WEBVTT
Kind: captions
Language: en
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Hello everyone welcome back, from today onwards
you are going to discuss analyses of
00:00:19.430 --> 00:00:29.410
flows that are no more isentropic. And before
we going to it we neglected a whole bunch
00:00:29.410 --> 00:00:35.720
of terms, when we derived the basic equations
of flow mass momentum energy
00:00:35.720 --> 00:00:42.539
consideration equations. So, will go revisit
them and then include those terms and see
00:00:42.539 --> 00:00:47.280
how the equation looks like, and what that
is telling us that is where we are going to
00:00:47.280 --> 00:00:53.480
start
today of course, this not much change in mass
00:00:53.480 --> 00:00:54.500
equation.
.
00:00:54.500 --> 00:01:10.170
Mass consideration equation we derived this
form, this still remains the same except for
00:01:10.170 --> 00:01:29.350
I
will write it in one more form, where this
00:01:29.350 --> 00:01:33.180
can be club together as total derivative this
is
00:01:33.180 --> 00:01:37.369
of course, fluid mechanics I will just avoid
going into details of this total derivative.
00:01:37.369 --> 00:01:40.020
So,
much this is called the total derivative will
00:01:40.020 --> 00:01:42.209
be using it in derivation because, it makes
our
00:01:42.209 --> 00:01:45.970
life simpler writing terms will decrease in
number.
00:01:45.970 --> 00:01:54.080
So, can be written in this form also, and
if we say the flow is incompressible I am
00:01:54.080 --> 00:01:55.959
going
to say density does not change at all and,
00:01:55.959 --> 00:01:59.550
so this term goes to 0 and that is how we
got to
00:01:59.550 --> 00:02:05.239
.incompressible flow divergence is 0 kind
of term. But, of course, we are not in
00:02:05.239 --> 00:02:12.610
incompressible world we are very much incompressible
world, so this need not be 0,
00:02:12.610 --> 00:02:21.790
next is momentum conservation where we got
an expression this is all I am referring to
00:02:21.790 --> 00:02:27.890
like say the 5'th lecture or 6'th lecture
somewhere at the beginning, where we derived
00:02:27.890 --> 00:02:34.860
all
the equations of motion.
00:02:34.860 --> 00:03:08.810
I will put this arrow for vector u everywhere,
I have written these extra terms now which
00:03:08.810 --> 00:03:13.840
we said 0 in the previous time and we derive,
I will just added these two terms. Basically
00:03:13.840 --> 00:03:20.499
this is telling you the total momentum conservation,
and the very first time when we
00:03:20.499 --> 00:03:26.909
derived it we had this term as dou by dou
t of rho u inside the derivative. And I have
00:03:26.909 --> 00:03:32.890
used this expression inside that to simplify
it to this form, I will not go and in details
00:03:32.890 --> 00:03:35.730
of
its fluid mechanics when you go derive equation
00:03:35.730 --> 00:03:39.560
for the first time it will come out to be
that and, so will keep only this form.
00:03:39.560 --> 00:03:45.360
Now, previously when we were deriving we said
that we do not have any body forces,
00:03:45.360 --> 00:03:50.150
we said this equal to 0. And then we said
that we do not have any viscosity, so every
00:03:50.150 --> 00:03:55.340
term in the shear force will have a mu inside
it, which is said to be 0, so this term was
00:03:55.340 --> 00:03:57.830
0,
so we neglected this. Now, we do not want
00:03:57.830 --> 00:04:02.859
to neglect it we will just keep it as is,
so
00:04:02.859 --> 00:04:13.599
basically with this becomes our momentum equation,
next is energy I will write what we
00:04:13.599 --> 00:04:56.280
had before and then from there we will add
two more terms similar to this.
00:04:56.280 --> 00:05:16.450
I have written this form here, the left hand
side is exactly same as what we derived at
00:05:16.450 --> 00:05:21.070
the
beginning. And we had this term already we
00:05:21.070 --> 00:05:26.370
did not have F b and tau terms before, we
said that it was 0 and we never wrote an expression
00:05:26.370 --> 00:05:31.670
for these, we just removed at the
beginning itself in the contribution stage
00:05:31.670 --> 00:05:36.850
itself. And later we said there is no heat
transfer
00:05:36.850 --> 00:05:45.800
and we neglected this also that is how we
came to this particular form of the equation.
00:05:45.800 --> 00:05:52.340
One more thing I need to tell you is this
F b is body force per unit mass it is roughly
00:05:52.340 --> 00:05:56.050
an
acceleration term F b is body force per unit
00:05:56.050 --> 00:06:00.440
mass, but everything else is per unit volume
basis. So, I am multiplying by density to
00:06:00.440 --> 00:06:03.210
get to per unit volume basis that is how we
get
00:06:03.210 --> 00:06:10.780
this form, now I want to rearrange this energy
equation, so that we can get more
00:06:10.780 --> 00:06:15.520
information out of this, what is it telling
is not very clear if I write it like this,
00:06:15.520 --> 00:06:17.060
this
perfectly correct. But, it does not give me
00:06:17.060 --> 00:06:20.110
full information we will try and rearrange
the
00:06:20.110 --> 00:06:21.110
terms here.
00:06:21.110 --> 00:06:27.910
.I will just take the left hand side, left
hand side of the equation I will expand it
00:06:27.910 --> 00:06:33.180
where I
will say h is e plus p by rho I will write
00:06:33.180 --> 00:06:51.530
it like that, I will see what I am getting
I will do
00:06:51.530 --> 00:07:04.720
one more thing I will use differentiation
by parts. So, that I will cut down one statement
00:07:04.720 --> 00:07:09.660
here, so I am taking rho as one quantity and
this bracket as another quantity and I am
00:07:09.660 --> 00:07:24.590
derivative taking derivative by parts from
here, this is coming from the first term alone
00:07:24.590 --> 00:07:30.400
plus the next term again, we have to do similar
thing I am going to group this rho u as
00:07:30.400 --> 00:07:35.370
together and the bracket is another term will
keep.
