Fluid Mechanics (Chapter 14)

8/10/2019 Fluid Mechanics (Chapter 14)

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Chapter 14.Fluid Mechanics

Dr. Yousef [email protected]


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Syllabus

14.1Density

14.2Pressure in a Fluid

14.3Buo anc

University Physics, ChapterDr. Y.Abou-Ali, IUST

14.4Fluid Flow

14.5Bernoulli's Equation

14.6Viscosity and Turbulence


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Goals for Chapter 14

To study density and pressure in a fluid. To apply Archimedes Principle of buoyancy. To describe surface tension and capillary action.


To study and solve Bernoulli's Equation for fluidflow.

To see how real fluids differ from ideal fluids(turbulence and viscosity).


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Introduction

Fluids play a vital role in many aspects of everyday life.

Theycirculatethrough our bodies and control ourweather.

Wedrinkthem,breathethem, andswimin them.

Airplanesflythrough them; shipsfloatin them.

A fluid is any substance that can flow; we use the term for bothliquidsandgases.



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Introduction

We begin our study withfluid statics, the study of fluids at rest ine uilibrium situations.

We usually think of a gasas easily compressedand a liquidasnearly incompressible.

Than will move to talk aboutfluid dynamics.



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14.1Density

An important property of any material is itsdensity, defined as itsmass per unit volume.

A homogeneous material such asiceorironhas the same densitythroughout.

We use the Greek letter (rho) for density. If a massmof homogeneous material has volume V, the density is:

m

V = (definition of density) (14.1)



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14.1Density

At right, liquids of differentdensities se arate with denser

liquids lower in the glass.



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14.1Density

The density of some materialsvariesfrom point to point within thematerial; someexamplesare the earths atmosphere,

which is less dense at high altitude.

and oceans (which are denser at greater depths).

For these materials, Eq. (14.1) describes the averagedensity. In general, the density of a material depends on environmentalfactors such astemperatureandpressure.

The SI unit of density is the kilogram per cubic meter (1 kg/m3).

3 31 g/cm 1000 kg/m=



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14.1Density

Thespecific gravityof a material is the ratio of its density to thedensity of water at 4.0o C, 1000 kg/m3.

It is a pure number without units.

Since it has nothing to do with gravity; relative density would

ave een e er.

Example14.1(The weight of a roomful of air):

Find the mass and weight of the air in a living room with a 4.0

m 5.0 m floor and a ceiling 3.0 m high. What is the mass and

weight of an equal volume of water?



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14.1Density

Solution:

Identify:We assume that air is homogeneous, so that the density is

the same throughout the room.

Set Up:We use Eq. (14.1) to relate the mass (the target variable) tothe volume (which we calculate from the dimensions of the room)

and the density (from Table 14.1, in your book).

Execute:The volume of the room is:

3(3 0m)(4 0m)(5 0m) 60 mV = =. . .

The massmairof air is given by Eq. (14.1):



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14.1Density

3 3air air (1 2 kg m )(60m ) 72 kgm V= = =. /

The weight of the air is: 2air air (72 kg)(9 8 m/s ) 700 Nw m g= = =.

3 3 4water water (1000 kg m ) (60 m ) 6 0 10 kgm V= = = / .

The weight is:4 2 5

water water (6 0 10 kg)(9 8 m/s ) 5 9 10 Nw m g= = = . . .



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We define the pressurepat the point as the normal force per unitarea, that is, the ratio dFto dA:

d Fp

d A

= (d e f i n i tio n o f p res su re ) (1 4 .2 )

If the pressure is the same at all points of a finite plane surfacewith areaA, then:

F

p A

= (14.3)

The SI unit of pressure in thepascal, where:

1 pascal = 1 Pa = N/m2University Physics, ChapterDr. Y.Abou-Ali, IUST


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Two related units, used principally in meteorology, are the bar,equal to105 Pa, andmillibar, equal to100 Pa.

Atmospheric pressurepais the pressure of the earths atmospheric,the pressure at the bottom of this sea of air which we live.

