WEBVTT
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In the last class, we had an introduction
to the three dimensional flow features in
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an axial flow compression machine. In the
last class we of course, talked more about
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the physics of the three dimensional flow,
how the three dimensional flow really develops
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inside the blade passage even if the flow
coming in is relatively uniform and then of
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course, it goes out with a full three dimensional
flow feature. In today's class, we will try
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to capture some of it in simple mathematical
forms. Later on of course, we will look at
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a little more comprehensive mathematical form
to capture the three dimensionality of the
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flow inside actual flow compressor machine,
more notably inside the rotor which is of
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course, the rotating machine.
Now, as we have seen, the flow inside the
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blade rows develops three dimensionality.
It may not have three dimensionality to begin
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with, especially in the first stage of an
axial flow compressor. In the subsequent stages,
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it normally goes in with a certain amount
of three dimensionality and further three
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dimensionality may develop depending on the
stage design, depending on the how the stages
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are stacked and then again if you have a multi
spool axial flow compressor arrangement. In
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a multi spool arrangement, the flow going
from the low pressure spool to the high pressure
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spool of a gastro vine engine quite often
has a small intervening duct in between which
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often is designed to straighten out the flow;
that means, the flow going into the high pressure
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spool is somewhat straightened out or made
uniform going into the first stage of the
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high pressure or H p spool. However, it immediately
again starts developing large amount of three
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dimensionality as it goes through successive
stages.
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So, three dimensionality of flow is a part
and parcel of the modern axial flow compressor.
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The basic theory that we are going to cover
today was indeed developed quite some time
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back, little more than fifty years back and
at time, the compressors were not very highly
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loaded. They were rather lightly loaded and
as a result of which certain very simple mathematical
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formulations were often sufficient to represent
what is happening and as I mentioned, later
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on in this lecture course, we will have another
lecture in which we will try to capture a
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little more modern version of axial flow compressor
three dimensionality. But in today's class,
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we will look at some what the simpler version
of the three dimensional flow in simple mathematical
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form.
So, let us take a look at what are the features
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of three dimensional flow and how do we go
about capturing it in simple mathematical
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form.
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The three dimensional flow features in an
axial flow compressor. It is understood that
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the flow we are talking about is air and not
any other fluid. For the movement, we will
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stick to the fact that we are dealing with
air and this air is as you know is very light.
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So, number of the assumptions can be made
with respect to flow of air through axial
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flow compressors and we will also make those
assumptions before we set forth on the mathematical
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formulations. But we will take a quick look
again at the physics of the flow and exactly
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how do we formulate the mathematical form.
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If we take a simple axial flow compressor
representation, we have the blades and those
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blades let us say are rotating in certain
directions and that direction is given by
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the arrows. Some are when these rotating blade
passages, we can say we have a fluid element
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and this fluid element is a representative
of the flow that is flowing in this entire
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passage. We have discussed this passage and
what happens inside this passage in some detail
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in the last class. So, much of that will of
course, be at the back of our mind when we
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go forward and hence we say that this fluid
element, a small element is representative
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of the fluid flow inside this passage. We
have seen that this passage consist of one
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side you have let us pressure surface, another
side you have a section surface, on the top
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you have casing and the bottom you have hub.
So, it is bounded by four dissimilar surfaces
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and so, we have a fluid element inside a bounded
passage of four dissimilar surfaces; however,
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we assume that this fluid element is representative
of the status of the entire flow inside the
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rotor passage. So, when we analyze what is
happening to this element, essentially we
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are trying to capture what is happening inside
the entire passage.
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So, we will have a look at this elemental
fluid or the fluid element and see what happens
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to this fluid element as a representation
of the entire fluid passage and try to figure
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out how this fluid passages flow or its path
through the blade passage can be captured
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in simple mathematical form. The next thing
we would like to remember is that this fluid
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element is also captured inside the curve
blade passage between two blades.
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Now the two blades as we just talked about
has this is a section surface and this is
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your pressure surface and as a result the
passage in between is actually a slightly
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defusing passage and of course, it is a curved
passage. So, let us say that this fluid element
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is situated on the mean of this passage which
as we have discussed before is indeed actually
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this curved line is indeed parallel to the
camber of the airfoil that is been used here.
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So, some are in the middle. Let us say that
we have a fluid element that is captured and
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is been subject to certain analysis.
You remember that in the cascade analysis
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that has been done before; we have a vector
diagram in front of the blade. Let us say
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this is a rotating blade. So, direction of
rotation is shown over here and in front,
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we have a vector diagram which shows all the
relative and the absolute velocities coming
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into the blade and of course at the exit side,
we have the absolute and the relative velocities
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and ` of course, the tangential components
shown here. This is what you have done already
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in the cascade analysis and this particular
blade passage is actually part of the domain
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that is been figured out enclosing one particular
blade and this bounded surface is of course,
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holding the fluid element in the mid passage
representing the fluid flow through this blade
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passage.
