WEBVTT
Kind: captions
Language: en
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So, welcome back. In the last class what we
were discussing is about the equation of the
00:00:29.440 --> 00:00:32.490
2-fluid model.
And we have discussed those equations. So,
00:00:32.490 --> 00:00:36.550
I will just briefly revise it again that what
are the equations we have discussed. So, we
00:00:36.550 --> 00:00:42.420
said that in 2 field model what you do you
solve individual equations for continuity,
00:00:42.420 --> 00:00:46.680
and momentum equation for each phases. So,
it means suppose if you have Two-phase, you
00:00:46.680 --> 00:00:49.609
solve 2 continuity equations and 2 momentum
equation.
00:00:49.609 --> 00:00:53.980
So, if both the phases are continuous, what
we do we solve continuity equation which is
00:00:53.980 --> 00:01:01.570
given by this this is dou by dou t of alpha
rho q plus del dot alpha rho qv v q equal
00:01:01.570 --> 00:01:06.670
to 0. This is the transient term, on a steady
state mass transfer term or transient term
00:01:06.670 --> 00:01:15.490
or the steady state term state and this is
convective term. So, convective transport
00:01:15.490 --> 00:01:19.730
so, that is equal to 0 and if suppose there
is some generation this will be actually equal
00:01:19.730 --> 00:01:22.570
to some generation.
But if we are not talking about any reactive
00:01:22.570 --> 00:01:26.840
system then it will be equal to 0. And for
a steady state system, what will happen that
00:01:26.840 --> 00:01:35.789
this 2 this term will be 0 for a steady state
ok , state this term will be 0. So, you will
00:01:35.789 --> 00:01:43.329
be getting this del dot alpha rho qv will
be equal to 0 ok. So, that is whatever the
00:01:43.329 --> 00:01:48.040
way you write this equation if rho is constant.
You can put it again for incompressible rho
00:01:48.040 --> 00:01:54.490
you can pull it out. So, this equation will
be rho del dot alpha v and that will be equal
00:01:54.490 --> 00:01:59.259
to 0, and this will be rho will be 0 .
So, finally, you will have a term del dot
00:01:59.259 --> 00:02:06.210
alpha into v will be equal to 0 for incompressible
steady state flow ok so, this way it will
00:02:06.210 --> 00:02:10.920
be simplified. Now you solve the same equation
for the both the phases. So, say I solved
00:02:10.920 --> 00:02:14.870
this for the q phase here the similar equation
you have to write for the s phase. So, it
00:02:14.870 --> 00:02:19.640
will be 2 continuity equation ok. So, one
will be for this alpha q this will be alpha
00:02:19.640 --> 00:02:27.660
q v q another equation will be del dot alpha
s vs that will be equal to 0. So, that is
00:02:27.660 --> 00:02:31.750
the one stage that will be the second.
So, one continuity equation one, continuity
00:02:31.750 --> 00:02:37.271
equation 2, this 2-continuity equation you
have to solve ok. Now for momentum again you
00:02:37.271 --> 00:02:40.970
have to solve the 2-momentum equation, now
this momentum equation is very close to the
00:02:40.970 --> 00:02:45.920
navier stoke equation ok, we have assumed
that mu is constant if you assume mu and rho
00:02:45.920 --> 00:02:50.220
is constant navier stokes equation. Or you
basically write a general momentum equation
00:02:50.220 --> 00:02:54.980
in terms of the tau. So, it is very close
to that momentum equation the only thing is,
00:02:54.980 --> 00:02:59.350
what we have done we have multiplied the each
term with alpha. It means the fraction volume
00:02:59.350 --> 00:03:03.790
fraction of that phase, and then you will
do the same thing for the second phase.
00:03:03.790 --> 00:03:08.220
So, we will multiply it is a by alpha s which
will be the volume fraction of the that phase.
00:03:08.220 --> 00:03:13.090
So, again this is the term this is the momentum
equation. The first term is under steady state
00:03:13.090 --> 00:03:18.700
acceleration or local acceleration. This is
convective acceleration term ok, this is pressure
00:03:18.700 --> 00:03:23.160
term which is multiplied by alpha q it means
that kind of contribution from that phase.
00:03:23.160 --> 00:03:27.960
Then tau del, dot tau; now because the tau
we are taking the velocity of that phase only,
00:03:27.960 --> 00:03:32.110
this is not multiplied with the alpha.
Then the overall body force term, which is
00:03:32.110 --> 00:03:38.441
being because of that phase. So, this is alpha
q rho into g. So, this is body force, let
00:03:38.441 --> 00:03:50.840
me write it again for the sake of simplicity.
This will be body force, viscous force , force,
00:03:50.840 --> 00:03:59.070
this is pressure and this is drag. And we
have already discussed that why the drag is
00:03:59.070 --> 00:04:03.950
written it it in this way, and the Ks q value
will be written for the different drag closures,
00:04:03.950 --> 00:04:08.620
the way we have already discussed in the during
the drag force discussion. And any issue of
00:04:08.620 --> 00:04:12.040
force if you multiply here.
Now, this is convective term, this is the
00:04:12.040 --> 00:04:15.390
local acceleration term. All the terms is
being multiplied with the alpha that in the
00:04:15.390 --> 00:04:19.919
viscous force term, because that will take
care of the velocity of that phase only. So,
00:04:19.919 --> 00:04:23.680
ideally speaking, if you write in terms of
the superficial velocity, again that alpha
00:04:23.680 --> 00:04:28.560
q term will be there. So, this is the way
we write, and if suppose the second phase
00:04:28.560 --> 00:04:32.389
is liquid ok or continuous fluid.
Suppose, if you are solving the 2-fluid model
00:04:32.389 --> 00:04:37.230
for bubble column, where both the phases are
actually a fluid whether it is a water and
00:04:37.230 --> 00:04:41.910
air both are fluid, you solve the exactly
same equation for the gas phase also. So,
00:04:41.910 --> 00:04:45.650
the same equation you will solve for the gas,
which same equation you will sell for the
00:04:45.650 --> 00:04:50.569
liquid phase. So, you both are fluid continuous
fluid you solve it it in this way; however,
00:04:50.569 --> 00:04:55.960
the equation modified slightly, if the one
phase is solid ok. It means discontinuous
00:04:55.960 --> 00:05:01.180
by nature itself. So, for the solid the equation
will be modified. If you will see that all
00:05:01.180 --> 00:05:06.939
the other terms are the same, only thing is
you introduce a new term of del ps which is
00:05:06.939 --> 00:05:16.130
the solid pressure term ok, solid pressure.
