WEBVTT
Kind: captions
Language: en
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Hello everyone, welcome back to our class,
we stopped at just writing the outline of
00:00:20.279 --> 00:00:25.990
conservation of Momentum, the Equation for
conservation of momentum last time. So,
00:00:25.990 --> 00:00:30.670
we will start from there hopefully we will
derive conservation of momentum and
00:00:30.670 --> 00:00:40.260
conservation of energy before end of today’s
class and probably go a little more stay on
00:00:40.260 --> 00:00:42.149
that.
.
00:00:42.149 --> 00:01:00.130
So, when we said conservation of momentum,
we said time rate of momentum, rate of
00:01:00.130 --> 00:01:25.560
increase
in volume V is equal to minus of net out flux
00:01:25.560 --> 00:01:42.320
across the surface S
00:01:42.320 --> 00:01:47.759
plus the net
force acting on volume V. If I say that there
00:01:47.759 --> 00:01:54.549
is no out flux, then I will say that the net
force is going to be directly increasing the
00:01:54.549 --> 00:01:58.149
net momentum at some particular rate that
is
00:01:58.149 --> 00:02:03.289
your Newton’s second law, which is directly,
gets you that. This is the extra term, which
00:02:03.289 --> 00:02:11.780
is the last term from the volume, we keep
it this way from in a simple expression, we
00:02:11.780 --> 00:02:48.700
will write for rate of change plus all the
remaining forces.
00:02:48.700 --> 00:03:15.450
.Now, we could have so many forces possible,
so we will just write it as the stress tensor
00:03:15.450 --> 00:03:22.840
is the full matrix of a 3 by 3 form of it,
and that is going to include along the diagonal,
00:03:22.840 --> 00:03:25.510
it
will be the normal force components and the
00:03:25.510 --> 00:03:31.560
half diagonal terms will be your sheer stress
terms. Right you would know tau x y the most
00:03:31.560 --> 00:03:41.280
common one used in Newton’s law of this
like that. So, this is the full equation,
00:03:41.280 --> 00:03:44.291
I am going to say that this f b is here body
force
00:03:44.291 --> 00:03:50.680
term, that is if there is any say magnetic
field across and that is going to cause some
00:03:50.680 --> 00:03:55.260
bulk
force or gravitational force. This is all
00:03:55.260 --> 00:04:05.440
force per unit volume the way, I have written
it,
00:04:05.440 --> 00:04:12.530
no it is actually absolute force, it is no
per unit volume, it is just absolute force
00:04:12.530 --> 00:04:14.470
no per
volume as if no.
00:04:14.470 --> 00:04:19.510
Now, we will say we will start simplifying
this and I will say this is going to be equal
00:04:19.510 --> 00:04:22.540
to
0 in most of my problems in gas dynamics,
00:04:22.540 --> 00:04:28.330
we will mostly say there is no gravitational
field or any other body forces electric or
00:04:28.330 --> 00:04:31.000
magnetic field. We do not have any of that
and
00:04:31.000 --> 00:04:46.270
this, we will simplify a little further. Now,
I will write this whole term by splitting
00:04:46.270 --> 00:04:51.834
this
into diagonal term and the half diagonal term,
00:04:51.834 --> 00:05:27.039
I will keep the that only the diagonal term
part.
00:05:27.039 --> 00:05:32.550
I am writing it as tau sheer it is means I
am having all the diagonal components of this
00:05:32.550 --> 00:05:35.800
0
and then I am taking, I have to take a dot
00:05:35.800 --> 00:05:43.479
product with this, dot product with the
perpendicular vector from there surface vector.
00:05:43.479 --> 00:05:46.419
And this is the force is acting along the
n
00:05:46.419 --> 00:05:50.990
direction normal force, if there is a surface
and the pressure is acting from outside on
00:05:50.990 --> 00:05:54.680
to
it, the normal is going to be out of the box
00:05:54.680 --> 00:05:58.430
and pressure is going to be inward.
So, that is giving you this minus sign in
00:05:58.430 --> 00:06:03.400
there and then in sheer force, it will
automatically take care of this sign based
00:06:03.400 --> 00:06:07.009
on this dot, It will take care of it right
side.
00:06:07.009 --> 00:06:13.159
Now, we substitute these 2 inside here and
we are going to say in our simple gas
00:06:13.159 --> 00:06:20.530
dynamics world, we will neglect this. We will
say there is no sheer force as in we are
00:06:20.530 --> 00:06:26.500
going to consider inviscid flow problem as
of now. Of course, we will come back and
00:06:26.500 --> 00:06:31.830
say friction can be taken into account at
a later stage as of now, we will say there
00:06:31.830 --> 00:06:35.319
is no
friction there is no sheer forces inside my
00:06:35.319 --> 00:06:42.689
flow. So, we will neglect it for now, only
normal force component will be there. So,
00:06:42.689 --> 00:06:52.210
now we just have to substitute all those inside.
00:06:52.210 --> 00:06:53.210
..
00:06:53.210 --> 00:07:01.759
Of course, I can just take this, I want to
convert all of them to volume term, volume
00:07:01.759 --> 00:07:24.919
integral term. So, this is going to be del
P not, del dot P that is just gradient of
00:07:24.919 --> 00:07:27.789
scalar
pressure is a scalar field. So, the gradient
00:07:27.789 --> 00:07:31.699
of that is a vector and that is your force
direction, we just got to this form that is
00:07:31.699 --> 00:07:36.139
just 1 term is still have other terms. We
want to
00:07:36.139 --> 00:07:57.029
look at the next term, the advection, term
this term when, we want to convert to the
00:07:57.029 --> 00:08:01.699
volume integrals, it is not very easy. So,
we will write a vector identity below this
00:08:01.699 --> 00:08:31.159
and go
by parallel that is easier for us.
