# Radiation Heat Transfer Figure (1)

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 Radiation Heat Transfer Figure (1) Radiation heat exchange between surfaces depends on the orientation of surfaces relative to each other; and this dependence on orientation is accounted for by the view factor. 1. The view factor Radiation heat transfer between surfaces depends on the orientation of the surfaces relative to each other as well as their radiation properties and temperatures, as illustrated in Figure (1). To account for the effects of orientation on radiation heat transfer between two surfaces, we define a new parameter called the view factor, which is a purely geometric quantity and is independent of the surface properties and temperature. It is also called the shape factor, configuration factor, and angle factor. The view factor based on the assumption that the surfaces are diffuse emitters and diffuse reflectors is called the diffuse view factor. we will consider radiation exchange between diffuse surfaces only, and thus the term view factor will simply mean diffuse view factor. The view factor from a surface i to a surface j is denoted by Fi → j or just Fij, and is defined as: Fij = the fraction of the radiation leaving surface i that strikes surface j directly Therefore, the view factor F12 represents the fraction of radiation leaving surface 1 that strikes surface 2 directly, and F21 represents the fraction of the radiation leaving surface 2 that strikes surface 1 directly. Note that the radiation that strikes a surface does not need to be absorbed by that surface. Also, radiation that strikes a surface after being reflected by other surfaces is not considered in the evaluation of view factors. Accordingly, the view factor F12 can be defined as: (1) where is the total rate of radiation leaves the entire (via emission and reflection) in all directions. And the view factor is defined as: (2) The relation between view factors and can be written as: (3) The foregoing relation, equation (3), is known as the reciprocity relation for view factor. It allows the calculation of a view factor from a knowledge of the other. Figure (2) The view factor from a surface to itself is zero for plane or convex surfaces and nonzero for concave surfaces. Noting that in the absence of strong electromagnetic fields radiation beams travel in straight paths, the view factor from a surface to itself will be zero unless the surface “sees” itself. Therefore, Fi→i = 0 for plane or convex surfaces and Fi → i 0 for concave surfaces, as illustrated in Figure (2). Figure (3) In a geometry that consists of two concentric spheres, the view factor F1 → 2 = 1 since the entire radiation leaving the surface of the smaller sphere will be intercepted by the larger sphere. The value of the view factor ranges between zero and one. The limiting case Fi → j = 0 indicates that the two surfaces do not have a direct view of each other, and thus radiation leaving surface i cannot strike surface j directly. The other limiting case Fi→j = 1 indicates that surface j completely surrounds surface i, so that the entire radiation leaving surface i is intercepted by surface j. For example, in a geometry consisting of two concentric spheres, the entire radiation leaving the surface of the smaller sphere (surface 1) will strike the larger sphere (surface 2), and thus F1 → 2 = 1, as illustrated in Figure (3). The view factor has proven to be very useful in radiation analysis because it allows us to express the fraction of radiation leaving a surface that strikes another surface in terms of the orientation of these two surfaces relative to each other. The underlying assumption in this process is that the radiation a surface receives from a source is directly proportional to the angle the surface subtends when viewed from the source. This would be the case only if the radiation coming off the source is uniform in all directions throughout its surface and the medium between the surfaces does not absorb, emit, or scatter radiation. That is, it will be the case when the surfaces are isothermal and diffuse emitters and reflectors and the surfaces are separated by a nonparticipating medium such as a vacuum or air. View factors for hundreds of common geometries are evaluated and the results are given in analytical, graphical, and tabular form in several publications. View factors for selected geometries are given in Tables (1) and (2) in analytical form and in Figures (4 to 7) in graphical form. The view factors in Table (1) are for three-dimensional geometries. The view factors in Table (2), on the other hand, are for geometries that are infinitely long in the direction perpendicular to the plane of the paper and are therefore two-dimensional. TABLE (1) TABLE (2)  Figure (5) View factor between two perpendicular rectangles with a common edge. Figure (4) View factor between two aligned parallel rectangles of equal size. Figure (6) View factor between two coaxial parallel disks. Figure (7) View factors for two concentric cylinders of finite length: (a) outer cylinder to inner cylinder; (b) outer cylinder to itself. 