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28 Introduction Table 1.2 §1.3 Forms of the electromagnetic wave spectrum Characterization Wavelength, λ Cosmic rays < 0.3 pm Gamma rays 0.3–100 pm X rays 0.01–30 nm Ultraviolet light 3–400 nm Visible light 0.4–0.7 µm Near infrared radiation 0.7–30 µm Far infrared radiation 30–1000 µm Millimeter waves 1–10 mm Microwaves 10–300 mm Shortwave radio & TV 300 mm–100 m Longwave radio 100 m–30 km            Thermal Radiation 0.1–1000 µm Black bodies. The model for the perfect thermal radiator is a so-called black body. This is a body which absorbs all energy that reaches it and reflects nothing. The term can be a little confusing, since such bodies emit energy. Thus, if we possessed infrared vision, a black body would glow with “color” appropriate to its temperature. of course, perfect radiators are “black” in the sense that they absorb all visible light (and all other radiation) that reaches them. It is necessary to have an experimental method for making a perfectly black body. The conventional device for approaching this ideal is called by the German term hohlraum, which literally means “hollow space”. Figure 1.13 shows how a hohlraum is arranged. It is simply a device that traps all the energy that reaches the aperture. What are the important features of a thermally black body? First consider a distinction between heat and infrared radiation. Infrared radiation refers to a particular range of wavelengths, while heat refers to the whole range of radiant energy flowing from one body to another. Suppose that a radiant heat flux, q, falls upon a translucent plate that is not black, as shown in Fig. 1.14. A fraction, α, of the total incident energy, called the absorptance, is absorbed in the body; a fraction, ρ, called the reflectance, is reflected from it; and a fraction, τ, called the Modes of heat transfer §1.3 29 Figure 1.13 Cross section of a spherical hohlraum. The hole has the attributes of a nearly perfect thermal black body. transmittance, passes through. Thus 1=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 a source at any temperature: 1 = αλ + ρλ + τλ (1.26) All radiant energy incident on a black body is absorbed, so that αb or αλb = 1 and ρb = τb = 0. Furthermore, the energy emitted from a black body reaches a theoretical maximum, which is given by the StefanBoltzmann law. We look at this next. Figure 1.14 The distribution of energy incident on a translucent slab. Introduction 30 §1.3 The Stefan-Boltzmann law. The flux of energy radiating from a body is commonly designated e(T ) W/m2 . The symbol eλ (λ, T ) designates the distribution function of radiative flux in λ, or the monochromatic emissive power: eλ (λ, T ) = de(λ, T ) or e(λ, T ) = dλ Thus e(T ) ≡ E(∞, T ) = λ 0 eλ (λ, T ) dλ (1.27) ∞ 0 eλ (λ, T ) dλ The dependence of e(T ) on T for a black body was established experimentally by Stefan in 1879 and explained by Boltzmann on the basis of thermodynamics arguments in 1884. The Stefan-Boltzmann law is eb (T ) = σ T 4 (1.28) where the Stefan-Boltzmann constant, σ , is 5.670400 × 10−8 W/m2 ·K4 or 1.714 × 10−9 Btu/hr·ft2 ·◦ R4 , and T is the absolute temperature. eλ vs. λ. Nature requires that, at a given temperature, a body will emit a unique distribution of energy in wavelength. Thus, when you heat a poker in the fire, it first glows a dull red—emitting most of its energy at long wavelengths and just a little bit in the visible regime. When it is white-hot, the energy distribution has been both greatly increased and shifted toward the shorter-wavelength visible range. At each temperature, a black body yields the highest value of eλ that a body can attain. The very accurate measurements of the black-body energy spectrum by Lummer and Pringsheim (1899) are shown in Fig. 1.15. The locus of maxima of the curves is also plotted. It obeys a relation called Wien’s law: (λT )eλ=max = 2898 µm·K (1.29) About three-fourths of the radiant energy of a black body lies to the right of this line in Fig. 1.15. Notice that, while the locus of maxima leans toward the visible range at higher temperatures, only a small fraction of the radiation is visible even at the highest temperature. Predicting how the monochromatic emissive power of a black body depends on λ was an increasingly serious problem at the close of the Modes of heat transfer §1.3 31 Figure 1.15 Monochromatic emissive power of a black body at several temperatures—predicted and observed. nineteenth century. The prediction was a keystone of the most profound scientific revolution the world has seen. In 1901, Max Planck made the prediction, and his work included the initial formulation of quantum mechanics. He found that eλb = 2π hco2 λ5 [exp(hco /kB T λ) − 1] (1.30) where co is the speed of light, 2.99792458 × 108 m/s; h is Planck’s constant, 6.62606876×10−34 J·s; and kB is Boltzmann’s constant, 1.3806503× 10−23 J/K. Radiant heat exchange. Suppose that a heated object (1 in Fig. 1.16a) radiates only to some other object (2) and that both objects are