123. Bernoulli’s EquationLearning Objectives
When a fluid flows into a narrower channel, its speed increases. That means its kinetic energy also increases. Where does that change in kinetic energy come from? The increased kinetic energy comes from the net work done on the fluid to push it into the channel and the work done on the fluid by the gravitational force, if the fluid changes vertical position. Recall the workenergy theorem, $${W}_{\text{net}}=\frac{1}{2}{\text{mv}}^{2}\frac{1}{2}{\text{mv}}_{0}^{2}\text{.}$$There is a pressure difference when the channel narrows. This pressure difference results in a net force on the fluid: recall that pressure times area equals force. The net work done increases the fluid’s kinetic energy. As a result, the pressure will drop in a rapidlymoving fluid, whether or not the fluid is confined to a tube. There are a number of common examples of pressure dropping in rapidlymoving fluids. Shower curtains have a disagreeable habit of bulging into the shower stall when the shower is on. The highvelocity stream of water and air creates a region of lower pressure inside the shower, and standard atmospheric pressure on the other side. The pressure difference results in a net force inward pushing the curtain in. You may also have noticed that when passing a truck on the highway, your car tends to veer toward it. The reason is the same—the high velocity of the air between the car and the truck creates a region of lower pressure, and the vehicles are pushed together by greater pressure on the outside. (See Figure 1.) This effect was observed as far back as the mid1800s, when it was found that trains passing in opposite directions tipped precariously toward one another. Making Connections: TakeHome Investigation with a Sheet of Paper:Hold the short edge of a sheet of paper parallel to your mouth with one hand on each side of your mouth. The page should slant downward over your hands. Blow over the top of the page. Describe what happens and explain the reason for this behavior. Bernoulli’s EquationThe relationship between pressure and velocity in fluids is described quantitatively by where $P$ is the absolute pressure, $\rho $ is the fluid density, $v$ is the velocity of the fluid, $h$ is the height above some reference point, and $g$ is the acceleration due to gravity. If we follow a small volume of fluid along its path, various quantities in the sum may change, but the total remains constant. Let the subscripts 1 and 2 refer to any two points along the path that the bit of fluid follows; Bernoulli’s equation becomes $${P}_{1}+\frac{1}{2}{\mathrm{\rho v}}_{1}^{2}+\rho {\mathrm{gh}}_{1}={P}_{2}+\frac{1}{2}{\mathrm{\rho v}}_{2}^{2}+\rho {\mathrm{gh}}_{2}\text{.}$$Bernoulli’s equation is a form of the conservation of energy principle. Note that the second and third terms are the kinetic and potential energy with $m$ replaced by $\rho $. In fact, each term in the equation has units of energy per unit volume. We can prove this for the second term by substituting $\rho =m/V$ into it and gathering terms: $$\frac{1}{2}{\mathrm{\rho v}}^{2}=\frac{\frac{1}{2}{\text{mv}}^{2}}{V}=\frac{\text{KE}}{V}\text{.}$$So $\frac{1}{2}{\mathrm{\rho v}}^{2}$ is the kinetic energy per unit volume. Making the same substitution into the third term in the equation, we find $$\rho \mathrm{gh}=\frac{\mathrm{mgh}}{V}=\frac{{\text{PE}}_{\text{g}}}{V},$$so $\rho \text{gh}$ is the gravitational potential energy per unit volume. Note that pressure $P$ has units of energy per unit volume, too. Since $P=F/A$, its units are ${\text{N/m}}^{2}$. If we multiply these by m/m, we obtain $\text{N}\cdot {\text{m/m}}^{3}={\text{J/m}}^{3}$, or energy per unit volume. Bernoulli’s equation is, in fact, just a convenient statement of conservation of energy for an incompressible fluid in the absence of friction. Making Connections: Conservation of Energy:Conservation of energy applied to fluid flow produces Bernoulli’s equation. The net work done by the fluid’s pressure results in changes in the fluid’s $\text{KE}$ and ${\text{PE}}_{\text{g}}$ per unit volume. If other forms of energy are involved in fluid flow, Bernoulli’s equation can be modified to take these forms into account. Such forms of energy include thermal energy dissipated because of fluid viscosity. The general form of Bernoulli’s equation has three terms in it, and it is broadly applicable. To understand it better, we will look at a number of specific situations that simplify and illustrate its use and meaning. Bernoulli’s Equation for Static FluidsLet us first consider the very simple situation where the fluid is static—that is, ${v}_{1}={v}_{2}=0$. Bernoulli’s equation in that case is $${P}_{1}+\rho {\mathrm{gh}}_{1}={P}_{2}+\rho {\mathrm{gh}}_{2}\text{.}$$We can further simplify the equation by taking ${h}_{2}=0$ (we can always choose some height to be zero, just as we often have done for other situations involving the gravitational force, and take all other heights to be relative to this). In that case, we get $${P}_{2}={P}_{1}+\rho {\mathrm{gh}}_{1}\text{.}$$This equation tells us that, in static fluids, pressure increases with depth. As we go from point 1 to point 2 in the fluid, the depth increases by ${h}_{1}$, and consequently, ${P}_{2}$ is greater than ${P}_{1}$ by an amount $\rho {\mathrm{gh}}_{1}$. In the very simplest case, ${P}_{1}$ is zero at the top of the fluid, and we get the familiar relationship $P=\rho \mathrm{gh}$. (Recall that $P=\mathrm{\rho gh}$ and $\text{\Delta}{\text{PE}}_{\text{g}}=\text{mgh}.$) Bernoulli’s equation includes the fact that the pressure due to the weight of a fluid is $\rho \text{gh}$. Although we introduce Bernoulli’s equation for fluid flow, it includes much of what we studied for static fluids in the preceding chapter. Bernoulli’s Principle—Bernoulli’s Equation at Constant DepthAnother important situation is one in which the fluid moves but its depth is constant—that is, ${h}_{1}={h}_{2}$. Under that condition, Bernoulli’s equation becomes $${P}_{1}+\frac{1}{2}{\mathrm{\rho v}}_{1}^{2}={P}_{2}+\frac{1}{2}{\mathrm{\rho v}}_{2}^{2}\text{.}$$Situations in which fluid flows at a constant depth are so important that this equation is often called Example 1: Calculating Pressure: Pressure Drops as a Fluid Speeds UpIn Section 122, we found that the speed of water in a hose increased from 1.96 m/s to 25.5 m/s going from the hose to the nozzle. Calculate the pressure in the hose, given that the absolute pressure in the nozzle is $1\text{.}\text{01}\times {\text{10}}^{5}\phantom{\rule{0.25em}{0ex}}{\text{N/m}}^{2}$ (atmospheric, as it must be) and assuming level, frictionless flow. Strategy Level flow means constant depth, so Bernoulli’s principle applies. We use the subscript 1 for values in the hose and 2 for those in the nozzle. We are thus asked to find ${P}_{1}$. Solution Solving Bernoulli’s principle for ${P}_{1}$ yields $${P}_{1}={P}_{2}+\frac{1}{2}{\mathrm{\rho v}}_{2}^{2}\frac{1}{2}{\mathrm{\rho v}}_{1}^{2}={P}_{2}+\frac{1}{2}\rho ({v}_{2}^{2}{v}_{1}^{2})\text{.}$$Substituting known values, $$\begin{array}{lll}{P}_{1}& =& 1\text{.}\text{01}\times {\text{10}}^{5}\phantom{\rule{0.25em}{0ex}}{\text{N/m}}^{2}\\ & & \text{}+\frac{1}{2}({\text{10}}^{3}\phantom{\rule{0.25em}{0ex}}{\text{kg/m}}^{3})\left[(\text{25.5 m/s}{)}^{2}(\text{1.96 m/s}{)}^{2}\right]\\ & =& 4.24\times {\text{10}}^{5}\phantom{\rule{0.25em}{0ex}}{\text{N/m}}^{2}\text{.}\end{array}$$Discussion This absolute pressure in the hose is greater than in the nozzle, as expected since $v$ is greater in the nozzle. The pressure ${P}_{2}$ in the nozzle must be atmospheric since it emerges into the atmosphere without other changes in conditions. Applications of Bernoulli’s PrincipleThere are a number of devices and situations in which fluid flows at a constant height and, thus, can be analyzed with Bernoulli’s principle. EntrainmentPeople have long put the Bernoulli principle to work by using reduced pressure in highvelocity fluids to move things about. With a higher pressure on the outside, the highvelocity fluid forces other fluids into the