WEBVTT
Kind: captions
Language: en
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Gas Dynamics
Prof. T.M. Muruganandam
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Department of Aerospace Engineering
Indian Institute of Technology, Madras
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Module - 14
Lecture - 27
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Quasi-1D flow with area variations, Choked
conditions
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Hello everyone, welcome back. I wanted to
move into area change during flow today,
00:00:18.770 --> 00:00:24.800
but just before that I will finish off with
something in expansions fan, just we will
00:00:24.800 --> 00:00:29.810
go
over the same piston analogy video again,
00:00:29.810 --> 00:00:32.860
but we would not do the video; we will just
look at it at the end.
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.
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If you go to the screen we have seen this
piston analogy already, I have definitely
00:00:38.450 --> 00:00:42.320
shown
this to you some four or five classes earlier.
00:00:42.320 --> 00:00:46.590
What I wanted to show now is this
streamline drawn here. We said that it is
00:00:46.590 --> 00:00:49.110
tracking of a particular particle especially
this
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one and this one along as a function of time,
and that becomes streamline if you look at
00:00:55.630 --> 00:01:02.560
in 2D flows, okay. Now what I want you to
see is they are starting very very close to
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each other; that is, the gap between them
is very small.
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If you go back to understanding fluid mechanics
the stream tubes, can be considered as
00:01:12.850 --> 00:01:18.861
the streamlines in this case in 2D flow, right.
The stream lines will perform the stream
00:01:18.861 --> 00:01:25.280
tube in 2D flow. The stream tube is really
small in area and a cross section here, and
00:01:25.280 --> 00:01:31.080
.after the expansion they have become really
wide. Basically telling that the flow has
00:01:31.080 --> 00:01:36.259
expanded; that is one way of looking at it.
Another way of looking at it is during the
00:01:36.259 --> 00:01:43.639
expansion fan, the stream tube area is changing,
okay, and the steam tube area is
00:01:43.639 --> 00:01:49.350
expanding if you look at it. Inside the expansion
fan the stream tube area is increasing
00:01:49.350 --> 00:01:55.719
with along the flow direction.
What this means is as it is increasing I can
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now say that if my flow is supersonic, you
remember this from long back we said that
00:02:00.649 --> 00:02:05.240
if there is area change; if the flow is incoming
and supersonic and the area is increasing,
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we are going to have higher Mach number.
That can also be seen from inside here if
00:02:11.650 --> 00:02:16.030
we think about the stream lines as stream
tube
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walls; that is the extra information I wanted
to give here in this particular expansion
00:02:20.670 --> 00:02:23.700
fan
example. We saw that similar thing is happening
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in compression wave the oblige shock
also, but I cannot show you what is happening
00:02:28.540 --> 00:02:30.830
inside a shock there. Here it is big region
I
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can show it you very nicely.
In the other case the area between the stream
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lines will be very large before and after
the
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oblige shock the stream line stunt they will
be much closer to each other. I cannot show
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you anything more than that in here I can
show that; in here I can show that the stream
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tube is expanding during the expansion fan,
after that it is staying constant cross section
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between them, okay. This is one final thing
I wanted to show before we jump into area
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variation. Since, I already discussed what
happens to area variation in supersonic flows,
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I
can now give this information already. This
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completes ours discussion on expansion fans
and shock expansion theory, all that together.
00:03:14.240 --> 00:03:20.480
Now we will move on to flow through ducts
or flow through some channels with area
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variation, typically compressible flows. So,
Mach numbers are going to be high enough;
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that is the condition we are going to work
with, okay. We will move on to a new
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problem. We will start with flow with area
variation. I would not go on derive all the
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equations again; we already derived most of
it. I in fact listed it out in your notes
00:03:39.850 --> 00:03:45.070
somewhere where it is all nicely written in
one page. So, just you just go back to that
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page and look at it whenever needed.
00:03:48.110 --> 00:03:49.110
..
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So, what we want to do now will be consider
some duct flow through a duct with some
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area variation. I am picking something like
a convergent divergent nozzle; flow through
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this is what we are going to study. And of
course, I have to revisit all my assumptions
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as
before. I am going to say that I am going
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to use quasi 1D assumption, where we say that
the flow is going to be going only along a
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particular direction, and I am not worried
about the perpendicular component velocity.
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Ideally, if I think about the real flow it
is
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going to flow along the wall; it cannot flow
any other direction. I am going to assume
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that we will not worry about those; we will
just say the flow is just going straight.
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It is going straight like this everywhere,
and when there is no space they will readjust
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to
go something like this, and in here it is
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going to go something like this. And when
I am
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out there, it is again going to go like this.
This is our assumption, quasi 1D assumption.
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We are going to use that so that our derivations
will become a little simpler, okay. When
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I say quasi 1D I am just going to say it as
stream normal direction velocity components
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are neglected; stream wise components only.
It is not really components; I am going to
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say all the velocity is along stream wise
direction which is for me in this direction
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for
now.
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I would not worry about area variation; one
side is more than the other side and all that.
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We will just assume. For me locally I am going
to consider some perpendicular line and
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flow is always perpendicular to it; that is
a simple way of thinking about it, easy to
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.convince myself that way; that is easier
to convince, okay. And of course, I am going
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to
also neglect some things. I am going to say
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heat transfer in or out of this flow is 0,
no
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heat transfer adiabatic I am going to assume
currently. And I am also going to say that
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there are no friction effects. We have been
assuming this all this time all of these
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assumptions till now. We want to start removing
them one by one after sometime.
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Eventually, we will consider a case where
these assumptions are not really valid; that
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is
towards the end.
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Now if I think about mass equation, if I look
at the integral form, the way I am going to
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start with this is I am going to say pick
two sections, section one, section two; it
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could be
any section. It could have been here and here
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or here and here whatever section, just
some two sections one and two. This is going
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to write the integral form. This is
something you should remember from before
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rho U A is a constant; that is the mass flow
rate through the duct momentum equation. We
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said that this will have a special term. It
is
00:07:18.510 --> 00:07:40.010
going to look some, I will write it in a nicer
fashion than this. It is going to be something
00:07:40.010 --> 00:07:44.980
like this, and we said that if it is a quasi
linear term then I can neglect this, currently
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I
cannot; it is no more a constant area duct.
