WEBVTT
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Hello and welcome to lecture number 33 of
this lecture series on turbo machinery aerodynamics.
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As promised in the last lecture, we shall
be having a tutorial session in today lecture.
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But before I start the tutorial, let me quickly
recap what we have been discussing in the
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last two lectures. As you know that last two
lectures have been devoted exclusively for
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centrifugal compressors, so during the first
lecture on the series that was on lecture
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number 31 where we introduce the aspect of
centrifugal compressors.
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We discussed in detail the thermodynamics
of centrifugal compressors, and we also discussed
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in some detail about why is it that centrifugal
compressors have certain amounts of benefits
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in some cases. And as well as the fact that,
there are a lot of disadvantages associated
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with centrifugal compressors and that is primarily
the reason why these compressors are not used
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in large sized engines.
So, the application of centrifugal compressors
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are primarily at this momentum limited to
smaller sized or smaller thrust class engines,
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because axial compressors have inherent advantages
over centrifugal compressors which makes them
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more suitable for larger sized engines.
But centrifugal compressors continue to be
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used in smaller engines. And of course, they
were used popularly in the early days of development
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of the jet engines, when axial compression
system was not that developed in those days.
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And then, we also discussed about different
components which constitute a centrifugal
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compressor like, the inlet part of the centrifugal
compressor, the inducer. Inducer is one which
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guides the flow into the impeller to ensure
that there is smooth entry of the flow into
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the impeller, then the impeller itself which
is the rotor of the centrifugal compressor,
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inducer is often attached to the impeller
and it forms the initial part of the impeller
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itself. Of course, in some of the older generation
engines and compressors, inducer was sometimes
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kept as a separate component and not necessarily
a part of the impeller itself.
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Impeller flow gets discharged into a vane
less space, which is where the diffusion continues
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from the impeller, it then continues in the
vane less space; we have seen the physics
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behind that. And then, the flow proceeds into
the diffuser which could be either vane type
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diffuser or it could be pipe diffuser, channel
diffuser and so on. From the diffuser the
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flow is collected through what are known as
collectors of volute, and then it is discharged
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from the compressor.
So, these are different components which constitute
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a centrifugal compressor, we have seen the
flow, as it passes through these different
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components how we can calculate different
parameters associated with the flow, so these
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were some of the things we had discussed in
the first lecture which was on during lecture
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number 31. In the previous lecture, we continued
our discussion on this, and many more advanced
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concepts like the carioles acceleration and
what its significance case in reference to
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centrifugal compressors.
We have seen that carioles acceleration basically
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leads to a tangential variation in the relative
velocity, and at the exit of the impeller
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this would mean that, there is what is known
as a slip. Slip is basically referring to
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a difference between the tangential components
of the absolute velocity to the tip blade
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speed, that is, c w 2 by u 2. And so, slip
factor basically affects the performance in
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terms of pressure rise of the centrifugal
compressor.
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We have seen that slip factor is a strong
function of the number of blades; therefore,
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there are empirical correlations from which
one can calculate slip factor for different
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types of impellers. And subsequently, we discussed
in detail about the performance characteristics
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of centrifugal compressor. We have seen that
is very similar to that of axial compressors,
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and we spent some time discussing about the
choking aspect of centrifugal compressor and
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also the fact that choking is different in
different components, that is, the way choking
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is calculated in the impeller or the that
is the rotating components or the stationery
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components like the inlet or the diffuser
is quite different.
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And also, the fact that one can continue to
operate the engine at a higher mass flow than
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the choking mass flow. If we can increase
the speed of the impeller and also the fact
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that this does not lead to choking of the
other components, that is, as long as all
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the other components are not choked, for a
given operating condition if the compressor
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is operating under choked conditions, if we
increase the rotational speed we can actually
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operated a slightly higher mass flow rates
at least theoretically.
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These were the different aspects that we had
discussed in the last two lectures, and therefore,
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it is about time that we have a tutorial session,
and I think I also mentioned why we are not
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emphasizing too much on the centrifugal or
the radial flow machines like centrifugal
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compressors or you will see shortly about
radial turbines is the fact that these components
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or these types of turbo machines, though have
been used extensively in the past, in the
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current day scenario their application is
very much limited. And most of the modern
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day jet engines that we are aware of the larger
sized ones, definitely use axial flow turbo
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machines like the compressors and turbines.
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So, today's lecture, we will devote towards
centrifugal compressors, basically trying
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to solve some problems on centrifugal compressors,
let us look at basically, have a tutorial
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session on centrifugal compressor. So, we
will have a tutorial on centrifugal compressors,
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let us take a look at the first problem that
we have today.
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So, the first problem statement is the following,
at the inlet of a centrifugal compressor eye,
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the relative Mach number is to be limited
to 0.97. The hub to tip radius ratio of the
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inducer is 0.4. The eye tip diameter is 20
centimeter and if the inlet velocity is axial,
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determine, part a - the maximum mass flow
rate for a rotational speed of 29160 rpm,
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part b - the blade angle at the inducer tip
for this mass flow. The inlet conditions can
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be taken as 101.3 kilopascals and 288 kelvin.
So, let us read this question little more
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carefully what is basically the problem statement
referring to, it tells us that there is a
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relative Mach number at the inlet eye and
that is limited to 0.97, that is, we have
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to keep the Mach number limited to 0.97 and
not let it attain supersonic speeds. The hub
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to tip ratio is given as 0.4, the eye tip
diameter is given as 20 centimeter, then we
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have the blade rotational speed, we need to
calculate maximum mass flow and the blade
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angle.
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So, let us take a look at the schematic of
this particular problem, we have this hub
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to tip ratio that is been given to us, the
radius ratio is 0.4 which means that r h by
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r t is 0.4. The inlet velocity is given as
axial, so C 1 is axial, and then the blade
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angle beta 1 is one of the parameters we need
to calculate, since U 1 is also specified,
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because we know the tip diameter and the rotational
speed, so from there we can calculate the
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blade angle beta 1.
