WEBVTT
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We are talking about axial flow turbines,
and in last class, last lecture, we talked
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about how to design and bring into the design
steps and design features, three dimensionality
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of flow through axial flow turbines. Now,
we know that the flow through the axial flow
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turbines, as in axial flow compressors, is
normally annulus in nature and that flow through
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the annulus space that is available to the
turbines is often not exactly uniform.
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So, some aspect of the non uniformity or some
aspect of the variation from the lower radius
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to the outer radius or inner radius to the
outer radius, and more specifically from the
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root to the tip of the blade of a rotor needs
to be factored into to begin within the design,
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and later on, in the computations, and of
course, later on in the analysis. We have
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seen how some of these things can be brought
into simple analysis. That we had done in
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the last class without getting into more complex
computational analysis.
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As I mentioned, we shall do computational,
an introduction to computational analysis
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through turbo machines towards the end of
this lecture series, but right now, we are
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looking at turbines specific certain theories
which factor in the three dimensionality of
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the flow and built it into the design. So,
most of it is indeed used for design, and
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then, immediate pose design analysis to find
out how the turbine is actually going to behave.
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So, in today's lecture, we will look at some
problems that actually use these theories
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that we have done in the last class, and then,
try to actually solve some problems which
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are prescribed problems, and from the problem
statement, we tried to figure out what kind
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of solutions can be arrived at using the theories
that we have done in the last lecture.
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And towards the end, I will leave you with
a few problems to solve by yourselves so that
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you can get the feel of the application of
these theories to actual problems, realistic
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problems and you also get a feel of the numbers.
It is essential for an engineer to get the
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feel of the numbers and that is why it is
essential that we look at a few problems which
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are probably a little textbook problems, simplicity
problems where, you know, typically you have
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all the data you require. In a real life problem,
you often may not actually have all the data
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that is require to solve, but we are dealing
with problems where the problem statement
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gives you all the data that you require.
And now, you need to use the theories that
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we have done to arrive at solutions, and in
a process, one learn how the solutions would
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look like and indeed as I said you get a feel
of the numbers. What the numbers would look
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like? So, that is very important for all practical
engineers. So, let us look at some of the
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problems related to three-dimensional flow
in axial flow turbines. Now, in axial flow
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turbines, that we are going to do some solved
problems, and then, I leave you with some
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exercise problems to solve for yourselves.
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Now, in the solved problems, we will first
look at the problem statement. In the first
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problem, the problem statement shows that
a constant nozzle exit angle which we have
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designated earlier is alpha 2 is being use
for axial flow turbine design. In which, it
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is prescribed that the temperature drop should
be 150 k, and at the hub, the flare velocity
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a blade speed u is 300 meters per second,
and at the tip, it is 400 meters per second.
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At it is prescribed that alpha 2 is 60 which
is constant as prescribed from a root to the
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tip hub to the tip and alpha 3 is 0, that
is, 0 whirl at the exit of the rotor. The
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radius ratio given for this particular problem
statement is 0.75, that is, hub radius to
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tip radius ratio is 0.75that is prescribed
here. The problem asks a solution in which
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he should complete the design velocity diagrams
at hub mean and tip of the stage, and thereafter,
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calculate the velocity components if the design
is a free vortex design for the turbine and
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compare the results of that with that of constant
nozzle exit angle. So, one can look at the
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whole thing from a free vortex point of view
and compare the results. So, this is a problem
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statement which we can now try to find a solution
to.
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Now, at the rotor illustration, we have done
in the last class - the equation for variation
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of whirl component is defined by the variation
of alpha 2 and that is given by C W 2 by C
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W m which is the mean and that is also equal
to C a 2 by C a 2 mean, which is also equal
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to C 2 by C 2 m which is mean and all these
velocity ratios are equal to radius ratio
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r by r mean r being at any radius to the power
sine square alpha 2. Now, this is what we
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had done in the last lecture following which,
what we get is, at the rotor exit, the actual
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velocity C a 3 square is equal to C a 3 mean
square plus twice U m into C W twice m whole
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thing multiplied by 1 minus radius ratio r
by r m to the power cost square alpha 2.
