The Roman natural philosopher, Pliny the Elder, purportedly had an intermittent fountain in his garden. As in Fig. 23.12, water enters a cylindrical tank at a constant flow rate [latex]Q_{in}[/latex] and fills until the water reaches [latex]y_{high}[/latex]. At this point, water siphons out of the tank through a circular discharge pipe, producing a fountain at the pipe’s exit. The fountain runs until the water level decreases to [latex]y_{low}[/latex], whereupon the siphon fills with air and the fountain stops. The cycle then repeats as the tank fills until the water reaches [latex]y_{high}[/latex], and the fountain flows again. When the siphon is running, the outflow [latex]Q_{out}[/latex] can be computed with the following formula based on Torricelli’s law: [latex]Q_{\text {out }}=C \sqrt{2 g y \pi r^2}[/latex] (23.28) Neglecting the volume of water in the pipe, compute and plot the level of the water in the tank as a function of time over 100 seconds. Assume an initial condition of an empty tank y(0) = 0, and employ the following parameters for your computation: [latex]\begin{array}{lll} R_T=0.05 \mathrm{~m} & r=0.007 \mathrm{~m} & y_{\text {low }}=0.025 \mathrm{~m} \\ y_{\text {high }}=0.1 \mathrm{~m} & C=0.6 & g=9.81 \mathrm{~m} / \mathrm{s}^2 \\ Q_{\mathrm{in}}=50 \times 10^{-6} \mathrm{~m}^3 / \mathrm{s} && \\ \end{array}[/latex]

Question 23.CS.5: The Roman natural philosopher, Pliny the Elder, purportedly ...

Applied Numerical Methods with MATLAB for Engineers and Scientists

The Roman natural philosopher, Pliny the Elder, purportedly had an intermittent fountain in his garden. As in Fig. 23.12, water enters a cylindrical tank at a constant flow rate $Q_{in}$ and fills until the water reaches $y_{high}$ . At this point, water siphons out of the tank through a circular discharge pipe, producing a fountain at the pipe’s exit. The fountain runs until the water level decreases to $y_{low}$ , whereupon the siphon fills with air and the fountain stops. The cycle then repeats as the tank fills until the water reaches $y_{high}$ , and the fountain flows again.
When the siphon is running, the outflow $Q_{out}$ can be computed with the following formula based on Torricelli’s law:

$Q_{\text {out }}=C \sqrt{2 g y \pi r^2}$ (23.28)

Neglecting the volume of water in the pipe, compute and plot the level of the water in the tank as a function of time over 100 seconds. Assume an initial condition of an empty tank y(0) = 0, and employ the following parameters for your computation: