Adrian Bejan(Editor), Allan D. Kraus (Editor). Heat transfer Handbok (776115), страница 26
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The surface heat flux q0 is2n−1 (n−1)/2n n−1xq0 = √ tkaΓ 1 +erfc √(3.293)i2α2 αt x=0Several special cases can be deduced from eqs. (3.292) and (3.293).BOOKCOMP, Inc. — John Wiley & Sons / Page 232 / 2nd Proofs / Heat Transfer Handbook / BejanTRANSIENT CONDUCTION123456789101112131415161718192021222324252627282930313233343536373839404142434445233Case 1: f (t) = T0 This is the case of constant surface temperature which occurswhen n = 0 and a = T0 − Ti :T (x, t) − Tix= erfc √(3.294)T0 − T i2 αtq0 =k(T0 − Ti )(παt)1/2(3.295)Case 2: f (t) = Ti + at This is the case of a linear variation of surface temperaturewith time which occurs when n = 2:xT (x, t) = Ti + 4ati 2 erfc √(3.296)2 αt2katq0 =(παt)1/2[233], (73)(3.297)Lines: 3285 to 3339———Case 3: f (t) = Ti + at 1/2 This is the case of a parabolic surface temperature-1.99896ptvariation with time with n = 1.———Long Page√xT (x, t) = Ti + a πt i erfc √(3.298) * PgEnds: Eject2 αtka πq0 =(3.299)[233], (73)2 αSpecified Surface Heat Flux For a constant surface heat flux q0 , the boundaryconditions of eq.
(3.291b) is replaced by−k∂T (0, t)= q0∂x(3.300)and the solution is√2q0 αtxT (x, t) = Ti +i erfc √k2 αtSurface Convection(3.301)The surface convection boundary condition−k∂T (0, t)= h [T∞ − T (0, t)]∂xreplaces eq. (3.291b) and the solution isT (x, t) − Tixh√x22= erfc √αt + √− e(hx/k)+(h αt/k ) erfcT∞ − T ik2 αt2 αtBOOKCOMP, Inc.
— John Wiley & Sons / Page 233 / 2nd Proofs / Heat Transfer Handbook / Bejan(3.302)(3.303)PgVar234123456789101112131415161718192021222324252627282930313233343536373839404142434445CONDUCTION HEAT TRANSFERConstant Surface Heat Flux and Nonuniform InitialTemperatureet al. (1995) considered a nonuniform initial temperature of the formT (x, 0) = a + bxwhere a and b are constants to find a modified version of eq. (3.301) as √q0x+ b i erfc √T (x, t) = a + bx + 2 αtk2 αtZhuang(3.304)(3.305)The surface temperature is obtained by putting x = 0 in eq.
(3.305), which gives αt q0+b(3.306)T (0, t) = a + 2π kZhuang et al. (1995) found that the predictions from eqs. (3.305) and (3.306) matchedthe experimental data obtained when a layer of asphalt is heated by a radiant burner,producing a heat flux of 41.785 kW/m 2. They also provided a solution when the initialtemperature distribution decays exponentially with x.Constant Surface Heat Flux and Exponentially Decaying Energy Generation When the surface of a solid receives energy from a laser source, the effect ofthis penetration of energy into the solid can be modeled by adding an exponentiallydecaying heat generation term, q̇0 e−ax /k (where a is the surface absorption coefficient), to the left side of eq. (3.290). The solution for this case has been reported bySahin (1992) and Blackwell (1990):xT = Ti + (T0 + Ti ) erfc √2 αt√xxq̇02+ 2 erfc √− 21 ea αt+ax erfc √ + a αtka2 αt2 αt√1 a 2 αt−axx−ax (a 2 αt−1)− e(3.307)erfc√ − a αt + e e22a αtBoth Sahin (1992) and Blackwell (1990) have also solved this case for a convectiveboundary condition (h, T∞ ) at x = 0.
Blackwell’s results show that for a givenabsorption coefficient a, thermal properties of α and k, initial temperature Ti , andsurface heat generation q̇0 , the location of the maximum temperature moves deeperinto the solid as time progresses. If h is allowed to vary, for a given time the greatervalue of h provides the greater depth at where the maximum temperature occurs.The fact that the maximum temperature occurs inside the solid provides a possibleexplanation for the “explosive removal of material” that has been observed to occurwhen the surface of a solid is given an intense dose of laser energy.BOOKCOMP, Inc. — John Wiley & Sons / Page 234 / 2nd Proofs / Heat Transfer Handbook / Bejan[234], (74)Lines: 3339 to 3386———5.48827pt PgVar———Normal PagePgEnds: TEX[234], (74)TRANSIENT CONDUCTION1234567891011121314151617181920212223242526272829303132333435363738394041424344453.8.3235Finite-Sized Solid ModelConsider one-dimensional transient conduction in a plane wall of thickness 2L, a longsolid cylinder of radius r0 , and a solid sphere of radius r0 , each initially at a uniformtemperature Ti .
