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BUF1113 (02P) Chapter5 Fluid Mechanics 1819I

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1




Mass, Density
Pressure
Buoyant Forces & Archimedes Principle
Pascal’s Principle
2
Introduction: Phases of matter
Solid
Liquid
Matter
Gas
3
Introduction: Fluids
Solid
• Fixed shape,
fixed size
Liquid
• does not
maintain fixed
shape (take the
shape of its
container)
• does not
• also not readily
readily change in compressible
shape or volume (volume can
when force is
change with
applied
large force)
Gas
• neither fixed
shape nor fixed
volume (it’s
expand to its
container)
• compressible
Fluids
4
Introduction: Fluids

Fluid’s definition:
“ a substance which has no definite shape and
has the ability to flow. Such as liquid and gas.”
5
Density


Every matter occupies space around it
Space is called “Volume (m3)”
6
Density
 Two objects made of the same material have the same density even
though they may have different masses and different volumes.
 That’s because the ratio of mass to volume is the same for both
objects
7
Density
8
Density



Closely pack  more mass
Iron has more mass per unit volume
The density ρ of a substance is its mass per unit volume:
m

v


The SI unit for density is kg/m3. Density is also
sometimes given in g/cm3.
Density is a characteristic property of any pure
substance.
9
Density
Object made of a particular pure
substance, such as gold can have
any size or mass, but the density
will the same for each.
 Why helium balloons float ?


Sometimes, we use the concept
of density to write mass:
m  v

..and hence the weight as:
mg   vg
10
Density
Example 1
What is the mass of a solid iron wrecking ball of
radius 18 cm? Given the density of iron is 7800
kg/m3.
Volume of ball
Answer:
4 3
m  v     r 
3

3
3 4
2
  7800 kg/m    18 10 m  
3

 190 kg
11
Density
Exercise 1
1. What volume does o.4 kg of alcohol occupy? What
is the weight of this volume?
2. What volume of water has the same mass of 100
cm3 of lead? What is the weight density of lead?
Ans : (1)5.06 104 m3 ,3.92 N (2)1.13 103 m3 ,1.1105 N/m3
12
Density
Exercise 2
1. Calculate the density of cubical solid if one of its
side is 0.02 m and its mass is 0.50 kg?
2. Air has density of 1.29 kg m/3 under the room
temperature. Calculate the mass of air in a cubical
room if one of the sides of the room in 8m?
Concept

Pressure is defined as the force per unit area.

Pressure is a scalar; the units of pressure in the SI
system are Pascal: 1 Pa = 1 N/m2.
14
15
Concept
Example 2
The two feet of a 60-kg person cover an area of 500
cm2.
(a) Determine the pressure exerted by the two feet
on the ground.
(b) If the person stands on one foot, what will the
pressure be under that foot?
16
Fluid pressure

Now, lets consider how the pressure in a liquid of
uniform density varies with depth;

The pressure at a depth h below the surface of the
liquid is due to the weight of the liquid above it. We
can quickly calculate:
20
Fluid pressure
21
Fluid pressure

However, for the everyday situation of a liquid in an
open container – such as water in the glass, a
swimming pool, an ocean and etc, there is a free
surface at the top exposed to the atmosphere.
Absolute
Pressure

Thus,
P = P₀ + ρgh
Atmospheric
Pressure
 The atmospheric pressure at sea level is about
1.013 x 105 N/m2 and decreases slowly with altitude.
25
Fluid pressure
Example 4
A swimming pool has a dimension of 28.0 m
by 8.5 m whose uniform depth is 1.8 m.
Calculate the absolute pressure and the total
force on the bottom of the pool.
27
Fluid pressure
Example 4: Answer
(a) Absolute pressure at the bottom of the pool:
P  P   gh
 1.103 105 Pa   1000 kg/m3  9.8 m/s 2  1.8 m 
 1.18 105 Pa
(b) Total force at the bottom of the pool:
F  PA  1.18 105 Pa   28.0 m  8.5 m   2.8 107 N
28
Exercise
Exercise 2
1. A Golf shoe has 10 cleats, each having an area of
6.5 ×10-6 m2 in contact with the floor. Assume that
in walking, there is one instant when all 10 cleats
support the entire weight of an 80 kg person, what
is the pressure exerted by the cleats on the floor?
2. A pipe contains water under a gauge pressure of
400 kPa. If you patch a 4mm diameter hole in the
pipe with a piece of tape, what force must the tape
be able to withstand?
Ans :(1)1.206 107 N/m2 (2)5.03 N
29
Exercise
Exercise 2
3. The water pressure in a certain house is 1.1×106
N/m2. How high must be the water level be above
the point to of release in the house?
4. What is the absolute pressure at the bottom of a
lake that is 30 m deep?
Ans : (3)112.24 m (4)3.95 105 Pa
30
Exercise
Exercise 2
5. A living room floor has floor dimensions of 4.50 m
and 3.20 m and a height of 2.40 m. The density of
air is 1.29 kg/m3. What does the air in the room
weigh? What force does the atmosphere exerts on
the floor of the room?
Ans : 437 N, 1.46 106 N
31
Buoyant force
32
Buoyant force

