Good afternoon. I welcome you all in this
session of fluid mechanics. Well, we are discussing
the continuity equation and deduced the equation
in different co-ordinate systems, Cartesian
co-ordinate system and also in cylindrical
polar co-ordinate system.
Well, let us see the equation that if we recollect
the continuity equation is del rho del t plus
divergence of rho into velocity vector, as
we recall, this was the differential equation
or this was the equation for continuity. That
means, the statement of conservation of mass
apply to a control volume what we have found
out that by expanding these in different frame
of reference, whether it is a rectangular
Cartesian co-ordinate system in conventional
x y z like this, x, y, z or in a cylindrical
polar co-ordinate system or even spherical
co-ordinate system, we can derive different
form of the equations, that equations with
respect to different co-ordinate system. Another
way of deriving it is to use the fundamental
principles of conservation of mass to a control
volume, a fluid appropriate to the co-ordinate
system.
So, in any case, this general equation in
vector form is the differential form of the
continuity equation. Since, this is a differential
equation, you say this is a differential operator.
So, this is known as differential form of
continuity equation. So, this gives the differential
equation. So, that is why it is known as differential
form of continuity equation. Now today, we
will find another or we will deduce another
form of continuity equation, which is known
as integral form of continuity equation That
means continuity equation.
Continuity equation means the same, and that
is equation derived from the same principle
of conservation of mass applied to a control
volume. So, that is the continuity equation
by its physical significance and this continuity
equation integrated form. So earlier, we discussed
the differential form. Now, integrated form.
Very simple. Let us consider a finite control
volume like that. So, to derive the integrate
form, we do not require any frame of difference.
Just we consider a control volume, when arbitrary
shape whose total volume is v and which is
bounded by a surface area A.
Now, let us consider the mass flux like this.
Let us consider the mass flux. This is the
mass flux, which crosses the control volume.
Let us consider these are the mass fluxes
crosses the control volume. This is the mass
flux. So, we define a control volume in a
flow field, where the mass flux crosses the
control surface. You know, the boundary of
the control volume is known as control surface.
This is the control surface.
Now, first of all we have to find out the
total mass efflux and the rate of mass increase
in the control volume. Now, if we specify
a small elemental area d a on the control
surface, where the velocity vector is free.
This is the velocity vector at this point,
enclosing this small elemental area d a. Then
we can tell that the mass efflux from this
elemental surface d a is what? Which is v
dot, that is, v is the velocity vector dot
n d a. What is this? This is the mass efflux.
I am not writing everything. I am telling
that mass, if this quantity represent the
mass efflux from this area d a.
What is n? This n is a unit vector in the
direction perpendicular to this area element
d a and taken positive when directed outwards
from this area. So that means, d a is the
scalar magnitude of the area and this is multiplied
with a unit vector to give a vector sense
of it, where unit vector is along the perpendicular
direction and taken positive when directed
outwards. So, physical significance of this
means, that the cross sectional area and the
velocity of flow normal to this cross sectional
area. As you know, to find out the flow across
any section, we multiply the cross sectional
area with the velocity of flow perpendicular
to this area. This is expressed by this dot
product in vector form.
So now, if we integrate this over the entire
area of the control volume; that means, the
entire area bounded by the control surface,
then we get this as the net rate on mass efflux
from the control volume. So, I can write this
as net rate of mass efflux from the c v control
volume, the total area. Now here, if we consider
an area, there is a mass influx. How it is
indicated? This is because in that case, v
the velocity vector and this, for example,
this is the normal direction outwards which
is taken positive outwards. Very important,
so that, this will be in the opposite direction.
So, there will be a negative sign; that means,
if we integrate this over the entire surface
area depending upon the mutual direction of
velocity vector and this vector in the net
efflux will be determined like that. That
means, efflux minus influx which will be taken
care of by the entire surface in this way.
So mathematically, one can tell that v dot
n bar d a n d a, where this n is an unit vector
as I have defined is the net rate of mass
efflux from control volume. rho, very good.
rho should be there. This could have been
the volume efflux. Very good. I am sorry.
