slide 1: FLUID MECHANICS
Module – 1
Fluid Mechanics is a branch of science which deals with the behavior of the fluids liquids or
gases at rest as well as in motion. Thus this branch of science deals with the static kinematics and
dynamic aspects of fluids.
Fluid Statics :  The study of fluids at rest is called Fluid Statics.
Fluid Kinematics :  The study of fluids in motion with out considering the
Pressure force is called Fluid Kinematics.
Fluid Dynamics :  The study of fluids in motion considering the
Pressure force is called Fluid Dynamics.
Liquid:
Liquid has definite volume and is incompressible compression is extremely small.
Gas:
A gas has no definite volume and is compressible.
Properties of fluids
Fluid Pressure at a Point :
Consider a small area dA in large mass of fluid. Let dF is the force exerted by the fluid on the
area dA in the normal direction.
Then the ratio dF/dA is known as called the intensity of pressure or simply pressure and this ratio is
represented by ‘p’. Mathematically the pressure at a point in a fluid at rest is p dF/dA. If ‘F’ is the
force uniformly distributed over the area ‘A’ then pressure at any point is given by
F Force
p
A Area
\ Force or pressure force F p A ·
The units of pressure is N/m
2
in S.I Units and kgf/m
2
in MKS Unit
1 kgf 9.81 N
Relationship Between Pressures
Absolute Pressure : is the pressure which is measured with reference to Absolute Vacuum
Pressure.
Gauge Pressure : is the reading of the pressure gauge. Gauge Pressure is the pressure
which is measured with help of a pressure measuring instrument in
which atmospheric pressure is taken as datumreference.
Eg. If Gauge Indicates 50 N/m
2
where Atmospheric Pressure is
100N/m
2
then the Pressure is Expressed as either
50 N/m
2
Gauge or
150 N/m
2
Absolute
Vacuum Pressure : is defined as the pressure below the Atmospheric Pressure. Eg. Gauge
tapped into a tank reads 31 kPa where atmospheric pressure is 100 kPa
Then the pressure is expressed as either 31 kPa Gauge or
70 kPa Absolure
slide 2: 2
Density or Mass Density :
Density or Mass Density of a fluid is defined as the ratio of the mass of a fluid to its volume.
ie. Mass per Unit Volume is called Density. It is denoted by the symbol r rho. The unit of Density in
S.I Unit is kg/m
3
. The density of fluids may be considered as constant while that of gases changes with
the variation of pressure and temperature.
Mathematically mass density is given by
r Mass of fluid
Volume of Fluid
Specific Weight or Weight Density :
Specific Weight or Weight Density of a fluid is the ratio between weight of the fluid to its
volume. ie. Weight per unit volume of a fluid is called Weight Density. It is denoted by the symbol
w.
Mathematically
. Wt of the fluid mass of the fluid acceleration dueto gravity
w
Volumeof fluid Volume of fluid
·
w g r ·
The unit of Specific Weight or Weight Density in S.I Unit is N/m
3
. Value of Specific Weight or
Weight Density for water is 1000 · 9.8 N/m
3
.
Specific Volume :
Specific Volume of a fluid is defined as the volume of a fluid occupied by a unit mass of fluid.
Mathematically
1/
Volumeof fluid
Specific volume
mass of the fluid
r
Specific Gravity or Relative Density :
Specific Gravity is defined as the ratio of the Weight Density of a fluid to the Weight Density of a
standard fluid. For liquids standard fluid taken is water and for gases the standard fluid taken is air. It
is denoted by the symbol S.
Mathematically
slide 3: 3
Density of liquid
S for liquids
Density of water
Bulk Modulus :
Consider a cylinder fitted with a piston as shown in figure below.
Let V Volume of gas enclosed in the
cylinder
p Pressure of gas when volume is V
Let the pressure is increased to p+dp the volume of gas decreases from V to VdV.
Then the increase in pressure dp
Decrease in Volume  dV Negative sign means the volume decreases
with increase in pressure.
Volumetric Strain  dV
V
Bulk Modulus K Increase of Pressure
Volumetric Strain
\ Bulk Modulus
dp
K
dV V

Compressibility is the reciprocal of Bulk Modulus. ie. Compressibility 1/K
Problems :
1. Find the Bulk modulus of elasticity of a liquid which is compressed from 70 N/m
2
to 130 N/m
2
and
volume of liquid decreases by 0.15.
Sol
n
:
Initial Pressure 70 N/m
2
Final Pressure 130 N/m
2
Increase in Pressure 13070N/m
2
Volumetric Strain  dV 0.015/100 0.0015 m
3
V
Bulk Modulus K Increase of Pressure 13070 4 · 10
4
N/m
2
Volumetric Strain 0.0015
2. What is the Bulk Modulus of Elasticity of a liquid which is compressed from a volume of 0.0125
m
3
at 80 N/m
2
pressure to a volume of 0.0124 m
3
at 150 N/m
2
pressure.
Sol
n
:
Initial Pressure 80 N//m
2
Final Pressure 150 N/m
2
Increase in Pressure 15080N/m
2
Initial Volume 0.0125 m
3
Final Volume 0.0124 m
3
Volumetric Strain  dV 0.01250.0124 0.08 m
3
V 0.0125
slide 4: 4
Bulk Modulus K Increase of Pressure 70.0 875 N/m
2
Volumetric Strain 0.08
Cohesive Property :
The attractive forces between molecules in a liquid can be viewed as residual electrostatic
forces and are sometimes called Vander Waals forces or Vander Waals bonds.
Liquid Molecules experience strong intermolecular attractive forces. When those forces are
between like molecules they are referred to as cohesive forces. For example the molecules of a water
droplet are held together by cohesive forces
A liquid is said to have more cohesive property when the molecules composing the liquid have
greater attraction towards each other than the surface in contact with it. eg. Mercury taken in a glass.
Adhesive Property :
When the attractive forces are between unlike molecules they are said to be adhesive forces.
The adhesive forces between water molecules and the walls of a glass tube are stronger than the
cohesive forces lead to an upward turning meniscus at the walls of the vessel and contribute to
capillary action.
A liquid is said to have more adhesive property when the molecules composing the liquid have
greater attraction towards the surface in contact with it than towards each other. eg. Water taken in a
glass.
Viscosity :
Viscosity is defined as the property of a fluid which offers resistance to the movement of one
layer of fluid over the adjacent layer of the fluid. Viscosity is fundamentally due to the cohesion
between liquid particles.
It is assumed that liquid is composed of so many fluid element layers. And when fluid is in
motion each layer of fluid is in relative motion with the adjacent fluid element layer. The top layer
causes shear stress on the adjacent lower layer of fluid while the lower layer causes a shear stress on
the adjacent top layer.
Newton’s Law of Viscosity states that the shear stress tt t t on fluid element is directly
proportional to the rate of shear strain.
Let two layers of fluid at a distance of ‘dy’ distance apart move over the other at different velocities
say u and u+du as shown in the figure below.
According to the law
.
du du
ie
dy dy
t a t m
tt t t Tau represents Shear Stress
du is the Rate of Shear Strain or Rate of Shear
dy Deformation or Velocity Gradient.
mm m m is the constant of proportionality and is known as
Coefficient of Dynamic Viscosity or only
Viscosity.
u
This equation can be rewritten as
u
y
t a
u
y
t m ·
slide 5: 5
Effect of viscosity on Motion
Consider a thin layer of a liquid sandwiched between two flat parallel plates. When a force ‘F’
is applied to the upper plate the upper plate will move with a velocity of u with respect to the lower
plate.
y
F
Force Applied t · Area
F m · u · A
y
Units of Viscosity
2
2
2
/ distance
/
/ 1/
sec
cos
1
1
10
1 1/100
Shear stress
changeof velocity change of
Force Area
Length time Length
Force Time
Length
Newton
SI unit of vis ity
m
N s
poise P
m
centiPoise cP poise
m
· ·


Kinematic Viscosity
It is defined as the ratio between the dynamic viscosity and density of the fluid. It is denoted by the
symbol ν ‘nu’.
2
2
2
cos
cos
/
Vis ity
Density
The unit of kinematic vis ity is obtained as
Force Time
mass
Length
length
But forceis mass length Time
substituting and simplifying
Length
Time
m u r
·
· ·
2
. . / i e cm s is known as stockes
Also One Stock 10
4
m
2
/sec
slide 6: 6
1 CentiStock 1 Stock
100
Types of Fluids :
1. Ideal Fluid : Ideal fluid is a fluid which does not possess viscosity.
2. Real Fluid : Real Fluid is a fluid which does possess viscosity.
3. Newtonian Fluid : Newtonian Fluid is a fluid which obeys Newton’s Laws of viscosity.
4. NonNewtonian Fluid : NonNewtonian Fluid is a fluid which does not obey the Newton’s
Laws of viscosity.
Problems :
1. Two horizontal plates are placed 1.25 cm apart the space between them being filled with oil of
viscosity 14 poises. Calculate the shear stress in oil if the upper plate is moved with a velocity of
2.5 m/s.
