WEBVTT
Kind: captions
Language: en
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Hello every one welcome back, we were discussing
Fanno Flow and most likely today
00:00:17.380 --> 00:00:25.269
will be the last class of Fanno Flow. And
next class on wards we will talk about heat
00:00:25.269 --> 00:00:29.470
transfer. I will just recap quickly and then
we will go to numerical example.
00:00:29.470 --> 00:00:30.470
.
00:00:30.470 --> 00:00:39.710
Lets us say, I have a duct and we are starting
with let us say M less than 1, M 1 less than
00:00:39.710 --> 00:00:55.530
1 the inlet and I am going to say this is
my duct length and some length L and for this
00:00:55.530 --> 00:01:07.790
mark number M 1 I am going to say that, the
for that particular friction factor and the
00:01:07.790 --> 00:01:15.380
hydraulic diameter of my duct, the L star
happens to be this long L max or L star
00:01:15.380 --> 00:01:23.200
happens to be this long and if I have only
this length. Then now, I can say that my flow
00:01:23.200 --> 00:01:28.320
since, it started subsonic it is going stay
subsonic up to here, but it is going towards
00:01:28.320 --> 00:01:31.040
M
equal to 1, from here it is going towards
00:01:31.040 --> 00:01:35.890
M equal to 1.
Now, we look at it is similar to the way we
00:01:35.890 --> 00:01:41.990
did the expansion fans, frankly meant curves
I am going to say from M 1 it is going to
00:01:41.990 --> 00:01:49.420
reach mark number 1 and somewhere along the
way it reaches M 2. Now, if think about M
00:01:49.420 --> 00:01:53.229
2 as my starting point, then it will take
00:01:53.229 --> 00:02:02.250
.exactly this length to reach this L star.
So, this length will be that particular L
00:02:02.250 --> 00:02:07.770
star for this
exit not number M 2 for this particular problem.
00:02:07.770 --> 00:02:11.700
That is the way I want to think about
this is the way to solve the problem.
00:02:11.700 --> 00:02:23.319
I am going to say, if this is my M 2 then
this is my L 1 star that is L max based on
00:02:23.319 --> 00:02:26.980
that f
that D h and this mark number M 1 that is
00:02:26.980 --> 00:02:41.510
this length. And this will be my L 2 star,
which is for that f for that D h if I use
00:02:41.510 --> 00:02:49.680
M 2 I will get this length. Now, the difference
happens to be the length that the float x
00:02:49.680 --> 00:02:53.309
to go from M 1 to M 2 till it reaches M equal
to
00:02:53.309 --> 00:02:59.330
1, this is the way I am going to think about
this Colum, this is one simple application.
00:02:59.330 --> 00:03:05.030
Now, I will make the case little more complicated
I will just say I extended the length, up
00:03:05.030 --> 00:03:10.449
to some length only up to here, not past my
l star and what will happen I will consider
00:03:10.449 --> 00:03:17.939
this as an M 3. If I start from M 2 and think
of this problem it is a duct of this length
00:03:17.939 --> 00:03:27.040
whatever, this length L prime x say, then
I will just add that L prime here L prime.
00:03:27.040 --> 00:03:29.439
And
now, adding one more section of this pipe,
00:03:29.439 --> 00:03:34.689
with same the enough and same area of
diameter I am adding one more section here.
00:03:34.689 --> 00:03:40.499
Then I can just think of this problem from
M 2 onwards as if it is same as the previous
00:03:40.499 --> 00:03:47.040
problem, just 1 is replaced by 2 and 3 is
replacing 2 here. That kind of problem just
00:03:47.040 --> 00:03:50.680
coming around here, that is all it should
be I am going to think about I am starting
00:03:50.680 --> 00:03:54.919
with
M 2 and that will have a L 2 star. And now,
00:03:54.919 --> 00:04:03.049
I have a M 3 such that if I go this extra
length, that will become a L 3 star, this
00:04:03.049 --> 00:04:07.829
is the way to think about this problem.
So, if I say this is the initial length, what
00:04:07.829 --> 00:04:12.089
is the mark number then if I add this much
length to way total will be the ma k number
00:04:12.089 --> 00:04:15.790
that kind of a problem, then I will approach
it this particular method I am just going
00:04:15.790 --> 00:04:19.739
to find L 3 star L 2 star etcetera. And then
based
00:04:19.739 --> 00:04:25.410
on that I can find the answer, will go through
numerical examples that is just one part,
00:04:25.410 --> 00:04:29.069
if
my length of my duct is exactly equal to L
00:04:29.069 --> 00:04:34.330
star then that is the simplest case you can
ever
00:04:34.330 --> 00:04:45.850
have. I have a duct f and D h are given and
I am going to say some M 1 less than 1 and
00:04:45.850 --> 00:04:52.340
I
am going to say this length L equal to L star.
00:04:52.340 --> 00:04:58.759
If I have this case, then I am going to say
fully it is going to become M equal to 1 at
00:04:58.759 --> 00:05:01.729
that
exit, simplest case really think about it
00:05:01.729 --> 00:05:04.830
for this L equal to L 1 star I should say,
for this
00:05:04.830 --> 00:05:10.280
mark number M 1 this is the length needed
and I am providing exactly that length. So,
00:05:10.280 --> 00:05:13.849
it
is just going to reach it, that is one simple
00:05:13.849 --> 00:05:25.280
case, where I will have M 2 equal to 1 here.
Now, the next case is if I have again M 1
00:05:25.280 --> 00:05:34.460
subsonic, but I am having a length greater
than
00:05:34.460 --> 00:05:47.699
.I call this again L while my L star is still
the same, this my L 1 star ends there from
00:05:47.699 --> 00:05:50.129
the
starting point if I think of that way.
00:05:50.129 --> 00:05:55.740
Now, I have more length, what will the flow
do it will decrease, it will send out extra
00:05:55.740 --> 00:06:00.240
compression waves from this excess length
such that, those compression waves go and
00:06:00.240 --> 00:06:02.250
it
subsonic. So, it will go faster it will go
00:06:02.250 --> 00:06:04.300
reach this point where thinking only steady
state
00:06:04.300 --> 00:06:11.009
as of time. So, all that compression waves
will go ahead and make this flow come slower
00:06:11.009 --> 00:06:18.099
into it. So, M 1 will now, change to M 1 prime
which will be less than M 1 value.
