Computational Fluid Mechanics and Heat Transfer Third Edition_7 potx

§5.6 Transient heat conduction to a semi-infinite region 223 and the one known b.c. is T(x = 0) = T ∞ or Θ ( ζ = 0 ) = 0 (5.47) If we call dΘ/dζ ≡ χ, then eqn. (5.45) becomes the first-order equa- tion dχ dζ =− ζ 2 χ which can be integrated once to get χ ≡ dΘ dζ = C 1 e −ζ 2 /4 (5.48) and we integrate this a second time to get Θ = C 1  ζ 0 e −ζ 2 /4 dζ + Θ(0)    = 0 according to the b.c. (5.49) The b.c. is now satisfied, and we need only substitute eqn. (5.49)inthe i.c., eqn. (5.46), to solve for C 1 : 1 = C 1  ∞ 0 e −ζ 2 /4 dζ The definite integral is given by integral tables as √ π,so C 1 = 1 √ π Thus the solution to the problem of conduction in a semi-infinite region, subject to a b.c. of the first kind is Θ = 1 √ π  ζ 0 e −ζ 2 /4 dζ = 2 √ π  ζ/2 0 e −s 2 ds ≡ erf(ζ/2) (5.50) The second integral in eqn. (5.50), obtained by a change of variables, is called the error function (erf). Its name arises from its relationship to certain statistical problems related to the Gaussian distribution, which describes random errors. In Table 5.3, we list values of the error function and the complementary error function, erfc(x) ≡ 1 − erf(x). Equation (5.50) is also plotted in Fig. 5.15. 224 Transient and multidimensional heat conduction §5.6 Table 5.3 Error function and complementary error function. ζ  2 erf(ζ/2) erfc(ζ/2)ζ  2 erf(ζ/2) erfc(ζ/2) 0.00 0.00000 1.00000 1.10 0.88021 0.11980 0.05 0.05637 0.94363 1.20 0.91031 0.08969 0.10 0.11246 0.88754 1.30 0.93401 0.06599 0.15 0.16800 0.83200 1.40 0.95229 0.04771 0.20 0.22270 0.77730 1.50 0.96611 0.03389 0.30 0.32863 0.67137 1.60 0.97635 0.02365 0.40 0.42839 0.57161 1.70 0.98379 0.01621 0.50 0.52050 0.47950 1.80 0.98909 0.01091 0.60 0.60386 0.39614 1.8214 0.99000 0.01000 0.70 0.67780 0.32220 1.90 0.99279 0.00721 0.80 0.74210 0.25790 2.00 0.99532 0.00468 0.90 0.79691 0.20309 2.50 0.99959 0.00041 1.00 0.84270 0.15730 3.00 0.99998 0.00002 In Fig. 5.15 we see the early-time curves shown in Fig. 5.14 have col- lapsed into a single curve. This was accomplished by the similarity trans- formation, as we call it 5 : ζ/2 = x/2 √ αt. From the figure or from Table 5.3, we see that Θ ≥ 0.99 when ζ 2 = x 2 √ αt ≥ 1.8214 or x ≥ δ 99 ≡ 3.64  αt (5.51) In other words, the local value of (T −T ∞ ) is more than 99% of (T i −T ∞ ) for positions in the slab beyond farther from the surface than δ 99 = 3.64 √ αt. Example 5.4 For what maximum time can a samurai sword be analyzed as a semi- infinite region after it is quenched, if it has no clay coating and h external ∞? Solution. First, we must guess the half-thickness of the sword (say, 3 mm) and its material (probably wrought iron with an average α 5 The transformation is based upon the “similarity” of spatial an temporal changes in this problem. §5.6 Transient heat conduction to a semi-infinite region 225 Figure 5.15 Temperature distribution in a semi-infinite region. around 1.5 × 10 −5 m 2 /s). The sword will be semi-infinite until δ 99 equals the half-thickness. Inverting eqn. (5.51), we find t δ 2 99 3.64 2 α = (0.003 m) 2 13.3(1.5)(10) −5 m 2 /s = 0.045 s Thus the quench would be felt at the centerline of the sword within only 1/20 s. The thermal diffusivity of clay is smaller than that of steel by a factor of about 30, so the quench time of the coated steel must continue for over 1 s before the temperature of the steel is affected at all, if the clay and the sword thicknesses are comparable. Equation (5.51) provides an interesting foretaste of the notion of a fluid boundary