WEBVTT
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We are talking about gas turbines, and we
just had a look at the basic theories of axial
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flow turbines, and radial flow turbines. In
today’s class, we will take a look at how
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simple problems related to these two kinds
of turbines can be solved using the theories
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that we have done. So, I will probably solve
a couple of problems for you, and then I will
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leave you with some problems for you to solve
for yourselves using the gas turbine theory
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that we have done in the last two or three
lectures. Now, these problems are set out
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for you, so that one you can use the theory
that we have done in the last lectures, and
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secondly you get a feel of the numbers.
In engineering, it is extremely important
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that you not only understand the equations
and the theories, that you also get a feel
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of the numbers. What are the possible correct
numbers under various situations for various
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parameters, that feel is extremely important
for good engineers. So, hopefully in today’s
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class, I will be able to expose to you some
of these engineering quantitative issues,
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and you get a feel for the numbers. So, in
today’s class, we will do the tutorial solved
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examples, and tutorial problems on axial and
radial flow turbines.
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Now, the radial and axial flow turbine that
we have done we know are slightly different
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kinds of turbines. Even though though they
do the same kind of job basically that they
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perform work. So, first we will take on the
axial flow turbines and take a look at a problem,
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which represents a typical axial flow turbine
problem and then we will discuss some of the
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tutorial problem that you could possibly attack
for yourselves and get a feel for the numbers.
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The first problem that we have is on axial
flow turbine and this problem statement shows
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that if you have a cooled axial flow turbine,
now we have discussed the cooling technology
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of axial flow turbines. So, if we have most
of the modern axial flow turbines do tend
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to be cool turbines, unless it happens to
be the last stage of a multistage turbine.
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So, let us take a look at a typical cooled
axial flow turbine problem and the problem
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shows that it has properties such that its
entry temperature is 1780 K and entry pressure
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is 1.4 mega pascals and carries a mass flow
of 40 kg per second, which means it is a medium
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sized engine. The mean radius data; that means,
mean is between the hub and the tip of a blade.
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So, the mean there is given as mach number.
The entry absolute mach number is given as
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0.3. The exit from the stator is given as
1.15, which means the flow has indeed gone
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a little supersonic at the exit of the stator
and the rotational speed at this mean radius
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is 400 meters per second. T 03 that is the
exit temperature form the turbine is gives
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as 1550 K and some simplistic angular parameters
are prescribed here; that is alpha 1 is equal
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to alpha 3 is equal to 0 degree; that means,
flow is coming in and going out axially that
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ofcourse simplifies a problem. But it is also
reasonably realistic. So, it is not just ideally
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simplified. But it does represent realistic
flow situation and then at the mean, the radius
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is 0.4 meters and as I mentioned, it represents
probably a medium sized engine.
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The axial flow velocity ratio across the rotor
C a 2 by C a 3 is equal to 1; that means axial
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velocity across the rotor is conserved. This
is again a simplifying assumption; but also
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reasonably realistic assumption. It is not
very ideal; it is a realistic assumption.
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The loss coefficients through the rotor and
the stator… The stator is of course we have
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been calling it nozzle. So, the loss coefficient
across the nozzle is given as 0.04 and that
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across the rotor is given as 0.08. So, these
two are prescriptions. These numbers are realistic
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numbers and represent a reasonably real compressor
turbine situation. Given are the thermo dynamic
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parameters. The specific rate heat ratio gamma
is given as 1.3 and the gas constant R is
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given as 287 Joules per kg K.
Now, those are standard parameters and they
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have been prescribed for this problem also.
Now, the problem asked for you to compute
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the flow properties along the mean line of
the stage, the degree of reaction, the total
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temperature delta T 0 based on stage loading
based stage loading. Now, this is something
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which we have discussed before. So, we will
try to compute this and then the isentropic
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efficiency which was been defined and the
flow areas at various axial stations and at
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the hub and tip radii at the nozzle inlet,
at the nozzle exit and at the rotor exit.
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So, let us take a look at these problems,
the parameters that are given and see what
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kind of solutions we can get out of this problem
that has been stated here.
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Now, we try to put the problems statement
in terms of at the mean radius as it is been
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prescribed in terms of the velocity diagram
that we have done before. And the problem
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statement now cast on to the velocity diagram
that we are familiar with shows that the entry
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temperature at station 1 has been prescribed
1780 K and pressure has been prescribed. Now
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at the station 1, alpha 1 there is now prescribed
as 0 degree and the entry mach number 0 3
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corresponds to c 1 which is shown in the diagram.
So, that c 1 corresponds to M 1 equal to 0.3
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at that particular station. Now, at the exit
from the stator nozzle, the mach number prescribed
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is 1.15 which is corresponding to the velocity
c 2; that is the absolute velocity.
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And that absolute velocity is prescribed in
the form of mach number as 1.15 depending
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on the local flow condition; that is local
temperature and the corresponding rotating
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speed U is given as 400 meters per second.
So, in that velocity diagram, you can see
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the value of U and that number quantitative
number is given as 400 meters per second.
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The other thing that is prescribed is across
the rotor, the axial velocity ratio now you
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can see C a 2 and C a 3 in the velocity diagram
here and that ratio is prescribed as 1 in
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this problem statement. Also prescribed as
that the exit, absolute flow angle alpha 3
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is 0. Now, this is something which is is similar
to the entry absolute flow angle.
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Now, corresponding to the to the prescription
that has been given here, we have the loss
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coefficients. The loss coefficients prescribed
for the nozzle or the stator nozzle blade
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is 0.04. The similar loss coefficient that
is prescribed for the rotor is 0.08. Normally,
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you would see that the rotor losses are somewhat
on the higher side; because it is a rotating
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blade. And in rotating blade, the losses are
normally of various origin and this is something
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we have discussed a little and you would probably
appreciate the fact that the number related
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to the rotor is on the higher side. Actually,
it is almost double of that of the loss in
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the nozzle.
The other thing that is given is that the
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exit total temperature is 1550 K, which means
it is undergone a total temperature drop of
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the order of 230 degree K across this turbine.
