WEBVTT
Kind: captions
Language: en
00:00:10.019 --> 00:00:20.010
Hello everyone, welcome back. We were looking
last class at trying to look at the 2-d
00:00:20.010 --> 00:00:27.129
problem as an analogy of a 1-d problem, like
we were trying to link 1-d x versus t as
00:00:27.129 --> 00:00:36.020
looking the same as x y is unsteady 1-d problem
is equivalent to a 2-d steady problem.
00:00:36.020 --> 00:00:37.670
That is what we were trying to look at.
.
00:00:37.670 --> 00:00:48.870
Today we have an animation to do that. Let
us go to the screen. What we are having here
00:00:48.870 --> 00:00:54.649
is a piston cylinder arrangement where there
is a duct, and the piston is moving and as
00:00:54.649 --> 00:00:57.920
it
moves, these dots are supposed to be representing
00:00:57.920 --> 00:01:03.960
gas and the gas is getting compressed,
that is the dots are getting closer to each
00:01:03.960 --> 00:01:04.960
other.
00:01:04.960 --> 00:01:05.960
..
00:01:05.960 --> 00:01:09.320
And this is the interface line. The red line
which is supposed to be are shock, it is shown
00:01:09.320 --> 00:01:15.880
as if it is outside, but it is really supposed
to be only inside. Can you see that, this
00:01:15.880 --> 00:01:17.909
is
physically possible right such a thing is
00:01:17.909 --> 00:01:23.590
possible to happen. So, this is our we already
solved a problem with this we said that if
00:01:23.590 --> 00:01:26.499
a piston moves at twenty meter per second,
and
00:01:26.499 --> 00:01:31.090
the shock will be going much faster than the
piston itself. We solve such a problem, this
00:01:31.090 --> 00:01:33.369
is supposed to be some such analogy.
.
00:01:33.369 --> 00:01:40.740
.Now, we will go to the next one, which is
supposed to be giving you the feel for what
00:01:40.740 --> 00:01:43.509
is
happening. Here, we are moving the piston
00:01:43.509 --> 00:01:51.070
parallel to itself in the direction in this
direction while the piston is moving the other
00:01:51.070 --> 00:01:56.439
direction. What we are seeing is we are
drawing the trajectory of the piston along
00:01:56.439 --> 00:01:59.400
with this line and this is the trajectory
of the
00:01:59.400 --> 00:02:04.659
shock. What I am doing here is x versus t
whatever I was plotting before it is a same
00:02:04.659 --> 00:02:09.890
thing I am getting here, a horizontal line
is roughly x versus t and the vertical line
00:02:09.890 --> 00:02:12.280
is here
position horizontal line is a time axis, vertical
00:02:12.280 --> 00:02:17.610
line is a position axis.
What we are seeing is if there is something
00:02:17.610 --> 00:02:23.900
some particle that is coming here only. When
it sees the shock from that point on it will
00:02:23.900 --> 00:02:26.710
start moving that is what is represented by
this
00:02:26.710 --> 00:02:32.600
yellow line here and it becomes your trajectory
we looked that it on the board, last time
00:02:32.600 --> 00:02:37.560
just wanted to put a animation in here and
if I pick another case here, if we follow
00:02:37.560 --> 00:02:41.070
any
point on this line up to this point till the
00:02:41.070 --> 00:02:43.220
shock touches. It is going to be standing
in the
00:02:43.220 --> 00:02:50.760
same spot after that it starts moving up that
is what you are seeing here, that is the idea.
00:02:50.760 --> 00:02:56.980
Now, if we look at the angles we are going
to label them the same way what we are
00:02:56.980 --> 00:03:05.150
going to use in the class. The angle of the
line representing the piston part is denoted
00:03:05.150 --> 00:03:09.430
by
theta and the angle representing the line
00:03:09.430 --> 00:03:14.820
denoted for the shock is given by beta is
our
00:03:14.820 --> 00:03:21.300
shock angle if it is x y coordinate system
currently we are having x versus t coordinate
00:03:21.300 --> 00:03:26.370
system and if it becomes x y coordinate system
theta is the angle of deflection of this
00:03:26.370 --> 00:03:30.880
incoming line.
The particle that comes here is deflected
00:03:30.880 --> 00:03:39.130
by the shock. Now, I am talking 2-d flow when
I am having a wall like this along theta direction
00:03:39.130 --> 00:03:44.010
and incoming flow is straight parallel to
the incoming, this yellow line then there
00:03:44.010 --> 00:03:47.730
will be a shock formed at angle beta such
that,
00:03:47.730 --> 00:03:52.790
the fluid particle that comes straight when
it touches that shock immediately, after that
00:03:52.790 --> 00:03:55.720
it
will be turned to go along this way, that
00:03:55.720 --> 00:04:01.170
is the idea and same thing happens in this
particular line also, that is what you are
00:04:01.170 --> 00:04:03.740
seeing and you can say that all these lines
any
00:04:03.740 --> 00:04:08.319
lines in between these two those particles
will be going in between these two, which
00:04:08.319 --> 00:04:13.110
means my overall fluid element that was here
which was having a big volume.
00:04:13.110 --> 00:04:17.709
Now, going to be compressed to have a small
volume, this is where we stopped last class
00:04:17.709 --> 00:04:22.009
we ended there and now I am just going to
show you the animation explanation for the
00:04:22.009 --> 00:04:28.529
same thing. Now, we do that is all we want
to show in animation, let us go to the board
00:04:28.529 --> 00:04:36.639
what we want to do is we will start with how
to look at this oblique shock problem,
00:04:36.639 --> 00:04:40.389
somehow we have to look at this public shock
problem such that it is easy to solve
00:04:40.389 --> 00:04:48.070
.analytically. So, happens that people have
already figured out how to look at it, is
00:04:48.070 --> 00:04:50.529
there is
a very nice way of looking at it which is
00:04:50.529 --> 00:04:54.370
again using a moving reference frame, we are
already comfortable with moving reference
00:04:54.370 --> 00:04:58.130
frame from one thing to the other. Now, we
are going to pick a special reference frame
00:04:58.130 --> 00:04:59.130
movement.
.
00:04:59.130 --> 00:05:03.479
Let us start with stationary normal shock
from now on when I say normal shock, it is
00:05:03.479 --> 00:05:08.270
stationary normal shock. We will go for moving
only later after that, I am having some
00:05:08.270 --> 00:05:17.360
velocity u 1 coming in and some smaller velocity
u 2 going out. Now, I want to have my
00:05:17.360 --> 00:05:27.229
observer, this particular person is going
to be moving straight along this normal shock
00:05:27.229 --> 00:05:36.580
with a velocity V. We are going to pick such
a case, if that is a case what will be the
00:05:36.580 --> 00:05:39.650
flow
field as seen by that moving observer that
00:05:39.650 --> 00:05:44.710
is what we want to see. So, we are going to
transform to the coordinate system as viewed
00:05:44.710 --> 00:05:50.770
by the flow field as viewed by this person,
this particular observer. So, I have this
00:05:50.770 --> 00:05:54.400
original velocity vector u 1. Now, I have
to add
00:05:54.400 --> 00:05:57.830
this relative velocity because the flow is
going to be going this way with respect to
00:05:57.830 --> 00:06:00.340
that
person I add that also.
