John H. Lienhard IV, John H. Lienhard V. A Heat Transfer Textbook (776116), страница 26
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When it is small, the temperature drop alongthe axis of the fin becomes small (see the lower graph in Fig. 4.9).174Analysis of heat conduction and some steady one-dimensional problems§4.5The group (mL)2 also has a peculiar similarity to the NTU (Chapter3) and the dimensionless time, t/T , that appears in the lumped-capacitysolution (Chapter 1).
Thus,h(P L)kA/Lis likeUACminis likehAρcV /tIn each case a convective heat rate is compared with a heat rate thatcharacterizes the capacity of a system; and in each case the system temperature asymptotically approaches its limit as the numerator becomeslarge. This was true in eqn. (1.22), eqn. (3.21), eqn. (3.22), and eqn. (4.50).The problem of specifying the root temperatureThus far, we have assmed the root temperature of a fin to be given information. There really are many circumstances in which it might be known;however, if a fin protrudes from a wall of the same material, as sketchedin Fig.
4.10a, it is clear that for heat to flow, there must be a temperaturegradient in the neighborhood of the root.Consider the situation in which the surface of a wall is kept at a temperature Ts . Then a fin is placed on the wall as shown in the figure. IfT∞ < Ts , the wall temperature will be depressed in the neighborhood ofthe root as heat flows into the fin. The fin’s performance should then bepredicted using the lowered root temperature, Troot .This heat conduction problem has been analyzed for several fin arrangements by Sparrow and co-workers.
Fig. 4.10b is the result of Sparrow and Hennecke’s [4.6] analysis for a single circular cylinder. TheygivehrTs − TrootQactual, (mr ) tanh(mL)(4.52)== fn1−Qno temp. depressionTs − T∞kwhere r is the radius of the fin. From the figure we see that the actualheat flux into the fin, Qactual , and the actual root temperature are bothreduced when the Biot number, hr /k, is large and the fin constant, m, issmall.Example 4.9Neglect the tip convection from the fin in Example 4.8 and supposethat it is embedded in a wall of the same material. Calculate the errorin Q and the actual temperature of the root if the wall is kept at 150◦ C.Figure 4.10 The influence of heat flow into the root of circularcylindrical fins [4.6].175176Analysis of heat conduction and some steady one-dimensional problems§4.5Solution. From Example 4.8 we have mL = 0.8656 and hr /k =120(0.010)/205 = 0.00586.
Then, with mr = mL(r /L), we have(mr ) tanh(mL) = 0.8656(0.010/0.080) tanh(0.8656) = 0.0756. Thelower portion of Fig. 4.10b then gives1−Ts − TrootQactual== 0.05Qno temp. depressionTs − T∞so the heat flow is reduced by 5% and the actual root temperature isTroot = 150 − (150 − 26)0.05 = 143.8◦ CThe correction is modest in this case.Fin designTwo basic measures of fin performance are particularly useful in a findesign. The first is called the efficiency, ηf .ηf ≡actual heat transferred by a finheat that would be transferred if the entire fin were at T = T0(4.53)To see how this works, we evaluate ηf for a one-dimensional fin with aninsulated tip:4(hP )(kA)(T0 − T∞ ) tanh mLtanh mL(4.54)=ηf =mLh(P L)(T0 − T∞ )This says that, under the definition of efficiency, a very long fin will givetanh(mL)/mL → 1/large number, so the fin will be inefficient.
On theother hand, the efficiency goes up to 100% as the length is reduced tozero, because tanh(mL) → mL as mL → 0. While a fin of zero lengthwould accomplish little, a fin of small m might be designed in order tokeep the tip temperature near the root temperature; this, for example, isdesirable if the fin is the tip of a soldering iron.It is therefore clear that, while ηf provides some useful informationas to how well a fin is contrived, it is not generally advisable to designtoward a particular value of ηf .A second measure of fin performance is called the effectiveness, εf :εf ≡heat flux from the wall with the finheat flux from the wall without the fin(4.55)Fin design§4.5177This can easily be computed from the efficiency:εf = ηfsurface area of the fincross-sectional area of the fin(4.56)Normally, we want the effectiveness to be as high as possible, But thiscan always be done by extending the length of the fin, and that—as wehave seen—rapidly becomes a losing proposition.The measures ηf and εf probably attract the interest of designers notbecause their absolute values guide the designs, but because they areuseful in characterizing fins with more complex shapes.
