Fluid Dynamics

Ms. Arra C. Quitaneg
6/30/2013
Prepared by: Ms. Paz Morales and Ms. Arra
Quitaneg

Pressure
 Which of the two exerts greater pressure?
 Why?
A. m= 100 g B. m= 200 g
6/30/2013
Quitaneg

Pressure
Which of the two exerts greater pressure?
Why?
A. m= 100 g
B. m= 100 g
6/30/2013
Quitaneg

PRESSURE
 Force per unit area
 When a fluid is at rest, it exerts a force perpendicular
to any surface in contact with it.
 The force exerted by the fluid is due to the molecules
colliding with its surroundings.
6/30/2013
Quitaneg

Factors affecting Pressure
 Force or Weight of the object
 Area of support
Pressure = Force/ Area
Pressure is the ratio of the force to the
area over which it is applied.
Unit: 1 Pascal (Pa) = 1 N/m2
6/30/2013
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Do you
want to
try this?
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6/30/2013
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LIQUID PRESSURE
 F = Ahg
 P = F/ A
 Liquid pressure = gh
 =density of the liquid
 g= acceleration due to
gravity
 h= depth
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Fluid Pressure Pressure:
 Units of measure: pounds per square inch
(psi), atmospheres (atm), or torr (which is
a millimeter of mercury).
 The S.I. unit for pressure is the pascal,
which is a Newton per square meter: 1 Pa
= 1 N / m 2. Atmospheric pressure is at
sea level is normally:
 1 atm = 1.01 ·10 5 Pa
= 760 torr = 14.7 psi.
 At the deepest ocean trench the pressure
is about 110 million Pascals.
Prepared by: Ms. Paz Morales and Ms.
Arra Quitaneg
6/30/2013

Fluid pressure
Liquid pressure Atmospheric pressure
 Fluid pressure in liquid varies
with depth.
 Due to the weight of the
water column above the
object submerged.
 Due to the weight of the
column of air above the
surface.
 Higher altitude – less
atmospheric pressure
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Which has greater liquid pressure?
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Liquid Pressure
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Pressure of the Earth’s
atmosphere.
1 atm = 1.013 x 105 Pa
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Body moves from higher pressure to
a lower pressure area.
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Water distribution
 If a water tower is only a storage device, why is it so
tall?
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WATER DISTRIBUTION
 Water accelerates only if pressure is out of balance.
 The deeper the water, the more weight there is
overhead and the greater the pressure.
 Water will not flow if there is no difference in
pressure.
6/30/2013
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Next time you take a shower…
Oh! Thanks to the
difference in
pressure that I am
able to take a
shower! Physics
works !!!! 
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Quitaneg

Think about these…
 Consider an air –filled balloon weighted so that it is
on the limit of sinking. That its over all density just
equals that of water. Now, if you push it beneath the
surface, it will
a) Sink
b) Return to the surface
c) Stay at the dept to which it is pushed
Explain your answer
6/30/2013
Quitaneg

 Farm silos have bands around
them to provide sturdiness to
the walls. Why are the bands
closer together near the bottom
of the silo than on top?
6/30/2013
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 Explain how level hose works.
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 Diver Jeff designs a new scuba
setup that is so profoundly
simple he is surprised that no
one has thought of this before.
He has attached one end of a
long garden hose to a large
Styrofoam to keep the hose
above the water level. He will
breathe through the other end
of the hose as he explores the
depths. What is wrong with his
simple design?
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Sample Problem:
Natsci 24: Topic 1 (Hydrostatics)
DLS-CSB-SMS-NATSCI AREA
Solution:
Thus, the equivalent force is 240 N.
 Problem 1:
 A child wants to pump up a
bicycle tire so that its pressure
is 2.5 x 105 Pa above that of
atmospheric pressure. If the
child uses a pump with a
circular piston 0.035 m in
diameter, what force must the
child exert?
Quitaneg
A
F
P PAF
NF 240
25
)
2
035.0
)()(105.2( xF
6/30/2013

