Adrian Bejan(Editor), Allan D. Kraus (Editor). Heat transfer Handbok (776115), страница 24
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The systemdescribed represents a two-dimensional (r, z) convecting spine discussed by Aziz andLunardini (1995). The equation governing the two-dimensional heat conduction inthe cylinder isLines: 2648 to 2686———-3.08784pt PgVar———Short Page(3.250) * PgEnds: Eject∂ 2θ1 ∂θ∂ 2θ+=0+∂R 2R ∂R∂Z 2where[218], (58)T − T∞θ=T1 = T∞rR=r0zZ=LLγ=r0and Bi is the Biot number, Bi = hr0 /k. The boundary conditions areθ(R,0) = 1(3.251a)∂θ(0,Z) = 0∂R(3.251b)∂θ(1,Z) = −Bi · θ(1,Z)∂R(3.251c)∂θ(R,1) = −Bi · γθ(R,1)∂R(3.251d)The solution obtained via the separation of the variables isθ=∞2λn J1 (λn )J0 (λn R) 2( coshλn γZ − Υ sinhλn γZ)λn + Bi2 [J0 (λn )]2n=1BOOKCOMP, Inc.
— John Wiley & Sons / Page 218 / 2nd Proofs / Heat Transfer Handbook / Bejan(3.252)TWO-DIMENSIONAL STEADY CONDUCTION123456789101112131415161718192021222324252627282930313233343536373839404142434445219whereΥ=λn sinhλn γ + Bi coshλn γλn coshλn γ + Bi sinhλn γand where J0 and J1 are the Bessel functions of the first kind (Section 3.3.5) and theeigenvalues λn are given byλn J1 (λn ) = Bi · J0 (λn )(3.253)The heat flow into the cylinder from the hot left face isq = 4πkr0 (T1 − T∞ )∞n=1λn [J1 (λn )]2Υλ2n + Bi2 [J0 (λn )]2(3.254)A three-dimensional plot of θ as a function of r and z is shown in Fig.
3.29 forr0 = 1, L = 1, and h/k = 1. This plot was generated using Maple V, Release 5.0. Asexpected, the temperature decreases along both the radial and axial directions. Ozisik(1993) has devoted a complete chapter to the method of separation of variables incylindrical coordinates and provides solutions for several other configurations.3.7.3 Solid Hemisphere with Specified Base and SurfaceTemperaturesr2 ∂φ∂r1 ∂∂φ+sin θ=0sin θ ∂θ∂θ(3.255)where φ = T − Tc . Two of the boundary conditions areθ(r0 , θ) = Ts − Tc = φs πφ r,=02(3.256a)(3.256b)Because the boundary condition at r = 0 falls on the θ = π/2 plane, which isthe base of the hemispherical droplet, it must meet the boundary condition of eq.(3.256b), that is, πφ r,=0(3.256c)2BOOKCOMP, Inc.
— John Wiley & Sons / Page 219 / 2nd Proofs / Heat Transfer Handbook / BejanLines: 2686 to 2752———12.95908pt PgVar———Short Page* PgEnds: Eject[219], (59)Poulikakos (1994) considers a hemispherical droplet condensing on a cold horizontalsurface as shown in Fig. 3.30. The heat conduction equation for the two-dimensional(r, θ) steady-state temperature distribution in the droplet is given by∂∂r[219], (59)220123456789101112131415161718192021222324252627282930313233343536373839404142434445CONDUCTION HEAT TRANSFER[220], (60)Lines: 2752 to 2780———3.46999pt PgVar———Normal PagePgEnds: TEX[220], (60)Figure 3.29 Three-dimensional plot of the temperature distribution in a solid cylinder.
(FromAziz, 2001.)The fourth boundary condition at θ = 0 is obtained by invoking the condition ofthermal symmetry about θ = 0, giving∂φ(r,0) = 0∂θ(3.256d)Use of the method of separation of the variables provides the solution for φ asφ = φs∞ Pn+1 (1) − Pn−1 (1) − Pn+1 (0) + Pn−1 (0) r nn=1r0Pn (cos θ)(3.257)where the P ’s are the Legendre functions of the first kind, discussed in Section 3.3.6.Ozisik (1993) may be consulted for a comprehensive discussion of the method ofseparation of the variables in spherical coordinates.BOOKCOMP, Inc.
— John Wiley & Sons / Page 220 / 2nd Proofs / Heat Transfer Handbook / BejanTWO-DIMENSIONAL STEADY CONDUCTION123456789101112131415161718192021222324252627282930313233343536373839404142434445221zTSr0rTCdy[221], (61)xLines: 2780 to 2809Figure 3.30 Hemispherical droplet condensing on an isothermal surface. (Adapted fromPoulikakos, 1994.)3.7.4———1.03503pt PgVar———Normal PagePgEnds: TEXMethod of SuperpositionThe configuration considered in the preceding section involved one nonhomogeneousboundary condition [either eq.
