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Welcome back to the video course on fluid
mechanics. In fluid kinematics, in the last
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lecture, we were discussing about the potential
flows. We can define the velocity potential
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as: u is equal to del phi by del x, v is equal
to del phi by del y and w is equal to del
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phi by del z. We have defined the consequence
rotationality of flow field for 3 dimensional
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flows and the potential flow where we have
defined with respect to the rotational flow
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fields. Further, we have discussed about the
potential flow, stream function and then we
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have derived the Laplace equation which governs
the inviscid incompressible irrigational flow
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fields.
In potential flows, we have seen how we can
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define a problem and how the boundary conditions
are defined. As we discussed potential flows
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which is the theories applicable for inviscid
incompressible irrotational flow fields given
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by the Laplace equation and lines of constant
potential is equipotential; the stream function
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is also defined and then the lines of constant
function is called stream line. We have also
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seen some examples related to the potential
flows. Today, we will discuss about the potential
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flow; we will see the basic potential flow;
then, super position of potential flow.
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Potential flow is as we have seen it is governed
by the Laplace equation. Potential flow solutions
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are always approximate since most of the fluid,
the assumption in potential flow is irrotational,
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the viscosity is neglected and hence we are
assuming as frictionless. Potential flow solutions
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are always approximates since fluid is assumed
to be frictionless. Exact solutions are got
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from the potential flow theory and represents
approximate solutions to real flow problems.
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Even we can derive the exact solutions for
potential flows but as far as real fluid flow
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is concerned this is an approximation for
the reality or the real fluid problem. The
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exact solution obtained in by using potential
flow theory gives only or represents approximate
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solutions to the real flow problem. So, the
potential flow which we have seen here is
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that we are assuming the flow as potential
but the reality is different; the solutions
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which are derived for the potential flow are
just approximation for the real fluid flow.
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We have already seen the Laplace equation
in the Cartesian coordinate system and now
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for potential flow in the cylindrical coordinate
system, we can describe this del phi as shown
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in this slide here: del phi is equal to del
phi by del r er plus 1 by r del phi by del
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theta etheta plus del phi by del z ez. In
the cylindrical coordinate system, we are
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defining in terms of r, theta and z. We can
use the unit vector er etheta and ez and then
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we can represent the radial velocity as del
phi by del r and then the tangential velocity
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vtheta we can represent as 1 by r del phi
by del theta and vz is represented as del
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phi by del z.
In the cylindrical coordinate system, the
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potential flow is represented with respect
to r, theta and z. This can be represented
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in terms, here the radial velocity. So, in
a cylindrical coordinate system the parameters
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are described as r, theta and z. The velocity
or the parameters can be described in terms,
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as we have seen for the Cartesian coordinate
system, here represents the velocity in xyz
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as uvw.
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Correspondingly, in cylindrical coordinate
system we can represent the radial velocity
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vr, the tangential velocity vtheta and the
velocity vz which are defined here as vr is
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equal to del phi by del r with respect to
the potential function; then, the tangential
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velocity is represent as 1 by r del phi by
del theta and the vz is represented as del
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phi by del z. Finally, the velocity can be
represented as vr er; the unit vectors plus
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vtheta e theta plus vz ez and the Laplace
equation in the cylindrical coordinate system
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can be represented as 1 by r del r of r del
phi by del r plus 1 by r square del square
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phi by del theta square plus del square phi
by del z square is equal to 0. So this equation
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represents the Laplace equation cylindrical
coordinate system. Some type of problems where
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we will be using theta in z coordinate system
or the cylindrical coordinate system we can
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use the equation in this particular form.
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Now, we have defined the potential function;
we have defined the stream function here.
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So here with respect to the stream function.
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We can represent the velocity for irrotational
function flow. We have already seen the stream
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function can be defined as u is equal to del
psi by del y and the velocity in y direction
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is minus del psi by del x, where phi is the
the stream function and then the velocities
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here are represented in terms of x and y direction.
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This is x this is y and psi is the stream
function. When we come as 2 dimensional flow
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with respect to stream function we can write
u is equal to del psi by del y and v is equal
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to minus del psi by del x which we have derived
earlier. From the condition of irrotationality
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we can write del v by del x is equal to del
v by del y. From the condition of irrotationality
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which we have seen earlier by substituting
by omegaz is equal to 0, we have show that
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del v by del x is equal to del u by del y.
If you substitute for u and v from these equations
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here as shown then we will get del square
psi by del x square plus del square psi by
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del y square is equal to 0s. This is another
form of the Laplace equation derived from
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the condition of irrotationality and the definition
of the stream function. Finally, we get a
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del square psi by del x square plus del square
psi by del y square is equal to 0. This is
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again Laplace equation in terms of stream
function.
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Depending up on the problem if we can represent
the flow field in terms of potential function
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or if we can represent the flow field in terms
of stream function we can write either of
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these two equations: if you represent phi
then we can write del square phi is equal
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to the 0 for the potential flow problems;
if we can represent the flow field as a stream
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function then we can use del square psi is
equal to 0. Both equations are valid for theorems
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depending up on potential flow problems depending
up on whether you are using the potential
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function or stream function. The stream function
similar to what we have seen in the case of
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a potential function, we can represent the
stream function in terms of cylindrical coordinate
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system, rtheta and z.