So, this is the blade element that we will
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be talking about and we will try to see how
this blade element behaves under certain assumptions
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or under certain circumstances which we will
set forth right in the beginning. So, let
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us take a look at what are these assumptions.
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Now these assumptions are created essentially
to create what we call a simple three dimensional
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flow analysis. As I mentioned, we will go
into a little more comprehensive three dimensional
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flow analysis later in this lecture series.
Now the initial assumptions are: the radial
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movement of the flow is governed by the radial
equilibrium of forces. Now that essentially
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means that the fluid element which is captured
there actually is in equilibrium in the radial
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direction; that means, the fluid element which
has been captured inside the blade passage
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has a certain balance of forces in the radial
direction and this balance of forces is what
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keeps the fluid element moving more or less
in the actual direction and this is what is
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meant by the first assumption; that means,
radial movement of the flow is governed by
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the radial equilibrium of forces. The forces
are balanced any radial equilibrium must necessarily
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be in consonants with passage shape that is
provided. So, the forces are adequately balanced
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not to create any radial movement of the fluid
element.
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So, the radial movement occurs when the blade
passage is creating a certain radial flow.
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For example, we shall see later on that in
the modern axial flow compressor, the blade
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passage because of the high compression ratio.
The blade passage is often somewhat converging
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in nature and that converging passage either
the tip is converging or the hub is converging.
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As a result of this converging, automatically
certain fluid element near either the hub
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or near the tip does acquire a radial component
of the flow. Now that is part of the blade
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passage shape and not part of the fluid mechanic
forces. So, at the movement, we are assuming
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that the fluid mechanic forces; the fluid
that is held inside the two blades, that fluid
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is in equilibrium of forces in the radial
direction. So, in the radial direction, all
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kinds of forces are to be balanced. Now this
is the first assumption that we will make
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that the fluid will be in radial equilibrium.
The other thing is that the gravitational
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forces can be neglected. Now as I mentioned,
we are dealing with air. Air is a very light
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fluid. So, we can say that, for all practical
purposes the gravitational force can be neglected
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and we can proceed without taking into account
the g force which in a very heavy fluid or
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heavy liquid would indeed be an important
issue. But as far as we are concerned, we
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are going to neglect the gravitational force.
Let us take a look at details of the fluid
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elements.
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If we look at this fluid element, we can see
that it is been created in a manner such that
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it fits into the blade passage and hence it
has a small angle of which is shown by the
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angle d theta. So, it is not rectangular or
a cubic element. It is an element that actually
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creates an angle d theta from the centre of
rotation, but it has a unit length. So, the
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length of the fluid element is considered
to be unity and it has a depth, radial depth
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of d r along which the fluid element is created
and this fluid element is at a distance r
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from the centre of rotation of the axial flow
compressor. So, it at a distance r and it
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itself has a small radial length of d r and
an actual length of unity and circumferential
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depth of d theta subtended by angle d theta.
So, that would be r d theta. So, that would
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be the width over here.
Now, we say that, if we us this co-ordinate
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system in which this direction is the actual
direction, the upward is of course, the radial
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direction and the direction in perpendicular
to the actual is the tangential direction.
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So, in this direction, we show that the fluid
has a tangential fluid velocity typically
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given by C w if it a stator blade or V w if
it is a rotor blade and the forward velocity
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is normally given by C a or V a and that is
a actual velocity and the forces or the pressure;
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the fluid pressure that is acting on the surfaces
is shown here.
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Now on the bottom side, the fluid pressure
is said to be or assumed to be p and it increases
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the pressure through the depth of the fluid
element which is d r and on the top surface,
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the fluid pressure is p plus d p. So, the
change of pressure from here to here is d
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p and on the sides it has therefore, an average
pressure of p plus d p by 2. So, that is the
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pressure that is acting on the two surfaces;
on this surface as well as on the surface
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on the other side and as a result of which
these other pressures acting on the fluid
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element from all its bounded surfaces of the
element and this fluid is in motion. It has
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a motion which is predominantly axial, given
by the actual velocity C a or V a and it may
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or quite often has velocity C w or V w and
the subscript w refers to the tangential or
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what in many books is referred to as whirled
component. That is how w comes in. So, it
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is a whirled component of the flow given normally
by C w or V w depending on whether it is stator
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or rotor.
We shall be mostly talking about the flow
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through the rotor which means the rotational
speed of the blades that contain the fluid
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element would also be brought into the picture.
