Topic: Heat Transfer
Determine $F_{12}$ and $F_{21}$ for the following configuration having long inclined plates and point B is directly above the center of $A_1$
- A. $F_{12}=0.5, F_{21}=0.707$
- B. $F_{12}=0.707, F_{21}=0.5$
- C. $F_{12}=0.707, F_{21}=1$
- D. $F_{12}=0.5, F_{21}=1$
Answer: A. $F_{12}=0.5, F_{21}=0.707$
Explanation: Using Hottel's crossed-string method for adjacent surfaces: $F_{12} = \frac{L_1 + L_2 - L_3}{2L_1}$.
$A_1$ length ($L_1$) = 200 mm. B is directly above the center of $A_1$ at a height of 100 mm.
$L_2$ (length of $A_2$) = $\sqrt{100^2 + 100^2} = 100\sqrt{2}$ mm.
$L_3$ (imaginary string closing the triangle from left edge to B) = $\sqrt{100^2 + 100^2} = 100\sqrt{2}$ mm.
$F_{12} = \frac{200 + 100\sqrt{2} - 100\sqrt{2}}{2 \times 200} = \frac{200}{400} = 0.5$.
By reciprocity theorem: $A_1 F_{12} = A_2 F_{21}$.
$200 \times 0.5 = 100\sqrt{2} \times F_{21} \implies 100 = 100\sqrt{2} F_{21} \implies F_{21} = \frac{1}{\sqrt{2}} = 0.707$.