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Now, this is also the theory that we had the
expression; we had shown in the last lecture.
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Now, from the prescribed radius ratio that
is given, we can see that the mean to tip
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radius ratio would be 0.875 and mean to hub
radius ratio would be 1.166, that is calculated
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from the hub to tip radius ratio that is prescribed
in the problem statement.
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Now, the work done by the rotor is given by
the rulers equation which you are all aware
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of. Now, in this particular problem statement,
alpha 3 would be equal to 0; exit whirl component
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is 0, which means C W 3 indeed would be 0,
and in which case, the specific enthalpy rise
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or, you know, specific work that we normally
use for aero thermodynamic relations, that
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is, a c p into delta t would be equal to U
m into C W 2 m C W 3 being 0, and from which,
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we can write that C W 2 m would indeed be
492 meters per second and corresponding C
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a 2 would be from the velocity triangles that
you can draw at station 2 that is before the
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rotor and that would come out to be 284 meters
per second. As per the prescription, that
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also would be equal to a actual velocity at
the exit of the rotors C a 3 m.
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Now, at the rotor hub, at the inlet of the
rotor, one can write a axial velocity C a
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2 h is equal to C a 2 m to the multiplied
by a radius ratio r m by r to the power sin
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square alpha 2 and this would yield a axial
velocity of 318.8 meters per second. On the
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other hand, the whirl component at hub, at
the inlet C W 2 h would be a C W 2 m into
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r m by r to the power sine square alpha 2
and that would be 552.2 meters per second.
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So, you now start getting the components of
the velocities at station 2, that is, before
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the rotor at the rotor inlet.
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Now, at the rotor tip of the inlet, one can
now find the actual velocity using the same
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relation, that is, C a 2 m into r m by r to
the power sin square alpha 2, and in which
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case, the actual velocity would now be 257
meters per second, and at the whirl component,
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C W 2 tip would be 447 meters per second.
So, as one can see that the whirl components
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has actually decreased from hub to the tip,
whereas, the actual velocity is also decreased
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a little from hub to the tip and this is a
consequence of the fact that alpha 2 has been
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held constant from root to the tip of this
prescribed problem. Soon as a result of which,
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we now have all the velocity components that
are required at the inlet to the rotor.
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Now, at the rotor tip outlet, we can find
what the actual velocity would be and this
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can be found from the axial velocity relation
that we had done earlier and that is given
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by C a 3 equal to whole thing root over C
a 3 square plus twice U m C W 2 m multiplied
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by 1 minus radius ratio r by r m to the power
cos square alpha 2, and if you do that, we
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can calculate the actual velocities at tip
and hub as at the outlet as 262 and 306 meters
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per second.
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So, once you apply the relations that we had
done in the last class for three-dimensional
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flow estimation, we now have the axial velocities
at tip and hub at the outlet to the rotor.
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On the other hand, if we calculate all the
values with reference to the free vortex design,
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we can see that the velocities that we get
in comparison to the nozzle angle can be now
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written down in a tabular form, and one can
see that these values are the ones in constant
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nozzle angle are given in red and the free
vortex values are given in black, and the
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actual velocities as you can see in, in, front
of the rotor and behind the rotor are constant
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for free vortex, only the whirl component
is changing as per free vortex variation from
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tip to hub. On the other hand, for constant
nozzle angle as one can see all the velocities
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are varying, and essentially only at the mean,
the values of free vortex and the constant
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nozzle angle are same at the tip as well as
at the mean, the axial velocity the whirl
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component velocity and the exit axial velocity
they are all same at the mean, but at the
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hub and the tip, the constant nozzle angle
gives completely different values compared
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to that of free vortex.
Now, this is something which you may like
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to take a look at and you may like to sit
down and draw all the velocity triangles that
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are born out of this two calculations, two
sets of design calculations and you will probably
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find that you get completely different blade
shapes. The blade shapes that come out of
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this two design exercises would indeed be
quite different from each other and that would
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tell you that, if you use different kind of
design law or design philosophy, you would
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in end up getting quite different blade shapes,
the three-dimensional blade shapes.
