WEBVTT
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Hello and welcome to lecture number fourteen
of this lecture series on jet propulsion.
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In the last several lectures, you have been
exposed to some of the basic aspects the thermodynamic
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aspects of the jet engines. We have analyzed
the ideal cycles, the component performance
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parameters like the intake, and the compressors
turbines combustion chambers, and nozzle and
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so on. And how we can incorporate these component
performance parameters into a real cycle analysis?
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So, during the component performance analysis,
we just assigned some performance parameters
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like isentropic efficiency or pressure laws
associated with various components, without
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really going into the working the basic aspects
of working of these components. So, what we
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will begin to do in the next several lectures
is to take up these components and understand
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the working of these components in greater
detail. You already understood the thermo
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dynamics behind or thermo dynamic operation
of compressors, and turbines.
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So in today’s lecture, we will begin to
understand we will we will start our understanding
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of the working of axial flow compressors in
little more detail. So, what we will be talking
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about in today’s lecture will be on axial
compressors; let us take a look at those topics
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that we shall be discussing in today’s lecture.
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So, we will start off our discussion today
with some brief description of the basic operation
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of axial compressors you have already been
exposed to some of these aspects in some of
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the earlier lectures; then we will also understand
about the construction of velocity triangles
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of axial compressors. We will derive some
expressions for the work, and the pressure
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ration of axial compressors and how they are
related.
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And then we will also discuss about some of
the design parameters which are involved in
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design of axial compressors. We will only
list a few of them this some of them which
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are important in the design of axial compressors
these are to do with flow coefficient the
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loading coefficient degree of reaction and
diffusion factor. So, we will have some very
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quick understanding, and discussion about
some of these different terminologies, which
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are used in design of axial compressors. So,
we will begin with our discussion the basic
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operation of axial compressors.
Now, axial compressors as your already your
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familiar with their basic function is to increase
the pressure of the incoming flow which is
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coming from the intake, and then deliver a
high pressure, and also high temperature relatively
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higher temperature air to the combustion chamber,
where fuel is added, and the actual heat addition,
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and combustion takes place in the combustion
chamber.
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So, axial compressors form one of the basic
components or aspects of a gas turbine engine,
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and in axial compressors as you might have
already noticed that they usually consist
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of a series of stages, and each stage basically
comprises or consists of a row of rotor blades
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followed by a row of stator blades.
So, a combination of a rotor and the stator
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constitute or comprise a stage of an axial
compressor, and the typical modern day axial
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compressors have several stages, and several
stages of rotors, and stators, and what basically
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happens in these stages is that the working
fluid is initially accelerated by the rotor
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blades. We will see how this is done later,
and subsequently it is decelerated in the
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stator passages.
And what happens in the stator is that the
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kinetic energy which was transferred to the
fluid in the rotor gets converted to increase
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in static pressure, and so what the rotor
basically does is that it derives its energy
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form the turbine; the turbine basically drives
the compressors. So, the energy that the rotor
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derives from the turbine is transferred to
the working fluid in the rotor.
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And then in the stator, stator has two functions
basically one is of course, that it will increase
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the static pressure by converting that kinetic
energy into pressure energy, the other function
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of course, is that at the exit of the rotor
the fluid also has a tangential component
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of velocity. So, the stator acts as a sort
of a guide where in some sense that it turns
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the flow back to its original direction partially.
So that the next stage of rotor can operate
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efficiently. So, stator has these two functions
that it not only increases the pressure but,
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also deflects the fluid in a particular desired
direction.
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So, and this process of the rotor transferring
energy to the working fluid; and then the
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stator doing its function is repeated in several
stages. So a multi stage axial compressor
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will have this process getting repeated over
several stages, which could be in a modern
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day compressor it could be as high as ten
to twelve stage or even more than that where
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in this process is repeated.
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So, in an axial compressor the compression
as I mentioned consists of a series of diffusions
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that is diffusion occurs both in the rotor
as well as in the stator now in the case of
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rotor because the rotor itself has a rotary
velocity that is there is a tangential velocity
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associated with the rotor blade itself, one
can identify two distinct velocity components;
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the absolute, and the relative velocities
that is if we were to observe the fluid particle
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as an observer from outside the compression
system that is the stationary frame of reference
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then we see one particular velocity, and if
the observer is actually seated on the blade
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itself that relative frame of reference then
we see another component of velocity.
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And therefore, rotation of the blade itself
leads to two distinct components of velocity
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the absolute, and relative components, and
these constitute basically the velocity triangle
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which we shall discuss shortly.
So, what happens in the rotor is that the
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absolute velocity is increased as a result
of the momentum or the energy that is transferred
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by the rotor on the working fluid, where as
there is a decrease in the relative velocity
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leading to diffusion, which is why I mentioned
that there is diffusion in both the rotor
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as well as the stator.
Now one of the factors that limits the amount
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of you may wonder why is that we need to have
so many stages of an axial compressor why
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cannot we have a limited number of stages
of axial compressor, the basic reason is that
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the per stage pressure ratio of pressure raise
in an axial compressor is limited, because
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an axial compressor as you might have already
guessed operates in an adverse pressure gradient
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environment that is the pressure downstream
is always higher than the upstream pressure,
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which means that the fluid has to overcome
this pressure gradient as it passes through
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an axial compressor stage therefore, if we
try to carry out too much of diffusion in
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one stage the adverse pressure gradient may
be so high that the flow will not be able
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to withstand these pressure gradients, and
there could be stalling of the flow from the
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blades.
And then that is one of the reasons that limits
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the per stage pressure ratio or per stage
pressure raise that can be achieved in an
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axial compressor, which is why we need to
have multiple stages of axial compressor each
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of them increasing the pressure by a certain
amount.
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So, on the other hand turbines do not really
have this problem, turbines always operate
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in a favorable pressure gradient mode and
therefore, it is possible to have a much larger
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pressure drop across a turbine stage as compared
to the pressure raise that we get from one
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stage of an axial compressor.
This means that in one stage of a turbine
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it is usually possible to achieve a substantial
pressure drop which can be achieved only in
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several stages of an axial compressor and
this is another reason if you next time when
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you take a look at a cutout diagram of a jet
engine you can notice that you will see several
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stages of axial compressors but, only very
few stages of turbine. So, typically a few
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stages of a in fact one stage of an axial
turbine can drive several stages of an axial
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compressor, and the reason for that is basically
that the per stage pressure ratio of compressors
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are limited due to adverse pressure gradient.
But, that problem is not really there for
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a turbine, because they always operate in
a favorable pressure gradient mode, and so
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you can actually expand in one stage much
more than what you would be able to compress
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in one stage of an axial compressor.
And so in a jet engine it is usually something
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you would notice that a few stages of turbines
would be driving several stages of a compressor,
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and that is one of the aspects that you need
to keep in mind. And in the case of compressor
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of course, which means that compressor design
in some sense would probably be little more
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challenging at least aerodynamically as compared
to a turbine.
