So, we will try to now combine some of the
basic solutions of the potential flow that
is the solution of laplace equations which
as you have already mentioned the basic solutions
are the uniform flow, point source, point
doublet they are all the basic solutions.
We will now try to combine them and see what
it happens. Of course, any super position
is also a solution of the laplace equation
because laplace equation is linear, so if
we combine the 2, any of these 2 solutions
or 3 solutions or even more solutions that
also is a new solution..
So, would now we will combine some of these
basic solutions and see what do we get. So,
the first 1 we will do it, a uniform flow
plus a point source, a point source placed
in a uniform flow. Let us call it two dimensional,
a two dimensional point source again remember
that two dimensional point source is actually
an infinite line source. Let us take the uniform
stream along the positive x direction that
is from left to right, uniform stream from
left to right. A uniform stream is always
represented by just a few parallel lines with
of course, arrow to give the direction of
the flow and the flow velocity in the x direction
we simply call it u infinity and we place
a point source in it. This is the point source
a two dimensional point source or an infinite
line source and we take the location of the
point source as our origin. This is just for
convenience, as you know that the potential
due to a point source is m by 2 pi log r where
m is the source strength and r is the distance
from the location of the point.
Now, the stream lines if we show some of the
stream lines of this point source they are
like this they are all radial, the flow which
is from the source the flow is coming out
in all direction radially. So, which is going
to the left that will of course, be opposed
by this uniform stream but, which is going
to the right that will be supplemented or
with the uniform stream increase. First of
all let us write the potential. What is the
potential of this? Consider this a steady
flow, forget about any time in it so let us
let phi at any point x y, call this plane
to be x y plane.
So, what will be phi x y? Yes, potential due
to uniform stream what it is?
U infinity x, and the point source.
M by 2 pi log r.
M by 2 pi log r. If we write you can write
even a constant plus a constant also you can
write that we will drop those constant all.
What is psi? The stream function. What is
stream function for this flow?
U infinity .
U infinity y is for the uniform stream eventually
these are the stream lines plus m by 2 pi
theta plus a constant
Find the velocity components? The 
velocity at any point u x y, the x component
of the velocity u x y is you can write d phi
d x or equal to d psi d y. How much is this?
yes?
u infinity plus?
U infinity plus m by?
Yeah
M by 2 pi r?
D r by r.
D r.
Wow why d r?
M x square x by root under x square plus x
2.
x by?
Under root x square plus 2.
M by 2 pi x by.
Under root x square plus r square.
Sir r square.
R
R and r r square.
So it becomes m by?
Two pi r square.
2 pi?
R square.
X by r or r square?
R square sir r square.
What about the other velocity component?
M by 2 pi. M by 2 pi.
Only m by 2 pi?
Into y by m square.
Y by r square.
So, this is the combined velocity field. This
is the velocity field due to a combination
of uniform stream and a point source. Is there
any point where the flow velocity is 0?
If we look to this problem from here, if there
is a point where the velocity is 0 that is
likely be to on the negative x axis.
At theta equal to, that is theta equal to
pi is there anything like that? What is that?
Say, we call point p where velocity is 0 equal
to 
minus m by 2 pi u infinity. Minus. M by 2
pi u infinity. M by 2 pi. U infinity. Comma
0 comma 0.
So, there is a point which is, which has no
velocity. Velocity at that point is 0, we
usually call this point a stagnation point.
The point at which the velocity is 0, that
is a stagnation point and all other properties
associated to that point they are called stagnation
properties. Like the pressure at the stagnation
point is called stagnation pressure, density
at that stagnation point is called stagnation
density, temperature at this stagnation point
called stagnation temperature.
However, remember in this case. This you should
remember that this stagnation property is
little more than what we are saying that velocity
where the velocity is 0, that is what is stagnation.
The stagnation properties can be associated
to any point irrespective of the flow velocity.
