Fluid Mechanics

DJJ20073 Fluid Mechanics Chapter 0: An Introduction to Fluid Mechanics

Chapter 0
An Introduction to Fluid Mechanics

Synopsis :

FLUID MECHANICS provides students with a strong understanding of the
fundamentals of fluid mechanics principles related to the fluid properties and behaviour
in static and dynamic situations. This course also exposes the students to the
demonstration at the real equipment of fluid mechanics.

Course Learning Outcome:

Upon completion of this course, students should be able to:

1. Explain the fundamentals of fluid.
2. Solve problems related to fluid properties, fluid statics and fluid dynamics.
3. Organize appropriate experiments in groups according to the standard operating

procedures.

Prepared by : Siti Safuan_June2021 Page 1

DJJ20073 Fluid Mechanics Chapter 0: An Introduction to Fluid Mechanics

Fluid mechanics ???? Why are we studying
fluid mechanics in a
Mechanical Engineering

programme?

Prepared by : Siti Safuan_June2021 Page 2

DJJ20073 Fluid Mechanics Chapter 0: An Introduction to Fluid Mechanics

WHY STUDY FLUID MECHANICS?

What is the necessity of studying the fluids? Fluids have become integral parts of our day-
to-day life. The study of the fluids helps us explore the use of potential fluids that can be
used for several new applications. For instances only after carrying out lots of study and
research on oil, humans have been successful in utilizing then as the fuel for the vehicles.
The fluids have huge potential to perform various functions. Some of these include:

a) There are several fluids that when burnt, produce lots of heat, which can be used
for various applications. Example of these fluids includes petrol and diesel for the
vehicles.

b) There are some fluids like oil that tend to exert very high pressure or force. These
fluids can be used for lifting various heavy loads. The fluids used in hydraulic
machines and hydraulic lifters are an example.

c) Some of the fluids have excellent flow properties which can be used for the
lubrication of various machines.

d) Fluids like water possess kinetic and potential energy, which is used for generation
of electricity as in hydroelectric power plants.

There are innumerable applications of the fluids. The study of fluid mechanics
helps us making the best use of fluids for different applications. The study helps
users understand the behaviour of fluid under various forces and at different
atmospheric conditions. The study also allows us to select proper fluid for various
applications.

Prepared by : Siti Safuan_June2021 Page 3

DJJ20073 Fluid Mechanics Chapter 0: An Introduction to Fluid Mechanics

TAKE CARE WITH THE SYSTEM OF UNITS

As any quantity are often expressed in whatever way you wish it's sometimes easy to

become confused on what exactly or what proportion is being mentioned. This is

particularly true within the field of hydraulics. Over the years many

various ways are wont to express the varied quantities involved. Even today different

countries use different terminology also as different units for an equivalent thing -

they even use an equivalent name for various things e.g. an American pint is 4/5 of a

British pint!

To avoid any confusion on this course we'll always use the SI (metric) system -
which you'll already be conversant in . It is essential that all quantities are expressed
in the same system or the wrong solutions will result.

Despite this warning you'll still find that this is often the foremost common
mistake once you attempt example questions.

Convert the questions below:
a. 100 cm2 to m2
b. 7 bar to N/m2
c. 5 kg/cm2 to N/m2
d. 1500 liter/minute to m3/ s
e. 300 dm3 to m3
f. 3.9 kg/cm2 to kN/m2

Prepared by : Siti Safuan_June2021 Page 4

DJJ20073 Fluid Mechanics Chapter 1: Introduction of Fluid

Introduction of Fluid CHAPTER 1

Fluid Mechanics is the section of Applied Mechanics concerned with the behavior of
fluids (liquids and gases) when they are in state of motion or rest. Whether the fluid is at
rest or motion, it is subjected to different forces and different climatic conditions and it
behaves in these conditions as per its physical properties. Fluid mechanics deals with
three aspects of the fluid: static, kinematics, and dynamics aspects.

After completing this chapter, you should be able to:

1. Explain fluid characteristics
2. Explain types of pressure
3. Explain the pressure in fluid

1.1 Fluid Characteristics
A substance that deforms continuously when acted on a shearing stress of any magnitude
is called as fluid. In a simple word, fluid is a substance that flow. Fluid flows under their
own weight and take the shape of any solid body with which they are in contact. All liquids
and gases substances are fluid. Water, oil, and others are very important in our day-to-
day life as they are used for various applications. For instances water is used for
generation of electricity in hydroelectric power plants and thermal power plants, water is
also used as the coolant in nuclear power plants, oil is used for the lubrication of
automobiles etc.

