Welcome back to the video course on fluid
mechanics. So in the last lecture, we were
discussing about the potentional flow of theories
and then we were discussing superposition
of elementary flows.
So in the last lecture we discussed about
the various kinds of superposition like direct
method, inverse method and then in superposition
process we have seen how we are superposing
uniform flow and then a source how we can
superpose to get a half body that we have
seen in the last lecture and also we have
seen how to superpose a two uniform flow with
source and sink of equal strength with uniform
flow a combination of source and sink of equal
strength with uniform flow to get rankine
ovals.
This also we have seen, from the uniform flow
equations and then the source and sink equations
how we are deriving the potential expression
for potential and then stream function and
then, how we are getting the velocities and
various other flow parameters.
Then we have seen after the rankine ovals,
we have also seen the flow past circular cylinders.
Flow past circular cylinder we can generate
by superposition of a doublet with a uniform
flow which we have seen earlier.
A doublet is superposed in a uniform flow.
Then, with respect to superposition of a doublet
and uniform flow, we have seen that the stream
function can be represented by si is equal
r sine theta minus k sine theta by r and velocity
potential phi is equal to U r cos theta plus
k cos theta by r and then the flow past circular
cylinders can be seen in this slide.
This is the cylinder and then, this is the
uniform flow and then with respect to the
doublet and uniform flow, we are analyzing
flow past through circular cylinder and then
we got the various parameters here for the
expressions for si and phi with respect to
the potential flow and then the uniform flow
and then the doublet. So, up to that we have
seen how to determine the pressure with respect
to the Bernoulli’s equation.
So with respect to the previous figures here
in the slide, say here as is the radius of
the cylinder and then say u is the velocity
of uniform flow. Then we can use to find the
pressure distribution , here since the maximum
velocity on surface of cylinder at r equal
to a that means theta is equal to plus or
minus phi by 2 V r is equal to 0 and the tangential
velocity on surface V theta is equal to minus
2 U sine theta and pressure distribution is
obtained from the Bernoulli’s equation p0
plus 1 by 2 rho u square is equal to ps half
rho v theta square where p0 is the pressure
far from cylinder and U is the velocity far
from the cylinder and ps is the surface pressure.
So surface pressure we can obtain as ps is
equal to p0 plus half rho U square into 1
minus 4 sine square theta with reference to
this figure here. So this you have seen in
the last lecture and now say as far as since
the cylinder is immersed in a fluid that means
flow past circular cylinder only we are analyzing.
So definitely there will be a drag force and
lift force acting upon the cylinder, if you
have analyzed the real fluid flow cases. So
then if you analyze a flow past circular cylinder
with respect to the superposition theories,
which you have seen now the drag force parallel
to the direction of uniform flow generally
we can obtain as Fx is equal minus integral
0 to 2 phi ps cos theta a d theta with reference
to this previous figure here.
A is the radius of the cylinder and then theta
is mentioned here in this shown here. So with
reference to this we can get the drag forces
which is parallel to the direction of uniform
flow is obtained as fx is equal to minus integral
0 to 2 phi ps cos theta a d theta and the
lift force perpendicular to direction of uniform
flow is obtained as F y is equal to minus
integral 0 to 2 phi ps sine theta a d theta
so now the substitution of ps from the previous
expression.
ps is equal to p0 plus half rho U square into
1 minus 4 sine square theta if you substitute
here for ps what you will get is this F x
is equal to 0 and F y is equal to 0. So with
respect to this figure here, Fx is in this
direction x direction and which is the drag
force Fy is the lift force in the normal vertical
direction y and then ps is the surface pressure
which is acted on the surface like this. So
if you substitute these values with respect
to our earlier analysis you get the drag force
Fx is equal to 0 and the lift force Fy is
equal to 0 with respect to the superposition
of this doublet and uniform flow.
But, we can see that in the real case in the
real fluid flow case, this is impossible.
Since definitely there is lift drag force
and also possibility of lift force with respect
to that there is a flow and surrounding the
cylinder or the cylinder is immersed in a
flow. So there is definitely drag force and
lift force, So the experience that there is
a drag on the cylinder when it is placed in
moving fluid due to the viscosity, so this
discrepancy, so when we analyze these kinds
of problem with respect to the potential flow
theory or the superposition of the elementary
flows with which we have seen so far, then
some parameters are not obtained properly
under predicted like drag force is 0.
