prova chap 13 sm

8/13/2019 Prova Chap 13 SM

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Chapter 13

Fluids

Conceptual Problems

1 If the gauge pressure is doubled, the absolute pressure will be(a) halved, (b) doubled, (c) unchanged, (d) increased by a factor greater than 2,(e) increased by a factor less than 2.

Determine the Concept The absolute pressure is related to the gauge pressure

according to atgauge PPP += .

Prior to doubling the gauge pressure,the absolute pressure is given by:

atgauge PPP += atgauge PPP =

Doubling the gauge pressure results inan absolute pressure given by:

atgauge2' PPP +=

Substituting for gaugeP yields: ( ) atatat 22' PPPPPP =+=

Because at2' PPP = , doubling the gauge pressure results in an increase in the

absolute pressure by a factor less than 2. ( )e is correct.

2 Two spherical objects differ in size and mass. Object A has a mass thatis eight times the mass of object B. The radius of object A is twice the radius of

object B. How do their densities compare? (a) BA > , (b) BA < ,

(c) BA = , (d) not enough information is given to compare their densities.

Determine the Concept The density ofan object is its mass per unit volume.The pictorial representation shown tothe right summarizes the informationconcerning the radii and masses of thetwo spheres. We can determine therelationship between the densities of Aand B by examining their ratio.

A

8

B

r m mr

2

Express the densities of the twospheres: 3

A34

A

A

AA

r

m

V

m

==

and

3

B34

B

B

BB

r

m

V

m

==


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Chapter 131272

Divide the first of these equations bythe second and simplify to obtain:

3

A

3B

B

A

3B3

4

B

3A3

4

A

B

A

r

r

m

m

r

m

r

m

==

Substituting for the masses and radiiand simplifying yields: ( )

12

83

3

B

A ==r

rmm

BA =

and ( )c is correct.

3 [SSM] Two objects differ in density and mass. Object A has a massthat is eight times the mass of object B. The density of object A is four times the

density of object B. How do their volumes compare? (a) B21

A VV = , (b) VA =VB ,

(c) VA = 2VB, (d) not enough information is given to compare their volumes.

Determine the Concept The density of an object is its mass per unit volume. Wecan determine the relationship between the volumes of A and B by examiningtheir ratio.

Express the volumes of the twoobjects:

A

AA

mV = and

B

BB

mV =

Divide the first of these equations bythe second and simplify to obtain:

B

A

A

B

B

B

A

A

B

A

m

m

m

m

V

V

==

Substituting for the masses anddensities and simplifying yields:

28

4 B

B

B

B

B

A ==m

m

V

V

BA 2VV =

and ( )c is correct.

4 A sphere is constructed by gluing together two hemispheres ofdifferent density materials. The density of each hemisphere is uniform, but thedensity of one is greater than the density of the other. True or false: The averagedensity of the sphere is the numerical average of the two different densities.

Clearly explain your reasoning.

Determine the Concept True. This is a special case. Because the volumes areequal, the average density is the numerical average of the two densities. This isnot a general result.


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Fluids 1273

5 In several jungle adventure movies, the hero and heroine escape thebad guys by hiding underwater for extended periods of time. To do this, theybreathe through long vertical hollow reeds. Imagine that in one movie, the wateris so clear that to be safely hidden the two are at a depth of 15 m. As a scienceconsultant to the movie producers, you tell them that this is not a realistic depthand the knowledgeable viewer will laugh during this scene. Explain why this is

so.

Determine the Concept Pressure increases approximately 1 atm every 10 m infresh water. To breathe requires creating a pressure of less than 1 atm in yourlungs. At the surface you can do this easily, but not at a depth of 10 m.

6 Two objects are balanced as in Figure 13-28. The objects haveidentical volumes but different masses. Assume all the objects in the figure aredenser than water and thus none will float. Will the equilibrium be disturbed if theentire system is completely immersed in water? Explain your reasoning.

Determine the Concept Yes. Because the volumes of the two objects are equal,the downward force on each side is reduced by the same amount (the buoyant

force acting on them) when they are submerged. The buoyant force is independent

of their masses. That is, if m1L1= m2L2andL1L2, then (m1c)L1(m2c)L2.

7 [SSM] A solid 200-g block of lead and a solid 200-g block of copperare completely submerged in an aquarium filled with water. Each block issuspended just above the bottom of the aquarium by a thread. Which of thefollowing is true?

(a) The buoyant force on the lead block is greater than the buoyant force on thecopper block.

(b) The buoyant force on the copper block is greater than the buoyant force onthe lead block.

(c) The buoyant force is the same on both blocks.(d) More information is needed to choose the correct answer.

Determine the Concept The buoyant forces acting on these submerged objects

are equal to the weight of the water each displaces. The weight of the displaced

water, in turn, is directly proportional to the volume of the submerged object.

Because Pb> Cu, the volume of the copper must be greater than that of the lead

and, hence, the buoyant force on the copper is greater than that on the lead.)(b is correct.

8 A 20-cm3block of lead and a 20-cm

3block of copper are completely

submerged in an aquarium filled with water. Each is suspended just above thebottom of the aquarium by a thread. Which of the following is true?


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Chapter 131274

(a) The buoyant force on the lead block is greater than the buoyant force on thecopper block..

(b) The buoyant force on the copper block is greater than the buoyant force onthe lead block.

(c) The buoyant force is the same on both blocks.(d) More information is needed to choose the correct answer.

Determine the Concept The buoyant forces acting on these submerged objects

are equal to the weight of the water each displaces. The weight of the displaced

water, in turn, is directly proportional to the volume of the submerged object.

Because their volumes are the same, the buoyant forces on them must be the

same. )(c is correct.

9 Two bricks are completely submerged in water. Brick 1 is made of

lead and has rectangular dimensions of 248. Brick 2 is made of wood andhas rectangular dimensions of 188. True or false: The buoyant force onbrick 2 is larger than the buoyant force on brick 1.

Determine the Concept False.The buoyant force on a submerged object depends

on the weight of the displaced fluid which, in turn, depends on the volume of the

displaced fluid. Because the bricks have the same volume, they will displace the

same volume of water and the buoyant force will be the same on both of them.

10 Figure 13-29 shows an object called a Cartesian diver.The diverconsists of a small tube, open at the bottom, with an air bubble at the top, inside aclosed plastic soda bottle that is partly filled with water. The diver normally

floats, but sinks when the bottle is squeezed hard. (a) Explain why this happens.(b) Explain the physics behind how a submarine can silentlysink verticallysimply by allowing water to flow into empty tanks near its keel. (c) Explain why afloating person will oscillate up and down on the water surface as he or shebreathes in and out.

Determine the Concept

(a) When the bottle is squeezed, the force is transmitted equally through the fluid,leading to a pressure increase on the air bubble in the diver. The air bubbleshrinks, and the loss in buoyancy is enough to sink the diver.

(b) As water enters its tanks, the weight of the submarine increases. When thesubmarine is completely submerged, the volume of the displaced water and,hence, the buoyant force acting on the submarine become constant. Because theweight of the submarine is now greater than the buoyant force acting on it, thesubmarine will start to sink.


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Fluids 1275

(c) Breathing in lowers ones average density and breathing out increases onesaverage density. Because denser objects float lower on the surface than do lessdense objects, a floating person will oscillate up and down on the water surface ashe or she breathes in and out.

11 A certain object has a density just slightly less than that of water sothat it floats almost completely submerged. However, the object is morecompressible than water. What happens if the floating object is given a slightdownward push? Explain.

Determine the Concept Because the pressure increases with depth, the object

will be compressed and its density will increase as its volume decreases. Thus, the

object will sink to the bottom.

12 In Example 13.11 the fluid is accelerated to a greater speed as it entersthe narrow part of the pipe. Identify the forces that act on the fluid at the entrance

to the narrow region to produce this acceleration.

Determine the Concept The acceleration-producing force acting on the fluid is

the product of the difference in pressure between the wide and narrow parts of the

pipe and the area of the narrow part of the pipe.

13 [SSM] An upright glass of water is accelerating to the right along aflat, horizontal surface. What is the origin of the force that produces theacceleration on a small element of water in the middle of the glass? Explain byusing a diagram.Hint: The water surface will not remain level as long as the

glass of water is accelerating. Draw a free body diagram of the small element ofwater.

Determine the Concept The pictorial representation shows the glass and an

element of water in the middle of the glass. As is readily established by a simple

demonstration, the surface of the water is not level while the glass is accelerated,

showing that there is a pressure gradient (a difference in pressure) due to the

differing depths (h1 > h2 and hence F1 > F2) of water on the two sides of the

element of water. This pressure gradient results in a net force on the element as

shown in the figure. The upward buoyant force is equal in magnitude to the

downward gravitational force.


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Chapter 131276

1h

2h

2F

1F

gF

bF

14 You are sitting in a boat floating on a very small pond. You take theanchor out of the boat and drop it into the water. Does the water level in the pond

rise, fall, or remain the same? Explain your answer.

Determine the Concept The water level in the pond will fall slightly.When the

anchor is in the boat, the boat displaces enough water so that the buoyant force on

it equals the sum of the weight of the boat, your weight, and the weight of the

anchor. When you drop the anchor into the water, it displaces just its volume of

water (rather than its weight as it did while in the boat). The total weight of the

boat becomes less and the boat displaces less water as a consequence.

