WEBVTT
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Hello and welcome to lecture number sixteen
of this lecture series on jet aircraft propulsion.
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We have been talking about axial compressors
over the last few lectures, and we are going
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to continue our discussion on axial flow compressors
today. And this would be the last lecture
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on the theory of axial flow compressors, and
subsequently in the next lecture we are going
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to discuss about we will basically have a
tutorial on axial compressors.
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Now, what in today’s lecture we are going
to cover two very important aspects of axial
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flow compressors. One of them is to do with
a very fundamental form of design of the axial
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flow compressor blades, and the that is what
we will initiate our discussion on and subsequently,
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we are going to discuss about the characteristics
performance characteristics of single and
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multi stage axial flow compressors. And the
performance characteristics is a very important
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part of axial flow compressor design, and
also it plays a very significant role in the
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performance of the engine as a whole, because
the compressor is to be matched with the turbine,
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and matching of the compressor and turbine
is in some way related to the performance
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characteristics.
And that also reflects the performance of
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the engine as a whole in the sense that performance
characteristics would tell us that what is
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the band of operation for this particular
compressor where in the operation is safe.
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And what happens, if we exceed these bands
of operation obviously, the performance is
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going to degrade. And in some cases as we
shall see today.
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The compressor might enter into might enter
into certain unstable modes of operation which
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can effect, which can drastically effect the
performance, and working of an engine as a
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whole. So, in this context it is very important
for us to understand the significance of performance
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characteristics. So, but before that let us
discuss about a very fundamental, form of
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or method of design of the axial compressors
in what is known as a free vortex deign, and
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we will see what is the principle behind a
free vortex design, and why we need to consider
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such a design methodology.
Subsequently, we will discuss about single
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and multi stage compressors, and we will also
towards the end of the lecture discuss in
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brief, about two instability modes of operation
of an axial compressor. These are known as
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rotating stall and surge, so that we will
take up towards the end of the lecture. So,
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we are going to discuss begin the lecture
today with discussion on what is known as
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free vortex design. And in subsequently we
will take up the performance characteristics.
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So, let us discuss about what is meant by
free vortex design, but before that let me
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give you a background about what is meant
by a radial equilibrium basically radial equilibrium
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is one of the conditions that needs to be
satisfied in the sense that all the flow parameters
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in terms of velocity pressure etcetera, need
in the radial direction also needs to be considered.
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In whatever design methodology or the velocity
triangle etcetera, which we have discussed
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over the last few lectures, we have not really
considered variations of properties in the
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radial direction.
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We have assumed that in a particular cross
section the blade speed is a constant, the
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tangential velocity is constant and many other
properties remain a constant. But in an actual
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blade you know that the blade speed is going
to vary, all the way from the hub to the tip.
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And so how do you consider or account for
this in a design methodology. So, radial variations
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of all these properties needs to be considered
and factored into when we take up a serious
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design of an axial compressor blade. And that
is one of the principles that will be made
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use of in, what we will be discussing about
in free vortex design. So, for a realistic
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design, what we need to do is to consider
radial variations of the following one is
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of course the blade speed, which we know will
vary any way, from hub to tip because it is
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a direct function of the radius. The other
parameter that we will need to be considered
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is the axial velocity. Then the radial variation
in tangential velocity and static pressure.
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So, if we have to maintain a reasonably uniform
flow at the compressor exit. And, why do we
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need to maintain that we need to maintain
a uniform flow at the compressor exit. Because
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the next stage or hum that is going to follow
a particular stage, will depend upon, what
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is coming into the stage from the previous
one. And so, if there is a uniform flow that
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exits one stage of an axial compressor. That
is also beneficial for these succeeding stages.
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So, one of the ways of ensuring that we can
have a fairly uniform flow exiting the compressor,
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is to ensure that there is a uniform distribution
of specific work input at each of the cross
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sections that is starting from the hub all
the way to the tip. If we can maintain, or
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at least try to maintain, relatively uniform
distribution of specific work in the radial
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direction, then it possible that we should
be able to get a fairly uniform flow at the
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compressor outlet, which is obviously good
thing for the succeeding stages.
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Now, in order to ensure that we have uniform
specific radial work distribution. Now, we
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know that the enthalpy difference across a
stage delta or even a compressor delta h not,
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is equal to enthalpy thrice across a compressor
stage. That is equal to the product of the
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blade speed that is U multiplied by delta
C W right that is something that we had discussed
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in one of the earlier lectures that the
Enthalpy rise, specific enthalpy rise will
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be equal to the product of blade speed times
the difference in the tangential velocity
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between the inlet and exit. So, delta C W
is basically the difference in the tangential
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velocity. So, if we equate this now we are
trying to maintain relatively uniform specific
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work input. So, we should be able to get some
hints from this as to how we can do that.
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So, we know that delta h not which is specific
enthalpy specific enthalpy change across a
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stage is equal to blade speed times delta
C W or C theta. And, which is equal to omega
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r, which basically the blade speed is the
product of the angular velocity times the
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radius this multiplied by delta C W is basically
equal to the specific work input.
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So, what does means is that for a given rotational
speed, that is if we fix omega then r times
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delta C W must be a constant. So, the product
of the radius, and the tangential velocity
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must be a constant. Now, this can ensure that
we can have for a given rotational speed;
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we can ensure that the specific work is a
constant. So one configuration or one design
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methodology, which can ensure that this is
satisfied is known as the free vortex design.
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And in a free vortex design, basically we
keep the product r times C W it is kept a
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constant that they exit for each of these
blade roles. And so, given the axial velocity
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and the blade speed at different cross section
from the hub to tip we have axial velocity
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and so on. And so, from hub to tip we know
all these parameters; and if we to if we were
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to ensure that the product r times C W is
kept a constant. We can actually solve the
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velocity triangle; all the way from hub to
tip.
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So, in vortex design methodology that is one
of the basic principles trying to keep the
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product r times C W A constant, which will
ensure that the specific work input, required
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for this particular blade would be a constant.
