Surviving under water in air bubble

Question

An incredible news story today is about a man who survived for two days at the bottom of the sea (~30 m deep) in a capsized boat, in an air bubble that formed in a corner of the boat. He was eventually rescued by divers who came to retrieve dead bodies. Details here. Since gases diffuse through water (and are dissolved in it) the composition of air in the bubble should be close to the atmosphere outside, if the surface of the bubble is large enough; so the excessive carbon dioxide is removed and oxygen is brought in to support life of a human.

Question: How large does the bubble have to be so that a person in it can have indefinite supply of breathable air?

Answer 1

Summary: I find a formula for the diameter of a bubble large enough to support one human and plug in known values to get $d=400\,{\rm m}$.

I'll have a quantitative stab at the answer to the question of how large an air bubble has to be for the carbon dioxide concentration to be in a breathable steady state, whilst a human is continuously producing carbon dioxide inside the bubble.

Fick's law of diffusion is that the flux of a quantity through a surface (amount per unit time per unit area) is proportional to the concentration gradient at that surface,

$$\vec{J} = - D \nabla \phi,$$

where $\phi$ is concentration and $D$ is the diffusivity of the species. We want to find the net flux out of the bubble at the surface, or $\vec{J} = -D_{\text{surface}} \nabla \phi$.

$D_{\text{surface}}$ is going to be some funny combination of the diffusivity of $CO_2$ in air and in water, but since the coefficient in water is so much lower, really diffusion is going to be dominated by this coefficient: it can't diffuse rapidly out of the surface and very slowly immediately outside the surface, because the concentration would then pile up in a thin layer immediately outside until it was high enough to start diffusing back in again. So I'm going to assume $D_{\text{surface}} = D_{\text{water}}$ here.

To estimate $\nabla \phi$, we can first assume $\phi(\text{surface})=\phi(\text{inside})$, fixing $\phi(\text{inside})$ from the maximum nonlethal concentration of CO2 in air and the molar density of air ($=P/RT$); then assuming the bubble is a sphere of radius $a$, because in a steady state the concentration outside is a harmonic function, we can find

$$\phi(r) = \phi(\text{far}) + \frac{(\phi(\text{inside})-\phi(\text{far}))a}{r},$$

where $\phi(\text{far})$ is the concentration far from the bubble, assumed to be constant. Then

$$\nabla \phi(a) = -\frac{(\phi(\text{inside})-\phi(\text{far}))a}{a^2} = -\frac{\phi(\text{inside})-\phi(\text{far})}{a}$$

yielding

$$J = D \frac{\phi(\text{inside})-\phi(\text{far})}{a}.$$

Next we integrate this over the surface of the bubble to get the net amount leaving the bubble, and set this $=$ the amount at which carbon dioxide is exhaled by the human, $\dot{N}$. Since for the above simplifications $J$ is constant over the surface (area $A$), this is just $JA$.

So we have $$\dot{N} = D_{\text{water}} A \frac{\phi(\text{inside})-\phi(\text{far})}{a} = D_{\text{water}} 4 \pi a (\phi(\text{inside})-\phi(\text{far})).$$

Finally assuming $\phi(\text{far})=0$ for convenience, and rearranging for diameter $d=2a$

$$d = \frac{\dot{N}}{2 \pi D_{\text{water}} \phi(\text{inside})}$$

and substituting

• $D = 1.6\times 10^{-9}\,{\rm m}^2\,{\rm s}^{-1}$ (from wiki)
• $\phi \approx 1.2\,{\rm mol}\,{\rm m}^{-3}$ (from OSHA maximum safe level of 3% at STP)
• $\dot{N}= 4\times 10^{-6}\,{\rm m}^3\,{\rm s}^{-1} = 4.8\times 10^{-6}\,{\rm mol}\,{\rm s}^{-1}$ (from $\%{\rm CO}_2 \approx 4\%$, lung capacity $\approx 500\,{\rm mL}$ and breath rate $\approx \frac{1}{5}\,{\rm s}^{-1}$)

I get $d \approx 400\,{\rm m}$.

It's interesting to note that this is independent of pressure: I've neglected pressure dependence of $D$ and human resilience to carbon dioxide, and the maximum safe concentration of carbon dioxide is independent of pressure, just derived from measurements at STP.

