WEBVTT
Kind: captions
Language: en
00:00:10.040 --> 00:00:17.480
Hello everyone welcome back. We were looking
last time about variation of various
00:00:17.480 --> 00:00:23.750
properties across a normal shock. And we were
looking at variation as a function of
00:00:23.750 --> 00:00:24.750
mock number.
.
00:00:24.750 --> 00:00:33.030
We will go to the screen, there where we saw
that, if I look at downstream mock number
00:00:33.030 --> 00:00:38.730
as a function of upstream mock number for
a static or normal shock that is stationary.
00:00:38.730 --> 00:00:44.800
Then as I increase my mock number as in the
incoming flow mock number with respect
00:00:44.800 --> 00:00:51.680
to the shock, if that is increasing we find
that the downstream mock number decreases
00:00:51.680 --> 00:00:57.360
from close to 1 to something very low. But,
as we go more and more downstream, as we
00:00:57.360 --> 00:01:02.290
go to very high mock numbers, we are finding
that the downstream mock number is
00:01:02.290 --> 00:01:07.100
tending to become a constant. It is becoming
asymptotic value of some different values
00:01:07.100 --> 00:01:13.040
for different gamma values, which we have
and if you look at u 2 by u 1, again that
00:01:13.040 --> 00:01:16.720
is
also going to a constant in each of these
00:01:16.720 --> 00:01:20.380
cases. So, we can find the exact asymptote.
00:01:20.380 --> 00:01:21.380
..
00:01:21.380 --> 00:01:32.320
If we look at analysis and if I look at t
2 by p t 1 or p 2 by p 1, the pressure ratio
00:01:32.320 --> 00:01:35.210
or the
temperature ratio, they are not going to any
00:01:35.210 --> 00:01:39.020
asymptote as we go to high enough mock
numbers.
00:01:39.020 --> 00:01:40.020
.
00:01:40.020 --> 00:01:44.800
And of course, p naught 2 by p naught on it
looks like it is going to asymptote. If I
00:01:44.800 --> 00:01:47.350
go
and look at it in long scale, it looks like
00:01:47.350 --> 00:01:50.369
it is continuously dropping. It is not going
to an
00:01:50.369 --> 00:01:51.369
asymptote.
00:01:51.369 --> 00:01:52.369
..
00:01:52.369 --> 00:01:55.660
Again that will be very clear if, I go take
delta s by or which is just log of p naught
00:01:55.660 --> 00:01:58.770
2 by
p naught 1. If you look at that it is continuously
00:01:58.770 --> 00:01:59.770
increasing.
.
00:01:59.770 --> 00:02:05.979
So, if I go back and look at the previous
one, where it is p naught 2 by p naught 1
00:02:05.979 --> 00:02:08.220
it is
dropping continuously. If I take log of these
00:02:08.220 --> 00:02:10.060
it will be some negative value, because it
is
00:02:10.060 --> 00:02:16.760
less than 1 in the axis. So, we have delta
s by r is negative of this, log of p naught
00:02:16.760 --> 00:02:19.269
2 by p
naught 1 is delta s by r.
00:02:19.269 --> 00:02:21.159
.
So, you are getting and you are seeing that
00:02:21.159 --> 00:02:23.730
it is continuously increasing. It is not going
to
00:02:23.730 --> 00:02:25.319
any asymptotic value.
.
00:02:25.319 --> 00:02:31.629
Now, will go back to the very first one M
2 alone it seems that it is going to a constant.
00:02:31.629 --> 00:02:36.930
If
I look at mock numbers if I go above 5 or
00:02:36.930 --> 00:02:42.879
so the downstream mock number across a
normal shock is almost the same value always
00:02:42.879 --> 00:02:50.480
irrespective of what my incoming mock
flow is. This is the resume where we call
00:02:50.480 --> 00:02:59.489
hypersonic, were things will go almost the
same. If I go above mock 7 that is more valid,
00:02:59.489 --> 00:03:04.220
it is better if it is mock 7 and above, you
want you can call it above 5 itself that it
00:03:04.220 --> 00:03:07.450
is doing that, and u 2 by u 2 is almost a
constant
00:03:07.450 --> 00:03:11.709
.after that point. They can actually find
the value for high mock number approximation
00:03:11.709 --> 00:03:18.109
which is what we are going to start with today.
So, now we will go to the board and find the
00:03:18.109 --> 00:03:19.989
asymptotic values for this.
.
00:03:19.989 --> 00:03:29.109
So, we are thinking about incoming mock number
tending to infinity that is what we are
00:03:29.109 --> 00:03:34.829
looking for. And we have this expression,
M 2 square we have derived all this normal
00:03:34.829 --> 00:03:55.250
shock relations, I am just writing, we are
having this. Now you want to say if I set
00:03:55.250 --> 00:03:58.959
M1
tending to infinity. M1 is being very large,
00:03:58.959 --> 00:04:01.939
then this will tend to I can neglect this
2 with
00:04:01.939 --> 00:04:07.999
respect to infinity, I will just ignore this
2. So, it will tend to something like gamma
00:04:07.999 --> 00:04:14.810
minus 1 M1 square divided by the denominator
again this M 1 square is very high
00:04:14.810 --> 00:04:19.199
compare this gamma minus 1, which is almost
like 0.4 0.3 that kind of numbers. I can
00:04:19.199 --> 00:04:26.620
neglect this compare to M1 square. So, it
will become 2 gammas M 1 square and we find
00:04:26.620 --> 00:04:31.730
that this is independent of mock number that
is why you are going to constant value of
00:04:31.730 --> 00:04:41.651
course, this is what you are getting.
Now, if I want to find the asymptotic value
00:04:41.651 --> 00:04:50.440
for various gamma values, let us say I put
gamma equal to 1.4. This is actually remember
00:04:50.440 --> 00:05:09.580
it is M 2 square, M 2 square tends to 0.4
by 2.8, which is 4 by 28, 1 by 7. So, I pick
00:05:09.580 --> 00:05:20.820
find my M 2, it is 1 by root 7, which is 0.378,
that is the value which our plot would have
00:05:20.820 --> 00:05:22.730
asymptotic to, if you go back and look at
the
00:05:22.730 --> 00:05:28.840
plot it will look like that but, it is not
go to the plot anyway you know it is true.
00:05:28.840 --> 00:05:42.050
.If I look at gamma equal to 1.66, which is
for more atomic gases, I can say this is exactly
00:05:42.050 --> 00:05:47.180
equal to 5 by 3, it is not really 1.66, it
is actually 1.66666 bar, if you want to think
00:05:47.180 --> 00:05:50.920
about
it right irrational numbers anyways. We will
00:05:50.920 --> 00:05:59.660
just keep it as 5 by 3, it is easier to work
with. Then I am going to say my M 2 square
00:05:59.660 --> 00:06:12.250
will become 2 by 3 divided by 10 by 3
which is 2 by 10, 1 by 5. So, you are going
00:06:12.250 --> 00:06:25.530
to get 1 by root 5 as your mock number for
that case again. So, M 2 becomes 0.447, this
00:06:25.530 --> 00:06:33.200
is what you are going to get. If I look at
gamma equal to 1.4, to gamma equal to 1.66,
00:06:33.200 --> 00:06:39.650
gamma equal to 1.4 gas will have lesser
speed of sound, which I said, in a way is
00:06:39.650 --> 00:06:46.310
related to more compressible gas. If I think
about it that way more compressible gas will
00:06:46.310 --> 00:06:52.390
go to much lower mock number than a less
compressible gas.
