Because flowing blood has mass and velocity it has kinetic energy (KE). This KE is proportionate to the mean velocity squared (V²; from KE = ½ mV²). Furthermore, as the blood flows inside a vessel, pressure is exerted laterally against the walls of the vessel; this pressure represents the potential or pressure energy (PE). The total energy (E) of the blood flowing within the vessel, therefore, is the sum of the kinetic and potential energies (assuming no gravitational effects) as shown below.

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(I1;I2;:::) is a sequence of Bernoulli trials, named for the Swiss mathemati-cian Jacob Bernoulli. Bernoulli trials form one of the simplest yet still most important of all random processes. If p = 0 then the gambler always loses and if p = 1 then the gambler always wins. These trivial cases are not interesting, so we will usually assume that 0. Bernoulli built his work off of that of Newton. Bernoulli (1700 – 1782) was a Dutch-born scientist who studied in Italy and eventually settled in Switzerland. Daniel Bernoulli was born into a. Family of renowned mathematicians. His father, Johann Bernoulli, was one of the early developers of calculus and his uncle Jacob Bernoulli.

E = KE + PE (where KE ∝ V²)Therefore, E ∝ V² + PE

There are two important concepts that follow from this relationship.

• Blood flow is driven by the difference in total energy between two points. Although pressure is normally considered as the driving force for blood flow, in reality it is the total energy that drives flow between two points (e.g., longitudinally along a blood vessel or across a heart valve). Throughout most of the cardiovascular system, KE is relatively low, so for practical purposes, it is stated that the pressure energy (PE) difference drives flow. When KE is high, however, adding KE to the PE significantly increases the total energy, E. To illustrate this, consider the flow across the aortic valve during cardiac ejection. Late during ejection, the intraventricular pressure (PE) falls slightly below the aortic pressure (PE), nevertheless, flow continues to be ejected into the aorta. The reason for this is that the KE of the blood as it moves across the valve at a very high velocity ensures that the total energy (E) in the blood crossing the valve is higher than the total energy of the blood more distal in the aorta.
• Kinetic energy and pressure energy can be interconverted so that total energy remains unchanged. This is the basis of Bernoulli's Principle. This principle can be illustrated by a blood vessel that is suddenly narrowed then returned to its normal diameter. In the narrowed region (stenosis), the velocity increases as the diameter decreases. Quantitatively, V ∝ 1/D² because flow (F) is the product of mean velocity (V) and vessel cross-sectional area (A) (F = V x A), and A is directly related to diameter (D) (or radius, r) squared (from A = π r²). If the diameter is reduced by one-half in the region of the stenosis, the velocity increases 4-fold. Because KE ∝ V², the KE increases 16-fold. Assuming that the total energy is conserved within the stenosis (E actually decreases because of resistance as shown in the figure), then the 16-fold increase in KE must result in a reciprocal decrease in in the magnitude of PE. Once past the narrowed segment, KE will revert back to its pre-stenosis value because the post-stenosis diameter is the same as the pre-stenosis diameter and flow (and therefore, velocity) is conserved. Because of the resistance of the stenosis, and the turbulence the likely occurs, the post-stenosis PE and E will both fall. To summarize this concept, blood flowing at higher velocities has a higher ratio of kinetic energy to potential (pressure) energy.

An interesting, yet practical application of Bernoulli's Principle is found when blood pressure measurements are made from within the ascending aorta. As described above, during ventricular ejection, the velocity and hence kinetic energy of the flowing blood is very high. The instantaneous blood pressure that is measured within the aorta will be significantly different depending upon how the pressure is measured. As illustrated to the right, if a catheter has an end-port (E) sensor that is facing the flowing stream of blood, it will measure a pressure that is significantly higher than the pressure measured by a side-port (S) sensor on the same catheter. The reason for the discrepancy in measured pressures is that the end-port measures the total energy of the flowing blood. As the flow stream 'hits' the end of the catheter, the kinetic energy (which is high) is converted to potential (or pressure) energy, and added to the potential energy to equal the total energy. The side-port will not be 'hit' by the flowing stream so kinetic energy is not converted to potential energy. The side-port sensor, therefore, only measures the potential energy, which is the lateral pressure acting on the walls of the aorta. The difference between the two types of pressure measurements can range from a few mmHg to more than 20 mmHg depending upon the peak velocity of the flowing blood within the aorta.

