WEBVTT
Kind: captions
Language: en
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Hello every one welcome back we just discussed
last class about some set of subroutines
00:00:17.090 --> 00:00:25.150
a interior points of subroutine and wall boundary
subroutine the next immediate thing
00:00:25.150 --> 00:00:34.339
you can go for I would say is free boundary
subroutine free boundary subroutine is
00:00:34.339 --> 00:00:40.690
related to pressure boundary condition we
already talked about in gas dynamics we either
00:00:40.690 --> 00:00:48.550
give velocity direction as boundary condition
or pressure in m o c it comes out to be
00:00:48.550 --> 00:00:52.680
velocity direction is your theta and pressure
comes out to be in terms of mark number
00:00:52.680 --> 00:00:56.090
those are your boundary conditions.
.
00:00:56.090 --> 00:01:07.490
If say I somehow know an already existing
point on boundary say this is my nozzle wall
00:01:07.490 --> 00:01:12.810
and I know how to solve this and suddenly
there is free jerk going outside and say I
00:01:12.810 --> 00:01:24.091
know this point a and I have already calculated
from this side interior point b now I want
00:01:24.091 --> 00:01:33.000
to find a point c on this that is my goal
I want to find the point c on a pressure boundary
00:01:33.000 --> 00:01:39.950
of this jet that is what we want and what
is given since it is a pressure boundary I
00:01:39.950 --> 00:01:43.200
know
my exit pressure the back pressure there that
00:01:43.200 --> 00:01:49.579
is given now I just have to go and find these
points how will I find I already told you
00:01:49.579 --> 00:01:53.180
the answer in some way mark number is decided
00:01:53.180 --> 00:01:59.900
.by the pressure that says that I already
know the p not in this region I know my
00:01:59.900 --> 00:02:02.970
stagnation pressure in.
This region I am assuming an isotropic flow
00:02:02.970 --> 00:02:04.750
all through right. So, I am keeping track
of
00:02:04.750 --> 00:02:10.200
my p not value all through and if I know my
p not value at this location then I can now
00:02:10.200 --> 00:02:21.019
find my mark number at c is going to be given
by p b by p not at c right that is giving
00:02:21.019 --> 00:02:29.799
your mark number which will in turn give you
nu at c once you know nu at c remaining
00:02:29.799 --> 00:02:36.380
thing is simple if I know this I am going
to send a positive characteristic from b towards
00:02:36.380 --> 00:02:50.040
c. So, I will have theta b minus minus nu
b equal to theta c minus nu c positive
00:02:50.040 --> 00:02:56.400
characteristic it will be a difference right.
So, and we know this we know this we know
00:02:56.400 --> 00:03:03.310
this. So, we can find this now theta c can
be obtained. So, the basic starting point
00:03:03.310 --> 00:03:06.920
will be
I have found theta c where in my wall boundary
00:03:06.920 --> 00:03:11.100
condition what did we know we know
we knew theta c that was given and we had
00:03:11.100 --> 00:03:19.230
to find nu c here it is the opposite problem
that is the only difference here k now after
00:03:19.230 --> 00:03:25.290
we have found this remaining thing is to find
the velocity direction there we know that
00:03:25.290 --> 00:03:30.000
the z boundary will be going with some
velocity vector value some some direction
00:03:30.000 --> 00:03:35.880
and that will be equal to your velocity vector
direction. So, I know that that location theta
00:03:35.880 --> 00:03:38.850
c is the angle of the velocity vector that
is
00:03:38.850 --> 00:03:44.500
found already that is tan of theta c will
be divided by d y by d x at that location,
00:03:44.500 --> 00:03:48.030
but what
I do not know is what is this curve here if
00:03:48.030 --> 00:03:50.049
I do not know the curve I cannot find the
b c
00:03:50.049 --> 00:03:56.040
line intersecting the this curve a c. So,
now, I have to find the point c as in the
00:03:56.040 --> 00:04:01.700
location x
y of point c to do that we again go for.
00:04:01.700 --> 00:04:10.519
Similar approximations [ noise] I am going
to say this is my point a this is my point
00:04:10.519 --> 00:04:18.750
c
and there is b here of course, we will start
00:04:18.750 --> 00:04:23.030
using this also as a curve characteristic
is also
00:04:23.030 --> 00:04:36.050
a curve and then now I am going to say d y
by d x for b c line is going to be tan of
00:04:36.050 --> 00:04:46.120
theta b
plus mu b plus theta c plus mu c whole by
00:04:46.120 --> 00:04:51.220
two similar to what we did before nothing
different it is same as positive characteristic
00:04:51.220 --> 00:04:54.480
for interior point exactly the same thing
now
00:04:54.480 --> 00:05:03.380
the other line it will be different from this
d y by d x for a c remember that a c is not
00:05:03.380 --> 00:05:05.220
a
characteristic line it is a stream line. So,
00:05:05.220 --> 00:05:08.880
it will go only with theta values it will
not go
00:05:08.880 --> 00:05:16.040
with theta plus mu or theta minus mu because
of that it will be tan of I will again do
00:05:16.040 --> 00:05:18.740
the
average business here it is theta a there
00:05:18.740 --> 00:05:24.800
it is theta c. So, it is theta a plus theta
c by two
00:05:24.800 --> 00:05:32.840
this is all it will be this is what you will
get k now with these two I have found an
00:05:32.840 --> 00:05:38.330
approximate line b c and approximate line
a c with that I am calculating the location
00:05:38.330 --> 00:05:41.550
of
point c we know lines two lines intersection
00:05:41.550 --> 00:05:43.390
we can calculate geometry simple geometry
00:05:43.390 --> 00:05:49.010
.I think it is high school even. So, we know
how to calculate the exact location x y from
00:05:49.010 --> 00:05:57.340
here remaining thing is just believe I will
write it once y c minus y b by x c minus x
00:05:57.340 --> 00:05:59.870
b is
equal to.
00:05:59.870 --> 00:06:06.590
D y by d x of b c this is one of the line
equations similarly I can replace this b by
00:06:06.590 --> 00:06:10.490
a I will
get another line equation for a c these two
00:06:10.490 --> 00:06:13.700
if I solve for y c together I will get value
of y c
00:06:13.700 --> 00:06:20.050
and then I will substitute that to get x c
thats all we will do typically k. So, we have
00:06:20.050 --> 00:06:34.800
another routine now which is what I call as
pressure boundary point that is I am finding
00:06:34.800 --> 00:06:39.840
a
value of as in the x y theta and mark number
00:06:39.840 --> 00:06:47.830
for a pressure boundary point now that
being said I actually did from wall corner
00:06:47.830 --> 00:06:53.610
two of pressure boundary will it work from
here at another point yes it will because
00:06:53.610 --> 00:07:00.960
I will go draw that separately we said that
this is
00:07:00.960 --> 00:07:06.460
some wall from here we did an interior point
somehow it came from here we got this
00:07:06.460 --> 00:07:11.200
point c.
Now how will I find this point next of course,
00:07:11.200 --> 00:07:18.270
there is one more characteristic going from
here right these two points will give you
00:07:18.270 --> 00:07:22.030
another interior point here that interior
point can
00:07:22.030 --> 00:07:28.310
now do exactly the same whatever we did with
this analysis these three points together.
