WEBVTT
Kind: captions
Language: en
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Hello everyone, welcome back. We are last
time solving some problems from Gas
00:00:17.750 --> 00:00:23.370
Dynamics from the point of view of stationary
shocks. This time we want to move on to
00:00:23.370 --> 00:00:27.790
moving shocks, we just want to solve problems
with shocks that are moving. And that
00:00:27.790 --> 00:00:32.110
will give you a little more physical feel,
which will directly lead us into going to
00:00:32.110 --> 00:00:36.120
oblique
shocks; that is where, we are going to…
00:00:36.120 --> 00:00:40.200
So the first problem we want to solve is this.
.
00:00:40.200 --> 00:00:55.589
I am given that, I have a 1 d duct and it
has stagnant air pressure 1 bar, temperature
00:00:55.589 --> 00:01:01.040
300
Kelvin and I said air, so gamma equal to 1.4,
00:01:01.040 --> 00:01:10.030
this is stagnant air in this duct. And I am
told that, there is a shock and it is moving,
00:01:10.030 --> 00:01:21.530
and behind the shock, pressure is 4.5 bar.
Now, I am supposed to find the shock speed
00:01:21.530 --> 00:01:28.280
and the speed of the gas behind, everything
with respect to the duct, the outside reference
00:01:28.280 --> 00:01:34.810
frame, so that is the basic problem.
So, to characterize the shock, what should
00:01:34.810 --> 00:01:37.770
I do? Basically I have to just start looking
at
00:01:37.770 --> 00:01:42.549
the problem and immediately say the information
given about the shock happens to be
00:01:42.549 --> 00:01:48.799
pressure P 2 divided by P 1; P 2 by P 1 is
given for us across the shock, no other
00:01:48.799 --> 00:01:52.619
.information is given across the shock, so
the only information about the shock is P
00:01:52.619 --> 00:02:01.990
2 by p
1 is given to be 4.5. Now, I will go to my
00:02:01.990 --> 00:02:11.650
normal shock tables and look at gamma equal
to 1.4 page and we will find that P 2 by P
00:02:11.650 --> 00:02:15.290
1 equal to 4.5 occurs for mach number equal
to
00:02:15.290 --> 00:02:21.920
2.0 basically mach 2 shock that is what we
are having.
00:02:21.920 --> 00:02:29.200
So now, we have shock strength, once we have
this information I do not want to really
00:02:29.200 --> 00:02:33.910
solve the problem in this coordinate system.
We know that, this is V equal to 0, V 1
00:02:33.910 --> 00:02:42.040
equal to 0, V 2 we are supposed to find out
and shock speed W S we are supposed to find
00:02:42.040 --> 00:02:52.970
out, that is the basic problem. So, I want
to find in the other coordinate system, I
00:02:52.970 --> 00:02:58.180
will
write here shock fixed coordinate system,
00:02:58.180 --> 00:03:05.840
I am going to have a u 1 and a u 2. And now,
I
00:03:05.840 --> 00:03:10.330
want to follow one convention for velocity
vector direction, I want to follow the
00:03:10.330 --> 00:03:17.260
convention for original problem.
So, I am going to say, this u 1, this is supposed
00:03:17.260 --> 00:03:22.500
to be negative velocity, will keep it that
way, we switched from that coordinate system
00:03:22.500 --> 00:03:29.110
to this coordinate system and I am going
to say u 1 is going to be negative. Because,
00:03:29.110 --> 00:03:34.750
it is going this way, I am saying V 2 is
positive, W S is positive that is the way
00:03:34.750 --> 00:03:37.290
we are going to solve the problem. Keep one
convention it will not confuse you, it will
00:03:37.290 --> 00:03:41.209
be very easy to solve, you will never get
into
00:03:41.209 --> 00:03:48.120
trouble.
Now, I want to find u 1, u 1 is going to be
00:03:48.120 --> 00:03:55.590
minus of M 1 a 1 because it is stagnant gas,
it
00:03:55.590 --> 00:04:00.430
is like a bus travelling against somebody
stationary, what is the speed of this person
00:04:00.430 --> 00:04:02.629
with
respect to the bus, it is the same velocity
00:04:02.629 --> 00:04:05.520
the other direction. So, I am just directly
minus
00:04:05.520 --> 00:04:17.190
M 1 a 1, this is going to be equal to minus
2 into square root of gamma R T, 1.4 into
00:04:17.190 --> 00:04:26.361
288.6 and 300. Just so we want to use this
number again and again, I will give you value
00:04:26.361 --> 00:04:39.560
for a 1, it is 348.2 meter per second and
so my u 1 is minus 2 times that, which is
00:04:39.560 --> 00:04:47.620
696.4
meter per second, this
00:04:47.620 --> 00:04:52.800
is my u 1 that is given. Innovators given
because P 2 by P 1 was
00:04:52.800 --> 00:04:58.640
given from there I found M 1, I do not want
to call it M 1 I will call it M of shock and.
00:04:58.640 --> 00:05:05.770
So, I am going to get to this point, we solved
up to this point now, we will go and write
00:05:05.770 --> 00:05:10.640
the equation, which we already wrote before.
00:05:10.640 --> 00:05:11.640
..
00:05:11.640 --> 00:05:21.050
u 1 equal to V 1 minus W S and u 2 equal to
V 2 minus W S this is the transformation
00:05:21.050 --> 00:05:25.740
between the 2 coordinate systems, these are
the expressions which are going to be used
00:05:25.740 --> 00:05:29.993
always in solving moving shock problems, very
easy to solve this way. And we know, in
00:05:29.993 --> 00:05:38.139
our case, V 1 equal to 0 and this information
has already been used, we already found u
00:05:38.139 --> 00:05:43.340
1. Now, all I have to do is substitute that
inside here and then now I am going to have
00:05:43.340 --> 00:05:47.759
set
of expressions to solve for it. Instead of
00:05:47.759 --> 00:05:51.020
that, I am doing it some other way, let us
just go
00:05:51.020 --> 00:06:02.800
for this V 2 minus V 1 equal to u 2 minus
u 1. For me to write this, I should have one
00:06:02.800 --> 00:06:05.570
coordinate system direction for all the velocities.
.