00:06:52.390 --> 00:06:57.460
Just a feeling you are supposed to get developed
out of this course. If it is more
00:06:57.460 --> 00:07:02.530
compressible, it will be more useful if you
are going for serious calculations with gamma
00:07:02.530 --> 00:07:06.920
changing during your flow. But, we are not
going to deal with that in our whole course,
00:07:06.920 --> 00:07:08.810
I
will just show you what happens if I change
00:07:08.810 --> 00:07:11.490
gamma that is all and we will let go here
if
00:07:11.490 --> 00:07:15.030
you go to in any case, if there is some course
if you want to have a course called high
00:07:15.030 --> 00:07:21.000
temperature gas dynamics, there we will think
about variation of gamma which
00:07:21.000 --> 00:07:24.480
temperature. If that is the case, then I will
have to worry about what happens to the
00:07:24.480 --> 00:07:30.490
compressibility as the gamma changes. This
is just give you a little feel for it. If
00:07:30.490 --> 00:07:32.300
you ever
have a course on high temperature gas dynamics,
00:07:32.300 --> 00:07:35.480
that will give you a lot more feel for it.
I will just leave it there if you want you
00:07:35.480 --> 00:07:48.159
can go read up more on high-temperature gas
dynamics. What will happen to pressure ratio?
00:07:48.159 --> 00:07:49.159
..
00:07:49.159 --> 00:07:55.780
We had this formula very easy to remember
one. So, we will just keep it this way, these
00:07:55.780 --> 00:08:02.300
is one of the forms of P 2 by P 1 we had.
If I, look at this, if I say M 1 is very big
00:08:02.300 --> 00:08:07.290
number
I am trying to go to asymptote. So, P 2 by
00:08:07.290 --> 00:08:18.060
P 1 for large, M1 will tend to I will neglect
this 1 with respect to M square, I will neglect
00:08:18.060 --> 00:08:19.950
this 1 also with respect to M square. So,
it
00:08:19.950 --> 00:08:26.870
is just going to become 2 gamma by gamma plus
1 M 1 square, if M 1 is very high P 2
00:08:26.870 --> 00:08:31.950
by P 1 is also very high. It is 1 M not independent
of mock number, it increases if I keep
00:08:31.950 --> 00:08:37.579
on increasing mock number. So, it is not going
to a constant that is all. So, it is not going
00:08:37.579 --> 00:08:43.649
to asymptote which is what we saw in our plot
also. Now, I will go rho 2 by rho 1
00:08:43.649 --> 00:08:51.829
density ratio rho 2 by rho 1, density ratio.
We had one more nice form easy to earth quiet
00:08:51.829 --> 00:09:03.800
in this case of course I have all ready tabulated
all the various formulae which you can
00:09:03.800 --> 00:09:07.420
us. I am just picking one of them which will
be very comfortable for this analysis.
00:09:07.420 --> 00:09:12.579
So, I am going to take this 1, and I am going
to neglect this 2 with respect to this M 1
00:09:12.579 --> 00:09:17.120
square of course, product multiplying by gamma
minus 1 should keep we would not
00:09:17.120 --> 00:09:22.009
remote that. So, this comes down too. I will
just neglect this 2 and if I neglected this
00:09:22.009 --> 00:09:24.620
2,
this looks like M 1 square will get cancelled
00:09:24.620 --> 00:09:28.370
and I will just have gamma plus 1 by
gamma minus 1. So, this tends to gamma plus
00:09:28.370 --> 00:09:34.720
1 by gamma minus 1, this is all you are
getting. Now will go and look at this is again,
00:09:34.720 --> 00:09:39.700
this says that this is independent of mock
number by rho 2 by rho 1 goes to a particular
00:09:39.700 --> 00:09:46.630
asymptotic value. We did not have a plot
for rho 2 by rho 1 specifically in the beginning
00:09:46.630 --> 00:09:48.830
of the class. But, if you look at this rho
2
00:09:48.830 --> 00:09:55.370
by rho 1 is same as u 1 by u 2, if rho 2 by
rho 1, goes to a constant u 1 by u 2 goes
00:09:55.370 --> 00:09:56.370
to a
00:09:56.370 --> 00:10:01.329
.constant, which means u 2 by u 1 also goes
to a constant, which is what we plotted and
00:10:01.329 --> 00:10:04.800
we showed that u 2 by u 1 goes to a constant,
anyways.
00:10:04.800 --> 00:10:11.370
Now, we will go look at what this number looks
like for different gamma values. If I put
00:10:11.370 --> 00:10:23.879
gamma equal to 1.4, 2.4 by 4, that is going
to be 6, if I go to gamma equal to 5 by 3
00:10:23.879 --> 00:10:32.449
which is officially 1.6666 and we will just
keep it that way to rho 2 by rho 1 will be
00:10:32.449 --> 00:10:40.029
8 by
3 divided by 2 by 3 that will be 4. What are
00:10:40.029 --> 00:10:44.649
we seeing here? I already told you when
gamma decreases it is more compressible gas.
00:10:44.649 --> 00:10:50.999
So, what we are seeing is, if I have an
infinite strength shock then I can compress
00:10:50.999 --> 00:10:58.089
the gas to 6 times its original density, While
if it is a less compressible gas, it is going
00:10:58.089 --> 00:11:01.170
to go lesser compression ratio, cannot go
to
00:11:01.170 --> 00:11:06.059
very high compression ratio, this is why I
am, saying gamma is in a way measure of
00:11:06.059 --> 00:11:11.160
compressibility. In a way, I will not tell
it is exact as of now, it is slightly in a
00:11:11.160 --> 00:11:13.680
way it is
correct, it is not always true.
00:11:13.680 --> 00:11:19.630
There is also molecular factor inside, if
I want to think about speed of sound. But,
00:11:19.630 --> 00:11:24.709
anyways gamma is in a way related to how compressible
the gas is to some extent draft
00:11:24.709 --> 00:11:32.959
measure. It works reasonably well if this
is a case, what will happen to my velocity
00:11:32.959 --> 00:11:36.400
this
is going to be equal to u 1 by u 2 velocity
00:11:36.400 --> 00:11:40.139
goes to 1 4th , here velocity goes to 1 6th
here
00:11:40.139 --> 00:11:46.100
that is what is happening here.
So, we have already looked at I will just
00:11:46.100 --> 00:11:49.110
write here, u 1 by u 2, then I will say I
have also
00:11:49.110 --> 00:11:56.420
looked at velocity ratio. I do not need to
deal with it separately, we will pick a particular
00:11:56.420 --> 00:12:10.009
relation. Actually I will, we can even pick,
I am not going by the nodes. Now I am, just
00:12:10.009 --> 00:12:18.689
going to go free-form writing 1 plus, this
is going to be more complex. So, I have, to
00:12:18.689 --> 00:12:21.480
go
by the nodes that all we will go back by the
00:12:21.480 --> 00:12:24.829
nodes because, I will get a m 2 square here,
then it will be m 2 and m 1 I have to link.
