WEBVTT
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We have been talking about various kinds of
flows through axial flow compressor. We have
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done lectures on two-dimensional flow features,
the two-dimensional flow theories and then
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we went on to extend that to three-dimensional
flows, the simple three-dimensional flow theories,
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and then, of course, we did a full three-dimensional
flow theory in a last lecture.
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Today we will try to take a look at using
the simple three-dimensional flow theories
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that we have done, essentially extension of
the two-dimensional flow theory that you have
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done earlier and how that extended flow theory
can be used to solve some simple problems.
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You see some of the simple problems that are
available in the textbooks or those which
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can be created to get the hang of the basics
of the subject, elements of the subjects can
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be solved with the help of, you know, simple
three-dimensional flow theories which are
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extended two-dimensional flow theories. The
more comprehensive 3-D flow theories that
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we have done indeed requires a large computing
capability, and hence strictly speaking cannot
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really be, you know, solve through simple
classroom lectures or simple problems that
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one encounters in textbooks or even the specialized
books on a turbo machinery. We will be covering
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some of those aspects later on in this lectures
series, where full 3-D flow theories are attempted
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to be solved. All the solution methods through
various computational techniques would be
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discussed in some detail.
So, today we will go back to our simple flow
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theories that we had done and those as I mentioned
are the extension of the 2-D flow theories,
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and those simple 3-D flow theories we will
try to invoke in solving some standard problems.
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Now, these problems are essentially designed
kind of problems. In this course, we are talking
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about turbo machineries which are specific
products; you know, we are not talking about
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basic physics or basic mathematics; we are
talking about specific products and that product
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are turbo machinery components, compressors
and turbines. If you need to analyze those
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things, you need to design them first, you
need to create them, only after the created
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you can really analyze them.
So today, the theories that we are going to
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do essentially are used for design and immediate
pose design, performance analysis to cater
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to the design that has been carried out. So,
the simple three 3-D flow theories are essentially
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used for various design purposes. So, let
us take a look at those flow theories very
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quickly. So, first we will do a couple of
problems, solutions of those problems, and
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then later on I will leave you with some two
or three unsolved problems which you can solve
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on your own using the theories that we have
done. Let us take a quick look at those theories.
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As I mentioned, these theories are essentially
used for design and immediate pose design
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analysis. So, the simple theory is essentially
that we have already put together can be now
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listed here. The first one of course is a
free vortex law which was derived from the
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simple radial equilibrium equation and that
leads as C w into r equal to constant. The
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corresponding opposite or theoretically the
opposite of that is the force vortex law which
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in some books is also referred to as a solid
body law, which means the flow essentially
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behave like a solid body and that is given
as C w by r equal to constant.
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The relaxed free vortex law or the generalized
free vortex law - a vortex law was put together
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as C w into r to the power n equal to constant,
where n is a parameter that could vary from
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minus 1 which is the forced vortex law to
1 which is free vortex law and any value in
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between the value of n could also be as we
have seen more than 1. Now, in many textbooks
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and many other literature, a generalized version
of the above laws is written down as for upstream
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flow C w1 equal to a R to the power n minus
b by R, and downstream of the rotor C w2 would
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be a R to the power n plus b by R. Now, R
here is the radius ratio which is small r
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by small r means, small r mean is the actual
radius at the mean radius of the blade, and
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r of course is radius at any location on the
blade from root to tip. Now, capital R is
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a ratio of these two parameters. And a and
b are the two constants to be used for the
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specific case, they would indeed vary from
one compressor rotor to another. So, before
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designing or applying these laws to a particular
compressor, the values of a and b need to
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be found out or prescribed before this comprehensive
version of the simple 3-D flow law can be
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applied.
Now, as we see here, the earlier stated laws
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in 1, 2, 3 have been generalized, and now
they have applied upstream and downstream
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that is the flow before going into the blade
can be subject to vortex law, and when it
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comes out at the rear of the blade, it can
be again subject to vortex law and as we see
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here, the two laws are slightly different
from each other; one is a R to the power n
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minus b by R and another is a R to the power
n plus b by R, and this of course stands to
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reason, because upstream the value of C w1
is normally rather small are or rather low,
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and it acquires a large value of C w2 after
going through the rotor, and hence that value
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is expected to be of a much higher order,
because work has been done work has been supplied
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and the downstream flow actually carries that
extra work, and indeed as we know C w 2 minus
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C w 1 is a measure of the work that has been
put in into the flow through the compressor
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blades.
So, these are the simplified generalized vortex
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laws that are applicable to compressors, which
we are going to use in the problems that we
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are going to encounter in today's lecture.
A fourth law which we have stated also is
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stated as exponential law and it simply says
that n that is written down in the earlier,
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you know simplified versions of the law is
0 and that gives us another law, earlier we
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are talking that n could be minus 1 to plus
1 or even more than 1 somewhere in between
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n is 0, and that value of a 0 actually caters
to another law and that law is referred to
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as exponential law. And we shall see that
this exponential law sometimes in some literature
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is also referred to as a law that creates
or caters to creation of constant reaction
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blading. Now, we have seen this earlier in
our lecture that some of the blades could
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have constant reaction from root to the tip
of the blade of a stage and that constant
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reaction could be some value like 0.5 which
is the more popular value of a constant reaction
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blading. However, if one goes into this theory
a little more in detail or one can try to
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get into the numeric's of a particular problem
solution, one would probably see that exponential
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law does not necessarily always aided to constant
reaction blading.