00:07:28.310 --> 00:07:32.860
Now this will become your new a this will
become your nu b this will become your nu
00:07:32.860 --> 00:07:37.490
c
k the nu values a b and c now I will just
00:07:37.490 --> 00:07:40.010
take this these two points as input we will
get
00:07:40.010 --> 00:07:44.639
this as output and that is the wall I just
keep continuing along this line and u will
00:07:44.639 --> 00:07:46.320
note is
that if the jet have has to turn like this
00:07:46.320 --> 00:07:49.450
then it also turns naturally the problem just
solves
00:07:49.450 --> 00:07:55.900
by itself no need to worry too much about
it k we they have given you a simple cases
00:07:55.900 --> 00:08:02.419
now I just want to discuss a little bit before
I go back and give you more routines.
00:08:02.419 --> 00:08:03.419
..
00:08:03.419 --> 00:08:11.960
Say I have a flow and the flow is supposed
to be expanding what do we expect in this
00:08:11.960 --> 00:08:16.170
case. I am drawing it as if it is a smooth
expansion does not matter what I want to do
00:08:16.170 --> 00:08:19.970
I
am bringing about some case here here mark
00:08:19.970 --> 00:08:26.620
number as m 1 and eventually it has to turn
and become some m 2 which is greater than
00:08:26.620 --> 00:08:34.380
m 1. This is what we expect somewhere up
there. Right now if I think about what will
00:08:34.380 --> 00:08:43.860
happen to my mark angles mark angle this is
mu one and this will be mu two will mu two
00:08:43.860 --> 00:08:55.620
will be higher or lower it will be lower this
will be my nu two this s what I will have
00:08:55.620 --> 00:08:58.310
finally. Why mark number is higher one by
m
00:08:58.310 --> 00:09:06.400
sine inverse. So, it is going to become that
now what is happening to my characteristic
00:09:06.400 --> 00:09:09.570
if
I think about the positive characteristics
00:09:09.570 --> 00:09:14.589
there might have been a positive characteristic
running like this and here there would have
00:09:14.589 --> 00:09:22.290
been a positive characteristic running like
this and then now it has turned this way what
00:09:22.290 --> 00:09:28.100
is really happening I am having these two.
If I think about they were initially same
00:09:28.100 --> 00:09:34.670
mark number. So, they were same mu one value
here they were running parallel suddenly because
00:09:34.670 --> 00:09:40.140
they are experiencing this expansion
fan this whole region at the end of it it
00:09:40.140 --> 00:09:49.240
is coming out quite diverging a characteristic
lines
00:09:49.240 --> 00:09:56.000
another way of thinking about in terms of
compression I have lets say a compression
00:09:56.000 --> 00:10:04.740
wall I have one.
mu one here, but here we know have the mark
00:10:04.740 --> 00:10:09.820
number decreases because of the
compression. So, it will be a higher value
00:10:09.820 --> 00:10:17.360
mu two will be higher now what will happen
these two they are both left running characteristics
00:10:17.360 --> 00:10:23.399
you look at it far away they want to
meet each other that is what we will end up
00:10:23.399 --> 00:10:28.050
with in case of compression happening in the
00:10:28.050 --> 00:10:34.070
.flow the characteristics will want to converge
this same family characteristics. Now they
00:10:34.070 --> 00:10:39.550
are talking about family characteristics if
I think about characteristics that are left
00:10:39.550 --> 00:10:43.899
running they are called one one family they
are all left running characteristic family
00:10:43.899 --> 00:10:45.560
and
there could be another family which is right
00:10:45.560 --> 00:10:50.320
running characteristics same family
characteristics if they want to meet they
00:10:50.320 --> 00:10:56.730
come together then there is compression
happening in that region if they diverge then
00:10:56.730 --> 00:10:59.350
there is expansion happening in that local
region k.
00:10:59.350 --> 00:11:05.180
But of course, I cannot say this when I am
thinking about one left running and one right
00:11:05.180 --> 00:11:09.270
running I am thinking about if there is a
case where right running is turning this way
00:11:09.270 --> 00:11:10.870
and
left running is turning this way. I cannot
00:11:10.870 --> 00:11:13.790
tell anything about the flow at that point
I need
00:11:13.790 --> 00:11:19.000
more characteristics to tell what will happen
there all I can tell is the flow is non-uniform
00:11:19.000 --> 00:11:23.279
nothing more. If they are same these are left
running characteristics and they are coming
00:11:23.279 --> 00:11:28.890
together then; that means, there is compression
happening here in here there is expansion
00:11:28.890 --> 00:11:35.390
happening here and if there is one of this
eventually they will all be parallel right.
00:11:35.390 --> 00:11:37.930
If I
think about this particular wave we think
00:11:37.930 --> 00:11:42.470
about this particular wave this will be parallel
to this wave there they are all left running
00:11:42.470 --> 00:11:44.040
waves. I am drawing right with respect to
the
00:11:44.040 --> 00:11:46.500
flow of stream line they are all left running
waves.
00:11:46.500 --> 00:11:53.170
So, I am seeing that these will become parallel
eventually to that also k which means
00:11:53.170 --> 00:11:57.870
there is no more change in mark number in
this region same thing will happen here I
00:11:57.870 --> 00:12:01.480
draw one more characteristic here this will
be parallel to this characteristic. I am not
00:12:01.480 --> 00:12:07.080
drawing it exactly parallel it should be something
like this and they will all be parallel
00:12:07.080 --> 00:12:12.390
after this point and what about before this
compression they will also be parallel to
00:12:12.390 --> 00:12:17.600
previous value this is what will happen we
will come back to what happens when things
00:12:17.600 --> 00:12:22.170
go wrong when they cross we will come back
to. This particular corner when things go
00:12:22.170 --> 00:12:26.350
wrong later, but we will just look at expansion
first now I will give you one more
00:12:26.350 --> 00:12:29.720
subroutine expansion fan corner expansion
subroutine.
00:12:29.720 --> 00:12:30.720
..
00:12:30.720 --> 00:12:41.820
I will just give it as expansion fan subroutine
expansion fan when I say expansion fan I
00:12:41.820 --> 00:12:50.450
am going to think about. So, many mac waves
in one small fan ideally it may have a
00:12:50.450 --> 00:13:00.830
sharp corner expansion typically I am given
this angle theta and now we want to say that
00:13:00.830 --> 00:13:05.410
the wall has turned. I want to solve this
problem the way to solve it I know the incoming
00:13:05.410 --> 00:13:12.590
mark number at that corner and I know the
outgoing theta. Of course, I know theta mark
00:13:12.590 --> 00:13:16.860
number at this point outgoing theta I know
I do not know the mark number as of now,
00:13:16.860 --> 00:13:21.631
but that can be found by our simple gas dynamics
calculations. I know the m one I can
00:13:21.631 --> 00:13:26.350
find m two based on this theta delta theta
is equal to delta nu you can solve that now
00:13:26.350 --> 00:13:27.931
that
delta theta equal to delta nu is what you
00:13:27.931 --> 00:13:32.410
want to use here. Anyway if I draw the first
mark waves let us say it is like this the
00:13:32.410 --> 00:13:36.110
last one will be like this with respect to
this local
00:13:36.110 --> 00:13:41.300
velocity vector this will be one angle here
this will be another angle. Of course, this
00:13:41.300 --> 00:13:46.850
angle is more than this angle these all we
already know. Now the way to solve it we
00:13:46.850 --> 00:13:51.440
already told that there are infinite waves
sitting inside here we will just pick a few
00:13:51.440 --> 00:13:55.750
of
them typically I pick ten k if you want more
00:13:55.750 --> 00:14:01.779
resolution in data you can pick hundred two
hundred whatever depending on your computer
00:14:01.779 --> 00:14:06.050
power.
