Faculty of Engineering
Engineering Physics I (BE121) Course
Fall 2012
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.

m
V
(kg/m3)
The density of solid and most liquids are almost constant but the density of
gases varies greatly with pressure and temperature.
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Fall 2012
Pressure:
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.,
P
F
A
(N/m2 or Pa)
The IS unit for pressure is Newton/Meter2 (N/m2) and also called Pascal (Pa)
1 Pa= 1 N/m2
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Variation of pressure 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
sample perpendicular to the surface.
The pressure exerted on the bottom face is (P) and the pressure exerted on
the top face is (Po). The cylinder is at equilibrium then,
F  0
P A  mg  PA
o
P A  Vg  PA
o
P A  Ahg  PA
o
P  P  gh
o
That is, the pressure P at a depth h below a point in the liquid at which the
pressure is P0 is greater by an amount 𝜌𝑔ℎ. If the liquid is open to the atmosphere
and P0 is the pressure at the surface of the liquid, then P0 is atmospheric pressure.
In our calculations and working of end-of-chapter problems, we usually take
atmospheric pressure to be
𝑷𝒐 = 𝟏. 𝟎𝟎 𝒂𝒕𝒎 = 𝟏. 𝟎𝟏𝟑 × 𝟏𝟎𝟓 𝑷𝒂
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Fall 2012
Pascal’s law:
If the pressure at any point in an enclosed fluid at rest is changed by P,
the pressure changes by an equal amount P at all points in the fluids.
An important application of Pascal’s law is the hydraulic lift shown in figure.
The pressure is the same in both sides, then:
F1 F2

A1 A2
A 
F2  F1  2 
 A1 
By proper design (A2 A1), a large o/p force can be obtained by a small i/p
force. The liquid volume pushed down on the left= A1x1, where x1 is the
displacement of the left piston. The volume pushed up on the right= A2x2, where
x2 is the displacement of the right piston.
 A1x1  A2 x2
x2 A1 F1


x1 A2 F2
F x  F2 x2
11

Work done by F1
Work done by F2
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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 a radius of
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.
A
F1  F2  1
 A2







  5  10  2 2
 13300
  15  10  2 2


  1.48  10 3 N


ii-The air pressure required to produce this force is:
P1 
F1
1.48  103

 1.88  105
2

2
A1
 5  10


Pa
Example
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 × 103 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
Your turn
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 h2=1 cm, determine the value of h1.
Given that the density of water and mercury are
1000, and 13600 Kg/m3, respectively.
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Solution
h
A
B
PA  PB
 w g h1  h  h2    m gh2   w gh
 w g h1  h2    m gh2
   w 
13600  1000 
2
h2  
h1   m
  1  10  0.126 m.
1000


 w 
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 liquid as shown in figure. The pressure at
the top and the bottom of the cube is Pt and Pb, respectively. Where
Pb  Pt   fluid gh . The pressure on the top face produces a force -Pt A, and the
pressure on the bottom face produces a force Pb A, the resultant of these two
forces is the buoyant force (B).
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𝐹𝑏 − 𝐹𝑡 = 𝐵
(𝑃𝑏 − 𝑃𝑡 )𝐴 = 𝐵
(𝜌𝑔ℎ𝑏 − 𝜌𝑔ℎ𝑡 )𝐴 = 𝐵
But
ℎ𝑏 − ℎ𝑡 = ℎ
𝜌𝑔ℎ𝐴 = 𝐵
But ℎ𝐴 = 𝑉𝑜𝑏𝑗𝑒𝑐𝑡 = 𝑉𝑑𝑖𝑠𝑝𝑙𝑎𝑐𝑒𝑑 𝑓𝑙𝑢𝑖𝑑
𝐵 = 𝜌𝑔𝑉𝑓𝑙𝑢𝑖𝑑
𝐵 = 𝑀𝑓𝑙𝑢𝑖𝑑 𝑔
𝐵 = 𝑊𝑓𝑙𝑢𝑖𝑑 𝑑𝑖𝑠𝑝𝑙𝑎𝑐𝑒𝑑
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/m3 is suspended
from a string and then completely immersed in a container of water. Calculate the
tension in the spring
a)
Before the metal is immersed.
b)
After the metal is immersed.
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Solution
a)
Before the metal is immersed:
T1  Mg  1 9.8  9.8N
b)
After the metal is immersed:
T2  Mg  B
 Mg   fluid gVobject
 Mg   fluid g
M
 object

