WEBVTT
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Hello and welcome to lecture number 9 of this
lecture series on Jet Aircraft Propulsion.
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So, as promised in the last lecture, today
we are going to solve some problems based
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on some of the discussions we have had in
the last two three lectures. So, as you might
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recall during one of our early lectures, we
started off with discussion on the Brayton
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cycle, we discussed about both the ideal and
the Brayton real Brayton cycle, we also understood
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what are the different sources of loses, which
lead to an actual Brayton cycle being different
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from that of an ideal cycle.
And basically the reason was that, the the
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presence of irreversibilities cause a Brayton
cycle, an actual Brayton cycle to be quite
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different from that of an ideal Brayton cycle.
And then subsequently, we have been discussing
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about application of the Brayton cycle in
terms of the actual jet engine cycles. And
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then, we have discussed various types of the
jet engine cycle like the turbojet, turbojet
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with afterburner, turbo fan and so on. So,
these are different modes or types of jet
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engine cycles which we have discussed. And
then, we have also looked at, what are the
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different components that constitute an a
jet engine cycle, and how we can estimate
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the performance of these components.
So, what we are going to do today is to have
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some numerical problems with us and see how
we can solve some of these problems based
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on what we have discussed. So, today we will
be discussing about application, well numerical
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application of the various Brayton cycle,
the real Brayton cycle and we will also be
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solving one or two problems based on component
performance analysis.
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So, we will be having a tutorial today on
ideal cycles, well real cycle as well and
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the component performance.
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So, let us take a look at the first problem
that we have. So, the first problem, the statement
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reads the following, it is a problem on a
Brayton cycle. A Brayton cycle operates with
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regenerator of 75 percent effectiveness, the
air at the inlet to the compressor is at 0.1
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mega Pascals and 30 degree Celsius, the pressure
ratio of the cycle is 6 and the maximum cycle
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temperature is 900 degree Celsius. If the
compressor and the turbine have efficiencies
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of 80 percent each, find the percentage increase
in cycle efficiency due to regeneration.
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So, here we have a problem which is related
to a Brayton cycle; obviously, it is not an
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ideal Brayton cycle, because we have effectiveness
of the regenerator coming into picture, we
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have also been given the compressor and turbine
efficiencies. So, put together it does not
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really constitute an ideal Brayton cycle.
So, it is a real Brayton cycle analysis that
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we would need to do, we have been given some
additional data like the cycle pressure ratio,
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the maximum cycle temperature and also the
inlet conditions of air, like it is pressure
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and temperature. So, what we are required
to find is the efficiency of the cycle with
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and without regeneration, it is basically
to see how much improvement or if at all we
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get by using regeneration.
So, to begin solving this problem, as ahead
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probably indicated in one of my earlier lectures,
and the earlier course that if you had a chance
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to go through it, I keep emphasizing the fact;
that it is necessary for us to first understand
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the problem statement very carefully and preferably
sketch a cycle diagram of the problem. Even
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though it may be a very trivial cycle, though
you you might have a feeling that it is very
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trivial, it is does need to sketch a cycle
diagram. I strongly recommend that you first
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sketch the cycle diagram on either PV or a
TS diagrams, and then mark those points where
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the data is known, and those points where
you need to find data.
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So, this will greatly have reduce the effort
of trying to figure out which data is given,
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which is not given and so on. And it is it
is a much more organized way of solving a
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problem. So, taking q from that, what I have
here is the cycle diagram for this problem.
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So, here we have a cycle diagram for a Brayton
cycle, I have indicated both the actual and
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the ideal Brayton cycles. So, those points
which are indicated with an s, that is 2 s
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indicates the isentropic point on that cycle,
similarly 4 s is for an isentropic process.
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Now, here in this problem, there is no mention
about pressure loss occurring during either
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heat addition or heat rejection process. So,
we can assume that there is no pressure loss
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occurring during these processes and so, it
is a constant pressure process. But since,
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the compressor and turbine efficiencies are
given; it means that those efficiencies, well
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those processes are not really isentropic.
So, if this is the Brayton cycle, Brayton
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cycle begins with the air at station 1, there
is a compression which is non isentropic which
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which takes the air to straight to a. And
and then there is a constant pressure heat
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addition between 2 a and 3; and that 3, we
have an expansion process which is again non
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isentropic, and this process goes all the
way up to 4 a.
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So, between 4 a and 1, we have the heat rejection
process and the, I have indicated two more
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point here 5 and 6 which could be somewhere
in between these heat addition and heat rejection
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processes, these two points indicate the regeneration
points. We will come back to regeneration
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little later, because that is the second half
of this problem, we will first solve this
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problem, find the efficiency of this cycle
without regeneration. Now, what is what are
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the data specified for us, we have the air
inlet conditions P 0 1 which is 0.1 mega Pascals,
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T 0 1 that is the stagnation temperature at
state 1, that is 303 Kelvin.
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We also have the maximum cycle temperature
which is T 0 3 that is 1173, 1173 Kelvin,
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these pressure ratio is 6, that is, so I have
indicated pressure ratio as pi c, then we
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have the compressor and turbine efficiency,
each of them equal to 0.8. So, this is the
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cycle diagram of Brayton cycle, both ideal
and real Brayton cycles have been indicated
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there, something we had discussed in one of
our earlier lectures.
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So, in this cycle, what we need to do is,
we know certain points, we know the pressure
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temperature, it is state 1, we know the pressure
ratio for this compression and expansion processes,
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the efficiencies are known, and the maximum
cycle temperature is known. And what for,
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what are we required to find? We are required
to basically, find the efficiency of this
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cycle. Now, if you recall from your thermodynamics,
if you have gone through the thermodynamic
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course, which I am sure you going to have,
then you would recall that efficiency.
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Thermal efficiency is basically defined as
the net work output divided by the heat input,
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that is how much work output do you get from
a cycle given a certain amount of heat, so
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that defines the thermal efficiency of a cycle.
So, if we need to do that, we need to find
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w net, and in this case, the net work output
as you know is the difference between the
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work generated by the turbine and the work
consumed by the compressor.
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So, w t minus w c will give us the net work
output, and what about heat input? Heat input
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is basically the enthalpy difference between
states 3 and state 2. So, difference between
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the enthalpy is there, will give us the heat
added per kilo gram of air, so ratio of this
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two will give us the efficiency. Now, the
question is how do we find the net work output
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of the turbine, and the work consumed by the
compressor, again I would suggest that we
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need to have a flashback, go back to thermodynamics,
you would recall that, the compressor and
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turbine, both of them are steady flow processes.
And for a steady flow process, you might recall
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that the net work output is, or work input
required is the difference in the enthalpies.
