Chapter 15B Chapter 15B -- Fluids in MotionFluids in MotionA PowerPoint Presentation byA PowerPoint Presentation by

Paul E. Tippens, Professor of PhysicsPaul E. Tippens, Professor of Physics

Southern Polytechnic State UniversitySouthern Polytechnic State University

© 2007

The lower falls at Yellowstone National Park: the water at the top of the falls passes through a narrow slot, causing the velocity to increase at that point. In this chapter, we will study the physics of fluids in motion.

Fluid Motion

Paul E. Tippens

Objectives: After completing this Objectives: After completing this module, you should be able to:module, you should be able to:•• Define the Define the rate of flowrate of flow for a fluid and for a fluid and

solve problems using velocity and crosssolve problems using velocity and cross-- section. section.

•• Write and apply Write and apply BernoulliBernoulli’’s equations equation for for the general case and apply for (a) a fluid the general case and apply for (a) a fluid at rest, (b) a fluid at constant pressure, at rest, (b) a fluid at constant pressure, and (c) flow through a horizontal pipe.and (c) flow through a horizontal pipe.

Fluids in MotionFluids in Motion

All fluids are assumed in this treatment to

exhibit streamline flow.

• Streamline flow is the motion of a fluid in which every particle in the fluid follows the same path past a particular point as that followed by previous particles.

• Streamline flow is the motion of a fluid in which every particle in the fluid follows the same path past a particular point as that followed by previous particles.

Assumptions for Fluid Flow:Assumptions for Fluid Flow:

Streamline flow Turbulent flow

• All fluids move with streamline flow.

• The fluids are incompressible.

• There is no internal friction.

• All fluids move with streamline flow.

• The fluids are incompressible.

• There is no internal friction.

Rate of FlowRate of FlowThe The rate of rate of flowflow RR is defined as the volume is defined as the volume VV of a fluid of a fluid that passes a certain crossthat passes a certain cross--section section AA per unit of time per unit of time tt..

The volume The volume VV of fluid is given by of fluid is given by the product of area the product of area A A and and vtvt:: V Avt

AvtR vAt

Rate of flow = velocity x area

vt

Volume = A(vt)

A

Constant Rate of FlowConstant Rate of FlowFor an incompressible, frictionless fluid, the velocity For an incompressible, frictionless fluid, the velocity increases when the crossincreases when the cross--section decreases:section decreases:

1 1 2 2R v A v A

A1

A2

R = A1 v1 = A2 v2

v1

v2

v2

2 21 1 2 2v d v d

Example 1:Example 1: Water flows through a rubber Water flows through a rubber hose hose 2 cm2 cm in diameter at a velocity of in diameter at a velocity of 4 m/s4 m/s. . What must be the diameter of the nozzle in What must be the diameter of the nozzle in order that the water emerge at order that the water emerge at 16 m/s16 m/s??

The area is proportional to the square of diameter, so:

2 21 1 2 2v d v d

2 22 1 12 2

2

(4 m/s)(2 cm)(20 cm)

v ddv

d2 = 0.894 cmd2 = 0.894 cm

Example 1 (Cont.):Example 1 (Cont.): Water flows through a Water flows through a rubber hose rubber hose 2 cm2 cm in diameter at a velocity of in diameter at a velocity of 4 m/s4 m/s. What is the . What is the rate of flowrate of flow in min m33/min?/min?

2 21

1 1(4 m/s) (0.02 m)

4 4dR v

R1 = 0.00126 m3/s

1 1 2 2R v A v A

21

1 1 1; 4dR v A A

3

1m 1 min0.00126min 60 s

R R1 = 0.0754 m3/minR1 = 0.0754 m3/min

Problem Strategy for Rate of Flow:Problem Strategy for Rate of Flow:

• Read, draw, and label given information.

• The rate of flow R is volume per unit time.

• When cross-section changes, R is constant.

•• Read, draw, and label given information.Read, draw, and label given information.

•• The rate of flow R is volume per unit time.The rate of flow R is volume per unit time.

•• When crossWhen cross--section changes, R is constant.section changes, R is constant.

1 1 2 2R v A v A

• Be sure to use consistent units for area and velocity.

• Be sure to use consistent units for area and velocity.

Problem Strategy (Continued):Problem Strategy (Continued):

• Since the area A of a pipe is proportional to its diameter d, a more useful equation is:

•• Since the area Since the area AA of a pipe is proportional to its of a pipe is proportional to its diameter diameter dd, a more useful equation is:, a more useful equation is:

• The units of area, velocity, or diameter chosen for one section of pipe must be consistent with those used for any other section of pipe.