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Since z coordinate system is not coming in
the case of stream function or since we are
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considering two dimensions so vtheta is equal
to minus del psi by del r; this is the definition
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of the tangential velocity with respect to
the stream functions psi. So vtheta is equal
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to minus del psi by del r and vr is equal
to 1 by r del psi by del theta. In cylindrical
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coordinate system rtheta with respect to stream
function we can write the tangential velocity
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as minus del psi by del r and the radial velocity
vr is equal to 1 by r del psi by del theta
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and change in value of the stream function
is related to the volume rate of flow.
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That means if you draw the stream function
with respect to stream line, if we consider
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a flow field like this and then if we can
draw the stream lines like this psi1, psi2
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and psi3 extra, the stream in change in value
thing function from one position to another
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that represent actually, the volume of ů.
This is psi1 and psi2, then the volume rate
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of flow q is equal to between psi2 and psi1.
One we can write the volume rate of flow is
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equal to psi2 minus psi1, that is, q1. Similarly,
between this psi2 and psi3 we can write q2
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is equal to psi3 minus the stream functions
psi3 minus psi2. Stream function q is the
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volume rate of flow can be represented as
the difference between this stream function.
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This potential flow with respect to the stream
function we can use to calculate the volume
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rate of flow between the fluid flows which
we will be considering. Now, with respect
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to this stream function let us consider a
small example.
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We can see an example for a fluid flow. With
fluid flow, radial velocity is given as vr
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is equal to lambda by r square cos theta and
if vtheta the tangential velocity is equal
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to 0, theta is equal to 0. Determine this
two stream functions psi and vtheta for the
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fluid flow?
The problem is the data given are in terms
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of the cylindrical coordinate system r and
theta; the radial velocity is already given
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vr is equal to lambda by r square cos theta
where lambda is constant and then a condition
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is given vtheta is equal to 0 and theta is
equal to 0 we have determine the stream functions
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psi and then the tangential velocity vtheta
for the fluid flow.
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Already given vr is the radial velocity equal
to lambda by r square cos theta. This with
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respect to over definition here vr is equal
to 1 by r del psi by del theta, this lambda
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by r square cos theta is equal to vr that
is equal to 1 by r del psi by del theta or
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we can write del theta is equal to lambda
by r cos theta. Now, we got an expression
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for psi with respect to theta as del psi by
del theta equal to lambda by r cos theta.
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To get the stream function we can just integrate
this with respect to theta. If we integrate
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theta del psi by del theta we get psi equal
to lambda by r sin theta f r. So, this stream
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function is obtained as psi is equal to lambda
by r sin theta that is equal to fr. If you
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want to determine the tangential velocity
vtheta definition is minus del psi by del
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r. Here, we have already derived the stream
function so we can just differentiate with
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respect to r to get the tangential velocity.
We differentiate the expression with respect
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to r we get vtheta is equal to lambda by r
square sin theta plus f dash r which is constant.
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One condition is given that the tangential
velocity vtheta is equal to 0 at theta is
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equal to 0. We can apply this condition here.
So, vtheta is equal to 0 so we get fr, f dash
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r is equal to 0. Finally, we can obtain vtheta
is equal to lambda by r square sin theta which
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is the expression asked in the question.
So have found the stream function as lambda
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by r sin theta and we got the tangential velocity
v component vtheta is equal to lambda by r
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square sin theta. As shown in this problem
by using these kinds of the polar coordinate
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system or cylindrical coordinate system we
can solve this kinds of problem in terms of
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the theta, the radial velocity or in times
of vr- the radial velocity or the tangential
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velocity vtheta and with respect to r psi
we can determine various functions like vtheta
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vr are distinct function. This is about the
representation of the potential flow with
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respect to the stream function.
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As I mentioned we can represent the streamlines
are the lines of constant psi. We can draw
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like this in the figure shown here, Psi1 psi2
psi3, these lines are constant and hence these
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lines are called stream lines. We can see
that there is no flow across a stream line,
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flow is in the direction of the stream line
once the stream line with respect to stream
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lines concept there cannot be a flow across
a stream line. So, flow across this
stream line is impossible. Since at every
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point stream line is tangent to the velocity
we can say that there is no flow across a
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stream line. This means that a streamline,
we can consider as a solid boundary so that
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flow cannot cross across a streamline. The
streamline concept is very useful to solve
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many of the fluid mechanics problems especially
when we can approximate it is next and as
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potential flows. The stream lines there cannot
be of any flow across a streamline and then
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the streamline can be considered as a solid
boundary where we can consider as a boundary
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between two various flow problems. So, streamline
is defined here.
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With respect to this streamline and also for
the plane of irrotational flow we have defined
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the potential function and stream function.
We have already seen that both satisfy the
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Laplace equation. So, in Laplace equation
which we have derived del square phi is equal
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to 0, del square psi is equal to 0. So, both
del square phi is equal to 0 function and
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both potential function satisfies the for
plane irrotational flow satisfy the Laplace
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equation.
