Adrian Bejan(Editor), Allan D. Kraus (Editor). Heat transfer Handbok (776115), страница 28
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— John Wiley & Sons / Page 244 / 2nd Proofs / Heat Transfer Handbook / Bejan(3.350)———Short PagePgEnds: TEX[244], (84)CONDUCTION-CONTROLLED FREEZING AND MELTING123456789101112131415161718192021222324252627282930313233343536373839404142434445245TABLE 3.14 InterfaceLocation Parameter λλSt0.00.20.40.60.81.01.20.00000.08220.35640.92051.99564.06018.1720and[245], (85)St =c(Tf − T0 )L(3.351)Lines: 3897 to 3930is the Stefan number, the ratio of the sensible heat to the latent heat. For water, Stis about 0.10, for paraffin wax about 0.90, for copper about 2.64, and for silicondioxide about 436.
Table 3.14 gives selected values of λ and St that satisfy eq. (3.350).Viskanta (1983) reports that the Neumann model accurately predicts the solidificationof n-octadecane on a horizontal plate.The solution presented here applies to the one-region melting problem if Tf isreplaced by the melting temperature Tm . With T0 > Tm , eq. (3.349) gives the temperature in the liquid region.3.10.2 Two-Region Neumann ProblemThe two-region Neumann problem allows for heat conduction in both the solid andliquid phases. For the configuration in Fig. 3.40, the mathematical description of theproblem is∂ 2 Ts1 ∂Ts=∂x 2αs ∂t(0 < x < xf and t > 0)(3.352)∂ 2 Tl1 ∂Tl=2∂xαl ∂t(xf < x < ∞ and t > 0)(3.353)with initial and boundary conditionsTl (x, 0) = Ti(3.354a)Ts (0, t) = T0(3.354b)Ts (xf , t) = Tl (xf , t) = TfBOOKCOMP, Inc.
— John Wiley & Sons / Page 245 / 2nd Proofs / Heat Transfer Handbook / Bejan(3.354c)———-2.90575pt PgVar———Short PagePgEnds: TEX[245], (85)246123456789101112131415161718192021222324252627282930313233343536373839404142434445CONDUCTION HEAT TRANSFERks∂Ts∂Tl− kl∂x∂x= ρLx=xfdxfdt(3.354d)where the subscripts s and l refer to the solid and liquid phases, respectively.The solutions for Ts and Tl are√Ts − T0erf(x/2 αs t)=(3.355)√Tf − T0erf(xf /2 αs t)√erfc(x/2 αl t)Ti − Tl=(3.356)√Ti − Tferfc(xf /2 αs t)√With xf /2 αs t denoted by λ, the interface energy balance given by eq. (3.354d)leads to the transcendental equation for λ:Ti − Tfe−λ−λ erf(λ) Tf − T02(kρc)l(kρc)s1/2√2e−λ (αs /αl )πL=1/2c(Tf − T0 )λ erfc λ(αs /αl )(3.357)Churchill and Evans (1971) noted that λ is a function of three parameters:Ti − Tfθ =Tf − T0∗(kρc)l(kρc)s1/2c(Tf − T0 )St =Lαsα =αl∗and solved eq.