00:06:00.340 --> 00:06:12.449
Let us say, this is my V vector, I will call
this capital V 1 this will be the velocity
00:06:12.449 --> 00:06:17.000
of the
incoming fluid as seen by the observer and
00:06:17.000 --> 00:06:18.840
then there is normal shock but it is still
going
00:06:18.840 --> 00:06:23.599
to be here. Normal shock does not move really,
it is in fact moving along this line. We
00:06:23.599 --> 00:06:33.479
would not see it. Now, the next thing is u
2, we said it has some u 2 velocity and the
00:06:33.479 --> 00:06:45.559
same V vector should be added to it and this
becomes the final net velocity. I have to
00:06:45.559 --> 00:06:54.270
.mark all the velocities u 2 is the same as
that plus this relative velocity V should
00:06:54.270 --> 00:06:57.839
be
added the net velocity is V 2 capital V 2
00:06:57.839 --> 00:07:01.309
there. So, overall what is happening is if
I draw
00:07:01.309 --> 00:07:12.150
a particular streamline it comes at this angle
and then there is a shock and then this
00:07:12.150 --> 00:07:17.620
velocity here. It is supposed to be a straight
line, just assume it is a straight line that
00:07:17.620 --> 00:07:21.599
is
fine. So, if I pick another streamline. Now,
00:07:21.599 --> 00:07:32.789
I have to be a very careful drawing a parallel
line that is going to do something like that.
00:07:32.789 --> 00:07:40.369
Now, in our flow we have assumed that, it
is invoice id which means I can call any
00:07:40.369 --> 00:07:45.800
streamline as a wall there is no affect of
boundary layer anywhere. I can call any
00:07:45.800 --> 00:07:50.950
streamline as a wall because no flow can go
past that streamline. So, I will transform
00:07:50.950 --> 00:07:54.659
this
to a case which I like. Now, I want to look
00:07:54.659 --> 00:07:57.080
at this problem like this till my head and
look
00:07:57.080 --> 00:08:03.869
at it. Now, it will look like I have just
rotated the problem slightly to the clockwise
00:08:03.869 --> 00:08:25.919
direction, this is what I will get. Now, basically
if I look at the normal shock problem
00:08:25.919 --> 00:08:32.750
with an observer moving along the shock direction
with the given velocity, then I am
00:08:32.750 --> 00:08:37.710
going to get some such case.
Now, I transformed a coordinate such that
00:08:37.710 --> 00:08:42.500
it is as observed by the particular observer
who is moving with respect to the stationary
00:08:42.500 --> 00:08:49.630
normal shock, this is what that particular
person will see it is a easy way of looking
00:08:49.630 --> 00:08:54.110
at the problem, even though I gave the analogy
like the piston cylinder arrangement, this
00:08:54.110 --> 00:08:57.690
is the best way of to transform. The thing,
the
00:08:57.690 --> 00:09:02.300
piston analogy which we saw in the animation
will have a little bit of a error, because
00:09:02.300 --> 00:09:06.890
there we assumed that the horizontal velocities
a constant before and after shock it is
00:09:06.890 --> 00:09:11.600
assumed to be constant which is not the case
in real life but this particular analogy it
00:09:11.600 --> 00:09:14.310
will
work perfectly, because we are moving along
00:09:14.310 --> 00:09:19.630
the line which separates the before and
after shock we are moving along the shock
00:09:19.630 --> 00:09:24.931
basically. Now, we will look at what happens
of course,, we have them. Now, mark those
00:09:24.931 --> 00:09:32.450
angles this is our theta and this is our shock
angle beta will keep those.
00:09:32.450 --> 00:09:42.052
Now, what if he is moving with a lesser velocity?
I will draw that case again I have to
00:09:42.052 --> 00:09:48.630
draw the same kind of lengths for velocity
vector. So that it is comparable. Now, he
00:09:48.630 --> 00:09:53.011
is
moving with a lesser velocity V. Let us say
00:09:53.011 --> 00:09:59.830
only this length, previously it was long.
Now, it is lesser V is lesser this is the
00:09:59.830 --> 00:10:04.370
movement of that particular observer. Now,
this I
00:10:04.370 --> 00:10:18.690
want to transform and that is going to go
to shock the same velocity here but a smaller
00:10:18.690 --> 00:10:27.440
relative velocity which will give me some
other V 1, same u 1, some other V. I will
00:10:27.440 --> 00:10:33.860
get
some other V 1 on the other side same u 2
00:10:33.860 --> 00:10:36.350
a lesser V.
00:10:36.350 --> 00:10:45.140
.So, I will get some other V 2 if I look at
this problem. The same way as before I
00:10:45.140 --> 00:10:57.760
transformed this to a flow field, it is going
to look like
00:10:57.760 --> 00:11:08.000
this is how the streamlines will go
which is equivalent to saying. There is a
00:11:08.000 --> 00:11:14.200
small angle of change here, there is a small
change here, theta is lesser in this case
00:11:14.200 --> 00:11:19.340
what is happening is my theta has decreased,
that
00:11:19.340 --> 00:11:24.550
is what you are noticing what happens to beta
would have increased a little. Those are
00:11:24.550 --> 00:11:31.570
the things that you will see from here, if
I change the relative velocity for the observer,
00:11:31.570 --> 00:11:37.760
then it is going to change the theta and beta
for my problem is there any other variable
00:11:37.760 --> 00:11:40.370
in
my problem of course, the strength of the
00:11:40.370 --> 00:11:44.650
normal shock itself which was our original
variable in our normal shock problem. So,
00:11:44.650 --> 00:11:52.041
I can change that also, if I say I increase.
Let
00:11:52.041 --> 00:11:58.790
us say I will keep the same V as this 1 but
I will increase my actually, I will keep the
00:11:58.790 --> 00:12:07.500
same V as this previous case.
And I am increasing my m 1 or u 1 in our case.
00:12:07.500 --> 00:12:12.320
It has to be u 1 will just keep it as u 1,
which is equivalent, if I say t 1 is the same
00:12:12.320 --> 00:12:16.360
for all the cases, it is a same thing. So,
which
00:12:16.360 --> 00:12:26.120
means my velocity vector upstream is going
to be longer and it is a same height. I have
00:12:26.120 --> 00:12:34.790
to draw the other case first and if this is
my u 2 sorry, u 1 sorry, u 2 will be much
00:12:34.790 --> 00:12:39.050
smaller
right. We know that already if we have high
00:12:39.050 --> 00:12:43.510
u 1, u 2 will be lesser, because my mark
number increases. It is much stronger it will
00:12:43.510 --> 00:12:51.370
be stopping the flow much more drastically.