Typically I will pick ten I will draw only
00:14:06.050 --> 00:14:10.490
five lines here k first one is the incoming
thing
00:14:10.490 --> 00:14:15.510
last one is the outgoing and other things
are somewhere in the middle. I am picking
00:14:15.510 --> 00:14:18.680
five
points let us say n equal to five right now
00:14:18.680 --> 00:14:23.250
what I want to do is I know my initial mark
number I know the final mark number I know
00:14:23.250 --> 00:14:25.720
the initial theta I know the final theta.
All I
00:14:25.720 --> 00:14:33.000
have to do is distribute this whole data inside
here ideally I do not distribute mark
00:14:33.000 --> 00:14:39.940
.number what I do is this I know n equal to
five right. If I have a I have to create five
00:14:39.940 --> 00:14:50.410
point with exactly same x y location that
is that corner all points have same x comma
00:14:50.410 --> 00:14:54.810
y
which is given by your wall geometry. I am
00:14:54.810 --> 00:14:57.640
going to put five points exactly one on top
of
00:14:57.640 --> 00:15:02.029
each other and I am going to process them
in a sequence in serial order n equal to one
00:15:02.029 --> 00:15:03.649
n
equal to two n equal to three n equal to four
00:15:03.649 --> 00:15:08.330
n equal to five. In that order only k now
how
00:15:08.330 --> 00:15:16.820
will I create those theta values first theta
value same as your theta 1 k and last theta
00:15:16.820 --> 00:15:21.050
value
is your theta end these two are known. Now
00:15:21.050 --> 00:15:23.860
what I do is I linearly interpolate inside
here
00:15:23.860 --> 00:15:34.880
linear interpolation I just do linear interpolation
in between these k. So, I will typically
00:15:34.880 --> 00:15:38.029
pick the twenty five percent fifty percent
seventy five percent hundred percent kind
00:15:38.029 --> 00:15:41.520
of
interpolation in the n equal to five case
00:15:41.520 --> 00:15:44.070
this will be happening now. Once I do this
I go
00:15:44.070 --> 00:15:49.149
back and ask expansion what should I do delta
theta equal to delta nu. So, nu will come
00:15:49.149 --> 00:16:00.269
out to be nu one nu one plus a first one is
nu one second one is known as nu one plus.
00:16:00.269 --> 00:16:10.740
Delta theta 1 nu one plus delta theta 2 etcetera
up to the last one nu end you can;
00:16:10.740 --> 00:16:16.250
obviously, think about this one right it is
not very difficult delta theta. Here is the
00:16:16.250 --> 00:16:19.250
delta
theta here and delta theta here will be the
00:16:19.250 --> 00:16:21.440
delta theta for from here to here like that
I can
00:16:21.440 --> 00:16:25.980
keep on going like this if I make it all uniform
then this will just be delta theta. I do not
00:16:25.980 --> 00:16:32.190
need to put delta theta 1 and delta theta
2 etcetera now once I find these numbers I
00:16:32.190 --> 00:16:34.800
can
get my mark number directly this is coming
00:16:34.800 --> 00:16:40.450
from here function of mu. I will get a mark
number array based on for individual nu values
00:16:40.450 --> 00:16:47.450
I will find individual mark numbers at
each of those points these all actually if
00:16:47.450 --> 00:16:50.490
I think about expansion of an subroutine.
I just
00:16:50.490 --> 00:16:55.860
had to create five points n equal to five
here I could have had hundred it is just most
00:16:55.860 --> 00:17:04.589
laborious work I have created five points
and I know x y location exactly and I know
00:17:04.589 --> 00:17:09.720
theta and m we told you for every point we
should know these four values x y theta and
00:17:09.720 --> 00:17:17.159
m we will know all four. So, I have created
some set of points here how will I use this
00:17:17.159 --> 00:17:21.939
if
I have an expansion corner in my flow and
00:17:21.939 --> 00:17:32.369
say I have solved like this. Till now
somehow
00:17:32.369 --> 00:17:40.700
I got this kind of solution that is I knew
some points already from there I got all points
00:17:40.700 --> 00:17:45.359
from there I got this point and from this
point I got this interior point from this
00:17:45.359 --> 00:17:48.269
interior
point I got this interior point these two
00:17:48.269 --> 00:17:50.970
interior points giving this these two interior
points
00:17:50.970 --> 00:17:57.970
giving this these two interior points giving
this this two interior points giving this
00:17:57.970 --> 00:18:01.330
like
that I can keep on finding like that right.
00:18:01.330 --> 00:18:06.760
And from here I got a wall point from this
interior point and this wall point I can
00:18:06.760 --> 00:18:11.970
calculate another interior point. I can find
this one from this interior point and this
00:18:11.970 --> 00:18:16.769
.interior point I can find another interior
point here from this and this together I can
00:18:16.769 --> 00:18:19.659
find
one here this and this together I can find
00:18:19.659 --> 00:18:22.480
one here like that I can keep on going. But
say
00:18:22.480 --> 00:18:28.129
there was a time I will extend this characteristic
also say I got this kind of situation here.
00:18:28.129 --> 00:18:34.840
Now of course, I will find all these interior
points now there is a point where my
00:18:34.840 --> 00:18:41.059
characteristic is coming close to this wall.
If I am a programmer I know that there is
00:18:41.059 --> 00:18:45.239
a
wall corner there that should be a given it
00:18:45.239 --> 00:18:50.450
cannot be solved arbitrarily you can solve
arbitrarily if u take this a sa function alone.
00:18:50.450 --> 00:18:52.379
Then this characteristic will go there and
hit
00:18:52.379 --> 00:19:00.210
slightly ahead of it then I will use that
value and just it will just go up the next
00:19:00.210 --> 00:19:02.179
the next
characteristic will go and hit some other
00:19:02.179 --> 00:19:06.859
wall right it will not I will draw one more
line
00:19:06.859 --> 00:19:09.890
here.
So, that it is closer I know all these interior
00:19:09.890 --> 00:19:19.271
points if my characteristic comes and hits
here now I am going to say this theta is different
00:19:19.271 --> 00:19:24.519
from this theta. So, this line alone will
diverge like this now the expansion fan looks
00:19:24.519 --> 00:19:26.549
like this automatically it will be taking
care
00:19:26.549 --> 00:19:31.779
of if I just use wall point. But if I know
for sure that there is an expansion corner
00:19:31.779 --> 00:19:34.700
and I
want to take care of that corner specially
00:19:34.700 --> 00:19:38.860
then I will create a bunch of points here
and
00:19:38.860 --> 00:19:45.490
then send out characteristics from there for
n equal to five its going to look something
00:19:45.490 --> 00:19:51.759
like this five points. So, what I have to
do is when the characteristic from the right
00:19:51.759 --> 00:19:57.330
running family comes here each of these points
in that sequence n equal to one two tree
00:19:57.330 --> 00:20:01.220
four five in that sequence n equal to one
will interact with this first giving an interior
00:20:01.220 --> 00:20:04.220
point.