00:12:24.829 --> 00:12:27.779
Let us not do it that way, I will just go
gamma
00:12:27.779 --> 00:12:37.839
plus 1 by gamma minus 1 plus P 2 by P 1 divided
by gamma plus 1 by gamma minus 1
00:12:37.839 --> 00:12:48.980
times P 2 by P 1 plus 1 we had such a relation.
No, I made a mistake here, this is not
00:12:48.980 --> 00:12:55.120
right this happens to be density ratio, whatever
I wrote now was density ratio. Pressure
00:12:55.120 --> 00:13:01.160
temperature ratio is P 1 by P 2 with a plus
sign in front of it. This is correct this
00:13:01.160 --> 00:13:04.649
is t 2 by t
1 and this formula already made a mistake
00:13:04.649 --> 00:13:10.249
and I was in a hurry.
All this time we have been doing, what happens
00:13:10.249 --> 00:13:13.589
when mock number turns to infinity
suddenly I am writing an expression in terms
00:13:13.589 --> 00:13:21.519
of P 2 by P 1, and P 1 by P 2, why because
I already know from here, if mock number turns
00:13:21.519 --> 00:13:23.750
to infinity P 2 by P 1 turns to infinity.
00:13:23.750 --> 00:13:28.120
.So, I will take the shortcut instead of substituting
all the expressions here it will be very
00:13:28.120 --> 00:13:30.149
complex.
So, I am, just taking a shortcut saying P
00:13:30.149 --> 00:13:34.850
2 by P 1 goes to infinity, if I say that is
very high
00:13:34.850 --> 00:13:42.709
compared to this number. So, I will neglect
this. So, this will tend to P 2 by P 1 divided
00:13:42.709 --> 00:13:49.490
by, now this is, there and this P 1 by P 2,
this is 1 by infinity. So, this is very small
00:13:49.490 --> 00:13:58.959
number compared to this. So, I will keep this
1 this is what I have. But, there is not a
00:13:58.959 --> 00:14:01.559
lot
P 2 by P 1 is going to turn to infinity. So,
00:14:01.559 --> 00:14:07.810
T 2 by T 1 will also attend to infinity.
But, not as fast as the other one because,
00:14:07.810 --> 00:14:09.800
this number is always more than 1, we just
did
00:14:09.800 --> 00:14:16.250
some calculations roughly. We found that,
that is 6 for gamma equal to 1.4. So, this
00:14:16.250 --> 00:14:18.929
will
be going up slower, temperature goes up slower
00:14:18.929 --> 00:14:23.300
across the shock compared to pressure
goes up faster. That is what you are seeing
00:14:23.300 --> 00:14:33.889
here anyway which is what we saw in the
plots alright. So, that brings us to the end
00:14:33.889 --> 00:14:42.249
of discussion on normal shocks that are
stationary. Ideally I have to go at this point
00:14:42.249 --> 00:14:48.649
and start solving numerical examples, which
I will do after we go to moving shocks after
00:14:48.649 --> 00:14:52.310
we finish moving shocks.
I want to give you a little more feel for
00:14:52.310 --> 00:14:54.470
moving shocks, then what is given in most
of the
00:14:54.470 --> 00:15:00.309
books and after that we will go and start
doing numerical examples, that is a little
00:15:00.309 --> 00:15:04.220
better
that gives you a different feel for what can
00:15:04.220 --> 00:15:08.709
be done. But, in case I want to solve a
problem. Typically what will be given will
00:15:08.709 --> 00:15:13.860
be there will be compressible flows tables
which we already said there are tables like
00:15:13.860 --> 00:15:19.060
this available, we said that we will use
isotropic low tables that is what we did before.
00:15:19.060 --> 00:15:22.629
Now I am, saying there are also normal
shock tables for compressible flows already
00:15:22.629 --> 00:15:28.589
available, they deal with only stationary
normal shocks, not moving normal shocks. Stationary
00:15:28.589 --> 00:15:31.220
normal shocks tables are available.
00:15:31.220 --> 00:15:32.220
..
00:15:32.220 --> 00:15:39.740
So, what they give will be, for each gamma
typically they will give 3 gamma’s 1.3,
00:15:39.740 --> 00:15:45.230
1.4
and 5 by 3, 1.666. These are the 3 values
00:15:45.230 --> 00:15:49.029
for which, they will give data typically.
What
00:15:49.029 --> 00:16:00.660
they will have be M 1 as the input variable
and they will give you M 2, P 2 by P 1, T
00:16:00.660 --> 00:16:09.709
2
by T 1, P naught 2 by P naught 1, P naught
00:16:09.709 --> 00:16:16.810
2 by P 1, these are the things they will
typically give you. There are some more things
00:16:16.810 --> 00:16:20.470
where they will give you rho 2 by rho 1.
But, I think it is unnecessary if I know P
00:16:20.470 --> 00:16:23.720
2 by P 1 and P 2 by T 1. I can find rho 2
by rho
00:16:23.720 --> 00:16:30.790
1, as long as my gas does not change its molecular
weight during the shock that is not the
00:16:30.790 --> 00:16:34.879
case in our ordinary gas dynamics. So, we
can very well say that there is no reaction
00:16:34.879 --> 00:16:40.180
during the shock. All right in that case I
can say that I can get rho 2 by rho 1 directly.
00:16:40.180 --> 00:16:43.380
So,
I do not need to have a table which has rho
00:16:43.380 --> 00:16:46.230
2 by rho 1 also because P equal to rho r T
is
00:16:46.230 --> 00:16:50.829
valid before and after.
So, P 2 by P 1 divided by T 2 by T 1 will
00:16:50.829 --> 00:16:58.730
be my rho 2 by rho 1 and there is one special
thing P naught 2 by P 1, It will be very helpful
00:16:58.730 --> 00:17:04.240
in solving problems will see that if I want.
When will I use this P naught 2 by P naught
00:17:04.240 --> 00:17:18.420
1 to find entropy jump. delta S is minus log
of this value. You have to of course, know
00:17:18.420 --> 00:17:20.950
in your tables whether they are giving it
as P
00:17:20.950 --> 00:17:24.290
naught 2 by P naught 1 or P naught 1 by P
naught 2, they will give you different books
00:17:24.290 --> 00:17:27.650
will give you different things you have to
know what is written there on the top.
00:17:27.650 --> 00:17:32.560
And you have to of course, get used to picking
the correct gamma page, each gamma
00:17:32.560 --> 00:17:36.930
will be one page typically. So, you have to
look the correct page otherwise you will
00:17:36.930 --> 00:17:40.370
.make a lot of mistakes. This is what happens
in most of the exams people will just take
00:17:40.370 --> 00:17:46.610
the wrong tables and because of that, they
will get wrong results. Anyways we will go
00:17:46.610 --> 00:17:50.520
to
numerical solutions later. Currently we will
00:17:50.520 --> 00:17:58.800
switch to moving shocks it gives you a little
more feel for waves in gas.
