Chapter 9 Fluids. MFMcGraw-PHY 1401Ch09e - Fluids-Revised: 7/12/20102 Chapter 9: Fluids Introduction...

Chapter 9

Fluids

MFMcGraw-PHY 1401 Ch09e - Fluids-Revised: 7/12/2010 2

Chapter 9: Fluids• Introduction to Fluids

• Pressure

• Measurement of Pressure

• Pascal’s Principle

• Gravity and Fluid Pressure

• Archimedes’ Principle

• Continuity Equation

• Bernoulli’s Equation

• Viscosity and Viscous Drag

• Surface Tension


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.


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.


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

24av

A

FP

242

2 m 100.1cm 100

m 1cm 0.1

A 500N person

weighs about 113 lbs.


Gravity’s Effect on Fluid Pressure

An imaginary cylinder of fluid

FBD for the fluid cylinder

P1A

P2Aw

x

y

Imaginary cylinder can be any size


Apply Newton’s 2nd Law to the fluid cylinder. Since the fluids isn’t moving the net force is zero.

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.


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

You noticed on the previous slide that the areas canceled out. Only the height matters since that is the direction of gravity.

Think of the pressure as a force density in N/m2


Example (text problem 9.19): 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


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


Measuring Pressure

A manometer is a U-shaped tube that is partially filled with liquid, usually Mercury (Hg).

Both ends of the tube are open to the atmosphere.


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.


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:

gauge meas atmP = P - P


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.


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

gdPA

From the figure atmBA PPP

and


The Many Units of Pressure

1 ATM equals 1.013x105 N/m2

14.7 lbs/in2

1.013 bar

76 cm Hg

760 mm Hg

760 Torr

34 ft H2O

29.9 in Hg


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.)


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.

In these problems neglect pressure due to columns of fluid.


Mathematically,

11

22

2

2

1

1

A

A

A

F

A

F

2point at 1point at

FF

PP


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.

50,000 N100

5000 N10

500 N1

F212 AA

Using Pascal’s Principle:


Archimedes’ Principle

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

wF2

F1

x

y


The magnitude of the buoyant force is:

APP

APAP

FFFB

12

12

12

gdPP 12From before:

gVgdAFB The result is

Buoyant force = the weight of the fluid displaced


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.


Example (text problem 9.28): 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?

x

y

w

FB

FBD for the barge

Apply Newton’s 2nd Law to the barge:

bw

bww

bwww

B

B

mAd

mV

gmgVgm

wF

wFF

0

m 5.1m 10.0*m 0.20kg/m 1000

kg 100.33

5

A

md

w

b


Example (text problem 9.40): 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.)

FBD for the metal mawFF B

VgFB water

The magnitude of the buoyant force equals the weight of the fluid displaced by the metal.

Solve for a:

1

ρ

ρ

ρ

ρ

objectobject

water

objectobject

water

V

Vgg

V

Vgg

m

Fa B

Apply Newton’s 2nd Law to the piece of metal:

x

y

w

FB


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:

The sign is minus because gravity acts down. BF causes a < g.


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


Steady flow is laminar; the fluid flows in layers. The path that the fluid in these layers takes is called a streamline.

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

Streamlines do not cross.

Crossing streamlines would indicate a volume of fluid with two different velocities at the same time.


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.

Faster Slower


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.


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 0.20 cm. How fast does the water move through the constriction?

m/s 50m/s 0.2cm 0.20

cm 0.12

122

21

12

12

2211

vr

rv

A

Av

vAvA

Simple ratios


Bernoulli’s Equation

Bernoulli’s equation is a statement of energy conservation.


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

This is the most general equation


Example (text problem 9.49): A nozzle is connected to a horizontal hose. The nozzle shoots out water moving at 25 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.


21

22atm1

22atm

211

2

1

2

12

1

2

1

vvPP

vPvP

Substituting:

v2 = 25 m/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:


Pa 101.3

m/s 25kg/m 10002

12

1

2

12

1

2

1

5

23

22

21

22

21

22atm1

vvv

vvPP

Example continued:

1Since v 0


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.


