WEBVTT
Kind: captions
Language: en
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Hello everyone, welcome back, last class we
stopped with the functional form of prandtl
00:00:17.610 --> 00:00:52.190
meyer function. I just write it once more,
this was the function we had last time and
00:00:52.190 --> 00:00:55.309
we
said that typically we work with angle and
00:00:55.309 --> 00:00:58.559
degrees. So, this tan inverse should be
converted to degrees units that is all it
00:00:58.559 --> 00:01:03.770
is nothing right. Now, we will look at this
in
00:01:03.770 --> 00:01:06.150
functional form on the screen there.
00:01:06.150 --> 00:01:07.150
..
00:01:07.150 --> 00:01:13.830
What I have plotted here is mach number on
the x axis all the way up to 100, and on the
00:01:13.830 --> 00:01:19.750
y axis it is angle and degrees starting from
M equal to 1 course you should know that
00:01:19.750 --> 00:01:26.410
below M equal to 1 this function is not valid.
It is not suppose to have any function value
00:01:26.410 --> 00:01:28.330
and you know that it will go complex numbers
anyways.
00:01:28.330 --> 00:01:35.810
So, from M equal to 1 onwards I have plotted
for three different gamma values, what we
00:01:35.810 --> 00:01:43.640
are saying here is specifically. Let us say
we will pick say gamma equal to 1.4 the green
00:01:43.640 --> 00:01:48.820
one. We are seeing that it starts here they
are all almost the same for very small mach
00:01:48.820 --> 00:01:53.671
numbers up to two or something there is not
much variation on the angle after that, when
00:01:53.671 --> 00:01:57.810
I go they split apart when you go to higher
and higher mach numbers.
00:01:57.810 --> 00:02:03.070
They split apart reasonably right, but if
I look at only gamma equal to 1.4 they tend
00:02:03.070 --> 00:02:05.950
to
become one particular value beyond some point.
00:02:05.950 --> 00:02:09.530
It tends to become an asymptote and we
gave the asymptotic value last time we get
00:02:09.530 --> 00:02:12.530
just to be 130.5 for gamma equal to 1.4 and
00:02:12.530 --> 00:02:18.510
.similarly, we can find such angles for the
other cases also, and I believe I gave that
00:02:18.510 --> 00:02:23.190
also
the last class anyways. So, now, I want to
00:02:23.190 --> 00:02:27.740
see the effect of gamma in here as usual we
always want to discuss compressibility.
00:02:27.740 --> 00:02:33.040
We said that gamma changing the gamma is one
way of changing the compressibility
00:02:33.040 --> 00:02:38.030
effects that is not the only way to change
compressibility. It is just one of the ways
00:02:38.030 --> 00:02:40.570
of
changing compressibility and I am going to
00:02:40.570 --> 00:02:44.320
say that gamma equal to 1.67 is least
compressible, and as I go more and more it
00:02:44.320 --> 00:02:49.610
is becoming more and more compressible
that is, what I am having here when I have
00:02:49.610 --> 00:02:52.880
gamma equal to 1.64 to 1.4 my gamma
decreases compressibility increases, what
00:02:52.880 --> 00:03:02.590
we are seeing is the prandtl meyer function
drops the functional value drops that is the
00:03:02.590 --> 00:03:10.060
change in the function need not be. So, high
to change the mach number mach number still
00:03:10.060 --> 00:03:13.210
changes.
Say, I want to go from mach number say 5 to
00:03:13.210 --> 00:03:20.670
10, if I have to do that in gamma equal to
1.3, I have to go from 60 degrees to something
00:03:20.670 --> 00:03:24.850
of the order of 73 degrees or. So, rough
estimate I am just picking up numbers from
00:03:24.850 --> 00:03:32.890
the plot the same thing, if I have to do from
1.67 and say I am going from 5 to 10, it has
00:03:32.890 --> 00:03:40.280
to start somewhere around say 95 degrees all
the way up to 125 degrees rough estimate.
00:03:40.280 --> 00:03:43.380
What we are seeing is I have to turn the flow
a
00:03:43.380 --> 00:03:49.910
lot for me to accelerate the flow, if I have
less compressible gas gamma equal to be very
00:03:49.910 --> 00:03:56.540
high value, when I have a easily compressible
gas, when I turn it when I change the flow
00:03:56.540 --> 00:04:00.560
direction that is I am sending an expansion
fan, when I send it I need to turn it by a
00:04:00.560 --> 00:04:03.930
small
angle to increase mach number very high, that
00:04:03.930 --> 00:04:09.220
is the special effect of compressibility
input, this is what we have to get out of
00:04:09.220 --> 00:04:15.780
these plots.
So, now, we will keep these in mind we will
00:04:15.780 --> 00:04:21.380
typically have only problems of gamma
equal to 1.4 in our simple gas dynamics world
00:04:21.380 --> 00:04:27.430
and with that we will get some numerical
problems and work on them today. Lets go to
00:04:27.430 --> 00:04:37.810
the board again, if I pick a case of we did
this a little bit last time, but I will just
00:04:37.810 --> 00:04:42.190
give some numerical examples. Let us say I
have a
00:04:42.190 --> 00:04:50.050
smooth corner smooth expansion corner, and
I am going to say I have mach number
00:04:50.050 --> 00:04:59.100
equal to M 1 equal to 2, and I am going to
call this situation M 2 and.
00:04:59.100 --> 00:05:08.660
I am given this angle to be 50 degrees, the
net turn I am giving for this flow is 50
00:05:08.660 --> 00:05:17.250
degrees. Let us say, if I give this and I
am telling you that this is air gamma equal
00:05:17.250 --> 00:05:23.060
to 1.4,
we want to find what is the final mach number.
00:05:23.060 --> 00:05:27.860
I am going to achieve in this we gave a
formula wise. Now, we just have to work through
00:05:27.860 --> 00:05:29.250
with the same formula, which we
00:05:29.250 --> 00:05:34.840
.discussed list time the way to think about
this is I have to find the nu of M equal to
00:05:34.840 --> 00:05:39.840
2
here, which is what I will call as nu 1, I
00:05:39.840 --> 00:05:44.090
have to find nu 1 from this function above,
but
00:05:44.090 --> 00:05:46.330
instead of this function we will go and use
tables.
00:05:46.330 --> 00:05:53.250
That is not easy function to work with we
just use tables nu of M equal to 2 comes out
00:05:53.250 --> 00:05:56.830
to
be 26.38 degrees for gamma equal to 1.4 of
00:05:56.830 --> 00:06:01.420
course, now you know you have seen the
plots, you have to look at for a specific
00:06:01.420 --> 00:06:03.920
gamma values for this gamma equal to 1.4 this
is
00:06:03.920 --> 00:06:12.710
the value we are getting now, with this you
remember the formula nu 2 equal to nu 1 plus
00:06:12.710 --> 00:06:19.400
delta theta, where theta is the angle expansion,
how much is there delta theta is the net
00:06:19.400 --> 00:06:23.070
expansion angle, if it is compressing then
it will have a minus sign automatically.