Here lift force is 0, which is not realistic
when we consider real fluid flow where the
viscosities are also considered. So this discrepancy
is called the d’Alembert’s law paradox
that means when they are using the potential
flow theory for the real fluid flow say like,
the flow surrounding a cylinder or which we
have seen here so the drag force becomes 0
or the lift force is 0 which is not realistic
as per the real fluid flow is concerned.
So this discrepancy is called the d Alembert’s
paradox so that means the potential theory
we can see here, it incorrectly predicts the
drag on a cylinder is 0 which is not realistic.
So that is what is called d’Alembert’s
paradox. So here the reason is obvious that
here we are considering as potential flow
and then viscosity is not considered then
effectively, when we superpose the uniform
flow and then the doublet in this particular
case just like flow past circular cylinder
it will be all the parameters will not be
predicted properly, but as we can see the
velocities and then the pressure can be predicted
with respect to this superposition of the
uniform flow and doublet as far flow past
a circular cylinder is considered.
So that means when we use this potential flow
theory for many of the real flow problems
we should be very careful which parameter
is obtained getting proper accurate which
parameter is not predicted properly. So we
should be very careful while using the potential
theory in the case of real fluid flow problems,
so that is very important.
So now with respect to this superposition
of potential flows before completing or before
closing this chapter we discussed numerical
examples here. We can see we will first see
superposition of potential flows. First example
is superposition of potential flow. So the
problem is a source with a flow volume rate
of 0.3 meter square per second and vertex
with 1.5 meter square per second are located
at the origin. Derive the equations for velocity
potential and stream function; calculate the
velocity resultant at 0.7 meter.
So here, we have got a source of strength
flow rate of 0.3 meter per square seconds
and then vertex. So a source and vertex combined
superposition superpose. So vertex strength
is 1.5 meter square per second, so we have
to derive the equation of velocity and potential
and stream function and calculate the resultant
velocity. So for source which we have already
analyzed earlier as far as the source is considered
concerned we can derive we obtained the stream
functions si is equal to m by two phi theta.
Where, m is the flow volume rate which is
given as 0.3 meter square per second and then
we have also seen earlier the velocity potential
can be expressed before say source as phi
is equal to m by 2 phi natural log r, where
r is the distance between the points which
we are considering. So here, for si and phi
source is obtained as si is equal to m by
2 phi theta and phi is equal to m by 2 phi
natural log r and then also we have seen for
vertex is concerned um earlier you have seen
from the vertex si this stream function can
be defined as si is equal to minus gamma by
2 phi natural log r and phi is equal to gamma
by 2 phi theta.
So here, this is gamma this is vertex strength
1. 5 meter square per second this is given.
So now for vertex we have seen si function
and phi function and m is already given as
0.3 meter square per second and gamma is equal
to 1.5 meter square per second. So now as
per our superposition theory, since now we
are dealing here with source and then vertex,
so we can use the combined equation or the
combined equation for the stream function
will be si is equal to so now which we have
already seen here m by 2 phi theta is for
si function so m is equal to point 3 by 2
phi into theta then we superpose the for the
vertex si function which is minus gamma by
2 phi natural log r.
So that means si is equal to point 3 by 2
phi theta, minus 1.52 phi natural log r which
can be simplified as 1 by phi 0.15 theta minus
0.75 natural log r. So now this is the expression
for the stream function for the superposed
the source and vertex which we can say in
this particular problem and now the velocity
potential is concerned phi is equal to m by2
phi natural log r for this source so m is
equal to 0.3.
So phi is equal to 0.3 by2 phi natural log
r and then say for vertex is concerned phi
is equal gamma by 2 phi theta. So, gamma is
1.5 plus 1.5 by 2 phi theta. So that is equal
to 1.1 by phi into 0.15 natural log r plus
0.75 theta.
So thus, we get the expression for stream
function and potential function after the
superposition of the source and the vertex
equation. So now we want to find the velocity
the resultant velocity. So for that we will
just calculate the radial velocity, the radial
velocity expression is V r is equal to del
phi del r so here if you differentiate this
phi function which is obtained here one differentiation
we will get V r is equal to del phi by del
r is equal to point 0.15 by phi r.
So similarly we can obtain the expression
for the tangential velocity V theta is equal
to1 o1 by pi1 by r del phi by del theta. So
the tangential velocity is equal to 1 by r
del phi by del theta. So if you differentiate
this function for phi you will get this is
equal to V theta is equal to 0.75 by phi r.
