Chapter 2.3, Problem 6P

Suppose that a tank containing a certain liquid has an outlet near the bottom. Let $h\left(t\right)$ be the height of the liquid surface above the outlet at time $t$. Torricelli’s principle states that the outflow velocity $\nu$ at the outlet is equal to the velocity of a particle falling freely (with no drag) from the height $h$.

Show that $\nu =\sqrt{2gh}$, where $g$ is the acceleration due to gravity.

By equating the rate of outflow to the rate of change of liquid in the tank, show that $h\left(t\right)$ satisfies the equation

$A\left(h\right)\frac{dh}{dt}=-\alpha a\sqrt{2gh}$

Where $A\left(h\right)$ is the area of the cross section of the tank at height $h$ and $a$ is the area of the outlet. The constant $\alpha$ is a contraction coefficient that accounts for the observed fact that the cross section of the (smooth) outflow stream is smaller than $a$. The value of $\alpha$ for water is about $0.6$.

Consider a water tank in the form of a right circular cylinder that is $3\text{m}$ high above the outlet. The radius of the tank is $1\text{m}$ and the radius of the circular outlet is $0.1\text{m}$. If the tank is initially full of water, determine how long it takes to drain the tank down to the level of the outlet.