INSTITUTE OF AERONAUTICAL ENGINEERING
(Autonomous)
Dundigal, Hyderabad -500 043
CIVIL ENGINEERING
COURSE LECTURE NOTES
Course Name FLUID MECHANICS
Course Code ACEB06
Programme B.Tech
Semester IV
Course
Coordinator Mr. Ch. V. S. S. Sudheer, Assistant Professor
Course Faculty Dr. P. Ram Mohan Rao
Mr. Ch. V. S. S. Sudheer
Lecture Numbers 1-60
Topic Covered All
COURSE OBJECTIVES (COs):
The course should enable the students to:
I Analyse and understand the broad principles of fluid statics, kinematics and dynamics
II Interpret the definitions of the basic terms used in fluid mechanics and their applications
III Summarize classifications of fluid flow and applications of continuity equation
IV Explore the knowledge of continuity, momentum and energy principles
V Obtain the wide applications of dimensional analysis in fluid flow problems
COURSE LEARNING OUTCOMES (CLOs):
Students, who complete the course, will have demonstrated the ability to do the
following:
ACEB06.01 Define fluid and its properties. Describe surface tension and relations in different
conditions.
ACEB06.02 Explain Newton’s law of viscosity. Classify fluids based on Newton’s law of
viscosity and solve problems on Viscosity.
ACEB06.03 Employ capillary principle to calculate capillary rise/fall in a given tube.
ACEB06.04 Understand the concept of surface tension and their applications.
ACEB06.05 Know the concept of capillarity and compressibility.
ACEB06.06 Employ principle of manometry to measure gauge and differential pressure surface.
ACEB06.07 Understand the Archimedes principle to solve numerical examples on fluid
pressure management.
PAGE 1
ACEB06.08 Application and measurement of pressure for different types of pipes such as
inclined, horizontal and vertical.
ACEB06.09 Examine stability of a floating body by determining its metacentric height
ACEB06.10 Understand the various types of flows in closed conduits.
ACEB06.11 Distinguish velocity potential function and stream function and solve for velocity
and acceleration of a fluid at a given location in a fluid flow
ACEB06.12 Establish Euler’s theorem and deduce Bernoulli’s equation for a ideal fluid and
comment on validation assumption made.
ACEB06.13 Examine Bernoulli’s equation for ideal and real fluids and evaluate the direction of
flow
ACEB06.14 Flow and velocity measuring instruments.
ACEB06.15 Employ Darcy-Weischbach and Chezy’s equation to calculate friction losses
ACEB06.16 Describe flow through pipes, and Distinguish between major loss and minor loss in
pipes.
ACEB06.17 Sketch HGL and TEL for a given pipe setting.
ACEB06.18 Distinguish between Drag force and lift force and Examine drag and lift force for a
given set
ACEB06.19 Flow and velocity measuring instruments.
ACEB06.20 Application and analysis of momentum principle in closed pipes to estimate the
force on pipe bends.
ACEB06.21 Understand the Dimensional quantities and analysis for various parameters.
ACEB06.22 Derive the problems based on Rayleigh’s method and Buckingham’s pi theorem
with applications.
ACEB06.23 Explain the concept of similitude with examples and different types of similitude
concepts.
ACEB06.24 Remember the concepts of dimensionless numbers to solve numerical problems
SYLLABUS
Module-I BASIC CONCEPTS AND DEFINITIONS
Distinction between a fluid and a solid; Density, Specific weight, Specific gravity, Kinematic
and dynamic viscosity; variation of viscosity with temperature, Newton law of viscosity; vapour
pressure, boiling point, cavitation; surface tension, capillarity, Bulk modulus of elasticity,
compressibility.
Module-II FLUID STATICS
Fluid Pressure: Pressure at a point, Pascal’s law, pressure variation with temperature, density and
altitude. Piezometer, U-Tube Manometer, Single Column Manometer, U -Tube Differential
Manometer, Micro manometers. Pressure gauges, Hydrostatic pressure and force: horizontal,
vertical and inclined surfaces. Buoyancy and stability of floating bodies.
Module-III FLUID KINEMATICS
Classification of fluid flow: steady and unsteady flow, uniform and non-uniform flow, laminar
and turbulent flow, rotational and irrotational flow, compressible and incompressible flow, ideal
and real fluid flow, one, two- and three-dimensional flows;
Stream line, path line, streak line and stream tube; stream function, velocity potential function.
One-, two and three - dimensional continuity equations in Cartesian coordinates.
Module-IV FLUID DYNAMICS
Surface and body forces; Equations of motion - Euler’s equation; Bernoulli’s equation –
derivation; Energy Principle; Practical applications of Bernoulli’s equation - Venturimeter,
orifice meter and pitot tube; Momentum principle; Forces exerted by fluid flow on pipe bend;
Vortex Flow – Free and Forced;
PAGE 2
Module-V DIMENSIONAL ANALYSIS
Dimensional Analysis and Dynamic Similitude - Definitions of Reynolds Number, Froude
Number, Mach Number, Weber Number and Euler Number; Buckingham’s π-Theorem.
Text Books:
1. C. S. P. Ojha, R. Berndtsson and P. N. Chadramouli, “Fluid Mechanics and Machinery”,
Oxford University Press, 2010.
2. P M Modi and S M Seth, “Hydraulics and Fluid Mechanics”, Standard Book House, 2014.
Reference Books:
1. K. Subramanya, “Theory and Applications of Fluid Mechanics”, Tata McGraw Hill.
2. R.L. Daugherty, J.B. Franzini and E.J. Finnemore, “Fluid Mechanics with Engineering
Applications”, International Student Edition, Tata Mc Graw Hill.
PAGE 3
Fluid Mechanics
Background
• There are three states of matter: solids, liquids and gases.
• Both liquids and gases are classified as fluids.
• Fluids do not resist a change in shape. Therefore fluids assume the shape of the
container they occupy.
• Liquids may be considered to have a fixed volume and therefore can have a
free surface. Liquids are almost incompressible.
• Conversely, gases are easily compressed and will expand to fill a container
they occupy.
• We will usually be interested in liquids, either at rest or in motion.
Liquid showing free surface Gas filling volume
Behaviour of fluids in containers
PAGE
MODULE – I
BASIC CONCEPTS AND DEFINATIONS
PAGE 4
Fluid Mechanics
Definition
The strict definition of a fluid is:
A fluid is a substance which conforms continuously under the action of
shearing forces.
To understand this, remind ourselves of what a shear force is:
Application and effect of shear force on a book
Definition Applied to Static Fluids
According to this definition, if we apply a shear force to a fluid it will deform and
take up a state in which no shear force exists. Therefore, we can say:
If a fluid is at rest there can be no shearing forces acting and therefore all
forces in the fluid must be perpendicular to the planes in which they act.
Note here that we specify that the fluid must be at rest. This is because, it is found
experimentally that fluids in motion can have slight resistance to shear force. This is
the source of viscosity.
PAGE
PAGE 5
Fluid Mechanics
Definition Applied to Fluids in Motion
For example, consider the fluid shown flowing along a fixed surface. At the surface
there will be little movement of the fluid (it will ‘stick’ to the surface), whilst further
away from the surface the fluid flows faster (has greater velocity):
If one layer of is moving faster than another layer of fluid, there must be shear forces
acting between them. For example, if we have fluid in contact with a conveyor belt
that is moving we will get the behaviour shown:
Ideal fluid Real (Viscous) Fluid
When fluid is in motion, any difference in velocity between adjacent layers has the
same effect as the conveyor belt does.
Therefore, to represent real fluids in motion we must consider the action of shear
forces.
PAGE
PAGE 6
Fluid Mechanics
Consider the small element of fluid shown, which is subject to shear force and has a
dimension s into the page. The force F acts over an area A = BC×s. Hence we have a
shear stress applied:
ForceStressAreaFA
τ
=
=
Any stress causes a deformation, or strain, and a shear stress causes a shear strain.
This shear strain is measured by the angle φ .
