Axial Flow Turbines
Hello and welcome to lecture number 23 of
this lecture series on turbo machinery
aerodynamics. We have been talking about actual
turbines in the last several lectures,
and we had a chance to discuss about different
aspects of axial turbines, starting from
very fundamental thermodynamics of axial turbines,
and moving towards twodimensional cascade
analysis, and also the 3D analysis, which
we are going to do
subsequently. So, based on what we have discussed
so far, on the basic thermodynamics
and the two-dimensional analysis; as I mentioned
in the last class, it is time that we, have
a tutorial session on understanding of axial
turbines, and how to solve problems, which
are related to axial flow turbines, but before
I go into the tutorial, let me just quickly
recap, what we have discussed in the last
several lectures.
We had an introductory lecture on axial turbines,
where we discussed about different
types of axial turbines, like well different
types of turbines in general, like the axial,
the
radial and the mixed flow turbines; of which
the axial turbine happens to be the one
which is the most commonly used in, especially
in aero engine applications and also in
marine as well as land based power plant applications;
for a variety of reason, basically
to do with the efficiency and and the convenience
in arranging axial turbines, as
compared to the other counter parts like the
radial or the mixed flow turbines.
Now, when we talk about the axial turbines
alone, we can further classify axial turbines
based on the nature of the flow through the
turbine itself. So, based on this classification,
we could have either an impulse turbine or
one could have reaction turbine; and so, the
turbine axial turbine may operate in either
of these modes, either an impulse turbine,
where in the entire pressure drop takes place
just in the nozzle; and there is no pressure
drop taking place in the rotor, the rotor
simply deflects the flow and does not contribute
in any way in the pressure drop that is taking
place in the turbine. So, this is true for
an
impulse turbine.
And the reaction turbine is one, where the
pressure drop is actually shared between both
the nozzle as well as the rotor. So, part
of the pressure drop takes place in the nozzle
and
the remaining part of that takes place in
the rotor. And so, we have defined what is
known as degree of reaction and how we can
calculate degree of reaction. We also had
a
session on, understanding the velocity triangles,
which are applicable for an axial turbine
and also these different types of axial turbine
like the impulse or the reaction type of
axial turbine. So, where in we discussed about,
how we can construct a velocity triangle
starting from the fundamental principle.
And then, we discussed about the various losses,
which occur in axial turbine and how
some of these losses can be quantified; one
may have two-dimensional losses or one may
have 3D losses like the secondary flows and
deep leakage losses and so on. We had a
rather detail discussion on these losses in
our discussion on axial compressors. And I
think, I mentioned that one could simply extend
the discussion, which was applicable for
axial compressors also for an axial turbine,
so which is why we restricted our discussion
on losses to the bare minimum, because it
is already been covered.
We also defined different mechanisms or methods
of calculating efficiency for axial
turbines. There are two commonly used efficiency
definitions; the total to static
efficiency and total to total efficiency;
each of them are used primarily, according
to the
applications, for which the turbine is being
used; for example, if if it is a land based
gas
turbine engine, which is not expected to produce
any nozzle thrust, one would like to
expand the flow to the maximum extent possible
without too much of kinetic energy
leaving the turbine. In this case, we would
define the total to static efficiency; we
would
want the flow to have or to reach the static
conditions without much dynamic head as it
leaves the turbine.
Now, in an aero engine kind of application,
where the turbine is meant only to drive the
compressor and some accessories; one would
still want from kinetic energy available at
the turbine exhaust, which can be further
expanded in in a nozzle and therefore, in
such
applications, we would define the total to
total efficiency. We have discussed some of
these topics in the last several lectures,
and we also of course, discussed about the
performance characteristics of axial turbines
very important. We have discussed how
what it is significance is in relation to
the engine as a whole and the significance
of
matching of turbine with compressors as well
as matching of turbine with the nozzle.
So, these were some of the topics that we
had discussed in the last few lectures; and
in
today’s lecture, which is basically a tutorial
session; what we going to do is, to try and
solve a few problems. And use our understanding
of the working of the axial turbines,
and put that to practice in terms of solving
problems. So, what I have for you today is
a
set of four problems on axial turbines, which
I will solve for you; and after this, I would
also give you a few exercise problems, which
I would like you to go ahead and solve
based on our discussion today, as well as
our understanding of these concepts in the
last
few lectures. So, this is, what we are going
to discuss in today’s class. We basically
be
having tutorial session on axial turbines.
So, let us go to the the first problem. So,
the first problem statement is the following,
its
states that a single stage gas turbine operates
at its design condition with an axial
absolute flow at entry and exit from the stage.
The absolute flow angle at the nozzle exit
is 70 degrees, and at the stage entry, the
total pressure and temperature are 311 kilo
Pascal and 850 degree celsius respectively;
the exhaust static pressure is 100 kilo pascal;
the total to static efficiency is 0.87 and
the mean blade speed is 500 meters per second.
