Fluid Mechanics - 203PHYS

203 PHYS - Physics 2Physics 2
(for Engineering)(for Engineering)
Instructor:
Dr. Sabar D. Hutagalung
Email: shutagalung@jazanu.edu.sa;
sdhutagalung@gmail.com

FLUID MECHANICSFLUID MECHANICS

Main ReferenceMain Reference
• Raymond A. Serway & John W. Jewett, Jr., Physics
for Scientists and Engineers with Modern Physics,
9th Edition, Brooks/Cole, 2014.

• Fluid is a collection of
molecules that are randomly
arranged and held together by
weak cohesive forcesweak cohesive forces and by
forces exerted by the walls of a
container.
• Both liquids and gases are
fluids.
What is a Fluid?What is a Fluid?

Common Fluids
• Liquids:
 Water, oil, mercury, gasoline, alcohol
• Gasses:
 Air, helium, hydrogen, steam
• Borderline:
 Jelly, asphalt, toothpaste, paint
Fluid mechanicsFluid mechanics is a study of the behavior ofis a study of the behavior of
fluids, either at rest (fluid statics) or in motionfluids, either at rest (fluid statics) or in motion
(fluid dynamics).(fluid dynamics).

Application Areas of Fluid MechanicsApplication Areas of Fluid Mechanics

Fluid StaticsFluid Statics::
 Fluid is at rest: no shear forces.
 Pressure is the only force acting.
Fluid DynamicsFluid Dynamics::
• Fluid in motion
Statics and Dynamics Fluid

DensityDensity
• The density of a fluid is the concentration of mass
• Mass = 100 g = 0.1 kg
• Volume = 100 cm3
= 10-4
m3
• Density = 1 g/cm3
= 1000 kg/m3
• Mass: 1 g = 0.001 kg = 10Mass: 1 g = 0.001 kg = 10-3-3
kgkg
• Volume: 1 cmVolume: 1 cm33
= 1 mL = 10= 1 mL = 10-6-6
mm33
• Density: 13.6 g/cmDensity: 13.6 g/cm33
= (13.6) (10= (13.6) (10-3-3
/10/10-6-6
))
kg/mkg/m33
= 13.6 x 10= 13.6 x 1033
kg/mkg/m33

Density of some common
substances

PressurePressure
• Units of pressure are N/m2
or Pascals (Pa) – 1 N/m2
= 1 Pa
• Atmospheric pressure = 1 atm = 101.3 kPa = 1 x 105
N/m2

Quiz 1
• Suppose you are standing directly behind
someone who steps back and accidentally
stomps on your foot with the heel of one
shoe. Would you be better off if that person
were (a) a large, male professional basketball
player wearing sneakers or (b) a petite woman
wearing spike-heeled shoes?

Quiz 1
Answer:
(a) P = F/A
Larger area A, lower pressure P.

Example 1 – The water bed
• The mattress of a water bed is 2.00 m long by 2.00 m wide
and 30.0 cm deep. (a) Find the weight of the water in the
mattress. (b) Find the pressure exerted by the water bed on
the floor when the bed rests in its normal position
• Ans: (a) The density of fresh water is 1000 kg/m3
, and the
volume of the water filling the mattress is V = (2.00 m)(2.00
m)(0.300 m) = 1.20 m3
.
• Hence, the mass of the water in the bed is

Example 1 – The water bed
• Ans: (b) Assume that the entire lower surface of the
bed makes contact with the floor.
• When the bed is in its normal position, the area in
contact with the floor is 4.00 m2
; thus,
F = Mg = 1.18 x 104
N; A = 4 m2

Quiz 2
• The pressure at the bottom of a filled glass of
water (ρ = 1 000 kg/m3
) is P. The water is
poured out and the glass is filled with ethyl
alcohol (ρ = 806 kg/m3
). What is the pressure
at the bottom of the glass? (a) smaller than P
(b) equal to P (c) larger than P (d)
indeterminate

Quiz 2
Answer: (a)
• The pressure at the bottom of the glass is (a)
smaller than P.
ρa < ρw
ma < mw
Fa < Fw
Aa = Aw a: alcohol
Pa < Pw w: water

Variation of Pressure with Depth
• Divers know that…
 Pressure increases with depth
• In Planes…
 Pressure decreases with height (pressurized cabins)
• Why?
 Density: mass by unit of volume (mass/volume)
 Incompressible fluid: the density is the uniform
throughout the liquid
The pressureThe pressure PP at aat a
depthdepth hh below a point inbelow a point in
the liquid is greater bythe liquid is greater by
an amountan amount ρρghgh..