Remember that a fluid continuously deforms when under the action of shear. This is
different to a solid: a solid has a single value of φ for each value of τ . So the longer
a shear stress is applied to a fluid, the more shear strain occurs. However, what is
known from experiments is that the rate of shear strain (shear strain per unit time) is
related to the shear stress:
Shear stress Rate of shear strainShear stress Constant Rate of shear strain
∝= ×
PAGE
PAGE 7
Fluid Mechanics
We need to know the rate of shear strain. From the diagram, the shear strain is:
xy
φ =
If we suppose that the particle of fluid at E moves a distance x in time t, then, using
S Rθ= for small angles, the rate of shear strain is:
1xx tyt t y
uy
φ ⎛ ⎞∆= = ⋅⎜ ⎟∆ ⎝ ⎠
=
Where u is the velocity of the fluid. This term is also the change in velocity with
height. When we consider infinitesimally small changes in height we can write this in
differential form, du dy . Therefore we have:
constant dudy
τ = ×
This constant is a property of the fluid called its dynamic viscosity (dynamic because
the fluid is in motion, and viscosity because it is resisting shear stress). It is denoted
µ which then gives us:
Newton’s Law of Viscosity:
dudy
τ µ=
PAGE
PAGE 8
Fluid Mechanics
Generalized Laws of Viscosity
We have derived a law for the behaviour of fluids – that of Newtonian fluids.
However, experiments show that there are non-Newtonian fluids that follow a
generalized law of viscosity:
n
duA Bdy
τ⎛ ⎞
= + ⎜ ⎟⎝ ⎠
Where A, B and n are constants found experimentally. When plotted these fluids
show much different behaviour to a Newtonian fluid:
Behaviour of Fluids and Solids
PAGE
PAGE 9
Fluid Mechanics
In this graph the Newtonian fluid is represent by a straight line, the slope of which is
µ . Some of the other fluids are:
• Plastic: Shear stress must reach a certain minimum before flow commences.
• Pseudo-plastic: No minimum shear stress necessary and the viscosity
decreases with rate of shear, e.g. substances like clay, milk and cement.
• Dilatant substances; Viscosity increases with rate of shear, e.g. quicksand.
• Viscoelastic materials: Similar to Newtonian but if there is a sudden large
change in shear they behave like plastic.
• Solids: Real solids do have a slight change of shear strain with time, whereas
ideal solids (those we idealise for our theories) do not.
Lastly, we also consider the ideal fluid. This is a fluid which is assumed to have no
viscosity and is very useful for developing theoretical solutions. It helps achieve
some practically useful solutions.
PAGE
PAGE 10
Fluid Mechanics
Units
Fluid mechanics deals with the measurement of many variables of many different
types of units. Hence we need to be very careful to be consistent.
Dimensions and Base Units
The dimension of a measure is independent of any particular system of units. For
example, velocity may be in metres per second or miles per hour, but dimensionally,
it is always length per time, or 1L T LT−= . The dimensions of the relevant base units
of the Système International (SI) system are:
Unit-Free SI Units Dimension Symbol Unit Symbol
Mass M kilogram kg Length L metre m Time T second s
Temperature θ kelvin K
Derived Units
From these we have some relevant derived units (shown on the next page).
Checking the dimensions or units of an equation is very useful to minimize errors.
For example, if when calculating a force and you find a pressure then you know
you’ve made a mistake.
PAGE
PAGE 11
Fluid Mechanics
SI Unit Quantity Dimension
Derived Base
Velocity 1LT− m/s 1m s−
Acceleration 2LT− m/s2 2m s−
Force 2MLT− Newton, N 2kg m s−
Pressure
Stress -1 2ML T
Pascal, Pa
N/m2 -1 2kg m s−
Density -3ML kg/m3 -3kg m
Specific weight -2 2ML T− N/m3 -2 2kg m s−
Relative density Ratio Ratio Ratio
Viscosity -1 1ML T− Ns/m2 -1 1kg m s−
Energy (work) 2 2ML T− Joule, J
Nm 2 2kg m s−
Power 2 3ML T−
Watt, W
Nm/s
2 3kg m s−
Note: The acceleration due to gravity will always be taken as 9.81 m/s2.
PAGE
PAGE 12
Fluid Mechanics
SI Prefixes
SI units use prefixes to reduce the number of digits required to display a quantity.
The prefixes and multiples are:
Prefix Name Prefix Unit Multiple
Tera
Giga
Mega
Kilo
Hecto
Deka
Deci
Centi
Milli
Micro
Nano
Pico
T
G
M
k
h
da
d
c
m µ
n
p
1012
109
106
103
102
101
10-1
10-2
10-3
10-6
10-9
10-12
Be very particular about units and prefixes. For example:
• kN means kilo-Newton, 1000 Newtons;
• Kn is the symbol for knots – an imperial measure of speed;
• KN has no meaning;
• kn means kilo-nano – essentially meaningless.
PAGE
PAGE 13
• Water: 1000 kg/m3; • Mercury: 13546 kg/m3; • Air: 1.23 kg/m3; • Paraffin: 800 kg/m3.
times the acceleration due to gravity: gγ ρ=
Relative Density (Specific Gravity)
A dimensionless measure of the density of a substance with reference to the density
of some standard substance, usually water at 4°C:
density of substancerelative densitydensity of water
specific weight of substancespecific weight of water
s s
w w
ρ γρ γ
=
=
= =
PAGE
Mass Density
The mass per unit volume of a substance, usually denoted as ρ . Typical values are:
Specific Weight
The weight of a unit volume a substance, usually denoted as γ . Essentially density
PAGE 14
Fluid Mechanics
Bulk Modulus
In analogy with solids, the bulk modulus is the modulus of elasticity for a fluid. It is
the ratio of the change in unit pressure to the corresponding volume change per unit
volume, expressed as:
Change in Volume Chnage in pressureOriginal Volume Bulk Modulus
dV dpV K
=
−=
Hence:
dpK VdV
= −
In which the negative sign indicates that the volume reduces as the pressure
increases. The bulk modulus changes with the pressure and density of the fluid, but
for liquids can be considered constant for normal usage. Typical values are:
• Water: 2.05 GN/m3; • Oil: 1.62 GN/m3.
The units are the same as those of stress or pressure.
Viscosity
The viscosity of a fluid determines the amount of resistance to shear force.
Viscosities of liquids decrease as temperature increases and are usually not affected
by pressure changes. From Newton’s Law of Viscosity:
shear stressrate of shear straindu dy
τµ = =
Hence the units of viscosity are Pa s⋅ or 2N s m⋅ . This measure of viscosity is
known as dynamic viscosity and some typical values are given:
PAGE
PAGE 15
Fluid Mechanics
Problems - Properties
a) If 6 m3 of oil weighs 47 kN, find its specific weight, density, and relative density.
(Ans. 7.833 kN/m3, 798 kg/m3, 0.800)
b) At a certain depth in the ocean, the pressure is 80 MPa. Assume that the specific
weight at the surface is 10 kN/m3 and the average bulk modulus is 2.340 GPa.
Find:
a) the change in specific volume between the surface and the large depth;
b) the specific volume at the depth, and;
c) the specific weight at the depth.
(Ans. -0.335×10-4 m3/kg, 9.475×10-4 m3/kg, 10.35 kN/m3)
c) A 100 mm deep stream of water is flowing over a boundary. It is considered to
have zero velocity at the boundary and 1.5 m/s at the free surface. Assuming a
linear velocity profile, what is the shear stress in the water?
(Ans. 0.0195 N/m2)
d) The viscosity of a fluid is to be measured using a viscometer constructed of two
750 mm long concentric cylinders. The outer diameter of the inner cylinder is 150
mm and the gap between the two cylinders is 1.2 mm. The inner cylinder is
rotated at 200 rpm and the torque is measured to be 10 Nm.
a) Derive a generals expression for
the viscosity of a fluid using this
type of viscometer, and;
b) Determine the viscosity of the
fluid for the experiment above.
(Ans. 6 × 10-4 Ns/m2)
PAGE
PAGE 16
Fluid Mechanics
Pressure
In fluids we use the term pressure to mean:
The perpendicular force exerted by a fluid per unit area.
This is equivalent to stress in solids, but we shall keep the term pressure.
Mathematically, because pressure may vary from place to place, we have:
0
lim FpA∆→
∆=
∆
As we saw, force per unit area is measured in N/m2 which is the same as a pascal
(Pa). The units used in practice vary:
• 1 kPa = 1000 Pa = 1000 N/m2
• 1 MPa = 1000 kPa = 1 × 106 N/m2
• 1 bar = 105 Pa = 100 kPa = 0.1 MPa
• 1 atm = 101,325 Pa = 101.325 kPa = 1.01325 bars = 1013.25 millibars
For reference to pressures encountered on the street which are often imperial:
• 1 atm = 14.696 psi (i.e. pounds per square inch)
• 1 psi = 6894.7 Pa ≈ 6.89 kPa ≈ 0.007 MPa
MODULE – II
FLUID STATICS
PAGE 17
Fluid Mechanics
Pressure Reference Levels
The pressure that exists anywhere in the universe is called the absolute pressure, absP .
