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Pressure on tank walls

  1. Jan 28, 2013 #1
    Hi all,

    A problem pops up while I was trying to calculate the pressure impacts on my aquarium tank walls.

    The first formula comes up to my mind was: p = d*h

    where d is pressure at one point inside the liquid
    d is density of the liquid (assume it's pure water)
    h is the height or distance from that point to liquid's surface.

    However, the formula DOES NOT take into account the WIDTH and LENGTH of the tank. To illustrate my problem, i make a comparison between my tank and a test tube.

    Assume I want to build a aquarium tank (10 x 10 x 10 meters ) and a test tube with the same height 10 meters but with normal radius (about 1 cm) just like normal test tubes. Measurement system does not matter here, the importance is that they have the same height. So if I want to build a tank that's (10x10x10) I probably need to use 3-centimeter-thick-toughened glass in order to withstand the pressure created by that amount of water. However, with the 10 meters high test tube, I just could use a normal glass with much less thin (let's say 1-centimeter-thick is enough right?).

    So my point is eventhough both have the same height but the aquarium tank requires much stronger glass, compared to the test tube does,to withstand the maximum capacity of water contained. Therefore, I believe the Width and Length ( or simply surface's area) should come in somewhere in the calculation.

    Can anyone help me point out where i went wrong or what i have missed out?

    Many thanks in advance!!!
     
  2. jcsd
  3. Jan 28, 2013 #2

    SteamKing

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    The pressure at any point in a static fluid depends only on the density of the fluid and the depth of the fluid above the point of interest. Remember, pressure has units of force / area.

    However, if you want to calculate the total force acting on one end of the aquarium, then you must know something about the area of the end of the aquarium, or the side, or whatever. The force can be determined by integrating the pressure gradient over the area of interest of the aquarium.

    In other words, the pressure at a depth of 10 meters does not differ whether the water is in a long thin tube or acting on a broad rectangular surface.
     
  4. Jan 28, 2013 #3
    In fact, my house has a aquarium tank about (3 x 3 x 1 meters) = (H x L x W) made of 3 cm thick toughened glass. When water is pump in at 75% capacity, the side glass is clearly curved. I just could not believe.
     
  5. Jan 28, 2013 #4
    Last edited by a moderator: May 6, 2017
  6. Jan 28, 2013 #5

    jbriggs444

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    This is a formula for pressure as a function of depth when using a "gravitational" system of units such as the U.S. traditional system. In such a system the distinction between force and mass is blurred.

    For instance, pressure (p) in pounds per square foot is the product of density (d) in pounds per cubic foot and height (h) in feet.

    This is a formula for pressure as a function of depth using a "coherent" system of units such as SI where a distinction is made between your mass and the force required to support your weight.

    For instance, pressure (p) in Pascals is the product of height (h) in meters, density (rho) in kilograms per cubic meter and gravity (g) in meters per second squared.


    http://00.edu-cdn.com/files/static/mcgrawhill-images/9780071393089/e0248-02.jpg [Broken]

    This is not a formula for "pressure". It is a formula for total force on the side of an aquarium in a gravitational system of units (note how the acceleration of gravity is not used).

    The force F on the [vertical] side of an aquarium is given by the integral of the force on each horizontal strip from top (a) to bottom (b).

    The force on each horizontal strip is given by the fluid pressure at the depth of that strip multiplied by the area of the strip.

    The pressure at the depth of the strip is computed by the first formula above [pressure = height (h) times density (rho)].

    The area of each strip is given by its width (w(h)) multiplied by its incremental height (dh).

    The article in which this appeared was using height and width in feet, density in pounds per cubic foot and was computing total force in pounds.
     
    Last edited by a moderator: May 6, 2017
  7. Jan 28, 2013 #6
    It's clear now. Thanks you a alot !!!
     
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