15 A horizontal pipe narrows from a diameter of 10 cm at location A to

5.0 cm at location B. For a nonviscous incompressible fluid flowing withoutturbulence from location A to location B, how do the flow speeds v(in m/s) at the

two locations compare? (a) BA vv = , (b) B21

A vv = , (c) B41

A vv = , (d) BA 2vv = ,

(e) BA 4vv =

Determine the Concept We can use the equation of continuity to compare the

flow rates at the two locations.

Apply the equation of continuity atlocations A and B to obtain: BBAA

vAvA = BA

BA v

A

Av =

Substitute forABandAAandsimplify to obtain: B

2

A

BB2

A41

2

B41

A vd

dv

d

dv

==


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Fluids 1277

Substitute numerical values andevaluate vA: B4

1B

2

Acm10

cm5vvv =

=

and )(c is correct.

16 A horizontal pipe narrows from a diameter of 10 cm at location A to5.0 cm at location B. For a nonviscous incompressible fluid flowing withoutturbulence from location A to location B, how do the pressuresP(in N/m2) at the

two locations compare? (a) BA PP = , (b) B21

A PP = , (c) B41

A PP = , (d) BA 2PP = ,

(e) PA= 4PB , (f) There is not enough information to compare the pressuresquantitatively.

Determine the Concept We can use the equation of continuity to compare the

flow rates at the two locations.

Apply Bernoullis equation forconstant elevations at locations Aand B to obtain:

2

B21

B

2

A21

A vPvP +=+ (1)

Apply the equation of continuity atlocations A and B to obtain: BBAA

vAvA = BA

BA v

A

Av =

Substitute forABandAAandsimplify to obtain: B

2

A

BB2

A41

2

B41

A vd

dv

d

dv

==

Substituting for vAin equation (1)yields: 2

B21

B

2

B

2

A

B21

A vPvd

dP +=

+

While the values of dB, dA, and are known to us, we need a value for vB(or vA)

in order to comparePAandPB. Hence )(f is correct.

17 [SSM] Figure 13-30 is a diagram of a prairie dog tunnel. Thegeometry of the two entrances are such that entrance 1 is surrounded by a moundand entrance 2 is surrounded by flat ground. Explain how the tunnel remains

ventilated, and indicate in which direction air will flow through the tunnel.


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Chapter 131278

Determine the Concept The mounding around entrance 1 will cause thestreamlines to curve concave downward over the entrance. An upward pressuregradient produces the downward centripetal force. This means there is a loweringof the pressure at entrance 1. No such lowering occurs aver entrance 2, so thepressure there is higher than the pressure at entrance 1. The air circulates inentrance 2 and out entrance 1. It has been demonstrated that enough air will

circulate inside the tunnel even with slightest breeze outside.

Estimation and Approximation

18 Your undergraduate research project involves atmospheric sampling.The sampling device has a mass of 25.0 kg. Estimate the diameter of a helium-filled balloon required to lift the device off the ground. Neglect the mass of the

balloon skinand the small buoyancy force on the device itself.

Picture the Problem We can use Archimedes principle and the condition forvertical equilibrium to estimate the diameter of the helium-filled balloon that

would just lift the sampling device.

Express the equilibrium conditionthat must be satisfied if the balloon-

payload is to just lift off:

mgFBF === nalgravitatiobouyant (1)

Using Archimedes principle,express the buoyant forceB:

gV

gmwB

balloonair

fluiddisplacedfluiddisplaced

=

==

Substituting forBin equation (1)

yields:

mggV =balloonair

ormV =balloonair (2)

The total mass m to be lifted is thesum of the mass of the payload mpand the mass of the helium mHe:

balloonHepHep Vmmmm +=+=

Substitute for min equation (2) toobtain:

balloonHepballoonair VmV +=

Solving for balloonV yields:

Heair

p

balloon =

m

V

The volume of the balloon is givenby:

3

613

34

balloon drV ==


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Fluids 1279

Substituting for balloonV yields:

Heair

p3

61

=

md

( )3

Heair

p6

=

md

Substitute numerical values andevaluate d:

( )( )

m50.3

kg/m179.0kg/m293.1

kg0.2563

33

=

=

d

19 Your friend wants to start a business giving hot-air balloon rides. Theempty balloon, the basket and the occupants have a total maximum mass of1000 kg. If the balloon has a diameter of 22.0 m when fully inflated with hot air,estimate the required density of the hot air. Neglect the buoyancy force on thebasket and people.

Picture the Problem We can use Archimedes principle and the condition forvertical equilibrium to estimate the density of the hot air that would enable the

balloon and its payload to lift off.

Express the equilibrium conditionthat must be satisfied if the balloon-

payload is to just lift off:

mgFBF === nalgravitatiobouyant (1)

Using Archimedes principle,express the buoyant forceB:

gV

gmwB

balloonair

fluiddisplacedfluiddisplaced

=

==

Substituting forBin equation (1)

yields:

mggV =balloonair mV =balloonair (2)

The total mass m to be lifted is thesum of the mass of the payload mpand the mass of the hot air:

balloonairhotpairhotp Vmmmm +=+=

Substitute for min equation (2) toobtain:

balloonairhotpballoonair VmV +=

Solving for hot airand simplifyingyields:

balloon

p

air

balloon

pballoonair

airhotV

m

V

mV=

=

The volume of the balloon is givenby:

3

613

34

balloon drV ==

Substituting for balloonV yields:3

p

air3

61

p

airairhot

6

d

m

d

m

==


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Chapter 131280

Substitute numerical values and

evaluate hot air:( )( )

3

3

3

airhot

kg/m.111

m0.22

kg10006kg/m293.1

=

=

Remarks: As expected, the density of the hot air is considerably less than the

density of the surrounding cooler air.

Density

20 Find the mass of a solid lead sphere with a radius equal to 2.00 cm.

Picture the Problem The mass of the sphere is the product of its density and

volume. The density of lead can be found in Figure 13-1.

Using the definition of density,

express the mass of the sphere:

( )334 RVm ==


evaluate m:( )( )

kg379.0

m102.00kg/m103.113233

34

=

= m

21 [SSM] Consider a room measuring 4.0 m 5.0 m 4.0 m. Undernormal atmospheric conditions at Earths surface, what would be the mass of theair in the room?

Picture the Problem The mass of the air in the room is the product of its densityand volume. The density of air can be found in Figure 13-1.

Use the definition of density to

express the mass of the air in the

room:

LWHVm ==


evaluate m:

)( )( )( )kg100.1

m4.0m5.0m0.4kg/m293.1

2

3

=

=m

22 An average neutron star has approximately the same mass as the Sun,but is compressed into a sphere of radius roughly 10 km. What would be theapproximate mass of a teaspoonful of matter that dense?

Picture the ProblemWe can use the definition of density to find the approximatemass of a teaspoonful of matter from a neutron star. Assume that the volume of ateaspoon is about 5 mL.


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Fluids 1281


express the mass of a teaspoonful of

matter whose density is :

teaspoonVm = (1)

The density of the neutron star is

given by: NSNSNS

V

m

=

Substituting for mNSand VNSyields:3

NS34

SunNS

r

m

=

Let = NSin equation (1) to obtain:3

NS

teaspoonSun

4

3

r

Vmm

=


evaluate m:

( )( )( )

Tg2

m10104

m105kg1099.1333

3630

=

m

23 A 50.0-g ball consists of a plastic spherical shell and a water-filledcore. The shell has an outside diameter equal to 50.0 mm and an inside diameterequal to 20.0 mm. What is the density of the plastic?

Picture the ProblemWe can use the definition of density to find the density ofthe plastic of which the spherical shell is constructed.

The density of the plastic is given

by:plastic

plastic

plasticV

m= (1)

The mass of the plastic is the

difference between the mass of the

ball and the mass of its water-filled

core:

waterballplastic mmm =


express the mass of the water:

waterwaterwater Vm =

Substituting for mwateryields: waterwaterballplastic Vmm =

Substitute for mplasticin equation (1)

to obtain: plastic

waterwaterballplastic

V

Vm

=


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Chapter 131282

Because 3inside34

water RV = and

( )3inside3outside34ball RRV = :( )

( )3inside3outside343

inside34

waterballplastic

RR

Rm

=

Substitute numerical values and evaluate plsstic:

( )( )

( ) ( )( )3

33

34

3

34

3

plastic g/cm.7480mm0.10mm0.25

mm0.10cm

g00.1g0.50

=

=

24 A 60.0-mL flask is filled with mercury at 0C (Figure 13-31). Whenthe temperature rises to 80C, 1.47 gof mercury spills out of the flask. Assumingthat the volume of the flask stays constant, find the change in density of mercuryat 80C if its density at 0C is 13 645 kg/m

3.

Picture the Problem We can use the definition of density to relate the change in

the density of the mercury to the amount spilled during the heating process.

The change in the density of the

mercury as it is warmed is given by:

= 0 (1)

where 0is the density of the mercury

before it is warmed.