This then is the amount of pressure greater than a pure vacuum. The atmosphere on
earth exerts atmospheric pressure, atmP , on everything in it. Often when measuring
pressures we will calibrate the instrument to read zero in the open air. Any measured
pressure, measP , is then a positive or negative deviation from atmospheric pressure.
We call such deviations a gauge pressure, gaugeP . Sometimes when a gauge pressure
is negative it is termed a vacuum pressure, vacP .
The above diagram shows:
(a) the case when the measured pressure is below atmospheric pressure and so is a
negative gauge pressure or a vacuum pressure;
(b) the more usual case when the measured pressure is greater than atmospheric
pressure by the gauge pressure.
PAGE 18
Fluid Mechanics
3.2 Pressure in a Fluid
Statics of Definition
We applied the definition of a fluid to the static case previously and determined that
there must be no shear forces acting and thus only forces normal to a surface act in a
fluid.
For a flat surface at arbitrary angle we have:
A curved surface can be examined in sections:
PAGE 19
Fluid Mechanics
And we are not restricted to actual solid-fluid interfaces. We can consider imaginary
planes through a fluid:
Pascal’s Law
This law states:
The pressure at a point in a fluid at rest is the same in all directions.
To show this, we will consider a very small wedge of fluid surrounding the point.
This wedge is unit thickness into the page:
PAGE 20
Fluid Mechanics
As with all static objects the forces in the x and y directions should balance. Hence:
0xF =∑ : sin 0y sp y p s θ⋅ ∆ − ⋅ ∆ ⋅ =
But ysins
θ ∆=
∆, therefore:
0y s
y s
y s
yp y p ss
p y p y
p p
∆⋅ ∆ − ⋅ ∆ ⋅ =
∆⋅ ∆ = ⋅ ∆
=
0yF =∑ : cos 0x sp x p s θ⋅ ∆ − ⋅ ∆ ⋅ =
But xcoss
θ ∆=
∆, therefore:
0x s
x s
x s
xp x p ss
p x p xp p
∆⋅ ∆ − ⋅ ∆ ⋅ =
∆⋅ ∆ = ⋅ ∆
=
Hence for any angle:
y x sp p p= =
And so the pressure at a point is the same in any direction. Note that we neglected the
weight of the small wedge of fluid because it is infinitesimally small. This is why
Pascal’s Law is restricted to the pressure at a point.
PAGE 21
Fluid Mechanics
Pressure Variation with Depth
Pressure in a static fluid does not change in the horizontal direction as the horizontal
forces balance each other out. However, pressure in a static fluid does change with
depth, due to the extra weight of fluid on top of a layer as we move downwards.
Consider a column of fluid of arbitrary cross section of area, A:
Column of Fluid Pressure Diagram
Considering the weight of the column of water, we have:
0yF =∑ : ( )1 2 1 2 0p A A h h p Aγ+ − − =
PAGE 22
Fluid Mechanics
Obviously the area of the column cancels out: we can just consider pressures. If we
say the height of the column is 2 1h h h= − and substitute in for the specific weight, we
see the difference in pressure from the bottom to the top of the column is:
2 1p p ghρ− =
This difference in pressure varies linearly in h, as shown by the Area 3 of the pressure
diagram. If we let 1 0h = and consider a gauge pressure, then 1 0p = and we have:
2p ghρ=
Where h remains the height of the column. For the fluid on top of the column, this is
the source of 1p and is shown as Area 1 of the pressure diagram. Area 2 of the
pressure diagram is this same pressure carried downwards, to which is added more
pressure due to the extra fluid.
To summarize:
The gauge pressure at any depth from the surface of a fluid is:
p ghρ=
PAGE 23
Fluid Mechanics
Summary
1. Pressure acts normal to any surface in a static fluid;
2. Pressure is the same at a point in a fluid and acts in all directions;
3. Pressure varies linearly with depth in a fluid.
By applying these rules to a simple swimming pool, the pressure distribution around
the edges is as shown:
Note:
1. Along the bottom the pressure is constant due to a constant depth;
2. Along the vertical wall the pressure varies linearly with depth and acts in the
horizontal direction;
3. Along the sloped wall the pressure again varies linearly with depth but also
acts normal to the surface;
4. At the junctions of the walls and the bottom the pressure is the same.
PAGE 24
Fluid Mechanics
Problems - Pressure
1. Sketch the pressure distribution applied to the container by the fluid:
2. For the dam shown, sketch the pressure distribution on line AB and on the
surface of the dam, BC. Sketch the resultant force on the dam.
PAGE 25
Fluid Mechanics
3. For the canal gate shown, sketch the pressure distributions applied to it. Sketch
the resultant force on the gate? If 1 6.0 mh = and 2 4.0 mh = , sketch the
pressure distribution to the gate. Also, what is the value of the resultant force
on the gate and at what height above the bottom of the gate is it applied?
PAGE 26
Fluid Mechanics
3.3 Pressure Measurement
Pressure Head
Pressure in fluids may arise from many sources, for example pumps, gravity,
momentum etc. Since p ghρ= , a height of liquid column can be associated with the
pressure p arising from such sources. This height, h, is known as the pressure head.
Example:
The gauge pressure in a water mains is 50 kN/m2, what is the pressure head?
The pressure head equivalent to the pressure in the pipe is just:
350 10
1000 9.815.1 m
p ghphg
ρ
ρ
=
=
×=
×≈
So the pressure at the bottom of a 5.1 m deep swimming pool is the same as the
pressure in this pipe.
Manometers
A manometer (or liquid gauge) is a pressure measurement device which uses the
relationship between pressure and head to give readings.
In the following, we wish to measure the pressure of a fluid in a pipe.
PAGE 27
Fluid Mechanics
Piezometer
This is the simplest gauge. A small vertical tube is connected to the pipe and its top is
left open to the atmosphere, as shown.
The pressure at A is equal to the pressure due to the column of liquid of height 1h :
1Ap ghρ=
Similarly,
2Bp ghρ=
PAGE 28
Fluid Mechanics
The problem with this type of gauge is that for usual civil engineering applications
the pressure is large (e.g. 100 kN/m2) and so the height of the column is impractical
(e.g.10 m).
Also, obviously, such a gauge is useless for measuring gas pressures.
U-tube Manometer
To overcome the problems with the piezometer, the U-tube manometer seals the fluid
by using a measuring (manometric) liquid:
Choosing the line BC as the interface between the measuring liquid and the fluid, we
know:
Pressure at B, Bp = Pressure at C, Cp
For the left-hand side of the U-tube:
PAGE 29
Fluid Mechanics
1B Ap p ghρ= +
For the right hand side:
2C manp ghρ=
Where we have ignored atmospheric pressure and are thus dealing with gauge
pressures. Thus:
1 2
B C
A man
p pp gh ghρ ρ
=
+ =
And so:
2 1A manp gh ghρ ρ= −
Notice that we have used the fact that in any continuous fluid, the pressure is the
same at any horizontal level.
PAGE 30
Fluid Mechanics
Differential Manometer
To measure the pressure difference between two points we use a u-tube as shown:
Using the same approach as before:
Pressure at C, Cp = Pressure at D, Dp
( )A B manp ga p g b h ghρ ρ ρ+ = + − +
Hence the pressure difference is:
( ) ( )A B manp p g b a hgρ ρ ρ− = − + −
PAGE 31
Fluid Mechanics
Problems – Pressure Measurement
1. What is the pressure head, in metres of water, exerted by the atmosphere?
(Ans. 10.3 m)
2. What is the maximum gauge pressure of water that can be measured using a
piezometer 2.5 m high?
(Ans. 24.5 kN/m2)
3. A U-tube manometer is used to measure the pressure of a fluid of density 800
kg/m3. If the density of the manometric liquid is 13.6 × 103 kg/m3, what is the
gauge pressure in the pipe if
(a) 1 0.5 mh = and D is 0.9 m above BC;
(b) 1 0.1 mh = and D is 0.2 m below BC?
(Ans. 116.15 kN/m2, -27.45 kN/m2)
4. A differential manometer is used to measure the pressure difference between
two points in a pipe carrying water. The manometric liquid is mercury and the
points have a 0.3 m height difference. Calculate the pressure difference when
0.7 mh = .
(Ans. 89.47 kN/m2)
5. For the configuration shown, calculate the weight of the piston if the gauge
pressure reading is 70 kPa.