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Now, so a careful design of a blade is very
much essential, because as I mentioned that
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you can increase the pressure ratio per stage
by only a limited amount and therefore, if
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you would like to minimize losses at the same
time try to maximize the pressure ratio that
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you can get from one stage of a compressor
that definitely calls for a very careful design
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of the compressor blades.
And in some cases you might notice in some
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engines that the compressor begins with one
set of a stationary blades, which are known
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as in that guide range usual compressors usually
compressors or actual compressors stage consists
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of a rotor followed by a stator that is rotor
imparts the kinetic energy to the fluid, and
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then stator increases the pressure by converting
that energy to pressure. But, in some engines
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the first stage of the compressor is preceded
by a row of stationary vanes which are known
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as inlet guide vanes.
And inlet guide vanes basically have a function
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in the sense that they permit the flow to
enter at a desired angle; especially, when
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the engine is operating at off design conditions
that is as the engine operates over several
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operating points, which may not necessarily
be the design operating point, then if the
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incoming flow angles are different then that
can affect the performance of the compressor,
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and therefore, the engine.
So, inlet guide vanes tend to deflect the
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flow at a certain desired angle. So, as so
that the off design performance of a the engines
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can be better than what it would normally
be. So, in an axial compressor the basic understanding
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of the working of an axial compressor begins
with understanding of what are known as the
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velocity triangles.
So, that is what we will be doing in the next
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few slides how to we construct a velocity
triangle or basically what do we mean by a
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velocity triangle. So, I think I already mentioned
that in a rotor because of the presence of
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the peripheral velocity of the blade speed
because of by virtue of its rotational speed
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one can identify to distinct velocity components
one is to do with the stationary frame of
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reference, and the other is the rotating frame
of reference, and so if you were to combine
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and all the three that is the blades speed
the relative velocity, and the absolute velocity
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or related to each other vectorially.
And so it is possible to construct a triangle,
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which relates which shows the relationship
between the blade speed the absolute velocity,
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and the relative velocity. And of course,
as we construct the velocity triangle it will
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also involve certain angles these are to do
with the inlet flow angles, and the blade
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angles which we will discuss shortly.
And so these angles also will be part of the
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so called velocity triangle. So, what we will
do in the next few slides is to see how we
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can construct given a certain rotor, and stator
combination how is it that. We can go about
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constructing a velocity triangle, because
that is where the basic design an understanding
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of the working of an axial compressor begins.
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So, what we do is that we will carry out or
the analysis that we are going to do is limited
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to a certain height of the blade. We will
take the main height of the blade, and we
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will assign the blade speed or the peripheral
velocity to a symbol U. So U denotes the peripheral
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velocity or the blade speed at that particular
station, at which the analysis is carried
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out. And we are going to carry out this analysis
for the main blade height.
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So, we will denote the absolute component
of velocity by symbol C, and relative velocity
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by V. So, absolute components of velocity
will be denoted by C and the relative component
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by V, and axial velocity the absolute component
of that is usually denoted by symbol C, subscript
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a whereas, a is for the axial, and the tangential
components will be denoted by a subscript
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w that means if we are denoting the tangential
component of the absolute velocity then it
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would be denoted by C subscript w; and the
tangential component of the relative velocity
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will be denoted by V subscript w.
And we will as I mentioned there are also
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certain angles which we will need to be familiar
with we will denote alpha for the angle between
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the absolute velocity and the axial direction.
So, the angle which the absolute velocity
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makes with the axial direction will be denoted
by alpha, and beta will be the angle corresponding
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to the relative velocity that is the direction
that is angle between the relative velocity,
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and the axial direction. So, alpha is for
the absolute velocity, and beta is for the
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relative velocity and absolute velocity is
to be denoted by C relative velocity or relative
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component by V.
So, what we will take up next is we will take
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up one row of a rotor blades, and followed
by a row of stator blades, and then we will
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see how we can think of beginning to understand,
and develop the velocity triangle for this
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kind of a configuration. So, what we will
do is to express the rotor blades on a two
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dimensional plane, you know that rotor blades
are arranged on an axis.
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So, let us assume that the rotor blades are
at arranged in a linear fashion which is what
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I will show in the next slide, and that is
basically for simplicity, and that is particularly
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true for a particular cross section. In this
case, we will assume that it is a the main
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blade height section.
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So, if we consider a row of rotor blades,
and stator blades so what is shown here by
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the these set of blades that are shown here
are the rotor blades let us say, and set of
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stator blades following them. And we will
denote these stations just upstream of the
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rotor by 1 station 1 and just downstream of
the rotor is 2 which is also the inlet for
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the stator 3 denotes the exit of the stator.
So, all these velocity triangles which we
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will construct is just at the leading edge
of the rotor. So, which is denoted by symbol
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or 1 or station 1 trailing edge of the rotor
is you know denoted by station 2 which is
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also the leading edge of the stator, and exit
of the stator is by a 3 station number 3.
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Now, this is the rotor, and stator configuration
that is a stage that we have, and so this
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rotor has a rotational direction as indicated
here, and so this is how the rotor would of
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a compressor would rotate.
And because this is imparting energy to the
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fluid that is a rotor blades do work on the
fluid and therefore, if the work has to be
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done on the fluid this obviously has to be
the direction in which the rotor blades rotate.
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In the case of turbines which we will see
latter on it would be the other way round
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because in turbines work is extracted from
the fluid and therefore, the direction of
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rotational velocity is opposite.
So, now we will begin with the inlet of the
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rotor and as I mentioned the angel between
the relative velocity and the axial direction
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is beta and since this is at the leading edge
of the rotor beta is have has a subscript
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of one. So, if beta one corresponds to the
angle between the relative velocity and the
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axial direction.
And so the relative velocity in this case
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is obviously V 1 because one corresponds to
the leading edge of the rotor. Now, why is
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it that we have indicated a direction of V
1 in this fashion. Now, you might notice that
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the blades both the rotor as well as stator
blades have already been set at a particular
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angle, and so the angle which is shown here
corresponds to the angle at which the relative
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velocity of the velocity of this particular
arrangement enters the blade in a tangential
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direction that is if you were to draw the
camber at of this particular rotor, and draw
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a tangent at the leading edge of the camber
line that would have an angle of beta one
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with the axial direction. Of course, this
is assuming that there is no incidents and
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things like that, which we will discuss in
later lectures.
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Now, for simplicity we are assume we are assumed
that the blade angle is represented by beta
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1. So, relative velocity is the angle which
is basically the angle of at the leading edge
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of the if you draw if you want to draw a tangent
at the leading edge on the camber of the blade.