(3.357) for a range of values of these parameters. Table 3.15 summarizes these results for λ.SolidTlks , ␣s, LTsLiquidTfkl , ␣lT00xxfFigure 3.40 Two-region Neumann problem (freezing).BOOKCOMP, Inc. — John Wiley & Sons / Page 246 / 2nd Proofs / Heat Transfer Handbook / Bejan[246], (86)Lines: 3930 to 4001———1.61536pt PgVar———Normal PagePgEnds: TEX[246], (86)CONDUCTION-CONTROLLED FREEZING AND MELTING123456789101112131415161718192021222324252627282930313233343536373839404142434445TABLE 3.15247Values of λθ∗Stα∗0.501.01.52.03.05.010.00.10.10.10.20.20.20.50.50.51.01.01.02.02.02.02.01.51.02.01.51.02.01.51.02.01.51.02.01.51.00.2020.2030.2040.2700.2710.2730.3740.3780.3830.4520.4600.4700.5170.5300.5460.1870.1880.1890.2410.2430.2450.3130.3200.3250.3580.3670.3780.3910.4030.4180.1730.1750.1760.2160.2190.2220.2680.2730.2800.2980.3050.3150.3170.3270.3390.1610.1630.1640.1950.1990.2020.2340.2400.2460.2540.2610.2700.2670.2750.2860.1410.1420.1440.1650.1670.1700.1860.1920.1970.1980.2030.2090.2040.2100.2170.1110.1130.1140.1230.1250.1280.1330.1350.1390.1370.1400.1440.1390.1420.1460.0710.0720.0730.0740.0750.0760.0770.0780.0790.0770.0790.0800.0780.0790.080[247], (87)Lines: 4001 to 4021———0.53903pt PgVar———Normal Page* PgEnds: Eject3.10.3Other Exact Solutions for Planar FreezingBesides the one- and two-region Neumann solutions, several other exact solutions forplanar freezing problems are available.
These include:1. The two-region problem with different solid- and liquid-phase densities (Lunardini, 1991)2. The two-region problem with phase change occurring over a temperature range(Cho and Sunderland, 1969)3. The one-region problem with a mushy zone separating the pure solid and pureliquid phases (Solomon et al., 1982)4. The two-region problem with temperature-dependent thermal conductivities ksand kl (Cho and Sunderland, 1974)5. The two-region problem with arbitrary surface temperature and initial conditions (Tao, 1978)3.10.4Exact Solutions in Cylindrical FreezingCarslaw and Jaeger (1959) give an exact solution for the freezing of a subcooled liquidwhile the solid phase remains at the freezing temperature.
The latent heat released isused to bring the subcooled liquid to its freezing temperature. The process is describedby the differential equationBOOKCOMP, Inc. — John Wiley & Sons / Page 247 / 2nd Proofs / Heat Transfer Handbook / Bejan[247], (87)248123456789101112131415161718192021222324252627282930313233343536373839404142434445CONDUCTION HEAT TRANSFER1 ∂r ∂rr∂T∂r=1 ∂Tα ∂t(r < rf )(3.358)and the initial boundary conditionsT (rf , t) = Tf(3.359a)lim T (r, t) = T0(3.359b)T (r, 0) = T0(3.359c)r−→∞where rf represents the radial growth of the solid phase and T0 < Tf is the subcooledliquid temperature.The solution of eq. (3.358) satisfying the conditions of eqs.
(3.359) is Ei − r 2 /4αtT = T0 + Tf − T0(3.360)Ei − rf2 /4αt[248], (88)Lines: 4021 to 4099———where λ = rf2 /4αt is given by 2λ2 · Ei − λ2 eλ + St = 00.51517pt PgVar———Normal PageIn eqs. (3.360) and (3.361), Ei is the exponential integral function discussed in Section * PgEnds: Eject3.3.4, and in eq. (3.361), St is the Stefan number. Table 3.16 provides the solution ofeq. (3.361).[248], (88)Another situation for which an exact solution is available is shown in Fig.