00:06:05.570 --> 00:06:12.030
.If I picked u 1 to be positive in the other
picture, let us go back here if I picked u
00:06:12.030 --> 00:06:16.990
1 to be
positive here then u 1 minus u 2 will be positive
00:06:16.990 --> 00:06:21.391
because u 2 should be less than u 1 in
the stationary shock condition. So, u 1 minus
00:06:21.391 --> 00:06:28.460
u 2 will be positive but in here, V 2 is
greater than V 1 because the shock is moving
00:06:28.460 --> 00:06:32.729
that is what is telling, the gas to move this
way. So, this will be positive velocity, V
00:06:32.729 --> 00:06:35.991
2 minus V 1 is positive and here, u 2 minus
u 1
00:06:35.991 --> 00:06:41.460
is negative, this kind of problems will come
up if you do not take care of the sign of
00:06:41.460 --> 00:06:47.270
velocity. So, always keep the sign information,
I put an arrow this way, so I am going to
00:06:47.270 --> 00:06:53.659
call it negative velocity, keep it that way
it will help you solve the problem easily.
00:06:53.659 --> 00:06:54.659
.
00:06:54.659 --> 00:07:04.710
Now, I can rewrite this as minus u 1 times
1 minus u 2 by u 1, u 2 by u 1 will not be
00:07:04.710 --> 00:07:11.810
effected why, both the velocities are negative,
ratio will not get affected. Both the
00:07:11.810 --> 00:07:15.710
velocities are negative in shock fixed coordinate
system so things will just work
00:07:15.710 --> 00:07:20.979
naturally, no problem for us. Now, we can
just start solving the problem, how will I
00:07:20.979 --> 00:07:27.830
know u 2 by u 1, I can get it from rho 1 by
rho 2, u 2 by u 1 is equal to rho 1 by rho
00:07:27.830 --> 00:07:29.990
2
from mass conservation across the shock.
00:07:29.990 --> 00:07:40.360
How will I get rho 1 by rho 2, P equal to
rho R T or rho equal to P by R T, I will write
00:07:40.360 --> 00:07:43.069
it
like this where, this R is specific gas constant
00:07:43.069 --> 00:07:48.419
of course, you do not need it in ratios but
anyway I am telling you. So, rho so I will
00:07:48.419 --> 00:07:53.650
write it as P 1 by P 2 into T 2 by T 1, this
is all
00:07:53.650 --> 00:08:00.539
I get it to be, most of the gas tables the
compressible flow tables or gas tables, none
00:08:00.539 --> 00:08:02.810
of
them will give you rho 2 by rho 1 across the
00:08:02.810 --> 00:08:05.139
shock. Because, they already have given
00:08:05.139 --> 00:08:10.199
.you P 2 by P 1 and T 2 by T 1, they do not
need to give this separately one column, just
00:08:10.199 --> 00:08:13.169
waste of paper if they give it, they will
never give it.
00:08:13.169 --> 00:08:19.069
So, all I have to do is, just go to that particular
condition, which we know it is not very
00:08:19.069 --> 00:08:23.191
difficult, we know that it is mach 2 now.
I have to go to that particular row, which
00:08:23.191 --> 00:08:27.849
has P
2 by P 1 equal to 44.5 or M 1 equal to 2,
00:08:27.849 --> 00:08:32.780
that condition. I will find T 2 by T 1 value,
that
00:08:32.780 --> 00:08:39.580
happens to be 1.687 and divided by P 2 by
P 1 value, which we know it is given in the
00:08:39.580 --> 00:08:52.530
problem 4.5 and this number comes out to be
0.375. So, basically it is coming down to
00:08:52.530 --> 00:08:57.651
1
minus 0.375 now, I know minus u 1, we just
00:08:57.651 --> 00:09:08.350
calculated u 1 to be 696.4.
So, V 2 minus 0 equal to, I put 0 of V 1 is
00:09:08.350 --> 00:09:19.190
0 for us, 696.4 into 1 minus 0.375, this
becomes your expressions. Now, you can just
00:09:19.190 --> 00:09:33.370
directly solve, V 2 comes out to be 435.3
meter per second now, have I found the shock
00:09:33.370 --> 00:09:44.960
speed, yes because from this expression,
minus u 1 happens to be the W S. Minus u 1,
00:09:44.960 --> 00:09:54.940
we already put a number here, shock speed
happens to be 696.4 meter per second, it is
00:09:54.940 --> 00:10:02.000
positive which says it is moving towards the
right, from left to right, V 2 is positive,
00:10:02.000 --> 00:10:03.660
velocity is from left to right.
.
00:10:03.660 --> 00:10:08.420
So, if I go back to the picture and say what
is happening, I am going to say the shock
00:10:08.420 --> 00:10:11.290
is
moving at 696 meter per second and the flow
00:10:11.290 --> 00:10:17.270
behind it, is going at 435 meter per second.
That is what, is the final outcome of this
00:10:17.270 --> 00:10:27.170
whole problem, we will go and look at another
problem with a little more modification.
00:10:27.170 --> 00:10:28.170
..
00:10:28.170 --> 00:10:32.690
I called this one problem a, I want to refer
to this problem again so we will call this
00:10:32.690 --> 00:10:43.110
one
problem b where, we want to say that, V 1
00:10:43.110 --> 00:10:49.630
is non zero, it is given to be 100 meter per
second, exactly same problem. V 1 equal to
00:10:49.630 --> 00:10:59.340
1 bar, T 1 equal to 300 Kelvin, gamma equal
to 1.4 it is air so R is also 288.6, there
00:10:59.340 --> 00:11:07.420
is a shock moving and P 2 is given to be 4.5
bar,
00:11:07.420 --> 00:11:13.070
same problem as before. Except for V 1 is
given to be 100 meter per second from left
00:11:13.070 --> 00:11:20.980
to
right that is, there is a velocity here already
00:11:20.980 --> 00:11:26.570
now we are supposed to find the same thing,
what is the value of V 2? And what is the
00:11:26.570 --> 00:11:31.840
value of shock speed.
These are the 2 things we want to find, just
00:11:31.840 --> 00:11:37.810
an extension to the previous problem now,
we have one variable non zero, that is all.
00:11:37.810 --> 00:11:40.280
I picked the same number so that, I can go
a
00:11:40.280 --> 00:11:45.490
little faster, I can use the same a 1 value
so I will just put a 1 otherwise, you just
00:11:45.490 --> 00:11:51.450
have to
calculate, square root of gamma R T 348.2
00:11:51.450 --> 00:11:56.440
meter per second. Now, I will go and write
u
00:11:56.440 --> 00:12:04.260
1 to V 1 minus W S, I told you this is the
expression we keep on using, we will use this.
00:12:04.260 --> 00:12:10.740
Now, we have a V 1 that is non zero. Previously
it was 0 in the previous problem now. I
00:12:10.740 --> 00:12:21.290
have to go and find out W S, W S will come
out to be, I will take it the other way V
00:12:21.290 --> 00:12:28.760
1
minus u 1 and now, this is going to be V 1.
