strictly following syllabus of KME 101T of AKTU for first yr 2021 fluid properties, bernoulli's equation with proof and numericals , pumps, turbine , hydraulic lift, continuity equation

1. Maharana pratap group of institutions
UNIT 2
Introduction to fluid mechanics and applications

2. DEFINE FLUID
 A fluid (or) liquid, which is capable of flowing.
It has no own shape, but confirms to the shape of
the containing vessels.
A fluid is a substance that
continually deforms under an applied shear stress
Liquids are like water, milk, air, steam.
MATTER EXISTS IN TWO STATES:
solids and the fluids.
fluids state being commonly divided into the
liquid and gaseous states.

3. DIFFERENCES BETWEEN SOLIDS
AND FLUIDS
 A solid is generally own shape and change in volume
under pure compressive load.
 It resistance to change in shape without a change in
volume under the application of tangential forces.
 The spacing and latitude of motion of molecules are
very small in solids, large in a liquid and extremely
large in gas.
 The intermolecular bonds are very strong in solids,
weak in liquids and very weak in gases.
 Solids are very compact and rigid. Solids materials
are steel, wood, plastics etc.

6. FLUID MECHANICS
 Fluid mechanics is that branch of science which deals
with the behavior of fluids (liquids or gases) at rest as
well as in motion.
 This branch of science deals with the static,
kinematics and dynamic aspects of fluids.
 The study of fluids at rest is called fluid statics.
 The study of fluids in motion, where pressure forces
are not considered, is called fluid kinematics.
 The pressure forces are also considered for the fluids
in motion, that branch of science is called fluid
dynamics.

11. UNITS AND DIMENSIONS
 The word dimensions are used to describe basic concepts like mass,
length, time, temperature and force. Units are the means of expressing
the value of the dimension quantitatively or numerically.
 All physical quantities are measured by units.
There are two types of units:
(i). Fundamental units.
(ii). Derived units.
FUNDAMENTALUNITS.
All physical quantities are expressed the following :
1.Length(L)
2.Mass(M)
3.Time(T)
DERIVED UNITS.
 Derived units are expressed in terms of fundamental units, this are area,
velocity, pressure etc.

12. SYSTEM OF UNITS
CGS Units
 The fundamental units of length, mass and time are taken as
centimeter, gram and second respectively.
FPS Units
 The fundamental units of length, mass and time are taken as
feet, pound and second respectively.
MKS Units
 In this system, the fundamental units of length, mass and time
are taken as meter, kilogram, and seconds respectively.
 The MKS units are called as gravitational units or engineers
units.
SI Units
 This system has six basic units, two supplementary units and
twenty seven derived units.

13. S.I. - Six Basic Units
Quantity SI Unit Dimension
Length Metre, m L
Mass Kilogram, kg M
Time Second, s T
Temperature Kelvin, K
Current Ampere, A I
Luminosity Candela Cd

14. Two Supplementary Units
One is for measuring the plane angle called
radian(rad).
Another for measuring solid angle called
stearadian(Sr).

15. Derived Units
Quantity SI Unit
Volume m3
Area m2
velocity m/s
Discharge m3 /s
acceleration m/s2
force N
Torque, energy,
work
Joule J (or) N m
power Watt W
pressure ( or stress) N/m2
density kg /m3
Dynamic viscosity N s/m2
surface tension N/m
Kinematic viscosity m2/s

16. DIFFERENT TYPES OFFLUIDS
Basically the fluids are classified into 5 types
and these are
1. Ideal fluid
2. Real fluid
3. Newtonian fluid
4. Non-Newtonian fluid, and
5. Ideal plastic fluid

18. Ideal Fluid
 A fluid which is incompressed and have no viscosity falls in the category of ideal
fluid.
 Ideal fluid is not found in actual practice but it is an imaginary fluid because all
the fluid that exist in the environment have some viscosity. there in no ideal fluid in
reality.
Real Fluid
 A fluid which has at least some viscosity is called real fluid.
 Actually all the fluids existing or present in the environment are called real fluids..
Newtonian Fluid
 If a real fluid obeys the Newton's law of viscosity (i.e the shear stress is directly
proportional to the rate of shear strain) then it is known as the Newtonian fluid.
 Example: water, kerosene
Non-Newtonian Fluid
 If real fluid does not obeys the Newton's law of viscosity then it is called Non-
Newtonian fluid.
 Example: paint, toothpaste
Ideal Plastic Fluid
 A fluid having the value of shear stress more than the yield value and shear stress
is proportional to the shear strain (velocity gradient) is known as ideal plastic fluid.

