John H. Lienhard IV, John H. Lienhard V. A Heat Transfer Textbook (776116), страница 11
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Always stop andask yourself, “Would I have access to a numerical value of thetemperature (or other data) that I specify at a given position ortime?” If the answer is no, then your result will be useless.Step 5. Substitute the general solution in the boundary and initial conditions and solve for the constants. This process gets very complicated in the transient and multidimensional cases. Fourierseries methods are typically needed to solve the problem. However, the steady one-dimensional problems are usually easy.
Inthe example, by evaluating at x = 0 and x = L, we get:Tw = −0 + 0 + C2Tw = −q̇L2+ C1 L + C22k =TwsoC2 = TwsoC1 =q̇L2k5960Heat conduction, thermal resistance, and the overall heat transfer coefficient§2.2Figure 2.6 Temperature distribution in the setting concreteslab Example 2.1.Step 6. Put the calculated constants back in the general solution to getthe particular solution to the problem. In the example problemwe obtain:T =−q̇q̇ 2x +Lx + Tw2k2kThis should be put in neat dimensionless form:T − Tw1 =2q̇L2 k 2 xx−LL(2.15)Step 7.
Play with the solution—look it over—see what it has to tell you.Make any checks you can think of to be sure it is correct. In thiscase we plot eqn. (2.15) in Fig. 2.6. The resulting temperaturedistribution is parabolic and, as we would expect, symmetrical.It satisfies the boundary conditions at the wall and maximizesin the center. By nondimensionalizing the result, we have succeeded in representing all situations with a simple curve.
Thatis highly desirable when the calculations are not simple, as theyare here. (Notice that T actually depends on five different things,yet the solution is a single curve on a two-coordinate graph.)Solutions of the heat diffusion equation§2.2Finally, we check to see if the heat flux at the wall is correct:∂T q̇Lq̇Lq̇x−=k=−qwall = −k∂x x=0k2k x=02Thus, half of the total energy generated in the slab comes outof the front side, as we would expect.
The solution appears tobe correct.Step 8. If the temperature field is now correctly established, you can,if you wish, calculate the heat flux at any point in the body bysubstituting T (r , t) back into Fourier’s law. We did this already,in Step 7, to check our solution.We shall run through additional examples in this section and the following one.
In the process, we shall develop some important results forfuture use.Example 2.2The Simple SlabA slab shown in Fig. 2.7 is at a steady state with dissimilar temperatures on either side and no internal heat generation. We want thetemperature distribution and the heat flux through it.Solution.
These can be found quickly by following the steps setdown in Example 2.1:Figure 2.7 Heat conduction in a slab (Example 2.2).6162Heat conduction, thermal resistance, and the overall heat transfer coefficient§2.3Step 1. T = T (x) for steady x-direction heat flowStep 2.d2 T= 0, the steady 1-D heat equation with no q̇dx 2Step 3. T = C1 x + C2 is the general solution of that equationStep 4.
T (x = 0) = T1 and T (x = L) = T2 are the b.c.sStep 5. T1 = 0 + C2 , so C2 = T1 ; and T2 = C1 L + C2 , so C1 =Step 6. T = T1 +T2 − T1LT − T1T2 − T1xx; or=LT2 − T1LStep 7. We note that the solution satisfies the boundary conditionsand that the temperature profile is linear.dT1 − T2dTT1 −= −kxStep 8. q = −kdxdxLso thatq=k∆TLThis result, which is the simplest heat conduction solution, calls tomind Ohm’s law.
Thus, if we rearrange it:Q=∆TL/kAis likeI=ERwhere L/kA assumes the role of a thermal resistance, to which we givethe symbol Rt . Rt has the dimensions of (K/W). Figure 2.8 shows how wecan represent heat flow through the slab with a diagram that is perfectlyanalogous to an electric circuit.2.3Thermal resistance and the electrical analogyFourier’s, Fick’s, and Ohm’s lawsFourier’s law has several extremely important analogies in other kinds ofphysical behavior, of which the electrical analogy is only one. These analogous processes provide us with a good deal of guidance in the solutionof heat transfer problems And, conversely, heat conduction analyses canoften be adapted to describe those processes.§2.3Thermal resistance and the electrical analogyFigure 2.8 Ohm’s law analogy to conduction through a slab.Let us first consider Ohm’s law in three dimensions:flux of electrical charge =I≡ J = −γ∇VA(2.16)I amperes is the vectorial electrical current, A is an area normal to thecurrent vector, J is the flux of current or current density, γ is the electricalconductivity in cm/ohm·cm2 , and V is the voltage.To apply eqn.
