Lecture notes on Radiation heat transfer

how radiation heat transfer works and what is radiation heat transfer coefficient. and how does radiation transfer heat from one object to another pdf free downlaod
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Part IV Thermal Radiation Heat Transfer 52310. Radiative heat transfer The sun that shines from Heaven shines but warm, And, lo, I lie between that sun and thee: The heat I have from thence doth little harm, Thine eye darts forth the fire that burneth me: And were I not immortal, life were done Between this heavenly and earthly sun. Venus and Adonis, Wm. Shakespeare, 1593 10.1 The problem of radiative exchange Chapter 1 described the elementary mechanisms of heat radiation. Be- fore we proceed, you should reflect upon what you remember about the following key ideas from Chapter 1: • Electromagnetic wave spectrum • The Stefan-Boltzmann law • Heat radiation & infrared radiation • Wien’s law & Planck’s law • Black body • Radiant heat exchange • Absorptance,α • Configuration factor,F 1–2 • Reflectance,ρ • Emittance,ε • Transmittance,τ • Transfer factor,F 1–2 • α+ρ+τ= 1 • Radiation shielding • e(T) ande (T) for black bodies λ The additional concept of a radiation heat transfer coefficient was devel- oped in Section 2.3. We presume that all these concepts are understood. The heat exchange problem Figure 10.1 shows two arbitrary surfaces radiating energy to one another. The net heat exchange, Q , from the hotter surface (1) to the cooler net 525526 Radiative heat transfer §10.1 Figure 10.1 Thermal radiation between two arbitrary surfaces. surface (2) depends on the following influences: • T andT . 1 2 • The areas of (1) and (2),A andA . 1 2 • The shape, orientation, and spacing of (1) and (2). • The radiative properties of the surfaces. • Additional surfaces in the environment, whose radiation may be reflected by one surface to the other. • The medium between (1) and (2) if it absorbs, emits, or “reflects” radiation. (When the medium is air, we can usually neglect these effects.) If surfaces (1) and (2) are black, if they are surrounded by air, and if no heat flows between them by conduction or convection, then only the§10.1 The problem of radiative exchange 527 first three considerations are involved in determiningQ . We saw some net elementary examples of how this could be done in Chapter 1, leading to 4 4 Q =A F σ T −T (10.1) net 1 1–2 1 2 The last three considerations complicate the problem considerably. In Chapter 1, we saw that these nonideal factors are sometimes included in a transfer factor F , such that 1–2 4 4 Q =A F σ T −T (10.2) net 1 1–2 1 2 Before we undertake the problem of evaluating heat exchange among real bodies, we need several definitions. Some definitions Emittance. A real body at temperatureT does not emit with the black 4 body emissive powere =σT but rather with some fraction,ε,ofe . b b The same is true of the monochromatic emissive power,e (T), which is λ always lower for a real body than the black body value given by Planck’s law, eqn. (1.30). Thus, we define either the monochromatic emittance,ε : λ e (λ,T) λ ε ≡ (10.3) λ e (λ,T) λ b or the total emittance,ε:   ∞ ∞ e (λ,T)dλ ε e (λ,T)dλ λ λ λ b e(T) 0 0 ε≡ = = (10.4) 4 4 e (T) σT σT b For real bodies, bothε andε are greater than zero and less than one; λ for black bodies,ε=ε = 1. The emittance is determined entirely by the λ properties of the surface of the particular body and its temperature. It is independent of the environment of the body. Table 10.1 lists typical values of the total emittance for a variety of substances. Notice that most metals have quite low emittances, unless they are oxidized. Most nonmetals have emittances that are quite high— approaching the black body limit of unity. One particular kind of surface behavior is that for