Draining a Tank

Water pours out of a hole faster when the tank is deep. As the level drops, the outflow slows, but not enough to stop: the tank still empties at a definite time. Separating the variables finds exactly when.

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How it works

Separable equations

In one sentence

A separable equation $y' = g(t)f(y)$ is solved by writing $\frac{dy}{f(y)} = g(t)\,dt$ and integrating both sides; here $f(h) = \sqrt h$ and $g = -k$ .

General formula

{\htmlData{sym=3}{\frac{dh}{dt}}} = -{\htmlData{sym=1}{k}}\sqrt{\htmlData{sym=0}{{h}}} \quad\Rightarrow\quad \int\frac{d{\htmlData{sym=0}{h}}}{\sqrt{\htmlData{sym=0}{{h}}}} = -\int {\htmlData{sym=1}{k}}\,dt \quad\Rightarrow\quad \sqrt{\htmlData{sym=0}{{h}}} = \sqrt{{\htmlData{sym=2}{h_0}}} - \frac{\htmlData{sym=1}{{k}}}{2}\,t

Deep water pushes harder, so it drains faster. As the level drops the outflow weakens, but only like $\sqrt h$ , which is why the tank still empties in finite time.

What the symbols mean

$h$
Water height
Depth of water above the hole, in meters.
$k$
Drain constant
Set by the hole's size and the tank's width; bigger drains faster.
$h_0$
Starting height
The depth when draining begins.
$\frac{dh}{dt}$
Rate of fall
How fast the level drops, in meters per minute.

Torricelli's law: water leaves the hole at speed $\sqrt{2gh}$ , so the level falls at a rate proportional to $\sqrt{h}$ .
The equation is separable: put every $h$ on one side and every $t$ on the other, then integrate both.
That gives $\sqrt{h} = \sqrt{h_0} - \frac{k}{2}t$ : the square root of the depth falls in a straight line.
The tank is empty at $t = 2\sqrt{h_0}/k$ , in finite time, unlike exponential decay.

Learn the math

Separable EquationsDiff Eq · Lesson 5, First Order Differential Equations