The document outlines a comprehensive syllabus for a B.Tech degree course in Fluid Mechanics at Mahatma Gandhi University, covering key concepts such as fluid properties, fluid motion kinematics, and equations of motion. It addresses topics like viscosity, pressure measurements, Bernoulli's theorem, and flow through pipes, including laminar and turbulent flow. The syllabus also delves into advanced concepts like non-Newtonian fluids, surface tension, capillarity, and the principles of hydrodynamics.

1. FLUID MECHANICS

2. 2
B.TECH. DEGREE COURSE
SCHEME AND SYLLABUS
(2002-03 ADMISSION ONWARDS)
MAHATMA GANDHI UNIVERSITY
KOTTAYAM
KERALA
Module 1
Introduction - Proprties of fluids - pressure, force, density, specific weight,
compressibility, capillarity, surface tension, dynamic and kinematic viscosity-
Pascal’s law-Newtonian and non-Newtonian fluids-fluid statics-measurement of
pressure-variation of pressure-manometry-hydrostatic pressure on plane and curved
surfaces-centre of pressure-buoyancy-floation-stability of submerged and floating
bodies-metacentric height-period of oscillation.
Module 2
Kinematics of fluid motion-Eulerian and Lagrangian approach-classification and
representation of fluid flow- path line, stream line and streak line. Basic
hydrodynamics-equation for acceleration-continuity equation-rotational and
irrotational flow-velocity potential and stream function-circulation and vorticity-
vortex flow-energy variation across stream lines-basic field flow such as uniform
flow, spiral flow, source, sink, doublet, vortex pair, flow past a cylinder with a
circulation, Magnus effect-Joukowski theorem-coefficient of lift.
Module 3
Euler’s momentum equation-Bernoulli’s equation and its limitations-momentum and
energy correction factors-pressure variation across uniform conduit and uniform
bend-pressure distribution in irrotational flow and in curved boundaries-flow through
orifices and mouthpieces, notches and weirs-time of emptying a tank-application of
Bernoulli’s theorem-orifice meter, ventury meter, pitot tube, rotameter.

3. 3
Module 4
Navier-Stoke’s equation-body force-Hagen-Poiseullie equation-boundary layer flow
theory-velocity variation- methods of controlling-applications-diffuser-boundary
layer separation –wakes, drag force, coefficient of drag, skin friction, pressure,
profile and total drag-stream lined body, bluff body-drag force on a rectangular plate-
drag coefficient for flow around a cylinder-lift and drag force on an aerofoil-
applications of aerofoil- characteristics-work done-aerofoil flow recorder-polar
diagram-simple problems.
Module 5
Flow of a real fluid-effect of viscosity on fluid flow-laminar and turbulent flow-
boundary layer thickness-displacement, momentum and energy thickness-flow
through pipes-laminar and turbulent flow in pipes-critical Reynolds number-Darcy-
Weisback equation-hydraulic radius-Moody;s chart-pipes in series and parallel-
siphon losses in pipes-power transmission through pipes-water hammer-equivalent
pipe-open channel flow-Chezy’s equation-most economical cross section-hydraulic
jump.

4. 4
DEFINITION OF FLUID
 A fluid is a substance that deforms continuously in the face of tangential or
shear stress, irrespective of the magnitude of shear stress .This continuous
deformation under the application of shear stress constitutes a flow.
 In this connection fluid can also be defined as the state of matter that cannot
sustain any shear stress.
Example : Consider Fig 1.1
Fig 1.1 Shear stress on a fluid body
If a shear stress τ is applied at any location in a fluid, the element 011' which is
initially at rest will move to 022', then to 033'. Further, it moves to 044' and
continues to move in a similar fashion.
In other words, the tangential stress in a fluid body depends on velocity of
deformation and vanishes as this velocity approaches zero. A good example is
Newton's parallel plate experiment where dependence of shear force on the velocity
of deformation was established.

5. 5
FLUID PROPERTIES:
Characteristics of a continuous fluid which are independent of the motion of the fluid
are called basic properties of the fluid. Some of the basic properties are as discussed
below.
Property Symbol Definition Unit
Density ρ
The density p of a fluid is its mass per unit volume . If a fluid
element enclosing a point P has a volume Δ and mass Δm
(Fig. 1.4), then density (ρ)at point P is written as
However, in a medium where continuum model is valid one
can write -
Fig 1.2 A fluid element enclosing point P
kg/m3

6. 6
Specific
Weight
γ
The specific weight is the weight of fluid per unit volume.
The specific weight is given
by γ= ρ g
Where ‘g’ is the gravitational acceleration. Just as weight
must be clearly distinguished from mass, so must the specific
weight be distinguished from density.
N/m3
Specific
Volume
v
The specific volume of a fluid is the volume occupied by unit
mass of fluid.
Thus m3 /kg
Specific
Gravity
s
For liquids, it is the ratio of density of a liquid at actual
conditions to the density of pure water at 101 kN/m2 , and at
4°C.
The specific gravity of a gas is the ratio of its density to that
of either hydrogen or air at some specified temperature or
pressure.
However, there is no general standard; so the conditions must
be stated while referring to the specific gravity
VISCOSITY (µ):
 Viscosity is a fluid property whose effect is understood when the fluid is in
motion.
 In a flow of fluid, when the fluid elements move with different velocities,
each element will feel some resistance due to fluid friction within the
elements.

7. 7
 Therefore, shear stresses can be identified between the fluid elements with
different velocities.
 The relationship between the shear stress and the velocity field was given by
Sir Isaac Newton.
Consider a flow (Fig. 1.3) in which all fluid particles are moving in the same
direction in such a way that the fluid layers move parallel with different velocities.
Fig 1.3 Parallel flow of a fluid
Fig.1.4 Two adjusent layers of moving fluid

8. 8
 The upper layer, which is moving faster, tries to draw the lower slowly
moving layer along with it by means of a force F along the direction of flow
on this layer. Similarly, the lower layer tries to retard the upper one,
according to Newton's third law, with an equal and opposite force F on it
(Figure 1.4).
 Such a fluid flow where x-direction velocities, for example, change with y-
coordinate is called shear flow of the fluid.
 Thus, the dragging effect of one layer on the other is experienced by a
tangential force F on the respective layers. If F acts over an area of contact A,
then the shear stress τ is defined as
τ = F/A
NEWTON’S LAW OF VISCOSITY:
 Newton postulated that τ is proportional to the quantity Δu/ Δy where Δy is
the distance of separation of the two layers and Δu is the difference in their
velocities.
 In the limiting case of, Δu / Δy equals du/dy, the velocity gradient at a point
in a direction perpendicular to the direction of the motion of the layer.
 According to Newton τ and du/dy bears the relation
Where, the constant of proportionality μ is known as the coefficient of viscosity or
simply viscosity which is a property of the fluid and depends on its state. Sign of τ
depends upon the sign of du/dy. For the profile shown in Fig. 1.4, du/dy is positive
everywhere and hence, τ is positive.

9. 9
Defining the viscosity of a fluid is known as Newton's law of viscosity. Common
fluids, viz. water, air, mercury obey Newton's law of viscosity and are known as
Newtonian fluids.
Other classes of fluids, viz. paints, different polymer solution, and blood do not obey
the typical linear relationship, of τ and du/dy and are known as non-Newtonian
fluids. In non-Newtonian fluids viscosity itself may be a function of deformation rate
as you will study in the next lecture.
KINEMATIC VISCOSITY:
The ratio dynamic viscosity to density of the given fluid is known as kinematic
viscosity.
IDEAL FLUID:
 Consider a hypothetical fluid having a zero viscosity ( μ = 0). Such a fluid is
called an ideal fluid and the resulting motion is called as ideal or inviscid
flow. In an ideal flow, there is no existence of shear force because of
vanishing viscosity.
 All the fluids in reality have viscosity (μ > 0) and hence they are termed as
real fluid and their motion is known as viscous flow.
 Under certain situations of very high velocity flow of viscous fluids, an
accurate analysis of flow field away from a solid surface can be made from
the ideal flow theory.
NON-NEWTONIAN FLUIDS
 There are certain fluids where the linear relationship between the shear stress
and the deformation rate (velocity gradient in parallel flow) as expressed by
 This not valid. For these fluids the viscosity varies with rate of deformation.

10. 10
 Due to the deviation from Newton's law of viscosity they are commonly
termed as non-Newtonian fluids. Figure 1.7 shows the class of fluid for
which this relationship is nonlinear.
Figure 1.7 Shear stress and deformation rate relationship of different fluids
 The abscissa in Fig. 1.7 represents the behaviour of ideal fluids since for the
ideal fluids the resistance to shearing deformation rate is always zero, and
hence they exhibit zero shear stress under any condition of flow.
 The ordinate represents the ideal solid for there is no deformation rate under
any loading condition.
 The Newtonian fluids behave according to the law that shear stress is linearly
proportional to velocity gradient or rate of shear strain . Thus for
these fluids, the plot of shear stress against velocity gradient is a straight line
through the origin. The slope of the line determines the viscosity.
 The non-Newtonian fluids are further classified as pseudo-plastic, dilatant
and Bingham plastic.

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COMPRESSIBILITY
 Compressibility of any substance is the measure of its change in volume
under the action of external forces.
 The normal compressive stress on any fluid element at rest is known as
hydrostatic pressure p and arises as a result of innumerable molecular
collisions in the entire fluid.
 The degree of compressibility of a substance is characterized by the bulk
modulus of elasticity E defined as


Where Δ and Δp are the changes in the volume and pressure respectively,
and is the initial volume. The negative sign (-sign) is included to make E
positive, since increase in pressure would decrease the volume i.e for Δp>0, Δ
<0) in volume.
 For a given mass of a substance, the change in its volume and density
satisfies the relation
m = 0, ρ ) = 0
using
we get

12. 12
 Values of E for liquids are very high as compared with those of gases (except
at very high pressures). Therefore, liquids are usually termed as
incompressible fluids though, in fact, no substance is theoretically
incompressible with a value of E as .
 For example, the bulk modulus of elasticity for water and air at atmospheric
pressure are approximately 2 x 106 kN/m 2 and 101 kN/m 2 respectively. It
indicates that air is about 20,000 times more compressible than water. Hence
water can be treated as incompressible.
 For gases another characteristic parameter, known as compressibility K, is
usually defined , it is the reciprocal of E
SURFACE TENSION OF LIQUIDS
 The phenomenon of surface tension arises due to the two kinds of
intermolecular forces
(i) Cohesion: The force of attraction between the molecules of a liquid by
virtue of which they are bound to each other to remain as one assemblage of
particles is known as the force of cohesion. This property enables the liquid
to resist tensile stress.
(ii) Adhesion: The force of attraction between unlike molecules, i.e. between
the molecules of different liquids or between the molecules of a liquid and
those of a solid body when they are in contact with each other, is known as
the force of adhesion. This force enables two different liquids to adhere to
each other or a liquid to adhere to a solid body or surface.

13. 13
Figure 1.8 the intermolecular cohesive force field in a bulk of liquid with
a free surface
A and B experience equal force of cohesion in all directions, C experiences a
net force interior of the liquid The net force is maximum for D since it is at
surface
 Work is done on each molecule arriving at surface against the action of an
inward force. Thus mechanical work is performed in creating a free surface or
in increasing the area of the surface. Therefore, a surface requires mechanical
energy for its formation and the existence of a free surface implies the
presence of stored mechanical energy known as free surface energy. Any
system tries to attain the condition of stable equilibrium with its potential
energy as minimum. Thus a quantity of liquid will adjust its shape until its
surface area and consequently its free surface energy is a minimum.
 The magnitude of surface tension is defined as the tensile force acting across
imaginary short and straight elemental line divided by the length of the line.
 The dimensional formula is F/L or MT-2. It is usually expressed in N/m in SI
units.
 Surface tension is a binary property of the liquid and gas or two liquids which
are in contact with each other and defines the interface. It decreases slightly

14. 14
with increasing temperature. The surface tension of water in contact with air
at 20°C is about 0.073 N/m.
 It is due to surface tension that a curved liquid interface in equilibrium results
in a greater pressure at the concave side of the surface than that at its convex
side.
CAPILLARITY:
 The interplay of the forces of cohesion and adhesion explains the
phenomenon of capillarity. When a liquid is in contact with a solid, if the
forces of adhesion between the molecules of the liquid and the solid are
greater than the forces of cohesion among the liquid molecules themselves,
the liquid molecules crowd towards the solid surface. The area of contact
between the liquid and solid increases and the liquid thus wets the solid
surface.
 The reverse phenomenon takes place when the force of cohesion is greater
than the force of adhesion. These adhesion and cohesion properties result in
the phenomenon of capillarity by which a liquid either rises or falls in a tube
dipped into the liquid depending upon whether the force of adhesion is more
than that of cohesion or not (Fig.1.9).
 The angle θ as shown in Fig. 1.9 is the area wetting contact angle made by the
interface with the solid surface.
 For pure water in contact with air in a clean glass tube, the capillary rise takes
place with θ = 0 . Mercury causes capillary depression with an angle of
contact of about 1300 in a clean glass in contact with air. Since h varies
 Inversely with D as found from Eq. (), an appreciable capillary rise or
depression is observed in tubes of small diameter only.

15. 15
Fig 1.9 Phenomenon of Capillarity
VAPOUR PRESSURE:
All liquids have a tendency to evaporate when exposed to a gaseous atmosphere. The
rate of evaporation depends upon the molecular energy of the liquid which in turn
depends upon the type of liquid and its temperature. The vapour molecules exert a
partial pressure in the space above the liquid, known as vapour pressure. If the space
above the liquid is confined (Fig. 1.10) and the liquid is maintained at constant
temperature, after sufficient time, the confined space above the liquid will contain
vapour molecules to the extent that some of them will be forced to enter the liquid.
Eventually an equilibrium condition will evolve when the rate at which the number
of vapour molecules striking back the liquid surface and condensing is just equal to
the rate at which they leave from the surface. The space above the liquid then
becomes saturated with vapour. The vapour pressure of a given liquid is a function of
temperature only and is equal to the saturation pressure for boiling corresponding to
that temperature. Hence, the vapour pressure increases with the increase in
temperature. Therefore the phenomenon of boiling of a liquid is closely related to the
vapour pressure. In fact, when the vapour pressure of a liquid becomes equal to the
total pressure impressed on its surface, the liquid starts boiling. This concludes that

16. 16
boiling can be achieved either by raising the temperature of the liquid, so that its
vapour pressure is elevated to the ambient pressure, or by lowering the pressure of
the ambience (surrounding gas) to the liquid's vapour pressure at the existing
temperature.
Fig 1.10
FORCES ON FLUID ELEMENTS:
Fluid Elements - Definition:
Fluid element can be defined as an infinitesimal region of the fluid continuum in
isolation from its surroundings.
Two types of forces exist on fluid elements
 Body Force: distributed over the entire mass or volume of the element. It is
usually expressed per unit mass of the element or medium upon which the
forces act.
Example: Gravitational Force, Electromagnetic force fields etc.
 Surface Force: Forces exerted on the fluid element by its surroundings
through direct contact at the surface.
Surface force has two components:

17. 17
 Normal Force: along the normal to the area
 Shear Force: along the plane of the area.
The ratios of these forces and the elemental area in the limit of the area tending to
zero are called the normal and shear stresses respectively.
The shear force is zero for any fluid element at rest and hence the only surface force
on a fluid element is the normal component.
PASCAL'S LAW OF HYDROSTATICS
PASCAL'S LAW
The normal stresses at any point in a fluid element at rest are directed towards the
point from all directions and they are of the equal magnitude.
Fig 1.11 State of normal stress at a point in a fluid body at rest
Derivation:
The inclined plane area is related to the fluid elements (refer to Fig 3.1) as follows

18. 18
Substituting above values in equation 3.1- 3.3 we get
CONCLUSION:
The state of normal stress at any point in a fluid element at rest is same and directed
towards the point from all directions. These stresses are denoted by a scalar quantity
p defined as the hydrostatic or thermodynamic pressure.
UNITS AND SCALES OF PRESSURE MEASUREMENT:
Pascal (N/m2) is the unit of pressure.
Pressure is usually expressed with reference to either absolute zero pressure (a
complete vacuum) or local atmospheric pressure.
 The absolute pressure: It is the difference between the value of the pressure
and the absolute zero pressure.
 Gauge pressure: It is the diference between the value of the pressure and the
local atmospheric pressure(patm)
 Vacuum Pressure: If p<patm then the gauge pressure becomes
negative and is called the vacuum pressure.But one should always remember
that hydrostatic pressure is always compressive in nature

19. 19
Fig 1.12 The Scale of Pressure
At sea-level, the international standard atmosphere has been chosen as Patm = 101.32
kN/m2
MANOMETERS FOR MEASURING GAUGE AND VACUUM PRESSURE
Manometers are devices in which columns of a suitable liquid are used to measure
the difference in pressure between two points or between a certain point and the
atmosphere.Manometer is needed for measuring large gauge pressures. It is basically
the modified form of the piezometric tube. A common type manometer is like a
transparent "U-tube" as shown in Fig.
Fig 1.13 A simple manometer to measure gauge pressure

20. 20
Fig 1.14 A simple manometer to measure vacuum pressure
One of the ends is connected to a pipe or a container having a fluid (A) whose
pressure is to be measured while the other end is open to atmosphere. The lower part
of the U-tube contains a liquid immiscible with the fluid A and is of greater density
than that of A. This fluid is called the manometric fluid.
The pressures at two points P and Q (Fig. 1.14) in a horizontal plane within the
continuous expanse of same fluid (the liquid B in this case) must be equal. Then
equating the pressures at P and Q in terms of the heights of the fluids above those
points, with the aid of the fundamental equation of hydrostatics, we have
Hence,
Where p1 is the absolute pressure of the fluid A in the pipe or container at its centre
line, and patm is the local atmospheric pressure. When the pressure of the fluid in the
container is lower than the atmospheric pressure, the liquid levels in the manometer
would be adjusted as shown in Fig. 4.5. Hence it becomes,

21. 21
(
MANOMETERS TO MEASURE PRESSURE DIFFERENCE
A manometer is also frequently used to measure the pressure difference, in course of
flow, across a restriction in a horizontal pipe.
Fig 1.15 Manometer measuring pressure difference
The axis of each connecting tube at A and B should be perpendicular to the direction
of flow and also for the edges of the connections to be smooth. Applying the
principle of hydrostatics at P and Q we have,
where, ρ m is the density of manometric fluid and ρw is the density of the working
fluid flowing through the pipe.
We can express the difference of pressure in terms of the difference of heads (height
of the working fluid at equilibrium).

22. 22
INCLINED TUBE MANOMETER:
 For accurate measurement of small pressure differences by an ordinary u-tube
manometer, it is essential that the ratio ρm/ρw should be close to unity. This is
not possible if the working fluid is a gas; also having a manometric liquid of
density very close to that of the working liquid and giving at the same time a
well defined meniscus at the interface is not always possible. For this
purpose, an inclined tube manometer is used.
 If the transparent tube of a manometer, instead of being vertical, is set at an
angle θ to the horizontal (Fig. 1.16), then a pressure difference corresponding
to a vertical difference of levels x gives a movement of the meniscus s =
x/sin𝜃 along the slope.
 Fig 1.16 An Inclined Tube Manometer
 If θ is small, a considerable mangnification of the movement of the meniscus
may be achieved.
 Angles less than 50 are not usually satisfactory, because it becomes difficult
to determine the exact position of the meniscus.
 One limb is usually made very much greater in cross-section than the other.
When a pressure difference is applied across the manometer, the movement

23. 23
of the liquid surface in the wider limb is practically negligible compared to
that occurring in the narrower limb. If the level of the surface in the wider
limb is assumed constant, the displacement of the meniscus in the narrower
limb needs only to be measured, and therefore only this limb is required to be
transparent.
INVERTED TUBE MANOMETER:
For the measurement of small pressure differences in liquids, an inverted U-tube
manometer is used.
Fig 1.17 An Inverted Tube Manometer
Here and the line PQ is taken at the level of the higher meniscus to equate
the pressures at P and Q from the principle of hydrostatics. It may be written that
where represents the piezometric pressure, (z being the vertical height of
the point concerned from any reference datum). In case of a horizontal pipe (z1= z2)
the difference in piezometric pressure becomes equal to the difference in the static
pressure. If is sufficiently small, a large value of x may be obtained for a

24. 24
small value of . Air is used as the manometric fluid. Therefore, is
negligible compared with and hence,
Air may be pumped through a valve V at the top of the manometer until the liquid
menisci are at a suitable level.
HYDROSTATIC THRUSTS ON SUBMERGED PLANE SURFACE:
Due to the existence of hydrostatic pressure in a fluid mass, a normal force is exerted
on any part of a solid surface which is in contact with a fluid. The individual forces
distributed over an area give rise to a resultant force.
INCLINED PLANE SURFACES:
Consider a plane surface of arbitrary shape wholly submerged in a liquid so that the
plane of the surface makes an angle θ with the free surface of the liquid. We will
assume the case where the surface shown in the figure below is subjected to
hydrostatic pressure on one side and atmospheric pressure on the other side.
Fig 1.18 Hydrostatic Thrust on Submerged Inclined Plane Surface

25. 25
Let p denotes the gauge pressure on an elemental area dA. The resultant force F on
the area A is therefore
Where h is the vertical depth of the elemental area dA from the free surface and the
distance y is measured from the x-axis, the line of intersection between the extension
of the inclined plane and the free surface. The ordinate of the centre of area of the
plane surface A is defined as
Hence from the above eqns, we get
Where is the vertical depth (from free surface) of centre c of area.
Implies that the hydrostatic thrust on an inclined plane is equal to the pressure at its
centroid times the total area of the surface, i.e., the force that would have been
experienced by the surface if placed horizontally at a depth hc from the free surface.

26. 26
Fig 1.19 Hydrostatic Thrust on Submerged Horizontal Plane Surface
The point of action of the resultant force on the plane surface is called the centre of
pressure . Let and be the distances of the centre of pressure from the y and
x axes respectively. Equating the moment of the resultant force about the x axis to
the summation of the moments of the component forces, we have
Solving for yp we can write
In the same manner, the x coordinate of the centre of pressure can be obtained by
taking moment about the y-axis. Therefore,

27. 27
From which,
The two double integrals in the numerators are the moment of inertia about the x-axis
Ixx and the product of inertia Ixy about x and y axis of the plane area respectively. By
applying the theorem of parallel axis
Where, and are the moment of inertia and the product of inertia of the
surface about the centroidal axes , and are the coordinates of the
center c of the area with respect to x-y axes.
With the help of above eqns it can be written as
1
2
The first term on the right hand side of the Eq. 1 is always positive. Hence, the centre
of pressure is always at a higher depth from the free surface than that at which the
centre of area lies. This is obvious because of the typical variation of hydrostatic
pressure with the depth from the free surface. When the plane area is symmetrical
about the y' axis, , and .

28. 28
HYDROSTATIC THRUSTS ON SUBMERGED CURVED
SURFACES
On a curved surface, the direction of the normal changes from point to point, and
hence the pressure forces on individual elemental surfaces differ in their directions.
Therefore, a scalar summation of them cannot be made. Instead, the resultant thrusts
in certain directions are to be determined and these forces may then be combined
vectorially. An arbitrary submerged curved surface is shown in Fig. 1.20. A
rectangular Cartesian coordinate system is introduced whose xy plane coincides with
the free surface of the liquid and z-axis is directed downward below the x - y plane.
Fig 1.20 Hydrostatic thrust on a Submerged Curved Surface
Consider an elemental area dA at a depth z from the surface of the liquid. The
hydrostatic force on the elemental area dA is
and the force acts in a direction normal to the area dA. The components of the force
dF in x, y and z directions are

29. 29
Where l, m and n are the direction cosines of the normal to dA. The components of
the surface element dA projected on yz, xz and xy planes are, respectively
Substituting Eqs we can write
Therefore, the components of the total hydrostatic force along the coordinate axes are
Where zc is the z coordinate of the centroid of area Ax and Ay (the projected areas of
curved surface on yz and xz plane respectively). If zp and yp are taken to be the
coordinates of the point of action of Fx on the projected area Ax on yz plane, , we can
write

30. 30
where Iyy is the moment of inertia of area Ax about y-axis and Iyz is the product of
inertia of Ax with respect to axes y and z. In the similar fashion, zp
' and x p
' the
coordinates of the point of action of the force Fy on area Ay, can be written as
Where Ixx is the moment of inertia of area Ay about x axis and Ixz is the product of
inertia of Ay about the axes x and z.
We can conclude from the above eqns that for a curved surface, the component of
hydrostatic force in a horizontal direction is equal to the hydrostatic force on the
projected plane surface perpendicular to that direction and acts through the centre of
pressure of the projected area., the vertical component of the hydrostatic force on the
curved surface can be written as
Where is the volume of the body of liquid within the region extending vertically
above the submerged surface to the free surface of the liquid. Therefore, the vertical
component of hydrostatic force on a submerged curved surface is equal to the weight
of the liquid volume vertically above the solid surface of the liquid and acts through
the center of gravity of the liquid in that volume.

31. 31
Buoyancy
 When a body is either wholly or partially immersed in a fluid, a lift is
generated due to the net vertical component of hydrostatic pressure forces
experienced by the body.
 This lift is called the buoyant force and the phenomenon is called buoyancy
 Consider a solid body of arbitrary shape completely submerged in a
homogeneous liquid as shown in Fig. 1.21. Hydrostatic pressure forces act on
the entire surface of the body.
Fig 1.21 Buoyant Forces on a Submerged Body

32. 32
To calculate the vertical component of the resultant hydrostatic force, the body is
considered to be divided into a number of elementary vertical prisms. The vertical
forces acting on the two ends of such a prism of cross-section dAz (Fig. 1.21) are
respectively
Therefore, the buoyant force (the net vertically upward force) acting on the elemental
prism of volume is -
Hence the buoyant force FB on the entire submerged body is obtained as
Where is the total volume of the submerged body. The line of action of the force
FB can be found by taking moment of the force with respect to z-axis. Thus
Substituting for dFB and FB, the x coordinate of the center of the buoyancy is
obtained as
Which is the centroid of the displaced volume. It is found that the buoyant force FB
equals to the weight of liquid displaced by the submerged body of volume . This
phenomenon was discovered by Archimedes and is known as the Archimedes
principle.

33. 33
ARCHIMEDES PRINCIPLE
THE BUOYANT FORCE ON A SUBMERGED BODY
 The Archimedes principle states that the buoyant force on a submerged body
is equal to the weight of liquid displaced by the body, and acts vertically
upward through the centroid of the displaced volume.
 Thus the net weight of the submerged body, (the net vertical downward force
experienced by it) is reduced from its actual weight by an amount that equals
the buoyant force.
THE BUOYANT FORCE ON A PARTIALLY IMMERSED
BODY
 According to Archimedes principle, the buoyant force of a partially immersed
body is equal to the weight of the displaced liquid.
 Therefore the buoyant force depends upon the density of the fluid and the
submerged volume of the body.
 For a floating body in static equilibrium and in the absence of any other
external force, the buoyant force must balance the weight of the body
STABILITY OF UNCONSTRAINED SUBMERGED BODIES IN
FLUID
 The equilibrium of a body submerged in a liquid requires that the weight of
the body acting through its cetre of gravity should be colinear with an equal
hydrostatic lift acting through the centre of buoyancy.
 In general, if the body is not homogeneous in its distribution of mass over
the entire volume, the location of centre of gravity G does not coincide with
the centre of volume, i.e., the centre of buoyancy B.
 Depending upon the relative locations of G and B, a floating or submerged
body attains three different states of equilibrium-
Let us suppose that a body is given a small angular displacement and then released.
Then it will be said to be in

34. 34
 STABLE EQUILIBRIUM: If the body returns to its original
position by retaining the originally vertical axis as vertical.
 UNSTABLE EQUILIBRIUM: If the body does not return to its original
position but moves further from it.
 NEUTRAL EQUILIBRIUM: If the body neither returns to its original
position nor increases its displacement further, it will simply adopt its new
position.
STABLE EQUILIBRIUM
Consider a submerged body in equilibrium whose centre of gravity is located below
the centre of buoyancy (Fig. 1.22a). If the body is tilted slightly in any direction, the
buoyant force and the weight always produce a restoring couple trying to return the
body to its original position (Fig. 1.22b, 1.22c).
Fig 1.22 A Submerged body in Stable Equilibrium
UNSTABLE EQUILIBRIUM
On the other hand, if point G is above point B (Fig. 1.23a), any disturbance from the
equilibrium position will create a destroying couple which will turn the body away
from its original position (1.23b, 1.23c).