Then that interior point will interact with
00:20:04.220 --> 00:20:08.340
n equal to two here to get the next interior
point
00:20:08.340 --> 00:20:12.239
from there it will interact to get next interior
point from there it will interact to get the
00:20:12.239 --> 00:20:16.549
next interior point. From there it will give
the last interior point I will go through
00:20:16.549 --> 00:20:26.669
this
whole sequence if I draw this in big
00:20:26.669 --> 00:20:30.549
this characteristic somehow is coming from
somewhere these two will interact and give
00:20:30.549 --> 00:20:37.429
me this point this and this then next n equal
to two point will go get me one more point
00:20:37.429 --> 00:20:41.809
here this point and n equal to three point
will
00:20:41.809 --> 00:20:47.000
go and give one more point. Here this point
and n equal to four will give me one more
00:20:47.000 --> 00:20:52.710
point here this point and this point n equal
to five point here and ne this point will
00:20:52.710 --> 00:20:57.619
give
me one more interior point here now I have
00:20:57.619 --> 00:21:04.029
increased number of points in this line alone.
And not on this line that it will automatically
00:21:04.029 --> 00:21:08.869
take care of itself no need to worry. So,
much about it what if it is not a sharp corner,
00:21:08.869 --> 00:21:13.509
but it is a smooth corner I do not need to
worry about anything there. If it is a smooth
00:21:13.509 --> 00:21:18.919
corner then I will just let this flow evolve
by
00:21:18.919 --> 00:21:24.610
.itself let this flow like this you just keep
solving for all these conditions let it evolve
00:21:24.610 --> 00:21:29.299
by
itself it will automatically turn k why this
00:21:29.299 --> 00:21:31.950
positive characteristic will have theta plus
mu,
00:21:31.950 --> 00:21:37.789
but theta is decreasing. So, automatically
your theta plus mu will also decrease and
00:21:37.789 --> 00:21:39.799
that
will take care of itself. So, you will see
00:21:39.799 --> 00:21:42.480
that the diverging is happening if you want
to be
00:21:42.480 --> 00:21:48.110
more accurate for a sharp corner then I will
have to do this expansion fan method.
00:21:48.110 --> 00:21:52.049
Otherwise I will just go and use my wall point
routine it will just work very nicely we
00:21:52.049 --> 00:21:57.909
still have not given you a full algorithm
for solving a flow field around any object
00:21:57.909 --> 00:22:03.549
yet k I
am just telling you partial version of it
00:22:03.549 --> 00:22:07.529
smooth corners.
We do not need to write a special coordinate
00:22:07.529 --> 00:22:14.109
for it k as of now we are at a point where
we can go and solve some standard problems
00:22:14.109 --> 00:22:16.880
without any shocks. In our flow if there is
a
00:22:16.880 --> 00:22:22.499
shock if there is subsonic flow behind the
shock we cannot solve it beyond that point.
00:22:22.499 --> 00:22:26.870
But if it is staying supersonic say it is
an oblique shock, weak oblique shock then
00:22:26.870 --> 00:22:32.390
definitely if the flow is subsonic behind
it I can solve it k that is still a special
00:22:32.390 --> 00:22:37.929
case but
that is the only thing we can solve with method
00:22:37.929 --> 00:22:38.929
of characteristics.
.
00:22:38.929 --> 00:22:44.090
If I think about how to solve this problem
let us say I want to solve flow through some
00:22:44.090 --> 00:22:51.580
random duct. So, this wall geometry is given
to me I know the exact wall function x y
00:22:51.580 --> 00:23:01.480
values of this wall I know exactly if I know
that we need to start somewhere. We know
00:23:01.480 --> 00:23:06.030
that we always solve towards downstream which
act I already talked about it ones last
00:23:06.030 --> 00:23:10.639
class we need to be given upstream conditions.
So, that we can solve the problem we are
00:23:10.639 --> 00:23:16.210
.given the wall conditions wall boundary we
should also been given the upstream
00:23:16.210 --> 00:23:23.559
conditions which let us say somehow I am given
these points initially. Say I know some
00:23:23.559 --> 00:23:32.779
end points initially if I know these all I
have to do is start using the routines most
00:23:32.779 --> 00:23:36.479
commonly used one is interior point routine,
but typically we have to solve in a logical
00:23:36.479 --> 00:23:41.860
fashion. So, I will start from top wall all
the way go to the bottom wall always if I
00:23:41.860 --> 00:23:44.970
think
about that the top wall and immediate next
00:23:44.970 --> 00:23:46.970
I will decrease the number of points. So,
it is
00:23:46.970 --> 00:23:58.559
easier to see what is happening.
Now, let us say I have only n equal to five
00:23:58.559 --> 00:24:00.649
initial number of points on this is n equal
to
00:24:00.649 --> 00:24:08.879
five now. I will first do integral point between
one and two then I will do interior point
00:24:08.879 --> 00:24:12.220
between two and three interior point between
three and four interior point between four
00:24:12.220 --> 00:24:23.330
and five that we already saw. This we have
marched one step forward from here from
00:24:23.330 --> 00:24:27.529
that point if you note the number of points
in the next step was lesser here there is
00:24:27.529 --> 00:24:31.849
five
points here there are only four points next
00:24:31.849 --> 00:24:36.639
point will next a step we will have five points
again from there. I will send one to the wall
00:24:36.639 --> 00:24:45.070
point I will create a wall point and then
interior points here interior points here
00:24:45.070 --> 00:24:49.100
and then another wall point here. So, here
in this
00:24:49.100 --> 00:24:55.849
step.
I have five points. So, we have to manage
00:24:55.849 --> 00:24:58.929
array such that you can store all this nicely
k it
00:24:58.929 --> 00:25:03.529
is not going to be all equal numbers you can
store sometimes it will be five sometimes
00:25:03.529 --> 00:25:05.549
it
will be four sometimes it will be 5 4 5 4
00:25:05.549 --> 00:25:09.590
5 n and n minus one always k. If I start with
n
00:25:09.590 --> 00:25:14.450
being on the wall to wall conditions this
is what you will get typically if this is
00:25:14.450 --> 00:25:18.169
the
situation now from here I have to keep on
00:25:18.169 --> 00:25:23.759
going easiest way to draw is not going this
point by point business I will just draw it
00:25:23.759 --> 00:25:34.779
like this since I am doing it on the board.
I can
00:25:34.779 --> 00:25:42.129
do these kind of things when I am solving
it in computers it may not been this simple
00:25:42.129 --> 00:25:45.749
I
can draw whatever I feel like here there we
00:25:45.749 --> 00:26:14.090
have to go and calculate each and every
thing.
00:26:14.090 --> 00:26:18.559
Some such thing if I now have to keep tracking
of this five four five four. I want to draw
00:26:18.559 --> 00:26:29.220
those lines here
something like this you can keep on tracking
00:26:29.220 --> 00:26:35.890
those k this is what I call
this m o c fabric characteristic fabric right.