00:17:58.800 --> 00:18:04.201
So, I am going to start with moving shocks
and I am going to assume that there is a
00:18:04.201 --> 00:18:12.070
shock moving and it is moving into still air
or still gas, no movement in the gas. I am
00:18:12.070 --> 00:18:19.970
going to assume the gas is stationary till
the shock comes in. So, I am going to draw,
00:18:19.970 --> 00:18:23.750
still
we are in 1 d world I am going to draw 1 d
00:18:23.750 --> 00:18:30.050
duct, where there is a shock moving with
some wave velocity. I am going to call it
00:18:30.050 --> 00:18:35.440
W s velocity of the shock wave velocity, and
I
00:18:35.440 --> 00:18:42.110
am going to say, I have to be careful about
using variables. I want two different reference
00:18:42.110 --> 00:18:46.130
frames. Because I have to compare, I have
to go from 1 reference frame to the other.
00:18:46.130 --> 00:18:49.460
Till
now we have been solving problems with reference
00:18:49.460 --> 00:18:53.690
frame fixed on the shock that is, I am
sitting on the shock. I see that the incoming
00:18:53.690 --> 00:18:57.930
flow is coming with some mock number and
it is going out with a lesser mock number
00:18:57.930 --> 00:19:02.420
that is what we have been seeing till now.
I am going to say I am sitting on the outside
00:19:02.420 --> 00:19:09.610
as in say on the duct on the tube or on the
ground and there is a shock moving stationary
00:19:09.610 --> 00:19:14.520
gas, shock is moving and this is a set of
compression waves that is what we said a shock
00:19:14.520 --> 00:19:19.680
is. So, what does that mean it is a strong
compression wave coming in, what will happen
00:19:19.680 --> 00:19:25.540
really behind it flow will follow right it
is like a snake example, there is a snake
00:19:25.540 --> 00:19:28.870
and everybody run. And that is the information
that is going they are all running along this
00:19:28.870 --> 00:19:34.420
way, there is too much crowd there is less
crowd here everybody wants to go this way
00:19:34.420 --> 00:19:38.810
that is the feel right.
So, I know that the velocity behind is going
00:19:38.810 --> 00:19:45.160
to be this way and here I am going to say,
now I am being careful using correct variables.
00:19:45.160 --> 00:19:56.910
I am using V 1 here, we will stick to U as
velocity in the reference frame, where shock
00:19:56.910 --> 00:20:09.380
is stationary, we will keep that as the
reference. I am, going to say shock stationary.
00:20:09.380 --> 00:20:12.950
In this reference frame, I will call the
velocity is U, in the other reference frame
00:20:12.950 --> 00:20:15.450
I will call the velocity is V. This is the
reason
00:20:15.450 --> 00:20:22.671
why we chose volume to be V cross and here
I am going to have a velocity V 2, some
00:20:22.671 --> 00:20:28.240
velocity V 2.
And just because I switch my reference frame,
00:20:28.240 --> 00:20:34.550
the P 2’s and T 2’s will not be any
different. Shock is just a shock. It does
00:20:34.550 --> 00:20:37.070
not care who is looking at it whether somebody
00:20:37.070 --> 00:20:40.370
.sitting on the shock is looking at the shock
or somebody standing on the ground is
00:20:40.370 --> 00:20:44.750
looking at the shock does not care, shock
is just a shock.
00:20:44.750 --> 00:20:50.540
So, the pressure ratios, density ratios, velocity
ratios they are all going to be the same as
00:20:50.540 --> 00:20:57.460
in this except for velocity I should not have
told. Pressure ratio, temperature ratio,
00:20:57.460 --> 00:21:02.680
density ratio those will be the same. Velocity
depends on reference frame that is a very
00:21:02.680 --> 00:21:07.660
important thing in engineering. Velocity depends
on reference frame. Now, I am
00:21:07.660 --> 00:21:11.960
choosing one reference frame here and this
is another reference frame, where I am saying
00:21:11.960 --> 00:21:18.690
I am stationary with respect to this incoming
gas. I am saying there is still air in the
00:21:18.690 --> 00:21:21.730
room
and there is a shock passing by, that is the
00:21:21.730 --> 00:21:24.480
condition. So, that is this case, I am keeping
V
00:21:24.480 --> 00:21:29.580
1, V 2 as velocities in that reference frame,
where it is with respect to lab our ground
00:21:29.580 --> 00:21:35.190
reference frame and this one is with respect
to the shock. Maybe the shock itself is
00:21:35.190 --> 00:21:39.800
moving currently.
All I know is I am sitting on the shock. I
00:21:39.800 --> 00:21:43.150
am sitting on the shock and because of that
I am
00:21:43.150 --> 00:21:50.200
going to see some velocity incoming this way.
I call that U 1 and we know from all the
00:21:50.200 --> 00:21:57.000
analysis last few weeks we are going to get
this to be U 2 and it is less than that value.
00:21:57.000 --> 00:22:07.410
We know all this already and T 2 by T 1, P
2 by P 1, rho 2 by rho 1, those ratios of
00:22:07.410 --> 00:22:12.880
thermo dynamics properties across the shock
does not change with reference frame. So,
00:22:12.880 --> 00:22:15.090
I
will just use the same ratios that are very
00:22:15.090 --> 00:22:20.830
important statement I am making here.
Now, I want to go from one reference frame
00:22:20.830 --> 00:22:30.070
to the other. How will I do it? I want to
write U 1 in terms of V 1, easy right, change
00:22:30.070 --> 00:22:35.510
in reference frame is just because of this
reference point is now moving with respect
00:22:35.510 --> 00:22:43.570
to my original reference. So, I am going to
say there if I add a negative W S to this
00:22:43.570 --> 00:22:49.880
whole system then the shock will become
stationary, W S to the left will make the
00:22:49.880 --> 00:22:52.200
shock stationary. So, I will add that to all
the
00:22:52.200 --> 00:22:57.520
velocities in this whole frame anywhere, any
analysis I do all the velocities I will add
00:22:57.520 --> 00:23:09.730
this. So, I am going to say U 1 here is equivalent
to V 1 minus W S
00:23:09.730 --> 00:23:18.500
and similarly, U 2 is
equal to V 2 minus W S.
00:23:18.500 --> 00:23:30.870
A quick check, if I change my reference frame,
any difference in a particular reference
00:23:30.870 --> 00:23:37.100
frame should not change, difference between
this to this should be the same as difference
00:23:37.100 --> 00:23:42.470
between this to that you will get directly.
Here I want to write it in a particular form.
00:23:42.470 --> 00:23:46.400
So,
I will put it is U 2 minus U 1 equal to V
00:23:46.400 --> 00:23:50.930
2 minus V 1 because I want to proceed from
this
00:23:50.930 --> 00:23:57.320
point. I am going to say yes, this is true,
a quick check. Now, I want to proceed from
00:23:57.320 --> 00:24:04.180
.here and I want to find the value of V 2
given V 1 is 0. So, I am going to set V 1
00:24:04.180 --> 00:24:07.360
equal to
0. I will just make it 0. So, my V 2 has a
00:24:07.360 --> 00:24:14.630
relation already U 2 minus U 1 in shock fixed
coordinates that will be my V 2.
