The following data was obtained at an asbestos site. Using the information given

Question

The following data was obtained at an asbestos site. Using the information given in the table below, determine whether the contractor can be released or held accountable for the amount of asbestos released at the work site. (Assume 95 percentile) The efficiency of a filter used in the duct supplying 100% outdoor air to a building is 85%. The average outdoor generation of particulates is 10^4 mug/sec and the volumetric flow of outdoor is 1060 ft^3/sec. Calculate the approximate total indoor concentration of particulates if the indoor concentration of particulates (due to indoor sources) is 10 mug/m^3 0.00006 g/m^3 0.0045 g/m^3 285 mug/m^3 335 mug/cm^3

Explanation / Answer

Volume of outdoor air=1060 ft3/sec

average generation of outdoor particulates=10^4 ug/sec

Total concentration of outdoor particulates= average generation of outdoor particulates/volume of outdoor air

=10^4 ug sec-1/1060 ft3s-1

=9.43ug/ft3

To convert ft3 into m3

We know that 1 ft3=0.028 m3

9.43ug/ft3=9.43/0.028=336.7ug/m3

Since 100% of outdoor air in building is converted 85% indoor air

therefore 85% of 336=285ug/m3

285ug/m3 is the total concentration of indoor particulates including the 10ugm3 particulates from indoor sources

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