Problems 189 a surface flux of 40 kW/m 2 . Evaluate the temperature of the rod in either side of the heated section if h = 150 W/m 2 K around the unheated rod, and T ambient = 27 ◦ C. 4.39 The heat transfer coefficient between a cool surface and a satu- rated vapor, when the vapor condenses in a film on the surface, depends on the liquid density and specific heat, the tempera- ture difference, the buoyant force per unit volume (g[ρ f −ρ g ]), the latent heat, the liquid conductivity and the kinematic vis- cosity, and the position (x) on the cooler. Develop the dimen- sionless functional equation for h. 4.40 A duralumin pipe through a cold room hasa4cmI.D. and a 5 cm O.D. It carries water that sometimes sits stationary. It is proposed to put electric heating rings around the pipe to protect it against freezing during cold periods of −7 ◦ C. The heat transfer coefficient outside the pipe is 9 W/m 2 K (including both convection and radiation). Neglect the presence of the water in the conduction calculation, and determine how far apart the heaters would have to be if they brought the pipe temperature to 40 ◦ C locally. How much heat do they require? 4.41 The specific entropy of an ideal gas depends on its specific heat at constant pressure, its temperature and pressure, the ideal gas constant and reference values of the temperature and pressure. Obtain the dimensionless functional equation for the specific entropy and compare it with the known equation. 4.42 A large freezer’s door has a 2.5 cm thick layer of insulation (k in = 0.04 W/m 2 K) covered on the inside, outside, and edges with a continuous aluminum skin 3.2 mm thick (k Al = 165 W/m 2 K). The door closes against a nonconducting seal 1 cm wide. Heat gain through the door can result from conduction straight through the insulation and skins (normal to the plane of the door) and from conduction in the aluminum skin only, going from the skin outside, around the edge skin, and to the inside skin. The heat transfer coefficients to the inside, h i , and outside, h o , are each 12 W/m 2 K, accounting for both con- vection and radiation. The temperature outside the freezer is 25 ◦ C, and the temperature inside is −15 ◦ C. a. If the door is 1 m wide, estimate the one-dimensional heat gain through the door, neglecting any conduction around 190 Chapter 4: Analysis of heat conduction and some steady one-dimensional problems the edges of the skin. Your answer will be in watts per meter of door height. b. Now estimate the heat gain by conduction around the edges of the door, assuming that the insulation is per- fectly adiabatic so that all heat flows through the skin. This answer will also be per meter of door height. 4.43 A thermocouple epoxied onto a high conductivity surface is in- tended to measure the surface temperature. The thermocou- ple consists of two each bare, 0.51 mm diameter wires. One wire is made of Chromel (Ni-10% Cr with k cr = 17 W/m·K) and the other of constantan (Ni-45% Cu with k cn = 23 W/m·K). The ends of the wires are welded together to create a measuring junction having has dimensions of D w by 2D w . The wires ex- tend perpendicularly away from the surface and do not touch one another. A layer of epoxy (k ep = 0.5 W/m·K separates the thermocouple junction from the surface by 0.2 mm. Air at 20 ◦ C surrounds the wires. The heat transfer coefficient be- tween each wire and the surroundings is h = 28 W/m 2 K, in- cluding both convection and radiation. If the thermocouple reads T tc = 40 ◦ C, estimate the actual temperature T s of the surface and suggest a better arrangement of the wires. 4.44 The resistor leads in Example 4.10 were assumed to be “in- finitely long” fins. What is the minimum length they each must have if they are to be modelled this way? What are the effec- tiveness, ε f , and efficiency, η f , of the wires? References [4.1] V. L. Streeter and E.B. Wylie. Fluid Mechanics. McGraw-Hill Book Company, New York, 7th edition, 1979. Chapter 4. [4.2] E. Buckingham. Phy. Rev., 4:345, 1914. [4.3] E. Buckingham. Model experiments and the forms of empirical equa- tions. Trans. ASME, 37:263–296, 1915. [4.4] Lord Rayleigh, John Wm. Strutt. The principle of similitude. Nature, 95:66–68, 1915. References 191 [4.5] J. O. Farlow, C. V. Thompson, and D. E. Rosner. Plates of the dinosaur stegosaurus: Forced convection heat loss fins? Science, 192(4244): 1123–1125 and cover, 1976. [4.6] D. K. Hennecke and E. M. Sparrow. Local heat sink on a convectively cooled surface—application to temperature measurement error. Int. J. Heat Mass Transfer, 13:287–304, 1970. [4.7] P. J. Schneider. Conduction Heat Transfer. Addison-Wesley Publish- ing Co., Inc., Reading, Mass., 1955. [4.8] A. D. Kraus, A. Aziz, and J.R. Welty. Extended Surface Heat Transfer. John Wiley & Sons, Inc., New York, 2001. 