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ME : MECHANICAL ENGINEERING
MODULE 3 : Fluid Mechanics & Turbomachinery

INDEX
FLUID MECHANICS
Contents Topics Pg. No.

Properties of Fluids 1

Fluid Statics 13

Buoyancy and Floatation 33

Flow Kinematics 42

Notes Viscous Flow 80

Boundary Layer Flow 86

Turbulent Flow 94

Flow Through Pipes 102

LMR (Last Minute Revision) 111

Assignment1 119

Assignment2 122
Assignments
Assignment3 125

Assignment4 129

Test Paper1 132

Test Paper2 135

Test Papers
Test Paper3 138

Test Paper4 140

Practice Problems 143

TURBOMACHINERY
Contents Topics Pg. No.

Introduction 155

Turbines 155

Pelton Wheel 156

Radial Flow Reaction Turbines 164

Francis Turbine 165

Axial Flow Reaction Turbine 171

Kaplan Turbine 172

Draft Tube 172

Unit Quantities 176

Notes Performance Characteristic Curves 177

Specific Speed of Turbine 178

Centrifugal Pumps 180

Specific Speed of Pump 184

Performance Characteristic Curves of Centrifugal
185
Pump
Cavitation 187

Net Positive Suction Head (NPSH) 188

Model Testing of Centrifugal pumps 189

List of Formulae 190

LMR (Last Minute Revision) 191

Assignment−1 193

Assignments Assignment−2 196

Assignment−3 198

Test Paper−1 200

Test Papers
Test Paper−2 203
SOLUTIONS
FLUID MECHANICS

Answer Key 206

Assignments
Model Solutions 207

Answer Key 213

Test Papers
Model Solutions 214

Answer Key 230

Practice Problems
Model Solutions 231

TURBOMACHINERY

Answer Key 239

Assignments
Model Solutions 240

Answer Key 243

Test Papers
Model Solutions 244
 Fluid Mechanics

PROPERTIES OF FLUIDS
1. Density or Mass Density :
It is defined as the ratio of mass of a fluid to its volume. Mathematically, mass density
is written as
Mass of fluid
ρ=
Volume of fluid

The value of density for water is 1gm / cm3 or 1000 kg/ m3

2. Specific Weight or Weight Density :

It is the ratio between the weight of a fluid to its volume. Mathematically

Weight of fluid Mass of fluid × g

w= = = ρg
Volume of fluid Volume of fluid

The value of specific weight (w) for water is 9810 N/m3 in SI units.

3. Specific Volume :
It is defined as the volume of a fluid occupied by a unit mass of a fluid.
Mathematically,
Volume of fluid 1
Specific Volume = =
Mass of fluid ρ
It is expressed as m3 / kg. It is commonly applied to gases.

4. Specific Gravity :
It is defined as the ratio of density of a fluid to the density of a standard fluid. For
liquids, the standard fluid is taken water and for gases, the standard fluid is taken as
air. Specific gravity is also called relative density. It is a dimensionless quantity and is
defined by S. Mathematically,
Density of liquid (or Weight density)
S (for liquids) =
Density of water (or Weight density)
Density of gas
S (for gases) =
Density of air

Thus weight density of a liquid = S × weight density of water

= S × 9810 N/m3
The density of a liquid = S × Density of water
= S × 1000 kg/m3

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Vidyalankar : GATE – ME

5. Viscosity :
It is defined as the property of a fluid which offers resistance to the movement of any
layer of fluid over another adjacent layer of the fluid. When two layers of a fluid, a
distance ‘dy’ apart, move one over the other at different velocities, say u and u + du
as shown in the figure, the viscosity together with relative velocity causes a shear
stress acting between the fluid layers.

Velocity variation near a solid boundary

The top layer causes a shear stress on the adjacent lower layer while the lower layer
causes a shear stress on the adjacent top layer. This shear stress is proportional to
the rate of change of velocity with respect to y.
du
The rate of change of velocity shear strain rate =
dy
1 ⎛ ∂v ∂u ⎞
For two dimensional flow, shear strain rate = ⎜ + ⎟
2 ⎝ ∂x ∂y ⎠
It is denoted by the symbol τ. Mathematically,
du du
τ∝ ⇒ τ =μ (Newton’s law of viscosity)
dy dy
where μ is the constant of proportionality and is known as coefficient of dynamic
du
viscosity or only viscosity. represents the rate of shear strain or rate of shear
dy
deformation or velocity gradient.
τ
Thus, μ=
⎛ du ⎞
⎜ dy ⎟
⎝ ⎠
Ns dyne − sec
Its unit is SI is2
while in CGS is ( or Poise )
m cm2
1 Ns
Also one Poise =
10 m2
1
1 centipoise = poise
100
The viscosity of water at 20°C is 0.01 poise or 1.0 centipoise.

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 Notes on Fluid Mechanics

6. Kinematic viscosity :
It is defined as the ratio between dynamic viscosity and density of fluid. It is denoted
by ν.
μ
ν=
ρ
Its SI unit is m2/s while in CGS it is cm2/s (or Stoke)
1 Stoke = 10−4 m2/s
1 centistoke = 0.01 stoke
Newton’s Law of Viscosity :
It states that the shear stress (τ) on a fluid element layer is directly proportional to the
rate of shear strain. Mathematically,
du
τ= μ
dy
Fluids which obey the above relation are known as Newtonain fluids and the fluids which
do not obey the above relation are called Non−Newtonain fluids.

Variation of Viscosity with Temperature :

The viscosity of liquids decreases with the increase of temperature while the viscosity of
gases increase with the increase of temperature. This is due to the reason that in luquids
the cohesive forces predominate the molecular momentum transfer, due to closely
packed molecules and with the increase in temperature, the cohesive forces decrease
with the result of decreasing viscosity. But in case of gases the cohesive forces are small
and molecular momentum transfer predominates. With the increase in temperature,
molecular momentum transfer increases and hence viscosity increases. The relation
between viscosity and temperature for liquids and gases is :
⎛ 1 ⎞
For liquids, μ = μ0 ⎜
2 ⎟
⎝ 1 + αt + β t ⎠
where μ = Viscosity of liquid at t°C, in poise
μ0 = Viscosity of liquid at 0°C, in poise
α, β are constants for the liquid
For water, μ0 = 1.79 × 10−3 poise, α = 0.03368 and β = 0.000221
For gases, μ = μ0 + αt − βt2
where for air; μ0 = 0.000017, α = 0.000000056, β = 0.1189 × 10−9

Types of Fluids :
1. Ideal Fluid :
A fluid, which is incompressible and is having no viscosity is known as an ideal fluid.
It is an imaginary fluid.
2. Real Fluid :
A fluid, which possesses viscosity, is known as real fluid. All the fluids, in actual
practice, are real fluids.
3. Newtonian Fluid :
A real fluid, which obeys Newton’s law of Viscosity is known as Newtonian Fluid.

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4. Non−Newtonian Fluid :
A real fluid, which does not obey Newton’s law of viscosity is known as a
Non−Newtonian Fluid.
5. Ideal Plastic Fluid :
A fluid, in which shear stress is more than the yield value and shear stress is
proportional to the rate of shear strain (or velocity gradient) is known as ideal plastic
fluid.

Types of fluids

Rheological Classification of Fluids :

There are two major types of Non−Newtonian fluids : Viscoelastic and Purely viscous
fluids. Viscoelastic fluids are characteristized by equation
du
τ= μ + αE
dy
where E is modulus of elasticity and α is a constant of fluid. A mixture of solid and liquid
flowing through a pipe is an example of viscoelastic fluid (slurry).
Purely Viscous fluids are characterized by the equation:
n
⎛ du ⎞
τ = τo + μ ⎜ ⎟
⎝ dy ⎠
i) If τo = 0, n < 1 it is pseudoplastic fluid (e.g. fine particle suspension, milk, blood)
ii) If τo = 0, n > 1 it is dilatant fluid (e.g. ultrafine irregular particle suspension,
concentrated solution of sugar in water)
iii) If τo = τo, n = 1 it is Bingham fluid (e.g. Hydrocarbon grease, clay suspended in water)
iv) If τo = τo, n > 1 it is yield pseudoplastic fluid (e.g. carboxypolymethylene)
Above four cases are time independent.
There are two time dependent purely viscous fluids govrned by the following equations:
n
⎛ du ⎞
i) Rheopectic fluids : τ = τo + μ ⎜ ⎟ + f(t)
⎝ dy ⎠
f(t) increases with time.
ii) Thixotropic fluids :
n
⎛ du ⎞
τ = τo + μ ⎜ ⎟ + f(t)
⎝ dy ⎠
f(t) decreases with time.

GATE/ME/SLP/Module_3/Ch.1_Notes /Pg.4
 Notes on Fluid Mechanics

Thermodynamic Properties :
Gases are compressible fluids and hence thermodynamic properties play an important
role. With the change of pressure and temperature, the gases undergo large variation in
density. The relationship between pressure (absolute), specific volume and absolute
temperature of a gas is given by equation of state as
p
pV = RT or = RT
ρ
where
p = Absolute pressure of a gas in N/m2
1
V = specific volume =
ρ
R = Gas constant = 287 J/kg−K
T = Absolute temperature in °K
ρ = Density of gas
1. Isothermal Process :
If the change in density occurs at constant temperature, then the process is called
isothermal and relationship between pressure and density is given by
P
= constant
ρ
2. Adiabatic Process :
If the change in density occurs with no heat exchange to and from the gas, the
process is called adiabatic. And if no heat generated within the gas due to friction,
the relationship between pressure and density is given by
P
= constant
ρk
where k = Ratio of specific heat of a gas at constant pressure and constant volume
= 1.4 for air
∗ Another form of equation of state ;
PV = nRT
where V = volume of gas
n = number of moles in a volume of gas
R = universal gas constant = 8.314 × 103 J/kg−mole K.

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Compressibility :
It is the reciprocal of bulk modulus of elasticity, K which is defined as the ratio of
compressive stress to volumetric strain.
dP 1
K= compressibility =
⎛ dV ⎞ K
⎜− V ⎟
⎝ ⎠
Relationship between k and pressure (p) of a Gas :
1. Isothermal Process : K=p
2. Adiabatic Process : K = pk
Surface Tension :
It is defined as the tensile force acting on the surface of a liquid in contact with a gas or
on the surface between two immiscible liquids such that the conact surface behaves like
a membrane under tension. The magnitude of this force per unit length of the free surface
will have the same value as the surface energy per unit area. It is denoted by σ.
Liquid Droplet :
Consider a small spherical droplet of a liquid of radius R. On the entire surface of the
droplet, the tensile force due to surface tension will be acting.
Let σ = surface tension of the liquid
p = Pressure intensity inside the droplet (in excess of the outside pressure intensity)
d = Diameter of droplet
By equilibrium of forces in spherical droplet
⎛π ⎞
p ⎜ d2 ⎟ = σ (2πd)
⎝4 ⎠
4σ
Then p =
d
Above equation shows that with the decrease of diameter of the droplet, presure intensity
inside the droplet increases.
Hollow Bubble:
e.g soap bubble : since soap bubble has two sufaces in contact with air,
⎛π ⎞
p ⎜ d2 ⎟ = 2σ(πd)
⎝4 ⎠
8σ
p=
d
Liquid Jet :
Pressure Intensity inside a Liquid Jet
Consider a jet of liquid of diameter d and length  and having internal pressure p in
excess of the outside pressure intensity. If jet is cut into two halves then forces acting on
one half will be those due to pressure intensity p on projected area d  and tensile force
due to surface tension σ acting along two sides(2  ). These two forces will be equal and
opposite for equilibrium and hence we have,
p × d ×  = σ 2
2σ
p=
d

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 Notes on Fluid Mechanics

Capillarity :
Capillarity is defined as a phenomenon of rise or fall of a liquid surface in a small tube
relative to the adjacent general level of liquid when the tube is held vertically in the liquid.
The rise of liquid surface is known as capillary rise while the fall of the liquid sruface is
known as capillary depression. It is expressed in terms of cm or mm of liquid. Its value
depends upon the specific weight of the liquid, diameter of the tube and surface tension
of liquid.
For equilibrium of vertical forces, acting on mass of liquid lying above (or below) the
general liquid level, the weight of the liquid column h (or total internal pressure in the ase
of capillary depression) must be balanced by the force, at surface of liquid, due to surface
tension σ.
Thus equating these two forces, we have
π 2
d h (ρg) = πd (σ cos θ)
4

4σ cos θ
h=
ρgd

Capillary Rise :
It happens when a glass tube of small diameter ‘d’ opened at both ends, is inserted in a
liquid such as water. In this case capillary rise is given by
4σ cos θ
h=
ρgd
where
σ = surface tension of liquid
ρ = Density of liquid
θ = angle of contact between liquid and glass tube
For water θ is approximately equal to zero.

Capillary rise
Capillary Depression :
It happens when a glass tube of diameter ‘d’ opened at both ends, is dipped in a liquid
such as mercury. In this case, capillary fall is given by,

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Vidyalankar : GATE – ME

4σcos θ
h=
ρgd
For mercury, θ is equal to 128°.

Example :
A flat plate of area 1.5 × 106 mm2 is pulled with a speed of 0.4 m/s relative to another
plate located at a distance of 0.15 mm from it. find the force and power required to
maintain this speed, if the fluid seperating them is having viscosity as 1 poise.
Solution :
Area of plate, A = 1.5 × 106 mm2 = 1.5 m2
speed of plate relative to another plate, du = 0.4 m/s
Distance between the plates, dy = 0.15 mm = 0.15 × 10−3 m
Ns
Viscosity μ = 1 poise = 0.1
m2
du 0.4 2
Now τ = μ = 0.1× = 266.67 N/m
dy 0.15 × 10−3
shear force, F = τA = 266.67 × 1.5 = 400 N
power, P = Fu = 400 × 0.4 = 160 W.

Example :
Calculate dynamic viscosity of oil, which is used for lubrication between a square plate of
size 0.8m × 0.8m and an inclined plane with an inclination 30°, as shown in the figure.
The weight of the square plate is 300N and it slides down the inclined plane with a
uniform velocity of 0.3 m/s. The thickness of oil film is 1.5 mm.

Solution :
Area of plate, A = 0.8 × 0.8 = 0.64 m2
Angle of plane, θ = 30°
Weight of plate, W = 300 N

GATE/ME/SLP/Module_3/Ch.1_Notes /Pg.8
 Notes on Fluid Mechanics

Velocity of plate, u = 0.3 m/s

Thickness, t = dy = 1.5 mm = 1.5 × 10−3 m
Let the viscosity of fluid between plate and inclined plane is μ.
Component of weight W, along the plane
= W cos 60° = 300 cos 60° = 150 N
F 150
and shear stress, τ = = N/m2
A 0.64
du
Now, τ = μ
dy
where, du = change of velocity = u − 0 = u = 0.3 m/s
dy = t = 1.5 × 10−3 m
150 0.3
∴ = μ×
0.64 1.5 × 10−3
⇒ μ = 11.7 poise

Example :
3
The velocity distribution for flow over a flat plate is given by u = y − y 2 in which u is the
4
velocity in metre per second at a distance y metre above the plate. Determine the shear
stress at y = 0.15 m. take dynamic viscosity of fluid as 8.5 poise.
Solution :
3
u= y − y2
4
du 3
= − 2y
dy 4
du 3
At y = 0.15, = − 2 × 0.15 = 0.75 − 0.30 = 0.45
dy 4
Viscosity, μ = 8.5 poise = 0.85 Ns/m2
du 2 2
τ=μ = 0.85 × 0.45 N/m = 0.3825 N/m
dy
Example :
The dynamic viscosity of anoil, used for lubrication between a shaft and sleeve is 6 poise.
The shaft is of diameter 0.4m and rotates at 190 r.p.m. Calculate the power lost in the
bearing for a sleeve length of 90 mm. The thickness of the oil film is 1.5 mm.
Solution :
Viscosity, μ = 6 poise = 0.6 Ns/m2
Diameter of shaft, D = 0.4 m
Speed of shaft, N = 190 r.p.m.
Sleeve length, L = 90 mm = 90 × 10−3 m
Thickness of oil film, t = 1.5 mm = 1.5 × 10−3 m
Tangential velocity of shaft,
πDN π × 0.4 × 190
u= = = 3.98 m/s
60 60

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Vidyalankar : GATE – ME

du
Using the relation, τ = μ
dy
where du = change of velocity = u − 0 = u = 3.98 m/s
dy = change of distance = t = 1.5 × 10−3 m
3.98
τ = 0.6 × = 1592 N/m2
1.5 ×10−3
This is shear stress on shaft.
∴ shear force on the shaft, F = τ × Area
= 1592 × π D × L
= 1592 × π × 0.4 × 90 × 10−3
= 180.05 N
D
torque on the shaft, T = Force ×
2
0.4
= 180.05 × = 36.01 Nm
2
2πN 2π × 190 × 36.01
⇒ Power lost = Tω = T × = = 716.48 W
60 60

Example :
What is the bulk modulus of elasticity of a liquid which is compressed in a cylinder from a
volume of 0.0125 m3 at 80N/cm2 pressure to a volume of 0.0124 m3 at 150 N/cm2
pressure ?
Solution :
Initial volume = 0.0125 m3
Final volume = 0.0124 m3
Decrease in volume, dv = 0.0125 − 0.0124 = 0.0001 m3
dV 0.0001
− =
V 0.0125
Initial pressure = 80 N/cm2
Final pressure = 150 N/cm2
Increase in pressure, dp = (150 − 80 ) = 70 N/cm2
dp 70 2
K= = = 8.75 × 103 N/cm
⎛ dV ⎞ ⎛ 0.0001 ⎞
⎜ − V ⎟ ⎜ 0.0125 ⎟
⎝ ⎠ ⎝ ⎠

Example :
The pressure outside the droplet of water of diameter 0.04 mm is 10.32 N/cm2
(atmospheric pressure). Calculate the pressure within the droplet if surface tension is
given as 0.0725 N/m of water.
Solution :
Diameter of droplet, d = 0.04 mm = 0.04 × 10−3 m
Pressure inside the droplet = 10.32 N/cm2
Surface tension, σ = 0.0725 N/m

GATE/ME/SLP/Module_3/Ch.1_Notes /Pg.10
 Notes on Fluid Mechanics

The pressure inside the droplet, in excess of outside pressure is given by;
4σ 4 × 0.0725
P= = = 7250 N/m2 = 0.725 N/cm2
d 0.04 × 10−3
Hence pressure inside the droplet = pressure outside the droplet
= 0.725 + 10.32 = 11.045/cm2
Example:
Calculate the capillary rise in a glass tube of 2.5 mm diameter when immersed vertically
in (a) water and (b) mercury. Take surface tension s σ = 0.0725 N/m for water and
σ = 0.52 N/m for mercury in contact with air. The specific gravity for mercury is given as
13.6 and angle of contact is 130°.
Solution:
Dia of tube ; d = 2.5 mm = 2.5 × 10−3 m
Surface tension, σ for water = 0.0725 N/m
σ for mercury = 0.52 N/m
sp. gr. of mercury = 13.6
⇒ Density = 13.6 × 1000 kg/m3
(a) capillary rise for water ( θ = 0°)
4σ 4 × 0.0725
h = = = 0.0118m = 1.18cm
ρgd 1000 × 9.81× 2.5 × 10−3
(b) For mercury
4σ cos θ 4 × 0.52 × cos130°
h = =
ρgd 13.6 × 1000 × 9.81× 2.5 × 10 −3
= −0.004m = −0.4cm
The negative sign indicates the capillary depression.
Example:
Two large plane surfaces are 2.4 cm apart. The space between the surfaces is filled with
glycerin. What force is required to drag a very thin plate of surface area 0.5m2 between
the two large plane surfaces at a speed of 0.6m/s if the thin plate is in the middle of the
two plane surfaces?
Take the dynamic viscosity of glycerin = 8.10 × 10−1Ns/m2
Solution:
Distance between two large surfaces = 2.4 cm
Area of thin plate, A = 0.5m2
Velocity of thin plate, u = 0.6m/s
Viscosity of glycerin, μ = 8.10 × 10−1Ns/m2
Let F1 = Shear force on the upper side of the thin plate
F2 = Shear force on the lower side of the thin plate
The shear stress (τ1) on the upper side of the thin plate is
⎛ du ⎞
τ1 = μ ⎜ ⎟
⎝ dy ⎠1
Where du = Relative velocity between thin plate and upper large plane surface
= 0.6m/s

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Vidyalankar : GATE – ME

dy = Distance between thin plate and upper large plane surface

= 1.2 cm
= 0.012 m (plate is a thin one and hence thickness of plate is neglected)
⎛ 0.6 ⎞
τ1 = 8.10× 10−1 × ⎜ ⎟ = 40.5N / m2
⎝ 0.012 ⎠
Now shear force, F1 = τ1 × A = 40.5 × 0.5 = 20.25N
Similarly shear stress (τ2) on the lower side of the thin plate is given by
⎛ du ⎞ ⎛ 0.6 ⎞
τ2 = μ ⎜ ⎟ = 8.10 × 10 −1 × ⎜ ⎟ = 40.5N / m
2

⎝ dy ⎠2 ⎝ 0.012 ⎠

F2 = τ2 × A = 40.5 × 0.5 = 20.25 N

Total F = F1 + F2 = 20.25 + 20.25 = 40.5 N

Example:
A cylinder of 0.6m3 in volume contains air at 50°C and 0.3 N/mm2 absolute pressure. The
air is compressed to 0.3m3. Find
(i) pressure inside the cylinder assuming isothermal process and
(ii) pressure and temperature assuming adiabatic process. Take k = 1.4
Solution:
Initial volume, V1 = 0.6 m 3
Temperature, t1 = 50°C
⇒ T = 273 + 50 = 323°k
pressure, p1 =0.3N/mm2 = 30 × 104 N/m2
Final volume, V2 = 0.3m3
k = 1.4
(i) Isothermal process
p
= constant or pV = constant
ρ
p1 V1 = p2 V2
V 30 × 104 × 0.6
⇒ p2 = p1 1 = = 0.6 × 106 N / m2
V2 0.3
= 0.6N/mm2
(ii) Adiabatic process.
p k
= constant or pV = constant
ρk
p1 V1 k = p2 V2 k
k 1.4
⎛V ⎞
⇒ p2 = p1 ⎜ 1 ⎟ = 30 × 104 × ⎛⎜ 0.6 ⎞⎟
⎝ V2 ⎠ ⎝ 0.3 ⎠
4 1.4
= 30× 10 × 2
= 0.791 × 106 N /mm2

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 Notes on Fluid Mechanics

Now; pV = nRT and also pVk = constant

nRT nRT
⇒ p= and × Vk = constant
V V
or nRTVk − 1 = constant
TVk−1 = constant
T1 V1 k − 1 = T2 V2 k − 1
k −1 1.4 −1.0
⎛V ⎞ ⎛ 0.6 ⎞
T2 = T1 ⎜ 1 ⎟ = 323 ⎜ ⎟
⎝ V2 ⎠ ⎝ 0.3 ⎠

= 323 × 2 0.4 = 426.2°k

t2 = 426.2−273 = 153.2°C

Example:
For water, the bulk modulus of elasticity is 200kN/cm2 what is the pressure required to
decrease its volume by 3% and what will be the change in density?
Solution:
⎛ ⎞
⎜ dp ⎟ dp
K= ⎜ ⎟ ⇒ 200 = ⇒ dp = 6 kN/cm2
⎜⎜ − dV ⎟⎟ 0.03
⎝ V ⎠
1
We know that ρ = ⇒ ρV = constant
V
⇒ ρdV +Vdρ = 0
dV dρ
⇒− =
V ρ
Hence decrease in volume will result in an increase in mass density by 3%.

FLUID STATICS
Fluid pressure at a point:
Consider a small area dA in large mass of fluid. If the fluid is stationary, then the forced
exerted by the surrounding fluid on the area dA will always be perpendicular to the
dF
surface dA. Let dF is the force acting on the area dA in normal direction. The ratio is
dA
know as the intensity of pressure or simply pressure and this ratio is represented by p.
Hence mathematically the pressure at a point in a fluid at rest is
dF
p=
dA
If the force (F) is uniformly distributed over the area (A), then pressure at any point is given by
F
p=
A
⇒ Force or pressure force, F = p × A

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Pascal’s Law :
It states that the pressure or intensity of pressure at a point in a static fluid is equal in all
directions.

Pressure Variation in a Fluid at Rest:

The pressure at any point in a fluid at rest is obtained by the hydrostatic law which states
that the rate of increase of pressure in a vertically downward direction must be equal to
the specific weight of the fluid at that point. Consider a small fluid element as shown in
fig.
Let ΔA = Cross sectional area of element
ΔZ = height of fluid element
p = pressure on the face AB
Z = distance of fluid element from free surface
Then according to Hydrostatic Law:
∂p
= ρ × g = w(weight desity)
∂Z
Above equation states that rate of increase of
pressure in a vertical direction is equal to weight
density of the fluid at that point Forces on a fluid element
Integrating we get, p = ρgZ
Where p is the pressure above atmospheric pressure and Z is the height of the point from
free surface. Z is called “pressure head”.

Absolute, Gauge, Atmospheric and Vacuum Pressures:

The pressure on a fluid is measured in two different systems. In one system, it is
measured above the absolute zero or complete vacuum and it is called absolute pressure
and in other system, pressure is measured above the atmospheric pressure and it is
called gauge pressure. Thus:
(1) Absolute pressure is defined as the pressure which is measured with reference to
absolute vacuum pressure.
(2) Gauge pressure is defined as the pressure which is measured with the help of
pressure measuring instrument, in which atmospheric pressure is taken as datum.
(3) Vacuum pressure is defined as the pressure below the atmospheric pressure. The
relationships between the absolute pressure, gauge pressure and vaccum pressure
are shown in Fig

Relationship between pressures

GATE/ME/SLP/Module_3/Ch.1_Notes /Pg.14
 Notes on Fluid Mechanics

Mathematically:
(i) Absolute pressure = Atmospheric pressure + Gauge pressure
or Pabs = Patm+ Pgauge
(ii) Vacuum pressure = Atmospheric pressure − absolute pressure
* The atmospheric pressure at sea level at 15°C is 101.3kN /m2
or 10.13 N/cm2
* The atmospheric pressure head is 760mm of mercury or 10.33 m of water

Measurement of pressure: The pressure of a fluid is measured by the following devices:

(1) Manometers
(a) Simple Manometers
(i) Piezometer
(ii) U − tube Manometer
(iii) Single Column Manometer

(b) Differential Manometer

(i) U − tube differential Manometer
(ii) Inverted U − tube Differential manometer
(2) Mechanical Gauges
(a) Diaphragm Pressure Gauge
(b) Bourdon tube pressure Gauge
(c) Dead − weight pressure Gauge
(d) Bellows pressure Gauge

Simple Manometer:
A simple manometer consists of a glass tube having one of its end connected to a point
where pressure it to be measured and other end remains open to atmosphere.