00:06:18.099 --> 00:06:25.120
Such that, it is equivalent to thinking about
let us say, I will pick this problem, this
00:06:25.120 --> 00:06:31.430
equivalent thinking about I am having a case
where this length is extended more here,
00:06:31.430 --> 00:06:39.469
such that I have M 1 prime here, M 1 prime
more in this side and now, that will be equal
00:06:39.469 --> 00:06:47.620
to that that M 1 prime giving L 1 star L 1
prime star let us say, going from this M 1
00:06:47.620 --> 00:06:54.520
prime to L 1 prime star. That will be equal
to your new value, that is how this problem
00:06:54.520 --> 00:06:57.960
goes at that point it is decreased such that,
that we have.
00:06:57.960 --> 00:06:58.960
.
00:06:58.960 --> 00:07:08.499
Now the next thing is I will do the whole
thing for supersonic. So, little different
00:07:08.499 --> 00:07:11.120
in
supersonic, simple cases are that exactly
00:07:11.120 --> 00:07:19.099
the same M 1 greater than 1 this is my L star
L
00:07:19.099 --> 00:07:29.150
1 star and this is my L if I think about M
2 greater than 1. It is just decreasing from
00:07:29.150 --> 00:07:34.430
M 1
towards M 1 M exit equal to 1, but before
00:07:34.430 --> 00:07:40.479
that we stopped the duct. So, it is not exactly
equal to 1 slightly greater than 1, it is
00:07:40.479 --> 00:07:42.780
decreasing monotonically towards there that
is the
00:07:42.780 --> 00:07:46.070
simplest case, similar to subsonic case.
00:07:46.070 --> 00:07:52.520
.If I make this equal to L 1 star then also
it simple case exit will be M equal to1. Now,
00:07:52.520 --> 00:07:59.150
I
will go for more than that length case, say
00:07:59.150 --> 00:08:04.860
I am just slightly more than that L star case,
if
00:08:04.860 --> 00:08:11.629
I have this case then typically there will
be a shock by the way it will be ahead of
00:08:11.629 --> 00:08:15.750
this L 1
star value that location in a shock let us
00:08:15.750 --> 00:08:19.539
be a head of this, what is the rational for
that,
00:08:19.539 --> 00:08:24.220
this extra length is what is going to cause
excess resistance. And, so that will send
00:08:24.220 --> 00:08:26.229
out the
set of compression waves and as they come
00:08:26.229 --> 00:08:30.229
in they find higher and higher mark number
flow resisting it.
00:08:30.229 --> 00:08:33.900
So, there will be a point where those set
of compression waves coming in coming to an
00:08:33.900 --> 00:08:39.820
equilibrium against the flow, it is like spread
well shock wants to run ahead and flows
00:08:39.820 --> 00:08:46.391
going this way, it will come and stay stabilise
somewhere typically. So, there will be a
00:08:46.391 --> 00:08:55.930
shock sitting here, it is subsonic flow now,
I will call this as my M 2, M 1 M 2, M 2
00:08:55.930 --> 00:09:00.060
prime I will call this there are just downstream
of the shock immediate downstream of
00:09:00.060 --> 00:09:05.720
stock is less than 1 and after that it takes
a subsonic part, we discuss this on the plot
00:09:05.720 --> 00:09:06.720
that
day.
00:09:06.720 --> 00:09:18.459
Because, in the subsonic section 4 f L star
by D versus mart number, we found that in
00:09:18.459 --> 00:09:27.700
subsonic region L star can be much higher.
So, instead of going to the supersonic wave
00:09:27.700 --> 00:09:34.180
where L star is only this length, instead
of this it takes subsonic solution, where
00:09:34.180 --> 00:09:38.410
it has
higher length. That is the idea for this same
00:09:38.410 --> 00:09:44.470
f and D this curve will just become like L
star curve it has higher L star. So, it jumps
00:09:44.470 --> 00:09:46.920
there, that is the idea behind this going
to
00:09:46.920 --> 00:09:51.779
subsonic side, after that it will still want
to reach to M equal to 1 and the exit this
00:09:51.779 --> 00:09:54.310
will be
the case.
00:09:54.310 --> 00:10:05.600
If I keep on extending this length, then will
go to situation where the shock goes more
00:10:05.600 --> 00:10:11.230
and more upstream, if I extended any further
then the shock will goes set outside this
00:10:11.230 --> 00:10:16.529
duct, set outside this duct after that if
I will look at what is the flow inside the
00:10:16.529 --> 00:10:19.360
duct it is
only subsonic flow I will go back to either
00:10:19.360 --> 00:10:23.120
more here, and the problem just becomes this.
00:10:23.120 --> 00:10:24.120
..
00:10:24.120 --> 00:10:26.210
Of course, there may be a shock sitting in
front of it and before that there could be
00:10:26.210 --> 00:10:29.330
a
supersonic flow even here, I am not showing
00:10:29.330 --> 00:10:33.259
you that part here that is all it should be.
(Refer Slide Time: 10: 41)
00:10:33.259 --> 00:10:39.079
With this idea exist wanted to re iterate
this whole thing in one sequence, want up
00:10:39.079 --> 00:10:44.190
I will
start with M l equal to 1 in here, M 1 equal
00:10:44.190 --> 00:10:52.060
to 1 I cannot send M 1 equal to 1 flow into
any duct because, immediately after that there
00:10:52.060 --> 00:10:55.769
is be a friction which will send
compression waves, which will slow down the
00:10:55.769 --> 00:11:00.850
flow and immediately we will get only a
subsonic case behind, that is why transonic
00:11:00.850 --> 00:11:02.579
problem is very, very difficult it could
00:11:02.579 --> 00:11:07.470
.become M equal to 1 immediately it forms
a shock and decreases somewhat now
00:11:07.470 --> 00:11:12.640
immediately after that.
And you will sudden have lower flow rates,
00:11:12.640 --> 00:11:15.260
that is the way it goes.
(Refer Slide Time: 11: 21)
00:11:15.260 --> 00:11:20.920
Now, we will go pick numerical example, I
will deal with only with only gamma equal
00:11:20.920 --> 00:11:27.899
to 1.4 typically, air as what I am considering
as my fluid, and I am having a simple case
00:11:27.899 --> 00:11:35.769
I
will consider this as problem a where I am
00:11:35.769 --> 00:11:46.019
having a duct with f equal to 0.002 hydraulic
diameter given to be 5 centimetres 0.05 m
00:11:46.019 --> 00:12:00.190
m or 0.05 meters. And L s given to be 2
meters, I am having M 1 equal to 0.2 mark
00:12:00.190 --> 00:12:07.709
number I want to find what does the exit
mark number that is the idea.
