WEBVTT
Kind: captions
Language: en
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And now, what we are going to see we are going
to see that how to write the equations for
00:00:30.980 --> 00:00:36.450
the multi phase flow for all these conditions
or all these flow regimes which we have learnt.
00:00:36.450 --> 00:00:43.480
Now, before going to that the flow equations,
I would first like to briefly revise some
00:00:43.480 --> 00:00:49.100
of the para things which we have already studied
maybe in the fluid mechanics, but to bring
00:00:49.100 --> 00:00:53.989
everyone at the same platform so that we can
understand the equation in more detail. I
00:00:53.989 --> 00:00:59.510
will do some revision and that revision is
very critical very necessary because we are
00:00:59.510 --> 00:01:05.320
going to build up our equations based on that
. Now first thing we always discussed in the
00:01:05.320 --> 00:01:12.780
multi phase flow or in any flow is that two
kind of measurement or two kind of the phenomenon
00:01:12.780 --> 00:01:20.650
or two kind of parameter; what we say is the
Eulerian flow and Lagrangian flow or a Eulerian
00:01:20.650 --> 00:01:25.230
measurement and Lagrangian measurement.
So, first what we are try going to understand
00:01:25.230 --> 00:01:30.540
that what is Eulerian what is Lagrangian.
So, that we should be very clear whether we
00:01:30.540 --> 00:01:35.830
are writing the equation in the Eulerian frame
of domain or we are writing in the Lagrangian
00:01:35.830 --> 00:01:40.960
frame of domain, whether we are doing the
measurement in Eulerian frame of domain or
00:01:40.960 --> 00:01:46.220
we are doing the measurement in Lagrangian
flow up domain . So, what we are going to
00:01:46.220 --> 00:01:58.890
do we are going to discuss about the Lagrangian
versus Eulerian .
00:01:58.890 --> 00:02:06.050
Now what is Lagrangian and what is Eulerian
. Now if we ; you might have studied about
00:02:06.050 --> 00:02:10.979
this in many places at the discrete places,
in your mass transfer, in your fluid mechanics
00:02:10.979 --> 00:02:15.060
and we have tried to derive the equations,
also we will again revise it because this
00:02:15.060 --> 00:02:20.950
is very very critical .
Now, what we are going to see is the equations,
00:02:20.950 --> 00:02:24.530
we are developing the measurement techniques,
we will be developing it will be either the
00:02:24.530 --> 00:02:30.220
Eulerian measurement or it will be the Lagrangian
measurement and we have to know, we should
00:02:30.220 --> 00:02:36.510
know; how to interchange the Eulerian information
to the Lagrangian information or the Lagrangian
00:02:36.510 --> 00:02:42.170
information to the Eulerian information. So,
that we can use the system or we can analyze
00:02:42.170 --> 00:02:46.870
the system in more detail.
Now, to understand this; what is Eulerian,
00:02:46.870 --> 00:02:53.790
what is Lagrangian , let us assume a system
in which a flow is taking place and the flow
00:02:53.790 --> 00:02:59.550
is changing with the position. Now we can
easily assume any diverging or converging
00:02:59.550 --> 00:03:05.370
section and if the section will be diverging,
definitely the area will change if the area
00:03:05.370 --> 00:03:10.110
will change my velocity is going to change.
So, suppose the fluid is flowing here in a
00:03:10.110 --> 00:03:16.209
diverging friction . So, what will happen?
Suppose, I am taking two places here; one
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is section one and another one is section
two . So, what will happen? Section one; suppose
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has a velocity V 1 , section two has a velocity
V 2 , why it is going to happen because my
00:03:29.660 --> 00:03:37.030
area has changed we know the equation of continuity
that A 1 V 1 will be equal to A 2 V 2, if
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the flow is incompressible or we can say that
overall mass transfer rate is going to be
00:03:42.270 --> 00:03:46.200
the same .
So, overall mass transfer is going to remain
00:03:46.200 --> 00:03:53.400
same. So, rho A 1 V 1 will be equal to rho
A 2 V 2 if rho is same rho; rho will be canceled
00:03:53.400 --> 00:03:59.980
out A 1 V 1 will be equal to A 2 V 2 , it
means I can write V 1 upon V 2 is nothing,
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but it will be equal to A 2 upon A 1 ok.
So, if your area will change your velocity
00:04:05.849 --> 00:04:12.349
will also change now it means if A 2 is higher
than the A 1, then what will happen the V
00:04:12.349 --> 00:04:18.549
1 will be higher than the V 2 ok. So, that
is going to happen. Now what we are going
00:04:18.549 --> 00:04:24.120
to see that if suppose, I am interested in
measuring the velocity at any of this point
00:04:24.120 --> 00:04:31.280
ok, say I am interested to measuring the velocity
at point one at any point here in between
00:04:31.280 --> 00:04:35.190
ok .
Now, what will happen suppose I have a probe
00:04:35.190 --> 00:04:40.650
which can measure the velocity say pitot tube
I can place the pitot tube at this place,
00:04:40.650 --> 00:04:44.810
this point and it will give me the velocity
at this point. So, suppose I am putting a
00:04:44.810 --> 00:04:54.400
pitot tube here , I hope everyone knows what
is pitot tube . So, I am putting a pitot tube
00:04:54.400 --> 00:05:00.370
along this point p and if I put a pitot tube
I will measure the velocity at that point.
00:05:00.370 --> 00:05:09.220
Now, if I assume that the flow is steady at
steady state
00:05:09.220 --> 00:05:17.320
it means what the V is not a function of time
ok. So, V will not change with the time. So,
00:05:17.320 --> 00:05:22.150
what will happen you will measure certain
velocity here? Now velocity at this point
00:05:22.150 --> 00:05:26.830
and velocity at this point is going to be
different why because the area is keep on
00:05:26.830 --> 00:05:32.910
changing now what you are measuring here is
the velocity at a fixed place ok.
00:05:32.910 --> 00:05:38.699
So, that is called Eulerian velocity. So,
what you will measure say pitot tube is giving
00:05:38.699 --> 00:05:44.720
a velocity V p which is the I am saying the
point velocity the velocity at a point p this
00:05:44.720 --> 00:05:56.710
V p is nothing, but is the Eulerian velocity
, it means what when we fix our frame of reference
00:05:56.710 --> 00:06:02.860
or when we fix our position or specify our
location and measure something at that location
00:06:02.860 --> 00:06:06.009
is called Eulerian set ok or Eulerian measurement
or Eulerian quantity.
00:06:06.009 --> 00:06:13.530
So, here we are measuring the velocity now
I have fixed my position. So, that what we
00:06:13.530 --> 00:06:20.470
are going to see is the this velocity how
it is changing with the time now if with the
00:06:20.470 --> 00:06:25.720
time this flow is at steady state, then dou
V by dou t will be 0 and you will be measuring
00:06:25.720 --> 00:06:30.960
the same velocity throughout the domain or
throughout the time, you are doing the measurement
00:06:30.960 --> 00:06:36.580
ok, if it is not efficient the flow is not
a steady state then with the time, you will
00:06:36.580 --> 00:06:39.970
see that how the velocity is changing at that
location.
00:06:39.970 --> 00:06:46.160
So, suppose a flow is not steady ok or is
the unsteady state flow then you will see
00:06:46.160 --> 00:06:54.100
at point p how the velocity is changing with
the time this is time and this is V p for
00:06:54.100 --> 00:07:06.060
unsteady state flow
for a steady state flow what you will see
00:07:06.060 --> 00:07:14.870
you will see this is V p and this is the time
, you will see there is no change in the velocity
00:07:14.870 --> 00:07:24.340
and this is for a steady state .
So, these all are called you Eulerian or Eulerian
00:07:24.340 --> 00:07:38.199
velocity
or Eulerian velocity where the position has
00:07:38.199 --> 00:07:44.080
been fixed ok, if I change the position if
I suppose, now if I am measuring instead of
00:07:44.080 --> 00:07:57.270
position p, I am measuring putting up new
pitot tube which is at a position s ok, I
00:07:57.270 --> 00:08:05.300
will see another velocity V s with time if
the flow is steady state this V s will be
00:08:05.300 --> 00:08:11.130
there , it can be more than V p, it can be
lower than the V p, it can be same to V p
00:08:11.130 --> 00:08:16.449
depending upon the flow condition ok whether
the flow is laminar or flow is turbulent we
00:08:16.449 --> 00:08:21.190
know, if the flow will be laminar, then what
you will see you will see a parabolic profile.
00:08:21.190 --> 00:08:25.680
So, definitely if you are measuring at the
center velocity will be higher if you are
00:08:25.680 --> 00:08:28.020
measuring near the wall, velocity will be
lower.
