WEBVTT
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Hello and welcome to lecture number 23 of
this lecture series on turbo machinery aerodynamics.
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We have been talking about actual turbines
in the last several lectures, and we had a
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chance to discuss about different aspects
of axial turbines, starting from very fundamental
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thermodynamics of axial turbines, and moving
towards two- dimensional cascade analysis,
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and also the 3D analysis, which we are going
to do subsequently. So, based on what we have
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discussed so far, on the basic thermodynamics
and the two-dimensional analysis; as I mentioned
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in the last class, it is time that we, have
a tutorial session on understanding of axial
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turbines, and how to solve problems, which
are related to axial flow turbines, but before
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I go into the tutorial, let me just quickly
recap, what we have discussed in the last
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several lectures.
We had an introductory lecture on axial turbines,
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where we discussed about different types of
axial turbines, like well different types
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of turbines in general, like the axial, the
radial and the mixed flow turbines; of which
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the axial turbine happens to be the one which
is the most commonly used in, especially in
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aero engine applications and also in marine
as well as land based power plant applications;
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for a variety of reason, basically to do with
the efficiency and and the convenience in
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arranging axial turbines, as compared to the
other counter parts like the radial or the
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mixed flow turbines.
Now, when we talk about the axial turbines
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alone, we can further classify axial turbines
based on the nature of the flow through the
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turbine itself. So, based on this classification,
we could have either an impulse turbine or
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one could have reaction turbine; and so, the
turbine axial turbine may operate in either
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of these modes, either an impulse turbine,
where in the entire pressure drop takes place
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just in the nozzle; and there is no pressure
drop taking place in the rotor, the rotor
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simply deflects the flow and does not contribute
in any way in the pressure drop that is taking
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place in the turbine. So, this is true for
an impulse turbine.
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And the reaction turbine is one, where the
pressure drop is actually shared between both
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the nozzle as well as the rotor. So, part
of the pressure drop takes place in the nozzle
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and the remaining part of that takes place
in the rotor. And so, we have defined what
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is known as degree of reaction and how we
can calculate degree of reaction. We also
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had a session on, understanding the velocity
triangles, which are applicable for an axial
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turbine and also these different types of
axial turbine like the impulse or the reaction
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type of axial turbine. So, where in we discussed
about, how we can construct a velocity triangle
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starting from the fundamental principle.
And then, we discussed about the various losses,
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which occur in axial turbine and how some
of these losses can be quantified; one may
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have two-dimensional losses or one may have
3D losses like the secondary flows and deep
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leakage losses and so on. We had a rather
detail discussion on these losses in our discussion
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on axial compressors. And I think, I mentioned
that one could simply extend the discussion,
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which was applicable for axial compressors
also for an axial turbine, so which is why
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we restricted our discussion on losses to
the bare minimum, because it is already been
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covered.
We also defined different mechanisms or methods
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of calculating efficiency for axial turbines.
There are two commonly used efficiency definitions;
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the total to static efficiency and total to
total efficiency; each of them are used primarily,
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according to the applications, for which the
turbine is being used; for example, if if
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it is a land based gas turbine engine, which
is not expected to produce any nozzle thrust,
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one would like to expand the flow to the maximum
extent possible without too much of kinetic
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energy leaving the turbine. In this case,
we would define the total to static efficiency;
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we would want the flow to have or to reach
the static conditions without much dynamic
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head as it leaves the turbine.
Now, in an aero engine kind of application,
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where the turbine is meant only to drive the
compressor and some accessories; one would
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still want from kinetic energy available at
the turbine exhaust, which can be further
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expanded in in a nozzle and therefore, in
such applications, we would define the total
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to total efficiency. We have discussed some
of these topics in the last several lectures,
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and we also of course, discussed about the
performance characteristics of axial turbines
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very important. We have discussed how what
it is significance is in relation to the engine
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as a whole and the significance of matching
of turbine with compressors as well as matching
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of turbine with the nozzle.
So, these were some of the topics that we
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had discussed in the last few lectures; and
in today's lecture, which is basically a tutorial
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session; what we going to do is, to try and
solve a few problems. And use our understanding
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of the working of the axial turbines, and
put that to practice in terms of solving problems.
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So, what I have for you today is a set of
four problems on axial turbines, which I will
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solve for you; and after this, I would also
give you a few exercise problems, which I
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would like you to go ahead and solve based
on our discussion today, as well as our understanding
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of these concepts in the last few lectures.
So, this is, what we are going to discuss
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in today's class. We basically be having tutorial
session on axial turbines.
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So, let us go to the the first problem. So,
the first problem statement is the following,
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its states that a single stage gas turbine
operates at its design condition with an axial
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absolute flow at entry and exit from the stage.
The absolute flow angle at the nozzle exit
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is 70 degrees, and at the stage entry, the
total pressure and temperature are 311 kilo
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Pascal and 850 degree celsius respectively;
the exhaust static pressure is 100 kilo pascal;
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the total to static efficiency is 0.87 and
the mean blade speed is 500 meters per second.
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Assuming constant axial velocity through the
stage, determine part (a) the specific work
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done, part (b) the Mach number leaving the
nozzle, and part (c) the axial velocity, part
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(d) the total to total efficiency, and last
part is to find the stage reaction. So, this
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problem statement is to do with single stage
gas turbine, which is operating under certain
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design conditions, the inlet stagnation pressure
and temperature at given the exhaust static
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pressure is given, the total to static efficiency
and the mean blade speed are given, based
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on this data we are required to calculate
a variety of other parameters and solve this
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particular problem.
Now, as we have done in the past the with
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reference to turbo machines, the very starting
point of solving a problem if you recall,
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when we had discussed about compressors, is
to get the velocity triangles right. So, the
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first point is to draw the velocity triangles;
and that is the starting point of solving
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any a such problems associated with turbo
machines. So, let us construct a generic velocity
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triangle, here we are not given, whether it
is a reaction turbine or an impulse turbine
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nothing is mentioned. So let us construct
a generic velocity of triangle like what we
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had done a few lectures earlier on; and then
from there we will see what are the parameters,
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which have been specified, and what is that
we need to calculate to be able to solve this
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problem.
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So, let us construct a velocity triangle in
general. So, this is a very generic velocity
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triangle, which is well, which is also true
for a reaction turbine stage; it is not really
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an impulse turbine as you can see, it is an
indeed a reaction stage. So, station 1 denotes
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the nozzle entry; station 2 denotes the nozzle
exit or the rotor entry; station 3 denotes
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the rotor exit. So, these are the three parameters
that have been the station numbers that we
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have defined.