So, this basically is a problem which refers
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to a typical centrifugal compressor problem,
which has the flow entering the inducer axially,
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in this case, the flow is indeed axial, and
that is one of the parameters which has been
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given. Rotational speed is given some of the
geometrical dimensions with reference to the
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inducer is also given that the tip diameter,
the hub to tip ratio etcetera is given to
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us. And one of the important statements in
the problem is that we need to ensure that
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the Mach number is limited to 0.97 at the
inlet. Actually start solving the problem
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from there onwards, so we know that the Mach
number is to be limited to that and from there
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we can calculate, let us say, the static temperature
because inlet stagnation temperature is given,
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the inlet velocities basically can be calculated.
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So, if you look at the rotational speed we
have been given that the rotational speed
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is 29160 rpm, diameter is 0.2 - 20 centimeters,
so we have phi N by 60 this is 305.36 meters
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per second. Now, from the velocity triangle
we can see that the relative Mach number is
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basically a relative velocity by square root
of gamma R T 1, and at the inlet V 1 is equal
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to square root of C 1 square plus U 1 square,
this divided by square root of gamma R T 1.
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So, we also know that T1 is equal to T 01
minus C 1 square by 2 Cp and T 01 is given
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as 288, C 1 of course is not known, C p we
will assume for air as 1005 Joules per kilogram
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Kelvin and therefore 2 Cp is 2010. So, here
we have two equations, one is for Mach number
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and the other is for static temperature. So,
let us substitute this in this equation, relative
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Mach number become square root of C 1 square
plus U 1 square divided by square root of
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gamma R into the temperature 288 minus C 1
square by 2010. Right hand side involves an
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a unknown that is C 1 and left hand side is
already given that we have to limit the Mach
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number to 0.97.
So, if you substitute all the values we have
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0.97 square is equal to C 1 square plus 305.63
square divide by gamma R into 288 that is
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115718.4 minus 0.2 C1 square. So, if we simplify
this, it is a simple quadratic equation which
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we can simplify, and then we can get C 1 is
114.62 meters per second.
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So, if we referred to the velocity triangle
we have basically determined C 1, and since
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U 1 is known you can now calculate beta 1,
because it is basically the tan of U 1 and
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C 1, of course, the first part of the question
is also to find the mass flow rate.
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So, let us calculate the mass flow rate, for
calculating mass flow rate we need the density,
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axial velocity is known, we also need the
area. Now, the static temperature, you can
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calculate here because, the stagnation temperature
is given, C 1 we have just now calculated,
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and therefore, static temperature would be
288 minus C 1 square by 2 Cp, C 1 square we
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have calculated and therefore, it becomes
281.46. And then we have, since no efficiency
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has actually been specified, we can use isentropic
relations here to calculate the corresponding
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static pressure, because stagnation pressure
has been specified, stagnation temperature
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and static temperature are known. So, if we
substitute these values here, we can calculate
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the static pressure P 1 which is 93.48 kilopascals.
Now, once we calculate the pressure and temperature,
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we can calculate the density which is P 1
by R T1 and that is 1.157 kilograms per meter
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cube.
So, this is one of the parameters which we
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need for calculating mass flow rate, the other
parameter is the area of a passage of the
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mass flow rate and that is the annulus area
at the inlet for which we have been given
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the tip diameter and the hub to tip ratio.
So, the annulus area would be phi square by
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4 into 1 minus r h by r t. So, that is 0.0264
meter square, so this is the annulus area
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through which the mass flow rate passes, and
we have calculated density. Now the third
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parameter for calculating mass flow rate is
the velocity and for mass flow rate we need
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to known the axial velocity, but at the inlet
we know that the absolute velocity itself
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is axial and therefore, C a1 should be equal
to C 1.
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So, mass flow rate is quite straight forward,
now we just multiply density, annulus area
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at inlet and the axial velocity, so rho 1
A 1 and C 1, the product of these three would
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give us the mass flow rate. And so we multiply
the density area and absolute velocity, we
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get the mass flow rate. Then, the second parameter
is the blade angle, so how does one calculate
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the blade angle? Blade angle at the inlet
is very straight forward again, because we
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have already calculated C 1 and we know the
blade speed U. So, tan inverse of C by U C
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1 by U 1 would give us the angle at the inlet
that is beta 1. So, if we substitute those
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values here, we get mass flow rate as rho
1 A 1 C 1 which is 1.157 into 0.0264 which
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is annulus area, C 1 is 114.62 meters per
second.
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So, the product of all the three will give
us the mass flow rate which comes out to be
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3.5 kilograms per second. Blade angle at the
inlet is at the tip is tan beta 1 which is
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C 1 by 1 therefore, beta 1 is tan inverse
C 1 by 1 which comes out to be 20.57 degrees.
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So, these are two parameters that we have
calculated the mass flow rate, which was the
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first part of the question; second part was
for the same mass flow what is the blade angle,
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so we have also calculated the blade angle
for this question.
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I think, I have mentioned this several times
in the past that the key to solving problems
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to do with turbo machines, the fundamental
problems to do with turbo machines is to get
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the velocity triangles. So, even for centrifugal
compressors as we have seen, velocity triangle
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is the starting point for any of this analysis,
so if you start solving the problem once you
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get the velocity triangle right. And so if
the velocity triangle can be set right, then
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solving the problem is quite easy, because
you know what to calculate and how to calculate
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them.
And so, I would urge you to keep this in mind,
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whenever you are solving a problem irrespective
of whether its axial compressors, axial turbine,
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centrifugal or radial turbines, whatever be
the problem that you are solving, it is very
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important that you understand the significance
of velocity triangles, and that is why you
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need to understand the physics very well to
be able to construct the velocity triangle
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and that would basically be the starting point
for solving any of these problems.
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So, let us move on to the next problem now
and let us see what this problem statement
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is, second problem statement is that a centrifugal
compressor has a pressure ratio of 4:1 with
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an isentropic efficiency of 80 percent when
running at 15000 rpm and inducing air at 293
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Kelvin. Curved vanes at the inlet give the
air a pre-whirl of 25 degrees to the axial
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direction at already eye. The tip diameter
of the eye of the impeller is 250 millimeters.
