A box has a hole which its diameter is 30 mm This hole emits

A box has a hole which its diameter is 30 mm. This hole emits diffusely, the total intensity associated with emission is 5000 (W/m^2.sr). The radiation emitted from the hole is intercepted by the surface area (which has the diameter of 50 mm). What is the rate at which radiation emitted by the hole is in by the other surface?

Solution

This is essentially a question of computing the View Factor:( see Incropera P 813)

The formula is   VFi= area integral 1/Ai[ cos(theta1)cos(theta2) dAi dAj/pi*R^2, where R is the distance between the surfaces

and similarly for j   where i and j refer to the two surfaces

Radiation intercepted by A2 from A1 is F12 = cos(theta1)cos(theta2)*A2 /R^2

From the geometry, theta1= 29.745 deg, theta2 = 60.255 deg

R^2=.65, A2 =pi (.05)^/4 m^2

R =.8062

F12= 178.426 *10^_6

(F12 proportional to A2 NOT A1)

if Emissivity of A1 is 5000 W/m^2

Since VF of 2 looking at 1 is the fraction of the surface leaving 1 intercepted by 2, b defn, the radiation absorbed is

5000*178.426*10^-6 =.892