2. View factor relations Radiation analysis on an enclosure consisting of N surfaces requires the evaluation of N 2 view factors, and this evaluation process is probably the most time-consuming part of a radiation analysis. However, it is neither practical nor necessary to evaluate all of the view factors directly. Once a sufficient number of view factors are available, the rest of them can be determined by utilizing some fundamental relations for view factors, as discussed next. 2.1 The reciprocity relation We have shown earlier the pair of view factors Fi → j and Fj → i are related to each other by: (4) This relation is referred to as the reciprocity relation or the reciprocity rule, and it enables us to determine the counterpart of a view factor from a knowledge of the view factor itself and the areas of the two surfaces. When determining the pair of view factors Fi→j and Fj→i, it makes sense to evaluate first the easier one directly and then the more difficult one by applying the reciprocity relation. Figure (8) Radiation leaving any surface i of an enclosure must be intercepted completely by the surfaces of the enclosure. Therefore, the sum of the view factors from surface i to each one of the surfaces of the enclosure must be unity. 2.2 The summation rule The conservation of energy principle requires that the entire radiation leaving any surface i of an enclosure be intercepted by the surfaces of the enclosure. Therefore, the sum of the view factors from surface i of an enclosure to all surfaces of the enclosure, including to itself, must equal unity. This is known as the summation rule for an enclosure and is expressed as, [Fig. (8)]: (5) where N is the number of surfaces of the enclosure. For example, applying the summation rule to surface 1 of a three-surface enclosure yields: The summation rule can be applied to each surface of an enclosure by varying i from 1 to N. Therefore, the summation rule applied to each of the N surfaces of an enclosure gives N relations for the determination of the view factors. Also, the reciprocity rule gives (1/2) N(N - 1) additional relations. Then the total number of view factors that need to be evaluated directly for an N-surface enclosure becomes: N 2 - [N + N(N - 1)] = N(N - 1) For example, for a six-surface enclosure, we need to determine only 6(6 - 1) = 15 of the 62 = 36 view factors directly. The remaining 21 view factors can be determined from the 21 equations that are obtained by applying the reciprocity and the summation rules. Figure (9) The view factor from a surface to a composite surface is equal to the sum of the view factors from the surface to the parts of the composite surface. 2.3 The superposition rule Sometimes the view factor associated with a given geometry is not available in standard tables and charts. In such cases, it is desirable to express the given geometry as the sum or difference of some geometries with known view factors, and then to apply the superposition rule, which can be expressed as the view factor from a surface i to a surface j is equal to the sum of the view factors from surface i to the parts of surface j. Note that the reverse of this is not true. That is, the view factor from a surface j to a surface i is not equal to the sum of the view factors from the parts of surface j to surface i. Consider the geometry in Figure (9), which is infinitely long in the direction perpendicular to the plane of the paper. The radiation that leaves surface 1 and strikes the combined surfaces 2 and 3 is equal to the sum of the radiation that strikes surfaces 2 and 3. Therefore, the view factor from surface 1 to the combined surfaces of 2 and 3 is: F1 → (2, 3) = F1 → 2 + F1 → 3 (6) With the aids of Table (2), one can obtain F1 → (2, 3) and F1 → 2 and hence from equation (6), one can obtain F1 → 3 as: F1 → 3 = F1 → (2, 3) - F1 → 2 (7) Also, one can obtain F (2, 3) →1 by applying both reciprocity and superposition rules. The result is: (8) Figure (10) Two surfaces that are symmetric about a third surface will have the same view factor from the third surface. 