00:26:35.890 --> 00:26:42.360
I call it fabric I do not see any book which
calls it a fabric just its like as if weaving
00:26:42.360 --> 00:26:45.330
of threads this way and this way k why do
I call
00:26:45.330 --> 00:26:50.830
it a fabric it will become more clear and
nicer to use fabric as an example that is
00:26:50.830 --> 00:26:55.729
why I
use this k this kind of woven net is what
00:26:55.729 --> 00:26:58.570
you will find finally.
00:26:58.570 --> 00:27:03.690
.Since I am having an expansion here I am
drawing it such that the characteristics are
00:27:03.690 --> 00:27:07.429
expanding except for I made one mistake like
here this is actually contracting. But ignore
00:27:07.429 --> 00:27:12.979
that for now other characteristics are all
looking like we are all expanding always I
00:27:12.979 --> 00:27:16.200
will
show you examples of all these next class
00:27:16.200 --> 00:27:24.179
anyway. So, this is the way I will solve flow
inside a duct this is one of the ways of holding
00:27:24.179 --> 00:27:29.259
the array values right. I can hold it like
five points four points n points n minus one
00:27:29.259 --> 00:27:31.270
points n points n minus one points like that.
I
00:27:31.270 --> 00:27:36.669
can store the array or this is a better way
of doing things I like the other one option
00:27:36.669 --> 00:27:43.840
which
is I will take another chalk color. I will
00:27:43.840 --> 00:27:48.369
pick this pink color initially when I am given
this
00:27:48.369 --> 00:28:00.789
initial line what I will do is from here I
will fill this whole triangle first
00:28:00.789 --> 00:28:03.690
k what do I mean
by fill this triangle I am going to find all
00:28:03.690 --> 00:28:12.129
the points inside this triangle. And on the
boundary I have found all these points already
00:28:12.129 --> 00:28:16.460
which means I am not solving the regular
way starting from one wall going to the other
00:28:16.460 --> 00:28:21.549
wall. I am solving it differently now I am
going to start from this this point is nothing
00:28:21.549 --> 00:28:28.149
to be done I will go to the next point from
there I will want to go to the bottom wall.
00:28:28.149 --> 00:28:32.759
I will start a characteristic from there I
am solving along my right running characteristics
00:28:32.759 --> 00:28:36.999
start go along my right running characteristic
till it meets the wall. That is how I am
00:28:36.999 --> 00:28:41.509
going to solve it now from here I will take
this interior points solve this point from
00:28:41.509 --> 00:28:45.409
there
go for wall point next one interior point
00:28:45.409 --> 00:28:49.269
interior point. Interior point wall point
next one
00:28:49.269 --> 00:28:53.889
one two three four five five interior points
and then wall point next one seven interior
00:28:53.889 --> 00:28:59.729
points then wall point just solve this whole
thing when I solve this whole thing now I
00:28:59.729 --> 00:29:07.499
will finally, have nine points along the last
line a I have started with five I will end
00:29:07.499 --> 00:29:12.440
up
with nine k one two three four five six seven
00:29:12.440 --> 00:29:18.659
eight nine k you will end up with two n
minus one actually it will be five plus four
00:29:18.659 --> 00:29:21.109
here n plus n minus one. That is what you
will
00:29:21.109 --> 00:29:27.190
end up with k we will end up with that after
that if I start solving only along negative
00:29:27.190 --> 00:29:36.019
characteristic there is always same number
of points. So, it is easier to store it in
00:29:36.019 --> 00:29:39.619
a …
array k and plotting also becomes easier if
00:29:39.619 --> 00:29:43.389
you start thinking this way I will just solve
this like this. From here again you can note
00:29:43.389 --> 00:29:48.070
this as nine points why I am losing the top
most point I am introducing new point at the
00:29:48.070 --> 00:29:53.309
bottom by creating one wall point k this is
the way I will be solving this whole thing
00:29:53.309 --> 00:29:56.500
now will always be nine point nine points
nine
00:29:56.500 --> 00:30:02.399
points nine points till till the end.
There is only one small problem when you come
00:30:02.399 --> 00:30:13.690
to the end if my nozzle say ends here
say my nozzle ends here. I will end up with
00:30:13.690 --> 00:30:17.279
a situation where I would have solved this
00:30:17.279 --> 00:30:23.759
.line note that there still be nine points
one two three four five six seven eight nine
00:30:23.759 --> 00:30:31.559
points
the next characteristic will be this one.
00:30:31.559 --> 00:30:36.110
Now I am going outside the nozzle on one end
while the other end I do not have any data
00:30:36.110 --> 00:30:41.590
k this is the problem with this now I need
to
00:30:41.590 --> 00:30:46.320
be given the pressure boundary point. So,
I we can solve this also I can solve for this
00:30:46.320 --> 00:30:52.720
outside region also pressure boundary point
yes k. So, I can solve up to here I have gone
00:30:52.720 --> 00:30:58.419
past the nozzle up to one side while the other
side I am still somewhere here I need data
00:30:58.419 --> 00:31:03.080
in this region.
That is the problem when this kind of analysis
00:31:03.080 --> 00:31:10.419
k what if I do the opposite I will pick
another color I can do not know green on green
00:31:10.419 --> 00:31:21.370
will work very well I will take yellow
here. So, I will pick another color and I
00:31:21.370 --> 00:31:24.520
want to solve all positive characteristics
the
00:31:24.520 --> 00:31:30.080
opposite one I will ignore the first one because
it is already on the top of wall anything
00:31:30.080 --> 00:31:35.649
that is not on the top wall I will send a
positive characteristic towards the top I
00:31:35.649 --> 00:31:37.789
can solve
it like this slightly off setting from the
00:31:37.789 --> 00:31:43.669
white line. So, that you can see what is happening
k again you are seeing that first one one
00:31:43.669 --> 00:31:47.539
point second one three five seven nine after
that
00:31:47.539 --> 00:31:52.799
everything else will have again nine nine
nine nine points and if I go this way I will
00:31:52.799 --> 00:32:04.229
go
up to their then the next one there is this
00:32:04.229 --> 00:32:07.139
characteristic coming from there and then
it
00:32:07.139 --> 00:32:14.379
goes out in to pressure boundary point on
the other side k again I will have one two
00:32:14.379 --> 00:32:17.789
three
four five six seven. I missed somewhere some
00:32:17.789 --> 00:32:21.399
other characteristics there should be nine
I
00:32:21.399 --> 00:32:30.100
missed one characteristic yes I missed a characteristic
here that characteristic should also
00:32:30.100 --> 00:32:36.549
be there k otherwise it will not work k one
two three four five six seven eight nine points
00:32:36.549 --> 00:32:44.619
k. So, I can go all the way up it will be
nine points the only problem now will be I
00:32:44.619 --> 00:32:47.529
have
gone passed the nozzle on that side and on
00:32:47.529 --> 00:32:51.970
this side I am still stuck here I do not know
what is happening here at the bottom depending
00:32:51.970 --> 00:32:57.740
on what you want to solve you can
choose there will be a there are three methods
00:32:57.740 --> 00:33:01.479
I gave you one thing is stepping along the
initial.