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So, for a compressor, net work output would
be equal to h 0 3 minus h 4 a, or h 4 in this
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case, h 0 4. And so, that is in turn equal
to C p times the temperature difference, similarly
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for a compressor it would be equal to h 0
3 minus h 0 2. And so, the difference between
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these two temperatures would give us the compressor
work output.
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So, with this in mind, let us begin our problem
solving here, we will first as I mentioned
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consider a Brayton cycle without any regeneration,
so without regeneration, let us look at how
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the cycle behaves. Now, for the first process
which is non isentropic, we know it is not
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isentropic, now if it were to be isentropic,
then we know the temperature at the end of
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the first process would be T 0 2 s which is
this stagnation temperature at the end of
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compression for an isentropic process.
So, this divided by T 0 1 should be equal
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to the pressure ratio raise to gamma minus
1 by gamma, this come from the isentropic
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analysis which we have discussed earlier.
So, temperature ratios, stagnation temperature,
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or static temperature ratios can be related
to the corresponding stagnation pressure or
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static pressure ratios, the raise to gamma
minus 1 by gamma, this is from, it it follows
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from isentropic relations, this is also equal
to T 0 3 by T 0 4 s. And what is that equal
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to? This is equal to cycle pressure ratio
raise to gamma minus 1 by gamma, and what
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is gamma? Gamma is 1.4; of course it is not
explicitly stated in the problem. So, what
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I will suggest is, in in a problem where gamma
is not been explicitly stated, you can safely
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take gamma as 1.4 which is the specific ratio
of specific heats for air. So, if gamma is
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equal to 1.4, then we have 6 raise to 1.4
minus 1 that is 0.4 divided by 1.4 which is
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1.668.
So, we get T 0 6, the T 0 2 s which has the
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stagnation temperature at the end of compression
for an isentropic process which is equal to
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T 0 1 multiplied by this , so we get T 0 3
into 1.668, so that is 505 Kelvin. Similarly,
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T 0 4 s can be determined by, because T 0
4 s will be equal to T 0 3 divided by 1.668,
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T 0 3 is already been specified in the problem
as 1173 Kelvin. So, T 0 4 s is equal to 1173
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divided by 1.668, so that is 705 Kelvin.
So, we now have two temperatures with us,
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each of them are the stagnation temperatures
for the process, if it were to have been isentropic.
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So, for an isentropic process, for an the
end of the compression, if the process was
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isentropic we would have a temperature of
T 0 2 s and similarly, for an expansion process
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that is the turbine process, if it were to
be isentropic we get a temperature of T 0
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4 s.
So, we have these two temperature with us,
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so what do we do with these temperatures,
we knew the actual temperatures, because that
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is what is the actual cycle about, it is not
Isentropic temperatures that we are concerned
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about, the work required by the compressor
will depend upon the actual temperatures and
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not the isentropic temperatures.
So, how do you find the actual temperatures,
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so for this, we have the efficiencies with
us, the compressor and the turbine efficiencies
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have been specified. And from our efficiency
definition if you recall, we can relate the
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isentropic temperatures to the actual temperatures
and since the efficiency is known, and we
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have also calculated the isentropic temperatures,
we now should be able to calculate the actual
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temperatures.
So, from the definition of isentropic efficiency,
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let us take up the compressor first now we
know that eta c which is isentropic efficiency
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of a compressor should be equal to T 0 2 s
minus T 0 1 divided by T 0 2 minus T 0 1 or
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T 0 2 minus T 0 1 is equal to T 0 2 s minus
t 0 1 divided by efficiency we have these
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both these temperatures with us which is 505
minus 303 divided by 0.8 which is the efficiency
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of the compressor, so this temperature difference
comes out to be 252 Kelvin.
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So, from the definition of isentropic efficiency,
let us take up the compressor first. Now,
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we know that eta c which is isentropic efficiency
of a compressor should be equal to T 0 2 s
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minus T 0 1 divided by T 0 2 minus T 0 1,
or T 0 2 minus T 0 1 is equal to T 0 2 s minus
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T 0 1 divided by efficiency, we have these
both these temperatures with us which is 505
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minus 303 divided by 0.8 which is the efficiency
of the compressor. So, this temperature difference
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comes out to be 252 Kelvin. Similarly, for
the turbine we have, turbine efficiency is
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T 0 3 minus T 0 4 divided by T 0 3 minus T
0 4 s.
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Therefore, T 0 3 minus T 0 4, which is the
temperature difference, the actual temperature
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difference of the turbine that would be equal
to efficiency times the isentropic temperature
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difference. So, that is equal to 0.8 multiplied
by 1173 minus 705, this is 375 Kelvin. So,
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we now have the actual temperature difference
for the compressor as well as for the turbine.
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So, we should now be able to find out the
work developed by the turbine and work required
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by the compressor. So, w t is equal to h 0
3 minus h 0 4 which is equal to c P into T
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0 3 minus T 0 4, that is again c P is not
explicitly given in the problem as I mentioned;
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if it not given we assume it to be air and
we we just take the specific heat at constant
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pressure for air which is 1.005 kilo joules
per kilo gram Kelvin.
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So, assuming c P to be 1.005 kilo joules per
kilo gram Kelvin, this multiplied by 375 which
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is the difference T 0 3 minus T 0 4, this
becomes 376.88 kilo joules per kilo gram.
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Similarly, we have w c which is work by the
compressor is equal to t h 0 2 minus h 0 1
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which is c P times T 0 2 minus T 0 1 which
is again equal to 253.26 kilo joules per kilo
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gram. So, we can also now find out the T 0
2, because the difference is known and T 0
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1 is known. So, T 0 2 is 252 plus 303 that
is 555 Kelvin and why do we need T T 0 2?
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Well we need T 0 2 if we have to calculate
the heat input.
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So, what we have now calculated is the work
developed by the turbine which is basically
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the enthalpy difference across the turbine
and so, for that we need the actual temperatures.
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How do we calculate actual temperatures? We
calculate actual temperatures based on the
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efficiency definition and for that, we need
the isentropic temperature, isentropic temperature
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comes from the pressure ratio raise to gamma
minus 1 by gamma. So, the next parameter that
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we need to find is the heat input.
So, heat input is basically the enthalpy difference
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between states 3 and 2 and so, that would
be equal to h 0 3 minus h 0 2 which is c P
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times T 0 3 minus T 0 2. And after we have
calculated the heat input, we can now easily
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calculate the efficiency, efficiency will
be equal to w t minus w c divided by Q 1 and
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which is basically net work output by heat
input.