00:12:28.760 --> 00:12:39.000
Now, you have to be little careful, in shock
fixed coordinates, I will draw another picture
00:12:39.000 --> 00:12:54.330
here, this is u 1 and it is having a value
M 1 a 1 in shock fixed coordinates. Now, I
00:12:54.330 --> 00:12:57.770
want
to make sure that, when I use this problem,
00:12:57.770 --> 00:13:03.220
I am not going to be messed up I want to use
it correctly. One way of doing it like this,
00:13:03.220 --> 00:13:04.980
after that I will tell you another method
both
00:13:04.980 --> 00:13:14.970
.should match exactly so I am looking at this
problem and I am going to say, I have to put
00:13:14.970 --> 00:13:22.770
u 1 as a minus of M 1 a 1.
Why am I putting this minus sign, I already
00:13:22.770 --> 00:13:25.321
did this in the last problem also but I did
not
00:13:25.321 --> 00:13:30.200
specifically tell this clearly there. But
here, I have to tell it, it is going from
00:13:30.200 --> 00:13:34.910
right to left
that velocity is negative, if it is left to
00:13:34.910 --> 00:13:37.250
right that is positive for us. So, I have
to be very
00:13:37.250 --> 00:13:42.500
careful, M 1 a 1 is the magnitude of the velocity,
it does not tell you the direction so I
00:13:42.500 --> 00:13:47.440
have to be very careful and put a direction
minus M 1 a 1. Now, I will substitute that
00:13:47.440 --> 00:13:53.660
so it
just becomes M 1 a 1, if this value comes
00:13:53.660 --> 00:13:55.690
out to be positive then the shock is moving
to
00:13:55.690 --> 00:13:59.700
the right, from left to right, that is the
way we are going to do it.
00:13:59.700 --> 00:14:03.540
Another way to look at the same problem, I
will get the same answer from another
00:14:03.540 --> 00:14:10.740
method, I will tell you that method also.
Imagine this, this 100 meter per second flow
00:14:10.740 --> 00:14:16.960
is
like a river flowing and the shock is moving
00:14:16.960 --> 00:14:20.850
on the river or it is like a boat going on
the
00:14:20.850 --> 00:14:26.850
river. If I think about it that way, there
is a boat moving with respect to the water
00:14:26.850 --> 00:14:30.090
in the
river and the water itself is moving. So,
00:14:30.090 --> 00:14:35.450
the actual velocity of the shock from outside,
actual velocity of the boat for somebody was
00:14:35.450 --> 00:14:40.410
standing outside, will be velocity of the
river plus the velocity of boat with respect
00:14:40.410 --> 00:14:48.050
to the river.
So, it will come out to be that M 1 a 1 plus
00:14:48.050 --> 00:14:50.870
the velocity of the river, that happens to
be
00:14:50.870 --> 00:14:57.370
this 100, that also will give you the same
answer V 1 plus M 1 a 1. You have to just
00:14:57.370 --> 00:14:59.460
a
crosscheck, I do not need to do it to 2 different
00:14:59.460 --> 00:15:02.410
methods, one is enough, if you just follow
the math, it will come out to be right as
00:15:02.410 --> 00:15:11.110
long as you do not make any mistakes. So now,
we just find this number, it is we know that
00:15:11.110 --> 00:15:17.380
M 1 is 2 because P 2 by P 1 is given to be
4.5, it is the same value. So, I am going
00:15:17.380 --> 00:15:33.940
to say it is 100 plus 2 into 348.2, the answer
comes out to be it is almost the same answer,
00:15:33.940 --> 00:15:43.050
696.4 was the previous answer plus 100,
796.4 meter per second this is the answer
00:15:43.050 --> 00:15:56.270
you are getting here. Now, I want to go and
solve V 2, now similar to this formula we
00:15:56.270 --> 00:15:59.350
have a V 2 formula u 2 equal to V 2 minus
W
00:15:59.350 --> 00:16:01.810
S.
00:16:01.810 --> 00:16:02.810
..
00:16:02.810 --> 00:16:22.960
So, how will I find my u 2, u 1 into u 2 by
u 1 at M equal to 2, easy way to do it, gas
00:16:22.960 --> 00:16:28.390
dynamics peoples like this a lot, canceling
ratios very nice way of doing things. It works
00:16:28.390 --> 00:16:33.880
great for us because in all compressible flows
tables, in all gas tables whatever you pick,
00:16:33.880 --> 00:16:38.330
they all give you only ratios so I have to
multiply all the ratios to get to whatever
00:16:38.330 --> 00:16:39.850
number
I get.
00:16:39.850 --> 00:16:54.030
So, I know this u 1 value, it happens to be
minus 696.4, this multiplied by u 2 by u 1
00:16:54.030 --> 00:16:57.670
I
want to find out, u 2 by u 1 we already found
00:16:57.670 --> 00:17:02.110
out from the previous problem it is the
same value, it has to be T 2 by T 1 divided
00:17:02.110 --> 00:17:18.390
by P 2 by P 1, that we found out to be 0.375.
If I do this, this is just u 2 now, I have
00:17:18.390 --> 00:17:22.480
to find V 2 so I will write it V 2 equal to
u 2 plus
00:17:22.480 --> 00:17:31.280
W S remember the minus sign, if you do not
use the minus sign, things will go wrong.
00:17:31.280 --> 00:17:46.360
Minus 696.4 into 0.375 plus 796.4, this is
what it comes out to be and the answer
00:17:46.360 --> 00:17:59.000
happens to be 535 meter per second, this is
what you are getting.
00:17:59.000 --> 00:18:11.861
Now, I want you to compare problem a, and
problem b, I will make a small table V 1 for
00:18:11.861 --> 00:18:33.180
problem a was 0, V 1 for problem b was 100,
W S 696.4 and here, it is 796.4, V 2 435
00:18:33.180 --> 00:18:46.680
here, it is 535, these are the numbers we
are getting. If you go look at u 1, u 2 those
00:18:46.680 --> 00:18:54.670
numbers will be different, other numbers if
you look at it, the same to be matching very
00:18:54.670 --> 00:18:58.360
nicely like this in fact, every number will
have this 100, it so happens that it will
00:18:58.360 --> 00:18:59.360
be
there.
00:18:59.360 --> 00:19:06.790
.Now is this logical, is this going to be
just this, if I have a V 1 some value, I just
00:19:06.790 --> 00:19:08.550
added to
all the values and it will come out to be
00:19:08.550 --> 00:19:11.680
this answer always, that is the question we
want
00:19:11.680 --> 00:19:20.230
to ask. It so happens that, for this type
of a problem, yes it is, why that is the question
00:19:20.230 --> 00:19:23.900
we
want to ask. Now, we will go to that side
00:19:23.900 --> 00:19:27.470
and start drawing pictures, more pictures
so I
00:19:27.470 --> 00:19:44.180
have all the pictures I need it looks like,
I have all the pictures I need.
