John H. Lienhard IV, John H. Lienhard V. A Heat Transfer Textbook (776116), страница 77
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(10.6) by dA and integrating over the entire hemisphere. For convenience we set r = 1, and we note (see Fig. 10.4) that dω = sin θ dθdφ. 2π π /2qoutgoing =φ=0θ=0i cos θ (sin θ dθdφ) = π i(10.7a)Kirchhoff’s law§10.2533In the particular case of a black body,ib =σT4eb== fn (T only)ππ(10.7b)For a given wavelength, we likewise define the monochromatic intensityiλ =10.2eλ= fn (T , λ)π(10.7c)Kirchhoff’s lawThe problem of predicting αThe total emittance, ε, of a surface is determined only by the physical properties and temperature of that surface, as can be seen fromeqn. (10.4).
The total absorptance, α, on the other hand, depends onthe source from which the surface absorbs radiation, as well as the surface’s own characteristics. This happens because the surface may absorbsome wavelengths better than others. Thus, the total absorptance willdepend on the way that incoming radiation is distributed in wavelength.And that distribution, in turn, depends on the temperature and physicalproperties of the surface or surfaces from which radiation is absorbed.The total absorptance α thus depends on the physical properties andtemperatures of all bodies involved in the heat exchange process.
Kirchhoff’s law2 is an expression that allows α to be determined under certainrestrictions.Kirchhoff’s lawKirchhoff’s law is a relationship between the monochromatic, directionalemittance and the monochromatic, directional absorptance for a surfacethat is in thermodynamic equilibrium with its surroundingsελ (T , θ, φ) = αλ (T , θ, φ)exact form ofKirchhoff’s law(10.8a)Kirchhoff’s law states that a body in thermodynamic equilibrium emitsas much energy as it absorbs in each direction and at each wavelength. If2Gustav Robert Kirchhoff (1824–1887) developed important new ideas in electricalcircuit theory, thermal physics, spectroscopy, and astronomy. He formulated this particular “Kirchhoff’s Law” when he was only 25.
He and Robert Bunsen (inventor of theBunsen burner) subsequently went on to do significant work on radiation from gases.534Radiative heat transfer§10.2this were not so, for example, a body might absorb more energy than itemits in one direction, θ1 , and might also emit more than it absorbs in another direction, θ2 . The body would thus pump heat out of its surroundings from the first direction, θ1 , and into its surroundings in the seconddirection, θ2 .
Since whatever matter lies in the first direction would berefrigerated without any work input, the Second Law of Thermodynamics would be violated. Similar arguments can be built for the wavelengthdependence. In essence, then, Kirchhoff’s law is a consequence of thelaws of thermodynamics.For a diffuse body, the emittance and absorptance do not depend onthe angles, and Kirchhoff’s law becomesελ (T ) = αλ (T )diffuse form ofKirchhoff’s law(10.8b)If, in addition, the body is gray, Kirchhoff’s law is further simplifiedε (T ) = α (T )diffuse, gray formof Kirchhoff’s law(10.8c)Equation (10.8c) is the most widely used form of Kirchhoff’s law. Yet, itis a somewhat dangerous result, since many surfaces are not even approximately gray.
If radiation is emitted on wavelengths much differentfrom those that are absorbed, then a non-gray surface’s variation of ελand αλ with wavelength will matter, as we discuss next.Total absorptance during radiant exchangeLet us restrict our attention to diffuse surfaces, so that eqn. (10.8b) isthe appropriate form of Kirchhoff’s law. Consider two plates as shownin Fig. 10.5. Let the plate at T1 be non-black and that at T2 be black. Thennet heat transfer from plate 1 to plate 2 is the difference between whatplate 1 emits and what it absorbs. Since all the radiation reaching plate1 comes from a black source at T2 , we may writeqnet =∞0∞ελ1 (T1 ) eλb (T1 ) dλ −αλ1 (T1 ) eλb (T2 ) dλ 0emitted by plate 1(10.9)radiation from plate 2absorbed by plate 1From eqn. (10.4), we may write the first integral in terms of total emittance, as ε1 σ T14 .
We define the total absorptance, α1 (T1 , T2 ), as the sec-Kirchhoff’s law§10.2535Figure 10.5 Heat transfer between twoinfinite parallel plates.ond integral divided by σ T24 . Hence,qnet =ε1 (T1 )σ T14 − α1 (T1 , T2 )σ T24 emitted by plate 1(10.10)absorbed by plate 1We see that the total absorptance depends on T2 , as well as T1 .Why does total absorptance depend on both temperatures? The dependence on T1 is simply because αλ1 is a property of plate 1 that maybe temperature dependent. The dependence on T2 is because the spectrum of radiation from plate 2 depends on the temperature of plate 2according to Planck’s law, as was shown in Fig. 1.15.As a typical example, consider solar radiation incident on a warmroof, painted black. From Table 10.1, we see that ε is on the order of0.94.
