Adrian Bejan(Editor), Allan D. Kraus (Editor). Heat transfer Handbok (776115), страница 37
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(1984). Various analytical and numericalmethods were employed to obtain short- and long-time solutions.4.5.1Isoflux Circular AreaBeck (1979) reported the following integral solution for a circular area of radius awhich is subjected to a uniform and constant flux q for t > 0:8 ∞ √ 2 dζ4kaRs =erf ζ Fo J1 (ζ) 2(4.63)π 0ζwhere erf is the error function, J1 (x) is a Bessel function of the first kind of order1 (Abramowitz and Stegun, 1965), and ζ is a dummy variable.
The dimensionlesstime is defined as Fo = αt/a 2 , where α is the thermal diffusivity of the half-space.The spreading resistance is based on the area-averaged temperature. Steady state isobtained when Fo → ∞, and the solution goes to 4kaRs = 32/(3π2 ).Beck (1979) gave approximate solutions for short and long times. For short timeswhere Fo < 0.6,BOOKCOMP, Inc.
— John Wiley & Sons / Page 285 / 2nd Proofs / Heat Transfer Handbook / Bejan286123456789101112131415161718192021222324252627282930313233343536373839404142434445THERMAL SPREADING AND CONTACT RESISTANCES84kaRs =πFo Fo Fo2Fo315Fo4−+++ππ8π32π512π(4.64)and for long times where Fo ≥ 0.6,4kaRs =322111−−+1−√3π23(4Fo) 6(4Fo)212(4Fo)3π3/2 Fo(4.65)The maximum errors of about 0.18% and 0.07% occur at Fo = 0.6 for the short- andlong-time expressions, respectively.4.5.2Isoflux HyperellipseThe hyperellipse is defined by (x/a)n + (y/b)n = 1 with b ≤ a, where n is theshape parameter and a and b are the axes along the x and y axes, respectively.The hyperellipse reduces to many special cases by setting the values of n and theaspect ratio parameter γ = b/a, which lies in the range 0 ≤ γ ≤ 1.
Therefore, thesolution developed for the hyperellipse can be used to obtain solutions for many othergeometries, such as ellipse and circle, rectangle and square, diamondlike geometries,√and so on. The transient dimensionless centroid constriction resistance k A R0 ,where R0 = T0 /Q, is given by the double-integral solution (Yovanovich, 1997) π/2 r0√2rk A R0 = √erfc √ √dr dω(4.66)π A 02 A Fo0with Fo = αt/A, and the area of the hyperellipse is given by A = (4γ/n)B(1 +1/n, 1/n) and B(x,y) is the beta function (Abramowitz and Stegun, 1965).
The upperlimit of the radius is given by r0 = γ/[(sin ω)n + γn (cos ω)n ]1/n and the aspect ratioparameter γ = b/a. The solution above has(1) for√ the following√√ characteristics:small dimensionless times, Fo ≤ 4 × 10−2 , k A R0 = (2/ π) Fo for all values ofn and γ; (2) for long dimensionless times, Fo ≥ 103 , the results are within 1% of thesteady-state values, which are given by the single integral π/2√dω2γk A R0 = √(4.67)n[(sin ω) + γn (cos ω)n ]1/nπ A 0which depends on the aspect ratio γ and the shape parameter n. The dimensionlessspreading resistance depends on the three parameters Fo, γ, and n in the transitionregion 4 × 10−2 ≤ Fo ≤ 103 in some complicated manner that can be deduced fromthe solution for the circular area.
For this axisymmetric shape we put γ = 1, n = 2into the hyperellipse double integral, which yields the following closed-form resultvalid for all dimensionless time (Yovanovich, 1997):√√111k A R0 = Fo √ − √ exp −4πFoππBOOKCOMP, Inc. — John Wiley & Sons / Page 286 / 2nd Proofs / Heat Transfer Handbook / Bejan[286], (26)Lines: 1195 to 1237———1.84224pt PgVar———Normal PagePgEnds: TEX[286], (26)TRANSIENT SPREADING RESISTANCE IN AN ISOTROPIC HALF-SPACE12345678910111213141516171819202122232425262728293031323334353637383940414243444511+ √ √ erfc √ √2 π Fo2 π Fo287(4.68)where the dimensionless time for the circle of radius a is Fo = αt/πa 2 .4.5.3Isoflux Regular PolygonsFor regular polygons having sides N ≥ 3, the area is A = N ri2 tan(π/N ), whereri is the radius of the inscribed circle.√ The dimensionless spreading resistance basedon the centroid temperature rise k A R0 is given by the following double integral(Yovanovich, 1997): π/N 1/ cos ω√Nrk A R0 = 2erfc √dr dω (4.69)√tan(π/N ) 02 N tan(π/N ) Fo0[287], (27)where the polygonal area is expressed in terms of the number of sides N , and forLines: 1237 to 1266convenience the inscribed radius has been set to unity.
