Physics 10: 7-14 Fluid Mechanics and Bernoulli's Theorem
Text Reading: Giancoli, Physics - Principles with Applications, Chapter 10: Sections 8-14
- 10.8: Because the particles are free to move in different directions, individual particles in flowing liquids and gases do not necessarily follow a nice trajectory, sliding easily past one another in layers. Any interference will cause one or more particles to vere off, creating a chain reaction that results in turbulence. Inherit friction or viscosity between the particles is always present, and becomes greater when turbulence occurs. The amount of fluid that can flow through a cross-sectional area A is determined by the density of the fluid and the rate of speed of the molecules: ρ * A * v. Flow of a incompressible fluid (ρ does not change) under constant pressure through a channel of changing diameter requires the speed to change: A1v1 = A2v2 [The equation of continuity.] As cross-section gets smaller, velocity increases, so that in any identical time interval, the same number of particles pass any point on the channel. If this weren't true, the particles would "back up" like cars in a traffic jam, increasing the density of the liquid -- which isn't allowed in an incompressible liquid.
- 10.9: Now consider an open system, where pressure is allowed to change. If we change the speed of the particles, density changes and pressure changes. Bernoulli's equation shows the relationship of pressure to altitude as air flows across a slanted surface: P1 + ½ρv12 + ρgy1 = P2 + ½ρv22 + ρgy2. We can now compare specific situations against a standard (unmoving air at sea level) or two different situations.
- 10.10: We can now apply Bernoulli's theorem to some common situations. One in particular is the flow of water trhough a small outlet from a large container (such as the flow of water from a reservoir through a dam spillway, or from a storage tank into the main pipe system). In this case, the only pressure is atmospheric pressure, which will be nearly the same at the top and bottom of the contain, so that P1 = P2. If the flow of water at the surface (point 2) is small compared to the flow of water through the spigot (point 1), we can assume v2 ~ 0. Since P1 and P2 cancel, ρ cancels, and v2 can be ignored, Bernoulli's equation simplifies to a form called Torricelli's theorem: ½v12 + gy1 = gy2. We can solve for v1 = sqrt(2g(y2 - y1)).
- 10.11-14: The remaining sections of this chapter discuss applications of these principles to the speed of flow (viscosity), flow within tubes where friction and drag occur, and blood flow in capillaries and through the heart.
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