BOOKCOMP, Inc. — John Wiley & Sons / Page 603 / 2nd Proofs / Heat Transfer Handbook / Bejan RADIATIVE EXCHANGE BETWEEN SURFACES 603 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 [603], (31) Lines: 1030 to 1030 ——— * 32.5553pt PgVar ——— Normal Page PgEnds: T E X [603], (31) TABLE 8.3 Important View Factors (Continued) 10. Outer surface of cylinder to annular diskat end of cylinder: R = r 1 r 2 L = l r 2 A = L 2 + R 2 − 1 B = L 2 − R 2 + 1 F 1−2 = B 8RL + 1 2π cos −1 A B − 1 2L (A + 2) 2 R 2 − 4 cos −1 AR B − A 2RL sin −1 R A 1 A 2 l r 2 r 1 11. Interior of finite-length, right-circular coaxial cylinder to itself: R = r 2 r 1 H = h r 1 F 2−2 = 1 − 1 R − √ H 2 + 4R 2 − H 4R + 1 π 2 R tan −1 2 √ R 2 − 1 H − H 2R √ 4R 2 + H 2 H sin −1 H 2 + 4(R 2 − 1) − 2H 2 /R 2 H 2 + 4(R 2 − 1) − sin −1 R 2 − 2 R 2 h A 2 r 2 r 1 12. Sphere to rectangle, r<d: D 1 = d l 1 D 2 = d l 2 F 1−2 = 1 4π tan −1 1 D 2 1 + D 2 2 + D 2 1 D 2 2 A 2 d r A 1 l 2 l 1 13. Sphere to coaxial disk: R = r a F 1−2 = 1 2 1 − 1 √ 1 + R 2 A 2 a A 1 r BOOKCOMP, Inc. — John Wiley & Sons / Page 604 / 2nd Proofs / Heat Transfer Handbook / Bejan 604 THERMAL RADIATION 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 [604], (32) Lines: 1030 to 1030 ——— * 32.50198pt PgVar ——— Normal Page PgEnds: T E X [604], (32) Figure 8.15 View factor between identical, parallel, directly opposed rectangles (configura- tion 7). Figure 8.16 View factor between perpendicular rectangles with common edge (configura- tion 8). BOOKCOMP, Inc. — John Wiley & Sons / Page 605 / 2nd Proofs / Heat Transfer Handbook / Bejan RADIATIVE EXCHANGE BETWEEN SURFACES 605 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 [605], (33) Lines: 1030 to 1030 ——— * 21.953pt PgVar ——— Normal Page PgEnds: T E X [605], (33) Figure 8.17 View factor between parallel coaxial disks of unequal radius (configuration 9). Figure 8.18 View factor evaluation between two surfaces. BOOKCOMP, Inc. — John Wiley & Sons / Page 606 / 2nd Proofs / Heat Transfer Handbook / Bejan 606 THERMAL RADIATION 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 [606], (34) Lines: 1030 to 1081 ——— 9.58319pt PgVar ——— Normal Page * PgEnds: Eject [606], (34) one’s disposal. Using an arbitrary coordinate origin, a vector pointing from the origin to a point on a surface may be written as r = x ˆ ı + yˆ + z ˆ k (8.46) where ˆ ı, ˆ, and ˆ k are unit vectors pointing into the x, y, and z directions, respectively. Thus the vector from dA i going to dA j is determined as s ij =−s ji = r j − r i = (x j − x i ) ˆ ı + (y j − y i )ˆ + (z j − z i ) ˆ k (8.47) The length of this vector is determined as s ji 2 = s ij 2 = S 2 ij = (x j − x i ) 2 + (y j − y i ) 2 + (z j − z i ) 2 (8.48) It will now be assumed that the local surface normals are known in terms of the unit vectors ˆ ı, ˆ, and ˆ k,or ˆ n = l ˆ ı + mˆ + n ˆ k (8.49) where l, m, and n are the direction cosines for the unit vector ˆ n. For example, l = ˆ n· ˆ ı is the cosine of the angle between ˆ n and the x axis, and m and n can be represented in a similar manner. Thus cos θ i and cos θ j may be evaluated as cos θ i = ˆ n i · s ij S ij = 1 S ij (x j − x i )l i + (y j − y i )m i + (z j − z i )n i (8.50) cos θ j = ˆ n j · s ji S ij = 1 S ij (x i − x j )l j + (y i − y j )m j + (z i − z j )n j (8.51) For contour integration the tangential vectors ds are found from eq. (8.46), that is, ds = dx ˆ ı + dy ˆ + dz ˆ k (8.52) where dx is the change in the x coordinate along the contour Γ, and dy and dz follow in a similar fashion. Examples for the application of area and contour integration to view factor evaluation may be found in textbooks (Modest, 2003; Siegel and Howell, 2002). Special Methods The mathematics of view factors follow certain rules, which may be exploited to simplify their evaluation. The two most important ones are the summation rule for a closed configuration consisting of N surfaces, N j=1 F i−j = 1 (8.53) BOOKCOMP, Inc. — John Wiley & Sons / Page 607 / 2nd Proofs / Heat Transfer Handbook / Bejan RADIATIVE EXCHANGE BETWEEN SURFACES 607 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 [607], (35) Lines: 1081 