CE374-Ch2a_ek [Compatibility Mode]

8/12/2019 CE374-Ch2a_ek [Compatibility Mode]

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CE 374 FLUID MECHANICS

Prepare y: Dr. Nuray Den i TokyayRevised b : Dr. El in Kentel


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.


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Introduction

H drostatics is the stud of fluids in which there is

no relative motion between the fluid particles.

ere no re a ve mo on, no ear ng re e o

be present.

The only stress that exists is a normal stress(theressure.

So it is the pressure that is primary interest in

.


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2.1. Scalar, Vector, Tensor Quantites and Fields

In order to define a point in space, we need 3

coordinates (x,y,z).

In eneral the h sical uantities are tensors of some

order:

with one characteristic, such as magnidute we call it as

sca ar:

(3)0

= 1

0-th order tensor or scalar. Ex: temperature


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2.1. Scalar, Vector, Tensor Quantites and Fields

If a physical quantity at a point can be defined with

three characteristics, such as magnidute, direction and

sense we call it as vector:

(3)1 = 3 1-st order tensor or vector. Ex: velocity

If a physical quantity at a point can be defined with

, -

(3)2 = 9 2-nd order tensor. Ex: stress tensor


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2.2. Surface & Bod Forces

There are two types of forces acting on the fluids:

1. Body forces

2. Surface Forces

Bod Forces, B: are all external forces which are

distributed over the mass of the fluid developedwithout h sical contact.

kBjBiBB zyx ++=)( kajaiamF zyxi ++= Inertia force can be

.


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Ex: When gravity is the only force and z is taken as

direction, then the body force is:

kmgkWkBB z ===

z

W

Surface forces are the forces that develop due to

.

Therefore they act on the contact surfaces.

Let us examine these surface forces in detail:


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Surface Forces and Stress at a oint

Let F be the resultant of all surface forces acting

on any surface A defined on the fluid surface.

z

B BCOA

ABOA

y

x

=

=

FA ACOA

z=

m O A

C

kAjAiAA zyx ++=


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Stress

The resultanf force F can be decomposed into components:

kFjFiFF zyx ++=

The area vector can also decomposed into components:

kAjAiAA zyx ++=

Then stress (i.e. the force per unit area) can be defined as:

A

FStress

A

0

lim)(


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Consider a point A in space. We can pass three planes

con a n ng po n , one s para e o xy-p ane, one s para e oxz-plane, & the other is paralel to yz-plane

xz z

yzAxx

xy

Ayy x

y y

xz zz

Azyzy

x


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In each plane, there will be one normal stress, and two

shear stress.

,

second one shows the direction of stress. Therefore,

.

Therefore in order to define stress (i.e. force per unit

area a a po n , we nee n ne componen s.z

A

zy

y

zx

x


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Stress Tensor

jF limRepresents nine scalar equations

iA Ai 0, , , , ,

plane direction

zz

z Stress tensor is a symmetricalsecond order tensor.

yz

zyzx yxxy

== xzxyxx

=yyyx

xy

xz

yzyyz = zzzyzx

xxFor all second order tensors, summation

:x

constantzzyyxx ==++ 3


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Thermodynamic Pressure z

On the fluid element nn

xA) Without shear stresses (=0):

a t e sur ace orces per unit

area &

y

yy

y

s

e on y o y orce, we g

are shown. sinsz =x cossy=zzW

= W

y-direction gives:

mayj=aaam zyx +=

ynnyy

ynnyy ay

2

=+

sinsz =


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Thermodynamic Pressure z

ynnyy ay

=+ nnx

If we shrink the volume to yyz s

obtain stress at a point,

i.e: 0 and is constant

y

nnyy =

x

zzW

Similarly Newtons second law in x and z directions would givennxx =and therefore:nnzz =

nnzzyyxx ===


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Pressure

===

What is the physical meaning of this equation?

The direction n is an arbitrary direction.

This equation means that the normal stress in any

on the direction. Therefore it is a scalar quantity.

Pressure in a fluid is constant at a point.


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Pascal Law ressure at a oint

In the absence of shear stresses, the normal stress

in a fluid is independent of orientation of the plane

thus can be represented by a single scalar quantity.Taking the fact that fluids can sustain only

compression, pressure p is set equal to negative of

this magnitude: pzzyyxx ===

B) With shear stresses (0):

1e u s ress: zzyyxx ==3


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The Gradient

Consider a pressure field acting on an infinitesimal frictionless

.

y-direction is

== += dpdxdydzpdxdzdypppdF yyy

The orce ector er unit o u e due to ressure ariations in

zand z directions is:

gradppkz

jy

ix

fd

==

+

+

==

y

p

dz

dyy

p

+

The Gradient operator

x dykzjyix

+

+

=


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2.3 Governing Equation

1 Consider a fluid element atrest or moving as a rigid body

(i.e. with no relative motion). = amF

2) Dividing by volume af =

,then the force per unit volume,

f, is due to pressure variations(Euler Equation)

and gravity only.


