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How fast will water drain from this area?

  1. Jun 17, 2011 #1

    I found this site searching for formulas that took into account gravity, volume, and pressure, to determine how fast water would drain out of a given size hole.

    In short, I'm into boats, trying to figure out given a certain size structure, how fast it would drain given certain sized water-letter-outers..

    This is my first post on this site. I'm a Computer Science guy, supposedly a math expert but when it comes to anything real world, certainly not..

    So, what I've got is an area that's (very irregular shape, but all put together) ~2 feet wide X ~2 feet deep X ~12 feet long.

    I'm trying to figure out, if filled with water, how fast it will drain with 2 X 3-inch diameter holes placed at the bottom of the "reservoir".

    Any idea where to start with this?


  2. jcsd
  3. Jun 17, 2011 #2
    Hey why don't you mail me at << e-mail address deleted by Mentor >>then we can discuss it.
    by the way you need to use the Bernoulli Equation, Equation of continuity from fluid dynamics and differential equations.But just mail me the shape of container as an image if you can bcs. it matters!!!
    Last edited by a moderator: Aug 30, 2011
  4. Jun 17, 2011 #3


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    It's best to discuss this problem here so many people can help answer your problem. Bhaskar is pretty much right in the general topics involved (execpt it uses calculus instead of D.E.), but what he didn't tell you is the problem is a common one with a known solution.

    If we assume your tank has a uniform cross-sectional area with respect to height, then the equation you're looking for is:

    [itex]t = \frac{2*A_{t}\left(\sqrt{z_{1}}-\sqrt{z_{2}}\right)}{C_{d}*A_{o}*\sqrt{2g}}[/itex]

    A_t is the cross-sectional area of the tank, A_o is the cross-sectional area of the orifice(s), and C_d is the orifice coefficient. z_1 and z_2 are the beginning and ending height of the fluid in the tank. You can probably assume C_d to be something like 0.6 t 0.7, it depends on what your holes look like.

    This should give you a good start on what you're after.
    Last edited by a moderator: Jun 17, 2011
  5. Jun 18, 2011 #4
    You will need differential equations because the pressure is directly proportional to instantaneous height.And shape matters, for different shapes pressure at bottom is different-ex. spherical,hemispherical,conical etc... and for straightforward shapes like cubical only common solutions are available.
    Last edited by a moderator: Aug 30, 2011
  6. Jun 18, 2011 #5
    Pressure at bottom depends on the shape? I don't think so...it just depends on the height of column of water directly on top of the hole. What depends on the shape is the volume of water remaining to be drained.
  7. Jun 19, 2011 #6
    Yes i know that but what if the height is varying due to hemispherical shape or any other curved shapes?Only at a single point pressure is =dgh.
  8. Jun 19, 2011 #7
    Well, 1994Bhaskar, you had me confused...on one sentence you do say that "pressure is directly proportional to the height" and in the next you say that "for different shapes pressure at the bottom is different".

    So, pressure is only proportional to the height and the equation above applies.
    Replace the letter z with the letter h, for height, and
    Lump all constant values into a single constant, K, and write the formula like this:

    t = K x At x ( sqrt(h1) - sqrt(h2) )

    (sorry, I don't know latex)

    Now, the area of the tank (At) is not constant (t is for tank not for time); instead, the tank has different cross-sectional areas at different heights...in other words, the area A is a function of the height...somebody needs to come up with a function A(h) depending on the profile along the wall of the tank.

    So, for lack of calculus abilities, I would simply take the formula given above and put it in a computer program...this is how I would approach it:

    Write a function A(h) that can be called from within a loop, decide on a dh (delta h) and evaluate the "differential" equation from initial height h1 all the way down to almost zero and add the delta times

    t = 0.0
    dh = 0.1
    for h from h1 down to dh, every dh:
    dt = K x A(h) x ( sqrt(h) - sqrt(h - dh)
    t = t + dt
    end loop

    I think that would do.


    For functions, I typically think y=f(x).
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