1) Piezometer:
It is the simplest form of manometer used for
measuring gauge pressures. One end of this
manometer is connected to the point where
pressure is to be measured and other end is
open to the atmosphere as shown in fig.

The rise of liquid gives the pressure head at that

point. If at point A, the height of liquid say water
is h in piezometer tube, then pressure at
A = ρgh N/m2

2) U−tube Manometer:
It consists of a glass tube in U− shape, one end of which is connecated to a point at
which pressure is to be measured and other end remains open to the aatmosphere
as shwon in fig. The tube generally constains mercury or any other liquid whose
specific gravity is greater thatn the specific gravity of the liquid whose pressure is to
be measured.

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Vidyalankar : GATE – ME

a) For Gauge Pressure:

p = ρ2 gh2 − ρ1 gh1
h1 = Height of light liquid above the datum line
h2 = Height of heavy liquid above the datum line
S1 = sp. gr. of light liquid
S2 = sp. gr. of heavy liquid.
ρ1 = Density of light liquid= 1000S1
ρ2 = Density of heavy liquid = 1000S2

b) For Vacuum Pressure:

p = − (ρ2gh2 + ρ1gh1)

3) Single Column Manometer:

It is a modified form of a U− tube manometer in which a reservoir having a large
cross sectional area (about 100 times) as compared to the area of the tube is
connected to one of the limbs of the manometer as shown in Fig. The other limb may
be vertical or inclined
a) Vertical Single Column Manometer:
Fig. shows the vertical single column manometer. Let x − x be the datum line in
the reservoir and in right limb of the manometer, when is not connected to the
pipe.

When the manometer is connected to the

pipe, due to high pressure at A, the heavy
liquid in the reservoir will be pushed
downward and will rise in the right limb.
Then pA, the pressure at A, which is to be
measured is given by
axh2
pA = h2 ρ2 g − h1ρ1 g + [ρ2g − ρ1g]
A
Where: h2 = Rise of heavy liquid in right limb
h1 = Height of center of pipe above x − x
A = Cross − sectional area of the reservoir

GATE/ME/SLP/Module_3/Ch.1_Notes /Pg.16
 Notes on Fluid Mechanics

a = Cross sectional area of the right limb

p1 = density of liquid in pipe
p2 = density of liquid in reservoir

b) Inclined Single Column Manometer :

Fig. shows the inclined single column manometer. This manometer is more sensitive.
Due to inclination the distance moved by the heavy liquid in the right limb will be
more

The pressure at A is ; pA = ρ2 g L sin θ − h1 ρ1g

Where L = Length of heavy liquid moved in right limb from x − x
θ = inclination of right limb with horizontal
4) Differential Manometers:
These are the devices used for measuring difference of pressures between two
points in a pipe or in two different pipes. A differential manometer consists of a U −
tube, containing a heavy liquid whose two ends are connected to the points, whose
difference of pressure is to be measured. Most common types of differential
manometers are:
a) U − tube differential manometer:
Fig shows the differential manometer of U − tube type.

U−tube differential manometers

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Vidyalankar : GATE – ME

i) For fig (a); pA − pB = hg (ρg −ρ1) + ρ2gy − ρ1 gx

Where h = Difference of mercury level in the U− tube
y = Distance of the center of B, from the mercury level in the right limb
x = Distance of the center of A, from the mercury level in the right limb.
ρ1 = Density of liquid at A
ρ2 = Density of liquid at B
ρg = Density of heavy liquid or mercury
Datum line is at x − x.
pA = pressure at A
pB = pressure at B
ii) For fig (b); pA − pB = gh (ρg − ρ1)

b) Inverted U − tube differential manometer:

It consists of an inverted U − tube, containing a light liquid. The two ends of the
tube are connected to the points whose difference of pressure is to be measured.
It is used for measuring difference of low pressures. Fig shows an inverted U −
tube differential manometer connected to the two points A and B let the pressure
at A is more than the pressure at B.
Here; pA − pB = ρ1 gh1 − ρ2gh 2 − ρs gh
Where
h1 = Height of liquid in the left limb
below the datum line x − x
h2 = Height of liquid in the right
limb
h = Difference in heights of light liquid.
ρ1 = Density of liquid at A
ρ2 = Density of liquid at B
ρs = Density of light liquid
pA = pressure at A
pB = pressure at B

Pressure at a Point in Compressible Fluid:

For compressible fluids, density (ρ) changes with pressure and temperature.
1) Isothermal Process : Case I: Temperature T is constant. Pressure at height Z is given
by
p = po e−gZ/RT
where po is the pressure where height is Zo . If the datum line is taken at Zo , then
Zo = 0 and po becomes the pressure at datum line.
2) Adiabatic Process : Here temperature T is not constant but the process follows
adiabatic law. Pressure at a height Z from ground level is given by
k
⎡ k − 1 ρo ⎤ k −1
p = po ⎢1 − gZ ⎥
⎣ k po ⎦
Where,
po = pressure at ground level where zo = 0
ρo = density of air at ground level
k = adiabatic index (Ratio of specific heats = CP/CV)

GATE/ME/SLP/Module_3/Ch.1_Notes /Pg.18
 Notes on Fluid Mechanics

Also,
k
⎡ k − 1 gZ ⎤ k −1
p = po ⎢1 − ⎥
⎣ k RTo ⎦
Where
To = temperature at ground level.

Temperature at any point in compressible fluid (For adiabatic process):

k
⎡ k − 1 gZ ⎤ k −1
T = To ⎢1 − ⎥
⎣ k RTo ⎦
Where
T = temperature at a height Z from ground level

Temperature Lapse Rate:

It is defined as the rate at which the temperature changes with elevation. The
temperature lapse rate is denoted by L and is given by
gΔz ⎛ k − 1⎞
L=
R ⎜⎝ k ⎟⎠
For air, R = 29.27 m−kg(f)/kg(m)°C abs and k = 1.405,
then for Δz = 100m, the variation in temperature ΔT may be obtained from above
equation.
9.81× 100 × 0.405
R ΔT =
1.405
R = 29.27 m−kg(f)/kg(m)°C abs
= 29.27 × 9.81 m2 /sec2 °C abs
9.81× 100 × 0.405
∴ ΔT = = 0.985°C.
29.27 × 9.81× 1.405
This shows that under adiabatic condition, the absolute temperature decreases by about
0.985°C for each 100 m increase in elevation. However, if the atmosphere is stable, the
temperature drop with altitude is generally somewhat less than that computed above.
• If k > 1, the lapse rate is negative, which means temperature decreases with the
increase of height.
• R is the “Gas Constant” and is equal to 287 J/kg − K.

Hydrostatic Forces on Surfaces:

This subject deals with the fluids at rest. This means that there will be no relative motion
between adjacent or neighbouring fluid layers. The velocity gradient will be zero.
du du
i.e. = 0 . The shear stress τ = μ will also be zero. Then the
dy dy
Force acting on the fluid particles will be
(i) Due to pressure of fluid normal to the surface.
(ii) Due to gravity (or self − weight of fluid particles)

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Vidyalankar : GATE – ME

Total Pressure:
It is defined as the force exerted by a static fluid on a surface either plane or curved when
the fluid comes in contact with the surface. This force always acts normal to the surface.
Centre of Pressure:
It is defined as the point of application of the total pressure on the surface. There are four
cases of submerged surfaces on which the total pressure forece and centre of pressure
is to be determined. The submerged surfaces may be
(1) Vertical plane surface
(2) Horizontal plane surface
(3) Inclined plane surface
(4) Curved surface
(1) Vertical Plane Surface Submerged in Liquid:
Consider a plane vertical surface of arbitrary shape immersed in a liquid as shown in fig.

A Total area of the surface

Let h = Distance of C.G. of the area from the free surface of liquid.
G = Centre of gravity of plane surface.
P = Centre of pressure
h* = Distance of center of pressure from free surface of liquid
Total pressure, F = ρg A h
For water, ρ = 1000kg/m3 and g = 9.81 m/s2
I
Centre of pressure, h* = G + h
Ah
Where IG = Moment of inertial of area about an axis passing through the C.G. of the area
and parallel to the free surface of the liquid.
* Centre of pressure (h*) lies below the C.G. of the vertical surface.
* Distance of centre of pressure from the free surface of liquid is independent of the
density of the liquid.

(2) Horizontal Plane Surface Submerged in Liquid:

Consider a plane horizontal surface immersed in a static fluid. As every point of the
surface is at the same depth from the free surface of the liquid, the pressure intensity
on the entire surface is equal to, p = ρgh, where h is depth of surface
Let A = Total area of surface

GATE/ME/SLP/Module_3/Ch.1_Notes /Pg.20
 Notes on Fluid Mechanics

Then total force, F on the surface

= p × Area = ρg h A
= ρ ghA
Where h = Depth of C. G. from free surface of liquid = h, also
h* = Depth of centre of pressure from free surface = h

(3) Inclined Plane Surface Submerged in Liquid:

Consider a plane surface of arbitrary shape immersed in such a way that the plane of
the surface makes on angle θ with the free surface of the liquid as shown in Fig.

Let A = Total area of inclined surface

h = depth of C.G. of inclined area from free surface
h* = Distance of centre of pressure from free surface of liquid
θ = Angle made by plane of the surface with free liquid surface

let the plane of the surface , if produce meet the free liquid surface
y = distance of the C.G. of the inclined surface from o − o
y* = distance of the centre of pressure from o − o

Total pressure, F = ρgA h

I Sin2 θ
Centre of pressure, h* = G +h
Ah
Where , IG = M. O. I of inclined surface about an axis passing thorough G and parallel
to o − o

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Vidyalankar : GATE – ME

(4) Curved Surface Submerged in Liquid:

Consider a curved surface AB submerged in a static fluid as shown in Fig.

Let dA is the area of a small strip at a depth h from water surface.

Total pressure on the curved surface, F = ∫ ρghdA
The total force on the curved surface is
F= Fx 2 + Fy 2
Where
Fx = ∫ dF = ρg∫ hdA sin θ
X

Fy = ∫ dFy = ρg∫ hdA cos θ

⎛ Fy ⎞
Here ; θ = tan F ⎟ −1 ⎜
⎝ x⎠
Also Fx = Total pressure force on the projected area of the curved surface on
vertical plane
Fy = Weight of liquid supported by the curved surface up to free surface of
liquid.

SOLVED EXAMPLES
Example:
A hydraulic press has a ram of 20cm diameter and plunger of 3cm diameter. It is used for
lifting a weight of 30kN. Find the force required at the plunger.
Solution:
Dia. of ram D = 20cm = 0.2m
π π
A = D2 = ( 0.2 ) = 0.0314m2
2
Area of ram,
4 4
π
Dia. of plunger, a = ( 0.03 ) = 7.068 × 10 −4 m2
2

4
Weight lifted, W = 30kN = 30 × 1000N = 300000N
Pressure intensity developed due to plunger
F
=
a

GATE/ME/SLP/Module_3/Ch.1_Notes /Pg.22
 Notes on Fluid Mechanics

By Pascal’s law, this pressure is transmitted equally in all directions.

F
Hence pressure transmitted at the ram =
a
⇒ Force acting on ram = pressure intensity × Area of ram
F F × 0.0314
= ×A = N
a 7.068 × 10−4
But force acting on ram = weight lifted = 30000N
F × 0.0314
⇒ 30000 =
7.068 × 10 −4
F = 675.2N.

Example:
The pressure intensity at a point in a fluid is given 3.924N/cm2. Find the corresponding
head of the fluid when the fluid is (a) water, and (b) oil of sp. gr. 0.9
Solution:
Pressure intensity, p = .0924 Ncm2 = 3.94 × 104 N/m2.. The corresponding head, Z, of
the fluid is given by
p
Z=
ρg
(a) For water, ρ = 1000kg/m3
p 3.924 × 104
⇒ Z= = = 4m of water
ρg 1000 × 9.81
(b) For oil, sp. gr = 0.9
ρo = 0.9 × 1000 = 900kg. /m3
p 3.924 × 104
Z= = = 4.44m of oil
ρg 1000 × 9.81
Example:
An open tank contains water up to a depth of 2m and
above it an oil of sp.gr. 0.9 for a depth of 1m . Find the
pressure intensity (i) at the interface of the two liquids,
and (ii) at the bottom of the tank.
Solution:
Height of water, Z1 = 2m
Height Oil, Z2 = 1m
Sp. gr. Oil, So = 0.9
Density of water , ρ1 = 1000 kg /m3
Density of oil, ρ2 = 0.9 × 1000 = 900 kg /m3
Pressure intensity of at any point is given by , p = ρgz
(i) At interface, i.e. at A
p = ρ2 × g × 1.0 = 900 × 9.81 × 1.0 = 8829 N/m2
(ii) At the bottom, i.e. at B
p = ρ2 gz2 + ρ1 gz1 = 900 × 9.8 ×1.0 + 1000 × 9.81 × 20
= 8829 + 19620 = 28449 N/m2

GATE/ME/SLP/Module_3/Ch.1_Notes /Pg.23
Vidyalankar : GATE – ME

Example:
What is the gauge pressure and absolute pressure at a point 3m below the free surface
of a liquid having a density of 1.53 × 103 kg/m3, specific gravity of mercury is 13.6 and
density of water = 1000kg/m3, if the atmospheric pressure is equivalent to 750mm of
mercury? The specific gravity of mercury is 13.6 and density of water = 1000kg /m3.
Solution:
Depth of liquid, Z1 = 3m
Density of liquid, ρ1 = 1.53 × 103kg /m3
Atmospheric pressure head, Z0 = 750 mm of Hg
= 0.75 m of Hg.
⇒ Atmospheric pressure, patm = ρo g Zo
Where po = density of Hg= 13.6 × 1000 kg/m3
And Zo = pressure head in terms of mercury
⇒ patm = (13.6 × 6 × 1000) × 9.81 × 0.75 (Zo = 0.75)
= 100062 N/m2
pressure at point, which is at a depth 3m from the free surface of the liquid is given
by,
p = ρ1 g Z1
= (1.53 × 1000) × 9.81 × 3 = 45028 N/m2

⇒ Gauge pressure, p = 45028 N/m2

Absolute pressure = Gauge pressure + Atmospheric pressure
= 45028 + 100062 = 145090 N/m2

Example:
A simple U − tube manometer containing mercury is
connected to a pipe in which a fluid of sp. gr. 0.8 and
having vacuum pressure is flowing. The other end of the
manometer is open to atmosphere. Find the vacuum
pressure in pipe, if the difference of mercury leveling the
two limbs is 40cm and the height of fluid in the left from
the centre of pipe is 15cm below.
Solution:
Sp. gr. of Fluid. S1= 0.8
Sp. gr. of mercury, S2 = 13.6
Density of fluid, ρ1 = 800
Density of mercury ρ2 = 13.6 × 1000
Difference of mercury level, h2 = 40cm = 0.4m.
Height of liquid, in left limb, h1 = 15cm = 0.15m.
Let the pressure in pipe = p
Equating pressure above datum line, A − A, we get
p = − [ ρ2 gh2 + ρ1 gh1 ]
= − [13.6 × 1000 × 9.81 × 0.4 + 800 ×9.81 ×0.15]
= − 54543.6 N/m2

GATE/ME/SLP/Module_3/Ch.1_Notes /Pg.24
 Notes on Fluid Mechanics

Example:
A single column manometer is connected to a pipe
containing a liquid of sp.gr.0.9 as shown in Fig. Find the
pressure in the pipe if the area of the reservoir is 100
times the area of the tube for the manometer reading as
shown in fig. The specific gravity of mercury is 13.6
Solution:
Sp.gr.of liquid in pipe , S1 = 0.9
⇒ Density, ρ1 = 900 kg/m3
sp. gr. of heavy liquid, S2 = 13.6
Density, ρ2 = 13.6× 1000 = 13600 kg/m3
Area of reservoir A
= =100
Area of right limb a
height of liquid, h1 = 20cm = 0.2m
Rise of mercury in right limb
h2 = 40cm = 0.4m
let pA = pressure in pipe
a
pA = h2 ⎡⎣ρ2 g − ρ1 g⎤⎦ + h2 ρ2 g − h1p1g
A
1
= × 04 [13600 × 9.81 − 900 × 9.81] + 0.4 × 13600 × 9.81 − 0.2 × 900 × 9.81
100
= 152134N/m2
Example:
A differential manometer is connected at the two points A and B as shown in Fig. At B
air pressure is 9.81 N/cm2 (abs), find the absolute pressure at A.

Solution:
Air pressure at B = 9.81 N/cm2
or pB = 9.81 × 104 N /m2
Density of oil = 0.9 × 1000 = 900 kg /m3
Density of mercury = 13.6 × 1000 = 13600kg/m3
Let the pressure at A is pA
Taking datum line at x − x
Pressure above x − x in the right limb
= 1000× 9.8 × 0.6 + pB
= 886 + 98100 = 103986

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Vidyalankar : GATE – ME

pressure above x − x in the left limb

= 13600 × 9.81 × 0.1 + 900× 9.81 × 0.2 + pA
= 13341.6 + 1765.8 + pA
Equating the two pressure head
⇒ 103986 = 13341.6 + 1765.8 + pA
⇒ pA = 88876.8 N/m2
= 8.887 N/cm2

Example:
Water is flowing through two different pipes in which an
inverted differential manometer having an oil sp.gr. 0.8 is
connected. The pressure head in the pipe A is 2m of
water, find the pressure in the pipe B for the manometer
readings as shown in fig
Solution:
pA
Pressure head at A = = 2m of water
ρg
⇒ pA = ρ × g × 2 = 1000 × 9.81 × 2 = 19620N/m2
Taking x − x as the left limb = pA = ρ1 gh1
= 19620 − 1000 × 9.81 × 0.3
= 16677 N /m2
pressure below x− x in the right limb
= pB − 1000 × 9.81 × 0.1 − 800 × 9.81 × 0.12
= pB − 1922.76
Equating the two pressures we get
16677 = pB − 1922.76
⇒ pB = 18599.76 N/m2

Example:
Calculate the pressure at height of 7500m above sea level if the atmospheric pressure is
10.143N/cm2 and temperature is 15°C at the sea level assuming (i) air is incompressible,
(ii) pressure variation follows isothermal law, and (iii) pressure variation follows adiabatic
law. Take the density of air at the sea−level as equal to 1.285 kg /m3. Neglect variation of
g with altitude.
Solution:
Height above sea −level, Z = 7500m
Pressure at sea −level, po = 10.143N/cm2 = 10.143 × 104 N /m2
Temperature at sea −level, to = 15°C
⇒ To = 273 + 15 = 288°k
Density of air, ρ = ρo = 1.285 kg./m3
i) Pressure when air is incompressible.
dp
= −ρg
dZ

GATE/ME/SLP/Module_3/Ch.1_Notes /Pg.26
 Notes on Fluid Mechanics

p z
⇒ ∫p1
dp = − ∫ ρgdZ ⇒ p = po = − ρg[ Z − Zo]
zo

⇒p = po − ρgZ { Zo = datum line = 0}

= 10.143 × 104 − 1.285 × 9.81 × 7500 = 6887 N/m2

ii) Pressure variation follows isothermal law.

P = po e−gZ./RT
= po e− gZρo /po
= 101430 e−7500 × 1.285 × 9.81 × / 101430 = 39939N/m2

iii) Pressure variation follows adiabatic law : [ k = 1.4]

k / ( k −1)
⎡ k − 1 ρo ⎤
p = po ⎢1 − gZ ⎥
⎣ k po ⎦
1.4
⎡ (1.4 − 1.0 ) ( 7500 × 1.285 ) ⎤ 1.4 −1.0
= 101430 ⎢1 − × 9.81 ⎥ = 34310N/m3
⎣ 1.4 101430 ⎦
Example:

Determine the total pressure on a circular plate of diameter

1.5m which is placed vertically in water in such a way that
the center of the plate is 3m below the free surface of
water. Find the position of center of pressure also.

Solution:

Dia of plate, d = 1.5 m

π
Area, A = (1.5 ) = 1.767m2
2

4
h = 3.0m
Total pressure, F = ρg A h = 1000 × 9.81 × 1.767 × 3
=52002.81N
position of centre of pressure ( h ) is given by
I
h* = G + h
Ah
πd4 π × 1.54
Where IG = = = 0.2485m4
64 64
0.2485
h* = + 3.0 = 3.0468m
1.767 × 3.0

Example:
A square aperture in the vertical side of a tank has one diagonal vertical and is
completely covered by a plane plate hinged along one of the upper sides of the aperture.
The diagonals of the aperture are 2m long and the tank contains a liquid of specific
gravity 1.15. The centre of aperture is 1.5m below the free surface. Calculate the thrust
exerted on the plate by the liquid and position of its centre of pressure

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Vidyalankar : GATE – ME

Solution:
Diagonals of aperture AC = BD = 2m
Area of square aperture = Area of Δ ACB + Area of Δ ACD
AC × BO AC × OD 2 × 1 2 × 1
= + = + = 2.0m2
2 2 2 2
sp.gr.of liquid = 1.15
⇒ Density of liquid (ρ) = 1.15 × 1000 = 150kg/m3
Depth of centre of aperture from free surface,
h = 1.5m
i) The thrust on the plate is given by
F = ρgA h = 1150 × 9.81 × 2 × 1.5 = 33844.5N
ii) Centre if pressure (h*) is given by
I
h* = G + h
Ah
Where IG = M.O.I. of ABCD about diagonal AC
= M.O.I. of triangle ABC about AC + M.O.I. of triangle ACD about AC
AC × OB3 AC × OD3 2 × 13 2 × 13 1 1 1 4
= + = + = + = m
12 12 12 12 6 6 3
1
⇒ h* = 3 + 1.5 = 1.611m
2 × 1.5
Example:
Fig. shows a tank full of water. Find
(i) Total pressure on the bottom of tank.
(ii) Weight of water in the tank.
(iii) Hydrostatic paradox between the results of (i) and (ii). Width of tank is 2m

GATE/ME/SLP/Module_3/Ch.1_Notes /Pg.28
 Notes on Fluid Mechanics

Solution:
Depth of water on bottom of tank
h1 = 3 + 0.6 = 3.6m
Width of tank = 2m
Length of tank at bottom, = 4m
⇒ Area of tank at the bottom, A = 4 ×2 = 8m2
(i) Total pressure F, on the bottom is
F = ρgA h = 1000 × 9.81 × 8 ×3.6 = 282528N
(ii) Weight of water in tank = ρg × Volume of tank
= 1000 × 9.81[3 × 0.4 × 2 × + 4 × 0.6 × 2] = 70632 N
(iii) For the results of (i) and (ii), it is observed that the total weight of water in the
tank is much less than the total pressure at the bottom of the tank. This is known
as “Hydrostatic paradox”.

Example:
A rectangular plane surface 2cm wide and 3m deep lies in water in such a way that its
plane makes an angle of 30° with the free surface of water. Determine the total pressure
and position of centre of pressure when the upper edge is 1.5m below the free water
surface.

Solution:
Width of plane surface, b = 2m
Depth, d = 3m
Angle, θ = 30°
Distance of upper edge from free water surface = 1.5m
(i) Total pressure force is given by
F = ρgA h
Where ρ = 1000 kg /m3
A = b × d = 3 × 2 = 6m2
h = Depth of C.G. from free water surface
1
= 1.5 + 1.5 sin30° = 1.5 +1.5 × = 2.25m
2
{ h = AE + EB = 1.5 + BC sin 30° = 1.5 + 1.5 Sin 30°}
⇒ F = 1000 × 9.81 × 6 ×2.25 = 132435N
(ii) Centre of pressure (h*)
I Sin2 θ bd3 2 × 33
h* = G + h;whereIG = = = 4.5m4
Ah 12 12

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Vidyalankar : GATE – ME

1
4.5 ×
*4.5 × sin2 30° 4
⇒h = + 2.25 = = 2.3333m
6 × 2.25 6 × 2.25
Example:
Find the total pressure and position of centre of pressure on a triangular plate of base 2m
and height 3m which is immersed in such a way that the plane of the plate makes an
angle of 60° with the free surface of water. The base of the plate is parallel to water
surface and at a depth of 2.5m from water surface.

Solution:
Base of plate, b = 2m
Height of plate, h = 3m
b×h 2×3
⇒ Area, A = = = 3m2
2 2
Inclination, θ = 60°
Depth of centre of gravity from free surface of water,
h = 2.5 + AG sin 60°
1 3
= 2.5 + × 3 ×
3 2
= 2.5 + 0.866 = 3.366m

(i) Total pressure Force (F)

F = ρgA h = 1000 × 9.81 × 3 ×3.366 = 99061.38N

(ii) Centre of Pressure (h*): Depth of centre of pressure from free surface of water is
given by
I Sin2 θ
h* = G +h
Ah

bh3 2 × 33 3
Where IG = = = = 1.5m4
36 36 2

1.5 × sin2 60
h* = + 3.366 = 0.1114 + 3.366 = 3.477m
3 × 3.366

GATE/ME/SLP/Module_3/Ch.1_Notes /Pg.30
 Notes on Fluid Mechanics

Example:
Compute the horizontal and vertical components
of the total force acting on a curved surface AB,
which is the form of a quadrant of a circle of
radius 2m as shown in Fig. Take the width of the
gate as unity.