layer. In the context of Fig. 1.9 and Fig. 1.10, we ob- serve that free stream flow around an object is disturbed in a thick layer near the object because the fluid adheres to it. It turns out that the thickness of this boundary layer of altered flow velocity increases in the downstream direction. For flow over a flat plate, this thickness is ap- proximately 4.92 √ νt, where t is the time required for an element of the stream fluid to move from the leading edge of the plate to a point of inter- est. This is quite similar to eqn. (5.51), except that the thermal diffusivity, α, has been replaced by its counterpart, the kinematic viscosity, ν, and the constant is a bit larger. The velocity profile will resemble Fig. 5.15. If we repeated the problem with a boundary condition of the third kind, we would expect to get Θ = Θ(Bi,ζ), except that there is no length, L, upon which to build a Biot number. Therefore, we must replace L with √ αt, which has the dimension of length, so Θ = Θ  ζ, h √ αt k  ≡ Θ(ζ, β) (5.52) 226 Transient and multidimensional heat conduction §5.6 The term β ≡ h √ αt  k is like the product: Bi √ Fo. The solution of this problem (see, e.g., [5.6], §2.7) can be conveniently written in terms of the complementary error function, erfc(x) ≡ 1 −erf(x): Θ = erf ζ 2 +exp  βζ + β 2   erfc  ζ 2 +β  (5.53) This result is plotted in Fig. 5.16. Example 5.5 Most of us have passed our finger through an 800 ◦ C candle flame and know that if we limit exposure to about 1/4 s we will not be burned. Why not? Solution. The short exposure to the flame causes only a very su- perficial heating, so we consider the finger to be a semi-infinite re- gion and go to eqn. (5.53) to calculate (T burn −T flame )/(T i −T flame ).It turns out that the burn threshold of human skin, T burn , is about 65 ◦ C. (That is why 140 ◦ For60 ◦ C tap water is considered to be “scalding.”) Therefore, we shall calculate how long it will take for the surface tem- perature of the finger to rise from body temperature (37 ◦ C) to 65 ◦ C, when it is protected by an assumed h  100 W/m 2 K. We shall assume that the thermal conductivity of human flesh equals that of its major component—water—and that the thermal diffusivity is equal to the known value for beef. Then Θ = 65 − 800 37 − 800 = 0.963 βζ = hx k = 0 since x = 0 at the surface β 2 = h 2 αt k 2 = 100 2 (0.135 × 10 −6 )t 0.63 2 = 0.0034(t s) The situation is quite far into the corner of Fig. 5.16. We read β 2  0.001, which corresponds with t  0.3 s. For greater accuracy, we must go to eqn. (5.53): 0.963 = erf 0    =0 +e 0.0034t  erfc  0 +  0.0034 t  Figure 5.16 The cooling of a semi-infinite region by an envi- ronment at T ∞ , through a heat transfer coefficient, h. 227 228 Transient and multidimensional heat conduction §5.6 By trial and error, we get t  0.33 s. In fact, it can be shown that Θ(ζ = 0,β) 2 √ π ( 1 − β ) for β  1 which can be solved directly for β = (1 − 0.963) √ π/2 = 0.03279, leading to the same answer. Thus, it would require about 1/3 s to bring the skin to the burn point. Experiment 5.1 Immerse your hand in the subfreezing air in the freezer compartment of your refrigerator. Next immerse your finger in a mixture of ice cubes and water, but do not move it. Then, immerse your finger in a mixture of ice cubes and water , swirling it around as you do so. Describe your initial sensation in each case, and explain the differences in terms of Fig. 5.16. What variable has changed from one case to another? Heat transfer Heat will be removed from the exposed surface of a semi-infinite