So, this is the fluid element that would indeed
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be subject to analysis in this simple mathematical
formulation. So, let us try to remember that
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we are analyzing a fluid element which has
pressure on all its bounded surfaces and this
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static pressure which is acting on this fluid
element on all its surfaces has to deal with
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the dynamic forces due to the motion of the
fluid particle or the fluid element which
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is shown by the velocity C a or V a or C w
or V w. So, this is a fluid element which
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we shall subject to mathematical analysis.
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Now if you resolve all the aerodynamic forces
and we are dealing essentially with only aerodynamic
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or fluid mechanic forces and we are not dealing
with any of the mechanical issues in this
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is machine, we are restricting ourselves essentially
to the aerodynamic issues. So, we will deal
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with only the aerodynamic forces and the aerodynamic
forces that are acting on this element and
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we say that the aerodynamic forces that are
acting in the radial direction because we
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had set forth right in the beginning, that
we would like to achieve radial balance of
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forces and hence we take the radial component
of all the forces that are acting.
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Now, the first term here is the static force
that is acting on the top surface of the fluid
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element which is p plus d p is the pressure
and hence that multiplied by the area and
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that area is r plus d r into d theta into
1 which is the unit length. So, that is the
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area and this is the pressure acting. So,
this whole thing is the force that is acting
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on the top surface of the fluid element.
Now this force is counted by the force acting
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on the bottom surface which is simply p into
r 1 d theta; r 1 d theta being the area of
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the bottom surface and p is the pressure acting
on the bottom surface. On the other hand,
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we have seen the two side phases also experience
pressure which we have said is the average
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is p plus d p by 2 and hence and there are
a two surfaces; one on this side, one on the
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other side and hence that is again multiplied
by the area d r d theta by 2 into 1; that
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is area.
So, the component of this is d theta by 2
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and that essentially gives us the area of
the fluid element which is acting in the radial
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direction.
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Now, if you look at the other surfaces; this
so called front surface and the so called
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back surface, they are essentially parallel
to each other and hence they are vertical
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and hence there is no radial component of
those they in those surfaces are themselves
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radial and hence they do not contribute any
more to the radial balance of forces. They
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are indeed already in the radial plane. So,
if we take these static forces, they are to
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be balanced by the dynamic force which is
the centrifugal force and that is the C w
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square by r is the acceleration in the centrifugal
direction or the radial direction and row
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d r r is of course, the volume of the fluid
element, a row of course, is the density and
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this and 1 of course, is the unit length.
So, this gives us the volume and this is into
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the density, that is a mass and this is of
course, the acceleration in the radial direction.
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So, that is the force due to the motion of
the fluid element in the blade passage which
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is executing radial force due to acceleration
in the tangential direction.
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There is no force due to the motion in the
actual direction because that is actually
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said to be in the purely in the actual direction
whereas, the one in the tangential direction
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or whirled component as we call it, would
indeed have a radial component of acceleration
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and that actually provides the dynamic force
due to the motion of the fluid element and
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that has to be balanced by the static forces
that are acting all over the fluid element.
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So, the static forces and the dynamic force
due to the motion are to be balanced and this
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balance is what we are talking about. This
has to be balanced to ensure that this fluid
21:25.340 --> 21:31.810
element is in equilibrium of forces so that
due to unbalance of this static and dynamic
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forces, there is no motion that accrues to
the fluid element. So, fluid element is not
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experience any radial motion due to unbalance
of these forces; that is the static forces
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and the dynamic forces. They are to be held
in balance by design, by intent of the blade
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that is created for axial flow compressors.
Let us move forward and let us see what happens
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to this equation.
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If we look at this equation, we will see that
it has a large number of terms that are second
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order or even higher order which means they
are products of small terms like d p d r or
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d p d theta or d r d theta which are to be
neglected. So, the combination of these terms,
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so, if you look at this in the first, we will
have product of d p d r or product of d p
22:33.700 --> 22:40.700
d theta or d r d theta; these terms are to
be neglected as small quantities.
22:40.910 --> 22:47.910
On the other hand, p r 1 d theta gets balanced
by a term which is created here that is p
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r d theta and they would neglect each other.
So, the result is that we arrive at a very
22:57.650 --> 23:04.650
simple equation and that is 1 by row d p d
r equal to 1 by r C w square. C w square of
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course, is the square of the whirled component
of the velocity of the fluid element. Now
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this is called the simple radial equilibrium
equation. In many of the books that you may
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study, you will probably find that this is
also simply referred to as radial equilibrium
23:27.010 --> 23:34.010
equation or sometimes radial equilibrium condition.
So, many of the text books that you may look
23:34.380 --> 23:40.780
at in axial flow compressors or in a chapter
of axial flow compressor, you might find this
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equation referred to as radial equilibrium
equation or radial equilibrium condition.
