WEBVTT
Kind: captions
Language: en
00:00:04.120 --> 00:00:29.489
So, welcome back.
Last class what we have done, we have found
00:00:29.489 --> 00:00:35.149
that if suppose a particle which is travelling
horizontally and only drag is opposing it
00:00:35.149 --> 00:00:40.219
is motion. So, how, so, this is the particle
which was travelling and suppose only drag
00:00:40.219 --> 00:00:45.649
if F D is only opposing the motion we have
derived the formula for the stokes regime
00:00:45.649 --> 00:00:50.999
and found that V, the how the velocity of
this particle will be changing if the initial
00:00:50.999 --> 00:00:55.609
velocity is V naught.
So, V naught into e raised to the power minus
00:00:55.609 --> 00:01:02.309
t upon tau where tau is nothing, but the response
time ok and similarly, we have defined that
00:01:02.309 --> 00:01:07.800
how the position of the particle will change
and that will be nothing, but it will be changing
00:01:07.800 --> 00:01:15.110
as per this formula that x will be equal to
V naught tau where V naught is nothing, but
00:01:15.110 --> 00:01:22.480
the initial velocity of the particle to 1
minus e raised to the power t upon tau ok
00:01:22.480 --> 00:01:28.110
and V naught can be written as also the initial
position of the particle. So, it can be x
00:01:28.110 --> 00:01:35.420
upon x s the starting position will be equal
to 1 minus e raised to the power t upon tau.
00:01:35.420 --> 00:01:41.230
So, this says that how the particle position
and velocity is going to change and this has
00:01:41.230 --> 00:01:46.190
numerous application I have already discussed
to track the motion of the particle. Now,
00:01:46.190 --> 00:01:51.500
what we have assumed here we have said that
only drag is acting. So, we assume that the
00:01:51.500 --> 00:01:54.830
particle is moving horizontally. If it is
moving horizontally, gravity component will
00:01:54.830 --> 00:01:59.710
be 0, if there is any buoyancy force that
is also not going to act because we are measuring
00:01:59.710 --> 00:02:03.570
the horizontal motion.
So, now, what we are going to do? We are going
00:02:03.570 --> 00:02:08.640
to complicate the problem little bit and now,
we will say that the same problem. But, suppose
00:02:08.640 --> 00:02:17.859
particle is moving in downward direction,
ok. So, now, if the particle is moving on
00:02:17.859 --> 00:02:22.030
the downward direction or I can say upward
direction only the directions if I will change
00:02:22.030 --> 00:02:25.379
the direction what will happen only the direction
of the forces will change.
00:02:25.379 --> 00:02:31.349
So, suppose if it is moving downward direction
the gravity is going to play F g that will
00:02:31.349 --> 00:02:38.150
be acting on the down side this is the momentum
ok if I assume that initial velocity was there
00:02:38.150 --> 00:02:43.379
as a V naught. So, this is momentum, drag
is going to oppose the relative motion if
00:02:43.379 --> 00:02:48.220
the air is stationary or the fluid in which
the particle is settling or moving down is
00:02:48.220 --> 00:02:55.299
stationary then definitely drag is going to
move at upward F d and if there is some buoyancy
00:02:55.299 --> 00:02:59.659
that is also going to act upward direction
buoyancy will always be upward.
00:02:59.659 --> 00:03:05.480
So, that will be the overall force balanced.
So, if I write the rate of change of equation
00:03:05.480 --> 00:03:13.400
the rate of change of momentum ok, that is
nothing, but m dV upon dt is equal to say
00:03:13.400 --> 00:03:19.650
V particle I will write it as a particle will
be equal to summation of the forces. Now,
00:03:19.650 --> 00:03:25.980
what would be the forces which will be acting.
So, this will be mp dV p upon dt and the summation
00:03:25.980 --> 00:03:32.220
of forces will be this will be equal to F
g, ok I am taking downward as a positive that
00:03:32.220 --> 00:03:39.329
upward as a negative minus it will be F b
minus F d. So, this will be the force which
00:03:39.329 --> 00:03:43.420
will be acting on the body and this is the
initial momentum that will be equal to the
00:03:43.420 --> 00:03:46.939
m dV upon dt.
Now, what we can do? We can do exactly same
00:03:46.939 --> 00:03:51.609
thing which we have done previously and what
we our target is to derive the formula for
00:03:51.609 --> 00:03:57.579
the velocity and the change in the position.
So, now, what we can do, in this case we can
00:03:57.579 --> 00:04:03.510
write it m p is nothing, but mass of the particle
that we can write it as a rho p density of
00:04:03.510 --> 00:04:14.080
the particle into volume of the particle.
So, this is volume. I am writing it as a V
00:04:14.080 --> 00:04:19.209
cross so that this velocity and volume does
not get one should not get confused. So, m
00:04:19.209 --> 00:04:24.470
dV p upon dt.
Now, what is F g? F g is what? Is the gravitational
00:04:24.470 --> 00:04:32.360
force. So, this will be again m of particle
into g. So, that is gravitational force. What
00:04:32.360 --> 00:04:39.259
will be the buoyancy force? Buoyancy force
will be rho of fluid into volume of particle
00:04:39.259 --> 00:04:45.680
ok into g. So, that will be the buoyancy force
and the drag force is nothing, but we have
00:04:45.680 --> 00:04:54.370
already discussed half rho of fluid into C
D into area into V square. So, this will be
00:04:54.370 --> 00:05:00.139
what that is the drag force, ok.
So, now if we just try to solve this equation
00:05:00.139 --> 00:05:05.900
we want to simplify this equation, what we
can do? We can write it as a rho p and m p
00:05:05.900 --> 00:05:14.900
again, here can be written as rho p into V
of particle into g minus rho f, V of particle
00:05:14.900 --> 00:05:27.990
into g minus half rho f, C D A into V square
now here what I can do it will be rho p minus
00:05:27.990 --> 00:05:40.010
rho f, I can take g and volume of particle
outside minus half rho f C D A V square, ok.
00:05:40.010 --> 00:05:49.830
Now, what is the we have already this rho
p into V p into dV p upon dt, ok. Now, what
00:05:49.830 --> 00:05:54.770
we want? we want to write the terms in terms
of the Reynold number because we know that
00:05:54.770 --> 00:05:59.830
the C D is a function of Reynold number. So,
we know this C D is a function of Reynold
00:05:59.830 --> 00:06:05.020
number. So, I will convert the whole equation
in terms of the Reynold number and to convert
00:06:05.020 --> 00:06:08.699
that the Reynold number will be defined for
the particle Reynold number.
00:06:08.699 --> 00:06:16.730
And, which is being defined R e will be equal
to what it will be d P it will be V P it will
00:06:16.730 --> 00:06:23.730
be rho of fluid upon mu of fluid this will
be what particle Reynold number, ok. So, particle
00:06:23.730 --> 00:06:27.909
Reynold number is defined it it in this way
we have already discussed it earlier. So,
00:06:27.909 --> 00:06:37.120
this will be the particle Reynold number.
So, V P we can write it as mu f upon rho f
00:06:37.120 --> 00:06:50.939
into d P into R e P ok. So, dV p will be equal
to what? mu f upon rho f d P into d R e f
00:06:50.939 --> 00:06:55.030
particle.