5a av( ) 1atm 1 013 10 Pa

= 1.013 bar = 1013 millibar

p = = .

Example14.2( The force of air):

In the room described in Example 14.1, what is the total downward

force on the surface of the floor due to air pressure of 1.00 atm?



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Solution:

IdentifyandSet Up:The pressure is uniform, so we use Eq. (14.3)to determine the force from the pressure and area.

Execute:The floor area is A = (4.0 m) (5.0 m) = 20 m2, so from

5 2 2

6

(1 013 10 N/m )(20 m )

= 2.0 10 N

F p A = =

.

Eq. (14.3) the total downward force is



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Pressure, Depth, and Pascals Law

If theweightof the fluid can beneglected, thepressurein a fluid isthesame throughout its volume. But often the fluids weight is notnegligible.

Atmospheric pressure is less at high altitude than at sea level,

which is why an airplane cabin has to pressurized when flying at35.000 feet.

When you dive in deep water, your ears tell you that the pressure

increasesrapidly with increasing depth below the surface.

We can derive a general relation between the pressure pat anypoint in a fluid at rest and the elevationyof the point.



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We will assume that the density and the acceleration due togravitygare the same throughout the fluid.

If the fluid in equilibrium, every volume element is in equilibrium.

Consider a thin element of fluid with height dy(figure below). Thebottom and top surface have area A, they are at elevations y &

y+ dy, above some reference level wherey= 0.

The volume of the fluid element is dV=A dy, its mass is dm= dV= A dy, and its weight is dw= dm g= gA dy.



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(p + dp) A

y

A

0

Element of fluid

thickness dy

(a)

dy

pA

dw

(b)



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What are the other forces on this fluid element?

The pressure at the bottom surface p; the total y-component ofupward force on this surface ispA.

y-component of (downward) force on the top surface is (p+ dp)A.

The fluid element is in equilibrium, so

0 so ( ) 0yF pA p dp A gA dy= + =dp

g

dy

= (14.4)



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Ifp1and p2are the pressure at elevations y1and y2, respectively,

and ifandgare constant, then

pressure in a fluid of uniform density2 1 2 1( )p p g y y = ( ) (14.5)

It is often convenient to express Eq. (14.5) in terms of the depthbelow the surface of a fluid (figure below).Fluid, density

P1=

P2=

y2 y1 =

1

2

y2

y1

y2y1



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o 2 1( )p p g y y gh = =

o pressure in a fluid of uniform densityp p g h= + ( ) (14.6)

The pressure pat a depth his greater than the pressure poat thesurface by an amount gh.

Note that the pressure is the same at any two points at the samelevel in the fluid. Theshapeof the container does not matter.



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The pressure in any fluid at the same elevation will be the same

regardless of the shape or size of the container.



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Pascals law:Pressure applied to an enclosed fluid is transmitted to

every portion of the fluid and the walls of the containing vessel.

The hydraulic lift shown schematically in figure below illustrates .

A piston with small cross-sectional areaA1exerts a forceF1on thesurface of a liquid such as oil. The applied pressure p = F1/A1transmitted through the connecting pipe to a larger piston of area

A2.



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The applied pressure is the same in both cylinders, so:

1 2 22 1

1 2 1

&F F A

p F FA A A

= = =

In a room with a ceiling height 3.0 m filled with air of uniformdensity 1.2 kg/m3, the difference in pressure between floor and

ceiling given by Eq. (14.6):

gh= (1.2 kg/m3) (9.8 m/s2) (3.0 m) = 35 Pa



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Absolute Pressure, Gauge Pressure, and Pressure Gauges

If the pressure inside a car tire is equal to atmospheric pressure,the tire is flat.

The pressure has to be greater than atmospheric to support the,

outside pressure.

When we say that the pressure in a car tire is (220 kPa or 2.2 105 Pa),we mean it is greaterthan atmospheric pressure (1.01 105 Pa) by

this amount.

The totalpressurein the tire is then 320 kPa.



