dT /dx|x+δx − dT /dx|xd2 Td2 (T − T∞ )→=δxdx 2dx 2(4.29)butFin design§4.5167sohPd2 (T − T∞ )(T − T∞ )=dx 2kA(4.30)The b.c.’s for this equation are(T − T∞ )x=0 = T0 − T∞d(T − T∞ ) −kA= hL A(T − T∞ )x=Ldx(4.31a)x=LAlternatively, if the tip is insulated, or if we can guess that hL is smallenough to be unimportant, the b.c.’s ared(T − T∞ ) =0(4.31b)(T − T∞ )x=0 = T0 − T∞ anddxx=LBefore we solve this problem, it will pay to do a dimensional analysis ofit.
The dimensional functional equation is(4.32)T − T∞ = fn (T0 − T∞ ) , x, L, kA, hP , hL ANotice that we have written kA, hP , and hL A as single variables. Thereason for doing so is subtle but important. Setting h(A/P )/k 1,erases any geometric detail of the cross section from the problem. Theonly place where P and A enter the problem is as product of k, h, orhL .If they showed up elsewhere, they would have to do so in a physicallyincorrect way. Thus, we have just seven variables in W, K, and m. Thisgives four pi-groups if the tip is uninsulated:⎛⎞2⎜⎟⎜xhP 2 hL AL ⎟T − T∞⎜⎟⎟= fn ⎜ ,L ,⎜L⎟T0 − T∞kAkA ⎠⎝=hL L kor if we rename the groups,Θ = fn (ξ, mL, Biaxial )(4.33a)3where we call hP L2 /kA ≡ mL because that terminology is common inthe literature on fins.If the tip of the fin is insulated, hL will not appear in eqn. (4.32). Thereis one less variable but the same number of dimensions; hence, there will168Analysis of heat conduction and some steady one-dimensional problems§4.5be only three pi-groups.
The one that is removed is Biaxial , which involveshL . Thus, for the insulated fin,Θ = fn(ξ, mL)(4.33b)We put eqn. (4.30) in these terms by multiplying it by L2 /(T0 − T∞ ). Theresult isd2 Θ= (mL)2 Θdξ 2(4.34)This equation is satisfied by Θ = Ce±(mL)ξ . The sum of these two solutions forms the general solution of eqn. (4.34):Θ = C1 emLξ + C2 e−mLξ(4.35)Temperature distribution in a one-dimensional fin with the tip insulated The b.c.’s [eqn. (4.31b)] can be written asdΘ Θξ=0 = 1 and=0(4.36)dξ ξ=1Substituting eqn.
(4.35) into both eqns. (4.36), we getC1 + C2 = 1andC1 emL − C2 e−mL = 0(4.37)Mathematical Digression 4.1To put the solution of eqn. (4.37) for C1 and C2 in the simplest form,we need to recall a few properties of hyperbolic functions. The four basicfunctions that we need are defined asex − e−x2xe + e−xcosh x ≡2sinh xtanh x ≡cosh xex + e−xcoth x ≡ xe − e−xsinh x ≡=ex − e−xex + e−x(4.38)Fin design§4.5169where x is the independent variable. Additional functions are definedby analogy to the trigonometric counterparts. The differential relationscan be written out formally, and they also resemble their trigonometriccounterparts.dsinh x =dxdcosh x =dx1 xe − (−e−x ) = cosh x21 xe + (−e−x ) = sinh x2(4.39)These are analogous to the familiar results, d sin x/dx = cos x andd cos x/dx = − sin x, but without the latter minus sign.The solution of eqns.
(4.37) is thenC1 =e−mL2 cosh mLand C2 = 1 −e−mL2 cosh mL(4.40)Therefore, eqn. (4.35) becomesΘ=e−mL(1−ξ) + (2 cosh mL)e−mLξ − e−mL(1+ξ)2 cosh mLwhich simplifies toΘ=cosh mL(1 − ξ)cosh mL(4.41)for a one-dimensional fin with its tip insulated.One of the most important design variables for a fin is the rate atwhich it removes (or delivers) heat the wall. To calculate this, we writeFourier’s law for the heat flow into the base of the fin:6d(T − T∞ ) Q = −kA(4.42)dxx=0We multiply eqn.
(4.42) by L/kA(T0 − T∞ ) and obtain, after substitutingeqn. (4.41) on the right-hand side,QLsinh mL= mL= mL tanh mLkA(T0 − T∞ )cosh mL(4.43)We could also integrate h(T − T∞ ) over the outside area of the fin to get Q. Theanswer would be the same, but the calculation would be a little more complicated.6170Analysis of heat conduction and some steady one-dimensional problems§4.5Figure 4.9 The temperature distribution, tip temperature, andheat flux in a straight one-dimensional fin with the tip insulated.which can be writtenQ3= tanh mLkAhP (T0 − T∞ )(4.44)Figure 4.9 includes two graphs showing the behavior of one-dimensional fin with an insulated tip.