2.4 The symmetry rule The symmetry rule can be expressed as two (or more) surfaces that possess symmetry about a third surface will have identical view factors from that surface [Fig. (10)]. The symmetry rule can also be expressed as if the surfaces j and k are symmetric about the surface i then Fi → j = Fi → k. Using the reciprocity rule, we can show that the relation Fj → i = Fk → i is also true in this case. View Factors between Infinitely Long Surfaces The Crossed-Strings Method Figure (11) Determination of the view factor F1 → 2 by the application of the crossed-strings method. Many problems encountered in practice involve geometries of constant cross section such as channels and ducts that are very long in one direction relative to the other directions. Such geometries can conveniently be considered to be two-dimensional, since any radiation interaction through their end surfaces will be negligible. These geometries can subsequently be modeled as being infinitely long, and the view factor between their surfaces can be determined by the simple crossed-strings method developed by H. C. Hottel in the 1950s. The surfaces of the geometry do not need to be flat; they can be convex, concave, or any irregular shape. To demonstrate this method, consider the geometry shown in Figure (11), and let us try to find the view factor F1 →2 between surfaces 1 and 2. The first thing we do is identify the endpoints of the surfaces (the points A, B, C, and D) and connect them to each other with tightly stretched strings, which are indicated by dashed lines. Hottel has shown that the view factor F1 → 2 can be expressed in terms of the lengths of these stretched strings, which are straight lines, as: (7) Note that L5 + L6 is the sum of the lengths of the crossed strings, and L3 + L4 is the sum of the lengths of the uncrossed strings attached to the endpoints. Therefore, Hottel’s crossed-strings method can be expressed verbally as: (8) The crossed-strings method is applicable even when the two surfaces considered share a common edge, as in a triangle. In such cases, the common edge can be treated as an imaginary string of zero length. The method can also be applied to surfaces that are partially blocked by other surfaces by allowing the strings to bend around the blocking surfaces. Example (1) Figure (12) The geometry considered in Example (1). Determine the view factors associated with an enclosure formed by two spheres, shown in Figure (12). Solution: The view factors associated with two concentric spheres are to be determined. Assumptions The surfaces are diffuse emitters and reflectors. Analysis The outer surface of the smaller sphere (surface 1) and inner surface of the larger sphere (surface 2) form a two-surface enclosure. Therefore, N = 2 and this enclosure involves N 2 = 22 = 4 view factors, which are F11, F12, F21, and F22. In this two-surface enclosure, we need to determine only: view factor directly. The remaining three view factors can be determined by the application of the summation and reciprocity rules. But it turns out that we can determine not only one but two view factors directly in this case by a simple inspection: F11 = 0, since no radiation leaving surface 1 strikes itself F12 = 1, since all radiation leaving surface 1 strikes surface 2 Actually it would be sufficient to determine only one of these view factors by inspection, since we could always determine the other one from the summation rule applied to surface 1 as: F11 + F12 = 1. The view factor F21 is determined by applying the reciprocity relation to surfaces 1 and 2: A1 F12 = A2 F21 which yields: Finally, the view factor F22 is determined by applying the summation rule to surface 2: F21 + F22 = 1 and thus: Discussion Note that when the outer sphere is much larger than the inner sphere (r2 r1), F22 approaches one. This is expected, since the fraction of radiation leaving the outer sphere that is intercepted by the inner sphere will be negligible in that case. Also note that the two spheres considered above do not need to be concentric. However, the radiation analysis will be most accurate for the case of concentric spheres, since the radiation is most likely to be uniform on the surfaces in that case. Example (2) Figure (13) The cylindrical enclosure considered in Example (2). EXAMPLE 12–2 Fraction of Radiation Leaving through an Opening Determine the fraction of the radiation leaving the base of the cylindrical enclosure shown in Figure 12–12 that escapes through a coaxial ring opening at its Determine the fraction of the radiation leaving the base of the cylindrical enclosure shown in Figure (13) that escapes through a coaxial ring opening at its top surface. The radius and the length of the enclosure are r1 = 10 cm and L = 10 cm, while the inner and outer radii of the ring are r2 = 5 cm and r3 = 8 cm, respectively. Solution: The fraction of radiation leaving the base of a cylindrical enclosure through a coaxial ring opening at its top surface is to be determined. Assumptions The base surface is a diffuse emitter and reflector. Analysis We are asked to determine the fraction of the radiation leaving the base of the enclosure that escapes through an opening at the top surface. Actually, what we are asked to determine is simply the view factor F1 → ring from the base of the enclosure to the ring-shaped surface at the top. We do not have an analytical expression or chart for view factors between a circular area and a coaxial ring, and so we cannot determine F1 → ring directly. However, we do have a chart for view factors between two coaxial