Radiation Exchange Between Surfaces

Radiation Exchange Between Surfaces
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Published Date:25-10-2017
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c13.qxd 3/6/06 11:07 AM Page 811 CHAPTER 13 Radiation Exchange Between Surfacesc13.qxd 3/6/06 11:07 AM Page 812 812 Chapter 13  Radiation Exchange Between Surfaces Having thus far restricted our attention to radiative processes that occur at a single surface, we now consider the problem of radiative exchange between two or more surfaces. This exchange depends strongly on the surface geometries and ori- entations, as well as on their radiative properties and temperatures. Initially, we assume that the surfaces are separated by a nonparticipating medium. Since such a medium neither emits, absorbs, nor scatters, it has no effect on the transfer of radia- tion between surfaces. A vacuum meets these requirements exactly, and most gases meet them to an excellent approximation. Our first objective is to establish geometrical features of the radiation exchange problem by developing the notion of a view factor. Our second objective is to develop procedures for predicting radiative exchange between surfaces that form an enclosure. We conclude our consideration of radiation exchange between surfaces by considering the effects of a participating medium, namely, an intervening gas that emits and absorbs radiation. 13.1 The View Factor To compute radiation exchange between any two surfaces, we must first introduce the concept of a view factor (also called a configuration or shape factor). 13.1.1 The View Factor Integral The view factor F is defined as the fraction of the radiation leaving surface i that is ij intercepted by surface j. To develop a general expression for F , we consider the arbi- ij trarily oriented surfaces A and A of Figure 13.1. Elemental areas on each surface, i j dA and dA , are connected by a line of length R, which forms the polar angles  and i j i  , respectively, with the surface normals n and n . The values of R,  , and  vary j i j i j with the position of the elemental areas on A and A . i j dA j n j θ j dA cos θ j j A , T j j R d n j–i i n i θ i dA i dA i A , T i i FIGURE 13.1 View factor associated with radiation exchange between elemental surfaces of area dA and dA . i jc13.qxd 3/6/06 11:07 AM Page 813 13.1  The View Factor 813 From the definition of the radiation intensity, Section 12.2.2, and Equation 12.6, the rate at which radiation leaves dA and is intercepted by dA may be i j expressed as dq  I cos  dA d ilj er, i i i ji where I is the intensity of radiation leaving surface i by emission and reflection er,i and d is the solid angle subtended by dA when viewed from dA. With ji j i 2 d  (cos  dA )/R from Equation 12.2, it follows that ji j j cos  cos  i j dq  I dA dA ilj er, i i j 2 R Assuming that surface i emits and reflects diffusely and substituting from Equation 12.22, we then obtain cos  cos  i j dq  J dA dA ilj i i j 2 R The total rate at which radiation leaves surface i and is intercepted by j may then be obtained by integrating over the two surfaces. That is, cos  cos  i j q  J dA dA   ilj i i j 2 A A R i j where it is assumed that the radiosity J is uniform over the surface A . From the def- i i inition of the view factor as the fraction of the radiation that leaves A and is inter- i cepted by A , j q ilj F  ij A J i i it follows that cos  cos  i j 1 F    dA dA (13.1) ij i j 2 A i A A R i j Similarly, the view factor F is defined as the fraction of the radiation that leaves A ji j and is intercepted by A . The same development then yields i cos  cos  i j 1 F    dA dA (13.2) ji i j 2 A j A A R i j Either Equation 13.1 or 13.2 may be used to determine the view factor associated with any two surfaces that are diffuse emitters and reflectors and have uniform radiosity. 