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The absolute velocity at inlet is 150 meters
per second and the impeller diameter is 600
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millimeters. Calculate the slip factor.
So, this is in fact, follow up question for
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this will also for will be given little later,
we will also solve that problems is quite
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similar to this, but with a small twist. So,
in this case, the question state that we have
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a centrifugal compressor which is developing
a pressure ratio of 4 : 1 with a certain efficiency
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and rotational speed. What is to be noted
here is that the air is no longer entering
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axially, unlike the previous question where
it was specifically mentioned that the inlet
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air is axial, here it is not axial it is coming
with a pre-whirl, that is, the pre-whirl is
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like a guide vane ahead of the inducer where
the guide vane set the fluid at a certain
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angle. In this case, the angle has been set
at 25 degrees and then we have been given
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some dimensions of the impeller and some speeds.
So, based on this we need to calculate the
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slip factor, so slip factor has as we have
learnt in the previous lecture, is basically
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the ratio of the tangential absolute velocity
at the impeller exit divided by the corresponding
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blade speed at the same location.
So, it is C w2 by U 2 which is the slip factor,
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we are required to find this. Now, since our
rotational speed is given and the diameter
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of the impeller is given, finding the blade
speed at the tip is very easy, because U 2
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is phi D 2 and by 60, and the D and N both
are given to us, so you can calculate the
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impeller diameter tip speed. But how do you
calculate C w2, this of course, will require
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us to solve the velocity triangle first at
the inlet and subsequently at the exit as
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well to be able to solve this problem.
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So, let us look at the velocity triangle at
the inlet of the inducer, so these are the
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inducer vanes which are shown and you can
see that the flow is not entering the inducer
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and axial direction. Is in fact, entering
at an angle of 25 degrees; that means, that
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C 1 is not axial, C 1 itself has a tangential
components C w1, then the inlet relative velocity
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is V 1 and the blade speed at the inlet is
U 1. So, this is which means that there would
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be a set of vanes, which will set this flow
angle of 25 degrees to ensure that the absolute
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velocity enters the inducer in this direction
and that is called the pre whirl, you can
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see this word mentioned here it is called
pre whirling.
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Pre whirl means that there is a whirl in component,
which is basically the tangential component
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of absolute velocity and that is why it is
called a pre whirl. In the absence of pre
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whirl, the flow would have entered axially,
but that may not really be an ideal condition
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for the inducer, because the relative velocity
might actually enter the inducer at a different
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angle of course, this has been designed for
a pre whirl.
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So, let us begin to solve this problem we
have the pressure ratio given to us and therefore
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we can calculate the exit stagnation temperature,
exit stagnation temperature is equal to T
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01 times the pressure ratio raise to gamma
minus 1 by gamma. And therefore, we get 293
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into 4 raise to 1.4 minus 1 by 1.4 that is
435.56 Kelvin. So, the isentropic temperature
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rise would be 435.56 minus T 01 that is 293,
we get 142.56 Kelvin, but the actual temperature
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rise is, isentropic temperature rise divided
by the efficiency, which in this case is specified
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as 80 percent, and therefore, we get delta
T not is equal to 1178.2 Kelvin which is 142.56
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divided by 0.8.
Therefore, from the temperature rise we can
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calculate the work done per unit mass, and
work done can be calculated in two ways, as
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we have seen in the past, one is by using
the temperature rise and the other method
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is by using velocity triangles that is U,
delta U, C w, will basically give us the work
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done per unit mass. So, here we know since
we know the temperature rise, we can calculate
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the work done per unit mass, which is basically
the enthalpy rise in the compressor and that
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is C p times delta T not, delta t not we have
calculated as 178, this multiplied by C p
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will give us the work done per unit mass.
So, work done here would be C p which is 1.005
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into delta T not, delta T not is 178.2, therefore,
we get 179 kilo joules per kilo gram.
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Now, at the exit, we can now calculate some
parameters at the exit, because our aim is
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to calculate the slip factor at the exit and
but of course, before that we need to calculate
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C w1, because that is required for calculating
the exit parameters as well. Now, at the inlet
21:17.980 --> 21:24.679
of the eye we have U 1 is phi d N by 60 that
is phi into the a tip of the eye of the impeller
21:24.679 --> 21:31.679
is 0.25 meters into 15000 divided by 60 and
that comes out to be 196.25 meters per second.
21:33.499 --> 21:40.499
C w1 is C 1 times sin 25, C 1 is given to
us and therefore, and also the angle is given
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alpha, C 1 sine alpha 1 therefore, that comes
out to be 63.4 meters per second.
21:47.840 --> 21:53.239
Now, peripheral velocity at the tip of the
impeller U 2 would be phi into capital D which
21:53.239 --> 21:59.230
is the tip diameter of the impeller that is
given as 600 millimeters multiplied by N by
21:59.230 --> 22:06.230
60, so phi into 0.6 into 1500 divided by 60
this is 471.2 meters per second. So, here
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we have calculated the conditions at the inlet
which refers to the C w1, and why we are calculating
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C w1 is because, we have just now calculated
the work done per unit mass, and work done
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per unit mass is also equal to U 1, well U
2 C w2 minus U 1 C w1, and from there U 1
22:31.460 --> 22:37.289
and U 2 are known C w1 we have just now calculated,
and therefore we can calculate C w2. Once
22:37.289 --> 22:43.840
you calculate C w2, ratio of C w2 to the blade
speed at the tip that is U 2 will give us
22:43.840 --> 22:45.090
the slip factor.
22:45.090 --> 22:52.090
So, let us now calculate C w2 which is the
whirl velocity at the tip of the impeller,
22:52.149 --> 22:59.149
and since the power work done per unit mass
is also equal to U 2 C w2 minus U 1 C w1 and
23:00.119 --> 23:05.840
this is equal to work done. We have just calculated
earlier 179 kilojoules per kilogram; this
23:05.840 --> 23:12.840
is equal to U 2 times C w2 that is 471.24
into C w2 minus U 1 C w1 that is 196.35 into
23:17.970 --> 23:24.499
63.4.