00:19:44.180 --> 00:19:45.180
.
00:19:45.180 --> 00:19:57.400
This was problem a and problem b picture,
I have to change, I have to draw new
00:19:57.400 --> 00:20:30.260
pictures, problem b I am putting 696 it was
696.4, that does not matter so much for us.
00:20:30.260 --> 00:20:37.480
If
I draw this picture, it becomes easier for
00:20:37.480 --> 00:20:42.680
us to understand the problem, if I look at
the
00:20:42.680 --> 00:20:51.300
picture from problem a, I have this case and
if I say, I have an observer standing here
00:20:51.300 --> 00:20:57.520
and
he is observing this particular problem. Now,
00:20:57.520 --> 00:21:00.360
if I say, he is not just standing there but
he
00:21:00.360 --> 00:21:08.190
is moving this way with 100 meter per second,
he is on a bus if he is moving like this,
00:21:08.190 --> 00:21:11.850
what will he see as this velocity 100 meter
per second.
00:21:11.850 --> 00:21:19.290
What will he see this problem as 696 plus
100, what will he see this plus 100, that
00:21:19.290 --> 00:21:26.500
happens to be this problem exactly. It is
just a shift in coordinate system, person
00:21:26.500 --> 00:21:28.763
who is
watching is moving, that is the only change
00:21:28.763 --> 00:21:32.660
if you look at it, moving shock problems
mostly it will come out to be something like
00:21:32.660 --> 00:21:38.190
this, it is easier to work with. For a given
strength, there is a particular delta u you
00:21:38.190 --> 00:21:41.620
will get, if the change just this variable,
all I
00:21:41.620 --> 00:21:44.680
have to do is add that number everywhere and
the problem is solved, we do not need to
00:21:44.680 --> 00:21:49.290
resolve the problem. That is, assuming we
already solved the problem once for one
00:21:49.290 --> 00:21:56.640
.particular V 1, if the change just V 1, very
easy to solve just add it, that is all it
00:21:56.640 --> 00:22:04.380
suppose
to be. Now, we will go to next problem, continuing
00:22:04.380 --> 00:22:06.850
the numbering I will call it problem
c.
00:22:06.850 --> 00:22:07.850
.
00:22:07.850 --> 00:22:17.710
We will continue the same problem in a way,
I am going to say the shock is going like
00:22:17.710 --> 00:22:25.440
this and it reach the wall, there is a wall
at the end, it should be a straight line,
00:22:25.440 --> 00:22:32.150
there is a
wall at the end. So, what is happening now,
00:22:32.150 --> 00:22:37.100
it is a shock it is going like this and there
is a
00:22:37.100 --> 00:22:44.260
flow coming here, what was the velocity there,
we will pick the problem a, 435 meter per
00:22:44.260 --> 00:22:52.410
second, this is V 1 equal to 0, V 2 is this
and this value was 696 meter per second
00:22:52.410 --> 00:22:54.950
whatever.
Now, this problem does not change till the
00:22:54.950 --> 00:22:59.640
shock goes and hits the wall, after that the
shock does not have to tell any more gas.
00:22:59.640 --> 00:23:02.170
What was the job of the shock, it has to go
and
00:23:02.170 --> 00:23:07.750
tell everybody move that way and it has told
all the gas that is present in the tube, it
00:23:07.750 --> 00:23:11.940
is
job is done, it vanishes. But now, there is
00:23:11.940 --> 00:23:15.680
a problem, what is the problem, will go to
the
00:23:15.680 --> 00:23:21.230
case where shock is reached the end so I will
draw one more picture. There is no V 1
00:23:21.230 --> 00:23:28.390
equal to 0, every condition is all this value,
all the gas is moving with 435 meter per
00:23:28.390 --> 00:23:36.690
second into a wall.
Now, there is a problem, things cannot go
00:23:36.690 --> 00:23:42.750
just through a wall, wall stops all the flow
which means, now there should be a new shock
00:23:42.750 --> 00:23:49.930
created here. And that has to be moving
against this flow, this is going like this
00:23:49.930 --> 00:23:59.000
such that, the velocity behind it is supposed
to
00:23:59.000 --> 00:24:04.381
.come to 0, that is the job of this shock.
This shock is going and telling this, what
00:24:04.381 --> 00:24:07.940
they
call the reflected shock, this shock’s job
00:24:07.940 --> 00:24:10.360
is to go and tell every fluid that is coming
this
00:24:10.360 --> 00:24:14.110
way that, there is a wall you cannot move
this way.
00:24:14.110 --> 00:24:17.760
So what will happen, all the people thought,
all the molecules actually, we will go back
00:24:17.760 --> 00:24:20.860
to molecules, all the molecules thought there
is a lot of space they are all coming this
00:24:20.860 --> 00:24:28.080
way that is what, this shock told them. When
the shock went here, till that point there
00:24:28.080 --> 00:24:34.650
was no information about this wall to this
fluid so they all kept coming. When the first
00:24:34.650 --> 00:24:39.500
layer of fluid that touched this, when the
shock that touched happen immediately after
00:24:39.500 --> 00:24:44.250
that, that fluid knows that there is a wall
there.
00:24:44.250 --> 00:24:49.490
So, it tells every other fluid this way coming
in, that do not come in this way there is
00:24:49.490 --> 00:24:52.060
a
problem here, that information is going this
00:24:52.060 --> 00:24:57.100
way. What will happen, this inducing a
velocity this way, in a way it is slowing
00:24:57.100 --> 00:25:00.090
down this fluid coming to the wall that is
what,
00:25:00.090 --> 00:25:03.620
is happening here, this is the physical feel.
Once you have the physical feel, problem is
00:25:03.620 --> 00:25:13.210
easier to solve and it has to come to V equal
to 0 that is what, we are trying to solve.
00:25:13.210 --> 00:25:16.740
Now, instead of looking at the problem with
shock going right to left with all velocities
00:25:16.740 --> 00:25:20.820
negative, I will just go look from the behind
of the board.