to 1127 ——— 4.5711pt PgVar ——— Normal Page PgEnds: T E X [607], (35) stating that the sum of fractions must total unity, and the reciprocity rule, A i F i−j = A j F j−i (8.54) which follows directly from eq. (8.44). The methods known as view factor algebra and the crossed-strings method are discussed briefly next. View Factor Algebra Many view factors for fairly complex configurations may be calculated without any integration simply by using the rules of reciprocity and summation and perhaps the known view factor for a more basic geometry. For exam- ple, suppose that the view factor for a corner piece from configuration 8 in Table 8.3 is given. Using this knowledge, the view factor F 3−4 between the two perpendicular strips shown in Fig. 8.19 can be evaluated. From the definition of the view factor and because the energy traveling to A 4 is the energy going to A 2 and A 4 minus the energy going to A 2 , it follows that F 3−4 = F 3−(2+4) − F 3−2 and using reciprocity, F 3−4 = 1 A 3 (A 2 + A 4 )F (2+4)−3 − A 2 F 2−3 Similarly, F 3−4 = A 2 + A 4 A 3 F (2+4)−(1+3) − F (2+4)−1 − A 2 A 3 F 2−(1+3) − F 2−1 All view factors on the right-hand side are for corner pieces and may be found by evaluating view factor 8 with appropriate dimensions. As a second example, the view factor from the inside surface of a finite-length cylinder to itself will be determined (as shown in the figure for view factor 11 for the A 3 A 4 A 2 A 1 Figure 8.19 View factor between two strips on a corner. BOOKCOMP, Inc. — John Wiley & Sons / Page 608 / 2nd Proofs / Heat Transfer Handbook / Bejan 608 THERMAL RADIATION 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 [608], (36) Lines: 1127 to 1178 ——— 0.12416pt PgVar ——— Long Page PgEnds: T E X [608], (36) limiting case of r 1 → 0). Calling the cylinder A 1 , and top and bottom openings of the cylinder A 2 and A 3 , respectively, the summation rule gives F 1−1 + F 1−2 + F 1−3 = 1orF 1−1 = 1 − 2F 1−2 because F 1−2 = F 1−3 . Now, using reciprocity, or A 1 F 1−2 = A 2 F 2−1 , and summation once again, F 1−1 = 1 − 2 A 2 A 1 F 2−1 = 1 − 2 A 2 A 1 (1 − F 2−3 ) because F 2−2 = 0. Then F 2−3 can be evaluated from configuration 9, with R 1 = R 2 = r/h = R, X = 2 + 1/R 2 ,or F 2−3 = 1 2 2 + 1 R 2 − 4 + 4 R 2 + 1 R 4 − 4 Finally, with A 2 /A 1 = πr 2 /2πrh = r/2h = R/2, F 1−1 = 1 − R − 1 2R 2 + 1 R 1 + 1 4R 2 = 1 + 1 2R − 1 + 1 4R 2 Crossed-Strings Method View factor algebra may be used to determine all the view factors in long enclosures with constant cross section. The method is called the crossed-strings method because the view factors can be determined experimentally by a person armed with four pins, a roll of string, and a yardstick. Consider the con- figuration in Fig. 8.20, which shows the cross section of an infinitely long enclosure, continuing into and out of the plane of the figure: The determination of F 1−2 is sought. Obviously, the surfaces shown are rather irregular (partly convex, partly concave), and the view between them may be obstructed. For such geometries, integration is out of the question; however, repeated application of the reciprocity and summation rules allows the evaluation of F 1−2 as F 1−2 = (A bc + A ad ) − (A ac + A bd ) 2A 1 (8.55) where, in general, A ab is the area (per unit depth) defined by the length of the string between points a and b. This relationship is easily memorized by looking at the configuration between any two surfaces as a generalized “rectangle,” consisting of A 1 , A 2 , and the two sides of A ac and A bd . Then F 1−2 = diagonals − sides 2 × originating area (8.56) As an example, suppose that the calculation of F 1−2 for configuration 1 in Table 8.3 via the crossed-strings method is desired. For that geometry, both “sides” would have BOOKCOMP, Inc. — John Wiley & Sons / Page 609 / 2nd Proofs / Heat Transfer Handbook / Bejan RADIATIVE EXCHANGE BETWEEN SURFACES 609 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 [609], (37) Lines: 1178 to 1199 ——— 4.68pt PgVar ——— Long Page PgEnds: T E X [609], (37) Figure 8.20 Crossed strings method for two-dimensional configurations. length h, both diagonals would have length √ h 2 + w 2 , and the originating area would have width w. Thus, F 1−2 = 2 √ h 2 + w 2 − 2h 2w = 1 + h w 2 − h w (8.57) as given in the table. 