Solution:
Width of gate = 1.0m
Radius of gate = 2m
⇒ Distance AO = OB = 2m
Horizontal Force, Fx exterted by water on gate is given by
Fx = Total pressure force on the projected area of curved surface AB on vertical
plane
= Total pressure force on OB { projected area of curved surface on vertical plane
= OB × 1}
= ρgA h
⎛ 2⎞
= 1000 × 9.81 × 2 × 1 × ⎜ 1.5 + ⎟
⎝ 2⎠
= 49050N
⎧ 2⎫
⎨Area of OB = A = BO × 1= 2 × 1 = 2,h = Depthof C.G.of OB from free surface = 1.5 + ⎬
⎩ 2⎭
I
The point of application of Fx is given by h* = G + h
Ah
bd3 1× 23 2 4
Where IG = M.O.I of OB about its C.G. = = = m
12 12 3
2
⇒ h* = 3 + 2.5 = 1 + 2.5 = 2.633m from free surface
2 × 2.5 7.5
Vertical force, Fy, exerted by water is given by
Fy = Weight of water supported by AB up to free surface
= Weight of portion DABOC
= Weight of DAOC + Weight of water in AOB
= ρg [Volume of DAOC + volume of AOB]
⎡ π ⎤
= 1000 × 9.81 ⎢ AD × AO ×1 + ( OB ) × 1⎥
2

⎣ 4 ⎦
⎡ π ⎤
= 1000 × 9.81 ⎢1.5 × 2 × 1 + 22 × 1⎥
⎣ 4 ⎦
= 60249.1N

GATE/ME/SLP/Module_3/Ch.1_Notes /Pg.31
Vidyalankar : GATE – ME

Example:

A cylinder 3 m in diameter and 4 m long retains

water on one side. The cylinder is supported as
shown in Fig. Determine the horizontal reaction
at A and the vertical reaction at B. The cylinder
weighs 196. 2kN. Ignore friction

Solution:
Dia of cylinder = 3m
Length of cylinder = 4m
Weight of cylinder, w = 196.2kN = 196200N
Horizontal force exerted by water
Fy = Weight of water enclosed in BDCOB
⎛π ⎞ π
= ρg × ⎜ R2 ⎟ ×  = 1000 × 9.81× × (1.5 ) × 4
2

⎝2 ⎠ 2
= 138684N
Force Fy is acting in the upward direction
For the equilibrium of cylinder
3 3
Horizontal Reaction of A = Fx = ρg(2πL) r = 1000 × 9.8 × 2 × × 4 × = 176580N
2 2
Vertical Reaction at B = Weight of Cylinder − Fy
= 196200 − 138684 = 57516N
Example:
The cylindrical gate of a canal head works, having a diameter of 3m & a length of 6m is
subjected to water pressure upto its top as shown. The gate resisting on the complete
floor of the head works is laterally supported on A where the coefficient of friction μ =
0.15. Assuming water tight conditions at B & no rotation of cylinder. Find the minimum
weight of the gate so that it may have no upward thrust motion resulting from water
pressure. C

3m D
RA A
Fy
F
B

Solution:
The reaction RA at point A due to horizontal component of the water pressure on
cylinder is
d 3
RA = (ρg) (Ld) = (1000 × 9.81) (6 × 3) × = 264870 N
2 2
As the gate tends to have upward motion, the frictional resistance F acting
downwards will be
F = μ . RA = 0.15 × 264870 = 39730 N

GATE/ME/SLP/Module_3/Ch.1_Notes /Pg.32
 Notes on Fluid Mechanics

The upward motion of the gate will be caused by the vertical component Fy of the
water pressure acting on the cylinder in the upward direction.
⎧ 1 ⎛ π ⎞⎫ 1 π
Fy = (ρg) ⎨ ⎜ d2 ⎟ ⎬ L = 9810 × × × 32 × 6
⎩ 2 ⎝ 4 ⎠⎭ 2 4
= 208028 N
The upward motion of the gate will be opposed by the weight W of the gate & the
frictional resistance F at A. Thus, considering limiting conditions of equilibrium of the
gate in vertical direction we have
W + 39730 = 208028
∴ W = 168298 N

BUOYANCY AND FLOATATION

Buoyancy:
When a body is immersed in a fluid, an upward force is exerted by the fluid on the body.
This upward force is equal to the weight of the fluid displaced by the body and is called
the force of buoyancy or simply buoyancy.
Centre of Buoyancy:
Is the defined as the point, through which the force of buoyancy is supposed to act. As
the force of buoyancy is a vertical force and is equal to the weight of the fluid displaced
by the body, the centre of buoyancy will be the centre of gravity of the fluid displaced.

Metacentre:
It is defined as the point about which a body starts oscillating when the body is tilted by a
small angle. The metacentre may also be defined as the point at which the line of action
of the force of buoyancy will meet the normal axis of the body when the body is given
small angular displacement.
Consider a body floating in a liquid as shown in fig. (a). let the body is in equilibrium and
G is the centre of gravity and B the centre buoyancy. For equilibrium, both the points lie
on the normal axis which is vertical.

Let the body is given a small angular displacement in the clockwise direction as shown in
fig. (b).The centre of buoyancy, which is the centre of gravity of the displaced liquid or
centre of gravity of the portion of the body sub − merged in liquid, will now be shifted
towards right from the normal axis. Let it is at B1 as shown in Fig. (b).

GATE/ME/SLP/Module_3/Ch.1_Notes /Pg.33
Vidyalankar : GATE – ME

The line of action of the force of buoyancy in this new position, will intersect the normal
axis of the body at some point say M. This point is called Metacentre.

Metacentric Height:
The distance MG, i.e. the distance between the metacentre of a floating body and the
centre of gravity of the body is called metacentric height.

Expression for Metacentric Height:

Fig. (a) shows the position of a floating body in equilibrium. The location of centre of
gravity and centric of buoyancy in this position is at G and B. The floating body is given a
small angular displacement in the clockwise direction. This is shown in fig. (b). The new
centre of buoyancy is at B1. The vertical line through B1 cuts the normal axis at M. Hence
M is the metacentre and GM is metacentre height.
I
Metacentric height = GM = − BG
V
Where;
V = Volume of the fluid displaced by the body.
= Volume of the body submerged in water.
I = Second moment of area of the plan of the body at water surface about the
axis y − y.

Stability of Floating and Submerged Bodies:

A submerged or a floating body is said to be stable if it comes back to its original position
after a slight disturbance. The relative position of the centre of gravity (G) and centre of
buoyancy (B1) of a body determines the stability of a submerged body

GATE/ME/SLP/Module_3/Ch.1_Notes /Pg.34
 Notes on Fluid Mechanics

Stability of a Submerged Body:

The position of centre of gravity and centre of buoyancy in case of a completely
submerged body are fixed. Consider a balloon, while is completely submerged in air. Let
the lower portion of the balloon contains heavier material, so that its centre of gravity is
lower than its centre of buoyancy shown in fig. (a). Let the weight of the balloon is W. The
weight W is acting through G, vertically in the downward direction, while the buoyant
force FB is acting vertically up, through B. For the equilibrium of the balloon W = FB. If the
balloon is given an angular displacement in the clockwise direction as shown in Fig. (a),
then W and FB constitute a couple acting in the anti−clockwise direction and bring the
balloon in original position. Thus the balloon in the position, shown by Fig. (a) is in stable
equilibrium.

a) Stable Equilibrium:
When W = FB and point B is above G, the body is said to be in stable equilibrium.

b) Unstable Equilibrium:
If W = FB, but the centre of buoyancy (B) is below centre of gravity (G), the body is in
unstable equilibrium as shown in Fig. (b). A slight displacement to the body, in the
clockwise direction, gives the couple due to W and FB also in the clock wise direction,
Thus the body does not return to its original position and hence the body is unstable
equilibrium.

c) Neutral Equilibrium:
If FB = W and B and G are at the same point, as shown in Fig. (c), the body is said to
be in neutral equilibrium.

Stability of Floating Body:

The stability of a floating body is determined form the position of Metacenter (M). In case
of floating body, the weight of the body is equal to is the weight of liquid displaced.
a) Stable Equilibrium:
If the point M is above G, the floating body will be in stable equilibrium as shown in
Fig. (a). If a slight angular displacement is given to the floating body in the clockwise
direction, the centre of buoyancy shifts from B to B1 such that the vertical line through
B1 cuts at M. Then the buoyant force FB through B1 and weight W through G
constitute a couple acting in the anti clockwise direction and thus bringing the floating
body in original position.

GATE/ME/SLP/Module_3/Ch.1_Notes /Pg.35
Vidyalankar : GATE – ME

b) Unstable Equilibrium:
If the point M is below G, the floating body will be in unstable equilibrium as shown in
Fig. (b). The disturbing couple is acting in the clockwise direction. The couple due to
buoyant force FB and W is also acting in the clockwise direction and thus overturning
the floating body.

c) Neutral Equilibrium:
If the point M is at the centre of gravity of the body, the floating body will be in neural
equilibrium.

Oscillation (Rolling) of a Floating Body:

If a floating body is tilted through an angle by an overturning couple and the over turning
couple is suddenly removed then the body will start oscillating. Thus, the body will be in a
state of oscillation as if suspended at the metacentre M. This is similar to the case of a
pendulum. The only force acting on the body is due to the restoring couple due to the
weight W of the body and force of buoyancy FB
Time period of oscillation is given by
K2
T = 2π
GM.g
Where
K = Radius of gyration
GM= MetaCentric height

SOLVED EXAMPLES
Example:
Find the volume of the water displaced and
position of centre of buoyancy for a wooden
block of width 2.5m and of depth 1.5m,
when it floats horizontally in water. The
density of wooden block is 650kg /m3 and its
length 6.0m.

GATE/ME/SLP/Module_3/Ch.1_Notes /Pg.36
 Notes on Fluid Mechanics

Solution:
Width = 2.5 m
Depth = 1.5 m
Length = 6.0 m
Volume of the Block = 2.5 × 1.5 × 6.0 = 22.50m3
Density of wood, ρ = 650 kg. /m3
⇒ Weight of block = p × g × volume = 650 × 9.81 × 22.50 = 143471N
For equilibrium: Weight of water displaced = weight of wooden block
Weight of water dispaced
⇒ Volume of water displaced =
Weight density of water
143471
= = 14.625m3
1000 × 9.81
{ Weight density of water = 1000× 9.81 N/m3}
Position of Centre of Buoyancy :
Volume of wooden block in water = Volume of water displaced
⇒ 2.5 ×h × 6.0 = 14.625m3; Where h is the depth of wooden block in water
⇒ h = 0.975m
0.975
⇒ Centre of Buoyancy = = 0.4875 frombase
2
Example :

Find the density of a metallic body which floats at the interface of mercury of sp. gr. 13.6
and water such that 40% of its volume is submerged in mercury and 60% in water.

Solution:
Let the volume of the body = Vm3
Then volume of body submerged in mercury V = 0.4 Vm3
Volume of body submerged in water = 0.6Vm3
For the equilibrium of the body,
Total buoyant force (upward force) = Weight of the body
But total buoyant force = Force of buoyancy due to water + Force of buoyancy due to
mercury
Force of buoyancy due to water = Weight of water displaced by body
= Density of water × g × Volume of water displaced
= 1000 ×g × Volume of body in water
= 1000 × g × 0.6 × VN

GATE/ME/SLP/Module_3/Ch.1_Notes /Pg.37
Vidyalankar : GATE – ME

Force of buoyancy due to mercury = Weight of mercury displaced by body

= g × density of mercury × volume of mercury
displaced
= g × 13.6 × 1000 × Volume of body in mercury
= g × 13.6 × 1000 × 0.4 VN
Weight of the body = Density × g × Volume of body = ρ × g × V
Where ρ is the density of the body
Hence for equilibrium, we have
Total buoyant force = Weight of the body
⇒ 1000 × g ×0.6 × V +13.6 × 1000 × g ×0.4V = ρ ×g ×V
⇒ ρ = 6040.00 kg /m3

Example:

A block of wood of specific gravity 0.7 floats in water. Determine the meta centric height
of the block if its size is 2m × 1m × 0.8m.

Solution:
Dimension of block = 2 × 1 × 0.8
Let depth of immersion = h m
Sp.gr. of wood = 0.7
Weight of wooden piece
= Density of wood × g× Volume.
= 0.7 × 1000 × 9.81 × 2 ×1 × 0.8 N
Weight of water displaced
= Density of water × g × Volume
= 1000 × 9.81 × 2 ×1 × h N
For equilibrium
Weight of wooden piece = Weight of water displaced
⇒ 700 × 9.81 × 2 ×1 × 1 × 0.8 = 1000 × 9.81 × 2 ×1 × h
⇒ h = 0.56 m
⇒ Distance of centre of buoyancy from bottom, i.e.
h 0.56
AB = = = 0.28m
2 2
and AG = 0.8/2 = 0.4m
⇒ BG = AG − AB = 0.4 − 0.28 = 0.12m

GATE/ME/SLP/Module_3/Ch.1_Notes /Pg.38
 Notes on Fluid Mechanics

The metacentric height is given by

I
GM = − BG
V
1 1
Where I = × 2 × 1.03 = m4
12 6
V = Volume of wood in water = 2 ×1 × h = 2 × 1 × 0.56 = 1.12m3
1 1
⇒ GM = × − 0.12 = 0.1488 − 0.12 = 0.0288m
6 1.12

Example:

A solid cylinder of diameter 4.0m has a height of 3m. Find the metacentric height of the
cylinder when it is floating in water with its axis vertical. The sp.gr.of the cylinder = 0.6

Solution:

Dia of cylinder, D = 4m
Height of cylinder = 3m
Depth of immersion of cylinder = 0.6 × 3.0 = 1.8m
⎧Let depth immersion is d. Then for equilibrium :
⎨
⎩Weight of cylinder in water = weight of water displaced
⇒ ρcylinder × g × Volume of cylinder
= ρ water × g × volume of cylinder in water
π π
 ρcylinder D2h = ρwater × D2 d
4 4
ρcylinder
⇒ d = ×h
ρwater
= Sp. gr. of cylinder × h}
1.8
⇒ AB = = 0.9m and AG = 1.5m
2
⇒ BG = AG − AB = 1.5 − 0.9 = 0.6m
I
now, metacentric height GM = − BG
V
But I = M.O.I. about y − y axis of the plan of the body

GATE/ME/SLP/Module_3/Ch.1_Notes /Pg.39
Vidyalankar : GATE – ME

π 4 π
× ( 4.0 )
4
= D =
64 64
and V = Volume of cylinder submerged in water
π
= D2 × Depthof immersion
4
π
= ( 4.0 ) × 1.8m3
2

4
π
× (4.0)4
⇒ GM = 64 − 0.6
π
× (4.0)2 × 1.8
4
1 4.02 1
= × − 0.6 = − 0.6 = −0.05m
16 1.8 1.8

*Negative sign means that metacentre (M) is below centre of gravity (G).

Example:
A solid cylinder of diameter 4.0m has a height of 4.0m. Find the metacentric height of the
cylinder if the specific gravity of the material of cylinder = 0.6 and it is floating in water
with its axis vertical. State whether the equilibrium is stable or unstable.

Solution:
D = 4m
Height,
h = 4m
sp.gr. = 0.6
Depth of cylinder in water = 0.6 × 4.0 = 2.4m
⇒ Distance of centre of buoyancy (B) from A;
2.4
AB = = 1.2m
2
h 4.0
AG = = = 2.0m
2 2
⇒ BG = AG − AB = 2.0 − 1.2 = 0.8m

GATE/ME/SLP/Module_3/Ch.1_Notes /Pg.40
 Notes on Fluid Mechanics

I
Now, metacentric height, GM = − BG
V
Where I = M.O.I. of the plan of the body about y − y
π 4 π
× ( 4.0 )
4
= D =
64 64
V = Volume of cylinder in water
π
= × D2 × Depthof cylinder in water
4
π
= × 42 × 2.4m3
4
π
× 44
I 64 1 42 1
⇒ = = × = = 0.416m
V π 16 2.4 2.4
× 42 × 2.4
4
I
⇒ GM = − BG = 0.4167 − 0.8 = −0.3833m
V
Negative sign means that the metacentre (M) is below the centre of gravity (G). Thus
the cylinder is in unstable equilibrium.

Example:
The time period of the rolling of a ship of weight 29430 kN in sea water is 10 seconds. the
centre of buoyancy of the ship is 1.5m below the centre of gravity. Find the radius of
gyration of ship if the moment of inertia of the ship at the water line about fore and aft
axis is 1000m4.
Take specific weight of sea water as 10100N/m3.
Solution:
Ti me period, , T = 10 s
Distance between centre of buoyancy and centre of gravity
BG = 1.5m
Moment of Inertia, I = 10000m4
Weight, W = 29430 × 1000N
Let the radius of gyration = K
I
Metacentric height GM = − BG
V
Where I = M.O.I
Weight of ship
V = Volume of water displaced =
sp.Weight of sea Water
29430 × 1000
= = 2912.6m3
10104
1000
⇒ GM = − 1.5 = 3.433 − 1.5 = 1.933m
2912.6
K2
Now, T = 2π
GM × g

GATE/ME/SLP/Module_3/Ch.1_Notes /Pg.41
Vidyalankar : GATE – ME

K2 2πΚ
⇒ 10 = 2π =
1.933 × 9.81 1.933 × 9.81
10 × 1.933 × 9.81
⇒ K= = 6.93m
2π

FLOW KINEMATICS
The fluid motion is described by two methods. They are (i) Lagrangian Method, and (ii)
Eulerian Method. In the Lagrangain method, a single fluid particle is followed during its
motion and its velocity acceleration, density etc, are descried. In case of Eulerian
method, the velocity, acceleration, pressure, density etc. are described at a point in a flow
field. The Eulerian method is commonly used in fluid mechanics.
Types of Fluid Flow: The fluid flow is classified as:
(i) Steady and unsteady flows.
(ii) Uniform and non − uniform flows.
(iii) Laminar and turbulent flows.
(iv) Compressible and incompressible flows
(v) Rotational and irrotational flow, and
(vi) One, two and three dimensional flows.
i) Steady and unsteady flows:

Steady flow is defined as that type of flow in which the fluid characteristics like
velocity, pressure, density, etc. at a point do not change with time. Thus for steady
flow,
⎛ ∂V ⎞ ⎛ ∂p ⎞ ⎛ ∂ρ ⎞
⎜ ∂t ⎟ = 0, ⎜ ⎟ = 0, ⎜ ⎟ =0
⎝ ⎠ xo ,yo ,zo , ⎝ ∂t ⎠ xo ,yo ,zo ⎝ ∂t ⎠ xo ,yo ,zo
Where (xo, yo, zo, ) is a fixed point in fluid field.

Unsteady, flow is that type of flow, in which the velocity pressure and density at a
point changes with respect to time. Thus, mathematically for unsteady flow,

⎛ ∂V ⎞ ⎛ ∂p ⎞
⎜ ∂t ⎟ ≠ 0, ⎜ ⎟ ≠ 0,etc.
⎝ ⎠ xo ,yo ,zo , ⎝ ∂t ⎠ xo ,yo ,zo

ii) Uniform and Non − Uniform Flows:

Uniform flow is defined as that type of flow in which the velocity at any given time
does not change with respect to space (i.e. length of direction of the flow). e.g. flow of
liquids under pressure through long pipeline constant diameter. Mathematically, for
uniform flow.
⎛ ∂V ⎞
⎜ ∂s ⎟ =0
⎝ ⎠ t = cons tan t
Where ∂V = Change of velocity
∂s = Length of flow in the direction s

GATE/ME/SLP/Module_3/Ch.1_Notes /Pg.42
 Notes on Fluid Mechanics

Non uniform flow is that type of flow in which the velocity at any given time changes
with respect to space. e.g. flow of liquids under pressure through long pipeline
varying diameter. Thus, mathematically, for non−uniform flow,
⎛ ∂V ⎞
⎜ ∂s ⎟ ≠0
⎝ ⎠ t=consstant
iii) Laminar and Turbulent Flows:
A flow is said to be laminar when the various fluid particles move in layers (or lamine)
with on layer of fluid sliding smoothly over an adjaction layer. Thus in the
development of a laminar flow, the viscosity of fluid plays a significant role. This type
of flow is also called stream− line flow or viscous flow.
Examples of laminar fuid motion are found in lubricating mechanis, in flow of fluids in
minute capillaries, blood veins and porous media, In flow of highly viscous fluids such
as oils in tube ad conduits.
A fluid motion is said to be turbulent when the fluid particles move in an entirely
haphazard or disorde manner, that results in a rapid and continuous mixing of the
fluid leading to momentum transfer as flow occurs. In such a flow eddies or vortices
of different sizes and shapes are present which move over large distances. The
occurrence of turbulent flow is more frequent than that of laminar flow. e.g. : flow in
natural stream, artifical channels, water supply pipes, sewers etc. are a few
examples of turbulent flow.
iv) Compressible and Incompressible Flows:
Compressible flow is that type of flow in which the density of the fluid changes form
point to point or in other words the density is not constant for the fluid.
In compressible flow is that type of flow in which the density is constant for the fluid
flow Liquids are generally incompressible while gases are compressible.
v) Rotational and Irrotational Flows:
Rotational flow is that type of flow in which the fluid particles while flowing along
stream lines also rotate about their own axis. eg. liquid in rotating time where velocity
of each partices varies directly as the distance from the centre of rotation. And if the
fluid particles while flowing along stream −lines, do not rotate about their own axis,
that type of flow is called irrotational flow. It exists only in case of flow of ideal fluids
for which no tangential or shear stresses occur. But the flow of practical fluids may
also be assumed to be irrotational if the viscosity of fluid has little significance.
VD
When the Reynolds number Re = of a pipe flow exceeds Recrit = 2000 the flow
ν
becomes turbulent.
vi) One, two and three dimensional flows:
a) One − dimensional flow:
It is that type of flow in which the flow parameter such as velocity is a function of
time and one space coordinate only. The variation of velocities in other two
mutually perpendicular direction is assumed negligible. Hence mathematically,
for one dimensional flow
u = f (x) , v = 0, w = 0
Where u, v and w are velocity components in x, y and z directions respectively.

GATE/ME/SLP/Module_3/Ch.1_Notes /Pg.43
Vidyalankar : GATE – ME

b) Two − dimensional flow:

It is that type of flow in which the velocity is a function of time and two rectangular
space co− ordinates say x and y. For a steady two − dimensional flow the
velocity is a function of two space co−ordinates only. The variation of velocity in
the third direction is negligible. Thus, mathematically, for two dimensional flow:
u = f1(x, y), v = f2 (x, y) and w = 0
c) Three − dimensional flow:
It is that type of flow in which the velocity is a function of time and three mutually
perpendicular directions. But, for a steady three dimensional flow the fluid
parameters are function of space co−ordinates (x, y and z) only. Thus,
mathematically, for three − dimensional flow:
u = f1 (x, y, z)
v = f2 (x, y, z)
w = f3 (x, y, z)

V = iu +ju + kw
where i, j, k are the unit vectors parallel to x, y, z axes respectively.

Discharge or Rate of Flow:

It is defined as the quantity of a fluid flowing per second through a section of a pipe or a
channel. For an incompressible fluid (or liquid) the rate of flow or discharge is expressed
as the volume of fluid flowing across the section per second. For compressible fluids, the
rate of flow is usually expressed as the weight of fluid flowing across the section. Thus
3
(i) For liquids the units of discharge are m / s or litres/s
(ii) For gases the units of discharge are kgf/s or Newton/s. Consider a liquid flowing
through a pipe in which
A = Cross − sectional area of pipe
V = Average velocity of fluid across the section
Then discharge Q = A × V
Continuity Equation (Mass Equation):
The equation based on the principle of conservation of mass is called continuity equation.
Thus for a fluid flowing through the pipe at all the cross − sections, the quantity of fluid
per second is constant. Consider two cross − sections of a pipe as shown in Fig.

GATE/ME/SLP/Module_3/Ch.1_Notes /Pg.44
 Notes on Fluid Mechanics

Let V1 = Average velocity at cross section 1 − 1

ρ1 = Density at section 1 − 1
A1 = Area of pipe at section 1 − 1
and V2, ρ2, A2 are corresponding values at section, 2 − 2
Then rate of flow at section 1−1 = ρ1 A1 V1
Rate of flow at section 2 − 2 = ρ2A2V2
According to law of conservation of mass
Rate of flow at section 1−1= Rate of flow at section 2 − 2
⇒ ρ1 A1 V1 = ρ2 A2 V2

Above equation is applicable to the compressible as well as incompressible fluids and is

called continuity equation. If fluid is incompressible, then ρ1 = ρ2 and continuity equation
becomes
A1 V1 = A2 V2

Differential Equation of Continuity:

∂ρ ∂ ∂ ∂
+ ( ρu) + ( ρv ) + ( ρw ) = 0
∂t ∂x ∂y ∂z

Where ρ is the density of the fluid. u, v and w are the inlet velocity components in x, y and
z directions respectively.

Above equation is the continuity equation in Cartesian co − ordinates in its most general
form. This equation is applicable to:
i) Steady and unsteady flow,
ii) Uniform and non − uniform flow, and
iii) Compressible and incompressible fluids
∂ρ
For steady flow, = 0 and hence continuity equation becomes
∂t
∂ ∂ ∂
( ρu) + ( ρv ) + ( ρw ) = 0
∂x ∂y ∂z
If the fluid is incompressible, then ρ is constant and the above equation becomes as
∂u ∂v ∂w
+ + =0
∂x ∂y ∂z

Above equation is the continuity equation in three dimensions. For a two dimensional
flow, the component w = 0 and hence continuity equation becomes as:
∂u ∂v
+ =0
∂x ∂y
Polar co − ordinates:
∂ρ ∂(ρ Vrr) ∂(ρ Vθ ) ∂(ρ Vz )
+ + +
∂t r ∂r r ∂θ ∂z
Where Vr is the velocity component in radial direction, and Vθ in tangential direction.
For steady flow:

GATE/ME/SLP/Module_3/Ch.1_Notes /Pg.45
Vidyalankar : GATE – ME

∂ ∂
(rVr ) + ( Vθ ) = 0
∂r ∂θ
∂Vr ∂Vθ
⇒ Vr + r + =0
∂r ∂θ

Velocity and Acceleration:

Let V is the resultant velocity at any point in a fluid flow. Let u, v and w are its
components in x , y and z direction. The velocity components are functions of space co −
ordinates and time. Mathematically, the velocity components are given as
u = f1 (x, y, z, t)
v = f2 (x, y, z, t)
w = f3 (x, y, z, t)
Resultant velocity V = u ˆi + vjˆ + wkˆ = u2 + v 2 + w 2
Let ax, ay and az, are the total accelerations in x, y and z direction respectively. Then;
du ∂u ∂u ∂u ∂u
ax = =u +v +w +
dt ∂x ∂y ∂z ∂t

dv ∂v ∂v ∂v ∂v
ay = =u +v +w +
dt ∂x ∂y ∂z ∂t

dw ∂w ∂w ∂w ∂w
az = =u +v +w +
dt ∂x ∂y ∂z ∂t

∂v
For steady flow, =0
∂t
∂u ∂v ∂w
⇒ = 0, = 0 and =0
∂t ∂t ∂t
Hence acceleration in x, y and z direction becomes
du ∂u ∂u ∂u
ax = =u +v +w
dt ∂x ∂y ∂z
dv ∂v ∂v ∂v
ay = =u +v +w
dt ∂x ∂y ∂z
dw ∂w ∂w ∂w
az = =u +v +w
dt ∂x ∂y ∂z

Acceleration vector A = a i + a ˆj + a kˆ
ˆ
x y z

Resultant acceleration A = a x 2 + a y 2 + a z2

Local Acceleration and Convective Acceleration:

Local acceleration is defined as the rate of increase of velocity with respect to time at a
given point in a flow field.
∂u ∂v ∂w
The expressions , and are knownas local acceleration.
∂t ∂t ∂t

GATE/ME/SLP/Module_3/Ch.1_Notes /Pg.46
 Notes on Fluid Mechanics

Convective acceleration is defined as the rate of change of velocity due to the change of
position of fluid particles in a fluid flow. The expressions for ax , ay and az for steady flow
are known as convective acceleration. For steady flow local acceleration is zero but the
connective acceleration is not necessarily zero. However in cse of uniform flow,
convective acceleration is also zero.