In such casesthe solutions are often so complex that ηf and εf plots serve as laborsaving graphical solutions. We deal with some of these curves later inthis section.The design of a fin thus becomes an open-ended matter of optimizing,subject to many factors. Some of the factors that have to be consideredinclude:• The weight of material added by the fin. This might be a cost factoror it might be an important consideration in its own right.• The possible dependence of h on (T − T∞ ), flow velocity past thefin, or other influences.• The influence of the fin (or fins) on the heat transfer coefficient, h,as the fluid moves around it (or them).• The geometric configuration of the channel that the fin lies in.• The cost and complexity of manufacturing fins.• The pressure drop introduced by the fins.Fin thermal resistanceWhen fins occur in combination with other thermal elements, it can simplify calculations to treat them as a thermal resistance between the rootand the surrounding fluid.
Specifically, for a straight fin with an insulatedtip, we can rearrange eqn. (4.44) as(T0 − T∞ )(T0 − T∞ )Q = 3−1 ≡RtfinkAhP tanh mL(4.57)178Analysis of heat conduction and some steady one-dimensional problems§4.5where1Rtfin = 3kAhP tanh mLfor a straight fin(4.58)In general, for a fin of any shape, fin thermal resistance can be written interms of fin efficiency and fin effectiveness. From eqns. (4.53) and (4.55),we obtainRtfin =1ηf Asurface h=1εf Aroot h(4.59)Example 4.10Consider again the resistor described in Examples 2.8 and 2.9, starting on page 76. Suppose that the two electrical leads are long straightwires 0.62 mm in diameter with k = 16 W/m·K and heff = 23 W/m2 K.Recalculate the resistor’s temperature taking account of heat conducted into the leads.Solution.
The wires act as very long fins connected to the resistor,so that tanh mL 1 (see Prob. 4.44). Each has a fin resistance of11=3= 2, 150 K/WRtfin = 3(16)(23)(π )2 (0.00062)3 /4kAhPThese two thermal resistances are in parallel to the thermal resistances for natural convection and thermal radiation from the resistorsurface found in Example 2.8. The equivalent thermal resistance isnow−11111+++Rtequiv =RtfinRtfinRtradRtconv−12=+ (1.33 × 10−4 )(7.17) + (1.33 × 10−4 )(13)2, 150= 276.8 K/WThe leads reduce the equivalent resistance by about 30% from thevalue found before.
The resistor temperature becomesTresistor = Tair + Q · Rtequiv = 35 + (0.1)(276.8) = 62.68 ◦ Cor about 10◦ C lower than before.Fin design§4.5179Figure 4.11 A general fin of variable cross section.Fins of variable cross sectionLet us consider what is involved is the design of a fin for which A andP are functions of x.
Such a fin is shown in Fig. 4.11. We restrict ourattention to fins for whichh(A/P )1kandd(a/P )1d(x)so the heat flow will be approximately one-dimensional in x.We begin the analysis, as always, with the First Law statement:Qnet = Qcond − Qconv =dUdtor7 dT dT −hP δx (T − T∞ )− kA(x)kA(x + δx)dx x=δxdx xdTdkA(x)δx=dxdxdT= ρcA(x)δxdt=0, since steady7Note that we approximate the external area of the fin as horizontal when we writeit as P δx. The actual area is negligibly larger than this in most cases.
