- Fluid height above the opening h
- 2m
- Gravitational acceleration g
- 9.81m/s²
6.2642m/s
Open with these values6.2642m/s
Result: 6.2642 m/sFluid 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.
6.2642m/s
Open with these values14.0071m/s
Open with these values8.0498m/s
Open with these valuesv = √(2 × g × h)
| 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 |
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.
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.
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.
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.
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.
Information, not professional advice.
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