This term you include, viscous forces because
00:05:16.130 --> 00:05:28.310
of the solid, forces body force again same
body force, this is drag, any additional force
00:05:28.310 --> 00:05:32.669
this is the local acceleration of the solid
convective acceleration. So, this is the acceleration
00:05:32.669 --> 00:05:36.749
completely so, local plus convective. So,
in that way you write for the solid the only
00:05:36.749 --> 00:05:41.139
thing you have to introduce a solid pressure
term extra. If there is continuous flow this
00:05:41.139 --> 00:05:45.229
term will not be there rest every term will
be there. As it is instead of the tau del
00:05:45.229 --> 00:05:50.880
dot tau s you will be writing that del dot
tau l say for the liquid or tau a for the
00:05:50.880 --> 00:05:55.189
air ok, or for the gas.
So, that is the way we write the equation
00:05:55.189 --> 00:06:00.999
into fluid model, and the basic assumption
is both the phases are assumed to be continuum
00:06:00.999 --> 00:06:05.969
interpenetrating. So, even if it is a solid,
we are assuming it to be continuous, and that
00:06:05.969 --> 00:06:11.409
actually bring lot of a junction in this model.
The very advantage of the model as discussed
00:06:11.409 --> 00:06:16.190
that you are solving only 4 equation ok, if
you do not solve other interactions. So, you
00:06:16.190 --> 00:06:21.360
are just solving 4 equations, and with the
4 equations you are getting the whole solution
00:06:21.360 --> 00:06:25.400
for the whole geometry, whole body or kind
of whole domain.
00:06:25.400 --> 00:06:31.110
Now, it is computationally much cheaper compared
to the other model which we are going to discuss.
00:06:31.110 --> 00:06:36.229
The only disadvantage is even the continuous
phase you make it interpenetrating continuum,
00:06:36.229 --> 00:06:40.749
even the discrete phase sorry you make it
interpenetrating continuum. And that is the
00:06:40.749 --> 00:06:45.169
major assumptions of this model. So, this
solid pressure and all those terms we will
00:06:45.169 --> 00:06:48.849
discuss, and as I said that the major problem
comes with the solid pressure, because solid
00:06:48.849 --> 00:06:53.099
do not put any pressure while flowing.
And the major problem is with this tau s,
00:06:53.099 --> 00:06:58.139
because the tau in the basic basic notion
has been defined as if this is proportional
00:06:58.139 --> 00:07:04.159
to dou v by dou x say or velocity gradient
and the proportionality constant is viscosity
00:07:04.159 --> 00:07:11.219
mu. Now the way the very notion that tau has
been defined cannot be do cannot be used here,
00:07:11.219 --> 00:07:15.340
because solid do not have any viscosity. Or
you can say solid hasn't finite viscosity.
00:07:15.340 --> 00:07:20.479
So, you have to write several equation which
will be written in terms of how the tau will
00:07:20.479 --> 00:07:25.120
be defined for the continuous phase, how the
tau will be defined for the solid phase.
00:07:25.120 --> 00:07:29.010
Now, for continuous phase, this equation is
pretty much same whatever you have done in
00:07:29.010 --> 00:07:34.069
your undergraduate courses or the lower level
courses of transport phenomena, the tau is
00:07:34.069 --> 00:07:40.270
nothing but mu dou v del v plus del v transpose
plus 2 by 3 mu minus k del dot v into I. Now
00:07:40.270 --> 00:07:46.770
this term is basically dilatational viscosity,
and that dilatation is viscosity comes into
00:07:46.770 --> 00:07:51.169
the picture, once there is any change in the
volume. So, for the incompressible flow, the
00:07:51.169 --> 00:07:55.099
change in volume will not be there, anyway
for incompressible flow from continuity equation
00:07:55.099 --> 00:08:06.819
del dot v equal to 0 for incompressible flow
this this term is going to be 0, it means
00:08:06.819 --> 00:08:11.110
whole this term will be removed.
And for compressible flow yes there can be
00:08:11.110 --> 00:08:14.689
change in the volume dilatational viscosity
will be there. So, this term will also be
00:08:14.689 --> 00:08:19.250
playing the role. But most of the time the
flow we deal with generally in chemical engineering
00:08:19.250 --> 00:08:23.430
the most of the fluid are incompressible in
nature rather than the gases. So, you can
00:08:23.430 --> 00:08:28.870
neglect this term easily. Then this is the
mu del v plus del v tau and I do not want
00:08:28.870 --> 00:08:33.500
to do go in detail of that, but just to revise
it a little bit, it has been done because
00:08:33.500 --> 00:08:37.690
tau has been found out asymmetric tensors,
symmetric second order tensor. So, that is
00:08:37.690 --> 00:08:41.340
why it will give the symmetric nature we write
this in this form.
00:08:41.340 --> 00:08:46.620
So, suppose if I write now tau xy, ideally
if they are symmetric tau xy will be equal
00:08:46.620 --> 00:08:54.460
to tau y x, it means, tau xy if you remember
will be defined as mu dou v y up on dou x
00:08:54.460 --> 00:09:01.400
and tau yx will be defined at mu dou z x upon
the y. So, they are not actually equal, but
00:09:01.400 --> 00:09:05.250
they are symmetric matrix they will be equal.
And that is why the another term has been
00:09:05.250 --> 00:09:10.340
added which is the transpose so, if I do the
transpose of this I will write it this, dou
00:09:10.340 --> 00:09:15.050
vx upon the y here it will be dou vy upon
dou x.
00:09:15.050 --> 00:09:18.570
Now, this becomes a transpose, it now this
2 becomes equal. So, that is why we write
00:09:18.570 --> 00:09:22.880
the tau it it in this way it is a 100 year
of research I do not want to go in detail
00:09:22.880 --> 00:09:27.770
of that, but if you want more on this, we
can discuss it in the forum or we can discuss
00:09:27.770 --> 00:09:31.970
it you can drop me a mail, and you can also
go through your transport phenomena books.
00:09:31.970 --> 00:09:36.490
So, that is the way it has been defined and
as I said that this will be like this, it
00:09:36.490 --> 00:09:41.500
will be defined it it in this way and it this
will come it it in this form.
00:09:41.500 --> 00:09:46.310
So, once this is tau x x actually this will
be 2 because dou vx plus dou v x will be there,
00:09:46.310 --> 00:09:52.680
tau xy it will be mu dou v upon dou x plus
dou v x upon dou y. So, that is the way continuous
00:09:52.680 --> 00:09:56.870
phase shear stress has been defined.
Now, coming to the solid shear stress. So,
00:09:56.870 --> 00:10:02.500
as I said this this is very critical and will
kind of left our discussion last time in this
00:10:02.500 --> 00:10:08.360
place only, because what you need to do now
you have to define mu of solid. Now as I said
00:10:08.360 --> 00:10:12.710
that solid actually do not deform if you put
it in the strain. So, if I put it like this
00:10:12.710 --> 00:10:16.130
if I rub my hand like this, I am not deforming.
So, that is the first definition.
