Adrian Bejan(Editor), Allan D. Kraus (Editor). Heat transfer Handbok (776115), страница 39
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Thelocations of the maximum errors were not given.———Long PagePgEnds: TEX4.7.2[295], (35)Circular Area on a Single Layer (Coating) on a Half-SpaceIntegral solutions are available for the spreading resistance for a circular source ofradius a in contact with an isotropic layer of thickness t1 and thermal conductivity k1which is in perfect thermal contact with an isotropic half-space of thermal conductivity k2 .
The solutions were obtained for two heat flux distributions correspondingto the flux parameter values µ = − 21 and µ = 0.Equivalent Isothermal Circular Contact Dryden (1983) obtained the solutionfor the equivalent isothermal circular contact flux distribution:q(r) =Q√2πa 2 1 − u20≤u≤1(W/m2)(4.87)The problem is depicted in Fig. 4.7.The dimensionless spreading resistance, based on the area-averaged temperature,is obtained from the integral (Dryden, 1983):4 k2 ∞ λ2 exp(ζt1 /a) + λ1 exp(−ζt1 /a) J1 (ζ) sin ζψ = 4k2 aRs =dζ(4.88)π k1 0 λ2 exp(ζt1 /a) − λ1 exp(−ζt1 /a)ζ2with λ1 = (1 − k2 /k1 )/2 and λ2 = (1 + k2 /k1 )/2.
The parameter ζ is a dummy variable of integration. The constriction resistance depends on the thermal conductivityBOOKCOMP, Inc. — John Wiley & Sons / Page 295 / 2nd Proofs / Heat Transfer Handbook / Bejan296123456789101112131415161718192021222324252627282930313233343536373839404142434445THERMAL SPREADING AND CONTACT RESISTANCESFigure 4.7 Layered half-space with an equivalent isothermal flux. (From Yovanovich et al.,1998.)[296], (36)Lines: 1525 to 1552ratio k1 /k2 and the relative layer thickness t1 /a.
Dryden (1983) presented simple———asymptotes for thermal spreading in thin layers, t1 /a ≤ 0.1, and in thick layers,-2.94893pt PgVar———t1 /a ≥ 10. These asymptotes were also presented as dimensionless spreading resisNormal Pagetances defined as 4k2 aRs . They are:Thin-layer asymptote:* PgEnds: Eject4 t1 k2k1(4k2 aRs )thin = 1 +(4.89)−[296], (36)π a k1k2Thick-layer asymptote:(4k2 aRs )thick =k22 a k22−lnk1π t1 k1 1 + k1 /k2(4.90)These asymptotes provide results that are within 1% of the full solution for relativelayer thickness: t1 /a < 0.5 and t1 /a > 2.The dimensionless spreading resistance is based on the substrate thermal conductivity k2 . The general solution above is valid for conductive layers where k1 /k2 > 1as well as for resistive layers where k1 /k2 < 1.
The infinite integral can be evaluatednumerically by means of computer algebra systems, which provide accurate results.4.7.3Isoflux Circular ContactHui and Tan (1994) presented an integral solution for the isoflux circular source.
Thedimensionless spreading resistance is 2 ∞J12 (ζ) dζ32 k2 2 8k24k2 aRs =+(4.91)1−23π2 k1πk10 [1 + (k1 /k2 ) tanh (ζt1 /a)]ζBOOKCOMP, Inc. — John Wiley & Sons / Page 296 / 2nd Proofs / Heat Transfer Handbook / BejanSPREADING RESISTANCE OF ISOTROPIC FINITE DISKS WITH CONDUCTANCE123456789101112131415161718192021222324252627282930313233343536373839404142434445297which depends on the thermal conductivity ratio k1 /k2 and the relative layer thickness t1 /a. The dimensionless spreading resistance is based on the substrate thermalconductivity k2 . The general solution above is valid for conductive layers, wherek1 /k2 > 1, as well as resistive layers, where k1 /k2 < 1.4.7.4Isoflux, Equivalent Isothermal, and Isothermal SolutionsNegus et al. (1985) obtained solutions by application of the Hankel transform methodfor flux-specified boundary conditions and with a novel technique of linear superposition for the mixed boundary condition (isothermal contact area and zero flux outsidethe contact area).
They reported results for three flux distributions: isoflux, equivalentisothermal flux, and true isothermal source. There results were presented below.Isoflux Contact Areafor ψq = 4k1 aRs :For the isoflux boundary condition, they reported the resultψq =∞328 +(−1)n αn Iq3π2π2 n=1(4.92)[297], (37)Lines: 1552 to 1602———-2.6009pt PgVarThe first term is the dimensionless isoflux spreading resistance of an isotropic half———space of thermal conductivity k1 , and the second term accounts for the effect of theNormal Pagelayer relative thickness and relative thermal conductivity.
The thermal conductivity * PgEnds: Ejectparameter α is defined asα=1−κ1+κwith κ = k1 /k2 . The layer thickness–conductivity parameter Iq is defined as1πIq =2 2(γ + 1)E2/(γ + 1) − √ Iγ − 2πnτ12π2 2γwithIγ = 1 +0.09375 0.0341797 0.00320435++γ2γ4γ6The relative layer thickness is τ1 = t/a and the relative thickness parameter isγ = 2n2 τ21 + 1The special function E(·) is the complete elliptic integral of the second kind(Abramowitz and Stegun, 1965).Equivalent Isothermal Contact Area For the equivalent isothermal fluxboundary condition, they reported the result for ψei = 4k1 aRs :BOOKCOMP, Inc.
