Question

Calculate the fraction of Energy Radiated by Inner walls of Spherical Shell falling on itself

• A. 1
• B. $\frac{r^2}{R^2}$
• C. $1 - \frac{r^2}{R^2}$
• D. $1 - \frac{R^2}{r^2}$

Answer 1

Answer: (C)

$1 - \frac{r^2}{R^2}$

Answer 2

The fraction of energy radiated by the inner walls of a spherical shell that falls back onto itself is determined by the view factor $F_{11}$. For a system with two surfaces, 1 (inner shell) and 2 (inner sphere), the sum of view factors from surface 1 to all other surfaces must equal 1. Therefore, $F_{11} + F_{12} = 1$, where $F_{12}$ is the view factor from the inner shell to the inner sphere. For two concentric spheres, the view factor from the outer sphere (radius $R$) to the inner sphere (radius $r$) is given by $F_{outer \to inner} = \frac{A_{inner}}{A_{outer}} = \frac{r^2}{R^2}$. In this case, $F_{12} = \frac{r^2}{R^2}$. Thus, the fraction of energy falling back on the shell is $F_{11} = 1 - F_{12} = 1 - \frac{r^2}{R^2}$.