WEBVTT
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Hello and welcome to lecture number 36, of
this lecture series on Turbo-machinery Aerodynamics.
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In the last class we had introduced a very
new topic, which is quite different from what
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we have discussed in some sense. That is,
on radial flow turbines. Of course, we have
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already discussed its counterpart centrifugal
compressor. But, in the context of turbines,
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radial flow turbines operate in a quite different
manner as compared to the axial flow turbines.
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So, last lecture was exclusively devoted towards
understanding of radial flow turbines, the
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basic working of radial flow turbines and
the thermodynamics of radial flow turbines.
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That is, as the flow passes through radial
flow turbine, what in a Thermodynamics sense
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happens to the flow, as it passes through
the different components of the turbine? We
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have discussed different types of radial flow
turbines, which are possible. The two broad
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classes are the outward flow radial turbines
and the inward flow radial turbines. We have
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seen that though the development, the original
development of radial flow turbines was in
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the outflow mode, very soon it was realized
that the inflow mode or the inflow radial
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flow turbine is a much more efficient way
of managing the flow, and achieving work output
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than the outflow type of turbine.
So, majority of the turbines, which are being
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used currently are inward flow radial turbines.
The radial turbines as a concept, was originally
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developed towards meeting the hydraulic power
requirement, that is, to convert hydraulic
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power into work output. And then, that continued
for a very long time. In fact, it even continues
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till date. And, the hydroelectric power plants
that we are aware of use one or more types
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of radial flow turbines. Of course, some of
them also have axial flow types, but apparently
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radial flow turbines are very commonly used.
And, the use of radial flow turbines for gas
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turbines was much later; that it was also
being considered as one of the options for
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use in gas turbine engine.
And, the modern day gas turbine technology
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has restricted the use of radial flow turbines.
It is primarily used in the smaller engines,
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smaller thrust class engines. And, it is the
main disadvantage, when it comes to use of
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radial flow turbines; for gas turbine applications,
is the fact that in order to achieve very
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high efficiencies and power output, the turbine
performance is a strong function of the inlet
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temperature. And in axial turbines, we have
already discussed that modern day gas turbines
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use different cooling technologies, which
can be used to, which can actually permit
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us to use turbine...temperatures, which are
much higher than their material limits.
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This is not. It is still possible in radial
flow turbines, but it is it is much more complicated
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to introduce and use some sort of cooling
mechanisms in radial flow turbines. There
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are complicated methods, which are being proposed
and probably being used in some types of engines.
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And, this is one major disadvantage. While
radial turbines are not really used in gas
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turbine technology, in gas turbine engines,
for larger thrust class engines because the
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turbine inlet temperature is sort of restricted.
We saw that there are two different types
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of inward flow turbines: the cantilever type
and the 90 degree inward flow turbines. And,
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we discussed in the previous class. We devoted
most of the time towards discussion on the
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90 degree inward flow turbine because that
is more commonly used.
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The cantilever type turbine is very similar
to the impulse turbine that we have already
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discussed in the axial turbine context. And
therefore, there are certain disadvantages
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associated with impulse turbine as we have
already seen. And, so the inward flow, 90
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degree inward flow turbine is like reaction
turbine. And, in geometrical terms, a 90 degree
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inward flow turbine looks exactly similar
to a centrifugal compressor. And, but of course,
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just at the direction of flow and the rotation
of the impeller or the rotor is exactly opposite
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in each cases.
We also discussed about the governing equations,
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which are used when we analyze the flow in
these different components. Starting from
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the outlet flow, the volute or the scroll,
and then the nozzle blades, and then it goes
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into the rotor or impeller. And then, there
is the exducer, which forms the later part
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of the impeller. Its counterpart and centrifugal
compressors was the inducer. And then, in
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typical turbine, we may also have a diffuser,
downstream of the rotor to recover part of
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the kinetic energy, which would otherwise
be lost.
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And, I think, we also discussed two other
aspects of radial flow turbines. They are
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the efficiency and performance parameters.
We defined different forms of efficiency,
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the total-to-static efficiency and how it
is related to total efficiency and the work
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output and so on. We also discussed or spent
quite some time discussing about the different
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loss parameters, which are used in radial
turbines, like the nozzle flow coefficient
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on the rotor flow coefficient. We also look
at the incidence losses, which is the flow
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entering the rotor, need not necessarily be
at the 0 incidence. Under off-design conditions,
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the incidence angles could be quite high,
which leads to a substantial increase in the
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losses; that is, an additional component of
loss that comes into picture when a turbine
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is operating in an off-design condition.
So, these are the different loss mechanisms
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that we discussed in a little bit detail.
We, then of course, spent lot of time discussing
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them because I think, it is fairly out of
the scope of this course to discuss the loss
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mechanisms and design optimization techniques
in detail; that it is a vast subject on it
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is own. So, I decided not to spend too much
time on that. We will discuss some aspects
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of design and performance in the next class.
But, today's class I thought it is a good
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idea to have a tutorial session. We shall
be discussing a few problems, which I will
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solve for you here. And then, I also have
a couple of exercise problems, which I think,
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you should be able to solve based on what
we had discussed today as well as in the previous
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class.
So, today's lecture is going to be a tutorial
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session. So, let us take a look at the first
tutorial problem that we have. Let us take
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a look at what the problem statement is.
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So, the problem number 1 that we have for
today is the following. The rotor of an inward
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flow radial turbine, which is designed to
operate at the normal condition is 23.76 centimeters
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in diameter and rotates at 38,140 revolutions
per minute. At the design point, the absolute
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flow angle at the rotor entry is 72 degrees.
The rotor mean exit diameter is one half of
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the rotor diameter and the relative velocity
at the rotor exit is twice the relative velocity
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at the rotor inlet. Determine the specific
work done.
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So, in this particular problem, it is basically
a very simple problem that, of course the
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first problem that I normally solve is a very
simple problem. We have some dimensions of
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the rotor; we have rotor diameter and we have
the rotational speed and the nozzle inlet
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angle. We also have been given that the rotor
mean diameter is one half of the rotor exit
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diameter and the relative velocity at rotor
exit is twice the relative velocity at the
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rotor inlet. So, based on this data that we
have, we need to find the specific work done.
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So, I would say this is a very simple problem
but, again as I keep emphasizing every time
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I have a tutorial session is that, we start
solving a problem with the velocity triangles.
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So, let us construct the velocity triangles
in this particular case. And then, we shall
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see and proceed towards solving this problem
because velocity triangle will help us to
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understand, what the known parameters are
and what are those parameters, which we need
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to estimate and calculate?