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The excess pressure above atmospheric pressure is usually calledgauge pressure.



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Example14.3( Finding absolute and gauge pressures):

A solar water-heating system uses solar panels on the roof, 12.0 m

above the storage tank. The water pressure at the level of the panels

.

gauge pressure?

Solution:


Identify:Water is nearly incompressible. Hence we can treat it as afluid of uniform density.


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Set Up:The level of the panels corresponds to point 2 in the figure

above, and the level of the tank corresponds to point 1. Hence ourtarget variable ispin Eq. (14.6); we are givenpo= 1 atm = 1.0110

-5

Pa andh= 12.0 m.

op p gh= +Execute:From Eq. (14.6), the absolute pressure is:


5 3 2

5

(1 01 10 Pa 1000 kg m 9 80 m s 12 0 m

2 19 10 Pa 2 16 atm

=

. . .

. .


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The gauge pressure is:

o5(2 19 1 01) 10 Pap p = . .

51 18 10 Pa 1 16 atm= =. .



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The simplest pressure gauge is the open-tube manometer (figurebelow). The U-shaped tube contains a liquid of density .

Pa=

y1

y2

y2- y1=

Pressurep

P + gy1 Pa+ gy2University Physics, ChapterDr. Y.Abou-Ali, IUST


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The pressure at the bottom of the tube due to the fluid in the left

column isp+ gy1, and the pressure at the bottom due to the fluid inthe right column ispa+ gy2.

These pressure are measured at the same point, so they must be

equal:

1 a 2p gy p gy + = +

14 8a 2 1( )p p g y y gh = = ( . ) In Eq. (14.8), pis the absolute pressure, and the difference p pabetween absolute and atmospheric pressure is thegauge pressure.



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Another common pressure gauge is the mercury barometer. It

consists of a long glass tube, closed at one end, that has to be filledwith mercury and then inverted in a dish of mercury.

p0 = 0

p = pay2 y1=

y2

y1


14 2 P i Fl id


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The pressure po at the top of the mercury column is practicallyzero. From Eq. (14.6),

14 9a 2 10 ( )p p g y y gh = = + = ( . )

Thus the mercury barometer reads the atmospheric pressure padirectly from the height of the mercury column.

A pressure of 1 mm Hg is called 1 torr.

In many applications, pressures are still commonly described interms of the height of the corresponding mercury column (mm Hg).


14 2 P i Fl id


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One common type of blood-pressure gauge, called a

sphygmomanometer, uses a mercury-filled manometer.

Blood-pressure readings, such as 130/80, give the maximum and,

torr.

We distinguish high-and low pressure sections of thecardiovascular system because the blood pressure varies

significantly as illustrated in figure below.

Blood pressure varies with height; the standard reference point isthe upper arm, level with the heart.


14 2 P i Fl id


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The high-pressure part includes the aorta, the arteries and

arterioles and the capillaries of the systemic circulation.


14 2 P i Fl id


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The blood pressure in the arteries varies and 16.0 KPa (systolic

pressure equal to 120 mm Hg).

The low-pressure part includes the veins and the pulmonary


circulation in this circulation the pressure varies only between 1.3

KPa and 3.3 KPa (10 - 25 mm Hg).

The pressure in our cardiovascular system exceeds the ambient air

pressure everywhere, this is correct only while lying down negativegauge pressures can occur while standing.

14 2 P i Fl id


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Example14.4( A tale of two fluids):

A manometer tube is partially filled with water. Oil (which does notmix with water, and has a lower density than water) is then poured

into the left arm of the tube until the oil-water interface is at the

midpoint of the tube (figure below). Both arms of the tube are open

to the air. Find a relationship between the heightshoilandhwater.

Solution:

Identify: the relationship between pressure and depth in a fluidapplies only to fluids of uniform density. Hence we cant write asingle equation for the oil and the water together. We can write a

pressure-depth relation for each fluid separately.