stream. This process is called entrainment. Entrainment devices have been in use since ancient times, particularly as pumps to raise water small heights, as in draining swamps, fields, or other lowlying areas. Some other devices that use the concept of entrainment are shown in Figure 2. Wings and SailsThe airplane wing is a beautiful example of Bernoulli’s principle in action. Figure 3(a) shows the characteristic shape of a wing. The wing is tilted upward at a small angle and the upper surface is longer, causing air to flow faster over it. The pressure on top of the wing is therefore reduced, creating a net upward force or lift. (Wings can also gain lift by pushing air downward, utilizing the conservation of momentum principle. The deflected air molecules result in an upward force on the wing — Newton’s third law.) Sails also have the characteristic shape of a wing. (See Figure 3(b).) The pressure on the front side of the sail, ${P}_{\text{front}}$, is lower than the pressure on the back of the sail, ${P}_{\text{back}}$. This results in a forward force and even allows you to sail into the wind. Making Connections: TakeHome Investigation with Two Strips of Paper:For a good illustration of Bernoulli’s principle, make two strips of paper, each about 15 cm long and 4 cm wide. Hold the small end of one strip up to your lips and let it drape over your finger. Blow across the paper. What happens? Now hold two strips of paper up to your lips, separated by your fingers. Blow between the strips. What happens? Velocity measurementFigure 4 shows two devices that measure fluid velocity based on Bernoulli’s principle. The manometer in Figure 4(a) is connected to two tubes that are small enough not to appreciably disturb the flow. The tube facing the oncoming fluid creates a dead spot having zero velocity (${v}_{1}=0$) in front of it, while fluid passing the other tube has velocity ${v}_{2}$. This means that Bernoulli’s principle as stated in ${P}_{1}+\frac{1}{2}{\mathrm{\rho v}}_{1}^{2}={P}_{2}+\frac{1}{2}{\mathrm{\rho v}}_{2}^{2}$ becomes $${P}_{1}={P}_{2}+\frac{1}{2}{\mathrm{\rho v}}_{2}^{2}\text{.}$$Thus pressure ${P}_{2}$ over the second opening is reduced by $\frac{1}{2}{\mathrm{\rho v}}_{2}^{2}$, and so the fluid in the manometer rises by $h$ on the side connected to the second opening, where $$h\propto \frac{1}{2}{\mathrm{\rho v}}_{2}^{2}\text{.}$$(Recall that the symbol $\text{\u221d}$ means “proportional to.”) Solving for ${v}_{2}$, we see that $${v}_{2}\propto \sqrt{h}\text{.}$$Figure 4(b) shows a version of this device that is in common use for measuring various fluid velocities; such devices are frequently used as air speed indicators in aircraft. Summary
Conceptual QuestionsExercise 1You can squirt water a considerably greater distance by placing your thumb over the end of a garden hose and then releasing, than by leaving it completely uncovered. Explain how this works. Exercise 2Water is shot nearly vertically upward in a decorative fountain and the stream is observed to broaden as it rises. Conversely, a stream of water falling straight down from a faucet narrows. Explain why, and discuss whether surface tension enhances or reduces the effect in each case. Exercise 3Look back to Figure 1. Answer the following two questions. Why is ${P}_{\text{o}}$ less than atmospheric? Why is ${P}_{\text{o}}$ greater than ${P}_{\text{i}}$? Exercise 4Give an example of entrainment not mentioned in the text. Exercise 5Many entrainment devices have a constriction, called a Venturi, such as shown in Figure 5. How does this bolster entrainment? Exercise 6Some chimney pipes have a Tshape, with a crosspiece on top that helps draw up gases whenever there is even a slight breeze. Explain how this works in terms of Bernoulli’s principle. Exercise 7Is there a limit to the height to which an entrainment device can raise a fluid? Explain your answer. Exercise 8Why is it preferable for airplanes to take off into the wind rather than with the wind? Exercise 9Roofs are sometimes pushed off vertically during a tropical cyclone, and