That is the amount of what you can expect
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that this turbine would have probably accomplished
in the process of functioning or in the process
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of transfer of energy from the gas to the
rotor and then mechanical work. Now, this
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is something which is what the turbine exists
for and the quantitative value prescribed
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here is 230 K; that is a pretty large number
and you could probably quickly notice that,
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that number is substantially higher than what
you can get in terms of temperature raise
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in any compressor stage. So, these are the
prescribed values for the problem that has
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been put in front of us. So, let us go along
and see whether we can complete the process
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of solution of this particular problem.
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Now at station 1, if we look at the numbers
that are given, and try to find the static
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parameters; if we can get value of T 1 and
if we apply the isotropic flow conditions
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at station 1, we get T 1 is equal to 1756.3
K using the isentropic flow conditions. And
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P 1 correspondingly using the same isentropic
conversion from total to static condition,
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we get 1321 kilo pascals converting from mega
pascal to kilo pascals. Now, we are using
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isentropic flow condition, because as we you
have done before in various thermo dynamics.
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That at any particular station, conversion
from static to total or total to static can
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be simply by using the ideal isentropic relations.
Because at that particular station, it is
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assumed that the flow is undergoing isentropic
conversion from total to static or static
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to total; that is an assumption; that is ofcourse
the theory and we can always invoke that theory
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at a particular station. There is no process
from 1 to 2; it is only at a particular station
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1. So, we can always invoke the isentropic
flow condition at that particular station
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in an ideal manner. So, that is what we have
done and we have got the static temperature
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and static pressure at station 1. Now, we
can find the velocity C 1. As I mentioned,
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M 1 that was given is corresponding to C 1
and that is dependent on the local temperature
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that we have just found.
And if we invoke the local temperature that
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has been just found, we get a velocity which
is 242.8 meters per second corresponding to
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mach 0.3 that has been prescribed there. Now,
since alpha 1 is given as 0, C 1 effectively
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becomes equal to C a 1 and corresponding C
w 1 would be 0. So, if you keep your eye on
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the velocity triangle, you would see that
we are trying to solve all the parameters
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that are shown in the velocity triangle as
far as possible. And the trick of solving
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such problems is that if you can find all
the parameters that are shown in this velocity
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triangle, you would be able to calculate the
performance of the turbine completely and
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fully.
So, one of the things that you may like to
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do while solving a problem is to compute all
the parameters; that are shown in this velocity
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triangle. So, let us do that and we found
at station 1, C w 1 would be equal to 0. Correspondingly
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then at station 1, we can say that the area
A 1 would be equal to the mass flow m dot
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gas divided by the local density rho 1 into
C a 1. Now, this ofcourse means that we are
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simply using the continuity condition at station
1 and simply using the continuity condition,
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we can find out the area at station 1. You
could notice that you would have to calculate
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the density rho 1 corresponding to T 1 and
P 1 that we have just calculated.
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So, if you calculate rho 1 and plug in that
value here, you would get an area that is
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a 0.0628 meters square at station 1; that
is at entry to the turbine stator nozzle blade.
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Now, applying the same isentropic method of
conversion from total to static, we can find
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the temperature at station 2. Because through
the stator nozzle, a lot of acceleration has
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taken place and the total temperature as remain
constant. But static temperature has undergone
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a large amount of change and that shows up
here that the static temperature now is 1485.4
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K substantially lower than 1756.3 K that we
have found at station 1. So, that is a drop
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due to the large change from potential to
kinetic energy through the stator nozzle.
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And correspondingly, we find a C 2 using the
same method again that is conversion from
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mach number to velocity using the local temperature,
which we have found as 1485.5. And we get
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a velocity now corresponding to mach 1.15
velocity of 856.1 meters per second. Now,
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this is a very high velocity jet and intended
purpose ofcourse is that this would now be
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made to impinge on the rotor blade to make
the rotor rotate. Now, as prescribed, 4 percent
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of the kinetic head is lost in the nozzle
blade. The nozzle loss coefficient was prescribed
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as 0 4 and that is corresponding to 4 percent
of the kinetic head as per the definition
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of the loss coefficient.
And as the result of which, the we can sort
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of do a reverse engineering and back calculate
and find that the ideal T 2 would have been
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T 2 prime would have been 1503.5 K and corresponding
to that, ideal C 2 prime would have been 877
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meters per second. Now, this is something
we have discussed before in our lecture that
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through the nozzle; because of the real flow
situation, the axial velocity C 2 is often
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a little less than what would have been the
ideal velocity C 2 prime. Now, this change
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from real to ideal is quite often a situation
that happens; because you see at the end of
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the nozzle, you are trying to achieve a sonic
velocity. Sometimes as we seen this particular
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problem, you are trying to achieve a slightly
supersonic velocity.
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So, what happens is you end of getting a little
less than what you prescribed or what you
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design for. And hence, the loss coefficient
comes in to the picture over here and this
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slight loss of a performance through the stator
nozzle needs to be accurately facted in your
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calculation. So, that you know exactly how
much you are losing in terms of the acceleration
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that the stator nozzle is expected to achieve.
So, that is what we see here that the ideal
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velocity could have been a little more through
the stator nozzle and as it happens, we are
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getting a little less of the order of almost
20 meters per second, 21 meters per second.
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Now, this is something what every turbine
design and needs to keep an eye on.
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Now, from the isentropic laws, we can also
find the value of static pressure P 2 at the
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end of stator nozzle and since P 0 2 is kind
of only factors in the 4 percent loss, it
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tells us that P 2 would be 625.5 kilo pascals.
Now, at station 3, it is given that exit angle
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is 0; alpha 3 is 0 and at station 3, it is
prescribed that alpha 2 could be corresponding
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to sine inverse psi into U divided by C 2.
Now, this is from the definition of the work
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done which we have done before; that is psi
is equal to delta of enthalpy change across
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the stage divided by U square, which is the
rotating speed.
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So, it is used as a normalizing parameter
and this is the definition of blade loading
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or the stage loading quite often used in case
of turbines and also sometimes in case of
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compressors. Now, if you do that, you get
a parameter that is C a by U into tan alpha
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2; from which, we wrote down as above that
you can get a parameter in terms of alpha
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2. Now, c p ofcourse from thermo dynamics
we know is gamma R divided by gamma minus
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1. And if we plug in all those values, the
stage loading psi can be computed as a 1.7878
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and this is ofcourse often as a non-dimensional
parameter and this number is often an indication
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of the amount of loading that you can prescribed
for a particular stage.