14 2 P i Fl id


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hoilhwater


14 2 Press re in a Fl id


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Set Up:Letpobe atmospheric pressure and letpbe the pressure at

the bottom of the tube. The densities of the two fluids are waterandoil(which is less than water). We use Eq. (14.6) for each fluid.

xecute:For the two fluids, Eq. (14.6) becomes,

o water waterp p g h= +

o oil oilp p g h= +


14 2 Pressure in a Fluid


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Since the pressurepat the bottom of the tube is the same for both

fluids,

wa eroil water

oilh h=




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Example:

Relative to the blood pressure in

supine position calculate the additionalblood pressure difference between the

brain and the feet in a standing

. = . .




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Solution:

To quantify the additional difference for the standing standardman we use pascals law:


For the pressure difference: brain feeth y y=

For the height of the person. this choice for peliminates the extraminus sign on the right hand:



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bloodp gh =

This difference is about 20% of the atmospheric pressure.


For physiological applications it is more useful to refer to pressure

that are measured relative to the pressure at the height of the heart.

For a standing person of 1.73 m height the heart is at a height of

1.22 m and the arterial venous pressure in the feet are increasedrelative to the pressure at the height of the heart by:

3

blood 1 06 10 9 80 1 22 12 7 KPa = 95 mmHgp gh ( . )( . )( . ) . = = =



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The pressure is increased to an average arterial value of 190mm Hg and an average venous value of 100 mm Hg.

In the scalp the pressures decrease for a standing person by:

31 06 10 9 80 0 51 5 3 KPa = -40 mmHg

bloodp gh ( . )( . )( . ) . = = =


The average arterial pressure drops to a value of 55 mm Hg andthe average venous pressure becomes -35 mm Hg.

Blood Pressure Measurement


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The figure above, illustrated what the health practitioner hearswith a stethoscope as a function of pressure.

Each sketch shows the pressure relative to systolic (p

s

) and

diastolic (pd) pressure at the right the normal variation of the blood

pressure (red curve) and the audible sound (orange curve).


Initially (Fig. a) the pressure in the cuff is increased until the flowof the blood through the brachial artery is below the cuff. A valve on

the bulb is then opened to lower the pressure in the cuff.

When the pressure in the brachial artery falls just below themaximum pressure generated by the heart (which is the systolic

pressure, where systole refers to the contraction of the heart muscle ,

the artery opens momentarily on each beat of the heart .



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The velocity of the blood in the artery is high and the blood flow isturbulent during these events this leads to a noisy blood flow easily

recognized by the health practitioner. The manometer now reads 120mm Hg for a standard man with a healthy heart.


intermittent sounds are heard until the pressure in the cuff fallsbelow the minimum heart pressure (which is the diastolic pressure

where diastolic refers to the expansion of the heart muscle) the a

continuous background sound is heard as illustrated in (Fig.d) the

transition to the continuous sound occurs at about 80 mm Hg for the

standard man with a healthy heart.

14.3 Buoyancy


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14.3Buoyancy

Buoyancyis a familiar phenomenon: a body immersed in waterseems to weight less than when it is in air.

When the body is less dense than the fluid, it floats.

The human body usually floats in water.


14.3 Buoyancy


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14.3Buoyancy

Archimedess principle states: When a body is completely or

partially immersed in a fluid, the fluid exerts an upward force on the

body equal to the weight of the fluid displaced by the body.

the body by the fluid is calledbuoyant force.


14.3 Buoyancy


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14.3Buoyancy


14.3 Buoyancy


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14.3Buoyancy

Example14.5( Buoyancy):

A 15.0-kg solid gold statue is being raised from a sunken ship (figure

below). What is the tension in the hoisting cable when the statue is at

rest and a) completely immersed; b) out of the water?


14.3 Buoyancy


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14.3Buoyancy

Identify:When the statue is immersed, it experiences an upwardbuoyant force = in magnitude to the weight of the fluid displaced.

Three forces are acting on the statue: weight, the buoyant force, and

the tension in the cable.

Solution:

Set Up:Our target variable is the tension T. We are givenmg, we

can calculate the buoyant forceBby using Archimedess principle.