Pressure and DepthPressure and Depth
• 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

Hydrostatic EquilibriumHydrostatic Equilibrium
• Pressure differences drive fluid flow
• If a fluid is in equilibrium, pressure forces must balance
• Pascal’s law: pressure change is transmitted through a fluid

Pascal’s lawPascal’s law
 The pressure is the same at all points
having the same depth, independent of
the shape of the container.
 Any increase in pressure at the surface
must be transmitted to every other
point in the fluid.
• 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.
2
2
1
1
A
F
A
F
P ==

• This concept was first recognized by the
French scientist Blaise Pascal (1623–1662) and
is called Pascal’s law.Pascal’s law.
• Pascal’s lawPascal’s law: A change in the pressure appliedA change in the pressure applied
to a fluid is transmitted undiminished to everyto a fluid is transmitted undiminished to every
point of the fluid and to the walls of thepoint of the fluid and to the walls of the
container.container.
• An important application of Pascal’s law is the
hydraulic presshydraulic press.

• Displacement ∆x1 equals the volume of liquid
pushed up on the right as the right piston
moves upward through a displacement ∆x2.
• That is, A1∆x1 = A2∆x2; thus, A2/A1 = ∆x1/∆x2 .
A2/A1 = F2/F1
• Thus, F2/F1 = ∆x1/∆x2 ,
• So, F1∆x1 = F2∆x2.
2
2
1
1
A
F
A
F
=

Example 2 – The car lift
• In a car lift used in a service station,
compressed air exerts a force on a small
piston that has a circular cross section and a
radius of 5.00 cm. This pressure is transmitted
by a liquid to a piston that has a radius of 15.0
cm. (a) What force must the compressed air
exert to lift a car weighing 13,300 N? (b) What
air pressure produces this force?

Solution
• (a) Force to lift a car:
• (b) The air pressure that produces this force is
2
2
1
1
A
F
A
F
=

Example 3 - A Pain in Your Ear
• Estimate the force exerted on your eardrum
due to the water when you are swimming at
the bottom of a pool that is 5.0 m deep.

Solution
• The air inside the middle ear is normally at
atmospheric pressure, P0.
• Let’s estimate the surface area of the eardrum to be
approximately, A = 1 cm2
= 1 x 10-4
m2
.
• Find pressure difference:
• Magnitude of the net force on the ear:

Example 4 - The Force on a Dam
• Water is filled to a height H
behind a dam of width w.
Determine the resultant force
exerted by the water on the
dam.

Solution
• Let us imagine a vertical y axis, with y = 0 at the
bottom of the dam
• Pressure at the depth h:
• Force exerted on the shaded strip of
area dA = w dy:
• Total force on the dam:

Solution without calculus
• The average pressure due to the water over
the face of the dam is the average of the
pressure at the top and the pressure at the
bottom:
• The total force on the dam is equal to the
product of the average pressure and the area:

Pressure Measurements:
(a) Barometer

(a) Barometer
• One instrument used to measure atmospheric
pressure is barometer, invented by Evangelista
Torricelli (1608–1647).
• Barometer is a long tube closed at one end filled
with mercury and then inverted into a dish of
mercury. The closed end of the tube is nearly a
vacuum, so the pressure at the top of the mercury
column can be taken as zero.
P0 = ρHggh,
where ρHg is the density of the mercury and h is the
height of the mercury column.
Mercury

(a) Barometer
• Let us determine the height of a
mercury column for one atmosphere of
pressure, P0 = 1 atm = 1.013 x 105
Pa:

(b) Manometer
• A device for measuring the pressure of a gas contained in agas contained in a
vesselvessel is the open tube manometer.
• One end of a U-shaped tube containing liquidliquid is open to the
atmosphere, and the other end is connected to a container of
gas at pressure P.
The difference in the pressures in each part
is, P – P0 = ρgh. The pressure P is called
the absolute pressure, and the
difference P - P0 is called the gauge
pressure.
For example, the pressure you measure in
your car tire is gauge pressure.
gas
liquid

Quiz 3
• Several common barometers are built, with a
variety of fluids. For which of the following
fluids will the column of fluid in the barometer
be the highest? (a) mercury (b) water (c) ethyl
alcohol (d) benzene.
• Answer: (c) because ethyl alcohol has the
lowest density. See Table 14.1, p. 419.