23:47.860 --> 23:54.860
This is born out of the simple balance of
forces of the fluid element that we have just
23:55.070 --> 24:02.070
looked at. Now this is a radial equilibrium
of forces that produces this simple radial
24:02.820 --> 24:07.740
equilibrium equation.
Let us go forward and see how we can make
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use of this radial equilibrium equation for
the sake of analysis and design of axial flow
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compressor flow.
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If we look at the fact that this flow is experiencing
a number of motions, we capture these motions
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typically in let us say, three equations which
are born out of the basic fluid mechanics
24:34.550 --> 24:41.550
and thermodynamics of the flow. The first
one is the energy equation which is the total
24:42.130 --> 24:49.120
enthalpy is equal to static enthalpy plus
the kinetic energy which is captured over
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here and this c square is the absolute velocity.
We can split it into two components; the actual
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component and the whirl component and we are
neglecting because we have assumed that the
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fluid is in radial balance of forces. Hence
there is no C r. If the fluid is not in radial
25:10.500 --> 25:17.500
balance of forces, this C over here would
indeed have a plus C r square which is the
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radial component of velocities.
So, at the movement we have assumed that that
25:23.310 --> 25:30.310
does not exist. There is no radial motion
of the fluid element. C p t of course, is
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the static enthalpy of the particular condition
of the state at which the fluid is in. The
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second equation comes from the equation of
state which is of course, as you know p equal
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to row r t and that produces a situation where
this can be written down, derived essentially
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from the equation of state and the third equation
that we will invoke here is the is the isentropic
26:01.440 --> 26:07.380
law; that is p divided by row to the power
gamma equal to constant this comes from the
26:07.380 --> 26:13.820
isentropic law. We are assuming indeed that
the flow through the axial flow compressor
26:13.820 --> 26:20.820
is prima facie isentropic in nature. There
is no heat transfer in or out of the flow
26:21.050 --> 26:28.050
during the process and hence it is essentially
to begin with, considered an isentropic process
26:29.460 --> 26:34.420
and hence it allows us to prima facie use
the isentropic law.
26:34.420 --> 26:41.420
So, these terms and notations are explained
over here that each is the total enthalpy,
26:43.600 --> 26:50.600
small H is the static enthalpy, pressure is
p, density is row which are the fluid properties
26:51.020 --> 26:56.410
or the properties of the state and then of
course, the thermo dynamic or thermal properties
26:56.410 --> 27:02.700
C p and gamma of the air under the given operating
condition.
27:02.700 --> 27:09.700
So, we invoke these fundamental laws of fluid
mechanics and thermo dynamics on the energy
27:11.010 --> 27:18.010
equation that we have looked at and let us
see where this transposition takes us to.
27:18.420 --> 27:23.820
So, invoking these laws of fluid mechanics
and thermo dynamics and then if you substitute
27:23.820 --> 27:30.820
them what we do is, we substitute this second
equation to begin with C p t in the first
27:31.090 --> 27:38.090
equation and get a new version of the energy
equation by differentiating with respect to
27:39.730 --> 27:46.730
r. Now, as I mentioned earlier, we are dealing
with what is happening in the radial direction
27:46.790 --> 27:53.620
or balance of forces, balance of energy in
the radial direction. That is what one of
27:53.620 --> 27:59.770
our prime concern at this moment and hence
we differentiate equation one with reference
27:59.770 --> 28:06.770
to r and that gives us d H d r equal to C
a into d C a d r plus c w into d C W d r plus
28:12.050 --> 28:19.050
gamma by gamma minus 1 into 1 by row d p d
r minus p by row square into d p d r.
28:22.800 --> 28:29.800
So, this is what we get by differentiating,
first we replace C p t from here and then
28:32.150 --> 28:38.110
we can say that this as a first term and this
goes into the second term and then H plus
28:38.110 --> 28:45.110
that whole thing is differentiated with reference
to r and if we do that this is what we get
28:45.120 --> 28:50.530
in terms of the variation of each parameter
in the radial direction.
28:50.530 --> 28:56.180
So, this is the enthalpy variation in the
radial direction. This is the C a variation
28:56.180 --> 29:02.540
in the radial direction, this is the C w variation
in the radial direction and this is the p
29:02.540 --> 29:07.660
variation pressure variation in the radial
direction and this is the density variation
29:07.660 --> 29:14.660
in the radial direction. Now this energy equation
indeed captures variation of all the important
29:15.200 --> 29:22.200
parameters in the radial direction. If you
do have access to or you have means to actually
29:22.470 --> 29:29.470
calculate the variation of each of these parameters
in the radial direction realistically, you
29:29.610 --> 29:36.610
can pluck them in here and solve this equation.