5. Transient and multidimensional heat conduction When I was a lad, winter was really cold. It would get so cold that if you went outside with a cup of hot coffee it would freeze. I mean it would freeze fast. That cup of hot coffee would freeze so fast that it would still be hot after it froze. Now that’s cold! Old North-woods tall-tale 5.1 Introduction James Watt, of course, did not invent the steam engine. What he did do was to eliminate a destructive transient heating and cooling process that wasted a great amount of energy. By 1763, the great puffing engines of Savery and Newcomen had been used for over half a century to pump the water out of Cornish mines and to do other tasks. In that year the young instrument maker, Watt, was called upon to renovate the Newcomen en- gine model at the University of Glasgow. The Glasgow engine was then being used as a demonstration in the course on natural philosophy. Watt did much more than just renovate the machine—he first recognized, and eventually eliminated, its major shortcoming. The cylinder of Newcomen’s engine was cold when steam entered it and nudged the piston outward. A great deal of steam was wastefully condensed on the cylinder walls until they were warm enough to accom- modate it. When the cylinder was filled, the steam valve was closed and jets of water were activated inside the cylinder to cool it again and con- dense the steam. This created a powerful vacuum, which sucked the piston back in on its working stroke. First, Watt tried to eliminate the wasteful initial condensation of steam by insulating the cylinder. But that simply reduced the vacuum and cut the power of the working stroke. 193 194 Transient and multidimensional heat conduction §5.2 Then he realized that, if he led the steam outside to a separate condenser, the cylinder could stay hot while the vacuum was created. The separate condenser was the main issue in Watt’s first patent (1769), and it immediately doubled the thermal efficiency of steam en- gines from a maximum of 1.1% to 2.2%. By the time Watt died in 1819, his invention had led to efficiencies of 5.7%, and his engine had altered the face of the world by powering the Industrial Revolution. And from 1769 until today, the steam power cycles that engineers study in their ther- modynamics courses are accurately represented as steady flow—rather than transient—processes. The repeated transient heating and cooling that occurred in New- comen’s engine was the kind of process that today’s design engineer might still carelessly ignore, but the lesson that we learn from history is that transient heat transfer can be of overwhelming importance. To- day, for example, designers of food storage enclosures know that such systems need relatively little energy to keep food cold at steady condi- tions. The real cost of operating them results from the consumption of energy needed to bring the food down to a low temperature and the losses resulting from people entering and leaving the system with food. The transient heat transfer processes are a dominant concern in the de- sign of food storage units. We therefore turn our attention, first, to an analysis of unsteady heat transfer, beginning with a more detailed consideration of the lumped- capacity system that we looked at in Section 1.3. 5.2 Lumped-capacity solutions We begin by looking briefly at the dimensional analysis of transient con- duction in general and of lumped-capacity systems in particular. Dimensional analysis of transient heat conduction We first consider a fairly representative problem of one-dimensional tran- sient heat conduction: ∂ 2 T ∂x 2 = 1 α ∂T ∂t with            i.c.: T(t = 0) = T i b.c.: T(t > 0,x = 0) = T 1 b.c.: −k ∂T ∂x     x=L = h ( T −T 1 ) x=L §5.2 Lumped-capacity solutions 195 The solution of this problem must take the form of the following dimen- sional functional equation: T −T 1 = fn  (T i −T 1 ), x, L, t, α, h, k  There are eight variables in four dimensions (K, s, m, W), so we look for 8−4 = 4 pi-groups. We anticipate, from Section 4.3, that they will include Θ ≡ (T −T 1 ) (T i −T 1 ) ,ξ≡ x L , and Bi ≡ hL k , and we write Θ = fn ( ξ,Bi, Π 4 ) (5.1) One possible candidate for Π 4 , which is independent of the other three, is Π 4 ≡ Fo = αt/L 2 (5.2) where Fo is the Fourier number. Another candidate that we use later is Π 4 ≡ ζ = x √ αt  this is exactly ξ √ Fo  (5.3) If the problem involved only b.c.’s of the first kind, the heat transfer coefficient, h—and hence the Biot number—would go out of the problem. Then the dimensionless function eqn. (5.1)is Θ = fn ( ξ,Fo ) (5.4) By the same token, if