The density of the mercury before it

is warmed is the ratio of its mass to

the volume it occupies:0

00

V

m= (2)

The volume of the mercury that

spills is the ratio of its mass to its

density at the higher temperature:

spilled

spilled

mV =

The density of the mercury at the

higher temperature is given by:

spilled

0

0

spilled0

0

mV

m

VV

m

+

=+

=

Solving for yields:

0

spilled

0

0

spilled0

V

m

V

mm=

= (3)

Substituting equations (2) and (3) in

equation (1) yields:0

spilled

0

spilled

00V

m

V

m=

=


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Fluids 1283


evaluate :3

36

3

kg/m5.24m1060.0

kg101.47 =

=

25 One sphere is made of gold and has a radius rAuand another sphere ismade of copper and has a radius rCu. If the spheres have equal mass, what is theratio of the radii, r

Au/r

Cu?

Picture the ProblemWe can use the definition of density to find the ratio of theradii of the two spheres. See Table 13-1 for the densities of gold and copper.


express the mass of the gold sphere:

3

AuAu34

AuAuAu rVm ==

The mass of the copper sphere is

given by:

3

CuCu34

CuCuCu rVm ==

Dividing the first of these equations

by the second and simplifying yields:

3

Cu

Au

Cu

Au

3

CuCu34

3

AuAu34

Cu

Au

==

r

r

r

r

m

m

Solve for rAu/rCuto obtain:3

Au

Cu

Cu

Au

Cu

Au

m

m

r

r=

Because the spheres have the same

mass:3

Au

Cu

Cu

Au

=

r

r


evaluate rAu/rCu:773.0

kg/m9.31

kg/m93.83

3

3

Cu

Au ==r

r

26 Since 1983, the US Mint has coined pennies that are made out of zincwith a copper cladding. The mass of this type of penny is 2.50 g. Model the pennyas a uniform cylinder of height 1.23 mm and radius 9.50 mm. Assume the coppercladding is uniformly thick on all surfaces. If the density of zinc is 7140 kg/m3and that of copper is 8930 kg/m

3, what is the thickness of the copper cladding?

Picture the Problem The pictorial representation shows a zinc penny with its

copper cladding. We can use the definition of density to relate the differencebetween the mass of an all-copper penny and the mass of a copper-zinc penny tothe thickness dof the copper cladding.


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Chapter 131284

Zn

Cu

r2

h

d

Express the mass 0f the cladded

penny:ZnCu mmm +=

In terms of the densities of copper

and zinc, our equation becomes:ZnZnCuCu VVm +=

Substituting for the volumes of zinc and copper gives:

( )[ ] ( ) ( )dhdrddhrdrm 22222

Zn2

Cu ++=

Factor h and r from (h 2d) and (r d)2to obtain:

+

+=

h

dh

r

drd

h

drhdrm

211

2122

2

2

Zn

2

Cu

Assuming that d


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Fluids 1285

Letting hrepresent the height of

the column of mercury, express

the pressure at its base:

ghgh

Hg

Hg

kPa101kPa101

==

Substitute numerical values andevaluate h: ( )( )

inHg8.29

m102.54

in1m0.7570

m/s9.81kg/m1013.6kPa101

2

233

=

=

=

h

28 The pressure on the surface of a lake isPat= 101 kPa. (a) At what

depth is the pressure 2Pat? (b) If the pressure at the top of a deep pool of mercury

isPat, at what depth is the pressure 2Pat?

Picture the Problem The pressure due to a column of height h of a liquid of

density is given byP= gh.The pressure at any depth in a liquid is the sum of

the pressure at the surface of the liquid and the pressure due to the liquid at a

given depth in the liquid.

(a) Express the pressure as a function

of depth in the lake:ghPP waterat +=

g

PPh

water

at

=

Substitute forPand simplify to

obtain: g

P

g

PPh

water

at

water

atat2

=

=


evaluate h: ( )( )m3.10

m/s81.9kg/m101.00

kPa101233

=

=h

(b) Proceed as in (a) with water

replaced byHgto obtain: g

P

g

PPh

Hg

at

Hg

atat2

=

=

Substitute numerical values andevaluate h: ( )( )

cm7.57

m/s9.81kg/m1013.6kPa101

233

=

=h

29 When at cruising altitude, a typical airplane cabin will have an airpressure equivalent to an altitude of about 2400 m. During the flight, ears oftenequilibrate, so that the air pressure inside the inner ear equalizes with the air


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Chapter 131286

pressure outside the plane. The Eustachian tubes allow for this equalization, butcan become clogged. If an Eustachian tube is clogged, pressure equalization maynot occur on descent and the air pressure inside an inner ear may remain equal tothe pressure at 2400 m. In that case, by the time the plane lands and the cabin isrepressurized to sea-level air pressure, what is the net force on one ear drum, dueto this pressure difference, assuming the ear drum has an area of 0.50 cm2?

Picture the ProblemAssuming the density of the air to be constant, we can usethe definition of pressure and the expression for the variation of pressure withdepth in a fluid to find the net force on ones ear drums.

The net force acting on one ear drum

is given by:

PAF = whereAis the area of an ear drum.

Relate the pressure difference to the

pressure at 2400 m and the pressure

at sea level:

hgPPP m2400levelsea ==

Substituting for Pin the expressionforFyields:

hAgF =

Substitute numerical values and evaluateF:

( )( )( )( ) N5.1cm50.0m2400m/s9.81kg/m293.1 223 ==F

30 The axis of a cylindrical container is vertical. The container is filled

with equal masses of water and oil. The oil floats on top of the water, and theopen surface of the oil is at a height habove the bottom of the container. What isthe height h, if the pressure at the bottom of the water is 10 kPa greater than thepressure at the top of the oil? Assume the oil density is 875 kg/m

3.


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Fluids 1287

Picture the Problem The pictorial

representation represents oil floating on

water in a cylindrical container that is

open to the atmosphere. We can

express the height of the cylinder as the

sum of the heights of the cylinders ofoil and water and then use the fact that

the masses of oil and water are equal to

obtain a relationship between hoil and

hwater. This relationship, together with

the equation for the pressure as

function of depth in a liquid will lead

us to expressions for hoiland hwater.

oilh

hoil

water

waterh

bottomP

atP

Express has the sum of the heights

of the cylinders of oil and water:

wateroil hhh += (1)

The difference in pressure between

the bottom of the water and the top

of oil is:

A

gmgmPPP oilwateratbottom +

==

whereAis the cross-sectional area of

the cylinder.

Substituting for mwaterand moil and

simplifying gives:oiloilwaterwater

hh

g

P += (2)

Because the masses of oil and waterare equal:

1oiloil

waterwater

oiloil

waterwater

oil

water ===hh

AhAh

mm

and

oiloilwaterwater hh =

Substituting for oiloilh in equation

(2) yields:

waterwater

waterwaterwaterwater

2

h

hhg

P

=

+=

and

gPhwater

water2

=

Proceed similarly to obtain:

g

Ph

oil

oil2

=


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Chapter 131288

Substituting for hwaterand hoilin

equation (1) gives:

+=

+=

wateroil

wateroil

11

2

2

2

g

P

g

P

g

Ph

Substitute numerical values and evaluate h:

( )m1.1

kg/m1000.1

1

kg/m875

1

m/s81.92

kPa103332

=

+=h

31 [SSM] A hydraulic lift is used to raise a 1500-kg automobile. Theradius of the shaft of the lift is 8.00 cm and the radius of the compressors pistonis 1.00 cm. How much force must be applied to the piston to raise the automobile?Hint: The shaft of the lift is the other piston.

Picture the Problem The pressure applied to an enclosed liquid is transmitted

undiminished to every point in the fluid and to the walls of the container. Hence

we can equate the pressure produced by the force applied to the piston to the

pressure due to the weight of the automobile and solve forF.

Express the pressure the weight of

the automobile exerts on the shaft of

the lift:shaft

autoauto

A

wP =

Express the pressure the forceapplied to the piston produces: pistonA

FP=

Because the pressures are the same,

we can equate them to obtain: pistonshaft

auto

A

F

A

w=

Solving forFyields:

shaft

piston

auto

shaft

piston

autoA

Agm

A

AwF ==

Substitute numerical values andevaluateF: ( )( )

N230

cm8.00

cm1.00m/s9.81kg1500

22

=

=F


19/76

Fluids 1289

32 A 1500-kg car rests on four tires, each of which is inflated to a gaugepressure of 200 kPa. If the four tires support the cars weight equally, what is thearea of contact of each tire with the road?

Picture the Problem The area of contact of each tire with the road is related to

the weight on each tire and the pressure in the tire through the definition of

pressure.

Using the definition of gauge pressure,

relate the area of contact to the

pressure and the weight of the car:gaugegauge

41

4P

mg

P

wA ==


evaluateA:

( )( )( )

22

cm184kPa2004

m/s9.81kg1500==A

33 What pressure increase is required to compress the volume of 1.00 kgof water from 1.00 L to 0.99 L? Could this compression occur in our oceans

where the maximum depth is about 11 km? Explain.

Picture the Problem The required pressure Pis related to the change in volumeV and the initial volume V through the definition of the bulk modulus B;

VV

PB

= .