00:07:35.370 --> 00:07:48.810
If I keep the rho u term together I am going
to have ideally I should have del dot of h
00:07:48.810 --> 00:07:55.250
plus u square by 2, which I want to write
I will write this right here e plus p by rho
00:07:55.250 --> 00:08:00.080
I will
put h as e plus p by rho I will take this
00:08:00.080 --> 00:08:03.250
e and this u plus u square by 2 together and
write
00:08:03.250 --> 00:08:07.450
it here. Will remember that there is one p
by rho term which is left out inside this
00:08:07.450 --> 00:08:19.550
whole
equation, this is one term plus the other
00:08:19.550 --> 00:08:28.410
term where derivative is on this rho u del
dot of
00:08:28.410 --> 00:08:42.140
rho u.
Now, will remember that there is a term del
00:08:42.140 --> 00:08:50.149
dot of rho u times p by rho u which is left,
which I will write separately plus del dot
00:08:50.149 --> 00:09:05.490
of we will just keep it like this I know I
can
00:09:05.490 --> 00:09:11.990
cancel the density will do that later. Now,
look at this expression I want to simplify
00:09:11.990 --> 00:09:17.019
it a
little bit, I will keep the terms with e plus
00:09:17.019 --> 00:09:20.279
u square by 2 as a common factor together,
if I
00:09:20.279 --> 00:09:28.769
look at that, that will be this term and this
term. If I look at that term I take this common
00:09:28.769 --> 00:09:34.889
factor out, what is remaining is dou by dou
rho by dou t plus del dot of rho u that
00:09:34.889 --> 00:09:41.540
happens to be equal to 0 from mass equation.
So, I will just remove that two terms from
00:09:41.540 --> 00:09:45.720
there, so my left hand side is only these
three terms.
00:09:45.720 --> 00:09:46.720
..
00:09:46.720 --> 00:09:58.230
So, I will continue from there by the way
that is the same exercise we did for getting
00:09:58.230 --> 00:10:03.449
the
momentum equation from the previously derived
00:10:03.449 --> 00:10:34.490
form to present form. Now, I will
rewrite this again I want to get to a point
00:10:34.490 --> 00:10:39.300
where I will go to h naught the stagnation
enthalpy. So, I will rearrange this now, now
00:10:39.300 --> 00:10:41.839
that it is in a better form I will rearrange
this
00:10:41.839 --> 00:10:47.459
such that it will be h instead of e in both
the places, so that I can write this whole
00:10:47.459 --> 00:10:50.269
thing as
total derivative.
00:10:50.269 --> 00:11:01.410
So, I will directly write it as h plus u square
by 2 will figure out what is missing after
00:11:01.410 --> 00:11:12.610
this, now will put all the missing terms I
added a p by rho here. So, that will come
00:11:12.610 --> 00:11:19.410
as the
next term sitting there minus rho into rho
00:11:19.410 --> 00:11:28.339
by dou t of p by rho I will put a plus inside
here
00:11:28.339 --> 00:11:42.200
plus rho dot del of p by rho plus the final
term there del dot, since I do not have space
00:11:42.200 --> 00:11:48.730
I
will cancel the density and write it there,
00:11:48.730 --> 00:11:58.220
I have these terms here. Now, I can club these
two terms together and I write it as rho into
00:11:58.220 --> 00:12:04.439
total derivative of h naught.
These two terms come together to form this
00:12:04.439 --> 00:12:16.309
one term minus, the remaining terms will
expand this such that I will have just dou
00:12:16.309 --> 00:12:19.819
p by dou t and dou rho by dou t. So, if I
do that
00:12:19.819 --> 00:12:32.699
it will be density into 1 by rho dou p by
dou t minus 1 by rho square p is also there
00:12:32.699 --> 00:12:37.069
dou
rho by rho t. These are the terms that comes
00:12:37.069 --> 00:12:40.230
from differentiating this p by rho that is
this
00:12:40.230 --> 00:12:55.730
first two terms plus I will keep the this
term as is u dot del of p by rho as is, now
00:12:55.730 --> 00:12:59.309
I will
expand this term actually I like the form
00:12:59.309 --> 00:13:02.709
this particular form I will expand this form
here
00:13:02.709 --> 00:13:07.720
next I will close this and I will open that
one next.
00:13:07.720 --> 00:13:14.029
.Where I want to split this term again rho
u together and p by rho together, so that
00:13:14.029 --> 00:13:16.170
will
come and cancel this particular term that's
00:13:16.170 --> 00:13:26.100
what am looking for finally. So, now, I will
write it as rho u dotted with del of p by
00:13:26.100 --> 00:13:30.790
rho that is one term plus p by rho outside
the
00:13:30.790 --> 00:13:40.369
derivative times del dot of rho u, this is
the other form. This is actually coming from
00:13:40.369 --> 00:13:43.470
this
term, but I am using the previous form which
00:13:43.470 --> 00:13:47.600
is nicer to expand and it comes to this
particular case.
00:13:47.600 --> 00:13:56.339
Now, I will look at this term this and this
they will get cancel, why this is having a
00:13:56.339 --> 00:13:59.481
minus
sign here, this minus sign minus rho u dot
00:13:59.481 --> 00:14:03.089
del of p by rho this is plus rho u dot del
of p
00:14:03.089 --> 00:14:11.839
by rho they get cancel. Now, I will look at
the remaining terms I will write it one more
00:14:11.839 --> 00:14:24.129
step rho D by D t of h naught minus dou p
by dou t, and this minus and this minus will
00:14:24.129 --> 00:14:32.160
become plus and density one of them will get
cancel plus p by rho dou rho by dou t that
00:14:32.160 --> 00:14:41.670
is only thing plus p by rho del dot of rho
u, this is my left hand side.
00:14:41.670 --> 00:14:50.709
Now, again I am seeing that this plus this
together is 0, why again continuity equation,
00:14:50.709 --> 00:14:55.970
continuity equation says this plus this is
equal to 0. So, p by rho multiplied by 0 these
00:14:55.970 --> 00:15:00.639
two terms will go away now, so I finally,
have left hand side to be just these two terms.
00:15:00.639 --> 00:15:07.559
Now, I will bring all the other terms and
write it together my full energy equation,
00:15:07.559 --> 00:15:26.879
now
be rho D h naught by D t, now I want to write
00:15:26.879 --> 00:15:33.929
it this on the other side.
