Fluid Mechanics

DJJ20073 Fluid Mechanics Chapter 3: Fluid Statics

Example 3.11
A rectangular pontoon has a width of 6 m, a length of 12 m, and a draught of 1.5 m in
fresh water (density 1000 kg/m3). Calculate:
a) the weight of the pontoon
b) its draught in sea water (density 1025 kg/m3)

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DJJ20073 Fluid Mechanics Chapter 3: Fluid Statics

Example 3.12
A barge has a base area of 240 m2. It is sinks to a depth of 0.45 m in sea water when it
is unloaded. When loaded with cargo, it sinks to a depth of 0.60 m in sea water.
Calculate:
a) The volume of sea water displaced by the barge when unloaded
b) The mass of the barge
c) The weight of the cargo carried by the barge

(density of sea water = 1025 kg/m3)

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DJJ20073 Fluid Mechanics Chapter 3: Fluid Statics

SELF ASSESSMENT

1. Determine the increase of pressure intensity per metre of depth in fresh water. The
answer should be in
a) Newtons per square metre
b) Milibars

2. At the foot of a mountain a mercury barometer reads 740 mm and a similar
barometer at the top of the mountain reads 590 mm. What is approximate height
of the mountain, assuming that the density of air is constant and equal to 1.225
kg/m3.

3. In Figure 1, fluid A is oil (s= 0.8), fluid B is brine (s=1.25). If a = 2.5 m and h = 0.3m
what is the pressure in N/m2 at X ?

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DJJ20073 Fluid Mechanics Chapter 3: Fluid Statics

4. What is the specific gravity of liquid B in a stable U tube manometer shown in
Figure 2.

P2 = 50N/m2

P1 = 70N/m2

s fluid B = ?

h2 = 10 cm

h1 = 10 cm

h3 = 5 cm
ρ fluid A = 700kg/m3

5. Consider a barge filled with rock. The barge is 7 m wide, 17 m long and 2.5 m
deep. If the barge and rock weight is 2.0 MN, determine the depth of
submergence, d of the barge in water as shown in the figure below.

rock

d

17 m

6. A rectangular scow 4.5m by 9.6m having vertical sides weight 36000 kg. What is
its draught in fresh water?

ans: Page 52
1. a) 9810 N/m2 b)98.1 mbars
2. 1665 m
3. -15941.25 N/m2
4. 0.37
5. 1.71 m
6. 0.83 m

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CHAPTER 4

Fluid Dynamics

Fluid dynamics is a sub-discipline of fluid mechanics that deals with fluid flow – the
natural science of fluids (liquids and gases) in motion. It has several sub-disciplines itself,
including aerodynamics (the study of air and other gases in motion) and hydrodynamics
(the study of liquids in motion). Fluid dynamics has a wide range of applications, including
calculating forces and moments on aircraft, determining the mass flow rate of petroleum
through pipelines, predicting weather patterns, understanding nebulae in interstellar
space and reportedly modelling fission weapon detonation. Some of its principles are
even used in traffic engineering, where traffic is treated as a continuous fluid.

After completing this chapter, you should be able to:
1. describe the different types of flow
2. describe flow rate
3. apply continuity equation law
4. apply Bernoulli Theorem

DJJ20073 Fluid Mechanics Chapter 4: Fluid Dynamics

4.1 Flow

Since fluid dynamics involves the study of the motion of fluid, one of the first concepts
that must be understood is how physicists quantify that movement. The term that
physicists use to describe the physical properties of the movement of liquid is flow. Flow
describes a wide range of fluid movement, such blowing through the air, flowing through
a pipe, or running along a surface. The flow of a fluid is classified in a variety of different
ways, based upon the various properties of the flow.

4.1.1 Types of Flow

a) Uniform Flow and Steady Flow
When a fluid is flowing in a pipe, the innumerable small particles get together and
form a flowing stream. These particles, while moving, group themselves in a variety
of ways, such as they may move in regular formation, just as disciplined soldiers do;
or they may swirl or jostle, like the individuals, in a disorderly mob. The type of flow
of a liquid depends upon the manner in which the particles unite and move. Though
there are many types of flows, yet the following are important from the subject point
of view:

i) Steady Flow:
A flow is said to be steady flow if different fluid properties and fluid velocity does not
change with respect to time.

If R is the set of fluid properties and V is the set of fluid particles, then for a steady
flow.

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DJJ20073 Fluid Mechanics Chapter 4: Fluid Dynamics

ii) Unsteady Flow:
A flow is said to be unsteady flow if different fluid properties and fluid velocity
changes with respect to time.

