fluid mechanics pt1

EY8152 FLUID MECHANICS & HEAT
TRANSFER
Unit I

A Control Volume approach for the Derivation of Euler’s Eqn.
(Fixed mass with
fixed identity ) (Fixed region in space)
(Fixed mass with same
identity & fixed energy)

Continuity Equation – Differential form
For a CV, the principle of conservation of mass can be stated as
Eq. A

Continuity Equation (Conservation of Mass)– Differential form
Anna University
To derive the continuity equation at a point in a fluid, the point is enclosed by an
elementary CV appropriate to the coordinate frame of reference and the influx and
efflux of mass across each surface as well as the rate of mass accumulation within the CV
is considered.

Therefore, according to the statement of conservation of mass (i.e. Eq. A) for a CV, it
can be written that

Therefore,
Continuity eqn.
for steady flow
In case of an incompressible flow, ρ = constant

Problem: Which of the following velocity fields pertain to the motion of steady, 2D flow of
an incompressible fluid?
Solution:
For a steady, 2D incompressible fluid flow the eqn. of continuity is
2 2 2 x x
e 2 2 2 2
(i) u=2x xy ; v=x 4xy y (iii)V=Ae i Ae yj
x x y
(ii)u A ; v=A log (xy) (iv)V= i+ j
y x y x y
   
 
 
2 2 2
u v
0
x y
(i) u=2x xy ; v=x 4xy y
u v
4x y; 4x 2y
x y
u v
(4x y) ( 4x 2y) 0
x y
 
 
 
  
 
    
 
 
      
 
The continuity eqn. is not satisfied and
hence the flow is not possible.

Conservation of Momentum: Equation of motion and
Momentum theorem
Anna University
Newton’s II Law of motion: The rate of change of momentum of a body is proportional to the
impressed action and takes place in the direction of the impressed action.
Euler’s Equation: The Equation of Motion of an Ideal Fluid
Using the Newton's second law of motion the relationship between the velocity
and pressure field for a flow of an inviscid fluid can be derived. The resulting
equation, in its differential form, is known as Euler’s Equation.
The equation is first derived by the scientist ‘Euler’.
Derivation:
Let us consider an elementary parallelopiped of fluid element as a control mass
system in a frame of rectangular cartesian coordinate axes as shown in Fig. The
external forces acting on a fluid element are the body forces (like gravity,
electromagnetic field, etc) and the surface forces (pressure forces).

Cont..
Anna University
A Fluid Element appropriate to a Cartesian Coordinate System
used for the derivation of Euler's Equation

Cont..
Anna University
Let Xx, Xy, Xz be the components of body forces acting per unit mass of the fluid element along
the coordinate axes x, y and z respectively.
The surface forces for an inviscid fluid will be the pressure forces acting on different surfaces
as shown in Fig.
Therefore, the net forces acting on the fluid element along x, y and z directions can be written
as
body force
surface force

Cont..
Anna University
Since each component of the force can be expressed as the rate of change of
momentum in the respective directions, we have
As the mass of system does not change with time, is constant with time and can be
taken common. Therefore we can write
Expanding the material accelerations in the above Eqns. in terms of their respective
temporal and convective components, we get
Rate of change of momentum = Body force + Surface force

After an infinitesimal time interval t , let the particle move to a new position given by
the coordinates (x + Δx, y +Δy , z + Δz).
Its velocity components at this new position be u + Δu, v + Δv and w +Δw.
Therefore we can write,
Material Derivate and Acceleration Cont..
Anna University
Let the position of a particle at any instant ‘t’ in a flow field be given by the space
coordinates (x, y, z) with respect to a rectangular cartesian frame of reference.
The velocity components u, v, w of the particle along x, y and z directions respectively can
then be written in Eulerian form as
u = u (x, y, z, t)
v = v (x, y, z, t)
w = w (x, y, z, t)
 
 
 
u u u x x, y y,z z,t t
v v v x x, y y,z z,t t
w w w x x, y y,z z,t t
          
          
          
The expansion of the RHS of the above eqns. In the form of Taylor’s series


Cont..
Anna University
The increment in space coordinates can be written as -
x u t; y v t; z w t        

