41 - Radiation

Radiation exchange (\(Q\)) between two surfaces can be stated as \[Q = Q_{1\text{--}2} = \left(\begin{array}{l}\text{radiation energy}\\ \text{leaving $A_1$ that}\\ \text{strikes $A_2$} \end{array} \right) - \left(\begin{array}{l}\text{radiation energy}\\ \text{leaving $A_2$ that}\\ \text{strikes $A_1$} \end{array} \right)\]

Radiation exchange (\(Q\)) between two surfaces \(A_1\) (at \(T_1\)) and \(A_2\) (at \(T_2\)) with emissivities \(\varepsilon_1\) and \(\varepsilon_2\) respectively, is \[Q = \frac{\sigma T_1^4 - \sigma T_2^4}{\dfrac{1-\varepsilon_1}{A_1\varepsilon_1} + \dfrac{1}{A_1F_{12}}+\dfrac{1-\varepsilon_2}{A_2\varepsilon_2}} \qquad (\text{where $T_1>T_2$})\] For \(A_1/A_2\rightarrow 0\), i.e., \(A_1<<A_2\), the above equation reduces to \[Q = A_1\varepsilon_1\sigma(T_1^4-T_2^4)\]

• For transfer between two large parallel plates, with \(A_1=A_2=A\), the above relation reduces to (as \(F_{12}=1\)): \[Q = \frac{A\sigma(T_1^4 - T_2^4)}{1/\varepsilon_1 + 1/\varepsilon_2 - 1}\]

• For long concentric cylinders or concentric spheres, the heat transfer rate is given by (inner=1; outer=2): \[Q = \frac{A_1\sigma(T_1^4 - T_2^4)}{1/\varepsilon_1 + (A_1/A_2)(1/\varepsilon_2 - 1)}\]