A Cylindrical furnance operating at a Temperature of 1200k and is emitting radiation uniformly in all Directions. The inside Diameter of the furnance is 2m and the length of the furnance is 5m. The walls of the furnance are at a temperature of 800k and are perfectly absorbing. -Calculate The Total Rate of Heat Transfer from the furnance walls. Solution Bo Heat transfer Through Radiation Rate of Heat transfer for the walls So of Surnance Q = Ain.. (T-T2") Q = Asu. σ . (T1^4 - T2^4) where Q: Rate of Heat transfer through Radiation. Asu: Surface Area of the furnace σ: Boltzmann constant = 5.67 x 10^-8 W / m^2 x K^4 T1: Tempreture of furnace wall T2: Tempreture inside furnace T1 = 800K° Furnace T2 = 1200K° L = 5m Dia = 2m Surface Area of Furnace Asu = (π.r^2) + (2.π.r.L) = π.(1m)^2 + (2π x 1m) = 3π = 9.42 m^2 Q = Asur x σ x (T1^4 - T2^4) = (9.42m²) × (5.67×10^-8 W/m²K^4) × (800K^4 - 1200K^4) = (53.4114 x 10^-8 W/K^4) x (4.096 x 10^9 - 20.736 x 10^9) = 53.4114 x 10^-8 x 10^9 x (-16.64 x 10^9) = 53.4114 x (-22.614) x 10^3 = -1207845.4 Q = -1,207,845.4 W = -1.207 MW Q = -1.207 MWa