DJJ20073 Fluid Mechanics Chapter 1: Introduction of Fluid

The following table lists the distinctions between fluid and solids.

Table 1.1: Properties of the three states of matter

Property Solid Liquid Gas
Particles are Particles are
Packing of particles Particles are packed less close packed very far
to each other. from each other.
packed very close Weaker than solids Negligible

to each another.

Forces of attraction Strongest

between the

particles

Shape and volume Fixed shape and No fixed shape but Neither fixed shape
fixed volume nor fixed volume
fixed volume Higher than solids Highest
Less than solids Least
Compressibility Negligible Higher than solids Highest

Density Highest

Kinetic Energy of Very small

particles

Example Ice Water Water vapour

1.2 Types of Pressure

Pressure could be phrased by referring to any different datum. Usually we referred in local
atmospheric pressure or absolute zero atmospheric pressure.

1.2.1 Atmospheric Pressure, Patm

The atmosphere is the layer of various mixture gases such as nitrogen, oxygen and
carbon dioxide that surrounds the Earth. The weight of this mass of the atmosphere exerts
pressure on the Earth surface and on everything on Earth. This pressure is called
atmospheric pressure. In other word, atmospheric pressure is the pressure due to
atmosphere at the surface of the earth depends upon the head of the air above the

Prepared by : Siti Safuan_June2021 Page 6

DJJ20073 Fluid Mechanics Chapter 1: Introduction of Fluid

surface. Atmospheric pressure at sea level is about 101.325 kN/m², equivalent to a head
of 10.35 m of water or 760 mm of mercury approximately and decreases with altitude.

1.2.2 Gauge Pressure, PG
Gauge pressure is the intensity of pressure measured above or below atmospheric
pressure.

1.2.3 Absolute Pressure, PA

Absolute pressure is the intensity of pressure measured above the absolute zero, which
is perfect vacuum.

PA = Patm + PG
- PV

1.2.4 Vacuum, Pv

Space in which there is no matter or in which the pressure is so low that any particles in
the space do not affect any processes being carried on there. It is a condition well below
normal atmospheric pressure and is measured in units of pressure (the pascal). A vacuum
can be created by removing air from a space using a vacuum pump or by reducing the
pressure using a fast flow of fluid.

Prepared by : Siti Safuan_June2021 Page 7

DJJ20073 Fluid Mechanics Chapter 1: Introduction of Fluid

Figure 1.1: Relationship between absolute, gauge and vacuum pressures

Example 1.1
What is the gauge pressure of air in a cylinder if the atmospheric pressure is 101.3 kN/m2
and the absolute pressure is 460 kN/m2.

Prepared by : Siti Safuan_June2021 Page 8

DJJ20073 Fluid Mechanics Chapter 1: Introduction of Fluid

Example 1.2

The pressure of a Bourdon Gauge is 7 bar. If the atmospheric pressure is 1.013 bar, find
the absolute pressure (in kN/m2).

Example 1.3

A pressure gauge connected to a pipe to drain the water. The reading on the pressure
gauge indicates 70.2 kN/m2. Calculate the absolute pressure inside the pipe.

Prepared by : Siti Safuan_June2021 Page 9

DJJ20073 Fluid Mechanics Chapter 1: Introduction of Fluid

1.3 Pressure in Fluid

Pressure is the force exerted by a fluid on the surfaces with which it is in contact or by
one part of a fluid on the adjoining part. The intensity of pressure at any point is the force
exerted on unit area at that point and is measured in newtons per square metre in SI
units. Also frequently used is bar, where 1 bar = 1 x 105 N/m2.

Pressure, P = Force,F
Area, A

P= F (SI unit : N/m2)
A

Consider a cylindrical container of height h and cross-sectional area A, which is filled with
a liquid of density, ρ.

 Volume of liquid in the container, V = Ah
 Mass of liquid in the container, m = Vρ

= Ah ρ
 Force on point X, F = weight of the liquid

= mg

= Ah ρg
 Pressure on point X,

P = F = Ahg
AA

Hence :

P = gh

Figure 1.2: Pressure and head

Prepared by : Siti Safuan_June2021 Page 10

DJJ20073 Fluid Mechanics Chapter 1: Introduction of Fluid

1.3.1 Pressure and Head

The pressure, P at a point in a fluid can be expressed in terms of the height, h of the
column of the fluid which causes the pressure, or which would cause an equal pressure
if the actual pressure is applied by other means. If  is the mass density of the fluid, P =
gh and the height, h is called the pressure head at the point. It is measured as a length
(e.g. in metres) of fluid. The name of the fluid must be given.