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We said its area may be changing. Initially
we said constant area duct I can say P plus
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rho U square is a constant. Now I cannot do
that anymore. I had to be more careful about
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which momentum equation I am using. This is
the correct one to use energy. This is
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going to be similar, because we said heat
transfer is not present. So, is going to be
00:08:11.889 --> 00:08:17.669
h
naught equal to constant, but writing in an
00:08:17.669 --> 00:08:21.970
expanded form it is going to be h 1 plus u
1
00:08:21.970 --> 00:08:31.900
square by 2 is equal to s 2 plus u 2 square
by 2; that is going to be this form. We will
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also
assume that the flow is isentropic, extra
00:08:36.930 --> 00:08:48.110
assumption, okay. Now if I want to solve this
flow problem from one section to another section
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and I want to see how flow properties
are changing.
00:08:52.450 --> 00:08:56.010
Now I am in trouble, because I have to do
this integral every time if I am using this
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set of
equations. It is not easy to use; of course,
00:09:00.350 --> 00:09:04.340
what is an equation when I say isentropic;
what
00:09:04.340 --> 00:09:16.890
is that? P equal to? Okay, we can write it
in several forms. Let us say I will use this
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form,
you guys hear this. This is one of the forms;
00:09:24.550 --> 00:09:31.040
you can use this form also, any of these
forms, okay. So, you can think about using
00:09:31.040 --> 00:09:34.710
any of these forms, but it is more difficult
to
00:09:34.710 --> 00:09:39.560
solve this problem from the integral form
of these equations, but these are the correct
00:09:39.560 --> 00:09:44.170
equations to use if you want to solve the
problem. Instead of this now we will start
00:09:44.170 --> 00:09:45.170
using
00:09:45.170 --> 00:09:50.410
.the differential form which is a little bit
easier, because I am gone to look at any section
00:09:50.410 --> 00:09:55.880
what happens just immediately next to it;
a very close line just next to it. So, the
00:09:55.880 --> 00:09:59.320
change
will be extremely small, very small d A which
00:09:59.320 --> 00:10:02.970
is what we will start looking at next, okay.
.
00:10:02.970 --> 00:10:16.390
And I am going to go back and use whatever
expressions we already derived. This was
00:10:16.390 --> 00:10:29.310
our mass equation in differential form, and
we had
00:10:29.310 --> 00:10:42.440
this and this mass momentum and
energy equations. We had different forms.
00:10:42.440 --> 00:10:50.660
And I wanted to write the isentropic relation
in this form; how will I write that? What
00:10:50.660 --> 00:11:03.230
will this be? d p by p, anybody directly
remembers it? If you do not remember this,
00:11:03.230 --> 00:11:08.010
the easy way to derive it is d p by d rho
is
00:11:08.010 --> 00:11:17.130
what? For isentropic a square which is gamma
RT. We will keep this form which is very
00:11:17.130 --> 00:11:26.730
nice form. Now I want d p by p. So, what do
I do? d P is d rho times gamma RT. I will
00:11:26.730 --> 00:11:35.300
keep it that way, okay. So, it will be 0 times
gamma RT divided by P.
00:11:35.300 --> 00:11:43.330
Now P can be rewritten as rho RT. I am using
ideal gas law already inside. Now RT will
00:11:43.330 --> 00:11:51.430
get cancelled. I am going to have a form gamma
d rho by rho. d p by p is gamma d rho
00:11:51.430 --> 00:11:58.801
by rho; that is the form we will use here,
okay. So, we have set of variables; what are
00:11:58.801 --> 00:12:04.480
all
the variables we have? Rho, u, p and t; these
00:12:04.480 --> 00:12:09.500
are our variables. Of course, a is also
changing, but that is the condition we are
00:12:09.500 --> 00:12:14.940
supplying to the flow; that is not a flow
variable really. As a condition we are giving
00:12:14.940 --> 00:12:19.230
to the flow. So, these are the equations,
four
00:12:19.230 --> 00:12:25.180
equations and four unknowns can be solved.
We just have to manipulate this to get to
00:12:25.180 --> 00:12:28.630
a
point where we can use it better.
00:12:28.630 --> 00:12:35.160
.And of course, I will just leave it to you
to think about it a little bit. I will just
00:12:35.160 --> 00:12:38.320
jump one
step, but we have derived something like this
00:12:38.320 --> 00:12:44.180
already in long time back; quasi 1D when I
introduce I derived some of this. So, I will
00:12:44.180 --> 00:12:51.709
just go and write expressions from there
directly. If I look at that then I will start
00:12:51.709 --> 00:13:05.461
with this was one of them. And of course,
this
00:13:05.461 --> 00:13:11.100
you cannot get without d rho by rho actually.
I should have given probably d rho by rho
00:13:11.100 --> 00:13:24.290
first. This M square in this d A by A expression
comes actually from d rho by rho in
00:13:24.290 --> 00:13:28.010
here.
That is why you go to that point, and this
00:13:28.010 --> 00:13:33.200
d A by A is coming from the original mass
expression. You can just link all of them
00:13:33.200 --> 00:13:35.920
together like this, okay; you can derive this.
We
00:13:35.920 --> 00:13:41.300
have derived this already; I do not want to
derive it again. We will just use this, and
00:13:41.300 --> 00:13:43.640
of
course, once I know d rho by rho I can use
00:13:43.640 --> 00:13:46.839
this isentropic relation, and I get d P by
p
00:13:46.839 --> 00:13:51.570
which will just be gamma multiplying this,
very simple, okay.
00:13:51.570 --> 00:13:52.570
.
00:13:52.570 --> 00:13:58.170
Now better expression we want to get will
be d T by T from energy equation which of
00:13:58.170 --> 00:14:04.370
course will come out to be, I will leave that
as an exercise for you to get to. It is not
00:14:04.370 --> 00:14:19.510
very
difficult to get, okay. We will have it in
00:14:19.510 --> 00:14:25.970
this particular form, and there is only one
more
00:14:25.970 --> 00:14:48.420
left which is probably the most used for analysis.
How does Mach number vary when I
00:14:48.420 --> 00:14:57.350
change area? That is what this gives. And
we have already used this in telling if I
00:14:57.350 --> 00:15:01.060
increase area what happens, if I decrease
area what happens and all that expression
00:15:01.060 --> 00:15:03.540
we
have already tried before.
00:15:03.540 --> 00:15:08.041
.We will anyway go over this once more because
this is directly applicable to flow
00:15:08.041 --> 00:15:14.060
through ducts with changing areas. So, if
I look at that again I am going to draw that
00:15:14.060 --> 00:15:20.880
table the way we have already done, M less
than 1, M greater than 1, and I am going to
00:15:20.880 --> 00:15:32.820
say d A greater than 0 and d A less than 0.
I believe you have already drawn this plot;
00:15:32.820 --> 00:15:37.220
I
have drawn this table sometime back definitely.