Very good. It is rho v n d a. So, rho should
come because otherwise this is the volume
flow rate. So, net rate of mass. Now, what
is net rate of increase in of mass in the
control volume? This is very simple. If we
take for example, at a point where the density
is rho and a small elemental volume d v enclosing
that point, then rho d v is the mass at that
point. So, simply we tell that it is the integral
of rho because rho is varying. rho can vary
within the control volume. So, simply rho
d v over the control volume.
All right. So therefore, we can write according
to our statement of conservation of mass applied
control volume, if you recall and it is very
simple. Very simple physical intuition that
the net rate of mass efflux from control volume
plus the net rate of increase of mass in control
volume is 0. That means, we can write that
this as the first term d v plus the net rate
of mass efflux rho v. This is one vector and
this is another vector.
So, this is the general from of the continuity
equation in integral form. So, this is the
continuity equation integrated form or integral
form. This is the general expression. Now,
you tell me, I am sorry. This is the rate
of change del del t del del t of rho. So,
there will be a del del theta del del t of
rho d v. I am sorry. del del t of rho d v.
So, this can be written of course, with another
because since the volume d v s del del t of
rho d v, let us write this way that this should
be the total volume del rho del t. So, del
rho del t I to d v plus it is this rho v dot
as it is. So, this is the volume integral
and this is the surface integral. Triple integral.
This is all right. So, this is the triple
integral. This is the integral form of the
continuity equation.
Now, you tell me under steady condition, what
will happen for steady flow? For steady flow,
what is this steady flow? This will be 0.
So, for steady flow, the equation is here.
I write surface integral rho v dot d a is
0. For incompressible flow, whether steady
or unsteady, incompressible flow means density
is constant. Incompressible flow, what will
be this term? That is over the area. Incompressible
flow also this will be 0 and more over, rho
will come out because rho is invariant in
this entire control volume.
So, rho will come out. So, simply for incompressible
flow, surface integral, sorry, v dot 0; that
means, it is the volume flow rate which is
conserved. Simply, physically in a compressible
flow, mass flow rate is conserve and for incompressible
flow, it is the volume flow rate which is
conserve. Now, for a steady flow, this is
the equation for incompressible flow. This
is the equation. So, this is a vector form.
Sometimes, it is difficult to recognize physically.
So, from engineering point of view, simply
you can guess this way. Let us consider a
channel formed by two stream lines. These
are the two stream lines. Sometimes, we describe
a stream tube. It is a tube formed by different
stream lines that the surface of the tube
is the stream surface, whose cross sectional
area is a circle. But we can think of a channel
also, which has got some depth in this direction
and which is bounded by two stream lines.
So, in that case, if you think from two dimensional
point of view, you see that if we define a
control volume like this for example, A B
C D. Then in this case, the flow will, let
us consider the velocity vector is such that
it is like this.
So, mass flux only crosses the control surface
or comes into the control volume. This is
formed by these two stream lines across this
surface A D and across the surface B C because
this surface or this line across which there
will be 0. That means 0. No mass flow; by
definition of no mass flow; by definition
of the stream line. So now, if we tell that
this cross sectional area is A 2 and this
cross sectional area is A 2 and this is A
1, and if we tell the velocity vector normal
or velocity normal to this area is v 1 and
velocity normal to this area is v 2. Then
this equation simply tells if rho 1 is the
density or rho 2 is the density, the integration
over the entire control volume will simply
yield rho 1 A1 V 1 is equal to rho 2 A 2 V
2.
So, this is a very simple case that when in
case of flow through a pipe, if we consider
the inlet section of area A 1, where the velocity
is V 1 perpendicular to the cross section,
then rho 1 A 1 V 1 is the mass flow coming
in. So, we do not go through all these detailed
equation from simple physics or physical concept.
We tell this is the mass flow coming into
the control volume. For example, in a pipe
there is no flow across the rigid boundary.
That is why a rigid boundary always behaves
like a stream surface or stream line because
across which there is no mass flow and that
mass be equal to that going out, rho 2 A 2
V 2. In case of incompressible flow, it is
A 1 V 1. That is very simple consequence of
continuity; that means, the volume flow rate
which is coming in to the control volume,
and mass going out the control volume. In
case of incompressible flow, it is the volume
flow rate which is conserved. While in case
of compressible flow, it is the mass flow
rate which is conserved. Well, I think this
is clear.
