Chapter 9: Fluids

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Fisica Generale - Alan Giambattista, Betty McCarty Richardson

Copyright © 2008 – The McGraw-Hill Companies s.r.l.

Chapter 9: Fluids

•Introduction to Fluids

•Pressure

•Pascal’s Principle

•Gravity and Fluid Pressure

•Measurement of Pressure

•Archimedes’ Principle

•Continuity Equation

•Bernoulli’s Equation

•Viscosity and Viscous Drag

•Surface Tension

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§9.1 Fluids

A liquid will flow to take the shape of the container that holds it. A gas will completely fill its container.

Fluids are easily deformable by external forces.

A liquid is incompressible. Its volume is fixed and is impossible to change.

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§9.2 Pressure

Pressure arises from the collisions between the particles of a fluid with another object (container walls for example).

There is a momentum change (impulse) that is away from the container walls. There must be a force exerted on the particle by the wall.

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Pressure is defined as .A

FP

The units of pressure are N/m2 and are called Pascals (Pa).

Note: 1 atmosphere (atm) = 101.3 kPa

By Newton’s 3rd Law, there is a force on the wall due to the particle.

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Example (text problem 9.1): Someone steps on your toe, exerting a force of 500 N on an area of 1.0 cm2. What is the average pressure on that area in atmospheres?

atm 49

Pa 10013.1

atm 1

N/m 1

Pa 1N/m 100.5

m 101.0

N 500

5226

24-av

A

FP

242

2 m 100.1cm 100

m 1cm 0.1

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§9.3 Pascal’s Principle

A change in pressure at any point in a confined fluid is transmitted everywhere throughout the fluid. (This is useful in making a hydraulic lift.)

7



Apply a force F1 here to a piston of cross-sectional area A1.

The applied force is transmitted to the piston of cross-sectional area A2 here.

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Mathematically,

11

22

2

2

1

1

A

A

A

F

A

F

2point at 1point at

FF

PP

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Example: Assume that a force of 500 N (about 110 lbs) is applied to the smaller piston in the previous figure. For each case, compute the force on the larger piston if the ratio of the piston areas (A2/A1) are 1, 10, and 100.

F2

1 500 N

10 5000 N

100 50,000 N

12 AA

Using Pascal’s Principle:

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The work done pressing the smaller piston (#1) equals the work done by the larger piston (#2).

2211 dFdF

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Example: In the previous example, for the case A2/A1 = 10, it was found that F2/F1 = 10. If the larger piston needs to rise by 1 m, how far must the smaller piston be depressed?

m 1021

21 dF

Fd

Using the result on the previous slide,

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§9.4 Gravity’s Effect on Fluid Pressure

A cylinder of fluid

FBD for the fluid cylinder

P1A

P2Aw

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Apply Newton’s 2nd Law to the fluid cylinder:

gdPP

gdPP

gdPP

gAdAPAP

wAPAPF

12

12

12

12

12

or

0

0

0

If P1 (the pressure at the top of the cylinder) is known, then the above expression can be used to find the variation of pressure with depth in a fluid.

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If the top of the fluid column is placed at the surface of the fluid, then P1=Patm if the container is open.

gdPP atm

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Example (text problem 9.15): At the surface of a freshwater lake, the pressure is 105 kPa. (a) What is the pressure increase in going 35.0 m below the surface?

atm 3.4kPa 343

m 35m/s 8.9kg/m 1000

23

atm

atm

gdPPP

gdPP

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Example: The surface pressure on the planet Venus is 95 atm. How far below the surface of the ocean on Earth do you need to be to experience the same pressure? The density of seawater is 1025 kg/m3.

m 950

N/m 109.5m/s 8.9kg/m 1025

N/m 109.5atm 94

atm 1atm 95

2623

26

atm

d

d

gd

gd

gdPP

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§9.5 Measuring Pressure

A manometer is a U-shaped tube that is partially filled with liquid.

Both ends of the tube are open to the atmosphere.

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A container of gas is connected to one end of the U-tube

If there is a pressure difference between the gas and the atmosphere, a force will be exerted on the fluid in the U-tube. This changes the equilibrium position of the fluid in the tube.