Execute:a)To findB, we first find the volume of the statue, using

the density of gold from table 14.1:

4 3

3 3gold

15 0 kg7 77 10 m

19 3 10 kg/m

mV

= = =

..

.University Physics, ChapterDr. Y.Abou-Ali, IUST

14.3Buoyancy


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ya y

Using table 14.1 again, we find the weight of this volume ofseawater:

sw sw sw

3 3 4 3 2(1 03 10 kg/m )(7 77 10 m )(9 80 m s )

w m g V g

= =

= . . . /

= .This equals the buoyant forceB.

The statue is at rest, so the net external force acting on it is zero.

2

( ) 0

(15 0kg)( 9 80 m s ) 7 84 N

147 N 7 84 N 139 N

yF B T mg

T mg B

= + + =

= =

= =

. . / .

- . University Physics, ChapterDr. Y.Abou-Ali, IUST

14.3Buoyancy


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y y

If a spring scale is attached to the upper end of the cable, it willindicate 7.84 N less than if the statue were not immersed in seawater.

3

Hence the submerged statue seems to weight 139 N, about 5% lessthan its actual weight of 147 N.

on the statue is3 4 3 2

3

(1 2 kg/m )( 7 77 10 m )( 9 80m s )

9 1 10 N

airB V g

= =

=

. . . /

. Thus the tension in the cable with the statue in air is equal to thestatues weight, 147 N.


14.3Buoyancy


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y y

Example:

The human brain is immersed in cerebrospinal fluid of density 1.007

g/cm3, density is slightly less than the average density of the brain

1.04 g/cm3. Thus most of the weight of the brain is supported by the

buoyant force of the surrounding fluid. What fraction of the weight


Solution:

This problem is an application of the Archimedes principle theweight of the brain is:

brain brainw V g= The buoyant force in turn for the brain fully impressed in thecerebrospinal fluid:

14.3Buoyancy


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y y

We know that these two force do not balance each medullaoblongata to the spinal cord which exerts an additional force on the

brain to determine the fraction of the weight of the brain that is not

cerebrospinal brainB V g=


Just 3.2% of the brains weight is not balanced by thecerebrospinal fluid, requiring only a small force to be exerted by thespinal cord on the brain.

brain cerebrospinal

brain

1 04 1 0070 032

1 04

w B

w

. ..

.

= = =

14.3Buoyancy


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y y

Surface Tension

The surface of the liquid behaves like amembraneunder tension. Surface tension arises because the molecules of the liquid exertattractive forces on each other.


14.3Buoyancy


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y y

Surface Tension

There is zero net force on a

molecule inside the volume of theliquid.

But a surface molecule isdrawn into the volume (figure

below). Thus the liquid tends to

minimize its surface area, just as

a stretched membrane does.


14.3Buoyancy


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Surface tension explains whyfreely falling raindropsarespherical(not teardrop-shaped): a sphere has a smaller surface area for its

volume than any other shape.

It also ex lains wh hot soa wateris used forwashin . To wash

Surface Tension

clothing thoroughly, water must be forced through the tiny spaces

between the fibers.

To do so requires increasing the surface area of the water, which is

difficult to achieve because of surface tension.

The job is made easier by increasing the temperature of the waterand adding soap, both of which decrease the surface tension.


14.3Buoyancy


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Surface tension is important for a millimeter-sized water drop,

which has a relatively large surface area for its volume.

Forlarge quantitiesof liquid, the ratio of surface area to volume is

Surface Tension

relatively small, and surface tension is negligible compared to

pressure forces.

For the remainder of this chapter, we will consider only fluids in

bulk and hence will ignore the effects of surface tension.


14.4Fluid Flow


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We will now consider motion of a fluid. Fluid flow can beextremely complex, as shown by the currents in river rapids.

Anideal fluidis a fluid that is incompressible(that is, its densitycant change) and has no internal frication (calledviscosity).

The path of an individual particle in a moving fluid is called aflow

line.

If the overall flow pattern does not change with time, the flow is

calledsteady flow.

