How to play Venturi Pipe
- Choose a test and read the areas or diameters, and the speed or the column difference, on the card.
- The same volume passes every section each second: A₁·v₁ = A₂·v₂, so narrower means faster.
- Faster water has lower pressure: Δp = ½·ρ·(v₂² − v₁²), and a column drops by Δp ÷ (ρ·g). Type your prediction and press Start.
- Watch the columns and the bars: what the water gains in motion, it loses in pressure.
Why does faster water push less?
Water doesn't compress, so every second the same volume must pass through every part of a pipe: Q = A·v. Where the pipe narrows, the water has to speed up. Something must push it to speed up, and that push comes from the pressure: the water is pushed harder from behind than from in front, so the pressure in the narrow part is lower. Bernoulli's principle puts numbers on it: along a flow, p + ½·ρ·v² + ρ·g·z stays the same, so pressure, motion, and height trade with one another. In a level pipe, Δp = ½·ρ·(v₂² − v₁²). Open columns on the pipe show it: the column over the narrow part stands Δp ÷ (ρ·g) lower. Turn this around and you have a Venturi meter: from the column difference and the areas you can work out how fast water flows. A nozzle uses the same idea to turn a hose's pressure into a fast jet, and a fountain turns that speed into height, h = v² ÷ (2·g).
A focused learning model
The pipe is level, the water doesn't compress, and friction is left out, so A₁·v₁ = A₂·v₂ and p + ½·ρ·v² is the same in both sections. In the pipes the flow builds up steadily over 1 s and then stays steady; the columns show the pressure above the air, p ÷ (ρ·g). The nozzle's jet leaves at air pressure. The fountain's water leaves at once and climbs like a thrown ball, without air resistance. A run lasts 2 s. The pipes are not to scale; the fountain is. Answers within 2% are right. Water is 1000 kg/m³; g = 9.8 m/s². Plays in real time.