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Posted By Ronofcam
Hi peeps,
This one'll test you lot!
What concentration of R410A gas in a (for example) 4m x 4m x 2.5m room is required to reduce the oxygen level to the level that would asphyxiate an average person?
Is there a calculation that one could use to work out such a thing? Also, what regulations are applicable (specific, not generic ones like HASAWA 74).
Thanks in advance.
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Posted By Chris Kelly
Assuming I am understanding you correctly, you need to be careful here as, although the gas is asphyxiant rather than toxic, the sudden change in oxygen content can lead to sudden loss of consciousness.
Accordingly it is safest to consider the atmosphere as if it contained toxic gas.
You need to consider 'lack of oxygen' rather than the quantity of the gas.
Get a gas meter and test your oxygen level - if it is below (or above) the % then it is dangerous.
Regards,
Chris
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Posted By J.Morrison
There is a BCGA guidance Note GN11,
“The management of risks associated with reduced oxygen atmospheres”,
which contains calculations which might help.
Provided you know the basic details, cylinder capacity and fill pressure,
you should be able to arrive at an answer to your problem.
As for the second part, sorry don’t know,
can’t think of anything outside of the regulations
for divers and CS entry which would state any
requirement for oxygen levels.
Regards
John
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Posted By Neil Pettitt
Not my field but suspect that the F Gas Regs might apply.
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Posted By Phil Grace
Ronofcam,
The answer to your Regs query seems pretty straightforward - it will be the Confined Spaces Regulations
As for safe levels there are many sources of detailed information that will clarify as follows:
23%/24% - Unsafe due to increased risk of combustion
21% - Normal average oxygen concentration in air (to be precise 20.9%)
19.5% - "Safe" minimum level
18% - Critical level which should never be approached in practice.
Hope this helps
Phil
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Posted By D. Hilton
Assuming no ventilation and instantaneous homogeneous mixing a concentration of 6.8% or 291.7 m3 of R410A is required to reduce the Oxygen concentration to 19.49%.
Room Vol 40 m3
Oxygen in room 836 m3
Nitrogen etc 3160 m3
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Posted By D. Hilton
that is
8.36 m3 Oxygen
31.6 m3 Nitrogen etc
2.91 m3 R410a
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Posted By Phil Rose
I have got to admit that I am impressed with the calculations. I suggest that confined spaces regs apply and also suggest that a reasonably simple precaution would be to use gas monitoring.
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Posted By Ronofcam
Thanks very much everyone for their input.
Darren, I'm interested how you arrived at those calculations. Can you enlighten me?
The more I look into this, the more I'm inclined to agree that some of the small rooms that are served by the VRF system are going to fall under the Confined Spaces Regs.
Thanks again to all.
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Posted By CFT
Why is the system leaking, or why is the gas present? Specifically F-gas regs are responsible for the blended gas, and CS as stated for the situation. This is usually for AC equipment. It is also incredibly quick to clear.
More concerned why it is happening though.
Curious.
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Posted By Robert K Lewis
Look at this from the other end.
Gram Molecular Weight for R10A is around 70 and thus every 70gm released creates 0.0224 cubic metre of gas at STP.
From here you can calculate the weight of R410A required to produce sufficient volume of gas to dilute the oxygen level to asphyxiation point. Forget the 18.0% level as this is for safe working. Mouth to mouth works at around 16% oxygen in exhaled breathe. 12% will give unconciouness which is recoverable but 10% is nearly always fatal.
Bob
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Posted By D. Hilton
Ronofcam
Use of Ideal Gas Law and Dalton's Law ;]
or in other terms:
The Oxygen concentration in air is 20.9%
If we take 19.5% Oxygen as the safe minimum limit then we know that 1.42% of Oxygen must be displaced.
to displace 1.42% of Oxygen we must displace an amount of air containing that Oxygen. In other words 1.42/20.9*100 = 6.83%.
As we are displacing the 6.83% of Air with R410A we can say that the concentration required to deplete the Oxygen content within the room is 6.83%.
Simples
Obviously if you want to calculate the conc of the chemical of concern required to result in an Oxygen concentration of 18% then
2.9/20.9*100 = 13.8%
Regards
D
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Posted By D. Hilton
NB
I calculated the Liquid to gas ratio as 298 and if that is correct then approx 11 litres of liquefied R410a would result in the 6.8% concentration
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Posted By D. Hilton
Also,
it's handy to know that 6.83%, 68300 ppm, 203922 mg/m3, 203 g/m3 is the concentration of any asphyxiate that will result in oxygen depletion below 19.5% regardless of the size of enclosure.