And then the flow enters the nozzle with a
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velocity of C 1, and it exits the velocity
absolute velocity of C 2; relative velocity
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at the nozzle exit is V 2, which is the velocity,
which the flow actually enters the rotor,
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and the corresponding blade angles also been
marked here; U is the mean blade speed the
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flow exits the rotor with the velocity of
V 3, and an absolute velocity of C 3 with
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the corresponding angles of alpha 3 and beta
3. So, this is a typical velocity triangle
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of an axial turbine stage; and, what we will
do is, we will take a look at what are the
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parameters that we know at this stage; so
from this velocity triangle, we basically
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know the blade speed that something, which
has been specified, and we also know the inlet
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conditions and exit conditions with static
pressure, and we are required to calculate
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a variety of other parameters.
So, let us, begin with the part (a) that is
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to calculate the first parameter, which is
to basically calculate the specific work done.
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Now, the first parameter that we need to calculate
is the specific work done; and yes we also
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been given the flow angle at the nozzle exit
that is alpha has been 2 given to us; so there
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are two parameters in this velocity triangle
specified, we have alpha two and the blade
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speed U, so these are the parameters, which
we know.
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So, let us calculate the specific work done,
we know that the total to static efficiency
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is basically, a function of the specific work
done with reference to the static pressure
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at the exit and isotropic conditions; and
therefore, the specific work done is basically
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total static efficiency into C P into T 0
1 1 minus P 3, where P 3 is a static pressure
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at the rotor exit divided by P 0 1 raise to
gamma minus 1 by gamma. So, these are parameters,
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out of this, all these parameters are known
to us; we know the total static efficiency
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it is given as 0.87 C P is known 1148, T 0
1 is 1123, P 3 by P 0 1 is 1 by 3.11, because
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the inlet condition is given as 311 kilo Pascal.
So, that is 1 bar by 3.11 raise to 0.248,
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which is basically gamma minus 1 by gamma
0.33 minus 1.33 minus 1 by 1.33 so we get
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0.248. So, the specific work done or the work
done by the turbine can be calculated, and
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it comes out to be 276 kilo joules per kilogram.
Now, at the nozzle exit the Mach number, the
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second part of the question is to calculate
the Mach number at the nozzle exit, now Mach
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number as we know is defined as ratio of absolute
velocity to the speed of sound; so M 2 is
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C 2 divided by gamma R T 2. So, here we need
to calculate two things; one is to calculate
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C 2 we also need to calculate the static temperature
at the nozzle exit that is T 2. Now velocity
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triangle, because it is stated that the flow
enters and leaves the stage axially, we have
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C W 3 is equal to 0; and since, we know that
W t is a product of mu times delta C W that
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is mu into C W 2 minus C W 3; and since, C
W 3 is 0; we can calculate C W 2, which is
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W t divided by U and that is 276 into 10 raise
to 3 divided by 500 that comes out to be 550
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meters per second.
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So, let us understand this a little better,
it is mentioned here that the flow enters
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the the well assuming a constant axial velocity
through the stage, and that the flow leaves
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the stage in an axial direction. So, we have
the turbine, which is operating in design
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conditions with an axial absolute flow entry
and exit from the stage. So, this actually,
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gets needs to be modified. In the sense that
C 1 is axial, C 3 is also axial, as per this
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question; so this general velocity triangle,
which I had drawn should have actually have
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been modified in the sense, that alpha 1 should
be 0 alpha 3 should also be 0; and which is
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why we get C W 3 as 0, because since, C 3
is axial, C W 3 is also 0.
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So, C 2 we know from the exit velocity triangle
is C w 2 divided by sin alpha 2. Let us, see
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that once again C 2 is C W 2, which is this
component and divided by sin alpha 2, and
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that is 588; alpha 2 is given as 70 degree,
so C 2 comes out to be 588 meters per second.
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We know that T 2 static temperature is T 0
2 minus C 2 square by 2 C p T 0 2 and T 0
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1 are same, because there is no change in
stagnation temperature in the rotor, in the
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stator. There is no work done in the stator,
and so stagnation temperature in the stator
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has to be unchanged.
So, T 2 is static pressure at rotor entry,
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which is T 0 1 minus C 2 square by 2 C p C
2. We have just now calculated, so we can
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calculate static temperature at the nozzle
exit; so once we calculate static temperature,
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we can now calculate the Mach number, because
Mach number is simply the ratio of absolute
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velocity to the speed of sound. So, we have
already calculated the absolute velocity that
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is 588 meters per second and the static temperature
973 kelvin; so M 2 is 588 divided by square
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root of 1.33 into 287, which is a gas constant,
multiplied by 973, so the Mach number comes
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out to be 0.97. So, that is that solved the
second part of the question.
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Third part of the question is to find the
axial velocity; axial velocity can be found
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from the inlet velocity triangle, because
C 2 is known alpha is known; so, C a is 2
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times cos alpha 2 and that is simply 200 meters
per second. Now, the fourth part of the question
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is to find the total to total efficiency.
I think, during our discussion on the efficiencies,
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I had derived an expression, which relates
the total to total efficiency, and the total
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to static efficiency. And, we will simply
make use of that equation, to calculate the
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total to total efficiency, total to static
efficiency has been given as 0.87, and so,
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let us, calculate now the total to total efficiency.
So, based on that equation, if you recall
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and go back to that lecture, you will find
that this derivation was shown. So, this is
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1 by eta tt, which is the total to total efficiency;
this is equal to 1 by eta ts minus C 3 square
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by 2 W t where W t is the specific work done.
So, all these parameters we have already calculated,
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so we get 1 by 0.87, which is total static
efficiency minus C 3 square; well C 3 is basically
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equal to C a, because the flow exits the stage
axially, so C 3 and C a, are same and therefore,
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this is 200 square divided by twice of W t
into 276 into 10 raise to 3.
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So, this comes out to be 1 by eta tt is 1.0775,
so the inverse of that eta tt is 0.93; total
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to total efficiency, we have calculated is
0.93. If you compare this with the total static
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efficiency, it is 0.87. So, you can see that
total to total efficiency is indeed greater
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than the total to static efficiency; this
is a comment, which I had made even during
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our discussion on efficiency is that in general,
the total to total efficiency is comes out
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to be higher than the total to static efficiency,
because of the very nature of the definition.
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I think, I had shown the T-S diagram to demonstrate,
why this efficiency total to total efficiency
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has to be higher than total to static efficiency?