13.1.2 View Factor Relations An important view factor relation is suggested by Equations 13.1 and 13.2. In par- ticular, equating the integrals appearing in these equations, it follows that A F  A F (13.3) i ij j ji This expression, termed the reciprocity relation, is useful in determining one view factor from knowledge of the other.c13.qxd 3/6/06 11:07 AM Page 814 814 Chapter 13  Radiation Exchange Between Surfaces T N T 1 J N J 1 T 2 J 2 J i FIGURE 13.2 T i Radiation exchange in an enclosure. Another important view factor relation pertains to the surfaces of an enclosure (Figure 13.2). From the definition of the view factor, the summation rule N F  1 (13.4)  ij j1 may be applied to each of the N surfaces in the enclosure. This rule follows from the conservation requirement that all radiation leaving surface i must be intercepted by the enclosure surfaces. The term F appearing in this summation represents the frac- ii tion of the radiation that leaves surface i and is directly intercepted by i. If the sur- face is concave, it sees itself and F is nonzero. However, for a plane or convex ii surface, F  0. ii 2 To calculate radiation exchange in an enclosure of N surfaces, a total of N view factors is needed. This requirement becomes evident when the view factors are arranged in the matrix form: … F F F 11 12 1N … F F F 21 22 2N      … F F F N1 N2 NN However, all the view factors need not be calculated directly. A total of N view factors may be obtained from the N equations associated with application of the J 1 J F = 1 F = 0 2 12 11 A A ___ 1 __1 _ F = F = 1 – 21 22 A A 2 2 1 FIGURE 13.3 View factors for the enclosure 2 formed by two spheres.c13.qxd 3/6/06 11:07 AM Page 815 13.1  The View Factor 815 summation rule, Equation 13.4, to each of the surfaces in the enclosure. In addition, N(N  1)/2 view factors may be obtained from the N(N  1)/2 applications of the reciprocity relation, Equation 13.3, which are possible for the enclosure. Accord- 2 ingly, only N  N  N(N  1)/2  N(N  1)/2 view factors need be determined directly. For example, in a three-surface enclosure this requirement corresponds to TABLE 13.1 View Factors for Two-Dimensional Geometries 4 Geometry Relation Parallel Plates with Midlines Connected by Perpendicular w i 2 1/2 2 1/2 (W  W )  4  (W  W )  4 i j j i i F  ij 2W i W  w /L, W  w /L L i i j j j w j Inclined Parallel Plates of Equal Width and a Common Edge j  F  1  sin ij w   2 α i w Perpendicular Plates with a Common Edge j 2 1/2 1  (w /w )  1  (w /w ) j i j i F  ij 2 w j i w i Three-Sided Enclosure w  w  w i j k F  w ij j 2w i w k j k i w i (continues)c13.qxd 3/6/06 11:07 AM Page 816 816 Chapter 13  Radiation Exchange Between Surfaces TABLE 13.1 Continued Geometry Relation Parallel Cylinders of Different Radii 1 2 2 1/2 F    C  (R  1) ij  2 r j r i 2 2 1/2  C (R  1) j i R 1 1  (R  1) cos      C C s R 1 1  (R  1) cos       C C R  r /r , S  s/r j i i C  1  R  S Cylinder and Parallel Rectangle s s r r 1 2 1 1 F  tan  tan ij s  s   1 2 L L j L i s 2 s 1 Infinite Plane and Row of Cylinders 2 1/2 D s F  1  1  D ij    s j 2 2 1/2 D s  D 1  tan    2   s D i only 3(3  1)/2  3 view factors. The remaining six view factors may be obtained by solving the six equations that result from use of Equations 13.3 and 13.4. To illustrate the foregoing procedure, consider a simple, two-surface enclosure involving the spherical surfaces of Figure 13.3. Although the enclosure is character- 2 ized by N  4 view factors (F , F , F , F ), only N(N  1)/2  1 view factor 11 12 21 22 need be determined directly. In this case such a determination may be made by inspection. In particular, since all radiation leaving the inner surface must reach the outer surface, it follows that F  1. The same may not be said of radiation leaving 12c13.qxd 3/6/06 11:07 AM Page 817 13.1  The View Factor 817 TABLE 13.2 View Factors for Three-Dimensional Geometries 4 Geometry Relation Aligned Parallel X  X/L, Y  Y/L Rectangles 2 2 1/2 (1  X ) (1  Y ) 2 (Figure 13.4) F  ln ij    2 2 X Y 1  X  Y j L X 2 1/2 1  X (1  Y ) tan 2 1/2 (1  Y ) i Y Y 2 1/2 1 1 1 X  Y (1  X ) tan  X tan X  Y tan Y 2 1/2  (1  X ) Coaxial Parallel Disks R  r /L, R  r /L