Now, we will switch on to the next section,
momentum equation. Now, what is momentum equation?
So far, we have derived the continuity equation.
That is the conservation of mass applied to
a fluid flow system. We have considered the
control volume approach, that we have considered
a control volume where the conservation of
mass we have applied. That is the very preliminary
and fundamental conservation statement and
we have derived the continuity equation.
Now, one thing we have seen that in deriving
the continuity equation, rather while applying
the conservation of mass to a fluid flow,
the concept of viscosity or question of viscosity
does not come into picture. Why? This is because
no force is coming into picture. It is only
the flow of fluid. So, when the flow field
is described, whether it is a viscous flow
or it is an ideal flow or inviscid flow does
not matter. If flow field is defined, it has
to satisfy continuity equation along with
the density field to make a possible flow
because the conservation of mass has to be
satisfied.
Now, we come to equation of motion. Now, equation
or statement of conservation of momentum to
derive the equation of motion for a fluid
flow. Conservation of momentum. What is the
conservation of momentum? The conservation
of momentum, if you recall, the principle
of conservation of momentum applied to a system.
First, consider the conservation of linear
momentum. That is rate of change of linear
momentum of a system is equal to the force
acting on the system in the same direction.
As you know, for a system mass is constant.
So, rate of change of momentum; that means,
mass into velocity is a vector can be written
as mass coming out.
So, mass into rate of change of velocity;
that means, mass into acceleration in a particular
direction is equal to the net force acting
on that system in that direction. The same
equation of motion, that Newton’s second
law, which we have read in the preliminary
physics, and that is applied to a fluid system
also. That means, if you consider a system
in a fluid flow of mass m, then this mass
times into acceleration, if it is accelerated
in a particular direction must equal to the
net force acting on that direction. So, this
is simply the momentum theorem or conservation
of momentum. Or you can tell the equation
of motion applied to system of solid or fluid
as well.
Now here, one question comes when you talk
about the total force acting on a fluid body
in a particular direction. You know, there
are two types of forces that act on a fluid
system. As we have read earlier, we have discussed
earlier that one force is the external force
acting on the fluid body. For example, the
gravitational force. There may be other external
forces, some magnetic field in which the fluid
flow is exposed or electrostatic field. These
are the external forces acting throughout
the mass. Another force which is acting, if
we take a fluid system in isolation, these
are the surface forces which appear in the
surface contact forces. As you know, when
it is taken as a free body in isolation from
its surrounding mass, it is the action or
reaction as the way you can think, that it
is because of this mutual interaction between
this fluid system with the surroundings. When
it has been kept in isolation, this appears
as the force at its surface.
So, these forces are of two types. One is
the normal force and another is the tangential
force. So, if the fluid viscosity is taken
into account, all fluids have viscosity. So,
fluid with viscosity, this surface force yields
to tangential forces also, apart from the
normal compressive forces. As you know, in
a fluid flow, the normal forces are only compressive
forces, which is the pressure. But if we consider
that the fluid to be ideal, this is a hypothetical
situation. Theoretically, if we consider that
fluid is frictionless, the only way we consider
a frictionless mechanical system and when
we analyze it, though no mechanical system
can be divided of friction. Similarly, if
we consider the fluid to be ideal fluid; that
means, 0 viscosity fluid, then the only surface
force is the pressure forces. That is normal
compressive forces on the surface.
So therefore, the shape of the equation or
the form of the equation will change, if we
take viscosity or if we neglect viscosity
of the fluid. So, depending on that, the final
expression which we derived from the application
of one of the basic laws of conservation of
momentum will differ.
So firstly, in your syllabus, we will discuss
the conservation of momentum or the principle
of conservation of momentum theorem applied
to an ideal fluid, and that is fluid without
viscosity. Then finally, the equations will
be derived by the application of this principle.
We will see the velocity field that is relationship
between the velocity field and the pressure
field and the resulting equation is what is
known as Euler’s equation, because Euler
was the scientist who first deduced this equation.
That means, they applied the equation of motions
Newton’s second law to the flow of an ideal
fluid. Let us see the Euler’s equation.
