Adrian Bejan(Editor), Allan D. Kraus (Editor). Heat transfer Handbok (776115), страница 21
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(3.136),the Kirchhoff transformation of eq. (3.138) can also be used for the hollow cylinderof Fig. 3.7. The final result for T ∗ is∗∗Ts,1− Ts,2r∗T ∗ = Ts,1ln+ln(r1 /r2 )r1(3.148)∗∗where Ts,1and Ts,2are as given in eq. (3.141). Once T ∗ for any radius r is found fromeq. (3.148), eq. (3.143) can be used to find the corresponding value of T .
The rate ofheat transfer then follows as2πkm L Ts,1 − Ts,2(3.149)q=ln(r2 /r1 )where km is the thermal conductivity at the mean temperature,Ts,m =Ts,1 + Ts,22Hollow Sphere The results for a hollow sphere (Fig. 3.8) whose thermal conductivity–temperature variation follows eq. (3.136) are∗∗ Ts,1− Ts,211∗+−(3.150)T ∗ = Ts,11/r2 − 1/r1 r1rBOOKCOMP, Inc. — John Wiley & Sons / Page 197 / 2nd Proofs / Heat Transfer Handbook / Bejan[197], (37)1.89345pt PgVar———Normal PagePgEnds: TEX[197], (37)198123456789101112131415161718192021222324252627282930313233343536373839404142434445CONDUCTION HEAT TRANSFER√1−1 + 1 + 2aT ∗a4πkm Ts,1 − Ts,2q=1/r1 − 1/r2T =(3.151)(3.152)Gebhart (1993) provides one-dimensional steady conduction analyses for singleand composite solids when the thermal conductivity varies simultaneously with location and temperature.
He also gives expressions for the conduction resistances of aplane wall, a hollow cylinder, and a hollow sphere for three cases of variable thermalconductivity: k = k(T ), k = k(x) or k(r), and k = k(x, T ) = k(T )f (x). Note thatthe last case assumes that k(x, T ) can be expressed as a product of two functions,k(T ) and f (x), each a function of a single variable.[198], (38)3.5.3Location-Dependent Energy GenerationPlane Wall Figure 3.18 presents a plane wall that experiences location-dependentenergy generation of the formx(3.153)q̇ = q̇0 1 −LThe temperature distribution in the wall is given byq̇0 Lxq̇0x3T = Ts,1 +−x2 −2k2k3L(3.154)Figure 3.18 Plane wall with linearly decaying, location-dependent internal energy generation.BOOKCOMP, Inc.
— John Wiley & Sons / Page 198 / 2nd Proofs / Heat Transfer Handbook / BejanLines: 1858 to 1890———0.66515pt PgVar———Normal PagePgEnds: TEX[198], (38)MORE ADVANCED STEADY ONE-DIMENSIONAL CONDUCTION123456789101112131415161718192021222324252627282930313233343536373839404142434445199and the maximum temperature occurs at x = L for an insulated face:T = Ts,1 +q̇0 L26k(3.155)Next, assume that the plane wall of Fig. 3.6 represents the shield of a nuclearreactor. The absorption of gamma radiation at the left surface (x = 0) triggers energyrelease into the shield which decays exponentially with the penetration distance xand can be represented by the relationq̇ = q0 ae−ax(3.156)where q0 (W/m 2) is the incident radiation heat flux and a m−1 is the absorptioncoefficient of the shield.The temperature distribution in the shield is q xq T = Ts,1 + 0 1 − e−ax + Ts,2 − Ts,1 + 0 e−aL − 1(3.157)akakLq0 aL1lna ak(Ts,1 − Ts,2 ) + q0 (1 − e−aL )0.7885pt PgVar(3.158)Solid Cylinder Reconsidering the solid cylinder of Fig.
3.16, q̇ will now beassumed to vary linearly with the radial distance r, that is,q̇ = ar(3.159)where a (W/m 4) is a constant. The temperature distribution in this case isT = T∞ +ar02a 3+r0 − r 33h9k(3.160)from which the centerline (r = 0) temperature Tc and the surface (r = r0 ) temperature Ts follow asT c = T∞ +ar02ar 3+ 03h9k(3.161)Ts = T∞ +ar023h(3.162)3.5.4 Temperature-Dependent Energy GenerationIn this section we present a collection of results for one-dimensional steady conduction in a plane wall, a solid cylinder, and a solid sphere when each experiences energygeneration that increases linearly with local temperature in accordance withBOOKCOMP, Inc.