Cont..
Anna University
The above equations (1a to 1c) tell that the operator for total differential with respect to
time, D/Dt in a convective field is related to the partial differential as:
; ;
Explanation of the above equation
The total differential D/Dt is known as the material or substantial derivative with respect to
time.
The first term in the right hand side of is known as temporal or local derivative which
expresses the rate of change with time, at a fixed position.
The last three terms in the right hand side of are together known as convective derivative
which represents the time rate of change due to change in position in the field.
t


1a, 1b, 1c

Cont..
Anna University
The terms in the left hand sides of Eqs (1a) to (1c) are defined as x, y and z components of
substantial or material acceleration.
The first terms in the right hand sides of Eqs (1a) to (1c) represent the respective local or
temporal accelerations, while the other terms are convective accelerations.
Thus we can write,
;
(Material or substantial acceleration) = (temporal or local acceleration) + (convective
acceleration)

Cont..
Anna University
FLOW CLASSIFICATION:

Cont..
Anna University
Important points:
1. In a steady flow, the temporal acceleration is zero, since the velocity at any point is
invariant with time.
2. In a uniform flow, on the other hand, the convective acceleration is zero, since the
velocity components are not the functions of space coordinates.
3. In a steady and uniform flow, both the temporal and convective acceleration vanish
and hence there exists no material acceleration.
Existence of the components of acceleration for different types of flow is
shown in the table below.

Problem Cont..
Anna University
Solution:

Eulers Eqn. Cont..
Anna University
These Eqns. are valid for both incompressible
and compressible flow
These Equations can be put into a single vector
form as
or
DV p
X
Dt

 

(2)
(3)
Eqn. 2 is the well known Euler’s eqn. in vector form, while eqn. 3 describe the Euler’s
eqns. in a rectangular cartesian coordinate system

Anna University
Mechanical Energy Equation
The above Eqn. is based on the assumption that no work or heat interaction between a fluid
element and the surrounding takes place.
The first term represents the flow work per unit mass, the second term represents the
kinetic energy per unit mass and the third term represents the potential energy per unit
mass.
Therefore the sum of three terms in the left hand side of the above Eqn. can be considered
as the total mechanical energy per unit mass which remains constant along a streamline for
a steady inviscid and incompressible flow of fluid. Hence this Eqn. is also known as
Mechanical energy equation or Bernoulli’s equation.
Bernoulli’s Equation (Conservation of Energy)

Cont..
Anna University
Each term in the above eqn. has the dimension of energy per unit mass. The eqn. can also
be expressed in terms of energy per unit weight as
In a fluid flow, the energy per unit weight is termed as head. Accordingly, the above
equation can be interpreted as
Pressure head + Potential head + Velocity head =Total head (total energy per
unit weight).
Bernoulli's Equation with Head Loss
The above Mechanical Energy eqn. is derived based upon the exclusion of frictional work
done by a moving fluid. This requires the description of shear stresses and their dependence
with shear rates in a flow field.
Piezometric head

Cont..
However, in many practical situations, problems related to real fluids can be analysed with
the help of a modified form of Bernoulli’s equation as
where, hf represents the frictional work done (the work done against the fluid friction) per
unit weight of a fluid element while moving from a station 1 to 2 along a streamline in the
direction of flow.
The term hf is usually referred to as head loss between 1 and 2, since it amounts to
the loss in total mechanical energy per unit weight between points 1 and 2 on a streamline
due to the effect of fluid friction or viscosity.
It physically signifies that the difference in the total mechanical energy between
stations 1 and 2 is dissipated into intermolecular or thermal energy and is expressed as loss
of head hf in the above eqn.
For an inviscid flow hL = 0, and the total mechanical energy is constant along a
streamline.