So, it will be a smaller u 2. Now, I will
00:12:51.370 --> 00:12:55.010
transform this same way as before. Now, it
is
00:12:55.010 --> 00:13:10.340
going to become I need more space. It looks
like the vector itself is too long
00:13:10.340 --> 00:13:23.010
and the
same V I have to add and this will be my V
00:13:23.010 --> 00:13:29.100
1, this is my u 1, this is my V and that is
my
00:13:29.100 --> 00:13:38.410
V 1 and if I pick the same U 2 from here across
here. It is a small value, I have drawn it
00:13:38.410 --> 00:13:54.310
too long and I will add this V, I am going
to get some particular V 2, if I turn my head
00:13:54.310 --> 00:13:57.320
slightly.
So that, this is horizontal in my eye, then
00:13:57.320 --> 00:14:07.700
I will get a particular picture, that is looking
like it is turning that nicely and the shock
00:14:07.700 --> 00:14:13.540
of angle is something like this is what I
will
00:14:13.540 --> 00:14:22.480
get, I just took a shortcut I should have
gone and drawn this kind of picture and then
00:14:22.480 --> 00:14:28.740
came back to this picture. I just took a shortcut
I am just telling you that the gap between
00:14:28.740 --> 00:14:34.200
beta and theta will become lesser if I increase
my u 1. The gap between beta and theta
00:14:34.200 --> 00:14:41.920
will be higher, if I increase my V, if I decrease
my V, that is what we have done till now
00:14:41.920 --> 00:14:49.600
basically, I am telling I need these two as
my variables for my fixing my problem, if
00:14:49.600 --> 00:14:51.910
I
want to say, I want to choose a particular
00:14:51.910 --> 00:14:56.200
oblique shock problem I have to pick a
00:14:56.200 --> 00:15:00.270
.particular u 1 and a particular velocity
with which my observer is moving along the
00:15:00.270 --> 00:15:03.610
normal shock.
So, if I pick these 2, then I will get to
00:15:03.610 --> 00:15:10.100
a nice condition which will match my flow
problem of course,, will this work we are
00:15:10.100 --> 00:15:14.210
just going through some particular analytical
approach will it work, if it is not working,
00:15:14.210 --> 00:15:19.870
I would not be teaching you this currently.
That is the only proof we know. It works because
00:15:19.870 --> 00:15:26.640
gas dynamics people having using this
for a long time why do we need two variables.
00:15:26.640 --> 00:15:31.680
Now, previously it was just one variable
normal shock, I want to give strength I needed
00:15:31.680 --> 00:15:36.460
only one variable. That is the strength of
the shock p 2 by p 1 or m 1 a mark numbers
00:15:36.460 --> 00:15:41.590
of the shock anyone variable given. I can
give every other property for the shock. Now,
00:15:41.590 --> 00:15:46.540
we need two variables why is that. It is 2-d
one more dimensions extra. So, I have to give
00:15:46.540 --> 00:15:50.870
some information about the second
dimension. So, I need one more dimension.
00:15:50.870 --> 00:15:56.140
So, I need one more data point on that.
That is the reason, why we need two variables.
00:15:56.140 --> 00:16:03.340
Now, of course,, it is not very nice to
look for this V and u 1 as the two variables
00:16:03.340 --> 00:16:06.970
instead of that we have to come up with
better variables. We will go and look at that
00:16:06.970 --> 00:16:12.440
soon beta and theta could be nice variables
but they are related probably I should pick
00:16:12.440 --> 00:16:18.440
m 1 and theta or m 1 and be tight will work.
We will go and look at that as time goes.
00:16:18.440 --> 00:16:26.560
Now, we will start solving the problem as
if it is a fresh problem, we would not assume
00:16:26.560 --> 00:16:32.440
that, I can solve the problem by using normal
shock analogy with relative velocity with
00:16:32.440 --> 00:16:36.330
respect to observer added and then it will
work, because it is a very complex thing to
00:16:36.330 --> 00:16:41.230
find
that V value to fit my problem. Instead, we
00:16:41.230 --> 00:16:45.980
will just go and pick, this is just to give
you a
00:16:45.980 --> 00:16:48.550
feel for what is actually happening.
00:16:48.550 --> 00:16:49.550
..
00:16:49.550 --> 00:16:55.529
Will just go start with solving as if it is
some new problem in gas dynamics. I am going
00:16:55.529 --> 00:17:02.660
to say it is there is a shock which has angle
beta with respect to my reference line which
00:17:02.660 --> 00:17:14.430
could be my horizontal axis and there is an
incoming velocity u 1. This u 1 is different
00:17:14.430 --> 00:17:18.570
from the u 1, we have been discussing in the
previous page of the board alright this u
00:17:18.570 --> 00:17:22.170
1 is
now in 2-d plane it is parallel to my reference
00:17:22.170 --> 00:17:31.670
line, that is my u 1. Now, we already know
that we want to look at it somewhat related
00:17:31.670 --> 00:17:37.090
to normal shock. So, I have to decompose
this into normal complement and tangential
00:17:37.090 --> 00:17:45.560
component along the shock, we want to get
to these two components when I do that naturally
00:17:45.560 --> 00:17:51.180
this angle will become beta, because I
am looking for tangential and I am going to
00:17:51.180 --> 00:18:03.040
label this as my u tangential ideally. I should
call it u tangential one and then this my
00:18:03.040 --> 00:18:11.980
u 1, normal ideally it should be u 1 t. You
will
00:18:11.980 --> 00:18:18.990
see that u 1 t and u 2 t are the same.
Now, I will draw the other picture from u
00:18:18.990 --> 00:18:21.690
1 normal I know that there will be a small
u 2
00:18:21.690 --> 00:18:30.160
normal, because it is a normal shock based
thing and we said that, this tangential velocity
00:18:30.160 --> 00:18:36.210
is like velocity along the shockwave, which
is related to my the small v in the previous
00:18:36.210 --> 00:18:45.350
discussion. It is related to my moment of
my observer my reference frames, if you want
00:18:45.350 --> 00:18:48.950
to think about. So, this cannot change across
the shock.
00:18:48.950 --> 00:18:55.020
So, I will just keep the same value here that
is why I did not put it as u t 1, I will just
00:18:55.020 --> 00:19:03.750
remove it. Now, it is just u t u tangential
will keep, just this is my u 2 normal and
00:19:03.750 --> 00:19:14.910
u
tangential is going to be the same I think
00:19:14.910 --> 00:19:18.180
this is the same and this is your final velocity
00:19:18.180 --> 00:19:30.059
.vector u 2 well this is u 2 u t. I have drawn
it, such that this line goes straight through
00:19:30.059 --> 00:19:36.720
this. So, it looks as if the velocity vector
has shifted ideally I have to move this triangle
00:19:36.720 --> 00:19:40.190
such that, this vortex comes and touches this
point and move this triangle such that, the
00:19:40.190 --> 00:19:45.150
vector touches this point then only it is
exactly correct but it is just to give a feel
00:19:45.150 --> 00:19:47.230
for what
is happening, this is the same thing happening
00:19:47.230 --> 00:19:52.790
at every point along this shock. So, I could
say that, it is the same. Now, what is my
00:19:52.790 --> 00:19:56.440
theta angle? It is the angle with respect
to my
00:19:56.440 --> 00:20:01.120
original reference line in my 2-d problem
made.