Assuming constant axial velocity through the
stage, determine part (a) the specific work
done, part (b) the Mach number leaving the
nozzle, and part (c) the axial velocity, part
(d) the total to total efficiency, and last
part is to find the stage reaction. So, this
problem
statement is to do with single stage gas turbine,
which is operating under certain design
conditions, the inlet stagnation pressure
and temperature at given the exhaust static
pressure is given, the total to static efficiency
and the mean blade speed are given, based
on this data we are required to calculate
a variety of other parameters and solve this
particular problem.
Now, as we have done in the past the with
reference to turbo machines, the very starting
point of solving a problem if you recall,
when we had discussed about compressors, is
to
get the velocity triangles right. So, the
first point is to draw the velocity triangles;
and
that is the starting point of solving any
a such problems associated with turbo machines.
So, let us construct a generic velocity triangle,
here we are not given, whether it is a
reaction turbine or an impulse turbine nothing
is mentioned. So let us construct a generic
velocity of triangle like what we had done
a few lectures earlier on; and then from there
we will see what are the parameters, which
have been specified, and what is that we need
to calculate to be able to solve this problem.
So, let us construct a velocity triangle in
general. So, this is a very generic velocity
triangle, which is well, which is also true
for a reaction turbine stage; it is not really
an
impulse turbine as you can see, it is an indeed
a reaction stage. So, station 1 denotes the
nozzle entry; station 2 denotes the nozzle
exit or the rotor entry; station 3 denotes
the
rotor exit. So, these are the three parameters
that have been the station numbers that we
have defined.
And then the flow enters the nozzle with a
velocity of C 1, and it exits the velocity
absolute velocity of C 2; relative velocity
at the nozzle exit is V 2, which is the velocity,
which the flow actually enters the rotor,
and the corresponding blade angles also been
marked here; U is the mean blade speed the
flow exits the rotor with the velocity of
V 3,
and an absolute velocity of C 3 with the corresponding
angles of alpha 3 and beta 3. So,
this is a typical velocity triangle of an
axial turbine stage; and, what we will do
is, we
will take a look at what are the parameters
that we know at this stage; so from this
velocity triangle, we basically know the blade
speed that something, which has been
specified, and we also know the inlet conditions
and exit conditions with static pressure,
and we are required to calculate a variety
of other parameters.
So, let us, begin with the part (a) that is
to calculate the first parameter, which is
to
basically calculate the specific work done.
Now, the first parameter that we need to
calculate is the specific work done; and yes
we also been given the flow angle at the
nozzle exit that is alpha has been 2 given
to us; so there are two parameters in this
velocity triangle specified, we have alpha
two and the blade speed U, so these are the
parameters, which we know.
So, let us calculate the specific work done,
we know that the total to static efficiency
is
basically, a function of the specific work
done with reference to the static pressure
at the
exit and isotropic conditions; and therefore,
the specific work done is basically total
static efficiency into C P into T 0 1 1 minus
P 3, where P 3 is a static pressure at the
rotor
exit divided by P 0 1 raise to gamma minus
1 by gamma. So, these are parameters, out
of
this, all these parameters are known to us;
we know the total static efficiency it is
given
as 0.87 C P is known 1148, T 0 1 is 1123,
P 3 by P 0 1 is 1 by 3.11, because the inlet
condition is given as 311 kilo Pascal. So,
that is 1 bar by 3.11 raise to 0.248, which
is
basically gamma minus 1 by gamma 0.33 minus
1.33 minus 1 by 1.33 so we get 0.248.
So, the specific work done or the work done
by the turbine can be calculated, and it
comes out to be 276 kilo joules per kilogram.
Now, at the nozzle exit the Mach number, the
second part of the question is to calculate
the Mach number at the nozzle exit, now Mach
number as we know is defined as ratio of
absolute velocity to the speed of sound; so
M 2 is C 2 divided by gamma R T 2. So, here
we need to calculate two things; one is to
calculate C 2 we also need to calculate the
static temperature at the nozzle exit that
is T 2. Now velocity triangle, because it
is stated
that the flow enters and leaves the stage
axially, we have C W 3 is equal to 0; and
since,
we know that W t is a product of mu times
delta C W that is mu into C W 2 minus C W
3; and since, C W 3 is 0; we can calculate
C W 2, which is W t divided by U and that
is
276 into 10 raise to 3 divided by 500 that
comes out to be 550 meters per second.
So, let us understand this a little better,
it is mentioned here that the flow enters
the the
well assuming a constant axial velocity through
the stage, and that the flow leaves the
stage in an axial direction. So, we have the
turbine, which is operating in design
conditions with an axial absolute flow entry
and exit from the stage. So, this actually,
gets needs to be modified. In the sense that
C 1 is axial, C 3 is also axial, as per this
question; so this general velocity triangle,
which I had drawn should have actually have
been modified in the sense, that alpha 1 should
be 0 alpha 3 should also be 0; and which
is why we get C W 3 as 0, because since, C
3 is axial, C W 3 is also 0.
So, C 2 we know from the exit velocity triangle
is C w 2 divided by sin alpha 2. Let us,
see that once again C 2 is C W 2, which is
this component and divided by sin alpha 2,
and that is 588; alpha 2 is given as 70 degree,
so C 2 comes out to be 588 meters per
second. We know that T 2 static temperature
is T 0 2 minus C 2 square by 2 C p T 0 2
and T 0 1 are same, because there is no change
in stagnation temperature in the rotor, in
the stator. There is no work done in the stator,
and so stagnation temperature in the stator
has to be unchanged.