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Now, we already know the direction of the
blade speed which is already shown here by
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this blue arrow, and so how are these three
components the relative velocity the absolute
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velocity, and the blade speed related. So,
vectorially they are related like this C which
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is the absolute velocity is equal to the vector
sum of U which is the blade speed, and the
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relative velocity V.
So, with that in mind let us construct the
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velocity triangle for the leading edge or
the inlet of the rotor so at the inlet of
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the rotor we now have all the 3 components
of velocity. We have the relative velocity
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which is V 1 and if we add vectorially add
V 1 and U we get the absolute velocity which
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is indicated here by the red arrow which is
C one.
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So, this represents the velocity triangle
at the inlet of the rotor or at the leading
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edge of the rotor. So, similarly we can now
construct the velocity triangles at the trailing
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edge as well. Now, before that let me also
mention that C 1 has an angle of alpha 1 with
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reference to the axis. So alpha 1 would be
this angle that is the angle which C 1 makes
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with the axial direction whereas, beta 1 is
angle which the relative velocity makes with
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the axis.
Now, at the trailing edge the same argument
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is valid that V 2 which is the relative velocity
exciting the rotor that is at the trailing
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edge it exceeds the trailing edge at an angle
beta 2 here again as I mentioned for simplicity
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we assume that the blade angle at the trailing
edge is what is shown here as beta 2 there
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is no deflection incidental deflection we
will discuss possible in the next class or
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little later.
So, V 2 is the relative velocity exciting
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the blade at the trailing edge and U remains
the same that is blade speed at a given cross
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section remains unchanged, and if that is
so we can now construct the velocity triangle
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at the trailing edge the same way, we constructed
it for the leading edge.
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And So, alpha 2 here corresponds to the angle
which is the angle which the absolute velocity
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makes with the axial direction. So, this is
how the velocity triangle at the trailing
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edge now looks; so, trailing edge velocity
triangle where in V 2 is added vectorially
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U 2 which U 2 and the U 1 are of course, the
same gives us the absolute velocity, which
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is C 2 C 2 making and angle of alpha 2 with
the axial direction.
23:09.419 --> 23:16.720
Now, unlike a rotor we which which had a blade
speed stator blades are stationary so there
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is no blade speed associated with stator blades,
which means that stator blades simply deflect
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the incoming flow at a desired angle depending
upon how the blades are set.
23:27.770 --> 23:36.279
So, C 2 is the angle or the velocity which
enters the stator, and it leaves the stator
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another velocity which obviously would be
lower than C 2 because stator blades defuse
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the flow, and therefore, C 3 will be less
than C 2 and it leaves the stator blades at
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an angle alpha 3.
And what we will see little later on is that
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by design normally we would like to keep alpha
3 equal to alpha 1, which means that another
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set of rotor blades can now follow this stator
blades, and this can be repeated over several
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stages, and that is how a multi stage axial
compressor is developed.
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So, let me just quickly go through what we
had done in this particular construction of
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velocity triangle. Now, we have rotor blades
set at a certain angle we know the direction
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of rotation of the rotor blades, and so depending
upon the blade angle at the leading edge,
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which in this case is beta 1. We can determine
the direction of the relative velocity, and
24:36.019 --> 24:40.759
blade speed is already known and therefore,
the absolute velocity can be determined, because
24:40.759 --> 24:45.440
these three components are related to 1 another
vectorially
24:45.440 --> 24:50.960
The same thing is followed for the trailing
edge, where we know the trailing edge the
24:50.960 --> 24:55.590
angle blade angle at the trailing edge, and
assuming that the blade leaves the trailing
24:55.590 --> 25:01.840
edge in the same direction we get the direction
of the relative velocity blade speed is known
25:01.840 --> 25:06.570
and therefore, the absolute velocity that
is C 2. In the case of stator, there is no
25:06.570 --> 25:12.899
more blade speed and therefore, there is no
relative velocity in a stator, and so C 2
25:12.899 --> 25:18.440
is the velocity which enters the stator blade
at its leading edge, and C 3 is the velocity
25:18.440 --> 25:23.070
living the blade stator blades at an angle
of alpha 3.
25:23.070 --> 25:31.919
So, in this particular slide we were trying
to discuss about how given a certain configuration
25:31.919 --> 25:40.230
of rotor, and blade stator combination one
can go about trying to make or construct the
25:40.230 --> 25:46.080
velocity triangle, because that is perhaps
one of the starting points of the design;
25:46.080 --> 25:53.120
where one would like to analyze a given configuration
of blade rotor, and stator blades, and so
25:53.120 --> 26:00.090
velocity triangles help us in trying to design
or begin the design of an axial compressor
26:00.090 --> 26:04.500
stage.
So, we will take a closer look at the velocity
26:04.500 --> 26:08.250
triangles you already constructed velocity
triangles at the leading edge and trailing
26:08.250 --> 26:15.039
edge. We can now super impose these two velocity
triangles, because one of the components of
26:15.039 --> 26:19.789
velocity is common for both that is the blade
speed U for leading edge and trailing edge
26:19.789 --> 26:25.429
both of them are same, which means that with
common U. We can now, supper impose these
26:25.429 --> 26:33.330
two velocity triangles and then try to infer
a few more aspects of axial compressor working
26:33.330 --> 26:39.799
from the supper imposed velocity triangle.
So, if you take a look at the velocity triangle
26:39.799 --> 26:44.820
the combined velocity triangle that has been
obtained by supper imposing the inlet velocity
26:44.820 --> 26:49.710
triangle with that of the outlet keeping the
blade speed that is peripheral velocity the
26:49.710 --> 26:56.519
same. So, what we see here is that in the
case of the velocity triangle of the leading
26:56.519 --> 27:03.370
edge, and as we moved from the leading edge
to the trailing edge. We see an increase in
27:03.370 --> 27:04.879
the absolute velocity.
27:04.879 --> 27:10.360
And so, there is an increase in absolute velocity
as you might have noticed at the inlet this
27:10.360 --> 27:16.100
is the velocity triangle consisting of C 1,
which is absolute velocity at leading edge
27:16.100 --> 27:22.789
at the inlet of the rotor b 1 is the relative
velocity, and U is the blade speed. Alpha
27:22.789 --> 27:30.549
1 corresponds to the angle of, which the absolute
velocity makes with the axis beta 1 is the
27:30.549 --> 27:34.039
relative velocity angle between relative velocity
and the axis.
27:34.039 --> 27:40.730
Similarly, at the exit we have C 2 which is
the absolute velocity at the exit V 2 which
27:40.730 --> 27:46.289
is the relative velocity at the exit, and
correspondingly there angles which are alpha
27:46.289 --> 27:56.559
2 and beta 2. Now, I have indicated a few
more of these components or if you were to
27:56.559 --> 28:04.879
take component of these velocities in different
directions. So, the axial velocity component
28:04.879 --> 28:11.379
is as I mentioned denoted by C a, which is
the velocity component of the absolute velocity
28:11.379 --> 28:20.159
in the axial direction that is C a.