— John Wiley & Sons / Page 297 / 2nd Proofs / Heat Transfer Handbook / Bejan[297], (37)298123456789101112131415161718192021222324252627282930313233343536373839404142434445THERMAL SPREADING AND CONTACT RESISTANCES∞ψei = 1 +8(−1)n αn Ieiπ n=1(4.93)where as discussed above, the first term represents the dimensionless spreading resistance of an isothermal contact area on an isotropic half-space of thermal conductivityk1 and the second term accounts for the effect of the layer relative thickness and therelative thermal conductivity.
The thermal conductivity parameter α is defined above.The relative layer thickness parameter Iei is defined as Iei =1 − β−2 β − β−1 + 21 sin−1 β−1 − 2nτ1with τ1 = t/a andβ = nτ1 +[298], (38)n2 τ21+1Lines: 1602 to 1654Isothermal Contact Area For the isothermal contact area, Negus et al. (1985)reported a correlation equation for their numerical results. They reported that ψT =4k1 aRs in the formψT = F1 tanh F2 + F3(4.94)———-0.16101pt PgVar———Normal PagePgEnds: TEXwhere[298], (38)F1 = 0.49472 − 0.49236κ − 0.0034κ2F2 = 2.8479 + 1.3337τ + 0.06864τ2withτ = log10 τ1F3 = 0.49300 + 0.57312κ − 0.06628κ2where κ = k1 /k2 .
The correlation equation was developed for resistive layers: 0.01 ≤κ ≤ 1 over a wide range of relative thickness 0.01 ≤ τ1 ≤ 100. The maximumrelative error associated with the correlation equation is approximately 2.6% at τ1 =0.01 and κ = 0.2. Numerical results for ψq , ψei , and ψT for a range of values ofτ1 and κ were presented in tabular form for easy comparison. They found that thevalues for ψq were greater than those for ψei and that ψei ≤ ψT . The maximumdifference between ψq and ψT was approximately 8%.
The values for ψT > ψei forvery thin layers, τ1 ≤ 0.1 and for κ ≤ 0.1; however, the differences were less thanapproximately 8%. For most applications the equivalent isothermal flux solution andthe true isothermal solution are simililar.4.8CIRCULAR AREA ON A SEMI-INFINITE FLUX TUBEThe problem of finding the spreading resistance in an semi-infinite isotropic circular flux tube has been investigated by many researchers (Roess, 1950; Mikic andBOOKCOMP, Inc. — John Wiley & Sons / Page 298 / 2nd Proofs / Heat Transfer Handbook / BejanCIRCULAR AREA ON A SEMI-INFINITE FLUX TUBE123456789101112131415161718192021222324252627282930313233343536373839404142434445299aqbk1t[299], (39)k1/k2 = 1Lines: 1654 to 1670 = t/b — ⬁——— = a/b ≤ 0.90.98595pt PgVarFigure 4.8 Isotropic flux tube with an isoflux area.
(From Yovanovich et al., 1998.)Rohsenow, 1966; Gibson, 1976; Yovanovich, 1976a, b; Negus and Yovanovich, 1984a,b; Negus et al., 1989). The system with uniform heat flux on the circular area is shownin Fig. 4.8.This problem corresponds to the case where κ = 1 and τ → ∞, and thereforethe spreading resistance depends on the system parameters (a, b, k) and the fluxdistribution parameter µ. The dimensionless spreading resistance defined as ψ =4kaRs , where Rs is the spreading resistance, depends on and µ. The results ofseveral studies are given below.4.8.1 General Expression for a Circular Contact Area with ArbitraryFlux on a Circular Flux TubeThe general expression for the dimensionless spreading (constriction) resistance4kaRs for a circular contact subjected to an arbitrary axisymmetric flux distributionf (u) (Yovanovich, 1976b) is obtained from the series4kaRs = 10 1∞J1 (δn )uf (u)J0 (δn u) du2uf (u) du n=1 δn J02 (δn ) 08/π(4.95)where δn are the positive roots of J1 (·) = 0 and = a/b is the relative size of thesource area.BOOKCOMP, Inc.
— John Wiley & Sons / Page 299 / 2nd Proofs / Heat Transfer Handbook / Bejan———Normal PagePgEnds: TEX[299], (39)300123456789101112131415161718192021222324252627282930313233343536373839404142434445THERMAL SPREADING AND CONTACT RESISTANCESFlux Distributions of the Form (1 − u2 )µ Yovanovich (1976b) reported thefollowing general solution for axisymmetric flux distributions of the form f (u) =(1 − u2 )µ , where the parameter µ accounts for the shape of the flux distribution. Thegeneral expression above reduces to the following general expression:∞161 J1 (δn )Jµ+1 (δn )4kaRs =(µ + 1)2µ Γ(µ + 1)π n=1 δ3n J0 (δn )(δn )µ(4.96)where Γ(·) is the gamma function (Abramowitz and Stegun, 1965) and Jν (·) is theBessel function of arbitrary order ν (Abramowitz and Stegun, 1965).The general expression above can be used to obtain specific solutions for variousvalues of the flux distribution parameter µ. Three particular solutions are considerednext.Equivalent Isothermal Circular Source The isothermal contact area requiressolution of a difficult mathematical problem that has received much attention bynumerous researchers (Roess, 1950; Kennedy, 1960; Mikic and Rohsenow, 1966;Gibson, 1976; Yovanovich, 1976b; Negus and Yovanovich, 1984a,b).Mikic and Rohsenow (1966) proposed use of the flux distribution correspondingto µ = − 21 to approximate an isothermal contact area for small relative contact areas0 < < 0.5.
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