— John Wiley & Sons / Page 199 / 2nd Proofs / Heat Transfer Handbook / BejanLines: 1890 to 1951———and the maximum temperature occurs atx=[199], (39)———Normal Page* PgEnds: Eject[199], (39)200123456789101112131415161718192021222324252627282930313233343536373839404142434445CONDUCTION HEAT TRANSFERq̇ = q̇s [1 + a(T − Ts )](3.163)where q̇S is the energy generation at the surface temperature Ts and a is a constant.Plane Wall For a plane wall of thickness 2L having identical surface temperaturesTs on both faces, the temperature distribution isT = Ts +1 cos nx−1a cos nL(3.164)√where n = a q̇s /k and nL < π/2 to ensure that the temperatures remain finite.If the convection cooling, characterized by the temperature T∞ and heat transfercoefficient h, is identical on both faces of the wall, the relationship between Ts andT∞ is given bymTs = T∞ + tan nLhwhere m =(3.165)Lines: 1951 to 2014√q̇s k/a.Solid Cylinder———For a solid cylinder of radius r0 , the temperature distribution is1 J0 (nr)T = Ts +−1(3.166)a J0 (nr0 )√where n = a q̇s /k and J0 is the Bessel function of the first kind of zero order (seeSection 3.3.5).
The parallel counterpart of eq. (3.165) isTs = T∞ +m J1 (nr0 )h J0 (nr0 )(3.167)√√where m = q̇s k/a, n = a q̇s /k, and J1 is the Bessel function of the first kind oforder 1. In eqs. (3.166) and (3.167), nr0 < 2.4048 to assure finite temperatures in thecylinder.Solid Sphere[200], (40)For a solid sphere of radius r0 , the temperature distribution is1 r0 sin nrT = Ts +−1(3.168)a r sin nr0where nr0 < π to assure finite temperatures in the sphere and the relationshipbetween Ts and the coolant temperature T∞ isT = Ts +k[1 − (nr0 ) cot nr0 ]hr0 awhere, here too, nr0 < π to assure finite temperatures in the sphere.BOOKCOMP, Inc. — John Wiley & Sons / Page 200 / 2nd Proofs / Heat Transfer Handbook / Bejan(3.169)0.49226pt PgVar———Normal PagePgEnds: TEX[200], (40)EXTENDED SURFACES1234567891011121314151617181920212223242526272829303132333435363738394041424344452013.5.5 Radiative–Convective Cooling of Solids with Uniform EnergyGenerationThe solutions obtained in Section 3.4.8 for a plane wall (the thermal symmetrycase), a solid cylinder, and a solid sphere are now extended to accommodate surfacecooling by simultaneous convection and radiation.
The surface energy balance foreach geometry gives 4dT 4h(Ts − T∞ ) + σ Ts − T∞ + k= 0 (plane wall)(3.170)dx x=LdT 4h(Ts − T∞ ) + σ Ts4 − T∞+k=0dr r=r0(solid cylinder and sphere) (3.171)where is the surface emissivity, σ the Stefan–Boltzmann constant, and T∞ representsthe surrounding or ambient temperature for both convection and radiation. In eqs.(3.170) and (3.171), the last terms can be evaluated using eqs. (3.105), (3.115), and(3.123), respectively.Because eqs. (3.170) and (3.171) require a numerical approach for their solutions,it is convenient to recast them in dimensionless form as 4dθ N1 (θs − 1) + N2 θs − 1 +=0(plane wall)(3.173)dX X=1 dθ 4N1 (θs − 1) + N2 θs − 1 +=0(cylinder and sphere)(3.173)dR R=1where θ = Ts /T∞ , N1 = hL/k for the plane wall, N1 = hr0 /k for the cylinder and33L/k for the plane wall, and N2 = σT∞r0 /k for the cylinder andsphere, N2 = σT∞sphere, X = x/L, R = r/r0 , and θ = T /T∞ .
The numerical values for θs are givenin Table 3.10 for a range of values of N1 and N2 and q̇L2 /kT∞ = 1 for the planewall and q̇r02 /kT∞ = 1 for the cylinder and sphere.3.6EXTENDED SURFACESThe term extended surface is used to describe a system in which the area of a surfaceis increased by the attachment of fins. A fin accommodates energy transfer by conduction within its boundaries, while its exposed surfaces transfer energy to the surroundings by convection or radiation or both.
Fins are commonly used to augment heattransfer from electronic components, automobile radiators, engine and compressorcylinders, control devices, and a host of other applications. A comprehensive treatment of extended surface technology is provided by Kraus et al. (2001).In this section we provide the performance characteristics (temperature distribution, rate of heat transfer, and fin efficiency) for convecting, radiating, and convectingradiating fins. Configurations considered include longitudinal fins, radial fins, andspines. The section concludes with a discussion of optimum fin designs.BOOKCOMP, Inc.