Sol
n
: Distance between plates y 1.25 cm 0.0125 m
Viscosity m 14 P 1.4 Ns/m
2
Change in Velocity u 2.5 m/s – 0 2.5 m/s
Shear Stress t
t m u
y
t 1.4 · 2.5 280 N/m
2
0.0125
2. The space between two square flat parallel plates is filled with oil. Each side of the plate is 60 cm.
The thickness of the oil film is 12.5mm. The upper plate which moves at 2.5 meter per sec
requires a force of 10N to maintain the speed. Determine
1. the Dynamic Viscosity of the oil and
2. the Kinematic viscosity of the oil if the specific gravity of the oil is 0.95.
Sol
n
:
Each side of the square plate 60 cm 0.60 m
\ Area 0.6 x 0.6 0.36 m
2
Thickness of the oil film y 12.5 · 10
3
m
Velocity of the plate u u 2.5 m/s
\ Change of velocity between the plates 2.5 m/s – 0 2.5 m/s
Force required on the upper plate F 10N
\ Shear Stress Force/Area 10/0.36 N/m
2
3
2
2
10 12.5 10
13.8 10
0.36 2.5
u
N s
m
y
t m m   · ·  · · ·
Sfor Liquids Weight Density Density of Liquid .95 r · 9.8
Weight Density of DensityWater 1000· 9.8
slide 7: 7
\r 950 kg/ m
3
ν Viscosity m 13.8 · 10
2
14.5 · 10
4
m
2
/sec
1.45 stoke.
Density r 950
Surface Tension :
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. Surface Tension is due to the cohesive force between the liquid molecules. It
is denoted by the symbol σ. Its unit in S.I unit is N/m.
The cohesive forces between liquid molecules are responsible for surface tension. The molecules at
the surface do not have other like molecules on all sides of them. Molecules below the surface act on
each other by forces that are equal in all directions. Thus the resultant force acting on those molecules
is zero. Molecules on the surface of the liquid do not have any molecules above it and so resultant
force acting on it is not zero. So the molecules near the surface have greater attraction for each other
than the molecules below the surface. So the surface behaves like a membrane under tension.
Surface Tension for water/air surface at 20º C is 0.073 N/m. ie. It would take a force of 0.073
N to break a surface film of water 1 m long. The surface tension of water decreases significantly with
temperature.
Examples of Surface Tension
The major reason for using hot water for washing is that its surface tension is lower and it is a
better wetting agent. But if the detergent lowers the surface tension the heating may be unneccessary.
If carefully placed on the surface a small needle can be made to float on the surface of water
even though it is several times as dense as water. If the surface is agitated to break up the surface
tension then needle will quickly sink.
Surface Tension on Liquid Droplet
slide 8: 8
Consider a small spherical droplet of a liquid of radius ‘r’. Bubble is visualized as two
hemispheres and noting that the internal pressure which tends to push the hemispheres apart is
counteracted by the surface tension acting around the circumference of the circle.
Let the droplet is cut into two halves. The forces acting on one half are
1. tensile force due to surface tension acting around the cut portion which is equal to
σ · πd
2. pressure force on the area πd
2
which is equal to p · πd
2
4 4
These two forces must be equal and opposite under equilibrium conditions.
ie. σ · πd πd
2
4
\ p 4σ
d
Surface tension is responsible for the shape of liquid droplets. Although easily deformed
droplets of water tend to be pulled into a spherical shape by the cohesive forces of the surface layer.
The spherical shape minimizes the "wall tension" of the surface layer.
Surface Tension on a Hollow Bubble :
A hollow bubble has two surfaces in contact with air one inside and other outside. These two surfaces
are subjected to surface tension.
1. Tensile force due to surface tension which is equal to 2 d s p·
2. Pressure force on the area
2
4
d p which is equal to
2
4
d
p
p ·
These two forces must be equal and opposite under equilibrium conditions.
slide 9: 9
2
. .2
4
8
d
i e d p
p
d
p s p s · · \
Surface Tension on a Liquid Jet :
Consider a liquid jet of diameter ‘d’ and length ‘L’ as shown above.
Force due to pressure p · Area of semi jet p · L · d
Force due to surface tension σ x 2 x L
Equating the forces we have
p · L · d σ · 2 · L
\ p σ · 2 · L 2 σ
L · d d
Problems :
1. A soap bubble o 62.5 mm diameter has an internal pressure in excess of the outside pressure of 20
N/m
2
. What is the tension in the soap film
Sol
n
:
Diameter of the soap bubble d 62.5 · 10
3
m
p 20 N/m
2
Surface Tension σ
p 8σ 20 8 σ
d 62.5 · 10
3
\ σ 0.156 N/m
Capillarity :
Capillarity is defined as a phenomenon of rise or fall of a liquid surface in a small tube
relative to adjacent general level of liquid when the tube is vertically held in the liquid. The rise of
liquid surface is known as Capillary rise and fall of liquid 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 the liquid.
Expression for Capillary Rise :
slide 10: 10
Consider a glass tube of small diameter ‘d’ opened at both ends and is inserted in a liquid say water.
The liquid will rise in the tube above the level of liquid.
Let h height of the liquid in the tube. Under a state of equilibrium the weight of liquid of height h is
balanced by the force at the surface of the liquid in the tube. But the force at the surface of the liquid
in the tube is due to surface tension.
Let σ Surface Tension of liquid
θ Angle of contact between liquid and glass tube
The weight of liquid of height h in the tube
Area of tube x h · w
π d
2
· h · w w is the weight density of liquid
4
Vertical Component of the surface tensile force
σ · π · d · cos θ
Equating the forces we have
σ · π · d · cos θ π d
2
· h · w
4
h 4 σ cos θ
wd
The value of θ between water and clean glass tube is approximately equal to zero and hence cos θ is
equal to unity. Then the rise of water is given by
h 4 σ
wd
Expression for Capillary Fall :
slide 11: 11
If the glass tube is dipped in mercury the level of mercury in the tube will be lower than the
general level of the outside liquid as shown in the figure below.
Let h Height of depression in the tube.
Then in equilibrium two forces are acting on the mercury inside the tube. First one is due to surface
tension acting in the downward direction and is equal to
σ · π · d · cos θ
Second one is due to hydrostatic force acting upward and is equal to intensity of pressure at a depth ‘h’
p · πd
2
4
w · h · πd
2
4
Equating the two we get
σ · π · d · cos θ w · h · πd
2
4
h 4σ cos θ
wd
The value of θ for mercury and glass tube is 128°.
Problems :
1. Determine the minimum size of a glass tubing that can be used to measure water level if the
capillary rise in the tube is not to exceed 0.3 mm. Take surface tension of water in contact with air
as 0.0735 N/m.
Sol
n
:
slide 12: 12
Capillary Rise h 0.3 mm 0.3 · 10
3
m
Surface Tension σ 0.0735 N/m
Specific Weight of water 9810 N/m
3
Capillary Rise h 4 σ cos θ
Wd
d 4 · 0.0735 0.1 m
9810 · 0.3 · 10
3
2. Calculate the capillarity effect in mm in a glass tube of 4 mm diameter when immersed in water
and Mercury. Surface Tension of water and mercury are 0.735 N/m and 0.5096 N/m. The angle of
contact of water is zero and that of mercury is 130°. Specific gravity of mercury is 13.6.
Sol
n
:
1 Water
Surface Tension σ 0.0735 N/m
Specific Weight of water 9810 N/m
3
Glass Tube Diameter 4 · 10
3
m
Capillary Rise h 4 σ cos θ
wd
h 4 · 0.0735 0.00749 m
9810 · 4 · 10
3
2 Mercury
Surface Tension σ 0.5096 N/m
θ 130°
Specific Weight of water 9810 N/m
3
Sfor Mercury Weight Density Density of Mercury 13.6 r · 9.8
Weight Density of DensityWater 1000· 9.8
w r · 9.8
\ w 13600 N/m
3
Capillary Rise h 4 σ cos θ
wd
h 4 · 0.5096 · cos 130°  0.024 m
13600 · 4 · 10
3
Pascal’s Law :
Pascal’s Law states that pressure or intensity of pressure at a point in a static fluid is equal in
all directions. ie. Pressure acting at a point in xy and z directions on a static fluid is same.
slide 13: 13
Consider an arbitrary fluid element of wedge shape in a fluid mass at rest as shown in the figure
above. Let the width of the element be unity and p
x
p
y
and p
z
are the pressures or intensity of
pressures acting on the face AB AC and BC. Let — ABC θ then the forces acting on the element are
1. Pressure force normal to the surfaces
2. Weight of the element
The forces on the faces are:
Force on face AB p
x
· Area of face AB p
x
· dy · 1
Force on face AC p
y
· Area of face AC p
y
· dx · 1
Force on face BC p
z
· Area of face BC p
z
· dz · 1
Weight of the element Volume x Specific Weight volume · w
Wedge shape can be assumed to be of a rectangular prism which is cut along the diagonal of the base
into two equal halves. Therefore volume wedge shape will be half the volume of rectangular prism
which is Base Area x Height
Volume of Wedge Shape dy · dx · 1
2
Weight of the Wedge shape dy · dx · 1 · w
2
Resolving force in the xdirection
p
x
· dy · 1 – p
z
· dz · 1 · sin90º  θ 0
p
x
· dy · 1 – p
z
· dz · 1 · cosθ 0
But cos θ dy
dz
ie. dy dz · cos θ
Substituting dz · cos θ for dy in the above equation
p
x
· dy · 1 – p
z
· dy 0
p
x
· dy p
z
· dy
\ p
x
p
z
1
Resolving force in the ydirection
p
y
· dx · 1 – p
z
· dz · 1 · cos90º  θ dy x dx x 1 x w 0
2
p
x
· dy · 1 – p
z
· dz · 1 · sinθ  dy x dx x 1 x w 0
2
The element is small and hence the weight of the element is negligible.