i i j j (Figure 13.5) 2 1  R j r j S  1  2 R j i 1 2 2 1/2 F  S  S  4(r /r ) ij j i 2 r L i i Perpendicular Rectangles H  Z/X, W  Y/X with a Common Edge 1 1 1 1 1 (Figure 13.6) F  W tan  H tan ij  W W H 1 2 2 1/2 1  (H  W ) tan j 2 2 1/2 (H  W ) Z 2 2 2 2 2 2 i W (1  W )(1  H ) W (1  W  H ) 1 X  ln Y  2 2  2 2 2 4 1  W  H (1  W )(W  H ) 2 2 2 2 H H (1  H  W )     2 2 2  (1  H )(H  W ) the outer surface, since this surface sees itself. However, from the reciprocity rela- tion, Equation 13.3, we obtain A A 1 1 F  F  21 12     A A 2 2 From the summation rule, we also obtain F  F  1 11 12 in which case F  0, and 11 F  F  1 21 22 in which case A 1 F  1  22   A 2c13.qxd 3/6/06 11:07 AM Page 818 818 Chapter 13  Radiation Exchange Between Surfaces 1.0 ∞ 10 4 j 0.7 L 2 0.5 0.4 i 1.0 Y 0.3 X 0.6 0.2 0.4 0.1 0.2 0.07 0.05 Y/L = 0.1 0.04 0.03 0.02 0.01 0.1 0.2 0.3 0.5 1.0 2 3 4 5 10 20 X/L FIGURE 13.4 View factor for aligned parallel rectangles. For more complicated geometries, the view factor may be determined by solv- ing the double integral of Equation 13.1. Such solutions have been obtained for many different surface arrangements and are available in equation, graphical, and tabular form 1–4. Results for several common geometries are presented in Tables 13.1 and 13.2 and Figures 13.4 through 13.6. The configurations of Table 13.1 are assumed to be infinitely long (in a direction perpendicular to the page) and are hence two-dimensional. The configurations of Table 13.2 and Figures 13.4 through 13.6 are three-dimensional. It is useful to note that the results of Figures 13.4 through 13.6 may be used to determine other view factors. For example, the view factor for an end surface of a 1.0 r j 8 j 6 r L i 0.8 5 i 4 0.6 3 r /L j 1.5 = 2 1.25 0.4 1.0 0.8 0.2 0.6 0.4 0.3 0 0.1 0.2 0.4 0.6 0.8 1 2 4 6 8 10 L/r i FIGURE 13.5 View factor for coaxial parallel disks. F F ij ijc13.qxd 3/6/06 11:07 AM Page 819 13.1  The View Factor 819 Y/X = j 0.5 0.02 Z i X Y 0.05 0.4 0.1 0.2 0.3 0.4 0.6 0.2 1.0 1.5 0.1 2.0 4 10 20 0 0.1 0.2 0.4 0.6 0.8 1 2 4 6 8 10 Z/X FIGURE 13.6 View factor for perpendicular rectangles with a common edge. cylinder (or a truncated cone) relative to the lateral surface may be obtained by using the results of Figure 13.5 with the summation rule, Equation 13.4. Moreover, Fig- ures 13.4 and 13.6 may be used to obtain other useful results if two additional view factor relations are developed. The first relation concerns the additive nature of the view factor for a sub- divided surface and may be inferred from Figure 13.7. Considering radiation from surface i to surface j, which is divided into n components, it is evident that n F  F (13.5) i( j)  ik k1 where the parentheses around a subscript indicate that it is a composite surface, in which case ( j) is equivalent to (1, 2, ... , k,... , n). This expression simply states that radiation reaching a composite surface is the sum of the radiation reaching its parts. Although it pertains to subdivision of the receiving surface, it may also be used to obtain the second view factor relation, which pertains to subdivision of the originating A i A n A k A 1 n FIGURE 13.7 A = A j Σ k k=1 Areas used to illustrate view factor relations. F ijc13.qxd 3/6/06 11:07 AM Page 820 820 Chapter 13  Radiation Exchange Between Surfaces surface. Multiplying Equation 13.5 by A and applying the reciprocity relation, Equa- i tion 13.3, to each of the resulting terms, it follows that n A F  A F (13.6) j ( j)i k ki  k1 or n A F k ki  k1 F  (13.7) ( j)i n A  k k1 Equations 13.6 and 13.7 may be applied when the originating surface is composed of several parts. For problems involving complicated geometries, analytical solutions to Equa- tion 13.1 may not be obtainable, in which case values of the view factors must be estimated using numerical methods. In situations involving extremely complex structures that may have hundreds or thousands of radiative surfaces, considerable error may be associated with the numerically calculated view factors. In