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Math Help - liquid draining from tank

  1. #1
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    liquid draining from tank

    A cylindrical tank with length of 5 ft. and radius 3 ft. is situated with its axis horizontal. If a circular bottom hole with a radius of 1 in. is opened and the tank is initially half full of xylene, how long will it take for the liquid to drain completely?

    Please help me, I am lost???
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    Quote Originally Posted by mandy123 View Post
    A cylindrical tank with length of 5 ft. and radius 3 ft. is situated with its axis horizontal. If a circular bottom hole with a radius of 1 in. is opened and the tank is initially half full of xylene, how long will it take for the liquid to drain completely?

    Please help me, I am lost???
    this is an application of Toricelli's principle ... the speed at which a fluid exits a hole is a function of the height of the fluid above the opening.

    v = \sqrt{2gh}

    further, the rate of change of volume w/r to time has the relationship

    \frac{dV}{dt} = -av , where a is the area of the opening.

    let y = 0 be at the tank's center. the volume of fluid in the tank as a function of the fluid's height above the opening ...

    V = \int_{-3}^{-3+h} 10\sqrt{9 - y^2} \, dy

    \frac{dV}{dt} = 10\sqrt{9 - (-3+h)^2} \cdot \frac{dh}{dt}

    -a\sqrt{2gh} = 10\sqrt{6h-h^2} \cdot \frac{dh}{dt}

    -\frac{a\sqrt{2g}}{10} \, dt = \sqrt{6-h} \, dh

    -\frac{a\sqrt{2g}}{10} t + C = -\frac{2}{3}(6-h)^{\frac{3}{2}}

    at t = 0, h = 3 ... C = -\frac{2}{3}(3)^{\frac{3}{2}}

    when h = 0 ...

    t = \frac{20}{3a\sqrt{2g}}\left[(6)^{\frac{3}{2}} - (3)^{\frac{3}{2}}\right]

    a = \frac{\pi}{144} \, ft^2

    g = 32 \, \frac{ft}{s^2}

    t \approx 363 \, s or a little over 6 min.

    hope I didn't commit any egregious errors ... I'm sure someone will let us both know if I did.
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  3. #3
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    Thank you so much, I thought it had something to do with torricelli's law, but I had no clue as where to start, but you explained it very well, and it makes more sense!!! Thanks!!!
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