John H. Lienhard IV, John H. Lienhard V. A Heat Transfer Textbook (776116), страница 79
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10.13. If the gray plate is diffuse, its radiation has the same geometrical distribution as that from a black body, and it will travel to otherobjects in the same way that black body radiation would. Therefore, wecan treat the radiation leaving the imaginary surface — the radiosity, thatis — as though it were black body radiation travelling to an imaginarysurface above the other plate.
Thus, by analogy to eqn. (10.13),B1 − B2Qnet1–2 = A1 F1–2 (B1 − B2 ) = 1(10.22)A1 F1–2where the final fraction shows that this is also a form of Ohm’s law:the radiosity difference (B1 − B2 ), can be said to drive heat through thegeometrical resistance, 1/A1 F1–2 , that describes the field of view betweenthe two surfaces.When two gray surfaces exchange radiation only with each other, thenet radiation flows through a surface resistance for each surface and ageometric resistance for the configuration.
The electrical circuit shownin Fig. 10.13 expresses the analogy and gives us means for calculatingQnet1–2 from Ohm’s law. Recalling that eb = σ T 4 , we obtainQnet1–2 = $eb1 − eb2resistances=1−εεA σ T14 − T24+11A1 F1–2+1−εεA(10.23)2For the particular case of infinite parallel plates, F1–2 = 1 and A1 = A2Heat transfer among gray bodies§10.4551(Fig.
10.6), and, with qnet1–2 = Qnet1–2 /A1 , we findqnet1–2 = 11ε1+1ε2 σ T14 − T24(10.24)−1Comparing eqn. (10.24) with eqn. (10.2), we may identifyF1–2 = 111+−1ε1ε2(10.25)for infinite parallel plates. Notice, too, that if the plates are both black(ε1 = ε2 = 1), then both surface resistances are zero andF1–2 = 1 = F1–2which, of course, is what we would have expected.Example 10.5One gray body enclosed by anotherEvaluate the heat transfer and the transfer factor for one gray bodyenclosed by another, as shown in Fig. 10.14.Solution.
The electrical circuit analogy is exactly the same as thatshown in Fig. 10.13, and F1–2 is still unity. Therefore, with eqn. (10.23), σ T14 − T24(10.26)Qnet1–2 = A1 qnet1–2 = 1 − ε111 − ε2++ε1 A1A1ε2 A2Figure 10.14 Heat transfer between anenclosed body and the body surroundingit.552Radiative heat transfer§10.4The transfer factor may again be identified by comparison to eqn.
(10.2):Qnet1–2 = A1 σ T14 − T24A1 11+−1ε1A2 ε21(10.27)=F1–2This calculation assumes that body (1) does not view itself.Example 10.6Transfer factor reciprocityDerive F2–1 for the enclosed bodies shown in Fig. 10.14.Solution.Qnet1–2 = −Qnet2–1A1 F1–2 σ T14 − T24 = −A2 F2–1 σ T24 − T14from which we obtain the reciprocity relationship for transfer factors:A1 F1–2 = A2 F2–1(10.28)Hence, with the result of Example 10.5, we haveF2–1 =Example 10.7A11F1–2 =11 A2A2+−1ε1 A1ε2(10.29)Small gray object in a large environmentDerive F1–2 for a small gray object (1) in a large isothermal environment (2), the result that was given as eqn. (1.35).Solution.
We may use eqn. (10.27) with A1 /A2 1:F1–2 =1 ε1A1 11+−1ε1A2 ε2(10.30)1Note that the same result is obtained for any value of A1 /A2 if theenclosure is black (ε2 = 1). A large enclosure does not reflect much radiation back to the small object, and therefore becomes like a perfectabsorber of the small object’s radiation — a black body.Heat transfer among gray bodies§10.4553Additional two-body exchange problemsRadiation shields. A radiation shield is a surface, usually of high reflectance, that is placed between a high-temperature source and its coolerenvironment.
Earlier examples in this chapter and in Chapter 1 show howsuch a surface can reduce heat exchange. Let us now examine the roleof reflectance (or emittance: ε = 1 − ρ) in the performance of a radiationshield.Consider a gray body (1) surrounded by another gray body (2), asdiscussed in Example 10.5.
Suppose now that a thin sheet of reflectivematerial is placed between bodies (1) and (2) as a radiation shield. Thesheet will reflect radiation arriving from body (1) back toward body (1);likewise, owing to its low emittance, it will radiate little energy to body(2). The radiation from body (1) to the inside of the shield and from theoutside of the shield to body (2) are each two-body exchange problems,coupled by the shield temperature. We may put the various radiationresistances in series to find (see Problem 10.46) σ T14 − T24(10.31)Qnet1–2 = 11 − ε21 − ε11 − εs1+++2+ε1 A1A1ε2 A2εs AsAsadded by shieldassuming F1–s = Fs–2 = 1. Note that the radiation shield reduces Qnet1–2more if its emittance is smaller, i.e., if it is highly reflective.Specular surfaces.
