WEBVTT
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Hello and welcome to lecture number 30 of
this lecture series on introduction in on
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jet aircraft propulsion. We have been discussing
about the components of aircraft engines and
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the last lecture, we had devoted exclusively
for discussion on one of the components which
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constitute an aircraft engine that is the
nozzle. And I think during the last lecture,
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we had some detailed discussion on different
types of nozzles. We have discussed about
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subsonic nozzles and supersonic nozzles and
various types of these nozzles like fixed
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geometry, variable geometry, axisymmetric
nozzles, and two-dimensional nozzles and so
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on.
We also had some chance to discuss about what
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are the functions of a nozzle. Besides the
fact that nozzles are primarily meant to generate
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thrust, modern day nozzles have various other
functions; Some of which are for thrust reversal
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which is used for breaking an aircraft, when
it is landing. Then thrust vectoring used
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popularly in compact aircraft and besides
that, it also has to keep noise under control
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and some of the aircraft have to like the
military aircraft also have to keep in mind
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that the nozzle exhaust should have as low
and IR signature; that is infrared signature
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as possible. So, that it can escape enemy
radar from being detected.
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So, these are some of the functions that a
modern day nozzle has to satisfy. With the
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fact that, it has to also work as a unit as
a single unit along with the rest of the other
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components like the intake, the compressor
the fan and the compressor, the combustion
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chamber, the turbine and the jet pipe and
so on. So, all these components put together
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constitute the engine and when an aircraft
is operational, it has to operate over a variety
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of operating ranges and all these components
need to be matched. And so, you would also
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be looking at matching process between various
components in the next few lectures.
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You would also be discussing about matching
of these components separately and matching
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is obviously a very important aspect. Even
though individual components may be designed
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for highest possible efficiency, it should
be ensured that when they are all put together
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and when they form an engine, then they all
work as optimum as possible and that is where,
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the importance of matching comes in to picture.
So, nozzle obviously is another component;
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the last component of an aircraft engine which
also needs to be matched with the other components.
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So, in today’s lecture, we are going to
discuss about the function. While we have
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already discussed the functions, we are going
to discuss about the working of these nozzles
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that different types of nozzles from a very
fundamental thermodynamic sense. If you have
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undergone thermodynamics or gas dynamics course,
you probably would have discussed about…
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you you would have already learnt little bit
about how different nozzles work; that is
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subsonic nozzle and a supersonic nozzle work
and how the pressure ratio across these nozzles
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can be controlled and shown. We will discuss
some of these aspects in today’s lecture.
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So, today’s lecture we shall be taking up
some of these topics. We shall talk about
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the working of subsonic and supersonic nozzles.
We will be very briefly discussing about the
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performance parameters of these nozzles. And
towards the end of the lecture, we also discuss
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some of the issues which nozzle expansion
will have like if it is not correctly expanded
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or fully expanded, then there could be some
problem with the flow that is coming out of
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the nozzle. So, before I start discussing
details about the subsonic or the convergent
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nozzle and subsequently the convergent-divergent
nozzle, let us review some of the concepts
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which I presumed.
You would have learnt in either the gas dynamics
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course or if you have undergone one of our
earlier courses on introduction to aerospace
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propulsion. Some of these topics have already
been covered there in detail. We will quickly
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review some of these concepts before proceeding
towards understanding of the working mechanism
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of these nozzles. Now, in the last class,
I think I mention that in a convergent nozzle,
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there is certain limitation to the Mach number
that can be achieved. That is, if we keep
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varying keep reducing the downstream pressure
the back pressure so called the back pressure,
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the flow will accelerate in a subsonic flow.
It will accelerate in the nozzle and what
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is the maximum Mach number that it can reach.
The maximum Mach number it can reach in a
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convergent nozzle as you probably know is
mach 1. That occurs when the flow or the flow
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rate has reached its maximum level at the
throat of the nozzle. Throat is the area that
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is the minimum cross sectional area of of
a convergent nozzle. When the mass flow reaches
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its maximum, then that is instance the Mach
number at the throat would be unity and under
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this condition, the flow or the nozzle is
said to be choked.
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Now, if you want to reduce the Mach number
further, one might intuitively want to reduce
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the area further; that is we are attach let
say another convergent nozzle to that. What
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you will notice obviously is that that is
not going to help; it is not going to reduce
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the… it is not going to increase the Mach
number further. If a Mach number increase
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has to indeed take place after the throat
of a convergent nozzle, we will need a divergent
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section. And that is why we have convergent-divergent
nozzles wherein we can achieve supersonic
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flow, which is not possible in conventional
simple convergent nozzle.
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So, in a convergent nozzle, as I am going
to illustrate here now this is an illustration
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of is convergent nozzle. So, in the first
one, we have a convergent nozzle which has
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upstream pressure stagnation pressure P 0
or p naught and stagnation temperature, T
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naught. As the flow accelerates through the
nozzle, it reaches sonic Mach number that
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is mach 1 at the throat. And ofcourse, this
occurs only for certain back pressures; it
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does not occur always. If there is a sufficient
pressure ratio, then the nozzle can be choked
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and you get sonic velocity here. Now, if you
want to increase the Mach number further and
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after; that is, after this we want to supersonic
Mach number, it will not help if you put in
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another convergent nozzle here.
Let us say we put another convergent nozzle
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with this as the inlet area of that nozzle
and an area which is much smaller than this
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at the exit. What will happen is that the
sonic point of the throat will shift. So,
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as you attach this additional portion here,
the throat will shift to a new location; which
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means that the location which was the throat
for the first nozzle, where you had Mach number
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unity will no longer be sonic Mach number.
You would have a subsonic mach number here.
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Sonic Mach number would have shifted. Ofcourse,
this will not keep happening if you keep increasing
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the area I mean the area ratio further, then
it will obviously result in loses.
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Then there will be increased loses in the
nozzle and it is not a feasible proposition.
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So, this requires that we attach a divergent
section after the throat. Attaching a divergent
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section is the only solution that we can get
a supersonic mach number in a nozzle that
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is convergent section ending in a throat followed
by a divergent section. Now, in a gas dynamics
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course, you must have derived what is known
as the area velocity Mach number relation
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right, where we must have seen that when mach
number is less than 1. Then when there is
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an increase in area, we have decrease in velocity;
that is we have a diffuser.
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And if there is a decrease in area, we have
an increase in velocity; whereas, if it is
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supersonic Mach number, then the trend is
reversed and increase in area will actually
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give an increase in velocity and vice versa.
That is, decrease in area will result in decrease
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in the velocity. This is for a supersonic
flow which means that if that is to be true,
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for this convergent nozzle which has attained
a sonic Mach number at its throat. You can
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achieve a supersonic Mach number further,
if we attach a divergent section; that is
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in at the end of a convergent section where
it has reached its throat, if you attach a
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divergent section there, then it is possible
that we can get supersonic Mach number in
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the divergent section.
And little later we will see that it does
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not always happen. It requires certain conditions
to be satisfied like the pressure ratio the
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back pressure has to be low enough for you
to achieve supersonic flow even in the divergent
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section; that we will discuss a little later.