35. 35
Fig 1.23 A Submerged body in Unstable Equilibrium
NEUTRAL EQUILIBRIUM
When the centre of gravity G and centre of buoyancy B coincides, the body will
always assume the same position in which it is placed (Fig 1.24) and hence it is in
neutral equilibrium.
Fig 1.24 A Submerged body in Neutral Equilibrium
Therefore, it can be concluded that a submerged body will be in stable, unstable or
neutral equilibrium if its centre of gravity is below, above or coincident with the
center of buoyancy respectively (Fig. 1.25).

36. 36
Fig 1.25 States of Equilibrium of a Submerged Body
(a) STABLE EQUILIBRIUM (B) UNSTABLE EQUILIBRIUM
(C) NEUTRAL EQUILIBRIUM
STABILITY OF FLOATING BODIES IN FLUID
 When the body undergoes an angular displacement about a horizontal axis,
the shape of the immersed volume changes and so the centre of buoyancy
moves relative to the body.
 As a result of above observation stable equlibrium can be achieved, under
certain condition, even when G is above B.
Figure 1.26a illustrates a floating body -a boat, for example, in its equilibrium
position.
Fig 1.26 A Floating body in Stable equilibrium

37. 37
IMPORTANT POINTS TO NOTE HERE ARE
a. The force of buoyancy FB is equal to the weight of the body W
b. Centre of gravity G is above the centre of buoyancy in the same vertical line.
c. Figure 1.26 b shows the situation after the body has undergone a small
angular displacement 𝜃 with respect to the vertical axis.
d. The centre of gravity G remains unchanged relative to the body (This is not
always true for ships where some of the cargo may shift during an angular
displacement).
e. During the movement, the volume immersed on the right hand side increases
while that on the left hand side decreases. Therefore the centre of buoyancy
moves towards the right to its new position B'.
Let the new line of action of the buoyant force (which is always vertical) through B'
intersects the axis BG (the old vertical line containing the centre of gravity G and the
old centre of buoyancy B) at M. For small values of the point M is practically
constant in position and is known as met centre. For the body shown in Fig. 1.26, M
is above G, and the couple acting on the body in its displaced position is a restoring
couple which tends to turn the body to its original position. If M were below G, the
couple would be an overturning couple and the original equilibrium would have been
unstable. When M coincides with G, the body will assume its new position without
any further movement and thus will be in neutral equilibrium. Therefore, for a
floating body, the stability is determined not simply by the relative position of B and
G, rather by the relative position of M and G. The distance of metacentre above G
along the line BG is known as metacentric height GM which can be written as
GM = BM -BG
Hence the condition of stable equilibrium for a floating body can be expressed in
terms of metacentric height as follows:

38. 38
GM > 0 (M is above G) Stable equilibrium
GM = 0 (M coinciding with G) Neutral equilibrium
GM < 0 (M is below G) Unstable equilibrium
The angular displacement of a boat or ship about its longitudinal axis is known as
'rolling' while that about its transverse axis is known as "pitching".
FLOATING BODIES CONTAINING LIQUID
If a floating body carrying liquid with a free surface undergoes an angular
displacement, the liquid will also move to keep its free surface horizontal. Thus not
only does the centre of buoyancy B move, but also the centre of gravity G of the
floating body and its contents move in the same direction as the movement of B.
Hence the stability of the body is reduced. For this reason, liquid which has to be
carried in a ship is put into a number of separate compartments so as to minimize its
movement within the ship.
PERIOD OF OSCILLATION
The restoring couple caused by the buoyant force and gravity force acting on a
floating body displaced from its equilibrium placed from its equilibrium position is
. Since the torque equals to mass moment of inertia (i.e., second
moment of mass) multiplied by angular acceleration, it can be written
Where IM represents the mass moment of inertia of the body about its axis of
rotation. The minus sign in the RHS of eqn arises since the torque is a retarding one
and decreases the angular acceleration. If θ is small, sin θ=θ and hence Eq. can be
written as

39. 39
The above equation represents a simple harmonic motion. The time period (i.e., the
time of a complete oscillation from one side to the other and back again) equals to
. The oscillation of the body results in a flow of the liquid around
it and this flow has been disregarded here. In practice, of course, viscosity in the
liquid introduces a damping action which quickly suppresses the oscillation unless
further disturbances such as waves cause new angular displacements.

40. 40
MODULE 2
KINEMATICS OF FLUID MOTION
“Fluid kinematics is the branch of fluid mechanics which deals with the
study of velocity and acceleration of the particles of fluid in motion and their
distribution in space without considering the force caring them.”
EULERIAN AND LAGRANGIAN APPROAH:
EULERIAN APPROAH:
This method describes the overall flow characteristics at various points as fluid
particles pass. Thus it gives the various quantities such as velocity, acceleration,
pressure, density etc. at any point in space and time, from a purely mathematical
view point. The eulerian method is preferable as the resulting equation can be solved
easily.
The velocities at any point(x, y, z) can be written as
U=f1(x, y, z, t)
V=f2(x, y, z, t)
W=f3(x, y, z, t)
Identities of the particles are made by specifying their initial position at a
given time. It is analytically represented as
 0s s s , t
r r
Where s
r
is the position vector of a particle at a time with its initial position
0s
r
.
LAGRANGIAN APPROACH:
In this method, the fluid motion is described by tracing the kinematic behavior of
each and every individual particle constituting the flow. Identities are made by
specifying their initial position at a given time. The position of the particle at any
instant of time then becomes a function of its identity and time.

41. 41
Analytical expression of the statement:
(1)
s Is the position vector of a particle (with respect to a fixed point of reference) at a
time t.
Is its initial position at a given time t =t0
Equation (1) can be written into scalar components with respect to a rectangular
Cartesian frame of coordinates as:
x = x(x0, y0, z0, t)
z = z(x0, y0, z0, t)
y = y(x0, y0, z0, t)
Where, x0, y0, z0 are the initial coordinates and x, y, z are the coordinates at a time t
of the particle.
Hence in can be expressed as
, , and are the unit vectors along x, y and z axes respectively.
CLASSIFICATION OF FLUIDS:
The fluid flow is classified as:
a) Steady and unsteady flow.
b) Uniform and non uniform flow.
c) Laminar and turbulent flow.
d) Compressible and incompressible flow.

42. 42
e) One, two, and three dimensional flow.
a) Steady and Unsteady Flow:
A steady flow is defined as a flow in which the various hydrodynamic parameters
such as velocity, pressure, density, etc and fluid properties at any point do not change
with time.
u v p
0; 0, 0
t t t t t
             
             
            
An unsteady Flow is defined as a flow in which the hydrodynamic parameters such
as velocity, pressure, density, etc and fluid properties changes with time.
v p
0 0
t t
    
    
    
b) Uniform and Non-uniform Flow:
Uniform Flow:
The flow is defined as uniform flow when in the flow field the velocity and other
hydrodynamic parameters do not change from point to point at any instant of time
For a uniform flow, the velocity is a function of time only, which can be expressed in
Eulerian description as
Implication:
1. For a uniform flow, there will be no spatial distribution of hydrodynamic and
other parameters.
2. Any hydrodynamic parameter will have a unique value in the entire field,
irrespective of whether it

43. 43
Non-uniform Flow:
When the velocity and other hydrodynamic parameters changes from one point to
another the flow is defined as non-uniform.
Important points:
1. For a non-uniform flow, the changes with position may be found either in the
direction of flow or in directions perpendicular to it.
2. Non-uniformity in a direction perpendicular to the flow is always encountered
near solid boundaries past which the fluid flows.
FOUR POSSIBLE COMBINATIONS
Type Example
1. Steady Uniform flow
Flow at constant rate through a duct of uniform
cross-section (The region close to the walls of
the duct is disregarded)
2. Steady non-uniform flow
Flow at constant rate through a duct of non-
uniform cross-section (tapering pipe)
3. Unsteady Uniform flow
Flow at varying rates through a long straight
pipe of uniform cross-section. (Again the
region close to the walls is ignored.)
4. Unsteady non-uniform
flow
Flow at varying rates through a duct of non-
uniform cross-section.

44. 44
C) laminar and turbulent flow:
Laminar flow:
This term is generally used in pipe flow. A laminar flow is one in which the fluid
particles move in layers or laminar with one layer slide over the other. Therefore
there is no exchange of fluid particles from one layer to the other and hence no
transfer of lateral momentum to the adjacent layers.
Fig.1 laminar flow
As the particles move in definite and observable paths or streamlines this flow is
sometimes known as streamline flow. it is a smooth and regular flow. The
velocity of flow is very small. The dimensionless Reynolds number is an
important parameter in the equations that describe whether flow conditions lead
to laminar or turbulent flow. In the case of flow through a straight pipe with a
circular cross-section, Reynolds numbers of less than 2300 are generally
considered to be of a laminar type .however, the Reynolds number upon which
laminar flows become turbulent is dependent upon the flow geometry. When the
Reynolds number is much less than 1, creeping motion or Stokes flow occurs.
This is an extreme case of laminar flow where viscous (friction) effects are much
greater than inertial forces.
Turbulent flow:
Fig.2 turbulent flow

45. 45
When the flow attains a certain velocity, it no longer remains steady and eddy
currents appear. The velocities in such a flow vary from point to point in
magnitude and direction as well as from instant to instant. All the fluid particles
are disturbed and they mix with each other.
On the basis of Reinhold’s number we can classify Laminar and turbulent
flow.
Re < 2000 flow in pipes is laminar
Re > 4000 flow in pipes is turbulent
Re between 2000 and 4000 flow may be laminar or turbulent.
d) Compressible and incompressible flow:
Compressible flow:
Compressible flow is that type of flow in which the density of the fluid changes from
point to point or in other words the density is not constant for the fluid. Thus
mathematically, for compressible flow
ρ ≠ constant
Incompressible flow:
Incompressible flow is that type of flow in which the density of the fluid is constant
for the fluid flow. Liquids are generally incompressible while gases are
compressible. Mathematically, for compressible flow,
ρ =constant
e) One, two, and three dimensional flow.
Fluid flow is three-dimensional in nature. This means that the flow parameters like
velocity, pressure and so on vary in all the three coordinate directions. Sometimes
simplification is made in the analysis of different fluid flow problems by selecting
the appropriate coordinate directions so that appreciable variation of the hydro
dynamic parameters take place in only two directions or even in only one.

46. 46
One-dimensional flow:
All the flow parameters may be expressed as functions of time and one space
coordinate only. The single space coordinate is usually the distance measured along
the centre-line (not necessarily straight) in which the fluid is flowing.
Mathematically, u = f(x), v = 0,  = 0
Where u, v,  are velocity components in x, y and z directions.
Example:
The flow in a pipe is considered one-dimensional when variations of pressure and
velocity occur along the length of the pipe, but any variation over the cross-section is
assumed negligible.
Fig.3 One-dimensional ideal flow along a pipe
In reality, flow is never one-dimensional because viscosity causes the velocity to
decrease to zero at the solid boundaries. If however, the non uniformity of the actual
flow is not too great, valuable results may often be obtained from “one dimensional
analysis” .The average values of the flow parameters at any given section
(perpendicular to the flow) are assumed to be applied to the entire flow at that
section.
Two-dimensional flow:
All the flow parameters are functions of time and two space coordinates (say x and
y). No variation in z direction. The same streamline patterns are found in all planes
perpendicular to z direction at any instant.

47. 47
Fig.4 Flow past a car antenna is approximately two-dimensional,
Two dimensional flow – flow parameter such as velocity is a function of
time and two rectangular space co-ordinates is called 2-D flow.
Mathematically u = f1 (x, y)
v = f2 (x, y)
w = 0
Three dimensional flow:
The hydrodynamic parameters are functions of three space coordinates and
time. It is that type of flow in which the velocity is a function of time and three
mutually perpendicular directions.
Mathematically
v = f(x, y, z, t) - unsteady flow
v = f(x, y, z) - steady flow
REPRESENTATION OF FLUID FLOW:
Path line:
A path line is the actual path traversed by a given (marked) fluid particle. A path line
is hence also an integrated pattern. A path line represents an integrated history of
where a fluid particle has been. In general it is a curve in a 3 – D space. If the flow is
2D, the curve becomes two dimensional

48. 48
Fig.5 path line
Stream line:
A streamline is a line that is everywhere tangent to the velocity vector at a given
instant of time. A streamline is hence an instantaneous pattern.
Fig.6 stream line
Equation for a streamline
Following points about streamlines are worth noting.
a) A streamline cannot intersect itself or two streamlines cannot cross. Because
the fluid is moving in the same direction as the streamlines, fluid cannot cross

49. 49
a streamline. Streamlines cannot cross each other. If they were to cross, this
would indicate two different velocities at the same point. This is not physically
possible. The above point implies that any particles of fluid starting on one
streamline will stay on that same streamline throughout the fluid
b) There cannot be any movement of the fluid mass across the streamlines.
c) Streamline sparing varies inversely as the velocity converging of streamlines
at any particular direction shows accelerated flow in that direction.
d) Where as path line indicates the path of one particular particle at successive
instants of time, streamline indicates the direction of flow of a number of
particles.
e) Series of streamlines represent the flow pattern at an instant.
In steady flow, streamlines remains invariant with time. The path lines and
streamlines will then be identical. In unsteady flow, pattern of streamline may not be
same.
The mathematical expression of a streamline can also be obtained from
Where V is the fluid velocity vector and dris a tangential vector along the
streamline. The above cross product is zero since the two vectors are in the same
direction. A useful technique in fluid flow analysis is to consider only a part of the
total fluid in isolation from the rest. This can be done by imagining a tubular surface
formed by streamlines along which the fluid flows. This tubular surface is known as
a stream tube, which is a tube whose walls are streamlines. Since the velocity is
tangent to a streamline, no fluid can cross the walls of a stream tube.
Stream tube:
Is a fluid mass bounded by a group of stream lines or is an imaginary tube formed
by a group of stream lines passing through a small closed curve which may or may
not be circular. The contents of stream tube is known as current filament.
Eg:- pipes and nozzles

50. 50
a) Stream tube has finite dimensions
b) Stream surface functions as if it were a solid wall.
c) Shape of a stream tube changes from one instant to another because of
the change in the position of streamlines.
Fig.7 Stream tube
Streak line:
FIG.8 Streak line:

51. 51
A streak line is an instantaneous line whose points are occupied by particles which
have earlier passed through a prescribed point in space. A streak line is hence an
integrated pattern. A streak line can be formed by injecting dye continuously into the
fluid at a fixed point in space. As time marches on, the streak line gets longer and
longer, and represents an integrated history of the dye streak.
Eg:- path taken by smoke coming out of chimney.
For steady flow, streamline, streak line and path line coincides.
For unsteady flow, streamline, streak line and path line are all different
Three basic principles are used in the analysis of the problems of fluid in
motion:
a) Principle of conservation of mass states that mass can neither be created nor
destroyed. On the basis of this principle, the continuity equation is derived.
b) Principle of conservation of energy – States that energy can neither be
destroyed nor be created. On the basis of this principle energy equation is
derived.
c) Principle of conservation of momentum or Impulse momentum principle
states that the impulse of the resultant force, or the product of the force and
time increment during which it cuts, is equal to the change in momentum of
the body. On the basis of this principle momentum equation is derived.
CONTINUITY EQUATION
The continuity equation is a statement of the conservation of mass in a system. A
most general expression on the basis of this principle may be obtained by considering
a fixed region within the flowing fluid. Since the fluid is neither created nor
destroyed in this region, it may be stated that “Rate of increase of fluid mass within
the region must be equal to the rate at which the fluid mass enters the region and the
rate at which the fluid mass leaves the region.”

52. 52
Consider a pipe which is uniform in diameter at both ends, but has a constriction
between the ends as in figure. This is called a venturi tube. Stations 1 and 2 have
cross-sectional areas A1 and A2, respectively. Let V1 and V2 be the average flow
speeds at these cross sections (one-dimensional flow). A further assumption is that
there are no leaks in the pipe nor is fluid being pumped in through the sides. The
continuity equation states that the fluid mass passing station 1 per unit time must
equal the fluid mass passing station 2 per unit time
Fig.9 venturi tube
(Mass rate)1 = (Mass rate)2 (1)
Mass rate = Density x Area x Velocity
This equation (1) reduces to
pl AlV1 = p2A2V2 (2)
Since the fluid is assumed to be incompressible, p [Greek letter rho] is a constant
and equation (2) reduces to
AlV1 = A2V2 (3)

53. 53
This is the simple continuity equation for inviscid, incompressible, steady, one-
dimensional flow with no leaks.
CONTINUITY EQUATION IS CARTESIAN CO-ORDINATES:
Consider a rectangular parallelopiped element with sides of length x, y,
z. Let p(x, y, z) the centre of parallelopiped, where the velocity components in x, y,
z direction v, v1 and  respectively and  be the mass density.
Fig.10 CARTESIAN CO-ORDINATES
Mass of fluid passing per unit time through the face with area y z normal
to x-axis through point  is
(y, y, z)
Mass of fluid flowing per unit time through face ABCD is
 
x
u y z u y z
x 2
   
            
.
Mass of fluid flowing per unit time through force A B C D    is
 
x
u y z u y z
x 2
   
            

54. 54
Net mass of the fluid remaining in the paralleopiped per unit time through
the force ABCD and A B C D    is
   
x x
u y z u y z u y z u y z
x 2 x 2
         
                           
=  u y z x
x

    

=  u y z x
x

    
 (1)
By applying the same method for the faces DCCD and ABBA and
ADDA and BCCB we obtain,
=  u x y z
y

    

(2)
=  y x y z
z

    

(3)
By adding (1), (2) and (3) net total mass of fluid that has remained in the
parallelopiped is
     u v
x y z
x y z
      
      
   
Since fluid is neither created nor destroyed in the paralleopiped, any
increase in the mass of the fluid contained in this space per unit time is equal to the
net total mass of fluid that remained in this parallelopiped per unit time.
The mass of the fluid in parallelopiped is x y z    and its rate of increase
with time is
 x y z x y z
t t
 
       
 
So
     u v 2 w
x y z x y z
x y z t
      
         
    

55. 55
     u v
x y z t
       
    
    
     y v v
0
t x y z
     
   
   
(4)
(4) Represents continuity equation in Cartesian co-ordinates which is
applicable for steady and unsteady flow, uniform and non-uniform flow and
compressible as well as incompressible flow.
For steady flow
t


= 0
So (4) reduces to
     u v w
0
x y z
     
  
  
(5)
For incompressible fluid, 0
x y z t
   
   
   
So (4) reduces to
u v w
0
x y z
  
  
  
(6)
CONTINUITY EQUATION IN CYLINDRICAL POLAR CO-
ORDINATES:
Consider any point p(r, , 2) in space. Let r,  and z be the small
increments in the direction r, , z respectively. So that 1PS r PQ r , PT z      .
Construct an elementary parallelopiped. Lt Vr, V and Vz be the components of
velocity V in the direction of r, , z at point . Let be the mass density of fluid at
point P.

56. 56
Fig.11 CYLINDRICAL POLAR CO-ORDINATES
Consider the forces PSTQ and PSTQ, the mass of fluid entering
paralleopiped per unit time through the face
 PST Q PVz r r    
Mass of the fluid leaving the paralleopiped per unit time through the face
P S TQ  
   2V r r r V r . r z
z

         

Wet mass of fluid that has remained in the parallelopiped per unit time
through this pair of faces.
=  2. V r, r z
z

    

Similarly the net mass of fluid that remains in the parallelopiped per unit
time through the pair of faces PTQ S and P T QS   .
=  r zV ,

    

and that through the pair of faces PQS T  and P Q ST 
=  r zV r , r

    


57. 57
Net total mass of fluid that has remained in the parallelopiped per unit time
through all the three pair of faces =
     r 2
r z
VV . r V
r
r r r z
     
      
   
The mass of fluid in parallelopiped,
=  r z r    
and its rate of increase with time
=  r r
t


  

=  r z r
t


  

Thus equating the two expressions
     r 2
r z
VV . r V
.
r r r z
     
      
   
=  r z r
t

  

Dividing both sides by volume of paralleopiped and taking the limit so as to
reduce the paralleopiped to point P, continuity equation is obtained as
     r 2
r
VV .r V
0
t r r z


    
   
   
EQUATION FOR ACCELERATION:
The Cartesian vector form of a velocity filed can be written as:
The flow filed is the most important variable in the fluid mechanics, i.e., knowledge
of the velocity vector filed is equivalent to solving a fluid flow problem.
The acceleration vector field can be calculated:

58. 58
Where the compact dot products is:
With a similar approach, we obtain the total acceleration vector:
The term is called the local acceleration and defined as the rate of increase
of velocity with respect to time at a given point in the flow field. It vanishes if the
flow is steady.
The three terms in the parentheses are called the convective acceleration and defined
as the rate of change of velocity due to the change of position of fluid particles in a
fluid flow. It rises when the particles move through regions of spatially varying
velocity, e.g. nozzle.

59. 59
ROTATIONAL FLOW:
In this type of flow the fluid element has net rotation about its own axis which is
perpendicular to the plane of motion. For example the fluid in the rotating tank.
Fig: 12 rotational flow
IRROTATIONAL FLOW:
Fig: 13irrotational flow
In this type of flow the fluid element has no net rotation about its own axis which is
perpendicular to the plane of motion. However the fluid element may be deformed
and distorted.
VELOCITY POTENTIAL:
The flow takes place in the pipe if there is a difference of pressure. The direction of
flow will take place from a higher to the lower pressure. Thus the velocity of flow in

60. 60
a certain direction will depend upon the potential difference which is known as
velocity potential.
STREAM FUNCTION:
It is defined as the scalar function of space and time such that its partial derivative
with respect to any direction gives the velocity component at right angles to that
direction. It is denoted by ψ (psi) and defined only for two dimensional flow.
Mathematically for steady flow it is defined as ψ=f (x, y) such that
Properties:
a) If the stream function (ψ) exists, it is a possible case of fluid flow which may
be rotational or irrotational.
b) If stream function (ψ) satisfies the Laplace equation it is possible case of an
irrotational flow.
CIRCULATION:
Circulation for a fluid element as shown in Fig is defined as the line integral of
the tangential component of velocity and is given by
Fig:13 circulation

61. 61
Where ds is an element of contour and the loop through the integral sign signifies
that the contour is closed.
VORTEX FLOW:
Vortex flow is defined as the flow of a fluid along a curved path or the flow of a
rotating mass of fluid .the vortex flow is of two types namely:
a) forced vortex flow
b) Free vortex flow.
FORCED VORTEX FLOW:
Forced vortex flow is defined as that type of vortex flow, in which some external
torque is required to rotate the fluid. The mass in this type of flow rotates at constant
angular velocity ω. The tangential velocity of any fluid particle is given by
V= ω*r
r=radius of particle from the axis of rotation.
Example:
Flow of liquid inside the impeller of a centrifugal pump.
Flow of water through the runner of the turbine.
FREE VORTEX FLOW:
When no external torque is required to rotate the fluid mass that type of flow is called
free vortex flow. Thus the liquid in case of free vortex is rotating due to the rotation
which is imparted to the fluid previously.
Examples:
A whirlpool in a river.
Flow of fluid in a centrifugal pump casing.

62. 62
VORTICITY:
It is defined as the value twice of rotation and hence it is given as=2 ω.
BASIC FLOW:
Uniform flow:
In uniform flow the velocity of the fluid remains constant. All the fluid particles are
moving with the same velocity. The uniform flow may be:
a) Parallel to x- axis.
b) Parallel to y axis.
Fig. 14 uniform flow
Source flow:
Fig.15 source flow
The source flow is the flow coming from a point (source) and moving out radially in
all directions of a plane at uniform rate. The strength of a source is defined as the
volume flow rate per unit depth. The unit of strength of the source is m2/s.it is
represented by q.
Let ur= radial velocity of flow at a radius r from the source.

63. 63
q= volume flow rate per unit depth.
r=radius.
The radial velocity ur at any radius r is given by,
ur =q/(2*π*r)
The above equation shows that with increase of r, the radial velocity decreases. And
at a large distance away from the source, the velocity will be approximately equal to
zero.
Sink flow:
The sink flow is the flow in which fluid moves radially inwards towards a point
where it disappears at a constant rate. The flow is just opposite to the source flow.
The fig shows a sink flow.
Fig.16 sink flow
The pattern of stream lines and equi potential lines of a sink is the same as that of a
source flow. all the equations derived for a source flow shall hold to good for sink
flow also except that in sink flow equations,q is to be replaced by(-q) .
Doublet:
The flow pattern due to a doublet is obtained when a source and a sink of equal
strength q are considered to approach one another under such conditions that their
strengths increase, as the distance between them approaches zero.

64. 64
Fig.17 doublet
FLOW PAST A CYLINDER WITH A CIRCULATION:
Fig.18 flow past a cylinder with a cylinder
A cylinder of infinite length kept across a uniform flow of fluid horizontally, the
stream lines are symmetrical on the top and bottom of the cylinder. For every point
on the upper side of the cylinder there is a corresponding point on the lower side
having the same pressure and velocity and so there can be no lift force on the
cylinder.
If the cylinder is rotated about its longitudinal axis anti-clock wise, the stream lines
will take up a path as shown in the fig. at the upper surface of the cylinder the force
will be slightly greater velocity than at the lower part. The addition of velocity on the
upper half is due to the fact that the rotating cylinder adds a little velocity to the air
adjacent to it. At the same time the rotating surface in the lower part retards the flow.
The velocity at the lower half has a magnitude given by the difference between the
velocity of flow and the velocity imparted due to rotation. The region of lower
velocity will exert more pressure than the region of higher velocity there by
providing a lifting force on the cylinder considered.

65. 65
MAGNUS EFFECT:
When a body (such as a sphere or circular cylinder) is spinning in a fluid, it creates
a boundary layer around itself, and the boundary layer induces a more widespread
circular motion of the fluid. If the body is moving through the fluid with a velocity V
the velocity of the fluid close to the body is a little greater than V on one side, and a
little less than V on the other. This is because the induced velocity due to the
boundary layer surrounding the spinning body is added to V on one side, and
subtracted from V on the other. In accordance with Bernoulli's principle, where the
velocity is greater the fluid pressure is less; and where the velocity is less, the fluid
pressure is greater. This pressure gradient results in a net force on the body, and
subsequent motion in a direction perpendicular to the relative velocity vector
Fig.18 Magnus effect
COEFFICENT OF LIFT:
The lift coefficient (CL or CZ) is a dimensionless coefficient that relates
the lift generated by an airfoil, the dynamic pressure of the fluid flow around the
airfoil, and the plan form area of the airfoil. It may also be described as the ratio of
lift pressure to dynamic pressure.

