Today is the tenth lecture in the series on
marine hydrodynamics. In the last few 2 classes,
we spent time on understanding several 2 dimensional
flows. In fact, in between I talked to about
how using the complex function theory, how
can represent various flows in the process.
We have talked about sources and sinks. We
have seen that a sink is nothing but a negative
source. In case of a source, the flow is always
in the radial direction. On the other hand,
we have seen that in case of the vertex, the
flow is always in the tangential direction.
Now, with this understanding, again we have
seen that when we have seen the additive characteristics
of the various sources in a flow that is possible,
the individual identities source characteristics
would maintain. On the other hand, when we
think of a boundary surface inside the flow;
particularly we have seen. In taking a case
of cylinder, in the presence of a source,
we have seen that the source additive property
will not on go when cylinder is introduced
to the flow.
Now, this understanding today, we will spend
some more time on various types approach,
particularly after source are sink what happens?
What is a double-t? Then, we will mention
few theorems, results, relative source sink
and doublets and also vertex. With this, we
will before going further. I will keep one
very important result that is circle theorem.
Although this result I would have told much
before, but let me mention this. What is a
circle theorem?
This theorem says that if there be a suppose,
we have a irrotational motion, we have a irrotational
motion of a 2 dimensional flow of the irrotational
motion of a 2 dimensional flow. Of course,
the fluid is an incompressible 
and in viscid. We are in the z plane, particular
in the compressed plane. So, we are talking
about a flow of irrotational flow of a fluid,
where the motion is irrotational. It is 2
dimensional in nature. The fluid is assumed
incompressible and in viscid. So, we are in
the z plane basically in the incompressible
plane.
If we introduce suppose, there are no region
boundaries. We usually assume, let there be
no rigid boundary. Let f z represent the complex
velocity potential of the flow complex velocity
potential potential of the flow. If f z represents
the complex velocity potential of the flow,
what we will do if you introduce a circular
cylinder; if a circular cylinder is cylinder
of radius where z is equal to a is introduced
in to the flow? Then, what will happen to
the new complex potential? Then, the new complex
potential becomes w is equal to f of z plus
f bar a square by z. So, this is what happens.
This is the circle theorem.
This result we apply, we can apply to several
several for the several applications. This
result, the simplest application is suppose,
I consider a flow w is equal to u z. That
means I have a uniform flow, which is p d
o. Then, if I apply for example, this is an
example I will take. So, this is a uniform
flow that has speeded as u naught. The speed
is the flow.
Then, if you apply circle theorem, so by circle
theorem, if I introduce a circular cylinder
the flow, then my new w z will be equal to
u naught z plus u a square by z. This is the
flow will come from past circular flow past
circular cylinder. This is the uniform flow
past of circular cylinder. Already, you we
have derived several several results on this
flow in one of our previous class. So, it
is directly coming in this. This result is
directly coming from the circle theorem.
With this now, I will go back to sources and
sinks another example. We know about source.
I will mark out another example of a source.
Suppose, suppose, I just consider a source
in a uniform stream. Seeing the source, let
me consider N as w is equal to u z plus m
n z. So, this is already called this as u
naught.
So, initially I had a stream velocity uniform
flow. I have added to that suppose, I have
a source of the stream present in this. What
will happen to this? If I from this say, I
can get what will be divided d w by d z? This
is u naught plus m by z. So, if d w by d z
is 0, it gives me z is equal to my, rather
I will say m by z equal to u naught. This
in place my z will be minus m by u naught.
So, that means when z is minus m by u naught,
this will be this floor.
So, at z is equal to minus m by u naught,
there will be a stagnation point. There will
be a stagnation point in the flow. We can
easily see that that this this point will
lie on x axis. Further, it can be as we seen
that what is the compressible velocity potential
particularly in the streamlines.
We have already seen w is equal to u z plus
m bar z, If you look at the complexity potential,
so from this, we can get phi plus i psi will
be u x plus i y plus m bar r e to power i
theta. If you put it u times x plus i y plus
m bar n plus i m theta, which we can write
as u x as started with u naught.
