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Chapter 14
Fluid Mechanics
(MEKANIK BENDALIR)
States of Matter (Keadaankeadaan Jirim)

Solid (pepejal)


Liquid (cecair)


Has a definite volume and shape
Has a definite volume but not a definite
shape
Gas – unconfined

Has neither a definite volume nor shape
States of Matter, cont


All of the previous definitions are
somewhat artificial
More generally, the time it takes a
particular substance to change its shape
in response to an external force
determines whether the substance is
treated as a solid, liquid or gas
Fluids (Bendalir)


A fluid is a collection of molecules that
are randomly arranged and held
together by weak cohesive forces
(daya-daya lekitan) and by forces
exerted by the walls of a container
Both liquids and gases are fluids
Statics and Dynamics with Fluids (Dinamik
dan statik bagi bendalir)

Fluid Statics (Statik bendalir)


Fluid Dynamics (dinamik bendalir)


Describes fluids at rest
Describes fluids in motion
The same physical principles that have
applied to statics and dynamics up to
this point will also apply to fluids
Forces in Fluids (Daya-daya dlm
bendalir)



Fluids do not sustain shearing stresses
(tegasan ricih) or tensile stresses (tegangan
tegasan)
The only stress that can be exerted on an
object submerged in a static fluid is one that
tends to compress (memampatkan) the
object from all sides
The force exerted by a static fluid on an
object is always perpendicular to the surfaces
of the object
Pressure (Tekanan)

The pressure P of
the fluid at the level
to which the device
has been
submerged is the
ratio of the force to
the area
F
P
A
Pressure, cont

Pressure is a scalar quantity



Because it is proportional to the magnitude
of the force
If the pressure varies over an area,
evaluate dF on a surface of area dA as
dF = P dA
Unit of pressure is pascal (Pa)
1 Pa  1 N/m
2
Contoh 1

(a)
(b)
(c)
Satu tilam air (waterbed) mempunyai ukuran 2.00
m panjang, 2.00 m lebar, dan 30.0 cm dalam.
Cari berat air di dalam tilam.
Cari tekanan yg dikenakan oleh air di dalam tilam
tersebut ke atas lantai apabila tilam berada dalam
kedudukan normal. Anggap keseluruhan
permukaan tilam terletak di atas lantai.
Katakan tilam air diganti dgn tilam biasa seberat
300 paun (1N=0.0225paun) yang disokong oleh 4
kaki. Keratan rentas kaki mempunyai jejari 3 cm.
Apakah tekanan yg dikenakan nke ats lantai?
Penyelesaian contoh 1
(a) Ketumpatan air  1, 000kg .m3
Isipadu air dlm tilam :
V  (2.00m)(2.00m)(0.300m)  1.20m3
Maka, jisim air dlm tilam adalah
M  V  (1, 000kg.m3 )(1.20m3 )  1.20 103 kg
Berat adalah
Mg  (1.20 103 kg )(9.80m.s 2 )  1.18 10 4 N
Berat ini lebih kurang 2, 650 paun. Berat tilam biasa adalah 300 paun.
(b) Luas permukaan tilam  A  4.00m 2 .
1.18 104 N
3
Maka, tekanan  P 

2.95

10
Pa
2
4.00m
Penyelesaian contoh 1 (contd)
(c) Berat tilam ditampung samarata oleh 4 kaki.
Maka, tekanan ke atas lantai adalah

F
mg
300 paun 
1N
P 


2 
2
A 4  r  4 (0.0200m)  0.225 paun 
 2.65  105 Pa
Perhatikan bahawa ini adalah 100 kali lebih besar
dari tekanan yg diberikan oleh tilam air.
Pressure vs. Force


Pressure is a scalar and force is a vector
The direction of the force producing a
pressure is perpendicular to the area of
interest
Measuring Pressure



The spring is
calibrated by a
known force
The force due to the
fluid presses on the
top of the piston
and compresses the
spring
The force the fluid
exerts on the piston
is then measured
Variation of Pressure with Depth
(Perubahan Tekanan dgn kedalaman)



Fluids have pressure that varies with depth
If a fluid is at rest in a container, all portions
of the fluid must be in static equilibrium
All points at the same depth must be at the
same pressure