So, for a 90 degree IFR turbine, we have the
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velocity triangles as shown.
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Here, we had already discussed this in the
last class. Let me, again quickly explain
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the construction of a 90 degree IFR turbine.
We have volute or a scroll, which sort of
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acts like chamber, through which the flow
enters stagnation chamber here, where the
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flow is stagnant and then that gets accelerated
through the nozzle blades. So, these are the
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nozzle blades.
Nozzle exit station is denoted by station
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two. The flow then passes through the rotor
blades or the impeller and exits at station
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three before going into the diffuser, which
is the exit of the diffuser denoted by station
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four. You can see that, as I mention that
is very similar to centrifugal turbine, a
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centrifugal compressor. In that case, of course,
the velocity, the flow direction is the other
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way round. It is reverse the flow, actually
proceeds in this direction. And, instead of
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nozzle blades, we have diffuser vanes. And
therefore, this aerofoil orientation will
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also be the reverse. Let us look at the velocity
triangles for this case. We normally have
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V 2 which is entering the rotor radically.
So, the nozzle flow leaves relative velocity;
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entering nozzle is in the 90 degrees and C
2 is at an angle of alpha 2. In this case,
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it is given as 72 degrees. So, alpha 2 in
this particular case is 72 degrees. U 2 is
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the blade speed at the station, that is, station
two.
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Now, as the flow leaves the rotor, the absolute
velocity leaving the rotor is axial. Under
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design condition, relative velocity that an
angle of beta 3 with the acceleration, that
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is, B 3 and U 3 is the rotor speed at station
three. So, this is typical inward flow radial
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turbine and the corresponding velocity triangles
at the rotor inlet and rotor exit. So, based
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on the data that we have, for this case we
basically have the rotational speed, we have
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the exit diameters. So, I think we should
able to find out U 2 and then subsequently,
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we also have been given some ratio of the
blade speed at the rotor exit to the mean
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diameter and the relative velocity at rotor
inlet and exit. So, with this data, we should
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be able to find the specific work done.
Now, let me recall what we had discussed in
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the last class, when we had derived a very
general expression for a 90 degree IFR turbine,
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where the specific work was, if you recall
a function of three distinct parameters. One
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is a difference between the blade speed at
the inlet and outlet is U 2 square minus U
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3 square. The second term was a function of
the relative velocity and third term function
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of the absolute velocities. So, we are going
to do exactly the same thing here to calculate
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the specific work done. Let us calculate these
three individual components, and then add
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up all of these and that gives us the specific
work done. So, specific work done was 1 by
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2 into three different terms, U 2 squares
minus U 3 square plus W 2 square V 2 square
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minus V 3 square. And, the third term was
the absolute velocity. So, let get these individual
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terms first, add them up and then we get the
specific work done.
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So, blade speed at the tip is U 2, which is
basically pi D into N divided by 60. And,
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so n has been given as 38,140 revolutions
per minute, the diameter is given as 23.76
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centimeters. So, it is 0. 2376. This divided
by 60. So, here U 2 comes out, may be if you
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substitute these values, we get U 2 as 474.50
meters per second.
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Now, this is where the velocity triangles
come into picture that, if we need to calculate
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V 2, if you have the velocity triangles in
front of us it is very straight forward, it
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is ratio between V 2 and U 2 is related by
tan alpha 2 or cotangent alpha 2 or cot alpha
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2. So, let us look at the velocity triangle
V 2 and U 2. So, tan alpha 2 is U 2 by V 2
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and therefore, V 2 is U 2 into cot alpha 2.
Alpha 2 is given as 7 2 degrees and therefore,
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V 2 the relative velocity at rotor inlet is
154.17 meters per second. Similarly, C 2 is
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U 2 sin alpha 2. C 2 is this, and this. So,
sin alpha 2 is U 2 by C 2 and therefore, C
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2 is U 2 sin alpha 2. And, since alpha 2 is
70 degrees, we get C 2 as 498.90 meters per
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second.
Now, C 3 square, that is, the absolute velocity
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at the exit, C 3 square is V 3 square minus
U 3 square. V 3 is given as twice of V 2,
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and, that is, 2 into 154.1 7 square. And,
U 3 is related to U 2 because the diameter
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at U 3, station U 3 is half that of U 2. Therefore,
U 3 is 0.5 times U 2. So, 2 into V 2 and 0.5
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times U 2.
So, what we get is that the square of this.
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We basically get C 3 square. C 3 square is
38, 786 meters squared per second squared.
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Similarly, let us find out the three individual
components, U 2 square minus U 3 square. This
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is U 2 square into 1 minus 1 by 4; this is
1 by 2 square. And, so it is 1 by 4. This
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should be 1, 68, 863 meter squared per second
squared. The second term is V 3 square minus
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V 2 square, that is, 3 into V 2 square because
V 3 is twice of V 2. So, this is 71, 305 meter
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squared per second squared. The third term
is C 2 square minus C 3 square. We already
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know C 2 and we know C 3 square. So, that
difference is 2, 10, 115 meter squared per
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second squared.
So, these are the three individual components,
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which contribute towards the specific work
done. So, the next specific work done would
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be 1 by 2 times the sum of these three components.
So, we add up all the three, divide that by
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two we get the specific work done.
Now, this is one way, while probably the direct
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way of calculating specific work done. we
can also approximate specific work done without
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having to undergo any of this, but of course,
this is an approximate estimate of the specific
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work done, that is, simply equal to the square
of the blade speed at the exit. So, delta
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W is U 2 square and why is it U 2 square?
That is because flow enters the rotor and
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leaves the rotor in the axial direction. So,
C w 3 is 0, C w 2 will basically be equal
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to U 2, specific work done is U 2 C w 2 minus
U 3 C w 3. The second term would become 0;
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the first term is equal to U 2 square. So,
delta C w should also be equal to simply U
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2 square. So, if we do that if we calculate
work done in both ways, let us see what happens.
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So, if we add up all the three different terms
and divide by 2, we get the specific work
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done as 225, 142 meter squared per second
squared. And, what we see is that, the fractional
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contribution of each of these three terms.
The first term which is U square is .375.
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V square is .158 and C square is .467, that
is, you can see that there is a fair share
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for all these three components in the specific
work done.