parallel disks, and we can always express a ring in terms of disks. Let the base surface of radius r1 = 10 cm be surface 1, the circular area of r2 = 5 cm at the top be surface 2, and the circular area of r3 = 8 cm be surface 3. Using the superposition rule, the view factor from surface 1 to surface 3 can be expressed as: F1 → 3 = F1 → 2 + F1 → ring since surface 3 is the sum of surface 2 and the ring area. The view factors F1→2 and F1 → 3 are determined from the chart in Figure (6). For : And for: Therefore, which is the desired result. Note that F1→2 and F1 → 3 represent the fractions of radiation leaving the base that strike the circular surfaces 2 and 3, respectively, and their difference gives the fraction that strikes the ring area. Example (3) Figure (14) The pyramid considered in Example (3). Determine the view factors from the base of the pyramid shown in Figure (14) to each of its four side surfaces. The base of the pyramid is a square, and its side surfaces are isosceles triangles. Solution The view factors from the base of a pyramid to each of its four side surfaces for the case of a square base are to be determined. Assumptions The surfaces are diffuse emitters and reflectors. Analysis The base of the pyramid (surface 1) and its four side surfaces (surfaces 2, 3, 4, and 5) form a five-surface enclosure. The first thing we notice about this enclosure is its symmetry. The four side surfaces are symmetric about the base surface. Then, from the symmetry rule, we have: F12 = F13 = F14 = F15 Also, the summation rule applied to surface 1 yields: However, F11 = 0, since the base is a flat surface. Then the two relations above yield: F12 = F13 = F14 = F15 = 0.25 Discussion Note that each of the four side surfaces of the pyramid receive one-fourth of the entire radiation leaving the base surface, as expected. Also note that the presence of symmetry greatly simplified the determination of the view factors. Example (4) Figure (15) The infinitely long triangular duct considered in Example (4). Determine the view factor from any one side to any other side of the infinitely long triangular duct whose cross section is given in Figure (15). Solution The view factors associated with an infinitely long triangular duct are to be determined. Assumptions The surfaces are diffuse emitters and reflectors. Analysis The widths of the sides of the triangular cross section of the duct are L1, L2, and L3, and the surface areas corresponding to them are A1, A2, and A3, respectively. Since the duct is infinitely long, the fraction of radiation leaving any surface that escapes through the ends of the duct is negligible. Therefore, the infinitely long duct can be considered to be a three-surface enclosure, N = 3. This enclosure involves N 2 = 32 = 9 view factors, and we need to determine: of these view factors directly. Fortunately, we can determine all three of them by inspection to be: F11 = F22 = F33 = 0 since all three surfaces are flat. The remaining six view factors can be determined by the application of the summation and reciprocity rules. Applying the summation rule to each of the three surfaces gives:   Noting that F11 = F22 = F33 = 0 and multiplying the first equation by A1, the second by A2, and the third by A3 gives:   Finally, applying the three reciprocity relations A1F12 = A2F21, A1F13 = A3F31, and A2F23 = A3F32 gives:   This is a set of three algebraic equations with three unknowns, which can be solved to obtain:   Discussion Note that we have replaced the areas of the side surfaces by their corresponding widths for simplicity, since A = Ls and the length s can be factored out and canceled. We can generalize this result as the view factor from a surface of a very long triangular duct to another surface is equal to the sum of the widths of these two surfaces minus the width of the third surface, divided by twice the width of the first surface. Example (5) Two infinitely long parallel plates of widths a = 12 cm and b = 5 cm are located a distance c = 6 cm apart, as shown in Figure (16). Determine the view factor F1 → 2 from surface 1 to surface 2 by using the crossed-strings method. Figure (16) The two infinitely long parallel plates considered in Example (5). Solution The view factors between two infinitely long parallel plates are to be determined using the crossed-strings method, and the formula for the view factor is to be derived. Assumptions The surfaces are diffuse emitters and reflectors. Analysis First we label the endpoints of both surfaces and draw straight dashed lines between the endpoints, as shown in Figure (16). Then we identify the crossed and uncrossed strings and apply the crossed-strings method to determine the view factor F1 → 2: where L1 = a = 12 cm L4 = 9.22 cm L2 = b = 5 cm L5 = 7.81 cm L3 = c = 6 cm L6 = 13.42 cm Substituting, The database is protected by copyright ©dentisty.org 2016
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