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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 $dh/dt=dr/dt$?

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Using the ideal gas law:

$P(h) = P_0 + \rho_w g h$, ${V_o P_0} = {V(h) P(h)}$ which gives $V(h) = {V_o P_0}{1 \over P_0 + \rho_w g h}$.

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  • $\begingroup$ I've tried that, but I don't have the initial conditions. $\endgroup$ – Mphysics Feb 10 '16 at 19:37
  • $\begingroup$ You need more information if you want to compute something. The question looked for a relationship... $\endgroup$ – copper.hat Feb 10 '16 at 19:38
  • $\begingroup$ @Mphysics If the bubble is released at a depth $h$ and a volume $V(h)$, this lets you compute $V_0$, the volume at zero depth. What were you looking for? $\endgroup$ – Ross Millikan Feb 10 '16 at 19:40
  • $\begingroup$ The problem is I don't have V(h) or V(0). I have to find the acceleration of the bubble as a(h) if I know the resistance force $F=k \rho Av$. That is all I have. It's a very confusing exercise, I've been solving it for a few days now, but I always end up nowhere. $\endgroup$ – Mphysics Feb 10 '16 at 19:58
  • $\begingroup$ The question as stated is really a physics question. The question asks about the relationship between depth and volume, and both current answers address this. You are adding extra information in comments and seem to be trying to answer a different question entirely. You should consider modifying the question and/or moving it to Physics SE. $\endgroup$ – copper.hat Feb 10 '16 at 20:09
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As you go deeper, the pressure increases, decreasing the volume of the gas in the bubble. In water, the pressure is appoximately $14.7(1+\frac d{33})$ psi where $d$ is the depth is measured in feet. The volume goes as the inverse of the pressure.

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  • $\begingroup$ I have a bubble that arises from the sea bottom and I'm looking for its acceleration. So I guess I need some formula that will represent rate of change of the volume, regarding the change of depth. dV/dt or dV/dh? $\endgroup$ – Mphysics Feb 10 '16 at 19:35
  • $\begingroup$ This will let you assess the volume and therefore the buoyancy force, but I don't know how you will get the resistance. That will dominate how fast it rises. $\endgroup$ – Ross Millikan Feb 10 '16 at 19:37
  • $\begingroup$ The resistance force is proportional to density of water, the cross-sectional area of the bubble and its velocity. So, I guess $F=k \rho Av$ $\endgroup$ – Mphysics Feb 10 '16 at 19:39

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