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So, let us do that and see how how much efficiency
do we get for this cycle . So, for this cycle
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the heat input comes out to be h 0 3 minus
h 0 2 that is 1.005 into 1173 which is the
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max cycle temperature minus h 0 2, that is
c P times 155. So, heat input is 620.09 kilo
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joules per kilo gram, therefore efficiency
is equal to the net work output which is w
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t minus w c 376.88 minus 253.26, the whole
thing divided by 621.09, so this is 19.9 percent.
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So, this Brayton cycle has an efficiency of
19.9 percent that is it can convert just about
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19.9 percent of the heat input to net work
output.
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This solves one part of the problem, now the
next part of the problem is to see what happens
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if we use a regenerator, do we get an increase
in efficiency; if so, how much improvement
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we get. So, regenerator as you know is is
is a mechanism by which we can transfer some
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amount of heat from one part of the cycle
to another. And so, the obvious thing to do
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in such a cycle is to transfer the heat which
is rejected, part of the heat which is rejected.
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Obviously, you know you cannot transfer the
entire heat which is rejected, because that
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would make heat rejected equal to 0, and that
it it violates the second law of thermodynamics.
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So, that is not possible, but we can definitely
transfer some part of the heat which is rejected
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back to the heat added process, thereby we
can save on the head added, and we also save
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on heat rejected. So, that seems to be a clever
idea of increasing the efficiency, because
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we are reducing Q 1 without effecting either
w net w t or w c and therefore, net work output.
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So, net work output does not change, but what
does change is the heat input, we can reduce
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heat input by using regeneration. So, let
us see if using a regeneration regenerator
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of certain effectiveness, given as 75 percent,
does that make a difference in the cycle efficiency,
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let us take a look at that.
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Now, if we consider a regenerator in in which
case we will need to find a few more temperatures,
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we will need T 0 4, T 0 4 is T 0 3 minus 375
that is 1173 minus 375 equal to 798 Kelvin.
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Regenerator effectiveness is given as 0.75
and this is defined as T 0 6 minus T 0 2 divided
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by T 0 4 minus T 0 2. And it means, it is
basically, effectiveness of regenerator tells
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us that if effectiveness was equal to 1, then
we will get T 0 4 is equal to T 0 6 is equal
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to T 0 4, that is we can heat the compressor
outlet all the way up to a temperature which
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is equal to the turbine exhaust. So, that
is a ideal process if the regenerator had
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no irreversibilities perhaps that would be
possible, but you know it is not, because
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regenerator effectiveness in this case is
0.75. And so, T 0 6 is a temperature which
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is less than T 0 4 and so, we need to find
how much it is.
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So, from the definition of effectiveness of
regenerator, we have T 0 6 minus T 0 4 divided
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by T 0 4 minus T 0 2, this is equal to 0.75.
So, all these temperatures are known, T 0
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2 is known, T 0 4 is known, effectiveness
is known, therefore we can calculate T 0 6.
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So, this comes out to be 737.3 Kelvin as compared
to T 0 4 which is 798 Kelvin. So, if the effectiveness
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was 1, then T 0 6 will be equal to T 0 4 which
is 798 Kelvin, now the heat input will now
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be equal to h 0 3 minus h 0 6.
So, you can see immediately the difference
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that, you need to only had this much amount
of heat here instead of h 0 3 minus h 0 2.
21:24.400 --> 21:32.419
So, this will be equal to c P times T 0 3
minus T 0 6 which comes out to be 437.88 kilo
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joules per kilo gram.
21:34.419 --> 21:40.879
So, heat input has changed, net work output
does not change, because we are not doing
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anything to the turbine or the compression
process. Therefore, the efficiency of this
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cycle would now be equal to 123.62 divided
by 437.9, this is equal to 0.2837 that is
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28.37 percent, and the cycle efficiency without
regeneration was 19.9. So, what is the percentage
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improvement increase due to regeneration,
it s equal to the difference 0.2837 minus
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0.199 divided by 0.199, so this is equal to
42.56 percent. So, what do you see here is
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that, using regeneration well how exactly
regeneration is to be carried out is is another
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issue all together.
But if we were indeed able to carry out regeneration,
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even if the regenerator was not 100 percent
effective, in this case it is 75, we still
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get an improvement in efficiency of the order
of 43 percent as 42.56 percent which is the
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huge improvement in efficiency. So, the the
moral of the story is that, regeneration is
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one of the ways of substantially improving
the efficiency of a cycle, which is something
23:00.960 --> 23:06.500
you would have understood if you had gone
through the thermodynamic course, that regeneration
23:06.500 --> 23:12.350
is definitely a way of increasing the efficiency.
In fact, if you recall there are two such
23:12.350 --> 23:20.179
cycles which have efficiencies as high as
that of Carnot cycle, and what are those cycles?
23:20.179 --> 23:28.029
The Stirling and Ericsson cycles, both these
cycles have two processes which are basically
23:28.029 --> 23:33.659
regeneration processes; one is at constant
volume, the other is at constant pressure.
23:33.659 --> 23:40.010
So, regeneration, using regeneration Stirling
and Ericsson cycles are able to achieve efficiencies
23:40.010 --> 23:46.920
which are as high or equal to that of a Carnot
cycle, which means the cycles have the maximum
23:46.920 --> 23:54.649
efficiency possible for that temperature limits.
So, regeneration, of course again during the
23:54.649 --> 23:59.370
discussion on thermo dynamics, you may have
come across description about Stirling and
23:59.370 --> 24:05.980
Ericsson cycles. Why why is that, these cycles
are not used? They are not used, because the
24:05.980 --> 24:13.110
regeneration makes the whole cycle extremely
bulky and complicated and so, it is not suitable
24:13.110 --> 24:18.899
for modern day applications. And so, they
have not really been used in any application,
24:18.899 --> 24:23.590
even though they have efficiencies equal to
Carnot cycle efficiency.
24:23.590 --> 24:31.009
So, regeneration, so this example that we
have just solved is one way of trying to see
24:31.009 --> 24:35.559
if regeneration makes a difference in the
overall efficiency or the thermal efficiency
24:35.559 --> 24:40.309
of the cycle. And we have seen that, it does
make a lot of difference, it makes about,
24:40.309 --> 24:45.060
and then this particular case we got about
43 percent improvements in the efficiency
24:45.060 --> 24:51.740
of a cycle with regeneration. So, this was
one problem, the first problem that we solved
24:51.740 --> 25:01.110
today by using the Brayton cycle analysis
for an real Brayton cycle with and without
25:01.110 --> 25:02.110
regeneration.
