Adrian Bejan(Editor), Allan D. Kraus (Editor). Heat transfer Handbok (776115), страница 38
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If the substrate thermal conductivityTABLE 4.8Three Heat Flux Distributionsµ− 21012BOOKCOMP, Inc. — John Wiley & Sons / Page 289 / 2nd Proofs / Heat Transfer Handbook / Bejanq(u)2πa 2Q√1 − u2Qπa 23Q 1 − u22πa 2[289], (29)Lines: 1295 to 1351———0.11618pt PgVar———Long PagePgEnds: TEX[289], (29)290123456789101112131415161718192021222324252627282930313233343536373839404142434445THERMAL SPREADING AND CONTACT RESISTANCESTABLE 4.9System Parameter Ranges0< <10 < τ1 < ∞0 < τ2 < ∞0< κ <∞0 < Bi < ∞is used, the right-hand side of the relationship must be multiplied by the thermalparameter κ.The coefficients An are functions of the heat flux parameter µ. They are −2 sin δ n2 2δn J0 (δn ) −2J1 (δn )An =δ2n J02 (δn )−2 sin δ 11 2 2 n−(δn ) tan δn δn J0 (δn ) (δn )2[290], (30)for µ =− 21for µ = 0Lines: 1351 to 1397(4.76)for µ =12The function Bn , which depends on the system parameters (τ1 , τ2 , κ, Bi), was definedasBn =φn tanh(δn τ1 ) − ϕn1 − φn———Long Page* PgEnds: Eject[290], (30)(4.77)and the two functions that appear in the relationship above are defined asφn =κ−1( cosh δn τ1 − ϕn sinh δn τ1 ) cosh δn τ1κ(4.78)ϕn =δn + Bi tanh δn τδn tanh(δn τ) + Bi(4.79)The eigenvalues δn are the positive roots of J1 (·) = 0 (Abramowitz and Stegun,1965).
For the special case of an isotropic disk where κ = 1, Bn = −ϕn , whichdepends on τ and Bi only. Since the general solution depends on several independentparameters, it is not possible to present the results in tabular or graphical form. Thefull solution can, however, be programmed easily into computer algebra systems.Characteristics of ϕn This function accounts for the effects of the parameters:δn , τ, and Bi. For extreme values of the parameter Bi, it reduces totanh δn τas Bi → ∞ϕn →(4.80)coth δn τas Bi → 0BOOKCOMP, Inc. — John Wiley & Sons / Page 290 / 2nd Proofs / Heat Transfer Handbook / Bejan———1.8284pt PgVarSPREADING RESISTANCE WITHIN A COMPOUND DISK WITH CONDUCTANCE123456789101112131415161718192021222324252627282930313233343536373839404142434445291For all values in the range 0 < Bi < ∞ and for all values τ > 0.72, tanh δn τ ≈ 1for all n ≥ 1. Therefore, φn ≈ 1 for values n ≥ 1.Characteristics of Bn When τ1 > 0.72, tanh δn τ = 1, φn = 1 for all 0 < Bi < ∞;therefore, Bn = 1 for n ≥ 1.4.6.1Special Cases of the Compound Disk SolutionThe general solution for the compound disk may be used to obtain spreading resistances for several special cases examined previously by many researchers.
Thesespecial cases arise when some of the system parameters go to certain limits. The special cases fall into the following two categories: isotropic half-space, semi-infiniteflux tube, and finite disk problems; and layered half-space and semi-infinite flux tubeproblems. Figures 4.5 and 4.6 show the several special cases that arise from the general case presented above. Results for several special cases are discussed in moredetail in subsequent sections.[291], (31)Lines: 1397 to 14244.6.2Half-Space Problems———0.0pt PgVarIf the dimensions of the compound disk (b, t) become very large relative to the radius———a and the first layer thickness t1 , the general solution approaches the solution forLong Pagethe case of a single layer in perfect thermal contact with an isotropic half-space.
In * PgEnds: Ejectthis case → 0, τ1 → 0 and the spreading resistance depends on the four systemparameters (a, t1 , k1 , k2 ) and the heat flux parameter µ. If we set t1 = 0 or k1 = k2 , the[291], (31)general solution goes to the special case of a circular heat source in perfect contactwith an isotropic half-space. In this case the spreading resistance depends on twosystem parameters (a, k2 ) and the heat flux parameter µ.
The dimensionless spreadingresistance is now defined as ψ = 4k2 aRs , and it is a constant depending on the heatflux parameter. The total resistance is equal to the spreading resistance in both casesbecause the one-dimensional resistance is negligible. The half-space problems areshown in Figs. 4.5d and 4.6d.4.6.3Semi-infinite Flux Tube ProblemsThe general solution goes to the semi-infinite flux tube solutions when the systemparameter τ2 → ∞.
In this case the spreading resistance will depend on the systemparameters (a, b, t1 , k1 , k2 ) and the heat flux parameter µ. The dimensionless spreading resistance will be a function of the parameters (, τ1 , κ) and µ. If one sets t1 = 0or k1 = k2 = k, the dimensionless spreading resistance ψ = 4kaRs depends on thesystem parameters (a, b, k) and µ or = a/b and µ only.