For unsteady flow,

Figure 4.1: Steady flow vs Unsteady flow

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DJJ20073 Fluid Mechanics Chapter 4: Fluid Dynamics

iii) Uniform flow:
In this type of flow, different fluid properties and fluid velocity does not varies with
respect to space (or directions).
Eg:- Constant discharge through a constant diameter pipe.

iv) Non-uniform flow:
In this type of flow, different fluid properties and fluid velocity varies with respect to
space (or directions).
Eg:- Constant discharge through a variable diameter pipe.

Figure 4.2: Uniform flow vs Non-uniform flow

b) Laminar Flow and Turbulent Flow
i) Laminar flow:
In a laminar flow, the flow of fluid is smooth and highly ordered. In this flow, the fluid
layers move parallel to each other and do not cross each other.

ii) Turbulent flow:
In a turbulent flow, the flow of fluid is chaotic and not in any order. In this flow, the
fluid layers cross each other and do not move parallel to each other.

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DJJ20073 Fluid Mechanics Chapter 4: Fluid Dynamics

In addition to these terms, there is another term called Transition Flow. It exists
between laminar and turbulent flow. In this region, the flow is very unpredictable and
often changes back and forth between laminar and turbulent states. Modern
experimentation has demonstrated that this type of flow may comprise short ‘burst’ of
turbulence embedded in a laminar flow.

Figure 4.3: Laminar vs Transition vs Turbulent Flow

c) Reynolds Number

Osborne Reynolds found that the types of flow is determined by the velocity,
density and viscosity of the fluid and the size of the conduit and depends on the
value of Reynolds number, Re

Re = vd


In which v = velocity, ρ = mass density and η = viscosity of the fluid, while d is

typical dimension which for a pipe is the diameter.

Re < 2100 : laminar flow

2100 < Re < 4000 : transition

Re > 4000 : turbulent flow

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DJJ20073 Fluid Mechanics Chapter 4: Fluid Dynamics

4.2 Flow Rate

4.2.1 Volume Flow Rate

The volume flow rate of a liquid (or commonly called flow rate or discharge) is a
measure of the volume of liquid that moves in a certain amount of time. The flow rate
depends on the area of the pipe or channel that the liquid is moving through, and the
velocity of the liquid.

Q = Av

Q = liquid flow rate (m3/s or L/s)
A = area of the pipe or channel (m2)
v = velocity of the liquid (m/s)

4.2.2 Mass Flow Rate

Mass flow rate is the rate of movement of a massive fluid through a unit area. Mass flow
depends on the density, velocity of the fluid and the area of the cross section. Meaning,
it is the movement of mass per unit time. The formula for mass flow rate is given:

• mass of fluid, m

Mass flow rate, m = time taken to collect fluid, t

•

m = Q

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DJJ20073 Fluid Mechanics Chapter 4: Fluid Dynamics

Example 4.1
An empty bucket weighs 2.0 kg. After 7 seconds of collecting water the bucket weighs 8
kg. What is the mass flow rate?

Example 4.2
If we know the mass flow is 1.7 kg/s, how long will it take to fill a container with 8 kg of
fluid?

Example 4.3
If the density of the fluid in the Example 4.2 is 850 kg/m3, what is the volume per unit time
(the discharge)?

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DJJ20073 Fluid Mechanics Chapter 4: Fluid Dynamics

4.3 Continuity Equation Law

The law of conservation of matter stipulates that matter can be neither created nor
destroyed, though it may be transformed (e.g. by a chemical process). Since this study
of the mechanics of fluids excludes chemical activity from consideration, the law reduces
to the principle of conservation of mass.

The principle is applied to fixed volumes, known as control volumes (or surfaces) as
shown in Figure 4.4.

Figure 4.4: A control volume

For any control volume, the principle of conservation of mass says

Mass entering = Mass leaving + Increase of mass in the control
per unit time per unit time volume per unit time

For steady flow, there is no increase in the mass within the control volume. Thus,
Mass entering per unit time = Mass leaving per unit time

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This can be applied to a stream tube such as shown below. No fluid flow across the
boundary made by the streamlines; so, mass only enters and leaves through the two ends
of this stream tube section.

Figure 4.5: A Stream tube section

Then, we can write:

Mass entering per unit time at end 1 = mass leaving per unit time at end 2
ρQ (entering) = ρQ (leaving)

But since flow is incompressible, the density is constants. Thus,

Q (entering) = Q (leaving)

This is the continuity equation for steady incompressible flow.