whichε is indepen- λ dent ofλ. We call such a surface a gray body. The monochromatic emis- sive power,e (T), for a gray body is a constant fraction,ε,ofe (T),as λ b λ indicated in the inset of Fig. 10.2. In other words, for a gray body,ε =ε. λTable 10.1 Total emittances for a variety of surfaces 10.1 Metals Nonmetals ◦ ◦ Surface Temp. ( C) ε Surface Temp. ( C) ε Aluminum Asbestos 40 0.93–0.97 Polished, 98% pure 200−600 0.04–0.06 Brick Commercial sheet 90 0.09 Red, rough 40 0.93 Heavily oxidized 90−540 0.20–0.33 Silica 980 0.80–0.85 Brass Fireclay 980 0.75 Highly polished 260 0.03 Ordinary refractory 1090 0.59 Dull plate 40−260 0.22 Magnesite refractory 980 0.38 Oxidized 40−260 0.46–0.56 White refractory 1090 0.29 Copper Carbon Highly polished electrolytic 90 0.02 Filament 1040−1430 0.53 Slightly polished to dull 40 0.12–0.15 Lampsoot 40 0.95 Black oxidized 40 0.76 Concrete, rough 40 0.94 Gold: pure, polished 90−600 0.02–0.035 Glass Iron and steel Smooth 40 0.94 Mild steel, polished 150−480 0.14–0.32 Quartz glass (2 mm) 260−540 0.96–0.66 Steel, polished 40−260 0.07–0.10 Pyrex 260−540 0.94–0.74 Sheet steel, rolled 40 0.66 Gypsum 40 0.80–0.90 Sheet steel, strong 40 0.80 Ice 0 0.97–0.98 rough oxide Cast iron, oxidized 40−260 0.57–0.66 Limestone 400−260 0.95–0.83 Iron, rusted 40 0.61–0.85 Marble 40 0.93–0.95 Wrought iron, smooth 40 0.35 Mica 40 0.75 Wrought iron, dull oxidized 20−360 0.94 Paints Stainless, polished 40 0.07–0.17 Black gloss 40 0.90 Stainless, after repeated 230−900 0.50–0.70 White paint 40 0.89–0.97 heating Lacquer 40 0.80–0.95 Lead Various oil paints 40 0.92–0.96 Polished 40−260 0.05–0.08 Red lead 90 0.93 Oxidized 40−200 0.63 Paper Mercury: pure, clean 40−90 0.10–0.12 White 40 0.95–0.98 Platinum Other colors 40 0.92–0.94 Pure, polished plate 200−590 0.05–0.10 Roofing 40 0.91 ◦ Oxidized at 590 C 260−590 0.07–0.11 Plaster, rough lime 40−260 0.92 Drawn wire and strips 40−1370 0.04–0.19 Quartz 100−1000 0.89–0.58 Silver 200 0.01–0.04 Rubber 40 0.86–0.94 Tin 40−90 0.05 Snow 10−20 0.82 Tungsten Water, thickness≥0.1 mm 40 0.96 Filament 540−1090 0.11–0.16 Wood 40 0.80–0.90 Filament 2760 0.39 Oak, planed 20 0.90 528§10.1 The problem of radiative exchange 529 Figure 10.2 Comparison of the sun’s energy as typically seen through the earth’s atmosphere with that of a black body hav- ing the same mean temperature, size, and distance from the earth. (Notice thate , just outside the earth’s atmosphere, is λ far less than on the surface of the sun because the radiation has spread out over a much greater area.) No real body is gray, but many exhibit approximately gray behavior. We see in Fig. 10.2, for example, that the sun appears to us on earth as an approximately gray body with an emittance of approximately 0.6. Some materials—for example, copper, aluminum oxide, and certain paints—are actually pretty close to being gray surfaces at normal temperatures. Yet the emittance of most common materials and coatings varies with wavelength in the thermal range. The total emittance accounts for this behavior at a particular temperature. By using it, we can write the emis- sive power as if the body were gray, without integrating over wavelength: 4 e(T)=εσT (10.5) We shall use this type of “gray body approximation” often in this chapter.530 Radiative heat transfer §10.1 Specular or mirror-like Reflection which is Diffuse radiation in which reflection of incoming ray. between diffuse and directions of departure are specular (a real surface). uninfluenced by incoming ray angle,θ. Figure 10.3 Specular and diffuse reflection of radiation. (Arrows indicate magnitude of the heat flux in the directions indicated.) In situations where surfaces at very different temperatures are in- volved, the wavelength dependence ofε must be dealt with explicitly. λ This occurs, for example, when sunlight heats