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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.

Respuesta :

Answer:

Time period  = 41654.08 s

Explanation:

Given data:

Internal volume is 210 m^3

Rate of air infiltration  [tex]9.4 \times 10^{-5} kg/s[/tex]

length of cracks 62 m

air density = 1.186 kg/m^3

Total rate of air infiltration [tex]= 9.4\times 10^{-5} \times 62 = 582.8\times 10{-5} kg/s[/tex]

total volume of air  infiltration[tex] = \frac{582.8\times 10{-5}}{1.156} = 5.04\times 10^{-3} m^3/s[/tex]

Time period [tex]= \frac{210}{5.04\times 10^{-3}} = 41654.08 s[/tex]

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