Using the definition of the bulk

modulus, relate the change in volumeto the initial volume and the required

pressure:

V

VBP

VV

PB

=

=


evaluate P:

atm197atm4.197

kPa101.325

atm1Pa1000.2

L1.00

L0.01Pa102.00

7

9

==

=

=P

Express the pressure at a depth h

in the ocean:

( ) ghPhP seawaterat +=

or

ghP seawater = g

Ph

seawater

=


20/76

Chapter 131290


evaluate h:( )( )

km0.2m/s81.9kg/m1025

atm

kPa101.325atm4.197

23

=h

Because a depth of only 2 km is required to produce a one-percent compression,

this does not occur in our oceans.

34 When a woman in high-heeled shoes takes a step, she momentarilyplaces her entire weight on one heel of her shoe. If her mass is 56.0 kg and if thearea of the heel is 1.00 cm2, what is the pressure exerted on the floor by the heel?Compare your answer to the pressure exerted by the one foot of an elephant on aflat floor. Assume the elephants mass is 5000 kg, that he has all four feet equallydistributed on the floor, and that each foot has an area of 400 cm2.

Picture the ProblemThe pressure exerted by the womans heel on the floor is

her weight divided by the area of her heel.

Using its definition, express the

pressure exerted on the floor by the

womans heel:

A

mg

A

w

A

FP ===


evaluatePwoman:

( )( )

atm2.45

kPa325.101

atm1N/m1049.5

m1000.1

m/s9.81kg56.0

26

24

2

woman

=

=

=

P

Express the pressure exerted by one of the elephants feet:

( )( )atm03.3

kPa325.101

atm1N/m10066.3

m10400

m/s9.81kg0005 2524

2

41

elephant ==

=

P

Express the ratio of the pressure

exerted by the womans heel to the

pressure exerted by the elephantsfoot:

18atm3.03

atm2.54

elephant

woman =P

P

Thus .18 elephantwoman PP

35 In the seventeenth century, Blaise Pascal performed the experimentshown in Figure 13-32. A wine barrel filled with water was coupled to a longtube. Water was added to the tube until the barrel burst. The radius of the lid was20 cm and the height of the water in the tube was 12 m. (a) Calculate the forceexerted on the lid due to the pressure increase. (b) If the tube had an inner radius


21/76

Fluids 1291

of 3.0 mm, what mass of water in the tube caused the pressure that burst thebarrel?

Picture the Problem The force on the lid is related to pressure exerted by the

water and the cross-sectional area of the column of water through the definition of

density. We can find the mass of the water from the product of its density and

volume.

(a) Using the definition of pressure,

express the force exerted on the lid:

PAF=

Express the pressure due to a column

of water of height h:

ghP water=

Substitute forPandAto obtain: 2water rghF =


evaluateF:

( )( )( ) ( )

kN15

m0.20m12

m/s9.81kg/m10001.

2

233

=

=

F

(b) Relate the mass of the water to

its density and volume:

2

waterwater rhVm ==

Substitute numerical values and evaluate m:

( )( ) ( ) kg0.34m103.0m12kg/m10001.2333

==

m

36 Blood plasma flows from a bag through a tube into a patients vein,where the blood pressure is 12 mmHg. The specific gravity of blood plasma at37C is 1.03. What is the minimum elevation of the bag so that the plasma flowsinto the vein?

Picture the Problem The minimum elevation of the bag h that will produce a

pressure of at least 12 mmHg is related to this pressure and the density of the

blood plasma through ghP blood= .

Using the definition of the pressure

due to a column of liquid, relate the

pressure at its base to its height:

ghP blood= g

Ph

blood=


22/76

Chapter 131292


evaluate h:

( )( )cm16

m/s9.81kg/m101.03

mmHg1

Pa133.32mmHg12

233

=

=h

37 Many people have imagined that if they were to float the top of aflexible snorkel tube out of the water, they would be able to breathe through itwhile walking underwater (Figure 13-33). However, they generally do not takeinto account just how much water pressure opposes the expansion of the chest andthe inflation of the lungs. Suppose you can just breathe while lying on the floorwith a 400-N (90-lb) weight on your chest. How far below the surface of thewater could your chest be for you still to be able to breathe, assuming your chesthas a frontal area of 0.090 m

2?

Picture the Problem The depth hbelow the surface at which you would be able

to breath is related to the pressure at that depth and the density of water wthrough ghP w= .

Express the pressure due to a column

of water of height h:ghP w=

g

Ph

w=

Express the pressure at depth hin

terms of the weight on your chest:chestyourof

chestyouron

A

wP=

Substituting forP yields:gA

wh

wchestyourof

chestyouron

=

Substitute numerical values and evaluate h:

cm452m/s9.813kg/m310001.2m0.090

N400=

=h

38 In Example 13-3 a 150-N force is applied to a small piston to lift a car

that weighs 15 000 N. Demonstrate that this does not violate the law ofconservation of energy by showing that, when the car is lifted some distance h,the work done by the 150-N force acting on the small piston equals the work doneon the car by the large piston.

Picture the Problem LetA1andA2represent the cross-sectional areas of the large

piston and the small piston, andF1andF2the forces exerted by the large and on

the small piston, respectively. The work done by the large piston is W1=F1h1and


23/76

Fluids 1293

that done on the small piston is W2= F2h2. Well use Pascals principle and the

equality of the volume of the displaced liquid in both pistons to show that W1and

W2are equal.

Express the work done in lifting the

car a distance h:

111 hFW =

whereF1is the weight of the car.

Using the definition of pressure,

relate the forcesF1 (= w) andF2to

the areasA1andA2:2

2

1

1

A

F

A

F=

2

121A

AFF =

Equate the volumes of the displaced

fluid in the two pistons:2211 AhAh =

1

221

A

Ahh =

Substitute in the expression for W1

and simplify to obtain:222

1

22

2

121 WhF

A

Ah

A

AFW ===

39 A 5.00-kg lead sinker is accidentally dropped overboard by fishermenin a boat directly above the deepest portion of the Marianas trench, near thePhilippines. By what percentage does the volume of the sinker change by the timeit settles to the trench bottom, which is 10.9 km below the surface?

Picture the ProblemThis problem is an application of the definition of the bulk

modulus. The change in volume of the sinker is related to the pressure change and

the bulk modulus by

VV

PB

= .

From the definition of bulk modulus:

B

P

V

V = (1)

Express the pressure at a depth h in

the ocean:

( ) ghPhP += at

The difference in pressure between

the pressure at depth hand the

surface of the water is:

( ) ghPhPP == at

Substitute for Pin equation (1) toobtain: B

gh

V

V =


24/76

Chapter 131294

Substitute numerical values and evaluate the fractional change in volume of the

sinker:

( )( )( )%4.1

GPa7.7

km9.10m/s81.9kg/m1025 23==

V

V

40 The volume of a cone of height h and base radius ris V= r2h/3. A jar

in the shape of a cone of height 25 cm has a base with a radius equal to 15 cm.The jar is filled with water. Then its lid (the base of the cone) is screwed on andthe jar is turned over so its lid is horizontal. (a) Find the volume and weight of thewater in the jar. (b) Assuming the pressure inside the jar at the top of the cone isequal to 1 atm, find the excess force exerted by the water on the base of the jar.Explain how this force can be greater than the weight of the water in the jar.

Picture the Problem The weight of the water in the jar is the product of its mass

and the gravitational field. Its mass, in turn, is related to its volume through thedefinition of density. The force the water exerts on the base of the jar can be

determined from the product of the pressure it creates and the area of the base.

(a) The volume of the water is:

( ) ( ) L9.5m10890.5m1025m1015 3322231 === V

Using the definition of density, relate

the weight of the water to the volume

it occupies:

Vgmgw ==

( ) N58N79.57s

m81.9m10890.5

m

kg1000.1

2

33

3

3 ==

= w

(b) Using the definition of pressure,

relate the force exerted by the water

on the base of the jar to the pressure

it exerts and the area of the base:

2rghPAF ==

Substitute numerical values and evaluateF:

( ) ( )( ) ( ) kN0.17m1015m1025m/s9.81kg/m1000.1 222233 == F


25/76

Fluids 1295

This occurs in the same way that the force on Pascals barrel is much greater than

the weight of the water in the tube. The downward force on the base is also the

result of the downward component of the force exerted by the slanting walls of

the cone on the water.

Buoyancy

41 A 500-g piece of copper with specific gravity 8.96, is suspended froma spring scale and is submerged in water (Figure 13-34). What force does thespring scale read?

Picture the Problem The scales reading is the difference between the weight of

the piece of copper in air and the buoyant force acting on it.

Express the apparent weight wof

the piece of copper:

Bww' =

Using the definition of density and

Archimedes principle, substitute for

wandBto obtain:( )Vg

VgVgw'

wCu

wCu

=

=

Express win terms of Cuand V

and solve for Vg: CuCu

wVgVgw ==

Substitute for Vg in the expression

for wto obtain: ( ) ww

w'

==Cu

w

Cu

wCu 1


evaluate w:( )( )

N4.36

m/s9.81kg0.5008.96

11 2

=

=w'

42 When a certain rock is suspended from a spring scale the scale-displayreads 60 N. However, when the suspended rock is submerged in water, the displayreads 40 N. What is the density of the rock?