objects here on earth. So- lar radiation (from a high temperature source) is on visible wavelengths, whereas radiation from low temperature objects on earth is mainly in the infrared range. We look at this issue further in the next section. Diffuse and specular emittance and reflection. The energy emitted by a non-black surface, together with that portion of an incoming ray of energy that is reflected by the surface, may leave the body diffusely or specularly, as shown in Fig. 10.3. That energy may also be emitted or reflected in a way that lies between these limits. A mirror reflects visible radiation in an almost perfectly specular fashion. (The “reflection” of a billiard ball as it rebounds from the side of a pool table is also specular.) When reflection or emission is diffuse, there is no preferred direction for outgoing rays. Black body emission is always diffuse. The character of the emittance or reflectance of a surface will nor- mally change with the wavelength of the radiation. If we take account of both directional and spectral characteristics, then properties like emit- tance and reflectance depend on wavelength, temperature, and angles of incidence and/or departure. In this chapter, we shall assume diffuse§10.1 The problem of radiative exchange 531 behavior for most surfaces. This approximation works well for many problems in engineering, in part because most tabulated spectral and to- tal emittances have been averaged over all angles (in which case they are properly called hemispherical properties). Experiment 10.1 Obtain a flashlight with as narrow a spot focus as you can find. Direct it at an angle onto a mirror, onto the surface of a bowl filled with sugar, and onto a variety of other surfaces, all in a darkened room. In each case, move the palm of your hand around the surface of an imaginary hemi- sphere centered on the point where the spot touches the surface. Notice how your palm is illuminated, and categorize the kind of reflectance of each surface—at least in the range of visible wavelengths. Intensity of radiation. To account for the effects of geometry on radi- ant exchange, we must think about how angles of orientation affect the radiation between surfaces. Consider radiation from a circular surface element, dA, as shown at the top of Fig. 10.4. If the element is black, the radiation that it emits is indistinguishable from that which would be emitted from a black cavity at the same temperature, and that radiation is diffuse — the same in all directions. If it were non-black but diffuse, the heat flux leaving the surface would again be independent of direc- tion. Thus, the rate at which energy is emitted in any direction from this diffuse element is proportional to the projected area ofdA normal to the direction of view, as shown in the upper right side of Fig. 10.4. If an aperture of areadA is placed at a radiusr and angleθ from a dA and is normal to the radius, it will seedA as having an area cosθdA. 1 The energy dA receives will depend on the solid angle, dω, it sub- a tends. Radiation that leavesdA within the solid angledω stays within dω as it travels todA . Hence, we define a quantity called the intensity a 2 of radiation,i (W/m ·steradian) using an energy conservation statement:  radiant energy fromdA dQ =(idω)(cosθdA)= (10.6) outgoing that is intercepted bydA a 1 The unit of solid angle is the steradian. One steradian is the solid angle subtended by a spherical segment whose area equals the square of its radius. A full sphere there- 2 2 2 fore subtends 4πr /r = 4π steradians. The aperturedA subtendsdω=dA r . a a532 Radiative heat transfer §10.1 Figure 10.4 Radiation intensity through a unit sphere. Notice that while the heat flux fromdA decreases withθ (as indicated on the right side of Fig. 10.4), the intensity of radiation from a diffuse surface is uniform in all directions. Finally, we computei in terms of the heat flux fromdA by dividing eqn. (10.6)bydA and integrating over the entire hemisphere. For conve- nience we setr = 1, and we note (see Fig. 10.4) thatdω= sinθdθdφ.   