1.3.2 Understanding Pressure in Liquids

Liquids have pressure because of their weights. For example, if you try to put your finger
over the end of a tap when it is turned on, you can feel the pressure of the water in the
pipe. This is just caused by the weight of water in the pipes all the way back up to the
cold water in the loft. The force of gravity acts on liquids, pulling them downwards into the
container.

Figure 1.3: Liquids always find their own level

Prepared by : Siti Safuan_June2021 Page 11

DJJ20073 Fluid Mechanics Chapter 1: Introduction of Fluid

When a liquid such as water is poured into the Pascal’s vase shown in Figure 1.3.2, the
water in each tube rises to the same height, h. This shows that a liquid always finds its
own level.

At the same level or depth, the liquid pressure must be the same, otherwise liquid
flow to equalize any pressure differences. This confirms the fact that the pressure in a
liquid

i) depends on vertical depth
ii) is independent of the shape and cross-sectional area of the containing vessel

Figure 1.4: Liquid pressure at the same depth acts equally in
all directions

Similar holes at the same height are made on the side of a bottle. These holes are plugged
up and the bottle is filled with water. On removing the plugs, it is found that water spurts
out equally fast and reaches the same distance away from each hole. This shows that
the pressure at any point in liquid acts equally in all directions.

Prepared by : Siti Safuan_June2021 Page 12

DJJ20073 Fluid Mechanics Chapter 1: Introduction of Fluid

Water jet

Figure 1.5: Liquid pressure increases with depth

The deeper the liquid, the faster the liquid spurts out. In conclusion, the pressure in a
liquid increase with depth.

Figure 1.6: Effect of density on liquid pressure

It is found that the jet of oil falls nearer to the tin and flows slower than the jet of water
although both tins contain equal amounts of oil and water. This shows that oil pressure is
less than water pressure.

Prepared by : Siti Safuan_June2021 Page 13

DJJ20073 Fluid Mechanics Chapter 1: Introduction of Fluid

Example 1.4
Find the head h of water corresponding to an intensity of pressure P of 340 000 N/m2.

Example 1.5

Calculate the pressure exerted by a column of water of height 0.85 m and kerosene of
the same height.
(The density of kerosene = 800 kg/m3)

Prepared by : Siti Safuan_June2021 Page 14

DJJ20073 Fluid Mechanics Chapter 1: Introduction of Fluid

Example 1.6

Figure above shows the cross section of a sea near a seaside. Find the difference of the
pressure between point A and point B.

[Density of seawater = 1050kg/m³]

Prepared by : Siti Safuan_June2021 Page 15

CHAPTER 2

Physical Properties of Fluid

Properties of fluids determine how fluids can be used in engineering and technology.
They also determine the behaviour of fluids in fluid mechanics. It is obvious that different
fluids can have grossly different characteristics. For example, gases are light and
compressible, whereas liquids are heavy and relatively incompressible. To quantify the
fluid behaviour differences certain fluid properties are used. The fluid properties are mass
density, specific weight, specific gravity, specific volume, viscosity and compressibility.

After completing this chapter, you should able to:

1. Apply physical properties of fluid
2. Solve problems related to the physical properties of fluid

2.1 Mass density, ρ (rho) is defined as the mass per unit volume.

Density,  = Mass, m (SI units, kg/m3)
Volume, V

=m
V

2.2 Specific weight, ω (omega) is defined as the weight per unit volume.

Specific Weight,  = Weight, W (SI units, N/m3)
Volume, V

DJJ20073 Fluid Mechanics Chapter 2: Physical Properties of Fluid

= W
V

Since weight, W = mass, m x acceleration of gravity, g

Or W = mg (g = 9.81 m/s2)

So  = mg = m g which m = 
VV V

Then  = g

2.3 Specific gravity or relative density, SG is the ratio of the weight of the substance
to the weight of an equal volume of water at 4 ºC.

