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University Physics Volume 1

14.7 Viscosity and Turbulence

University Physics Volume 114.7 Viscosity and Turbulence
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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 what viscosity is
  • Calculate flow and resistance with Poiseuille's law
  • Explain how pressure drops due to resistance
  • Calculate the Reynolds number for an object moving through a fluid
  • Use the Reynolds number for a system to determine whether it is laminar or turbulent
  • Describe the conditions under which an object has a terminal speed

In Applications of Newton’s Laws, which introduced the concept of friction, we saw that an object sliding across the floor with an initial velocity and no applied force comes to rest due to the force of friction. Friction depends on the types of materials in contact and is proportional to the normal force. We also discussed drag and air resistance in that same chapter. We explained that at low speeds, the drag is proportional to the velocity, whereas at high speeds, drag is proportional to the velocity squared. In this section, we introduce the forces of friction that act on fluids in motion. For example, a fluid flowing through a pipe is subject to resistance, a type of friction, between the fluid and the walls. Friction also occurs between the different layers of fluid. These resistive forces affect the way the fluid flows through the pipe.

Viscosity and Laminar Flow

When you pour yourself a glass of juice, the liquid flows freely and quickly. But if you pour maple syrup on your pancakes, that liquid flows slowly and sticks to the pitcher. The difference is fluid friction, both within the fluid itself and between the fluid and its surroundings. We call this property of fluids viscosity. Juice has low viscosity, whereas syrup has high viscosity.

The precise definition of viscosity is based on laminar, or nonturbulent, flow. Figure 14.34 shows schematically how laminar and turbulent flow differ. When flow is laminar, layers flow without mixing. When flow is turbulent, the layers mix, and significant velocities occur in directions other than the overall direction of flow.

Figure A is the schematic of a laminar flow that occurs in layers without mixing. Fluid velocity is different for the different layers. Figure B is the schematic of a turbulent flow caused by the obstruction. Turbulent flow mixes the fluid resulting in the uniform fluid velocity.
Figure 14.34 (a) Laminar flow occurs in layers without mixing. Notice that viscosity causes drag between layers as well as with the fixed surface. The speed near the bottom of the flow (vbvb) is less than speed near the top (vtvt) because in this case, the surface of the containing vessel is at the bottom. (b) An obstruction in the vessel causes turbulent flow. Turbulent flow mixes the fluid. There is more interaction, greater heating, and more resistance than in laminar flow.

Turbulence is a fluid flow in which layers mix together via eddies and swirls. It has two main causes. First, any obstruction or sharp corner, such as in a faucet, creates turbulence by imparting velocities perpendicular to the flow. Second, high speeds cause turbulence. The drag between adjacent layers of fluid and between the fluid and its surroundings can form swirls and eddies if the speed is great enough. In Figure 14.35, the speed of the accelerating smoke reaches the point that it begins to swirl due to the drag between the smoke and the surrounding air.

Figure is a photo of smoke that rises smoothly at the bottom and forms swirls and eddies at the top.
Figure 14.35 Smoke rises smoothly for a while and then begins to form swirls and eddies. The smooth flow is called laminar flow, whereas the swirls and eddies typify turbulent flow. Smoke rises more rapidly when flowing smoothly than after it becomes turbulent, suggesting that turbulence poses more resistance to flow. (credit: “Creativity103”/Flickr)

Figure 14.36 shows how viscosity is measured for a fluid. The fluid to be measured is placed between two parallel plates. The bottom plate is held fixed, while the top plate is moved to the right, dragging fluid with it. The layer (or lamina) of fluid in contact with either plate does not move relative to the plate, so the top layer moves at speed v while the bottom layer remains at rest. Each successive layer from the top down exerts a force on the one below it, trying to drag it along, producing a continuous variation in speed from v to 0 as shown. Care is taken to ensure that the flow is laminar, that is, the layers do not mix. The motion in the figure is like a continuous shearing motion. Fluids have zero shear strength, but the rate at which they are sheared is related to the same geometrical factors A and L as is shear deformation for solids. In the diagram, the fluid is initially at rest. The layer of fluid in contact with the moving plate is accelerated and starts to move due to the internal friction between moving plate and the fluid. The next layer is in contact with the moving layer; since there is internal friction between the two layers, it also accelerates, and so on through the depth of the fluid. There is also internal friction between the stationary plate and the lowest layer of fluid, next to the station plate. The force is required to keep the plate moving at a constant velocity due to the internal friction.