Now, in the next slide what I have shall show
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you is probably something you would have learnt
in gas dynamics or in the previous course
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which is a direct outcome of the area Mach
number velocity relation, which is what I
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had just described that what happens for subsonic
flow and what happens for supersonic flow
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are entirely opposite, when it comes to nozzles
and diffusers.
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So, let us take a look at what happens there.
Now, in a subsonic flow, let us say the first
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two diagrams the top two pictures or illustrations
are true for a subsonic flow. Subsonic flow
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mach number upstream Mach number is less than
1. Decrease in area results in decrease in
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pressure and temperature, since static pressure
and temperature; yet at the same time results
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in increase in velocity and hence the Mach
number. If we put the nozzle the other way
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round, that’s Mach number is still less
than 1 upstream; but area is now increasing.
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Then the pressure and temperature static pressure
and temperature increases, velocity and therefore
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the mach number decreases.
Therefore, the first one is known as a subsonic
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nozzle; second one is known as a subsonic
diffuser. Now, the reverse exact reverse of
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this happens, when the flow is supersonic.
If Mach number is greater than 1, then an
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increase in area as you can see here. We will
result in decrease in static pressure and
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temperature and therefore velocity and Mach
number and therefore, this is supersonic nozzle.
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And Mach number greater than 1 that is in
supersonic flow, if there is a decrease in
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area, it results in increase in static pressure
and temperature and therefore, it will it
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results in increase decrease in velocity and
hence mach number.
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So, this is basically supersonic diffuser.
Say nozzle in a subsonic flow acts like a
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diffuser in a supersonic flow. A diffuser
in a subsonic flow acts like a nozzle in supersonic
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flow. So, this one important aspect that you
need to understand here that what a nozzle
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for a subsonic flow is will act like a diffuser
for a supersonic flow and what acts like a
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diffuser in a subsonic flow acts like a nozzle
in a supersonic flow. So, this gives us some
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idea on how we can proceed now. We are struck
with the fact that in a in a convergent section.
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We can only achieve sonic Mach number. So,
what we do if we have to get a supersonic
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Mach number beyond the throat?
So, after the throat now if we need to get
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a supersonic flow, the best way to do that
is or well rather the only way to do that
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is to attach a divergent section. So, a divergent
section as we have seen now will get us a
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supersonic flow; because a divergent section
in a supersonic flow is a nozzle for supersonic
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flow. So, if we have to get achieve supersonic
flow, we will need a divergent section. So,
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all supersonic nozzles will have a convergent
section which will end in a throat, which
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is the minimum area of that entire nozzle
followed by a divergent section, where we
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can get supersonic flow.
So, in a supersonic nozzle, we normally referred
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to supersonic nozzles as convergent-divergent
nozzle; because it as a convergent section,
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it also has a separate divergent section and
in between these two, there is a throat where
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the area is minimum. So, supersonic nozzles
will be basically convergent-divergent nozzle.
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So, today’s lecture, we will discuss in
detail about how a supersonic nozzle can work.
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And we will see very soon that it cannot work
in under any circumstance that needs to be
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certain pressure ratio across the nozzle which
has to be maintained. So, let us begin our
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discussion with a discussion on the subsonic
nozzle first. Let us see what happens in the
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case of a convergent nozzle.
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So, in a convergent nozzle, what we will discuss
begin with is that convergent nozzle as we
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have seen can give as the maximum of Mach
number of 1. So, beyond that, obviously we
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cannot increase the Mach number; that is a
convergent nozzle. So, before we look at the
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convergent nozzle, let us look at one set
of equations which will be used in both the
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cases where whether it is subsonic or supersonic
flow. So, we will look at one of the governing
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equations for this; where we will derive an
expression for the mass flow rate which can
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pass through these geometries whether it is
convergent or convergent-divergent nozzle.
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So, mass flow rate as we know it from continuity
is equal to the product of the density, the
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velocity and the area.
So, from product of these three, we can calculate
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mass flow rate. So, let us express mass flow
rate in terms of parameters like the upstream
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parameters, the stagnation pressure and temperature
and some of the other parameters including
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Mach number. Let us express or derive an expression
for mass flow rate which we can use in our
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subsequent discussion on convergent as well
as C-D nozzles; that is convergent divergent
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nozzle. So, let us basically consider flow
of calorically perfect gas through a nozzle,
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which could be either be a convergent nozzle
or convergent-divergent nozzle. So, mass flow
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rate as we know it is product of density,
velocity and area. Density we know is pressure
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divided by R T; P by R T.
Velocity is Mach number time square root of
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gamma R T and area. So, here we have we are
going to simplify this. We have now Mach number
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times area P by P naught, which is P naught
is stagnation pressure multiplied by P naught
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in to square root of gamma divided by square
root of R T in to square root of T by T naught.
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This basically comes from this expression
here; because we have as a square root of
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R T here and R T. So, this becomes square
root of R T and T by T naught. So, this again
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we can simplify. We now have square root of
gamma in to P naught which is stagnation pressure
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divided by square root of T naught in to R
multiplied by Mach number in to area. In the
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numerator, we have 1 plus gamma minus 1 by
2 M square and where does this come?
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This comes from the temperature ratio; that
is T T naught by T is 1 plus gamma minus 1
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by 2 M square. And that is why, we have the
square root 1 plus gamma minus 1 by 2 M square
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rise to 1 by 2; because it is square root
of this. Similarly, pressure we have expressed
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in terms of Mach number. We have 1 plus gamma
minus 1 by 2 M square rise to gamma by gamma
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minus 1. This we can again simplify. This
after we simplified we be we get an expression
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mass flow rate is equal to area time stagnation
pressure divided by stagnation time pressure
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square root square root of gamma by R which
are properties of the gas multiplied by Mach
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number M divided by 1 plus gamma minus 1 by
2 M square whole rise to gamma plus 1 divided
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by 2 in to gamma minus 1.
So, this is a simple expression which relates
16:19.180 --> 16:25.079
mass flow rate and some of the other parameters
which are involved; mass flow rate, then the
16:25.079 --> 16:30.019
inlet stagnation pressure temperature and
Mach number. Besides ofcourse, the gas properties
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like the ratio of specific heats gamma and
the gas constant R. So, what we have here
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is an expression which tells us that what
the mass flow rate which a given area can
16:41.510 --> 16:47.230
give. So, here again the area comes in to
picture, once we know the stagnation pressure
16:47.230 --> 16:52.749
temperature and the Mach number or vice versa.
If we know the mass flow rate and ofcourse
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the area, we can actually calculate what Mach
number we are operating at and so from this
16:59.050 --> 17:03.709
expression which can be used for both subsonic
and supersonic nozzles.
17:03.709 --> 17:08.130
We can basically relate mass flow rate with
the other parameters like stagnation pressure
17:08.130 --> 17:14.449
temperature, Mach number and so on. So, with
this background in mind, we will now try to
17:14.449 --> 17:20.970
analyze flow through convergent nozzle. But
we will look at an isentropic flow through
17:20.970 --> 17:27.030
a convergent nozzle. When we talk about isentropic
flow, we are inherently assuming that the
17:27.030 --> 17:34.039
flow is adiabatic and ofcourse, it is reversible.