So, T 2 is static pressure at rotor entry,
which is T 0 1 minus C 2 square by 2 C p C
2.
We have just now calculated, so we can calculate
static temperature at the nozzle exit; so
once we calculate static temperature, we can
now calculate the Mach number, because
Mach number is simply the ratio of absolute
velocity to the speed of sound. So, we have
already calculated the absolute velocity that
is 588 meters per second and the static
temperature 973 kelvin; so M 2 is 588 divided
by square root of 1.33 into 287, which is
a
gas constant, multiplied by 973, so the Mach
number comes out to be 0.97. So, that is
that solved the second part of the question.
Third part of the question is to find the
axial velocity; axial velocity can be found
from
the inlet velocity triangle, because C 2 is
known alpha is known; so, C a is 2 times cos
alpha 2 and that is simply 200 meters per
second. Now, the fourth part of the question
is
to find the total to total efficiency. I think,
during our discussion on the efficiencies,
I had
derived an expression, which relates the total
to total efficiency, and the total to static
efficiency. And, we will simply make use of
that equation, to calculate the total to total
efficiency, total to static efficiency has
been given as 0.87, and so, let us, calculate
now
the total to total efficiency.
So, based on that equation, if you recall
and go back to that lecture, you will find
that this
derivation was shown. So, this is 1 by eta
tt, which is the total to total efficiency;
this is
equal to 1 by eta ts minus C 3 square by 2
W t where W t is the specific work done. So,
all these parameters we have already calculated,
so we get 1 by 0.87, which is total static
efficiency minus C 3 square; well C 3 is basically
equal to C a, because the flow exits the
stage axially, so C 3 and C a, are same and
therefore, this is 200 square divided by twice
of W t into 276 into 10 raise to 3.
So, this comes out to be 1 by eta tt is 1.0775,
so the inverse of that eta tt is 0.93; total
to
total efficiency, we have calculated is 0.93.
If you compare this with the total static
efficiency, it is 0.87. So, you can see that
total to total efficiency is indeed greater
than
the total to static efficiency; this is a
comment, which I had made even during our
discussion on efficiency is that in general,
the total to total efficiency is comes out
to be
higher than the total to static efficiency,
because of the very nature of the definition.
I
think, I had shown the T-S diagram to demonstrate,
why this efficiency total to total
efficiency has to be higher than total to
static efficiency?
So, if you look at the expansion process here,
which is something I had mentioned in the
during our definition of the efficiency; total
to total efficiency is defined, based on the
temperature stagnation temperature here, with
reference to the stagnation temperature at
the exit isotropic; where as total to static
efficiency is defined, based on the stagnation
condition, here with this static condition
here and the corresponding isotropic
parameters, so we can see that this difference
is always less than this other difference’
and therefore, it is inherent that total to
total efficiency has to be higher than that
of total
to static efficiency. So, that is coming from
the very basic definition and therefore, that
should be indeed true; so there is something
that which I have also seen in this particular
problem that if we calculate that this actually,
it should come out to be true.
Let us, move on to the last part of the question,
which is to calculate the degree of
reaction. Now degree of reaction again; we
have defined derived an expression for
degree of reaction in the one of the earlier
lectures. So, this is basically, equal to
1 minus
C a by 2 U into tan beta 3 minus tan beta
2. So, from the velocity triangle let us see,
what
these parameters are; and how do we calculate
tan beta 3. Let us, go back to the velocity
triangle here; tan beta 3 is equal to this
component divided by the axial velocity that
is U
divided by C a. And similarly, tan beta 2
is equal to the component given by this that
is C
W 2 minus u divided by C a.
So, from these two equations, we can calculate
tan beta 2 as well as tan beta 3; and
therefore, what you can see here is that tan
beta 2, beta 3 is U by C a, tan beta 2 can
be
equated to tan alpha 2 minus U by C a, so
if you substitute both these expressions here,
we can simplify the expression for the degree
of reaction, and that is 1 minus 1 by 2 C
a
by U into tan alpha 2. So, all these parameters,
we have already calculated, and so let us,
just substitute that here, and then we get
the degree of reaction as 0.451
So, we have now calculated all the five parameters,
which were required to be calculated
for this question, we have calculated specific
work done and axial velocity, the total to
total efficiency and now the degree of the
reaction as well; so that completes the first
question, which was to do with very simple
single stage axial turbine with certain
parameters, which has been specified, and,
how do you proceed towards calculating the
other parameters and which obviously started
from the velocity triangles, trying to solve
the velocity triangles to calculate the flow
angles, and therefore, the other parameters
which we were require to be calculated.
So, let us, now move on to the next question.