Revised 01/27/20

Learning Objectives

By the end of this section, you will be able to:

• Explain the terms in Bernoulli’s equation.
• Explain how Bernoulli’s equation is related to conservation of energy.
• Explain how to derive Bernoulli’s principle from Bernoulli’s equation.
• Calculate with Bernoulli’s principle.
• List some applications of Bernoulli’s principle.

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 work-energy theorem,

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 rapidly-moving fluid, whether or not the fluid is confined to a tube.

There are a number of common examples of pressure dropping in rapidly-moving fluids. Shower curtains have a disagreeable habit of bulging into the shower stall when the shower is on. The high-velocity 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 mid-1800s, when it was found that trains passing in opposite directions tipped precariously toward one another.

Figure 1. An overhead view of a car passing a truck on a highway. Air passing between the vehicles flows in a narrower channel and must increase its speed (v₂ is greater than v1), causing the pressure between them to drop (P_i is less than P_o). Greater pressure on the outside pushes the car and truck together.

Making Connections: Take-Home Investigation with a Sheet of Paper

The relationship between pressure and velocity in fluids is described quantitatively by Bernoulli’s equation, named after its discoverer, the Swiss scientist Daniel Bernoulli (1700–1782). Bernoulli’s equation states that for an incompressible, frictionless fluid, the following sum is constant:

[latex]P+frac{1}{2}rho v^{2}+rho gh=text{constant}[/latex],

where P is the absolute pressure, ρ 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

[latex]{P}_{1}+frac{1}{2}{{rho v}_{1}}^{2}+rho {gh}_{1}={P}_{2}+frac{1}{2}{{rho v}_{2}}^{2}+rho {gh}_{2}[/latex].

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 ρ. In fact, each term in the equation has units of energy per unit volume. We can prove this for the second term by substituting ρ = m/V into it and gathering terms:

[latex]frac{1}{2}{rho v}^{2}=frac{frac{1}{2}{text{mv}}^{2}}{V}=frac{text{KE}}{V}[/latex]

So [latex]frac{1}{2}{rho v}^{2}[/latex] is the kinetic energy per unit volume. Making the same substitution into the third term in the equation, we find

[latex]rho {gh}=frac{mgh}{V}=frac{{text{PE}}_{text{g}}}{V}[/latex],

so ρgh 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 N/m². If we multiply these by m/m, we obtain N ⋅ m/m³ = J/m³, 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

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.

Let us first consider the very simple situation where the fluid is static—that is, v₁ = v₂ = 0. Bernoulli’s equation in that case is

P₁ + ρgh₁ = P₂ + ρgh₂.

We can further simplify the equation by taking h₂ = 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

[latex]{P}_{2}={P}_{1}+rho {gh}_{1}[/latex]

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 h₁, and consequently, P₂ is greater than P₁ by an amount ρgh₁. In the very simplest case, P₁ is zero at the top of the fluid, and we get the familiar relationship P = ρgh. (Recall that P = ρgh and [latex]Delta{text{PE}}_{text{g}}= mgh[/latex]) Bernoulli’s equation includes the fact that the pressure due to the weight of a fluid is ρgh. 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 Depth

Another important situation is one in which the fluid moves but its depth is constant—that is, h₁ = h₂. Under that condition, Bernoulli’s equation becomes

[latex]{P}_{1}+frac{1}{2}{{rho v}_{1}}^{2}={P}_{2}+frac{1}{2}{{rho v}_{2}}^{2}[/latex].

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Situations in which fluid flows at a constant depth are so important that this equation is often called Bernoulli’s principle. It is Bernoulli’s equation for fluids at constant depth. (Note again that this applies to a small volume of fluid as we follow it along its path.) As we have just discussed, pressure drops as speed increases in a moving fluid. We can see this from Bernoulli’s principle. For example, if v₂ is greater than v₁ in the equation, then P₂ must be less than P₁ for the equality to hold.