00:10:16.130 --> 00:10:21.260
So, it means there is no viscosity in this,
or in other way you can say that in finite
00:10:21.260 --> 00:10:27.400
viscosity huge resistance and any stress or
any strain rate is not going to change. Anything
00:10:27.400 --> 00:10:32.370
ideally the mu is definition itself is pretty
much critical, and if you want to use in the
00:10:32.370 --> 00:10:37.780
2-phase model in this form, you need to define
the mu s, and there the problem starts and
00:10:37.780 --> 00:10:43.390
many people like the drag different researcher
has different different relation. Here also
00:10:43.390 --> 00:10:47.830
different researcher has given different relation
and different kind of correlation for the
00:10:47.830 --> 00:10:53.750
mu s calculation. And most of those mu s calculation
which has been said is actually the mu s is
00:10:53.750 --> 00:10:57.430
going to be the part of the 3-mu collision,
mu kinetic and mu friction.
00:10:57.430 --> 00:11:03.170
So, in that way the mu s has been breaking,
that why you are going to see why the solid
00:11:03.170 --> 00:11:08.510
will see a resistance in their motion, because
the mu is nothing but has been defined, it
00:11:08.510 --> 00:11:13.420
is a resistance in their motion. Now the resistance
in the motion can be because of for the solids,
00:11:13.420 --> 00:11:17.690
it can be because of the collision because
they are having collision with one other particle
00:11:17.690 --> 00:11:22.980
their motion is waiting restricted. Then this
kinetic energy because of that there is kind
00:11:22.980 --> 00:11:27.320
of motion is random and if their motion will
be random there again they can hit the other
00:11:27.320 --> 00:11:31.050
particle, their motion can be restricted and
definitely the friction will restrict the
00:11:31.050 --> 00:11:34.820
motion of the solid.
So, this is the 3 way the mu as said that
00:11:34.820 --> 00:11:40.890
the mu of solid is actually the summation
of these 3 resistances. And now this resistances
00:11:40.890 --> 00:11:46.120
are actually being defined by the different
researcher, and given the different correlation,
00:11:46.120 --> 00:11:52.340
that collisional shear stress has been given
by this 4 by 5, alpha s rho s d s g 1 plus
00:11:52.340 --> 00:11:57.470
epsilon and this s into phi s theta s upon
pi. Where, theta is nothing but the granular
00:11:57.470 --> 00:12:03.360
temperature. This is nothing but the granular
temperature. And we will discuss about that
00:12:03.360 --> 00:12:08.071
what is granular temperature.
Later on, this is restitution coefficient
00:12:08.071 --> 00:12:15.990
, again we will discuss about the restitution
coefficient after sometime. This is radial
00:12:15.990 --> 00:12:28.460
distribution function ,
rho s is the density of this solid, alpha
00:12:28.460 --> 00:12:32.920
s is the fraction of the solid which is present,
ds is the diameter of the solid. So, in this
00:12:32.920 --> 00:12:37.680
way it has been defined, but this is not the
only correlation several correlation is available
00:12:37.680 --> 00:12:42.650
lun et al syamlal gidaspow lot of correlation
you will if you go and see the books, you
00:12:42.650 --> 00:12:46.940
see the literature you will see all these
names will be featured that different correlations
00:12:46.940 --> 00:12:51.950
are there by the different researcher.
But what is the notion if you see? The notion
00:12:51.950 --> 00:12:56.570
is how well the particles are flat packed,
how close the particles are there that is
00:12:56.570 --> 00:13:02.490
why the alpha s is being taken into the account.
Then, your gravitational acceleration of the
00:13:02.490 --> 00:13:05.961
particle is being taken into the account,
that how the particle is accelerating, how
00:13:05.961 --> 00:13:10.340
the particles are packed near about, that
is why the radial distribution function has
00:13:10.340 --> 00:13:14.760
been there how they are readily distributed,
the restitution coefficient which says that
00:13:14.760 --> 00:13:19.790
after this the collision whether the collision
is elastic or un elastic, elastic means they
00:13:19.790 --> 00:13:24.380
have they will hit each other and they will
go back to the same place. Completely inelastic
00:13:24.380 --> 00:13:28.910
they will hit each other the velocity will
be dead. So, they will not move or if the
00:13:28.910 --> 00:13:34.280
restitution coefficient value between the
0 to 1, if one means completely elastic 0,
00:13:34.280 --> 00:13:36.810
completely inelastic in between they will
move certain distance.
00:13:36.810 --> 00:13:40.580
You can find it out what will be the probability
of the collision, and after each collision
00:13:40.580 --> 00:13:44.950
how far they will move, that will define the
probability of the other collision, theta
00:13:44.950 --> 00:13:49.340
s is the granular temperature, it is being
defined as a function of kinetic energy of
00:13:49.340 --> 00:13:54.180
the fluctuations it means how much kinetic
energy it tells how fast it is moving or fluctuating
00:13:54.180 --> 00:13:59.650
from it is location. So, those all combined
together define that what is the mean collision.
00:13:59.650 --> 00:14:03.840
So, these parameters remain same only the
powers and the empirical coefficient values
00:14:03.840 --> 00:14:08.690
used to be changed for the different correlation.
Similarly, the kinetic viscosity has been
00:14:08.690 --> 00:14:13.230
defined again, if you will see that this is
the function of granular temperature, this
00:14:13.230 --> 00:14:18.230
is the function of e s which is the ristitution
coefficient, this is function of g naught
00:14:18.230 --> 00:14:22.080
which is the radial distribution function
and the volume fraction also. These are our
00:14:22.080 --> 00:14:26.940
empirically developed correlation based on
several experiments this has been developed,
00:14:26.940 --> 00:14:31.840
and they found that how this coefficients
or this this the kinetic energy viscosity
00:14:31.840 --> 00:14:35.970
will be correlated, or how you can define
the viscosity for the solids.
00:14:35.970 --> 00:14:40.840
Similarly, the friction viscosity the frictional
viscosity is also being defined, and that
00:14:40.840 --> 00:14:45.580
this very close to the collisional viscosity
in the way is bulk viscosity has been defined
00:14:45.580 --> 00:14:49.850
similarly, and that is also very close to
the collisional viscosity. Only the power
00:14:49.850 --> 00:14:56.010
differences are there. So, overall the viscosities
has been defined for the solid, and this viscosity
00:14:56.010 --> 00:15:01.300
is being used actually for the solids calculation
of the solid viscosity all these individuals
00:15:01.300 --> 00:15:05.930
as being used, and then you calculate mu s
by summing of all these 3. So, you get that
00:15:05.930 --> 00:15:11.890
mu s value, and you define all these 3.