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So, constant reaction blading is not synonymous
with exponential law. One can have a constant
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reaction blading, that is a separate issue
and one can design one quite carefully to
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create constant reaction, but that does not
necessarily cater to exponential law which
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is derived from the vortex law where all kinds
of assumptions were need and those assumptions
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would not be valid, if you go for a pure constant
reaction blading. So, constant reaction blading
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are not exactly in line with all the assumptions
made in creating the vortex laws. So let us
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keep that in mind that the exponential law
which have stated here at the end does not
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necessarily lead to constant reaction blading.
Having stated the various laws that cater
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to creation of compressor blades and immediate
post design analysis, let us take a look at
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a couple of problems, that would use a some
of these laws and essentially through the
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numerical numbers would show you what happens
if you apply this law or that law, what happens
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to the various parameters through the blades,
how what happens to the blade shape, the blade
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geometry and the flow parameters through the
blade. So, let us do a couple of problems
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to exemplify the laws that we have just stated
here.
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The first problem that we take as an example
is a application of free vortex design that
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is being advocated for design of an axial
compressor rotor, and this has a high hub
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to tip ratio of 0.9, we have talked about
this earlier before and 0.9 is definitely
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high hub to tip ratio blade. And this high
hub to tip ratio is taking to the constant
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through the stage that means the the high
hub to tip ratio is constant from leading
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edge of the rotor till that the trailing edge
of the rotor and indeed till the trailing
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edge of the stator. So, it is a constant radius
hub and tip stage that we are advocating.
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Typically, this would be valid for a last
stage of an axial flow compressor or one before
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last stage, and such a stage indeed have a
high hub to tip ratio of the order of 0.9,
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and they indeed do have constant hub and tip
radius. So, the problems being stated here
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is reasonably a realistic problem.
Now, at the rotor tip the diameter is one
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meter, and the flow angles given are alpha
1 equal to 30 degree, beta 1 equal to which
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is related flow angle 60 degree at the exit
to the rotor, alpha 2 the absolute flow angle
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is 60 degree and the related flow angle beta
2 is 30 degree. Now, we can see here of course
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that the blade is essentially of a symmetric
velocity triangle through which, would immediately
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mean that the tip of the blade has a degree
of reaction 50 percent or 0.5.
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Now, this compressor is been designed for
rotating speed of 6000 rpm and the density
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of the working medium is 1.5 which indicates
that 1.5 kg per meter cube which indicates
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that this is probably somewhere towards the
rear of a compressor stage and definitely
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not the first stage in the rear where the
density of the air has already gone up to
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somewhat higher values. Or it could be indicative
of a working medium which is not air, but
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something heavier than air. Also stated in
the problem are that the enthalpy H is constant
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along the radius from the root to the tip.
And this is typically what we had also assume
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for free vortex theory if you remember. The
entropy change across the rotor from root
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to tip is also held constant. So, the losses
that we have done in our earlier lectures,
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essentially are cater to the entropy change,
and this can also cater to the efficiency
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of the or isentropic efficiency of the compressor
rotor and the compressor stage. And this entropy
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change is now being held constant from root
to tip along the blade length.
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Now, for this rotor the problem requires you
to determine at the design point at which
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all those values are prescribed the performance
parameters, that means you try to find out
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what is actual velocity, which is again prescribed
to be constant from root to tip. Then, the
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mass flow rate which is passing through the
blade and that needs to be found out what
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should be the mass flow rate that the compressor
should process, quite often mass flow rate
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is indeed a primary parameter for which a
compressor is designed. So, mass flow rate
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here is being asked to be found out given
the blade geometry. Then, the ideal minimum
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power to be supplied to this rotor, this means
no efficiency is coming into picture. We are
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assuming which is 100 percent efficiency and
as a result the turbine that is being used
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to supply power or any other prime mover has
to supply a minimum amount of power to do
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this compression job.
So, what is the minimum power or what is the
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power that the compression is doing. So, that
is the minimum power that you need to supply.
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So, that it needs to be also found out in
many actual designs that power may be available
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that power may be available or sometimes you
need to find out how much power you need to
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be supplied with to do a prescribed amount
of compression. And the next thing you need
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to find out of course, other flow angles at
the blade root, flow angles have been prescribed
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here only at the tip. Now, you need to find
out the flow angles at the root then at the
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mean, and then of course you need to find
out the degree of reaction. We have defined
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degree of reaction and if you remember degree
of reaction essentially a two-dimensional
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parameter. So, this problem and as I mentioned,
in today's lecture we are doing problem we
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are which in which we are using extended two-dimensional
flow theories; three-dimensional theories
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- simple three-dimensional theories, in which
many of the two-dimensional concepts are also
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being used and used to solve the problems
at hand. So, degree of reaction is a two-dimensional
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definition, but this problem wants you to
find the value of degree of reaction at the
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blade root. So, let us see whether we can
find a solution to the problem given the problem
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statement as it is.
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Now, given the problem that the rpm is 6000
rpm, first thing is we can find out the angular
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velocity and that is 628.4 radiance per second.
Now, this angular velocity normally is to
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be held constant from root to tip. So, with
with the use of this angular velocity, we
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can now find the actual blade speed - the
solid body blade speed at tip root and mean.
