Fluid Mechanics Theory S

INTRO & FUNDAMENTALS

FLUID : A fluid may be defined as substance which is
capable of flowing. A fluid deforms continuously when
subjected to a tangential or shear stress, however small
the shear stress may be.

Note : Solids can resist tangential stress under static
condition, fluids can do it only under dynamic condition
Moreover, when tangential stress disappears solids regain
either fully or partly their original shape, whereas fluids
can never regain it’s original shape.

Note : Liquid have considerable Cohesion. Gases have
negligible cohesion.

FLUIDS PROPERTIES :-

i) Density () : Mass per unit volume

 = m Kg/m3
v

ii) Specific weight () : Weight per unit volume
 = .g. N/m3

iii) Specific Volume : Volume per unit mass

 1 = m m3/kg
v


v) Specific Gravity : Sg

Sg = Density of fluid
Density of standard fluid

For Liquid standard is pure water at 101 KN/m2 & 40c.
In gas standard fluid is hydrogen or air

Sg  Hg = 13.6
g = 13.6 x 1000

= 13,600 Kg/m3
  Hg = 13,600 x 9.81 = 133416 N/m3

Sp. gravity of sea water Ssw = 1.02

 sw = 1.02 x 9.81 =
Sp. gravity of kerosene Sk = 0.8

k = 0.8 x 1000 = 800 Kg/m3

Note :

For water

v) Viscosity : Viscosity is the property by virtue of which
fluid offers resistance to deformation under the influence
of a shear force

Newton’s Law of Viscosity

   du --- (1)
dy

= shear stress

 = Coeffi. of Viscosity or Dynamic Viscosity

du = Velocity gradient or strain rate or Rate of shear
dy

strain or Rate of angular deformation

Equation (1) is known as Newton’s law of viscosity.

* The fluids which obeys Newton’s law of viscosity are
known as Newtonian fluids.
Example : Water, air, Hg etc.

* Fluids which do not obey this law are called Non
Newtonian fluids
Example : Paints, blood, different polymer solutions etc.

Dimensional formula & units of 



   dy  F / L2  ML1T 2 [M L1T 1]
1/T 1/ T
du /

Units of 
NS/m2 or Kg/m.s.

1 N sec/m2 = 1 Pa.Sec.
(1N = 105 dynes)
= 105 dyne sec/104 cm2
= 10 dyne sec/cm2

= 10 poise , [1 poise = 1 dyne sec/cm2]

1 Poise = 0.1 Pa. Sec.

1 Poise = 0.1 Ns/m2

Note : For Newtonian fluids  depends strongly on
temperature & varies very little with pressure.
* For liquids  decreases with increase in temperature
* For Gases  increase with increases in temperature

CAUSES OF VISCOSITY

i) Inter molecular force of cohesion

ii) Molecular momentum exchange

Note : In liquids, intermolecular cohesion is the
predominant cause of viscosity. Because cohesion
decrease with temperature, the liquid viscosity does
likewise.

* In gases, cohesive force is very small and viscosity is
dictated by molecular momentum exchange. As random
molecular motion increase with increase in temperature,
viscosity increase.

Ideal Fluids
* Hypothetical fluid having zero viscosity (  = 0)
(Invicid, Non viscous, Frictionless)

* Incompressible (Non dilatant)
( V = 0) v = 0, K = 

*  = 0.5

* No Surface tension
Note : In fact ideal fluid does not exist in nature

Non-Newtonian Fluids

Which does not obey Newton’s law of viscosity (i.e.  is
not proportional to)
Such fluid follows Power Law

  A dv n + B
 dy



 - Shear stress
A - Multiplying constant (depends upon type of fluid)
n - Power index
B - Additive constant or y intercept or shear intercept

(i) A =  , n = 1, B = 0  Newtonian fluids

  dv  Bewton's law of viscosity
dy

or   dv  Newton's fluid
dy

(ii) A  0, n < 1 , B = 0

- Non linear curve (convex)

- Pseudo plastic fluids (Eg. : Blood)

(iii) A  0, n > 1 , B = 0
- Dialatent fluid
(Eg. : Butter, Quick sand

(iv) A  0, n > 1 , B > 0
- Bingham Plastic
(Eg. : Drilling mud, Sewage mud)

(v) A  0, n < 1 , B > 0
- Thixotropic
(Eg. : Printers ink, wall paints)

(vi) X- axis ( Y coordinate = 0)
Ideal fluids (zero shear stress)

(vii) Y- axis ( X coordinate = 0)
Rigid body or Perfect solid

(ddyv = 0 or angular deformation = 0)

Note : Study of Non Newtonian fluid is called
RHEOLOGY).

Kinematic Viscosity ()

V   {  = Dynamiv Viscosity}


Dimensional formula - L2T-1
unit - m2/s or cm2/sec

cm2/sec - stoke
m2/s.1 stoke = 10-4 m2/s

Significance of Kinematic Viscosity ()
* Kinematic Viscosity is important for gases not for
liquids
* w > air
w < air at any temperature

 is less for gas   is used (as  is very low   is
considerable)
Note : Sayvolt viscometer is used to measure kinematic
viscosity.