25:02.110 --> 25:06.860
So, let us take a look at what is a second
problem that we have for us today, second
25:06.860 --> 25:12.231
problem is again a gas turbine problem, again
in some sense it is basically a Brayton cycle
25:12.231 --> 25:23.070
problem. So, here we have a gas turbine operating
at a pressure ratio of 11.314 produces zero
25:23.070 --> 25:32.090
net work output when 473.35 kilo joule of
heat is added per kilo gram of air. If the
25:32.090 --> 25:39.260
inlet air temperature is 300 Kelvin and the
turbine efficiency is 71 percent, find the
25:39.260 --> 25:44.299
compressor efficiency.
So, here this is again a Brayton cycle problem,
25:44.299 --> 25:50.990
but we have slightly altered it for a gas
turbine. So, we have the pressure ratio specified
25:50.990 --> 25:57.999
and it is given that this gas turbine produces
no net work output, which means that the work
25:57.999 --> 26:03.010
output of the turbine should be equal to the
work input required for the compressor. So,
26:03.010 --> 26:07.670
there is no net work output, the heat input
is given, the efficiency of the turbine is
26:07.670 --> 26:12.640
given, we are required to find efficiency
of the compressor.
26:12.640 --> 26:20.909
So, in this particular problem, we have which
is again a simple Brayton cycle analysis,
26:20.909 --> 26:27.019
the net work output is given as 0 which is
true for most of the air craft engines. If
26:27.019 --> 26:34.559
if you recall, aircraft engines do not regenerate
net work output, turbine in an aircraft exits
26:34.559 --> 26:39.630
only for driving the compressor and a few
other accessories. So, in most of the jet
26:39.630 --> 26:44.429
engines, the there is no net work output from
the jet engine, because turbine drives the
26:44.429 --> 26:48.139
compressor and that is it is. So, there is
no net work output that way, but of course,
26:48.139 --> 26:53.731
there are shaft power outputs required for
turbo props and turbo shafts and so on, but
26:53.731 --> 26:59.970
for pure jet engines the net work output will
be 0. So, in this case, we have the cycle
26:59.970 --> 27:06.169
pressure ratio, we have certain temperatures
given to us and efficiency of the turbine;
27:06.169 --> 27:09.600
we need to find what is the efficiency of
the compressor.
27:09.600 --> 27:15.730
So, let us try to solve this problem, so here
we have the net work output as zero which
27:15.730 --> 27:25.100
means that w c will be equal to w t, or T
0 2 minus T 0 1 which is the temperature difference
27:25.100 --> 27:31.539
across the compressor will be equal to T 0
3 minus T 0 4 which is the temperature drop
27:31.539 --> 27:39.679
across the turbine, this in turn is equal
to T 0 3 minus T 0 2 is equal to T 0 4 minus
27:39.679 --> 27:49.000
T 0 1. Now, cycle pressure ratio is given
as 11.314 and that should be equal to the
27:49.000 --> 27:56.889
isentropic stagnation temperature ratios,
that is T 0 2 s divided by T 0 1 is equal
27:56.889 --> 28:05.220
to P 0 2 by P 0 1 raise to gamma minus 1 by
gamma, this is equal to 11.314 raise to gamma
28:05.220 --> 28:10.159
minus 1 is 1.4 minus 1 that is 0.5 divided
by 1.4.
28:10.159 --> 28:18.100
Therefore, T 0 2 s that is this stagnation
temperature isentropic will be equal to T
28:18.100 --> 28:26.350
0 1 multiplied by 11.314 raise to gamma minus
1 by gamma. So, this comes to be 600 Kelvin,
28:26.350 --> 28:34.860
now heat added is given as 466.35 kilo joules
per kilo gram. So, since heat added is the
28:34.860 --> 28:40.660
process between states 2 and 3, heat added
occurs during that particular process. So,
28:40.660 --> 28:52.860
c P times T 0 3 minus T 0 2 is equal to 476.35,
or T 0 3 minus T 0 2 is 474, because that
28:52.860 --> 29:02.470
is 476.354 divided by 1.005. So, we are assuming
that specific heat at constant pressure is
29:02.470 --> 29:08.379
the same as that of air which is 1.005 kilo
joules per kilo gram Kelvin.
29:08.379 --> 29:17.090
So, we have calculated that is the difference
that is T 0 3 minus T 0 2 which is 474 Kelvin.
29:17.090 --> 29:24.119
So, the next step in this is to calculate
the corresponding temperature so that we can
29:24.119 --> 29:30.419
determine the efficiency of the compressor.
Now, from our first equation that we wrote
29:30.419 --> 29:35.850
that was equating the work output of the turbine
to that of the compressor, we can also write
29:35.850 --> 29:45.259
T 0 4 as equal to T 0 1 plus T 0 3 minus T
0 2, T 0 1 is known as 300 Kelvin it is given
29:45.259 --> 29:51.470
in the problem plus this difference T 0 3
minus T 0 2 we just calculated in the previous
29:51.470 --> 29:58.350
step that was 474, so we get a total temperature
of 774 Kelvin.
29:58.350 --> 30:06.149
Now, turbine efficiency is is given as 71
percent and the definition of turbine efficiency
30:06.149 --> 30:16.210
is T 0 3 minus T 0 4 divided by T 0 3 minus
T 0 4 s. So, we can simplify that in a maneuvers
30:16.210 --> 30:24.740
shown here, T 0 3 multiplied by 1 minus T
0 4 by T 0 3 divided by T 0 4 as multiplied
30:24.740 --> 30:33.899
by T 0 3 by T 0 4 s minus 1, also we know
that this ratio that is T 0 3 by T 0 4 s should
30:33.899 --> 30:39.910
be equal to the cycle pressure ratio raise
to gamma minus 1 by gamma, because the net
30:39.910 --> 30:44.960
work output is anyway 0. So, the turbine pressure
ratio will also be equal to the compressor
30:44.960 --> 30:51.340
pressure ratio, therefore we get this ratio
from this expression .
30:51.340 --> 30:59.830
So, here we have this ratio T 0 3 by T 0 4
s which we can determine from this expression
30:59.830 --> 31:06.879
and we also know the turbine efficiency. So,
if you simplify that we get T 0 4 by T 0 3
31:06.879 --> 31:18.899
is equal to 1 minus 0.71 by 2 which is 0.645,
therefore T 0 3 is equal to 774 which is T
31:18.899 --> 31:28.140
0 4 divided by 0.645 that is 1200 Kelvin.
And therefore, T 0 2 will be equal to 1200
31:28.140 --> 31:36.340
that is T 0 3 minus the difference between
T 0 3 and T 0 2, that is 474, 726 Kelvin.