If velocity of flow across the entry to the control volume is measured, and that the velocity
is constant at v1 m/s; then, if cross sectional area of the stream tube at entry is A1, we get

Q (entering) = v1A1

Thus, if the velocity of flow leaving the volume is v2, and the area of the stream tube at
exit is A2, then

Q (leaving) = v2A2

Therefore, the continuity equation may also be written as v1A1 = v2A2 Page 61

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DJJ20073 Fluid Mechanics Chapter 4: Fluid Dynamics

4.3.1 Application of Continuity Equation
We can apply the principle of continuity to pipes with cross sections which change along
their length. Refer to the following samples.

Sample 1
Consider the diagram of a pipe with contraction below.

Figure 4.6: Pipe with a contraction

A liquid is flowing from left to right and the pipe is narrowing in the same direction. By
continuity principle, the mass flow rate must be the same at each section. We can write
as:

ρ1V1A1 = ρ2V2A2

As we considering a liquid, usually water, which is not very compressible, the density
changes very little so we can say ρ1 = ρ2 = ρ. This also says that the volume flow rate is
constant or that Q1 = Q2

V1A1 = V2A2

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Sample 2

This is diffuser, a pipe which expands or diverges as on in Figure 4.7. Derivations of
equation are the same as Sample 1 above.

Figure 4.7: A pipe with diverges
Sample 3

The continuity principle can also used to determine the velocities in pipes coming into a
junction.

Figure 4.8: A pipe with a junction
Total mass flow into the junction = total mass flow out of the junction

ρ1Q1 = ρ2Q2 + ρ3Q3
When the flow is incompressible (e.g. if it is water) ρ1 = ρ2 = ρ3 = ρ

Q1 = Q2 + Q3
A1V1 = A2V2 + A3V3

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DJJ20073 Fluid Mechanics Chapter 4: Fluid Dynamics

Example 4.4

If the areas in Figure 4.6; A1 = 10 x 10-3 m2 and A2 = 3 x 10-3 m2 and the upstream mean
velocity, v1 = 2.1 m/s, what is the downstream mean velocity?

Example 4.5

If the diameter of a diffuser (Figure 4.7) at section 1 is d1 = 30 mm and at section 2 d2 =
40 mm and the mean velocity at section 2 is v2 = 3.0 m/s. Calculate the velocity entering
the diffuser.

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DJJ20073 Fluid Mechanics Chapter 4: Fluid Dynamics

Example 4.6

Oil flows through a pipeline which contracts from 450 mm diameter at A to 300 mm
diameter at B and then forks, one branch being 150 mm diameter discharging at C and
the other branch 225 mm diameter discharging at D. If the velocity at A is 1.8 m/s and the
velocity at D is 3.6 m/s, what will be the discharges at C and D and the velocities at B and
C?

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DJJ20073 Fluid Mechanics Chapter 4: Fluid Dynamics

4.4 Bernoulli’s Equation

A general approach to obtaining the parameters of motion (for both solids and fluids) is
to apply the principle of conservation of energy. When friction is negligible, the sum of
kinetic energy and gravitational potential energy is constant. Thus,

Kinetics energy = 1mv2
2

Gravitational potential energy = mgh

(Note: m is the mass, v is the velocity and h is the height above the datum).

In order to apply this to a falling body where we have an initial velocity of zero, and it falls
through a height of h, the followings applied:

Initial kinetic energy = 0

Initial potential energy = mgh
Final kinetics energy = 1mv2

2

Final potential energy = 0

From here, we know that

Kinetic energy + potential energy = constant

So,

Initial kinetic Initial potential Final kinetic Final kinetic

energy + energy = energy + energy

Thus, mgh = 1mv2
2

Or v = 2gh Page 66

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DJJ20073 Fluid Mechanics Chapter 4: Fluid Dynamics

4.4.1 Energy of a flowing fluid
A liquid may possess three forms of energy:

• Potential energy
If a liquid of weight W is at a height of Z above datum line
Potential energy = WZ
Potential energy per unit weight = Z

The potential energy per unit weight has dimensions of Nm/N and is measured as a
length or head Z and can be called the potential head.

• Pressure energy
When a fluid flows in a continuous stream under pressure it can do work. If the area
of cross-section of the stream of fluid is A, then force due to pressure P on cross-
section is PA. If a weight W of liquid passes the cross-section

Volume passing cross-section = W


Distance moved by liquid = W
A

Work done = force × distance = PA  W = PW
ωA 

Pressure energy per unit weight = P = P
 g

Similarly the pressure energy per unit weight P/W is equivalent to a head and is
referred to as the pressure head.