00:12:07.709 --> 00:12:12.649
To start this problem always if it is a friction
problem immediately I want to find L star
00:12:12.649 --> 00:12:18.980
because, based on L verses L star whether
L is less than or greater than L star, my
00:12:18.980 --> 00:12:26.220
approach to the problem changes. So, first
thing I need to do is find L star. So, M 1
00:12:26.220 --> 00:12:31.089
equal
to 0.2 from tables gamma equal to 1.4 tables.
00:12:31.089 --> 00:12:47.370
So, if I go look at it 4 f L star by D h is
given to be 14.533. So, based on f and D h
00:12:47.370 --> 00:12:57.180
given for my friction factor for my particular
duct I am going to get L star as 90.83 meters.
00:12:57.180 --> 00:13:07.269
Now, your numbers look ridiculous at L is
2 meters and L star is 90 meters, only when
00:13:07.269 --> 00:13:10.259
it
goes 90 meters long my mark number will ever
00:13:10.259 --> 00:13:15.100
reach M equal to 1, I am starting with
very, very low mark number that is why this
00:13:15.100 --> 00:13:18.600
case is happened. If I started with 0.9 it
may
00:13:18.600 --> 00:13:23.880
.not be very long, it may be just 2 meters
also you remembered that curve that was
00:13:23.880 --> 00:13:27.230
looking that actually I had a curve in this
port. A curve is looking like this
00:13:27.230 --> 00:13:28.230
.
00:13:28.230 --> 00:13:33.610
and mark 0.2 is somewhere their high out there
that is why it is looking like this, the
00:13:33.610 --> 00:13:36.220
numbers may look ridiculous at beginning we
know it immediately.
00:13:36.220 --> 00:13:37.529
.(Refer Slide Time: 13: 36)
00:13:37.529 --> 00:13:42.209
If this is the case what should I do I just
have to go and draw some dotted line and L
00:13:42.209 --> 00:13:46.269
star
somewhere out there, this will become L 2
00:13:46.269 --> 00:13:53.129
star, where this is my M 2. So, this is the
remaining length I know this is 2 meters from
00:13:53.129 --> 00:13:59.560
M 1 from the inlet to the star is L 1 star
that is given here, I will make it L 1 star
00:13:59.560 --> 00:14:04.260
this my L 2 star. So, L equal to L 2 star
minus L
00:14:04.260 --> 00:14:19.339
1 star which means I can find my L 2 star,
L 2 star value is to 88.83 meters.
00:14:19.339 --> 00:14:32.640
So, I will go and find 4 f L by d 4 f L 2
star by D h that comes out to be 14.213 meters.
00:14:32.640 --> 00:14:42.980
Previously, it was 14.33 for M equal to 0.2
from there it has become slightly lesser this
00:14:42.980 --> 00:14:49.639
mark number increased little bit that is what
we have, from here I should not be putting
00:14:49.639 --> 00:14:57.519
meters here it is not meters sorry no unit
is it is not dimensioned L by D. So, from
00:14:57.519 --> 00:15:03.199
here I
am going to find my mark number M 2 this will
00:15:03.199 --> 00:15:18.300
give me a M 2 of 0.028, we will look it
does change only from 0.208 because, this
00:15:18.300 --> 00:15:21.259
length is far small compare to my L 1 star
that
00:15:21.259 --> 00:15:29.040
I have here L is 2 meters L 1 star is 90 meters
it is not that much of change when I look
00:15:29.040 --> 00:15:34.190
at the overall needed length.
So, there is not much of change. So, this
00:15:34.190 --> 00:15:39.970
is the first problem, let us pick some other
length let say L equal to 45 meters the same
00:15:39.970 --> 00:15:49.449
problem as before. Let us want to see what
if I go half way to L 1 star perfectly. So,
00:15:49.449 --> 00:15:51.459
since I know the problem I just go through
the
00:15:51.459 --> 00:15:56.000
same process nothing changes, I know that
length is less than L 1 star. So, if I go
00:15:56.000 --> 00:16:02.910
through
this one L 2 star is 45.83 meters.
00:16:02.910 --> 00:16:26.629
.So, 4 f L 2 star by D h is going to be 0.265
sorry I jumped 0.265 is my mark number this
00:16:26.629 --> 00:16:46.120
comes out to be 7.333. And this gives me a
mark number of 0.265 you still find that even
00:16:46.120 --> 00:16:53.979
if I go half way through L star it does not
gone too much away from mark 0.2.
00:16:53.979 --> 00:16:54.979
.
00:16:54.979 --> 00:16:58.730
This is the idea of this particular curve
I will go back to this curve. I am having
00:16:58.730 --> 00:17:03.040
4 f L star
by D verses mark number, this curve is very
00:17:03.040 --> 00:17:08.189
steep at the beginning and then it turns to
become shallow, only when it is shallow there
00:17:08.189 --> 00:17:14.089
will be too much change in mark number
till that time it changes only in L star L
00:17:14.089 --> 00:17:16.580
basically, changes mainly in the length of
the
00:17:16.580 --> 00:17:21.980
duct not much change in M 1 that is what is
currently happen.
00:17:21.980 --> 00:17:28.510
So, when you get an answer like this you should
think about either it is ridiculous or not
00:17:28.510 --> 00:17:36.790
currently it is not. So, you can leave with
this particularly problem. Now, I will change
00:17:36.790 --> 00:17:41.350
the problem in another way around solve it
and mark number the way, suddenly I want to
00:17:41.350 --> 00:17:47.850
solve the problem by saying how much length
should I have. So, that my pressure at the
00:17:47.850 --> 00:17:51.800
exit is half of pressure at the inlet I am
changing the problem.
00:17:51.800 --> 00:17:52.800
..
00:17:52.800 --> 00:18:02.570
I am going to say, I am having the duct f
and D h are given for my cube, I am going
00:18:02.570 --> 00:18:07.670
to
say there is some M 1 let us say this again
00:18:07.670 --> 00:18:19.710
0.2 and I am given some P 1 I want to know
what should be the length L such that P 2
00:18:19.710 --> 00:18:23.850
equal to half of P 1 this is a problem I want
to
00:18:23.850 --> 00:18:31.130
solve. So, how I will approach the problem
it is not very difficult I am again going
00:18:31.130 --> 00:18:36.200
to say
for this M 1 I know my L 1 star from problem
00:18:36.200 --> 00:18:42.060
a I know the L star it is 90.83 meters and
I
00:18:42.060 --> 00:18:48.470
know this number.