(2.16) to a one-dimensional current flow, as pictured inFig. 2.9, we write eqn. (2.16) asJ = −γ∆VdV=γ,dxL(2.17)but ∆V is the applied voltage, E, and the resistance of the wire is R ≡L γA. Then, since I = J A, eqn. (2.17) becomesI=ER(2.18)which is the familiar, but restrictive, one-dimensional statement of Ohm’slaw.Fick’s law is another analogous relation. It states that during massdiffusion, the flux, j1 , of a dilute component, 1, into a second fluid, 2, is6364Heat conduction, thermal resistance, and the overall heat transfer coefficient§2.3Figure 2.9 The one-dimensional flow ofcurrent.proportional to the gradient of its mass concentration, m1 . Thusj1 = −ρD12 ∇m1(2.19)where the constant D12 is the binary diffusion coefficient.Example 2.3Air fills a thin tube 1 m in length.
There is a small water leak at oneend where the water vapor concentration builds to a mass fraction of0.01. A desiccator maintains the concentration at zero on the otherside. What is the steady flux of water from one side to the other ifD12 is 2.84 × 10−5 m2/s and ρ = 1.18 kg/m3 ?Solution.kgjwater vapor = 1.18 3m2−5 m2.84 × 10= 3.35 × 10−7s0.01 kg H2 O/kg mixture1mkgm2 ·sContact resistanceOne place in which the usefulness of the electrical resistance analogy becomes immediately apparent is at the interface of two conducting media.No two solid surfaces will ever form perfect thermal contact when theyare pressed together.
Since some roughness is always present, a typicalplane of contact will always include tiny air gaps as shown in Fig. 2.10§2.3Thermal resistance and the electrical analogyFigure 2.10 Heat transfer through the contact plane betweentwo solid surfaces.(which is drawn with a highly exaggerated vertical scale). Heat transferfollows two paths through such an interface. Conduction through pointsof solid-to-solid contact is very effective, but conduction through the gasfilled interstices, which have low thermal conductivity, can be very poor.Thermal radiation across the gaps is also inefficient.We treat the contact surface by placing an interfacial conductance, hc ,in series with the conducting materials on either side.
The coefficient hcis similar to a heat transfer coefficient and has the same units, W/m2 K. If∆T is the temperature difference across an interface of area A, then Q =Ahc ∆T . It follows that Q = ∆T /Rt for a contact resistance Rt = 1/(hc A)in K/W.The interfacial conductance, hc , depends on the following factors:• The surface finish and cleanliness of the contacting solids.• The materials that are in contact.• The pressure with which the surfaces are forced together.
This mayvary over the surface, for example, in the vicinity of a bolt.• The substance (or lack of it) in the interstitial spaces. Conductiveshims or fillers can raise the interfacial conductance.• The temperature at the contact plane.The influence of contact pressure is usually a modest one up to around10 atm in most metals. Beyond that, increasing plastic deformation of6566Heat conduction, thermal resistance, and the overall heat transfer coefficient§2.3Table 2.1 Some typical interfacial conductances for normalsurface finishes and moderate contact pressures (about 1 to 10atm). Air gaps not evacuated unless so indicated.SituationIron/aluminum (70 atm pressure)Copper/copperAluminum/aluminumGraphite/metalsCeramic/metalsStainless steel/stainless steelCeramic/ceramicStainless steel/stainless steel(evacuated interstices)Aluminum/aluminum (low pressureand evacuated interstices)hc (W/m2 K)45, 00010, 000 − 25, 0002, 200 − 12, 0003, 000 − 6, 0001, 500 − 8, 5002, 000 − 3, 700500 − 3, 000200 − 1, 100100 − 400the local contact points causes hc to increase more dramatically at highpressure.
Table 2.1 gives typical values of contact resistances which bearout most of the preceding points. These values have been adapted from[2.1, Chpt. 3] and [2.2]. Theories of contact resistance are discussed in[2.3] and [2.4].Example 2.4Heat flows through two stainless steel slabs (k = 18 W/m·K) that arepressed together. The slab area is A = 1 m2 . How thick must theslabs be for contact resistance to be negligible?Solution. With reference to Fig. 2.11, the total or equivalent resistance is found by adding these resistances, which are in series:L11 L1LL++=++Rtequiv =kA hc A kAA 18 hc18Since hc is about 3,000 W/m2 K,12Lmust be = 0.00033183000Thus, L must be large compared to 18(0.00033)/2 = 0.003 m if contactresistance is to be ignored. If L = 3 cm, the error is about 10%.Thermal resistance and the electrical analogy§2.367Figure 2.11 Conduction through twounit-area slabs with a contact resistance.Resistances for cylinders and for convectionAs we continue developing our method of solving one-dimensional heatconduction problems, we find that other avenues of heat flow may also beexpressed as thermal resistances, and introduced into the solutions thatwe obtain.
We also find that, once the heat conduction equation has beensolved, the results themselves may be used as new thermal resistances.Example 2.5Radial Heat Conduction in a TubeFind the temperature distribution and the heat flux for the long hollowcylinder shown in Fig.
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