So, this is the problem statement for this
question; this is basically, 50 percent reaction
stage question; so the problem statement is
the following; combustion gases enter the
first stage of a gas turbine at a stagnation
temperature and pressure of 1200 kelvin and
4 bar; the rotor blade tip diameter is 0.75
meter; the blade height is 0.12 meter, and
the shaft speed is 10,500 rpm. At the mean
radius, the stage operates with a reaction
of 50 percent; a flow coefficient of 0.7 and
a
stage loading coefficient of 2.5. Determine,
part (a) ,the relative and absolute flow angles
for the stage; part (b), the velocity at nozzle
exit; part (c), the static temperature and
pressure at nozzle exit assuming a nozzle
efficiency of 0.96 and the mass flow.
So, here this is a question, which contains
to 50 percent reaction stage, and we have
the
temperature and pressure at the inlet the
degree of reaction is given to us, the rotational
speed, and also the mean diameter is specified;
and based on this, we are required to
calculate basically, solve the velocity triangle
and calculate the flow angles the relative
and absolute flow angles; part (b) was to
calculate velocity at nozzle exit that is
ah the
absolute velocity, and the static pressure
and temperature at nozzle exit, with certain
nozzle efficiency, which has been specified
and also the mass flow rate; so these are
the
parameters that we will need to calculate
for this particular question. So, as always
we
will start with the velocity triangle, let
us construct the velocity triangle first,
and then we
will proceed towards solving the question.
So, this is what we had drawn for 50 percent
reactions stage during our discussing on this
couple of lectures .early, on so this is a
velocity triangle for a typical 50 percent
reaction
stage, and the the mean feature that you can
observe for 50 percent reactions stage is
that
the velocity triangles are symmetrical or
mirror images; so what you have at the inlet
of
the rotor is, what happens at the exit as
well it just a mirror image; and which means
that
the angles should also be equal alpha 2 will
be equal to beta 3 and beta 2 will be equal
to
alpha alpha 3 and so on. Similarly, the velocity
components if the axial velocity is
constant, then we have C 2 is equal to v 3
and V 2 is equal to C 3.
So, in this question we have been given the
blade height at the mean diameter and the
speed, which means that we can calculate the
blade speed that is U flow coefficient is
given as 0.7, which means we can also calculate
the axial velocity from their, because
flow coefficient is the ratio of the axial
velocity to the blade speed, so we have the
blade
speed as well as the axial velocity. We are
now required to solve the velocity triangle
and
calculate few other parameters.
So, let us start with the first part of it,
which is to calculate all the angles involved
in the
velocity triangle. Now for this case, we also
have been given the loading coefficient psi
which is delta h not by U square; and for
a 50 percent reaction stage, delta h not equal
to
U times delta C W, which is also same as delta
V w. And therefore, the stage loading
coefficient reduces to delta h not by U square,
which is V W 3 plus V W 2 divided by U;
and so this if we express in terms of the
angles, let us take a look at what is V w
2 and V
w 3; V w 2 corresponds to this component here,
which is in terms of beta 2 here and the
axial velocity; similarly, V w 3 is for the
exit in terms of the axial component and the
blade angle that is beta 3.
So, this we have expressed in terms of axial
components V w 3 is C a times tan beta 3;
and V w 2 is c a times tan beta 2. So, from
our discussion on degree of reaction, the
specifically for 50 percent degree of reaction
case, we can basically, equate degree of
reaction as C a by U into tan beta 3 minus
tan beta 2 divided by 2; and that is, because
the angles are similar so if you if you replace
alpha 3 by beta 2 and beta alpha 2 by beta
3, we can express degree of reaction in terms
of this.
So, if we look at these two equations and
we try to simplify them, what we basically
get
is an expression, in terms of the degree of
reaction, the loading coefficient and the
flow
coefficient, C a by U is the coefficient,
which we know is equal to 0.7 in this question;
so
what we get tan beta 3 is equal to psi by
2 plus degree of reaction divided by the flow
coefficient phi similarly, tan beta 2 is equal
to psi by 2 minus degree of reaction divided
by phi; this basically, happens if we simply
add and subtract these two equations, we can
simplify them and get this equation.
So, since we have been we already know the
loading coefficient psi, the flow coefficient
C a by U and degree of reaction; we substitute
for those values here and we get the
angles beta 3, which is 68.2 degrees and beta
2, which comes out to be 46.98 degrees.
Now so for 50 percent reaction stage, we do
not need to actually calculate the other
angles, because other angles are equal to
what we already calculated; alpha 2 will be
equal to beta 3, which is in term equal to
68.2 degrees; and alpha 3 is equal to beta
2,
which is 46.98 degrees. So in this question
that we have, because it happens to be 50
percent reaction stage, we could make lot
of simplifications in calculation of degree
of
reaction or equating that to the loading coefficient
and so on. And, also the fact that we
do not need to really calculate the other
angles, because beta 2 is equal to alpha 3
and
beta 3 is equal to alpha 2.
So, we just calculated beta 2 and beta 3 and
the absolute angles or indeed equal to these
angles as well; so that completes the first
part of the question. Now, let us look at
what is
the second part? Second part is to calculate
the velocity at nozzle exit; subsequently
we
need to calculate static pressure and temperature
at nozzle exit with a certain nozzle
efficiency specified. So let us calculate
the velocity at the nozzle exit, and for which
we
will make use of the fact that the dimensions
have been specified, the mean, the radius
has been given the blade speed is known. And,
so we can basically, calculate the axial
velocity and also the absolute velocity.