(3.247a), (3.251a), or (3.256a)]. When two or morenonhomogeneous boundary conditions occur, the analysis can be split into two subanalyses each containing one nonhomogeneous boundary condition. Each subanalysis can then be solved using the method of separation of the variables, and the sum ofthe solutions to the two subanalyses will provide the solution to the overall problem.This approach is illustrated in Fig.
3.31 for a rectangular plate with two homogeneousboundary conditions.The mathematical description of the problem is∂ 2θ∂ 2θ+=0∂x 2∂y 2(3.258)where θ = T − T3 . The boundary conditions areθ(0,y) = θ1 = T1 − T3(3.259a)θ(x,0) = θ2 = T2 − T3(3.259b)θ(L,y) = 0(3.259c)θ(x,H ) = 0(3.259d)BOOKCOMP, Inc. — John Wiley & Sons / Page 221 / 2nd Proofs / Heat Transfer Handbook / Bejan[221], (61)222123456789101112131415161718192021222324252627282930313233343536373839404142434445CONDUCTION HEAT TRANSFERFigure 3.31 Rectangular plate with two nonhomogeneous boundary conditions.[222], (62)Lines: 2809 to 2828———-0.10812pt PgVar———Normal Page* PgEnds: Eject[222], (62)Figure 3.32 Splitting the problem of Fig.
3.31 into two subproblems with known solutions.The problem is split into two subproblems, as indicated in Fig. 3.32, and the twosolutions can be obtained from eq. (3.248) with appropriate adjustment to accountfor the definition of θ and the coordinates, x and y. The sum of the two solutions isθ=∞4θ1 sinh [(2n + 1)π(L − x)/H )] sin [(2n + 1)(πy/H )]π n=0sinh [(2n + 1)πL/H )]2n + 1+3.7.5∞4θ2 sinh [(2n + 1)π(H − y)/L)] sin [(2n + 1)(πx/L)]π n=0sinh [(2n + 1)(πH /L)]2n + 1(3.260)Conduction Shape Factor MethodAlthough the conduction shape factor method does not give the temperature distribution, it provides a simple equation for the rate of heat transfer:BOOKCOMP, Inc. — John Wiley & Sons / Page 222 / 2nd Proofs / Heat Transfer Handbook / BejanTWO-DIMENSIONAL STEADY CONDUCTION123456789101112131415161718192021222324252627282930313233343536373839404142434445q = kS ∆T223(3.261)where k is the thermal conductivity of the conducting medium, ∆T the temperature difference driving the heat flow, and S the conduction shape factor.
Table 3.12provides expressions for the conduction shape factor for various two-dimensionalconfigurations.The conduction resistance for a two-dimensional system follows from eq. (3.261) asRcond =1Sk(3.262)3.7.6 Finite-Difference MethodCartesian Coordinates In the finite-difference approach, the conducting regionis covered with a grid consisting of intersecting lines. The points of intersectionare called nodes. For a rectangular region, the grid lines are drawn parallel to theboundaries.
For simplicity, the spacings ∆x and ∆y are chosen so that ∆x = ∆y.Nodes are identified by double-subscript notation, ij , where i and j count the gridlines along the x- and y-coordinate directions, respectively. The node i, j , whichrepresents a particular subvolume, is presumed to be isothermal at the temperature,Ti,j . In Fig. 3.33, five different types of nodes are identified together with their controlvolumes, which are shown as dashed enclosures.
The finite-difference approximationfor each type of node is given here with the control volume assumed to have no energy *generation.14Ti+1,j + Ti,j +1 + Ti−1,j + Ti,j −1 = 0(3.263)2. Node on plane convecting surface:2Ti−1,jLines: 2828 to 2883———-0.5499pt PgVar———Normal PagePgEnds: Eject[223], (63)1. Internal node:− Ti,j +[223], (63)2h ∆xh ∆x+ Ti,j −1 + Ti,j +1 +T∞ − 2+ 2 Ti,j = 0kk(3.264)3.
