PhysicsLAB Resource Lesson
Fluids In Motion

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When real fluids flow through pipes, two distinct forces act on them. One is the frictional forces exerted on the fluid by the walls of the pipe and the other is the viscous forces within the fluid. The fluid layers next to the walls of the pipe "stick" slightly to the pipe. As you move further from the walls towards the center of the fluid, this boundary layer ends and the fluid moves faster and more coherently. Viscous forces within the fluid produce a shearing action that results in tiny layers of fluid of ever-increasing speed which eventually reach the speed of the free stream in the center of the pipe. Energy is lost within the fluid to both of these forces.
An ideal fluid is one that meets the following specifications: steady flow, irrotational flow, nonviscous flow, and incompressible flow. Steady flow is laminar flow which means that the particles flow along streamlines - that is, every particle moves along the same path as previous particles followed. Every particle at the same place in a fluid will have the same velocity. Steady flow only occurs at low velocities. When streamlines are forced closer together, the velocity in the fluid is greater. Irrotational flow means that no fluid elements (small volume packets) have angular velocity - that is, there is no turbulence in the form of  whirlpools or eddy currents. Nonviscous flow means that viscosity can be neglected - that is, there are no shearing forces within the fluid which subsequently result in the production of heat as the fluid flows. Incompressible flow means that the density of the fluid remains constant.
If a fluid system has no sources providing additional fluid or sinks draining off fluid, the volume of fluid entering the first cross-sectional area must equal the volume of fluid flowing out the later cross-sectional area. 
Since density, rho = m/V (or rhoV = m) is a constant in an incompressible fluid, we say that mass is conserved in a closed fluid system.
The Continuity Equation
states that the cross-sectional area of the pipe and the velocity of the fluid are inversely proportional - that is, fluids flow faster through narrower pipes. We can see this by the fact that the streamlines are forced closed together whenever the pipe narrows. Next time you watch water flowing from a faucet note how the water stream narrows as the water falls. This reduction in cross-sectional area is required by the Continuity Equation since the water is increasing in speed as it falls.
Ideal liquids also obey a special statement of conservation of energy first developed by Daniel Bernoulli in 1738
which is often expressed as
Before investigating a derivation of Bernoulli's Equation, note that each term is actually an expression of energy/volume and has units of J/m3.
  •  has units of N/m2 which can be also expressed as work per unit volume, Nm/m3 = J/m3
  •  is just potential energy per unit volume, mgh/V, which is measured in J/m3
  •  is just kinetic energy per unit volume, ½mv2/V, which is measured in J/m3

If there is no change in potential energy along the length of the pipe, then this equation can be rephrased as
and we see that the kinetic energy of the fluid will decrease if the pressure increases. Combining the Continuity Equation and Bernoulli's Equation, we have the result that when the cross-sectional area of a pipe decreases, the velocity - and hence the kinetic energy - of the fluid increases, and the pressure decreases. This is called the Bernoulli or Venturi Effect.
Suppose you have an open tank of water (with an extremely large upper cross-sectional area having no water currents) has a spigot (with a tremendously small cross-sectional area) located just above its bottom surface. The spigot has a nozzle which creates a vertical jet of water, or fountain.
How high will the water jet upwards?
Since the area of spigot is so small compared to the area of the tank, the Continuity Equation tells us that the exiting speed of the water v2 will be very large. We can consider the rate at which the water level in the tank drops with its very large cross-sectional area to approximate zero.
Since P1 and P2 are both nearly equal to atmospheric pressure, we can consider them equal to each other.

Using conservation of energy, the water would jet back to the surface of the tank. The result derived above is called Torricelli's Result.

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