We are talking about axial flow turbines, and in last class, last lecture, we talked
about how to design and bring into the design steps and design features, three dimensionality
of flow through axial flow turbines. Now, we know that the flow through the axial flow
turbines, as in axial flow compressors, is normally annulus in nature and that flow through
the annulus space that is available to the turbines is often not exactly uniform.
So, some aspect of the non uniformity or some aspect of the variation from the lower radius
to the outer radius or inner radius to the outer radius, and more specifically from the
root to the tip of the blade of a rotor needs to be factored into to begin within the design,
and later on, in the computations, and of course, later on in the analysis. We have
seen how some of these things can be brought into simple analysis. That we had done in
the last class without getting into more complex computational analysis.
As I mentioned, we shall do computational, an introduction to computational analysis
through turbo machines towards the end of this lecture series, but right now, we are
looking at turbines specific certain theories which factor in the three dimensionality of
the flow and built it into the design. So, most of it is indeed used for design, and
then, immediate pose design analysis to find out how the turbine is actually going to behave.
So, in today's lecture, we will look at some problems that actually use these theories
that we have done in the last class, and then, try to actually solve some problems which
are prescribed problems, and from the problem statement, we tried to figure out what kind
of solutions can be arrived at using the theories that we have done in the last lecture.
And towards the end, I will leave you with a few problems to solve by yourselves so that
you can get the feel of the application of these theories to actual problems, realistic
problems and you also get a feel of the numbers. It is essential for an engineer to get the
feel of the numbers and that is why it is essential that we look at a few problems which
are probably a little textbook problems, simplicity problems where, you know, typically you have
all the data you require. In a real life problem, you often may not actually have all the data
that is require to solve, but we are dealing with problems where the problem statement
gives you all the data that you require. And now, you need to use the theories that
we have done to arrive at solutions, and in a process, one learn how the solutions would
look like and indeed as I said you get a feel of the numbers. What the numbers would look
like? So, that is very important for all practical engineers. So, let us look at some of the
problems related to three-dimensional flow in axial flow turbines. Now, in axial flow
turbines, that we are going to do some solved problems, and then, I leave you with some
exercise problems to solve for yourselves.
Now, in the solved problems, we will first look at the problem statement. In the first
problem, the problem statement shows that a constant nozzle exit angle which we have
designated earlier is alpha 2 is being use for axial flow turbine design. In which, it
is prescribed that the temperature drop should be 150 k, and at the hub, the flare velocity
a blade speed u is 300 meters per second, and at the tip, it is 400 meters per second.
At it is prescribed that alpha 2 is 60 which is constant as prescribed from a root to the
tip hub to the tip and alpha 3 is 0, that is, 0 whirl at the exit of the rotor. The
radius ratio given for this particular problem statement is 0.75, that is, hub radius to
tip radius ratio is 0.75that is prescribed here. The problem asks a solution in which
he should complete the design velocity diagrams at hub mean and tip of the stage, and thereafter,
calculate the velocity components if the design is a free vortex design for the turbine and
compare the results of that with that of constant nozzle exit angle. So, one can look at the
whole thing from a free vortex point of view and compare the results. So, this is a problem
statement which we can now try to find a solution to.
Now, at the rotor illustration, we have done in the last class - the equation for variation
of whirl component is defined by the variation of alpha 2 and that is given by C W 2 by C
W m which is the mean and that is also equal to C a 2 by C a 2 mean, which is also equal
to C 2 by C 2 m which is mean and all these velocity ratios are equal to radius ratio
r by r mean r being at any radius to the power sine square alpha 2. Now, this is what we
had done in the last lecture following which, what we get is, at the rotor exit, the actual
velocity C a 3 square is equal to C a 3 mean square plus twice U m into C W twice m whole
thing multiplied by 1 minus radius ratio r by r m to the power cost square alpha 2.
Now, this is also the theory that we had the expression; we had shown in the last lecture.
Now, from the prescribed radius ratio that is given, we can see that the mean to tip
radius ratio would be 0.875 and mean to hub radius ratio would be 1.166, that is calculated
from the hub to tip radius ratio that is prescribed in the problem statement.