VELOCITY POTENTIAL FUNCTION AND STREAM FUNCTION

Velocity Potential Function:
It 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. It is defined by φ (phi.).
Mathematically, the velocity potential is defined as
φ = f (x, y, z)
For steady flow such that
φ = f(x, y, z, t) for unsteady flow
−∂φ
u =
∂x
−∂φ
v = −−−−−−− (1)
∂y
−∂φ
w =
∂z
were u, v, and w are components of velocity in x, y and z directions respectively.
The negative sign shows that φ decreases with an increase in value of x, y, z i.e. flow is
always in the direction of decreasing φ.
The velocity components in cylindrical polar co− ordinates in terms of velocity potential
function are given by
∂φ
ur = −−−−−− (2)
∂r
1 ∂φ
uθ =
r ∂θ

Where ur = velocity component in radial direction (i.e. in r direction) and uθ = Velocity

component in tangential direction (i.e. in θ direction)
The continuity equation may be expressed in vector notation as ∇ 2 φ = 0. This above
equation is known as laplace equation.

The continuity equation for an incompressible steady flow becomes

∂2φ ∂2φ ∂2φ
+ + =0
∂x 2 ∂y 2 ∂z2
Above equation in known as Laplace equation.
For two dimensional case the equation reduces to
∂2φ ∂2φ
+ =0
∂x 2 ∂y 2

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Vidyalankar : GATE – ME

If any function of φ that satisfies the Laplace equation, will correspond to some case of
fluid flow.
The properties of potential function are:
(1) If velocity potential (φ) exists, the flow should be irrotational.
(2) If velocity potential (φ) exists, and satisfies the Laplace equation, it represents the
possible steady incompressible irrotational flow.
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 components at right angles to that direction. it
is denoted by Ψ and defined only for two dimensional flow. Mathematically, for steady
flow it is defined as Ψ = f(x, y) such that
∂Ψ
=v
∂x
∂Ψ
= −u
∂y
The velocity components in cylindrical polar co − ordinates in terms of stream function
are given as
1 ∂Ψ ∂Ψ
ur = anduθ = −
r ∂θ ∂r
Where ur = radial velocity and uθ = tangential velocity
For irrotational flow:
∂ 2 Ψ ∂Ψ
+ =0
∂x 2 ∂y 2
which is Laplace Equation for Ψ .
The properties of stream function ( Ψ ) are:
1) If stream function ( Ψ ) exists, it is a possible case of fluid flow which may be
rotational or irrotational
2) If stream function ( Ψ ) satisfies the Laplace equation, it is a possible case of an
irrotational flow.
3) The difference of its values at two points represents the flow across any line joining
the points.
4) When Ψ1 − Ψ 2 = 0, the two points lie on the same stream line, there is no flow
across the stream line. It means that stream line is given by Ψ = constant.
Equipotential Line:
A property of the stream function is that the difference of its values at two points
represents the flow across any line joining the points. Therefore, when two points lie on
the same stream line, then since there is no flow across a streamline, the difference
between the stream functions ψ1 and ψ 2 at these two points is equal to zero, i.e.
( ψ1 − ψ 2 ) = 0. In other words, it means that streamline is given by ψ = constant.

Similarly φ = constant, represents a curve for which the velocity potential is same at every
point, and hence it represens an equipotential line.

GATE/ME/SLP/Module_3/Ch.1_Notes /Pg.48
 Notes on Fluid Mechanics

Consider two curves, viz., φ = constant and ψ = constant, intersecting each other at any
point. The slopes of these curves at the point of intersection may be determined as
indicate below.

For the curve φ = constant,

⎛ ∂φ ⎞
∂y ⎜⎝ ∂x ⎟⎠ −u u
slope = = = =
∂x ⎛ ∂φ ⎞ −v v
⎜ ∂y ⎟
⎝ ⎠
Similarly for the curve ψ = constant,
⎛ dψ ⎞
∂y ⎜⎝ ∂x ⎟⎠ v v
slope = = = =−
∂x ⎛ ∂ψ ⎞ −u u
⎜ ∂y ⎟
⎝ ⎠
Therefore the product of the slopes of these two curves at the point of intersection = −1,
which indicates that these two sets of curves, viz., streamlines and equipotential lines
intersect each other orthogonally at all points of intersection.
A grud obtained by drawing a series of streamlines and equipotential lines is known as a
flow net. A flow net may be drawn for a two dimensional irrotational flow and it provides a
simple, yet valuable indication of the flow pattern. Fig. below shows the elements of a
flow net which has been obtained by drawing a set of cues corresponding to ψ = C1, C2,
C3 etc., and φ = C1, C2, c3 etc.
At any point in the direction n along the equipotential line since φ = constant, C2
∂φ
=0
∂n

Elements of a flow net

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Vidyalankar : GATE – ME

But according to the definition of the velocity potential

∂φ
− = Vn
∂n
∴ Vn = 0
∂φ
Further − = Vs
∂s
The relations therefore indicate that there is no flow along the diretion tangential to the
equipotential lines, but the flow always takes place in the direction at right angles to the
equipotential lines.
Similarly at the point shown in fig. above, by definition
∂ψ
= Vn = 0
∂s
∂ψ
and = − Vs
∂n
These relations also indicate that there is no flow in the direction normal to the
streamlines but the flow is always along the direction tangential to the streamlines.
Relation between Stream Function and Velocity Potential Function:
∂φ ∂Ψ
= = −u
∂x ∂y
∂φ ∂Ψ
− = =v
∂y ∂x

Rotation:
It is defined as the movement of a fluid element in such a way that both of its axes
1 ⎛ ∂v ∂u ⎞
(horizontal as well as vertical) rotate in same direction. It is equal to ⎜ − ⎟ for a
2 ⎝ ∂x ∂y ⎠
two − dimensional element in x − y plane. The rotational components are:
1 ⎛ ∂v ∂u ⎞
ωz = ⎜ − ⎟
2 ⎝ ∂x ∂y ⎠
1 ⎛ ∂w ∂v ⎞
ωx = ⎜ − ⎟
2 ⎝ ∂y ∂z ⎠
1 ⎛ ∂u ∂w ⎞
ωy = ⎜ −
2 ⎝ ∂z ∂x ⎟⎠

* If velocity potential function exists, then ω x = ω y = ω z = 0. In terms of stream

function we can write by substituting values of u, v, w in terms of φ i.e.
∂φ ∂φ ∂φ
u=− , v= − , w= −
∂x ∂y ∂z

∂ψ ∂ψ
u= − , v= −
∂y ∂x

GATE/ME/SLP/Module_3/Ch.1_Notes /Pg.50
 Notes on Fluid Mechanics

We have
1 ⎡ ∂2 Ψ ∂2 Ψ ⎤
ωz = ⎢ + 2 ⎥
2 ⎣ ∂x 2 ∂y ⎦

The above equation is called Poisson’s equation.

∂2 Ψ ∂2 Ψ
For irrotational flow, ω z = 0 and hence + 2 =0
∂x 2 ∂y
This is Laplace’s equation for ψ .

* It stream function satisfies Laplace equation, the flow is irrotational.

Equations of Motion:
According to Newton’s second law of motion, the net force Fx acting on a fluid element in
the direction of x is equal to mass m of the fluid element multiplied by the acceleration ax
in the x − direction. Thus, mathematically Fx = m ax
In the fluid flow, the following forces are present:
1) Fg, gravity force
2) Fp, the pressure force
3) F ν , the force due to viscosity
4) Fc, force due to compressibility
5) Ft, fore due to turbulence of flow

Thus , net force Fx, = (Fg )x + (Fp)x + (Fν)x + (Ft)x+(Fc)x

* If the force due to compressibility, FC is negligible, the resulting net force
Fx =(Fg)x + (Fp)x + (Fν)x + (Ft)x
and equations of motion are called “Reynold’s equations of motion”

* For flow where (Ft) is negligible, the resulting equations of motion are known as
“Navier − Stokes equation”.

* Fx = (Fg )x + (Fp )x + (Fv )x (Navier Stoke’s equation.)

This equation is useful in analysis of viscous flow.

* If the flow is assumed to be ideal, viscous force (Fν) is zero and equations of motion
are known as “Euler’s equations of motion”.
Fx = (Fg )x + (Fp )x

Euler’s Equation of Motion:

This is Equation of motion in which the force due to gravity and pressure are taken into
consideration
dp
Euler’s Equation is + gdz + vdv = 0
ρ
Bernouli’s equation is obtained by integrating the Euler’s equation of motion as

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Vidyalankar : GATE – ME

dp
∫ ρ ∫
+ gdz + ∫ vdv = constant

This equation is applicable for steady, irrotational flow of incompressible fluids.

If flow is incompressible, ρ is constant i.e. independent of pressure (hence p is treated as

constant).
p v2
+ gz + = Constant
ρ 2
p v2
⇒ + + z = Constant
ρg 2g
Above equation is Bernoulli’s equation in which:
p
= pressure energy per unit weight of fluid or pressure head
ρg
v2 / 2g = Kinetic energy per unit weight or kinetic head
z = potential energy per unit weight or potential head
Assumptions:
The following are the assumptions made in the derivation of Bernoulli’s equation:
(1) The fluid is ideal, i.e. viscosity is zero
(2) The flow is steady
(3) The flow is incompressible
(4) The flow is irrotational

Bernoulli’s Equation for incompressible Real Fluid:

The Bernoulli’s equation was derived on the assumption that fluid is inviscid (non −
viscous ) and therefore frictionless. But all the real fluids are viscous and hence offer
resistance to flow. Thus there are always some losses in fluid flows and hence in the
applications of Bernoulli’s equation, these losses have to be taken into consideration.
Thus the Bernoulli’s equation for real fluids between points 1 and 2 is given as
p1 v12 p v2
+ + z1 = 2 + 2 + z 2 + hL
ρg 2g ρg 2g
Where hL is loss of energy per unit weight of fluid between points 1 and 2.

Practical Applications of Bernoulli’s Equation:

Bernoulli's equation is applied in all problems of incompressible fluid flow where energy
considerations are involved e.g.
1) Venturimeter
2) Orifice meter
3) Pitot − tube
The Momentum Equation:
It is based on the law of conservation of momentum, which states that the net force
acting on fluid mass is equal to the change in momentum of flow per unit time in that
direction. hence;
d ( mv )
F= − − − − − − − −(1)
dt

GATE/ME/SLP/Module_3/Ch.1_Notes /Pg.52
 Notes on Fluid Mechanics

Where F is the force on a fluid mass m.

Above equation is known as the “momentum principle”
Equation (1) may also be written as
F. dt = d (mv)
Which is known as the “impulse − momentum equation”.
Force exerted by a flowing fluid on Pipe − Bend:
The impulse momentum equation is used to determine the resultant force exerted by a
flowing fluid on a pipe bend. Consider two sections (1) and (2) as shown in fig.

Forces on bend
Let, V1 = Velocity of flow at section (1)
p1 = pressure intensity at section (1)
A1 = area of cross section of pipe at section (1) and
V2, p2.,A2 = corresponding values of velocity, pressure and area at section (2)
Let Fx and Fy be the components of the forces exerted by the flowing fluid on the bend in
x and y direction respectively. Then
Fx = ρQ (V1 − V2 cosθ ) + p1 A1 − p1 A2 cosθ
Fy = ρ Q (−V2 sin θ) − p2 A2 sin θ
The resultant force (FR) acting on the bend = Fx2 + Fy2
And the angle made by the resultant force with horizontal direction is given by
Fy
tanθ =
Fx
Moment of Momentum Equation:
Moment of momentum equation is derived from moment of momentum principle which
states that the resulting torque acting on a rotating fluid is equal to the rate of change of
moment of momentum.
let V1 = Velocity of fluid at section 1
r1 = radius of curvature of section 1
Q = rate of flow of fluid
ρ = density of fluid
V2 and r2 = velocity and radius of curvature at section 2.
Hence resultant torque; T = ρQ (V2r2 − V1 r1)
Above equation is known as moment of momentum equation. This equation is applied:
1) For analyzing flow problems in turbines and centrifugal pumps.
2) For finding torque exerted by water on sprinkler.

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Vidyalankar : GATE – ME

Free Liquid Jets:

Free liquid jet is defined as the jet of water coming out from the nozzle in atmosphere.
The path travelled by the free jet is parabolic. Consider a jet coming from the nozzle as
shown in fig. Let the jet at A, makes an angle θ with the horizontal direction If U is the
velocity of jet of water, then the horizontal component and vertical component of this
velocity at A are U cos θ and U sinθ. Consider another point P (x, y) on the centre line of
the jet. The co−ordinates of P from A are x and y.

Free liquid jet

gx 2
Equation of motion y = x tan θ − sec 2 θ
2U2
U2 sin2 θ
Maximum height attained by the jet S =
2g
2Usin θ
Time of flight T =
g
T Usin θ
Time to reach highest point T ′ = =
2 g
U2 sin2θ
Horizontal range of the jet =R=
g
Maximum horizontal range
U2
Rmax = when θ = 45°
g
Hydraulic Coefficients :
The hydraulic Coefficients are:
(1) Coefficient of velocity, Cv
(2) Coefficient of contraction, Cc
(3) Coefficient of discharge, Cd
1) Coefficient of velocity (Cv):
It is defined as the ratio between the actual velocity of jet of liquid at vena − contracta
and the theoretical velocity of jet. It is denoted by CV and mathematically CV is given
as:
Actual velocity of jet at vena − contracta
CV =
Theoritical velocity
V
= (H is the height from which water falls)
2gH

GATE/ME/SLP/Module_3/Ch.1_Notes /Pg.54
 Notes on Fluid Mechanics

The value of CV varies from 0.95 to 0.99 for different orifices, depending upon the
shape, size of the orifice and on the head under which the flow takes place. CV is
generally taken as 0.98 for sharp edged orifices

2) Coefficient of Contraction (Cc):

It is defined as the ratio of the area of the jet at vena − contracta to the area of orifice.
It is denoted by Cc .
area of jet at vena - contracta a c
Cc = =
area of orfice a
The value of Cc varies form 0.61 to 0.69 depending on shape and size of the orifice
and head of liquid under which flow takes place. In general, the value of Cc may be
taken as 0.64,
3) Coefficient of discharge (Cd):
It is defined as the ratio of actual discharge from an orifice to the theoretical
discharge from the orifice. It is denoted by Cd
Q ActualDischarge
Cd = =
Q th TheoriticalDischarge
Also Cd = Cv × Cc
The value of Cd varies from 0.61 to 0.65. For general purpose, the value of Cd is
taken as 0.62
Venturimeter :
A Venturimeter is a device used for measuring the rate of flow of a fluid flowing through a
pipe. It consists of three parts:
(i) A short converging part, (ii) throat, and (iii) diverging part. It is based on the principle
of Bernoulli’s equation.
Expression for Rate of Flow through Venturimeter:
Consider a venturimeter fitted in a horizontal pipe through which a fluid is flowing (say
water) as shown in fig.
Let d1 = diameter at inlet or at section (1)
p1 = pressure at section (1)
V1 = Velocity of fluid at section (1)
π
A1 = area of section (1) = d12
4

Venturimeter
And d2, p2, V2, a2, are corresponding values at section (2).
a1a 2
Discharge, Q = × 2gh
a12 − a 22
p1 − p2
Where h = = Difference of pressure heads at sections (1) and (2).
ρg
Above equation gives the discharge under ideal conditions and is called, theoretical
discharge. Actual discharge will be less than theoretical discharge.

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Vidyalankar : GATE – ME

a1a 2
⇒ Qact = Cd × 2gh
a12 − a 22
Where Cd is the coefficient of discharge.
Value of ‘h’ givenby differential U −tube manometer:
Case I:
Let the differential manometer contains a liquid which is heavier than the liquid flowing
through the pipe. Let
Sh = sp.gr. of the heavier liquid
So = sp.gr. of the liquid flowing through pipe
x = difference of the heavier liquid column in U− tube

⎡S ⎤
Then h = x ⎢ h − 1⎥
⎣ So ⎦
Case II:
If the differential manometer contains a liquid which is lighter than the liquid flowing
through the pipe, the value of h is given by
⎡ S ⎤
h = x ⎢1 −  ⎥
⎣ So ⎦
Where S  = sp. gr. of lighter liquid in U − tube.

Orifice Meter or Orifice Plate:

It is a cheaper device as compared to venturimeter. It consists of a flat circular plate
which has circular shape edged hole called orifice, which is concentric with the pipe. The
orifice diameter is kept generally 0.5 times the diameter of the pipe, through it may vary
from 0.4 to 0.8 times the pipe diameter.
A differential manometer is connected at section (1), which is at a distance of about 1.5
time to 2.0 times the pipe diameter upstream from the orifice plate, and at section (2),
which is at a distance about half the diameter of the orifice on the down stream side from
the orifice plate.

GATE/ME/SLP/Module_3/Ch.1_Notes /Pg.56
 Notes on Fluid Mechanics

Let p1 = pressure at section (1),

V1 = velocity at section (1),
a1 = area of pipe at section (1),
p2, V2, a2, are corresponding values at section (2)
C a a 2gh
Discharge, Q = d o 1
a12 − a 02
Section (2) at vena − contracts and a2 represents the area of the vena − contract. If ao is
the area of orifice then,
C
ao = c
a2
Where Cc = coefficient of contraction
* The coefficient of discharge for orifice meter is much smaller than that for a
venturimeter.

Pitot Tube:
It is a device used for measuring the velocity of flow at any point in a pipe or a channel. It
is based on the principle that if the velocity of flow at a point becomes zero, the pressure
there is increased due to conversion of kinetic energy into pressure energy. In its
simplest form the pitot tube consists of a glass tube, bent at right angles as shown in fig.

The lower end, which is bent through 90° is directed in the upstream direction as shown
in fig. The liquid rise up in the tube due to the conversion of kinetic energy into pressure
energy. The velocity is determined by measuring the rise of liquid in the tube.
The velocity at any point in flow is given by
V = Cv 2gh
Where
Cv = coefficient of velocity
h = Height of liquid in the pitot tube above water surface

Experimental Determination of Hydraulic Coefficients:

1) Determination of Cd:
The water is allowed to flow through an orifice fitted to a tank under a constant head,
H, as shown in Fig. The water is collected in a measuring tank for a known time, t.
The height of water in the measuring tank is noted down.
Then actual discharge through orifice,

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Vidyalankar : GATE – ME

Area of measuring tank × Height of water in measuring tank

Q =
Time(t)
And theoretical discharge = area of orifice × v

where v = 2gH = velocity of free fall from height H

Q
⇒ Cd =
a × 2gΗ

2) Determination of coefficient of Velocity(Cv):

Let C − C represents the vena − contracta of a jet of water coming out from an orifice
under constant head H as shown in fig. (above). Consider a liquid particle which is at
vena − contracta at any time and takes the position at P along the jet in time ‘t’.
Let x = horizontal distance travelled by the particles in time ‘t’
y = vertical distance between P and C − C
V = actual velocity of jet at vena contracata
Then
x
Cv =
4yH

Flow Through Large Orifices:

If the head of liquid is less than 5 times the depth of the orifice, the orifice is called large
orifice. In case of small orifice, the velocity in the centre cross − section of the jet is
considered to be constant and discharge can be calculated by Q = Cd × a × 2gh . But, in
case of a large orifice, the velocity is not constant over the entire cross section of the jet
and hence Q cannot be calculated by above formula.
1) 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 as shown in fig.

Let H1 = height of liquid above top edge of orifice

H2 = height of liquid above bottom edge of orifice
b = breadth of orifice

GATE/ME/SLP/Module_3/Ch.1_Notes /Pg.58
 Notes on Fluid Mechanics

d = depth of orifice = H2 −H1

Cd = coefficient of discharge
2
Then discharge, Q = Cd × b 2g ⎡⎣H32 / 2 − H13 / 2 ⎤⎦
3
2) Discharge Through Fully Submerged Orifice:
Fully submerged orifice is one which has its
whole of the outlet side sub merged under liquid
so that it discharges a jet of liquid into the liquid
of the same kind. It is also called totally drowned
orifice. Consider two points (1) and (2), point (1)
being in the reservoir on the upstream side of
the orifice and point (2) being at the vena
contracta as shown in fig.
Let H1 = Height of water above the top of the orifice on the upstream side.
H2 = Height of water above the bottom of the orifice
H = Difference in water level
b = width of orifice
Cd = Coefficient of discharge
Discharge through orifice, Q = Cd × b (H2 − H1) × 2gH

3) Discharge Through Partially Submerged Orifice:

Partially submerged orifice is one which has its
outlet side partially submerged under liquid as
shown in fig. It is also known as partially downed
orifice.
Discharge ; Q =
2
Cd × b × (H2 − H1) × 2gH + Cd × b × 2g ⎡⎣H3 / 2 − H13 / 2 ⎤⎦
3
4) Time of Emptying a Cylindrical Tank Through an Orifice at its Bottom:
Consider a tank containing some liquid upto a height 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 form the
height H1 to a height H2.
Let A = Area of the tank
a = Area of the orifice
H1 = Initial height of the liquid
H2 = Final height of the liquid
T = Time in seconds for the liquid to fall
from height H1 to H2
2A ⎡ H1 − H2 ⎤
Hence T = ⎣ ⎦
Cd. a 2g
For emptying the tank completely; H2 = 0 and
2A H1
T =
Cd. a 2g

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Vidyalankar : GATE – ME

5) Time of Emptying a Hemispherical Tank:

Consider a hemispherical tank of radius R fitted
with an orifice of area ‘a’ at its bottom as shown
in Fig. The tank contains some liquid whose
initial height is H1 and in time T, the height of
liquid falls to height H2

π ⎡4 2 5/2 5/2 ⎤
Here, T =
Cd .a. 2g ⎣
( 3/2
)
⎢ 3 R H1 − H2 − 5 H1 − H2 ⎥
3/2
(⎦
)
For completely emptying the tank, H2 = 0 and hence
π ⎡4 2 5/2 ⎤
⎢ 3 RH1 − 5 H1 ⎥
3/2
T =
Cd .a. 2g ⎣ ⎦

6) Time of Emptying a Circular Horizontal Tank:

Consider a circular horizontal tank of length L and radius R, containing liquid upto a
height of H1. Let an orifice of area ‘a’ is fitted at the bottom of the tank. Let T is the
time required to bring the liquid level from H1 to H2.

4L ⎡( 2R − H )3 / 2 − ( 2R − H )3 / 2 ⎤
Then, T =
3Cd .a. 2g ⎣ ⎦
2 1

For completely emptying the tank, H2 = 0 and hence

4 ⎡( 2R )2 / 3 − ( 2R − H )3 / 2 ⎤
T =
3C .a. 2g ⎣ ⎦
1
d

FLOW THROUGH MOUTHPIECES:

1) Flow through an External Cylindrical Mouthpiece:

A mouthpiece is a short length of a pipe which is two or three times its diameter in
length. If this pipe is fitted externally to the orifice, the mouthpiece is called external
cylindrical mouthpiece, and the discharge through orifice increases.
Consider a tank having an external cylindrical mouthpiece of cross sectional area a1,
attached to one of its sides as shown in fig. The jet of liquid entering the mouthpiece
contracts to form a vena− contracta at a section C −C. Beyond this section, the jet
again expands and fills the mouthpiece completely.

GATE/ME/SLP/Module_3/Ch.1_Notes /Pg.60
 Notes on Fluid Mechanics

Let H = Height of liquid above the centre of mouthpiece

Vc = Velocity of liquid at C − C section
ac = Area of flow at vena − contracta
V1 = Velocity of liquid at outlet
a1 = Area of mouthpiece at outlet
Cc = Coefficient of contraction

Thus V1 = Cv 2gH
Where Cv = coefficient of velocity = 0.855
For mouthpiece; Cc = 1
Thus Cd = Cc × Cv = 0.855
*The value of Cd for mouthpiece is more than the value of Cd for orifice, and so
discharge through mouthpiece will be more.
Q = Cc Cv . a . 2gH
= Cd . a . 2gH

2) Flow Through a Convergent − Divergent Mouthpiece:

If a mouthpiece converges upto vena − contracta and then divervges as shown in fig.,
that type of mouthpiece is called Convergent − Divergent Mouthpiece. For this
mouthpiece Cd = 1. Let H is the head of liquid over the mouthpiece.

Velocity at outlet is given by V1 = 2gH

The discharge Q is given as; Q = ac× 2gH
Where ac = area at vena − contracta and is given by
a1
ac =
H − Hc
1+ a
H
Where a1 = area at outlet
p p
Ha = , Hc = c
ρg ρg
p is atmospheric pressure, pc is pressure at vena − contracta.

GATE/ME/SLP/Module_3/Ch.1_Notes /Pg.61
Vidyalankar : GATE – ME

Notch:
A notch is a device used for measuring the rate of flow of a liquid through a small
channel or a tank. It may be defined as an opening in the side of a tank or a small
channel in such a way that the liquid surface in the tank or channel is below the top
edge of the opening.

Weir:
A weir is concrete or masonry structure, placed in an open channel over which the
flow occurs. It is generally in the form of vertical wall, with a sharp edge at the top,
running all the way across the open channel. The notch is of small size while the weir
is of a bigger size. The notch is generally made of metallic plate while weir is made of
concrete or masonry structure.

1) Discharge over Rectangular sharp crested Notch or Weir:

The expression for discharge over a rectangular notch or weir is same.

Consider a rectangular notch or weir provided in a channel carrying water as shown in fig.
Let H = Head of water over the crest
L = Length of the notch or weir
2
Discharge, Q = CdL 2gH3 / 2
3

2) Discharge over a Triangular Notch or Weir:

Let H = head of water above the V− notch

θ = angle of notch

8 θ
Total discharge, Q = Cd tan 2gH5 / 2
15 2

3) Discharge Over a Trapezoidal Notch or Weir:

As shown in fig. a trapezoidal notch or weir is a combination of a 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 discharge through a triangular
notch or weir.

GATE/ME/SLP/Module_3/Ch.1_Notes /Pg.62
 Notes on Fluid Mechanics

Discharge is maximum when nd nd

1) Hydraulic mean depth = d/2 O
= 2 ( n2 + 1 − n)
L
2)
d d
3) Perpendiculars drawn from centre of top
width O on three sides are equal.