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So, at the tip its omega into r tip and that
would be 314.2 meters per second, at the hub
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it is omega into r hub and there is 282.5
meters per second and at the mean that is
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exactly the mean between hub and the tip,
the U mean is omega into r mean and that would
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be to equal to 298.5 meters per second. So,
these are the three solid body blade speeds
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with which the blades are actually rotating
and with them you are not have to find out
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what kind of work they can actually do.
Now, from the standard velocity diagram that
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we have done in all lectures, indeed you can
have compressors with some unusual kind of
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velocity diagrams, but we are not doing those
things at this moment, though specially the
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transonic ones we are not bothered about those
things in this particular problem. So, we
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will take some standard velocity diagrams
of a rotor inlet. In which U tip would typically
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be C w plus V w, now at the inlet it would
be U tip 1 or U 1 and that would be C w 1
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plus V w 1. Now, this would be then C a in
to tan alpha 1 plus tan beta 1 at the tip.
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Now, we have already prescribed in this problem
that the value of C a is constant from root
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to tip. They already found U tip above which
is 314.2 meters per second, and at the tip
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the value of alpha and beta are already prescribed.
And if you can use those values of 30 and
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60 degrees and put them over here, you can
directly find the value of C a and the direct
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calculation gives you C a equal to 136 meters
per second. So, that is the actual velocity,
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constant from root to tip that the flow through
this compressor blade is being designed for.
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So, that is actual velocity.
Now, the mass flow through the compressor
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rotor or compressor that we have at hand is
indeed directly found from using the continuity
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and that is the annulus area into the actual
velocity which gives the volume flow rate
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and that into density is the mass flow rate.
Now, the annulus area of this particular compressor
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is phi equal to r tip square minus r hub square
and that is the area annulus area, all compressors
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is essentially have annulus area subtended
between the tip and the hub, and that into
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the actual velocity C a is the volume flow
rate through this compressor, and that into
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the density prescribed value is indeed would
be the mass flow rate, and that then turns
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around to be if you put in all the values
are available, the mass flow rate turns out
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to be 30.4 kilograms per second. So, that
is a mass flow rate that has been asked for
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and now we can we see that it can be directly
found, having found the actual velocity in
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the earliest step.
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The next things to be found are the whirl
components of the tangential components of
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the velocities. Now, at the inlet at the tip
C w1 tip can be found out from C a into tan
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alpha 1 at the tip. And that comes out to
be C w1 tip 78.6 meters per second. Now, if
21:47.830 --> 21:54.830
you apply the free vortex law, which is simplified
3-D flow theory. If you apply the free vortex
21:57.899 --> 22:04.899
law, you can find the value of C w1 mean,
and as a result of which you can see that
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the value at C w1 mean would be 82.73 meters
per second, applying C w into r equal to constant.
22:14.340 --> 22:21.340
At the exit to the tip, the C w2 is C a into
tan alpha 2, and this would be 235.6 meters
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per second. Now, this can be used to find
again using free vortex law. The C w 2 mean
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that is at the mean radius and C w 2 mean
would then be again applying the free vortex
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law of C w into r equal to constant, and that
could come out to be 248 meters per second.
22:47.600 --> 22:54.240
So, we can find C w 1 mean and C w 2 mean
that means before the rotor and after the
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rotor in both the stations applying free vortex
law. A free vortex law can be applied between
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this at the inlet to the rotor at the exit
to the rotor or some people sometimes use
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it in the middle of the rotor at the mean
passage. But in this problem we we are not
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doing that we are applying it at the inlet
and at the exit, and having applied that we
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have got the values of C w 1 mean and C w
2 mean.
23:25.730 --> 23:32.480
From this, we can now find that the minimum
power to be supplied that means with hundred
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percent efficiency of the compressor turbine,
and the supply shaft, intermediate shaft between
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the turbine and compressor if all that is
hundred percent, then the power absorbed by
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the rotor is indeed that the power is required
for doing the compression job, and that is
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typically as for free vortex law is is supposed
to be constant from root to tip. So, if you
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find the value at the tip or at the mean anywhere,
it would be a valid everywhere, and that work
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done is m dot into U mean into C w 2 mean
minus C w 1 mean and that gives us a value
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of 1.513 mega watts. So, that is the kind
of a power you would need to activate this
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particular compressor for that kind of a mass
flow, and given the prescribed rotating speed
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of 6000 rpm. So, given all the parameters
we can now see that the amount of work that
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would be necessary to activate this compressor
is 1.513 mega watts.
24:49.870 --> 24:56.870
Now, if you apply the free vortex law, you
can find the value of C w1 hub and C w2 hub
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that means the C w values before the rotor
and after the rotor using the free vortex
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law of C w r equal to constant, and that comes
out to be 87.3 meters per second, and C w
25:16.669 --> 25:23.669
2 would be 262 meters per second. Now, using
all these values, we can now use the velocity
25:27.200 --> 25:34.200
triangles that you have done in quite a great
detail in the earlier lectures. And those
25:34.299 --> 25:41.299
velocity triangles simply yield that the flow
angles at the hub tan alpha 1 equal to C w1
25:44.100 --> 25:51.100
hub by C a equal to 87.3 by 136 which is the
actual velocity, and that would be point 0.642
25:53.580 --> 26:00.580
and hence the alpha 1 is 32.75 degrees; alpha
2 which is the absolute flow angle at the
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exit to the rotor would then be C w2 hub divided
by C a, and that 262 as we are found earlier
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divided by 136, and that value of tan alpha
2 is 1.928 and that yields alpha 2 of 62.6
26:23.510 --> 26:28.179
degrees.