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Also

atmc PP At point C

B'B PP

The pressure at point B is the pressure of the gas.

gdP

gdPPPP

gdPPP

BCB

CBB

gauge

atm

'

From the figure:

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A Barometer

The atmosphere pushes on the container of mercury which forces mercury up the closed, inverted tube. The distance d is called the barometric pressure.

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Atmospheric pressure is equivalent to a column of mercury 76.0 cm tall.

gdPA

From the figure atmBA PPP

and

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Example (text problem 9.22): An IV is connected to a patient’s vein. The blood pressure in the vein has a gauge pressure of 12 mm of mercury. At least how far above the vein must the IV bag be placed in order for fluid to flow into the vein? Assume that the density of the IV fluid is the same as blood.

Blood:

mm 121

1gauge

h

ghP Hg

IV: 2bloodgauge ghP

The pressure is equivalent to raising a column of mercury 12 mm tall.

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At a minimum, the gauge pressures must be equal. When h2 is large enough, fluid will flow from high pressure to low pressure.

mm 154

mm 12kg/m 1060

kg/m 600,133

3

1blood

blood

12

2blood1gauge

h

g

ghh

ghghP

Hg

Hg

Hg

Example continued:

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§9.6 Archimedes’ Principle

wF2

F1

An FBD for an object floating submerged in a fluid.

The total force on the block due to the fluid is called the buoyant force.

12

12

where FF

FFF

B

25



The magnitude of the buoyant force is:

APP

APAP

FFFB

12

12

12

gdPP 12From before:

gVgdAFB The result is

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gVFB

Archimedes’ Principle: A fluid exerts an upward buoyant force on a submerged object equal in magnitude to the weight of the volume of fluid displaced by the object.

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m 5.1m 10.0*m 0.20kg/m 1000

kg 100.3

0

3

5

A

md

mAd

mV

gmgVgm

wF

wFF

w

b

bw

bww

bwww

B

B

Example (text problem 9.30): A flat-bottomed barge loaded with coal has a mass of 3.0105 kg. The barge is 20.0 m long and 10.0 m wide. It floats in fresh water. What is the depth of the barge below the waterline?

w

FB

FBD for the barge

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Example (text problem 9.34): A piece of metal is released under water. The volume of the metal is 50.0 cm3 and its specific gravity is 5.0. What is its initial acceleration? (Note: when v=0, there is no drag force.)

w

FB

FBD for the metal

mawFF B

VgFB water

The buoyant force is the weight of the fluid displaced by the object

Solve for a:

1

ρ

ρ

ρ

ρ

objectobject

water

objectobject

water

V

Vgg

V

Vgg

m

Fa B

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Since the object is completely submerged V=Vobject.

water

gravity specific

where water = 1000 kg/m3 is the density of water at 4 °C.

Given 0.5gravity specificwater

object

2

objectobject

water m/s 8.710.5

11

..

11

ρ

ρ

g

GSg

V

Vga

Example continued:

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§9.7 Fluid Flow

A moving fluid will exert forces parallel to the surface over which it moves, unlike a static fluid. This gives rise to a viscous force that impedes the forward motion of the fluid.

A steady flow is one where the velocity at a given point in a fluid is constant.

V1 = constant

V2 = constant

v1v2

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Steady flow is laminar; the fluid flows in layers. The path that the fluid in these layers takes is called a streamline. Streamlines do not cross.

An ideal fluid is incompressible, undergoes laminar flow, and has no viscosity.

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The continuity equation—Conservation of mass.

The amount of mass that flows though the cross-sectional area A1 is the same as the mass that flows through cross-sectional area A2.

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is the mass flow rate (units kg/s)Avt

m

Avt

V

is the volume flow rate (units m3/s)

222111 vAvA The continuity equation is

If the fluid is incompressible, then 1= 2.

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Example (text problem 9.41): A garden hose of inner radius 1.0 cm carries water at 2.0 m/s. The nozzle at the end has radius 0.2 cm. How fast does the water move through the constriction?

m/s 50m/s 0.2cm 0.2

cm 0.12

122

21

12

12

2211

vr

rv

A

Av

vAvA

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§9.8 Bernoulli’s Equation

Bernoulli’s equation is a statement of energy conservation.

