Playground
A tank with a hole at the bottom; water stream exit velocity depends on height. Adjust water level to see velocity change.
Variables
| Symbol | Name | SI | Dimension | Range |
|---|---|---|---|---|
| Efflux Velocityoutput Speed of fluid leaving the orifice | m/s | L·T⁻¹ | 0 – 50 | |
| Gravity Gravitational acceleration | m/s² | L·T⁻² | 9.8 – 9.8 | |
| Head Height Height of fluid surface above orifice | m | L | 0 – 100 |
Deep dive
Derivation
Apply Bernoulli between the free surface (point 1, large area, v₁ ≈ 0, P₁ = P_atm, height h above orifice) and the orifice (point 2, P₂ = P_atm, height 0, velocity v): P_atm + 0 + ρgh = P_atm + ½ρv² + 0. Pressure cancels, gh = ½v², giving v = √(2gh) — identical to free-fall from height h.
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 ~62% of orifice area) and viscous losses — the discharge coefficient C_d ≈ 0.6.
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 ~60% of A·v
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
Worked examples
Hole 1 m below surface
Given:
- g:
- 9.8
- h:
- 1
Find: v
Solution
v = √(2 × 9.8 × 1) = √19.6 ≈ 4.43 m/s
Drain hole at the bottom of a 5 m tank
Given:
- g:
- 9.8
- h:
- 5
Find: v
Solution
v = √(2 × 9.8 × 5) = √98 ≈ 9.9 m/s
Scenarios
What if…
- scenario:
- What if the hole is actually quite large (not small)?
- answer:
- The free-surface velocity is no longer ≈ 0 and you must keep both terms in Bernoulli, giving v = √(2gh / (1 - (A_2/A_1)²)). For tiny holes the correction is negligible.
- scenario:
- What if the tank is pressurized above the water?
- answer:
- Add the gauge pressure as an extra head: v = √(2(gh + ΔP/ρ)). A pressurized tank shoots water out faster than gravity alone would predict.
- scenario:
- What if the fluid is very viscous (like syrup)?
- answer:
- Torricelli over-predicts. Viscous drag at the orifice walls reduces v significantly — use Poiseuille-type flow instead of inviscid Bernoulli.
Limiting cases
- condition:
- h → 0
- result:
- v → 0
- explanation:
- No head, no flow.
- condition:
- h = 5 m
- result:
- v ≈ 9.9 m/s
- explanation:
- Same as the speed of an object dropped from 5 m — Torricelli is free-fall in disguise.
- condition:
- h → ∞
- result:
- v → ∞
- explanation:
- Idealized — viscosity, vena contracta, and air drag eventually limit real efflux.
Context
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.
Hook
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.
Dimensions: [√(gh)] = √((L·T⁻²)(L)) = √(L²·T⁻²) = L·T⁻¹ = [v] ✓
Validity: Inviscid flow, large tank (surface velocity ≈ 0), atmospheric pressure on both sides, hole small compared to tank cross-section.