• The units of area, velocity, or diameter chosen for one section of pipe must be consistent with those used for any other section of pipe.

2 21 1 2 2v d v d

The Venturi MeterThe Venturi Meter

The higher velocity in the constriction B causes a The higher velocity in the constriction B causes a difference of pressure between points A and B.difference of pressure between points A and B.

PA - PB = ghPA - PB = gh

h

AB

C

Demonstrations of the Venturi PrincipleDemonstrations of the Venturi Principle

The increase in air velocity produces a difference of pressure that exerts the forces shown.

Examples of the Venturi Effect

Work in Moving a Work in Moving a Volume of FluidVolume of Fluid

P1

A1

P1

A1

P2

A2

A2

P2

h

Volume V

Note differences in pressure P and area A

Fluid is raised to a height h.

22 2 2 2

2

; FP F P AA

11 1 1 1

1

; FP F PAA

F1

, F2

Work on a Fluid (Cont.)Work on a Fluid (Cont.)

Net work done on fluid is sum of work done by input force Fi less the work done by resisting force F2 , as shown in figure.

Net work done on fluid is sum of work done by input force Fi less the work done by resisting force F2 , as shown in figure.

Net Work = P1 V - P2 V = (P1 - P2 ) V

F1 = P1 A1

F2 = P2 A2

v1

v2

A1

A2

h2

h1 s1

s2

Conservation of EnergyConservation of Energy

Kinetic Energy K:2 22 1½ ½K mv mv

Potential Energy U:

2 1U mgh mgh

Net Work = K + U

2 21 2 2 1 2 2( ) (½ ½ ) ( )P P V mv mv mgh mgh

also Net Work = (P1 - P2 )V

F1 = P1 A1

F2 = P2 A2

v1

v2

A1

A2

h2

h1 s1

s2

Conservation of EnergyConservation of Energy2 2

1 2 2 1 2 2( ) (½ ½ ) ( )P P V mv mv mgh mgh

Divide by V, recall that density m/V, then simplify: 2 2

1 1 1 2 2 2½ ½P gh v P gh v

Bernoulli’s Theorem:2

1 1 1½P gh v Const

v1

v2

h1

h2

BernoulliBernoulli’’s Theorem (Horizontal Pipe):s Theorem (Horizontal Pipe):2 2

1 1 1 2 2 2½ ½P gh v P gh v

h1 = h2

v1 v2

Horizontal Pipe (h1 = h2 )

2 21 2 2 1½ ½P P v v

h

Now, since the difference in pressure P = gh,

2 22 1½ ½P gh v v Horizontal

Pipe

Example 3:Example 3: Water flowing at Water flowing at 4 m/s4 m/s passes through passes through a a VenturiVenturi tube as shown. If tube as shown. If h h = 12 cm= 12 cm, what is the , what is the velocity of the water in the constriction?velocity of the water in the constriction?

v1 = 4 m/s

v2h

h = 6 cm2 22 1½ ½P gh v v

Bernoulli’s Equation (h1 = h2 )

2gh = v22 - v1

2Cancel , then clear fractions:

2 2 22 12 2(9.8 m/s )(0.12 m) (4 m/s)v gh v

v2 = 4.28 m/sv2 = 4.28 m/s Note that density is not a factor.

BernoulliBernoulli’’s Theorem for Fluids at Rest.s Theorem for Fluids at Rest.

For many situations, the fluid remains at rest so that v1 and v2 are zero. In such cases we have:

2 21 1 1 2 2 2½ ½P gh v P gh v

P1 - P2 = gh2 - gh1 P = g(h2 - h1 )P = g(h2 - h1 )

h

= 1000 kg/m3

This is the same relation seen earlier for finding the pressure P at a given depth h = (h2 - h1 ) in a fluid.

TorricelliTorricelli’’s Theorems Theorem

2v gh

h1

h2h

When there is no change of pressure, P1 = P2.

2 21 1 1 2 2 2½ ½P gh v P gh v

Consider right figure. If surface v2 and P1 = P2 and v1 = v we have:

Torricelli’s theorem:

2v gh

v2

Interesting Example of Interesting Example of TorricelliTorricelli’’s Theorem:s Theorem:

vv

v

Torricelli’s theorem:

2v gh

• Discharge velocity increases with depth.

• Holes equidistant above and below midpoint will have same horizontal range.

• Maximum range is in the middle.

