# Chapter 2

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24 Οκτ 2013 (πριν από 4 χρόνια και 8 μήνες)

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Faculty of Engineering

Engineering Physics I (BE121) Course

Fall 2012

1

Chapter
2

Fluid mechanics

Reference: PHYSICS for Scientists and Engineers with Modern Physics
Eighth

Edition
Raymond A. Serway, John W. Jewett, Jr.

Matter is normally classified as
being in one of the three states: solid, liquid
or gases.

Fluid:

both liquid and gases are called fluids.

The study of fluids will be treated from two different approaches. First, we
will consider the mechanics of fluids at rest (
fluid statics
), then we
will treat the
fluids in motion (
fluid dynamics
).

Part one: Fluid statics

Density:

The density of a substance is defined as the amount of mass
of

a unit
volume of the substance.

(kg/m
3
)

The density of solid and most liquids are
almost constant but the density of
gases varies greatly with pressure and temperature.

Faculty of Engineering

Engineering Physics I (BE121) Course

Fall 2012

2

Pres
s
ure:

The
only stress that can be exerted on an object submerged in static fluid is
one that tends to compress the object from all sides, i.e., the force exerted

by static
fluid on an object is always perpendicular to the surface of the object.

The pressure is defined as the magnitude of the normal force acting per unit
surface area, i.e.,

(
N
/m
2

or Pa
)

The IS unit for pressure is
Newton/Meter
2

(N/m
2
) and also
called Pascal

(Pa)

1 Pa= 1 N/m
2

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Variation of p
res
s
ure

with depth
:

Consider a liquid of density (

) at rest as shown in figure. We assume that

is uniform throughout the liquid. Let us select a sample of liquid contained within
an imaginary cylinder of cross section area (A) extending from depth (d) to (d+h).
The liquid external to our sample exerted forces at all points on the surface of the
sam
ple perpendicular to the surface.

The pressure exerted on the bottom face is (P) and the pressure exerted on
the top face is (P
o
). The cylinder is at equilibrium then,

That is, the pressure P at a depth h below a point in the liqui
d at which the
pressure

is P
0

is greater by an amount

If the liquid is open to the atmosphere
and

P
0

is the pressure at the surface of the liquid, then P
0

is atmospheric pressure.
In

our calculations and working of end
-
of
-
chapter problems, we
usually take
atmospheric

pressure to be

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Fall 2012

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Pascal’s law
:

If the pressure at any point in an enclosed fluid at rest is changed by

P,
thepressrechangesbyaneqalamont

Patallpointsintheflids.

An
important application of Pascal’s law is the hydraulic lift shown in figure.

The pressure is the same in both sides, then:

By proper design (A
2

A
1
), a large o/p force can be obtained by a small i/p
force. The liquid volume pushe
d down on the left= A
1

x
1
, where

x
1

is the
displacement of the left piston. The volume
pushed up on the right= A
2

x
2
, where

x
2

is the displacement of the right piston.

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Example:

In a car lift used in a service station, compressed air exerts a force
on small piston that has a circular cross section and a radius of 5cm.
This pressure is transmitted by a liquid to a piston that has

15 cm.

i.

What force must the compressed

air exert to lift a car
weighting 13300N?

ii.

What is the air pressure produces this force?

Solution

i.

ii
-
The air pressure required to produce this force is:

E
xample

A U
-
tube contains some mercury. Water is
poured into one arm of the U
-
tube and
oil is

poured into the other arm, as shown in Fig.

The column of water, density

1.0 × 10
3

kgm

3
, is 53 cm high. The column of oil is 71 cm high.

Calculate the
density of the oil.

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Solution

Example:

A U
-

tube of a uniform cross section area
opened to atmosphere is partially filled with
mercury. Water is then poured into both arms. If
the equilibrium configuration of the tube is as
shown with h
2
=1 cm, determine the value of h
1
.

Given that the density of water and mercury are
1000, and 13600
Kg/m
3
, respectively.

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Solution

Buoyant forces and Archimedes’s principle
:

Buoyant force is the upward force exerted by a fluid on any immersed
object. Archimedes’s principle states that
the magnitude of the buoyant force
always equals to the weight of the fluid displaced by the object
.