66. 66
QUESTIONS
1. Find the direction of stream lines and equipotential lines in uniform flow
when uniform flow is along – (1) x – axis (2) y – axis (3) an angle to x-axis.
Case I. Uniform flow along x-axis with constant velocity u.
u U
y x
 
   
 
v 0
x x
 
   
 
Uy and Ux   
Hence stream lines are parallel to x-axis and equipotential lines are parallel
to y-axis.
Case 2. Uniform flow parallel to y axis ie, u = 0 and v = U
u 0
y x
 
  
 
v U
y y
 
   
 
Ux    and Uy 
Hence stream lines are parallel to y-axis and equipotential lines are parallel
to x-axis.
Case 3. A uniform flow inclined at angle ‘’ to x-axis. Such flow can be
considered to flow at velocity U cos  in x-direction and U sin  in y-direction. In
case cylindrical co-ordinates are used, then x = r cos  and y = r sin .
 rU sin     
and  rU cos    
Stream lines and equipotential lines in uniform flow (u = U cos  & v = U sin )

67. 67
2. What is a source flow? What is the source strength? Find the direction of
stream lines and equipotential lines in a source.
Source: In two-dimensional flow, a source is a point from which fluid
discharge radially outwards in all directions.
A point source is plane source (two-dimensional). A line source as
compared to a plane source is three –dimensional. In fact a cross-section of a line
source is a plane point source. The point source remains at the origin ‘O’ and fluid
flows radially outwards from the source. The radial velocity (ur) of the flow at a
distance ‘r’ from the origin is given by:
r
q k
u
2 r r
 

where q = discharge per unit length and k = source strength.
or
q
k
2


Source strength (k). The source strength is a measure of the radial velocity
at unit radius. It is defined as volume flow per unit depth
q
k
2
 
 
 
.
r
k
u
r

when r = 1, then ur = k
Stream function for source flow
r
I
u
r



and u



but r
q
u
2 r


1 q
r 2 r


 
or
q
2
  

or 1
q
C
2
   


68. 68
But 0  when  = 0, hence C1 = 0
q
2

  

The stream function is a function of angle ‘’.
Equipotential function of source flow
rU
r



and
1
U
r




r
q
u
2 r r

  
 
q 1
or . r
2 r
  

q r
2 r

 
 
q
logr
2
 

Hence equipotential function is the function of ‘r’ (the distance from the
source origin)
3. Derive an expression of pressure distribution in source flow
We consider two points in the source flow to derive an expression of
pressure distribution in source flow. First point at r = 1 and other point at large
distance away from the source where radial velocity ur = 0. Now apply Bernoulli’s
equation –
2 2 2
1 1 2 2
1 2
p u p u
z z
g 2g g 2g
    
 
Now as flow is in one plane, hence 1 2z z , P1 = P. ie, pressure at point
r = 1 and u1 = ur. P2 = P0 ie, pressure at large distance from source where u2 = ur = 0.
2
0 rP P u
g 2g




69. 69
2
r
0
u
P P
2

  
Now r
q
u
2 r


2
0
q
P P
2 2 r
  
    
 
2
2 2
q
8 r



Hence the pressure increases inversely proportional to the square of the
radius from the source.
4. What is a sink flow? What is the pressure variation in a sink flow?
A sink is a source with a negative strength. Fluid flows radially inwards in a
plane towards a point where the fluid disappears at a constant rate. For a sink flow –
r
q
u kr
2 r

 

where
q
k
2



= sink strength
The pressure variation in a plane sink flow is given by the same expression
as applicable in the source flow –
2 2
0 2 2 2
k q
P P
2r 8 r
 
  

5.What is the Langrangian method of describing fluid motion?
In Langrangian method, the motion of each particle is defined with respect
to its location in space and time from a fixed position before the start of the motion.
The movement of single particle is observed which gives the path followed by the
particle. The position of the fluid particle in space (x, y, z) at any time from its fixed
position (a, b, c) at t = 0, shall be
x = f1(a, b, c, t)
y = f2 (a, b, c, t)
z = f3 (a, b, c, t)

70. 70
From the position, velocities, u, v and w and acceleration ax, ay, and az can
be given as -
(1) Velocities
x y x
u ,v & w
t t t
  
  
  
(2) Accelerations
2 2 2
x y z2 2 2
x y x
a ,a &a
t t t
  
  
  
The resultant of velocities and acceleration can be calculated. The resultant
of pressure and densities can also be worked out.
6. What is the Eulerian method of describing fluid motion?
In this method, instead of concentrating on single fluid particle and its
movement, concentration is made on all fluid particles passing through a fixed point.
The changes of velocity, acceleration, pressure and density of the fluid particles as
they pass through the fixed point are observed. Hence the Eulerian method describes
the overall flow characteristics at various points as fluid particles pass through them.
The velocity at any point (x, y, z) can be given as
u = f1(x, y, z, t), v = f2(x, y, z, t), w = f3(x, y, z, t)
7. What is the difference between Langrangian and Eulerian method?
Langrangian Eulerian
1. Observer concentrates on the
movement of single particle.
2. Observer has to move with the fluid
particle to observe its movement.
3. The path and changes in velocity,
acceleration, pressure and density of a
single particle are described.
4. Not commonly used.
1. Observer concentrates on various
fixed point particles
2. Observer remains stationary and
observes changes in the fluid
parameters at the fixed point only.
3. The method describes the overall flow
characteristics at various points as
fluid particles pass.
4. Commonly used.

71. 71
8. What are various lines of flow to describe the motion of the fluid?
The various lines of flow are: (1) path line, (2) stream line, (3) stream tube,
(4) streak line or filament line and (5) potential lines.
9. What is a path line in fluid motion? What happens to path line during steady
and unsteady flow?
The path traced or taken by a single fluid particle in motion over a period of
time is called its path line. Hence path line is a locus of a fluid particle as it moves
along. The path line shows the direction of velocity of the particle as it moves ahead.
During steady flow, path line concides with the stream line as there is no fluctuation
in velocity. However the path line fluctuates between different stream lines during an
unsteady flow.
10. What is a stream line?
A stream line is an imaginary line drawn through a flowing fluid such that
the tangent at each point on the line indicates the direction of the velocity of the fluid
particle at that point. For example, if ABC is a stream line, then the direction of
velocities of A, B and is C as shown
11. Explain the following (1) the value of velocity at right angle to the stream
line (2) can there be any flow across the stream line and (3) can two stream lines
cross each other?
Since velocity of the fluid particle at any point on the stream line is
tangential to the stream line, there cannot by any component of velocity normal or
right angle to the stream line i.e. the component of velocity at right angle to the
stream line is zero.
There cannot be any flow across the stream line as the flow is always
tangential to the stream line.
Two stream lines cannot cross each other as otherwise there would be two
velocities at that point, one each tangential to the stream lines. This is inconsistent
with the definition of a stream line.

72. 72
12. How does a stream line behave in the vicinity of a solid surface?
The stream line in the vicinity of a solid surface conforms to the outline of
the boundary surface. For example, the stream lines ariund a solid cylinder and
within a closed conduit are as shown in the figure
13. What happens to stream lines during steady and steady flow conditions?
In steady flow, the pattern of stream lines remains unchanging with time.
However the pattern of stream lines may or may not remain same with time for
unsteady flow. If unsteady flow is due to the change of magnitude of velocity, the
stream line pattern remains invariant (unchanging) with time. However if the
unsteadiness due to the change in the direction of the velocity, the stream line pattern
cannot remain same.
14. What is a stream tube?
The stream tube consists of stream lines forming its boundary surface. The
stream tube is defined as a circular space formed by the collection of stream lines
passing through the perimeter of a closed curve in a steady flow. As stream tube is
bounded on all sides by stream lines therefore no fluid can enter or leave the stream
tube from the sides except from the ends. Hence stream tube behaves as a solid
surface tube. The general equation of continuity can be applied on stream tube
though it has no solid boundaries. Stream tube may be of regular or irregular shape.
The velocity is uniform throughout the cross-section of the stream tube which
necessitates small cross-section for the stream tube. In steady flow with uniform
velocity, all stream lines are straight and parallel. The contents of stream tube are
called current-filaments.
15. What do you understand from streak line or filament line?
It is an instantaneous picture of the positions of all fluid particles in the flow
which have passed or emerged from a given point. The line of smoke from a
cigarette or from a chimney is nothing but a streak line.
16. What is a potential line?
The lines of equal velocity potential are called potential lines. These

73. 73
potential lines cut stream lines orthogonally i.e. a stream line and a potential line are
at right angle to each other.
17. Distinguish between stream lines, streak lines and path lines
Stream lines Streak lines Path lines
1. Imaginary lines
showing positions of
various fluid particles
2. Particle may change
stream line depending on
type of flow
3. Stream lines cannot
intersect each other, they
are always parallel
4. No flow across stream
line
1. Real line showing
instantaneous positions of
various particles
2. May change from
instant to instant
3. Streak line changes with
time. Two streak lines
may intersect each other
4. Flow across streak line
is possible
1. Real line showing
successive position of one
particle
2. Particle may cross its
path line
3. Two path lines for two
particles may intersect
each other
4. Flow across a path line
is possible by other
particles
18. What are different types of fluid flow?
The fluid flows can be: (1) steady or unsteady (2) uniform or non-uniform
(3) laminar or turbulent (4) compressible or incompressible (5) rotational or
irrotational (6) one-, two- or three-dimensional
19. Define steady and unsteady flow with one particle example
The flow in which fluid characteristics like velocity, acceleration, pressure
or density do not change with time at any point in the fluid is called steady flow. A
flow of water with constant discharge through a pipeline is an example of steady
flow. For steady flow, we have the following:
2
2
v v dP
0, 0, 0, 0
t t tt
  
   
  
The flow in which fluid characteristics like velocity, acceleration, pressure

74. 74
and density change with time at any point is called unsteady flow. The water with
varying discharge through a pipe is an unsteady flow. For unsteady flow, we have:
2
2
v v dP
0, 0, 0,& 0
t t tt
  
   
  
In steady flow, the path line and stream line will concide. In unsteady fluid
flow, the path line of successive particle will be different and stream line pattern of
the flow will be changing at every instant.
20. Define laminar and turbulent flow with one practical example.
A laminar flow is one in which the fluid particles move in layers with each
layer sliding over the other. There is no movement of fluid particles from one layer
to another. Flow of blood in small veins and oil flow in bearings are examples of
laminar flow. It is a smooth regular flow where velocity of flow is very small
The flow in which adjacent layers cross each other and the layers do not
move along the well defined path is called turbulent flow. The flow through rivs or
canals, smoke from chimney or cigarette are turbulenter flow.
21. How can flows be classified laminar or turbulent or transit flow?
The flow can be classified by Reynolds number. The Reynolds number is
the ratio of inertia force to viscous force. The viscous force tends to make the motion
of fluid in parallel layers while inertia force (mass x acceleration) tends to diffuse the
fluid particles. Higher viscous force (higher increase in velocity or inertial force) are
turbulent flow which have higher values of Reynolds number.
Reynolds number =
Inertiaforce VL
Viscousforce



For flow in pipes, the flow can be:
(a) Laminar flow if Reynolds number < 2100
(b) Transit flow if Reynolds number is in between 2100 to 4000
(c) Turbulent flow if Reynolds number > 4000

75. 75
22. What do you understand by compressible and incompressible flows?
The flow in which the volume and thereby the density of fluid does not
remain constant during the flow is called compressible flow. Gases are most
compressible. The compressibility related problems are observed when gases flow
through orifice, nozzle, turbine, compressor and in flight of planes.
The flow in which the changes in volume and thereby in density of the fluid
are insignificant, the flow is said is to be incompressible. Flow of liquids are
incompressible flow. For gases, in subsonic aerodynamic conditions air flow can be
considered incompressible.
23. Differentiate one-, two- and three-dimensional flow.
1. One-dimensional flow: One -dimensional flow is a flow in which the velocity of
the flow is a function of time and one space coordinate (x, y, or z). The flow
through a pipe is one-dimensional flow. For one-dimensional flow, we have for
(a) Steady flow: V = f(x)
(b) Unsteady flow: V = f(x, t)
Fig.
2. Two-dimensional flow: Two-dimensional flow is a flow in which the velocity of
the flow is a function of two space coordinates (xy, xz or yz) and time. The flows
between two parallel plates of large extent, flow over a long spillway and flow at
the middle part of the wing of an aircraft are two-dimensional flows. For two-
dimensional flow, we have for:
(a) Steady flow: v = f(x, y)
(b) Unsteady flow: v = f(x, y, t)
Fig.
3. Three-dimensional flow: Three-dimensional flow is a flow in which the velocity of
the flow is a function of three space coordinates (x, y, z) and time. Flow in
converging or diverging pipe section is a three-dimensional flow.
For three-dimensional flow, we have for -

76. 76
(a) Steady flow: v = f(x, y, z)
(b) Unsteady flow: v=f(x, y, z, t)
24. What is discharge of rate of flow (Q)?
The discharge is the amount of fluid flow per unit time. Hence discharge is
Q =
Volume
Time
=
Area length
Time

= Area  velocity
= A  V
The discharge can also be defined as cross-sectional area of the flow
multiplied by the velocity of flow.
25. What is the principle of conservation of mass?
The principle of conservation of mass states that matter cannot be created or
destroyed in non-nuclear processes.
26. What is the equation of continuity of flow?
The equation of continuity of flow is based on the principle of conservation
of mass. It states that the mass of fluid entering the stream tube from one end must be
equal to the mass of fluid leaving the stream tube at the other end per unit time and
there is no accumulation of fluid in the stream tube.
27. What do you understand from rotation and Vorticity of fluid particles?
A flow is said to be rotational if fluid particles are rotating about their own
mass centres. The rotation of a fluid particle at a point is specified by funding the
average angular velocity of two perpendicular linear elements of fluid particle.
Rotation of the fluid particle takes place about an axis which is perpendicular to the
plane formed by these two elements of the fluid. The sense of rotation is given by the
right hand rule for the motion of a right-hand screw. Vorticity of a rotation flow has
twice the value of the rotation of the flow. Both rotation and Vorticity are vectors.

77. 77
28. What are the conditions for irrotational flow?
The irrotational flow will have both rotation and vorticity as zero.
w = 0 and 0 
Or x y zw w w 0,   and x y z 0     
The above gives the conditions of irrotation of the flow as:
w v
y z
 

 
u w
z x
 

 
w u
x y
 

 
Note: The conditions in polar coordinates is 0  which gives
rv V 1 V
0
r r r
  
  
 
29. What is circulation? How is it related to vorticity?
Circulation is defined as the line integral of the tangential velocity of the
fluid around a closed curve
 Circulation =
c
V ds  Ñ
Where V = tangential velocity & ds = small length of curve.
Fig.
Consider a number of stream lines in a closed curve. Let tangential and
normal components of velocity V are V = V cos  and rV = V sin . If we calculate
circulation around a rectangular rotating element of sizedx dy , then we get -
c
Vcos ds   Ñ

78. 78
v u
udx v dx dy u dy dx vdy
y y
    
        
    
v u
dx.dy
x y
  
  
  
z dx dy   
z   Area enclosed by closed curve in plane at right angle to z-axis
Hence vorticity is the circulation per unit surface area.

A

  Where A = are enclosed by closed curved
Or
v u
dxdy x y
   
    
  
30. Explain the physical significance and use of the term ‘stream function’
The flow rate (q) between two stream lines per unit thickness and time is
known as stream function ().
Physical significance of stream function
Consider a flow between two stream lines AA and BB which is two-
dimensional steady and incompressible. The flow (q) per unit time and thickness
between these stream lines AA and BB is stream function . The flow across
points A and B will remain same =  = q irrespective of their joining line which
may be AB or ACB or ADB. If we assume  = 0 at point A, then  at point B will
be flow per unit time and thickness = q = 
Stream function is also defined as the scalar function of space and time
whose partial derivative with respect to any direction gives the velocity component at
right angles to that direction. Hence in steady two-dimensional flow,  f x,y 
which gives v
x



and u
y




79. 79
.
31. Show that stream function satisfies the equation of continuity.
The equation of continuity for a two-dimensional flow is
u v
0
x y
 
 
 
(1)
As per stream function, we have
u
y

 

v
x



Put the values of u and v in the continuity equation (eqn. (1))
0
x y y x
    
    
    
Or
2 2
0
x y y x
   
  
   
The above is satisfying as LHS = RHS. Hence stream function satisfies the
equation of continuity.
32. How and  in polar coordinates is used to find velocity components in
radial and tangential directions.
 = f (r, )
Radial velocity ru = r
1
u
r



Tangential velocity u =
r


 f r,  
Radial velocity ru
r




80. 80
Tangential velocity u =
1
r


Note: Following may be used as simple way to remember -
(1) ru u ,v u 
(2) dx rd ,dy dr  
33. Enumerate the properties of stream function .
The properties of stream function are:
(1)  is constant at all places at a stream line.
(2) The flow around any path in the fluid is zero
(3) The velocity vector at any point on the stream line can be found out by
differentiating its stream function i.e. u
y

 

and v
x



(4) The discharge between the stream lines is equal to difference of their
stream functions i.e.
2 1q    
(5) The rate of change of stream function with distance is proportion to the
component of velocity normal to that direction
(6) If two stream lines superimposed, then
 1 21 2
s s s
    
 
  
34. What do you understand from the velocity potential?
The flow takes place as in pipeline when there is a difference of pressure.
The flow always takes place from higher to lower pressure side. This potential
difference resulting into the flow of fluid is known as velocity potential and it is
denoted by  . The velocity potential is defined as a scalar function of space and time
such that its negative derivative with respect to any direction gives the fluid velocity
in that direction. Hence for steady flow, if velocity potential is given by
 f x,y,z  , then

81. 81
u ,v
x y
 
  
 
And w
z

 

(Note: negative sign indicates that flow is taking place in direction in which velocity
potential decreases.)
35. Derive the Laplace equation for the velocity potential.
For steady flow, the continuity equation is:
u v w
0
x y z
  
  
  
Now u ,v
x y
 
  
 
and w
z

 

. Put these values in continuity equation.
0
x x y y z z
         
         
         
This is Laplace equation
2 2 2
2 2 2
0
x y z
     
   
  
36. What are the properties of potential function?
The properties of potential function are:
(1) Flow having potential function  are irrotational
(2) Flows are steady, incompressible and irrotation if velocity potential
satisfies the Laplace equation.
37. What is the relation between stream function   and velocity potential  
?
As per definition, the derivatives of stream and velocity potential give
components of velocities as under:
u ,v
y x
 
 
 
Also u ,v
x y
 
 
 

82. 82
 u ,
y x
 
  
 
And v ,
x y
 
  
 
Hence, we can write -
x y
 

 
and
y x
 

 
38. Prove equipotential line and stream line are orthogonal to each other.
 f x,y constant  
d .dx dy 0
x y
 
   
 
u
x

 

and v
y

 

, we get
 udx vdy 0  
or
dy u
dx v
 
It is slope 1m of equipotentail line
Now stream function along a stream line is constant.
  f x,y constant  
Or d .dx .dy 0
x y
 
   
 
As v
x



and u
y

 

, we get
vdx udy 0 

83. 83

dy v
dx u

Hence the above is slope 2m of stream line
Now 1 2
u v
m m 1
v u
     
Hence equipotentail line and stream line are orthogonal.
39. What is a flow net? What is the condition of flow net to be a set of square?
A flow net is a graphical representation of stream lines and equipotential
lines. The stream lines in flow net show the direction of flow which the equipotential
lines join the equal velocity potential   points in the flow. Flow net provides a
simple graphical technique for studying flows which are two dimensional and
irrotational as compared to mathematical calculation techniques which are difficult
and tedious. In flow net, the stream lines are spaced such that the rate of flow q is
same between each successive pair of streamline. The flow net can be drawn for
irrotational flows. is constant along stream line and  is
If q is the flow and 1 2V &V are velocities along streamline 1 2& 
 1 1 2 2q V n V n    = constant
Where 1n and 2n are the distances between two streamlines
 1 2
2 1
V n
V n



But 1 1 2 2V s V s     = constant
 1 2
2 1
V s
V s



2 2
1 1
n s
n s
 

 
or 2 1
2 1
n n
s s
 

 

84. 84
or
n
s


= constant
In case
n
s


= 1, then n = s . We get a set of square in the flow net
40. What are the methods used for drawing the flow nets?
Three methods are used for drawing the flow nets which are: (1) analytical
method (2) graphical method (3) electrical analogy method. In analytical method,
flow net can be drawn be finding the expression for  and  in terms of x and y
which can be plotted. In graphical method, the flow passage is equally devided to
draw stream lines and then equally potential lines are drawn orthogonally to the
stream line. In electric analogy method, equipotential lines are drawn using null
method. Stream lines are later drawn orthogonally.
41. Explain graphical method of drawing flow net.
The graphical method is carried out as under:
(1) The flow passage between the boundaries is derived into suitable
number of equal parts and a set of stream lines are drawn enclosing these equal parts.
(2) Equipotential lines are drawn orthogonally to those stream lines such
that each block of flow net is almost a square.
42. At a point on a stream line, the velocity is 3 m/s and the radius of curvature
is 9 m. If the rate of increase of velocity along the stream line at this point is 1
m/s2, then the total acceleration at this point would be:
(a) 1 m/s2 (b) 3 m/s2 (c) 21
m/s
3
(d) 2
2 m /s
2
2
n
V 9
a 1 m/s
r 9
  
2
a 1 m/s 
2 2 2
na a a 1 1 2 m/s    
Answer (d) is correct

85. 85
43. If u and v are the components of velocity in the x and y directions of flow
given by u = ax + by, v = cx + dy. Then the condition to be satisfied is:
(a) a + c = 0 (b) b + d = 0 (c) a + b + c + d = 0 (d) a + d = 0
u = ax + by v = cx + dy
u
a
x



v
d
y



Flow has to satisfy continuity equation
u v
0
x y
 
 
 
a + d = 0
Answer (d) is correct
44. x component of velocity in a two-dimensional incompressible flow is given by
u = y2 + 4xy. If Y-component of velocity v equals zero at y = 0, the expressions
for v is given by –
(a) uy (b) 2y2 (c) –2y2 (d) 2xy
u = y2 + 4xy
u
4y
x



From continuity equation
u v
0
x y
 
 
 
v u
4y
y x
 
   
 
2
24y
v c 2y c
2
      
Condition y = 0, v = 0.
c 0 
2
v 2y  

86. 86
Answer (c) is correct.
45. Given the x-component of velocity u = 6xy –x2, the y-component of flow v is
given by,
(a) 6y2 – 5xy (b) –6xy + 2x2 (c) –6x2 – 2xy (d) 4xy – 3y2
The continuity equation is:
u v
0
x y
 
 
 
2
u 6xy 2x 
u
6y 4x
x

 

v u
6y 4x
y x
 
     
 
If we carryout integration, we get –
2
y
v 6 4xy c
2
   
2
3y 4xy c   
Answer (d) is correct.
46 The velocity component representing the irrotational flow is:
(a) u = x + y, v = 2x – y (b) u = 2x + 3y, v = – 2y2 + x
(c) u = x2, v = –2xy (d) u = –2x, v = 2y
For irrotation flow,
u v
0
y x
 
 
 
For choice (a) :
u
1
y



, and
v
2
x



u v
1 2 1 0
y x
 
      
 
For choice (b)
u v
3 and 1
y x
 
  
 

87. 87
u v
3 1 2 0
y x
 
     
 
For choice (c)
u v
0, 2y
y x
 
  
 
u v
2y 0
y x
 
    
 
For choice (d)
u v
0, 0
y x
 
 
 
u v
0
y x
 
 
 
Also
v u
2 2 0
x y
 
    
 
Hence flow is irrotational.
Answer (d) is correct.
47. Which of the following is rotational flow?
(a) u = y, v = 3/2 x (b) u = xy2, v = x2y (c) Both (a) and (b)(d) none of the other
For choice (a), we have
u v
1, and 3/ 2
y x
 
 
 
u v
1 3/ 2 0
y x
 
    
 
It is rotational flow
For choice (b), we have
u
2xy
y



and
v
2xy
x



u v
2xy 2xy 0
y x
 
    
 
It is an irrotational flow.
Answer (a) is correct.

88. 88
MODULE 3
Forces acting on Fluid in motion:
The various forces that may influence the motion are due to gravity, pressure,
viscosity, turbulence, surface tension and compressibility.
Alike mechanics of solids, dynamics of fluid is also governed by Newton’s second
law of motion. It states that the resultant force on any fluid element must be equal to
the product of mass and the acceleration of the element. In Mathematical form the
law is expressed as
F Ma  (1)
Therefore if the mass of fluid is influenced by all the above mentioned forces then
we can write the equation as:-
g p v s t cF F F F F F Ma      (2)
By resolving the various forces and acceleration in x, y and z directions, the
following equations of motion may be obtained
x x x x x xax g p v s t eM F F f F F F     
y y y y y yay g p v s t eM F F F F F F      (2.1)
z z z z z z za g p v s t eM F F F F F F     
In most of the fluid flow cases, the surface tension forces & the compressibility
forces are not significant. Hence they may be neglected. Then equations 2.0 and 2.1
becomes
a g p v tM F F F F    (3)
and
x x x xax g p v tM F F F F   
y y y yay g p v tM F F F F    (3.1)
z z z zaz g p v tM F F F F   

89. 89
Equations 3.1 are known as Reynold’s equation of motion which is useful in
turbulent flow analysis.
For Laminar or viscous, turbulent forces become less significant and hence these
may be neglected.
Thus equation 3 and 3.1 can be modified as
a g p rM F F F   (4.0)
and
ax gx px vxM F F F  
ay gy py vyM F F F   (4.1)
az gz pz vzM F F F  
The equation 4.0 and 4.1 are known as Navier stoke’s equation which are useful in
viscons flows.
Further if viscous forces are also little significant, then the equations can be modified
as
a g pM F F  (5. 0)
and
x xax g pM F F 
y yay g pM F F  (5.1)
z z za g pM F F 
The equation 5.0 and 5.1 are known as Euler’s equation of motion.

90. 90
Euler’s Equation of Motion:
Fig.1 cylindrical fluid element
In Euler’s equation of motion, the forces due to gravity and pressure are considered.
Consider a cylindrical element of cross section dA and length dS. The forces acting
in the cylindrical element are
1. Presence force PdA in the direction of flow.
2. Preserve force
P
P dA
S
 
 
 
opposite to direction of flow.
3. Weight of element gd AdS
Let  be the angle between the direction of flow and the line of action of weight of
element.
P
PdA P dS dA gdA dScos dAds as
S
 
      
 
Where as – acceleration in the direction of s
Now
dV
as
dt
 , where v = f(s, t)
v ds v v v ds
v v
s dt t s t dt
     
    
     
Q

91. 91
If the flow is steady,
v
t


= 0
s
v
a v
s

 

ie,
P V
pdA p dS dA g dA ds cos dA ds .v.
S S
  
      
  
Dividing by dA ds
1 p v
gcos v
s s
 
   
  
dz v 1 p dz
cos , v g 0
ds s s ds
 
    
  
ie,
p
gdz vdv 0

  

(I)
(I) is known as Euler’s equation of motion.
Integrating (I),
dp
g dz vdv 
   = constant
If the flow is in compressible,  = constant.
ie,
2
p v
gz
2
 

= constant
ie,
2
p v
z
g g
 
 
= constant (II)
Where
p
g
- Pressure energy per unit weight of fluid.
z – Potential energy per unit weight of fluid.
2
v
g
- K.E. per unit weight of fluid.
(II) Is known as Bernoulli’s equation.
Following are the assumptions made in the derivation of Bernoulli’s Equation.