So, I will put it as rho D h naught by D t
00:15:33.929 --> 00:15:35.670
the I have taken the dou p by dou t to the
other
00:15:35.670 --> 00:15:46.869
side, and I have the remaining terms I will
write it as plus rho F b vector dotted with
00:15:46.869 --> 00:15:59.329
u
plus del dot of tau tensor dotted with u vector,
00:15:59.329 --> 00:16:12.800
this one more term minus del dot of u.
Finally, have this particular form we would
00:16:12.800 --> 00:16:17.070
not do any more changes to this, this gives
you some more physical field.
00:16:17.070 --> 00:16:22.930
All this time we said, total enthalpy is a
net energy content of the fluid, this includes
00:16:22.930 --> 00:16:28.910
kinetic energy as well as internal energy,
thermal energy including the pressure energy
00:16:28.910 --> 00:16:30.959
it
is having all forms of energy, thermal energy,
00:16:30.959 --> 00:16:35.829
pressure energy, and kinetic energy
complete set is inside this. So, this is the
00:16:35.829 --> 00:16:43.069
net content of energy in that particular fluid
element, now we want to see the change in
00:16:43.069 --> 00:16:46.329
that particular quantity, if I look at it
may
00:16:46.329 --> 00:16:54.290
change the simple term is this one.
It may change, because of if I call it a del
00:16:54.290 --> 00:16:59.660
dot q is divergence of heat, this is a heat
conduction, heat flukes q basically talking
00:16:59.660 --> 00:17:04.540
about divergence of heat away from this point.
So, if there is divergence of heat away from
00:17:04.540 --> 00:17:08.020
this point, that is going to have energy loss,
00:17:08.020 --> 00:17:14.799
.so it is negative or you can think about
it as minus of divergence is the convergence
00:17:14.799 --> 00:17:17.730
of
heat, opposite of divergence right it is converging
00:17:17.730 --> 00:17:22.520
of heat into that point that is increasing
energy, whichever way you think about it that
00:17:22.520 --> 00:17:27.890
is one term.
The next term is here shear stress tensor
00:17:27.890 --> 00:17:31.669
doing work with that particular fluid is going
to
00:17:31.669 --> 00:17:36.370
finally, generate some amount of energy because,
it is increasing energy of the system by
00:17:36.370 --> 00:17:42.240
adding work to that particular fluid that
is going to increase energy and that is this.
00:17:42.240 --> 00:17:45.299
This
is the same kind of work, but done by the
00:17:45.299 --> 00:17:48.669
body force the bulk on the bulk fluid itself
and
00:17:48.669 --> 00:17:53.900
that is given by this term, work done per
unit time that is coming out to be this,
00:17:53.900 --> 00:17:56.080
everything is in terms of per unit volume
basis.
00:17:56.080 --> 00:18:03.340
And finally, there is one more term here which
is talking about if I suddenly have
00:18:03.340 --> 00:18:11.789
unsteady pressure rise that can also add energy
to my fluid, where have you seen this we
00:18:11.789 --> 00:18:16.750
have seen one example, where we said there
is no heat transfer, there is no body force,
00:18:16.750 --> 00:18:23.230
there is no shear, but we have a moving shock.
If we had a moving shock, then when the
00:18:23.230 --> 00:18:28.080
shock crosses my current point of interest,
then when the shock crosses I am going to
00:18:28.080 --> 00:18:33.220
have pressure rising, and the pressure rises
I am going to have t naught increasing that
00:18:33.220 --> 00:18:34.600
is
how it is show right.
00:18:34.600 --> 00:18:38.710
We found that when a moving shock crosses
my point t naught increases that is directly
00:18:38.710 --> 00:18:43.169
come in front. These are the various mechanisms
by which we may change energy of the
00:18:43.169 --> 00:18:50.720
fluid, till now we thought about only simple
things, now will go and look at little more
00:18:50.720 --> 00:18:56.770
details of what all other things can change
the net energy of course, in particular we
00:18:56.770 --> 00:18:59.049
are
going to think about, we will still neglect
00:18:59.049 --> 00:19:00.750
this body force term of course, you can add
it
00:19:00.750 --> 00:19:06.360
its simply in of term, you are going to think
about heat transfer and viscous forces.
00:19:06.360 --> 00:19:11.930
We will not deal with this complicated viscous
force term, I did not explain what this tau
00:19:11.930 --> 00:19:17.010
tensor is really, I am just going around it
is not really important for this course will
00:19:17.010 --> 00:19:18.630
just
go around it. Because, we are going to work
00:19:18.630 --> 00:19:26.480
in typically one dimension it will come out
to be much simpler than this, if I send an
00:19:26.480 --> 00:19:33.000
expansion wave what happens, this will
automatically become negative pressure is
00:19:33.000 --> 00:19:35.260
going to drop that will automatically pull
out
00:19:35.260 --> 00:19:40.250
energy from the system.
That can also happen I may have if there is
00:19:40.250 --> 00:19:44.820
a moving expansion it is going to cool down
the gas, it is going to pull out net energy
00:19:44.820 --> 00:19:48.080
from the gas also that can also happen any
of
00:19:48.080 --> 00:19:52.780
these can be happen in there.
00:19:52.780 --> 00:19:53.780
..
00:19:53.780 --> 00:20:02.450
Next thing we want to do is entropy, so I
will start from the standard thermo dynamics
00:20:02.450 --> 00:20:09.250
form of entropy, when I write this small s
I am going to say it is per mass basis. We
00:20:09.250 --> 00:20:11.300
just
work with that form, we had the same convention
00:20:11.300 --> 00:20:17.769
from the beginning will maintain that
d e which is the internal energy plus p d
00:20:17.769 --> 00:20:26.340
v where d v I will write it as d of 1 by rho
because, that's per mass basis volume. So,
00:20:26.340 --> 00:20:30.520
it is 1 by rho will rewrite this now this
can be
00:20:30.520 --> 00:20:41.549
re written from calculus as minus p by rho
square d rho, it can be re written like this.