00:24:14.630 --> 00:24:25.470
That is what I am getting. So, I am going
to have P 2 equal to U 2 minus U1. I want
00:24:25.470 --> 00:24:28.360
to
write it in terms of mock number, how will
00:24:28.360 --> 00:24:31.570
I do it? I want to write it as U 2 by U 1.
Then
00:24:31.570 --> 00:24:37.880
I have expressions for U 2 by U 1, U1 by U2
everything we have expressions. So, I will
00:24:37.880 --> 00:24:51.810
pull out a U 1 and I will have U 2 by U 1
minus 1. Now, I want to find the value of
00:24:51.810 --> 00:24:56.910
U 1
what will it be. From here I am shifting to
00:24:56.910 --> 00:25:02.950
here. Simple thing I can just go to math and
say V 1 was said to be 0.
00:25:02.950 --> 00:25:10.030
Then I will get U 1 is W minus of W S minus
is important here, minus W S or I can say
00:25:10.030 --> 00:25:15.299
this is stationary, this is moving, Now I
am switching reference frame to this. So,
00:25:15.299 --> 00:25:17.090
this
will be moving this way with the same velocity
00:25:17.090 --> 00:25:20.140
and this is stationary. So, it is minus, it
is
00:25:20.140 --> 00:25:26.600
just coming out directly. So I will rewrite
this as W S times. I will take the minus sign
00:25:26.600 --> 00:25:29.110
in
to the bracket it will become 1 minus U 2
00:25:29.110 --> 00:25:35.220
by U 1 that will become I expression. Where
U
00:25:35.220 --> 00:25:42.010
2 by U 1 is shock fixed coordinates.
We know everything about shock fixed coordinate
00:25:42.010 --> 00:25:45.940
properties. We have already thrashed
it out like last for classes 5 classes. Now,
00:25:45.940 --> 00:25:48.990
all we have to do is just start substituting
things
00:25:48.990 --> 00:26:00.900
inside. Only one question W S is the velocity
of the shock, do we know the value, in a
00:26:00.900 --> 00:26:06.880
way we know the value because, I have been
dealing with U 1 all this times. How will
00:26:06.880 --> 00:26:11.980
I
write U 1 in terms of say mock number, U 1
00:26:11.980 --> 00:26:21.890
can be written as M 1a 1. Now, I know U 1
is equal to minus W S and I have already used
00:26:21.890 --> 00:26:28.440
up the minus sign. So, it should just
become M 1 a 1, I have already used up the
00:26:28.440 --> 00:26:34.060
minus sign and the minus sign went into this
to flip the terms the other way. So, I am
00:26:34.060 --> 00:26:37.770
going to say this is equal to M 1 a 1. Now,
I will
00:26:37.770 --> 00:26:47.110
go and write a better expression.
00:26:47.110 --> 00:26:48.110
..
00:26:48.110 --> 00:27:17.800
I am picking up one of the relations which
we wrote before, it should be 2 gamma, this
00:27:17.800 --> 00:27:22.890
expression I never wrote separately. But this
comes from whatever we wrote today for P
00:27:22.890 --> 00:27:28.060
2 by P 1 when we did the asymptotic analysis.
1plus 2 gamma by gamma, plus 1 times M
00:27:28.060 --> 00:27:32.310
square minus 1 will give you this form. I
just rearranged that term to get to this,
00:27:32.310 --> 00:27:34.970
nothing
great you can get to this form it is not very
00:27:34.970 --> 00:27:40.680
difficult to get.
So, I have rewritten M 1, M1 square I have
00:27:40.680 --> 00:27:42.690
rewritten and then put a half power on top
of
00:27:42.690 --> 00:27:49.820
it. So, I got this expression. Why did I go
to p 2 by p 1? In reality when we want to
00:27:49.820 --> 00:27:54.510
create
a moving shock, the way to create it, is to
00:27:54.510 --> 00:28:00.240
create a pressure differential, if the pressure
differential is strong enough then I am going
00:28:00.240 --> 00:28:06.720
to form a moving compression wave which
I am going past myself, I have to wait for
00:28:06.720 --> 00:28:09.630
this to happen shock tubes case. But anyway
I
00:28:09.630 --> 00:28:12.940
will go ahead and tell you that if there is
a whole bunch of compression waves come
00:28:12.940 --> 00:28:17.810
together, they form a strong thin-layer called
the shock.
00:28:17.810 --> 00:28:22.080
We will go into details of this again later,
I will just tell you now that you whole bunch
00:28:22.080 --> 00:28:30.750
of compression waves together is what forms
this thin-layer called shock. And now if I
00:28:30.750 --> 00:28:38.040
know my P 2 by P 1 that is caused because
of the shock, then I can tell what will be
00:28:38.040 --> 00:28:42.010
the
speed at which the shock will move given a
00:28:42.010 --> 00:28:50.810
1 and the gas in which it is moving. I should
know what is the temperature of the whatever
00:28:50.810 --> 00:28:57.890
stationary gas and it the stationary gas is
gamma value, I need to know these 2 and the
00:28:57.890 --> 00:29:03.750
P 2 by P 1 which is what is typically
referred to as a strength of the shock. How
00:29:03.750 --> 00:29:05.550
much compression can the shock give, that
is
00:29:05.550 --> 00:29:11.940
.called the strength of the shock so, if I
give the strength of the shock and in what
00:29:11.940 --> 00:29:15.350
medium
is it travelling, we can tell at what speed
00:29:15.350 --> 00:29:20.480
it will travel, that is the idea that is why
we
00:29:20.480 --> 00:29:23.250
write it like this.
So, most of the moving shock expressions I
00:29:23.250 --> 00:29:29.330
am going to show you are all going to be in
P 2 by P 1, very useful in P 2 by P 1 form.
00:29:29.330 --> 00:29:31.690
So, the next one, we already left it half
done
00:29:31.690 --> 00:29:44.880
V 2 is W s times 1 minus u 2 by u 1 which
can be written as W s times 1 minus rho 1
00:29:44.880 --> 00:29:48.900
by
rho 2. We already wrote an expression for
00:29:48.900 --> 00:29:53.220
rho 2 by rho 1, in terms of P 2 by P 1. Now,
I
00:29:53.220 --> 00:29:57.860
just go and invert that expression and I am
going to write it. So, my expression W s,
00:29:57.860 --> 00:30:40.890
I am
going to substitute the top 1, now we want
00:30:40.890 --> 00:30:47.920
to go again to these two extremes of the
analysis, 2 extremes of the formulae. I am
00:30:47.920 --> 00:30:54.550
going to say what if my shock is very, very
weak, it is going to cause a very small pressure
00:30:54.550 --> 00:31:00.120
jump, a weak compression wave,
extremely weak compression wave. If I pick
00:31:00.120 --> 00:31:04.690
that particular case, if it is a weak
compression wave I can say that my P 2 by
00:31:04.690 --> 00:31:10.640
P 1 is approximately 1.
