Torricelli's Law

Equivalent forms

v = \sqrt{2 g (h_1 - h_2)}

Bernoulli's equation collapsed into a free-fall analogy.

Symbol	Name	SI unit	Dimension	Range
$v$	Efflux Velocityoutput Speed of fluid leaving the orifice	m/s	L·T⁻¹	0 – 50
$g$	Gravity Gravitational acceleration	m/s²	L·T⁻²	9.8 – 9.8
$h$	Head Height Height of fluid surface above orifice	m	L	0 – 100

Unit systems

Symbol	SI	CGS	Imperial
$v$	m/s	cm/s	ft/s
$g$	m/s²	cm/s²	ft/s²
$h$	m	cm	ft

Where it holds

Inviscid flow, large tank (surface velocity

\approx 0)

, atmospheric pressure on both sides, hole small compared to tank cross-section.

Dimensional analysis

$[\sqrt{gh}] = \surd ((L\cdot T^{-2})(L)) = \sqrt{L^{2}\cdot T^{-2}} = L\cdot T^{-1} = [v] \checkmark$

Discovery

Evangelista Torricelli · 1643

Derived as a special case of energy conservation almost a century before Bernoulli formalized the general principle. Torricelli also invented the mercury barometer.

Try this

How fast does water shoot from a hole in a tank?

A water tank has a small hole 1 m below the water surface. Find the speed at which water exits the hole.

Research status: stable

Real-world applications

Draining tanks and reservoirs
Water clocks (clepsydras) used by ancient civilizations
Garden sprinkler design
Dam spillway calculations
Estimating hull-breach flooding rates in ships

Common misconceptions

The efflux speed does NOT depend on the fluid's density (it cancels out)
It does NOT depend on the orifice area (only the head height)
Real jets are narrower than the hole (vena contracta) — the actual flow rate $is \sim 60$ % of $A\cdot v$

Experimental verification

Verified by Torricelli using water jets from holes at varying depths. Modern PIV measurements show real efflux velocities are slightly less due to vena contracta (jet narrows

to \sim 62

% of orifice area) and viscous losses — the discharge coefficient

C_d \approx 0.6

.

Derivation

Apply Bernoulli between the free surface (point 1, large area,

v_{1} \approx 0

,

P_{1} = P_atm

, height h above orifice) and the orifice (point 2,

P_{2} = P_atm

, height 0, velocity v):

P_atm + 0 + \rho gh = P_atm + \tfrac{1}{2}\rho v^{2} + 0

.

Pressure cancels,

gh = \tfrac{1}{2}v^{2}

, giving

v = \sqrt{2gh}

— identical to free-fall from height h.

Limiting cases

h \to 0

⟶

v \to 0

No head, no flow.

h = 5\,\mathrm{m}

⟶

v \approx 9.9\,\mathrm{m/s}

Same as the speed of an object dropped from 5 m — Torricelli is free-fall in disguise.

h \to \infty

⟶

v \to \infty

Idealized — viscosity, vena contracta, and air drag eventually limit real efflux.

What if…

What if the hole is actually quite large (not small)?

The free-surface velocity is no longer $\approx 0$ and you must keep both terms in Bernoulli, giving $v = \surd (2gh / (1$ - (A_2/A_ $1)^{2}))$ . For tiny holes the correction is negligible.

What if the tank is pressurized above the water?

Add the gauge pressure as an extra head: $v = \surd (2(gh + \Delta P/\rho ))$ . A pressurized tank shoots water out faster than gravity alone would predict.

What if the fluid is very viscous (like syrup)?

Torricelli over-predicts. Viscous drag at the orifice walls reduces v significantly — use Poiseuille-type flow instead of inviscid Bernoulli.

1

Hole 1 m below surface

Given ·

g:: 9.8
h:: 1

Find · v

Steps

Apply $v = \sqrt{2gh}$
$v = \sqrt{2 \times 9.8 \times 1} = \sqrt{19.6}$
$v \approx 4.43\,\mathrm{m/s}$

Result ·

v = \sqrt{2 \times 9.8 \times 1} = \sqrt{19.6} \approx 4.43\,\mathrm{m/s}

2

Drain hole at the bottom of a 5 m tank

Given ·

g:: 9.8
h:: 5

Find · v

Steps

$v = \sqrt{2gh} = \sqrt{98}$
$v \approx 9.9\,\mathrm{m/s}$
Same as a ball dropped from 5 m

Result ·

v = \sqrt{2 \times 9.8 \times 5} = \sqrt{98} \approx 9.9\,\mathrm{m/s}

Bernoulli's Equation→Continuity Equation→