Cont..Bernoulli’s Eqn. - Definition
The sum of the kinetic, potential, and flow energies
of a fluid particle is constant along a streamline during
steady flow when the compressibility and frictional effects
are negligible.
Bernoulli equation can be viewed as the “conservation of
mechanical energy principle”
In the light of Newton’s second law of motion, the Bernoulli
equation can also be viewed as the work done by the pressure and
gravity forces on the fluid particle is equal to the increase in the
kinetic energy of the particle.
Assumptions in Bernoulli’s Eqn.:
Inviscid flow (ideal fluid, frictionless)
Steady flow
Along a streamline
Constant density (incompressible flow)
No shaft work or heat transfer

Cont..Bernoulli’s Eqn. - Limitation
Care must be exercised when applying the Bernoulli equation since it is an
approximation that applies only to inviscid regions of flow.
In general, frictional effects are always important very close to solid
walls (boundary layers) and directly downstream of bodies (wakes). Thus, the Bernoulli
approximation is typically useful in flow regions outside of boundary layers and wakes,
where the fluid motion is governed by the combined effects of pressure and gravity forces.

Cont..Bernoulli’s Eqn. - Limitation
Frictional effects and components that disturb the streamlined structure of flow in a
flow section make the Bernoulli equation invalid.

Cont..Bernoulli’s Eqn. - Application
1) Pitot tube
If a stream of uniform velocity flows into a blunt body, the streamlines take a pattern
similar to this:
Note how some move to the top and some to the bottom. But one, in the centre, goes to
the tip of the blunt body and stops. It stops because at this point the velocity is zero - the
fluid does not move at this one point. This point is known as the stagnation point.

Cont..
The blunt body stopping the fluid does not have to be a solid. It could be a static column
of fluid. Two pressure measuring device, one as normal and one as a Pitot tube within the
pipe can be used in an arrangement shown below to measure velocity of flow.
Using the above theory, we have the eqn. for p2
which is an expression for velocity obtained from two pressure measurements and the
application of the Bernoulli equation. This equation is for ideal flow only. To
account for real fluid effects, the equation can be modified into
where is the Cv - coefficient of velocity to be determined empirically.

Cont..
Anna University
The necessity of one piezometer and one Pitot
tube and thus two readings make this
arrangement a little awkward. Connecting the
two tubes to a manometer would simplify
things but there are still two tubes. The Pitot
static tube (Pitot Darcy tube)
combines the tubes, and they can then be
easily connected to a differential manometer.
The holes on the side of the tube connect to
one side of a manometer and register the
static head, (h1), while the central hole is
connected to the other side of the manometer
to register, as before, the stagnation head (h2).
The difference of the two heads, being the
dynamic head, is now measured directly by the
differential manometer.

Anna University
3Weight mg
Specific weight( ) = g (N/m )
Volume V
   
Energy Grade Line (EGL)
Piezometric Line (Hydraulic Grade Line - HGL)
2
1 1
1
p v
The Total Energy = z
2g
 

HGL & EGL
Ideal Total Energy Line

Cont..
EGL declines continually along the flow direction due to friction and other
irreversible losses in the flow.
EGL cannot increase in the flow direction unless energy is supplied to the fluid.
HGL can rise or fall in the flow direction, but can never exceed EGL.
The downward slope of both grade lines is larger for the smaller diameter section
of pipe since the frictional head loss is greater.
Finally, HGL decays to the liquid surface at the outlet since the pressure there is
atmospheric
Free discharge from a reservoir through a horizontal pipe with a diffuser
Dynamic head = EGL - HGL

Cont..
Idealized Bernoulli-type flow

Cont..
A steep jump occurs in EGL and
HGL whenever mechanical energy is
added to the fluid (by a pump, for
example). Likewise, a steep drop
occurs in EGL and HGL whenever
mechanical energy is removed from
the fluid (by a turbine, for example),
as shown in Fig.
steep jump
steep drop

Problem
A pump draws a solution having specific gravity of 1.85 from a storage tank through
a 7.5 cm steel pipe in which the flow velocity is 1.0 m/s. The pump discharges
through a 5 cm steel pipe to an overhead tank, the end of the discharge pipe is 15 m
above the level of the solution in the feed tank, and the friction losses in the entire
piping system are 5 m. What pressure must the pump develop and what should be
its power? Take pump efficiency as 60%.