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So, if you look at the expansion process here,
which is something I had mentioned in the
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during our definition of the efficiency; total
to total efficiency is defined, based on the
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temperature stagnation temperature here, with
reference to the stagnation temperature at
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the exit isotropic; where as total to static
efficiency is defined, based on the stagnation
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condition, here with this static condition
here and the corresponding isotropic parameters,
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so we can see that this difference is always
less than this other difference' and therefore,
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it is inherent that total to total efficiency
has to be higher than that of total to static
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efficiency. So, that is coming from the very
basic definition and therefore, that should
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be indeed true; so there is something that
which I have also seen in this particular
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problem that if we calculate that this actually,
it should come out to be true.
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Let us, move on to the last part of the question,
which is to calculate the degree of reaction.
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Now degree of reaction again; we have defined
derived an expression for degree of reaction
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in the one of the earlier lectures. So, this
is basically, equal to 1 minus C a by 2 U
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into tan beta 3 minus tan beta 2. So, from
the velocity triangle let us see, what these
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parameters are; and how do we calculate tan
beta 3. Let us, go back to the velocity triangle
19:35.750 --> 19:42.750
here; tan beta 3 is equal to this component
divided by the axial velocity that is U divided
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by C a. And similarly, tan beta 2 is equal
to the component given by this that is C W
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2 minus u divided by C a.
So, from these two equations, we can calculate
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tan beta 2 as well as tan beta 3; and therefore,
what you can see here is that tan beta 2,
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beta 3 is U by C a, tan beta 2 can be equated
to tan alpha 2 minus U by C a, so if you substitute
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both these expressions here, we can simplify
the expression for the degree of reaction,
20:21.169 --> 20:28.169
and that is 1 minus 1 by 2 C a by U into tan
alpha 2. So, all these parameters, we have
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already calculated, and so let us, just substitute
that here, and then we get the degree of reaction
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as 0.451
So, we have now calculated all the five parameters,
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which were required to be calculated for this
question, we have calculated specific work
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done and axial velocity, the total to total
efficiency and now the degree of the reaction
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as well; so that completes the first question,
which was to do with very simple single stage
20:59.960 --> 21:06.120
axial turbine with certain parameters, which
has been specified, and, how do you proceed
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towards calculating the other parameters and
which obviously started from the velocity
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triangles, trying to solve the velocity triangles
to calculate the flow angles, and therefore,
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the other parameters which we were require
to be calculated.
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So, let us, now move on to the next question.
So, this is the problem statement for this
21:27.760 --> 21:34.740
question; this is basically, 50 percent reaction
stage question; so the problem statement is
21:34.740 --> 21:41.130
the following; combustion gases enter the
first stage of a gas turbine at a stagnation
21:41.130 --> 21:48.130
temperature and pressure of 1200 kelvin and
4 bar; the rotor blade tip diameter is 0.75
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meter; the blade height is 0.12 meter, and
the shaft speed is 10,500 rpm. At the mean
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radius, the stage operates with a reaction
of 50 percent; a flow coefficient of 0.7 and
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a stage loading coefficient of 2.5. Determine,
part (a) ,the relative and absolute flow angles
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for the stage; part (b), the velocity at nozzle
exit; part (c), the static temperature and
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pressure at nozzle exit assuming a nozzle
efficiency of 0.96 and the mass flow.
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So, here this is a question, which contains
to 50 percent reaction stage, and we have
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the temperature and pressure at the inlet
the degree of reaction is given to us, the
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rotational speed, and also the mean diameter
is specified; and based on this, we are required
22:42.230 --> 22:47.570
to calculate basically, solve the velocity
triangle and calculate the flow angles the
22:47.570 --> 22:52.500
relative and absolute flow angles; part (b)
was to calculate velocity at nozzle exit that
22:52.500 --> 22:58.909
is the absolute velocity, and the static
pressure and temperature at nozzle exit, with
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certain nozzle efficiency, which has been
specified and also the mass flow rate; so
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these are the parameters that we will need
to calculate for this particular question.
23:08.149 --> 23:13.010
So, as always we will start with the velocity
triangle, let us construct the velocity triangle
23:13.010 --> 23:17.940
first, and then we will proceed towards solving
the question.
23:17.940 --> 23:24.940
So, this is what we had drawn for 50 percent
reactions stage during our discussing on this
23:26.070 --> 23:33.070
couple of lectures .early, on so this is a
velocity triangle for a typical 50 percent
23:34.440 --> 23:41.440
reaction stage, and the the mean feature that
you can observe for 50 percent reactions stage
23:41.779 --> 23:47.620
is that the velocity triangles are symmetrical
or mirror images; so what you have at the
23:47.620 --> 23:54.620
inlet of the rotor is, what happens at the
exit as well it just a mirror image; and which
23:54.830 --> 24:00.750
means that the angles should also be equal
alpha 2 will be equal to beta 3 and beta 2
24:00.750 --> 24:06.190
will be equal to alpha alpha 3 and so on.
Similarly, the velocity components if the
24:06.190 --> 24:12.580
axial velocity is constant, then we have C
2 is equal to v 3 and V 2 is equal to C 3.
24:12.580 --> 24:18.990
So, in this question we have been given the
blade height at the mean diameter and the
24:18.990 --> 24:25.990
speed, which means that we can calculate the
blade speed that is U flow coefficient is
24:26.370 --> 24:31.360
given as 0.7, which means we can also calculate
the axial velocity from their, because flow
24:31.360 --> 24:38.360
coefficient is the ratio of the axial velocity
to the blade speed, so we have the blade speed
24:40.509 --> 24:44.919
as well as the axial velocity. We are now
required to solve the velocity triangle and
24:44.919 --> 24:46.940
calculate few other parameters.
24:46.940 --> 24:53.940
So, let us start with the first part of it,
which is to calculate all the angles involved
24:54.279 --> 25:01.279
in the velocity triangle. Now for this case,
we also have been given the loading coefficient
25:01.409 --> 25:08.409
psi which is delta h not by U square; and
for a 50 percent reaction stage, delta h not
25:10.480 --> 25:17.480
equal to U times delta C W, which is also
same as delta V w. And therefore, the stage
25:17.559 --> 25:24.559
loading coefficient reduces to delta h not
by U square, which is V W 3 plus V W 2 divided
25:25.409 --> 25:32.409
by U; and so this if we express in terms of
the angles, let us take a look at what is
25:33.269 --> 25:40.269
V w 2 and V w 3; V w 2 corresponds to this
component here, which is in terms of beta
25:41.309 --> 25:48.309
2 here and the axial velocity; similarly,
V w 3 is for the exit in terms of the axial
25:50.200 --> 25:54.919
component and the blade angle that is beta
3.