00:25:20.820 --> 00:25:26.360
What will it look like, I will just transform
this problem to I turn the board the other
00:25:26.360 --> 00:25:29.350
way
and look at it from the other side. Then,
00:25:29.350 --> 00:25:39.940
it will look like, this is the problem we
are going
00:25:39.940 --> 00:25:49.550
to solve, here I am given V 1 and V 2 both
the values are given, I am renumbering things
00:25:49.550 --> 00:25:54.960
it is no more connected with problem a. I
am renumbering it here V 1 and V 2, this old
00:25:54.960 --> 00:26:00.570
numbering I will remove, would not get confused
with these numbers. This is the
00:26:00.570 --> 00:26:04.720
problem we want to solve now, we are supposed
to find what is the speed, at which the
00:26:04.720 --> 00:26:13.890
shock goes.
So, let us go and find out how to do this
00:26:13.890 --> 00:26:18.843
so again we are going to write the same set
of
00:26:18.843 --> 00:26:26.570
equations, we will say u 1 equal to V 1 minus
W S, u 2 equal to V 2 minus W S
00:26:26.570 --> 00:26:34.820
remember, there is no compressible flow tables
for moving shocks. So, we have to do
00:26:34.820 --> 00:26:41.480
this transformation of coordinates everytime
to go to stationary shock case and then use
00:26:41.480 --> 00:26:46.080
the tables come back to moving shock problem,
transform to one side come back
00:26:46.080 --> 00:26:52.330
transform again.
We are now given V 2, this happen to be 0
00:26:52.330 --> 00:26:58.250
and V 1 has a value 435 meter per second,
that is given to us. So, I will again go and
00:26:58.250 --> 00:27:02.930
write V 2 minus V 1 equal to u 2 minus u 1
00:27:02.930 --> 00:27:08.710
.remember, to keep the velocity directions
important. If it goes negative if it goes
00:27:08.710 --> 00:27:18.890
leftwards, it is minus sign, will keep that
into account. So, V 1 minus 435 meter per
00:27:18.890 --> 00:27:24.930
second, I will write it here I have put an
arrow on 435 so that, we do not want to touch
00:27:24.930 --> 00:27:29.390
that, I will change it only here.
V 1 equal to minus 435 meter per second which
00:27:29.390 --> 00:27:34.860
means, it is going right to left, will keep
it that way and V 2 is zero. So, I am finally
00:27:34.860 --> 00:27:40.190
getting it to be this side 0 minus, minus
435
00:27:40.190 --> 00:27:48.720
will be 435 equal to now, I do not know u
2, I do not know u 1 and I am supposed to
00:27:48.720 --> 00:27:53.840
guess mach number of the shock or P 2 by P
1 of the shock. So, we have to work through
00:27:53.840 --> 00:28:11.400
this problem only, how will I do it, let us
say, I will write u 2 as minus M 2 a 2
00:28:11.400 --> 00:28:16.750
remember, minus is coming because in my shock,
fixed coordinate it is going from right
00:28:16.750 --> 00:28:19.970
to left.
Similarly, I will do for u 1, it will become
00:28:19.970 --> 00:28:30.220
a plus M 1 a 1, do I know a 1, yes I know
a 1
00:28:30.220 --> 00:28:37.170
because I found the temperature here. T 2
by T 1 was known, for the problem a, T 2 by
00:28:37.170 --> 00:28:42.140
T
1 was I think 1.68 something like that, I
00:28:42.140 --> 00:28:59.550
did not give you that part of the problem
anyway, I will just do it right now, 300 into
00:28:59.550 --> 00:29:05.980
1.687. We wrote this T 2 by T 1 value before
in fact, in this page in fact, it is slightly
00:29:05.980 --> 00:29:14.030
shown here, anyway. So from there, I will
get a 1
00:29:14.030 --> 00:29:19.900
value I have to find square root of gamma
R T and that comes out to be 452.2 meter per
00:29:19.900 --> 00:29:27.300
second.
You get it to this value, you know up to here,
00:29:27.300 --> 00:29:39.740
I do not know M 1 but if I know M 1, I will
know M 2 now a 2, can I write in terms of
00:29:39.740 --> 00:29:47.500
a 1. Yes I can, will write it that way M 1
a 1
00:29:47.500 --> 00:29:58.180
minus M 2 a 1 into square root of T 2 by T
1, can I write it like this. Yes because a
00:29:58.180 --> 00:30:02.450
2 by a
1 is just square root of gamma R T 2 divided
00:30:02.450 --> 00:30:07.790
by square root of gamma R T 1 so I can
simplify to this. So, what I did is, a 2 is
00:30:07.790 --> 00:30:14.350
equal to a 2 by a 1 times a 1, again the favorite
multiplication of ratios, gas dynamics deal
00:30:14.350 --> 00:30:16.830
with.
So now, I have this, I do not know M 1, I
00:30:16.830 --> 00:30:21.000
know a 1, I know a 1, I do not know M 2, I
do
00:30:21.000 --> 00:30:27.380
not know T 2 by T 1 but if I know M 1, I will
know M 2 from normal shock tables, I
00:30:27.380 --> 00:30:34.740
know T 2 by T 1 from the same row normal shock
tables. So, if I know M 1, I know this
00:30:34.740 --> 00:30:42.600
whole expression and then now I have to see,
whether 435 is matched so it is now a
00:30:42.600 --> 00:30:51.250
problem of iteration. I have to guess an M
1 and see, if 435 is matched by this side
00:30:51.250 --> 00:30:54.400
and I
keep on doing this, till I solve the problem.
00:30:54.400 --> 00:31:00.230
.I will start guessing m 1 value, next value,
next value till it solves, instead of writing
00:31:00.230 --> 00:31:03.350
it
like this, we will rearrange this and write
00:31:03.350 --> 00:31:16.270
it a little better, will go to the next section.
.
00:31:16.270 --> 00:31:27.809
It so happens that, for most of the problems
it will be this, delta V by a 1 equal to M
00:31:27.809 --> 00:31:34.110
1
minus M 2 square root of T 2 by T 1. I will
00:31:34.110 --> 00:31:37.390
keep getting this always in most of the
moving shock problems where, I do not know
00:31:37.390 --> 00:31:42.380
the shock strength. The previous problem
section a and b when be solved, we knew the
00:31:42.380 --> 00:31:48.700
P 2 by P 1 for the shock. So, it is very easy
to solve. Here, we need iteration but this
00:31:48.700 --> 00:31:51.561
is the general form where, delta V is V 2
minus
00:31:51.561 --> 00:31:57.510
V 1 that is what, you will have.
Now, will get to numbers so I will take this
00:31:57.510 --> 00:32:01.610
also to that side, I will write a function
that
00:32:01.610 --> 00:32:10.720
looks like M 1 minus M 2 square root of T
2 by T 1 minus, I know delta V was 435 and
00:32:10.720 --> 00:32:14.300
I
know a 1 is 452.2 I will get that number to
00:32:14.300 --> 00:32:22.660
be 0.962 equal to 0. So, I have to form a
table
00:32:22.660 --> 00:32:34.940
for iteration M 1, M 2, T 2 by T 1 and the
whole thing, I will call it left hand side.