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So, we can calculate specific work done also
by the second method, which is basically because
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in this particular case, beta 2 is 0 and alpha
3 is also 0. In which case, delta W simply
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becomes U 2 square. So, if you simply square
474.5 whole squares, you get 2 25,150 meter
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squared per second squared. You can see that,
they are very close. Specific work done calculated
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in either of these ways, should give you the
correct answer. Both these methods give you
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identical specific work done. So, one way
is to calculate the individual components
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and add up all of them, which is a more general
method because irrespective of how the velocity
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triangles are, you could, that is, still valid;
whereas delta W is equal to U 2 square is
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valid, only if in this case. Like in this
case, the incidence is 0, the deviation also
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0, in which case you can directly calculate
delta W. And, one would get the same delta
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W expecting the round of errors, which is
seen in the both this calculation.
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So, this first problem as you can, already
as we have seen is a very simple problem,
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which involves simple application of mine
to solving the velocity triangle to calculate
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the different components or constituents of
the specific work done. One is blade speed;
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the other is relative velocity and the absolute
velocity. We add up all three, divide them
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by 2; we should get the specific work done.
So, let us now, move on to the next problem
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that I have for you. And, it is a slightly
more involved problem, but of course, again
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I normally keep these problems limited to
very fundamental aspects of the particular
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aspect that we have designed working on. In
this case it is the radial turbines. So, we
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are just looking at very fundamentally thermodynamics
of radial turbine and how we can apply some
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of these principles to calculate some parameters
associated with radial flow turbines.
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Let us take a look at the second question
we have for today.
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Problem statement number 2 is: a radial inflow
turbine develops 60 kilowatts power when running
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at 60,000 revolutions per minute. The pressure
ratio P 0 1 by P 3 of the turbine is 2.0.
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The inlet total temperature is 1,200 Kelvin.
The rotor has an inlet tip diameter of 12
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centimeters and an exit tip diameter of 7.5
centimeters. The hub to tip ratio at the exit
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is 0. 3. The mass flow rate is 0.35 kilograms
per second. The nozzle angle is 70 degrees
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and the rotor exit blade angle is 40 degrees.
If the nozzle loss coefficient is 0.07, determine
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the total-to-static efficiency of the turbine
and the rotor loss coefficient.
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So, here we can see, of course that the problem
involves lot more data than we had for first
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case. We had the power input or power developed
by the turbine is 16 kilowatts, the rotational
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speed is 60000 revolutions per minute. The
pressure ratio is 2. Turbine inlet temperature
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is 1200. The dimensions of the rotor the tip
diameter is 12 centimeter, exit tip diameter
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of 7.5 centimeter. Hub to tip ratio is 0.3.
Mass flow rate and the angles, and additionally,
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the fact that the nozzle has a loss coefficient
of 0.07.
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We need to find total-to-static efficiency
and the rotor loss coefficient. So, this is
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the problem statement for this second question
that we have. As always, we will first start
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with the velocity triangle. It is exactly
the same, as we have seen in the first problem.
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Nevertheless, let us just quickly look at
the velocity triangles and understand, what
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are the data provided for using this question
and what is that we need to find.
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So, in this case, the velocity triangle again
is the same as we have seen in the previous
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question. This angle is given to us as 70
degrees and the exit blade angle beta 3 is
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given as 40 degrees. We have the dimensions
at station two and station three and hub to
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tip diameter ratio is also given to us. We
have the power output and the rotational speed.
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So, quite a bit of data is given to us. Rotational
speed is given; means U 2 minus U 3 already
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known, since angles are specified the other
components can also be calculated.
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So, let us start calculating some of these
parameters. The rotor tip rotational speed
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is we can calculate from pi D N by 60, which
is 377 meters per second. And here, D 2 is
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given in this question. It is given as 12
centimeters; rotational speed is 60,000 revolutions
23:08.590 --> 23:14.559
per minute. So, if we substitute pi and the
rotational speed and the diameter, is divided
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that by 60, we get 377 meters per second.
Now, from the velocity triangle we know that,
23:22.970 --> 23:29.970
at the rotor inlet beta 2 is 0. And therefore,
sin alpha 2 is simply U 2 by C 2, where C
23:32.490 --> 23:39.490
2 is U 2 into cosecant alpha 2. Alpha 2 is
given as 70 degrees. And, U 2 we have already
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calculated as 377 meters per second. And therefore,
C 2 is equal to 377 into cosecant 70 degrees,
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that is, 401.185 meters per second.
T 0 2 is given to us; turbine inlet temperature
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is 1200 kelvin. Since C 2 is now calculated,
we can calculate T 2 static temperature, that
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is, T 0 2 minus C 2 square by 2 C p. That
should come out to be 1130 kelvin.
24:10.200 --> 24:17.200
So, this is the preamble of a question, that
is, we solve some simple parameters, which
24:17.289 --> 24:21.559
anyways is required for solving the rest of
the problem. So, first part of the question
24:21.559 --> 24:27.059
is to find the total- to- static efficiency.
And, that is where, we will make use of the
24:27.059 --> 24:33.009
pressure ratio that we have been given to,
we have given as 2 .0.
24:33.009 --> 24:38.029
From the pressure ratio, we should able to
use that data to calculate the total-to-static
24:38.029 --> 24:42.059
efficiency. And, so that the next part of
the question, we are going to solve is to
24:42.059 --> 24:49.059
find the stagnation temperature drop across
the turbine. And, from the power output, we
24:49.159 --> 24:56.059
can actually find the isentropic or the actual
power output, actual temperature drop across
24:56.059 --> 25:03.059
the turbine. We should take the ratio of the
two; we get the total-to-static efficiency.
25:03.110 --> 25:10.110
So, the stagnation temperature drop, which
is given by T 0 1 minus T 3 s is equal T 0
25:11.179 --> 25:18.179
1 into 1 minus T 3 s by T 0 1. We can convert
this ratio into the pressure ratio because
25:18.820 --> 25:25.820
this ratio has been given as 1 by 2, that
is, 0.5. T 0 1 is known to us. And therefore,
25:26.899 --> 25:30.990
this temperature drop is required for calculating
the total-to-static efficiency because total-to-
25:30.990 --> 25:37.990
static efficiency is T 0 1 minus T 0 3, divided
by T 0 1 minus T 3 s.
25:38.610 --> 25:45.610
So, T 0 1 is 1200. This multiplied by 1 minus
0. 5, rise to gamma minus 1 by gamma. You
25:47.330 --> 25:53.799
can calculate this. And, this should come
out to be 190.92 Kelvin. Turbine power output,
25:53.799 --> 26:00.799
as we know is m dot C p into T 0 1 minus T
0 3. Power output is given as 60 kilo watts;
26:01.419 --> 26:07.200
mass flow rate is 0.35 kilograms per second.