This double-integral solution———has characteristics identical to those of the double-integral solution given√above for-0.04442ptPgVarthe √hyperellipse;that is, for small dimensionless time, Fo ≤ 4 × 10−2 , k A R0 =———√(2/ π) F0 for all polygons N ≥ 3; and for long dimensionless times, Fo ≥ 103 ,Normal Pagethe results are within 1% of the steady-state values, which are given by the following* PgEnds: Ejectclosed-form expression (Yovanovich, 1997):√N1 + sin(π/N )1[287], (27)k A R0 =ln(4.70)π tan(π/N )cos(π/N )√The dimensionless spreading resistance k A R0 depends on the parameters: Fo andN in the transition region: 4 × 10−2 ≤ Fo ≤ 103 in some complex manner which,as described above, may be deduced from√ the solution for the circular area. Thesteady-state solution gives the values k ARo = 0.5617, 0.5611, and 0.5642 forthe equilateral triangle, N = 3, the square, N = 4, and the circle, N → ∞.
Thedifference between the values for the triangle and the circle is approximately 2.2%,whereas the difference between the values for the square and the circle is less than0.6%. The following procedure is proposed for computation of the centroid-basedtransient spreading resistance for the range 4 × 10−2 ≤ Fo ≤ 106 . The closed-formsolution for the circle is the basis of the proposed method.
For any planar singlyconnected contact area subjected to a uniform heat flux, take√ψ01= 2 Fo 1 − exp −ψ0 (Fo → ∞)4π · Fo11+ √ erfc √ √(4.71)2 Fo2 π FoBOOKCOMP, Inc. — John Wiley & Sons / Page 287 / 2nd Proofs / Heat Transfer Handbook / Bejan288123456789101112131415161718192021222324252627282930313233343536373839404142434445THERMAL SPREADING AND CONTACT RESISTANCES√where ψ0 = k A R0 . The right-hand side of eq. (4.71) can be considered to bea universal dimensionless time function that accounts for the transition from smalltimes to near steady state. The procedure proposed should provide quite accurateresults for any planar, singly connected area.
A simpler expression that is based onthe Greene (1989) approximation of the complementary error function is proposed(Yovanovich, 1997):√ψ01 22(4.72)= √ 1 − e−z + a1 π ze−a2 (z+a3 )ψ0 (Fo → ∞)z π√ √where z = 1/(2 π Fo) and the three correlation coefficients are a1 = 1.5577, a2 =0.7182, and a3 = 0.7856. This approximation will provide values of ψ0 with maximum errors of less than 0.5% for Fo ≥ 4 × 10−2 .[288], (28)4.6 SPREADING RESISTANCE WITHIN A COMPOUND DISKWITH CONDUCTANCEThe spreading, one-dimensional flow and total resistances for steady conductionwithin compound disks is important in many microlectronic applications.
The heatenters a compound disk of radius b through a circular area of radius a located in thetop surface of the first layer of thickness t1 and thermal conductivity k1 , which is inperfect thermal contact with the second layer (called the substrate) of thickness t2and thermal conductivity k2 .The lateral boundary r = b is adiabatic, the face at z = t = t1 + t2 is either cooledby a fluid through the film conductance h or it is in contact with a heat sink through acontact conductance h. In either case, h is assumed to be uniform. A compound diskwith uniform heat flux and uniform conductance along the lower surface is shown inFig.
4.4.Figure 4.4 Compound disk with uniform heat flux and conductance. (From Yovanovich etal., 1980.)BOOKCOMP, Inc. — John Wiley & Sons / Page 288 / 2nd Proofs / Heat Transfer Handbook / BejanLines: 1266 to 1295———0.62859pt PgVar———Long PagePgEnds: TEX[288], (28)SPREADING RESISTANCE WITHIN A COMPOUND DISK WITH CONDUCTANCE123456789101112131415161718192021222324252627282930313233343536373839404142434445289The boundary condition over the source area may be modeled as either uniformheat flux or isothermal.
The complete solution for these two boundary conditionshas been reported by Yovanovich et al. (1980). The spreading resistance Rs and onedimensional system resistance R1D are related to the total system resistance R total inthe following manner:R total = Rs + R1D = Rs +t1t21++k1 A k2 A hA(K/W)(4.73)where A = πb2 . The general solution for the dimensionless spreading resistance parameter ψ = 4k1 aRs depends on several dimensionless geometric and thermophysical parameters: = a/b, τ = t/b, τ1 = t1 /b, τ2 = t2 /b, κ = k1 /k2 , Bi = hb/k2 ,and µ, a parameter that describes the heat flux distribution over the source area.
Thegeneral relationship between the total heat flow rate Q and the axisymmetric heat fluxdistribution q(u) isq(u) =Q(1 + µ)(1 − u2 )µπa 20≤u≤1(4.74)The heat flux distributions corresponding to three values of the parameter µ arepresented in Table 4.8, where Q/πa 2 is the average flux on the area.
When µ = 0,the heat flux is uniform, and when µ = − 21 , the heat flux distribution is calledthe equivalent isothermal distribution because it produces an almost isothermal areaprovided that a/b < 0.6.The independent system parameters have the ranges given in Table 4.9. If the firstlayer conductivity is lower than the substrate conductivity (i.e., 0 < κ < 1), the layer(coating) is said to be thermally resistive, and if the conductivity ratio parameter liesin the range 1 < κ < ∞, the layer is said to be thermally conductive.The general dimensionless spreading resistance relationship was given as (Yovanovich et al., 1980):∞4k1 aRs =8(µ + 1) J1 (δn )An ()Bn (τ, τ1 , κ, Bi)πδnn=1(4.75)The first layer thermal conductivity and the radius of the heat source were used tonondimensionalize the spreading resistance.
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