John H. Lienhard IV, John H. Lienhard V. A Heat Transfer Textbook, страница 4

If T2 → T1 , the process will approach beingquasistatic and reversible. But the rate of heat transfer will also approach2T = absolute temperature, S = entropy, V = volume, p = pressure, and “rev” denotesa reversible process.Relation of heat transfer to thermodynamics§1.2Figure 1.2 Irreversible heat flowbetween two thermal reservoirs throughan intervening wall.zero if there is no temperature difference to drive it. Thus all real heattransfer processes generate entropy.Now we come to a dilemma: If the irreversible process occurs atsteady state, the properties of the wall do not vary with time.

We knowthat the entropy of the wall depends on its state and must therefore beconstant. How, then, does the entropy of the universe increase? We turnto this question next.Entropy productionThe entropy increase of the universe as the result of a process is the sumof the entropy changes of all elements that are involved in that process.The rate of entropy production of the universe, ṠUn , resulting from thepreceding heat transfer process through a wall isṠUn = Ṡres 1 +Ṡwall+Ṡres 2(1.5)= 0, since Swallmust be constantwhere the dots denote time derivatives (i.e., ẋ ≡ dx/dt). Since the reservoir temperatures are constant,Ṡres =Q.Tres(1.6)Now Qres 1 is negative and equal in magnitude to Qres 2 , so eqn.

(1.5)becomes 11.(1.7)−QṠUn = res 1 TT219Introduction10§1.3The term in parentheses is positive, so ṠUn > 0. This agrees with Clausius’s statement of the Second Law of Thermodynamics.Notice an odd fact here: The rate of heat transfer, Q, and hence ṠUn ,is determined by the wall’s resistance to heat flow. Although the wallis the agent that causes the entropy of the universe to increase, its ownentropy does not change.

Only the entropies of the reservoirs change.1.3Modes of heat transferFigure 1.3 shows an analogy that might be useful in fixing the conceptsof heat conduction, convection, and radiation as we proceed to look ateach in some detail.Heat conductionFourier’s law. Joseph Fourier (see Fig.

1.4) published his remarkablebook Théorie Analytique de la Chaleur in 1822. In it he formulated a verycomplete exposition of the theory of heat conduction.Hebegan his treatise by stating the empirical law that bears his name:the heat flux,3 q (W/m2 ), resulting from thermal conduction is proportionalto the magnitude of the temperature gradient and opposite to it in sign. Ifwe call the constant of proportionality, k, thenq = −kdTdx(1.8)The constant, k, is called the thermal conductivity. It obviously must havethe dimensions W/m·K, or J/m·s·K, or Btu/h·ft·◦ F if eqn. (1.8) is to bedimensionally correct.The heat flux is a vector quantity. Equation (1.8) tells us that if temperature decreases with x, q will be positive—it will flow in the x-direction.If T increases with x, q will be negative—it will flow opposite the xdirection. In either case, q will flow from higher temperatures to lowertemperatures.

Equation (1.8) is the one-dimensional form of Fourier’slaw. We develop its three-dimensional form in Chapter 2, namely: = −k ∇Tq3The heat flux, q, is a heat rate per unit area and can be expressed as Q/A, where Ais an appropriate area.Figure 1.3 An analogy for the three modes of heat transfer.11Figure 1.4 Baron Jean Baptiste Joseph Fourier (1768–1830). JosephFourier lived a remarkable double life. He served as a high government official in Napoleonic France and he was also an applied mathematician of great importance. He was with Napoleon in Egypt between1798 and 1801, and he was subsequently prefect of the administrative area (or “Department”) of Isère in France until Napoleon’s firstfall in 1814. During the latter period he worked on the theory ofheat flow and in 1807 submitted a 234-page monograph on the subject.

