Presets
Playback
Tank, throat and valve
Narrow the throat or close the valve and watch the green static-head line and the piezometer columns move.
Model checks
The science
Continuity and Bernoulli
Water is incompressible, so the flow Q = A v is the same in every cross-section and the speed rises where the tube narrows. Along a streamline p/(ρg) + v2/(2g) + z = const minus losses, so where the speed head grows the static head must fall. A piezometer reads that static head.
Losses
Entrance loss Ke v2/2g (Ke = 0.5), Darcy friction f/(2r) · v2/2g integrated along the tube, a diffuser expansion loss (K = 0.15 on the speed-head drop) and the outlet valve, (Kv + 1) v32/2g. The flow follows from the tank head: H = R Q2.
Venturi meter and cavitation
Two taps give the flow, Q = Cd A2 √(2g(h1 - h3)/(1 - (A2/A1)2)); the discharge coefficient is a little below 1 because of losses. The static head cannot fall below the vapour-pressure limit (about -10.1 m of water at sea level), so when the throat reaches it the flow chokes and the water cavitates.
Honest simplifications
- Loss coefficients and the friction factor are constants (no Reynolds-number dependence); the flow is steady and one-dimensional.
- The jet and tracer dots are illustrations; the tank is large so its level is held constant. The bench picture is a Blender render of the 3D page's model.
What to try
- Pick Strong constriction and compare piezometer h1 and h3.
- Pick Valve nearly open: h3 goes negative, so an open tube there would suck air in.
- Pick Cavitation limit and see the flow stop growing in the second graph.
- Switch the head lines off and read the columns alone.