00:20:41.549 --> 00:20:51.600
Now, will use this particular definition and
we want to write D s by D t total derivative
00:20:51.600 --> 00:20:56.610
of entropy, right entropy change in my fluid
element, if I write that using this particular
00:20:56.610 --> 00:21:01.220
derivative notation I can immediately jump
to this particular derivative, and I am going
00:21:01.220 --> 00:21:16.549
to say it is D e by D t minus p by rho square
D rho by D t. Now, I will look at this minus
00:21:16.549 --> 00:21:25.049
D rho by D t, I will go and use my continuity
equation, which is out here this is why I
00:21:25.049 --> 00:21:30.640
wrote this here D rho by D t if I put a minus
sign, then I it is equivalent to taking this
00:21:30.640 --> 00:21:32.360
part
to the other side of the equation.
00:21:32.360 --> 00:21:37.370
So, it will just come out to be rho times
del dot u, I will use that in there, so I
00:21:37.370 --> 00:21:45.710
will
simplify this to D e by D t plus p by rho
00:21:45.710 --> 00:21:55.470
square into rho del dot u one rho gets cancel.
I
00:21:55.470 --> 00:22:02.650
have this particular form here, now I need
to write a D e by D t which is just talking
00:22:02.650 --> 00:22:08.299
about the internal energy of the fluid and
not the total enthalpy, which is what we being
00:22:08.299 --> 00:22:09.320
writing till now.
00:22:09.320 --> 00:22:16.850
.So, I now I need to split this total enthalpy
from my full energy equation, for this I have
00:22:16.850 --> 00:22:22.669
to think about what are the components of
my energy equation, I am going to say I have
00:22:22.669 --> 00:22:28.500
my enthalpy plus kinetic energy as my total
energy equation, where enthalpy is having
00:22:28.500 --> 00:22:35.490
internal energy plus p by rho. So, if I write
conservation equation for kinetic energy
00:22:35.490 --> 00:22:41.769
alone, which I can obtain by multiplying the
momentum equation by another u vector dot
00:22:41.769 --> 00:22:45.730
product of it.
And then finding out the what the final expression
00:22:45.730 --> 00:22:49.090
is that comes out to be the kinetic
energy conservation equation, we would not
00:22:49.090 --> 00:22:53.870
do that here I will just give you the final
expression of course, this is again been simplified
00:22:53.870 --> 00:22:59.539
by using continuity equation. So, that
density is outside even though it is used
00:22:59.539 --> 00:23:03.320
for compressible flows, I am putting u square
by
00:23:03.320 --> 00:23:09.899
2 because, it is that particular form of kinetic
energy form per unit mass basis.
00:23:09.899 --> 00:23:29.779
This term comes out to be rho F b dot u plus
del dot of tau tensor dot u vector plus
00:23:29.779 --> 00:23:40.179
pressure times del dot u minus there is a
free term, which I did not want to go into
00:23:40.179 --> 00:23:42.000
details
of it. But, since it is in the equation I
00:23:42.000 --> 00:23:44.980
have to talk about it a little bit, it is
call the
00:23:44.980 --> 00:23:49.610
dissipation of energy due to viscosity, dissipation
of kinetic energy due to viscosity in
00:23:49.610 --> 00:23:55.450
this case. A written a equation for kinetic
energy conservation alone, where we did not
00:23:55.450 --> 00:23:59.810
talk about internal energy of the gas we just
that kinetic energy part.
00:23:59.810 --> 00:24:06.740
What we have finding is kinetic energy change
is due to external force doing work on the
00:24:06.740 --> 00:24:12.070
system, on the particular fluid element or
shear forces doing work on the fluid element
00:24:12.070 --> 00:24:15.679
or
pressure forces doing work on the fluid element,
00:24:15.679 --> 00:24:21.700
a this is actually pressure work on the
fluid element, this is the divergence of the
00:24:21.700 --> 00:24:23.809
fluid element, and this is the expansion of
the
00:24:23.809 --> 00:24:30.030
fluid element multiply by the pressure that
is like p d v work that is forming here. And
00:24:30.030 --> 00:24:36.269
then this is here energy removed from kinetic
energy, so that it enters into thermal
00:24:36.269 --> 00:24:40.520
energy will see that soon.
This is the part, which causes heating of
00:24:40.520 --> 00:24:43.080
the gas when there is viscosity, we would
not go
00:24:43.080 --> 00:24:49.450
into details of it will also have some terms
like I believe it is u dot del dot tau that
00:24:49.450 --> 00:24:51.649
kind of
form will be there, I do not want to get into
00:24:51.649 --> 00:24:55.370
the details of it will just ignore it for
now, it
00:24:55.370 --> 00:25:01.620
is not needed except for this one particular
place. So, if I subtract this from our original
00:25:01.620 --> 00:25:07.720
energy equation which we had before, then
it is going to come to point where I can write
00:25:07.720 --> 00:25:25.720
it as rho D e by D t is equal to minus del
dot q minus p times del dot u plus p.
00:25:25.720 --> 00:25:32.980
.Only these will be remaining, will find that
the viscous work is going to be internal and
00:25:32.980 --> 00:25:36.049
it
will just sit in the total energy also. So,
00:25:36.049 --> 00:25:38.809
only thing remaining is this particular part,
what
00:25:38.809 --> 00:25:45.950
we are now seeing is this is only the thermal
part of the energy, and we look at the
00:25:45.950 --> 00:25:50.320
thermal part of the energy we are finding
that if there is net convergence of heat into
00:25:50.320 --> 00:25:53.140
the
fluid element, then it increases. If there
00:25:53.140 --> 00:25:58.299
is net contraction of fluid element del dot
u is
00:25:58.299 --> 00:26:02.400
related to rate of expansion of the fluid
element minus of that is like compression
00:26:02.400 --> 00:26:04.380
of the
fluid element that is external work done on
00:26:04.380 --> 00:26:10.360
my fluid that can cause heating of the fluid.
