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Crazy water pressure formula implication?

  1. Feb 6, 2009 #1
    Hey everyone, I just thought of a mind-boggling "thought experiment":


    As a reminder, the formula for water pressure acting on an object due to weight is:

    (Water density)*(Acceleration of gravity)*(Depth of object underwater)

    Note that the shape of the container doesn't matter.


    Imagine a water-filled tank the size of the Pacific Ocean (exact measurements won't be necessary). The tank is completely filled with water and is sealed shut, so any pressure other than water pressure is absent. Now imagine that at one corner of the tank there is an incredibly narrow "chimney" about ten nanometers in diameter and extending about ten kilometers above the top of the tank. This tube is connected directly to the big tank and is sealed shut, though the water level can be changed at will. I didn't do the math, but I'm pretty sure a few drops of water would raise the water level by at least hundreds of meters.

    The weird part: According to the formula, water pressure throughout a container is dependent only upon the height of the container, assuming gravity and density remain constant. Technically, our imaginary tank is over ten kilometers tall, since the chimney is part of the tank. Does this mean that adding just a few drops of water can create enormous pressure everywhere in the ocean-sized tank? Seems to violate the law of conservation of energy, doesn't it?

    Yes, I know that pressure is force/area, so the weight of the drops is concentrated like a needle. But can anyone tell me how--preferably on a molecular level--that force is distributed to every part of a body of water as big as the PACIFIC OCEAN?

    Thanks
     
  2. jcsd
  3. Feb 6, 2009 #2
    Nice question but the increased pressure at the bottom of the narrow tube will cause water to be forced out until equilibrium is reached .In fact for a tube so narrow surface tension effects will play a major part and only some of the water will flow out.
     
  4. Feb 6, 2009 #3

    stewartcs

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    The pressure at any point in the container is equal to [itex] \rho g h [/itex]. However, the h refers to the height of the water column directly above the point. That is, the water pressure is only the same at points that have the same fluid column heights above them, and not the tank's total height.

    CS
     
  5. Feb 6, 2009 #4
    Surface tension is irrelevant to my question, but thanks for the reality check anyway.


    So basically, the weight of the water in the tube pushes the water below it aside? Then the water that was just pushed aside pushes other molecules aside, then those molecules push others aside, etc., spreading the pressure like a chain of dominoes? Maybe!

    At first, I disregarded the domino analogy because I thought that the molecules would eventually run out of steam (pun not intended), while dominoes would keep going because of gravity. However, thinking twice, I was reminded of two things:


    1. The molecules are pushing through a vacuum, so friction is absent (hydrogen bonding and van der waal's forces might complicate things though).

    2. More importantly, the molecules DO have gravity on their side, since they're being propelled by the weight of the tube's water!


    These might seem like simple insights, but they're good enough for me.


    One more question. If what I said is true, then I wonder how fast the pressure spreads through our ocean-sized tank? Instantaneously, or slower?


    If my understanding is correct, pressure is scalar and applies equally to all parts of a container, regardless of shape or size.


    Thank for replying guys
     
  6. Feb 6, 2009 #5

    stewartcs

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    Pressure is transmitted equally and undiminished in all directions, but the actual value of the pressure due to the hydrostatic head pressure of the fluid at that point varies with the height of the fluid column.

    For example if the pressure at h1 = 1 psi and h2 = 2 psi, then an increase in pressure at h1 of 3 psi (h1 now is 4 psi), h2 will be increased equally to 5 psi.

    Does that help?

    CS
     
  7. Feb 6, 2009 #6

    stewartcs

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    Pressure waves in fluid travel at different speeds depending on the fluid properties. The maximum they will travel is the speed of sound for that particular medium.

    [tex]c = \sqrt{\frac{E}{\rho}} [/tex]

    where,

    E is the bulk modulus of elasticity of the fluid
    rho is the fluid density

    CS
     
  8. Feb 6, 2009 #7

    russ_watters

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    I think you may have said that in a way you didn't intend - the "...directly above them..." part doesn't apply. The fluid height is measured from the level of the chosen point to the top of the tank, regardless of where horizontally those two points are wrt each other.
     
  9. Feb 6, 2009 #8

    Oh! Yes, I knew that. I must have misinterpreted what you said earlier. For example, the pressure at the top of the tube is zero, and at the bottom it'd be [tex]{\rho}{g}[/tex](10000m) if surface tension didn't play a role. Thanks anyway.