So this gives the expression for the tangential
velocity. So thus, we got the expression for
the radial velocity and expression for the
tangential velocity.
So now the next part of the problem is, we
want to find the resultant velocity at say
0.7 meter square so x is equal to 0.7 and
y is equal to 0.7.So r with respect to the
x and y we can write the position r is equal
to square root of 0.7 square plus 0.7 square
that is equal to 0.89 meter.
So that now we can obtain the value of the
radial velocity at 0.7 as Vr is equal to 0.7
is equal to since the expression for radial
velocity we obtained Vr is equal to 0.15 by
phi r. So this is equal to 0.15 by phi into
0.9899 which r is obtained. So this is equal
to V r is equal to 0.0482 meter per second
and then similarly we can find the expression
for V we obtain the value for the tangential
velocity at 0.7.
So the expression for V theta as obtained
is 0.75 by phi r so 0.75 by phi into r is
0.9899. So this is equal to 0.241., So then
once we obtained the radial velocity and tangential
velocity we can obtain the resultant velocity
is equal to V resultant is equal to square
root of Vr square plus V theta square so that
V is here, root of 0.0482 square plus 0.241
square.
So if you take square root of these, you will
get the resultant velocity is equal to point
two four five nine meter per second. So like
this here in this particular problem, we have
superposed a source and then a vertex we superposed
and then we have found the velocity for the
expression for the potential velocity potential
and then the expression for the stream function
and then finally, we found the radial velocity
and tangential velocity and we got the resultant
velocity at a particular point from the radial
velocity and tangential velocity. So this
is one example with respect to superposition
of a source and then vertex.
Now as a second numeric example here we consider
a half body.
So as we have seen the half body which is
a superposition of source and then the uniform
flow so, here the problem is a half body is
formed by superposition of a two dimensional
source of volume flow rate two meter square
per second in a uniform flow of velocity one
meter per second. Calculate the stagnation
point and maximum thickness of resulting half
body? So this is the problem.
So here half body is as we have seen in the
seen earlier half body is concerned here we
get a half body by superposition of a source
and then the uniform flow in this slide here.
So here we have superposed here the uniform
flow velocities of one meter per second then
the volume flow rate is for the source is
concerned it is two meter square two meter
square per second.
So we want to find the stagnation point and
maximum thickness of the resulting half body.
So as far as the half body is concerned various
parameters, these things we also discussed
the stagnation point and then the thickness
and all other parameters so with respect to
the figure here.
So we have diagnosed in this problem. So now
you use the Cartesian coordinate system with
respect to this figure. This figure shows
this is the origin x axis is this direction
and here this is the y axis here, there is
uniform flow of velocity 1meter per second
and here we have a source of 2 meter square
per second.
So now with respect to this figure, we can
obtain in Cartesian coordinate system, this
stream function. So there is uniform flow
and then the source is there, so with respect
to this si is equal to stream function can
be written as si is equal to U0 y plus m by
2 phi tan inverse y by x. So here y and x
is already defined y is this direction x is
this direction and m is the strength of the
source and U0 is the velocity uniform flow
velocity.
So now this stream function is obtained by
superposing the stream function for the uniform
flow and the stream function for the source
and then this we can write as si is equal
to U0 y plus, if m by 2 phi is represented
as K then plus K tan inverse y by x. So this
is we can write in this way with respect to
this figure. So this is the velocity potential
velocity potential K 1 K2 K3 etc. and then
we have the stream lines and the potential
line is drawn here for various values.
So now the velocity, the expression for velocity
is U the velocity in x direction velocity
in x direction is u is equal to minus del
si by del y u is equal to del si by del y,
so we can differentiate this expression for
the expression for si is U0 y plus K tan inverse
y by x.
So if you differentiate u is equal to del
si by del y is equal to U0 plus K x by x square
plus y square so we get u is equal to del
si by del y is equal to U0 plus K into x by
x square plus y square. So similarly the velocity
in y direction u is equal to minus del si
by del x so that is equal to K into y by x
square plus y square. So thus, we got the
expression for velocity in x direction and
velocity in y direction u and v from the obtained
expression of stream function with respect
to the superposition of the uniform flow and
source for the half body which we considered
here.