25:54.919 --> 26:01.919
So, this we have expressed in terms of axial
components V w 3 is C a times tan beta 3;
26:03.080 --> 26:10.080
and V w 2 is c a times tan beta 2. So, from
our discussion on degree of reaction, the
26:11.750 --> 26:18.190
specifically for 50 percent degree of reaction
case, we can basically, equate degree of reaction
26:18.190 --> 26:24.630
as C a by U into tan beta 3 minus tan beta
2 divided by 2; and that is, because the angles
26:24.630 --> 26:31.630
are similar so if you if you replace alpha
3 by beta 2 and beta alpha 2 by beta 3, we
26:32.409 --> 26:35.470
can express degree of reaction in terms of
this.
26:35.470 --> 26:40.169
So, if we look at these two equations and
we try to simplify them, what we basically
26:40.169 --> 26:47.169
get is an expression, in terms of the degree
of reaction, the loading coefficient and the
26:49.029 --> 26:55.789
flow coefficient, C a by U is the coefficient,
which we know is equal to 0.7 in this question;
26:55.789 --> 27:01.909
so what we get tan beta 3 is equal to psi
by 2 plus degree of reaction divided by the
27:01.909 --> 27:08.639
flow coefficient phi similarly, tan beta 2
is equal to psi by 2 minus degree of reaction
27:08.639 --> 27:13.730
divided by phi; this basically, happens if
we simply add and subtract these two equations,
27:13.730 --> 27:20.730
we can simplify them and get this equation.
So, since we have been we already know the
27:22.460 --> 27:28.919
loading coefficient psi, the flow coefficient
C a by U and degree of reaction; we substitute
27:28.919 --> 27:35.919
for those values here and we get the angles
beta 3, which is 68.2 degrees and beta 2,
27:35.970 --> 27:42.970
which comes out to be 46.98 degrees. Now so
for 50 percent reaction stage, we do not need
27:43.879 --> 27:48.539
to actually calculate the other angles, because
other angles are equal to what we already
27:48.539 --> 27:55.539
calculated; alpha 2 will be equal to beta
3, which is in term equal to 68.2 degrees;
27:56.110 --> 28:03.110
and alpha 3 is equal to beta 2, which is 46.98
degrees. So in this question that we have,
28:05.210 --> 28:11.460
because it happens to be 50 percent reaction
stage, we could make lot of simplifications
28:11.460 --> 28:17.259
in calculation of degree of reaction or equating
that to the loading coefficient and so on.
28:17.259 --> 28:21.149
And, also the fact that we do not need to
really calculate the other angles, because
28:21.149 --> 28:26.460
beta 2 is equal to alpha 3 and beta 3 is equal
to alpha 2.
28:26.460 --> 28:32.259
So, we just calculated beta 2 and beta 3 and
the absolute angles or indeed equal to these
28:32.259 --> 28:37.509
angles as well; so that completes the first
part of the question. Now, let us look at
28:37.509 --> 28:41.779
what is the second part? Second part is to
calculate the velocity at nozzle exit; subsequently
28:41.779 --> 28:47.210
we need to calculate static pressure and temperature
at nozzle exit with a certain nozzle efficiency
28:47.210 --> 28:53.889
specified. So let us calculate the velocity
at the nozzle exit, and for which we will
28:53.889 --> 29:00.009
make use of the fact that the dimensions have
been specified, the mean, the radius has been
29:00.009 --> 29:05.620
given the blade speed is known. And, so we
can basically, calculate the axial velocity
29:05.620 --> 29:07.909
and also the absolute velocity.
29:07.909 --> 29:13.080
Now, at the mean radius we can for, which
the tip diameter and the blade height will
29:13.080 --> 29:19.230
be used. So, the tip diameter is given as
75 centimeters so 0.75 minus blade height
29:19.230 --> 29:26.230
0.12 divided by 2, so this is the mean radius
that is 0.315. So, at the mean radius, we
29:28.490 --> 29:35.399
can calculate the blade speed, the speed of
rotation is given here, and the so phi D n
29:35.399 --> 29:42.399
by 60 will give us the blade speed mean, blade
speed that is 346.36 meters per second. Since,
29:43.029 --> 29:48.710
the flow coefficient has already been specified,
we can calculate the axial velocity as 5 times
29:48.710 --> 29:55.710
U mean that is 70 percent of this, so that
is 242.45 meters per second.
29:56.440 --> 30:03.240
So, the absolute velocity at the nozzle exit
is C a by cos alpha; and alpha is given, we
30:03.240 --> 30:10.240
already calculated in the previous part of
the question that was 68.2 degrees; so C 2
30:10.690 --> 30:17.690
is equal to C a by cos alpha 2 and that is
652.86 meters per second. So, having calculated
30:19.580 --> 30:25.490
the absolute velocity, we can now calculate
the static temperature based on the absolute
30:25.490 --> 30:31.549
velocity, because stagnation temperature is
known. It is gives as 1200 kelvin. So, in
30:31.549 --> 30:36.840
this question, we have been given the stagnation
temperature at the inlet of the turbine as
30:36.840 --> 30:41.350
1200 kelvin, and we know that in the nozzle
,there is no change in the stagnation temperature,
30:41.350 --> 30:48.350
so T 0 1 should be equal to T 0 2. And so,
T 2 is basically, equal to T 0 1 minus C 2
30:48.490 --> 30:53.909
square by 2 C P and same that is, because
T 0 1 and T 0 2 are same; so, from there you
30:53.909 --> 30:59.620
can calculate static temperature; static temperature
at the nozzle exit will be equal to T 0 2,
30:59.620 --> 31:06.620
which is T 0 1 minus C 2 square by 2 C p.
So, we get 1200 minus C 2 square 652.86 square
31:10.490 --> 31:17.490
divided by 2 into C p, so this can be calculated
at 1016.3 kelvin.
31:19.539 --> 31:25.159
Next part is, calculate the stagnation static
pressure at the nozzle exit, for which we
31:25.159 --> 31:31.240
will make use of the nozzle efficiency definitions;
so this is again from the fundamental cycle
31:31.240 --> 31:38.240
analysis that I assume you would be aware
of for rate and cycles, where we could, where
31:38.779 --> 31:45.110
we have defined nozzle efficiency, in terms
of enthalpies h 0 1 being the inlet enthalpy
31:45.110 --> 31:52.110
of the nozzle, h 2 is the inlet static exit
static enthalpy divided by h 0 1 minus h 2
31:52.720 --> 31:59.720
S. So, we can express this in terms of temperatures
1 minus T 2 by T 0 1 divided by 1 minus P
32:02.279 --> 32:09.269
2 by P 0 1 raise to gamma minus 1 by gamma,
this is because we can express the denominator
32:09.269 --> 32:15.740
in terms of 1 minus T 2 S by T 0 1, which
is basically, in terms of pressure ratios
32:15.740 --> 32:22.740
for an isotropic condition.
