Topic: Heat Transfer
For infinite parallel planes with emissivities $\epsilon_{1}$ and $\epsilon_{2}$, the interchange factor for radiation from surface 1 to surface 2 is given by-
- A. $\frac{\epsilon_{1}\epsilon_{2}}{\epsilon_{1}+\epsilon_{2}-\epsilon_{1}\epsilon_{2}}$
- B. $\frac{1}{\epsilon_{1}}+\frac{1}{\epsilon_{2}}$
- C. $\epsilon_{1}+\epsilon_{2}$
- D. $\epsilon_{1}\epsilon_{2}$
Answer: A. $\frac{\epsilon_{1}\epsilon_{2}}{\epsilon_{1}+\epsilon_{2}-\epsilon_{1}\epsilon_{2}}$
Explanation: For infinite parallel planes, the interchange factor for radiation is $f_{12} = \frac{1}{\frac{1}{\epsilon_{1}} + \frac{1}{\epsilon_{2}} - 1} = \frac{1}{\frac{\epsilon_{2}+\epsilon_{1}-\epsilon_{1}\epsilon_{2}}{\epsilon_{1}\epsilon_{2}}} = \frac{\epsilon_{1}\epsilon_{2}}{\epsilon_{2}+\epsilon_{1}-\epsilon_{1}\epsilon_{2}}$.