Case 1
- Fluid height above the opening h
- 2m
- Gravitational acceleration g
- 9.81m/s²
6.2642m/s
Open with these valuesTorricelli Velocity Calculator
Change any value to update the result.
6.2642m/s
Efflux velocity: 6.2642 m/sExamples
How the result moves
Gravitational acceleration g (m/s²) → Efflux velocity (m/s)Your current valid inputs, with one value varied across the range.
How the result moves
Held fixed: Fluid height above the opening h 2.000000 m.
| Gravitational acceleration g (m/s²) | Efflux velocity (m/s) |
|---|---|
| 5.00000 | 4.4721 |
| 7.50000 | 5.4772 |
| 9.81000Your value | 6.2642 |
| 10.00000 | 6.3246 |
| 12.50000 | 7.0711 |
Fluid leaving a hole h below the surface jets out at v = √(2gh) — exactly the speed a stone reaches falling that same height. A 2 m head on Earth gives 6.26 m/s. Density does not appear: water and mercury leave the same head at the same speed.
Worked examples
Case 2
- Fluid height above the opening h
- 10m
- Gravitational acceleration g
- 9.81m/s²
14.0071m/s
Open with these valuesCase 3
- Fluid height above the opening h
- 20m
- Gravitational acceleration g
- 1.62m/s²
8.0498m/s
Open with these valuesHow it's calculated
v = √(2 × g × h)
- StepMeasure straight down from the free surface to the centre of the opening.Fluid height above the opening h=2m
- StepLeave gravity at 9.81 m/s² unless you need another planet.
- ResultRead the ideal exit speed; a real opening delivers somewhat less.Fluid height above the opening h=2m→6.2642m/s
Reference table
| h, g | Setting | v (m/s) |
|---|---|---|
| 1, 9.81 | Earth | 4.4294 |
| 2, 9.81 | Earth | 6.2642 |
| 5, 9.81 | Earth | 9.9045 |
| 10, 9.81 | Earth | 14.0071 |
| 2, 9.80665 | Earth, standard gravity | 6.2631 |
| 20, 1.62 | Moon | 8.0498 |
Questions
What is Torricelli's law?
It states that an ideal fluid leaving an opening a height h below its free surface exits at v = √(2gh). It follows from energy conservation: the pressure head above the hole turns into kinetic energy.
How do I calculate the efflux velocity?
Multiply 2 by gravity and by the height, then take the square root. For a 2 m head on Earth: √(2 × 9.81 × 2) = √39.24, about 6.264 m/s. Only the depth below the surface and gravity enter the formula.
Does the fluid's density change the exit speed?
No — density cancels out of the energy balance, so water, oil and mercury all leave a given head at the same speed. The size of the opening does not change the speed either, only how much volume flows out per second.
Why is it the same as a free fall?
Both come from √(2gh). An object dropped from rest through height h arrives at v = √(2gh), and the jet leaves the hole at exactly that speed — a neat sanity check on the number.
How accurate is this for a real tank?
It is an ideal-fluid result. A real opening loses speed to viscosity and edge turbulence and delivers roughly 60 to 97 per cent of it, depending on the shape. The figure is also instantaneous: as the tank drains, h falls and the jet slows.