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B Water pressure in a sphere

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  1. Mar 22, 2017 #1
    I've been wondering how I can find the pressure water exerts on a sphere it's in.
    As for cyllinders and pyramids, and prisms - the thing is extremely easy - you just take the one formula and put in numbers.

    What about a situation when we have a sphere whose radius is R, and it's half full with water.
    Could you explain to me in simple terms how to find this pressure?
    If it's not possible to explain it in an easy way, then please use calculus.
     

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  3. Mar 22, 2017 #2
    What's in the other half of the sphere? Air? That's going to matter a lot.

    Try to list out all the parameters that you think might be important.
     
  4. Mar 22, 2017 #3

    russ_watters

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    Since fluid pressure does not depend on the container shape, I'm curious as to what that "one formula" is and how you are or aren't using it...
     
  5. Mar 22, 2017 #4
    Khashishi,
    Yes, it's air.
    I think that this pressure will depend on the radius of the sphere, the density of the liquid that is in the sphere and the weight of the liquid.

    russ_watters,
    The formula I've been using is this one:
    [tex]p = \rho gh[/tex] , where [tex]\rho[/tex] is the density of the liquid.
     
  6. Mar 22, 2017 #5
    What about the pressure of the air?
     
  7. Mar 22, 2017 #6
    [tex]p = \frac{m_{water}g}{2\pi R^{2}} +\frac{m_{air}g}{2\pi R^{2}} [/tex]
    Is it close to any extent?
     
  8. Mar 22, 2017 #7
    Many errors there.
    Air pressure doesn't work like that. Take a look at the ideal gas law.
    Why did you change ##h## in your equation to ##\frac{1}{2\pi R^2}##?

    It's not even clear if you are asking the right question. Are you looking for the wall stress? That's not the same as the pressure.
     
  9. Mar 22, 2017 #8
    Pressure = Force / Area, thus I assumed the force we need to take into consideration is the weight of the water in the sphere and the area is half of the sphere's area.
    And yes, I am looking for the wall stress. Sorry for having confused you with pressure.
     
  10. Mar 22, 2017 #9
    Is this a real world problem? (Is this a problem on Earth?)

    If your sphere is airtight, then you could have a different air pressure inside and outside the sphere. In which case, the difference in air pressure is probably going to be the dominant source of pressure, unless your sphere is quite large (so that the water is quite deep inside the sphere). For instance, (Earth) surface air pressure is equal to the pressure of about 10 meters depth of H2O (assuming Earth gravity). So, if your sphere is 20 meters tall, and half filled with air at surface pressure, then the pressure in the middle of the sphere is about half of the pressure at the bottom (under 10 meters of water).

    Now, if you have some kind of valve that allows air pressure to equalize inside and outside, then you can mostly ignore the air for the purposes of calculating sphere wall stresses, because the internal and external pressures cancel out. The water pressure at any point in the sphere is just ##\rho gh## where ##h## is the depth of the point below the surface.

    The average wall stress can be estimated by integrating over the 3/2 times the pressure of the interior over the full volume of the sphere, and then dividing by the volume of the wall. The wall stress is larger than the interior pressure, because there is tension in the wall.

    See also https://en.wikipedia.org/wiki/Cylinder_stress#Thin-walled_assumption

    edit: you have to multiply the pressure by 3/2, since the pressure is force in 3 directions, and the wall tension is mainly in two directions.
     
    Last edited: Mar 22, 2017
  11. Mar 22, 2017 #10
    So, for the case where air pressure cancels out, we have
    ##p = \rho gh##
    Integrate over volume by cutting the sphere into horizontal slices. The slice at depth ##h## has a volume of ##\pi h^2 dh##
    ##\frac{3}{2} \int p dV = \frac{3}{2}mg \int_{-R}^{0} \pi h^2 (-h) dh##
    ##=\frac{3}{8} \rho g \pi R^4##
    Now divide by volume of the wall, which is just ##4\pi R^2 t##, where t is the thickness.
    ##\sigma = \frac{3}{32t} \rho g R^2##
    where ##\sigma## is the average wall tension.
     
  12. Mar 22, 2017 #11
    Thank you, Khashishi.
    You have solved my problem entirely.
     
  13. Mar 23, 2017 #12

    Nidum

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    Sorry but there seem to be some serious flaws in this analysis .

    Perhaps @Khashishi would like to have another look at the problem ?
     
  14. Mar 23, 2017 #13

    russ_watters

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    I agree. This looks to me like the OP is looking for water pressure in the tank. It's just air pressure plus hydrostatic pressure.
     
  15. Mar 23, 2017 #14

    russ_watters

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    Why do you think pressure depends on geometry for a sphere if it doesn't for other shapes?
    [/quote]
    russ_watters,
    The formula I've been using is this one:
    [tex]p = \rho gh[/tex] , where [tex]\rho[/tex] is the density of the liquid.[/QUOTE]
    That is the correct formula. You just need to add the air pressure over the water, if applicable.
     
  16. Mar 23, 2017 #15
    In post 8 they said they want the wall stress.
     
  17. Mar 23, 2017 #16

    russ_watters

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    Nothing of what the OP said implied he needed wall stress until that, and they didn't volunteer it, you did. Since you are the one who introduced the term and the OP seemed pretty clear to me to be about fluid pressure, I'm not convinced the OP knows what "wall stress" means. They are very different things, but I could see someone unfamiliar with the term incorrectly thinking that the pressure exerted against the wall is "wall stress".
     
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