Quite often to begin with, you may not have
29:37.190 --> 29:43.440
access to the variation in the radial direction
of each of these parameters. These are the
29:43.440 --> 29:49.490
important parameters; the enthalpy variation,
the axial velocity variation, the whirl component
29:49.490 --> 29:55.740
variation, the pressure variation and the
density variation. Now in a modern axial flow
29:55.740 --> 30:01.960
compressor, I might as well mentioned here,
in a modern axial flow compressor, it is entirely
30:01.960 --> 30:08.930
possible that all the terms on the right hand
side would indeed have a tendency to vary
30:08.930 --> 30:14.970
in the radial direction and some of the terms
may vary quite a lot quite significantly certainly
30:14.970 --> 30:21.970
not negligible. With all those variations,
can we still keep the variation of enthalpy
30:23.130 --> 30:29.850
in the radial direction negligible or to the
minimum? That is the question that the axial
30:29.850 --> 30:36.850
flow designer or the initial designer analyst
would have to contend with and decide to begin
30:36.940 --> 30:40.700
with.
So, these are the issues that a typical designer
30:40.700 --> 30:47.700
analyst, we will have to decide during the
process of the design itself before the designing
30:48.110 --> 30:54.240
the blade shapes. So, let us look at the simplest
possible thing because as I mentioned, we
30:54.240 --> 30:59.600
are doing the simplest possible radial analysis
today.
30:59.600 --> 31:06.600
So, we will look at it in the simple possible
fashion. So, we say that in this equation,
31:07.620 --> 31:14.080
we shall consider the variations of these
parameters probably in the most simple way.
31:14.080 --> 31:20.960
Let us look at what can be done. To begin
with, if we differentiate the equation three
31:20.960 --> 31:27.960
that was given earlier, that is the isentropic
law, in the radial direction again, we get
31:28.400 --> 31:34.230
a variation of density in the radial direction
and that comes out, in terms of the pressure
31:34.230 --> 31:39.700
in the radial direction with reference to
the other terms. If we substitute that, the
31:39.700 --> 31:46.700
energy equation takes the form d H d r equal
to C a into d C a d r plus C w into d C w
31:48.390 --> 31:54.670
d r plus 1 by row into d p d r.
So, this is of course, a similar version compared
31:54.670 --> 32:01.670
to this one which has a little more variations.
So, those variations have been simplified
32:02.740 --> 32:08.760
out by invoking the isentropic law and that
is what isentropic law allows us to do.
32:08.760 --> 32:14.620
So, this is the first simplification that
the flow is isentropic and by invoking by
32:14.620 --> 32:20.710
isentropic flow, we have simplified the situation;
that is the third term and we are able to
32:20.710 --> 32:27.710
put this third term here as 1 by row d p d
r. Now this tells us that we have now three
32:29.470 --> 32:34.850
variations; the axial velocity in the radial
direction, the whirl component in the radial
32:34.850 --> 32:41.850
direction and a static pressure in the radial
direction and as a result, it assumed that
32:42.520 --> 32:48.050
to begin with, density is sort of assumed
to be more or less constant. We will do that
32:48.050 --> 32:55.050
formally as we go along. Now in the modern
axial flow compressor, we will do later on.
32:55.760 --> 33:01.880
We shall see that in the first stage, it is
possible that the C a variation in the radial
33:01.880 --> 33:08.250
direction could be quite uniform to begin
with, but in the subsequent stages we shall
33:08.250 --> 33:15.250
see later on in fact, we have look at in the
earlier last lecture that the C a profile
33:16.260 --> 33:22.870
going into the stages can vary substantially
in the radial direction. So, that variation
33:22.870 --> 33:29.870
of C a would need to be invoked here appropriately
to tell you what is happening of the radial
33:31.210 --> 33:35.320
balance of forces of the static and the dynamic
forces.
33:35.320 --> 33:42.320
So, those are realties that an analyst will
have to bring in as and when those realistic
33:43.160 --> 33:50.160
numbers or realistic values are indeed made
available to him. That may happen in the second
33:50.530 --> 33:57.530
round of analyses or the third round of analysis
when the realistic values start coming through
33:57.820 --> 34:04.040
from the first cut design or analysis. Now
let us take a look at what happens to this
34:04.040 --> 34:11.040
equation which we have simplified already.
If we look at it, we can now invoke the radial
34:13.190 --> 34:18.899
equilibrium equation or the simple radial
equilibrium equation which was of course,
34:18.899 --> 34:24.300
1 by row d p d r equal to 1 by r into c w
square.