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Higher this number is higher the stage is
being us to perform. And so, keeping an eye
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on this number tells you whether the stage
has been sufficiently loaded or the stage
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loading has been maximized. And in case of
aircraft gas turbine engines as we have discussed
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before, we would like to maximize this number
without causing and U gas dynamic are structural
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problems. Now, from all these calculations
what we get is alpha 2 is equal to 56.6 degree.
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This is the angle at which the flow is coming
out from the stator nozzle. Correspondingly,
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the C a 2 the axial component of this velocity
is before 170 meters per second.
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And C w 2, that is the old component or tangential
component of the absolute velocity that would
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be 715 meters per second. And ofcourse, your
V w 2 that is the relative tangential component
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would be 315 meters per second. So, some of
these are numbers which you would probably
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like to get used to. Because as you see these
numbers are substantially different from what
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you found in station 1, where C w 1 was ofcourse
0 and hence the change across the stator nozzle
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only. Even when no work has been done, a substantial
change in the velocity field has been achieved
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through a large acceleration and this is what
was ofcourse intended by there so called stator
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nozzle.
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If we move forward with these numbers and
move on to the rotors, what we find there
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is that at the entry to the rotor, the relative
flow angle beta 2 can be now found from the
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tangential velocity V w 2 and the axial velocity
C a 2 and if we plug in those numbers, we
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get beta 2 is equal to 33.8 degrees. Now,
it can be also shown that corresponding to
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this, that the M 2 relative at the entry to
the rotor that is relative mach number to
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the entry to the rotor would be 0.76, which
is subsonic. Now, we just saw that the exit
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from the stator M 2 is actually supersonic;
that was prescribed as 1.15 and we see by
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conversion through velocity triangle, the
M 2 relative is now subsonic.
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Let us take a quick look at the velocity triangle;
you see corresponding to our calculations,
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c 2 has been prescribed clearly as supersonic.
We got the velocity value and we got the v
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2 velocity value. When we converted v 2 to
M 2 relative, this is clearly subsonic by
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the local flow conditions and this is exactly
what is been achieved through this velocity
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vector transformation that the exit velocity
here can be clearly supersonic. But the entry
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here would be subsonic and as I have been
mentioned earlier, this is done deliberately
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that in most of the modern turbines today.
Still admit flow in to the rotor which is
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subsonic.
It is indeed possible to admit flow into the
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rotor which is supersonic; that means you
can indeed have supersonic rotors. But normally
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in aircraft gas turbine, even today it is
not a done thing. Supersonic rotors have been
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used in other applications notably in termission
the applications in space craft; but they
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are normally not used in aircraft gas turbines.
So, in aircraft gas turbines, we still keep
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the v 2 value subsonic. This is deliberate
and very done very consciously for the turbine
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designs; so that, the losses through these
turbines are somewhat under control. As you
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can see, the loss to the rotor is already
little on the higher side and through the
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shocks, it would have gone even higher and
hence, most of the aircraft gas turbine even
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today are essentially subsonic rotors.
Let us get back to the calculations that we
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were doing. Through the rotor, we can now
find the area at the station 2 which is at
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the entry to the rotor and that is again found
by using the continuity condition. And the
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continuity condition tells us that value of
A 2 can be written down in terms of the various
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flow parameters there and it comes out to
the of the order of 0.58 meters square. Now,
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axial velocity acts at the exit of the stage
can be found simply by using the velocity
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triangle that we were looking at and this
comes out to be 470 meters per second. Now,
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we have seen that this axial velocity is constant
across the rotor. So, the value of C a 2 would
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be exactly same. So, that is how this velocity
has been computed.
25:15.679 --> 25:22.880
Now, at the exit again it is been prescribed
that alpha 3 is 0. So, C w 3 would again be
25:22.880 --> 25:32.070
0. And therefore, the value of V w 3 would
be equal to the rotating speed of the rotor
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and that is 400 meters per second and corresponding
relative velocity leaving the rotor would
25:39.020 --> 25:47.159
be 617 meters per second. Now, this is ofcourse
the relative velocity. So, as we can see that
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the relative velocity here is also essentially
subsonic and we will see that it is a little
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on the higher side compared to V 2. Now, the
exit flow angle beta 3 can be computed and
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this comes out to be 40.3. This is invoking
the fact that alpha 3 is indeed 0 and static
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temperature at station 3 now can be computed
again invoking the isentropic laws as we have
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done in station 1 and 2.
And if we do that, we find the temperature
26:21.059 --> 26:29.070
to be 1461 K. The total temperature ofcourse
was prescribed. So, it is it is easy to use
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the isentropic laws and find the static temperature
that is 1461 K and the sonic speed at that
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particular station is 738.3 meters per second
and we can very well see that this velocity
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V 3 would be subsonic. So, the exit velocity
from the rotor also is subsonic. So, both
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the entry and the exit velocity from the rotor
are deliberately kept subsonic to avoid shock
27:00.110 --> 27:07.310
losses, which again would reduce the efficiency
of the turbine; increase the losses and reduce
27:07.310 --> 27:14.150
the efficiency. And most of the designer presently
designers are extremely sensitive to efficiency
27:14.150 --> 27:19.730
of the turbines and hence they try to keep
the turbines subsonic even today.
27:19.730 --> 27:25.679
And the other thing is most of the work done
is being accomplished higher level of work
27:25.679 --> 27:31.200
done is being accomplished by increasing the
turbine entry temperature. And this increase
27:31.200 --> 27:37.340
of turbine entry temperature is quite often
the focus of attention of modern research
27:37.340 --> 27:42.130
and more and more cooling technologies applied.
And we have prescribed that this particular
27:42.130 --> 27:47.540
problem actually applies blade cooling and
that is how, that high entry temperature was
27:47.540 --> 27:55.659
admitted and hence we see that we have a turbine
here, which is a cool turbine. The entry temperature
27:55.659 --> 28:04.330
is very high and that allows the designer
to keep the rotor subsonic and hence, the
28:04.330 --> 28:11.470
rotors are not under any compulsion to go
supersonic to get more work done.