A streamline is a curve whose tangent at any point is in thedirection of the fluid velocity at that point.


14.4Fluid Flow


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We will consider only steady-flow situations, for which flow linesand streamlines are identical.


14.4Fluid Flow


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The flow lines passing through the edge of an imaginary element ofareaA, form a tube called aflow tube(see figure 14.19/466).

From the definition of a flow line, in steady flow no fluid can crossthe side walls of a flow tube; the fluids in different flow tubes cant

.

Laminar flow, in which adjacent layers of fluid slide smoothly pasteach other and the flow is steady.

At sufficiently high flow rates, or when boundary surface causeabrupt changes in velocity, the flow can become irregular andchaotic. This is calledturbulent flow.


14.4Fluid Flow


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Laminar flow Turbulent flow


14.4Fluid Flow


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The mass of a moving fluid does not change as it flows. This leadsto an important quantitative relationship called the continuity

equation.

The Continuity Equation

v

v11 1 1dV A v dt =

2 2 2dV A v dt =


14.4Fluid Flow


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Let us consider the case of an incompressible fluid so that the

density has the same value at all point.


The mass dm flowing into the tube across in time dtis:

1 1 1dm A v dt =

2 2 2dm A v dt =

The mass dm2flowing into the tube acrossA2in time dtis:

In steady flow the total mass in the tube is constant, so dm1= dm2,


14.4Fluid Flow


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1 1 2 2A v d t A v d t =

c o n t i n u i t y e q u a t i o n , i n c o m p r e s s i b l e f l u i d

1 1 2 2A v A v= ( 1 4 . 1 0 )

( )

The product Av is the volume flow ratedV/dt, the rate at whichvolume crosses a section of the tube:

volum e f low ra ted V

A vd t

= ( ) (1 4 .11 )

Themass flow rateis the mass flow per unit time through a crosssection. This is equal to density times the volume flow rate dV/dt.


14.4Fluid Flow


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We can generalise Eq. (14.10):

1 1 1 2 2 2A v A v = (continuity equation, compressible fluid) (14.12)

.

As part of a lubricating system for heavy machinery, oil of density

850 kg/m3 is pumped through a cylindrical pipe of diameter 8.0 cm

at a rate of 9.5 liters per second. a) What is the speed of the oil?

What is the mass flow rate? b) If the pipe diameter is reduced to 4.0cm, what are the new values of the speed and volume flow rate?

Assume that the oil is incompressible.


14.4Fluid Flow


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Identify and Set Up:we use Eq. (14.11), to determine the speed v1in the 8.0 cm-diameter section. Using Eq. (14.10), to find the speed v2

in the 4.0 cm-diameter section.

xecute:a)The volume flow rate dV dt equals the product v,

Solution:

where A1is the cross-sectional area of the pipe of diameter 8.0 cm

and radius 4.0 cm. Hence:

3 3

12 21

( 9 5 L/s)( 10 m /L1 9 m/s

(4 0 10 m)

dV dt v

A

= = =

/ . ).

.

The mass flow rate is: dV/dt= (850 kg/m3)(9.5 10-3 m3/s) = 8.1

kg/s.


14.4Fluid Flow


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b)Since the oil is incompressible, the volume flow rate has the same

value (9.5 L/s) in both sections of pipe. From Eq. (14.10),

2 21

2 1 2 2

(4 0 10 m)(1 9 m/s) 7 6 m/s

Av v

= = =

.. .

.


Speed of Blood in Capillaries


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Example:

The heart of the standard man pumps 5 liters of blood per minute into the aorta:

A) What is the volume flow rate in the cardiovascular system

B) What is the speed of the blood in the aorta

C) If we assume that the blood passes through all systemic capillaries in our body

,

D) What is the speed of the blood in parallel through the system capillaries in thehuman body.




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Solution:

A)The amount of the blood flowing:

B)A typical inner diameter of the aorta is daorta= 2.2 cm this leads to

-

m3/s


The speed of the blood in aorta is obtained from its inner cross-

sectional area and the volume passing per second using:

m2



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We obtain:

This is the frequently used results: blood flows through the aorta at

an avera e s eed of about 20 cm/sec.