Well, adiabatic flow is true in this case;
17:34.039 --> 17:42.140
because the amount of change in pressure that
occurs across the nozzle is much larger than
17:42.140 --> 17:47.710
the amount of heat transfer, which may take
place across the walls of the nozzle. So,
17:47.710 --> 17:53.679
with that assumption in mind, we can keep
we can safely assume that the flow through
17:53.679 --> 17:57.510
these nozzles are adiabatic.
There is no heat transfer really. Even though
17:57.510 --> 18:01.940
there would be heat transfer, that is very
small compared to the pressure drop which
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occurs in the nozzle. Therefore it safe to
assume that the flow is adiabatic and so with
18:07.500 --> 18:13.730
for an isentropic nozzle flow or isentropic
flow across a nozzle, we will now look at
18:13.730 --> 18:19.210
what happens in a convergent nozzle; how the
pressure changes, how the mass flow rate changes
18:19.210 --> 18:24.620
and so on in convergent nozzle, which is basically
a subsonic nozzle; which we will very soon
18:24.620 --> 18:30.830
seen that the max Mach number which we can
get and reasons behind that is unity. You
18:30.830 --> 18:34.419
can get a Mach number of 1 at the throat of
the nozzle.
18:34.419 --> 18:39.409
So, we will look at a convergent nozzle first
in a subsonic flow, which means the convergent
18:39.409 --> 18:44.270
nozzle in a subsonic flow will have to have
a decreasing area along the flow direction.
18:44.270 --> 18:50.320
We will now consider the effect of back pressure
on the exit velocity, mass flow rate, pressure
18:50.320 --> 18:56.900
distribution etcetera. And we will also assume
that the flow is entering the nozzle from
18:56.900 --> 19:02.549
a reservoir in such a way that, its stagnation
pressure there therefore inlet velocity will
19:02.549 --> 19:09.090
be zero and stagnation pressure and temperature
will not change in the nozzle. That is stagnation
19:09.090 --> 19:13.559
temperature cannot change; because it is adiabatic.
There is no heat transfer across the nozzle
19:13.559 --> 19:18.419
valves and the stagnation pressure we will
assume does not change; because we will assume
19:18.419 --> 19:19.800
that there are no pressure loses.
19:19.800 --> 19:25.650
Let us take a look at what happens. So, I
have here a very simple schematic of a convergent
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nozzle. Though this is a convergent nozzle
in subsonic flow, it has to be increasing
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area. We will now see the effect of back pressure
which I have denoted by P subscript b on the
19:37.820 --> 19:42.120
flow properties in the nozzle. The upstream
pressure that the reservoir pressure is P
19:42.120 --> 19:47.410
naught, which is stagnation pressure; it also
has stagnation temperature of T naught. The
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nozzle exit pressure is denoted by P subscript
e or P e. Let us look at the ratio P by P
19:53.909 --> 20:01.270
naught that is a pressure at any instant versus
the reservoir pressure for any value of back
20:01.270 --> 20:04.419
pressure.
The first instance is if back pressure is
20:04.419 --> 20:09.610
equal to this reservoir pressure, what happens.
Well, if the pressures at both ends are same,
20:09.610 --> 20:13.430
obviously there is no flow taking place across
the nozzle. So, there will not be any flow.
20:13.430 --> 20:20.101
Let us see what happens if we decrease the
back pressure. We have decrease the back pressure
20:20.101 --> 20:25.529
now; but it is still greater than the critical
pressure. Critical pressure is denoted by
20:25.529 --> 20:31.460
P star. Critical pressure tells us or denotes
that pressure at which we get a sonic flow
20:31.460 --> 20:37.429
or choked flow at the exit of the nozzle.
So, second instance we have P b which is greater
20:37.429 --> 20:41.330
than p star.
But it is still less than this reservoir pressure.
20:41.330 --> 20:47.120
So, P b is less than p naught as you can see
here. This is the location where P b by P
20:47.120 --> 20:52.130
naught is equal to 1. Now, we have P b by
P naught less than 1 there. But it is still
20:52.130 --> 20:58.320
greater than the critical pressure. Now, if
you further reduce it, we reduce back pressure
20:58.320 --> 21:02.700
and now it is equal to the critical pressure.
Critical pressure refers to the pressure,
21:02.700 --> 21:09.419
where we get sonic fluid at the throat. Now,
when P b is equal to P star; that is when
21:09.419 --> 21:16.710
the flow has attain the maximum mass flow
rate that it can pass through, we get choked
21:16.710 --> 21:23.309
flow here. Mach number is equal to 1 and mass
flow rate reaches the maximum value possible.
21:23.309 --> 21:29.799
So, that is why P by P naught; when p P b
is equal to P star that pressure ratio P star
21:29.799 --> 21:37.201
by P naught corresponds to a choked flow.
Any further reduction in P b, if it is P b
21:37.201 --> 21:44.429
less than P star, it will still lead us to
the same flow rate that we are talking about.
21:44.429 --> 21:50.590
Whatever be the flow rate here, we continue
to get the same mass flow that we get at this
21:50.590 --> 21:56.131
particular point. So, it does not change the
pressure characteristic; it is the same. Because
21:56.131 --> 22:01.820
the way the pressure drops in the nozzle will
continue to remain the same, even after we
22:01.820 --> 22:08.630
reduce back pressure any further than the
critical pressure. So, this is how pressure
22:08.630 --> 22:10.980
ratio varies across the nozzle.
22:10.980 --> 22:16.020
Let us look at how mass flow rate varies.
We have discussed about pressure ratio already.
22:16.020 --> 22:21.110
The first chart here is for the mass flow
rate; the second one is the pressure ratio
22:21.110 --> 22:31.059
exit to the inlet stagnation pressure. So,
we has beginning our flow right here, where
22:31.059 --> 22:36.100
mass flow rate is 0 which occurs when P b
is equal to P naught; that is ratio P b by
22:36.100 --> 22:42.149
P naught is equal to 1; that is when we have
the zero mass flow rate. Now, as the ratio
22:42.149 --> 22:47.570
decreases as P b by P naught is reducing,
mass flow rate is increasing. As you can see
22:47.570 --> 22:55.260
here from 1, it goes to 2. When P b is equal
to P star which is the critical pressure,
22:55.260 --> 23:00.740
then we have the mass mass flow rate which
reaches its maximum value, which is m dot
23:00.740 --> 23:04.039
max.
If you look at the corresponding pressure
23:04.039 --> 23:10.440
ratio plot here, we starting from here where
P by P naught is zero is equal to 1 and as
23:10.440 --> 23:15.970
that ratio continuously decreases, we reach
here which is P star by P naught. Any further
23:15.970 --> 23:20.390
reduction does not change the mass flow rate
and it does not change the pressure ratio
23:20.390 --> 23:26.370
as well. p P star by P naught remains the
same; mass flow rate has reached its maximum
23:26.370 --> 23:32.840
value. So, this particular condition of the
flow which we have seen just now discussed
23:32.840 --> 23:38.789
that is at state 3 and beyond is known as
the choked flow. We are saying it is choked
23:38.789 --> 23:44.179
will literally choked means that whatever
we could pass the maximum that one could pass
23:44.179 --> 23:49.220
has reached and that is known as choked. We
cannot take up anything more than that; that
23:49.220 --> 23:52.779
is known as choking.
