Worked exercise — Siphon from a bottle

Problem

A bottle of water is turned upside down into a flexible little tube resting on the table. The free surface of the water in the bottle is at $h_1 = 30\;\text{cm}$ above the outlet of the tube. At what speed does the water leave the tube? (The bottle is large enough that the speed at the surface is negligible; viscous friction is ignored.)

Solution

Step 1 — Bernoulli’s theorem. We apply it between the free surface (point 1, speed $\approx 0$, atmospheric pressure) and the outlet of the tube (point 2, speed $v$, atmospheric pressure). The atmospheric pressures cancel and what remains is the balance between potential and kinetic energy per unit volume:

$\rho g h_1 = \tfrac{1}{2}\,\rho v^2$

Step 2 — Solving for the speed. The density cancels out:

$v = \sqrt{2 g h_1} = \sqrt{2\cdot 9{,}8\cdot 0{,}3} \approx 2{,}4\;\text{m/s}$

$\ev{v \approx 2{,}4\;\text{m/s}}$

This is the same speed a body would have if dropped from $30\;\text{cm}$: it is Torricelli’s theorem, a special case of Bernoulli’s. Gravity converts potential energy into kinetic energy for both a stone and a fluid.

Linked atoms

Argomenti: Fluid statics and dynamics Concetti: Bernoulli’s theorem