So, let us say we will set the ultimate weak
00:31:10.640 --> 00:31:17.370
shock will be P 2 by P 1 equal to 1 or tending
to 1 is what I have to do really. Let us say
00:31:17.370 --> 00:31:20.220
we will make it equal to 1. I want to see
what
00:31:20.220 --> 00:31:30.300
happens to my W s? Actually, we will look
at V 2 first because the formula is just right
00:31:30.300 --> 00:31:38.180
above V 2 will tend to, what will happen that
will become 1, this will become 1. It so
00:31:38.180 --> 00:31:44.540
happens that P 2 by P 1 will become 1, gamma
plus 1 by gamma minus 1 plus 1
00:31:44.540 --> 00:31:48.059
numerator also gamma plus 1 by gamma minus
1 plus 1.
00:31:48.059 --> 00:31:56.880
So, these 2 will just become 1 overall, 1
minus 1 becomes 0. So, my V 2 tends to 0 that
00:31:56.880 --> 00:32:05.710
means, if it is a very weak compression wave
there is no flow behind it. Now, what is the
00:32:05.710 --> 00:32:11.550
speed at which this way will travel that is
the next question W s. I want to find what
00:32:11.550 --> 00:32:15.480
it
tends to, I am going to look at this top expression
00:32:15.480 --> 00:32:22.240
and again I am going to set P 2 by P 1
equal to 1. What is going to happen, P 2 by
00:32:22.240 --> 00:32:26.309
P 1 becomes 1, gamma plus 1 both the
denominator are the same 2 gamma. So, it will
00:32:26.309 --> 00:32:32.100
be just add the numerators gamma plus 1,
plus gamma minus 1 minus 1 plus 1 cancels
00:32:32.100 --> 00:32:38.730
2 gamma by 2 gamma becomes 1, square
root of 1. So, it becomes a 1.
00:32:38.730 --> 00:32:47.430
So, What happens is if I have a very weak
compression wave, it is going to travel at
00:32:47.430 --> 00:32:55.450
speed of sound in that medium and it may not
change any flow as in no pressure jump,
00:32:55.450 --> 00:33:03.150
no velocity jump that is what you are going
to get if it is very, very weak, if it is
00:33:03.150 --> 00:33:07.710
anything
more than P 2 by P 1 equal to 1, then velocity
00:33:07.710 --> 00:33:12.620
will not be 0 and I will have slightly
00:33:12.620 --> 00:33:20.809
.stronger than speed of sound as the shock
strength. So, if I have already told you this
00:33:20.809 --> 00:33:23.170
is
ead once I will tell you again.
00:33:23.170 --> 00:33:27.630
If I have, a case where 2 3 shock waves, 2
3 compression waves, weak compression
00:33:27.630 --> 00:33:33.900
waves come together, then they will get together
I will not tell you how currently it gets
00:33:33.900 --> 00:33:37.330
together, we will just keep it like that.
We will go and do this analysis again. When
00:33:37.330 --> 00:33:40.890
we
go and deal with shock tubes, when the shocks
00:33:40.890 --> 00:33:48.240
come together, when the compression
waves come together, they will tend to overlap
00:33:48.240 --> 00:33:54.260
1 on top of the other and become 1 strong
wave, there are some 10 waves, they all will
00:33:54.260 --> 00:34:00.870
coalesce to form 1 strong wave. It will still
be that thin shockwave as far as our gas dynamics
00:34:00.870 --> 00:34:07.760
is constant assumption. It will be
reasonably thin. If that happens, then my
00:34:07.760 --> 00:34:12.869
P 2 by P 1 is the overall, it should actually
be P
00:34:12.869 --> 00:34:19.290
2 by P 1 multiplied by P 3 by P 2, multiplied
by P 4 by P 3 like that, overall it will
00:34:19.290 --> 00:34:26.760
become P 4 by P 1 or something or P n by P
1, if there are n waves whatever.
00:34:26.760 --> 00:34:32.810
So, overall that P 2 by P 1 across that whole
set of waves together will not be 1, it will
00:34:32.810 --> 00:34:37.929
be
strong, when that happens this is not really
00:34:37.929 --> 00:34:44.730
true, it will be moving with some reasonably
high-speed. At what speed, we will go look
00:34:44.730 --> 00:34:47.780
at it, if P 2 by P 1 is anything more than
1,
00:34:47.780 --> 00:34:55.770
what will happen? This number will not be
1 but, higher. If this number is higher, then
00:34:55.770 --> 00:35:03.150
my speed of the shock is going to be more
than a 1, it is going to travel supersonic.
00:35:03.150 --> 00:35:05.730
Any
compression wave, if it has any reasonable
00:35:05.730 --> 00:35:12.380
compression anything more than 1, P 2 by P
1 will move supersonic, that is what we are
00:35:12.380 --> 00:35:18.650
seeing here, opposite is true for expansion
waves we would not deal with it right now.
00:35:18.650 --> 00:35:22.590
But, we will go to expansion when we have
to, first we will deal with all compressions
00:35:22.590 --> 00:35:27.720
then we will go deal with expansions.
So, this is one extreme of the case, we said
00:35:27.720 --> 00:35:32.680
P 2 by P 1 tending to 1, very weak
compression waves. Now, we will look at strong
00:35:32.680 --> 00:35:36.620
compression waves, really strong
compression waves. What will happen there?
00:35:36.620 --> 00:35:42.420
We have to do the same kind of analysis as
before we are going to do the asymptotic analysis,
00:35:42.420 --> 00:35:48.940
we are going to set that P 2 by P 1
tends to infinity, very large values and we
00:35:48.940 --> 00:35:52.150
want to see what is the effect?
00:35:52.150 --> 00:35:53.150
..
00:35:53.150 --> 00:36:17.369
The first one we want to do is W s, I will
write it again here, I have this and I am
00:36:17.369 --> 00:36:20.240
going
to say this tends to infinity which means
00:36:20.240 --> 00:36:22.840
I will neglect the other term. So, I am going
to
00:36:22.840 --> 00:36:36.970
get to only the first-term remaining, roughly
rough estimate, I will just say it tends to
00:36:36.970 --> 00:36:40.330
that
is more mathematically correct, we will just
00:36:40.330 --> 00:36:43.780
say it tends to this for large values of P
2 by
00:36:43.780 --> 00:36:51.180
P 1 which means if I keep on increasing my
P 2 by P 1 my shock mock number, if I want
00:36:51.180 --> 00:36:57.190
to call it is a shock mock number, shock velocity
divided by a 1, I will use it that way.
00:36:57.190 --> 00:37:00.510
Now, I will put 1 more term here, I will call
the actually I will write it separately, I
00:37:00.510 --> 00:37:10.770
do
not want to rewrite everything. This one tends
00:37:10.770 --> 00:37:19.260
to square root of gamma plus 1 by 2
gamma times P 2 by P 1, this is what it will
00:37:19.260 --> 00:37:27.369
come down to and we are seeing that if my
compression gets stronger and stronger, my
00:37:27.369 --> 00:37:31.470
shock mock number will keep on increasing,
there is no limit to this, it is not asymptotic
00:37:31.470 --> 00:37:37.780
to any values. It will keep on increasing.