31:36.340 --> 31:41.879
So, we now have all the temperatures that
are required for estimating the compressor
31:41.879 --> 31:49.999
efficiency, the compressor efficiency as we
know it is defined as equal to the isentropic
31:49.999 --> 31:58.169
temperature difference T 0 2 s minus T 0 1
divided by T 0 2 minus T 0 1.So, all these
31:58.169 --> 32:04.360
temperatures we have already calculated.
So, we get 300 T 0 2 s which is 600 minus
32:04.360 --> 32:13.470
T 0 1 which is 300 divided by T 0 2 726 minus
300 which is equal to 0.704 or 70.4 percent.
32:13.470 --> 32:21.770
So, we have determined the efficiency of the
compressor, because the efficiency of the
32:21.770 --> 32:26.900
turbine is given and how did we solve this
problem, basically it is given that the net
32:26.900 --> 32:34.249
work output is 0. So, we can equate the work
done by the turbine to that required by the
32:34.249 --> 32:39.130
compressor. And if we were to assume that,
the specific heat is the same for both the
32:39.130 --> 32:44.879
compressor and the turbine, then we get, we
can equate basically a temperature difference
32:44.879 --> 32:50.200
across the compressor and the turbine, we
also know the pressure ratio of the cycle.
32:50.200 --> 32:55.919
So, we can find out the actual isentropic
temperatures and since turbine efficiency
32:55.919 --> 33:01.159
is known, if we substitute for the cycle pressure
ratios and the isentropic temperatures, we
33:01.159 --> 33:06.940
can actually find all the temperatures which
are needed to calculate the compressor efficiency.
33:06.940 --> 33:13.070
So, compressor efficiency in this case, we
have calculated as 70.4 percent. So, this
33:13.070 --> 33:19.730
second problem that we have solved today here
is again to do with the Brayton cycle as in
33:19.730 --> 33:24.820
the case of first cycle, first cycle of course
we did Brayton cycle with and without regeneration.
33:24.820 --> 33:30.570
In this case, we have estimated the efficiency
of one of the components which is the compressor,
33:30.570 --> 33:36.059
given some of the other cycle parameters.
So, the next problem that we are going to
33:36.059 --> 33:43.360
solve here today would be on an intake. And
we have already analyzed in our last lecture,
33:43.360 --> 33:48.369
an intake and its performance, we have seen
that there are two distinct perform parameters;
33:48.369 --> 33:54.100
one is the total pressure ratio which is also
known as the pressure recovery, and the second
33:54.100 --> 33:58.470
parameter is the diffuser efficiency, isentropic
efficiency of a diffuser.
33:58.470 --> 34:04.499
So, these are two parameters which which tell
us something about the performance of air
34:04.499 --> 34:10.291
intake. So, in the next problem, we are going
to talk about an air intake and how we can
34:10.291 --> 34:14.230
estimate some of the performance parameters
of an air intake.
34:14.230 --> 34:21.230
So, the problem number 3 that we have for
today is an aircraft flies at a Mach number
34:21.230 --> 34:28.960
of 0.75 ingesting an air flow of 80 kilo grams
per second at an altitude where the ambient
34:28.960 --> 34:38.309
temperature and pressure are 222 Kelvin and
10 kilo Pascals respectively. The inlet design
34:38.309 --> 34:44.800
is such that the Mach number at the entry
to the inlet is 0.6 and that at the compressor
34:44.800 --> 34:52.540
phase is 0.4, the inlet has an isentropic
efficiency of 0.95.
34:52.540 --> 34:59.609
Find part a, the area of the inlet, entry
part b, the inlet pressure recovery, and part
34:59.609 --> 35:07.180
c the compressor phase diameter, so in this
particular problem, we have been given that
35:07.180 --> 35:11.170
an aircraft is flying at a certain Mach number
at an altitude where the pressure and temperature
35:11.170 --> 35:17.740
is given. And then we have also been given
what is a Mach number at the inlet entry and
35:17.740 --> 35:22.570
the compressor phase and also, of course the
isentropic efficiency of this particular intake.
35:22.570 --> 35:27.319
So, given these parameters we have been asked
to find the area of the inlet entry, and the
35:27.319 --> 35:33.780
pressure recovery and also the diameter of
the compressor phase. So, based on some of
35:33.780 --> 35:40.420
these parameters which are known, we should
be able to find out the required parameters.
35:40.420 --> 35:47.619
So, first thing that we will do is to try
to find out the properties at state 1, that
35:47.619 --> 35:55.359
is right at the intake entry that is state
1 and then based on that, we can find out
35:55.359 --> 36:04.109
the area of the inlet entry, because area
is equal to the velocity divided by the mass
36:04.109 --> 36:10.020
flow rate, mass flow rate divided by velocity
times density and then subsequently, we will
36:10.020 --> 36:12.760
move to station 2 which is the compressor
phase.
36:12.760 --> 36:19.640
So, Mach number in this case is here given
by is given as 0.75 and therefore, we can
36:19.640 --> 36:25.190
find these flight speed which is u a, u subscript
a is the flight speed which is Mach number
36:25.190 --> 36:31.320
times square root of gamma R T a, this is
in turn equal to 0.75 into square root of
36:31.320 --> 36:37.450
1.4 into 287 into the temperature is given
as 222.
36:37.450 --> 36:45.220
So, the flight speed comes out to be 224 meters
per second. And the density, the free stream
36:45.220 --> 36:50.510
density rho a will be equal to P a by R T
a, which is from the state equation, this
36:50.510 --> 37:01.020
is equal to 0.156 kilo grams per meter cube.
And so, once the Mach number is already known
37:01.020 --> 37:05.660
to us, the static temperature is known, T
a is known to us. So, we can find out the
37:05.660 --> 37:12.609
total temperature which is T naught a, this
is T a into 1 plus gamma minus 1 by 2 M square
37:12.609 --> 37:20.500
and that is basically from the energy equation.
So, T a is given as 222, Mach number is 0.75,
37:20.500 --> 37:27.500
so we get T 0 a is equal to 246 Kelvin.
Similarly, we can find the total pressure,
37:27.500 --> 37:32.920
total pressure is P 0 a which is P a into
1 plus gamma minus 1 by 2 M square raise to
37:32.920 --> 37:39.210
gamma by gamma minus 1, the pressure is already
given to us, P a is specified. Therefore,
37:39.210 --> 37:51.049
we can find P 0 a, this is 14522.8 Pascal
or 114.52 kilo Pascals. Therefore, the static
37:51.049 --> 37:59.609
temperature at the inlet entry that is T 1
is equal to T 0 a divided by 1 plus gamma
37:59.609 --> 38:03.710
minus 1 by 2 M square.