92. 92
(i) The fluid is ideal, viscosity is zero.
(ii) The flow is steady
(iii) The flow is incompressible
(iv) The flow is irrotational
The Bernoulli’s equation was derived on the assumption that fluid is invisicid (non-
viscous) and therefore frictionless. But all the real fluids are viscous and hence after
resistance to flow. Hence there are always some losses in fluid flows and leave in the
application of Bernoulli’s equation there losses have to be taken into consideration.
Thus the Bernoulli’s equation for real fluids between any two points 1 and 2 is given
as
2 2
1 1 2 2
1 2 L
p v p v
z z h
g 2g g 2g
     
 
Where hL is the loss energy between point 1 and 2.
Kinetic Energy Corrections factor:
Kinetic Energy correction factor is defined as the ratio of the kinetic energy of the
flow per second based on actual velocity across the section to the kinetic energy of
the flow per second based on average velocity across the section. It is denoted by 2.
EE/sec bsed on actual velocity
KE/sec based an average velocity
 
Momentum Correction Factor:
It is defined as the ratio of momentum of flow per second based on actual velocity to
the momentum of flow per second based an average velocity across a section. It is
denoted by .
Momentum per second based an actual velocity
Momentum/ second based an average velocity
 
Statement of Bernoulli’s theorem:
It states that in a steady, ideal flow of an in compressible fluid, the total energy at any

93. 93
point of the fluid is constant.
Application of Bernoulli’s theorem:
Bernoulli’s equation finds wide application in the solution of many problems of fluid
flow. Some of the simple applications of Bernoulli’s theorem are
Venturimeter:
It is a device used for measuring the rate of flow of fluid through a pipe.
The basic principle on which a venturimeter works is that by reducing the cross
sectional area of the flow passage, a pressure difference is created and the
measurement of pressure difference enables the determination of the discharge
through pipe.
Fig:2 venturimeter
Parts:
A venturimeter consists of (1) inlet section followed by convergent cone (2) a
cylindrical threat and (3) a gradually divergent cone. The convergent cone is a short
pipe which tapers from the original size of the pipe to that of the threat of the
venturimeter. The throat of a venturimeter is a short parallel sided tube having its
cross sectional area smaller than that of the pipe. The divergent cone of venturimeter
is a gradually diverging pipe with its cross sectional area increasing that of the throat
to the original size of the pipe.
Principle:

94. 94
The basic principle on which a venturi meter works is that by reducing the cross-
sectional area of the flow passage, a pressure difference is created and the
measurement of the pressure difference enables the determination of the discharge
through the pipe.
Discharge:
Let a1 and a2 be the cross sectional at the inlet section and the threat. Let p1 and p2 be
pressures and v1 and v2 be velocities at section (1) and (2) respectively. Assuming
the fluid to be incompressible and there is no less of energy and applying the
Bernoulli’s equation, we get,
2 2
1 1 2 2
1 2
p v p v
z z
w 2g w 2g
     (1)
If the venturimeter is connected to a horizontal pipe then z1 = z2.
2 2
1 1 2 2p v p v
w g w g
  
 
2 2
1 2 2 1p p v v
w w g g
  
 
(2)
In the above expression, 1 2p p
w w
 
 
 
is the difference between pressure heads
l and 2 which is known as neutrihead and is denoted by h. That is,
2 2
2 1v v
h
g g
 
 
(3)
Further the Qth represents the discharge through the pipe then by continuity
equation
th 1 1 2 2Q a v a v 
or th
1
1
Q
v
a
 and th
2
2
Q
v
a
 (4)
    2
2 2
th 2 th 1 th
2 2
2 1
QQ /a Q /a 1 1
h
2g 2g 2g a a
 
     
  

95. 95
or 1 2
th
2 2
1 2
a a 2gh
Q
a a


(5)
But in actual practice there is always some loss of energy as the fluid flows through
the venturimeter, on account of which the actual discharge will be less than the
theoretical discharge. The actual discharge may therefore be obtained by multiplying
the theoretical discharge by a factor cd. (Coefficient of discharge.)
a
th
Q
Cd
Q

1 2
a
2 2
1 2
a a 2gh
Q Cd .
a a


Since for a given venturimeter the cross sectional areas of the inlet section and the
threat is a1 and a2 are fixed. Therefore we introduce constant of venturimeter c.
1 2
2 2
1 2
a a 2g
c
a a


aQ Cd . c . h
The value of Cd depends on rate of flow, viscosity of fluid and surface roughness,
but in general for fluids of low viscosity a value of about 0.98 is usually adopted for
Cd of venturimeter.
ORIFICE METER:

96. 96
Fig.3 orificemeter
Orifice meter is another simple device used for measuring the discharge through
pipes. The working principle is same as that of venturimeter. Its advantage is that it
is a cheaper arrangement and its installation requires a smaller length as compared
with venturimeter.
An orifice meter consists of a flat plate (circular) with a circular hole called orifice
which is concentric with pipe axis. The thickness of plate t is less than or equal to
0.05 times the diameter of pipe.
Applying Bernoulli’s equation between (1) and (2)
2 2
1 1 2 2
1 2
p v p v
z z
w 2g w 2g
    
If 1 2z z
2 2
1 1 2 2p v p v
w 2g w 2g
  
2 2
1 2 2 1p p v v
w w 2g 2g
  
2 2
2 1v v
n
2g 2g
 
 
1/ 22
2 1v 2gh v  (1)
If a2 is the area of vena contract a and a0 is the area at orifice then
Coefficient of contraction, 2
c
0
a
c
a

2 0 ca a c  (2)
By continuity equations,
1 1 2 2a v a v
2
1 2
1
a
v v
a


97. 97
Substituting (2) for a2
0 c
1 2
1
a c
v .v
a
 (3)
Substituting (3) in (1)
2 2 2
0 e 2
2 2
1
a c v
v 2gh
a
 
2
2 2 20
2 e 2
1
a
v 2gh c v
a
 
   
 
2
2 2 20
2 e 2
1
a
v c v 2gh
a
 
  
 
2
2 20
2 e
13
a
v 1 c 2gh
a
  
   
   
2
2
20
e
1
2gh
v
a
1 c
a

 
  
 
Q = 2 2 2 0 ev a v a c  
0 e
2
20
e
1
a c 2gh
Q
a
1 c
a

 
  
 
(4)
We know that
2
0
1
d e
2
20
e
1
a
1
a
c c
a
1 c
a
 
  
 

 
  
 

98. 98
2
20
c
1
c d
2
0
1
a
1 c
a
c c
a
1
a
 
  
 

 
  
 
(5)
Substituting (5) on (4)
2
20
c
1
0 d
2 2
20 0
c
1 1
a
1 c
a 2gh
Q a c
a a
1 1 c
a a
 
  
 
  
   
    
   
d 0 d 0 1
2 2 2
1 00
1
c a 2gh c a a 2gh
Q
a aa
1
a
 
 
  
 
where cd = coefficient of discharge of orifice meter. The coefficient of discharge of
orifice meter is much smaller than venturimeter.
Pitot tube:
Fig.4 Pitot tube
It is a simple device used for measuring the velocity of flow at any point in a
pipe or a channel. The basic principle used in this device is that the velocity of flow
at a particular point is reduced to zero, the pressure there is increased due to the

99. 99
conversion of kinetic energy into pressure energy and by measuring the increase in
the pressure energy at this point the velocity of flow may be determined. It is named
in the honour of its originator Herixi de Pilot, a feemh engineer, who in 1732 adopted
this principle for measuring the velocities in given scene.
In its simplest form, it consists of a glass tube, bent at right angles. The liquid rises
in the tube due to conversion of KE into pressure energy.
Consider two points (1) and (2) at the same level in such a way that point (2) is just
at the inlet of the pilot tube and (1) is far away from the tube.
2 2
1 1 2 2
1 2
p v p v
z z
W 2g W 2g
     (1)
1 2z z and v2 = 0
2
1 1 2p v p
w 2g w
   (2)
We know 1p
H
w
 and 2p
H h
w
 
2
1v
H H h
2g
    (3)
2
1v
h
2g
  or 1v 2gh (4)
Actual velocity is given by
actV Cv 2gh
Cr – coefficient of pilot tube.
Rotameter:
The Rotameter also known as variable-area meter. It consists of a vertical transparent
conical tube in which there is a rotor or float having a sharp circular edge. The rotor
has grooves on its head which ensures that as the liquid flows past, it causes the rotor
to rotate about its axis. The rotor is heavier than the liquid and hence it will sink to
the bottom of the tube when the liquid is at rest. But as the liquid beings to flow

100. 100
through the meter, it lifts the rotor until it reaches the steady level corresponding to
the discharge. This rate of flow of liquid can then be read from graduations engraved
on the tube by prior calibration, the sharp edge of the float serving as a pointer. The
rotating motion of the float helps to keep it steady. In this condition of equilibrium,
the hydrostatic and dynamic thrusts of the liquid on the upper side of the rotor will be
equal to the hydrostatic thrust on the upper side, plus the apparent weight of the
rotor.
ENERGY OR HEAD LOSSES OF FLOWING LIQUID DUE TO
SUDDEN CHANGE IN VELOCITY:
When the velocity of the fluid flowing changes either in magnitude or
direction, there is a large scale turbulence generated due to the formation of the
eddies in which a portion of the fluid’s energy is dissipated as heat, which is
considered to be a loss. Both in gradual and sudden change in velocity, these are
losses, but more loss happens when there is a sudden change in velocity. Some of the
loss of energy caused due to the change of velocity are:-
a) Loss of energy due to sudden enlargement.
b) Loss of energy due to sudden contraction.
c) Loss of energy due to gradient contraction or enlargement.
d) Loss of energy at the entrance to a pipe from large vessel.
e) Loss of energy at the exit from a pipe
f) Loss of energy due to an obstruction in the flow passage
g) Loss of energy in bends
h) Loss of energy in pipe fittings
The above mentioned losses are termed as ‘minor’ losses, as its value is less
compared to the loss due to friction in long pipes which is termed as ‘major’ loss.
a) Loss due to sudden enlargement: Consider a pipe of C.S. area A1 and
carrying a liquid of sp. wgt. W, connected to another pipe of larger C.S. area A2.

101. 101
Let V1 and V2 be velocities of flow of liquid in narrower and wider pipes
respectively. From continuity since Q = A1V1 = A2V2. The velocity V2 is smaller
than V1 and hence a change of momentum takes place as liquid flows from narrower
pipe to a wider pipe.
This change of momentum per sec will be equal to the net force acting in the
direction of flow on the liquid. Now if P1 and P2 are the pressure intensities at the
end sections BC and FE of the control volume and p is the mean pressure of the
eddying fluid over the annual face GD, then the force acting on the liquid in the
control volume in the dim of flow is
 1 1 2 1 2 2p A p A A p A      (1)
Fig.5 sudden enlargement
Experimentally it is found that 1p p  . Thus the net force is equal to  1 2 2p p A .
Momentum of liquid passing through 1wQV
BC
g
 (2)
Momentum of liquid passing through 2wQV
EF
g
 (3)
Change of momentum per sec
 2 1wQ V V
g

 (4)

102. 102
Then from impulse momentum equation
   1 2 2 2 1
wQ
p p A v v
g
   (5)
or
 
 1 2
2 1
2
p p 1 Q
v v
W g A

   2 2 1
1
V V V
g
  (6)
If hL is the head loss between the sections BC and EF due to sudden enlargement,
then applying Bernoulli’s equation between BC & EF.
2 2
1 1 2 2
1 2 L
p V p V
z z h
w 2g w 2g
     
Let 1 2z z pipe in horizontal, then,
2 2
1 2 1 2
L
p p v v
h
w w 2g 2g
  
          
(7)
 
2 2
1 2
2 2 1
v v1
v v v
g 2g 2g
    (8)
 2
1 2
L
v v
h
2g

 (9)
or
2 22 2
1 1 2 2
L
2 1
v A v A
h 1 1
2g A 2g A
   
     
   
(10)
b) Loss of energy due to sudden contraction: Let us consider a pipe
carrying liquid of sp. weight w, whose cross sectional area abruptly changes from
A1 to A2 (A1 > A2)

103. 103
Fig.6 sudden contraction
Just upstream of the junction between two pipes, in the wider pipe the streamlines are
curved and liquid is accelerated, due to which the pressure at the annular face varies
in an unknown manner which cannot be determined easily. Moreover in the region
just upstream of the junction there being accelerating flow no major loss of energy
occurs. However immediately downstream of the junction as the liquid flows from
the wider pipe to the narrower pipe a vena contracta is formed, after which the
stream of liquid widens again to fill completely the narrower pipe. In between the
vena contracta and the wall of the pipe, a lot of eddies are formed which cause a
considerable dissipation of energy.
Between vena-contracta and sec 2-2 at a certain distance away from it where the
velocity has again become almost uniform, the flow pattern is similar to that after a
sudden enlargement. As such equation (9) may be applied between the sections at
venacontracta and 2-2 to obtain approx. value of head loss due to sudden contraction.
2
L 2
L
V V
h
2g
 
  
 
(9.1)
Applying continuity equation 2
c c 2 2 c
c
A
A V A V , V
A
 
22 2
2
L
c
v A
h 1
2g A
 
    
 
(9.2)

104. 104
But we know c
c
2
A
C
A
 hence (9.2) can be written as,
22
2
L
c
v 1
h 1
2g C
 
  
 
Often loss of head due to sudden contraction is expressed as,
2
2
L
v
h k .
2g

The value of Cc or k is not constant; however it depends on the ratio 2
1
A
A
. Below
table gives some values of k for different values of 2
1
D
D
, the ratio of diameters of
narrower and wider pipes.
D2/D1 0 0.2 0.4 0.6 0.8 1.0
k 0.5 0.45 0.38 0.28 0.14 0
If Cc is assumed to be 0.62,
22 2
2 2
L
v v1
h 1 0.375
2g 0.62 2g
 
   
 
ie, k = 0.375
However in general the head loss due to sudden contraction is taken as
2
2v
0.5
2g
ie, k = 0.5
c) Loss of energy at the entrance to a pipe: depends on the form of
entrance. Thus for rounded or bell-mouthed entrance, loss is relatively less
compared to sharp – cornered entrance. In general, for a sharp-cornered entrance the
loss of head of head at the entrance is equal to
2
v
0.5
2g
where v is the mean vel. of
flow of liquid in the pipe. Flow pattern is similar to the case of sudden contraction of
pipe.

105. 105
d) Loss of energy at the exit from a pipe:- is equal to
2
v
2g
where V is the
mean vel. of flow in the pipe. Flow pattern is similar to the case of sudden
enlargement.
e) Loss of energy in various pipe fittings:- All pipe fittings such as values
couplings etc. cause a loss of energy. The loss of head in various pipe fittings may
also be represented as
2
L
v
h k
2g

The value of k depends on the type of pipe fitting.
FLOW THROUGH ORIFICES AND MOUTHPIECES:
An orifice is an opening having a closed perimeter, made in the walls or bottom of a
tank or vessel containing liquid through which the fluid may be discharged.
A mouthpiece is a short tube of length not more than 2-3 times its diameter, which is
filled to a circular opening or orifice of the same diameter, provided in a tank or
vessel containing fluid, such that it is an extension of the orifice and through which
also the fluid may be discharged.
Both are used for measuring the rate of flow of fluid.
Classifications of
1) Orifices:
a) According to the size – small and large
b) According to the shape – circular, square and triangular
c) According to the shape of upstream edge – sharp edged and bell
mouthed orifices.
d) Orifices discharging free and drowned or submerged orifice.
The drowned or submerged orifices are divided as fully submerged and
partially submerged orifices.

106. 106
2) Mouth pieces:
a) Shape – cylindrical, convergent, divergent and convergent –
divergent.
b) Position – external and internal
External can be any shape – But internal is generally of cylindrical
only.
SHARP EDGED ORIFICE DISCHARGING FREE:
The liquid emerging out of orifice as a free get is under the influence of gravity force
only.
Vena contractor:
The section of jet where streamlines are straight, parallel to each other and
perpendicular to plane of orifice and the jet has minimum cross sectional area.
Since at the vena contractor, as the streamlines are parallel and straight the pressure
in the jet at this section is uniform and is equal to the pressure of fluid surrounding
the jet, which in case of a free jet is equal to the atmospheric pressure.
Flow through an orifice:
Assume the flow to be steady through the orifice, apply Bernoulli’s equation between
(1) and (2), neglecting the loss of energy.
2 2
1 1 2 2
1
p v p v
z 0
W 2g W 2g
     (1)
We know p2 = pa (2)

107. 107
Fig:6 flow through a orifice
Assume the flow to be steady through the orifice, apply Bernoulli’s equation between
(1) and (2), neglecting the loss of energy.
2 2
1 1 2 2
1
p v p v
z 0
W 2g W 2g
     (1)
We know p2 = pa (2)
And p1 = pa + w (h – z1) (3)
Substituting (2) and (3) on (1)
2 2
2 1v v
h
2g 2g
  (4)
v1 – is called velocity of approach.
Applying continuity equation,
1 1 c 2Q a v a v 
c
1 2
1
a
V v
a
 (5)
Sub. (5) in (4).
22 2
c2 1
1
av v
h
2g a 2g
 
  
 

108. 108
2
2 c
2
1
a
v 1 2gh
a
  
   
   
2 2
c
1
2gh
v
a
1
a

 
  
 
(6)
ac – area of cross section at vena contractor
a1 – area of cross section of reservoir.
If the reservoir is assumed to be sufficient by large and the point 1 is assumed to be
sufficiently far from the orifice then velocity of approach v1 will be very small in
comparison with v2 hence it may be neglected in equation (4). So we get
2
2v
h
2g

2v 2gh (7)
(7) Is known as Torricelli’s formula in honor of Evangelista Torricelli who
in 1643 demonstrated experimentally.
(7) Gives velocity of vena – contracta which is always more than the
average velocity of point at a section in the plane of the orifice.
In the above analysis the loss of energy that takes place as the liquid flows through
the orifice has been neglected. But there is always some loss of energy due to
friction and Surface Tension. Thus (7) gives the ideal or theoretical velocity of the
jet. The actual velocity of jet at venacontracta may therefore be determined by
multiplying the ideal velocity by a factor called coeff of velocity.
HYDRAULIC COEFFICIENTS:
The hydraulic coefficients are:
1) Co-efficient of velocity,Cv
2) Co-efficient of contraction, Cc
3) Co-efficient of discharge,Cd

109. 109
Co-efficient of velocity (Cv):
Co-efficient of velocity is the ratio of actual velocity of jet at venacontracta to the
theoretical velocity of the jet.
vv c . 2gh 
The loss of energy is due to friction mainly, at orifice. The value Cv varies from 0.95
to 0.99 for different orifices, depending on the shape and size of the orifice and on
the head of liquid under which the flow takes place. An avg. value of 0.97 – 0.98
can be assumed.
In the sharp edged orifices the friction loss is very small and the value of Cv is
slightly less than unity for water and other similar liquids.
Co-efficient of contraction (Cc):
Cc in the ratio of the area of the jet at vena contracta to the area of the orifice. The
area of c.s. of jet is minimum at vena-contracta. In other words the jet of liquid
issuing from an orifice undergoes a contraction. The actual area of the jet at vena
contracta may therefore be determined by multiplying the area of the orifice by a
factor called coefficient of contraction (Cc).
Cc=
cont
orifice
area of jet at vena racta
area of
Cc in the ratio of the area of the jet at vena contracta to the area of the orifice. The
value varies from 0.61 – 0.69 depending on the size and shape of the orifice and head
of liquid under which the flow takes place. For general purpose, for a sharp edged
orifice the value is taken is 0.64 – 0.65. For a bell mouthed orifice, contraction after
the orifice is completely eliminated. The value of Cc in such case is taken as unity.
Co-efficient of discharge, (Cd):
Co-efficient of discharge in the ratio of actual discharge to the theoretical discharge.
unity. The cross sectional area of the jet of liquid is suing from an orifice will be
equal to the area of the orifice which is the theoretical area of jet. The product of
ideal area of jet and ideal velocity of jet will give ideal discharge of jet. But due to

110. 110
the effect of friction, the actual velocity of jet is reduced and due to the contraction of
jet, the actual discharge is always less than ideal discharge. The actual discharge of
liquid thro’ an orifice therefore can be determined by multiplying the ideal discharge
by a factor called coeff. of discharge (Cd)
Cd =
discharge
discharge
actual
therotical
The Cd varies from 0.61 – 0.65 depending on size and shape of orifice and head of
liquid under which the flow takes place. For general purpose for small sharp edged
orifices discharging water or liquids of similar viscosity an arg. value of 0.62 – 0.63
may be assumed.
Coefficient of resistance Cr is defined as the ratio of the loss of K.E. as the liquid
flows this an orifice and the actual K.E. possessed by the flowering liquid.
Theroetical K.E. per unit weight of liquid
 
2
2
th
2ghV
h
2g 2g
   (10)
Actual K.E. per unit weight of liquid
 
222
v 2
v
C 2ghV
hC
2g 2g
   (11)
 Loss of K.E.  2 2
v vh hC h 1 C    (12)
 2 2
v v
r 2 2 2
v v v
h 1 C 1 C 1
C 1
hC C C
  
      
 
FLOW THROUGH LARGE ORIFICES:

111. 111
Fig7 large orifice
Discharge through large rectangular orifice:- consider a large rectangular orifice in
one side of the tank discharging freely into atmosphere under a constant head, H.
Let H1 = height of liquid above top edge of orifice
H2 = height of liquid above bottom edge of orifice.
b = breadth of orifice
d = depth of orifice = H2 – H1
Cd = coeff. of discharge.
Consider an elementary strip of depth ‘dh’ at a depth of h below free surface of the
liquid in the tank.
Area of the strip = b × dh
Theoretical velocity of water through strip = 2gh
 Discharge through elementary strip  dQ Cd b dh . 2gh  .
Integrating,
2
1
H
H
Q Cd . b 2gh dh 
2
1
H3/ 2
H
h
Cd.b 2g
3/ 2
 
  
 
3/ 2 3/ 2
2 1
2
Cd.b 2g H H
3
   
Discharge through fully submerged orifice:
Fully submerge orifice is one which has its whole of its outlet side submerged under

112. 112
liquid so that it discharges a jet of liquid into the liquid of same kind. It is also called
totally drowned orifice.
Discharge through the orifice=Cd × Area ×Velocity
= Cd× b (H2-H1) ×√2𝑔ℎ
Where
H1= height of the water above the top of the orifice on the upstream side.
H2= height of the water above the bottom of the orifice on the upstream side.
h= difference in water level.
b= width of orifice.
Cd= Co-efficient of discharge.
Discharge through partially submerged orifice:
Fig.8 partially submerged orifice:
Partially submerged orifice is the one which has its outlet side partially submerged in
the liquid. It is also known as partially drowned orifice. Thus the partially submerged
orifice has two Parts. The upper portion behaves as an orifice discharging free while
the lower portion behaves as submerged orifice. Only a large orifice can behave as a
partially submerged orifice. The total discharge Q through partially submerged
orifice is equal to the discharges through free and the submerged portions.

113. 113
Discharge through the submerged portion
Q1 = Cd× b (H2-H1) ×√2𝑔ℎ
Discharge through free portion
Q2= 3/ 2 3/ 2
2 1
2
Cd.b 2g H H
3
   
Total discharge Q=Q1+Q2
= Cd× b (H2-H1) ×√2𝑔ℎ+ 3/ 2 3/ 2
2 1
2
Cd.b 2g H H
3
   
TIME OF EMPTYING A TANK THROUGH AN ORIFICE AT ITS
BOTTOM:
Fig:9 orifice
Consider a tank containing liquid up to a height of H1. Let an orifice is fitted at the
bottom of the tank. It is required to find the time for the liquid surface to fall from
height H1 to height H2. Let A be the area of tank, a be the area of orifice and T be
the time in seconds for the liquid to fall from H1 to H2.
Let at any time, the height of liquid from orifice is h and let the liquid surface fall by
a small height dh in time dT then,
Vol. of liquid leaving the tank in time dT = A × dh
Theoretical velocity through the orifice v = (2gh) ½
Discharge through orifice/sec. dQ = Cd. (Area of orifice) (Theo. vel.)

114. 114
= Cd. a 2gh
Discharge through orifice in dT, dQ = cd . a 2gh . dT.
We know, vol. of liquid leaving the tank = vol. flowing throw orifice.
ie,   dA dh C . a 2gh dT 
d
Adh
dT
C .a 2gh

 - ve sign is inserted because with the increase of time,
head on orifice decreases.
Integrating between limits H1 to H2 the total time is
T
0
dT =
2
1
H
½
d H
A
h dh
C .a 2g


2
1
H½
d H
A h 1
½ 1C .a 2g
  
  
 
2 1
d
2A
H H
C .A 2g
    
T 1 2
d
2A
H H
C .a 2g
   
For emptying completely, H2 = 0 hence
1
d
2A H
T
C .a 2g

TIME OF EMPTYING A HEMISPHERICAL TANK:
Consider a hemispherical tank of radius R fitted with an orifice of area ‘a’ at the
bottom of the tank. The tank contain some liquid up to a height H1 and in time T the
liquid falls to a height of H2

115. 115
Fig:10 hemispherical tank
T=
2cd a g

 
×
4
3
× 𝑅{ (H1
1
2⁄
–H2
1
2⁄
) −
2
5
× (H1
5
2⁄
–H2
5
2⁄
)}
For completely emptying the tank
T=
2cd a g

 
×
4
3
𝑅 (H1
1
2⁄
) −
2
5
(H1
5
2⁄
)
TIME OF EMPTYING A CIRCULAR HORIZONTAL TANK:
Consider a circular horizontal tank of length L and radius R, containing liquid up to a
height of H1. An orifice of area ‘a’ at the bottom of the tank. The tank contain some
liquid up to a height H1 and in time T the liquid falls to a height of H2
T=
3 2
L
cd a g  
× (2R-H2)1/2-(2R-H1)3/2
For emptying the tank (2R)1/2-(2R)3/2
T=
3 2
L
cd a g  
×(2R)
1
2 − (2R– H1 )
3
2
FLOW THROUGH A MOUTH PIECE:
A mouth piece is a short tube fitted to a circular orifice provide in a tank or a
reservoir. By fitting a mouth piece the discharge through the orifice can be increased.
The increase in discharge would, however result only if the mouth piece is running
full so that the jet of liquid emerging from the mouthpiece is same as the diameter as
that if the mouthpiece. A mouthpiece will be running full if its length is equal to
about two or three times its diameter and the head does not exceed a certain level.

116. 116
Flow through an external cylindrical mouthpiece:
Fig.11 external cylindrical mouthpiece
Consider a tank having a cylindrical mouthpiece of c.s. area a, attached
externally.
Sp. wt. of liquid = w
Head of liquid flowing above the centre of mouth piece = H.
Since the entrance section is exactly same as sharp edged orifice, the jet of
liquid entering contracts to form vena-contracta at a section c – c.
Beyond c – c, the jet expands filling the mouth piece completely.
ac – area of c.s. of jet of vena-contracta.
Cc – coeff. of contraction of orifice.
ac = Cc . Assume Cc = 0.62 ie, ac = 0.62 a (1)
Further if vc is the velocity of jet at venacontracta and V is the vel. of the jet
at the outlet them by continuity.
Vc . ac = V.a.
c
c
a V
V v.
a 0.62
  (2)
Applying Bernoulli’s equation between (1) and section b – b;
2
a a L
v
H H H h
2g
    (3)

117. 117
hL is the head loss between sections b – b and c – c. The flow pattern is
similar to that after sudden enlargement and hL is similar to that in sudden
enlargement.
 
2
2 2
c
L
V
v
V V V0.62
h 0.375
2g 2g 2g
 
       (4)
 (3) Becomes,
2 2
V V
H 0.375
2g 2g
 
2
V
H 1.375
2g

2gH
V 0.855 2gH
1.375
   (5)
Again thV 2gH , therefore v
th
V
C
V
 =
0.855 2gh
0.855
2gh

For mouthpiece Cc = 1. So d c vC C C 1 0.855 0.855    
In actual practice friction reduces value of C.v and is approx. equal to 0.82
Flow through a convergent divergent mouth piece:
If a mouth piece converges up to the venacontracta and then diverges that type of
mouth piece is called a convergent divergent mouth piece. As in this mouth piece
there is no sudden enlargement of the jet, the loss of energy due to sudden
enlargement is eliminated. The co-efficient of discharge for this mouthpiece is unity.
Let H is the head of the liquid over the mouthpiece.
The discharge Q= ac×√2𝑔𝐻
Where ac= area of venacontracta

118. 118
Fig.12 convergent divergent mouth piece
Flow through internal or re- entrant or borda’s mouth piece:
A short cylindrical tube attached to an orifice in such a way that the tube projects
inwardly to a tank, is called an internal mouth piece. It is also called re- entrant or
borda’s mouth piece. If the length of the mouthpiece is equal to the diameter the jet
of liquid comes out of the mouthpiece with out touching the sides of the tube. Then
the mouthpiece is running free. But the length of the diameter is three times the
diameter; the jet comes out with its diameter equal to the diameter of the mouthpiece.
Then the mouthpiece is said to be running full.
Borda’s mouth piece running free:
Discharge Q= a × Cd × √2𝑔𝐻
Here Cd=0.5
Fig.13 Borda’s mouth piece running free

119. 119
Borda’s mouth piece running full:
Fig.14 Borda’s mouth piece running full
Discharge Q= a × Cd × √2𝑔𝐻
Here Cd=0.707
FLOW OVER NOTCHES AND WEIRS:
Definition
A notch made of metallic plate are provided in narrow channels (particularly in
laboratory channels) to measure the rate of flow of liquid from a tank or in a channel.
A notch is an opening provided in the side of a tank or vessel such that the liquid
surface in the tank is below the top edge of the opening.
A weir is the name of the concrete or masonry structure built across a river or stream
in order to raise water level on the upstream side and to allow the excess water to
flow over its entire length to downstream side. A Dam is similar to weir, but the
difference is that in dam, the excess water flow to the down stream only through a
small portion called spill way. Weirs way also be used for measuring the discharge
or rate of flow of liquid in rivers or streams.
The sheet of water flowing thro’ a notch or a weir is nappe or vein. The bottom edge
of a notch or the top of a weir over which the water flows is known as sill or crest
and its height above the bottom of the channel or tank is crest height.

120. 120
Classification of Notches: - According to the
1) Shape of opening:-
a) Rectangular notch
b) triangular (V-notch)
c) trapezoidal notch
d) parabolic notch
e) Stepped notch.
2) Effect of the sides on the nappe:
a) Notch with end contraction
b) Notch without end contraction or suppressed notch.
Classification of weir:- According to the
1) shape:
a) Rectangular,
b) triangular
c) trapezoidal weir & Cipolletti weir
2) Shape of crest:
a) Thin plate or sharp edged weir
b) narrow crested weir
c) broad crested weir
d) ogee shaped weir
3) Discharge conditions:
a) Freely discharging weir
b) Submerged or (drowned) weir.
4) Effect of the sides on the nappe emerging from weir:
a) weir with end contraction
b) weir with out end contraction

121. 121
Discharge over a rectangular Notch or Weir
Fig.15 rectangular Notch
Fig.16 rectangular Weir
Consider a rectangular notch or weir provided in a channel carrying water as
shown in figure.
H = Head of water over the crest.
L = Length of the notch or weir
For finding the discharge of water over the weir or notch consider an
elementary horizontal strip of water of thickness dh and length L at a depth h from
the free surface of water as shown in figure.
The area of strip = L × dh
And theoretical velocity of water flowing through strip = 2gh .
The discharge dQ through the strip is
dQ = Cd × Area of strip × Theoretical velocity.