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DJJ20073 Fluid Mechanics Chapter 4: Fluid Dynamics

• Kinetic energy
If a weight W of liquid has a velocity v,

Kinetic energy = 1 W v2
2g
v2

Kinetic energy per unit weight =
2g

The kinetic energy per unit weight is also measured as a length and referred to as
the velocity head.

The total energy of the fluid is the sum of these three forms of energy
Total head = pressure head + velocity + head potential head

Total energy per unit weight = P + v2 +Z
g 2g

4.4.2 Limitations of Bernoulli’s Theorem

Bernoulli’s Theorem states that the total energy of each particle of body of fluid is the
same provided that no energy enters or leaves the system at any point. The division of
the energy between pressure, kinetic and potential energy, may vary, but the total
remains constant. In symbols:

H = P + v2 + Z = constant
g 2g

The Bernoulli’s equation has been derived on certain assumptions, which are rarely
possible. Thus, the Bernoulli’s theorem has following limitations:

• The Bernoulli’s equation has been derived under the assumption that the velocity
of every liquid particle, across any cross-section of a pipe, is uniform. But, in actual
practice, it is not so. The velocity of liquid particle in the centre of a pipe is
maximum and gradually decreases towards the walls of the pipe due to the pipe

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DJJ20073 Fluid Mechanics Chapter 4: Fluid Dynamics

friction. Thus, while using the Bernoulli’s equation, only the mean velocity of the
liquid should be considered.

• The Bernoulli’s equation has been derived under the assumption that no external
force, except the gravity force, is acting on the liquid. But, in actual practice, it is
not so. There are always some external forces (such as pipe friction etc.) acting
on the liquid, which effect the flow of the liquid. Thus, while using the Bernoulli’s
equation, all such external forces should be neglected. But, if some energy is
supplied to, or, extracted from the flow, the same should be taken into account.

• The Bernoulli’s equation has been derived under the assumption that there is no
loss of energy of the liquid particle while flowing. But, in actual practice, it is rarely
so. In a turbulent flow, some kinetic energy is converted into heat energy. And in
a viscous flow, some energy is lost due to shear forces. Thus, while using the
Bernoulli’s equation, all such losses should be neglected.

• If the liquid is flowing in a curved path, the energy due to centrifugal force should
also be taken into account.

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DJJ20073 Fluid Mechanics Chapter 4: Fluid Dynamics

4.4.3 Application for Bernoulli’s Theorem
Bernoulli’s equation can be applied to many situations including the great ones. In the
following sections, we will see more examples of its application. First of all, we need to
establish an understanding of the following fundamentals and concepts.

a) Horizontal Pipe

Example 4.7

Figure 4.9 below shows a venturi tube whose radii at sections X and Y are 2.0 cm and
1.0 cm respectively. If the velocity of water flow at X and Y are 5.0 m/s and 20.0 m/s
respectively, calculate the

(a) pressure difference between X and Y
(b) rate of water flow

Figure 4.9

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DJJ20073 Fluid Mechanics Chapter 4: Fluid Dynamics

b) Incline Pipe

Example 4.8
Water flows through a pipe system as shown in Figure 3.8. In section 1, the velocity of
the water is 1.42 m/s while the pressure is 68.7 kN/m2. Calculate the flow rate if the
pressure and diameter section 2 is 34.3 kN/m2 and 50 cm. Ignore any head loss.

Figure 4.10

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DJJ20073 Fluid Mechanics Chapter 4: Fluid Dynamics

c) Venturi meter

The Venturi effect is the reduction in fluid pressure that results when a fluid flows through
a constricted section of pipe. The venturi effect is named after Giovanni Battista Venturi
(1746–1822), an Italian physicist.

A venturi meter is used to measure the flow through a pipeline. It can be found in
carburettors, fluid flow meters and aircraft airspeed indicators. It consists of a rapidly
converging section which increases the velocity of flow and hence reduces the pressure.
It then returns to the original dimensions of the pipe by a gently diverging 'diffuser' section.
By measuring the pressure differences the discharge can be calculated. This is a
particularly accurate method of flow measurement as energy loss are very small.

Figure 4.11: Venturi meter

In order to understand the principles of a venturi meter, we must first review two basic
concepts. Based on the principle of Bernoulli’s equation, the total amount of head at point
1 must equal the total amount of head at point 2.