00:15:37.220 --> 00:15:44.019
Now we just want to look at those four
variables we just listed, density, velocity,
00:15:44.019 --> 00:15:50.090
pressure, temperature; what happens to all
of
00:15:50.090 --> 00:16:00.470
them? So, let us speak a case; it is just
to remind you what is suppose to happen. Oh,
00:16:00.470 --> 00:16:03.470
I
will also put M so that it is easier.
00:16:03.470 --> 00:16:12.541
I will keep M on top of all of them; of course,
we have expressions for all of them. We
00:16:12.541 --> 00:16:21.300
just wrote expression for every one of them.
We will just look at this. M square always
00:16:21.300 --> 00:16:29.040
positive, and it is just going to be some
number for any particular M. But here if M
00:16:29.040 --> 00:16:33.240
square is more than one this is going to be
one sign positive. If it is M square is less
00:16:33.240 --> 00:16:34.820
than
one it is going to be opposite sign which
00:16:34.820 --> 00:16:37.720
means if my Mach number is supersonic, it
is
00:16:37.720 --> 00:16:42.959
going to be positive sign. If it is subsonic,
it is going to be negative sign; this is what
00:16:42.959 --> 00:16:46.160
is
the critical thing. If I look at area increase,
00:16:46.160 --> 00:16:52.130
my Mach number will increase if it is
supersonic; that is M greater than one that
00:16:52.130 --> 00:16:56.560
will make it positive. If this is less than
0, then
00:16:56.560 --> 00:17:01.350
this will also be less than 0.
So, I am going to pick a case d A greater
00:17:01.350 --> 00:17:04.600
than 0 which means this is positive, and I
will
00:17:04.600 --> 00:17:09.769
pick mach number greater than one; that will
make this positive. So, the overall thing
00:17:09.769 --> 00:17:12.949
is
positive. Mach number will increase; d M is
00:17:12.949 --> 00:17:16.971
positive; that is this case. What if it is
M
00:17:16.971 --> 00:17:21.230
less than one? This will just change sign.
So, I can immediately write that this will
00:17:21.230 --> 00:17:25.530
be
decreasing; Mach number decreases there, okay.
00:17:25.530 --> 00:17:30.530
Now let us say I am considering a case
where Mach number is more than one. This is
00:17:30.530 --> 00:17:37.220
positive, and my area d A is less than 0;
this will also be less than 0. So, I am going
00:17:37.220 --> 00:17:43.500
to have Mach number decreasing for
supersonic area decreasing case; supersonic
00:17:43.500 --> 00:17:50.530
area decreasing case that is this. I am going
to have Mach number decreasing, and of course,
00:17:50.530 --> 00:17:54.500
you can prove that it will be the other
way in here.
00:17:54.500 --> 00:17:57.910
It so happens that when Mach number increases
velocity increases. We have already
00:17:57.910 --> 00:18:03.160
seen it as a numerical example recently, and
we have also seen it from looking at the
00:18:03.160 --> 00:18:07.840
expressions also you can tell the same thing.
I will just go through the remaining ones
00:18:07.840 --> 00:18:17.800
a
little faster. You can also use d u by u expression
00:18:17.800 --> 00:18:26.160
which I believe I have here. I can use
this also to tell the same thing. If supersonic
00:18:26.160 --> 00:18:29.440
if area increases velocity increases. I can
do
00:18:29.440 --> 00:18:36.610
.that also. And supersonic area increases
velocity increases; it can be told from here
00:18:36.610 --> 00:18:43.970
also.
Now since I am adiabatic I can now immediately
00:18:43.970 --> 00:18:48.040
tell that if velocity increases
temperature drops; t naught is constant.
00:18:48.040 --> 00:19:02.960
So, whatever velocity is this will be the
opposite arrow, okay. Now we can also show
00:19:02.960 --> 00:19:10.650
from momentum equation, and in fact even we
also already derived this for whatever be
00:19:10.650 --> 00:19:16.760
the Mach number this M square is always positive,
which means velocity if there is a
00:19:16.760 --> 00:19:20.110
change, density change will be opposite of
that, because there is a minus sign. If this
00:19:20.110 --> 00:19:26.169
increases, this decreases; you can always
tell that. If velocity increases, density
00:19:26.169 --> 00:19:39.700
decreases. So, I can draw the same thing for
this. And now we also did this d P by P and
00:19:39.700 --> 00:19:47.220
d rho by rho; d P by P is directly proportional
to d rho by rho by a constant. So, I can
00:19:47.220 --> 00:20:01.210
also tell it will be the same direction.
So, I can populate this whole table. The main
00:20:01.210 --> 00:20:03.490
idea is you have to just get used to it. I
will
00:20:03.490 --> 00:20:08.620
just tell you a simple way to think about
it; p rho RT will always go the same direction,
00:20:08.620 --> 00:20:10.730
u
will be the opposite direction. U will go
00:20:10.730 --> 00:20:13.820
the same as M; u and M will go one way, p
rho
00:20:13.820 --> 00:20:18.390
RT will go the other way in any of these blocks
if you look at; that is what will be
00:20:18.390 --> 00:20:23.750
happening there. It is a easy way to remember;
p rho and t whatever is in your p equal to
00:20:23.750 --> 00:20:27.150
rho RT they are all going to go on one way.
And flow related thing Mach number and
00:20:27.150 --> 00:20:32.190
velocity will go the opposite side; that is
what will happen.
00:20:32.190 --> 00:20:35.430
Now you just need to remember only one thing;
how does Mach number vary when I
00:20:35.430 --> 00:20:39.710
change area? If you remember that you can
tell anything else, because you are
00:20:39.710 --> 00:20:45.450
remembering the other statement p rho RT go
opposite of u and M; it is easy way to
00:20:45.450 --> 00:20:53.430
remember this. Now all we need to do is go
to a problem flow situation and look at what
00:20:53.430 --> 00:20:59.440
happens there. I will redraw this picture
again with some other picture next to it.
00:20:59.440 --> 00:21:00.440
..
00:21:00.440 --> 00:21:09.870
Oh, before that I wanted to revisit this d
A by A expression
00:21:09.870 --> 00:21:18.480
M square minus 1 times d u
by u, okay. Now if I say that there is no
00:21:18.480 --> 00:21:30.340
change in velocity for a given Mach number,
then my duct should have constant area; that
00:21:30.340 --> 00:21:36.820
is what this statement says now, as one
inference I am going to have. Another inference
00:21:36.820 --> 00:21:48.480
I can say now is if d A equal to 0, what
all can happen? If d A is 0 what all can happen?
