John H. Lienhard IV, John H. Lienhard V. A Heat Transfer Textbook, страница 6

Thus5d(T − T∞ )hA=−(T − T∞ )dtρcV(1.20)5Is it clear why (T −Tref ) has been changed to (T −T∞ ) under the derivative? Remember that the derivative of a constant (like Tref or T∞ ) is zero. We can therefore introduce(T − T∞ ) without invalidating the equation, and get the same dependent variable onboth sides of the equation.Modes of heat transfer§1.323Figure 1.10 The cooling of a body for which the Biot number,hL/kb , is small.The general solution to this equation isln(T − T∞ ) = −t+C(ρcV hA)(1.21)The group ρcV hA is the time constant, T . If the initial temperature isT (t = 0) ≡ Ti , then C = ln(Ti − T∞ ), and the cooling of the body is givenbyT − T∞= e−t/TTi − T∞(1.22)All of the physical parameters in the problem have now been “lumped”into the time constant.

It represents the time required for a body to coolto 1/e, or 37% of its initial temperature difference above (or below) T∞ .Introduction24§1.3The ratio t/T can also be interpreted astcapacity for convection from surfacehAt (J/◦ C)==TρcV (J/◦ C)heat capacity of the body(1.23)Notice that the thermal conductivity is missing from eqns. (1.22) and(1.23). The reason is that we have assumed that the temperature of thebody is nearly uniform, and this means that internal conduction is notimportant. We see in Fig. 1.10 that, if L (kb / h) 1, the temperature ofthe body, Tb , is almost constant within the body at any time.

ThushL 1 implies that Tb (x, t) T (t) Tsurfacekband the thermal conductivity, kb , becomes irrelevant to the cooling process. This condition must be satisfied or the lumped-capacity solutionwill not be accurate.We call the group hL kb the Biot number 6 , Bi. If Bi were large, ofcourse, the situation would be reversed, as shown in Fig. 1.11. In thiscase Bi = hL/kb 1 and the convection process offers little resistanceto heat transfer. We could solve the heat diffusion equation1 ∂T∂2T=2∂xα ∂tsubject to the simple boundary condition T (x, t) = T∞ when x = L, todetermine the temperature in the body and its rate of cooling in this case.The Biot number will therefore be the basis for determining what sort ofproblem we have to solve.To calculate the rate of entropy production in a lumped-capacity system, we note that the entropy change of the universe is the sum of theentropy decrease of the body and the more rapid entropy increase ofthe surroundings.

The source of irreversibility is heat flow through theboundary layer. Accordingly, we write the time rate of change of entropyof the universe, dSUn /dt ≡ ṠUn , asṠUn = Ṡb + Ṡsurroundings =6−QrevQrev+TbT∞Pronounced Bee-oh. J.B. Biot, although younger than Fourier, worked on the analysis of heat conduction even earlier—in 1802 or 1803. He grappled with the problemof including external convection in heat conduction analyses in 1804 but could not seehow to do it.

Fourier read Biot’s work and by 1807 had determined how to analyze theproblem. (Later we encounter a similar dimensionless group called the Nusselt number, Nu = hL/kfluid . The latter relates only to the boundary layer and not to the bodybeing cooled. We deal with it extensively in the study of convection.)Modes of heat transfer§1.325Figure 1.11 The cooling of a body for which the Biot number,hL/kb , is large.orṠUndTb= −ρcVdt11−T∞Tb.We can multiply both sides of this equation by dt and integrate the righthand side from Tb (t = 0) ≡ Tb0 to Tb at the time of interest:∆S = −ρcV Tb Tb011−T∞TbdTb .(1.24)Equation 1.24 will give a positive ∆S whether Tb > T∞ or Tb < T∞ becausethe sign of dTb will always opposed the sign of the integrand.Example 1.4A thermocouple bead is largely solder, 1 mm in diameter.

It is initiallyat room temperature and is suddenly placed in a 200◦ C gas flow. Theheat transfer coefficient h is 250 W/m2 K, and the effective valuesof k, ρ, and c are 45 W/m·K, 9300 kg/m3 , and c = 0.18 kJ/kg·K,respectively. Evaluate the response of the thermocouple.Introduction26§1.3Solution. The time constant, T , isTρc π D 3/6ρcD=2hAh πD6hm2·K 1000 W(9300)(0.18)(0.001) kg kJm=6(250)m3 kg·KWkJ/s= 1.116 s=ρcV=Therefore, eqn.