19. PROPERTIES OF FLUIDS
Density (or) Mass Density:(ρ)
 Density or mass density of a fluid is defined as the ratio
of the mass of a fluid to its volume.
 Thus mass per unit volume of a fluid is called density.
ρ = Mass of fluid / Volume of fluid
 Its units ,kg/m3
 Temperature increase with density decrease
 Pressure increase with density increase
Example - Water = 1000 kg/m3, Mercury = 13600 kg/m3,
Air = 1.23 kg/m3, Paraffin Oil = 800 kg/m3(at pressure
=1.013 N/m2, and Temperature = 288.15K.)

20. Specific weight or weight density:(w)
 Specific weight or weight density of a fluid is the ratio
between the weight of a fluid to its volume.
 The weight per unit volume of a fluid is called Specific
weight or weight density
 It various from place to place because of acceleration due
to gravity changing from place to place.
 Specific weight, w = Weight of fluid / Volume of fluid
(w=W/V = mg/V = ρg)
 w = ρg
 Its units, N/m3
 Temperature increase with specific weight decrease
 Pressure increase with specific weight increase
 Water =9810 N/m3, Mercury = 132943 N/m3, Air =12.07
N/m3, Paraffin Oil =7851 N/m3

21. Specific Gravity (or) Relative Density :(S)
 Specific gravity is defined as the ratio of the density of a
fluid to density of a standard fluid.
 S = density of a fluid / density of a standard fluid
( mostly water in case of liquid).
For example -Specific gravity of mercury is 13.6

24. Viscosity
 Viscosity is the property of a fluid, due to cohesion and
interaction between
molecules, which offers resistance to sheer deformation.
 Different fluids deform at different rates under the same shear
stress.
 Fluid with a high viscosity such as syrup, deforms more slowly
than fluid with a low viscosity such as water.
 Shear stress,

25. Viscosity is defined as the property of a fluid which offers resistance to the
movement of one layer of fluid over adjacent layer of the fluid.
When two layers of a fluid, a distance ‘dy’apart, move one over the other at different
velocities, say u and u+du.
The viscosity together with relative velocity causes a shear stress acting between the
fluid layers.
The top layer causes a shear stress on the adjacent lower layer while the lower layer
causes a shear stress on the adjacent top layer.
This shear stress is proportional to the rate of change of velocity with respect to y.

26. Dynamic Viscosity (µ):
 Its defined as the Shear stress(τ), required causing
unit rate of shear deformation(du/dy).

 Its units, N-s/m2
µ = τ /(du/dy).
(or) kg/m-s (or) poise
Kinematic Viscosity (ν):
 Its defined as the ratio of dynamic viscosity to mass
density.
 Its units, m2/s (or) stoke. 1 m2/s = 104 stokes
1 N-s/m2 = 10 poise

27. BULK MODULUS
 It define as the ratio of change in pressure to the rate of
change of volume is called as bulk modulus of the material.
 Bulk modulus (K) = (change in pressure) /(volumetric strain)
 K = -(dp/(dV/V))
 V
olumetric strain is the change in volume divided by the
original volume. (dV/V)
 Negative sign for dV indicates the volume decreases as
pressure increases.
 K = dp/(dρ/ρ) [dV/V = - dρ/ρ]
 Typical values of Bulk Modulus:
• K = 2.05 x 109 N/m2 forwater
• K = 1.62 x 109 N/m2 foroil.

28. COMPRESSIBILITY
 The compressibility of a fluid is the reduction of the volume of the
fluid due to an external pressure acting on it.
 A compressible fluid will reduce (or) change in volume in the
presence of external pressure.
 Compressibility is the reciprocal of the bulk modulus of elasticity, K
which is defined as the ratio of compressive stress to volumetric
strain.
 Compressibility is given by = 1/K
 Its unit in N/m2
 In nature all the fluids are compressible. Gases are highly
compressible but liquid s are not highly compressible.

29. Pascal’s law

31. Hydraulic system
 Pascal’s s law is applicable in hydraulic system.

33. Proof of pascal’s law in 2D

38. Mass of fluid entering tube in a given time
must equal mass of fluid exiting tube…
m1 m2
The Continuity Equation –
Conservation of Mass

39. A2v2 t
V1 V2
A1 X1 A2 X2
A1v1 t
A1 v1 A2 v2 OR… Av Constant
(V = Volume in this step)
(Substitute x = …..?)
(v = Velocity in this step)

40. • Av Constant is often called the
units of
• As A ↑, v
AnApplication:
“Volume Flow Rate” equation because the
term Av is equivalent to Volume/time, with SI
m3
s
↓.
As velocity
increases,
streamlines
get closer.