00:21:48.480 --> 00:21:57.600
One thing could be d u is 0; other thing
could be Mach number could go to 1. These
00:21:57.600 --> 00:22:01.261
are the two possibilities. When I say d A
is
00:22:01.261 --> 00:22:08.580
equal to 0, what do I mean? I am talking about
a case where I am saying there is no area
00:22:08.580 --> 00:22:13.789
change in my duct; does it mean that I am
always going to be Mach one in a constant
00:22:13.789 --> 00:22:17.900
area duct? Not really.
So, I cannot use this meaning all the time,
00:22:17.900 --> 00:22:27.790
but I can tell one thing for sure if M equal
to 1
00:22:27.790 --> 00:22:37.210
definitely d A is 0 irrespective of what happens
to u; that I can say for sure, okay. This
00:22:37.210 --> 00:22:41.669
is
a just one way thing. So, this is one thing
00:22:41.669 --> 00:22:47.690
I can say for sure. If Mach number is one
definitely d A is 0. That does not mean if
00:22:47.690 --> 00:22:53.740
d A is 0 Mach number is 1; one of the common
mistake in simple books. If you go pick some
00:22:53.740 --> 00:22:57.750
simple books not well written books, you
will see these kinds of mistakes happening
00:22:57.750 --> 00:23:02.560
there. I can give you a crazy example for
it; I
00:23:02.560 --> 00:23:08.840
liked the example from one of the books so
I can use that in here. If I pick an example
00:23:08.840 --> 00:23:18.100
like this
and another case where it is this; I am going
00:23:18.100 --> 00:23:24.660
to have d A is 0 at these two places.
Both these places, the slope of area changes
00:23:24.660 --> 00:23:29.179
0, right, because the slope is 0 here. It
is as if
00:23:29.179 --> 00:23:32.679
locally it is a constant area duct here, same
thing in this place.
00:23:32.679 --> 00:23:43.770
.So, now I am going to label them one, two
and three. Here also we will do the same
00:23:43.770 --> 00:23:52.010
thing one, two and three. If I have this let
us pick different cases; I can have a case
00:23:52.010 --> 00:24:02.990
of let
us say A, I can have a case where M 1 less
00:24:02.990 --> 00:24:09.410
than 1; what can happen now if M 1 is less
than 1? M 2 is more than M 1 is all I can
00:24:09.410 --> 00:24:18.250
say really, because subsonic flow we already
wrote the table; subsonic flow this will accelerate.
00:24:18.250 --> 00:24:27.130
So, I can tell that M 2 greater than M
1. Now what all can happen? Let us pick a
00:24:27.130 --> 00:24:34.880
case where M 2 is greater than M 1 and goes
to Mach one let us assume. I could have two
00:24:34.880 --> 00:24:42.440
cases. So, I will say M 2 less than 1, and
I
00:24:42.440 --> 00:24:51.679
can also have a case where M 2 greater than
1, I can have a case M 2 equal to 1.
00:24:51.679 --> 00:24:59.860
I can have different cases like this; something
must go wrong in this. What will happen
00:24:59.860 --> 00:25:07.450
here? One of them should not happen. These
are three possibilities. I can say that M
00:25:07.450 --> 00:25:11.010
1 is
say 0.5; it can go to 0.8; that will be this
00:25:11.010 --> 00:25:13.830
case. Here I am saying it is going to 1.2,
here I
00:25:13.830 --> 00:25:21.450
am saying it is 1, okay. One of them should
not happen, which one? The second one
00:25:21.450 --> 00:25:29.030
cannot happen, why? Because directly going
from subsonic to supersonic and that will
00:25:29.030 --> 00:25:33.870
be violated in your table if you look at it.
Say, I am having 0.5, it is going to 1.2;
00:25:33.870 --> 00:25:38.840
let us
pick this case; this cannot happen, why? We
00:25:38.840 --> 00:25:45.299
will go to that table. Let us say I am having
0.5 and I want to go to 1.2, and I am having
00:25:45.299 --> 00:25:52.821
only a converging duct d A less than 0.
Someone look at only this column, and I am
00:25:52.821 --> 00:25:56.730
going to say my Mach number is less than 1
when I start. Mach number keeps on increasing
00:25:56.730 --> 00:26:02.190
which is a good thing; I am going from
0.5 towards 1.2, but when it reaches 1 I should
00:26:02.190 --> 00:26:06.790
not look at this row, but I have to jump to
this row and look at this one. Because I want
00:26:06.790 --> 00:26:09.831
to go from 1 to 1.2 I have to use this row.
If
00:26:09.831 --> 00:26:16.570
I have the same area variation, it is going
to decrease Mach number again. So, that
00:26:16.570 --> 00:26:24.660
particular case cannot happen. So, I have
written it there, but we would not think about
00:26:24.660 --> 00:26:31.650
that particular case at all. Now let us go
back and look at what happens to three; this
00:26:31.650 --> 00:26:34.539
we
will not consider. This will not be a possible
00:26:34.539 --> 00:26:37.789
case.
What about when M 2 is less than 1 which is
00:26:37.789 --> 00:26:43.730
only case that is possible? What will
happen to M 3 in terms of M 2? M 3 less than
00:26:43.730 --> 00:26:52.950
M 2 and what about M 3 with respect to M
2 in terms of this one, this particular case
00:26:52.950 --> 00:27:00.679
and M 2 equal to 1, cannot say. Now you are
at
00:27:00.679 --> 00:27:04.350
a situation where I do not know which table
I have to look at; which row of the table
00:27:04.350 --> 00:27:06.080
I
have to look at? I do not know, because this
00:27:06.080 --> 00:27:11.870
is M 2 equal to 1. We have only rows that
correspond to M 2 less than 1 or greater than
00:27:11.870 --> 00:27:16.159
1. All I can say is M 3 will not be equal
to
00:27:16.159 --> 00:27:22.010
1; can be less than 1 or greater than 1.
00:27:22.010 --> 00:27:30.440
.I cannot tell which one it is going to go;
information is not sufficient as of now, because
00:27:30.440 --> 00:27:37.900
we are on the border between the two rows
data there the table. We have to figure out
00:27:37.900 --> 00:27:44.750
what can we done. This is one system I have.
What happens in the other system? Let us
00:27:44.750 --> 00:27:51.559
go there and see what happens there. I will
again consider the case A for this duct if
00:27:51.559 --> 00:27:57.260
I
pick this duct, again we will pick same M
00:27:57.260 --> 00:28:03.929
1 less than 1. We have not done case B M 1
greater than 1 yet; we will go back to that
00:28:03.929 --> 00:28:08.900
M 1 less than 1. What will happen to M 2?