PAGE 32
Fluid Mechanics
(Ans. 61.6 kN)
6. A hydraulic jack having a ram 150 mm in diameter lifts a weight W = 20 kN
under the action of a 30 mm plunger. What force is required on the plunger to
lift the weight?
(Ans. 800 N)
PAGE 33
Fluid Mechanics
3.4 Fluid Action on Surfaces
Plane Surfaces
We consider a plane surface, PQ, of area A, totally immersed in a liquid of density
ρ and inclined at an angle φ to the free surface:
Side Elevation
Front Elevation
PAGE 34
Fluid Mechanics
If the plane area is symmetrical about the vertical axis OG, then 0d = . We will
assume that this is normally the case.
Find Resultant Force:
The force acting on the small element of area, Aδ , is:
R p A gy Aδ δ ρ δ= ⋅ = ⋅
The total force acting on the surface is the sum of all such small forces. We can
integrate to get the force on the entire area, but remember that y is not constant:
R gy A
g y A
ρ δ
ρ δ
= ⋅
= ⋅
∫∫
But y Aδ⋅∫ is just the first moment of area about the surface. Hence:
R gAyρ=
Where y is the distance to the centroid of the area (point G) from the surface.
Vertical Point Where Resultant Acts:
The resultant force acts perpendicular to the plane and so makes an angle 90 φ° − to
the horizontal. It also acts through point C, the centre of pressure, a distance D below
the free surface. To determine the location of this point we know:
Sum of moments of forces
Moment of about on all elements about
R OO
=
PAGE 35
Fluid Mechanics
Examining a small element first, and since siny s φ= , the moment is:
( )
( )2
Moment of about sin
sin
R O g s A s
g s A
δ ρ φ δ
ρ φ δ
= ⋅⎡ ⎤⎣ ⎦
= ⋅
In which the constants are taken outside the bracket. The total moment is thus:
2Moment of about sinR O g s Aρ φ δ= ⋅ ⋅∫
But 2s Aδ⋅∫ is the second moment of area about point O or just OI . Hence we have:
2
Moment of about sinsin
sinsin
sin
O
O
O
O
R O g IgAy OC g I
DAy I
IDAy
ρ φρ ρ φ
φφ
φ
= ⋅
× = ⋅
× = ⋅
= ⋅
If we introduce the parallel axis theorem:
( )2
2
sin
O G
G
I I A OG
yI Aφ
= + ×
⎛ ⎞= + ⋅ ⎜ ⎟
⎝ ⎠
Hence we have:
2 2
2
sinsin
G
G
I AyDAy
IyAy
φφ
+= ⋅
= +
Hence, the centre of pressure, point C, always lies below the centroid of the area, G.
PAGE 36
Fluid Mechanics
Plane Surface Properties
PAGE 37
Fluid Mechanics
Plane Surfaces – Example
Problem
Calculate the forces on the hinges supporting the canal gates as shown. The hinges
are located 0.6 m from the top and bottom of each gate.
Plan
Elevation
PAGE 38
Fluid Mechanics
Solution
We will consider gate AB, but all arguments will equally apply to gate BC.
The length of the gate is 3.0 sin30 3.464 mL = = . The resultant pressure on the gate
from the high water side is:
( )1 1 1
3 4.510 9.81 3.464 4.52
344 kN
P gA yρ=
= × × × ×
=
Similarly for the low water side:
( )2 2 2
3 3.010 9.81 3.464 3.02
153 kN
P gA yρ=
= × × × ×
=
The net resultant force on the gate is:
1 2 344 153 191 kNP P P= − = − =
To find the height at which this acts, take moments about the bottom of the gate:
1 1 2 2
4.5 3344 153 363 kNm3 3
Ph Ph P h= +
= × − × =
Hence:
363 1.900 m191
h = =
Examining a free-body diagram of the gate, we see that the interaction force between
the gates, BR , is shown along with the total hinge reactions, AR and the net applied
hydrostatic force, P. Relevant angles are also shown. We make one assumption: the
PAGE 39
Fluid Mechanics
interaction force between the gates acts perpendicular on the contact surface between
the gates. Hence BR acts vertically downwards on plan.
From statics we have Moments about 0A =∑ :
( )sin30 02
12 2
B
B
B
LP R L
PR
R P
⋅ + =
⋅ =
=
Hence 191 kNBR = and the component of BR perpendicular to the gate is 95.5 kN.
By the sum of forces perpendicular to the gate, the component of AR perpendicular to
the gate must also equal 95.5 kN. Further, taking the sum of forces along the gate, the
components of both AR and BR must balance and so 191 kNA BR R= = .
The resultant forces AR and BR must act at the same height as P in order to have
static equilibrium. To find the force on each hinge at A, consider the following figure:
PAGE 40
Fluid Mechanics
Taking moments about the bottom hinge:
( ) ( )
( ),
,
0.6 6 0.6 0.6 0
191 1.900 0.651.7 kN
4.8
A A top
A top
R h R
R
− − − − =
−= =
And summing the horizontal forces:
, ,
, 191 51.7 139.3 kNA A top A btm
A btm
R R R
R
= +
= − =
It makes intuitive sense that the lower hinge has a larger force. To design the bolts
connecting the hinge to the lock wall the direct tension and shear forces are required.
Calculate these for the lower hinge.
(Ans. T = 120.6 kN, V = 69.7 kN)
PAGE 41
Fluid Mechanics
Curved Surfaces
For curved surfaces the fluid pressure on the infinitesimal areas are not parallel and
so must be combined vectorially. It is usual to consider the total horizontal and
vertical force components of the resultant.
Surface Containing Liquid
Consider the surface AB which contains liquid as shown below:
• Horizontal Component
Using the imaginary plane ACD we can immediately see that the horizontal
component of force on the surface must balance with the horizontal force ACF .
Hence:
Force on projection of surfaceonto a vertical planexF =
PAGE 42
Fluid Mechanics
xF must also act at the same level as ACF and so it acts through the centre of
pressure of the projected surface.
• Vertical Component
The vertical component of force on the surface must balance the weight of liquid
above the surface. Hence:
Weight of liquid directly above the surfaceyF =
Also, this component must act through the centre of gravity of the area ABED,
shown as G on the diagram.
• Resultant
The resultant force is thus:
2 2x yF F F= +
This force acts through the point O when the surface is uniform into the page, at
an angle of:
1tan y
x
FF
θ −=
to the horizontal. Depending on whether the surface contains or displaces water
the angle is measured clockwise (contains) or anticlockwise (displaces) from the
horizontal.
PAGE 43
Fluid Mechanics
Surface Displacing Liquid
Consider the surface AB which displaces liquid as shown below:
• Horizontal Component
Similarly to the previous case, the horizontal component of force on the surface
must balance with the horizontal force EBF . Hence again:
Force on projection of surfaceonto a vertical planexF =
This force also acts at the same level as EBF as before.
• Vertical Component
In this case we imagine that the area ABDC is filled with the same liquid. In this
case yF would balance the weight of the liquid in area ABDC. Hence:
PAGE 44
Fluid Mechanics
Weight of liquid which would lie above the surfaceyF =
This component acts through the centre of gravity of the imaginary liquid in area
ABDC, shown as G on the diagram.
The resultant force is calculated as before.
Both of these situations can be summed up with the following diagram:
PAGE 45
Fluid Mechanics
Curved Surfaces – Example
Problem
Determine the resultant force and its direction on the gate shown:
Solution
The horizontal force, per metre run of the gate, is that of the surface projected onto a
vertical plane of length CB:
( )3 610 9.81 6 12
176.6 kN
x CB CBF gA yρ=
⎛ ⎞= × × × × ⎜ ⎟⎝ ⎠
=
And this acts at a depth 2 6 4 m3
h = ⋅ = from the surface. The vertical force is the
weight of the imaginary water above AB:
23 610 9.81 1
4277.4 kN
yF π⎛ ⎞= × ×⎜ ⎟
⎝ ⎠=
PAGE 46
Fluid Mechanics
In which 2 4Rπ is the area of the circle quadrant. The vertical force is located at:
4 4 6 2.55 m3 3
Rxπ π
×= = =
to the left of line BC. The resultant force is thus:
2 2
2 2176.6 277.4328.8 kN
x yF F F= +
= +=
And acts at an angle:
1
1
tan
277.4tan176.6
57.5
y
x
FF
θ −
−
=
=
=
measured anticlockwise to the horizontal. The resultant passes through point C. Also,
as the force on each infinitesimal length of the surface passes through C, there should
be no net moment about C. Checking this:
Moments about 0176.6 4 277.4 2.55 0
706.4 707.4 0
C =
× − × =− ≈
∑
The error is due to rounding carried out through the calculation.