00:08:28.020 --> 00:08:32.789
If your flow profile is turbulent then what
will happen that near the wall you will see
00:08:32.789 --> 00:08:37.620
a tilted profile where the velocity will not
be the same, but at the central portion the
00:08:37.620 --> 00:08:42.270
velocity will almost be same everywhere in
that case, if the 2 point is between this
00:08:42.270 --> 00:08:47.050
flat region your V p and V s will be the same
if you are a laminar flow you are going to
00:08:47.050 --> 00:08:51.680
have a parabolic profile V p and V s will
be different. So, what we will see we will
00:08:51.680 --> 00:08:56.880
see that how the velocity is changing with
the location. So, if I change the location
00:08:56.880 --> 00:09:03.710
the velocity is changed for a steady state
for unsteady state flow, the velocity can
00:09:03.710 --> 00:09:09.750
change even at the same location with the
time ok for a steady state flow the velocity
00:09:09.750 --> 00:09:13.850
at the same location with the time will be
this same, but if you change the location
00:09:13.850 --> 00:09:18.380
it may change ok.
So, these all are called Eulerian field or
00:09:18.380 --> 00:09:23.880
Eulerian measurement or Eulerian velocity,
but the problem is our most of the conservation
00:09:23.880 --> 00:09:30.230
law whether the conservation of mass conservation
of momentum conservation of energy conservation
00:09:30.230 --> 00:09:36.760
of angular momentum all those equations are
written in terms of the Lagrangian field ok
00:09:36.760 --> 00:09:42.890
what do you mean by the Lagrangian domain
it means if I am measuring the velocity of
00:09:42.890 --> 00:09:49.130
a molecule. So, what will happen if the flow
is taking place in this system then the Lagrangian
00:09:49.130 --> 00:09:54.450
system if I am talking about if I do to the
same diverging section ?
00:09:54.450 --> 00:10:02.560
Say if one section section two , if I am talking
about a Lagrangian system then what will happen;
00:10:02.560 --> 00:10:08.570
if Lagrangian means you are sitting on a molecule
of a fluid element and you are seeing, how
00:10:08.570 --> 00:10:12.850
the velocity of that fluid element is changing
with the time.
00:10:12.850 --> 00:10:17.850
It means; suppose there is a molecule of the
fluid element, I am sitting on top of it,
00:10:17.850 --> 00:10:23.190
what will happen? I will see that though the
flow is steady state my velocity is changing
00:10:23.190 --> 00:10:55.520
with that time ok, please remember, this though
the flow is steady state ok ok and this is
00:10:55.520 --> 00:11:01.540
very critical. So, though the flow is steady
state the velocity will keep on changing with
00:11:01.540 --> 00:11:06.220
the time if I am sitting on the molecule.
Why because the area is changing my velocity
00:11:06.220 --> 00:11:17.050
is going to change and this is called the
Lagrangian velocity . So, what is happening
00:11:17.050 --> 00:11:23.500
my domain is not fixed and domain is changing
with the time and because the domain is changing
00:11:23.500 --> 00:11:28.510
my environment is also changing with the time
and that is the reason that I am seeing a
00:11:28.510 --> 00:11:34.510
different velocity field altogether I am not
measuring the velocity, I am not seeing the
00:11:34.510 --> 00:11:39.459
velocity at a fixed domain I am changing the
velocity with the time.
00:11:39.459 --> 00:11:45.440
It means what what is changing the velocity
is changing with the time and the location
00:11:45.440 --> 00:11:51.230
of the molecule is also changing, it means
here, if I am aging the velocity, this velocity
00:11:51.230 --> 00:11:58.040
will be the function of x, y, z and time t
, it means if I have to write in terms of
00:11:58.040 --> 00:12:07.180
the derivative I will write it as DV by Dt
it will be what it will be dou V by dou x
00:12:07.180 --> 00:12:16.600
into dou x upon dou t ok because position
is changing with the time dou V by dou y into
00:12:16.600 --> 00:12:27.640
dou y upon dou t plus dou V upon dou z into
dou z upon dou t plus dou V upon dou t , how
00:12:27.640 --> 00:12:32.709
this is changing velocity is changing with
the time ok. So, if that is also happening.
00:12:32.709 --> 00:12:40.810
So, now this is called actually the substantial
derivative or material derivative which is
00:12:40.810 --> 00:12:46.250
being widely used to understand the Lagrangian
velocity field that; what is the Lagrangian
00:12:46.250 --> 00:12:52.880
velocity. This Dx by Dt will be equal to what
it will be the velocity if V is a vector we
00:12:52.880 --> 00:13:02.490
know that and has a 3 component V x V y V
z , then what will we can write it as a vx
00:13:02.490 --> 00:13:15.380
dou V upon dou x plus vy dou V upon dou y
plus vz dou V upon dou z plus dou V upon dou
00:13:15.380 --> 00:13:20.180
t .
Where V x V y and V z at V is nothing, but
00:13:20.180 --> 00:13:31.440
it is a velocity of the fluid element
ok when it is velocity of the fluid element,
00:13:31.440 --> 00:13:37.120
this DV by Dt will be equal to the substantial
derivative or the material derivative in which
00:13:37.120 --> 00:13:43.760
this is what is called the Eulerian part where
we are seeing that how velocity is changing
00:13:43.760 --> 00:13:49.420
with the position and position is changing
with the time in all these 3 and how the velocity
00:13:49.420 --> 00:13:55.529
is also changing with the time.
So, this part if suppose you are doing the
00:13:55.529 --> 00:13:59.579
Eulerian measurement, if I am measuring the
pitot tube, what I am measuring is this part
00:13:59.579 --> 00:14:04.250
that how my velocity is changing with the
time if I put a pitot tube at a particular
00:14:04.250 --> 00:14:09.380
location, I am not changing the position here
in this part. So, what is happening at a fixed
00:14:09.380 --> 00:14:14.910
location, I am seeing that how the particle
velocity or how the fluid velocity is changing
00:14:14.910 --> 00:14:26.530
with the time this is Eulerian velocity and
this is what is the Lagrangian velocity DV
00:14:26.530 --> 00:14:32.139
quantity or the material derivative and whatever
the conservation equation.
00:14:32.139 --> 00:14:37.800
We see we all see in terms of the substantial
derivative where the fluid element velocity
00:14:37.800 --> 00:14:43.360
comes into the picture or the fluid velocity
also comes into the picture and that fluid
00:14:43.360 --> 00:14:48.959
velocity is being measured by how the fluid
element is changing the position with the
00:14:48.959 --> 00:14:55.020
time that is nothing, but Dx upon Dt which
is nothing, but the x velocity or x element
00:14:55.020 --> 00:15:02.610
of the velocity of the fluid V DV Dy by Dt
or dou y by Dt which is the y element velocity
00:15:02.610 --> 00:15:07.709
of the fluid dou z by dou t which is the z
element velocity of the fluid ok.
00:15:07.709 --> 00:15:14.220
So, that is the component similarly any quantity
any parameter, you can write it in this form.
00:15:14.220 --> 00:15:23.571
So, I can say that dou phi by Dt where phi
is any quantity it can be written as vx dou
00:15:23.571 --> 00:15:35.399
phi by dou x plus vy dou phi by dou y plus
vz dou phi by dou z plus dou phi by dou t
00:15:35.399 --> 00:15:40.699
, but this part is the Eulerian part if you
do the Eulerian measurement you actually get
00:15:40.699 --> 00:15:45.440
only this ok while writing the conservation
equation we see all this.
00:15:45.440 --> 00:15:51.600
So, what we need to learn how to convert these
two parameters and how to start the measurement
00:15:51.600 --> 00:15:58.260
in such a way that the Lagrangian information
or the Eulerian information can be interchangeably
00:15:58.260 --> 00:16:04.600
used to understand about the system and this
whole equation this whole equation whatever
00:16:04.600 --> 00:16:16.000
in terms of the velocity we know it is also
called as Euler's acceleration equation and
00:16:16.000 --> 00:16:21.490
you have already have used in your fluid mechanics
once you do the fluid mechanics you derive
00:16:21.490 --> 00:16:25.850
the Navier stoke equation, you derive the
Eulerian equation and you mean this equations
00:16:25.850 --> 00:16:30.279
like if the Eulerian acceleration is 0, how
the equations will be converted you do all
00:16:30.279 --> 00:16:35.830
sort of things. So, this is called Eulerian
acceleration equation in which this is stilled
00:16:35.830 --> 00:16:39.459
that how the fluid element is being actually
accelerating.
00:16:39.459 --> 00:16:46.180
So, this is; what is the Lagrangian domain.
What is the Eulerian domain to summarize again
00:16:46.180 --> 00:16:53.010
Eulerian domain means if your location is
fixed, if your frame of reference is fixed
00:16:53.010 --> 00:16:59.430
and you are seeing the element of the fluid
velocity or how the fluid velocity is changing
00:16:59.430 --> 00:17:06.029
in that element, in that fixed element is
called Eulerian. So, what is happening the
00:17:06.029 --> 00:17:12.869
fluid element is keep on changing, but the
location is fixed in the Lagrangian field,
00:17:12.869 --> 00:17:18.579
you are seeing that how the velocity of a
fluid element is changing with the time. So,
00:17:18.579 --> 00:17:24.989
what is happening fluid element is fixed,
but the location of the fluid element or the
00:17:24.989 --> 00:17:27.620
position of the fluid element is changing
with the time .