So, C w2, we can calculate as 406.27, and
23:24.499 --> 23:31.499
therefore, we calculate slip factor as ratio
of C w2 to U 2 and therefore, 406.27 divided
23:32.679 --> 23:39.679
by 471.24 that is 0.862. So, here we get a
slip factor of 0.86. What are the implications
23:42.720 --> 23:49.720
of this well, the implications depends upon
the kind of application we have, but the basic
23:50.179 --> 23:55.789
implication of having a slip factor much lower
than 1, is that it is leads to lower and lower
23:55.789 --> 24:01.509
pressure ratios, that is higher the slip factor
lower is a pressure ratio for which it is
24:01.509 --> 24:07.109
actually been design for.
This is because the tangential velocity at
24:07.109 --> 24:13.330
the exit that is basically the swirl velocity
at the exit is directly related to the pressure
24:13.330 --> 24:20.330
ratio; you can actually derive an expression
which relates the pressure rise P O2 by P
24:20.409 --> 24:25.539
01 in terms of C w and U and so on. Therefore,
there is a direct correlation between that,
24:25.539 --> 24:32.539
so lower the slip factor which means lower
is C w2 is comparison T U 2, and therefore,
24:33.840 --> 24:40.100
that will affect the pressure rise achieved
in a centrifugal compressor.
24:40.100 --> 24:47.100
That is why modern day designers would like
to maximize, increase the value of slip factor
24:47.330 --> 24:54.289
to keep it as close as possible to 1. We have
seen the strong dependence of slip factor
24:54.289 --> 25:01.289
on the number of the blades, and that is one
of the key optimization challenges because,
25:01.450 --> 25:06.629
one can keep increasing number of blades,
but then the problem with that would be the
25:06.629 --> 25:11.929
fact that increasing number of blades also
leads to an increase in these king friction
25:11.929 --> 25:16.909
losses, and therefore, that is going to affect
the efficiency in some way. The designer does
25:16.909 --> 25:21.710
not want to have a poor efficiency with the
higher pressure ratio that does not make sense.
25:21.710 --> 25:28.710
So, one need to have a choice of basically
a mix of high efficiency as well as higher
25:28.820 --> 25:35.600
pressure ratio, and that requires the very
intelligent way of trying to optimize this
25:35.600 --> 25:41.279
case, where we have on hand one option of
increasing number of blades with the risk
25:41.279 --> 25:47.919
of in lower efficiency because of losses.
Other option is to reduce number of blades
25:47.919 --> 25:53.409
and have higher law efficiency, possible higher
efficiency because of lower frictional losses,
25:53.409 --> 25:57.820
but one may end up with the lower pressure
ratio, so there is a trade of required to
25:57.820 --> 26:04.259
attain an optimum condition. So, let us now
again look at another problem, third problem,
26:04.259 --> 26:08.129
also involves slip, let us take a look at
what the problem statement is.
26:08.129 --> 26:14.970
The problem number 3 states that air at a
stagnation temperature of 22 degree Celsius
26:14.970 --> 26:21.609
enters the impeller of a centrifugal compressor
in the axial direction. The rotor, which has
26:21.609 --> 26:28.509
17 radial vanes, rotates at 15000 rpm. The
stagnation pressure ratio between the diffuser
26:28.509 --> 26:35.509
outlet and the impeller inlet is 4.2 and the
total to total efficiency is 0.83 percent
26:37.190 --> 26:44.190
or 0.83. Determine the impeller tip radius,
assuming that the air density at the impeller
26:44.669 --> 26:49.940
outlet is 2 kilograms per meter cube and the
axial width at the entrance to the diffuser
26:49.940 --> 26:54.429
is 11 millimeters, you also need to determine
the absolute Mach number at this point.
26:54.429 --> 26:59.570
We assume that the slip factor is 1 minus
2 by N, where N is the number of vanes. So,
26:59.570 --> 27:05.460
we have been permitted to use this standards
slip factor, which states that the slip factor
27:05.460 --> 27:11.690
is equal to 1 minus 2 by N, N being the number
of vanes. So, in this question we have been
27:11.690 --> 27:18.690
given the number of blades here 17 radial
vanes, rotational speed 15000 rpm and the
27:18.749 --> 27:25.059
stagnation pressure ratio 4.2, total to total
efficiency 83 percent and then, the density
27:25.059 --> 27:29.629
at the impeller outlet and the axial width
at the entrance of the diffuser.
27:29.629 --> 27:34.139
So, there are the parameters which are given
to us based on which we need to calculate
27:34.139 --> 27:41.139
two parameters, one is the impeller tip radius,
and the second is, we also need to calculate
27:41.619 --> 27:46.799
the Mach number at the exit of the impeller,
so we need to of course, calculate the absolute
27:46.799 --> 27:52.700
Mach number. So, this is the question which
is very similar to one of the questions we
27:52.700 --> 27:58.139
have solved earlier, so I will skip the velocity
triangle part leaving that to you to figure
27:58.139 --> 28:02.289
out the velocity triangle and triangle construct
the velocity triangle for that kids.
28:02.289 --> 28:08.590
So, this question involves or requires us
to calculate two parameters, one of course
28:08.590 --> 28:15.590
the hint given to us is that we can calculate
slip factor using the Steinitz formula, they
28:15.720 --> 28:19.210
have been giving the number of blades. So,
we can actually calculate the slip factor
28:19.210 --> 28:25.190
from there, and then we know the pressure
ratio and total to total efficiency, and so
28:25.190 --> 28:32.190
we can actually calculate the power required
from what data has been given to us, the other
28:32.529 --> 28:38.220
key information is that the flow enters the
impeller in the axial direction, which means
28:38.220 --> 28:42.979
that there is no tangential component to the
absolute velocity at the inlet. So, C w1 should
28:42.979 --> 28:49.169
be 0, and so you can calculate work done based
on just U 2 times C w2.