Now, at the mean radius we can for, which
the tip diameter and the blade height will
be
used. So, the tip diameter is given as 75
centimeters so 0.75 minus blade height 0.12
divided by 2, so this is the mean radius that
is 0.315. So, at the mean radius, we can
calculate the blade speed, the speed of rotation
is given here, and the so phi D n by 60
will give us the blade speed mean, blade speed
that is 346.36 meters per second. Since,
the flow coefficient has already been specified,
we can calculate the axial velocity as 5
times U mean that is 70 percent of this, so
that is 242.45 meters per second.
So, the absolute velocity at the nozzle exit
is C a by cos alpha; and alpha is given, we
already calculated in the previous part of
the question that was 68.2 degrees; so C 2
is
equal to C a by cos alpha 2 and that is 652.86
meters per second. So, having calculated
the absolute velocity, we can now calculate
the static temperature based on the absolute
velocity, because stagnation temperature is
known. It is gives as 1200 kelvin. So, in
this
question, we have been given the stagnation
temperature at the inlet of the turbine as
1200 kelvin, and we know that in the nozzle
,there is no change in the stagnation
temperature, so T 0 1 should be equal to T
0 2. And so, T 2 is basically, equal to T
0 1
minus C 2 square by 2 C P and same that is,
because T 0 1 and T 0 2 are same; so, from
there you can calculate static temperature;
static temperature at the nozzle exit will
be
equal to T 0 2, which is T 0 1 minus C 2 square
by 2 C p. So, we get 1200 minus C 2
square 652.86 square divided by 2 into C p,
so this can be calculated at 1016.3 kelvin.
Next part is, calculate the stagnation static
pressure at the nozzle exit, for which we
will
make use of the nozzle efficiency definitions;
so this is again from the fundamental cycle
analysis that I assume you would be aware
of for rate and cycles, where we could, where
we have defined nozzle efficiency, in terms
of enthalpies h 0 1 being the inlet enthalpy
of
the nozzle, h 2 is the inlet static exit static
enthalpy divided by h 0 1 minus h 2 S. So,
we
can express this in terms of temperatures
1 minus T 2 by T 0 1 divided by 1 minus P
2 by
P 0 1 raise to gamma minus 1 by gamma, this
is because we can express the denominator
in terms of 1 minus T 2 S by T 0 1, which
is basically, in terms of pressure ratios
for an
isotropic condition.
So, from this we can simplify this and calculate
the pressure ratios P 2 by P 0 1, in terms
of the temperature ratio as well as the nozzle
efficiency and so, the static pressure at
the
nozzle exit can be simply calculated in terms
of inlet pressure, stagnation that is 4 bar
multiplied by 0.84052 rise to 4.03; so this
is 1.986 bar; and the last part of the question
is
to calculate the mass flow rate; mass flow
rate is expressed, in terms of density, the
area
and the axial velocity. Now density is P 2
by R T 2, and how do you calculate the
annulus area; so, for calculating this area
we have the tip diameter as well as the blade
height; so from there you can calculate the
annulus area and multiply that with the axial
velocity, one can calculate the mass flow
rate. So, density is P 2 by R T 2, annulus
area
in terms of the blade tip diameter and the
blade height and the axial velocity, which
is
already something calculated. So, if you multiplied
all of them the mass flow rate as 39.1
kilo gram per second.
So, this completes the second question, which
was very similar to what we have solved
in the first question except the fact that
here, we had also firstly, reaction turbine
50
percent reaction turbine stage, the first
question was on an impulse turbine stage;
and
second difference being the fact that here
we had nozzle efficiency and correspondingly,
we had calculated the static temperature and
pressure at the exit of the nozzle using the
nozzle efficiency definitions. So, having
understood two distinct problems, one to do
with an impulse turbine and this second question
was 50 percent reaction turbine, we will
now proceed to solving two other problems,
which are slightly different from what we
have already solved.
So, the third question, the problem statements
is the following; the inlet a single stage
axial flow turbine operates with an inlet
temperature of 1100 kelvin, and total
temperature of 3.4 bar, the total temperature
drop across the stage is 144 kelvin, and the
isotropic efficiency of the turbine is 0.9,
the mean blade speed is 298 meters per second,
and the mass flow rate is 18.75 kilo grams
per second, the turbine operates with a
rotational speed of 12,000 rpm. If the convergent
nozzle is operating under choked
condition, determine the part (a) blade loading
coefficient part (b) the pressure ratio of
the stage and part (c) flow angles.
So, in this question, we have been given that
the nozzle is operating under a choked
condition that is one of the things that we
need to keep in mind, we also have been given
isotropic efficiency of the turbine, besides
few other parameters like the temperature
and
pressure; the temperature drop in the stage,
blade speed and the mass flow rate. So, as
we
have been doing in the past being construct
the velocity triangle for generic turbine,
axial
turbine; and so that we are familiar, with
the parameters, which are specified and those
we should be need to the calculate. And so
in this question we have the blade speed,
which is given 298 meters per second and the
mass flow rate is given to us, and the
pressure and temperature at the inlet have
been specified.