So, if that is the case we can ensure that
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r times C W is kept a constant.
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Now, for example let us, take a look at one
example where in we can see what has happens,
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if we were to maintain the velocity triangles
if we were to solve the velocity triangles,
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keeping r times C W a constant. If we were
to do that; so let me repeat what I was saying
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so, basically if we have r times delta C W,
which we are trying to maintain a constant
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for different cross sections all the way from
hub to the tip. Then from axial velocity,
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which is known at the inlet and blade speed
we can solve the velocity triangle.
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So, let us take one example of, what would
happen if we were to take to do this, what
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is shown here are three cross sections at
the hub the mean and the tip. And their corresponding
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velocity triangles and if we were to use the
free vortex method for designing such a blade.
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This is how the velocity triangles are likely
to look like.
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This is just one example now let us, take
a look at the hub section, where we have the
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velocity triangle at the inlet. And the exit
of the blade; so this particular velocity
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triangle, which is shown here. C 1 is absolute
velocity entering the blade at the leading
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edge at the hub v 1 is the corresponding relative
velocity. U is the blade speed; and at the
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exit we have C 2, which is the velocity absolute
velocity exiting the blade and V 2, which
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is the relative velocity exiting the blade.
And if you try to maintain r times C W constant;
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we can see that the blade cross section well
the orientation of the blade itself changes
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drastically as we move from the hub towards
the tip, which means that and we can see velocity
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triangles at the mean; and also at the tip.
So, what you notice is that if you were to
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is were to stag these different cross sections
from the hub all the way to the tip. You can
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clearly see that the blade is going to be
twisted it is no longer going to be a straight
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blade as we have been discussing so far, where
we can at least the velocity triangles we
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have discussed in the last few lectures we
were seeing that the blades were little bit
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straight. Now, here with the free vortex design
methodology we can see that the blade is no
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longer going to remain straight.
It is going to have a fairly significant twist.
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The other significant thing that you can try
and notice is that during our discussion on
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degree of reaction. We have seen that when
degree of reaction is close to 0.5 or is equal
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to 0.5. The velocity triangles, become symmetric.
So, that is something that you can probably
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try to see here at the mean cross section
of the blade. We can see that the velocity
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triangles are more or less symmetric. And
this is indication that the degree of reaction
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here is likely to be close to 0.5, whereas
at the hub and the tip. It is quite different.
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And, what comes out is that in most of the
design a processed that one would encounter.
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The degree of reaction approaches zero close
to the hub; it becomes very low. And towards
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the tip it becomes much higher than 0.5 it
might approach one, and even though we are
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maintaining a degree of reaction of around
0.5 at the main section.
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So, this is just an example of one case of
how we can use free vortex design methodology
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to design a blade cross section, and there
of course, the other variants of this methodology,
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which we shall discuss now. And by using certain
modifications to the free vortex design methodology
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we can of course, arrive at other methods
of designing a blade.
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Now, what is important in all the these methods
is that we need to satisfy the conditions
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of radial equilibrium. That is the equations
of motion the three-dimensional equations
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of motion needs to be satisfied.
That is one fundamental requirement in all
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these design methodologies. So based on this
besides free vortex design, where in we have
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r times C W is equal to constant there are
other methods like forced vortex, which is
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r times C W is equal to a r square exponential
method which could be r times C W is equal
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to a r plus b. And constant reaction, which
is r C W is a r square plus b, where a and
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b are of course constants, and needs to be
fixed (( )).
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So, these are other variants of the methods,
where in we can try to ensure a constant specific
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radial work. And that is required, because
we would like to maintain a fairly uniform
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set of properties in the exit of a particular
rotor blade. So, free vortex design basically
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is one of the attempts or one of the methods
by, which we can try to ensure a constant
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specific radial work criteria.
So, what we have discussed in brief for now
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is one of the ways of one of the methods or
popular methods of design of an axial compressor
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blade. And what we shall discuss next would
be as I discussed had initially. We should
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be talking about the different performance
characteristics of single and multi stage
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axial compressors.
So we will begin our discussion, with consideration
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of a single stage axial flow compressor. We
will first discuss about performance characteristics
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of single stage compressor, followed by this
we will be talking about the multi stage characteristics.
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Now stage of an axial compressor as you probably
know by now compressors comprises of a rotor
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and a stator. So let us, take up a characteristic
rotor and the stator and then we will discuss
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about, how we can characterize this particular
rotor and see what happens as you change different
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properties.
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Now, we have a typical axial compressor stage
here, which consists of a set rotor blades,
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followed by a set of stator blades. We have
already seen this velocity triangle (( )) on.
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At the rotor inlet we have absolute velocity
entering at C 1 relative velocity at V 1 blade
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speed is zero. And at the exit of the rotor
we have, absolute velocity is equal to C 2
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and relative velocity is V 2. And blade speed
of course, remains the same its U. And this
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velocity triangle is, what is basically goes
into the stator basically the absolute velocity
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that enters the stator. And exits at velocity
of C 3 living at angle alpha.
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So, here we can see as the as the flow enters
the rotor and exits the rotor. There is a
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deceleration in terms of the relative velocity
and that is, what leads to the diffusion,
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and similarly, as the flow enters the stator.
The absolute velocity decelerates from C 3
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from C 2 to C 3 leading to diffusion.
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So, the combined velocity triangle is something,
which we have discussed earlier as well. If
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we were to overlap the two velocity triangles
at the inlet and exit of the rotor then we
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get a combined velocity triangle with all
the angles indicated here as well. Let me,
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just quickly go through these velocity triangles
we have, absolute velocity at an angle alpha
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1 at the inlet, of the rotor and relative
velocity V 1 at an angle beta one at the inlet,
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at the exit we have the absolute velocity
leaving the rotor at an angle alpha 2. And
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relative velocity leaving the rotor at angle
beta two.