\ p
x
· dy · 1 – p
z
· dz · 1 · sinθ 0
slide 14: 14
But sin θ dx
dz
ie. dx dz · cos θ
Substituting dz · sin θ for dx in the above equation
p
y
· dx – p
z
· dx 0
p
y
· dx p
z
· dx
\ p
y
p
z
2
From equation 1 and 2
p
x
p
y
p
z
3
Hence it is proved that pressure at a point is the same is all the directions.
Pressure Head :
The pressure at any point in the fluid at rest is obtained by the Hydrostatic Law which states that the
rate of increase of pressure in vertical downward direction must be equal to the specific weight at that
point. This is proved as:
Consider a small fluid element as shown in the figure below. Let
ΔA Crosssectional Area of the element
ΔZ Height the fluid element
p Pressure on the face AB
Z Distance of the fluid element from the free surface.
The forces acting on the fluid element are :
1. Pressure force on AB p · ΔA acting perpendicular to face AB in the downward direction
2. Pressure force on CD p + ∂p ΔZ ΔA acting perpendicular to face CD in vertically upward
∂z
direction
3. Weight of the fluid element Weight Density · Volume w · ΔA · ΔZ
4. Pressure forces on face BC and AD are equal and opposite.
For equilibrium of the fluid element
p · ΔA  p + ∂p ΔZ ΔA + w · ΔA · ΔZ 0
∂z
p · ΔA  p ΔA  ∂p ΔZ ΔA + w · ΔA · ΔZ 0
∂z
 ∂p ΔZ ΔA + w · ΔA · ΔZ 0
slide 15: 15
∂z
 ∂p ΔZ ΔA w · ΔA · ΔZ
∂z
∂p w 1
∂z
Equation 1 states that rate of increase of pressure in vertical downward direction must be equal to the
specific weight at that point.
By integrating the above equation for liquids
∫ dp ∫ w dZ
or p wZ
or Z p
w
Here Z is called Pressure Head.
Problems :
1. Find the pressure at a point which is 4 m below the free surface of water. Take specific gravity of
water as 9.8 KN/m
3
.
Sol
n
:
Given
Z 4 m
w 9800 N/m
3
Pressure at the point wZ 9800 · 4 39.2 kN/ m
2
2. Find the depth of oil of specific gravity 0.8 which will produce a pressure of 50 kN/m
2
.
Sol
n
:
Given
Specific Gravity 0.8 Specific Weight of the Liquid w
l
Specific Weight of the Standard Liquidw
w
\ w
l
0.8 · 1000 · 9.8 7.84 kN/ m
3
.
Pressure at the point p 50000 N/ m
2
p 50000 w
l
Z 7.84 · 1000 · Z
\ Z 6.38m.
3. A hydraulic press has a ram of 30 cm diameter and a plunger of 4.5 cm
diameter. Find the weight lifted by hydraulic press when the force applied at the plunger is 500 N.
Sol
n
:
Diameter of Ram D 30 cm 30 · 10
2
m
Diameter of Ram D 4.5cm 4.5 · 10
2
m
Force on Plunger F 500 N
Area of Ram A π D
2
π · 0.3
2
0.07068 m
2
4 4
Area of Plunger a π d
2
π · 0.045
2
0.00159 m
2
4 4
Pressure intensity due to plunger Force on Plunger F 500.00 314465.4 N/m
2
Area of Plunger a 0.00159
According to Pascal’s Law the pressure intensity will be equally transmitted in all directions.
Therefore the pressure intensity at the Plunger will be same as the pressure intensity at the Ram.
slide 16: 16
314465.4 N/m
2
But the pressure intensity at the Ram
Weight W 314465.4
Area of Ram A
W 314465.4 · 0.07068 22226.4 N
4 A tank filled with water for a depth of 7.5 m above a layer of mud of specific gravity 1.5. The
pressure at the bottom of the tank is found to be 450 kN/m
2
. Determine the thickness of the mud
layer.
Solution:
Pressure at the bottom of the tank is due to pressure exerted by the water layer and pressure
exerted by the layer of mud.
Let
P
Total
P
water
+ P
mud
450 kN/m
2
450 × 1000 N/m
2
450 × 1000 P
water
+ P
mud
P
water
w
w
h
w
1000 × 9.8 × 7.5
73500 N/m
2
P
mud
w
mud
h
mud
Given the Specific Gravity of the Mud w
mud
1.5
w
water
w
mud
1.5 × 1000 × 9.8
P
mud
1.5 × 1000 × 9.8 × h
mud
P
Total
73500 + 1.5 × 1000 × 9.8 × h
mud
450 × 1000 73500 + 1.5 × 1000 × 9.8 × h
mud
h
mud
26.5 m
Vapour Pressure:
The pressure at which a liquid will boil is called Vapour Pressure. All liquids have the
tendency to evaporate or vaporize. When a liquid enclosed with in a space is heated the space above
the liquid will be filled with these vapours and these then exert a pressure P
v
called Vapour Pressure
on the liquid. Also some of the molecules get reabsorbed into the liquid. When the number of
molecules getting reabsorbed is equal to number of molecules being released from the liquid the
Vapour Pressure exerted becomes a constant and this Vapour Pressure is called Saturated Vapour
Pressure. Increase of temperature hastens evaporation. When the pressure above the liquid becomes
the Vapour Pressure of the liquid ‘boiling’ occurs.
Eg. Water boils at a lesser temperature at lower atmospheric pressure Regional pressure regions like
hills.
Measurement of Pressure:
The pressure of a fluid is measured by the following devices:
1. Manometers
2. Mechanical Gauges
slide 17: 17
Manometers :
Manometers are defined as the devices used for measuring the pressure at a point in a fluid by
balancing the column of fluid by the same or another column of fluid.
Advantages of Manometers :
1. Inexpensive and easy to fabricate
2. Good accuracy
3. Require little maintenance
4. High Sensitivity
Disadvantages of Manometers :
1. Usually bulk and very large.
Manometers are classified as
1. Simple Manometers
2. Differential Manometers
Simple Manometers:
A Simple Manometer consists of a glass tube having one end connected to a point where
pressure is to be measured and other end remains open to the atmosphere. Common Types of Simple
Manometer are
1. Piezometer
2. UTube Manometer
3. Single Column Manometer
Piezometer :
Piezometer is the simplest form of manometers used for measuring gauge pressures. One end
of this manometer is connected to the point where the pressure is to be measured and other end is open
to atmosphere as shown in the figure below.
Working :
Due to the pressure at point A the liquid will rise in the tube to a height say ‘h’. The rise of
liquid is balanced by the weight of the liquid in the tube acting in the downward direction. If ‘w’ is the
weight density of the liquid the force due to weight of liquid in the tube
w · h . which is the pressure at point A.
slide 18: 18
Note:
If the liquid whose pressure is to be measured is replaced by ‘h’ height of water so as to have the same
pressure intensity then
w
water
· h
water
w
liquid
· h
liquid
h
water
w
liquid
· h
liquid
Specific Gravity of the liquid · h
liquid
w
water
UTube Manometer :
UTube Manometer consists of a glass tube bent in Ushape one end is connected to a point
at which the pressure is to be measured and other end remains open to the atmosphere. The tube
generally contains mercury or any other liquid whose specific gravity is greater than the specific
gravity of the liquid whose pressure is to be measured.
a For Gauge Pressure
Let
B is the point at which pressure is to be measured
‘h’ be the pressure head when the liquid is replaced by water
AA is the datum line.
h
1
Height of light liquid above the datum line.
h
2
Height of heavy liquid above the datum line.
S
1
Specific Gravity of the light liquid.
S
2
Specific Gravity of the heavy liquidsay Mercury.
Pressure at the datum line AA in the right column when the heavy liquidmercury is replaced by
water
h
2
· S
2
w
water
Pressure at the datum line AA in the left column when the heavy liquid is replaced by water
h + h
1
· S
1
w
water
Pressure at the datum line AA in the left column is same as pressure at the datum line AA in the right
column.
slide 19: 19
ie.
h + h
1
· S
1
w
water
h
2
· S
2
w
water
\ h h
2
S
2
 h
1
S
1
a For Vacuum Pressure
Let
B is the point at which pressure is to be measured
‘h’ be the pressure head when the liquid is replaced by water
AA is the datum line.
h
1
Height of light liquid above the datum line in the left column.
h
2
Height of heavy liquid above the datum line in the left column.
S
1
Specific Gravity of the light liquid.
S
2
Specific Gravity of the heavy liquidsay Mercury.