Now, the only thing I had to think about is,
00:18:48.470 --> 00:18:54.060
what should be my mark number such that I
will get half the pressure. So, it is very
00:18:54.060 --> 00:18:57.830
difficult to do, it is not very, very difficult
to do,
00:18:57.830 --> 00:19:05.260
but as just look through that problem. It
is, so happens that there is a another column
00:19:05.260 --> 00:19:11.760
called P by P star given in your tables. If
you have that it is easy otherwise it is little
00:19:11.760 --> 00:19:16.790
more complicated to solve this problem. It,
so happens that in my tables I have it. So,
00:19:16.790 --> 00:19:20.480
we
will just use this directly for this example.
00:19:20.480 --> 00:19:34.340
P 2 by P star we know it is equal to P 1 P
star looks like I am switching star up and
00:19:34.340 --> 00:19:40.430
down
this keep it as up now. So, we know this because,
00:19:40.430 --> 00:19:44.730
P star is going to be the same whatever
my initial condition is fixed, I am going
00:19:44.730 --> 00:19:47.320
to end up with this I am going to say this
is
00:19:47.320 --> 00:20:01.260
equal to 2.728 that is I already divided by
2 to get this also I am having this also.
00:20:01.260 --> 00:20:05.740
Now,
this from my tables gamma equal to 1.4 and
00:20:05.740 --> 00:20:17.550
this P by P star, will give me my M 2 as
0.396 this is the value I get, now my actual
00:20:17.550 --> 00:20:19.500
objective is to find length of the duct.
00:20:19.500 --> 00:20:25.350
.So, what should I do I should go and find
L 2 star again from there. So, from my tables
00:20:25.350 --> 00:20:32.330
again for this mark number my L 2 star actually,
what will get from tables is 4 f L star
00:20:32.330 --> 00:20:44.550
etcetera. So, if my L 2 star will be actually,
I will put 4 f by D h this number will be
00:20:44.550 --> 00:21:01.341
2.388. So, from here I will find my L 2 star,
L 2 star happens to be 14.925 meters. Think
00:21:01.341 --> 00:21:08.400
about this when I started with M equal to
0.2 my number was 90 meters, and it became
00:21:08.400 --> 00:21:16.550
roughly 0.4 I have only 14.9 meters. So, the
curve is starting to turn, that particular
00:21:16.550 --> 00:21:17.550
curve
here.
00:21:17.550 --> 00:21:18.550
.
00:21:18.550 --> 00:21:24.160
This particular curve is starting to turn
around 0.3 plus it will start turn steeply,
00:21:24.160 --> 00:21:26.550
initially it
was out there now, I slowly turning, turning,
00:21:26.550 --> 00:21:32.360
turning that turning point start seriously
only after 0.3 mark number, above 0.3 mark
00:21:32.360 --> 00:21:37.330
number there is more sensitivity to mark
number. And when we are going very close to
00:21:37.330 --> 00:21:40.630
M 1 there is lot more change to it.
00:21:40.630 --> 00:21:41.630
..
00:21:41.630 --> 00:21:50.060
So, we know that from here my star value is
90 meters away, and from this point my star
00:21:50.060 --> 00:21:59.360
value is only 14.9 meters away. So, the difference
must be my L, L happens to be 90.83
00:21:59.360 --> 00:22:15.220
minus 14. I will make it 14.93 because, it
gives me nearest number, 75.9 meters is the
00:22:15.220 --> 00:22:22.320
distance I need to extend this duct before
it goes half the pressure at the end line
00:22:22.320 --> 00:22:30.060
that is
the current situation, we have that is the
00:22:30.060 --> 00:22:35.700
idea of that particular problem.
Now, we still did not do a case where I have
00:22:35.700 --> 00:22:42.640
a subsonic entry and my length is more than
my L star. So, if we pick a case where I have
00:22:42.640 --> 00:22:50.920
a subsonic entry and my length is more
than L star let us say I have only 100 meters,
00:22:50.920 --> 00:22:55.910
100 is a very next number slightly above
90. So, I will just pick 100.
00:22:55.910 --> 00:22:56.910
..
00:22:56.910 --> 00:23:05.620
So, I am going to say my L equal to 100 meters
M 1 equal to 0.2 which gives my L 1 star
00:23:05.620 --> 00:23:19.300
is equal to 91.83 meters these are the numbers
we have. Now, we know that 90. sorry
00:23:19.300 --> 00:23:27.710
90.83 meters, we know that now, there will
be a compression waves travelling upstream
00:23:27.710 --> 00:23:36.600
and adjusting this M 1 to M 1 prime I do not
know what value it is, but I know my M 2
00:23:36.600 --> 00:23:42.750
the exit of the duct will be M equal to 1.
So, it will be getting adjusted to some M
00:23:42.750 --> 00:23:43.750
1
prime.
00:23:43.750 --> 00:23:52.730
And what it will get adjusted to M 1 prime
will now, have a L star of 100 that is how
00:23:52.730 --> 00:23:55.390
it
will get adjusted. So, only I have to do is
00:23:55.390 --> 00:24:00.300
4 f into 100 divided by D h, if I do this
of
00:24:00.300 --> 00:24:08.710
course, I am still using same f and same D
h for all my problems if I use that it will
00:24:08.710 --> 00:24:15.680
be 4
into 0.002 into 100 divided by 0.05 meters,
00:24:15.680 --> 00:24:27.260
this number comes out to be 16. So, my M 1
prime comes out to be 0.193.
00:24:27.260 --> 00:24:38.900
So, I started with 0.2, but since the length
is more than this length the flow immediately
00:24:38.900 --> 00:24:46.450
got adjusted to M at the inlet happens to
the 0.193 and after that it is just a simple
00:24:46.450 --> 00:24:50.080
fan of
flow the condition, where it is just chopped
00:24:50.080 --> 00:24:55.140
at the exit it is M equal to 1 very nice
problem. So, it is just go slowly increasing
00:24:55.140 --> 00:25:00.370
math number and eventually it will go a point
where it is M equal to 1. if you want to find
00:25:00.370 --> 00:25:02.300
any other variable I am sure you can find
out
00:25:02.300 --> 00:25:10.490
from just going to supersonic cases from now.
Of course, I can now, tell that find temperature
00:25:10.490 --> 00:25:14.300
at some location or tell me when the
temperature goes triple the initial temperature
00:25:14.300 --> 00:25:16.430
all those kinds of problems we can solve.