And this is the energy added from kinetic
00:26:10.360 --> 00:26:17.159
energy to thermal energy due to viscous
effects. If we look at this previous equation
00:26:17.159 --> 00:26:21.570
and this equation, in the kinetic energy
equation there is a minus v I did not write
00:26:21.570 --> 00:26:23.679
the term, but it is going to be negative it
is
00:26:23.679 --> 00:26:27.789
pulling out energy from kinetic energy. And
here it is plus v it is adding that energy
00:26:27.789 --> 00:26:31.060
into
this thermal energy.
00:26:31.060 --> 00:26:35.120
If we look at that total energy equation which
we had before I will go back to it once
00:26:35.120 --> 00:26:39.650
more that is this one, we find that there
is no phi in this particular thing why it
00:26:39.650 --> 00:26:42.210
is just a
internal rearrangement, it is taking from
00:26:42.210 --> 00:26:43.570
one pocket and putting it to the other pocket.
So,
00:26:43.570 --> 00:26:47.880
it is not there at all, this is the net energy
over all how much do I contain that kind of
00:26:47.880 --> 00:26:56.679
thing is this. In here if we look at this
part this is sitting only in my internal energy
00:26:56.679 --> 00:27:00.710
equation, this part is sitting only in my
kinetic energy equation this also sitting
00:27:00.710 --> 00:27:02.720
only in
my kinetic energy equation.
00:27:02.720 --> 00:27:09.770
This term is hidden somewhere inside, it is
sitting in that p dot del u kind of term we
00:27:09.770 --> 00:27:17.259
would not worry about that part will see it
will come up now. So, this is my internal
00:27:17.259 --> 00:27:26.690
energy equation, now I want to put this D
e by D t inside here, and I have a rho here
00:27:26.690 --> 00:27:29.472
will
multiply this entropy equation with rho, so
00:27:29.472 --> 00:27:38.990
that I will get rho D e by D t. So, I am going
to finally have, form rho into temperature
00:27:38.990 --> 00:27:44.000
D s by D t is equal to I will put rho into
D e by
00:27:44.000 --> 00:27:55.299
D t instead of that I will write this expression,
it should be minus del dot q minus p into
00:27:55.299 --> 00:28:02.460
del dot u plus phi I have this particular
term.
00:28:02.460 --> 00:28:16.029
Now, I will write this particular thing p
by rho times del dot u, this is my net entropy
00:28:16.029 --> 00:28:27.309
equation what is that last term should have
a multiply by a rho also. Because, I
00:28:27.309 --> 00:28:32.590
multiplied the whole expression with rho while
rho D e by D t was this term that one
00:28:32.590 --> 00:28:37.510
should also be multiplied by rho that p will
get rho will get cancel. So, I have this term
00:28:37.510 --> 00:28:43.899
now you will find that this term and this
term gets out, they are getting cancel, so
00:28:43.899 --> 00:28:46.600
I am
00:28:46.600 --> 00:29:05.840
.finding that for my entropy to increase,
it has to be through convergence of heat into
00:29:05.840 --> 00:29:11.019
the
fluid element or through frictional heating
00:29:11.019 --> 00:29:14.909
of the fluid element.
These are the two modes where I can have entropy
00:29:14.909 --> 00:29:21.440
increase in my fluid flow, all this
point I always used compressible density was
00:29:21.440 --> 00:29:25.110
changing it is, so happens that I used my
continuity equation. So, that a density always
00:29:25.110 --> 00:29:31.000
comes out of my derivatives, but I know
that it is compressible flow, so the essential
00:29:31.000 --> 00:29:37.420
part of this particular statement is that
if I do
00:29:37.420 --> 00:29:46.230
not have heating of my fluid, either by heat
conduction or by viscous forces I will not
00:29:46.230 --> 00:29:53.679
have any entropy change in my fluid flow.
These are my predominant changes in entropy,
00:29:53.679 --> 00:29:58.680
entropy changes are mainly due to this of
course, we also know that there is one more
00:29:58.680 --> 00:30:05.130
way we can change entropy that is through
shocks in the flow. Which is just sitting
00:30:05.130 --> 00:30:08.570
inside this which we could not talk about
here,
00:30:08.570 --> 00:30:17.740
that will come as delta p naught that will
not be from here that should come from, the
00:30:17.740 --> 00:30:22.430
direct expansion of this term.
If I expand it we did this long back where
00:30:22.430 --> 00:30:28.220
we I do not know where I have that
expression right now, but we found that delta
00:30:28.220 --> 00:30:43.470
s by r was equal to log of p naught 2 by t
naught 1 to the power gamma by gamma minus
00:30:43.470 --> 00:30:55.270
plus 1, multiplied by p naught 1 by p
naught 2. We had one such expression this
00:30:55.270 --> 00:31:01.260
is going to take care of that fluid mechanical
way of increasing entropy, this is one more
00:31:01.260 --> 00:31:04.720
form of writing it this is actually derived
from this top equation we would not go into
00:31:04.720 --> 00:31:07.370
the details of that right, now it was done
I
00:31:07.370 --> 00:31:19.870
believe just after moving shocks discussion.
So, we want to look at those particular two
00:31:19.870 --> 00:31:21.980
characteristics that is one is heating of
the
00:31:21.980 --> 00:31:28.590
fluid element or viscous forces causing entropy
change. So, we are going to deal with
00:31:28.590 --> 00:31:35.440
two types of flow fields, now where there
is entropy change important first one will
00:31:35.440 --> 00:31:37.870
pick
up friction and friction flow next one will
00:31:37.870 --> 00:31:41.330
pick heating. But, before that I just wanted
to
00:31:41.330 --> 00:31:50.749
give you this small snipped of analytical
work by Italian engineer called croccos. It
00:31:50.749 --> 00:31:55.990
is a
nice point to give it here it has to be given,
00:31:55.990 --> 00:32:04.169
so that you will understand the connection
between vorticity and entropy. Since, we are
00:32:04.169 --> 00:32:09.200
deriving gradients of entropy in my flow
etcetera it is a good point to give it this
00:32:09.200 --> 00:32:10.220
point.
00:32:10.220 --> 00:32:11.220
..