28:49.169 --> 28:56.169
So, the specific work required is U 2 C w2
minus U 1 C w1, C w1 is 0 and therefore, work
28:57.489 --> 29:04.489
is U 2 C w2 which is also equal to sigma times
U 2 square, because sigma is C w2 by U 2 and
29:07.220 --> 29:13.369
so we can express work done in terms of sigma,
which is the slip factor and the blade speed
29:13.369 --> 29:19.009
at the tip of the impeller U 2, so sigma U
2 square would be the work done - specific
29:19.009 --> 29:22.019
work done.
So, we can now express the efficiency in terms
29:22.019 --> 29:29.019
of well, U 2 in terms of efficiency and pressure
ratio, this comes from the total to total
29:29.960 --> 29:34.479
efficiency we have already defined that earlier
on with the context of a turbine, so this
29:34.479 --> 29:39.850
similar aspect we can also apply for this
case. So, when we use the efficiency definition,
29:39.850 --> 29:46.850
we have the outlet temperature T 03 minus
T 01; T 03 being the isentropic temperature
29:48.690 --> 29:55.690
minus T 01 divided by T 03 minus T 01. And
then, the pressure ratio is given to us, so
29:56.179 --> 30:01.299
the numerator gets converted can be expressed
in terms of the pressure ratio, pressure ratio
30:01.299 --> 30:08.299
is been given to us as 4.2 and the denominator
which is 3 minus T 01 is basically the work
30:09.359 --> 30:16.159
done divided by specific heats. So, you can
express that in terms of w times w divided
30:16.159 --> 30:23.159
by C p, w is sigma into U 2 square, and therefore,
we can express U 2 in terms of these parameters
30:26.019 --> 30:30.629
which are known to us the inlet stagnation
temperature, sigma, the temperature and pressure
30:30.629 --> 30:35.070
ratio as well as the efficiency.
So, if we express U 2 in terms of efficiency
30:35.070 --> 30:42.070
and pressure ratio, we have U 2 square is
C p T 01 multiplied by phi c raise to gamma
30:42.129 --> 30:49.129
minus 1 by gamma minus 1 divided by sigma
into efficiency. Efficiency is given as 83
30:49.129 --> 30:54.809
percent and sigma given can be calculated
based on the number of blades. So, sigma is
30:54.809 --> 31:01.809
1 minus 2 by N, N is 17, we have 17 numbers
of vanes or blades, so sigma the slip factor
31:01.889 --> 31:06.239
is 0.8824.
So, we know that temperature, we know pressure
31:06.239 --> 31:13.239
ratio and the efficiency; if we substituted
that we get the blades speed as 452 meters
31:13.379 --> 31:20.379
per second and then omega, because the speed
is given to us as a 15000 rpm, based on that
31:20.859 --> 31:27.529
we can now calculate the tip radius, because
tip radius is U 2 or the tip speed is basically
31:27.529 --> 31:34.529
equal to omega into r therefore, r t is equal
to U 2 by omega and omega here is the rotational
31:35.059 --> 31:41.229
speed in radiance per second 15000 into 2
phi by 60, we get 1570 radiance per second.
31:41.229 --> 31:48.229
So, if you substitute for those 452 divided
by 1570, we get the tip radius as 0.288 meters.
31:50.129 --> 31:55.700
So, this is the first part of the question
where we are required to find tip diameter
31:55.700 --> 32:02.590
of the impeller, which in this case comes
out by 288 millimeters or 0.288 meters. The
32:02.590 --> 32:07.229
second part of the question is to find the
Mach number, well, the absolute Mach number
32:07.229 --> 32:14.229
at the impeller exit, which means that we
have to take the ratio of C 2 to the speed
32:14.259 --> 32:19.799
of sound at that location that is a 2. So,
C 2 by A 2 will gives us the Mach number at
32:19.799 --> 32:25.340
station 2 which is impeller exit. So, we need
to now find out the absolute Mach number,
32:25.340 --> 32:31.429
we also need to know the temperature - static
temperature - at the exit of the impeller
32:31.429 --> 32:33.609
to be able to calculate the Mach number.
32:33.609 --> 32:40.609
Now, at the exit of the impeller we have C
2, which is root of C w2 square plus C r2
32:41.970 --> 32:47.759
square where C r is the radial velocity, the
absolute component of radial velocity, and
32:47.759 --> 32:53.149
how do we calculate C r2, C r2 would basically
be coming from the mass flow rate, mass flow
32:53.149 --> 32:59.830
rate divided by the density times the annulus
area. Now, in this case, we have mass flow
32:59.830 --> 33:05.359
rate which has been given to us as 2 kilograms
per second, density is also given as 2 kilograms
33:05.359 --> 33:12.359
per meter cube. Tip radius we have just now
calculated 288 millimeters or 0.288 meters,
33:12.789 --> 33:19.159
and the axial width is given to us in the
question is 11 millimeters, so this 0.011.
33:19.159 --> 33:26.159
So, C r2 which is the radial velocity at the
exit of the impeller is mass flow rate divided
33:26.220 --> 33:33.220
by density times the annulus area 2 phi into
tip radius times the axial width, it gives
33:34.849 --> 33:41.849
the annular area, that comes out to be 50.3
meters per second. Similarly, C w2, we can
33:42.909 --> 33:49.029
calculate because we know U 2, we also know
the slip factor sigma product of those 2 would
33:49.029 --> 33:55.499
give us the tangential velocity at the exit
C w2, so that is 400 meters per second. So,
33:55.499 --> 34:02.499
C 2 is equal to square root of 50.3 squares
plus 400 square that is 402.5 meters per second.
34:03.039 --> 34:09.290
So, for calculating Mach number, we now have
absolute velocity, we now need to calculate
34:09.290 --> 34:14.629
the static temperature as well. So, for calculating
the static temperature we will make use of
34:14.629 --> 34:20.859
rothalpy, we have already discussed about
that earlier on. So, will make use of that
34:20.859 --> 34:26.800
concept here, the conservation of rothalpy
in the impeller, and based on that we will
34:26.800 --> 34:30.669
calculate the static temperature.