41. Bernoulli’s equation

45. pressure head = p/ρg
Velocity head =
𝑈2
2𝑔
Potential head = z
Bernoulli’s equation

46. Restrictions/ assumptions in application of Bernoulli’s
equation:
1.Flow is steady
2.Density is constant (incompressible)
3.Friction losses are negligible
4. It relates the states at two points along a single streamline, (not conditions on two
different streamlines)
All these conditions are impossible to satisfy at any
instant in time so Bernoulli’s equation is applicable to ideal fluid only. Therefore we
use Bernoulli’s equation for real fluid. And we use modified equation for real fluids
𝑃1
𝜌𝑔
+
𝑣1
2
2𝑔
+ 𝑧1 =
𝑃2
𝜌𝑔
+
𝑣2
2
2𝑔
+ 𝑧2 + 𝑙𝑜𝑠𝑠 𝑜𝑓 ℎ𝑒𝑎𝑑

47. A1 X1
• Wnet = W1 – W2 =
• F1 does positive work…W1 =F1 x1 P
1 A1 x1
x2 P2 A2 x2
• F2 does negative work..W2=F2
A2 X2 V
P
1 V P2 V
Bernoulli’s Equation – Conservation of
Energy

48. The net work on the fluid, then, must equal the
Wnet =
change in energy of the fluid.
K U

49. DERIVATION-

53. Problem on Bernoulli’s equation for real fluids
Solution -
Ans

54. APPLICATION OF BERNOULLI’S
EQUATION
• Venturi meter
• Orifice meter
• Pitot tube

55. VENTURIMETER

57. VENTURI METER

59. ORIFICE METER

61. ORIFICE METER

62. PITOT TUBE

65. Hydraulic turbines

67. CLASSIFICATION OF TURBINES
1. According to the type of energy available at inlet.
(i). Impulse Turbine: If the energy available at the inlet of the turbine is only kinetic
energy,
the turbine is known as impulse turbine.Ex. Pelton turbine
(ii). Reaction Turbine: If the energy available at the inlet of the turbine is kinetic
energy as
well as pressure energy, the turbine is known as reaction turbine. Ex. -Francis
turbine,
Kaplan turbine.
2. According to the Head at the Inlet of Turbine
(i). High Head Turbine: The net head varies in this turbine is from 150 m to 2000 m
or even
more. It requires small quantity of water. Eg: pelton turbine.
(ii). Medium Head Turbine: in this turbine, the net head varies from 30 m to 150 m.
It
requires moderate quantity of water. Eg: Francis turbine.
(iii). Low Head Turbine: In low head turbines, the net head is less than 30 m. it
requires
large quantity of water. Eg: Kaplan turbine.
3. According to the direction of flow
(i). Tangential Flow Turbine: If the water flows along the tangent of the runner, the
Turbine
is known as tangential flow turbine. For Example: Pelton turbine.
(ii). Radial Flow Turbine: If the water flows in the radial direction through the runner,
the
turbine is called radial flow turbine. Ex- Francis turbine

69. Working Principle of a Turbine
A fast moving fluid (it may be water, gas, steam or wind) is made to strike on the
blades of
the turbine. As the fluid strikes the blades, it rotates the runner. Here the energy of
the
moving fluid is converted into rotational energy. A generator is coupled with the
shaft of the turbine. With the rotation of the runner of the turbine, the shaft of the
generator also rotates. The generator converts the mechanical energy of the
runner into electrical energy.
Main Parts of a pelton wheel Turbine
The main parts of a turbine are
1. Nozzle: It guides the steam to flow in designed direction and velocity.
2. Runner: it is the rotating part of the turbine and blades are attached to the
runner.
3. Blades: It is that part of the turbine on which the fast moving fluid strikes and
rotates the runner.
4. Casing: It is the outer air tight covering of the turbine which contains the runner
and blades. It protects the internal parts of the turbine.
5. Pen stock : Pen stock is a solid pipe which take water from the dam(structure which
store water on its one side) and take it to the turbine at very high speed.
6. Spear -A spear is provided with in the pen stock and its main function is to increase
and
decrease the speed of water entering into the turbine. If more water is needed spear
move
back and allow more to enter into the turbine and if less water is needed spear moves

70. ROTODYNAMIC MACHINES
• The device in which the fluid is in continuous motion
and imparts energy conversion is known as fluid
machines.
• The fluid machines are pumping the fluid from lower
level to higher level.
• The hydraulic machines which converts the
mechanical energy into hydraulic energy are called
pumps.
• These fluid machines are called rotodynamic
machines (or) turbomachines.
• The energy transfer takes place between fluid and
rotodynamic element.
PUMPS

71. CLASSIFICATION OF ROTODYNAMIC
MACHINES
The energy transfer takes place between fluid
and rotodynamic element, the main category of
fluid machines are
Hydraulic turbines.
Pumps.