00:32:11.220 --> 00:32:17.929
So, will derive it the simplest way first
then will give you the most complicated
00:32:17.929 --> 00:32:23.929
expression, that is the final result. It is
also going to introduce you stream wise
00:32:23.929 --> 00:32:41.019
coordinate system, I am calling this as my
stream line I am using this kind of s small
00:32:41.019 --> 00:32:44.570
s
while if I use the other s this is entropy
00:32:44.570 --> 00:32:55.120
per mass, this is my stream wise distance
coordinate system. So, I am having this is
00:32:55.120 --> 00:33:03.149
s let us say this is my reference point am
picking this is my n, the normal stream normal
00:33:03.149 --> 00:33:09.029
coordinate system.
So, I will say that if this is my point n
00:33:09.029 --> 00:33:13.230
comma s this is going to be a delta n there,
so this
00:33:13.230 --> 00:33:22.159
point will be a delta s here. This will be
n comma s plus delta s while here it will
00:33:22.159 --> 00:33:26.260
be n
plus delta n this will be my delta n, n plus
00:33:26.260 --> 00:33:32.000
delta n and s this will be n plus delta n
s plus
00:33:32.000 --> 00:33:40.429
delta n, that kind of coordinate system I
am going to have. Now, let us say I have some
00:33:40.429 --> 00:33:45.249
particular velocity direction with respect
to my previous original rectilinear coordinate
00:33:45.249 --> 00:33:48.210
system.
Let us say this is having a theta x y coordinate
00:33:48.210 --> 00:33:51.639
system this where theta with respect to x
y
00:33:51.639 --> 00:34:00.429
coordinate system, that is my velocity vector
that is the case then I go from here to the
00:34:00.429 --> 00:34:10.520
next n point. It is going to be again velocity
vector will be parallel to that local stream
00:34:10.520 --> 00:34:20.190
line which is this line another s curve and
that angle is going to be theta plus dou theta
00:34:20.190 --> 00:34:24.540
by
dou n times delta n, just a generalize form
00:34:24.540 --> 00:34:27.419
of writing it for a small change in n how
much
00:34:27.419 --> 00:34:33.669
will be the change in theta is what this represents
and will just keep it, that way very
00:34:33.669 --> 00:34:37.819
generalized form and here this gap was delta
s.
00:34:37.819 --> 00:34:53.450
.I am going to say this is delta s plus dou
by dou n of delta s times delta n change in
00:34:53.450 --> 00:34:57.030
delta
s with delta n is what is shown there, that
00:34:57.030 --> 00:35:00.530
is the this point to this point the gap is
there.
00:35:00.530 --> 00:35:04.810
Just taking care of full details of course,
we may not use all of them and what will
00:35:04.810 --> 00:35:10.500
happen to velocity here. This velocity will
be u which is the value here this is my
00:35:10.500 --> 00:35:21.130
velocity u.
It will become u plus dou u by dou n times
00:35:21.130 --> 00:35:27.210
delta n this will be the velocity at this
point
00:35:27.210 --> 00:35:32.060
should I think about components of velocity,
there is no components here. Because, it is
00:35:32.060 --> 00:35:36.359
a stream wise coordinate system which means
all the velocity is going to be only along
00:35:36.359 --> 00:35:41.390
this s, there is no velocity component perpendicular
to it because this is the stream line.
00:35:41.390 --> 00:35:46.040
So, I can just talk about velocity always
along this direction there is no perpendicular
00:35:46.040 --> 00:35:48.839
component.
So, I have basically brought my problem down
00:35:48.839 --> 00:35:54.970
I do not need to think about u as a vector,
I will just think about u and it looks as
00:35:54.970 --> 00:35:58.089
if it is a one d problem finally, because,
it is going
00:35:58.089 --> 00:36:06.099
to be varying only inside this stream tube,
between any s and s plus delta n, n and n
00:36:06.099 --> 00:36:08.660
plus
delta n that particular thing is one set of
00:36:08.660 --> 00:36:14.160
stream tube and my problem is one d problem
inside this which means, now I can use all
00:36:14.160 --> 00:36:18.860
the one d equations.
So, will write the one d conservation equation
00:36:18.860 --> 00:36:27.750
for this whatever we are used all the time
will just write it here, rho u times delta
00:36:27.750 --> 00:36:33.900
n this suppose to be rho u a is constant and
I say
00:36:33.900 --> 00:36:39.970
delta n here, this is the delta n is this
length and unit depth inside. I am picking
00:36:39.970 --> 00:36:43.819
two d
problem its unit depth inside, so delta n
00:36:43.819 --> 00:36:52.180
times one is what is given as area here. Similarly
momentum, now you have to be a little bit
00:36:52.180 --> 00:36:55.420
careful forces can act along this n direction
also.
00:36:55.420 --> 00:37:02.369
So, now I have think about s momentum first
only along the s direction s momentum
00:37:02.369 --> 00:37:06.300
equation like, we had x momentum and y momentum
way see here it is s momentum
00:37:06.300 --> 00:37:16.880
equation. Now, it can be written as rho u
dou u by dou s equal to minus dou p by dou
00:37:16.880 --> 00:37:21.790
s, if
you look at this is your acceleration of your
00:37:21.790 --> 00:37:26.280
fluid element inertial forces here, that has
be
00:37:26.280 --> 00:37:29.540
compensated by the pressure force from here
to here.
00:37:29.540 --> 00:37:34.329
The fluid is accelerating this way because,
I am going to say pressure here is higher
00:37:34.329 --> 00:37:36.690
than
pressure here. So, there is d p by d s that
00:37:36.690 --> 00:37:39.170
is negative and what is a net force acting,
that
00:37:39.170 --> 00:37:44.960
will be minus of d p by d s times that area.
I have divided by area throughout that is
00:37:44.960 --> 00:37:46.990
this,
this is basically here force balance, momentum
00:37:46.990 --> 00:37:49.089
increase is equal to the force that is what
00:37:49.089 --> 00:37:55.050
.is written here or you think about it just
f equal to m a kind of formula, next one is
00:37:55.050 --> 00:37:57.770
a
normal momentum equation.
00:37:57.770 --> 00:38:02.530
Where, now I have to think about this stream
line is curved and it is curving if it is
00:38:02.530 --> 00:38:07.350
curving then, there is force perpendicular
to the velocity vector that is when a flow
00:38:07.350 --> 00:38:13.780
stream line will turn it is like a centrifugal
force centripetal force, so I will write an
00:38:13.780 --> 00:38:21.050
expression for that that will be rho u square
by r, this is your centripetal component this
00:38:21.050 --> 00:38:29.240
should be given by minus of dou p by dou n.
