John H. Lienhard IV, John H. Lienhard V. A Heat Transfer Textbook (776116), страница 29
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We anticipate, from Section 4.3, that they will includeΘ≡(T − T1 ),(Ti − T1 )ξ≡x,Land Bi ≡hL,kand we writeΘ = fn (ξ, Bi, Π4 )(5.1)One possible candidate for Π4 , which is independent of the other three,isΠ4 ≡ Fo = αt/L2(5.2)where Fo is the Fourier number. Another candidate that we use later isξx√√(5.3)this is exactlyΠ4 ≡ ζ =αtFoIf the problem involved only b.c.’s of the first kind, the heat transfercoefficient, h—and hence the Biot number—would go out of the problem.Then the dimensionless function eqn.
(5.1) isΘ = fn (ξ, Fo)(5.4)By the same token, if the b.c.’s had introduced different values of h atx = 0 and x = L, two Biot numbers would appear in the solution.The lumped-capacity problem is particularly interesting from the standpoint of dimensional analysis [see eqns.
(1.19)–(1.22)]. In this case, neither k nor x enters the problem because we do not retain any featuresof the internal conduction problem. Therefore, we have ρc rather thanα. Furthermore, we do not have to separate ρ and c because they onlyappear as a product. Finally, we use the volume-to-external-area ratio,V /A, as a characteristic length. Thus, for the transient lumped-capacityproblem, the dimensional equation is(5.5)T − T∞ = fn (Ti − T∞ ) , ρc, V /A, h, t196Transient and multidimensional heat conduction§5.2Figure 5.1 A simpleresistance-capacitance circuit.With six variables in the dimensions J, K, m, and s, only two pi-groupswill appear in the dimensionless function equation.
hAtt(5.6)Θ = fn= fnρcVTThis is exactly the form of the simple lumped-capacity solution, eqn. (1.22).Notice, too, that the group t/T can be viewed ash(V /A)hk(V /A)tαtt==·= Bi Fo2Tρc(V /A) kk(V /A)2(5.7)Electrical and mechanical analogies to thelumped-thermal-capacity problemThe term capacitance is adapted from electrical circuit theory to the heattransfer problem. Therefore, we sketch a simple resistance-capacitancecircuit in Fig.
5.1. The capacitor is initially charged to a voltage, Eo . Whenthe switch is suddenly opened, the capacitor discharges through the resistor and the voltage drops according to the relationEdE+=0dtRC(5.8)The solution of eqn. (5.8) with the i.c. E(t = 0) = Eo isE = Eo e−t/RC(5.9)and the current can be computed from Ohm’s law, once E(t) is known.I=ER(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)§5.2Lumped-capacity solutions197is uniform in the body and we can lump the capacitance into a singlecircuit element.
The thermal resistance is 1/hA, and the temperaturedifference (T − T∞ ) is analogous to E(t). Thus, the thermal response,analogous to eqn. (5.9), is [see eqn. (1.22)]hAtT − T∞ = (Ti − T∞ ) exp −ρcVNotice that the electrical time constant, analogous to ρcV /hA, is RC.Now consider a slightly more complex system. Figure 5.2 shows aspring-mass-damper system. The well-known response equation (actually, a force balance) for this system isd2 xdxc+ k x = F (t)m dt 2 + dt(5.11)where k is analogous to 1/C or to hAthe damping coefficient is analogous to R or to ρcVWhat is the mass analogous to?A term analogous to mass would arise from electrical inductance, but weFigure 5.2 A spring-mass-dampersystem with a forcing function.did not include it in the electrical circuit.
Mass has the effect of carryingthe system beyond its final equilibrium point. Thus, in an underdampedmechanical system, we might obtain the sort of response shown in Fig. 5.3if we specified the velocity at x = 0 and provided no forcing function.Electrical inductance provides a similar effect. But the Second Law ofThermodynamics does not permit temperatures to overshoot their equilibrium values spontaneously. There are no physical elements analogousto mass or inductance in thermal systems.198Transient and multidimensional heat conduction§5.2Figure 5.3 Response of an unforcedspring-mass-damper system with aninitial velocity.Next, consider another mechanical element that does have a thermal analogy—namely, the forcing function, F . We consider a (massless)spring-damper system with a forcing function F that probably is timedependent, and we ask: “What might a thermal forcing function looklike?”Lumped-capacity solution with a variable ambient temperatureTo answer the preceding question, let us suddenly immerse an object ata temperature T = Ti , with Bi 1, into a cool bath whose temperature isrising as T∞ (t) = Ti + bt, where Ti and b are constants.
Then eqn. (1.20)becomesT − T∞T − Ti − btd(T − Ti )=−=−dtTTwhere we have arbitrarily subtracted Ti under the differential. Thenbtd(T − Ti ) T − Ti+=dtTT(5.12)To solve eqn. (5.12) we must first recall that the general solution ofa linear ordinary differential equation with constant coefficients is equalto the sum of any particular integral of the complete equation and thegeneral solution of the homogeneous equation. We know the latter; itis T − Ti = (constant) exp(−t/T ).
A particular integral of the completeequation can often be formed by guessing solutions and trying them inthe complete equation. Here we discover thatT − Ti = bt − bTLumped-capacity solutions§5.2199satisfies eqn. (5.12). Thus, the general solution of eqn. (5.12) isT − 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 5.1The flow rates of hot and cold water are regulated into a mixing chamber.