Now, the problem is if you are using it for
00:15:11.890 --> 00:15:16.960
the solid, the accuracy of your prediction
is going to depend that how accurately you
00:15:16.960 --> 00:15:21.250
are choosing these models. So, what you are
doing? You are increasing more and more modeling
00:15:21.250 --> 00:15:27.200
part. And modeling which is based on the coefficients
are calculated based on some empirically developed
00:15:27.200 --> 00:15:32.000
correlations. So, drag was already there now
we have introduced more models ok, more empirically
00:15:32.000 --> 00:15:37.160
developed models or more models. Now that
actually make the model use of the model little
00:15:37.160 --> 00:15:41.780
bit complicated or tricky and one need to
be very careful while using the 2 fluid models,
00:15:41.780 --> 00:15:46.500
you should understand that what are the closures
which you are using?
00:15:46.500 --> 00:15:51.480
What are the equations you are using, what
is the limitation of each equation, and if
00:15:51.480 --> 00:15:56.800
you use this equation how your predictability
will be there. And because there are lot of
00:15:56.800 --> 00:16:03.080
undefined function, if you get the solution
your experimental validation becomes a must.
00:16:03.080 --> 00:16:08.240
If you are not validating experimentally,
these models are not matured enough to give
00:16:08.240 --> 00:16:10.930
you the right solution particularly for multi-phase
flow.
00:16:10.930 --> 00:16:16.050
So, what you need to do? The experimental
validation becomes must. So, all this simulation
00:16:16.050 --> 00:16:21.220
at least you need to experimentally validate
to find that all the closures parameter all
00:16:21.220 --> 00:16:26.110
these equations or all this the closure equations
which we have used, you have used correctly
00:16:26.110 --> 00:16:31.980
and has a validity for this the experimental
law for the setup or you have used or for
00:16:31.980 --> 00:16:36.220
the condition for which you are operating.
And that validation should not be at a one
00:16:36.220 --> 00:16:40.870
label it should be a different label for different
operating conditions, it should not be only
00:16:40.870 --> 00:16:46.110
volume fraction, it should not be only mean
velocity, because everywhere the kinetic energy
00:16:46.110 --> 00:16:51.100
kind of a granular temperature is taking play
a role, it should also be validated at the
00:16:51.100 --> 00:16:55.200
fluctuation label. We discuss how the granular
temperature is being defined we have already
00:16:55.200 --> 00:16:59.200
discussed it earlier we will discuss the equation
of the granular temperature completely.
00:16:59.200 --> 00:17:04.209
But because everything is a function of granular
temperature, you need to also validate it
00:17:04.209 --> 00:17:08.759
with the fluctuation velocity. So, if you
are using 2 fluid model, you are using algebraic
00:17:08.759 --> 00:17:13.730
slip model, you are using lot of empirical
correlation or assumed values, and because
00:17:13.730 --> 00:17:19.829
of that the experimental validation becomes
very, very must very necessary, and it should
00:17:19.829 --> 00:17:23.990
not be just validated at the mean velocity
level or mean condition level, it should be
00:17:23.990 --> 00:17:29.150
validated at mean level and at least at the
fluctuation level. You should match the fluctuation
00:17:29.150 --> 00:17:33.429
velocities too or you should match the kinetic
energy of the fluctuations.
00:17:33.429 --> 00:17:38.120
So, that is the way it has been defined similarly
the solid pressure has been defined again
00:17:38.120 --> 00:17:43.610
if you will see, it is a function of theta
s which is the granular temperature, restitution
00:17:43.610 --> 00:17:48.749
coefficient and radial distribution function.
All this is the function of that and if you
00:17:48.749 --> 00:17:53.259
will see this this is nothing but the way
it has been defined is exactly same way as
00:17:53.259 --> 00:17:57.049
the pressure has been defined how the pressure
is being defined in the kinetic theory of
00:17:57.049 --> 00:18:01.539
gases, that once the particle particle of
molecules goes and hit the walls.
00:18:01.539 --> 00:18:06.919
So, number of collisions per unit time and
based on that we define that how much force
00:18:06.919 --> 00:18:11.940
will be acting on that wall per unit area,
we calculate the pressure. Exactly same way
00:18:11.940 --> 00:18:16.149
it has been defined here that how many number
of particles are available, what is their
00:18:16.149 --> 00:18:21.179
probability of hitting towards any wall or
hitting towards the each other, in that way
00:18:21.179 --> 00:18:25.950
it has been defined here also solid pressure
is also being defined. Over all this the solid
00:18:25.950 --> 00:18:29.370
pressure is there this is again one of the
equation which is given by lun et al, but
00:18:29.370 --> 00:18:33.379
this is not the only equation several other
equation is there, but all will be the function
00:18:33.379 --> 00:18:38.100
of granular temperature radial distribution
function and the ristitution coefficient.
00:18:38.100 --> 00:18:43.100
Again, you have to validation becomes must,
and validation in which you cannot you cannot
00:18:43.100 --> 00:18:47.999
only validate at mean level you have to also
validate at fluctuation level. And this granular
00:18:47.999 --> 00:18:53.360
temperature will discuss that how it has been
done, these kind of discuss all this the radial
00:18:53.360 --> 00:18:58.100
distribution function is being defined, in
this way that this is 1 minus alpha s volume
00:18:58.100 --> 00:19:03.450
fraction at that place divided by the maximum
volume fraction which is possible raised to
00:19:03.450 --> 00:19:06.429
the power 1 by 3 whole inverse.
So, that is the way the radial distribution
00:19:06.429 --> 00:19:12.370
function g naught has been defined. Now we
have used drag coefficient to solve it, we
00:19:12.370 --> 00:19:16.690
have discussed about the drag a lot during
our previous discussion, but what we have
00:19:16.690 --> 00:19:23.070
done instead of writing in terms of the fd,
we have written in terms of the Ks q and multiplied
00:19:23.070 --> 00:19:28.669
by the slip velocity. Now if this is 2 way
coupling the particle will the kind of particle
00:19:28.669 --> 00:19:33.809
will put a drag, on fluid fluid will put a
drag on particle the equation will be same.
00:19:33.809 --> 00:19:39.250
So, Ks q will be equal to Kq s, it means both
drag will be equal only the directions will
00:19:39.250 --> 00:19:44.649
be different, and that will be defined at
alpha s rho into friction factor upon tau
00:19:44.649 --> 00:19:50.909
s, in that way it will be defined. And then
this tau s is nothing but the particle relaxation
00:19:50.909 --> 00:19:54.940
times we have already discussed that way earlier
also, that once you write it in terms of the
00:19:54.940 --> 00:19:59.580
f the only change is we are writing in terms
of the particle relaxation time also. So,
00:19:59.580 --> 00:20:05.149
tau s is being defined as rho dp square upon
18 mu, that we have already discussed. The
00:20:05.149 --> 00:20:09.870
mu will be for the continuous phase. D s will
and rho will be for the discrete feature.