It is
00:28:08.900 --> 00:28:16.900
a diverging duct subsonic; it is going to
decrease Mach number further M 2 less than
00:28:16.900 --> 00:28:23.279
M
1, and here it is converging duct. This is
00:28:23.279 --> 00:28:28.150
already subsonic; it is going much more
subsonic, say 0.5 became 0.2 further subsonic.
00:28:28.150 --> 00:28:35.580
Now it is going back to converging duct. All
I can say currently is M 3 is what greater
00:28:35.580 --> 00:28:37.559
or
less than M 2?
00:28:37.559 --> 00:28:40.490
Student: Greater than M2.
Greater than M 2; unless I am given exact
00:28:40.490 --> 00:28:42.750
area values I cannot go and calculate what
is
00:28:42.750 --> 00:28:47.360
the exact Mach number. Currently, we would
not worry about how to calculate it; this
00:28:47.360 --> 00:28:49.530
is
what will happen. Can I have any other options
00:28:49.530 --> 00:28:55.220
here for M 2? Like we had in the other
case we cannot have; there is only one option
00:28:55.220 --> 00:29:00.360
there. This is the only thing that one can
happen there. Now we will go to the next case.
00:29:00.360 --> 00:29:13.070
I do not have space here I will go and
write here
00:29:13.070 --> 00:29:14.070
.
00:29:14.070 --> 00:29:30.240
.Have this, and I am considering case B where
I am going to choose M 1 greater than 1
00:29:30.240 --> 00:29:36.279
already. I have M 1 greater than 1, and I
am having a converging duct here. If you go
00:29:36.279 --> 00:29:39.340
and
look at the table it is going to go in the
00:29:39.340 --> 00:29:43.370
direction of M 1 equal to 1 or Mach number
equal
00:29:43.370 --> 00:29:47.690
to 1. By the way a simple thing; if it is
the converging ducts it is going to go towards
00:29:47.690 --> 00:29:49.970
M
equal to 1, if it is a diverging duct it is
00:29:49.970 --> 00:29:53.490
going to go away from M equal to 1. But this
may
00:29:53.490 --> 00:29:56.730
get confusing after sometime if you do not
remember which is towards which is away,
00:29:56.730 --> 00:30:01.610
you will get confuse. Instead remember the
other statement also. I like this statement;
00:30:01.610 --> 00:30:06.010
anyways you had to get used to something in
your compressible flows.
00:30:06.010 --> 00:30:11.299
I am going to say it is going to go supersonic;
it is going to go towards M equal to 1
00:30:11.299 --> 00:30:17.789
which means 1.5 will go towards 1; maybe it
will become 1.1, I do not know. All I can
00:30:17.789 --> 00:30:25.650
tell is M 2 less than M 1. Let us consider
a case where it is this. Another case I will
00:30:25.650 --> 00:30:29.850
say
M 2 equal to. I will specifically say that
00:30:29.850 --> 00:30:32.990
it is something like 1.1 in this case, and
here it is
00:30:32.990 --> 00:30:40.330
exactly equal to 1. Can I have a case again
M 2 less than 1? This particular case is not
00:30:40.330 --> 00:30:45.539
possible; because again similar argument as
what we said here one case is not possible.
00:30:45.539 --> 00:30:50.210
Similarly, I can say that there also it will
not be possible. It has to cross and suddenly
00:30:50.210 --> 00:30:54.880
area change should change the other away for
it go cross Mach one.
00:30:54.880 --> 00:31:06.409
Now I have these two cases. If it is M 2 greater
than 1, what happens to M 3? Supersonic
00:31:06.409 --> 00:31:13.370
flow area increases; what happens to it? This
will also increase which means M 3 is
00:31:13.370 --> 00:31:24.090
greater than M 2. Now if I look at M 2 equal
to 1, again similar situation arises. I cannot
00:31:24.090 --> 00:31:28.700
tell what it is going to do, because I am
not anywhere on the rows of the table. I am
00:31:28.700 --> 00:31:33.340
in
between two rows of the table. I can have
00:31:33.340 --> 00:31:39.539
M 3 greater than or less than 1. This probably
not a mathematical symbol; am just using it
00:31:39.539 --> 00:31:50.091
any way, could be any of those cases. I
cannot tell for sure what can happen. What
00:31:50.091 --> 00:32:00.110
about the other system? If I have case B here
M 1 we said greater than 1 I am going to pick.
00:32:00.110 --> 00:32:11.630
Am picking M 1 greater than 1, what
happens to M 2? Supersonic flow diverging
00:32:11.630 --> 00:32:17.710
is going to increase M 2 greater than M 1.
And of course, that means it is definitely
00:32:17.710 --> 00:32:22.690
more than 1; we do not need to worry. It is
going to be 1.5 becoming 3, something like
00:32:22.690 --> 00:32:25.190
that, and it is converging now. It is going
to
00:32:25.190 --> 00:32:32.190
go in the reverse direction M 3 less than
M 2. We would not worry about what is
00:32:32.190 --> 00:32:36.909
happening from 1 to 3. We would not compare
M 1 and M 3 as of now. It is not a serious
00:32:36.909 --> 00:32:44.610
problem that way; we would not worry about
it. Any other possibilities here; no other
00:32:44.610 --> 00:32:51.500
possibilities exist, okay. The nice thing
about this particular example is I am going
00:32:51.500 --> 00:32:58.230
to say
here also d A is going to be equal to 0, here
00:32:58.230 --> 00:33:00.590
also d A is going to be equal to 0.
00:33:00.590 --> 00:33:10.220
.In this case whichever be the condition I
pick, I am never going to get M equal to 1
00:33:10.220 --> 00:33:15.010
in
this kind of a duct, but in this kind of duct
00:33:15.010 --> 00:33:21.049
I have possibility of M equal to 1 happening;
that is what we need to think about. It is
00:33:21.049 --> 00:33:24.870
something more than just d A equal to 0. It
has
00:33:24.870 --> 00:33:32.809
to be area constriction. It cannot be area
opening more to something else; that is one
00:33:32.809 --> 00:33:37.770
more thing you have to look at here. This
is called chocking of the flow, because they
00:33:37.770 --> 00:33:40.960
are
constrained of duct; we are chocking the duct
00:33:40.960 --> 00:33:43.890
that is the idea. If you have a pipe and I
am
00:33:43.890 --> 00:33:45.630
crushing it in the center that is called the
chocking.
