001a (PPT) Introduction & Properties of fluids.pdf

Dr. Vijay G. S.
Professor
Dept. of Mech. & Mfg. Engg.
9980032104
vijay.gs@manipal.edu
Cabin: Room No. 6, Faculty Chambers 3,
Adjacent to CAD Lab.,
AB1, First Floor

Fundamental Concepts
• In general, matter can be distinguished by the physical forms known as
solid, liquid, and gas.
• The liquid and gaseous phases are usually combined and given a
common name of fluid.
Definition of fluid:
A fluid is a substance which deforms continuously under the action of
shearing forces, however small they may be.
• Conversely, it follows that: If a fluid is at rest, there can be no shearing
forces acting and, therefore, all forces in the fluid must be
perpendicular to the planes upon which they act.

How are shear stresses developed in a fluid??
Fluid at rest:
• If the velocity of the fluid is same at every point within the fluid, no
shear stresses will be produced, since the fluid particles are at rest
relative to each other.
Moving fluid:
• If the particles of the fluid move relative to each other so that they have
different velocities, causing the original shape of the fluid to become
distorted, the fluid is said to be in motion.
• Shear stresses are developed when the fluid is in motion.
Fundamental Concepts-Contd…

• Fluid Statics deals with action of forces on fluids at rest or in
equilibrium.
• Fluid Kinematics deals with geometry of motion of fluids
without considering the cause of motion.
• Fluid dynamics deals with the motion of fluids considering the
cause of motion.

• Fluid mechanics is the study of fluids at rest or in motion.
• It has traditionally been applied in:
The design of pumps, compressor, turbines, etc.
The design of dams and canals, etc.
The design of piping and ducting in chemical plants
The aerodynamics of airplanes and automobiles
• In recent years fluid mechanics is truly a ‘high-tech’ discipline and many
exciting areas have been developed:
The aerodynamics of multi-story buildings
Fluid mechanics of atmosphere, sports, and micro fluids.

• Intermolecular force is very
large; hence, molecules
are not free to move.
• exhibit definite shape and
volume.
• undergo certain amount of
deformation and then
attain state of equilibrium
when subjected to tensile,
compressive and shear
forces.
• Intermolecular force
is comparatively
small;
• exhibit definite
volume, but they
assume the shape of
the container
• Intermolecular force is
very small.
• The molecules are free to
move along any direction.
Therefore gases will
occupy or assume the
shape of the container as
well as the volume of the
container.

Why liquids and gases together are called fluids??
Flow property of liquids and gases:
• Liquids and gases undergo continuous or prolonged angular
deformation or shear strain when subjected to tangential force or
shear force.
• This property of the liquid is called flow of liquid and such property
of the gas is called flow of gas.
• Any substance which exhibits the ‘property of flow’ is a fluid.
• Therefore liquids and gases are considered as fluids.

Compressible and incompressible fluids
• Gases offer little resistance against compressive forces.
Therefore gases are called compressible fluids.
• Liquids offer maximum resistance against compressive
forces; Hence liquids are also called incompressible fluids.

1. Mechanics : Deals with action of forces on bodies at rest or in motion.
2. State of rest and Motion:
• They are relative and depend on the frame of reference.
• If the position with reference to frame of reference is fixed with time, then the body
is said to be in a state of rest. Otherwise, it is said to be in a state of motion.
3. Scalar and vector quantities:
• Quantities which require only magnitude to represent them are called scalar
quantities.
• Quantities which require magnitude and direction to represent them are called
vector quantities.
e.g., Scalars: Mass, time interval, distance traveled, speed, etc.
Vectors: Weight, Displacement, Velocity, etc.
Preliminary Concepts

4. Displacement and Distance
Unit: m
Preliminary Concepts-Contd…
5. Velocity and Speed:
• Rate of displacement is called Velocity.
• Rate and distance travelled is called Speed.
Unit: m/s

6. Acceleration: Rate of change of velocity is called acceleration. Negative
acceleration is called retardation.
Unit: m/s2
7. Mass: A measure of the amount of matter contained by the body. It is a scalar
quantity. (Unit: kg)
8. Volume: It is the measure of space occupied by the body. (Unit: m3; 1 m3 =
1000 litres)
9. Momentum: The capacity of a body to impart motion to other bodies is called
momentum.
• The momentum of a moving body is measured by the product of mass and
velocity of the moving body
• Momentum = Mass × Velocity
• Unit: kg.m/s

10. Newton’s first law of motion: Every body continues to be in its state of rest or
of uniform motion unless compelled by an external agency.
11. Inertia: It is the inherent property of the body to retain its state of rest or
uniform motion.
12. Force: It is an external agency which overcomes or tends to overcome the
inertia of a body. (Unit: N)
13. Newton’s second law of motion: The rate of change of momentum of a body
is directly proportional to the magnitude of the applied force and takes place
in the direction of the applied force.