00:06:23.070 --> 00:06:31.300
It will get a negative delta now, I will just
put numbers here 26.38 plus the net change
00:06:31.300 --> 00:06:35.200
in
stream line direction is 50 degrees. So, I
00:06:35.200 --> 00:06:48.240
will add 50 degrees here. So, my final prandtl
meyer angle happens to be 76.38 degrees. Now,
00:06:48.240 --> 00:06:53.330
I have I know that this flow will reach a
mach number, such that it will have 76.38
00:06:53.330 --> 00:06:58.630
degrees as the prandtl meyer function. So,
what I have to do is go back to that plot
00:06:58.630 --> 00:07:03.270
and read out 76.38.
What is the mach number and instead of doing
00:07:03.270 --> 00:07:09.139
that from the plot, I am doing it from my
tables in compressible flow tables book, and
00:07:09.139 --> 00:07:18.160
I am getting that mach number to be 4.95
what has essentially happened is my mach two
00:07:18.160 --> 00:07:24.830
flow, when it expands over this corner
has gone to a point where it is having mach
00:07:24.830 --> 00:07:29.960
number of 4.95 here, that is what has
happened. I am assuming that after this is
00:07:29.960 --> 00:07:38.290
going just straight. It is not changing anywhere
now, this being said I have to inculcate the
00:07:38.290 --> 00:07:41.770
habit of drawing expansion fan the correct
way.
00:07:41.770 --> 00:07:49.229
So, let us say till here the line is all straight
and this is the first point of change and
00:07:49.229 --> 00:07:51.509
let,
us say this is the last point of change and
00:07:51.509 --> 00:07:56.570
since it is a smooth corner. There is expansion
fan will be spread over this whole region
00:07:56.570 --> 00:07:59.200
this is, what we discussed last time now,
I have
00:07:59.200 --> 00:08:07.289
to say that the very first wave will be at
an angle with respect to that local stream
00:08:07.289 --> 00:08:12.720
line
direction, such that this will be equal to
00:08:12.720 --> 00:08:17.410
mu locally why the expansion fan each
individual wave travels at speed of sound
00:08:17.410 --> 00:08:23.160
locally. So, with respect to that local velocity
vector, it is going to have that mu value
00:08:23.160 --> 00:08:25.470
remember that it is always local velocity
vector
00:08:25.470 --> 00:08:30.810
local mach number. So, for M equal to 2 this
whole region is uniform flow up to this first
00:08:30.810 --> 00:08:31.810
line.
00:08:31.810 --> 00:08:39.620
.So, M equal to 2 mu angle will be sin inverse
1 by 2, which is 30 degrees. So, it should
00:08:39.620 --> 00:08:44.610
be drawn such that it is exactly 30 degrees.
Since, it is not very nice I will draw it
00:08:44.610 --> 00:08:46.929
again
we will make it look like it is close to correct.
00:08:46.929 --> 00:08:52.029
So, it is going to more like this for this
particular problem. It should actually be
00:08:52.029 --> 00:08:54.959
probably slightly lesser, but anyways this
is
00:08:54.959 --> 00:09:01.660
your 30 degrees and it is going to keep on
going there.
00:09:01.660 --> 00:09:08.279
What should be the final mach final not mach
number final angle of my expansion
00:09:08.279 --> 00:09:15.139
wave, if I look at it every expansion wave
is like a sound wave. So, for that local mach
00:09:15.139 --> 00:09:20.439
number, which happens to be 4.95. I have to
find the mu for that particular mach number
00:09:20.439 --> 00:09:24.490
and find the local velocity direction, which
will be parallel to the wall here.
00:09:24.490 --> 00:09:33.069
So, I will draw this local direction of velocity
vector and I have to find the local mach
00:09:33.069 --> 00:09:40.709
angle for that mach number, which happens
to be mu of 4.95, happens to be 11.66
00:09:40.709 --> 00:09:52.100
degrees. I will use this number this is sin
inverse 1 by 4.95 that happens to be this.
00:09:52.100 --> 00:09:56.360
So,
with respect to this flow velocity direction.
00:09:56.360 --> 00:10:04.420
I am going to have 11 degrees from here,
which is a very small angle like this. So,
00:10:04.420 --> 00:10:09.429
from the last point of change it is going
to have
00:10:09.429 --> 00:10:12.240
an angle like this, now I will erase whatever
is unnecessary.
00:10:12.240 --> 00:10:28.410
So, that it is not confusing I draw this again
and this actually, I will remove this also
00:10:28.410 --> 00:10:36.230
now you know it is 50 degrees, if you want
we will mark it here this is 50 degrees with
00:10:36.230 --> 00:10:42.749
respect to horizontal line that is this wall
angle. Now, my initial expansion wave line
00:10:42.749 --> 00:10:50.579
goes like this final one goes like this this
is the correct way to draw it actually crosses
00:10:50.579 --> 00:10:53.170
the
horizontal and goes below also for this particular
00:10:53.170 --> 00:10:59.209
case it does what matters is with
respect to the local velocity vector direction
00:10:59.209 --> 00:11:06.720
in there, I am going to have that 11.66
degrees this is the last wave.
00:11:06.720 --> 00:11:18.420
So, overall this final angle with respect
to my horizontal will be what 50 minus 11.66
00:11:18.420 --> 00:11:30.399
right what is that 40 30 930 8.34 38.3 degrees
that is what will be the angle below
00:11:30.399 --> 00:11:37.130
horizontal that should be the final angle
here think about it that way this. Now, let
00:11:37.130 --> 00:11:39.649
us say
I want to draw intermediate expansion fans
00:11:39.649 --> 00:11:44.220
a few of them there are. So, many ways I said
its infinite number of expansion waves here
00:11:44.220 --> 00:11:53.769
in, which case I can draw few more when I
draw others. I have to draw it, such that
00:11:53.769 --> 00:11:57.829
there is a slope change in a linear fashion
to
00:11:57.829 --> 00:11:59.819
some extent.
I am assuming that this change is linear it
00:11:59.819 --> 00:12:03.240
need not be, but to some extent this angle
and
00:12:03.240 --> 00:12:09.110
this angle must be around this middle angle
or the middle one should be having an angle
00:12:09.110 --> 00:12:13.300
.in between these two angles. It cannot be
changing just think about it that way every
00:12:13.300 --> 00:12:20.019
angle should be like that it is a smooth change
in the slope of this expansion this is what
00:12:20.019 --> 00:12:27.970
you have to think about this is one particular
case of smooth expansion lets go for
00:12:27.970 --> 00:12:29.720
smooth compression ok.