00:17:27.620 --> 00:17:33.149
So, that is called when the position of the
fluid element changes with the time it is
00:17:33.149 --> 00:17:39.720
called Lagrangian measurement or Lagrangian
velocity or Lagrangian quantity when the element
00:17:39.720 --> 00:17:45.529
is fixed observation element is fixed and
the fluid element in that observation circle
00:17:45.529 --> 00:17:52.809
or in that observation frame is keep on changing
and we see with the time, how the quantity
00:17:52.809 --> 00:17:57.450
is changing in that fixed frame is called
Eulerian measurement or Eulerian quantity.
00:17:57.450 --> 00:18:06.649
So, these are the two quantities which we
need to understand while doing most of the
00:18:06.649 --> 00:18:12.820
measurements we have, we measure in the Eulerian
field domain like thermocouple like pitot
00:18:12.820 --> 00:18:20.039
tube like pressure gauge, what we do in thermocouple?
We fix the thermocouple probe or the that
00:18:20.039 --> 00:18:25.269
a particular location. So, we measure the
temperature at that location.
00:18:25.269 --> 00:18:31.769
In case of pitot tube, you fix the probe at
a particular location you measure the velocity
00:18:31.769 --> 00:18:39.299
at that particular location and what you measure
is the Eulerian field most of our conservation
00:18:39.299 --> 00:18:44.720
equation is written in terms of the Lagrangian
field and that is why there is always a confusion
00:18:44.720 --> 00:18:49.919
between the Eulerian field measurement and
Lagrangian field measurement and I hope that
00:18:49.919 --> 00:18:54.320
now it will be little bit clarified.
So, what we are going to do we are now going
00:18:54.320 --> 00:18:59.309
to see that how to interchangeably use the
equations how to write the equation for the
00:18:59.309 --> 00:19:04.239
different multi phase flow domain now before
that what we are going to do we are going
00:19:04.239 --> 00:19:12.019
to derive the equation and that is very common
equation is called Navier stoke equation and
00:19:12.019 --> 00:19:17.029
the equation of conservation of mass
So, first we will see that how that equations
00:19:17.029 --> 00:19:23.139
are derived. So, that once for all our basics
are clear and then by using that equation
00:19:23.139 --> 00:19:29.139
of momentum, we will try to see for different
multi phase flow regime how that the same
00:19:29.139 --> 00:19:35.289
equation can be modified what we will try
to do we will try to fix our discussion in
00:19:35.289 --> 00:19:41.789
the single direction or in the one dimensional
domain you can convert that the same equation
00:19:41.789 --> 00:19:47.679
very easily in two dimensional and 3 dimensional
domain; however, it will be very numerically
00:19:47.679 --> 00:19:54.149
very challenging to get the solution its almost
impossible to get an analytical solution for
00:19:54.149 --> 00:19:59.090
that that is the reason that why we will be
limiting our discussion for this course in
00:19:59.090 --> 00:20:04.379
A 1 dimensional domain. So, that we can solve
the problem, but one can easily extend the
00:20:04.379 --> 00:20:10.869
same discussion in two dimensional and 3 dimensional
domains, but then you need numerical approaches
00:20:10.869 --> 00:20:16.840
numerical methods or advanced computational
methods or computational methods to solve
00:20:16.840 --> 00:20:20.559
those equations.
Second thing; what we are going to do we will
00:20:20.559 --> 00:20:26.059
be limiting our discussion most of the time
in the laminar flow regime, again for the
00:20:26.059 --> 00:20:31.730
sake of simplicity of the problem. So, that
we can understand the things because turbulent
00:20:31.730 --> 00:20:37.179
will be completely different ball field and
if you go in the turbulent domain your number
00:20:37.179 --> 00:20:42.010
of equations will be increased your number
of variable will be increased and it would
00:20:42.010 --> 00:20:47.961
be very difficult to get an analytical solution
for that. So, one is interested can do that
00:20:47.961 --> 00:20:54.179
we will also try to look into the computational
fluid dynamics method, how to solve the computationally
00:20:54.179 --> 00:21:00.249
the equations; we will briefly try to see
towards that, but that is not the main focus
00:21:00.249 --> 00:21:05.830
of this course, but the main focus of this
course is to write the equation in one dimensional
00:21:05.830 --> 00:21:12.100
field in one dimensional domain for laminar
flow equation if you get the turbulent flow;
00:21:12.100 --> 00:21:18.059
do not be worried, it is exactly same you
are solving just few more equations to account
00:21:18.059 --> 00:21:21.889
the turbulent field.
We will also try to have a small discussion
00:21:21.889 --> 00:21:26.590
on the turbulent field, but most of the time
we will be limiting our discussion on the
00:21:26.590 --> 00:21:32.129
laminar flow field. So, these all I am classifying
or clarifying. So, that in future classes
00:21:32.129 --> 00:21:36.769
there will be no confusion and we will be
keep on carrying our discussion on the laminar
00:21:36.769 --> 00:21:42.429
flow field ok, one dimensional laminar flow
field that is what we are going to cover turbulent
00:21:42.429 --> 00:21:47.760
flow field, if you read the paper if you read
the book some books discuss about the turbulent
00:21:47.760 --> 00:21:53.239
flow field do not get confused you can easily
solve it only thing is the analytical solution
00:21:53.239 --> 00:21:57.960
will not be possible and you have to depend
on the computational fluid dynamics method
00:21:57.960 --> 00:22:03.429
or advanced numerical approach methods to
solve those problem people use alias people
00:22:03.429 --> 00:22:09.580
use dns people use some computational field
in the finite volume method or finite difference
00:22:09.580 --> 00:22:15.110
method or computational approach to solve
those equations, but we will limit ourselves
00:22:15.110 --> 00:22:21.019
in one dimensional laminar flow.
So, before going to the multi phase flow,
00:22:21.019 --> 00:22:26.480
as I said that what we are going to see is
how this Lagrangian field Eulerian field is
00:22:26.480 --> 00:22:34.149
being kind of generated how to write the
equations properly and how the basic notion
00:22:34.149 --> 00:22:42.009
of how to develop and how to write first the
general momentum equation or general mass
00:22:42.009 --> 00:22:47.559
conservation equation. So for that; the very
basic thing which has been started is called
00:22:47.559 --> 00:23:03.889
the Reynolds transport theorem ok.
So, what we are going to do is theorem also
00:23:03.889 --> 00:23:10.349
known as widely RRT ok. So, what Reynolds
transport theorem says, suppose, there is
00:23:10.349 --> 00:23:17.909
a fluid element which is moving a fluid is
flowing here and if I take a small element
00:23:17.909 --> 00:23:26.669
here a small say volume which is being there
and if suppose at time t equal to t plus delta
00:23:26.669 --> 00:23:32.649
t. this volume is being kept, then what will
happen that you will see that it will move
00:23:32.649 --> 00:23:38.110
to a certain distance and it is the same element,
I am saying that it has moved to the certain
00:23:38.110 --> 00:23:42.460
distance and what will happen you will see
this is again moving .
00:23:42.460 --> 00:23:49.289
Now, what we can do we can divide this element
in 3 part; one is the part one , this is the
00:23:49.289 --> 00:23:54.989
part two which is actually the intercept and
this is the part 3 . So, what is happening
00:23:54.989 --> 00:24:00.299
a fluid element is there which is being placed
in a moving flow and this will be swept with
00:24:00.299 --> 00:24:05.349
a velocity of the fluid and because of that
swept some of the property will actually move
00:24:05.349 --> 00:24:11.759
outside of this control surface from the outer
side and some of the property, from this control
00:24:11.759 --> 00:24:17.009
surface fund will move to the inside of the
control volume two. So, there is a control
00:24:17.009 --> 00:24:22.889
volume and these are two; one and two is the
two control surfaces which is moving in the
00:24:22.889 --> 00:24:28.690
property is moving in from control surface
one and moving out from control surface two.
00:24:28.690 --> 00:24:34.159
So, now if I have to write here; so, if I
see that suppose if I am talking about any
00:24:34.159 --> 00:24:39.039
extensive property which is a property of
mass and I am denoting it with a number and
00:24:39.039 --> 00:24:52.989
so, this is extensive property ok. So, this
can be energy mass this can be momentum this
00:24:52.989 --> 00:25:02.190
can be energy and I am defining one more quantity
eta which is in intensive property or I will
00:25:02.190 --> 00:25:13.419
say the per unit mass quantity ,
it means if I am defining and as a mass, then
00:25:13.419 --> 00:25:20.340
eta will be mass upon mass it means the value
will be equal to 1 , if I am defining the
00:25:20.340 --> 00:25:29.499
momentum it will be momentum per unit mass
momentum is defined by mV divided by m it
00:25:29.499 --> 00:25:34.279
means this will be cancelled out it will be
the velocity and. So, on if you define the
00:25:34.279 --> 00:25:38.720
energy and all is all. So, on
So, if you define kinetic energy t will be
00:25:38.720 --> 00:25:45.139
half mV square it will be only half V square.