The top graph shows how the heat removal increases with mL to a virtual maximum at mL 3. This meansthat no such fin should have a length in excess of 2/m or 3/m if it is being used to cool (or heat) a wall. Additional length would simply increasethe cost without doing any good.Also shown in the top graph is the temperature of the tip of such afin. Setting ξ = 1 in eqn. (4.41), we discover thatΘtip =1cosh mL(4.45)Fin design§4.5171This dimensionless temperature drops to about 0.014 at the tip when mLreaches 5. This means that the end is 0.014(T0 − T∞ ) K above T∞ at theend. Thus, if the fin is actually functioning as a holder for a thermometeror a thermocouple that is intended to read T∞ , the reading will be in errorif mL is not significantly greater than five.The lower graph in Fig. 4.9 hows how the temperature is distributedin insulated-tip fins for various values of mL.Experiment 4.1Clamp a 20 cm or so length of copper rod by one end in a horizontalposition.
Put a candle flame very near the other end and let the arrangement come to a steady state. Run your finger along the rod. How doeswhat you feel correspond to Fig. 4.9? (The diameter for the rod shouldnot exceed about 3 mm. A larger rod of metal with a lower conductivitywill also work.)Exact temperature distribution in a fin with an uninsulated tip. Theapproximation of an insulated tip may be avoided using the b.c’s givenin eqn. (4.31a), which take the following dimensionless form:dΘ = Biax Θξ=1(4.46)Θξ=0 = 1 and −dξ ξ=1Substitution of the general solution, eqn.
(4.35), in these b.c.’s yieldsC1 + C2−mL(C1emL− C2e−mL )=1= Biax (C1 emL + C2 e−mL )(4.47)It requires some manipulation to solve eqn. (4.47) for C1 and C2 and tosubstitute the results in eqn. (4.35). We leave this as an exercise (Problem4.11). The result isΘ=cosh mL(1 − ξ) + (Biax /mL) sinh mL(1 − ξ)cosh mL + (Biax /mL) sinh mL(4.48)which is the form of eqn. (4.33a), as we anticipated. The correspondingheat flux equation is(Biax /mL) + tanh mLQ4=1 + (Biax /mL) tanh mL(kA)(hP ) (T0 − T∞ )(4.49)172Analysis of heat conduction and some steady one-dimensional problems§4.5We have seen that mL is not too much greater than unity in a welldesigned fin with an insulated tip. Furthermore, when hL is small (as itmight be in natural convection), Biax is normally much less than unity.Therefore, in such cases, we expect to be justified in neglecting termsmultiplied by Biax .
Then eqn. (4.48) reduces toΘ=cosh mL(1 − ξ)cosh mL(4.41)which we obtained by analyzing an insulated fin.It is worth pointing out that we are in serious difficulty if hL is solarge that we cannot assume the tip to be insulated. The reason is thathL is nearly impossible to predict in most practical cases.Example 4.8A 2 cm diameter aluminum rod with k = 205 W/m·K, 8 cm in length,protrudes from a 150◦ C wall.
Air at 26◦ C flows by it, and h = 120W/m2 K. Determine whether or not tip conduction is important in thisproblem. To do this, make the very crude assumption that h hL .Then compare the tip temperatures as calculated with and withoutconsidering heat transfer from the tip.Solution.2mL =hP L2=kABiax =2120(0.08)2= 0.8656205(0.01/2)120(0.08)hL== 0.0468k205Therefore, eqn. (4.48) becomescosh 0 + (0.0468/0.8656) sinh 0cosh(0.8656) + (0.0468/0.8656) sinh(0.8656)1== 0.68861.3986 + 0.0529Θ (ξ = 1) = Θtip =so the exact tip temperature isTtip = T∞ + 0.6886(T0 − T∞ )= 26 + 0.6886(150 − 26) = 111.43◦ CFin design§4.5173Equation (4.41) or Fig. 4.9, on the other hand, givesΘtip =1= 0.71501.3986so the approximate tip temperature isTtip = 26 + 0.715(150 − 26) = 114.66◦ CThus the insulated-tip approximation is adequate for the computationin this case.Very long fin.If a fin is so long that mL 1, then eqn.
(4.41) becomesemL(1−ξ)emL(1−ξ) + e−mL(1−ξ)=mL→∞emL + e−mLemLlimit Θ = limitmL→∞orlimit Θ = e−mLξmL→largeSubstituting this result in eqn. (4.42), we obtain [cf. eqn. (4.44)]4Q = (kAhP ) (T0 − T∞ )(4.50)(4.51)A heating or cooling fin would have to be terribly overdesigned for theseresults to apply—that is, mL would have been made much larger thannecessary. Very long fins are common, however, in a variety of situationsrelated to undesired heat losses.
In practice, a fin may be regarded as“infinitely long” in computing its temperature if mL 5; in computingQ, mL 3 is sufficient for the infinite fin approximation.Physical significance of mL. The group mL has thus far proved to beextremely useful in the analysis and design of fins. We should thereforesay a brief word about its physical significance. Notice that(mL)2 =L/kA1/h(P L)=internal resistance in x-directiongross external resistanceThus (mL)2 is a hybrid Biot number. When it is big, Θ|ξ=1 → 0 and wecan neglect tip convection.
![](https://s.studizba.com/z.php?f=/templates/si/i/noavatar.png&w=100&h=100&t=1"){kind=avatar}