region, with a b.c. of either the first or the third kind, in accordance with Fourier’s law: q =−k ∂T ∂x     x=0 = k(T ∞ −T i ) √ αt dΘ dζ      ζ=0 Differentiating Θ as given by eqn. (5.50), we obtain, for the b.c. of the first kind, q = k(T ∞ −T i ) √ αt  1 √ π e −ζ 2 /4  ζ=0 = k(T ∞ −T i ) √ παt (5.54) Thus, q decreases with increasing time, as t −1/2 . When the temperature of the surface is first changed, the heat removal rate is enormous. Then it drops off rapidly. It often occurs that we suddenly apply a specified input heat flux, q w , at the boundary of a semi-infinite region. In such a case, we can §5.6 Transient heat conduction to a semi-infinite region 229 differentiate the heat diffusion equation with respect to x,so α ∂ 3 T ∂x 3 = ∂ 2 T ∂t∂x When we substitute q =−k∂T/∂x in this, we obtain α ∂ 2 q ∂x 2 = ∂q ∂t with the b.c.’s: q(x = 0,t >0) = q w or q w −q q w      x=0 = 0 q(x  0,t = 0) = 0or q w −q q w      t=0 = 1 What we have done here is quite elegant. We have made the problem of predicting the local heat flux q into exactly the same form as that of predicting the local temperature in a semi-infinite region subjected to a step change of wall temperature. Therefore, the solution must be the same: q w −q q w = erf  x 2 √ αt  . (5.55) The temperature distribution is obtained by integrating Fourier’s law. At the wall, for example:  T w T i dT =−  0 ∞ q k dx where T i = T(x →∞) and T w = T(x = 0). Then T w = T i + q w k  ∞ 0 erfc(x/2  αt) dx This becomes T w = T i + q w k  αt  ∞ 0 erfc(ζ/2)dζ    =2/ √ π so T w (t) = T i +2 q w k  αt π (5.56) 230 Transient and multidimensional heat conduction §5.6 Figure 5.17 A bubble growing in a superheated liquid. Example 5.6 Predicting the Growth Rate of a Vapor Bubble in an Infinite Superheated Liquid This prediction is relevant to a large variety of processes, ranging from nuclear thermodynamics to the direct-contact heat exchange. It was originally presented by Max Jakob and others in the early 1930s (see, e.g., [5.10, Chap. I]). Jakob (pronounced Yah  -kob) was an im- portant figure in heat transfer during the 1920s and 1930s. He left Nazi Germany in 1936 to come to the United States. We encounter his name again later. Figure 5.17 shows how growth occurs. When a liquid is super- heated to a temperature somewhat above its boiling point, a small gas or vapor cavity in that liquid will grow. (That is what happens in the superheated water at the bottom of a teakettle.) This bubble grows into the surrounding liquid because its bound- ary is kept at the saturation temperature, T sat , by the near-equilibrium coexistence of liquid and vapor. Therefore, heat must flow from the superheated surroundings to the interface, where evaporation occurs. So long as the layer of cooled liquid is thin, we should not suffer too much error by using the one-dimensional semi-infinite region solu- tion to predict the heat flow. §5.6 Transient heat conduction to a semi-infinite region 231 Thus, we can write the energy balance at the bubble interface:  −q W m 2   4πR 2 m 2     Q into bubble =  ρ g h fg J m 3   dV dt m 3 s     rate of energy increase of the bubble and then substitute eqn. (5.54) for q and 4πR 3 /3 for the volume, V . This gives k(T sup −T sat ) √ απt = ρ g h fg dR dt (5.57) Integrating eqn. (5.57) from R = 0att = 0uptoR at t, we obtain Jakob’s prediction: R = 2 √ π k∆T ρ g h fg √ α  t (5.58) This analysis was done without assuming the curved bubble interface to be plane, 24 years after Jakob’s work, by Plesset and Zwick [5.11]. It was verified in a more exact way after another 