Picture the Problem We can use the definition of density and Archimedes

principle to find the density of the rock. The difference between the weight of the

rock in air and the weight of the rock in water is the buoyant force acting on the

rock.


26/76

Chapter 131296


density of the rock: rock

rockrock

V

m= (1)

Apply Archimedes principle to

obtain:gV

gmwB

fluid

displaced

fluid

displaced

fluiddisplaced

fluiddisplaced

=

==

Solve forfluiddisplacedV :

g

BV

fluiddisplacedfluid

displaced

=

Because rockfluiddisplaced VV = and

waterfluiddisplaced = :

g

BV

water

rock

=

Substituting in equation (1) and

simplifying yields:water

rockwater

rockrock

B

w

B

gm ==


evaluate rock:( )

33

33

rock

kg/m100.3

kg/m1000.1N40N60

N60

=

=

43 [SSM] A block of an unknown material weighs 5.00 N in air and4.55 N when submerged in water. (a) What is the density of the material?

(b) From what material is the block likely to have been made?


principle to find the density of the unknown object. The difference between the

weight of the object in air and in water is the buoyant force acting on the object.

(a) Using its definition, express the

density of the object:object

object

objectV

m= (1)


obtain: gV

gmwB

fluiddisplaced

fluiddisplaced

fluiddisplaced

fluiddisplaced

=

==


g

BV

fluiddisplacedfluid

displaced

=


27/76

Fluids 1297

Because objectfluiddisplaced VV = and


g

BV

water

object

=

Substitute in equation (1) and

simplify to obtain: water

object

water

object

object

B

w

B

gm==

Substitute numerical values and evaluate object:

( ) 3333object kg/m1011kg/m1000.1N4.55N00.5

N00.5=

=

(b) From Table 13-1, we see that the density of the unknown material is close to

that of lead.

44 A solid piece of metal weighs 90.0 N in air and 56.6 N whensubmerged in water. Determine the density of this metal.


principle to find the density of the unknown object. The difference between the

weight of the object in air and the weight of the object in water is the buoyant

force acting on the object.


density of the metal: metal

metalmetal

V

m= (1)


obtain:gV

gmwB

fluiddisplaced

fluiddisplaced

fluiddisplaced

fluiddisplaced

=

==


g

BV

fluiddisplacedfluid

displaced

=

Becausemetalfluiddisplaced

VV = and


g

BV

water

metal=

Substitute in equation (1) and

simplify to obtain:water

metalwater

metalmetal

B

w

B

gm==


28/76

Chapter 131298

Substitute numerical values and evaluate metal:

( ) 3333metal kg/m1069.2kg/m1000.1N56.6N0.90

0.0N9=

=

45 A homogeneous solid object floats on water, with 80.0 percent of itsvolume below the surface. When placed in a second liquid, the same object floatson that liquid with 72.0 percent of its volume below the surface. Determine thedensity of the object and the specific gravity of the liquid.

Picture the ProblemLet Vbe the volume of the object and Vbe the volume that

is submerged when it floats. The weight of the object is Vg and the buoyant

force due to the water is wVg. Because the floating object is in translational

equilibrium, we can use = 0yF to relate the buoyant forces acting on theobject in the two liquids to its weight.

Apply = 0yF to the objectfloating in water:

0ww == VgV'gmgV'g (1)

Solving for yields:

V

V'w=


evaluate :( )

3

33

kg/m800

800.0kg/m1000.1

=

=V

V

Apply = 0yF to the objectfloating in the second liquid and

solve for mg:

gVmg L720.0 =

Solve equation (1) for mg: Vgmg w800.0 =

Equate these two expressions to

obtain:wL 800.0720.0 =

Substitute in the definition of

specific gravity to obtain:11.1

720.0

800.0gravityspecific

w

L ===

46 A 5.00-kg iron block is suspended from a spring scale and issubmerged in a fluid of unknown density. The spring scale reads 6.16 N. What isthe density of the fluid?


29/76

Fluids 1299

Picture the Problem We can use Archimedes principle to find the density of the

unknown fluid. The difference between the weight of the block in air and the

weight of the block in the fluid is the buoyant force acting on the block.


obtain: gV

gmwB

fluiddisplaced

fluiddisplaced

fluiddisplaced

fluiddisplaced

=

==

Solve forfluiddisplaced :

gV

B

fluiddisplacedfluid

displaced=

Because blockFefluiddisplaced VV = :

Fe

blockFeblockFe

f gm

B

gV

B==

Substitute numerical values and evaluate f:

( )( )( )( )

( ) 33332

2

f kg/m100.7kg/m1096.7m/s81.9kg00.5

N16.6m/s81.9kg00.5=

=

47 A large piece of cork weighs 0.285 N in air. When held submerged

underwater by a spring scale as shown in Figure 13-35, the spring scale reads

0.855 N. Find the density of the cork.

Picture the Problem The forces acting on the cork areB, the upward force due to

the displacement of water, mg, the weight of the piece of cork, and Fs, the force

exerted by the spring. The piece of cork is in equilibrium under the influence of

these forces.

Apply = 0yF to the piece of cork: 0s = FwB (1)or

0scork = FVgB (2)

Express the buoyant force as a

function of the density of water:VgwB

wfluiddisplaced

== w

BVg=

Substitute for Vgin equation (2) to

obtain:0s

w

cork = FB

B

(3)

Solve equation (1) forB: sFwB +=


30/76

Chapter 131300

Substitute in equation (3) to obtain:0s

w

scorks =

++ F

FwFw

or

0w

scork =

+

Fww

Solving for corkyields:

s

wcorkFw

w

+=

Substitute numerical values and evaluate cork:

( ) 333cork kg/m250N0.855N0.285

N0.285kg/m1000.1 =

+=

48 A helium balloon lifts a basket and cargo of total weight 2000 N understandard conditions, at which the density of air is 1.29 kg/m3and the density ofhelium is 0.178 kg/m3. What is the minimum volume of the balloon?

Picture the Problem Under minimum-volume conditions, the balloon will be in

equilibrium. LetBrepresent the buoyant force acting on the balloon, wtotrepresent

its total weight, and Vits volume. The total weight is the sum of the weights of its

basket, cargo, and helium in its balloon.

Apply = 0yF to the balloon: 0tot=wB

Express the total weight of the

balloon:

Vgw Hetot N2000 +=

Express the buoyant force due to the

displaced air:

VgwB airfluiddisplaced ==

Substitute forB and wtotto obtain: 0N2000 Heair = VgVg

Solving for Vyields:

( )gV

Heair

N2000

=

Substitute numerical values and evaluate V:

( )( )3

233m183

m/s9.81kg/m0.178kg/m1.29

N2000=

=V


31/76

Fluids 1301

49 [SSM] An object has neutral buoyancywhen its density equals thatof the liquid in which it is submerged, which means that it neither floats nor sinks.If the average density of an 85-kg diver is 0.96 kg/L, what mass of lead should thedive master suggest be added to give the diver neutral buoyancy?

Picture the Problem Let V= volume of diver, Dthe density of the diver, VPbthe

volume of added lead, and mPbthe mass of lead. The diver is in equilibrium under

the influence of his weight, the weight of the lead, and the buoyant force of the

water.

Apply = 0yF to the diver: 0PbD = wwB

Substitute to obtain: 0PbDDPbDw =+ gmgVgV

or

0PbDDPbwDw =+ mVVV

Rewrite this expression in terms

of masses and densities:0Pb

D

DD

Pb

Pbw

D

Dw =+ m

mmm

Solving for mPbyields: ( )( )wPbD

DDwPbPb

=

mm

Substitute numerical values and evaluate mPb:

( )( )( )

( )( ) kg9.3kg/m1000.1kg/m1011.3kg/m100.96kg85kg/m100.96kg/m1000.1kg/m1011.3

333333

333333

Pb =

=m

50 A 1.00-kg beaker containing 2.00 kg of water rests on a scale. A

2.00-kg block of aluminum (density 2.70 103kg/m3) suspended from a springscale is submerged in the water, as in Figure 13-36. Find the readings of bothscales.

Picture the Problem The hanging scales reading w is the difference betweenthe weight of the aluminum block in air wand the buoyant force acting on it. The

buoyant force is equal to the weight of the displaced fluid, which, in turn, is the

product of its density and mass. We can apply a condition for equilibrium to relate

the reading of the bottom scale to the weight of the beaker and its contents and the

buoyant force acting on the block.


32/76

Chapter 131302

Express the apparent weight wofthe aluminum block. This apparent

weight is the reading on the hanging

scale and is equal to the tension in

the string:

Bww' = (1)

LettingFsbe the reading of the

bottom scale and choosing upward

to be the +y direction, apply

0= yF to the system consisting ofthe beaker and the water to obtain:

( ) 0wbs =+ BgMMF

or

( )

( ) 'wb

wbs

wwgMM

BgMMF

++=

++= (2)

Using Archimedes principle,

substitute forBin equation (1) to

obtain:

Vgww' w= (3)

Express win terms of Aland V

and solve for Vg: AlAl

wVgVgw ==

Substitute for Vg in equation (3) and

simplify to obtain:w

www'

==

Al

w

Al

w 1

Substitute numerical values and evaluate w:

( )( ) N4.12m/s9.81kg2.00kg/m102.70kg/m1000.11 2

33

33

=

=w'

Substitute numerical values in equation (2) and evaluateFs:

( )( ) ( )( )N7.36

N4.12m/s81.9kg00.2m/s81.9kg2.00kg00.1 22s

=

++=F

51 When cracks form at the base of a dam, the water seeping into thecracks exerts a buoyant force that tends to lift the dam. As a result, the dam can

topple. Estimate the buoyant force exerted on a 2.0-m thick by 5.0-m wide damwall by water seeping into cracks at its base. The water level in the lake is 5.0 mabove the cracks.