The corresponding value of tan beta 1 comes
26:28.179 --> 26:35.179
out to be again using the values available
at U hub and that is 1.436 which is a value
26:39.159 --> 26:46.159
of beta 1 of 55.15 degrees; corresponding
time beta 2 if you calculate the same way
26:47.220 --> 26:54.000
comes out to be a small value of 8.64 degrees.
Now, you can see here that at the exit to
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the rotor the value of beta 2 is rather small;
at the entry the value of beta 1 is rather
27:01.110 --> 27:06.029
high. Now, that is the angle at which that
is a relative flow angle with which the flow
27:06.029 --> 27:13.029
is going into the rotor blade, and that is
what is often referred to as the blade stagger
27:14.710 --> 27:21.710
angle. That is angle at which the blade would
need to be oriented to get the blade in working
27:22.820 --> 27:28.730
order. So, beta 1, beta 2 are the relative
flow angles at which the blade rotor blades
27:28.730 --> 27:35.460
would have to be oriented to get the work
done. So, those are the values that we get
27:35.460 --> 27:41.340
from using the free vortex law that has been
prescribed in this problem.
27:41.340 --> 27:47.799
The next step in this problem is to find out
the degree of reaction at the hub as required.
27:47.799 --> 27:51.840
And degree of reaction and the hub can be
found out by using the degree of reaction
27:51.840 --> 27:57.639
relation that we have done in the lecture
series, lectures earlier, and that is tan
27:57.639 --> 28:04.639
beta 2 hub minus tan alpha 1 hub into C a
by 2 U hub. All these values have already
28:05.539 --> 28:12.220
been calculated in the earliest steps, and
if you just put in all the values, you get
28:12.220 --> 28:19.220
a degree of reaction R x hub equal to 0.382.
As you can see here, the degree of reaction
28:20.299 --> 28:26.299
at the hub is much lower than that at the
tip where it was prescribed as 0.5.
28:26.299 --> 28:31.840
So, at the hub it is expected that the degree
of reaction in a free vortex design would
28:31.840 --> 28:38.630
be lower and as we have stated in our lectures
earlier that one needs to keep an eye on this
28:38.630 --> 28:44.850
value, because the value of degree of reaction
at the hub, if one is not careful especially
28:44.850 --> 28:51.850
in a free vortex design could indeed go pretty
close to 0 or indeed if one is not a very
28:53.419 --> 28:58.240
careful it could go below 0 that means it
could be negative, and in negative degree
28:58.240 --> 29:03.740
of reaction compressor blade is quite useless
as a compressor, because it is now starting
29:03.740 --> 29:10.740
to behave like a turbine. So, degree of reaction
negative is certainly not acceptable and in
29:11.659 --> 29:17.269
free vortex design, there is a tendency for
the degree of reaction near the hub to become
29:17.269 --> 29:24.269
very low quite close to 0, and if one is not
careful, it could become less than 0. So,
29:25.429 --> 29:31.559
that is why in any design very quickly one
needs to find out what is the degree of reaction
29:31.559 --> 29:38.559
at the hub. Because that is where it could
go rather low, rather dangerously low and
29:38.710 --> 29:44.590
that is normally not acceptable. So, these
are some of the things that one needs to be
29:44.590 --> 29:50.200
careful about when one using the free vortex
design.
29:50.200 --> 29:57.200
The other issue is that the free vortex design
actually shows that it can be symmetric at
29:57.350 --> 30:04.039
only one radial location along the blade length.
In this problem, it was prescribed that degree
30:04.039 --> 30:11.039
of reaction R x is equal to 0.5 or 50 percent
at the tip which means the diffusion split
30:11.919 --> 30:18.120
between rotor and stator is 50-50 at the tip.
There are many designers who would like to
30:18.120 --> 30:24.970
do that at the mean, if you do that at the
mean as we have discussed before if you remember,
30:24.970 --> 30:29.990
at the tip the value of degree of reaction
would be much higher than 0.5, but at the
30:29.990 --> 30:36.299
hub it could go rather dangerously close to
0. This particular design proposes that it
30:36.299 --> 30:42.779
is 0.5 at the tip, and as a result of which
at the hub it is very safe, and it is about
30:42.779 --> 30:48.049
it is 0.382.
So, the value of degree of reaction is another
30:48.049 --> 30:55.049
issue which needs to be kept an eye on even
one is using free vortex design degree of
30:55.620 --> 31:01.990
reaction, as we know is essentially a two-dimensional
parameter, but this is where it helps the
31:01.990 --> 31:08.490
three-dimensional blade design that it tells
you when the design may possibly be going
31:08.490 --> 31:14.070
towards the wrong direction, if you do not
keep an eye on the degree of reaction. And
31:14.070 --> 31:21.070
of course a symmetric blading, the ideal the
original symmetric blading design or 50 percent
31:21.690 --> 31:28.059
reaction is possible in free vortex design
only at one place. If you want symmetric blading
31:28.059 --> 31:33.799
from root to tip and people have done such
design earlier, you have to go for a constant
31:33.799 --> 31:39.100
reaction blading. So, in this problem that
was not prescribed, it was prescribed as a
31:39.100 --> 31:40.610
free vortex design.