00:20:09.870 --> 00:20:17.249
So, this is for discrete phase say solid,
this will be for solid, and this will be for
00:20:17.249 --> 00:20:23.620
fluid. So, that is the way this particle relaxation
time has been defined, f is the drag factor
00:20:23.620 --> 00:20:28.970
which is nothing but c d upon 24 upon re the
only thing we have done, we are multiplied
00:20:28.970 --> 00:20:33.240
with the volume fraction because, we know
that the fraction of the solid present is
00:20:33.240 --> 00:20:37.629
going to play a role. So, volume fraction
we have multiplied of that phase, and then
00:20:37.629 --> 00:20:43.019
divided by v square.
Then again one of the typical drag correlations
00:20:43.019 --> 00:20:46.941
syamlal o brien we have discussed several
drag correlation, I am not going to use discuss
00:20:46.941 --> 00:20:51.080
all these things here again, but these are
certain correlation you can use syamlal, you
00:20:51.080 --> 00:20:56.769
can use gidaspow, you can use ergun you can
use wen yu any correlation depending upon
00:20:56.769 --> 00:21:00.920
what is valid, which drag coefficient is valid
in your case.
00:21:00.920 --> 00:21:06.320
So, you can use this equations there. V square
r is nothing but the terminal velocity of
00:21:06.320 --> 00:21:11.559
the solid it has been defined with that and
terminal velocity of the solid can be calculated
00:21:11.559 --> 00:21:15.299
the way we have already learned that how to
calculate the terminal velocity or settling
00:21:15.299 --> 00:21:20.299
velocity of the solid. So, that is what we
can calculate here. Reynold number is always
00:21:20.299 --> 00:21:24.770
based on the slip velocity as I already discussed
in the running the discussion of the drag,
00:21:24.770 --> 00:21:29.090
that once we talk about the drag Reynold number
we always define based on the slip velocity.
00:21:29.090 --> 00:21:34.960
We did it here also in the slip velocity,
these diameter rho is the density of the fluid,
00:21:34.960 --> 00:21:39.370
mu is the density of the fluid, d is the diameter
of the solid and it will be the slip velocity.
00:21:39.370 --> 00:21:43.950
So, that is the way Reynold number has been
defined. And you can calculate the drag coefficient.
00:21:43.950 --> 00:21:48.830
So, you can calculate all these parameters,
again in the drag what you need to do these
00:21:48.830 --> 00:21:53.440
are the correlation which are empirically
developed. And because these are correlations
00:21:53.440 --> 00:22:00.430
which are empirically developed their validation
is necessary their use is limited; their accuracy
00:22:00.430 --> 00:22:05.490
is limited and again your accuracy of the
prediction is going to depend on the accuracy
00:22:05.490 --> 00:22:09.909
of the models or the closure equations you
are using to predict the flow.
00:22:09.909 --> 00:22:15.230
So, that is the way it has been defined granular
temperature I have discussed a lot earlier
00:22:15.230 --> 00:22:19.830
also, that what is the granular temperature
it is actually a quantity which is analogous
00:22:19.830 --> 00:22:25.159
to the thermodynamic temperature, and that
comes because of the fluctuation motion, fluctuation
00:22:25.159 --> 00:22:29.950
motion of the solid, and that is being actually
validated this will be equal to the kinetic
00:22:29.950 --> 00:22:35.999
energy of the solid is equivalent is being
validated with the analogous temperature that
00:22:35.999 --> 00:22:42.370
is actually 3 by 2 k t and this tk is k and
t, this will be equivalent to half rho v square
00:22:42.370 --> 00:22:48.139
and from there, you actually calculate that
what will be your kinetic in your temperature
00:22:48.139 --> 00:22:52.299
or your grandeur temperature.
So, it means if you just remember 3 by 2 is
00:22:52.299 --> 00:23:00.509
equal to half rho v square, half rho v square,
it means 3 by 2 kt is half rho v square, that
00:23:00.509 --> 00:23:06.100
is the way the molecular in the any gas kinetic
theory of gases we use that t is the thermodynamic
00:23:06.100 --> 00:23:11.899
temperature k is the constants Boltzmann constant,
and rho and v square is the particle velocities
00:23:11.899 --> 00:23:16.850
or particle fluctuating velocities. So, this
half rho v square, we can write it in terms
00:23:16.850 --> 00:23:20.270
of the kinetic energy of the fluctuation if
you are writing in terms of the fluctuation
00:23:20.270 --> 00:23:28.409
velocities, this is called Ks and actually
it is being calculated as half rho v x prime
00:23:28.409 --> 00:23:35.030
square, plus v y prime square, plus v z prime
square. It means kinetic energy of fluctuation
00:23:35.030 --> 00:23:40.080
in the x direction fluctuation in the y direction
fluctuation in the z direction all the kinetic
00:23:40.080 --> 00:23:43.659
energy.
You add together, and that is what is the
00:23:43.659 --> 00:23:48.639
k s is the total kinetic energy of the solids.
Now that we are representing with the solids,
00:23:48.639 --> 00:23:53.580
this is Ks half rho v square, now instead
of mv square because it is a continuous fluid
00:23:53.580 --> 00:23:58.059
we are writing in terms of the rho in not
in terms of the m; so, it will be per unit
00:23:58.059 --> 00:24:03.379
volume. So, that is what we are going to do,
and we will do it into the terms of per unit
00:24:03.379 --> 00:24:08.610
mass. So, that the rho will go out and we
will do it per unit mass and volume we will
00:24:08.610 --> 00:24:13.159
see that for each cell if you do it, then
Ks will be defined as if you are writing per
00:24:13.159 --> 00:24:19.450
unit mass that will be half v prime square.
Once I came v prime square it is v x prime
00:24:19.450 --> 00:24:24.750
square plus v y prime square plus vz prime
square. So, Ks is kinetic energy of the solid
00:24:24.750 --> 00:24:29.399
fluctuation per unit mass. Because we have
divided it by rho. So, this will be becomes
00:24:29.399 --> 00:24:33.830
now per unit watts, if you multiply with the
rho, it will be per unit volume if you divide
00:24:33.830 --> 00:24:38.789
by the rho it will be per unit mass. That
is the way we define the kinetic energy of
00:24:38.789 --> 00:24:44.139
fluctuations this will be Ks. Now this is
equivalent to the granular temperature, kt
00:24:44.139 --> 00:24:48.799
which is proportional to the kt a quantity
we have given theta, which is analogous to
00:24:48.799 --> 00:24:53.460
the thermodynamic temperature.
So, if you write it 3 by 2 theta will be equal
00:24:53.460 --> 00:25:02.169
to ks. So, theta is nothing but 2 by 3 of
Ks . That is the way we define, we use the
00:25:02.169 --> 00:25:06.700
fluctuation we calculate the fluctuation motion
we calculate the motion fluctuation in all
00:25:06.700 --> 00:25:12.340
the 3 directions, and then the summit together
multiply it by 2 by 3, we get the granular
00:25:12.340 --> 00:25:16.909
temperature quantity which is analogous to
the thermodynamic temperature. We have already
00:25:16.909 --> 00:25:21.389
discussed about this in the previous classes
that how that granular temperature is being
00:25:21.389 --> 00:25:24.059
done.