So, we know that h 02 which is stagnation
34:30.669 --> 34:35.810
temperature at the exit of the impeller would
be equal to inlet stagnation temperature plus
34:35.810 --> 34:42.810
the work done, that is W c, so h 01 plus W
c will be h 02, and therefore, h 01 plus sigma
34:44.570 --> 34:51.570
U 2 square, because W c. We know in this case
is sigma U 2 square, and therefore, h 2 would
34:52.490 --> 34:59.490
be equal to h 01 plus sigma U 2 square minus
half C 2 square, because h 2 plus half C 2
35:00.980 --> 35:07.980
square is h 02, this we will convert in terms
of temperatures. Since h 2 is equal to h 01
35:08.730 --> 35:14.090
plus sigma U 2 square minus half C 2 square
correspondingly in terms of temperature we
35:14.090 --> 35:21.000
get T 2 is equal to T 01 plus sigma U 2 square
minus half C 2 square divide by C p.
35:21.000 --> 35:26.090
So, we substitute all these values and we
get temperature, static temperature at the
35:26.090 --> 35:32.120
impeller exit as 394.5 Kelvin, therefore,
the Mach number at the exit of the impeller
35:32.120 --> 35:39.120
is 402, which is the absolute velocity divided
by square root of 1.4 into r 287 into 394.5,
35:42.960 --> 35:49.360
Mach number comes out be 1.01. So, we can
see that the absolute Mach number is just
35:49.360 --> 35:54.440
about sonic, it has just crossed the sonic
Mach number, and in fact, in the relative
35:54.440 --> 35:59.970
frame of reference the Mach number can in
fact be higher than what you have calculated
35:59.970 --> 36:06.970
for the absolute case here.
So, the third problem we just now solved concerned
36:08.320 --> 36:15.000
was basically about calculating, Firstly the
Mach number we have seen how to calculate
36:15.000 --> 36:21.640
the Mach number as well as how do you calculate,
based on the parameters like the slip factor
36:21.640 --> 36:28.640
in this case to be approximated as 1 minus
2 by N. We can also calculate the mass flow
36:28.970 --> 36:35.270
rate as we have done in this question. Now,
let us take up one more problem which is in
36:35.270 --> 36:41.260
some sense identical to what we have solved
in the second question and partly also identical
36:41.260 --> 36:45.670
to what we have just now solved for the third
question, so we will require understanding
36:45.670 --> 36:50.420
of how those two problems were solved to be
able to solve this question. So, we will be
36:50.420 --> 36:57.060
calculating the slip factor as well as the
Mach number in this question that we were
36:57.060 --> 36:57.570
going to solve.
36:57.570 --> 37:02.480
Let us take a look at the problem statement
for this question, a centrifugal compressor
37:02.480 --> 37:09.480
with a backward leaning blades develops a
pressure ratio of 5:1 with an isentropic efficiency
37:10.170 --> 37:16.510
of 83 percent. The compressor runs at 15000
rpm. Inducers are provided at the inlet of
37:16.510 --> 37:22.600
the compressor. So, that air enters at an
absolute velocity of 120 meters per second.
37:22.600 --> 37:27.910
The inlet stagnation temperature is 250 Kelvin
and the inlet air is given a pre whirl of
37:27.910 --> 37:34.100
22 to the axial direction at all radii. The
mean diameter of the eye of the impeller is
37:34.100 --> 37:41.080
250 millimeters and the impeller tip diameter
is 600 millimeters. Determine the slip factor
37:41.080 --> 37:44.170
and the relative Mach number at the impeller
tip.
37:44.170 --> 37:49.900
So, you can immediately see that it has components
of both the second question as well as third
37:49.900 --> 37:55.200
question, second question we actually calculated
the slip factor, and in the third question
37:55.200 --> 37:59.320
we calculated the absolute Mach number of
course, in this case we are required to calculate
37:59.320 --> 38:01.040
the a relative Mach number.
38:01.040 --> 38:05.790
Let us look at the velocity triangles first,
both at the inlet as well as the exit, this
38:05.790 --> 38:10.840
is the velocity triangle that you should get
at the inlet and you have the schematic of
38:10.840 --> 38:16.950
the inducer and also fixed inlet guide when
which gives a pre whirl to the inlet flow
38:16.950 --> 38:21.550
which is entering the inducer.
So, these guide vanes would which are set
38:21.550 --> 38:28.550
at an angle of alpha 1 gives a pre-whirl which
is given in this case as 22 degrees causing
38:28.630 --> 38:34.480
the absolute velocity not to be axial, and
therefore, the absolute velocity C 1 is not
38:34.480 --> 38:40.910
axial unlike some of the questions we have
solved earlier on, and that causes the velocity
38:40.910 --> 38:46.210
triangle at the inlet to look like what is
shown here, this is C 1 the relative velocity
38:46.210 --> 38:52.830
V 1 and the blades speed V 1. At the exit
on the other hand, these are backward leaning
38:52.830 --> 38:58.260
blades, and so the velocity triangle for a
backward leaning blade should look like this.
38:58.260 --> 39:02.750
Since the blades are leaning like this, the
relative velocity leaves the blades in this
39:02.750 --> 39:08.990
direction with an angle of beta 2, and this
is the absolute velocity C 2 and the blade
39:08.990 --> 39:13.470
speed U 2 at the tip of the impeller, the
axial velocity component is shown here as
39:13.470 --> 39:18.910
C a and the rotational speed is omega which
is given in this case as 15000 rpm. Let us
39:18.910 --> 39:24.160
first try to solve the inducer part of the
question and then, we will move towards solving
39:24.160 --> 39:25.090
the second part.