Now, the next thing is, we have to do this
00:37:37.780 --> 00:37:43.490
analysis for V 2 by V 1, V 2 by a 1, I want
to
00:37:43.490 --> 00:37:50.380
call this V 2 by a 1, I could do it as V 2
by a 2 also. We will currently do it as V
00:37:50.380 --> 00:37:58.450
2 by a 1
and then we will look at a 2 by V 2 by a 2.
00:37:58.450 --> 00:38:04.930
From whatever we wrote before, I can say
that I will just directly write the tending
00:38:04.930 --> 00:38:14.870
2 thing, gamma plus 1 by 2 gamma times P 2
by
00:38:14.870 --> 00:38:23.190
P 1, this 1 just comes from W s whatever is
here is exactly here. Now, the remaining
00:38:23.190 --> 00:38:33.360
term is 1 minus u 2 by u 1, that is going
to tend to 1 minus gamma minus 1 by gamma
00:38:33.360 --> 00:38:43.830
plus 1, this is what it should become. So,
if I set P 2 by P 1, P 1 is very large you
00:38:43.830 --> 00:38:46.460
should
get to this particular formula.
00:38:46.460 --> 00:38:55.040
.Now, once I have this. Now, I just have to
simplify this it is going to, this whole bracket
00:38:55.040 --> 00:39:02.010
is going to become 2 by gamma plus 1, it is
going to become 2 by gamma plus 1,
00:39:02.010 --> 00:39:06.170
because this gamma will get cancelled with
minus gamma there, minus gamma plus 1 is
00:39:06.170 --> 00:39:09.900
what is there. So, it will become 2 by gamma
plus 1. So, I will get to this particular
00:39:09.900 --> 00:39:12.320
form.
Now, we just have to write it this gamma plus
00:39:12.320 --> 00:39:23.190
1 and square root of gamma plus 1 inside,
simplify this you will get to 2 by gamma times
00:39:23.190 --> 00:39:36.250
gamma plus 1, P 2 by P 1, this is what
you are getting. Again, we are seeing that
00:39:36.250 --> 00:39:38.700
the induced velocity, I am dividing by a 1,
this
00:39:38.700 --> 00:39:44.680
is not like mock number behind the shock,
it is not really that, I am dividing by a
00:39:44.680 --> 00:39:47.060
1. We
have to go and divide by a 2 next we will
00:39:47.060 --> 00:39:51.730
do it again, if I divided by a 1, V 2 by a
1 then,
00:39:51.730 --> 00:39:57.531
I am getting that particular value will keep
on increasing forever. For a given a 1, if
00:39:57.531 --> 00:40:00.600
I
keep on increasing my compression, my V 2
00:40:00.600 --> 00:40:06.200
keeps on increasing. If I have a stronger
shock, the flow induced by the stronger shock
00:40:06.200 --> 00:40:13.410
will be higher, that is what we are having.
In all this analysis we have assumed V 1 equal
00:40:13.410 --> 00:40:20.670
to 0. If V 1 is not 0, really nothing goes
wrong, we just have to keep track of what
00:40:20.670 --> 00:40:23.040
is the value of u 1 all the time. We will
go and
00:40:23.040 --> 00:40:28.840
do a numerical example next class where we
will see what if V 1 is not equal to 0. Then
00:40:28.840 --> 00:40:34.420
we just have to keep track of it, the values
will be much higher in that is all will happen
00:40:34.420 --> 00:40:41.870
nothing special. Now, we will look at I want
to make this become M 2 as in mock
00:40:41.870 --> 00:40:50.000
number. So, I have to look at V1 by a 2. So,
I have to multiply this with a1 by a 2. So,
00:40:50.000 --> 00:40:53.310
let
us find a 1 by a 2.
00:40:53.310 --> 00:40:54.310
.
00:40:54.310 --> 00:41:07.380
.I will go to the next page, a 1 by a 2 it
is the same gas, same gamma, same R, I will
00:41:07.380 --> 00:41:09.940
just
remove it. So, it will just become square
00:41:09.940 --> 00:41:16.520
root of T 1 by T 2. I do not need to do much
about this. Now, we have to write this T 1
00:41:16.520 --> 00:41:19.580
by T 2 in terms of P 2 by P 1 because, we
are
00:41:19.580 --> 00:41:24.820
writing everything in terms of P 2 by P 1.
T 1 by T 2 is whatever we wrote before for
00:41:24.820 --> 00:41:29.140
T 2
by T 1, we have right it inverted. So, it
00:41:29.140 --> 00:41:35.840
is going to be, this is the only place where
it is
00:41:35.840 --> 00:41:46.460
different, it will be P 1 by P 2 divided by
gamma plus 1 by gamma minus 1, plus P 2 by
00:41:46.460 --> 00:41:52.050
P 1, whole power half, this is what you are
going to get.
00:41:52.050 --> 00:42:01.120
So, for large values of P 2 by P 1 similar
analysis, we just want to see what happens
00:42:01.120 --> 00:42:11.980
when P2 by P1 tends to infinity, a 1 by a
2 will become, I am saying this is very high,
00:42:11.980 --> 00:42:16.880
this I will neglect, I will have on the denominator
only P 2 by P 1, in the numerator this
00:42:16.880 --> 00:42:22.560
is very small 1 by infinity very small. So,
I will have only this term. So, I am going
00:42:22.560 --> 00:42:28.630
to get
gamma plus 1 by gamma minus 1 divided by P
00:42:28.630 --> 00:42:38.010
2 by P 1 whole power half. Now, I want
to go and write it with the V 2 by a 1, already
00:42:38.010 --> 00:42:51.880
we have V 2 by a 1 into a 1 by a 2 is my M
2, which is my mock number of the flow induced
00:42:51.880 --> 00:42:57.770
due to the shock, that is this value.
So, if I take the previous formulae which
00:42:57.770 --> 00:43:01.400
is here this one and multiply with whatever
we
00:43:01.400 --> 00:43:05.880
derived just now I will take this particular
of course, remember that this is only for
00:43:05.880 --> 00:43:16.820
very
large values of P 2 by P 1 multiplied by a
00:43:16.820 --> 00:43:22.090
1 by a 2 which is here for large values, again
I
00:43:22.090 --> 00:43:39.210
will put it as square root of. So, we are
finding that P 2 by P 1 gets cancelled which
00:43:39.210 --> 00:43:44.860
means irrespective of the strength, if the
large value if this very strong shock, it
00:43:44.860 --> 00:43:49.940
goes to
one particular final mock number, that is
00:43:49.940 --> 00:43:51.510
why I am doing the all this analysis if it
does
00:43:51.510 --> 00:43:55.450
not go to any asymptotic I would not even
be dealing with such a thing, anyways we will
00:43:55.450 --> 00:44:01.430
simplify this further, inside the square root
there is gamma plus 1 which I can cancel.
