Physlets run in a Java-enabled browser, except Chrome, on the latest Windows & Mac operating systems. If Physlets do not run, click here for help on updating Java and setting Java security.

# Illustration 15.2: Bernoulli's Principle at Work

Please wait for the animation to completely load.

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:

P_{1} + (1/2)ρv_{1}^{2} + ρgy_{1} = P_{2} + (1/2)ρv_{2}^{2} + ρgy_{2} = 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 P_{1} = P_{2} = P_{atm}. At the top of the reservoir the water is essentially stationary, so v_{1} = 0 m/s there. This means that Bernoulli's equation simplifies to

ρgy_{1} = (1/2)ρv_{2}^{2} + ρgy_{2} or that v_{2}^{2} = 2 g (y_{1} - y_{2}) = 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.

« previous

next »