— John Wiley & Sons / Page 201 / 2nd Proofs / Heat Transfer Handbook / Bejan[201], (41)Lines: 2014 to 2081———-2.41492pt PgVar———Normal PagePgEnds: TEX[201], (41)202123456789101112131415161718192021222324252627282930313233343536373839404142434445CONDUCTION HEAT TRANSFERTABLE 3.10 Dimensionless Surface Temperature in Solids with Uniform EnergyGeneration and Radiative–Convective Surface CoolingθsN1N2Plane WallSolid CylinderSolid Sphere0.250.500.751.000.250.500.751.000.250.500.751.000.250.500.751.000.250.250.250.250.500.500.500.500.750.750.750.751.001.001.001.001.45971.42701.39701.36981.29931.28381.26931.25591.22581.21641.20751.19911.18241.17591.16981.16401.28381.25591.23201.21151.17591.16401.15341.14391.12881.12211.11591.11031.10201.09761.09351.08971.20751.18401.16461.14841.12541.11591.10761.10041.09051.08531.08071.07641.07101.06771.06471.06193.6.1Longitudinal Convecting FinsThe five common profiles of longitudinal fins shown in Fig.
3.19 are rectangular,trapezoidal, triangular, concave parabolic, and convex parabolic. The analytical expressions obtained are based on several assumptions.1.2.3.4.The heat conduction in the fin is steady and one-dimensional.The fin material is homogeneous and isotropic.There is no energy generation in the fin.The convective environment is characterized by a uniform and constant heattransfer coefficient and temperature.5.
The fin has a constant thermal conductivity.6. The contact between the base of the fin and the primary surface is perfect.7. The fin has a constant base temperature.Rectangular Fin For the rectangular fin (Fig. 3.19a), the temperature distribution,rate of heat transfer, and fin efficiency are given for five cases of thermal boundaryconditions.1.
Constant base temperature and convecting tip:θcoshm(b − x) + H sinhm(b − x)=θbcoshmb + H sinhmbBOOKCOMP, Inc. — John Wiley & Sons / Page 202 / 2nd Proofs / Heat Transfer Handbook / Bejan(3.174)[202], (42)Lines: 2081 to 2116———-2.70891pt PgVar———Normal PagePgEnds: TEX[202], (42)203EXTENDED SURFACES123456789101112131415161718192021222324252627282930313233343536373839404142434445qf = kmAθbsinhmb + H coshmbcoshmb + H sinhmb(3.175)qid = (hP b + ht A)θbη=(3.176)qfqid(3.177)TbTbLqf␦kbxAkinsulatedtipT⬁ , h xT⬁ , hb———0.25815pt PgVarT⬁ , hL␦bxkqfxbT⬁ , hT⬁ , h(c)(d)TbT⬁ , hL␦bqfkxb(e)T⬁ , hFigure 3.19 Longitudinal fins of (a) rectangular, (b) trapezoidal, (c) triangular, (d) concaveparabolic, and (e) convex parabolic profiles.BOOKCOMP, Inc. — John Wiley & Sons / Page 203 / 2nd Proofs / Heat Transfer Handbook / Bejan[203], (43)Lines: 2116 to 2127(b)TbLqfxebP = fin perimeter= 2(L ⫹ δ)Tbk␦bT⬁ , htqfT⬁ , hT⬁ , h(a)␦bLT⬁ , h———Normal PagePgEnds: TEX[203], (43)204123456789101112131415161718192021222324252627282930313233343536373839404142434445CONDUCTION HEAT TRANSFERwhere θ = T − T∞ , θb = Tb − T∞ , m2 = hP /kA = 2h/kδ, H = ht /km, andTb is the fin base temperature, T the fin temperature at location x, T∞ the convectiveenvironmental temperature, b the fin height, A the fin cross-sectional area, P the finperimeter, k the fin thermal conductivity, h the convective heat transfer coefficient forsurfaces other than the fin tip, ht the tip convective heat transfer coefficient, qf thefin heat dissipation, and qid the ideal fin heat dissipation.2.
Constant base temperature and insulated tip (H = 0):coshm(b − x)θ=θbcoshmb(3.178)qf = kmAθb tanhmb(3.179)tanhmbmb(3.180)η=[204], (44)3. Constant base and tip temperatures:Lines: 2127 to 2171(θt /θb ) sinhmx + sinhm(b − x)θ=θbsinhmb(3.181)coshmb − (θt /θb )qf = kmAθbsinhmb———Normal Page(3.182)* PgEnds: Ejectwith qid and η given by eqs. (3.176) and (3.177), respectively, Tt taken as the prescribed tip temperature, and θt = Tt − T∞ .4. Convective heating at the base and insulated tip:Bi cosh(mb − x)θ=θfBi coshmb + mb sinhmbqf = kmAθfBi sinhmbBi coshmb + mb sinhmb(3.183)(3.184)where Bi = hf b/k, θf = Tf −T∞ , and hf and Tf characterize the convection processat the fin base. Equations (3.176) and (3.177) can be used to find qid and η, but θbmust be found first from eq.
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