So, from this we can simplify this and calculate
32:23.629 --> 32:29.669
the pressure ratios P 2 by P 0 1, in terms
of the temperature ratio as well as the nozzle
32:29.669 --> 32:36.149
efficiency and so, the static pressure at
the nozzle exit can be simply calculated in
32:36.149 --> 32:43.149
terms of inlet pressure, stagnation that is
4 bar multiplied by 0.84052 rise to 4.03;
32:44.899 --> 32:51.899
so this is 1.986 bar; and the last part of
the question is to calculate the mass flow
32:52.220 --> 32:59.220
rate; mass flow rate is expressed, in terms
of density, the area and the axial velocity.
33:01.909 --> 33:08.909
Now density is P 2 by R T 2, and how do you
calculate the annulus area; so, for calculating
33:09.789 --> 33:16.090
this area we have the tip diameter as well
as the blade height; so from there you can
33:16.090 --> 33:23.090
calculate the annulus area and multiply that
with the axial velocity, one can calculate
33:23.320 --> 33:29.100
the mass flow rate. So, density is P 2 by
R T 2, annulus area in terms of the blade
33:29.100 --> 33:35.169
tip diameter and the blade height and the
axial velocity, which is already something
33:35.169 --> 33:41.929
calculated. So, if you multiplied all of them
the mass flow rate as 39.1 kilo gram per second.
33:41.929 --> 33:48.139
So, this completes the second question, which
was very similar to what we have solved in
33:48.139 --> 33:55.139
the first question except the fact that here,
we had also firstly, reaction turbine 50 percent
33:55.340 --> 34:00.559
reaction turbine stage, the first question
was on an impulse turbine stage; and second
34:00.559 --> 34:04.799
difference being the fact that here we had
nozzle efficiency and correspondingly, we
34:04.799 --> 34:11.250
had calculated the static temperature and
pressure at the exit of the nozzle using the
34:11.250 --> 34:17.300
nozzle efficiency definitions. So, having
understood two distinct problems, one to do
34:17.300 --> 34:24.300
with an impulse turbine and this second question
was 50 percent reaction turbine, we will now
34:25.300 --> 34:30.980
proceed to solving two other problems, which
are slightly different from what we have already
34:30.980 --> 34:31.829
solved.
34:31.829 --> 34:38.829
So, the third question, the problem statements
is the following; the inlet a single stage
34:40.069 --> 34:46.559
axial flow turbine operates with an inlet
temperature of 1100 kelvin, and total temperature
34:46.559 --> 34:53.559
of 3.4 bar, the total temperature drop across
the stage is 144 kelvin, and the isotropic
34:56.440 --> 35:03.440
efficiency of the turbine is 0.9, the mean
blade speed is 298 meters per second, and
35:03.920 --> 35:10.920
the mass flow rate is 18.75 kilo grams per
second, the turbine operates with a rotational
35:11.520 --> 35:18.520
speed of 12,000 rpm. If the convergent nozzle
is operating under choked condition, determine
35:19.010 --> 35:25.000
the part (a) blade loading coefficient part
(b) the pressure ratio of the stage and part
35:25.000 --> 35:30.930
(c) flow angles.
So, in this question, we have been given that
35:30.930 --> 35:36.790
the nozzle is operating under a choked condition
that is one of the things that we need to
35:36.790 --> 35:42.290
keep in mind, we also have been given isotropic
efficiency of the turbine, besides few other
35:42.290 --> 35:47.930
parameters like the temperature and pressure;
the temperature drop in the stage, blade speed
35:47.930 --> 35:54.930
and the mass flow rate. So, as we have been
doing in the past being construct the velocity
35:56.579 --> 36:03.579
triangle for generic turbine, axial turbine;
and so that we are familiar, with the parameters,
36:03.869 --> 36:09.329
which are specified and those we should be
need to the calculate. And so in this question
36:09.329 --> 36:16.220
we have the blade speed, which is given 298
meters per second and the mass flow rate is
36:16.220 --> 36:20.349
given to us, and the pressure and temperature
at the inlet have been specified.
36:20.349 --> 36:26.910
So, let us solve the first part of the question,
which was to calculate the loading coefficient,
36:26.910 --> 36:30.800
which is very straight forward, because we
know the temperature drop and we know the
36:30.800 --> 36:36.829
blade speed. So, psi is simply C p times delta
T not divided by U square, so that comes out
36:36.829 --> 36:43.829
to be 1.8615. Now, second part of the question
is to find the pressure ratio of the stage,
36:45.170 --> 36:52.170
so we need to find P 0 1 by P 0 3, so to calculate
that we know the temperature drop in this
36:52.569 --> 36:59.569
stage, it is given us 144 kelvin. The inlet
temperature is given T 0 2 equal to T 0 1
37:00.190 --> 37:05.609
that is 1100 kelvin, based on that we can
calculate the exit stagnation temperature
37:05.609 --> 37:12.609
T 0 3, which will be T 0 1 minus delta T not
and that is 1100 minus 144, so that is 956
37:15.369 --> 37:19.130
kelvin.
So, having calculated the exit stagnation
37:19.130 --> 37:24.780
temperature, we can we need to also calculate
the pressure ratio, because that is the objective
37:24.780 --> 37:29.540
of this second part of the question; for this
we will make use of the nozzle efficiency
37:29.540 --> 37:36.540
definition; so isotropic efficiencies, I mean
isotropic efficiency of a turbine, which is
37:36.589 --> 37:42.190
defined in terms of these the actual drop
in stagnation temperature divided by isotropic
37:42.190 --> 37:49.190
drop, so isotropic efficiency is defined as
one T 0 1 minus T 0 3 divided by T 0 1 minus
37:51.900 --> 37:58.900
T 0 3 as numerator is already known that is
delta T not divided by T 0 1 into 1 minus
37:59.950 --> 38:04.760
P 0 3 by P 0 1 rise to gamma minus 1 by gamma
.
38:04.760 --> 38:10.760
So, if we simplify this, because the temper
the efficiency is already given to us, we
38:10.760 --> 38:16.020
can substitute for the efficiency, the temperature
drop is known, inlet stagnation temperature
38:16.020 --> 38:23.020
is known, and therefore, P 0 3 by P 0 1 comes
out to be 0.53; and therefore, the pressure
38:24.390 --> 38:31.390
ratio of the turbine is an inverse of that
is 1.875. So, that solves the second part
38:32.329 --> 38:37.960
of the question were we are required to calculate
the pressure ratio of this particular turbine
38:37.960 --> 38:44.960
stage single stage axial turbine. The next
part is to calculate the flow angles, we need
38:46.010 --> 38:50.280
to calculate all the angles that are involved
in the velocity triangle; so we are going
38:50.280 --> 38:56.609
to use the fact that the nozzle is operating
under a choked conditions, so that is the
38:56.609 --> 39:02.740
one of the information that we have; and that
means that mach number at the nozzle exit
39:02.740 --> 39:08.420
is one and therefore, the absolute flow at
the nozzle exit C 2 will be equal to square
39:08.420 --> 39:15.420
root of gamma R T 2.