31:40.610 --> 31:47.610
Let us go to the next problem, the example
2. And in this the axial flow compressor is
31:49.039 --> 31:55.690
originally designed with free vortex law,
and this free vortex law has a degree of reaction
31:55.690 --> 32:02.690
0.6 at the mean, with hub to tip ratio of
0.6, with flow angles at the mean radius given
32:05.539 --> 32:12.539
are alpha 1 equal to 30 degree and beta 1
equal to 60 degrees. What is required first
32:12.880 --> 32:18.950
is that you calculate the relative absolute
flow angles at the hub and the tip, that completes
32:18.950 --> 32:25.950
the so called blade is geometry blade design,
and both at the inlet and the exit, and to
32:26.860 --> 32:32.309
complete the whole thing you calculate the
degree of reaction at both hub and tip. As
32:32.309 --> 32:37.299
we have seen those are the critical values
that need to be checked out at the beginning
32:37.299 --> 32:41.750
of the design itself.
So, this problem states that you should find
32:41.750 --> 32:48.279
out those values as per free vortex law to
begin with. Now, if this axial compressor
32:48.279 --> 32:55.279
is to be redesigned with exponential law which
we have discussed earlier, then it is prescribed
32:56.110 --> 33:02.779
that you recalculate the relative and the
absolute flow angles at the hub and at the
33:02.779 --> 33:09.779
tip both at the inlet and at the exit of the
rotor, and then corresponding degrees of reaction
33:10.789 --> 33:17.789
at both hub and tip for this exponential law,
it is prescribed at the value of a is 100
33:18.009 --> 33:25.009
and the value of b is four 40 that is prescribed
for this exponential law to be applied for
33:26.120 --> 33:31.799
redesign of this original compressor blade
which was designed with free vortex law. Let
33:31.799 --> 33:34.820
us see how we can proceed to solve this problem.
33:34.820 --> 33:41.820
Now, the original free vortex law designed
can be directly applied, as we had done in
33:42.669 --> 33:49.580
the first problem, and the procedure that
was adopted in the first problem can be adopted
33:49.580 --> 33:56.580
again to find the original free vortex design
of this problem, that gives us alpha 1 is
33:56.950 --> 34:03.950
equal to 37.6 degrees, beta 1 at the hub is
24.8 degrees, alpha 2 at the hub is equal
34:06.269 --> 34:13.269
to 66.6 degrees and beta 2 at the hub is equal
to minus 30 degrees. Now, this is something
34:13.780 --> 34:19.570
that can happen that beta 2 at the hub can
become minus which means it is gone on the
34:19.570 --> 34:26.300
other side of the actual direction, whereas
the other flows are on the other side of the
34:26.300 --> 34:33.300
actual direction. The values of the tip are
alpha 1 tip is equal to 43.9 degrees, beta
34:34.379 --> 34:41.379
1 tip is equal to 67.5 degrees, alpha 2 at
the tip is 54.2 degrees that is at the rear
34:43.970 --> 34:50.750
of the rotor and beta 2 at the tip reality
flow angle is 56.3 degrees.
34:50.750 --> 34:57.750
Now, we can see here that beta 1 in the hub
is very low that is 24.8 degrees, beta 1 at
34:59.490 --> 35:06.060
the tip is 67.5 degrees and those are the
relative flow angles, and also essentially
35:06.060 --> 35:11.579
are close to the stagger angle of the blades.
So, at the tip the stagger angle is very high,
35:11.579 --> 35:18.579
and the blade is set at a high angle, and
at the hub the blade is more actual actually
35:20.619 --> 35:27.420
set. So, those are the typical flow angles
that you can get of typical flow free vortex
35:27.420 --> 35:33.349
design and those angular settings are also
indicative of the kind of values, you may
35:33.349 --> 35:40.349
expect out of a typical free vortex design.
At the hub, the values as you can see at beta
35:40.839 --> 35:47.240
2 at the hub has gone negative, it can indeed
go negative, because the hub airfoil at the
35:47.240 --> 35:53.950
is typically of a very high camber and it
takes the flow from positive flow angle to
35:53.950 --> 36:00.950
negative angle; whereas at the tip the you
can see here the beta 1 tip is 67.5, beta
36:01.740 --> 36:08.740
2 tip is 56.3 which indicates that delta beta
at the tip is only about 11 degrees 11.2 degrees.
36:12.359 --> 36:19.359
That means it is a low camber airfoil, whereas
at the tip it was 54 degrees of camber. So,
36:19.560 --> 36:25.420
at the hub it was 54 degrees of camber.
So, typically you would in a free vortex design,
36:25.420 --> 36:30.290
you would have a hub camber that is highly
camber blade, at the tip it is a flattered
36:30.290 --> 36:37.290
blade. So here in this problem, the tip camber
is 11 degrees and hub camber is about 54 degrees.
36:38.660 --> 36:45.460
So, also it tells us that the beta 1 and beta
2 values have changed a lot from root to the
36:45.460 --> 36:50.859
tip of the blade which is indicative that
the blade would be highly twisted. And that
36:50.859 --> 36:56.720
is also typical of free vortex that blades
do tend to be highly twisted blades. That
36:56.720 --> 37:02.349
is something which the early designers found
out and that is one of the reasons, the other
37:02.349 --> 37:08.990
vortex laws were created. So, that the blades
are not always very highly loaded in a highly
37:08.990 --> 37:15.260
twisted, because highly twisted blades do
get structurally very highly loaded. The high
37:15.260 --> 37:21.420
twist creates structural loading and that
is a bit of a problem in modern compressor.