00:33:45.630 --> 00:33:50.710
Now you are chocking the duct in here; that
is why it is called chocked flow. If it goes
00:33:50.710 --> 00:33:53.740
to
a situation where Mach number at that point
00:33:53.740 --> 00:34:00.809
is sonic, it does not ever happen in the other
configuration. Both are going to have d A
00:34:00.809 --> 00:34:08.350
equal to 0, and as per this I cannot tell
anything for sure, but we have to look at
00:34:08.350 --> 00:34:13.099
more than just what it is. It has to be a
chocked
00:34:13.099 --> 00:34:24.510
flow for it to go to M equal to 1. Now after
this we want to be able to go to a situation
00:34:24.510 --> 00:34:35.509
where I can start solving whatever flow field
in different situations; for that I should
00:34:35.509 --> 00:34:40.279
be
ready to extract more information out of this.
00:34:40.279 --> 00:34:41.279
.
00:34:41.279 --> 00:34:49.450
What we have going to say now is I have a
duct with some chocked area. Since, it is
00:34:49.450 --> 00:34:57.269
some chocked area something special I am going
to call this is A star. I know that it is
00:34:57.269 --> 00:35:05.140
chocked area which means my Mach number here
is 1, because that is by definition
00:35:05.140 --> 00:35:09.369
chocking for me. You want to say the least
area point and it is chocked I am going to
00:35:09.369 --> 00:35:12.430
say.
We will go and redefine that chocking more
00:35:12.430 --> 00:35:15.640
after we have done some derivation. As of
00:35:15.640 --> 00:35:21.500
.now we will just assume that it has to be
the least area or the minimum area possible,
00:35:21.500 --> 00:35:24.059
and
we will assume that Mach number is 1 there.
00:35:24.059 --> 00:35:29.680
There is also a possibility like I will go
back here. There is also a possibility that
00:35:29.680 --> 00:35:33.529
from here I start with say Mach 0.2, it went
to a
00:35:33.529 --> 00:35:38.130
higher Mach number, but it did not reach.
It need not always take this path; it can
00:35:38.130 --> 00:35:43.369
also take this path. Let us say it did not
take this
00:35:43.369 --> 00:35:50.579
path currently. We are considering a duct
which has this particular path going through;
00:35:50.579 --> 00:35:56.980
that is the path you are going to consider.
Now our job is to go and find let us say section
00:35:56.980 --> 00:36:03.999
one. I want to know the mach number for this
area, whatever; area A 1 is given let us say
00:36:03.999 --> 00:36:10.420
I want to find the mark number M 1; that is
the overall idea, how will I find it? So,
00:36:10.420 --> 00:36:14.479
now I
will start using the integral equations and
00:36:14.479 --> 00:36:17.060
trying to get to some point where it will
look
00:36:17.060 --> 00:36:21.779
like I know how to solve this problem. I have
already assumed that it is an isentropic
00:36:21.779 --> 00:36:27.630
flow, and I am starting to solve the problem,
okay.
00:36:27.630 --> 00:36:42.119
I am going to pick this M dot mass flow rate
equal to, I am going to write it like this;
00:36:42.119 --> 00:36:45.170
I
am gone to pick two conditions, one and star
00:36:45.170 --> 00:36:52.599
condition. I am going to say I know that the
flow is chocked here, and I am going to say
00:36:52.599 --> 00:36:59.549
also know isentropic flow T naught and P
naught do not change. I know that also; for
00:36:59.549 --> 00:37:03.960
this whole flow fields somewhere I have
measured it. I know somethingâ€™s about the
00:37:03.960 --> 00:37:08.859
flow. I know that the flow is chocked here,
and I know T naught and P naught for the flow.
00:37:08.859 --> 00:37:12.499
These are the information I know
currently, and we want to solve every other
00:37:12.499 --> 00:37:19.970
point in the flow; can we do it? Let us see
how we can solve this problem.
00:37:19.970 --> 00:37:25.440
So, I am going to look at only this set of
expressions. I want to rewrite them in such
00:37:25.440 --> 00:37:28.369
a
way that I will eliminate all the variables.
00:37:28.369 --> 00:37:34.799
I know Mach number here. Let us say I want
to call it M star. M star will be 1, and here
00:37:34.799 --> 00:37:43.229
this is some M 1. I want to write all these
variables or the ratios of them in terms of
00:37:43.229 --> 00:37:46.990
M equal to 1 or M star and M 1; I want to
write
00:37:46.990 --> 00:37:55.130
these two. So, let us start with what we know
already rho naught by rho is equal to what;
00:37:55.130 --> 00:38:03.869
in terms of Mach number what it will be? 1
plus gamma minus 1 by 2 M square to the
00:38:03.869 --> 00:38:14.059
power 1 by gamma minus 1; this is what I have
already. So, now, if I say I have a special
00:38:14.059 --> 00:38:21.509
condition rho naught by rho star, this is
going to be equal to M equal to 1 will simplify
00:38:21.509 --> 00:38:29.769
to
1 plus gamma divided by 2; this is what I
00:38:29.769 --> 00:38:34.599
have.
So, now if I take the ratios of these two,
00:38:34.599 --> 00:38:37.420
I will get rho by rho star which is what I
have
00:38:37.420 --> 00:38:46.630
here, okay. If I use M 1 here this will become
rho 1; that is the way I look at it. And here
00:38:46.630 --> 00:38:55.200
.I have used M equal to 1; that becomes the
star condition, this is what I have. Similarly
00:38:55.200 --> 00:39:03.190
I
can do for u 0, okay; u 1 in terms of Mach
00:39:03.190 --> 00:39:15.390
number is what? U 1 in terms of Mach
number, M into speed of sound; so, it will
00:39:15.390 --> 00:39:26.960
be M 1 times A 1 which is written as M 1
square root of gamma RT 1. Now I still have
00:39:26.960 --> 00:39:37.420
to write T 1 in terms of Mach number. That
is going to be M 1 square root of gamma R.
00:39:37.420 --> 00:39:42.369
I will go and write a separate expression
here T naught by T equal to 1 plus gamma minus
00:39:42.369 --> 00:39:49.140
1 by 2 M square. I have this here.
I want to use that inside here. I will take
00:39:49.140 --> 00:39:51.209
T there and the whole expression below. It
is
00:39:51.209 --> 00:39:59.020
going to be T naught divided by I will extend
the square root further 1 plus gamma
00:39:59.020 --> 00:40:06.839
minus 1 by 2 M 1 square. This is the expression
I will have. Now if I want a u star
00:40:06.839 --> 00:40:13.690
expression that is just substituting M 1 equal
to 1 in this. So, that is going to come down
00:40:13.690 --> 00:40:23.229
to square root of 2 gamma RT naught by gamma
plus 1, okay. You can simplify to
00:40:23.229 --> 00:40:35.440
something like this. Now you want to simplify
it much further. I will substitute the ratios
00:40:35.440 --> 00:40:51.380
of these such that I will get A 1 by A star.