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And C W 1 corresponds to the tangential component
of the absolute velocity C W 2 corresponds
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to the tangential component of the absolute
velocity at the exit of the rotor. And similarly,
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V W 1 and V W 2 are the corresponding relative
velocity components. Delta C W is the difference
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between C W 2 and C W 1; and C a is that axial
component of the absolute velocity.
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So, with this in mind from the velocity triangles
we can infer that C W 2 is equal to U minus
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C a tan beta two. So, let see where that comes
from C W 2 is this component this should be
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equal to U, which is the blade speed minus
the this particular component, which is what
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is given by C a tan beta 2. So C a is this
component; C a times tan beta 2 is this component.
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So, U minus C a tan beta 2 is C W 2 similarly,
C W 1 is C a tan alpha, C a C W 1 is this
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so, C a tan alpha 1 is this component. Now
we also know that delta h not that is the
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specific enthalpy rise across a stage should
be equal to U times delta C W. And delta C
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W is C W 2 minus C W 1.
So, if you go to substitute for C W 2 and
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C W 1 here, we get delta h not is equal to
U multiplied by U minus C a times tan alpha
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1 plus tan beta 2 or delta C W by U is equal
to delta h not divided by U square, which
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is 1 minus C a divided by U into tan alpha
1 plus tan beta 2. So, what we have here is
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the specific work ratio or the stage loading,
which is expressed in terms of two important
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parameters. One is the axial velocity; and
the blade speed and of course the angles the
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blade the inlet angle alpha 1 and the blade
outlet angle. So, the specific work ratio
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or the blade loading has expressed in terms
of two distinct parameters.
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So, what this means is that if we go to change
the design mass flow rate; mass flow rate
19:20.850 --> 19:28.320
is directly proportional to the axial velocity
C a. And so, if we change design mass flow
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rate or the blade speed either of them is
going to effect the loading characteristics.
19:35.149 --> 19:42.659
Because delta C W by U is equal to 1 minus
C a by U into tan alpha 1 plus tan beta 2.
19:42.659 --> 19:48.260
So the stage loading is directly a function
of either the mass flow or the blade speed,
19:48.260 --> 19:53.230
besides of course, the angle so if we were
to assume that the inlet angle angles are
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fixed.
This means that the performance of the stage
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will directly depend upon the mass flow, which
is in turn equal to the axial velocity or
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the blade speed U. That is this ratio C a
by U plays a significant role in the stage
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performance characteristic. So the blade performance
or stage performance is a direct function
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of the ratio C a by U. And so, we have already
defined c a by U if you recall earlier on
20:23.879 --> 20:32.501
we have defined C a by U as the blade loading
in terms of C a by U as we have defined in
20:32.501 --> 20:33.501
the last lecture.
20:33.501 --> 20:40.210
So, let us look at what happens as you change
C a by U. That is basically the flow co efficient
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thus defined last slide. So, what we have
plotted here are two parameters one is the
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stage loading. In terms of what we just now
derived delta h not by U square, which is
20:51.259 --> 21:00.750
delta C W by U. And these stage efficiency;
so, delta C W by U is equal to delta h not
21:00.750 --> 21:06.029
by U square that is 1 minus C a by U into
tan alpha 1 plus tan beta 2. So, if you were
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to plot this for different values of the flow
co efficient C a by U. As we keep changing
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the flow co efficient what happens to this
loading characteristics.
21:17.779 --> 21:25.600
So, what we will see is the one we indicated
by this thick blue line refers to the measured
21:25.600 --> 21:31.110
loading characteristics as C a by U is changed,
which could be either by changing the mass
21:31.110 --> 21:41.279
flow or by the speed. So, we observe a characteristic
as shown here, which means that as we reduce
21:41.279 --> 21:48.530
the mass flow the pressure rise or the stage
loading increases up to a certain point. And
21:48.530 --> 21:54.230
then subsequently if we continue to reduce
the mass flow it will decrease. And this has
21:54.230 --> 21:58.549
certain implications, which we will discuss
during the multi stage performance characteristic
21:58.549 --> 22:05.970
as well. And so we see that at this point
that is indicated by C a by U which is corresponding
22:05.970 --> 22:11.760
to the design condition. We have this particular
characteristic and if we were to look at the
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corresponding stage efficiency. We get the
maximum efficiency around that point. That
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is corresponding to C a by U design, which
is of course less than hundred percent
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of there is a certain efficiency associated
with the stage. And if we were to draw a tangent
22:30.779 --> 22:36.779
at this point were C a by U is given by this
point if you if you draw a tangent at that
22:36.779 --> 22:40.519
point.
The slope of that is basically given by tan
22:40.519 --> 22:45.830
alpha one plus tan beta two corresponding
to the design condition. So, this the slope
22:45.830 --> 22:51.820
of this line basically is, what is given here
by tan alpha 1 plus tan beta 2 corresponding
22:51.820 --> 22:59.769
to the design condition. So, what is shown
here is basically, how a particular stage
22:59.769 --> 23:06.549
of an axial compressor is going to behave
as we change the mass flow or the speed.
23:06.549 --> 23:11.740
So, in terms of either of them we have represented
them by the flow co efficient that is C a
23:11.740 --> 23:18.539
by U. So as c a by U changes how is it that
the stage performance changes in terms of
23:18.539 --> 23:24.480
the loading of the stage as well as the efficiency
of the stage. So what we see here is that
23:24.480 --> 23:33.929
basically the measured performance and, how
it performs in terms of the efficiency corresponding
23:33.929 --> 23:41.160
to C a by U design. So, what this means is
that the as one deviates from C a by U design.
23:41.160 --> 23:46.080
The stage is not going to perform as it is
supposed to be for the design conditions.
23:46.080 --> 23:52.460
So, what we will see next is for off design
conditions, where in C a by U is either greater
23:52.460 --> 23:58.679
than C a by U design or C a by U is less than
C a by U design. How the performance of the
23:58.679 --> 24:04.240
stage would get effected. So, let us take
a look at what happens in terms of the velocity
24:04.240 --> 24:09.429
triangles as well as what happens on the blade
as we change the flow co efficient from its
24:09.429 --> 24:10.429
design value.