Example 1. Calculating Pressure: Pressure Drops as a Fluid Speeds Up

In Example 1 from Flow Rate and Its Relation to Velocity, 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 1.0 × 10⁵ N/m² (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 P₁.

Solution

Solving Bernoulli’s principle for P₁ yields

[latex]{P}_{1}={P}_{2}+frac{1}{2}{{rho v}_{2}}^{2}-frac{1}{2}{{rho v}_{1}}^{2}={P}_{2}+frac{1}{2}rho left({{v}_{2}}^{2}-{{v}_{1}}^{2}right)[/latex]

Substituting known values,

[latex]begin{array}{c}{P}_{1} = 7 1.01times 10^{5} text{ N/m}^{2} +frac{1}{2}left(10^{3}text{ kg/m}^{3}right)left[left(25.5 text{ m/s}right)^{2}-left(1.96 text{ m/s}right)^{2}right] = 4.24times {10}^{5}text{ N/m}^{2}end{array}[/latex]

Discussion

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This absolute pressure in the hose is greater than in the nozzle, as expected since v is greater in the nozzle. The pressure P₂ in the nozzle must be atmospheric since it emerges into the atmosphere without other changes in conditions.

There are a number of devices and situations in which fluid flows at a constant height and, thus, can be analyzed with Bernoulli’s principle.

People have long put the Bernoulli principle to work by using reduced pressure in high-velocity fluids to move things about. With a higher pressure on the outside, the high-velocity 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 low-lying areas. Some other devices that use the concept of entrainment are shown in Figure 2.

Figure 2. Examples of entrainment devices that use increased fluid speed to create low pressures, which then entrain one fluid into another. (a) A Bunsen burner uses an adjustable gas nozzle, entraining air for proper combustion. (b) An atomizer uses a squeeze bulb to create a jet of air that entrains drops of perfume. Paint sprayers and carburetors use very similar techniques to move their respective liquids. (c) A common aspirator uses a high-speed stream of water to create a region of lower pressure. Aspirators may be used as suction pumps in dental and surgical situations or for draining a flooded basement or producing a reduced pressure in a vessel. (d) The chimney of a water heater is designed to entrain air into the pipe leading through the ceiling.

The airplane wing is a beautiful example of Bernoulli’s principle in action. Figure 2(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 2(b).) The pressure on the front side of the sail, P_front, is lower than the pressure on the back of the sail, P_back. This results in a forward force and even allows you to sail into the wind.

Making Connections: Take-Home Investigation with Two Strips of Paper

Figure 3 shows two devices that measure fluid velocity based on Bernoulli’s principle. The manometer in Figure 3(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 (v₁=0) in front of it, while fluid passing the other tube has velocity v₂. This means that Bernoulli’s principle as stated in [latex]{P}_{1}+frac{1}{2}{{rho v}_{1}}^{2}={P}_{2}+frac{1}{2}{{rho v}_{2}}^{2}[/latex] becomes [latex]{P}_{1}={P}_{2}+frac{1}{2}{{rho v}_{2}}^{2}[/latex].

Figure 3. (a) The Bernoulli principle helps explain lift generated by a wing. (b) Sails use the same technique to generate part of their thrust.

Thus pressure P₂ over the second opening is reduced by [latex]frac{1}{2}{{rho v}_{2}}^{2}[/latex], and so the fluid in the manometer rises by h on the side connected to the second opening, where

[latex]hpropto frac{1}{2}{{rho v}_{2}}^{2}[/latex]

(Recall that the symbol ∝ means “proportional to.”) Solving for v₂, we see that

[latex]{v}_{2}propto sqrt{h}[/latex].

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.

Figure 4. Measurement of fluid speed based on Bernoulli’s principle. (a) A manometer is connected to two tubes that are close together and small enough not to disturb the flow. Tube 1 is open at the end facing the flow. A dead spot having zero speed is created there. Tube 2 has an opening on the side, and so the fluid has a speed v across the opening; thus, pressure there drops. The difference in pressure at the manometer is [latex]frac{1}{2}{{rho v}_{2}}^{2}[/latex] , and so [latex]h[/latex] is proportional to [latex]frac{1}{2}{{rho v}_{2}}^{2}[/latex] . (b) This type of velocity measuring device is a Prandtl tube, also known as a pitot tube.