So, we can replace now this above equation
00:06:55.030 --> 00:07:01.660
and we can write this dV P in terms of the
Reynold number. So, this will be rho or you
00:07:01.660 --> 00:07:09.840
just write it. So, that the step of simplicity
rho P V P dV P upon dt will be equal to we
00:07:09.840 --> 00:07:22.020
have derived it rho f rho P minus rho f into
g into V P minus half rho of fluid A into
00:07:22.020 --> 00:07:30.150
C D into V square of d P square. Now, we will
replace the V P with this and if you do it,
00:07:30.150 --> 00:07:38.460
it will be just rho P into V P this will be
d re upon d R e P upon dt and then you have
00:07:38.460 --> 00:07:50.020
to multiply it by mu f upon rho f into d P
this will be equal to rho P minus rho f into
00:07:50.020 --> 00:08:03.030
g V P minus half rho f A C D and V P can be
written as mu f square upon rho f square into
00:08:03.030 --> 00:08:10.259
d P square into R e P.
Now, we can simplify it we can replace the
00:08:10.259 --> 00:08:17.949
d P; d P is equal to what? this is the particle,
is spherical. It will be pi by 6 d P cube
00:08:17.949 --> 00:08:23.909
area if the particle is spherical it is pi
by 4, d P square will replace A and V P here.
00:08:23.909 --> 00:08:38.019
So, we will get mu f upon rho f into d P into
rho P and d R e P upon dt, this V P again
00:08:38.019 --> 00:08:47.851
this will be rho P minus rho f into g this
will be equal to minus half rho f, A is pi
00:08:47.851 --> 00:08:59.720
by 4, d P square ok this will be C D, mu f
square upon rho f square into d P square into
00:08:59.720 --> 00:09:09.370
R e P square upon V P and V P we can write
it as this term rho P minus rho f into g this
00:09:09.370 --> 00:09:19.820
will be minus half rho P into pi by 4, d P
square into C D this will be mu f square rho
00:09:19.820 --> 00:09:30.410
f square into d p square into R e P and this
will be pi by 6, d P cube, the V P.
00:09:30.410 --> 00:09:35.170
Now, we will just try to simplify it. So,
this just 4 pi-pi will be cancelled out, this
00:09:35.170 --> 00:09:40.510
d P square this d P 5. So, d P square d P
square here will be cancelled out we get the
00:09:40.510 --> 00:09:44.200
d P cube.
And, if you just simplify this equation what
00:09:44.200 --> 00:09:52.220
we are going to get we are going to get it
as if we just simplify it it will be mu f
00:09:52.220 --> 00:10:05.820
into upon rho f, d P into d R e P upon dt
this will be equal to here rho P is also there
00:10:05.820 --> 00:10:16.800
rho P minus rho f into g and this will be
equal to minus half actually it will be 8
00:10:16.800 --> 00:10:23.829
and just 3. So, 6, so, this will be 3 by 4.
I can write it 3 by 4 because this will be
00:10:23.829 --> 00:10:32.490
8 and 6. So, this will be cancelled out. So,
it will be 3 by 4 into C D into rho P into
00:10:32.490 --> 00:10:45.690
mu f square upon rho f square into d P cube
into R e P square. So, this also square, so,
00:10:45.690 --> 00:10:49.630
R e P square.
So, I hope I have written this correct, rho
00:10:49.630 --> 00:10:54.070
P and you will get this now we will further
simplify it we will write it as a d R e P
00:10:54.070 --> 00:10:59.589
upon dt term. So, what we are going to do
we are just writing it in terms of the C D
00:10:59.589 --> 00:11:04.480
into R e square. So, this if you try to simplify
what we are going to do we are just going
00:11:04.480 --> 00:11:12.459
to divide by this term over all. So, this
will be 4 by 3. Now, if you divide by the
00:11:12.459 --> 00:11:16.940
rho P; rho p - rho P here will be cancelled
out. So, this I will write it as as it is
00:11:16.940 --> 00:11:30.079
mu f rho P upon rho f d P, ok. Now, I am dividing
by this terms 4 by 3, it will be rho f square
00:11:30.079 --> 00:11:41.680
d P cube upon mu f square, ok upon rho P.
This will be equal to again here 4 by 3 rho
00:11:41.680 --> 00:11:56.251
P minus rho f ok into g upon this will be
rho P into mu f square into rho f into d P
00:11:56.251 --> 00:12:03.839
cube, ok. So, this will be minus C D into
R e P square. So, that is what we are going
00:12:03.839 --> 00:12:07.060
to get.
Now, what we are going to do we are just going
00:12:07.060 --> 00:12:12.519
to simplify this and this will be sorry here
dR e p upon dt is missing. Now, we are going
00:12:12.519 --> 00:12:17.140
to simplify it. If we will simplify it you
will see that this will be what this is rho
00:12:17.140 --> 00:12:23.589
of fluid ok. So, if you simplify it you see
rho of fluid will be cancelled with here,
00:12:23.589 --> 00:12:28.610
mu f square will be cancelled from this place
rho P rho P will be cancelled out, what you
00:12:28.610 --> 00:12:36.060
are going to get you are going to get it 4
upon 3 into rho of particle sorry rho of fluid
00:12:36.060 --> 00:12:41.960
into d P square into d P square just d P and
one this will be cancelled out.
00:12:41.960 --> 00:12:52.640
So, rho of fluid d P square upon mu f into
d R e P upon dt this will be equal to 4 upon
00:12:52.640 --> 00:13:11.209
3 rho P minus rho f into d P cube into g into
rho of fluid upon rho P into mu f square minus
00:13:11.209 --> 00:13:17.880
C D R e square R e P square, now this rho
P should be cancelled out actually. So, rho
00:13:17.880 --> 00:13:23.910
f rho f and rho of particle this is rho of
particle failure there is something somewhere
00:13:23.910 --> 00:13:28.110
something is wrong this is also rho of fluid
actually, not rho of particles this is rho
00:13:28.110 --> 00:13:31.860
of fluid.
So, that is why it is making strong here rho
00:13:31.860 --> 00:13:39.240
of fluid will be there. So, if you do the
rho of fluid this will be if I cancelled it
00:13:39.240 --> 00:13:44.630
rho of fluid rho of fluid will be cancelled
out. So, rho P will not be cancelled out here.
00:13:44.630 --> 00:13:54.240
So, pi by 6 rho of fluid. So, this will be
C D ok, here rho of fluid will be there instead
00:13:54.240 --> 00:13:59.980
of rho P.
C D 3 by 4 rho of fluid C D rho f square rho
00:13:59.980 --> 00:14:05.920
f square d P cube upon R e square R e P square
f. So, this is rho of fluid. Now, if it will
00:14:05.920 --> 00:14:10.190
be rho of fluid it will be just divided by
the here rho of fluid. So, this will be rho
00:14:10.190 --> 00:14:15.649
of fluid again. So, rho of fluid will come
here one rho of fluid because rho of fluid
00:14:15.649 --> 00:14:19.930
rho of fluid this will be cancelled out. So,
only one rho of fluid will be there that will
00:14:19.930 --> 00:14:25.839
be here this rho P will go. There will be
no rho P here, sorry, please check it, just
00:14:25.839 --> 00:14:34.480
do it and here again it will be once you will
divide it it will be rho of fluid square it
00:14:34.480 --> 00:14:40.660
will not be the square it will be only the
rho of fluid. So, it will be rho of fluid
00:14:40.660 --> 00:14:47.380
into d P cube mu f square ok and there will
be no rho P here.