such situa- tions, Equation 13.3 should be used to check the accuracy of individual view fac- tors, and Equation 13.4 should be used to determine whether the conservation of energy principle is satisfied 5. EXAMPLE 13.1 Consider a diffuse circular disk of diameter D and area A and a plane diffuse sur- j face of area A  A . The surfaces are parallel, and A is located at a distance L from i j i the center of A . Obtain an expression for the view factor F . j ij SOLUTION Known: Orientation of small surface relative to large circular disk. Find: View factor of small surface with respect to disk, F . ij Schematic: D dr dA j A j θ j L R θ i A i Assumptions: 1. Diffuse surfaces. 2. A  A . i jc13.qxd 3/6/06 11:07 AM Page 821 13.1  The View Factor 821 Analysis: The desired view factor may be obtained from Equation 13.1. cos  cos  i j 1 F  dA dA   ij i j 2 A i A A R i j Recognizing that  ,  , and R are approximately independent of position on A , this i j i expression reduces to cos  cos  i j F  dA  ij j 2 A R j or, with    , i j 2 cos  F  dA  ij j 2 A R j 2 2 2 With R  r  L , cos   (L/R), and dA  2r dr, it follows that j D/2 2 r dr D 2 F  2L   (13.8)  ij 2 2 2 2 2 0 (r  L ) D  4L Comments: 1. Equation 13.8 may be used to quantify the asymptotic behavior of the curves in Figure 13.5 as the radius of the lower circle, r , approaches zero. i 2. The preceding geometry is one of the simplest cases for which the view factor may be obtained from Equation 13.1. Geometries involving more detailed inte- grations are considered in the literature 1, 3. EXAMPLE 13.2 Determine the view factors F and F for the following geometries: 12 21 D A A 1 A 1 1 L = D L A A 2 L = D 2 A A 3 3 A 2 (1) (2) (3) 1. Sphere of diameter D inside a cubical box of length L  D. 2. One side of a diagonal partition within a long square duct. 3. End and side of a circular tube of equal length and diameter. SOLUTION Known: Surface geometries. Find: View factors.c13.qxd 3/6/06 11:07 AM Page 822 822 Chapter 13  Radiation Exchange Between Surfaces Assumptions: Diffuse surfaces with uniform radiosities. Analysis: The desired view factors may be obtained from inspection, the reci- procity rule, the summation rule, and/or use of the charts. 1. Sphere within a cube: By inspection, F  1  12 2 A D  1 By reciprocity, F  F   1   21 12 2 A 6 6L 2 2. Partition within a square duct: From summation rule, F  F  F  1 11 12 13 where F  0 11 By symmetry, F  F 12 13 Hence F  0.50  12 A 2 L 1 By reciprocity, F  F   0.5  0.71  21 12 L A 2 3. Circular tube: From Table 13.2 or Figure 13.5, with (r /L)  0.5 and (L/r )  2, F  0.172 3 1 13 From summation rule, F  F  F  1 11 12 13 or, with F  0, F  1  F  0.828  11 12 13 2 A D /4 1 From reciprocity, F  F   0.828  0.207  21 12 DL A 2 13.2 Radiation Exchange Between Opaque, Diffuse, Gray Surfaces in an Enclosure In general, radiation may leave an opaque surface due to both reflection and emis- sion, and on reaching a second opaque surface, experience reflection as well as absorption. In an enclosure, such as that of Figure 13.8a, radiation may experience multiple reflections off all surfaces, with partial absorption occurring at each. Analyzing radiation exchange in an enclosure may be simplified by making certain assumptions. Each surface of the enclosure is assumed to be isothermal and to be characterized by a uniform radiosity and a uniform irradiation. Opaque, dif- fuse, gray surface behavior is also assumed, and the medium within the enclosure is taken to be nonparticipating. The problem is generally one in which either the tem- perature T or the net radiative heat flux q  associated with each of the surfaces is i i known. The objective is to use this information to determine the unknown radiative heat fluxes and temperatures associated with each of the surfaces.c13.qxd 3/6/06 11:07 AM Page 823 13.2  Radiation Exchange Between Opaque, Diffuse, Gray Surfaces 823 13.2.1 Net Radiation Exchange at a Surface The term q , which is the net rate at which radiation leaves surface