24:10.429 --> 24:16.749
So, as C a by U let us take a look at three
different cases here; the first case is, when
24:16.749 --> 24:22.960
C a by U is actually equal to the design condition.
Then we have the stage performing normally
24:22.960 --> 24:29.549
that is normal operation under design condition
of this particular stage. So the velocity
24:29.549 --> 24:36.730
triangle with the relative velocity at entering
the stage at an angle beta one and absolute
24:36.730 --> 24:43.789
velocity at C 1 C 1 entering at alpha one.
Now, let us consider the first off design
24:43.789 --> 24:51.820
condition, that is if C a by U is less than
C a by U design. And if you assume that this
24:51.820 --> 24:58.139
is only by reducing the axial velocity or
changing the mass flow alone keeping the blade
24:58.139 --> 25:06.750
speed same. That means if we fix U and just
change the mass flow it means that for a lower
25:06.750 --> 25:15.220
mass flow, how the stage is going to perform.
So, as C a is less than C a by C a design.
25:15.220 --> 25:20.159
This means the axial velocity is now less
than the design axial velocity. As we reduce
25:20.159 --> 25:24.889
the axial velocity what, happens is we see
that the velocity triangle getting altered.
25:24.889 --> 25:30.860
So, this was the original velocity triangle;
and as we reduce the axial velocity the velocity
25:30.860 --> 25:35.679
triangle has got altered. Because now, C a
by U is now less than C a design, which means
25:35.679 --> 25:42.029
that the relative velocity will now enter
the rotor at angle, which now greater than
25:42.029 --> 25:49.110
what it was at the design condition. So, here
beta 1 is actually greater than the beta 1
25:49.110 --> 25:54.789
at the design.
So, therefore, it means that the flow approaches
25:54.789 --> 26:02.110
the stage at a very high angle of incidence,
which is basically a positive incidence here.
26:02.110 --> 26:06.870
There is a chance that the flow would separate
from the suction surface of this rotor.
26:06.870 --> 26:12.259
So, there is flow separation occurring here
on the suction surface of this blade. Therefore,
26:12.259 --> 26:19.620
this is basically a positive incidence flow
separation. The other extreme of this is C
26:19.620 --> 26:26.379
a by U is greater than C a by U design, which
means C a is greater than C a design. If U
26:26.379 --> 26:32.100
were to be fixed the blade (( )) is fixed
then there is a possibility that the incidence
26:32.100 --> 26:38.220
is now negative. And there could be flow separation
from the pressure surface and that basically
26:38.220 --> 26:43.299
means a negative incidence flow separation
flow might separate from the pressure surface
26:43.299 --> 26:48.650
of the blade.
So, these are two extreme cases of operation
26:48.650 --> 26:56.039
of the stage when the flow co efficient is
different from the design flow co efficient.
26:56.039 --> 27:01.679
It is either less than design co efficient
or greater than if it is less than if the
27:01.679 --> 27:08.039
blade fixed then C a is less than the design.
Axial velocity and the flow separates from
27:08.039 --> 27:11.669
the suction surface might separate from the
suction surface leading to positive incidence
27:11.669 --> 27:18.240
separation. And if C a is greater than C a
of design then it could lead to negative incidence
27:18.240 --> 27:21.980
separation.
So, having understood some of the aspects
27:21.980 --> 27:27.210
of a single stage performance characteristics
of an axial compressor. Let’s now move on
27:27.210 --> 27:34.230
to multi stage axial compressor. That is if
there multiple number of these stages one
27:34.230 --> 27:38.879
after another how does the performance, how
can we evaluate the performance of such an
27:38.879 --> 27:45.190
axial compressor. Now, to do that we need
to understand, what are the different parameters
27:45.190 --> 27:49.970
or variants or variables based on which we
can evaluate this performance.
27:49.970 --> 27:54.490
So, at the outlet of the compressor we are
interested in two things one is the pressure
27:54.490 --> 28:00.640
ratio; stagnation pressure ratio across the
compressor. And the other parameter is the
28:00.640 --> 28:05.260
efficiency of this compressor. So, these are
two parameters, which we will be interested
28:05.260 --> 28:11.900
in. And so, we will need to see what are the
other variables on which these two parameters
28:11.900 --> 28:19.070
depend that is efficiency and pressure ratio
of the compressor expressed in terms of other
28:19.070 --> 28:25.730
variables. So, that we can then go ahead and
then look at the performance characteristics.
28:25.730 --> 28:31.870
So, for a multi stage compressor now we are
going to denote the inlet of the multi stage
28:31.870 --> 28:37.720
compressor by station one. And exit by of
the compressor by two that means the overall
28:37.720 --> 28:43.630
pressure ratio of the compressor is P 0 2
by P 0 1. So, the compressor outlet pressure
28:43.630 --> 28:50.759
and efficiency isentropic efficiency is a
function of several properties several variables.
28:50.759 --> 28:57.669
So it could be mass flow rate, the inlet stagnation
pressure, inlet temperature, the rotational
28:57.669 --> 29:05.499
speed, the ratio of specific heats gas, constant
for the working fluid, viscosity of the working
29:05.499 --> 29:11.820
fluid, the design of the blades themselves
and the diameter d. So, these are the different
29:11.820 --> 29:17.070
parameters on, which the outlet pressure outlet
stagnation pressure and outlet efficiency
29:17.070 --> 29:18.659
depend upon.
29:18.659 --> 29:24.600
So, if you were to express this in terms of
non dimensional parameters, so if we carry
29:24.600 --> 29:30.779
out dimensional analysis of these terms that
we have listed here. Then non dimensional
29:30.779 --> 29:36.059
parameters come out to be on the left hand
side we have the stagnation pressure ratio
29:36.059 --> 29:40.710
of the compressor, P 0 2 by P 0 1 and the
efficiency.