So, if anybody asks what is Euler’s equation,
do not try to; Euler’s equations, d u d
t d u d t, like that you must say that Euler’s
equation is finally, an equation relating
the velocity and pressure field, for an ideal
fluid which is deduced from the application
of the conservation of momentum. So now, let
us see that conservation of momentum, as you
know that mass into acceleration is equal
to the force impressed on the system. Let
us first consider the Euler’s equation in
a x y Cartesian co-ordinate system z and let
us define the fluid system compatible to this.
Let us define a fluid system like this, A
B; it is a parallel as you know, A B C D compatible
to the appropriate or compatible to the co-ordinate
system.
Now, this is a fluid system now. Be careful,
this is a system and not a control volume.
There is the difference. You will see afterwards
how the statement relates for both system
and control volume. First of all, we are taking
a fluid system because we are utilizing this
expression, mass into acceleration A in a
particular direction is force in that direction.
Simply as you have read earlier in school
that is the Newton’s second law of motion.
So therefore, acceleration only a system can
have. Control volume may be fixed. Control
volume may move with fluid which will come
afterwards. But control volume may be fixed.
But it is system; that means, the Newton’s
second law is basically defined for a system
that will be modified for a control volume
afterwards with the help of transport. It
denotes transport theory which I will tell
afterwards.
So, now we consider a system. System means
which is moving in the flow or moving with
the flow field. System has got a definite
mass. So, volume may not be constant, but
mass is constant with fixed identity. You
can think your system is composed of different
Lagrangian particles of fixed identity. That
means, system is a fixed mass with fixed identity,
which is moving in the flow. Just like the
Lagrangian particles move because it is composed
by Lagrangian particles of fixed identity.
So, they are called a system master acceleration.
So, please think. This is not a control volume
as we did in case of continuity system.
Now, in this system, let us consider these
are the phases. What is the body? What are
the pressure forces? First, surface forces
acting on this. Then what are the forces on
the surface? If we consider the fluid to be
an ideal fluid or ideal flow, whatever you
call, flow of ideal fluid is the ideal flow.
Then pressure forces, which is the only force
acting here as P, on this surface A E H D.
You know how to define or specify this surface,
the x surface. So, the pressure force in another
x surface parallel to it. B F G C will be
simply p plus del p del x into d x. Before
that, I must define that this is d x. So,
this is d y; that means, in y direction, this
is the dimension of the system and this is
d z; that means, this is d z dimension. Any
dimension within the system. Neglecting the
higher order term, we can tell the pressure
is changed over a distance d x for these two
planes p plus del p del x d. Extremely simple.
Now, if we define the pressure here, on this
A B C D plane, the y plane, then the pressure
on the another y plane which is separated
by d y distance in the positive y direction;
that means, E F G H will be p plus del p del
y d y. My first job is to fix the force because
acceleration, I know now from the kinematics.
Similarly, if the pressure is p, if we specify
at this plane A B E F, then the pressure at
another z plane, that is D C G H will be p
plus del p del z into d z d x d y d z at the
dimensions of the system. So therefore, after
recognizing these pressure forces, let us
consider a body force acting on this system.
Let us consider a body force like this, which
is denoted by x bar at any direction. Let
x bar is the body force per unit mass. As
I have told, the body force is the external
force which is acting throughout the mass.
So, it can be expressed in terms of the mass
per unit mass. So, body force per unit mass.
x bar is the body force per unit mass, which
can be written in terms of its component x
y and x z, in terms of its component in x
direction, y direction and z direction with
i j k as the unit vectors in x y z direction.
Now, next task is what? What is m? m of the
body is the system is the rho d x d y d z.
Now, if we write separately the equation of
motion, it is equal to m into a x is net force
in the x direction.
So, m into a y is the net force in the y direction.
Let us define in terms of vector. So, m a
z is the net force in the z direction. So
now, you see what are the net forces in x
y and z direction. So, this if you recall,
m a, now let us see that what are the forces.
Just a minute. Just this well, I think you
can better see this.
So now, so if we now, you see that what are
the forces now. If we recall that this m in
x direction a x is F x, m a y is F y and m
a z is F z. Now, what is a x? a x is d u d
t. So, when if I write this for x direction,
then I can write rho d x d y d z into d u
d t. What is the net force in the x direction?
