Radiation Exchange Between Surfaces Questions and Answers

Answer: A

The view factor is \(1.53 \times 10^{-5}\).

For two small surfaces far apart compared with their size, the view factor reduces to the differential-area form:

\[ \begin{aligned} F_{12} &= \frac{A_{2}\cos\theta_{1}\cos\theta_{2}}{\pi r^{2}} \\ &= \frac{(16 \times 10^{-6})(0.866)(0.866)}{\pi (0.5)^{2}} \\ &= \frac{1.2 \times 10^{-5}}{0.7854} \\ &= 1.53 \times 10^{-5} \end{aligned} \]

Leaving out the cosines gives \(2.04 \times 10^{-5}\), and keeping only one cosine factor gives the distractor \(1.76 \times 10^{-5}\). The values of order \(10^{-3}\) are far too large for two 4 mm squares half a metre apart, which can intercept only a tiny fraction of each other's radiation.

Answer: B

The view factor from the large cylinder to either annular end is 0.29.

Call the small cylinder 1 and the large cylinder's inner surface 2. The given 0.34 is \(F_{12}\), from the small cylinder to the large one. Reciprocity converts it, using the area ratio \(A_{1}/A_{2} = r_{1}/r_{2} = 5/10\) for equal lengths:

\[ \begin{aligned} F_{21} &= \frac{A_{1}}{A_{2}}F_{12} = 0.5 \times 0.34 = 0.17 \\ F_{21} + F_{22} + 2F_{2,\text{end}} &= 1 \\ 2F_{2,\text{end}} &= 1 - 0.17 - 0.25 = 0.58 \\ F_{2,\text{end}} &= 0.29 \end{aligned} \]

The summation rule over the enclosure, with two identical annular ends, gives the answer. 0.17 is \(F_{21}\) itself, and using 0.34 directly without reciprocity gives \((1 - 0.34 - 0.25)/2 \approx 0.21\), not offered. 0.33 and 0.42 match no consistent working.

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.

Answer: D

The view factor is \(1 - \dfrac{(\pi dh + \pi d^{2}/2)}{6L^{2}}\).

Call the rod's whole surface r. Its area is the curved surface plus two end discs, and the cube's inner area is six faces:

\[ \begin{aligned} A_{r} &= \pi dh + 2 \times \frac{\pi d^{2}}{4} \\ &= \pi dh + \frac{\pi d^{2}}{2} \\ A_{S} &= 6L^{2} \end{aligned} \]

The rod is convex and fully enclosed, so everything leaving it reaches the cube: \(F_{r-S} = 1\). Reciprocity gives \(A_{S}F_{S-r} = A_{r}F_{r-S}\), so \(F_{S-r} = A_{r}/(6L^{2})\). By the summation rule, \(F_{S-S} + F_{S-r} = 1\), so

\[ F_{S-S} = 1 - \frac{\pi dh + \pi d^{2}/2}{6L^{2}} \]

0 would hold for a flat or convex surface, but the inside of a cube sees itself. 1 would be correct only with no rod inside. \(\dfrac{(\pi dh + \pi d^{2}/2)}{6L^{2}}\) is \(F_{S-r}\), the fraction that reaches the rod, not the fraction that returns to the cube.

Answer: A

The answer is \(\pi/6\).

All radiation leaving the sphere (surface 1) reaches the box (surface 2), so \(F_{12} = 1\). The non-trivial shape factor is the one from the box to the sphere, found by reciprocity, \(A_{1}F_{12} = A_{2}F_{21}\):

\[ \begin{aligned} F_{21} &= \frac{A_{1}}{A_{2}} = \frac{\pi d^{2}}{6 l^{2}} \\ &= \frac{\pi}{6}\quad (l = d) \end{aligned} \]

Since \(F_{12} = 1\) is not offered, the key reads the question as asking for this factor. \(\pi/3\) would follow from taking only three faces of the cube, and \(4\pi/3\) and \(3\pi/2\) exceed one, which no shape factor can.

Answer: D

The view factor is \(F_{22} = 1 - \left(\frac{D_{1}}{D_{2}}\right)\).

All radiation leaving the inner tube reaches the outer tube, so \(F_{12} = 1\). By reciprocity, \(A_{2}F_{21} = A_{1}F_{12}\), and per unit length the areas are in the ratio of the diameters, so \(F_{21} = D_{1}/D_{2}\). Since the outer tube's inside surface is concave, it also sees itself, and the factors from surface 2 must add up to one:

\[ F_{22} = 1 - F_{21} = 1 - \frac{D_{1}}{D_{2}} \]

\(D_{1}/D_{2}\) alone is \(F_{21}\), not \(F_{22}\). Zero would be right only for a flat or convex surface, and \(D_{2}/D_{1} - 1\) can exceed one.