Now, today we will go to the 3rd lecture in
a series, the first two lectures we have talked
about first we talked about the basic Marine
Hydrodynamics, it is introduction, motivation.
In the second lecture, we have talked about
basically the continuity equation. And while
talking about the continuity equation, we
come across talk about what is the vector,
then irrotational motion and from that we
could get that existence of potential phi
which exist in case of a irrotational motion
in a fluid flow problem. And today again now
we will go to, what about the flow direction?
In which direction how the flow, how to know
the direction of the fluid flow, so in this
context we will talk about stream lines and
path lines.
Stream lines - 
for a fixed instant of time, let us draw a
path in the fluid a space curve at a particular
point, let us draw a tangent to this. And
if it q bar velocity vector, if a velocity
vector q bar is same as the direction of the
tangent; that means, q bar cross d r bar equal
to 0 and then this line, this curve we call
the stream line, because that means along
a stream line, the flow it gives the direction
of the flow of the fluid. And then in as a
result there is no fluid usually across a
stream line there will be no flow, no fluid
will pass through the streamline, until or
unless gives the direction of the flow.
So, if u simplify this one, then you have
d r bar is equal to d x i hat d y j hat, these
are the components d z k hat, then we have
already q bar is equal to u i hat plus v j
hat plus w k hat from q dot d r bar is 0,
this will give us i j k you have u v w, you
have d x d y d z. And if this is 0, because
this is what you have q bar cross d r bar
means and if this is 0, then and gives us
from this you have u d z, rather if I go to
this then v d z v d z minus w d y is 0. Then
you have j then you have u d z that is u d
z minus w d y w d x is 0 and then we have
k k component u d y minus v d x is 0, from
this we can easily get d y by v is d z by
w. And from here, we will get d x by u equal
to d y d w by sorry this is called d z by
w and then you have from here, you will get
from the third one you will get d x by u is
equal to d y by v.
So, if you combine that three things, you
will get d x by u d y by v is equal to d z
by w. So, this is what. So, this is what we
call the they are just equation of the streamline.
So, you can take any two of them and solve
it and, so that will give us two sets of constant,
if i say that I will take either any two either
this two this two I can take, I will get two
sets of equations and I can take this two
or I can this take one and third. So, that
will give you often. So, because of that we
say that, it gives two infinity set of constraints
and this is the equation of the streamline.
Now, let me just take a simple example to
understand what exactly how it means.
So, if I say I will take a simple example,
like if I say my velocity vector q bar x i
hat minus j sorry y j hat, if this is the
velocity vector, then what will happen to
my streamlines equation of the streamline
will be d x by x will be d y y minus y, once
this gives me. So, it gives me l n x minus
l n y plus constant which gives me l n x y
is equal to constant and that which gives
me x y is equal to constant and since, if
x y is constant then what will happen, if
I draw these lines.
So, this is for each constant suppose, we
call it k 
and this constant say since it is. So, what
will happen then, what if I draw these lines
let me say in that straight line this is x
axis, this is my y axis and if I draw it,
this will be my curves lines is bar there
is well is a constant. And again because,
similarly we can get in a now, I will go to
this is one example for x positive, y positive
we can get when k is a constant or k is equal
to 1, 2, 3 it can have any values and also,
so we can have this is one of the example.
So, now some of the points to be noted note,
so in case of a streamline in case of a streamline
stream line stream line flow the flow direction
of flow is in one direction is in one direction,
as I have already mentioned there is no flow
across if there is no flow across a streamline
across the streamline. And I have mentioned
the the differential equation the equation
of the streamline, yields double infinitive
solution 
and in this case you have, this q bar is x
i hat this is a two dimensional flow, this
is the one. Now, once you know what is a streamline,
what will happen like in case of a vertex
line you have got a vertex stream what will
happen, if we take up any closed curve.