C p, we are assuming for gasses as 1.148.
26:07.200 --> 26:14.200
Therefore, T 0 1 minus T 0 3 is 149.33 kelvin.
So, if we simply take of ratio, this total-to-static
26:18.269 --> 26:25.269
efficiency is 149.33 kelvin divided by 190.
92. And, that is 0.782. So, total-to- static
26:29.380 --> 26:36.380
efficiency in this case is 0.782.
So, that solves the first part of the question
26:36.450 --> 26:42.070
where we are required to calculate the total-to-static
efficiency. Now, the next part of the question
26:42.070 --> 26:49.070
involves or requires us to calculate the rotor
loss coefficient. We have been given the nozzle
26:49.740 --> 26:56.740
loss coefficient as 0. 07. And, so we will
need to make use of that, rather complex formula
26:57.509 --> 27:03.149
that we had derived in the last class. If
you remember it was a very long formula, which
27:03.149 --> 27:08.330
was relating the last coefficient to the radius
ratio and the static temperature ratios.
27:08.330 --> 27:14.960
So, we will make use of that formula to calculate
the nozzle, well the rotor loss coefficient
27:14.960 --> 27:20.200
and that formula involves both the nozzle
coefficient, as well as rotor loss coefficient
27:20.200 --> 27:27.200
multiplied by the angles and so on. So, we
know that in this case, the radius ratio because
27:28.029 --> 27:35.029
this will be required in that formula. Radius
ratio r 3 by r 2 is basically the hub diameter
27:37.419 --> 27:42.769
at station plus the shroud diameter divided
by 2, that is, the mean diameter divided by
27:42.769 --> 27:49.769
diameter station two. And, this we know the
hub to tip ratio is given as 0.4. So, we substitute
27:51.370 --> 27:58.370
zeta here and that is basically 0.4 in this
question. This is multiplied by the d 3 s
27:58.389 --> 28:02.659
plus d 3 s divided by 2 into d 2. So, if we
substitute these diameters, which have been
28:02.659 --> 28:09.659
given, as well as the hub to tip ratio we
can get the radius ratio as 0. 406.
28:12.919 --> 28:19.919
Now for efficiency this was the long expression,
I was mentioning total-to-static efficiency
28:20.779 --> 28:27.779
is 1 plus 1 by 2 into zeta n, the nozzle coefficient,
T 3 by T 2 into cosecant square alpha 2 plus
28:29.850 --> 28:36.850
r 3 by r 2 square into zeta r, which is the
rotor loss coefficient to cosecant square
28:36.919 --> 28:42.059
beta 3 plus cot square beta 3, this whole
raise to minus one.
28:42.059 --> 28:49.059
Now, this, though it looks rather complex
formulae, we have derived this expression
28:50.179 --> 28:56.080
right from the first principle. So, if we
know, if we can actually relate the first
28:56.080 --> 29:00.960
principle formulae of total-to-static efficiency,
which is T naught 1 minus T naught 3 by T
29:00.960 --> 29:07.490
naught 1 minus T 3 s, the denominator get
expressed in terms of the nozzle and the stator
29:07.490 --> 29:13.620
or the rotor loss coefficients. Numerator
is expressed in terms of static temperature
29:13.620 --> 29:16.940
ratios and so on.
So, from that, we can actually derive this
29:16.940 --> 29:23.940
without much difficulty. The numerator actually
becomes C p times U into C p times C w delta
29:26.570 --> 29:32.240
C w. And then, delta C w we express in terms
of the angles and so on. So, it is a very
29:32.240 --> 29:37.710
simple two to three step derivation from which
we can derive this long, rather longest expression,
29:37.710 --> 29:43.470
which we see here for total-to-static efficiency.
Now, in this expression, we still have an
29:43.470 --> 29:49.200
unknown that is T 3 by T 2. T 2 we have already
calculated, but we do not know the value of
29:49.200 --> 29:56.200
T 3, which also can be. In fact calculate,
provided we know the exit stagnation temperature
29:58.320 --> 30:05.019
and the exit absolute velocity. So, if that
is known, we can actually, we should be able
30:05.019 --> 30:10.750
to calculate because in this case, beta 3
is given and from the velocity triangle, since
30:10.750 --> 30:17.750
beta 3 is given, we can. And, U 3 is known,
we can calculate C 3. What about T naught
30:19.000 --> 30:22.179
3?
For T naught 3, the power output is given,
30:22.179 --> 30:28.149
inlet stagnation temperature is given. So,
we can calculate T naught 3 from there. T
30:28.149 --> 30:35.149
naught 3 minus C 3 square by 2 C p will give
us T 3. And then, we can take ratio T 3 by
30:35.320 --> 30:41.090
T 2. That would be deriving the whole thing
from the first principle or the whole thing
30:41.090 --> 30:47.950
can be expressed in a single definition term,
which also I think I had mentioned in the
30:47.950 --> 30:52.330
last class. Which is basically, in terms of
some of these parameters, which we know and
30:52.330 --> 30:59.070
T 3 by T 2 is actually defined terms of the
velocities and loss coefficients. 1 minus
30:59.070 --> 31:06.070
U 2 square by 2 C p, T 2 into 1 plus r 3 by
r 2 whole square multiplied by 1 plus zeta
31:06.240 --> 31:10.399
r cosecant square beta 3 minus 1 into cot
square alpha 2.
31:10.399 --> 31:16.379
And, so if you substitute for these values
here, the only unknown is zeta r. So, we should
31:16.379 --> 31:23.379
get 0.9396 minus 0.02187 zeta r. So, this,
if you substitute in this expression where
31:24.590 --> 31:30.970
zeta n is also known, you get a quadratic
equation and you can solve that zeta r, which
31:30.970 --> 31:35.399
is the rotor loss coefficient can be calculated
as 0.62.
31:35.399 --> 31:42.399
So, this is one way of calculating zeta r.
The other way, of course is to calculate T
31:43.519 --> 31:47.929
3 by T 2 using what I had mentioned. T 2,
we have already calculated. We know what the
31:47.929 --> 31:54.749
value of T 2 is. For calculating T 3, it involves
two to three steps. One is to calculate the
31:54.749 --> 32:00.039
stagnation temperature at exit T naught 3,
which can be calculated from the power expression.
32:00.039 --> 32:06.039
Power is equal to mass flow rate into C p
into delta T. T naught 1 minus T naught 3.