.
00:12:29.720 --> 00:12:36.839
We will keep this just go for a smooth compression
case, where I am going to have a
00:12:36.839 --> 00:12:53.399
wall and it is going to turn to a 60 degree
angle, 60 degree angle and its turning into
00:12:53.399 --> 00:12:55.670
the
flow, what does that mean it is going to block
00:12:55.670 --> 00:12:57.170
the flow it is going to decrease the mach
number.
00:12:57.170 --> 00:13:02.829
So, it is a compression wave sitting there,
and I am going to say the incoming mach
00:13:02.829 --> 00:13:10.290
number M 1 equal to 4 gamma equal to 1.4.
We have not currently worried about finding
00:13:10.290 --> 00:13:13.449
the pressure and temperature, if needed we
can do it we will do it in some other example
00:13:13.449 --> 00:13:23.259
for lines how will I work on this problem
exactly the same way nu 1 will be nu at M
00:13:23.259 --> 00:13:35.800
.equal to 4 this value, you get from tables
that value happens to be 65.79 degrees nu
00:13:35.800 --> 00:13:41.790
2
equal to nu 1 plus delta theta, but by our
00:13:41.790 --> 00:13:46.119
convention delta theta is the amount of
expansion in this case it is compressing.
00:13:46.119 --> 00:13:59.290
So, I have to give a negative value for this
65.79 minus that 60 degrees we said, which
00:13:59.290 --> 00:14:09.350
happens to be 5.79 degrees this is what I
am going to get 5.79 degrees. Now, if I go
00:14:09.350 --> 00:14:13.550
back
and look at my table such that nu of that
00:14:13.550 --> 00:14:18.439
particular M 2 let us call this as my M 2
nu of
00:14:18.439 --> 00:14:28.930
that particular M 2 happens to be 5.79 that
M 2 value will be 1.3 mach number. So,
00:14:28.930 --> 00:14:33.230
essentially if there is.
A smooth compression of my flow from mach
00:14:33.230 --> 00:14:41.529
4 up to 60 degrees, then I will end up with
the flow 1.3 mach number this is, what I am
00:14:41.529 --> 00:14:46.029
saying again I have to go back and think
about how to draw these compression waves
00:14:46.029 --> 00:14:49.559
it is weak compression waves. So, I can still
think about it as isentropic and. So, I can
00:14:49.559 --> 00:14:52.319
say that the angle at, which it is going to
go
00:14:52.319 --> 00:14:57.639
with respect to the flow will be mach angle
again, if I say it is mach angle. Now, I have
00:14:57.639 --> 00:15:06.239
to find the nu value for this, I will put
mu 1 for M equal to 4 is sin inverse 1 by
00:15:06.239 --> 00:15:15.119
4 that is
75.21 degrees, and for this case mu 2 happens
00:15:15.119 --> 00:15:29.339
to be 50.29 degrees.
I think there is a problem somewhere I made
00:15:29.339 --> 00:15:37.620
a mistake somewhere, I think it cannot be
that you guys have a calculator you can check
00:15:37.620 --> 00:15:40.321
whether this 75.21 is right. I think there
is
00:15:40.321 --> 00:15:46.569
some mistake there anyways I know that, when
mach number increases the mu angle
00:15:46.569 --> 00:15:56.129
should decrease. I am seeing that there is
a mistake there this value sin inverse 1 by
00:15:56.129 --> 00:16:00.569
4 is
14.47 degrees. We will use that number and
00:16:00.569 --> 00:16:11.610
tell me the same for 1.3, what is this value
this is correct good now, I am more comfortable
00:16:11.610 --> 00:16:16.209
with this.
So, what I am seeing is locally with respect
00:16:16.209 --> 00:16:25.059
to this flow vector mach 4 flow has mach
angle 14.47, which is a very small angle subtending
00:16:25.059 --> 00:16:28.370
an angle very small with respect to
this, I am actually drawing it slightly big.
00:16:28.370 --> 00:16:32.179
So, that it is easy to visualize, and then
as it
00:16:32.179 --> 00:16:38.149
goes mach number decreases, which means my
mu angle should increase. So, when I am
00:16:38.149 --> 00:16:43.009
drawing my local velocity vector like this
with respect to that my mu would have
00:16:43.009 --> 00:16:47.730
increased. I am drawing too high probably
here and I do not want to draw it this high,
00:16:47.730 --> 00:16:51.690
because the final angle is 60 degrees plus
50 degrees.
00:16:51.690 --> 00:17:00.189
So, it is going to go somewhere like this
and with respect to this final angle. I am
00:17:00.189 --> 00:17:04.049
going
to say my mu is 50 degrees with respect to
00:17:04.049 --> 00:17:07.490
that, which means with respect to this, I
have
00:17:07.490 --> 00:17:11.570
.to draw a 50 degree thing there, which means
with respect to horizontal it has crossed
00:17:11.570 --> 00:17:18.600
90,
and it is coming inward going to go like this
00:17:18.600 --> 00:17:27.070
this will be your situation.
If I think about a case it is more like this
00:17:27.070 --> 00:17:31.490
it is going to look something like this. We
would not worry about how they interact right.
00:17:31.490 --> 00:17:35.480
Now, this is a very complicated case,
when such a thing happens there is probably
00:17:35.480 --> 00:17:39.140
a normal shock sitting in front of it.
Typically, we do not see compression was running
00:17:39.140 --> 00:17:44.140
up stream we will currently keep it
like this, we are interested in.
00:17:44.140 --> 00:17:47.880
Studying what is happening only very close
to the wall this region alone. Nowhere, else
00:17:47.880 --> 00:17:52.980
just this region were whatever theory. I have
said will be applicable, because it is all
00:17:52.980 --> 00:17:55.630
very
small change at any time very small changes
00:17:55.630 --> 00:18:01.360
close to isentropic. So, this will work when
I go more out these compression waves will
00:18:01.360 --> 00:18:04.870
come together, and they will form a strong
compression wave, which will become a shock
00:18:04.870 --> 00:18:09.140
eventually when that point happens it is
non isentropic. I cannot use any of these
00:18:09.140 --> 00:18:15.810
relations things will go wrong there. So,
this
00:18:15.810 --> 00:18:22.500
will be the case here hopefully. You will
get comfortable with drawing these mach angle
00:18:22.500 --> 00:18:47.790
that is the idea now we will go for another
example, which is all sharp corners.
00:18:47.790 --> 00:18:48.790
.
00:18:48.790 --> 00:19:05.020
.This is most common example in expansion
fan in gas dynamics. I am having a sharp
00:19:05.020 --> 00:19:13.370
corner change, and it is 20 degrees and the
incoming mach number happens to be 2 it is
00:19:13.370 --> 00:19:17.641
here.