With the kinetic theory of granular flow,
00:25:24.059 --> 00:25:28.769
you can actually derive the transport of the
solid, and then you can find that the transport
00:25:28.769 --> 00:25:33.909
equation for the solid fluctuation kinetic
energy. And that transport equation for the
00:25:33.909 --> 00:25:39.549
solid fluctuation kinetic energy can be converted
in terms of the granular temperature equation,
00:25:39.549 --> 00:25:44.340
I am not going in detail of this, but if you
want you can go and see this discussion, this
00:25:44.340 --> 00:25:49.830
is a kind of derivation this is chapman cowling
book is there where this derivation is being
00:25:49.830 --> 00:25:53.470
given you can find it out that how this things
has been derived.
00:25:53.470 --> 00:25:58.809
So, you can do that derivation. So, this is
if you do that, you will find that, this is
00:25:58.809 --> 00:26:04.710
actually the granular temperature equation
is being very close to the momentum equation.
00:26:04.710 --> 00:26:09.340
The only thing is the momentum equation you
take velocity, in the granular temperature
00:26:09.340 --> 00:26:14.759
you take the the temperature and again please
remember, that is nothing but the kinetic
00:26:14.759 --> 00:26:19.510
energy of fluctuation, it means it is nothing
but the summation of fluctuation in all the
00:26:19.510 --> 00:26:23.769
3 directions. So, that is the way it comes
and then this equation if you want you can
00:26:23.769 --> 00:26:27.779
derive, we can also discuss over the forum
if you have any question about this.
00:26:27.779 --> 00:26:34.690
So, if you see this this is 3 by 2, dou by
dou t, del rho alphas in to theta s. So, if
00:26:34.690 --> 00:26:40.980
you see this term then plus del dot rho alpha
s vs into theta s. So, this is the local un
00:26:40.980 --> 00:26:46.129
steady state term, this is the convective
fluctuation term. So, this is un steady state
00:26:46.129 --> 00:26:50.120
this is the convective way how the flux whole
fluctuation is being transferred from one
00:26:50.120 --> 00:26:57.440
location to another location that is the convective
term . So, you can say convective acceleration
00:26:57.440 --> 00:27:08.119
term of the fluctuation, this is the unsteady
state term, the first term is del ps into
00:27:08.119 --> 00:27:12.700
I and ps into I is the generation of energy
by the solid stress tensor.
00:27:12.700 --> 00:27:18.059
So, this is the first term which is being
earned with all of the solid stress tensor,
00:27:18.059 --> 00:27:24.399
that how much energy is being kind of generated
because of this. So, viscous dissipation,
00:27:24.399 --> 00:27:28.320
because of that the solid stress term, how
much energy is generated. So, that is the
00:27:28.320 --> 00:27:31.970
term is being given here. So, this is actually
we are doing the complete energy balance.
00:27:31.970 --> 00:27:36.250
So, how much energy under steady state term
is there, how much energy has been transported.
00:27:36.250 --> 00:27:40.370
So, total energy acceleration fluctuation
energy, then how it is being consumed.
00:27:40.370 --> 00:27:45.019
So, it can be because of the energy dissipation
or generation how much generation will be
00:27:45.019 --> 00:27:49.100
there because of the viscous dissipation.
So, this is this term is being given. Now
00:27:49.100 --> 00:27:53.539
the second term is actually going to talk
about the energy diffusion. So, whatever energy
00:27:53.539 --> 00:27:58.679
has been transported, how it is diffusing
out, and then the third term is actually the
00:27:58.679 --> 00:28:03.080
collision dissipation, that how much is the
collisional dissipation of the solid is there.
00:28:03.080 --> 00:28:08.700
And the 4th term is the interface energy exchange.
So, how much energy is being exchanged between
00:28:08.700 --> 00:28:13.340
the interfaces. That is the way the kinetic
theory of the granular flow is being defined.
00:28:13.340 --> 00:28:17.580
This is a big derivation, I am not doing in
detail of this, because that is not the scope
00:28:17.580 --> 00:28:22.330
of this course we want to just show you the
equation. So, that I can just tell you this
00:28:22.330 --> 00:28:26.710
or know that once we are solving the kinetic
theory of granular flow, momentum equation
00:28:26.710 --> 00:28:30.299
with the kinetic theory of granular flow what
equation, you are solving then you are solving
00:28:30.299 --> 00:28:33.519
actually this equation.
So, what we do? For the gas solid flow in
00:28:33.519 --> 00:28:38.619
2, fluid model equation or Euler Euler equations
once you solve for the gas solid flow, generally
00:28:38.619 --> 00:28:43.670
we solve the kinetic theory of granular flow.
So, 2 fluid model or Euler Euler equation
00:28:43.670 --> 00:28:47.629
with kinetic theory of granular flow. So,
once you solve with the kinetic theory of
00:28:47.629 --> 00:28:52.690
granular flow, you actually solve this equation.
And for theta s calculation you do 2 by 3
00:28:52.690 --> 00:28:57.490
Ks, for the ps and tau I again you use the
correlation which we have discussed earlier,
00:28:57.490 --> 00:29:01.809
just before one or 2 slides.
The way we saw. So now, if you solve the kinetic
00:29:01.809 --> 00:29:05.629
theory of granular flow the 2-fluid model
with the kinetic theory of granular flow what
00:29:05.629 --> 00:29:10.909
you are going to solve, you are going to solve
now 5 equation, 2 equation for the continuity
00:29:10.909 --> 00:29:15.790
of individual phases. 2 for momentum of individual
phases and one for the kinetic theory of the
00:29:15.790 --> 00:29:19.659
granular flow. So, 5 equation now you are
going to solve together. I am not counting
00:29:19.659 --> 00:29:22.960
any other closure equation definitely you
have to solve to get this solution.
00:29:22.960 --> 00:29:27.239
But majorly you are solving this 5 equation.
And that is the way the kinetic theory of
00:29:27.239 --> 00:29:31.700
granular flow has been defined and has been
derivated, that this is the way it has come
00:29:31.700 --> 00:29:37.320
out. So, you solve along with all these equations,
and that is called 2 fluid model or Euler
00:29:37.320 --> 00:29:41.679
Euler approach. Now the major advantage of
the approach again I am saying that, you are
00:29:41.679 --> 00:29:46.509
just solving even if you solve the kinetic
theory of granular flow which is being used
00:29:46.509 --> 00:29:52.640
against a closure to model a mean motion of
the solids to the fluctuation motion of the
00:29:52.640 --> 00:29:56.779
solid, it gives the correlation in that way.