buildings sometimes explode outward when hit by a tornado. Use Bernoulli’s principle to explain these phenomena. Exercise 10Why does a sailboat need a keel? Exercise 11It is dangerous to stand close to railroad tracks when a rapidly moving commuter train passes. Explain why atmospheric pressure would push you toward the moving train. Exercise 12Water pressure inside a hose nozzle can be less than atmospheric pressure due to the Bernoulli effect. Explain in terms of energy how the water can emerge from the nozzle against the opposing atmospheric pressure. Exercise 13A perfume bottle or atomizer sprays a fluid that is in the bottle. (Figure 6.) How does the fluid rise up in the vertical tube in the bottle? Exercise 14If you lower the window on a car while moving, an empty plastic bag can sometimes fly out the window. Why does this happen? Problems & ExercisesExercise 1Verify that pressure has units of energy per unit volume. Show/Hide Solution Solution$\begin{array}{lll}P& =& \frac{\text{Force}}{\text{Area}},\\ (P{)}_{\text{units}}& =& {\text{N/m}}^{2}=\text{N}\cdot {\text{m/m}}^{3}={\text{J/m}}^{3}\\ & =& \text{energy/volume}\end{array}$ Exercise 2Suppose you have a wind speed gauge like the pitot tube shown in Section 122 Example 2(b). By what factor must wind speed increase to double the value of $h$ in the manometer? Is this independent of the moving fluid and the fluid in the manometer? Exercise 3If the pressure reading of your pitot tube is 15.0 mm Hg at a speed of 200 km/h, what will it be at 700 km/h at the same altitude? Show/Hide Solution Solution184 mm Hg Exercise 4Calculate the maximum height to which water could be squirted with the hose in Section 122 example if it: (a) Emerges from the nozzle. (b) Emerges with the nozzle removed, assuming the same flow rate. Exercise 5Every few years, winds in Boulder, Colorado, attain sustained speeds of 45.0 m/s (about 100 mi/h) when the jet stream descends during early spring. Approximately what is the force due to the Bernoulli effect on a roof having an area of $\text{220}\phantom{\rule{0.25em}{0ex}}{\text{m}}^{2}$? Typical air density in Boulder is $1\text{.}\text{14}\phantom{\rule{0.25em}{0ex}}{\text{kg/m}}^{3}$, and the corresponding atmospheric pressure is $8\text{.}\text{89}\times {\text{10}}^{4}\phantom{\rule{0.25em}{0ex}}{\text{N/m}}^{2}$. (Bernoulli’s principle as stated in the text assumes laminar flow. Using the principle here produces only an approximate result, because there is significant turbulence.) Show/Hide Solution Solution$2\text{.}\text{54}\times {\text{10}}^{5}\phantom{\rule{0.25em}{0ex}}\text{N}$ Exercise 6(a) Calculate the approximate force on a square meter of sail, given the horizontal velocity of the wind is 6.00 m/s parallel to its front surface and 3.50 m/s along its back surface. Take the density of air to be $\text{1.29 kg}{\text{/m}}^{3}$. (The calculation, based on Bernoulli’s principle, is approximate due to the effects of turbulence.) (b) Discuss whether this force is great enough to be effective for propelling a sailboat. Exercise 7(a) What is the pressure drop due to the Bernoulli effect as water goes into a 3.00cmdiameter nozzle from a 9.00cmdiameter fire hose while carrying a flow of 40.0 L/s? (b) To what maximum height above the nozzle can this water rise? (The actual height will be significantly smaller due to air resistance.) Show/Hide Solution Solution(a) $1\text{.}\text{58}\times {\text{10}}^{6}\phantom{\rule{0.25em}{0ex}}{\text{N/m}}^{2}$ (b) 163 m Exercise 8(a) Using Bernoulli’s equation, show that the measured fluid speed _{$v$} for a pitot tube, like the one in Figure 4(b), is given by $$v={\left(\frac{2\rho \prime \mathrm{gh}}{\rho}\right)}^{1/2},$$where $h$ is the height of the manometer fluid, $\rho \prime $ is the density of the manometer fluid, $\rho $ is the density of the moving fluid, and $g$ is the acceleration due to gravity. (Note that $v$ is indeed proportional to the square root of $h$, as stated in the text.) (b) Calculate $v$ for moving air if a mercury manometer’s $h$ is 0.200 m.
