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descarte8  
#1 Posted : 21 July 2021 09:50:45(UTC)
Rank: Super forum user
descarte8

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

Evans38004  
#2 Posted : 21 July 2021 09:58:33(UTC)
Rank: Forum user
Evans38004

Not sure you can correctly assume that all fibres will be uniform in size 5um x 0.2um! 

descarte8  
#3 Posted : 21 July 2021 10:00:09(UTC)
Rank: Super forum user
descarte8

thats why its purely a theorhetical and academic process, im just wondering if my math is correct

CptBeaky  
#4 Posted : 21 July 2021 11:09:26(UTC)
Rank: Super forum user
CptBeaky

Purely theorhetically, yes your maths is correct, although with so many decimal places in the equation I may have missed one at some point.

descarte8  
#5 Posted : 21 July 2021 11:24:08(UTC)
Rank: Super forum user
descarte8

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 :)

​​​​​​​
CptBeaky  
#6 Posted : 21 July 2021 11:58:47(UTC)
Rank: Super forum user
CptBeaky

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

peter gotch  
#7 Posted : 21 July 2021 12:11:26(UTC)
Rank: Super forum user
peter gotch

Captain - go to the top of the class.......I think.

Security image reflects text speak for the weather. SUNi

descarte8  
#8 Posted : 21 July 2021 12:38:02(UTC)
Rank: Super forum user
descarte8

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.

peter gotch  
#9 Posted : 22 July 2021 10:05:25(UTC)
Rank: Super forum user
peter gotch

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?

bill4000  
#10 Posted : 28 April 2022 21:46:47(UTC)
Rank: Forum user
bill4000

Originally Posted by: CptBeaky Go to Quoted Post
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
bill4000  
#11 Posted : 01 May 2022 23:33:48(UTC)
Rank: Forum user
bill4000

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.
Kate  
#12 Posted : 02 May 2022 15:58:30(UTC)
Rank: Super forum user
Kate

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 ...

Kate  
#13 Posted : 02 May 2022 16:06:07(UTC)
Rank: Super forum user
Kate

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.

thanks 1 user thanked Kate for this useful post.
bill4000 on 03/05/2022(UTC)
bill4000  
#14 Posted : 03 May 2022 08:03:36(UTC)
Rank: Forum user
bill4000

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
Kate  
#15 Posted : 03 May 2022 09:59:46(UTC)
Rank: Super forum user
Kate

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

bill4000  
#16 Posted : 04 May 2022 05:25:12(UTC)
Rank: Forum user
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
thanks 1 user thanked bill4000 for this useful post.
Kate on 04/05/2022(UTC)
Gerry Knowles  
#17 Posted : 04 May 2022 07:16:47(UTC)
Rank: Forum user
Gerry Knowles

Its not Friday afternoon is it!!!!

bill4000  
#18 Posted : 04 May 2022 07:28:02(UTC)
Rank: Forum user
bill4000

Originally Posted by: bill4000 Go to Quoted Post
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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