Change in momentum in time ‘t’ = mv – mu
Rate of change of momentum
mv mu
t


F = 1 N,
when m = 1 kg and a = 1 m/s2
then K = 1
F = ma
(Unit of force is N)
(1 N = 1 kg × 1 m/s2)

14. Weight:
• It is the gravitational force acting on the body.
• It is defined as the force required to attract a mass of 1 kg with an
acceleration of 9.81 m/s2 towards the center of the earth.
• It is a vector quantity.
Weight = mass × acceleration due to gravity
W = mg (g = 9.81 m/s2)
Unit: N

15. Work:
Work = Force × Displacement ; Unit: Nm (Linear motion)
Work = Torque × Angular displacement; Unit: Nm-rad (Rotory motion)
16. Energy: Capacity of doing work is called energy.
Unit: Nm or J
Potential energy = mgh
Kinetic energy = ½ mv2

17. Power: Rate of doing work is called Power.
Work Force Displacement
Power
Time Time

 
(Unit: Nm/s or J/s or W)
For linear motion, Power = (Force × velocity) W
For rotary motion, Power = (Torque × angular velocity)
2 2
60 60
N NT
Power T T W
 

   

Systems of Units:
• The SI units (System International Units) is the official international system of
units.
• But older systems, particularly the CGS and FPS engineering gravitational
systems are still in use and probably will be around for some time.
• Thus it becomes necessary to be expert in the use of all the three systems.

Derived quantities in CGS System:

Dimensions: Dimensions of the primary quantities:
T
K

Dimensions of derived quantities can be expressed in terms
of the fundamental dimensions.
T-1
ML/T2
L2/T
L/T

1. Mass density or Specific mass ():
Mass density or specific mass is defined as the mass per unit volume of the
fluid.
Properties of fluids
• With the increase in temperature, volume of fluid increases and hence mass
density decreases.
• With the increase in pressure, volume of the fluid decreases and hence
mass density increases.

Properties of fluids-Contd…
2. Weight density or Specific weight ():
Weight density or Specific weight of a fluid is defined as the weight per unit
volume.
• With increase in temperature, volume increases and hence the specific
weight decreases.
• With increase in pressure, volume decreases and hence specific weight
increases.

3. Relationship between mass density and weight density:

4. Specific gravity or Relative density (S):
It is defined as the ratio of specific weight of the fluid to the specific weight of
a standard fluid.
OR
It is defined as the ratio of mass density of the fluid to the mass density of a
standard fluid.

In case of liquids, water at 4oC is considered as the standard liquid.
water = 1000 kg/m3 (In S.I. units)
water = water × g = 1000 × 9.81 = 9810 N/m3 (In S.I. units)
Liq = S × water = S × 9810 N/m3 (In S.I. units)
Liq = S × water = S × 1000 kg/m3 (In S.I. units)
( )
( )
( )
Liq Liq
Liq
Water Water
Weightdensity density of liquid
S for liquids
Weightdensity density of water
S
 
 

 

In case of gases, air at 15oC is considered as the standard gas at sea level.
Air = 1.22 kg/m3 at 15oC
 Air = Air × g = 11.97 N/m3
Gas = S × Air = S × 11.97 N/m3 (In S.I. units)
 Gas = S × Air = S × 1.22 kg/m3 (In S.I. units)
( )
( )
( )
Gas Gas
Gas
Air Air
Weightdensity density of gas
S for gases
Weightdensity density of air
S
 
 

 

5. Specific volume (): It is defined as the volume per unit mass of the fluid
Unit: m3/kg
• As the temperature increases volume increases and hence specific
volume increases.
• As the pressure increases volume decreases and hence specific volume
decreases.

Problem 1: Calculate specific weight, density, specific volume and specific
gravity and of one liter of Petrol which weighs 6.867N.
Problem 2: Specific gravity of a liquid is 0.7 Find (i) Mass density (ii) specific
weight. Also find the mass and weight of 10 liters of liquid.
Problems

6. Viscosity:
• Viscosity is the property by virtue of which a fluid offers resistance to the
motion of one layer of fluid over another layer of the fluid.
• In other words, it is the reluctance of the fluid to flow or it is the resistance
against the flow or shear deformation.
• Viscous force is that force of resistance offered by a layer of fluid for the
motion of another layer over it.
• In case of liquids, viscosity is due to cohesive force between the molecules of
adjacent layers of liquid.
• In case of gases, molecular activity between adjacent layers is the cause of
viscosity.