8.3.2 Radiative Exchange between Black Surfaces Consider an enclosure consisting of N opaque, black, isothermal surfaces. One or more of these surfaces may not be actual material but a hole through which radiation escapes, and through which external radiation may enter the enclosure. A hole is usually best modeled as a cold black surface, because a hole does not reflect internal radiation (black) and because usually no diffuse energy is entering through the hole (no emission or T = 0 K). If external radiation is entering through an opening, this energy tends to be directional and is best accounted for in the energy balances for individual surfaces that receive it. An energy balance for any isothermal surface A i in the enclosure yields Q i A i = E bi − H i H i = N j=1 E bj F i−j + H oi i = 1, 2, ,N (8.58) where Q i /A i is the average radiative heat flux on A i , E bi = σT 4 i is the surface’s emissive power, and H i is the total irradiation onto A i , consisting of the fractions of emitted radiation from all surfaces in the enclosure (including itself if A i is concave) BOOKCOMP, Inc. — John Wiley & Sons / Page 610 / 2nd Proofs / Heat Transfer Handbook / Bejan 610 THERMAL RADIATION 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 [610], (38) Lines: 1199 to 1252 ——— -1.86508pt PgVar ——— Normal Page PgEnds: T E X [610], (38) that are intercepted by A i , plus, possibly, external radiation H oi that enters through a hole and hits A i (per unit area of A i ). If the temperatures for all N surfaces making up the enclosure are known, eq. (8.58) constitutes a set of N explicit equations for the unknown radiative fluxes (Q i /A i ). Suppose that for surfaces i = 1, 2, ,n the heat fluxes are prescribed (and temperatures are unknown), whereas for surfaces i = n +1, ,N the temperatures are prescribed (heat fluxes unknown). Unlike for the heat fluxes, no explicit relations for the unknown temperatures exist. Placing all unknown temperatures on one side, eq. (8.58) may be written as E bi − n j=1 F i−j E bj = Q i A i + H oi + N j=n+1 F i−j E bj i = 1, 2, ,n (8.59) where everything on the right-hand side of the equation is known. In matrix form this is written as A · e b = b (8.60) where A = 1 − F 1−1 −F 1−2 ··· −F 1−n −F 2−1 1 − F 2−2 ··· −F 2−n . . . . . . . . . −F n−1 −F n−2 ··· 1 − F n−n (8.61) e b = E b1 E b2 . . . E bn b = Q 1 A 1 + H o1 + N j=n+1 F 1−j E bj Q 2 A 2 + H o2 + N j=n+1 F 2−j E bj . . . Q n A n + H on + N j=n+1 F n−j E bj (8.62) The n × n matrix A is readily inverted on a computer (generally with the aid of a software library subroutine), and the unknown temperatures are calculated as e b = A −1 · b (8.63) 8.3.3 Radiative Exchange between Diffuse Gray Surfaces It will now be assumed that all surfaces are gray and that they are diffuse emitters, absorbers, and reflectors. Under these conditions, = λ = α λ = α = 1 − ρ. The BOOKCOMP, Inc. — John Wiley & Sons / Page 611 / 2nd Proofs / Heat Transfer Handbook / Bejan RADIATIVE EXCHANGE BETWEEN SURFACES 611 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 [611], (39) Lines: 1252 to 1293 ——— 7.37305pt PgVar ——— Normal Page PgEnds: T E X [611], (39) radiative heat flux leaving surface A i now consists of emission plus the reflection of incoming radiation, J i = i E bi + ρ i H i (8.64) which is called the surface radiosity. In the same way as eq. (8.58) was formulated, one can make an energy balance at surface A i , but there are now two ways to make this balance, Q i A i = J i − H i = i E bi − α i H i i = 1, 2, ,N (8.65) the first stating that the net flux is equal to outgoing minus incoming radiation, the other stating that net flux is equal to the emitted minus the absorbed radiation. Eliminating H i between them gives an expression for radiosity in terms of local temperature and flux, J i = E bi − 1 i − 1 Q i A i (8.66) Replacing emissive power by radiosities in eq. (8.58) and using eq. (8.65) leads to N simultaneous equations for the