Cont..
Solution:
Anna University

Problem
A pipe 300 m long has a slope of 1 in 100 and tapers from 1 m
diameter at the high end to 0.5 m at the low end. Quantity of water
flowing is 5400 lpm. If the pressure at the high end is 70 kPa, find the
pressure at the low end.

Basic kinematic modes of a fluid
The movement of a fluid element in space has three distinct features simultaneously.
Translation
Rate of deformation
Rotation (or spin)
Pure translation in absence of
rotation and deformation of a
fluid element in a 2D flow
described by a rectangular
cartesian coordinate system.
Translation and rotation without
deformation represent rigid-body
displacements which do not induce any
strain in the body
Translation is essentially the gross motion of the parcel of fluid. The other two phenomenon
represent relative motions and distortions superimposed upon the gross movement. All three
phenomena are, in general, independent of each other, in the sense that one can occur in the
absence of the others

Cont..
Anna University
In absence of deformation and rotation,
a) There will be no change in the length of the sides of the fluid
element.
b) There will be no change in the included angles made by the
sides of the fluid element.
c) The sides are displaced in parallel direction.
This is possible when the flow velocities u (the x component velocity)
and v (the y component velocity) are neither a function of x nor of y,
i.e., the flow field is totally uniform.

Translation with Linear deformation Cont..
Now consider a situation where a component of flow velocity becomes the function
of only one space coordinate along which that velocity component is defined.
For example,
•if u = u(x) and v = v(y), the fluid element ABCD suffers a change in its linear
dimensions along with translation
• there is no change in the included angle by the sides as shown in Fig.
Fluid element in
translation with
continuous linear
deformation

Cont..
The changes in lengths along the coordinate axes per unit time per unit original lengths
are defined as the components of linear deformation or strain rate in the respective
directions.
Therefore, linear strain rate component in the x and y direction is given below
Translation with Linear Deformations
Observations from the figure:
Since u is not a function of y and v is not a function of x
•All points on the linear element AD move with same velocity in the x direction.
•All points on the linear element AB move with the same velocity in y direction.
•Hence the sides move parallel from their initial position without changing the included
angle.
This situation is referred to as translation with linear deformation.
Strain rate:
&

Cont..
Fluid element in translation with simultaneous linear and angular deformation
Let us consider both the velocity component u and v are functions of x and y, i.e.,
u = u(x,y) ; v = v(x,y)
* Point B has a relative displacement in y direction with respect to the point A.
* Point D has a relative displacement in x direction with respect to point A. Hence the
included angle between AB and AD changes.
* The fluid element suffers a continuous angular deformation along with the linear
deformations in course of its motion

Cont..
Definition: Rate of Angular deformation ( )
It’s defined as the rate of change of angle between the linear segments AB and AD
which were initially perpendicular to each other.
xy
xy
d d
dt dt
BE DF
tan ;tan
AE AF
for small angles, tan and tan
uv y tx t
BE DF yxtan ; tan
uAE AF v
x 1 t y 1 t
x y
 
  
   
     
   
     
   
            

Cont..
1.The transverse displacement of B with respect to A and the lateral displacement
of D with respect to A can be considered as the rotations of the linear segments AB
and AD about A.
2.This brings the concept of rotation in a flow field.
Definition of rotation( ) at a point:
The rotation at a point is defined as the arithmetic mean of
the angular velocities of two perpendicular linear segments
meeting at that point.
The angular velocities of AB and AD about A are respectively, but in
the opposite sense.
d d
&
dt dt
 
Considering the anticlockwise direction as positive, the rotation at A can be
written as,


Cont..
xy
z
0
0
 
 
In another special case

Cont..
Rotation in a flow field can be expressed in a vector form as
x y z
1
i j k
2
       
Irrotational Flow

Vorticity
Definition: The vorticity Ω in its simplest form is defined as a vector
which is equal to two times the rotation vector
Vorticity Ω, may be expressed as   x y z
i j k
V i j k
x y z
u v w
 
 
            
   
 
 
2 V    
For an irrotational flow, vorticity components are zero
The vorticity components are separately given by
x x y y
z z
w v u w
2 ; 2
y z z x
v u
2
x y
     
                 
  
     
  
Note: When the components of
rotation or vorticity zero, then the
motion is described as irrotational.