Now, the work done by the rotor is given by the rulers equation which you are all aware
of. Now, in this particular problem statement, alpha 3 would be equal to 0; exit whirl component
is 0, which means C W 3 indeed would be 0, and in which case, the specific enthalpy rise
or, you know, specific work that we normally use for aero thermodynamic relations, that
is, a c p into delta t would be equal to U m into C W 2 m C W 3 being 0, and from which,
we can write that C W 2 m would indeed be 492 meters per second and corresponding C
a 2 would be from the velocity triangles that you can draw at station 2 that is before the
rotor and that would come out to be 284 meters per second. As per the prescription, that
also would be equal to a actual velocity at the exit of the rotors C a 3 m.
Now, at the rotor hub, at the inlet of the rotor, one can write a axial velocity C a
2 h is equal to C a 2 m to the multiplied by a radius ratio r m by r to the power sin
square alpha 2 and this would yield a axial velocity of 318.8 meters per second. On the
other hand, the whirl component at hub, at the inlet C W 2 h would be a C W 2 m into
r m by r to the power sine square alpha 2 and that would be 552.2 meters per second.
So, you now start getting the components of the velocities at station 2, that is, before
the rotor at the rotor inlet.
Now, at the rotor tip of the inlet, one can now find the actual velocity using the same
relation, that is, C a 2 m into r m by r to the power sin square alpha 2, and in which
case, the actual velocity would now be 257 meters per second, and at the whirl component,
C W 2 tip would be 447 meters per second. So, as one can see that the whirl components
has actually decreased from hub to the tip, whereas, the actual velocity is also decreased
a little from hub to the tip and this is a consequence of the fact that alpha 2 has been
held constant from root to the tip of this prescribed problem. Soon as a result of which,
we now have all the velocity components that are required at the inlet to the rotor.
Now, at the rotor tip outlet, we can find what the actual velocity would be and this
can be found from the axial velocity relation that we had done earlier and that is given
by C a 3 equal to whole thing root over C a 3 square plus twice U m C W 2 m multiplied
by 1 minus radius ratio r by r m to the power cos square alpha 2, and if you do that, we
can calculate the actual velocities at tip and hub as at the outlet as 262 and 306 meters
per second.
So, once you apply the relations that we had done in the last class for three-dimensional
flow estimation, we now have the axial velocities at tip and hub at the outlet to the rotor.
On the other hand, if we calculate all the values with reference to the free vortex design,
we can see that the velocities that we get in comparison to the nozzle angle can be now
written down in a tabular form, and one can see that these values are the ones in constant
nozzle angle are given in red and the free vortex values are given in black, and the
actual velocities as you can see in, in, front of the rotor and behind the rotor are constant
for free vortex, only the whirl component is changing as per free vortex variation from
tip to hub. On the other hand, for constant nozzle angle as one can see all the velocities
are varying, and essentially only at the mean, the values of free vortex and the constant
nozzle angle are same at the tip as well as at the mean, the axial velocity the whirl
component velocity and the exit axial velocity they are all same at the mean, but at the
hub and the tip, the constant nozzle angle gives completely different values compared
to that of free vortex. Now, this is something which you may like
to take a look at and you may like to sit down and draw all the velocity triangles that
are born out of this two calculations, two sets of design calculations and you will probably
find that you get completely different blade shapes. The blade shapes that come out of
this two design exercises would indeed be quite different from each other and that would
tell you that, if you use different kind of design law or design philosophy, you would
in end up getting quite different blade shapes, the three-dimensional blade shapes.
Now, if we go onto the second example in which, the problem statement reads that it is propose
that for design of an actual flow turbine two design methods are to be explored. Now,
first design method that is prescribed is C W 2 m is equal to C W 2 h that is equal
to C W 2 t; that means the whirl components at hub mean and tip are equal to each other.
Now, this type of design is what we have called earlier solid body essentially design; that
means the fluid behaves more like a solid body and correspondingly the whirl component
everywhere are same. The second design that is prescribed is where
C a 2 is equal to C a 2 hub into the radius ratio hub to tip radius ratio to the power
sin square alpha 2, which somehow tries to make use of the constant alpha 2 prescription
and that is your Case B. The second design that is suggested here in Case B, and incase
C, we have C W ratio C W 2 t to C W 2 h has equal to the radius ratio r h by r t.