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

4th power


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. Neglect the extra pressure required to accelerate the fluid from the syringe into the entrance needle.

(a) What must the pressure of the fluid in the syringe be in order to inject the solution at a rate of 0.250 mL/sec?

Solve Poiseuille’s Law for the pressure difference:

Pa 4716

seccm 250.0cm 103.0

cm 00.3sec Pa 1000.288 341

3

4

t

V

r

LP


Example continued:

This pressure difference is between the fluid in the syringe

and the fluid in the vein. The pressure in the syringe is

Pa 6860Pa 4720Pa 2140

PPP

PPP

vs

vs

51.013x10 Pa16 (mm Hg)× = 2132 Pa

760 mm Hg

Conversion:


(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 = 0.686 N.


Viscous Drag

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

rvFD 6

Where is the viscosity of the fluid that the sphere is falling through, r is the radius of the sphere, and v is the velocity of the sphere.


Example (text problem 9.62): 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

wx

y

FB

Apply Newton’s Second Law to the sphere

Drag Buoyant


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


Surface Tension

The surface of a fluid acts like 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.


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

r

rPAF


(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


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

Quick Questions


Consider a boat loaded with scrap iron in a swimming pool. If the iron is thrown overboard into the pool, will the water level at the edge of the pool rise, fall, or remain unchanged?

1. Rise2. Fall3. Remain unchanged


1. Rise2. Fall3. Remain unchanged

Consider a boat loaded with scrap iron in a swimming pool. If the iron is thrown overboard into the pool, will the water level at the edge of the pool rise, fall, or remain unchanged?


In the presence of air, the small iron ball and large plastic ball balance each other. When air is evacuated from the container, the larger ball

1. rises.2. falls.3. remains in place.


The weight of the stand and suspended solid iron ball is equal to the weight of the container of water as shown above. When the ball is lowered into the water the balance is upset. The amount of weight that must be added to the left side to restore balance, compared to the weight of water displaced by the ball, would be

1. half. 2. the same.3. twice. 4. more than twice.


A pair of identical balloons are inflated with air and suspended on theends of a stick that is horizontally balanced. When the balloon on the left is punctured, the balance of the stick is

1. upset and the stick rotates clockwise.

2. upset and the stick rotates counter-clockwise.

3. unchanged.


Consider an air-filled balloon weighted so that it is on the verge of sinking—that is, its overall density just equals that of water.Now if you push it beneath the surface, it will

1. sink.2. return to the surface.3. stay at the depth to

which it is pushed.

Extras


The density of the block of wood floating in water is

1. greater than the density of water.2. equal to the density of water.3. less than half that of water.4. more than half the density of water. 5. … not enough information is given.


1. greater than the density of water.2. equal to the density of water.3. less than half that of water.4. more than half the density of water. 5. … not enough information is given.

The density of the block of wood floating in water is


1. rises.2. falls.3. remains in place.

In the presence of air, the small iron ball and large plastic ball balance each other. When air is evacuated from the container, the larger ball


1. half. 2. the same.3. twice. 4. more than twice.

The weight of the stand and suspended solid iron ball is equal to the weight of the container of water as shown above. When the ball is lowered into the water the balance is upset. The amount of weight that must be added to the left side to restore balance, compared to the weight of water displaced by the ball, would be


1. upset and the stick rotates clockwise.

2. upset and the stick rotates counter-clockwise.

3. unchanged.

A pair of identical balloons are inflated with air and suspended on theends of a stick that is horizontally balanced. When the balloon on the left is punctured, the balance of the stick is


1. sink.2. return to the surface.3. stay at the depth to

which it is pushed.

Consider an air-filled balloon weighted so that it is on the verge of sinking—that is, its overall density just equals that of water.Now if you push it beneath the surface, it will

Chapter 9 Fluids. MFMcGraw-PHY 1401Ch09e - Fluids-Revised: 7/12/20102 Chapter 9: Fluids Introduction...

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