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Find the time until a tank with a hole is empty

by RoRi

Consider a water tank shaped like a right circular cone with its point up. Find the amount of time necessary for the tank to completely drain if there is a hole in its base. Express the time as a function of the dimension of the water tank and the area A_0 of the hole in the tank.

Incomplete.

2 comments

  1. Evangelos says:

    I could have used the aid of some good figures to explain the equations, but I’m not finding any good plotting software for the purpose of these types of exercises. Anyway, if you’re reading, I hope you can use your imagination, and possibly some pen and paper.

    We can describe the volume of the cone as the volume swept out by rotating the right triangle with (y, x) vertices.

        \begin{align*} (0, 0)\\ (0, R)\\ (H, 0)\\ \end{align*}

    around the y axis. The area of this triangle is the area of the ordinate set underneath the function x = my + b from y = 0 to y = H. We know that x(0) = R and x(H) = 0 since the base is at height 0 and the tip is at height H.

    Plugging in the above values and solving for m and b gives us the following line

        \begin{align*} x &= R - \frac{R}{H}y\\ &= R(1 - \frac{y}{H}) \end{align*}

    And the cross-sectional area at a given y from 0 to H, A(y) is the circle swept out by the height of the function at y

        \begin{align*} A(y) &= \pi [R(1 - \frac{y}{H})]^2 \\ &= \pi R^2 (1 - \frac{2y}{H} + \frac{y^2}{H^2}) \end{align*}

    Now that we know A(y), we can use equation (8.61) from section 8.27 to determine the time it would take to empty the cone (or in other words, to take the height of the water from H to 0).

        \begin{align*} \int_{H}^{0} \frac{A(y)}{\sqrt{y}} dy &= -4.8 A_{0} \int dt + C \\ \end{align*}

        \begin{align*} \int_{H}^{0} \frac{\pi R^2 (1 - \frac{2y}{H} + \frac{y^2}{H^2})}{\sqrt{y}} dy &= -4.8 A_{0} \int dt + C \\ \end{align*}

        \begin{align*} \frac{\pi R^2}{4.8 A_{0}} \int_{0}^{H} \frac{1}{\sqrt{y}} - \frac{2\sqrt{y}}{H} + \frac{y^{3/2}}{H^2} dy &= \int dt + C \\ \end{align*}

        \begin{align*} \frac{\pi R^2}{4.8 A_{0}} [2\sqrt{y} - \frac{4\sqrt{y}}{3H} + \frac{2y^{5/2}}{5H^2}]_{0}^{H} &= \frac{\pi R^2}{4.8 A_{0}} [2\sqrt{H} - \frac{4}{3}\sqrt{H} + \frac{2}{5}\sqrt{H} \\ &= \frac{\pi R^2}{4.8 A_{0}} [\frac{16}{15}\sqrt{H}] \\ & = \frac{2 \pi R^2}{9 A_{0}} \sqrt{H} \end{align*}

    As requested.

    • Evangelos says:

      I should mention, the integral on the right-hand side

          \begin{align*} \int dt + C \end{align*}

      Can be written as a bounded integral, getting rid of the constant C (since we start the clock at t = 0)

          \begin{align*} \int_{0}^{t_0} dt &= [t]_{0}^{t_0} \\ & = t_0 \end{align*}

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