i, represents the i net effect of radiative interactions occurring at the surface (Figure 13.8b). It is the rate at which energy would have to be transferred to the surface by other means to maintain it at a constant temperature. It is equal to the difference between the sur- face radiosity and irradiation and may be expressed as q  A ( J  G ) (13.9) i i i i From Equation 13.9 and the definition of the radiosity J , i J  E   G (13.10) i i i i The net radiative transfer from the surface may also be expressed in terms of the surface emissive power and the absorbed irradiation: q  A (E   G ) (13.11) i i i i i where use has been made of the relationship   1   for an opaque surface. This i i relationship is illustrated in Figure 13.8c. Substituting from Equation 12.35 and rec- ognizing that   1    1  for an opaque, diffuse, gray surface, the radios- i i i ity may also be expressed as J   E  (1   )G (13.12) i i bi i i T , A , ε T , A , ε 1 1 1 j j j q J (a) i i G i T , A , ε i i i G A ρ G A i i i i i J i E A i i G A J A i i i i 1 – ε ____i ε A α G A i i i i i E bi q q i i q i (b) (c) (d) FIGURE 13.8 Radiation exchange in an enclosure of diffuse, gray surfaces with a nonparticipating medium. (a) Schematic of the enclosure. (b) Radiative balance according to Equation 13.9. (c) Radiative balance according to Equation 13.11. (d) Network element representing the net radiation transfer from a surface.c13.qxd 3/6/06 11:07 AM Page 824 824 Chapter 13  Radiation Exchange Between Surfaces Solving for G and substituting into Equation 13.9, it follows that i J   E i i bi q  A J  i i i   1   i or E  J bi i q  (13.13) i (1   )/ A i i i Equation 13.13 provides a convenient representation for the net radiative heat transfer rate from a surface. This transfer, which may be represented by the network element of Figure 13.8d, is associated with the driving potential (E  J ) and a sur- bi i face radiative resistance of the form (1 )/ A . Hence, if the emissive power that i i i the surface would have if it were black exceeds its radiosity, there is net radiation heat transfer from the surface; if the inverse is true, the net transfer is to the surface. It is sometimes the case that one of the surfaces is very large relative to the other surfaces under consideration. For example, the system might consist of multi- ple small surfaces in a large room. In this case, the area of the large surface is effec- tively infinite (A l ), and we see that its surface radiative resistance, (1   )/ A , i i i i is effectively zero, just as it would be for a black surface (  1). Hence, J  E , i i bi and a surface which is large relative to all other surfaces under consideration can be treated as if it were a blackbody. This important conclusion was reached in Sec- tion 12.6 based on a physical argument and has now been confirmed from our treat- ment of gray surface radiation exchange. Again, the physical explanation is that, even though the large surface may reflect some of the irradiation incident upon it, it is so big that there is a high probability that the reflected radiation reaches another point on the same large surface. After many such reflections, all the radiation that was originally incident on the large surface is absorbed by the large surface, and none ever reaches any of the smaller surfaces. 13.2.2 Radiation Exchange Between Surfaces To use Equation 13.13, the surface radiosity J must be known. To determine this i quantity, it is necessary to consider radiation exchange between the surfaces of the enclosure. The irradiation of surface i can be evaluated from the radiosities of all the sur- faces in the enclosure. In particular, from the definition of the view factor, it follows that the total rate at which radiation reaches surface i from all surfaces, including i, is N A G  F A J i i ji j j  j1 or from the reciprocity relation, Equation 13.3, N A G  A F J i i i ij j  j1 Canceling the area A and substituting into Equation 13.9 for G , i i N q  A J  F J i i i ij j    j1c13.qxd 3/6/06 