thermally black. All heat leaving object 1 arrives at object 2, and all heat arriving at object 1 comes from object 2. Thus, the net heat transferred from object 1 to object 2, Qnet , is the difference between Q1 to 2 = A1 eb (T1 ) Introduction 32 Figure 1.16 another. §1.3 The net radiant heat transfer from one object to and Q2 to 1 = A1 eb (T2 ) Qnet = A1 eb (T1 ) − A1 eb (T2 ) = A1 σ T14 − T24 (1.31) If the first object “sees” other objects in addition to object 2, as indicated in Fig. 1.16b, then a view factor (sometimes called a configuration factor or a shape factor ), F1–2 , must be included in eqn. (1.31): Qnet = A1 F1–2 σ T14 − T24 (1.32) We may regard F1–2 as the fraction of energy leaving object 1 that is intercepted by object 2. Example 1.5 A black thermocouple measures the temperature in a chamber with black walls. If the air around the thermocouple is at 20◦ C, the walls are at 100◦ C, and the heat transfer coefficient between the thermocouple and the air is 15 W/m2 K, what temperature will the thermocouple read? Modes of heat transfer §1.3 33 Solution. The heat convected away from the thermocouple by the air must exactly balance that radiated to it by the hot walls if the system is in steady state. Furthermore, F1–2 = 1 since the thermocouple (1) radiates all its energy to the walls (2): 4 4 − Twall hAtc (Ttc − Tair ) = −Qnet = −Atc σ Ttc or, with Ttc in ◦ C, 15(Ttc − 20) W/m2 =   5.6704 × 10−8 (100 + 273)4 − (Ttc + 273)4 W/m2 since T for radiation must be in kelvin. Trial-and-error solution of this equation yields Ttc = 51◦ C. We have seen that non-black bodies absorb less radiation than black bodies, which are perfect absorbers. Likewise, non-black bodies emit less radiation than black bodies, which also happen to be perfect emitters. We can characterize the emissive power of a non-black body using a property called emittance, ε: enon-black = εeb = εσ T 4 (1.33) where 0 < ε ≤ 1. When radiation is exchanged between two bodies that are not black, we have Qnet = A1 F1–2 σ T14 − T24 (1.34) where the transfer factor, F1–2 , depends on the emittances of both bodies as well as the geometrical “view”. The expression for F1–2 is particularly simple in the important special case of a small object, 1, in a much larger isothermal environment, 2: F1–2 = ε1 for A1  A2 (1.35) Example 1.6 Suppose that the thermocouple in Example 1.5 was not black and had an emissivity of ε = 0.4. Further suppose that the walls were not black and had a much larger surface area than the thermocouple. What temperature would the thermocouple read? Introduction 34 §1.3 Solution. Qnet is now given by eqn. (1.34) and F1–2 can be found with eqn. (1.35): 4 4 hAtc (Ttc − Tair ) = −Atc εtc σ Ttc − Twall or 15(Ttc − 20) W/m2 =   (0.4)(5.6704 × 10−8 ) (100 + 273)4 − (Ttc + 273)4 W/m2 Trial-and-error yields Ttc = 35◦ C. Radiation shielding. The preceding examples point out an important practical problem than can be solved with radiation shielding. The idea is as follows: If we want to measure the true air temperature, we can place a thin foil casing, or shield, around the thermocouple. The casing is shaped to obstruct the thermocouple’s “view” of the room but to permit the free flow of the air around the thermocouple. Then the shield, like the thermocouple in the two examples, will be cooler than the walls, and the thermocouple it surrounds will be influenced by this much cooler radiator. If the shield is highly reflecting on the outside, it will assume a temperature still closer to that of the air and the error will be still less. Multiple layers of shielding can further reduce the error. Radiation shielding can take many forms and serve many purposes. It is an important element in superinsulations. A glass firescreen in a fireplace serves as a radiation shield because it is largely opaque to radiation. It absorbs heat radiated by the fire and reradiates that energy (ineffectively) at a temperature much lower than that of the fire. Experiment 1.4 Find a small open flame that produces a fair amount of soot. A candle, kerosene lamp, or a cutting torch with a fuel-rich mixture should work well. A clean blue flame will not work well because such gases do not radiate much heat. First, place your finger in a position about 1 to 2 cm to one side of the flame, where it becomes uncomfortably hot. Now take a piece of fine mesh screen and dip it in some soapy water, which will fill up the holes. Put it between your finger and the flame. You will see that your finger is protected from the heating until the water evaporates. Water is relatively transparent to light. What does this experiment show you about the transmittance of water to infrared wavelengths? A look ahead §1.5 1.4 A look ahead What we have done up to this point has been no more than to reveal the tip of the iceberg. The basic mechanisms of heat transfer have been explained and some quantitative relations have been presented. However, this information will barely get you started when you are faced with a real heat transfer problem. Three