P1 + v12 + Z1 = P2 + v22 + Z2
g 2g g 2g

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DJJ20073 Fluid Mechanics Chapter 4: Fluid Dynamics

Since the venturi meter is its horizontal position, then Z1 = Z2

We obtained v22 − v12 = P1 − P2 …. (1)
2g  … (2)

For uniform flow, Q1 = Q2

A1v1 = A2v2

v2 = A1v1
A2

Substitute equation (2) into equation (1)

v12  A 2 − 1 = 2g P1 − P2 
1  ω 

A22

 2g P1 - P2  
 
   
v1 =  
  A12 − 1 
 A22 

From Q = A1v1
So
v1 = Q
A1

Assume that H = P1 − P2
ω

So equation for v1 becomes Q= 2gH
A1
 A12 − 1
A22

By taking m as the ratio of the cross sectional area, namely:

m = A1 or m =  d1 2
A2 d2

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Therefore, the formula for the theoretical flow rate can be reduced to :

Qth = A1 2gH
m2 −1

The above formula can be corrected to obtain the actual flow rate by multiplying it with
the coefficient called the discharge coefficient, Cd.

So, the actual flow rate is:

Qactual = Cd x Qtheory

Horizontal venturi meter: H = P1 − P2
ρg

Inclined or Vertical venturi meter: H = P1 − P2 + Z1 − Z2
ρg

Horizontal, Inclined or Vertical venturi meter: H = x ωmanometer − 1
ωpipe

Inclined venturi meter Vertical venturi meter

Figure 4.12: Inclined and vertical venturi meter

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Example 4.9

A venturi meter with a 150 mm diameter at inlet and 100 mm at throat is laid with its axis
horizontal and is used for measuring the flow of oil specific gravity 0.9. The oil-mercury
differential manometer shows a gauge difference of 200 mm. Assume coefficient of the
metre as 0.98. Calculate the discharge in litres per minute.

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Example 4.10

Sea water has a density of 1025 kg/m3 was channeled through a pipe diameter of 1200
mm at a rate of 8.0 m3/s. Venturi meter with a diameter of 825 mm in the neck, mounted
on the pipe. Calculate the coefficient of discharge of the venturi meter if the head
difference in pressure between the sections in the mouth and throat venturi meter is equal
to 10 m sea water column. (Z1 = 300 mm, Z2 = 600 mm)

Figure 4.13: Inclined venturi meter

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Example 4.11

Oil flowed vertically in a pipeline and through the venturi meter. The diameter at the inlet
and throat of the meters are 40 cm and 20 cm, respectively. Inlet section located 30 cm
below the throat section of the venturi meter. Manometer U-tube which is connected to
the venturi meter shows the difference readings as high as 15 cm. Determine the flow
rate of oil if relative density of oil and mercury are 0.9 and 13.6, respectively. (Given Cd =
0.98)

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d) Orifice meter
An orifice meter is used to measure the discharge in a pipe. An orifice meter, in its
simplest form, consists of a plate having a sharp edged circular hole known as an orifice.
This plate is fixed inside a pipe as shown in the figure below.

A mercury manometer is inserted to know the Figure 4.14: Orifice meter
difference of pressures between the pipe and
the throat. Let

x = reading of mercury manometer
P1 = pressure at inlet

v1 = velocity of liquid at inlet,
A1 = area of pipe at inlet, and

P2, v2, A2 = corresponding values at the

orifice plat

Qth = A1 2gH
m2 −1

Qactual = Cd x Qtheory

H = P1 − P2
ρg

Or

H = x ωmanometer − 1
ωpipe

m = A1
A2

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Example 4.12

An orifice meter consisting of 100 mm diameter orifice in a 250 mm diameter pipe has
coefficient equal to 0.65. The pipe delivers oil (SGoil = 0.8). The pressure difference on
the two sides of the orifice plate is measured by a mercury oil differential manometer. If
the differential gauge reads 80 mm of mercury, calculate the rate of flow in litre/s.

Example 4.13

Water flows at the rate of 0.147 m3/s through a 150 mm diameter orifice inserted in a 300
mm diameter pipe. If the pressure gauge fitted upstream and downstream of the orifice
plate have shown readings of 176.58 kN/m2 and 88.29 kN/m2 respectively, find the
coefficient of discharge Cd of the orifice meter.

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e) Pitot tube

A Pitot tube is an instrument to determine the velocity of flow at the required point in a
pipe or a stream. The pitot tube was invented by the French engineer Henri Pitot (1695 –
1771) in the early 18th century and was modified to its modern form in the mid-19th
century by French scientist Henry Darcy.