Figure is a schematic drawing of the set-up for the measurement of viscosity for laminar flow of fluid between two plates of area A. L is the separation between two plates. The bottom plate is fixed. When the top plate is pushed to the right, it drags the fluid along with it.
Figure 14.36 Measurement of viscosity for laminar flow of fluid between two plates of area A. The bottom plate is fixed. When the top plate is pushed to the right, it drags the fluid along with it.

A force F is required to keep the top plate in Figure 14.36 moving at a constant velocity v, and experiments have shown that this force depends on four factors. First, F is directly proportional to v (until the speed is so high that turbulence occurs—then a much larger force is needed, and it has a more complicated dependence on v). Second, F is proportional to the area A of the plate. This relationship seems reasonable, since A is directly proportional to the amount of fluid being moved. Third, F is inversely proportional to the distance between the plates L. This relationship is also reasonable; L is like a lever arm, and the greater the lever arm, the less the force that is needed. Fourth, F is directly proportional to the coefficient of viscosity, ηη. The greater the viscosity, the greater the force required. These dependencies are combined into the equation

F=ηvAL.F=ηvAL.

This equation gives us a working definition of fluid viscosity ηη. Solving for ηη gives

η=FLvAη=FLvA
(14.17)

which defines viscosity in terms of how it is measured.

The SI unit of viscosity is Nm/[(m/s)m2]=(N/m2)s or PasNm/[(m/s)m2]=(N/m2)s or Pas. Table 14.4 lists the coefficients of viscosity for various fluids. Viscosity varies from one fluid to another by several orders of magnitude. As you might expect, the viscosities of gases are much less than those of liquids, and these viscosities often depend on temperature.

Fluid Temperature
(°C)(°C)
Viscosity
ηη ×103×103
Air 0 0.0171
20 0.0181
40 0.0190
100 0.0218
Ammonia 20 0.00974
Carbon dioxide 20 0.0147
Helium 20 0.0196
Hydrogen 0 0.0090
Mercury 20 0.0450
Oxygen 20 0.0203
Steam 100 0.0130
Liquid water 0 1.792
20 1.002
37 0.6947
40 0.653
100 0.282
Whole blood 20 3.015
37 2.084
Blood plasma 20 1.810
37 1.257
Ethyl alcohol 20 1.20
Methanol 20 0.584
Oil (heavy machine) 20 660
Oil (motor, SAE 10) 30 200
Oil (olive) 20 138
Glycerin 20 1500
Honey 20 2000–10000
Maple syrup 20 2000–3000
Milk 20 3.0
Oil (corn) 20 65
Table 14.4 Coefficients of Viscosity of Various Fluids

Laminar Flow Confined to Tubes: Poiseuille’s Law

What causes flow? The answer, not surprisingly, is a pressure difference. In fact, there is a very simple relationship between horizontal flow and pressure. Flow rate Q is in the direction from high to low pressure. The greater the pressure differential between two points, the greater the flow rate. This relationship can be stated as

Q=p2p1RQ=p2p1R

where p1p1 and p2p2 are the pressures at two points, such as at either end of a tube, and R is the resistance to flow. The resistance R includes everything, except pressure, that affects flow rate. For example, R is greater for a long tube than for a short one. The greater the viscosity of a fluid, the greater the value of R. Turbulence greatly increases R, whereas increasing the diameter of a tube decreases R.

If viscosity is zero, the fluid is frictionless and the resistance to flow is also zero. Comparing frictionless flow in a tube to viscous flow, as in Figure 14.37, we see that for a viscous fluid, speed is greatest at midstream because of drag at the boundaries. We can see the effect of viscosity in a Bunsen burner flame [part (c)], even though the viscosity of natural gas is small.

Figure A is a schematic drawing of the non-viscous flow of fluid in a tube. All layers of fluid move with the same speed. Figure B is a schematic drawing of the nonviscous flow of fluid in a tube. Layers at the center of the tube move at a higher speed. Figure C is a photo of a Bunsen burner with the conical – shaped flame above it.
Figure 14.37 (a) If fluid flow in a tube has negligible resistance, the speed is the same all across the tube. (b) When a viscous fluid flows through a tube, its speed at the walls is zero, increasing steadily to its maximum at the center of the tube. (c) The shape of a Bunsen burner flame is due to the velocity profile across the tube. (credit c: modification of work by "jasonwoodhead23"/Flickr)

The resistance R to laminar flow of an incompressible fluid with viscosity ηη through a horizontal tube of uniform radius r and length l, is given by

R=8ηlπr4.R=8ηlπr4.
(14.18)

This equation is called Poiseuille’s law for resistance, named after the French scientist J. L. Poiseuille (1799–1869), who derived it in an attempt to understand the flow of blood through the body.