(1.22) becomesT − 200◦ C= e−t/1.116 or T = 200 − 180 e−t/1.116 ◦ C(20 − 200)◦ CThis result is plotted in Fig. 1.12, where we see that, for all practicalpurposes, this thermocouple catches up with the gas stream in lessthan 5 s. Indeed, it should be apparent that any such system willcome within 95% of the signal in three time constants. Notice, too,that if the response could continue at its initial rate, the thermocouplewould reach the signal temperature in one time constant.This calculation is based entirely on the assumption that Bi 1for the thermocouple. We must check that assumption:Bi ≡(250 W/m2 K)(0.001 m)/2hL== 0.00278k45 W/m·KThis is very small indeed, so the assumption is valid.Experiment 1.2Invent and carry out a simple procedure for evaluating the time constant of a fever thermometer in your mouth.RadiationHeat transfer by thermal radiation.

All bodies constantly emit energyby a process of electromagnetic radiation. The intensity of such energyflux depends upon the temperature of the body and the nature of itssurface. Most of the heat that reaches you when you sit in front of a fireis radiant energy. Radiant energy browns your toast in an electric toasterand it warms you when you walk in the sun.Modes of heat transfer§1.3Figure 1.12 Thermocouple response to a hot gas flow.Objects that are cooler than the fire, the toaster, or the sun emit muchless energy because the energy emission varies as the fourth power of absolute temperature. Very often, the emission of energy, or radiant heattransfer, from cooler bodies can be neglected in comparison with convection and conduction.

But heat transfer processes that occur at hightemperature, or with conduction or convection suppressed by evacuatedinsulations, usually involve a significant fraction of radiation.Experiment 1.3Open the freezer door to your refrigerator. Put your face near it, butstay far enough away to avoid the downwash of cooled air. This way youcannot be cooled by convection and, because the air between you and thefreezer is a fine insulator, you cannot be cooled by conduction. Still yourface will feel cooler. The reason is that you radiate heat directly into thecold region and it radiates very little heat to you.

Consequently, yourface cools perceptibly.2728Introduction§1.3Table 1.2 Forms of the electromagnetic wave spectrumCharacterizationWavelength, λCosmic rays< 0.3 pmGamma rays0.3–100 pmX rays0.01–30 nmUltraviolet light3–400 nmVisible light0.4–0.7 µmNear infrared radiation0.7–30 µmFar infrared radiation30–1000 µmMillimeter waves1–10 mmMicrowaves10–300 mmShortwave radio & TV300 mm–100 mLongwave radio100 m–30 km⎫⎪⎪⎪⎪⎬⎪⎪⎪⎪⎭Thermal Radiation0.1–1000 µmThe electromagnetic spectrum.

Thermal radiation occurs in a rangeof the electromagnetic spectrum of energy emission. Accordingly, it exhibits the same wavelike properties as light or radio waves. Each quantum of radiant energy has a wavelength, λ, and a frequency, ν, associatedwith it.The full electromagnetic spectrum includes an enormous range ofenergy-bearing waves, of which heat is only a small part. Table 1.2 liststhe various forms over a range of wavelengths that spans 17 orders ofmagnitude.

Only the tiniest “window” exists in this spectrum throughwhich we can see the world around us. Heat radiation, whose main component is usually the spectrum of infrared radiation, passes through themuch larger window—about three orders of magnitude in λ or ν.Black bodies.

The model for the perfect thermal radiator is a so-calledblack body. This is a body which absorbs all energy that reaches it andreflects nothing. The term can be a little confusing, since such bodiesemit energy. Thus, if we possessed infrared vision, a black body wouldglow with “color” appropriate to its temperature. of course, perfect radiators are “black” in the sense that they absorb all visible light (and allother radiation) that reaches them.§1.3Modes of heat transfer29Figure 1.13 Cross section of a spherical hohlraum.

The holehas the attributes of a nearly perfect thermal black body.It is necessary to have an experimental method for making a perfectlyblack body. The conventional device for approaching this ideal is calledby the German term hohlraum, which literally means “hollow space”.Figure 1.13 shows how a hohlraum is arranged. It is simply a device thattraps all the energy that reaches the aperture.What are the important features of a thermally black body? Firstconsider a distinction between heat and infrared radiation. Infrared radiation refers to a particular range of wavelengths, while heat refers tothe whole range of radiant energy flowing from one body to another.Suppose that a radiant heat flux, q, falls upon a translucent plate thatis not black, as shown in Fig.

1.14. A fraction, α, of the total incidentenergy, called the absorptance, is absorbed in the body; a fraction, ρ,Figure 1.14 The distribution of energyincident on a translucent slab.Introduction30§1.3called the reflectance, is reflected from it; and a fraction, τ, called thetransmittance, passes through. Thus1=α+ρ+τ(1.25)This relation can also be written for the energy carried by each wavelength in the distribution of wavelengths that makes up heat from asource at any temperature:1 = αλ + ρλ + τλ(1.26)All radiant energy incident on a black body is absorbed, so that αb orαλb = 1 and ρb = τb = 0.