That is p plus del p del x into this one,
first p into d y d z minus p plus del p del
x d x into d y d z. So, net force in the x
direction is p d y d z minus p plus del p
del x d x into d y d z. Well, d y d z. Well,
this is the net force. p into d y d z, you
can see p plus del p del x into d y d z d
x into d y d z, this is the area plus the
x component of the body force.
So therefore, this is rho per unit mass. This
is the part. x bar nomenclature is the body
force per unit mass. So, mass times, its x
component. All right. So, if you make this
cancel d x d y d z from both these sides,
you get an expression like this, x sides.
Take first minus 1 upon rho del p del x. Any
problem? This is the x direction. Final equation,
we get by equating the forces. Similarly,
if we do for this, a y is d v d t and mass
remains the same. What is F y? F y is, let
us write F y. What is F y? F y is d x d z
p into d x d z minus p plus this force del
p del y d y times d x d z p plus del p del
y. This is the pressure there times d x d
z.
So, this becomes equal to minus del p del
y d x d y. That means in any direction, the
net pressure forces either in x y z will come
out to be the corresponding gradient. It was
del p del x d x d y d z and it is del p del
y d x d y d z. Similarly, if you consider
the net forces in z direction, it will be
minus del t del z d x d y d z. This is because
in z direction p into d x d y minus p plus
del p del z d z into d x d y and it will ultimately
yield to that. If we use this as F y and a
y is D v D t and a z is D w D t.
So finally, we get the same type of thing
that we get for y component D v D t is equal
to x y, in consideration of the body force,
per unit mass, y component of the body force
per unit mass and D w D t is equal to x z
minus 1 upon rho del p del z .So, this is
the x component and this is the y component.
All right. Now again, I write this thing.
So therefore, what we get in x direction,
we get D u D t equation of motion in x direction
is x minus 1 upon rho del p del x. Similarly,
for y direction I write, what is y direction?
D v D t is equal to x y minus 1 upon rho del
p del y and for z direction, D w D t is equal
to x z minus 1, where p is the pressure in
the flow, u is the x component of velocity,
v is the y component of velocity and w is
the z component of velocity in a typical x
y z co-ordinate system and, xx x y x z at
the x component body force per unit mass,
y component of the body force per unit mass,
and z component of the body force per unit.
About this, we do not have any control from
fluid mechanics point of view. This is described
by the physics of the body force field.
Now, we can write this in a different form
by splitting of this total acceleration. That
means, this can be written as del u del t
plus u del u del x. This is another form.
We can write explicitly u del u del x v del
u del y plus w. That means, splitting of the
total derivative or substantial derivative
in terms of temporal and convective; that
means, explicitly showing the acceleration
in its components temporal and convective
is equal to x minus 1 upon rho del p del x.
So, any one of this is the x component equation
of motion. Similarly, this is del v del t,
the temporal acceleration plus the convective
acceleration, del v del x associated with
u, and del v del y associated with v. This
you know earlier, w del v del z from the kinematics
is equal to x y minus 1 upon rho del p del
y. So, this is the y direction motion; equation
of motion in y direction and this will be
del w del t plus u del w del x plus v del
w del y well plus w del w del z is equal to
x z minus 1 upon rho del p del z. So therefore,
we see this is the z direction. So, this is
x direction equation of motion rather than
y direction and z direction. These three sets
of equation, this, this and this or this,
this and this, they constitute the Euler’s
equation. This is known as Euler’s equations.
That is the Euler’s equation, E u l e r
s, the scientist, big scientist, Euler’s
equation. Euler’s equation or equations
of motions for the flow of an ideal fluid.
That means, it is the conservation of momentum
for the flow of an ideal fluid.
Now, you see that by the method of induction,
as we did in case of continuity equation,
these three equations can be put in a vector
form. Can anybody tell what is the vector
form? These three equations D u D t x minus
1 upon rho del p del x, D v D t is x y minus
corresponding gradient it is the y component
del p del y, it is the z component del p del
z and the body force z component, this is
the y component of the body force per unit
mass and this is the x component of the body
and gradient with respect to x direction.
So, can you tell this can be written like
that, d of v D v D t, where v is the vector
is equal to x minus gradient of v gratitude.
This is the vector form. Why v is given by
three scalar components in three co-ordinate
directions? Well, any problem?
