Answer:
20 W/m², 20 W/m², -20 W/m²
Yes, the wall is under steady-state conditions.
Explanation:
Air temperature in room = 20°C
Air temperature outside = 5°C
Wall inner temperature = 16°C
Wall outer temperature = 6°C
Inner heat transfer coefficient = 5 W/m²K
Outer heat transfer coefficient = 20 W/m²K
Heat flux = Concerned heat transfer coefficient × (Difference of the temperatures of the concerned bodies)
q = hΔT
Heat flux from the interior air to the wall = heat transfer coefficient of interior air × (Temperature difference between interior air and exterior wall)
⇒ Heat flux from the interior air to the wall = 5 (20-6) = 20 W/m²
Heat flux from the wall to the exterior air = heat transfer coefficient of exterior air × (Temperature difference between wall and exterior air)
⇒Heat flux from the wall to the exterior air = 20 (6-5) = 20 W/m²
Heat flux from the wall to the interior air = heat transfer coefficient of interior air × (Temperature difference between wall and interior air)
⇒Heat flux from the wall and interior air = 5 (16-20) = -20 W/m²
Here the magnitude of the heat flux are same so the wall is under steady-state conditions.