00:14:47.380 --> 00:14:54.290
So, in that case if you do that this will
be what rho f rho f will be cancelled out
00:14:54.290 --> 00:15:00.190
and rho will be as remain as it is. So, this
will be not rho f it will be rho P. So, this
00:15:00.190 --> 00:15:07.870
will be rho P d P square upon mu f d R e P
upon dt into this. Now, this number is called
00:15:07.870 --> 00:15:18.009
this number is called the Galileo number,
G a; represented with G a and G a is nothing,
00:15:18.009 --> 00:15:26.220
but Galileo number.
And, the Galileo number G a is nothing, but
00:15:26.220 --> 00:15:40.170
it is the ratio of gravitational force
to the viscous force. So, ratio of gravitational
00:15:40.170 --> 00:15:44.839
force to the viscous force; it means what,
where the fluid is settling what is the importance
00:15:44.839 --> 00:15:49.110
if the fluid is getting settling. So, how
much gravity is pulling it up and how much
00:15:49.110 --> 00:15:54.190
the viscosity of the fluid is actually resisting
the motion of the particle. So, that is what
00:15:54.190 --> 00:15:58.879
is the Galileo number. It is very important
once the settling is going on are anything
00:15:58.879 --> 00:16:03.110
any object is moving under the gravity.
So, it shows that how much is the gravity
00:16:03.110 --> 00:16:08.110
is playing role to pull the particle or the
particle and how much is the viscosity of
00:16:08.110 --> 00:16:12.970
the system or viscosity of the fluid is resisting
the motion of the particle. So, that is the
00:16:12.970 --> 00:16:18.060
Galileo number. Now, we already know that
tau which is response time is nothing, but
00:16:18.060 --> 00:16:26.250
rho P into d P square upon 18 mu f. So, what
we can do? This is what we can write it in
00:16:26.250 --> 00:16:29.730
terms of the tau this quantity we can write
it in terms of the tau.
00:16:29.730 --> 00:16:41.300
Now, if we do the replacement, this will be
4 by 3 into 18 tau ok into d R e P upon dt
00:16:41.300 --> 00:16:48.260
this will be equal to Galileo number minus
C D into R e square R e of particle square.
00:16:48.260 --> 00:16:54.810
So, this will be the equation you will get
to see that how the particle motion is changing
00:16:54.810 --> 00:17:01.779
with the time. Now, if you just simplify it
this will be nothing, but 24 tau into d R
00:17:01.779 --> 00:17:09.669
e P upon dt will be equal to G a minus C D
into R e P square.
00:17:09.669 --> 00:17:15.289
So, this is the motion under the gravity this
will be the equation which will tell you that
00:17:15.289 --> 00:17:20.660
under the gravity how this R e P and this
will be changed. Now, you can further integrate
00:17:20.660 --> 00:17:25.770
it you can solve this problem again just by
replacing suppose for the stokes law you can
00:17:25.770 --> 00:17:31.360
say this is nothing, but 24 upon R e P you
can just cancelled it out it will be 1 minus
00:17:31.360 --> 00:17:36.409
x 1 upon x, ln problem and you can solve it
you will get the equation which will be similar
00:17:36.409 --> 00:17:41.760
to whatever we have got earlier. So, you will
get the equation for the velocity, you will
00:17:41.760 --> 00:17:46.580
get the equation for the position. So, you
can calculate everything whatever the way
00:17:46.580 --> 00:17:50.409
you want, ok.
Now, based on that you can find it out that
00:17:50.409 --> 00:17:54.840
how the particle velocity will change or particle
Reynold number will change with the time,
00:17:54.840 --> 00:17:58.970
how the particle position will change with
the time. So, that all calculation we can
00:17:58.970 --> 00:18:08.850
do again ok just suppose if I say that in
stokes regime for stoke regime C D is equal
00:18:08.850 --> 00:18:18.690
to what 24 upon R e, now this will be 24 tau
d R e P upon dt this will be equal to G a
00:18:18.690 --> 00:18:23.990
minus 24 upon R e if I do I will say 24 R
e P.
00:18:23.990 --> 00:18:30.990
So, what I will get I will get this value
or we can say that d R e P upon G a minus
00:18:30.990 --> 00:18:44.760
24 upon R e P is equal to dt into 1 upon 24
tau ok. So, we can integrate it in terms of
00:18:44.760 --> 00:18:49.620
the d A and we can see that how the time is
changing suppose we can say that initial is
00:18:49.620 --> 00:18:54.810
R e naught if it is has certain velocity finally,
say R e P then we can integrate it and we
00:18:54.810 --> 00:18:59.000
can find it out that how the Reynold number
will change with the time. The Reynold number
00:18:59.000 --> 00:19:03.590
can be converted in terms of the velocity
by using the same R equal to V rho d upon
00:19:03.590 --> 00:19:08.409
mu. We can calculate in terms of the velocity,
we can find it out how velocity is changing
00:19:08.409 --> 00:19:13.440
with the time, velocity can be converted in
terms of dx upon dt, we can find it out how
00:19:13.440 --> 00:19:17.250
the position is changing with the time.
So, under this influence again you can track
00:19:17.250 --> 00:19:20.700
the particle trajectory you can track the
particle velocity.
00:19:20.700 --> 00:19:27.160
But, what we are more interested is in is
suppose the term which we called as a terminal
00:19:27.160 --> 00:19:37.580
velocity or settling velocity. Now, what is
terminal velocity or if the particle is moving
00:19:37.580 --> 00:19:44.000
down or settling down I will say it as the
same as a settling velocity. So, how it has
00:19:44.000 --> 00:19:49.160
been defined terminal velocity? Terminal velocity
is being defined when all the forces are balanced
00:19:49.160 --> 00:20:00.299
balanced. It means what that dV upon dt term
is going to be 0.
00:20:00.299 --> 00:20:05.500
So, it means if I neglect that the initial
transition period where the particle velocity
00:20:05.500 --> 00:20:11.159
is changing with the time and before it achieve
a constant velocity this equation if I neglect
00:20:11.159 --> 00:20:16.290
that part if I put this dV by dt, the same
equation what will happen db by dt is equal
00:20:16.290 --> 00:20:26.059
to 0, it means if dV by dt is equal to 0 then
d R e P upon dt is also going to be 0. So,
00:20:26.059 --> 00:20:31.750
particle velocity then this equation will
be modified and this equation will be G a
00:20:31.750 --> 00:20:41.760
minus C D R e square R e P square will be
equal to 0 or you can say that R e P square
00:20:41.760 --> 00:20:46.390
C D is equal to genetic algorithm.
So, you can find it out the correlation here
00:20:46.390 --> 00:20:56.120
directly and for stokes regime again if I
do that; so, before going to the stokes regime
00:20:56.120 --> 00:21:05.090
what I will do I will write just normal equation.