00:08:31.159 --> 00:09:03.270
So, this is a vector identity, we just take
it from mathematicians, we will just use it,
00:09:03.270 --> 00:09:06.790
if
you look at it really. It is just differentiation
00:09:06.790 --> 00:09:09.070
by parts that is all we are doing inside this
is
00:09:09.070 --> 00:09:13.980
nothing rate differentiation by chain rule,
I guess that is what it is call, we just doing
00:09:13.980 --> 00:09:17.280
that
inside here, it just look very complex. Of
00:09:17.280 --> 00:09:20.640
course, we still do not know what to do with
this term, we will go back to that and write
00:09:20.640 --> 00:09:26.560
another identity for that. If we using tensor
notation as in the Einstein notation, index
00:09:26.560 --> 00:09:32.080
notation then this life will become a little
easier, I will just do not want to teach you
00:09:32.080 --> 00:09:34.750
this in this course.
So, we will just live with vector notation
00:09:34.750 --> 00:09:37.970
and just get over with it, we will just use
it only
00:09:37.970 --> 00:09:43.630
once in this whole course and that is this
point, we want to worry about it too much.
00:09:43.630 --> 00:09:45.850
So,
now, I have to just see the parallel between
00:09:45.850 --> 00:09:53.500
this and this my B vector is a velocity vector,
A vector is another velocity vector. This
00:09:53.500 --> 00:09:56.440
phi is a scalar and that is my density in
this
00:09:56.440 --> 00:10:01.930
.case, now you just have to go and find the
parallel and write the equation that will
00:10:01.930 --> 00:10:40.680
become. So, I use this to get to this form.
So, now, we can simplify it a little bit just
00:10:40.680 --> 00:10:47.750
per unit to change this to flower bracket,
now
00:10:47.750 --> 00:10:59.750
it is consistent, now this looks simple enough
to work with, I still want to think about
00:10:59.750 --> 00:11:02.900
this
term, it is not very easy to work with. So,
00:11:02.900 --> 00:11:05.530
we will go and simplify that a little more
again,
00:11:05.530 --> 00:11:10.650
we will go and ask mathematicians, they will
tell me another identity for it. So, we will
00:11:10.650 --> 00:11:13.000
go to this point and start writing from the
beginning.
00:11:13.000 --> 00:11:14.000
.
00:11:14.000 --> 00:11:55.570
I will pick the identity write now, this is
an identity of just the operator operating
00:11:55.570 --> 00:11:58.750
on a
scalar multiplied by a vector, that is what
00:11:58.750 --> 00:12:04.920
this is it is A dot del as an operator
differentiation operator acting on the product
00:12:04.920 --> 00:12:11.720
of a scalar and a vector that comes out to
be this. So, now, we will just have to look
00:12:11.720 --> 00:12:20.420
at it in terms of our rho’s and u’s, we
had an
00:12:20.420 --> 00:12:50.190
expression that was this, now this can be
written as this form remember that here. This
00:12:50.190 --> 00:12:53.610
is
not integral over surface to integral surface
00:12:53.610 --> 00:12:55.270
volume, that is not what we did here, this
is
00:12:55.270 --> 00:12:58.840
just expressing this in terms of this whole
expression.
00:12:58.840 --> 00:12:59.840
..
00:12:59.840 --> 00:13:07.260
If you go back and look at where, it was sitting
here inside this volume integral, that is
00:13:07.260 --> 00:13:12.500
still here, inside the volume integral. So,
now, I have to take that expression substitute
00:13:12.500 --> 00:13:18.520
inside here along with this expression that
will be 3 times inside this, all the 3 terms
00:13:18.520 --> 00:13:22.010
are
going to be representing this 1 surface integral,
00:13:22.010 --> 00:13:23.070
now I will go back 1 more level.
.
00:13:23.070 --> 00:13:29.200
And I will say in my original equation I have
this surface integral is going to be replaced
00:13:29.200 --> 00:13:34.470
by those 3 terms, which we just talked about
plus this term, which became these two
00:13:34.470 --> 00:13:41.769
terms of which, this term we wrote as one
simple expression in here.
00:13:41.769 --> 00:13:42.769
..
00:13:42.769 --> 00:13:47.500
So, there is whole set of equations, we will
put all of them together and write it once,
00:13:47.500 --> 00:14:30.190
the
full equation in volume integral form.
00:14:30.190 --> 00:14:31.190
.
00:14:31.190 --> 00:14:38.620
I will fold and write here, I want to give
some gap a little bit here, it helps after
00:14:38.620 --> 00:15:02.170
some
time. This whole thing is 1 equation of which,
00:15:02.170 --> 00:15:09.260
we know that there is your del P term, that
coming from tau tensor dot n d s integral
00:15:09.260 --> 00:15:14.440
converted to volume integral. These are the
three terms, which we just talked about of
00:15:14.440 --> 00:15:21.370
which 2 of them are just sitting here directly
above. This is the first un study term, now
00:15:21.370 --> 00:15:24.560
the time derivative is operating on individual
00:15:24.560 --> 00:15:29.460
.terms instead of what, we had before as a
group dou by dou t of rho u together we have,
00:15:29.460 --> 00:15:35.311
it is operating on individual, that is the
whole thing we have.