At time t = 0, the exposed surface in each geometry is exposed to ahot convective environment (h, T∞ ). The single parabolic partial differential equationdescribing the one-dimensional transient heating of all three configurations can bewritten as1 ∂1 ∂Tn ∂Ts=(3.308)s n ∂s∂sα ∂twhere s = x, n = 0 for a plane wall, s = r, n = 1 for a cylinder, and s = r, n = 2for a sphere. In the case of a plane wall, x is measured from the center plane. Theinitial and boundary conditions for eq. (3.308) areT (s, 0) = Ti∂T (0, t)=0∂sk(3.309a)(thermal symmetry)∂T (L or r0 , t)= h [T∞ − T (L or r0 , t)]∂sLines: 3386 to 3439(3.309b)θ=n=1R ν J−ν (λn R) −λ2n τeλ2n + Bi2 + 2ν · Bi J−ν (λn )2 Bi(3.309c)(3.311)The cumulative energy received over the time t is2 ∞2(1 + n) Bi2 1 − e−λn τQ = ρcV (T∞ − Ti )λ2 λ2n + Bi2 + 2ν · Bin=1 n(3.312)Solutions for a plane wall, a long cylinder, and a sphere can be obtained from eqs.(3.310) and (3.312) by putting n = 0, (ν = 21 ), n = 1(ν = 0), and n = 2(ν = − 21 ),respectively.
It may be noted thatJ−1/2 (z) =2πz1/2BOOKCOMP, Inc. — John Wiley & Sons / Page 235 / 2nd Proofs / Heat Transfer Handbook / Bejancos z———Normal PagePgEnds: TEX[235], (75)(3.310)where θ = (T∞ − T )/(T∞ − Ti ), R = x/L for a plane wall, R = r/r0 for bothcylinder and sphere, Bi = hL/k or hr0 /k, τ = αt/L2 or τ = αt/r02 , ν = (1 − n)/2,and the λn are the eigenvalues given byλn J−(ν−1) (λn ) = Bi · J−ν (λn )———0.65923pt PgVarAccording to Adebiyi (1995), the separation of variables method gives the solutionfor θ as∞[235], (75)236123456789101112131415161718192021222324252627282930313233343536373839404142434445CONDUCTION HEAT TRANSFERJ1/2 (z) =J3/2 (z) =2πz2πz1/2sin z1/2 sin z− cos zzWith these representations, eqs.
(3.310) and (3.312) reduce to the standard formsappearing in textbooks. Graphical representations of eqs. (3.310) and (3.312) arecalled Heisler charts.3.8.4Multidimensional Transient ConductionFor some configurations it is possible to construct multidimensional transient conduction solutions as the product of one-dimensional results given in Sections 3.8.2 and3.8.3. Figure 3.38 is an example of a two-dimensional transient conduction situationin which the two-dimensional transient temperature distribution in a semi-infiniteplane wall is the product of the one-dimensional transient temperature distribution inan infinitely long plane wall and the one-dimensional transient temperature distribution in a semi-infinite solid.
Several other examples of product solutions are given byBejan (1993).3.8.5Finite-Difference MethodExplicit Method For two-dimensional transient conduction in Cartesian coordinates, the governing partial differential equation isT⬁ ,hT⬁ ,hT⬁ ,hx2=x1T⬁ ,hT⬁ ,hx1⫻x22L2LT⬁ ,hT(x1, x2, t) ⫺ T⬁=T1 ⫺ T⬁semiinfiniteplanewallT( x2, t) ⫺ T⬁⫻T1 ⫺ T⬁infinitelylong planewallT( x1, t) ⫺ T⬁T1 ⫺ T⬁semiinfinitesolidFigure 3.38 Product solution for a two-dimensional transient conduction problem.BOOKCOMP, Inc. — John Wiley & Sons / Page 236 / 2nd Proofs / Heat Transfer Handbook / Bejan[236], (76)Lines: 3439 to 3461———-2.89299pt PgVar———Normal Page* PgEnds: Eject[236], (76)TRANSIENT CONDUCTION123456789101112131415161718192021222324252627282930313233343536373839404142434445∂ 2T∂ 2T1 ∂T+=∂x 2∂y 2α ∂t237(3.313)which assumes no internal heat generation and constant thermal properties.