So, gamma equal to 1.4 P 1 equal to 1 atmosphere
00:19:17.641 --> 00:19:29.550
t 1 equal to 300 kelvin. We have this
whole thing now, we are told to find what
00:19:29.550 --> 00:19:37.000
will be the final pressure temperature mach
number standard things every property that
00:19:37.000 --> 00:19:42.140
is needed P naught 2 T naught 2 everything
we want to find all the properties.
00:19:42.140 --> 00:19:49.030
So, how will I do this it is exactly same
procedure as what we did just before this
00:19:49.030 --> 00:19:53.730
board,
which was find the local prandtl meyer angle
00:19:53.730 --> 00:20:07.920
nu at M equal to 2 this is my nu 1, which
from tables will be 26.38 degrees and nu 2
00:20:07.920 --> 00:20:12.320
equal to nu 1 this is expansion angle. So,
it is
00:20:12.320 --> 00:20:21.280
plus 20 degrees. So, the final value is 46.38
degrees this, if I go and use my prandtl
00:20:21.280 --> 00:20:33.860
meyer function tables from compressible data
books. I will get M 2 equal to 2.83 this is
00:20:33.860 --> 00:20:34.920
the first phase.
00:20:34.920 --> 00:20:40.610
.The next thing I want to do is draw that
shape their expansion fan shape exact shape,
00:20:40.610 --> 00:20:47.940
how will I do that what is the first angle
going to be 30 degrees sin n was 1 by M 1
00:20:47.940 --> 00:20:52.280
sin n
was 1 by 2. So, with respect to my flow vector
00:20:52.280 --> 00:20:57.780
going to be a 30 degree line that is my
first it is really not 30.
00:20:57.780 --> 00:21:06.870
Assume, it is close to 30 this is a first
line now, the next one for this I have to
00:21:06.870 --> 00:21:15.890
find mu
2.83 that comes out to be 20.7 degrees. So,
00:21:15.890 --> 00:21:20.080
the final stream line angle is going to be
parallel to this wall, which is what is the
00:21:20.080 --> 00:21:27.470
use of this expansion fan and. So, now, with
respect to that I have to draw a 20 degree
00:21:27.470 --> 00:21:30.560
line it. So, happens that this change itself
is 20
00:21:30.560 --> 00:21:35.400
degrees. So, I know what my 20 degree looks
like 20.7 will be slightly above this 20
00:21:35.400 --> 00:21:38.770
degrees.
So, my final expansion line happens to be
00:21:38.770 --> 00:21:44.080
only straight above horizontal, since I am
exaggerating it looks like too big it should
00:21:44.080 --> 00:21:46.360
not be. It should be almost that horizontal
line
00:21:46.360 --> 00:21:53.720
may be, I will draw one more line very close.
So, this is my set of expansion fans this
00:21:53.720 --> 00:21:59.880
whole region is my expansion fan. This is
the way to draw for this particular picture
00:21:59.880 --> 00:22:01.780
that
is the easy part now, you have to go and find
00:22:01.780 --> 00:22:10.100
the pressures and temperatures, how will I
do that very simple expansion is all isentropic
00:22:10.100 --> 00:22:18.220
So, I know my p naught and t naught remains
the same. So, I have to find P 2 equal to
00:22:18.220 --> 00:22:26.680
this is the gas dynamics way of writing this
term. So, get use to this P naught 2 by P
00:22:26.680 --> 00:22:36.070
naught 1 into P naught 1 by P 1 into P 1 it
is the nice way of writing pressure equal
00:22:36.070 --> 00:22:39.050
to
something pressure multiplied by a set of
00:22:39.050 --> 00:22:44.610
ratios. Now, we know that there is 1
atmosphere remaining numbers we have to find,
00:22:44.610 --> 00:22:49.580
since it is an isentropic expansion I can
say that p naught does not change we did this
00:22:49.580 --> 00:22:53.330
long back.
It is also T naught does not change anyway
00:22:53.330 --> 00:22:59.140
here I will make this 1 P naught does not
change. So, this P naught 2 is equal to P
00:22:59.140 --> 00:23:01.820
naught 1 reaming things is just numbers. I
have
00:23:01.820 --> 00:23:08.970
to find from isentropic flow tables for M
2 P by P naught is given in my tables, and
00:23:08.970 --> 00:23:17.900
I
used that that is 0.0352 divided by my tables
00:23:17.900 --> 00:23:21.000
has P by P naught.
So, I will do divide it by the other value
00:23:21.000 --> 00:23:28.690
for M 1 that is 0.128 multiply it by pressure
P 1
00:23:28.690 --> 00:23:35.270
for us it is one atmosphere. So, I will just
multiply by one final answer is 0.275
00:23:35.270 --> 00:23:48.430
atmospheres this is what I have now the same
thing. I have to do for temperature of
00:23:48.430 --> 00:23:52.340
course, I can of course, just say to you that
T 2 instead of all P’s replace it with T’s
00:23:52.340 --> 00:23:56.100
that
is all you need to do really. Let us say I
00:23:56.100 --> 00:24:01.600
will just go and do that T 2 will be I will
just do
00:24:01.600 --> 00:24:11.580
.it once more T naught 2 into T naught 2 by
T naught 1 into 1 by T 1 by T naught 1 into
00:24:11.580 --> 00:24:14.180
T
1 this is a nice way of writing it for me
00:24:14.180 --> 00:24:19.040
and.
I am going to find this to be 0.384 divided
00:24:19.040 --> 00:24:33.310
by 0.555 multiplied by 300 kelvin and that
answer happens to be 207.6 kelvin that is
00:24:33.310 --> 00:24:41.640
what I have there. Now, the remaining thing
is
00:24:41.640 --> 00:24:46.910
finding T naught 2 and P naught 2 that is
what is left not very difficult. It should
00:24:46.910 --> 00:24:50.250
be same
as t naught 1 and P naught 1, we have numbers
00:24:50.250 --> 00:24:57.350
for all the ratios. I have T 2 by T naught
2 or P 2 by P 2 by P naught 2 or the same
00:24:57.350 --> 00:25:00.240
thing for case one anyone of them I have to
use.
00:25:00.240 --> 00:25:01.240
.
00:25:01.240 --> 00:25:11.840
Let us say I use P 1. So, P naught 1 is equal
to P 1 divided by P 1 by P 01 1 divided by
00:25:11.840 --> 00:25:28.720
that number was 0.128, that was to be 7.81
atmospheres and similar thing T naught 1 T
00:25:28.720 --> 00:25:32.270
1
divided by that ratio is 0.555.