37:21.420 --> 37:28.380
So.. But that is typical of free vortex design
that you get a highly twisted blade. The camber
37:28.380 --> 37:33.970
is very high at the hub rather low at the
tip and you end up getting a highly twisted
37:33.970 --> 37:34.569
blade.
37:34.569 --> 37:39.869
Let us now look at the degree of reaction.
In this particular problem, the degree of
37:39.869 --> 37:46.869
reaction prescribed was at the mean, and using
those prescribed values we get the free vortex
37:46.880 --> 37:53.880
degree of reaction at the hub 0.29 which is
a good value, and at the tip it is 0.744 which
37:55.349 --> 38:01.740
is again a good value. So, in the two degrees
of reaction that we get are safe values at
38:01.740 --> 38:08.740
the hub and at the tip. So, these are the
free vortex values of the original blade design.
38:08.930 --> 38:15.010
What we need to do now is recalculate these
values using the exponential law.
38:15.010 --> 38:22.010
Now, if we go for the exponential redesign,
the law can be stated as C w1 equal to a minus
38:23.250 --> 38:30.250
b by R upstream of the blade, and C w2 equal
to a plus b by R downstream of the blade where
38:33.530 --> 38:40.530
capital R here is the radius ratio which is
r by r mean that is related to the mean radius
38:41.930 --> 38:48.930
of the blade, and the values a and b given
are 140, these can actually be a dimensional
38:52.300 --> 38:58.490
values are expressed in terms of meters per
second as per the law that is stated here,
38:58.490 --> 39:05.490
which directly gives us the values of C w1
and C w2 in meters per second, because R is
39:06.300 --> 39:11.569
a non-dimensional parameter, it is a radius
ratio really.
39:11.569 --> 39:16.599
Using this law now from the prescribed values
that were given to us in the problem statement
39:16.599 --> 39:23.599
C w1 hub would be equal to 46.7 meters per
second, C w1 tip is 68 meters per second which
39:27.450 --> 39:34.450
are comparatively low values at the inlet
to the rotor. C a1 hub is 121 meters per second
39:36.050 --> 39:43.050
and C a1 tip is 94.1 meters per second by
solving the velocity triangles. Now, as we
39:44.550 --> 39:51.550
can see here C a1 is not constant, in free
vortex design it is normally assumed and it
39:52.119 --> 39:58.109
is prescribed, and it is held constant from
hub to tip. The moment you are weird away
39:58.109 --> 40:05.109
from the free vortex law and using other kinds
of law C 1 C a is not constant anymore. It
40:05.750 --> 40:12.510
varies from hub to the tip and you can solve
the velocity triangles to find those values.
40:12.510 --> 40:19.510
Now, using these values and using the prescribed
law in front and behind the rotor, one can
40:20.960 --> 40:27.960
find at the hub C a2 hub that is behind the
rotor 142 meters per second, C w2 hub is now
40:31.150 --> 40:38.150
higher than C w1 is 153 meters per second.
Now, using all these values of velocities
40:38.480 --> 40:45.480
the whirl component and the actual components,
we can find the angles. Now, tan alpha 1 hub
40:45.930 --> 40:52.930
is equal to C w1 hub divided by C a1, and
this would yield an alpha 1 hub of 21 degrees.
40:55.839 --> 41:02.839
And tan alpha 2 hub is C w2 hub minus C a
at by C a2 and this yields alpha 2 hub of
41:06.020 --> 41:13.020
43 degrees; tan beta 1 hub is equal to U hub
by C a1 minus tan alpha 1 and this yields
41:16.559 --> 41:23.559
a beta 1 hub of 49.1 degree; and tan beta
2 hub yields a value of 21.4 degree using
41:28.940 --> 41:34.480
similar relation.
So, as we can see here now, the relative flow
41:34.480 --> 41:41.480
angles beta 1 and beta 2, beta 1 is very high
at the inlet, but it is comparatively low
41:42.260 --> 41:48.660
at the exit. And now we can see here beta
2 at the hub is actually positive, it is no
41:48.660 --> 41:55.660
more negative and this is what it has done.
So, one can say that the flow turning here
41:56.130 --> 42:03.130
is of the order of 27 degrees a little more
than 27 degrees not as high as we had seen
42:03.309 --> 42:10.309
free vortex design, where it was almost double
of the order of 54 degrees. Hence using the
42:12.710 --> 42:19.710
exponential law, the degree of reaction at
the hub R x hub is equal to 0.59 which is
42:20.650 --> 42:27.650
a very safe and good degree of reaction, and
correspondingly we can use the prescribed
42:29.690 --> 42:35.420
exponential law and find the values at the
tip.
