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Thread: DFQ: "A vertical conical tank..."

  1. #1
    Nov 2008

    DFQ: "A vertical conical tank..."

    Hey guys.

    I've spent probably 6 hours today trying to solve a basic differential equation, without success, and I'm about to go crazy.

    The problem:

    "A vertical conical tank, point down, with a radius of 0.5 meters, and a height of 3 meters, is filled with water. How long does it take to drain the tank through a small hole in the bottom of the tank, when the water level has sunk by 1.5 meters after 45 minutes?"

    The answer is supposed to be ~55 minutes.

    I'm not gonna do the entire thing here, but here is general way of how I tried solving it:

    1) Volume of a cone is $\displaystyle V = \frac{\pi * r^2 * h}{3}$ => $\displaystyle V(t) = \frac{\pi * r^2}{3} * h(t)$
    2) General solution to a "draining tank through a small hole" is $\displaystyle \frac{dV}{dt} = -k * \sqrt{h}$
    3) Rearranging and finding the derivative, getting $\displaystyle \frac{\pi * r^2}{3} * \frac{dh}{\sqrt{h}}= -k dt$
    4) Integrating both sides = $\displaystyle \frac{2\pi * r^2}{3} * \sqrt{h} = -kt+C$
    5) Finding an expression for h(t): $\displaystyle h(t) = ((-kt+C) * \frac{3}{2\pi*r^2})^2$
    6) Using t = 0 to find C, and then the information about the water level after 45 minutes, to find k.
    7) Finding the time when h(t) = 0 (tank is empty). Standard second degree equation.
    8) Getting the wrong answer. All but one time, I've gotten t = 2.56 hours, which is completely wrong.

    Anyone able to see if there's anything wrong with that? Or am I just making some sort of fluke error while calculating? Anyone able to get it right (~55 minutes)?

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  2. #2
    MHF Contributor
    skeeter's Avatar
    Jun 2008
    North Texas
    your mistake is treating r as a constant ... it's not.

    I arrived at 54.663 min, but hear this ... this problem has some ugly constants.

    $\displaystyle r = \frac{h}{6}$

    $\displaystyle V = \frac{\pi}{3}\left(\frac{h}{6}\right)^2 h$

    $\displaystyle V = \frac{\pi}{108} h^3$

    $\displaystyle \frac{dV}{dt} = \frac{\pi}{36} h^2 \frac{dh}{dt}$

    $\displaystyle -k\sqrt{h} = \frac{\pi}{36} h^2 \frac{dh}{dt}$

    $\displaystyle -\frac{36k}{\pi} \, dt = h^{\frac{3}{2}} \, dh$

    $\displaystyle -\frac{36k}{\pi} t + C = \frac{2}{5} h^{\frac{5}{2}}$

    when t = 0 , h = 3 ...

    $\displaystyle C = \frac{2}{5} 3^{\frac{5}{2}}$

    when t = 45 , h = 1.5

    solve for k

    I'll leave you to work out the rest ... like I said C and k are two ugly constants. I stored their values in my calculator and found that h = 0 at about t = 55 min.
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  3. #3
    Nov 2008
    Ah...! Thanks a lot man, you just saved my hair from being viciously ripped out.
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