122. 122
dQ = Cd × L × dh × 2gh (1)
where Cd – coeff. of discharge
The total discharge Q, for the whole notch or weir is determined by
integrating Eqn. (1) from 0 to H.
H H
½
0 0
Q Cd L 2gh . dh Cd L 2g h dh     
H3/ 2
0
h
Cd L 2g
3/ 2
 
    
 
3/ 22
Q CdL 2g H
3

Discharge once a triangular notch or weir
Let H = Head of water above V – witch, O – angle of notch.
Consider horizontal strip of water of thickness ‘dh’ at a depth of h from the
free surface of water as shown in figure.
From the figure we have
Fig.17 triangular notch
Q AC Ac
tan
2 Oc H h
 

 AC H h tan
2

 

123. 123
AB = AC × 2 = 2(H – h) tan
2

 Area of strip = 2(H – h) tan
2

× dh
Theoretical velocity of water through strip = 2gh
 Discharge, dQ through the strip is
dQ = Cd × Area of strip × Velocity (theoretical)
= 2Cd (H – h) tan dh 2gh
2


Total discharge Q =  
H
0
2Cd H h tan 2gh dh
2

  
 
H
½
0
2Cd tan 2g H h h dh
2

 
 
H
½ 3/ 2
0
2Cd tan 2g Hh h dh
2

 
H3/ 2 5/ 2
0
Hh h
2Cd tan 2g
2 3/ 2 5/ 2
 
  
 
332 5/ 22 2
Q 2cd tan 2g HH H
2 3 5
  
    
 
5/ 24
2cd tan 2g H
2 15
  
   
 
5/ 28
Q Cd tan 2g H
15 2

 
For a right angled V – notch, if Cd = 0.6
o
90  tan 1
2

 
Discharge 5/ 28
Q 0.6 1 2 9.81 H
15
     
Q = 1.417 H5/2
Advantage of triangular notch or weir over rectangular notch or weir:
A triangular notch is preferred over a rectangular notch due to the following reasons:
1) The expression for right angled V notch or weir is very simple.

124. 124
2) For measuring low discharge, a triangular notch gives more accurate results
than a rectangular notch.
3) In the case of a triangular notch only one reading i.e. (H) is required for the
computation of discharge.
4) Ventilation of a triangular notch is not necessary.
Discharge over a trapezoidal notch or weir
As shown in figure a trapezoidal notch or weir is a combination of
rectangular and triangular notch or weir. Thus the total discharge will be equal to the
sum of discharge through a rectangular weir or notch and the discharge through
triangular notch or weir.
Fig.18 trapezoidal notch or weir
Let H - height of water over the notch.
L – Length of crest of the notch.
Cd1 – Coefficient of discharge for rectangular portion (ABCD)
Cd2 – Coefficient of discharge for triangular portion (FAD and BCE)
The discharge through rectangular portion ABCD is given by
3/ 2
1 1
2
Q Cd L 2g H
3
  
The discharge over the two triangular notches FDA and BCE is equal to the
discharge through a single triangular notch of angle  and it is given by

125. 125
5/ 2
2 2
8
Q Cd tan 2g H
15 2

    
 Discharge through trapezoidal notch or weir FDCEF = Q1 + Q2
3/ 2 5/ 2
1 2
2 8
Q Cd L 2g H Cd tan 2g H
3 15 2

     
Discharge over a stepped notch:
A stepped notch is a combination of rectangular notches. The discharge through a
stepped notch is the sum of the discharges through the different rectangular notches.
Consider a stepped notch as shown in fig:
Fig.19 stepped notch
Let H1= height of water above the creast of notch(1)
L1=length of notch (1)
H2, L2 and H3, L3 are corresponding values for notches 2and 3 respectively.
Cd= Co-efficient of discharge
Total discharge Q=Q1+Q2+Q3
Q=
2
3
×Cd×L1 × √2𝑔 (H1
3/2- H2
3/2) +
2
3
×Cd×L2 × √2𝑔 (H2
3/2- H3
3/2) +
2
3
×Cd×L1
× √2𝑔 × H3
3/2

126. 126
Discharge over a cipolletti weir or notch:
Fig.20 cipolletti weir or notch
Cipolletti weir is a trapezoidal weir which has side slopes of 1 horizontal to 4
vertical as shown in fig. thus in ∆ABC
tan
𝜃
2
=
𝐴𝐵
𝐵𝐶
= / 4H
H
=
1
4
by giving this slope to the sides, an increase in discharge through the triangular
portions ABC and DEF of the weir is obtained.if this slope is not provided the weir
would be a rectangular one and due to end contraction, the discharge would decrease.
Thus in case of cipolletti weir the factor of end contraction is not required.
The discharge of V notch of angle  is given by
= 58
2 tan
15 2
Cd g H

   
tan
2

=
2 15 1
15 8 4
14.2
2
or

 

Q= 58
2 14.2
15
Cd g H   

127. 127
Time required to empty a reservoir or a tank with a rectangular weir or notch:
Consider a reservoir or a tank with uniform cross-sectional area A. A
rectangular or notch is provided in one of its sides.
Let L – Length of crest of weir or notch
Cd – Coefficient of discharge
H1 – Initial height of liquid above the crest of notch.
H2 – Final height of liquid above the crest of notch.
T = Time required in seconds to lower the height of liquid from H1 to H2.
Let at any instant the height of liquid surface above the crest of weir or
notch be h and in a small time dT, let the liquid falls by dh. Then,
–Adh = Q × dt
Q = 3/ 22
cd L 2g h
3
  
–Adh = 3/ 22
cd L 2g h dT
3
   
or
dT = 3/ 2
Adh
2/3Cd L 2g h

  
The total time T is obtained by integrating the above equation between the
limits H1 to H2.
T =
2
1
HT
3/ 2
0 H
Adh
dT
2
Cd L 2g h
3


  
 
2
1
H
3/ 2
H
A
h dh
2
Cd L 2g
3


 

2
1
H½
H
3A h
–½2Cd L 2g
 
  
   

128. 128
2 1
3A 1 1
T
Cd L 2g H H
 
  
    
Time required to empty a reservoir or a tank with a triangular weir or notch.
Consider a reservoir or a tank of uniform cross-sectional area A, having a
triangular weir or notch in one of its sides.
Let q = Angle of notch
Cd = Coefficient of discharge
H1 = Initial height of liquid above the apex of notch.
H2 = Final height of liquid above the apex of notch.
T = Time required in seconds, to lower the height from H1 to H2
above the apex of notch.
Let at any instant, the height of liquid surface above the apex of weir or
notch be h and in a small time dT, let the liquid surface falls by ‘dh’. Then
–Adh = Q × dT
–ve sign is taken, as with increase of T, decreases and Q for a triangular
notch is
5/ 28
Q Cd tan 2g h
15 2

   
5/ 28
Adh Cd tan 2g dT
15 2

      
5/ 2
Adh
dT
8
Cd tan 2g h
15 2



   
The total time is obtained by integrating the above equation between the
limits H1 and H2
2
1
HT
5/ 2
0 H
Ah
dT
8
Cd tan 2g h
15 2



  
 

129. 129
2
1
H
5/ 2
H
A
T h dh
8
Cd tan 2g
15 2



 

2
1
H3/ 2
H
15A h
3/ 28 Cd tan 2g
2
 
      
3/ 2 3/ 2
2 1
5A 1 1
T
H H4 Cd tan 2g
2
 
       
Velocity of approach:
Velocity of approach is defined as the velocity with which the water approaches or
reaches the weir or notch before it flows over it. Thus Va is the velocity of approach,
then an additional head ha equal to Va
2/2g due to the velocity of approach, is acting
on the water flowing over the notch. Then the initial height of the water over the
notch becomes (H+h) and final height becomes ha. Then all the formulas are
changed taking into consideration of the velocity of approach. The velocity of
approach, Va is determined by finding the discharge over the notch or weir
neglecting velocity of approach. Then dividing the discharge by cross sectional area
of the channel on the upstream side of the weir or notch, the velocity of approach is
obtained. Mathematically,
Va=
𝑄
𝑎𝑟𝑒𝑎 𝑜𝑓 𝑡ℎ𝑒 𝑐ℎ𝑎𝑛𝑛𝑒𝑙
EFFECT ON DISCHARGE OVER A NOTCH OR WEIR DUE TO
ERROR IN THE MEASUREMENT OF HEAD
For an accurate value of the discharge over a weir or notch, an accurate
measurement of head over the weir or notch is very essential as the discharge over a
triangular notch is proportional to H5/2 and in case of rectangular notch it is
proportional to H3/2. A small error in the measurement of head, will affect the
discharge considerably. The following cases of error in the measurement of head will
be considered:

130. 130
(i) For Rectangular Weir or Notch
(ii) For Triangular Weir or Notch.
For Rectangular Weir or Notch. The discharge for a rectangular weir or notch
is given by equation (8.1) as
Q = 3/ 2
d
2
C L 2g H
3
   
= KH3/2 ... (i)
where K = d
2
C L 2g
3
  
Differentiating the above equation, we get
dQ = 1/ 23
K H dH
2
 ... (ii)
Dividing (ii) by (i),
dQ
Q
=
1/ 2
3/ 2
3
K H dH
3 dH2
2 HKH

 ... (8.6)
Equation (8.6) shows that an error of 1% in measuring H will produce 1.5%
error in discharge over a rectangular weir or notch.
For Triangular Weir or Notch. The discharge over a triangular weir or notch is
given by equation (8.2) as
Q = d
8
C . tan 2g
15 2

× H5/2
= KH5/2 ... (iii)
Where K = d
8
C . tan 2g
15 2

Differentiating equation (iii), we get
dQ = 3/ 25
K H dH
2
 ... (iv)

131. 131
Dividing (iv) by (iii), we get
dQ
Q
=
3/ 2
5/ 2
5
K H dH
5 dH2
2 HKH

 ... (8.7)
Equation (8.7) shows that an error of 1% in measuring H will produce 2.5%
error in discharge over a triangular weir or notch.
EMPIRICAL FORMULAE FOR DISCHARGE OVER
RECTANGULAR WEIR
The discharge over a rectangular weir is given by
Q = 3/ 22
Cd 2g L H
3
     without velocity of approach ... (i)
=  3/ 2 3/ 2
a a
2
Cd 2g L H h h
3
    
 
with velocity of approach
...(ii)
Equation (i) and (ii) are applicable to the weir of notch fro which the crest
length is equal to the width of the channel. This type of weir is called Suppressed
weir. But if the weir is not suppressed, the effect of end contraction will be taken into
account.
(a) Francis’ Formula. Francis on the basis of his experiments established
that end contraction decreases the effective length of the crest of weir and hence
decreases the discharge. Each end contraction reduces the crest length by 0.1 × H,
where H is the head over the weir. For a rectangular weir there are two end
contractions only and hence effective length
L = (L - 0.2 H)
And Q =   3/ 2
d
2
C L 0.2 H 2gH
3
    
If Cd = 0.623, g = 9.81 m/s2, then
Q =   3/ 22
.623 2 9.81 L 0.2 H H
3
      

132. 132
= 1.84 [L - 0.2 × H] H3/2 ... (8.11)
If end contractions are suppressed, then
H = 1.84 LH3/2 ...(8.12)
If velocity of approach is considered, then
Q = 1.84 L [(H + ha)3/2 - ha
3/2] ... (8.13)
(b) Bazin’s Formula. On the basis of results of a series of experiments,
Bazin’s proposed the following formula for the discharge over a rectangular weir as
Q = m × L 2g × H3/2 ... (8.14)
where m = d
2 .003
C 0.405
3 H
  
H = height of water over the weir
If velocity of approach is considered, then
Q = m1 × L × 2g [(H + ha)3/2] ... (8.15)
Where m1 =
 a
.003
0.405
H h


.
Discharge over a broad crested weir:
A weir having a wide crest is known as broad-created weir.
Let H = height of water, above the crest
L = length of the crest
.

133. 133
Fig.21 broad crested weir
If 2L > H, the weir is called broad crested weir
If 2L < H, the weir is called a narrow crested weir
Fig. 8.10 shows a broad-crested weir.
Let h = head of water at the middle of weir which is constant
v = velocity of flow over the weir
Applying Bernoulli’s equation to the still water surface on the upstream side
and running water at the end of weir,
0 + 0 + H = 0 +
2
v
2g
+ h

2
v
2g
= H - h
 v =  2g H h
The discharge over weir Q = Cd × Area of flow × Velocity
= Cd × L × h ×  2g H h
= Cd × L ×  2 3
2g Hh h ... (8.18)
The discharge will be maximum, if (Hh2 - h2) is maximum
or  2 3d
Hh h 0
dh
  or 2h × H - 3h3 = 0 or 2H = 3h
 h =
2
3
H
Qmax will be obtained by substituting this value of h in equation (8.18) as
Qmax = Cd × L ×
2 3
2 2
2g H H H
3 3
    
     
     

134. 134
= Cd × L × 2g
2 34 8
H H H
9 27
  
= Cd × L × 2g
3 34 8
H H
9 27
 = Cd × L × 2g
  3
12 8 H
27

= Cd × L × 2g
34
H
27
 = Cd × L × 2g ×0.3849 × H3/2
= .3849 × 2 9.81 × Cd × L × H3/2 = 1.7047 × Cd × L × H3/2
= 1.705 × Cd × L × H3/2
DISCHARGE OVER A NARROW-CRESTED WEIR
For a narrow-crested weir, 2L < H. It is similar to a rectangular weir or
notch hence, Q is given by
fig.22 NARROW-CRESTED WEIR
Q =
2
3
 Cd × L × 2g × H3/2
DISCHARGE OVER AN OGEE WEIR
an Ogee weir, in which the crest of the weir rises up to maximum height of 0.115 ×
H (where H is the height of water above inlet of the weir) and then falls as shown in
Fig. 8.11. The discharge for an Ogee weir is the same as that of a rectangular weir,
and it is given by

135. 135
Q =
2
3
 Cd × L × 2g × H3/2
DISCHARGE OVER SUBMERGED OR DROWNED WEIR
When the water level on the downstream side of a weir is above the crest of
the weir, then the weir is called to be a submerged or drowned weir. Fig. 8.12 shows
a submerged weir. The total discharge, over the weir is obtained by dividing the weir
into two parts. The portion between upstream and downstream water surface may be
treated as free weir and portion between downstream water surface and crest of weir
as a drowned weir.
Fig.23 SUBMERGED OR DROWNED WEIR
Let H = height of water on the upstream of the weir
h = height of water on the downstream side of the weir
Then Q1 = discharge over upper portion
=  1
3/ 2
d
2
C L 2g H h
3
   
Q2 = discharge through drowned portion
= 2dC × Area of flow × Velocity of flow
=  2dC L h 2g H h   
 Total discharge, Q = Q1 + Q2
=    1 2
3/ 2
d d
2
C L 2g H h C L h 2g H h
3
       

136. 136
QUESTIONS:
1) State the assumptions made in Bernoulli’s equation
In Euler’s equation of motion, forces due to gravity and pressure are taken
into consideration. The assumptions are
(1) Fluid is ideal i.e. non-viscous
(2) Flow is incompressible
(3) Flow is non-turbulence
(4) Flow is steady
(5) Flow is irrotational
(6) Flow is one dimensional i.e.along stream line and
(7) Velovity is uniform over the cross-section.
2) What is a pitot static tube?
The Pitot tube is suitable for measuring velocity of the flow in open channel
or used for measuring speed of flying aircraft. Special type of Pitot tube known as
pitot static tube is used with aircraft. This consists of two concentric tubes. Inside
tube faces the flow of air and it acts similar to Pitot tube i.e. converting the kinetic
energy of air into high pressure energy. The outer tube has openings perpendicular to
the flow of air and it measures the static pressure of air (low pressure side). These
two tubes are connected individually to each limb of the manometer so that the
manometer level difference is the dynamic head (h)
Speed of aircraft = V = 2gh
3.) Define the term Cd,Cc,Cv and derive the equation d c vC C C  where Cd =
coefficient of discharge, Cc = coefficient of contraction and Cv = coefficient of
velocity.
Cd, Cc and Cv are hydraulic coefficients which are used to find the actual
discharge, contraction and velocity of the liquid passing out of a orifice in the form a
jet. the jet goes on contracting from the mouth of orifice upto a distance of about one

137. 137
half of the orifice diameter. Later it begins to expand. The place where jet has least
cross-sectional area is known as vena contracta. The velocity at vena contracta is
maximum.
Coefficient of velocity (Cv). It is the ratio of the actual velocity of the liquid
jet at vena contracta to the theoretical value of the liquid jet. Therefore the coefficient
of velocity is always less than one.
v
Actualvelocityof jetat venacontracta
C
Theoreticalvelocityof the jet

=
V
2gh
The value of Cv varies from 0.95 to 0.99.
Coefficient of contraction (Cc). It is the ratio of the area of the jet at vena-
contracta to the area of the orifice. Since the jet keeps on converging after passing
orifice upto vena-contracta, the cross-section area of the vena contracta is minimum.
The coefficient of contraction is always less than one.
c
Areaof jetat venacontracta
C
Areaof theorifice

=
a
A
The value of coefficient of contraction varies from 0.61 to 0.69.
Coefficient of discharge (Cd). It is the ratio of the actual discharge to the
theoretical discharge. Since theoretical velocity and area of orifice is more than the
actual velocity of the jet and area at vena contracta, the theoretical discharge is
always larger than the actual discharge. Hence the value of coefficient of discharge is
always less than one.
d
th th
Q V a
C
Q V A
  
       
d v cC C .C

138. 138
The value of Cd varies from 0.61 to 0.65.
4) Define vena contracta.
Vena contracta. The liquid flowing past through an orifice in the form of a jet of
liquid is c converging type. The jet has a minimum area of cross-section at a distance
of about half the diameter of the orifice and this section is called vena contracta. At
this section, the stream lines are straight and parallel to each other besides being
perpendicular to the plane of the orifice. Beyond this section of vena contracta, the
liquid jet diverges and it bends towards downward direction due to the effect of
gravity.
5) An external cylindrical mouthpiece of diameter 120 mm is discharging with a
constant head of 6 m. The coefficient of contraction = 0.62 and coefficient of
discharge = 0.86 and atmospheric pressure head = 10.0 m of water. Find (1)
discharge and (2) absolute pressure head of vena contracta.
Area of mouthpiece (a) =
2
d
4

or 2 2 2
a 0.12 1.13 10 m
4

   
Discharge dQ C a 2gh  
= 2
0.86 1.13 10 2 9.81 6
    
= 0.105
2
m /s
Pressure head at vena contracta is
2
c
c 0
V
H H H
2g
  
Now 1 1
c
c
V V
V
C 0.62
  and
2
11.375V
H
2g

Therefore c 0H H H 1.89H  

139. 139
= 0H 0.89H
= 10 - 0.89  6
= 10 - 5.39
= 4.66 m
6) The diameter of the throat and exit of a convergent-divergent mouthpiece are
50 and 110 mm respectively is fitted to the side of a tank full of water. If the
maximum vacuum pressure is 6 m of water, find the maximum head of water
for steady flow. The atmospheric pressure head is 10.3 m of water. Find also the
discharge from the mouthpiece.
The maximum vacuum pressure will be at throat only. Hence absolute
pressure at throat = 6 m of water
Now Hc = atmospheric pressure - absolute pressure at throat
= 10.3 - 6 = 4.3 m
Now 0 c1
c
H Ha
1
a H

 
2
1
2
c
d 10.3 4.3
1
Hd

 
2
0.1 6
1
0.05 H
 
  
 
4 =
6
1
H

16 = 1+
6
H
or H =
6
15
= 0.4 m of water
Q = ac 2gh

140. 140
=
 2
0.05
2 9.81 0.4
4

  
=
4
3.14 25 10
2.8
4

 

= 54.97 4 3
10 m /s

7) Find the expression for hydraulic coefficient and discharge for Borda’s
mouthpiece running free.
In this a short cylindrical tube is attached to the orifice in such a way that
the tube is projected inwardly in the liquid tank. Hence it is also known as internal
mouthpiece. In running free Borda’s mouthpiece the length of tube is kept equal to
the diameter of the mouthpiece. In this condition, the jet of liquid comes out from the
mouthpiece without touching sides of the mouthpiece. The cross-section area of the
jet (ac) is lesser than the cross-sectional area of the mouthpiece (a) i.e. a>ac. The flow
through the mouthpiece is taking place due to pressure intensity at the mouthpiece
which is equal to gH .
Force = area  pressure intensity
= a gH (1)
If Vc is the velocity of the jet, then mass of fluid flowing per unit time is -
M = c ca V 
Force = Mass  acceleration
=  c c ca V V 0  
= 2
c ca V
or a. .g.H = 2
c c.a .V from equation (1)
Applying Bernoulli’s equation between point ‘A’ on free surface and section 1-1, we
get -

141. 141
2
c
22
CA A
A 1
VV
z z
g 2g g 2g

    
 
Now
2
cA
A 1 A
pp
z z H, 0, V 0
g g
    
 

2
CV
H
2g

cV 2gH (ii)
Now we have found out -
2
c ca. .g.H p.a .V 
Put the value of Vc from eqn. (ii), we get -
a. .g.H = c.a .2gH
or a = 2 ac
or c
c
a
C 0.5,
a
  Cc = coefficient of contraction
As there is no loss of head due to running free of the liquid jet, therefore coefficient
of velocity Cv = 1
Coefficient of discharge Cd = Cc  Cv
= 0.5  1
= 0.5
Discharge Q = Cd.a. 2gH
= 0.5 a 2gH
8) An internal mouthpiece of 1.0 m diameter discharging water at constant head
of 9 m. Find the discharge when (1) running free and (2) running full
2
2d
a 1
4 4
 
  

142. 142
Running free
Q = 0.5  a  2gH
= 0.5
2
1
2 9.81 9
4

   
= 30.5 3.14 13.29
5.22m /s
4
 

Running full
Q = 0.707  a  2gH
= 0.707
2
1
2 9.81 9
4

   
= 3
7.37m /s
9) A nozzle is situated at a distance of 1.0 m above the ground level and o
45 to
the horizontal. The diameter of nozzle is 100 mm and the jet strikes the ground
at 4 m from the nozzle. Find the discharge from the nozzle.
x = 4 m
Ux  t = 4 where Ux = horizontal velocity
h = Uyt - gt2 Uy = vertical velocity
Now h = -1 m when it touches ground at B w.rt point A
 y 2
x x
4 1 16
1 U . g
U 2 U
   
Now x
U
U Ucos45
2
 
y
U
U Ucos45
2
 
 2
1 16 2
1 4 9.81
2 U

    

143. 143

2 9.81 16
U
5


or U = 5.6 m/s
Discharge Q = 4 area of nozzle
Q = 5.6 
 2
0.1
4

= 4.396 2 3
10 m /s

10) What is a weir? What is the difference between a weir & notch?
Flow through an open channel takes place under atmospheric pressure. The
motive force to create the flow is derived from the slope of the bed/ground of the
channel and the kinetic energy possessed by the flowing liquid. A weir is an
obstruction placed in an open channel over which the flow occurs. The weir is
generally in the form of a vertical wall with a sharp edge at the top, running all the
way across the open channel. When the liquid flows over the weir, the height of the
liquid above the sharp edge of the weir bears a relationship with the amount of
discharge across it. This is the reason why weirs are widely used for discharge
measuring purpose.
A notch is a sharp edge device which permits the liquid to go through it
while liquid downstream exposed to the atmospheric pressure. The water may be
flowing out from a tank or reservoir by providing a notch in the way of a depression
in the side of a tank or a reservoir. The only difference between a weir and a
rectangular notch is that a weir runs all the way across the channel but a notch needs
not run full way across the channel.
Theoretically a weir or notch may be regarded as special form of wide
orifice in which the free water surface does not touch the upper edge of the orifice.

144. 144
MODULE 4
BOUNDARY LAYER THEORY
Fig.1 Boundary layer
When a real fluid flows past a solid body or a solid wall, the fluid particle adheres to
the boundary and the condition of no slip occurs. This means that the velocity of
fluid close to the boundary will be same as that of the boundary. If the boundary is
stationary, the velocity will be higher and as a result of the variations of velocity
gradient
du
dy
will exist. The velocity of fluid increases from zero velocity on the
stationary boundary to free stream velocity (U) of the fluid in the direction normal to
the boundary. This variation of velocity from zero to free stream velocity in the
direction normal to the boundary takes place in a narrow region in the vicinity of
solid boundary. This narrow region of the fluid is called boundary layer.
According to boundary layer theory, the flow fluid in the neighbourhood of the solid
boundary may be divided into two regions as shown in fig1.
1) A very thin layer of the fluid, called the boundary layer, in the immediate
neighourhood of the solid boundary, where the variation of velocity from
zero at the solid boundary to free stream velocity in the direction normal to
the boundary take place. In this region, the velocity gradient
du
dy
exists and

145. 145
hence the fluid exerts a shear stress on the wall in the direction of motion.
The value of shear stress is given by
τ =μ
du
dy
2) The remaining fluid which is out side the boundary layer. The velocity
outside the boundary layer is constant and equal to the free stream velocity.
As there is no variation of velocity i8n this region velocity gradient
du
dy
becomes zero. As a result of this the shear stress is zero.
BOUNDARY LAYER SEPARATION:
fig: 2 boundary layer separation:
The boundary layer thickness is considerably affected by the pressure gradient in the
direction of flow. If the pressure gradient is zero then the boundary layer continues to
grow in thickness along a flat plate. With the decreasing pressure in the direction of
flow, with negative pressure gradient, the boundary layer tends to be reduced in
thickness. However with the negative pressure gradient, the boundary layer tends to
be reduced in thickness. However with the pressure increasing in the direction of
flow, with positive pressure gradient, the boundary layer thickens rapidly. the
adverse pressure gradient plus the boundary shear decreases the momentum in the
boundary layer and if they both act over a sufficient distance they cause the fluid in
the boundary layer to come to rest, the retarded fluid particle in general cannot in

146. 146
general penetrate too far into the region of increased pressure owing to their small
kinetic energy. thus the boundary layer is deflected sideways from the boundary,
separates from it and moves into the main stream. This phenomenon is called
boundary layer separation.
Separation occurs with both laminar and turbulent boundary layers. Laminar
boundary layer is more prone to separation than turbulent boundary layer. Thus the
turbulent boundary layer can flow farther around the cylinder before it separates than
can the laminar boundary layer. Therefore the wake size will be narrower if the flow
is turbulent at the separation point than if it is laminar.
METHODS OF CONTROLLING THE BOUNDARY LAYER:
a) motion of the solid body:
Fig.3 flow past a rotating cylinder
The formation of the boundary layer is due to the difference between the velocity of
the flowing fluid and that of the solid boundary. As such it is possible to eliminate
the formation of a boundary layer by causing the solid boundary to move with the
flowing fluid. Such a motion of the boundary may be achieved in the simplest ways
by rotating a circular cylinder lying in a stream of fluid. so that on the upper side of
the cylinder, where the fluid as well as the cylinder move in the same direction the
boundary layer does not form and hence the separation is completely eliminated.
However on the lower side of the cylinder where the fluid motion is opposite to that
of the cylinder, separation would occur.
b) acceleration of the fluid in the boundary layer:
This method consists of supplying additional energy to the particles of the fluid
which are being retarded in the boundary layer. This may be achieved either injecting
the fluid into the region of the boundary layer from the interior of the body with the

147. 147
help of some suitable devices or by diverting a portion of the fluid of the main stream
from the region of high pressure to region
Fig.4 injecting fluid into the boundary layer
of the boundary layer through a slot provide in the body as in the case of the slotted
wing. However a disadvantage of this method is that if the fluid is injected into a
laminar boundary layer it undergoes a transition to turbulent boundary layer which
results in an increased skin friction.
c) suction of the fluid from the boundary layer:
Fig.6 slotted wing
Fig.7 suction of fluid through boundary layer
In this method the slow moving fluid in the boundary layer is removed by suction
through slots or through a porous surface so that on the down stream of the point of
suction a new boundary layer starts developing which is able to with stand an
adverse pressure gradient and hence separation is prevented. More over the suction