39:25.090 --> 39:31.010
Now, the inlet temperature is given as 300
Kelvin and therefore, based on the pressure
39:31.010 --> 39:37.780
ratio and isentropic relation we can calculated
the exit stagnation temperature isentropic
39:37.780 --> 39:44.730
and that is T 02 s, which is equal to T 01
into phi C raise to gamma minus 1 by gamma
39:44.730 --> 39:51.730
and this is 250 multiplied by 5 raise to 0.4
by 1.4, that is, gamma minus 1 is 0.4 divided
39:52.700 --> 39:59.700
by 1.4, this basically 395.95 Kelvin. Therefore,
the stagnation temperature isentropic is 395.95
40:04.670 --> 40:11.390
minus 300 that is 95.95 Kelvin, but the actual
temperature rise is this divided by the efficiency,
40:11.390 --> 40:17.770
in this case, efficiency is given as 0.83
and therefore, T 0 s actual basically would
40:17.770 --> 40:24.770
be equal to this divided by efficiency that
is 95.95 divided by 0.83 that is 115.6 Kelvin.
40:27.340 --> 40:34.340
So, the specific work required would be C
p times delta t naught that is 1005 into 115.6,
40:35.880 --> 40:42.880
this is 116.886 kilojoules per kilogram, now
since it is given that C 1 is 150 we can calculate
40:45.210 --> 40:52.100
C w1 which is the whirl component of the absolute
velocity that is C 1 times sine alpha 1, alpha
40:52.100 --> 40:53.830
1 is 22.
(Refer Slide Time: 40:54)
40:53.830 --> 40:59.150
Let me go back to the velocity triangle here,
C 1 times sine alpha 1 would give us C w1
40:59.150 --> 41:05.230
which is this component, so that comes out
to be 56.2. Now, here what we going to do
41:05.230 --> 41:11.240
is, since we know the specific work done from
the temperature rise, we also know that specific
41:11.240 --> 41:18.240
work done is equal to the product of U 2 times
C w2 minus U 1 times C w1. C w1 we have calculated,
41:20.100 --> 41:27.100
we can now calculate U 1 and U 2, therefore,
we get C w2 and once we know C w2 and we can
41:27.220 --> 41:30.630
calculate the slip factor which is C w2 by
U 2.
41:30.630 --> 41:37.630
So, let us first calculate U 1 and U 2, U
1 is phi mean of the eye of the impeller into
41:37.800 --> 41:44.800
the rotational speed by 60 that is given as
phi into 0.25 into 15000 by 60, this is 196.3
41:47.620 --> 41:54.620
meters per second. And U 2 is the blade speed
at the tip of the impeller, phi t into N by
41:55.050 --> 42:02.050
60 that is phi into 0.6 into 15000 by 60 is
471.24 meters per second. Now, specific work
42:02.700 --> 42:08.100
done as we have seen, we calculated that already
from the previous calculations specific work
42:08.100 --> 42:15.100
is C p delta T not that is 116.186, specific
work is 116.186 into 10 raise to 3 is equal
42:16.400 --> 42:23.400
U 2 which is 471.24 into C w2 minus U 1 which
is 196.3 into C w1 56.2.
42:24.850 --> 42:31.850
So, from this we can calculate C w2 that is
269.6 meters per second, therefore, slip factor
42:33.250 --> 42:40.250
is the ratio of this C w2 divided by U 2 that
is 269.96 divided by 471.24 that is 0.573.
42:44.500 --> 42:49.160
We can see that the slip factor is a very
low number here, one would normally expect
42:49.160 --> 42:54.950
relatively higher slip factor of the order
of 0.7, 0.8 or even higher than that, this
42:54.950 --> 42:59.070
is the very low slip factor. And we have already
discussed the disadvantages of having very
42:59.070 --> 43:05.250
low by use of slip factor, basically affecting
the pressure rise of the centrifugal compressor
43:05.250 --> 43:11.430
for a given rotational speed. So, the next
part of the question is to calculate the Mach
43:11.430 --> 43:14.160
number at the tip in the relative frame.
43:14.160 --> 43:18.860
So, the relative Mach number at the tip of
the compiler for which we will refer to the
43:18.860 --> 43:24.100
velocity triangles once again, let us look
at just the velocity exit velocity triangle,
43:24.100 --> 43:27.510
for the backward leaning blade we have seen
that the velocity triangle would look like
43:27.510 --> 43:33.890
this. We have the absolute velocity C 2 and
the relative V 2 which is leaving the blades
43:33.890 --> 43:38.820
tangentially, and so this is however, the
exit velocity triangle would look like as
43:38.820 --> 43:43.760
we need to calculate V 2 and also the temperature
at the exit of the impeller to calculate the
43:43.760 --> 43:44.580
Mach number.
43:44.580 --> 43:50.200
So, from the velocity triangle we see that
V 2 is equal to square root of C a square
43:50.200 --> 43:57.130
plus U 2 minus C w2 the whole square. So,
this C a at the inlet of course, we are assuming
43:57.130 --> 44:02.580
that axial velocity does not really change,
as it passes through this impeller, this is
44:02.580 --> 44:08.400
equal to C a square that is C 1 cos alpha
1 the whole square minus, well, plus U 2 minus
44:08.400 --> 44:13.330
C w2 the whole square.
Since all these parameters are known we substituted
44:13.330 --> 44:20.330
that we get the relative velocity 222.9 meters
per second. And then, we need to also calculate
44:21.610 --> 44:28.470
the static temperature at the tip, T 2 which
is T 02 minus C 2 square by 2 C p, T 02 we
44:28.470 --> 44:34.460
can calculate because, we know the efficiency
and the pressure ratio. So, from there we
44:34.460 --> 44:41.460
calculate T 02 which is T 01 plus T 02 s minus
T 01 by efficiency and that comes out to be
44:42.430 --> 44:49.060
365.61.
And we also need to calculate C 2, from the
44:49.060 --> 44:56.060
velocity triangle we see that C 2 is C 2 square
is equal to the this component C w2 square
44:57.620 --> 45:04.620
plus C a square. So, C w2 we have already
calculated that is 269.9, that is, square
45:05.700 --> 45:12.700
plus C a square which we can calculate from
this C 1 cos alpha 1 that is 139.08 square.
