Adrian Bejan(Editor), Allan D. Kraus (Editor). Heat transfer Handbok (776115), страница 27
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However, at each timet, the implicit method requires that the node equations be solved simultaneously ratherthan sequentially.Other Methods Several improvements of the explicit and implicit methods havebeen advocated in the numerical heat transfer literature. These include the three-timelevel scheme of Dufort and Fankel, the Crank–Nicholson method, and alternatingBOOKCOMP, Inc. — John Wiley & Sons / Page 238 / 2nd Proofs / Heat Transfer Handbook / Bejan———Normal PagePgEnds: TEX[238], (78)• Node at exterior corner with convection:p+1p+1p+1p1 + 4Fo(1 + Bi)Ti,j − 2Fo Ti−1,j + Ti,j −1 = Ti,j + 4Bi · Fo · T∞ (3.322)(1 + 4Fo)Ti,j[238], (78)239PERIODIC CONDUCTION123456789101112131415161718192021222324252627282930313233343536373839404142434445direction explicit methods.
For a discussion of these methods as well as stabilityanalysis, the reader should consult Pletcher et al. (1988).3.9PERIODIC CONDUCTIONExamples of periodic conduction are the penetration of atmospheric temperaturecycles into the ground, heat transfer through the walls of internal combustion engines,and electronic components under cyclic operation. The periodicity may appear in thedifferential equation or in a boundary condition or both. The complete solution toa periodic heat conduction problem consists of a transient component that decaysto zero with time and a steady oscillatory component that persists. It is the steadyoscillatory component that is of prime interest in most engineering applications. Inthis section we present several important solutions.[239], (79)3.9.1 Cooling of a Lumped System in an Oscillating TemperatureEnvironmentLines: 3565 to 3609Revisit the lumped thermal capacity model described in Section 3.8.1 and considera scenario in which the convective environmental temperature T∞ oscillates sinusoidally, that is,———Normal Page(3.324) * PgEnds: EjectT∞ = T∞,m + a sin ωtwhere a is the amplitude of oscillation, ω = 2πf the angular frequency, f thefrequency in hertz, and T∞,m the mean temperature of the environment.The method of complex combination described by Arpaci (1966), Myers (1998),Poulikakos (1994), and Aziz and Lunardini (1994) gives the steady periodic solutionasθ= √11 + B2sin(Bτ − β)(3.325)whereθ=T − T∞,maB=ρV cωhAτ=hAtρV cβ = arctan B(3.326)A comparison of eq.
(3.325) with the dimensionless environmental temperature variation (sin Bτ) shows that the temperature of the body oscillates with the same frequency as that of the environment but with a phase lag of β. As the frequency ofoscillation increases, the phase angle β = arctan B increases, but the amplitude ofoscillation 1/(1 + B 2 )1/2 decreases.3.9.2Semi-infinite Solid with Periodic Surface TemperatureConsider the semi-infinite solid described in Section 3.8.2 and let the surface temperature be of the formBOOKCOMP, Inc.
— John Wiley & Sons / Page 239 / 2nd Proofs / Heat Transfer Handbook / Bejan———-5.78989pt PgVar[239], (79)240123456789101112131415161718192021222324252627282930313233343536373839404142434445CONDUCTION HEAT TRANSFERT (0, t) = Ts = Ti + a cos ωt(3.327)In this case, eqs. (3.291) are still applicable, although the initial condition of eq.(3.291a) becomes irrelevant for the steady periodic solution, which is ω 1/2 −[(ω/2α)1/2 x ]T (x, t) = Ti + aecos ωt −x(3.328)2αThree conclusions can be drawn from this result. First, the temperatures at all locations oscillate with the same frequency as the thermal disturbance at the surface.Second, the amplitude of oscillation decays exponentially with x. This makes the solution applicable to the finite thickness plane wall.