So, it is u naught x plus m bar n r plus i
times u naught y plus m theta. See, if I look
at the streamlines, psi will be u naught y
plus m theta, which is u naught u naught y
plus m tan inverse y by x where my r is equal
to x square plus y square. I call it z is
equal to x plus i y. So, if a psi equal to
constant that means the streamlines will be
tan inverse y by x that would be constant.
These are the streamlines. These will be the
streamlines. We can see what will happen when
we put y is equal to 0. If you take y is equal
to 0, then that itself will give us a constant
that means we can always say y is equal to
0 will be the line x is equal to y is equal
to 0. It means it is the line x axis on that
will be there would be no flow, which will
be x axis means y is equal to 0 will be streamline.
All we can say no flow across the x axis.
This is no flow across x axis. So, this is
an example where we see that when there is
a stream and the uniform flow, you can have
a stagnation point in the flow. That stagnation
point is on the x axis. They will not be any
flow that is about there is no flow, which
will be crossing the x axis. This is because
x axis itself gives like a streamline. Now,
I will take a 
different example. Suppose, I have a source,
which is can we consider a source? You may
consider a source.
Now, they have wall and middle it is along
the x axis. Let there be point A that is a
coma 0. Now, let there be source of strength
m at a. This is on the x axis. There is a
wall. If that is a source, what will be the,
with much will be at minus a 0 that will be
on the other side.
Suppose that there is a mirror image. So,
then what will be the complex potential? If
I look at the complex potential m log z minus
a plus m logs n z plus a, so that means the
point z is equal to; if z is equal to point
ratio r, I have a 1, then this z is equal
to minus a. It is just on an opposite side
of the 1. That means it is automated. The
distance is not the same distance on the wall
and appears as if it is an opposite side.
Then, it is like a mirror image. So, what
will happen to this? So, this also can be
written.
So, from here what a happen d w by d here
if d w by d z, I calculate that will give
you m this can be from here. It will be m
logs z square minus a square. So, when I said
d w by d z that will give me 2 m by z square
minus a square into; so this will d d w by
d z. Then, only of wall what will happen near
the wall? We have that the wall means x is
equal to 0. So, what will happen to d w by
d z near the wall? d w by d z is this will
be 2 m z is x is 0.
So, this will be i y my z is x is 0. So, this
is minus y square minus a square. So, I can
call it minus 2 i times 2 m y by y square
plus a square. Again, if you look at d w bar
by d z bar from here itself, we will get 2
i m y square plus a square. So, from this,
I can get q square, which is d w by d z into
d w bar by d z. That gives me 4 m square y
square by y square plus a square, square.
Now, that means on the wall, x is equal to
0. This is my wall. Of course, the speed will
direct towards the wall in this direction.
This is the speed. At this speed, the fluid
will be flowing from the source once we have
q. But, since this wall is an also streamline,
then we can have, we also can see that what
will happen to the force pressure only 1.
So, if I say while the pressure is the pressure
at infinity, pi is the pressure at infinity.
From other lesson, q we have seen, q square
is equal to 4 m square y square by y square
plus a square, square. So, that is q at infinity,
this it will be 0. So, at any point on the
wall, that will be 0 at infinity, pi is the
pressure. q square is this. y is equal to
infinity. It should be 0 because there is,
if it is 0 that means I will have p by rho
is equal to p by rho plus q square by 2. If
I go to Bernoulli’s equation, I have pi
by rho.
That means my q my p is equal to whether also
p by rho equal to pi by rho minus q square
by 2. This is 2 m y the q square by 2. So,
q square is a 2 half is there. So, we will
get 2 m square y square by y square plus a
square, square. So, this will give me the
pressure at any point the wall. If I know,
if liquid is the rho is at rest, so if q is
again I will say if q is 0 that means if the
liquid is at rest that means these are at
any point on a wall. It will be same as the
pressure at infinity.
So, if the liquid is at rest, pressure on
the wall is same as, same as the pressure
at infinity. Now, with this understanding,
if I want to calculate what is the force that
is excited on the wall due to the source?
Then, I can always do because I know p. Then,
I can calculate the pressure. From the pressure,
pressure from pressure, I can calculate the
force.
