The CRC Handbook of Mechanical Engineering. Chapter 4. Heat and Mass Transfer (776127), страница 15
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A special type of view factor algebra may be used to determine all the viewfactors in long enclosures with constant cross section. The method is called the crossed-strings methodsince the view factors can be determined experimentally with four pins, a roll of string, and a yardstick.Consider the configuration in Figure 4.3.10, which shows the cross section of an infinitely long enclosure,continuing into and out of the plane of the figure.
Repeatedly applying reciprocity and summation rulesallows the evaluation of F1–2 asF1-2 =(Abc+ Aad ) - ( Aac + Abd )2 A1(4.3.14)FIGURE 4.3.10 The crossed-strings method for arbitrary two-dimensional configurations.where Aab is the area (per unit depth) defined by the length of the string between points a and b, etc.This formula is easily memorized by looking at the configuration between any two surfaces as ageneralized ”rectangle,” consisting of A1, A2, and the two sides Aac and Abd. ThenF1-2 =diagonals - sides2 ´ originating areaExample 4.3.3Calculate F1–2 for the configuration shown in Figure 4.3.11.Solution.
From the figure it is obvious that2s12 = (c - d cos a ) + d 2 sin 2 a = c 2 + d 2 - 2cd cos a© 1999 by CRC Press LLC(4.3.15)4-67Heat and Mass TransferFIGURE 4.3.11 Infinitely long wedge-shaped groove for Example 4.3.3.Similarly, we have22s22 = (a + c) + (b + d ) - 2(a + c) (b + d ) cos a2d12 = (a + c) + d 2 - 2(a + c) d cos a2d22 = c 2 + (b + d ) - 2c(b + d ) cos aandF1-2 =d1 + d2 - (s1 + s2 )2aRadiative Exchange between Opaque Surfaces (Net Radiation Method)Consider an enclosure consisting of N opaque surfaces.
The enclosure is closed, or, if not, no surfaceexternal to the surface reflects or emits radiation into the enclosure (i.e., the open configuration may beartificially closed by replacing openings with cold, black surfaces); any external radiation entering theenclosure is dealt with individually for each surface [see Equation (4.3.17) below]. All surfaces areassumed to be gray, and emit and reflect diffusely.
Traditionally, the radiosity J of the surfaces isdetermined, defined as the total diffuse radiative energy leaving a surface (by emission and reflection),Ji = e1 Ebi + ri Hi ,i = 1, N(4.3.16)where Hi is the incoming radiative flux (irradiation) onto surface Ai. This leads to N simultaneousequations for the unknown radiosities, specifically,éJi = e i Ebi + (1 - e i ) êêëNåj =1ùJi Fi - j + Hoi úúû(4.3.17a)orNJ i = qi +åJ Fj i- j+ Hoi(4.3.17b)j =1depending on whether surface temperature or surface flux are known on surface Ai.
In Equation (4.3.17)Hoi is irradiation on surface Ai from outside the enclosure, if any; Hoi is always zero for closed configurations, but is useful in the presence of external light sources (such as solar energy, lasers, etc.). The© 1999 by CRC Press LLC4-68Section 4radiosity neither is a useful quantity to determine, nor is there a need to determine it. Eliminating theradiosities from Equations (4.3.17a) and (4.3.17b) leads to N simultaneous equations in temperature (Ebi)and heat flux (qi):qieiNåj =1öæ 1ç e - 1÷ Fi - j q j + Hoi = Ebi øè jNåFi- jEbj(4.3.18)j =1Note that no artificial closing surfaces (j > N) appear in Equation (4.3.18), since for these surfaces ej =1 and Ebj = 0.
Thus, such closing surfaces may simply be ignored in the analysis.Since Equation (4.3.18) is a set of N equations, this requires that N values of emissive power Ebiand/or flux qi must be given as boundary conditions, in order to solve for the remaining N unknowns.For computer calculations Equation (4.3.18) may be recast in matrix form(4.3.19a)C × q = A × eb - howhereCij =öæ 1- ç - 1÷ Fi - jej è ejød ij(4.3.19b)Aij = d ij - Fi - j(4.3.19c)ïì1 if i = jd ij = íïî0 if i ¹ j(4.3.20)dij is Kronecker’s delta, i.e.,and q, eb, and ho are vectors of the surface heat fluxes qi, emissive powers Ebi, and external irradiationsHoi (if any). For example, if the temperatures are given for all the surfaces, and the heat fluxes are tobe determined, Equation (4.3.19) is solved by matrix inversion, and()(q = C -1 × A × e b - C -1 × h o)(4.3.21)Example 4.3.4A right-angled groove, consisting of two long black surfaces of width a, is exposed to solar radiationqsol (Figure 4.3.12).