1 by rho.
1 by rho, very good. Always I am missing something.
Which side.
Very good. grad p by rho, very good. Very
good. So, I am happy that today you are very
careful. k x z and grad p. Just by the simple
preliminary vector that gradient of scalar
field is just their gradient in the respective
co-ordinate direction. So, it is the definition
of, please any other thing. This is all right?
So therefore, this is the definition or this
is the vector form of the Euler’s; D v D
t is equal to x bar minus grad p by rho or
somebody or some book or somewhere there,
I draw here D v D t is x bar by rho minus
x bar into rho minus rho x bar into grad.
This is another rho. Does not matter here.
This is the acceleration. Physical significance,
you see. As you define the equation of motion,
each and every term is the force. This is
the acceleration; that means, force per unit
mass. This is the negative of the inertia
force per unit mass and this is the body force
unit mass and this is the pressure force per
unit mass. If you write in this way, rho D
v D t is the acceleration time the density.
That means, inertia force per unit volume.
Negative of that because inertia force defined
with a negative sign and this is the body
force per unit volume and this is the pressure
force per unit volume. Only the units are
different. Any one of the two ways we can
express.
So, this is precisely the Euler’s equation
of motion in a vector form. So, why I am writing
in the vector form? Again, similar to continuity
equation, if I have to derive the Euler’s
equation of motion in a different frame of
reference, not in a Cartesian frame of reference
like x y z, rather if we have to define it
in a cylindrical polar co-ordinate system
of spherical polar co-ordinate system, one
way is to follow the mathematical approach.
That means, you have to write this thing in
the corresponding, so view will be all right
and v bar, it will be defined with its scalar
components in the respective co-ordinate access.
So, only thing is that the grad p has to be
expressed in terms of the respective co-ordinates.
For example, if we think of a cylindrical
co-ordinate system for example, this is x
and this is y. That means, r theta and this
z co-ordinate, then we define v as I v r plus
j v theta plus k v z with the v r v theta
v z, with the r theta z component, a velocity
and I j k are the unit vectors along that.
So, what is the grad p in that case? I del
t del r plus j 1 upon r del p del theta plus
k, this probably you know, del p del z.
So therefore, one can write this D v D t is
equal to what? x minus p by rho. Now, if I
write the r component, what will be this?
D v r D t is equal to x r. If I define x r
is the body force per unit mass in r direction
minus 1 upon rho del p del r, is it all right?
Can anybody tell is it all right? I have written
the equation with a gap. Is it all right?
What is that? There will be another component,
that is the beauty that v theta square by
r. Because of the v theta, I told you earlier
there is an inward radial acceleration; that
means, in the negative at this minus v theta
square by r is the acceleration because of
v theta. That component comes.
So, even if you follow this D v D t, so when
you convert the total differential with the
curvature of the, in consideration of the
curvature of the co-ordinate system, mathematically
also you arrive to this. So, you better write
D v or D t as the total change in the, when
you make a vector transformation. Some of
you may be very strong in mathematics. You
can do it, but you can follow this way also
without going for the mathematical complication.
When you make the D v D t scalar component
in r direction, it will be the total differential
of v r with respect to t with an additional
term. So, it comes from the mathematics considering
in the curvature of the co-ordinate system.
But one can think in a physical way also,
that it is the rate of change of radial velocity
with time. It is the radial acceleration and
acceleration at the radial direction along
with another radial acceleration which arises
wholly from the tangential velocity. That
means, in a flow field where the radial velocity
is 0, having only tangential velocity, there
also a radial acceleration is there. This
is known as centripetal acceleration as you
have already seen.
So similarly, in theta direction, it will
be D v theta D t plus v r v theta by r is
equal to x theta minus 1 upon rho del p del
theta. Similarly, D v z D t, there only D
v z D t; z direction acceleration is responsible
for the change of v z z component of velocity
with time; x z minus, please.
Now, these three equations, this one, this
one and this one, this is r direction, this
is theta direction and this is z direction
can also be derived by a similar fashion.
That is by taking a control volume appropriate
to r theta z co-ordinate and making its force
balance and equating between the mass time
acceleration, according to Newton’s second
law of motion.
Thank you.