 fluid 

 Mg 1 
 

object


 1000 
 1  9.8  1 
  6.17 N
 2700 
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Floating Object
Now consider an object of volume Vobj and density ⍴𝑜𝑏𝑗 < ⍴𝑓𝑙𝑢𝑖𝑑 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 Vdisp 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 the surface of the fluid), the buoyant force has a magnitude,
𝐵 = ⍴𝑓𝑙𝑢𝑖𝑑 𝑔 𝑉𝑑𝑖𝑠𝑝
Because the weight of the object is 𝐹𝑔 = 𝑀𝑔 = ⍴𝑜𝑏𝑗 𝑔 𝑉𝑜𝑏𝑗
and because 𝐹𝑔 = 𝐵, we see that
⍴𝑓𝑙𝑢𝑖𝑑 𝑔 𝑉𝑑𝑖𝑠𝑝 = ⍴𝑜𝑏𝑗 𝑔 𝑉𝑜𝑏𝑗 ,
or
𝑉𝑑𝑖𝑠𝑝
⍴𝑜𝑏𝑗
=
𝑉𝑜𝑏𝑗
⍴𝑓𝑙𝑢𝑖𝑑
This equation shows that 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.
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:
Fg  B
mg   water gVsubmerged
 sphereV   water
 sphere 
 water
2
V
2
 500kg / m 3
When the sphere floats in glycerin:
Fg  B
mg   glycerin gVsubmerged
 sphereV   glycerin 0.4V 
 sphere  0.4  glycerin
 glycerin 
500
 1250kg / m 3
0.4
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Faculty of Engineering
Engineering Physics I (BE121) Course
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Part two: Fluid dynamics
When fluid is in motion, its flow can be characterized as:
i.
Steady flow or non-steady flow:
In steady flow, velocity of fluid particles passing by any point
remains constant with time. But in non-steady flow, 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 turbulent
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.
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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 x1=v1t, if A1 is the cross section area in this region, then the mass of
the shaded region at (1) is:
m1  A1 x1
 A1v1 t
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:
m2  A2 x 2
 A2 v 2 t
The mass that crosses A1 in a time interval t must equal to the mass of the
fluid that crosses A2 in the same time interval, i.e.,
A1v1 t  A2 v2 t
A1v1  A2 v 2
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=Av) is called the volume flow rate (m3/s). So, the
continuity equation means that “The volume of fluid entering one end of the
tube in a given time interval equals the volume of fluid leaving the other end
of the tube in the same time interval”.
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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=A1v1=A2v2
A
v 2  v1  1
 A2
ii.
Q  A1v1 

2
 2.5  10

  1.5   4
  0.9  10 2


4





2.5  10 
4
2 2
2
2


  11.57 m / s.



 1.5  0.736  10 3
m3 / s
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 0.5cm2 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 bucketfilling 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 v2:
Substitute numerical values:
Bonus Problem
Over what horizontal distance can the water be projected?
Answer: 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) moves a distance x1 while the
fluid at point (2) moves a distance x2. The force exerted by fluid at (1) is
F1  P1 A1 where A1 is the cross section area of the pipe at (1). The work done by
this force in time interval t is W1  P1 A1x1  P1V , 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 W2  P2 A2 x2  P2V . So, the net work done by the fluid is: W = W1 W2
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W  P1  P2 V
(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:
K .E 
1
1
mv22  mv12
2
2
(2)
where m is the fluid mass in portion (1) and (2), v1 is the fluid velocity at point (1)
and v2 is the fluid velocity at point (2). The change in potential energy is:
(3)
P.E  mgy2  mgy1
But W  K.E  P.E
From equations (1), (2), and (3), we got:
P1  P2 V

1
1
mv22  mv12  mgy2  mgy1
2
2
Dividing both sides by V, then
1m 2 1m 2 m
m
v2 
v1  gy 2  gy1
2V
2V
V
V
1 2 1 2
P1  P2  v 2  v1  gy 2  gy1
2
2
1 2
1
P1  v1  gy1  P2  v22  gy2
2
2
P1  P2 
This could be written as:
1
P  v 2  gy  cona tan t
2
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104 Pa and the
pressure in the smaller pipe is 6104 Pa, at what rate does the water flow through
the pipes?
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Solution
D1= 10 cm P1=8104 Pa
D2= 5 cm P2=6104 Pa
1 2
1
v1  gy1  P2  v 22  gy 2
2
2
1
1
8 10 4  1000  v12  0  6 10 4  1000  v22  0
2
2
P1 
But from the continuity equation:
radius = D/2
Q  A1v1  A2 v2

4
(2)
D12 v1 

D22 v 2
4
D
v 2  v1  1
 D2



 10 
v 2  v1  
5
v2  4v1
2
2
(3)
From equation (1) and equation (3):
1
1
2
8  10 4   1000  v12  6  10 4   1000  4v1 
2
2
v1  1.63m / s
But from equation 2, the flow rate is:
Q  A1v1 

4
D12 v1 

4
(1)
0.12 1.63  0.0128m 3 / s
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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 P1 - P2 is known.
Apply Bernoulli’s equation to points 1 and 2,
noting that y1 = y2 because the pipe is
horizontal:
Solve the equation of continuity for v1:
Substitute this expression into Equation (1):
Solve for v2:
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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:
P1 
1 2
1
v1  gy1  P2  v 22  gy 2
2
2
But from the continuity equation:
Q  A1v1  A2 v2
A 
 v2  v1  1   zero
 A2 
because A1  A2
The velocity of the fluid at point 2 is almost zero, then:
 Po 
1 2
v1  gy1  P0  0  gy 2
2
1
 v12  g  y 2  y1 
2
1
 v12  gh
2
So, the fluid velocity at the orifice is:
v1  2 gh
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. Without 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 into
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 m2. 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
P1 
1 2
1
v1  P2  v22
2
2
1
1
∆𝑃 = 𝑃2 − 𝑃1 = ⍴(𝑉12 − 𝑉22 ) = × 1.2 × (852 − 752 ) = 960
2
2
𝐹 = 𝐴 × ∆𝑃 = 25 × 960 = 24000𝑁
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