If we think of any closed curve and draw a
stream line, in the inside the fluid you have
a closed curve, let it a curve like in case
of vertex lines, vertex tubes we have seen
that we have taken any closed curve inside
the fluid and we have drawn vertex lines,
we got a vertex tubes. Similarly, suppose
I say I had draw I have any closed curve and
that each point of the curve, if I draw a
line, the stream line, then this will give
a tube and this tube we call this stream tube
and let us look at the flow in a stream tube,
if I look at the flow in a stream tube. Then
let us say that lets look at the stream tube
of various cross section, this is cross section
A 1 this side is A 2, let this A 3 is the
boundary of the tube and then these are the
two cross sections. So, there are three one
is A 1, A 2 and A 3.
So, let us say that fluid is flowing along
this direction and it is going out in this
direction and let me say that, at this point
if n hat is the normal unit normal outer drawn
direction and here also, we have outer drawn
direction normal n hat. So, if that is the
case and the vessel that the fluid is entering
through the cross section A 1 and it is going
out through the cross section A 2. Because,
it is and there is a flow what will happen,
if we will say from the continuity equation,
we have for a in compressive fluid we have
got this is equal to 0, we have seen in case
of continuity equation.
If this is the case then which is same as,
integral over the surface q bar dot n hat
d s that is 0. Now, if I look at now, cross
section now, this surface we have three surface
we have A 1 we have A 1, A 2 and A 3. Since,
we have three surface, since we have a stream
line flow, this is a stream tube and there
is no flow which will cross through A 3. So,
because of that we will have, integral over
this is called the surface s 1, this is my
surface s 2, this is my surface s 3. So, there
is no flow across s 3. So, what will happen
my this will be same as, integral over s 1
q bar dot n hat d s plus integral over s 2
q bar dot n hat d s will be 0 because, there
is no flow which is crossing the surface s
3 and if that is the case, then because, we
have one is the inward direction.
Because, this side you have the stream tube,
this side your flow is entering, this is the
way flow is entering and this is the flow
so outer direction is upward. So, we can have
this is the this is the outer direction, here
it is outer normal. So, if then in that case,
so we will have integral over s 1; that means,
A 1 q 1 dot n hat d A 1 plus integral over
A 2 q 2 dot n hat d A 2 this is 0 and which
implies q 1 A 1 because, the normal’s are
in opposite direction. So, we will have a
negative sign so, that will give you q 1 A
1 is q 2 A 2 and this is what in case of a
stream tube, the amount of fluid that will
enter through this and in that that same amount
of fluid will go out through this.
So, now, if I say A 1 is A 2 which implies
from this relation, we will have q 1 is q
2; that means, when you have a tube, stream
tube of uniform cross section the speed remains
the same. Now, to demonstrate this let me
take an example, let us consider a stream
tube a circular pipe.
Circular pipe, so this is a tube, it has fluid
will enter through this one and it will exit
in this way, it will enter this way till fluid
release, let this point be A, B, C and the
D. If i say if I say my diameter D A is 6
inches and here, D B the diameter at the point
B is equal to I will set 12 inch at C I will
set this is 4 inch D C that is 4 inch and
diameter at this point D D, this is I will
set 2 inch, so it is clear.
Here, this is diameter is 6 inch here, the
diameter is 12 inch from there, there is a
bifurcation and here, diameter at point C
it is a 4 inch and here it is 2 inch. Now,
if I say the velocity, the fluid velocity
at B is 2 feet per second, if it is at 2 feet
per second and again velocity at C and we
say, this is 9 feet per second, my question
is, find the fluid velocity at A and D. That
means, at which rate the fluid will be entering
through this tube and a, which rate a fluid
will be going out through the D tube, through
the face D.
Now, since we have seen that incase of a stream
tube, we have seen that the q 1 A 1. So, q
A D A q A rather I will call this as s 1 is
q B s B s 1 s 2 are the surface area and this
is a cylinder here. So, my q A into s 1 is
pi into 
D square by 4, 36 this is 36 by 4 and this
is q b 
q B, here it is 144 into pi by 4 into q b
q B at this point is 2 feet to this gives
me, pi will cancel 4, 4, 4, 4 then I have
q A will be if I cut this with this then it
will be 4. So, 4 into 2 is equals to 8. So,
q is 8 feet per second.
So, my speed at here, becomes 8 feet per second.
Now, in the same way, what will I will do
that I will be because, I know this speed
here, I have the fluid speed if I say that
at q B at this will be q B s B will be same
as q C my s C plus my q D s D and if I utilize
this substitute for this, then I will easily
get I am not going to the detail I can get
my q D as 36 feet per second, this is what
I will get, it can easily check that, just
a simple calculation I am leaving it to you.
So, this two example suggest that how the
flow, fluid flows inside a tube of certain
cross section, if the cross sectional area
changes then, the velocity of fluid changes,
fluid velocity changes and here that is why
sometimes when have a very the flow through
unusual the speed because, very high it is
because, that the unusual, the area squeeze
we squeeze the area and in the process, the
speed of the flow speed will be high, fluid
will flow water higher rate, layer the speed
will be high. Now, this I will go to, what
is path lines particularly, path of a particle.
Basically, what we talk about here the path
of a particle is a curve again it is a curve
the path of a particular a curve which a particular
fluid will describe during its motion. So,
the differential equation of the path line
will be d r by d t that this will represent
the fluid motion. So, d r by d t will give
you q bar and what is q bar q bar is u i hat
plus v j hat plus w k hat. So, that is; that
means, and if you has I put it in r as write
it for r as x i hat plus y j hat plus z k
hat, then we have d x by d t equal to u d
y by d t that is equal to v d z by d t that
is w, these these are the equation for the
path lines.
Now, the question comes we have already seen
that what how how far the path lines and stream
lines are difference, basic difference between
path lines and stream lines. So, the tangent
to stream lines, we have already we know that
the tangent if we draw a tangent to the stream
line it gives us the direction of velocity,
tangent to a stream line, if we draw a tangent
to a stream line it gives us the fluid particle,
it gives us the velocity of the fluid particle,
of the particle basically the fluid particle
at various point on the other hand.
The tangent to a path line, if we draw a tangent
tangent to a path line gives the direction
of velocity, it gives the velocity direction
of velocity of a given particle that, there
is at various at a various time. So, basic
difference is that here, at here it gives
the that when you draw a tangent to the stream
line, it gives us the velocity of the particle
at various points at a same instance of time
on the other hand, if you draw tangent to
a path line direction of the velocity of a
given particle this is the difference at various
times.
Here, various points we have at various times
here, the given particle for a particular
particle at various times whereas, in case
of stream lines, it gives the same particle
at various points. Basically, your time is
fixed and in this case your time is varying
space position is fixed. On the other hand,
if I will say that when the motion is steady,
if the motion is steady then stream line do
not vary with the time and it coincide with
the path line. Now, with this background of
stream lines and path lines, I will just go
to two dimensional flow.
Why the need of two dimensional flow? Because,
everything nature is three dimensional why
should we think of a two dimensional flow.
When it comes to particularly we have the
three components of velocity u, v, w suppose,
one of the component is very small or negligible.
So, in such a situation what we will do, then
we can consider if one of the component is
small, then we can consider the flow as two
dimensional and in fact, it has been found
that in many situations when the flow is symmetric
along the particular direction then it is
easy to analyze consider the flow as two dimensional.
And the from experiments and observations
it has been seen that while physical modeling
in many situation, the two two dimensional
flow is a very accurate realization of the
flow situation. Even if there are axis symmetric
flows, here the flow can consider as a two
dimensional flow. So, because of that there
are two aspect, one is either the certain
symmetric characteristics of the flow, if
certain symmetric character characteristic
of the flow is there or when the one of the
components of the velocity is negligible.