SG = substance = subs tan ce (No unit)
 water  water

2.4 Specific volume,  (upsilon) is defined as the reciprocal of mass density. It is used
to mean volume per unit mass.

Specific Volume,  = Volume, V (SI units, m3/kg).
Mass, m

= V
m

Or  = 1


Prepared by : Siti Safuan_June2021 Page 17

DJJ20073 Fluid Mechanics Chapter 2: Physical Properties of Fluid

Example 2.1

Given the volume and mass of crude oil is 7.3 m3 and 6500 kg. Determine:

i. Mass density
ii. Relative density/specific gravity
iii. Specific weight

Example 2.2
A liquid weighs 3.7 N on a volume of 100 ml. Find:

a) the liquid density
b) specific gravity of the fluid.

(given the density of water = 1000 kg/m3 and g = 9.81 m/s2)

Prepared by : Siti Safuan_June2021 Page 18

DJJ20073 Fluid Mechanics Chapter 2: Physical Properties of Fluid

Example 2.3
A beaker is filled with a liquid up to the mark of one litre and weighed. The weight of the
liquid is found to be 6.5 N. Determine:

a) The specific weight
b) The specific gravity
c) The specific volume

Example 2.4
A golf ball has a diameter of 42 mm and a mass of 45 g. Calculate the density of the
ball.

Prepared by : Siti Safuan_June2021 Page 19

DJJ20073 Fluid Mechanics Chapter 2: Physical Properties of Fluid

2.5 Viscosity

A fluid at rest cannot resist shearing forces but once it is in motion, shearing forces are
set up between layers of fluid moving at different velocities. The viscosity of the fluid
determines the ability of the fluid in resisting these shearing stresses. The Coefficient of
Dynamic Viscosity, ƞ (eta) is defined as the shear force per unit area required to drag one
layer of fluid with unit velocity past another layer unit distance away from it in the fluid. SI
unit, pascal-second (Pa.s). Kinematic Viscosity,  (nu) is the ratio of dynamic viscosity to
mass density,  = /. The SI unit for kinematic viscosity is m2/s.

In the Figure 1.1 above :
F : shear force (N)
A : area (m2)
h : distance between layers (m)
v : velocity of layer (m/s)

Dynamic viscosity , = Shear stress ,  Kinematic viscosity , = Dynamic viscosity , 
Density, 
.

Shear rate, 

Prepared by : Siti Safuan_June2021 Page 20

DJJ20073 Fluid Mechanics Chapter 2: Physical Properties of Fluid

Example 2.5

The density of an oil is 850 kg/m3. Find its relative density and kinematic viscosity if the
dynamic viscosity is 5 ×10-3 kg/ms.

Example 2.6

A fluid is flowing between two layers of 2 mm2 area. Calculate the shearing force if the
shear velocity is 0.25 m/s and has length 2 m and dynamic viscosity is 2 Ns/m2.

Example 2.7

A fluid moves along length 0.75 m with velocity 2 m/s and has shearing stress of 2 N/m2.
Calculate its dynamic viscocity.

Prepared by : Siti Safuan_June2021 Page 21

DJJ20073 Fluid Mechanics Chapter 2: Physical Properties of Fluid

2.6 Fluid Compressibility

All materials whether solids or fluids (liquids/gases) can change the shape under
application of external force or stress. In case of fluid, the force is generally termed as
'pressure' as it is uniform from all the directions. But how do we judge the solid/fluid is
susceptible to change in the volume upon application of forces/pressure. Thus, in a simple
word, Compressibility is a measure of change in the volume of fluid under the effect of
force/pressure or stress.

When fluid is acted upon by force, volume gets reduced. This effect is also judge by
'Bulk Modulus of elasticity'. The Compressibility coefficient can also be expressed as
reciprocal of Bulk Modulus of elasticity.

Thus, - (Change in Volume/Original Volume = Change in Pressure/Bulk Modulus) as
given by Equation (1)

The negative (-ve) sign suggests the volume gets decreased when pressure increases.
When we consider the unit mass of fluid, Volume is reciprocal of density (as given by
Equation (2)). Now by differentiating the and substituting the value of V and dV, final

Prepared by : Siti Safuan_June2021 Page 22

DJJ20073 Fluid Mechanics Chapter 2: Physical Properties of Fluid

equation is equation (3) which gives relation between change in pressure and change in
density of fluid and density also affected by temperature change. Thus during
compression, rise in temperature causes the change in density as well as change in
pressure.

If the temperature remains constant, conditions are called 'isothermal' and Isothermal
Bulk modulus and if no heat is added or removed during compression, conditions are
called 'adiabatic' and Adiabatic Bulk modulus. One of the most important outcome from
the above discussion is the outcome of index of process which is resulted when taking
ratio of adiabatic bulk modulus to the isothermal bulk modulus. The process index is the
ratio of specific heat at constant pressure to the specific heat at constant volume.