I
00:25:16.430 --> 00:25:20.750
.am just avoiding those you know you can calculated
actually, temperature decreases at
00:25:20.750 --> 00:25:23.950
subsonic to M equal to 1. So, temperature
should goes to one third of I want to ask
00:25:23.950 --> 00:25:28.490
a
question. So, now we will look at supersonic
00:25:28.490 --> 00:25:32.110
inlet, I just add here.
00:25:32.110 --> 00:25:42.960
So, I am going to say it is all subsonic cases
and this I believed was problem d a b and
00:25:42.960 --> 00:25:48.070
this was problem c just did not number there.
.
00:25:48.070 --> 00:26:00.020
Now, we will go for next problem, I will call
this 2 a now, I am going to say M 1 is 2.0
00:26:00.020 --> 00:26:11.070
instead of 0.2 same f, same D h just for simplicity
I did not want to change this numbers.
00:26:11.070 --> 00:26:18.030
Of course, I will give you another exercise
with this, with all this numbers varying not
00:26:18.030 --> 00:26:23.320
very difficult to calculate, but it is just
something you want to get use to.
00:26:23.320 --> 00:26:27.170
These numbers you can make a mistake typically,
in looking up a tables the gamma can
00:26:27.170 --> 00:26:32.510
be different in, different books some books
will give you, so many gamma values,
00:26:32.510 --> 00:26:35.950
different tables for each gamma values should
not be looking at the wrong gamma value
00:26:35.950 --> 00:26:41.010
you will get all wrong solutions, typically
the mistake made by the most students. So,
00:26:41.010 --> 00:26:47.580
I
will pick a case and being careful and I am
00:26:47.580 --> 00:26:55.380
going to say L equal to 1.5 meters, previously
we had case a was L equal to 2 meters I am
00:26:55.380 --> 00:26:58.760
picking 1.5 meters you will see while in a
minute.
00:26:58.760 --> 00:27:09.550
.I already calculated some things
M 1 equal to 2 if I start here of course,
00:27:09.550 --> 00:27:11.920
the very first
thing I want to find the math number at the
00:27:11.920 --> 00:27:18.130
exit that is the overall objective of my
problem. Now, to find that half curve of course,
00:27:18.130 --> 00:27:22.291
I had known how close to M equal to 1
am I at the exit. So, I will first fine L
00:27:22.291 --> 00:27:34.950
star, L 1 star, if I find 4 f L 1 star by
D h for this
00:27:34.950 --> 00:27:44.800
mark number for that gamma equal to 1.4 I
am are going to get a number 0.305 which is
00:27:44.800 --> 00:28:03.580
going to give me L 1 star as sorry 1.906 meters
this is the reason I did not choose 2
00:28:03.580 --> 00:28:06.260
meters.
Because, I already calculated this one before
00:28:06.260 --> 00:28:13.550
coming to class now. We are finding that
now, I chosen a case L less than L 1 star
00:28:13.550 --> 00:28:16.340
which is very simple now, it is supersonic
it will
00:28:16.340 --> 00:28:20.610
end with supersonic about lesser mark number
than the initial. Similar, thing I need to
00:28:20.610 --> 00:28:24.440
do
now I will just find that it has 0.406 meters
00:28:24.440 --> 00:28:27.880
from this point to L star, that is the distance
it
00:28:27.880 --> 00:28:39.940
has. So, I have to find L 2 star happens to
be 0.406 meters, this 0.406 meters I just
00:28:39.940 --> 00:28:54.010
had to
find 4 f L 2 star by D h and 4 f L 2 star
00:28:54.010 --> 00:29:00.530
by D h for that f and D h given happens to
be
00:29:00.530 --> 00:29:07.660
0.065.
Now, I have to go to gamma equal to 1.4 tables
00:29:07.660 --> 00:29:14.070
look for where this occurs in that table
and choose that mark number, that mark number
00:29:14.070 --> 00:29:25.870
comes out to be 1.30 I started with
mark 2 and I am ending with 1.3 marks. And
00:29:25.870 --> 00:29:29.730
within that 0.4 meters left it is going go
to
00:29:29.730 --> 00:29:34.440
M equal to 1 that is what we are seeing, here
you are seeing that there is a lot of
00:29:34.440 --> 00:29:39.710
sensitivity to length in terms of mark number
change for small length change mark
00:29:39.710 --> 00:29:48.910
number changes a lot, that is about supersonic
case. Now, we will choose a case where L
00:29:48.910 --> 00:29:50.260
is greater than L star.
00:29:50.260 --> 00:29:51.260
..
00:29:51.260 --> 00:30:00.220
Now, I will choose L greater than L star.
So, I am going to pick L equal to 3 same M
00:30:00.220 --> 00:30:14.100
equal to 2.0 as my mark number I begin L equal
to 3 meters. So, we except a shock let us
00:30:14.100 --> 00:30:24.540
say this was my L 1 star I am going fast that
point and M equal to 2.0 as the beginning.
00:30:24.540 --> 00:30:32.560
Now, I know my shock should sitting slightly
inside of this, I told you already. So, I
00:30:32.560 --> 00:30:35.040
am
excepting a shock somewhere here and I am
00:30:35.040 --> 00:30:41.840
going to say mark number before as M 2
and mark number after M 2 prime just after
00:30:41.840 --> 00:30:47.520
the shock, shock has happening at that
infinite as minute as can in location.
00:30:47.520 --> 00:30:57.370
So, I want to find that particular length
at which the shock occurs, I will call this
00:30:57.370 --> 00:31:01.770
as
location or shock L s a length at which shock
00:31:01.770 --> 00:31:08.980
occurs, such that I will get M 3 equal to
1
00:31:08.980 --> 00:31:15.150
that is the overall thing. I am just doing
mark number variation extra in my tube of
00:31:15.150 --> 00:31:18.370
course, I know that once know my mark number
distribution in my tube I can find any
00:31:18.370 --> 00:31:23.370
other property because, I have fan flow of
tables with me, for a given mark number I
00:31:23.370 --> 00:31:25.720
can
just go and find any other property.
00:31:25.720 --> 00:31:31.100
If want to find say somewhere in the middle
I can just go and calculate from M 2 prime
00:31:31.100 --> 00:31:33.280
I
went, so much down steam with this f and D
00:31:33.280 --> 00:31:37.400
h I can solve that problem, it is not very
difficult to solve and if I want P y by P
00:31:37.400 --> 00:31:39.490
star whatever I can get every number. So,
I can
00:31:39.490 --> 00:31:43.660
change pressures also across. So, I am not
solving those that probably I will give those
00:31:43.660 --> 00:31:47.510
as
exercise problems. So, how will I solve this.
