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University Physics Volume 1

14.6 Bernoulli’s Equation

University Physics Volume 114.6 Bernoulli’s Equation
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  1. Preface
  2. Unit 1. Mechanics
    1. 1 Units and Measurement
      1. Introduction
      2. 1.1 The Scope and Scale of Physics
      3. 1.2 Units and Standards
      4. 1.3 Unit Conversion
      5. 1.4 Dimensional Analysis
      6. 1.5 Estimates and Fermi Calculations
      7. 1.6 Significant Figures
      8. 1.7 Solving Problems in Physics
      9. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
    2. 2 Vectors
      1. Introduction
      2. 2.1 Scalars and Vectors
      3. 2.2 Coordinate Systems and Components of a Vector
      4. 2.3 Algebra of Vectors
      5. 2.4 Products of Vectors
      6. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
    3. 3 Motion Along a Straight Line
      1. Introduction
      2. 3.1 Position, Displacement, and Average Velocity
      3. 3.2 Instantaneous Velocity and Speed
      4. 3.3 Average and Instantaneous Acceleration
      5. 3.4 Motion with Constant Acceleration
      6. 3.5 Free Fall
      7. 3.6 Finding Velocity and Displacement from Acceleration
      8. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
    4. 4 Motion in Two and Three Dimensions
      1. Introduction
      2. 4.1 Displacement and Velocity Vectors
      3. 4.2 Acceleration Vector
      4. 4.3 Projectile Motion
      5. 4.4 Uniform Circular Motion
      6. 4.5 Relative Motion in One and Two Dimensions
      7. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
    5. 5 Newton's Laws of Motion
      1. Introduction
      2. 5.1 Forces
      3. 5.2 Newton's First Law
      4. 5.3 Newton's Second Law
      5. 5.4 Mass and Weight
      6. 5.5 Newton’s Third Law
      7. 5.6 Common Forces
      8. 5.7 Drawing Free-Body Diagrams
      9. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
    6. 6 Applications of Newton's Laws
      1. Introduction
      2. 6.1 Solving Problems with Newton’s Laws
      3. 6.2 Friction
      4. 6.3 Centripetal Force
      5. 6.4 Drag Force and Terminal Speed
      6. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
    7. 7 Work and Kinetic Energy
      1. Introduction
      2. 7.1 Work
      3. 7.2 Kinetic Energy
      4. 7.3 Work-Energy Theorem
      5. 7.4 Power
      6. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
    8. 8 Potential Energy and Conservation of Energy
      1. Introduction
      2. 8.1 Potential Energy of a System
      3. 8.2 Conservative and Non-Conservative Forces
      4. 8.3 Conservation of Energy
      5. 8.4 Potential Energy Diagrams and Stability
      6. 8.5 Sources of Energy
      7. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
    9. 9 Linear Momentum and Collisions
      1. Introduction
      2. 9.1 Linear Momentum
      3. 9.2 Impulse and Collisions
      4. 9.3 Conservation of Linear Momentum
      5. 9.4 Types of Collisions
      6. 9.5 Collisions in Multiple Dimensions
      7. 9.6 Center of Mass
      8. 9.7 Rocket Propulsion
      9. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
    10. 10 Fixed-Axis Rotation
      1. Introduction
      2. 10.1 Rotational Variables
      3. 10.2 Rotation with Constant Angular Acceleration