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Now, if we go onto the second example in which,
the problem statement reads that it is propose
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that for design of an actual flow turbine
two design methods are to be explored. Now,
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first design method that is prescribed is
C W 2 m is equal to C W 2 h that is equal
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to C W 2 t; that means the whirl components
at hub mean and tip are equal to each other.
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Now, this type of design is what we have called
earlier solid body essentially design; that
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means the fluid behaves more like a solid
body and correspondingly the whirl component
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everywhere are same.
The second design that is prescribed is where
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C a 2 is equal to C a 2 hub into the radius
ratio hub to tip radius ratio to the power
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sin square alpha 2, which somehow tries to
make use of the constant alpha 2 prescription
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and that is your Case B. The second design
that is suggested here in Case B, and incase
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C, we have C W ratio C W 2 t to C W 2 h has
equal to the radius ratio r h by r t.
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Now that, if you remember is nothing but you
free vortex design. So, we have three design
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that is suggested for design of an actual
flow turbine - one in which the flow behaves
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like a solid body; the second one in which
one may use the alpha 2 equal to constant
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that is constant nozzle angle prescription
and Case C is which resembles of that of a
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free vortex design.
What is prescribed are some common data and
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these are the actual velocity at the mean
is prescribed as 200 meters per second; entry
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alpha 2 is 60 degrees; exit alpha 3 is 0.
The degree of reaction is 0.5 at mean and
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radius ratio prescribed is 0.8. What is asked
for is complete the velocity diagrams for
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all the cases. The velocity diagrams are asked
for because those are the velocity diagrams
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based on which final blade shapes would be
created. So, the velocity diagrams would give
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a fairly good idea about the blade shapes
that are being created by three cases - a
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b and c.
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Let us look at how to go about finding a solution
to this particular problem statement. From
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the prescribed data of the example 2, one
can calculate that radius ratio r m by r t
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would be 0.89 and r t by r m would be 1.11.
This is calculated from the hub to tip radius
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ratio that is prescribed in the problem, and
then, the whirl component at mean C W 2 m
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can be found from the mean velocity diagram
that is a C a 2 into tan alpha 2 and that
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gives C W 2 m equal to 346.5 meters per second
and it is prescribed that alpha 3 is 0. So,
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C W 3 m would be 0.
Now, C W 3 m equal to 0 is what we had done
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in the first problem also, where it is prescribed
that C W 3 m could be given as 0 or alpha
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is given as 0. This is a fairly standard prescription
for many of the designs, because what happens
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is, when the flow is going out of the turbine
quite often, the prescription often encourages
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that the flow going out of the turbine does
not have any whirl component, because the
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whirl component going to going out of a turbine
typically of a single turbine or a multistage
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turbine, last stage would be quite useless.
So, typically of a turbine it is quite often
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unless you know it is one of the earlier turbines
are one of the middle stage turbines. The
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prescription quite often comes with alpha
3 equal to 0 which leaves 0 whirl component,
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and if the flow is going in to a nozzle or
going into exhaust, any whirl component present
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is quite useless, only component that is useful
for nozzle effect for thrust making is the
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axial component. So, giving a prescription
of whirl component 0 is pretty much a practical
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and standard prescription for turbine design
as I mentioned, unless you are designing a
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turbine, which is one of the middle stages
of a multistage turbine.
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So, as we have seen in both the problems C
W 3 and alpha 3 of the prescribed to be a
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0, and as a result of which, the problems
does become a little simple, but the prescription
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is realistic it is not really idealistic.
Let us get back at the problem. For this particular
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problem statement, it is given the degree
of reaction r x is 0.5. Now, that means, it
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brings us back to the symmetrical balding
concept that you may have done earlier and
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certainly done in some detail in axial flow
compressors.
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So, the moment you put a degree of reaction
0.5, the symmetrical bleeding concept comes
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in, and then, you have at the mean, alpha
2 m equal to beta 3 m that would be equal
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to 60 degrees as prescribed and alpha 3 m
equal to beta 2 m equal to 0 degrees as prescribed.