Example 4:Example 4: A dam springs a leak A dam springs a leak at a point at a point 20 m20 m below the surface. below the surface. What is the emergent velocity?What is the emergent velocity?

2v ghhTorricelli’s theorem:

2v gh

Given: h = 20 m g = 9.8 m/s2

22(9.8 m/s )(20 m)v

v = 19.8 m/s2v = 19.8 m/s2

Strategies for BernoulliStrategies for Bernoulli’’s Equation:s Equation:

• Read, draw, and label a rough sketch with givens.

• The height h of a fluid is from a common reference point to the center of mass of the fluid.

• In Bernoulli’s equation, the density is mass density and the appropriate units are kg/m3.

• Write Bernoulli’s equation for the problem and simplify by eliminating those factors that do not change.

• Read, draw, and label a rough sketch with givens.

• The height h of a fluid is from a common reference point to the center of mass of the fluid.

• In Bernoulli’s equation, the density is mass density and the appropriate units are kg/m3.

• Write Bernoulli’s equation for the problem and simplify by eliminating those factors that do not change.

2 21 1 1 2 2 2½ ½P gh v P gh v

Strategies (Continued)Strategies (Continued)2 2

1 1 1 2 2 2½ ½P gh v P gh v

• For a stationary fluid, v1 = v2 and we have:

P = g(h2 - h1 )P = g(h2 - h1 )

• For a horizontal pipe, h1 = h2 and we obtain:

h

= 1000 kg/m3

2 21 2 2 1½ ½P P v v

• For no change in pressure, P1 = P2 and we have:

Strategies (Continued)Strategies (Continued)2 2

1 1 1 2 2 2½ ½P gh v P gh v

2v gh

Torricelli’s Theorem

General Example: General Example: Water flows through the pipe at the rate of 30 L/s. The absolute pressure at point A is 200 kPa, and the point B is 8 m higher than point A. The lower section of pipe has a diameter of 16 cm and the upper section narrows to a diameter of 10 cm. Find the velocities of the stream at points A and B.

8 m

A

BR=30 L/s

AA = (0.08 m)2 = 0.0201 m3

AB = (0.05 m)2 = 0.00785 m3

2; 2DA R R

3 3

22 22

0.030 m /s 0.030 m /s1.49 m/s; 3.82 m/s 0.0201 m 0.00785 mA

A

R Rv vA A

vA = 1.49 m/s vB = 3.82 m/s

R = 30 L/s = 0.030 m3/s

General Example (Cont.):General Example (Cont.): Next find the absolute pressure at Point B.

8 m

A

BR=30 L/s

Consider the height hA = 0 for reference purposes.

Given: vA = 1.49 m/s vB = 3.82 m/s PA = 200 kPa hB - hA = 8 m

PA + ghA +½vA2 = PB + ghB + ½vB

20

PB = PA + ½vA2 - ghB - ½vB

2

PB = 200,000 Pa + ½1000 kg/m3)(1.49 m/s)2

– (1000 kg/m3)(9.8 m/s2)(8 m) - ½1000 kg/m3)(3.82 m/s)2

PB = 200,000 Pa + 1113 Pa –78,400 Pa – 7296 Pa

PB = 115 kPaPB = 115 kPa

SummarySummary

Bernoulli’s Theorem:2

1 1 1½P g h v C o n s ta n t

1 1 2 2R v A v A 2 21 1 2 2v d v d

Streamline Fluid Flow in Pipe:

PA - PB = ghHorizontal Pipe (h1 = h2 )

2 21 2 2 1½ ½P P v v

Fluid at Rest:

Torricelli’s theorem:

2v gh

Summary: BernoulliSummary: Bernoulli’’s Theorems Theorem

2 21 1 1 2 2 2½ ½P gh v P gh v

• Read, draw, and label a rough sketch with givens.

• The height h of a fluid is from a common reference point to the center of mass of the fluid.

• In Bernoulli’s equation, the density r is mass density and the appropriate units are kg/m3.

• Write Bernoulli’s equation for the problem and simplify by eliminating those factors that do not change.

• Read, draw, and label a rough sketch with givens.

• The height h of a fluid is from a common reference point to the center of mass of the fluid.

• In Bernoulli’s equation, the density r is mass density and the appropriate units are kg/m3.

• Write Bernoulli’s equation for the problem and simplify by eliminating those factors that do not change.

CONCLUSION: Chapter 15BCONCLUSION: Chapter 15B Fluids in MotionFluids in Motion