0
$\begingroup$

Pressure

On average, a column of air one square centimeter in cross-section, measured from sea level to the top of the atmosphere, has a mass of about 1.03 kg and weight of about 10.1 N (2.28 lbf) (A column one square inch in cross-section would have a weight of about 14.7 lbs, or about 65.4 N).

The standard atmosphere (symbol: atm) is a unit of pressure equal to 101325 Pa or 1013.25 hectopascals or millibars. Equivalent to 760 mmHg (torr), 29.92 inHg, 14.696 psi. (The pascal is a newton per square meter or in terms of SI base units, kilogram per meter per second-squared.)

Atmospheric Pressure, Wikipedia

Therefore I conclude that Atmospheric pressure ≈ 1 bar.

The deeper you go under the sea, the greater the pressure of the water pushing down on you. For every 33 feet (10.06 meters) you go down, the pressure increases by 14.5 psi (1 bar). In the deepest ocean, the pressure is equivalent to the weight of an elephant balanced on a postage stamp, or the equivalent of one person trying to support 50 jumbo jets!

Pressure, The National Oceanic and Atmospheric Administration

Therefore I conclude that at 10 meters of depth in seawater, the pressure on an object is 2 bar.

Rigid, air-tight boxes

Imagine with have a rigid, air-tight box containing normal air at normal atmospheric pressure. With the assistance of weights, we lower the box to a steady 33 feet of depth. This is Box A.

Now imagine we take an identical box, but construct it in a vacuum chamber. As a result it contains no air. This box is then exposed to normal atmospheric pressure. This is Box B.

Question

As Box A has 1 bar internal pressure (outwards) and 2 bar external pressure (inwards), does this result in a 1 bar pressure differential similar to Box B, but without the need of expensive vacuum chambers? Would this be a viable way to test a design for structural weak-points?

Improve this question
$\endgroup$
5
  • 1
    $\begingroup$ THe buoyant force would be the same, but the weight of the box would be different, so the difference between the weight and the buyant force woudl be different, creating a different net external force on the two boxes. $\endgroup$
    – Zo the Relativist
    Dec 19, 2014 at 17:46
  • $\begingroup$ Do you mean rather than 2-1=1 it would be 1-0=1? The forces are larger, but the resultant is the same? $\endgroup$
    – Lewis Goddard
    Dec 19, 2014 at 17:50
  • $\begingroup$ @JerrySchirmer: The buoyant force would definitely be different, because water is much denser than air. $\endgroup$
    – Jan Hudec
    Dec 19, 2014 at 22:31
  • $\begingroup$ @JanHudec: the buoyant force only depends on the volume of displaced water. The net force on the thing is the difference between it's actual weight and the buoyant force. Though I think we're arguing semantics now. $\endgroup$
    – Zo the Relativist
    Dec 19, 2014 at 22:44
  • $\begingroup$ @JerrySchirmer: Buoyant force depends on volume of displaced fluid and density of that fluid. In one case water is being displaced, in the other case air is. Therefore the buoyant forces are different. $\endgroup$
    – Jan Hudec
    Dec 19, 2014 at 22:47

2 Answers 2

Reset to default
1
$\begingroup$

The shear and tension stress on the material would be the same, because only the difference matters for that. So for testing strength of the structure the cases are equivalent.

The material would still be under the total pressure though, so if the material itself can't withstand that pressure, it would degrade (it may change crystalline structure, compounds may undergo chemical reaction and such with corresponding change in modulus of strength; or it may just break on inhomogeneities and crumble). However solid material generally withstand significantly higher pressures.

And under water there is also higher difference between pressure on the top and bottom surface and corresponding difference in buoyancy.

Improve this answer
$\endgroup$
1
$\begingroup$

It seems your question asks simply is the effect on the box the same with 1 bar differential pressure of either air or water. Yes, the effect would be the same for air and water. Both are fluids exerting uniform differential pressure of 1 bar.

Improve this answer
$\endgroup$
2
  • $\begingroup$ So the fact the forces are different do not matter, just the resultant force? $\endgroup$
    – Lewis Goddard
    Dec 19, 2014 at 22:31
  • 1
    $\begingroup$ Yes, static equilibrium of forces of fluids are independent of the composition of the fluid if the question asks simply for the comparative effect of a differential pressure. $\endgroup$
    – G. R. Wilbourn
    Dec 19, 2014 at 22:34

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.