11:07 AM Page 825 13.2  Radiation Exchange Between Opaque, Diffuse, Gray Surfaces 825 or, from the summation rule, Equation 13.4, N N q  A F J  F J i i ij i ij j     j1 j1 Hence N N q  A F (J  J )  q (13.14) i i ij i j ij   j1 j1 This result equates the net rate of radiation transfer from surface i, q , to the sum of i components q related to radiative exchange with the other surfaces. Each component ij may be represented by a network element for which (J  J ) is the driving potential i j 1 and (A F ) is a space or geometrical resistance (Figure 13.9). i ij Combining Equations 13.13 and 13.14, we then obtain N J  J E  J i j bi i  (13.15)  1 (1   )/ A j1 (A F ) i i i i ij As shown in Figure 13.9, this expression represents a radiation balance for the radiosity node associated with surface i. The rate of radiation transfer (current flow) to i through its surface resistance must equal the net rate of radiation transfer (current flows) from i to all other surfaces through the corresponding geometrical resistances. Note that Equation 13.15 is especially useful when the surface temperature T i (hence E ) is known. Although this situation is typical, it does not always apply. bi In particular, situations may arise for which the net radiation transfer rate at the surface q , rather than the temperature T , is known. In such cases the preferred i i form of the radiation balance is Equation 13.14, rearranged as N J  J i j q  (13.16) i  1 j1 (A F ) i ij q i1 J q 1 i2 J 2 –1 (A F ) i i1 q –1 i3 (A F ) i i2 J 3 –1 (A F ) E J i i3 bi i J q N–1 i q 1– ε i(N–1) ____ i –1 (A F ) ε i i(N–1) A i i Node corresponding –1 (A F ) i iN to the surface i J N q iN FIGURE 13.9 Network representation of radiative exchange between surface i and the remaining surfaces of an enclosure.c13.qxd 3/6/06 11:07 AM Page 826 826 Chapter 13  Radiation Exchange Between Surfaces Use of network representations was first suggested by Oppenheim 6. The net- work is built by first identifying nodes associated with the radiosities of each of the N surfaces of the enclosure. The method provides a useful tool for visualizing radia- tion exchange in the enclosure and, at least for simple enclosures, may be used as the basis for predicting this exchange. An alternative direct approach to solving radiation enclosure problems involves writing Equation 13.15 for each surface at which T is known, and writing i Equation 13.16 for each surface at which q is known. The resulting set of N linear, i algebraic equations is solved for J , J , ..., J . With knowledge of the J , Equation 1 2 N i 13.13 may then be used to determine the net radiation heat transfer rate q at each i surface of known T or the value of T at each surface of known q . For any number i i i N of surfaces in the enclosure, the foregoing problem may readily be solved by the iteration or matrix inversion methods of Chapter 4. EXAMPLE 13.3 In manufacturing, the special coating on a curved solar absorber surface of area A  2 2 15 m is cured by exposing it to an infrared heater of width W  1 m. The absorber and heater are each of length L  10 m and are separated by a distance of H  1 m. The upper surface of the absorber and the lower surface of the heater are insulated. Room walls, T sur Absorber surface, A , T , ε 2 2 2 H Heater, A , T , ε 1 1 1 W The heater is at T  1000 K and has an emissivity of   0.9, while the absorber is 1 1 at T  600 K and has an emissivity of   0.5. The system is in a large room whose 2 2 walls are at 300 K. What is the net rate of heat transfer to the absorber surface? SOLUTION Known: A curved, solar absorber surface with a special coating is being cured by use of an infrared heater in a large room. Find: Net rate of heat transfer to the absorber surface. Schematic: Collector, L = 10 m 2 T = 600 K, A = 25 m , 2 2 A' 2 ε = 0.5 2 T = T = 300 K 3 sur ε = 1 3 H = 1 m A' 3 2 T = 1000 K, A = 10 m , 1 1 ε = 0.9 1 W = 1 m Heater, L = 10 mc13.qxd 3/6/06 11:07 AM Page 827 13.2  Radiation Exchange Between Opaque, Diffuse, Gray Surfaces 827 Assumptions: 1. Steady-state conditions exist. 