34:24.300 --> 34:30.159
Now this can be now be substituted in the
energy equation which we are put in the last
34:30.159 --> 34:37.159
slide and in the third term, it is a direct
substitution and it is not involve a great
34:37.490 --> 34:44.490
deal of problem and if we do that, the equation
takes the form now d h d r equal to c a into
34:46.450 --> 34:53.450
d C a by d r plus c w into d C W d r plus
C W square by r. Now C W square by r of course,
34:55.590 --> 35:02.530
is the acceleration term which we had talked
about earlier and hence we what we see is
35:02.530 --> 35:09.530
that all the three are essentially representative
of the dynamics of the fluid and the statics
35:11.350 --> 35:17.400
of the fluid have essentially now been eliminated
from the energy equation.
35:17.400 --> 35:24.400
Now this of cause tell us that the variation
of d C a d r and d C d w d r are the prime
35:26.580 --> 35:32.890
forces which would have to be brought into
the focus if we are to get more realistic
35:32.890 --> 35:38.670
analysis, but that as I mentioned, we shall
do in one of the lectures later on.
35:38.670 --> 35:45.670
Now, all this together tells us what should
be the variation of enthalpy in the radial
35:46.060 --> 35:52.920
direction because this enthalpy is what is
going is to undergo change as the flow goes
35:52.920 --> 35:58.940
through an axial flow compressor because an
axial compressor remember is a working machine
35:58.940 --> 36:05.020
and this working machine has to work. When
it works, the enthalpy of the fluid would
36:05.020 --> 36:11.830
change. The very purpose of doing the work
is to change the enthalpy or capital h which
36:11.830 --> 36:18.170
is the total enthalpy would indeed change.
What we are looking at the left hand term
36:18.170 --> 36:25.170
is the variation of total enthalpy; that is
static plus to dynamic in the radial direction
36:25.580 --> 36:31.490
from root to the tip of the blade. How does
it change? That is what or how should it be
36:31.490 --> 36:38.490
allowed to change? That is what we are contending
with at this moment. How should the enthalpy
36:38.970 --> 36:45.970
change occur as it passes through the blade
from root to the tip of the blade.
36:46.140 --> 36:53.140
So, this is something which the designer analyst
would have to decide a priori before he sets
36:53.590 --> 37:00.590
forth his design and creation of axial flow
compressor. Later on of course, it will have
37:00.600 --> 37:06.540
to be analyzed what happens to it under various
operating conditions, but those are the issues
37:06.540 --> 37:10.290
we should deal with a little later in this
lecture series.
37:10.290 --> 37:17.290
So, let us take a look at the most simplified
version; that means, we say to begin with,
37:17.860 --> 37:24.860
the radial variation of enthalpy is to be
deliberately held 0; that means enthalpy in
37:28.000 --> 37:32.810
the radial direction would essentially remain
constant.
37:32.810 --> 37:39.810
If we do that, we see that the flow actually
has a constant enthalpy characteristic and
37:44.960 --> 37:51.960
hence the work done distribution also; that
means, d H along the length of the blade from
37:54.580 --> 38:01.580
root to tip in the radial direction we can
say is 0. So, d H d r is to be held 0. This
38:04.150 --> 38:11.150
is a deliberate decision that is been taken
for let us say, a simplified axial flow compressor
38:12.450 --> 38:16.810
design and analysis.
So, that is the first thing we do. We simplify
38:16.810 --> 38:23.810
the whole thing and say that the left hand
term d H d r is being held 0; that means,
38:25.490 --> 38:32.490
the work done is uniform from the root to
the tip along the blade length in the radial
38:33.050 --> 38:39.540
direction. So, this allows us a certain freedom
or certain simplification; that means the
38:39.540 --> 38:46.540
work done from root to tip is to be held constant.
So, somehow we have to manage to do equal
38:47.280 --> 38:54.010
amount of work at all the radial section from
root to the tip of the blade. How do you do
38:54.010 --> 38:58.330
that? That is a separate issue, we will come
to that and we will do that a little in the
38:58.330 --> 39:04.830
next lecture, but today we just make an assumption
that the work done is to be held constant
39:04.830 --> 39:09.620
along the length of the blade. So, that is
the first assumption. That gives us the left
39:09.620 --> 39:13.790
hand term of the energy equation now goes
0.
39:13.790 --> 39:20.790
Let us see what happens to that. So, we have.
So, we have brought it on the inside. So,
39:20.960 --> 39:27.960
d C a d r C w d C w d r and C w square by
r is equal to 0 and if C a is held constant
39:31.760 --> 39:38.160
already here, then the first term also goes
0 so; that means, the constancy of c a is
39:38.160 --> 39:45.150
now to be invoked and if we do that, the first
term goes away and we are left with only this
39:45.150 --> 39:52.150
term and this term is taken to the other side.
So, we have minus C w square by r and hence
39:54.050 --> 40:00.210
this becomes now the energy equation or the
governing equation of the fluid element that
40:00.210 --> 40:06.840
we are looking at.