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Now as per our definition if we use the definition
of the stream function and potential function
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we can write its dy by dx along psi is equal
to constant and is equal to v by u, that is,
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the lines of constant psi or which we have
discussed the streamlines and then dy by dx
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along the potential psi is constant that means
equal to, as per definition, that is equal
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to minus u by v, where u is the velocity in
the x direction and u the velocity in the
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y direction. These lines are constant where
the potential psi is constant where these
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lines are called a equipotential lines. With
respect to the concept of the potential velocity
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potential and the stream function we can draw
the stream lines where dy by dx along psi
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is equal to constant which is equal to v by
u which are called the stream lines. Then
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we can draw lines where the potentials are
constant or dy by dx along phi is equal to
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constant as minus v by u. These are lines
of constant potential which are called a equipotential
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lines.
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This equipotential lines are when we draw
with respect to fluid flow we can see that
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these equipotential lines are orthogonal to
the streamline. So if I draw here you can
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see that these are the streamlines here. If
you plot the potential lines we can see that
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these are orthogonal to the 90 degree angle.
So this is phi1, phi2 and phi3, like that
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phi4. We can draw the potential equipotential
lines and then the streamlines already drawn
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here. We can see equipotential lines are orthogonal
to the streamlines at all point where they
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intersect or equipotential lines intersect
with respect to the streamlines at 90 degree
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or they are orthogonal since the product of
slope is minus 1. When we draw the streamlines
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and then when we draw the equipotential lines
these lines together consists a family of
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streamlines and equipotential lines called
flow net. This flow net concept is very useful
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in many of the fluid flow problems. So, flow
net consists of family of streamlines and
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equipotential lines.
We have seen how to draw streamlines with
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respect to the direction of the fluid flow
and we have also draw the equipotential lines;
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together they are orthogonal or the product
of slope is minus 1 and then the family occurs
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with respect to streamlines and equipotential
lines from the flow net.
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So, here you can see the flow net. Flow net
consists of streamlines and equipotential
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lines; the streamlines are drawn here just
like this red color and yellow color and equipotential
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lines phi is constant is also drawn. So they
form the flow net. This flow net concept is
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very useful in many of the flow net.
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This flow net we can use for visualization
of flow patterns as shown here. It can be
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also used to get graphical solution for fluid
flow problems. They can also determine the
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velocity and discharge can be obtained from
the flow net since velocity is inversely proportional
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to streamline spacing. Here this figure shows
a flow net for a fluid flow in a bend so there
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is a flow through a bend.
The streamlines are plotted here like this
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and then equipotential lines are also plotted.
If we can draw the streamlines and then equipotential
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lines finally get the flow net we can get
a visualization of the pattern as per the
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fluid flow is concerned especially for potential
flows or the flows which we can the approximate
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as potential flow. Many of the problems like
flow through an earth dam we can use this
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concept.
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For example, you consider an earth dam like
this. In this earth dam, let us assume that
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it is impermeable. Here, if the head or the
potential is 10 meter and here is 2 meter
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you can see that there is fluid surface and
then with respect to this for the earth dam
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problem. We can draw the streamlines for this
and then correspondingly we can plot the equipotential
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lines which give the flow net. The flow net
for an earth dam is drawn here. With respect
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to this, once the flow net is drawn we can
use this flow net pattern to calculate the
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velocities or discharge since the velocity
is inversely proportional to this streamline
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spacing so this concept of the streamlines
or the flow net with respect to streamline
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and equipotential lines. There are large numbers
of applications like in earth dam so flow
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through a bend, wherever the flow problems
can be approximated as a potential flow we
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can use the concept of the flow net.