The entire groove surface is kept isothermal at temperature T. Determine the netradiative heat transfer rate from the groove.Solution. We may employ Equation (4.3.19). However, the enclosure is not closed, and we must closeit artificially. We note that any radiation leaving the cavity will not come back (barring any reflectionfrom other surfaces nearby). Thus, our artificial surface should be black. We also assume that, with theexception of the (parallel) solar irradiation, no external radiation enters the cavity.
Since the solarirradiation is best treated separately through the external irradiation term Ho, our artificial surface isnonemitting. Both criteria are satisfied by covering the groove with a black surface at 0 K (A3). Eventhough we now have three surfaces, the last one does not really appear in Equation (4.3.18) (since Eb3= 0 and 1/e3 – 1 = 0):© 1999 by CRC Press LLC4-69Heat and Mass TransferFIGURE 4.3.12 Right-angled groove exposed to solar irradiation, Example 4.3.4.q1 = Eb1 - F1-2 Eb 2 - Ho1 = sT 4 (1 - F1-2 ) - qsol cos aq2 = Eb 2 - F2-1 Eb1 - Ho 2 = sT 4 (1 - F2-1 ) - qsol sin aFrom the crossed-strings method, Equation (4.3.15), we findF1-2 =a+a-(2a + 0) = 1 2 - 2 = 0.293 = F()22 -12aand[]Q ¢ = a(q1 + q2 ) = a 2sT 4 - qsol (cos a + sin a )Example 4.3.5Consider a very long duct as shown in Figure 4.3.13.
The duct is 30 ´ 40 cm in cross section, and allsurfaces are covered with gray, diffuse surface material. Top and bottom walls are at T1 = T3 = 1000 Kwith e1 = e3 = 0.3, while the side walls are at T2 = T4 = 600 K with e2 = e4 = 0.8 as shown. Determinethe net radiative heat transfer rates for each surface.FIGURE 4.3.13 Two-dimensional gray, diffuse duct for Example 4.3.5.© 1999 by CRC Press LLC4-70Section 4Solution. Using Equation (4.3.18) for i = 1 and i = 2 and noting that F1–2 = F1–4 and F2–1 = F2–3,i = 1:ööæ1æ 1q1- 2ç - 1÷ F1-2 q2 - ç - 1÷ F1-3 q1 = 2 F1-2 ( Eb1 - Eb 2 )e1øè e1 øè e2i = 2:ööæ 1æ1q2- 2ç - 1÷ F2-1q1 - ç - 1÷ F2- 4 q2 = 2 F2-1 ( Eb 2 - Eb1 )e2øè e2è e1 øThe view factors are readily evaluated from the crossed-strings method as F1–2 =1/4, F1–3 = 1 – 2F1–2 =1/2,F2–1 = 4/3 F1–2 =1/3 and F2–4 = 1 – 2F2–1 = 1/3.
Substituting these, as well as emissivity values, into therelations reduces them to the simpler form of1é 1 æ 1æ 1ö1ö 1ùê 0.3 - è 0.3 - 1ø 2 ú q1 - 2è 0.8 - 1ø 4 q2 = 2 ´ 4 ( Eb1 - Eb 2 )ûë111ù1é 1 æ 1-2æ- 1ö q +- 1ö q = 2 ´ ( Eb 2 - Eb1 )è 0.3 ø 3 1 êë 0.8 è 0.8 ø úû 3 23or1311q - q = ( E - Eb 2 )6 1 8 2 2 b1-1472q + q = - ( Eb1 - Eb 2 )9 1 6 23Thus,æ 13 ´ 7 - 14 ´ 1 ö q = æ 1 ´ 7 - 2 ´ 1 ö E - Eb2 )è 6 6 9 8 ø 1 è 2 6 3 8 ø ( b1q1 =3 13´ ( E - Eb 2 ) =s T14 - T247 2 b114()andæ - 1 ´ 14 + 7 ´ 13 ö q = æ 1 ´ 14 - 2 ´ 13 ö E - Eb2 )è 8 9 6 6 ø 2 è 2 9 3 6 ø ( b1q2 =3 22´ ( E - Eb 2 ) = - s T14 - T247 3 b17()Finally, substituting values for temperatures,Q1¢ = 0.4 m ´© 1999 by CRC Press LLC3W´ 5.670 ´ 10 -8 2 4 1000 4 - 600 4 K 4 = 4230 W / m14m K()4-71Heat and Mass TransferQ2¢ = - 0.3 m ´2W´ 5.670 ´ 10 -8 2 4 1000 4 - 600 4 K 4 = - 4230 W / m7m K()Note that, for conservation of energy, both heat transfer rates must add up to zero.Small Body Inside Isothermal Enclosure.