5 years by Scriven [5.12]. These calculations are more complicated, but they lead to a very similar result: R = 2 √ 3 √ π k∆T ρ g h fg √ α  t = √ 3 R Jakob . (5.59) Both predictions are compared with some of the data of Dergarabe- dian [5.13] in Fig. 5.18. The data and the exact theory match almost perfectly. The simple theory of Jakob et al. shows the correct depen- dence on R on all its variables, but it shows growth rates that are low by a factor of √ 3. This is because the expansion of the spherical bub- ble causes a relative motion of liquid toward the bubble surface, which helps to thin the region of thermal influence in the radial direction. Con- sequently, the temperature gradient and heat transfer rate are higher than in Jakob’s model, which neglected the liquid motion. Therefore, the temperature profile flattens out more slowly than Jakob predicts, and the bubble grows more rapidly. Experiment 5.2 Touch various objects in the room around you: glass, wood, cork- board, paper, steel, and gold or diamond, if available. Rank them in 232 Transient and multidimensional heat conduction §5.6 Figure 5.18 The growth of a vapor bubble—predictions and measurements. order of which feels coldest at the first instant of contact (see Problem 5.29). The more advanced theory of heat conduction (see, e.g., [5.6]) shows that if two semi-infinite regions at uniform temperatures T 1 and T 2 are placed together suddenly, their interface temperature, T s , is given by 6 T s −T 2 T 1 −T 2 =  (kρc p ) 2  (kρc p ) 1 +  (kρc p ) 2 If we identify one region with your body (T 1  37 ◦ C) and the other with the object being touched (T 2  20 ◦ C), we can determine the temperature, T s , that the surface of your finger will reach upon contact. Compare the ranking you obtain experimentally with the ranking given by this equation. Notice that your bloodstream and capillary system provide a heat 6 For semi-infinite regions, initially at uniform temperatures, T s does not vary with time. For finite bodies, T s will eventually change. A constant value of T s means that each of the two bodies independently behaves as a semi-infinite body whose surface temperature has been changed to T s at time zero. Consequently, our previous results— eqns. (5.50), (5.51), and (5.54)—apply to each of these bodies while they may be treated as semi-infinite. We need only replace T ∞ by T s in those equations. [...]... left-hand side of Fig 5.7 × Θ2 x L 2 = 1 , Fo2 = 0.565, Bi−1 = 4.75 2 2 = 0.91 from interpolation between lower lefthand side and upper righthand side of Fig 5.7 Thus, at the axial line of interest, Θ = (0.93)(0.91) = 0.846 so T − 20 = 0.846 100 − 20 or T = 87.7◦ C Transient multidimensional heat conduction 251 Product solutions can also be used to determine the mean temperature, Θ, and the total heat. .. get C1 + C 2 = Ti ri and C1 + C 1 = To ro Therefore, C1 = Ti − To ri ro ro − r i and C2 = Ti − Ti − T o ro ro − r i Putting C1 and C2 in the general solution, and calling Ti − To ≡ ∆T , we get T = Ti + ∆T ro r i ro − r (ro − ri ) ro − ri Then 4π (ri ro ) dT = k∆T dr ro − r i 4π (ri ro ) m S= ro − r i Q = −kA where S now has the dimensions of m 243 244 Transient and multidimensional heat conduction §5.7... meter Problems 5.19 and 2.15 meter [5.16] none [5.16] meter [5.16, 5.17] 4 The boundary of a spherical hole of radius R conducting into an infinite medium 5 Cylinder of radius R and length L, transferring heat to a parallel isothermal plane; h L 2π L cosh−1 (h/R) 6 Same as item 5, but with L → ∞ (two-dimensional conduction) 2π cosh −1 (h/R) 7 