Approximating the second-order derivatives in x and y and the first-order derivative in t byforward differences, the explicit finite-difference approximations (using ∆x = ∆y)for various nodes (see Fig. 3.33) can be expressed using the Fourier modulus, Fo =α ∆t/(∆x)2 , Bi = h ∆x/k, and t = p ∆t.• Internal node: With Fo ≤ 41 ,p+1pppppTi,j = Fo T1+1,j + Ti−1,j + Ti,j +1 + Ti,j −1 + (1 − 4Fo)Ti,j(3.314)[237], (77)• Node at interior corner with convection: With Fo(3 + Bi) ≤ 43 ,p+1ppppTi,j = 23 Fo Ti+1,j + 2Ti−1,j + 2Ti,j +1 + Ti,j −1 + 2BiT∞ p+ 1 − 4Fo − 43 Bi · Fo Ti,j• Node on a plane surface with convection: With Fo(2 + Bi) ≤p+1pppTi,j = Fo 2Ti−1,j + Ti,j +1 + Ti,j −1 + 2Bi · T∞Lines: 3461 to 3514(3.315)———Normal PagePgEnds: TEX1,2p+ (1 − 4Fo − 2Bi · Fo) Ti,j(3.316)• Node at exterior corner with convection: With Fo(1 + Bi) ≤ 41 ,p+1ppTi,j = 2Fo Ti−1,j + Ti,j −1 + 2Bi · T∞p+ (1 − 4Fo − 4Bi · Fo) Ti,j = 0(3.317)• Node on a plane surface with uniform heat flux: With Fo ≤ 41 ,p+1Ti,j∆xpppp(3.318)= (1 − 4Fo)Ti,j + Fo 2Ti−1,j + Ti,j +1 + Ti,j −1 + 2Fo · q kThe choice of ∆x and ∆t must satisfy the stability constraints, introducing eachof the approximations given by eqs.
(3.314)–(3.318) to ensure a solution free ofnumerically induced oscillations. Once the approximations have been written for eachnode on the grid, the numerical computation is begun with t = 0(p = 0), for whichthe node temperatures are known from the initial conditions prescribed. Because eqs.(3.313)–(3.318) are explicit, node temperatures at t = ∆t (p = 1) can be determinedfrom a knowledge of the node temperatures the preceding time, t = 0(p = 0). ThisBOOKCOMP, Inc. — John Wiley & Sons / Page 237 / 2nd Proofs / Heat Transfer Handbook / Bejan———4.84828pt PgVar[237], (77)238123456789101112131415161718192021222324252627282930313233343536373839404142434445CONDUCTION HEAT TRANSFER“marching out” in time type of computation permits the transient response of thesolid to be determined in a straightforward manner.
However, the computational timenecessary to cover the entire transient response is excessive because extremely smallvalues of ∆t are needed to meet the stability constraints.Implicit Method In the implicit method, the second derivatives in x and y areapproximated by central differences but with the use of temperatures at a subsequenttime, p + 1, rather than the current time, p, while the derivative in t is replaced bya backward difference instead of a forward difference. Such approximations lead tothe following equations:• Internal node:p+1(1 + 4Fo)Ti,jp+1p+1p+1p+1p− Fo Ti+1,j + Ti−1,j + Ti,j +1 + Ti,j −1 = Ti,j(3.319)• Node at interior corner with convection: p+1p+1p+1(1 + 4Fo) 1 + 13 Bi Ti,j − 23 Fo Ti+1,j + 2Ti−1,jp+1p+1p+ 2Ti,j +1 + Ti,j −1 = Ti,j + 43 Fo · Bi · T∞Lines: 3514 to 3565———-5.41692pt PgVar(3.320)• Node on a plane surface with convection:p+1p+1p+1p+1[1 + 2Fo(2 + Bi)] Ti,j − Fo 2Ti−1,j + Ti,j +1 + Ti,j −1=pTi,j+ 2Bi · Fo · T∞(3.321)• Node on a plane surface with uniform heat flux:p+12Fo · q ∆xp+1p+1p+1p+ Fo 2Ti−1,j + Ti,j +1 + Ti,j −1 = Ti,j +k(3.323)The implicit method is unconditionally stable and therefore permits the use ofhigher values of ∆t, thereby reducing the computational time.
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