      4. 10.3 Relating Angular and Translational Quantities
      5. 10.4 Moment of Inertia and Rotational Kinetic Energy
      6. 10.5 Calculating Moments of Inertia
      7. 10.6 Torque
      8. 10.7 Newton’s Second Law for Rotation
      9. 10.8 Work and Power for Rotational Motion
      10. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
    11. 11 Angular Momentum
      1. Introduction
      2. 11.1 Rolling Motion
      3. 11.2 Angular Momentum
      4. 11.3 Conservation of Angular Momentum
      5. 11.4 Precession of a Gyroscope
      6. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
    12. 12 Static Equilibrium and Elasticity
      1. Introduction
      2. 12.1 Conditions for Static Equilibrium
      3. 12.2 Examples of Static Equilibrium
      4. 12.3 Stress, Strain, and Elastic Modulus
      5. 12.4 Elasticity and Plasticity
      6. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
    13. 13 Gravitation
      1. Introduction
      2. 13.1 Newton's Law of Universal Gravitation
      3. 13.2 Gravitation Near Earth's Surface
      4. 13.3 Gravitational Potential Energy and Total Energy
      5. 13.4 Satellite Orbits and Energy
      6. 13.5 Kepler's Laws of Planetary Motion
      7. 13.6 Tidal Forces
      8. 13.7 Einstein's Theory of Gravity
      9. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
    14. 14 Fluid Mechanics
      1. Introduction
      2. 14.1 Fluids, Density, and Pressure
      3. 14.2 Measuring Pressure
      4. 14.3 Pascal's Principle and Hydraulics
      5. 14.4 Archimedes’ Principle and Buoyancy
      6. 14.5 Fluid Dynamics
      7. 14.6 Bernoulli’s Equation
      8. 14.7 Viscosity and Turbulence
      9. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
  3. Unit 2. Waves and Acoustics
    1. 15 Oscillations
      1. Introduction
      2. 15.1 Simple Harmonic Motion
      3. 15.2 Energy in Simple Harmonic Motion
      4. 15.3 Comparing Simple Harmonic Motion and Circular Motion
      5. 15.4 Pendulums
      6. 15.5 Damped Oscillations
      7. 15.6 Forced Oscillations
      8. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
    2. 16 Waves
      1. Introduction
      2. 16.1 Traveling Waves
      3. 16.2 Mathematics of Waves
      4. 16.3 Wave Speed on a Stretched String
      5. 16.4 Energy and Power of a Wave
      6. 16.5 Interference of Waves
      7. 16.6 Standing Waves and Resonance
      8. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
    3. 17 Sound
      1. Introduction
      2. 17.1 Sound Waves
      3. 17.2 Speed of Sound
      4. 17.3 Sound Intensity
      5. 17.4 Normal Modes of a Standing Sound Wave
      6. 17.5 Sources of Musical Sound
      7. 17.6 Beats
      8. 17.7 The Doppler Effect
      9. 17.8 Shock Waves
      10. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
  4. A | Units
  5. B | Conversion Factors
  6. C | Fundamental Constants
  7. D | Astronomical Data
  8. E | Mathematical Formulas
  9. F | Chemistry
  10. G | The Greek Alphabet
  11. Answer Key
    1. Chapter 1
    2. Chapter 2
    3. Chapter 3
    4. Chapter 4
    5. Chapter 5
    6. Chapter 6
    7. Chapter 7
    8. Chapter 8
    9. Chapter 9
    10. Chapter 10
    11. Chapter 11
    12. Chapter 12
    13. Chapter 13
    14. Chapter 14
    15. Chapter 15
    16. Chapter 16
    17. Chapter 17
  12. Index

