Question

Two concentric spheres of diameters d1 = 0.8 m and d2 = 1.2 m, have surface temperatures T1 = 450 K and T2 = 300 K 

respectively. If the surface emissivities are 0.5 and 0.05 respectively, determine the net radiation heat exchange between 

the two spheres.

Answer 1

net radiation heat exchange between the two spheres.

$\dot Q=\frac{\sigma A_1(T_1^{4}-T_2^{4})}{\frac{1-\epsilon_1 }{\epsilon_1}+\frac{1}{F_{12}}+\frac{A_1}{A_2}+\frac{1-\epsilon_2 }{\epsilon_2}}$

Here, $\sigma = 5.67\times10^{-8}\frac{W}{m^2K^4}$ and F₁₂=1

$\dot Q=\frac{5.67\times10^{-8} \times 8.04\times (450^{4}-300^{4})}{\frac{1-0.5 }{0.5}+\frac{1}{1}+\frac{8.04}{0.0314}+\frac{1-0.05 }{0.05}}=51.14$ W