00:25:32.270 --> 00:25:48.600
.So, I have to multiply this with 300. So,
300 by 0.555 that number is 540.5 kelvin and
00:25:48.600 --> 00:25:50.600
of
course, P naught 2 is equal to P naught 1
00:25:50.600 --> 00:25:53.490
T naught 2 equal to T naught 1. So, now, I
have
00:25:53.490 --> 00:25:58.000
solved this full problem of expansion any
other property. I want I can find what are
00:25:58.000 --> 00:26:00.220
the
properties important in this problem anything
00:26:00.220 --> 00:26:09.070
else that, we have not talked about till now
only one more happens to be velocity. So,
00:26:09.070 --> 00:26:17.240
I will just find the velocity also, why do
I have
00:26:17.240 --> 00:26:25.400
to talk about this just in case, if somebody
has a doubt that mach number increases.
00:26:25.400 --> 00:26:33.640
I cannot just say velocity increases, because
temperature decreases. I have u equal to M
00:26:33.640 --> 00:26:40.280
into square root of gamma RT. We know that
T decreases M increasing what happens to
00:26:40.280 --> 00:26:46.620
velocity it can go the other way, but we will
just check it has to accelerate, then only
00:26:46.620 --> 00:26:48.730
you
can call that the flow is accelerating, when
00:26:48.730 --> 00:26:56.400
it is expanding we did this long time back,
when we were discussing stagnation points
00:26:56.400 --> 00:27:00.440
and stagnation enthalpy there.
We said if my temperature drops my velocity
00:27:00.440 --> 00:27:06.620
increases that should be satisfied even
here. So, we will just check it, u 1 I want
00:27:06.620 --> 00:27:11.931
to calculate. So, I will put M 1 was 2 for
our
00:27:11.931 --> 00:27:20.730
problem into 1.4 into 288.7 into temperature
was 300 kelvin and this answer happens to
00:27:20.730 --> 00:27:30.090
be 696.4 meter per second this exact number
we have been using a lot 696.4, because we
00:27:30.090 --> 00:27:37.990
typically it started mach two problem in supersonic.
So, we will always get that u 2 we found that
00:27:37.990 --> 00:27:48.059
the mach number was 2.83 into square root
of 1.4 288.72 into temperature we found was
00:27:48.059 --> 00:28:01.240
207.6. So, if I
find this number it is 289.7
00:28:01.240 --> 00:28:11.780
meter per second this multiplied by 2.83 is
my final answer, which is 819 .8 meter per
00:28:11.780 --> 00:28:18.100
second. If I go look at this number here,
if I just see the speed of sound here, what
00:28:18.100 --> 00:28:21.270
will be
the speed of sound I will divide this by 2
00:28:21.270 --> 00:28:28.230
it will be a 348.2. So, my speed of sound
has
00:28:28.230 --> 00:28:32.850
gone down yes, because my temperature is lesser
it is colder.
00:28:32.850 --> 00:28:38.510
So, it has gone down, but still a mach number
effect is higher. So, we are finding that
00:28:38.510 --> 00:28:41.940
the
velocity is increasing. Even though, I am
00:28:41.940 --> 00:28:47.870
getting colder my velocity is still increasing
eventually we will go to a point, where if
00:28:47.870 --> 00:28:51.630
I keep on increasing if I want to increase
velocity I have to change my mach number by
00:28:51.630 --> 00:28:56.580
a huge amount, because temperatures will
become close to 0, then you have to change
00:28:56.580 --> 00:29:00.950
your mach number by a huge amount there
will be a point where you cannot change anymore.
00:29:00.950 --> 00:29:05.570
What is that point called the v max? right
We did this v max, when we introduced T
00:29:05.570 --> 00:29:13.140
naught T naught equal to T plus u square by
2 c p we said and, if I put T equal to 0 you
00:29:13.140 --> 00:29:17.620
will get T naught equal to v max square by
2 c p that v max what you will reach
00:29:17.620 --> 00:29:24.500
.eventually it will go a steady state. We
have seen plots of this also right. So, this
00:29:24.500 --> 00:29:29.690
is one
problem which we solved like this. I will
00:29:29.690 --> 00:29:37.790
give you another case a little more complex
version of this and that will be the last
00:29:37.790 --> 00:29:41.180
numerical example, I think I will solve in
my
00:29:41.180 --> 00:29:44.800
expansion alone world.
.
00:29:44.800 --> 00:29:56.800
Let us, say somehow I have a flow that is
like this and here mach number is 3 gamma
00:29:56.800 --> 00:30:01.600
equal to 1.4 of course, I could have done
problems with gamma equal to 1.67 or 1.3.
00:30:01.600 --> 00:30:04.380
It is
just a change in one number it does not change
00:30:04.380 --> 00:30:11.980
much. So, much for us now I am telling
you that this angle is 20 degrees, 20 degrees
00:30:11.980 --> 00:30:14.710
is the tramp angle, and I am telling that
this
00:30:14.710 --> 00:30:18.060
is parallel to this.
So, it is a straight section and outside is
00:30:18.060 --> 00:30:25.210
vacuum that is the setting we have I want
to find
00:30:25.210 --> 00:30:33.100
what is the flow field here the absolute flow
field let us say this is my nozzle exist from
00:30:33.100 --> 00:30:37.970
my rocket and it is sitting in space currently
the flow coming out is M equal to 1 M equal
00:30:37.970 --> 00:30:42.850
.to 3 at gamma equal to 1.4 now, I want to
see what is the flow field here that is the
00:30:42.850 --> 00:30:46.250
idea,
if I think about it that way what should I
00:30:46.250 --> 00:30:49.210
do I will go just faster this time mu of M
equal
00:30:49.210 --> 00:30:55.559
to 3 I will mark this region 1, this region
as 2 and somewhere outside is 3 we mark this
00:30:55.559 --> 00:31:00.640
something like this.
So, I will just go nu 1 is nu at M equal to
00:31:00.640 --> 00:31:16.170
3 happens to be 55.76 degrees and mu 1
happens to be 19.47 degrees. Now, nu 2 must
00:31:16.170 --> 00:31:24.950
be nu 1 plus 20 degrees, because it is
expanding its plus 20 degrees which is 75.76
00:31:24.950 --> 00:31:34.010
degrees. So, from here I will get a M 2,
which happens to be approximately equal to
00:31:34.010 --> 00:31:39.940
4.87 I do not have clean resolution at this
high mach numbers in my tables.
00:31:39.940 --> 00:31:46.690
So, I just gave roughly 4.87 I did not interpolate
very well, and this will give you mu 2
00:31:46.690 --> 00:31:53.630
value of 11.77 degrees. Once, I know this
I should be able to draw the expansion fan
00:31:53.630 --> 00:32:00.950
there. So, I am having 19 degrees roughly
20 degrees is the angle. We know that this
00:32:00.950 --> 00:32:05.390
angle we said as 20 degrees.
So, it should be roughly that that is the
00:32:05.390 --> 00:32:14.510
first expansion line with respect to the incoming
vector it is 19.5 degrees above the last one
00:32:14.510 --> 00:32:21.460
going to be 11 degrees from this wall, which
means it should be roughly midpoint is 10
00:32:21.460 --> 00:32:26.400
degrees. So, it should be slightly above that.