Proof:

Consider a cube immersed in a liq
uid as shown in figure. The pressure at
the top and the bottom of the cube is P
t

and P
b
, respectively.
W
here
.
The pressure on the top face produces a force
-
P
t

A, and the
pressure on the bottom face produces a force P
b

A, the resultant of these two
forces is the buoyant force (B).

h

A

B

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(

)

(

)

But

But

The

buoyant force equals the weight of the fluid displaced by the
object
.

Example:

A piece of aluminum with mass 1kg and density 2700 kg/m
3

is suspended
from a string and then completely immersed in a container of water. Calculate the
tension in the spring

a)

B
efore the metal is immersed.

b)

After the metal is immersed.

Faculty of Engineering

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Solution

a)

Before the metal is immersed:

b)

After the metal is immersed:

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Floating Object

Now consider an object of volume
V
obj

and density

in static
equilibrium floating on the surface of a fluid, that is, an object that is

only
partially
submerged. In this case, the upward buoyant force

is balanced by the
downward gravitational force acting on the object. If
V
disp

is the

volume of the
fluid displaced by the object (this volume is the same as the volume

of that part of
the object beneath th
e surface of the fluid), the buoyant force has a

magnitude
,

Because the weight of the object is

and because

, we see that

,

or

This equation shows that
t
he 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.

Example:

A plastic sphere floats in water with 50%

of its volume submerged. The
same sphere floats in glycerin with 40% of its volume submerged. Determine the
densities of the sphere and glycerin

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Solution

When
the sphere floats
in water:

When the sphere floats in glycerin:

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Part
two: Fluid dynamics

When fluid is in motion, its flow can be characterized as:

i.

Steady flow or non
-
:

, velocity of fluid particles passing by any point
remains

constant with time. But
in non
-
velocity of fluid
particles passing any point changes with time.

ii.

Laminar flow or Turbulent flow:

In laminar flow, particles of the fluid follow a smooth path such that
the paths of different particles never cross each other while turb
ulent
flow is irregular flow.

Laminar flow

Because the motion of the real fluid is very complex, we will consider the ideal
flow with the following assumptions:

i.

The fluid is non
-
viscous.

ii.

The flow is steady and laminar.

iii.

The fluid is incompressible,
i.e., the density of the fluid is constant.

Stream lines and tube of flow:

The path taken by a fluid particle is called a streamline. The velocity of a
particle is always tangent to the streamline as shown in the above figure. A set of
streamlines form
a tube of flow. Fluid particles can not flow into or out of the
sides of the tube of
flow;

otherwise the streamlines would cross each other.

Faculty of Engineering

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Fall 2012

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Continuity equation
:

Consider an ideal fluid flowing through a pipe of non
-
uniform size as
shown in figure.

In a time interval

t, the fluid at the bottom end of the pipe moves a
distance

x
1
=v
1

t
, if A
1

is the cross section area in this region, then the mass of
the shaded region at (1) is:

where

is the density of the fluid. The fluid that moves through the upper end of
the pipe in a time interval

t has a mass:

The mass that crosses A
1

in a time interval

t must equal to the mass of the
fluid that crosses A
2

in the same tim
e interval, i.e.,

This equation is called the continuity equation

for fluids
“The

product of
the area and fluid speed at all points along a pipe is constant for
incompressible fluids”.

The product (Q=A
v
) is called the volume flow
rate (m
3
/s). So, the
continuity equation means that “
T
he volume of fluid entering one end of the
tube in a give
n time interval

equals
the volume of fluid leaving the other end
of the tube in the same time interval

.

Faculty of Engineering

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Example:

A pipe with a diameter of
2.5 cm is connected to another pipe with a
diameter of 0.9cm. If the velocity of the fluid

in the 2.5 cm pipe is 1.5 m/s:

i.

What is the velocity in the 0.9cm pipe?

ii.

How much water flows per seconds from the 0.9 cm pipe?

Solution

i.

Q=A
1
v
1
=A
2
v
2

ii.

Example:

A water hose 2.5cm in diameter is used by a gardener to fill 30L bucket.
The gardener notes that it takes 1min to fill the bucket. A nozzle with an opening
of cross section area of 0.5cm
2

is then attached to the hose. The nozzle is held so
that water is projected horizontally from a point 1m above the ground.

i.

What is the speed of water in the hose?

ii.