PAGE 47
Fluid Mechanics
Problems – Fluid Action on Surfaces
1. You are in a car that falls into a lake to a depth as shown below. What is the
moment about the hinges of the car door (1.0 × 1.2 m) due to the hydrostatic
pressure? Can you open the door? What should you do?
(Ans. 50.6 kNm, ?, ?)
2. A sluice gate consist of a quadrant of a circle of radius 1.5 m pivoted at its
centre, O. When the water is level with the gate, calculate the magnitude and
direction of the resultant hydrostatic force on the gate and the moment required
to open the gate. The width of the gate is 3 m and it has a mass of 6 tonnes.
(Ans. 61.6 kN, 57˚, 35.3 kNm)
PAGE 48
Fluid Mechanics
3. The profile of a masonry dam is an arc of a circle, the arc having a radius of 30
m and subtending an angle of 60˚ at the centre of curvature which lies in the
water surface. Determine: (a) the load on the dam in kN/m length; (b) the
position of the line of action to this pressure.
(Ans. 4280 kN/m, 19.0 m)
4. The face of a dam is curved according to the relation 2 2.4y x= where y and x
are in meters, as shown in the diagram. Calculate the resultant force on each
metre run of the dam. Determine the position at which the line of action of the
resultant force passes through the bottom of the dam.
(Ans. 1920 kN, 14.15 m)
PAGE 49
Fluid Mechanics
Introduction
Hydrostatics involves only a few variables: ρ , g, and h, and so the equations
developed are relatively simple and experiment and theory closely agree. The study
of fluids in motion is not as simple and accurate. The main difficulty is viscosity.
By neglecting viscosity (an ideal fluid), we do not account for the shear forces which
oppose flow. Based on this, reasonably accurate and simple theories can be derived..
Using experimental results, these theories can then be calibrated by using
experimental coefficients. They then inherently allow for viscosity.
As we will be dealing with liquids, we will neglect the compressibility of the liquid.
This is incompressible flow. This is not a valid assumption for gases.
Classification of Flow Pattern
There are different patterns of fluid flow, usually characterized by time and distance:
• Time: A flow is steady if the parameters describing it (e.g. flow rate, velocity,
pressure, etc.) do not change with time. Otherwise a flow is unsteady.
• Distance: A flow is uniform if the parameters describing the flow do not
change with distance. In non-uniform flow, the parameters change from point
to point along the flow.
From these definitions almost all flows will be one of:
MODULE – III
FLUID KINEMATICS
PAGE 50
Fluid Mechanics
Steady uniform flow
Discharge (i.e. flow rate, or volume per unit time) is constant with time and the cross
section of the flow is also constant. Constant flow through a long straight prismatic
pipe is an example.
Steady non-uniform flow
The discharge is constant with time, but the cross-section of flow changes. An
example is a river with constant discharge, as the cross section of a river changes
from point to point.
Unsteady uniform flow
The cross-section is constant but the discharge changes with time resulting in
complex flow patterns. A pressure surge in a long straight prismatic pipe is an
example.
Unsteady non-uniform flow
Both discharge and cross section vary. A flood wave in a river valley is an example.
This is the most complex type of flow.
Visualization
To picture the motion of a fluid, we start by examining the motion of a single fluid
‘particle’ over time, or a collection of particles at one instant. This is the flow path of
the particle(s), or a streamline:
PAGE 51
Fluid Mechanics
At each point, each particle has both velocity and acceleration vectors:
A streamline is thus tangential to the velocity vectors of the particles. Hence:
• there can be no flow across a streamline;
• therefore, streamlines cannot cross each other, and;
• once fluid is on a streamline it cannot leave it.
We extend this idea to a collection of paths of fluid particles to create a streamtube:
Streamlines and streamtubes are theoretical notions. In an experiment, a streakline is
formed by injecting dye into a fluid in motion. A streakline therefore approximates a
streamline (but is bigger because it is not an individual particle).
PAGE 52
Fluid Mechanics
Dimension of Flow
Fluid flow is in general three-dimensional in nature. Parameters of the flow can vary
in the x, y and z directions. They can also vary with time. In practice we can reduce
problems to one- or two-dimensional flow to simplify. For example:
One dimensional flow
A two-dimensional streamtube
Flow over an obstruction
PAGE 53
Fluid Mechanics
Fundamental Equations
To develop equations describing fluid flow, we will work from relevant fundamental
physical laws.
The Law of Conservation of Matter
Matter cannot be created nor destroyed (except in a nuclear reaction), but may be
transformed by chemical reaction. In fluids we neglect chemical reactions and so we
deal with the conservation of mass.
The Law of Conservation of Energy
Energy cannot be created nor destroyed, it can only be transformed from one form to
another. For example, the potential energy of water in a dam is transformed to kinetic
energy of water in a pipe. Though we will later talk of ‘energy losses’, this is a
misnomer as none is actually lost but transformed to heat and other forms.
The Law of Conservation of Momentum
A body in motion remains in motion unless some external force acts upon it. This is
Newton’s Second Law:
Rate of change
Force =of momentum
( )d mvF
dtdvmdt
ma
=
=
=
To apply these laws to fluids poses a problem, since fluid is a continuum, unlike rigid
bodies. Hence we use the idea of a ‘control volume’.
PAGE 54
Fluid Mechanics
Control Volume
A control volume is an imaginary region within a body of flowing fluid, usually at
fixed location and of a fixed size:
It can be of any size and shape so we choose shapes amenable to simple calculations.
Inside the region all forces cancel out, and we can concentrate on external forces. It
can be picture as a transparent pipe or tube, for example.
PAGE 55
Fluid Mechanics
4.2 The Continuity Equation
Development
Applying the Law of Conservation of Mass to a control volume, we see:
Rate of mass Rate of mass Rate of mass
= +entering leaving increase
For steady incompressible flow, the rate of mass increase is zero and the density of
the fluid does not change. Hence:
Rate of mass Rate of mass
=entering leaving
The rate of mass change can be expressed as:
Rate of mass Fluid Volume
=change density per second
×
Using Q for flow rate, or volume per second (units: m3/s, dimensions: L3T-1):
in outQ Qρ ρ=
And as before, assuming that the flow is incompressible:
in outQ Q=
PAGE 56
Fluid Mechanics
Consider a small length of streamtube:
The fluid at 1-1 moves a distance of s vt= to 2-2. Therefore in 1 second it moves a
distance of v. The volume moving per second is thus:
Q Av=
Thus, for an arbitrary streamtube, as shown, we have:
1 1 2 2Av A v=
PAGE 57
Fluid Mechanics
A typical application of mass conservation is at pipe junctions:
From mass conservation we have:
1 2 3
1 1 2 2 3 3
Q Q QAv A v A v
= +
= +
If we consider inflow to be positive and outflow negative, we have:
No. of Nodes
1
0i ii
Av=
=∑
PAGE 58
Fluid Mechanics
Mass Conservation – Example
Problem
Water flows from point A to points D and E as shown. Some of the flow parameters
are known, as shown in the table. Determine the unknown parameters.
Section Diameter
(mm)
Flow Rate
(m3/s)
Velocity
(m/s)
AB 300 ? ?
BC 600 ? 1.2
CD ? 3 42Q Q= 1.4
CE 150 4 30.5Q Q= ?
PAGE 59
Fluid Mechanics
Solution
From the law of mass conservation we can see:
1 2Q Q=
And as total inflow must equal total outflow:
1
3 4
3 3
3
0.51.5
outQ QQ QQ Q
Q
=
= += +
=
We must also work out the areas of the pipes, 2
4i
idA π
= . Hence:
A1 = 0.0707 m3 A2 = 0.2827 m3 A4 = 0.0177 m3
Starting with our basic equation, Q Av= , we can only solve for 2Q from the table:
( )( )2
3
0.2827 1.2
0.3393 m /
Q
s
=
=
We know that 1 2Q Q= and so we can now calculate 3Q from previous:
1 3
313
1.50.3393 0.2262 m /s
1.5 1.5
Q QQQ
=
= = =
PAGE 60
Fluid Mechanics
And so,
334
0.2262 0.1131 m /s2 2
QQ = = =
Thus we have all the flows. The unknown velocities are:
11
1
0.3393 4.8 m/s0.0707
QvA
= = =
44
4
0.1131 6.4 m/s0.0177
QvA
= = =
And lastly, the diameter of pipe CD is:
233
3
0.2262 0.1616 m1.4
QAv
= = =
33
4 0.454 mAdπ
= =
And so it is likely to be a 450 mm ∅ pipe.