So, and. So, on you can define the quantity
00:25:45.139 --> 00:25:50.639
eta which is intensive property per unit mass
quantity and is the extensive property which
00:25:50.639 --> 00:25:55.749
is mass energy momentum any quantity we are
defining. So, we are defining a general equation
00:25:55.749 --> 00:26:01.570
. So, suppose a quantity N which is being
swept and it is being from control surface
00:26:01.570 --> 00:26:06.490
one, it is entering to the control through
volume two and then it is going out again
00:26:06.490 --> 00:26:11.989
through the control surface 3. So, what will
happen at any time t quantity N suppose at
00:26:11.989 --> 00:26:21.309
time t the quantity N will be what the quantity
N will be in what will be the quantity N in
00:26:21.309 --> 00:26:31.539
control surface two at time t minus how much
is come in from control surface one at time
00:26:31.539 --> 00:26:35.409
t .
Now, at time t plus delta t if I just think
00:26:35.409 --> 00:26:40.110
about at t plus delta t how the things will
happen this has been changed the location
00:26:40.110 --> 00:26:47.210
has been moved. So, what will happen this
will be moved again and this quantity two
00:26:47.210 --> 00:26:55.799
will be moved and it will be seeing that how
much of this quantity has moved at t plus
00:26:55.799 --> 00:27:02.639
delta t from control surface 3 and how much
has come in in the control surface in the
00:27:02.639 --> 00:27:10.369
control volume two t plus delta t . So, that
will be at time t equal to at an time equal
00:27:10.369 --> 00:27:16.909
to t, the quantity and or the system on this
N, this will be the total change in the quantity
00:27:16.909 --> 00:27:22.489
which will be coming in at t plus delta t,
how much quantity will be going out that will
00:27:22.489 --> 00:27:26.339
be this quantity.
So, rate of change of this quantity will be
00:27:26.339 --> 00:27:35.129
dN upon Dt and if I assume that this delta
t is very small then I can say that delta
00:27:35.129 --> 00:27:44.139
N delta N and upon delta t . So, this will
be equal to delta N will be nothing, but N
00:27:44.139 --> 00:27:54.740
t plus delta t N t plus delta t minus N t
upon delta t , I can write it in this way,
00:27:54.740 --> 00:28:08.739
now if I write it N t plus delta t is what.
N t plus delta t is N t 3 ok into t plus delta
00:28:08.739 --> 00:28:31.210
t ok minus ; it will be N 2 t plus delta t
ok, then this will be this minus again N t
00:28:31.210 --> 00:28:46.470
at two at N t ok , then again it will be plus
N of one into t sorry this will be N 2 and
00:28:46.470 --> 00:28:52.599
this will be N one t, this will be in this
case ok upon delta t .
00:28:52.599 --> 00:28:58.450
So, what you can say you can say that the
system final minus initial I can write it
00:28:58.450 --> 00:29:13.070
as N to t minus N 2 t plus delta t upon delta
t this will be the one then I can write it
00:29:13.070 --> 00:29:25.549
N , sorry, I can write it this will be again
, there is something this is minus, this is
00:29:25.549 --> 00:29:36.870
plus . Now this will be plus N 3 t plus delta
t upon delta t which is moving out of the
00:29:36.870 --> 00:29:46.639
control surface known minus and one t upon
delta t these are the 3 quantities.
00:29:46.639 --> 00:29:52.460
Now, this is what this is the control volume
quantity that how much is the quantity is
00:29:52.460 --> 00:29:59.399
moving out of the control volume in per unit
time. So, I can write it out this quantity
00:29:59.399 --> 00:30:08.370
as dou by dou t of integral control volume
integral of the same quantity rho in terms
00:30:08.370 --> 00:30:13.369
of the intrinsic property which is eta into
DV ok.
00:30:13.369 --> 00:30:20.309
So, this is how much of that quantity this
quantity is moving per unit volume because
00:30:20.309 --> 00:30:26.229
section two is a control volume. So, I have
multiplied with by the rho and I have just
00:30:26.229 --> 00:30:31.360
made it the in quantity N and in terms of
the eta. So, it means eta is what quantity
00:30:31.360 --> 00:30:36.200
per unit mass I have multiplied kg per unit
volume. So, mass mass will be cancelled out
00:30:36.200 --> 00:30:41.820
it will be carried as N capital N ok if you
write it it in this way and then per unit
00:30:41.820 --> 00:30:46.390
volume how much is the quantity is changing
. So, I can write this whole term is dou by
00:30:46.390 --> 00:30:52.099
dou t of this term.
Now, what about the surface 3; it means how
00:30:52.099 --> 00:30:57.539
much of the same quantity is going out from
the surface 3 and how much of this quantity
00:30:57.539 --> 00:31:04.029
is coming in from the surface one. So, we
can write it out as how much quantity is going
00:31:04.029 --> 00:31:14.809
out from the surface separately. So, I can
write it out N 3 t plus delta t is equal to
00:31:14.809 --> 00:31:21.529
that how much is the quantity going out of
the surface it will be nothing, but the integral
00:31:21.529 --> 00:31:26.330
surface integral , it will be the control
surface integral of the quantity which is
00:31:26.330 --> 00:31:31.320
going out of the system.
Now, that will be nothing, but the eta which
00:31:31.320 --> 00:31:38.799
is the intrinsic property and the way it has
been swept out from that surface now the surface
00:31:38.799 --> 00:31:44.369
velocity, suppose is V, if V is the velocity
of the surface at which the surface is being
00:31:44.369 --> 00:31:49.519
swept or it means you can say the velocity
of the fluid then the surface will be swept
00:31:49.519 --> 00:31:56.289
as the velocity V in the direction normal
to the area. So, this we will write as a V
00:31:56.289 --> 00:32:02.519
dot N which is the unit vector in the normal
direction of the dA .
00:32:02.519 --> 00:32:10.479
So, that will be the amount of the quantity
which will be swept out from the surface 3
00:32:10.479 --> 00:32:22.860
similarly the similar amount will be swept
in from the surface one into the similar amount
00:32:22.860 --> 00:32:29.580
will be swept in from the surface one it means
the overall change will be written as this
00:32:29.580 --> 00:32:36.099
minus this can be written as this will be
the total swept or total change will be the
00:32:36.099 --> 00:32:42.700
change at the relative change in the velocity
it means V in minus V out and I am writing
00:32:42.700 --> 00:32:50.909
that as a V. So, that will be the V dot N
into dA where this d V is nothing, but the
00:32:50.909 --> 00:32:56.690
change in the velocity ok that how much is
the velocity of coming swept in and how much
00:32:56.690 --> 00:33:00.969
mass is being swept out.
So, that is the overall total change in the
00:33:00.969 --> 00:33:05.450
velocity how it is changing with the time
will be written here. So, this whole quantity
00:33:05.450 --> 00:33:17.390
can be written as dN by dt.
Dn by Dt of the system will be equal to how
00:33:17.390 --> 00:33:26.099
the quantity is changing with the time ok;
with the time how the quantity is changing
00:33:26.099 --> 00:33:33.629
in the control volume dou by dou t ok plus
in the control surface, how the quantity is
00:33:33.629 --> 00:33:42.090
being changed . So, this V dot N which is
a unit vector into t now this is called the
00:33:42.090 --> 00:34:00.029
overall Reynold transport theorem . Now any
transfer of the quantity can be derived through
00:34:00.029 --> 00:34:04.149
by using this Reynold transport.
So, what we have done we have done a very
00:34:04.149 --> 00:34:09.500
simplified approach we have taken we have
taken a moving fluid element in which I have
00:34:09.500 --> 00:34:14.200
taken an element here now element is moving
with the flow. So, what will happen some of
00:34:14.200 --> 00:34:18.780
the quantity, suppose any quantity whether
it is a mass, whether it is a energy, whether
00:34:18.780 --> 00:34:25.230
it is a momentum, in its angular momentum
any of the quantity will also be swept because
00:34:25.230 --> 00:34:29.440
of the fluid velocity element.
So, what will happen the swept will be because
00:34:29.440 --> 00:34:35.200
of the change in the property at the surface;
how much, it is being swept in from the left
00:34:35.200 --> 00:34:40.579
surface to the control volume how much it
is being swept from the control volume to
00:34:40.579 --> 00:34:47.000
the right surface ok and that is the total
change will be what we will see the effective
00:34:47.000 --> 00:34:52.679
change in the quantity. So, effective change
in the quantity with the time you will see
00:34:52.679 --> 00:34:59.349
how the quantity is changing within the control
volume and how the quantity is changing from
00:34:59.349 --> 00:35:03.109
the control surface ok.
Now, if you have a different six elements
00:35:03.109 --> 00:35:08.440
all the surface element will be changing to
quantity and you can have a proper equation
00:35:08.440 --> 00:35:14.190
that how any quantity is changing with the
time from the control surface and from the
00:35:14.190 --> 00:35:21.730
control volume. So, this is Reynold transport
theorem and you can derive all your conservation
00:35:21.730 --> 00:35:27.079
equations by using the Reynold transport theorem
ok. So, what we are going to first see that
00:35:27.079 --> 00:35:33.460
how to derive the mass conservation equation
by using the Reynold transport theorem. Now
00:35:33.460 --> 00:35:39.680
for mass conservation equation what will be
my property my property will be mass. So,
00:35:39.680 --> 00:35:47.430
N will be what N will be equal to mass that
is say written by m.