Now that, if you remember is nothing but you free vortex design. So, we have three design
that is suggested for design of an actual flow turbine - one in which the flow behaves
like a solid body; the second one in which one may use the alpha 2 equal to constant
that is constant nozzle angle prescription and Case C is which resembles of that of a
free vortex design. What is prescribed are some common data and
these are the actual velocity at the mean is prescribed as 200 meters per second; entry
alpha 2 is 60 degrees; exit alpha 3 is 0. The degree of reaction is 0.5 at mean and
radius ratio prescribed is 0.8. What is asked for is complete the velocity diagrams for
all the cases. The velocity diagrams are asked for because those are the velocity diagrams
based on which final blade shapes would be created. So, the velocity diagrams would give
a fairly good idea about the blade shapes that are being created by three cases - a
b and c.
Let us look at how to go about finding a solution to this particular problem statement. From
the prescribed data of the example 2, one can calculate that radius ratio r m by r t
would be 0.89 and r t by r m would be 1.11. This is calculated from the hub to tip radius
ratio that is prescribed in the problem, and then, the whirl component at mean C W 2 m
can be found from the mean velocity diagram that is a C a 2 into tan alpha 2 and that
gives C W 2 m equal to 346.5 meters per second and it is prescribed that alpha 3 is 0. So,
C W 3 m would be 0. Now, C W 3 m equal to 0 is what we had done
in the first problem also, where it is prescribed that C W 3 m could be given as 0 or alpha
is given as 0. This is a fairly standard prescription for many of the designs, because what happens
is, when the flow is going out of the turbine quite often, the prescription often encourages
that the flow going out of the turbine does not have any whirl component, because the
whirl component going to going out of a turbine typically of a single turbine or a multistage
turbine, last stage would be quite useless. So, typically of a turbine it is quite often
unless you know it is one of the earlier turbines are one of the middle stage turbines. The
prescription quite often comes with alpha 3 equal to 0 which leaves 0 whirl component,
and if the flow is going in to a nozzle or going into exhaust, any whirl component present
is quite useless, only component that is useful for nozzle effect for thrust making is the
axial component. So, giving a prescription of whirl component 0 is pretty much a practical
and standard prescription for turbine design as I mentioned, unless you are designing a
turbine, which is one of the middle stages of a multistage turbine.
So, as we have seen in both the problems C W 3 and alpha 3 of the prescribed to be a
0, and as a result of which, the problems does become a little simple, but the prescription
is realistic it is not really idealistic. Let us get back at the problem. For this particular
problem statement, it is given the degree of reaction r x is 0.5. Now, that means, it
brings us back to the symmetrical balding concept that you may have done earlier and
certainly done in some detail in axial flow compressors.
So, the moment you put a degree of reaction 0.5, the symmetrical bleeding concept comes
in, and then, you have at the mean, alpha 2 m equal to beta 3 m that would be equal
to 60 degrees as prescribed and alpha 3 m equal to beta 2 m equal to 0 degrees as prescribed.
This of course, makes the problem a little simple to handle. However, many turbines in
the past and the early days of turbine design have been used, using these kind of a somewhat
simpler design prescriptions and those turbines were a operating quite well.
It makes the designers job definitely a simple and a probably analysis also become simple
in those days and long back thirty forty years back if you may remember, the aid from a computational
fluid dynamics was not available, and as a result, the refinement that is possible in
today's turbine design was not really available in those days, and as a result, somewhat simpler
design prescriptions were often used for design and those were functional, they worked fine
cell. So, a similar somewhat simpler design prescriptions has been also prescribed here.
Now, the blade velocity U m would come out to be same as a C W 2 m because u r beta 2
m is 0, and as a result of which, 346.5 meters per second as calculated just a little above,
and at any radius, we can now calculate the blade velocities from the radius ratios that
we have just written down. So, the hub blade velocity would be 308 meters per second and
the tip blade velocity U t would be 385 meters per second. So, the blade velocity is vary
all the way from root to tip as a per omega into r concept.
Now, for case A, we have three cases to be actually looked at, and as I indicated, this
is a fluid which is a prescribed to be behaving like a solid body for a cases which we have
done in the last lecture n equal to 0 for the equation C W 2 equal to r to the power
n. Now, when you put n equal to 0, it is a Cases which fluid behaves like a solid body.