00:20:01.120 --> 00:20:09.710
By the final velocity vector after the shock,
if I extend this line that u 2 line if I extend
00:20:09.710 --> 00:20:16.460
that is the theta I am talking about. So,
what can I tell about the relation between
00:20:16.460 --> 00:20:20.559
u 1 n
and u 1 that is just direct trigonometry here.
00:20:20.559 --> 00:20:31.320
It will be u 1 sin beta is my u 1, normal
u 1,
00:20:31.320 --> 00:20:45.220
sin beta and u t equal to you 1 cos beta.
Now, we want to write something more of
00:20:45.220 --> 00:20:58.570
course,, I can also say that m 1 normal equal
to u 1 normal by a 1, which is equal to u
00:20:58.570 --> 00:21:05.590
1
by a 1 which is m 1, m 1 sin beta. It just
00:21:05.590 --> 00:21:08.630
comes out to be that we will just keep it
there
00:21:08.630 --> 00:21:10.320
we would not.
Use this relation immediately we will use
00:21:10.320 --> 00:21:12.900
it after sometime. Now, I want to look at
this
00:21:12.900 --> 00:21:25.550
triangle again in little more detail I will
take this triangle draw it a little bigger
00:21:25.550 --> 00:21:28.630
I change
the angles. Let us say we would not worry
00:21:28.630 --> 00:21:48.030
about this change in angle from the actually
this is correct. Now, it may be close and
00:21:48.030 --> 00:21:52.420
we know that. This is our 90 degree angle
because this is normal and this is tangential
00:21:52.420 --> 00:21:58.650
this is your 90 degree angle.
Now, I want to relate this u 2 with this u
00:21:58.650 --> 00:22:03.990
2 normal, which means I should know either
this angle or this angle, what I currently
00:22:03.990 --> 00:22:07.390
know is this angle. This angle is known to
be
00:22:07.390 --> 00:22:17.070
theta, that is what I know and I also know
that this u tangential is parallel to my shock
00:22:17.070 --> 00:22:21.760
which means with respect to my reference line.
This is having an angle beta these are the
00:22:21.760 --> 00:22:27.870
things I know already. Now, I need to find,
let us say I want to find this angle. So that,
00:22:27.870 --> 00:22:33.481
I
will be get a sin theta relation. So, I want
00:22:33.481 --> 00:22:37.070
to find this angle, how I will find it? This
is
00:22:37.070 --> 00:22:43.300
some 8th standard trigonometry I think.
So, I am drawing parallel line to this original
00:22:43.300 --> 00:22:47.130
reference line. Now, you look at a slant line
with parallel line you know this angle will
00:22:47.130 --> 00:22:53.570
be beta right included opposite angle, I think
that is what it is called. So, this whole
00:22:53.570 --> 00:22:58.179
thing is beta and now I know there is one
more
00:22:58.179 --> 00:23:06.110
parallel line here, which means this angle
is theta, this triple thing angle is theta
00:23:06.110 --> 00:23:10.210
from
here to here while the whole thing is beta
00:23:10.210 --> 00:23:12.870
which means, this particular angle which I
am
00:23:12.870 --> 00:23:23.559
.talking about this particular angle. I am
talking about will become beta minus theta,
00:23:23.559 --> 00:23:25.490
that
is what it should come out to be this particular
00:23:25.490 --> 00:23:32.140
angle between u 2 vector and u t vector,
that is what I should get once, I get this
00:23:32.140 --> 00:23:34.890
triangle is very simple to solve. I can relate
any
00:23:34.890 --> 00:23:39.130
two vectors with the third one, because I
know trigonometry.
00:23:39.130 --> 00:23:49.330
Now, I can go underwrite u 2 normal is equal
to u 2 times sin of beta minus theta I
00:23:49.330 --> 00:23:53.700
wanted sin. So, I took this angle, if I wanted
cos I would have used this angle which will
00:23:53.700 --> 00:24:02.020
be 90 minus beta minus theta anyways. So,
it is going to be u 2 times sin of this angle
00:24:02.020 --> 00:24:06.940
will be this which is from trigonometry simple
steps. Now, u t I have another expression
00:24:06.940 --> 00:24:18.220
u 2 cos beta minus theta remember, this will
be used after sometime u t has 2 values but
00:24:18.220 --> 00:24:20.670
I
know u t is the same across both.
00:24:20.670 --> 00:24:27.160
So, I can link this u 1 cos beta and u 2 cos
of beta minus theta later we will do it, that
00:24:27.160 --> 00:24:33.070
is
one relation between beta and theta. So, if
00:24:33.070 --> 00:24:38.140
I want to find m 2 normal all. I have to do
is
00:24:38.140 --> 00:24:50.800
divide by a two similar to that, I will get
it to be m 2 cos m 2 sin m 2 sin beta minus
00:24:50.800 --> 00:25:00.710
theta
this is what I will get. Now, of course,,
00:25:00.710 --> 00:25:03.260
it is so tempting to just go and start using
normal
00:25:03.260 --> 00:25:09.610
shock tables with m 1 normal, because we know
it already because we said that only
00:25:09.610 --> 00:25:15.660
thing that matters across this u 1 normal
and u 1, u 2 normal with respect to my observer.
00:25:15.660 --> 00:25:20.630
It is just a normal shock flow which means
u 1 normal and u 2 normal are related by my
00:25:20.630 --> 00:25:25.929
normal shock. Let us say I do not want to
do that I want to analysis from scratch as
00:25:25.929 --> 00:25:30.380
if I
do not know normal shock relations and then
00:25:30.380 --> 00:25:35.340
after sometime will say yes, I know normal
shock relations, if I want to do that I have
00:25:35.340 --> 00:25:39.750
to start with mass conservation momentum
conservation energy conservation across the
00:25:39.750 --> 00:25:40.990
shock for a 2-d problem.
00:25:40.990 --> 00:25:41.990
..
00:25:41.990 --> 00:25:49.860
So, I will go start again with a shock at
angle beta. Now, I am going to pick a control
00:25:49.860 --> 00:25:57.711
volume like, this is my control volume, I
am picking assume it is a rectangle there
00:25:57.711 --> 00:26:01.549
seems
to be a bend there and. I am saying there
00:26:01.549 --> 00:26:06.799
is a velocity incoming, it is u 1 normal and
it
00:26:06.799 --> 00:26:15.670
also has a tangential velocity incoming u
t on the other side, there is u 2 normal outgoing,
00:26:15.670 --> 00:26:24.490
it is also has a tangential velocity u t outgoing,
I am drawing it differently. Now,
00:26:24.490 --> 00:26:27.470
previously I wanted to add vector. Now, I
want to show that at this point this is the
00:26:27.470 --> 00:26:36.940
vector velocity I want to solve this problem
by shifting my coordinate system to parallel
00:26:36.940 --> 00:26:42.300
and perpendicular to my shock.
