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 updating Java & setting Java security.
Exploration 15.2: Bernoulli's Equation
Please wait for the animation to completely load.
Bernoulli's equation describes the conservation of energy in an ideal fluid system. Assume an ideal fluid (position is given in meters and pressure is given in pascals). The dark blue in the animation is a section of water as it flows into the region marked by the horizontal line and the corresponding water that must move out of the region in the top right. We will explore the connection between Bernoulli's equation and conservation of energy. Restart.
Note: The format of the pressure is written in shorthand. For example, atmospheric pressure, 1.01 x 105 Pa, is written as 1.01e+005.
The relationship between the speed and dimensions of the water going in compared with the water leaving is governed by the continuity equation (what flows in must flow out unless there is a leak in the pipes!): Av = constant, where A is the cross-sectional area and v is the speed of the liquid. Assume the pipes are cylindrical.
- What is the volume of both blue regions (should be the same)?
- What is the speed of the water in the left pipe?
- What is the cross-sectional area of the left pipe?
- What is the speed of the water leaving the region (in the right pipe)?
- What is the cross-sectional area of the right pipe?
- Does the continuity equation hold?
As the water travels through this pipe system, work must be done on the fluid to raise it up and to increase its speed. The work done must be equal to the change in kinetic plus potential energy.
- Given the pressure (you can move the red pressure indicators), find the force (from the water behind it) on the lower left dark blue region.
- What is the work done by that force for the duration of the animation?
- Similarly, find the force on the upper right dark blue region that opposes the motion.
- What is the work done by that force for the duration of the animation (note that the displacement and force are in opposite directions, so this is negative work)?
- What is the net work done, then, during the duration of the animation, on the water in the middle region?
- Calculate the difference in kinetic energy of the dark blue regions. Note that since the volume is the same, the mass is the same. (The density of water is 1000 kg/m3.)
- Calculate the difference in potential energy of the center of mass of the dark blue regions. Does the net work equal the difference in kinetic energy plus the difference in potential energy?
This is all described by Bernoulli's equation.
- Show that the net work is equal to (Pleft - Pright) Avt.
- Show that the net change in kinetic energy is (1/2) ρAvt (vright2 - vleft2).
- Show that the net change in potential energy is ρgAvt (yright - yleft).
P is the pressure, ρ is the density of the fluid, v is the speed of the fluid flow, A is the cross-sectional area, t is the time, and y is the height of the fluid. Combining these three terms, we get
(Pleft - Pright) = (1/2)ρ(vright2 - vleft2) + ρg(yright - yleft),
or Bernoulli's equation, P + (1/2)ρv2 + ρgy = constant, so that Bernoulli's equation is simply another way to restate conservation of energy.
Exploration authored by Anne J. Cox.