And where does it comes from, this is basically,
38:03.710 --> 38:11.420
because there is no change in total temperature
from the ambient that is at station a, all
38:11.420 --> 38:16.930
the way up to 0.1. And in fact, all the way
up to called the compressor phase, there is
38:16.930 --> 38:23.099
no change in the total temperature, because
firstly we assume that this compression process
38:23.099 --> 38:29.140
is adiabatic, there is no heat input or heat
out rejected by this process. So, total temperature
38:29.140 --> 38:35.230
does not change, total pressure does change,
because of frictional effects, but total temperature
38:35.230 --> 38:42.740
cannot change. So, T 1 is T 0 a divided by
1 plus gamma minus 1 by 2 M square. So, if
38:42.740 --> 38:46.980
you substitute for all these values, we get
the static temperature at the inlet entry
38:46.980 --> 38:52.410
as 230.4 Kelvin.
38:52.410 --> 38:56.460
Having found the static temperature, we can
find static pressure as well, P 1 which is
38:56.460 --> 39:02.540
P 0 8 divided by 1 plus gamma minus 1 by 2
M square raise to gamma by gamma minus 1.
39:02.540 --> 39:10.460
So, this is 11386 Pascal, therefore density
rho 1 is P 1 by R T 1 which is, now since
39:10.460 --> 39:17.450
we have calculated P one and T 1, we can find
density that is 0.1722 kilo grams per meter
39:17.450 --> 39:23.770
cube. So, having found the density and the
mass flow rate is already been given, we can
39:23.770 --> 39:28.890
now find out the area at the inlet entry.
So, A 1 will be equal to mass flow rate divided
39:28.890 --> 39:35.440
by u 1 into rho 1, because mass flow rate
is rho a v that is velocity times density
39:35.440 --> 39:41.630
times area. And how do you find u 1, u 1 is
found from the Mach number, Mach number at
39:41.630 --> 39:48.049
inlet entry is given as 0.6. So, M 1 times
square root of gamma R T 1 gives us u 1 that
39:48.049 --> 39:54.250
multiplied by rho 1. And so, if you substituted
for all these values, we have mass flow rate
39:54.250 --> 40:03.089
as 80 k g s per second divided by M 1 which
is 0.6 square root of gamma R T 1,1.4, R is
40:03.089 --> 40:11.790
287 for air and T 1 is 230.4, just multiplied
by density that is 0.1722, area at the inlet
40:11.790 --> 40:18.130
entry comes out to be 2.54 meter square.
So, this solves the first part of the problem
40:18.130 --> 40:23.710
where we are required to find out the area
at the inlet entry, what we basically did
40:23.710 --> 40:30.440
was to find out the static temperature and
pressure at the inlet of the diffuser, and
40:30.440 --> 40:35.280
that would help us in finding out the density.
And since Mach number is given, we can find
40:35.280 --> 40:40.799
out the velocity and therefore, from from
the mass flow rate relation, mass flow rate
40:40.799 --> 40:47.559
divided by u 1 times rho 1, we can find out
the area at the inlet entry.
40:47.559 --> 40:52.579
The second part of the problem is to find
out the pressure recovery, now pressure recovery
40:52.579 --> 40:59.000
as you know it is the ratio of the stagnation
pressures or at the outlet of the intake to
40:59.000 --> 41:05.250
the inlet of the intake. I already mentioned
that stagnation temperature does not change
41:05.250 --> 41:10.030
all the way from inlet to outlet. So, T 0
a should be equal to T 0 1 should be equal
41:10.030 --> 41:15.590
to T 0 2, but that is not true for the pressure,
stagnation pressure is not the same, there
41:15.590 --> 41:19.490
will be a stagnation pressure loss. So, we
need to find out, how much is the pressure
41:19.490 --> 41:20.490
recovery.
41:20.490 --> 41:25.490
Now, in order to find out the pressure recovery,
we will make use of the diffuser efficiency
41:25.490 --> 41:33.099
definition. Now, T 0 2 as we know it is equal
to T 0 a, and the diffuser efficiency is defined
41:33.099 --> 41:42.010
as T 0 2 s minus T a divided by T0 a minus
T a, diffuser efficiency is given as 0.95,
41:42.010 --> 41:47.070
T 0 a is known you already calculated that,
T a is also known.
41:47.070 --> 41:55.140
So, if you substitute for all these values,
we get T 0 2 s. So, T 0 2 s comes out to be
41:55.140 --> 42:06.290
245.75 Kelvin and therefore, pressure recovery
is equal to P 0 2 by P 0 1 and that is equal
42:06.290 --> 42:13.480
to T 0 2 s divided by T 0 1 raise to gamma
by gamma minus 1, pressure recovery of the
42:13.480 --> 42:22.900
diffuser is 0.982. This means that, from the
inlet at station 1 to the outlet which is
42:22.900 --> 42:28.480
station 2 that is the compressor phase, there
is a certain loss in total pressure that is
42:28.480 --> 42:34.069
ratio of these pressures, total pressures
is 0.982. So, there is about 2 percent loss
42:34.069 --> 42:41.170
in total pressure as the flow moves from station
1 to 2.
42:41.170 --> 42:47.940
Now, the next, the third part of problem is
to find out the area at the compressor phase,
42:47.940 --> 42:53.730
you will use the same principle as we did
for calculating the area at the inlet entry.
42:53.730 --> 42:59.940
We need static pressure and temperature at
station 2, now static temperature can be easily
42:59.940 --> 43:05.930
found out, because T 0 2 is equal to T 0 a,
which is already known. So, T 2 is equal to
43:05.930 --> 43:14.099
T 0 2 divided by 1 plus gamma minus 1 by 2
M 2 square, and M 2 is given as 0.4. So, if
43:14.099 --> 43:21.349
you substituted for all these values, we get
T 2 is equal to 239.3 Kelvin. And what about
43:21.349 --> 43:28.030
P 2, P 2 is equal to P 1 into the static temperature
ratios; T 2 by T 1 raise to gamma by gamma
43:28.030 --> 43:33.750
minus 1, T 1 has already been calculated in
the previous part, so static pressure comes
43:33.750 --> 43:38.300
out to be 13001 Pascal.
43:38.300 --> 43:45.920
Therefore, density can be calculated, density
is equal to P 2 by R T 2 which is 0.1893 k
43:45.920 --> 43:52.630
g per meter cube and velocity of the compressor
phase is Mach number times square root of
43:52.630 --> 43:59.000
gamma R T 2. So, that comes to be 124.03 meters
per second. Therefore, area at the compressor
43:59.000 --> 44:06.280
phase is mass flow rate is divided by u 2
times rho 2 that is 3.407 meter square. And
44:06.280 --> 44:11.480
once we know the area, we can also find out
the ratio, the diameter which is pi d square
44:11.480 --> 44:18.240
by 4 is equal to the area and therefore, diameter
can be calculated as 2.08 meters.