So, we have simplified at three levels, we
40:06.840 --> 40:13.190
have said that d H d r is equal to 0. We are
also now saying that c a is held constant.
40:13.190 --> 40:20.190
So, d C a d r is also 0 and then we are left
with only two terms now and we take the acceleration
40:23.290 --> 40:29.930
term to the right hand side. So, it becomes
minus C W square by r and as a result of which,
40:29.930 --> 40:36.930
we get a very simple situation where we can
say that the equation becomes d C W d r equal
40:39.390 --> 40:46.390
to minus d r by r. Now this is the simplified
version subsequent to number of assumption
40:48.580 --> 40:54.630
that we have made. Now remember, those are
assumption, simplifying assumptions we are
40:54.630 --> 41:01.230
dealing with rather simplified axial flow
compressor in which number of assumption have
41:01.230 --> 41:08.230
been made, the density is held constant, the
gravity has been neglected. We have assumed
41:09.590 --> 41:14.830
that the work done or the enthalpy is held
constant from root to the tip of the blade
41:14.830 --> 41:20.070
along the length of the blade. We are also
assuming that while flowing through the blade
41:20.070 --> 41:27.070
passage right from the entry to the exits,
C a is held constant. It remain constant from
41:27.550 --> 41:32.180
root to the tip of the blade at the entry
as well as the exist and through the blade
41:32.180 --> 41:39.180
passage and when we put all these assumption
put together, a simplified fluid mechanic
41:40.090 --> 41:46.440
equation that comes through from the energy
equation. These are derived with all those
41:46.440 --> 41:53.440
assumptions and we get a very simple equation
d C w d r equal to minus d r by r.
41:54.530 --> 42:01.530
Now this on integration, yields a very simple
relation which is C w r equal to constant.
42:04.820 --> 42:09.480
Now this of course, is a very famous equation,
you would probably see it in many of the books
42:09.480 --> 42:16.480
that you may look at and this condition is
very commonly known as free vortex law. This
42:19.180 --> 42:26.180
is a free vortex law which the axial flow
compressor designers arrived at a little more
42:26.510 --> 42:33.510
than fifty years back and allow them to design
axial flow compressors invoking those simplification.
42:33.910 --> 42:40.910
Remember we had a large number of simplifications;
thermo dynamic, fluid mechanic, gravity is
42:41.400 --> 42:47.090
neglected, density is kind of held constant
which is true for you know low pressure compressors
42:47.090 --> 42:52.840
which are not very high speed.
So, comparatively low speed. So, all those
42:52.840 --> 42:59.720
things taken into account, the reality of
the flow has been somewhat simplified and
42:59.720 --> 43:06.270
we have taken a fluid element in which certain
fluid static forces and dynamic forces were
43:06.270 --> 43:13.270
allowed to hold forth along its surfaces.
As a result, the balance of those forces in
43:14.550 --> 43:21.550
the radial direction finally, tells us that
we can say that whirl component c w into r
43:24.170 --> 43:27.980
is equal to constant along the length of the
blade.
43:27.980 --> 43:34.980
Now, this is a very simple, but very important
and useful relation. It tells us what should
43:35.220 --> 43:40.900
be the nature of variation of C w along the
length of the blade. Otherwise there is no
43:40.900 --> 43:47.210
other way you can find out what should be
the value of C w at any radius of the blade
43:47.210 --> 43:53.820
from root to the tip. C a we have allowed
it to be constant, but how about C w? We have
43:53.820 --> 43:59.830
no way of figuring out what the value of the
C w should be unless we set forth a law like
43:59.830 --> 44:06.350
this. We need a law like this to figure out
what should be the value of C w and of course,
44:06.350 --> 44:11.930
if you do not have C w, you cannot go forward
or analyze your axial flow compressor.
44:11.930 --> 44:17.720
So, this is a very important law which was
arrived at by the pioneers of the axial flow
44:17.720 --> 44:24.490
compressor and it allowed them to actually
design and analyze the axial flow compressor
44:24.490 --> 44:30.360
in very simple fashion, but that is all right,
but without this we cannot go forward.
44:30.360 --> 44:37.360
So, C w r tells us that with the increase
of radius, the value of C w can reduce. So,
44:40.230 --> 44:47.230
at the tip, the value of C w would be small.
At the hub the value of C w W world be somewhat
44:47.420 --> 44:54.050
large. So, did the product of c w and r is
constant from root to tip and this is what
44:54.050 --> 45:01.050
is called the free vortex law. Now let me
explain very briefly, why it is called a free
45:01.290 --> 45:03.410
vortex law.
45:03.410 --> 45:10.410
The term free vortex actually comes from the
fact that it denotes in some way the strength
45:13.130 --> 45:20.130
of the free vortex which as you all know from
basic fluid mechanics or aerodynamics, essentially
45:23.420 --> 45:29.120
represents the lift per unit length of the
aerofoil sections used at that particular
45:29.120 --> 45:36.120
radial station.