Finally, a bubble this large will probably rapidly break up due to buoyancy and Plateau-Rayleigh instabilities.

Answer 2

Maybe I should turn the comment to an answer.

The physics of the situation is the same as when one can upturn a water glass with the water not falling out. The atmospheric pressure keeps it in.

There exist the diving bells with open bottoms . As they are lowered the pressure in the air goes up to balance the water pressure, because the lower in the water the higher the pressure. I think one atmosphere is ten meters of water. The person in the link was at 30 meters and that is why he had to be decompressed in order not to get the bends.

The oxygen versus CO2 and volume to survive one day is solved in this link:

The atmosphere is about 20 percent oxygen. People breathe this in, and breathe 15 percent oxygen out, making the air that's left lower in Good Ol' Oh Two, but still quite breathable. Each minute a person at rest takes in roughly seven to eight liters of air, which adds up to about 11,000 liters of air a day. That sounds like a lot, volume-wise it's only 388 cubic feet. A ten by ten by ten foot room has 1000 cubic feet of air.Add that to the fact that you'll be breathing out a lot of oxygen, and you only need about 19 cubic feet of pure oxygen a day. Your breathing may get labored by the end of the second day, but a relatively small room should be fine for about three days. right?

wrong because there is CO2 exhaled which at large concentrations is poisonous:

Once the carbon dioxide levels in the room rise to two percent, carbon dioxide poisoning starts to happen. That occurs when the overall oxygen level falls to 19 percent, about half way through the first day.

For the person to survive three days the compressed air he was breathing must have been from a lot of volume. From these numbers an order of magnitude estimate for non renewable volume to survive for 3 days is 6000 cubic feet. An enclosure of 10*25*25 in feet is a reasonable one on a tugboat. He was just lucky the air siphoned where he was trapped.

..........

Answer 3

This is to summarize some of the excellent comments made previously by participants of this discussion, and to emphasize a couple of important points.

1. The original question implied that gas exchange between the bubble and surrounding water may be enough to sustain indefinitely breathing organisms inside. However this does not seem to be possible if the bubble is below the sea level because the gas in the bubble will eventually dissolve in the water and the bubble will cease to exist. The reason is that the concentration of dissolved gas near the bubble surface has to be higher than the ambient concentration. The ambient concentration is not higher than its value at the ocean surface (it is convective mixing rather than diffusion what transports dissolved air into the ocean). However the bubble is at the pressure $\sim$4 times the atmospheric pressure, and according to the Henry law the dissolved gas concentration next to the bubble should be $\sim$4 times higher than the ambient (the possible variation of temperature does not change the overall conclusion). So eventually the dissolved gas will diffuse away and the bubble will disappear. Thus, for any size of the bubble, with or without breathing organisms inside, there is no steady-state solution, if the bubble is below the sea level. This is illustrated in the figure: for either regular or upside-down siphon, the equilibrium state has water level equal in both legs. If one leg end in Fig. A is below the equilibrium level then eventually there will be no gas on this side of the siphon; this is similar to having no water in one leg in Fig. B if the end of this leg is above the equilibrium level.

   

2. The good news is that it would take a long time for the bubble to diffuse away. Estimate the diffusion time as

   $ \tau \sim \frac{n_{g}}{c} \frac{a^2}{D}, $

   where $c/n_g$ is the ratio of concentrations of dissolved gas molecules and gas molecules in the bubble, $a$ is the size of the bubble, $D$ is the diffusion coefficient of dissolved gas.

   According to Wikipedia, at 25 C, $c/n_g$ values are: for N$_2$ $1.492\times10^{-2}$; for O$_2$ $3.181\times10^{-2}$; for CO$_2$ $0.8317$.

   Also, according to Wikipedia, the values of $D_g$ are: for C0$_2$ at 5-25 C $1.07-1.91 \times 10^{-5}$ cm$^2$/s; for N$_2$ at 25 C $1.88 \times 10^{-5}$ cm$^2$/s; for O$_2$ at 25 C $2.1\times 10^{-5}$ cm$^2$/s.