Note that in a problem such as this the individual calculations do not pose a problem.
It is the strategy used to solve it that is key. In this problem, we started from some
knowns and one equation. Even though we couldn’t see all the way to the end from
Step 1, with each new calculation another possibility opened up. This is the ‘art of
problem solving’ and it can only be learned by practice!
PAGE 61
Fluid Mechanics
The real behaviour of fluids flowing is well described by an experiment carried out
by Reynolds in 1883. He set up the following apparatus:
The discharge is controlled by the valve and the small ‘filament’ of dye (practically a
streamline) indicates the behaviour of the flow. By changing the flow Reynolds
noticed:
• At low flows/velocities the filament remained intact and almost straight. This
type of flow is known as laminar flow, and the experiment looks like this:
MODULE – IV
FLUID DYNAMICS
PAGE 62
Fluid Mechanics
• At higher flows the filament began to oscillate. This is called transitional flow
and the experiment looks like:
• Lastly, for even higher flows again, the filament is found to break up
completely and gets diffused over the full cross-section. This is known as
turbulent flow:
Reynolds experimented with different fluids, pipes and velocities. Eventually he
found that the following expression predicted which type of flow was found:
Re vlρµ
=
In which Re is called the Reynolds Number; ρ is the fluid density; v is the average
velocity; l is the characteristic length of the system (just the diameter for pipes), and;
µ is the fluid viscosity. The Reynolds Number is a ration of forces and hence has no
units.
Flows in pipes normally conform to the following:
• Re 2000< : gives laminar flow;
PAGE 63
Fluid Mechanics
• 2000 Re 4000< < : transitional flow;
• Re 4000> : turbulent flow.
These values are only a rough guide however. Laminar flows have been found at
Reynolds Numbers far beyond even 4000.
For example, if we consider a garden hose of 15 mm diameter then the limiting
average velocity for laminar flow is:
( ) ( )3
3
Re
10 0.0152000
0.55 100.073 m/s
vl
v
v
ρµ
−
=
=×
=
This is a very low flow and hence we can see that in most applications we deal with
turbulent flow.
The velocity below which there is no turbulence is called the critical velocity.
PAGE 64
Fluid Mechanics
Characteristics of Flow Types
For laminar flow:
• Re < 2000;
• ‘low’ velocity;
• Dye does not mix with water;
• Fluid particles move in straight lines;
• Simple mathematical analysis possible;
• Rare in practical water systems.
Transitional flow
• 2000 < Re < 4000
• ‘medium’ velocity
• Filament oscillates and mixes slightly.
Turbulent flow
• Re > 4000;
• ‘high’ velocity;
• Dye mixes rapidly and completely;
• Particle paths completely irregular;
• Average motion is in the direction of the flow;
• Mathematical analysis very difficult - experimental measures are used;
• Most common type of flow.
PAGE 65
Fluid Mechanics
4.3 The Energy Equation
Development
We apply the Law of Conservation of Energy to a control volume. To do so, we must
identify the forms of energy in the control volume. Consider the following system:
The forms of energy in this system are:
• Pressure energy:
The pressure in a fluid also does work by generating force on a cross section
which then moves through a distance. This is energy since work is energy.
• Kinetic energy:
This is due to the motion of the mass of fluid.
• Potential energy:
This is due to the height above an arbitrary datum.
PAGE 66
Fluid Mechanics
Pressure Energy
The combination of flow and pressure gives us work. The pressure results in a force
on the cross section which moves through a distance L in time tδ . Hence the pressure
energy is the work done on a mass of fluid entering the system, which is:
1 1m A Lρ=
And so the pressure energy at the entry is:
1 1PrE pAL p A L= =
Kinetic Energy
From classical physics, the kinetic energy of the mass entering is:
2 21 1 1
1 1KE2 2
mv A Lvρ= =
Potential Energy
The potential energy of the mass entering, due to the height above the datum is:
1 1 1PE mgz A Lgzρ= =
Total Energy
The total energy at the entry to the system is just the sum:
* 21 1 1 1 1 1 1 1 1
12
H p A L A Lv A Lgzρ ρ= + +
Be careful to
distinguish the
density ρ and the
pressure p.
PAGE 67
Fluid Mechanics
Final Form
It is more usual to consider the energy per unit weight, and so we divide through by
1 1mg gA Lρ= :
*1
11 1
21 1 1 1 1 1 1 1
1 1 1 1 1 12
1 11
1
12
2
HHgA L
p A L A Lv A LgzgA L gA L gA L
p v zg g
ρ
ρ ρρ ρ ρ
ρ
=
= + +
= + +
Similarly, the energy per unit weight leaving the system is:
2
2 22 2
2 2p vH z
g gρ= + +
Also, the energy entering must equal the energy leaving as we assume the energy
cannot change. Also, assuming incompressibility, the density does not change:
1 2
2 21 1 2 2
1 22 2
H H
p v p vz zg g g gρ ρ
=
+ + = + +
And so we have Bernoulli’s Equation:
2 2
1 1 2 21 2 constant
2 2p v p vz z Hg g g gρ ρ
+ + = + + = =
PAGE 68
Fluid Mechanics
Comments
From Bernoulli’s Equation we note several important aspects:
1. It is assumed that there is no energy taken from or given to the fluid between
the entry and exit. This implies the fluid is frictionless as friction generates
heat energy which would be an energy loss. It also implies that there is no
energy added, say by a pump for example.
2. Each term of the equation has dimensions of length, L, and units of metres.
Therefore each term is known as a head:
• Pressure head: pgρ
;
• Kinetic or velocity head: 2
2vg
;
• Potential or elevation head: z .
3. The streamtube must have very small dimensions compared to the heights
above the datum. Otherwise the height to the top of a cross-section would be
different to the height to the bottom of a cross-section. Therefore, Bernoulli’s
Equation strictly only applies to streamlines.
We have derived the equation from energy considerations. It can also be derived by
force considerations upon an elemental piece of fluid.
PAGE 69
Fluid Mechanics
Energy Equation – Example
Problem
For the siphon shown, determine the discharge and pressure heads at A and B given
that the pipe diameter is 200 mm and the nozzle diameter is 150 mm. You may
neglect friction in the pipe.
PAGE 70
Fluid Mechanics
Solution
To find the discharge (or flow) apply Bernoulli’s Equation along the streamline
connecting points 1 and 2. To do this note:
• Both 1p and 2p are at atmospheric pressure and are taken to be zero;
• 1v is essentially zero.
2 21 1 2 2
1 20 00
22
1 2
2 2
2
p v p vz zg g g g
vz zg
ρ ρ= ==
⎛ ⎞⎛ ⎞ ⎛ ⎞+ + = + +⎜ ⎟⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠⎝ ⎠
− =
Hence, from the figure:
22
2
1.22 0.152 9.815.18 m/s
v
v
+ =×
=
And using continuity:
( )2 2 2
2
3
0.155.18
40.092 m /
Q A v
s
π
=
= ⋅
=
For the pressure head at A, apply Bernoulli’s equation from point 1 to A:
2 2
1 11
0 02 2A A
Ap v p vz zg g g gρ ρ= =
⎛ ⎞⎛ ⎞+ + = + +⎜ ⎟⎜ ⎟
⎝ ⎠ ⎝ ⎠
PAGE 71
Fluid Mechanics
Hence:
( )2
1 2A A
Ap vz zg gρ
= − −
Again using continuity between point 2 and A and the diameter of the pipe at A:
( )
2
2
0.092
0.20.092
42.93 m/s
A
A A
A
A
Q QA v
v
v
π
==
⋅ =
=
Hence the kinetic head at A is just 2
0.44 m2
Avg
= , and so:
2.44 0.44
2.88 m
Apgρ
= − −
= −
This is negative gauge pressure indicating suction. However, it is still a positive
absolute pressure.
Similarly to A, at B we have, B Av v= and 1 1.22 mBz z− = and so:
( )2
1 21.22 0.440.78 m
B BB
p vz zg gρ
= − −
= −=
PAGE 72
Fluid Mechanics
Discharge Measurement in Pipelines
We consider two kinds of meters based on constricting the flow: the Venturi meter
and the Orifice meter, as shown.
The constriction in these meters causes a difference in pressure between points 1 and
2, and it is this pressure difference that enables the discharge to be measured.
