How to calculate the oxygen
level drop in a room if any gas cylinder leaks in to the atmosphere.
The air around us is a mixture of gases,
mainly nitrogen and oxygen, but containing much smaller amounts of water vapor,
argon, and carbon dioxide, and very small amounts of other gases. According to
the Occupational Safety and Health Administration, OSHA, determined the optimal
range of oxygen in the air for humans runs between 19.5 and 23.5 percent. As we
aware that lower oxygen content (depending upon %) may have various effects in
our human body.
So, when gas cylinder is stored in area/gases
are used in an area, it is very important to predict the concentration of gas
present in that area for carrying out adequate safety precaution. The
calculation are as follows
Assumption/Conditions:
1.
Volume
of the room = 165 Cu.m
2.
Gas
stored : Nitrogen in cylinder
3.
No air
exchange rate in the room : 0
4.
Weight
of N2 in
cylinder =
16 kg
Calculation
Density of air at 30°C(our environment temp) = 1.1644 kg/Cu.m
Mass of air in the
room =
165 * 1.1644 = 192.126 kg
Molecular
weight of air = 29
So,
Number of kg-moles of air = 192.126/29
= 6.625034
Since
the volume ratio is the same as mole ratio for gases,
Therefore
Number
of kg-moles of N2 = 6.625034*.78
(78 % of N2
in
atmosphere)
= 5.167527
Number
of kg-moles that would add if leaked from the cylinder is = 16(weight of N2 in
cylinder)/28
= 0.57.
Total
kg-moles of Nitrogen in the room =
5.16752 +
0.57 = 5.737527
If
the released volume of N2 is added to denominator, then the % of
N2
in
air (after Nitrogen release is)
= (5.737527 /
6.625034+0.57) *100 = 79.7429 %
The
oxygen % in air is reduces to =
19.287 % which below 19.5%. so we can go for any of the below said
recommendation
Note : More over the above
said will come if equal dispersion happens & in case of any jet leakage the
oxygen level near the leakage area will be low which increases the risk.