And we also have these tangential components
28:20.159 --> 28:26.970
for example, the tangential component of a
V 1 which is the relative velocity at the
28:26.970 --> 28:34.100
inlet is denoted by V w 1 which is basically
this component of velocity which is this,
28:34.100 --> 28:42.779
and V w 2 corresponds to the tangential component
of the relative velocity. So, relative velocity
28:42.779 --> 28:49.090
component in the tangential direction which
is U is denoted by V w 2 and this is the direction
28:49.090 --> 28:56.250
of these two velocity components. Similarly,
we have the corresponding components of the
28:56.250 --> 29:03.799
absolute velocity C w 1 is the tangential
component of the absolute velocity at inlet
29:03.799 --> 29:12.489
C w 2 is the tangential component of the absolute
velocity at the exit of the rotor.
29:12.489 --> 29:19.350
And the difference between C w 2 and C w 1
is denoted by delta C w, and that we shall
29:19.350 --> 29:26.489
see little later is delta C w does play a
significant role in the performance analysis
29:26.489 --> 29:33.929
of these compressor. So, delta C w is basically
the difference between C w 2 and C w 1 which
29:33.929 --> 29:40.659
are basically the tangential component of
the absolute velocities, and taken in the
29:40.659 --> 29:48.299
tangential direction.
So, this particular slide gives us an idea
29:48.299 --> 29:54.230
of how these velocity triangles are related
in the sense that, we can immediately see
29:54.230 --> 30:00.100
from these two velocity triangles that between
C 1 and C 2; C 2 obviously is higher than
30:00.100 --> 30:05.789
C 1 that means that there is an increase in
absolute velocity. But, at the same time we
30:05.789 --> 30:11.100
can see that relative velocity has reduced
or decreased so, V 1 is greater than V 2.
30:11.100 --> 30:16.210
So, there is a deceleration in terms of the
relative velocity and that is why we see an
30:16.210 --> 30:21.700
increase in pressure even in the rotor, because
the relative velocity has decreased. So, there
30:21.700 --> 30:26.369
is a diffusion of occurring in the rotor and
obviously there is also a diffusion occurring
30:26.369 --> 30:33.129
in the stator, because a we have seen in the
stator C 3 is less than C 2, and there is
30:33.129 --> 30:37.330
no longer any more relative velocity components
there.
30:37.330 --> 30:43.029
So, this is one of the configurations of velocity
triangles of course, there depending upon
30:43.029 --> 30:47.580
the blade design you might see different types
of velocity triangles some of them we will
30:47.580 --> 30:53.370
discuss a little later. On how the velocity
triangles can look like if certain design
30:53.370 --> 30:58.840
parameters are changed, so that we will discuss
a little later.
30:58.840 --> 31:05.029
So, having understood some basic aspects of
velocity triangle, and how we can construct
31:05.029 --> 31:11.940
velocity triangle we have now understood that
in the rotor basically there is an increase
31:11.940 --> 31:16.629
in the absolute velocity. But at the same
time there is a deceleration in terms of their
31:16.629 --> 31:24.169
relative velocity which is by we get an increase
in the pressure even at the rotor and the
31:24.169 --> 31:27.679
diffusion process is split between the rotor
and the stator.
31:27.679 --> 31:34.040
So, what we will see next is to just get some
idea of how the properties vary as we move
31:34.040 --> 31:40.139
from rotor to stator that is in a stage of
an axial compressor how different properties
31:40.139 --> 31:47.389
is like enthalpy or pressure varies, as we
move from inlet of the rotor towards the exit
31:47.389 --> 31:48.389
of the stator.
31:48.389 --> 31:54.010
So, in a stage what basically happens is that
the properties obviously undergo a certain
31:54.010 --> 32:00.269
change. For example, if you were to look at
the enthalpy b the stagnation enthalpy or
32:00.269 --> 32:06.960
total enthalpy what is indicated here just
an illustration that is shown here. We have
32:06.960 --> 32:13.179
the rotor and the stator both of them put
together constitute a stage of an axial compressor.
32:13.179 --> 32:22.320
So, if you were to express the variation of
total enthalpy across a stage what we see
32:22.320 --> 32:28.359
is that in the rotor there is an energy addition
occurring that is energy is added on the fluid
32:28.359 --> 32:33.710
rotor does work on the fluid, and so there
obviously is an increase in the total enthalpy
32:33.710 --> 32:39.299
or stagnational enthalpy. So, x 0 2 is greater
than x 0 1.
32:39.299 --> 32:44.999
On the other hand, in the case of stator there
is no more energy addition taking place, and
32:44.999 --> 32:51.759
therefore, if you have to assume that the
flow passing through the stator is adiabatic
32:51.759 --> 32:56.750
in which case this stagnation enthalpy does
not change. So, stagnation enthalpy remains
32:56.750 --> 33:02.220
fixed between stations 2 and 3 which is the
inlet of stator 2 inlet of the exit of the
33:02.220 --> 33:08.230
stator.
On the other hand, we look at the absolute
33:08.230 --> 33:14.620
velocity in the rotor there is an increase
in absolute velocity. So, C 1 is less than
33:14.620 --> 33:19.559
C 2 that is there is an increase in absolute
velocity from the inlet of the rotor to the
33:19.559 --> 33:24.429
exit of the rotor and in the case of stator
there is a deceleration occurring that is
33:24.429 --> 33:31.429
what we had observed earlier as well that
C 3 is less than C 2.
33:31.429 --> 33:38.820
And static pressure on the other hand, static
pressure as I mentioned the diffusion is split
33:38.820 --> 33:43.010
between the rotor, and stator diffusion begins
in the rotor and continues in the stator.
33:43.010 --> 33:47.909
So, there is a continuous increase in static
pressure all the way from inlet of the rotor
33:47.909 --> 33:56.519
till the exit of the stator.
Similarly, we can also map out other properties
33:56.519 --> 34:02.809
like the relative velocity relative velocity
know we know is yeah decreases in the rotor
34:02.809 --> 34:08.320
and there is no relative velocity as such
in the stator. Total pressure also can be
34:08.320 --> 34:14.430
expressed in a similar way like total enthalpy,
because if you were to assume that there are
34:14.430 --> 34:19.290
no frictional losses in the stator total pressure
in the stator does not change. So, total pressure
34:19.290 --> 34:24.100
increase basically occurs in the rotor, because
that is where you are adding energy.
34:24.100 --> 34:31.090
So, total in the same way as total enthalpy
total pressure basically raises in the rotor
34:31.090 --> 34:39.290
and remains constant at least in the ideal
case in the stator. So, this is just to give
34:39.290 --> 34:45.900
an illustration or of how these properties
vary and in a stage of an axial compressor,
34:45.900 --> 34:54.630
and so what we have discussed is to one of
the basic ways of beginning design of axial
34:54.630 --> 34:58.040
compressors basically starting from the velocity
triangles.