Problem- Rotational
The velocity components of a 3D, incompressible fluid flow are prescribed as:
Make calculations for the third component of velocity. Further check whether flow is
irrotational?
2 2 2 2
u x z 5; v y z 3     

Stream Function(ψ)
dy
dx
e
dψ
c
vx
vy
B
B
b
b
a
a
A
A
x
y
o
Let aa and bb two streamlines in a flow bounded by
solid boundaries AA and BB
If the streamline aa is defined by ψa, which will be labelled by a numerical value
representing the flow rate per unit depth between AA and streamline aa, then, a ocQ 
and, similarly, if thereforeb oeQ  b a ced Q     
x y
x y
d v dy v dx
v dy v dx
f(x,y)
  
  
 
 
Hence, the total derivaive
x y
d dx dy
x y
comparing eqns c and d
v ; v
y x
 
  
 
 
  
 
c d

Stream Function(ψ)
The concept of stream function is a direct consequence of the principle of continuity.
Let us consider a 2D steady incompressible flow parallel to the x-y plane in a rectangular
cartesian coordinate system. The flow field in this case is defined by
u=u(x,y) & v=v(x,y)
It ‘s differentiation w.r.t x gives the velocity in y-direction (generally taken as –ve) and its
differentiation w.r.t y gives the velocity in x-direction.
Since ψ is a function of x and y, its total differential is
Stream function is constant along a streamline. The streamlines are thus lines of constant ψ.
However, the ψ varies from one streamline to another. Each streamline of flow pattern can be
represented as
ψ=C1 ; ψ=C2 etc.
u ;v
y x
 
  
 
d dx dy
x y
 
  
 

Cont..
ψ is constant along a streamline, i.e., dψ=0
Therefore, udy-vdx=0 ψ(x,y)=constant
• Substitute for u and v in terms of ψ in the expression of vorticity
For irrotational flow, the vorticity is zero and hence, (a)
Eqn (a) is the Laplace eqn. Thus the ψ for irrotational motion satisfies the Laplace eqn.
v dy
u dx

z z
2 2
2 2
v u
2
x y x x y y
x y
         
                    
    
   
   2 2
2 2
0
x y
   
 
 

Cont..
Substitute for u and v in terms of ψ in the continuity eqn. for an incompressible fluid
which is true
Hence the stream function satisfies the eqn. of continuity
• The algebraic addition of the stream functions of the individual flow fields gives the another ψ
•The stream function can also be defined as the flux or flow rate between two streamlines. The
units of ψ are m2/s; discharge per unit thickness of flow.
2 2
u v
0 or 0
x y x y y x
or
x y x y
0
        
  
 
                
 
 
    
 

1 2
1 2( ) ( )
x x x x
   
      
   

Problem
A 2-D flow is described by the velocity components:
u=5x3 and v=-15x2y
Evaluate the stream function, velocity and acceleration at point P (x=1m and y=2m)
Solution:
(i) At the point P, i.e., at x=1, y=2
u=5(1)3 = 5 m/s ; v=-15 (1)2 x 2= -30 m/s
resultant velocity, V =
(ii) For steady flow
2 2
5 30 925 30.41m/s  
3 2 2 5 2
x
5 2
x
y
3 2 2
u u
a u v 5x (15x ) ( 15x y)(0) 75x m / s
x y
Acceleration in x direction at x=1, y=2
a 75(1 ) 75m / s
v v
Furthur a u v
x y
5x ( 30xy) ( 15x y)( 15x )
1
 
     
 

 
 
 
 
    
 4 4
50x y 225x y

Cont..
Acceleration in y direction at x=1, y=2
ay= - 150(1)4(2)+ 225 (1)4(2) =150 m/s2
Thus the resultant acceleration
2 2 2 2 2
x y
2 3 3
3
3 2
a a a 75 150 167.7m / s
(iii) w.k.t., the stream function is
d dx dy vdx udy
x y
=15x ydx 5x dy d(5x y)
upon integration, it gives
=5x y
(1,2) 5(1) (2) 10m / s
    
 
     
 
 

  

