Discussion
Consider two infinitely long thin concentric tubes of circular cross section as shown in the figure. If D1 and D2 are the diameters of the inner and outer tubes respectively, then the view factor F22 is given by
Answer: D
The view factor is \(F_{22} = 1-\frac{D_{1}}{D_{2}}\).
The inner tube is convex, so it cannot see itself, \(F_{11}=0\), and everything it emits reaches the outer tube, \(F_{12}=1\). Reciprocity, per unit length, gives:
\[ \begin{aligned} A_{1}F_{12} &= A_{2}F_{21} \\ F_{21} &= \frac{\pi D_{1}}{\pi D_{2}} = \frac{D_{1}}{D_{2}} \end{aligned} \]
The summation rule for surface 2 then gives \(F_{22} = 1 - F_{21} = 1 - D_{1}/D_{2}\).
\(D_{1}/D_{2}\) is F21, not F22. Zero would be true only for a convex surface, but the inside of the outer tube is concave and sees part of itself. \(D_{2}/D_{1}-1\) is greater than 1 for a large outer tube, which no view factor can be.
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