6a. Newton’s Law of Viscosity:

6a. Newton’s Law of Viscosity-Contd…:

Consider a liquid between a pair of plates separated by a distance ‘Y’.
Let the bottom plate be fixed plate and the upper plate be movable.
Let
‘A’ be the contact area (Wetted area) of the movable plate,
‘F’ be the force required to move the plate with a velocity ‘U’.

According to Newton’s law of viscosity,
i.e., Shear force is directly proportional to area
i.e., Shear force is inversely proportional to distance
i.e., Shear force is directly proportional to velocity
The constant of proportionality  is known as
‘Coefficient of Dynamic Viscosity’ or simply ‘Viscosity’
or ‘Absolute Viscosity’

• ‘’ is the force required per unit area and known as the ‘Shear Stress’.
• The above equation is called ‘Newton’s law of viscosity’.
• Newton’s law of viscosity states that the shear stress () on a fluid
element layer is directly proportional to the rate of shear strain or velocity
gradient (du/dy)

Velocity gradient or rate of shear strain:
• It is the difference in velocity per unit distance between any two layers.
• If the velocity profile is linear, then the velocity gradient is given by
• If the velocity profile is non – linear, then the velocity gradient is given by
• The velocity gradient (du/dy) = rate of shear strain (d/dt)
U
Y
du
dy
U
Y
 

du
dy
 


  
 
du d
dy dt

dy
u+du
u

  
 
du d
dy dt

Proof for (du/dy) = (d/dt)

Unit of dynamic viscosity
du
dy
du
dy
 



 
 
 
 
 
  
 
 
 
 
2
2
2
2 2
/
/
. /
N m Ns
Pa s
m s m
m
OR
kg m s s
Ns kg
m m ms
SI unit of 
CGS unit of  is poise
2
1
dyne s
poise
cm


 
2 2
1
1 1 10
10
Ns Ns
poise OR poise
m m




2
2
1 10
. ., 1 10
centipoise poise
i e cP P

6b. Kinematic Viscosity:
The Kinematic Viscosity () is defined as the ratio of
dynamic viscosity () of the fluid to its mass density ().




Unit of kinematic viscosity
NOTE: Unit of kinematic viscosity in
CGS system is cm2/s and is called
‘stoke’
1 stoke = 1 cm2/s = 10-4 m2/s
1 centistoke (cS)=10-2stoke=10-6m2/s

Effect of Pressure on Viscosity of fluids:
• Pressure has very little or no effect on the viscosity of fluids.
Effect of Temperature on Viscosity of fluids:
• Temperature affects the viscosity.
• Viscous forces in a fluid are due to cohesive forces and molecular momentum
transfer.
• In liquids, the cohesive forces predominates the molecular momentum
transfer, due to closely packed molecules.
• Whereas in gases the cohesive forces are small and molecular momentum
transfer predominates.

1. Effect of temperature on viscosity of liquids:
As the temperature increases cohesive force decreases and hence viscosity
decreases.
• For liquids, the viscosity (t) (in units of poise) as a function of temperature
‘t’ (in units of oC) is given by:
where,
o = Viscosity of liquid at 0oC
 and  = Constants for the liquid
Note: For water,
o = 1.7910-3 poise,  = 0.03368 and  = 0.000221
2
1
( )
1
o
t
t t
 
 
 
  
 
 

2. Effect of temperature on viscosity of gases:
As the temperature increases momentum transfer in molecules increases and
hence viscosity increases.
• For gases, the viscosity (t) (in units of poise) as a function of temperature
‘t’ (in units of oC) is given by:
where,
o = Viscosity of the gas at 0oC
 and  = Constants for the gas
Note: For air ,
o = 0.000017 poise,  = 56 10-9 and  = 0.1189 10-9
2
( ) o
t t t
   
  

• As temperature
increases,
– Viscosity decreases
for liquids
– Viscosity increases
for gases