28:11.470 --> 28:17.190
Rotors can be kept subsonic. Another reason,
why modern designers would like to keep the
28:17.190 --> 28:24.580
rotors subsonic is because many of the modern
rotors are also subjected to cooling technology.
28:24.580 --> 28:29.250
Some of these cooling technologies as we have
seen are extremely intricate and if we have
28:29.250 --> 28:35.281
supersonic flow over them, it will be an extremely
complicated fluid mechanic situation. And
28:35.281 --> 28:41.080
most of the turbine designers would like to
avoid that kind of complication in their turbine
28:41.080 --> 28:46.840
design; because it would impact on the efficiency
or aero thermodynamic efficiency or what we
28:46.840 --> 28:53.710
call isentropic efficiency of the turbine
quite adversely. So, as we see by the numbers
28:53.710 --> 28:56.940
that they are deliberately kept subsonic.
28:56.940 --> 29:03.650
Let us proceed with this problem and see whether
we can complete it. The degree of reaction
29:03.650 --> 29:09.789
that we have prescribed before, as we see
now if it is computed using all the parameters,
29:09.789 --> 29:16.150
comes out to be only 0.106. And as we have
discussed in many of the modern turbines,
29:16.150 --> 29:22.289
the degree of reaction value are of that order
much lower compared to what you have done
29:22.289 --> 29:28.220
probably in your compressor chapter why the
degree of reaction values are somewhat much
29:28.220 --> 29:33.450
higher than these numbers. So, you got to
get the feel of these numbers. The degree
29:33.450 --> 29:38.270
of reaction in turbine is substantially lower
than that of compressors and these are the
29:38.270 --> 29:44.600
numbers you would need to get used to.
The mach number at the exit then the absolute
29:44.600 --> 29:52.690
mach number is 0.6375and the relative exit
mach number as we just saw in subsonic and
29:52.690 --> 29:59.700
it is 0.836. It is slightly higher than what
it came in with in to the rotors. So, the
29:59.700 --> 30:07.169
rotor is clearly a reaction turbine. It has
accelerate the flow a little and has produced
30:07.169 --> 30:13.610
a reaction and hence it is a reaction turbine.
The relative total temperature and pressure
30:13.610 --> 30:21.440
at the exit can also be computed to compute
the calculation and we get T 0 3 relative
30:21.440 --> 30:31.070
as 1614 K and P 0 2 relative as 897.3 kilo
pascals by applying the rotor loss. I need
30:31.070 --> 30:36.080
to mention here very clearly here that the
rotor loss coefficient that was prescribed
30:36.080 --> 30:42.480
should be applied across the rotor relative
frame.
30:42.480 --> 30:49.370
So, the 8 percent loss that was prescribed
should be applied to P 0 2 relative and if
30:49.370 --> 30:57.110
you do that, you get P 0 3 relative at as
872.7 kilo pascals. Now, this is something
30:57.110 --> 31:03.150
you need to again very clearly understand.
That the loss prescription that is normally
31:03.150 --> 31:09.060
given should be applied in the absolute scale
in the stator nozzle and should be applied
31:09.060 --> 31:15.760
in the relative frame in the rotor and when
you are calculating the pressure losses respectively
31:15.760 --> 31:23.390
across these two rows of blades. So, losses
in the rotor take place in the relative frame;
31:23.390 --> 31:30.830
in the stator, it takes place in the absolute
frame. So, you need to take note of this and
31:30.830 --> 31:38.320
do your future problems accordingly. Using
the isentropic relation, we can now find the
31:38.320 --> 31:50.080
stator static value that is P 3 as 566 kilo
Pascals and the total pressure is 731 kilo
31:50.080 --> 31:54.210
Pascals.
Correspondingly, now we can find the exit
31:54.210 --> 32:00.830
area at station 3 and this exit area is now
A 3 equal to… Again using the continuity
32:00.830 --> 32:09.820
condition and using m dot which is constant
across the stage; the local density rho 3
32:09.820 --> 32:17.289
and C a 3, the local axial velocity C a 3.
If we apply all those, we get at station 3
32:17.289 --> 32:24.929
the area is equal to 0.63 meters per second.
You can see that the area across the stage
32:24.929 --> 32:31.850
is going slightly higher; because the flow
is axially losing its pressure and temperature
32:31.850 --> 32:37.240
and it is losing its density and hence the
area requirement would be higher, as you go
32:37.240 --> 32:45.330
through the stage. So, these are the numbers
you need to understand and made and try to
32:45.330 --> 32:51.360
make it clear that when you get these numbers
when you are solving a problem, certain numbers
32:51.360 --> 32:56.340
would tell you whether your solution is proceeding
along the right path and you get the feel
32:56.340 --> 33:02.080
of the number that you are probably okay,
when you get the numbers in hand.
33:02.080 --> 33:09.510
Let us finally put them altogether; the performance
parameters that we get from this in in terms
33:09.510 --> 33:15.899
of temperature ratio. The total temperature
ratio is 1.148; the total pressure ratio or
33:15.899 --> 33:25.149
the pressure drop across the stage is 1.915,
which is a modest total pressure ratio across
33:25.149 --> 33:31.440
turbine stage. Many of the modern turbines
could possibly have much higher pressure ratio
33:31.440 --> 33:40.970
across its stage. The efficiency of this particular
turbine can now be computed using the thermodynamic
33:40.970 --> 33:48.490
laws that we have done before and it is simply
is 92.9 percent, which is a fairly good efficiency.
33:48.490 --> 33:55.510
Many of the modern turbines can have efficiency
slightly higher than these; but 92.9 nearly
33:55.510 --> 34:02.649
93 percent is definitely good efficiency.
