In a home, air infiltrates from the outside through cracks around doors and windows. Consider a residence where the total length of cracks is 62 m and the total internal volume is 210 m3 . Due to the wind, 9.4 x 10-5 kg/s of air enters per meter of crack and exits up a chimney. Assume air temperature is the same inside and out and air density is constant at 1.186 kg/m3 . If windows and doors are not opened or closed, estimate the time required for one complete air change in the building.


Answer 1


Time period  = 41654.08 s


Given data:

Internal volume is 210 m^3

Rate of air infiltration  9.4 * 10^(-5) kg/s

length of cracks 62 m

air density = 1.186 kg/m^3

Total rate of air infiltration = 9.4* 10^(-5) * 62 = 582.8* 10{-5} kg/s

total volume of air  infiltration= \frac{582.8* 10{-5}}{1.156} = 5.04* 10^(-3) m^3/s

Time period = (210)/(5.04* 10^(-3)) = 41654.08 s

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