A series of full-period consecutive dust samples are collected during an 8 hour

Question

A series of full-period consecutive dust samples are collected during an 8 hour work shift. Samples were collected on filters and weighed gravimetrically, with results as follows:

Sample no . Flowrate( liters per min) Sample Duration (min) Mass Collected (micrograms ug) 1) 2.0 60 110 2) 2.0 120 745 3) 2.0 90 350 4) 2.0 120 700 5) 2.0 90 255 What is the 8 hour TWA concentration in milligrams per cubic meter (mg/m^3)? The final units are in mg/m^3. You have mass in ug. You also have minutes and liters per minute. That should tell you how many liters of air were sampled. Airborne concentration is in mass per volume of air. You can calculate ug/l. Convert to mg/m^3 and then calculate the 8 hour TWA (show all work please)

Thanks !

Explanation / Answer

flowrate( liters per min) Sample Duration (min) Mass Collected (micrograms ug)

1) 2.0 60 110

2) 2.0 120 745

3) 2.0 90 350

4) 2.0 120 700

5) 2.0 90 255

-------------------------------------

as we can see from the data that flowrate( liters per min) is consant , so there is no relation of flowrate with other two variables

.........................

so, lets find relation between sample duration and mass collected

so,

Sample Duration (min)(x) Mass Collected (micrograms ug) (y)

1) 60 110

2) 120 745

3) 90 350

4) 120 700

5) 90 255

so, regression equation will be ...(using calculator to find regression equation .... we can do it maually as well, but it's very lengthy and painful work... use calculators for such calculations)

so,

y=10.75x-600

here, y = Mass Collected (micrograms ug)

x = Sample Duration (min)

for

x = 8 hours = 8*60 min = 480 min

Mass Collected (micrograms ug) = y = 10.75*480-600 = 4560 ug = 4.56 mg

flowrate = 2.0 L/min = 0.002 m^3/min

----------------------------------------

so,

8 hour TWA concentration = mass/(volume)

= (mass collected) / (flowrate * sample duration)

= 4.56/(0.002*480) mg/m^3

= 4.75 mg/m^3

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