00:15:35.311 --> 00:15:42.980
Now, we want to group it in a nice fashion,
we want to say, I will group these 3 times,
00:15:42.980 --> 00:15:52.060
that is why I wanted that gap here. I want
to group these terms all of them have a u
00:15:52.060 --> 00:15:57.660
vector multiplied by something, now we will
group them that way just take that bracket
00:15:57.660 --> 00:16:20.680
alone, it will look like u vector multiplied
by it will look like this. Now this is just
00:16:20.680 --> 00:16:23.930
your
mass equation, which we just derived last
00:16:23.930 --> 00:16:31.300
class, this is u vector multiplied by mass
equation and if I say my flow is obeying mass
00:16:31.300 --> 00:16:38.990
equation, this will automatically be 0. So,
my expression can be simplified to just this
00:16:38.990 --> 00:16:47.540
term plus 0 plus these 2 terms, that is what
we have going to use from now on. And now,
00:16:47.540 --> 00:16:53.430
we will say again what are logics, we used
last time we said the integral should be valid
00:16:53.430 --> 00:17:01.130
for any small volume d V. So, ideally my
flower bracket inside the integral that must
00:17:01.130 --> 00:17:06.569
be equal to 0. So, we get to the differential
form of the expression remember the integral
00:17:06.569 --> 00:17:11.169
form which, we started with and the
differential form, we will use all of them.
00:17:11.169 --> 00:17:13.060
So, what is the integral form, we will go
back
00:17:13.060 --> 00:17:15.069
here.
.
00:17:15.069 --> 00:17:22.329
This original thing is my integral form is
this whole expression of which you should
00:17:22.329 --> 00:17:30.001
remember this form along with the change to
derivative form, which will be in here, this
00:17:30.001 --> 00:17:31.001
whole expressions.
00:17:31.001 --> 00:17:32.001
..
00:17:32.001 --> 00:17:37.190
Up to here, you want to keep both these expressions,
we will use 1 or the other
00:17:37.190 --> 00:17:45.090
depending on the situation, whichever is convenient.
So, I will get a final expression, I
00:17:45.090 --> 00:18:24.580
will write it in a next board.
.
00:18:24.580 --> 00:18:32.179
This happens to be your differential form
of full momentum equation, of course, it is
00:18:32.179 --> 00:18:34.850
3
equations together I have to put this vector
00:18:34.850 --> 00:18:41.190
symbol here on top, it is actually, 3 equations
together. If I look at it there is x component
00:18:41.190 --> 00:18:42.880
1 full equation is there, y component that
is a
00:18:42.880 --> 00:18:49.020
.full equation, z component is a full equation,
this is grad P is going to have 3 components
00:18:49.020 --> 00:18:51.570
in there.
So, each of them will go to one particular
00:18:51.570 --> 00:18:55.981
equation specifically, If you want to think
about it dou P by dou X is going to be in
00:18:55.981 --> 00:19:01.909
X momentum equation, dou P by dou Z will be
in Z momentum equation or the W momentum equation,
00:19:01.909 --> 00:19:10.500
which I am using. Now is this
expression for compressible or incompressible
00:19:10.500 --> 00:19:15.899
flow.
. Incompressible.
00:19:15.899 --> 00:19:21.450
Why is it incompressible flow.
. Compressible.
00:19:21.450 --> 00:19:27.850
He already changes the answers. So, he say
it is compressible flow, why is it valid for
00:19:27.850 --> 00:19:30.759
compressible flow.
. .
00:19:30.759 --> 00:19:37.921
Del dot u equal to 0, we used somewhere, where
did, we use it.
00:19:37.921 --> 00:19:38.921
.
00:19:38.921 --> 00:19:44.649
I want to go say it is here, this is a mass
equation, we said this whole term is equal
00:19:44.649 --> 00:19:49.399
to 0,
in mass equation, that is we saying right.
00:19:49.399 --> 00:19:54.730
So, if we look at the original expression
for
00:19:54.730 --> 00:20:00.519
momentum equation, if I do not use this simplification,
the actual momentum equation
00:20:00.519 --> 00:20:05.090
.happens to be this, whatever is inside this
flower bracket equal to 0. That is your original
00:20:05.090 --> 00:20:12.529
momentum equation, if I look at it that way,
I am having this dou rho by dou t term. So,
00:20:12.529 --> 00:20:17.059
in theory I have to say the density variation
is also taken into talk on to in this excepts
00:20:17.059 --> 00:20:20.399
for
I am simplifying this expression using mass
00:20:20.399 --> 00:20:26.289
equation. Because, mass equation is also
satisfied in my flow problem, mass equation
00:20:26.289 --> 00:20:28.549
and momentum equations are satisfied. So,
I
00:20:28.549 --> 00:20:33.610
will simplify this, so even though I write
my expression in this form.
00:20:33.610 --> 00:20:34.610
.
00:20:34.610 --> 00:20:40.299
It is valid for compressible flow that is
an important point to have to remember, it
00:20:40.299 --> 00:20:43.269
is
valid for compressible flows. Even though
00:20:43.269 --> 00:20:50.200
they does not seem to be a direct time
variation term for density, that is the typical
00:20:50.200 --> 00:20:54.380
thing people use in interviews they will tell
you to write this expression. And then they
00:20:54.380 --> 00:21:00.360
will ask whether it is for compressible flow
or incompressible flow standard drop just
00:21:00.360 --> 00:21:02.740
tell you that, do not want to fall in that
trap
00:21:02.740 --> 00:21:11.049
anyway. So, now, we will move on to consideration
of energy.