So, what we are doing? If you solve the using
00:29:56.779 --> 00:30:00.830
drag, both way drag, it means you are using
the drag in both the equation individually,
00:30:00.830 --> 00:30:06.970
like if you are using the drag in both the
equation, let me go back quickly, let me go
00:30:06.970 --> 00:30:12.700
quickly back. So, if you are solving this
equation drag equation, there in both the
00:30:12.700 --> 00:30:16.690
equation it means now you are using the we
are coupling one fluid is affecting the motion
00:30:16.690 --> 00:30:21.220
of another, and solid or discrete phase is
affecting the motion of the fluid fluid phase
00:30:21.220 --> 00:30:25.269
is affecting the motion of the discrete phase,
your solving 2 way coupling and you are joining
00:30:25.269 --> 00:30:29.159
it with the drag.
If you are solving the kt gf also, it means
00:30:29.159 --> 00:30:33.590
what? You are solving how the mean motion
of the discrete phase or solid phase is correlated
00:30:33.590 --> 00:30:38.600
with the fluctuating motion of the solid phase,
you are solving that coupling to, and if you
00:30:38.600 --> 00:30:43.230
want to solving in the fluid phase how the
mean motion of the fluid phase is correlated
00:30:43.230 --> 00:30:47.220
with the mean with fluctuating motion of the
fluid phase, you have to solve any turbulence
00:30:47.220 --> 00:30:53.350
model, like k epsilon model r ng model. So,
all these things you have to or k mean model
00:30:53.350 --> 00:30:56.870
any of these models you have to use. And I
am not going in detail of those turbulence
00:30:56.870 --> 00:31:01.259
model because that is not the scope of this
course. That is more towards the turbulence
00:31:01.259 --> 00:31:04.029
course.
So, you can solve 3-way coupling, all the
00:31:04.029 --> 00:31:08.999
3-way coupling. The only thing which we have
not discussed in solving this is that, how
00:31:08.999 --> 00:31:13.940
the fluctuating motion of the solid is correlated
with the fluctuating motion of the fluid.
00:31:13.940 --> 00:31:18.749
So, that equation we are not solving here,
but all other 3 way coupling we can easily
00:31:18.749 --> 00:31:22.450
solve. What you need to do you have to just
keep on adding the equation. So, if it is
00:31:22.450 --> 00:31:26.440
of only one way coupling or 2 way coupling
your you have to just solve 4 equation, if
00:31:26.440 --> 00:31:31.460
you want a 3-way coupling you have to solve
the kinetic theory of granular flow also equations
00:31:31.460 --> 00:31:35.210
so, it means 5 equation.
If you want that all the turbulence also,
00:31:35.210 --> 00:31:38.720
the turbulence for the continuous phase you
have to solve one more equation say k epsilon
00:31:38.720 --> 00:31:43.739
equation or k omega equation for the continuous
load. So, you will have to solve the 6 equations.
00:31:43.739 --> 00:31:47.840
I am not again telling you not talking about
the drag closures and all they were definitely
00:31:47.840 --> 00:31:52.619
several closures equation you have to solve,
but majorly you are solving this 6 equation.
00:31:52.619 --> 00:31:58.979
And that is the major advantage because, you
are increasing you are reducing the amount
00:31:58.979 --> 00:32:05.480
of computational power which is required to
get the solution. You are not doing any inherent
00:32:05.480 --> 00:32:09.879
assumptions like the slip velocity you are
not using any algebraic equation and all,
00:32:09.879 --> 00:32:15.359
which is restricting the use of the asm model,
for very low volume fraction and volume with
00:32:15.359 --> 00:32:18.539
the co current flow all those assumptions
we have removed.
00:32:18.539 --> 00:32:23.049
And what we have found, we have found the
equation which can be have a very wide applicability.
00:32:23.049 --> 00:32:28.659
So, that is the major advantage the applicability
is very wide, you can use it for co current
00:32:28.659 --> 00:32:33.779
you can use it for counter current cross flow,
for high volume fraction system for low volume
00:32:33.779 --> 00:32:40.389
fraction system anything you can use it.
The only thing is you are assuming discrete
00:32:40.389 --> 00:32:46.879
phase also as a continuous phase, you are
assuming solid as interpenetrating continuum,
00:32:46.879 --> 00:32:52.070
and you are using several closure equations
which are empirically developed and therefore,
00:32:52.070 --> 00:32:57.330
the accuracy of your model is limited to the
accuracy of those closure equations which
00:32:57.330 --> 00:33:01.779
you are going to use or the constants in the
closure equation which you are going to use.
00:33:01.779 --> 00:33:08.489
And therefore, it requires serious validation.
Experimental validation, becomes must at least
00:33:08.489 --> 00:33:13.669
at the current state of the art whatever it
is. Maybe a day will come when we will use
00:33:13.669 --> 00:33:17.669
the multi scale modeling approach the way
I discussed that you define the drag from
00:33:17.669 --> 00:33:22.739
the Euler LaGrange you develop the close other
closure equations from other equations, and
00:33:22.739 --> 00:33:28.159
then you use those equations which is developed
based on the fundamental here and maybe then
00:33:28.159 --> 00:33:30.690
the validation requirement will not be that
much serious.
00:33:30.690 --> 00:33:35.759
But at current state or that validation is
must in 2 fluid model, if you are using 2
00:33:35.759 --> 00:33:40.090
fluid model equation, definitely you need
to validate your simulation data. So, with
00:33:40.090 --> 00:33:44.779
this we have completed the other leg also
of the 2-fluid model or continuum model, now
00:33:44.779 --> 00:33:51.519
the next of modeling technique is Euler LaGrange
model. Or we will say DPM model discrete phase
00:33:51.519 --> 00:33:56.559
model or discrete element model, now discrete
element model is one of the model of the discrete
00:33:56.559 --> 00:33:59.369
phase.
So, what the Euler LaGrange approach says.
00:33:59.369 --> 00:34:04.619
As I said that one phase is Euler continuum,
another phase is LaGrange discrete. And you
00:34:04.619 --> 00:34:09.820
are solving continuous equation or continuum
equation for the fluid phase, you are solving
00:34:09.820 --> 00:34:14.800
discrete phase equation you stand with the
LaGrange way, what is the LaGrange way of
00:34:14.800 --> 00:34:20.139
tracking we have already discussed that in
the module 3, that lagrangain way of tracking
00:34:20.139 --> 00:34:25.399
means you are solving newtons second law of
motion for individual particles, we have done
00:34:25.399 --> 00:34:28.270
it for one particle.
Now, what you need to do if you are solving
00:34:28.270 --> 00:34:32.990
for a multi-phase say fluidized bed or bubble
column, you have to solve the newtons second
00:34:32.990 --> 00:34:37.899
law of motion for each bubble. And you have
to also see that how they are interacting
00:34:37.899 --> 00:34:42.309
with each other. While solving the module
phase 3 module 3 where we solve the interaction
00:34:42.309 --> 00:34:48.800
equations, model for acting we were not taking
we said that, we are not solving it for when
00:34:48.800 --> 00:34:54.250
you are solving it for only one particle,
and interaction between the 2 particles or
00:34:54.250 --> 00:34:59.089
2 this bubbles is missing, we are not solving
that we are not solving the collisional thing.