the b.c.’s had introduced different values of h at x = 0 and x = L, two Biot numbers would appear in the solution. The lumped-capacity problem is particularly interesting from the stand- point of dimensional analysis [see eqns. (1.19)–(1.22)]. In this case, nei- ther k nor x enters the problem because we do not retain any features of the internal conduction problem. Therefore, we have ρc rather than α. Furthermore, we do not have to separate ρ and c because they only appear as a product. Finally, we use the volume-to-external-area ratio, V/A, as a characteristic length. Thus, for the transient lumped-capacity problem, the dimensional equation is T −T ∞ = fn  ( T i −T ∞ ) , ρc, V/A, h, t  (5.5) 196 Transient and multidimensional heat conduction §5.2 Figure 5.1 A simple resistance-capacitance circuit. With six variables in the dimensions J, K, m, and s, only two pi-groups will appear in the dimensionless function equation. Θ = fn  hAt ρcV  = fn  t T  (5.6) This is exactly the form of the simple lumped-capacity solution, eqn. (1.22). Notice, too, that the group t/T can be viewed as t T = hk(V/A)t ρc(V/A) 2 k = h(V/A) k · αt (V/A) 2 = Bi Fo (5.7) Electrical and mechanical analogies to the lumped-thermal-capacity problem The term capacitance is adapted from electrical circuit theory to the heat transfer problem. Therefore, we sketch a simple resistance-capacitance circuit in Fig. 5.1. The capacitor is initially charged to a voltage, E o . When the switch is suddenly opened, the capacitor discharges through the re- sistor and the voltage drops according to the relation dE dt + E RC = 0 (5.8) The solution of eqn. (5.8) with the i.c. E(t = 0) = E o is E = E o e −t/RC (5.9) and the current can be computed from Ohm’s law, once E(t) is known. I = E R (5.10) Normally, in a heat conduction problem the thermal capacitance, ρcV, is distributed in space. But when the Biot number is small, T(t) §5.2 Lumped-capacity solutions 197 is uniform in the body and we can lump the capacitance into a single circuit element. The thermal resistance is 1/ hA, and the temperature difference (T − T ∞ ) is analogous to E(t). Thus, the thermal response, analogous to eqn. (5.9), is [see eqn. (1.22)] T −T ∞ = ( T i −T ∞ ) exp  − hAt ρcV  Notice that the electrical time constant, analogous to ρcV/ hA,isRC. Now consider a slightly more complex system. Figure 5.2 shows a spring-mass-damper system. The well-known response equation (actu- ally, a force balance) for this system is m  What is the mass analogous to? d 2 x dt 2 + c  the damping coefficient is analogous to R or to ρcV dx dt + k  where k is analogous to 1/C or to hA x = F(t) (5.11) A term analogous to mass would arise from electrical inductance, but we Figure 5.2 A spring-mass-damper system with a forcing function. did not include it in the electrical circuit. Mass has the effect of carrying the system beyond its final equilibrium point. Thus, in an underdamped mechanical system, we might obtain the sort of response shown in Fig. 5.3 if we specified the velocity at x = 0 and provided no forcing function. Electrical inductance provides a similar effect. But the Second Law of Thermodynamics does not permit temperatures to overshoot their equi- librium values spontaneously. There are no physical elements analogous to mass or inductance in thermal systems. 198 Transient and multidimensional heat conduction §5.2 Figure 5.3 Response of an unforced spring-mass-damper system with an initial velocity. Next, consider another mechanical element that does have a ther- mal analogy—namely, the forcing function, F. We consider a (massless) spring-damper system with a forcing function F that probably is time- dependent, and we ask: “What might a thermal forcing function look like?” Lumped-capacity solution with a variable ambient temperature To answer the preceding question, let us suddenly immerse an object at a temperature T = T i , with Bi  1, into a cool bath whose temperature is rising as T ∞ (t) = T i +bt, where T i and b are constants. Then eqn. (1.20) becomes d(T −T i ) dt =− T −T ∞ T =− T −T i −bt T where we have arbitrarily subtracted T i under the differential. Then d(T −T i ) dt + T −T i T = bt T (5.12) To solve eqn. (5.12) we must first recall that the general solution of a linear ordinary differential equation with constant coefficients is equal to the sum of any particular integral of the complete equation and the general solution of the homogeneous equation. We know the latter; it is T − T i = (constant) exp(−t/T ). A particular integral of the complete equation can often be formed by guessing solutions and trying them in the complete equation. Here we discover that T −T i = bt − bT [...]