   So, with these numbers, for an air bubble of the size $a\sim$ 1 m the lifetime is $\sim 10^{11}$ s $\sim$10,000 years! This is just unrealistic (if such things were possible we would've known about that!). The problem with this estimate is that the gas diffusion coefficient in water, D$_g\sim$10$^{-5}$cm$^2$/s = 10$^{-9}$m$^2$/s applies only to phenomena on small spatial scales, e.g., transport in porous media. We know that we can make carbonated water at home in a matter of minutes rather than months! On macroscopic scales transport is dominated by turbulence and convective flows, and, rather than diffusion, macroscopic transport models use the notion of effective ``exchange velocity'' $V_{ex}$, such that the flux of dissolved gas is

   $ \Gamma = A c V_{ex}, $

   where $A$ is the area, $c$ the concentration of dissolved gas.

   The idea is that dissolved gas is removed by background convection and/or turbulence at the rate $V_{ex}$. Typical magnitude of $V_{ex}$ in the ocean, near the surface, can be 1 m/day [reference needed].

   The lifetime of an air pocket is then

   $\tau \sim \frac{a^3}{a^2 (c/n_g) V_{ex}} = \frac{n_g}{c} \frac{a}{V_{ex}},$

   which, for $V_{ex}$=1 m/day, yields $\tau\sim$100 days, much more reasonable that 10,000 years!

3. As pointed out earlier, probably one does not need to make complex assumptions about water flows, turbulent diffusion etc if the available bubble of air contained the equivalent of $\sim$25 $m^3$ at atmospheric pressure, now compressed into $25/4\sim$6 m$^3$ volume, theoretically that would be enough for 2.5 days, without degassing excessive CO$_2$ into the water, or importing O$_2$ from surrounding water (the latter is impossible anyway as discussed in (1) above). The way it is described in the media, the man ``...survived, breathing inside a four-foot high bubble of air as it slowly shrank from the waters rising from the ceiling of the tiny toilet and adjoining bedroom'', but probably that four-foot bubble communicated with a larger volume or air under the hull of the boat - and that's the most reasonable explanation of this miraculous survival.

4. One can note that due to different solubility of CO$_2$, N$_2$, and O$_2$ these gases will be leaving the bubble at different rates, and this is perhaps good news. Due to the breathing human present inside, the CO$_2$ level in the bubble will be increasing and the O$_2$ level will be accordingly decreasing. Once this goes to the dangerous level, say 5$\%$ CO$_2$ and 15$\%$ O$_2$, it will be mostly CO$_2$ that will be leaving the bubble, transport of other gases will be much slower. As pointed out by other people in this discussion, with the ``classical'' diffusion the gas removal rate would not be enough to counterbalance the buildup of CO$_2$ (unrealistic surface area needed), but background flows and turbulence in water can significantly increase the rate of transport, and perhaps the available surface area as well (foam formation).

   To make a quantitative estimate, let's take that the human rate of O$_2$ consumption and CO$_2$ production is $\sim 10^{25}$ molecules/day; and for the concentration of dissolved carbon dioxide use

   $ c_{CO2} \sim 0.8 \times 0.05 \times 4 \times 6.02 \times 10^{23} / (22.4 \times 10^{-3} m^3) \sim 4 \times 10^{24} m^{-3} $

   For the rate of removal of carbon dioxide take

   $ \Gamma_{CO2} \sim a^2 V_{ex} c_{CO2}, $

   assuming transport of dissolved CO$_2$ with the exchange velocity $V_{ex}$=1 m/day, and one can see that just a couple of square meters of surface area would be enough to balance the buildup of exhaled CO$_2$ concentration. This effect can probably extend the survival time by a factor of $\sim$2. Normally only a quarter of available oxygen in the air can be used, reducing it from 20$\%$ on inhale to 15$\%$ on exhale. With excessive CO$_2$ removed from the air by water, however, probably surviving on the fraction of oxygen in the air as low as 10$\%$ is possible. For example, the diving gas mix Heliox uses 10$\%$ oxygen (and the rest helium).

5. And last, it is interesting to note that in principle in this kind of situation one could possibly survive free ascent on a single breath from this depth; in the history of submarine accidents some (well trained) people managed to escape to the surface from sunken submarines even from greater depths by free ascent.