00:44:01.430 --> 00:44:07.940
So, it will come down to square root of 2
by this gamma minus 1 will go to the
00:44:07.940 --> 00:44:14.480
denominator, there is a gamma already in the
denominator. This is the expression you
00:44:14.480 --> 00:44:23.430
will get finally for I have to a closed bracket
square root of 2 by gamma times gamma
00:44:23.430 --> 00:44:34.880
minus 1. If I say my gamma is 1.4, if I say
I am having air kind of gas, diatomic
00:44:34.880 --> 00:44:44.280
molecules then M 2 max I have to also say
induced, I will keep on using this induced
00:44:44.280 --> 00:44:52.390
this
is a induced flow mock number, maximum possible
00:44:52.390 --> 00:45:01.510
is going to be equal to what is this
square root of 2 by 1.4 times 0.4. I am just
00:45:01.510 --> 00:45:04.220
substituting gamma equal to 1.4 there and
that
00:45:04.220 --> 00:45:05.220
comes out to be 1.89.
00:45:05.220 --> 00:45:17.371
.If I want to create a supersonic flow by
sending a shockwave into still medium, I can
00:45:17.371 --> 00:45:25.550
maximum achieve only 1.89, that is what this
is telling you finally. Irrespective of
00:45:25.550 --> 00:45:30.600
whatever be the initial gas temperature initial
whatever be the gas present I can
00:45:30.600 --> 00:45:37.830
maximum reach only this, if my gamma for the
gas happens to be 1.4, the same thing if it
00:45:37.830 --> 00:45:46.700
is 5 by 3, monotonic gases this is going to
give I do not want to write the whole thing
00:45:46.700 --> 00:45:55.970
again, I will just write the final value that
comes out to be 1.34, If it is a less
00:45:55.970 --> 00:46:03.740
compressible gas, I cannot induce very high
mock number, it is going to go only slowly
00:46:03.740 --> 00:46:09.130
that is what we are seeing here.
If it is less compressible gas, it will have
00:46:09.130 --> 00:46:12.130
only lesser velocity, lesser mock number,
I am
00:46:12.130 --> 00:46:17.930
not talking about velocity here. Velocity
is not this simple, velocity is more complex
00:46:17.930 --> 00:46:27.020
than this. It is not very easy, we will go
look at what the plots look like, I just want
00:46:27.020 --> 00:46:28.760
to
write one more expression will start the next
00:46:28.760 --> 00:46:31.970
class with all the plots. I will just write
an
00:46:31.970 --> 00:46:37.530
expression for this whole thing, V 2 by a
1 multiplied by a 1 by a 2, the full expression
00:46:37.530 --> 00:46:44.240
not just the large P 2 by P 1 value, just
write the full thing and probably that, with
00:46:44.240 --> 00:46:47.660
that
will close today will deal with how the function
00:46:47.660 --> 00:46:49.810
looks like on plot next time.
.
00:46:49.810 --> 00:47:01.430
So, I
am going to call it M 2 induced, this is induced
00:47:01.430 --> 00:47:06.330
in V 1 equal to 0 gas, remember
that, if V 1 is not equal to 0, all the formula
00:47:06.330 --> 00:47:08.619
here are all wrong. I have to keep track of
V
00:47:08.619 --> 00:47:48.190
1, we have made V1 equal to 0. I will get
to something like, multiplied by, I am folding
00:47:48.190 --> 00:47:58.590
and writing here. So, I am multiplying this
with gamma plus 1 by gamma minus 1 plus P
00:47:58.590 --> 00:48:10.680
.2 by P 1 divided by gamma plus 1, gamma minus
1, P2 by P1 plus 1, this is my whole
00:48:10.680 --> 00:48:15.200
expression.
Now, I am going to observe something. I am
00:48:15.200 --> 00:48:24.890
going to say when my P 2 by P 1 is close to
1, my V 2 was 0, which is what we saw first.
00:48:24.890 --> 00:48:29.770
Alright V2 is 0, which is if it is a weak
compression wave it does not induce any velocity
00:48:29.770 --> 00:48:39.680
that is the first thing we said which
means, my M 2 induced must be 0 and we saw
00:48:39.680 --> 00:48:48.740
that, if it is very very strong shock,
infinitely strong shock, then it is going
00:48:48.740 --> 00:48:57.010
to induced some finite mock numbers finally,
which is supersonic we saw for gamma equal
00:48:57.010 --> 00:49:03.359
to 1.4, it was 1.89 value. If that is a case,
there was a point where it should have crossed
00:49:03.359 --> 00:49:10.091
from subsonic to supersonic. There must
be some P 2 by P 1 at which it will cross
00:49:10.091 --> 00:49:13.119
from subsonic to supersonic, that is if I
want to
00:49:13.119 --> 00:49:18.060
create a supersonic flow behind my shock,
my compression ratio for that shock should
00:49:18.060 --> 00:49:20.349
at
least be some value.
00:49:20.349 --> 00:49:27.950
How will I find that I have to set this equal
to 1, and find the P 2 by P 1 for that. I
00:49:27.950 --> 00:49:31.170
tried
solving that from here, it is very complex
00:49:31.170 --> 00:49:37.920
because it is a P 1 by P2, P2 by P1 to the
power half again to the power half here, P
00:49:37.920 --> 00:49:41.580
2 by P 1, P 1 by P2, P2 by P1 everywhere,
not
00:49:41.580 --> 00:49:47.630
easy to solve. So, I reserved at to graphical
way of doing things, this is the old-style
00:49:47.630 --> 00:49:55.010
longtime back when mathematics was to complex,
they just went to plots, easier just go 1
00:49:55.010 --> 00:50:00.390
direction, now find out the inverse easy to
do that is what people followed. I will just
00:50:00.390 --> 00:50:03.590
follow that, simple enough for our course
it is nice we do not need to inverse this,
00:50:03.590 --> 00:50:07.650
it is
not going to be much use anyway. So, I just
00:50:07.650 --> 00:50:14.960
want to say that if I create a strong enough
shock wave I maybe creating supersonic flow,
00:50:14.960 --> 00:50:18.800
most likely I will be creating subsonic
flow behind.
00:50:18.800 --> 00:50:23.910
We will look at all this next time with plots
of these expressions, every expression I have
00:50:23.910 --> 00:50:28.620
put up here for moving shock. We will look
at plots of it and then we will see where
00:50:28.620 --> 00:50:30.850
it
goes, it looks like have already crossed my
00:50:30.850 --> 00:50:36.190
time. We will look at the plots and then we
will start inferring items from there and
00:50:36.190 --> 00:50:42.980
will say what does a moving shock do to a
gas
00:50:42.980 --> 00:50:48.300
that is stationary, after that we will go
to numerical examples where we will first
00:50:48.300 --> 00:50:53.490
solve a
simple problems of shocks inflow that is whatever
00:50:53.490 --> 00:50:58.850
we said we will do later and then we
will go to moving shocks where we will say,
00:50:58.850 --> 00:51:03.850
shock is moving into stationary gas, after
that we will say shock is moving into moving
00:51:03.850 --> 00:51:10.990
gas, moving one way and moving the other
way and we will see what the effects are.
00:51:10.990 --> 00:51:13.180
We will do all this hopefully next class if
not
00:51:13.180 --> 00:51:18.600
we will go to one more class, any questions?
See you people next time.
00:51:18.600 --> 00:51:19.600
.