tasks, in particular, must be completed to solve actual problems: • The heat diffusion equation must be solved subject to appropriate boundary conditions if the problem involves heat conduction of any complexity. • The convective heat transfer coefficient, h, must be determined if convection is important in a problem. • The factor F1–2 or F1–2 must be determined to calculate radiative heat transfer. Any of these determinations can involve a great deal of complication, and most of the chapters that lie ahead are devoted to these three basic problems. Before becoming engrossed in these three questions, we shall first look at the archetypical applied problem of heat transfer–namely, the design of a heat exchanger. Chapter 2 sets up the elementary analytical apparatus that is needed for this, and Chapter 3 shows how to do such design if h is already known. This will make it easier to see the importance of undertaking the three basic problems in subsequent parts of the book. 1.5 Problems We have noted that this book is set down almost exclusively in S.I. units. The student who has problems with dimensional conversion will find Appendix B helpful. The only use of English units appears in some of the problems at the end of each chapter. A few such problems are included to provide experience in converting back into English units, since such units will undoubtedly persist in the U.S.A. for many more years. Another matter often leads to some discussion between students and teachers in heat transfer courses. That is the question of whether a problem is “theoretical” or “practical”. Quite often the student is inclined to 35 36 Chapter 1: Introduction view as “theoretical” a problem that does not involve numbers or that requires the development of algebraic results. The problems assigned in this book are all intended to be useful in that they do one or more of five things: 1. They involve a calculation of a type that actually arises in practice (e.g., Problems 1.1, 1.3, 1.8 to 1.18, and 1.21 through 1.25). 2. They illustrate a physical principle (e.g., Problems 1.2, 1.4 to 1.7, 1.9, 1.20, 1.32, and 1.39). These are probably closest to having a “theoretical” objective. 3. They ask you to use methods developed in the text to develop other results that would be needed in certain applied problems (e.g., Problems 1.10, 1.16, 1.17, and 1.21). Such problems are usually the most difficult and the most valuable to you. 4. They anticipate development that will appear in subsequent chapters (e.g., Problems 1.16, 1.20, 1.40, and 1.41). 5. They require that you develop your ability to handle numerical and algebraic computation effectively. (This is the case with most of the problems in Chapter 1, but it is especially true of Problems 1.6 to 1.9, 1.15, and 1.17). Partial numerical answers to some of the problems follow them in brackets. Tables of physical property data useful in solving the problems are given in Appendix A. Actually, we wish to look at the theory, analysis, and practice of heat transfer—all three—according to Webster’s definitions: Theory: “a systematic statement of principles; a formulation of apparent relationships or underlying principles of certain observed phenomena.” Analysis: “the solving of problems by the means of equations; the breaking up of any whole into its parts so as to find out their nature, function, relationship, etc.” Practice: “the doing of something as an application of knowledge.” Problems 37 Problems 1.1 A composite wall consists of alternate layers of fir (5 cm thick), aluminum (1 cm thick), lead (1 cm thick), and corkboard (6 cm thick). The temperature is 60◦ C on the outside of the for and 10◦ C on the outside of the corkboard. Plot the temperature gradient through the wall. Does the temperature profile suggest any simplifying assumptions that might be made in subsequent analysis of the wall? 1.2 Verify eqn. (1.15). 1.3 q = 5000 W/m2 in a 1 cm slab and T = 140◦ C on the cold side. Tabulate the temperature drop through the slab if it is made of • Silver • Aluminum • Mild steel (0.5 % carbon) • Ice • Spruce • Insulation (85 % magnesia) • Silica aerogel Indicate which situations would be unreasonable and why. 1.4 Explain in words why the heat diffusion equation, eqn. (1.13), shows that in transient conduction the temperature depends on the thermal diffusivity, α, but we can solve steady conduction problems using just k (as in Example 1.1). 1.5 A 1 m rod of pure copper 1 cm2 in cross section connects a 200◦ C thermal reservoir with a 0◦ C thermal reservoir. The system has already reached steady state. What are the rates of change of entropy of (a) the first reservoir, (b) the second reservoir, (c) the rod, and (d) the whole universe, as a result of the process? Explain whether or not your answer satisfies the Second Law of Thermodynamics. [(d): +0.0120 W/K.] 1.6 Two thermal energy reservoirs at temperatures of 27◦ C and −43◦ C, respectively, are separated by a slab of material 10 cm thick and 930 cm2 in cross-sectional area. The slab has