Now, as per our definition of the stagnation
points this U0 that is the velocity should
be 0 at stagnation point as defined here,
the stagnation point is shown here. So for
stagnation point u is equal to 0, v is equal
to 0. Now if you use this equation now with
respect to this if you substitute for u is
equal to 0, v is equal to 0 you will get x
is equal to minus m by 2 phi U0 where K is
taken as m by 2 phi. So x is equal to - m
by 2 phi U0, so that is equal to -m is already
given as 2 meters per second, so this is equal
to - 2 by2 phi U is equal to the velocity
is equal to 1, So -2 by2 phi into 1so this
is equal to -1 by phi.
So now we want to determine the coordinates
of the stagnation point. So from this you
will get the coordinates of the stagnation
points as a,0 or it will be 1 by phi 0, so
the coordinates of the stagnation point is
a,0 or 1 by phi 0 and this stream function
at stagnation point with respect to this now
y is equal to here say 0 here.
So we will get si stream function at stagnation
point is 0 plus m by2 phi tan inverse y 0.
So tan inverse 0. So this is equal to 2 by
m is 2, so 2 by2 phi into phi so this is equal
to 1. You obtained si stream function at stagnation
point is equal to 1.
So now the half body is described by dividing
stream lines. So as you can see this is the
half body with respect to this figure here
or this figure here, this is the half body.
So the half body is described by the dividing
stream line. So that we can write si is equal
to m by 2 so si is equal to m by 2 that is
equal to phi into m by 2 phi so that we will
get si is equal to phi K.
So that is with respect to this figure it
is obvious si is equal to phi K which we have
already seen when we derived all the equation
so we can write U0 y plus m by 2 phi tan inverse
y by x is equal to m by 2 with respect to
the half body which is shown here.
So then we can write U0 y plus m theta by
2 phi now with respect to this figure this
is equal to m by 2 or you will get y is equal
to m into 1 minus theta by phi divided by
2 U0 and at theta is equal to 0 we get y max
is equal to m by 2 U0 which is the maximum
ordinate and at theta by is equal to phi by
2. We get y is equal to m by 4 U0 that is
the upper ordinate at origin and theta is
equal to phi y is equal to 0 which is the
stagnation point.
So here with respect to this figure, everything
is clear this is stagnation point and y max
y this is y max. So, all these parameters
for this particular problems are clear with
respect to the figure here. This figure you
can see that the y max is defined here and
then si is equal to minus phi K here and si
is equal phi K here and all other parameters
are already defined so maximum width is equal
to 2 y max. So a theta is equal to 3 phi by2
y is equal to -m by 4 U0 which is the lower
ordinate at origin so here lower ordinate
at origin y is equal to - m by 4 U0.
So now the equation of half body becomes U0
y plus m by 2 phi tan inverse y by x is equal
to m by 2 and then maximum thickness occurs
as x tends to infinity is the maximum thickness
that means here the maximum thickness with
respect to figure is maximum this two times
y max. So that is equal to 1 into y since
U0 is 1 into y plus1 by phi m is to 1 by phi
tan inverse 0 that is equal to 1. So we get
y is equal to 0.5 meter so maximum thickness
with respect to this particular problem is
1 meter. Here since it is two times y max
y max is 0.5 so two times 0.5 is equal to
1 meter.
So, with respect to figure this figure. So
now we have obtained this particular problem
the superpose the half body is obtained by
superposition of a source and then a uniform
flow and then we calculated the stagnation
points and also the maximum thickness of the
resulting half body is calculated. So like
this we have say in two examples for a as
far as one is concerned in super position
of vertex and source and then another one
is superposition of a uniform flow and then
the one source is superposed, so that we go
the half body.
So like this we can superpose various flows
various elementary flows of uniform flow or
we can superpose the source sink doublet or
the vertex, so these are the elementary flows
and then we can obtain the complex flow phenomena
like which we have seen half body or the flow
past circular cylinder.
So like that or rankine oval like that we
can obtain various complex flow condition
by superposition of the elementary flows and
then this can be applied with for various
practical cases just like say in a homogenous
isotropic media like ground water flow is
concerned. Somehow the particular cases like
we can apply this superposition theories to
solve some of the complex problem.
But as I mentioned we should be careful that
here we are assuming the potential flow theory.
So we should be very careful in the use of
this superposition techniques to express determine
its certain parameters like drag lift etc.
may not predict properly.
So we should be careful in the usage of this
potential theory and then this kinds of superposition.
So now this chapter on kinematics fluid flow,
now we will summarise what we have done so
far in this particular chapter we have discussed
the kinematics of fluid motion so as we have
seen here we are here say not considering
as for the particular force is considered.