An especially simple — but important — case occurs if asmall, convex body A1 (i.e., a surface that cannot “see” itself, or F1–1 = 0) is totally enclosed by anisothermal enclosure A2. Then, with N = 2 and F1–2 = 1, Equation (4.3.18) reduces toq1 =()s T14 - T24Eb1 - Eb 2=ööA æ 1A æ 111+ 1 ç - 1÷+ 1 ç - 1÷e1 A2 è e 2e1 A2 è e 2øø(4.3.22)If the enclosure is large, i.e., A1 ! A2, then Equation (4.3.22) simplifies further to(q1 = e1s T14 - T24)(4.3.23)Radiation Shields. If it is desired to minimize radiative heat transfer between two surfaces, it is commonpractice to place one or more radiation shields between them (usually thin metallic sheets of lowemissivity).
If two surfaces Ai and Aj are close together, so that Ai @ Aj and Fi–j @ 1, then the radiativeexchange between them is, from Equation (4.3.22),q=Ebi - EbjRij,Rij =11+-1ei e j(4.3.24)FIGURE 4.3.14 Placement of radiation shields between two large, parallel plates.where Rij is termed the radiative resistance.
Equation (4.3.24) is seen to be analogous to an electricalcircuit with “current” q and “voltage potential” Ebi – Ebj. Therefore, expressing radiative fluxes in termsof radiative resistances is commonly known as network analogy. The network analogy is a very powerfulmethod of solving one-dimensional problems (i.e., whenever only two isothermal surfaces see each other,such as infinite parallel plates, or when one surface totally encloses another). Consider, for example,two large parallel plates, A1 and AN, separated by N – 2 radiation shields, as shown in Figure 4.3.14. Leteach shield have an emissivity eS on both sides. Then, by applying Equation (4.3.24) to any twoconsecutive surfaces and using the fact that q remains constant throughout the gap,© 1999 by CRC Press LLC4-72Section 4q=Eb1 - Eb 2E- EbkE- EbNE - EbN= L = bk -1= L = bN -1= Nb1R12Rk -1,kRN -1, NR j -1, j(4.3.25)åj -2whereR j -1, j =1e j -1+1-1ej(4.3.26)and, if e2 = e3 L = eN–1 = eS,NåRj -1, jæ 2ö11+- 1 + ( N - 2) ç - 1÷e1 e Nè eSø=j =2(4.3.27)Equations (4.3.24) to (4.3.27) are also valid for concentric cylinders, concentric spheres, and similarconfigurations, as long as rN – r1 ! r1.
Also, the relations are readily extended to shields with nonidenticalemissivities.While the network analogy can (and has been) applied to configurations with more than two surfacesseeing each other, this leads to very complicated circuits (since there is one resistance between any twosurfaces). For such problems the network analogy is not recommended, and the net radiation method,Equation (4.3.18), should be employed.Radiative Exchange within Participating MediaIn many high-temperature applications, when radiative heat transfer is important, the medium betweensurfaces is not transparent, but is “participating,” i.e., it absorbs, emits, and (possibly) scatters radiation.In a typical combustion process this interaction results in (1) continuum radiation due to tiny, burningsoot particles (of dimension <1 mm) and also due to larger suspended particles, such as coal particles,oil droplets, fly ash; (2) banded radiation in the infrared due to emission and absorption by moleculargaseous combustion products, mostly water vapor and carbon dioxide; and (3) chemiluminescence dueto the combustion reaction itself.
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