Pressure at the datum line AA in the left column when the liquids are replaced by water
h + h
1
· S
1
+
h
2
· S
2
w
water
Pressure at the datum line AA in the right column 0
Pressure at the datum line AA in the left column is same as pressure at the datum line AA in the right
column.
ie.
h + h
1
· S
1
+
h
2
· S
2
w
water
0
\ h  h
2
S
2
+ h
1
S
1
Problems :
1.The right limb of a simple UTube manometer containing mercury is open to the atmosphere while
the left limb is connected to a pipe in which a fluid of specific gravity 0.9 is flowing. The center of
the pipe is 12 cm below the level of mercury in the right limb. Find the pressure of fluid in the pipe
if the difference of mercury level n the two limbs is 20 cm. Take Specific Gravity of Mercury as
13.6.
Sol
n
:
slide 20: 20
Specific Gravity of fluid 0.9
Specific Gravity of mercury 13.6
Difference of mercury level h
2
20 cm 20 · 10
2
m
Height of the fluid from AA h
1
2012 8 cm 8 · 10
2
m
h Pressure head of fluid in terms of water head
h h
2
S
2
 h
1
S
1
20 · 10
2
· 13.6  8 · 10
2
· 0.9
2.648
m
Pressure wh 20 · 10
2
· 9810 25976.88 N/m
2
2.A simple UTube Manometer containing mercury is connected to a pipe in which a fluid of specific
gravity 0.8 and having vacuum pressure is flowing. The other end of the manometer is open to
atmosphere. Find the vacuum pressure in the pipe if the difference of mercury level in the two limbs
is 40 cm and the height of the fluid in the left limb from the center of the pipe is 15 cm below. Take
Specific Gravity of Mercury as 13.6.
Sol
n
:
Specific Gravity of fluid 0.8
Specific Gravity of mercury 13.6
Difference of mercury level h
2
40 cm 40 · 10
2
m
Height of the liquid in the left limb h
1
15 · 10
2
m
h Pressure head of fluid in terms of water head
h  h
2
S
2
+ h
1
S
1
40 · 10
2
· 13.6 + 15 · 10
2
· 0.8
556 · 10
2
m
Pressure wh 556 · 10
2
· 9810 54543.6 N/m
2
Single Column Manometer:
Single Column Manometer is a modified form of UTube Manometer in which a reservoir of
large cross sectional area as compared to the area of the tube is connected to the one of the limbs. The
right limb may be vertical or inclined. Thus there are two types of single column manometer. They are
slide 21: 21
1. Vertical Single Column Manometer
2. Inclined Single Column Manometer
Vertical Single Column Manometer:
Vertical Single Column Manometer consists of a reservoir which is connected to a point at
which the pressure is to be measured and other end connected to a limb. Other end of the limb remains
open to the atmosphere. The Vertical Single Column Manometer generally contains mercury or any
other liquid whose specific gravity is greater than the specific gravity of the liquid whose pressure is to
be measured.
Let XX be the datum line in the reservoir and in the right limb of the manometer. 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.
Let
∆h
Fall of heavy liquid in the reservoir
h
2
Rise of heavy liquid in the right limb
h
1
Height of center of pipe above XX
h Pressure Head at A which is to be measured
A CrossSectional Area of the reservoir
a Cross Sectional Area of the right limb
S1 Specific Gravity of the liquid in pipe
S2 Specific Gravity of the liquid in pipe
Fall of heavy liquid in the reservoir will cause a rise of heavy liquid level in the right limb.
A · ∆h
a · h
2
\ ∆h
a · h
2
A
Now consider the datum line YY. Pressure head in the right limb above YY
∆h
+ h
2
S
2
Pressure head in the left limb above YY
∆h
+ h
1
S
1
+ h
Equating two pressures
∆h
+ h
2
S
2
∆h
+ h
1
S
1
+ h
h ∆h
+ h
2
S
2
 ∆h
+ h
1
S
1
∆hS
2
 S
1
+ h
2
S
2
 h
1
S
1
But ∆h
a · h
2
slide 22: 22
A
h a · h
2
S
2
 S
1
+ h
2
S
2
 h
1
S
1
A
But a ratio is very small and so can be neglected.
A
Then
h h
2
S
2
 h
1
S
1
.
Inclined Single Column Manometer
Inclined Single Column Manometer consists of a reservoir which is connected to a point at
which the pressure is to be measured and other end connected to an inclined limb. Other end of the
limb remains open to the atmosphere. The Vertical Single Column Manometer generally contains
mercury or any other liquid whose specific gravity is greater than the specific gravity of the liquid
whose pressure is to be measured.
Let
∆h
Fall of heavy liquid in the reservoir
h
2
Vertical Rise of heavy liquid in the right limb
θ Inclination of right limb with the horizontal
h
1
Height of center of pipe above XX
h Pressure Head at A which is to be measured
A CrossSectional Area of the reservoir
a Cross Sectional Area of the right limb
S1 Specific Gravity of the liquid in pipe
S2 Specific Gravity of the liquid in pipe
L Length of heavy liquid moved in right limb from XX
Fall of heavy liquid in the reservoir will cause a rise of heavy liquid level in the right limb.
A · ∆h
a · h
2
\ ∆h
a · h
2
A
Now consider the datum line YY. Pressure head in the right limb above YY
∆h
+ h
2
S
2
Pressure head in the left limb above YY
slide 23: 23
∆h
+ h
1
S
1
+ h
Equating two pressures
∆h
+ h
2
S
2
∆h
+ h
1
S
1
+ h
h ∆h
+ h
2
S
2
 ∆h
+ h
1
S
1
∆hS
2
 S
1
+ h
2
S
2
 h
1
S
1
But ∆h
a · h
2
A
h a · h
2
S
2
 S
1
+ h
2
S
2
 h
1
S
1
A
But a ratio is very small and so can be neglected.
A
Then
h h
2
S
2
 h
1
S
1
.
But sin θ h
2
L
\ h L sin θ · S
2
 h
1
S
1
Differential Manometer :
Differential Manometers are devices used for measuring the difference of pressures between
two points in a pipe or in two different pipes. A Differential Manometer consists of a UTube
containing a heavy liquid whose two ends are connected to the points whose difference of pressures is
to be measured. Most common types of Differential Manometers are:
1. UTube Differential Manometer
2. Inverted UTube Differential Manometer
UTube Differential Manometer
A UTube Differential Manometer consists of a UTube containing a heavy liquid whose two
ends are connected to the points whose difference of pressures is to be measured. Let two points A and
slide 24: 24
B are at two levels and also contains oils of different specific gravities. These points are connected
to UTube Differential Manometer. Let the pressure at A and B are p
A
and p
B
respectively.
Two Pipes are at different levels containing two different liquids
Let
h Difference of mercury level in the UTube
y Distance of 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
h
A
Pressure Head at A
h
B
Pressure Head at B
S
1
Specific Gravity of liquid at A
S
2
Specific Gravity of liquid at B
S
g
Specific Gravity of the heavy liquid
Consider the datum line at XX
Pressure Head above XX in the left limb
h + x S
1
+
h
A
Pressure Head above XX in the right limb
h · S
g
+ y · S
2
+
h
B
Equating the two pressure head
h + x S
1
+
h
A
h · S
g
+ y · S
2
+
h
B
h
A

h
B
h · S
g
+ y · S
2
h + x S
1
hS
g
– S
1
+ y S
2
 xS
1
\ Difference of pressure head at A and B
h
A

h
B
hS
g
– S
1
+ y S
2
 xS
1
A and B are at the same level and contains the same liquid of Specific Gravity S
1
Let
h Difference of mercury level in the UTube
y Distance of center of B from the mercury level in the right limb
x Distance of the center of A and B from the mercury level in the limb.
h
A
Pressure Head at A
h
B
Pressure Head at B
S
1
Specific Gravity of liquid at A and B
S
g
Specific Gravity of the heavy liquid
Consider the datum line at XX
Pressure Head above XX in the right limb
h· S
g
+ x· S
1
+
h
B
Pressure head above XX in left limb
h + x S
1
+
h
A
slide 25: 25
Equating the two pressure head
h· S
g
+ x· S
1
+
h
B
h + x S
1
+
h
A
h
A

h
B
hS
g
+ xS
1
+h + x S
1
hS
g
 S
1
Meters Of Water
Inverted UTube Differential Manometer:
Inverted UTube Differential Manometers are used for measuring the difference of low
pressures. It consists of an inverted UTube containing a light liquid. The two ends of the tube are
connected to the points whose difference of pressures is to be measured.