2. Convection effects are negligible. 3. Absorber and heater surfaces are diffuse and gray. 4. The surrounding room is large and therefore behaves as a blackbody. Analysis: The system may be viewed as a three-surface enclosure, with the third surface being the large surrounding room, which behaves as a blackbody. We are interested in obtaining the net rate of radiation transfer to surface 2. We solve the problem using both the radiation network and direct approaches. Radiation Network Approach The radiation network is constructed by first identifying nodes associated with the radiosities of each surface, as shown in step 1 in the schematic below. Then each radiosity node is connected to each of the other radiosity nodes through the appropriate space resistance, as shown in step 2. We will treat the surroundings as having a large but unspecified area, which introduces 1 1 difficulty in expressing the space resistances (A F ) and (A F ) . Fortunately, 3 31 3 32 from the reciprocity relation (Equation 13.3), we can replace A F with A F and 3 31 1 13 A F with A F , which are more readily obtained. The final step is to connect the 3 32 2 23 blackbody emissive powers associated with the temperature of each surface to the radiosity nodes, using the appropriate form of the surface resistance. 4 E = T b3 3 1 – ε 3 ε A 3 3 J 3 J 3 J 3 –1 –1 –1 –1 (A F ) (A F ) (A F ) (A F ) 1 13 2 23 1 13 2 23 J J 1 2 J J J J 1 – ε 1 – ε 1 2 1 2 1 2 –1 –1 (A F ) (A F ) A A 1 12 ε 1 12 ε 1 1 2 2 4 4 E = T E = T b1 1 b2 2 Step 1 Step 2 Step 3 In this problem, the surface resistance associated with surface 3 is zero accord- 4 2 ing to assumption 4; therefore, J  E  T  459 W/m . 3 b3 3 Summing currents at the J node yields 1 4 4 T  J J  J J  T 1 1 1 2 1 3   (1) (1   )/ A 1/A F 1/A F 1 1 1 1 12 1 13 while summing the currents at the J node results in 2 4 4 T  J J  J J  T 2 2 2 1 2 3   (2) (1   )/ A 1/A F 1/A F 2 2 2 1 12 2 23  The view factor F may be obtained by recognizing that F  F , where A is 12 12 12 2 shown in the schematic as the rectangular base of the absorber surface. Then, fromc13.qxd 3/6/06 11:07 AM Page 828 828 Chapter 13  Radiation Exchange Between Surfaces Figure 13.4 or Table 13.2, with Y/L  10/1  10 and X/L  1/1  1, F  0.39 12 From the summation rule, and recognizing that F  0, it also follows that 11 F  1  F  1  0.39  0.61 13 12 The last needed view factor is F . We recognize that, since radiation propagating 23  from surface 2 to surface 3 must pass through the hypothetical surface A , 2  A F  A F 2 23 2 23 and from symmetry F  F . Thus 23 13 2  A 2 10 m F  F   0.61  0.41 23 13 2 A 15 m 2 4 We may now solve Equations 1 and 2 for J and J . Recognizing that E  T  1 2 b1 1 2 56,700 W/m and canceling the area A , we can express Equation 1 as 1 56,700  J J  J J  459 1 1 2 1   (1  0.9)/0.9 1/0.39 1/0.61 or 10J  0.39J  510,582 (3) 1 2 2 4 Noting that E  T  7348 W/m and dividing by the area A , we can express b2 2 2 Equation 2 as 7348  J J  J J  459 2 2 1 2   2 2 (1  0.5)/0.5 1/0.41 15 m /(10 m  0.39) or 0.26J  1.67J  7536 (4) 1 2 2 Solving Equations 3 and 4 simultaneously yields J  12,487 W/m . 2 An expression for the net rate of heat transfer from the absorber surface, q , 2 may be written upon inspection of the radiation network and is 4 T  J 2 2 q  2 (1 )/ A 2 2 2 resulting in 2 (7348  12,487)W/m q   77.1 kW 2 2 (1  0.5)/(0.5  15 m ) Hence, the net heat transfer rate to the absorber is q q  77.1 kW.  net 2 Direct Approach Using the direct approach, we write Equation 13.15 for each of the three surfaces. We use reciprocity to rewrite the space resistances in terms of the known view factors from above and to eliminate A . 