34:58.040 --> 35:03.410
So, what we will do next is that given this
understanding of the velocity triangle and
35:03.410 --> 35:08.590
there construction we will try to now derive
an expression or relate the pressure rise
35:08.590 --> 35:15.690
in an axial compressor to some of the other
parameters, and see on what parameters does
35:15.690 --> 35:19.690
pressure rise across a stage of an axial compressor
would depend upon.
35:19.690 --> 35:27.480
So, basically what we are going to do is to
assume that the axial velocity will remain
35:27.480 --> 35:34.680
the same. So, we will assume that the axial
velocity C a 1, and C a 2 is basically equal
35:34.680 --> 35:44.370
to C a that is axial velocity remains unchanged.
So, if that is so from the velocity triangles
35:44.370 --> 35:50.360
we will let me go back to the velocity triangle
so from the both these velocity triangles
35:50.360 --> 35:56.590
let’s take up the inlet velocity triangle
which has C 1 and B 1 and U.
35:56.590 --> 36:05.900
So, if we were to take this ratio U, U divided
by C a that is basically equal to the tan
36:05.900 --> 36:14.010
of alpha 1 plus tan of beta 1. So, tan beta
1 is basically this component of U this part
36:14.010 --> 36:21.390
of U divided by C a and tan of alpha 1 is
the remaining component of U divided by C
36:21.390 --> 36:27.350
a, which means U by C a should be equal to
tan alpha 1 plus tan beta 1.
36:27.350 --> 36:33.380
Similarly, for the second velocity triangle
that is the exit velocity triangle we have
36:33.380 --> 36:41.680
tan of alpha 2 plus tan of beta 2 is equal
to U by C a. So, U by C a, we have expressed
36:41.680 --> 36:48.270
in terms of the angles tan alpha 1 plus tan
beta 1 similarly, it is also equal to tan
36:48.270 --> 36:55.800
alpha 2 plus tan beta 2.
Now, if you look at the change in angular
36:55.800 --> 37:03.020
momentum of the air that is passing through
the rotor. We know that as the fluid passes
37:03.020 --> 37:09.730
through the rotor there is a change in the
velocity components, and you have also already
37:09.730 --> 37:17.480
identified that the component of the absolute
velocity in the tangential direction are denoted
37:17.480 --> 37:23.820
by C w 1 and C w 2.
So, as the rotor does work on the fluid its
37:23.820 --> 37:28.150
basically the tangential component of the
velocity that plays a significant role here,
37:28.150 --> 37:34.170
because it is the tangential motion of the
rotor which basically does work, and so we
37:34.170 --> 37:42.440
can basically express the work done per unit
mass flow as a product of the blade velocity,
37:42.440 --> 37:49.620
and the difference between the tangential
component of these velocities between the
37:49.620 --> 37:56.120
inlet of the rotor, and exit of the rotor;
that means work done per unit mass would be
37:56.120 --> 38:02.620
basically the product of the blade speed the
peripheral velocity and the difference in
38:02.620 --> 38:07.600
the tangential component of these absolute
velocities.
38:07.600 --> 38:15.920
So, U multiplied by C w 2 minus C w 1 basically
gives us tells us how much work is done on
38:15.920 --> 38:22.811
the fluid per unit mass. So, work done per
unit mass of the air passing through the compressor
38:22.811 --> 38:30.120
can basically be expressed in terms of product
of U times C w 2 minus C w 1, which basically
38:30.120 --> 38:37.610
U times delta C w, and here C w 1 and C w
2 are the tangential component of velocities
38:37.610 --> 38:42.030
of the fluid before and after the rotor respectively.
38:42.030 --> 38:50.930
So, if you were to look at the first equation
we wrote it was U by C a is equal to tan alpha
38:50.930 --> 38:58.220
1 plus tan beta 1 and U by C a is also equal
to tan alpha 2 plus tan beta 2. We combine
38:58.220 --> 39:04.650
that with this worked on equation, then we
get work done per unit mass is equal to U
39:04.650 --> 39:09.900
times C a into tan alpha 2 minus tan alpha
1.
39:09.900 --> 39:16.400
Now, tan alpha 2 minus tan alpha 1 is also
equal to tan beta 1 minus tan beta 2, which
39:16.400 --> 39:21.670
you can let’s go back to the velocity triangle
again. So, the difference between tan alpha
39:21.670 --> 39:28.350
2 and tan alpha 1 is also equal to the difference
between tan beta 1 and tan beta 2.
39:28.350 --> 39:35.360
So, work done per unit mass can also be expressed
as U time C a into tan beta 1 minus tan beta
39:35.360 --> 39:40.690
2. So, you could use either of these expressions
to determine how much work is done per unit
39:40.690 --> 39:45.840
mass which means that one of the advantages
of having the velocity triangle is that you
39:45.840 --> 39:51.480
can calculate how much work is done per unit
mass if we know the angles and obviously the
39:51.480 --> 39:56.050
peripheral loss T would be known because the
speed at which the rotor rotates is kind of
39:56.050 --> 40:02.300
a design parameter. So, you would know a clearly
how much is the speed of rotation of the rotor.
40:02.300 --> 40:08.640
And knowing that and the axial velocity one
can express these components, that is the
40:08.640 --> 40:13.580
blade speed, and axial velocity and the blade
angles in terms of the work done per unit
40:13.580 --> 40:19.580
mass. And what it means is that work done
per unit mass is basically a function of the
40:19.580 --> 40:28.340
blade speed, and delta C w. And so how will
it affect the performance so, basically this
40:28.340 --> 40:35.660
work done per unit mass will manifest itself
in the form of an increase in stagnation temperature
40:35.660 --> 40:41.030
because you are adding energy to the flow.
We have seen that stagnation enthalpy changes
40:41.030 --> 40:46.280
that h 0 2 is greater than h 0 1, because
we are adding energy to the flow therefore,
40:46.280 --> 40:52.730
whatever work is being done on the flow that
is U times delta C w is basically equal to
40:52.730 --> 40:59.520
or will manifest itself as increase in stagnation
enthalpy, which is basically increase in stagnation
40:59.520 --> 41:01.240
temperature.
41:01.240 --> 41:07.640
That means U times delta C w should be equal
to h 0 2 minus h 0 1. So, this is per unit
41:07.640 --> 41:14.240
mass so work done have so here as we have
calculated U times delta C w should be equal
41:14.240 --> 41:22.610
to h 0 2 minus h 0 1. So, if that is the case
then we can simplify this further that is
41:22.610 --> 41:30.660
multiplying this by C P and T in delta temperature
so T 0 2 minus T 0 1 is equal to U times delta
41:30.660 --> 41:38.310
C w by C P, which again can be simplified
as delta T not divided by T 0 1 is U times
41:38.310 --> 41:45.120
delta C w by C P T 0 one.