00:35:47.430 --> 00:35:52.570
So, what will be eta? Eta will be the intrinsic
property which will be per unit mass. So,
00:35:52.570 --> 00:35:58.869
eta will be equal to m upon m it means we
will write it as a one. So, in the same system
00:35:58.869 --> 00:36:03.960
same Reynold transport equation I am going
to replace the N and eta. So, what is going
00:36:03.960 --> 00:36:15.369
to be dm upon Dt . So, rate of change of mass
of the system is equal to dou by dou t of
00:36:15.369 --> 00:36:21.369
the control volume ok you will write eta will
be equal to one. So, this will be rho of the
00:36:21.369 --> 00:36:27.690
control volume DV dv is nothing, but the control
volume integral plus you are going to have
00:36:27.690 --> 00:36:36.480
a control surface eta will be equal to 1 V
dot N into dA .
00:36:36.480 --> 00:36:42.750
Now, what we can do we can this is independent
of the time. So, I can put the time inside
00:36:42.750 --> 00:36:49.619
the control volume is anywhere of the at arbitrary
. So, we can write it dou rho upon dou t in
00:36:49.619 --> 00:36:59.990
the control volume V and this will be the
control surface V dot N and is the direction
00:36:59.990 --> 00:37:04.780
normal to the surface dA. Now, we can use
the gauss divergence theorem .
00:37:04.780 --> 00:37:17.960
So, what does theorem do the theorem changes
the control surface integral to the control
00:37:17.960 --> 00:37:24.089
volume and if you apply the gauss divergence
theorem here this quantity will be written
00:37:24.089 --> 00:37:32.430
as exactly same control volume dou rho upon
dou t into DV plus this can be changed to
00:37:32.430 --> 00:37:39.440
the control volume integral and it will be
written as del dot V and rho will be also
00:37:39.440 --> 00:37:45.809
coming in here the rho will be there here
the rho will be there.
00:37:45.809 --> 00:37:59.890
So, this del dot V or it will be del dot rho
V here we have done this. So, this will be
00:37:59.890 --> 00:38:08.609
rho will also be here there will be a rho
here ok. So, that will be the control volume.
00:38:08.609 --> 00:38:13.080
So, this will be the control surface here.
So, because you are changing in the intrinsic
00:38:13.080 --> 00:38:21.049
property here the rho will be there it will
be there rho. So, it will be rho V into n.
00:38:21.049 --> 00:38:26.300
So, that is the whole Reynold transport theorem.
So, here the rho also will come here and you
00:38:26.300 --> 00:38:32.720
will see that what is del dot V rho now if
you do that I can write it under both the
00:38:32.720 --> 00:38:42.569
control volume integral. So, del rho upon
del t plus del dot rho V and this will be
00:38:42.569 --> 00:38:45.960
d v.
Now, this is control volume because we are
00:38:45.960 --> 00:38:51.010
using an arbitrary control volume system we
can do that integral this will be Dx Dy dz,
00:38:51.010 --> 00:38:56.730
we will integrate it, it will come at a delta
x delta y delta z and we can simplify it as
00:38:56.730 --> 00:39:07.500
a del rho upon del t plus del dot rho V will
be equal to 0 and this is nothing, but the
00:39:07.500 --> 00:39:16.700
mass conservation equation .
If the flow is incompressible then del rho
00:39:16.700 --> 00:39:24.910
by del t will be 0 rho will come out and del
dot V will come to 0 for incompressible flow
00:39:24.910 --> 00:39:38.840
ok. So, you can derive that how the mass conservation
equation can be derived from the Reynolds
00:39:38.840 --> 00:39:43.650
transport theorem again going back to this
eta how it rho comes because we are changing
00:39:43.650 --> 00:39:48.920
the property and to the intrinsic property
and that is why the rho has been multiplied
00:39:48.920 --> 00:39:54.819
here again ok. So, that because you are changing
the N to the eta ok that is why I forget here
00:39:54.819 --> 00:39:59.590
rho I introduce it and please be there and
I hope now there will be no confusion why
00:39:59.590 --> 00:40:02.980
the rho comes here.
So, what you have to do you have to just replace
00:40:02.980 --> 00:40:08.670
it if you replace it you will get and just
use that gauss divergence theorem which actually
00:40:08.670 --> 00:40:14.730
changes the surface integral to the control
volume integral gauss divergence theorem and
00:40:14.730 --> 00:40:20.309
it will be nothing, but del dot rho v. So,
by using that you can derive the mass conservation
00:40:20.309 --> 00:40:25.040
equation general mass conservation equation
you can simplify it for the incompressible
00:40:25.040 --> 00:40:30.010
flow where del rho by del t will be 0 and
then you can write it as del dot V will be
00:40:30.010 --> 00:40:35.290
equal to 0 because rho will be constant it
will come out and that you can take it as
00:40:35.290 --> 00:40:41.970
a 0 . So, right hand side. So, you will get
the del dot V equal to 0 which is the incompressible
00:40:41.970 --> 00:40:47.050
mass conservation equation for the incompressible
fluid or velocity continuity equation.
00:40:47.050 --> 00:40:52.800
So, this can be derived similarly one can
derive the momentum equation Navier stokes
00:40:52.800 --> 00:40:58.720
equation and for Navier stoke equation derivation
what you need to do you have to take the momentum
00:40:58.720 --> 00:41:04.420
.
It means the system property, N will be equal
00:41:04.420 --> 00:41:13.539
to mV and eta will be equal to V . So, what
we will do we will say dN upon Dt Reynold
00:41:13.539 --> 00:41:22.309
transport theorem of the system will be equal
to control volume dou by dou t of the control
00:41:22.309 --> 00:41:37.630
volume of rho eta d z plus control surface
rho eta V this is eta dot N into dA.
00:41:37.630 --> 00:41:48.010
Now, we will replace it for the momentum it
means this will be d upon Dt of mV will be
00:41:48.010 --> 00:41:58.910
equal to dou upon dou t of integral rho eta
is equal to V into DV plus control system
00:41:58.910 --> 00:42:09.869
rho V into V dot N into dA.
Now, again what I will do because this is
00:42:09.869 --> 00:42:14.960
the control volume this dou by dou t I can
take it inside. So, I can write it control
00:42:14.960 --> 00:42:23.769
volume dou by dou t of rho z into DV where
V is nothing, but the control volume integral
00:42:23.769 --> 00:42:35.400
plus control surface rho V into V dot N into
dA . So, again I can put that gauss divergence
00:42:35.400 --> 00:42:41.359
theorem and if I will put the gauss divergence
theorem this will be control volume dou by
00:42:41.359 --> 00:42:51.980
dou t of rho V into DV and this will be again
converted to the control volume and del dot
00:42:51.980 --> 00:43:04.569
rho V c DV ok, again the gauss divergence
theorem the control surface to the control
00:43:04.569 --> 00:43:11.990
volume now again what we can do.
We can say that d upon Dt of mg is nothing,
00:43:11.990 --> 00:43:26.849
but this will be equal to control volume integral
of dou rho V upon dou t plus del dot rho vv
00:43:26.849 --> 00:43:33.460
ok of the control volume DV .
Now, this is the right hand term term whatever
00:43:33.460 --> 00:43:39.839
we have seen here and this will be nothing,
but capital DV upon Dt that will be the substantial
00:43:39.839 --> 00:43:44.190
derivative of the velocity or you can say
that it will be the material derivative this
00:43:44.190 --> 00:43:49.890
whole term the right hand term we can write
it as rho this this term we can write it as
00:43:49.890 --> 00:43:57.650
rho DV upon Dt if rho is constant ok if rho
is not constant we can write it dou rho V
00:43:57.650 --> 00:44:02.900
upon dou t ok.
So, what we are going to do if rho is rho
00:44:02.900 --> 00:44:09.221
is incompressible we are writing going to
write it it is this way d rho DV upon Dt . Now
00:44:09.221 --> 00:44:12.730
DV upon Dt now what is the left hand side
term left hand side term if you will see I
00:44:12.730 --> 00:44:22.599
can write it d m upon Dt into V plus m DV
upon Dt the left hand side term this term
00:44:22.599 --> 00:44:25.529
.
Now, we know that rate of change of the mass
00:44:25.529 --> 00:44:30.910
this is V equal to 0 . So, what will happen
you will see m DV upon Dt which is nothing,
00:44:30.910 --> 00:44:38.119
but the force ok . So, left hand side term
is nothing, but the summation of all the forces
00:44:38.119 --> 00:44:43.849
which is acting on the element ok will be
equal to whatever the material derivative
00:44:43.849 --> 00:44:50.109
is DV upon Dt of the control volume interior.
So, now, on any element whatever the element
00:44:50.109 --> 00:44:56.130
we have thought any element what are the total
forces acting the total forces acting can
00:44:56.130 --> 00:45:04.579
be written in terms of the body forces plus
the surface forces Fg plus Fs . Now, let us
00:45:04.579 --> 00:45:09.200
simplify that what will be the body force
or instead of g, I will write it though let
00:45:09.200 --> 00:45:19.220
it right Fb which is the body force to the
surface force ok.