At each of these stations, the blade height
34:02.649 --> 34:09.399
can also be calculated simply using the geometry
at that particular station; that is h i equal
34:09.399 --> 34:16.190
to a i divided by twice pi or a mean at the
particular station. And if we put them all
34:16.190 --> 34:25.329
in table in station 1, 2 and 3, we get the
areas which we have computed before and these
34:25.329 --> 34:32.009
areas are now shown over here. And then across
the rotor, as you can see there is you know
34:32.009 --> 34:37.730
increase in area and across the stage indeed,
there is been a slight increase in area. The
34:37.730 --> 34:45.139
height as you can see again has change across
the stations. Correspondingly, the tip radius
34:45.139 --> 34:52.790
and the hub radius can be found using the
geometry that is available with us. So, this
34:52.790 --> 35:01.329
table gives us an idea about the changes that
typically occur across a stator and the rotor.
35:01.329 --> 35:05.039
This may vary from one kind of turbine to
another.
35:05.039 --> 35:10.789
If you go from HP to LP, these variations
could be slightly different. If you were crossed
35:10.789 --> 35:16.430
to last stage of the LP, it will again slightly
different. It depends on the turbine design
35:16.430 --> 35:23.539
and as I just mentioned that the pressure
ratio 1.95915 is a modest pressure ratio.
35:23.539 --> 35:28.660
Many of the modern turbines could indeed have
pressure ratios quite higher than this. And
35:28.660 --> 35:35.400
if they are, then all the values that you
see here in terms of area, height and the
35:35.400 --> 35:43.339
tip radius and so on would indeed change from
station 1 to 2 to 3. So, it depends on turbine
35:43.339 --> 35:49.710
and every turbine would have unique such numbers
except that if you get the feel of the numbers,
35:49.710 --> 35:55.089
you would know whether the numbers are coming
out in a correct trend and that trend is what
35:55.089 --> 36:00.910
you can get the feel of when you are solving
a problem. But the numbers for every turbine
36:00.910 --> 36:05.209
would be quite different from each other.
36:05.209 --> 36:10.829
What I will do now is I will prescribe a few
problems for you to solve for yourself using
36:10.829 --> 36:16.140
the methodology that we have just solve the
problem. So, the first problem is actually
36:16.140 --> 36:23.400
that of a impulse turbine, where reaction
is 0. And if we do that, we prescribe that
36:23.400 --> 36:32.450
it has a entry pressure of 414 kilo pascals,
static pressure of 207 and exit pressure of
36:32.450 --> 36:39.749
P 0 from the stator nozzle of 400 kilo pascal.
So, it is lost about 14 kilo Pascals across
36:39.749 --> 36:47.210
the stator nozzle. The exit static pressure
from the whole stage is given as 200 kilo
36:47.210 --> 36:55.200
Pascals. Now, when operating with a rotating
speed U mean at the mean of the rotor blade,
36:55.200 --> 37:00.349
291 meters per second.
The entry temperature given is a modest 1100
37:00.349 --> 37:09.089
K and alpha 2 that is exit from the stator
nozzle is prescribed as 70 degrees; that is
37:09.089 --> 37:15.630
from the exit of the stator nozzle. Again
it is assumed or prescribed that C 1 that
37:15.630 --> 37:23.910
is entry velocity is equal to the C 3, which
is exit velocity from the stage. What is asked
37:23.910 --> 37:31.319
for is compute the total-to-total efficiency
of this particular stage prescribed or the
37:31.319 --> 37:38.729
values of C p and gamma, which are the normal
values. The C p is 1148 k g per kilo joules
37:38.729 --> 37:47.430
k g per k and gamma is 1.333. So, if you use
those numbers, you would probably get an answer
37:47.430 --> 37:54.869
close to 91 percent as efficiency of the stage.
I hope you can sit down and solve the problem
37:54.869 --> 37:58.650
using the method that we have adopted before.
37:58.650 --> 38:06.950
Let us go on to the next problem which corresponds
actually now to not a impulse turbine. Let
38:06.950 --> 38:12.039
us look at the problem and we will see that
the axial velocity as prescribed through the
38:12.039 --> 38:19.249
axial turbine is held constant by design;
that means, C a 1 is equal to C a 2 is equal
38:19.249 --> 38:25.289
to C a 3. And this entry and exit velocities
are also axial; that means, C a 1 is equal
38:25.289 --> 38:33.269
to C 1; C a 3 is equal to C 3. It is a simplified
problem and the flow coefficient that is C
38:33.269 --> 38:41.349
a by U through the rotor is 0.6. Now, if you
do that, the gas leaves the nozzle with alpha
38:41.349 --> 38:46.660
2 equal to 68.2; that is at the exit of the
stator nozzle.
38:46.660 --> 38:53.900
Now, at whole mean diameter that means, across
the stage if you if you just travels through
38:53.900 --> 39:01.130
the mean diameter from the stator entry to
the rotor exit, you are required to find the
39:01.130 --> 39:08.229
stage loading coefficient, psi; the relative
flow angles at rotor mean diameter, beta 2
39:08.229 --> 39:15.140
and beta 3; those are the relative flow angles;
the degree of reaction, R x and the total-to-total
39:15.140 --> 39:20.190
and total-to-state static efficiencies that
we have prescribed in our lecture earlier.
39:20.190 --> 39:26.130
And you are required to find both the efficiencies
and the answers given are that your stage
39:26.130 --> 39:34.049
loading coefficient is 1.5, the relative flow
angles are 40 and 59 degrees, the degree of
39:34.049 --> 39:41.739
reaction is 0.25 and the two efficiencies
the total efficiency is 90.5 and the total-to-static
39:41.739 --> 39:48.049
efficiency is 81.6. As we have discussed during
the lecture, the total-to-static efficiency
39:48.049 --> 39:55.380
is normally lower than the total total-to-total
efficiency. So, these are the answers you
39:55.380 --> 39:58.210
would get by solving the second problem.
