A 5m × 5m × 3m room has four 230 mm thick external brick walls. Total wall fenestration is 10 sqm

Q. A 5m × 5m × 3m room has four 230 mm thick external brick walls. Total wall fenestration is 10 sqm. The temperature difference between indoor and outdoor is 2 degC. The air to air transmittance values for 230 mm thick brick wall and 200 mm thick aerated concrete block wall are 2.4 and 1.7 W/sqm degC respectively. If the brick walls are replaced with the aerated concrete block walls, then the change in conductive heat flow through the walls is___________W.

Ans: 69.5 to 70.5

Sol:

Here, Area of 4 walls of the room = 4 × (5 × 3) = 60 m², Height = 3m.
Total area of wall fenestration (Openings) = 10 m².
Thus, net external wall area= 50 m²
Thermal transmittance value of 230 mm thick brick wall (U-value of wall) = 2.4 W/sq.m. deg C.
Thermal transmittance value of 200 mm thick aerated concrete block wall (U-value of aerated concrete wall) 1.7 W/sq.m. deg C
Difference between out door and indoor temperature (At) = 2°
Conductive heat flow through brick wall
= (U-value of brick wall × area of wall × Δt) = (2.4 × 50 × 2) = 240 watts. Conductive heat flow through aerated concrete block wall = (U-value of ACC wall x area of wall × Δt) = (1.7 × 50 × 2) = 170 watts.
Thus, change in conductive heat flow through the wall due to change of materials = 240 – 170 = 70 Watts.