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(theoretical) conversion of asbestos fibre limit to mg/m3
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Morning all, if you would please indulge me, I know why this is not practical nor realistic, but purely for academic purposes... Based on a limit level of 0.1fibres per cm3 Assuming each fibre is 5um long by 0.2um diameter And an density of 2.5g/cm3 the conversion of 0.1 fibres per cm3 would equate to 0.03925mg/m3
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Rank: Forum user
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Not sure you can correctly assume that all fibres will be uniform in size 5um x 0.2um!
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Rank: Super forum user
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thats why its purely a theorhetical and academic process, im just wondering if my math is correct
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Rank: Super forum user
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Purely theorhetically, yes your maths is correct, although with so many decimal places in the equation I may have missed one at some point.
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Rank: Super forum user
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Thanks cap, I do sometimes like these kind of impractical logic puzzles, like if someone was stuck in a lift how long it would take through normal respiration to accumulate sufficient concentrations of CO2 to create an IDLH atmosphere, there are not useful but keep my mind active :)
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Rank: Super forum user
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Actually you might be out once I tried a simpler equation Calculation is (5*0.1*3.14)/100,000,000 = total volume of 0.0000000157 cm³ for each strand 0.0000000157*2.5 = 0.00000003925g per strand
0.1 * 1,000,000 = 100,000 strands per m³
100,000 * 0.00000003925 = 0.003925g per m³ or 3.925mg I think I forgot the "0.1" per cm³ last time
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Rank: Super forum user
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Captain - go to the top of the class.......I think. Security image reflects text speak for the weather. SUNi
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Rank: Super forum user
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Indeed
0.1 fibres per cm3 = 100,000 fibres per m3
1fibre = 0.0157 um3 volume
100,000 fibres = 1570um3 or 1.570 mm3 or 0.00157 cm3
1cm3 = 2.5g
0.00157cm3 = 0.003925g (per m3)
= 3.925 mg/m3
I missed the conversion from g to mg at the end i think.
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Rank: Super forum user
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descarte - there is the other problem. Usually the fibres will be mixed in with something else, some of which is going to be drawn into an air sample. So do you [theoretically] want to calculate the mg/m3 for the asbestos alone or for the mix?
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Rank: Forum user
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Originally Posted by: CptBeaky  Actually you might be out once I trieda simpler equationCalculation is (5*0.1*3.14)/100,000,000 = total volume of0.0000000157 cm³ for each strand0.0000000157*2.5 =0.00000003925g per strand
0.1 * 1,000,000 = 100,000 strands per m³
100,000 *0.00000003925 =0.003925g per m³ or 3.925mgI think I forgot the "0.1" per cm³ last time
Sorry to come in late...but I am doing a similar cal.
Isn't the volume of a cylinder pi (r)^2 × h.
Therefore Fibre volume = 1.57x10(-9) m^3 when working in microns.
And thus the m3 to ml conversion is to x by 1000000. Thus, i get a vastly different strand volume of 1.57x10^(-13) ml.
This greatly impacts the calculation. In fact I get for 100000 fibres in 1 m3 is equal to only 3.92x10^(-5)mg
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Rank: Forum user
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Hi everyone. Would appreciate a check of my calls if anyone is interested. Purely foe theoretical purposes only as I think the earlier calls done may be very inaccurate.
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Rank: Super forum user 
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The formula for volume of a cylinder is correct but I don't think the radius is squared correctly there. How about:
For a cylindrical fibre of diameter 0.2 micron (so radius 0.1 micron) and length 5 micron: Volume = (0.1 E-6) x (0.1 E-6) x (5 E-6) x pi cubic metres = (0.01 E-12) x (5 E-6) x pi cubic metres = (5 E-20) x pi cubic metres = (1.57 E-19) cubic metres I had thought there was a superscript format here but I don't find it when I need it ...
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Rank: Super forum user
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Looking again I wonder if you just made a typo and meant to put -19 not -9 in the first place? Because when you proceed to multiply as you say by E6 that is what gives you E-13 as you end up with.
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 1 user thanked Kate for this useful post.
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Rank: Forum user
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Hi Kate. Really appreciate the feedback and check.
I do believe the volume of a cylinder is correct. Simply put it the area of a circle multiplied by the height so pi r^2×h.
But u did pick up a typo (thankyou)...but it was inconsequential though. In my comment above, it should have been "Therefore Fibre volume = 1.57x10^(-9)"
Many thanks
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Rank: Super forum user
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I don't see how you get 1.57 x 10^(-9) cubic metres I make it 1.57 x 10^(-19) cubic metres
Hence multiplying by 10^(-6) you get 1.57 x 10^(-13) ml as you say So teh typo I am suggesting is that the -9 should be -19
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Rank: Forum user
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Hi Kate. My apologies, that should have been a -19 and is a typo. But at the end...it seems as though u agree with the ml volume.
Thanks agian
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1 user thanked bill4000 for this useful post.
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Its not Friday afternoon is it!!!!
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Rank: Forum user
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Originally Posted by: bill4000 Hi Kate. My apologies, that should have been a -19 and is a typo. But at the end...it seems as though u agree with the ml volume.
Thanks agian
Also...in case the OP was interested. For 3.92mg of pure asbestos (assumeing fibre dimensions as per above) distributed in 1m3 volume...equates to 10000fb/ml. Note that 1x10^(10) fibres are in 3.92 mg.
Incredible right
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(theoretical) conversion of asbestos fibre limit to mg/m3
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