So, the nozzle has choked means that it has
23:52.779 --> 23:59.169
pass the maximum mass flow that it can. And
any further increase will not change the mass
23:59.169 --> 24:05.200
flow rate that the nozzle can handle; that
is known as the choked flow. So, for a subsonic
24:05.200 --> 24:10.210
nozzle or convergent nozzle, choked flow is
the limit to which a nozzle can operate; beyond
24:10.210 --> 24:14.570
which nothing can happen. If we continued
to, it will infact lead to more losses or
24:14.570 --> 24:20.110
if one tries to increase the mass flow; because
mass flow cannot increase beyond that level.
24:20.110 --> 24:26.559
So, nozzle when operating under choked condition
will have mass flow rate is equal to maximum;
24:26.559 --> 24:32.100
Mach number is equal to unity, sonic Mach
number; pressure will be equal to the critical
24:32.100 --> 24:34.760
pressure, which we have denoted by P star.
24:34.760 --> 24:40.320
So, from the above figure that we have just
discussed now, the back the exit pressure
24:40.320 --> 24:47.320
P e will be equal to the back pressure, for
all back pressures greater than p star. Exit
24:47.320 --> 24:53.610
pressure will be equal to the critical pressure,
for any back pressure which is less than the
24:53.610 --> 24:59.500
critical pressure itself. So, for all back
pressures lower than the critical pressure,
24:59.500 --> 25:04.409
exit pressure will be equal to critical pressure;
Mach number is unity and mass flow rate is
25:04.409 --> 25:08.899
maximum; that is we get a choked flow. Back
pressure lower than the critical pressure
25:08.899 --> 25:13.490
cannot be sensed in the nozzle upstream flow
and it does not affect the mass flow rate.
25:13.490 --> 25:18.380
So, any back pressure which is lower than
the critical will not be sensed by the upstream
25:18.380 --> 25:22.110
flow and therefore, it does not change the
mass flow rate.
25:22.110 --> 25:29.080
So, before we discussed a little more on how
we can analyze the flow, let us look at quickly
25:29.080 --> 25:32.330
look at nozzle efficiency. We have already
derived this in one of the earlier lectures.
25:32.330 --> 25:37.610
But it has been quickly reviewed that nozzle
efficiency we are talking about a process.
25:37.610 --> 25:41.880
This process which is shown by the dotted
line, where the nozzle inlet pressure is been
25:41.880 --> 25:48.870
noted by P o that is P naught stagnation inlet
and exit condition is denoted by P e which
25:48.870 --> 25:53.940
is static pressure. So, this is the process
actual process. The solid line here shows
25:53.940 --> 25:59.429
the isentropic process which will be denoted
by P e s, where s stands for the isentropic
25:59.429 --> 26:03.149
flow.
Now, the corresponding exit stagnation pressure
26:03.149 --> 26:10.299
would be P o e or p P naught is the stagnation
pressure; subscript e is for the exit. Now,
26:10.299 --> 26:16.750
what is to be also kept in mind is that there
is no change in stagnation temperature; because
26:16.750 --> 26:21.370
it is an adiabatic flow. Stagnation temperature
cannot change, unless there is heat transfer.
26:21.370 --> 26:26.820
So, T o T naught i which is stagnation temperature
inlet will be equal to T naught t, which is
26:26.820 --> 26:31.490
stagnation temperature at the throat that
will also be equal to stagnation temperature
26:31.490 --> 26:36.850
at the exit T naught e. So, this is how the
process will look like.
26:36.850 --> 26:44.330
Now, with that in mind, we have defined efficiency
earlier. Nozzle efficiency is h naught i which
26:44.330 --> 26:48.750
is stagnation enthalpy at the inlet minus
h e static enthalpy at the exit divided by
26:48.750 --> 26:55.740
h naught i minus h e s, which is static enthalpy
for an isentropic process; where h e s is
26:55.740 --> 27:01.000
static enthalpy for an isentropic process.
We express this in terms of temperatures.
27:01.000 --> 27:06.590
So, efficiency will be equal to T naught i
minus T e divided by T naught i minus T e
27:06.590 --> 27:14.659
s which is 1 minus T e by T naught e; because
T naught i and T naught t are the same divided
27:14.659 --> 27:21.620
by T e s by T naught i. So, this we will now
express in terms of the Mach number and the
27:21.620 --> 27:23.120
pressure ratios.
27:23.120 --> 27:29.799
So, for a choked flow what happens is that
Mach number is equal to 1. So, this temperature
27:29.799 --> 27:36.929
ratio that you see here T naught e by T e
is 1 plus gamma minus 1 by 2 M e square. And
27:36.929 --> 27:43.179
since M e is equal to 1, when we equate that
equal to 1, we get 1 minus 2 by gamma plus
27:43.179 --> 27:52.139
1 divided by 1 minus T c or critical pressure
divided P naught i rise to gamma minus 1 by
27:52.139 --> 27:58.960
gamma. So, from this if we simplify, we get
the critical pressure ratio or just the pressure
27:58.960 --> 28:06.320
ratio P naught i by P c is equal to 1 by 1
minus 1 by eta n the nozzle efficiency multiplied
28:06.320 --> 28:11.380
by gamma minus 1 by gamma plus 1 rise to gamma
by gamma minus 1.
28:11.380 --> 28:18.139
So, if we see that in this case, when Mach
number is equal to unity, the pressure ratio
28:18.139 --> 28:22.420
the critical pressure ratio is only a function
of the gas and ofcourse the nozzle pressure
28:22.420 --> 28:29.820
ratio. So, if we get this number P naught
i by P c less than the actual pressure ratio
28:29.820 --> 28:36.529
P naught i by P a or the ambient where P a
is the ambient pressure, the nozzle is operating
28:36.529 --> 28:42.830
under choked condition. So, if this pressure
ratio that we have seen with which we can
28:42.830 --> 28:47.900
calculate based on the gas properties and
nozzle efficiency. If that is less than the
28:47.900 --> 28:52.149
pressure ratio corresponding to the ambient
pressure, the nozzle is operating under choked
28:52.149 --> 28:57.080
condition.
And so, given a certain nozzle geometry and
28:57.080 --> 29:03.029
some of its properties, the type of gas which
is normally either air or combustion products.
29:03.029 --> 29:07.360
From the pressure ratio, we can actually find
out whether the nozzle is going to operate
29:07.360 --> 29:12.880
under choked condition or not. So, if that
pressure ratio which we have just now derived
29:12.880 --> 29:18.480
P naught i by P c comes out to be less than
P naught i by P ambient, then it means that
29:18.480 --> 29:27.830
the nozzle is choked. And most of the aircraft
which use subsonic nozzle or convergent nozzles,
29:27.830 --> 29:32.389
nozzle will most of the time atleast under
the design operating condition be operating
29:32.389 --> 29:36.160
under choked condition.