Here, we said the stagnation pressure is a
pressure were the at the stagnation point,
where the flow velocity is 0 but, in practice
stagnation temperature, stagnation pressure,
all these properties are associated to each
and every point irrespective of the flow velocity.
It is not necessary the flow velocity to be
0 to define a stagnation pressure at any point
it is not necessary that the velocity at that
point has to be 0. We can, the way we can
define a pressure temperature at any point
similarly, we can also define a stagnation
pressure stagnation temperature at each and
every point.
This is by assuming or imagining what would
be the pressure or what would be the temperature
if the velocity becomes 0 there. Say at point
there is some velocity say 10 meter per second.
Imagine and it has some pressure some temperature.
Imagine what would be the pressure and temperature
if that 10 meter per second becomes 0? It
need not be really 0, just imagine then what
would be the pressure and temperature that
is actually called the stagnation pressure
and stagnation temperature but, then comes
something else, you are assuming or you are
imagining that your speed is becoming 0 from
say 10 meter per second, how it is becoming
0?
How we are making the flow velocity 0 in your
imagination? You are imagining, all right
but, how you are imagining that it is becoming
0, to make it 0 it has to follow certain process.
So, it depends what that process is how we
are making the velocity 0 there it will depend,
the pressure at that point will be depend
how we are making it 0. So, actual stagnation
quantities need that this process must be
isentropic, for temperature and enthalpy adiabatic
is sufficient but, for all other properties
the process should be isentropic.
That means you have to imagine that you are
making the flow velocity 0 there isentropically,
then the pressure at that point whatever pressure
at that point will become that is what is
stagnation pressure.
So, just if are asked that define stagnation
pressure simply saying that where the pressure
velocity is 0 the pressure at that point is
stagnation pressure is not actually a correct
definition. The correct definition is if the
flow velocity at any point become 0 isentropically
then the pressure at that point would be the
stagnation pressure. However ,at present we
are discussing incompressible inviscid irrotational
flow and for such flows all processes are
isentropic. So, we need not specifically mention
every time that it is stopped isentropically,
it is stopped isentropically but, you must
remember that this process of stopping, of
course, the imaginary process of stopping
must be isentropic then only you will get
the stagnation pressure, correct stagnation
pressure otherwise not.
Look we have, if you look to this only this
vertical component of the velocity v component
of the velocity you can see that the v component
of velocity are constant on a circle, the
v components of velocities are constant on
a circle. Take this v component, what it gives?
(r square we have simply written x square
plus y square. What is this? This is an equation
of a circle with radius or center at 0 m by
2 pi v, circle with center at 0 m by 2 pi
v sorry m by 4 pi v.
Circle with center at m by four pi v. So,
for constant v, if v is fixed then all these
points are on a circle so on each circle the
vertical component of velocity is constant.
Find the stream function at the stagnation
point or on a stream line which passes through
the stagnation point? You know any streamline
is simply psi equal to constant. psi equal
to constant gives a streamline, so find the
stream line which passes through that stagnation
point. How much it is? Stream function is
a function of x and y and you know x and y
for the stagnation point.
So, how much? Let us call that to be psi 0.
How much it is?
m by 2 final.
M by 2. So, the streamline passing through
the stagnation point what will be the streamline
passing through the stagnation point? (Any
streamline is given by psi x y equal to constant
or u infinity y plus m by 2 pi theta equal
to constant and since this is passing through
the stagnation point the constant is already
known m by 2. See, this equal to constant,
this equal to constant is equation for all
the streamlines, different streamlines for
different value of the constant. The streamline
which is passing through this for that we
have already found a constant m by 2. So,
in this case this equal to m by 2 the constant
equal to m by 2 or we can write y equal to
how much.
M by 2 u infinity into 1 minus theta by pi.
Say, this is the equation of the streamline
passing through the stagnation point. What
type of line is this? Can you plot this line?
Can you plot this line? A plotting a line
in x y plane only.