00:32:34.940 --> 00:32:39.241
So, we
have to start guessing, each column will be
00:32:39.241 --> 00:32:44.320
one guess, I will start guessing with mach
number 2.
00:32:44.320 --> 00:32:50.380
And I go to normal shock tables, gamma equal
to 1.4 and I look for M 2 that is, 0.577,
00:32:50.380 --> 00:32:54.200
T
2 by T 1 happens to be, we already know this
00:32:54.200 --> 00:33:01.730
number 1.687. Now, if I do this calculation
on the left and side, that number comes out
00:33:01.730 --> 00:33:13.020
to be 0.285, I do not know how to guess the
next number. I will generally guess 3, I could
00:33:13.020 --> 00:33:19.390
be having to guess less than 2 or more than
2 I do not know, I say I choose 3, will see
00:33:19.390 --> 00:33:23.960
what happens. This number is 4.475, again
00:33:23.960 --> 00:33:34.370
.normal shock tables gamma equal to 1.4 for
mach 3, this number is 2.679 and this is 1.26
00:33:34.370 --> 00:33:41.440
so it looks like I went the wrong direction.
I guessed wrong, I should not be guessing
00:33:41.440 --> 00:33:46.740
from 2, I should have guessed lesser, I guess
the more, the number went the same direction.
00:33:46.740 --> 00:33:51.200
Most of the functions in gas dynamics are
monotonic, it is a good thing for us that
00:33:51.200 --> 00:33:58.920
is what, is used here. Now, I will go guess
something lesser, this is very close to 0
00:33:58.920 --> 00:34:01.410
and when I go from one number increase, it
is
00:34:01.410 --> 00:34:06.690
too high. So, I will guess something very
close to this but lesser, something nice I
00:34:06.690 --> 00:34:11.179
will
start with 1.5, that number is 0.701, 1.32.
00:34:11.179 --> 00:34:27.220
And now, my left and side comes out to be
minus 0.267 now, I know the answer is
00:34:27.220 --> 00:34:33.530
between this and this, I just have to guess
a number in between. It so happens that, in
00:34:33.530 --> 00:34:38.850
my
gas tables there is no 1.75, my gas tables
00:34:38.850 --> 00:34:45.800
has 1.7, 1.72, 1.74, 1.76. So, I thought I
will
00:34:45.800 --> 00:34:56.950
pick 1.7, I am still guessing I am not interpolating,
I could have interpolated and got to
00:34:56.950 --> 00:35:12.140
the answer faster. I got this answer which
says, I should go more towards 2 even now,
00:35:12.140 --> 00:35:15.610
I
could interpolate but it is so close that,
00:35:15.610 --> 00:35:18.350
we will just try the next number, in the next
row
00:35:18.350 --> 00:35:27.460
in my gas tables, that is easier.
So, I will go 1.72, this is 1.70, the next
00:35:27.460 --> 00:35:37.080
one is 1.72 and that number is 0.635, 1.473
and
00:35:37.080 --> 00:35:46.560
this number comes out to be minus 0.013 it
is even close but still the same side. So,
00:35:46.560 --> 00:35:48.730
if I
took a interpolation probably, I would have
00:35:48.730 --> 00:35:55.990
been a closer match, I do not know I am
guessing so it comes out to be this, I could
00:35:55.990 --> 00:35:58.170
have picked 1.74 instead of, 1.75 I picked
1.7.
00:35:58.170 --> 00:36:13.090
Now, we will look at the next one 1.74, I
am moving the correct direction, 0.631, 1.487
00:36:13.090 --> 00:36:22.430
and this number is 0.008, this is the closest
I could get and I have crossed.
00:36:22.430 --> 00:36:27.330
If I want at this point, I could interpolate
between these two to get to a closer answer,
00:36:27.330 --> 00:36:32.300
which will probably be 1.735, which if I approximate
to second decimal the answer is
00:36:32.300 --> 00:36:43.200
1.74 so this happens to be my mach number.
Typically, in gas dynamics, if you guess
00:36:43.200 --> 00:36:48.500
right at the initial conditions, you will
solve in the 6 th or 7 th iteration, most
00:36:48.500 --> 00:36:52.740
of the
problems. Unless you start with say, mach
00:36:52.740 --> 00:36:57.910
20 and slowly decrease every 2 then you will
take some so many iterations to get to the
00:36:57.910 --> 00:37:02.650
answer.
So, once I know M 1 equal to 1.74, I will
00:37:02.650 --> 00:37:10.590
go and find W S, we had the formula on the
board before so I will just write it. This
00:37:10.590 --> 00:37:18.500
you know, how to derive it from two different
methods, this number happens to be 351.8 meter
00:37:18.500 --> 00:37:23.960
per second, this is the speed with which
the shock moves with respect to outside observer
00:37:23.960 --> 00:37:29.550
with respect to the duct. And P 2 by P 1
00:37:29.550 --> 00:37:46.990
.for this shock is 3.336, now we will again
observe stuffs, I have numbers anyway, I will
00:37:46.990 --> 00:37:53.210
erase and do it.
.
00:37:53.210 --> 00:38:07.520
Problem a, we had a case where, a shock is
moving this way P 1 equal to 0, this was 435
00:38:07.520 --> 00:38:19.730
and this was 696 and now we are saying, it
went and it found a wall and so that becomes
00:38:19.730 --> 00:38:26.750
your what was this problem, problem c. Problem
b I said, I did not erase b probably here,
00:38:26.750 --> 00:38:35.690
it is not problem b we are working on problem
c I think. So, I just did not erase that part,
00:38:35.690 --> 00:38:47.940
anyway problem a set this, problem c it says,
now this is the velocity here now, I am
00:38:47.940 --> 00:38:59.410
thinking about shock going this way and the
flow incoming is 435 meter per second, V 2
00:38:59.410 --> 00:39:05.290
is 0.
And when the shock goes this way, it is going
00:39:05.290 --> 00:39:12.540
with velocity we just found 351 meter per
second, it is close to 352 meter per second.
00:39:12.540 --> 00:39:14.480
We are not interested in the actual number
I
00:39:14.480 --> 00:39:34.240
just want to tell you the trend, here P 2
by P 1 was 4.5 and here, P 2 by P 1 is 3.336
00:39:34.240 --> 00:39:38.060
some
such number, I could have written as 3.34
00:39:38.060 --> 00:39:46.810
that is enough. So, what we are finding is,
when my shock goes first, it is having some
00:39:46.810 --> 00:39:49.730
particular pressure ratio and then it is created
a supersonic flow.