3 Surface 1 4 T  J J  J J  J 1 1 1 2 1 3 (5)   (1   )/ A 1/A F 1/A F 1 1 1 1 12 1 13c13.qxd 3/6/06 11:07 AM Page 829 13.2  Radiation Exchange Between Opaque, Diffuse, Gray Surfaces 829 Surface 2 4 T  J J  J J  J J  J J  J 2 2 2 1 2 3 2 1 2 3     (6) (1   )/ A 1/A F 1/A F 1/A F 1/A F 2 2 2 2 21 2 23 1 12 2 23 Surface 3 4 T  J J  J J  J J  J J  J 3 3 3 1 3 2 3 1 3 2     (7) (1   )/ A 1/A F 1/A F 1/A F 1/A F 3 3 3 3 31 3 32 1 13 2 23 Substituting values of the areas, temperatures, emissivities, and view factors into Equations 5 through 7 and solving them simultaneously, we obtain J  51,541 1 2 2 2 W/m , J  12,487 W/m , and J  459 W/m . Equation 13.13 may then be written 2 3 for surface 2 as 4 T  J 2 2 q  2 (1 )/ A 2 2 2 This expression is identical to the expression that was developed using the radiation network. Hence, q 77.1 kW. 2 Comments: 1. In order to solve Equations 5 through 7 simultaneously, we must first multiply both sides of Equation 7 by (1   )/ A  0 to avoid division by zero, result- 3 3 3 4 ing in the simplified form of Equation 7, which is J  T . 3 3 4 2. If we substitute J  T into Equations 5 and 6, it is evident that Equations 5 3 3 and 6 are identical to Equations 1 and 2, respectively. 3. The direct approach is recommended for problems involving N  4 surfaces, since radiation networks become quite complex as the number of surfaces increases. 4. As will be seen in Section 13.3, the radiation network approach is particularly useful when thermal energy is transferred to or from surfaces by additional means, that is, by conduction and/or convection. In these multimode heat trans- fer situations, the additional energy delivered to or taken from the surface can be represented by additional current into or out of a node.  5. Recognize the utility of using a hypothetical surface (A ) to simplify the evalu- 2 ation of view factors. 6. We could have approached the solution in a slightly different manner. Radia- tion leaving surface 1 must pass through the openings (hypothetical surface 3) in order to reach the surroundings. Thus, we can write. F  F 13 13  A F  A F  A F 1 13 1 13 3 31 A similar relationship can be written for exchange between surface 2 and the sur-  roundings, that is, A F  A F . Thus, the space resistances which connect to 2 23 3 32 radiosity node 3 in the radiation network above can be replaced by space resis- tances pertaining to surface 3. The resistance network would be unchanged, and the space resistances would have the same values as those determined in the fore- going solution. However, it may be more convenient to calculate the view factors by utilizing the hypothetical surfaces 3. With the surface resistance for surface 3 equal to zero, we see that openings of enclosures that exchange radiation with large surroundings may be treated as hypothetical, nonreflecting black surfaces ()   1 whose temperature is equal to that of the surroundings (T  T ). 3 3 surc13.qxd 3/6/06 11:07 AM Page 830 830 Chapter 13  Radiation Exchange Between Surfaces 13.2.3 Blackbody Radiation Exchange In Example 13.3, we saw that large surroundings may be treated as hypothetical black surfaces. Real surfaces, such as those that are charred or coated with high- emissivity finishes, also exhibit near-black behavior. Matters are simplified signifi- cantly when all the surfaces of the enclosure are black. Since the absorptivity of a black surface is unity, there is no reflection and the radiosity is composed solely of the emitted energy. Hence, Equation 13.14 reduces to N 4 4 q  A F (T  T ) (13.17) i  i ij i j j1 EXAMPLE 13.4 A furnace cavity, which is in the form of a cylinder of 75-mm diameter and 150- mm length, is open at one end to large surroundings that are at 27°C. The sides and bottom may be approximated as blackbodies, are heated electrically, are well insu- lated, and are maintained at temperatures of 1350 and 1650°C, respectively. D Side, T 1 Heater wire L Insulation Bottom, T 2 How much power is required to maintain the furnace conditions? SOLUTION Known: Surface temperatures of cylindrical furnace and surroundings. Find: Power required to maintain prescribed temperatures. Schematic: q T = 300 K sur A , T = T 3 3 sur A , T = 1350°C 1 1 L = 0.15 m A , T = 1650°C 2 2 D = 0.075 m

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