As you can see from x equal to minus infinity
to x equal to, how much is that m by m by
2 pi u infinity, from x equal to minus infinity
to x equal to minus m by 2 pi u infinity,
this line will coincides with the x axis.
So, coming from minus infinity to plus infinity
this is this streamline and let us say then
this is that, this is that point which is
minus m by 2 pi u infinity and what will happen
after this? Looking to this equation as you
can see as theta increases from 0 at theta
equal to 0, this has a value of m by 2 to
u infinity at theta equal to 0, as theta increases
from m by sorry 0 to pi the value decreasing
at theta equal to pi it becomes 0, that becomes
this point only. Then as it again increase
from pi to 2 pi this y become negative y become
negative but, it is symmetric. The value at
theta equal to pi by 2 is same as the value
at theta equal to 3 pi by 2 but, with negative
sign.
So, it is something like this. This geometry
is symmetric 
and this is the location of source, this is
the location of the source. So, this is the
usually referred to as the zero streamline
and the maximum what should we call? Thickness
or half thickness? When theta equal to 0 the
y is m by 2 u infinity or 2 phi u infinity,
what it is? 2 u infinity m by 2 u infinity.
You can call it half thickness because the
on the lower half also there is a m by 2 u
infinity.
So, the maximum half thickness. (Now, what
is a streamline?
Excuse me sir sir that point on x axis whether
it is m by 2 pi u infinity or m 2 u infinity.
What it is coming.
M by 2 u infinity.
When theta equal to 0.
Also this stagnation point was minus m by
2 pi u infinity.
Stagnation point is minus m by 2 pi u infinity.
This is not the stagnation point this is half
thickness, this thickness that is the coordinate
of that stagnation point this, coordinate
of the stagnation point. (So, this is 
maximum half thickness this we will call it
say h, let us call this to be h .
Which one?
That is a stagnation point minus m by 2 phi
u infinity.
With the phi m by 2 infinity.
This is the height is m by 2 u infinity.
Now, what is a streamline? We have already
defined or by definition streamlines are lines
to which velocity is perpendicular parallel
or tangential velocity vector is always tangential
to streamline. So, what is the normal component
of velocity to the streamline?
0.
Now, if we have a solid body in a flow, then
an incompressible inviscid flow, not incompressible
in any inviscid flow the boundary condition
is normal component of the relative velocity
is 0 or the normal component of velocity is
0 if the body is at rest that is the boundary
condition.
Now, you see that all streamlines satisfy
that. Streamlines are lines on which normal
component of velocity is 0, if you have solid
body in an inviscid flow then the boundary
condition on the inviscid on the solid body
is that the normal component of normal component
of the velocity is 0 on the solid body. So,
the solid body are same similar to streamlines.
As far as inviscid flow is concerned solid
bodies and streamlines are same or any streamline
can be thought of as a solid body. You can
write that in any inviscid flow.
Yes?
Now, in this case this is a streamline. We
can have other streamlines let us say some
other streamlines we can show approximately
(these are also a streamlines these interior
dotted lines.
Now, let us say that this body what we are
calling the 0 streamline if we replace it
by a solid body then what it is? This problem
now then become uniform flow over a body solid
body of this set. Of course, a semi infinite
body.
So, if we replace this. this particular streamline
what we are calling the 0 streamline, what
we have called the 0 streamline this part
we can forget only this, only this much, only
this much.
And this body is usually called a semi-infinite
pairing of maximum thickness of m by 2 infinite
into 2 then this problem now becomes what
we have started with a combination of uniform
stream dot and a source.
Now, become uniform stream passed a semi-infinite
pairing. The problem now becomes and these
particular streamlines these dotted streamlines
that I have shown here they now become streamlines
within the body have no significance, this
source that is also now inside the body nothing
it is not in the flow.