The semi-infinite flux tubeproblems are shown in Figs. 4.5c and 4.6c.4.6.4Isotropic Finite Disk with ConductanceIn this case, one puts k1 = k2 = k or κ = 1. The dimensionless spreading resistance ψ = 4kaRs depends on the system parameters (a, b, t, k, h) and µ or theBOOKCOMP, Inc. — John Wiley & Sons / Page 291 / 2nd Proofs / Heat Transfer Handbook / Bejan123456789101112131415161718192021222324252627282930313233343536373839404142434445292BOOKCOMP, Inc. — John Wiley & Sons / Page 292 / 2nd Proofs / Heat Transfer Handbook / Bejanq = ⫺1/2=0baqqbqak1k1t1k2t1k2(c) — 0, — ⬁, 0 < < ⬁(a) 0 < < ⬁arbk1t1k2t2tqaqabzh = k1/k2Bi = hb/k1 = t1/b = a /bk1k1t1k2k2T=0T=0 = t /bt1(b) Bi — ⬁, — ⬁, 0 < < ⬁(d) — 0, — ⬁, 0 < < ⬁Figure 4.5 Four problems with a single layer on a substrate. (From Yovanovich et al., 1998.)[292], (32)Lines: 1424 to 1433———* 528.0pt PgVar———Normal Page* PgEnds: PageBreak[292], (32)123456789101112131415161718192021222324252627282930313233343536373839404142434445BOOKCOMP, Inc.
— John Wiley & Sons / Page 293 / 2nd Proofs / Heat Transfer Handbook / Bejant1k2t2T=0T=0 = k1/k2Bi = hb/k11 = t1/b = a /bk1k1zh(d ) — 0, = 1, — ⬁(b) = 1, Bi — ⬁293Figure 4.6 Four problems for isotropic systems. (From Yovanovich et al., 1998.)aqaqtk1rbak1q(c) = 1, — ⬁(a) = 1k1b=0qbaq = ⫺1/2aqb[293], (33)Lines: 1433 to 1440———* 528.0pt PgVar———Normal Page* PgEnds: PageBreak[293], (33)294123456789101112131415161718192021222324252627282930313233343536373839404142434445THERMAL SPREADING AND CONTACT RESISTANCESdimensionless system parameters ( = a/b, τ = t/b, Bi = hb/k) and µ. Thisproblem is shown in Fig.
4.6b. These special cases are presented in the followingsections.4.7 SPREADING RESISTANCE OF ISOTROPIC FINITE DISKSWITH CONDUCTANCEThe dimensionless spreading resistance for isotropic (κ = 1) finite disks (τ1 < 0.72)with negligible thermal resistance at the heat sink interface (Bi = ∞) is given by thefollowing solutions (Kennedy, 1960; Mikic and Rohsenow, 1966; Yovanovich et al.,1998): For µ = − 21 :∞4kaRs =8 J1 (δn ) sin δn tanh δn τπ n=1 δ3n J02 (δn )[294], (34)(4.81)Lines: 1440 to 1476For µ = 0:———∞16 J12 (δn )tanh δn τ4kaRs =π n=1 δ3n J02 (δn )(4.82)If the external resistance is negligible Bi → ∞, the temperature at the lower face ofthe disk is isothermal.
The solutions for isoflux µ = 0 heat source and isothermalbase temperature were given by Kennedy (1960) for the centroid temperature and thearea-averaged contact area temperature.4.7.1Correlation EquationsA circular heat source of radius a is attached to one end of a circular disk of thicknesst, radius b, and thermal conductivity k. The opposite boundary is cooled by a fluidat temperature Tf through a uniform heat transfer coefficient h. The sides of the diskare adiabatic and the region outside the source area is also adiabatic.
The flux overthe source area is uniform. The heat transfer through the disk is steady. The externalresistance is defined as R ext = 1/ hA, where A = πb2 .The solution for the isoflux boundary condition and with external thermal resistance was recently reexamined by Song et al. (1994) and Lee et al. (1995). Theynondimensionalized the constriction resistance based on the centroid and areaaveraged temperatures using the square root of the source area (as recommendedby Yovanovich, 1976b, 1991, 1997; Yovanovich and Burde, 1977; Yovanovich andSchneider, 1977; Chow and Yovanovich, 1982; Yovanovich et al., 1984; Yovanovichand Antonetti, 1988) and compared the analytical results against the numerical resultsreported by Nelson and Sayers (1992) over the full range of independent parameters:Bi = hb/k, = a/b, τ = t/b.
Nelson and Sayers (1992) also chose the squareroot of the source area to report their numerical results. The agreement between theanalytical and numerical results were reported to be in excellent agreement.BOOKCOMP, Inc. — John Wiley & Sons / Page 294 / 2nd Proofs / Heat Transfer Handbook / Bejan5.87788pt PgVar———Long PagePgEnds: TEX[294], (34)SPREADING RESISTANCE OF ISOTROPIC FINITE DISKS WITH CONDUCTANCE123456789101112131415161718192021222324252627282930313233343536373839404142434445295Lee et al. (1995) recommended a simple closed-form expression for the dimensionless constriction resistance based on the area-averaged and centroid temperatures.√They defined the dimensionless spreading resistance parameter as ψ = π kaRc ,where Rc is the constriction resistance, and they recommended the following approximations: For the area-averaged temperatureψave = 21 (1 − )3/2 φc(4.83)and for the centroid temperature:1ψmax = √ (1 − )ϕcπ(4.84)with[295], (35)ϕc =Bi tanh (δc τ) + δcBi + δc tanh δc τ(4.85)Lines: 1476 to 15251δc = π + √π(4.86)———-2.5917pt PgVarThe approximations above are within ±10% of the analytical results (Song et al.,1994; Lee et al., 1995) and the numerical results of Nelson and Sayers (1992).
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