In its simplest form, a pitot tube consists of a glass tube bent a through 90° as shown in
Figure 4.15 (a).

The lower end of the tube faces the direction of the flow. The liquid rises up in the tube
due to the pressure exerted by the flowing liquid. By measuring the rise of liquid in the
tube, we can find out the velocity of the liquid flow.

Let h = height of the liquid in the
d pitot tube above the
surface,

= depth of tube in the liquid

v = velocity of the liquid
Applying Bernoulli’s equation for A and B, we found that :

v act = C 2gh
which C is coefficient of pitot tube

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Figure 4.15: (a) Simplest form of pitot tube (b) Pitot tube used in a pipe

Figure 4.15 (b) shows a pitot tube used in a pipe. Po - P is the difference between the
static pressure and the pressure at the impact hole and is measured by a differential
gauge.

Example 4.14
A pitot-static tube used to measure air velocity along a wind tunnel is coupled to a
manometer which shows a difference of head of 4 mm water. The density of the air is 1.2
kg/m3. Obtain the air velocity, assuming that the coefficient for the pitot tube is unity.

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Example 4.15

A submarine moves horizontally in a sea and has its axis below water surface. A Prandtl
pitot tube placed in front of the submarine and along its axis is connected to the two limbs
of a U-tube containing mercury. The difference in mercury level is found to be 170 mm.
Find the speed of submarine knowing that specific gravity of sea water is 1.025

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SELF ASSESSMENT

1. Water is flowing through a tapered pipe having end diameters of 150 mm and 50
mm respectively. Find the discharged at the larger end and velocity head at the
smaller end, if the velocity of water at the larger end is 2.5 m/s.

2. Water is flowing along a pipe with a velocity of 7.2 m/s. Express this as a velocity
head in metres of water. What is corresponding pressure in N/m2.

3. A pipe 300 m long has a slope of 1 in 100 and tapers from 1 m diameter at the
higher end to 0.5 m at the lower end. The quantity of water flowing is 900
liters/second. If the pressure at the higher end is 70 kPa, find the pressure at the
lower end.

4. A venturi meter has an area ratio of 9 to 1, the larger diameter being 300 mm.
During the flow, the recorded pressure head in the large section is 6.5 metres
and that at the throat 4.25 metres. If the meter coefficient is 0.99, compute the
discharge through the meter.

5. A horizontal venturi meter 160 mm x 80 mm is used to measure the flow of an oil
of specific gravity 0.8. Determine the deflection of the oil-mercury gauge, x if the
discharge of the oil is 50 litres/s. Take coefficient of venturi meter as 1.

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6. A 300 mm x 150 mm venturi meter is provided in a vertical pipeline carrying oil of
specific gravity 0.9, the flow being upwards. The difference in elevations of the
throat section and entrance section of the venturi meter is 300 mm. The
differential U-tube mercury manometer shows a gauge deflection of 250 mm.
Calculate
i) discharge of the oil, and

ii) Pressure difference between the entrance and throat section

Take the coefficient of meter as 0.98 and the specific gravity of the mercury as
13.6.

ans

1. 1.0.044 m3/s, 22.41m/s
2. 2.64, 25.9kN/m2
3. 89.23kPa
4. 0.052m3/s
5. 296mm
6. 0.149m3/s, 33.8kN/m2

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DJJ20073 Fluid Mechanics Chapter 5: Energy Loss in Pipelines

CHAPTER 5
Energy Loss in Pipelines

Losses of energy in a pipeline are due to shock from the disturbance of the normal flow
due to bends or sudden changes of section and frictional resistance to flow. These losses
are conveniently expressed as energy lost in Nm/N, as the head lost in terms of fluid in
the pipe and related to velocity head. If v = velocity in the pipe, velocity head = v2/2g and
head lost = k(v2/2g) where k is a constant.

Losses of energy in a pipeline cannot be ignored. When the shock losses and friction loss
have been determined, they are inserted in Bernoulli’s equation in the usual way.

P + v2 + Z
ρg 2g

After completing this chapter, you should be able to:

a) Explain the round pipe system.
b) apply the types of head loss
c) apply head loss in pipeline systems

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DJJ20073 Fluid Mechanics Chapter 5: Energy Loss in Pipelines

5.1 Velocity Profile in Circular Pipe

Not all fluid particles travel at the same velocity within a pipe. The shape of the velocity
curve (the velocity profile across any given section of the pipe) depends upon whether
the flow is laminar or turbulent. If the flow in the pipe is laminar, the velocity distribution
at a cross section will be parabolic in shape with the maximum velocity at the center being
about twice the average velocity in the pipe. In turbulent flow, a fairly flat velocity
distribution exists across section of pipe, will the result that entire fluid flows at a given
single value. Figure 4.1 helps illustrate the above ideas. The velocity of the fluid in contact
with the pipe wall is essentially zero and increases the further away from the wall.

