Adrian Bejan(Editor), Allan D. Kraus (Editor). Heat transfer Handbok (776115), страница 25

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— John Wiley & Sons / Page 226 / 2nd Proofs / Heat Transfer Handbook / BejanTWO-DIMENSIONAL STEADY CONDUCTION123456789101112131415161718192021222324252627282930313233343536373839404142434445227[227], (67)Lines: 2996 to 3001———-4.07896pt PgVar———Normal Page* PgEnds: Eject[227], (67)Figure 3.35 Finite-difference grid for radial and axial conduction in a solid cylinder with sixdifferent types of nodes.∂ 2T∂ 2T1 ∂T+ 2 =0+2∂rr ∂r∂z(3.268)Let the outer surface temperature of the cylinder be Ts . The face at z = 0 is insulatedwhile the face at z = L experiences a constant heat flux q .

This description givesthe boundary conditionsBOOKCOMP, Inc. — John Wiley & Sons / Page 227 / 2nd Proofs / Heat Transfer Handbook / Bejan228123456789101112131415161718192021222324252627282930313233343536373839404142434445CONDUCTION HEAT TRANSFER∂T (0,z)=0∂r(thermal symmetry)(3.269a)T (r0 ,z) = Ts(3.269b)∂T (r,0)=0∂z(3.269c)∂T (r,L)= q ∂z(3.269d)Six different nodes (a, b, c, d, e, and f ) have been identified in Fig. 3.35. The finitedifference approximations for these nodes are as follows:1. Node a (internal node):1∆r∆rTi,j −1 + 1 +Ti,j +1 + Ti+1,j + Ti−1,j = 0 (3.270)1−− Ti,j +42rj2rj2. Node b (node on an insulated surface, z = 0):1∆r∆rTi,j −1 + 1 +Ti,j +1 = 02Ti+1,j + 1 −− Ti,j +42rj2rjq ∆r=02k4. Node d (node at the corner of two insulated surfaces):− Ti,j + + 13 Ti+1,j + 2Ti,j +1 = 0(i = 1, j = 1)———(3.271)(3.272)(3.273)(3.275)Taking r0 = 1m, L = 1m, Ts = 25°C, k = 20W/m · K, and q = 1000 W/m 2,Aziz (2001) produced the following temperatures at the 20 nodes shown in Fig.

3.35:BOOKCOMP, Inc. — John Wiley & Sons / Page 228 / 2nd Proofs / Heat Transfer Handbook / Bejan———Normal Page* PgEnds: Eject[228], (68)5. Node e (node at the corner of an insulated surface and a constant heat fluxsurface):1q ∆rTi−1,j + 2Ti,j +1 +=0(3.274)− Ti,j + +3k6. Node f (node on the longitudinal axis):− Ti,j + 16 Ti−1,j + Ti+1,j + 4Ti,j +1 = 0Lines: 3001 to 30761.44237pt PgVar3. Node c (node on a constant heat flux surface, z = L):∆r1∆rTi,j −1 + 1 +Ti,j +1 + 2Ti−1,j1−− Ti,j +42rj2rj+[228], (68)TRANSIENT CONDUCTION123456789101112131415161718192021222324252627282930313233343536373839404142434445T1,1 = 31.03°CT3,3 = 32.33°CT1,2 = 30.54°CT3,4 = 28.95°CT1,3 = 29.14°CT4,1 = 41.42°CT1,4 = 27.13°CT4,2 = 40.43°CT2,1 = 32.02°CT4,3 = 37.38°CT2,2 = 31.46°CT4,4 = 32.15°CT2,3 = 29.87°CT5,1 = 51.59°CT2,4 = 27.53°CT5,2 = 50.42°CT3,1 = 35.24°CT5,3 = 46.69°CT3,2 = 34.50°CT5,4 = 39.35°C229A thorough discussion of the finite-difference method appears in Ozisik (1994).The book also features a large number of examples.[229], (69)Lines: 3076 to 3127———3.8 TRANSIENT CONDUCTION2.71507pt PgVar———The term transient conduction is used when the temperature in a heat conductionNormal Pageprocess depends on both time and the spatial coordinates.

Three models that are commonly used to study transient conduction are the lumped thermal capacity model, the * PgEnds: Ejectsemi-infinite solid model and the finite-sized model. The finite-difference methodprovides one of many numerical methods that are used to analyze complicated con[229], (69)figurations.3.8.1Lumped Thermal Capacity ModelThe lumped thermal capacity model assumes that spatial temperature variationswithin the body are negligible and the temperature variation is solely a function oftime. Consider a body of arbitrary shape of volume V , surface area As , density ρ,and specific heat c, initially at a temperature Ti , as indicated in Fig. 3.36.