3.41.A line heat sink of strength Qs (W/m) located at r = 0 and activated at time t = 0causes the infinite extent of liquid at a uniform temperature Ti (Ti > Tf ) to freeze.The interface grows radially outward. The mathematical formulation for the solid andliquid phases leads to1 ∂∂Ts1 ∂Ts(3.362)r=(0 < r < rf )r ∂r∂rαs ∂t1 ∂∂Tl1 ∂Tlr=(rf < r < ∞)(3.363)r ∂r∂rαl ∂t(3.361)with initial and boundary conditionsTl (∞, t) = Ti(3.364a)Tl (r, 0) = Ti(3.364b)Ts (rf , t) = Tl (rf , t) = Tfdrf∂Ts∂Tl= ρL− klks∂r∂r r=rfdtBOOKCOMP, Inc. — John Wiley & Sons / Page 248 / 2nd Proofs / Heat Transfer Handbook / Bejan(3.364c)(3.364d)CONDUCTION-CONTROLLED FREEZING AND MELTING123456789101112131415161718192021222324252627282930313233343536373839404142434445249TABLE 3.16 Stefan Numberand Interface Location ParameterStλ0.10.20.30.40.50.60.70.80.18460.31430.44910.60060.78111.00951.32371.8180[249], (89)Ozisik (1993) gives the solution as rf2r2QsEi −Ts = Tf +− Ei −4πks4αs t4αs tTi − Tfr2 Ei −Tl = Ti − 4αl tEi − rf2 /4αl t(0 < r < rf )(3.365)———10.1232pt PgVar(rf < r < ∞)(3.366)√where λ = rf /2 αs t is obtained from the transcendental equationkl (Ti − Tf ) −λ2 αs /αlQ s − λ2+= λ2 αs ρLee4πEi(−λ2 αs /αl )(3.367)The solution presented by eqs.
(3.365) through (3.367) has been extended byOzisik and Uzzell (1979) for a liquid with an extended freezing temperature.Figure 3.41 Cylindrical freezing due to a line strength of fixed strength.BOOKCOMP, Inc. — John Wiley & Sons / Page 249 / 2nd Proofs / Heat Transfer Handbook / BejanLines: 4099 to 4132———Normal PagePgEnds: TEX[249], (89)2501234567891011121314151617181920212223242526272829303132333435363738394041424344453.10.5CONDUCTION HEAT TRANSFERApproximate Analytical SolutionsBecause of the mathematical complexity and the restrictive nature of exact analytical solutions, several approaches have been employed to generate approximate analytical solutions that provide rapid results in a number of practical situations. Themethods used are the quasi-steady solution, the heat balance integral approach ofGoodman (1958) and Lunardini (1991), and the perturbation method of Aziz andNa (1984), Aziz and Lunardini (1993), and others.
A collection of such solutions isprovided next.One-Region Neumann Problemfor this problem isThe quasi-steady-state solution where St = 0T = T0 + (Tf − T0 )xf2 =xxf2k(T0 − Tf )tρLOne-Region Neumann Problem with Surface Convectioncooling, the boundary condition of eq. (3.347a) is replaced by∂T k= h [T (0, t) − T∞ ]∂x x=0(3.368)(3.369)With convectiveh(Tf − T∞ )(x − xf )k + hxfρLxfhxft=1+h(Tf − T∞ )2k(3.371)(3.372)as the results.Outward Cylindrical Freezing Consider a saturated liquid at the freezing temperature Tf , surrounding a cylinder of radius r0 whose outer surface is kept at a subfreezing temperature, T0 < Tf .
If a quasi-steady-state assumption (St = 0) is used,the solution isTf − T0rlnln(rf /r0 ) r0ρL1 21 2 rft=rf ln −rf − r02k(Tf − T0 ) 2r04T = T0 +BOOKCOMP, Inc. — John Wiley & Sons / Page 250 / 2nd Proofs / Heat Transfer Handbook / BejanLines: 4132 to 4174———3.06346pt PgVar———Short Page(3.370) * PgEnds: Ejectwhere T∞ is now the coolant temperature. The quasi-steady-state approach withSt = 0 yieldsT = Tf +[250], (90)(3.373)(3.374)[250], (90)CONDUCTION-CONTROLLED FREEZING AND MELTING123456789101112131415161718192021222324252627282930313233343536373839404142434445251An improvement on the quasi-steady-state solution can be achieved with the regular perturbation analysis provided by Aziz and Na (1984).
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