A 1 by A star I want to substitute. If I do
00:40:51.380 --> 00:40:55.010
this, I
have to take the remaining things there rho
00:40:55.010 --> 00:41:06.910
star by rho multiplied by u star by u. Actually
I have to write it as u rho 1 and u 1, and
00:41:06.910 --> 00:41:09.299
we just have to substitute these functions
inside
00:41:09.299 --> 00:41:15.469
there.
And it is going to be rho star by rho 1 will
00:41:15.469 --> 00:41:19.359
be this divided by this will give you rho
star
00:41:19.359 --> 00:41:27.309
by rho 1. So, I am going to write it as 1
plus gamma minus 1 by 2 M square divided by
00:41:27.309 --> 00:41:31.519
1
plus gamma divided by 2; the whole thing to
00:41:31.519 --> 00:41:37.600
the power 1 by gamma minus 1. This is the
first thing multiplied by u star by u 1 which
00:41:37.600 --> 00:41:42.640
is this divided by this where some of them
will get cancelled. We would not cancel any
00:41:42.640 --> 00:41:48.589
of them as of now. We will write it as such
and cancel it later square root of 2 gamma
00:41:48.589 --> 00:42:00.469
RT naught by gamma plus 1 divided by M 1
square root of gamma RT naught by 1 plus gamma
00:42:00.469 --> 00:42:10.460
minus 1 by 2 M 1 square; this is what
I have. Now of course, I can directly cancel
00:42:10.460 --> 00:42:20.589
gamma RT naught, gamma RT naught. I am
going to get something like this.
00:42:20.589 --> 00:42:21.589
..
00:42:21.589 --> 00:42:33.609
Now I just have to simplify this whole thing.
If I look at it, the expressions are almost
00:42:33.609 --> 00:42:36.470
the
same in this bracket and in this square root
00:42:36.470 --> 00:42:39.819
terms. There is just a 1 by M 1 which will
be
00:42:39.819 --> 00:42:44.009
separate which we will put it in the front.
The remaining things if I look at it, they
00:42:44.009 --> 00:42:47.650
are
almost the same thing. If I slightly rewrite
00:42:47.650 --> 00:42:52.489
this, I will get it as 2 plus gamma minus
1 M
00:42:52.489 --> 00:42:58.760
square whole divided by 2;; then that by 2
will get cancel it this divided by 2. If I
00:42:58.760 --> 00:43:01.569
look at
here, it will be the reciprocal of this in
00:43:01.569 --> 00:43:05.770
the numerator, and here the reciprocal of
this in
00:43:05.770 --> 00:43:12.589
the denominator with the square root here.
I can rewrite this expression as this whole
00:43:12.589 --> 00:43:18.269
thing to the power half; that is what will
happen here.
00:43:18.269 --> 00:43:28.490
I will go and write it like this 1 plus gamma
minus 1 by 2 M square by 1 plus gamma by
00:43:28.490 --> 00:43:36.469
2 to the power 1 by gamma minus 1. I also
have a 1 by M 1, and the remaining terms
00:43:36.469 --> 00:43:42.640
inside the square root I will combine them
as one and rewrite it in a nicer fashion 1
00:43:42.640 --> 00:43:47.710
plus
gamma minus 1 by 2 M 1 square divided by 1
00:43:47.710 --> 00:43:56.719
plus gamma by 2. I write it like this. Now
if I look at it, this and this are exactly
00:43:56.719 --> 00:43:59.140
the same and they are multiplied. So, the
powers
00:43:59.140 --> 00:44:06.289
will be adding. So, I will just do that 1
by gamma minus 1 plus 1 by 2, and it is going
00:44:06.289 --> 00:44:13.059
to
be 2 plus gamma minus 1 by 2 times gamma minus
00:44:13.059 --> 00:44:21.170
1. This is going to be gamma plus 1
by 2 times gamma minus 1; this is what it
00:44:21.170 --> 00:44:28.569
comes out to be.
So, I will write my A 1 by A star expression.
00:44:28.569 --> 00:44:35.260
It is going to be 1 plus gamma minus 1 by
2
00:44:35.260 --> 00:44:47.099
M square divide by 1 plus gamma divided by
2 whole to the power gamma plus 1 by 2
00:44:47.099 --> 00:44:56.890
times gamma minus 1 with a 1 by M 1 in the
front. Now I will have to put M 1 square
00:44:56.890 --> 00:45:04.119
.here. Now I have our expression where I am
going to say area is a function of Mach
00:45:04.119 --> 00:45:08.249
number. This is not an ideal expression for
us, but this is the only thing we can get
00:45:08.249 --> 00:45:13.430
right
now. I can say that area with respect to critical
00:45:13.430 --> 00:45:19.859
area; I am using the word critical for the
first time, this is the area at chocked condition.
00:45:19.859 --> 00:45:24.559
In compressible flows they use the word critical
a lot. I am going to say the area divided
00:45:24.559 --> 00:45:33.010
by the critical area; will this be more than
one or less than one? It will always be more
00:45:33.010 --> 00:45:39.109
than one, because A star is defined to be
the least area in your whole duct. So, this
00:45:39.109 --> 00:45:40.819
will
always be more than one; this can never be
00:45:40.819 --> 00:45:46.589
less than one this function. So, that being
said
00:45:46.589 --> 00:45:53.279
will it automatically be so? Even if it is
subsonic, I can say when it is supersonic
00:45:53.279 --> 00:45:56.539
Mach
number square yield by Mach number, it is
00:45:56.539 --> 00:45:58.259
going to be very high power; that is easy
to
00:45:58.259 --> 00:46:01.849
say that it is going to increase. Can I tell
that that is the same when it is decreasing
00:46:01.849 --> 00:46:06.229
when
it is subsonic? It is so we will look at the
00:46:06.229 --> 00:46:12.260
plot after sometime, but that is what will
happen there.