Then then we can always consider the flow
as two dimensional and that gives a very good
accurate realization of the flow characteristics.
Suppose, if we consider one cross section,
particularly what happened in the large ocean
when you do with large problems related to
ocean waves, are often we consider the flow
as if the flow characteristics is similar
in each section, each cross section, particular
cross section we always consider the flow
symmetric assuming the symmetric characteristic
of the flow, we always consider the flow as
two dimensional.
So, and the it has been seen in many cases
that, this gives a very good realization,
particularly for large class of marine hydrodynamics
problems and it is easy because, the moment
you reduce the dimension by one, the complexity
of the problem reduces and that helps us to
understand a get a very accurate realization
of the situation, in a very simplify manner,
often we try with a two dimensional problem,
once we get a proper feeling about the problem,
then we go to the general case that is what
we do.
Now, with this I will now go to few problems
related to the two dimensional flow problems,
particularly we have already talked about
the what will happen to the continuity equation
in two dimensional, continuity equation particularly
here I want to say, emphasize the often the
flow is incompressible because, as I have
mentioned earlier that here, we will be considering
for a large section of this force as the basically
marine hydrodynamics, we will concentrate
on incompressible fluid and for a. So, for
a incompressible fluid if that would have
continuity equation will be del u by del x
plus del v by del y is 0 you see the continuity
equation.
And now, let us consider the differential
because, and then we have the streamlines,
we have come across the streamlines that is
the equation of the stream lines is stream
lines are d x by u is d y by v, this will
give us the stream lines for a two dimensional
flow. If that is the case, then what will
happen let us consider the differential u
d u v d x minus u d y if this, consider this
as 0, if this is 0 that gives us from this
we can always get this v d x minus u d y is
0.
If v d x minus u d y is a exact differential
d psi then what will happen. So, that will
be 0, then psi is a constant and that psi
because, this psi is equal to constant again
what will happen, if I relate with. So, further
we can see d psi as del psi by del x because,
if psi is a function of x y, if I say psi
is a function of x and y then d psi is del
psi by del x d x plus del psi by del y d y
and this is same as v d x minus u d y see,
if you compare this component wise, we will
just compare the x part d x part and d y part,
then we can say del psi by del x is equals
to v whereas, u is minus del psi by del y.
So, in fact this psi is called the stream
function, psi is equal to constant, psi is
the stream function, this psi is called the
stream function and psi is equal to constant
as we have seen that they will give the stream
lines that will give the stream lines. Another
point of view here, now I will relate phi
with the, the stream function with the velocity
potential.
We have already seen in in case of the two
dimensional flow, we have already seen u is
equal to phi x v is equal to phi y. Now, from
here again, if you relate with the stream
function. So, we have u is equal to phi x
and phi x is nothing, but we have u is nothing,
but minus del psi by del y minus psi y similarly,
if we say v that is a phi y and v is again
del psi by del x.
So, which gives us del phi by del x is equal
to minus del psi by del y and again del psi
del phi by del y is del psi by del x, this
is the sometimes say complex function theory,
this is the Cauchy Cauchy Riemann equation,
particularly if f is a function of a compression
f of z that is given as phi x y plus i psi
x y then f is analytic, we say f is analytic
in z provided, if it satisfy the satisfy the
Cauchy Riemann, the Cauchy Riemann equation.
So, this is the Cauchy Riemann equation, in
short we call it as Cauchy Riemann equation.
So; that means, for a two dimensional flow,
we will have if the flow is potential. So,
here it is very clear; that means, I have
already mentioned that a stream lines exist
for all flow exist for all flow on the other
hand, we have velocity potential will exist
only when the flow is irrotational, it will
exist for irrotational flow, on the other
hand for stream lines they are only for two
dimensional flow, stream lines they are associated
with with two dimensional flow, when the flow
is two dimensional.