Let us examine Poiseuille’s expression for R to see if it makes good intuitive sense. We see that resistance is directly proportional to both fluid viscosity ηη and the length l of a tube. After all, both of these directly affect the amount of friction encountered—the greater either is, the greater the resistance and the smaller the flow. The radius r of a tube affects the resistance, which again makes sense, because the greater the radius, the greater the flow (all other factors remaining the same). But it is surprising that r is raised to the fourth power in Poiseuille’s law. This exponent means that any change in the radius of a tube has a very large effect on resistance. For example, doubling the radius of a tube decreases resistance by a factor of 24=1624=16.

Taken together, Q=p2p1RQ=p2p1R and R=8ηlπr4R=8ηlπr4 give the following expression for flow rate:

Q=(p2p1)πr48ηl.Q=(p2p1)πr48ηl.
(14.19)

This equation describes laminar flow through a tube. It is sometimes called Poiseuille’s law for laminar flow, or simply Poiseuille’s law (Figure 14.38).

Figure is the schematics of a tube of length l and radius r. Fluid flows through the tube in the direction from greater pressure p2 to lower pressure p1. Flow is laminar and is higher at the center of a tube.
Figure 14.38 Poiseuille’s law applies to laminar flow of an incompressible fluid of viscosity ηη through a tube of length l and radius r. The direction of flow is from greater to lower pressure. Flow rate Q is directly proportional to the pressure difference p2p1p2p1, and inversely proportional to the length l of the tube and viscosity ηη of the fluid. Flow rate increases with radius by a factor of r4r4.

Example 14.8

Using Flow Rate: Air Conditioning Systems An air conditioning system is being designed to supply air at a gauge pressure of 0.054 Pa at a temperature of 20°C.20°C. The air is sent through an insulated, round conduit with a diameter of 18.00 cm. The conduit is 20-meters long and is open to a room at atmospheric pressure 101.30 kPa. The room has a length of 12 meters, a width of 6 meters, and a height of 3 meters. (a) What is the volume flow rate through the pipe, assuming laminar flow? (b) Estimate the length of time to completely replace the air in the room. (c) The builders decide to save money by using a conduit with a diameter of 9.00 cm. What is the new flow rate?

Strategy Assuming laminar flow, Poiseuille’s law states that

Q=(p2p1)πr48ηl=dVdt.Q=(p2p1)πr48ηl=dVdt.

We need to compare the artery radius before and after the flow rate reduction. Note that we are given the diameter of the conduit, so we must divide by two to get the radius.

Solution

  1. Assuming a constant pressure difference and using the viscosity η=0.0181mPasη=0.0181mPas,
    Q=(0.054Pa)(3.14)(0.09m)48(0.0181×10−3Pas)(20m)=3.84×10−3m3s.Q=(0.054Pa)(3.14)(0.09m)48(0.0181×10−3Pas)(20m)=3.84×10−3m3s.
  2. Assuming constant flow Q=dVdtΔVΔtQ=dVdtΔVΔt
    Δt=ΔVQ=(12m)(6m)(3m)3.84×10−3m3s=5.63×104s=15.63hr.Δt=ΔVQ=(12m)(6m)(3m)3.84×10−3m3s=5.63×104s=15.63hr.
  3. Using laminar flow, Poiseuille’s law yields
    Q=(0.054Pa)(3.14)(0.045m)48(0.0181×10−3Pas)(20m)=2.40×10−4m3s.Q=(0.054Pa)(3.14)(0.045m)48(0.0181×10−3Pas)(20m)=2.40×10−4m3s.

    Thus, the radius of the conduit decreases by half reduces the flow rate to 6.25% of the original value.

Significance In general, assuming laminar flow, decreasing the radius has a more dramatic effect than changing the length. If the length is increased and all other variables remain constant, the flow rate is decreased:

QAQB=(p2p1)πrA48ηlA(p2p1)πrB48ηlB=lBlA QB=lAlBQA.QAQB=(p2p1)πrA48ηlA(p2p1)πrB48ηlB=lBlA QB=lAlBQA.

Doubling the length cuts the flow rate to one-half the original flow rate.

If the radius is decreased and all other variables remain constant, the volume flow rate decreases by a much larger factor.

QAQB=(p2p1)πrA48ηlA(p2p1)πrB48ηlB=(rArB)4 QB=(rBrA)4QAQAQB=(p2p1)πrA48ηlA(p2p1)πrB48ηlB=(rArB)4 QB=(rBrA)4QA

Cutting the radius in half decreases the flow rate to one-sixteenth the original flow rate.

