Home » Blog » Find the time required for the water level in a tank to drop 1 foot

Find the time required for the water level in a tank to drop 1 foot

Consider a water tank with vertical sides and with cross-section a square of area 4 feet. Water exits the tank through a hole which as area equal to \frac{5}{3} square inches. The water level is initially 2 feet above the level of the hole. Find the time required for the water level to drop 1 foot.


One comment

  1. Evangelos says:

    Tom gave us a nice little formula to use for these leaky tank problems, as long as we get our setup down right, it’s simply a matter of evaluating the integrals.

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

    Or as I like to set it up

        \begin{align*} \int_{y_0}^{y_1} \frac{A(y)}{\sqrt{y}} dy &= -4.8 A_{0} \int_{t_0}^{t_1} dt \end{align*}

    With y being the height of the water, A(y) being the cross-sectional area of the tank at height y, and A_0 being the area of the hole in the bottom of the tank. We are given the areas A(y) and A_0, along with the values of y_0, y_1, and t_0 as follows

        \begin{align*} A(y) &= 4 ft^2 \\ A_0 &= \frac{5}{3} in^2\\ y_0 &= 2 ft\\ y_1 &= 1 ft\\ t_0 &= 0 s\\ \end{align*}

    And the problem asks us to find the value of t_1 that satisfies the equation.

    But, before we set up the integral, we need to do a unit conversion to match the unit of area on both sides. As such, I changed the units of A_0 from square inches to square feet

        \begin{align*} A_0 &= \frac{5}{3} in^2\\ & = \frac{5}{432} ft^2 \end{align*}

    Now, to set up the integrals

        \begin{align*} 4 ft^2 \int_{2}^{1} \frac{dy}{\sqrt{y}} &= -4.8 (\frac{5}{432} ft^2) \int_{0}^{t_1} dt \end{align*}

    Some rearranging and simplifying gives us

        \begin{align*} 4 ft^2 \int_{1}^{2} \frac{dy}{\sqrt{y}} &= \frac{24}{432} ft^2 \int_{0}^{t_1} dt \end{align*}

    Evaluating the integrals on both sides gives us

        \begin{align*} 8 ft^2 [\sqrt{y}]_{1}^{2} = \frac{24}{432} ft^2 [t]_{0}^{t_1} s \end{align*}

        \begin{align*} 8 ft^2 [\sqrt{2} - 1] = \frac{24}{432} ft^2 [t_1] s \end{align*}

        \begin{align*} 144 [\sqrt{2} - 1] =  [t_1] s \end{align*}

    Plugging these numbers into a calculator gives us

        \begin{align*} t_1 \approx 59.6 s \end{align*}

Point out an error, ask a question, offer an alternative solution (to use Latex type [latexpage] at the top of your comment):