Importance of Compressibility
The concept of the bulk modulus is mainly applied to the liquids, since for the gases the
Compressibility is so great that the value of K is not constant but can vary according to
the conditions, but proportional to the pressure and changes very rapidly. For liquids,
the changes in pressure is not so significant as gases have, and thus liquid mostly
treated as 'Incompressible' and gases are called, 'Compressible'. Normally gases are
treated as 'Incompressible' if the pressure changes are so small (as per equation (3)).
And also if Bulk modulus of elasticity of the fluid is large, such as for water, flow may be
treated as 'Incomressible'. For a gas where the bulk modulus of elasticity can be as
much as 20,000 times smaller than water, there is no sharp ine of demarcation
betweeen compressible or incompressible flow.

For example, if we consider a typical regenerative cryogenic refrigerator with a working
fluid as helium at a pressure of 20 bar, the Compressibility coefficient is 0.5 E -7 at a
pressure change of 1.2 bar would lead to change density by 5 %.

Prepared by : Siti Safuan_June2021 Page 23

DJJ20073 Fluid Mechanics Chapter 2: Physical Properties of Fluid

A property that is commonly used to characterize compressibility is bulk modulus, 
(kappa). Bulk modulus is defined as the ratio of the change in pressure to the rate of
change of volume due to the change in pressure.

Bulk Modulus,  = − Change in pressure
Rate of change of volume

= − Change in pressure = −dP  dV
Change in volume/Original volume V

 = −V dP
dV

Example 2.8

Determine the bulk modulus of a liquid whose volume decreases by 4% for an increase
in pressure of 500 x 105 Pa.

Prepared by : Siti Safuan_June2021 Page 24

DJJ20073 Fluid Mechanics Chapter 2: Physical Properties of Fluid

SELF ASSESSMENT

1. Assume the density of water to be 1000 kg/m3 at atmospheric pressure 101 kN/m2.
What will be:
a) the gauge pressure
b) the absolute pressure of water at a depth of 2000 m below the free surface?

2. Determine in Newton per square metre, the increase in pressure intensity per
metre depth in fresh water. The mass density of fresh water is 1000 kg/m3.

3. Given specific weight of fluid is 6.54 kN/m3 and its mass is 8.3 kg, calculate the
following:
a) volume of fluid
b) specific volume of fluid
c) density of fluid

4. Given oil specific gravity is 0.89, find :
a) density of oil
b) specific weight of oil
c) specific volume of oil

ans: 1 a) 117.72 kN/m2 , b) 218.72 kN/m2

2 9810 N/m2

3 a) 0.012 m3 b) 0.0015 m3/kg c) 691.67 kg/m3

4 a) 0.89 x 103 kg/m3 b) 8730.9 N/m3 c) 0.00112 m3/kg

Prepared by : Siti Safuan_June2021 Page 25

CHAPTER 3
Fluid Statics

Fluid statics is the field of physics that involves the study of fluids at rest. Because
these fluids are not in motion, that means they have achieved a stable equilibrium state,
so fluid statics is largely about understanding these fluid equilibrium conditions. When
focusing on incompressible fluids (such as liquids) as opposed to compressible fluids
(such as most gases), it is sometimes referred to as hydrostatics.
A fluid at rest does not undergo any sheer stress, and only experiences the influence of
the normal force of the surrounding fluid (and walls, if in a container), which is the
pressure. This form of equilibrium condition of a fluid is said to be a hydrostatic
condition. Fluids that are not in a hydrostatic condition or at rest, and are therefore in
some sort of motion, fall under the other field of fluid mechanics, fluid dynamics.

After completing this chapter, you should be able to:
1. Apply concept of Pascal’s Law into the hydraulic jack.
2. Apply concept of piezometer, barometer and manometer.
3. Explain concept of a bourdon gauge.
4. Apply concept of buoyancy.

3.1 Pascal’s Law

DJJ20073 Fluid Mechanics Chapter 3: Fluid Statics

Blaise Pascal, a seventeenth-century scientist, has described two important principles
about pressure:

• Pressure acts uniformly in all directions on a small volume of fluid;

Figure 3.1: Pressure acting uniformly in all directions

• In a fluid confined by solid boundaries, pressure acts perpendicular to the
boundary.

Figure 3.2: Direction of fluid pressures on boundries

These principles are called Pascal’s Law.