Third, the amplitude of oscillationdecays exponentially with the square root of the frequency ω. Thus, higher-frequencydisturbances damp out more rapidly than those at lower frequencies. This explainswhy daily oscillations of ambient temperature do not penetrate as deeply into theground as annual and millenial oscillations. The surface heat flux variation followsdirectly from eq. (3.328):q (0, t) = −k ω 1/2∂T (0, t)= ka∂xαcos ωt −π45.8702pt PgVar———Normal PagePgEnds: TEXand this shows that q (0, t) leads T (0, t) by π/4 radians.Semi-infinite Solid with Periodic Surface Heat Flux[240], (80)In this case, the boundary condition of eq. (3.327) is replaced withq (0, t) = −k∂T(0, t) = q0 cos ωt∂x(3.330)and the solution takes the form ω 1/2q0 α 1/2 −[ω/2α)1/2 x ]πT (x, t) = Ti +ex−cos ωt −k ω2α4(3.331)It is interesting to note that the phase angle increases as the depth x increases with theminimum phase angle of π/4 occurring at the surface (x = 0).
A practical situationin which eq. (3.331) becomes useful is in predicting the steady temperature variationsinduced by frictional heating between two reciprocating parts in contact in a machine.This application has been described by Poulikakos (1994).3.9.4Lines: 3609 to 3664———(3.329)3.9.3[240], (80)Semi-infinite Solid with Periodic Ambient TemperatureThe surface boundary condition in this case isT∞ = T∞,m + a cos ωtBOOKCOMP, Inc. — John Wiley & Sons / Page 240 / 2nd Proofs / Heat Transfer Handbook / Bejan(3.332)PERIODIC CONDUCTION123456789101112131415161718192021222324252627282930313233343536373839404142434445241and the steady periodic temperature distribution is given byθ=√Bi(Bi + 2Bi + 2)1/22e−πX√cos 2πτ − π X − β(3.333)whereT − T∞,ma ω 1/2X=x2παωtτ=2π h 2 1/2Bi =ωk αθ=β = arctan11 + Bi(3.334a)(3.334b)(3.334c)(3.334d)(3.334e)[241], (81)Lines: 3664 to 3745———1.74728ptNote that as h −→ ∞, Bi −→ ∞, and β −→ 0, eqs.
(3.333) reduce to eq. (3.328)———with T∞,m = Ti . The presence of the factor Bi/(Bi + 2Bi + 2)1/2 in eq. (3.333) showsNormal Pagethat convection enhances the damping effect and that it also increases the phase angle* PgEnds: Ejectby an amount β = arctan 1/(1 + Bi).3.9.5[241], (81)Finite Plane Wall with Periodic Surface TemperatureConsider a plane wall of thickness L with the face at x = 0 insulated and the face atx = L subjected to a periodic temperature change of the formT (L, t) = Ti + a cos ωt(3.335)where Ti is the initial temperature of the wall and the insulated boundary conditionat x = 0 gives∂T (0, t)=0∂x(3.336)The steady periodic solution is xxω 1/2ω 1/2T (x, t) = Ti + aφ1L cos ωt + φ2L(3.337),,L 2αL 2αwhere the numerical values of φ1 as a function of x/L and (w/2α)1/2 L are suppliedin Table 3.13, and φ2 , which is also a function of x/L and (w/2α)1/2 L, is given byφ2 = arctanBOOKCOMP, Inc.
— John Wiley & Sons / Page 241 / 2nd Proofs / Heat Transfer Handbook / Bejanφa φd − φb φcφa φb − φ c φd(3.338)PgVar242123456789101112131415161718192021222324252627282930313233343536373839404142434445CONDUCTION HEAT TRANSFERTABLE 3.13 Values of the Amplitude Decay Function, φ as a Function of x/Land σ = (ω/2α)1/2 Lσx/L = 00.1250.2500.3750.5000.6250.7500.8751.0000.00.51.01.52.04.08.0∞1.000.980.770.470.270.040.000.001.000.980.770.470.270.040.000.001.000.980.770.470.280.050.010.001.000.980.780.480.300.080.010.001.000.980.790.520.360.130.020.001.000.980.810.580.450.220.050.001.000.980.850.680.580.370.140.001.000.990.910.830.770.640.360.001.001.001.001.001.001.001.00[242], (82)whereφaφbφcφd ω 1/2ω 1/2= cosL coshL2α2α ω 1/2ω 1/2= cosx coshx2α2α ω 1/2ω 1/2= sinL sinhL2α2α ω 1/2ω 1/2= sinx sinhx2α2α(3.339a)(3.339b)(3.339c)(3.339d)Consider an infinitely long cylinder of inside radius, ri , extending to infinity in theradial direction.