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2222

2111 2

1

2

1vgyPvgyP

Potential energy per unit volume

Kinetic energy per unit volume

Work per unit volume done by the fluid

Points 1 and 2 must be on the same streamline

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Example (text problem 9.50): A nozzle is connected to a horizontal hose. The nozzle shoots out water moving at 25.0 m/s. What is the gauge pressure of the water in the hose? Neglect viscosity and assume that the diameter of the nozzle is much smaller than the inner diameter of the hose.

2222

2111 2

1

2

1vgyPvgyP

Let point 1 be inside the hose and point 2 be outside the nozzle.

The hose is horizontal so y1=y2. Also P2 =Patm.

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21

22atm1

22atm

211

2

1

2

12

1

2

1

vvPP

vPvP

Substituting:

v2 = 25m/s and v1 is unknown. Use the continuity equation.

2

2

1

222

1

2

2

21

21

2

2v

d

dv

d

d

vA

Av

Since d2<<d1 it is true that v1<<v2.

Example continued:

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Pa 101.3

m/s 0.25kg/m 10002

12

1

2

12

1

2

1

5

23

22

21

22

21

22atm1

vvv

vvPP

Example continued:

40



§9.9 Viscosity

A real fluid has viscosity (fluid friction). This implies a pressure difference needs to be maintained across the ends of a pipe for fluid to flow.

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Viscosity also causes the existence of a velocity gradient across a pipe. A fluid flows more rapidly in the center of the pipe and more slowly closer to the walls of the pipe.

The volume flow rate for laminar flow of a viscous fluid is given by Poiseuille’s Law.

4

8r

LP

t

V

where is the viscosity

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Example (text problem 9.55): A hypodermic syringe attached to a needle has an internal radius of 0.300 mm and a length of 3.00 cm. The needle is filled with a solution of viscosity 2.0010-3 Pa sec; it is injected into a vein at a gauge pressure of 16.0 mm Hg.

(a) What must the pressure of the fluid in the syringe be in order to inject the solution at a rate of 0.150 mL/sec?

Solve Poiseuille’s Law for the pressure difference:

Pa 2830

scm15.0

cm 103.0

cm 00.3sec Pa 1000.288 3

41

2

4

t

V

r

LP

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Example continued:

This pressure difference is between the fluid in the syringe and the fluid in the vein; it is the given gauge pressure.

Pa 4960Pa 2830Pa 2130

PPP

PPP

vs

vs

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(b) What force must be applied to the plunger, which has an area of 1.00 cm2?

Example continued:

The result of (a) gives the force per unit area on the plunger so the force is just F = PA = 0.496 N.

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§9.10 Viscous Drag

The viscous drag force on a sphere is given by Stokes’ law.

rvFD 6

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Example (text problem 9.64): A sphere of radius 1.0 cm is dropped into a glass cylinder filled with a viscous liquid. The mass of the sphere is 12.0 g and the density of the liquid is 1200 kg/m3. The sphere reaches a terminal speed of 0.15 m/s. What is the viscosity of the liquid?

FBD for sphere

mawFFF BD

FD

w

x

y

FBApply Newton’s Second Law to the sphere

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Example continued:

When v = vterminal, a = 0 and

sec Pa 4.26

06

06

06

0

t

sls

sslt

sllt

slt

BD

rv

gVgm

gmgVrv

gmgVrv

gmgmrv

wFF

Solving for

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§9.11 Surface Tension

The surface of a fluid acts like a a stretched membrane (imagine standing on a trampoline). There is a force along the surface of the fluid.

The surface tension is a force per unit length.

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Example (text problem 9.70): Assume a water strider has a roughly circular foot of radius 0.02 mm. The water strider has 6 legs.

(a) What is the maximum possible upward force on the foot due to the surface tension of the water?

The water strider will be able to walk on water if the net upward force exerted by the water equals the weight of the insect. The upward force is supplied by the water’s surface tension.

N 1092 62

rr

PAF

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(b) What is the maximum mass of this water strider so that it can keep from breaking through the water surface?

Example continued:

To be in equilibrium, each leg must support one-sixth the weight of the insect.

kg 1056

or 6

1 6g

FmwF

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Summary

•Pressure and its Variation with Depth

•Pascal’s Principle

•Archimedes Principle

•Continuity Equation (conservation of mass)

•Bernoulli’s Equation (conservation of energy)

•Viscosity and Viscous Drag

•Surface Tension

Chapter 9: Fluids

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