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00:24:42,520 --> 00:24:49,520
Now, we also consider here another problem
with respect to a flow beneath a concrete
196
00:24:50,890 --> 00:24:57,890
dam. Here, in this slide we can see there
is a concrete dam; this is the dam position
197
00:25:06,810 --> 00:25:13,810
and then we are considering a domain 60 meter
length, this is 0,0 then 60,0; the depth is
198
00:25:22,810 --> 00:25:29,810
60 by 15 and here 0,20. We are now considering
there is a concrete dam on a permeable foundation
199
00:25:39,250 --> 00:25:45,930
like this. You can see that here the flow
condition, the boundary condition here it
200
00:25:45,930 --> 00:25:52,930
is 5 meter depth and here it is the downstream
end; it is 5 meter. There is a level difference,
201
00:25:55,640 --> 00:26:00,850
here the boundary conditions and here the
potential if we consider with respect to this
202
00:26:00,850 --> 00:26:06,930
line as datum, you can see that there is a
potential meter of 10 meter. So on the upstream
203
00:26:06,930 --> 00:26:13,930
side, the flow comes and the potential function
phi is equal to 10 meter and downstream side
204
00:26:15,440 --> 00:26:22,440
this is the water level. The downstream side
phi is equal to 5 meter and since we are considering
205
00:26:24,550 --> 00:26:31,550
this boundary at the bottom as impermeable
we can say that del phi by del n that means
206
00:26:33,250 --> 00:26:38,050
no flux can cross this boundary. So, del phi
by del n is equal to 0; here also del phi
207
00:26:38,050 --> 00:26:44,580
by del n is equal to 0 and this side also
del phi by del n is equal to 0. Also we can
208
00:26:44,580 --> 00:26:51,380
assume that stream functions psi is equal
to 0, psi is equal to 0 here and here also
209
00:26:51,380 --> 00:26:58,380
psi is equal to 0. Then, similarly, here there
is a concrete bed and the stream function
210
00:27:01,060 --> 00:27:08,060
here, let us assume psi is equal to 10 and
on this also del phi by del n is concrete
211
00:27:10,360 --> 00:27:14,370
dam; then in upstream there is a concrete
bed; in downstream also there is a concrete
212
00:27:14,370 --> 00:27:20,470
bed. So del phi by del n is equal to 0 on
this phase also. With respect to this problem
213
00:27:20,470 --> 00:27:27,470
we can solve now. The equations of the del
square phi is equal to 0 we can solve and
214
00:27:28,950 --> 00:27:35,950
del square psi is equal to 0. The Laplace
equation in terms of phi the potential function
215
00:27:36,550 --> 00:27:41,650
and then the Laplace equation terms of stream
function also we can solve. In this particular
216
00:27:41,650 --> 00:27:47,310
domain, the boundary conditions are given
here and then we can determine the phi and
217
00:27:47,310 --> 00:27:53,590
psi at various points like this, at various
points we can find the potential function
218
00:27:53,590 --> 00:28:00,590
and stream functions. Then we can interpolate
between two to get the stream lines and the
219
00:28:00,720 --> 00:28:03,400
potential lines like in the figure.
220
00:28:03,400 --> 00:28:08,990
So here we have drawn with respect to the
problem given here and the boundary conditions
221
00:28:08,990 --> 00:28:15,590
and then the use the Laplace equation in terms
of phi and then use the Laplace equation in
222
00:28:15,590 --> 00:28:19,870
terms of psi del square phi is equal to 0
and del square psi is equal to 0.With respect
223
00:28:19,870 --> 00:28:26,800
to this boundary conditions we can get the
potential function and the stream function
224
00:28:26,800 --> 00:28:31,780
in various locations and finally we can call
this stream lines like this.
225
00:28:31,780 --> 00:28:38,420
You can see here these stream lines are plotted
like this and then the potential functions
226
00:28:38,420 --> 00:28:45,420
are also plotted. So, a flow net is formed
here for the flow beneath the concrete dam
227
00:28:47,770 --> 00:28:54,770
and this can be used to calculate how much
will be the discharge. These are the equipotential
228
00:28:58,640 --> 00:29:05,640
lines and these are stream lines. So, finally,
we get a flow net for this particular problem.
229
00:29:11,680 --> 00:29:18,130
This can be used to find the discharge or
the flow which can go through the pores media
230
00:29:18,130 --> 00:29:23,390
from this place to this place and then finally
it will exist at this place. This can be used
231
00:29:23,390 --> 00:29:29,550
to find the velocity; these are discharge
between with respect to the potential flow
232
00:29:29,550 --> 00:29:33,050
equations and potential flow theories.
233
00:29:33,050 --> 00:29:40,050
Potential flow theory has got lag in applications
as described in the earth dam here or the
234
00:29:40,400 --> 00:29:47,400
flow beneath concrete dam as described here.
Further this will be this aspect will be discussed
235
00:29:48,890 --> 00:29:55,890
later. Now, after the flow net we will see
some of the basic potential flows. The potential
236
00:29:59,850 --> 00:30:05,550
flow where the applications directly we can
apply Laplace equation and where the simple
237
00:30:05,550 --> 00:30:12,550
flow surface and we can call some basic potential
flows like uniform flow, like a source since
238
00:30:13,940 --> 00:30:19,700
and double x. Since the potential flows are
governed by the Laplace equation which is
239
00:30:19,700 --> 00:30:26,700
a linear equation we can have number of particular
solution and with this particular solution
240
00:30:29,780 --> 00:30:33,680
can be added to yield solution for complex
problem.
241
00:30:33,680 --> 00:30:40,680
If we have a problem like where the potential
functions for example potential function phi1
242
00:30:42,350 --> 00:30:49,350
phi2 phi3 extra are known then since the Laplace
equation which we are considered del square
243
00:30:51,100 --> 00:30:57,740
phi is equal to 0 or del square psi is equal
to 0; this is the linear form of the linear
244
00:30:57,740 --> 00:31:04,740
equation. We can superpose or we can add to
yield the solution of complex problem. If
245
00:31:05,340 --> 00:31:12,340
phi1 phi2 phi3 extra the solution obtained
then for various problem we can superpose
246
00:31:12,750 --> 00:31:19,750
the problems to the elementary flows like
uniform flow, like a flow with source and
247
00:31:21,450 --> 00:31:28,450
sink and then we can superpose whether to
get complex flow problem. This concept we
248
00:31:28,590 --> 00:31:35,590
can use to solve many of the complex problem
which can be kindly approximated with respect
249
00:31:37,140 --> 00:31:43,840
to the potential theories. This we will discuss
before going to the complex problems. You
250
00:31:43,840 --> 00:31:50,430
will see the basic potential flows.