It turns out that α is just about the same. If we repaint the roofwhite, ε will not change noticeably. However, much of the energy arriving from the sun is carried in visible wavelengths, owing to the sun’svery high temperature (about 5800 K).3 Our eyes tell us that white paintreflects sunlight very strongly in these wavelengths, and indeed this isthe case — 80 to 90% of the sunlight is reflected. The absorptance of3Ninety percent of the sun’s energy is on wavelengths between 0.33 and 2.2 µm (seeFigure 10.2). For a black object at 300 K, 90% of the radiant energy is between 6.3 and42 µm, in the infrared.Radiative heat transfer536§10.3white paint to energy from the sun is only 0.1 to 0.2 — much less thanε for the energy it emits, which is mainly at infrared wavelengths. Forboth paints, eqn.
(10.8b) applies. However, in this situation, eqn. (10.8c)is only accurate for the black paint.The gray body approximationLet us consider our facing plates again. If plate 1 is painted with whitepaint, and plate 2 is at a temperature near plate 1 (say T1 = 400 K andT2 = 300 K, to be specific), then the incoming radiation from plate 2 hasa wavelength distribution not too dissimilar to plate 1.
We might be verycomfortable approximating ε1 α1 . The net heat flux between the platescan be expressed very simplyqnet = ε1 σ T14 − α1 (T1 , T2 )σ T24 ε1 σ T14 − ε1 σ T24= ε1 σ T14 − T24(10.11)In effect, we are approximating plate 1 as a gray body.In general, the simplest first estimate for total absorptance is the diffuse, gray body approximation, eqn. (10.8c). It will be accurate either ifthe monochromatic emittance does not vary strongly with wavelength orif the bodies exchanging radiation are at similar absolute temperatures.More advanced texts describe techniques for calculating total absorptance (by integration) in other situations [10.2, 10.3].One situation in which eqn. (10.8c) should always be mistrusted iswhen solar radiation is absorbed by a low temperature object — a spacevehicle or something on earth’s surface, say.
In this case, the best first approximation is to set total absorptance to a value for visible wavelengthsof radiation (near 0.5 µm). Total emittance may be taken at the object’sactual temperature, typically for infrared wavelengths. We return to solarabsorptance in Section 10.6.10.3Radiant heat exchange between two finiteblack bodiesLet us now return to the purely geometric problem of evaluating the viewfactor, F1–2 . Although the evaluation of F1–2 is also used in the calculation§10.3Radiant heat exchange between two finite black bodiesFigure 10.6 Some configurations for which the value of theview factor is immediately apparent.of heat exchange among diffuse, nonblack bodies, it is the only correctionof the Stefan-Boltzmann law that we need for black bodies.Some evident results. Figure 10.6 shows three elementary situations inwhich the value of F1–2 is evident using just the definition:F1–2 ≡ fraction of field of view of (1) occupied by (2).When the surfaces are each isothermal and diffuse, this corresponds toF1–2 = fraction of energy leaving (1) that reaches (2)A second apparent result in regard to the view factor is that all theenergy leaving a body (1) reaches something else.
Thus, conservation ofenergy requires1 = F1–1 + F1–2 + F1–3 + · · · + F1–n(10.12)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 surrounded by three other bodies. It sees all three bodies, but it also views537538Radiative heat transfer§10.3Figure 10.7 A body (1) that views three other bodies and itselfas well.itself, in part. This accounts for the inclusion of the view factor, F1–1 ineqn. (10.12).By the same token, it should also be apparent from Fig. 10.7 that thekind of sum expressed by eqn. (10.12) would also be true for any subsetof the bodies seen by surface 1.
Thus,F1–(2+3) = F1–2 + F1–3Of course, such a sum makes sense only when all the view factors arebased on the same viewing surface (surface 1 in this case). One might betempted to write this sort of sum in the opposite direction, but it wouldclearly be untrue,F(2+3)–1 ≠ F2–1 + F3–1 ,since each view factor is for a different viewing surface—(2 + 3), 2, and3, in this case.View factor reciprocity. So far, we have referred to the net radiationfrom black surface (1) to black surface (2) as Qnet . Let us refine ournotation a bit, and call this Qnet1–2 :Qnet1–2 = A1 F1–2 σ T14 − T24(10.13)Likewise, the net radiation from (2) to (1) isQnet2–1 = A2 F2–1 σ T24 − T14(10.14)Radiant heat exchange between two finite black bodies§10.3Of course, Qnet1–2 = −Qnet2–1 .
It follows thatA1 F1–2 σ T14 − T24 = −A2 F2–1 σ T24 − T14orA1 F1–2 = A2 F2–1(10.15)This result, called view factor reciprocity, is very useful in calculations.Example 10.1A jet of liquid metal at 2000◦ C pours from a crucible. It is 3 mm in diameter. A long cylindrical radiation shield, 5 cm diameter, surroundsthe jet through an angle of 330◦ , but there is a 30◦ slit in it. The jetand the shield radiate as black bodies. They sit in a room at 30◦ C, andthe 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.
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