00:21:11.049 --> 00:21:12.049
..
00:21:12.049 --> 00:21:17.279
This same thing can be used here rate of increase
in volume V, I am going to say how
00:21:17.279 --> 00:21:23.039
much is being collected inside my packet,
we talked about this accumulation term, that
00:21:23.039 --> 00:21:39.710
is
going to be there plus the net out flux. I
00:21:39.710 --> 00:21:43.429
am writing little less in here ideally, I
have to
00:21:43.429 --> 00:21:49.380
write net out flux is net rate of out flux
of energy across surface S all that I have
00:21:49.380 --> 00:21:52.260
to write
here really, I am just writing here, net out
00:21:52.260 --> 00:22:03.519
flux across s. That is how much am I keeping
inside how much, how I am sending outward
00:22:03.519 --> 00:22:09.779
out of my control surface, those have 2
things.
00:22:09.779 --> 00:22:14.779
And then other forms of work done by me on
the surroundings, that is a next thing
00:22:14.779 --> 00:22:23.830
remaining thing. I will just call it W out
for simplicity ideally, I have to write net
00:22:23.830 --> 00:22:27.240
rate of
work done on the surroundings, that is what
00:22:27.240 --> 00:22:32.600
I have to here ideally, all these let us say
are
00:22:32.600 --> 00:22:42.240
going out from the system. Now what is the
input, the heat transfer in that is only things
00:22:42.240 --> 00:22:49.659
or some other source term that is producing
energy inside, all these are just rate of
00:22:49.659 --> 00:22:55.279
increase of energy inside the volume, but
what increased it. Everything seems to be
00:22:55.279 --> 00:22:57.889
using
of energy is for something, what increase
00:22:57.889 --> 00:23:00.890
the energy that is your net in flux, I will
just
00:23:00.890 --> 00:23:09.990
call this your actually. It is not just Q
in this is not just W out ideally, I have
00:23:09.990 --> 00:23:18.509
to write rate
of you are thinking about time rate of change
00:23:18.509 --> 00:23:24.789
of quantities, in this case it is energy
equation. So, it is energy quantities energy.
00:23:24.789 --> 00:23:36.909
Now, let us look at Q in term, what is a net
rate of Q in, what are the various ways by
00:23:36.909 --> 00:23:49.899
which, I can give energy into my control volume,
Q is heat transfer. So, what are all the
00:23:49.899 --> 00:23:55.639
.various things that is pass, some kind of
what is a process, we are looking for the
00:23:55.639 --> 00:23:58.600
process
not heat exchanger say as that is a mechanism
00:23:58.600 --> 00:24:10.090
by which give heat to gas yes convexion
not really convexion. It is not convexion,
00:24:10.090 --> 00:24:15.159
we are looking for conduction basically, heat
is
00:24:15.159 --> 00:24:20.749
being conducted, because the walls are hot
and the gas is cold something like that.
00:24:20.749 --> 00:24:32.489
So, there is heat transport due to temperature
gradient, what log law of conduction
00:24:32.489 --> 00:24:39.710
named after who, Fourier’s law of heat conduction,
it should have been sometime in high
00:24:39.710 --> 00:24:47.490
school you are on be on this for first year
at least in engineering. So, now I want to
00:24:47.490 --> 00:24:51.200
look
at how much is the net heat into my volume.
00:24:51.200 --> 00:24:57.999
So, I have to look at my control volume and
say every surface, this is my surface of my
00:24:57.999 --> 00:25:04.279
control volume as and I am going to say
every surface there is some amount of heat
00:25:04.279 --> 00:25:09.559
coming in.
Which means, temperature here is higher than
00:25:09.559 --> 00:25:15.230
temperature here, what is a direction of
my temperature gradient, gradient has a direction
00:25:15.230 --> 00:25:29.799
by the way, what will be the direction
of my temperature gradient, del t has a direction.
00:25:29.799 --> 00:25:40.460
It is in the direction of nobody here
thought about del let say del P, we look at
00:25:40.460 --> 00:25:54.070
that just now momentum equation, what is the
direction of del P. Normal to iso surface.
00:25:54.070 --> 00:25:57.489
So, let us use that terminology here, I do
not
00:25:57.489 --> 00:25:58.489
know whether.
.
00:25:58.489 --> 00:26:04.309
It will work very well, let us say I have
my space x y coordinate system and I am looking
00:26:04.309 --> 00:26:15.230
at some random gas volume. And I have some
pressure distribution let us say wherever
00:26:15.230 --> 00:26:23.619
I
put lot of doubts pressure high there something.
00:26:23.619 --> 00:26:27.179
Pressure is high here, low here, if I take
00:26:27.179 --> 00:26:35.489
.a gradient over this volume here, let say
if I take grad P del P. This happens to be
00:26:35.489 --> 00:26:38.860
a
vector, it will have one direction finally,
00:26:38.860 --> 00:26:44.190
because it is a vector, it has to have a particular
direction, what will that direction be it
00:26:44.190 --> 00:26:53.889
will be towards the highest change.