What is the speed of water at the exit of the nozzle?

Solution

We first find the speed of
the
water in the hose from the bucket
-
filling information.

Find the cross
-
sectional area of the
hose:

Evaluate the volume flow rate:

Solve for the speed of the water in
the hose:

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Solve the continuity equation for
fluids for
v
2
:

Substitute numerical

values:

Bonus Problem

Over what horizontal distance can the water be projected
?

4.52 m

Bernoulli’s equation
:

Bernoulli’s equation gives the relation between fluid speed, elevation and
pressure. Consider the flow of a segment of an ideal fluid through a non
-
uniform
pipe in a time interval

t as shown in figure.

In a time interval

t, the fluid at point (1) m
oves a distance

x
1

while the
fluid at point (2) moves a distance

x
2
. The force exerted by fluid at (1) is
where A
1

is the cross section area of the pipe at (1). The work done by
this force in time interval

t is
, where V is the volume of fluid at
portion (1). Similarly, the work done by the fluid at
portion (2) in the same time
interval is
. So, the net work done by the fluid is:

W = W
1

-
W
2

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(1)

Part of this work

goes into changing the kinetic energy of the fluid and the
other part goes into changing the potential energy of the fluid. The change in
kinetic energy is:

(2)

where m is the fluid mass in portion (1) and (2),
v
1

is the fluid vel
ocity at point (1)
and
v
2

is the fluid velocity at point (2). The change in potential energy is:

(3)

But

From equations (1),
(2), and (3), we got:

Dividing both sides by
V
, then

This

could
be
written as:

Example:

A horizontal pipe 10 cm in diameter has a smooth reduction to a pipe 5cm
in diameter. If the pressure of the water in the larger pipe is 8

10
4

Pa

and the
pressure in the smaller pipe is 6

10
4

Pa, at what rate does the water flow through
the pipes?

Faculty of Engineering

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Fall 2012

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Solution

D
1
= 10 cm

P
1
=8

10
4

Pa

D
2
= 5 cm

P
2
=6

10
4

Pa

(1)

But
from the continuity equation:

(2)

(3
)

From equation (1) and equation (
3
):

But
from equation 2
, the flow rate is:

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Fall 2012

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Applications of Bernoulli’s
equation

The Venturi Tube

The horizontal constricted pipe illustrated in

the
Figure,

known as a
Venturi tube,
can be used to measure the flow speed

of an incompressible fluid. Determine the
flow speed at point 2

of Figure if the pressure difference
P
1

-

P
2

is known.

Apply
Bernoulli’s equation

to points 1 and 2,
noting

that
y
1

=

y
2

because the pipe is
horizontal
:

Solve the equation of continuity for
v
1:

Substitute this expression into Equation (1):

Solve for
v
2
:

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2
-

The flow of a
fluid through an orifice

Consider the large tank of water shown in figure. Applying Bernoulli’s
equation at point (1) and (2), then:

But from the continuity equation:

The velocity of the fluid at point 2
is almost zero, then:

So, the fluid velocity at the orifice is:

Other Applications on Fluid Mechanics

The curvature of the wing surfaces causes

the pressure above the wing to be lower
than that below the wing due to the

Bernoulli
Effect
. This pressure difference
assists with the lift on the wing.

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Fig.1

Another
example, 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. Figure
2

shows air
adhering to the ball and being deflected downward as a result. Because

the ball
pushes the air down, the air must push up on the ball. Wi
thout the dimples,

the
friction force is lower and the golf ball does not travel as far.

Fig.
2

Fig.
3

A number of devices operate by means of the pressure differentials that result

from differences 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 as illustrated in Figure
3
. This reduction in pressure

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causes the liquid to rise into the air stream. The liquid is then dispersed int
o
a fine

spray of droplets. You might recognize that this so
-
called atomizer is used
in perfume

bottles and paint sprayers.

Example

An aerofoil has an effective area of 25 m
2
. Air of density 1.2 Kgm
-
3

flows over
the aerofoil at a speed of 85 ms
-
1

and under the aerofoil at 75 ms
-
1
.

Calculate the lift force on the aerofoil.

Solution

2

1

1
2

(

1
2

2
2
)

1
2

1

2

(
85
2

75
2
)

960

25

960

24000
𝑁

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