The normal force if my pressure here is higher
00:38:29.240 --> 00:38:34.920
than pressure here then my stream line
will curve up, as the pressure goes from lower
00:38:34.920 --> 00:38:40.480
n to higher n pressure decreases which
means my stream line will turn up, that is
00:38:40.480 --> 00:38:43.109
what is this is showing here minus of dou
p by
00:38:43.109 --> 00:38:47.849
dou n is the net force upward, that is going
to causes centripetal acceleration that is
00:38:47.849 --> 00:38:53.520
this
terms, now I will just write since we have
00:38:53.520 --> 00:39:02.281
this u by r, r is here radius of curvature.
I will write this as in terms of theta which
00:39:02.281 --> 00:39:09.160
we are given in the drawing, rho u square
went
00:39:09.160 --> 00:39:15.390
to 1 by r can be represented as dou theta
by dou s that, happens to be the curvature.
00:39:15.390 --> 00:39:21.430
1 by r
is given as curvature which is equal to dou
00:39:21.430 --> 00:39:28.800
theta by dou s, how does theta vary as I go
this way that talks about the curvature of
00:39:28.800 --> 00:39:34.839
this, this particular curve s coordinate system
now.
00:39:34.839 --> 00:39:43.440
The only other thing left is energy will write
the simplified form simplest form where
00:39:43.440 --> 00:39:47.980
will say that there is no heat transfer, there
is no viscosity just for this croccoâ€™s theorem
00:39:47.980 --> 00:39:57.540
derivation alone. When we say that we know
that h naught is constant this is the old
00:39:57.540 --> 00:40:01.250
1 d
derivation I am just writing it again. In
00:40:01.250 --> 00:40:05.109
fact, of in today us derivation also you can
say
00:40:05.109 --> 00:40:08.970
when you said those terms v equal to 0 you
just automatically get to this of course,
00:40:08.970 --> 00:40:12.520
I can,
now introduce it may have a non stationary
00:40:12.520 --> 00:40:18.250
flow where pressure may be changing with
time, I could get that kind of dou p by dou
00:40:18.250 --> 00:40:22.069
t term here.
If I want to currently we will keep it simplified
00:40:22.069 --> 00:40:24.650
when we give you the generalized
expression I will give you the full thing,
00:40:24.650 --> 00:40:27.890
one more thing I need to tell here is this
telling h
00:40:27.890 --> 00:40:34.770
naught is constant for every stream line s
it may. So, happen that this stream line can
00:40:34.770 --> 00:40:39.380
have a different h naught from this stream
line, I did not tell anything about h naught
00:40:39.380 --> 00:40:41.950
is
constant for all s I just said h naught is
00:40:41.950 --> 00:40:44.420
constant along a particular s, that all we
know 1 d
00:40:44.420 --> 00:40:45.420
gas dynamics.
00:40:45.420 --> 00:40:48.780
.We just know that inside a stream tube h
naught does not change it may change from
00:40:48.780 --> 00:40:53.130
one
stream tube to the next one that can still
00:40:53.130 --> 00:40:54.180
happen.
.
00:40:54.180 --> 00:41:01.740
So, now I will go and write again entropy
relation where this s is entropy per unit
00:41:01.740 --> 00:41:17.720
mass, I
will just directly go into p d v v d p sorry
00:41:17.720 --> 00:41:27.140
d h minus v d p. Which can be express in terms
of d h naught from the expression written
00:41:27.140 --> 00:41:34.329
there, d h naught minus u d u will be your
d h
00:41:34.329 --> 00:41:44.010
minus 1 by rho d p I can write it like this.
So, from here I want go and look at individual
00:41:44.010 --> 00:41:49.010
variation of entropy along s direction and
n direction.
00:41:49.010 --> 00:41:55.710
So, I will take this definition this is for
any particular change in s, so now, I will
00:41:55.710 --> 00:41:58.630
go make
it change only one particular direction, t
00:41:58.630 --> 00:42:09.069
dou as by dou stream wise coordinate dou small
s is equal to I know that along stream wise
00:42:09.069 --> 00:42:12.599
direction dou s dou by dou h naught as, I
will
00:42:12.599 --> 00:42:25.440
just write it also h naught by dou s which
is 0. Because, that is what we said is constant
00:42:25.440 --> 00:42:40.470
minus u dou u by dou s minus 1 by rho dou
p by rho s this, what we get.
00:42:40.470 --> 00:42:58.530
And, now I will write the other one stream
normal direction dou s by dou s this going
00:42:58.530 --> 00:43:05.369
to
be equal to dou h naught by dou n minus u
00:43:05.369 --> 00:43:20.460
dou u by dou n minus 1 rho p by dou n. I have
this, now I will just go back to this dou
00:43:20.460 --> 00:43:24.750
s by dou stream line this any problem, this
is dou
00:43:24.750 --> 00:43:37.950
s by dou n. Now, we look at this dou s by
dou s what we see is from my momentum
00:43:37.950 --> 00:43:41.630
equation s momentum equation, that is going
to become 0.
00:43:41.630 --> 00:43:48.579
.If I take this rho 1 by rho here this is
equal to that other term and the one is negative
00:43:48.579 --> 00:43:50.740
of
the other these two will give out to be equal
00:43:50.740 --> 00:44:00.940
to 0. So, I just say equal to 0 from s
momentum equation, from s momentum equation
00:44:00.940 --> 00:44:08.200
that is 0 that cannot be said about the
other term. If we look at the other term,
00:44:08.200 --> 00:44:15.549
I can write something for dou p by dou n that
will be rho u square by r, even if I write
00:44:15.549 --> 00:44:18.930
it that is not going to be 0 that term will
just
00:44:18.930 --> 00:44:24.960
take.
So, this is going to be equal to dou h naught
00:44:24.960 --> 00:44:42.630
by dou n minus u dou u by dou n plus u
square by r, just have this particular form.