So that, it is easy to solve, because I already
00:26:42.300 --> 00:26:46.411
know the velocity vectors in these
dimensions, it is easier to solve that. Let
00:26:46.411 --> 00:26:54.420
us say the area on this line for my control
volume is some area for the flow for the normal
00:26:54.420 --> 00:27:08.220
flow alone the other region. Let us say I
have a upstream and a downstream and I am
00:27:08.220 --> 00:27:14.470
picking a rectangle such that, a up the same
here and here and a down is same here and
00:27:14.470 --> 00:27:19.690
here, that is nice to look that particular
case.
00:27:19.690 --> 00:27:24.890
So that, your equation will becomes simpler.
Now, first thing I want to do is conservation
00:27:24.890 --> 00:27:33.370
of mass momentum energy along the shock
direction. First we will pick, that is along
00:27:33.370 --> 00:27:38.669
the tangential direction. So, let us pick
mass, if
00:27:38.669 --> 00:27:44.030
I pick mass conservation, if I pick along
this direction as it is, if it is a 1-d problem.
00:27:44.030 --> 00:27:46.850
So, I
can start using my 1-d relations rho, u is
00:27:46.850 --> 00:27:50.980
constant for my control volume. It is a big
control volume I am going to say rho u is
00:27:50.980 --> 00:27:54.500
constant. So, I have to find out how much
is
00:27:54.500 --> 00:27:59.720
.the mass entering through this section, how
much is leaving this section? This section,
00:27:59.720 --> 00:28:04.810
I
have to link all of them together. So, my
00:28:04.810 --> 00:28:20.970
mass expression will be rho 1 time u tangential
times a upstream plus rho 2 u tangential,
00:28:20.970 --> 00:28:31.179
a downstream these are the masses. That are
entering this way which is equal to rho 1,
00:28:31.179 --> 00:28:39.080
u tangential a upstream again there plus rho
2 u
00:28:39.080 --> 00:28:47.280
tangential a downstream. So, happens that
this is automatically satisfied always for
00:28:47.280 --> 00:28:51.500
any
value of u t it is a silly equation it looks
00:28:51.500 --> 00:28:56.500
like it is identically satisfied for any value
of u t
00:28:56.500 --> 00:29:03.840
that is all we see. So, u t cannot be governed
by this equation really it is automatically.
00:29:03.840 --> 00:29:09.030
Satisfied for any value of course,, this is
happening because I have pictured my control
00:29:09.030 --> 00:29:14.940
volume such that a upstream is the same for
both the sections here and here it is same
00:29:14.940 --> 00:29:17.530
as
down. It is same here and here if I picked
00:29:17.530 --> 00:29:21.180
something else, then I have to take into
account that there is a change and then it
00:29:21.180 --> 00:29:27.380
will become more complex problems. Let us
not worry about that. Now, I want to do momentum
00:29:27.380 --> 00:29:35.270
equation momentum conservation.
Now, momentum conservation I have to be worried
00:29:35.270 --> 00:29:39.450
about it is a 2-d problem. So, I have
to worry about two different momentums going
00:29:39.450 --> 00:29:45.580
along this way of course,, when it is
going along the shock, it can be my along
00:29:45.580 --> 00:29:49.559
the shock momentum that is going along the
shock or the normal momentum being carried
00:29:49.559 --> 00:29:55.169
by the mass flow, this way along the shock
both are possible. Let us say, we will pick
00:29:55.169 --> 00:30:01.540
the tangential momentum, first if I pick the
tangential momentum.
00:30:01.540 --> 00:30:12.890
That is u t being carried by the mass flow,
this is way, that will be rho 1 u t A up times
00:30:12.890 --> 00:30:16.510
u
t, this is my mass flow rate times momentum
00:30:16.510 --> 00:30:26.400
times velocity, because my momentum rate
of course,, I have to have this is normal
00:30:26.400 --> 00:30:28.500
component no. So, I should have a pressure
also
00:30:28.500 --> 00:30:37.570
along with this P 1. I will call this plus
on the other section P 1 A up sorry, A up
00:30:37.570 --> 00:30:41.530
is also
there with this, I am looking at only this
00:30:41.530 --> 00:30:50.350
section and plus this downstream section P
2 a
00:30:50.350 --> 00:31:07.049
down plus rho 2 u t A down u t this is all
the momentum entering this way. Now, this
00:31:07.049 --> 00:31:09.040
is
to be equal to the momentum leaving on the
00:31:09.040 --> 00:31:12.059
other side, which is equal to if I do the
same
00:31:12.059 --> 00:31:18.620
thing here. I will get exactly the same relations
a up P 1, because it is still on the same
00:31:18.620 --> 00:31:28.860
side of the shock plus rho 1 u t A up is my
mass flow rate multiplied by u t again plus,
00:31:28.860 --> 00:31:38.971
the other section exactly the same thing will
come up again P 2 A down plus rho 2 u t A
00:31:38.971 --> 00:31:46.160
down times u t it.
So, happens that, this expression is also
00:31:46.160 --> 00:31:52.960
identically satisfied, tangential momentum
equation is identically satisfied. Let us
00:31:52.960 --> 00:32:00.140
look at normal momentum, this the u 1 normal
being carried by the mass flow going this
00:32:00.140 --> 00:32:04.350
way, that is also happening if I do that since
it
00:32:04.350 --> 00:32:09.370
.is this way momentum, the pressure term will
not be there in that, it will just be without
00:32:09.370 --> 00:32:20.160
pressure normal momentum equation. There will
be no P 1 related and there is no shear
00:32:20.160 --> 00:32:23.330
stress we assumed it to be 0.
So, there is no other stress on this area,
00:32:23.330 --> 00:32:32.610
it will just be rho 1u t A up times u 1 normal
plus
00:32:32.610 --> 00:32:44.500
rho 2 u t A down times u 2 normal, these are
the net to momentum of normal momentum
00:32:44.500 --> 00:32:52.299
being carried by this mass in this way, This
is equal to on the other end, again you will
00:32:52.299 --> 00:33:10.320
get the same thing A up times u 1 normal plus
rho 2 u t a down times u 2 normal again it
00:33:10.320 --> 00:33:15.429
comes up that, it looks like some expression
equal to the same expression, which means
00:33:15.429 --> 00:33:21.440
it is again identically satisfied. So, we
are coming up with some set of expressions
00:33:21.440 --> 00:33:24.780
which
looks as if, the tangential components will
00:33:24.780 --> 00:33:29.640
work irrespective of whatever u t I pick.