Potential Function(Ф)
Its derivative in any direction gives the velocity in that direction,
(b)
The function Ф is called the velocity potential function, and lines of constant potential
function are termed equipotential lines.
Characteristics of a Ф are:
• Since Ф is a function of x and y alone, its total differential is:
For an equipotential line, the Ф is constant, i.e., d Ф=0. Therefore for an equipotential line
which prescribes the slope of equipotential line at any point.
u ;v
x y
 
 
 
d dx dy udx vdy
x y
 
    
 
dy u
udx vdy 0;
dx v
   

Cont..
• Substitute for u and v in terms of potential function in the expression for vorticity
Hence the velocity potential satisfies the condition for irrotational flow.
The existence of velocity potential function means that a possible flow must be
irrotational. Consequently the 2D irrotational flow is often called the potential flow
• Combining eqn.b with the continuity eqn.
•The potential function decreases along the flow direction and accordingly some authors
prefer to define
z z
v u
2 0
x y x y y x
         
                     
2 2
2 2
u v
0 or 0
x y x x y y
or 0
x y
         
    
    

          
 
  

u ;v
x y
 
 
 
Therefore it satisfies the Laplace eqn.

Cauchy Riemann Equations
The above discussion with reference to stream and velocity potential function leads us to
conclude that:
(i) Stream function applies to both rotational and irrotational flows. The flow has only to be
steady and incompressible,
(ii) Potential function exists only for irrotational flow
(iii) For irrotational flow , both ψ and Ф satisfy Laplace eqn; consequently they are
interchangeable. The relationship between the ψ and Ф for a steady, irrotational
incompressible flow
In hydromechanics, these eqns. are sometimes called the Cauchy-Riemann eqns.
u ;v
y x x y
   
    
   

2D flow in terms of ψ and φ function - Comparison
y
r
r
xv u ;v v
y x x y
1 1
v ;v
r r r r
u ; v
y
u v
0
x y
v(rv )
0
r
( u) ( )
x y
x
x
v
0
y


   
   
      
   
   
    
   

 
 
 

  
   
 
 
   
 
  
  



Incompressible Continuity eqn. in cartesian coordinate
Incompressible Continuity eqn. in cylindrical polar coordinate
Compressible Continuity eqn. in cartesian coordinate

Problem
Check whether the following functions satisfy continuity and are valid potential functions; A is a
numerical constant.
Solution:
For the function to satisfy both continuity and the requirements of potential function, it must
satisfy the Laplace eqn.
2 2
e
A
(i) (x y ),(ii) A(cosx siny)
2
(iii) Alog xy
     
 
2 2
2 2
0
x y
    
  
  
2 2
2
2
2
2
A
(i) (x y )
2
Ax; A
x x
Ay; A
y y
  
  
 
 
  
   
 
2 2
2 2
A ( A) 0
x y
    
     
  
The function is possible for continuity and is a potential function.

Cont..
(ii) A(cosx siny)  
2 2
2 2
A(cosx siny) 0
x y
    
     
  
It is not valid potential function.
e(iii) Alog xy 
2 2
2 2 2 2
A A
0
x x yy
      
       
    
It is not valid potential function.

Problem
Does the velocity potential exist for the 2D incompressible flow prescribed by
u=x-4y; v=-(y+4x)
If so determine its form as well as that of stream function
Solution:
From the given velocity components
u v
1; 1;
x y
u v
1 1 0
x y
 
  
 
 
   
 
The continuity eqn. is satisfied and the flow is possible.
 
i j k
V 0
x y z
u v w
 
 
       
   
 
 
Further,

Cont..
 
     
   
i 0 (y 4x) j 0 (x 4y
y z x z
+k y 4x (
i j k
V
x y z
x 4y (y 4x) 0
x 4y ) 0
x y
     
     
 
        
  
      
 
      
   
 
  
 
 
The zero vorticity implies irrotational flow and hence the velocity potential does exist.