29:40.710 --> 29:46.210
This is a function of mass flow rate multiplied
by square root of gamma R T 0 1 divided by
29:46.210 --> 29:53.769
P 0 1 D square. The other parameter is omega
D by square root of gamma R T 0 1; then we
29:53.769 --> 30:00.389
have the omega D square by viscosity mu. Then,
the ratio of specific heats and the design.
30:00.389 --> 30:05.919
Now, for a particular design suppose we have
frozen a particular design and we can assume
30:05.919 --> 30:11.070
that gamma, which is of this particular fluid
and the viscosity do not really affect the
30:11.070 --> 30:17.179
performance significantly or do not change
much and for a given diameter of the engine.
30:17.179 --> 30:21.559
Because, the diameter has been frozen the
diameter has also been frozen; and the gas
30:21.559 --> 30:28.659
constant fixed. So this particular set of
non-dimensional parameters will now reduce
30:28.659 --> 30:37.740
to P 0 2 by P 0 1. And the efficiency as a
function of m dot root t 0 1 by P 0 1; and
30:37.740 --> 30:43.309
n times T 0 square root of T 0 1, where n
is the rotational speed.
30:43.309 --> 30:48.220
So, the pressure ratio and the efficiency
of the compressor depends up on basically
30:48.220 --> 30:53.990
two parameters here. One is corresponding
to mass flow rate and the other corresponding
30:53.990 --> 31:00.409
to the rotational speed. And so, let us further
reduce it in terms of standard conditions,
31:00.409 --> 31:05.580
which we will see that is known as corrected
mass flow and the corrected speed. So, pressure
31:05.580 --> 31:11.470
ratio and efficiency as a function of mass
flow rate times the root of inlet stagnation
31:11.470 --> 31:17.690
temperature divided by P 0 1, and the speed
that is n times square root of T 0 1.
31:17.690 --> 31:24.919
So, we will further express them in terms
of standard conditions. That is standard ambient
31:24.919 --> 31:32.220
pressure and temperature; so, that this performance
characteristics can be used in any other ambient
31:32.220 --> 31:37.029
conditions. So, that once it is standardized
and corrected it can be used under other conditions
31:37.029 --> 31:42.090
as well. So, if you process this further in
terms of standard day pressure and temperature.
31:42.090 --> 31:50.730
When we have P 0 2 by P 0 1, and the efficiency
are functions of m dot square root of theta
31:50.730 --> 31:55.350
divided by delta and N divided by square root
of theta.
31:55.350 --> 32:03.059
Here theta is equal to T 0 1 divided by T
0 1 standard day. And delta is equal to P
32:03.059 --> 32:11.909
0 1 divided by P 0 1 standard day, where T
0 1 standard day is 288.15 kelvin. And P 0
32:11.909 --> 32:16.669
1 standard day is 101.32 kilo pascal.
32:16.669 --> 32:23.450
So, if you go to express these performance
parameters in terms of the characteristic
32:23.450 --> 32:30.450
of a compressor, multi stage compressor. So,
what I have shown here is one typical performance
32:30.450 --> 32:37.470
characteristics of a particular axial compressor.
We have on the Y axis the pressure ratio P
32:37.470 --> 32:45.259
0 2 by P 0 1 and expressed in terms of the
X axis that is mass flow rate m dot square
32:45.259 --> 32:51.940
root of theta divided by delta. We also have
the efficiency; as a function of the mass
32:51.940 --> 32:58.409
flow rate m dot square root of theta by delta.
So let us, look at the pressure ratio characteristics
32:58.409 --> 33:04.830
first. So, typical compressor multi stage
axial compressor characteristic would comprise
33:04.830 --> 33:11.029
of pressure ratio versus mass flow at different
speeds. So, these lines that are shown here
33:11.029 --> 33:17.559
are for different speed ratios that is N by
square root of theta.
33:17.559 --> 33:23.879
That is for different speeds of rotation how
the characteristic change. Let us, take up
33:23.879 --> 33:30.210
one particular speed let us say .7. So, what
does this line mean is that as we change the
33:30.210 --> 33:35.950
mass flow rate? As the mass flow reduces the
pressure ratio increases. And it reaches a
33:35.950 --> 33:41.129
particular peak beyond, which the pressure
ratio, which is not show here. It will droop
33:41.129 --> 33:47.820
or it will fall drastically. The reason by
I not showing the points on the left hand
33:47.820 --> 33:53.789
side of the line, which is indicated as the
surge line is because after this point the
33:53.789 --> 33:58.451
compressor operation is unstable. Because
of what is known as surge, which I will discuss
33:58.451 --> 34:03.770
shortly what is meant by surge.
So, there is an instability in the compressor
34:03.770 --> 34:11.240
performance, which prohibits the compressor
operation on the left hand side of this line.
34:11.240 --> 34:16.480
So, compressor operation is possible; stable
compressor operation is possible; only on
34:16.480 --> 34:20.669
towards the right hand side of the compressor
of this particular line. So the surge line
34:20.669 --> 34:26.609
dictates one of the limits of the compressor
operation. And so, what performance characteristic
34:26.609 --> 34:31.819
here shows how the performance in terms of
pressure ratio varies as we change the mass
34:31.819 --> 34:35.220
flow.
We also have the efficiency characteristics
34:35.220 --> 34:42.710
shown here how the efficiency changes as we
change the mass flow. So, we have peak efficiency
34:42.710 --> 34:48.290
corresponding to the mass flow, which is shown
here, so far a particular mass flow at 0.85
34:48.290 --> 34:55.059
speeds. We have the peak efficiency of occurring
at a mass flow somewhere around here. And
34:55.059 --> 34:57.430
that is true for other mass flows as well.
34:57.430 --> 35:03.240
So, mass flow versus efficiency, mass flow
versus total pressure ratio both of these
35:03.240 --> 35:10.470
put together. Define the performance of a
multi stage axial flow compressor. Now, let
35:10.470 --> 35:17.950
me take a closer look at the pressure ratio
versus the mass flow characteristic. So, you
35:17.950 --> 35:22.940
might have noticed here that the peak efficiency
is occurring at a point, which is slightly
35:22.940 --> 35:31.150
away from the surge line. And at the same
time I mentioned that operation on the left
35:31.150 --> 35:36.940
hand side of the surge line is not possible,
because of the presence of instabilities.