2π π/2 q = i cosθ(sinθdθdφ)=πi (10.7a) outgoing φ=0 θ=0§10.2 Kirchhoff’s law 533 In the particular case of a black body, 4 e σT b i = = = fn(T only) (10.7b) b π π For a given wavelength, we likewise define the monochromatic intensity e λ i = = fn(T,λ) (10.7c) λ π 10.2 Kirchhoff’s law The problem of predictingα The total emittance, ε, of a surface is determined only by the phys- ical properties and temperature of that surface, as can be seen from eqn. (10.4). The total absorptance, α, on the other hand, depends on the source from which the surface absorbs radiation, as well as the sur- face’s own characteristics. This happens because the surface may absorb some wavelengths better than others. Thus, the total absorptance will depend on the way that incoming radiation is distributed in wavelength. And that distribution, in turn, depends on the temperature and physical properties of the surface or surfaces from which radiation is absorbed. The total absorptanceα thus depends on the physical properties and temperatures of all bodies involved in the heat exchange process. Kirch- 2 hoff’s law is an expression that allowsα to be determined under certain restrictions. Kirchhoff’s law Kirchhoff’s law is a relationship between the monochromatic, directional emittance and the monochromatic, directional absorptance for a surface that is in thermodynamic equilibrium with its surroundings exact form of ε (T,θ,φ)=α (T,θ,φ) (10.8a) λ λ Kirchhoff’s law Kirchhoff’s law states that a body in thermodynamic equilibrium emits as much energy as it absorbs in each direction and at each wavelength. If 2 Gustav Robert Kirchhoff (1824–1887) developed important new ideas in electrical circuit theory, thermal physics, spectroscopy, and astronomy. He formulated this par- ticular “Kirchhoff’s Law” when he was only 25. He and Robert Bunsen (inventor of the Bunsen burner) subsequently went on to do significant work on radiation from gases.534 Radiative heat transfer §10.2 this were not so, for example, a body might absorb more energy than it emits in one direction,θ , and might also emit more than it absorbs in an- 1 other direction,θ . The body would thus pump heat out of its surround- 2 ings from the first direction,θ , and into its surroundings in the second 1 direction,θ . Since whatever matter lies in the first direction would be 2 refrigerated without any work input, the Second Law of Thermodynam- ics would be violated. Similar arguments can be built for the wavelength dependence. In essence, then, Kirchhoff’s law is a consequence of the laws of thermodynamics. For a diffuse body, the emittance and absorptance do not depend on the angles, and Kirchhoff’s law becomes diffuse form of ε (T)=α (T) (10.8b) λ λ Kirchhoff’s law If, in addition, the body is gray, Kirchhoff’s law is further simplified diffuse, gray form ε T =α T (10.8c) ( ) ( ) of Kirchhoff’s law Equation (10.8c) is the most widely used form of Kirchhoff’s law. Yet, it is a somewhat dangerous result, since many surfaces are not even ap- proximately gray. If radiation is emitted on wavelengths much different from those that are absorbed, then a non-gray surface’s variation ofε λ andα with wavelength will matter, as we discuss next. λ Total absorptance during radiant exchange Let us restrict our attention to diffuse surfaces, so that eqn. (10.8b)is the appropriate form of Kirchhoff’s law. Consider two plates as shown in Fig. 10.5. Let the plate atT be non-black and that atT be black. Then 1 2 net heat transfer from plate 1 to plate 2 is the difference between what plate 1 emits and what it absorbs. Since all the radiation reaching plate 1 comes from a black source atT , we may write 2   ∞ ∞ q = ε (T )e (T )dλ− α (T )e (T )dλ (10.9) net λ 1 λ 1 λ 1 λ 2 1 1 b b 0 0       emitted by plate 1 radiation from plate 2 absorbed by plate 1 From eqn. (10.4), we may write the first integral in terms of total emit- 4 tance, asε σT .We define the total absorptance,α (T ,T ), as the sec- 1 1 1 2 1§10.2 Kirchhoff’s law 535 Figure 10.5 Heat transfer between two infinite parallel plates. 