Because that gives as the maximum mass flow
29:36.160 --> 29:42.130
which can which is the nozzle can handle.
And ofcourse, this choking condition can be
29:42.130 --> 29:48.730
changed by changing the area ratio which is
one advantage which adjustable or variable
29:48.730 --> 29:53.980
nozzles have. So, if the nozzle exit area
can be changed, then the choking condition
29:53.980 --> 30:00.090
can be changed. So, the nozzle can be unchoked
and then depending upon the operating condition,
30:00.090 --> 30:05.780
the nozzle area can be changed to get again
a choked flow at the exit. So, that is obviously
30:05.780 --> 30:12.530
one of the advantages of a variable area nozzle.
So, for a convergent nozzle, let me summarize
30:12.530 --> 30:14.230
what we have just now discussed.
30:14.230 --> 30:19.590
So, in a convergent nozzle, if it is operating
under choked condition, the exit Mach number
30:19.590 --> 30:26.870
is unity as we have seen. The exit flow parameters
will then be defined by the critical parameters;
30:26.870 --> 30:30.770
that is the exit pressure will be equal to
critical pressure; exit temperature will be
30:30.770 --> 30:35.570
equal to critical temperature; exit density
is critical density and so on. All exit properties
30:35.570 --> 30:39.820
will be equal to critical parameters. This
also we have discussed when we were talking
30:39.820 --> 30:45.450
about the cycle analysis real cycle analysis
or even ideal cycle, where we have actually
30:45.450 --> 30:50.610
calculate the pressure ratio and then say
just see if the nozzle is choked or not. If
30:50.610 --> 30:56.770
it is indeed choked, then we have to equate
the exit conditions to be choked parameters.
30:56.770 --> 31:01.320
So, if the pressure ratio is greater than
the critical pressure ratio, the nozzle is
31:01.320 --> 31:07.649
set to be choked. In which case, the parameters
at the exit of the nozzle will be equal to
31:07.649 --> 31:14.520
the critical parameters that we have which
we can calculate. Because critical temperature
31:14.520 --> 31:22.159
for example, T e will basically be a function
of gamma; that is it will be 2 by gamma plus
31:22.159 --> 31:27.690
1 in to T naught and so on. Because T naught
by T is equal to 1 plus gamma minus 1 by 2
31:27.690 --> 31:34.889
M square; for critical condition, M is equal
to 1. So, you get T naught by T e is equal
31:34.889 --> 31:39.960
to 2 by gamma plus 1 and so on or gamma plus
1 by 2. So, based on this, we can actually
31:39.960 --> 31:44.820
calculate the exit conditions of the nozzle,
which will only be purely a function of the
31:44.820 --> 31:52.440
upstream stagnation conditions and the ratio
of specific heats that is gamma.
31:52.440 --> 32:00.110
So, having understood operation of a subsonic
nozzle, let us now take up a supersonic nozzle;
32:00.110 --> 32:05.190
that is a convergent-divergent nozzle. Let
us now look at how an isentropic flow behaves
32:05.190 --> 32:10.990
as it passes through a C-D nozzle, a convergent-divergent
nozzle. We will analyze the flow in a very
32:10.990 --> 32:15.080
similar manner as we have done for subsonic
flow; that is we keep changing the exit back
32:15.080 --> 32:19.620
pressure and see how the flow varies in the
nozzle. Ofcourse, in this case the nozzle
32:19.620 --> 32:24.080
the flow is going to be slightly more complicated.
Because of the fact that we are going to deal
32:24.080 --> 32:28.930
with supersonic flows, there will be presence
of shocks. So, some of these aspects we are
32:28.930 --> 32:31.649
going to discuss very soon.
32:31.649 --> 32:38.500
So, in a C-D nozzle, the basic advantage is
that we can achieve supersonic Mach number,
32:38.500 --> 32:44.340
which can be achieved in the divergent section.
Well but divergent section alone cannot guarantee
32:44.340 --> 32:49.900
a supersonic flow. We need to ensure certain
back pressure. So, that is what we are going
32:49.900 --> 32:51.240
to discuss now.
32:51.240 --> 32:57.640
So, for that let us again take up a very simple
C-D nozzle. It is a linear C-D nozzle just
32:57.640 --> 33:02.419
for the sake of simplicity. We have a convergent
section as you can see here, which is very
33:02.419 --> 33:08.340
similar to the convergent nozzle we have seen
in few slides earlier. Followed by this, this
33:08.340 --> 33:13.690
throat this is a throat section, we have the
divergent section; convergent nozzle ending
33:13.690 --> 33:19.000
in a throat followed by the divergent section.
This constitutes a typical convergent-divergent
33:19.000 --> 33:23.190
nozzle. Upstream pressure that stagnation
pressure, this reservoir stagnation pressure
33:23.190 --> 33:30.590
is P naught; exit pressure static pressure
is P e; we have a back pressure P b. So, let
33:30.590 --> 33:34.840
us now look at how the flow would vary as
we keep changing the back pressure.
33:34.840 --> 33:41.909
Initially, when P a is equal to P b which
is the equal to the stagnation pressure, there
33:41.909 --> 33:49.090
is no flow taking place. As we decrease the
back pressure, we have now new pressure which
33:49.090 --> 33:55.870
is P b which is less than P naught; but it
is still greater than the critical pressure.
33:55.870 --> 34:01.309
In a convergent section, the flow will accelerate
which means there is a drop in static pressure
34:01.309 --> 34:07.580
from inlet to the throat. From the throat
till the exit, it is a divergent section in
34:07.580 --> 34:13.080
a subsonic flow. So, there has to be diffusion
or deceleration leading to increase in static
34:13.080 --> 34:17.500
pressure. So, static pressure increases; but
it reaches a value which is lower than the
34:17.500 --> 34:24.790
stagnation pressure upstream. We further decrease
it; we reach a point where which is basically
34:24.790 --> 34:31.470
corresponding to the choked condition.
We still have subsonic flow; because when
34:31.470 --> 34:36.550
P b is equal to P star which is the critical
pressure at the throat, we get sonic flow
34:36.550 --> 34:42.340
at the throat and that is indicated by P star.
So, you can see that for all these three cases,
34:42.340 --> 34:47.690
we still have a subsonic flow at the nozzle
exit. Now our objective here is to get supersonic
34:47.690 --> 34:55.179
flow which means that we have to now further
decrease the back pressure. Now, what happens
34:55.179 --> 35:00.600
is for a back pressure which is lower than
the critical pressure, there are different
35:00.600 --> 35:05.069
conditions which can exits. That is, adds
the back pressure is lowered to value which
35:05.069 --> 35:09.900
is lower than the critical pressure, which
means that the flow is no longer… well it
35:09.900 --> 35:13.730
becomes sonic at the throat.