Buoyant Forces andBuoyant Forces and
Archimedes’s PrincipleArchimedes’s Principle
• The pressure at the bottom (Pbot) of the cube higher
than the pressure at the top (Ptop), Pbot > Ptop by an
amount ρfluidgh, where h is the height of the cube and
ρfluid is the density of the fluid.
• Upward force = PbotA, and
• Downward force = PtopA; where A is area.
• Magnitude of buoyant force,buoyant force, BB:
wherewhere VVdispdisp == AhAh is the volume of the fluid displaced by the cube.is the volume of the fluid displaced by the cube.

Totally Submerged Object
• When an object is totally submergedtotally submerged in a fluid,
the volume Vdisp = Vobj; so, the magnitude of the
upward buoyant force is:
BB == ρρfluidfluidgVgVobjobj
• The net force on the object is
B - Fg =(ρfluid - ρobj)gVobj.
• Hence, if the density of the object is less than the density ofHence, if the density of the object is less than the density of
the fluid, the downward gravitational force is less than thethe fluid, the downward gravitational force is less than the
buoyant force and the unsupported object acceleratesbuoyant force and the unsupported object accelerates
upward.upward.

Floating Object
• Consider an object of volume Vobj and density
ρobj < ρfluid in static equilibrium floating on the
surface of a fluid, that is, an object that is only
partially submerged.
Fg = B, then ρfluidgVdisp = ρobjgVobj
ρfluidVdisp = ρobjVobj

Lighter than air?Lighter than air?
• The atmosphere is a fluid
• Scientific instruments are carried into the
atmosphere by balloon filled with Helium, in
this case ρair > ρballoon, the balloon will lift-up.
• If ρair = ρballoon and the balloon will stop rising (FB
up = FGdown)

Quiz 4
(a) (b) (c)
Answer: (c). Maximum Buoyant force obtained when the chest isAnswer: (c). Maximum Buoyant force obtained when the chest is
fully submerged in the water and help to reduce downward force.fully submerged in the water and help to reduce downward force.
Treasur
e chest

Example 5
T1 = Fg = 7.84 N T2 = 6.84 N

Solution
• The scale reads the true weight,
• When immersed in water, the buoyant force B
reduces the scale reading to an
apparent weight,
• In equilibrium model of the crown in water:

Solution
• The volume of the crown (Vc) is equal to
the volume of the displaced water (Vdisp),
Find the density of the crown:
Note: Density of gold is 19.3 x 103
kg/m3
. Therefore, Archimedes
should have reported that the king had been cheated. Either the
crown was hollow, or it was not made of pure gold.
T1 = Fg = 7.84 N
T2 = 6.84 N

Solution
• What If ? Suppose the crown has the same weight but
is indeed pure gold and not hollowpure gold and not hollow. What would the
scale reading be when the crown is immersed in
water?
• Answer: Find the buoyant force:
Find the tension in the string hanging from the scale:
T1 = Fg = 7.84 N
T2 = 6.84 N

Example - Floating ObjectsExample - Floating Objects
Q. If the density of an iceberg is 0.86
that of seawater, how much of an
iceberg’s volume is below the sea?

Fluid DynamicsFluid Dynamics
Laminar and Turbulent flowLaminar and Turbulent flow
• A flow in motion can be
Laminar or steady: each particle follows
a smooth path
• velocity of fluid particles passing any
point remains constant in time
Turbulent: turbulent flow is irregular
flow characterized by small whirlpool-
like regions
• occurs above a certain speed
The smoke first moves in
laminar and then in
turbulent flow.

Ideal FluidsIdeal Fluids
• Steady: velocity, density and pressure not
change in time; no turbulence
• Incompressible: constant density
• Nonviscous: no internal friction between
adjacent layers
• Irrotational: no particle rotation about center
of mass

• The product of the area and the fluid speed at
all points along a pipe is constant for an
incompressible fluid:
constant2211 == vAvA
The Continuity EquationThe Continuity Equation
rateFlow2211 == vAvA
Volume flux = Flow rate (volume/unit time)Volume flux = Flow rate (volume/unit time)

Conservation of Mass: The Continuity Eqn.Conservation of Mass: The Continuity Eqn.
The rate a fluid enters a pipe must equal the rate the fluidThe rate a fluid enters a pipe must equal the rate the fluid
leaves the pipe. i.e. There can be no sources or sinks of fluid.leaves the pipe. i.e. There can be no sources or sinks of fluid.
“The water all has to go somewhere”

Example 7
yf = 1 m
xf
Similar to Projectile motionSimilar to Projectile motion
at zero launch angleat zero launch angle