Sample Problem:
 Problem 2:
 In a classroom demonstration, a
73.5-kg physics professor lies on a
“bed of nails.” The bed consists of
a large number of evenly spaced,
relatively sharp nails mounted in a
board so that the points extend
vertically outward from the board.
While the professor is lying down,
approximately 1900 nails make
contact with his body. What is the
average force exerted by each nail
on the professor’s body?
Quitaneg6/30/2013

Sample Problem:
 Problem 2:
 In a classroom demonstration, a
73.5-kg physics professor lies on a
“bed of nails.” The bed consists of
a large number of evenly spaced,
relatively sharp nails mounted in a
board so that the points extend
vertically outward from the board.
While the professor is lying down,
approximately 1900 nails make
contact with his body. What is the
average force exerted by each
nail on the professor’s body?
 Solution:
 This gives about 0.379 N
Quitaneg
A
F
P
NF 379.0
nails
S
mkg
F
1900
)8.9)(5.73( 2
A
mg
A
W
P
6/30/2013

Sample Problem:
 Problem 3:
 A swimming pool has the
dimensions 15.0 m x 20.0 m. It
is filled with water to a
uniform depth of 8.00 m. The
density of water = 1.00 × 103
kg/m3.
Quitaneg6/30/2013

Sample Problem:
Solution:
Thus, the equivalent pressure is 1.8 x 105 Pa
 Problem 3:
 A swimming pool has the
dimensions 15.0 m x 20.0 m. It
is filled with water to a
uniform depth of 8.00 m. The
density of water = 1.00 × 103
kg/m3. What is the total
pressure exerted on the
bottom of the swimming pool?
Quitaneg
ghP
PaxP
5
108.1
)8.9)(8)(100.1( 33
3
s
mm
m
kg
xP
6/30/2013