But since the back pressure is lower than
35:13.730 --> 35:19.339
critical pressure and we have a divergent
section here which will act like a nozzle
35:19.339 --> 35:24.750
in a supersonic flow. So, we have a divergent
section. So, the flow becomes supersonic which
35:24.750 --> 35:31.010
is why you can see the static pressure continues
to drop; because the flow is accelerating
35:31.010 --> 35:37.250
and that happens in a supersonic flow. So,
after the throat, it become flow becomes supersonic
35:37.250 --> 35:42.790
and under certain conditions if the back pressure
is not low enough, you might have a normal
35:42.790 --> 35:49.180
shock within the divergent section of the
nozzle itself. So, there could be a normal
35:49.180 --> 35:53.480
shock here, which means that downstream of
the normal shock the flow is subsonic.
35:53.480 --> 35:59.619
So, after the normal shock, we have a subsonic
flow which means that a divergent section
35:59.619 --> 36:04.920
will now behave like subsonic flow diffuser.
So, you can see there is again an increase
36:04.920 --> 36:13.530
of static pressure. So, this occurs till a
point where the back pressure is low enough
36:13.530 --> 36:17.720
to ensure that there is a supersonic flow.
That is, they have to continuously decrease
36:17.720 --> 36:23.380
even we are now at the back pressure which
is equal to P d. We have to decrease further
36:23.380 --> 36:28.869
even if it is P e, we continue to have subsonic
flow at the nozzle exit, which is not the
36:28.869 --> 36:36.190
objective. So, the back pressure has now is
now supposed to be lower than all these pressures.
36:36.190 --> 36:41.170
So, if back pressure is further reduced, we
now we can ensure that the normal shock that
36:41.170 --> 36:46.490
was existing here can be pushed out of the
nozzle. So, that we get a supersonic flow
36:46.490 --> 36:51.720
at the nozzle exit, which is basically our
objective that we need a supersonic flow at
36:51.720 --> 36:56.960
the nozzle exit. So, for back pressures which
are not as low as this; but it is lower than
36:56.960 --> 37:02.820
critical pressure, we might under having a
shock within the divergent section of the
37:02.820 --> 37:10.109
nozzle which is obviously not a good thing.
So, if you look at the Mach number plot for
37:10.109 --> 37:17.380
these conditions, we had the first case. That
is for subsonic flow, we have Mach number
37:17.380 --> 37:19.340
increasing from the inlet all the way to the
throat.
37:19.340 --> 37:25.080
But it is not Mach number equal to 1 at the
throat; after which, again the Mach number
37:25.080 --> 37:29.640
decreases. For critical pressure, when we
have sonic flow at the throat, Mach number
37:29.640 --> 37:36.890
increases which is Mach 1 and then again it
decreases or decelerates; because it is subsonic.
37:36.890 --> 37:41.930
For back pressure is equal to P d, we have
Mach number equal to 1 at the throat. It continues
37:41.930 --> 37:46.430
to increase, but only up to the shock; after
which, it becomes subsonic and it decreases
37:46.430 --> 37:53.290
below 1. For Mach numbers for continuous supersonic
Mach number all the way up up to the nozzle
37:53.290 --> 37:59.180
exit, we need to ensure that the back pressures
are low enough; So that, there is no shock
37:59.180 --> 38:06.770
within the divergent section of the nozzle.
So, this is a basic operation of C-D nozzle;
38:06.770 --> 38:14.410
a convergent-divergent nozzle, where we can
achieve supersonic flow under certain operating
38:14.410 --> 38:18.890
conditions that is depending upon the nozzle
pressure ratio. One can ensure that they can
38:18.890 --> 38:25.550
achieve supersonic flow in a C-D nozzle. So,
simply fitting a divergent section at the
38:25.550 --> 38:31.010
throat of a convergent nozzle will not suffice.
We need to ensure that there is sufficient
38:31.010 --> 38:36.560
pressure ratio, which can lead to supersonic
flow within the nozzle. So, with this background
38:36.560 --> 38:40.920
in mind, we shall now derive some expression
as we get for convergent nozzle also to evaluate
38:40.920 --> 38:45.470
the performance or some of the parameters
like mass flow rate through a supersonic nozzle
38:45.470 --> 38:46.470
and so on.
38:46.470 --> 38:50.589
So, what we will do is we are going to assume
as we have done for convergent nozzle that
38:50.589 --> 38:54.980
the flow is adiabatic because heat transfer
per unit mass is much smaller than the difference
38:54.980 --> 39:00.951
in enthalpy between inlet and outlet. The
flow can be assumed to be isentropic up to
39:00.951 --> 39:05.960
the throat. But from the throat to the exit,
it may not really the isentropic; because
39:05.960 --> 39:10.630
there could be possible shocks especially
under certain operating conditions.
39:10.630 --> 39:17.750
So, nozzle efficiency we have already derived
for the subsonic nozzle which is h naught
39:17.750 --> 39:23.640
i the inlet stagnation enthalpy minus h e
divided by h naught i minus h e s, which we
39:23.640 --> 39:28.410
can expressed in terms of temperature ratio
is 1 minus T e by T naught e divided by 1
39:28.410 --> 39:33.230
minus T e s by T naught i. This will now be
expressed in terms of the pressure ratio.
39:33.230 --> 39:39.960
So, we get 1 minus P e by P naught e raise
to gamma minus 1 by gamma divided by 1 minus
39:39.960 --> 39:46.240
P e by P naught i raise to gamma minus 1 by
gamma. So, here we will now express P e by
39:46.240 --> 39:50.500
P naught e in terms of the nozzle efficiency
and the other pressure ratio.
39:50.500 --> 39:55.890
So, the exit pressure to the exit stagnation
pressure ratio is equal to 1 minus eta n which
39:55.890 --> 40:01.940
is nozzle efficiency in to 1 minus P e by
P naught i raise to gamma minus 1 by gamma,
40:01.940 --> 40:07.920
the whole thing raise to gamma by gamma minus
1. Now, we can now express P naught i by P
40:07.920 --> 40:14.380
naught e as P e by P naught e multiplied by
P naught i by P e. So, from this, we get P
40:14.380 --> 40:21.640
naught i by P e P naught e is equal to P 0
i which is string stagnation pressure divided
40:21.640 --> 40:30.079
by P e in to this term; that is, 1 minus eta
n in to 1 minus P e by P naught i raise to
40:30.079 --> 40:34.640
gamma minus 1 by gamma rise to gamma by gamma
minus 1.
40:34.640 --> 40:42.000
So, the exit velocity; So, which we have also
derived for the cycle analysis is u e which
40:42.000 --> 40:47.200
is square root of 2 in to h naught i minus
h e, which in terms of efficiency becomes
40:47.200 --> 40:53.270
2 in to eta n in to h naught i minus h e s
and this again in terms of temperature is
40:53.270 --> 41:01.260
2 c p eta n in to T naught i minus T e s.
This is on simplification 2 gamma R by gamma
41:01.260 --> 41:08.220
minus 1 in to eta n in to T 0 i multiplied
by 1 minus P e by P 0 i raise to gamma minus
41:08.220 --> 41:09.530
1 by gamma.