Let the pressure at A is more than the pressure at B.
h Difference of the light liquid
h
1
Height of liquid in the left limb below the datum line XX
h
2
Height of the liquid in the right limb
h
A
Pressure Head at A
h
B
Pressure Head at B
S
1
Specific Gravity of liquid at A
S
2
Specific Gravity of liquid at B
S
Specific Gravity of the light liquid
Consider the datum line at XX
Pressure Head in the left limb below XX
h
A
 S
1
h
1
Pressure Head in the right limb below XX
h
B
 S
2
h
2
Sh
Equating the two pressure heads
h
A
 S
1
h
1
h
B
 S
2
h
2
Sh
h
A
 h
B
S
1
h
1
 S
2
h
2
Sh
Problems :
1. A pipe contains an oil of specific gravity 0.8. A differential manometer connected at the two points
A and B of the pipe shows a difference in mercury level as 20 cm. Find the difference of pressure
head at the two points.
slide 26: 26
Sol
n
:
Given Specific Gravity of oil 0.8
Difference of mercury level 20 cm 0.20 m
Specific Gravity of Mercury 13.6
Pressure Head above XX in the right limb
h · S
g
· w
w
+ x · S
1
· w
w
+
H
B
Pressure Head above XX in left limb
h + x S
1
· w
w
+
H
A
Equating the two pressure heads
h + x S
1
· w
w
+
H
A
h · S
g
· w
w
+ x · S
1
· w
w
+
H
B
H
A
 H
B
h S
g
 S
1
0.2013.6 – 0.8 2.56 meters of water.
2. A pipe contains an oil of specific gravity 0.9. A differential manometer connected at the two points
A and B shows a difference in mercury level as 15 cm. Find the difference of pressure head at the
two points.
Sol
n
:
Given Specific Gravity of oil 0.9
Difference of mercury level 15 cm 0.15 m
Specific Gravity of Mercury 13.6
Pressure above XX in the right limb
h · S
g
· w
w
+ x · S
1
· w
w
+
P
B
slide 27: 27
Pressure above XX in left limb
h + x S
1
· w
w
+
P
A
Equating the two pressures
h + x S
1
· w
w
+
P
A
h · S
g
· w
w
+ x · S
1
· w
w
+
P
B
P
A
 P
B
h S
g
 S
1
0.1513.6 – 0.9 1.905 meters of water.
3. A differential UTube manometer is connected at two points A and B of two pipes as shown in the
figure below. The pipe A contains a liquid of specific gravity 1.5 while pipe B contains a liquid of
specific gravity 0.9. The pressure at A and B are 9.8 · 10
4
N/m
2
and 17.64 · 10
4
N/m
2
. Find the
difference in mercury level in the differential manometer.
Sol
n
:
Given
Specific gravity of Liquid at A S
a
1.5
Specific gravity of Liquid at B S
b
0.9
Pressure at A P
a
9.8 · 10
4
N/m
2
Pressure at B P
b
17.64 · 10
4
N/m
2
Taking XX as datum line.
Pressure at XX in the left limb
h · 13.6 · w
w
+ 2 · 1.5 · w
w
+ 3 · 1.5 · w
w
+ P
a
h · 13.6 · w
w
+ 2 · 1.5 · w
w
+ 3 · 1.5 · w
w
+ 9.8 · 10
4
Pressure at XX in the right limb
h · 0.9 · w
w
+ 2 · 0.9 · w
w
+ P
b
h · 0.9 · w
w
+ 2 · 0.9 · w
w
+ 17.64 · 10
4
Equating the two pressures
h · 13.6· w
w
+2 · 1.5· w
w
+3· 1.5· w
w
+9.8· 10
4
h · 0.9· w
w
+2· 0.9 · w
w
+17.64 · 10
4
h 0.181 m
4. A differential manometer is connected at two points A and B as shown in figure below. At B air
pressure is 9.8 · 10
4
N/m
2
absolute find the absolute pressure at A.
slide 28: 28
Sol
n
:
Air pressure at B 0.98 · 10
4
N/m
2
Let the pressure at A be P
a
and pressure at B be P
b
Taking XX as datum line.
Pressure at XX in the right limb
0.60 · w
w
+ P
b
0.60 · w
w
+ 0.98 · 10
4
Pressure at XX in the left limb
0.10 · 13.6 · w
w
+ 0.20 · 0.9 · w
w
+ P
a
Equating the two pressures
0.60 · w
w
+ 0.98 · 10
4
0.10 · 13.6 · w
w
+ 0.20 · 0.9 · w
w
+ P
a
P
a
588 N/m
2
absolute
5. The pressure between two points M and N in a pipe conveying oil of specific gravity 0.9 is
measured by an inverted UTube and the column connected to point N stands 1.5 m higher than
that at point M. A pressure gauge attached directly to the pipe at M reads 20 N/cm
2
. Determine the
pressure at N.
Sol
n
:
Let the pressure at M be P
m
and pressure at N be P
n
Air pressure at P
m
20 N/cm
2
20 · 10
4
N/m
2
Taking XX as datum line.
Pressure at XX in the left limb attached to point N
 0.40 · 0.9 · w
w
+ P
N
slide 29: 29
Pressure at XX in the right limb attached to point M
 0.40 · 13.6 · w
w
+ P
M
 0.40 · 13.6 · w
w
+ 20 · 10
4
Equating the two pressures
 0.40 · 0.9 · w
w
+ P
N
 0.40 · 13.6 · w
w
+ 20 · 10
4
P
N
 0.40 · 13.6 · w
w
+ 0.40 · 0.9 · w
w
P
N
 53312
+ 3528  49784 N/m
2
6. An inverted UTube manometer containing an oil of specific gravity 0.9 is connected to find the
pressure difference of two points of a pipe containing water. If the manometer reading is 40 cm
find the difference of pressures.
Sol
n
:
Let the pressure at left limb be P
L
and pressure at right limb be P
R
Taking XX as datum line.
Pressure at XX in the left limb
 0.40 · 0.9 · w
w
+ P
L
Pressure at XX in the right limb
 0.40 · 13.6 · w
w
+ P
R
Equating the two pressures
 0.40 · w
w
+ P
L
 0.40 · 0.9 · w
w
+ P
R
P
L
 P
R
 0.40 · 0.9 · w
w
+ 0.40 · w
w
P
L
 P
R
392 N/m
2
Mechanical Gauges :
Mechanical Gauges are devises used for measuring the pressure by balancing the fluid column
by spring or dead weight. Mechanical Gauges are suitable for measuring medium and high pressures.
The commonly used mechanical pressure gauges are :
slide 30: 30
1. Diaphragm Pressure Gauge
2. Bourdon Pressure Gauge
3. Dead Weight Pressure Gauge
4. Bellows Pressure Gauge
Bourdon Pressure Gauge:
When atmospheric pressure prevails in the gauge the tube is undeflected and in this condition the
pointer is calibrated to read zero pressure. When pressure is applied to the gauge the tube tends to
straighten thereby actuating the pointer to read the corresponding pressure.
Pressure on Immersed Surfaces :
Total Pressure :
Total Pressure is defined as the force exerted by a fluid on a surface either plane or curved
when the fluid comes in contact with the surfaces. This force always acts normal to the surfaces.
Center of Pressure :
Center of Pressure is defined as the point of application of total pressure on the surface.
There are four cases of submerged surfaces on which the total pressure force and center of pressure is
to be determined. They are:
1. Vertical Plane Surface
2. Horizontal Plane Surface
3. Inclined Plane Surface
4. Curved Surface
Vertical Plane Surface :
Consider a plane vertical surface of arbitrary shape immersed in liquid as shown in figure.
slide 31: 31
Let
A Total Area of the surface
_
h Distance of the Center of Gravity of the Area from the free surface of the liquid.
p Center of Pressure
h Distance of center of pressure from the free surface of liquid
a Total Pressure
Total Pressure on the surface may be determined by dividing the entire surface into a number
of small parallel strips. The force on the parallel strip is then calculated and the total pressure force on
the whole area is calculated by integrating the force on the small strip.
Consider a strip of thickness dh and width of b at a depth of h from the free surface.
Pressure intensity on the strip p wh
Area of the strip dA b · dh
Total Pressure Force on the strip dF p · Area
wh · b· dh
\ Total Pressure on the whole surface
F ∫ dF ∫ wh · b · dh w ∫ h · b · dh
But
∫ h · b· dh ∫ h dA
Moment of surface area about the free surface of the liquid
Area of the surface · Distance of Center of Gravity form the free surface
_
A · h
_
\ F wAh
b Center Pressure
Center Pressure is calculated using the ‘Principal of Moments’ which states that the moment of
the resultant force about an axis is equal to the sum of moments of the components about the same
axis.
The resultant force is acting at P at a distance h from free surface of the liquid. Moment of force F
about the free surface of the liquid F · h 1
slide 32: 32
Moment of force dF acting on a strip about the free surface of the liquid
dF · h dF wh · b · dh
Sum of all such moments about free surface of the liquid
∫ wh · b · dh · h
w ∫ h · b · dh · h
w ∫ bh
2
dh
w ∫ h
2
dA bdh dA
But
∫ bh
2
dh ∫ h
2
dA ∫ h · dA · h
Moment of Inertia of the surface about the free surface of the liquid.
I
0
\ Sum of moment s about free surface
wI
0
2
Equating equation 1 and 2 we get
F · h h · dA · h
But _
F wAh
_
wAh · h wI
0
\ h wI
0
3
wA h¯
_
I
0
I
G
+ A · h
2
I
G
Moment of Inertia of the area about an axis passing through the Center Of Gravity of the area and
parallel to the free surface of liquid.