So, let us solve the first part of the question,
which was to calculate the loading
coefficient, which is very straight forward,
because we know the temperature drop and
we know the blade speed. So, psi is simply
C p times delta T not divided by U square,
so
that comes out to be 1.8615. Now, second part
of the question is to find the pressure ratio
of the stage, so we need to find P 0 1 by
P 0 3, so to calculate that we know the
temperature drop in this stage, it is given
us 144 kelvin. The inlet temperature is given
T
0 2 equal to T 0 1 that is 1100 kelvin, based
on that we can calculate the exit stagnation
temperature T 0 3, which will be T 0 1 minus
delta T not and that is 1100 minus 144, so
that is 956 kelvin.
So, having calculated the exit stagnation
temperature, we can we need to also calculate
the pressure ratio, because that is the objective
of this second part of the question; for this
we will make use of the nozzle efficiency
definition; so isotropic efficiencies, I mean
isotropic efficiency of a turbine, which is
defined in terms of these the actual drop
in
stagnation temperature divided by isotropic
drop, so isotropic efficiency is defined as
one
T 0 1 minus T 0 3 divided by T 0 1 minus T
0 3 as numerator is already known that is
delta T not divided by T 0 1 into 1 minus
P 0 3 by P 0 1 rise to gamma minus 1 by
gamma .
So, if we simplify this, because the temper
the efficiency is already given to us, we
can
substitute for the efficiency, the temperature
drop is known, inlet stagnation temperature
is known, and therefore, P 0 3 by P 0 1 comes
out to be 0.53; and therefore, the pressure
ratio of the turbine is an inverse of that
is 1.875. So, that solves the second part
of the
question were we are required to calculate
the pressure ratio of this particular turbine
stage single stage axial turbine. The next
part is to calculate the flow angles, we need
to
calculate all the angles that are involved
in the velocity triangle; so we are going
to use
the fact that the nozzle is operating under
a choked conditions, so that is the one of
the
information that we have; and that means that
mach number at the nozzle exit is one and
therefore, the absolute flow at the nozzle
exit C 2 will be equal to square root of gamma
R T 2.
And similarly, that is also fixes the temperature
pressure at the nozzle axial, because the
of the temperature and pressure would be the
critical condition and therefore, we can
actually, calculate static pressure and temperature
from the isotropic relations, because
operating under choked condition with Mach
number equal to 1. So, once we calculate
the absolute velocity, we can also calculate
some of the flow angles that are involved,
because blade speed is known and therefore,
we can proceed towards solving the
velocity triangle calculated all the angle.
So, the nozzle since, is it operating under
choked condition mach number is exit mach
number is unity, so C 2 is a equal to square
root of gamma R T 2 and therefore, T 2 by
T
0 2 by T 2 is equal to gamma plus 1 by 2 and
that is 1.165. Since, T 0 2 is equal to T
0 1,
that is known the static temperature at the
nozzle exit is T 2, which is 944.2 kelvin;
and
therefore, the absolute velocity of the gases
leaving the choked nozzle will be square root
of gamma R T 2 that is 600.3 meters per second.
And, now we can calculate the axial
velocity, because the blade speed is known,
the flow coefficient is known, so U times
C a
equal to U times that is 298 multiplied by
0.95; and at that is 283 meters per second;
and
from the velocity triangle, we can now calculate
cos alpha 2, which is C a divided by C 2
that is 283 by 600, and therefore, alpha 2
is 62 degrees.
We now need to calculate beta 2, which is
the blade angle at the inlet of the rotor
and for
which we will make use of the velocity triangle
again; and that is from the velocity
triangle, you can see that this ratio U by
C a is tan alpha 2 minus tan beta 2, which
is
inverse of the flow coefficient. Since, tan
alpha 2 is known and phi is also known we
can
calculate tan beta 2, which is tan alpha 2
minus 1 by phi that is 0.828 or beta 2 is
39.6degrees.
Now, the third part of the question well,
where in we are calculating all angle, we
need to
now calculate the angle circle, the rotor
exit; now this specific work of turbine, which
is
W T is C P is delta T not, which can be express
in terms of U into delta C W that is U
into C a tan alpha 2 plus tan alpha 3. In
which case, we know the angles at the inlet
tan
alpha 2 is known U is known C a is known and
delta T not is also known therefore, tan
alpha 3 is C P times delta T not by U C a
minus tan alpha 2, so all these parameters
are
known to us simply substitute that in we get
alpha 3 as 4.54 degrees.
Similarly, U by C a is also equal to tan beta
3 minus tan alpha 3 and therefore, tan beta
3
is equal to 1, by phi plus tan alpha 3 and
therefore, beta 3 is 48.54 degrees. So, we
are
calculated all the angles that are involved
starting from the rotor entry, alpha 2 and
alpha
beta 2 and rotor exit, the alpha 3 and beta
3; so you see that to be able to solve in
fact, all
three problems that we have now solved, requires
you to have good understanding of the
velocity triangles, because of that is where
that is basically, the starting point of solving
these questions, if necessary that, you can
constant the velocity triangle, clearly
understanding the different components of
velocity, which has been specified, and then
use that information to proceed and solve
the problem, based on what is already given
to
even in the problem like some questions, you
have mean blade speed; some questions
may specify axial velocity or some angles
so on.