Section Summary

• Bernoulli’s equation states that the sum on each side of the following equation is constant, or the same at any two points in an incompressible frictionless fluid:

[latex]{P}_{1}+frac{1}{2}{{rho v}_{1}}^{2}+rho {gh}_{1}={P}_{2}+frac{1}{2}{{rho v}_{2}}^{2}+rho {gh}_{2}[/latex].

• Bernoulli’s principle is Bernoulli’s equation applied to situations in which depth is constant. The terms involving depth (or height h) subtract out, yielding
  [latex]{P}_{1}+frac{1}{2}{{rho v}_{1}}^{2}={P}_{2}+frac{1}{2}{{rho v}_{2}}^{2}[/latex].
• Bernoulli’s principle has many applications, including entrainment, wings and sails, and velocity measurement.

Conceptual Questions

1. You 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.

2. Water 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.

3. Refer to Figure 1. Answer the following two questions. Why is P_o less than atmospheric? Why is P_o greater than P_i ?

Figure 1. An overhead view of a car passing a truck on a highway. Air passing between the vehicles flows in a narrower channel and must increase its speed (v₂ is greater than v₁), causing the pressure between them to drop (P_i is less than P_o). Greater pressure on the outside pushes the car and truck together.

4. Give an example of entrainment not mentioned in the text.

5. Many entrainment devices have a constriction, called a Venturi, such as shown in Figure 5. How does this bolster entrainment?

6. Some chimney pipes have a T-shape, 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.

7. Is there a limit to the height to which an entrainment device can raise a fluid? Explain your answer.

8. Why is it preferable for airplanes to take off into the wind rather than with the wind?

9. Roofs 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.

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11. It 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.

12. Water 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.

13. A 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?

Figure 6. Atomizer: perfume bottle with tube to carry perfume up through the bottle. (credit: Antonia Foy, Flickr)

14. If you lower the window on a car while moving, an empty plastic bag can sometimes fly out the window. Why does this happen?

Problems & Exercises

1. Verify that pressure has units of energy per unit volume.

2. Suppose you have a wind speed gauge like the pitot tube shown in Example 2 from Flow Rate and Its Relation to Velocity. 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?

3. If 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?

4. Calculate the maximum height to which water could be squirted with the hose in Example 2 from Flow Rate and Its Relation to Velocity if it: (a) Emerges from the nozzle. (b) Emerges with the nozzle removed, assuming the same flow rate.

5. Every 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 220 m²? Typical air density in Boulder is 1.14 kg/m³, and the corresponding atmospheric pressure is 8.89 × 10⁴ N/m². (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.)

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 1.29 kg/m³. (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.

7. (a) What is the pressure drop due to the Bernoulli effect as water goes into a 3.00-cm-diameter nozzle from a 9.00-cm-diameter 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.)

8. (a) Using Bernoulli’s equation, show that the measured fluid speed vfor a pitot tube, like the one in Figure 4(b), is given by

[latex]v={left(frac{2rho′gh}{rho }right)}^{1/2}[/latex],

where h is the height of the manometer fluid,

[latex]rho′[/latex] is the density of the manometer fluid, [latex]rho[/latex] 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.

Glossary

Bernoulli’s equation:

the equation resulting from applying conservation of energy to an incompressible frictionless fluid: P + 1/2pv² + pgh = constant , through the fluid

Bernoulli’s principle:

Bernoulli’s equation applied at constant depth: P₁ + 1/2pv₁² = P₂ + 1/2pv₂²

Selected Solutions to Problems & Exercises

Bernoulli Roulette System Review

1. [latex]begin{array}{c} {P}&=&frac{text{Force}}{text{Area}}, (P)_{text{units}}&=&text{N/m}^{2}=text{N}cdottext{m/m}^{3}=text{J/m}^{3} &=& text{energy/volume}end{array}[/latex]

3. 184 mm Hg

5. 2.54 × 10⁵N

7. (a) 1.58 × 10⁶N/m² (b) 163 m