251
00:31:50,430 --> 00:31:57,430
As mentioned elementary flows are uniform
flow, source, sink and the vortex. These are
252
00:32:01,200 --> 00:32:07,340
the elementary flows which we consider in
the potential flows theory, uniform flow source
253
00:32:07,340 --> 00:32:14,340
and sink and vortex so each one of this discuss
in detail. Now we will discuss the basic potential
254
00:32:15,370 --> 00:32:21,640
flows.
First, we will discuss the uniform flow. This
255
00:32:21,640 --> 00:32:28,640
is the simplest plane flow described by either
stream function or velocity potential. Some
256
00:32:31,550 --> 00:32:38,550
of the flows like the ground water flow without
the pumping or velocities very low then it
257
00:32:40,800 --> 00:32:46,910
can be approximated as uniform flow sometimes
depending up on the flow conditions. The uniform
258
00:32:46,910 --> 00:32:53,910
flow concept becomes most simple or simplest
plane flow where there is no complexity; there
259
00:32:54,240 --> 00:33:01,240
are no sources or nothing.
In the uniform flow we can approximate using
260
00:33:03,980 --> 00:33:09,020
potential flow theory. This we can either
represent using the stream function equation
261
00:33:09,020 --> 00:33:16,020
or the potential equation and the stream lines
are straight and parallel as far as the potential
262
00:33:19,520 --> 00:33:22,720
flow is concerned and the magnitude of the
velocities is constant.
263
00:33:22,720 --> 00:33:29,720
Here, we can see that it can be horizontal
or implant but as you can see here the stream
264
00:33:32,809 --> 00:33:39,160
lines are straight and parallel so that is
the peculiarity of uniform flow and magnitude
265
00:33:39,160 --> 00:33:46,160
of velocity is constant. Here you can see
the psi is equal to psi1 or psi is equal to
266
00:33:46,450 --> 00:33:51,530
psi2 and then the streamlines are psi is equal
to psi1 or psi is equal to psi2 or psi is
267
00:33:51,530 --> 00:33:58,530
equal to psi3 or psi is equal to psi4 like
that. We can represent the uniform flow with
268
00:33:58,840 --> 00:34:05,840
equipotential lines and streamlines are drawn
as shown in this uniform flow or it can be
269
00:34:06,440 --> 00:34:10,300
implanted like with respect to an angle alpha.
270
00:34:10,300 --> 00:34:17,300
Here also phi1 and phi2 represent the equipotential
lines and psi1 psi2 psi3 psi4 represent the
271
00:34:19,559 --> 00:34:22,929
streamline.
272
00:34:22,929 --> 00:34:29,929
With respect to this now we will use the potential
flow theory. As we have represented the velocity
273
00:34:32,109 --> 00:34:39,109
in the x direction it can be represented as
u is equal to del phi by del x and the for
274
00:34:39,329 --> 00:34:46,329
this particular problem del phi by del y is
equal to 0 for the case of the uniform flow.
275
00:34:47,369 --> 00:34:54,369
So, this is the flow in one direction; we
can represent del phi by del x is equal to
276
00:34:55,599 --> 00:35:02,599
u and then we can write phi is equal to ux
plus c, where c is an arbitrary constant and
277
00:35:03,240 --> 00:35:09,220
this can be said to 0. So that for the uniform
flow we can write phi is equal to Ux, where
278
00:35:09,220 --> 00:35:16,220
U is the velocity the x direction, phi is
equal to Ux plus c and c is an arbitrary constant
279
00:35:16,680 --> 00:35:23,250
which we said to 0, phi is equal to Ux. So
that we can write now del psi by del y is
280
00:35:23,250 --> 00:35:30,250
equal to u as per the definition of the stream
function; del psi by del y is equal to 0 and
281
00:35:31,230 --> 00:35:36,390
also the uniform flow del psi by del x is
equal to 0. Since we assume that v is equal
282
00:35:36,390 --> 00:35:43,029
to as per the definition of the uniform flow,
the stream lines are straight and parallel
283
00:35:43,029 --> 00:35:48,059
and the magnitude of the velocity is constant.
So, with respect to this we can write that
284
00:35:48,059 --> 00:35:55,059
phi is equal to Ux and then psi is equal to
Uy. So that del phi by del x is equal to U
285
00:35:57,670 --> 00:36:04,670
and del psi by del y is equal to U. Finally,
we get the expression for the potential as
286
00:36:05,710 --> 00:36:12,710
phi is equal to Ux plus c and the expression
for Ux plus c, we said c is equal to 0, we
287
00:36:13,549 --> 00:36:20,549
get phi is equal to U into x and psi is equal
to U into y. This represents the uniform flow.