00:45:19.220 --> 00:45:26.730
So, Fb is the body force Fs is the surface
force now what is the body force which will
00:45:26.730 --> 00:45:32.309
be acting in absence of any other force it
will be nothing, but Fg which will be the
00:45:32.309 --> 00:45:37.079
gravitational force which will be acting on
the element now what are the surface forces.
00:45:37.079 --> 00:45:42.319
So, let us assume a simplified case and I
am assuming the cuboid .
00:45:42.319 --> 00:45:56.319
So, let us assume it is being kept inside
a moving fluid element ok and this is the
00:45:56.319 --> 00:46:01.690
direction x this is the direction y this is
the direction z ok.
00:46:01.690 --> 00:46:10.660
So, what is the body force acting on this
that Fg it will be nothing, but it will be
00:46:10.660 --> 00:46:22.850
rho into g into Dx , Dy Dz is this is the
Dx Dy and Dz a very small element, I am choosing
00:46:22.850 --> 00:46:29.980
Dx, Dy, Dz or delta x delta y delta z then
this will be equal to actually rho g delta
00:46:29.980 --> 00:46:37.730
x delta y delta z , if Dx divided is it is
very small infinitesimally small then we can
00:46:37.730 --> 00:46:42.530
simply write it in terms of the delta x delta
y delta z. So, that will be nothing, but the
00:46:42.530 --> 00:46:46.400
body force which will be acting on it ok on
the element.
00:46:46.400 --> 00:46:52.160
Now, what are the surface forces now if you
see that the surface forces the surface forces
00:46:52.160 --> 00:46:58.070
will acting on all the surfaces let us assume
a sink one surface the surface which is in
00:46:58.070 --> 00:47:07.369
the direction of state x the surface which
is in the direction of x this is Dz this is
00:47:07.369 --> 00:47:15.150
Dy and this is in the direction of x ok or
I can write it delta x delta y and delta z
00:47:15.150 --> 00:47:19.519
delta y and this is in the direction.
So, what are the forces which will be acting
00:47:19.519 --> 00:47:24.560
one will be acting which will be the normal
to this surface and that force is being given
00:47:24.560 --> 00:47:30.710
by sigma. So, I will write it as sigma at
x which is the force acting normal to the
00:47:30.710 --> 00:47:36.099
surface, then there will be shear forces will
be acting now the shear forces is what which
00:47:36.099 --> 00:47:42.539
is tau and the tau notation whatever we are
going to follow it is tau is what; it is a
00:47:42.539 --> 00:47:49.999
second order tensor .
What do you mean by second order tensor? It
00:47:49.999 --> 00:47:58.200
means we need two direction to specify distance
and number of component will be what how many
00:47:58.200 --> 00:48:03.799
number of component it will be the number
of dimensional space it means I will write
00:48:03.799 --> 00:48:11.529
it as . So, tau xy is what two , it is a second
order tensor it means you require two directions
00:48:11.529 --> 00:48:16.749
to specify the second order tensor and the
number of element will be equal to how much
00:48:16.749 --> 00:48:21.380
is the number of space you are using. So,
suppose if you are using in 3 dimensional
00:48:21.380 --> 00:48:26.460
space. So, number of the space will be 3 and
raise to the power N where N is nothing, but
00:48:26.460 --> 00:48:34.079
order of the tensor tensor ok.
So, it means if you have a 3 dimensional space
00:48:34.079 --> 00:48:39.030
order of tensor is two. So, you will have
nine component of the shear stress if you
00:48:39.030 --> 00:48:45.240
have A 2 dimensional domain and order is 2.
So, you will have 4 direct 4 component ok.
00:48:45.240 --> 00:48:51.259
So, tau will have 4 component and the notation
which we are going to follow is tau xy suppose
00:48:51.259 --> 00:48:57.809
if I am writing say tau xy , it means what
I am talking about I am talking about this
00:48:57.809 --> 00:49:12.650
is y momentum in x direction ok, it means
tau xy will be y direction momentum, it means
00:49:12.650 --> 00:49:17.410
the velocity will be in the y direction y
directional velocity it means vy component
00:49:17.410 --> 00:49:23.349
and effect of the vy component in the x direction.
So, that is the way we are going to distribute
00:49:23.349 --> 00:49:26.039
it.
Now, we are talking about in the x direction.
00:49:26.039 --> 00:49:32.790
So, there will be sigma xx which is the normal
stress component sigma xx, ok, sigma xx will
00:49:32.790 --> 00:49:37.150
be acting now that in terms of the force if
you have to write you have to multiply it
00:49:37.150 --> 00:49:41.539
with the area and area will be what which
will be the perpendicular to it. So, in terms
00:49:41.539 --> 00:49:47.970
of the force the area perpendicular to the
surface this will be delta y delta z ok. So,
00:49:47.970 --> 00:49:52.720
that will be the area perpendicular. So,
how much it will be this , it will be coming
00:49:52.720 --> 00:49:58.920
into the surface x now there will be the force
which will be in the x direction again which
00:49:58.920 --> 00:50:03.790
will be effect of the y momentum.
So, it will be tau xy which will be acting
00:50:03.790 --> 00:50:10.289
on it and tau xy suppose the direction will
be somewhere here that will be tau xy say
00:50:10.289 --> 00:50:17.980
tau xy and tau zx the direction will be somewhere
here that will be tau xz actually which will
00:50:17.980 --> 00:50:23.769
be the z momentum in the x direction y momentum
in the x direction these are the forces which
00:50:23.769 --> 00:50:27.839
will be acting in the x direction
Similarly, the forces will be acting in the
00:50:27.839 --> 00:50:32.660
y direction the forces will be acting in the
z direction, but I am just writing in one
00:50:32.660 --> 00:50:39.079
directional forces which is in the x direction.
So, what I can say I can say that total force
00:50:39.079 --> 00:50:44.340
which is acting in the direction of x that
will be written as sigma xx that would be
00:50:44.340 --> 00:50:50.980
multiplied by delta y delta z
Then it will be the force which will be acting
00:50:50.980 --> 00:50:57.249
in the again x direction which will be the
y momentum . So, x into y now that is the
00:50:57.249 --> 00:51:01.829
force which is the y directional force which
acting in the x direction. So, what will be
00:51:01.829 --> 00:51:06.279
the surface which will be acting here you
have to multiply with that surface and that
00:51:06.279 --> 00:51:13.730
surface will be what this is the y momentum
force it will be delta x into delta z similarly
00:51:13.730 --> 00:51:22.579
it will be tau xz that will be multiplied
by delta y into delta x .
00:51:22.579 --> 00:51:28.009
Similar; so, these are the forces which are
acting on the x side of the surface similar
00:51:28.009 --> 00:51:32.349
forces will be acting on the other side. So,
this is on this side of the surface whatever
00:51:32.349 --> 00:51:37.560
we have said. So, this side of the surface
similarly the forces will be the similar forces
00:51:37.560 --> 00:51:43.430
will be acting on this side. So, total net
change in the x direction force will be you
00:51:43.430 --> 00:51:49.599
have to just subtract it if you subtract it.
You will write it as sigma xx minus at say
00:51:49.599 --> 00:51:58.470
x minus sigma xx at x plus delta x ok you
multiply it by delta y delta z , the total
00:51:58.470 --> 00:52:09.259
forces surface forces I am adding it, it will
be tau xy at surface y minus tau xy at surface
00:52:09.259 --> 00:52:20.450
y plus delta y and this will be multiplied
by delta x delta z plus tau xz at xz minus
00:52:20.450 --> 00:52:31.490
tau xz at z j plus delta z upon multiply by
delta y delta x ok plus that is the total
00:52:31.490 --> 00:52:34.549
force.
So, total force this is the surface forces
00:52:34.549 --> 00:52:41.559
plus the body force body force is nothing,
but the Fg Fg can be written as Fg can be
00:52:41.559 --> 00:52:50.470
written as as we have defined it is nothing,
but Fg is nothing, but rho into g delta x
00:52:50.470 --> 00:52:57.740
delta y delta z now you can divide it by delta
y delta z delta x delta z that will be the
00:52:57.740 --> 00:53:06.319
F and you just divide it by the delta x delta
y delta z, this will be nothing, but you can
00:53:06.319 --> 00:53:14.500
write it sigma xx minus sigma xx this is at
x, this is as x plus delta x upon delta x
00:53:14.500 --> 00:53:30.910
plus tau xy at y minus tau xy at y plus delta
y upon delta y plus tau x z at z minus tau
00:53:30.910 --> 00:53:35.900
x z at z plus delta z upon delta j plus rho
g .
00:53:35.900 --> 00:53:42.829
So, you can write it as by using if delta
x delta y delta z are very small infinitesimally
00:53:42.829 --> 00:53:46.089
small.
You can write it in terms of the differential
00:53:46.089 --> 00:53:52.109
equations, we can write as d sigma xx , I
will write it in terms of the dou upon dou
00:53:52.109 --> 00:54:06.250
x minus dou tau xy upon dou y minus tau x
z upon dou z plus rho g ok. So, whole minus
00:54:06.250 --> 00:54:17.750
will come out it will be dou sigma xx upon
dou x plus dou tau xy upon tho y plus dou
00:54:17.750 --> 00:54:23.579
tau zz upon dou z now because we are writing
it in terms of the x direction gravity also
00:54:23.579 --> 00:54:28.220
I will write it it in the x direction and
instead of g I will write it as rho gx.