Now, if we assume that the flow is adiabatic
41:45.120 --> 41:50.320
and since there is no work done in the fluid
as it passes through the stator it means that
41:50.320 --> 41:59.010
T 0 3 is equal to T 0 2, and so let us now
define a stage efficiency we have already
41:59.010 --> 42:04.140
seen what is meant by isentropic efficiency
that is defined for the whole compressor at
42:04.140 --> 42:08.350
that time I have also mentioned that in addition
to isentropic efficiency, and polytrophic
42:08.350 --> 42:14.780
efficiency we can also define stage efficiency;
which is basically the isentropic efficiency
42:14.780 --> 42:22.750
applied for one stage. So, stage efficiency
is h 0 3 s minus h 0 1 divided by h 0 3 minus
42:22.750 --> 42:31.360
h 0 1 which can be simplified as T 0 3 s divided
by T 0 1 is equal to 1 plus stage efficiency
42:31.360 --> 42:35.360
times delta T not by T 0 1.
42:35.360 --> 42:42.610
So, in the above equation we can we can basically
substitute for T 0 1 delta T not as T 0 3
42:42.610 --> 42:49.820
minus T 0 1, because there is no change in
stagnation temperature in the stator. So,
42:49.820 --> 42:56.320
in terms of pressure ratio we now have P 0
3 divided by P 0 1, which is because we have
42:56.320 --> 43:03.570
this temperature ratio which is isentropic
T 0 3 s by T 0 1 can be equated to pressure
43:03.570 --> 43:09.521
ratios rise to the gamma minus 1 by gamma.
So, P 0 3 by P 0 1 that is pressure ratio
43:09.521 --> 43:16.710
per stage will be equal to 1 plus stagnation
stage efficiency into delta T not by T 0 1
43:16.710 --> 43:24.600
raise to gamma by gamma minus 1. We have already
calculated what is delta T not by T 0 1 this
43:24.600 --> 43:32.100
is basically equal to U times delta C w divided
by C P T 0 1. Therefore, pressure ratio per
43:32.100 --> 43:39.500
stage is equal to 1 plus stage efficiency
into U times delta C w divided by C P T 0
43:39.500 --> 43:46.760
1 rise to gamma by gamma minus 1.
So, here we have an expression which is basically
43:46.760 --> 43:52.520
the pressure ratio the stagnation pressure
ratio across a stage being related to a few
43:52.520 --> 43:57.960
other parameters like the temperature raise
or in terms of the velocity triangle components
43:57.960 --> 44:05.120
like the blade speed delta C w, and of course,
the stage efficiency. So, once we know once
44:05.120 --> 44:11.620
we can determine the velocity triangle for
particular stage, and of course, if we know
44:11.620 --> 44:16.590
the stage efficiency we can actually calculate
how much pressure ratio this particular stage
44:16.590 --> 44:20.270
would be able to develop.
So, pressure ratio per stage as you can see
44:20.270 --> 44:28.210
depends upon a few parameters one of them
is the blade speed that means higher the blade
44:28.210 --> 44:34.880
speed better is the velocity pressure ratio
per stage. And of course, there is other parameter
44:34.880 --> 44:42.760
that is delta C w, and delta C w is basically
related to the blade angles beta 1, and beta
44:42.760 --> 44:50.260
2 that is deflection of the fluid also determines
or increases the pressure ratio per stage.
44:50.260 --> 44:54.900
So increasing the blade speed increasing the
deflection these are different ways of improving
44:54.900 --> 45:00.140
the pressure ratio per stage but, that is
not always a possible as we will see little
45:00.140 --> 45:01.140
later.
45:01.140 --> 45:07.070
So, what we can see that the it basically
the equation relates the temperature raise
45:07.070 --> 45:13.200
per stage to the pressure ratio, and so to
obtain a high temperature ratio for a given
45:13.200 --> 45:19.870
pressure ratio that is to basically minimize
the number of stages. We could either increase
45:19.870 --> 45:27.990
the blade speed or axial velocity or fluid
deflection but, increasing the blade speed
45:27.990 --> 45:32.640
has certain limitations that is if we if we
were to rotate the blades at very high speeds
45:32.640 --> 45:36.250
it can obviously lead to very large amounts
of stress.
45:36.250 --> 45:41.870
So, blade stresses will limit the blade speed
high axial velocity, and fluid deflection
45:41.870 --> 45:46.820
also are limited because of the increase in
adverse pressure gradients which may lead
45:46.820 --> 45:52.750
to these stalling of these blades. So, there
are certain limitations because of which we
45:52.750 --> 45:58.410
cannot increase these parameters, and therefore,
as we had discussed in the initial part of
45:58.410 --> 46:06.190
the lecture that per stage pressure ratio
is very much limited. So, we have now identified
46:06.190 --> 46:11.660
what are the parameters or based on which
the pressure ratio can be increased or improved
46:11.660 --> 46:17.560
and we have also seen what limits these parameters.
Let us now take a look at some of the important
46:17.560 --> 46:23.670
design parameters which are often used in
design of axial compressor, and then we will
46:23.670 --> 46:28.110
we will take a closer look at some of these
parameters like degree of reaction and the
46:28.110 --> 46:29.280
diffusion factor.
46:29.280 --> 46:36.740
So, there are I have listed here four important
design parameters one of them is the flow
46:36.740 --> 46:43.890
coefficient; flow coefficient is usually denoted
by five, and this is a equal to the ratio
46:43.890 --> 46:49.980
of the axial velocity to the blade speed.
So, C a by U is basically the flow coefficient
46:49.980 --> 46:57.130
besides that often one would use the stage
loading coefficient, which is denoted by psi,
46:57.130 --> 47:03.340
and that is basically equal to the enthalpy
rise across the stage divided by U square.
47:03.340 --> 47:09.251
So, delta h not by U square which is in turn
equal to delta C w divided by U. So, this
47:09.251 --> 47:16.030
gives us an idea of how much work is done
per stage.
47:16.030 --> 47:22.990
The other parameter which is often used in
fact always used in the design of a these
47:22.990 --> 47:30.290
compressors is the degree of reaction, which
is denoted by R subscript x, and also the
47:30.290 --> 47:32.740
diffusion factor which is denoted by d star.
47:32.740 --> 47:38.400
So, we will take a closer look at what is
meant by degree of reaction; and the diffusion
47:38.400 --> 47:47.260
factor. Now, degree of reaction is basically
an indication of the split of work or pressure
47:47.260 --> 47:53.520
raise between the rotor and the stator as
I mentioned that pressure rise is shared by
47:53.520 --> 47:59.380
both the rotor, and the stator. So, how much
of this total pressure rise that is done by
47:59.380 --> 48:05.370
the stage is being done by the rotor is what
degree of reaction indicates it indicates
48:05.370 --> 48:10.170
how much of a pressure rise of course, in
the rotor as compared to the whole stage.