And similarly, that is also fixes the temperature
39:15.839 --> 39:21.290
pressure at the nozzle axial, because the
of the temperature and pressure would be the
39:21.290 --> 39:26.250
critical condition and therefore, we can actually,
calculate static pressure and temperature
39:26.250 --> 39:31.440
from the isotropic relations, because operating
under choked condition with Mach number equal
39:31.440 --> 39:38.440
to 1. So, once we calculate the absolute velocity,
we can also calculate some of the flow angles
39:39.200 --> 39:44.609
that are involved, because blade speed is
known and therefore, we can proceed towards
39:44.609 --> 39:48.170
solving the velocity triangle calculated all
the angle.
39:48.170 --> 39:53.500
So, the nozzle since, is it operating under
choked condition mach number is exit mach
39:53.500 --> 40:00.230
number is unity, so C 2 is a equal to square
root of gamma R T 2 and therefore, T 2 by
40:00.230 --> 40:07.230
T 0 2 by T 2 is equal to gamma plus 1 by 2
and that is 1.165. Since, T 0 2 is equal to
40:10.280 --> 40:16.940
T 0 1, that is known the static temperature
at the nozzle exit is T 2, which is 944.2
40:16.940 --> 40:22.799
kelvin; and therefore, the absolute velocity
of the gases leaving the choked nozzle will
40:22.799 --> 40:29.799
be square root of gamma R T 2 that is 600.3
meters per second. And, now we can calculate
40:30.680 --> 40:36.930
the axial velocity, because the blade speed
is known, the flow coefficient is known, so
40:36.930 --> 40:43.930
U times C a equal to U times that is 298 multiplied
by 0.95; and at that is 283 meters per second;
40:47.160 --> 40:53.049
and from the velocity triangle, we can now
calculate cos alpha 2, which is C a divided
40:53.049 --> 41:00.049
by C 2 that is 283 by 600, and therefore,
alpha 2 is 62 degrees.
41:03.750 --> 41:10.150
We now need to calculate beta 2, which is
the blade angle at the inlet of the rotor
41:10.150 --> 41:15.329
and for which we will make use of the velocity
triangle again; and that is from the velocity
41:15.329 --> 41:22.329
triangle, you can see that this ratio U by
C a is tan alpha 2 minus tan beta 2, which
41:23.000 --> 41:29.470
is inverse of the flow coefficient. Since,
tan alpha 2 is known and phi is also known
41:29.470 --> 41:36.470
we can calculate tan beta 2, which is tan
alpha 2 minus 1 by phi that is 0.828 or beta
41:37.270 --> 41:43.099
2 is 39.6degrees.
Now, the third part of the question well,
41:43.099 --> 41:47.630
where in we are calculating all angle, we
need to now calculate the angle circle, the
41:47.630 --> 41:54.630
rotor exit; now this specific work of turbine,
which is W T is C P is delta T not, which
41:55.510 --> 42:02.510
can be express in terms of U into delta C
W that is U into C a tan alpha 2 plus tan
42:02.900 --> 42:09.900
alpha 3. In which case, we know the angles
at the inlet tan alpha 2 is known U is known
42:11.910 --> 42:18.670
C a is known and delta T not is also known
therefore, tan alpha 3 is C P times delta
42:18.670 --> 42:25.670
T not by U C a minus tan alpha 2, so all these
parameters are known to us simply substitute
42:26.450 --> 42:33.450
that in we get alpha 3 as 4.54 degrees.
Similarly, U by C a is also equal to tan beta
42:36.880 --> 42:43.880
3 minus tan alpha 3 and therefore, tan beta
3 is equal to 1, by phi plus tan alpha 3 and
42:46.049 --> 42:52.319
therefore, beta 3 is 48.54 degrees. So, we
are calculated all the angles that are involved
42:52.319 --> 42:57.950
starting from the rotor entry, alpha 2 and
alpha beta 2 and rotor exit, the alpha 3 and
42:57.950 --> 43:04.609
beta 3; so you see that to be able to solve
in fact, all three problems that we have now
43:04.609 --> 43:11.319
solved, requires you to have good understanding
of the velocity triangles, because of that
43:11.319 --> 43:16.450
is where that is basically, the starting point
of solving these questions, if necessary that,
43:16.450 --> 43:21.940
you can constant the velocity triangle, clearly
understanding the different components of
43:21.940 --> 43:28.790
velocity, which has been specified, and then
use that information to proceed and solve
43:28.790 --> 43:33.930
the problem, based on what is already given
to even in the problem like some questions,
43:33.930 --> 43:39.500
you have mean blade speed; some questions
may specify axial velocity or some angles
43:39.500 --> 43:42.569
so on.
So, in this question we need basically, solved
43:42.569 --> 43:49.569
we calculated the pressure ratio, we are also
calculated all the angles that are involved
43:50.950 --> 43:55.150
and in this case of course, it is also given
towards that in nozzle is operating under
43:55.150 --> 44:02.150
choked conditions. So, I have one more question
for you in this is different from the all
44:03.480 --> 44:08.450
the questions we have solved, in the sense
that this question requires as to find the
44:08.450 --> 44:14.190
number of stages that are required for operating
a certain turbines, it is a multi stage turbine,
44:14.190 --> 44:17.579
how do you calculate the number of stages;
if you know the pressure drop across the entire
44:17.579 --> 44:22.230
turbine, how can you estimate the number of
stages that are involved.
44:22.230 --> 44:28.740
So, this is a problem statement is the following;
a multi stage axial turbine is to be designed
44:28.740 --> 44:34.299
with impulse stages and is to operate with
an inlet pressure and temperature of 6 bar
44:34.299 --> 44:40.900
and 900 kelvin and outlet pressure of 1 bar.