72. PUMPS
• The rotating element does work on the fluid, it is
called pump. These elements add energy to the fluid.
• Pump converts the mechanical energy to hydraulic
energy by lifting water to a higher level.
• The hydraulic energy refers both potential and kinetic
energies of a liquid.
• The lifting of water to a higher level is carried out by
the various action of pumps such as centrifugal action
and reciprocating action.

73. TYPES OF PUMP
Rotodynamic pump:
Increase in energy level is due to combination of
Centrifugal energy, pressure energy and kinetic
energy.
Positive displacement pump:
Liquid is sucked and then displaced due to the thrust
Exerted on it by a moving member that results in the
lifting of liquid to a desired height.

74. TYPES OF PUMP
There are two main categories of pump:
 Rotodynamic pumps.
 Positive displacement pumps.
Reciprocating
Rotary
Rotodynamic
Centrifugal
Axial flow
Mixed flow
Positive displacement
PUMP

75. CENTRIFUGAL PUMP
The mechanical energy is converted into pressure
energy by centrifugal force acting on the fluid, the
hydraulic machine is called centrifugal pumps.
The centrifugal pumps acts as a reverse of an inward
radial flow reaction turbine.
The flow in centrifugal pumps is in the radial outward
directions.

76. MAIN PARTS OF CENTRIFUGAL
PUMP
1. Impeller.
2. Casing.
3. Suction pipe with a foot valve and a strainer.
4. Delivery pipe.

77. IMPELLER
The rotating part of a centrifugal pump is called
impeller.
It consists of a series of backward curved vanes.
The impeller is mounted on a shaft which is
connected to the shaft of an electrical motor.

78. TYPES OF IMPELLER

79. CASING
It is an airtight chamber, which accommodates the
rotating impeller.
The area of flow of the casing gradually increases in
the direction of flow of water to convert kinematic
energy into pressure energy.
The following three types of casing are commonly
adopted:
1.Volute casing.
2.Vortex casing.
3.Diffuser (or) casing with guide blades

80. WORKING OF CENTRIFUGAL
PUMP

81. WORKING OF CENTRIFUGAL PUMP
Works on the principle that when a certain mass of
fluid is rotated by an external source, it is thrown
away from the central axis of rotation and a
centrifugal head is impressed which enables it to rise
to a higher level.
In order to start a pump, it has to be filled with water
so that the centrifugal head developed is sufficient to
lift the water from the sump. The process is called
priming.
Pump is started by electric motor to rotate the
impeller.
Rotation of impeller in casing full of water produces
forced vortex which creates a centrifugal head on the
liquid.
The delivery valve is opened as the centrifugal head
is impressed.

82. Pascal’s
principle
82
Hydraulic lift

83. Problem 1
83
 The area of A1 = 10 cm2, The area of A2 = 100 cm2
 Force 2 (F2) = 100 Newton, Calculate F1 by using
pascal’s law.
Solution - As per pascal’s law p1 = p2
F1/A1 = F2/A2
F1/10 = 100/100
F1 = 10 N
Problem- 1

84. Problem 2
 If the area of A1 = 0.001 m2 and the area of A2 = 0.1 m2,
external input force F1 = 100 N, then the external output
force F2 ?
84
Problem -2
Solution - P1 = P2
F1/A1 = F2/A2
100/0.001 = F2/0.1
F2 = 10000 N

85. Problem 3
 Car’s weight = 16,000 N. What is the external input force
F?
85
Solution - P1 = P2
F1/A1 = F2/A2
F/50 = 16000/4000
F = 200 N
Problem -3

86. Problem 4
 Area of A is 60 cm2 and
area of B is 4,200 cm2,
determine the external input
force of F
.
86
Problem -4
Answer - F = 50 N

87. Problem 5
 The hydraulic lift has a large cross section and a small
cross section. Large cross-sectional area is 20 times the
small cross-sectional area. If on the small cross section is
given an input force of 25 N, then determine the output
force.
87
Solution - P1 = P2
F1/A1 = F2/A2
25/A = F2/20A
F2 = 500 N

88. Prepared by department of mechanical engineering.
For any querries contact to your subject faculty
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