An exceptionwould be the tip of the fin in Fig. 4.11.180Analysis of heat conduction and some steady one-dimensional problems§4.5Figure 4.12 A two-dimensional wedge-shaped fin.Therefore,d(T − T∞ )dhPA(x)−(T − T∞ ) = 0dxdxk(4.60)If A(x) = constant, this reduces to Θ −(mL)2 Θ = 0, which is the straightfin equation.To see how eqn. (4.60) works, consider the triangular fin shown inFig. 4.12. In this case eqn. (4.60) becomes d(T − T∞ )2hbxd2δb−(T − T∞ ) = 0dxLdxkorξhL2d2 Θ dΘ+Θ=0−dξ 2dξ kδ(4.61)a kindof (mL)2This second-order linear differential equation is difficult to solve becauseit has a variable coefficient. Its solution is expressible in Bessel functions: 4Io 2 hLx/kδΘ= 4(4.62)2Io 2 hL /kδFin design181where the modified Bessel function of the first kind, Io , can be looked upin appropriate tables.Rather than explore the mathematics of solving eqn.
(4.60), we simplyshow the result for several geometries in terms of the fin efficiency, ηf ,in Fig. 4.13. These curves were given by Schneider [4.7]. Kraus, Aziz, andWelty [4.8] provide a very complete discussion of fins and show a greatmany additional efficiency curves.Example 4.11A thin brass pipe, 3 cm in outside diameter, carries hot water at 85◦ C.It is proposed to place 0.8 mm thick straight circular fins on the pipeto cool it. The fins are 8 cm in diameter and are spaced 2 cm apart.
Itis determined that h will equal 20 W/m2 K on the pipe and 15 W/m2 Kon the fins, when they have been added. If T∞ = 22◦ C, compute theheat loss per meter of pipe before and after the fins are added.Solution. Before the fins are added,Q = π (0.03 m)(20 W/m2 K)[(85 − 22) K] = 199 W/mwhere we set Twall = Twater since the pipe is thin. Notice that, sincethe wall is constantly heated by the water, we should not have a roottemperature depression problem after the fins are added. Then wecan enter Fig. 4.13a with222LhL315(0.04 − 0.15)3r2=== 0.306= 2.67 and mLr1PkA125(0.025)(0.0008)and we obtain ηf = 89%.
Thus, the actual heat transfer given by0.02 − 0.0008Qwithout fin0.02 119 W/mfraction of unfinned areafins+ 0.89 [2π (0.04 − 0.015 )] 50m22W15 2[(85 − 22) K]m Karea per fin (both sides), m2soQnet = 478 W/m = 4.02 Qwithout finsFigure 4.13 The efficiency of several fins with variable cross section.182Problems183Problems4.1Make a table listing the general solutions of all steady, unidimensional constant-properties heat conduction problemns inCartesian, cylindrical and spherical coordinates, with and without uniform heat generation.
This table should prove to be avery useful tool in future problem solving. It should include atotal of 18 solutions. State any restrictions on your solutions.Do not include calculations.4.2The left side of a slab of thickness L is kept at 0◦ C. The rightside is cooled by air at T∞ ◦ C blowing on it. hRHS is known. Anexothermic reaction takes place in the slab such that heat isgenerated at A(T − T∞ ) W/m3 , where A is a constant. Find afully dimensionless expression for the temperature distribution in the wall.4.3A long, wide plate of known size, material, and thickness L isconnected across the terminals of a power supply and servesas a resistance heater. The voltage, current and T∞ are known.The plate is insulated on the bottom and transfers heat outthe top by convection.
The temperature, Ttc , of the bottonis measured with a thermocouple. Obtain expressions for (a)temperature distribution in the plate; (b) h at the top; (c) temperature at the top. (Note that your answers must depend onknown information only.) [Ttop = Ttc − EIL2 /(2k · volume)]4.4The heat tansfer coefficient, h, resulting from a forced flowover a flat plate depends on the fluid velocity, viscosity, density, specific heat, and thermal conductivity, as well as on thelength of the plate. Develop the dimensionless functional equation for the heat transfer coefficient (cf.
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