On the other hand, when it comes to the sorry
stream lines particularly I will say that
a stream function this is not stream lines,
stream function, exist for all flow whereas,
this is again stream function, exist for two
dimensional whereas, velocity potential velocity
potentials here, both two it exist for both
two or three dimensional flow. So, this is
a very, very this is a very clear difference
between as I say the stream lines, stream
function and velocity potential.
Now, another thing here that what is the another
basic difference now let us just say that
if we say, phi what will happen to phi x psi
x plus phi y psi y this will give phi x is
u we have seen we have seen phi x is u and
psi x we have seen it is v whereas, phi y
is v we have seen psi y is minus u. So, this
is same as u v minus u v that is equal to
0.
So, which even one must say that grad phi
dot grad psi is nothing, but grad phi is 0
which implies that the stream lines, when
phi is equal to constant gives the stream
lines psi equals constant it gives the phi
is equal to constant is equal to constant,
it gives the equip potentials equip potentials
I have already mentioned where as the psi
is equals to constant gives the stream line,
it gives the stream lines.
So, this suggest me that the stream lines
and the equip potential lines, cut orthogonally
this is another. So, this is another very
important result. Here another point I work
I would like to highlight that I have already
mentioned that the stream function is independent
of the fact whether the flow is irrotational
or not, on the other hand the velocity potential
phi they exist only for irrotational, when
the flow is irrotational and that is of potential
kind.
So, irrotational flow is are often called
as the potential flow. Now, I will come to
two examples, couple of examples rather to
understand that how the flow characteristics
are defined in case of various examples are
few examples rather I will concentrate on
to understand the path lines, stream lines
then flow irrotational flow stream functions
and how they are related. So, to do that I
will go to this one by one let us work out
few examples.
And, so here I will first four five examples
let us work out. Let me say q bar equal to
k square x j hat minus y i hat divided by
x square plus y square, here I assume k is
a constant. So, now, whether this q bar represents,
whether it has a it relates with some fluid
motion or not then if, so if a motion is possible,
fluid flow is possible is possible find the
stream lines.
And then that is the first part, second part
test whether the flow is irrotational, test
for irrotational motion. Again if the flow
is irrotational, then find the velocity potential.
So, here I have preferred this given the velocity
vector, I have two questions, first question
I am emphasizing whether to test whether the
flow is irrotational flow is possible fluid
flow is possible or not, once we say that
the fluid flow is possible.
Then, we will find that what is the stream
line because, that will give as a direction
of the flow and then, we will check it for
whether the flow is irrotational, if the flow
is irrotational, then we will find the corresponding
velocity potential. So, here we will find
the stream lines, we find the velocity potential
at the end we will check whether the cut orthogonally
or not that is what we will do, then it will
be clarify some of our whatever we have theoretically
discussed, let us check for it.
First this is a two dimensional flow and for
the two dimensional flow we have divergent
of q bar is 0 that is del u by del x plus
del v by del y is 0, this we have to check
because, it is two dimensional and it will,
so del u by del x what will happen here, here
your u is minus k square y by x square plus
y square and your v is k square that is x
y x square plus y square and then, what will
happen your which implies del dot q bar will
be k square into minus del by del x y by x
square plus y square plus del by del y into
this is x by x square plus y square.
And this gives us k square and if you take
the derivative with this is minus 2 x y by
x square plus y square, square this will gives
plus 2 x y divided by x square plus y square,
square and that gives us 0. Hence, fluid motion
is possible in compressive fluid, we have
a possible fluid motion, motion is possible
and once the fluid motion is possible. Now,
whether we say that, what will be the stream
lines now check for irrotational let us see
first, whether the flow is irrotational.
If it is irrotational, we have to check it
for curl of q bar is 0 curl of q bar is 0
because, for irrotational and that is same
as because, the flow is two dimensional here,
we have to check it as if del u by del y is
equal to del v by del x this we have to check
and here, we can see that, easily here del
u by del y minus del v by del x this equals
to y square minus x square del v by del y,
this will give you y square minus x k square
into by x square plus y square, square then
plus x square minus y square divided by check
it and that is this as 0.