Pascal’s Law states that if pressure is applied to a non-flowing fluid in a container,
then that pressure is transmitted equally in all directions within the container.

3.1.1 The Hydraulic Jack Page 27

Prepared by : Siti Safuan_June2021

DJJ20073 Fluid Mechanics Chapter 3: Fluid Statics

A Hydraulic Jack is used to lift a heavy load with the help of a light force. A diagram of a
hydraulic jack is shown in Figure 3.3. An external force, F is applied to the piston of the
small cylinder and forces oil or water out into the large cylinder thus raising the piston
supporting the load, W. The force, F acting on area A1 produces a pressure P1 which is
transmitted equally in all directions through the liquid. If the two pistons are at the same
level, the pressure P2 acting on the larger piston must equal P1.

A1 A2

Figure 3.3: A Hydraulic Jack System

Based on the Figure 3.3 above:
F = external force exerted on the small piston (N)
W = the load to be lifted (N)
A1 = cross-sectional area of small piston (m2)
A2 = cross-sectional area of large piston (m2)

Pressure on small piston, P1 = F
A1

Pressure on large piston, P2 = W
A2

Based on Pascal’s Law, pressure in liquid is transmitted from small piston to large piston.

Prepared by : Siti Safuan_June2021 Page 28

DJJ20073 Fluid Mechanics Chapter 3: Fluid Statics

Thus, P1 = P2
F =W
A1 A2

There are three cases of hydraulic jack system depends on the position of small piston
and large piston.

hh

F =W F + gh = W F = W + gh
A1 A2 A1 A2 A1 A2

Figure 3.4: Different piston positions and the corresponding equation

Example 3.1 Page 29

Prepared by : Siti Safuan_June2021

DJJ20073 Fluid Mechanics Chapter 3: Fluid Statics

A force F of 850 N is applied to the small cylinder of a hydraulic jack. The area A1 of the
small piston is 15 cm2 and the area A2 of the larger piston is 150 cm2. What load W can
be lifted on the large piston:
a) if the pistons are at the same level?
b) if the large piston is 0.75 m below the smaller?
c) if the small piston is 0.40 m below the larger?

Example 3.2 Page 30

Prepared by : Siti Safuan_June2021

DJJ20073 Fluid Mechanics Chapter 3: Fluid Statics

A hydraulic jack containing water has a small piston diameter of 80 mm and the diameter
of the large piston is eight times larger than the diameter of the smaller piston. The small
piston surface 6 m above the large piston. What is the force required at the small piston
to lift a mass of 3500 kg at the large piston?

3.2 Fluid Pressure Measurement Page 31

Prepared by : Siti Safuan_June2021

DJJ20073 Fluid Mechanics Chapter 3: Fluid Statics

3.2.1 Mercury Barometer

Atmospheric pressure is usually measured by a mercury barometer. A simple barometer
consists of a tube more than 760 mm long inserted in an open container of mercury with
a closed and evacuated end at the top and open end at the bottom with mercury extending
from the container up into the tube. A void is produced at the top of the tube which is very
nearly a perfect vacuum. Figure 2.7 below shows an example of a barometer.

Mercury rises in the tube to a height of approximately 760 mm at sea level. The level of
mercury will rise and fall as atmospheric pressure changes; direct reading of the mercury
level gives prevailing atmospheric pressure as a pressure head (of mercury), which can
be converted to pressure using the relation Patm = ρgh.

Figure 3.5: Barometer

Although a barometer can be used to measure atmospheric pressure, it is often
necessary to measure pressures of other fluids. There are several ways to accomplish
this task.

Example 3.3

Prepared by : Siti Safuan_June2021 Page 32

DJJ20073 Fluid Mechanics Chapter 3: Fluid Statics

The level of mercury in a barometer tube is 760 mm above the level of the mercury in the
bowl. What is the atmospheric pressure in N/m2?

3.2.2 Piezometer Tube

A simple vertical tube open at the top, which is attached to the system containing the
liquid where the pressure (higher than atmospheric pressure) to be measured. As the
tube is open to the atmosphere, the pressure measured is the gauge pressure.

h
A•

Figure 3.6: Piezometer Tube
The pressure at point A can be determined by measuring the height of head, h.

Prepared by : Siti Safuan_June2021 Page 33

DJJ20073 Fluid Mechanics Chapter 3: Fluid Statics

PA = gh

where  is the density of the fluid in the container? Piezometer is a measurement tool that
is very easy and accurate, but it is appropriate if the pressure in the container is higher
than the atmospheric pressure (if not, the air will be sucked into the system). In addition,
the pressure to be measured must be relatively small so that height of head is reasonable.
To avoid capillary effects, a piezometer’s tube should be 12 mm or larger.