00:31:47.510 --> 00:31:55.620
.So, I am going to first read this part of
the problem, as supersonic flow going some
00:31:55.620 --> 00:32:02.150
length less than L star which is just now,
we did a of second 2 a right that was this
00:32:02.150 --> 00:32:08.950
problem. And this of 2 b I keep forgetting
to label these problems these 2 b. And, so
00:32:08.950 --> 00:32:12.600
I
will solve this as it is 2 a, after that I
00:32:12.600 --> 00:32:14.490
will have a normal shock problems, I will
just go to
00:32:14.490 --> 00:32:21.160
normal shock tables and jump from here to
here across this, once I do this from here
00:32:21.160 --> 00:32:24.750
to
here remaining portion. Now, I have to see
00:32:24.750 --> 00:32:30.390
if I get this mark number to be 1 at the exit.
So, again I will solve this as 1 a problem.
00:32:30.390 --> 00:32:36.265
And I will see if this much is 0 how will
I check that simplest way to check, what if
00:32:36.265 --> 00:32:39.420
it is
more than the length of this particular M
00:32:39.420 --> 00:32:43.620
2 star, M 2 prime star that is the case. So,
all I
00:32:43.620 --> 00:32:49.360
need to do find the remaining length and see
if that is more than L star corresponding
00:32:49.360 --> 00:32:52.780
to
M 2 prime, that is all I need to do if it
00:32:52.780 --> 00:32:56.390
is equal to that L star then I have solved
my
00:32:56.390 --> 00:33:04.630
problem. So, I have to go through some iteration
here, I cannot directly guess what my L
00:33:04.630 --> 00:33:09.950
s will be. So, I had to go through a set of
iterations I have gone through iterations.
00:33:09.950 --> 00:33:17.660
So, I will just give it to you I will tell
you how it works I am going to have a full
00:33:17.660 --> 00:33:28.390
column
L s L 1 star minus L s 4 f by D h times L
00:33:28.390 --> 00:33:33.440
1 star minus L s and that is going to give
me
00:33:33.440 --> 00:33:49.190
my M 2 just to be clear I have written this
L s L 1 star minus L s that happens to be
00:33:49.190 --> 00:33:52.180
my L
2 star really, if you go and look at problem
00:33:52.180 --> 00:33:55.250
2 a that is was those are L 2 star. Now, I
will
00:33:55.250 --> 00:34:01.480
say L 4 f L 2 star by D h from here I will
go read out M 2 from there.
00:34:01.480 --> 00:34:06.460
Once I know M 2 I will go to normal shock
tables and find M 2 prime from there, from
00:34:06.460 --> 00:34:15.460
there I will go to 4 f by D h times L 2 prime
star from this mark number I will go to fine
00:34:15.460 --> 00:34:20.009
flow of tables find this of course, now you
have to pick only subsonic solutions. Now,
00:34:20.009 --> 00:34:24.730
I
will pick only subsonic solution from here,
00:34:24.730 --> 00:34:29.829
and from here I will get a L 2 prime star
I
00:34:29.829 --> 00:34:36.119
will just have to multiply appropriately D
h by 4 f you will get that, after all this
00:34:36.119 --> 00:34:42.060
finally, I
have to check whether this distance is more
00:34:42.060 --> 00:34:49.539
than or less than my that L 2 prime star.
So, the way I want to do it I will find L
00:34:49.539 --> 00:34:52.960
2 prime star that will be distance from this
shock
00:34:52.960 --> 00:35:02.100
till where M equal to 1 occurs, that plus
this L s must be equal to full duct length.
00:35:02.100 --> 00:35:06.960
So, I
will do it that way, I am going to say I have
00:35:06.960 --> 00:35:12.500
L s plus L 2 prime star, as my last column
I
00:35:12.500 --> 00:35:18.019
want to make this last column equal to L my
3 meters, that is the overall objective of
00:35:18.019 --> 00:35:24.349
my
solving this that is my iterative process.
00:35:24.349 --> 00:35:34.540
So, now I am going to pick let us say 1.3
because, I already have answer for 1.3, 1.30
00:35:34.540 --> 00:35:40.691
I
am picking. So, if I pick 1.30 again you get
00:35:40.691 --> 00:35:50.040
numbers and I said that L 1 star was 1.906
00:35:50.040 --> 00:36:06.160
.meters. So, I can get the L 1 star minus
L s that will be 0.606 meters from here, I
00:36:06.160 --> 00:36:09.450
am
going to get 4 f L 2 star by D h that number
00:36:09.450 --> 00:36:19.829
happens to be 0.097 from this I can back
calculate I can go to my fan of tables get
00:36:19.829 --> 00:36:25.019
a mark number for gamma equal to 1.4 1.393
is
00:36:25.019 --> 00:36:31.799
my mark number.
And M 2 prime and now, going to normal shock
00:36:31.799 --> 00:36:38.950
tables across this 2 columns alone
0.743, again draw the normal shock tables
00:36:38.950 --> 00:36:44.319
when you flirting pages you should go to the
correct gamma of tables. Now, I will go back
00:36:44.319 --> 00:36:51.900
to fan of flow tables, from here to next one
that number is 0.137 from here I will calculate
00:36:51.900 --> 00:37:05.559
my L 2 star 0.865. Now, I will add L s
plus L 2 star that comes out to be 2.156 meters.
00:37:05.559 --> 00:37:17.690
So, I am finding that in my shock is here,
my duct becomes L equal to 1 somewhere here
00:37:17.690 --> 00:37:22.790
itself, that is what I find in this case,
if this is where it becomes M equal to 1 and
00:37:22.790 --> 00:37:26.630
I have a
extra length or have not there will be more
00:37:26.630 --> 00:37:30.839
compression waves generated and it will go
upstream and push this shock further up stair.
00:37:30.839 --> 00:37:36.920
So, my logical way to go is I have to
decrease my L s. So, I will go decrease my
00:37:36.920 --> 00:37:41.510
L s I will make it 1.0 I initially when I
iterate
00:37:41.510 --> 00:37:46.190
call I p call as nice numbers eventually when
you are getting close to the answer we will
00:37:46.190 --> 00:37:58.009
think about something else.
So, from here 1 minus 10.906 minus 1 0.906
00:37:58.009 --> 00:38:09.049
from here I will write the next number
0.145. Now, this if I go to the tables I do
00:38:09.049 --> 00:38:11.500
not get exact number I have to interpolate.
