Irodov 1.323: How Long Until the Tank Drains – Amit & Deepali Sinha Foundation

Problem Irodov 1.323

A cylindrical vessel of height $h$ and base area $S$ is completely filled with water. A small orifice of area $s$ , with $s \ll S$ , is opened in the bottom of the vessel and water drains out through it. Neglecting viscosity, find the time $\tau$ for all the water to drain from the vessel.

The cylindrical vessel of base area S, water height h, draining through the small base orifice of area s at the instantaneous speed v. — The cylindrical vessel of base area $S$ , water height $h$ , draining through the small base orifice of area $s$ at the instantaneous speed $v$ .

Given

$h$
$S$
$s \ll S$

Find

The time

\tau

for the vessel to empty completely

Solution

Step 1The instantaneous efflux speed

Let $x$ be the height of the water remaining at time $t$ , with $x(0)=h$ . Because $s\ll S$ , the water surface falls far more slowly than the jet moves, so at every instant the flow is effectively the steady Torricelli jet that fixed depth $x$ would drive:

v(x) = \sqrt{2gx}.

Step 2Continuity between the falling surface and the jet

The volume the falling surface vacates per unit time, $S(-dx/dt)$ , equals the volume rate leaving through the orifice, $sv(x)$ , so

-S\frac{dx}{dt} = s\sqrt{2gx}.

Separating the variables,

\dfrac{dx}{\sqrt x} = -\dfrac{s}{S}\sqrt{2g}\,dt

(1)

Step 3Integrate from full to empty

The height runs from $h$ at $t=0$ down to $0$ at the drain time $t=\tau$ , so integrating (1) over those limits,

\int_h^0 x^{-1/2}\,dx = -\frac{s}{S}\sqrt{2g}\int_0^\tau dt \;\Longrightarrow\; \Bigl[2\sqrt{x}\Bigr]_h^0 = -\frac{s}{S}\sqrt{2g}\,\tau.

The left side evaluates to $-2\sqrt h$ , giving

-2\sqrt h = -\frac{s}{S}\sqrt{2g}\,\tau \;\Longrightarrow\; \tau = \frac{2\sqrt h\,S}{s\sqrt{2g}}.

Step 4Simplify

Combining the constants under one square root,

\tau = \frac{S}{s}\cdot\frac{2\sqrt h}{\sqrt{2g}} = \frac{S}{s}\sqrt{\frac{4h}{2g}}.

\tau = \dfrac{S}{s}\sqrt{\dfrac{2h}{g}}

(2)

Check

Dimensionally $\sqrt{h/g}$ is a time and $S/s$ is dimensionless, so $\tau$ is a time, as required. And $\tau\to\infty$ as $s\to0$ (a vanishing hole never finishes draining) while $\tau\to0$ as $h\to0$ (nothing to drain), both as they must.

Remark

The drain time is the time a single particle would take to fall the height $h$ from rest under gravity, $\sqrt{2h/g}$ , scaled up by the area ratio $S/s$ . That ratio is the whole story: the vessel drains as if the entire volume $Sh$ had to be pushed, drop by drop, through the narrow orifice at the (falling) Torricelli speed, so a vessel a hundred times larger in cross section than its own hole takes a hundred times longer to empty than that single free-fall time. There is a factor of two in this problem, but it sits elsewhere: draining at the initial rate $s\sqrt{2gh}$ all the way down would clear the volume $Sh$ in $(S/s)\sqrt{h/2g}$ , exactly $\tau/2$ , so the slowing of the jet as the head falls costs the vessel precisely a doubling.