Since, this is 0. So, the flow is irrotational.
So, this is satisfy and we have we have already
seen, this satisfies dimension q is 0 fluid
motion is possible and here, we have seen
fluid is irrotational. So, now, we will go
for what are the stream lines and what are
the velocity potential, to find the stream
lines what we will do.
We have a stream lines will give us equation
of stream line is d x by u is d y by v and
that that gives us that is why two dimensional
flow. So, that will give us d x by minus y
d z y by x square that will give you d y by
x which gives us x square plus y square equals
to constant. So, this gives the stream lines.
So, it is concentric circles, this all will
give us concentric circles on the other hand,
if we look at the velocity potential.
So, in that case, we will have u is equal
to phi x and v is equal to phi y already,
we know u v. So, we have phi x is equal to
k square y by x square plus y square and phi
y minus k square x by x square plus y square,
if you integrate it then you get phi from
this one, we can get phi is equal to k square
y integral d x by x square plus y square and
that gives us k square inverted tan inverse
x by y tan inverse x by y plus f of y.
Now, what will happen to if I take del phi
by del y of this, this gives me k square del
phi by del y if I go far del phi by del y
minus k square x by x square plus y square
plus f dash y and this is same as, already
I have been given phi y is this minus k square
x by x square plus y square. So, which implies
because, this implies f dash y is 0 because,
this term will get cancelled into this term,
when we have f dash y is 0 and; that means,
f y is constant and this constant, we can
always say it will 0 without loss of generality
because, a constant, we can always take it
as 0.
So, we have taken arbitrary constant we can
take it as 0 that gives us which gives. So,
phi we get it phi x y we have got it as k
square tan inverse x by y. Now, if I say that
phi x y is constant will give me the equip
potential lines, if I put this as constant
which gives me x is equal to c y. So, if these
are the concentrated stream lines now, x is
equal to c y is a centroid lines, if x equals
to c y then, what will happen they will just
pass through the origin any line will pass
through the origin sorry it will go through
this origin and you can have, number of lines
and this is a line passing through the origin
these are all circles.
So, if you look at at draw at any point a
tangent you will see that, these two lines
the stream lines and the equip potentials
they will cut each other orthogonally. So,
this is what it shows. The another point is
that here, we have seen the flow the stream
lines they are circles whereas, the flow is
rotational, the stream lines are circles the
following circular path the fluid follows
the circular path whereas, the flow is irrotational.
Now, I am showing in another example that,
the flow is rotational they are circular as
well as the flow is here rotational in nature,
they are not irrotational, let us look at
this example.
Suppose, my q bar is minus w y, w x, 0 let
us say omega y then we can see, easily that
del dot q bar is 0 and again. So, which shows
that flow is incompressible fluid flow, flow
is incompressible; that means, there is a
fluid motion if it is possible and if you
look at curl of q is 0 which is same as, which
implies i j k if you take del by del x del
by del y del by del z and then u is minus
omega y omega x this is 0 and then this you
calculate, it to 2 omega k hat and which is
not equal to 0.
So, curl of q sorry it is not 0 for irrotational
flow, this is for irrotational flow here I
am finding here, curl of q is not 0 hence,
this flow flow is not irrotational. Now, look
at what is the stream lines, if you look at
the stream lines, we have d x by minus y is
d y by x and again which gives you x square
plus y square is equal to constant. So, if
this is the stream line for each constant
you have a circle passing through origin and
centre is at origin on the other hand here
the, curl is not 0.
So; that means, we have a fluid flow that
the fluid particle follow circular path whereas,
the flow can be remain irrotational whereas,
in the previous example we have shown that
the flow can be irrotational, but still the
flow can follow circular path, with this I
will stop today.
Thank you.
