Question

What is the heat loss coefficient that has a symbol Uair and is found from (volumetric flow * density * specific heat capacity) of a building with a volume of 19456 cubic feet, if there is a natural air change per hour of 0.4

Answer 1

Uair = 0.0749 KW/k = 74.9 W/k

Step-by-step explanation:

The natural air change per hour is given by the formula:

Natural Air Change per Hour = ACPH = 60*Volume Flow/Volume

where,

ACPH = 0.4

Volume Flow = ? in ft³/min

Volume = 19456 ft³

Therefore,

0.4 = (60 min)(Volume Flow)/(19456 ft³)

Volume Flow = (0.4)(19456 ft³)/(60 min) = (129.7 ft³/min)(1 min/60 s)

Volume Flow = (2.16 ft³/s)(0.3048 m/1 ft)³ = 0.061 m³/s

Now, we find heat loss coefficient:

Uair = Volumetric Flow*Density of air*Specific Heat Capacity of air

Uair = (0.061 m³/s)(1.225 kg/m³)(1 KJ/kg.k)

Uair = 0.0749 KW/k = 74.9 W/k