We measure the temperature of the water as it leaves, using athermometer with a time constant, T . On a particular day, the system started with cold water at T = Ti in the mixing chamber. Thenhot water is added in such a way that the outflow temperature riseslinearly, as shown in Fig. 5.4, with Texit flow = Ti + bt. How will thethermometer report the temperature variation?Solution.
The initial condition in eqn. (5.13), which describes thisprocess, is T − Ti = 0 at t = 0. Substituting eqn. (5.13) in the i.c., weget0 = C1 − bTsoC1 = bTand the response equation isT − (Ti + bt) = bT e−t/T − 1(5.14)This result is graphically shown in Fig. 5.4. Notice that the thermometer reading reflects a transient portion, bT e−t/T , which decaysfor a few time constants and then can be neglected, and a steadyportion, Ti + b(t − T ), which persists thereafter. When the steady response is established, the thermometer follows the bath with a temperature lag of bT . This constant error is reduced when either T orthe rate of temperature increase, b, is reduced.Second-order lumped-capacity systemsNow we look at situations in which two lumped-thermal-capacity systemsare connected in series. Such an arrangement is shown in Fig.
5.5. Heat istransferred through two slabs with an interfacial resistance, h−1c betweenthem. We shall require that hc L1 /k1 , hc L2 /k2 , and hL2 /k2 are all much200Transient and multidimensional heat conduction§5.2Figure 5.4 Response of a thermometer to a linearly increasingambient temperature.less than unity so that it will be legitimate to lump the thermal capacitance of each slab.
The differential equations dictating the temperatureresponse of each slab are thendT1= hc A(T1 − T2 )dtdT2= hA(T2 − T∞ ) − hc A(T1 − T2 )slab 2 : −(ρcV )2dtslab 1 : −(ρcV )1(5.15)(5.16)and the initial conditions on the temperatures T1 and T2 areT1 (t = 0) = T2 (t = 0) = Ti(5.17)We next identify two time constants for this problem:1T1 ≡ (ρcV )1 hc A and T2 ≡ (ρcV )2 hAThen eqn.
(5.15) becomesT2 = T11dT1+ T1dt(5.18)Notice that we could also have used (ρcV )2 /hc A for T2 since both hc and h act onslab 2. The choice is arbitrary.Lumped-capacity solutions§5.2201Figure 5.5 Two slabs conducting in series through an interfacial resistance.which we substitute in eqn. (5.16) to gethcdT1dT1d2 T1dT1+ T1 − T∞ += T1 T2− T2T1T12dtdtdtdthord2 T1+dt 211hc++TT2hT2 1≡bT1 − T∞dT1=0+dtT T 1 2 (5.19a)c(T1 − T∞ )if we call T1 − T∞ ≡ θ, then eqn.
(5.19a) can be written asd2 θdθ+ cθ = 0+b2dtdt(5.19b)Thus we have reduced the pair of first-order equations, eqn. (5.15) andeqn. (5.16), to a single second-order equation, eqn. (5.19b).The general solution of eqn. (5.19b) is obtained by guessing a solutionof the form θ = C1 eDt . Substitution of this guess into eqn. (5.19b) givesD 2 + bD + c = 0(5.20)3from which we find that D = −(b/2) ± (b/2)2 − c. This gives us twovalues of D, from which we can get two exponential solutions.
By adding202Transient and multidimensional heat conduction§5.2them together, we form a general solution:⎡⎤⎡⎤22 2 2bbbb− c ⎦ t + C2 exp ⎣− −− c ⎦tθ = C1 exp ⎣− +2222(5.21)To solve for the two constants we first substitute eqn. (5.21) in thefirst of i.c.’s (5.17) and getTi − T∞ = θi = C1 + C2(5.22)The second i.c.
can be put into terms of T1 with the help of eqn. (5.15):hc AdT1 =(T1 − T2 )t=0 = 0−dt t=0(ρcV )1We substitute eqn. (5.21) in this and obtain⎤⎡⎤⎡22 2 2bbbb− c ⎦ C1 + ⎣− −− c ⎦ C20 = ⎣− +2222 = θi − C 1soC1 = −θiandC2 = θi3−b/2 − (b/2)2 − c32 (b/2)2 − c3−b/2 + (b/2)2 − c32 (b/2)2 − cSo we obtain at last:3⎡T1 − T∞−cθb/2 +b3≡=exp ⎣−Ti − T∞θi22 (b/2)2 − c⎡3−b/2 + (b/2)2 − cb3+exp ⎣−222 (b/2) − c(b/2)2⎤2 2b+− c⎦ t2⎤2 2b−− c⎦ t2(5.23)This is a pretty complicated result—all the more complicated whenwe remember that b involves three algebraic terms [recall eqn.
(5.19a)].Yet there is nothing very sophisticated about it; it is easy to understand.A system involving three capacitances in series would similarly yield athird-order equation of correspondingly higher complexity, and so forth.§5.3Transient conduction in a one-dimensional slab203Figure 5.6 The transient cooling of aslab; ξ = (x/L) + 1.5.3Transient conduction in a one-dimensional slabWe next extend consideration to heat flow in bodies whose internal resistance is significant—to situations in which the lumped capacitanceassumption is no longer appropriate. When the temperature within, say,a one-dimensional body varies with position as well as time, we mustsolve the heat diffusion equation for T (x, t).
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