288
00:36:22,819 --> 00:36:29,400
These results can be generalized to provide
the velocity potential and stream function
289
00:36:29,400 --> 00:36:36,400
for a uniform flow at an angle alpha with
the x axis. As shown in the previous slide,
290
00:36:36,759 --> 00:36:43,759
these lines are present: phi is equal to Ux
and psi is equal to Uy and if you consider
291
00:36:49,739 --> 00:36:56,739
certain angle as shown here then we can represent
the psi and phi like this. For phi is equal
292
00:37:00,369 --> 00:37:07,369
to U into x cos alpha plus y sin alpha and
psi is equal to U into y cos alpha minus y
293
00:37:08,009 --> 00:37:14,140
x sin alpha So the results are not generalized,
provided the velocity potential and stream
294
00:37:14,140 --> 00:37:20,720
function are uniform flow pattern angle alpha.
This is called uniform flow and the uniform
295
00:37:20,720 --> 00:37:26,380
flow represented with respect to the potential
function and then the stream function as phi
296
00:37:26,380 --> 00:37:33,380
is equal to Ux and psi is equal to Uy as shown
here. phi is equal to Ux and psi is equal
297
00:37:40,279 --> 00:37:46,299
to Uy for the horizontal type of flow like
this and for inclined type flow it is phi
298
00:37:46,299 --> 00:37:52,640
is equal to U into x cos alpha where alpha
is the angle plus y sin alpha and psi is equal
299
00:37:52,640 --> 00:37:59,640
to Y into y cos alpha minus x sin alpha, this
is about the uniform potential flow with respect
300
00:38:00,059 --> 00:38:06,690
to the uniform flow which we have discussed.
This is the simplest plane flow. As discussed
301
00:38:06,690 --> 00:38:11,690
here uniform flow is the simplest flow described
by either stream function and we have seen
302
00:38:11,690 --> 00:38:14,339
the stream function and velocity potential.
303
00:38:14,339 --> 00:38:21,339
Now, the second kind of the basic or elementary
potential flow which we represent is called
304
00:38:22,839 --> 00:38:29,839
the source and sinks; the source and sink
is purely radial flow type. So the fluid flow
305
00:38:33,710 --> 00:38:40,710
is radialy outward from the origin perpendicular
to the xy plane only we have to consider the
306
00:38:42,369 --> 00:38:49,369
vr- the radial velocity here and psi is equal
to constant. This line represents psi is equal
307
00:38:52,269 --> 00:38:59,230
to constant and this dash line represents
the phi is equal to constant.
308
00:38:59,230 --> 00:39:06,230
If we consider the flow with respect to an
angle theta, we can derive various values
309
00:39:09,380 --> 00:39:16,380
of the potential function and the stream function.
Let us consider m as the volume rate of flow
310
00:39:20,930 --> 00:39:27,930
emanating from the line per unit length as
shown in this figure. If m is the volume rate
311
00:39:29,499 --> 00:39:36,259
of flow emanating from the line per unit length,
if we use the conservation of mass we can
312
00:39:36,259 --> 00:39:43,259
write this as 2phi r into vr is equal to m.
With respect to this figure, we can write
313
00:39:43,680 --> 00:39:50,680
m is equal 2phi r into vr, r is shown here.
If we consider the radial distance which we
314
00:39:54,799 --> 00:40:01,799
consider here 2phi r into vr is equal to m
where vr is the radial velocity. Finally,
315
00:40:02,299 --> 00:40:07,069
for the considered source sink we can write
the volume rate of flow emanating from the
316
00:40:07,069 --> 00:40:11,119
line per unit length vr is equal to m divided
by 2phi r. So with respect to this conservation
317
00:40:11,119 --> 00:40:18,119
of mass we can write vr- radio velocity is
equal to m by 2phi r and for the potential
318
00:40:20,549 --> 00:40:27,549
flow, the tangential velocity is equal to
0. As defined here it is purely radial flow;
319
00:40:29,989 --> 00:40:36,989
there is no tangential component for the velocity.
So, the radial velocity is equal to m divided
320
00:40:40,069 --> 00:40:45,940
by 2phi r and Vtheta is equal to 0. So that
we can write del phi by del r which is the
321
00:40:45,940 --> 00:40:52,549
radial velocity component del phi by del r
is equal to m by 2phi r. Finally, we can get
322
00:40:52,549 --> 00:40:59,549
an expression for phi. The potential function,
phi is defined as phi is equal to m divided
323
00:41:01,239 --> 00:41:06,359
by 2phi by r. The integration of expression
del phi by del r is equal to m by 2phi by
324
00:41:06,359 --> 00:41:13,359
r. We get an expression for the velocity potential
phi as 2phi by r. So this represent as far
325
00:41:16,460 --> 00:41:23,460
the source flow is concerned as shown in this
figure the potential function phi is represented
326
00:41:23,759 --> 00:41:29,789
as m is equal to m divided by 2phi natural
log r, where m is the volume rate of flow
327
00:41:29,789 --> 00:41:36,789
emanating from the line per unit length as
shown in this figure. So, if volume rate of
328
00:41:39,430 --> 00:41:45,739
flow, m is positive the flow is radially outward
and the flow is considered to be a source
329
00:41:45,739 --> 00:41:48,599
flow.