The electrical circuit analogy that we have developedis for diffuse surfaces. If the surface reflection or emission has directional characteristics, different methods of analysis must be used [10.2].One important special case deserves to be mentioned. If the two graysurfaces in Fig. 10.14 are diffuse emitters but are perfectly specular reflectors — that is, if they each have only mirror-like reflections — thenthe transfer factor becomesF1–2 = 111+−1ε1ε2for specularlyreflecting bodies(10.32)This result is interestingly identical to eqn. (10.25) for parallel plates.Since parallel plates are a special case of the situation in Fig.
10.14, itfollows that eqn. (10.25) is true for either specular or diffuse reflection.554Radiative heat transfer§10.4Example 10.8A physics experiment uses liquid nitrogen as a coolant. Saturatedliquid nitrogen at 80 K flows through 6.35 mm O.D. stainless steelline (εl = 0.2) inside a vacuum chamber. The chamber walls are atTc = 230 K and are at some distance from the line. Determine theheat gain of the line per unit length. If a second stainless steel tube,12.7 mm in diameter, is placed around the line to act as radiationshield, to what rate is the heat gain reduced? Find the temperatureof the shield.Solution.
The nitrogen coolant will hold the surface of the line atessentially 80 K, since the thermal resistances of the tube wall and theinternal convection or boiling process are small. Without the shield,we can model the line as a small object in a large enclosure, as inExample 10.7: Qgain = (π Dl )εl σ Tc4 − Tl4= π (0.00635)(0.2)(5.67 × 10−8 )(2304 − 804 ) = 0.624 W/mWith the shield, eqn. (10.31) applies. Assuming that the chamber areais large compared to the shielded line (Ac Al ), σ Tc4 − Tl4Qgain = 11 − εc1 − εl1 − εs1+++2+εl AlAlε Aεs AsAs 2 cneglect=π (0.00635)(5.67 × 10−8 )(2304 − 804 ) 1 − 0.21 − 0.20.00635+1 +2+10.20.01270.2= 0.328 W/mThe radiation shield would cut the heat gain by 47%.The temperature of the shield, Ts , may be found using the heatloss and considering the heat flow from the chamber to the shield,with the shield now acting as a small object in a large enclosure: Qgain = (π Ds )εs σ Tc4 − Ts4 0.328 W/m = π (0.0127)(0.2)(5.67 × 10−8 ) 2304 − Ts4Solving, we find Ts = 213 K.Heat transfer among gray bodies§10.4The electrical circuit analogy when more than two gray bodiesare involved in heat exchangeLet us first consider a three-body transaction, as pictured in at the bottom and left-hand sides of Fig.
10.15. The triangular circuit for threebodies is not so easy to analyze as the in-line circuits obtained in twobody problems. The basic approach is to apply energy conservation ateach radiosity node in the circuit, setting the net heat transfer from anyone of the surfaces (which we designate as i)Qneti =ebi − Bi1 − εiεi Ai(10.33a)equal to the sum of the net radiation to each of the other surfaces (callthem j)⎛⎞$Bi − Bj⎝ B⎠(10.33b)Qneti =1AFi i−jjFor the three body situation shown in Fig.
10.15, this leads to three equationsQnet1 , at node B1 :eb1 − B1B1 − B2 B1 − B3=+1 − ε111ε1 A1A1 F1–2A1 F1–3Qnet2 , at node B2 :eb2 − B2B2 − B1 B2 − B3=+1 − ε211ε2 A2A1 F1–2A2 F2–3Qnet3 , at node B3 :eb3 − B3B3 − B1 B3 − B2=+1 − ε311ε3 A3A1 F1–3A2 F2–3(10.34a)(10.34b)(10.34c)If the temperatures T1 , T2 , and T3 are known (so that eb1 , eb2 , eb3 areknown), these equations can be solved simultaneously for the three unknowns, B1 , B2 , and B3 . After they are solved, one can compute the netheat transfer to or from any body (i) from either of eqns. (10.33).Thus far, we have considered only cases in which the surface temperature is known for each body involved in the heat exchange process.
Letus consider two other possibilities.555Radiative heat transfer556§10.4Figure 10.15 The electrical circuit analogy for radiationamong three gray surfaces.An insulated wall. If a wall is adiabatic, Qnet = 0 at that wall. Forexample, if wall (3) in Fig. 10.15 is insulated, then eqn. (10.33b) showsthat eb3 = B3 . We can eliminate one leg of the circuit, as shown on theright-hand side of Fig.
10.15; likewise, the left-hand side of eqn. (10.34c)equals zero. This means that all radiation absorbed by an adiabatic wallis immediately reemitted. Such walls are sometimes called “refractorysurfaces” in discussing thermal radiation.The circuit for an insulated wall can be treated as a series-parallelcircuit, since all the heat from body (1) flows to body (2), even if it doesso by travelling first to body (3). ThenQnet1 =eb1 − eb21 − ε1+ε1 A1111 /(A1 F1–3 ) + 1 /(A2 F2–3 )+1+1 − ε2ε2 A21 /(A1 F1–2 )(10.35)§10.4Heat transfer among gray bodiesA specified wall heat flux.
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