Gradient vector gives you the direction of
00:26:53.889 --> 00:26:58.210
highest change at that point, which your
direction pressure changing the highest, it
00:26:58.210 --> 00:27:00.879
will be putting that direction for you, which
is
00:27:00.879 --> 00:27:07.669
in a way equivalent, what he was saying. It
is perpendicular to the iso surface,
00:27:07.669 --> 00:27:13.190
perpendicular to that, there will be a line,
where I can tell pressure is almost the same.
00:27:13.190 --> 00:27:17.629
I
will get to that, that is your iso surface,
00:27:17.629 --> 00:27:20.320
but I can have a crazy situation in my flow
field
00:27:20.320 --> 00:27:34.630
where, pressure is very high, in this annulus
then pressure gradient will be this way here.
00:27:34.630 --> 00:27:41.730
This way here, this way here, this way here,
like that, but still iso surface is still
00:27:41.730 --> 00:27:46.950
present.
Depending on the resolution, we will see different
00:27:46.950 --> 00:27:52.230
things, am I interested in such small
differences in distances, if that is the case
00:27:52.230 --> 00:27:56.950
then we will such effects. If I say I am
interested in my minimum delta x happens to
00:27:56.950 --> 00:28:01.830
be that is big block, there is no pressure
variation for me. I will just take 1 average
00:28:01.830 --> 00:28:07.259
pressure for this, of course engineering
problems, we always, we will say I am interested
00:28:07.259 --> 00:28:13.450
in lengths of the order of something let
say 1 centimeter, lengths of the order microns
00:28:13.450 --> 00:28:19.460
are not useful for me.
So, like that we will set our own limits,
00:28:19.460 --> 00:28:23.149
depending on the problem it may change, you
would is below over an air craft then I am
00:28:23.149 --> 00:28:31.070
not bothered about lengths of less than 1
centimeter probably. If it is flower my hand
00:28:31.070 --> 00:28:36.039
then I may be interested in lengths of the
order of 1 millimeter, but anything lesser,
00:28:36.039 --> 00:28:38.950
if it is flow over this chalk probably, I
am
00:28:38.950 --> 00:28:44.580
interested in less than 0.1 millimeter also.
Because, the overall size itself is some are
00:28:44.580 --> 00:28:48.700
small the overall size itself is roughly 10
m m or something.
00:28:48.700 --> 00:28:53.070
So, we may be interested in something much
more finer that can also happen depending
00:28:53.070 --> 00:28:57.470
on the problem, you may want to resolve more
that is for people, who want to go
00:28:57.470 --> 00:29:04.370
computational mode other, it does not matter
so much. But, in any case grad P, you
00:29:04.370 --> 00:29:08.529
should know is going to be perpendicular to
the iso surface perfect mathematical
00:29:08.529 --> 00:29:11.299
expression.
But the actual way, we want to we will think
00:29:11.299 --> 00:29:18.080
about it, it will point towards the maximum
change direction, it will go towards change
00:29:18.080 --> 00:29:22.789
in the for upper direction. We will go from
low pressure to high pressure direction that
00:29:22.789 --> 00:29:26.450
will be the direction that is the way it will
be.
00:29:26.450 --> 00:29:34.539
.Now, same thing for temperature, we wanted
temperature gradient, so that, we will have
00:29:34.539 --> 00:29:38.009
heat conduction, that is what we were looking
for.
00:29:38.009 --> 00:29:39.009
.
00:29:39.009 --> 00:29:49.169
Now, if I say my normal vector is like this
and my q vector, the heat flux vector is like
00:29:49.169 --> 00:30:01.059
this then if I put q dot n that will have
a negative sign. Because, n vector is like
00:30:01.059 --> 00:30:05.669
this q is
vector is like this and you will get q dot
00:30:05.669 --> 00:30:08.100
n will be a negative sign, we want to cancel
that
00:30:08.100 --> 00:30:13.971
because we want to say the q is positive,
this q corresponds to positive. So, I have
00:30:13.971 --> 00:30:20.241
to say
my q in, I will put dot saying it is rate
00:30:20.241 --> 00:30:23.350
of I do not want to go and write every time
rate of
00:30:23.350 --> 00:30:28.460
q in, I will just put a dot on top and say
that is time rate of change of q n.
00:30:28.460 --> 00:30:40.480
Now write this as minus integral over the
surface
00:30:40.480 --> 00:30:52.710
q dot n d s, now what is this q, that is
where your Fourier’s law coming conduction
00:30:52.710 --> 00:31:05.059
coefficient times del T, this is your
Fourier’s law of heat conduction. Now, I
00:31:05.059 --> 00:31:10.929
did this gradient of pressure now. Now gradient
of temperature, what is this say if I go back
00:31:10.929 --> 00:31:13.429
to this picture, I am saying heat is going
this
00:31:13.429 --> 00:31:20.129
way in which, means this is higher temperature,
this is lower temperature right. So, my
00:31:20.129 --> 00:31:27.880
grad T will look going outward, but we already
defined it to be going inward.
00:31:27.880 --> 00:31:35.489
So, there has to be a minus sign, that is
one way of looking at it, we just have to
00:31:35.489 --> 00:31:37.279
be
consists, why we put this minus why, we put
00:31:37.279 --> 00:31:41.679
that minus, they are 2 different minus signs.
They are going to cancel each other soon,
00:31:41.679 --> 00:31:47.160
if I substitute this q inside here, that will
cancel that minus sign, we will keep it this
00:31:47.160 --> 00:31:49.200
way. I will use this form sometimes, I will
use
00:31:49.200 --> 00:31:59.489
this only once, anyways finally will just
say q equal to 0 overall, that is also there
00:31:59.489 --> 00:32:00.489
by the
00:32:00.489 --> 00:32:05.669
.way the q in need not be always through conduction,
what are the other possibilities of
00:32:05.669 --> 00:32:14.039
giving heat to a volume of gas radiation.