33/76

Fluids 1303

Picture the Problem We can apply Archimedes principle to estimate the

buoyant force exerted by the water on the dam. We can determine whether the

buoyant force is likely to lift the dam by comparing the buoyant and gravitational

forces acting on the dam.

The buoyant force acting on the damis given by Archimedes principle:

gV

gmwB

waterdisplacedwater

waterdisplacedwaterdisplaced

=

==

The volume of the displaced water is

the same as the volume of the dam:

LWHggVB waterdamwater ==

whereL, W, andHare the length and

width of the dam, andH is the height of

the water above the cracks.

Substitute numerical values and evaluateB:

( )( )( )( )( ) N109.4m/s81.9m0.5m0.5m0.2kg/m1000.1 5233 ==B

52 Your team is in charge of launching a large helium weather balloonthat is spherical in shape, and whose radius is 2.5 m and total mass is 15 kg(balloon plus helium plus equipment). (a) What is the initial upward accelerationof the balloon when it is released from sea level? (b) If the drag force on the

balloon is given byFD=12r

2v

2, where ris the balloon radius, is the density of

air, and v the balloons ascension speed, calculate the terminal speed of theascending balloon.

Picture the ProblemThe forces actingon the balloon are the buoyant force B,its weight mg, and a drag force FD. Wecan find the initial upward accelerationof the balloon by applying Newtonssecond law at the instant it is released.We can find the terminal speed of theballoon by recognizing that whenay = 0, the net force acting on theballoon will be zero.

B

gm

DF

y

(a) Apply = yy maF to the balloonat the instant of its release to obtain:

yamgmB balloonballoon = (1)


34/76

Chapter 131304

Using Archimedes principle, expressthe buoyant forceBacting on theballoon:

grgV

gV

gmwB

3

air34

balloonair

fluiddisplaced

fluiddisplaced

fluiddisplaced

fluiddisplaced

==

=

==

Substitute in equation (1) to obtain: yamgmgr balloonballoon3air34 =

Solving for ayyields:g

m

ray

= 1

balloon

3

air34

Substitute numerical values and evaluate ay:

( )( ) ( ) 2233

34

m/s45m/s81.91kg15

m5.2kg/m29.1=

=

ya

(b) Apply = yy maF to the balloonunder terminal-speed conditions toobtain:

02t2

21 = vrmgB

Substitute forB: 02t2

213

air34 = vrmggr

Solving for vtyields: ( )

2

3

air34

t

2

r

gmrv

=

Substitute numerical values and evaluate v:

( )( )[ ]( )( )( )

m/s3.7m/s33.7kg/m29.1m5.2

m/s81.9kg15m5.2kg/m29.1232

233

34

t ==

=

v

53 [SSM] A ship sails from seawater (specific gravity 1.025) intofreshwater, and therefore sinks slightly. When its 600,000-kg load is removed, itreturns to its original level. Assuming that the sides of the ship are vertical at thewater line, find the mass of the ship before it was unloaded.

Picture the Problem Let V = displacement of ship in the two cases, m be the

mass of ship without load, and mbe the load. The ship is in equilibrium under

the influence of the buoyant force exerted by the water and its weight. Well

apply the condition for floating in the two cases and solve the equations

simultaneously to determine the loaded mass of the ship.


35/76

Fluids 1305

Apply = 0yF to the ship in freshwater:

0w = mgVg (1)

Apply = 0yF to the ship in saltwater:

( ) 0sw =+ gmmVg (2)

Solve equation (1) for Vg:

w

mgVg=

Substitute in equation (2) to obtain: ( ) 0w

sw =+ gmmmg

Solving for myields:

wsw

w

=

mm

Add mto both sides of the equationand simplify to obtain:

wsw

sw

wsw

w

wsw

w

1

=

+

=

+

=+

m

m

mm

mm


evaluate m + m:

( )( )

( )( )

kg105.2

1025.1

025.1kg1000.6

025.1

025.1kg1000.6

7

5

ww

w5

=

=

=+

mm

Continuity and Bernoulli's Equation

54 Water flows at 0.65 m/s through a 3.0-cm-diameter hose thatterminates in a 0.30-cm-diameter nozzle. Assume laminar nonviscous steady-state

flow. (a) At what speed does the water pass through the nozzle? (b) If the pump atone end of the hose and the nozzle at the other end are at the same height, and ifthe pressure at the nozzle is 1.0 atm, what is the pressure at the pump outlet?

Picture the Problem LetA1represent the cross-sectional area of the hose, A2the

cross-sectional area of the nozzle, v1the speed of the water in the hose, and v2the

speed of the water as it passes through the nozzle. We can use the continuity


36/76

Chapter 131306

equation to find v2 and Bernoullis equation for constant elevation to find the

pressure at the pump outlet.

(a) Using the continuity equation,

relate the speeds of the water to the

diameter of the hose and thediameter of the nozzle:

2211 vAvA =

or

2

2

21

2

1

44vdvd = 12

2

2

12 v

ddv =


evaluate v2: ( )

m/s65

m/s0.65m/s0.65cm0.30

cm3.02

2

=

=

=v

(b) Using Bernoullis equation for

constant elevation, relate the

pressure at the pumpPPto the

atmospheric pressure and the

velocities of the water in the hose

and the nozzle:

2

221

at

2

121

P vPvP +=+

Solve for the pressure at the pump: ( )212221atP vvPP +=

Substitute numerical values and evaluatePP:

( )( ) ( )atm22

kPa101.325

atm1Pa1021.2

m/s0.65m/s65.0kg/m1000.1kPa325.1016

2233

21

P

==

+=P

55 [SSM] Water is flowing at 3.00 m/s in a horizontal pipe under apressure of 200 kPa. The pipe narrows to half its original diameter. (a) What isthe speed of flow in the narrow section? (b) What is the pressure in the narrowsection? (c) How do the volume flow rates in the two sections compare?

Picture the Problem Let A1 represent the cross-sectional area of the larger-

diameter pipe, A2 the cross-sectional area of the smaller-diameter pipe, v1 thespeed of the water in the larger-diameter pipe, and v2the velocity of the water in

the smaller-diameter pipe. We can use the continuity equation to find v2and

Bernoullis equation for constant elevation to find the pressure in the smaller-

diameter pipe.


37/76

Fluids 1307

(a) Using the continuity equation,

relate the velocities of the water to

the diameters of the pipe:

2211 vAvA =

or

2

2

21

2

1

44v

dv

d = 12

2

2

12 v

d

dv =


evaluate v2: ( ) m/s0.12m/s3.002

121

12 =

=

d

dv

(b) Using Bernoullis equation for


pressures in the two segments of the

pipe to the velocities of the water in

these segments:

2

2w21

2

2

1w21

1 vPvP +=+

Solving forP2yields:

( )2221w211

2

2w212

1w21

12

vvP

vvPP

+=

+=

Substitute numerical values and evaluateP2:

( )( ) ( )[ ] kPa133m/s0.12m/s00.3kg/m1000.1kPa200 223321

2 =+=P

(c) From the continuity equation; V2V1 II = .

56 The pressure in a section of horizontal pipe with a diameter of 2.00 cmis 142 kPa. Water flows through the pipe at 2.80 L/s. If the pressure at a certainpoint is to be reduced to 101 kPa by constricting a section of the pipe, whatshould the diameter of the constricted section be?

Picture the Problem Let A1 represent the cross-sectional area of the 2.00-cm

diameter pipe, A2the cross-sectional area of the constricted pipe, v1the speed of

the water in the 2.00-cm diameter pipe, and v2 the speed of the water in the

constricted pipe. We can use the continuity equation to express d2in terms of d1

and to find v1and Bernoullis equation for constant elevation to find the speed ofthe water in the constricted pipe.