Compressibility
Susceptibility to change in volume

Bulk modulus K v
dp dp
K  dv / v 
d/

Compressibility = 1
K

or Compressibility = 1 dp

 d

Surface Tension () :-

Surface Tension is the unbalanced normal force acting at
the interface of two different media.

* Surface tension is line force per unit length

  F
L

* It is also defined as surface energy per unit area

  Surface energy  F.L  F
Aera L2 L

* It arises due to two kinds of intermolecular forces
cohesion & Adhesion

* As temperature increase, Cohesion decrease therefore
surface tension decreases.

* The spherical shape of bubbles or droplets is due to
surface tension.

Reason :- Surface tension is a normal force on the surface.
It will tend to convert that irregular shape to that of a
geometrical shape having least surface area which is
nothing but a sphere.

HYDROSTATIC EXCESS PRESSURE OR
GAGE PRESSURE INSIDE A BUBBLE

Capillarity
The rise & fall of liquids in tubes of smaller dia (d < 10
mm)

Capillarity is due to both Cohesion & adhesion

Note :- Wetting liquid rise & Non welding liquid falls

h  4  Cos 

d

* As temperature increase, Cohesion decrease, surface

tension () decrease,  h decrease

* For Hg, as T increase, h decrease

*  for pure water at clean glass  = 0

h  4
d

h 4
d

 h1 = d2
h2 d1

As dia. increase, capillarity decrease
Note : In all pressure measuring instrument
 = H
Here H = Rise of fluid due to pressure only & not due to
capillarity
Minimum dia. of piezometer etc. should not be less
than 10 mm.

VAPUR PRESSURE

Evaporability means a change from liquid to gaseous
phase. The evaporation rate varies for different liquids
and depends upon the prevailing pressure and temperature
conditions.

Consider a liquid enclosed in a closed space. The
liquid molecules having energy leave the liquid space in
the vapour state. The vapour molecules are in gaseous
nature and exert their own pressure, called vapour
pressure.

Note : Hg is used as Barometric fluid .

Reason :
1) Hg is only liquid which has either zero or negligible
vapour pressure, so measured heat is almost correct.

2) Since Hg has higher density hence heat measureable is
reasonable.

P = gh.

Cavitation
When liquid is passing through low pressure region,
bubbles forms. It is called cavitation.

(As pressure reduces to it is vapor pressure, liquid boils &
vaporization takes place. If liquid contains dissolved
gases, they are liberated at such low pressure.

Effects of Cavitation :
i) Lots of noise.
ii)  will have a sharp fall.
iii) Damaging of adjoining pipe boundaries.

Ques. : Cavitation is due to
a) Vapour Pressure
b) Low Pressure
c) Low Vapour Pressure
d) High Pressure

Ans (b)

Ques. : Arrange vapour pressure of Kerosene, Hg, H2O &
petroleum in ascending order
Ans. : Hg < H2O < Kerosene < Petroleum.

Continuum Concept

* The concept of continuum assumes a continuous
distribution of mass within the matter or system with no
empty space.

* Liquid satisfy continuum concept due to predominant
cohesion force in liquid molecular

* Knudsen number

Kn    Mean free path fluid
Characterstic length of
L

Kn describes degree of departure from continuum.
(Just before collision of two molecules distance between
two molecules is molecular mean free path).

* Concept of continuum holds good when Kn < 0.01
* For liquid , molecular mean free path = 0
So, Knudson no. = 0

Que. : For a fluid  = 0.048 Pa-S, Sg = 0.913 for flow
over solid plate, the velocity. of a point 75 mm away from
surface is 1.125 m/s. Calculate shear stresses at the solid
boundary & also at points25 mm, 50 mm & 75 mm away
from surface in normal direction, if velocity distribution
across the surface is
(i) Linear (ii) Parabolic with vertex at point 75 mm away
from surface.

Sol. : (i) Linear velocity distribution

du is smae at all pionts
dy

so, du = 1.125
dy 0.075

 = du = 0.048 x 1.125  0.72N / m2
dy 0.075

(  is same at all locations)

(ii) Parabolic velocity distribution
Let, u = ly2 + my + n

At,

a) At y = 0 , u = 0
n=0

b) At y = 0.075 m, u = 1.125 m/s
1.125 = 0.075 2 l + 0.075

c) At y = 0.075 m , du =0
dy

du =2ly +m
dy

 0 = 2 l x 0.075 + m

Solving we get,
l = -200. m = 30

So, u = -200 y2 + 30 y, du = 30 - 400 y
dy

y(m) 0 0.025 0.05 0.075
10 0
du (/sec) 30 20 0.48 0
dy

   du 1.44 0.96

dy

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