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 the conservation of energy
  • Describe how to derive Bernoulli’s principle from Bernoulli’s equation
  • Perform calculations using Bernoulli’s principle
  • Describe some applications of Bernoulli’s principle

As we showed in Figure 14.27, when a fluid flows into a narrower channel, its speed increases. That means its kinetic energy also increases. The increased kinetic energy comes from the net work done on the fluid to push it into the channel. Also, if the fluid changes vertical position, work is done on the fluid by the gravitational force.

A pressure difference occurs when the channel narrows. This pressure difference results in a net force on the fluid because the pressure times the area equals the force, and this net force does work. Recall the work-energy theorem,

Wnet=12mv212mv02.Wnet=12mv212mv02.

The net work done increases the fluid’s kinetic energy. As a result, the pressure drops in a rapidly moving fluid whether or not the fluid is confined to a tube.

There are many common examples of pressure dropping in rapidly moving fluids. For instance, shower curtains have a disagreeable habit of bulging into the shower stall when the shower is on. The reason is that the high-velocity stream of water and air creates a region of lower pressure inside the shower, whereas the pressure on the other side remains at the standard atmospheric pressure. This pressure difference results in a net force, pushing the curtain inward. Similarly, when a car passes a truck on the highway, the two vehicles seem to pull toward each other. The reason is the same: The high velocity of the air between the car and the truck creates a region of lower pressure between the vehicles, and they are pushed together by greater pressure on the outside (Figure 14.29). 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 is an overhead view of a car passing a truck on a highway. Air passing between the vehicles flows in a narrower channel and increases the speed from v1 to v2, causing the pressure between vehicles to drop from Po to Pi.
Figure 14.29 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 (v2v2 is greater than v1v1), causing the pressure between them to drop (pipi is less than po).po). Greater pressure on the outside pushes the car and truck together.

Energy Conservation and Bernoulli’s Equation

The application of the principle of conservation of energy to frictionless laminar flow leads to a very useful relation between pressure and flow speed in a fluid. This relation is called Bernoulli’s equation, named after Daniel Bernoulli (1700–1782), who published his studies on fluid motion in his book Hydrodynamica (1738).

Consider an incompressible fluid flowing through a pipe that has a varying diameter and height, as shown in Figure 14.30. Subscripts 1 and 2 in the figure denote two locations along the pipe and illustrate the relationships between the areas of the cross sections A, the speed of flow v, the height from ground y, and the pressure p at each point. We assume here that the density at the two points is the same—therefore, density is denoted by ρρ without any subscripts—and since the fluid in incompressible, the shaded volumes must be equal.

Figure is the schematics of a fluid flowing in an “S”-shaped pipeline with the cross-section area reducing from A1 (left bottom part) to A2 (right top part). The left bottom part is at the height y1 above the ground; right top part is at the height h2 above the ground. Fluid moves with the velocity v1 at the bottom part and v2 at the top part. Volume of fluid dv takes dx1 in the part of pipeline and dx2 in the top part of pipeline.
Figure 14.30 The geometry used for the derivation of Bernoulli’s equation.

We also assume that there are no viscous forces in the fluid, so the energy of any part of the fluid will be conserved. To derive Bernoulli’s equation, we first calculate the work that was done on the fluid:

dW=F1dx1F2dx2dW=F1dx1F2dx2
dW=p1A1dx1p2A2dx2=p1dVp2dV=(p1p2)dV.dW=p1A1dx1p2A2dx2=p1dVp2dV=(p1p2)dV.

The work done was due to the conservative force of gravity and the change in the kinetic energy of the fluid. The change in the kinetic energy of the fluid is equal to

dK=12m2v2212m1v12=12ρdV(v22v12).dK=12m2v2212m1v12=12ρdV(v22v12).

The change in potential energy is

dU=mgy2mgy1=ρdVg(y2y1).dU=mgy2mgy1=ρdVg(y2y1).

The energy equation then becomes

dW=dK+dU (p1p2)dV=12ρdV(v22v12)+ρdVg(y2y1) (p1p2)=12ρ(v22v12)+ρg(y2y1).dW=dK+dU (p1p2)dV=12ρdV(v22v12)+ρdVg(y2y1) (p1p2)=12ρ(v22v12)+ρg(y2y1).

Rearranging the equation gives Bernoulli’s equation:

p1+12ρv12+ρgy1=p2+12ρv22+ρgy2.p1+12ρv12+ρgy1=p2+12ρv22+ρgy2.

This relation states that the mechanical energy of any part of the fluid changes as a result of the work done by the fluid external to that part, due to varying pressure along the way. Since the two points were chosen arbitrarily, we can write Bernoulli’s equation more generally as a conservation principle along the flow.