N unknown radiosities, J i = i E bi + ρ i N j=1 F i−j J j + H oi (8.67a) or J i = Q i A i + N j=1 F i−j J j + H oi (8.67b) dependent on whether temperature (E bi ) or flux (Q i /A i ) is known for surface i. While commonly solved for in the older literature, there is rarely ever any need to determine radiosities. Eliminating them through the use of eq. (8.66) yields a set of N simultaneous algebraic equations, 1 i Q i A i − N j=1 1 j − 1 F i−j Q j A j + H oi = E bi − N j=1 F i−j E bj i = 1, 2, ,N (8.68) Equation (8.68) contains N different E bi and N different Q i /A i . Therefore, if the temperatures are specified for all N surfaces, all fluxes can be calculated. It is also possible to specify an arbitrary mix of N surface temperatures and heat fluxes, and eq. (8.68) allows determination of the remaining N unknowns (the one exception being BOOKCOMP, Inc. — John Wiley & Sons / Page 612 / 2nd Proofs / Heat Transfer Handbook / Bejan 612 THERMAL RADIATION 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 [612], (40) Lines: 1293 to 1329 ——— 2.73325pt PgVar ——— Normal Page PgEnds: T E X [612], (40) that it is not proper to specify all N heat fluxes: because from conservation of energy, N i=1 Q i = 0, this amounts to only N − 1 specifications). As for black enclosures, holes are modeled as cold black surfaces; because for such surfaces E bi = J i = 0, they do not appear in eq. (8.68) (except for external irradiation, entering through holes, accounted for at the surfaces receiving this irradiation). Convex Surface Exposed to Large Isothermal Enclosure In many impor- tant engineering applications a flat or convex surface (i.e., a surface that cannot “see itself”) is radiating into (and receiving radiation from) a large isothermal enclosure. In such a case, N = 2(A i being the convex surface and A e the large enclosure), and, if F i−e = 1, eq. (8.68) reduces to Q i A i = σ(T 4 i − T 4 e ) (1/ i ) + (A i /A e )[(1/ e ) − 1] i σ T 4 i − T 4 e (8.69) because for a large enclosure, A i A e . Equation (8.69) provides a simple, yet accurate set of radiation boundary conditions for engineering problems, which are dominated by conduction and/or convection. 8.3.4 Radiation Shields If it is desired to minimize radiative heat transfer between two surfaces, it is common practice to place one or more radiation shields between them (usually, thin metallic sheets of low emittance). In these situations any two shields A i and A j often enclose one another, or are very close together, such that F i−j 1. The radiative heat transfer between two diffusely reflecting plates is then, from eq. (8.69), Q = E bi − E bj R ij R ij = 1 i A i + 1 j − 1 1 A j (8.70) where R ij is termed the radiative resistance. Equation (8.70) is seen to be analogous to an electrical circuit with current Q and voltage potential E bi − E bj . Therefore, expressing radiative fluxes in terms of radiative resistances is commonly known as the network analogy (Oppenheim, 1956). The network analogy is a very powerful method for solving one-dimensional problems (e.g., 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, or concentric, plates, A 1 and A N , separated by N − 2 radiation shields, as shown in Fig. 8.21. Let each shield j have an emittance j on both sides. Then by applying eq. (8.70) to any two consecutive surfaces and using the fact that Q remains constant throughout the gap, Q = E b1 − E b2 R 12 =···= E bk−1 − E bk R k−1,k =···= E bN−1 − E bN R N−1,N = E b1 − E bN N j=2 R j−1,j (8.71) . temperatures and heat fluxes, and eq. (8.68) allows determination of the remaining N unknowns (the one exception being BOOKCOMP, Inc. — John Wiley & Sons / Page 612 / 2nd Proofs / Heat Transfer Handbook. − 1 √ 1 + R 2 A 2 a A 1 r BOOKCOMP, Inc. — John Wiley & Sons / Page 604 / 2nd Proofs / Heat Transfer Handbook / Bejan 604 THERMAL RADIATION 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 [604],. edge (configura- tion 8). BOOKCOMP, Inc. — John Wiley & Sons / Page 605 / 2nd Proofs / Heat Transfer Handbook / Bejan RADIATIVE EXCHANGE BETWEEN SURFACES 605 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 [605],