Another possibility anything else, convexion
00:32:14.039 --> 00:32:21.639
will not be working, conduction is the
physical process, convexion is the gas molecule
00:32:21.639 --> 00:32:27.249
let say, I have a hot plate the gas
molecule here will come take the heat by conduction.
00:32:27.249 --> 00:32:31.869
And then it is moves away, some
other gas comes here, takes the heat and it
00:32:31.869 --> 00:32:38.210
moves away that, whole process is what
people call convexion engineering wise, it
00:32:38.210 --> 00:32:43.940
is useful.
They will just define 1 coefficient of heat
00:32:43.940 --> 00:32:49.539
convexion and problem can be solved faster.
So, they have a separate thing call convexion,
00:32:49.539 --> 00:32:52.970
really it is just conduction with flow that
is
00:32:52.970 --> 00:32:58.809
what is convexion, it is not anything special.
So, you would not go into that I am just
00:32:58.809 --> 00:33:02.429
going to say there is no separate convexion
term, that is why I want to introduce this,
00:33:02.429 --> 00:33:06.499
it is
there is only conduction term, there could
00:33:06.499 --> 00:33:12.899
be radiation term other than that. There could
be 1 more thing, chemical reaction can be
00:33:12.899 --> 00:33:18.389
exothermic producing heat or endothermic.
It can be q in negative, it may be taking
00:33:18.389 --> 00:33:21.659
away heat, it could be q out from the volume
that
00:33:21.659 --> 00:33:27.250
could also be there, I could have chemical
reactions. But, currently in our gas dynamics,
00:33:27.250 --> 00:33:33.950
we are simply in a flow of system, we will
say no chemical reactions, no radiation. But,
00:33:33.950 --> 00:33:42.409
they are just approximations, I could have
a case where it matters, this is just 1 term,
00:33:42.409 --> 00:33:46.929
we
have just written that last term in this 4
00:33:46.929 --> 00:33:52.690
term expression. Now, we will go and look
at the
00:33:52.690 --> 00:34:18.030
rate of energy accumulation, that is a easiest
term time rate of change of stuff.
00:34:18.030 --> 00:34:23.230
This is e is internal energy per unit mass,
we already had this convention and this is
00:34:23.230 --> 00:34:28.849
kinetic energy per unit mass. This is a total
energy of the gas per unit mass, this is the
00:34:28.849 --> 00:34:31.260
gas
is energy and this is the flow energy all
00:34:31.260 --> 00:34:34.220
together. This is all per mass multiplied
by
00:34:34.220 --> 00:34:39.440
density will make it the whole, energy per
volume into small volume will mean that
00:34:39.440 --> 00:34:44.679
small box, how much energy am I having integral
over the whole volume. That will be
00:34:44.679 --> 00:34:50.000
the net energy I will have inside my control
volume time rate of change of that, this is
00:34:50.000 --> 00:34:55.909
your the first term rate of increase in wall
inside the volume. Rate of increase of energy
00:34:55.909 --> 00:35:02.900
content inside the volume, that is this term
this is 1 term. Now, the next term is energy
00:35:02.900 --> 00:35:08.760
out flux that is this net out flux across
S, what out flux, energy out flux rate of
00:35:08.760 --> 00:35:16.780
energy
going out across the surface.
00:35:16.780 --> 00:35:17.780
..
00:35:17.780 --> 00:35:42.640
I have to say net out flux. Similar to all
the other terms, we did for out flux, it will
00:35:42.640 --> 00:35:46.900
be u
dot n times d s multiplied by that quantity,
00:35:46.900 --> 00:35:52.260
that quantity happens to be energy per unit
volume rho times e plus u square by 2. That
00:35:52.260 --> 00:35:59.430
is the whole quantity, we are talking about,
we get to this form. Now the only term left
00:35:59.430 --> 00:36:04.000
in this equation, we have an expression for
this that is here, we have an expression for
00:36:04.000 --> 00:36:08.930
this that is out here, then the next thing
is this
00:36:08.930 --> 00:36:13.590
right hand side, which is what is the first
thing we did, that is this expression.
00:36:13.590 --> 00:36:19.950
Now only thing left is this is rate of work
out. Now rate of work out what are all the
00:36:19.950 --> 00:36:25.380
various mechanism by which my fluid can do
work on something around, something
00:36:25.380 --> 00:36:35.430
surroundings, turbine work yes that is a possibility.
That is called the shaft work
00:36:35.430 --> 00:36:46.920
typically, yes that is a possibility anything
else, pressures doing work p d v related work.
00:36:46.920 --> 00:36:53.910
But, it is flow, we want to find rate of work
done, how do you find rate of work done, I
00:36:53.910 --> 00:37:05.450
want to what is has the units high school
physics. I am pushing a block of metal, what
00:37:05.450 --> 00:37:07.150
is
a net work done, what is the rate of work
00:37:07.150 --> 00:37:16.500
done?