00:33:01.479 --> 00:33:07.279
Line just one step at a time all points used
if I go that way I will have n points n minus
00:33:07.279 --> 00:33:13.019
one points n points n minus one points like
that if I do this I will solve at up to the
00:33:13.019 --> 00:33:15.229
end
even here it will be almost uniform at the
00:33:15.229 --> 00:33:22.359
exit which is one nice thing that need not
always be the case there may be situations
00:33:22.359 --> 00:33:30.429
where you may want data only on one side not
the other side for example, if I have a nozzle
00:33:30.429 --> 00:33:41.269
flow
and I am interested in the flow only in
00:33:41.269 --> 00:33:47.549
this small region in the center and I want
to solve for method of characteristics based
00:33:47.549 --> 00:33:53.260
method somehow I started with some initial
set of lines and I am going to solve only
00:33:53.260 --> 00:33:54.260
top
00:33:54.260 --> 00:33:58.330
.half it acts a its a symmetric problem. So,
I will solve only a top of its only a two
00:33:58.330 --> 00:34:00.619
d
symmetric problem so.
00:34:00.619 --> 00:34:32.680
From here I can solve like this this is my
nozzle I could have solved like tis using
00:34:32.680 --> 00:34:34.830
only
positive characteristic family this is the
00:34:34.830 --> 00:34:37.610
most expensive way of calling for region in
here
00:34:37.610 --> 00:34:41.830
now I know several points inside this region
this is the most expensive way of solving
00:34:41.830 --> 00:34:44.590
it
why I needed to solve all these extra points
00:34:44.590 --> 00:34:48.780
to solve for this flow field instead if I
had
00:34:48.780 --> 00:35:02.760
used only my negative characteristic then
I would have solved these
00:35:02.760 --> 00:35:07.000
and by the time I am
whatever region is interested is crossed I
00:35:07.000 --> 00:35:09.380
haven’t even gone passed this region there
may
00:35:09.380 --> 00:35:13.420
be times when I want we use this method there
may be times when I want to use the
00:35:13.420 --> 00:35:17.790
other method because I am interested only
in z boundary not interested in what is
00:35:17.790 --> 00:35:22.310
happening in the middle then I will use my
left running characteristic depending on what
00:35:22.310 --> 00:35:29.590
I am interested then I can choose if I am
doing a analysis of whether y nozzle will
00:35:29.590 --> 00:35:33.380
be
useful for creating of low field of particular
00:35:33.380 --> 00:35:39.930
number in this region of some particular size
then I will use my red lines the right running
00:35:39.930 --> 00:35:45.600
characteristics if I am interested in finding
the z boundary shape for the same nozzle for
00:35:45.600 --> 00:35:51.650
a given pressure outside then I will use left
running characteristics or if I want to know
00:35:51.650 --> 00:35:57.300
the exit pressure exit plain pressure variation
then I will use this other method n n minus
00:35:57.300 --> 00:36:00.980
one n n minus one like this advancing the
initial everywhere together.
00:36:00.980 --> 00:36:06.910
So, there are different ways of solving. So,
you can write your code in different modes
00:36:06.910 --> 00:36:13.350
depending on what your objective is application
is k typically if you are just writing flow
00:36:13.350 --> 00:36:18.860
through a nozzle I would say go towards axis
that is right running characteristic typically
00:36:18.860 --> 00:36:26.660
if you are solving for the top half k that
is one way of looking at this now in all these
00:36:26.660 --> 00:36:30.400
we
needed this initial set of data points of
00:36:30.400 --> 00:36:31.800
course, there will be problems where we will
be
00:36:31.800 --> 00:36:36.280
given the initial flow condition and then
we have to solve from there that is simple
00:36:36.280 --> 00:36:38.060
there
is no more trouble there I can just put how
00:36:38.060 --> 00:36:42.260
many ever points I feel like there, but there
will be times when nobody knows what the data
00:36:42.260 --> 00:36:49.140
is that is one of the one such case
happens in the nozzle throat nozzle throat
00:36:49.140 --> 00:36:53.030
we know its somewhere there n equal to one
k.
00:36:53.030 --> 00:37:06.570
So, if I expand re just this region if I think
about just that region
00:37:06.570 --> 00:37:11.500
something like this now
typically if you are looking at the transonic
00:37:11.500 --> 00:37:20.370
flow around this region k mark line contours
look like this
00:37:20.370 --> 00:37:24.600
something like this this is how your mark
line contours look like for
00:37:24.600 --> 00:37:31.140
transonic flow near the throat region I am
assuming it is choking somewhere there if
00:37:31.140 --> 00:37:34.820
this
is the case and you know that as the a flow
00:37:34.820 --> 00:37:36.760
goes through this then I am assuming flow
is
00:37:36.760 --> 00:37:41.080
.going towards supersonic mark number should
be increasing along these lines typically
00:37:41.080 --> 00:37:49.000
m equal to one happens here and this one is
m less than one this one is m greater than
00:37:49.000 --> 00:37:57.400
one how do I know this people have analytically
solved this problem already k I do not
00:37:57.400 --> 00:38:04.420
remember the exact a.
Paper right now reference I will give you
00:38:04.420 --> 00:38:13.900
the reference tomorrow k I searched for today,
but I will get to you tomorrow if I just a
00:38:13.900 --> 00:38:19.740
the author s name is hall I just do not remember
the journal currently k author s name is hall
00:38:19.740 --> 00:38:24.490
it is something very old paper probably
nineteen sixty five or something k completely
00:38:24.490 --> 00:38:28.650
analytical work what they really did was
they imagined one cylinder this way another
00:38:28.650 --> 00:38:34.360
cylinder this way flow in between two
cylinders chocked flow in between two cylinders
00:38:34.360 --> 00:38:41.010
that is the problem they solved for they
have given analytical expression for velocity
00:38:41.010 --> 00:38:46.271
actually they have solved for u and v k. So,
from there we can get theta and m if you want
00:38:46.271 --> 00:38:52.970
k. So, we have analytical expressions for
those its not simple its they have done some
00:38:52.970 --> 00:38:57.660
series expansion based mathematical
methods to solve for this region they have
00:38:57.660 --> 00:39:00.140
solved that the solutions have already
published on the web.
00:39:00.140 --> 00:39:06.860
We have used this data and created nozzles
and they work well we have used that kind
00:39:06.860 --> 00:39:10.020
of
starting lines then used m o c to create nozzles
00:39:10.020 --> 00:39:17.410
and that nozzle works well. So, I will say
that data is reasonably close. So, we typically
00:39:17.410 --> 00:39:21.820
do not use the m equal to one line we will
typically pick one point one or one point
00:39:21.820 --> 00:39:25.130
zero five and the solution is not very valid
for
00:39:25.130 --> 00:39:29.530
more than m equal to one point two its transonic
the errors will become higher for m
00:39:29.530 --> 00:39:34.130
more than one point one. So, I typically use
one one point zero five one point one kind
00:39:34.130 --> 00:39:37.380
of
numbers for that particular condition I will
00:39:37.380 --> 00:39:40.940
find this initial line k if find the initial
line
00:39:40.940 --> 00:39:46.480
after that the problem is extremely simple
somehow from there contours I have created
00:39:46.480 --> 00:39:54.200
this initial line after that I will go and
follow this initial technique now I have given
00:39:54.200 --> 00:39:56.910
this
initial line from there I can just go on solve
00:39:56.910 --> 00:40:02.370
for anyone of these methods the white line
thing marching forward on the initial line
00:40:02.370 --> 00:40:06.360
or going along left characteristic or right
characteristic any of these methods.