35:36.940 --> 35:43.190
So, we would like the compressor to be operating
at point at a safe operating point, which
35:43.190 --> 35:49.470
is away from the surge line. So, what is shown
here is in terms of this dotted line corresponds
35:49.470 --> 35:55.941
to the actual operating line of an engine,
which means we are really not operating very
35:55.941 --> 36:00.890
(( )) we would not want to operate very close
to the surge line. Because there is always
36:00.890 --> 36:07.560
a risk that the compressor might go into surge.
If there is a certain fluctuation in the mass
36:07.560 --> 36:11.820
flow rate.
So, the engine the compressor is usually operated
36:11.820 --> 36:19.349
or designed for operation along an operating
line, which is away from the surge line. And
36:19.349 --> 36:23.700
the difference between these two points is
basically referred to as surge margin. So,
36:23.700 --> 36:28.740
there is always a certain margin provided
between the engine operation and the surge
36:28.740 --> 36:35.510
line, which indicates or denotes that there
is a certain margin provided for a safe operation
36:35.510 --> 36:41.970
of the compressor. So, surge lines is one
of the extremes of operations of the compressor.
36:41.970 --> 36:46.900
That the engine actually (( )) slightly away
from the surge line, which is known as the
36:46.900 --> 36:53.960
engine operation line. So, the multi stage
performance characteristic as defined here
36:53.960 --> 37:00.680
is in terms of the efficiency as well as the
pressure ratio. So, in some books and literature
37:00.680 --> 37:07.400
you might also find that the efficiency plots
are shown in the same pressure ratio plot
37:07.400 --> 37:13.549
as well so you will also find contours of
constant efficiency plotted on the same graph.
37:13.549 --> 37:22.420
Now, I mentioned that the axial compressor
performance is limited by surge on one of
37:22.420 --> 37:29.700
the sides. And that there is one of the instability
modes of operation of the compressor. And
37:29.700 --> 37:34.170
so before I discuss surge let me talk about
yet another mode of instability called, which
37:34.170 --> 37:40.070
is known as the rotating stall.
So, axial compressor performance is hindered
37:40.070 --> 37:45.450
by two instability modes there of course other
modes as well. But primarily two modes of
37:45.450 --> 37:51.880
instability one is known as the rotating stall,
and the other is known as surge.
37:51.880 --> 37:59.520
Now rotating stall is a non axis symmetric
mode of instability. And on the other hand
37:59.520 --> 38:04.721
surge is the axis symmetric mode of instability
and it is periodic, where as rotating stalled
38:04.721 --> 38:10.780
is not necessarily periodic it is a periodic
in nature. But both of these modes of operation
38:10.780 --> 38:17.690
eventually can hamper the performance of a
compressor drastically. So rotating stalled
38:17.690 --> 38:23.250
basically involves progression around the
blade unless of a stall pattern. That is the
38:23.250 --> 38:31.119
stall pattern progressively moves around the
annulus of the axial compressor, where in
38:31.119 --> 38:38.460
one or more adjacent blade passages are instantaneously
stalled. And then the stall cell continuously
38:38.460 --> 38:44.400
moves or progresses around the annulus of
the compressor. So, this means that the rotating
38:44.400 --> 38:50.240
stall can lead to alternate loading, and unloading
of the of the blade, which means that there
38:50.240 --> 38:56.880
is a possibility of fatigue failure eventually
if rotating stall mode was to be present continuously.
38:56.880 --> 39:03.011
So, let us see what actually happens during
rotating stall. Now, during rotating stall
39:03.011 --> 39:11.020
let us say there is a certain pocket of non
uniformity in the incoming flow. And this
39:11.020 --> 39:17.710
pocket of non uniformity may cause a certain
one either one blade or a set of adjacent
39:17.710 --> 39:24.109
blades to operate under stall condition. So
here we have a non uniform flow; that means
39:24.109 --> 39:30.900
there is a flow entering the compressor
at a very high incidence than the design condition.
39:30.900 --> 39:38.440
So, high non uniformity entering into this
particular set of blades causes this blade
39:38.440 --> 39:46.339
let us say to get to be stalling, which means
that as this blade stalls. It also causes
39:46.339 --> 39:54.760
the flow entering the second the adjacent
blade to enter at a higher incidence, because
39:54.760 --> 39:58.150
the blades are actually rotating. So, there
is also a relative motion here the blades
39:58.150 --> 40:02.640
rotate.
And, because of this the non uniformity now
40:02.640 --> 40:09.290
progresses, and move towards the adjacent
blade. Causing the adjacent blade, also to
40:09.290 --> 40:15.210
now stall you has this blade stalls it unstalls
this particular, which is earlier under stall.
40:15.210 --> 40:20.280
Because the flow is now deflected in this
fashion here. And also because of the rotation
40:20.280 --> 40:24.700
of the blades themselves.
So, this stall blade, which was undergoing
40:24.700 --> 40:30.480
a stall. And this stall cell as you see here
moves towards the adjacent blades and the
40:30.480 --> 40:35.880
new stall is now on this particular blade.
And eventually it unloads the previous blade,
40:35.880 --> 40:41.329
which was undergoing stall. So, this continuously
happens and this stall cell moves progressively
40:41.329 --> 40:46.910
moves from one end to another around the annulus.
And, what you can immediately see here is
40:46.910 --> 40:52.839
that the direction of the rotation of the
stall cell is opposite to that of the rotation
40:52.839 --> 40:59.270
of the blades themselves. So rotating stalls
cells rotate in a direction, which is opposite
40:59.270 --> 41:04.289
to that of the rotor rotation. And you can
see the rotor rotation obviously for this
41:04.289 --> 41:10.359
compressor is in the direction from right
to left. Whereas, the stall cell moves from
41:10.359 --> 41:15.800
left to right. So similarly on a annulus you
can imagine that the rotating stall cells
41:15.800 --> 41:21.690
will actually move in a direction opposite
to that of the rotor rotation.