330
00:41:48,599 --> 00:41:55,599
If m is negative the flow is toward the origin
and the flow is considered to be a sink flow.
331
00:42:00,759 --> 00:42:05,119
This is either with respect to the domain
which we are considering what is coming and
332
00:42:05,119 --> 00:42:12,119
what is going out; with respect to that particular
point we can define if m is positive then
333
00:42:12,220 --> 00:42:19,220
it can be the source flow and if m is negative
it can be the sink flow. So the flow rate,
334
00:42:19,609 --> 00:42:26,609
m is called the strength of the source or
sinks. We have considered the source or sink.
335
00:42:28,670 --> 00:42:35,670
As shown here whether it can be coming to
the domain or going out of the domain. It
336
00:42:35,869 --> 00:42:41,509
can be a sink or a source depending up on
the case whether m is negative or positive,
337
00:42:41,509 --> 00:42:48,109
where m is defined as the volume rate of flow
emanating from the line per unit length. This
338
00:42:48,109 --> 00:42:54,970
m is called with respect to the potential
function phi is equal to m by 2phi natural
339
00:42:54,970 --> 00:43:01,970
of r. Here, this m is called the strength
of the source or sink. With respect to stream
340
00:43:05,460 --> 00:43:12,460
function, we can define this in polar coordinate
system in cylindrical coordinate system define
341
00:43:12,579 --> 00:43:19,099
theta is equal to the tangential velocity.
vtheta is equal to minus del psi by del r
342
00:43:19,099 --> 00:43:25,460
and the radial velocity vr is equal to 1 by
r del psi by del theta.
343
00:43:25,460 --> 00:43:32,460
With respect to this for the source or sink
it is purely radial flow we can that vtheta
344
00:43:32,979 --> 00:43:39,979
is equal to 0 so that here del psi by del
r is equal to 0 or now find the V r is equal
345
00:43:40,410 --> 00:43:45,210
to one by r del psi by del theta so this is
the radial velocity which we have already
346
00:43:45,210 --> 00:43:51,400
seen so this is equal to m by 2phi r so which
will give the stream function is equal to
347
00:43:51,400 --> 00:43:58,400
m by 2 phi theta.
So finally, we can derive here del psi by
348
00:44:00,299 --> 00:44:07,299
del theta. This we can integrate that we will
get psi is equal to m by 2 phi theta, this
349
00:44:09,950 --> 00:44:16,950
gives the stream function. For a source or
sink type flow which is described here, source
350
00:44:17,869 --> 00:44:24,869
or sink which are purely radial flow we have
derived the velocity potential as m by 2phi
351
00:44:27,269 --> 00:44:34,269
natural log r and then the stream function
psi is equal to m divided by 2phi into theta,
352
00:44:40,549 --> 00:44:47,549
where m is the strength of the source or sink
which we considered. Now we got the stream
353
00:44:49,420 --> 00:44:54,739
function and the potential function.
354
00:44:54,739 --> 00:45:01,739
As far as source and sink is concerned which
we represent, the stream lines are just radial
355
00:45:18,160 --> 00:45:25,160
lines.
356
00:45:27,670 --> 00:45:34,670
The source the stream lines are radial lines
and the equipotential lines are concentric
357
00:45:38,670 --> 00:45:44,999
circles centered at the origin as shown in
figure. So point source or sink is a point
358
00:45:44,999 --> 00:45:51,999
of singularity in the flow field. Since the
radial velocity, r is defined like this is
359
00:45:53,920 --> 00:46:00,920
the origin so from origin r tends to 0 the
sources strength is put on the origin. So,
360
00:46:01,680 --> 00:46:08,680
when r tends to 0 we can see that the radial
velocity Vr radial velocity v will be tending
361
00:46:10,869 --> 00:46:17,869
to become infinity as per the definition.
We can see singularity may occur at the point
362
00:46:23,029 --> 00:46:30,029
source of singularity so Vr is defined as
m by 2phi r, when r tends to 0 as shown here
363
00:46:35,650 --> 00:46:42,650
you can see that Vr becomes Vr tending to
infinity. Then source or sink point source
364
00:46:45,890 --> 00:46:52,890
become a singularity, the point of singularity
where r is tending to 0. Vr becomes infinity
365
00:46:54,660 --> 00:47:01,660
which is called a singular point or sink is
we can represent as a singularity. So this
366
00:47:03,910 --> 00:47:10,910
concept of source and sink on applications
especially we consider the point media flow
367
00:47:10,960 --> 00:47:17,960
so sometimes there are not much complexity.
For the flow in media is homogeneous, isotopic
368
00:47:18,369 --> 00:47:25,369
cases we can consider. If there is a recharge
well or there is a plumbing well application
369
00:47:26,940 --> 00:47:33,940
of the potential theory we can approximate
and then we will try to solve the problem
370
00:47:37,680 --> 00:47:43,839
with respect to the pores media flow. These
are some of the applications which will be
371
00:47:43,839 --> 00:47:50,839
discussing later. With respect to these now
point sources is singularity as discussed
372
00:47:55,229 --> 00:47:58,489
and now the third one is so called the vortex
flow.