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This of course, makes the problem a little
simple to handle. However, many turbines in
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the past and the early days of turbine design
have been used, using these kind of a somewhat
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simpler design prescriptions and those turbines
were a operating quite well.
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It makes the designers job definitely a simple
and a probably analysis also become simple
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in those days and long back thirty forty years
back if you may remember, the aid from a computational
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fluid dynamics was not available, and as a
result, the refinement that is possible in
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today's turbine design was not really available
in those days, and as a result, somewhat simpler
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design prescriptions were often used for design
and those were functional, they worked fine
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cell. So, a similar somewhat simpler design
prescriptions has been also prescribed here.
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Now, the blade velocity U m would come out
to be same as a C W 2 m because u r beta 2
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m is 0, and as a result of which, 346.5 meters
per second as calculated just a little above,
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and at any radius, we can now calculate the
blade velocities from the radius ratios that
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we have just written down. So, the hub blade
velocity would be 308 meters per second and
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the tip blade velocity U t would be 385 meters
per second. So, the blade velocity is vary
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all the way from root to tip as a per omega
into r concept.
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Now, for case A, we have three cases to be
actually looked at, and as I indicated, this
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is a fluid which is a prescribed to be behaving
like a solid body for a cases which we have
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done in the last lecture n equal to 0 for
the equation C W 2 equal to r to the power
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n. Now, when you put n equal to 0, it is a
Cases which fluid behaves like a solid body.
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Now, the actual speed is calculated from the
actual velocity pressure derived from the
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energy equation for the cases n equal to 0
and this comes out to be C a 2 is equal to
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C a 2 m whole thing root over 1 minus 2 tan
square alpha 2 m into l n radius ratio r by
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r m. Now, this can be derived; you can said
an and derived in the same way we were done
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earlier for the cases n equal to 0 from the
energy equation that we are written down involving
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the various velocity components.
And you have to put the value of n equal to
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0 there and he would arrive at this a solution
which we are looking at for a axial velocity
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at any station with relation to axial velocity
at a mean radius along the blade length.
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Now, using this, you can also calculate the
angles across the rotor from the above considerations.
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If you do that, the solutions that you get
for the case A tells you that the actual velocity
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prescribed at the mean was a 200 and C W 2
at the mean was found to be 346.5. On the
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other hand, we found the axial velocity is
varying from tip to hub and both at the rotor
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entry as well as at the rotor exit. At the
rotor exit, the C W 3 is prescribed to be
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actually 0, and correspondingly the value
of alpha 3 also has been prescribed to be
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0. The alpha 2 variation is shown here as
part of our results. It varies all the way
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from hub to the tip. Correspondingly the beta
3 also varies exactly in the same manner from
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hub to the tip, and the beta 2 values are
shown here. It is 0 at the mean as we have
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calculated, as off we prescribed the degree
of reaction being 0. However, there is a small
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value beta 2 at the hub and a small vale of
beta 2 at the mean, at the tip that is been
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shown here.
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Now, if we move to the results of a case B,
the prescribed condition is C a 2 t is equal
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to C a 2 h into radius ratio r h by r t prescribed
here in the problem to the power sine square
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alpha 2. Now, this of course, brings us to
the fact that all the velocities at the tip
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and hub and mean can be the ratios of them
can be put down as equal to each other. All
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of them would be equal to some relation to
alpha 2.
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Now, this allows us to write down that far
constant nozzle angle, which is the prescribed
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case B that we were looking at. C a 2 can
be now written down as C a 2 m into radius
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ratio r m by r to the power sine square alpha
2, and in case of C W 2, that would be C W
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2 m to the into radius ratio r m by r to the
power sine square alpha 2 and C 2 is equal
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to C 2 m to the multiplied by radius ratio
r m by r to the power sine square alpha 2.
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So, far constant nozzle angle case, all the
velocity C h 2, C W 2 and C 2 which is the
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absolute velocity can be found from the mean
values, that is, at the mean radius to hub
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to tip at any radius by using this radius
ratio concept.
