A radiation source is to be built, as shown in the diagram, for an experimental study of radiation. The base of the hemisphere is to be covered by a circular plate having a centered hole of radius R/2. The underside of the plate is to be held at 555 K by heaters embedded in its surface. The heater surface is back. The hemispherical surface is well-insulated on the outside. Assume gray diffuse processes and uniform distribution of radiation. (a) Find the ratio of the radiant intensity at the opening to the intensity of emission at the surface of the heated plate. (b) Find the radiant energy loss through the opening in watts for R = 0.3 m. (c) Find the temperature of the hemispherical surface.

Question

GIVEN

• A radiation source as shown above

• Radius of hole = R/2

• Temperature of underside of plate (T2) = 555 K

• Underside of plate is black

• Hemispherical surface is well insulated on the outside

FIND

(a) The ratio of the radiant intensity at the opening to the intensity at the surface of the heated plate (G1/ Eb2)

(b) The radiant energy loss through the opening (q1) in watts for R = 0.3 m

(c) The temperature of the hemispherical surface (T3)

ASSUMPTIONS

• Gray diffuse processes

• Uniform distribution of radiation

• Radiation entering A1 is negligible, i.e., A1 as a black body at 0 K Heat loss through insulation is negligible

PROPERTIES AND CONSTANTS

The Stephan-Boltzmann constant (?) = 5.67 × 10–8 W/(m2 K4)

Answer 1

The problem consists of radiative exchange between two black surfaces and a gray surface. It can be solved by simplifying which applies to gray surfaces



The radiosities, are



Let the opening the surface 1, the heater surface be surface 2, and the hemisphere be surface 3. Since A1 and A2 cannot see themselves or each other: F11 = F22 = F12 = F21 = 0 Since A1 and A2 are black ?1 = ?2 = 1 and ?1 = ?2 = 0 Neglecting radiation entering A1 Eb1 = 0 ? J1 = 0

In steady state, surface A3 has no net heat gain or loss q3 = 0.



Incorporating these simplifications into the above equations



(a) Combining



The shape factors must sum to unity: F31 + F32 + F33 = 1



From examination of the geometry, it is clear that F13 = 1 and F23 = 1





(b) The radiation energy loss at the opening is given by the irradiance of surface 1



(c)



Since J3 = G3, yields



But A3 is opaque, so ?3 = 0



Combining these two equations