148. 148
of the fluid from the boundary layer also greatly delays its transition from laminar to
turbulent due to which skin friction drag is reduced.
d) streamlining of the body shapes:
By the use of suitably shaped bodies the point of transition of the boundary layer
from laminar to turbulent can be moved downstream which results in the reduction of
the skin friction drag. Furthermore by streamlining of body shapes the separation
may be eliminated.
WAKES:
A wake is the region of recirculating flow immediately behind a moving solid body,
caused by the flow of surrounding fluid around the body. In fluid dynamics,
a wake is the region of disturbed flow (usually turbulent) downstream of a solid body
moving through a fluid, caused by the flow of the fluid around the body. In
incompressible fluids (liquids) such as water, a bow wake is created when a
watercraft moves through the medium; as the medium cannot be compressed, it must
be displaced instead, resulting in a wave. As with all wave forms, it spreads outward
from the source until its energy is overcome or lost, usually by friction or dispersion.
The formation of these waves in liquids is analogous to the generation of shockwaves
in compressible flow, such as those generated by rockets and aircraft traveling
supersonically through air. For a blunt body in subsonic external flow, for example
the Apollo or Orion capsules during descent and landing, the wake is massively
separated and there exists a reverse flow region behind the body, in which the flow is
actually moving toward the body. This phenomenon is often observed in wind
tunnel testing of aircraft, and is especially important when parachute systems are
involved, because unless the parachute lines extend the canopy beyond the reverse
flow region, the chute can fail to inflate and thus collapse. Parachutes deployed into
wakes suffer dynamic pressure deficits which reduce their expected drag forces.
High-fidelity computational fluid dynamics simulations are often undertaken to
model wake flows, although such modeling has uncertainties associated
with turbulence modeling (for example RANSversus LES implementations), in
addition to unsteady flow effects. Example applications include rocket stage

149. 149
separation and aircraft store separation This pattern consists of two wake lines that
form the arms of a V, with the source of the wake at the point. Each wake line is
offset from the path of the wake source by around 17° and is made up with feathery
wavelets that are angled at roughly 53° to the path. The interior of the V is filled with
transverse curved waves, each of which is an arc of a circle centered at a point lying
on the path at a distance twice that of the arc to the wake source
Fig ;8 Wake pattern of a boat
This pattern is independent of the speed and size of the wake source over a
significant range of values.
FORCE EXERTED BY A FLOWING FLUID ON A STATIONARY
BODY
Consider a body held stationary in a real fluid, which is flowing at a uniform
velocity U as shown in Fig.
Fig.9

150. 150
The fluid will exert a force on the stationary body. The total force (FR)
exerted by the fluid on the body is perpendicular to the surface of the body. Thus the
total force is inclined to the direction of motion. The total force can be resolved in
two components, one in the direction of motion and other perpendicular to the
direction of motion.
Drag.
The component of the total force (FR) in the direction of motion is called ‘drag’. This
component is denoted by FD. Thus drag is the force exerted by the fluid in the
direction of motion.
Lift.
The component of the total force (FR) in the direction perpendicular to the direction
of motion is known as lift. This is denoted by FL. Thus lift is the force exerted by the
fluid in the direction perpendicular to the direction of motion. Lift force occurs only
when the axis of the body is inclined to the direction of fluid flow. If the axis of the
body is parallel to the direction of fluid flow, lift force is zero. In that case only drag
force acts.
If the fluid is assumed ideal and the body is symmetrical such as a sphere or
cylinder, both the drag and lift will be zero.
EXPRESSION FOR DRAG AND LIFT:
Consider an arbitrary shaped soil body placed in a real fluid, which is
flowing with a uniform velocity U in a horizontal direction as shown in Fig. Consider
a small elemental area dA on the surface of the body. The forces acting on the
surface area dA are :

151. 151
Fig.11 arbitrary shaped soil body
1. Pressure force equal to p × dA, acting perpendicular to the surface and
2. Shear force equal to 0 dA  , acting along the tangential direction to the
surface.
Let  = Angle made by pressure force with horizontal direction.
(a) Drag Force (FD). The drag force on elemental area
= Force due to pressure in the direction of fluid motion
+ force due to shear stress in the direction of fluid motion.
= pdA cos  + 0dA cos (90o-  ) = pdA cos + 0dA sin
Total drag, FD = Summation of pdA cos + Summation of 0dA sin
= 0pcos dA sin dA     . ... (1)
The term pcos dA is called the pressure drag or form drag while the term
0 sin dA  is called the friction drag or skin drag or shear drag.
(b) Lift Force (FL). The lift force an elemental area
= Force due to pressure in the direction perpendicular to the direction of
motion
+ Force due to shear stress in the direction perpendicular to the direction of
motion.

152. 152
= -pdA sin + 0dA sin (90o-  ) = -pdA sin + 0dA cos
Negative sign is taken with pressure force as it is acting in the downward
direction while shear force is acting vertically up.
Total lift, FL = 0dA cos - pdAsin   
= 0cos dA - psin dA    ... (2)
The drag and lift for a body moving in a fluid of density, at a uniform
velocity U are calculated mathematically, as
FD =
2
D
U
C A
2

... (3)
FL =
2
L
U
C A
2

... (4)
Where CD = Co-efficient of drag,
CL = Co-efficient of lift,
A = Area of the body which is the projected area of the body perpendicular
to the direction of flow
or
= Largest projected area of the immersed body.
Then resultant force on the body, FR =
2 2
D LF F ... (5)
The equations (14.3) and (14.4) which give the mathematical expression for
drag and lift are derived by the method of dimensional analysis.
Dimensional Analysis of Drag and Lift. In the chapter Dimensional and Model
Analysis it is shown in problem12.6 that the force exerted by a fluid on a supersonic
plane is given by:
2 2
2
k
F L U ,
UL U
 
   
  
... (i)

153. 153
Also in problem 12.7, it is shown that the force exerted by a fluid on a
partially sub-merged body is given by :
2 2
2
Lg
F L U ,
UL U
 
    
... (ii)
Thus the general expression for the force exerted by a fluid (air or water) on
a body (completely sub-merged or partially sub-merged) is given as
2 2
2 2
Lg k
F L U ,
UL U U
 
   
  
... (6)
Drag = Pressure Drag + Skin Friction Drag
TYPES OF DRAG:
Attempts have been made to rationalize the definitions and terminology associated
with drag. On the whole the new terms have not been widely adopted. Here we will
use the widely accepted traditional terms and indicate alternatives in parentheses.
Total drag
This is formally defined as the force corresponding to the rate of decrease in
momentum in the direction of the undisturbed external flow around the body, this
decrease being calculated between stations at infinite distances upstream and
downstream of the body. Thus it is the total force or drag in the direction of the
undisturbed flow. It is also the total force resisting the motion of the body through
the surrounding fluid. There are a number of separate contributions to total drag. As
a first step it may be divided into pressure drag and skin-friction drag.
Skin- friction drag (or surface- friction drag)
This is the drag that is generated by the resolved components of the traction due to
the shear stresses acting on the surface of the body. This traction is due directly to
viscosity and acts tangentially at all points on the surface of the body. At each point
it has a component aligned with but opposing the undisturbed flow (i.e. opposite to
the direction of flight). The total effect of these components, taken (i.e. integrated)

154. 154
over the whole exposed surface of the body, is the skin-friction drag. It could not
exist in an invisicid flow.
.Normal Pressure drag
This is the drag that is generated by the resolved components of the forces due to
pressure acting normal to the surface at all points. It may itself be considered as
consisting of several distinct contributions:
(i) Induced drag (sometimes known as vortex drag);
(ii) Wave drag; and
(iii) Form drag (sometimes known as boundary-layer pressure drag).
Induced drag (or vortex drag)
For a wing of finite span, the mere existence of lift leads to an additional source of
drag, known as induced drag. This drag is due to the high pressure fluid on the lower
wing surface moving outward and around the wingtip to meet the low pressure fluid
on the upper surface of the wing. When viewed from the front or rear of the aircraft,
we see that this flow produces vortices on the wingtips, which are often
called trailing vortices.
These trailing vortices connect with the starting vortex, or downwash, to form a U-
shaped vortex connected to the wingtips. Now a vortex can contain a great deal of
kinetic energy, and it requires energy to produce a vortex. Since the wings are
constantly supplying energy to the fluid to create the vortices, the fluid is doing work
on the wings, resulting in a drag on the wings.
Wave drag
This is the drag associated with the formation of shock waves in high-speed flight.
Shock waves radiate away a considerable amount of energy, energy that is
experienced by the aircraft as drag. Although shock waves are typically associated
with supersonic flow, they can form at much lower speeds at areas on the aircraft
where local airflow accelerates to supersonic speeds. The effect is typically seen
at transonic speeds above about Mach 0.8, but it is possible to notice the problem at
any speed over that of the critical Mach of that aircraft's wing

155. 155
Form drag (or boundary-layer pressure drag)
This can be defined as the difference between the profile drag and the skin-friction
drag where the former is defined as the drag due to the losses in total pressure and
total temperature in the boundary layers. Form drag and pressure drag are virtually
the same type of drag. Form or pressure drag is caused by the air that is flowing over
the aircraft or airfoil. The separation of air creates turbulence and results in pockets
of low and high pressure that leave a wake behind the airplane or airfoil (thus the
name pressure drag). This opposes forward motion and is a component of the total
drag. Since this drag is due to the shape, or form of the aircraft, it is also called form
drag.
STREAM LINED BODY:
A stream lined body is defined as that body whose surface coincides with the
streamlines, when the body is placed in a flow. In that case the separation of the flow
will take place only at the trailing edge. Though the boundary layer will start at the
leading edge, will become turbulent from laminar, yet it does not separate up to the
rear most part of the body in the case of stream lined body. Thus behind a
streamlined body, wake formation zone will be very small and consequently the
pressure drag will be very small. Then the total drag on the streamlined body will be
due tom friction only. A body may be stream lined at
1) At low velocities but may not be so at high velocities.
2) When placed in a particular position in the flow but may not be so when
placed in another position.
BLUFF BODY:
A bluff body is defined as that body whose surface does not coincides with the
stream lines, when placed in a flow. Then the flow is separated from the body much
ahead of the trailing edge with the result of a very large wake formation zone. Then
the drag due to pressure will very large compared to the drag due to friction on the
body. Thus the bodies of such a shape in which the pressure drag is very large as
compared to the friction drag are called bluff body.

156. 156
DRAG ON A FLAT PLATE
(a) Plate held parallel to the direction of flow of fluid. When a thin flat
plate is held parallel to the action of flow of fluid (i.e., the plate is held at zero
incidence) the total drag by the fluid on the plate is equal to the friction drag which is
due to the formation of the boundary on the surface of the plate. The magnitude of
the friction drag exerted on the plate however depends whether the boundary layer
formed over the surface of the plate is laminar or turbulent or partly and partly
turbulent.
(b)Plate held perpendicular to the direction of flow of fluid. Consider an
infinitely long plate held perpendicular to the direction of flow of fluid in a stream of
infinite extent. For an ideal fluid flowing past the plate the flow pattern will be
symmetrical on the upstream and the down stream sides of the plate further, because
of symmetrical flow pattern the pressure distribution will also be symmetrical and
hence there will be no drag. However, for a real fluid flowing past a flat plate the
flow pattern as well as the pressure distribution are markedly different form those for
an ideal fluid. The flow separates at the top and the bottom edges of the plate,
thereby forming a fairly wide wake on the downstream side... As expected the
pressure on the upstream face of the plate is positive for the entire width, with a
maximum at the middle section along which the stagnation points are located. On the
downstream face of the plate in the wake zone a constant pressure drop takes place as
indicated by a more or less constant negative pressure over the entire width of the
plate.
Fig: 12 Plate held perpendicular to the direction of flow of fluid

157. 157
for a flat plate held perpendicular to the flow the skin friction drag is negligible as
compared with the pressure drag, and so the effects of inertia forces become
predominant event at much lower values of Re than for a cylinder. Moreover, unlike
a sphere or a cylinder, in the case of a flat plate, except at low Reynolds number (say
Re <100), separation of the boundary layer always occurs at the same place-the sharp
edges of the plate-whatever be the value of the Reynolds number. As such in the case
of a flat plate the drag coefficient CD is a function of Re only at low and moderate
values of Re. However, as the value of Re exceeds 103, CD assumes a constant value
of about 2.0,. A reduction in the value of CD however occurs if the ratio of the length
L of the plate to its width B is not very large. The reduction in the value of CD in this
case is on account of the fact that the flow in not truly two-dimensional in character.
The flow of fluid round the ends tends to reduce the pressure at the upstream face of
the plate and increase at the rear face. The value of CD therefore decreases as the
length of the plate is reduced and these end effects become more significant. This is
demonstrated by the values of CD, obtained for a thin flat plate held perpendicular to
the flow, for different values of the ratio of length to breadth (L/B) of the plate which
are tabulated below:
L/B 1 2 4 10 18 
CD
1.10 1.15 1.19 1.29 1.40 2.01
The coefficient of drag, although remains independent of Reynolds number
for Re > 1000, it varies markedly with the (L/B) ratio of the plate.
The variation of CD with Re, for an infinitely long flat plate follows the
trend similar to that for a circular disc, both held perpendicular to the flow. However,
because of the flow in the case of circular disc being three-dimensional in character
the limiting value of CD is only about 1.1, which is much smaller than the limiting
value of CD equal to about 2, for the flat plate having flow which is two-dimensional
in character.

158. 158
where L = Length of body
 = Viscosity of fluid
F = Force exerted
g = Acceleration due to gravity
U = Velocity of body
 = Density of fluid
k = Bulk modules of fluid
If the body is completely sub-merged in the fluid, the force exerted by the
fluid on the body due to gravitational effect is negligible. Hence the non-dimensional
term containing ‘g’ i.e., 2
Lg
U
is neglected. if the velocity of the body is comparable
with velocity of sound, the effect due to compressibility is to be considered. But if
the ratio of the velocity of the body to the velocity of the sound is less than 0.3, the
force exerted by the fluid on the body due to compressibility is negligible.
2 2 2 2 UL
F L U L U
UL
    
            
Where
UL

= Reynold number = Re
F =  2 2
eL U R  .
Now F is the total force exerted by the fluid on the body. The total force
having two components, one in the direction of motion called drag force and other
component in the direction perpendicular to the direction of motion, called lift force.
The two components of F are expressed as
2 2
D D
L U
F C
2

 
Where CD is a function of Re and is called co-efficient of drag

159. 159
=
2
D
U
C A
2

Q L2 = Area = A
And
2 2
L L
L U
F C
2

 
Where CL is a function of Re and is called co-efficient of lift
=
2
L
U
C .A
2

.
DRAG ON A SPHERE
Consider the flow of a real fluid past a sphere.
Let U = Velocity of the flow of fluid over sphere,
D = Diameter of sphere,
 = Mass density of fluid, and
 = Viscosity of fluid
If the Reynold number of the flow is very small less than 0.2
e
UD
i.e., R 0.2
 
  
 
, the viscous forces are much more important that the inertial
forces as in this case viscous forces are much more predominate than the inertial
forces, which may be assumed negligible. G.G. Stokes developed a mathematical
equation for the total drag on a sphere immersed in a flowing fluid for which
Reynold number is upto 0.2, so that inertia forces may be assumed negligible.
According to his colution, total drag is
DF 3 DU 
(i) Expression for Cd for Sphere for Reynold Number less than 0.2.
From equation (14.3), the total drag is given by
2
D D
U
F C A
2

  
For sphere, FD = 3 DU

160. 160
A = Projected area of the sphere = 2
D
4

 3 DU =
2
2
D
U
C D
4 2
 

CD = 2
2 e
3 DU 24 24
UD RU
D
4 2
 
 
 

eR
DU
 
 
 
Q
This equation is called ‘Stoke’s law’.
(ii) Value of CD for sphere when Re is between 0.2 and 5. With the
increase of Reynold number, the inertia forces increase and must be taken into
account. When Re lies between 0.2 and 5, Oseen, a Swedish physicist, improved
Stoke’s law as
D
e e
24 3
C 1
R 16R
 
  
 
This Equation is called Oseen formulae and is valid for Re between 0.2 and
5.
(iii) Value of CD for Re from 5.0 to 1000. The drag co-efficient for the
Reynold of number from 5 to 1000 is equal to 0.4.
(iv) Value of CD for Re 1000 to 1000,000. In this range, CD is independent
of the Reynold number and its value is apptoximately equal to 0.5.
(v) Value of CD for Re more than 105. The value of CD is approximately
equal to 0.2 for the Reynold number more than 105.
TERMINAL VELOCITY OF A BODY
Terminal velocity is defined as the maximum constant velocity of a falling
body (such as sphere or a composite body such as parachute together with man) with
which body will be travelling. When the body is allowed to fall from rest in the
atmosphere, the velocity of the body increases due to acceleration of gravity. With

161. 161
the increase of the velocity, the drag force, opposing the motion of body also
increases. A stage is reached when the upward drag force acting on the body will be
equal to the weight of the body. Then the net external force acting on the body will
be zero and the body will be travelling at constant speed. This constant speed is
called terminal velocity of the falling body. If the body drops in a fluid, at the instant
it has acquired terminal velocity the net force acting on the body will be zero. The
forces acting on the body at this state will be:
1. Weight of body (W), acting downward,
2. Drag force (FD), acting vertically upward, and
3. Buoyant force (FB), acting vertically up.
The net force on the body should be zero, i.e., W = FD + FB
DRAG ON A CYLINDER
Fig.13 drag on a stationary cylinder
Consider a real fluid flowing over a circular cylinder of diameter D and
length L when the cylinder is placed in the fluid such that its length is perpendicular
to the direction of flow. If the Reynolds number of the flow is less than 0.2
U d
i.e., 0.2
 
 
 
, the inertia force is negligibly small as compared to viscous force
and hence the flow pattern about the cylinder will be symmetrical. As the Reynold
number is increased, inertia forces increase and hence they must be taken into
consideration for analysis of flow over cylinder. With the increase of the Reynold
number, the flow pattern becomes unsymmetrical with respect to an axis

162. 162
perpendicular to the direction of flow. The drag force, i.e., the force exerted by the
following fluid on the cylinder in the direction of flow depends upon the Reynold
number of the flow. From experiments, it has been observed that:
(i) When Reynold number (Re) < 1, the drag force is directly proportional to
velocity and hence the drag co-efficient (CD) is inversely proportional to Reynold
number.
(ii) With the increase of the Reynold number from 1 to 2000, the drag co-
efficient decreases and reaches a minimum value of 0.95 at Re = 2000.
(iii) With the further increase of the Reynold number from 2000 to 3 × 104,
the co-efficient of drag increases and attains maximum value of 1.2 at Re = 3 × 104.
(iv) The value of co-efficient of drag decreases if the Reynold number is
increased from 3 × 104 to 3 × 105. At Re = 3 × 105, the value of CD = 0.3.
(v) If the Reynold number is increased beyond 3 × 106, the value of CD
increases and it becomes equal to 0.7 in the end.
Flow of Ideal Fluid over Stationary Cylinder.
Consider the flow of an ideal fluid over a cylinder, which is stationary
Let U = Free stream velocity of fluid
R = Radius of the cylinder
 = Angle made by any point say C on the circumference of the
cylinder with the direction of flow.
The flow pattern will be symmetrical and the velocity at any point say C on
the surface of the cylinder is given by u = 2U sin
The velocity distribution over the upper half and lower half of the cylinder
from the axis AB of the cylinder are identical and hence the pressure distributions
will also be same. Hence the lift acting on the cylinder will be zero.

163. 163
Flow Pattern Around the Cylinder when a Constant Circulation  is imparted
to the Cylinder.
Circulation is defined as the flow along a closed curve. Mathematically, the
circulation is obtained if the product of the velocity component along the curve at
any point and the length of the small element containing that point is integrated
around the curve.
Fig.14
Consider a fluid flowing with a free stream velocity equal to U. Within the
fluid consider a closed curve as shown in Fig. 14.10. Let E is any point on the closed
curve and ‘dS’ is a small length of the closed curve containing point E.
Let 1 = Angle made by the tangent at E with the direction of flow,
1
u = Component of free stream velocity along the tangent at E and is
given as
= U cos 1
 By definition, circulation along the closed curve is
 = Ñvelocity component along curve × Length of element
= ÑU cos 1 × dS
where Ñ= Integral for the complete closed curve.

164. 164
EXPRESSION FOR LIFT FORCE ACTING ON ROTATING
CYLINDER.
Fig.15 Rotating Cylinder
The flow over a cylinder to which a constant circulation (Ѓ) is imparted is obtained
as shown in fig. the velocity a t the point C is obtained by
v= 2 sin
2
U
R




Where U is the velocity of the flow of the fluid.R is the radius of the cylinder.for the
upper half portion of the cylinder, varies from 00 to 1800 and hence component of
velocity,2Usin𝜃 is positive. But for the lower half portion of the cylinder,𝜃varies
from1800 to 3600.as sin𝜃for the value of 𝜃more then 1800 and less than 3600 is
negative and hence component of velocity 2Usin𝜃will be negative. This means, the
velocity on the upper half portion of the cylinderwill be more than the velocity on the
lower half portion of the cylinder.but from Bernoulli’s theorm we know that at a
surface where velocity is less pressure will be more there and viceversa. Hence on
the lower half portion of the cylinderwhere velocity is less pressure will be morethan
the pressure on the upper half portion of the cylinder.due to this difference of
pressure on the two portions of the cylinder, a force will be acting on the cylinder in
a dirction perpendicular to the direction of flow.this force is nothing but a lift
force.thus by rotating a cylinder at a constant velocity in a uniform flow field,a lift
force can be developed.

165. 165
Let a cylinder is rotating is a uniform flow field. The resultant flow pattern will be
as shown in Consider a small length of the element of the surface of the cylinder.
Let ps = Pressure on the surface of the element on cylinder
ds = Length of element
R = Radius of cylinder
p = Pressure of the fluid far away from the cylinder
U = Velocity of fluid far away from the cylinder
us = Velocity of fluid on the surface of the cylinder
Applying Bernoulli’s equation to a point far away from cylinder and to a
point lying on the surface of cylinder such that both the points are on the same
horizontal line, we have
22
s sp up U
g 2g g 2g
  
 

22
s sp up U
g g 2g 2g
  
 
=
22
s
2
up U
1
g 2g U
 
  
  
su u 2Usin
2 R

   

Substituting this value of us in (i), we get
2
2
s
2
2Usin
p p U 2 R
1
g g 2g U
  
   
    
  
 
  

166. 166
=
2
2
2 22
2
4U sin 4Usin
2 R2 Rp U
1
g 2g U
   
     
    
 
  
=
2 2
2
2 2 2
p U 4sin
1 4sin
g 2g U 2 R4 R U
   
            
or ps =
2 2
2
2 2 2
gU 4sin
p 1 4sin
2g U 2 R4 R U
   
     
  
we have the lift force acting on the small length ds on the element, due to
pressure ps as
= Component of ps in the direction perpendicular to flow × Area of the
element
= (-ps sin) × (ds × L)
Negative sign is taken as the component of ps perpendicular to the flow is
acting in the downward direction.
Now L = length of the cylinder
ds = R × d
 Lift force on the element = -ps sin × Rd × L  ds = RdQ
The total force is obtained by integrating equation (iii) over the centre
surface of the cylinder.
 Total lift,
2 2
L s s0 0
F p sin Rd L p R L sin d
 
           
Substituting the value of ps from equation (ii), we get
2
2 2
L 2 2 20
gU 4sin
F p 1 4sin RL sin d
2g U 2 R4 R U
     
           
   


167. 167
=
2 2
2 3
2 2 20
gU sin 4sin
RL psin sin 4sin d
2g U 2 R4 R U
      
               

But
2 2 3
0 0
sin d sin d 0
 
      

2 2
2
L 0
gU 4sin
F R L d
2g U 2 R
   
         

=
2
2 22 2
0 0
gU 4 L gU
R L sin d sin d
2g U 2 R g
    
       
   
But
2
2 2
0
0
sin2 2 sin4
sin d
2 4 2 4

       
         
   

 L
L g L g
F U gU LUT LUT
g g g
 
        

Equation is known as Kutta-Joukowski equation.
DRAG FORCE ACTING ON A ROTATING CYLINDER.
The resultant flow pattern for a rotating cylinder in a uniform flow field is shown in
Fig. 15. The resultant flow pattern is symmetrical about the vertical axis of the
cylinder. Hence the velocity distribution and also pressure distribution is symmetrical
about the vertical axis and as such there will be no drag on the cylinder.
Expression for Lift Co-efficient for Rotating Cylinder.
The lift co-efficient is defined by the equation as
2
L L
U
F C A
2


where CL = Lift co-efficient, A = Projected area
U = Free stream velocity or uniform velocity of flow.
For a rotating cylinder, the lift force is given by equation

168. 168
LF LUT 
A = Projected area of cylinder = 2RL
Substituting these values we get
2
L
U
LUT C 2RL
2

    or L 2
LUT
C
RURL U
 
 

From equation (14.14), we have 1
u
2 R



or
R

=
1
2 u
Substituting this value of
R

in equation (14.17), CL is also expressed
1
L
2 u
C
U

where 1
u =Velocity of rotation of the cylinder in the tangential direction.
Location of Stagnation Points for a Rotating Cylinder in a Uniform Flow-field.
Stagnation points are those points on the surface of the cylinder, where velocity is
zero. For a rotating cylinder, the resultant velocity is given by equation
u = 2U sin +
2 R


.
For stagnation point, u = 0
 2U sin +
2 R


= 0 or 2U sin = -
2 R


or sin = -
4 UR


The solution of equation given the location of stagnation points on the surface of the
cylinder. There are two values of , which satisfy the above equation. As sin  is
negative in equation it means  is more than 180o but less than 360o. The two values
of  are such that one value is between 180o and 270o and other value is between
270o and 360o.

169. 169
For a single stagnation point,  = 270o and then equation becomes as
sin 270o = -
4 UR


or -1 =
4 UR


(Q sin 270o = -1)
 = 4 UR
DEVELOPMENT OF LIFT ON AN AIROFOIL
The airfoil is characterized by its chord length C, angle of attack  (which
is the angle between the direction of the fluid flowing and chord line) and span L of
the airfoil. The lift on the airfoil is due to negative pressure created on the upper part
of the airfoil. The drag force on the airfoil is always small due to the design of the
shape of the body, which is stream-lined.
fig.16 airofoil
From the theoretical analysis, the circulation  developed on the airfoil so
that the stream-line at the trailing edge of the airfoil is tangential to the airfoil, is
given as
CU sin   
Where C = Chord length, U = Free stream velocity of airfoil,  = Angle of
attack
Lift force FL is given by equation as
LF UL UL CUsin        (Q
CUsin   )

170. 170
= 2
CU L sin 
The lift force is also given by equation as
2
L L
U
F C A
2

  
where CL = Co-efficient of lift
A = Projected area = C × L for airfoil

2
L L
U
F C C L
2

   
Equating the two value of lift force we get
2
2
L
U
CU L sin =C C L
2

    

2
L
2 CU L sin
C 2 sin
C L UH2
 
   
 
Thus it is clear from equation that co-efficient of lift depends upon the angle
of attack.
Steady-state of a Flying Object.
When a flying object for example airplane is in a steady-state, the weight of the
airplane is equal to the lift force and thrust developed by the engine is equal to the
drag force. Hence
W = Lift force =
2
L
AU
C
2

where W = Weight of the airplane and
2
L
AU
C
2

= Lift force.

171. 171
CHARACTERSTICS OF AIROFOIL:
1) Chord and chord length:
Fig: 17 Chord
Chord is an imaginary reference line in the airofoil section above and sometimes
below which the curvature of the upper and lower surfaces of the airofoil
respectively is described.if the aerofoil section is a flat botttmed one as chown in fig.
the cord is the bottom line itself. Ift he aerofoil section have convex curvature on
both the upper and lower surfaces or the aerofoil section has convex curvature at the
bottom surface, the cord can be found out gy joining the most forward point of the
leading edge to the rear most point on the trailing edge. The cord length is the length
of the projection of the aerofoil on the chord and is denoted by the letter “c”.
2) Span:
Fig.18 Span
this is the distance measured from the left wing tip to the right wing tip of the
aeroplane and is denoted by the letter “b”

172. 172
3) Frontal area:
This is the area of the wing as seen from the plan of the airplane snd is the area of the
projection of the actual outline of the wing on a horizontial plane through the cord.
The symbol used is “S”
4) Aspect ratio:
This is a non dimensional constant for a given wing and given by span/cord length
for a rectangular wing.aspect ratio is the ratio of the square of the span to the frontal
area for wings other than rectangular in shape in the plan view.
5) Camber:
Fig.19 camber
Camber is the name given to describe the curvature of the upper and lower surfaces
of an aerofoil section. Camber is defined as the ratio of the vertical intercept and
chord length, the vertical intercept being measured as thre length between surface
concerded and the chord. The length of the intercept between the cord and the upper
surface expressed as the percentage of the chord length is calkled lower camber. The
mean camber is the verticaslintercept between thew geometrical central lines of the
aerofoil section at the centre of the cord expressed as the percentage of the chord
length. Mean camber line is also called the median lin e. the maximum camger refer
to the greatest departure of the surface of the curvature of the aerofoil section from
the cord line.