Otherwise, the fluid would not be in equilibrium
Pressure and Depth (Tekanan dan
Kedalaman)


Examine the darker
region, a sample of
liquid within a
cylinder
 It has a crosssectional area A
 Extends from
depth d to d + h
below the surface
Three external
forces act on the
region
Pressure and Depth, cont

The liquid has a density of 



Assume the density is the same throughout the
fluid
This means it is an incompressible liquid (cecair
tak termampat)
The three forces are:



Downward force on the top, P0A
Upward on the bottom, PA
Gravity acting downward, Mg

The mass can be found from the density:
M  V   Ah
Pressure and Depth, final

Since the net force must be zero:
ˆ
ˆ
ˆ
F

PA
j

P
A
j

M
gj

o
This chooses upward as positive
Solving for the pressure gives
 P = P0 + gh
The pressure P at a depth h below a point in
the liquid at which the pressure is P0 is
greater by an amount gh



Density Notes (Nota bagi
ketumpatan)



Density is defined as the mass per unit
volume of the substance
The values of density for a substance
vary slightly with temperature since
volume is temperature dependent
The various densities indicate the
average molecular spacing in a gas is
much greater than that in a solid or
liquid
Density Table
Atmospheric Pressure (Tekanan
atmosfera)


If the liquid is open to the atmosphere,
and P0 is the pressure at the surface of
the liquid, then P0 is atmospheric
pressure
P0 = 1.00 atm = 1.013 x 105 Pa
Pascal’s Law (Hukum Pascal)



The pressure in a fluid depends on
depth and on the value of P0
An increase in pressure at the surface
must be transmitted to every other
point in the fluid
This is the basis of Pascal’s law
Pascal’s Law, cont


Named for French scientist Blaise Pascal
A change in the pressure applied to
a fluid is transmitted undiminished
to every point of the fluid and to
the walls of the container
P1  P2
F1 F2

A1 A2
Pascal’s Law, Example



Diagram of a
hydraulic press
(penekan hidraulik)
(see figure)
A large output force
can be applied by
means of a small
input force
The volume of liquid
pushed down on the
left must equal the
volume pushed up on
the right
Pascal’s Law, Example cont.


Since the volumes are equal,
A1x1  A2x2
Combining the equations,
 F1x1  F2 x2 which means W1 = W2

This is a consequence of Conservation of
Energy (keabadian tenaga)
Contoh 2

Pengangkat kereta di dalam satu woksyop
menggunakan udara termampat untuk
menolak satu omboh (piston) berjejari
5.00cm dgn mengenakan daya ke atasnya.
Tekanan yg wujud disebar oleh cecair ke satu
omboh berjejari 15.0cm. Apakah daya yg
mesti dikenakan oleh udara termampat untuk
mengangkat kereta sebeart 13,3000N?
Apakah tekanan udara yg menghasilkan daya
ini?
Penyelesaian contoh 2
 A1 
  (5.00 102 m) 2 
4
F1    F2  
(1.33

10
N)
2
2 
  (15.00 10 m) 
 A2 
 1.48 103 N
Tekanan udara yg menghasilkan daya ini adalah
F1
1.48 103 N
5
P


1.88

10
Pa
2
2
A1  (5.00 10 m)
Tekanan ini adalah dua kali ganda tekanan udara.
Pascal’s Law, Other
Applications




Hydraulic brakes (brek hidraulik)
Car lifts (lif kereta)
Hydraulic jacks (hidraulik jek)
Forklifts
Contoh 3

Anggarkan daya yg dikenakan ke atas
gegendang telinga semasa kita
berenang di dalam kolam pada
kedalaman 5.0 m.
Penyelesaian Contoh 3