42:35.420 --> 42:42.410
The C w2 tip it would be 132 meters per second
that is a whirl component. Correspondingly
42:42.410 --> 42:49.410
now, tan alpha 1 tip would be C w1 tip by
C a1, and that would yield a value of alpha
42:51.800 --> 42:58.800
1 tip equal to 35.85 degrees, tan alpha 2
tip is equal to C w2 tip divided by C a2 and
43:02.200 --> 43:09.200
that would yield a value of 60.32 degrees,
tan beta 1 tip U tip by C a1 minus tan alpha
43:13.270 --> 43:20.270
1 that gives a value of 2.6, and beta 1 tip
of 69 degrees, that is a very high beta 1,
43:22.790 --> 43:28.750
but that is typical of compressor rotor blade
where the relative flow angle at the tip is
43:28.750 --> 43:35.750
the highest flow angle. Beta 2 tip correspondingly
as we see here now comes out to be 67.4 degrees
43:39.109 --> 43:46.040
which indicates that the camber at the the
tip is very very small of the order of just
43:46.040 --> 43:53.040
about 2 degrees all it less than that.
So, the degree of reaction at the tip is Rx
43:53.390 --> 44:00.390
tip is equal to 0.734 which essentially means
that you have much more of a diffusion going
44:02.329 --> 44:09.329
on in the tip. The work done at the tip is
decided by the camber is decided by the U
44:10.589 --> 44:17.589
tip which is rather high. And more of that
work is now being converted to pressure inside
44:18.970 --> 44:25.910
the tip blade passage, and much less would
be left for the stated to diffuse. So, degree
44:25.910 --> 44:32.910
of reaction of the order of 0.7 or higher
indicates that rotor is participating in the
44:34.059 --> 44:41.059
diffusion of the flow more than that in the
stator. So, the degree of reaction at the
44:41.339 --> 44:44.760
tip has been found to be 0.735.
44:44.760 --> 44:51.760
So, the values obtained for of original free
vortex design and the redesign exponential
44:54.800 --> 45:01.800
law permits us to now to make a few remarks
- concluding remarks of this particular problem
45:03.700 --> 45:08.470
solution. The degree of reaction at the hub
for the exponential design is much higher
45:08.470 --> 45:15.470
than that of the free vortex design and that
makes it very safe design. There is no way
45:16.210 --> 45:23.210
that particular blade would encounter negative
degree of reaction under any operating condition.
45:24.270 --> 45:29.920
The rotor twist that is the values of beta
1 and beta 2, and the variation from root
45:29.920 --> 45:36.920
to tip is much less for the exponential design.
Now, this means that the blade is much less
45:38.030 --> 45:42.839
twisted. Now, this is what I was talking about
that free vortex design tends to be a high
45:42.839 --> 45:49.660
twisted blade. People resort to exponential
design and other kinds of vortex laws to bring
45:49.660 --> 45:56.660
the twist down, because highly twisted blade
can get very highly stressed under the aerodynamic
45:56.849 --> 46:03.770
loading of the on the blade surfaces.
So, the structural loading of the blade is
46:03.770 --> 46:09.140
now substantially reduced and this is true
of the modern compressor blades where the
46:09.140 --> 46:16.089
pressure rise is higher, blades are indeed
aerodynamically highly loaded, and high aerodynamic
46:16.089 --> 46:23.089
load in combination with blade twist would
create very high stress levels on the blade
46:25.589 --> 46:32.589
solid body, and that would essentially could
be prohibitive from the design point of view.
46:33.200 --> 46:40.200
So, exponential law essentially relieves those
stresses by reducing the twist of the blade.
46:41.000 --> 46:48.000
So, this is how the second problem has been
solved and we have given a redesign blade,
46:48.150 --> 46:52.680
and its blade geometry as a part of this solution.
46:52.680 --> 46:58.020
Now, I will leave you with a few problems
which you can attempt to solve on your own
46:58.020 --> 47:04.890
on the basis of the theories that we have
done, following the examples that we are given
47:04.890 --> 47:11.890
in this lecture. So, I will leave you with
a few tutorial problems for your own exercise.
47:12.410 --> 47:19.410
The first problem is a 3-D design of a rotor
blade of axial flow compressor following the
47:20.400 --> 47:27.400
following vortex laws may may be used; first
is C w1 equal to a R minus b by R that is
47:30.859 --> 47:37.410
upstream of the blade, and the other law is
downstream of the blade would be C w2 equal
47:37.410 --> 47:44.410
to a R plus b divided by R where R is indeed
the radius ratio with respect to the mean
47:46.309 --> 47:51.670
radius.
The following data may be used. The hub to
47:51.670 --> 47:58.670
tip radius ratio is prescribed as 0.6, which
is a moderate hub to tip ratio. C a which
48:00.180 --> 48:07.180
is the constant across the rotor from the
inlet to the exit at mean radius, and the
48:10.329 --> 48:17.329
mean radius C w 1 is 60 meters per second,
C w2 is 150 meters per second, the specific
48:22.240 --> 48:29.240
work that is prescribed and is being supplied
or is going to be supplied is 21.6 kilo joules
48:30.369 --> 48:37.369
per kg, and that is the work that you would
be supplied with. With these prescriptions
48:39.140 --> 48:46.140
calculate the following parameters, at mean
radius delta t equal delta t that is the temperature
48:46.740 --> 48:53.740
rise through this compressor, the degree of
reaction Rx, the values of a and b that are
48:56.210 --> 49:03.210
given in the laws, they are to be now found
out. At the root and the tip find out the
49:04.299 --> 49:11.299
degree of reaction, at the root and the tip
find out the inlet and exit axial velocities
49:11.359 --> 49:18.359
is C a1 and C a2; that is at the inlet and
at the exit of the rotor. And the inlet and
49:19.190 --> 49:26.190
the exit flow angles at the root mean and
tip. That means all the alphas and betas at
49:27.380 --> 49:34.380
the root mean and tip at the exit and at the
inlet to the blades. So, those are all the
49:35.640 --> 49:42.030
parameters that would complete the entire
blade design, and you have to find out given
49:42.030 --> 49:48.260
here the work that is being supplied and the
vortex laws that have been prescribed. So,
49:48.260 --> 49:51.740
that is the first problem. Let us take a look
at the second problem.
