Subtopic: Steady state heat transfer in conduction, convection and radiation
Question Statement:
Two small parallel plane square surfaces, each 4 mm × 4 mm, are placed 0.5 m apart (centre to centre), with 30° angle between the radial distance and both surface normals. The view factor between the two surfaces is:
Options:
(A) 1.53×10⁻⁵
(B) 1.76×10⁻⁵
(C) 3.82×10⁻³
(D) 4.41×10⁻³
Official Answer Key:
A
Step-by-Step Solution & Explanation
For two small differential/near-differential parallel coplanar-normal surfaces separated by distance r with both normals at angle θ to the line joining their centres: F1-2 = (A2 cos θ1 cos θ2)/(π r²). A2 = (0.004)² = 1.6×10⁻⁵ m², cosθ1=cosθ2=cos30°=0.866, r=0.5 m, r²=0.25. F1-2 = 1.6×10⁻⁵×0.866×0.866/(π×0.25) = 1.6×10⁻⁵×0.75/0.7854 = 1.2×10⁻⁵/0.7854 = 1.528×10⁻⁵. Answer: A.
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