Cont..
(a) The differential dФ for the velocity potential function is
2 2
d dx dy udx vdy
x y
= (x-4y)dx-(y+4x)dy = xdx-ydy-4(ydx+xdy)
=xdx ydy 4d(xy)
which on integration gives
(b)The differential d for str
x y
4xy cons
eam funct
tant
2
ion is
d dx dy
x y
2
 
    
 
 

 
  
 
 
  
 
2 2
vdx udy 4xdx 4ydy d(xy)
which on inte
2x 2y xy constant
gration gives
 
  
 
 

Boundary layer (BL) concept
•A major breakthrough in fluid mechanics occurred in 1904 when Ludwig
Prandtl (1875-1953), German Scientist introduced the concept of boundary
layer and derived the equations for boundary layer flow by correct reduction of
Navier-Stokes equations.
•BL approximation given by the Prandtl’s is used to bridge the gap between
Euler and Navier-Stokes eqn.

Cont…
•He hypothesized that for fluids having relatively small viscosity, the effect of
internal friction in the fluid is significant only in a narrow region surrounding
solid boundaries or bodies over which the fluid flows.
•Thus, close to the body is the boundary layer where shear stresses exert an
increasingly larger effect on the fluid as one moves from free stream towards
the solid boundary.
•However, outside the boundary layer where the effect of the shear stresses on
the flow is small compared to values inside the boundary layer (since the
velocity gradient is negligible), thus
* the fluid particles experience no vorticity and therefore,
* the flow is similar to a potential flow.
•Hence, the surface at the boundary layer interface is a rather fictitious one,
that divides rotational and irrotational flow.
u
y



Cont…
•The below figure shows the Prandtl's model regarding boundary layer flow.
•Hence with the exception of the immediate vicinity of the surface, the flow is
frictionless (inviscid) and the velocity is U (the potential velocity).
•In the region, very near to the surface (in the thin layer), there is friction in the
flow which signifies that the fluid is retarded until it adheres to the surface (no-
slip condition).
•The transition of the mainstream velocity from zero at the surface (with
respect to the surface) to full magnitude takes place across the boundary layer.

Cont…
Additional features of BL:
•The condition is true for the BL zone; whilst the conditions for flow
beyond the BL and at its outer border are:
•The shear stress acting at the plate is defined by . This stress set up a
shear force which opposes the fluid motion and fluid close to the wall is
decelerated.
Due to increasing plate surface area along the flow direction, more and
more of the fluid is retarded and so the thickness of the BL also increases.
•In addition to the distance along the flat plate (x), the BL growth is also
governed by the magnitude of the incoming fluid velocity and the kinematic
viscosity of the flowing fluid. (BL thickness less for high velocity and greater for
high kinematic viscosity)
•For smooth plate laminar BL exists till Re = 3x105 and the transition may be
delayed until Re= 5x105. Beyond which the layer treated as turbulent BL.
u
0
y



o
u
0 & u = U or U
y 


 u
y

  


Cont…
•However, for rough plates, turbulent approach flows, vibrations, acoustic
noise, flow unsteadiness and curvature of the wall contribute to an earlier
transition at much lower values.
•The below figure depicts the distribution of shear stress along the plate in the
flow direction.
u
y

  


Cont…
•The turbulent BL does not extend to the solid surface. An extremely thin layer
called laminar sublayer, is formed wherein the flow is essentially of laminar
character.
•In a laminar BL, the velocity gradient becomes less steep and parabolic in
nature. Consequently, a turbulent BL has a fuller velocity profile and a much
steeper velocity gradient at the plate surface.
Fig. Comparison of Laminar & Turbulent BL

Cont…
•Theoretically, for an infinite plate, the BL goes on thicknening indefinitely.
However, in practice, the growth is curtailed by other surfaces in the vicinity.
Re<2000
Re>2000

BL Separation
The boundary layer thickness is considerably affected by the pressure gradient in the
direction of flow. If the pressure gradient is zero then the boundary layer continues to
grow in thickness along a flat plate.

