Chapter 5, Problem 15P

As depicted in Fig. P5.15, the velocity of water, $v\left(\text{m/s}\right)$, discharged from a cylindrical tank through a long pipe can be computed as

$v=\sqrt{2gH}\text{tanh}\left(\frac{\sqrt{2gH}}{2L}t\right)$

FIGURE P5.15

where $g=9.81$ ${\text{m/s}}^2,H=$ initial head $\left(\text{m}\right)$, $L=$ pipe length $\left(\text{m}\right)$, and $t=$ elapsed time $\left(\text{s}\right)$. Determine the head needed to achieve $v=5\text{ m/s in 2}\text{.5 s for a 4-m-long pope}$(a) graphically, (b) by bisection, and (c) with false position. Employ initial guesses of $x_l=0\text{ and }{x}_u=2$ m with a stopping criterion of ${\epsilon }_s=1\%$. Check you results.