Problem 3: Find the kinematic viscosity of an oil having density 981 kg/m3. The
shear stress at a point in oil is 0.2452 N/m2 and velocity gradient at that point is
0.2 per second.
Problem 4: A flat plate 0.025 m distance from a fixed plate moves at 60 cm/sec
and requires a force of 2 N/m2 to maintain this speed. Determine the fluid
viscosity between the plates.
Problem 5: If velocity distribution of fluid over a plate is given by u = (2/3)y –y2 in
which ‘u’ is the velocity in m/sec at a distance ‘y’ meter above the plate,
determine the shear stress at y = 0 and y = 0.15 m. Take the dynamic viscosity  =
8.63 poise.
Problems on Viscosity

Problem 6: Determine the intensities of shear of oil having viscosity equals 1 poise.
The oil is used for lubricating the clearance between a shaft of diameter 10 cm and
its bearing. The clearance is 1.5 mm and shaft rotates at 150 rpm.
Problem 7: Calculate the dynamic viscosity of oil which is used for lubrication
between a square plate of size 0.8 x 0.8 m2 and an inclined plane with the angle of
inclination 300. The weight of square plate is 300 N and it slides down with uniform
velocity 0.3 m/sec. The thickness of the oil film is 1.5 mm.
Problem 8: A shaft of diameter 0.4 m is supported in a journal bearing of length 90
mm and clearance 1.5 mm. The viscosity of lubricant is 6 poise. The shaft rotates at
190 rpm. Calculate the power lost in the bearing.

Problem 9: Two large plane surfaces are 2.4 cm apart. The space between the
surfaces is filled with fluid. What force is required to drag a very thin plate of
surface area 0.5 square meter between two large plane surfaces at a speed of 0.6
m/sec, if
(i) The thin plate is in the middle of two plane surfaces
(ii) The thin plate is at a distance of 0.8 cm from one of the plane surfaces. Take
dynamic viscosity of fluid as 8.1 x 10-1 Ns/m2
Problem 10: A vertical gap 2.2 cm wide of infinite extent contain a fluid of viscosity
2 Ns/m2 and specific gravity 0.9. A metallic plate 1.2 m x 1.2 m x 0.2 cm is to be
lifted up with a constant velocity of 0.15 m/sec through the gap. If the plate is in
the middle of the gap, find the force required. The weight of the plate is 40 N.

7. 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 of two immiscible liquids such
that the contact surface behaves like a membrane under tension.
• Surface tension is due to cohesion between the molecules of liquid and
weak adhesion between the molecules on the free surface of the liquid and
molecules of air.

• A molecule inside the liquid gets attracted by equal forces from the
surrounding molecules, i.e., the resultant force acting on it is zero.
• A molecule on the surface gets attracted by the molecules below it, i.e., the
resultant force is acting downwards.
• All the molecules at the free surface will experience a downward pull; the
entire free surface of the liquid will have a tendency to move downwards
• The free surface of the liquid will behave like a very thin film (or elastic
membrane) under tension.
• The property of the liquid surface to offer resistance against tension is called
surface tension.
Surface Tension-Contd…

Consequences of Surface tension:
• Liquid surface supports small loads
• Formation of spherical droplets of liquid
• Formation of spherical bubbles of liquid
• Formation of cylindrical jet of liquids.

Measurement (Unit) of surface tension:
• Surface tension is measured as the force exerted by the film on a line of
unit length on the surface of the liquid.
• It can also be defined as the force required for maintaining unit length of
the film in equilibrium.
,
:
F
SurfaceTension
L
N
Unit
m
 

Surface tension on
a doubly curved
membrane

• Consider the surface ABCD of a
fluid region with curvature on
both sides
• Curved edges AB and DC
subtend angle d1 at their center
of curvature, with radius r1
• Curved edges BC and AD
subtend angle d2 at their center
of curvature, with radius r2
• Surface tension force acting on
edges AB and CD = r1d1
• Surface tension force acting on
edges AD and BC = r2d2
Pressure on
inside = pi
Pressure on
outside = po
Relationship between pressure and surface tension:

Relationship between pressure and surface tension – Contd…
 
1 2
2 2 1 1
2 sin 2 sin
2 2 i o
d d
r d rd p p dA
 
   
   
  
   
   
   
1 2
2 2 1 1 1 1 2 2
2 2
2 2 i o
d d
r d rd p p rd r d
 
     
   
  
   
   
where, dA = (r1d1)(r2d2) is the area of the curved surface ABCD

   
2 2 1 1 1 2 1 1 2 2
i o
r d d rd d p p rd r d
       
  
 
1 2
1 1
i o
p p
r r

 
  
 
 
Dividing throughout by r1d1r2d2
• If the fluid region (shown in fig) is hollow, then two interfaces are formed
• i.e., surface tension forces will be doubled (because of two interfaces), and
the final equation will be:
 
1 2
1 1
2
i o
p p
r r

 
  
 
 

Case (a): For a liquid droplet:
r1 = r2 = r
Substituting in  
1 2
1 1
i o
p p
r r

 
  
 
 
 
2 4
i o
p p
r D
 
   where, D is the diameter of droplet = 2r
Case (b): For a liquid jet:
r1 = r and r2 = 
1 2
1 1
i o
p p
r r

 
  
 
 
 
2
i o
p p
r D
 
   where, D is the diameter of droplet = 2r

Case (c): For a hollow bubble of liquid-gas interface:
r1 = r2 = r
1 2
1 1
2
i o
p p
r r

 
  
 
 
 
4 8
i o
p p
r D
 
  

Effect of temperature on surface tension of liquids:
• In case of liquids, surface tension decreases with increase in temperature. This
is due the decrease in cohesive forces between the liquid molecules.
• Pressure has no or very little effect on surface tension of liquids.

Problem 11: The pressure outside the droplet of water of diameter 0.04
mm is 10.32 N/cm2 (atm. pr.). Calculate the pressure within the droplet if
surface tension is given as 0.0725 N/m of water.
Problems on Surface Tension

8. Capillarity:
• Capillarity is defined as the phenomena of rise or fall of a liquid surface in a
tube of small diameter relative to the adjacent general level of liquid, when the
tube is dipped and held vertically in the liquid.
• If there is a rise in the liquid surface in the tube, it is known as ‘capillary rise’
• If there is a fall in the liquid surface in the tube, it is known as ‘capillary
depression’

Capillarity-Contd…
• Any liquid between contact surfaces attains curved shaped surface known as
the Miniscus
• If adhesion is more than cohesion then the meniscus will be concave.
• If cohesion is greater than adhesion meniscus will be convex.

Derivation for capillary rise of a liquid

 

h
Liquid
 D
= Surface tension of the liquid
= Angle of contact between liquid and glass tube
D = Diameter of glass tube
h = Capillary rise of the liquid
Then,
Weight of liquid column of height h = Vertical
component of surface tension force

   
   
 
    
  
2
. ., cos
4
4 cos 4 cos
D
i e h D
h
D gD

Derivation for capillary depression of a liquid




h
Liquid
 D
Upward Hydrostatic force at A = Downward
vertical component of surface tension force

  


   
   
 
   
 
     
  
2
2
. ., cos
4
,
cos
4
4 cos 4 cos
A
A
D
i e p D
But p h
D
h D
h
D gD
A

Problem 12: Calculate the capillary effect in mm in a glass tube of 4 mm
diameter, when immersed in a) water b) mercury. The temperature of the
liquid is 20oC and the value of surface tension of water and mercury at 20oC
in contact with air are 0.0735 N/m and 0.51 N/m respectively. The angle of
contact is zero for water and 130o for mercury. Density of water at 20oC is
998 kg/m3.
Problem 13: Determine the minimum size of the glass tubing that can be
used to measure water level if capillary rise is not to exceed 2 mm. Take  =
0.0736 N/m at 20oC.
Problem 14: If a tube is made so that one limb is 20 mm in diameter and
other 2 mm in diameter and water is poured in the tube. What is the
difference in the level of surface of liquid in the two limbs? Take  = 0.073
N/m for water.
Problems on Capillarity

9. Classification of fluids or types of fluids:
The fluids may be classified into the following types:
• Ideal fluid and Real fluid
• Newtonian fluid and Non-newtonian fluid
• Ideal (Bingham) plastic and Thixotropic fluid
Pseudoplastic
fluid
Dailatant
fluid

Classification of fluids or types of fluids-Contd…
Ideal fluid v/s Real Fluid
• Any fluid which is
incompressible and having no
viscosity (i.e., viscosity = zero)
is called an Ideal fluid.
•  = 0 for all values of du/dy
• Any fluid which possesses viscosity
is called a real fluid.
•   0 for all values of du/dy

Newtonian fluid v/s Non-Newtonian fluid
• A real fluid which obeys
Newton’s law of viscosity is
called Newtonian fluid.
• In such fluids, the shear stress
varies directly as shear strain
• The stress-strain curve is a
straight line passing through
origin; the slope of the line gives
dynamic viscosity of the fluid.
• e.g., water, kerosene, air, etc.
• A real fluid which does not obeys
Newton’s law of viscosity is called Non-
Newtonian fluid.
• In such fluids, the shear stress is not
proportional to the shear strain
• Its stress-strain curve is a curved line
passing through origin
• e.g., salt solutions and molten
polymers like: paint, blood, ketchup,
shampoo, etc.

