In an apartment the interior air temperature is 20°C and exterior air temperatures is 5°C. The wall has inner and outer surface temperatures of 16°C and 6°C, respectively. The inner and outer convection heat transfer coefficients are 5 and 20 W/m2. K, respectively. Calculate the heat flux from the interior air to the wall, from the wall to the exterior air, and from the wall to the interior air. Is the wall under steady-state conditions?

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.