So, in this question we need basically, solved
we calculated the pressure ratio, we are
also calculated all the angles that are involved
and in this case of course, it is also given
towards that in nozzle is operating under
choked conditions. So, I have one more
question for you in this is different from
the all the questions we have solved, in the
sense that this question requires as to find
the number of stages that are required for
operating a certain turbines, it is a multi
stage turbine, how do you calculate the number
of stages; if you know the pressure drop across
the entire turbine, how can you estimate
the number of stages that are involved.
So, this is a problem statement is the following;
a multi stage axial turbine is to be
designed with impulse stages and is to operate
with an inlet pressure and temperature of
6 bar and 900 kelvin and outlet pressure of
1 bar. The isotropic efficiency of the turbine
is 85 percent, all the stages are to have
a nozzle angle outlet of 75 degrees, and equal
inlet and outlet rotor blade angles; the mean
blade speed is 250 meters per second and
the axial velocity is 150 per second and is
a constant across the turbine; estimate the
number of stages required for this turbine;
so this is an impulse turbine, and so it is
needless to say that the rotor inlet and outlet
angles are going to be equal. It is also given
that we have the inlet pressure in temperature
for the nozzle entry, we have been given
the isotropic efficiency and the nozzle exit
angle, we also have the mean blade speed and
axial velocity; so lot of parameters that
are involved in the velocity triangles are
given to
us, but we need to calculate the number of
stages that are required for getting this
kind of
pressure ration that is involved in this particular
turbine.
So, let us take a quick look at the velocity
triangle, because we will need to come back
to
this little later; an impulse turbine would
involve a velocity triangle like what is shown
here; C 2 is the flow exiting nozzle at an
angle of alpha 2, in this case, it is given
to us; V
2 is the relative velocity entering the rotor
at an angle of beta 2 and it leaves the rotor
at
velocity V 3, which is equal to 2, and an
angle beta 3, which is equal to beta 2, the
absolute velocities at the rotor exit are
is equal to C 3, which is in fact equal to
V 3 and V
2 for constant axial velocity. So, in this
case it is indeed specified that the axial
velocity
is constant, so it means that C 3 is equal
to V 3 which is in turn equal to V 2.
So, we will come back to this velocity triangle
as we proceed to solve this equation; so
first part of the question, we will what we
will try to do is that since, we know the
pressure drop across overall pressure drop
is known; let us calculate the overall
stagnation temperature drop across the turbine;
and then what we will do is to estimate
the stagnation temperature drop in one stage
therefore, the overall temperature drop
divided by temperature drop in one stage will
give us the number of stages, an estimate
of the number of stages required.
So, the overall pressure ratio is given as
6 therefore, T 0 1 by T not E S, where T not
E S
is the stagnation temperature, isotropic at
the exit of the turbine is equal to P 0 1
by P 0 E
raise to gamma minus 1 by gamma. Therefore,
the stagnation temperature at the exit of
the turbine is equal to 576.9 kelvin, this
is isotropic temperature; so the actual
temperature overall will be equal to the isotropic
temperature difference multiplied by
the isotropic efficiency, which is 85 percent
in this case; so delta T not overall, which
is
the stagnation temperature across the entire
turbine is equal to eta T, which is the
isotropic efficiency multiplied by T 0 1 minus
T naught E S, T 0 1 is given as 900, T
naught E S is calculated as 576.9 kelvin,
so the overall temperature drop in the turbine
is
274.6 kelvin.
So, now we come back to the velocity triangles;
we know from the inlet velocity
triangles C 2 equal to C a divided by cos
alpha 2, C 2 is this, cos alpha 2 is C a divided
by C 2 and therefore, C 2 is C a by cos alpha
2, C a is known, alpha 2 is also known
therefore, C 2 is equal to 150 divided by
cos 75 that is 579.5 meters per second. Now,
this stagnation temperature at the inlet is
known T 0 2, which is equal to T 0 1; and
therefore, static temperature at the nozzle
exit T 2 is equal to T 0 2 minus C 2 square
by
2 C P; therefore, T 0 1, which is equal to
T 0 2 is 900 kelvin, C 2 we just now calculated
579.5 and therefore, we can calculate T 2
that is 753.7 kelvin.
Now, it is given specifically that this is
an impulse turbine stage, which means that
degree of reaction is basically 0; so when
the degree of reaction is 0, it implies that
the
degree of reaction, we have defined as H 2
minus H 3 divided by H 0 1 minus H 0 3;
since, the degree of reaction is 0, we have
H 2 is equal to H 3, implies the static
temperature should be same T 2 and T 3 should
be the same and that is equal to 753.7
kelvin, because there is no change in static
conditions across rotor, the entire pressure
drop has actually take place in the nozzle
nothing changes in the rotor T 2 is equal
to T 3.
