Heat Transfer Handbook part 26 ppt

BOOKCOMP, Inc. — John Wiley & Sons / Page 241 / 2nd Proofs / Heat Transfer Handbook / Bejan PERIODIC CONDUCTION 241 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 [241], (81) Lines: 3664 to 3745 ——— 1.74728pt PgVar ——— Normal Page * PgEnds: Eject [241], (81) and the steady periodic temperature distribution is given by θ = Bi (Bi 2 + 2Bi + 2) 1/2 e − √ πX cos  2πτ − √ π X − β  (3.333) where θ = T − T ∞,m a (3.334a) X =  ω 2πα  1/2 x (3.334b) τ = ωt 2π (3.334c) Bi = h k  2 α  1/2 ω (3.334d) β = arctan 1 1 + Bi (3.334e) Note that as h −→ ∞, Bi −→ ∞, and β −→ 0, eqs. (3.333) reduce to eq. (3.328) with T ∞,m = T i . The presence of the factor Bi/(Bi+2Bi+2) 1/2 in eq. (3.333) shows that convection enhances the damping effect and that it also increases the phase angle by an amount β = arctan 1/(1 +Bi). 3.9.5 Finite Plane Wall with Periodic Surface Temperature Consider a plane wall of thickness L with the face at x = 0 insulated and the face at x = L subjected to a periodic temperature change of the form T(L,t)= T i + a cos ωt (3.335) where T i is the initial temperature of the wall and the insulated boundary condition at x = 0gives ∂T(0,t) ∂x = 0 (3.336) The steady periodic solution is T(x,t)= T i + aφ 1  x L ,  ω 2α  1/2 L  cos  ωt + φ 2  x L ,  ω 2α  1/2 L  (3.337) where the numerical values of φ 1 as a function of x/L and (w/2α) 1/2 L are supplied in Table 3.13, and φ 2 , which is also a function of x/L and (w/2α) 1/2 L,isgivenby φ 2 = arctan φ a φ d − φ b φ c φ a φ b − φ c φ d (3.338) BOOKCOMP, Inc. — John Wiley & Sons / Page 242 / 2nd Proofs / Heat Transfer Handbook / Bejan 242 CONDUCTION HEAT TRANSFER 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 [242], (82) Lines: 3745 to 3783 ——— 0.51117pt PgVar ——— Normal Page * PgEnds: Eject [242], (82) TABLE 3.13 Values of the Amplitude Decay Function, φ as a Function of x/L and σ = (ω/2α) 1/2 L σ x/L = 0 0.125 0.250 0.375 0.500 0.625 0.750 0.875 1.000 0.0 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 0.5 0.98 0.98 0.98 0.98 0.98 0.98 0.98 0.99 1.00 1.0 0.77 0.77 0.77 0.78 0.79 0.81 0.85 0.91 1.00 1.5 0.47 0.47 0.47 0.48 0.52 0.58 0.68 0.83 1.00 2.0 0.27 0.27 0.28 0.30 0.36 0.45 0.58 0.77 1.00 4.0 0.04 0.04 0.05 0.08 0.13 0.22 0.37 0.64 1.00 8.0 0.00 0.00 0.01 0.01 0.02 0.05 0.14 0.36 1.00 ∞ 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 where φ a = cos   ω 2α  1/2 L  cosh   ω 2α  1/2 L  (3.339a) φ b = cos   ω 2α  1/2 x  cosh   ω 2α  1/2 x  (3.339b) φ c = sin   ω 2α  1/2 L  sinh   ω 2α  1/2 L  (3.339c) φ d = sin   ω 2α  1/2 x  sinh   ω 2α  1/2 x  (3.339d) 3.9.6 Infinitely Long Semi-infinite Hollow Cylinder with Periodic Surface Temperature Consider an infinitely long cylinder of inside radius, r i , extending to infinity in the radial direction. The inner surface is subjected to a periodic temperature variation of the form T(r i ,t)= T i + a cos ωt (3.340) where T i is the initial temperature of the cylinder. The equation governing the tem- perature distribution is ∂ 2 T ∂r 2 + 1 r ∂T ∂r = 1 α ∂T ∂t (3.341) The other boundary condition is T(∞,t) = T i , ∂T(∞,t) ∂r = 0 (3.342) BOOKCOMP, Inc. — John Wiley & Sons / Page 243 / 2nd Proofs / Heat Transfer Handbook / Bejan CONDUCTION-CONTROLLED FREEZING AND MELTING 243 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 [243], (83) Lines: 3783 to 3836 ——— -2.0078pt PgVar ——— Normal Page PgEnds: T E X [243], (83) and the initial condition is T(r,0) = T i (3.343) An application of the method of complex combination gives the steady-state periodic solution as θ = T − T i a = Λ 1 cos ωt − Λ 2 sin ωt (3.344) where Λ 1 = ker  √ ω/α r i  ker  √ ω/α r  + kei  √ ω/α r i  kei  √ ω/α r  ker 2  √ ω/α r i  + kei 2  √ ω/α r i  (3.345a) Λ 2 = ker  √ ω/α r i  kei  √ ω/α r  − kei  √ ω/α r i  ker  √ ω/α r  ker 2  √ ω/α r i  + kei 2  √ ω/α r i  (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 by Arpaci (1966), Myers (1998), Poulikakos (1994), and Aziz and Lunardini (1994). The method may be extended to numerous other periodic heat problems of engineering interest. 3.10 CONDUCTION-CONTROLLED FREEZING AND MELTING Heat conduction with freezing (melting) occurs in a number of applications, such as ice formation, permafrost melting, metal casting, food preservation, storage of latent energy, and organ preservation and cryosurgery. Books and review articles on the subject include those of Lunardini (1991), Cheng and Seki (1991), Rubinsky and Eto (1990), Aziz and Lunardini (1993), Viskanta (1983, 1988), and Alexiades and Solomon (1993). Because of the vastness of the literature, only selected results that are judged to be of fundamental importance are discussed in this section. 3.10.1 One-Region Neumann Problem The one-region Neumann problem deals with a semi-infinite region of liquid initially at its freezing temperature, T f . At time t>0, the face at x = 0 is suddenly reduced and kept at T 0 such that T 0 <T f , as shown in Fig. 3.39. This initiates the extraction of heat by conduction from the saturated liquid to the surface and the liquid begins to freeze. As the cooling continues, the interface (assumed sharp) between the solid and liquid phases penetrates deeper into the liquid region. The prediction of the location of the interface calls for determination of the one-dimensional transient temperature distribution in the solid assuming that the liquid continues to remain at T f at all times. The governing partial differential equation is BOOKCOMP, Inc. — John Wiley & Sons / Page 244 / 2nd Proofs / Heat Transfer Handbook / Bejan 244 CONDUCTION HEAT TRANSFER 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 [244], (84) Lines: 3836 to 3897 ——— 2.97874pt PgVar ——— Short Page PgEnds: T E X [244], (84) Figure 3.39 One-region Neumann problem (freezing). ∂ 2 T ∂x 2 = 1 α ∂T ∂t (3.346) with the boundary conditions T(0,t) = T 0 (3.347a) T(x f ,t)= T f (3.347b) where x f denotes the location of the interface, which is not known a priori and must be determined as part of the solution. An energy balance at the interface gives k ∂T ∂x     x=x f = ρL ∂x f dt (3.348) where k andρ are the thermal conductivity and density ofthesolidphase,respectively, and L is the latent heat. The temperature distribution in the solid is given as T f − T T f − T 0 = 1 − erf  x/2 √ αt  erf  x f /2 √ αt  (3.349) where x f /2 √ αt, denoted by λ, is a root of the transcendental equation √ πλerf  λe λ 2  = St (3.350) BOOKCOMP, Inc. — John Wiley & Sons / Page 245 / 2nd Proofs / Heat Transfer Handbook / Bejan CONDUCTION-CONTROLLED FREEZING AND MELTING 245 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 [245], (85) Lines: 3897 to 3930 ——— -2.90575pt PgVar ——— Short Page PgEnds: T E X [245], (85) TABLE 3.14 Interface Location Parameter λ λ St 0.0 0.0000 0.2 0.0822 0.4 0.3564 0.6 0.9205 0.8 1.9956 1.0 4.0601 1.2 8.1720 and St = c(T f − T 0 ) L (3.351) is the Stefan number, the ratio of the sensible heat to the latent heat. For water, St is about 0.10, for paraffin wax about 0.90, for copper about 2.64, and for silicon dioxide 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 solidification of n-octadecane on a horizontal plate. The solution presented here applies to the one-region melting problem if T f is replaced by the melting temperature T m . With T 0 >T m , eq. (3.349) gives the temper- ature in the liquid region. 3.10.2 Two-Region Neumann Problem The two-region Neumann problem allows for heat conduction in both the solid and liquid phases. For the configuration in Fig. 3.40, the mathematical description of the problem is ∂ 2 T s ∂x 2 = 1 α s ∂T s ∂t (0 <x<x f and t>0) (3.352) ∂ 2 T l ∂x 2 = 1 α l ∂T l ∂t (x f <x<∞ and t>0) (3.353) with initial and boundary conditions T l (x, 0) = T i (3.354a) T s (0,t) = T 0 (3.354b) T s (x f ,t)= T l (x f ,t)= T f (3.354c) BOOKCOMP, Inc. — John Wiley & Sons / Page 246 / 2nd Proofs / Heat Transfer Handbook / Bejan 246 CONDUCTION HEAT TRANSFER 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 [246], (86) Lines: 3930 to 4001 ——— 1.61536pt PgVar ——— Normal Page PgEnds: T E X [246], (86)  k s ∂T s ∂x − k l ∂T l ∂x  x=x f = ρL dx f dt (3.354d) where the subscripts s and l refer to the solid and liquid phases, respectively. The solutions for T s and T l are T s − T 0 T f − T 0 = erf(x/2 √ α s t) erf(x f /2 √ α s t) (3.355) T i − T l T i − T f = erfc(x/2 √ α l t) erfc(x f /2 √ α s t) (3.356) With x f /2 √ α s t denoted by λ, the interface energy balance given by eq. (3.354d) leads to the transcendental equation for λ: e −λ 2 λ erf(λ) − T i − T f T f − T 0  (kρc) l (kρc) s  1/2 e −λ 2 (α s /α l ) λ erfc  λ(α s /α l ) 1/2  = √ π L c(T f − T 0 ) (3.357) Churchill and Evans (1971) noted that λ is a function of three parameters: θ ∗ = T i − T f T f − T 0  (kρc) l (kρc) s  1/2 α ∗ = α s α l St = c(T f − T 0 ) L and solved eq. (3.357) for a range of values of these parameters. Table 3.15 summa- rizes these results for λ. Solid Liquid T f T l x f T 0 T s 0 x k ss ,␣ k ll , ␣ ␳,L Figure 3.40 Two-region Neumann problem (freezing). BOOKCOMP, Inc. — John Wiley & Sons / Page 247 / 2nd Proofs / Heat Transfer Handbook / Bejan CONDUCTION-CONTROLLED FREEZING AND MELTING 247 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 [247], (87) Lines: 4001 to 4021 ——— 0.53903pt PgVar ——— Normal Page * PgEnds: Eject [247], (87) TABLE 3.15 Values of λ θ ∗ St α ∗ 0.50 1.0 1.5 2.0 3.0 5.0 10.0 0.1 2.0 0.202 0.187 0.173 0.161 0.141 0.111 0.071 0.1 1.5 0.203 0.188 0.175 0.163 0.142 0.113 0.072 0.1 1.0 0.204 0.189 0.176 0.164 