Understanding Black Body Radiation Exchange

This problem involves calculating the fraction of radiative energy received between two black bodies using the concept of view factors and the reciprocity theorem.

Given Information

• Surface 1 Area: $A_1 = 1 \, m^2$
• Surface 2 Area: $A_2 = 4 \, m^2$
• Fraction of energy emitted by $S_1$ received by $S_2$ = 40%, which equals 0.4. This fraction represents the view factor $F_{12}$. Therefore, $F_{12} = 0.4$.

Applying View Factor Reciprocity

For radiation heat transfer between surfaces, the view factor reciprocity theorem is essential. It states that the product of the area of one surface and the view factor from that surface to another is equal to the product of the second surface's area and the view factor from the second surface to the first.

The formula is:

$A_1 F_{12} = A_2 F_{21}$

Where:

• $A_1$ and $A_2$ are the surface areas of the bodies.
• $F_{12}$ is the fraction of energy leaving surface $S_1$ that is incident upon surface $S_2$.
• $F_{21}$ is the fraction of energy leaving surface $S_2$ that is incident upon surface $S_1$.

Calculating the Fraction Received by $S_1$

We need to determine $F_{21}$, the fraction of energy emitted by $S_2$ that is received by $S_1$. We have the values:

• $A_1 = 1 \, m^2$
• $A_2 = 4 \, m^2$
• $F_{12} = 0.4$

Substitute these values into the reciprocity equation:

$ (1 \, m^2) \times 0.4 = (4 \, m^2) \times F_{21} $

Simplify the equation:

$ 0.4 = 4 \times F_{21} $

Now, solve for $F_{21}$:

$ F_{21} = \frac{0.4}{4} $

$ F_{21} = 0.1 $

Conclusion

The fraction of energy emitted by $S_2$ that is received by $S_1$ is $0.1$.