So, this we can say that R e P is G a upon
00:21:05.090 --> 00:21:12.230
C D and V t if you do this R e P in terms
of the velocity of the particle you can write
00:21:12.230 --> 00:21:20.080
it as a V of terminal into rho of particle
ok into sorry rho of fluid into d of particle
00:21:20.080 --> 00:21:29.210
upon mu of fluid, so, will be equal to G a
upon C D. So, you can write it as V t is nothing,
00:21:29.210 --> 00:21:41.190
but G a into mu f upon rho f into d P upon
C D. So, that will be the terminal velocity
00:21:41.190 --> 00:21:44.760
of the particle.
In general case, you can write it in this
00:21:44.760 --> 00:21:50.080
way; the only thing is you have to find that
the C D verses R e correlation and C D versus
00:21:50.080 --> 00:21:55.010
R e correlation you have to find it out with
the terms that ok how the C D is changing
00:21:55.010 --> 00:21:58.570
with the Reynold number. So, you have to find
the Reynold number, it will be iterative solution.
00:21:58.570 --> 00:22:03.170
So, you have to find the Reynold number or
you have to guess a Reynold number for that
00:22:03.170 --> 00:22:07.230
Reynold number you will calculate the value
of C D. Once you will calculate the value
00:22:07.230 --> 00:22:11.900
of C D you will get the value of V t.
Now, calculate again with the new Reynold
00:22:11.900 --> 00:22:17.440
number you V t you calculate the Reynold number
and see that if the assumed Reynold number
00:22:17.440 --> 00:22:22.300
and this calculate Reynold number has certain
difference. If they are same it means your
00:22:22.300 --> 00:22:26.679
assumption is correct, your Reynold number
assumption is correct. If they are not same,
00:22:26.679 --> 00:22:31.330
you choose the new V t Reynold number the
Reynold number calculated by this V t and
00:22:31.330 --> 00:22:36.330
again keep on doing that iteration till you
are not getting the number of value or Reynold
00:22:36.330 --> 00:22:41.610
number within a certain percentage and that
percentage error generally we take as a less
00:22:41.610 --> 00:22:47.560
than 5 percent. So, with the iterative solution
you can get the value of V t, ok.
00:22:47.560 --> 00:22:57.100
For stokes regime you do not need to do anything
you can further we can simplify and that will
00:22:57.100 --> 00:23:03.550
be what the stoke regime C D will be equal
to 24 upon R e. So, now, what we will do C
00:23:03.550 --> 00:23:13.409
D R e P square was equal to G a. So, we will
say 24 upon R e P into R e P square will be
00:23:13.409 --> 00:23:20.260
equal to G a. So, this will be R e P will
be equal to what, this will be G a upon 24.
00:23:20.260 --> 00:23:34.330
Now, you can open the G a; G a was nothing,
but rho P minus rho f into d P cube into g
00:23:34.330 --> 00:23:44.470
upon mu f square, ok and it was this, ok.
So, this will be this R e or this was 4 by
00:23:44.470 --> 00:23:55.940
3. I think this 4 by 3 term we have included
here. So, this will be 4 by 3, this was what
00:23:55.940 --> 00:24:02.100
the Galileo number was.
Now, if you do that this is rho f now you
00:24:02.100 --> 00:24:07.710
do this upon 24 you will get that R e P is
nothing, but V if you want to find the terminal
00:24:07.710 --> 00:24:14.110
velocity it will be this into 24 here will
be coming. Now, V P R e P is nothing, but
00:24:14.110 --> 00:24:23.650
V of terminal into rho of particle sorry rho
of fluid into d of particle upon mu of fluid
00:24:23.650 --> 00:24:30.080
this will be equal to 4 upon 3 into this will
be actually 18, I will write just this is
00:24:30.080 --> 00:24:40.640
4, 6. So, I will write it as a 1 upon 18 or
I will write rho P minus rho f into d P cube
00:24:40.640 --> 00:24:49.190
into g into rho f upon 18 mu f square.
Now, if you do that if I have to find the
00:24:49.190 --> 00:24:54.080
V t what I will do I will cut this mu and
mu this rho of fluid and rho of fluid will
00:24:54.080 --> 00:24:59.460
be cancelled out and this d P will make a
d P square. So, V t will be equal to what
00:24:59.460 --> 00:25:09.950
it will be rho P minus rho f into g into d
P square upon 18 mu f. So, this will be the
00:25:09.950 --> 00:25:24.370
terminal velocity in case of stokes regime
00:25:24.370 --> 00:25:29.899
terminal velocity. So, this you have already
done, but how this terminal velocity has been
00:25:29.899 --> 00:25:36.000
derived that is what I have just shown you
that this is nothing, but a stokes regime
00:25:36.000 --> 00:25:42.299
how this terminal velocity will be defined
or derived. So, this is rho P minus rho f
00:25:42.299 --> 00:25:46.809
g d P square upon 18 mu f.
So, if you have a turn you have to find the
00:25:46.809 --> 00:25:51.450
particle terminal velocity in stokes regime
you can use this formula directly. Everything
00:25:51.450 --> 00:25:56.010
is known if you know the particle diameter
if you know the fluid under which it is being
00:25:56.010 --> 00:25:59.929
settled down. If you know the particle density
you can calculate the terminal velocity very
00:25:59.929 --> 00:26:07.440
easily, but what will happen in case of non
this stokes regime you need to do the iteration.
00:26:07.440 --> 00:26:10.490
Now, again I would like to explain the procedure
of this.
00:26:10.490 --> 00:26:17.200
So, this what we have done we have defined
that V t is nothing, but this is equal to
00:26:17.200 --> 00:26:29.850
the non-stokes regime this is G a mu f G a
mu f upon rho f d P C D mu f upon rho f into
00:26:29.850 --> 00:26:36.779
d P into C D this is in the non-stokes regime
or any other regime. So, what we need to do
00:26:36.779 --> 00:26:50.050
we need to guess a Reynold number R e P, for
that R e P calculate C D value ok from the
00:26:50.050 --> 00:26:54.350
chart or from the correlation which you are
using for the C D value there are different
00:26:54.350 --> 00:26:58.350
correlation has been given in the literature
and we will discuss that correlation for the
00:26:58.350 --> 00:27:06.159
different drag for different C D values correlation
is given. Calculate the C D value, calculate
00:27:06.159 --> 00:27:15.590
V t . I will write here more elaborately calculate
the C D value for guessed R e P.
00:27:15.590 --> 00:27:26.330
Now, calculate the V t value calculate R e
R e P now this I will say calculated R e P
00:27:26.330 --> 00:27:34.860
C which is calculated this is R e P g which
is the guessed one. So, this is C, now, calculate
00:27:34.860 --> 00:27:46.860
that R e P C minus R e g R e P g upon say
R e P g this you take as the mod. If this
00:27:46.860 --> 00:27:56.460
error is less than 5 percent ok, error is
less than 5 percent then you assume that whatever
00:27:56.460 --> 00:28:03.029
the C D you have done is calculated is fine.
If it is more than 5 percent then now you
00:28:03.029 --> 00:28:14.230
replace if less than 5 percent say I will
write it here then terminate, terminate your
00:28:14.230 --> 00:28:18.519
program suppose if you are writing the code
and you can write try to write a code for
00:28:18.519 --> 00:28:29.029
this that then it will be terminate. If this
is no, if this is not correct then R e P will
00:28:29.029 --> 00:28:36.810
be equal to you say R e P g will be equal
to R e P C and then you again go back and
00:28:36.810 --> 00:28:44.279
calculate this . So, you have to keep on repeating
this loop till your V t your R e P which was
00:28:44.279 --> 00:28:49.280
guessed or an R e P which is being calculated
does not come within the 5 percent range.