The inner surface is subjected to a periodic temperature variation ofthe form(3.340)where Ti is the initial temperature of the cylinder. The equation governing the temperature distribution is∂ 2T1 ∂T1 ∂T=+∂r 2r ∂rα ∂t(3.341)The other boundary condition isT (∞, t) = Ti ,BOOKCOMP, Inc. — John Wiley & Sons / Page 242 / 2nd Proofs / Heat Transfer Handbook / Bejan∂T (∞, t)=0∂r———0.51117pt PgVar3.9.6 Infinitely Long Semi-infinite Hollow Cylinder with PeriodicSurface TemperatureT (ri , t) = Ti + a cos ωtLines: 3745 to 3783(3.342)———Normal Page* PgEnds: Eject[242], (82)CONDUCTION-CONTROLLED FREEZING AND MELTING123456789101112131415161718192021222324252627282930313233343536373839404142434445243and the initial condition isT (r, 0) = Ti(3.343)An application of the method of complex combination gives the steady-state periodicsolution asθ=T − Ti= Λ1 cos ωt − Λ2 sin ωta(3.344)where√√√√ω/α ri ker ω/α r + kei ω/α ri kei ω/α r√√Λ1 =ker 2 ω/α ri + kei2 ω/α ri√√√√ker ω/α ri kei ω/α r − kei ω/α ri ker ω/α r√√Λ2 =ker 2 ω/α ri + kei2 ω/α riker(3.345a)[243], (83)(3.345b)and ker and kei are the Thomson functions discussed in Section 3.3.5.As indicated in Section 3.9.1, the method of complex combination is described byArpaci (1966), Myers (1998), Poulikakos (1994), and Aziz and Lunardini (1994).
Themethod may be extended to numerous other periodic heat problems of engineeringinterest.3.10CONDUCTION-CONTROLLED FREEZING AND MELTINGHeat conduction with freezing (melting) occurs in a number of applications, such asice formation, permafrost melting, metal casting, food preservation, storage of latentenergy, and organ preservation and cryosurgery. Books and review articles on thesubject include those of Lunardini (1991), Cheng and Seki (1991), Rubinsky andEto (1990), Aziz and Lunardini (1993), Viskanta (1983, 1988), and Alexiades andSolomon (1993).
Because of the vastness of the literature, only selected results thatare judged to be of fundamental importance are discussed in this section.3.10.1One-Region Neumann ProblemThe one-region Neumann problem deals with a semi-infinite region of liquid initiallyat its freezing temperature, Tf .
At time t > 0, the face at x = 0 is suddenly reducedand kept at T0 such that T0 < Tf , as shown in Fig. 3.39. This initiates the extractionof heat by conduction from the saturated liquid to the surface and the liquid begins tofreeze.
As the cooling continues, the interface (assumed sharp) between the solid andliquid phases penetrates deeper into the liquid region. The prediction of the locationof the interface calls for determination of the one-dimensional transient temperaturedistribution in the solid assuming that the liquid continues to remain at Tf at all times.The governing partial differential equation isBOOKCOMP, Inc. — John Wiley & Sons / Page 243 / 2nd Proofs / Heat Transfer Handbook / BejanLines: 3783 to 3836———-2.0078pt PgVar———Normal PagePgEnds: TEX[243], (83)244123456789101112131415161718192021222324252627282930313233343536373839404142434445CONDUCTION HEAT TRANSFER[244], (84)Figure 3.39 One-region Neumann problem (freezing).Lines: 3836 to 3897———2.97874pt PgVar∂ 2T1 ∂T=2∂xα ∂t(3.346)T (0, t) = T0(3.347a)T (xf , t) = Tf(3.347b)with the boundary conditionswhere xf denotes the location of the interface, which is not known a priori and mustbe determined as part of the solution.
An energy balance at the interface gives∂xf∂T k= ρL∂x x=xfdt(3.348)where k and ρ are the thermal conductivity and density of the solid phase, respectively,and L is the latent heat.The temperature distribution in the solid is given as√ erf x/2 αtTf − T=1−√ Tf − T0erf xf /2 αt(3.349)√where xf /2 αt, denoted by λ, is a root of the transcendental equation 2√π λ erf λeλ = StBOOKCOMP, Inc.
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