Cari dahulu tekanan yg tak seimbang.
Anggarkan luas gegendang telinga.
Kemudian tentukan daya yg dikenakan oleh
air ke atas gegendang.
Udara di dlm telinga tengah selalunya berada
pada tekanan udara, P0. Maka, untuk
mendapatkan paduan daya kita perlu
timbangkan perbezaan di antara jumlah
tekanan di bawah kolam dan tekanan udara.
Penyelesaian Contoh 3 (contd)
Pbot-P0=gh
=(1.00x103kg.m-3)(9.80m.s-2)(5.0m)
=4.9x104Pa
Luas permukaan gegendang telinga
adalah 1cm2=1x10-4m2.
Maka, daya F=(Pbot-P0)A ≈ 5N.
Magnitud daya sebanyak ini membuatkan
telinga berdengung.
Contoh 4

Air diisi ke ketinggian H dibelakang satu
empangan selebar w (lihat Fig.14.5).
Tentukan paduan daya yg dikenakan
oleh air ke atas empangan.
Penyelesaian contoh 4
P   gh   g ( H  y )
dF  PdA   g ( H  y ) wdy
Maka, jumlah daya ke atas empangan adalah
H
1
F   PdA    g ( H  y ) wdy   gwH 2
0
2
Pressure Measurements:
Barometer


Invented by Torricelli
A long closed tube is filled
with mercury and inverted in
a dish of mercury



The closed end is nearly a
vacuum
Measures atmospheric
pressure as P0   Hg gh
One 1 atm = 0.760 m (of Hg)
Pressure Measurements:
Manometer




A device for measuring the
pressure of a gas
contained in a vessel
One end of the U-shaped
tube is open to the
atmosphere
The other end is
connected to the pressure
to be measured
Pressure at B is P0+ρgh
Absolute (Mutlak) vs. Gauge Pressure
(tekanan tolok)



P = P0 + gh
P is the absolute pressure (tekanan
mutlak)
The gauge pressure is P – P0


This is also gh
This is what you measure in your tires
Buoyant Force (daya apongan)



The buoyant force
is the upward force
exerted by a fluid on
any immersed object
The parcel is in
equilibrium
There must be an
upward force to
balance the
downward force
Buoyant Force, cont



The upward force, B, must equal (in
magnitude) the downward gravitational
force
The upward force is called the buoyant
force
The buoyant force is the resultant force
due to all forces applied by the fluid
surrounding the parcel
Archimedes’s Principle (Prinsip
Archimedes)



The magnitude of the buoyant force
always equals the weight of the fluid
displaced by the object
This is called Archimedes’s Principle
Archimedes’s Principle does not refer to
the makeup of the object experiencing
the buoyant force
 The object’s composition is not a
factor since the buoyant force is
exerted by the fluid
Archimedes’s Principle, cont



The pressure at the
top of the cube
causes a downward
force of Pt A
The pressure at the
bottom of the cube
causes an upward
force of Pb A
B = (Pb – Pt) A = Mg
Archimedes's Principle: Totally
Submerged Object (Objek yg tenggelam
sepenuhnya)




An object is totally submerged in a fluid
of density fluid
The upward buoyant force is
B  fluid gV  fluid gVobj
The downward gravitational force is
w = mg =  obj gVobj
The net force is B - Fg = ( fluid   obj )gVobj
Archimedes’s Principle: Totally
Submerged Object, cont



If the density of the object is
less than the density of the
fluid, the unsupported object
accelerates upward
If the density of the object is
more than the density of the
fluid, the unsupported object
sinks
The motion of an object in
a fluid is determined by the
densities of the fluid and
the object
Archimedes’s Principle:
Floating Object (Objek terapong)



The object is in static equilibrium
The upward buoyant force is balanced
by the downward force of gravity
Volume of the fluid displaced
corresponds to the volume of the object
beneath the fluid level
 obj Vfluid

fluid Vobj
Archimedes’s Principle:
Floating Object, cont

The fraction of
the volume of a
floating object
that is below the
fluid surface is
equal to the ratio
of the density of
the object to that
of the fluid
Archimedes’s Principle, Crown
Example




Archimedes was (supposedly) asked, “Is the
crown made of pure gold?”
Crown’s weight in air = 7.84 N
Weight in water (submerged) = 6.84 N
Buoyant force will equal the apparent weight
loss


Difference in scale readings will be the buoyant
force
Lihat contoh 14.5 Serway m.s. 429
Archimedes’s Principle, Crown Example,
cont. (contoh mahkota raja)