An isothermal sphere of radius R transfers heat to an isothermal... It has a length 2L and a radius ro Finding the temperature field in this situation is inherently complicated 248 Transient and multidimensional heat conduction §5.8 It requires solving the heat conduction equation for T = fn(r , z, t) with b.c.’s of the first, second, or third kind However, Fig 5.27a suggests that this can somehow be viewed as a combination of an infinite cylinder and an infinite slab... plot shown.) Steady multidimensional heat conduction §5.7 Figure 5.25 Heat transfer through a thick, hollow sphere Example 5.10 Calculate S for a thick hollow sphere, as shown in Fig 5.25 Solution The general solution of the heat diffusion equation in spherical coordinates for purely radial heat flow is: C1 + C2 r when T = fn(r only) The b.c.’s are T = T (r = ri ) = Ti and T (r = ro ) = To substituting... Fo = 0.2, Bi = 10, and x/L = 0 analytically 5.9 Prove that when Bi is large, eqn (5.34) reduces to eqn (5.33) 5.10 Check the point at Bi = 0.1 and Fo = 2.5 on the slab curve in Fig 5.10 analytically Chapter 5: Transient and multidimensional heat conduction 254 Figure 5.28 Problem 5.6 5.11 Configuration and temperature response for Sketch one of the curves in Fig 5.7, 5.8, or 5.9 and identify: • The... warm) into cold water before heating starts or by placing warm eggs directly into simmering water [5.20] A copper block 1 in thick and 3 in square is held at 100◦ F on one 1 in by 3 in surface The opposing 1 in by 3 in surface is adiabatic for 2 in and 90◦ F for 1 inch The remaining surfaces are adiabatic Find the rate of heat transfer [Q = 36.8 W.] Two copper slabs, 3 cm thick and insulated on the outside,... electrical resistance strip heater is fastened to a firebrick wall, unformly at 15◦ C When it is suddenly turned on, it releases heat at the uniform rate of 4000 W/m2 Plot the temperature of the brick immediately under the heater as a function of time if the other side of the heater is insulated What is the heat flux at a depth of 1 cm when the surface reaches 200◦ C 5.29 Do Experiment 5.2 and submit a report... Situation Shape factor, S 8 An isothermal sphere of radius R, near an insulated plane, transfers heat to a semi-infinite medium at T∞ (see items 4 and 7) Dimensions 4π R 1 + R/2h meter Source [5.18] 9 Parallel cylinders exchange heat in an infinite conducting medium −1 cosh 10 Same as 9, but with cylinders widely spaced; L R1 and R2 11 Cylinder of radius Ri surrounded by eccentric cylinder of radius Ro > Ri... every 14 cm They can be assumed to stay at the temperature of that wall Find the heat flux through the wall if the first wall is at 40◦ C and the one with ribs is at 0◦ C Find the temperature in the middle of the wall, 2 cm from a rib, as well Figure 5.23 Heat transfer through a wall with isothermal ribs Steady multidimensional heat conduction §5.7 Solution The flux plot for this configuration is shown in . 0.8 875 4 1.30 0.93401 0.06599 0.15 0.16800 0.83200 1.40 0.95229 0.0 477 1 0.20 0.22 270 0 .77 730 1.50 0.96611 0.03389 0.30 0.32863 0. 671 37 1.60 0. 976 35 0.02365 0.40 0.42839 0. 571 61 1 .70 0.98 379 0.01621 0.50. 0.01621 0.50 0.52050 0. 479 50 1.80 0.98909 0.01091 0.60 0.60386 0.39614 1.8214 0.99000 0.01000 0 .70 0. 677 80 0.32220 1.90 0.99 279 0.0 072 1 0.80 0 .74 210 0.2 579 0 2.00 0.99532 0.00468 0.90 0 .79 691 0.20309 2.50. isothermal lines and the adiabatic, 7 7 These are lines in the direction of heat flow. It immediately follows that there can §5 .7 Steady multidimensional heat conduction 2 37 or heat flow, lines