... require that hc L1 /k1 , hc L2 /k2 , and hL2 /k2 are all much 200 Transient and multidimensional heat conduction §5.2 Figure 5.4 Response of a thermometer to a linearly increasing ambient temperature less than unity so that it will be legitimate to lump the thermal capacitance of each slab The differential equations dictating the temperature response of each slab are then slab 1 : slab 2 : dT1 = hc A( T1 −... The choice is arbitrary Lumped-capacity solutions §5.2 2 01 Figure 5.5 Two slabs conducting in series through an interfacial resistance which we substitute in eqn (5 .16 ) to get T1 dT1 dT1 d2 T 1 dT1 hc T1 + T1 − T ∞ + = T 1 T2 − T2 2 dt dt dt dt h or d 2 T1 + dt 2 1 1 hc + + T1 T2 hT2 ≡b T1 − T∞ dT1 + =0 dt T1 T2 (5 .1 9a) c(T1 − T∞ ) if we call T1 − T∞ ≡ θ, then eqn (5 .1 9a) can be written as dθ d2 θ +... spherical apples, 10 cm in diameter are taken from a 30◦ C environment and laid out on a rack in a refrigerator at 5◦ C They have approximately the same physical properties as water, and h is approximately 6 W/m2 K as the result of natural convection What will be the temperature of the centers of the apples after 1 hr? How long will it take to bring the centers to 10 ◦ C? How much heat will the refrigerator... the bath with a temperature lag of bT This constant error is reduced when either T or the rate of temperature increase, b, is reduced Second-order lumped-capacity systems Now we look at situations in which two lumped-thermal-capacity systems are connected in series Such an arrangement is shown in Fig 5.5 Heat is transferred through two slabs with an interfacial resistance, h 1 between c them We shall... the outer center of the slab—at x = 0 or ξ = 1 Then 4 × Θ (0, Fo) = π    3π 1 π 2 Fo − exp − exp −  2 2 3  = 0.085 at Fo = 1 = 0.7 81 at Fo = 0 .1 = 0 .97 6 at Fo = 0. 01 2 Fo + 10 10 at Fo = 1 = 0.036 at Fo = 0 .1 = 0.267 at Fo = 0. 01 5π 1 exp − 2 5 2    Fo + · · ·   10 −27 at Fo = 1 = 0.0004 at Fo = 0 .1 = 0 .10 8 at Fo = 0. 01 Thus for values of Fo somewhat greater than 0 .1, only the first term in... = hA(T2 − T∞ ) − hc A( T1 − T2 ) −(ρcV )2 dt −(ρcV )1 (5 .15 ) (5 .16 ) and the initial conditions on the temperatures T1 and T2 are T1 (t = 0) = T2 (t = 0) = Ti (5 .17 ) We next identify two time constants for this problem :1 T1 ≡ (ρcV )1 hc A and T2 ≡ (ρcV )2 hA Then eqn (5 .15 ) becomes T 2 = T1 1 dT1 + T1 dt (5 .18 ) Notice that we could also have used (ρcV )2 /hc A for T2 since both hc and h act on slab 2...Lumped-capacity solutions §5.2 19 9 satisfies eqn (5 .12 ) Thus, the general solution of eqn (5 .12 ) is T − Ti = C1 e−t/T + b(t − T ) (5 .13 ) The solution for arbitrary variations of T∞ (t) is given in Problem 5.52 (see also Problems 5.3, 5.53, and 5.54) Example 5 .1 The flow rates of hot and cold water are regulated into a mixing chamber We measure the temperature of the water as it leaves, using a thermometer... The transient cooling of a slab; ξ = (x/L) + 1 5.3 Transient conduction in a one-dimensional slab We next extend consideration to heat flow in bodies whose internal resistance is significant—to situations in which the lumped capacitance assumption is no longer appropriate When the temperature within, say, a one-dimensional body varies with position as well as time, we must solve the heat diffusion equation... is a pretty complicated result—all the more complicated when we remember that b involves three algebraic terms [recall eqn (5 .1 9a) ] Yet there is nothing very sophisticated about it; it is easy to understand A system involving three capacitances in series would similarly yield a third- order equation of correspondingly higher complexity, and so forth §5.3 Transient conduction in a one-dimensional slab... • Extrapolate the given curves using a straightedge • Evaluate Θ using the first term of eqn (5.34), as discussed in Sect 5.5 • If Bi is small, use a lumped-capacity result Figure 5.8 and Fig 5 .9 are similar graphs for cylinders and spheres Everything that we have said in general about Fig 5.7 is also true for these graphs They were simply calculated from different solutions, and the numerical values . briefly at the dimensional analysis of transient con- duction in general and of lumped-capacity systems in particular. Dimensional analysis of transient heat conduction We first consider a fairly. our attention, first, to an analysis of unsteady heat transfer, beginning with a more detailed consideration of the lumped- capacity system that we looked at in Section 1. 3. 5.2 Lumped-capacity. as t T = hk(V /A) t ρc(V /A) 2 k = h(V /A) k · αt (V /A) 2 = Bi Fo (5.7) Electrical and mechanical analogies to the lumped-thermal-capacity problem The term capacitance is adapted from electrical circuit theory to the heat transfer