00:40:06.360 --> 00:40:14.440
I could be using k and solve this this will
give you nice contoured nozzles a sorry nice
00:40:14.440 --> 00:40:21.660
flow estimation inside a nozzle that whose
geometry is already known k, but the main
00:40:21.660 --> 00:40:28.250
reason why method of characteristics became
famous was because of its ability to create
00:40:28.250 --> 00:40:34.640
nozzle shapes with whatever flow conditions
we want k typically we wanted uniform
00:40:34.640 --> 00:40:39.300
because you want to create supersonic tunnels
where the mark numbers is exactly the
00:40:39.300 --> 00:40:44.970
.same although that is the idea of this. So,
we will go and look at how we can make
00:40:44.970 --> 00:40:59.770
nozzles yes o I dint talk about axis point
a thank you I will I will talk about that
00:40:59.770 --> 00:41:02.010
also
there is one or subroutine which is not very
00:41:02.010 --> 00:41:09.740
different from wall subroutine that has
slipped my mind axis how will I work at axis
00:41:09.740 --> 00:41:13.270
I want to solve only the top half of my
problem.
00:41:13.270 --> 00:41:14.270
.
00:41:14.270 --> 00:41:19.800
So, there is a right running characteristic
that is coming to the axis let us call this
00:41:19.800 --> 00:41:25.400
as my
axis point by the way axis could be axi symmetric
00:41:25.400 --> 00:41:30.710
that kind of axis or symmetry line may
be we will call it symmetry point symmetry
00:41:30.710 --> 00:41:40.190
is more clear we are solving only two d
remember only two d. So, it is only symmetry
00:41:40.190 --> 00:41:50.670
point k. So, I have point here a and I come
here I want to find this b we know at b theta
00:41:50.670 --> 00:41:59.390
is zero theta b is equal to zero is as known.
So, now, I can write this a right running
00:41:59.390 --> 00:42:08.300
characteristic.
So, it is a theta a plus nu a equal to theta
00:42:08.300 --> 00:42:14.540
b plus nu b. So, now, I can find this nu b
once I
00:42:14.540 --> 00:42:19.070
know nu b I know mark number at that point
remaining things are similar to that to wall
00:42:19.070 --> 00:42:25.270
point except for the t is s I can actually
use a wall point routine here exactly if u
00:42:25.270 --> 00:42:28.450
want k
except for wall condition will be y equal
00:42:28.450 --> 00:42:33.120
to zero always that is your symmetric line
or
00:42:33.120 --> 00:42:36.780
you can just have another routine it is not
very difficult we know that y equal to zero
00:42:36.780 --> 00:42:40.440
at
that point directly all I have to do is find
00:42:40.440 --> 00:42:43.500
when will this line come and meet here again
I
00:42:43.500 --> 00:42:57.280
will go for d y by d x of a b is given by
tan of average of theta s and nu s and it
00:42:57.280 --> 00:42:59.350
s a
negative characteristic. So, it is theta a
00:42:59.350 --> 00:43:01.340
minus nu a.
00:43:01.340 --> 00:43:13.380
.Plus theta b minus nu b whole by two. So,
it is intersection of this line with y equal
00:43:13.380 --> 00:43:15.590
to
zero line this is the other equation. So,
00:43:15.590 --> 00:43:19.390
all I have to do is go and substitute y c
is equal to
00:43:19.390 --> 00:43:25.530
zero and get x c from here directly right
this is equal to one more c here a and b.
00:43:25.530 --> 00:43:32.930
So, here
this a why a b minus y a by x b minus x a
00:43:32.930 --> 00:43:37.960
that is here d y by d x a b now we know that
y
00:43:37.960 --> 00:43:42.580
b is equal to zero we are given these two
values. So, I just have to find x b from there
00:43:42.580 --> 00:43:53.630
direct. So, y b is zero k x b is known x and
y are calculated theta b is zero nu b can
00:43:53.630 --> 00:43:56.520
be
calculated. So, I know the full point k this
00:43:56.520 --> 00:44:02.110
is the way you solve from there k isc slipped
this I should have included that and that
00:44:02.110 --> 00:44:05.930
is the point that is the subroutine you will
have
00:44:05.930 --> 00:44:09.200
to use at these symmetry points every time
when there is negative characteristic comes
00:44:09.200 --> 00:44:17.940
and hits the axis you have to use this symmetry
point routine here for all those points. So,
00:44:17.940 --> 00:44:50.430
now, let s go and discuss how to create a
nozzle.
00:44:50.430 --> 00:44:51.430
.
00:44:51.430 --> 00:44:58.170
Let us say somehow the initial points after
the throat is given to you this is my axis
00:44:58.170 --> 00:45:03.110
symmetry line whatever that is given to you.
Somehow the initial line is given to you as
00:45:03.110 --> 00:45:21.100
of now I will use the same five points I will.
In fact, use four points on my initial line
00:45:21.100 --> 00:45:31.850
now let us say m exit is known to me that
is my target mark number I want to get from
00:45:31.850 --> 00:45:37.910
this nozzle at the exit and and I want theta
exit equal to zero this is the standard thing
00:45:37.910 --> 00:45:41.410
we
want for supersonic tunnels any ways k. So,
00:45:41.410 --> 00:45:50.241
I want to get such a situation. So, for this
special condition what we have to do is think
00:45:50.241 --> 00:45:59.970
about this radius here by the way upstream
radius can be different from downstream radius.
00:45:59.970 --> 00:46:07.080
.Ok, but let us say they are same currently
or just worry about downstream radius right
00:46:07.080 --> 00:46:15.910
now I am going to have the wall turning around
as if it is from some point just a circle
00:46:15.910 --> 00:46:25.660
with center some where there just a circle
with radius r downstream r d now I will
00:46:25.660 --> 00:46:37.050
imagine the wall being on that curve for some
time for how long up to a point where this
00:46:37.050 --> 00:46:54.080
angle comes out to be new exit by two why
we will see it in a minute say if that is
00:46:54.080 --> 00:46:57.240
new
exit by two that is my last point on my wall
00:46:57.240 --> 00:47:04.210
k this is ive created a wall along this where.
I have. So, many points on this wall now of
00:47:04.210 --> 00:47:07.810
course, I can go on start calculating
characteristics from here and go back and
00:47:07.810 --> 00:47:10.000
everything that whole set of things can be
done
00:47:10.000 --> 00:47:19.760
from here
this whole thing can be done, but now I am
00:47:19.760 --> 00:47:29.390
going to say after this point I want
to do something different. So, I will if I
00:47:29.390 --> 00:47:32.490
consider a characteristic that is a right
running
00:47:32.490 --> 00:47:38.240
characteristic that is coming from the last
point on this circular wall when it hits the
00:47:38.240 --> 00:47:43.100
axis
or the symmetry point I know that its a negative
00:47:43.100 --> 00:47:54.390
characteristic. So, theta let us label this
point as a theta a plus nu a is equal to lets
00:47:54.390 --> 00:48:01.780
label this point b theta b plus nu b this
is
00:48:01.780 --> 00:48:13.360
known now we also know that b is a symmetry
point. So, this is equal to zero. So, my nu
00:48:13.360 --> 00:48:23.570
b is equal to theta a plus nu a.