The isotropic efficiency of the turbine is
44:40.900 --> 44:47.900
85 percent, all the stages are to have a nozzle
angle outlet of 75 degrees, and equal inlet
44:49.410 --> 44:54.980
and outlet rotor blade angles; the mean blade
speed is 250 meters per second and the axial
44:54.980 --> 45:01.390
velocity is 150 per second and is a constant
across the turbine; estimate the number of
45:01.390 --> 45:08.119
stages required for this turbine; so this
is an impulse turbine, and so it is needless
45:08.119 --> 45:14.430
to say that the rotor inlet and outlet angles
are going to be equal. It is also given that
45:14.430 --> 45:19.049
we have the inlet pressure in temperature
for the nozzle entry, we have been given the
45:19.049 --> 45:25.369
isotropic efficiency and the nozzle exit angle,
we also have the mean blade speed and axial
45:25.369 --> 45:30.420
velocity; so lot of parameters that are involved
in the velocity triangles are given to us,
45:30.420 --> 45:36.339
but we need to calculate the number of stages
that are required for getting this kind of
45:36.339 --> 45:39.410
pressure ration that is involved in this particular
turbine.
45:39.410 --> 45:44.740
So, let us take a quick look at the velocity
triangle, because we will need to come back
45:44.740 --> 45:51.200
to this little later; an impulse turbine would
involve a velocity triangle like what is shown
45:51.200 --> 45:58.200
here; C 2 is the flow exiting nozzle at an
angle of alpha 2, in this case, it is given
45:59.809 --> 46:06.599
to us; V 2 is the relative velocity entering
the rotor at an angle of beta 2 and it leaves
46:06.599 --> 46:11.960
the rotor at velocity V 3, which is equal
to 2, and an angle beta 3, which is equal
46:11.960 --> 46:18.400
to beta 2, the absolute velocities at the
rotor exit are is equal to C 3, which is in
46:18.400 --> 46:25.329
fact equal to V 3 and V 2 for constant axial
velocity. So, in this case it is indeed specified
46:25.329 --> 46:30.859
that the axial velocity is constant, so it
means that C 3 is equal to V 3 which is in
46:30.859 --> 46:34.690
turn equal to V 2.
So, we will come back to this velocity triangle
46:34.690 --> 46:40.960
as we proceed to solve this equation; so first
part of the question, we will what we will
46:40.960 --> 46:47.960
try to do is that since, we know the pressure
drop across overall pressure drop is known;
46:48.030 --> 46:54.040
let us calculate the overall stagnation temperature
drop across the turbine; and then what we
46:54.040 --> 47:00.329
will do is to estimate the stagnation temperature
drop in one stage therefore, the overall temperature
47:00.329 --> 47:06.130
drop divided by temperature drop in one stage
will give us the number of stages, an estimate
47:06.130 --> 47:08.430
of the number of stages required.
47:08.430 --> 47:15.299
So, the overall pressure ratio is given as
6 therefore, T 0 1 by T not E S, where T not
47:15.299 --> 47:21.920
E S is the stagnation temperature, isotropic
at the exit of the turbine is equal to P 0
47:21.920 --> 47:27.990
1 by P 0 E raise to gamma minus 1 by gamma.
Therefore, the stagnation temperature at the
47:27.990 --> 47:34.990
exit of the turbine is equal to 576.9 kelvin,
this is isotropic temperature; so the actual
47:38.630 --> 47:45.630
temperature overall will be equal to the isotropic
temperature difference multiplied by the isotropic
47:46.760 --> 47:52.520
efficiency, which is 85 percent in this case;
so delta T not overall, which is the stagnation
47:52.520 --> 47:59.420
temperature across the entire turbine is equal
to eta T, which is the isotropic efficiency
47:59.420 --> 48:06.420
multiplied by T 0 1 minus T naught E S, T
0 1 is given as 900, T naught E S is calculated
48:06.770 --> 48:13.770
as 576.9 kelvin, so the overall temperature
drop in the turbine is 274.6 kelvin.
48:16.740 --> 48:23.329
So, now we come back to the velocity triangles;
we know from the inlet velocity triangles
48:23.329 --> 48:30.289
C 2 equal to C a divided by cos alpha 2, C
2 is this, cos alpha 2 is C a divided by C
48:30.289 --> 48:36.660
2 and therefore, C 2 is C a by cos alpha 2,
C a is known, alpha 2 is also known therefore,
48:36.660 --> 48:43.660
C 2 is equal to 150 divided by cos 75 that
is 579.5 meters per second. Now, this stagnation
48:47.270 --> 48:53.599
temperature at the inlet is known T 0 2, which
is equal to T 0 1; and therefore, static temperature
48:53.599 --> 49:00.599
at the nozzle exit T 2 is equal to T 0 2 minus
C 2 square by 2 C P; therefore, T 0 1, which
49:00.780 --> 49:07.780
is equal to T 0 2 is 900 kelvin, C 2 we just
now calculated 579.5 and therefore, we can
49:09.049 --> 49:16.049
calculate T 2 that is 753.7 kelvin.
Now, it is given specifically that this is
49:17.299 --> 49:22.380
an impulse turbine stage, which means that
degree of reaction is basically 0; so when
49:22.380 --> 49:27.210
the degree of reaction is 0, it implies that
the degree of reaction, we have defined as
49:27.210 --> 49:34.150
H 2 minus H 3 divided by H 0 1 minus H 0 3;
since, the degree of reaction is 0, we have
49:34.150 --> 49:40.970
H 2 is equal to H 3, implies the static temperature
should be same T 2 and T 3 should be the same
49:40.970 --> 49:47.839
and that is equal to 753.7 kelvin, because
there is no change in static conditions across
49:47.839 --> 49:54.299
rotor, the entire pressure drop has actually
take place in the nozzle nothing changes in
49:54.299 --> 49:57.940
the rotor T 2 is equal to T 3.
49:57.940 --> 50:04.119
Now, at the exit of the velocity triangle
at rotor entry well, let us look at rotors
50:04.119 --> 50:09.380
entry first; we have calculated alpha 2, which
is known to us; let us calculate beta 2 as
50:09.380 --> 50:16.380
well, tan beta 2 we know is C 2 sin alpha
2 minus U by C a, beta 2 is C 2 sin alpha
50:21.130 --> 50:28.130
2 that is this component minus U divided by
C a; so that is C 2 sin alpha 2 minus U divided
50:30.970 --> 50:37.970
by C a, and that comes out to be beta 2 give
comes out to be 64.16 degrees. And therefore,
50:40.539 --> 50:47.539
from this, we can calculate the relative velocity
V 2, and that is C a cos beta 2, it is 344.15
50:48.500 --> 50:53.750
meters per second; for constant axial velocity
in an impulse turbine, we have seen that V
50:53.750 --> 50:59.280
2 is equal to V 3 which is equal to C 3; and
therefore, we can calculate the stagnation
50:59.280 --> 51:06.280
temperature at the rotor exit T 0 3 in terms
of, because T 3 and T 2 are same T 0 3 is
51:09.039 --> 51:16.039
T 2 plus C 3 square by 2 Cp that is 753.7
plus 344.14 square by 2 in to C p that is
51:17.710 --> 51:22.890
805.28 kelvin.