But we are considering various other parameters
like velocity field acceleration without giving
much attention to the force which is causing
the flow. So that the analysis becomes much
simple and then we have also derived with
respect to various theories.
We derived the law of conservation of mass,
the continuity equation in differential form
and integration form. We have derived the
equations and then we have seen how we can
apply for various kinds of practical problems
as far as law of conservation of mass is concerned.
Then in this chapter the kinematics of fluid
flow, we discussed the rotational and irrigational
flow and then as far as irrotational flow
is concerned which is potential is concerned
various applications we have seen potential
theory we have seen the Laplace equation how
we can utilize with respect to stream function
and potential function and then we have also
discussed the elementary potential flow uniform
flow source sink doublet and the vertex and
finally we have seen the how we can superpose
various this elementary flows to obtain complex
flows and then how we can utilize to solve
various complex problem we have discussed.
So this is chapter on kinematics of fluid
flow further the next chapter we will be discussing
dynamics of fluid flow so far we have not
given much attention to the force causing
the flow but the dynamics of fluid flow. Consider
in next chapter we will be giving much attention
to the force causing the flow. So that we
can have the problem will become much more
complex and then we will be analyzing it with
various fundamental principal of the physics
and the fluid mechanics. So on the in the
video course on fluid mechanics now we will
discuss the dynamics of fluid flow.
So the last chapter we have seen the kinematics
of fluid flow where we did not give much attention
to the force causing the flow, but with respect
to potential flow and then with respect to
various other approximations only we have
considered.
So now we will see the dynamics of fluid flow.
So as I have mentioned the dynamics of fluid
flow is concerned since we are considering,
we have to consider all the forces acting
upon it and then on the flow which is causing
the fluid flow. So now in dynamics of fluid
flow the main objectives which we have said
here, this particular video course is concerned.
First we will introduce the concepts necessary
to analyze fluids in motion. So as far as
dynamics of fluid flow is concerned and then
secondly we discussed the finite control volume
analysis with respect to which we will be
deriving the momentum equation energy equation
and various other fundamentals equations and
then we will be discussing the Euler and Bernoulli’s
theorem.
So with respect to this control volume analysis
and then we will be discussing the practical
applications of Bernoulli’s equations and
then the inverse momentum theory and finally
we will be discussing more about the momentum
and energy equation’s and its applications
with respect to the dynamics of fluid flow.
So now as I mentioned say the dynamics of
fluid flow is concerned say here we are considering
the forces causing the flow and then we are
say as far as the fluid dynamics is concerned
it is the analysis of the fluid motion.
So as we have seen the fluids statics earlier
there the motion is not considered the fluid
in is considered at rest so here the fluid
is in motion. So we are trying to analyze
with respect to say various forces acting
due to which the flow fluid movement or flow
takes place. So we are trying to analyze these
fluids in motion with respect to various fundamental
principles fundamental theories and then we
are trying to obtain the various fluid flow
properties like velocity pressure depth and
other parameters with respect to when the
particular section is considered or particular
point is considered with respect to fluid
movement.
So the motion of fluids here predicted same
way as motion of solids as I mentioned earlier
so the fluid mechanics is most of the theories
which we use in engineering mechanics or mechanics
of solids or very much valued for fluid mechanics
also.
So just like we deal with the solids in motion
or the moving solids, so like that how we
predict the various parameters like velocity
and other parameters like that only as far
as motion of fluid is concerned we are trying
to predict the various parameters very similar
to the motion of solid and then the various
fundamental laws of physics and physical properties
of fluids.
We can directly utilize just like the fundamental
laws which we use as far as motion of fluid
is concerned those laws are also very much
valued as far as fluid mechanics or fluid
dynamics of fluids is concerned then other
than this fundamental laws of physics, we
will be using various physical properties
of the fluid which say particular case which
we are concerned and then the physical properties
also will be considered so that we would be
able to solve the particular problem or we
will be able to predict the various fluid
flow properties that are considered.
So now as I mentioned the fluid flow analysis
is complex just like complex flow if you consider
the spray behind a car or wave on beaches
hurricanes or tornados or various atmospheric
phenomena like the movement of clouds or rain
fall. All these fluid flows are concerned
all these system is very complex to the movement
with respect to.
We are trying to analyze with respect to the
fluid flow or fluid motion the analysis becomes
much more complex. So now for this all these
particular problems are concerned say we want
insight into this fluid motion.