48:10.170 --> 48:16.970
So, degree of reaction gives us an indication
of what is the ratio of the pressure rise
48:16.970 --> 48:22.450
in the rotor to the pressure rise across the
stage. So, it basically provides a measure
48:22.450 --> 48:27.791
of the extent to which the rotor assists or
contributes towards the overall pressure ratio
48:27.791 --> 48:30.330
in this stage.
48:30.330 --> 48:37.320
So, diffusion factor is a defined as the static
enthalpy rise in the rotor to the stagnation
48:37.320 --> 48:45.230
enthalpy rise in the stage, which is basically
h 2 minus h 1 divided by h 0 3 minus h 0 1;
48:45.230 --> 48:51.550
which we can also approximate as h 2 minus
h 1 divided by h 0 2 minus h 0 1, because
48:51.550 --> 48:59.500
h 0 2 is equal to h 0 3. So for so we are
going to simplify this for a relatively in
48:59.500 --> 49:05.261
compressible for nearly incompressible flow,
we can approximate h 2 minus h 1 from the
49:05.261 --> 49:11.720
energy equation as 1 by row into P 2 minus
P 1, and for the stage h 0 3 minus h 0 1 as
49:11.720 --> 49:19.040
1 by row into P 0 3 minus P 0 1, if that was
the case then the diffusion degree of reaction
49:19.040 --> 49:25.670
would be h 2 minus h 1 divided by h 0 2 minus
h 0 1 which is approximated as P 2 minus P
49:25.670 --> 49:29.730
1 by P 0 2 minus P 0 1.
49:29.730 --> 49:35.780
Now, from the energy equation if we were to
look at the apply the energy equation in the
49:35.780 --> 49:43.200
relative frame of reference, then we have
in the rotor h 1 plus V 1 square by 2 is equal
49:43.200 --> 49:48.420
to h 2 plus V 2 square by 2, which is that
V 1 and V 2 are the relative velocities at
49:48.420 --> 49:55.790
the inlet, and exit of the rotor. So, we can
express the degree of reaction as V 1 square
49:55.790 --> 50:03.480
by 2 V 1 square minus V 2 square divided by
2 U into C w 2 minus C w 1, because delta
50:03.480 --> 50:08.470
h not is a expressed in terms of U times delta
C w.
50:08.470 --> 50:14.180
So, for a particular stage if you were to
assume axial velocity to be constant then
50:14.180 --> 50:19.800
we can express these velocity components as
V 1 square minus V 2 square is equal to V
50:19.800 --> 50:28.790
w 1 square minus V w 2 square and V w 1 minus
V w 2 is C w 1 minus C w 2. So, this we can
50:28.790 --> 50:33.370
derive as we look at the velocity triangle
and if you assume constant axial velocity.
50:33.370 --> 50:40.060
So, if we were to assume these, and we substitute
this in the degree of reaction on simplification
50:40.060 --> 50:47.270
we get degree of reaction as 1 by 2 minus
C a by 2 U into tan alpha 1 minus tan beta
50:47.270 --> 50:53.650
2 or from the velocity triangle degree of
reaction is also equal to C a by 2 U into
50:53.650 --> 50:58.490
tan beta 1 plus tan beta 2.
And of course, there are many other expressions
50:58.490 --> 51:06.600
for degree of reaction based on simplification,
and the understanding of the velocity triangle;
51:06.600 --> 51:11.880
so degree of reaction as we can see here can
be expressed in terms of the axial velocity
51:11.880 --> 51:18.820
the blade speed, and the angles as we get
it from the velocity triangles. So, it basically
51:18.820 --> 51:26.300
tells us what is the amount of work sharing
between the rotor as compared to the whole
51:26.300 --> 51:27.300
stage.
51:27.300 --> 51:33.170
So, let us take a look at a few special cases
of the degree of reaction degree of reaction
51:33.170 --> 51:40.030
if it were to be 0 if we if we put R as 0
here, then we get beta 2 is equal to minus
51:40.030 --> 51:45.480
beta 1 this means that there is no pressure
rise occurring in the rotor the entire pressure
51:45.480 --> 51:49.510
raise is due to the stator of the rotor basically
deflects the flow incoming flow.
51:49.510 --> 51:53.780
So, this is something like an impulse bleeding
which we will discuss a little later when
51:53.780 --> 52:01.180
we take up the turbines; degree of reaction
is equal to 0.5 gives us alpha 1 equal to
52:01.180 --> 52:05.920
beta 2, and alpha 2 equal to beta 1 which
means the velocity triangles are symmetric
52:05.920 --> 52:11.240
or mirror images, and there would be equal
pressure rise taken place in the rotor and
52:11.240 --> 52:16.950
the stator. And degree of reaction is equal
to 1 means that alpha 2 is equal to minus
52:16.950 --> 52:22.160
alpha 1, and the entire pressure rise takes
place in the rotor where as the stator does
52:22.160 --> 52:24.650
not contribute to the pressure rise.
52:24.650 --> 52:29.870
So, the velocity triangles would look something
like this for all these three cases if degree
52:29.870 --> 52:37.010
of reaction was 0 then we get beta 2 is minus
beta 1. So, this is beta 2 here this is exactly
52:37.010 --> 52:42.940
opposite equal and exactly opposite to that
of beta 1; this is how it would be if we have
52:42.940 --> 52:49.050
an impulse balding for an axial compressor
degree of reaction equal to 0.5 means alpha
52:49.050 --> 52:56.330
1 is equal to beta 2, and alpha 2 is equal
to beta 1, so that is the velocity triangles
52:56.330 --> 53:01.120
are mirror images of one and other.
So, this the inlet velocity triangle if you
53:01.120 --> 53:05.830
take a mirror image of that we get the exit
velocity triangle. So, this is how it would
53:05.830 --> 53:13.470
be if degree of reaction was 0.5 for degree
of reaction of one, we get alpha 2 is equal
53:13.470 --> 53:19.370
to minus alpha 1. So, here we have alpha 2
which is equal, and opposite to that of alpha
53:19.370 --> 53:24.970
1.
So, here there is the pressure rise is occurring
53:24.970 --> 53:32.370
only because of the rotor. Now, let us now
look at the last parameter that we will discussed
53:32.370 --> 53:37.050
today that is called the diffusion factor,
and we have already seen that pressure ratio
53:37.050 --> 53:43.380
per stage pressure ratio is significantly
dependent on the fluid deflection that is
53:43.380 --> 53:49.950
difference beta 1 minus beta 2 that is a strong
parameter that influences the pressure ratio.
