Weight of air

[Gordon]

thought this might be interesting/useful for some of you.

Have worked out how much lighter cylinders are when theyre nearly empty.

from 232 - 50 bar in a:

15 = 3.3kg
12 = 2.7
10 = 2.2

24 (2x12) = 5.2
20 (2x10) = 4.4

from 300 - 50 bar

15 = 4.5
12 = 3.5
10 = 2.7

thought it might be useful because I struggled on ascents for a while until someone suggested adding a bit more weight to allow for the air Id breathed.

Edit: All sorted these figures are near enough right now - cheers Nigel

[Paul Watts]

It's worth remembering that a weight check should be done with <50bar

[Ben Thompson]

Hi Gordon,

Out of curiosity, did you have a source for air weight? I (and I haven't looked this up yet) always though air was a bit over a gram a litre (I think about 1.3 or something like that) which is about double what you're using. Am I just remembering it wrong?

[Woz]

Could have saved you all that maths:
http://www.subaqua.co.uk/cgi-bin/cylinder-buoyancy.cgi

[Nigel Hewitt]

15 = 1.65kg
12 = 1.35
10 = 1.1
232-50 bar?
Well the total weight of air at 232 bar in each is 4.1, 3.3 and 2.7Kgs respectivly so may I suggest a maths error somewhere.

May I also agree with another reply to say that a weight check should be done at 'almost empty' as it's no good carrying a 50bar reserve if you can't stay down and breath it but bob up to the surface.

[Tony Dwyer]

Hi Gordon,

Out of curiosity, did you have a source for air weight? I (and I haven't looked this up yet) always though air was a bit over a gram a litre (I think about 1.3 or something like that) which is about double what you're using. Am I just remembering it wrong?

have a look here:

http://www.cashflo.co.uk/Air.html

[Tony Dwyer]

232-50 bar?
Well the total weight of air at 232 bar in each is 4.1, 3.3 and 2.7Kgs respectivly so may I suggest a maths error somewhere.

May I also agree with another reply to say that a weight check should be done at 'almost empty' as it's no good carrying a 50bar reserve if you can't stay down and breath it but bob up to the surface.

I agree that there appears to be some error in the numbers.
See: http://www.cashflo.co.uk/Air.html for the source value of the weight of air.

[quote:The weight of dry air (no moisture content) at 0 deg C and under a normal atmospheric pressure of 1013 mbar is 1.293 Kg/cu metre. :endquote]

with air @ 0.001293 kg per ltr

15 x 232 = 3480 ltrs - 15 x 50 = 750

3480 x 0.001293 = 4.49964 kgs (round it to 4.5)
750 x 0.001293 = 0.96975 (round it to 0.97)

difference is 3.53 kgs

12 x 232 = 2784 ltrs - 12 x 50 = 600

2784 x 0.001293 = 3.599712 kgs (round it to 3.6)
600 x 0.001293 = 0.7758 (round it to 0.78)

difference is 2.82 kgs

10 x 232 = 2320 ltrs

2320 x 0.001293 = 2.99976 kgs (round it to 3)
500 x 0.001293 = 0.6465 (round it to 0.65)

difference is 2.35 kgs

[PeteM]

The weight of dry air (no moisture content) at 0 deg C and under a normal atmospheric pressure of 1013 mbar is 1.293 Kg/cu metre.



Trouble is we don't tend to dive in water 0 degrees as it tends to be a bit stiff :)

A better figure to take would be the 10 degree figure of 1.247 Kg/cu m or maybe the 20 degree figure of 1.205, which is almost the same as the figures as those from Nigel were based.

[Nick Kay]

I've always used 1.2g/Litre

I note the comments that weight-checks should be sone with <50B...

However, knowing the weight of air, size of cylinders and gauge pressure means you can calculate your "correct" weighting with any gauge pressure, i.e. you don't have to empty your cylinders down to 50B to do a weight check

[Tony Dwyer]

Trouble is we don't tend to dive in water 0 degrees as it tends to be a bit stiff :)

A better figure to take would be the 10 degree figure of 1.247 Kg/cu m or maybe the 20 degree figure of 1.205, which is almost the same as the figures as those from Nigel were based.