00:28:49.280 --> 00:28:54.700
So, that will be the iterative, you can write
a program and for any particle you can calculate
00:28:54.700 --> 00:28:59.720
that what will be the terminal velocity. The
only thing is the determination of the C D
00:28:59.720 --> 00:29:04.140
and how to find that C D we will discuss it
later to the different classes that how the
00:29:04.140 --> 00:29:09.179
C D correlations are given by the many papers
and depending on the correlation you can take
00:29:09.179 --> 00:29:13.919
a C D correlation and you can keep on doing
the same iteration again and again till you
00:29:13.919 --> 00:29:18.170
are not conversed to this value, once you
terminal through this value your program will
00:29:18.170 --> 00:29:21.350
be terminated.
So, you can write your code you can we can
00:29:21.350 --> 00:29:25.039
discuss it in the assignment or I will give
a assignment within which you can write a
00:29:25.039 --> 00:29:30.210
code if you want we will see that we will
discuss it further to understand that how
00:29:30.210 --> 00:29:35.070
to do it. If you do not want to write a code
you can still do it numerically by your own
00:29:35.070 --> 00:29:38.710
hand, but it will take time because every
time you have to do the computation.
00:29:38.710 --> 00:29:44.179
So, you can still do it without writing a
code, you can write a code both are ok both
00:29:44.179 --> 00:29:49.789
up to you how you want to do it the overall
idea is that how to calculate the C D value.
00:29:49.789 --> 00:29:54.919
Now, what we are going to do which is the
most popular formula which we discussed which
00:29:54.919 --> 00:30:03.091
we use here when the particle is being only
settled in presence of F g and F d. So, if
00:30:03.091 --> 00:30:08.460
I am neglecting the buoyancy say if the rho
f value is very very low and if I am neglecting
00:30:08.460 --> 00:30:13.929
the buoyancy effect then this F g will be
equal to we can derive the formula for that
00:30:13.929 --> 00:30:18.529
and why I am deriving because this is what
you have done in your undergraduate, F d will
00:30:18.529 --> 00:30:28.340
be equal to F g and in that case half rho
F d is what half rho of fluid into area into
00:30:28.340 --> 00:30:35.240
C D into V square now because this is settling
we have neglected the dV upon dt term it means
00:30:35.240 --> 00:30:38.950
all the forces are balanced. So, this will
be the the settling velocity or the terminal
00:30:38.950 --> 00:30:44.559
velocity and this will be m g into g, m of
particle sorry, into g.
00:30:44.559 --> 00:30:51.960
Now, m of particle can be written as rho of
particle into volume of particle into g and
00:30:51.960 --> 00:30:59.960
this can be written as rho of particle into
pi by 6, d P cube into g. Now, we can solve
00:30:59.960 --> 00:31:07.090
this V t. So, V t will be what? V t will be
equal to this values pi by 6 this. So, you
00:31:07.090 --> 00:31:12.380
can calculate this value this area will be
equal to what. So, this will be rho P pi by
00:31:12.380 --> 00:31:25.730
6 d P cube into g and this will be 2 times
of rho f, sorry this will be 2 times of rho
00:31:25.730 --> 00:31:33.260
f upon A into C D.
Now, you can further simplify it and you can
00:31:33.260 --> 00:31:45.059
write it as this will be 3. So, rho P into
pi d P cube into g upon 6, 2 upon rho f and
00:31:45.059 --> 00:31:53.340
area can be written as pi by 4 d P square
into C D, ok. Now, if you further simplify
00:31:53.340 --> 00:32:00.419
it this we will get what, this will be 8 and
this so, this will get a 4 upon 3 and pi-pi
00:32:00.419 --> 00:32:07.799
will be cancelled out d P cube will be d P
only. So, this will be d P, ok. Now, this
00:32:07.799 --> 00:32:20.510
will be rho P this will be g upon rho f into
C. So, this will be what you will get V t
00:32:20.510 --> 00:32:27.309
square value here. So, V t will be equal to
what this will be under root of this value.
00:32:27.309 --> 00:32:34.570
So, what you will get? You will get that formula
for the V t versus C D again you have to do
00:32:34.570 --> 00:32:38.809
the if you do not know the C D value. So,
what you have to do you have to guess the
00:32:38.809 --> 00:32:43.389
Reynold number again the same look will be
there. You have to guess the Reynold number
00:32:43.389 --> 00:32:48.389
if you have to find that Reynold number guess
the Reynold number find the C D value; for
00:32:48.389 --> 00:32:53.730
that C D value whatever you have guessed you
can calculate that V t value.
00:32:53.730 --> 00:32:59.070
Again, from this V t value you can calculate
the Reynold number. If both the Reynold number
00:32:59.070 --> 00:33:04.100
is same then it is fine, you can terminate
your calculation or your program. If they
00:33:04.100 --> 00:33:09.880
are not same you can take that newly calculated
value of Reynold number as a guessed value.
00:33:09.880 --> 00:33:14.789
Again, you can find the C D value for this
you can again find the V t value and you can
00:33:14.789 --> 00:33:20.130
keep on running this loop till your error
is not coming within a margin and margin generally
00:33:20.130 --> 00:33:24.880
we take as a 5 percent .
Suppose, if there is a density difference
00:33:24.880 --> 00:33:32.610
if we have to do that. So, this will be 4
upon 3, it will be d P generally rho P minus
00:33:32.610 --> 00:33:42.510
rho f into g upon rho f into C D. So, you
can use this formula again to calculate the
00:33:42.510 --> 00:33:47.649
terminal velocity. So, what you can do? What
we have done? Now, in this part we have seen
00:33:47.649 --> 00:33:54.429
that how the value of particle terminal velocity
of particle settling velocity will change
00:33:54.429 --> 00:34:00.080
under the influence of different forces, how
it will change for the stokes regime and for
00:34:00.080 --> 00:34:05.840
the other than the stokes regime, how to calculate
the particle terminal velocity in case of
00:34:05.840 --> 00:34:11.330
any flow and how to calculate the position
of the particle, how it is changing with the
00:34:11.330 --> 00:34:15.720
time, how the velocity of the particle is
changing with the time if we do not want to
00:34:15.720 --> 00:34:20.330
neglect the transient term if we want to calculate
we can do that.
00:34:20.330 --> 00:34:25.810
So, what we can do, we can calculate the particle
position, we can calculate the particle velocity,
00:34:25.810 --> 00:34:30.880
we can change the find the particle position
with the time, we can find the particle velocity
00:34:30.880 --> 00:34:37.660
with the time. So, you can track the particle
in every domain if only drag is acting once
00:34:37.660 --> 00:34:43.510
we have seen and another case we have seen
when the drag as well as the other forces
00:34:43.510 --> 00:34:49.450
like buoyancy and gravity is also acting.