External corner node with convection:Ti,j −1 + Ti−1,j +2h ∆xh ∆xT∞ − 2 1 +Ti,j = 0kk(3.265)4. Internal corner node with convection:2h ∆x2 Ti,j −1 + Ti,j +1 + Ti+1,j + Ti,j −1 +T∞kh ∆x−2 3+Ti,j = 0kBOOKCOMP, Inc. — John Wiley & Sons / Page 223 / 2nd Proofs / Heat Transfer Handbook / Bejan(3.266)224123456789101112131415161718192021222324252627282930313233343536373839404142434445CONDUCTION HEAT TRANSFERTABLE 3.12 Conduction Shape Factors for Selected Two-Dimensional Systems[q = Sk(T1 − T2 )]SystemSchematicCase 1:RestrictionsShape Factorz > D/22πD1 − D/4zT2zIsothermal sphere buriedin a semi-infinite mediumT1DT2Case 2:LDT12πLLDzHorizontal isothermalcylinder of length Lburied in a semi-infinitemediumcosh−1 (2z/D)2πLln(4z/D)LDz > 3D/2[224], (64)T2Case 3:Vertical cylinder in asemi-infinite mediumLT12πLln(4L/D)LDD———2.00122pt PgVarCase 4:Conduction betweentwo cylinders of lengthL in infinite mediumD2D1T1T2wL D1 , D2Lwcosh−12πL4w 2 − D12 − D222D1 D1Case 5:Horizontal circularcylinder of length Lmidway between parallelplanes of equal lengthand infinite widthT2⬁z T1⬁DT1dwT1DT2zL2πLln(8z/πD)w>DLw2πLln(1.08w/D)D>dLDcosh−12πLD 2 + d 2 − 4z22DdT2DT1z D/2Lz⬁T2Case 7:Conduction through theedge of adjoining wallsDT2Circular cylinder oflength L centered in asquare solid of equallengthCase 8:⬁zCase 6:Eccentric circularcylinder of length L ina cylinder of equal lengthLines: 2883 to 2923LBOOKCOMP, Inc.
— John Wiley & Sons / Page 224 / 2nd Proofs / Heat Transfer Handbook / BejanD > L/50.54D———Normal Page* PgEnds: PageBreak[224], (64)225TWO-DIMENSIONAL STEADY CONDUCTION123456789101112131415161718192021222324252627282930313233343536373839404142434445TABLE 3.12 Conduction Shape Factors for Selected Two-Dimensional Systems[q = Sk(T1 − T2 )] (Continued)SystemSchematicCase 9:Conduction throughcorner of three walls witha temperature difference∆T1−2 across the wallLLShape FactorL length andwidth of wall0.15LNone2DLCase 10:DDisk of diameter D andT1 on a semi-infinitemedium of thermalconductivity k and T2kCase 11:LT1[225], (65)T2T1T2Square channel oflength LRestrictionswWLines: 2923 to 2996W< 1.4w2πL0.785 ln(W/w)W> 1.4w2πL0.930(W/w) − 0.0505.
Node on a plane surface with a uniform heat flux: 2q ∆x2Ti+1,j + Ti,j +1 + Ti,j −1 +− 4Ti,j = 0k———Normal PagePgEnds: TEX[225], (65)(3.267)In eqs. (3.264)–(3.266), h = 0 applies to a node on an adiabatic boundary andh = ∞ appiles to a node on an isothermal boundary.By writing an appropriate finite-difference approximation for each node in thegrid, a set of n-linear algebraic equations (one for each of the n nodes) in the unknownnode temperatures can be produced. A standard numerical procedure or a computercode can be used to solve the system of equations giving the temperatures at all thenodes.Consider the square plate shown in Fig.
3.34. For ∆x = ∆y = 0.1/16 = 0.00625m, the finite-difference solution generated by Aziz (2001) gives the temperatures onthe convecting surface. For j = 1, 3, 5, 7, . . . 17,T17,j = 67.85, 68.24, 70.19, 73.24, 77.13, 81.67, 86.81, 92.66, and 100.00 all in °C.Cylindrical Coordinates Consider the solid cylinder of radius r0 and lengthL (Fig. 3.35) in which steady conduction occurs along the r and z directions. Theconduction equation isBOOKCOMP, Inc.
— John Wiley & Sons / Page 225 / 2nd Proofs / Heat Transfer Handbook / Bejan———-0.51883pt PgVar226123456789101112131415161718192021222324252627282930313233343536373839404142434445CONDUCTION HEAT TRANSFERjFluidT⬁ , hexternal cornernode (i, j)FluidT⬁ , hq⬙node on aplane withuniform heatflux (i, j)internalnode (i, j)node on a planesurface (i, j)y[226], (66)internal cornernode (i, j)FluidT⬁ , h⌬y⌬xxLines: 2996 to 2996———iFigure 3.33 Types of nodes and the corresponding control volumes.*63.854pt PgVar———Normal PagePgEnds: TEX[226], (66)Figure 3.34 Grid for two-dimensional conduction in a square plate.BOOKCOMP, Inc.
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