39:58.210 --> 40:05.509
Let us look at the third problem. This is
following the design data applied to an uncooled
40:05.509 --> 40:13.200
axial flow turbine. You get P 0 1 equal to
400 kilo pascals; T 0 1 equal to 859 K and
40:13.200 --> 40:21.349
at mean radius alpha 2 equal to 63.8 degree;
degree of reaction is 0.5 and phi, the flow
40:21.349 --> 40:28.150
coefficient as 0.6; P 1, static temperature
at the exit at the entry is 200 kilo pascals
40:28.150 --> 40:35.059
and total-to-static efficiency is now prescribed
in this problem as 85 percent. If the axial
40:35.059 --> 40:40.119
velocity is held constant as in the last problem,
through the whole stage you are required to
40:40.119 --> 40:47.039
compute the specific work done by the gas,
the blades speed and the stage exit static
40:47.039 --> 40:52.589
temperature. Now, these are the numbers that
you would get. You would get 131 kilo joules
40:52.589 --> 41:00.880
per kg and the blade speed would be 301 meters
per second and the stage exit static temperature
41:00.880 --> 41:04.979
would be 707.6 K.
41:04.979 --> 41:10.440
If you look at the fourth problem, it is an
axial flow turbine now again with a cooled
41:10.440 --> 41:15.809
rotor. Now, as I mentioned, you could have
cooled rotor. So, you have a problem here
41:15.809 --> 41:23.280
which says that the mass flow is 20 kg per
second; the entry temperature total temperature
41:23.280 --> 41:32.900
is 1000 K and P 0 1 is 4 bar; C a is 260 meters
per second and this is assumed to be constant
41:32.900 --> 41:40.369
through the entire stage. Again it is prescribed
that C 1 is equal to C 3; U mean is 360 meters
41:40.369 --> 41:47.069
per second; alpha 2 that is exit from the
stator nozzle is 65 degree; and alpha 3 at
41:47.069 --> 41:54.440
the exit to the stage is 10 degree and the
nozzle loss coefficient is prescribed as 0.5.
41:54.440 --> 42:03.280
You are required to compute the relative flow
angles at the rotor mean diameter beta 2 and
42:03.280 --> 42:10.440
beta 3 at the mean, the degree of reaction
R x, the stage loading coefficient psi, the
42:10.440 --> 42:19.089
power output of this working turbine and the
nozzle exit throat area, neglecting the real
42:19.089 --> 42:25.199
flow effect; that means, you assume that is
an isentropic ideal flow and you can still
42:25.199 --> 42:33.969
find the nozzle exit throat area. And if you
adopt all these prescriptions, the answers
42:33.969 --> 42:40.750
that you would be getting as the relative
low angles would be 37.2 degrees and 57.4
42:40.750 --> 42:47.130
degrees respectively. The degree of reaction
would be 0.29; the stage loading co efficient
42:47.130 --> 42:57.430
would be very good 3.35; power output correspondingly
would be 43.40 kilo watts and the nozzle exit
42:57.430 --> 43:04.819
throat area would be 0.04 meter per square.
This is the throat area at the exit of the
43:04.819 --> 43:14.949
nozzle, which is typically at the end of the
converging channel that is prescribed or normally
43:14.949 --> 43:18.319
used in the stator nozzle.
43:18.319 --> 43:23.969
The fifth problem that is prescribed to you
is the axial flow turbine with cooled nozzle
43:23.969 --> 43:29.900
and rotor blades; that means, both nozzle
and rotor are prescribed to be cooled operates
43:29.900 --> 43:35.809
with the following flow parameters at the
reference diameter, which is the mean. The
43:35.809 --> 43:42.569
entry temperature is a respectable good modern
temperature that is 1800 K. The entry pressure
43:42.569 --> 43:49.859
is 1000 kilo pascals; C a 3 by C a 2 as we
have done in all problem is equal to 1; that
43:49.859 --> 43:56.489
is axial velocity across the rotor is conserved
mach number at the exit of the stator nozzle
43:56.489 --> 44:05.589
is 1.1; that is it is allowed to go slightly
supersonic and U mean is 360 meters per second.
44:05.589 --> 44:14.500
Alpha 2 prescribed is 45 and alpha 3 at the
exit of the rotor is prescribed as 5 degree.
44:14.500 --> 44:20.770
You are required to compute the following
that is C 2 at the exit of the stator nozzle;
44:20.770 --> 44:28.520
C a 2, the axial velocity between the stator
and the rotor; C w 2, the absolute tangential
44:28.520 --> 44:35.680
component of the velocity at station 2 between
the stator and the rotor; C 3, the absolute
44:35.680 --> 44:42.700
velocity at the exit of the rotor; C a 3,
the axial velocity and C w 3, the tangential
44:42.700 --> 44:47.479
velocity at the exit of the rotor. So, all
these velocities that is you are required
44:47.479 --> 44:53.410
to complete the velocity triangle that we
have seen before; the total temperature drop
44:53.410 --> 44:59.779
across the stage and correspondingly ofcourse
the total temperature ratio across the stage,
44:59.779 --> 45:07.780
the pressure ratio across the stage and then
P 0 3 for a polytrophic efficiency of 89 percent.
45:07.780 --> 45:13.499
And if you adopt all the prescriptions that
are given and the velocities that are another
45:13.499 --> 45:20.410
parameter that asked for, you get those numbers.
The velocities are 830 meters per second;
45:20.410 --> 45:30.089
585 meters per second and 587 meters per second
for C 3 and 585 meters per second for C a
45:30.089 --> 45:38.229
3 and C w 3. You get a temperature drop of
the order of the 184 K; corresponding temperature
45:38.229 --> 45:45.799
ratio is 1.148 and the corresponding pressure
ratio or pressure drop across the turbine
45:45.799 --> 45:55.869
is 1.792, which gives you exit pressure of
591 kilo Pascals. So, those are the answers
45:55.869 --> 46:03.029
you are likely to get, if you adopt the prescriptions
that are given in this problem. Let us now
46:03.029 --> 46:10.160
take a look at a problem of radial flow turbine;
this radial flow turbine which we have done
46:10.160 --> 46:15.390
in the last class. Using the simple theories
that we have done in the last class, we can
46:15.390 --> 46:17.359
try to solve this problem.
46:17.359 --> 46:23.890
What is prescribed here is that the mass flow
is 2 kg s per second, which means it is a
46:23.890 --> 46:30.920
small turbine. The entry pressure is 400 kilo
pascals; the entry temperature is 1100 K and
46:30.920 --> 46:36.979
the entry pressure undergoes 1 percent loss;
that means, the pressure at the exit of the
46:36.979 --> 46:47.519
stator nozzle is 0.99 times P 0 1. The nozzle
exit angle alpha 2 is 70 degrees; the polytrophic
46:47.519 --> 46:54.869
efficiency is prescribed as 0.85 for the turbine.
