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I would like to get some insight into a component with a sealed chamber of air which is submerged in water and forced down to a depth x. As the air has no communication with the outside of the cavity, how would this effect the buoyancy of the component?

Reasoning for this, I would like to determine the buoyancy force with increasing depth (ignoring the air chamber compressing). If this is possible, I would appreciate some guidance on how to calculate it.

Following this, I would like to recalculate the problem but using helium instead of air to determine the effects with depth.

Hope this makes sense! It's been a while since I looked at some physics!

Many thanks

CJU

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  • $\begingroup$ Most likely, the weight of the air or helium would be negligible compared to the weight of the chamber itself. Therefore almost no effect. $\endgroup$
    – LLlAMnYP
    Jul 4, 2016 at 8:27
  • $\begingroup$ The buoyancy is the weight of the displaced liquid, i.e. it doesn't matter what's inside, as long as the volume will be the same. Having said that, the volume of pressure tanks does change quite a bit for these high pressure dives, so you have to look at the data sheet for this equipment. $\endgroup$
    – CuriousOne
    Jul 4, 2016 at 9:04
  • $\begingroup$ Perfect, that answers the question for me. I know it's easy to calculate buoyancy but I was questioning my sanity the more I thought about an air chamber and pressure differentials as high depths. Thank you! $\endgroup$
    – CJU
    Jul 4, 2016 at 10:01

1 Answer 1

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Since you want some guidance on how to calculate the buoyancy, here is the equation: $B = \rho g V$, with $B$ the bouyant force, $\rho$ the density of the water at the depth x and $V$ the displaced volume.

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