Lessons · fluids

A draining tank: why the jet speed depends only on the depth

Regime 2 — calculus does what algebra cannot. The acceleration isn't constant, so the algebra formulas don't apply; calculus is the only road in, and SymPy proves the closed form solves the equation of motion.

Regime 2 · calculus does more Math machine-derived & checked by SymPy 2 modeling assumptions (author-asserted)

Water drains from an orifice a depth $h = 5$ m below the surface of a tank. A parcel of water falling from the surface trades potential energy for kinetic energy, and leaves the orifice at $v = \sqrt{2gh}$ — exactly the speed it would reach falling freely through $h$ . Drag the cursor down from the surface: the KE and PE bars (per unit volume) trade while the Total bar stays flat. The efflux speed is set by the depth alone, not by the width of the tank or how much water sits above.

Efflux speed

v = \sqrt{2gh}

10 m/s

Total energy density

\rho g h

50000 Pa

KE density at half depth

25000 Pa

Speed at half depth

\sqrt{gh}

7.071 m/s

Energy conservation along a streamline (Bernoulli) is the first integral of

F = ma

\tfrac12\rho v^2 + \rho g y = \text{const}

For a parcel of fluid following a streamline, Newton's second law integrates to the statement that pressure, kinetic energy density, and potential energy density trade so their sum is constant. With the surface and the orifice both at atmospheric pressure, the kinetic energy gained equals the potential energy lost, $\tfrac12\rho v^2 = \rho g h$ .

check $\tfrac{d}{dd}(\text{KE}+\text{PE})=0$ ; check $\tfrac12\rho v^2+\rho g y=\rho g h$ ; show $v=\sqrt{2gd}$ ; orifice speed $\sqrt{2gh}$

✓ The total energy per unit volume does not change with depth: $\tfrac{d}{dd}(\text{KE}+\text{PE}) = 0$ . [structural]
✓ Bernoulli along the streamline: $\tfrac12\rho v^2 + \rho g y = \rho g h$ is constant as the parcel falls. [structural]
✓ The speed falls out of the kinetic energy: $\tfrac12\rho v^2 = \text{KE}$ , so $v = \sqrt{2gd}$ . [structural]
✓ At the orifice ( $d = h$ ) the efflux speed is Torricelli's law $v = \sqrt{2gh}$ — the same as a body that fell freely through $h$ . [structural]

Dimensional homogeneity: checked by SymPy (holds).

✗ Common misconception: “A taller column of water, or a wider tank, pushes the water out of the orifice faster.”

Only the depth of the orifice below the surface sets the jet speed: $v = \sqrt{2gh}$ . A wide tank and a narrow pipe at the same depth eject water at the same speed, and the amount of water sitting above the orifice does not matter — just as a block's speed at the bottom of a ramp depends only on the height dropped, not the ramp's shape. It is the same energy conservation: the potential energy density $\rho g h$ a parcel loses falling to the orifice becomes kinetic energy density $\tfrac12\rho v^2$ , and the density $\rho$ cancels. Drag the cursor and read the speed off the kinetic-energy bar — it tracks the depth, nothing else.

Modeling assumptions — author-asserted, disclosed not discharged

An ideal fluid (incompressible, no viscosity) draining slowly, so the surface level barely moves and the surface and jet are both at atmospheric pressure; energy is conserved along a streamline (Bernoulli).
$g = 10$ m/s² as a magnitude, with depth $d$ measured downward from the surface; the parcel starts essentially at rest at the surface.

The stacked graph, fully annotated

A static rendering (Matplotlib) at the default parameters — the interactive version is in the Graph tab above.

Formulas used

Hover a formula to preview its reference entry; click to open it in the reference (or the concept graph):