Now, the actual speed is calculated from the actual velocity pressure derived from the
energy equation for the cases n equal to 0 and this comes out to be C a 2 is equal to
C a 2 m whole thing root over 1 minus 2 tan square alpha 2 m into l n radius ratio r by
r m. Now, this can be derived; you can said an and derived in the same way we were done
earlier for the cases n equal to 0 from the energy equation that we are written down involving
the various velocity components. And you have to put the value of n equal to
0 there and he would arrive at this a solution which we are looking at for a axial velocity
at any station with relation to axial velocity at a mean radius along the blade length.
Now, using this, you can also calculate the angles across the rotor from the above considerations.
If you do that, the solutions that you get for the case A tells you that the actual velocity
prescribed at the mean was a 200 and C W 2 at the mean was found to be 346.5. On the
other hand, we found the axial velocity is varying from tip to hub and both at the rotor
entry as well as at the rotor exit. At the rotor exit, the C W 3 is prescribed to be
actually 0, and correspondingly the value of alpha 3 also has been prescribed to be
0. The alpha 2 variation is shown here as part of our results. It varies all the way
from hub to the tip. Correspondingly the beta 3 also varies exactly in the same manner from
hub to the tip, and the beta 2 values are shown here. It is 0 at the mean as we have
calculated, as off we prescribed the degree of reaction being 0. However, there is a small
value beta 2 at the hub and a small vale of beta 2 at the mean, at the tip that is been
shown here.
Now, if we move to the results of a case B, the prescribed condition is C a 2 t is equal
to C a 2 h into radius ratio r h by r t prescribed here in the problem to the power sine square
alpha 2. Now, this of course, brings us to the fact that all the velocities at the tip
and hub and mean can be the ratios of them can be put down as equal to each other. All
of them would be equal to some relation to alpha 2.
Now, this allows us to write down that far constant nozzle angle, which is the prescribed
case B that we were looking at. C a 2 can be now written down as C a 2 m into radius
ratio r m by r to the power sine square alpha 2, and in case of C W 2, that would be C W
2 m to the into radius ratio r m by r to the power sine square alpha 2 and C 2 is equal
to C 2 m to the multiplied by radius ratio r m by r to the power sine square alpha 2.
So, far constant nozzle angle case, all the velocity C h 2, C W 2 and C 2 which is the
absolute velocity can be found from the mean values, that is, at the mean radius to hub
to tip at any radius by using this radius ratio concept.
If we do that at the station 3, at the exit of the rotor, it is prescribed that alpha
3 is equal to 0 and C W 3 is also equal to 0. The expression for axial velocity comes
out as we have done in the last lecture a C a 3 square is equal to C a 3 square 3 C
a 3 m class twice U m C W 2 m to multiplied by 1 minus radius ratio to the power cos square
alpha 2. Now, this allows us to calculate C a 3 at exit of the rotor. If we now use
the relation that we have done which is essentially for the case B, which is a exit angle from
the rotor is held constant from root to the tip of the blade.
Now, this is something which we have discussed in last lecture that, holding the exit angle
from the rotor constant from root to the tip of the blade makes the rotor untwisted or
very likely twisted. Now, this is important for a stator nozzle blade cooling purpose.
We shall be doing a cooling technology from next lecture onwards, but this particular
design philosophy of holding alpha 2, 2 equal to constant from hub to tip or root to tip
of the blade of the stator essentially caters to cooling technology, and if we apply this,
in this present problem what we see is the result that we get for case B.
Let us look at the results of Case B. What we see here is the axial velocity is now vary
all the way from hub to the tip. At the hub, it is 218.5. At the mean, it is prescribed
as 200. So, the variation of axial velocity at the entry as well as at the exit is quite
pronounced from hub to tip. The values of C W 2 are also variables from hub to tip point.
Substantially C W 3 is being held constant; alpha 2 by prescription is held constant;
alpha 3 by prescription is 0. We get a variation of beta 2 from 17.9 2 minus
19.4 and we get a variation of beta 3 from 54 to 60 nine from hub to tip. So, these are
the results of the Case B which is of a constant alpha 2 from hub to the tip of the stator
nozzle. Now, we can move to the Case C. Now, Case C is what we had seen was actually the
free vortex design. Now, the free vortex design as we have mentioned before is not the most
popular design for actual flow turbine even though it is a very popular design for axial
flow compressors. For turbine, it is not the most popular design,
but it is the simpler design it of course works. If you make actual flow turbine with
a free vortex design, it will surely work; there is no reason why it should not work,
but it is not the most popular design today, and ever since the cooling technology came
into the market, the free vortex design has been essentially replaced by the constant
alpha 2 design which is the more popular design philosophy for turbines essentially as I mentioned
to cater to a cooling technology. But let us look at the results that we get for this
problem a statement problem to Case C for a free vortex design philosophy.
