Ahoy hive mind!

The rough scenario I'm looking for some help over is;

  • Picture a tub of $\mathrm{10 \ (l) × 5 \ (w) × 2 \ (h)}$ floating in a body of water
  • The full mass is $50,000 \ \mathrm{kg}$, so the displaced $50\ \mathrm{m^3}$ of water reaches half way up the tub sides
  • Add a mirrored cavity, equivalent to the top, except with only $\mathrm{1\ m}$ vertical walls (structure included in the $50,000\ \mathrm{kg}$ of mass)
  • This occurs in freshwater, and at seal level.


Now bubble air into the inverted cavity, filling the entire $50\ \mathrm{m^3}$ and raising the double-tub $1\ \mathrm{m}$ higher in the water, to the point where the top of the trapped air pocket is level with the water surface.


My question :

How would one go about calculating the pressure of the trapped air?

Any help appreciated! The idea is to hypothesise how vast a submerged tank would be needed to qualify as a realistic, low-pressure, energy stoage reservoir (any further thoughts would also be welcome!)


1 Answer 1


The pressure of the air at the bottom is the pressure on top plus the pressure due to the change in height: $p=W/A+\rho_{air}gh$. You can also calculate that pressure by considering the pressure of the water there: $p=p_{atmosphere}+\rho_{water}gh$.

From these two you can get both, $p$ and $h$

  • $\begingroup$ Super thanks for that, I was out of my depth (h)! $\endgroup$ Nov 10, 2021 at 23:33
  • $\begingroup$ @AlexanderForster I am glad it helped! $\endgroup$
    – user65081
    Nov 11, 2021 at 0:02

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