Now, at the exit of the velocity triangle
at rotor entry well, let us look at rotors
entry
first; we have calculated alpha 2, which is
known to us; let us calculate beta 2 as well,
tan beta 2 we know is C 2 sin alpha 2 minus
U by C a, beta 2 is C 2 sin alpha 2 that is
this component minus U divided by C a; so
that is C 2 sin alpha 2 minus U divided by
C
a, and that comes out to be beta 2 give comes
out to be 64.16 degrees. And therefore,
from this, we can calculate the relative velocity
V 2, and that is C a cos beta 2, it is
344.15 meters per second; for constant axial
velocity in an impulse turbine, we have seen
that V 2 is equal to V 3 which is equal to
C 3; and therefore, we can calculate the
stagnation temperature at the rotor exit T
0 3 in terms of, because T 3 and T 2 are same
T
0 3 is T 2 plus C 3 square by 2 Cp that is
753.7 plus 344.14 square by 2 in to C p that
is
805.28 kelvin.
Therefore, the per stage temperature drops,
stagnation temperature drops is T 0 1 minus
T 0 3, and that is 900 minus 805.28 94.7 kelvin.
We have already calculated the overall
temperature drop, which was 274.6 that divided
by 94.7comes out to be 2.89 and that is
to be approximated to the next integer that
is three stages, so we have we are required
three stages for achieving this kind of a
pressure drop in an generating this work output
from this kind of impulse turbine, which has
this efficiency and mean blade speed and so
on. So, this is one of ways of estimating
the number of stages that are required, basically
they are calculating the overall temperature
drop and calculating the temperature drop
per stage diving the 2 will give us the number
of stages required.
So, this completes this particular problem
as well, we have solved so far 4 distinct
problems, which have to do with different
types of turbines, how to analyze velocity
triangle and solved the problem using these
velocity triangles. So, what I will do now,
is
to leave you with exercise problems, which
you can solved based on what we have
solved in today’s lecture, as well as the
discussions we had in the last few lectures,
I will
also give the final answers of these questions,
so you can check after you have solved
these problems.
So, the first exercise problem is a following
an axial flow turbine operating with an
overall stagnation of 8 is to 1 has a polytrophic
efficiency of 0.85, determine, the total to
total efficiency of the turbine, if the exhaust
Mach number of the turbine is 0.3,
determine the total to static efficiency.
If in addition, the exhaust velocity of the
turbine
is 160 meter per seconds, determine the inlet
total temperature the answers to these
questions are 88 percent that is total to
total efficiency total to static efficiency
is 86.17
percent, inlet total temperature is 1170.6.
The second exercise problem is the mean blade
radii of the rotor of a mixed flow turbine
is 0.3 meter at the inlet and 0.1 at the outlet;
the rotor rotates at 20000 revolutions per
minute and the turbine is required to produce
430 kilo watts; the flow velocity at nozzle
exist is 700 meters per second and the flow
direction is at 70 degrees to the meridional
plane. Determining the absolute and relative
flow angles and the absolute exist velocity.
If the gas flow rate is 1 kg per second and
also the velocity of the through flow is
constant through the router, so the angles
are alpha 2 is 70 degrees beta 2 is 7.02 degrees
alpha 3 18.4 degrees and beta 3 is 50.37 degrees
The third question is an axial flow gas turbine
stage develops 3.36 mega watts at a mass
flow rate of 27.2 kg per second; at the stage
entry, the stagnation pressure and
temperature are 772 kilo Pascal and 727 degree
Celsius respectively; the static pressure
at exit from the nozzle is 482 kilo Pascal
and the corresponding absolute flow direction
is
72 degrees to the axial direction; assuming
axial velocity is constant across the stage,
and
the gas enters and leaves the stage without
any absolute swirl velocity. Determine, part
(a) the nozzle exit velocity; part (b) the
blade speed part (c) the total to static efficiency
and (d) the stage reaction; answers are 488
meters per second, ythe blade speed is 266.1
meter per second, total to static efficiency
0.83 and the stage reaction is0.128.
And last question is the single stage axial
turbine has a mean radius of 30 centimeters
and a blade height of 6 centimeters; the gas
enters the turbine stage at 1900 kilo Pascal
and 1200 kelvin, and the absolute velocity
leaving the stator of nozzle 600 meters per
second and inclined at 65 degree to the axial
direction; the relative angles at the inlet
and
outlet of the rotor are 25 degrees and 60
degrees respectively; if the stage efficiency
is
0.88, calculate part (a) the rotor rotational
speed part (b), stage pressure ratio part
(c)
flow coefficient (d) degree of reaction and
part (e) the power delivered by the turbine;
here, the answers are rotational speed is
13550 rpm, the pressure ratio is 2.346, flow
coefficient is 0.6, degree of reaction is0.41,
power delivered is 34.6 megawatts.
So, these are four different problems that
I have put up as an exercise for you, and
I hope
with what we have discussed in last few classes
as well as today’s tutorial, you will be
able to solve these problems, based on these
discussions. So, we will continue our
discussion on axial turbines further in basically,
looking at the 3D flows in further
lectures. And, I hope you will be able to
understand the basic looking of a threedimensional
flow in axial turbine based on the fundamental
that we have discussed based
on two-dimensional flow in axial flow turbines.
So, we will continue discussion on some
of these topics in the coming lectures.