44:18.240 --> 44:26.520
So, in this particular problem which was basically
related to an air intake, we have been given
44:26.520 --> 44:31.369
some of the parameters like Mach number at
various stations and also the diffuser efficiency,
44:31.369 --> 44:38.109
given those parameters we could find out the
areas at different locations and also the
44:38.109 --> 44:43.750
pressure recovery, amount of losses, total
pressure losses occurring in the diffuser
44:43.750 --> 44:49.500
which is basically specified by the pressure
recovery. So, in this problem, we have basically
44:49.500 --> 44:55.490
tried to use the diffuser efficiency to calculate
the pressure recovery across the diffuser.
44:55.490 --> 45:02.380
Now, the last problem that we will be solving
in today’s tutorial section is something
45:02.380 --> 45:08.170
to do with in a nozzle and that is again component
of a gas turbine engine which we were discussing
45:08.170 --> 45:14.049
in the last lecture. And nozzle as you know,
it have been well basically have one of the
45:14.049 --> 45:20.069
parameters used to specify a nozzle is the
nozzle efficiency. So, we will see how we
45:20.069 --> 45:25.079
can use the nozzle efficiency in assessing
the performance of a nozzle.
45:25.079 --> 45:33.619
So, the problem statement number 4 is a turbo
jet engine operates at an altitude where the
45:33.619 --> 45:42.089
ambient temperature and pressure are 216.7
Kelvin and 24.444 four kilo Pascals.
45:42.089 --> 45:49.119
So, the ambient temperature and pressure are
given, the flight Mach number is 0.9, and
45:49.119 --> 45:55.550
the inlet conditions to the convergent nozzle
are 1000 Kelvin and 60 kilo Pascals. If the
45:55.550 --> 46:03.420
nozzle efficiency is 0.98, the ratios of specific
heats is 1.33, determine whether the nozzle
46:03.420 --> 46:09.440
is operating under chocked condition or not,
also determined the nozzle exit pressure.
46:09.440 --> 46:16.130
So, in this case, we have the ambient conditions,
the ambient temperature and pressure and then
46:16.130 --> 46:22.110
the nozzle entry conditions for the convergent
nozzle. So, in this case the nozzle is convergent,
46:22.110 --> 46:28.320
and we have the inlet conditions for the convergent
nozzle, the nozzle efficiency is also given,
46:28.320 --> 46:32.829
and based on this data we have to find out
whether this nozzle is operating under chocked
46:32.829 --> 46:39.190
condition and also we need to find out the
exit pressure of the nozzle.
46:39.190 --> 46:47.020
Now, let us go back to the definition of the
nozzle, we have defined efficiency of a nozzle
46:47.020 --> 46:54.920
as efficiency of a nozzle is equal to h 0
6 which is the nozzle entry stagnation enthalpy
46:54.920 --> 47:03.200
minus h 7 which is nozzle exit static enthalpy
divided by h 0 6 minus h 7 s where h 7 s is
47:03.200 --> 47:12.140
the static enthalpy for an isentropic process.
This can be simplified as T 0 6 minus T 7
47:12.140 --> 47:21.069
divided by T 0 6 minus T 7 s, which is in
turn equal to 1 minus T 7 divided by T 0 6
47:21.069 --> 47:29.230
divided by 1 minus T 7 s divided by T 0 6.
Now, T 0 6 is also equal to T 0 7, because
47:29.230 --> 47:34.339
there is no change in stagnation temperature
across the nozzle. So, we have 1 minus T 7
47:34.339 --> 47:42.900
by T 0 7 divided by 1 minus T 7 s by T 0 6.
Now, under chocked condition, we know that
47:42.900 --> 47:48.260
the Mach number at the exit of the nozzle
will become 1. So, if if we if we substitute
47:48.260 --> 47:55.240
for M is equal to 1, we know that T 0 7 by
T 7 will be equal to 1 plus gamma minus 1
47:55.240 --> 48:02.539
by 2 M 7 square. So, M 7 is equal to 0 as
is equal to 1 and therefore, that expression
48:02.539 --> 48:11.049
reduces to 1 minus 2 by gamma plus 1. Similarly,
the pressure under chocked condition is basically
48:11.049 --> 48:17.400
the critical pressure. And so, this temperature
ratio that we have T 7 s by T 0 6 will become
48:17.400 --> 48:25.200
equal to the critical pressure divide by the
P 0 6 that is total pressure, P c by P 0 6
48:25.200 --> 48:29.520
raise to gamma minus 1 by gamma.
So, efficiency of the nozzle reduces to 1
48:29.520 --> 48:38.230
minus 2 by gamma plus 1 divided by 1 minus
P c by P 0 6 raise to gamma minus 1 by gamma,
48:38.230 --> 48:45.829
or this pressure ratio that we have P 0 6
by P c can be reduced as 1 by 1 minus 1 by
48:45.829 --> 48:52.680
eta and into gamma minus 1 by gamma plus 1
raise to gamma by gamma minus 1. So, this
48:52.680 --> 48:57.060
pressure ratio which is basically relating
the nozzle entry total pressure to the critical
48:57.060 --> 49:02.430
pressure, critical pressure is the pressure
which the nozzle attains when the Mach number
49:02.430 --> 49:08.030
at the exit is equal to 1 if the nozzle is
operating under chocked condition. And if
49:08.030 --> 49:14.130
it is indeed operating under chocked condition,
the the parameters are easily calculated,
49:14.130 --> 49:19.510
because Mach number is 1. So, if we know the
ratio of specific heats and the nozzle efficiency,
49:19.510 --> 49:21.070
we can calculate the pressure ratio.
49:21.070 --> 49:27.059
So, in this case we can now calculate P 0
6 by P c, because rest of the parameters are
49:27.059 --> 49:32.530
known, efficiency is known, the gamma is known.
So, if we substitute for all those values,
49:32.530 --> 49:42.780
we have P 0 6 by P c is equal to 1.878, also
we can calculate P 0 6 by P a as equal to
49:42.780 --> 49:53.060
60 by 24.4 that is 2.45 kilo Pascal.
