How much Oxygen is there for a person to survive in an air-tight enclosure?
Introduction:
Have you ever wondered how long a person could survive
in a closed, air-tight space with the Oxygen that is available?
I have.
This relates to the other web page on Oxygen: "Is
There Enough Oxygen in the Atmosphere?"
Some of the numbers I will use here are from that page.
If you don't want to go through the details of calculation, you will see the
results in larger red print as you go down the page.
(1) How much Oxygen does an average person
need in a day to live?
3 sources:
(a) According to NASA
the average
person needs 0.84 kg of
O_{2} per day (day and night activity).
(b) According to MadSci: The amount
of air breathed by an average adult varies from 6 Litre/minute (when at rest) to
50 Litre/minute after hard exercise. So, at rest the amount of Oxygen
absorbed by the blood in the lungs is 300 milliLitre/minute to 3,500 mL/minute.
My
Calculations for a person at rest:
In 1 day the volume of O_{2} absorbed by a person at rest is 300mL/minute x 60
minute/hour x 24 h/day = 432,000 mL/day or 432 L/day. Since the mass of 1
mole of O_{2} is 32 gram and 1 mole of any gas has a volume of 22.4 L, the number
of mole/day is 432 L/day / 22.4 L/mole = 19.28 mole.
The mass of O_{2} needed per day is
then 19.28 mole x 32 gram/mole = 617
gram or 0.617 kg of Oxygen per day, which is less than
the NASA figure.
My Calculations
for a person doing hard exercise all day:
In 1 day the volume of O_{2} absorbed by a person at rest is 3,500
mL/minute x 60 minute/hour x 24 h/day =5,040,000 mL/day or 5,040 L/day.
Since the mass of 1 mole of O_{2} is 32 gram and 1 mole of any gas has a volume of
22.4 L, the number of mole/day is 5,040 L/day / 22.4 L/mole = 225 mole.
The mass of O_{2} needed per day is
then 225 mole x 32 gram/mole = 7,200
gram or 7.2 kg of Oxygen per day, which is much more
than the NASA figure. This is an unrealistic figure because a person
doesn't do hard exercise day and night, but it does give an upper limit to
Oxygen consumption for the average person.
An aside: About 5L of blood passes through your lungs
every minute. Adult does from 12 to 18 breaths/minute.
According to Dr. Garfield Kirchner, neighbour, 'one
does not breathe the entire lungs volume with every average breath, but only a
small bit, I think around 500 ml -- called the tidal volume. A
yawn or a sigh will force one to breath more as will deep sleep'. Lungs
can hold up to 4 to 6 L -- called the Vital Capacity.
'Inspiratory reserve: the room for extra air beyond a normal tidal breath is
about 3 L.
Expiratory reserve: the amount that can be forcefully expired beyond a normal
tidal breath (exhalation) is 1.1 L.
Residual Volume: that which is left in the lungs after the most forceful
exhalation is 1.2 L.'
Very roughly 5% of Volume of air is consumed in each breath (about 21% O_{2}
in and about 16% O_{2} out).
CO_{2} output is another issue not covered here.
(c) Body tissues need
14.7 L of Oxygen/hour. So, in 1 day they need 14.7 L/h x 24h/day = 352.8
L/day.
This is 352.8 L/day/22.4L/mole = 15.75 mole/day.
Which is 15.75 mole/day x 32 gram/mole = 504 gram/day or 0.504 kg/day of
Oxygen. This is the absolute minimum required. This confirms we are
in the right ball park.
We will use the 3 following figures:
0.617 kg for an average person at rest
0.84 kg for an average person going about his regular duties, including sleeping
7.2 kg for an average person that does continuous heavy exercise day and
night (just to make a comparison).
(2) How long would it take to use up all the O_{2} in 1 cubic metre (1 m^{3} of space at sea level) -- this is not realistic, but what the heck -- see a more realistic situation in part (3) below
The volume in Litre (L) in cubic metre (m^{3})is:
1000L/m^{3}.