It was given to such luminaries as Lagrange and Laplace forreview. They found fault with his adaptation of a series expansionsuggested by Daniel Bernoulli in the eighteenth century. Fourier’stheory of heat flow, his governing differential equation, and the nowfamous “Fourier series” solution of that equation did not emerge inprint from the ensuing controversy until 1822. (Etching from Portraits et Histoire des Hommes Utiles, Collection de Cinquante Portraits,Société Montyon et Franklin 1839-1840).12Modes of heat transfer§1.313Example 1.1The front of a slab of lead (k = 35 W/m·K) is kept at 110◦ C and theback is kept at 50◦ C.

If the area of the slab is 0.4 m2 and it is 0.03 mthick, compute the heat flux, q, and the heat transfer rate, Q.Solution. For the moment, we presume that dT /dx is a constantequal to (Tback − Tfront )/(xback − xfront ); we verify this in Chapter 2.Thus, eqn. (1.8) becomes50 − 110q = −350.03= +70, 000 W/m2 = 70 kW/m2andQ = qA = 70(0.4) = 28 kWIn one-dimensional heat conduction problems, there is never any realproblem in deciding which way the heat should flow.

It is therefore sometimes convenient to write Fourier’s law in simple scalar form:q=k∆TL(1.9)where L is the thickness in the direction of heat flow and q and ∆T areboth written as positive quantities. When we use eqn. (1.9), we mustremember that q always flows from high to low temperatures.Thermal conductivity values. It will help if we first consider how conduction occurs in, for example, a gas. We know that the molecular velocity depends on temperature. Consider conduction from a hot wall toa cold one in a situation in which gravity can be ignored, as shown inFig.

1.5. The molecules near the hot wall collide with it and are agitatedby the molecules of the wall. They leave with generally higher speed andcollide with their neighbors to the right, increasing the speed of thoseneighbors. This process continues until the molecules on the right passtheir kinetic energy to those in the cool wall. Within solids, comparableprocesses occur as the molecules vibrate within their lattice structureand as the lattice vibrates as a whole. This sort of process also occurs,to some extent, in the electron “gas” that moves through the solid. TheIntroduction14§1.3Figure 1.5 Heat conduction through gasseparating two solid walls.processes are more efficient in solids than they are in gases.

Notice that−q1dT=∝dxkk(1.10)since, in steadyconduction, q isconstantThus solids, with generally higher thermal conductivities than gases,yield smaller temperature gradients for a given heat flux. In a gas, bythe way, k is proportional to molecular speed and molar specific heat,and inversely proportional to the cross-sectional area of molecules.This book deals almost exclusively with S.I. units, or Système International d’Unités. Since much reference material will continue to be available in English units, we should have at hand a conversion factor forthermal conductivity:hft1.8◦ FJ···0.0009478 Btu 3600 s 0.3048 mKThus the conversion factor from W/m·K to its English equivalent, Btu/h·ft·◦ F, is1=1 = 1.731W/m·KBtu/h·ft·◦ F(1.11)Consider, for example, copper—the common substance with the highestconductivity at ordinary temperature:W/m·K= 221 Btu/h·ft·◦ FkCu at room temp = (383 W/m·K) 1.731Btu/h·ft·◦ F15Figure 1.6 The approximate ranges of thermal conductivity of various substances.(All values arefor the neighborhood of room temperature unless otherwise noted.)Introduction16§1.3The range of thermal conductivities is enormous.

As we see fromFig. 1.6, k varies by a factor of about 105 between gases and diamond atroom temperature. This variation can be increased to about 107 if we include the effective conductivity of various cryogenic “superinsulations.”(These involve powders, fibers, or multilayered materials that have beenevacuated of all air.) The reader should study and remember the orderof magnitude of the thermal conductivities of different types of materials. This will be a help in avoiding mistakes in future computations, andit will be a help in making assumptions during problem solving. Actualnumerical values of the thermal conductivity are given in Appendix A(which is a broad listing of many of the physical properties you mightneed in this course) and in Figs. 2.2 and 2.3.Example 1.2A copper slab (k = 372 W/m·K) is 3 mm thick.