So, I
00:38:11.500 --> 00:38:18.340
will just pick a curve of estimate of the
order of 1.52 I will just pick the closest
00:38:18.340 --> 00:38:22.420
1 to
interpolate, from here I go to normal shock
00:38:22.420 --> 00:38:30.300
tables across here, 0.694. Now, I from here,
to fan of flow tables and get the next number
00:38:30.300 --> 00:38:45.740
0.220 which will give me my L 2 star as
1.375. So, my L s plus L 2 star is 2.375 meters
00:38:45.740 --> 00:38:53.930
that is what I have finally, 2.375.
So, of course, here 2.15 begin 2.375 still
00:38:53.930 --> 00:38:59.109
not enough, my shock has to go further
upstream. So, I will change more drastically
00:38:59.109 --> 00:39:13.630
I will make it 0.5 meters it has gone well
ead 0.5 meters. Now, I will go through the
00:39:13.630 --> 00:39:28.119
same process 1.406 meters 4 f L by d will
give me 0.225 M 2 is approximately 1.75 mark
00:39:28.119 --> 00:39:37.890
number from their normal shock tables
gives me 0.628 mark number, downstream of
00:39:37.890 --> 00:39:40.109
the shock from here I will go back to fan
of
00:39:40.109 --> 00:39:48.720
flow subsonic solution gamma equal to 1.4
I am going to get 0.391 from there I can
00:39:48.720 --> 00:40:00.779
calculate my L 2 star 2.444 and that gives
me 2.944 meters getting close to the answer.
00:40:00.779 --> 00:40:08.900
So, I still have to go further upstream, but
I am close to the answer. So, now I will start
00:40:08.900 --> 00:40:28.990
changing slowly I will go to 0.40 that number
is 1.506 that I do 4 f L by D 0.241 and this
00:40:28.990 --> 00:40:34.660
.is approximately 1.80 till I am doing, approximately
I will do more better I have
00:40:34.660 --> 00:40:48.289
interpolation at the last step. The next across
the normal shock that number is 0.616 after
00:40:48.289 --> 00:40:59.880
that I will do 4 f L by D 0.431 which will
give me my L 2 star as 2.694 which now, has
00:40:59.880 --> 00:41:08.640
crossed 3.094 meters.
So, now, I find that if put my shock at 0.4
00:41:08.640 --> 00:41:12.140
meters from inlet I am going to have mark
1
00:41:12.140 --> 00:41:16.930
slightly downstream of this exit which means
now, my answer is between the last 2
00:41:16.930 --> 00:41:24.089
iterations. So, now, I will do more careful
iteration. So, now onwards I will not do this
00:41:24.089 --> 00:41:29.260
approximates shock I will try to end the correct
interpolation in the everything, I will
00:41:29.260 --> 00:41:39.049
pick of course, I want to pick the midpoint
0.45 I will pick the midpoint, that gives
00:41:39.049 --> 00:41:49.200
me
1.456 that gives me the next column 0.233.
00:41:49.200 --> 00:41:57.190
Now, from here I want to go find mark number
ead of shock that value now I will do
00:41:57.190 --> 00:42:06.720
more carefully 1.744 even more approximate
methods doing it little more carefully 774
00:42:06.720 --> 00:42:13.839
sorry 774 of course, the way I check my answer
was because, 744 was not in between
00:42:13.839 --> 00:42:20.809
these 2 numbers when you are iterating just
go through this simple checks 774. Now, I
00:42:20.809 --> 00:42:29.791
will go look at normal shock tables and that
gives me 0.623 from here I go back to the
00:42:29.791 --> 00:42:42.680
fan of flow tables 0.407 and that is giving
me 2.544 meters.
00:42:42.680 --> 00:42:57.600
So, I am getting 2.994 of course, I can go
one more iteration and try 0.46 should I go
00:42:57.600 --> 00:43:05.430
down or up 44 or some again try a 0.44 and
I may get a answer even closer to 3 when I
00:43:05.430 --> 00:43:13.099
want to stop at this 2.994 I will stop, this
where it goes at this point, this is the iterative
00:43:13.099 --> 00:43:19.010
procedure where you have to follow to get
to problem solution of this type. Once I
00:43:19.010 --> 00:43:23.359
solved up to this point this is the starting
point if I want to know more properties.
00:43:23.359 --> 00:43:30.079
Now, I know that my shock sitting at 0.45
meters up to that it is supersonic after that
00:43:30.079 --> 00:43:32.559
it is
also subsonic, that is what I figured out
00:43:32.559 --> 00:43:35.119
after that if I want to know the final x it
exit
00:43:35.119 --> 00:43:39.190
pressure or my P naught at the exit all those
things I have to go through the calculation
00:43:39.190 --> 00:43:42.460
all the way P naught 2 by P naught 2 star
at etcetera has to go through and find out
00:43:42.460 --> 00:43:45.130
all
these details. And remember that there is
00:43:45.130 --> 00:43:46.990
P naught change across the shock also you
have
00:43:46.990 --> 00:43:52.651
to take that also into occur.
Of course, I am not doing that right now,
00:43:52.651 --> 00:43:59.299
I will give you another problem with pressure
change. So, I think this will be your last
00:43:59.299 --> 00:44:03.359
numerical example I would not give you any
more of numerical examples.
00:44:03.359 --> 00:44:04.359
..
00:44:04.359 --> 00:44:12.820
So, now, this becomes here 2 c I am starting
with M 1 equal to 2.0 same f same D h.
00:44:12.820 --> 00:44:20.720
Now, I am telling I have that exact situation
like in problem b I am having length equal
00:44:20.720 --> 00:44:32.381
to 3 meters. And we started at M 1 equal to
2.0. So, we know my shock location is L s
00:44:32.381 --> 00:44:39.890
equal 0.45 meters, may be I am right may be
it has to be 442 something I will just
00:44:39.890 --> 00:44:44.500
assume it 45 right. Now, I did not go through
one more iteration to make it any more
00:44:44.500 --> 00:44:55.559
accurate. So, we will keep it as 0.45 meters.
So, if the question is this if P 1 is equal
00:44:55.559 --> 00:45:04.220
to 25 bar find P exit, that is my P exit find
my P
00:45:04.220 --> 00:45:12.749
exit that is the question I want to ask. It
is easier to work with P 0. So, I will switch
00:45:12.749 --> 00:45:15.779
to P
0 from here itself I know P 1 and I know M
00:45:15.779 --> 00:45:20.859
1 I can find P 0 where should I go I said
graphic tables gamma equal to 1.4 I said graphic
00:45:20.859 --> 00:45:25.200
tables there is, so many tables in
compressible flow tables books. So, you have
00:45:25.200 --> 00:45:28.089
to choose the correct table and correct
gamma value.