Note from Figure 4.1 that the velocity profile depends upon the surface condition of the
pipe wall. A smoother wall results in a more uniform velocity profile than a rough pipe
wall.

Figure 5.1: Velocity profile in a circular pipe.

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DJJ20073 Fluid Mechanics Chapter 5: Energy Loss in Pipelines

5.2 Loss of Head of A Liquid Flowing in A Pipe

A pipe is defined as a closed conduit of circular section through which the fluid flows filling
the complete cross-section. The fluid in the pipe has no free surface. It will be at a
pressure above or below atmospheric and this pressure may vary along the pipe.

It has been experimentally found that when a liquid is flowing in a pipe, it loses its energy
(or sometimes termed as head) due to friction of the wall, change of cross-section or
obstruction in the flow. Head loss can be divided into two categories: major loss and
minor loss. All problems in the pipeline can be solved by entering the head loss into the
Bernoulli’s Equation:

P1 + v 2 + Z1 = P1 + v22 + Z2 +h
ρg 1 ρg 2g

2g

where h is the head loss due to the major loss and minor losses. All such losses are
expressed in terms of velocity head.
Though there are many types of losses of heads, yet the following losses, which occur in
a flowing liquid, are important from the subject point of view :
- Loss of head due to friction in a pipe.
- Loss of head due to sudden enlargement,
- Loss of head due to sudden contraction,
- Loss of head at the entrance in a pipe,
- Loss of head at the exit in a pipe,
All these losses of heads will be discussed in the following pages one-by-one.

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DJJ20073 Fluid Mechanics Chapter 5: Energy Loss in Pipelines

5.2.1 Loss of head due to friction in a pipe

When there is a loss of head due to friction in a pipeline in terms of the velocity head, we
assume that the frictional resistance per unit area of the pipe wall is proportional to the
square of the mean velocity of flow. Consider a cylinder of fluid of length L completely
filling the pipe of cross sectional area A and moving with a mean velocity v (Figure 4.2).
The force acting on the cylinder is the force due to pressure difference and the force due
to frictional resistance. Since the velocity is constant and there is no acceleration, the
resultant of these two forces in the direction of motion must be zero.

Figure 5.2: Loss due friction

Darcy Wesbach’s formula allows loss due to friction of the fluid through the pipe calculated

:

hf = 4fLv 2
2gd.

where hf is the head loss due to friction (m)

L = the length of the pipe (m)

d = the internal diameter of the pipe (m)

v = the flow velocity within the pipe (m/s)

g = the acceleration due to gravity (9.81 m/s2)

f = the coefficient of friction.

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DJJ20073 Fluid Mechanics Chapter 5: Energy Loss in Pipelines

Example 5.1

Find the loss of head due to friction in a pipe 300 m long and 150 mm diameter when the
discharge is 2.73 m3/min and the resistance coefficient f = 0.01.

Example 5.2

Determine the diameter of pipe if the major loss due to friction in the pipe is 13820 mm.
Pipe length is 360 m. If the flow velocity is 2.38 m/s, determine the fluid flow rate. Take f
as 0.005.

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DJJ20073 Fluid Mechanics Chapter 5: Energy Loss in Pipelines

5.2.2 Loss of head due to sudden enlargement

Figure 5.3: Sudden enlargement causes the vortex occurs

As fluid flows from a smaller pipe into larger pipe through sudden enlargement, its velocity
abruptly decreases; causing turbulence that generates an energy loss. The amount of
turbulence and the amount of energy is dependent on the ratio of the sizes of the two
pipes. The loss is calculated from

he = (v1 - v 2 )2
2g

which v1 = flow velocity in small pipe
v2 = flow velocity in large pipe

Example 5.3

Calculate a loss due to enlargement when water flow through pipes with a diameter of
153 mm and suddenly enlarged to a diameter of 305 mm. Water flows at a rate of 0.3
m3/s.

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DJJ20073 Fluid Mechanics Chapter 5: Energy Loss in Pipelines

5.2.3 Loss of head due to sudden contraction

When fluid flows through the pipes is reduced dramatically, vena contracta will occur in
early attenuation of the second section.