Newtonian fluid v/s Non-Newtonian fluid

A power law model
More information on Non-Newtonian fluids
Ostwald-de-Waelemodel
n
du
m
dy
m flow behaviour index
n flow consistency index

 
  
 



Ideal-plastic fluid v/s Thixotropic fluid
• A real fluid in which the strain
starts after certain initial
stress (o) and then the stress-
strain relationship will be
linear.
• o is called initial yield stress.
• Sometimes they are also called
Bingham’s Plastics.
• e.g., toothpaste which will not
be extruded until a
certain pressure is applied to
the tube. It then is pushed out
as a solid plug.
• A real fluid in which the strain starts
after certain initial stress (o) and
then the stress-strain relationship
will be curved.
• e.g., gels, colloids, printers ink,
cytoplasm,etc.

Ideal-plastic fluid v/s Thixotropic fluid

Real Fluids
Ideal
Solid

• Gases are compressible fluids, and hence their thermodynamic properties play
an important role.
• With the change of pressure and temperature, the gases undergo a large
variation in density.
• The relationship between the absolute pressure, absolute temperature and
the specific volume is given by the equation of state as:
10. Thermodynamic properties:
p
p RT OR RT

  
where, p is the absolute pressure in N/m2;
 is the specific volume = 1/ in m3/kg
R is the gas constant = 287 J/kg K
T is the absolute temperature in K

Isothermal Process:
• If the changes in density of the gas
occurs at constant temperature,
then the process is called
isothermal.
• The relationship between the
pressure (p) and density () is given
as:
Thermodynamic properties-Contd…
Adiabatic Process:
• If the changes in density of the gas
occurs with no heat exchange from/to
the gas, and if no heat is generated
within the gas due to friction, then the
process is called adiabatic.
• The relationship between the pressure
(p) and density () is given as:
tan
p
Cons t

 tan
k
p
Cons t


k = 1.4 for air, It is the ratio of the
specific heat of the gas at constant
pressure and volume (k = Cp/Cv)

• It is the property by virtue of which there will be a change in the volume of
the fluid due to the change in pressure.
• It is the reciprocal of the ‘Bulk modulus of elasticity’ (K) of the fluid
• Bulk modulus of elasticity of the fluid is defined as the ratio of the change in
pressure (compressive stress) to the volumetric strain.
11. Compressibility

• Let  be the volume of the gas
enclosed in the cylinder when
the pressure is p.
• When the pressure of the gas is
increased to p+dp, the volume
of the gas decreases to -d
Compressibility-Contd…

  

Change in volume
Volumetric strain
Original volume
The negative sign means that the volume decreases with increase of pressure

 


 
  

 
  
0
,
lim
/
V
Change in pressure OR Compressive stress
Bulk Modulus K
Volumetric strain
P
K
V V
dP
K V
dV
constant
0
V m
dV V d
dV V d
dV d
V

 
 


 
 
 
 
dP dP
K V
dV d


   

Example: For water, K = 2106 kN/m2 [(1/K)water = 0.510-9m2/N]
For air at STP, K = 101 kN/m2 [(1/K)Air = 1000010-9 m2/N]
Therefore, air is 20000 times more compressible than water



 
 

 
      
1
d
dV d
Compressibility
K dP V dP dP
Unit of Bulk Modulus (K) is N/m2 or Pa
Unit of Compressibility (1/K) is m2/N

Relationship between Bulk Modulus (K) and Pressure (p) for a gas:
(a) For Isothermal process:
tan
. ., tan [ 1/ ]
p
Cons t
i e p Cons t



   
Differentiating,
0
. .,
pd dp
dp dp
i e p K
d
d
K p
 

   

  
 

 
 

Relationship between Bulk Modulus (K) and Pressure (p) for a gas-Contd…
(b) For adiabatic process:
 
tan
. ., tan [ 1/ & 1/ 1/ ]
k
k
k k k
p
Cons t
i e p Cons t

  

      
Differentiating,
1
1
0
. ., 0 ( )
k k
k
p k d dp
i e pkd dp dividing throughout by
dp dp
pk K
d
d
K pk


   
  

   

  
 

 
 

Problem on Bulk modulus
Problem 15: Determine the Bulk Modulus of Elasticity of a liquid, if the
pressure of the liquid is increased from 70 N/cm2 to 130 N/cm2. The
volume of the liquid decreases by 0.15%.