So, the force will be F is equal to rho. This
will be you can easily see. This will be minus
infinity to infinity that is 2 m square y
square by y square plus a square, square d
y. So, the source will if this one would and
this if you will calculate this, this will
give us by rho m square by a, because basically
what we are doing here. We are substituting
further pressure and integrating. This pressure
is per unit breadth of the wall. This force
this is force per unit breadth. This becomes
the force.
Again, if I have the relation q, my q is on
the wall. My q is 2 m y by y square plus a
square. If this, if I substitute, if I take
y is equal to a tan theta, then this will
give me y is well tan theta. That will give
me 2 m a tan theta by a square sec square
theta. That will give me m by a sin 2 theta.
That will give me the velocity. So, if I am,
my wall is here, the point is here a, that
is the point a, 0. Then, if this makes an
angle is theta with this x axis, this point
is here. This angle is theta. Then, q is m
by a sin 2 theta. It can be seen theta will
be maximum, and then q is maximum. q is maximum
if theta is equal to phi by 4 plus minus pi
by 4. This is pi by 4. It can be seen if q
is maximum. Then, you can see that at that
point, the speed is maximum.
We can see that p is minimum at that point.
Theta is equal to plus minus pi by 4 because
from Bernoulli’s equation, once I have one
form boundary, it has speed which is maximum.
p has to be minimum. This can be checked from
the relations. So, this is the way we calculate.
Suppose, due to a source, how the force is
calculated on a wall in a fluid flow? Now,
this I will go to doublet because we have
seen in case of a source or a sink, we have
a source. It is radial in flow, is radial.
Now, if we think about doublet where as if
you are thinking of 2 sources, so let us look
at what I mean by a doublet.
So, now let us look at two things. I have
a source of strength. This is the point is
minus a. This point is a. Here, I put a source
and here, I will put a sink. So, what will
be the complex velocity potential? The 2 are
at a distance to way source and sink they
are at distance to a. so, the complex velocity
potential will be m logs z minus a z plus
a minus m log z minus a. I can write it as
m log of, I will 
put it z into 1 plus a by z plus m logs, minus
m into logs z into 1 minus a by z. This gives
me m logs first time.
It would b m logs z, either I will say m log
1 plus a by z plus or minus m log 1 minus
a by z. This is because I have a m log z term
here, m log z term here. They will cancel.
So, what I will do? I like expand it. So,
m, if I expand it, therefore this is so. Let
us see this.
Then, I will get it a by z plus where they
are at the second terminal minus a square
by 2 z square 
plus higher order comes here. Then, plus a
by z minus a square by 2 z square plus the
higher times this and this coming that will
give 2 m a by z plus 1 higher terms. This
is because I am assuming a is small. If I
assume a is small, this gives me. So, I look
at 2 things.
If a is tending to 0, the strength of source
is turning to infinity in such a manner, so
that m a tends to a constraint, 2 m a turns
to constant. What is called a constant? So,
let me repeat. Well, there is there is a source;
there is a sink, which is of strength m. Now
I will just say that the source and sink,
there are very, they become very close to
each other, whereas they become close to each
other. The strength m will tend to infinity.
So, in such a manner that 2 m a will give
m a constant and that y becomes a constant.
w will be w mu by z. This is called a doublet.
So, this w represents the complex velocity
potential of 
a doublet of strength mu. If you look calculate
what are the streamlines, w can be written
as phi plus i psi. This can be written as
mu x by x square plus y square plus i times
u y by x square plus y square. It gives me
psi is equal to mu y by x square plus y square.
Then, this is constant. It gives me a constant
that means this will give me the streamlines
that is x square plus y square plus 2 k y
is equal to constant. We have seen this with
lot of examples in the last class. This is
like this. It will touch the x axis. The flow
factor would be like this. Of course, this
is cylinder circular flow. So, again the flow
will be the center will be, it will touch
the x axis. Again, the circular flow and flow
will be and this is the flow that will due
to a doublet.
So, from a source and sink, when they rest
together, particularly close to each other,
we will get a doublet and a flow again will
be in fact. In in our last class, if mu is
equal to u a square, you have seen that when
mu is equal to mu a square, we have seen a
simple singular example. That means with mu
is equal to u a square, then w is equal to
u a square by z that we have seen in this
example that a it is a like we have speed
u naught test. There is a cylinder of radius
a, other speed called strength speed u naught.