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Posted By Robert K Lewis
These thoughts may be useful
Oxygen Depletion
-Any atmosphere containing measurably less than 20.9 % of oxygen can be considered depleted.
-Humans should not be exposed during prolonged periods to concentrations below 19.5 %. Note however this is about routine work
-The first symptoms and effects of oxygen depletion below normal concentrations and down to 14 % are an increased heart rate and tiredness.
-Between 14 % and 11 % the agility of mind and body diminish significantly,
-where as between 11 % and 8 % headaches, dizziness and unconsciousness occur quickly.
-Between 8 % and 6 % unconsciousness occurs in a few minutes and reanimation is only possible if it occurs rapidly.
-At concentrations below 6 % unconsciousness takes place almost immediately and severe brain damage and death are almost certain.
For me your critical point is around 12% as an answer to your question.
Bob
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Posted By D. Hilton
It's not as simple as that Bob, it depends on the level of physical activity in conjunction with the ambient concentration of oxygen inhaled.
The ambient O2 concentration effects the level of saturated blood within the body.
A decrease in the saturated blood concentration (SaO2) of 10% provides a reasonable chance of recovery without physical harm or fatality.
A decrease of 15% SaO2 will lead to physical harm and or fatality.
At rest ambient O2 levels will be lower than during moderate physical effort.
So during escape or carrying out moderate work, an ambient O2 conc of 18% will result in a reduction of SaO2 of 10%. While an ambient O2 conc of 17% during the same scenario will result in a reduction of SaO2 of 15%.
a difference of 1% conc would make the difference between life or death in such a situation.
For the above reason 19.5% O2 conc is selected as a safe limit.
NB Decrease in SaO2 can be calculated as
EXP(10.5-0.455*(Amb O2 Conc Inh).
Please note this does not include the effects of inhalation of CO or other fire effluent.
Regards
D
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Posted By Ronofcam Thanks again, all. Most enlightening! :) Darren, I understand the logic of what you are saying. Like most things, obvious once you know the answer! The million dollar question is, though, to displace 6.83% air of a 4m x 4m x 2.5m room, how much R410A would need to be introduced to the room to do that?! The MSDS sheet is here: http://www.bocsds.com/uk/sds/special/r410a.pdf Thanks again to all those brighter than me! :)
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Posted By D. Hilton
8.66 litres or 8.73 Kg liquefied gas expansion factor at 293K = 336 (Rho Liquid/Rho Gas)
2910 litres gas in room to give 6.83 % conc
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Posted By Ronofcam
Thanks again, Darren. Most helpful.
Out of curiosity, how did you arrive at those figures? Just interested to know. :)
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Posted By D. Hilton
Good thing you asked. The 2.91 m3 is based on the sum of the partial pressure of 3 gases into a theoretical sealed box which results in a slightly higher volume of gas required to provide the 6.83% concentration within the room.
So to clarify:
Room volume 40 m3
Concentration of gas required is 6.83% of 40 m3 (2.73 m3)
From BOC data:
Liquid density 1.09 water @1000 kg/m3 gives (1090 kg/m3)
Gas density 2.5 Air @ 1.2 kg/m3 gives (3 kg/m3)
Liquid to gas expansion calculated as Rho liquid/Rho gas (1090/3)giving an expansion factor of 363.3 (note error in previous reply)
We know that 2730 litres of R410a is required to result in 6.83% concentration.
2730 litres gas / 363.3 expansion factor gives 7.51 litres of liquified gas within cylinder.
7.51 litres of liquid x liquid density 1.09 kg/litre gives 8.19 kg liquid gas.
Regards
D
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Posted By Ronofcam
Darren, been away for a few days.
Thanks very much for your valuable input. Based on your information, I have been able to knock a few calculations out.
Once you see the working out it's fairly self explanatory.
Thanks again.
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Posted By D. Hilton
happy to help
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Posted By Robert K Lewis
DH
I was not intending any of the figures as meaning that any work at all should continue. Escape is the name of the game for occupants. The equipment design should prevent any of the figures below 16% ever being achieved without emergency automatic ventilation being available.
If work is intended it is a totally different situation and one enters the realm of confined spaces!
Bob
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