4 ond integral divided byσT . Hence, 2 4 4 q = ε (T )σT −α (T ,T )σT (10.10) net 1 1 1 1 2 1 2       emitted by plate 1 absorbed by plate 1 We see that the total absorptance depends onT , as well asT . 2 1 Why does total absorptance depend on both temperatures? The de- pendence onT is simply becauseα is a property of plate 1 that may 1 λ 1 be temperature dependent. The dependence onT is because the spec- 2 trum of radiation from plate 2 depends on the temperature of plate 2 according to Planck’s law, as was shown in Fig. 1.15. As a typical example, consider solar radiation incident on a warm roof, painted black. From Table 10.1, we see that ε is on the order of 0.94. It turns out thatα is just about the same. If we repaint the roof white, ε will not change noticeably. However, much of the energy ar- riving from the sun is carried in visible wavelengths, owing to the sun’s 3 very high temperature (about 5800 K). Our eyes tell us that white paint reflects sunlight very strongly in these wavelengths, and indeed this is the case — 80 to 90% of the sunlight is reflected. The absorptance of 3 Ninety percent of the sun’s energy is on wavelengths between 0.33 and 2.2 µm (see Figure 10.2). For a black object at 300 K, 90% of the radiant energy is between 6.3 and 42 µm, in the infrared.536 Radiative heat transfer §10.3 white paint to energy from the sun is only 0.1 to 0.2 — much less than ε for the energy it emits, which is mainly at infrared wavelengths. For both paints, eqn. (10.8b) applies. However, in this situation, eqn. (10.8c) is only accurate for the black paint. The gray body approximation Let us consider our facing plates again. If plate 1 is painted with white paint, and plate 2 is at a temperature near plate 1 (sayT = 400 K and 1 T = 300 K, to be specific), then the incoming radiation from plate 2 has 2 a wavelength distribution not too dissimilar to plate 1. We might be very comfortable approximatingε α . The net heat flux between the plates 1 1 can be expressed very simply 4 4 q =ε σT −α (T ,T )σT net 1 1 1 2 1 2 4 4 ε σT −ε σT 1 1 1 2 4 4 =ε σ T −T (10.11) 1 1 2 In effect, we are approximating plate 1 as a gray body. In general, the simplest first estimate for total absorptance is the dif- fuse, gray body approximation, eqn. (10.8c). It will be accurate either if the monochromatic emittance does not vary strongly with wavelength or if the bodies exchanging radiation are at similar absolute temperatures. More advanced texts describe techniques for calculating total absorp- tance (by integration) in other situations 10.2, 10.3. One situation in which eqn. (10.8c) should always be mistrusted is when solar radiation is absorbed by a low temperature object — a space vehicle or something on earth’s surface, say. In this case, the best first ap- proximation is to set total absorptance to a value for visible wavelengths of radiation (near 0.5 µm). Total emittance may be taken at the object’s actual temperature, typically for infrared wavelengths. We return to solar absorptance in Section 10.6. 