This is the velocity potential.
At the same time, here we are seeing that
the similar function complex velocity potential
of a w is representing a doublet. Thus, it
is sometimes this this example such as that
this constant value changes the nature of
the problem that can be different problem.
By physical characterization, we define on
how we look at the problem and what is the
value associated, the constant associated
with it. So, that is that is why, in many
situations, we always have a similar problem
that can be expressed, rather a similar function
can be used to understand the various flow
characteristics. All that will depend on that
constant vector, the physical parameter that
is involved in the flow.
So, that is one of the examples. This is one
example. We showed that physical characteristics
will change the nature of the problem. Often,
even often the complex velocity potential
is similar in nature. With this, I will go
to another part of the problem that is I will
go to the blasius theorem 
and the blasius theorem.
So, basically this theorem talks about the
forces, component of forces that is acting
on a body or on a boundary when there is a
flow. Suppose, let us think of let a fixed
cylinder be placed in a liquid, which is flowing.
The fluid is flowing irrationally and steadily.
Let w is the complex velocity potential associated
to flow. A fixed cylinder is placed in a liquid,
which is flowing irrationally and steadily.
w is the complex velocity potential of the
flow.
Let x and y be the component of forces. m
is the moment on the cylinder. These are the
force on moments in cylinder, moments acting
on a cylinder. Then, neglecting, if you neglect
the external forces 
due to the fluid, the forces acting on the
cylinder will be x minus i y. that will be
given by 1 by 2 i times 2 integral over c.
I told you d w by d z square d z whether it
is called moment, m is the real part real
part of minus rho by 2 integral over c that
is z d w by d z square d z.
So, this theorem, which we will have, which
cylinder is include, which is moving irrotationally,
if you want to calculate the force on moments.
So, let us see. Here, I can easily say that
x is the force and y is the left force m a,
m is a moment that is acting on the cylinder.
Now, I will illustrate through an example.
We take the case of an example.
Suppose I say my w is u z plus a square by
z plus i k log z. This will prevents this
velocity potential. It prevents. I have a
infinite cylinder placed in a uniform flow
in the presence of a. Here, I have a vertex
and the circulation k, the strength of the
vertex. Then, if I have to find the vertical
force, find the forces and moment that is
acting on the cylinder.
If I have to do that, look at here, w is given.
What will happen to my d w by d z? My d w
by d z in which u 1 minus a square by z square
plus i k by z, if this is, then what will
happen to my x plus i y? If I go to this in
other way, my x minus i y will be rho by 2
into rho by d w by d z square d z. Again,
we check an answer at if i y plus i x as a
rho by 2 per minus rho by 2 d w by d z square
d z. So, if I explain this, d w by d z means
minus rho by 2 d w by d z is u minus a square
by z square plus i k by z square d z. So,
this is a i by circle; is a closed circle.
So, if I apply Cauchy residue theorem. It
looks like Cauchy residue theorem. So, this
will give me y plus i x will give me minus
rho by 2, 2 pi i in to because I have at z
is equal to 0. I have a is equal to 0. There
are flows. So, you can say sum of the residuals
and only since we have, we can see that in
this problem the only residues and z is equal
to 0 that is a residue. It can be calculated
2 pi into 2 i u k.
That will give me 2, 2 cancel i, i minus pi
2 pi rho a. From this, I can always say that
my y is 2 pi rho k. Again, I do not have any
here x what is 0. Further, if I calculate
the moment, we are applying the same process
minus rho by 2 ideal part of minus rho by
2 integral c z d w by d z square d z. I can
easily find that this is 0 and not going to
because this again can be applied by the Cauchy
residue theorem rather says by Cauchy residue
theorem, this will be 0.
So, here we have a; this y part with this
del j. In the present surface, there is a;
but if x components they represent able force
vertical force. If the lift force and x component
in fact the horizontal force that is a drag
force that means there is no drag in this
case, whereas the lift force is gained by
this. On the other hand, we have to calculate
the moment and that is given 0.