So, it is not creating any rate of expansion
anywhere within the flow. It now does not
exist only what you have a uniform flow and
a semi infiniting semi-infinite fairing emerged
in it or in other way we have solved the flow
over a semi-infinite fairing. Only thing that
when you started of course, you did not know
that what body we are going to get or whether
we are going to get a body really that was
also not known to us but, you see the idea
is this that we started with this and now
we see that this is what is the practical
case this is a practical problem of semi-infinite
uniform flow over a semi-infinite fairing
and we have solved it by indirect approach.
Of course, it gives an idea that perhaps some
other combination will give flow over something
else .
We now know the potential function stream
function associated with this flow with this
problem the velocity at any point within the
flow field, the important flow field is now
only this, only this much which is outside
this. This becomes a solid body. At any point
here we know the flow velocity already you
have found out what are u v.
So, you can find the flow velocity at any
point the unknown and you can apply the Bernoulli's
equation to find the pressure. Looking from
the symmetry of the problem since the velocity
at any point on the upper half and the lower
half will be same the pressure will also be
same. So, pressure on the upper half of this
body on this upper surface of the body and
pressure on the lower surface of the body
will be same. This we can say without even
calculating the pressure but, calculating
the pressure is quite simple you just satisfy
the Bernoulli's equation.
Take a point at the infinity or far away from
here where we have flow velocity only u infinity
far away we have flow velocity only u infinity
pressure p infinity so at that point from
Bernoulli's equation p infinity plus half
rho u infinity square and take any point here
where the pressure is p the velocity is square
of the velocity is u square plus v square
that u and v we have found.
And we can find the pressure there. So, just
we can write that phi psi u v (from here you
can find p at any point, in particular if
you find the pressure on the body surface
on the body surface you will see the pressure
on the upper part and on the lower part are
exactly identical meaning that there will
be no net force in the y direction the body
is symmetric about x.
So, the pressure distribution is also symmetric
about x. So, there will be net no net force
in the y direction that means there will be
no lift force in this case. Even you can say
that the x component of the force that will
also be 0. So there will be no net force acting
on the body. This can also be thought of as
flow over a clip if you consider just only
half of this body say upper half looks like
a clip, it looks like a clip.
So, it is a flow over a clip also and as you
see that in that case at the stagnation point
the flow velocity is 0. So, that is the most
sheltered place. So, if you are near a clip
and then suddenly a very strong storm is coming
up and you want to be sheltered you can know
that the stagnation point is the most sheltered
position near about stagnation point and it
is near about is the sheltered position .
So, that is the place where you can go and
take your refuge. You can 
extend this problem to three dimension also.
You can extend this problem to three dimension
also of course, in that case then you will
find a body of revolution like this, you revolve
it, that is the body you will find and in
three dimension of course, you have the real
point source potential given by m by 4 pi
r. Anyhow, we will not do that. We will rather
take up another case which is all most similar
but, now instead of a point source we will
take a point doublet again two dimensional,
two dimensional point doublet 
or an infinite line doublet.
Next, case we consider. What is the potential
due to a point doublet?
Mu by.
4 pi
4 pi actually becomes for three dimensional
case that is the 
point or either perhaps infinite line doublet,
you call it infinite line doublet. potential
function mu by 2 pi yes? mu by 2 pi cos theta
by r. This x r theta we can replace this x
also, we can make it r cos theta so that we
have only (the velocity components u and v,
what are u and v? r r theta. The radial component
of velocity it is d phi d r or 1 by r d psi
d theta, the radial component it is should
I write that u r u theta, this u is now the
radial component not the x component. I think
perhaps better to write that subscript. Is
d phi d r equal to 1 by r d psi d theta. How
much is this?
Infinity cos theta
Minus.
Mu by 2 pi cos theta by r square.
Mu by 2 pi r square cos theta, we should have
written the cos theta once. What about that
theta component? It is 1 by r d phi d theta
or minus d psi d r. This is what both the
term on negative minus u infinity plus. Inside
the bracket there should minus . Now look
to the expression for u r where is u r 0 other
than that cos theta equal to 0 and so theta
equal to 0 or pi which is just the x axis
other than that.