If you apply free vortex design philosophy, the whirl components at tip and hub the ratios
are directly related to the hub to tip radius ratio. Now, if we apply the same at the outlet
also, C W 2 also is held constant as per free vortex principle across the blade at mean
various. So, C W, C a 2 is equal to C a 3, and if you apply all these, in the, as per
the free vortex law, we get a set of results directly as per very well known free vortex
law prescriptions.
The results that we get that the axial velocities are held constant from hub to tip as well
as across the rotor. C W 2 varies substantially from hub to the tip as per free vortex law
and which would tell you that you would end up getting a substantially twisted blade.
C W 3 is being held constant. So, the trailing edge would be rather leaner. On the other
hand, the value of alpha 3 is 0 corresponding to the prescription. Alpha 2 varies from hub
to tip and the C W 2 variation shows that, and then, of course, you have the variation
of beta 2 which goes minus at the tip and this is something that comes out of the free
vortex design, and as a result of which, you get a variation of beta 3 which also varies
from hub to the tip of the blade.
So, you have the results of the Case C Tabulated here, and then, we have three Cases -- a,
b and c these variations can now be all put together into one table which are essentially
tries to compare the 3 Cases -- a, b and C, and as you can see the Case A is given in
red, the Case B that is a constant nozzle angle is given in b and the free vortex design
is given in Case C and all three of them are brought together. If you sit down, use the
velocity is that I given the angles that are given, and if you draw the velocity triangles
of all the cases, you will probably get a very clear picture of what kind of blades
actually come out of the three-dimensional blade shapes that should come out of the three
Cases that are prescribed here. The three cases that are prescribed here had certain
commonalities, that is, the mean actual velocity prescribed by 200. The exit the angles were
prescribed to be 0. The whirl components at the exit were prescribed to be 0.
So, with those common prescriptions, we try to put together; the blades that would come
out even with those commonalities, and we see that three completely different blade
shapes are likely to result from three cases that are prescribed here for axial flow turbine
design. So, as I mentioned, you can probably sit down and actually draw the velocity triangles
and you would find that three different cases or three different blade shapes and he would
indeed need to choose different aerofoil shapes, different blade sections from hub to tip for
the each of these three cases. So, you end up having three completely different blade
shapes for three different design philosophies even though we started out with a common data
prescription for all three cases. So, this is an example which tells you with certain
simplifications or simplified problem. It still tells you that, if you have three different
design philosophies, you end up with three completely different blade shapes. I will
now leave you with a few problems which you can sit down and solve for yourself and get
a feel of the numbers that come out of solving of examples problems that are prescribed with
numerical values. So, let us look at some of the problems you can solve for yourselves.
The first exercise, exercise, problem, the problem statement reads that an axial turbine
rotor that is prescribed with rotor inlet and outlet flow in radial equilibrium, which
means the static pressure and the dynamic pressure are balanced at the inlet and outlet
of the rotor. The whirl component of the flow is designed to vary radically as per this
prescription as C W at the inlet as a r minus b by r and at the outlet as a r plus b by
r. Now, a and b are the constants, and what is required for you to find is find the inlet
and outlet exit velocities and the expressions for those velocities.
In this case, you can see the answers given here and the actual velocity is of course
would remain constant across the rotor. So, we are dealing with rotor only. So, the problem
statement is essentially for rotor. Part b of the problem is - it is prescribed that
at mean various given value is point 3 meters. The actual velocity is 10 meters per second
and the degree of reaction is 0.5. The blade loading coefficient is prescribed as, as,
per the definition, psi rotor is equal to work done, specific work done 8 0 by U tip
square and the r p m in 7640 rpm. The hub to tip radius ratio is given as 0.5,
and at 80 percent of the rotor radius, it is required for you to find the rotor relative
flow inlet and outlet angles. So, what you are required to find the beta values of a
beta 2 and beta 3 for the particular problem statement that is given here. Now, the answers
given are 43.3 degree and 10.4 degree for beta 2 and beta 3. So, you can try to sit
down and see whether you can arrive at those answers.