41:21.690 --> 41:28.970
So the rotating stall is in an instability,
which often precedes surge. And we will see
41:28.970 --> 41:33.960
that a little later; and the stall patterns
will obviously move that in a direction, which
41:33.960 --> 41:38.740
is opposite to that of rotor revolution. And
the frequency of the rotation of a stall cell
41:38.740 --> 41:43.230
is of course, not fixed it can keep changing,
but it can be as high as about fifty percent
41:43.230 --> 41:48.050
of rotor frequency.
And rotating stall may usually be initiated
41:48.050 --> 41:57.000
by the presence of a non uniform flow (( )) rotor
blades. So, what I have shown here are three
41:57.000 --> 42:02.630
different modes of operation of the stall.
One is that rotating stall may be affecting
42:02.630 --> 42:07.690
only a part of span. This is the annulus of
the compressor; this is the hub, and this
42:07.690 --> 42:15.470
is the casing. So, rotating stall may be affect
part of the span or it may be a full span
42:15.470 --> 42:21.869
stall. That it affects all the way from the
hub to the tip. And eventually it may lead
42:21.869 --> 42:29.720
to surge, where in the entire annulus undergoes
stall. So, this means that the rotating
42:29.720 --> 42:36.140
stall is one of the precursors is likely to
be one of the precursors of the surge. That
42:36.140 --> 42:43.540
is one of the rotating stall were to continue
and progress leading to full annulus stall.
42:43.540 --> 42:45.770
It may eventually lead to surge.
42:45.770 --> 42:52.000
So, which brings us to what is meant by what
is actually meant by surge, we have already
42:52.000 --> 42:57.320
seen surge line is seen in compressor characteristic,
which was on the left hand side. On one of
42:57.320 --> 43:03.520
the limits of compressor operation and I mentioned
that you cannot really operate the compressor
43:03.520 --> 43:10.050
beyond the surge line. Because of the occurrence
of surge itself. So, let us understand what
43:10.050 --> 43:18.200
we mean by surge. Now, surge is surge line
is obtained as by joining all the points on
43:18.200 --> 43:23.349
the speed lines of compressor, where surge
is likely to occur. So, surge line is the
43:23.349 --> 43:28.049
locus of all the points of a multi stage axial
compressor.
43:28.049 --> 43:35.869
Surge basically, involves fluctuation of fluid
back and forth along the annulus. Axial fluctuation
43:35.869 --> 43:43.500
of the fluid of the working fluid. And the
entire annulus of the compressor is affected
43:43.500 --> 43:48.210
by surge, which is why I mentioned that surge
is axis symmetric. That is the entire annulus
43:48.210 --> 43:55.680
of the compressor is affected by a separating
flow. And it is violent oscillation of the
43:55.680 --> 44:05.369
fluid back and forth. So, the onset of surge
can lead to very drastic effects on the compressor,
44:05.369 --> 44:09.310
and as the engine as a whole. Because compressor
is feeding in to the compressor chain bar
44:09.310 --> 44:13.059
(( )) and the turbine and so on, which means
that if there is something drastic happening
44:13.059 --> 44:18.230
compressor. It will immediately affect the
combustion chamber and therefore, the turbine
44:18.230 --> 44:22.130
which means the whole engine gets affected
as a result of surge.
44:22.130 --> 44:27.520
So surge is characterized by violent and periodic
oscillations in the flow. It might lead to
44:27.520 --> 44:31.579
flame blow out in the combustion chamber.
Because if there is a back and forth movement
44:31.579 --> 44:35.060
of fluid in the combustion in the compressor.
44:35.060 --> 44:39.901
It can lead to a flame out a flame blow out
in the combustion chamber. And obviously it
44:39.901 --> 44:45.000
can lead to substantial damage of compressors,
which is why surge is something that always
44:45.000 --> 44:51.890
needs to be avoided. Designers always try
to keep a safe margin between the surge line
44:51.890 --> 44:56.220
and the operating line, which is basically
known as the surge margin.
44:56.220 --> 45:01.750
I mentioned that there is a certain margin
provided between the surge line and the engine
45:01.750 --> 45:07.930
operating line. And this is basically referring
to this surge margin. And so designers would
45:07.930 --> 45:16.370
always like to keep a certain margin there.
So now, let us try to understand what is actually
45:16.370 --> 45:23.930
meant by a surge; and let me take up the performance
characteristics once again we have the mass
45:23.930 --> 45:28.730
flow in terms of the flow co efficient I have
taken at only one stage of the compressor
45:28.730 --> 45:31.869
here.
We have flow co efficient and the pressurized
45:31.869 --> 45:39.741
co efficient, what happens to the stage performance
as we keep changing the mass flow. Now, let
45:39.741 --> 45:45.660
us say we have the stage characteristics,
which have been plotted here and in solid
45:45.660 --> 45:52.549
line. And the throttle characteristics in
terms of these dashed lines, which means that
45:52.549 --> 45:57.700
as we throttle the compressor as we reduce
the mass flow.
45:57.700 --> 46:04.710
The throttle characteristics keeps changing.
And this particular stage let us say operating
46:04.710 --> 46:10.890
at point a. As we throttle the compressor
will continuously move towards the left. And
46:10.890 --> 46:18.670
it keeps moving along these dotted lines.
A to B and to C D and so on, but beyond a
46:18.670 --> 46:26.660
certain point what we see what are indicated
by the points here with dash. That is D dash
46:26.660 --> 46:32.900
C dash and B dash are those points, which
corresponds to unstable operation of the compressor
46:32.900 --> 46:40.589
And why is it exactly unstable, because what
happens here is that as we reduce mass flow
46:40.589 --> 46:45.470
in on the right hand side of the particular
line. Let us say, after on the right hand
46:45.470 --> 46:52.299
side of E. As we reduce the mass flow it is
accompanied by an increase in the pressure
46:52.299 --> 46:58.369
ratio. Whereas, on the left hand side we have
as we reduce the mass flow rate. It also result
46:58.369 --> 47:02.920
in reduction in pressure ratio which means
as the pressure ratio reduces the pressure
47:02.920 --> 47:05.690
downstream is lower than the pressure in the
upstream.