Example 3.4

A pressure tube is used to measure the pressure of oil (mass density ρ = 640 kg/m3) in a
pipeline. If the oil rises to a height of 1.2 m above the center of the pipe, what is the gauge
pressure in N/m2 at that point?

3.2.3 Manometer U- tube Page 34

Prepared by : Siti Safuan_June2021

DJJ20073 Fluid Mechanics Chapter 3: Fluid Statics

a) Simple Manometer U-tube

One end of the U-tube is connected to pressure that is to be measured, while the other
end is left open to atmosphere. The tube contains a liquid, which is called the manometric
fluid, which does not mix with the fluid whose pressure is to be measured.

The fluid whose pressure is being measured should have a lesser density than the
manometric fluid (ρ < ρman). Refer Figure 3.7 for the illustration.

Fluid density, ρ

Manometric fluid density, ρman

Figure 3.7: U-tube Manometer

Pressure in a continuous static fluid is the same level at any horizontal level so,

PB = PC …. (1)

For left-hand limb; Page 35
Pressure at B = Pressure at A + pressure due to h1

Prepared by : Siti Safuan_June2021

DJJ20073 Fluid Mechanics Chapter 3: Fluid Statics

PB = PA + ρgh1 … (2)
... (3)
For right-hand limb;
Pressure at C = Pressure at D + pressure due to h2

PC = PD + ρ gh

man 2

But PD = patm (atmospheric pressure due to open end)

As we are measuring gauge pressure, Patm = 0

PC = ρmangh2

But PC = PB. By equating (2) and (3), gives:

PA + ρgh1 = ρmangh2

PA = ρmangh2 - ρgh1

Example 3.5 Page 36

Prepared by : Siti Safuan_June2021

DJJ20073 Fluid Mechanics Chapter 3: Fluid Statics

A U tube manometer is used to measure the pressure at point A, which is below
atmospheric pressure. If the relative density of mercury is 13.6 and the atmospheric
pressure is 101.3 kN/m2, determine the absolute pressure at point A when h1 = 15 cm
and h2 = 30 cm. Specific weight of water is 9.81 kN/m3.

b) Differential U-tube Manometer Page 37

Prepared by : Siti Safuan_June2021

DJJ20073 Fluid Mechanics Chapter 3: Fluid Statics

In some cases, the different between the pressures at two points is desired rather than
the actual value of the pressure at each point. A manometer to determine this pressure
difference is called the differential manometer. The liquids in manometer will rise or fall
as the pressure at either end (or both ends) of the tube changes.

Figure 3.8: Differential Manometer

In the above figure:

But PC = PD PC = PA + wgh
Thus PD = PB + wgh2 + Hggh1
(same horizontal level)
PA + wgh = PB + wgh2 + Hggh1

Pressure different between point A and point B is
PA - PB = wgh2 + Hggh1 - wgh

Example 3.6 Page 38

Prepared by : Siti Safuan_June2021

DJJ20073 Fluid Mechanics Chapter 3: Fluid Statics

According to Figure 3.9, a Figure 3.9
differential U-tube manometer is
used to measure the pressure
difference of water between point
A and point B. If relative density of
mercury is 13.6 whereas h = 15 cm
and h1 = h2 = 5 cm, find the
pressure difference.

Example 3.7 Page 39

Prepared by : Siti Safuan_June2021

In Figure 3.10, if h1 = 0.11 m and
h2 = 0.75 m, find the pressure

DJJ20073 Fluid Mechanics Chapter 3: Fluid Statics

c) Inverted U-tube Manometer Page 40

Prepared by : Siti Safuan_June2021

DJJ20073 Fluid Mechanics Chapter 3: Fluid Statics

Figure 3.11: Inverted U-tube manometer

Inverted U-tube manometer is used to measure the difference in fluid pressures between
two points on a pipe or other pressure source. The manometer is usually filled with air.
Air can be inserted or removed through a valve at the top to control fluid level in
manometer. The other alternative is filled the top portion of manometer with another fluid
that has a specific gravity lower than the fluid at A and B.

