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Pressure underwater and equivalent weight

  1. Sep 28, 2015 #1
    Hello,

    I am trying to calculate the equivalent amount of weight that a submarine would be under at 1000m below the sea. Apologies if this isn't that well explained!!!

    At 1000m - 101 atmospheres - which is 14.5 psi per atmosphere - so at 1000m it would be 1464.5 psi.

    Do I then multiple 1464.5 by the entire surface area of the submarine or just the top part that is facing the surface?

    I am aiming to try and get a figure that will enable me to say 'at 1000m deep the submarine is supporting the equivalent weight of X jumbo jets / cars etc'

    Any help would be so much appreciated!!!
     
  2. jcsd
  3. Sep 28, 2015 #2

    Bystander

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    The "journalistic/pop-sci" approach isn't that well defined. Your body's surface area is 2500 -3000 square inches and under a load of (omigosh!) 30,000- 40,000 pounds just here at the earth's surface --- and pearl divers triple and quadruple that load.
    Fifteen hundred pounds per square inch is fifteen hundred pounds per square inch --- it is not a concept that requires translation.
     
  4. Sep 28, 2015 #3
    Neither. Have you ever heard of Archimedes Principle?
     
  5. Sep 28, 2015 #4
    yes I have read a little - but I am not sure how to apply to what I am trying to do. It's quite confusing!!!
     
  6. Sep 28, 2015 #5

    sophiecentaur

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    If you want to know the 'upthrust' then all you need to know is the density of the water, g and the volume i.e the Weight of water displaced.
    The difficulty would be in knowing the actual volume of the submarine and the actual volume of the compressed water. Of course, if the submarine is not going up or down, the upthrust is equal to its weight.
     
  7. Sep 28, 2015 #6
    Hello,

    If you are at 1000m doesn't the pressure mean that you are negatively buoyant?
     
  8. Sep 28, 2015 #7
    Please tell us your understanding of Archimedes Principle so that we can help you understand it better and apply it.

    Actually, the pressure at 1000 m has very little to do with what you want to determine.

    Chet
     
  9. Sep 28, 2015 #8

    russ_watters

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    I suppose people might be interested in how much water is sitting on top of a sub, so you could multiply by the horizontal cross sectional area...
     
  10. Sep 28, 2015 #9
    From the OP it looks like he is not interested in the buoyancy but rather the forces acting to "crush" the body of the submarine.
    But it's not completely clear what he is after.
     
  11. Sep 29, 2015 #10

    sophiecentaur

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    In which case, it should be pointed out to him that it isn't the 'total' of the forces acting that counts but the pressure on the weakest part of the hull that will determine whether the sub will be crushed or not.
     
  12. Sep 29, 2015 #11

    DaveC426913

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    While factually true, it does not get the OP what he is looking for.
     
  13. Sep 29, 2015 #12

    DaveC426913

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    Avoid the phrase "supporting", since that suggests the pressure is from above. It is actually from all sides.

    You could say "resisting a crushing force of" - and yes, you'd multiply the surface area of the sub by 1464.

    So, a 100ft sub with a 20ft diameter will be resisting (2x106)*1464 pounds, or about 3200 kilotons. That's about 10,000 Jumbo jets.
     
  14. Sep 29, 2015 #13

    sophiecentaur

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    Should we really be answering such questions as if they are meaningful? The jumbo jet thing is for popular journalists and not appropriate for PF. There is a difference between intensive and extensive variables.
     
  15. Sep 29, 2015 #14
    If you have the sub on the ground and put a jumbo jet on top, there will be the weight of the jet pushing down and a normal force pushing up, with a magnitude equal to the weight of the jumbo jet. Would it be meaningful to say that the sub supports (or resists) the weight of 2 jumbo jets?

    The result of multiplying pressure by area gives a meaningful force only of the surface is flat. Otherwise the changing direction of force should be taken into account.
    But anyway, the crushing effect depends on pressure (stress) and not on net force.

    It will be more accurate to compare the pressure on the sub with the pressure applied by x jumbo jets stacked on in top of the other, on teh landing gear of the bottom one. Will be a cool picture to imagine. :smile:
     
  16. Sep 29, 2015 #15
    Pascal's princple states that the pressure exerted by a fluid (water) is equal over the entire surface of the walls containing the fluid (in this case the sub hull keeping the water out). Since pressure = force/area, it's correct to multiply the pressure times the total area of the sub, to get the total force acting on the sub from outside. Obviously, the crushing effect of this force applies to all sides, so it would make more sense to calculate how much force is supported by a significant surface element, say a square foot, which would be 144 sq.in.*1464.5 lb/sq.in = 210,888 lb or 105.4 long tons. This is the weight each square foot of the sub needs to be able to support.105.4 tons is about the weight of a mid-size airliner, or 50 passenger cars. Compare this with the force usually supported by building floors of about 100 lb. per sq. foot. .
     
  17. Sep 29, 2015 #16
    This is not correct. At each location on the hull, the pressure acts perpendicular to the surface, so one must integrate the pressure over the surface area vectorially to get the total force acting on the sub hull. Also, Pascal's principle states that the pressure at a given location acts equally in all directions, not that it is equal over the entire surface of the walls containing the fluid. Some parts of the hull are at greater depth than others, so the pressure acting on the parts of the surface at greater depth is higher.

    Chet
     
  18. Sep 29, 2015 #17

    sophiecentaur

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    "one must integrate the pressure over the surface area vectorially to get the total force acting on the sub hull."
    But what on earth does that mean and how is it relevant to anything? It's just an arbitrary summing of values.
     
  19. Sep 29, 2015 #18
    It will mean the buoyant force, won't it?. Calculated the hard way. :smile:
     
  20. Sep 29, 2015 #19
    Yes, exactly.

    I should add that I didn't think that any of my post was relevant to the OP's question, unless he was interested in the buoyant force. The primary reason for my response in post #16 was just to correct the misinformation that the responder in post #15 had dispersed.

    As far as issues associated with deformation and crushing of submarine hulls under external pressure are concerned, we have the computational modelling tools to handle such problems. But certainly, a simplistic analysis (such as in some of the posts in this thread) is not adequate. Crushing of a hull, in particular, is a complicated buckling problem that takes into account the stress-strain behavior of the hull.

    Chet
     
  21. Sep 29, 2015 #20

    DaveC426913

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    I confess, after doing the back-of-napkin calcs, and coming up with the answer of 10,000 jumbo jets, I must concur.

    Suggesting a sub is withstanding the weight of 10,000 jumbo jets is going to do more harm than good. Saying every square inch must withstand the weight-equivalent of a VW Beetle is more meaningful.
     
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