Illustration 1: The Continuity Equation
A continuity equation is simply a way to express that "what goes in must come out." This simply means that as a fluid goes from a pipe of one diameter to another, the fluid flow changes. Restart. 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 through the pipes from left to right (assume they are cylindrical, that is, the vertical distances in the animation correspond to the diameter of a circular cross section). Notice that as the water enters the narrower pipe, it goes faster. How long does it take the dark blue region to cross a line in the wide region and a line in the narrow region? The volume of the dark blue region divided by this time is the volume flow rate for each region. It should be the same for both regions because whatever goes in (per second) must come out (per second). We express this with the continuity equation \(Av =\) constant, where \(A\) is the cross-sectional area and \(v\) is the speed of the fluid flow (What are the units of \(Av\)? They should be volume/time). When we couple this with Bernoulli's equation (conservation of energy),
\[P+(1/2)\rho v^{2}+\rho gy=\text{ constant}\nonumber\]
where \(P\) is the pressure, \(\rho\) is the density of the fluid, \(v\) is the speed of the fluid flow, and \(y\) is the height of the fluid (you can, of course, pick any point to be \(y = 0\text{ m}\)), we find a change in pressure as well. In this case, because the pipe is horizontal, \(y\) is the same, so we simply use \(P + (1/2)\rho v^{2} =\text{ constant}\), so as the speed increases, the pressure decreases. Note the pressure readings by sliding the pressure indicator along the center of the pipe.
Note
The format of the pressure is written in shorthand. For example, atmospheric pressure, \(1.01\times 10^{5}\text{ Pa}\), is written as \(1.01e+005\).
Illustration authored by Anne J. Cox.
Illustration 2: Bernoulli's Principle at Work
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:
\[P_{1}+(1/2)\rho v_{1}^{2}+\rho gy_{1}=P_{2}+(1/2)\rho v_{2}^{2}+\rho gy_{2}=\text{ constant}\nonumber\]
where \(P\) is the pressure, \(\rho\) is the density of the fluid, \(v\) is the speed of the fluid flow, and \(y\) is the vertical position from \(y = 0\text{ m}\) (you can, of course, pick any point to be \(y = 0\text{ m}\)---this is equivalent to picking any spot to be the zero of potential energy, but once you have picked the \(y = 0\text{ m}\) 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 \(P_{1} = P_{2} = P_{\text{atm}}\). 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
\[\rho gy_{1}=(1/2)\rho v_{2}^{2}+\rho gy_{2}\quad\text{or that}\quad v_{2}^{2}=2g(y_{1}-y_{2})=2g\Delta h\nonumber\]
where \(\Delta h\) 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.
Illustration 3: Ideal and Viscous Fluid Flow
Bernoulli's equation describes the conservation of energy in an ideal fluid system such as the one shown in this animation (position is given in tenths of meters and pressure is given in pascals). Restart. The vertical tubes are open to the atmosphere. Notice that the water level is lower to the right, indicating a lower pressure. Why is the pressure lower in the narrower tube? Notice also that the pressure only changes when going from a wider to a narrower tube. The pressure indicator measures the gauge pressure, not the absolute pressure (gauge pressure is pressure above atmospheric pressure). When there is viscosity (that is the fluid sticks together a bit so there is some friction), but still a smooth (laminar) flow, the pressure drops along the length of a pipe. Try it. For viscous flow, notice that, to get the same volume per unit time (\(Av =\) volume/time, where \(A\) is the cross-sectional area and \(v\) is the speed of the fluid flow), the pressure drops more in the narrower tube than in the wider tube. The equation governing the flow is Poiseulle's equation, \(Av = \pi R^{4}\Delta P/8\eta L\), where \(R\) is the radius of the tube, \(L\) is the length of the tube, \(\Delta P\) is the pressure difference and \(\eta\) is the viscosity of the fluid.
Note
The format of the pressure is written in shorthand. For example, atmospheric pressure, \(1.01\times 10^{5}\text{ Pa}\), is written as \(1.01e+005\).
Illustration authored by Anne J. Cox.
Illustration 4: Airplane lift
The animation shows the cross section (side view) of a model airplane wing with air moving past the wing (position is given in centimeters and time is given in seconds). Restart. Where is the speed of the air greatest? Where will the pressure be higher? How does this explain the lift on an airplane? Using Bernoulli's principle we can find the pressure difference between the top and the bottom wing and it is
\[\Delta P=\rho (v_{\text{above}}^{2}-v_{\text{below}}^{2})/2\nonumber\]
First, find the speed of the air above (once above the wing, the air speed is constant) and below the wing. We can find the average speed easily as the displacement over the time interval, and we find that \(v_{\text{below}} = 950\text{ cm/s} = 9.5\text{ m/s}\) and \(v_{\text{above}} = 990\text{ cm/s} = 9.9\text{ m/s}\).
Now we can calculate the pressure difference using our results for the air speed and the density of air, \(\rho = 1.3\text{ kg/m}^{3}\). We find that in this case \(\Delta P = 5\text{ Pa}\). If the surface area of the wing is \(0.1\text{ m}^{2}\), what is the net force (lift) on this wing? Since \(P = F/A\), we find that the net force will be the pressure difference times the area or \(0.5\text{ N}\).
The reason an air flow pattern develops that yields different speeds on the top and the bottom is that the air flowing around the wing moves into nonideal fluid flow. Initially, since the air on top has farther to travel, the air on the bottom of the wing gets to the back of the wing and moves up to "fill" this space, but this instability causes a turbulent wake that eventually allows a new, more stable, air-flow pattern such as the one shown, where air-particles that travel across the top go faster. For a greater difference in pressure, the wing is tilted up (the angle of tilt is called the angle of attack), and this increases the lift.
Note
The format of the time is written in shorthand. For example, a time of \(6.00\times 10^{-3}\text{ s}\), is written as \(6.00e-003\).
Illustration authored by Anne J. Cox.
Physlets were developed at Davidson College and converted from Java to JavaScript using the SwingJS system developed at St. Olaf College.