So, its somewhere here this whole region is
00:32:26.400 --> 00:32:35.400
my expansion fan that whole region is
expansion fan. So, the flow comes here turns
00:32:35.400 --> 00:32:38.290
and then goes out straight like this that
is
00:32:38.290 --> 00:32:44.451
what is happening this is the first case of
course, now you can go and find p 2 t 2 every
00:32:44.451 --> 00:32:49.390
other property you want u 2 whatever right.
Now, we will go out of that I just I believe
00:32:49.390 --> 00:32:55.760
you can do remaining part, now the next
thing is we have not given any angle change,
00:32:55.760 --> 00:33:00.700
but you have given pressure outside. We
said in gas dynamics, there are two types
00:33:00.700 --> 00:33:04.450
of boundary condition velocity direction and
pressure. We have been using velocity direction
00:33:04.450 --> 00:33:08.971
till now, suddenly we are giving
pressure boundary condition, if I gave pressure
00:33:08.971 --> 00:33:14.060
boundary condition here what should it
be I have to just go and find what is the
00:33:14.060 --> 00:33:18.720
pressure here it. So, happens that p is 0,
if it is 0
00:33:18.720 --> 00:33:25.040
what should be my mach number keeping same
P naught isentropic should be M tends to
00:33:25.040 --> 00:33:30.870
infinity right.
So, I know that my final mach number M 3 actually,
00:33:30.870 --> 00:33:39.500
I will just put tends to infinity not
equal to it tends to infinity, which means
00:33:39.500 --> 00:33:45.040
now I know my M 3, I know my M 2, I just
have to draw the expansion fan problem is
00:33:45.040 --> 00:33:52.410
different. So, I will find my nu 3 what is
nu 3
00:33:52.410 --> 00:33:59.020
.for gamma equal to 1.4 this number. We discussed
last class that number happens to be
00:33:59.020 --> 00:34:11.580
130.5 degrees and nu 2 we calculated just
now 75.76. So, my delta theta the change in
00:34:11.580 --> 00:34:27.700
angle of the stream line comes out to be nu
3 minus nu 2 happens to be 54.74 degrees,
00:34:27.700 --> 00:34:35.730
which means the flow is already turned 20
degrees from horizontal, it can turn only
00:34:35.730 --> 00:34:37.309
54
degrees more.
00:34:37.309 --> 00:34:48.289
So, I will erase these things for clarity
it can turn only 54 degrees more from here,
00:34:48.289 --> 00:34:52.349
this is
my 90 degrees its 60 degrees. So, it can turn
00:34:52.349 --> 00:35:00.950
only 54 degrees. So, the final velocity
vector direction is this. It cannot go anymore
00:35:00.950 --> 00:35:15.240
turning around the body it can turn only till
here and this change is 54.74 degrees that
00:35:15.240 --> 00:35:26.799
is this change. Now, I just have to draw the
expansion fan angle my first wave here, what
00:35:26.799 --> 00:35:32.670
will be the angle for that my first wave of
the expansion fan what should be the angle
00:35:32.670 --> 00:35:40.690
for that the same thing here 11.77, but we
already have an expansion there. I do not
00:35:40.690 --> 00:35:43.160
need to go and look at it anymore, I know
that
00:35:43.160 --> 00:35:45.849
the mach number here is all the same velocity
vector direction is all the same.
00:35:45.849 --> 00:35:50.690
So, I just have to draw a parallel line to
that that is my first expansion wave this
00:35:50.690 --> 00:35:55.499
and this
are parallel right the same condition and
00:35:55.499 --> 00:36:05.089
the last expansion wave. What is the mu value
for that sin inverse 1 by infinity sin inverse
00:36:05.089 --> 00:36:13.039
0 is 0. So, you do not have a mu angle 0 with
respect to the flow velocity vector direction
00:36:13.039 --> 00:36:21.499
my last expansion wave is going almost on
it. So, it comes to a point where it is not
00:36:21.499 --> 00:36:24.509
physical for us. We already told you that
M
00:36:24.509 --> 00:36:29.660
tends to infinity is not a real situation
by then my continuum assumption breaks down.
00:36:29.660 --> 00:36:35.480
So, I cannot be using this analysis here by
here by here pressure is 0 here and non 0
00:36:35.480 --> 00:36:39.910
here
that situation will never arise and at this
00:36:39.910 --> 00:36:42.020
particular situation, where pressure is close
to 0
00:36:42.020 --> 00:36:47.130
I cannot apply continuum any more it is no
more continuum there is no molecule it
00:36:47.130 --> 00:36:52.180
cannot be continuum anymore. So, it is just
a hypothetical analysis, but one thing it
00:36:52.180 --> 00:36:54.989
tells
is, if I have a case like this the flow will
00:36:54.989 --> 00:36:57.010
just go straight like this it will never try
and turn
00:36:57.010 --> 00:37:01.670
around the body that is the very important
thing. You should know if, I start with high
00:37:01.670 --> 00:37:05.680
mach number it cannot be it turned too much
already, because it is already turned a lot
00:37:05.680 --> 00:37:06.680
ok.
.
00:37:06.680 --> 00:37:10.049
.That is the way you have to think of and
you should of course, remember the plot, which
00:37:10.049 --> 00:37:18.430
we just saw nu of M versus M it looks something
like this, which means if I start
00:37:18.430 --> 00:37:24.210
somewhere at reasonably high mach number say
five after that I have I can turn only. So,
00:37:24.210 --> 00:37:29.779
much especially if it is a very compressible
gas I can turn only lesser.
00:37:29.779 --> 00:37:37.369
So, if it is a less if it is a more compressible
gas. It will turn faster the curve will become
00:37:37.369 --> 00:37:43.190
flat very fast this kind of effect will happen,
which means I want to turn my flow more it
00:37:43.190 --> 00:37:48.019
is probably not possible if it is very compressible,
but still my mach number can go to
00:37:48.019 --> 00:37:56.900
infinity that kind of situation can arise.
So, the next thing we can do is once you
00:37:56.900 --> 00:38:01.440
understand these it is just the simple extension
of this ok.
00:38:01.440 --> 00:38:02.440
.
00:38:02.440 --> 00:38:14.500
.I will consider a first flow problem, which
is I have some jet coming out with a non
00:38:14.500 --> 00:38:22.599
mach number and velocity direction. Let, us
say M equal to 2 and I have to tell you P
00:38:22.599 --> 00:38:29.069
naught value let us say I will give you P
1 equal to 1 atmosphere, if such a situation
00:38:29.069 --> 00:38:32.940
is
coming out and let us say I do not have P
00:38:32.940 --> 00:38:38.569
exist is equal to let us say 0.1 atmosphere,
if I
00:38:38.569 --> 00:38:45.799
have such a situation, and let us say I will
give a T 1 also 300 kelvin.