Substituting I
0
in equation 3
_ _
\ h I
G
+ A · h
2
I
G
+ h
Ah¯ Ah¯
From the equation it is clear that
1. The Center of Pressure lies below the Center of Gravity of the Vertical Surface.
2. The Distance of Center Of Pressure from the free surface of liquid is independent of the specific
weight of the liquid.
slide 33: 33
Problems:
1.A rectangular plane surface is 2m wide and 3m deep. It lies in a vertical plane in water. Determine
the total pressure and position of center of pressure on the plane surface when its upper edge is
horizontal and a coincide with water surface b 2.5m below the free water surface.
Sol
n
:
Width of the plane surface b 2m
Depth of plane surface d 3m
a Upper Coincide with water surface
slide 34: 34
_
F wAh
w 9810 N/m
3
A 3 · 2 6 m
2
_
h Center Of Gravity from the base for a rectangular plane surface d/2
0.5 · 3 1.5m
F 9810 · 6 · 1.5
88920 N.
Depth of Center Of Pressure
_
h I
G
+ h
Ah¯
I
G
bd
3
12
2 · 3
3
4.5 m
4
12
h 4.5 + 1.5 2.0 m
6 · 1.5
bUpper Edge is 2.5m below the water surface
_
F wAh
w 9810 N/m
3
A 3 · 2 6 m
2
_
h Center Of Gravity from the base for a rectangular plane surface 2.5 + d/2
2.5 + 0.5 · 3 4m
slide 35: 35
F 9810 · 6 · 4
235440 N.
Depth of Center Of Pressure
_
h I
G
+ h
Ah¯
I
G
bd
3
12
2 · 3
3
4.5 m
4
12
h 4.5 + 4 4.1875 m
6 · 4
2.Determine the total pressure on a circular plate of diameter 1.5 m 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.
Sol
n
:
Diameter of the plate d 1.5m
Area A π1.5
2
1.767 m
2
4
_
h 3.0 m
_
Total Pressure F wAh 9810 · 1.767 · 3.0 52002.81 N.
Position of Center Of Pressure
h I
G
+ h
Ah¯
I
G
πd
4
64
π · 1.5
4
0.2485m
4
64
h 0.2485 + 3.0 3.0468m
1.767 · 3.0
Horizontal Plane Surface Immersed in Liquid :
Consider a horizontal plane surface immersed in a static fluid. Every point on the surface is at the same
depth from the free surface of the liquid the pressure intensity will be equal on the entire surface and
equal to p wh where h is depth of the surface.
slide 36: 36
Let A Total area of the surface
Then total force F on the surface
wh · Area whA
_
wAh
_
h Depth of Center of Gravity from the free surface h
h Depth of center of pressure form the free surface h
Problem:
1.Figure below shows a tank full of water. Find:
1 Total pressure on the bottom of the tank
2 Weight of water in the tank.
3 Hydostatic Paradox between results of 1 and 2
Width of tank is 2m.
Sol
n
:
Depth of tank on the bottom of the tank h
1
3 + 0.6 3.6m
Width of the tank 2m
Length of the tank at the bottom 4m
A 4 · 2 8m
2
1 Total Pressure F on the bottom is
_
F wAh
9810 · 8 · 3.6
282528 N
2 Weight of water in tank
w · Volume of tank
9810 · 3 · 0.4 · 2 + 4 · .6 · 2
70632 N
slide 37: 37
From the results of 1 and 2 it can be seen that the total weight of the water in the tank is much
less than the total pressure at the bottom of the tank. This is known as Hydrostatic Paradox.
Inclined Plane Surface Immersed in Liquid :
Consider a plane surface of arbitrary shape immersed in liquid in such a way that the plane of the
surface makes an angle θ with the free surface of the liquid.
A Total Area of the surface
_
h Distance of the Center of Gravity of the Area from the free surface of the liquid.
p Center of Pressure
h Distance of center of pressure from the free surface of liquid
θ Angle made by the plane of the surface with the free liquid surface.
Let the plane of the surface if produced meet the free liquid surface at O. Then OO is the axis
perpendicular to the plane of the surface.
Let
_
y Distance of the Center Of Gravity of the inclined surface from OO.
y Distance of Center Of Pressure from OO.
a Total Pressure
Total Pressure on the surface may be determined by dividing the entire surface into a number
of small parallel strips. The force on the parallel strip is then calculated and the total pressure force on
the whole area is calculated by integrating the force on the small strip.
Consider a strip of thickness dh and width of b at a depth of h from the free surface and at a
distance y from the axis OO.
pressureintensity on thestrip p wh
Area of the strip dA b dh
Total pressure force on the strip dF p Area
wh b dh
Total pressure on the wh
· · · · \
ole surface F dF wh b dh w h b dh
But sin
h h
y y
q · · · ·
∫ ∫ ∫
slide 38: 38
_
h h h sin θ
y y¯ y
h y sin θ
\ F ∫ dF ∫ wh · b · dh w ∫ y sin θ · b · dh w sin θ ∫ y · b · d
But
∫ y · b· dh ∫ y dA
Moment of surface area about the free surface of the liquid
Area of the surface · Distance of Center of Gravity form the free surface
_ _
A · y y Distance of Center Of Gravity from Axis OO
_ _ _
\ F wAh h y sin θ
b Center Pressure
Center Pressure is calculated using the ‘Principal of Moments’ which states that the moment of
the resultant force about an axis is equal to the sum of moments of the components about the same
axis.
The resultant force is acting at P at a distance y about the axis OO.
Moment of force F about the axis OO F · y 1
Pressure Force on the strip
dF
wh · b · dh dF wh · b · dh
wy · sin θ · dA h y sin θ bdh dA
Moment of force dF acting on a strip about the axis OO
dF · y
wy · sin θ · dA · y
wy
2
sin θ · dA
Sum of all such moments about free surface of the liquid
∫ wy
2
sin θ · dA
w sin θ ∫ y
2
· dA
w ∫ y · dA · y
But
∫ y · dA · y
Moment of Inertia of the surface about OO
I
0
\ Sum of moment s about free surface
w · sin θ · I
0
2
Equating equation 1 and 2 we get
slide 39: 39
F · y w · sin θ · I
0
But _
F wAh and y h
_ sin θ
wAh · h w sin θ I
0
sin θ
\ h wI
0
sin
2
θ 3
wA h¯
_
I
0
I
G
+ A · y
2
I
G
Moment of Inertia of the area about an axis passing through the Center Of Gravity of the area and
parallel to the free surface of liquid.
Substituting I
0
in equation 3
_ _
_ y h
\ h wsin
2
θ I
G
+ A · y
2
sin θ
wA h¯
_
w · sin
2
θ· I
G
+ A · h
2
sin
2
θ
w Ah¯ w· A h¯ sin
2
θ
_
h I
G
sin
2
+ h
Ah¯
Buoyancy and Floatation :
When a body is immersed in a fluid an upward force is exerted by the fluid 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 denoted by the letter F
b
.
If Weight of the Body is greater than Force of Buoyancy the body will sink.
If Weight of the Body is equal to Force of Buoyancy the body will just float in the Liquid. ie. Top
surface of the body will coincide with the fluid’s free surface.
If Weight of the Body is less than Force of Buoyancy the body will float in the Liquid.
slide 40: 40
Centre of Buoyancy:
It is defined as the point through which force of buoyancy is supposed to act. The center of
buoyancy will be the center of gravity of the fluid displaced as the force of buoyancy is a vertical force
and is equal to the weight of the fluid displaced by the body.
Problems:
1.Find the volume of water displaced and position of centre of buoyancy for a wooden block of width
2.5m and a depth 1.5m when it floats horizontally in water. The density of wooden block is
650kg/m
3
and its length is 6.0m.
Sol
n
:
Width 2.5m
Depth 1.5m
Length 6.0 m
Volume of the block 2.5 · 1.5 · 6.0 22.50 m
3
Density of wood 650
Weight of the block r · g · volume
650 · 9.81 · 22.50
143471 N
For equilibrium the weight of the water displaced Weight of the wooden Block 143471 N
Volume of water displaced
Weight of water displaced 143471 14.625 m
3
Weight Density of water 9810
Position of Centre of Buoyancy
Volume of wooden block in water Volume of water displaced
h Depth of wooden block in water
2.5 · h · 6.0 14.625 m
3
h 0.975 m
\ Centre of buoyancy 0.975/2 0.4875 m from base.
2. A wooden block 2 m · 1 m · 0.5 m and a specific gravity of 0.76 is floating in water. What load
may be placed on the block so that it may completely immerse in water.
Sol
n
:
Given Volume of Block 2 · 1 · 0.5 1 m
3
Specific Gravity of Wood 0.76
Let W Weight placed on the block of wood.
Weight of Wooden Block
w
wood
0.76
w
water
\ w
wood
0.76 · w
water
0.76 · 1000 · 9.8
slide 41: 41
w
wood
W
wood
0.76 · 1000 · 9.8
V
\ W
wood
0.76 · 1000 · 9.8 · V
0.76 · 1000 · 9.8 · 1
7.45 kN
Total Weight Acting Downwards
7.45 kN + W kN
and the volume of water displaced when the block is completely immersed in it 1 m
3
Upwards thrust when the block is completely immersed
w
water
1000 · 9.8 W
water
W
water
Volume of water 1
\ W
water
9.8 kN
Now equating the total downward weight and upwards thrust
7.45 kN + W kN 9.8 kN
\ W 9.8 kN – 7.45 kN 2.35 kN
MetaCentre :
It is defined as the point about which a body starts oscillating when the body is tilted by a small
angle. MetaCenter is also 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 a small angular displacement.
