Determine F12 and F21 for the following configurations

Step 1 of 5

The area of sphere is \(A_{1}\).

The area of concentric hemisphere is \(A_{2}=2 A_{1}\).

(a)

The schematic diagram of long duct is as follows:

In order to determine the shape factor, we have to:

Apply the reciprocity theorem.

\(F_{21}=\frac{A_{1}}{A_{2}} F_{12}\)

For \(A_{1}=2 R L, A_{2}=\frac{3}{4} 2 \pi R L\) and by inspection \(F_{12}=1\).

Therefore,

\(\begin{array}{l}
F_{21}=\left(\frac{2 R L}{\frac{3}{4} 2 \pi R L}\right)(1) \\
F_{21}=0.424
\end{array}\)

Shape factor is \(F_{21}=0.424\) and \(F_{12}=1\).

(b)

The schematic diagram of small sphere under concentric hemisphere is as follows:

In order to determine the shape factor, we have to:

Apply the summation rule.

\(F_{11}+F_{12}+F_{13}=1\)

For \(F_{12}=F_{13}\) by symmetry.

Therefore,

\(\begin{array}{l}
F_{12}=\frac{1}{2} \\
F_{12}=0.5
\end{array}\)

The expression from reciprocity theorem can be given as:

\(F_{21}=\frac{A_{1}}{A_{2}} F_{12}\)

Substitute the given values \(A_{2}=2 A_{1}\) and \(F_{12}=0.5\) in the above expresion.

\(\begin{array}{l}
F_{21}=\frac{A_{1}}{2 A_{1}}(0.5) \\
F_{21}=0.25
\end{array}\)

Shape factor is \(F_{21}=0.25\) and \(F_{12}=0.5\).