Now,
00:33:29.640 --> 00:33:39.600
let us I do not want to do the energy again
you will again see that m dot times h if I
00:33:39.600 --> 00:33:41.350
put h
plus u square.
00:33:41.350 --> 00:33:45.330
That will again give exactly the same kind
of result. So, I do not want to do it. Let
00:33:45.330 --> 00:33:50.190
us say
that tangential component transfer or transport
00:33:50.190 --> 00:33:56.230
along tangential component, it is not very
useful for us. It is not giving me any governing
00:33:56.230 --> 00:33:59.890
equation really, it is just telling me a
equal to a or b equal to b something like
00:33:59.890 --> 00:34:07.740
that not very useful. Let us go to the normal
direction. The same picture, I am going to
00:34:07.740 --> 00:34:09.700
solve for normal direction across area A.
.
00:34:09.700 --> 00:34:23.050
Now, I will get rho 1 u 1 normal times A equal
to, this is my mass conservation rho 2 u 2
00:34:23.050 --> 00:34:28.700
normal times A. This is one expression yes;
this is very different it is not A equal to
00:34:28.700 --> 00:34:29.700
A
00:34:29.700 --> 00:34:41.159
.kind of relation. So, we need to keep it,
next one normal momentum component. It
00:34:41.159 --> 00:34:46.800
is
going to have P 1 times a plus rho 1 u 1 normal
00:34:46.800 --> 00:34:53.700
A times u 1 normal equal to again I am
using 1-d relations, because I am having flow,
00:34:53.700 --> 00:34:59.211
only along I am considering only that
side. We have decoupled the problem P 2 A
00:34:59.211 --> 00:35:11.190
plus rho 2 u 2 normal A times u 2 normal.
This is what, I have here does not seem like
00:35:11.190 --> 00:35:12.620
anything will get simplified, we will keep
it
00:35:12.620 --> 00:35:22.150
as is tangential momentum
being transported along the normal direction,
00:35:22.150 --> 00:35:29.849
that is going to
look like rho 1 u 1 normal A times, the tangential
00:35:29.849 --> 00:35:43.160
velocity this is going to be equal to rho
2 u 2 normal A times the tangential velocity.
00:35:43.160 --> 00:35:54.260
Now, in here I could have assumed that u t
1 is different from u t 2, if I assumed that
00:35:54.260 --> 00:35:56.230
then
all these expressions none of them will go
00:35:56.230 --> 00:36:00.430
to 0 actually I should go back here. So, if
I had
00:36:00.430 --> 00:36:09.450
gone and said that u t 1 is different from
u t 2, if I ever said that then I will have
00:36:09.450 --> 00:36:10.480
a
situation where none of these expressions
00:36:10.480 --> 00:36:14.570
will become identically 0, actually even then
it
00:36:14.570 --> 00:36:17.910
will all become identically 0. So, I do not
need to worry about this whole set of
00:36:17.910 --> 00:36:23.050
expressions tangential transport is always
0. But, even if I said u t 1 is different
00:36:23.050 --> 00:36:24.599
from u t
2.
00:36:24.599 --> 00:36:32.111
I will go here now tangential momentum where,
I will get u t 1 here u t 2 here. Now, if
00:36:32.111 --> 00:36:34.780
I
go and use the mass relation inside here,
00:36:34.780 --> 00:36:40.420
we have put rho 2 u 2 normal A is equal to
rho
00:36:40.420 --> 00:36:49.099
1, u 1 normal A. Then I will get rho 1 u 1
normal A times u t 2, you will get from here
00:36:49.099 --> 00:36:56.640
that u t 1 equal to u t 2 that is what you
all get finally, but if I do not do that,
00:36:56.640 --> 00:37:00.170
if I already
know that u t is constant, because I am saying
00:37:00.170 --> 00:37:05.700
it is the velocity with which my observer
is moving along my normal shock. Once I say
00:37:05.700 --> 00:37:10.680
that u t is the same, that is what I have
used and that is the understanding.
00:37:10.680 --> 00:37:16.340
I used to solve this if I did not use that
then I will get u t 1 equal to u t 2 at this
00:37:16.340 --> 00:37:21.730
point.
Now, since I used it this equation, just becomes
00:37:21.730 --> 00:37:25.360
mass equation. It is not anything new, it
is a dependent equation, it is in fact the
00:37:25.360 --> 00:37:28.410
same equation multiplied by u t, it is not
giving
00:37:28.410 --> 00:37:50.160
anything new will go to energy equation h
1 plus half u 1 normal square 2 plus half
00:37:50.160 --> 00:37:55.771
u 2
normal square. Now, is this correct or not
00:37:55.771 --> 00:37:59.390
really, because kinetic energy when I call
it
00:37:59.390 --> 00:38:05.340
kinetic energy it does not know direction.
So, I cannot just write you 1 normal square
00:38:05.340 --> 00:38:08.240
or
u 2 normal squares alone. I should also take
00:38:08.240 --> 00:38:13.960
into account u t square that is what, it should
be really.
00:38:13.960 --> 00:38:20.450
.So, this is not really true but wait a minute
will keep this expression but I will write
00:38:20.450 --> 00:38:25.360
the
more correct expression which is half u 1
00:38:25.360 --> 00:38:32.520
square, half u 2 square I will write it this
way,
00:38:32.520 --> 00:38:38.400
this is the perfect equation, this is correct
as long as there is no heat transfer across
00:38:38.400 --> 00:38:40.900
there
is no work done by the fluid and those kind
00:38:40.900 --> 00:38:44.500
of other expectations. This is correct, this
is
00:38:44.500 --> 00:38:50.560
your basic h naught 1 equals to h naught 2
relations which we had in normal shock now,
00:38:50.560 --> 00:38:57.290
since I know from before that u t 1 equal
to u t 2, when I decompose this expression
00:38:57.290 --> 00:39:02.180
u 1
square u, u will become u square will become
00:39:02.180 --> 00:39:07.780
u t square plus u normal square of course,,
I can put subscript 1 or 2 for both, it will
00:39:07.780 --> 00:39:12.030
work that is why I did not put 1 or 2, if
I do that
00:39:12.030 --> 00:39:19.099
and I say that u t 1 equal to u t 2. This
expression reduces to this expression.
00:39:19.099 --> 00:39:22.060
This is the more correct expression h naught
1 equal to h naught 2, this is the more
00:39:22.060 --> 00:39:27.750
correct expression it. So, it happens that,
this becomes this, If I use u t 1 equal to
00:39:27.750 --> 00:39:33.430
u t 2.
So, we will keep this expression also. Now,
00:39:33.430 --> 00:39:45.650
we want to say something special, I want to
say that, if I pick equation is 1, 2 and 3,
00:39:45.650 --> 00:39:51.430
if I pick these three equations, they look
exactly
00:39:51.430 --> 00:39:58.900
same as what we had for stationary normal
shock some 8 or 9 classes before, when we
00:39:58.900 --> 00:40:05.510
derived the normal shock, we had this exact
same set of equations without the subscript
00:40:05.510 --> 00:40:11.700
n, that is the only change, that is why I
wanted to write this expression here.