Consider a floating body in a liquid as shown in figure. Let the body is in equilibrium and G is
the Centre of Gravity and B the Centre of 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 clock wise direction. The center of
buoyancy which is the Center of Gravity of the portion of the body submerged in liquid will now be
shifted towards right from the normal axis. Let it be at B
1
. The line of action of the force of buoyancy
in this new position will intersect the normal axis of the body at same point say M. This point M is
called MetaCenter.
MetaCentric Height:
The distance GM ie. the distance between the metacenter of a floating body and the center of
a floating body and the Centre of Gravity of the body is called MetaCentric Height.
Meta Centric Height GM BM + BG
slide 42: 42
+ sign is used if G is lower than B and – sign is to be used if G is higher than B.
BM Distance between MetaCenter and Center of Buoyancy BG Distance between Meta
Center and Center of Gravity
BM I/V Moment of Inertia Of the Plane
Volume of water displaced
Note:
Consider a block of Specific Gravity S
b
and Height H
b
immersed in a liquid of Specific Gravity S
l
.
Then the depth of immersion of the block in the liquid h
Weight of Liquid Displaced Weight of Block
r liquid
· Volume of Liquid Displaced · 9.8 r block
· Volume of Block · 9.8
r liquid
· Base Area · Height of Liquid Displaced · 9.8 r block
· Base Area · Height of Block · 9.8
r liquid
· Base Area · h of Liquid Displaced · 9.8 r block
· Base Area · H
b
of Block · 9.8
Base Area of Liquid Displaced and Base Area of the Block are same
\ r liquid
· h r block
· H
b
h r block
· H
b
r liquid
r block
· r water
· H
b
r liquid
r water
r block
· r water
· H
b
r water
r liquid
h S
b
· H
b
S
l
If the block is immersed in water then the depth of immersion
h S
b
· H
b
S
w
h S
b
· H
b
Problems:
1. A block of wood of specific gravity 0.8 and size 1.2m · 0.4m · 0.3m floats in water. Determine its
metacentric height for tilt about its longitudinal axis.
Sol
n
:
slide 43: 43
Specific Gravity of wood 0.8
l 1.2 m
b 0.4 m
d 0.3 m
Depth of immersion of the block
h S
b
· H
b
0.8 · 0.3 0.24 m
Distance of Centre of Buoyancy from the bottom of the block
OB 0.24/2 0.12 m
Distance of Centre of Gravity from the Bottom of the Block
OG 0.3/2 0.15 m
BG OG – OB 0.15 – 0.12 0.03 m
Moment of Inertia of the rectangular section about the central axis and parallel to the long side
I lb
3
12
1.2 · 0.4
3
0.0064 m
4
Volume of water Displaced 1.2 · 0.4 · 0.24 0.1152 m
3
BM I/V 0.0064 0.056 m
3
0.1152
MetaCentric Height BM  BG 0.056 – 0.03 0.026 m
Conditions of Equilibrium of a 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 GravityG and Centre Of
BuoyancyB
1
of a body determines the stability of a submerged body.
Stability of a submerged body :
slide 44: 44
The position of a Centre of GravityG and Centre Of BuoyancyB
1
in case of a completely
submerged body are fixed.
Consider a balloon which is completely submerged in air. Let the lower portion of the balloon
contains a heavy material so that its Centre of Gravity is lower than its Centre of Buoyancy.
Let the weight of the balloon is W. The weight W is acting through G vertically in the downward
direction while the buoyant force F
B
upward direction through B.
For the equilibrium of the balloon W F
B
.
If the balloon is given a small angular displacement in the clockwise direction then W and F
B
constitute a couple acting in the anticlockwise direction and brings the balloon in the original
position. The balloon in this position is said to be in stable equilibrium.
a Stable Equilibrium
When W F
B
and Centre Of BuoyancyB is above Centre Of GravityG the body is said to
be in stable equilibrium.
b Unstable Equilibrium
When W F
B
and Centre Of BuoyancyB is below Centre Of GravityG the body is said to
be in stable equilibrium.
A slight displacement to the body in the clockwise direction gives the couple due to
W and F
B
also
in the clockwise direction and hence the body is in Unstable equilibrium.
c Neutral Equilibrium
If W F
B
and B and G are in the same point the body is said to be in Neutral Equilibrium.
Stability of a Floating Body :
a Stable Equilibrium
If the point M is above G the floating body will be in Stable Equilibrium. If a slight
displacement is given to the body in the clockwise direction the Centre Of Buoyancy shifts from B to
B
1
such that vertical line through B1 such that vertical line through B
1
cuts at M. Then the buoyant
slide 45: 45
force F
B
through M and weight W through G constitute a couple acting in the anticlock direction
and thus bringing the floating body in the original position.
b Unstable Equilibrium
If the point M is below G the floating body will be in unstable equilibrium. The disturbing
couple is acting in the clockwise direction. The couple due to buoyant force F
B
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 Neutral
Equilibrium.
Problems:
1. A cylinder has a diameter of 0.3m and specific gravity of 0.8. What is the maximum permissible
length in order that it may float in water with its axis vertical
Sol
n
:
Given that
Cylinder Diameter d 0.3m
Specific Gravity of Cylinder S
c
0.8
Let l be the height of the cylinder then
Depth of immersion of the cylinder
h S
c
· H
0.8 · l
Distance of Center of Buoyancy from the bottom of the cylinder
OB 0.8 · l
2
Distance of Center of Gravity from the bottom of the cylinder
OG l/2
BG OG – OB
l  0.8l 0.1l
2 2
Moment of Inertial of the Circular Section about its center of gravity
I πd
4
64
Volume of Water Displaced
4
2
d2
V 0.8
4
I d /64
BM
V
0.8
4
d2
16 0.8
l
d
l
l
p p p ·
· ·
slide 46: 46
For Stable Equilibrium the MetaCenter should be above the Center of Gravity G or may coincide
with G.
i.e
BG BM
0.1l d
2
16· 0.8l
l
2
d
2
16 · 0.8 · 0.1
l d
1.28
l 0.8838 · d
This means that the cylinder cannot float with its longitudinal axis vertical when the length
exceeds 0.8838 times of its diameter.
2. A wooden cylinder of a circular section and uniform density with specific gravity of 0.6 is required
to float in an oil film of specific gravity 0.9. If the cylinder has a diameter d and length of l
show that l cannot exceed 0.75d for the cylinder to float with its longitudinal axis vertical.
Sol
n
:
Given that
Cylinder Diameter d 0.3m
Specific Gravity of Cylinder S
c
0.6
Specific Gravity of Oil S
o
0.9
Let l be the height of the cylinder then
Depth of immersion of the cylinder
h S
c
· H
S
o
0.6 · l 2l
0.9 3
Distance of Center of Buoyancy from the bottom of the cylinder
OB 1 · 2l l
2 3 3
Distance of Center of Gravity from the bottom of the cylinder OG l/2
BG OG  OB
2 3 6
l l l

slide 47: 47
Moment of Inertial of the Circular Section about its center of gravity
I πd
4
64
Volume of Water Displaced
V πd
2
· 2l
4 3
BM I/V πd
4
64
πd
2
· 2l
4 3
32d
2
32l
For Stable Equilibrium the MetaCenter should be above the Center of Gravity G or may coincide
with G.
i.e
BG BM
l 32d
2
6 32l
l
2
18d
2
32
l
2
9d
2
16
l 3 d
4
l 0.75 · d
This means that the cylinder cannot float with its longitudinal axis vertical when the length
exceeds 0.75 times of its diameter.
3. A solid cylinder of 3 meters diameter has a height of 3 meters. It is made up of a material whose
specific gravity is 0.8 and is floating in water with its axis vertical. Find its metacenter and state
whether its equilibrium is stable or unstable.
Solution:
Given Diameter 3m
Height H
c
3m
Specific Gravity of Cylinder S
c
0.8
Depth of immersion of the cylinder h S
c
× H
c
0.8 × 3 2.4 m
slide 48: 48
Distance of Center of Center of
Buoyancy From the bottom of the
Cylinder OB 2.4/2 1.2 m
Distance of Center of Center of
Gravity from the bottom of the
Cylinder OG 3/2 1.5 m
BG OG – OB 1.5 – 1.2 0.3 m
Moment of Inertia of the Circular
Section I π d
4
/64 π 3
4
/64
1.27 π m
4
Volume of water Displaced π d
2
× h π 3
2
× 2.4 5.4 π m
3
4 4
BM I/V 1.27 π / 5.4 π 0.235 m
And MetaCentric Height
GM BM – BG 0.235 – 0.3 0.065 m
Minus Sign means that the Meta Center is below the Center of Gravity. Therefore the cylinder is in
unstable equilibrium.
4 A wooden cone of Specific Gravity 0.8 is required to float vertically in water. Determine the least
apex angle which shall enable the cone to float in stable equilibrium.