Object submerged in a fluid appear to weigh less
when its outside the fluid.
For e.g.: a large rock that you would have difficulty
lifting off the ground can often be easily lifted from
the bottom of a stream.
 When its break through the surface of the water,
suddenly its seems to be heavier.


This is the e.g. of the buoyant force.
33
Buoyant force

The buoyant force is an upward force exerted by the
fluid.
34
Buoyant force

The BF occurs because the
pressure in a fluid increases
with depth.

Thus, the upward pressure
on the bottom surface of a
submerged object is
greater than the downward
pressure on its top surface.
35
Buoyant force
Buoyant force  weight of the fluid displaced
FB  mF g
  FVF g
Where:
F = density of the fluid
VF = volume of displaced fluid / volume
of submerged object
36
Buoyant force
Consider:
At the top cylinder:
Pressure, P1 = ρFgh1
and Force,
F1 = P1A = ρFgh1A
(downward)
At the bottom cylinder:
Pressure, P2 = ρFgh2
and,
Force, F2 = P2A = ρFgh2A (upward)
38
Buoyant force

Thus, the net force on the cylinder exerted by the
fluid pressure is known as the buoyant force, FB and
act upward.

The magnitude of FB is;
Weight of the
Fluid displaced
Volume of
the
cylinder
39
Archimedes Principle

This result is valid no matter what the shape of
the object.

Archimedes’s Principle:
“ The buoyant force on an object immersed in a fluid
is equal to the weight of the fluid displaced
by that object.”
40
Archimedes Principle

“Fluid displaced” mean a volume of the fluid equal to
submerged volume of the object ( or that part of the
object that is submerged )

For e.g.: if the object is placed in a glass or tube
initially filled to the brim with water, the water that
flows over the top represents the water displaced by
the object.
41
Archimedes Principle
Example 5
Consider two identical pails of water filled to the
brim. One pail contains only water, the other has a
piece of wood floating in it. Which pail has the
greater weight?
42
Archimedes Principle
Example 5
Consider two identical pails of water filled to the
brim. One pail contains only water, the other has a
piece of wood floating in it. Which pail has the
greater weight?
Answer:
Both pail weigh the same.
According to Archimedes’ principle, the wood
displaced a volume of water with equal weight of
the wood.
43
Archimedes Principle
Example 6
A 70-kg ancient statue lies at the bottom
of the sea. Its volume is 3.0 x 104 cm3.
(a) What is the weight of the statue?
(b) What is the buoyant force exerted on
the statue?
(c) How much force is needed to lift it?
Given the density for seawater, ρ = 1.1025
x 103 Kg/m3 .
44
Archimedes Principle
Example 6: Answer
(a) Weight of the statue:
w  mg   70 kg   9.8 m/s 2   6.9 102 N
(b) The buoyant force (equal to weight of
displaced seawater):
FB  ms g
  s vs g
Volume of seawater =
volume of statue
 1.1025 103 kg  3.0 102 m3  9.8 m/s 2 
 3.0 102 N
45
Archimedes Principle
Example 6: Answer
(c) Force to lift the statue:
F  mg  FB
 6.9 102 N  3.0 102 N
 3.9 102 N
46
Archimedes Principle
Example 7
A typical hot air balloon has a volume of 2200 m3.
The density of air at temperature 20°C is 1.205
kg/m3. The density of the hot air inside the balloon
at a temperature 100°C is 0.946 kg/m3.
How much weight can the hot air balloon lift?
47
Archimedes Principle
Example 7: Answer
FB
The weight of the extra load can be
found by: (refer to the FBD)
FB  mB g  wL
 wL  FB  mB g
Thus, find the buoyant force exerted
on the balloon:
FB  mB g   AvB g
 1.205 kg   2200 m3  9.8 m/s 2 
 2.6 104 N
WL
m Bg
48
Archimedes Principle
Example 7: Answer
The mass of the balloon (equal to
mass of the hot air inside):
FB
mB   H v   0.946 kg/m3  2200 m3   2081 kg
and the weight:
mB g   2081 kg   9.8 m/s 2   2.0  104 N
WL
m Bg
Hence, the extra load:
wL  FB  mB g  2.6 104 N  2.0 104 N  6000 N
49
Archimedes Principle
Case 1: Totally submerged object
 An object is totally submerged in a fluid if its density
is greater than the density of the fluid (ρ₀ > ρF).
 When an object is totally submerged, the upward FB
is:
FB  Fluid vFluiddispl g  Fluid vobject g
50
Archimedes Principle