That would account for why I'm so tired lately. The stiff stuff is hard to fin through. Unfortunately I have frequently found myself diving in sub 10 degree water. Not too often at zero degrees, I have to admit. :)

My intent was to demo the math, not take issue with Nigel's numbers.

[Nigel Hewitt]

A better figure to take would be the 10 degree figure of 1.247 Kg/cu m or maybe the 20 degree figure of 1.205, which is almost the same as the figures as those from Nigel were based.The problem is that the deviations from the 'Ideal' Gas Law give us even more errors at this sort of pressure.
http://www.nigelhewitt.co.uk/diving/maths/vdw3.jpg
The purple is idea law pressure and the blue is real world for a set weight of gas in a fixed volume.
My figures were not based on density, that is a rather nebulous concept for gases under pressure, but the Van der Waals constants adjusted for air.

Sorry about this. I'm a physicist who now seems to have become a computer programmer so a bit of real maths is a treat.

[Ben Thompson]

Sorry about this. I'm a physicist who now seems to have become a computer programmer so a bit of real maths is a treat.

Well.. you're a naughty physicist, as there aren't any labels on the graph! Which bit's which on it? :p

[Tony Dwyer]

Ah well! If you're gonna get picky! :)

I don't introduce newbies to Van der Waals constants. They (am me) get confused enough as it is.

The numbers aren't that far apart and perhaps it's better to err a little on the heavy side.

I'm adhering to the KISS principle here.

[Paul Renucci]

:) Nice one to Tony Dwyer - if we are to believe the link I came across on yesterdays forum about, in my words, rebreathers and cylinders becoming obsolete, then no 'buddy' may have to worry about the weight of air nor that of our cylinders :eek: see http://www.likeafish.biz/ :D

I still can't believe this invention buds :)

[Gordon]

Sorry guys, seem to have made a mistake but cant quite suss out where.

think you guys were right bout the weight of air being twice what I said but cant figure why Ive got it wrong.

[Nigel Hewitt]

Well.. you're a naughty physicist, as there aren't any labels on the graph! Which bit's which on it? :pTo those that care some things are obvious.:rolleyes:

Pressure up the left in bar.

[Andy Wade]

The problem is that the deviations from the 'Ideal' Gas Law give us even more errors at this sort of pressure.
http://www.nigelhewitt.co.uk/diving/maths/vdw3.jpg
The purple is idea law pressure and the blue is real world for a set weight of gas in a fixed volume.
My figures were not based on density, that is a rather nebulous concept for gases under pressure, but the Van der Waals constants adjusted for air.

Sorry about this. I'm a physicist who now seems to have become a computer programmer so a bit of real maths is a treat.

No need to apologise Nigel, I'm a thicky who likes to read this stuff and rather winsomely imagine that I might one day understand a small part of it. :D

[PeteM]

Sorry guys, seem to have made a mistake but cant quite suss out where.

think you guys were right bout the weight of air being twice what I said but cant figure why Ive got it wrong.

Show us the calculations

[Gordon]

taking a 232bar 10l cylinder:

232x10 = 2320litres
1mole of a gas occupies 24litres
0.8x14+0.2x16=14.4g/mole of air
2320/24=96.667moles
14.4x96.667=1392g=1.4kg

[PeteM]

taking a 232bar 10l cylinder:

232x10 = 2320litres
1mole of a gas occupies 24litres
0.8x14+0.2x16=14.4g/mole of air
2320/24=96.667moles
14.4x96.667=1392g=1.4kg

The grammes per mole figures look different from those I have looked up, your figure appears reduced by 50% - have you taken atomic weights rather than g/mole?

http://www.newton.dep.anl.gov/askasci/phy00/phy00844.htm (search for Mellendorf)
http://www.apogee-inst.com/pdf_files/Molefractionandpartialpressure.pdf (at end)
http://www.seva.net/~smithsch/Weight_of_Gases.html

[Nigel Hewitt]

0.8x14+0.2x16=14.4g/mole of air
Molecular weight of O2 is 32 and N2 is 28.
Air comes out as 28.8

[Gordon]

Cheers. Have editted first post accordingly...

Feeling a little bit of a fool - trying to be too clever.