Force into displacement gives you the work
00:37:16.500 --> 00:37:22.450
not the rate of work for center velocity,
were
00:37:22.450 --> 00:37:28.660
both, there both vectors. So, what should
I do that was cannot be generally multiplied
00:37:28.660 --> 00:37:32.710
should be dot product, you have to know which
1, because we are looking for energy,
00:37:32.710 --> 00:37:38.630
which is scalar not a vector. That is 1 simple
way of looking at and continue mechanics
00:37:38.630 --> 00:37:42.819
if you give this answer, they would not accept,
you have to give more logical answer for
00:37:42.819 --> 00:37:43.819
that.
00:37:43.819 --> 00:37:49.119
.But, in here we can give this answer that
is you will say, you are looking for a scalar
00:37:49.119 --> 00:37:51.390
how
to get a scalar from 2 vectors dot product
00:37:51.390 --> 00:38:01.330
easy answer that is enough for us. So, what
is
00:38:01.330 --> 00:38:13.710
my force really, my force happens to be this
stress times d s. Now I have to multiply this
00:38:13.710 --> 00:38:23.109
with velocity along a particular direction.
So, I have to dotted it with my velocity,
00:38:23.109 --> 00:38:28.349
this is
my rate of work done, this is overall term,
00:38:28.349 --> 00:38:35.980
now I have to be careful about this, why
because I want to put a minus sign in front.
00:38:35.980 --> 00:38:46.020
So, if I have stress say I am fluid element
and I am moving forward, but the next fluid
00:38:46.020 --> 00:38:50.910
element is not moving, what we will be feel
that is a velocity gradient. So, I have to
00:38:50.910 --> 00:38:53.790
full
that person along, the next fluid element,
00:38:53.790 --> 00:39:01.700
I have to full along it is means, I am applying
force this way. And I am moving this way,
00:39:01.700 --> 00:39:06.690
what is the actual force applied on me by
the
00:39:06.690 --> 00:39:11.890
surroundings, it should be the opposite direction
right.
00:39:11.890 --> 00:39:21.000
Because, the next fluid element is slowing
me down, I am doing work against that force
00:39:21.000 --> 00:39:25.619
this correct way to look at it, you have to
think about it and then come up with you do
00:39:25.619 --> 00:39:31.240
not explanation for it finally. I will tell
you this particular way of looking at it,
00:39:31.240 --> 00:39:34.390
but you
may have some other version of it, you can
00:39:34.390 --> 00:39:39.800
think about it and which, are way you feel
like I will just tell you sheer force. It
00:39:39.800 --> 00:39:41.550
need not be just sheer force, we will look
at normal
00:39:41.550 --> 00:39:48.030
force also after this at sheer force is what
is the example? I gave you right now, I could
00:39:48.030 --> 00:39:51.520
go for normal force also, we will go for normal
force after this.
00:39:51.520 --> 00:40:00.240
I am going to have this particular force,
which is minus this is my W dot out again,
00:40:00.240 --> 00:40:02.860
I am
going to say net rate of work done on the
00:40:02.860 --> 00:40:05.894
surroundings this is what, we are looking
at we
00:40:05.894 --> 00:40:13.279
will get to this point. So, now, again I will
write this stress tensor in diagonal and half
00:40:13.279 --> 00:40:31.579
diagonal terms separately then I can write
this expression as tau sheer dot u d s, I
00:40:31.579 --> 00:40:33.770
missed
here equal to sign here which, I do not think
00:40:33.770 --> 00:40:36.660
it is a serious trouble. I missed this equal
to
00:40:36.660 --> 00:40:55.470
sign form and then I have to use a correct
sign again here n dot u, here the negative
00:40:55.470 --> 00:40:58.360
sign
is automatically taken and talk on, you may
00:40:58.360 --> 00:41:05.530
should think about it. Again I would say now
I have this everywhere d s.
00:41:05.530 --> 00:41:11.730
So, already I have to do is the integral overall
the surface to get my full force not just
00:41:11.730 --> 00:41:14.770
full
force full rate of work done on the surroundings
00:41:14.770 --> 00:41:21.900
that is, what I need to do finally. So, I
have to just put integral over this over the
00:41:21.900 --> 00:41:28.339
whole surface that is what I need to do for
both
00:41:28.339 --> 00:41:36.069
the terms integral over the whole surface.
Now, you taken into account all the terms
00:41:36.069 --> 00:41:37.069
all
00:41:37.069 --> 00:41:45.180
.have to be just look at the whole expression,
the whole expression is not very difficult
00:41:45.180 --> 00:41:49.559
to
write to anywhere. I will go to the next section.
00:41:49.559 --> 00:41:57.559
We have in converted the surface integral
to volume integrals. And we will write it
00:41:57.559 --> 00:42:02.140
without the convention remember that form
and then we will convert it and then the
00:42:02.140 --> 00:42:09.349
whole volume integral, we will just say all
the integrand the inside the integral should
00:42:09.349 --> 00:42:11.460
be
0. We will tell that also and then we will
00:42:11.460 --> 00:42:16.259
remember the differential form, we will keep
both the forms. So, I will write both.
00:42:16.259 --> 00:42:17.259
.
00:42:17.259 --> 00:43:28.170
The first 1 just the whole term, I am going
to fold the equation, in case there is shaft
00:43:28.170 --> 00:43:32.950
work, I will just write this immediately after
this point. I will say this is equal to 0
00:43:32.950 --> 00:43:35.480
and
my problem, in case there is other forms of
00:43:35.480 --> 00:43:41.309
work out then I will keep that this, whole
thing is the left hand side, now that is going
00:43:41.309 --> 00:44:10.410
to be equal to. The remaining terms again
I
00:44:10.410 --> 00:44:16.540
wrote this term, If I want I can even write
chemical reaction here, I just said it is
00:44:16.540 --> 00:44:20.730
q in rate
of heat n through radiation and we will again
00:44:20.730 --> 00:44:24.520
say, we will make it equal to 0 in our
problem same thing as shaft work.