So, with this calculation whatever we have
00:34:49.450 --> 00:34:54.780
done till now, any particle which is flowing
or the group of the particle is flowing we
00:34:54.780 --> 00:34:58.571
can track the motion of the particle for the
group of the particle you have to solve the
00:34:58.571 --> 00:35:04.020
equation for individual particles. The only
change will be the C D value will be modified
00:35:04.020 --> 00:35:08.070
and we will discuss all these things that
how the C D value will be modified for a particle
00:35:08.070 --> 00:35:11.230
cloud and based on that you can calculate
it.
00:35:11.230 --> 00:35:16.050
So, it means what whether it is a you have
a particle which you have a hundred particles
00:35:16.050 --> 00:35:20.860
which are flowing together, you have to solve
the hundred Newton second law of motion to
00:35:20.860 --> 00:35:26.800
track the particle position, ok. So, whether
it is a fluidized bed, whether it is a pneumatic
00:35:26.800 --> 00:35:32.380
conveying, whether it is a general solid is
suspended in the air and they are moving,
00:35:32.380 --> 00:35:36.890
you can track the velocity of those particles
you can track the position of those particles.
00:35:36.890 --> 00:35:41.300
And, if you know the velocity if you know
the position you can also if you know the
00:35:41.300 --> 00:35:46.100
position you can also calculate the fraction
of the solids present there. So, you can understand
00:35:46.100 --> 00:35:50.300
about the system.
So, all the gas solid system where the Lagrangian
00:35:50.300 --> 00:35:55.860
approach is there you can solve the Navier-Stokes
equation to get the terminal velocity to get
00:35:55.860 --> 00:36:01.810
the settling velocity to find how the position
of the particle is changing with the time,
00:36:01.810 --> 00:36:05.870
how the velocity of the particle is changing
with the time, ok.
00:36:05.870 --> 00:36:11.120
So, now, the typical value I am just seeing
that giving you that about the terminal velocity;
00:36:11.120 --> 00:36:21.350
so, if suppose if I have a 100 micro meter
dust particle which is settled in air ok,
00:36:21.350 --> 00:36:29.870
suspended in air then its terminal velocity
suppose, once the everything will be percent
00:36:29.870 --> 00:36:34.450
the terminal velocity will be around 1 meter
per second. I have assumed the density of
00:36:34.450 --> 00:36:38.060
the dust is equal to the 2900 which is the
density of the sand.
00:36:38.060 --> 00:36:44.100
So, you will get it as a 100 meter 1 meter
per second and if suppose the rainfall which
00:36:44.100 --> 00:36:51.040
we have always seen the drop if suppose I
am assuming that the droplet size of the rain
00:36:51.040 --> 00:37:00.070
is 2 mm which is very typical then the u t
in air, in air will be around 3 meter per
00:37:00.070 --> 00:37:05.010
second, ok. So, that will be the settling
velocity of the rain droplet if it is size
00:37:05.010 --> 00:37:10.400
is 2 mm you will see that it is moving at
the 3 meter per second velocity. If you have
00:37:10.400 --> 00:37:15.620
a suspended particle 100 micrometer dust particle
which is suspended in the air, you want to
00:37:15.620 --> 00:37:19.950
find that the terminal velocity you will see
that the terminal velocity will be in the
00:37:19.950 --> 00:37:24.640
range of 1 meter per second.
So, that is the way you can do that if suppose
00:37:24.640 --> 00:37:29.710
again if suppose if we assume that the particle
whatever we have assumed right now, either
00:37:29.710 --> 00:37:34.470
particle is moving horizontally or particle
is moving vertically suppose if the particle
00:37:34.470 --> 00:37:38.300
is moving at a particular angle.
So, what you can do the drag will act it it
00:37:38.300 --> 00:37:44.570
in that angle, you can say angle is theta
the drag F D will be here you can say this
00:37:44.570 --> 00:37:55.650
will be F D cos theta this will be component
F D sin theta and you can do the balance you
00:37:55.650 --> 00:38:00.060
can solve the horizontal component force,
you can solve the vertical component force
00:38:00.060 --> 00:38:05.951
you will get a particle horizontal velocity
by using this motion with dV by dt. You can
00:38:05.951 --> 00:38:10.780
get the water particle vertical velocity by
using this F D sin theta force and what you
00:38:10.780 --> 00:38:15.790
will have you will have the particle x velocity
you will have particle y velocity, you can
00:38:15.790 --> 00:38:21.420
find it out the resultant velocity magnitude
and direction of the vector of the velocity
00:38:21.420 --> 00:38:23.700
and you can find that how the particle will
move.
00:38:23.700 --> 00:38:28.630
So, using the 2 dimensional domain, we are
right now solving the 1 dimensional domain
00:38:28.630 --> 00:38:33.810
you can easily convert this problem to 2 dimensional
domain to find it out, how the particle is
00:38:33.810 --> 00:38:38.080
moving with the time how the particle position
will change with the time. So, in the most
00:38:38.080 --> 00:38:44.020
realistic cases the domain is actually 3 dimensional,
you can further do it in the third dimension
00:38:44.020 --> 00:38:48.720
you can find you can just resolve it in terms
of the vector another angle will come this
00:38:48.720 --> 00:38:54.330
is of the theta 5 will also come.
You can resolve the F D value in r theta and
00:38:54.330 --> 00:39:01.550
z coordinate or in xy and z coordinate and
you can find it out the velocity in the x
00:39:01.550 --> 00:39:06.870
direction velocity in the y direction velocity
in the z direction and the combined motion
00:39:06.870 --> 00:39:11.560
of the particle that where the particle is
moving how the particle position are changing.
00:39:11.560 --> 00:39:15.800
So, these are the more realistic cases which
I am just giving the examples.
00:39:15.800 --> 00:39:21.580
So, with this what you can do you can track
the position of the particle, with the time
00:39:21.580 --> 00:39:26.150
you can track the velocity of the particle
with the time in 2 dimensional domain, in
00:39:26.150 --> 00:39:30.860
3 dimensional domain, in one dimensional domain.
So, what you need to do, you just need to
00:39:30.860 --> 00:39:36.971
solve this now the only problem is the it
is computational very costly. Computationally
00:39:36.971 --> 00:39:41.180
it will be very costly, we will discuss it
again later the computational cost why it
00:39:41.180 --> 00:39:45.460
will be computationally very costly whatever
we have done we have done for one particle,
00:39:45.460 --> 00:39:52.690
but even if I take a small fluidized bed say
a diameter of 2 inch a height of 20 centimetre
00:39:52.690 --> 00:39:56.760
there will be millions of the particle will
be present inside if the particle diameter
00:39:56.760 --> 00:40:02.640
is say around 50 micron or 75 micron I am
talking about a group a particle then millions
00:40:02.640 --> 00:40:06.360
of the particles will be present there.
And, then what you need to do you have to
00:40:06.360 --> 00:40:11.240
solve these equations for the million of the
particle. We have neglected the collision
00:40:11.240 --> 00:40:14.690
between the particle we have not seen the
particle-particle interaction this is the
00:40:14.690 --> 00:40:18.770
one way interaction. We are just seeing that
the drag is actually interacting with the
00:40:18.770 --> 00:40:23.430
mean motion of the particle the particle interaction
the particle fluctuations we have not taken
00:40:23.430 --> 00:40:26.320
into the account.
