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What is Fluid?????
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A fluid may be liquid, vapour or gas. It has no permanent
shape but takes up the shape of a containing vessel or
channel or is shaped by external forces (eg the atmosphere).
A fluid consists of atoms/molecules in random motion
(translation) and in continual collision with the
surroundings.
Fluids are readily deformable, and flow.
Solids have ‘frozen’ molecules that vibrate and do not
translate. Solids resist change of shape.
What is Fluid?????
FT
Attached
plates
Solid
(Rectangular
Block)
B
t
t
FT
For a solid, application of a shear stress causes a deformation
which, if modest, is not permanent and solid regains original
position.
B
What is Fluid?????
Fluid
at rest
Sliding (shearing
occur between
fluid layer)
FT
For a fluid, continuous deformation takes place with an
infinite number of layers sliding over each other.
Deformation continues until the force is removed.
FT
What is Fluid?????
FT
to
FT
to
0<t<a
t1 t2
FT
to<t1<t2
FT
a) Solid
a) Fluid
A fluid is a substance for which a shear stress tends to
produce unlimited deformation.
Dimensions and Unit
1. Primary Dimensions
Primary dimension SI Unit
BG Unit
Conversion Factor
Mass (M)
Kilogram(kg)
Slug
1slug = 14.5939kg
Length (L)
Meter(m)
Foot (ft)
1 ft = 0.3048 m
Time (T)
Second (s)
Second (s) 1 s = 1 s
Temperature (Q)
Kelvin (K)
Rankine ® 1 K = 1.8 R
Dimensions and Unit
Secondary Dimension
SI Unit
velocity
acceleration
m/s
m/s2
N
kg m/s2
force
energy (or work)
power
In primary
dimension
.
ms-1
ms-2
LT-1
LT-2
kg ms-2
M LT-2
Joule J
N m,
kg m2s-2
kg m2/s2
Watt W
Nms-1
N m/s
kg m2s-3
2
3
kg m /s
ML2T-2
ML2T-3
Dimensions and Unit
Secondary Dimension
SI Unit
pressure ( or stress)
Pascal P,
N/m2,
kg/m/s2
density
specific weight
relative density
In primary
dimension
.
Nm-2
kg m1s-2
kg/m3
kg m-3
N/m3
kg m2s-2
kg/m2/s2
a ratio
.
no units
ML-1T-2
ML-3
ML-2T-2
1
no dimension
Viscosity
A fluid offers resistance of motion due to its viscosity or
internal friction. Viscosity arises from movement of
molecules from one layer to another moving at a different
velocity. Slower layer tend to retard faster layers hence
resistance
Viscosity
Shear stress
F
t
A
x
Shear strain  
y
Viscosity
Rate of Shear strain

x x1
 

ty t y
t
u

y
Viscosity
For most fluids used in engineering it is found that
the shear stress is directly proportional to rate of
shear when straight and parallel flow is involved
u
t
y
u
t  cons tan t 
y
Viscosity
u
where
is velocity change in y direction
y
du
So, at any point is the true velocity gradient
dy
du
t 
dy
Newton's law of viscosity
The constant of proportionality;  is called the dynamic viscosity
or just viscosity
EXAMPLE 0
1) Determine the unit of dynamic viscosity; 
2) Kinematic viscosity;u is defined as the ration of
dynamic viscosity to fluid density; Determine the
unit of Kinematic viscosity
EXAMPLE 1
Circular plate slides over the larger flat surface on a thin film of
liquid that has a thickness of 0.02 mm. The plate has an diameter
of d=6.3cm and mass of 20 g. If the plate is given an initial velocity
of 5m/s, calculate the force required to move the plate at a steady
velocity.Assume dynamic viscosity of liquid =1.805 kg/m-s
u=5 m/s
=1.805kg/m-s
y=0.02 mm
Surface
EXAMPLE 1.1
A board 1m by 1m that weight 25 N slides down an inclined ramp
(slope = 20O) with a velocity of 2.0 cm/s. The board is separated
from the ramp by a thin film of oil with a velocity of 0.05 N.s/m2.
Neglect the edge effect, calculate the spacing between the board
and the ramp
EXAMPLE 1
A board 1m by 1m that weight 25 N slides down an inclined ramp
(slope = 20O) with a velocity of 2.0 cm/s. The board is separated
from the ramp by a thin film of oil with a velocity of 0.05 N.s/m2.
Neglect the edge effect, calculate the spacing between the board
and the ramp
EXAMPLE 1.2
A piston moves inside a cylinder at a velocity of 5 m/s, as shown
in figure. The 150 mm diameter piston is centrally located within
the 150.2 mm inside diameter cylinder. The film of oil separating
the piston from the cylinder has an an absolute viscosity of 0.40
N.s/m2. Assuming a linear velocity profile, find the
a) Shear stress in the oil
b) Force F required to maintain the given motion
c) Force by which the required force would change if the
velocity increased by a factor of 2
EXAMPLE 1.2
Viscosity
Fluids which do not obey the Newton's law of viscosity are called
as non-Newtonian fluids. Generally non-Newtonian fluids are
complex mixtures: slurries, pastes, gels, polymer solutions etc.,
Viscosity
Variation of Viscosity with Temperature
For gases
 T 
  
 o  To 
n
Power law
3/ 2
 T  (T  S )


o
T
  o

o
T S
Sutherland law
Viscosity
Variation of Viscosity with Temperature
For Liquid
  ae
 bT

 T0   T0 
ln
 a  b   c 
0
T  T 
For water To = 273.16K, mo = 0.001792 kg/(m.s)
and a = -1.94, b = -4.80 , c = 6.74
2
CONTIMUUM
From a microscopic point of a view a fluid
is not continuous and homogenous
substance but consists of atom or molecules
in random motion and with relatively large
space between them. Under such
circumstances it has no meaning refer to the
velocity at a point in a fluid because that
point may be empty space at particular
instant. When we refer to the velocity of a
fluid, we usually imply a quantity of fluid
consisting of an enormous number of atoms
of molecules-fluid velocity is a
macroscopic concept.
macroscopic
microscopic
DENSITY
X
Total Mass = m1
Total Volume = V1
Fluid
mi
Average Density =  
vi
DENSITY
m
Density at point =   lim  
v 0 V
 
dm/dV
Average Density
dV
Specific Weight
The weight per unit volume of a fluid is called its specific
volume and equal to g, the product of its density and the
acceleration of gravity
gair = (1.204 kg/m3) (9.807m/s2) = 11.8N/m3 or = 0.0752lbf/ft3
gwater = (998kg/m3) (9.807m/s2) = 9790N/m3 or = 62.4lbf/ft3
Specific Gravity
Specific gravity (SG) which is the ratio of density to the
standard density of some reference fluid at 20oC and 1 atm
SGair
 gas
 gas


3
 air 1.204kg / m
SGwater
 liquid
 liquid


 water 998kg / m 3
Compressibility and the Bulk modulus
P
V
P  dP
V  dV
Bulk modulus (K) = (change in pressure) / (volumetric strain)
Compressibility and the Bulk modulus
Bulk modulus (K) = (change in pressure) / (volumetric strain)
Where the volumetric strain is the ratio of the change in
volume to the initial volume
 dV / V  dP / K
or
K  VdP / dV
Compressibility and the Bulk modulus
From the mass conservation;
dm  Vd  dV  0
or
V
dV  ( )d

So that
 VdP
K
V
d

or
dP
K
d
Compressibility and the Bulk modulus
Typical values of Bulk Modulus:
K = 2.05x109 N/m2 for water
K = 1.62 x 109 N/m2 for oil.