32:06.039 --> 32:10.129
So, all the three parameters are there. All
the parameters are known except T 0 3 we can
32:10.129 --> 32:17.129
calculate stagnation temperature. Then, to
calculate T 3, we also need to know the velocity
32:17.840 --> 32:24.840
at the exit, that is, C 3. And, to calculate
velocity at the exit C 3, we know beta 3,
32:27.129 --> 32:32.480
we also know U 3. And, how do we know U 3?
Because rotational speed is given to us and
32:32.480 --> 32:36.649
the diameter at the hub or at the exit of
the rotor is also known.
32:36.649 --> 32:43.649
So, from that we can calculate U 3. Since
U 3 is known, beta 3 is known and we can calculate
32:44.490 --> 32:51.490
C 3 that is basically, U 3 cot alpha 3 should
be equal to C3. So, once C 3 is also known,
32:53.720 --> 32:59.570
the static temperature as at the exit T 3
is equal to stagnation temperature T 0 3 minus
32:59.570 --> 33:05.429
C 3 square by 2 C p. And, so that, it is a
rather easier way rather than this longest
33:05.429 --> 33:09.019
formula, I just shown.
And then, substitute for that in the expression
33:09.019 --> 33:16.019
and we should able to calculate zeta r, which
is the rotor loss coefficient. Nozzle loss
33:16.360 --> 33:22.809
coefficient is already known to us. It is
0.07. So, what you can see is that the loss
33:22.809 --> 33:29.389
coefficient for the nozzle. This case has
been given as 0.07. And, for the rotor it
33:29.389 --> 33:36.389
is 0.62. And, so I think, I have also given
some range of these values in the last class.
33:37.129 --> 33:43.299
I had mentioned typically, the rotor loss
coefficient are on the higher side because
33:43.299 --> 33:47.759
of the fact that there are rotational effects
come into picture losses, associated with
33:47.759 --> 33:53.480
the rotor are much more than the rotor losses
associated with the stationary component like
33:53.480 --> 33:59.759
a nozzle; especially when the flow is accelerating.
So, in this case the rotor loss coefficient
33:59.759 --> 34:05.919
is, in fact close to one order magnitude higher
than the nozzle loss coefficient.
34:05.919 --> 34:12.610
So, this solves our second problem, which
required us to calculate the total-to-static
34:12.610 --> 34:19.610
efficiency as well as the rotor loss coefficient.
So, you can clearly see that this is slightly
34:21.580 --> 34:25.899
more involved question. In this, of course
I have taken the easier route of directly
34:25.899 --> 34:30.869
substituting this in the formulae. What I
would strongly urge you to do and probably
34:30.869 --> 34:35.710
leave that as an exercise for you to derive
these equations from the first principles
34:35.710 --> 34:41.929
and not simply use the direct longest formulae.
It is very easy to derive the equations from
34:41.929 --> 34:48.710
the first principles. In the total-to-static
efficiency definition term, it is basically
34:48.710 --> 34:55.710
T 0 1 minus T 0 3 divided by T 0 1 minus T
3 s. Or, let us express that in terms of enthalpy,
34:56.159 --> 35:03.159
h 0 1 minus h 0 3 divided by h 0 1 minus h
3 s. The denominator gets expressed in two
35:03.440 --> 35:10.440
separate forms. One is to do with nozzle loss
coefficient; second is the rotor loss coefficient.
35:11.530 --> 35:17.790
Numerator gets expressed in terms of mass
flow rate C p and delta T and so on. So, from
35:17.790 --> 35:24.790
this, you can actually say denominator has
a nozzle and rotor loss coefficient term.
35:25.700 --> 35:29.809
Numerator is already known to us.
So, this can be simplified and you can actually
35:29.809 --> 35:36.809
calculate the rotor loss coefficient, given
the nozzle loss coefficient in a much simpler,
35:40.460 --> 35:45.650
less confusing manner than simply substituting
them in the formulae. I picked up this method
35:45.650 --> 35:50.220
because if we take up any textbook, you would
normally see this kind of method where they
35:50.220 --> 35:57.220
would refer to the derivation, which was discussed
earlier on what I have done. And, in the problem
35:57.220 --> 36:01.700
we just simply substitute, plug in the values
and calculate the corresponding efficiency
36:01.700 --> 36:07.050
and other terms required.
Now, that brings us to the third question.
36:07.050 --> 36:13.069
Let us, now proceed towards the third problem
that we have for us to solve. And then, we
36:13.069 --> 36:17.630
will see how this problem is different from
the previous problems.
36:17.630 --> 36:24.630
An inward flow turbine with 12 vanes is required
to develop 230 Kilowatts at an inlet stagnation
36:25.619 --> 36:32.619
temperature of 1,050 kelvin and a flow rate
of 1 kilogram per second. Using the optimum
36:34.160 --> 36:41.160
efficiency design method and assuming a total-to-static
efficiency of 0.81, determine the absolute
36:41.970 --> 36:47.800
and relative flow angles at the rotor inlet;
part b is the overall pressure ratio, P 0
36:47.800 --> 36:54.800
1 by P 3; part c is the rotor tip speed and
the inlet absolute Mach number.
36:55.069 --> 37:00.270
So, this is a three part question where we
are required to calculate three different
37:00.270 --> 37:04.490
part aspects. One is the angles, absolute
and relative flow angles and the pressure
37:04.490 --> 37:11.369
ratio, and of course, the rotors tip speed
and the Mach number. And of course, you can
37:11.369 --> 37:16.260
see what is mentioned here is that we can
assume optimum efficiency design method. I
37:16.260 --> 37:21.520
mentioned some preliminary aspects of this
in the last class. I would urge you to take
37:21.520 --> 37:28.010
up slightly more detail reading of this in
any of the text books that we have mentioned.
37:28.010 --> 37:33.819
You can take up, pickup any book on the turbo
machines and we will find a small section
37:33.819 --> 37:40.059
and optimum efficiency design in methodology,
wherein a few formulae would basically be
37:40.059 --> 37:45.530
derived again from the first principles. And,
you can see what it is basically trying to
37:45.530 --> 37:49.540
tell us.
So, in this question, it is question which
37:49.540 --> 37:56.040
involves optimum efficiency design methodology,
which can be assumed. And then, we can, we
37:56.040 --> 38:01.809
are required to calculate the angles, the
Mach number and so on. So, the first part
38:01.809 --> 38:07.809
of the question is to find the flow angles;
the relative and absolute flow angles.
38:07.809 --> 38:14.809
For optimum design, it is known that the absolute
angle at the nozzle exit, that is, alpha 2
38:16.780 --> 38:22.359
is simply related to the number of blades.
