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A reservoir of water has a hole in it at a height that you can adjust (position is given in meters). Notice what happens to the water flow out of the opening. Restart. Assume an ideal fluid. How do you describe what happens to the initial speed of the water depending on the height of the hole? The velocity of the water out of the opening can be determined according to Bernoulli's equation, which compares two points in the fluid:

P₁ + (1/2)ρv₁² + ρgy₁ = P₂ + (1/2)ρv₂² + ρgy₂ = constant,

where P is the pressure, ρ is the density of the fluid, v is the speed of the fluid flow, and y is the vertical position from y = 0 m (you can, of course, pick any point to be y = 0 m---this is equivalent to picking any spot to be the zero of potential energy, but once you have picked the y = 0 m spot you must be consistent).

Consider point 1 to be the top of the fluid and point 2 to be the point where the fluid leaves the hole. Given this assignment, we can easily see that P₁ = P₂ = P_atm. At the top of the reservoir the water is essentially stationary, so v₁ = 0 m/s there. This means that Bernoulli's equation simplifies to

ρgy₁ = (1/2)ρv₂² + ρgy₂      or that      v₂² = 2 g (y₁ - y₂) = 2gΔh,

where Δh is the height of the water above the hole (not the height of the opening, although the two are related).

Illustration authored by Anne J. Cox and Mario Belloni.