Our aim here is we want insight into this
fluid motion by using fundamental principles
like law of conservation of mass conservation
of momentum, conservation of energy and other
fundamental principles like Newton’s laws.
So that is the way which we are going to approach
the dynamics of fluid flow.
So as far as dynamics of fluid flow is concerned
any fluid element obey normal laws of mechanics
as I mentioned, the laws of mechanics which
we are using for solid mechanics is considered
most of the laws are valid for fluid flow
also concerned. So since most of the time
compared to the solid is concerned we have
a definite shape and that definite shape only
we will be considering in the analysis. But
as far as fluid is concerned we have to, it
is continuously flowing we have to consider
instead of considering the total flow we may
be considering particular element or particular
control volume with respect to that only what
happens we will be analyzing. So that is the
difference between the solid analysis of motion
of solids and the analysis of motion of fluids.
So now when a force is applied say we can
obtain the behavior with respect to the application
of forces using Newton’s three fundamental
laws. This Newton’s laws, which are applicable
to solids movement or motion of solid, these
are also valid for fluid also. So as far as
the force is applied the behavior we can obtain
from Newton’s law three laws.
First law is a body will remain at rest or
in a state of uniform motion in a straight
line until acted upon by an external force.
So this is one of the fundamental principles
which we use always in physics and also solid
mechanics and now this we will be also using
fluid mechanics theories of fluid mechanics
principles the development of fluid mechanics
principles are concerned.
So the first law is body will remain at rest
or in a state of uniform motion in a straight
line until acted upon by an external force.
So if there is only external force acting
upon the body. So here the body means here
the fluid then only if there is no external
force the body will remain at rest or in a
uniform flow in a state of uniform motion
in a straight line there are no external force
it will be a rest or it will be keep on moving
And second Newton second law is rate of change
of momentum of body is proportional to the
force applied and takes place in direction
of action of that force. So this is the Newton’s
second law.
So here, the rate of change of momentum of
a body is always say when a force is acting
upon the momentum will be we have to consider
the momentum of the body. So the rate of change
of momentum is proportional to the force applied
and then takes place in the direction of that
force which is acting upon the body. So this
is Newton’s second law.
So this Newton’s second law is very much
used in fluid mechanics principles fluid mechanics
theories most of the fluid mechanics theories
we have derived based upon this Newton’s
second law, based upon which we can obtain
force is equal to mass into acceleration and
other related theories which we will be discussing
further and third Newton’s law is action
and reaction are equal and opposite so these
also we can use as far as dynamics of fluid
flow is concerned.
So the basic fundamental principles just like
Newton’s laws are we can definitely utilize
as far as dynamics of fluid flow is concerned
we will be using these principles to derive
various fundamentals equations as far as dynamics
of fluid flow is concerned.
So now we will be looking into what are the
different forces acting as far as fluid flow
is concerned. So as I mentioned say this Newton’s
second law is one of the fundamental law which
we will be utilizing to derive most of the
equations concerned to dynamics of fluid flow.
So as I mentioned force is equal to mass into
acceleration. So this we will be utilizing
to derive like conservation of mass conservation
of energy and all these principles so here
the various forces say as far as fluid.
So when the fluid motion is concerned the
various forces which have influenced the fluid
motion are: first one is the gravity force.
Gravity force has got a definite impact upon
the fluid motion. So first one which we will
be generally considering is gravity force
and second one is fluid pressure, based upon
which the fluid is moving from one position
to another. So second force influence the
fluid motion is the fluid pressure and third
force is due to the molecular viscosity within
the fluid itself and then fourth one is the
surface tension say which is acted upon fluid
surface, so surface tension is important and
then if the fluid is compressible or if it
is incompressible so the compressibility of
the fluid that affects compressible force
that we have to consider as far as the fluid
motion is concerned.
And then lastly the Reynolds stresses force.
Forces due to turbulent fluctuations the flow
can be either laminal or turbinal we will
be discussing further. So with respect to
turbulence there will be Reynolds stresses
will be produced these are some of the important
forces which influence the fluid motions just
like gravity force, fluid pressure, then force
due to molecular viscosity, then surface tension
compressibility Reynolds stresses. So all
these forces we have to consider as far as
dynamics of fluid flow.