BL Separation
With the decreasing pressure in the direction of flow (- ve pressure gradient), the BL
tends to be reduced in thickness.
However with the pressure increasing in the direction of flow, with positive pressure
gradient, the BL thickens rapidly.
The adverse pressure gradient plus the boundary shear decreases the momentum in the
BL, so the BL is deflected sideways from the boundary, separates from it and moves into
the main stream. This phenomenon is called boundary layer separation.
As a result of the reverse flow, large irregular eddies are formed in which much energy is
dissipated as heat.
Separation occurs with both laminar and turbulent BLs. Laminar boundary layer is more
prone to separation than turbulent BL.
Since the separation of the BL gives rise to additional resistance to flow, attempts should
be made to avoid separation by some means.

Development of BL over airfoil

Methods of controlling the BL
injecting fluid into the BL – Injection method
slotted wing – By Pass Method
suction of fluid through BL - Suction method
& streamlining of the body shapes

BL Parameters
BL thickness (δ):
BL thickness is defined as that distance from the surface where the
local velocity equals 99% of the free stream velocity
(u 0.99U )y  
Displacement thickness (δ*):
y
y+δ*
Ideal fluid flow over
solid boundary
(absence of BL)
Real fluid flow over
solid boundary
(formation of BL)
It is the distance measured
perpendicular to the solid
boundary, by which the free
stream is displaced due to
formation of BL

Cont…
The displacement thickness for the BL may be defined as the
distance measured perpendicular to the solid boundary, by which
the boundary is displaced to compensate the reduction in mfr of
the flow fluid due to formation of BL.
*
s0
u
1 dy
U

 
   
 

Momentum thickness (θ):
s s0
u u
1 dy
U U

 
   
 

The momentum thickness for the BL may be defined as the
distance measured perpendicular to the solid boundary, by which
the boundary is displaced to compensate the reduction in
momentum of the flow fluid due to formation of BL.

Cont…
Energy thickness (δ**):
2
**
s s0
u u
1
U U
   
     
   

Shape factor (H):
*
H



The energy thickness for the BL may be defined as the distance
measured perpendicular to the solid boundary, by which the
boundary is displaced to compensate the reduction in KE of the
flow fluid due to formation of BL.

Problem
2 1/ m
o o
u y y u y
(a) 2 ; (b)
U U
     
              
2
*
o0 0
2 3 2 3
2 2
0
*
u y y
1 dy 1 2 dy
U
2 y 1 y 2 1
y
2 3 2 3
3
 

      
                   
        
             
          

 
 
Determine the displacement, momentum thickness in terms of the
nominal boundary layer thickness ( ) in respect of the following
velocity profiles:
where ‘u’ is the velocity at a height ‘y’ above the surface and Uo is
the free stream velocity. Also find shape factor
Solution:
(a)


Problem cont..
2 2
o o0 0
2 3 4
0
2 3 4 5
2 3 4
0
u u y y y y
1 dy 2 1 2 dy
U U
y y y y
= 2 5 4 dy
2 y 5 y 4 y 1 y
2 3 4 5
2
 


           
                                 
        
                      
       
          
          

 

2 3 4 5
2 3 4
5 4 1
2 3 4 5
5 1 2
1 1
3 5 15
          
         
          
 
        
 
*
3 5
H
2 2
15
 
  
 

Problem cont..
(b)
1/ m
*
o0 0
m 1 m 1
m m
1/ m 1/ m
0
*
u y
1 dy 1 dy
U
1 m m 1
y y
m 1 m 1
m m
1
m 1 m 1 m 1
 
 
    
             
      
   
   
           
     
 
1/ m 1/ m
o o0 0
1/ m 2/ m
0
m 1 m 2
m m
1/ m 2/ m
0
u u y y
1 dy 1 dy
U U
y y
= dy
1 m 1 m
y y
m 1 m 2
 

 
      
                  
     
          
 
   
 


Problem cont..
m 1 m 2
m m
1/ m 2/ m
*
1 m 1 m
m 1 m 2
m m m
m 1 m 2 (m 1)(m 2)
(m 1) m 2
H
m (m 1)(m 2) m
 
   
   
      
   
   
  
   

fluid mechanics pt1

Recommended

Recommended

More Related Content

What's hot

What's hot (20)

Similar to fluid mechanics pt1

Similar to fluid mechanics pt1 (20)

More from Shanu Jp

More from Shanu Jp (6)

Recently uploaded

Recently uploaded (20)

fluid mechanics pt1