41:09.530 --> 41:15.430
So, this combine with the previous equation
that we have seen we will now express that
41:15.430 --> 41:20.760
in terms of the Mach number. So, the exit
Mach number as we know is M e square of that
41:20.760 --> 41:27.950
is u e square by a e square that is u e square
by gamma R T e. Now, we know that T e by T
41:27.950 --> 41:36.079
naught e by T e is 1 plus gamma minus 1 by
2 M e square which is T naught i by T e. Therefore,
41:36.079 --> 41:42.250
we can express M e from what we have seen
in the previous equation that is u e square
41:42.250 --> 41:49.240
and this equation we get 2 in to eta n by
gamma minus 1 multiplied by 1 minus gamma
41:49.240 --> 41:56.920
minus 1 by 2 M i square where M i is the inlet
Mach number multiplied by 1 minus P i by P
41:56.920 --> 42:02.970
naught i raise to gamma minus 1 by gamma.
This ofcourse can be simplified and written
42:02.970 --> 42:09.460
here in terms of the pressure ratio 2 by gamma
minus 1 in to the nozzle efficiency 1 minus
42:09.460 --> 42:16.359
P e by P naught i raise to gamma minus 1 by
gamma divided by 1 minus eta n in to 1 minus
42:16.359 --> 42:21.550
P e by P naught i raise to gamma minus 1 by
gamma. So, here we have expressed the Mach
42:21.550 --> 42:27.960
number in terms of the pressure ratio that
is the exit pressure to the inlet pressure.
42:27.960 --> 42:34.740
And so, this will be useful in determining
whether we get a sonic Mach number or supersonic
42:34.740 --> 42:39.410
Mach number especially a supersonic nozzle
one would want to have supersonic Mach number
42:39.410 --> 42:45.210
at the exit. So, we can find out what is the
pressure ratio that will be required, if we
42:45.210 --> 42:49.421
have to achieve a supersonic Mach number.
Let us say we want to achieve a Mach number
42:49.421 --> 42:57.000
of 1.2. So, given the exit Mach number is
1.2, we can now find out what will be the
42:57.000 --> 43:02.880
pressure ratio that the nozzle will have to
develop, if one one has to achieve the supersonic
43:02.880 --> 43:08.310
Mach number with the certain nozzle efficiency.
So, nozzle efficiency in mind, one can actually
43:08.310 --> 43:13.010
find out what will be the kind of pressure
ratio that is needed to achieve that Mach
43:13.010 --> 43:18.619
number. So, the second part that we going
to look at now is to see how we can also act
43:18.619 --> 43:24.789
derive expression for the area ratio. We now
just derived the equation for the Mach number.
43:24.789 --> 43:28.190
We will now look at how we can derive an expression
for the area ratio.
43:28.190 --> 43:33.330
So, we have already discussed about the mass
flow rate of the governing equation little
43:33.330 --> 43:39.530
earlier. We will assume again isentropic flow
up the up to the throat. So, we can express
43:39.530 --> 43:46.530
the throat area and the exit area as A t by
A e which is throat area by exit area expressed
43:46.530 --> 43:51.830
in terms of the corresponding pressures and
Mach number P naught e by P naught t which
43:51.830 --> 43:58.440
is the stagnation pressure at the throat multiplied
by M e by M t Mach number at the throat in
43:58.440 --> 44:04.329
to 1 plus gamma minus 1 by 2 M t square divided
by 1 plus gamma minus 1 by 2 M e square raise
44:04.329 --> 44:10.740
to gamma plus 1 in to divided by 2 in to gamma
minus 1. This also can be expressed in terms
44:10.740 --> 44:15.470
of the inlet pressure.
Now, if the throat is choked where we have
44:15.470 --> 44:21.900
Mach number M t as equal to 1, the throat
area becomes A star, which is critical area.
44:21.900 --> 44:29.819
A star by A e would now be equal to P naught
e which is exit stagnation pressure multiplied
44:29.819 --> 44:37.020
by exit Mach number divided by inlet stagnation
pressure, which leads to the choking multiplied
44:37.020 --> 44:43.569
by gamma plus 1 by 2 by 1 minus gamma minus
1 by 2 M e square the whole raise to gamma
44:43.569 --> 44:49.870
plus 1 by 2 in to gamma minus 1. So, from
this, we can actually find out what is the
44:49.870 --> 44:56.270
critical area ratio. Area ratio between the
throat and the exit of the nozzle, which can
44:56.270 --> 45:01.359
lead to critical flow that is choked flow
at the exit of the nozzle.
45:01.359 --> 45:06.580
So, this is one of the expressions that one
can use. We have also seen how we can calculate
45:06.580 --> 45:13.160
the pressure ratio that is required to develop
a certain supersonic Mach number at the exit
45:13.160 --> 45:17.140
of the convergent nozzle. We will also look
at what is the mass flow rate that one can
45:17.140 --> 45:21.569
achieve. Especially, if it is under choked
condition, what how we can calculate the mass
45:21.569 --> 45:26.050
flow rate? Because you have already derived
general equation for mass flow rate in terms
45:26.050 --> 45:33.070
of the inlet stagnation pressure temperature,
the Mach number and the area. And we now know
45:33.070 --> 45:39.099
that for chocking, the Mach number at the
throat becomes 1. So, from that, we can actually
45:39.099 --> 45:43.940
calculate the mass flow rate in terms of the
throat area and the inlet parameters.
45:43.940 --> 45:49.859
So, mass flow rate we can calculate using
the equation we have discussed earlier, where
45:49.859 --> 45:55.099
we substitute M is equal to 1. We now get
A star which is the throat area multiplied
45:55.099 --> 46:01.780
by P naught i inlet stagnation pressure divided
by T naught i square root; the gas properties
46:01.780 --> 46:08.190
gamma by R square root 1 by 1 plus gamma plus
1 by 2 raise to gamma plus 1 by 2 in to gamma
46:08.190 --> 46:12.110
minus 1. So, here we see mass flow rate is
function of the inlet stagnation pressure
46:12.110 --> 46:20.210
temperature and ofcourse the throat area.
Now, another aspect concerning the designer
46:20.210 --> 46:27.230
is that he would like to always try to keep
the exit area the area ratio that is A i by
46:27.230 --> 46:33.900
A e as close as possible to unity.
Because if the exit area is very large, then
46:33.900 --> 46:40.400
it has other implications like it might lead
to high levels of external drag; because obviously
46:40.400 --> 46:46.579
the entire nozzle and the engine is housed
in casing in a nasal. So, if the nozzle area
46:46.579 --> 46:51.869
is increased very large; if we have an exit
area which is much larger than the inlet area,
46:51.869 --> 46:56.950
then the possibilities of external drag increasing
would be very high. But if we does not do
46:56.950 --> 47:02.490
that; if we tries to keep the area ratio as
close as possible to unity, then the nozzle
47:02.490 --> 47:07.250
may not be operating under what is known as
the fully expanded condition.
47:07.250 --> 47:11.870
That is fully expanded condition is one, where
the exit pressure is equal to the ambient
47:11.870 --> 47:18.190
pressure; P e is equal to P a. If P e is not
equal to P a, then the nozzle could be either
47:18.190 --> 47:25.069
be operating as an under expanded nozzle or
a over expanded nozzle. Both of which has
47:25.069 --> 47:29.140
certain issues which we are going to discuss
now. That is, if a nozzle is under expanded,
47:29.140 --> 47:32.800
what are the problems; if it is over expanded,
what are the problems. So, will take a look
47:32.800 --> 47:38.650
at how we can how a designer can keep this
in mind in under these conditions. So, an
47:38.650 --> 47:45.030
incomplete expansion is a result of a difference
between the nozzle exit pressure and the ambient
47:45.030 --> 47:47.380
pressure.