373
00:47:58,489 --> 00:48:04,019
The vortex flow: it is the flow in which the
streamlines are concentric circle. First we
374
00:48:04,019 --> 00:48:08,369
discussed the uniform flow; second one, we
discussed the source and sink and third type
375
00:48:08,369 --> 00:48:14,069
of basic or elementary potential flow is called
the vortex flow. The vortex flow is the flow
376
00:48:14,069 --> 00:48:19,680
in which the streamlines are concentric circles
and the velocity along each streamline is
377
00:48:19,680 --> 00:48:24,390
inversely proportional to the distance from
the center. This kind of flow is called a
378
00:48:24,390 --> 00:48:31,390
vortex flow and the vortex motion can be either
rotational or irrotational.
379
00:48:33,079 --> 00:48:40,079
This represents the rotational type which
we have discussed here; the velocity, the
380
00:48:46,380 --> 00:48:52,869
flow, streamlines are concentric circles and
velocity along each streamline is inversely
381
00:48:52,869 --> 00:48:58,739
proportional to the distance as shown in this
figure. So, phi constant is the dotted lines
382
00:48:58,739 --> 00:49:02,809
here and these lines represent the psi constant
lines.
383
00:49:02,809 --> 00:49:09,249
Then we can define the potential function
with respect to as phi is equal to K theta,
384
00:49:09,249 --> 00:49:16,249
where K is a constant, theta is show in this
figure. So phi is equal to K theta and psi
385
00:49:17,039 --> 00:49:24,039
can be represented as psi is equal to minus
K natural log r. Finally, in this case, vortex
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00:49:27,029 --> 00:49:32,690
flows Vr is equal to the radial velocity component
is equal to 0 and vtheta is defined as 1 by
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00:49:32,690 --> 00:49:38,119
r del phi by del theta which is equal to minus
del psi by del r. This is equal to vtheta
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00:49:38,119 --> 00:49:44,670
is equal to minus K by r. For vortex flow
we define phi is equal to K theta; the potential
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00:49:44,670 --> 00:49:51,219
function for vortex flow is defined as phi
is equal to K theta and psi is equal to minus
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00:49:51,219 --> 00:49:57,079
K natural log r, where K is the constant.
If the fluid were rotating as a rigid body
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00:49:57,079 --> 00:50:04,079
has already shown earlier vtheta is equal
to K1r, where K1 is a constant.
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00:50:05,200 --> 00:50:09,150
This type of vortex motion is rotation and
cannot be discussed with respect to velocity
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00:50:09,150 --> 00:50:13,759
potential; the vortex flow can be either rotational
or irrotational.
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00:50:13,759 --> 00:50:20,759
We call the rotational vertex as forced vortex,
for example, the motion of a fluid or the
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00:50:25,049 --> 00:50:30,160
motion of a liquid contained in a tank that
is rotated about its axis with angular velocity
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00:50:30,160 --> 00:50:34,960
omega. This is called a rotational vortex.
Then irrotational vortex, free vortex, for
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00:50:34,960 --> 00:50:41,710
example the swirling motion of the water as
it drains from a bathtub. This represents
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00:50:41,710 --> 00:50:44,589
the irrigational vortex.
The irrotational vortex is called the free
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00:50:44,589 --> 00:50:50,469
vortex and for example swirling motion of
the water as it drain from a bathtub is called
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00:50:50,469 --> 00:50:54,809
irrotational vortex and rotational vortex
is called the forced vortex so the motion
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00:50:54,809 --> 00:50:59,489
of the fluid contained in a tank that is rotated
about its axis with angular velocity omega
402
00:50:59,489 --> 00:51:04,779
so this is called rotational vortex and also
we can have combined vortex. The combined
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00:51:04,779 --> 00:51:10,380
vortex is the forced vortex as a central core
and a velocity distribution corresponding
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00:51:10,380 --> 00:51:16,769
to that of a free vortex outside the core.
We can write vtheta is equal to omega into
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00:51:16,769 --> 00:51:23,769
r, where r is less than are equal to r0 and
vtheta is equal to K by r as represented earlier;
406
00:51:26,349 --> 00:51:33,349
vtheta is the tangential velocity and K is
the constant which we considered. The vortex
407
00:51:35,640 --> 00:51:42,319
flow can be rotational vortex which is called
forced vortex or it can be irrotational vortex
408
00:51:42,319 --> 00:51:49,210
called free vortex or we can also have combined
vortex which is called forced vortex as defined.
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00:51:49,210 --> 00:51:53,509
So this is about the vortex flows.
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00:51:53,509 --> 00:52:00,509
Further, we will be discussing about the circulation
with respect to the vortex flows and then
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00:52:01,749 --> 00:52:08,749
we will be discussing the doublet; then the
combination of all the elementary basic flows
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00:52:10,440 --> 00:52:16,710
to represent complex flow system which can
be in certain places we can apply for the
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00:52:16,710 --> 00:52:20,710
real fluid flow problem. So, this will be
discussing in the next lecture.
414