Using the continuity equation, relate

the volume flow rate in the 2.00-cm

diameter pipe to the volume flow

rate in the constricted pipe:

2211 vAvA =

or

2

2

21

2

1

44v

dv

d =

2

112

v

vdd = (1)


38/76

Chapter 131308

Using the continuity equation,

relate v1to the volume flow rateIV:2

1

V

2

141

V

1

V1

4

d

I

d

I

A

Iv

===

Using Bernoullis equation for


pressures in the two segments of thepipe to the velocities of the water in

these segments:

2

2w21

2

2

1w21

1 vPvP +=+

Solve for v2to obtain: ( ) 21

w

12

2v

PPv +

=

Substituting for v2in equation (1) and simplifying yields:

( ) ( ) ( )4

w

2

1

1

1

w

2

1

1

1

2

1

w

1

112

12

12

1

2+

=+

=+

=

v

PP

d

v

PPdv

PP

vdd

Substitute for v1and simplify to obtain:

( ) ( )4

w

2

V

1

4

1

2

1

4

w

2

2

1

V

1

12

18

14

2+

=

+

=

I

PPd

d

d

I

PP

dd

Substitute numerical values and evaluate d2:

( ) ( )( )( )

cm68.1

1kg/m1000.1L/s80.28

kPa101.325kPa142m0200.0

cm00.2

4332

422 =

+

=

d

57 [SSM] Blood flows at at 30 cm/s in an aorta of radius 9.0 mm.(a) Calculate the volume flow rate in liters per minute. (b) Although the cross-

sectional area of a capillary is much smaller than that of the aorta, there are manycapillaries, so their total cross-sectional area is much larger. If all the blood fromthe aorta flows into the capillaries and the speed of flow through the capillaries is1.0 mm/s, calculate the total cross-sectional area of the capillaries. Assumelaminar nonviscous, steady-state flow.


39/76

Fluids 1309

Picture the Problem We can use the definition of the volume flow rate to find

the volume flow rate of blood in an aorta and to find the total cross-sectional area

of the capillaries.

(a) Use the definition of the volume

flow rate to find the volume flowrate through an aorta:

AvI =V


evaluateIV:( ) ( )

L/min6.4L/min58.4

m10

L1

min

s60

s

m10634.7

m/s30.0m100.9

33

35

23

V

==

=

=

I

(b) Use the definition of the volume

flow rate to express the volumeflow rate through the capillaries:

capcapV vAI = cap

Vcap

v

IA =


evaluateAcap:22

35

cap m107.6m/s0010.0

/sm107.63

=

=A

58 Water flows through a 1.0 m-long conical section of pipe that joins acylindrical pipe of radius 0.45 m, on the left, to a cylindrical pipe of radius0.25 m, on the right. If the water is flowing into the 0.45 m pipe with a speed of1.50 m/s, and if we assume laminar nonviscous steady-state flow, (a) what is the

speed of flow in the 0.25 m pipe? (b) What is the speed of flow at a positionxinthe conical section, ifxis the distance measured from the left-hand end of theconical section of pipe?

Picture the Problem The pictorial representation summarizes what we knowabout the two pipes and the connecting conical section. We can apply thecontinuity equation to find v2. We can also use the continuity equation in (b)provided we first express the cross-sectional area of the transitional conicalsection as a function ofx.

m90.01=d m50.02 =d

?2 =vm/s50.11=v

x

?=v

0 L

(a) Apply the continuity equation tothe flow in the two sections ofcylindrical pipes:

2211 AvAv =


40/76

Chapter 131310

Solving for v2, expressing the cross-sectional areas in terms of theirdiameters, and simplifying yields:

2

2

112

241

2

141

1

2

112

===

d

dv

d

dv

A

Avv

Substitute numerical values andevaluate v2: ( ) m/s9.4

m50.0

m90.0m/s50.1

2

2 =

=v

(b) Continuity of flow requires that: ( ) ( ) ( ) ( )xAxvAv =00

Solve for v(x) to obtain:( ) ( )

( )( )xA

Avxv

00= (1)

The diagram to the right is anenlargement of the upper half of thetransitional conical section. Use thecoordinates shown on the line to

establish the following proportion:

( )0

m45.0m25.0

0

m45.0

=

Lx

xr

0 L

( )m45.0,0

( )m25.0,L

( )( )xrx,

x

Solving the proportion for r(x)yields:

( ) ( )L

xxr m20.0m45.0 =

whereLis the length of the conicalsection.

The cross-sectional area of theconical section varies withxaccording to:

( ) ( )( )

( )2

2

m20.0m45.0

=

=

L

x

xrxA

Substituting forA(x) in equation (1)and simplifying yields: ( ) ( )

( )

( )

( )

2

2

1

2

2

1

m20.0m45.0

0

m20.0m45.0

0

=

=

L

x

rv

L

x

rvxv


41/76

Fluids 1311

Substitute numerical values andsimplify further to obtain: ( )

( )( )

( )

( )23

2

2

20.0m45.0

s/m304.0

m.01m20.0m45.0

m45.0m/s50.1

x

xxv

=

=

To confirm the correctness of thisequation, evaluate v(0) and v(1.0 m): ( ) ( )

m/s5.1m45.0

s/m304.00

2

3

==v

and

( )( )( )

m/s9.4

m0.120.0m45.0

s/m304.0m0.1

2

3

=

=v

59 The $8-billion, 800-mile-long Alaskan Pipeline has a maximum

volume flow rate of 240,000 m3

of oil per day. Most of the pipeline has a radiusof60.0 cm. Find the pressurePat a point where the pipe has a 30.0-cm radius.Take the pressure in the 60.0-cm-radius sections to beP = 180 kPa and thedensity of oil to be 800 kg/m3. Assume laminar nonviscous steady-state flow.

Picture the Problem Let the subscript 60 denote the 60.0-cm-radius pipe and the

subscript 30 denote the 30.0-cm-radius pipe. We can use Bernoullis equation for

constant elevation to express Pin terms of v60and v30, the definition of volumeflow rate to find v60, and the continuity equation to find v30.

P

P'

cm0.6060 =rcm0.3030 =r

30v60v

Using Bernoullis equation for


pressures in the two pipes to the

velocities of the oil:

2

30212

6021 vP'vP +=+

Solving forPyields: ( )23026021 vvPP' += (1)

Use the definition of volume flow

rate to express v60:2

60

V

60

V60

r

I

A

Iv

==


42/76

Chapter 131312

Using the continuity equation, relate

the speed of the oil in the half-

standard pipe to its speed in the

standard pipe:

30306060 vAvA = 6030

6030 v

A

Av =

Substituting for v60 andA30yields:2

30

V

30

V

60

V

30

6030

rI

AI

AI

AAv

===

Substitute for v60and v30and simplify to obtain:

+=

+=

4

30

4

60

2

2V

2

2

30

V

2

2

60

V21

11

2 rr

IP

r

I

r

IPP'

Substitute numerical values in equation (1) and evaluateP:

( )

( ) ( )

kPa144

m300.0

1

m600.0

1

2

s3600

h1

h24

d1

d

m1040.2kg/m800

kPa180

44

2

23

53

=

+=

P'

60 Water flows through a Venturi meter like that in Example 13-11 with apipe diameter of 9.50 cm and a constriction diameter of 5.60 cm. The U-tubemanometer is partially filled with mercury. Find the volume flow rate of the waterif the difference in the mercury level in the U-tube is 2.40 cm.

Picture the Problem Well use its definition to relate the volume flow rate in the

pipe to the speed of the water and the result of Example 13-11 to find the speed of

the water.


volume flow rate:1

2

11V vrvAI ==

Using the result of Example 13-11,

find the speed of the water upstream

from the Venturi meter:

=

1

2

22

2

1w

Hg

1

R

R

ghv


43/76

Fluids 1313

Substituting for v1yields:

=

1

2

2

2

2

1w

Hg2

V

R

R

ghrI

Substitute numerical values and evaluateIV:

( ) ( )( )( )

( )L/s1.13

1m0.0560

m0.0950kg/m1000.1

m0.0240m/s9.81kg/m1013.62m0950.0

4 233

2332

V =

=

I

61 [SSM] Horizontal flexible tubing for carrying cooling water extendsthrough a large electromagnet used in your physics experiment at Fermi NationalAccelerator Laboratory. A minimum volume flow rate of 0.050 L/s through thetubing is necessary in order to keep your magnet cool. Within the magnet volume,

the tubing has a circular cross section of radius 0.500 cm. In regions outside themagnet, the tubing widens to a radius of 1.25 cm. You have attached pressuresensors to measure differences in pressure between the 0.500 cm and 1.25 cmsections. The lab technicians tell you that if the flow rate in the system dropsbelow 0.050 L/s, the magnet is in danger of overheating and that you shouldinstall an alarm to sound a warning when the flow rate drops below that level.What is the critical pressure difference at which you should program the sensorsto send the alarm signal (and is this a minimum, or maximum, pressuredifference)? Assume laminar nonviscous steady-state flow.

Picture the Problem The pictorial representation shows the narrowing of the

cold-water supply tubes as they enter the magnet. We can apply Bernoullisequation and the continuity equation to derive an expression for the pressure

differenceP1P2.

cm50.21=d cm000.12 =d2v1v

1P

2P

Apply Bernoullis equation to thetwo sections of tubing to obtain:

222

12

212

11 vPvP +=+

Solving for the pressure difference

21 PP yields: ( )2122212

1212

221

21

vv

vvPPP

=

==


44/76

Chapter 131314

Factor 21v from the parentheses to

obtain:

=

= 11

2

1

2212

12

1

2

2212

1

v

vv

v

vvP

From the continuity equation wehave:

2211 vAvA =

Solving for the ratio of v2to v1,expressing the areas in terms of thediameters, and simplifying yields:

2

2

1

2

241

214

1

2

1

1

2

===

d

d

d

d

A

A

v

v

Use the expression for the volumeflow rate to express v1: 2

1

V

2

141

V

1

V1

4

d

I

d

I

A

Iv

===

Substituting for v1and v2/v1in the

expression for Pyields:

= 1

4

4

2

1

2

2

1

V21

d

d

d

IP

Substitute numerical values and evaluate P:

( )( )

kPa20.01cm.0001

cm.502

m0250.0

L

m10

s

L050.04

kg/m10.001

4

2

2

33

33

21 =

=

P

Because 2

V

IP (P as a function ofIVis a parabola that opens upward), this

pressure difference is the minimum pressure difference.

62 Figure 13-37 shows a Pitot-static tube, a device used for measuring thespeed of a gas. The inner pipe faces the incoming fluid, while the ring of holes inthe outer tube is parallel to the gas flow. Show that the speed of the gas is given

by( )

g

gL2

=

ghv , where Lis the density of the liquid used in the

manometer and gis the density of the gas.

Picture the ProblemLet the numeral 1 denote the opening in the end of the innerpipe and the numeral 2 to one of the holes in the outer tube. We can applyBernoullis principle at these locations and solve for the pressure differencebetween them. By equating this pressure difference to the pressure difference dueto the height h of the liquid column we can express the speed of the gas as a

function of L, g,g, and h.


45/76

Fluids 1315

Apply Bernoullis principle atlocations 1 and 2 to obtain:

22g2

12

21g2

11 vPvP +=+

where weve ignored the difference inelevation between the two openings.

Solve for the pressure difference

P =P1P2:

21g2

122g2

121 vvPPP ==

Express the speed of the gas at 1: Because the gas is brought to a halt(that is, is stagnant) at the opening to

the inner pipe, 01=v .

Express the speed of the gas at 2: Because the gas flows freely past the

holes in the outer ring, vv =2 .

Substitute to obtain: 2g2

1 vP =

Letting A be the cross-sectional areaof the tube, express the pressure at adepth h in the column of liquid

whose density is 1:

A

B

A

wPP +=

liquiddisplaced

21

where AhgB g= is the buoyant force

acting on the column of liquid ofheight h.

Substitute to obtain:

( )ghPA

ghA

A

ghAPP

gL2

gL21

+=

+=

or ( )ghPPP gL21 ==

Equate these two expressions for P: ( )ghv gL2g21 =

Solving for vyields: ( )g

gL2

=

ghv

Note that the correction for buoyant force due to the displaced gas is very small

and that, to a good approximation, .2g

L

ghv=

Remarks: Pitot tubes are used to measure the airspeed of airplanes.


46/76

Chapter 131316

63 [SSM] Derive the Bernoulli Equation in more generality than donein the text, that is, allow for the fluid to change elevation during its movement.Using the work-energy theorem, show that when changes in elevation are

allowed, Equation 13-16 becomes 2221

22

2

121

11 vghPvghP ++=++ (Equation

13-17).

Picture the ProblemConsider a fluid flowing in a tube that varies in elevation aswell as in cross-sectional area, as shown in the pictorial representation below. Wecan apply the work-energy theorem to a parcel of fluid that initially is contained

between points 1 and 2. During time t this parcel moves along the tube to theregion between point 1 and 2. Let V be the volume of fluid passing point 1during time t.The same volume passes point 2 during the same time. Also, letm= Vbe the mass of the fluid with volume V. The net effect on the parcelduring time t is that mass m initially at height h1 moving with speed v1 istransferredto height h2with speed v2.

1A

2

A

2v

1v

1 '1

'22

1h

2h111 APF =

222 APF =

2x

1x

0g =U

Express the work-energy theorem: KUW total += (1)

The change in the potential energy ofthe parcel is given by:

( ) ( )( ) ( )

( )12

2

12

hhVg

hhgm

ghmghmU

=

=

=

The change in the kinetic energy ofthe parcel is given by:

( ) ( )

( )( )( )212221

21

222

1

212

1222

1

vvV

vvm

vmvmK

=

=

=

Express the work done by the fluidbehind the parcel (to the parcels leftin the diagram) as it pushes on theparcel with a force of magnitude

111 APF = , whereP1is the pressure at

point 1:

VPxAPxFW 1111111 ===


47/76

Fluids 1317

Express the work done by the fluidin front of the parcel (to the parcelsright in the diagram) as it pushes onthe parcel with a force of magnitude

222 APF = , whereP2is the pressure

at point 2:

VPxAPxFW 2222222 ===

where the work is negative because theapplied force and the displacement arein opposite directions.

The total work done on the parcel is: ( ) VPPVPVPW 2121total ==

Substitute for U, K, and Wtotalin equation (1) to obtain:

( ) ( ) ( )2122211221 vvVhhVgVPP +=

Simplifying this expression by

dividing out V yields:( ) ( )2122211221 vvhhgPP +=

Collect all the quantities having asubscript 1 on one side and thosehaving a subscript 2 on the other toobtain:

2

221

22

2

121

11 vghPvghP ++=++

Remarks: This equation is known as Bernoullis equation for the steady,

nonviscous flow of an incompressible fluid.

64 A large keg of heightH and cross-sectional areaA1is filled with rootbeer. The top is open to the atmosphere. There is a spigot opening of areaA2,which is much smaller thanA1, at the bottom of the keg. (a) Show that when theheight of the root beer is h, the speed of the root beer leaving the spigot is

approximately gh2 . (b) Show that ifA2


48/76

Chapter 131318

(a) Apply Bernoullis equation to the

beer at the top of the keg and at the

spigot:

2

2beer21

2beer2

2

1beer21

1beer1

v

ghPvghP

+

+=++

or, because v10, h2= 0,P1=P2=Pat,

and h1= h,2

22

1 vgh= ghv 22

=

(b) Use the continuity equation to

relate v1and v2:2211 vAvA =

Substitute dh/dtfor v1and gh2

for v2to obtain:ghA

dt

dhA 221 =

Solving for dh/dtyields:gh

A

A

dt

dh2

1

2=

(c) Separate the variables in the

differential equation to obtain:dt

h

dh

g

AA=

2

21

Express the integral fromHto hand

0 to t: =th

H

dth

dh

g

AA

0

21

2

Evaluate the integral to obtain: ( ) thHg

AA=

2

2 21

Solving for hgives: 2

1

2 22

= tg

A

AHh

(d) Solve h(t) for the time-to-drain

t: g

H

A

At'

2

2

1=


evaluate t

( )

min46h1s1039.6

m/s81.9

m00.22

1000.13

21

4

1

==

=

A

At'

65 A siphon is a device for transferring a liquid from container tocontainer. The tube shown in Figure 13-38 must be filled to start the siphon, butonce this has been done, fluid will flow through the tube until the liquid surfacesin the containers are at the same level. (a) Using Bernoullis equation, show that


49/76

Fluids 1319

the speed of water in the tube is gdv 2= . (b) What is the pressure at the highest

part of the tube?

Picture the Problem Let the letter adenote the entrance to the siphon tube andthe letter bdenote its exit and assume that bin at the surface of the liquid inthe container to the right. Assuming streamline flow between these points, we canapply Bernoullis equation to relate the entrance and exit speeds of the waterflowing in the siphon to the pressures at either end, the density of the water, andthe difference in elevation between the entrance and exit points. Well also usethe equation of continuity to argue that, provided the surface area of the beaker islarge compared to the area of the opening of the tube, the entrance speed of thewater is approximately zero.

(a) Assuming that the height of the containers isH, apply Bernoullis equationat the surface of the left container and at point b:

( ) ( )dhHgvPhHgvP ++=++ 2

b2

1

at

2

surface2

1

at (1)

Apply the continuity equation to apoint at the surface of the liquid inthe container to the left and to pointb:

surfacesurfacebb AvAv = surface

bbsurface

A

Avv =

or, becauseAsurface>>Ab,

0surface =v

With va= 0, equation (1) becomes: ( ) ( )dhHgvhHg += 2b21

Solving for vbgives: gdv 2b =

(b) Relate the pressure at the highest part of the tube Ptopto the pressure atpoint b:

( ) ( ) 2b21at2

top21

top vdhHgPvhHgP ++=++

The equation of continuity requires that vtop= vb= va:

( ) ( ) 2top21at2

top21

top vdhHgPvhHgP ++=++

or, simplifying,

( ) ( )dhHgPhHgP +=+ attop

Solve forPtopto obtain: gdPP = attop

Remarks: If we letPtop= 0, we can use this result to find the maximum

theoretical height a siphon can lift water.


50/76

Chapter 131320

66 A fountain designed to spray a column of water 12 m into the air has a1.0-cm-diameter nozzle at ground level. The water pump is 3.0 m below theground. The pipe to the nozzle has a diameter of 2.0 cm. Find pump pressurenecessary if the fountain is to operate as designed. (Assume laminar nonviscoussteady-state flow.)

Picture the Problem Let the letter P denote the pump and the 2.0-cm diameterpipe and the letter N the 1-cm diameter nozzle. Well use Bernoullis equation to

express the necessary pump pressure, the continuity equation to relate the speed

of the water coming out of the pump to its speed at the nozzle, and a constant-

acceleration equation to relate its speed at the nozzle to the height to which the

water rises.

Using Bernoullis equation,

prova chap 13 sm

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