Therefore, the per stage temperature drops,
51:22.890 --> 51:29.569
stagnation temperature drops is T 0 1 minus
T 0 3, and that is 900 minus 805.28 94.7 kelvin.
51:29.569 --> 51:36.569
We have already calculated the overall temperature
drop, which was 274.6 that divided by 94.7comes
51:40.289 --> 51:47.289
out to be 2.89 and that is to be approximated
to the next integer that is three stages,
51:48.970 --> 51:54.960
so we have we are required three stages for
achieving this kind of a pressure drop in
51:54.960 --> 52:00.500
an generating this work output from this kind
of impulse turbine, which has this efficiency
52:00.500 --> 52:05.839
and mean blade speed and so on. So, this is
one of ways of estimating the number of stages
52:05.839 --> 52:10.349
that are required, basically they are calculating
the overall temperature drop and calculating
52:10.349 --> 52:15.940
the temperature drop per stage diving the
2 will give us the number of stages required.
52:15.940 --> 52:21.960
So, this completes this particular problem
as well, we have solved so far 4 distinct
52:21.960 --> 52:25.750
problems, which have to do with different
types of turbines, how to analyze velocity
52:25.750 --> 52:31.849
triangle and solved the problem using these
velocity triangles. So, what I will do now,
52:31.849 --> 52:38.829
is to leave you with exercise problems, which
you can solved based on what we have solved
52:38.829 --> 52:45.549
in today's lecture, as well as the discussions
we had in the last few lectures, I will also
52:45.549 --> 52:50.140
give the final answers of these questions,
so you can check after you have solved these
52:50.140 --> 52:50.670
problems.
52:50.670 --> 52:55.690
So, the first exercise problem is a following
an axial flow turbine operating with an overall
52:55.690 --> 53:02.079
stagnation of 8 is to 1 has a polytrophic
efficiency of 0.85, determine, the total to
53:02.079 --> 53:07.710
total efficiency of the turbine, if the exhaust
Mach number of the turbine is 0.3, determine
53:07.710 --> 53:13.440
the total to static efficiency. If in addition,
the exhaust velocity of the turbine is 160
53:13.440 --> 53:18.940
meter per seconds, determine the inlet total
temperature the answers to these questions
53:18.940 --> 53:25.940
are 88 percent that is total to total efficiency
total to static efficiency is 86.17 percent,
53:26.680 --> 53:31.349
inlet total temperature is 1170.6.
53:31.349 --> 53:38.349
The second exercise problem is the mean blade
radii of the rotor of a mixed flow turbine
53:38.700 --> 53:45.700
is 0.3 meter at the inlet and 0.1 at the outlet;
the rotor rotates at 20000 revolutions per
53:46.410 --> 53:52.220
minute and the turbine is required to produce
430 kilo watts; the flow velocity at nozzle
53:52.220 --> 53:58.319
exist is 700 meters per second and the flow
direction is at 70 degrees to the meridional
53:58.319 --> 54:05.000
plane. Determining the absolute and relative
flow angles and the absolute exist velocity.
54:05.000 --> 54:11.980
If the gas flow rate is 1 kg per second and
also the velocity of the through flow is constant
54:11.980 --> 54:18.980
through the router, so the angles are alpha
2 is 70 degrees beta 2 is 7.02 degrees alpha
54:19.520 --> 54:26.520
3 18.4 degrees and beta 3 is 50.37 degrees
54:26.520 --> 54:32.710
The third question is an axial flow gas turbine
stage develops 3.36 mega watts at a mass flow
54:32.710 --> 54:38.950
rate of 27.2 kg per second; at the stage entry,
the stagnation pressure and temperature are
54:38.950 --> 54:45.950
772 kilo Pascal and 727 degree Celsius respectively;
the static pressure at exit from the nozzle
54:46.309 --> 54:52.410
is 482 kilo Pascal and the corresponding absolute
flow direction is 72 degrees to the axial
54:52.410 --> 54:58.410
direction; assuming axial velocity is constant
across the stage, and the gas enters and leaves
54:58.410 --> 55:04.319
the stage without any absolute swirl velocity.
Determine, part (a) the nozzle exit velocity;
55:04.319 --> 55:11.319
part (b) the blade speed part (c) the total
to static efficiency and (d) the stage reaction;
55:11.660 --> 55:17.529
answers are 488 meters per second, ythe blade
speed is 266.1 meter per second, total to
55:17.529 --> 55:23.809
static efficiency 0.83 and the stage reaction
is0.128.
55:23.809 --> 55:30.039
And last question is the single stage axial
turbine has a mean radius of 30 centimeters
55:30.039 --> 55:37.039
and a blade height of 6 centimeters; the gas
enters the turbine stage at 1900 kilo Pascal
55:37.170 --> 55:43.770
and 1200 kelvin, and the absolute velocity
leaving the stator of nozzle 600 meters per
55:43.770 --> 55:49.700
second and inclined at 65 degree to the axial
direction; the relative angles at the inlet
55:49.700 --> 55:55.279
and outlet of the rotor are 25 degrees and
60 degrees respectively; if the stage efficiency
55:55.279 --> 56:02.279
is 0.88, calculate part (a) the rotor rotational
speed part (b), stage pressure ratio part
56:02.520 --> 56:08.480
(c) flow coefficient (d) degree of reaction
and part (e) the power delivered by the turbine;
56:08.480 --> 56:15.480
here, the answers are rotational speed is
13550 rpm, the pressure ratio is 2.346, flow
56:17.880 --> 56:24.880
coefficient is 0.6, degree of reaction is0.41,
power delivered is 34.6 megawatts.
56:25.130 --> 56:32.039
So, these are four different problems that
I have put up as an exercise for you, and
56:32.039 --> 56:37.240
I hope with what we have discussed in last
few classes as well as today's tutorial, you
56:37.240 --> 56:42.230
will be able to solve these problems, based
on these discussions. So, we will continue
56:42.230 --> 56:47.539
our discussion on axial turbines further in
basically, looking at the 3D flows in further
56:47.539 --> 56:53.250
lectures. And, I hope you will be able to
understand the basic looking of a three- dimensional
56:53.250 --> 56:59.450
flow in axial turbine based on the fundamental
that we have discussed based on two-dimensional
56:59.450 --> 57:03.859
flow in axial flow turbines. So, we will continue
discussion on some of these topics in the
57:03.859 --> 57:10.859
coming lectures.