The second exercise problem that is given is the gas exits from the turbine stator or
nozzle at a radically constant angle alpha 2. So, it is a constant nozzle exit angle
problem. The gas is prescribed to be in radial equilibrium. The axial velocity variation
at that station is given as C a 2 in to r to the power into sine square alpha 2 equal
to constant. This is what we have done in our lectures also, and for a turbine in which
the axial velocity at the radius, 0.3 is a again prescribed 100 meters per second, and
if the turbine has stated above is designed with a constant alpha 2 equal to 45 degree,
find the actual velocity at that station at 0.6 meters radius.
Now, the answers given here is very simple and that is 70.7 meters per second. So, it
is a constant exit, stator exit angle problem and you have to apply the relations that we
have done in the lecture or in the earlier problem that has been solved for you.
The third and the last problem that is prescribed for you to solve is an axial turbine is designed
with free vortex at stator nozzle exit and 0 whirls at stator at rotor exit. So, it is
a free vortex problem applied at stator and nozzle exit and a rotor inlet. So, the station
between the stator and the rotor carries free vortex prescription. For the following operating
condition, that is, at the entry t 0, 1 is equal to 1000 k mass flow is given as the
thirty 2 kegs per second. The hub radius is 0.56; the tip radius is 0.76 meters; rpm prescribed
is 8 thousand; the degree of reaction is point 5 and a actual velocity is constant that is
1 into 3 meters per second, and it is prescribed that the inlet and exit absolute velocities
are equal to each other, that is, c 1 is equal to c 3.
You are required to find C 2 that is the nozzle exit velocity. As you know, it expands ha
hugely from C 1 to 2, and then, you are required to find the mach number, maximum mach number
at the stage. In this particular stage, typically it is most likely to be a somewhere in the
nozzle exit, and then, the reaction at the root, the power output of this particular
working turbine and t 0 3 and t 3 at that stage exit. So, these are the numerical values
that you need to find out of this prescribed problem.
The answers solutions that are given here is that the nozzle exit velocity C 2 is would
be for in 80 meter per second. The maximum mach number in this stage is so solved as
point 0.818. The reaction at the root is 0.08, and you remember at the root, it is quite
often specially free, free, vortex. It is likely to be very close to 0 and 0 of course
would mean an impulse turbine, and we are looking at a problem in which, the solution
actually comes pretty close to giving you an impulse station, impulse section at the
root of this particular turbine, and then, the power output are work done for this particular
rotor given the mass flow is 3.42 mega watts, and the temperatures at the exit t 3 is 907
and the static temperature t 3 is 8 892 k. Now, you can sit down and try to solve this
problem and see whether you can come. You can standard values of gas constant r,
that is, 247 joules per kg k and value of C p as thousand 147 joules per kg k. So, you
can use those standard values to solve this problem in which, numerical are given and
you are required to find the certain prescribed velocities, mach numbers, work done and the
exit temperature. So, I leave you with these problems for you to solve yourself so that
you can get a feel of the numbers the typically come out of axial flow turbine design. So,
some of these problems would give you an idea how the turbine designs indeed proceeded with,
and what kind of numbers you get? What kind of variations? You get a you get a feel of
the numbers by solving these problems. In the next class, we will be looking at turbine
blade cooling, because in this class and in the earlier lecture, we had looked at the
design philosophy of alpha 2 is constant from root to the tip and we have stated again and
again that this particular design philosophy essentially caters to turbine blade cooling.
In the next lecture onwards, we will devote ourselves to looking at this turbine blade
cooling technology and how it impacts the turbine blade, the turbine blade shape, and
essentially, the aerodynamics or the aero thermodynamics of the flow over the turbine
blades is very strongly impacted, and essentially, the aerodynamics of the blade changes hugely
by application of cooling technology.
We will look at various cooling technologies and how do they actually impact the turbine
design of modern actual flow turbines specifically these cooling technologies are very widely
used in aero engines, and we will look at some of the typical examples of these applied
cooling technologies in turbine, actual turbine, rotor and stator. So, we shall be doing turbine
cooling technologies from next lecture onwards.