So, at any radial station, you are deploying
45:36.230 --> 45:42.220
normally aerofoil. You have discussed this
in the earlier lectures in some detail and
45:42.220 --> 45:48.430
these aero foils are essentially lift producing
entities. Each of those aero foils creates
45:48.430 --> 45:54.390
lift. How do they create lift? They create
lift from aerodynamic understanding or aerodynamic
45:54.390 --> 46:00.510
theory by creating a vortex system around
itself and this vortex system has strength
46:00.510 --> 46:04.310
of its own, without the strength it cannot
create lift.
46:04.310 --> 46:10.300
So, lift is essentially a direct result of
the strength of the vortex system if the vortex
46:10.300 --> 46:17.300
system is 0, the lift also would be 0. So,
strength of this vortex system is now to be
46:18.690 --> 46:25.690
held constant from root to the tip of the
blade and hence it is called free vortex so
46:26.200 --> 46:33.200
that all the vortex from root to the tip of
the blade are of the same strength. Now if
46:33.400 --> 46:40.400
you write down in terms of lift from the basic
aerodynamics that all of you have done, normally
46:40.560 --> 46:47.560
it is written as lift per unit length and
that is l equal to row V and a capital of
46:49.690 --> 46:56.430
omega and that is a row is density, V is in
the inlet velocity or the velocity of the
46:56.430 --> 47:03.430
fluid and capital omega is the length of the
circulation and this tells us that if the
47:03.790 --> 47:08.240
strength of the circulation is held constant,
if the density is constant and if the inlet
47:08.240 --> 47:15.240
velocity is constant, the lift produced would
also be constant from the root to the tip
47:15.880 --> 47:20.370
of the blade.
Now, the strength of the circulation is what
47:20.370 --> 47:27.370
we are talking about and that is what gives
rise to the vortex and this is to be held
47:28.870 --> 47:35.870
constant and hence the term was coined as
I said more than 50 years back, free vortex;
47:36.110 --> 47:43.110
that means, it is free of any change of vortex
strength from root to the tip of the blade.
47:43.560 --> 47:50.290
This also means that the twilling edge of
the blade which typically produces a twilling
47:50.290 --> 47:56.260
edge vortex sheet coming out of the blade
or any three dimensional body made up of aero
47:56.260 --> 48:02.300
foils or lifting elements creates a twilling
edge vortex sheet and this vortex sheet would
48:02.300 --> 48:08.110
have constant strength from root to the tip
or hub to the tip of the blade.
48:08.110 --> 48:14.190
So, it will not have any variation or strength
along the length of the blade. Now this is
48:14.190 --> 48:21.190
what is meant by free vortex law or free vortex
condition and this allows us to create a simple
48:23.360 --> 48:30.360
blades which are of immense use to begin with
and this free vortex law was one of the first
48:30.950 --> 48:37.950
things that needed to be created to analyze
and design axial flow compressor rotor blades.
48:39.890 --> 48:46.890
So, today we have set forth a very simply
relation born out of the basic energy equation
48:49.030 --> 48:54.830
and deployment of the basic thermo dynamic
conditions invoking the isentropic law and
48:54.830 --> 49:00.730
making a number of assumptions; all of them
put together finally, has given us two varies
49:00.730 --> 49:06.000
simply things: one is the radial equilibrium
equation or what at the moment what we call
49:06.000 --> 49:12.900
as simple radial equilibrium equation and
consequent to that, we get the free vortex
49:12.900 --> 49:19.520
law. This free vortex is something we will
go forward with and in the next class, we
49:19.520 --> 49:26.520
will look at this free vortex law as a design
principle and how various design principles
49:27.240 --> 49:34.240
have been created based on this free vortex
law and it is derivatives of various laws
49:34.630 --> 49:41.240
which are simply called vortex laws again
derived from the word free vortex, a terminological
49:41.240 --> 49:47.530
free vortex and various vortex laws have been
created there off which tells us how the compressor
49:47.530 --> 49:53.450
should be designed and further analyzed. So,
this is what we will be doing in the next
49:53.450 --> 49:54.010
class.
49:54.010 --> 50:00.430
We will be looking at free vortex law as a
design law and the other blade design laws
50:00.430 --> 50:07.430
which are derived from those free vortex laws
and how do they help us essentially in analyzing
50:07.930 --> 50:13.850
the axial flow compressor blades. Later on,
we will talk about the various design principles.
50:13.850 --> 50:20.850
So, in the next lecture, we will follow it
up with the design laws which govern the axial
50:21.380 --> 50:23.290
flow compressor, flow through the axial flow
compressors.