At timet ≥ 0, the body is immersed in a convective environment (T∞ , h), where T∞ < Ti ,and allowed to cool. The differential equation describing the cooling process isρV cdT= −hAs (T − T∞ )dt(3.276)with the initial conditionT (t = 0) = Ti(3.277)T − T∞= e−hAs t/ρV cTi − T∞(3.278)The solution isBOOKCOMP, Inc. — John Wiley & Sons / Page 229 / 2nd Proofs / Heat Transfer Handbook / Bejan230123456789101112131415161718192021222324252627282930313233343536373839404142434445CONDUCTION HEAT TRANSFER␧AsRadiation sinkat Ts .h, T∞V, ␳, cqconvectionorqradiationT(0) = TiFigure 3.36 Cooling of a lumped thermal capacity body.and the cumulative energy transfer to the coolant Q over a period of time t isQ = ρV c(Ti − T∞ ) 1 − e−hAs t/ρV c(3.279)The lumped thermal capacity model is valid for Bi = hV /kAs < 0.10, a conditionthat is met in many engineering applications, such as the annealing of metals.Attention now turns to some refinements of the basic lumped thermal capacitymodel.Internal Energy Generation If the body experiences uniform thermal energygeneration Ėg (W) at time t = 0, the temperature variation in the body is given byĖg − hAs (T − T∞ )= e−hAs t/ρV cĖg − hAs (Ti − T∞ )(3.280)This situation occurs when an electronic component is suddenly energized.Temperature-Dependent Specific Heattemperature, that is,If the specific heat varies linearly withc = c∞ [1 ± β(T − T∞ )](3.281)the temperature variation given by Aziz and Hamad (1977) islnT − T∞T − TihAs t± β (Ti − T∞ )=−Ti − T ∞Ti − T aρV c∞(3.282)Pure Radiation Cooling If the body is cooled solely by radiation, the termhAs (T − T∞ ) is replaced in eq.

(3.276) by σAs (T 4 − Ts4 ), where is the surfaceemissivity and Ts is the effective sink temperature for radiation. The solution for Tin this case is given by Aziz and Hamad (1977):BOOKCOMP, Inc. — John Wiley & Sons / Page 230 / 2nd Proofs / Heat Transfer Handbook / Bejan[230], (70)Lines: 3127 to 3169———0.56714pt PgVar———Normal Page* PgEnds: Eject[230], (70)TRANSIENT CONDUCTION123456789101112131415161718192021222324252627282930313233343536373839404142434445lnTs + T Ts − TiTs − T Ts + T iT4σAs Ts3 tTi+ 2 arctan=− arctanTsTsρV c231(3.283)Equation (3.283) is useful, for example, in designing liquid droplet radiation systemsfor heat rejection on a permanent space station.Simultaneous Convective–Radiative Cooling In this case, the radiative term,−σAs (T 4 −Ts4 ) appears on the right-hand side of eq.

(3.276) in addition to −hAs (T −T∞ ). An exact solution for this case does not exist except when T∞ = Ts = 0. Forthis special case the exact solution is1 + σTi3 / h(T /Ti )31hAs t=ln(3.284)3ρV c(1 + σTi3 / h)(T /Ti )3[231], (71)Lines: 3169 to 3226Temperature-Dependent Heat Transfer Coefficient For natural convectioncooling, the heat transfer coefficient is a function of the temperature difference, andthe functional relationship ish = C(T − T∞ )n———Normal Page(3.285) * PgEnds: Ejectwhere C and n are constants.

Using eq. (3.285) in (3.276) and solving the resultingdifferential equation givesT − T∞nhi As t −1/n= 1+(3.286)Ti − T ∞ρV cwhere n = 0 and hi = C(Ti − T∞ )n .Heat Capacity of the Coolant Pool If the coolant pool has a finite heat capacity,the heat transfer to the coolant causes T∞ to increase.

Denoting the properties of thehot body by subscript 1 and the properties of the coolant pool by subscript 2, thetemperature–time histories as given by Bejan (1993) areT1 (t) = T1 (0) −T1 (0) − T2 (0)(1 − e−nt )1 + ρ1 V1 c1 /ρ2 V2 c2(3.287)T2 (t) = T2 (0) +T1 (0) + T2 (0)(1 − e−nt )1 + (ρ2 V2 c2 /ρ1 V1 c1 )(3.288)where t1 (0) and T2 (0) are the initial temperatures andn = hAsρ1 V1 c1 + ρ2 V2 c2(ρ1 V1 c1 )(ρ2 V2 c2 )BOOKCOMP, Inc. — John Wiley & Sons / Page 231 / 2nd Proofs / Heat Transfer Handbook / Bejan———-0.79182pt PgVar(3.289)[231], (71)232123456789101112131415161718192021222324252627282930313233343536373839404142434445CONDUCTION HEAT TRANSFERq0⬙Tsxh, TaxxT(0, t) = Ts = Ti ⫹ at nT(x,0) = Ti⳵T⫺k ⫺⫺|x= 0 = q0⬙⳵XT(x,0) = Ti(a)⳵T⫺k ⳵⫺X⫺|x= 0 = h[Ta ⫺ T(0, t)](b)(c)Figure 3.37 Semi-infinite solid with (a) specified surface temperature, (b) specified surfaceheat flux, and (c) surface convection.3.8.2[232], (72)Lines: 3226 to 3285Semi-infinite Solid Model———As indicated in Fig.

3.37, the semi-infinite solid model envisions a solid with oneidentifiable surface and extending to infinity in all other directions. The parabolicpartial differential equation describing the one-dimensional transient conduction is∂ 2T1 ∂T=2∂xα ∂t0.00218pt PgVar———Long PagePgEnds: TEX(3.290)[232], (72)Specified Surface Temperature If the solid is initially at a temperature Ti , andif for time t > 0 the surface at x = 0 is suddenly subjected to a specified temperature–time variation f (t), the initial and boundary conditions can be written asT (x, 0) = Ti(x ≥ 0)T (0, t) = f (t) = Ti + at n/2T (∞, t) = Ti(t ≥ 0)where a is a constant and n is a positive integer.Using the Laplace transformation, the solution for T is obtained asnxn/2 nT = Ti + aΓ 1 +(4t) i erfc √22 αt(3.291a)(3.291b)(3.291c)(3.292)where Γ is the gamma function (Section 3.3.2) and i n erfc is the nth repeated integralof the complementary error function (Section 3.3.1).

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