So, what you have to do you solve several
00:40:26.320 --> 00:40:31.350
equations together to get that how the position
of each particle is going to change with the
00:40:31.350 --> 00:40:36.420
time, how the velocity of each particle is
going to change with the time if you know
00:40:36.420 --> 00:40:40.900
that you can calculate how the pressure is
going to be modified with the time and you
00:40:40.900 --> 00:40:45.840
can have the complete statistics you can have
a complete hydrodynamics of flow dynamics
00:40:45.840 --> 00:40:50.380
of the system. We will discuss again about
these things later on, right now the idea
00:40:50.380 --> 00:40:53.710
is the one dimensional motion that how to
solve this.
00:40:53.710 --> 00:40:58.940
So, I hope now you can solve the particle
motion in one-dimensional whatever the forces
00:40:58.940 --> 00:41:05.910
is acting on it you can solve it. Now, what
is left is that external forces. So, external
00:41:05.910 --> 00:41:11.310
forces can be in many of the application what
we see through the for the particle separation
00:41:11.310 --> 00:41:16.490
particularly, we use either the electric field
or we use the magnetic field to separate the
00:41:16.490 --> 00:41:25.880
particle from the main fluid. So, suppose
like a particle is suspended from like typical
00:41:25.880 --> 00:41:31.830
industries where kind of power plants where
the cool particles are burned to get the energy.
00:41:31.830 --> 00:41:36.640
Now, the cool particles are burn they form
the flue gases the flue gases is being used
00:41:36.640 --> 00:41:43.020
to heat the water to generate the steam, but
at the end of the day some of the soot particles
00:41:43.020 --> 00:41:48.070
or ash particle is being carried away with
the flue gases you cannot discharge your fluid
00:41:48.070 --> 00:41:53.300
gases to the the fluid gases to the environment
because they have the suspended particles.
00:41:53.300 --> 00:41:57.680
You have to remove that suspended particle
and to remove that suspended particles what
00:41:57.680 --> 00:42:03.120
we use we use electrostatic precipitator.
First we use the back filter, but after a
00:42:03.120 --> 00:42:06.980
certain particle size then the back filter
is not able to do cyclone separator is not
00:42:06.980 --> 00:42:10.080
able to separate then we need to use the electric
field.
00:42:10.080 --> 00:42:16.200
And, what ESP do, it have a plate. It has
a plate where the electrical charge is being
00:42:16.200 --> 00:42:21.140
generated gas flows from there now because
of this the particle has a certain charge
00:42:21.140 --> 00:42:25.560
and it moves either through the anode or the
positive or negative depending upon what is
00:42:25.560 --> 00:42:29.740
the charge generated on the particle it will
move on this side.
00:42:29.740 --> 00:42:35.340
Now, in this case if external force is there
similar thing is true with the magnetic particle
00:42:35.340 --> 00:42:41.330
magnetic force if the particle has some magnetic
field or suppose it is iron or something I
00:42:41.330 --> 00:42:44.970
can put a magnetic field I can put the magnet
and then what will happen the particle will
00:42:44.970 --> 00:42:49.570
be move towards the magnet. So, the motion
of the particle will change. So, what we need
00:42:49.570 --> 00:42:55.200
to see now what we are going to do now next
that in case in presence of the external field
00:42:55.200 --> 00:43:01.390
in external force field how the particle trajectory
will change with the time, how to calculate
00:43:01.390 --> 00:43:07.120
the particle settling velocity of particle
velocity that when the external force field
00:43:07.120 --> 00:43:10.200
is also present.
So, what we are going to do we are going to
00:43:10.200 --> 00:43:16.120
case take a typical case of the ESP we will
see that how in the ESP the particle is going
00:43:16.120 --> 00:43:23.200
to be changing its motion. So, now, what will
happen again in case of this this ESP what
00:43:23.200 --> 00:43:28.580
we are going to do we are going to take the
electrostatic charge or the coulomb forces
00:43:28.580 --> 00:43:36.530
inside and coulomb forces F C is being defined
as q into E where q is nothing, but is the
00:43:36.530 --> 00:43:42.330
coulomb and E is nothing, but volt per meter
or you can say the q is nothing, but the charge
00:43:42.330 --> 00:44:01.100
on the particle on the particle and E is electric
field intensity. So, that will be the coulomb
00:44:01.100 --> 00:44:05.060
forces. If suppose if I put an electric field
then this will be the force which will be
00:44:05.060 --> 00:44:10.780
acting on the particle. So, what we can do
we can write the same equation again mp into
00:44:10.780 --> 00:44:14.980
dV p upon dt is equal to summation of the
force.
00:44:14.980 --> 00:44:19.980
Now, the summation of the forces can be written
as this will be what are the forces which
00:44:19.980 --> 00:44:29.630
is going to act that is going to be F D that
is going to act here plus F C which is the
00:44:29.630 --> 00:44:36.870
F C the coulomb forces or electric forces
plus F g ok, because once the particles will
00:44:36.870 --> 00:44:41.190
move vertically downward also.
So, if we do that we can write it as md P
00:44:41.190 --> 00:44:52.330
upon dt m P dV P upon dt will be equal to
C D this will be half rho of fluid into C
00:44:52.330 --> 00:45:02.010
D into A into V of particle square plus q
into E, ok. Now, that q into E will be there
00:45:02.010 --> 00:45:09.200
which will be plus or minus depending on the
charge of the system plus mg, ok. Now, we
00:45:09.200 --> 00:45:14.960
can write it as we can solve this as bps equation
we can solve these equations again further
00:45:14.960 --> 00:45:19.290
and we can write it in terms of the Reynold
number. We can calculate the Reynold number,
00:45:19.290 --> 00:45:23.770
how it is changing with the time.
We can calculate the this values because this
00:45:23.770 --> 00:45:29.080
q is constant. You can calculate you can solve
it the way we have solved it earlier the only
00:45:29.080 --> 00:45:34.270
thing one more constant value will be added
and you can have all those things you can
00:45:34.270 --> 00:45:39.120
find it out that how the particle position
will change with the time, how the particle
00:45:39.120 --> 00:45:42.920
velocity will change with the time. You can
write it in in terms of the Reynold number
00:45:42.920 --> 00:45:46.900
also you can calculate that how the Reynold
number is changing with the time.
00:45:46.900 --> 00:45:51.560
I am not doing this calculation you can do
it very easily we have already done q is a
00:45:51.560 --> 00:45:55.690
constant force which is acting there. So,
that value is just that is going to act as
00:45:55.690 --> 00:46:01.730
a constant ok. So, now, we can have all this
calculation we can find it out. Now, in the
00:46:01.730 --> 00:46:06.680
more simplified case if suppose I am assume
that the particle is moving horizontally the
00:46:06.680 --> 00:46:12.280
q E force is dominating over the mg it is
just moving horizontally then in that case
00:46:12.280 --> 00:46:20.800
the same equation will be modified as half
rho f C D A V P square plus minus q E and
00:46:20.800 --> 00:46:25.350
this is the more realistic case because in
the electrostatic precipitator particle first
00:46:25.350 --> 00:46:29.820
move towards the wall gravity does not play
much role once it is stick to the wall then
00:46:29.820 --> 00:46:35.070
it will fall down because of the gravity.
