How does the depth affect the volume (the radius) of an air bubble in the water, if the temperature and density of the water are constant. Is there any relation combining this?
Can I say that $dh/dt=dr/dt$?
[Physics] volume of the air bubble in the water
airfluid dynamicsthermodynamicswater
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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.
The (very small) air bubbles you see sticking on your hand when you immerse your hand in tap water, are of two types:
1- Bubbles which contain air outside the water. This air is trapped along your fingerprint lines when you insert your hand inside water. Typically, the lower the surface tension, the lower the probability and size of these bubbles. You can effectively decrease the surface tension of water by dissolving washing powder or soap to it. Interestingly, this will cause a lot of bubbles to form ON the water, but very few UNDER the surface.
2- Bubbles which form due to dissolved gases in the water itself. Contrary to solids, gases dissolve better in a liquid at it's temperature decreases. Boiling the tap water and then letting it cool in very low pressure conditions will remove most of these gases and keep them from forming again.
So, as a conclusion, boil the water, cool it under low pressure conditions and then add washing powder to it. Do not stir the water after adding it and just give it time to dissolve by itself. Now slowly insert your hand inside the water. There would be far fewer bubbles on your hand when it gets in.
Best Answer
Using the ideal gas law:
$P(h) = P_0 + \rho_w g h$, ${V_o P_0} = {V(h) P(h)}$ which gives $V(h) = {V_o P_0}{1 \over P_0 + \rho_w g h}$.