10.3 Radiant heat exchange between two finite black bodies Let us now return to the purely geometric problem of evaluating the view factor,F . Although the evaluation ofF is also used in the calculation 1–2 1–2§10.3 Radiant heat exchange between two finite black bodies 537 Figure 10.6 Some configurations for which the value of the view factor is immediately apparent. of heat exchange among diffuse, nonblack bodies, it is the only correction of the Stefan-Boltzmann law that we need for black bodies. Some evident results. Figure 10.6 shows three elementary situations in which the value ofF is evident using just the definition: 1–2 F ≡ fraction of field of view of (1) occupied by (2). 1–2 When the surfaces are each isothermal and diffuse, this corresponds to F = fraction of energy leaving (1) that reaches (2) 1–2 A second apparent result in regard to the view factor is that all the energy leaving a body (1) reaches something else. Thus, conservation of energy requires 1=F +F +F +···+F (10.12) 1–1 1–2 1–3 1–n where (2), (3),…,(n) are all of the bodies in the neighborhood of (1). Figure 10.7 shows a representative situation in which a body (1) is sur- rounded by three other bodies. It sees all three bodies, but it also views538 Radiative heat transfer §10.3 Figure 10.7 A body (1) that views three other bodies and itself as well. itself, in part. This accounts for the inclusion of the view factor,F in 1–1 eqn. (10.12). By the same token, it should also be apparent from Fig. 10.7 that the kind of sum expressed by eqn. (10.12) would also be true for any subset of the bodies seen by surface 1. Thus, F =F +F 1–(2+3) 1–2 1–3 Of course, such a sum makes sense only when all the view factors are based on the same viewing surface (surface 1 in this case). One might be tempted to write this sort of sum in the opposite direction, but it would clearly be untrue, F ≠F +F , 2–1 3–1 (2+3)–1 since each view factor is for a different viewing surface—(2+ 3), 2, and 3, in this case. View factor reciprocity. So far, we have referred to the net radiation from black surface (1) to black surface (2) as Q . Let us refine our net notation a bit, and call thisQ : net 1–2 4 4 Q =A F σ T −T (10.13) net 1 1–2 1–2 1 2 Likewise, the net radiation from (2) to (1) is 4 4 Q =A F σ T −T (10.14) net 2 2–1 2–1 2 1§10.3 Radiant heat exchange between two finite black bodies 539 Of course,Q =−Q . It follows that net net 1–2 2–1 4 4 4 4 A F σ T −T =−A F σ T −T 1 1–2 2 2–1 1 2 2 1 or A F =A F (10.15) 1 1–2 2 2–1 This result, called view factor reciprocity, is very useful in calculations. Example 10.1 ◦ A jet of liquid metal at 2000 C pours from a crucible. It is 3 mm in di- ameter. A long cylindrical radiation shield, 5 cm diameter, surrounds ◦ ◦ the jet through an angle of 330 , but there is a 30 slit in it. The jet ◦ and the shield radiate as black bodies. They sit in a room at 30 C, and ◦ the shield has a temperature of 700 C. Calculate the net heat transfer: from the jet to the room through the slit; from the jet to the shield; and from the inside of the shield to the room. Solution. By inspection, we see thatF = 30/360= 0.08333 jet–room andF = 330/360= 0.9167. Thus, jet–shield 4 4 Q =A F σ T −T net jet jet–room jet–room jet room   2 π(0.003) m −8 4 4 = (0.08333)(5.67× 10 ) 2273 − 303 m length = 1, 188 W/m Likewise, 4 4 Q =A F σ T −T net jet jet–shield jet–shield jet shield   2 π(0.003) m −8 4 4 = (0.9167)(5.67× 10 ) 2273 − 973 m length = 12, 637 W/m The heat absorbed by the shield leaves it by radiation and convection to the room. (A balance of these effects can be used to calculate the shield temperature given here.) To find the radiation from the inside of the shield to the room, we needF . Since any radiation passing out of the slit goes to the shield–room540 Radiative heat transfer §10.3 room, we can find this view factor equating view factors to the room with view factors to the slit. The slit’s area isA =π(0.05)30/360= slit 2 0.01309 m /m length. Hence, using our reciprocity and summation rules, eqns. (10.12) and (10.15), A π(0.003) jet F = F = (0.0833)= 0.0600 slit–jet jet–room A 0.01309 slit F = 1−F −F = 1− 0.0600− 0= 