173. 173
6) Sweep back angle:
Fig.20 sweep back angle
The angle between a line drawn perpendicular to the longitudinal plane of symmetry
of the aeroplane and the horizontial projection of the reference line of the wing.
7) Dihedral angle:
Fig.21 dihedral angle
This is the acute angle to which the wing inclines in front view of the aeroplane.
8) Relative wind and angle of attack:
The relative motion of an aerofoil immersed in fluid can happen on two ways.
a) The fluid moves relative to the stationary aerofoil.
b) Movement of the aerofoil in a still fluid.

174. 174
Fig.22 relative wind and angle of attack
The velocity of the fluid relative to the aerofoil is same in both the above cases. For
any such motion of aerofoil in airthe relative velocity is termed as relative wind. The
speed of an aeroplane is the speed relative to air or called air speed while ground
speed the trem applied to the speed of the aerofoil relative to the ground. The angle
of incidence or the angle of attack of an aerofoil is the angle made by the direction of
the relative wind on the cord of the aerofoil.
9) Wing loading:
This is the ratio of the total weight of the airplane and the wing area.
10) Plan form:
The shape of the wing of an air craft as viewed in plan.
POLAR DIAGRAM:
The various drag and lift cofficents of an aerofoil having a particular aspect ratio
may be determined by testing the airofoil having a particular aspect ratio may be
determined by testing the airfoil at different angles of attack in a wind tunnel. The
values of these coefficients abtained by such tests may be represented graphically by
plotting these coffecients against the angle of attack α .These plots thus indicate the
Variation of the copefficents with the angle of attack α. However the variation of
these coefficients may also be shown by a single curve known as the polar diagram
in this diagram the lift coefficient CL is plotted against thre drag coefficient CD.
The corresponding angle of attack are represented by points on the curve.The slope
of the line drawn from the orgin to a point on the curve will thus give the ratio of lift

175. 175
to drag and the maximum value of this ratio will be equal to the slope of the line
tangent to the curve and drawn from the orgin.The point corresponding to zero lift
and minium drag are indicated on the curve.The stall point as well as the stalling
angle is also indicated.
FLOW OF VISCOUS FLUID THROUGH CIRCULAR PIPE:
Consider a horizontal pipe of radius R. The viscous fluid is flowing fro left
to right in the pipe as shown in fig.
Fig. 23horizontal pipe of radius R
Consider a fluid element of radius r, sliding in the cylindrical fluid element
of radius (r+dr).let the length of the fluid element be x . If p is the intensity of
pressure on the face AB of the fluid element then intensity of pressure on the face
CD will be
p
p x
x
 
  
 
Shear stress distribution
Let  is the shear stress acting on the fluid element. Then the forces acting
on the fluid element are.
1 The pressure force, 2
p r on face AB.
2. The pressure force,
p
p x
x
 
  
 
2
r on face CD.
3. The shear force, 2 r x   on the fluid element..
For steady and uniform flow, there is no acceleration and hence the resultant
force in the direction of flow is zero.

176. 176

2
p r -
p
p x
x
 
  
 
2
r - 2 r x   =0
Or
2p
x r 2 r x 0
x x
 
      
 
p r
x 2

  

----------------------------------(1)
The shear stress τ across a section varies with ‘r’ as
p
x


a cross a section is
constant. Hence shear distribution is shown as
Velocity distribution
To obtain the velocity distribution across the section the value of the shear stress
1 2
[ ]
g
p p


Is substituted in eq (1).but y is measured from the pipe wall. Hence
y=R-r and dy dr 
du
dr
  
2
p r du
x dr



  
Or
1
2
p r du
x dr

 
 (2)
Integrating equation(2) w.r.t ‘r’ we get
21
4
p
u r C
x

 
  (3)
Where C is the constant of integration and its value is obtained from the boundary
condition at r=R, u=0

177. 177
21
0
4
p
R C
x

 
 
21
4
p
C R
x

 
 
Substituting this value of C in the equation (3)
2 21 1
4 4
p p
u r R
x x
 
   
 
2 21
( )
4
p
u R r
x

 
   (4)
In equation (3) μ,
p
x


amd R are constant. Which means the velocity u, varies with
the square of r.thus equation is an equation of parabola. This shows that the velocity
distribution across the section of a pipe is parabolic
Fig.24 shear stress and velocity distrubution

178. 178
Drop of pressure for a given length (L) of a pipe.
Fig.25 horizontal pipe
Average velocity U= 21
( )
8
p
R
x

 

2
8
( )
p U
x R
 

 
Integrate the above equation w.r.t.x, we get
1 1
2
2 2
8 U
p x
R

   
1 2
[ ]p p 2 2 1
8
[ ]
U
R x x

 
= 2
8 U
L
R

= 2
8
2
U
L
d

 
 
 
1 2
[ ]p p = 2
32 U
L
gd


2 1 21 2
8
[ ] [ ]
U
R
p p x x

   

179. 179
1 2
[ ]
g
p p


= 2
32 U
L
gd


QUESTIONS:
Examine the following velocity profile to find if the flow is attached or detached:
(a)
2
u y y
2
U
 
   
  
(b)
3 4
u y y y
2 2
U
   
     
     
(c)
2 3 4
u y y y
2 2
U
     
       
       
Case (a)
2
2
u y y
2
U
 
 
or
2
2
y y
u U 2
 
     
or
du 2 2y
U
dy
 
  
  
or
y 0
du 2U
dy 
 
  
 
positive value
Hence flow will remain attached
Case (b)
3 4
2y y y
u U 2
     
      
       

2 3
3 4
du 2 3y 8y
U
dy
 
   
   

180. 180
Now
y 0
du 2U
dy 
  
  
 
negative value
The flow is separated
Case(C)
2 3 4
y y y
u U 2 2
      
        
         

2 3
2 3 4
du 4y 3y 8y
U
dy
 
   
   

y 0
du
0
dy 
 
 
 
Hence flow is about to be detached
2.What are the effects of boundary layer?
The effects of boundary layer are –
(1) Boundary layer adds on to the effective thickness of the solid body
which is called displacement thickness. This displacement thickness increases the
pressure drag on the solid body.
(2) As velocity gradient exists in the boundary layer, shear force acts on
the boundary surface, resulting in skin friction drag.
3. What is no slip condition at the boundary? Why is the velocity of fluid
particles equal to zero at the static boundary? How does this velocity vary away
from the boundary?
The fluid particle on the boundary surface stick to the surface. The sticking
of the fluid particles to the surface is due to the viscosity of the fluid. Hence the fluid
particles on the surface will have same velocity as that of the surface. There is no
relative velocity between fluid particles and the boundary surface i.e. there is no
slippage of fluid particles on the boundary surface. This state is called no slip
condition at the boundary.

181. 181
In no slip condition, the fluid particles will have same velocity as that of the
boundary surface. The static boundary surface has zero velocity. Hence fluid
particles in this condition will have zero velocity.
On static boundary surface, the velocity of fluid particles at the boundary is
zero in the no slip condition. Further away the boundary surface, the fluid velocity
gradually increases as shown in the figure. The velocity changes from zero at the
boundary surface to the free stream velocity outside the boundary layer.
4. Why is the flow in the boundary layer analyzed on the principles of viscous
flow theory?
In the boundary layer, the viscous forces are predominant. The resistance is
offered by the boundary surface to the fluid flow which is mainly due to viscous
effects. All real fluids are viscous. The potential flow theory is applicable for non-
viscous fluids (ideal fluids). The potential flow theory can also be applied in the free
stream flow of the fluid outside the boundary layer where shear stresses are
practically negligible. The flow in the boundary layer can only be anlayzed on the
principles of viscous flow theory.
5. What are the factors affecting, the growth of boundary layer on a flat plate?
Factors affecting the growth of boundary layer on a flat plate are:
1. The thickness of the boundary layer grows from leading edge to trailing
edge of the plate. More is the distance ‘x’ from the leading edge, more
is the thickness of the boundary layer.
2. Free stream velocity. The thickness of the boundary layer at any point
‘x’ from the leading edge is inversely proportional to the free stream
velocity. If free stream velocity increases, the thickness of the
boundary layer decreases.
3. Density of the fluid. The thickness boundary layer varies inversely
with the density of the fluid. Hence lower density fluids give thicker
boundary layer as compared to high density fluids.
4. Viscosity of the fluids. The boundary layer thickness varies directly

182. 182
proportional to the viscosity of the fluids. If viscosity of fluid is high, it
will give thicker boundary layer.
6. What is shape factor?
The shape factor is the ratio of displacement thickness to the momentum
displacement. Hence shape factor is:
displacement thickness
H
momentum displacement

=
*


= 0
0
u
1 dy
U
u u
1 dy
U U


 
 
 
 
 
 


7. What do you understand by Navier-Stroke’s equations? What are the
applications of Naiver-Stroke’s equations?
Navier-Stoke’s equations are the fundamental equations for the analysis of
viscous flows. In incompressible flow without turbulence, there are four unknown
characteristics of the flow viz u, v, w, and P. Hence three Navier-Stroke’s equations
with continuity equation
u v w
0
x y z
   
   
   
are the sufficient conditions for the
determination of the flow characteristics. The Navier-Stroke’s equations are:
2 2 2
x 2 2 2
1 P du u u u
B
x dt x y z
     
     
      
2 2 2
y 2 2 2
1 P du v v v
B
y dt x y z
     
     
      
2 2 2
z 2 2 2
1 P dw w w w
B
z dt x y z
     
     
      

183. 183
xB , yB and zB are components of body force per unit mass. The solutions of
Navier Stoke’s equations are possible for such flow conditions where the fluid
characteristics such as viscosity and density are constant and boundary conditions are
simple.
The applications of Navier-Stoke’s equations are generally used for the analysis of :
(1) Laminar flow in circular pipes
(2) Laminar unit directional flow between stationary parallel plates
(3) Laminar flow between concentric rotating cylinders
(4) Laminar unit directional flow between parallel plates having relative
motion.
8. Examine the following velocity profile to find if the flow is attached or
detached:
(a)
2
u y y
2
U
 
   
  
(b)
3 4
u y y y
2 2
U
   
     
     
(c)
2 3 4
u y y y
2 2
U
     
       
       
Case (a)
2
2
u y y
2
U
 
 
or
2
2
y y
u U 2
 
     
or
du 2 2y
U
dy
 
  
  

184. 184
or
y 0
du 2U
dy 
 
  
 
positive value
Hence flow will remain attached
Case (b)
3 4
2y y y
u U 2
     
      
       

2 3
3 4
du 2 3y 8y
U
dy
 
   
   
Now
y 0
du 2U
dy 
  
  
 
negative value
The flow is separated
Case(C)
2 3 4
y y y
u U 2 2
      
        
         

2 3
2 3 4
du 4y 3y 8y
U
dy
 
   
   

y 0
du
0
dy 
 
 
 
Hence flow is about to be detached

185. 185
MODULE 5
DESCRIPTION OF THE BOUNDARY LAYER:
From fig 1 On-coming flow is irrotational and has a uniform velocity U. The
boundary layer starts out as a laminar boundary layer, in which fluid particles move
in smooth layers and the velocity distribution is approximately parabolic. As the flow
moves on, the continual action of shear stress tends to slow down additional fluid
particles, causing the boundary layer thickness to increase with distance downstream
from the leading edge. See below for a definition of the boundary layer thickness the
flow within the boundary layer is subject to wall shear, and dominated by viscous
forces. The velocity gradient (hence the rotation of fluid particles) is the largest at the
wall, and decreases with distance away from the wall, and tends to zero on matching
with the main stream flow.
Fig 1 boundary layer.
Roughly speaking, the flow is said to be rotational within the boundary layer, but is
irrotational outside the boundary layer. As the thickness of laminar boundary layer
increases, it becomes unstable and some eddying commences. These changes take
place over a short length known as the transition zone.
It finally transforms into a turbulent boundary layer, in which particles move in
haphazard paths. Due to the turbulent mixing, the velocity distribution is much more
uniform than that in the laminar boundary layer. The increase of thickness along the

186. 186
plate continues indefinitely but with a diminishing rate. If the plate is smooth (i.e.,
negligible roughness size), laminar flow persists in a very thin film called the viscous
sub-layer in immediate contact with the plate and it is in this sub-layer that the
greater part of the velocity change occurs.
THICKNESSES OF THE BOUNDARY LAYER
a) Boundary Layer Thickness δ
The velocity within the boundary layer increases to the velocity of the main stream
asymptotically. It is conventional to define the boundary layer thickness δ as the
distance from the boundary at which the velocity is 99% of the main stream velocity.
Fig.3 Boundary Layer Thickness
There is other ‘thicknesses’, precisely defined by mathematical expressions, which
are measures of the effect of the boundary layer on the flow.
b) Displacement Thickness δ*
It is defined by δ
*
is the distance by which the boundary surface would have
to be shifted outward if the fluid were frictionless and carried at the same mass flow
rate as the actual viscous flow. It also represents the outward displacement of the
streamlines caused by the viscous effects on the plate. Conceptually one may ‘add’
this displacement thickness to the actual wall and treat the flow over the ‘thickened’
body as an in viscid flow.

187. 187
Expression for δ*
Consider the flow of a fluid having free stream velocity equal to U over a thin
smooth plate as shown in fig4
Fig.4 thin smooth plate
At a distance ‘x’ from the leading edge consider a section 1-1. The velocity of the
fluid at B is zero and the velocity at C, which lies an the boundary layer, is U. thus
the velocity varies from zero at B to U at C, where BC is the thickness of the
boundary layer i.e
Distance BC= δ
At section 1-1 consider an elemental strip.
y= distance of the elemental strip from the plate
dy = thickness of the elemental strip from the plate.
u= velocity of the fluid at the elemental strip.
b= width of the plate
Area of the plate dA = b× dy
Mass of the fluid flowing per second flowing through the elemental strip
=ρ× Velocity× Area of elemental strip.

188. 188
=ρ×u× b× dy (1)
If there had been no plate, then the fluid would have been flowing with a constant
velocity equal to the free stream velocity (U) at he section 1-1.then the mass of the
fluid flowing per second through the elemental strip would have been
=ρ×U×b× dy (2)
As U is more than u , hence due to the presence of the plate and consequently due to
the formation of boundary layer, there will be a reduction in the mass flowing per
second through the elemental strip.
This reduction in mass/sec flowing through the elemental strip
=eq(2)-eq(1)
= (ρ×U×b× dy )-( ρ×u× b× dy )
=ρb(U-u) dy
Total reduction in mass of fluid/sec flowing through BC due to the plate
=
0
bu(U-u)dy

 (3)
Let the pklate is displaced by a distance δ* and then velocity of flow for the distance
δ* is equal to free stream velocity .loss of mass of the fluid /sec flowing through the
distance δ*
=ρ×Velocity×Area
= ρ×U× δ*×b (4)
Equating (4) and (3)
0
b(U-u)dy

 = ρ×U× δ*×b

189. 189
c) Momentum Thickness θ
Momentum Thickness is defined as the distance ,measured perpendicular to the boundary of
the solid body, by which the boundary should be displaced to compensate for the reduction
in momentum of the flowing fluid on account of boundary layer formation. it is denoted by θ
Expression for 
Consider the flow of a fluid having free stream velocity equal to U over a thin
smooth plate as shown in fig.4Consider the section 1-1 at a distance x from the
leading edge. The velocity of the fluid at B is zero and the velocity at C, which lies
an the boundary layer, is U. thus the velocity varies from zero at B to U at C, where
BC is the thickness of the boundary layer i.e
Distance BC= δ
At section 1-1 consider an elemental strip.
y= distance of the elemental strip from the plate
dy = thickness of the elemental strip from the plate.
u= velocity of the fluid at the elemental strip.
b= width of the plate
Area of the plate dA = b× dy
Mass of the fluid flowing per second flowing through the elemental strip
=ρ× Velocity× Area of elemental strip.
=ρ×u× b× dy (1)
Momentum of this fluid=Mass× Velocity = (ρ×u× b× dy ) ×u (2)

190. 190
Momentum of this fluid in the absence of boundary layer= (ρ×U×b× dy ) ×U (3)
Loss of momentum through the elemental strip= (ρ×U×b× dy ) ×U-(ρ×u× b×dy ) ×u
=ρ b u (U-u) dy (4)
Total loss in momentum/sec through BC=
0
bu

 (5)
Let = distance by which the plate is displaced when flowing with a constant
velocity U
Loss of momentum/sec of fluid flowing through distance  with a velocity U
= mass of the fluid through  ×velocity
= (ρ× ×b×U) ×U
=ρ bU2 (6)
Equating (6) and (5)
ρ bU2=
0
bu(U-u)dy


d) Energy thickness (δ**):
It is defined as the distance measured perpendicular to the boundary of the solid
body, by which the boundary should be displaced to compensate for the reduction in
kinetic energy of the flowing fluid on account of the boundary layer formation, it is
denoted by δ**.

191. 191
Expression for(δ**)
Consider the flow of a fluid having free stream velocity equal to U over a thin
smooth plate as shown in fig.4.At a distance ‘x’ from the leading edge consider a
section 1-1. The velocity of the fluid at B is zero and the velocity at C, which lies an
the boundary layer, is U. thus the velocity varies from zero at B to U at C, where BC
is the thickness of the boundary layer i.e
Distance BC= δ
At section 1-1 consider an elemental strip.
y= distance of the elemental strip from the plate
dy = thickness of the elemental strip from the plate.
u= velocity of the fluid at the elemental strip.
b= width of the plate
Area of the plate dA = b× dy
Mass of the fluid flowing per second flowing through the elemental strip
=ρ× Velocity× Area of elemental strip.
=ρ×u× b× dy (1)
Kinetic energy of the fluid=
1
2
× mass×velocity2=
1
2
× (ρ×u× b× dy ) ×u2 (2)
Kinetic energy of the fluid in the absence of boundary layer
=
1
2
× (ρ×U× b× dy ) ×U2 (3)
Loss of K.E energy through the elemental strip eq(3)-eq(2)
=
1
2
× (ρ×U× b× dy ) ×U2 -
1
2
× (ρ×u× b× dy ) ×u2

192. 192
Total loss of K.E of fluid passing through BC
=
0
bu

 (U2-u2) dy (4)
Let δ**= distance by which the plate is displaced to compensate for the reduction in
K.E
Loss of K.E through δ**of the fluid flowing with velocity U
=
1
2
× mass×velocity2=
1
2
× (ρ × area ×velocity) × velocity2
=
1
2
× (ρ × b× δ**×U) U2 (5)
Equating eq (5) and eq (4)
1
2
× (ρ × b× δ**×U) U2=
0
bu

 (U2-u2) dy
δ**=
0
u
U

 (1-u2/U2) dy
FRICTIONAL LOSS IN PIPE FLOW
When a liquid is flowing through a pipe, the velocity of the liquid layer adjacent to
the pipe wall is zero. The velocity of liquid goes on increasing from the wall and thus
velocity gradient and hence shear stresses are produced in the whole liquid due to
viscosity. This viscous action causes loss of energy which is usually known as
frictional loss.
On the basis of his experiments, William Froude gave the following laws of fluid
fraction for turbulent flow.
The friction resistance for turbulent flow is:
(i) proportional to Vn, where n varies from 1.5 to 2.0

193. 193
(ii) proportional to the density of fluid,
(iii) proportional to the area of surface in contact,
(iv) independent of pressure,
(v) dependent on the nature of the surface in contact.
Expression for loss of head due to Friction in pipes.
Consider a uniform horizontal pipe, having steady flow as shown in Fig. 5. Let 1-1
and 2-2 are two sections of pipe.
Let p1 = pressure intensity at section 1-1,
V1 = velocity of flow at section 1-1,
L = length of the pipe between sections 1-1 and 2-2,
Fig.5 uniform horizontal pipe
d = diameter of pipe,
f= frictional resistance per unit wetted area per unit velocity,
hf = loss of head due to friction,
p2,V2 = are values of pressure intensity and velocity at section 2-2.
Applying Bernoulli’s equations between sections 1-1 and 2-2,
Total head at 1-1 = Total head at 2-2 + loss of head due to friction between
1-1 and 2-2
or
2 2
1 1 2 2
1 2 f
p V p V
z z h
g 2g g 2g
     
 

194. 194
But z1 = z2 as pipe is horizontal
V1 = V2 as dia. of pipe is same at 1-1 and 2-2
 1 2
f
p p
h
g g
 
 
or 1 2
f
p p
h
g g
 
 
(i)
But fh is the head lost due to friction and hence intensity of pressure will be reduced
in the direction of flow by frictional resistance.
Now frictional resistance = frictional resistance per unit wetted area per unit velocity
 wetted area  velocity2
or 2
1F f dL V    1 2wettedarea d Lvelocity V V V     Q
= 2
f P L V    d Perimeter P  Q (ii)
The forces acting on the fluid between sections 1-1 and 2-2 are:
1. pressure force at section 1-1 = p1  A where A = Area of pipe
2. pressure force at section 2-2 = p2  A
3. frictional force F1 as shown in Fig.10.3.
Resolving all forces in the horizontal direction, we have
1 2p p 1A A F 0   ...(10.1)
or   2
1 2 1p p A F f P L V        2
1From ii ,F f PLV  Q
or
2
1 2
f P L V
p p
A
  
 
But from equation (i), 1 2 fp p pgh 
Equating the value of  1 2p p , we get
2
f
f P L V
pgh
A
  

195. 195
or 2
f
f P
h L V
g A

   

(iii)
In equation (iii),
2
P Wettedperimeter d 4
A Area dd
4

  


2
2
f
f 4 f 4LV
h L V
g d g d
 
     
 
(iv)
putting
f f
g 2



, where f is known as co-efficient of friction
Equation (iv), becomes as
2 2
f
4.f LV 4f.L.V
h .
2g d d 2g
 

...(a)
Equation (a) is known as Darcy-Weisbach equation. This equation is
commonly used for finding loss of head due to friction in pipes.
Sometimes equation (a) is written as
2
f
f.L.V
h
d 2g


...(b)
Then f is known as friction factor.
Expression for co-efficient of friction in terms of shear stress.
The forces acting on a fluid between sections 1-1 and 2-2 of Fig. 5 in horizontal
direction as
1p p2 1A A F 0  
or  1 2 1p p A F  force due to shear stress 0
= shear stress  surface area
= 0 d L   
or   2
1 2 0p p d d L
4

      2
A d
4
 
 
 
Q

196. 196
Cancelling d to both sides, we have
 1 2 0
d
p p L
4
   
or   0
1 2
4 L
p p
d
 
  ...(c)
Equation (c) can be written as
2
1 2
f
p p 4f.L.V
h
g d 2g

 
 
or  1 2p p 
4f.L.V
g
d 2g


...(d)
Equating the value of  1 2p p in equations (c) and (d),
2
4f.L.V
g
d 2g


or
2 2
0
fV g fV
g
2g 2g

   
or
2
0
V
f
2

  ...(e)
 0
2
2
f .
V



...(f)
(a) Darcy-Weisbach Formula.
The loss of head (or energy) in pipes due to friction is calculated from Darcy-
Weisbach equation
2
f
4.f.L.V
h
d 2g


...(a)
Where fh = loss of head due to friction
f = Co-efficient of friction which is a function of Reynold number
=
e
16
R
for eR < 2000 (viscous flow)

197. 197
= 1/ 4
e
0.079
R
for eR varying from 4000 to 106
L = length of pipe,
V = mean velocity of flow,
d = diameter of pipe.
(b) Chezy’s Formula for loss of head due to friction in pipes.
Expression for loss of head due to friction in pipes is derived.
2
f
f P
h L V
g A

   

(a)
Where fh = loss of head due to friction,
P = wetted perimeter of pipe,
A = area of cross-section of pipe,
L = Length of pipe,
And V = mean velocity of flow.
Now the ratio of
 
A Areaof flow
P Perimeter wetted
 
  
 
is called hydraulic mean depth
or hydraulic radius and is denoted by m.
 Hydraulic mean depth,
2
d
A d4m
P d 4

 
 (b)
Substituting
A
m
P
 or
P
A
=
1
m
in equation (b), we get
2
f
f
h L V
g

  

Or 2 f
f
g 1 g h
V h m m
f L f L
 
      
 

f fg h g h
V m m
f L f L
 
   
 
(c)

198. 198
Let
g
f


= C, where C is a constant known as Chezy’s constant and fh
L
=
i, where i is loss of head per unit length of pipe.
Substituting the values of
g
f


and fh
L
in equation (c), we get
V C mi (d)
Equation (d) is known as Chezy’s formula. Thus the loss of head due to
friction in pipe from Chezy’s formula can be obtained if the velocity of flow through
pipe and also the value of C is known. The value of m for pipe is always equal to d/4.
HYDRAULIC GRADIENT AND TOTAL ENERGY LINE
The concept of hydraulic gradient line and total energy line is very useful in
the study of flow of fluids through pipes. They are defined as:
Hydraulic Gradient Line.
It is defined as the line which gives the sum of pressure head
p
w
 
 
 
and datum head
(z) of a flowing fluid in a pipe with respect to some reference line or it is the line
which is obtained by joining the top of all vertical ordinates, showing the pressure
head (p/w) of a flowing fluid in a pipe from the centre of the pipe. It is briefly written
as H.G.L. (Hydraulic Gradient Line).
Total Energy Line.
It is defined as the line which gives the sum of pressure head, datum head and kinetic
head of a flowing fluid in a pipe with respect to some reference line. It is also defined
as the line which is obtained by joining the tops of all vertical ordinates showing the
sum of pressure head and kinetic head from the centre of the pipe. It is briefly written
as T.E.L. (Total Energy Line).
FLOW THROUGH SYPHON
Syphon is a long bent pipe which is used to transfer liquid from a reservoir at a
higher elevation to another reservoir at a lower level when the two reservoirs are

199. 199
separated by a hill or high level ground as shown in Fig. 6 The point C which is at
the highest of the syphon is called the summit. As the point C is above the free
surface of the water in the tank A, the pressure at C will be less than atmospheric
pressure. Theoretically, the pressure at C may be reduced to - 10.3 m of water but in
actual practice this pressure is only - 7.6 m of water or 10.3 - 7.6 = 2.7 m of water
absolute If the pressure at C becomes less than 2.7 m of water absolute, the dissolved
air and other gases would come out from water and collect at the summit. The flow
of water will be obstructed. Syphon is used in the following cases:
1. To carry water from one reservoir to another reservoir separated by a hill
or ridge.
2. To take out the liquid from a tank which is not having any outlet.
3. To empty a channel not provided with any outlet sluice.
f i g
.
Fig.6 syphon
FLOW THROUGH PIPES IN SERIES OR FLOW THROUGH
COMPOUND PIPES
Pipes in series or compound pipes is defined as the pipes of different lengths and
different diameters connected end to end (in series) to form a pipe line as shown in
Fig. 7

200. 200
Fig.7 pipe in series
Let, L1, L2, L3 = length of pipes 1, 2 and 3 respectively
d1, d2, d3 = diameter of pipes 1, 2, 3 respectively
V1, V2, V3 = velocity of flow through pipes 1, 2, 3
f1, f2, f3 = co-efficient of frictions for pipes 1, 2, 3
H = difference of water level in the two tanks.
The discharge passing through each pipe is same.
 1 1 2 2 3 3Q A V A V A V  
The difference in liquid surface levels is equal to the sum of the total head loss in the
pipes.