Let H = Height of water over the notch

L = Length of the crest of the notch
Cd1 = Coefficient of discharge for rectangular portion ABCD
Cd2 = Coefficient of discharge for triangular portion [FAD and BCE]
2 8 θ
Discharge Q = Cd1L 2gH3 / 2 + Cd2 tan 2gH5 / 2
3 15 2
* A triangular notch or weir is preferred to a rectangular notch or weir due to
following reasons
(1) The expression for discharge of right angled V − notch is very simple
(2) For measuring low discharge, a triangular notch gives more accurate results than
a rectangular notch.
(3) In case of triangular notch, only one reading, i.e. (H) is required for the
computation of discharge
(4) Ventilation of a triangular notch is not necessary.

Effect on Discharge Over a Notch or Weir due to Error in the Measurement of Head:
(1) For Rectangular Weir or Notch:
dQ 3 dH
=
Q 2 H
Above equation shows that an error of 1% in measuring H will produce 1.5% error in
discharge over a rectangular notch or weir.
(2) For Triangular Weir or Notch:
dQ 5 dH
=
Q 2 H
Above equation shows that an error of 1% in measuring H will produce 2.5% error in
discharge over a triangular notch or weir.

Time Required to Empty a Reservoir or a Tank with a Rectangular Weir or Notch:

3A ⎡ 1 1 ⎤
T = ⎢ − ⎥
Cd .L. 2g ⎣⎢ H2 H1 ⎦⎥
Where
L = Length of crest of the 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
A = Cross − sectional area

GATE/ME/SLP/Module_3/Ch.1_Notes /Pg.63
Vidyalankar : GATE – ME

Time Required to Empty a Reservoir or a Tank With a Triangular weir or Notch:

5A ⎡ 1 1 ⎤
T = ⎢ − 3/2 ⎥
θ 3/2
4Cd tan 2g ⎣ H2 H1 ⎦
2
Where
θ = Angle of the 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
A = Cross − sectional area

Velocity of Approach:
Velocity of approach is defined as velocity with which the water approaches or reaches
the weir or notch before it flows over it. Thus if Va is the velocity of approach, then an
V2
additional head ha equal to a due to velocity of approach, is acting on the water flowing
2g
over the notch. Then initial height of water over the notch becomes (H+ha) and final
height becomes equal to ha.
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 the cross− sectional area
of the channel on the upstream side of the weir or notch, the velocity of approach is
obtained. Mathematically.
Q
Va =
Area of Channel
Example:
Discharge over a rectangular weir, with velocity of approach
2
Cd ,L 2g ⎡(H1 + ha ) − ha 3 / 2 ⎤
3/2
=
3 ⎣ ⎦
Empirical Formulae for Discharge over a Rectangular Weir:
1) Francis’s Formula:
Q = 1.84 [L − 0.2H] H3/ 2
If end − contractions are suppressed then
Q = 1.84 LH3/2
If velocity of approach is considered then
Q = 1.84 L ⎡(H + ha ) − ha 3/ 2 ⎤
3/2
⎣ ⎦
2) Bazin’s Formula:
2 0.003
Q = mL 2gH3 / 2 Where m = Cd = .405 +
3 H
H = height of water over the weir
If velocity of approach is considered then
0.003
2g ⎡(H + ha ) ⎤
3/2
Q = m1 L Where m1 = 0.405 +
⎣ ⎦ (H + ha )

GATE/ME/SLP/Module_3/Ch.1_Notes /Pg.64
 Notes on Fluid Mechanics

Cipolletti Weir or Notch :

This weir is a trapezoidal weir, which has side slopes of 1 horizontal to 4 vertical at
shown in fig.
Thus in ΔABC
θ AB H / 4 1
tan = = =
2 BC H 4
θ ⎛ 1 ⎞
⇒ = tan−1 ⎜ ⎟ = 14°2′
2 ⎝4⎠
Discharge through cipolletti weir is
2
Q = CdL 2g H3 / 2
3
If velocity of approach, Va is to be taken into consideration:
2
Q = CdL 2g ⎡(H + ha ) − ha 3/ 2 ⎤
3/ 2

3 ⎣ ⎦

Discharge Over Broad − Crested Weir:

A weir having a wide crest is known as broad crested weir
Let H = height of water, above the crest
L = Length of the crest

If 2L > H, the weir is called broad crested weir.

If 2L < H, the weir is called a narrow crested weir
Let h = head of water at the middle of weir which is constant
V = Velocity of flow over the weir
The discharge over weir,
Q = Cd L (
2g Hh2 − h3 )
2
For maximum discharge , h = H and , Qmax = 1.705 CdLH3/2
3
Discharge Over a Narrow − Crested Weir:
2
Q = CdL 2gH3 / 2
3

GATE/ME/SLP/Module_3/Ch.1_Notes /Pg.65
Vidyalankar : GATE – ME

Discharge Over an Ogee Weir:

Fig shows an ogee weir, in which the crest of the weir rises up to a maximum height of
0.11 × H (where H is the height of water above the inlet of weir) and then falls as shown
in fig.

The discharge for an ogee weir is the same as that of a rectangular weir, and is given by
2
Q = CdL 2gH3 / 2
3
Discharge Over Submerged or Drowned Weir:
When the water level on the downstream side of a weir is above the crest of weir, then
the weir is called to be submerged or drowned weir. Fig. shows a submerged weir. The
total discharge, over the weir is obtained by dividing the weir into two parts. The portion
between upstream and down stream water surface may be treated as free weir and
portion between down stream water surface and crest of weir as drowned weir.

Let H = height of water on the upstream of the weir

h = height of water on the downstream side of weir

2 2
Cd 1L 2g [H − h] + CdLh 2g (H − h )
3/2
Discharge, Q =
3 3
Energy Equation :
The general equation for conservation of energy for an incompressible fluid can be
written as
⎛ p1 V12 ⎞ ⎛p V2 ⎞
⎜ + + Z1 ⎟ + qw + W = ⎜ 2 + 2 + Z2 ⎟ + (U2 − U1 )
⎝ ρg 2g ⎠ ⎝ ρg 2g ⎠

GATE/ME/SLP/Module_3/Ch.1_Notes /Pg.66
 Notes on Fluid Mechanics

Where,
qw = Heat added per unit weight of fluid flow
U1, U2 = Internal energy per unit weight of fluid at the respective states and
W = External work done (i.e.) shaft work added per unit weight of fluid.

SOLVED EXAMPLES
Example:
A 25cm diameter pipe carries oil of sp. gr. 0.9 at a velocity of 3m/s. At another section the
diameter is 20cm. Find the velocity at this section and also mass rate of flow of oil.
Solution:

At section 1, D1 = 25cm = 0.25m

π π
A1 = D2 = × ( 0.25 ) = 0.049m2
2

4 4
V1 = 3m / s
At section 2, D2 = 20cm = 0.2m
π
( 0.2 ) = 0.0314m2
2
A2 =
4
Applying continuity equation at section 1 and 2
A1 V1 = A2 V2
⇒ 0.049 × 3.0 = 0.0314 × V2
⇒ V2 = 4.68m/s

Mass flow rate of oil = sp.gr.of oil × Density of water × A1V1

= 0.9 × 1000 × 0.049 ×3.0
= 132.23 kg/s
Example:
The following case represents the two velocity components such that they satisfy the
continuity equation:
u = x2 + y2 + z2 ; v = xy2 − yz2 + x ŷ
Solution:
The continuity equation for incompressible fluid is given by
∂u ∂v ∂w
+ + =0
∂x ∂y ∂z
u = x 2 + y2 + z 2
∂u
⇒ = 2x
∂x
v = xy2 − yz2 + xy
∂v
⇒ = 2xy − z2 + x
∂y
∂u ∂v
Substituting the values of and in continuity equation
∂x ∂y
∂w
2x + 2xy − z2 + z + =0
∂z

GATE/ME/SLP/Module_3/Ch.1_Notes /Pg.67
Vidyalankar : GATE – ME

∂w
⇒ = −3x − 2xy + z2
∂z
⇒ ∂ w = (−3x − 2xy + z2)dz
z3
⇒ w = (−3xz − 2xyz + ) + Constant of integration
3
Where constant of integration cannot be a function of z. But it can be a function of x
and y i.e. f (x, y)
z3
⇒ w = (−3xz − 2xyz + ) + f (x,y)
3
Example:
The velocity potential function (φ) is given by an expression
− xy 3 x3 y
φ= − x2 + + y2
3 3
(i) Find the velocity components in x and y direction
(ii) Show that φ represents a possible case of flow
Solution:
− xy 3 x3 y
φ = − x2 + + y2
3 3
∂φ − y 3 3x 2 y
= − 2x +
∂x 3 3
∂φ −3xy 2 x 3
= + + 2y
∂y 3 3
−∂φ y 3
(i) u = = + 2x − x 2 y
∂x 3

−∂φ x3
v = = xy 2 − − 2y
∂y 3
(ii) The given value of φ, will represent a possible case of flow if it satisfies the
∂2φ ∂2φ
Laplace equation i.e. + =0
∂x 2 ∂y 2
Now
∂φ − y 3
= − 2x + x 2 y
∂x 3
2
∂ φ
⇒ = −2 + 2xy
∂x 2

∂φ x3
and = − xy 2 + + 2y
∂y 3

∂2φ
⇒ = −2xy + 2
∂y 2

GATE/ME/SLP/Module_3/Ch.1_Notes /Pg.68
 Notes on Fluid Mechanics

∂2φ ∂2φ
⇒ + = ( −2 + 2xy ) + ( −2xy + 2 ) = 0
∂x 2 ∂y 2

Hence Laplace equation is satisfied and hence φ represents a possible case of

flow.
Example
If, for a two dimensional potential flow, the velocity potential is given by φ = (2y − 1 )x,
determine the velocity at the point P (4, 5). Determine also the value of stream function at
the point P.
Solution :

φ = x (2y − 1 )
∂φ
(i) u = − = −2y + 1
∂x
∂φ
v =− = −2x
∂y
At P (4, 5)
u = 1 − 2×5 = −9
v = − 2 × 4 = −8
Velocity at P = −9 î −8 ĵ
Resultant Velocity at P = 92 + 82 = 12.04

∂Ψ
(ii) = −u = 2y − 1 ....(1)
∂y
∂Ψ
= − v = 2x ….(2)
∂x
Integrating equation (1) w.r.t.y we get;
∂y 2
Ψ =2 −y+k ….(3)
2
Differentiating above equation w.r.t. x we get
∂Ψ ∂k
= {k may be a functionof x}
∂x ∂x
∂Ψ
But = 2x
∂x
∂k
⇒ = 2x
∂x
∫ 2xdx = x
2
⇒ k=
Hence Ψ = y2 − y + x2
At P (4,5) ; Ψ = 52 − 5 + 42 = 36 units

GATE/ME/SLP/Module_3/Ch.1_Notes /Pg.69
Vidyalankar : GATE – ME

Example:
A pipe, through which water is flowing, is having diameters, 20cm and 10cm at the cross
− sections 1 and 2 respectively. The velocity of water at section 1 is given 4.0m/s. Find
the velocity head at sections 1 and 2 and also rate of discharge.
Solution:
D1 = 20cm = 0.2m
π π
⇒ Area A1 = D12 = ( 0.2 ) = 0.0.314m2
2

4 4
V1 = 4.0m / s
D2 = 0.1m
π
A2 = ( 0.1) = 0.00785m2
2

4
(i) Velocity head at section 1
( 4.0 )
2
V12
= = = 0.815m
2g 2 × 9.81
V22
(ii) Velocity head at section 2 =
2g
From equation of continuity
A1V1 = A2 V2
AV 0.0314 × 4.0
⇒ V2 = 1 1 = = 16.0m / s
A2 0.00785
(16.0 )
2

⇒ Velocity head at section 2 = = 83.047m

2 × 9.81

(iii) Rate of discharge = A1 V1 = A2 V2

= 0.0314 × 4.0 = 0.1256 m3/s
Example:
The water is flowing through a pipe having diameters 20cm and 10cm at sections 1 and 2
respectively. The rate of flow through pipe is 35litres/s. The section 1 is 6m above datum
and section 2 is 4m above datum. If the pressure at section 1 is 39.24 N /cm2, find the
intensity of pressure at section 2.

Solution:
At section 1, D1 = 20cm = 0.2m
π
A1 = ( 0.2 ) = 0.0314m2
2

GATE/ME/SLP/Module_3/Ch.1_Notes /Pg.70
 Notes on Fluid Mechanics

p1 = 39.24 N/cm2
= 39.24 × 104 N/m2
Z1 = 6.0m
At section 2, D2 = 0.10m
π
A2 = ( 0.1) = 0.00785m2
2

4
Z2 = 4m
Rate of flow, Q = 35 lit/s = 0.035m3 /s
Q = A1 V1 = A2 V2
Q 0.035
⇒ V1 = = = 1.114m / s
A1 0.00314
Q 0.035
V2 = = = 4.456m / s
A 2 0.00785
Applying Bernoulli’s equation at sections 1 and 2, we get
p1 V12 p V2
+ + Z1 = 2 + 2 + Z2 …. neglecting losses
ρg 2g ρg 2g
39.24 × 104 (1.114 ) ( 4.456 )
2 2
p2
⇒ + + 6.0 = + + 4.0
1000 × 9.81 2 × 9.81 1000 × 9.81 2 × 9.81
⇒ p2 = 41.05 × 9810N/m2
= 40.27N/cm2
Example:
A pipe of diameter 400mm carries water at a velocity of 25m/s. The pressures at the
points A and B are given as 29.43 N/cm2 and 2.563N/cm2 respectively while the datum
heads at A and B are 28 m and 30m. Find the loss of head between A and B.

Solution:
Dia of pipe, D = 400mm = 0.4m
Velocity, V= 25m /s
At point A, pA = 29.43 N /cm2 = 29.43 ×104 N/m2
ZA = 28m
VA = V = 25m/s

GATE/ME/SLP/Module_3/Ch.1_Notes /Pg.71
Vidyalankar : GATE – ME

⇒Total head at A
p V2
HA = A + A + Z A
ρg 2g
29.43 × 104 252
= + + 28 = 89.85m
1000 × 9.81 9.81
At point B, pB = 22.563 N/cm2 = 22.563 × 104 N/m2
ZB = 30m
VB = V = VA= 25m /s
⇒ Total Head at B
p V2
HB = B + B + ZB
ρg 2g
22.563 × 104 252
= + + 30 = 84.85m
1000 × 9.81 2 × 9.81
⇒ Loss of Head = HA − HB = 89.85−84.85 = 5.0m
Example:
A horizontal venturimeter with inlet and throat diameters 30cm and 15cm respectively is
used to measure the flow of water. The reading of differential manometer connected to
the inlet and the throat is 20cm of mercury. Determine the rate of flow. Take Cd = 0.98
Solution:
Dia. Of inlet, d1 = 30cm
π π
⇒ Area at inlet, a1 = d12 = ( 30 ) = 706.85cm2
2

4 4
Dia. at throat, d2 = 15cm
π
⇒ a2 = × 152 = 176.7cm2
4
Cd = 0.98
Reading of differential manometer = x = 20 cm of mercury
⇒ difference of pressure head is given by
⎡S ⎤
h = x ⎢ h − 1⎥
⎣ So ⎦
Where Sh = Sp. Gravity of mercury = 13.6 , So = Sp. Gravity of water = 1
⎡ 13.6 ⎤
⇒ h = 20 ⎢ − 1⎥ = 252.0cm of water
⎣ 1 ⎦
a1a 2
Now, Q = Cd × 2gh
a12 − a 22
706.85 × 176.7
= 0.98 × × 2 × 981× 252
( 706.85 ) − (176.7 )
2 2

= 125756cm3 /s = 125.756 lit /s

GATE/ME/SLP/Module_3/Ch.1_Notes /Pg.72
 Notes on Fluid Mechanics

Example:

Find the discharge of water flowing through a pipe 30cm

diameter placed in an inclined position where
venturimeter is inserted, having a throat diameter of
15cm. The difference of pressure between the main and
throat is measured by a liquid of sp. gr. 0.6 in an inverted
U− tube which gives a reading of 30cm. The loss of head
between the main and throat is 0.2 times the kinetic
head of pipe.
Solution:

Dia. At inlet; d1 = 30cm

π
a1 = ( 30 ) = 706.85cm2
2

4
Dia. At throat; d2 = 15cm
π
(15 ) = 176.7cm2
2
a2 =
4
Reading of differential manometer, x = 30cm
Difference of pressure head (for inclined venturimeter) is given by;
⎡ S ⎤ ⎛p ⎞ ⎛p ⎞
h = x ⎢1 −  ⎥ = ⎜ 1 + Z1 ⎟ − ⎜ 2 + Z2 ⎟
⎣ S o ⎦ ⎝ ρg ⎠ ⎝ ρg ⎠
Where S  = 0.6 and So = 1.0
⎡ 0.6 ⎤
= 30 ⎢1 − ⎥ = 12.0cm of water
⎣ 1.0 ⎦
Loss of Head, hL = 0.2 × kinetic had for pipe
0.2V12
=
2g
Applying Bernoull’s equation at sections (1) and (2) we get
p1 V12 p V2
+ + Z1 = 2 + 2 + Z2 + hL
ρg 2g ρg 2g

⎛p ⎞ ⎛p ⎞ V2 V2 V2
⇒ ⎜ 1 + Z1 ⎟ − ⎜ 2 + Z2 ⎟ + 1 − 2 = hL = 0.2 1
⎝ ρg ⎠ ⎝ ρg ⎠ 2g 2g 2g

V12 V22
⇒ 12.0 + 0.8 − =0
2g 2g
Now using continuity equation;
a1V1 = a2V2

GATE/ME/SLP/Module_3/Ch.1_Notes /Pg.73
Vidyalankar : GATE – ME

π
a 2 V2 4 ( ) 2 V2
2
15 V
⇒ V1 = = =
a1 π 4
( 30 )
2

4
2
0.8 ⎛ V2 ⎞ V22
⇒ 12.0 + − =0
2g ⎜⎝ 4 ⎟⎠ 2g
⇒ V2 = 157.4cm/s
⇒ Discharge = a2 V2 = 176.7 × 157.4 = 2780cm3 /s = 27.8 lit/s

Example:
An orifice meter with orifice diameter 15cm is inserted in a pipe of 30 cm diameter. The
pressure difference measured by a mercury oil differential manometer on the two sides of
the orifice meter gives a reading of 50cm of mercury. Find the rate of flow of oil of sp.gr.
0.9 when the coefficient of the meter = 0.64.
Solution:
Dia. of orifice, do = 15cm
π
ao = (15 ) = 176.7cm2
2
⇒ Area,
4
Dia. of pipe, d1 = 30cm
π
⇒ Area, a1 = ( 30 ) = 706.85cm2
2

4
sp.gr.of oil, So = 0.9
Reading of Diff. manometer, x = 50cm of mercury
⎡ Sg ⎤ ⎡ 13.6 ⎤
Differential head, h = x ⎢ − 1⎥ = 50 ⎢ − 1⎥
S
⎣ o ⎦ ⎣ 0.9 ⎦
= 705.5cm of oil
Cd= 0.64
a o a1
⇒ Q = Cd . × 2gh
a12 − a 20
176.7 × 706.85
= 0.64 × × 2 × 981× 705.5
( 706.85 ) − (176.7 )
2 2

= 137414.25 cm3/s = 137.414 litres/s

Example:
A Pitot − static tube placed in the centre of a 300mm pipe line has one orifice pointing
upstream and other perpendicular to it. The mean velocity in the pipe is 0.80 of the
central celocity. Find the discharge through the pipe if the pressure difference between
the two orifices is 60mm of water. Take the coefficient of pitot tube as Cv = 0.98
Solution :
Dia. Of pipe, d = 300mm = 0.30m
Diff. of pressure head, h = 60mm of water = 0.06 m of water
Cv = 0.98

GATE/ME/SLP/Module_3/Ch.1_Notes /Pg.74
 Notes on Fluid Mechanics

Mean velocity = V = 0.80 × central velocity

Central velocity = Cv 2gh = 0.98 × 2 × 9.81× 0.06 = 1.063m / s
⇒ V = 0.80 × 1.063 = 0.8504 m/s
Discharge, Q = A V
π π
= d2 × V = ( 0.30 ) × 0.8504 = 0.06m3 / s
2

4 4
Example:
A nozzle of diameter 20mm is fitted to a pipe of diameter 40mm. Find the force exerted
by the nozzle on the water which is flowing through the pipe at the rate of 1.2 m3 /minute.

Solution:
Dia. of pipe, D1 = 40mm = 0.04m
π
A1 = ( 0.04 ) = 0.001256m2
2
⇒ Area,
4
Dia. of nozzle, D2, = 20 mm = 0.02m
π
⇒ Area, A2 = ( 0.02 ) = 0.000314m2
2

4
Discharge, Q = 1.2m3/min = 0.02m3/s
Applying continuity equation at sections (1) and (2) we get,
A1V1 = A2V2 = Q
Q 0.02
⇒ V1 = = = 15.92m / s
A1 0.001256
Q 0.02
and V2 = = = 63.69m / s
A 2 0.000314
Applying Bernoulli’s equation at sections (1) and (2) we get
(Neglecting losses)
p1 V12 p V2
+ + Z1 = 2 + 2 + Z2
ρg 2g ρg 2g
p
Now Z1 = Z2, 2 = atmospheric pressure = 0
ρg
p1 V12 V22
⇒ + =
ρg 2g 2g
p1 V22 − V12 ( 63.69 ) − (15.92 )
2 2

⇒ = = = 193.83m of water
ρg 2g 2 × 9.81

GATE/ME/SLP/Module_3/Ch.1_Notes /Pg.75
Vidyalankar : GATE – ME

⇒ p1 = 193.83 × 1000 × 9.81 = 1901472 N/m2

Let the force exerted by the nozzle on water = F
Net force in the direction of x = rate of change of momentum in the direction of x
⇒ p1 A1 − p2 A2 + Fx = ρQ (V2 − V1)
Where p2 = atmospheric pressure = 0 and ρ = 1000 (for water)
⇒ 1901472 ×0.001256 − 0 + Fx = 1000 × 0.02 (63.69 − 15.92)
⇒ Fx = −1472.09N
−ve sign indicates that the force exerted by nozzle on water is acting from right to left.

Example:
A nozzle is situated at a distance of 1m
above the ground level and is inclined at an
angle of 45° to horizontal. The diameter of
the nozzle is 50mm and the jet of water from
the nozzle strikes the ground at a horizontal
distance of 4 m. Find the rate of flow of
water.
Solution :
Distance of nozzle above ground = 1m
Angle of inclination, θ = 45°
Dia. of nozzle, d = 50mm = 0.05m
π
⇒ Area, A = ( 0.05 ) = 0.001963m2
2

4
The horizontal distance x = 4m
The co−ordinates of the point B, which is on the centerline of the jet of water and is
situated on the ground, with respect A (origin) are
x = 4m and y = −1.0m { From A, point B is vertically down by 1m}

The equation of jet is given by

gx 2
y = x tan θ − − 2 sec 2 θ
2U
9.81× 42
⇒ −1.0 = 4 tan 45° − × sec 2 45°
2U2
⇒ U = 5.60m/s
Now, rate of flow of fluid = A × U = 0.001963 × 5.6 = 0.011m3/s
Example:
The head of water over the centre of an orifice of diameter 20mm is 1m. The actual
discharge through the orifice is 0.85lit/s. Find the coefficient for discharge.
Solution:
Diameter of orifice, d = 20mm = 0.02m
π
⇒ Area, a = ( 0.02 ) = 0.000314m2
2

4
Head, H = 1m
Actual discharge, Q = 0.85liters /s= 0.00085m3/s

GATE/ME/SLP/Module_3/Ch.1_Notes /Pg.76
 Notes on Fluid Mechanics

Theoretical Velocity, Vth = 2gH = 2 × 9.8 × 1 = 4.429m / s

⇒ Theoretical discharge, Q th = Vth × Area of orifice
= 4.429 × 0.000314 = 0.0939 m3 /s
ActualDischarge
⇒ Coefficient of discharge =
TheoriticalDischarge
0.00085
= = 0.61
0.00139
Example:
The head of water over an orifice of diameter 100mm is 10m. The water coming out from
orifice is collected in a circular tank of diameter 1.5m. The rise of water level in this tank
is 1.0m in 25 seconds. Also, the co−ordinates of a point on the jet, measured from vena −
contracta are 4.3m horizontal and 0.5m vertical. Find Cd ,Cv and Cc.
Solution :
Head, H = 10m
Dia. of orifice, d = 100mm = 0.1m
π
⇒ Area, a = ( 0.1) = 0.007853m2
2

4
Dia. of measuring tank, D = 1.5m
π
⇒ Area, A = (1.5 ) = 1.767m2
2

4
Rise of water, h = 1m
In time, t = 25 seconds
Horizontal distance, x = 4.3m
Vertical distance, y = 0.5m
Now theoretical velocity, Vth = 2gH = 2 × 9.81× 10 = 14.0m / s

⇒ Theoretical discharge, Qth = Vth × Area of orifice

= 14.0 × 0.007854 = 0.1099m3 /s
A × h 1.767 × 1.0
Actual discharge, Q = = = 0.07068m3 / s
t 25
Q 0.07068
⇒ Cd = = = 0.643
Q th 0.1099
x 4.3
Cv = = = 0.96
4yH 4 × 0.5 × 10
Cd 0.643
Cc = = = 0.669
Cv 0.9
Example:
A rectangular orifice 0.9m wide and 1.2m deep is discharging water from a vessel. The
top edge of the orifice is 0.6m below the water surface in the vessel. Calculate the
discharge through the orifice if Cd = 0.6 and percentage error if the orifice is treated as a
small orifice.