If the object has a mass, m and density,ρ₀ thus,
Fg  mobject g  object vobject g

Hence, the net force,
FB  Fg   Fluid vobject g  object vobject g    Fluid  object  vobject g

When ρ₀ > ρF , then FB < Fg and the supported
object sink!
51
Archimedes Principle
Case 2: Floating object
 In general, an object floats on a fluid if its density is
less than that of fluid (ρ₀ < ρF).
 In this case, the upward buoyant force is balanced by
the downward gravitational force,
FB  Fg
 Fluid vF g  object vobject g
displ
vFdispl
vobject
object

 Fluid
52
Archimedes Principle
Example 8
A 100 g block of wood has a volume of 120 cm3. Will
it float in water? Will it float in gasoline?
53
Archimedes Principle
Example 8: Answer
Lets compare the buoyant force with the weight:
Weight of the block:
mB g  100 103 kg  9.8 m/s 2   0.98 N
Buoyant force by the water:
FB   wvB g  1000 kg/m3 120 106 m3  9.8 m/s 2   1.176 N
Since FB > mg, it will float in water.
54
Archimedes Principle
Example 8: Answer
Buoyant force by the gasoline:
FB   g vB g   680 kg/m3 120 106 m3  9.8 m/s 2   0.78 N
Since FB < mg, it will sink in gasoline.
55
Archimedes Principle
Example 9
A scuba diver and her gear displace a volume of 65.0
L and have a total mass of 68.0 kg.
(a) What is the buoyant force on the diver in
seawater?
(b) Will the diver sink or float?
Given the density for seawater, ρ = 1.025 x 103 Kg/m3
.
56
Archimedes Principle
Example 9: Answer
(a) Buoyant force on the diver:
FB   s vd g
 1.025 103 kg/m3  0.065 m3  9.8 m/s 2 
 653 N
(b) Will the diver sink/float?
md g   68 kg   9.8 m/s 2   666 N
Thus, since the weight > buoyant force, the diver
will sink!
57
Archimedes Principle
Exercise 3
A crane lift the 16,000 kg steel hull of a sunken ship out
of the sea. Determine:
(a) the buoyant force exerted by seawater on the hull
(b) the tension in the crane’s cable when the hull is
fully submerged in the water
(c) the tension when the hull is completely out of the
water?
[Given the density for seawater, ρ = 1.1025 x 103Kg/m3
and density of steel, ρ = 7.8 x 103 Kg/m3]
Ans : (a) 2.2 104 N (b) 1.34 105 N (c) 1.56 105 N
58
Archimedes Principle
Exercise 4
A tin coffee can floating in water has an internal
volume of 180 cm3 and a mass of 112 g. How many
grams of metal can be added to the can without
causing it to sink in the water?
Ans : 68 g
59
Pascal’s Principle
60
Pascal’s Principle

In fact that the pressure in a fluid depends on depth
and the atmospheric pressure P₀, any increase in
pressure at the surface must be transmitted to every
other points in fluid.

Pascal’s principle:
“If an external pressure is applied to a confined fluid,
the pressure at every point within the fluid increases by
that amount”.
A change in pressure at any point in an enclosed fluid is
transmitted equally throughout the fluid.
61
Pascal’s Principle

Some application of Pascal’s Principle:
a) Hydraulic lift
which a small input force is used to exert large
output force
62
Pascal’s Principle
 The pressure must be same for
both piston.
Pin  Pout
 Since the pressure is define as force per unit area:
Fin Fout

Ain Aout
 The mechanical advantage: Fout  Aout
Fin
Ain
63
Pascal’s Principle
Example 7
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 5.0 cm. This
pressure is transmitted by a liquid to a piston
that has a radius of 15.0 cm.
What force must the compressed air exert to lift
a car weighing 13,300 N? What air pressure
produced this force?
65
Pascal’s Principle
Example 7: Answer
input force: Fin  Fout
Ain
Aout
   5 102 m 2 

 1.33 104 N  
2
  15 102 m  


 1.48 103 N
pressure due this force:
Fin
1.48 103 N
5
Pin 


1.88

10
Pa
2
Ain   5 102 m 
66
Pascal’s Principle
Exercise 5
A force of 400 N is applied to a small piston of
hydraulic press whose diameter is 4 cm. What must
be the diameter of the large piston if it is to lift a 200
kg load?
0.088 m
67
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