F  B  T2  Fg  0
B = Fg – T2
(Weight in air –
“weight” in water)
 Archimedes’s
principle says
B = gV
 Then to find the
material of the
crown, crown =
mcrown in air / V
Archimedes’s Principle, Iceberg Example




What fraction of the iceberg is below
water?
The iceberg is only partially submerged
and so Vfluid / Vobject = 
/ fluid
object
applies
The fraction below the water will be the
ratio of the volumes (Vwater / Vice)
Lihat contoh 14.6 Serway m.s. 430.
Archimedes’s Principle,
Iceberg Example, cont


Vice is the total
volume of the
iceberg
Vwater is the volume
of the water
displaced


This will be equal to
the volume of the
iceberg submerged
About 89% of the
ice is below the
water’s surface
Types of Fluid Flow – Laminar
lamina)

(Pengaliran
Laminar flow




Steady flow (pengaliran mantap)
Each particle of the fluid follows a smooth
path
The paths of the different particles never
cross each other
The path taken by the particles is called a
streamline (garis strim)
Types of Fluid Flow –
Turbulent (alir bergelora)


An irregular flow (alir tak sekata)
characterized by small whirlpool like
regions
Turbulent flow occurs when the
particles go above some critical speed
Viscosity (kelikatan)



Characterizes the degree of internal friction in
the fluid
This internal friction, viscous force (daya
likat), is associated with the resistance that
two adjacent layers of fluid have to moving
relative to each other
It causes part of the kinetic energy of a fluid
to be converted to internal energy
Ideal Fluid Flow (alir bendalir unggul)
There are four simplifying assumptions
made to the complex flow of fluids to
make the analysis easier
(1) The fluid is nonviscous (bendalir
tak likat)– internal friction is neglected
(2) The flow is steady (alir mantap)–
the velocity of each point remains
constant

Ideal Fluid Flow, cont
(3) The fluid is incompressible
(bendalir tak termampat)
– the density remains constant
(4) The flow is irrotational (bendalir
tak berputar) – the fluid has no angular
momentum about any point
Streamlines (Garis strim)



The path the
particle takes in
steady flow is a
streamline
The velocity of the
particle is tangent
to the streamline
A set of
streamlines is
called a tube of
flow
Equation of Continuity (persamaan
keselanjaran)



Consider a fluid moving
through a pipe of
nonuniform size
(diameter)
The particles move
along streamlines in
steady flow
The mass that crosses
A1 in some time interval
is the same as the mass
that crosses A2 in that
same time interval
Equation of Continuity, cont



m1 = m2 A1v1 = A2v2
Since the fluid is incompressible,  is a
constant
A1v1 = A2v2


This is called the equation of continuity for
fluids
The product of the area and the fluid
speed at all points along a pipe is constant
for an incompressible fluid
Equation of Continuity,
Implications




The speed is high where the tube is
constricted (small A)
The speed is low where the tube is wide
(large A)
The product, Av, is called the volume flux or
the flow rate
Av = constant is equivalent to saying the
volume that enters one end of the tube in a
given time interval equals the volume leaving
the other end in the same time

If no leaks are present
Contoh 5

Air dari satu hos berdiameter 2.50cm telah
digunakan oleh seorang tukang kebun untuk
mengisi baldi 30.0-L. Pekebun ini mendapati
bahawa dia memerlukan 1.00 min untuk
mengisi baldi itu dgn air sepenuhnya. Satu
nozzle dgn keratan rentas 0.500cm2
disambungkan ke hos. Nozzle ini dipegang
pada ketinggian 1.00m dari tanah. Berapa
projection mengufuk air yg dipancut keluar.
Penyelesaian contoh 5
Titik 1 adalah hos dan titik 2 pula adalah nozzle.
 (2.50cm) 2 
d 
2
Keratan rentas hos : A1       

4.91
cm

2
4
 


Kadar pengaliran air ke baldi  30.0 L / min
2
30.0 103 cm3
A1v1  30.0 L / min 
 500cm3 / s
60.0s
500cm3 / s 500cm3 / s
v1 

