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GATE 2023 ME – Question 43

Heat Transfer · Radiative heat transfer: laws, view factors, radiation network · 2 marks · Multiple choice

A cylindrical rod of length $h$ and diameter $d$ is placed inside a cubic enclosure of side length $L$. $S$ denotes the inner surface of the cube. The view-factor $F_{S-S}$ is

  1. 0
  2. 1
  3. $\frac{(\pi dh + \pi d^2/2)}{6L^2}$
  4. $1 - \frac{(\pi dh + \pi d^2/2)}{6L^2}$

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Correct answer: (D) $1 - \frac{(\pi dh + \pi d^2/2)}{6L^2}$

Explanation

The rod is convex, so everything leaving its surface goes to the enclosure: $F_{rod-S} = 1$. By reciprocity, $A_SF_{S-rod} = A_{rod}F_{rod-S}$, so $F_{S-rod} = \frac{A_{rod}}{A_S} = \frac{\pi dh + \pi d^2/2}{6L^2}$, with the area of the rod as given in the options and the inner area of the cube as $6L^2$. The remaining radiation from S falls on S itself, so $F_{S-S} = 1 - F_{S-rod} = 1 - \frac{\pi dh + \pi d^2/2}{6L^2}$.