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Hydrostatics

akp =

pkz

jy

ix

fd

=

+

+

==

fakp =+=r

r

r

rr

Equation of

/volumebody forceresultantp=

s equat on s ows t at var at on o pressure s ue to t e

body force per unit volume in that direction.


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Fluid at rest a = 0

0=

+

+

+

k

z

pj

y

pi

x

pakp =

0=x

p

0=y

povern ng eren a

equation for the

=

z

p

distribution


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If gravity is the only body force, then

kfr

rr

=

kp =

equipotential surfaces.

e s e e sp acemen vec or on an equ po en a

surface:

kdzjdyidxsd rrrr ++=If we take the dot product of ds with the equation of


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kdzjdyidxsdr

rrr

++= ppp

fp

r

r

r

r

r

r

rr

==

=

=

zyx

On t e ot er an dp=0 on an equipotentia sur ace.

Therefore 0= sdf r

r

If dot product of two vectors is zero, these vectors are

perpendicular to each other. Therefore

equipotential surfaces are perpendicular to the direction ofresultant body force.

If gravity is the only body force, then equipotential

surfaces are horizontal planes.


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2.4 Pressure distribution(in an incompressible fluid at rest or moving as a rigid body withno acceleration)

zpatm =

s

A

atm

A zP

dzdp =z

p

zs

hhzzpp AsAatm == )(

zA pAatmabsoluteA ==

hppp atmabsolutegage ==

gagehpA =

Therefore, in an incompressible fluid, the pressure changes

.

with depth only. It remains constant on horizantal planes.


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Equi otential surfaces

The surfaces on which pressure is constant are called

equ o en a ur ace .

If gravity is the only body force, thenequ po en a sur aces are or zon a p anes.

z

zspatm

h hp =


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H drostatic Condition: Incom ressible Fluids

I we are wor ing exc usive y wit a iqui , t en t ere is a

free surface at the liquid-gas interface. For mostapplications, the pressure exerted at the surface is

atmos heric ressure, = Then the e uation is,

written as follows:

p = h + patm

p = h + po


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H drostatic Condition: Incom ressible Fluids

The pressure in a homogenous, incompressible fluid at rest

depends on the depth of the fluid relative to some reference

& is not influenced by the shape of the container.

lines of constantressure

p = pop = p1p = p2

1

For p1 = p = h1 + po

2 o


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Standard Atmos here

pT


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Absolute & Ga e Pressure

z

patm atmabsoluteau e ppp =absolutepressure

gaugepressure

w

p

.

Manometers measure gauge presssure.


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Absolute & Ga e Pressures

Local atmospheric

Absolute zero reference


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The basic principles in measuring the pressure are:

1. On equipotential surfaces, pressure is constant.

2. If the gravity is the only body force, then equipotentialsurfaces are horizontal planes.

3. In an inco ressib e uid ressure increases in o in

down, and decreases in moving up.

p p


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2.5. Measurement of Pressure

Barometers measures local atmospheric pressure

hppp

pp

mAatmB

BB

+==

= '

Evangelista Torricelli

(1608-1647)h

hppppp mvatmBA

==+===

0

equa po en a neB'

'BB pp =


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The various forms of manometers are:

Piezometers

U Tube ManometersDifferential Manometers

Inclined Manometers


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Piezometer Tube

hpp atmA +=atm

h Disadvantages:1. The pressure in the

greater than atmosphericpressure.

2. Pressure must be relativelysmall to maintain a small

.3. The measurement of

pressure must be of aliquid.


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U-tube manometer

patmThe fluid in the U-tube is

w

Aknown as the gage fluid.

The a e fluid t eh1

h2 mdepends on the application,

. . ,

whether the fluid equalpotentialline

1 2

=

.

21 hphp matmwA +=+


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Differential manometer

pp FY

=

=

BA phhhp =++ 332211

332211 BA

B

A

Final notes:

h1h31 3

1) Common gage fluids are Hg and H2O,some oils, and must be immiscible (i.e.

h2

that do not mix).

2) Temp. must be considered in very2

potential

line

accurate measurements, as t e gagefluid properties can change.

ap ar y can p ay a ro e, u n many

cases each meniscus will cancel.