45:13.730 --> 45:20.730
So, C 2 comes out to be 303.68 meters per
second, therefore the static temperature is
45:21.220 --> 45:28.220
equal to 365.61 minus C 2 square by 2 C p,
therefore T 2 comes out to be 319.73 Kelvin.
45:31.060 --> 45:37.710
Therefore, the relative Mach number is, the
relative velocity divided by square root of
45:37.710 --> 45:43.540
gamma R T 2, and this if we substitute we
get relative Mach number of 0.62. So, relative
45:43.540 --> 45:50.540
Mach number at the impeller tip in this case,
is calculated as 0.62. So, this completes
45:51.850 --> 45:57.140
four problems that we have solved in today
class, we will started off with a very simple
45:57.140 --> 46:02.770
problem which just involved solving the velocity
triangle and calculating the velocities and
46:02.770 --> 46:06.810
also the angles involved.
Second question was to calculate basically
46:06.810 --> 46:12.610
the slip, third question involved calculating
the Mach number in the absolute frame and
46:12.610 --> 46:17.260
the last question was combination of the second
and third calculating the slip factor as well
46:17.260 --> 46:22.200
as the Mach number at the tip. So, these were
four question that we have solved in today
46:22.200 --> 46:29.110
lecture, I now have a few exercise problems
which you can take up and solve based on our
46:29.110 --> 46:33.700
discussion today as well as what we have discussed
during the lectures.
46:33.700 --> 46:38.130
Let us take a look at the first exercise problem,
the first exercise problem states that the
46:38.130 --> 46:44.660
design mass flow rate of a centrifugal compressor
is 7.5 kilograms per second with inlet stagnation
46:44.660 --> 46:50.610
temperature of 300 Kelvin and pressure of
100 kilopascal. The compressor has straight
46:50.610 --> 46:56.530
radial blades at the outlet, the blade angle
at the inducer inlet tip is 50 degrees and
46:56.530 --> 47:01.720
the inlet hub to tip ratio is 0.5, the impeller
is designed to have a relative Mach number
47:01.720 --> 47:08.720
of 0.9 at the inducer inlet tip.
If the tip speed is 450 meters per second,
47:08.970 --> 47:14.800
determine part a -the air density at inducer
inlet, part b - inducer inlet diameter, part
47:14.800 --> 47:20.590
c - the rotor rpm and part d - the impeller
outlet diameter. The answers to these four
47:20.590 --> 47:25.750
different parts of the questions are density
should come out to be 0.988 kilograms per
47:25.750 --> 47:32.750
meter cube, inducer inlet diameter is 0.258,
rotor rpm is 17100 rpm and impeller outlet
47:33.290 --> 47:35.200
diameter is 0.502 meters.
47:35.200 --> 47:42.200
The second exercise question is, a centrifugal
compressor runs at 10000 rpm and delivers
47:42.630 --> 47:48.650
600 meter cube per minute of air at a pressure
ratio of 4 is to 1. The isentropic efficiency
47:48.650 --> 47:54.150
of the compressor is 0.82. The outlet radius
of the impeller is twice the inner radius.
47:54.150 --> 48:01.150
The axial velocity is 60 meters per second
if the ambient conditions are 1 bar and 293
48:01.210 --> 48:08.040
Kelvin, determine part a - the impeller diameter
at inlet and outlet the power input, and the
48:08.040 --> 48:13.180
impeller angles at the inlet. The answer to
the questions are impeller diameter at the
48:13.180 --> 48:20.180
inlet should be 0.92 meters and outlet 0.461,
and the power input is 2044 kilowatts impeller
48:24.160 --> 48:29.760
and diffuser angles at the inlet are 13.9
degrees and 7.1 degrees.
48:29.760 --> 48:34.760
Third exercise problem is 30 kilograms of
air per second is compressed in a centrifugal
48:34.760 --> 48:40.000
compressor at a rotational speed of 15000
rpm. The air enters the compressor axially.
48:40.000 --> 48:46.070
The compressor has a tip radius of 30 centimeters.
The air leaves the tip with a relative velocity
48:46.070 --> 48:51.800
of 100 meters per second at an angle of 80
degrees. Assuming an inlet stagnation pressure
48:51.800 --> 48:56.820
and temperature of 1 bar and 300 Kelvin, respectively,
find part a - the torque required to derive
48:56.820 --> 49:02.810
the compressor, the power required and the
compressor delivery pressure. So, the torque
49:02.810 --> 49:09.810
in this case will be 4085 Newton meters, power
required 6.417 megawatts and compressor delivery
49:10.590 --> 49:13.980
pressure is 6.531 bar.
49:13.980 --> 49:19.930
And the last problem is a centrifugal compressor
has an impeller tip speed of 366 meters per
49:19.930 --> 49:23.870
second. Determine the absolute Mach number
of the flow leaving the radial vanes of the
49:23.870 --> 49:30.010
impeller when the radial component of velocity
at impeller exit is 30.5 meters per second
49:30.010 --> 49:35.250
and the slip factor is 0.90. Given that the
flow area at the impeller exit is 0.1 meters
49:35.250 --> 49:41.860
square and the total to total efficiency is
90 percent, determine the mass flow rate.
49:41.860 --> 49:48.800
So, in this case we have the absolute Mach
number as 0.875 and mass flow rate as 5.61
49:48.800 --> 49:52.420
kilograms per second.
So, these are four exercise problems that
49:52.420 --> 49:57.950
you can solve based on what we have discussed
in the last three lectures including today's,
49:57.950 --> 50:03.370
and I hope based on this discussions you will
be able to solve these four exercise problems
50:03.370 --> 50:10.200
that we have for you today. And we will continue
our discussion on some of these topics especially,
50:10.200 --> 50:15.450
some of the radial flow machines we will probably
be taking up the radial flow turbines in the
50:15.450 --> 50:20.240
coming lectures. So, we will discuss more
to do with radial flow machines in some of
50:20.240 --> 50:23.020
the coming lectures, in the next few lectures.