 2 22 2 2 2 2
2 3 3 3 31 1 1 1 2 2 2 2 2
1 2 3
V V 4f L V0.5V 4f L V 0.5V 4f L V V
I
2g d 2g 2g d 2g 2g d 2g 2g

      
  
...(1)
If minor losses are neglected, then above equation becomes as
22 2
3 3 31 1 1 2 2 2
1 2 3
4f L V4f L V 4f L V
H
d 2g d 2g d 2g
  
  
...(2)
If the co-efficient of friction is same for all pipes
i.e., f1 = f2 = f3 = f, then equation (3) becomes as
22 2
3 3 31 1 1 2 2 2
1 2 3
4f L V4f L V 4f L V
H
d 2g d 2g d 2g
  
  

201. 201
=
22 2
3 31 1 2 2
1 2 3
L V4f L V L V
2g d d d
 
  
 
. ... (4)
FLOW THROUGH PARALLEL PIPES
Consider a main pipe which divides into two or more branches as shown in
Fig. 7 and again join together downstream to form a single pipe, then the branch
pipes are said to be connected in parallel. The discharge through the main is
increased by connecting pipes in parallel. The rate of flow in the main pipe is equal
to the sum of rate of flow through branch pipes. Hence from Fig.8 we have
Q = Q1 + Q2 ...(1)
In this, arrangement, the loss of head for each branch pipe is same.
Loss of head for branch pipe 1 = Loss of head for branch pipe 2
Fig.8 parallel pipes
or
2
1 1 1
1
4f L V
d 2g
=
2
2 2 2
2
4f L V
d 2g
...(1)
If f1 = f2, then
2
1 1
1
L V
d 2g
=
2
2 2
2
L V
d 2g
...(2)
POWER TRANSMISSION THROUGH PIPES
Power is transmitted through pipes by flowing water or other liquids flowing through
them. The power transmitted depends upon (i) the weight of liquid flowing through

202. 202
the pipe and (ii) the total head available at the end of the pipe. Consider a pipe AB
connected to a tank as shown in Fig. 9. The power available at the end B of the pipe
and the condition for maximum transmission of power will be obtained as mentioned
below.
Let L = length of the pipe,
d = diameter of the pipe,
H = total head available at the inlet of pipe,
V = velocity of flow in pipe,
hf = loss of head due to friction, and f = co-efficient of friction.
Fig.9 power transmission through pipes
The head available at the outlet of the pipe, if minor losses are neglected
= Total head at inlet - loss of head due to friction
=
2
f
4f L V
H h H
d 2g
 
  

2
f
4f L V
h
d 2g
   
 
  
Q
Weight of water flowing through pipe per sec,
W = g × volume of water per sec = g × Area × Velocity
= 2
g d V
4

  
The power transmitted at the outlet of the pipe
= weight of water per sec × head at outlet

203. 203
=
2
2 4f L V
g d V H
4 d 2g
    
           
Watts
Power transmitted at outlet of the pipe,
2
2g 4fLV
P d V H
1000 4 d 2g
  
      
kW
Efficiency of power transmission,
=
Power available at outlet of the pipe
Power supplied at the inlet of the pipe
=
Weight of water per sec Head available at outlet
Weight of water per sec Head at inlet


=
 f fW H h H h
W H H
  


Condition for Maximum Transmission of Power. The condition for
maximum transmission of power is obtained by differentiating equation with respect
to V and equating the same to zero.
Thus  
d
P 0
dV

or
2
2d g 4fLV
d HV 0
dV 1000 4 d 2g
   
        
or
2
2g 4 3 f L V
d H 0
1000 4 d 2g
      
     
or
2
4fLV
H 3 0
d 2g
  

or fh 3 h 0  
2
f
4fLV
h
d 2g
 
   
Q
 H=3hf or hf =
H
3
This is the condition for maximum transmission of power. It states that
power transmitted through a pipe is maximum when the loss of head due to friction is
one-third of the total head at inlet.

204. 204
Maximum Efficiency of Transmission of power.
Efficiency of power transmission through pipe is given by equation
fH h
H

 
For maximum power transmission through pipe the condition is given by
equation) as
f
H
h
3

Substituting the value of hf in efficiency, we get maximum,
max
H H/3
H

  =1-
1 2
3 3
 or 66.7%
WATER HAMMER IN PIPES
Consider a long pipe AB as shown in Fig.10 connected at one end to a tank
containing water at a height of H from the centre of the pipe. At the other end of the
pipe, a valve to regulate the flow of water is provided. When the valve is completely
open, the water is flowing with a velocity, V in the pipe. If now the valve is suddenly
closed, the momentum of the flowing water will be destroyed and consequently a
wave of high pressure will be set up. This wave of high pressure will be transmitted
along the pipe with a velocity equal to the velocity of sound wave and may create
noise called knocking. Also this wave of high pressure has the effect of hammering
action on the walls of the pipe and hence it is known as water hammer.
Fig.10 water hammer in pipes

205. 205
The pressure rise due to water hammer depends upon: (i) the velocity of
flow of water in pipe, (ii) the length of pipe, (ii) time taken to close the valve, (iii)
Elastic properties of the material of the pipe. The following cases of water hammer in
pipes will be considered:
1. Gradual closure of valve,
2. Sudden closure of valve and considering pipe rigid, and
3. Sudden closure of valve and considering pipe elastic.
Gradual closure of Valve.
Let the water is flowing through the pipe AB shown in Fig. 10and the valve
provided at the end of the pipe is closed gradually.
Let- A = area of cross-section of the pipe AB,
L = length of pipe,
V = velocity of flow of water through pipe,
t = time in second required to close the valve, and
p = intensity of pressure wave produced.
Mass of water in pipe AB = volumeof water A L   
The valve is closed gradually in time ‘t’ seconds and hence the water is brought from
initial velocity V to zero velocity in time seconds.
Retardation of water =
Changeof velocity V 0 V
Time t t

 
Retarding force = Mass ×Retardation = AL
V
t
 (i)
If p is the intensity of pressure wave produced due to closure, of the valve, the force
due to pressure wave,
= p areaof pipe p A   (ii)
Equating the two forces, given by equation (i) and (ii),

206. 206
V
AL p A
t
   
LV
p
t


Head of pressure,
p LV LV
H
g g t g t
 
  
    
or
LV
H
gt
 (a)
(i) The valve closure is said to be gradual if t >
2L
C
Where t = time in sec, C = velocity of pressure wave
(ii) The valve closure is said to be sudden if t <
2L
C
Where C = velocity of pressure wave.
Sudden Closure of Valve and Pipe is Rigid.
Equation (a) gives the relation between increase of pressure due to water hammer in
pipe and the time required to close the valve. If t = 0, the increase in pressure will be
infinite. But from experiments, it is observed that the increase in pressure due to
water hammer is finite, even for a very rapid closure of valve. Thus equation (a) is
valid only for (i) Incompressible fluids and (ii) when pipe is rigid. But when a wave
of high pressure is created, the liquids get compressed to some extent and also pipe
material gets stretched. For a sudden closure of valve [the valve of t is small and
hence a wave of high pressure is created] the following two cases will be considered:
(i) Sudden closure of valve and pipe is rigid, and
(ii) Sudden closure of valve and pipe is elastic.
Consider a pipe AB in which water is flowing as shown in Fig. 10 Let the pipe is
rigid and valve fitted at the end B is closed suddenly.
Let A = Area of cross-section of pipe AB,
L = Length of pipe,
V = Velocity of flow of water through pipe,

207. 207
p = Intensity of pressure wave produced,
K = Bulk modules of water.
When the valve is closed suddenly, the kinetic energy of the flowing water is
converted into strain energy of water if the effect of friction is neglected and pipe
wall is assumed perfectly rigid.
 Loss of kinetic energy = 21
massof waterinpipe V
2
 
= 21
.AL V
2
  (Q mass = × volume = × A × L)
Gain of strain energy =
2 2
1 p 1 p
volume AL
2 K 2 K
 
    
 
Equating loss of kinetic energy to gain of strain energy

2
21 1 p
AL V AL
2 2 K
   
or 2 2 21 2K
p AL V KV
2 AL
     

2
2 K
p KV V K V

    

= V ×C  K/ C Q ...(b)
Where C = velocity * of pressure wave.
Sudden Closure of Valve and Pipe is Elastic.
Consider the pipe AB in which water is flowing as shown in Fig. 10. Let the
thickness‘t’ of the pipe wall is small compared to the diameter D of the pipe and also
let the pipe is elastic.
Let E = Modulus of Elasticity of the pipe material,
1
m
= Poisson’s ratio for pipe material,

208. 208
p = Increase of pressure due to water hammer,
t = Thickness of the pipe wall,
D = Diameter of the pipe.
When the valve is closed suddenly, a wave of high pressure of intensity p will be
produced in the water. Due to this high pressure p, circumferential and longitudinal
stress in the pipe wall will be produced.
Let f1 = longitudinal stress in pipe
fc = circumferential stress in pipe,
The magnitude of the stress are given as l
pD
f
4t
 and c
pD
f
2t

Now from the knowledge of strength of material we know, strain energy
stored in pipe material per unit volume
= 2 2 l c
l c
2f f1
f f
2E m
 
  
 
=
2 2 2 2 2 2 2 2
2 2 2
pD pD
2
1 pD pD 1 p D p D p D4t 2t
2E 4t 2t m 2E 16t 4t 4mt
 
       
         
      
 
Taking
1 1
m 4

1
i.e.,poisson ratio
4
 
 
 
Strain energy stored in pipe material per unit volume
2 2 2 2 2 2 2 2 2 2
2 2 2 2 2
1 p D p D p D 1 p D p D
2E 2E16t 4t 4t 4 4t 8Et
 
      
 
Total volume of pipe material = D t L  
 Total strain energy stored in pipe material
= Strain energy per unit Volume × total volume

209. 209
=
2 2 2 3
2
p D p D L
D t L
8Et8Et

    
=
2 2 2
p D DL p A DL
8Et 2Et
   

2
D
Areaof pipe A
4
 
   
 
Q
Now loss of kinetic energy of water = 2 21 1
mV AL V
2 2
    m AL Q
Gain of strain energy in water =
2 2
1 p 1 p
volume AL
2 K 2 K
 
    
 
Then Loss of kinetic energy of water = Gain of strain energy in water + Strain energy
stored in pipe material.

2 2
21 1 p p A DL
AL V AL
2 2 K 2Et
  
      
 
Divide by AL,
2 3 2 2
V 1 p p D p 1 D
2 2 K 8Et 2 K Et
  
    
 
or 2 2
V p 
1 D
K Et
 
 
 

2
2 V
p
1 D
K Et



or
2
V
p V
1 D 1 D
K Et K Et
 
  
 
...(c)
Time taken by pressure wave to travel from the value to the tank and
from tank to the valve
Let T = the required time taken by pressure wave
L = Length of the pipe
C = Velocity of pressure wave
Then total distance = L + L = 2L
Time, T =
Distance 2L
.
Velocityof pressure wave C


210. 210
DISCHARGE THROUGH OPEN CHANNEL BY CHEZY’S
FORMULA
Consider uniform flow of water in a channel as shown in Fig. 11. As the flow is
uniform, it means the velocity, depth of flow and area of flow will be constant for a
given length of the channel. Consider sections 1-1 and 2-2.
fig.11 open channel
Let L = Length of channel
A = Area of flow of water,
i = Slope of the bed
V = Mean velocity of flow of water,
P = Wetted perimeter of the cross-section,
f = Frictional resistance per unit velocity per unit area.
The weight of water between sections 1-1 and 2-2.
W = Specific weight of water × volume of water
= w× A× L
Component of W along direction of flow = W × sin i = wAL sin i
Frictional resistance against motion of water = f × surface area × (velocity)n
The value of n is found experimentally equal to 2 and surface area = P × L.

211. 211
Frictional resistance against motion = f × P × L × V2
The forces acting on the water between sections 1-1 and 2-2 are:
1. Component of weight of water along the direction of flow,
2. Friction resistance against flow of water,
3. Pressure force at section 1-1,
4. Pressure force at section 2-2.
As the depths of water at the sections 1-1 and 2-2 are the same the pressure
forces on these two sections are same and acting in the opposite direction. Hence
they channel each other. In case of uniform flow, the velocity of flow is constant for
the given length of the channel. Hence there is no acceleration acting on the water.
Hence the resultant force acting in the direction of flow, we get
wAL sin i - f × P × L × V2 = 0
wAL sin i = f × P × L × V2
2 wALsini w A
V sini
f P L f P
   
 
or
w A
V sini
f P
   ...(i)
But
A
P
= m
= hydraulic mean depth or hydraulic radius,
w
f
= C = Chezy’s constant
Substituting the values of
A
P
and
w
f
in equation (i), V = C msini
For small values of i, sin i tan i i V=C m sin i ; (a)
 Discharge, Q = Area × Velocity = A × V

212. 212
= A× C mi (b)
MOST ECONOMICAL SECTION OF CHANNELS
A section of a channel is said to be most economical when the cost of construction of
the channel is minimum. But the cost of construction of a channel depends upon the
excavation and the lining. To keep the cost down or minimum, the wetted perimeter,
for a given discharge, should be minimum. This condition is utilized for determining
the dimensions of a economical sections of different form of channels.
Most economical section is also called the best section or most efficient section as
the discharge, passing through a most economical section of channel for a given
cross-sectional area (A), slope of the bed (i) and a resistance co-efficient, is
maximum. But the discharge, Q is given by equation
Q = AC mi = AC
A i
P
 A
m
P
 
 
 
Q
For a given A, i and resistance co-efficient C, the above equation is written as
Q = K
1
P
, where K = AC Ai = constant
Hence the discharge, Q will be maximum, when the wetted perimeter P is minimum.
This condition will be used for determining the best section of a channel i.e, best
dimensions of a channel for a given area.
The conditions to be most economical for the following shapes of channels will be
considered:
1. Rectangular Channel, 2. Trapezoidal Channel,.
Most Economical Rectangular Channel.
The condition for most economical section is that for a given area, the perimeter
should be minimum. Consider a rectangular channel as shown in Fig.12

213. 213
Fig.12 Rectangular Channel
Let b = width of the channel,
d = depth of the flow.
Area of flow, A = b×d ...(i)
Wetted perimeter, P = d + b + d = b + 2d ...(ii)
From equation (i), b =
A
d
Substituting the value of b in (ii),
A
P b 2d 2d
d
    ...(iii)
For most economical section, P should be minimum for a given srea.
or
 
dP
0
d d

Differentiating the equation (iii) with respect to d and equating the same to
zero, we get
 
d A
2d 0
d d d
 
  
 
or 2
A
0
d 2
 

or
2
A 2d
But A = b×d,  b×d= 2d2 or b = 2d
Now hydraulic mean depth,
A b d
m
P b 2d

 

(Q A = bd, P = b + 2d)

214. 214
=
2d d
2d 2d


(Q b = 2d)
=
2
2d d
4d 2
 ...(16.10)
it is clear that rectangular channel will be most economical when:
(i) Either b = 2d means width is two times depth of flow.
(ii) Or m =
d
2
means hydraulic depth is half the depth of flow.
Most Economical Trapezoidal Channel.
The trapezoidal section of a channel will be most economical, when its wetted
perimeter is minimum. Consider a trapezoidal section of channel as shown in Fig. 13
Let b = width of channel at bottom,
d = depth of flow,
= angle made by the sides with horizontal.
Fig.13 Trapezoidal Channel
The side slop is given as 1 vertical to n horizontal.
 Area of flow,
   BC AD b b 2nd
A d d
2 2
  
    (Q AD = b+2nd)
=  
2b 2nd
d b nd d
2

    (i)

215. 215

A
b nd
d
 

A
b nd
d
  ...(ii)
Now wetted perimeter, P = AB + BC + CD = BC + 2CD (Q AB = CD)
= 2 2 2 2 2 2
b 2 CE DE b 2 n d d b 2d n 1        (iia)
Substituting the value of b from equation (ii), we get
2A
P nd 2d n 1
d
    (iii)
For most economical section, P should be minimum or
 
dP
0
d d

 Differentiating equation (iii) with respect to d and equating it equal to
zero, we get
 
2d A
nd 2d n 1 0
d d d
 
    
 
or 2
2
A
n 2 n 1 0
d
    (Q n is constant)
or 2
2
A
n 2 n 1
d
  
Substituting the value of A from equation (i) in the above equation,
  2
2
b nd d
n 2 n 1
d

   or 2b nd
n 2 n 1
d

  
or 2b nd nd b 2nd
2 n 1
d b
  
   or 2b 2nd
2 n 1
d

  (a)
But from Fig. 16.8,
b 2nd
d

half of top width
And 2
d n 1 = CD = one of the sloping side

216. 216
Equation (a) is the required condition for a trapezoidal section to be most
economical, which can be expressed as half of the top width must be equal to one of
the sloping sides of the channel.
(ii) Hydraulic mean depth,
 
 
b nd dA d
m
P 2 b nd 2

  

(b)
Hence for a trapezoidal section to be most economical hydraulic mean depth must be
equal to half the depth of flow.
MOODY CHART:
The Moody chart is a graph in non-dimensional form that relates the friction factor,
Reynolds number and relative roughness for fully developed flow in a circular pipe.
It can be used for working out pressure drop or flow rate down such a pipe.
Description:
This dimensionless chart is used to work out pressure drop, ΔP (Pa) (or head
loss, hf (m)) and flow rate through pipes. Head loss can be calculated using
the Darcy–Weisbach equation:
And pressure drop can then be evaluated as:
Or directly from
Where ρ is the density of the fluid, V is the average velocity in the pipe, f is the
friction factor from the Moody chart, l is the length of the pipe and d is the pipe
diameter. The basic chart plots Darcy–Weisbach friction factor against Reynolds
number for a variety of relative rough nesses and flow regimes. The relative
roughness being the ratio of the mean height of roughness of the pipe to the pipe
diameter or .
Moody chart showing friction factor plotted against Reynolds number for various
rough nesses The Moody chart can be divided into two regimes of
flow: laminar and turbulent. For the laminar flow regime, the Darcy–

217. 217
Weisbach friction factor was determined analytically by Poiseuille and is used.
In this regime roughness has no discernible effect. For the turbulent flow regime, the
relationship between the friction factor and the Reynolds number is more complex
and is governed by the Colebrook equation which is implicit in f:
In 1944, Lewis F. Moody plotted the Darcy–Weisbach friction factor into what is
now known as the Moody chart
HYDRAULIC JUMP OR STANDING WAVE:
Consider the flow of water over a dam as shown in fig. 14
Fig.14 hydraulic jump
The height of the water at the section1-1 is very small. As we moves towards the
down stream the height of the water increases rapidly over a short length of the
channel. This is because at section 1, the flow is a shooting flow as the depth of the
water at this section 1-1 is less than the critical depth. Shooting flow is unstable flow
and does not continue on the downstream side. Then this shooting will convert itself
to a streaming or tranquil flow and hence depth of water will increase. This sudden
increase of depth of water is called a hydraulic jump.
QUESTIONS
1. Find the displacement thickness, momentum thickness and energy thickness
for the velocity distribution in the boundary layer given by –

218. 218
2
u y y
2
U
   
    
    
Displacement thickness (*)
* =
0
u
1 dy
U

 
 
 

=
2
0
y y
1 2 dy
     
     
      

=
2 3
2
0
2 y 1 y
y . .
2 3

 
  
  
=
2 3
2
1
3
 
  
 
=
1
1 1
3
 
   
 
=
3

Momentum thickness ()
 =
0
u u
1 dy
U U
  
 
 

=
2 2
2 20
y y 2y y
2 1 dy
    
     
     

=
2 3 3 3 4
2 3 3 40
2y 4y 2y y 2y y
dy
  
     
     

=
2 3 4
2 3 40
2y 5y 4y y
dy
  
   
    


219. 219
=
2 3 4 5
2 3 4
0
2 y 5 y 4 y 1 y
. . . .
2 3 4 5

 
   
    
=
5 1
. .
3 5
   
=
15 25 15 3
15
   
  
=
2
. .
15

Energy thickness ()
a =
2
20
u u
1 dy
U U
  
  
 

=
2 2
2 20
2y y 2y y
1 dy
     
                

=
2 2 4 3
2 2 4 30
2y y 4y y 4y
1 dy
  
            

=
3 5 4 2 4 6 2
3 5 4 2 4 6 50
2y 8y 2y 8y y 4y y 4y
dy
  
       
        

=
2 3 4 5 6 7
2 3 4 5 60
0
2 y 1 y 8 y 12 y 6 y 1 y
. . . . . .
2 3 4 5 6 7

  
     
      

=
1 12 1
. 2. .
3 5 7
    
=
105 35 210 252 105 15
105
     
   
=
2
105

2. The velocity distribution in the boundary layer over a plate is given by –

220. 220
3
u 3 y 1 y
U 2 2
 
   
  
Find the Displacement thickness (*)
* =
0
u
1 dy
U

 
 
 

= 30
3 y 1 y
1 . . dy
2 2
   
      

=
2
20
3 y 1 y
1 . dy
2 2
  
     

=
2 4
2
0
3 y 1 y
y . .
2 5.2 42.

 
  
 
=
3 1
4
 
     
 
=
8 6 1
8
  
   
=
3
.
8

3. The velocity distribution in the boundary layer is given by
2
2
u 3 y 1 y
. . .
U 2 2
 
 
Find (1) Ratio of displacement thickness (*) to thickness of boundary layer ()
and (2) Ratio of momentum thickness () to boundary layer thickness () -
(*) =
0
u
1 dy
U
 
 
 

=
2
20
3 y 1 y
1 . dy
2 2
   
        


221. 221
=
2 3
2
0
3 y 1 y
y . .
2. 2 32.

 
  
  
=
3 1
.
4 6
    
=
12 9 2
12
  
 
 
*
 =
5
.
12

*


=
5
12
Momentum thickness  =
0
u u
1 dy
U U
   
  
  

or  =
2 2
2 20
3 y y 3 y y
. 1 . dy
2 22 2
     
                

=
2 2
2 20
3 y y 3 y y
. 1 . dy
2 22 2
  
          

=
2 3 2 3 4
2 3 2 3 40
3 y 9 y 3y y 3 y y
. . dy
2 4 24 2 4
 
            

=
2 3 4
2 3 40
3 y 11 y 3 y y
. dy
2 4 2 4
 
        

=
2 3 4 5
2 3 4
0
3 y 11 y 3 y 1 y
. . . .
2.8 2 3 4 54. 2. 4

 
   
   
=
3 11 3 1
4 12 8 20
 
    
 
=
90 110 45 6
120
   
 
 

222. 222
=
19
.
120




=
19
120
4. In a laminar boundary layer over a flat plate, velocity distribution is given by
u = a + by + cy2. Using the boundary conditions, find the values of the constants
a, b and c.
u = a + by + cy2
Boundary conditions are -
(1) u = 0 at y = 0
(2) u = U at y = 
(3)
du
dy
= 0 at y = 
Now if y = 0 then u = 0, we get -
0 = a a = 0
Now u = U at y = 
U = b +c2 (i)
Now
du
dy
= 0 at y = 

du
dy
= b + 2 cy
Now
y
du
0
dy 
 
 
 
or b + 2 c = 0 (ii)
Using eqn (i) & (ii)
2 2 2
U 2c c c       

223. 223
or 2
U
c  

 2
U
b 2c 2

       

U
2

Hence velocity distribution is -
2
y y
u 2U U
   
    
    
5. The velocity distribution in a turbulent boundary layer is as per power law -
1/ 7
u y
U
 
  
 
Find (1) displacement thickness (2) momentum thickness (3) energy thickness
(4) shape factor and (5) energy loss if U = 10 m/s, = 20 mm,  = 12 kg/m3
Displacement thickness (*)
* =
0
u
1 dy
U
 
 
 

=
1/7
0
y
1 dy
   
  
   

=
8/ 7
1/ 7
0
y
y
8
7

 
 
 
  
 
=
7
1
8
 
  
 
=
8

Momentum thickness

224. 224
 =
0
u
1 dy
U
 
 
 

=
1/ 7
1/ 7
0
y y
1 dy
     
    
      

=
1/7 2/7
1/7 2/70
y y
dy
 
    

=
8/ 7 9/ 7
1/ 7
2/ 7
0
1 y y
.
8 9
.
7 7

 
 
 
 
 
=
7 7
8 9
 
  
 
=
7
.
72

Energy thickness
 =
2
20
u u
1 dy
U U
  
  
 

=
1/7 2/7
1/7 2/70
y y
1 dy
  
    

=
1/7 3/7
1/7 3/70
y y
dy
  
    

=
8/ 7 10/ 7
1/ 7 3/ 7
0
y y
8 10
. .
7 7

 
 
 
  
 
=
7 7
.
8 10
 
  
 

225. 225
=
5 4
.7
40
 
  
 
=
7
.
40

Shape factor
Shape factor H =
*


= 8
7
.
72


=
9
7
Energy loss
37 7
. 20 10
40 40

     
Energy loss per unit width =   21
U U
2

= 3 21
1.2 3.5 10 10 10
2

    
= 2.1 Nm/s
7.What are the effects of boundary layer?
The effects of boundary layer are –
(1) Boundary layer adds on to the effective thickness of the solid body
which is called displacement thickness. This displacement thickness increases the
pressure drag on the solid body.
(2) As velocity gradient exists in the boundary layer, shear force acts on
the boundary surface, resulting in skin friction drag.

226. 226
8. What is no slip condition at the boundary? Why is the velocity of fluid
particles equal to zero at the static boundary? How does this velocity vary away
from the boundary?
The fluid particle on the boundary surface stick to the surface. The sticking
of the fluid particles to the surface is due to the viscosity of the fluid. Hence the fluid
particles on the surface will have same velocity as that of the surface. There is no
relative velocity between fluid particles and the boundary surface i.e. there is no
slippage of fluid particles on the boundary surface. This state is called no slip
condition at the boundary.
In no slip condition, the fluid particles will have same velocity as that of the
boundary surface. The static boundary surface has zero velocity. Hence fluid
particles in this condition will have zero velocity.
On static boundary surface, the velocity of fluid particles at the boundary is
zero in the no slip condition. Further away the boundary surface, the fluid velocity
gradually increases as shown in the figure. The velocity changes from zero at the
boundary surface to the free stream velocity outside the boundary layer.
9. Why is the flow in the boundary layer analyzed on the principles of viscous
flow theory?
In the boundary layer, the viscous forces are predominant. The resistance is
offered by the boundary surface to the fluid flow which is mainly due to viscous
effects. All real fluids are viscous. The potential flow theory is applicable for non-
viscous fluids (ideal fluids). The potential flow theory can also be applied in the free
stream flow of the fluid outside the boundary layer where shear stresses are
practically negligible. The flow in the boundary layer can only be analyzed on the
principles of viscous flow theory.
10. What are the factors affecting, the growth of boundary layer on a flat plate?
Factors affecting the growth of boundary layer on a flat plate are:
1. The thickness of the boundary layer grows from leading edge to trailing
edge of the plate. More is the distance ‘x’ from the leading edge, more

227. 227
is the thickness of the boundary layer.
2. Free stream velocity. The thickness of the boundary layer at any point
‘x’ from the leading edge is inversely proportional to the free stream
velocity. If free stream velocity increases, the thickness of the
boundary layer decreases.
3. Density of the fluid. The thickness boundary layer varies inversely
with the density of the fluid. Hence lower density fluids give thicker
boundary layer as compared to high density fluids.
4. Viscosity of the fluids. The boundary layer thickness varies directly
proportional to the viscosity of the fluids. If viscosity of fluid is high, it
will give thicker boundary layer.
11. What is shape factor?
The shape factor is the ratio of displacement thickness to the momentum
displacement. Hence shape factor is:
displacement thickness
H
momentum displacement

=
*


= 0
0
u
1 dy
U
u u
1 dy
U U


 
 
 
 
 
 


12. What do you understand by Navier-Stroke’s equations? What are the
applications of Naiver-Stroke’s equations?
Navier-Stoke’s equations are the fundamental equations for the analysis of
viscous flows. In incompressible flow without turbulence, there are four unknown
characteristics of the flow viz u, v, w, and P. Hence three Navier-Stroke’s equations

228. 228
with continuity equation
u v w
0
x y z
   
   
   
are the sufficient conditions for the
determination of the flow characteristics. The Navier-Stroke’s equations are:
2 2 2
x 2 2 2
1 P du u u u
B
x dt x y z
     
     
      
2 2 2
y 2 2 2
1 P du v v v
B
y dt x y z
     
     
      
2 2 2
z 2 2 2
1 P dw w w w
B
z dt x y z
     
     
      
xB , yB and zB are components of body force per unit mass. The solutions of
Navier Stoke’s equations are possible for such flow conditions where the fluid
characteristics such as viscosity and density are constant and boundary conditions are
simple.
The applications of Navier-Stoke’s equations are generally used for the analysis of :
(1) Laminar flow in circular pipes
(2) Laminar unit directional flow between stationary parallel plates
(3) Laminar flow between concentric rotating cylinders
(4) Laminar unit directional flow between parallel plates having relative
motion.
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