GATE/ME/SLP/Module_3/Ch.1_Notes /Pg.77
Vidyalankar : GATE – ME

Solution:
Width of orifice, b = 0.9m
Depth of orifice, d = 1.2m
H1 = 0.6m
H2 = H1 +d =0.6 + 1.2 = 1.8m
Cd = 0.6
2
Q = Cdb 2g ⎡⎣H23 / 2 − H13 / 2 ⎤⎦
3
2
= × 0.6 × 0.9 × 2 × 9.81 ⎡⎣1.83 / 2 − 0.63 / 2 ⎤⎦ = 3.1097 m3 /s
3
Discharge for a small orifice
Q1 = Cd a 2gh

d 1.2
Where h = H1 + = 0.6 + = 1.2m and a = b× d = 0.9 ×1.2
2 2

Q1 = 0.6 × 0.9 × 1.2 2 × 9.81× 1.2 = 3.1442m3 / s

Q1 − Q 3.1442 − 3.1097
% error = = = 0.01109 = 1.109%
Q 3.1097
Example:
Find the discharge through a totally drowned orifice 2.0 m wide and 1 m deep, if the
difference of water levels on both the sides of the orifice be 3m. Take Cd= 0.62 .
Solution:
b = 2.0m
d = 1.0m
Difference of water level on both the sides
H = 3m
Cd = 0.62
Q = Cd a 2gH = Cd × b × d × 2gH
= 0.62 × 2 × 1.0 × 2 × 9.81× 3 = 9.513 m3 /s
Example:
A convergent divergent mouthpiece having diameter of 4.0cm is discharging water under
a constant head of 2.0m. Determine the maximum outer diameter for maximum
discharge. Find maximum discharge also. Take Ha = 10.3 m of water and Hsep = 2.5 m of
water (absolute)
Solution:
Dia. of throat, dc = 4.0cm
π
⇒ Area, ac = ( 4 ) = 12.566cm2
2

4
Constant head, H = 2.0 m = 200 cm

GATE/ME/SLP/Module_3/Ch.1_Notes /Pg.78
 Notes on Fluid Mechanics

Find max. dia at outlet, d1 and Qmax

Ha = 10.3m
Hset = 2.5m (absolute)
The discharge, Q in convergent − divergent mouthpiece depends on the area of throat.
⇒ Qmax = ac × 2gH = 12.566 × 2 × 981× 200
3
= 7871.5cm /s
a1 H − Hc 10.3 − 2.5
now
ac
= 1+ a
H
= 1+
2.0
{Hc = Hsep = 2.5} = 2.2135
π 2
d1 ⎛d ⎞
2

now 4 = 2.2135 ⇒ ⎜ 1 ⎟ = 2.2135

π 2 ⎝ dc ⎠
d
4 c
d1
⇒ = 1.4877
dc
⇒ d1 = 1.4877 × dc =1.4877 × 4.0 = 5.95cm
Example:
The head of water over a rectangular notch is 900mm. The discharge is 300 liters/s. Find
the length of the notch when Cd = 0.62.
Solution :
Head over notch, H = 90cm = 0.9m
Discharge, Q = 300 lit/s = 0.3m3/s
Cd = 0.62
Let length of notch = L
2
Q = CdL 2gH3 / 2
3
2
⇒ 0.3 = × 0.62 × L × 2 × 9.81 × ( 0.9 )
3/2

3
⇒ L = 0.192m = 192mm
Example:
A rectangular notch 40cm long is used for measuring a discharge of 30 liters per second.
An error of 1.5mm was made, while measuring the head over the notch. Calculate the
percentage error in the discharge. Take Cd = 0.60.
Solution:
L = 40cm Q = 30 lit/s = 3000 cm3 /s
dH = 1.5 mm = 0.15cm Cd = 0.60
Let height of water over rectangular notch = h
2
Now Q = Cd L 2gH3 / 2
3
2
⇒3000= × 0.60 × 40 × 2 × 981 × H3 / 2
3
⇒H = 12.16cm

GATE/ME/SLP/Module_3/Ch.1_Notes /Pg.79
Vidyalankar : GATE – ME

dQ 3 dH 3 0.15
= = × = 0.0185 = 1.85%
Q 2 H 2 12.16

Example:
Water is flowing in a rectangular channel of 1m wide and 0.75m deep. Find the discharge
over a rectangular weir of crest length 60cm if the head of water over the crest of weir is
20cm and water from channel flows over the weir. Take Cd = 0.62. Take velocity of
approach into consideration.
Solution:
A = 1.0 × 0.75 = 0.75m2 L = 60cm = 0.6m2
H1 = 20cm = 0.2m Cd = 0.62
2
Q = CdL 2gH13 / 2
3
2
× 0.62 × 0.6 × 2 × 9.81 ( 0.2 ) m3 / s = 0.0982m3 / s
3/ 2
=
3
Q 0.0982
Velocity of approach, Va = = = 0.1309m / s
A 0.75

Va2 ( 0.1309 )
2

⇒ Additional head, ha = = = 0.0008733m

2g 2 × 9.81

2
CdL 2g ⎡(H1 + ha ) − ha 3 / 2 ⎤
3/2
Q =
3 ⎣ ⎦

2
× 0.62 × 0.6 × 2 × 9.81 ⎡( 0.2 + 0.00087 ) − ( 0.00087 ) ⎤ = 0.09881m3/s
3/ 2 3/ 2
=
3 ⎣ ⎦

VISCOUS FLOW
This topic deals with the flow of fluids which are viscous and flowing at very low velocity.
At low velocity the fluid moves in layers. Each layer of fluid slides over the adjacent layer.
du
Due to relative velocity between two layers, the velocity gradient exists and hence a
dy
shear stress
du
τ = μ acts onthelayers.
dy

Flow of Viscous Fluid Through Circular Pipe:

For the flow of viscous fluid through circular pipe, the velocity distribution across a
section, the ratio of maximum velocity to average velocity, the shear stress distribution
and drop of pressure for a given length is to be determined. The flow through the circular
pipe will be viscous or laminar, if the Reynolds number (Re) is less than 2000. The
expression for Reynold number is given by

GATE/ME/SLP/Module_3/Ch.1_Notes /Pg.80
 Notes on Fluid Mechanics

ρVD
Re =
μ
Where
ρ = Density of fluid flowing through pipe
V = Average velocity of fluid.
D = Diameter of pipe and
μ = Viscosity of fluid

Consider a horizontal pipe of radius R. The viscous fluid is flowing from left to right in the
pipe as shown in fig. (a). Consider a fluid element of radius r, sliding in a cylindrical fluid
element of radius (r +dr). Let the length of fluid element be Δx. If ‘p’ is the intensity of
⎛ ∂p ⎞
pressure, on face CD will be ⎜ p + Δx
⎝ ∂x ⎟⎠

∂p
The shear stress τ across a section varies with ‘r’ as across a section is constant.
∂x
Hence shear stress distribution across a section is linear as shown in fig. below

The Reynold’s number for different velocity distribution is as shown below.

Curve A represents Re = 1000
Curve B represents Re = 4000 D C
A
Curve C represents Re = 6000 B
Curve D represents Re = 10,000

GATE/ME/SLP/Module_3/Ch.1_Notes /Pg.81
Vidyalankar : GATE – ME

(i) Velocity Distribution:

−1 ∂p 2 2
u = ⎡R − r ⎤⎦
4μ ∂x ⎣
Above equation shows that the velocity distribution across the section of a pipe is
parabolic.

(ii) Ratio of Maximum Velocity to Average Velocity:

Umax
= 2.0 (Umax at r = 0 )
U
−1 ∂p 2
Umax = R
4μ ∂x

1 ⎛ −∂p ⎞ 2 ⎧ π ⎛ −∂p ⎞ 4 Q Q ⎫
U = R ⎨Q = R ;U = = ⎬
8μ ⎜⎝ ∂x ⎟⎠ ⎩
⎜
8μ ⎝ ∂x ⎠⎟ A πR2 ⎭

(iii) Drop of pressure for a given length of pipe:

32μUL
p1 − p2 = (HagenPoiseuille formula)
ρgD2

Flow of Viscous Fluid Between Two Parallel Plates:

Consider two parallel plates fixed and kept at a distance ‘t’ apart as shown in fig. A
viscous fluid is flowing between these two plates from left to right. Consider a fluid
element of length Δ x and thickness Δy at distance y from the lower fixed plate. If p is the
intensity of pressure on the face AB of the fluid element then intensity of pressure on the
⎛ ∂p ⎞
face CD will be ⎜ p + Δx . Let τ is the shear stress acting on the face BC then the shear
⎝ ∂x ⎟⎠
⎛ ∂τ ⎞
stress on the face AD will be ⎜ τ + Δy ⎟ . The width of the element in direction
⎝ ∂y ⎠
perpendicular to the paper is assumed as unity.

GATE/ME/SLP/Module_3/Ch.1_Notes /Pg.82
 Notes on Fluid Mechanics

(i) Velocity Distribution :

−1 ∂p
u = ⎡ ty − y 2 ⎤⎦
2μ ∂x ⎣
Above equation is the equation
of parabola. This distribution is
shown in Fig (a) below.
Velocity distribution and shear
stress distribution across a
section of parallel plates
(ii) Ratio of Maximum Velocity to Average Velocity:
−1 ∂p 2
Umax = t (Umax at y = t/2 i.e. mid−way between the plates)
8μ ∂x

−1 ∂p 2 ⎧ −1 ∂p 3 Q Q ⎫
U = t ⎨Q = t ;U = = ⎬
12μ ∂x ⎩ 12μ ∂x A t ×1 ⎭

Umax 3
=
U 2
(iii) Drop of pressure head for a given length :
12μUL
p1 − p2 =
t2
p − p2 12μUL
pressure drop hf = 1 =
ρg ρgt 2
(iv) Shear stress Distribution:
1 ∂p
τ =− [ t − 2y ]
2 ∂x
τ varies linearly with y, as shown in fig (b)
−1 ∂p
max shear stress. τo = t (at y = 0 or t)
2 ∂x
Kinetic Energy correction Factor:
It is defined as the ratio of the kinetic energy of the flow per second based on actual
velocity across a section to the kinetic energy of the flow per second based on average
velocity across the same section. It is denoted by α.
K.E / s based on actual veloicty
α =
K.E / sbased onaverage velocity

Momentum Correction Factor:

It is defined as the ratio of momentum of the flow per second based on actual velocity to
the momentum of the flow per second based on average velocity across a section. It is
denoted by β.
Momentum / s based on actual veloicty
β =
Momentum / sbased onaverage velocity

GATE/ME/SLP/Module_3/Ch.1_Notes /Pg.83
Vidyalankar : GATE – ME

* Momentum correction factor and energy correction factor for laminar flow through
4
flow a circular pipe are and 2.0 respectively.
3
Viscous Resistance on a shaft Rotating between Bearings:

μπ2D2NL
Viscous resistance =
60t
Where
μ = Coefficient of dynamic viscosity
D = Diameter of Shaft
N = speed of shaft in r.p.m.
t = thickness of oil film
L = length of oil film
Torque Required to over come the Frictional Resistance:
μπ2D3NL
=T=
120t
Power absorbed in overcoming frictional Resistance
μπ3D3N2L
=P= Watts
3600t
SOLVED EXAMPLES
Example:
A crude oil of viscosity 0.97 poise and relative density 0.9 is flowing through a horizontal
circular pipe of diameter 100mm and of length 10m. Calculate the difference of pressure
at the two ends of the pipe, if 100kg of the oil is collected in a tank in 30seconds.
Solution:
μ = 0.97 poise = 0.097 Ns/m2
Density of oil, ρ = 0.9 × 1000 = 900kg/m2
Dia of pipe, D = 100mm = 0.1m
L = 10m
Mass of oil collected, M = 100kg.
In time t = 30s
32μUL
p1 − p2 = N/m2
D2
100
mass of oil/s = kg / s
30
= ρ × Q = 900Q
100 1
⇒ Q = × = 0.0037m3 / s
30 900
Q 0.0037 0.0037
U = = = = 0.471m / s
A π 2 π
( 0.1)
2
D
4 4
32μUL 32 × 0.097 × 0.47 × 10
p1 − p2 = = = 1462.28N/m2
D2 ( 0.1)
2

GATE/ME/SLP/Module_3/Ch.1_Notes /Pg.84
 Notes on Fluid Mechanics

Example:
Determine the pressure gradient and the shear stress at the two horizontal parallel plates
and the discharge per meter width for the laminar flow of oil with a maximum velocity of
2m/s between two horizontal parallel fixed plates which are 100mm apart. Given μ =
2.4525 Ns/m2
Solution:
Umax = 2m/s
t = 100mm = 0.1m
μ = 2.4525 Ns/m2
Now,
−1 ∂P 2
Umax = t
8μ ∂x
1 ∂p
× ( 0.1)
2
⇒ 2.0 = − ×
8 × 2.4525 ∂x
∂p
⇒ = −3924N / m2per m
∂x
−1 ∂p −1
τo = t = ( −3924 ) × 0.1 = 196.2N / m2
2 ∂x 2
Q = Mean velocity × Area
2 2
= Umax × ( t × 1) = × 2.0 × 0.1× 1 = 0.133m2 / sper m {width is taken as unity}
3 3

Example:
Water at 15°C flows between two large parallel plates at a distance of 1.6 mm apart.
Determine the maximum velocity and the pressure drop per unit length and the shear
stress at the walls of the plates if the average velocity is 0.2 m/s. The viscosity of water at
15°C is given by as 0.01 poise.
Solution:
t = 1.6 mm = 0.0016m

U = 0.2 m/s
μ = 0.01 poise = 0.001 Ns/m2
3
Umax = U = 1.5 × 0.2 = 0.3m / s
2
12μU
p1 − p2 =
t2
12μU ∂p
pressure drop per unit length = =
t 2
∂x
∂p 12 × 0.01 0.2
⇒ = × = 937.44N / m2per m
∂x ( )
2
10 0.0016

GATE/ME/SLP/Module_3/Ch.1_Notes /Pg.85
Vidyalankar : GATE – ME

Maximum shear stress,

−1 ∂p 1
τo = t = × 937.44 × 0.0016 = 0.749N / m2
2 ∂x 2
Example:
A shaft having a diameter of 50mm diameter rotates centrally in a journal bearing having
a diameter of 50.15 mm and length 100mm. The angular space between the shaft and
the bearing is filled with oil having viscosity of 0.9 poise. Determine the power absorbed
in the bearing when the speed of rotation is 60 r.p.m
Solution:

Dia. of shaft, D = 50mm = 0.05m

Dia. of bearing, D1 = 50.15 mm = 0.05015m
L = 100mm = 0.1m
μ = 0.9 poise = 0.09 Ns/m2
N = 600 r.p.m.
D − D 50.15 − 50
Thickness of oil film, t = 1 = = 0.075mm = 0.075 × 10−3 m
2 2
πDN π × 0.05 × 600
Tangential speed of shaft, V = = = 0.5 × π m / s
60 60
du V 0.5 × π
Shear stress, τ = μ = μ = 0.09 × = 1883.52N / m2
dy t 0.075 × 10−3
Shear Force, F = τ × A
= 1883.52 × πDL
= 1883.52 × π × 0.05 × 0.1
= 29.586N
D 0.05
Resisting Torque; T=F = 29.586 × = 0.7378Nm
2 2

2πNT 2 × 3.14 × 600 × 0.7387

Power, P = = = 46.41W
60 60

BOUNDARY LAYER FLOW

Introduction :
When a real fluid flows past a solid body or a solid wall, the fluid particles adhere to the
boundary and 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
of fluid at the boundary will be zero. Farther away from the boundary , the velocity will be
du
higher and as a result of this variation of velocity the velocity gradient will exist. The
dy

GATE/ME/SLP/Module_3/Ch.1_Notes /Pg.86
 Notes on Fluid Mechanics

velocity of fluid increases from zero velocity on the stationary boundary to free stream
velocity (U) of the fluid in the direction normal to 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 the solid boundary. This narrow region of the fluid is called
boundary layer.
According to boundary layer theory, the flow of fluid in the neighborhood of the solid
boundary may be divided into two regions as shown in fig.

1) A very thin layer of the fluid, called the boundary layer, in the immediate
neighborhood 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 takes
du
place. In this region, the velocity gradient exists and hence the fluid exerts a
dy
shear stress on the wall in the direction of motion. The value of shear stress is given
du
by τ = μ
dy
2) The remaining fluid, which is outside the boundary layer. The velocity outside the
boundary layer is constant and equal to free stream velocity. As there is no variation
du
of velocity in this region, the velocity gradient becomes zero. As a result of this,
dy
the shear stress is zero.
Laminar Boundary Layer:
Consider the flow of fluid, having free stream velocity (U), over a smooth thin plate which
is flat and placed parallel to the direction for free stream of fluid as shown in fig. Let us
consider the flow with zero pressure gradient on one side of the plate, which is stationary.

Flow over a Plate

GATE/ME/SLP/Module_3/Ch.1_Notes /Pg.87
Vidyalankar : GATE – ME

The velocity of fluid on the surface of the plate should be equal to the velocity of the
plate. Hence velocity of fluid on the surface of the plate is zero. But at a distance away
from the plate the fluid is having a certain velocity. Thus a velocity gradient is set up in
the fluid near the surface of the plate. This velocity gradient develops shear resistance,
which retards the fluid. Thus the fluid with a uniform free stream velocity (U) is retarded in
the vicinity of the solid surface of the plate and the boundary layer region begins at sharp
leading edge.
At subsequent points downstream the leading edge,the boundary layer region increases
because the retarded fluid is further retarded. This is also referred to as growth of
boundary layer. Near the leading edge of the surface of the plate, where the thickness is
small the flow in the boundary layer is laminar though the main flow is turbulent. This
layer of fluid is said to be laminar boundary layer. This is shown by AE in Fig. The
thickness of the laminar boundary layer at a distance x from the leading edge of the plate
varies as x . The length of the plate from the leading edge, up to which laminar
boundary layer exists, is called laminar zone. This is shown by distance AB. The distance
of B from leading edge is obtained from Reynolds’s number the boundary layer is
Ux
laminar. The Reynolds number is given by (Re) x =
ν
Where:
x = Distance form the leading edge
U = Free stream velocity of fluid.
v = kinematics viscosity of fluid.
Hence for laminar boundary layer,
Ux
= 2 ×105
ν

Turbulent Boundary Layer:

If the length of the plate is more than the distance x, calculated from above equation, the
thickness of boundary layer will go on increasing in the down stream direction. Then the
laminar boundary layer becomes unstable and motion of fluid within it, is disturbed and
irregular which leads to a transition from laminar to turbulent boundary layer. This short
length over which the boundary layer flow changes from laminar to turbulent is called
transition zone. This is shown by distance BC in Fig. Further downstream the transition
zone the boundary layer is turbulent and continues to grow in thickness. This layer of
boundary is called turbulent boundary layer, which is shown by portion FG in Fig. The
thickness of the tabulant boundary layer at a distance x from the leading edge of the plate
varies as x 4 / 5 .

Laminar Sub− Layer:

This is the region in the turbulent boundary layer zone, adjacent to the solid surface of
the plate as shown in fig. In this zone, the velocity variation is influenced only by viscous
effects. Here velocity variation is linear and so the velocity gradient is constant.
Therefore, the shear stress in the laminar sub−layer would be constant and equal to the
boundary shear stress τo.
⎛ ∂u ⎞ u⎧ ∂u u ⎫
τo = μ ⎜ ⎟ = μ ⎨for linear var iation = ⎬
∂y
⎝ ⎠y = 0 y ⎩ ∂y y ⎭

GATE/ME/SLP/Module_3/Ch.1_Notes /Pg.88
 Notes on Fluid Mechanics

Boundary Layer Thickness (δ):

It is defined as the distance from the boundary of the solid body measured in y − direction
to the point, where the velocity of the fluid is approximately equal to 0.99 times the free
stream velocity (U) of the fluid. It is denoted by δ

Displacement 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 flow rate on
account of boundary layer formation. It is also defined as the distance, perpendicular to
the boundary, by which the free stream is displaced due to the formation of boundary
layer
δ⎛ u⎞
δ* = ∫ ⎜ 1 − ⎟ dy
o
⎝ U ⎠
Where u is the velocity at a distance y, perpendicular to boundary layer.
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 boundary layer formation.
δ u ⎡ u⎤
θ = ∫ ⎢1 − ⎥ dy
o U
⎣ U⎦
Energy Thickness (δ**):
It is 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 boundary layer formation.
δ u ⎡ u2 ⎤
δ** = ∫ ⎢1 − 2 ⎥ dy
o U
⎣ U ⎦
Drag Force on a Flat Plate due to Boundary Layer:
L
FD = ∫o τobdx
Where L = length of plate.
b = width of plate
⎛ du ⎞
τo is the boundary shear stress and equal to μ ⎜ ⎟
⎝ dy ⎠y = 0
and is given by “Von Karman” momentum integral equation as:
τo ∂θ
=
pU 2
∂x

Local co−efficient of Drag (CD*):

1
It is defined as the ratio of shear stress τo to the quantity ρU2
2
τo
CD* =
1 2
ρU
2

GATE/ME/SLP/Module_3/Ch.1_Notes /Pg.89
Vidyalankar : GATE – ME

Average Coefficient of Drag (CD):

1
It is defined as ratio of total drag force to the quantity ρAU2 . It is also called Drag
2
Coefficient.
FD
CD =
1 2
ρU A
2
Where
A = Area of the surface (or plate)
U = Free stream velocity
ρ = Mass density of fluid

Boundary Condition for Velocity Profiles:

du
(1) At y = 0, u = 0 and has some finite value
dy
(2) At y = δ, u = U
du
(3) At y = δ, =0
dy

Nominal Boundary Layer Thickness:

It is given by
5x
δ =
Rex
and for the maximum thickness of boundary layer,
5.48x
δmax =
Rex
* CD is inversely proportional to square root of Reynold Number.
μU
* τo ∝ Rex
x
2
u 3 ⎛ y ⎞ 1⎛ y ⎞
* For a profile given by = −
U 2 ⎜⎝ δ ⎟⎠ 2 ⎜⎝ δ ⎟⎠
4.64x
δ =
Re x
μU
τo = 0.323 Rex
x
1.92
CD =
ReL

Turbulent Boundary Layer on a Flat Plate:

The velocity profile for turbulent boundary layer is given by Blasius as
n
u ⎛y⎞
=
U ⎜⎝ δ ⎟⎠

GATE/ME/SLP/Module_3/Ch.1_Notes /Pg.90
 Notes on Fluid Mechanics

1
Where n = for Re < 107 but more than 5 × 105
7
1/ 4
⎛ u ⎞
And τo = 0.0225ρU2 ⎜ ⎟
⎝ ρδU ⎠
Analysis of Turbulent Boundary Layer:

For 5 × 10 5 < Re < 107

0.37x 0.072
δ = andCD =
(Rex ) (ReL )
1/ 5 1/ 5

Where x = Distance from the leading edge

Rex = Reynold number for length x
ReL = Reynold number at the end of plate
For 107 < Re <109
0.455
CD = (Schlichting Formula)
(log10 ReL )
2.58

Separation of Boundary Layer:

When a solid body is immersed in a flowing fluid, a thin layer of fluid called the boundary
layer is formed adjacent to the solid body. In this thin layer of fluid, the velocity varies
from zero to free stream velocity in the direction normal to the solid body. Along the
length of the solid body, the thickness of the boundary layer increases. The fluid layer
adjacent to the solid surface has to do work against surface friction at the expense of its
kinetic energy. This loss of kinetic energy is recovered from the immediate fluid layer in
contact with the layer adjacent to solid surface through momentum exchange process.
Thus the velocity of the layer goes on decreasing. Along the length of the solid body, at a
certain point a stage may come when the boundary layer may not be able to keep
sticking to the solid body if it cannot provide kinetic energy to overcome the resistance
offered by the solid body. In other words, the boundary layer will be separated form the
surface. This phenomenon is called boundary layer separation. The point on the body at
which the boundary layer is on the verge of separation from the surface is called point of
separation.
Location of separation point:
The separation point is determined from the condition,
⎛ ∂u ⎞
⎜ ⎟ =0
⎝ ∂y ⎠ y = 0
For a given velocity profile, it can be determined whether the boundary layer has
separated, or on the verge of separation or will not separate form the following
conditions:
⎛ ∂u ⎞
(1) If ⎜ ⎟ is −ve − − − − − − the flow has separated.
⎝ ∂y ⎠ y = 0

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Vidyalankar : GATE – ME

⎛ ∂u ⎞
(2) If ⎜ ⎟ = 0 − − − − − − the flow is on the verged of separation
⎝ ∂y ⎠ y = 0
⎛ ∂u ⎞ the flow will not separate or flow
(3) If ⎜ ⎟ is +ve − − − − − −
⎝ ∂y ⎠ y = 0 will remain attached with the surface.

SOLVED EXAMPLES
Example:
Find the displacement thickness, the momentum thickness and energy thickness for the
u y
velocity distribution in the boundary layer given by = where u is the velocity at a
U δ
distance y from the plate and u = U at y = δ where δ = boundary layer thickness. Also
calculate the value of δ*/θ
Solution:
δ ⎛ u⎞ δ ⎛ y⎞
Displacement thickness δ* = ∫o ⎜ 1 − U ⎟ dy =
⎝ ⎠
∫
o ⎜ 1 − δ ⎟ dy
⎝ ⎠
δ
⎡ y2 ⎤ δ
= ⎢y − ⎥ =
⎣ 2 δ ⎦o 2
δ u⎛ u⎞ δ y⎛ y⎞
Momentum thickness θ = ∫ ⎜ 1 − ⎟ dy = ∫ ⎜ 1 − ⎟ dy
o U
⎝ U⎠ o δ⎝ δ⎠
δ
⎛ y y2 ⎞
δ ⎡ y2 y3 ⎤ δ
= ∫o ⎜⎝ δ δ2 ⎟⎠ ⎢⎣ 2δ 3δ2 ⎥⎦ = 6
− dy = −
o

δ u⎡ u2 ⎤
Energy thickness δ** = ∫o U ⎢⎣1 − U2 ⎥⎦ dy
δ y ⎡ y2 ⎤
= ∫o ⎢1 − ⎥ dy
δ ⎣ δ2 ⎦
δ
⎛ y y3 ⎞
δ ⎡ y2 y4 ⎤ δ
∫o ⎜⎝ δ − δ3 ⎟⎠ dy =
= ⎢ − 3⎥ =
⎣ 2 δ 4δ ⎦o 4
δ* (δ / 2)
= = 3.0
θ (δ / 6)
Example:
For the velocity profile in laminar boundary layer as
3
u 3 ⎛ y ⎞ 1⎛ y ⎞
= −
U 2 ⎜⎝ δ ⎟⎠ 2 ⎜⎝ δ ⎟⎠
Find the thickness of the boundary layer and the shear stress 1.5m from the leading edge
of a plate. The plate is 2m long and 1.4 m wide and is placed in water which is moving
with a velocity of 200 mm per second. Find the total drag force on plate if μ for
water = 0.01 poise

GATE/ME/SLP/Module_3/Ch.1_Notes /Pg.92
 Notes on Fluid Mechanics

Solution:
Velocity profile is
2
u 3 ⎛ y ⎞ 1⎛ y ⎞
= −
U 2 ⎜⎝ δ ⎟⎠ 2 ⎜⎝ δ ⎟⎠
Distance of x from leading edge, x = 1.5m
Length of plate, L = 2m
Width of plate, b = 1.4m
Velocity of water, U = 200mm /s = 0.2m/s
0.01
Viscosity of water, μ = 0.01 poise = = 0.001Ns / m2
10
For the given velocity profile, thickness of boundary layer is
4.64x
δ =
Re x
ρUx 1000 × 0.2 × 1.5
Rex = = = 300000
μ 0.001
4.64 × 1.5
⇒ δ = = 0.0127m = 12.7mm
300000
μU
τo = 0.323 Rex
x
0.2
= 0.323 × 0.001 × × 300000 = 0.0235N / m2
1.5
L L
μU
FD = ∫ τobdx = ∫ 0.323
o o
x
Rex bdx

L L
μU ρUx ρU 1
= 0.323 ∫ bdx = 0.323μU × b∫ dx
o
x μ μ o x
L
⎡ ⎤
ρU ⎢ x1/ 2 ⎥
= 0.323 μU ×b⎢ ⎥
μ ⎢ 1 ⎥
⎣ 2 ⎦o
ρUL
= 0.646μU b
μ
1000 × 0.2 × 2.0
= 0.646 × 0.001 ×0.2 × × 1.4 = 0.1138N
0.001
⇒ Total drag force = Drag force on both sides of the plate
= 2 × 0.1138 = 0.2276N

Example:
Determine the thickness of boundary layer at the trailing edge of smooth plate of length
4m and of width 1.5m, when the plate is moving with a velocity of 4m/s in stationary air.
Take kinematic viscosity of air as as 1.5×10−5 m2/s.