The rotor maximum diameter that is at the
46:54.869 --> 47:06.859
tip of the radial turbine is 0.4 meters. The
radial flow entry in to the rotor is prescribed
47:06.859 --> 47:13.869
and that is assumed to be or allowed to be
assumed to be equal to the exit of the rotor,
47:13.869 --> 47:20.599
which is C a through and it is supposed to
be axial. So, the axial velocity at the exit
47:20.599 --> 47:27.150
is to be taken as equal to the radial velocity
relative radial velocity at the entry to the
47:27.150 --> 47:32.230
rotor.
The hub to tip radius ratio is 0.4 and T 0
47:32.230 --> 47:41.329
3 at the exit is 935 K. The thermo dynamic
parameters prescribed here gamma equal to
47:41.329 --> 47:50.729
1.33; R equal to 287.287 kilo joules per kg
K and C p as 1.158 kilo joules per kg K. The
47:50.729 --> 47:58.029
problem asked you to compute rotors tip speed,
rotation speeds and rpm of the rotor; the
47:58.029 --> 48:06.189
mach number velocities, the rotor width at
the tip of the rotor and T 0 2 relative; the
48:06.189 --> 48:12.420
stagnation pressure, mach number, hub and
tip radii at rotor exit; at the rotor exit
48:12.420 --> 48:21.079
plane, the relative velocity V 3, T 0 3 relative,
relative flow angle beta 3 and the relative
48:21.079 --> 48:27.219
mach number M 3 at mean radius; ofcourse,
they are all connected to each other; the
48:27.219 --> 48:34.059
values of beta 3 are the relative flow angle
and M 3 relative at different radii at rotor
48:34.059 --> 48:42.489
exit. Now, these are the answers that are
expected out of this problem statement.
48:42.489 --> 48:47.960
Let us look at the radial flow turbine diagram
that we have done before. What is prescribed
48:47.960 --> 48:54.881
is the tip diameter of the rotor is prescribed
as 0.4. At the entry to the turbine, what
48:54.881 --> 49:02.980
is prescribed is .0 P 0 1 equal to 400 kilo
pascals, T 0 1 equal to 1100 K and alpha 2
49:02.980 --> 49:13.459
over here is prescribed as 70 degree. V 2
r is that means this V 2 r here is to be equal
49:13.459 --> 49:20.059
to this C a 3. So, this radial component which
is equal to V 2 here in this diagram as is
49:20.059 --> 49:26.420
normally often done in many gas turbines would
be taken as equal to C a 3; that is this C
49:26.420 --> 49:34.609
a 3 that is coming out of the rotor; the absolute
component of the velocity and its axial direction.
49:34.609 --> 49:43.219
The T 0 3 prescribed is 935 K and the hub
to tip radius ratio of the rotor at the exit
49:43.219 --> 49:48.819
station, you see there is a hub here station
and there is a tip of the rotor at the exit
49:48.819 --> 49:54.509
and this is prescribed as 0.4. So, these are
the prescriptions with which we go forward
49:54.509 --> 49:58.380
the loss across the stator nozzle is 1 percent.
49:58.380 --> 50:03.369
So, with this we move forward to solve the
whole problem. The rotor tip speed can be
50:03.369 --> 50:11.119
simply found by using the isentropic laws
and that tells us that if you use the enthalpy
50:11.119 --> 50:20.410
drop, you get a rotor tip speed of 437 meters
per second. Now, this rotational speed then
50:20.410 --> 50:30.689
can be found simply by dividing this U by
r and that gives us to 2185 radians per second.
50:30.689 --> 50:37.460
Corresponding to which, you would get the
rpm equal to 20870 rpm. So, as you can see
50:37.460 --> 50:44.119
here, typical radial turbine would be running
at very high rpm compared to an axial flow
50:44.119 --> 50:52.699
turbine. The corresponding at the rotor tip,
the value of C 2 would be 465 meters per second
50:52.699 --> 50:58.089
using the velocity triangles that we have
just seen. The corresponding V 2 r would be
50:58.089 --> 51:04.460
159 meters per second. The local speed of
sound using the local temperature T 2, you
51:04.460 --> 51:11.970
have to do that and if you do that, you get
T 2 equal to 707 K and corresponding sonic
51:11.970 --> 51:17.190
speed as 620 meters per second.
51:17.190 --> 51:25.069
And then the nozzle exit mach number M 2 comes
out to be 0.75. So, as you see here it is
51:25.069 --> 51:30.930
not sonic; it is still subsonic. The area
at the rotor tip now is can be found using
51:30.930 --> 51:39.819
the continuity condition, as we have always
done before and this comes out to be .0 0.0164.
51:39.819 --> 51:45.950
You have to find the local density and before
that, the local pressure. So, if you do that,
51:45.950 --> 51:51.980
you get this area at station 2. Using this
area at station 2 which is the entry to the
51:51.980 --> 51:57.759
rotor, you can find the width of the rotor
tip; that is the axial width of the rotor
51:57.759 --> 52:05.689
tip. Using the geometry, you can find the
rotor tip width to be 0.013 meters. This is
52:05.689 --> 52:13.049
simple geometry and then you can find the
relative total temperature using the flow
52:13.049 --> 52:19.410
conditions that we have used before; that
is T 0 2. You take out the absolute flow kinetic
52:19.410 --> 52:26.430
energy and plug in the relative kinetic energy
and if you do that, the T 0 2 relative is
52:26.430 --> 52:31.819
1017 K.
52:31.819 --> 52:37.950
The expansion ratio of the turbine at this
operating point using the prescribed polytrophic
52:37.950 --> 52:44.359
efficiency can be found by using the relation
that we have done before and it gives you
52:44.359 --> 52:54.670
a velocity a value of 2.16, which ofcourse
gives us a P 0 3 of 185 kilo pascals. Now,
52:54.670 --> 53:03.150
given that V 2 r is equal to C a 3 which would
be then equal to 159 meters per second and
53:03.150 --> 53:11.569
this ofcourse is same as V 3. That we can
therefore get M 3 relative that is the relative
53:11.569 --> 53:20.420
mach number at the exit of the rotor as 0.267.