 102cm / s  1.02m / s
2
A1
4.91cm
Penyelesaian contoh 5 (contd)
Gunakan persamaan keselanjaran untuk
mencari v2  vxi . Laju air yg kelaur dari nozzle.
A1v1  A2 v2  A2 vxi
vxi 
A1
v1
A2
4.91cm 2
vxi 
(1.02 m / s )
2
0.500cm
 10.0m / s
1 2
gt
2
1
1.00m  0  0  (9.80m / s 2 )t 2
2
2(1.00m)
t
 0.452 s
2
9.80m / s
x f  xi  vxi t  0  (10.0m / s )(0.452 s )  4.52m
y f  yi  v y t 
Bernoulli’s Equation


As a fluid moves through a region
where its speed and/or elevation above
the Earth’s surface changes, the
pressure in the fluid varies with these
changes
The relationship between fluid speed,
pressure and elevation was first derived
by Daniel Bernoulli
Bernoulli’s Equation, 2




Consider the two shaded
segments
The volumes of both
segments are equal
The net work done on the
segment is W =(P1 – P2) V
Part of the work goes into
changing the kinetic energy
and some to changing the
gravitational potential
energy
Bernoulli’s Equation, 3

The change in kinetic energy:



K = ½ mv22 - ½ mv12
There is no change in the kinetic energy of
the unshaded portion since we are
assuming streamline flow
The masses are the same since the
volumes are the same
Bernoulli’s Equation, 4

The change in gravitational potential
energy:



U = mgy2 – mgy1
The work also equals the change in
energy
Combining:
W = (P1 – P2)V =½ mv22 - ½ mv12 + mgy2 – mgy1
Bernoulli’s Equation, 5



Rearranging and expressing in terms of density:
P1 + ½ v12 + mgy1 = P2 + ½ v22 + mgy2
This is Bernoulli’s Equation and is often
expressed as
P + ½ v 2 + gy = constant
When the fluid is at rest, this becomes P1 – P2 =
gh which is consistent with the pressure
variation with depth we found earlier
Bernoulli’s Equation, Final

The general behavior of pressure with
speed is true even for gases

As the speed increases, the pressure
decreases
Contoh 6

Satu tangki yg tertutup mengandungi cecair
yg mempunyai ketumpatan . Ia mempunyai
lubang pada ketinggian y1 dari bawah tangki.
Lubang ini terdedah ke atmosfera dan
diameternya adalah lebih kecil dari diameter
tangki. Udara di atas cecair ditetapkan pada
tekanan P. Tentukan laju cecair apabila
keluar lubang tersebut jika ketinggian cecair
adalah h dari lubang. (Rujuk Fig. 14.21
Serway m.s. 435)
Penyelesaian Contoh 6
A2
A1. Guna persamaan Bernoulli pada titik 1 dan 2.
P1  P0 .
1 2
 v1   gy1  P   gy2 ;
2
2( P  P0 )
v1 
 2 gh ; Jika P
P0 

y2  y1  h
P0 , maka v1 ( P ).
Jika tangki terdedah kepada udara, P  P0 danv1  2 gh .
Laju cecair kelaur dari lubang sama dgn laju jatuh bebas
dari ketinggian h. Fenomena ini dipanggil hukum Torricelli.
Applications of Fluid Dynamics




Streamline flow around
a moving airplane wing
Lift is the upward force
on the wing from the air
Drag is the resistance
The lift depends on the
speed of the airplane,
the area of the wing, its
curvature, and the
angle between the wing
and the horizontal
Lift – General


In general, an object moving through a fluid
experiences lift as a result of any effect that
causes the fluid to change its direction as it
flows past the object
Some factors that influence lift are:


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The shape of the object
The object’s orientation with respect to the fluid flow
Any spinning of the object
The texture of the object’s surface
Golf Ball
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The ball is given a
rapid backspin
The dimples
increase friction
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Increases lift
It travels farther
than if it was not
spinning
Atomizer (Pengabus)
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A stream of air passes
over one end of an
open tube
The other end is
immersed in a liquid
The moving air reduces
the pressure above the
tube
The fluid rises into the
air stream
The liquid is dispersed
into a fine spray of
droplets