A treated aluminum plate is irradiated with G-1000 Wm2 while at a temperature of 400 K The plate is treated with a surface coating that provides an absorptivity of 0.14 and an emissivity of 0.76. Find:

(a) The radiosity, J, of the plate surface
(b) The net radiation heat flux at the top surface

(a) The radiosity, J, of the plate surface is 1343.2 W/m²

(b) The net radiation heat flux at the top surface is 1863.2 W/m²

Explanation:

J = εσT⁴ + (1 - ε)G

where;

J is the  radiosity of the plate surface

ε is the emissivity  

σ is the stefan-Boltzmann constant

T is the temperature of the surface

G is the irradiance of the surface

J = 0.76 x 5.67 x 10⁻⁸ x (400)⁴ + (1 - 0.76) x 1000

J = 1343.2 W/m²

Part (b) net radiation heat flux at the top surface

this is equivalent to sum of reflected and emitted heat.

Rnet = G - (Rσ + Rt) + Rε

Rnet = G + Rε - (Rσ + Rt)

Rnet = G + εσT⁴ -  (1 - ε)G

Rnet = 1000 + 1103.2 - 240

Rnet = 1863.2 W/m²

Therefore, the net radiation heat flux at the top surface is 1863.2 W/m²

answer: chicken nuggets.

explanation: