Consider an enclosure formed by three surfaces having the following values of shape factors, emissivities and temperatures Surface 1 i.e. curved cylindrical has an emissivity 0.75 and temperature 800 K Surface 2 i.e. closing disc has an emissivity 0.8 and temperature 700 K Surface 3 i.e. closing disc has an emissivity 0.8 and temperature 700 K The closing flat discs are 25 mm in diameter and they have interspacing distance equal to 100 mm. If the shape factor between these two identical discs is 0.05, calculate the net rate of radiant heat flow from the curve surface to each of the closing end surface

F 21 + F 23 + 1 and A 1 = 0.00785 m2, A2 = 0.000491 m2. Then from reciprocity theorem, F 12 = 0.0594. Q12 = F 12 A 1 σ b (T 14 – T 2 4) = 3.561 W.