The CRC Handbook of Mechanical Engineering. Chapter 4. Heat and Mass Transfer (776127), страница 26
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In fact, in some cases,especially with a convective boundary condition, they may produce incorrect results. It is thus necessaryto examine the physical viability of the results, such as overall energy balances, when using theseapproximations.© 1999 by CRC Press LLC4-112Section 4All of the examples are for melting, but freezing problems have the same solutions when the propertiesare taken to be those of the solid and hsl is replaced everywhere by –hsl.
It is assumed here that theproblems are one-dimensional, and that the material is initially at the fusion temperature Tf .Examples of the Quasi-Static Approximation for Cartesian Coordinate Geometries. Given a semi-infinite solid (Figure 4.4.16), on which a time-dependent temperature T0(t) > Tf is imposed at x = 0, theabove-described quasi-static approximation of Equations (4.4.42) to (4.4.46) easily yields the solutioné kX (t ) = ê2 lë rhsltò[0[Tl ( x, t ) = T0 (t ) - T0 (t ) - TfùT0 (t ) - Tf dt úû]] Xx(t)12for t ³ 0in 0 £ x £ X (t ) for t ³ 0(4.4.54)(4.4.55)The heat flux needed for melting, q(x,t), can easily be determined from the temperature distribution inthe liquid (Equation 4.4.55), which is linear because of the steady-state form of the heat conductionequation in this quasi-static approximation, so thatq( x, t ) = - klT0 (t ) - TfdTl ( x, t )= kldxX (t )(4.4.56)For comparison of this approximate solution to the exact one (Equations (4.4.47) and (4.4.50)),consider the case where T0(t) = T0 = constant.
Rearranging to use the Stefan number, Equations (4.4.54)and (4.4.55) become12X (t ) = 2(Ste l 2)[T ( x , t ) = T - [T - T ](Stel12x 2(a l t )0012f2)l12(a t )]for t > 0in 0 £ x £ X (t ) for t ³ 0(4.4.57)(4.4.58)It is easy to show that l¢ in the exact solution (Equation 4.4.48) approaches the value (Stel/2)1/2 whenStel ® 0, and that otherwise l¢ < (Stel/2)1/2. The approximate solution is therefore indeed equal to theexact one when Stel ® 0, and it otherwise overestimates the values of both X(t) and T(x,t).
While theerrors depend on the specific problem, they are confined to about 10% in the above-described case(Alexiades and Solomon, 1993).For the same melting problem but with the boundary condition of an imposed time-dependent heatflux q0(t),æ dT ö- kl ç l ÷ = q0 (t ) for t > 0è dx ø 0,t(4.4.59)the quasi-static approximate solution isX (t ) º© 1999 by CRC Press LLC1rhsltò q (t) dt00for t > 0(4.4.60)4-113Heat and Mass TransferTl ( x, t ) = Tf +q0klùé q0t - x ú in 0 £ x £ X (t ) for t > 0êrhûë sl(4.4.61)For the same case if the boundary condition is a convective heat flux from an ambient fluid at thetransient temperature Ta(t), characterized by a heat transfer coefficient h,æ dT ö- kl ç l ÷ = h Ta (t ) - Tl (0, t )è dx ø 0,t[]for t ³ 0(4.4.62)the quasi-static approximate solution isX (t ) = -2kl ìïæ kl ök+ íç ÷ + 2 lh ïîè h ørhsl[Tl ( x, t ) = Tf (t ) Ta (t ) - Tf12tò[0üïTa (t ) - T f dt ýïþ] h[X(t) + k ]h X (t ) - x]for t ³ 0in 0 £ x £ X (t ) for t > 0(4.4.63)(4.4.64)lExamples of the Quasi-Static Approximation for Cylindrical Coordinate Geometries.
It is assumed inthese examples that the cylinders are very long and that the problems are axisymmetric. Just as in theCartesian coordinate case, the energy equation (4.4.51) is reduced by the approximation to its steadystate form. HereTl (ri , t ) = T0 (t ) > Tf(4.4.65)for t > 0Consider the outward-directed melting of a hollow cylinder due to a temperature imposed at theinternal radius ri. The solution is[Tl (r, t ) = Tf + T0 (t ) - Tf] ln[[rln r R(t )]i]R(t )in ri £ r £ R(t ) for t > 0(4.4.66)and the transient position of the phase front, R(t), can be calculated from the transcendental equation22 R(t ) ln4kR(t )2= R(t ) - ri2 + lrhslritò [T (t) - T ] dt00f(4.4.67)If the melting for the same case occurs due to the imposition of a heat flux q0 at r pæ dT ö- kl ç l ÷ = q0 (t ) > 0 for t > 0è dx ø r ,t(4.4.68)ithe solution isTl (r, t ) = Tf -© 1999 by CRC Press LLCq0 (t ) rirlnin ri £ r £ R(t ) for t > 0klR(t )(4.4.69)4-114Section 4ærR(t ) = ç ri2 + 2 irhslèòt0öq0 (t ) dt÷ø12(4.4.70)for t > 0If the melting for the same case occurs due to the imposition of a convective heat flux from a fluidat the transient temperature Ta(t), with a heat transfer coefficient h, at riæ dT ö- kl ç l ÷ = h Ta (t ) - T f (ri , t ) > 0 for t > 0è dr ø r ,t[](4.4.71)iThe solution is[Tl (r, t ) = Tf + Ta (t ) - Tf] ln[riln[r R(t )]]R(t ) - kl h riin ri £ r £ R(t ) at t > 0(4.4.72)with R(t) calculated from the transcendental equation22 R(t ) ln2k ö4kR(t ) æ2= ç1 - l ÷ R(t ) - ri2 + lrhslrih ri øè[]tò [T (t) - T ] dt0af(4.4.73)The solutions for inward melting of a cylinder, where heating is applied at the outer radius ro, are thesame as the above-described ones for the outward-melting cylinder, if the replacements ri ® ro, q0 ® –q0,and h ® – h are made.
If such a cylinder is not hollow, then ri = 0 is used.Estimation of Freezing and Melting TimeThere are a number of approximate formulas for estimating the freezing and melting times of differentmaterials having a variety of shapes. The American Society of Heating, Refrigerating, and Air-Conditioning Engineers (ASHRAE) provides a number of such approximations for estimating the freezingand thawing times of foods (ASHRAE, 1993). For example, if it can be assumed that the freezing orthawing occurs at a single temperature, the time to freeze or thaw, tf , for a body that has shape parametersP and R (described below) and thermal conductivity k, initially at the fusion temperature Tf , and whichis exchanging heat via heat transfer coefficient h with an ambient at the constant Ta, can be approximatedby Plank’s equationtf =æ Pd Rd 2 ö+ç÷k øTf - Ta è hhsl r(4.4.74)where d is the diameter of the body if it is a cylinder or a sphere, or the thickness when it is an infiniteslab, and where the shape coefficients P and R for a number of body forms are given in Table 4.4.2.Shape coefficients for other body forms are also available.
To use Equation 4.4.74 for freezing, k and rshould be the values for the food in its frozen state. In thawing, they should be for the unfrozen food.Other simple approximations for melting and thawing times can be found in Cleland et al. (1987).Example of Using Plank’s Equation (4.4.74) for Estimating Freezing Time. Estimate the time neededto freeze a fish, the shape of which can be approximated by a cylinder 0.5 m long having a diameter of0.1 m.
The fish is initially at its freezing temperature, and during the freezing process it is surroundedby air at Ta = –25°C, with the cooling performed with a convective heat transfer coefficient h = 68W/m2 K. For the fish, Tf = –1°C, hsl = 200 kJ/kg, rs = 992 kg/m3, and sks = 1.35 W/m K.© 1999 by CRC Press LLC4-115Heat and Mass TransferTABLE 4.4.2Shape Factors for Equation (4.4.74)FormsPRSlabCylinderSphere1/21/41/61/81/161/24From ASHRAE, in Fundamentals, ASHRAE, Atlanta,1993, chap. 29.
With permission.By using Table 4.4.2, the geometric coefficients for the cylindrical shape of the fish are P = 1/2 andR = 1/16, while d is the cylinder diameter, = 0.1 m. Substituting these values into Equation (4.4.74) givestf =2200, 000 × 992 æ 1 4 (0.1) 1 16 (0.1) ö= 6866 sec = 1.9 hr+ç1.35 ÷ø-1 - (-25) è 68In fact, freezing or melting of food typically takes place over a range of temperatures, and approximatePlank-type formulas have been developed for various specific foodstuffs and shapes to represent realitymore closely than Equation (4.4.74) (ASHRAE, 1993).Alexiades and Solomon (1993) provide several easily computable approximate equations for estimating the time needed to melt a simple solid body initially at the fusion temperature Tf.
It is assumed thatconduction occurs in one phase (the liquid) only, that the problems are axi- and spherically symmetricfor cylindrical and spherical bodies, respectively, and that the melting process for differently shapedbodies can be characterized by a single geometric parameter, r, in the body domain 0 £ r £ L, using ashape factor, w, defined byw=LA-1V(4.4.75)where A is the surface area across which the heat is transferred into the body and V is the body volume,to account for the specific body shape:0 for a slab insulated at one endw = 1 for a cylinder2 for a sphere(4.4.76)0 £ w £ 2 always, and w may be assigned appropriate values for shapes intermediate between the slab,cylinder, and sphere.
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