Question

A hemispherical tank of radius 2 m is initially full of water and has an outlet of 12${cm}^2$ cross-sectional area at the bottom. The outlet is opened at some instant. The flow through the outlet is according to the law v (t) = $0.6\sqrt{2gh\left(t\right)}$, where v (t) and h (t) are respectively the velocity of the flow through the outlet and the height of water level above the outlet at time t and g is the acceleration due gravity. Fine the time it takes to empty the tank.

Answer 1

Let at time $t$ the depth of water is $h$, the radius of water surface is $r$
Then, $r^2=R^2-(R-h{)}^2$
$\Rightarrow r^2=2Rh-h^2$
Now, if in time $dt$ the decrease in water level is $dh$, then
$-\pi r^{2}dh=0.6\sqrt{2gh}.adt$
($a$ is cross-sectional area of the outlet)
$\Rightarrow \frac{-\pi }{\left(0.6\right)a\sqrt{2}g}(2Rh-h^2)\frac{dh}{\sqrt{h}}=dt$
$\Rightarrow \frac{\pi }{\left(0.6\right)a\sqrt{2}g}{\int }_R^0(h^{3/2}-2R{h}^{1/2})dh={\int }_0^{1}dt$
$\Rightarrow \frac{\pi }{\left(0.6\right)a\sqrt{2}g}{[\frac{2}{5}{h}^{3/2}-\frac{4}{3}R{h}^{3/2}]}_R^0=t$
$\Rightarrow t=\frac{\pi }{\left(0.6\right)a\sqrt{2}g}[0-R^{5/2}(\frac{2}{4}-\frac{4}{3})]=\frac{7\pi \times {10}^5}{135\sqrt{g}}$