00:40:11.700 --> 00:40:17.520
Ideally it should be written this way and
with this, it should become this, I wanted
00:40:17.520 --> 00:40:19.900
to
write it like this. So that, I can just say
00:40:19.900 --> 00:40:23.600
1, 2 and 3 directly, since I am getting exactly
the
00:40:23.600 --> 00:40:30.970
same differential equations the solution should
be exactly the same. It is not differential
00:40:30.970 --> 00:40:34.740
equation sorry; differential equation can
have different solutions based on boundary.
00:40:34.740 --> 00:40:42.369
This is algebraic expression will give exactly
the same answer, which means I do not
00:40:42.369 --> 00:40:46.280
need to go and solve this whole thing and
spend two more extra lectures for getting
00:40:46.280 --> 00:40:49.870
to
you, m 2 to m 1 relation it is going to be
00:40:49.870 --> 00:40:55.791
the exactly same derivation.
I will just skip that part and say that, if
00:40:55.791 --> 00:41:03.040
I put you 1 normal alone in my normal shock
calculations, I will get that particular set
00:41:03.040 --> 00:41:06.820
of solutions directly from there. That is
the
00:41:06.820 --> 00:41:12.320
important aspect we have to just convert to
normal component from original coordinates
00:41:12.320 --> 00:41:16.070
and then we will start solving the whole problem.
If I go to that picture, it is easier to
00:41:16.070 --> 00:41:21.859
look at. Not this picture. I think I will
take this picture, if I go to this picture,
00:41:21.859 --> 00:41:25.450
if I am
given this u 1 and this beta. Let us say,
00:41:25.450 --> 00:41:28.500
and then if I want to find P 2 by p 1 across
the
00:41:28.500 --> 00:41:35.849
shock, all I have to do is go find the u 1
normal after that, if I go to the equations
00:41:35.849 --> 00:41:38.720
it will
look like normal shock equations. So, I will
00:41:38.720 --> 00:41:41.080
get my solution same as normal shock tables
00:41:41.080 --> 00:41:46.070
.solutions. I will go find the m 1 normal
from there, I will go find the m 2 normal
00:41:46.070 --> 00:41:48.100
actually
I do not need m 2 normal currently, we wanted
00:41:48.100 --> 00:41:52.670
only P 2 by P 1.
So, from m 1 normal in my normal shocks stationary
00:41:52.670 --> 00:41:56.270
normal shock tables, I will get P 2
by P 1 across and that is my solution for
00:41:56.270 --> 00:42:02.570
the problem. If I wanted the velocity direction,
then I have to go find u 2 normal and find
00:42:02.570 --> 00:42:05.589
this u t. Put this triangle complete it and
get
00:42:05.589 --> 00:42:12.940
the V 2 or of course, you can cheat and say
I will go use this function. I will go use
00:42:12.940 --> 00:42:15.310
this
function and I will get my u 2 from beta and
00:42:15.310 --> 00:42:18.200
theta I can get the relations anyway I want
I
00:42:18.200 --> 00:42:26.810
can do. It is all the same it is all trigonometrics.
So, we have different ways of looking at this
00:42:26.810 --> 00:42:34.140
problem and we found that t naught 2 by t
naught 1 is equal to 1 even for an oblique
00:42:34.140 --> 00:42:37.230
shock even if there is extra velocity there
on
00:42:37.230 --> 00:42:41.870
the tangential component, why is that ? It
is because the tangential velocity is the
00:42:41.870 --> 00:42:45.000
same
before and after, that is the special reason
00:42:45.000 --> 00:42:48.260
P naught 2 by P naught 1 will be less than
1
00:42:48.260 --> 00:42:54.130
which you will see, that is similar to normal
shock P naught 2 by P naught 1, that is what
00:42:54.130 --> 00:43:06.221
you will get. Now, you want to start solving
for beta, theta in terms of each other. If
00:43:06.221 --> 00:43:10.050
I
ever go and write an expression for P 2 by
00:43:10.050 --> 00:43:16.350
P 1 for an oblique shock, how will I write?
It
00:43:16.350 --> 00:43:24.589
is not very difficult to write, it is going
to be the same expression as in normal shock.
00:43:24.589 --> 00:43:25.589
.
00:43:25.589 --> 00:43:38.030
So, P 2 by P 1 equal to 1 plus 2 gamma by
gamma plus 1 times m 1 square minus 1 this
00:43:38.030 --> 00:43:43.350
was one relation just check whether this relation
is correct. I think it is correct. I do not
00:43:43.350 --> 00:43:50.650
have that expression here. So, here instead
of it being m 1 all I have to do is make it
00:43:50.650 --> 00:43:52.380
m 1
00:43:52.380 --> 00:43:59.820
.normal, then suddenly I solve this problem
to get P 2 by P 1 all I have to do is this
00:43:59.820 --> 00:44:02.900
but
will I know my m 1 normal already that is
00:44:02.900 --> 00:44:11.530
related to my beta.
We said that m 1 normal is equal to m 1 sin
00:44:11.530 --> 00:44:14.920
beta. So, as long as I do not know my beta
I
00:44:14.920 --> 00:44:21.089
cannot tell my P 2 by P 1 if beta is 90 degrees
of course,, I will get the same thing then
00:44:21.089 --> 00:44:26.220
the shock becomes normal shock. It is perpendicular
to the flow direction and u
00:44:26.220 --> 00:44:34.760
tangential becomes 0 naturally. So, to solve
these, I need to basically it is coming down
00:44:34.760 --> 00:44:39.470
to I need to give you m 1 and beta for me
to solve the problem, which is what we said
00:44:39.470 --> 00:44:45.420
sometime back, I need 2 variables to define
this problem fully. Now, once I know this
00:44:45.420 --> 00:44:48.360
I
can find theta ideally. I should be able to
00:44:48.360 --> 00:44:51.360
find theta how will I do it? Let us say we
will
00:44:51.360 --> 00:44:59.810
go back to this board.
If I am given m 1 and beta and of course,
00:44:59.810 --> 00:45:03.950
I am given all the properties in state 1 then
I
00:45:03.950 --> 00:45:09.610
can say that, I can find m 1 and beta from
u 1 and beta. Once I know m 1 and beta, I
00:45:09.610 --> 00:45:11.720
can
find u 1 normal or m 1 normal from there,
00:45:11.720 --> 00:45:18.910
I will get m 2 normal. I already know u
tangential, because I am given beta u 1 cos
00:45:18.910 --> 00:45:25.100
beta will be my u tangential. It is the same
value here, which means I know u t here, once
00:45:25.100 --> 00:45:31.121
I know u t here and u 2 normal.
Here, I will know the final vector u 2 and
00:45:31.121 --> 00:45:35.589
I have u 2 normal already, if I have these
two I
00:45:35.589 --> 00:45:41.940
will go to this expression u 2 normal equal
to u 2 times sin of beta minus theta in this.