Flow and Resistance as Causes of Pressure Drops

Water pressure in homes is sometimes lower than normal during times of heavy use, such as hot summer days. The drop in pressure occurs in the water main before it reaches individual homes. Let us consider flow through the water main as illustrated in Figure 14.39. We can understand why the pressure p1p1 to the home drops during times of heavy use by rearranging the equation for flow rate:

Q=p2p1Rp2p1=RQ.Q=p2p1Rp2p1=RQ.

In this case, p2p2 is the pressure at the water works and R is the resistance of the water main. During times of heavy use, the flow rate Q is large. This means that p2p1p2p1 must also be large. Thus p1p1 must decrease. It is correct to think of flow and resistance as causing the pressure to drop from p2p2 to p1p1. The equation p2p1=RQp2p1=RQ is valid for both laminar and turbulent flows.

Figure is the schematic drawing of few small water lines leading to the individual houses that merge into the main water line.
Figure 14.39 During times of heavy use, there is a significant pressure drop in a water main, and p1p1 supplied to users is significantly less than p2p2 created at the water works. If the flow is very small, then the pressure drop is negligible, and p2p1p2p1.

We can also use p2p1=RQp2p1=RQ to analyze pressure drops occurring in more complex systems in which the tube radius is not the same everywhere. Resistance is much greater in narrow places, such as in an obstructed coronary artery. For a given flow rate Q, the pressure drop is greatest where the tube is most narrow. This is how water faucets control flow. Additionally, R is greatly increased by turbulence, and a constriction that creates turbulence greatly reduces the pressure downstream. Plaque in an artery reduces pressure and hence flow, both by its resistance and by the turbulence it creates.

Measuring Turbulence

An indicator called the Reynolds number NRNR can reveal whether flow is laminar or turbulent. For flow in a tube of uniform diameter, the Reynolds number is defined as

NR=2ρvrη(flow in tube)NR=2ρvrη(flow in tube)
(14.20)

where ρρ is the fluid density, v its speed, ηη its viscosity, and r the tube radius. The Reynolds number is a dimensionless quantity. Experiments have revealed that NRNR is related to the onset of turbulence. For NRNR below about 2000, flow is laminar. For NRNR above about 3000, flow is turbulent.

For values of NRNR between about 2000 and 3000, flow is unstable—that is, it can be laminar, but small obstructions and surface roughness can make it turbulent, and it may oscillate randomly between being laminar and turbulent. In fact, the flow of a fluid with a Reynolds number between 2000 and 3000 is a good example of chaotic behavior. A system is defined to be chaotic when its behavior is so sensitive to some factor that it is extremely difficult to predict. It is difficult, but not impossible, to predict whether flow is turbulent or not when a fluid’s Reynold’s number falls in this range due to extremely sensitive dependence on factors like roughness and obstructions on the nature of the flow. A tiny variation in one factor has an exaggerated (or nonlinear) effect on the flow.

Example 14.9

Using Flow Rate: Turbulent Flow or Laminar Flow In Example 14.8, we found the volume flow rate of an air conditioning system to be Q=3.84×10−3m3/s.Q=3.84×10−3m3/s. This calculation assumed laminar flow. (a) Was this a good assumption? (b) At what velocity would the flow become turbulent?

Strategy To determine if the flow of air through the air conditioning system is laminar, we first need to find the velocity, which can be found by

Q=Av=πr2v.Q=Av=πr2v.

Then we can calculate the Reynold’s number, using the equation below, and determine if it falls in the range for laminar flow

R=2ρvrη.R=2ρvrη.

Solution

  1. Using the values given:
    v=Qπr2=3.84×10−3m3s3.14(0.09m)2=0.15ms R=2ρvrη=2(1.23kgm3)(0.15ms)(0.09m)0.0181×10−3Pas=1835.v=Qπr2=3.84×10−3m3s3.14(0.09m)2=0.15ms R=2ρvrη=2(1.23kgm3)(0.15ms)(0.09m)0.0181×10−3Pas=1835.

    Since the Reynolds number is 1835 < 2000, the flow is laminar and not turbulent. The assumption that the flow was laminar is valid.
  2. To find the maximum speed of the air to keep the flow laminar, consider the Reynold’s number.
    R=2ρvrη2000 v=2000(0.0181×10−3Pas)2(1.23kgm3)(0.09m)=0.16ms.R=2ρvrη2000 v=2000(0.0181×10−3Pas)2(1.23kgm3)(0.09m)=0.16ms.

Significance When transferring a fluid from one point to another, it desirable to limit turbulence. Turbulence results in wasted energy, as some of the energy intended to move the fluid is dissipated when eddies are formed. In this case, the air conditioning system will become less efficient once the velocity exceeds 0.16 m/s, since this is the point at which turbulence will begin to occur.

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