So depending upon the type of problem which
we are considering or type of equation which
we are going to derive some of the forces
will be which will be more important for that
particular case and then others can be the
effect will be negligible and that negligible
forces can be neglected as far as the particular
derivation is concerned or the particular
problem is concerned important forces will
be considering as far as the derivation of
the equation or the particular problem is
concerned.
Now, based upon these as since, we are now
dealing with dynamics of fluid flow. So the
fluid motion we can classify depending upon
the how the fluid properties are changing
with respect to motion, we can classify the
fluid motion like uniform flow, non uniform
flow, steady flow, or unsteady flow.
So this we have already seen earlier. Uniform
flow as we discussed earlier means the velocity
is same magnitude and direction at a very
point. So if you consider for example, a channel
flow say that if it is uniform flow if you
consider with respect to various sections
say here section one, section two, section
three, here the flow is in this direction
say this piece of channel which we consider
here.
For uniform flow means we will say that flow
is uniform for this particular case when the
velocity remains same velocity and other fluid
parameters like depth remains same in magnitude
and direction at every point of the flow is
concerned. So this is what is called uniform
flow and that means the.
Variation with respect to space if there is
spatial variation is not there that means
not there is a flow parameters remains same
in magnitude and direction. So that is the
uniform flow. But as far as non uniform flow
is concerned we can see a flow is said to
be non-uniform.
Whenever we go from section one to section
two the flow parameters like velocity pressure
and all these parameters will be changing
with respect to space. So the flow with respect
to fluid motion when we consider a spatial
wise so the parameters the velocity pressure
and other depth parameters are changing that
is what is called non uniform flow. The third
category is concerned steady flow.
So if we say that the flow parameters are
not changing with respect time means only
the spatial variation is there in that case,
we say that the flow is steady that means
with respect to time there is no change for
the particular fluid flow is concerned and
whenever there is change with respect to the
velocity is changing with respect to time,
if it is one d one dimensional flow with respect
to space and time or with respect to x y z
and time then we say that it is unsteady flow
that means or we call it as transient flow.
So with respect to this uniform flow, non
uniform flow, steady flow, we can again classify
the fluid motion to various characteristics
like steady uniform, so we say that there
is no change with respect to space and time
that means it is steady and uniform.
So for example for a pipe of constant diameter
it is steady uniform it can be considered
as a steady uniform flow and then we can also
say classify the fluid motion. As steady non
uniform so that means if you consider the
flow in name tapered pipe, so if you consider
the flow with respect to a tapered pipe like
this, so here you can see the diameter is
changing from one section to another, so it
is tapered pipe. So in this case we can say
that the flow is steady non uniform flow with
the definition of the steady flow and non
uniform flow.
And then we can also classify the unsteady
flow as unsteady uniform flow that means for
example, a pipe connected to a pump pumping
constant rate is 0. So this is a case of unsteady
uniform flow or we can have an unsteady uniform
flow. For example, waves in a channel. So
these are some of the typical fluid motion
classification like with respect to how the
changes takes place with respect to space
x y z axes the space and then with respect
to time how the changes takes place. So the
dynamics of fluid flow or the fluid motion
we can classify accordingly.
So now as far as fluid motion is concerned
as I mentioned there can also be the effect
of compressing if any compressible force is
acting then, the fluid is compressible then
definitely the motion ,the fluid motion is
affected accordingly, we can say that the
fluids in motion, we have to consider the
rate is compressible or incompressible depending
upon the property of the fluid and then if
it is compressible then with respect to various
forces acting then properties will change
and then we have to consider.
Whether the flow is compressible or incompressible
and then obviously we have to also consider
with respect to the dynamics fluid flow dynamics
or dynamics of fluid flow, we have to consider
the flow is whether it is 3 dimension as I
mentioned earlier most of the time the fluid
flow is varying with respect to x y and z
but that means 3 dimension. So most of times
we can say approximate the flow instead of
if you consider the flow as three dimension
then it will be analysis especially since
the fluid motion is we are dealing with so
the analysis becomes much complex.
So we can just simplify the 3 dimension flow
to 2 dimension or 1 dimension depending upon
the case. so we have to also see as far as
dynamics of fluid flow is concerned whether
we are dealing with 3 dimension flow or whether
we are dealing with 2 dimension flow or 1
dimensions flow.