Bernoulli’s Equation

For an incompressible, frictionless fluid, the combination of pressure and the sum of kinetic and potential energy densities is constant not only over time, but also along a streamline:

p+12ρv2+ρgy=constantp+12ρv2+ρgy=constant
(14.16)

A special note must be made here of the fact that in a dynamic situation, the pressures at the same height in different parts of the fluid may be different if they have different speeds of flow.

Analyzing Bernoulli’s Equation

According to Bernoulli’s equation, if we follow a small volume of fluid along its path, various quantities in the sum may change, but the total remains constant. Bernoulli’s equation is, in fact, just a convenient statement of conservation of energy for an incompressible fluid in the absence of friction.

The general form of Bernoulli’s equation has three terms in it, and it is broadly applicable. To understand it better, let us consider some specific situations that simplify and illustrate its use and meaning.

Bernoulli’s equation for static fluids

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

p1+ρgh1=p2+ρgh2.p1+ρgh1=p2+ρgh2.

We can further simplify the equation by setting h2=0.h2=0. (Any height can be chosen for a reference height of zero, as is often done for other situations involving gravitational force, making all other heights relative.) In this case, we get

p2=p1+ρgh1.p2=p1+ρgh1.

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 h 1 , and consequently, p2p2 is greater than p1p1 by an amount ρgh1ρgh1. In the very simplest case, p1p1 is zero at the top of the fluid, and we get the familiar relationship p=ρghp=ρgh. (Recall thatp=ρgh(Recall thatp=ρgh and ΔUg=mgh.)ΔUg=mgh.) Thus, Bernoulli’s equation confirms the fact that the pressure change due to the weight of a fluid is ρghρgh. Although we introduce Bernoulli’s equation for fluid motion, it includes much of what we studied for static fluids earlier.

Bernoulli’s principle

Suppose a fluid is moving but its depth is constant—that is, h1=h2h1=h2. Under this condition, Bernoulli’s equation becomes

p1+12ρv12=p2+12ρv22.p1+12ρv12=p2+12ρv22.

Situations in which fluid flows at a constant depth are so common that this equation is often also called Bernoulli’s principle, which is simply 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.) Bernoulli’s principle reinforces the fact that pressure drops as speed increases in a moving fluid: If v2v2 is greater than v1v1 in the equation, then p2p2 must be less than p1p1 for the equality to hold.

Example 14.6

Calculating Pressure In Example 14.5, 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 1.01×105N/m21.01×105N/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 p1p1.

Solution Solving Bernoulli’s principle for p1p1 yields

p1=p2+12ρv2212ρv12=p2+12ρ(v22v12).p1=p2+12ρv2212ρv12=p2+12ρ(v22v12).

Substituting known values,

p1=1.01×105N/m2+12(103kg/m3)[(25.5 m/s)2(1.96 m/s)2]=4.24×105N/m2.p1=1.01×105N/m2+12(103kg/m3)[(25.5 m/s)2(1.96 m/s)2]=4.24×105N/m2.

Significance This absolute pressure in the hose is greater than in the nozzle, as expected, since v is greater in the nozzle. The pressure p2p2 in the nozzle must be atmospheric, because the water emerges into the atmosphere without other changes in conditions.

Applications of Bernoulli’s Principle

Many devices and situations occur in which fluid flows at a constant height and thus can be analyzed with Bernoulli’s principle.

Entrainment

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 as pumps to raise water to small heights, as is necessary for draining swamps, fields, or other low-lying areas. Some other devices that use the concept of entrainment are shown in Figure 14.31.

Figure A is a drawing of a Bunsen burner: air and natural gas enter at the bottom and move upwards. Figure B is a drawing of an atomizer: a squeeze bulb creates a horizontally flowing jet of air that pulls moving drops of perfume upwards and out of the bottle. Figure C is a drawing of a common aspirator that has water moving from the top to the bottom and combines with the flow of air that enters from a side. Figure D is a drawing of a chimney in which a flow of hot air moves upwards combining with the cool air entering from the side.
Figure 14.31 Entrainment devices 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.