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If we do that at the station 3, at the exit
of the rotor, it is prescribed that alpha
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3 is equal to 0 and C W 3 is also equal to
0. The expression for axial velocity comes
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out as we have done in the last lecture a
C a 3 square is equal to C a 3 square 3 C
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a 3 m class twice U m C W 2 m to multiplied
by 1 minus radius ratio to the power cos square
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alpha 2. Now, this allows us to calculate
C a 3 at exit of the rotor. If we now use
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the relation that we have done which is essentially
for the case B, which is a exit angle from
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the rotor is held constant from root to the
tip of the blade.
29:27.190 --> 29:32.679
Now, this is something which we have discussed
in last lecture that, holding the exit angle
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from the rotor constant from root to the tip
of the blade makes the rotor untwisted or
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very likely twisted. Now, this is important
for a stator nozzle blade cooling purpose.
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We shall be doing a cooling technology from
next lecture onwards, but this particular
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design philosophy of holding alpha 2, 2 equal
to constant from hub to tip or root to tip
30:03.559 --> 30:10.559
of the blade of the stator essentially caters
to cooling technology, and if we apply this,
30:14.500 --> 30:20.370
in this present problem what we see is the
result that we get for case B.
30:20.370 --> 30:27.370
Let us look at the results of Case B. What
we see here is the axial velocity is now vary
30:30.779 --> 30:36.659
all the way from hub to the tip. At the hub,
it is 218.5. At the mean, it is prescribed
30:36.659 --> 30:43.659
as 200. So, the variation of axial velocity
at the entry as well as at the exit is quite
30:46.679 --> 30:53.679
pronounced from hub to tip. The values of
C W 2 are also variables from hub to tip point.
30:54.750 --> 31:00.549
Substantially C W 3 is being held constant;
alpha 2 by prescription is held constant;
31:00.549 --> 31:07.549
alpha 3 by prescription is 0.
We get a variation of beta 2 from 17.9 2 minus
31:10.610 --> 31:17.610
19.4 and we get a variation of beta 3 from
54 to 60 nine from hub to tip. So, these are
31:19.029 --> 31:26.029
the results of the Case B which is of a constant
alpha 2 from hub to the tip of the stator
31:27.230 --> 31:34.230
nozzle. Now, we can move to the Case C. Now,
Case C is what we had seen was actually the
31:35.240 --> 31:42.240
free vortex design. Now, the free vortex design
as we have mentioned before is not the most
31:42.669 --> 31:49.379
popular design for actual flow turbine even
though it is a very popular design for axial
31:49.379 --> 31:54.059
flow compressors.
For turbine, it is not the most popular design,
31:54.059 --> 32:01.059
but it is the simpler design it of course
works. If you make actual flow turbine with
32:01.399 --> 32:07.909
a free vortex design, it will surely work;
there is no reason why it should not work,
32:07.909 --> 32:13.570
but it is not the most popular design today,
and ever since the cooling technology came
32:13.570 --> 32:19.259
into the market, the free vortex design has
been essentially replaced by the constant
32:19.259 --> 32:26.259
alpha 2 design which is the more popular design
philosophy for turbines essentially as I mentioned
32:26.840 --> 32:33.149
to cater to a cooling technology. But let
us look at the results that we get for this
32:33.149 --> 32:40.149
problem a statement problem to Case C for
a free vortex design philosophy.
32:44.259 --> 32:51.259
If you apply free vortex design philosophy,
the whirl components at tip and hub the ratios
32:52.039 --> 32:59.039
are directly related to the hub to tip radius
ratio. Now, if we apply the same at the outlet
33:01.860 --> 33:08.860
also, C W 2 also is held constant as per free
vortex principle across the blade at mean
33:10.059 --> 33:17.059
various. So, C W, C a 2 is equal to C a 3,
and if you apply all these, in the, as per
33:19.330 --> 33:26.330
the free vortex law, we get a set of results
directly as per very well known free vortex
33:26.799 --> 33:29.679
law prescriptions.
