ch13 - PROBLEM 13.1 KNOWN Various geometric shapes...

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PROBLEM 13.1 KNOWN: Various geometric shapes involving two areas A 1 and A 2 . FIND: Shape factors, F 12 and F 21 , for each configuration. ASSUMPTIONS: Surfaces are diffuse. ANALYSIS: The analysis is not to make use of tables or charts. The approach involves use of the reciprocity relation, Eq. 13.3, and summation rule, Eq. 13.4. Recognize that reciprocity applies to two surfaces; summation applies to an enclosure. Certain shape factors will be identified by inspection. Note L is the length normal to page. (a) Long duct (L): By inspection, 12 F 1.0 = < By reciprocity, ( ) 1 21 12 2 A 2 RL 4 F F 1.0 0.424 A 3 / 4 2 RL 3 π π = = × = = < (b) Small sphere, A 1 , under concentric hemisphere, A 2 , where A 2 = 2A Summation rule 11 12 13 F F F 1 + + = But F 12 = F 13 by symmetry, hence F 12 = 0.50 < By reciprocity, 1 1 21 12 2 1 A A F F 0.5 0.25. A 2A = = × = < (c) Long duct (L): By inspection, 12 F 1.0 = < By reciprocity, 1 21 12 2 A 2RL 2 F F 1.0 0.637 A RL π π = = × = = < Summation rule, 22 21 F 1 F 1 0.64 0.363. = = = < (d) Long inclined plates (L): Summation rule, 11 12 13 F F F 1 + + = But F 12 = F 13 by symmetry, hence F 12 = 0.50 < By reciprocity, ( ) 1 21 12 1/ 2 2 A 20L F F 0.5 0.707. A 10 2 L = = × = < (e) Sphere lying on infinite plane Summation rule, F 11 + F 12 + F 13 = 1 But F 12 = F 13 by symmetry, hence F 12 = 0.5 < By reciprocity, 1 21 12 2 A F F 0 A = since 2 A . → ∞ < Continued …..

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PROBLEM 13.1 (Cont.) (f) Hemisphere over a disc of diameter D/2; find also F 22 and F 23 . By inspection, F 12 = 1.0 < Summation rule for surface A 3 is written as 31 32 33 F F F 1. + + = Hence, 32 F 1.0. = By reciprocity, 3 23 32 2 A F F A = ( ) 2 2 2 23 D/ 2 D D F / 1.0 0.375. 4 4 2 π π π = = By reciprocity, 2 2 1 21 12 2 A D D F F / 1.0 0.125. A 4 2 2 π π = = × = < Summation rule for A 2 , 21 22 23 F F F 1 or + + = 22 21 23 F 1 F F 1 0.125 0.375 0.5. = = = < Note that by inspection you can deduce F 22 = 0.5 (g) Long open channel (L): Summation rule for A 1 11 12 13 F F F 0 + + = but F 12 = F 13 by symmetry, hence F 12 = 0.50. < By reciprocity, ( ) 1 21 12 2 A 2 L 4 F F 0.50 0.637. A 2 1 / 4 L π π × = = = × = × COMMENTS: (1) Note that the summation rule is applied to an enclosure. To complete the enclosure, it was necessary in several cases to define a third surface which was shown by dashed lines. (2) Recognize that the solutions follow a systematic procedure; in many instances it is possible to deduce a shape factor by inspection.
PROBLEM 13.2 KNOWN: Geometry of semi-circular, rectangular and V grooves. FIND: (a) View factors of grooves with respect to surroundings, (b) View factor for sides of V groove, (c) View factor for sides of rectangular groove. SCHEMATIC: ASSUMPTIONS: (1) Diffuse surfaces, (2) Negligible end effects, “long grooves”. ANALYSIS: (a) Consider a unit length of each groove and represent the surroundings by a hypothetical surface (dashed line). Semi-Circular Groove: ( ) 2 21 12 21 1 A W F 1; F F 1 A W / 2 π = = = × 12 F 2/ . π = < Rectangular Groove: ( ) ( ) ( ) 4 4 1,2,3 1,2,3 4 4 1,2,3 1 2 3 A W F 1; F F 1 A A A H W H = = = × + + + + ( ) ( ) 1,2,3 4 F W / W 2H .

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