So, this is what is the more realistic case
00:46:35.070 --> 00:46:40.380
and you can again do the calculations you
now, integration will be very simple this
00:46:40.380 --> 00:46:44.720
is constant you can convert everything in
terms of the Reynold number or you can write
00:46:44.720 --> 00:46:50.280
it in terms of the V P and dt and you can
find it out m P you can calculate again as
00:46:50.280 --> 00:46:56.440
a rho P into V P. V P you can write it in
terms of the d P. The way we have done I do
00:46:56.440 --> 00:47:01.120
not want to do it the same thing again and
again. So, you can do this exercise take it
00:47:01.120 --> 00:47:06.960
as a assignment, I will put it as actually
as assignment in the or next assignment which
00:47:06.960 --> 00:47:11.030
will be uploaded and if there is any problem
we can discuss it further.
00:47:11.030 --> 00:47:17.990
So, what we are more interested is in once
the velocity or this kind of the dvb d V P
00:47:17.990 --> 00:47:23.630
upon dt is equal to 0. It means all the forces
are balanced and then the particle is moving
00:47:23.630 --> 00:47:30.530
with a particular thing in that case you can
write it as half rho of fluid area of particle
00:47:30.530 --> 00:47:40.530
is pi by 4 d P square into C D into V P square
will be equal to q into E, ok.
00:47:40.530 --> 00:47:47.930
We can find it out or V instead of V P we
called it as a electric drift velocity or
00:47:47.930 --> 00:48:00.520
E S which is called electric drift velocity,
because now the particle is moving in influence
00:48:00.520 --> 00:48:06.510
of the electric field and the drag is opposing
the motion of that. So, that is the electric
00:48:06.510 --> 00:48:15.160
drift velocity V is called. So, this will
be half rho. So, V E S square will be equal
00:48:15.160 --> 00:48:27.840
to what q E upon 2 into rho f actually this
will be 8 upon pi d P square into C D.
00:48:27.840 --> 00:48:35.200
So, if you want to find it out in the normal
regime again, what you need to do? You have
00:48:35.200 --> 00:48:40.660
to guess a Reynold number, you have to find
the C D value, you have to find the V S value,
00:48:40.660 --> 00:48:45.620
with the V S again you find the Reynold number.
If both are same then your guessed value of
00:48:45.620 --> 00:48:50.770
Reynold number is correct and your calculation
of V E S is correct if not the new Reynold
00:48:50.770 --> 00:48:55.200
number which we have calculate you again use
the Reynold number to calculate the C D, keep
00:48:55.200 --> 00:49:00.460
on running the same loop again and you can
calculate that what will be the electric drift
00:49:00.460 --> 00:49:04.240
velocity of the particle.
So, if you write a same program or you make
00:49:04.240 --> 00:49:09.090
a excel sheet where the same calculation is
being done then you can simply calculate very
00:49:09.090 --> 00:49:17.980
easily that what will be the V E S. Now, in
stokes regime
00:49:17.980 --> 00:49:25.890
what will happen the C D will be replaced
by 24 upon R e and R e is nothing, but 24
00:49:25.890 --> 00:49:33.590
upon u E S because we are writing it in terms
of this into d of particle into rho of fluid
00:49:33.590 --> 00:49:40.420
into mu of fluid. So, I will replace it here
and if you do it it will be V E S square will
00:49:40.420 --> 00:49:52.830
be equal to 8 into q into E upon this was
rho of fluid, ok. So, this will be rho of
00:49:52.830 --> 00:50:02.110
fluid into pi into d P square and C D value
will be what this will be 24, this will be
00:50:02.110 --> 00:50:11.290
mu f, this will be u E S, d P into rho f ok.
So, now, this rho f-rho f will be cancelled
00:50:11.290 --> 00:50:18.060
out. This u E S this will be cancelled out,
this will be 3 and this d P and d P will be
00:50:18.060 --> 00:50:30.780
cancelled out. So, you will get V E S will
be q E upon 3 pi mu f into d P.
00:50:30.780 --> 00:50:38.160
So, you will calculate the ESP value, V E
S value, the electric drift velocity values
00:50:38.160 --> 00:50:42.750
in terms of the stokes regime if the particle
Reynold number is less than 1. You can calculate
00:50:42.750 --> 00:50:47.850
in the general regime in general regime the
solution will be iterative in the stokes regime
00:50:47.850 --> 00:50:52.900
the solution will be straightforward and what
you have to do you can just find that how
00:50:52.900 --> 00:50:58.170
the electrical mobility will take place. So,
how the particle velocity will be there, so,
00:50:58.170 --> 00:51:03.810
that is the way the things can be defined
in most of the cases in ESP whatever the way
00:51:03.810 --> 00:51:15.450
we have discussed the ESP is used only for
micron or below particle size ok.
00:51:15.450 --> 00:51:21.020
So, the particle size should be less than
1.1 micron then ESP is used and most of the
00:51:21.020 --> 00:51:26.650
time we are in the stokes regime because particle
size is very very small, it pulls the pull
00:51:26.650 --> 00:51:32.190
Reynold number very low and you can use this
value to calculate that what is the V E S.
00:51:32.190 --> 00:51:37.370
If it is not in the stokes regime you can
this use this value you can do the iterative
00:51:37.370 --> 00:51:41.900
solution and you can find it out that what
is the E S value.
00:51:41.900 --> 00:51:48.160
Now, the only problem whatever we have discussed
till now is the value of C D and I said that
00:51:48.160 --> 00:51:54.180
the C D value for the stoke regime it is very
clear it is 24 by R e; other than the stokes
00:51:54.180 --> 00:51:58.450
regime it is difficult to calculate either
you have to do on the chart which will be
00:51:58.450 --> 00:52:02.980
changing with the particle diameter which
will be changing with the particle size or
00:52:02.980 --> 00:52:09.000
particle shape. So, many authors have done
lot of work to find the C D value and several
00:52:09.000 --> 00:52:12.850
C D correlation has been developed.
Now, in the next time what we are going to
00:52:12.850 --> 00:52:18.110
do we are going to discuss those C D correlations
that what are those C D correlations, how
00:52:18.110 --> 00:52:22.810
those C D correlation has been developed,
what is the advantage of each correlation
00:52:22.810 --> 00:52:27.180
and when to use which correlation. So, if
you know the C D correlation what you need
00:52:27.180 --> 00:52:32.810
to do? You can do all these calculations if
you have the C D correlation, you can just
00:52:32.810 --> 00:52:37.700
do the iteration find the C D value from the
correlation use the Reynold number guess the
00:52:37.700 --> 00:52:42.840
Reynold number find the C D value for that
Reynold number by using the correlation find
00:52:42.840 --> 00:52:48.280
the u V E S value or u t value for that and
then again keep on doing the iteration.
00:52:48.280 --> 00:52:53.410
So, we will see that the different drag loss
which is being developed by many researcher
00:52:53.410 --> 00:52:58.910
for the different applications and how that
drag loss can be implemented or can be integrated
00:52:58.910 --> 00:53:04.660
with this C D values to calculate the velocity
of the particle, with the time to calculate
00:53:04.660 --> 00:53:09.660
the position of the particle with the time
or to calculate the settling velocity or terminal
00:53:09.660 --> 00:53:26.350
velocity. So, that we will discuss in the
next class.
00:53:26.350 --> 00:53:30.760
Thank you .