John H. Lienhard IV, John H. Lienhard V. A Heat Transfer Textbook, страница 5

It is protected fromcorrosion on each side by a 2-mm-thick layer of stainless steel (k = 17W/m·K). The temperature is 400◦ C on one side of this composite walland 100◦ C on the other. Find the temperature distribution in thecopper slab and the heat conducted through the wall (see Fig. 1.7).Solution. If we recall Fig. 1.5 and eqn. (1.10), it should be clear thatthe temperature drop will take place almost entirely in the stainlesssteel, where k is less than 1/20 of k in the copper. Thus, the copper will be virtually isothermal at the average temperature of (400 +100)/2 = 250◦ C.

Furthermore, the heat conduction can be estimatedin a 4 mm slab of stainless steel as though the copper were not eventhere. With the help of Fourier’s law in the form of eqn. (1.8), we getq = −k400 − 100dTK/m = 1275 kW/m2 17 W/m·K ·dx0.004The accuracy of this rough calculation can be improved by considering the copper. To do this we first solve for ∆Ts.s. and ∆TCu (seeFig. 1.7). Conservation of energy requires that the steady heat fluxthrough all three slabs must be the same. Therefore,∆Tq= kLs.s.∆T= kLCuModes of heat transfer§1.317Figure 1.7 Temperature drop through acopper wall protected by stainless steel(Example 1.2).but(400 − 100)◦ C ≡ ∆TCu + 2∆Ts.s.(k/L)Cu= ∆TCu 1 + 2(k/L)s.s.= (30.18)∆TCuSolving this, we obtain ∆TCu = 9.94 K.

So ∆Ts.s. = (300 − 9.94)/2 =145 K. It follows that TCu, left = 255◦ C and TCu, right = 245◦ C.The heat flux can be obtained by applying Fourier’s law to any ofthe three layers. We consider either stainless steel layer and getq = 17W 145 K= 1233 kW/m2m·K 0.002 mThus our initial approximation was accurate within a few percent.One-dimensional heat diffusion equation. In Example 1.2 we had todeal with a major problem that arises in heat conduction problems. Theproblem is that Fourier’s law involves two dependent variables, T andq.

To eliminate q and first solve for T , we introduced the First Law ofThermodynamics implicitly: Conservation of energy required that q wasthe same in each metallic slab.The elimination of q from Fourier’s law must now be done in a moregeneral way. Consider a one-dimensional element, as shown in Fig. 1.8.Introduction18§1.3Figure 1.8 One-dimensional heat conduction through a differential element.From Fourier’s law applied at each side of the element, as shown, the netheat conduction out of the element during general unsteady heat flow isqnet A = Qnet = −kA∂2Tδx∂x 2(1.12)To eliminate the heat loss Qnet in favor of T , we use the general FirstLaw statement for closed, nonworking systems, eqn. (1.3):−Qnet =d(T − Tref )dTdU= ρcAδx = ρcAδxdtdtdt(1.13)where ρ is the density of the slab and c is its specific heat capacity.4Equations (1.12) and (1.13) can be combined to give1 ∂T∂2Tρc ∂T≡=∂x 2k ∂tα ∂t4(1.14)The reader might wonder if c should be cp or cv .

This is a strictly incompressibleequation so cp = cv = c. The compressible equation involves additional terms, andthis particular term emerges with cp in it in the conventional rearrangements of terms.Modes of heat transfer§1.319Figure 1.9 The convective cooling of a heated body.This is the one-dimensional heat diffusion equation. Its importance isthis: By combining the First Law with Fourier’s law, we have eliminatedthe unknown Q and obtained a differential equation that can be solvedfor the temperature distribution, T (x, t). It is the primary equation uponwhich all of heat conduction theory is based.The heat diffusion equation includes a new property which is as important to transient heat conduction as k is to steady-state conduction.This is the thermal diffusivity, α:α≡Jm3 kg·Kk= α m2/s (or ft2/hr).ρc m·s·K kg JThe thermal diffusivity is a measure of how quickly a material can carryheat away from a hot source.

Since material does not just transmit heatbut must be warmed by it as well, α involves both the conductivity, k,and the volumetric heat capacity, ρc.Heat ConvectionThe physical process. Consider a typical convective cooling situation.Cool gas flows past a warm body, as shown in Fig. 1.9. The fluid immediately adjacent to the body forms a thin slowed-down region called aboundary layer.

Heat is conducted into this layer, which sweeps it awayand, farther downstream, mixes it into the stream. We call such processesof carrying heat away by a moving fluid convection.In 1701, Isaac Newton considered the convective process and suggested that the cooling would be such thatdTbody∝ Tbody − T∞dt(1.15)where T∞ is the temperature of the oncoming fluid. This statement suggests that energy is flowing from the body. But if the energy of the bodyIntroduction20§1.3is constantly replenished, the body temperature need not change.

