Answer: -1543371.65837 W
= - 1543.372 kW.
Step-by-step explanation:
Using the equation;
Q= EσA -------------------------------------------------------------------------------------------(1).
Where Q= net rate of radiation heat transfer between the floor and the ceiling of the furnace, σ = Boltzmann's constant, A= area of the cube, E = emissitivity.
Recall that the emissitivity of a black body is equals to one(1).
From the question, the parameters given are; The view factor from the ceiling to the floor of the furnace,F12 = 0.2, σ = 5.67 × 10-8., A= (4.45×4.45) m.
Slotting in the parameters into the equation;
Q= EσA[T(2)^4 - (T(1)^4] ---------------------------------------------------------------------(2).
Therefore, Q= (1)× (5.67×10^-8) × (4.45×4.45) m × [(547)^4 - (1100)^4]
= 0.0000011228 × (89526025681 - 1.4641×10^12).
= 0.0000011228×(-1.374574×10^12)
= -1543371.65837 W
= -1543.372 kW.