00:38:45.799 --> 00:38:49.940
If I have something like this, I will just
go for how will I calculate this problem
00:38:49.940 --> 00:38:54.589
remaining thing is just going through the
tables that much of use for us. We will just
00:38:54.589 --> 00:38:56.130
let
us assume that you can do it. I will give
00:38:56.130 --> 00:39:00.640
you exercise later on the website. Now, we
know
00:39:00.640 --> 00:39:05.420
that mach number is pressure is 1 atmosphere
and it has to go decrease to 0.1
00:39:05.420 --> 00:39:11.779
atmosphere, which means the flow has to expand
more if it has to expand more, because
00:39:11.779 --> 00:39:16.869
this is the new boundary condition.
Then it has to expansion fans the very first
00:39:16.869 --> 00:39:22.319
fan will be with respect to this angle 30
degrees something like this this angle is
00:39:22.319 --> 00:39:26.359
30 degrees that is the first fan angle last
one will
00:39:26.359 --> 00:39:30.360
.be decided by what is the final mach number?
Let us say you can calculate all that, I will
00:39:30.360 --> 00:39:36.900
just draw some angle roughly currently let
us assume this is correct for now, but how
00:39:36.900 --> 00:39:41.350
will I calculate this final mach number.
The flow is turning like this right you can
00:39:41.350 --> 00:39:43.430
imagine that flow is going like this and it
is
00:39:43.430 --> 00:39:50.759
turning out, if I think about the same thing
on the bottom it is going to go like this,
00:39:50.759 --> 00:39:54.460
and
its going to turn down like this, this is
00:39:54.460 --> 00:39:57.630
what should happen let us start getting use
to this,
00:39:57.630 --> 00:40:01.960
because we are going to draw more and more
of this as the classes come up next week or
00:40:01.960 --> 00:40:08.920
the weak after next. We will draw more of
this a lot more of this and how will I find
00:40:08.920 --> 00:40:11.200
it, I
know the pressure here and I am going to assume
00:40:11.200 --> 00:40:16.309
this isentropic. So, I know P naught
value is same as P naught 1. So, I will just
00:40:16.309 --> 00:40:21.109
go and use P e by P naught 1 this will give
me
00:40:21.109 --> 00:40:27.619
my M at the exit M at the outside boundary.
Let us call this ambient pressure I do not
00:40:27.619 --> 00:40:31.369
want to call it exist it will confuse you
later,
00:40:31.369 --> 00:40:37.710
and jet boundary mach number at the jet boundary
the final line will be the jet boundary
00:40:37.710 --> 00:40:42.760
to find the mach number. We will go and deal
with jet separately later, I am just giving
00:40:42.760 --> 00:40:47.360
you this one example, I will find the mach
number how will I find this mach number. I
00:40:47.360 --> 00:40:53.029
am having P a by P naught that from isentropic
tables for gamma equal to 1.4, it is going
00:40:53.029 --> 00:40:58.700
to give me some mach number. Once, I know
the mach number I know the delta nu right.
00:40:58.700 --> 00:41:06.779
So, I know how much it will turn by this will
give me a nu at jet boundary. I know that
00:41:06.779 --> 00:41:18.799
nu 1 oh sorry nu minus nu 1 is my delta theta.
So, I have to find the delta theta now, I
00:41:18.799 --> 00:41:24.749
will tell that at the end of it.
It has turned by so, much angle that is what
00:41:24.749 --> 00:41:28.680
it will come out to be it will turn by whatever
angle it is I have not calculated this, but
00:41:28.680 --> 00:41:32.239
you can calculate it is not it is the same
procedure. I have to go follow it again and
00:41:32.239 --> 00:41:38.240
again and I will go to a point, where I know
this once I know mach number I can find temperature
00:41:38.240 --> 00:41:44.650
also even, otherwise you can find
temperature how we said it is isentropic.
00:41:44.650 --> 00:41:53.720
So, I know my P a by P 1 is equal to T a by
T 1 to the power gamma by gamma minus 1,
00:41:53.720 --> 00:41:58.780
I can use this also to get my temperature
at the exit at the ambient condition. That
00:41:58.780 --> 00:42:03.680
is also
a possibility two ways of doing this, this
00:42:03.680 --> 00:42:07.650
is the way we will approach this problem why
did I pick this problem separately all this
00:42:07.650 --> 00:42:11.900
time. We have been considering boundary
conditions as velocity vector direction. This
00:42:11.900 --> 00:42:14.780
is one problem were pressure at the exit is
00:42:14.780 --> 00:42:19.369
.given pressure is the boundary condition
given for the flow direction two boundary
00:42:19.369 --> 00:42:22.970
conditions are typically used in gas dynamics
world.
00:42:22.970 --> 00:42:33.279
Once, this is done the next step is just go
and use this expansion fan concept along with
00:42:33.279 --> 00:42:38.319
all the previous compression wave concepts.
You have all that mix together in flow
00:42:38.319 --> 00:42:46.109
situation applications, that is the only thing
left and the way to start is by picking the
00:42:46.109 --> 00:42:49.839
simplest problem even possible in supersonic
flow.
00:42:49.839 --> 00:42:50.839
.
00:42:50.839 --> 00:42:58.410
A flat plate 2 d flat plate at an angle of
attack, let us say this angle is some alpha
00:42:58.410 --> 00:43:03.000
as in
aerospace terms and I am having uniform supersonic
00:43:03.000 --> 00:43:09.980
flow coming in, and I want to tell
what will happen here? It is of course, considered
00:43:09.980 --> 00:43:13.440
to be inviscid they are still in inviscid
world.
00:43:13.440 --> 00:43:19.520
.So, I am going to draw this one line that
is going to touch this now, this is going
00:43:19.520 --> 00:43:23.049
to help
us to understand the problem the way to approach
00:43:23.049 --> 00:43:28.890
this problem is to say I want to draw
what is the wave nature at the front, if I
00:43:28.890 --> 00:43:34.069
look at this dotted line and above then I
will just
00:43:34.069 --> 00:43:40.430
block this region with paper. It looks like
this flow with a sudden expansion corner
00:43:40.430 --> 00:43:46.079
stream line can be equivalent to a wall in
inviscid flow we are in inviscid gas dynamics.
00:43:46.079 --> 00:43:52.739
So, all I have to do is, it will have an expansion
fan like this flow will come like this, and
00:43:52.739 --> 00:43:59.009
turn around like this that is the easy part
now a little more difficult part. I will block
00:43:59.009 --> 00:44:01.750
the
other side of the dotted line.