Thenwith the help of eqn. (1.3) we get, from eqn. (1.15) (see Problem 1.2),Q ∝ Tbody − T∞(1.16)This equation can be rephrased in terms of q = Q/A asq = h Tbody − T∞(1.17)This is the steady-state form of Newton’s law of cooling, as it is usuallyquoted, although Newton never wrote such an expression.The constant h is the film coefficient or heat transfer coefficient. Thebar over h indicates that it is an average over the surface of the body.Without the bar, h denotes the “local” value of the heat transfer coefficient at a point on the surface. The units of h and h are W/m2 K orJ/s·m2·K.

The conversion factor for English units is:1=0.0009478 BtuK3600 s (0.3048 m)2···J1.8◦ Fhft2or1 = 0.1761Btu/h·ft2 ·◦ FW/m2 K(1.18)It turns out that Newton oversimplified the process of convectionwhen he made his conjecture. Heat convection is complicated and hcan depend on the temperature difference Tbody − T∞ ≡ ∆T . In Chapter 6 we find that h really is independent of ∆T in situations in whichfluid is forced past a body and ∆T is not too large. This is called forcedconvection.When fluid buoys up from a hot body or down from a cold one, hvaries as some weak power of ∆T —typically as ∆T 1/4 or ∆T 1/3 .

This iscalled free or natural convection. If the body is hot enough to boil a liquidsurrounding it, h will typically vary as ∆T 2 .For the moment, we restrict consideration to situations in which Newton’s law is either true or at least a reasonable approximation to realbehavior.We should have some idea of how large h might be in a given situation. Table 1.1 provides some illustrative values of h that have beenModes of heat transfer§1.321Table 1.1 Some illustrative values of convective heat transfercoefficientsSituationNatural convection in gases• 0.3 m vertical wall in air, ∆T = 30◦ CNatural convection in liquids• 40 mm O.D. horizontal pipe in water, ∆T = 30◦ C• 0.25 mm diameter wire in methanol, ∆T = 50◦ CForced convection of gases• Air at 30 m/s over a 1 m flat plate, ∆T = 70◦ CForced convection of liquids• Water at 2 m/s over a 60 mm plate, ∆T = 15◦ C• Aniline-alcohol mixture at 3 m/s in a 25 mm I.D.

tube, ∆T = 80◦ C• Liquid sodium at 5 m/s in a 13 mm I.D. tube at 370◦ CBoiling water• During film boiling at 1 atm• In a tea kettle• At a peak pool-boiling heat flux, 1 atm• At a peak flow-boiling heat flux, 1 atm• At approximate maximum convective-boiling heat flux, underoptimal conditionsCondensation• In a typical horizontal cold-water-tube steam condenser• Same, but condensing benzene• Dropwise condensation of water at 1 atmobserved or calculated for different situations.

They are only illustrativeand should not be used in calculations because the situations for whichthey apply have not been fully described. Most of the values in the table could be changed a great deal by varying quantities (such as surfaceroughness or geometry) that have not been specified. The determinationof h or h is a fairly complicated task and one that will receive a greatdeal of our attention. Notice, too, that h can change dramatically fromone situation to the next.

Reasonable values of h range over about sixorders of magnitude.h, W/m2 K4.335704, 000805902, 60075, 0003004, 00040, 000100, 00010615, 0001, 700160, 000Introduction22§1.3Example 1.3The heat flux, q, is 6000 W/m2 at the surface of an electrical heater.The heater temperature is 120◦ C when it is cooled by air at 70◦ C.What is the average convective heat transfer coefficient, h? What willthe heater temperature be if the power is reduced so that q is 2000W/m2 ?Solution.h=q6000== 120 W/m2 K∆T120 − 70If the heat flux is reduced, h should remain unchanged during forcedconvection.

Thus∆T = Theater − 70◦ C =qh=2000 W/m2= 16.67 K120 W/m2 Kso Theater = 70 + 16.67 = 86.67◦ CLumped-capacity solution. We now wish to deal with a very simple butextremely important, kind of convective heat transfer problem. The problem is that of predicting the transient cooling of a convectively cooledobject, such as we showed in Fig. 1.9. With reference to Fig. 1.10, weapply our now-familiar First law statement, eqn. (1.3), to such a body:Q−hA(T − T∞ )=dUdt(1.19)d[ρcV (T − Tref )]dtwhere A and V are the surface area and volume of the body, T is thetemperature of the body, T = T (t), and Tref is the arbitrary temperatureat which U is defined equal to zero.