00:39:49.730 --> 00:39:57.110
Actually, it was not supersonic, was it M
2 we did not calculate, I think I calculated
00:39:57.110 --> 00:39:59.690
I did
not do that part for you people, M 2 was 0.577
00:39:59.690 --> 00:40:04.460
it was not supersonic really, the
temperature is also high, velocity is high
00:40:04.460 --> 00:40:08.500
and temperature is high, it is still subsonic
flow.
00:40:08.500 --> 00:40:16.910
Because, the speed of sound for this case,
a 1 we picked was 450 or something like that,
00:40:16.910 --> 00:40:23.950
.something higher. So, it is still subsonic
case somewhere here, anyways it is subsonic
00:40:23.950 --> 00:40:28.470
flow created, it is going like this and it
is suddenly meeting a wall and that information
00:40:28.470 --> 00:40:32.280
has to be transferred to the whole fluid by
the second shock.
00:40:32.280 --> 00:40:39.930
It is not moving very fast as fast as this
one, it is moving slower but with respect
00:40:39.930 --> 00:40:44.359
to the
fluid, it is actually moving very fast. With
00:40:44.359 --> 00:40:47.910
respect to the fluid if I think about it,
it is
00:40:47.910 --> 00:40:52.350
going extremely fast. The fluid is going this
way but still the wave is going this way,
00:40:52.350 --> 00:40:55.060
it is
like a treadmill you know, it is like a treadmill
00:40:55.060 --> 00:41:01.040
what is the thing the belt, the belt is going
this way with 435 meter per second and the
00:41:01.040 --> 00:41:05.730
person running on it is running this way for
352 with respect outsider.
00:41:05.730 --> 00:41:13.670
Which means, is actually running the total
velocity added these two, which is 787 meter
00:41:13.670 --> 00:41:21.520
per second, the shock is actually travelling
faster in the gas that is heated, the main
00:41:21.520 --> 00:41:25.190
thing
is the gas is hot, it has been heated by the
00:41:25.190 --> 00:41:28.520
first shock. So, it is traveling very fast
but
00:41:28.520 --> 00:41:35.869
when it reflects, it does not need such a
high strong shock for it to tell the fluid
00:41:35.869 --> 00:41:38.690
to stop,
that is the main thing here. Previously, it
00:41:38.690 --> 00:41:44.240
was going with 4.5 now, it is going with 3.336,
that is enough.
00:41:44.240 --> 00:41:51.270
With that, it can say stop there is no more
movement needed, that is the information
00:41:51.270 --> 00:41:58.740
taken by this shock, that is all it is carrying.
Overall problem if you look at it, I had a
00:41:58.740 --> 00:42:07.320
duct, I had zero stagnant gas initially, there
was a shock sent in P 2 by P 1 4.5, it created
00:42:07.320 --> 00:42:15.240
a flow of 435 meter per second this way and
then it stopped and that information is
00:42:15.240 --> 00:42:21.230
carried by this. And then every gas that is
processed by this second shock will come to
00:42:21.230 --> 00:42:25.160
stop, it will not move any more.
What happened to the overall gas that was
00:42:25.160 --> 00:42:28.230
initially sitting in this whole region now,
it is
00:42:28.230 --> 00:42:33.820
compressed and it is sitting in a very small
region here, that is the other thing that
00:42:33.820 --> 00:42:36.240
you
have to think about. If I think about let
00:42:36.240 --> 00:42:41.950
us say, I will pick this whole volume, that
whole
00:42:41.950 --> 00:42:50.109
volume is probably now sitting in this small
area here, that small volume here why, that
00:42:50.109 --> 00:42:58.010
gas was compressed twice once by this, once
by this. So, overall if I want to call it
00:42:58.010 --> 00:43:01.560
P 3 by
P 2 then I can say the net final pressure
00:43:01.560 --> 00:43:07.790
is 4.5 times 3 of the order of 14, of the
order of
00:43:07.790 --> 00:43:12.530
14 times compressed.
So, the volume will be 1 by 14 of that roughly,
00:43:12.530 --> 00:43:19.000
it need not be that because temperature is
also changing, it is roughly that. So, all
00:43:19.000 --> 00:43:21.730
that volume is now only here, all that gas
which
00:43:21.730 --> 00:43:26.620
was initially there let us say, if I had this
gas or some color gas, all that color will
00:43:26.620 --> 00:43:27.620
be only
00:43:27.620 --> 00:43:33.650
.inside this small region, what is here then
the gas that was in this volume here, like
00:43:33.650 --> 00:43:36.040
that it
gets compressed on the other side, you have
00:43:36.040 --> 00:43:48.510
compressed it twice with two different
shocks. I guess I will not be able to complete
00:43:48.510 --> 00:43:52.040
problem d fully, we will see how much
ever we can go, if needed we will stop with
00:43:52.040 --> 00:43:58.340
the iteration part. We will go to problem
d, I
00:43:58.340 --> 00:44:11.540
can take a shortcut a little bit with using
the expressions from forth, we will see.
00:44:11.540 --> 00:44:12.540
.
00:44:12.540 --> 00:44:20.410
It is almost problem c except for I am picking
a particular velocity, this is a case where,
00:44:20.410 --> 00:44:24.240
I
am having a flow through a duct 20 meter per
00:44:24.240 --> 00:44:32.170
second. I am having a simple flow, flow
through a pipe 20 meter per second and suddenly,
00:44:32.170 --> 00:44:39.620
somebody puts a lid on it, they closed
the flow. I want to find what happens inside
00:44:39.620 --> 00:44:44.900
the duct now, this is the unsteady problem,
we are trying to solve an unsteady problem
00:44:44.900 --> 00:44:51.040
already. What is going to happen, all this
flow has to be stopped because there is a
00:44:51.040 --> 00:44:54.380
lid that is closed and says no more flow through
it.
00:44:54.380 --> 00:45:01.070
So, there has to be a shock that is going
this way and V equal to 0 should be enforced
00:45:01.070 --> 00:45:06.690
there by that shock that is the goal that
is what, the goal of this particular shock,
00:45:06.690 --> 00:45:10.030
it is
going to do that job. Let us say, we know
00:45:10.030 --> 00:45:19.530
this T 1 equal to 300 Kelvin temperature here,
V 1 equal to this and this means that, a 1
00:45:19.530 --> 00:45:30.000
equal to 348.2 meter per second, I am using
same number so that, it is easier. Now, we
00:45:30.000 --> 00:45:31.620
do not like solving the problem from right
to
00:45:31.620 --> 00:45:34.490
left, we are always using left to right.