00:45:28.089 --> 00:45:37.120
So, now I will go find P 0 1 that will be
25 divided by P by P naught that is for M
00:45:37.120 --> 00:45:48.490
equal
to 2 that is 0.128 that gives me 195 bar of
00:45:48.490 --> 00:45:50.999
course, I could work with just P by P star
all
00:45:50.999 --> 00:45:57.641
through I just wanted to work with P 0 by
P 0 star just for one of it. So, now, I want
00:45:57.641 --> 00:46:04.460
to
find we know this is a P 0 1 we will keep
00:46:04.460 --> 00:46:13.400
this. Now, I will look at shock we know the
shock mark number right, from the previously
00:46:13.400 --> 00:46:19.740
problem we know that shock mark
number is 1.774.
00:46:19.740 --> 00:46:34.660
.So, I will go and find that value P 0 to
prime by P 0 2 we labelled this as 2 this
00:46:34.660 --> 00:46:39.069
as 2
prime then I have exit, these are the values
00:46:39.069 --> 00:46:47.049
I have P 0 2 by P 0 2 prime I really tried
that
00:46:47.049 --> 00:47:01.640
value 0.824 I have this number. So, I want
to find what is be the P 0 star at this point,
00:47:01.640 --> 00:47:06.220
by
the way since it is all subsonic flow here
00:47:06.220 --> 00:47:13.099
P 0 star here is same as P 0 star here, p
0 star
00:47:13.099 --> 00:47:18.249
does not change P 0 star will be the same
value and we know that at exit M equal to
00:47:18.249 --> 00:47:21.440
1.
So, P 0 here is equal to P 0 star.
00:47:21.440 --> 00:47:28.259
So, all I need to plan P 0 only up to here
that is enough of course, if I started at
00:47:28.259 --> 00:47:30.690
working
with P star then I have to track only P star
00:47:30.690 --> 00:47:33.049
up to here and that will be the pressure at
the
00:47:33.049 --> 00:47:51.300
exit I wanted it to take a longer path. So,
now actually I want to find P 0 2 by P 0 1
00:47:51.300 --> 00:47:56.109
is
equal to P 0 2 by P 0 star into P 0 star by
00:47:56.109 --> 00:48:07.489
P 0 1 I told you gas most people like ratios
multiplication multiplying it. So, this is
00:48:07.489 --> 00:48:10.769
equivalent to I am in my tables given as P
0 by P
00:48:10.769 --> 00:48:19.029
0 star. So, I will go find at M 2 what is
this value divided that with M 1 what is that
00:48:19.029 --> 00:48:24.150
value.
And that comes out to be 1.412 divided by
00:48:24.150 --> 00:48:40.849
1.687 that value is 0.837. So, now look at
these numbers P 0 2 prime P 0 2 I have, I
00:48:40.849 --> 00:48:46.039
have P 0 2 by P 0 1 only thing left is P 0
2
00:48:46.039 --> 00:48:53.580
prime divided by P 0 2 star P 0 prime star
downstream star value. So, I have to just
00:48:53.580 --> 00:49:03.780
find
that value P 0 2 prime divided by P 0 star
00:49:03.780 --> 00:49:10.660
prime, that value happens to be 1.163. Now,
if
00:49:10.660 --> 00:49:19.819
I multiply all these appropriately I can get
P 0 star prime divided by P 0 1 that what
00:49:19.819 --> 00:49:25.230
you
need to do I will just do that right here
00:49:25.230 --> 00:49:32.749
P 0 star prime by P 0 1.
So, of course, I had take reciprocal of this
00:49:32.749 --> 00:49:44.210
1 by 1.163. So, now, I have P 0 prime at the
denominator I go multiply with this I get
00:49:44.210 --> 00:49:47.640
P 0 2 at the denominator and it comes out
to be
00:49:47.640 --> 00:49:56.819
0.824 and I will take this number then I will
have P 0 1 at the denominator which is what
00:49:56.819 --> 00:50:10.319
I want. So, I multiply with this 0.837 I actually,
then find number for this I just use this
00:50:10.319 --> 00:50:15.289
I
know P 0 1 is given to be 195 bar we calculated.
00:50:15.289 --> 00:50:28.970
So, I am going to multiply this ratio
with P 0 1 value to get P 0 star prime happens
00:50:28.970 --> 00:50:38.960
to be 115.6 bar.
Now from here of course, I can go to isotropic
00:50:38.960 --> 00:50:46.950
tables and find M equal to 1 what will be
the P value, that comes out to be P at exit
00:50:46.950 --> 00:50:53.670
happens to be 0.528 of that you should know
gamma equal to 1.4 is 0.58 is the ratio that
00:50:53.670 --> 00:51:02.999
comes out to be 61 bar this what I end of
it.
00:51:02.999 --> 00:51:07.849
Now of course, it look like I went through
as little bit longer path because, it is P
00:51:07.849 --> 00:51:11.069
naught
I wanted to solve a problem in class with
00:51:11.069 --> 00:51:13.069
P naught variations, so I went through it
we
00:51:13.069 --> 00:51:17.339
.could have solved it with P star then I would
have avoided going to my isotropic tables
00:51:17.339 --> 00:51:24.049
completely that is the only difference.
Now, I started with 25 bar and I am ending
00:51:24.049 --> 00:51:28.329
with 61 bar is that logical, flow is going
from
00:51:28.329 --> 00:51:34.940
low pressure to high pressure it looks like
is that right. It happens to be correct what
00:51:34.940 --> 00:51:40.049
will
be the reason I give my momentum is decreasing
00:51:40.049 --> 00:51:46.369
I am going from M equal to 2 to M
equal 1 finally, that is why it is matching
00:51:46.369 --> 00:51:50.960
otherwise, the flow will never go from low
pressure to high pressure think about it,
00:51:50.960 --> 00:51:54.589
but I had other energy not just pressure energy
I
00:51:54.589 --> 00:51:59.150
had also other energy here momentum or kinetic
energy I had excess of it.
00:51:59.150 --> 00:52:04.490
And that is now, decreasing and that is becoming
pressure energy during the process that
00:52:04.490 --> 00:52:11.109
is what happened here, I will probably show
some animation tomorrow if not we will
00:52:11.109 --> 00:52:18.650
just start heat transparent wave duct and
what happens to it has turn x during duct,
00:52:18.650 --> 00:52:21.579
inside
the duct that we will start with tomorrow
00:52:21.579 --> 00:52:22.890
see people in next class.
00:52:22.890 --> 00:52:23.890
.