GATE/ME/SLP/Module_3/Ch.1_Notes /Pg.93
Vidyalankar : GATE – ME

Solution:
L = 4m
b = 1.5m
U = 4m/s
v = 1.5 × 10−5 m2 /s
UL 4.0 × 4.0
ReL = = = 10.66 × 105
v 1.5 × 10−5
As the Reynold number is more than 5 × 105 and hence the boundary layer at the
trailing edge is turbulent.
The boundary layer thickness for turbulent boundary layer is given by
0.37x
δ = {x = L and Rex = ReL}
(Rex )
1/ 5

0.37 × 4.0
= = 0.0921m = 92.1mm
(1.66 × 105 )
1/ 5

Example:
In above problem determine the total drag on one side of the plate assuming that the
boundary layer is turbulent from the beginning. Take ρ for air = 1.226kg/m3
Solution:
0.072 0.072
CD = = = 0.00448
(ReL ) (10.66 × 10 )
1/ 5 1/ 5
5

1
FD = ρAU2 × CD
2
1
= × 1.226 × 6.0 × 42 × 0.00448 {A = b × L= 1.5 × 4 = 6m2}
2
= 0.2637N

TURBULENT FLOW
Introduction:
In laminar flow the fluid particles move along straight parallel paths in layers or laminae,
such that the paths of individual fluid particles do not cross those of neighboring particles.
Laminar flow is possible only at low velocities and when the fluid is highly viscous. But
when the velocity is increased or fluid is less viscous, the fluid particles do not move in
straight paths. The fluid particles move in random manner resulting in general mixing of
particles. This type of flow is called turbulent flow.
A laminar flow changes to turbulent flow when (i) velocity is increased or (ii) diameter of a
pipe is increased or (iii) viscosity of fluid is decreased. The transition from laminar to
turbulent depends on the mean velocity as well as Reynold’s number given by,
ρVD
Re = . In case of circular pipe if Re < 2000, the flow is said to be laminar and if
μ
Re > 4000, the flow changes from laminar to turbulent.

GATE/ME/SLP/Module_3/Ch.1_Notes /Pg.94
 Notes on Fluid Mechanics

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 experiments, William Froude gave the following laws of fluid friction for
turbulent flow:
The frictional resistance for turbulent flow is:
(i) proportional to Vn, 1.5 ≤ n ≤ 2.0
(ii) proportional to the density of fluid.
(iii) proportional to the area of surface in contacat
(iv) independent of pressure
(v) dependent on the nature of 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. 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 pipe between section 1−1 and 2−2
d = diameter of pipe
f′ = frictional resistance per unit wetted area per unit velocity
hf = loss of head due to friction.
p2, V2 = values of pressure intensity and velocity at section 2−2
f ′ 4LV 2
Then hf = ×
ρg d
f′ f 4fLV 2
Putting = , where ‘f’ is known as coefficient of friction, We get hf =
pg 2 2gd
Above equation is known as “Dracy Weisbach Equation”.

Expression for coefficient of frication in terms of shear stress:.

2 τo
f = 2
ρU
Where τo = wall shear stress, U = average velocity

GATE/ME/SLP/Module_3/Ch.1_Notes /Pg.95
Vidyalankar : GATE – ME

Shear Stress in Turbulent Flow:

Boussinesq expressed the turbulent shear in mathematical form as
du
τt = η
dy
Where τt = Shear stress due to turbulence
η = eddy viscosity
u = average velocity at a distance y from boundary
The ratio of η (eddy viscosity) and ρ ( mass density) is known as kinematic eddy viscosity
and is denoted by ε. Mathematically,
η
ε =
p
If the shear stress due to viscous flow is also considered, then the total shear stress
becomes
du du
τ = μ +η
dy dy
The value of η = 0 for laminar flow. For other cases the value of η may be several
thousand times the value of μ .
Reynold’s Expression for Turbulent Shear Stress:
Expression for turbulent shear stress between two layers of a fluid at a small distance
apart is
τ = ρu′v ′
Where u′,v ′ = fluctuating components of velocity in the direction of x and y due to
turbulence. This shear stress is also known as “Reynold Stress”.
As u′ and v ′ are varying and hence τ will also very. Hence to find the shear stress, the
time average on both the sides of the above equation is taken. Then:
τ = ρu′v ′

Prandtl Mixing Length Theory for Turbulent Shear Stress:

According to this theory
du du
u′ =  , v′ = 
dy dy
2
⎛ du ⎞
u′v ′ =  ⎜⎝ dy ⎟⎠
2

2
⎛ du ⎞
hence τ = 2 ⎜ ⎟
⎝ dy ⎠
Where
 = mixing length, is that distance between two layers in the transverse direction
such that the lumps of fluid particles from one layer should reach the other layer
and the particles are mixed in the other layer in such a way that the momentum
of the particles in the direction of x is same.
= K y; where K is karman constant = 0.4

GATE/ME/SLP/Module_3/Ch.1_Notes /Pg.96
 Notes on Fluid Mechanics

Velocity Distribution in Turbulent Flow in pipes:

Umax − u ⎛R ⎞
= 5.75Log10 ⎜ ⎟
u* ⎝y⎠
Where
U max is the maximum velocity at the centre of pipe
y = distance from the pipe wall, R is the radius of pipe
u = velocity at any distance y from pipe wall
τ
u* = o , τo = wall shear stress
ρ
(u* is also called shear velocity)
ρ = density of fluid

(Umax − u) is known as ‘velocity defect’.

Hydrodynamically Smooth and Rough Boundaries:

Let k is the average height of the irregularities projecting from the surface of boundary as
shown in fig.

If the value of k is large form a boundary then the boundary is called rough boundary and
if the value of the k is less, then boundary is known as smooth boundary.
For turbulent flow analysis along a boundary, the flow is divided into two portions. The
first portion consists of a thin layer of fluid in the immediate neibhourhood of the boundary
where viscous shear stress predominates while the shear stress due to turbulence is
negligible. This protion is known as laminar sub layer. The height upto which the effect of
viscosity predominates in this zone is denoted by δ′ . The 2nd portion of flow, where shear
stress due to turbulence is large compared to viscous stress is known is turbulent zone.
If the average height k, of the irregularities, projecting from the surface of a boundary is
much less than δ’, the thickness of laminar sub − layer as shown in fig. (a), the boundary
is called smooth boundary. This because, outside the laminar − sub layer the flow is
turbulent and eddies of various size present in turbulent flow try to penetrate the laminar
sub−layer and reach the surface of boundary. But due to great thickness of laminar sub
layer the eddies are unable to reach the surface irregularities and hence the boundary
behaves as a smooth boundary. This type of boundary is called hydro − dynamically
smooth boundary.
Now, if the Reynold number of flow is increased then the thickness of laminar sub − layer
will decrease. If the thickness of laminar sub − layer becomes much smaller than the
average height k of irregularities of the surface as shown in fig. (b), the boundary will act

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Vidyalankar : GATE – ME

as rough boundary. This is because the irregularities of the surface above the laminar
sub− layer and the eddies present in turbulent zone will come in contact with the
irregularities of the surface and lot of energy will be lost.
Nikuradse’s experiment:
k
(1) If is less than 0.25, the boundary is called smooth boundary
δ′
k
(2) If is greater than 6.0, the boundary is rough.
δ′
⎛k⎞
(3) If 0.25 < ⎜ ⎟ < 6.0, the boundary is in transition.
⎝ δ′ ⎠
u*k
In terms of roughness Reynold number :
ν
u*k
(1) If < 4, boundary is smooth.
ν
u*k
(2) If 4 < <100, boundary is in transition state.
ν
u*k
(3) If > 100, the boundary is rough.
ν

Velocity Distribution for Turbulent Flow in Smooth pipes:

u ⎛y⎞ δ′
= 5.57Log10 ⎜ ⎟ Where y ′ = ( at y = y′,u = 0 )
u* ⎝ y′ ⎠ 107
11.6ν
Also δ′ =
u*
(v = kinematics viscosity)
u u* y
Hence; = 5.75Log10 + 5.55
u* ν
Velocity Distribution for Turbulent flow in Rough Pipes
k
Here y ′ =
30
u ⎛y⎞
= 5.57L og10 ⎜ ⎟ + 8.5
u* ⎝k⎠

Velocity Distribution for Turbulent Flow in Terms of Average Velocity:

(a) For smooth pipes
U u*R
= 5.57Log10 + 1.75
u* ν
Where U = average velocity
R = pipe radius

GATE/ME/SLP/Module_3/Ch.1_Notes /Pg.98
 Notes on Fluid Mechanics

(b) For rough pipes

U R
= 5.57Log10 + 4.75
u* k
(c) Difference of the velocity at any point and average velocity for smooth and rough
pipes
u−U ⎛y⎞
= 5.75L og10 ⎜ ⎟ + 3.75
u* ⎝R ⎠

Velocity Distribution for Turbulent Flow in Smooth Pipes by Power Law:

1/n
u ⎛y⎞
=⎜ ⎟
umax ⎝R ⎠
⎛ 1⎞
Where exponent ⎜ ⎟ dependsuponReynold'snumber .
⎝n⎠
1 1
For Re = 4 × 103, =
n 6
1 1
Re = 1.1 × 105 =
n 7
1 1
Re ≥ 2 × 106 =
n 10
1 1
Thus if = , the velocity distribution law becomes as :
n 7
1/ 7
u ⎛y⎞
=
umax ⎜⎝ R ⎟⎠
Above equation is known as (1/7)th power law of velocity distribution for smooth pipes.

Resistance of Smooth and Rough Pipes:

(a) Variation of ‘f’ for Laminar Flow:
16
f =
Re

(b) Variation of ‘f’ for Turbulent Flow:

For turbulent flow the coefficient of frication is a function of Re and k / D ration. For
relative roughness (k/D), in the turbulent flow the boundary may be smooth or rough
and hence the value of ‘f’ will be different for those boundaries.
(i) ‘f’ for smooth pipes
According to Blasius
0.0791
f = for 4000 < Re < 105
(R e )
1/ 4

For 105 < Re < 4 × 106

1
4f
( )
= 2.03L og10 Re 4f − 0.91

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Vidyalankar : GATE – ME

For Re upto 4 × 107

0.05525
f = 0.0008 +
(R e )
0.237

(ii) ‘f’ for Rough Pipes

1 ⎛R ⎞
= 2L og10 ⎜ ⎟ + 1.74
4f ⎝k⎠
where R is the radius of pipe.

(c) Value of ‘f’ for commercial pipes

(i) Smooth pipes
1 ⎛R ⎞ ⎛ R 4f ⎞
− 2L og10 ⎜ ⎟ = 2Log10 ⎜ e ⎟⎟ − 0.8
4f ⎝k⎠ ⎜ R/k
⎝ ⎠
(ii) Rough pipes
1 ⎛R ⎞
− 2L og10 ⎜ ⎟ = 1.74
4f ⎝k⎠

SOLVED EXAMPLES

Example:
Determine the wall shearing stress in a pipe of diameter 100mm which carriers water.
The velocities at the pipe centre and 30mm from the pipe centre are 2m/s and 1.5m/s
respectively. The flow in the pipe is given as turbulent
Solution: R

Dia. of pipe = D = 100mm = 0.1m y

∴ Radius = R = 0.05m
Velocity at 30 mm or 0.03m from centre = 1.5m/s
30 mm
∴ Velocity (at r = 0.03 m ), u = 1.5m/s
Let the wall shearing stress = τo

For turbulent flow, the velocity distribution in terms of centerline velocity (umax) is given by;
umax − u ⎛R ⎞
= 5.75Log10 ⎜ ⎟
u *
⎝y⎠
Where u = 1.5m/s at y = (R−r) = 0.05 − 0.03= 0.02m
2.0 − 1.5 0.05
⇒ *
= 5.57Log10 = 2.288
u 0.02
⇒ u* = 0.2185m/s
τo
now u* =
ρ
τo
⇒ 0.2185 = ⇒τo = 47.676N/m2
1000

GATE/ME/SLP/Module_3/Ch.1_Notes /Pg.100
 Notes on Fluid Mechanics

Example:
For turbulent flow in a pipe of diameter 300mm, find the discharge when the centre line
velocity is 2.0 m/s and the velocity at a point 100mm from the centre as measure by pitot
tube is 1.6m/s.
Solution:
Dia of pipe , D = 300mm = 0.3m
∴ Radius, R = 0.15m
Velocity of centre, umax = 2.0 m/s, velocity (at r = 100 mm = 0.1m) u = 1.6 m/s
Now y = R − r = 0.15 − 0 10 = 0.05m
∴ Velocity at (r = 0.1 m or at y; = 0.05m) , u = 1.6 m/s
The velocity in terms of centre − line velocity is given by
umax − u ⎛R ⎞
*
= 5.75Log10 ⎜ ⎟
u ⎝y⎠
2.0 − 1.6 ⎛ 0.15 ⎞
⇒ = 5.75Log10 ⎜ ⎟
u* ⎝ 0.05 ⎠
⇒ u* = 0.1458 m/s
The relation between velocity of any point and average velocity for smooth and rough
pipes is:
u−U ⎛y⎞
⇒ *
= 5.75Log10 ⎜ ⎟ + 3.75
u ⎝R ⎠
at y = R, u = umax where U is the average velocity.
umax − U ⎛R ⎞
⇒ *
= 5.75Log10 ⎜ ⎟ + 3.75 = 3.75
u ⎝R ⎠
2.0 − U
⇒ = 3.75 ⇒ U = 0.4533m / s
u*
π π
( )
⇒ Discharge, Q = A × U = D2U = 0.32 × 1.4533 = 0.1027m3 / s
4 4
Example:
For above problem, find the coefficient of friction and average height of rough pipe.
Solution:
We have
R = 0.15m
u* = 0.1458m/s
U = 1.4533m/s
2 τo τ
Now, f = andu* = o
ρU 2
ρ
From these we get
f
u* = U
2
f
⇒ 0.1458 = 1.4533 2 ⇒ f = 0.02

GATE/ME/SLP/Module_3/Ch.1_Notes /Pg.101
Vidyalankar : GATE – ME

Height of rough pipe is obtained from

1 ⎛R ⎞
= 2Log10 ⎜ ⎟ + 1.74
4f ⎝k⎠
1 ⎛ 0.15 ⎞
⇒ = 2Log10 ⎜ ⎟ + 1.74
4 × 0.2 ⎝ k ⎠
⇒ k = average hight of roughness = 0.01898m = 18.98mm

FLOW THROUGH PIPES

Loss of Energy in Pipes due to Friction :
(a) Dracy − Weisbach Formula:
4fLV 2 16
hf = ; f= forRe < 2000(Viscous Flow)
2gd Re
0.079
f = for4000 < Re < 106
Re1/ 4
(b) Chezy’s Formula:
V = C mi
ρg
Where; C = = Chezy' s constant
f′
d f′ f
m = d = pipe dia. also =
4 pg 2
hf
i =
L
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.
Minor Energy (Head) Losses:
The loss of head or energy due to friction in a pipe is known as major loss while the loss
of energy due to change of velocity of the flowing fluid in magnitude or direction is called
minor loss of energy. The minor loss of energy includes the following cases:
1) Loss of head due to sudden enlargement:
Consider a liquid flowing through a pipe which has sudden enlargement as shown in fig.

V2

GATE/ME/SLP/Module_3/Ch.1_Notes /Pg.102
 Notes on Fluid Mechanics

Consider two sections 1−1 and 2−2 before and after the enlargement.
Let p1 = pressure intensity of section 1−1
V1 = velocity of flow at section 1−1
A1 = area of pipe at section 1−1
p2, V2 and A2 = Corresponding values at section 2−2
( V − V2 )
2
⎡⎛ A ⎞ V2 ⎤
2

Then head loss, he = 1 = ⎢⎜ 1 − 1 ⎟ 1 ⎥

2g ⎢⎣⎝ A 2 ⎠ 2g ⎥⎦

2) Loss of Head due to Sudden Contraction:

Consider a liquid flowing in a pipe which has a sudden contraction in area as shown
in fig.

2
kV22 ⎡ 1 ⎤
Then; head loss; hc = , Where k = ⎢ − 1⎥
2g ⎣ Cc ⎦
Cc = Coefficient of contraction
When Cc = 0.62
V2
Then hc = 0.375 2
2g
If the value of Cc is not given, then the head loss due to contraction is taken as;
V2
hc = 0.5 2
2g
3) Loss of Head at the entrance of a pipe:
V2
hi = 0.5
2g
4) Loss of Head at the exit of pipe:
V2
ho =
2g
5) Loss of Head due to Bend in pipe:
kV 2
hb =
2g
Where k = coefficient of bend
The value of k depends upon
(i) Angle of bend
(ii) Radius of curvature of bend
(iii) Diameter of pipe

GATE/ME/SLP/Module_3/Ch.1_Notes /Pg.103
Vidyalankar : GATE – ME

6) Loss of Head in various Pipe Fittings (e.g. values, couplings):

kV 2
h =
2g
Where; k = coefficient of pipe fitting.

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:
1) Hydraulic Gradient Line:
⎛ p ⎞
It is defined as the line which gives the sum of pressure head ⎜ ⎟ and datum head
⎝ ρg ⎠
(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
⎛ p ⎞
head ⎜ ⎟ of a flowing fluid in a pipe, from the centre of pipe. It is briefly written as
⎝ ρg ⎠
H.G.L.
2) Total Energy Line:
It is defined as the line, which gives the sum of pressure head, datum head and
kinetic head of an 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 pipe. It is
briefly written as T.E.L.
Flow Through Pipes in Series:
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 pipeline as shown in fig.

The discharge passing through each pipe is same.

⇒ Q = A1V1 = A2V2 = A3V3
Total head loss = Difference in liquid surfaces
4f1L1 V12 4f2L 2 V22 4f3L 3 V32
⇒ H = + +
2gd1 2gd2 2gd3

GATE/ME/SLP/Module_3/Ch.1_Notes /Pg.104
 Notes on Fluid Mechanics

Equivalent Pipe:
This is defined as the pipe of uniform diameter having loss of head and discharge equal
to the loss of head and discharge of compound pipe consisting of several pipes of
different lengths and diameters.
Hence for an equivalent pipe
L = L1+ L2 + L3 + −−−−−−−−−−
L L L L
And 5 = 51 + 52 + 53 + −−−−−−−−−−−
d d1 d2 d3
Above equation is known as “Dupuit’s Equation”.

Flow Through Parallel Pipes:

Consider a main pipe which divides into two or more branches as shown in fig. 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;
Q = Q1 + Q2
Also ; loss of head for branch pipes is equal.
4f1L1 V12 4f2L 2 V22
⇒ =
2gd1 2gd2
Power Transmission Through Pipes:
Consider a pipe AB connected to tank as shown in Fig.

The power available at the end B of the pipe will be;

ρg π ⎛ 4fLV 2 ⎞
P = × d2 × V ⎜ H − ⎟ kW
1000 4 ⎝ 2gd ⎠
Efficiency of power transmission;
H − hf 4fLV 2
η = ; hf =
H 2gd

GATE/ME/SLP/Module_3/Ch.1_Notes /Pg.105
Vidyalankar : GATE – ME

For maximum transmission of power;

H 2
hf = , ηmax = or 66.7%
3 3
Flow Through Nozzles:
Fig. below shows a nozzle fitted at the end of a long pipe. The total energy by fitting the
nozzle at the end of the pipe consists of pressure energy and kinetic energy .By fitting the
nozzle at the end of the pipe, the total energy is converged into kinetic energy. Thus
nozzles are used, where higher velocities of flow are required.

Outlet velocity of nozzle is given by

2gH
V =
⎛ 4fL a 2 ⎞
⎜1+ × ⎟
⎝ D A2 ⎠
Where
π 2
a = area of nozzle at outlet = d
4
π 2
A = area of the pipe = D
4
f = co−efficient of friction for pipe
Discharge through nozzle = a × V
Power in kW at the outlet of nozzle
0.5ρaV 3 ρg ⎛ V2 ⎞
= = aV ⎜ ⎟ (if losses are neglected)
1000 1000 ⎝ 2g ⎠
ρgaV ⎡ 4fLV 2 ⎤
= ⎢H − ⎥ (if losses are considered)
1000 ⎣ 2gD ⎦
Efficiency of power transmission through nozzle:
V2 1
η = =
2gH 4fLa 2
1+
DA 2
For maximum power transmission:
H ⎡ 4fLV 2 ⎤
hf = ⎢hf = = headlossinpipe ⎥
3 ⎣ 2gD ⎦

GATE/ME/SLP/Module_3/Ch.1_Notes /Pg.106
 Notes on Fluid Mechanics

Diameter of nozzle for maximum power transmission: It can be calculated from either of
the equations given below;
1/ 4
⎛ D5 ⎞
d =⎜ ⎟
⎝ 8fL ⎠
A 8fL
=
a D

Example :
Find the head loss due to friction in a pipe of diameter 300mm and length 50m, through
with water is flowing at velocity of 3m/s using (i) Dracy formula, (ii) Chezy’s formula for
which C = 60. Take ν for water = 0.01 stoke.
Solution:
Dia of pipe; D = 300mm = 0.30
Length of pipe; L = 50m
Velocity of flow; V = 3m/s
Chezy’s constant; C = 60
Kinematic Viscosity ν = 0.01 stoke
= 0.01 × 10−4 m2/s
(i) Dracy formula is given by:
4fLV 2
hf =
2gd
Vd 3.0 × 0.3
Re = = = 9 × 105
ν 0.01× 10−4
(for 40000 < Re < 106)
0.079 0.079
Hence f = = = 0.00256
( )
1/ 4 1/ 4
Re 9 × 105
4 × 0.00256 × 50 × 32
⇒ Head loss = hf = = 0.7828m
0.3 × 2.0 × 9.81
(ii) Chezy’s Formula:
V = C mi
d 0.30
C = 60, m = = = 0.075m
4 4
⇒ 3 = 60 0.075 × i ⇒ i = 0.0333
h h
But i = f = f ⇒ hf = 50 × 0.0333 = 1.665m
L 50
Example:
An oil of sp. gr. 0.7 is flowing through a pipe of diameter 300mm at the rate of 500 liters/s.
Find the head loss due to friction and power required to maintain the flow for a length of
1000m. Take ν = 0.29 stokes.

GATE/ME/SLP/Module_3/Ch.1_Notes /Pg.107
Vidyalankar : GATE – ME

Solution :
Sp. gr.of oil, S = 0.7
Dia. of pipe, D = 300mm = 0.3m
Discharge Q = 500litres/s = 0.5m3 /s
ν = 0.29 × 10 −4 m2 / s
Length of pipe, L =1000 m
Q 0.5 0.5 × 4
Velocity V = = = = 7.073m / s
Area π 2 π × 0.32
d
4
V × D 7.0373 × 0.3
⇒ Reynold number, Re = = = 7.316 × 104
ν 0.29 × 10 −4
(for 4000 < Re < 106)
0.079 0.079
⇒ Coefficient of friction, f = = = 0.0048
Re1/ 4 ( )
1/ 4
7.316 × 104
4fLV 2 4 × 0.0048 × 1000 × 7.0732
⇒ Head lost due to friction, hf = = = 163.18m
2gd 0.3 × 2 × 9.81
ρgQhf
Power required = kW
1000
Where ρ = density of oil = 0.7 × 1000 = 700kg/m3
700 × 9.81× 0.5 × 163.18
⇒ Power required = = 560.28kW
1000
Example:

A pipe line AB of diameter 300mm and of length 400m carries water at the rate of
50litres/s. The flow takes place from A to B where point B is 30 metres above A. Find the
pressure at A if the pressure at B is 19.62 Ncm2. Take f = 0.008.