Note: Center of Buoyancy of a cone is at ¾ distance of the length of the cone from the Apex.
Solution:
Given the Specific Gravity of the Cone w
c
0.8
w
w
w
c
ρ
c
× 9.8 0.8 × 1000 × 9.8
ρ
c
0.8 × 1000
Length of the Cone L
Length of the cone immersed in water l
Radius of the Cone R
Diameter of the Cone D
Radius of the Cone at the liquid level r
Diameter of the Cone at the liquid level d
Apex Angle 2α
slide 49: 49
tan α R r
L l
R L × tan α
Volume of the Cone 1 π R
2
L
3
1 × π × L × tan α
2
× L
3
Weight of Cone ρ
c
× Volume of the Cone × 9.8
0.8 × 1000 × 1 × π × L
3
tan
2
α
× 9.8
3
Weight of Water Displaced ρ
w
× Vol. Cone Displaced by
water × 9.8
1000 × 1 × π × l
3
tan
2
α × 9.8
3
Weight of the cone Weight of water displaced
0.8 × 1000 × 1 × π × L
3
tan α
3
× 9.8 1000 × 1 × π × l
3
tan
2
α
× 9.8
3 3
0.8 × L
3
l
3
l 0.8
1/3
× L
Distance of Center of Buoyancy from
the Apex OB 0.75 l
0.75 × 0.8
1/3
× L
Distance of Center of Gravity from
the apex OG 0.75L
Volume of Liquid Displaced V 1 × π × l
3
tan
2
α
3
Moment of Inertia of the Circular
Section about the liquid level I π × d
4
64
But r l × tan α
So d 2 × l × tan α
I π × 2 × l × tan α
4
64
π × l
4
tan
4
α
4
BM I / V
slide 50: 50
π × l
4
tan α
4
4_________
1 × π × l
3
tan
2
α
3
0.75 × l × tan
2
α
For stable equilibrium the Meta Centric should be above Center of Gravity.
i.e
BG BM
OG – OB BM
0.75L  0.75 × 0.8
1/3
× L 0.75 × l × tan
2
α
L × 0.75 1 – 0.8
1/3
0.75 × 0.8
1/3
× L × tan
2
α
tan
2
α 1 – 0.8
1/3
0.8
1/3
tan
2
α 0.08
α 15º 37’
5 A conical buoy 1 meter long and of base diameter 1.2 meter floats in water with its apex
downwards. Determine the minimum height of the buoy for stable equilibrium. Take Specific
weight of water as 1000 × 9.8 N/m
2
Note: Center of Buoyancy of a cone is at ¾ distance of the length of the cone from the Apex.
Solution:
Length of the Cone L 1 m
Length of the cone immersed in water l
Radius of the Cone at the liquid level r
Diameter of the Cone D 1.2 m
tan α R r
L l
tan α 0.6 r
1 l
r 0.6l
slide 51: 51
d 1.2l
Volume of water displaced 1 × π × r
2
l
3
1 × π × 0.6l
2
l
3
0.377 l
3
m
2
Moment of Inertia of the Circular
Section about the liquid level I π × d
4
64
I π × 1.2l
4
64
0.01018 l
4
BM I/V
0.01018 l
4
0.377 l
3
0.27l
Distance of Center of Buoyancy from
the Apex OB 0.75 l
Distance of Center of Gravity from
the apex OG 0.75 × 1
For stable equilibrium the Meta Centric should be above Center of Gravity.
i.e
BG BM
OG – OB BM
0.75 – 0.75 l 0.27 l
l 0.735m
Now Volume of water Displaced 0.377 l
3
0.377 0.735
3
0.15 m
3
This should be equal to the weight of the buoy therefore weight of the buoy
w
w
W
w
Mass × 9.8
V
w
Volume
Mass × 9.8 W
w
0.15 × 1000 × 9.8 1470 N.
Weight of water displaced W
w
W
b
.
slide 52: 52
6 A conical buoy floating with its apex pointing downward is 3.5m high and 2 m in base diameter.
Calculate its weight if it is just stable when floating in seawater weighing 10055 N/m
3
.
Solution:
Note: Center of Buoyancy of a cone is at ¾ distance of the length of the cone from the Apex.
Length of the Cone L 3.5 m
Length of the cone immersed in water l
Radius of the Cone at the liquid level r
Diameter of the Cone D 2 m
tan α R r
L l
tan α 1 r
3.5 l
r l/3.5
d 2l/3.5
Volume of water displaced 1 × π × r
2
l
3
1 × π × l/3.5
2
l
3
0.08544 l
3
m
2
Moment of Inertia of the Circular
Section about the liquid level I π × d
4
64
I π × 2l/3.5
4
64
0.005233 l
4
BM I/V
0.005233 l
4
0.08544 l
3
0.06124l
Distance of Center of Buoyancy from
the Apex OB 0.75 l
Distance of Center of Gravity from
the apex OG 0.75 × 3.5
slide 53: 53
For stable equilibrium the Meta Centric should be above Center of Gravity.
i.e
BG BM
OG – OB BM
0.75 × 3.5 – 0.75 l 0.06124l
0.81124 l 0.75 × 3.5
l 0.75 × 3.5
0.81124
l 3.235 m
Now Volume of water Displaced 0.08544 l
3
0.08544 3.235
3
2.89468 m
3
This should be equal to the weight of the buoy therefore weight of the buoy
w
w
W
w
Mass × 9.8
V
w
Volume
Mass × 9.8 W
w
2.89468 × 1000 × 9.8 28367.89 N.
Weight of water displaced W
w
Weight of Body W
b
.
7 A wooden cylinder of specific gravity 0.6 and circular cross section is required to float in oil of
specific gravity 0.90. Find the L/D ratio of the cylinder to float with its longitudinal axis vertical
in oil where L is the height of the cylinder and D its diameter.
Solution:
Given that
Cylinder Diameter D
Height of the cylinder L
Specific Gravity of Cylinder S
c
0.6
Specific Gravity of Oil S
o
0.9
Depth of immersion of the cylinder
h S
c
· L
S
o
0.6 · L 2L
0.10 3
Distance of Center of Buoyancy from the bottom of the cylinder
OB 1 · 2L L
2 3 3
Distance of Center of Gravity from the bottom of the cylinder
OG L/2
BG OG – OB
slide 54: 54
L  L L
2 3 6
Moment of Inertial of the Circular Section about its center of gravity
I πd
4
64
Volume of Water Displaced
V πd
2
· 2L
4 3
BM I/V πd
4
64
πd
2
· 2L
4 3
32d
2
32L
For Stable Equilibrium the MetaCenter should be above the Center of Gravity G or may coincide
with G.
i.e
BG BM
L 32d
2
6 32L
L
2
18d
2
32
L
2
9d
2
16
L 3 d
4
L 0.75 · d
8 A cone of specific gravity S floats in water with its apex downwards. It has a diameter D and
vertical Height L. Show that for stable equilibrium of the cone
L 1 D
2
S
1/3
1/2
2 1  S
1/3
1/2
Note: Center of Buoyancy of a cone is at ¾ distance of the length of the cone from the Apex.
Solution:
slide 55: 55
Given
Diameter of the Cone D
Length of the Cone L
Specific Gravity of the Cone S
Let
G Centre of Gravity of Cone
B Centre of Buoyancy of Cone
2 α Apex Angle of Cone
h Depth of immersion
d Diameter of cone at water surface
Distance of Center of Gravity from
the apex OG 0.75L
Distance of Center of Buoyancy from
the Apex OB 0.75 l
Weight of the cone Weight of water displaced
S × 1000 × 1 × π × R
2
× L × 9.8 1000 × g × 1 × r
2
× l
3 3
SR
2
L r
2
l
l SR
2
L
r
2
tan α R r
L l
R L × tan α
r l × tan α
l S × L
2
× tan
2
α × L
l
2
× tan
2
α
l S × L
3
l
2
l
3
S × L
3
l S
1/3
× L
OG – OB 0.75L  0.75l
slide 56: 56
OG – OB 3 L  l
4
OG – OB 3 L  S
1/3
× L
4
OG – OB 3 L1  S
1/3
4
Volume of Liquid Displaced V 1 × π × d
2
× l
3 × 4
Volume of Liquid Displaced V 1 × π × d
2
× S
1/3
× L
3 × 4
Moment of Inertia of the Circular
Section about the liquid level I π × d
4
64
I π × d
4
64
BM I/V
π × d
4
64___________
1 × π × d
2
× S
1/3
× L
3 × 4
3 d
2
16LS
1/3
tan α R r
L l
tan α R L
r l
tan α R D L
r d l
D L
d l
d D l
L
d D S
1/3
× L
L
d D S
1/3
For stable equilibrium the Meta Centric should be above Center of Gravity.
slide 57: 57
i.e
BG BM
OG – OB BM
3 L1  S
1/3
3 d
2
4 16LS
1/3
L
2
4 × d
2
_______
161  S
1/3
S
1/3
L
2
4 × D
2
× S
1/3
× S
1/3
161  S
1/3
S
1/3
L
2
D
2
× S
1/3
41  S
1/3
L 1 D
2
S
1/3
1/2
2 1  S
1/3
1/2