And, this comes from what is known as the
38:22.359 --> 38:29.359
Whitfield's formula, which basically equates
or which basically relates the alpha 2 to
38:29.950 --> 38:34.690
the number of blades. So, cos square alpha
2 is equal to 1 by N; so, that is, where n
38:34.690 --> 38:36.020
is the number of vanes.
38:36.020 --> 38:43.020
So, for optimum design, it actually has been
shown by Whitfield that, cos square alpha
38:43.809 --> 38:50.230
2 is 1 by n. So, that is, where n is the number
of vanes. So, in this equation we have 12
38:50.230 --> 38:56.670
vanes and since we have been told to assume
optimum design, we can simply substitute the
38:56.670 --> 39:03.670
number of vanes here and calculate alpha 2.
So, alpha 2 would be 73.22 degrees. Also,
39:04.190 --> 39:11.190
another consequence of the Whitfield's equation
is that, we have beta 2 is equal to 2 into
39:11.589 --> 39:18.589
90 minus alpha 2. Beta 2 is the blade angle
at the inlet of the rotor. And, that is equal
39:20.140 --> 39:26.200
to 2 into 90 minus alpha 2. And, so this comes
to be 33.56 degrees.
39:26.200 --> 39:33.200
So, the next part of the question is to find
the pressure ratio, well, basically the total-to-static
39:34.069 --> 39:41.069
efficiency. And, for which, of course we will
use the total-to-static efficiency basically
39:42.990 --> 39:49.990
is T 0 1, 1 minus T 0 3 by 3, T 0 1 minus
T 3 or T 3 s. And, from there we can express
39:51.569 --> 39:58.530
the denominator in terms, P 3 by P 0 1. And,
in this question, we know that the power developed
39:58.530 --> 40:04.890
is given as 230 kelvin, inlet stagnation temperature
is given, mass flow rate is given and the
40:04.890 --> 40:08.730
total-to-static efficiency is given. So, we
need to find the pressure ratio in this case.
40:08.730 --> 40:15.730
So, all we have do is, we just substitute
the power output here, which is m dot into
40:16.250 --> 40:23.250
C p into delta T. C p is known, assumed and
the stagnation temperature is also can be
40:23.680 --> 40:28.450
calculated or it is given at the inlet.
Since the efficiency is known, we can calculate
40:28.450 --> 40:35.450
the pressure ratio, P 3 by P 0 1 as 0.32165.
The inverse of this is the turbine pressure
40:35.520 --> 40:42.520
ratio, which is P 0 1 by P 3 and that is 3.109.
So, this basically comes from the efficiency
40:44.480 --> 40:51.480
definition eta T s is T 0 1 minus T 0, T 0
3 divided by T 0 1 minus T 3 s. The denominator
40:54.640 --> 41:00.329
gets expressed in terms of the pressure ratio,
from the isentropic relation numerated is
41:00.329 --> 41:06.520
simply C p times delta T. And therefore, that
is the power output of the turbine.
41:06.520 --> 41:12.410
Now, the third part of the question is to
find the Mach number and then, we should also
41:12.410 --> 41:18.700
need to find the blade speed at the tip of
the rotor. So, we will first find the absolute
41:18.700 --> 41:24.059
Mach number at the inlet, for which we will
find first the Mach number corresponding to
41:24.059 --> 41:25.680
stagnation conditions.
41:25.680 --> 41:32.680
So, M 0 2, this again is coming from the very
basic definition for delta M 0 2 square is
41:34.020 --> 41:41.020
equal to delta W by gamma minus 1 into 2 cos
beta 2 by 1 plus cos beta 2. Now, where does
41:41.440 --> 41:46.579
this equation come from? Again I will urge
you to try to derive this equation from the
41:46.579 --> 41:53.579
fundamentals. Delta W is h 0 1 minus h 0 3,
which is m dot C p into T 0 1 minus T 0 3.
41:56.740 --> 42:01.970
And, there the temperatures can actually be
expressed in terms of the corresponding velocities.
42:01.970 --> 42:08.970
And therefore, we can express the Mach number
in terms of the, well, part of the temperature
42:10.079 --> 42:15.309
get expressed in terms of Mach number, wherein
this flow angles alpha 2 and beta 2 will also
42:15.309 --> 42:18.290
come into picture.
So, from those fundamental equation we can
42:18.290 --> 42:24.089
actually derive the stagnation Mach number,
which is M 0 2 Mach number at the inlet of
42:24.089 --> 42:31.089
the rotor, M 0 2 square is delta W by gamma
minus 1 into 2 cos beta 2 by 1 plus cos beta
42:33.210 --> 42:38.180
2. So, all the parameters on the right hand
side are known to us. We just, substitute
42:38.180 --> 42:45.180
these different values and we get M 0 2 is
0.7389.
42:45.339 --> 42:51.720
The absolute Mach number which is basically
based on this static conditions, M 2 is related
42:51.720 --> 42:58.720
to M 0 2 as we know. So, M 2 square is M 0
2 square by 1 plus gamma minus 1 by 2 M 0
42:59.309 --> 43:04.780
2 square. This again follows from the isentropic
relations. We have already seen that, it is
43:04.780 --> 43:09.630
the relation between stagnation temperatures
to static temperature, stagnation pressure
43:09.630 --> 43:14.770
to static pressure and so on. We can also
relate the corresponding Mach numbers in this
43:14.770 --> 43:19.549
way.
So, since we have calculated M 0 2, we substitute
43:19.549 --> 43:25.079
that here and then, we get the static Mach
number based on static conditions that the
43:25.079 --> 43:32.079
absolute Mach number as 0.775. So, this is
again, this can be calculated in multiple
43:32.790 --> 43:38.430
ways. The other way, which I suggest you can
try to calculate would be to take the ratio
43:38.430 --> 43:45.430
of T, absolute velocity; that is, C 2 divided
by square root of gamma or T 2. Stagnation
43:46.990 --> 43:53.990
temperature is known at the inlet and to find
static temperature, we need T naught minus
43:54.440 --> 43:59.670
C squared by 2 C p. So, is basically we need
to calculate C square at the inlet. And, to
43:59.670 --> 44:06.630
calculate C square we will of course, need
the blade speed because you need to calculate
44:06.630 --> 44:13.170
because blade angle is known. And so, if you
know the blade speed at tip, that is U 2,
44:13.170 --> 44:16.630
you can solve C 2, and then, therefore, calculate
Mach number.
44:16.630 --> 44:19.680
In this case, of course we are doing at other
way round. We are calculating the Mach number
44:19.680 --> 44:24.500
first and then, we are now going to calculate
the blade speed at the tip. But, it can also
44:24.500 --> 44:29.270
be done the other way round.