00:44:42.630 --> 00:44:54.390
Now, since dou s by dou s is 0 and s can vary
only along s or n, entropy can vary along
00:44:54.390 --> 00:44:59.020
stream wise stream normal direction and we
say it does not vary along stream not stream
00:44:59.020 --> 00:45:04.869
wise direction. I can rewrite this partial
derivative as total derivative, now same thing
00:45:04.869 --> 00:45:11.360
can be done for h naught also this can also
be written as total derivative.
00:45:11.360 --> 00:45:33.980
If I do that t d s by d n is equal to d h
naught by d n plus u into u by r minus dou
00:45:33.980 --> 00:45:48.261
u by dou
n, I will just say here that this term can
00:45:48.261 --> 00:45:56.079
be shown to be equal to minus of omega. The
virtuosity vector this is actually should
00:45:56.079 --> 00:45:58.920
be a vector, but since it is 2 d problem,
the vector
00:45:58.920 --> 00:46:02.970
is also always going to be pointing out, of
the board and its going to be having a negative
00:46:02.970 --> 00:46:06.350
value because, in this case it will be pointing
inside the board into the board.
00:46:06.350 --> 00:46:11.819
So, I will keep it as is this is from definition
from virtuosity in stream wise stream
00:46:11.819 --> 00:46:16.730
normal coordinate system, you will get it
to be this will just write it like this. So,
00:46:16.730 --> 00:46:26.800
I finally
end up with the form I may check whether I
00:46:26.800 --> 00:46:44.059
made a mistakes or not sure of this whether
it is plus or minus
00:46:44.059 --> 00:46:47.450
as of, now I will assume this plus it could
be minus just a minute will
00:46:47.450 --> 00:47:02.380
just go through this d s by d n t d s by d
n, I will just keep it as minus there.
00:47:02.380 --> 00:47:14.770
So, it will become plus u omega is equal to
d h naught by d n we get to this expression
00:47:14.770 --> 00:47:19.140
its
look like I made one small mistake in my derivation,
00:47:19.140 --> 00:47:26.369
but I am not sure. So, I believe this
should be minus omega here, but expression
00:47:26.369 --> 00:47:29.890
seems to be correct finally. So, maybe I
made a mistake here or not made a mistake
00:47:29.890 --> 00:47:35.030
I am not sure about this, whether it is minus
or plus I am not sure, but this is the final
00:47:35.030 --> 00:47:38.830
expression I am getting here.
But, the physics I want to talk will not change
00:47:38.830 --> 00:47:44.530
here as of now there might be a this could
be a minus or a plus currently will just leave
00:47:44.530 --> 00:47:48.160
it there physics, I want to talk about is
if I
00:47:48.160 --> 00:47:54.310
have enthalpy gradients in my flow from one
stream line to other stream line that is in
00:47:54.310 --> 00:48:02.500
.the normal stream, normal direction then
I may have entropy variation that is one way
00:48:02.500 --> 00:48:07.170
of
looking it or if I starting with constant
00:48:07.170 --> 00:48:12.130
enthalpy in all the stream lines.
If I have constant enthalpy in all the stream
00:48:12.130 --> 00:48:16.980
lines then if I have entropy gradient, then
I
00:48:16.980 --> 00:48:21.490
will definitely produce virtuosity in my flow.
This is a important statement of croccos
00:48:21.490 --> 00:48:27.890
theorem this is called the croccos theorem
typically most of the fluid flows which we
00:48:27.890 --> 00:48:35.589
start or work with will always have d h naught
by d n 0 and we also have d h naught by d
00:48:35.589 --> 00:48:40.020
s 0, that we already saw it is h naught does
not change along the flow as stream line.
00:48:40.020 --> 00:48:45.339
So, I am going to say that since we started
with enthalpy constant at the beginning it
00:48:45.339 --> 00:48:46.890
is
going to be remaining the same all through
00:48:46.890 --> 00:48:52.120
except, when there is a if that is again I
made
00:48:52.120 --> 00:48:59.309
this 0 then I will produce virtuosity in the
flow. If there is entropy gradient across
00:48:59.309 --> 00:49:04.089
stream
lines d by d n of s not equal to 0, if I have
00:49:04.089 --> 00:49:09.390
this not equal to 0 then I will definitely
produce virtuosity this is the main statement
00:49:09.390 --> 00:49:11.119
of croccos theorem.
.
00:49:11.119 --> 00:49:19.339
Now, I will just write the full or more complicated
expression for croccos theorem then
00:49:19.339 --> 00:49:43.670
will stop here and we look at the implications
of this next class. The full equation
00:49:43.670 --> 00:49:51.750
of
croccos theorem is t gradient of s plus u
00:49:51.750 --> 00:49:58.750
cross del cross u is equal to gradient and
h
00:49:58.750 --> 00:50:10.859
naught plus dou u by dou t this is, the full
croccos theorem expression. Where what we
00:50:10.859 --> 00:50:16.920
derived did not have any non stationary term
we removed that part and this is the most
00:50:16.920 --> 00:50:23.240
complicated form, where this is your virtuosity
omega vector this case.
00:50:23.240 --> 00:50:30.700
.We have this particular term what we are
seen again is if I start with the isenthalpic
00:50:30.700 --> 00:50:34.060
flow
or home enthalpy flow you are having referring
00:50:34.060 --> 00:50:40.690
into some other books, isenthalpic flow
then and I am having stationary flow then,
00:50:40.690 --> 00:50:43.849
I do not have these terms that 0 then if there
is
00:50:43.849 --> 00:50:50.079
a entropy gradient in my flow I will produce
virtuosity in the flow. This is the main
00:50:50.079 --> 00:50:54.190
statement from croccos theorem will keep,
it will look at examples of virtuosity
00:50:54.190 --> 00:51:02.580
produced in the flow next class and then start
going into frictional force and how to
00:51:02.580 --> 00:51:08.480
analyze flow with friction. This one small
part I wanted to discuss also by today as
00:51:08.480 --> 00:51:10.320
not
enough time and I do not want to hurry will
00:51:10.320 --> 00:51:17.049
do it next time. So, any question see you
people next class.
00:51:17.049 --> 00:51:18.049
.