21 % of the atmosphere consists of O_{2} by volume. So, 1000 L of
air consists of 0.21 x 1000 L = 210 L of O_{2}.
1 mole of any gas at STP has a volume of 22.4 L. Therefore 210 L of O_{2}
is 210L/22.4L/mole = 9.38 mole.
The mass of 9.38 mole of Oxygen is: 9.38 mole x 32 gram/mole = 300 gram or 0.3
kg.
How long would the 3 persons under different exercise modes last in an
air-tight enclosure of 1 m^{3} that has 0.3 kg of Oxygen?
For a person at rest | For a person doing NASA type activity |
For a person exercising hard all day | |
Oxygen needed by 1 person/day |
0.617 kg/day | 0.84 kg/day | 7.2 kg/day |
Length of time needed to use all the Oxygen in 1 m^{3} |
0.3 kg / 0.617 kg/day = 0.48 day |
0.3 kg / 0.84 kg/day = 0.36 day |
0.3 kg / 7.2 kg/day = 0.042 day |
(3) According to psu.edu the lowest %
Volume of O_{2} a person can survive at standard pressure is
about 17.5%. How long would it take to use up just the O_{2} in 1
m^{3} for life? [realistic]
The volume in Litre (L) in cubic metre (m^{3})is:
1000L/m^{3}.
21 % of the atmosphere consists of O_{2} by volume. So, 1000 L of
air consists of 0.21 x 1000 L = 210 L of O_{2}.
If 17.5% of the atmosphere consists of O2 by volume, then 1000L of air would
consist of 0.175 x 1000 L = 175 L of O2.
Useable Oxygen is therefore 210 L - 175 L = 35 L for a person.
1 mole of any gas at STP has a volume of 22.4 L. Therefore 35 L of O_{2}
is 35L/22.4L/mole = 1.6 mole.
The mass of 1.6 mole of Oxygen is: 1.6 mole x 32 gram/mole = 51 gram or 0.051
kg.
How long would the 3 persons under different exercise modes last in an
air-tight enclosure of 1 m^{3} that has 0.3 kg of Oxygen, but only .051
kg is useable?
For a person at rest | For a person doing NASA type activity |
For a person exercising hard all day | |
Oxygen needed by 1 person/day |
0.617 kg/day | 0.84 kg/day | 7.2 kg/day |
Length of time needed to use all the useable Oxygen in 1 m^{3}. |
0.051 kg / 0.617 kg/day = 0.083 day or 2 hours |
0.051 kg / 0.84 kg/day = 0.061 day or 1.5 hours |
0.051 kg / 7.2 kg/day = 0.0071 day or 10 minutes |
(4) What about other spaces such as rooms,
houses, etc.? How long would one person last?
We're assuming a completely enclosed place - no more Oxygen in or
out.
We're also assuming 1 person.
We will use a conversion factor of 0.3 m/foot and a conversion factor of 0.09 m^{2}/ft^{2}
Length of time before the Oxygen level drops to 17.5% of the air |
|||
For a person at rest | For a person doing NASA type activity |
For a person exercising hard all day | |
Room 12 ft wide x 12 ft long x 8 ft ceiling 3.6 m x 3.6 m x 2.4 m = 31.1m^{3} |
31.1 m^{3} x 0.083 day/m^{3} = 2.6 days |
31.1 m^{3} x 0.061 day/m^{3} = 1.9 days |
31.1 m^{3} x 0.0071 day/m^{3} = 0.22 days or 5.3 h |
House 1500 square feet x 8 ft ceiling 135 m2 x 2.4 m = 324 m^{3} |
324 m^{3} x 0.083 day/m^{3} = 27 days |
324 m^{3} x 0.061 day/m^{3} = 19.8 days |
324 m^{3} x 0.0071 day/m^{3} = 2.3 days |
CO_{2} build-up would also happen.
O. Hooge, B.C. Canada
tfrisen@shaw.ca
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