00:44:01.750 --> 00:44:06.789
I am having only this dotted line and flow
below, what is this saying there is a wall,
00:44:06.789 --> 00:44:12.849
which is coming into the flow this is like
a compression and it is a sudden change. So,
00:44:12.849 --> 00:44:15.380
it
is going to have an oblique shock, which is
00:44:15.380 --> 00:44:18.570
some few classes before I have to go find
the
00:44:18.570 --> 00:44:22.890
angle. We hope that currently it is not a
detached shock, it has a shock the case is
00:44:22.890 --> 00:44:25.400
a
complicated case for this problem. We would
00:44:25.400 --> 00:44:28.440
not deal with it right now, let us say it
is
00:44:28.440 --> 00:44:33.960
attached shock and I can find the beta value
from there, this is my beta value. I can find
00:44:33.960 --> 00:44:36.839
it
let us hope we can just go find the mach number
00:44:36.839 --> 00:44:45.570
and gamma, I can get a beta value.
So, I have drawn this let us just change the
00:44:45.570 --> 00:44:53.499
color a little bit for the plate. Let us say
that
00:44:53.499 --> 00:44:59.099
green thing is my plate this is my shock this
is my expansion. I have drawn this my flow
00:44:59.099 --> 00:45:07.640
field here is also parallel like this. Now
the next thing to do I have to find what will
00:45:07.640 --> 00:45:12.630
happen on the back this flow field is going
to go like this. This flow field is going
00:45:12.630 --> 00:45:16.530
to go
like this, if there is no problem they will
00:45:16.530 --> 00:45:19.400
just keep going straight like this, but that
is not
00:45:19.400 --> 00:45:24.099
going to be the case why, because the pressure
here is going to be different from the
00:45:24.099 --> 00:45:29.260
pressure here. How do we know that this below
this dotted line region is going to go
00:45:29.260 --> 00:45:34.830
through a oblique shock, which is a compression
it is going to increase pressure here. Let
00:45:34.830 --> 00:45:42.769
us mach region let us call this infinity region
this is called 1, this is called 2.
00:45:42.769 --> 00:45:50.770
We can tell now that P 1 less than P infinity
and P 2 is greater than P infinity. I can
00:45:50.770 --> 00:45:53.720
tell
that for these two stream lines. Now, because
00:45:53.720 --> 00:46:00.769
of this I am going to have a case were this
pressure is higher than this. When these two
00:46:00.769 --> 00:46:06.470
packets of stream lines packets of fluids
come along the closest to the wall, when they
00:46:06.470 --> 00:46:10.089
touch what they are going to see
immediately is there is a pressure jump, what
00:46:10.089 --> 00:46:15.299
will happen flow will want to go from high
pressure to low pressure. Now, once that is
00:46:15.299 --> 00:46:17.390
said I will erase these things.
00:46:17.390 --> 00:46:24.190
.Once, that is said we can also tell that
the stream line wants to turn that way, because
00:46:24.190 --> 00:46:29.410
that region is lower pressure the same reason.
I can tell that the stream line has to go
00:46:29.410 --> 00:46:35.740
away from high pressure region for the low
pressure stream line that will also turn,
00:46:35.740 --> 00:46:39.979
if I
say this when will they stop turning it will
00:46:39.979 --> 00:46:46.940
go to a point, where the pressures will become
equal consider a case where it is turning
00:46:46.940 --> 00:46:53.210
too much, what will happen then now that side
will be finding that oh these guys are pushing
00:46:53.210 --> 00:47:00.720
too much without any force a better way of
explaining this is after I put the waves here
00:47:00.720 --> 00:47:03.239
right, if I think about what will happen here.
I
00:47:03.239 --> 00:47:09.779
will bock the top side I am having only the
bottom side. I am having flow along this wall
00:47:09.779 --> 00:47:13.710
and it is now turning I do not know by how
much let us just say it is turning, because
00:47:13.710 --> 00:47:15.579
it is
turning away from the flow direction, which
00:47:15.579 --> 00:47:21.369
means it is an expansion corner.
So, it has gone to be an expansion wave here
00:47:21.369 --> 00:47:28.440
expansion fan here this region is my
expansion fan. let us say if I look at the
00:47:28.440 --> 00:47:32.719
top section alone flow is coming like this
and
00:47:32.719 --> 00:47:37.910
wall is moving in, why do I call this a wall
it is actually the stream line center line
00:47:37.910 --> 00:47:43.589
will be
as, if it is a wall in a inviscid flow stream
00:47:43.589 --> 00:47:45.720
liens are equivalent to a wall in inviscid
flow.
00:47:45.720 --> 00:47:49.800
So, I can just say that the wall is moved
in this is equality to our problem, which
00:47:49.800 --> 00:47:54.080
we did
some days back this problem. So, it is going
00:47:54.080 --> 00:47:58.880
to have a shock in this region with respect
to
00:47:58.880 --> 00:48:07.519
this line. It is going to have a beta value
there will be a shock there now, because of
00:48:07.519 --> 00:48:10.940
this
what is going to happen this shock will cause
00:48:10.940 --> 00:48:15.920
the pressure, which is low to become
higher while this pressure, which is high
00:48:15.920 --> 00:48:18.819
will decrease now across this expansion fan
to
00:48:18.819 --> 00:48:22.219
become lower.
So, ideally these two waves should be such
00:48:22.219 --> 00:48:28.720
that they will come to an optimum point,
where they will reach the same pressure is
00:48:28.720 --> 00:48:33.020
that the only condition. There is one more
condition needed I told you already in gas
00:48:33.020 --> 00:48:37.279
dynamics always. There are two conditions
what are the two conditions one is the velocity
00:48:37.279 --> 00:48:42.180
vector direction, other is the pressure. We
keep on repeating this all through we will
00:48:42.180 --> 00:48:48.489
see how it goes you have to make sure that
these two vector directions are the same at
00:48:48.489 --> 00:48:56.140
the same time. These two pressures are the
same two conditions two constants.
00:48:56.140 --> 00:49:01.509
I have to get what are those what is the full
delta theta here, which is same as delta nu
00:49:01.509 --> 00:49:06.630
I
have to find this how much is the change and
00:49:06.630 --> 00:49:13.240
the second thing is this beta angle two
variables two conditions. I can definitely
00:49:13.240 --> 00:49:17.499
solve this problem. Now, it takes a lot more
effort to solve this problem we would not
00:49:17.499 --> 00:49:20.219
solve it now we will solve it after some time
it
00:49:20.219 --> 00:49:25.630
needs iterative procedure and we do not have
enough time anyway. So, this is a simplest
00:49:25.630 --> 00:49:29.130
.problem I have done now. I will do one more
and then we will switch on to some other
00:49:29.130 --> 00:49:34.849
case. we will go and solve more problems after
that see you people next class .
00:49:34.849 --> 00:49:35.849
.
.
00:49:35.849 --> 00:49:36.849
.