49:51.740 --> 49:58.650
Now, the in the second problem, the table
a table is being given here, it shows a few
49:58.650 --> 50:05.650
data of an axial flow compressor rotor which
is been designed with free vortex theory.
50:06.720 --> 50:13.720
Given a few data, you are to calculate all
the data to complete the table. So, few data
50:14.299 --> 50:19.670
are given with those data can you complete
the entire table, actually the problem is
50:19.670 --> 50:26.390
very similar to the earlier problem, calculating
all the alphas and betas, and C a 1 and U
50:26.390 --> 50:33.390
1 and so on, and so far. So, all the flow
velocities, all the flow angles, and then
50:33.490 --> 50:39.910
the work done, and then of course the degree
of reaction; so, all those values that we
50:39.910 --> 50:44.930
have been calculating essentially has now
been put in a tabular form, and you are asked
50:44.930 --> 50:51.210
to complete the table. You can also plot the
entry and exit flow velocity angles at the
50:51.210 --> 50:56.900
root, mean and tip and that will give you
an idea, what kind of blade you are getting
50:56.900 --> 50:59.680
out of this particular problem.
50:59.680 --> 51:05.250
The third and the last problem that I will
leave you with is an axial flow compressor
51:05.250 --> 51:12.250
with hub to tip ratio of 0.4 which you will
notice is a somewhat lower hub to tip ratio
51:12.890 --> 51:18.549
as we have discussed earlier in our lectures.
And the maximum diameter for this compressor
51:18.549 --> 51:25.020
is 0.6 meters, and this has been designed
with a constant reaction of 50 percent of
51:25.020 --> 51:31.500
0.5 from root to tip. So, this is a constant
reaction design which we have discussed before
51:31.500 --> 51:38.220
in the lecture and a little bit today also.
And this has the blade tip speed of U tip
51:38.220 --> 51:45.170
of 300 meters per second. The stagnation temperature
pressure rises 16 degrees centigrade which
51:45.170 --> 51:52.170
means it is a comparatively lowly loaded compressor.
The actual velocity prescribed for the flow
51:52.619 --> 51:58.430
near the casing upstream of the rotor not
exactly at the leading edge, but it will upstream
51:58.430 --> 52:05.430
is 120 meters per second. For air you can
take the value of C p as 1.005 kilo joules
52:06.970 --> 52:13.970
per kg K as we have probably done in all the
earlier lectures and the earlier tutorial
52:14.859 --> 52:19.790
that you have done.
The problem asked you to determine the actual
52:19.790 --> 52:26.790
velocity before and after the rotor in such
a manner that the radial equilibrium is maintained
52:28.609 --> 52:35.390
or obtain. So, that is how the 3-D flow is
brought into this particular problem. That
52:35.390 --> 52:40.869
you are to maintain radial equilibrium and
if you are to maintain radial equilibrium,
52:40.869 --> 52:47.150
can you get the actual velocity before and
after the rotor, and as I was stating earlier,
52:47.150 --> 52:54.150
if the axial velocity is constant across the
rotor as it is assumed in a free vortex design,
52:54.400 --> 53:01.400
you cannot get a constant reaction from hub
to tip. So, the constant reaction design is
53:01.619 --> 53:08.619
an intrinsically a contradictory to constant
axial flow across the blade. So, this problem
53:09.540 --> 53:16.540
exempli exemplifies that particular understanding
and asks you to actually calculate those values
53:16.790 --> 53:23.720
to essentially formalize that particular understanding.
So, I hope you will be able to sit down and
53:23.720 --> 53:29.630
do the calculations using the same steps that
we have done in the earlier examples, and
53:29.630 --> 53:35.069
you should have no problem getting the answers.
The accuracy of the answers depends on the
53:35.069 --> 53:42.069
accuracy of the some of the earlier parameters
like C p and all that you use. So, I hope
53:43.089 --> 53:47.170
you will be able to solve the problems and
get your answers; the trend of the answers
53:47.170 --> 53:51.900
are already been given in the two examples
that we have taken. So, you should be able
53:51.900 --> 53:58.599
to quickly make out whether you are getting
the answers that are reasonably correct answers.
53:58.599 --> 54:04.270
With this we come to the end of today's lecture
that is the solution of the problems and problems
54:04.270 --> 54:09.849
for your exercise.
In the next class, we will be moving towards
54:09.849 --> 54:16.849
some of the other issues of compressors which
are related to compressor instability, flow
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distortion in the compressor which are indeed
extremely vexing issues and we shall look
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at the physics of the problem in some detail
in the next few lectures, trying to look at
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issues that are three-dimensional and much
more than three-dimensional, their distortions
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and sometimes they are born out of or leading
to instability in the compressors. Those are
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the issues we shall be doing in the next few
lectures.