Now, I said we are starting from parallel
00:48:23.570 --> 00:48:28.350
to the axis where theta equal to zero and
rotated
00:48:28.350 --> 00:48:40.580
by nu exit by two angle which means my this
a the theta at this point since it is a
00:48:40.580 --> 00:48:46.430
expansion isotropic expansion around the corner
is going to be same theta and same nu I
00:48:46.430 --> 00:48:52.590
started with m equal to one theta equal to
zero m equal to one corresponds to nu equal
00:48:52.590 --> 00:48:55.310
to
zero I started with nu is equal to theta equal
00:48:55.310 --> 00:49:00.120
to zero finally, I went with nu equal to nu
exit by two which means my theta will be theta
00:49:00.120 --> 00:49:06.360
exit by two actually theta will be equal to
nu exit by two also because delta theta is
00:49:06.360 --> 00:49:09.950
equal to delta nu because of that ill finally,
end
00:49:09.950 --> 00:49:20.430
up with nu exit by two plus nu exit by two
which is nu exit. So, nu at b is equal to
00:49:20.430 --> 00:49:25.140
nu exit
which means at b I already reached mark mark
00:49:25.140 --> 00:49:32.730
number which I want at the exit once I
get that mark number I do not want any expansion
00:49:32.730 --> 00:49:40.350
or compression to happen. So, after
this I want this characteristic to go straight
00:49:40.350 --> 00:49:47.330
no changes that is what I ideally want k once
I think about that remaining things are easy
00:49:47.330 --> 00:49:52.580
after that every characteristic should be
parallel then I have achieved whatever I wanted
00:49:52.580 --> 00:50:04.020
I have not completed the full fabric yet.
I will just leave it like this they are all
00:50:04.020 --> 00:50:07.440
parallel lines exactly same theta theta equal
to zero
00:50:07.440 --> 00:50:13.890
and mark number exactly same from this point
on that is what I get I have achieved my
00:50:13.890 --> 00:50:20.390
objective remaining thing I need to do is
once I have got this expansion up to this
00:50:20.390 --> 00:50:24.210
point I
do not want any more change from here till
00:50:24.210 --> 00:50:28.890
here. So, I need to have a parallel line going
00:50:28.890 --> 00:50:39.060
.there such that along this line finally,
when I reach there this wall should reflect
00:50:39.060 --> 00:50:43.450
back
exactly the same parallel line if I do that
00:50:43.450 --> 00:50:47.600
then there is no more change they are all
parallel once I get these parallel there is
00:50:47.600 --> 00:50:53.770
no more expansion this way which is what we
wanted or no compression. So, because of that
00:50:53.770 --> 00:50:56.150
what I will do is the last characteristic
I
00:50:56.150 --> 00:51:03.220
get from here I will just put.
The same theta value and nu value at this
00:51:03.220 --> 00:51:09.460
point on that wall point that is the key idea
of
00:51:09.460 --> 00:51:14.760
method of characteristics to create a wall
point to start solving it the logical way
00:51:14.760 --> 00:51:17.410
I will
start from this characteristic first one that
00:51:17.410 --> 00:51:21.480
is going outside of a I will pick this point
I will
00:51:21.480 --> 00:51:30.680
take that wall point put this same theta and
nu here on that point I still have to find
00:51:30.680 --> 00:51:36.620
the x
y the way to think about it this is my point
00:51:36.620 --> 00:51:43.051
a which is already known to me now I have
a
00:51:43.051 --> 00:51:47.760
point b this is this is different from the
other b there let us say I will use one and
00:51:47.760 --> 00:51:54.340
two
point one is already known point two is known.
00:51:54.340 --> 00:52:02.450
Now, from some other interior line I want
to create a wall point three where my curve
00:52:02.450 --> 00:52:08.400
will go some other way this is my new creation
this is my nozzle design. So, how will I
00:52:08.400 --> 00:52:15.770
create three I am going to say theta 2 theta
three equal to theta 2 nu three equal to nu
00:52:15.770 --> 00:52:19.770
two
for the last point on this characteristic
00:52:19.770 --> 00:52:23.520
whatever this is a after this the next point
is this at
00:52:23.520 --> 00:52:30.390
that situation I will do this remaining things
is just geometry similar to what we did for
00:52:30.390 --> 00:52:35.760
pressure boundary point exactly the same way
I will say that this is stream line direction.
00:52:35.760 --> 00:52:41.720
So, I will take tan of theta 1 plus tan of
theta three by two a sorry tan of theta 1
00:52:41.720 --> 00:52:44.940
plus theta
three by two will be my d y by d x this way
00:52:44.940 --> 00:52:48.630
this way it will be average of theta plus
mu
00:52:48.630 --> 00:52:55.670
of these two points similar to pressure boundary
point that exact way we will solve. So, I
00:52:55.670 --> 00:53:00.570
will get my x y here and I already know theta
nu. So, I have created one point there.
00:53:00.570 --> 00:53:06.120
We are interested in x y because we are creating
the wall how will I solve the next one
00:53:06.120 --> 00:53:13.060
hit the next characteristic this is the last
point here I will put this value here if I
00:53:13.060 --> 00:53:19.890
call this
one as four five will have theta 5 equal to
00:53:19.890 --> 00:53:25.810
theta 4 and a nu five equal to nu four I will
get
00:53:25.810 --> 00:53:34.870
to this this is the way you will create this
whole curve and once you get this sur[face]we
00:53:34.870 --> 00:53:43.320
have solved and we have created a new nozzle
which will give me that decided exit
00:53:43.320 --> 00:53:49.410
conditions theta e equal to zero and m e exactly
that value how will I make sure that I
00:53:49.410 --> 00:53:54.490
have done the correct thing the last characteristic
that is going from here is going to have
00:53:54.490 --> 00:54:00.430
the value of this is a positive characteristic.
So, theta minus nu is equal to.
00:54:00.430 --> 00:54:09.910
.Theta zero and nu is nu b minus nu exit will
be the value nu b is here k I will get theta
00:54:09.910 --> 00:54:15.100
minus nu equal to constant is equal to minus
nu exit. So, when it reaches the final point
00:54:15.100 --> 00:54:22.300
at that point we know we want theta equal
to zero I will automatically get nu equal
00:54:22.300 --> 00:54:27.000
to nu
exit there k. So, it is matching this is the
00:54:27.000 --> 00:54:29.400
reason why we wanted to expand only up to
nu
00:54:29.400 --> 00:54:35.160
e by two that is the special thing if you
want if you want to go for some other theta
00:54:35.160 --> 00:54:39.361
e then
you have to choose something else that I will
00:54:39.361 --> 00:54:45.000
leave it to you if you want to think about
more that is beyond course subjects k course
00:54:45.000 --> 00:54:50.150
curriculum remaining things I will discuss
in next class I will also bring some example
00:54:50.150 --> 00:55:00.150
solutions next time I will show how I have
solved cases anything else see you people
00:55:00.150 --> 00:55:01.150
next class.
00:55:01.150 --> 00:55:02.150
.