You would ofcourse need to find T 3 over there;
53:20.420 --> 53:29.559
so, which is taken to be constant from root
to tip at the rotor exit. Now at station 3,
53:29.559 --> 53:39.059
A 3 can be found as using the continuity condition
as m dot divided by rho 3 in to C a 3 and
53:39.059 --> 53:48.390
this yields rotor exit area to be 0.02363
meters square. So, that is the area you would
53:48.390 --> 53:52.670
get at the exit of the radial turbine rotor.
53:52.670 --> 53:59.499
At the rotor exit area, if you find the radius
using the area that we have found, we get
53:59.499 --> 54:11.839
the tip area using 0.4 as the ratio; tip radius
to be 0.0946 meters and hub radius to be 0.0378
54:11.839 --> 54:21.049
meters. And hence, the mean radius would be
0.06624 meters; corresponding to which, they
54:21.049 --> 54:30.259
are C w 3 the tangential velocity which is
equal to U 3 would be 144.8 meters per second.
54:30.259 --> 54:38.809
The corresponding velocity triangle at the
rotor exit would yield that C 3 at the mean
54:38.809 --> 54:45.069
of the rotor exit between; that is mean between
the hub and the tip of the rotor exit is 215
54:45.069 --> 54:54.329
meters per second and the T 0 3 at that mean
would be 944 K. The corresponding mean mach
54:54.329 --> 55:01.650
number would can be now computed and you would
get a value of 0.362 as the mean mach number
55:01.650 --> 55:10.839
at the exit. The corresponding exit flow angle
would be 42.3. The radial variation of these
55:10.839 --> 55:18.700
two parameters that is beta 3 and M 3 relative
can now be found and a mach number and the
55:18.700 --> 55:24.079
exit flow angle is go up with radius in a
marginally non-linear manner.
55:24.079 --> 55:31.119
So, let us look at the solution in a graphical
manner. Beta 3 is shown over here and M 3
55:31.119 --> 55:37.869
relative on the other y axis and the change
of radius across as we can see M 3 varies
55:37.869 --> 55:44.619
along this line path and beta 3 varies in
a slightly different manner. Both of them
55:44.619 --> 55:52.410
ofcourse going up from hub to the tip of the
exit of the turbine; so, this is what the
55:52.410 --> 55:55.960
solution of a radial flow turbine is.
55:55.960 --> 56:01.179
I will leave you with tutorial problem which
you can solve on your own. The radial turbine
56:01.179 --> 56:09.130
problem prescribed here is that the operating
point data are given to you P 0 1 is 699 kilo
56:09.130 --> 56:23.640
pascals, T 0 1 equal to 1145 K; the prescribed
nozzle exit pressure is 527.2 kilo pascals
56:23.640 --> 56:32.359
and the nozzle exit temperature is 1029 K.
The stage exit static pressure is given as
56:32.359 --> 56:42.619
384.7 kilo pascals and static temperature
there is 914.5 K. The T 0 3 value correspondingly
56:42.619 --> 56:53.099
is prescribed as 924.7 K. If it is given that
the impeller exit area mean diameter to the
56:53.099 --> 57:00.689
impeller tip diameter is chosen as 0.49 that
is ratio between tip and mean, the design
57:00.689 --> 57:08.059
rpm is prescribed as 24000 rpm.
Assuming that the relative velocity at the
57:08.059 --> 57:13.700
rotor inlet is radial as we have done before
and the absolute flow at the rotor exit is
57:13.700 --> 57:20.179
axial again as we have done before, you are
asked to compute the total-to-static efficiency
57:20.179 --> 57:25.939
of the radial turbine; the impeller rotor
tip diameter and the loss coefficients in
57:25.939 --> 57:31.279
the nozzle and the rotor using the definition
of the loss coefficient that we have done
57:31.279 --> 57:37.609
in the lecture. If you do those things, you
would get answers; the total-to-static efficiency
57:37.609 --> 57:47.049
would be 90.5; the impeller rotor tip diameter
would be 0.27 meters and the loss coefficients
57:47.049 --> 57:57.510
would be for the nozzle stator nozzle 0.05316
and for the rotor 0.2. As we see, the rotor
57:57.510 --> 58:02.329
loss coefficient is substantially higher than
that of nozzle loss coefficient.
58:02.329 --> 58:07.219
So, these are the numbers you would get and
as you can see, these numbers are quite different
58:07.219 --> 58:13.679
from that of axial flow turbine. So, we have
done two solved two problems. So, one on axial
58:13.679 --> 58:18.869
flow turbine; one on radial flow turbine and
we have give prescribed some problem for you
58:18.869 --> 58:25.579
to solve. And as you can see, you can get
the feel of the numbers that the numbers corresponding
58:25.579 --> 58:32.219
to radial flow turbine are quite different
from that of axial flow turbine. If you talk
58:32.219 --> 58:39.040
in terms of efficiency; if you talk in terms
of rpm; you talk in terms of the mass flows,
58:39.040 --> 58:44.729
the numbers of axial flow and radial flow
turbine are quite different from each other.
58:44.729 --> 58:50.299
And by solving the problems, I hope you would
get the feel of these numbers and that is
58:50.299 --> 58:57.089
very important for engineering people to get
the feel of the numbers. We will leave the
58:57.089 --> 59:02.729
turbines with these problems and we will proceed
in the next class.
59:02.729 --> 59:08.099
We will proceed on to the combustion chamber,
and that is what we will be doing in the next
59:08.099 --> 59:15.109
class. So, we have finished the turbine chapter,
and from the next class we will proceed on
59:15.109 --> 59:20.809
to the combustion chamber, which actually
spatially in a gas turbine you remember comes
59:20.809 --> 59:26.920
before the turbine. So, we will now look at
how the hot gas is created that is applied
59:26.920 --> 59:30.739
to the turbine, and that is what we will be
doing in the next class.