00:54:28.220 --> 00:54:34.380
So, what we have done total force which is
acting in the x direction similarly one can
00:54:34.380 --> 00:54:38.720
write it in for the y direction for the z
direction and that will be the total summation
00:54:38.720 --> 00:54:45.390
of the force and that is the force which we
are writing is integral control volume of
00:54:45.390 --> 00:54:54.720
d of V upon Dt , I am writing it for incompressible
flow of dou z ok now this divided by the total
00:54:54.720 --> 00:55:01.450
volume, there the total volume has been divided.
Now if you do that here delta V delta xy this
00:55:01.450 --> 00:55:10.480
is delta x delta y delta z is here and that
you are writing as minus of dou sigma xx upon
00:55:10.480 --> 00:55:21.280
dou x plus dou tau xy upon dou y plus dou
tau z z upon dou z plus rho gx .
00:55:21.280 --> 00:55:27.829
Now, if you do that if this is infinitesimally
small DV can be written as Dx Dy Dz if you
00:55:27.829 --> 00:55:32.830
integrate it it will be delta x delta y delta
z that can be cancelled out and you can write
00:55:32.830 --> 00:55:46.109
it as rho DV upon Dt is equal to minus of
dou sigma xx upon dou x plus dou tau xy upon
00:55:46.109 --> 00:55:54.759
dou y plus dou tau zz upon dou z plus rho
gx . Now the normal stress term we know that
00:55:54.759 --> 00:55:59.309
the normal stress is the combination of the
pressure plus the stress in the normal direction.
00:55:59.309 --> 00:56:14.779
So, I can write it as minus sigma t plus tau
xx ok, upon dou x plus tau xy upon dou y plus
00:56:14.779 --> 00:56:21.999
dou tau zz tau xz sorry.
You know this is xz everywhere sorry for this
00:56:21.999 --> 00:56:32.670
tau xz upon dou z plus rho gx .
So, you can write it as rho DV upon Dt will
00:56:32.670 --> 00:56:45.890
be equal to minus dou p upon dou x minus dou
tau x x upon dou x plus dou tau x y upon dou
00:56:45.890 --> 00:56:56.910
y plus dou tau xz upon dou z plus rho gx and
this term can written as minus dou p upon
00:56:56.910 --> 00:57:08.359
dou x minus del dot tau plus rho gx and this
is the x component of the tau. So, this is
00:57:08.359 --> 00:57:16.589
rho Dz upon Dt .
Now, the whole Navier stoke equation can be
00:57:16.589 --> 00:57:27.859
written as dou V upon Dt minus dou p minus
del dot tau plus rho g and this is nothing,
00:57:27.859 --> 00:57:35.249
but it is the momentum equation. Now if we
write it in terms of mu also constant then
00:57:35.249 --> 00:57:46.180
this whole equation can be written as rho
DV upon Dt will be minus dou p plus mu del
00:57:46.180 --> 00:57:54.960
two z because tau is equal to nothing, but
minus mu dou V . So, delta 2 V plus rho g
00:57:54.960 --> 00:58:08.750
and this is called Navier stokes equation
So, this has been derived from the RRT by
00:58:08.750 --> 00:58:14.509
doing the property balance and property is
nothing, but the momentum . So, that is the
00:58:14.509 --> 00:58:22.499
general equation which is called momentum
equation or general momentum equation and
00:58:22.499 --> 00:58:27.380
from momentum equation if you assume rho is
constant mu is constant you can come to the
00:58:27.380 --> 00:58:30.190
Navier stoke equation ok which will be in
this form.
00:58:30.190 --> 00:58:35.970
So, this is the 3 dimensional Navier stokes
equation has been derived from the basic
00:58:35.970 --> 00:58:41.829
Reynold transport theory where you assuming
a control volume in which some property and
00:58:41.829 --> 00:58:47.460
the properties momentum which is coming in
which is being there and then is going out
00:58:47.460 --> 00:58:52.019
from a control volume. So, how much property
is coming into the control volume how much
00:58:52.019 --> 00:58:57.849
property is going outside of the control volume
the net change is this or written it in this
00:58:57.849 --> 00:59:01.450
form.
Now, we are going to use this equation in
00:59:01.450 --> 00:59:06.470
terms of now this is DV by Dt again if you
open this DV by Dt that will go in terms of
00:59:06.470 --> 00:59:12.450
the Eulerian as well as the Lagrangian measurement
variables. Now what we are going to do we
00:59:12.450 --> 00:59:17.539
are going to use this equation we are going
to see that how the equation will be modified
00:59:17.539 --> 00:59:21.759
in case of multi phase flow.
But before doing that what I am going to do
00:59:21.759 --> 00:59:26.400
I am going to simplify the equation in the
one dimensional form this is the 3 dimensional
00:59:26.400 --> 00:59:32.030
equation if you want you can get the solution
for this, but you again need a numerical approach
00:59:32.030 --> 00:59:37.220
or computational fluid dynamics method to
solve this equation in the 3 dimensional domain
00:59:37.220 --> 00:59:42.040
now what I am going to do I am going to do
simplify the equation in the one dimensional
00:59:42.040 --> 00:59:45.000
form.
Now, if I have to simplify the equation in
00:59:45.000 --> 00:59:47.670
the one dimensional form what I am going to
do.
00:59:47.670 --> 00:59:59.569
This rho DV by Dt will be equal to minus dou
p by dou x minus I will write force in terms
00:59:59.569 --> 01:00:06.999
of the tau. So, this will be dou say tau upon
dou x I am writing it in the one dimensional
01:00:06.999 --> 01:00:13.390
plus rho g ok let us assume that this is the
my equation now what I am going to do I am
01:00:13.390 --> 01:00:17.640
writing it in terms of the one dimensional
again I am going to write it in terms of one
01:00:17.640 --> 01:00:24.670
dimensional. So, say in the direction of x
I am writing. So, dou zx into DV x upon Dt
01:00:24.670 --> 01:00:31.999
or Dx ok assuming that the flow is steady
state vx upon vt term I am neglecting the
01:00:31.999 --> 01:00:39.490
steady state term.
So, the steady state flow . So, the flow is
01:00:39.490 --> 01:00:43.940
steady state it means dou vx upon dou t we
are neglecting this is the one dimension.
01:00:43.940 --> 01:00:51.230
So, rho dou vx upon to vx upon dou x will
be equal to minus dou p upon dou x minus dou
01:00:51.230 --> 01:00:58.290
tau upon dou x plus rho gx ok I am writing
it in x direction ok.
01:00:58.290 --> 01:01:03.390
Now, what I am going to do I am going to multiply
with the area which is the cross sectional
01:01:03.390 --> 01:01:10.319
area everywhere I can do that into it now
if I do that area multiplication this is what
01:01:10.319 --> 01:01:15.940
this is nothing, but the mass flow rate and
I am going to denote it with m naught ok.
01:01:15.940 --> 01:01:21.180
So, this is how much of the solid is flowing
in the x almost the fluid is flowing in the
01:01:21.180 --> 01:01:38.759
x direction m naught d V x upon Dx area into
dou p upon dou x minus now area into gx.
01:01:38.759 --> 01:01:50.319
Now, A into Dx if I try to reduce it it will
be in terms of the perimeter if I am assuming
01:01:50.319 --> 01:01:55.829
that this x is very small element and I am
just doing the changes in a very small element,
01:01:55.829 --> 01:02:04.130
I can just cancel it out I can write it minus
a dou p upon dou x minus area upon Dx it will
01:02:04.130 --> 01:02:16.210
be perimeter which is p into dou tau A gx
ok. So, into m naught into DV x upon Dx many
01:02:16.210 --> 01:02:21.640
books write it in terms of the g in a standard
or multi phase flow books this some book also
01:02:21.640 --> 01:02:26.130
write it in terms of the g. So, do not get
confused, if you see the g symbol it is just
01:02:26.130 --> 01:02:30.740
nothing, but rho V into x.
Now, we can simplify it what we want again
01:02:30.740 --> 01:02:35.339
I am going to write it in terms of the dou
p by dou x because that is what we want to
01:02:35.339 --> 01:02:45.349
calculate. So, I will write it as minus m
naught into dvx upon Dx divided by area this
01:02:45.349 --> 01:02:55.460
will be minus p upon a into d tau plus rho
into gx . So, that will be what is the total
01:02:55.460 --> 01:03:00.660
pressure drop you are going to see and that
total pressure drop is going to be the function
01:03:00.660 --> 01:03:05.970
of all this ok.
Now, what if you will see here ok what we
01:03:05.970 --> 01:03:15.099
are going to do we are going to say this is
as dou p by Dx and because the pressure drop
01:03:15.099 --> 01:03:19.940
is going to take place it is going to be negative
and instead of writing final minus initial
01:03:19.940 --> 01:03:26.700
I am going to write initial final it means
everywhere I will take the minus sign out
01:03:26.700 --> 01:03:31.130
.