• Vapour pressure: The pressure exerted by a
vapour on the liquid phase with which it is in
equilibrium.
• Vapour pressure is a function of temperature
Vp = f(T)
12. Vapour pressure and Cavitation:
For water

• At pressures lower than the vapour pressure, more molecules of the liquid
vapourize and escape from the surface of the liquid than are absorbed from the
vapour, resulting in evaporation.
• At the vapour pressure, the exchange is equal and there is no net evaporation.
Hence, it is also called evaporation pressure.
• A liquid with a high vapour pressure at normal temperatures is often referred to
as volatile.
• If the pressure over the liquid surface goes below the vapour pressure, then,
there will be vapourisation.
• But if the pressure above the liquid surface is more than the vapour pressure,
then there will not be vapourisation, unless there is heating.
• Note: In Carburetors and sprayers the vapours of liquid are created by reducing
the pressure below vapour pressure of the liquid.
Vapour pressure and Cavitation-Contd…

Cavitation: In case of Hydraulic turbines sometimes pressure goes below the
vapour pressure of the liquid. This leads to vapourisation and formation of
bubbles of liquid.
• When the bubbles are carried to a high pressure zone, they burst causing
partial vacuum.
• The surrounding liquid enters this space with very high velocity, exerting large
force on the local turbine parts, forming cavities on them.
• This phenomenon is called cavitation. Therefore, the turbines are designed
such that there is no cavitation.
Vapour pressure and Cavitation-Contd…

• While describing any matter, we define the property of the matter with respect to
space coordinates. The properties may be pressure, density, temperature, etc. or
some other point functions. We define these properties to be continuous functions of
space coordinates and/or time within the matter.
• This means that at each and every point in the matter, there is a molecule. In other
words the molecules are very closely packed such that the distance between the
adjacent molecules is almost zero. This holds good for solid and liquid matter as the
molecules are very closely packed, and thus can be considered as an assemblage of
single substance. Thus such matter can be considered to be “continuous media” or
“continuum”.
13. Concept of continuous medium/continuum:

• Definition of continuum: A continuum is a continuous medium of a substance in which
the various properties or point functions can be defined to be continuous functions of
space coordinates and time.
• But what about gases? In gases the molecules are less densely packed. Then can we
consider the gaseous medium as a continuum? If the gas pressure is very low, then it
may not be a continuous medium. But under normal conditions we find that the gas
molecules are also very closely packed.
• The number of molecules within a certain volume of a gas is given by the Avogadro’s
hypothesis.
Concept of continuous medium/continuum-Contd..

• At standard temperature and pressure, there are 6.0231023 molecules per 22.4 liters
of gas. That is the molecular density is 2.71025 molecules/m3, which is very high. So
under normal conditions the gas molecules are also so closely packed that the gas can
be considered as a continuous medium.
• But at very low gas pressures, the average distance between the molecules increases
and at every point in the gas we may not find a molecule, thereby we may not be
sure of considering it as a continuum.
• This problem was solved by a scientist Knudsen who established a criteria (known as
Knudsen criteria) for defining a continuum.

Knudsen considered two distance parameters:
•  as the mean free path between the molecules, and
• L as the characteristic dimension of the problem under consideration.
Note: The mean free path is the average distance travelled by a moving particle (such
as an atom, a molecule, a photon) between successive impacts (collisions) which
modify its direction or energy or other particle properties.
• If these two distances  and L are comparable (of the same order), i.e., if the ratio
/L is large, then the concept of continuum is not valid.
• This ratio /L is called as the Knudsen number.

Value of Knudsen
number
Continuum
validity
Flow condition
/L < 0.01 Yes
All fluid flow
problems
0.01 < /L < 0.1 No Slip flow
0.1 < /L < 10 No
Flow regime is
transitional
/L > 10 No Free molecular flow

Additional problems
1. Viscosity of water is 0.01 poise. Find its kinematic viscosity if specific
gravity is 0.998.
2. A plate at a distance 0.0254 mm from a fixed plate moves at 0.61 m/s and
requires a force of 1.962 N/m2 area of plate. Determine dynamic viscosity
of liquid between the plates.
3. A plate having an area of 1 m2 is dragged down an inclined plane at 45o to
horizontal with a velocity of 0.5m/s due to its own weight. There is a
cushion of liquid 1mm thick between the inclined plane and the plate. If
viscosity of oil is 0.1 Pa-s find the weight of the plate.