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Inclined Manometers

This type of manometer is used to measure small

pressure changes.

patm

Lx

h

F

==

potential

line

atmatmA PLPhp +=+= sin


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Measurement of Pressure:

Mechanical & Electrical Devices

Spring

Bourdon Gage:

Pressure Transducer:

Motion


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2.6. Hydrostatic Forces on Plane Surfaces

Consider a plane surface which is immersed completely in an incompressible liquid. Wewish to evaluate the resultant hydrostatic force on the upper face. Since there can be

, .calculation, the plane of the submerged surface is extended so as to intersect with theplane of the free surface. The trace of intersection is shown as the x axis in the

figure. Note that the y axis is coplanar with the top surface of the plate.

dAhdApdF ==

o

dAhFA

=

=

hph

c

dF

F y yc o Distance toa centroid

A

=

A

dAyF sin

p

x

=

Ac

dAyy 1

AhAF == sindA

ccc

AhApF cc ==

cpc

xp

y


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This equation shows that these two systems are

equivalent, but not equal. Because, the point ofa lication of F is not at the centroid, but at a lower

distance, cp, called center of pressure.

hcFEquivalent hc

cp=center ofpressure hc

c=centroid

y

p cc ==


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Center of Pressure

Application point of the resultant hydrostatic force iscalled the center of pressure.

Let F act at (xp, yp)

The value of yp can be obtained by equating moments

= ypdAFy

a out t e x-axis

dAydAyyypdA == 2sinsin

=== AAdAyypda sin 2

=Ix IAyI xccxc +=

+ 2

AydAF sin Ayc

AyAy cc

=A

x dAyI 2secon momen o

area about x-axis 2cxcx AyII +=

second moment of an area is realated to thesecond moment of an area, Ixc about the

centroidal axis by the parallel-axis-theorem

f


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Center of Pressure

Similarly for xp

IAyxIIxydAxpdA xycccxycxy + sin AyAyAyydAF ccccA

p =====sin

=the product of

A

xyntert a o t earea A

ccxycxy yAxII +=using the transfer


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The Pressure Prism Conce t

Pressure prism is a geometric representation of the

hydrostatic force on a plane surface.

h1

=A

p

h1

of thepressure prism

FRcpFR

h2

bh2

Th it d f th lt t h d t ti f i l t th


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The magnitude of the resultant hydrostatic force is equal to the

vo ume o e re ure r m a e roug cen ro .

h1

h1

FRcpFR

h2

b

h2

F1

+

F2

h1(h2-h1)

bhhF )(

2 12

122

= bhhhF )( 1211 = ApFFF cR =+= 21 +


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Force on a plane area with top edge in a free surface


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Pressure Prism

pressure distributionrectangular area pressure prism.

on the area


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2.7. H drostatic Forces on Curved Surfaces

Example: Arch Dams

Picture: Gkekaya Dam

H d t ti F C d S f


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H drostatic Force on Curved Surfaces

Forces on curved surfaces submerged in any static fluid can be

artiall determined b methods used on lane sur aces.

A curved surface is shown in figure below, submerged in a static

fluid. The force on any area element dA of this surface isdirected along the normal to to the area element and is given as:

dApdF =

The (-) sign indicates that pressure andarea are in opposite directions.

z

zs

Taking the dot product of each side of

the equation above with unit vectors i, jh dAp

an , we ge e componen s x, yand dFz:y

dAx

dAy

x

zxx dApdF = yy dApdF = zz dApdF =

Hydrostatic Forces on Curved Surfaces:


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Horizontal Componentsz

p

dAcos

xx pdApdAdF == cos

xcA

xx AppdAF ==

x

ycA

yy ppdAF ==

where Ax and Ay are the projected areas of the curved surfaceon planes perpendicular to the x- and y- axes.

en:The horizontal component of the hydrostatic force on a curvedsurface is the roduct of the ressure at the centroid of thearea projected on a vertical plane perpendicular to the x- and y-axis.



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Vertical Componentz

zz dApdApdF == sin

ph === ddAhdApF zAA zz

dA = zF

x

e ver ca componen o e y ros a c orce on acurved surface is the weight of the liquid volume between

surface.

2 8 Buoyancy & Flotation


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2.8. Buoyancy & Flotation

distribution on submerged volumes is called the

uoyant orce.

The buoyant force acts through the center of gravityof the submer ed volume in a direction o osite to

that of gravity.

Forces on a submerged body


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Forces on a submerged body

a submer ed bod b free-bod dia ram c free-bod shown F

Forces on a floatin ob ect


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Forces on a floatin ob ect


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Buo anc and Stabilit


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Buo anc and Stabilit

Buoyancy force FB is equal only to

sp ace

Three scenarios are possible:

1.bodyfluid Sinking body

=submergedB

liquiddisplacedB

F =


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,

when displaced, it returns to its original equilibrium

.

cg WW

cgcp

FBcp FB

Restoring couple


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A body is in unstable equilibrium if, when displaced, it

moves to a new equilibrium position.

cg

WW

cg

cpFB

cp FB

Overturning

coup e

Stabilit of a submer ed bod


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Stabilit of a submer ed bod

(a) unstable (b) neutral (c) stable

Stabilit of a floatin bod


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Stabilit of a floatin bod

(a) equilibrium position (b) rotated position

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