33:29.679 --> 33:36.679
The results that we get that the axial velocities
are held constant from hub to tip as well
33:36.919 --> 33:43.919
as across the rotor. C W 2 varies substantially
from hub to the tip as per free vortex law
33:45.259 --> 33:52.259
and which would tell you that you would end
up getting a substantially twisted blade.
33:53.470 --> 34:00.470
C W 3 is being held constant. So, the trailing
edge would be rather leaner. On the other
34:00.620 --> 34:07.620
hand, the value of alpha 3 is 0 corresponding
to the prescription. Alpha 2 varies from hub
34:09.639 --> 34:15.810
to tip and the C W 2 variation shows that,
and then, of course, you have the variation
34:15.810 --> 34:22.810
of beta 2 which goes minus at the tip and
this is something that comes out of the free
34:23.000 --> 34:30.000
vortex design, and as a result of which, you
get a variation of beta 3 which also varies
34:30.440 --> 34:33.390
from hub to the tip of the blade.
34:33.390 --> 34:40.390
So, you have the results of the Case C Tabulated
here, and then, we have three Cases -- a,
34:40.910 --> 34:47.910
b and c these variations can now be all put
together into one table which are essentially
34:50.150 --> 34:57.150
tries to compare the 3 Cases -- a, b and C,
and as you can see the Case A is given in
34:57.190 --> 35:04.190
red, the Case B that is a constant nozzle
angle is given in b and the free vortex design
35:04.220 --> 35:11.220
is given in Case C and all three of them are
brought together. If you sit down, use the
35:13.800 --> 35:19.930
velocity is that I given the angles that are
given, and if you draw the velocity triangles
35:19.930 --> 35:26.930
of all the cases, you will probably get a
very clear picture of what kind of blades
35:27.380 --> 35:34.070
actually come out of the three-dimensional
blade shapes that should come out of the three
35:34.070 --> 35:39.890
Cases that are prescribed here. The three
cases that are prescribed here had certain
35:39.890 --> 35:46.890
commonalities, that is, the mean actual velocity
prescribed by 200. The exit the angles were
35:47.420 --> 35:53.140
prescribed to be 0. The whirl components at
the exit were prescribed to be 0.
35:53.140 --> 36:00.140
So, with those common prescriptions, we try
to put together; the blades that would come
36:00.450 --> 36:06.780
out even with those commonalities, and we
see that three completely different blade
36:06.780 --> 36:13.780
shapes are likely to result from three cases
that are prescribed here for axial flow turbine
36:13.880 --> 36:20.880
design. So, as I mentioned, you can probably
sit down and actually draw the velocity triangles
36:21.080 --> 36:26.740
and you would find that three different cases
or three different blade shapes and he would
36:26.740 --> 36:33.740
indeed need to choose different aerofoil shapes,
different blade sections from hub to tip for
36:35.220 --> 36:40.200
the each of these three cases. So, you end
up having three completely different blade
36:40.200 --> 36:47.200
shapes for three different design philosophies
even though we started out with a common data
36:48.830 --> 36:55.830
prescription for all three cases. So, this
is an example which tells you with certain
36:57.360 --> 37:03.290
simplifications or simplified problem. It
still tells you that, if you have three different
37:03.290 --> 37:09.690
design philosophies, you end up with three
completely different blade shapes. I will
37:09.690 --> 37:16.690
now leave you with a few problems which you
can sit down and solve for yourself and get
37:18.290 --> 37:25.290
a feel of the numbers that come out of solving
of examples problems that are prescribed with
37:29.240 --> 37:36.240
numerical values. So, let us look at some
of the problems you can solve for yourselves.
37:37.730 --> 37:43.480
The first exercise, exercise, problem, the
problem statement reads that an axial turbine
37:43.480 --> 37:50.340
rotor that is prescribed with rotor inlet
and outlet flow in radial equilibrium, which
37:50.340 --> 37:57.340
means the static pressure and the dynamic
pressure are balanced at the inlet and outlet
37:59.600 --> 38:06.020
of the rotor. The whirl component of the flow
is designed to vary radically as per this
38:06.020 --> 38:13.020
prescription as C W at the inlet as a r minus
b by r and at the outlet as a r plus b by
38:17.960 --> 38:24.960
r. Now, a and b are the constants, and what
is required for you to find is find the inlet
38:27.530 --> 38:34.530
and outlet exit velocities and the expressions
for those velocities.