0.940 slit–shield slit–jet slit–slit    0 A slit F = F shield–room slit–shield A shield 0.01309 = (0.940)= 0.08545 π(0.05)(330)/(360) Hence, for heat transfer from the inside of the shield only, 4 4 Q =A F σ T −T net shield shield–room room shield–room shield π(0.05)330 −8 4 4 = (0.08545)(5.67× 10 ) 973 − 303 360 = 619 W/m Both the jet and the inside of the shield have relatively small view factors to the room, so that comparatively little heat is lost through the slit. Calculation of the black-body view factor,F . Consider two elements, 1–2 dA anddA , of larger black bodies (1) and (2), as shown in Fig. 10.8. 1 2 Body (1) and body (2) are each isothermal. Since elementdA subtends 2 a solid angledω , we use eqn. (10.6) to write 1 dQ =(i dω )(cosβ dA ) 1to2 1 1 1 1 But from eqn. (10.7b), 4 σT 1 i = 1 π Note that because black bodies radiate diffusely,i does not vary with 1 angle; and because these bodies are isothermal, it does not vary with position. The element of solid angle is given by cosβ dA 2 2 dω = 1 2 s§10.3 Radiant heat exchange between two finite black bodies 541 Figure 10.8 Radiant exchange between two black elements that are part of the bodies (1) and (2). wheres is the distance from (1) to (2) and cosβ enters becausedA is 2 2 not necessarily normal tos. Thus,   4 σT cosβ cosβ dA dA 1 2 1 2 1 dQ = 1to2 2 π s By the same token,   4 σT cosβ cosβ dA dA 2 1 2 1 2 dQ = 2to1 2 π s Then   cosβ cosβ 1 2 4 4 Q =σ T −T dA dA (10.16) net 1 2 1–2 1 2 2 πs A A 1 2 The view factorsF andF are immediately obtainable from eqn. 1–2 2–1 4 4 (10.16). If we compare this result withQ =A F σ(T −T ),we net 1 1–2 1–2 1 2 get   1 cosβ cosβ 1 2 F = dA dA (10.17a) 1–2 1 2 2 A πs 1 A A 1 2542 Radiative heat transfer §10.3 From the inherent symmetry of the problem, we can also write   1 cosβ cosβ 2 1 F = dA dA (10.17b) 2–1 2 1 2 A πs 2 A A 2 1 You can easily see that eqns. (10.17a) and (10.17b) are consistent with the reciprocity relation, eqn. (10.15). The direct evaluation of F from eqn. (10.17a) becomes fairly in- 1–2 volved, even for the simplest configurations. Siegel and Howell 10.4 provide a comprehensive discussion of such calculations and a large cat- alog of their results. Howell 10.5 gives an even more extensive tabula- tion of view factor equations, which is now available on the World Wide Web. At present, no other reference is as complete. We list some typical expressions for view factors in Tables 10.2 and 10.3. Table 10.2 gives calculated values of F for two-dimensional 1–2 bodies—various configurations of cylinders and strips that approach in- finite length. Table 10.3 givesF for some three-dimensional configu- 1–2 rations. Many view factors have been evaluated numerically and presented in graphical form for easy reference. Figure 10.9, for example, includes graphs for configurations 1, 2, and 3 from Table 10.3. The reader should study these results and be sure that the trends they show make sense. Is it clear, for example, thatF → constant, which is 1 in each case, 1–2 as the abscissa becomes large? Can you locate the configuration on the right-hand side of Fig. 10.6 in Fig. 10.9? And so forth. Figure 10.10 shows view factors for another kind of configuration— one in which one area is very small in comparison with the other one. Many solutions like this exist because they are a bit less difficult to cal- culate, and they can often be very useful in practice. Example 10.2 A heater(h) as shown in Fig. 10.11 radiates to the partially conical shield(s) that surrounds it. If the heater and shield are black, calcu- late the net heat transfer from the heater to the shield. Solution. First imagine a plane(i) laid across the open top of the shield: F +F = 1 h−s h−i But F can be obtained from Fig. 10.9 or case 3 of Table 10.3, h−i

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