So, this will be DV x upon Dx plus p upon
01:03:31.130 --> 01:03:39.970
a d tau minus rho g x ok because this p is
now initial minus finest a final minus initial
01:03:39.970 --> 01:03:45.220
initial minus final. So, it will be in this
form, we can write it now if we write it;
01:03:45.220 --> 01:03:51.580
it in this form what we can say that the total
delta p across the pipeline in the single
01:03:51.580 --> 01:03:58.769
phase flow is going to be the function of
3 things one is the frictional parameter second
01:03:58.769 --> 01:04:04.720
one is the acceleration field because of the
acceleration because of the frictional and
01:04:04.720 --> 01:04:13.059
because of the gravitational.
So, dp by dx can be written as m naught upon
01:04:13.059 --> 01:04:26.519
a into dvx upon dx plus p upon a tau w minus
rho gx .
01:04:26.519 --> 01:04:42.549
So, this is acceleration
this is due to friction or viscous forces
01:04:42.549 --> 01:04:52.519
and this is due to gravitational forces or
gravity forces . So, the dp by dx in any pipeline
01:04:52.519 --> 01:04:56.710
is function of these 3 things whatever we
have discussed for the limiting conveying
01:04:56.710 --> 01:05:03.000
in the previous classes that this is the function
of acceleration fluid acceleration it is the
01:05:03.000 --> 01:05:06.539
function of the friction of the fluid it is
a function of the gravity.
01:05:06.539 --> 01:05:11.349
Now, if you have a multi phase flow what you
are going to have this is the phase one you
01:05:11.349 --> 01:05:18.030
will have the delta p because in the overall
delta p in a pipeline because of the acceleration
01:05:18.030 --> 01:05:25.130
of one phase acceleration of other phase or
second phase friction because of the phase
01:05:25.130 --> 01:05:31.329
one friction because of phase two gravity
force because of phase one gravity force because
01:05:31.329 --> 01:05:35.059
of phase two in single phase flow this is
going to be the function this.
01:05:35.059 --> 01:05:46.680
So, I can say dp upon dx of frictional component
is written as p upon a tau w ok now d tau
01:05:46.680 --> 01:05:53.200
w you can write or tau w you can write if
the change is very small and p upon a in a
01:05:53.200 --> 01:05:57.619
cylindrical pipe if we are thinking about
then the perimeter is nothing, but pi by d
01:05:57.619 --> 01:06:08.839
area is pi by 4 d square to tau w or you can
write it equal to 4 tau w upon d in single
01:06:08.839 --> 01:06:15.250
direction one dimensional .
Then dp upon dx due to acceleration is nothing,
01:06:15.250 --> 01:06:22.440
but m naught upon a into dvx upon dx that
is the Lagrangian acceleration term how it
01:06:22.440 --> 01:06:30.799
is changing with the velocity is changing
with the position x ok then dp upon dx which
01:06:30.799 --> 01:06:35.519
is the gravitational forces is equal to rho
g x
01:06:35.519 --> 01:06:42.089
So, this is the combination of all these 3
now even in the single phase flow if you want
01:06:42.089 --> 01:06:48.680
to find the pressure drop the problem is it
is forced to measure the Lagrangian velocity
01:06:48.680 --> 01:06:55.109
you should know how the V is changing with
the x second and most important is how the
01:06:55.109 --> 01:07:01.000
tau is being defined and the tau is being
defined tau w is defined in terms of the fanning
01:07:01.000 --> 01:07:15.170
friction factor it is F rho u square by two
where F is fanning friction factor
01:07:15.170 --> 01:07:20.859
and we know that finding the fanning friction
factor itself is a difficult job and you have
01:07:20.859 --> 01:07:26.170
to go to the moody chart or some correlation
which is being developed empirically developed
01:07:26.170 --> 01:07:31.960
correlation to find the F value for laminar
flow it is relatively simpler.
01:07:31.960 --> 01:07:48.229
So, if the flow is laminar F is nothing, but
16 by re and if flow is turbulent then what
01:07:48.229 --> 01:07:53.499
you need to do there are several empirical
based equation you have to either use that
01:07:53.499 --> 01:08:01.380
or you have to use the moody chart ok and
where the F is being plotted whether in Reynold
01:08:01.380 --> 01:08:06.930
number , this is for the laminar flow and
this is for the different turbulent flow and
01:08:06.930 --> 01:08:18.359
this is for different k upon d where k is
nothing but the roughness factor coefficient
01:08:18.359 --> 01:08:24.690
or roughness value this k upon d is called
roughness coefficient and k is roughness ok.
01:08:24.690 --> 01:08:31.469
So, the simplest equation for the turbulent
flow via the pipe is smooth is equal to it
01:08:31.469 --> 01:08:41.480
is 0.625 upon re raised to the power 0.25
ok. This is the way F has been defined if
01:08:41.480 --> 01:08:57.150
the roughness is placed the roll. There is
different correlation correlation is there
01:08:57.150 --> 01:09:02.420
which can be used to find that what will be
the value of destruction factor and the one
01:09:02.420 --> 01:09:06.619
form of the Cole book. The correlation for
the fanning friction term is one upon under
01:09:06.619 --> 01:09:22.900
root F is equal to minus 4 log of twice ki
upon d plus 9.35 upon re under root F plus
01:09:22.900 --> 01:09:29.420
3.48 . This is the one form of the correlation
there are different cause both correlations
01:09:29.420 --> 01:09:33.830
have a form of this correlation available
which is valid for the different in Reynold
01:09:33.830 --> 01:09:37.839
number there are other correlations also available
to find the F values.
01:09:37.839 --> 01:09:43.109
So, what I am trying to say that even for
the single phase flow calculating the delta
01:09:43.109 --> 01:09:49.779
p is a challenge and that is mainly because
of the acceleration term and because of the
01:09:49.779 --> 01:09:54.610
frictional term and the frictional term the
problem comes with the friction factor value
01:09:54.610 --> 01:10:01.130
that how to calculate the suitable friction
factor value for the one dimensional flow
01:10:01.130 --> 01:10:06.550
in the single phase flow and different correlations
are available, if the flow is laminar finding
01:10:06.550 --> 01:10:11.130
the values is relatively simpler because you
have a straightforward correlation F equal
01:10:11.130 --> 01:10:15.900
to upon re.
If the flow is turbulent; then, if there is
01:10:15.900 --> 01:10:20.900
a roughness parameter, then, the becomes the
solution becomes iterative. And if you see
01:10:20.900 --> 01:10:25.780
that if you have to solve this equation; then,
you have to do the iteration to find the value
01:10:25.780 --> 01:10:30.239
of F for the different Reynold number or for
the same little number. How the value of F
01:10:30.239 --> 01:10:35.630
will change you have to do the iterative solution
to get the F value or you have to move to
01:10:35.630 --> 01:10:41.030
the moody chart you have to see that how the
winner in all number F is changing and what
01:10:41.030 --> 01:10:45.810
is your roughness parameter and depending
upon the roughness value, you can find the
01:10:45.810 --> 01:10:51.239
k upon d for a particular k upon d for a particular
in all number you can find the F value.
01:10:51.239 --> 01:10:56.199
So, it is still a challenge in doing in the
single phase flow the challenges will get
01:10:56.199 --> 01:11:01.900
double will increase at least. So, I will
not double I will say once the multi phase
01:11:01.900 --> 01:11:09.139
flow comes into the picture why because now
the problem is even to find that what is the
01:11:09.139 --> 01:11:15.460
area. So, first we have to find what is the
area, what is the velocity inside of the fluid
01:11:15.460 --> 01:11:21.230
because now it is not filled with that single
phase it is filled with both the phases available
01:11:21.230 --> 01:11:27.639
a particular fraction in the pipeline is being
occupied by that particular phase, you have
01:11:27.639 --> 01:11:33.090
to find that fraction, you have to calculate
the velocity inside you have to calculate
01:11:33.090 --> 01:11:38.440
the Reynold number inside and based on that
Reynold number you have to calculate the F
01:11:38.440 --> 01:11:42.660
value and then you can calculate the delta
p in the pipeline.
01:11:42.660 --> 01:11:48.300
So, what we are going to do from the next
class is we are going to calculate the use
01:11:48.300 --> 01:11:53.659
the regimes. First we will focus on the gas
liquid flow, we will take the different regime
01:11:53.659 --> 01:12:00.139
homogeneous regime separated regime annular
regime all those regimes we will take we will
01:12:00.139 --> 01:12:05.820
see how this equations need to be modified
in those regimes to calculate the delta p.
01:12:05.820 --> 01:12:11.670
We will also try to see some of the empirically
developed correlation like Lockhart Martinelli
01:12:11.670 --> 01:12:18.210
correlation which is widely used in all the
industries to find the delta p in a pipeline.
01:12:18.210 --> 01:12:23.619
So, you will understand those systems and
we will see that in each regimes; how the
01:12:23.619 --> 01:12:29.460
equation will get modified what are the challenges
we have to see in the each regime and then
01:12:29.460 --> 01:12:34.760
we will try to do some assignment problem
on each regimes to understand to have more
01:12:34.760 --> 01:12:53.800
clarity on it .