With reference to the Figure 3.12 on the right, the
balance of the left :
PA = PC + wgh1 + ogh2
PC = PA - wgh1 - ogh2
To balance on the right side
PB = PD + wg(h1 + h2)
PD = PB - wg(h1 + h2)

Prepared by : Siti Safuan_June2021 Figure 3.12
Page 41

DJJ20073 Fluid Mechanics Chapter 3: Fluid Statics

Since PC = PD Fluid Q
Fluid P
PA - wgh1 - ogh2 = PB - wg(h1 + h2) Figure 3.13
PA - PB = wgh1 + ogh2 - wg(h1 + h2)

= gh2 (o - w)

Example 3.8
In Figure 3.13, fluid P is water and fluid Q
is oil (SG = 0.9). If h = 69 cm, y = 75 cm
and z = 23 cm, what is the pressure
difference (in kN/m2) between A and B.

Example 3.9 Page 42

AcPcreoprdariendgbyto: SFitiigSuarfeuan3_.1Ju4n,e2c0a2lc1ulate the
pressure difference between B and A (in
kN/m2) if p = 22 cm, q = 80 cm and r = 10

DJJ20073 Fluid Mechanics Chapter 3: Fluid Statics

d) Combined U-tube Manometer Page 43

Prepared by : Siti Safuan_June2021

DJJ20073 Fluid Mechanics Chapter 3: Fluid Statics

Figure 3.15: Combine manometer

Datum: N, O and P, so PN = PO = PP …(1)
Left-hand limb: ...(2)
PN = PM – ρAgh1
Right-hand limb: …(3)
PP = PQ + ρBgh2
PQ = Patm Page 44

= Patm + ρBgh2
Put (2) & (3) into (1),
PM – ρAgh1 = Patm + ρBgh2
PM = Patm + ρBgh2 + ρAgh1
Example 3.10

Prepared by : Siti Safuan_June2021

DJJ20073 Fluid Mechanics Chapter 3: Fluid Statics

A manometer containing mercury is used to measure the pressure of an oil of specific
gravity 0.89 as shown in the Figure 3.16 below. Calculate the pressure at M, PM if the
difference level of mercury is 45 cm.

Figure 3.16

Prepared by : Siti Safuan_June2021 Page 45

DJJ20073 Fluid Mechanics Chapter 3: Fluid Statics

3.3 Bourdon Gauge

The pressure to be measured is applied to a curved tube, oval in cross section. Pressure
applied to the tube tends to cause the tube to straighten out, and deflection of the end of
the tube is communicated through a system of levers to a recording needle.

This gauge is widely used for steam and compressed gases. The pressure indicated is
the difference between that communicated by the system to the external (ambient)
pressure, and is usually referred to as the gauge pressure.

(a) Front view (b) Rear view with back of case removed

Figure 3.17: Bourdon tube pressure gauges

Prepared by : Siti Safuan_June2021 Page 46

DJJ20073 Fluid Mechanics Chapter 3: Fluid Statics

3.4 Buoyancy

Buoyancy is the force that causes objects to float. It is the force exerted on an object that
is partly or wholly immersed in a fluid. Buoyancy is caused by the differences in pressure
acting on opposite sides of an object immersed in a static fluid. It is also known as
the buoyant force. Buoyancy is the phenomena due to Buoyant Force. This buoyancy
force is always acting vertically upward, and has the following characteristics:

i) The buoyancy force is equal to the weight of the fluid displaced by the solid
body.

ii) The buoyancy force acts through the centroid of the displaced volume of fluid,
called center of buoyancy.

iii) A floating body displaces a volume of fluid whose weight is equal to the weight
of the body.

The above principle is known Archimedes’ Principle and can be defined mathematically
as demonstrated below.

Figure 3.18: Buoyancy force

For equilibrium: +∑Fy = 0

Fb – W = 0 or Fb = W

Prepared by : Siti Safuan_June2021 Page 47

DJJ20073 Fluid Mechanics Chapter 3: Fluid Statics

Therefore, we can write:
Fb = weight of fluid displaced by the body

or Fb = W = mg = ρgV
where Fb = buoyant force

V = displaced volume of fluid
W = weight of fluid
The buoyant force will act ↑ through the centre of buoyancy (B), while the weight will act
↓ through the centre of gravity (G) of the body.

Archimedes’ Principle state that an object which is partially or wholly immersed
in a fluid (liquid or gas) is acted upon by an upward buoyant force equal to the
weight of the fluid it displaces.

Buoyant force = Weight of fluid displaced by the object = Weight of the object
Fb = Wdisplaced fluid = Wobject
Vdisplaced fluid = Vimmersed object

Prepared by : Siti Safuan_June2021 Page 48