Bernoulli’s Equation
( )VPPW 21 −=
2
1
2
2
2
1
2
1
mvmvK −=∆
12 mgymgyU −=∆
UKW ∆+∆=
• An increase in the speed of the fluid occurs simultaneously with aAn increase in the speed of the fluid occurs simultaneously with a
decrease in pressure or a decrease in the fluid´s potential energy.decrease in pressure or a decrease in the fluid´s potential energy.
• The work done:The work done:

Bernoulli’s Equation
where: m = ρV
By rearranging the terms gives
Bernoulli’s equation:
When fluid at rest v1 = v2 = 0:

Quiz 5
Answer:Answer: (a)(a)
Because when we blow air thru small space between them means speed of
air at space between baloons is higher than speed of air at opposite side of
baloons, v >vo. So that pressure at the space between them is lower, P < Po
therefore, the balloons will pushed move toward each othersthe balloons will pushed move toward each others.
v
P
PoPo
vovo
Based on Bernoulli’s equation:Based on Bernoulli’s equation:
v P
v P

Problem
• What happens to
the car and truck?
• They move toward
each other or they
move away from
each other?
Explain, why.
TruckTruck
CarCar

Example 8
Pressure P1 is greater than
pressure P2 because v1 < v2.
This device can be used to
measure the speed of fluid
flow.

Example 8 - Solution
y1 = y2 (because the pipe is horizontal)
=
0
(Bernoulli’s equation)(Bernoulli’s equation)

Example 9: Torricelli’s LawExample 9: Torricelli’s Law
• An enclosed tank containing a liquid of density ρ has a hole in its side at
a distance y1 from the tank’s bottom. The hole is open to the atmosphere,
and its diameter is much smaller than the diameter of the tank. The air
above the liquid is maintained at a pressure P.
• Determine the speed of the liquid as it leaves the hole when the liquid’s
level is a distance h above the hole.
If P>>Po If tank is open P=Po

If P>>Po If tank is open P=Po
Example 9 ……Example 9 ……

Torricelli Law – cont.Torricelli Law – cont.
WHAT IF? What if the position of the hole in Figure 14.20 could
be adjusted vertically? If the tank is open to the atmosphere and
sitting on a table, what position of the hole would cause the
water to land on the table at the farthest distancefarthest distance from the
tank?
Answer: Model a parcel of water exiting the hole as a projectile.

Torricelli Law – cont.Torricelli Law – cont.
Maximum horizontal position if hole halfwayMaximum horizontal position if hole halfway
between botton and top tank.between botton and top tank.

Other Applications of
Fluid Dynamics
Streamline flow around a moving airplane wing.Streamline flow around a moving airplane wing.
• Let’s assume the airstream approaches the wing horizontally
from the right with a velocity v1.
• The tilt of the wing causes the airstream to be deflected
downward with a velocity v2.
• Because the airstream is deflected by the wing, the wing must
exert a force on the airstream.
• According to Newton’s third law, the airstream exerts a force F
on the wing that is equal in magnitude and opposite in
direction.
• This force has a vertical component called lift (or aerodynamic
lift) and a horizontal component called drag.

Spinning golf ballSpinning golf ball
• A golf ball struck with a club is
given a rapid backspin due to the
slant of the club. The dimples on
the ball increase the friction force
between the ball and the air so
that air adheres to the ball’s
surface.
• Because the ball pushes the air
down, the air must push up on the
ball.

Bend it like Beckham
Dynamic lift
http://www.tudou.com/programs/view/qLaZ-A0Pk_g/

Beckham, Applied Physicist
~ 5m
Distance 25 m
Initial v = 25 m/s
Flight time 1s
Spin at 10 rev/s
Lift force ~ 4 N
Ball mass ~ 400 g
a = 10 m/s2
A swing of 5 m.
Goal!!Goal!!

Atomizer
• A number of devices operate by means of
the pressure differentialspressure differentials that result from
differences in a fluid’s speeddifferences in a fluid’s speed.
• For example, a stream of air passing over one
end of an open tube, the other end of which
is immersed in a liquid, reduces the pressure
above the tube.
• This reduction in pressure causes the liquidcauses the liquid
to rise into the airstreamto rise into the airstream. The liquid is then
dispersed into a fine spray of droplets.
• It is atomizeratomizer that used in perfume bottlesthat used in perfume bottles
and paint sprayersand paint sprayers.

Fluid Mechanics - 203PHYS

Recommended

Recommended

More Related Content

What's hot

What's hot (20)

Similar to Fluid Mechanics - 203PHYS

Similar to Fluid Mechanics - 203PHYS (20)

Recently uploaded

Recently uploaded (20)

Fluid Mechanics - 203PHYS