So as we have seen earlier for example, if
you consider a river flow is concerned say
a large river is concerned. So now if you
consider the 2 sections say at distance of
1 kilometer so this particular case is concerned
if you are dealing if you are interested only
what happens on the direction of the flow
if the direction of flow is in this direction,
what happens in this direction? Even though
this river flow is concerned it is 3 dimension
flow depending upon the particular problem,
we are solving we can analyze as 1 dimension
flow that means in this direction of flow
we can consider so that using 1dimension how
the variation of 1 dimension parameter.
Since it is more important for this particular
problem is concerned so we can analyze it
1 dimension or when we consider a particular
section of the river like these, here say
what happens in this particular say section
then we can take this as 2 D that means it
will be say with respect to vertical and with
respect to the lateral direction, so what
may be z and y so that will be the y and z
will be considered here or say if you want
a complete analysis you can go for a 3 dimension
analysis.
As far as dynamics of fluid flow is concerned
depending upon the problem which we are dealing
we will be considering 3 dimensional flow
or 2 dimensional flow or 1 dimensional flow
whenever we need to say approximate or we
want to simplify the problem we will be considering
the 3 D flow as 2 D flow or 1 D flow.
And then some of the other important parameters
as far as dynamics of fluid flow is concerned
we will be considering since fluid is flowing,
fluid we are considering fluid motion. So
the mass flow rate how much mass is flowing
that is an important parameter so that we
may have to consider fluids or fluids in motion
and then volume flow rate that means discharge,
discharge cube that means how much discharging
is passing at a particular section or between
two sections, how much is the change in discharge.
So another important parameter is so called
volume flow rate or the discharge. If you
know the cross sectional area we can say,
most of the time we will be interested in
the mean velocity. So mean velocity can be
expressed as V is equal to Q by A which is
the volume flow rate divided by area of cross
section. So now further we will discuss about
the analysis of fluid that means with respect
to motion.
So as I mentioned here, when we deal with
the fluid flow compared to the solid movement,
here fluid is continuously moving say, if
you consider river flow or a pipe flow or
any kind of this flow you can see that continuous
movement is endless stream of fluid is there.
So it is very important that we have to see
what part of this stream that we will be analyzing
or the stream starts constitute the system
to analyze. So generally we are dealing with
system so we have to see which part of that
which one we are analyzing.
So as far as solid motion is concerned particular
solid how it is moving we will be dealing,
but as far as fluid motion is concerned we
have to see, since we are dealing with an
endless stream of fluid we have to see what
part of that will be considering the analysis.
So we have two alternatives generally - first
one is we can see the behavior of a specific
element of fluid. So here is you consider
this particular element what happens? That
is one approach. So behavior of a specific
element of fluid of fixed mass that means
generally we will be considering a closed
system just like a pipe this particular system
what happens? So that is one way approach.
So within that fixed mass what happens with
respect to fluid motion and second one is
define the system to be studied as a fixed
region in space known as control volume through
which fluid flows.
So generally, this will be an open system
so this here we can see a channel flow or
river flow. So if you consider a control volume
like this between section one and section
two what happens? So that is we can define
a system of fixed region in space and then
what happens in that is so called control
volume and then this is this is an open system.
So many of the important fluid mechanics problems,
we can solve using this finite control volume.
So this is we are just defining a control
volume and then what happens to this control
volume that is what we are analyzing. So many
of the fluid dynamics problem we can analyze
using finite control volume analysis.
Here this finite control volume analysis is
much easy to interpret physically and the
usage is also vary we can use this concept
is simple to interpret also the formula derived
from basic laws we can apply directly to the
system. So since we are considering a finite
control volume with respect to the fluid motion
of dynamics of flow, it is the formula derived
we can easily apply to the system which we
are dealing with.
Then as we have seen the flow description
is concerned we can either use the Eulerian
description of Lagrangian description, but
here as far as the finite control volume analysis
is concerned we will be dealing with the Eulerian
description and then we will be using the
Reynold’s transport theorem which we have
seen earlier.
So now with respect to this finite control
volume approach we will be deriving the conservation
of momentum equation earlier. We have already
discussed how to derive the conservation of
mass equation the continuity equation with
respect to the differential approach and integral
approach that we have already discussed earlier
in the kinematics of fluid flow.
So now in this dynamics of fluid flow first
we will be discussing how to derive the conservation
of linear the conservation of momentum equation
and then we will be discussing about the conservation
of energy equation and further we will be
going for the with respect to the dynamics
of fluid flow of the fluid motion is concerned,
we will be analyzing various problems with
respect to fundamental principles and its
basic applications.