    Hmm! So I take it my domino model was pretty much right?
     
  10. Feb 6, 2009 #9

    stewartcs

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    Yeah that probably was a very poor way for me to word that! Long day at work! :smile:

    The total column height above that point (like in the attachement) is what I meant. Thanks for catching that Russ. I hate adding confusion to threads!

    Photo courtesy of: http://www.ac.wwu.edu/~vawter/PhysicsNet/Topics/Pressure/HydroStatic.html [Broken]

    CS
     

    Attached Files:

    Last edited by a moderator: May 4, 2017
  11. Feb 8, 2009 #10
    I'm no expert, but wouldn't a chimney that small cause capillary action? Due to adhesive forces I think.
     
  12. Feb 8, 2009 #11
    You are right Idoubt.This point has already been mentioned with reference to surface tension but capillary action is probably a better way of describing it.
     
  13. Nov 23, 2009 #12
    Given most other factors as assumed irrelevent for this application.

    It would be prudent to take the the column of water for that chimney as a single column of many columns with in the whole tank.

    You would then take the the area of that chimney column and find how many others of that same area would fit in the tank than take an average of all to find the net increase in tank pressure.
     
  14. Nov 23, 2009 #13

    russ_watters

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    Sorry, but that's completely wrong. Pressure works as we said: it is just p=rho*g*h.
     
    Last edited: Nov 24, 2009
  15. Nov 24, 2009 #14
    I calculate that the mass of water in the column is @ 0.078539816339744830961566084581988 grams
    at the base of the column (the surface of the enclosed sea), then, the pressure will be increased by that amount of mass divided by the total square inches of the surface covering the sea.
    (please forgive any misplaced decimals if they don't really affect the practical outcome)

    Wouldn't that put the highest pressure somewhere around half way up the column?
     
  16. Nov 24, 2009 #15
    I don't think so. If I'm not mistaken, you're using the area of the ocean to calculate pressure when you should be using the area of the chimney (pi*25nm^2).


    The purpose of this thread was to illustrate through hyperbole the intuitive paradox that pressure at a given point depends exclusively on the weight density of the fluid and the depth below the surface. I find it hard to imagine that the ocean's pressure change created by the water in this unbelievably minuscule chimney is exactly the same as that created by water in a gargantuan chimney of equal height. (Discounting forces like capillary action.)

    This is the crux of my experiment:
    I can just picture a scuba diver swimming peacefully along in the Pacific Ocean, when suddenly, someone adds a single, tiny drop of water to the tube, crushing the diver to bits.:surprised Pardon my sadism...
     
  17. Nov 24, 2009 #16
    I did the area and volume and 1 g/cc.

    You can't get PSI without putting Ps on top of SIs.
    When you add a tiny amount of mass, the sealed, enclosed horizontal surface where the little column empties represents the SI you added the minuscule fraction of a P to. At that level, every SI of the surface shares the load of the mass above, which is practically nothing in the first place.
    I bet Archimedes would eureke on that one.

    I understand that much but I don't understand where the point of highest pressure must be in that system.

    Or if I'm off, I hope somebody smarter will clear it up.
     
    Last edited: Nov 24, 2009
  18. Nov 24, 2009 #17

    russ_watters

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    You are completely wrong. The weight of the water column does not get spread out over the area of the surface of the tank.

    Again, p=rho*g*h. That's it. There is no more to it than that. The pressure is therefore highest at the bottom of the tank and the pressure is equal to the density of water times g times the vertical distance from the bottom of the tank to the top of the colum. The pressure at any point is equal to rho*g* the vertical distance between that point and the top of the colum.
     
    Last edited: Nov 24, 2009
  19. Nov 25, 2009 #18
    I think I get it now.
    It really does take that kind of pressure to lift that column.
    Thanks.
     
  20. Nov 25, 2009 #19
    Ya russ i am aware of that formula perhaps you misunderstood the statment i made.

    I use that formula for each column then apply an average to find the new pressure from the old supposedly know pressure.

    The single small tube of great height would not pruduce a massive amount of overall pressure increase it would apply to only that single column.

    And since water pushes out in all directions it would be spread throughout the whole.

    If water didnt push to the side it wouldnt run off a table.
     
  21. Nov 25, 2009 #20
    How do you figure this seems to violate conservation of energy?
     
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