Applying the energy equation between the inlet and the constriction:
2 2
1 1 2 2
2 2p v p vg g g gρ ρ
+ = +
Thus the difference in height in the piezometer is:
2 2
1 2 1 2
2p p v vh
g gρ− −
= =
And from continuity 1 1 2 2Q Av A v= = , and using 1 2k A A= we get:
PAGE 73
Fluid Mechanics
( )2 2 2 2 2 22 1 2 1 1v A A v k v= =
And also:
2 21 22v gh v= +
Which after substituting for 22v and rearranging gives:
1 2
21
ghvk
=−
Hence the discharge is:
1 1
1 2
21
Q Av
ghAk
=
=−
This equation neglects all losses. The actual discharge requires the introduction of the
coefficient of discharge, dC :
1 2
21actual d
ghQ C Ak
=−
For properly designed Venturi meters, dC is about 0.97 to 0.99 but for the Orifice
meter it is much lower at about 0.65.
PAGE 74
Fluid Mechanics
4.4 The Momentum Equation
Development
We consider again a general streamtube:
In a given time interval, tδ , we have:
1 1
2 2
momentum enteringmomentum leaving
Q t vQ t v
ρ δρ δ
==
From continuity we know 1 2Q Q Q= = . Thus the force required giving the change in
momentum between the entry and exit is, from Newton’s Second Law:
( )d mvF
dt=
( )
( )
2 1
2 1
Q t v vF
tQ v v
ρ δδ
ρ
−=
= −
This is the force acting on a fluid element in the direction of motion. The fluid exerts
an equal but opposite reaction to its surroundings.
PAGE 75
Fluid Mechanics
Application – Fluid Striking a Flat Surface
Consider the jet of fluid striking the surface as shown:
The velocity of the fluid normal to the surface is:
cosnormalv v θ=
This must be zero since there is no relative motion at the surface. This then is also the
change in velocity that occurs normal to the surface. Also, the mass flow entering the
control volume is:
Q Avρ ρ=
Hence:
PAGE 76
Fluid Mechanics
( )
( )( )2
cos
cos
d mvF
dtAv v
Av
ρ θ
ρ θ
=
=
=
And if the plate is perpendicular to the flow then:
2F Avρ=
Notice that the force exerted by the fluid on the surface is proportional to the velocity
squared. This is important for wind loading on buildings. For example, the old wind
loading code CP3: Chapter V gives as the pressure exerted by wind as:
20.613 sq v= (N/m2)
In which sv is the design wind speed read from maps and modified to take account of
relevant factors such as location and surroundings.
PAGE 77
Fluid Mechanics
Application – Flow around a bend in a pipe
Consider the flow around the bend shown below. We neglect changes in elevation
and consider the control volume as the fluid between the two pipe joins.
The net external force on the control volume fluid in the x-direction is:
1 1 2 2 cos xp A p A Fθ− +
In which xF is the force on the fluid by the pipe bend (making it ‘go around the
corner’). The above net force must be equal to the change in momentum, which is:
( )2 1cosQ v vρ θ −
Hence:
( )( )
( ) ( )
1 1 2 2 2 1
2 1 1 1 2 2
2 2 2 1 1 1
cos cos
cos cos
cos
x
x
p A p A F Q v v
F Q v v p A p A
Qv p A Qv p A
θ ρ θ
ρ θ θ
ρ θ ρ
− + = −
= − − +
= + − +
PAGE 78
Fluid Mechanics
Similarly, for the y-direction we have:
( )( )
( )
2 2 2
2 2 2
2 2 2
sin sin 0
sin 0 sin
sin
y
y
p A F Q v
F Q v p A
Qv p A
θ ρ θ
ρ θ θ
ρ θ
− + = −
= − +
= +
The resultant is:
2 2x yF F F= +
And which acts at an angle of:
1tan y
x
FF
θ −=
This is the force and direction of the bend on the fluid. The bend itself must then be
supported for this force. In practice a manhole is built at a bend, or else a thrust block
is used to support the pipe bend.
PAGE 79
Fluid Mechanics
Application – Force exerted by a firehose
Problem
A firehose discharges 5 l/s. The nozzle inlet and outlet diameters are 75 and 25 mm
respectively. Calculate the force required to hold the hose in place.
Solution
The control volume is taken as shown:
There are three forces in the x-direction:
• The reaction force RF provided by the fireman;
• Pressure forces PF : 1 1p A at the left side and 0 0p A at the right hand side;
• The momentum force MF caused by the change in velocity.
So we have:
M P RF F F= +
The momentum force is:
PAGE 80
Fluid Mechanics
( )2 1MF Q v vρ= −
Therefore, we need to establish the velocities from continuity:
( )
3
1 21
5 100.075 4
1.13 m/s
QvA π
−×= =
=
And
( )
3
2 25 100.025 4
10.19 m/s
vπ
−×=
=
Hence:
( )( )( )
2 1
3 310 5 10 10.19 1.13
45 N
MF Q v vρ−
= −
= × −
=
The pressure force is:
1 1 0 0PF p A p A= −
If we consider gauge pressure only, the 0 0p = and we must only find 1p . Using
Bernoulli’s Equation between the left and right side of the control volume:
PAGE 81
Fluid Mechanics
2 2
1 1 0 0
02 2p v p vg g g gρ ρ =
⎛ ⎞+ = +⎜ ⎟
⎝ ⎠
Thus:
( )
( )
2 21 1 0
32 2
2
2
10 10.19 1.132
51.28 kN/m
p v vρ⎛ ⎞= −⎜ ⎟⎝ ⎠⎛ ⎞
= −⎜ ⎟⎝ ⎠
=
Hence
( ) ( )1 1 0 0
23 0.075
51.28 10 04
226 N
PF p A p A
π
= −
⎛ ⎞= × −⎜ ⎟⎜ ⎟
⎝ ⎠=
Hence the reaction force is:
45 226181 N
R M PF F F= −= −= −
This is about a fifth of an average body weight – not inconsequential.
PAGE 82
INTRODUCTION:
Many practical real flow problems in fluid mechanics can be solved by using equations and analytical procedures. However, solutions of some real flow problems depend heavily on experimental data.
Sometimes, the experimental work in the laboratory is not only time-consuming, but also expensive.
So, the main goal is to extract maximum information from fewest experiments. In this regard,
dimensional analysis is an important tool that helps in correlating analytical results with experimental data and to predict the prototype behavior from the measurements on the model.
Dimensions and Units:
Dimensional analysis is a mathematical technique which makes use of the study of dimensions as an
aid to the solution of several engineering problems. It deals with the dimensions of the physical
quantities are measured by comparison, which is made with respect to an arbitrarily fixed value.
• Length L, mass M and Time T are three fixed dimensions which are of importance in fluid mechanics. If in any problem of fluid mechanics, heat is involved then temperature is also
taken as fixed dimension.
• These fixed dimensions are called fundamental dimensions or fundamental quantity. The
following are the Fundamental Dimensions (MLT)
• Mass - kg - M
• Length - m - L
• Time - s - T
• Secondary dimensions are those quantities which possess more than one fundamental
dimension. 1. Geometric dimensions
• Area - m2 - L2
• Volume - m3 - L3
2. Kinematic dimensions
• Velocity - m/s - L
T or L.T-1
• Acceleration - m/s2 - L
T2 or L.T-2
3. Dynamic dimensions
• Force -N - ML
T or M.L.T-1
• Density -kg/m3 - M/L3 or M.L-3
Use of Dimensional Analysis:
1. Conversion from one dimensional unit to another
2. Checking units of equations (Dimensional Homogeneity) 3. Defining dimensionless relationship using
• Rayleigh’s Method
• Buckingham’s π-Theorem
4. Model Analysis
MODULE – V
DIMENSIONAL ANALYSIS
PAGE 83
Dimensional Homogeneity:
Dimensional Homogeneity means the dimensions in each equation on both sides equal.
Consider the following equation – 𝑉 = √2 𝑋 𝑔 𝑋 ℎ
Dimensions of L. H. S = 𝑉 = 𝐿
𝑇= 𝐿 𝑋 𝑇−1
Dimensions of R. H. S = √2 𝑋 𝑔 𝑋 ℎ = √𝐿
𝑇−2 𝑋 𝐿 = 𝐿 𝑋 𝑇−1
⸫ Dimensions of L. H. S = Dimensions of R. H. S
Hence the above equation is said to be dimensionally homogenous as Dimensions of L. H. S and R. H. S are equal.
Methods of dimensional analysis:
Rayleigh’s Method:
To define relationship among variables. This method is used for determining the expression for a variable which depends upon maximum three or four variables only.