Velocity measurement

Figure 14.32 shows two devices that apply Bernoulli’s principle to measure fluid velocity. The manometer in part (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=0v1=0) in front of it, while fluid passing the other tube has velocity v2v2. This means that Bernoulli’s principle as stated in

p1+12ρv12=p2+12ρv22p1+12ρv12=p2+12ρv22

becomes

p1=p2+12ρv22.p1=p2+12ρv22.

Thus pressure p2p2 over the second opening is reduced by 12ρv2212ρv22, so the fluid in the manometer rises by h on the side connected to the second opening, where

h12ρv22.h12ρv22.

(Recall that the symbol means “proportional to.”) Solving for v 2 v 2 , we see that

v2h.v2h.

Part (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 A is a drawing of a manometer that 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 while Tube 2 has an opening on the side. Figure B is a drawing of a manometer that is connected to two tubes one of which (Tube 1) is inserted into another (Tube 2). Tube 1 is open at the end facing the flow while Tube 2 has an opening on the side.
Figure 14.32 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, so the fluid has a speed v across the opening; thus, pressure there drops. The difference in pressure at the manometer is 12ρv2212ρv22, so h is proportional to 12ρv22.12ρv22. (b) This type of velocity measuring device is a Prandtl tube, also known as a pitot tube.

A fire hose

All preceding applications of Bernoulli’s equation involved simplifying conditions, such as constant height or constant pressure. The next example is a more general application of Bernoulli’s equation in which pressure, velocity, and height all change.

Example 14.7

Calculating Pressure: A Fire Hose Nozzle Fire hoses used in major structural fires have an inside diameter of 6.40 cm (Figure 14.33). Suppose such a hose carries a flow of 40.0 L/s, starting at a gauge pressure of 1.62×106N/m21.62×106N/m2. The hose rises up 10.0 m along a ladder to a nozzle having an inside diameter of 3.00 cm. What is the pressure in the nozzle?

Figure is a drawing of the fire truck with the extended ladder. Fireman on the top of the ladder uses hose to extinguish the fire. The flow of water from the hose is parallel to the ground and is 10 meters above it.
Figure 14.33 Pressure in the nozzle of this fire hose is less than at ground level for two reasons: The water has to go uphill to get to the nozzle, and speed increases in the nozzle. In spite of its lowered pressure, the water can exert a large force on anything it strikes by virtue of its kinetic energy. Pressure in the water stream becomes equal to atmospheric pressure once it emerges into the air.

Strategy We must use Bernoulli’s equation to solve for the pressure, since depth is not constant.

Solution Bernoulli’s equation is

p1+12ρv12+ρgh1=p2+12ρv22+ρgh2p1+12ρv12+ρgh1=p2+12ρv22+ρgh2

where subscripts 1 and 2 refer to the initial conditions at ground level and the final conditions inside the nozzle, respectively. We must first find the speeds v1v1 and v2v2. Since Q=A1v1Q=A1v1, we get

v1=QA1=40.0×10−3m3/sπ(3.20×10−2m)2=12.4m/s.v1=QA1=40.0×10−3m3/sπ(3.20×10−2m)2=12.4m/s.

Similarly, we find

v2=56.6m/s.v2=56.6m/s.

This rather large speed is helpful in reaching the fire. Now, taking h1h1 to be zero, we solve Bernoulli’s equation for p2p2:

p2=p1+12ρ(v12v22)ρgh2.p2=p1+12ρ(v12v22)ρgh2.

Substituting known values yields

p2=1.62×106N/m2+12(1000kg/m3)[(12.4 m/s)2(56.6 m/s)2](1000kg/m3)(9.80m/s2)(10.0m)=0.p2=1.62×106N/m2+12(1000kg/m3)[(12.4 m/s)2(56.6 m/s)2](1000kg/m3)(9.80m/s2)(10.0m)=0.

Significance This value is a gauge pressure, since the initial pressure was given as a gauge pressure. Thus, the nozzle pressure equals atmospheric pressure as it must, because the water exits into the atmosphere without changes in its conditions.

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