47:05.690 --> 47:13.930
And so, that can lead to sudden fluctuation
in the mass flow rate. And that is, why operation
47:13.930 --> 47:20.760
on the left hand side of what is given here
as the throttle characteristic can lead to
47:20.760 --> 47:26.829
an unstable operation of the compressor. So,
what happens really here is that beyond this
47:26.829 --> 47:32.310
point, which is given here as the tangent
of the stage characteristic any point on the
47:32.310 --> 47:40.690
left hand side of this, where this stage characteristic
will have a positive slope. We mean that reduction
47:40.690 --> 47:46.880
in mass flow will be accompanied by reverse
flow, which can fluctuate back and forth.
47:46.880 --> 47:53.329
Because momentarily would have a lower pressure
of upstream than the downstream, which means
47:53.329 --> 47:59.020
that there would be a reverse flow.
And as the flow moves from the downstream
47:59.020 --> 48:03.700
stages to the upstream stages. There is a
decrease in pressure downstream as compared
48:03.700 --> 48:09.680
to upstream and then this continues in a cycle
leading to rapid fluctuations in the flow
48:09.680 --> 48:15.849
back and forth. So, this is what leads to
what is meant by surge, where in there is
48:15.849 --> 48:22.240
violent oscillation of mass flow back and
forth. And something that can be explained
48:22.240 --> 48:26.320
using the stage characteristic and the throttle
characteristic.
48:26.320 --> 48:31.869
So, as we change the throttle characteristic
from extreme left hand side right hand side
48:31.869 --> 48:37.349
as shown here. As we keep throttling, which
means it indicates some increase in flow resistance.
48:37.349 --> 48:45.150
We obtain different performance characteristic
as denoted here by these points. But all these
48:45.150 --> 48:50.790
points as you can see have a negative slope.
Whereas, as the stage characteristic reaches
48:50.790 --> 48:55.220
a point, where in the slope is positive.
48:55.220 --> 49:00.619
That indicates that these are the points where
basically the operation of the compressor
49:00.619 --> 49:05.980
is unstable. And this is what is reflected
in the multi stage characteristic as well.
49:05.980 --> 49:13.410
So, what I had shown was one particular line
here. And beyond this as you can see the slope
49:13.410 --> 49:19.460
will have will be basically positive. Whereas,
on this side will be negative. So, operation
49:19.460 --> 49:26.410
of the compressor is safe on all these lines
and all these points, where the slope is positive.
49:26.410 --> 49:31.830
And the locus of all these points together
constitute what is known as the surge line.
49:31.830 --> 49:36.500
So, operation of the compressor is unstable
on the left hand side of the surge line. Whereas,
49:36.500 --> 49:44.020
it is safe to operate the compressor on the
right hand side of this surge line. So,
49:44.020 --> 49:51.950
surge is basically the one of the modes of
instability of an axial compressor, which
49:51.950 --> 49:57.470
is something that the designer would always
want to eliminate and avoid.
49:57.470 --> 50:03.630
And rotating stall is one of the possible
one of possibly one of the precursors of surge.
50:03.630 --> 50:08.600
And that is something that would eventually
lead to the occurrence of surge. So, let me
50:08.600 --> 50:16.079
wind up today’s lecture, where in we were
discussing the about two important aspects.
50:16.079 --> 50:20.579
One was the stage characteristics, the single
stage characteristic as well as the multi
50:20.579 --> 50:25.640
stage performance characteristics of the axial
compressor. And the other aspect that we were
50:25.640 --> 50:31.640
discussing was the free vortex theory we started
the lecture today with some preliminary discussion
50:31.640 --> 50:36.579
about the free vortex theory.
And how it can be used for in preliminary
50:36.579 --> 50:44.809
design of the axial compressors, subsequently,
we started discussing about single stage characteristic.
50:44.809 --> 50:52.260
And how we can how we can basically evaluate
the performance of a single stage by relating
50:52.260 --> 50:57.359
the flow co efficient to the pressure rise
or the stage loading.
50:57.359 --> 51:02.350
And subsequently, we discussed about multi
stage performance characteristic in brief.
51:02.350 --> 51:07.579
And then we also discussed about two modes
of instabilities. One is known as the rotating
51:07.579 --> 51:14.799
stall, where in the stall cell progressively
moves around the annulus. We then saw that
51:14.799 --> 51:20.470
rotating stall can eventually or if allowed
to continue may lead to occurrence of surge
51:20.470 --> 51:27.950
as well, where in the entire annulus of the
compressor undergoes flow separation or flow
51:27.950 --> 51:31.329
reversal.
And there could be rapid oscillations or fluctuations
51:31.329 --> 51:39.720
of the flow, which could also lead to extensive
damage to the engine as a whole. So, single
51:39.720 --> 51:44.069
stage multi stage performance characteristic
the significance of that and also the free
51:44.069 --> 51:49.260
vortex theory. These were some of the topics
we had discussed in today’s lecture. And
51:49.260 --> 51:55.180
what we will take up for tomorrow’s well
for the next lecture would be basically a
51:55.180 --> 51:58.500
tutorial.
On axial compressors and we will wind up our
51:58.500 --> 52:03.470
discussion on axial compressor with the next
lecture, which will basically be a tutorial.
52:03.470 --> 52:09.640
We will try to solve some problems pertaining
to axial compressors. And also we will have
52:09.640 --> 52:14.750
a few exercise problems, which we can solve
based on the discussion in the lecture. So,
52:14.750 --> 52:17.660
let us take up these topics for discussion
in the next lecture.