Now, this is a very classical example to calculate
the course based on blasius theorem. We have
the blasius theorem to calculate this. Now,
I will take few more examples in this class
on this source and sink. Now, suppose I have
a source. We have seen that if I have source
and sink, which are placed, we have taken
2 to 3 cases.
The first cases is if I say that I have a
source placed at that main a have sink, a
source placed at z is equal to minus a, we
will have this sink which is placed at z is
equal to a. Both are of strength m. Then,
we have seen that that will relate to a doublet.
Now, on the other hand provided m tends to
infinity, a tends to 0, on the other hand
if I suppose I say that I do not say that
the situation. Suppose, there is a. What will
happen if I have just a source or sink, both
2 equal sources? Here, it is a source and
the sink source plus sink. It has giving a
doublet.
What will happen if we have 2 sources? We
think of 2 sources. 2 sources, both are of
a same strength, 2 sources of same strength.
Then, my w will be m log z minus m plus m
log z plus a. That gives me m log z minus
a into z plus a. That gives me m block. That
gives me z square minus a square. That is
x square minus y square minus a square plus
2 i x y z square. That gives me this from
which I can get my psi I will get my psi as
because same. So, this will give me 2 x y
by x square m times 2 x y by x square minus
y square minus a square. So, that means 2
m x y by rather it will give it tan 
inverse this.
So, my psi is tan inverse 2 m x y divided
by x square minus y square minus a square.
If psi is equal to constant are the equations
of the streamlines that will give me the streamlines.
If I further simplify this, then I will get
x square minus y square plus this will give
me 
plus 2 x y. We would have taken as 1, then
I will get 2 x y cot psi, but other 2, 2 x
y psi by m, I do not have to take this. That
will be equal to a square. If I simplify this,
I can get this is same as x plus y cot psi
by m into x minus y tan psi by m. This is
equal to a square.
So, if you look at this, they are nothing
but we will assume that these again give me
rectangular hyperboles. So, in the rectangular
hyperboles, so what we have seen in the 2
examples? We have a where you have a source
and sink. We have a source and sink of strength,
rather 2 sources 2 sources of strength m placed
at this tends 2 a. Then, we are getting hyperboles,
rectangular hyperboles.
On the other hand, we have seen that we get
a doublet if you take a source and the sink
placed against same distance 2 a, whereas
the distance is small. So, that is a. So,
what here I mean to say again that just look,
if you look at the authentically compress
potential w is equal to m log z plus a plus
m log z minus a, at the same time you are
looking at the the compress potential w is
equal to m log z minus a minus m log z plus
a.
So, if these 2 potentials, this to compress
potential, there difference is mathematically
looking, there is only a change of sign in
both the cases. Only one psi it has, but that
leads to a flow, which is quite different
in both the cases. So, what we have here?
So, that always it suggest that simple representation
of the flow like complex number gives us very
simple change in the flow complex number pattern.
It particularly gives us very different types
of flow. Physically, it is quite; it is very
interesting physically to understand such
kind of flows.
This has in fact because of this in understanding
of 2 dimensional flows. The role of complex
velocity potential of the role of complex
function theory cannot be ignored. This is
because without much of the difficulty, we
are able to analyze several complex flows.
That is possible because of not only we are
able to analyze the flow pattern, we are also
able to get the streamlines. Even if we have
seen that using the blasius theorem, we are
able to analyze various flows. Particularly,
we are able to calculate the force and moments
on the boundary of the flow. This is what
blasius theorem has given.
So, in the next class again, we will look
into a detail about complex flow, various
types of complex flow. We will come to that
particularly afterwards. We will come to little
more about conform and mapping where how can
we represent a simple flow. We can change
them simple flow, a complex flow to have in
one plane to very simple plane in other plane.
In the process, we will be able to analyze
more complex flow patterns by using simple
complex velocity potential.
This is because we can transfer from one plane,
a complex flow in one plane can be transferred,
transform to another plane where we can analyze
the flow in a very simpler manner. Again,
we can go back to our new plane where the
flow is well complex. That is known as confirm
and mapping. In the next class, we will talk
a little about confirm and mapping, its application
to several flows.
Thank you all. Today I will stop here.