0.144 0.114 0.073 0.2 2.0 0.270 0.241 0.216 0.195 0.165 0.123 0.074 0.2 1.5 0.271 0.243 0.219 0.199 0.167 0.125 0.075 0.2 1.0 0.273 0.245 0.222 0.202 0.170 0.128 0.076 0.5 2.0 0.374 0.313 0.268 0.234 0.186 0.133 0.077 0.5 1.5 0.378 0.320 0.273 0.240 0.192 0.135 0.078 0.5 1.0 0.383 0.325 0.280 0.246 0.197 0.139 0.079 1.0 2.0 0.452 0.358 0.298 0.254 0.198 0.137 0.077 1.0 1.5 0.460 0.367 0.305 0.261 0.203 0.140 0.079 1.0 1.0 0.470 0.378 0.315 0.270 0.209 0.144 0.080 2.0 2.0 0.517 0.391 0.317 0.267 0.204 0.139 0.078 2.0 1.5 0.530 0.403 0.327 0.275 0.210 0.142 0.079 2.0 1.0 0.546 0.418 0.339 0.286 0.217 0.146 0.080 3.10.3 Other Exact Solutions for Planar Freezing Besides the one- and two-region Neumann solutions, several other exact solutions for planar freezing problems are available. These include: 1. The two-region problem with different solid- and liquid-phase densities (Lu- nardini, 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 pure liquid phases (Solomon et al., 1982) 4. The two-region problem with temperature-dependent thermal conductivities k s and k l (Cho and Sunderland, 1974) 5. The two-region problem with arbitrary surface temperature and initial condi- tions (Tao, 1978) 3.10.4 Exact Solutions in Cylindrical Freezing Carslaw and Jaeger (1959) give an exact solution for the freezing of a subcooled liquid while the solid phase remains at the freezing temperature. The latent heat released is used to bring the subcooled liquidtoits freezing temperature. The process is described by the differential equation BOOKCOMP, Inc. — John Wiley & Sons / Page 248 / 2nd Proofs / Heat Transfer Handbook / Bejan 248 CONDUCTION HEAT TRANSFER 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 [248], (88) Lines: 4021 to 4099 ——— 0.51517pt PgVar ——— Normal Page * PgEnds: Eject [248], (88) 1 r ∂ ∂r  r ∂T ∂r  = 1 α ∂T ∂t (r < r f ) (3.358) and the initial boundary conditions T(r f ,t)= T f (3.359a) lim r−→∞ T(r,t) = T 0 (3.359b) T(r,0) = T 0 (3.359c) where r f represents the radial growth of the solid phase and T 0 <T f is the subcooled liquid temperature. The solution of eq. (3.358) satisfying the conditions of eqs. (3.359) is T = T 0 +  T f − T 0  Ei  − r 2 /4αt  Ei  − r 2 f /4αt  (3.360) where λ = r 2 f /4αt is given by λ 2 · Ei  − λ 2  e λ 2 + St = 0 (3.361) In eqs. (3.360) and (3.361), Ei is the exponential integral function discussed in Section 3.3.4, and in eq. (3.361), St is the Stefan number. Table 3.16 provides the solution of eq. (3.361). Another situation for which an exact solution is available is shown in Fig. 3.41. A line heat sink of strength Q s (W/m) located at r = 0 and activated at time t = 0 causes the infinite extent of liquid at a uniform temperature T i (T i >T f ) to freeze. The interface grows radially outward. The mathematical formulation for the solid and liquid phases leads to 1 r ∂ ∂r  r ∂T s ∂r  = 1 α s ∂T s ∂t (0 <r<r f ) (3.362) 1 r ∂ ∂r  r ∂T l ∂r  = 1 α l ∂T l ∂t (r f <r<∞) (3.363) with initial and boundary conditions T l (∞,t)= T i (3.364a) T l (r, 0) = T i (3.364b) T s (r f ,t)= T l (r f ,t)= T f (3.364c)  k s ∂T s ∂r − k l ∂T l ∂r  r=r f = ρL dr f dt (3.364d) BOOKCOMP, Inc. — John Wiley & Sons / Page 249 / 2nd Proofs / Heat Transfer Handbook / Bejan CONDUCTION-CONTROLLED FREEZING AND MELTING 249 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 [249], (89) Lines: 4099 to 4132 ——— 10.1232pt PgVar ——— Normal Page PgEnds: T E X [249], (89) TABLE 3.16 Stefan Number and Interface Location Parameter St λ 0.1 0.1846 0.2 0.3143 0.3 0.4491 0.4 0.6006 0.5 0.7811 0.6 1.0095 0.7 1.3237 0.8 1.8180 Ozisik (1993) gives the solution as T s = T f + Q s 4πk s  Ei  − r 2 4α s t  − Ei  − r 2 f 4α s t  (0 <r<r f ) (3.365) T l = T i − T i − T f Ei  − r 2 f /4α l t  Ei  − r 2 4α l t   (r f <r<∞) (3.366) where λ = r f /2 √ α s t is obtained from the transcendental equation Q s 4π e −λ 2 + k l (T i − T f ) Ei(−λ 2 α s /α l ) e −λ 2 α s /α l = λ 2 α s ρL (3.367) The solution presented by eqs. (3.365) through (3.367) has been extended by Ozisik 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 250 / 2nd Proofs / Heat Transfer Handbook / Bejan 250 CONDUCTION HEAT TRANSFER 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 [250], (90) Lines: 4132 to 4174 ——— 3.06346pt PgVar ——— Short Page * PgEnds: Eject [250], (90) 3.10.5 Approximate Analytical Solutions Because of the mathematical complexity and the restrictive nature of exact analyti- cal solutions, several approaches have been employed to generate approximate an- alytical solutions that provide rapid results in a number of practical situations. The methods used are the quasi-steady solution, the heat balance integral approach of Goodman (1958) and Lunardini (1991), and the perturbation method of Aziz and Na (1984), Aziz and Lunardini (1993), and others. A collection of such solutions is provided next. One-Region Neumann Problem The quasi-steady-state solution where St = 0 for this problem is T = T 0 + (T f − T 0 ) x x f (3.368) x 2 f = 2k(T 0 − T f ) ρL t (3.369) One-Region Neumann Problem with Surface Convection With convective cooling, the boundary condition of eq. (3.347a) is replaced by k ∂T ∂x     x=0 = h [ T(0,t)−T ∞ ] (3.370) where T ∞ is now the coolant temperature. The quasi-steady-state approach with St = 0 yields T = T f + h(T f − T ∞ )(x − x f ) k + hx f (3.371) t = ρLx f h(T f − T ∞ )  1 + hx f 2k  (3.372) as the results. Outward Cylindrical Freezing Consider a saturated liquid at the freezing tem- perature T f , surrounding a cylinder of radius r 0 whose outer surface is kept at a sub- freezing temperature, T 0 <T f . If a quasi-steady-state assumption (St = 0) is used, the solution is T = T 0 + T f − T 0 ln(r f /r 0 ) ln r r 0 (3.373) t = ρL k(T f − T 0 )  1 2 r 2 f ln r f r 0 − 1 4  r 2 f − r 2 0   (3.374) . & Sons / Page 242 / 2nd Proofs / Heat Transfer Handbook / Bejan 242 CONDUCTION HEAT TRANSFER 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 [242],. times. The governing partial differential equation is BOOKCOMP, Inc. — John Wiley & Sons / Page 244 / 2nd Proofs / Heat Transfer Handbook / Bejan 244 CONDUCTION HEAT TRANSFER 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 [244],. & Sons / Page 246 / 2nd Proofs / Heat Transfer Handbook / Bejan 246 CONDUCTION HEAT TRANSFER 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 [246],