So, let us now calculate the blade speed at
44:29.270 --> 44:36.270
the exit of the nozzle or, that is the entry
rotor, entry tip of the rotor. Delta W by
44:37.089 --> 44:44.089
C p T 0 1 as we know is equal to gamma minus
1 into coos beta 2 into U 2 square by a 0
44:44.849 --> 44:49.099
1 square.
So, here a 0 1 is speed of sound based on
44:49.099 --> 44:56.099
stagnation temperature, square root of gamma
r T 0 1. This can be calculated as 633.8 meters
44:56.630 --> 45:01.880
per second. Assuming, T 0 1 is equal to T
0 2. Since in this case, we know the power
45:01.880 --> 45:07.470
output as stagnation temperature and beta
2, we can simply substitute for these values
45:07.470 --> 45:14.470
and then, calculate the blade speed at the
tip U 2. And, that comes out as 538.1 meters
45:14.790 --> 45:19.460
per second. So, the other approach to calculate
Mach number would be to actually calculate
45:19.460 --> 45:25.309
the blade speed first and then, since blade
speed is known and alpha 2 is known, you can
45:25.309 --> 45:32.309
calculate C 2. And, from C 2 you can calculate
the static temperature T 2. T 0 2 minus C
45:33.210 --> 45:39.030
2 square by 2 C P. And therefore, Mach number
would be C 2 by square root of gamma or T
45:39.030 --> 45:43.200
2. So, this is the other way of calculating
the Mach number. We have calculated that in
45:43.200 --> 45:47.950
the slightly different way.
So, that completes the third problem that
45:47.950 --> 45:54.200
we had set aside for today's tutorial. So,
I have one more problem to solve, which is
45:54.200 --> 46:00.819
a very simple problem, basically not involving
in the calculation. But, this is just to compare
46:00.819 --> 46:06.329
performance or operation of two different
types of turbines that we have discussed about.
46:06.329 --> 46:11.359
One is the axial turbine, which we had rather
detailed discussion, several lectures earlier
46:11.359 --> 46:17.900
on and of course, the radial turbine. So,
let us compare under certain given operating
46:17.900 --> 46:23.559
conditions, how the work output of these two
different turbines can be calculated and how
46:23.559 --> 46:25.640
do they compare with each other.
46:25.640 --> 46:32.089
So, the fourth problem statement is the following:
Compare the specific power output of axial
46:32.089 --> 46:37.910
and radial turbines in the following cases.
Axial turbine, in this case is given as alpha
46:37.910 --> 46:44.910
2 is equal to beta 3 is 60 degrees and alpha
3 is equal to beta 2 is 0 degrees.
46:45.450 --> 46:50.970
For the radial turbine, alpha 2 is given as
60 degrees and beta 3, alpha 3, beta 2; all
46:50.970 --> 46:56.579
of them are 0. If the rotational speed is
the same in both this cases, we are required
46:56.579 --> 47:01.799
to calculate this specific power output. So,
what we can see is that immediately for the
47:01.799 --> 47:08.799
axial turbine, we have, we can see that alpha
is equal to beta 3 and alpha 3 is equal to
47:09.299 --> 47:15.599
beta 2. Immediately, tells us that this is
a 50 percent reaction turbine, which means
47:15.599 --> 47:21.059
that the velocity triangles would be symmetrical.
And, for the radial turbine we have already
47:21.059 --> 47:27.220
seen the velocity triangle, which is for the
nominal operation condition, where alpha 2
47:27.220 --> 47:34.220
is given, beta 2 is 0, alpha 3 and beta 3
are respectively 0 at the exit.
47:34.549 --> 47:39.960
Let us take a look at the velocity triangles
first. And, so here, we have the velocity
47:39.960 --> 47:45.140
triangle for both these cases, the axial as
well as radial turbine. Let us look at the
47:45.140 --> 47:51.609
axial turbine first. Velocity triangle at
the inlet is given by this. Alpha 2 is given
47:51.609 --> 47:58.609
as 60 degrees. And, so alpha 2 at 60 degrees
and that is equal to beta 3, that is, also
47:59.520 --> 48:06.520
60 degrees and beta 2 is 0, as you can see
and alpha 2 is 0, that is, C 3 makes 0 degrees
48:07.450 --> 48:11.530
with the axial direction.
And, since it is an axial turbine, we will
48:11.530 --> 48:18.530
assume U 2 is equal to U 3 is equal to U,
blade speed. It is for the same circumferential
48:19.170 --> 48:24.640
radial location. So, U 3 and U 2 are the same
and that is equal to U.
48:24.640 --> 48:30.839
And, these are the corresponding velocities,
C 2. At the inlet, the absolute velocity V
48:30.839 --> 48:36.510
2 is related and at the rotor exit, we have
C 3 and correspondingly V 3, which is the
48:36.510 --> 48:41.980
relative velocity.
For the radial turbine, we have been given
48:41.980 --> 48:48.980
that alpha 2 is again 60 degrees and beta
3 is also 60 degrees. And, this is typical
48:49.670 --> 48:56.380
velocity triangle for an inward flow radial
turbine. Since alpha 2 is 60 degrees, we have
48:56.380 --> 49:03.380
C 2, which is at 60 degrees to the radial
location here. V 2 is the relative velocity
49:04.190 --> 49:09.010
which is entering in the radial direction.
U 2 and U 3 they are not the same. They are
49:09.010 --> 49:15.960
different in this case. and, the exit of the
rotor we have beta 3, which is again 60 degrees
49:15.960 --> 49:22.000
and axial velocity which is C 3 and C a 3,
they are the same. V 3 is the relative velocity
49:22.000 --> 49:24.359
at the exit of the rotor.
49:24.359 --> 49:31.359
So, for the axial turbine, we have alpha 2
and beta 3 as 60 degrees; alpha 3 and beta
49:32.430 --> 49:37.760
2 as 0 degrees; specific work for this case,
in this fifty percentage reaction case, would
49:37.760 --> 49:44.760
be U times delta C w; U times C w 2 and C
w 3. Here as you can see, C w 3 would be equal
49:46.369 --> 49:51.740
to 0 for the axial turbine because the flow
is leaving in the axial direction. So, C w
49:51.740 --> 49:58.740
3 is 0 and C w 2 is equal to U 2. C w 2 is
the tangential component of C 2, which is
50:01.130 --> 50:07.880
equal to U 2. And therefore, the specific
work done for the axial turbine would be simply
50:07.880 --> 50:12.490
U square.