00:44:24.520 --> 00:44:29.530
We will again say it is equal to 0 and we
want our consider that, now of course I can
00:44:29.530 --> 00:44:37.299
write this in terms of gradient in t using
Fourier’s law. If I use Fourier’s law,
00:44:37.299 --> 00:44:39.839
this will
become just this 1 term, I will write separately
00:44:39.839 --> 00:44:44.800
here, this is integral over surface, which
I
00:44:44.800 --> 00:44:52.869
missed here, integral over the surface there.
Now I can write this as integral over the
00:44:52.869 --> 00:45:01.980
volume, this is the first change.
00:45:01.980 --> 00:45:14.089
.Now, I will use my heat conduction law here
and I should have minus sign here, now
00:45:14.089 --> 00:45:25.871
that will become
this term of course, I just tell you, that
00:45:25.871 --> 00:45:31.730
this is a possibility, we would
not most likely use this. We will use only
00:45:31.730 --> 00:45:34.839
bulk quantities, we would not go for at a
point,
00:45:34.839 --> 00:45:38.920
what is a temperature, what is the next temperature
point nearby and find out the
00:45:38.920 --> 00:45:44.940
gradient, we will never do that in our simple
gas dynamics world. It is needed when we
00:45:44.940 --> 00:45:48.240
want to think about how much is the heat conduction
across a shock which, we will
00:45:48.240 --> 00:45:54.119
never go to in our course here for now we
are in simple world.
00:45:54.119 --> 00:46:07.069
Now, we just want to go write this in terms
of everything in terms of volume integral,
00:46:07.069 --> 00:46:13.849
which is not difficult to do like, we did
this before. If this can be rewritten in terms
00:46:13.849 --> 00:46:19.500
of del
P dot u kind of term like that, you can write
00:46:19.500 --> 00:46:25.190
this term also there will be a grad of this
whole term, it will come out. We will just
00:46:25.190 --> 00:46:30.220
write the final form I want worry about how,
we got to this, we have the same method used
00:46:30.220 --> 00:46:35.660
in mass equation, it will look the same. So,
I will just write the final form here.
00:46:35.660 --> 00:46:36.660
.
00:46:36.660 --> 00:46:44.990
It will be and I am removing the integral
and just integral over the volume is for whole
00:46:44.990 --> 00:46:49.289
thing and I will just say there, whatever
is inside the integral mass be 0. I will just
00:46:49.289 --> 00:47:42.860
write
that form. We will leave it in this form actually
00:47:42.860 --> 00:47:45.749
I just told that it is a grad somewhere.
00:47:45.749 --> 00:47:46.749
..
00:47:46.749 --> 00:47:50.470
Here in this term I just told when, we make
it a volume integral just a few minutes back,
00:47:50.470 --> 00:47:55.900
I told it is going to become grad P, it would
not grad P, because this dot u here.
00:47:55.900 --> 00:47:56.900
.
00:47:56.900 --> 00:48:01.781
It will just become del dot P u, it will become
this term all the other terms was similar
00:48:01.781 --> 00:48:04.269
to
what you did before in mass equation, this
00:48:04.269 --> 00:48:09.000
1 term alone and then we get to this form,
this
00:48:09.000 --> 00:48:16.099
is your energy equation, energy conservation
equation in differential form. We will keep
00:48:16.099 --> 00:48:22.050
all these forms. Now we, are at a point where,
we have all the govern equations for the
00:48:22.050 --> 00:48:30.269
flow along with this, we should keep the equation
for entropy that is t d s equal to d h
00:48:30.269 --> 00:48:37.390
.minus V d p, we are keep that form or d u
plus p d v are not d u for you. It will be
00:48:37.390 --> 00:48:40.690
d e
internal energy change in internal energy
00:48:40.690 --> 00:48:46.869
plus p d v, that form and P equal to rho r
t gas e
00:48:46.869 --> 00:48:55.519
state equation. So, ideally we have five equations.
We will look at what happens when there are
00:48:55.519 --> 00:49:06.700
5 equations, how many variables are, there
are 5 equations, how many variables are there,
00:49:06.700 --> 00:49:18.559
what are the 5 variables we have.
Velocity volume not relay, because it is all
00:49:18.559 --> 00:49:20.840
differential equation is at a point there
is no
00:49:20.840 --> 00:49:30.950
volume for your fluid. It is point pressure,
temperature, density, velocity, internal energy
00:49:30.950 --> 00:49:42.369
is just temperature
enthalpy still temperature c p times temperature,
00:49:42.369 --> 00:49:50.260
we will think about
this. We will get back to this next time,
00:49:50.260 --> 00:49:51.720
there is still one more variable, which I
we just
00:49:51.720 --> 00:49:57.590
missed anyways.
So, next class onwards, we will get in to
00:49:57.590 --> 00:50:03.380
actual gas dynamics, we just laid the foundation
out, we know thermodynamics very well. Now
00:50:03.380 --> 00:50:07.359
we know, we have used mechanics next
class onwards, we will just starts solving
00:50:07.359 --> 00:50:11.780
problems of different types in gas dynamics,
see you guys next time.
00:50:11.780 --> 00:50:12.780
.