53:49.950 --> 53:54.730
But, we also seen that there are limitations
as you keep increasing the deflation angle
53:54.730 --> 54:01.150
that is deflection beta 1 minus beta 2, then
there is an increasing adverse pressure gradient,
54:01.150 --> 54:07.350
and there is a threat of flow separation or
blade stalling that can influence the performance
54:07.350 --> 54:10.950
of the compressor.
So, we will now define a parameter which will
54:10.950 --> 54:17.270
tell us how close you is how close is the
rotor towards what are the chances that the
54:17.270 --> 54:20.370
flow will separate on the section surface
of the blade.
54:20.370 --> 54:27.870
So, fluid deflection is a basically what decides
the pressure ratio one of the parameters which
54:27.870 --> 54:32.870
decides the pressure ratio but, there are
certain limitations to that. So, diffusion
54:32.870 --> 54:37.390
factor is a parameter that associates the
blade stall with deceleration on the section
54:37.390 --> 54:44.770
surface of the airflow. So, it’s basically
defined as the difference between the maximum
54:44.770 --> 54:50.320
velocity, and the minimum velocity that is
at the trailing edge to the inlet velocity
54:50.320 --> 54:56.930
at the leading edge. So V max minus V 2 by
V 1 basically defines or tells us, what is
54:56.930 --> 55:02.310
the chance that there could be a blade stall
occurring on the section surface?
55:02.310 --> 55:08.750
So, let us understand how this V max minus
V 2 by V 1 come from so here we have the velocity
55:08.750 --> 55:14.520
distribution on the blade surface. So, this
is the pressure surface shown here and this
55:14.520 --> 55:19.540
is the section surface of the blade the velocity
on the section surface. So, we know that at
55:19.540 --> 55:24.390
from the leading edge to the max pressure
minimum pressure point the flow accelerates
55:24.390 --> 55:30.630
velocity reaches its speed, and that is where
we get V max V 2 on the other hand is a velocity
55:30.630 --> 55:35.270
at the trailing edge.
And so that is the minimum velocity that basically
55:35.270 --> 55:42.080
is that velocity at the trailing edge, and
we have this pressure surface velocity distribution.
55:42.080 --> 55:49.130
So, the higher we keep V max by V 2 which
is also true as we keep increasing beta 1
55:49.130 --> 55:54.430
minus beta 2 this difference also increases
beyond a certain level we are trying to decelerate
55:54.430 --> 55:59.620
the flow too much and the obviously the flow
will not be able to sustain itself, and with
55:59.620 --> 56:05.550
stand the pressure gradient, and it may separate
so diffusion factor relates this V max to
56:05.550 --> 56:13.140
V 2, and so that is why we get the diffusion
factor V max minus V 2 by 1.
56:13.140 --> 56:22.580
And so in the earlier days when this was sort
of identified of course, it was in go back
56:22.580 --> 56:28.190
in 1950’s Lieblein basically propose an
empirical parameter, because V max minus V
56:28.190 --> 56:33.580
to by V 1 is something that is in those days
it was difficult to calculate and measure
56:33.580 --> 56:39.270
from experiments, and they were not have computational
tools to measure to estimate these numbers.
56:39.270 --> 56:46.180
So, he came with an empirical parameter for
diffusion factor basically tried to express
56:46.180 --> 56:51.840
diffusion factor in terms of parameters, which
can be measured or it can be known and as
56:51.840 --> 56:57.190
we can see as we will see it depends strongly
upon the solidity which is basically the cord
56:57.190 --> 57:03.760
to the spacing ratio, and over the years it
has been proven to be a sort of a dependable
57:03.760 --> 57:10.300
parameter indicator of the approach to separation
for a variety of blade shades.
57:10.300 --> 57:16.860
So, diffusion factor according to Lieblein
is defined as 1 minus V 2 by V 1 plus V w
57:16.860 --> 57:24.630
1 minus V w 2 divided by 2 C by s into V 1.
So, here we can see that all these parameters
57:24.630 --> 57:32.100
can be known based on velocity triangles the
velocity components can be known and therefore,
57:32.100 --> 57:38.520
it is possible for us to calculate the diffusion
factor given velocity triangles, and if the
57:38.520 --> 57:45.240
cord and is spacing are known.
So, this is much more handier expression to
57:45.240 --> 57:52.500
calculate the diffusion factor, and also estimate
how much it is for a given design, and so
57:52.500 --> 57:58.830
what has been observed is that diffusion factor
around 0.5 is considered to be a safe number,
57:58.830 --> 58:06.060
and as we exceed diffusion factor of 0.5 there
is a I mean a threat of a the chance of blade
58:06.060 --> 58:11.970
stall that might occur. But, of course, modern
day design or tools like computational tools
58:11.970 --> 58:17.350
etcetera have helped us in trying to exceed
these limits to the extent possible.
58:17.350 --> 58:24.860
So, diffusion factors is we parameter that
we have seen is one of the other design parameters,
58:24.860 --> 58:29.460
which will tell us how close our design is
or what are the chances that the blade might
58:29.460 --> 58:33.630
stall given a certain design conditions.
58:33.630 --> 58:41.560
So, let us take a relook at what we had discussed
in today’s class; today’s class was basically
58:41.560 --> 58:48.350
an introductory class on lecture on axial
flow compressor basic operation of axial compressors.
58:48.350 --> 58:53.500
We have discussed about velocity triangles,
and how we can construct a velocity triangle
58:53.500 --> 59:00.200
for a given configuration. And then we have
also derived expression for work done on the
59:00.200 --> 59:07.070
fluid by the compressor and pressure ratio
as related to the geometric parameters, and
59:07.070 --> 59:11.110
like the plate speed, and delta C w and so
on.
59:11.110 --> 59:16.550
And so towards the end of the lecture we have
also discussed about a few design parameters
59:16.550 --> 59:22.440
like the flow coefficient the loading coefficient
the degree of reaction, and the diffusion
59:22.440 --> 59:26.870
factor which are basically parameters which
are used during the initial stages of design
59:26.870 --> 59:28.680
of axial compressors.
59:28.680 --> 59:35.070
So, we will continue discussion of our study
of axial compressors in the next lecture as
59:35.070 --> 59:41.610
well, where we will basically we will be talking
about what is meant be a cascade, and cascade
59:41.610 --> 59:47.630
analysis which will involve a nomenclature
of different parameters of cascade. We will
59:47.630 --> 59:53.860
also be talking about certain loss parameters
which are used in performance analysis, and
59:53.860 --> 59:59.370
we will take up some of these topics for discussion
during the next lecture; subsequently we will
59:59.370 --> 1:00:04.220
also take up discussion of performance of
single stage, multi stage characteristics
1:00:04.220 --> 1:00:08.960
in later lectures. So, let us discuss these
topics during the next lecture.