00:45:41.940 --> 00:45:44.210
I
know u 2 normal u 2. I know beta only thing
00:45:44.210 --> 00:45:51.580
I do not know is theta. So, I can get theta
from here but this is such a big process to
00:45:51.580 --> 00:45:54.150
get to theta. So, instead we want to write
an
00:45:54.150 --> 00:46:00.790
expression for directly theta equal to some
big expression in terms of m 1 and beta that
00:46:00.790 --> 00:46:04.290
is
the goal for us. The next few moments, we
00:46:04.290 --> 00:46:10.849
are trying to get to that point.
So, the way to do it we already said that
00:46:10.849 --> 00:46:23.000
u t is equal to u 1 cos beta from the downstream
section. It is equal to u 2 cos of beta minus
00:46:23.000 --> 00:46:31.310
theta, we know this already. Now, I am going
to use the other expression from the next
00:46:31.310 --> 00:46:39.710
page again which was u 1 in terms of u 1
normal. Now, I am going to use just these
00:46:39.710 --> 00:46:45.290
2 as my expression u 1 in terms of u 1 normal
will be u 1 normal divided by sin beta multiplied
00:46:45.290 --> 00:46:55.930
by this cos beta equal to again u 2
normal u 2 in terms of u 2 normal will be
00:46:55.930 --> 00:47:06.810
u 2 normal by sin beta minus theta multiplied
by this cos beta cos beta minus theta. I have
00:47:06.810 --> 00:47:20.470
this expression from here I can write u 1
normal by u 2 normal equal to this tan beta.
00:47:20.470 --> 00:47:23.980
This will become tan beta on the
denominator which I will take it to that site
00:47:23.980 --> 00:47:37.300
become tan beta by tan of beta minus theta,
this is one relation between u 1 normal and
00:47:37.300 --> 00:47:38.300
u 2 normal.
00:47:38.300 --> 00:47:47.800
.We know one more, what is that in terms of
m 1 normal sorry, the normal shock relation
00:47:47.800 --> 00:47:52.100
but let us say we would not write the full
expression. Now, I will just call it this
00:47:52.100 --> 00:47:56.900
is equal
to rho 2 by rho 1 which is true from mass
00:47:56.900 --> 00:48:00.800
equation in the normal direction. It will
be true
00:48:00.800 --> 00:48:07.200
you are going to get to this form. Of course,
I can write in terms of m 1 which is our old
00:48:07.200 --> 00:48:15.320
formula gamma plus 1 times. It should be m
1 square but in our case it is m 1 square
00:48:15.320 --> 00:48:19.660
sin
square beta. I have used m 1 normal square
00:48:19.660 --> 00:48:28.300
and I have substituted in terms of m 1. This
divided by 2 plus gamma minus 1 times m 1
00:48:28.300 --> 00:48:38.190
square again I am going to put m 1 sin beta
square m 1 square sin square beta, this is
00:48:38.190 --> 00:48:47.940
what I get.
Now, what I have to do is take this equal
00:48:47.940 --> 00:48:53.990
to this and manipulate it such that, I will
get tan
00:48:53.990 --> 00:49:00.520
theta equal to something I have to expand
this and make it tan theta equal to something
00:49:00.520 --> 00:49:04.490
instead of directly starting with this. I
want to write this expression for so many
00:49:04.490 --> 00:49:08.440
times. I
will just put rho 2 by rho 1 for a few seconds
00:49:08.440 --> 00:49:09.970
and then we will go on substitute this to
get
00:49:09.970 --> 00:49:16.680
to some big expressions. I would not finish
this today I will go to a point where rho
00:49:16.680 --> 00:49:19.680
2 by
rho 1, if I substitute I will get a good answer
00:49:19.680 --> 00:49:27.260
I would not substitute rho 2 by rho 1 today.
.
00:49:27.260 --> 00:49:44.710
So, I have tan beta by tan of beta minus theta
equal to rho 2 by row 1. Now, you should
00:49:44.710 --> 00:49:50.780
know my expansion for tan of beta minus theta
is equal to tan beta minus tan theta
00:49:50.780 --> 00:49:59.869
divided by 1 plus tan beta tan theta. You
should know this if I use that expansion,
00:49:59.869 --> 00:50:04.030
then I
am going to get to a form where this is equal
00:50:04.030 --> 00:50:13.220
to tan beta multiplied by 1 plus tan beta
tan
00:50:13.220 --> 00:50:23.200
theta divided by tan beta minus tan theta.
00:50:23.200 --> 00:50:31.630
.Now, what I want to do is club this denominator
with this tan beta take it to my left-hand
00:50:31.630 --> 00:50:38.650
side of this expression just this equal to
this expression and I do that I will end up
00:50:38.650 --> 00:50:47.570
with
rho 2 by rho 1 multiplied by 1 minus tan theta
00:50:47.570 --> 00:51:02.940
by tan beta and this is equal to 1 plus tan
beta tan theta, this is what I will have.
00:51:02.940 --> 00:51:05.740
Now, I want it tan theta equal to something
I have
00:51:05.740 --> 00:51:09.970
to group all the tan theta terms together.
So, I will take this tan theta term to the
00:51:09.970 --> 00:51:12.970
other
side and I will write tan theta terms on left
00:51:12.970 --> 00:51:19.050
of my new equation tan theta multiplied by
it
00:51:19.050 --> 00:51:26.430
is going to be tan beta in here, tan beta
this minus sign when it goes there.
00:51:26.430 --> 00:51:39.420
It will become plus. So, plus rho 2 by rho
1 times 1 by tan beta, this is 1 term. Now,
00:51:39.420 --> 00:51:42.000
I
will take this 1 to the other side, this is
00:51:42.000 --> 00:51:48.119
equal to minus 1 here and rho 2 by rho 1 here,
rho
00:51:48.119 --> 00:51:59.130
2 by rho 1 minus 1, this will become i expression.
So, I will end up with tan theta equal
00:51:59.130 --> 00:52:21.600
to rho 2 by rho 1 minus 1 times tan beta times
tan beta divided by tan square beta plus
00:52:21.600 --> 00:52:34.820
rho 2 by row 1. Now, I have an expression
for tan theta which is from which, if I invert
00:52:34.820 --> 00:52:42.410
tan, I will get theta in terms of rho 2 by
rho 1 and beta where. Now, my rho 2 by rho
00:52:42.410 --> 00:52:47.640
1
depends on my m 1 and beta m 1 sin beta. It
00:52:47.640 --> 00:52:54.570
is related to m 1 normal. So, next class we
will substitute that m 1 normal in here and
00:52:54.570 --> 00:52:59.930
simplify this expression everything in terms
of m 1 which is the original mark number and
00:52:59.930 --> 00:53:07.500
beta. So, we will continue that part and it
will give you a very nice expression will
00:53:07.500 --> 00:53:10.609
do that next class. See you people next class.
00:53:10.609 --> 00:53:11.609
.