00:45:34.490 --> 00:45:40.130
.So, I will transform this to a problem of
this type and I am going to say there is a
00:45:40.130 --> 00:45:43.619
shock
that is moving left to right and it is job
00:45:43.619 --> 00:45:49.200
is to make V 2 equal to 0 while V 1 was having
a
00:45:49.200 --> 00:45:55.599
minus 20 meter per second velocity. Now, I
am putting correct direction minus 20
00:45:55.599 --> 00:46:10.330
because it is moving the other way, so V 2
minus V 1 equal to 0 minus of minus 20, 20
00:46:10.330 --> 00:46:30.780
meter per second, this is also equal to u
2 minus u 1. So, we want to solve this, we
00:46:30.780 --> 00:46:39.290
also
know this, we wrote the expressions in different
00:46:39.290 --> 00:46:43.770
form now, I am writing another form
that is all.
00:46:43.770 --> 00:46:50.090
Actually, I do not need to do all these, I
will just we do not want to derive the whole
00:46:50.090 --> 00:46:54.760
thing, I will take the expression from below.
From the previous time, we wrote this know
00:46:54.760 --> 00:47:05.180
delta V by a 1 is equal to M 1 minus M 2 square
root T 2 by T 1, I can use this directly,
00:47:05.180 --> 00:47:06.520
I
do not want to go derive this whole thing
00:47:06.520 --> 00:47:10.160
for you. It is same problem as before except
for
00:47:10.160 --> 00:47:14.280
we gave a new dimension to the problem, we
said it is an unsteady problem suddenly,
00:47:14.280 --> 00:47:19.310
there is a wall that appeared in the duct
after that, the problem is exactly the same,
00:47:19.310 --> 00:47:21.500
that
way I will cut short and finish it off in
00:47:21.500 --> 00:47:25.660
this 2 minutes left.
So, what do we do exactly same procedure,
00:47:25.660 --> 00:47:29.960
delta V is given to be 20 meter per second,
a
00:47:29.960 --> 00:47:43.210
1 is given to be 348.2 so I will get a number
for this, that number is 0.057. So, I just
00:47:43.210 --> 00:47:46.560
have
to solve this problem with this expression
00:47:46.560 --> 00:47:58.510
M 1 minus M 2 square root T 2 by T 1 minus
0.057 equal to 0, this is my left hand side
00:47:58.510 --> 00:48:00.921
and I just want to solve this problem. Again
I
00:48:00.921 --> 00:48:05.920
have to go on guess that is the same thing,
I will go through the same guessing process,
00:48:05.920 --> 00:48:22.345
will again go form the same table I do not
want to write below so I will go to next page.
00:48:22.345 --> 00:48:23.345
..
00:48:23.345 --> 00:48:28.200
Again, I always like starting with M 1 equal
to 2 so we will start with that in fact, I
00:48:28.200 --> 00:48:38.290
remember this number 0.577 also, this gives
you left hand side of 1.19, while it is
00:48:38.290 --> 00:48:43.880
supposed to be 0. So, I am guessing too bad,
now I have experience from the previous
00:48:43.880 --> 00:48:48.839
problems I know, if I increase mach numbers
left hand side increases so I will decrease
00:48:48.839 --> 00:48:56.940
immediately. Let us say, I will guess something
low 1.4 because it is too high I am going
00:48:56.940 --> 00:49:04.830
more, I cannot guess 1 if I guess 1, things
will go wrong, it would not go wrong very
00:49:04.830 --> 00:49:05.830
much.
.
00:49:05.830 --> 00:49:12.240
.Go back here, if I guess 1, what will happen
what will be M 2 value, it will also be 1,
00:49:12.240 --> 00:49:17.950
what will be the value of T 2 by T 1, no change
it is a mach wave T 2 by T 1 is also 1 so
00:49:17.950 --> 00:49:26.790
1 minus 1 will be 0 and it is minus 0.057.
Now, you look at it, it is very close to 0
00:49:26.790 --> 00:49:31.130
so I
should not be guessing 1.4 really, I should
00:49:31.130 --> 00:49:32.830
be guessing something close to 1 so that,
I
00:49:32.830 --> 00:49:41.160
will get to the answer. We will do that instead
of 1.4, I was not thinking like this when
00:49:41.160 --> 00:49:42.970
I
was doing the problem in one paper.
00:49:42.970 --> 00:49:43.970
.
00:49:43.970 --> 00:49:56.609
So, I will start guessing 1.1 here, I skip
one column in my guess 0.911, 1.065 and this
00:49:56.609 --> 00:50:08.421
comes out to be 0.103 this is what, we have
here. And I know if it is 1.0, we know the
00:50:08.421 --> 00:50:17.490
values it is we just did this know, 0.057
you know this also so the answer is somewhere
00:50:17.490 --> 00:50:23.940
in between this, probably very close to 1.
Now, I just have to guess let us say, in my
00:50:23.940 --> 00:50:29.180
tables I have only a set of numbers I will
get 1.02, that is the closest I can get to
00:50:29.180 --> 00:50:37.920
1.
I have 0.981, 1.013 that gives minus 0.024,
00:50:37.920 --> 00:51:00.520
I want to go even closer so I have to increase
from here a little bit, go 1.04, 0.962, 1.026
00:51:00.520 --> 00:51:08.099
and this number comes out to be 0.008, which
we will say is approximately 0. So, I am getting
00:51:08.099 --> 00:51:16.300
a mach number to be this, for if I stop at
20 meter per second flow in a duct with a
00:51:16.300 --> 00:51:21.800
sudden lid completely closed no mass flow
through it then I will produce a 1.04 mach
00:51:21.800 --> 00:51:27.300
number flow, mach number shock going
against the flow that is what, we know as
00:51:27.300 --> 00:51:31.109
of now.
We will go and find the implications of it
00:51:31.109 --> 00:51:36.990
next class, because I have one more full page
of implications, because of this. We cannot
00:51:36.990 --> 00:51:39.530
go through that, now we will stop here, we
00:51:39.530 --> 00:51:44.349
.will just stop with M 1 equal to 1.04, will
come and continue the same problem next
00:51:44.349 --> 00:51:49.570
class at the beginning. After that, we will
do one more problem and then we will go to
00:51:49.570 --> 00:51:55.470
2D problems, any questions you have. So, easy
way of solving, if you just remember that
00:51:55.470 --> 00:51:59.920
one formula and just iterate, see you people
next class.
00:51:59.920 --> 00:52:00.920
.