38:34.890 --> 38:41.890
In this case, you can see the answers given
here and the actual velocity is of course
38:42.420 --> 38:49.180
would remain constant across the rotor. So,
we are dealing with rotor only. So, the problem
38:49.180 --> 38:55.270
statement is essentially for rotor. Part b
of the problem is - it is prescribed that
38:55.270 --> 39:02.270
at mean various given value is point 3 meters.
The actual velocity is 10 meters per second
39:03.120 --> 39:10.120
and the degree of reaction is 0.5. The blade
loading coefficient is prescribed as, as,
39:12.080 --> 39:18.720
per the definition, psi rotor is equal to
work done, specific work done 8 0 by U tip
39:18.720 --> 39:25.720
square and the r p m in 7640 rpm.
The hub to tip radius ratio is given as 0.5,
39:27.620 --> 39:34.620
and at 80 percent of the rotor radius, it
is required for you to find the rotor relative
39:36.130 --> 39:43.130
flow inlet and outlet angles. So, what you
are required to find the beta values of a
39:43.430 --> 39:50.430
beta 2 and beta 3 for the particular problem
statement that is given here. Now, the answers
39:52.280 --> 39:59.280
given are 43.3 degree and 10.4 degree for
beta 2 and beta 3. So, you can try to sit
40:00.910 --> 40:04.660
down and see whether you can arrive at those
answers.
40:04.660 --> 40:10.450
The second exercise problem that is given
is the gas exits from the turbine stator or
40:10.450 --> 40:17.130
nozzle at a radically constant angle alpha
2. So, it is a constant nozzle exit angle
40:17.130 --> 40:24.130
problem. The gas is prescribed to be in radial
equilibrium. The axial velocity variation
40:24.190 --> 40:30.280
at that station is given as C a 2 in to r
to the power into sine square alpha 2 equal
40:30.280 --> 40:37.280
to constant. This is what we have done in
our lectures also, and for a turbine in which
40:37.290 --> 40:44.290
the axial velocity at the radius, 0.3 is a
again prescribed 100 meters per second, and
40:45.000 --> 40:52.000
if the turbine has stated above is designed
with a constant alpha 2 equal to 45 degree,
40:53.070 --> 40:58.300
find the actual velocity at that station at
0.6 meters radius.
40:58.300 --> 41:05.300
Now, the answers given here is very simple
and that is 70.7 meters per second. So, it
41:06.360 --> 41:13.360
is a constant exit, stator exit angle problem
and you have to apply the relations that we
41:13.710 --> 41:19.670
have done in the lecture or in the earlier
problem that has been solved for you.
41:19.670 --> 41:26.670
The third and the last problem that is prescribed
for you to solve is an axial turbine is designed
41:28.080 --> 41:35.080
with free vortex at stator nozzle exit and
0 whirls at stator at rotor exit. So, it is
41:37.730 --> 41:44.500
a free vortex problem applied at stator and
nozzle exit and a rotor inlet. So, the station
41:44.500 --> 41:50.060
between the stator and the rotor carries free
vortex prescription. For the following operating
41:50.060 --> 41:56.850
condition, that is, at the entry t 0, 1 is
equal to 1000 k mass flow is given as the
41:56.850 --> 42:03.850
thirty 2 kegs per second. The hub radius is
0.56; the tip radius is 0.76 meters; rpm prescribed
42:06.000 --> 42:13.000
is 8 thousand; the degree of reaction is point
5 and a actual velocity is constant that is
42:13.750 --> 42:19.760
1 into 3 meters per second, and it is prescribed
that the inlet and exit absolute velocities
42:19.760 --> 42:24.150
are equal to each other, that is, c 1 is equal
to c 3.
