# Time for the tank to be empty

A hemispherical tank of radius 2 metres is initially full of water and has an outlet of 12 cm$^2$ cross-sectional are at the bottom. The outlet is opened at some instant.

The flow through the outlet is according to the law $v(t) = 0.6 \sqrt{2gh(t)}$, 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 to gravity. Find the time it takes to empty the tank.

I know to write a differential equation by relating the decrease of water level to the outflow in order to solve. But don't know how and as we are not not given the initial height how can we solve it.

The volume $V$ of a sperical cap of height $h$ in a sphere of radius $r$ is $$V=\pi \left(h^2r-\frac{h^3}{3}\right).$$ The flow satisfies the differential equation $$\frac{dV}{dt}=-S\cdot v(t)$$ where $S$ is the area of the hole at the bottom and $v(t)=0.6 \sqrt{2gh(t)}$ the velocity of the flow through the hole. Hence, we obtain $$\frac{dV}{dt}=\pi \left(h^2r-\frac{h^3}{3}\right)'=\pi(2hr-h^2)h'=-0.6 S\sqrt{2gh}.$$ Now we separate the variables and integrate $$\pi\int_0^{r} \frac{2hr-h^2}{\sqrt{h}}\,dh=0.6 S\sqrt{2g}\cdot T \Rightarrow T=\frac{14\pi r^{5/2}}{15\cdot 0.6 S\sqrt{2g}}$$ where we used the conditions $h(0)=r$ (the hemisphere is full) and $h(T)=0$ (the hemisphere is empty).
• @ user123733 You mean $\frac{dV}{dt}=-S\cdot v(t)$? The term $S\cdot v(t)$ is how much water flow out in one unit of time. Therefore is the negative rate of change of the volume of water inside the tank. Is it clear now? – Robert Z Oct 31 '16 at 8:00
• $V(t)$ is the volume of water inside the cap, $v(t)$ is the velocity of the flow. For the equation after 'Hence" see my edited answer. – Robert Z Nov 1 '16 at 7:05