00:46:12.260 --> 00:46:18.440
Now instead of having this divided by 2 and
stuff I can rewrite this as one more
00:46:18.440 --> 00:46:26.650
statement 2 plus gamma minus 1 times M square
divided by 1 plus gamma whole to the
00:46:26.650 --> 00:46:36.709
power gamma plus 1 by 2 times gamma minus
1. This is one more form you will see.
00:46:36.709 --> 00:46:43.990
This is probably easier; they are doing some
Huron coding for this function, or I will
00:46:43.990 --> 00:46:46.640
take
this divided by 2 as a square root in here
00:46:46.640 --> 00:46:49.039
for this whole thing; that is another way
people
00:46:49.039 --> 00:46:54.309
write it, different ways of writing. Whichever
form you remember is good for us.
00:46:54.309 --> 00:46:59.539
This is called the area Mach number relationship,
very very useful in calculating Mach
00:46:59.539 --> 00:47:07.089
numbers at different points in your duct.
This is the area Mach number relationship.
00:47:07.089 --> 00:47:10.109
The
next thing we want to do is go back and substitute
00:47:10.109 --> 00:47:15.900
it in this expression; I will just use this
expression here. I will erase this. I want
00:47:15.900 --> 00:47:24.660
to find the mass flow rate through my duct.
I
00:47:24.660 --> 00:47:39.880
will erase this, not very difficult to do.
I will not use this rho 1, u 1, A 1, more
00:47:39.880 --> 00:47:40.880
difficult.
00:47:40.880 --> 00:47:41.880
..
00:47:41.880 --> 00:47:53.390
I will go use this. M dot is going to be equal
to I will write everything only in terms of
00:47:53.390 --> 00:47:55.400
P
naught and T naught, the stagnation conditions.
00:47:55.400 --> 00:48:07.390
I will just pick this rho star by rho
naught times rho naught into u star which
00:48:07.390 --> 00:48:12.390
is nothing but for M equal to 1, right. So,
it is
00:48:12.390 --> 00:48:21.671
just going to be square root of gamma RT star.
Now I have to expand T star further and
00:48:21.671 --> 00:48:31.699
multiply by A star I will keep it right now,
one more line I will write. Rho not is P
00:48:31.699 --> 00:48:37.890
naught by RT naught into rho star by rho naught
we already had an expression. I just
00:48:37.890 --> 00:48:55.859
erased it. It is going to be 1 plus gamma
minus 1; for M equal to 1 it is just 1 plus
00:48:55.859 --> 00:49:02.329
gamma
divided by 2 to the power 1 by 1 minus gamma;
00:49:02.329 --> 00:49:07.650
that is the function I have.
Of course, I put 1 minus gamma here instead
00:49:07.650 --> 00:49:12.579
of 1 by gamma minus 1, because it has to
be reciprocal of that. I have already taking
00:49:12.579 --> 00:49:18.509
care of that; that is my rho naught multiplied
by rho star by rho naught. Now I have to do
00:49:18.509 --> 00:49:25.549
this square root of gamma R. I will keep that
as it is square root of T star; T star in
00:49:25.549 --> 00:49:30.150
terms of T naught. T naught by T star is equal
to 1
00:49:30.150 --> 00:49:35.809
plus gamma divided by 2. This we already wrote
some time back. So, I want T star; that
00:49:35.809 --> 00:49:47.990
will come out to be 2 T naught by 1 plus gamma
square root of that multiplied by A star.
00:49:47.990 --> 00:49:56.049
This is the expression I have. Now if you
see nothing really gets canceled except for
00:49:56.049 --> 00:49:59.460
this
RT naught and RT naught here inside the square
00:49:59.460 --> 00:50:02.319
root.
Everything else stays as it is, and probably
00:50:02.319 --> 00:50:07.059
I can simplify this 1 plus gamma to the power
1 by gamma 1 minus gamma and 1 plus gamma
00:50:07.059 --> 00:50:14.029
to the power half. Let us say I will leave
it to you an exercise, and I will just write
00:50:14.029 --> 00:50:19.069
the final expression. If I write the final
00:50:19.069 --> 00:50:25.299
.expression in a form which is very common,
I will take the A star and put it in the
00:50:25.299 --> 00:50:33.259
denominator here M dot by A star; that is
going to be given as P naught by square root
00:50:33.259 --> 00:50:35.470
T
naught. I am writing it in a particular form
00:50:35.470 --> 00:50:44.329
multiplied by square root of gamma by r 2
by
00:50:44.329 --> 00:50:53.279
gamma plus 1 to the power gamma plus 1 by
gamma minus 1. You can end up with this
00:50:53.279 --> 00:50:58.759
form; it is not very difficult. You will end
up with this form after some time. It is not
00:50:58.759 --> 00:51:04.559
very difficult; you will just go through the
math. I will just say you know how to do that
00:51:04.559 --> 00:51:07.329
math.
Now what is the mass flow rate at section
00:51:07.329 --> 00:51:12.749
one? If you think about it must be the same
as
00:51:12.749 --> 00:51:18.160
that at this section, right, it is a duct.
It is said it is one duct and it is just area
00:51:18.160 --> 00:51:20.569
changing;
mass flow rate here must be same as mass flow
00:51:20.569 --> 00:51:25.059
rate here; another proof this sentence
here. The expression written here M dot is
00:51:25.059 --> 00:51:27.500
equal to rho 1 u 1 A 1 is equal to rho star
u
00:51:27.500 --> 00:51:35.279
star A stat. This is going to help us solve
problem faster. I have this expression. I
00:51:35.279 --> 00:51:39.569
can of
course go and do the same kind of derivation
00:51:39.569 --> 00:51:46.979
for rho 1 u 1 A 1. I can do that and there
will be a better expression than this. There
00:51:46.979 --> 00:51:52.339
will be 1 by M 1 somewhere and then there
will be 1 by 1 plus gamma minus 1 by 2 M square
00:51:52.339 --> 00:51:56.839
kind of terms will be sitting inside
here; that can also be done.
00:51:56.839 --> 00:52:01.019
And I will give an exercise of that form,
and I will put it up on the web for that purpose.
00:52:01.019 --> 00:52:06.440
That can also be derived, and that means I
can get mass flow rate. If I know the Mach
00:52:06.440 --> 00:52:12.299
number and the critical area, I can get mass
flow rate through any duct. Of course, you
00:52:12.299 --> 00:52:16.690
should also know the stagnation conditions
pressure and temperature, and of course, the
00:52:16.690 --> 00:52:20.920
gas the gamma value you should know. These
are the things you need to know to get to
00:52:20.920 --> 00:52:29.549
mass flow rate through the duct. We will go
look at more expressions and the functional
00:52:29.549 --> 00:52:36.359
form of A by A star, what happens to Mach
number when my area A star changes and all
00:52:36.359 --> 00:52:39.869
that in the coming classes. See you people
next class.
00:52:39.869 --> 00:52:40.869
.