Solution:

Dia of pipe, D = 300mm = 0.3m

Length of pipe, L = 400m
Discharge, Q = 50 liters /s = 0.05 m3/s
Q 0.05
⇒ Velocity, V`= = = 0.7074m / s
Area π
( 0.3 )
2

GATE/ME/SLP/Module_3/Ch.1_Notes /Pg.108
 Notes on Fluid Mechanics

Pressure at B, pB = 19.62N/cm2= 19.62 ×104 N/m2

f = 0.008
Applying Bernoulli’s equation at points A and B and taking datum line passing
through A we have
pA VA2 p V2
+ + Z A = B + B + ZB + hf
ρg 2g ρg 2g
But VA = VB [∵ Dia is same]
ZA = 0, ZB = 30
4fLV 2 pA 19.62 × 104 4 × 0.008 × 400 × 0.70742
And hf = ⇒ +0 = + 30 + = 51.088 m
2gd ρg 1000 × 9.81 0.3 × 2 × 9.81
⇒ pA = 51.088 ×1000 × 9.81N /m2 = 50.12 N/cm2

Example:
A pipe line of length 2000m is used for power transmission. If 110.3625 kW power is to
be transmitted through the pipe in which water having a pressure of 490.5 N/cm2 at inlet
is flowing. Find the diameter of the pipe and efficiency of transmission if the pressure
drop over the length of pipe is 98.1N/cm2. Take f = 0.0065.
Solution:
Length of pipe, L = 2000m
Power transmitted, = 110.3625 kW
Pressure at inlet, p = 490.5N/cm2 = 490.5× 104N/m2
p 490.5 × 104
⇒ pressure head at inlet, H = = = 500m [∵ ρ = 1000]
ρg 1000 × 9.81
pressure drop = 98.1 N /cm2 = 98.1 × 104 N/m2
98.1× 104
⇒ Loss of head = hf = = 100m
1000 × 9.81
Coefficient of friction, f = 0.0065
Head available at the end of the pipe = H − hf = 500 −100 = 400m
Let diameter of pipe = d
Now, power transmitted is given by
ρg × Q × (H − hf )
P = kW
1000
1000 × 9.81× Q × 400
⇒110.3625 =
1000
⇒Q = 0.02812
π
But discharge, Q = Area × Velocity = d2 × V
4
π 2
⇒ d × V = 0.02812
4
0.358
⇒ V =
d2

GATE/ME/SLP/Module_3/Ch.1_Notes /Pg.109
Vidyalankar : GATE – ME

The head loss due to friction

4 fLV 2 4 × 0.0065 × 2000 × V 2
hf = =
2gd 2 × 9.81× d
2
2.65 V 2 2.65 ⎛ 0.358 ⎞
100 = = ⎜ ⎟
d d ⎝ d2 ⎠
0.003396
100 =
d5
⇒ d = 0.1277m = 127.7m
H − hf 500 − 100
η = = = 0.80 = 80%
H 500
Example :
The head of water at the inlet of a pipe 2000m long and 500 mm diameter is 60m. A
nozzle of diameter 100mm at its outlet is fitted to the pipe. Find the velocity of water at
the outlet of the nozzle if f = 0.01 for the pipe.
Solution:
Head of water at inlet of pipe, H = 60mm
Length of pipe, L = 2000m
Dia. of pipe, D = 500mm = = 0.50m
Dia of nozzle at outlet, d = 100mm = 0.1m
Coefficient of friction. f = 0.01

The velocity at outlet of nozzle is given by

2gH 2 × 9.81× 60
V = = 2
⎛ 4fLa 2 ⎞ ⎛π 2 ⎞
⎜1+ 2 ⎟ 4 × 0.1× 2000 ⎜ 4d ⎟
⎝ DA ⎠ 1+ ⎜ ⎟
0.5 ⎜⎜ π D2 ⎟⎟
⎝4 ⎠
2 × 9.81× 60
= 2
= 30.61m / s
4 × 0.1× 2000 ⎛ 0.1× 0.1 ⎞
1+ ⎜ ⎟
0.5 ⎝ 0.5 × 0.5 ⎠

Stagnation:
Consider the streamline steady flow of fig. in which stream A along a horizontal path is
brought to rest at O so that Vo = 0. Applying Bernoulli equation to points 1 and O on
streamline A.

GATE/ME/SLP/Module_3/Ch.1_Notes /Pg.110
 Notes on Fluid Mechanics

V12 p1 V2 p
+ + gh1 = o + o + gho
2 ρ 2 ρ
But h1 = ho , and Vo = 0 hence
1 2
po = ρV1 + p1
2
The point O is called stagnation point , and pressure po is called stagnation pressure.

LMR (LAST MINUTE REVISION)

• The weight density or specific weight of a fluid is equal to weight per unit volume.
• Specific volume is the reciprocal of mass density.
• The shear stress is proportional to velocity gradient.
• Poise and stokes are the units of viscosity and kinematic viscosity respectively.
• Compressibility is the reciprocal of Bulk Modulus of Elasticity.
• The pressure at any point in a fluid is defined as the force per unit area.
• The viscosity of liquids decreases with the increase of temperature while the viscosity
of gases increase with the increase of temperature.
• Types of Fluids :
1. Ideal Fluid 2. Real Fluid 3. Newtonian Fluid
4. Non−Newtonian Fluid 5. Ideal Plastic Fluid

• Rheological Classification of Fluids :

There are two major types of Non−Newtonian fluids : Viscoelastic and Purely viscous
fluids.
• Surface Tension is defined as the tensile force acting on the surface of a liquid in
contact with a gas or on the surface between two immiscible liquids such that the
contact surface behaves like a membrane under tension.
• Capillarity is defined as a phenomenon of rise or fall of a liquid surface in a small
tube relative to the adjacent general level of liquid when the tube is held vertically in
the liquid.
• Pascal’s Law states that the pressure or intensity of pressure at a point in a static
fluid is equal in all directions.

• Pressure Variation in a Fluid at Rest:

The pressure at any point in a fluid at rest is obtained by the hydrostatic law which
states that the rate of increase of pressure in a vertically downward direction must be
equal to the specific weight of the fluid at that point.
• Absolute pressure is defined as the pressure which is measured with reference to
absolute vacuum pressure.
• Gauge pressure is defined as the pressure which is measured with the help of
pressure measuring instrument, in which atmospheric pressure is taken as datum.

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Vidyalankar : GATE – ME

• Vacuum pressure is defined as the pressure below the atmospheric pressure.

• The pressure of a fluid is measured by the following devices:
(1) Manometers
(a) Simple Manometers
(i) Piezometer
(ii) U − tube Manometer
(iii) Single Column Manometer
(b) Differential Manometer
(i) U − tube differential Manometer
(ii) Inverted U − tube Differential manometer
(2) Mechanical Gauges
(c) Diaphragm Pressure Gauge
(d) Bourdon tube pressure Gauge
(e) Dead − weight pressure Gauge
(f) Bellows pressure Gauge
• Simple manometers are used to measure pressure at a point while differential
manometers are used for measuring difference of pressures between two points in a
pipe or in two different pipes.
• A single column manometer is used for measuring small pressures, where accuracy
is required.
• For compressible fluids, density (ρ) changes with pressure and temperature.
• Temperature Lapse Rate is defined as the rate at which the temperature changes
with elevation.
• Total Pressure is defined as the force exerted by a static fluid on a surface either
plane or curved when the fluid comes in contact with the surface. This force always
acts normal to the surface.
• Centre of Pressure is defined as the point of application of the total pressure on the
surface.
• When the fluid is at rest the shear stress is zero.
• The centre of pressure of a plane vertical surface lies at a depth of two−third the
height of the immersed surface.
• Buoyancy:
When a body is immersed in a fluid, an upward force is exerted by the fluid on the
body. This upward force is equal to the weight of the fluid displaced by the body and
is called the force of buoyancy or simply buoyancy.
• Centre of Buoyancy is defined as the point, through which the force of buoyancy is
supposed to act.
• Meta − Centre is defined as the point about which a body starts oscillating when the
body is tilted by a small angle.
• Meta − Centric Height:
The distance between the meta − centre of a floating body and the centre of gravity
of the body is called meta − centric height.

GATE/ME/SLP/Module_3/Ch.1_Notes /Pg.112
 Notes on Fluid Mechanics

• Conditions of equilibrium of a floating and submerged body are:

Equilibrium Floating Body Sub−merged Body

(1) Stable Equilibrium M is above G B is above G

(2) Unstable Equilibrium M is below G B is below G
(3) Neutral Equilibrium M and G coincide B and G coincide

• The fluid motion is described by two methods. They are (i) Lagrangian Method, and
(ii) Eulerian Method.
• Types of Fluid Flow: The fluid flow is classified as:
(i) Steady and unsteady flows.
(ii) Uniform and non − uniform flows.
(iii) Laminar and turbulent flows.
(iv) Compressible and incompressible flows
(v) Rotational and irrotational flow, and
(vi) One, two and three dimensional flows.
• Discharge or Rate of Flow is defined as the quantity of a fluid flowing per second
through a section of a pipe or a channel.
The equation based on the principle of conservation of mass is called continuity
equation. This equation is applicable to:
(i) Steady and unsteady flow,
(ii) Uniform and non − uniform flow, and
(iii) Compressible and incompressible fluids
• Local acceleration is defined as the rate of increase of velocity with respect to time at
a given point in a flow field.
• Convective acceleration is defined as the rate of change of velocity due to the
change of position of fluid particles in a fluid flow.
• Velocity Potential Function 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.
(1) If velocity potential (φ) exists, the flow should be irrotational.
(2) If velocity potential (φ) exists, and satisfies the Laplace equation, it represents the
possible steady incompressible irrotational flow.
• Stream Function is defined as the scalar function of space and time, such that its
partial derivative with respect to any direction gives the velocity components at right
angles to that direction.
(1) If stream function ( Ψ ) exists, it is a possible case of fluid flow which may be
rotational or irrotational
(2) If stream function ( Ψ ) satisfies the Laplace equation, it is a possible case of an
irrotational flow.
• A line along which the velocity potential (φ) is constant, is called equipotential line
• The equipotential lines are orthogonal to the stream lines at all points of intersection.
• A grid obtained by drawing a series of equipotential lines and stream lines is called a
flow net. the flow net is an important tool in analyzing two− dimensional irrotational
flow problems.

GATE/ME/SLP/Module_3/Ch.1_Notes /Pg.113
Vidyalankar : GATE – ME

• It stream function satisfies Laplace equation, the flow is irrotational.

• In the fluid flow, the following forces are present:
(1) Fg, gravity force
(2) Fp, the pressure force
(3) F ν , the force due to viscosity
(4) Fc, force due to compressibility
• If the force due to compressibility, FC is negligible, equations of motion are called
“Reynold’s equations of motion”
• For flow where (Ft) force due to turbulence is negligible, the resulting equations of
motion are known as “Navier − Stokes equation”.
• If the flow is assumed to be ideal, viscous force (Fν) is zero and equations of motion
are known as “Euler’s equations of motion”.
• Euler’s Equation of Motion is Equation of motion in which the force due to gravity
and pressure are taken into consideration.
• Bernoulli’s equation is obtained by integrating the Euler’s equation of motion.
• The following are the assumptions made in the derivation of Bernoulli’s equation:
(1) The fluid is ideal, i.e. viscosity is zero
(2) The flow is steady
(3) The flow is incompressible
(4) The flow is irrotational
• Bernoulli's equation is applied in all problems of incompressible fluid flow where
energy considerations are involved e.g.
(1) Venturimeter
(2) Orifice meter
(3) Pitot − tube
• The Momentum Equation is based on the law of conservation of momentum, which
states that the net force acting on fluid mass is equal to the change in momentum of
flow per unit time in that direction.
• Moment of Momentum Equation is derived from moment of momentum principle
which states that the resulting torque acting on a rotating fluid is equal to the rate of
change of moment of momentum.
This equation is applied:
(1) For analyzing flow problems in turbines and centrifugal pumps
(2) For finding torque exerted by water on sprinkler.

• Free liquid jet is defined as the jet of water coming out from the nozzle in
atmosphere. The path travelled by the free jet is parabolic.
• Coefficient of velocity (Cv) is defined as the ratio between the actual velocity of jet
of liquid at vena −contracta and the theoretical velocity of jet. The value of CV varies
from 0.95 to 0.99 for different orifices, depending upon the shape, size of the orifice
and on the head under which the flow takes place. CV is generally taken as 0.98 for
sharp edged orifices

GATE/ME/SLP/Module_3/Ch.1_Notes /Pg.114
 Notes on Fluid Mechanics

• Coefficient of Contraction (Cc) is defined as the ratio of the area of the jet at vena −
contracta to the area of orifice. The value of Cc varies form 0.61 to 0.69 depending on
shape and size of the orifice and head of liquid under which flow takes place. In
general, the value of Cc may be taken as 0.64.
• Coefficient of discharge (Cd) is defined as the ratio of actual discharge from an
orifice to the theoretical discharge from the orifice. The value of Cd varies from 0.61
to 0.65. For general purpose, the value of Cd is taken as 0.62
• A Venturimeter is a device used for measuring the rate of flow of a fluid flowing
through a pipe.
• Orifice Meter or Orifice Plate consists of a flat circular plate which has circular
shape edged hole called orifice, which is concentric with the pipe. The orifice
diameter is kept generally 0.5 times the diameter of the pipe, through it may vary
from 0.4 to 0.8 times the pipe diameter.
• Pitot Tube is a device used for measuring the velocity of flow at any point in a pipe
or a channel. It is based on the principle that if the velocity of flow at a point becomes
zero, the pressure there is increased due to conversion of kinetic energy into
pressure energy.
• If the head of liquid is less than 5 times the depth of the orifice, the orifice is called
large orifice.
• Fully submerged orifice is one which has its whole of the outlet side sub merged
under liquid so that it discharges a jet of liquid into the liquid of the same kind. It is
also called totally drowned orifice.
• A mouthpiece is a short length of a pipe which is two or three times its diameter in
length.
• A notch is a device used for measuring the rate of flow of a liquid through a small
channel or a tank.
• A weir is concrete or masonry structure, placed in an open channel over which the
flow occurs.
• A triangular notch or weir is preferred to a rectangular notch or weir due to following
reasons:
(1) The expression for discharge of right angled V − notch is very simple
(2) For measuring low discharge, a triangular notch gives more accurate results than
a rectangular notch.
(3) In case of triangular notch, only one reading, i.e. (H) is required for the
computation of discharge
(4) Ventilation of a triangular notch is not necessary.
• Velocity of approach is defined as velocity with which the water approaches or
reaches the weir or notch before it flows over it.
• The flow through the circular pipe will be viscous or laminar, if the Reynolds number
(Re) is less than 2000.
• Kinetic Energy correction Factor is defined as the ratio of the kinetic energy of the
flow per second based on actual velocity across a section to the kinetic energy of the
flow per second based on average velocity across the same section.

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Vidyalankar : GATE – ME

• Momentum Correction Factor is defined as the ratio of momentum of the flow per
second based on actual velocity to the momentum of the flow per second based on
average velocity across a section.
• Momentum correction factor and energy correction factor for laminar flow through
4
flow a circular pipe are and 2.0 respectively.
3
• When a solid body is immersed in a flowing fluid, there is a narrow region of the fluid
in the neighborhood of the solid body, where the velocity of the fluid varies from zero
to free−stream velocity. This narrow region of fluid is called boundary layer.
• If the Reynold’s number is more than 5 ×105, the boundary layer is called turbulent
boundary layer.
• Laminar Sub− Layer is the region in the turbulent boundary layer zone, adjacent to
the solid surface of the body. In this zone, the velocity variation is influenced only by
viscous effects. Here velocity variation is linear and so the velocity gradient is
constant.
• Boundary Layer Thickness (δ) is defined as the distance from the boundary of the
solid body measured in y − direction to the point, where the velocity of the fluid is
approximately equal to 0.99 times the free stream velocity (U) of the fluid.
• Displacement Thickness (δ*) 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 flow rate on account of boundary layer formation.
• Momentum Thickness (θ) 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 boundary layer formation.
• Energy Thickness (δ**) is 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 boundary layer formation.
• If the pressure gradient is positive, the boundary layer separates from the surface
and back flow and eddies formation take place due to which a great loss of energy
occur.
• When the velocity is increased or fluid is less viscous, the fluid particles do not move
in straight paths. The fluid particles move in random manner resulting in general
mixing of particles. This type of flow is called turbulent flow.
• A laminar flow changes to turbulent flow when (i) velocity is increased or (ii) diameter
of a pipe is increased or (iii) viscosity of fluid is decreased.
• In case of circular pipe if Re < 2000, the flow is said to be laminar and if Re > 4000,
the flow changes from laminar to turbulent.
• 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.

GATE/ME/SLP/Module_3/Ch.1_Notes /Pg.116
 Notes on Fluid Mechanics

• If the average height k, of the irregularities, projecting from the surface of a boundary
is much less than δ’, the thickness of laminar sub − layer as shown in fig. (a), the
boundary is called smooth boundary.
• If the thickness of laminar sub − layer becomes much smaller than the average
height k of irregularities of the surface as shown in fig. (b), the boundary will act as
rough boundary.
• Nikuradse’s experiment:
k
(1) If is less than 0.25, the boundary is called smooth boundary
δ′
k
(2) If is greater than 6.0, the boundary is rough.
δ′
⎛k⎞
(3) If 0.25 < ⎜ ⎟ < 6.0, the boundary is in transition.
⎝ δ′ ⎠
u*k
• In terms of roughness Reynold number :
ν
u*k
(1) If < 4, boundary is smooth.
ν
u*k
(2) If 4 < <100, boundary is in transition state.
ν
u*k
(3) If > 100, the boundary is rough.
ν
• The loss of head or energy due to friction in a pipe is known as major loss whiles the
loss of energy due to change of velocity of the flowing fluid in magnitude or direction
is called minor loss of energy. The minor loss of energy includes the following cases:
(1) Loss of head due to sudden enlargement
(2) Loss of Head due to Sudden Contraction
(3) Loss of Head at the entrance of a pipe
(4) Loss of Head at the exit of pipe
(5) Loss of Head due to Bend in pipe
(6) Loss of Head in various Pipe Fittings (e.g. values, couplings)

• Hydraulic Gradient Line is defined as the line which gives the sum of pressure head
⎛ p ⎞
⎜ ⎟ and datum head (z) of a flowing fluid in a pipe with respect to some reference
⎝ ρg ⎠
line or it is the line which is obtained by joining the top of all vertical ordinates,
⎛ p ⎞
showing the pressure head ⎜ ⎟ of a flowing fluid in a pipe, from the centre of pipe.
⎝ ρg ⎠
• Total Energy Line is defined as the line, which gives the sum of pressure head,
datum head and kinetic head of an 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 pipe.

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• Equivalent Pipe is defined as the pipe of uniform diameter having loss of head and
discharge equal to the loss of head and discharge of compound pipe consisting of
several pipes of different lengths and diameters.
• The total energy by fitting the nozzle at the end of the pipe consists of pressure
energy and kinetic energy. By fitting the nozzle at the end of the pipe, the total
energy is converged into kinetic energy. Thus nozzles are used, where higher
velocities of flow are required.



GATE/ME/SLP/Module_3/Ch.1_Notes /Pg.118
 Assignment on Fluid Mechanics

ASSIGNMENT − 1
Duration : 45 Min. Max. Marks : 30
Q1 to Q6 carry one mark each

1. Choose the wrong statement

(A) A fluid deforms continuously under a shear force
(B) A fluid does not stay in shape
(C) A gas is a fluid which expands continuously till it occupies the space
completely
(D) A liquid is a fluid which essentially behaves as a gas

2. The density of fluid varies with

(A) Change of temperature
(B) Change of pressure
(C) Change of temperature and pressure
(D) None of the above

3. The force of attraction between molecules of fluid and surface of contact is

(A) cohesion (B) adhesion
(C) viscosity (D) friction

4. Surface energy per unit surface area of liquid is

(A) surface tension
(B) atmospheric pressure
(C) force offered by a flowing liquid
(D) force of cohesion among the molecules of the liquid

5. In which fluids shearing stress increases uniformly with velocity gradient?

(A) Bingham and Newtonian (B) Newtonian and Inviscid
(C) Inviscid and Dilatant (D) Dilatant and Bingham

6. The velocity of flow changes from one point to the other at any instant of time,
the flow is
(A) unsteady (B) rotational
(C) turbulent (D) non − uniform

Q7 to Q18 carry two marks each

7. The density of water at 0°C in kg/m3 is

(A) 999.8 (B) 981.7
(C) 958.8 (D) 900

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Vidyalankar : GATE – ME

8. A soap bubble has a spherical shape because

(A) pressure inside it is uniform
(B) a spherical shape will have maximum energy
(C) a sphere will have minimum surface area for a given volume
(D) a sphere will have minimum volume for a given surface energy

9. A liquid in atmosphere of air begins to boil when

(A) vapour pressure of liquid is slightly less than air pressure
(B) vapour pressure of liquid slightly exceeds the air pressure
(C) vapour pressure of liquid is equal to the air pressure
(D) vapour pressure of liquid attains a critical value

10. The pressure in pipeline is read through the height of a liquid column in a
manometer whose one end is open to atmosphere. If the column in manometer is
1m and density of liquid is 0.72 times that of water, the pressure in the pipeline is
2
(A) 7.265 kN/m (B) 108.06 kN/m2
2
(C) 0.72 kN/m (D) 101.473 kN/m2

11. For a plane which if symmetrical about y − axis, the centre of pressure
(A) coincides with the centroid
(B) coincides with the centre of gravity
(C) lies on the y −axis but below the centroid
(D) lies on the y −axis but above the centroid

12. A metal piece having density equal to the density of a fluid is placed in the liquid.
The metal piece will
(A) sink to the bottom (B) float on the surface
(C) will be partially immersed (D) will be wholly immersed.

13. Which of the flow measuring device through which fluid does not flow?
(A) Orifice plate (B) Venturimeter
(C) Pitot tube (D) Elbow meter
14. A fluid whose viscosity does not change with rate of deformation is known as
(A) ideal fluid (B) real fluid
(C) non − Newtonian (D) Newtonian fluid
15. Dynamic viscosity remains constant with respect to velocity gradient for
(A) Newtonian and Bingham fluids
(B) Bingham and Dilatant fluids
(C) Dilatant and pseudoplastic fluids
(D) Pseudoplastic and Newtonian fluids

GATE/ME/SLP/Module_3/Ch.1_Assign /Pg.120
 Assignment on Fluid Mechanics

16. The atmospheric pressure measured as 760 mm of mercury column is equal to

(A) 0.1033N/mm2 (B) 10.33 m of water column
2
(C) 10.33N/mm (D) 1.033 m of water column

17. A 40m long wall is one side of a water

tank storing water to a depth of 3.5 m.
What is the resultant force acting on 3.5m
wall?
(Take density of water = 999.8 kg/m3)
(A) 2.6 MN
(B) 3.4 MN
(C) 3.8 MN
(D) 2.4 MN

18. In previous question, what is the location of centre of pressure yf?

(A) 2.33m (B) 1.33m
(C) 3.33m (D) 0.333m



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Vidyalankar : GATE – ME

TEST PAPER − 1
Duration : 30 Min. Max. Marks : 25
Q1 to Q5 carry one mark each

1. The specific weight of water at 0°C is

(A) 5009.9N/m3 (B) 9808.4N/m3
(C) 1509N/m3 (D) 20000N/m3
2. A small diameter tube is inserted in a beaker filled with mercury
(A) The mercury in the tube will drop below the level and the meniscus will
be convex
(B) The mercury in the tube will drop below the level in the beaker and the
meniscus will be concave
(C) The mercury will rise in the tube and meniscus will be convex
(D) The mercury will rise in the tube and meniscus will be concave

3. The pressure above an oil surface, whose specific gravity is 0.9, is 9810N/m2.
The pressure at a depth of 3m would be
(A) 0.026N/mm2 (B) 0.09813N/mm2
2
(C) 0.0363N/mm (D) 0.07113N/mm2
4. The height of mercury barometer column is measured at a place a 757 mm. the
atmospheric pressure at that place is
(A) 101kN/m2 (B) 10.3 kN/m2
2
(C) 17.95kN/m (D) 55.7 kN/m2

5. A fluid flowing through a tube has a density of 788 kg/m3. The piezometer tube in
the pipe records a height of 760mm above the centre line of the tube of 320mm
diameter. The pressure of fluid in the tube is (taking atmospheric pressure as
101kN/m2)
(A) 106.875 kN/m2 (B) 105.638 kN/m2
2
(C) 101.600 kN/m (D) 101.473 kN/m2

Q6 to Q13 carry two marks each

6. Match the list I and II using code given below

I (characteristics) II (type of fluid)
n
⎛ ∂u ⎞
a. τxy = μ ⎜ ⎟ ,n < 1 1. Bingham
⎝ ∂y ⎠
n
⎛ ∂u ⎞
b. τxy = μ ⎜ ⎟ ,n > 1 2. Yield pseudoplastic
⎝ ∂y ⎠

c. τxy = τo + ⎛⎜ ∂u ⎞⎟ 3. pseudoplastic
⎝ ∂y ⎠
n
⎛ ∂u ⎞
d. τxy = τo +μ ⎜ ⎟ ,n > 1 4. dilatant
⎝ ∂y ⎠

GATE/ME/SLP/Module_3/Ch.1_Test /Pg.132
 Test on Fluid Mechanics

Code.
a b c d
(A) 3 4 1 2
(B) 3 1 4 2
(C) 2 4 1 3
(D) 4 1 2 3

7. A vertical sluice gate 1m deep × 2m wide is immersed in water such that its c.g.
is at a depth of 1m. Total force acting on the gate is
(A) 1999.6N (B) 1000N
(C) 19616.N (D) 2000N

8. A wooden block of rectangular base of 4m × 2m and depth of 1 m is floated with

depth as vertical in water. The specific gravity of wood is 0.7. The portion of
depth out of water is
(A) 300mm (B) 560mm
(C) 650mm (C) 700mm

9. Match the lists I and II. Use the code given below:
I (Quantity to be measured) II (Instrument)
a. Flow through pipe 1. piezometer
b. Flow through channel 2. Venturimeter
c. Pressure in a pipe 3. Pitot tube
d. Velocity of flow 4. Manometer
5. Orifice plate
6. V− notch.

Code a b c d
(A) 2,5 6,4 1 2,3
(B) 5 2,4 1,3 6
(C) 2,5 6 1,4 3
(D) 1,3 2,6 1,4 3,5

10. Fig. depicts an oil tank whose one side is open to atmosphere and the other side
sealed with air above the oil. Specific gravity of oil is 0.88. What is the pressure
at A and B?

(A) 0, 200N/m2
(B) 0. 25893 N/m2
(C) 200, 25893N/m2
(D) 25893, 0 N/m2
(Take density of water = 999.8 kg/m3)

11. In previous Q10, what are pressures at points C and D?

(A) 25893 N/m2, 51786.22 N/m2
(B) 51786.22 N/m2, 0
(C) 51786.22N/m2, 25893 N/m2
(D) 0,51786.22N/m2

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Vidyalankar : GATE – ME

12. In Q10, what are pressure at points E and F?

(A) 0, −12946.5 N/m2 (B) 0, 12946.5 N/m2
(C) 12946.5 N/m2, 25893 N/m2 (D) 25893N/m2, 12946.5 N/m2

13. In fig. what is the gauge pressure at point A if specific weight of mercury is
132.8 k N/m3?

(A) 20.28kN/m2
(B) 19.28kN/m2
(C) 29.28kN/m2
(D) 39.28kN/m2

Q14(a) & (b) carry two marks each

Linked Answer Question

14(a). A circular plate 2m in diameter is immersed in water, its greatest and least
depths below the surface being 3m and 2 m respectively. What is the position of
centre of pressure below the water surface?
(Take density of water = 999.8 kg/m3)
(A) 2.625m (B) 2.525m
(C) 1.625m (D) 3.525m

14(b). In previous part (part(a)), what is the total pressure on one face of the plate?
(A) 67.042kN (B) 57.042 kN
(C) 77.042 kN (D) 47.042 kN



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