Additional problems – Contd…
4. A flat plate is sliding at a constant velocity of 5 m/s on a
large horizontal table. A thin layer of oil (of absolute
viscosity = 0.40 N-s/m2) separates the plate from the
table. Calculate the thickness of the oil film (mm) to limit
the shear stress in the oil layer to 1 kPa.
5. A shaft of diameter 20mm and mass 15 kg slides
vertically in a sleeve with a velocity of 5 m/s. The gap
between the shaft and the sleeve is 0.1 mm and is filled
with oil. Calculate the viscosity of oil if the length of the
shaft is 500 mm.

6. If the equation of velocity profile over a plate is V = 2y2/3 in which ‘V’ is the
velocity in m/s and ‘y’ is the distance in ‘m’. Determine the shear stress at (i)
y = 0 (ii) y = 75 mm. Take μ = 835 centipoise.
7. A circular disc of 0.3m diameter and weight 50 N is kept on an inclined
surface with a slope of 45o. The space between the disc and the surface is 2
mm and is filled with an oil of dynamics viscosity 1 N-s/m2. What force will
be required to pull the disk up the inclined plane with a velocity of 0.5 m/s.

8. Dynamic viscosity of oil used for lubrication between a shaft and a sleeve is 600
centipoise. The shaft is of diameter 0.4 m and rotates at 190 rpm. Calculate the
power lost in the bearing for a sleeve length of 0.09 m. Thickness of the oil is
1.5 mm.
9. Two large surfaces are 2.5 cm apart. This space is filled with glycerin of absolute
viscosity 0.82 Ns/m2. Find what force is required to drag a plate of area 0.5 m2
between the two surfaces at a speed of 0.6 m/s. (i) when the plate is
equidistant from the surfaces, (ii) when the plate is at 1 cm from one of the
surfaces.

10. Through a very narrow gap of ‘h’ a thin plate of large extent is pulled at a
velocity `V’. On one side of the plate is an oil of viscosity μ1 and on the other
side there is an oil of viscosity μ2. Determine the position of the plate for the
following conditions. (i) Shear stress on the two sides of the plate is equal. (ii)
The pull required, to drag the plate is minimum.
11. What is the pressure inside the droplet of water 0.05 mm in diameter at
20oC, if the pressure outside the droplet is 103 kPa Take  = 0.0736 N/m at
20oC
12. A liquid bubble 2 cm in radius has an internal pressure (in excess of the
outside) of 13 Pa. Calculate the surface tension of liquid film.

13. A capillary tube having an inside diameter 5 mm is dipped in water at 20oC.
Determine the rise of water in the tube. Take  = 0.0736N/m at 20oC.
14. Calculate capillary rise in a glass tube when immersed in Hg at 20oC. Assume 
for Hg at 20oC as 0.51 N/m. The diameter of the tube is 5mm. Take  = 130o.
15. If a U-tube is made so that one limb is 20 mm in diameter and the other 2 mm
in diameter and water is poured in the tube, what is the difference in the
levels of the surface of the liquid in the two limbs.  = 0.073 N/m for water.

16. The change in volume of certain mass of liquids is observed to be 1/500th of
original volume when the pressure on it is increased by 5 MPa. Determine
the Bulk modulus and the compressibility of the liquid.
17. Find the pressure that must be applied to water at atmospheric pressure to
reduce its volume by 1% . Take K = 2 GPa.

Surface tension on a Liquid droplet.
• Consider a spherical droplet of liquid of diameter D.
• On the entire surface of the droplet, the tensile force due to surface tension
will be acting. Let  be the surface tension of the liquid
• Let ‘p’ be the pressure inside the droplet in excess of outside pressure (p =
pinside – poutside).

The above equation shows that the pressure intensity inside the droplet
increases with decrease in its diameter.
Note: For a hollow bubble, there are two surfaces: one outside and another
inside. Then, the equilibrium equation is
 
2
8
2
4
D
D p p
D
 
 
     

Surface tension on a jet of liquid.
• Consider a cylindrical jet of diameter D.
L
D

p

• On the entire surface of the jet, the tensile force due to surface tension will be
acting. Let  be the surface tension of the liquid
• Let ‘p’ be the pressure inside the droplet in excess of outside pressure (p =
pinside – poutside).
• Then we can write,
Force due to pressure = Force due to surface tension
i.e., p × area of semi-jet =  × length of surface
p × L × D =  × 2L
2
p
D

 

001a (PPT) Introduction & Properties of fluids.pdf

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