For the radial turbine, alpha 2 and beta 3
50:12.490 --> 50:19.490
are 60 degrees; beta 2 and alpha 3 are equal
to 0. And, specific work done is U 2 C w 2
50:20.559 --> 50:27.559
minus U 3 C w 3. In this case of course, C
w 3 is again 0, like in the case of axial
50:28.710 --> 50:35.710
turbine. And, C w 2 is at the inlet, is equal
to U 2. And, of course at the exit U 2 and
50:38.799 --> 50:45.799
U 3 are not the same. But, since C w 3 is
0, the radial turbine also develops a work
50:46.109 --> 50:51.799
which is equal to U square. In this case it
is, U 2 square. So, specific work done in
50:51.799 --> 50:58.190
this particular case one of them is for an
axial turbine and other is for radial turbine.
50:58.190 --> 51:03.900
For given these conditions for the same rotational
speed, both of these turbines generate the
51:03.900 --> 51:09.099
same work output. They both are functions
of square of the blade speed.
51:09.099 --> 51:15.589
So, in this specific case that for example,
that we have looked at because of the conditions
51:15.589 --> 51:21.780
that have been specified to us for the same
blade angles and rotational speeds, both these
51:21.780 --> 51:28.339
turbines generate the same work output. So,
just to give you an idea of how the work done
51:28.339 --> 51:32.730
can be calculated for different types of turbine
configurations, of course, we have seen axial
51:32.730 --> 51:39.270
turbines in greater detail, including a tutorial
session earlier on. And, this is for just
51:39.270 --> 51:45.940
to make a comparison between the work outputs
of these two different types of turbine configurations.
51:45.940 --> 51:52.940
So, that completes the fourth problem, as
well that we have solved today. And, what
51:52.980 --> 51:58.530
I have for you are two exercise problems,
which I would leave it for you to solve. And,
51:58.530 --> 52:04.740
of course I had also left few exercises in
between the couple of problems where I had
52:04.740 --> 52:09.619
requested that you should try to solve it
in a different way. The method I had used
52:09.619 --> 52:13.599
to solve is one of the ways of solving the
problems. You can also attempt to solve the
52:13.599 --> 52:17.809
problem in a different way, for which I had
given some hints. So, I suggest that you would
52:17.809 --> 52:23.839
also solve those problems using the other
alternative approach that I had suggested.
52:23.839 --> 52:30.619
So, let us take a look at the first exercise
problem that I have: a small inward radial
52:30.619 --> 52:37.619
flow gas turbine, comprising a ring of nozzle
blades, radial-vaned rotor and an axial diffuser,
52:37.950 --> 52:44.950
operates at nominal design point with a total-to-total
efficiency of 0.90. At the turbine entry the
52:45.130 --> 52:52.130
stagnation pressure and temperature of the
gas is 400 kilopascals and 1140 kelvin respectively.
52:55.359 --> 53:00.140
The flow leaving the turbine is diffused to
a pressure of 100 kilopascals and has negligible
53:00.140 --> 53:07.140
final velocity. Given that the flow is just
choked at the nozzle exit, determine the impeller
53:07.829 --> 53:12.299
peripheral speed and the flow outlet angle
from the nozzles.
53:12.299 --> 53:18.400
So, this question has an additional component,
that is, diffuser and it is given that at
53:18.400 --> 53:23.170
the exit of the diffuser the flow has negligible
velocity. So, we can assume that the flow
53:23.170 --> 53:30.170
exits the diffuser with almost 0 velocity,
and the flow at the nozzle exit is just choked,
53:30.859 --> 53:35.980
which means that Mach number at nozzle exit
is 1, velocity there would be equal to square
53:35.980 --> 53:42.980
root of gamma r T. And, stagnation temperature
is given to us. And, so that should help you
53:43.750 --> 53:49.280
in finding out the parameters at the rotor
entry. And, since rotor exit conditions are
53:49.280 --> 53:56.280
lot of fix. You can also calculate the conditions
at the rotor exit using the fact that the
53:57.130 --> 54:04.039
diffusion pressure is given and the fact that
velocity at the exit is close to 0. So, in
54:04.039 --> 54:11.039
this case the tip speed comes out 586 meters
per second and the angle alpha 2 is 73.75
54:12.339 --> 54:13.380
degrees.
54:13.380 --> 54:18.440
Now, the second question exercise problem
is: if the mass flow rate of gas through the
54:18.440 --> 54:24.140
turbine given in this problem 1, the previews
problem is 3.1 kilogram per second, the ratio
54:24.140 --> 54:31.140
of rotor axial width to rotor tip radius is
0. 1 and the nozzle isentropic velocity ratio
54:33.150 --> 54:40.150
is 0.96. Assuming that the space between the
nozzle exit and nozzle rotor entry is negligible
54:40.930 --> 54:47.210
and ignoring the effects of blade blockage,
determine the static pressure and static temperature
54:47.210 --> 54:54.210
at the nozzle exit; the rotor tip diameter
and rotational speed; the power transmitted
54:55.480 --> 55:02.220
assuming a mechanical efficiency of 93.5 percentage.
So, we need to use part of data which is given
55:02.220 --> 55:07.130
in the first question, to be able to solve
this question as well. So, answer to part
55:07.130 --> 55:14.130
1 is the pressure static pressure is 205.
80 kilopascal; temperature is 977 kelvin;
55:15.829 --> 55:22.829
the rotor tip diameter is 125.44 millimeters
and rotational speed is 89,200 revolutions
55:22.859 --> 55:29.859
per minute, power transmitter with this mechanical
efficiency would come out to be 1 megawatts.
55:30.059 --> 55:36.619
So, these are two exercise problems that I
have for you. You can solve this based on
55:36.619 --> 55:41.630
what we have discussed in the last couple
of lectures including today's. And, I would
55:41.630 --> 55:46.069
also suggest that you solve couple of those
problems, which we solved in today's lecture
55:46.069 --> 55:51.789
that is problem number 2 and 3 in a different
approach, from what we have solved in the
55:51.789 --> 55:56.670
tutorials. So, I would suggest that you also
solve those problems using a slightly different
55:56.670 --> 56:00.579
approach.
So, we would have just one more lecture on
56:00.579 --> 56:05.369
radial turbines, where we would discuss some
aspects of performance and some preliminary
56:05.369 --> 56:10.980
design aspects related to radial flow turbines.
So, these we will take up in the next class,
56:10.980 --> 56:13.380
which would be lecture number 37.