And therefore, we can see that from both these
49:53.060 --> 49:57.099
expressions, we can see that P c is greater
than P a, this is critical pressure is greater
49:57.099 --> 50:03.580
than the static pressure ambient which means
that the nozzle is operating under chocked
50:03.580 --> 50:10.859
condition. And so, that is if the nozzle exit
pressure exceeds the ambient pressure, this
50:10.859 --> 50:16.940
means that the nozzle is indeed chocking that
it the nozzle is operating under chocked condition,
50:16.940 --> 50:21.660
exit Mach number is equal to 1, the maximum
mass flow rate that the nozzle can handle
50:21.660 --> 50:28.329
has already been basing through the nozzle.
And once we can find out whether it is chocking
50:28.329 --> 50:32.960
or not, then the exit pressure is equal to
the chocking pressure which will be equal
50:32.960 --> 50:40.950
to the inlet pressure divided by the pressure
ratio 60 divided by 1.878 that is 31.95 kilo
50:40.950 --> 50:44.589
Pascals.
So, nozzle exit pressure can be calculated
50:44.589 --> 50:52.540
from this expression which is basically equal
to the critical pressure that is P 0 6, 60
50:52.540 --> 51:00.740
kilo Pascal divided by 1.878, 31.95 kilo Pascals.
So, in this problem we have we have been,
51:00.740 --> 51:07.599
we we could use the nozzle efficiency definition
and in cooperate that in the pressure ratio
51:07.599 --> 51:12.630
for calculating the critical pressure ratio
to determine whether the nozzle is operating
51:12.630 --> 51:17.859
under chocked condition or not. And once it
is chocked, then the exit pressure will be
51:17.859 --> 51:22.930
equal to the chocking pressure which again
can be calculated from the critical pressure
51:22.930 --> 51:28.529
ratio.
So, what we are solved in in today’s tutorial
51:28.529 --> 51:33.609
are four different problems, two of them related
to Brayton cycle and two other problem related
51:33.609 --> 51:39.020
to the components, one to the intake and one
to the nozzle. And of course, as I discussed
51:39.020 --> 51:44.390
in the last class, we will be talking up detailed
analysis and discussion of all these individual
51:44.390 --> 51:51.680
components in different lectures, in few lectures
from now. We will take up intake in detail,
51:51.680 --> 51:56.180
compressor and fan, combustion chamber, turbine,
nozzle, all these components will be detailed
51:56.180 --> 52:00.720
discussed in detailed subsequently, their
geometry instruction, etcetera also will be
52:00.720 --> 52:08.230
part of that discussion. So, I have few excises
problem for you to solve based on our discussion
52:08.230 --> 52:11.130
and the tutorial that we had today.
52:11.130 --> 52:17.230
So, you have four different problems as excises
a problem, the first one is for a Brayton
52:17.230 --> 52:23.010
cycle. So, a Brayton cycle with two stages
of compression, when two stages of expansion
52:23.010 --> 52:28.880
has an overall pressure ratio of 8.0, air
enters each stage of compressor at 300 Kelvin
52:28.880 --> 52:34.770
and each stage of the turbine at 1300 Kelvin.
Determine the thermal efficiency, part a with
52:34.770 --> 52:41.260
no regenerator, part b with an ideal regenerator,
part c if the compressor and turbine have
52:41.260 --> 52:49.119
80 percent efficiency each with no regenerator.
So, answer to part a is 35 percent, part b
52:49.119 --> 52:57.260
is 69.6 percent, and part c is if compressors
and turbine, both are efficiency of 80 percent
52:57.260 --> 53:02.760
and no regeneration, then the efficiency is
26 percent.
53:02.760 --> 53:09.010
The second problem is in a gas turbine plant,
air at the inlet is at 27 degree Celsius and
53:09.010 --> 53:16.960
0.1 MPa, the pressure ratio is 6.25, the maximum
temperature is 800 degree Celsius. Compressor
53:16.960 --> 53:21.960
and turbine efficiencies are in 80 percent
each, find part a, compressor work per kilo
53:21.960 --> 53:28.810
gram of air, part b turbine work per k g of
air, part c heat supplied per k g of air,
53:28.810 --> 53:36.310
and part d the cycle efficiency.
Answer to this is part a 259.4 kilo joule
53:36.310 --> 53:44.910
per kilo gram, part b 351.68 kilo joules per
kilo gram, part c 569.43 kilo joules per kilo
53:44.910 --> 53:49.590
gram, and part d 16.2 percent.
53:49.590 --> 53:54.530
The third problem is to do with an intake,
an aircraft is flying at an Mach number of
53:54.530 --> 54:00.089
0.8 at an altitude where the ambient static
pressure is 40 kilo Pascal. If the diffuser
54:00.089 --> 54:06.940
pressure recovery is 0.9, determine the isentropic
efficiency of the diffuser, in this case the
54:06.940 --> 54:13.160
isentropic efficiency comes out to be 0.738.
So, here we have a Mach number, flight Mach
54:13.160 --> 54:18.670
number, flight the static pressure is given
and the diffuser pressure recovery is specified,
54:18.670 --> 54:23.270
we need to find isentropic efficiency, answer
is 0.738.
54:23.270 --> 54:32.480
And the last problem is a nozzle of a turbo
jet engine, it develops a thrust of 519 Newton’s
54:32.480 --> 54:37.820
second per kilo gram, air craft is flying
at 240 meters per second, the pressure and
54:37.820 --> 54:44.440
temperature at the nozzle entry are 1.284
kilo Pascals and 993 Kelvin respectively.
54:44.440 --> 54:51.819
If the ratio of specific heats is 0.33, determine
the nozzle efficiency, the nozzle can be assumed
54:51.819 --> 54:56.460
to be operating under chocked condition. So,
here its specified at the nozzle is operating
54:56.460 --> 55:01.430
under chocked condition and which means that,
exit pressure is the critical pressure and
55:01.430 --> 55:07.470
so on, we need to find the nozzle efficiency.
If the thrust is given, the flight velocities
55:07.470 --> 55:14.200
are given and nozzle entry conditions are
also specified, so the nozzle efficiency is
55:14.200 --> 55:20.250
a 0.95.
So, here we have four different exercise problems
55:20.250 --> 55:25.650
and based on this particular tutorial, we
had discussed today, I am sure you would be
55:25.650 --> 55:31.480
able to find out the, solve these problems
and be able to find out the answers which
55:31.480 --> 55:36.400
have been given in these exercise problems.
So, that brings us to the end of today’s
55:36.400 --> 55:43.210
tutorial section, we will continue our discussion
on cycle analysis of jet engines in the next
55:43.210 --> 55:50.049
class, we will basically be taking up real
cycle analysis of different types of jet engine.
55:50.049 --> 55:55.550
And subsequently, we also have another tutorial
section on the real cycle analysis of jet
55:55.550 --> 55:55.619
engines.