Question

Consider a 4.45-m × 4.45-m × 4.45-m cubical furnace whose floor and ceiling are black and whose side surfaces are reradiating. The floor and the ceiling of the furnace are maintained at temperatures of 547 K and 1100 K, respectively. Determine the net rate of radiation heat transfer between the floor and the ceiling of the furnace. (Given: The view factor from the ceiling to the floor of the furnace F12 = 0.2, σ = 5.67 × 10-8.)

Answer 1

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.