C) Lets assume that blood passes through each single systemiccapillary at the rate found in part (a). For the outer diameter a

capillary we use 9 m this value leads to an inner diameter of

dcapillary= 7 m the cross section area is:



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We use the equation of the continuity to derive the speed for blood:

Then it would no longer be possible to exchange oxygen and

nutrients with the surrounding tissue. The physiological purpose of


D)A slow flow of blood in the systemic capillaries is achieved byarranging them parallel to each other with a combined cross-section

that is large than the cross- section of the aorta. The equation of

continuity allows us to determine the actual speed of blood in thecapillaries once we have know the compined cross-section and the is

2100\cm2 so:

e car ovascu ar sys em wou e os .



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Again this is a frequently used value blood flows very slowly

smm

A

t

V

v

capillary

aortacapillary /10421.0/103.8

425 ==

=


through the capillaries at less than 1 mm/s.

14.5Bernoullis Equation


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We can derive an importantrelationship called Bernoulli's

equa on a re a es e pressure,

flow speed, and height for flow of anideal, incompressible fluid.




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To derive Bernoullis equation, we apply the work-energy theoremto the fluid in a section of a flow tube.

2 21 11 1 1 2 2 22 2p gy v p gy v + + = + +

The subscripts 1 & 2 refer to any two points along the flow tube, sowe can also write:

212

constantp gy v + + = (14.18)




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Blood flow characteristics are changed dramatically by plaque.




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Example14.7( Water pressure in the home):

Water enters a house through a pipe with an inside diameter of 2.0

cm at an absolute pressure of 4.0 105 Pa (about 4 atm). A 1.0 cm

diameter pipe leads to the second-floor bathroom 5.0 m above (figure14.24). When the flow speed at the inlet pipe is 1.5 m/s, find the flow

, , .

Solution:

Identify and Set Up:Let points 1 & 2 be at the inlet pipe and at thebathroom, respectively. We take y1= 0(at the inlet) and y2= 5.0m

(at the bathroom). Our first two targets are the speed v1 and

pressurep1.




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Execute:From continuity equation, Eq. (14.10):

21

2 1 22

(1 0 cm)(1 5 m s) 6 0 m s

(0 50 cm)

Av v

A

= = =.

. / . / .

1 1, 2

2 2 512 1 2 1 2 12

3 3 2 2 2 21

23 3 2

( ) ( ) 4 0 10 Pa

1 0 10 kg m )( 36 m s 2 25 m s )

(1 0 10 kg m )( 9 8 m s )( 5 0 m)

p p v v g y y = =

.

( . / / . /

. / . / .




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The volume flow rate is:

5 5 5

5

4 0 10 Pa 0 17 10 Pa 0 49 10 Pa

3 3 10 Pa 3 3 atm

=

= =

. - . - .

. .

2 2 22 2

4 3

(0 50 10 m ) (6 0 m s)

4 7 10 m s 0 47 L s

dV A vdt

= =

= =

. . /

. / . /

Read Example:14.8/470&14.9/471.




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Viscosity is the reason why it takes effort to paddle a canoe

Viscosityis internalfrictionin a fluid. Viscous forces oppose the

motion of one portion of a fluid relative to another.

Viscosity

through calm water.

Viscous effects are important in the flow of fluids inpipes, the flowof blood, and the lubrication of engine parts.

Fluids that flow readily, such as water or gasoline, have smallerviscosities than do thick liquids such as honey or motor oil.




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Instead, the flow pattern becomes extremely irregular and

When the speed of a flowing fluid exceeds a certaincritical value,

the flow is no longer laminar.

Turbulence

comp ex, an c anges con nuous y w me; ere s no s ea y-

state pattern..

This irregular, chaotic flow is calledturbulence.




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Fluid Mechanics (Chapter 14)

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Transcript of Fluid Mechanics (Chapter 14)