42:24.150 --> 42:31.150
You are required to find C 2 that is the nozzle
exit velocity. As you know, it expands ha
42:32.400 --> 42:39.400
hugely from C 1 to 2, and then, you are required
to find the mach number, maximum mach number
42:39.530 --> 42:46.530
at the stage. In this particular stage, typically
it is most likely to be a somewhere in the
42:48.480 --> 42:55.480
nozzle exit, and then, the reaction at the
root, the power output of this particular
42:56.190 --> 43:03.190
working turbine and t 0 3 and t 3 at that
stage exit. So, these are the numerical values
43:05.360 --> 43:09.290
that you need to find out of this prescribed
problem.
43:09.290 --> 43:16.290
The answers solutions that are given here
is that the nozzle exit velocity C 2 is would
43:17.600 --> 43:24.600
be for in 80 meter per second. The maximum
mach number in this stage is so solved as
43:26.440 --> 43:33.440
point 0.818. The reaction at the root is 0.08,
and you remember at the root, it is quite
43:36.870 --> 43:42.630
often specially free, free, vortex. It is
likely to be very close to 0 and 0 of course
43:42.630 --> 43:49.630
would mean an impulse turbine, and we are
looking at a problem in which, the solution
43:49.770 --> 43:56.770
actually comes pretty close to giving you
an impulse station, impulse section at the
43:57.850 --> 44:04.850
root of this particular turbine, and then,
the power output are work done for this particular
44:06.730 --> 44:13.730
rotor given the mass flow is 3.42 mega watts,
and the temperatures at the exit t 3 is 907
44:19.160 --> 44:26.160
and the static temperature t 3 is 8 892 k.
Now, you can sit down and try to solve this
44:26.760 --> 44:33.760
problem and see whether you can come. You
can standard values of gas constant r,
44:34.450 --> 44:41.450
that is, 247 joules per kg k and value of
C p as thousand 147 joules per kg k. So, you
44:45.950 --> 44:52.690
can use those standard values to solve this
problem in which, numerical are given and
44:52.690 --> 44:59.690
you are required to find the certain prescribed
velocities, mach numbers, work done and the
45:02.480 --> 45:09.020
exit temperature. So, I leave you with these
problems for you to solve yourself so that
45:09.020 --> 45:16.020
you can get a feel of the numbers the typically
come out of axial flow turbine design. So,
45:16.530 --> 45:23.430
some of these problems would give you an idea
how the turbine designs indeed proceeded with,
45:23.430 --> 45:29.120
and what kind of numbers you get? What kind
of variations? You get a you get a feel of
45:29.120 --> 45:36.120
the numbers by solving these problems.
In the next class, we will be looking at turbine
45:36.750 --> 45:43.750
blade cooling, because in this class and in
the earlier lecture, we had looked at the
45:44.760 --> 45:51.240
design philosophy of alpha 2 is constant from
root to the tip and we have stated again and
45:51.240 --> 45:58.070
again that this particular design philosophy
essentially caters to turbine blade cooling.
45:58.070 --> 46:04.000
In the next lecture onwards, we will devote
ourselves to looking at this turbine blade
46:04.000 --> 46:10.680
cooling technology and how it impacts the
turbine blade, the turbine blade shape, and
46:10.680 --> 46:17.680
essentially, the aerodynamics or the aero
thermodynamics of the flow over the turbine
46:18.720 --> 46:25.720
blades is very strongly impacted, and essentially,
the aerodynamics of the blade changes hugely
46:28.320 --> 46:31.820
by application of cooling technology.
46:31.820 --> 46:38.820
We will look at various cooling technologies
and how do they actually impact the turbine
46:39.200 --> 46:46.180
design of modern actual flow turbines specifically
these cooling technologies are very widely
46:46.180 --> 46:53.180
used in aero engines, and we will look at
some of the typical examples of these applied
46:54.120 --> 47:01.120
cooling technologies in turbine, actual turbine,
rotor and stator. So, we shall be doing turbine
47:02.050 --> 47:05.900
cooling technologies from next lecture onwards.