Assignment
 Bring the following by group:
2 identical drinking glasses
2 straws
1 pin
½ crosswise short bond paper
2 pingpong balls
1 yard length of a string
Adhesive tape
2 empty aluminum cans (identical)
hair dryer
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Assignment
 Clay
 Aluminum foil
 Coins
 basin
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Problem-solving:
 The surface of the water
in a storage tank is 30 m
above a water faucet in
the kitchen of a house,
calculate the water
pressure at the faucet.
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Pressure Pressure
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 Estimate the pressure exerted on a floor by a) a 50 kg
model standing momentarily on a single spiked heel
(area = 0.05 cm2) and compare it b) to the pressure
exerted by a 1500 kg elephant standing on one foot
(area = 800 cm2).
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Pascal’s principle
 Blaise Pascal
 Pressure applied to a confined fluid increases
the pressure throughout by the same amount.
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 Any change in the pressure of an incompressible fluid
is transmitted uniformly in all directions throughout
the fluid.
 Pressure is transferred undiminished in a fluid.
P P
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Application: Hydraulic Jack
Changes in pressure are transmitted uniformly
through a fluid and the pressure pushes outward in all
directions.
F
A
F
A
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oil
A2
F1
A1
F2h1
h2
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 A piston of cross-sectional area "a" is used in a
hydraulic press to exert a small force of magnitude "f"
on the enclosed liquid. A connecting pipe leads to a
larger piston of cross-sectional area A. If the piston
diameters are 3.80 cm and 53.0 cm, what force
magnitude on the small piston will balance a 20.0 kN
force on the large piston?
6/30/2013
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Surface tension
 Tendency of the liquid
surface to contract.
 Surface behaves as elastic
film.
 Caused by molecular
attractions.
 Soapy water has less surface
tension than pure water.
Spherical water droplet
Water strider
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 Why don’t boats made of
steel SINK?
 Is it possible for a material
that is MORE DENSE than
water to float?
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 A body immersed in water seems to weigh
less than when it is in air.
 Buoyant force- force exerted by liquid to an
object immersed on it
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Buoyant force
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BUOYANT FORCE
FB = F2 – F1
FB = g A (h2 – h1 )
FB = gV
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Buoyant force
FB = gV
 =density of the liquid
g=acceleration due to gravity
V = volume of the object
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Buoyancy
 Accordingly, if the weight of the displaced water equals the
weight of the object then it floats.
 Buoyant force is now equal to the weight of the object.
 This is known as the principle of flotation:
 A floating object displaces a weight of fluid equal
to its own weight.
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Archimedes principle
The buoyant force
acting on an object
fully or partially
submerged in a
fluid is equal to the
weight of the fluid
displaced by the
object.
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Floating of an object
 Object floats if density of object is less than
the density of the fluid.
 An object is in equilibrium if the net force is
zero.
Buoyant force = Weight of the object
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Fraction of the object submerged:
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FLOATING RESTAURANTS
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 A 15.0 kg solid gold statue is being raised from a
sunken ship. What is the tension in the hoisting cable
when the statue is at rest. A) completely immersed b)
out of water
 Density of gold= 19.3 x 103 kg/m3
 Density of air = 1.2 kg/m3
 Density of seawater- 1.025 x 103 kg/m3
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 A wooden block of wood with a volume of 125 m3 floats
on seawater. What fraction of object is submerged on
seawater? Density of seawater- 1.025 x 103 kg/m3
Density of wood = 0.9 x 103 kg/m3
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Seatwork
 A block of steel measuring 2m on each side was
accidentally dropped in seawater reaching a depth
of 10 m. How much force is needed to lift the steel
block if its density is 7.8 x 103 kg/m3 ?
Density of seawater- 1.025 x 103 kg/m3
 A 0.5 kg block of wood is floating in water. What
is the magnitude of the buoyant force acting on
the block?
6/30/2013
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 A 70 kg ancient statue lies at the bottom of the sea. Its
volume is 3.0 x 104 cm3. How much force is needed to
lift it?
6/30/2013
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 A small piston with a diameter of 0.5 m is used in a
hydraulic lift. The larger piston has a diameter of 2m,
how much force is needed to lift a car weighing 10 000
N ?
 How much pressure is felt by woman standing on a
heel with an area of 0.025 cm 2 , if her mass is 55 kg?
6/30/2013
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The strongest aluminum boat!!!
 Prepare 144 in2 aluminum
foil.
 Make your own boat design
using the cut aluminum
foil.
 Prepare for the battle.
 Which boat design will
support more coins?
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POINTING SYSTEM
NO . OF COINS POINTS
10 5
15 10
20 15
25 20
30 25
35 30
40 35
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Activities…….
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When the speed
of the fluid
increases,
pressure in the
fluid decreases.
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Bernoulli's Principle
 For Bernoulli's Principle to
apply, the fluid is assumed
to have these qualities:
◦ fluid flows smoothly
◦ fluid flows without any swirls
(which are called "eddies")
◦ fluid flows everywhere through
the pipe (which means there is
no "flow separation")
◦ fluid has the same density
everywhere (it is
"incompressible" like water)
Quitaneg6/30/2013

Bernoulli's Principle
 Only true for a non-viscous fluid flowing
at a constant height. It follows directly
from the Bernoulli equation:
 P + ½ v2 + gy = constant.
 If y is a constant, then
P + ½ v2 = constant.
 This shows that if pressure increases,
then speed decreases, and versa visa.
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Airplane wing
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Bernoulli’s principle Applications
 Atmospheric pressure
decreases in a strong
wind, tornado or
hurricane.
Faster air
LOWER PRESSURE
HIGHER PRESSURE
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The curved
shape of an
umbrella can be
disadvantageous
on a windy day!
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Think…think…think…
 Roofs of houses are sometimes blown off
(or are they pushed off?) during a storm.
Explain using Bernoulli’s principle.
 An unvented building with airtight closed
windows is in more danger of losing its
roof than a well vented building.
6/30/2013
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 Children are told to avoid standing too
close to a rapidly moving train because
they might get sucked under it. Is it
possible? Why?
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House roofing
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Assignment
 Which is the
best roof design
to prevent roof
destruction due
to difference in
pressure above
and below the
roof.
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Reminders
 Quiz on Monday
 Part 1: Written exam
 Part 2: Practical exam(floating restaurant design)
 Bring all the materials you need for the design.
Research on the weights of the objects you want to
place in your floating restaurant. Practice
computation.
6/30/2013
Quitaneg

Fluid Dynamics

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