Figure 5.4: Sudden contraction causes vena contracta

Head loss due to sudden contraction, hc = (v c - v2 )2
2g

From equation of continuity, Acvc = A2v2

vc = A2v2
Ac

If the coefficient of contraction, Cc = Ac
A2

So vc = 1 v2
Cc

Then by knowing the CC, the head loss due to sudden contraction can be calculated by

the formula :

hc =  1 - 1 2 v22
Cc 2g

where v2 = flow velocity in small pipe
Cc = coefficient of contraction

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DJJ20073 Fluid Mechanics Chapter 5: Energy Loss in Pipelines

Example 5.4

A horizontal pipeline carrying water at a rate of 0.236 m3/s is reduced suddenly from 450
mm diameter to 300 mm diameter. Calculate :

(i) The head loss due to sudden contraction
(ii) Pressure difference between the two pipes.
(Take Cc = 0.67)

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DJJ20073 Fluid Mechanics Chapter 5: Energy Loss in Pipelines

5.2.4 Loss of head at the entrance in a pipe

When a fluid enters a pipe from a large vessel (or tank or reservoir) some loss of energy
occurs at the entrance to the pipe which is sometimes known as inlet loss of energy. The
loss of energy actually depends on the form of the entrance. Thus for a rounded or bell-
mouthed entrance the loss energy is relatively much less as compared with a sharp-
cornered entrance. For a sharp-cornered entrance :

 v1  2
2g
hin = 0.5

where v1 = flow velocity in the pipe

Figure 5.5: Fluid from the large tank flow into the pipe

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DJJ20073 Fluid Mechanics Chapter 5: Energy Loss in Pipelines

5.2.5 Loss of head at the exit in a pipe

The outlet end of a pipe carrying a liquid may be either left free so that the liquid is

discharged freely in the atmosphere or it may be connected to large reservoir so that the

pipe outlet becomes submerged and the liquid is discharged into a large body of static

liquid. The liquid leaving the pipe at its outlet end still-possesses a kinetic energy

corresponding to the velocity of flow of liquid in pipe, which is ultimately dissipated either

in the form of a free jet or it is lost in turbulence in the reservoir, depending on the condition

of the outlet. Therefore :

hex = v12
2g

where v1 = flow velocity in the pipe

Figure 5.6: Fluid from a pipe into a large tank

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DJJ20073 Fluid Mechanics Chapter 5: Energy Loss in Pipelines

5.3 Pipeline Problems

All pipeline problems should be solved by applying Bernoulli’s Theorem between points
for which the total energy is known and including expressions for any loss of energy due
to shock or to friction, thus

P1 + v 12 + Z1 = P1 + v22 + Z2 +h
ρg 2g ρg 2g

When two or more pipes of different diameters or roughness are connected in such a way
that the fluid follows a single flow path throughout the system, the system represents a
‘series pipeline’. In a series pipeline, the total energy loss is the sum of the individual
minor losses and all pipe friction losses.

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DJJ20073 Fluid Mechanics Chapter 5: Energy Loss in Pipelines

Example 5.5

Water from a large reservoir is discharged to atmosphere through a 100 mm diameter
pipe 550 m long. The entry from the reservoir to the pipe is sharp and the outlet is 8 m
below the surface level in the reservoir. Taking f = 0.01, calculate the discharge.

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DJJ20073 Fluid Mechanics Chapter 5: Energy Loss in Pipelines

Example 5.6

Water from a large reservoir is discharged to atmosphere through a pipe 1200 m long.
The diameter for the first 600 m length of the pipe is 400 mm and 250 mm for the
remaining 600 m long. The entry from the reservoir to the pipe is sharp and the outlet is
30 m below the surface level in the reservoir. Taking f = 0.004 for 400 mm pipe and f =
0.006 for 250 mm pipe while Cc = 0.66, calculate the discharge.

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DJJ20073 Fluid Mechanics Chapter 5: Energy Loss in Pipelines

Example 5.7

Water is discharged from a reservoir into the atmosphere through a pipe 39 m long. There
is a sharp entrance to the pipe and the diameter is 50 mm for 15 m from the entrance.
The pipe then enlarges suddenly to 75 mm in diameter for the remaining of its length.
Taking into account the loss of head at entry and at the enlargement, calculate the
difference of level between the surface of the reservoir and the pipe exit which will
maintain a flow of 2.8 dm3/s. Take f as 0.0048 for the 50 mm pipe and 0.0058 for the 75
mm pipe.

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