Methodology:
1. Let X is a function of X1, X2, and X3 and mathematically it can be written as
X = f (X1, X2, X3)
2. This can be also written as X = K (X1a , X2b , X3c ) where K is constant and a, b and c are arbitrarily powers.
3. The values of a, b and c are obtained by comparing the powers of the fundamental dimension
on both sides.
Buckingham’s π-Theorem:
This method of analysis is used when number of variables are more.
Theorem:
If there are n variables in a physical phenomenon and those n variables contain m dimensions, then
variables can be arranged into (n – m) dimensionless groups called Φ terms.
Explanation:
1. If f (X1, X2, X3, ……… Xn) = 0 and variables can be expressed using m dimensions then, f (π1, π2, π3, ……… πn - m) = 0, where, π1, π2, π3, … are dimensionless groups.
2. Each π term contains (m + 1) variables out of which mare of repeating type and one is of non
– repeating type. 3. Each π term being dimensionless, the dimensional homogeneity can be used to get each π
term. π denotes a non-dimensional parameter.
Selecting Repeating Variables:
1. Avoid taking the quantity required as the repeating variable.
2. Repeating variables put together should not form dimensionless group.
3. No two repeating variables should have same dimensions. 4. Repeating variables can be selected from each of the following properties.
• Geometric property → Length, height, width, area
• Flow property → Velocity, Acceleration, Discharge
• Fluid property → Mass density, Viscosity, Surface tension
PAGE 84
Model Analysis:
For predicting the performance of the hydraulic structures (such as dams, spillways etc.) or hydraulic
machines (such as turbines, pumps etc.) before actually constructing or manufacturing, models of the
structures or machines are made and tests are conducted on them to obtain the desired information. Model is a small replica of the actual structure or machine. The actual structure or machine is called
as Prototype. Models can be smaller or larger than the Prototype. Model Analysis is actually an
experimental method of finding solutions of complex flow problems.
Similitude or Similarities:
Similitude is defined as the similarity between the model and prototype in every aspect, which means that the model and prototype have similar properties.
Types of Similarities:
• Geometric Similarity →Length, Breadth, Depth, Diameter, Area, Volume etc.,
• Kinematic Similarity → Velocity, Acceleration etc.,
• Dynamic Similarity → Time, Discharge, Force, Pressure Intensity, Torque, Power
Geometric Similarity:
The geometric similarity is said to be exist between the model and prototype if the ratio of all
corresponding linear dimensions in the model and prototype are equal. Mathematically,
Kinematic Similarity
The kinematic similarity is said exist between model and prototype if the ratios of velocity and
acceleration at corresponding points in the model and at the corresponding points in the prototype are
the same. Also the directions of the velocities in the model and prototype should be same.
Dynamic Similarity
The dynamic similarity is said exist between model and prototype if the ratios of corresponding forces
acting at the corresponding points are equal.
It means for dynamic similarity between the model and prototype, the dimensionless numbers should be same for model and prototype.
P P
m
V a and V ar rV am
where is Velocity RatioVr
where is Acceleration Ratioa r
= =
P
m
r
FFr
F
where is Force RatioF
=
P P
m
P P P
r
m m m m
2 3VB AL D and and L L Lr rVL B D A
where is Scale RatioLr
= = = = =
PAGE 85
Forces acting on moving fluid:
Following are the different types of forces to be considered when the liquid is in motion.
• Inertia force, Fi
• Viscous force, Fv
• Gravity force, Fg
• Pressure force, Fp
• Surface tension force, Fs
• Elastic force, Fe
1. Inertia Force, Fi :
• It is the product of mass and acceleration of the flowing fluid and acts in the direction opposite to the direction of acceleration.
• It always exists in the fluid flow problems.
2. Viscous Force, Fv:
• It is equal to the product of shear stress due to viscosity and surface area of the flow.
• It is important in fluid flow problems where viscosity is having an important role to play.
3. Gravity Force, Fg:
• It is equal to the product of mass and acceleration due to gravity of the flowing fluid.
• It is present in case of open surface flow.
4. Pressure force, Fp:
• It is equal to the product of pressure intensity and cross-sectional area of flowing fluid.
• It is present in case of pipe-flow.
5. Surface Tension Force, Fs:
• It is equal to the product of surface tension and length of surface of the flowing fluid.
6. Elastic Force, Fe:
• It is equal to the product of elastic stress and area of the flowing fluid
Dimensionless Numbers:
Dimensionless numbers are obtained by dividing the inertia force by viscous force or gravity force or
pressure force or surface tension force or elastic force. Following are some of the dimensionless numbers described below. The detailed applications of these dimensionless numbers shall be
discussed detailed in further sections.
e
FR
oeyn
cold’
is num
ober, R
Inertia r e VL VDor
V sc us Force
=
=
u
iE
ule
or’s
cnumbe
rr, E
Inert a F r e V
P ess r
ure Fo ce p /==
e
Inertia Force V
Surface Tension Force / LWeber’s number, W ==
e
Inertia Force V
Gravity Force LgFroude’s number, F ==
PAGE 86
Model Laws:
The laws on which the models are designed for dynamic similarity are called model laws or laws of
similarity.
• Reynold’s Model Law
• Froude Model Law.
• Euler Model Law.
• Weber Model Law
• Mach Model Law
1. Reynold’s Model Law:
If the viscous forces are predominant, the models are designed for dynamic similarity based on
Reynold’s number. Models based on Reynolds’s Number includes:
• Pipe Flow
• Resistance experienced by Sub-marines, airplanes, fully immersed bodies etc.
• Mathematically, it can be expressed as:
•
2. Froude Model Law:
If the gravity force is predominant, the models are designed for dynamic similarity based on Froude number. Froude Model Law is applied in the following fluid flow problems:
• Free Surface Flows such as Flow over spillways, Weirs, Sluices, Channels etc.,
• Flow of jet from an orifice or nozzle
• Where waves are likely to formed on surface
• Where fluids of different densities flow over one another
3. Euler Model Law:
Euler Model Law is applied in the following cases:
• Closed pipe in which case turbulence is fully developed so that viscous forces are
negligible and gravity force and surface tension is absent where phenomenon of
cavitation’s takes place.
4. Weber Model Law:
Weber Model Law is applied in the following cases:
• Capillary rise in narrow passages
• Capillary movement of water in soil
• Capillary waves in channels
• Flow over weirs for small heads
Inertia Force V
Elastic Force CMach’s number, M = =
m Pm Pm P
m p
R Re e m p
ρ ρV VL L
=
=
e rm p
pm
m pm p
FF
VV
X L X Lg g
=
=
PAGE 87
5. Mach Model Law:
Mach Model Law is applied in the following cases:
• Flow of aero plane and projectile through air at supersonic speed ie., velocity more than
velocity of sound.
• Aero dynamic testing
• Underwater testing of torpedoes, and
• Water-hammer problems
Problems:
1. Find Dimensions for the following:
• Stress / Pressure
• Work
• Power
• Kinetic Energy
• Dynamic Viscosity
• Kinematic Viscosity
• Surface Tension
• Angular Velocity
• Momentum
• Torque
2. Check Dimensional Homogeneity of the following:
• Q = AV
• EK = v2/2g 3. Water flowing through a pipe of diameter 30 cm at a velocity of 4 m/s. Find the velocity of oil
flowing in another pipe of diameter 10cm, if the conditions of dynamic similarity is satisfied
between two pipes. The viscosity of water and oil is given as 0.01 poise and 0.025 poise. The
specific gravity of oil is 0.8. 4. In 1 in 40 model of a spillway, the velocity and discharge are 2 m/s and 2.5 m3/s. Find
corresponding velocity and discharge in the prototype
5. In a 1 in 20 model of stilling basin, the height of the jump in the model is observed to be 0.20m. What is height of hydraulic jump in the prototype? If energy dissipated in the model is 0.1kW,
what is the corresponding value in prototype?
6. A 7.2 m height and 15 m long spillway discharges 94 m3/s discharge under a head of 2m. If a 1:9 scale model of this spillway is to be constructed, determine the model dimensions, head over
spillway model and the model discharge. If model is experiencing a force of 7500 N, determine
force on the prototype.
7. A Dam of 15 m long is to discharge water at the rate of 120 cumecs under a head of 3 m. Design a model, if supply available in the laboratory is 50 lps
8. A 1:50 spillway model has a discharge of 1.5 cumecs. What is the corresponding discharge in
prototype?. If a flood phenomenon takes 6 hour to occur in the prototype, how long it should take in the model.
PAGE 88