47:47.380 --> 47:51.800
Incomplete expansion could be either over
expanded nozzle, where the exit pressure is
47:51.800 --> 47:57.530
greater than the ambient pressure. If P e
is greater than P a, it means that the flow
47:57.530 --> 48:03.700
is capable of additional expansion. That is,
the flow has potential for expanding further
48:03.700 --> 48:08.319
and this is now carried out using expansion
waves, which will originate from the lip of
48:08.319 --> 48:14.150
the nozzle. That is, because the flow is having
enough energy to further expand and that is
48:14.150 --> 48:18.790
why, it will carry out that expansion; because
the pressure has to eventually become equal
48:18.790 --> 48:22.630
to the ambient pressure.
So, that expansion takes place through nozzles;
48:22.630 --> 48:28.829
through the formation of expansion waves.
The other extreme case would be over expanded
48:28.829 --> 48:34.589
nozzle; that is if P e is less than P a. That
is, the nozzle is expanding more than what
48:34.589 --> 48:40.770
it should and how can that be taken in to
account. Well that can happen if there are
48:40.770 --> 48:46.069
I mean in in the instance where the exit pressure
is greater is less than ambient pressure,
48:46.069 --> 48:51.079
there would be shock waves which would again
be originating from the nozzle. So, across
48:51.079 --> 48:55.210
the shock wave, the pressure would change
increase the static pressure would increase.
48:55.210 --> 48:59.880
So, that you have exit pressure finally equal
to the ambient pressure.
48:59.880 --> 49:04.750
Well, the ideal case one would like to have
ofcourse is fully expanded nozzle, where the
49:04.750 --> 49:08.580
exit pressure is exactly equal to ambient
pressure. So, there are no shock waves or
49:08.580 --> 49:14.700
expansion waves here. Now, if the exit pressure
is much less than the ambient pressure, then
49:14.700 --> 49:18.280
the shock waves will actually occur within
the divergent section of the nozzle. We have
49:18.280 --> 49:24.230
already seen that when we discussing about
the nozzle performance and behavior that if
49:24.230 --> 49:28.819
P e much less than P a, shock waves may actually
occur within the divergent section of the
49:28.819 --> 49:29.890
nozzle.
49:29.890 --> 49:35.010
So, these are the illustrations of all these
four cases; fully expanded nozzle P e is equal
49:35.010 --> 49:43.630
to P a. So, there are no issues within the
nozzle itself. If P e is less than P a which
49:43.630 --> 49:49.190
means that it is under expanded, so there
would be expansion waves as you can see here
49:49.190 --> 49:53.750
which originate from the lip of the nozzle.
These are the expansion waves through which
49:53.750 --> 50:00.680
the flow will further expand and become equal
to the ambient pressure. Now, if P e is greater
50:00.680 --> 50:08.231
than P a that means it has it is actually
expanded more than what it should be, then
50:08.231 --> 50:13.359
we have an over expanded nozzle. In which
case, the pressure gets finally equalized
50:13.359 --> 50:17.150
through what are known as oblique shocks.
Because now the pressure has to eventually
50:17.150 --> 50:21.660
become equal to the ambient pressure and that
happens through what are known as oblique
50:21.660 --> 50:23.670
shocks.
You may have oblique shocks originating from
50:23.670 --> 50:29.740
the lip of the nozzle and the pressure gets
finally equalized there. If P e is much less
50:29.740 --> 50:36.140
than P a, then we have an occurrence of an
normal shock or which which could occur within
50:36.140 --> 50:42.359
the nozzle itself in the divergent section
and the pressure gets across the normal shock,
50:42.359 --> 50:47.230
we know there is an increase in static pressure.
So, across the normal shock, the exit pressure
50:47.230 --> 50:52.119
finally becomes equal to that of the ambient
pressure. But that is through a normal shock,
50:52.119 --> 50:58.651
which obviously means there are going to be
loses across such nozzle geometry. So, these
50:58.651 --> 51:07.049
are different operating conditions of supersonic
nozzle where the depending upon the exit pressure
51:07.049 --> 51:10.619
in relation to the ambient pressure, the nozzle
may operate under different conditions.
51:10.619 --> 51:16.670
So, let me now quickly summarize what we have
discussed in this lecture. We started our
51:16.670 --> 51:23.069
lecture today with discussion on subsonic
nozzle convergent nozzle. And we have seen
51:23.069 --> 51:29.690
that there are certain problems with subsonic
nozzle in the sense that there is a limitation
51:29.690 --> 51:34.380
to the maximum Mach number, which a subsonic
nozzle can deliver and that sonic Mach number
51:34.380 --> 51:39.280
that occurs right at the throat. So, we have
seen how we can as we change the back pressure,
51:39.280 --> 51:44.760
how the flow behaves through a subsonic nozzle.
We also derived an equation for analyzing
51:44.760 --> 51:50.790
this. We then took up the convergent divergent
nozzle. We have discussed in detail about
51:50.790 --> 51:57.230
how the flow flow behavior can analyze right
from the inlet to the exit as the back pressure
51:57.230 --> 52:01.310
keeps changing.
And then we have also derived equations for
52:01.310 --> 52:06.460
calculating the pressure ratio pressure ratio
across the nozzle required for getting a desired
52:06.460 --> 52:10.950
supersonic Mach number at the exit of the
nozzle, what will be the mass flow rate corresponding
52:10.950 --> 52:17.410
to choked flow and so on. And towards the
end, we have discussed about different operating
52:17.410 --> 52:23.450
conditions of supersonic nozzle; where depending
upon the exit pressure in relation to the
52:23.450 --> 52:30.100
ambient pressure, the nozzle may undergo either
over expansion or under expansion. Or if the
52:30.100 --> 52:34.950
exit pressure is much lower than the ambient,
one might also have normal shock within the
52:34.950 --> 52:40.079
divergent section of the nozzle. So, these
are some of the topics that we have discussed
52:40.079 --> 52:42.460
in today’s lecture.
52:42.460 --> 52:49.170
And in the next lecture that we going to discuss,
we will basically be having a tutorial wherein
52:49.170 --> 52:56.700
we will solve some problems on intakes as
well as nozzles. So, we are we have already
52:56.700 --> 53:03.850
discussed about intakes. Today’s lecture,
we conclude our discussion on nozzles and
53:03.850 --> 53:09.510
so in the next class, we will take up some
simple problems on intakes and nozzles and
53:09.510 --> 53:13.890
we will solve them in the class. I will also
have a few exercise problems for you which
53:13.890 --> 53:18.839
you can solve based on our discussion during
the last few lectures as well as our discussion
53:18.839 --> 53:23.470
during the tutorial. So, we will take up some
of these topics for discussion in the next
53:23.470 --> 53:23.619
class.