00:34:59.089 --> 00:35:02.710
We are not solving the interaction that 2
particles are interacting also with each other.
00:35:02.710 --> 00:35:07.030
We have not thought that that is one particle
which is moving or multiple particle moving,
00:35:07.030 --> 00:35:10.240
but they are not interacting with each other.
That is a very ideal case.
00:35:10.240 --> 00:35:15.309
In the real case of the reactor scale definitely
the fluidized bed the particles will be interacting
00:35:15.309 --> 00:35:19.299
with each other. So, we have to also solve
the interaction between the particles. So,
00:35:19.299 --> 00:35:23.470
that is what we do in the Euler LaGrange.
For the fluid phase we call the continuum
00:35:23.470 --> 00:35:27.720
equation which is nothing but the momentum
equation, we have discussed earlier for the
00:35:27.720 --> 00:35:32.250
dispersed phase tracking we do it in the LaGrange
way, it means you solve the newtons second
00:35:32.250 --> 00:35:37.230
law of motion for the discrete phase, that
is what it is given here. Then particle collisions
00:35:37.230 --> 00:35:41.349
are included and model through a spring dashpot
model.
00:35:41.349 --> 00:35:46.049
Now, this is not necessary you can use any
other collisional model also, but this is
00:35:46.049 --> 00:35:50.880
the most famous model which is a spring dashpot
model. That is what is the eulerian LaGrange
00:35:50.880 --> 00:35:55.440
approach is there, it means what you are going
to solve, you are going to solve one continuum
00:35:55.440 --> 00:36:00.260
equation. It mean one continuum equation for
the fluid phase, it means one continuity one
00:36:00.260 --> 00:36:07.440
momentum equation for the fluid phase. This
will be the 2 equations , here LaGrange way
00:36:07.440 --> 00:36:12.049
you are solving the newton second law of motion
for each particle, and that please mind it
00:36:12.049 --> 00:36:21.569
each particle.
It means, suppose, if I have 50,000 particle,
00:36:21.569 --> 00:36:34.460
then you will have to solve 50,000 equation
equation. And that enormously increase your
00:36:34.460 --> 00:36:40.140
computational line, why? Because you are now
solving huge number of equations, even with
00:36:40.140 --> 00:36:44.109
the 3-way coupling you are solving only 6,
now you are solving even without having any
00:36:44.109 --> 00:36:48.260
coupling any collisional equation, I have
not included that, you have to solve say if
00:36:48.260 --> 00:36:53.430
you have 50,000 particle 50,002. And then
on top of that we have solved the collisional
00:36:53.430 --> 00:36:57.569
equations so that you can model that how the
particles are having collision.
00:36:57.569 --> 00:37:01.680
And based on the collisional equation, we
will discuss that it has been further divided
00:37:01.680 --> 00:37:06.500
in the different classes. So, that is called
Euler LaGrange model, and in the approach
00:37:06.500 --> 00:37:10.410
definitely it is more accurate, because you
are solving very fundamental equations without
00:37:10.410 --> 00:37:15.549
having any assumption that the solid is becomes
continuum, they are interpenetrating we are
00:37:15.549 --> 00:37:20.440
neglected all those assumptions. So, it is
more towards the real model or real physics,
00:37:20.440 --> 00:37:24.971
but the computational time requirement is
enormous in this case. And that is the major
00:37:24.971 --> 00:37:29.000
drawback major advantage your move towards
the reality.
00:37:29.000 --> 00:37:33.390
Definitely the prediction accuracies are much
better, but still we will discuss we use lot
00:37:33.390 --> 00:37:38.990
of empirical correlation and therefore, experimental
validation again is must in this case also.
00:37:38.990 --> 00:37:43.010
So, discrete element model actually how it
has been developed, it has been developed
00:37:43.010 --> 00:37:49.030
as an outgrowth of a molecular dynamics
simulation used in computational statistics
00:37:49.030 --> 00:37:53.310
physics. So, it comes actually with the molecular
dynamics simulation, that how the molecules
00:37:53.310 --> 00:37:56.079
are moving together how they are interacting
with each other.
00:37:56.079 --> 00:38:02.070
That is the way discrete element model has
been developed. Later on, in the Cundall and
00:38:02.070 --> 00:38:06.120
Strack, what they have done? They have actually
clubbed this discrete element model, with
00:38:06.120 --> 00:38:10.420
the fluid phase, the discrete element model
originally developed is only for the granular
00:38:10.420 --> 00:38:15.480
phase there was no fluid involved they clubbed
it with the fluid phase, and that is become
00:38:15.480 --> 00:38:20.319
Euler LaGrange model, that how the flow both
the continuous place and the discrete phase
00:38:20.319 --> 00:38:25.520
are moving together. And he has developed
it independently not as a molecular dynamic
00:38:25.520 --> 00:38:28.839
suggest.
But independently developed in 1970's that
00:38:28.839 --> 00:38:33.549
how the dm model will be working with the
fluid phase. So, that is the history of the
00:38:33.549 --> 00:38:37.750
fluid this da model, some people get confused
that it is very close to the molecular simulation.
00:38:37.750 --> 00:38:42.460
Yes, it is really close to molecular simulation,
but then the fluid part we will have to add
00:38:42.460 --> 00:38:47.460
here. And definitely the collisional way we
model collision is little bit different whatever
00:38:47.460 --> 00:38:51.760
you do in the molecular dynamics.
So, what is the basic idea of the DEM simulation?
00:38:51.760 --> 00:38:57.799
That has been given here, that approximate
the collisional interactions between the particles
00:38:57.799 --> 00:39:04.190
using idealized force model that dissipate
energy. And we will come that why the experimental
00:39:04.190 --> 00:39:08.799
this things is needed, and why the DEM is
still not matured enough to give the production
00:39:08.799 --> 00:39:13.490
on it is own. So, what you need to first do?
We approximate a collisional interactions,
00:39:13.490 --> 00:39:18.470
that how the particles will interact with
each other by idolizing the force field we
00:39:18.470 --> 00:39:23.170
say that ok this will be the force field around
the particle, and how the energy will dissipate
00:39:23.170 --> 00:39:28.250
because of the collisional collision.
We try to have a approximate this and we find
00:39:28.250 --> 00:39:33.079
the collisional interactions. Then we integrate
the equation of motion with it is the first
00:39:33.079 --> 00:39:37.270
we see that how the collisional interactions
will be there, then we integrate the equation
00:39:37.270 --> 00:39:42.099
system equation of motion, that equation of
motion we actually add, now even to do the
00:39:42.099 --> 00:39:46.609
equation of motion it determines the individual
position of the particle And the velocity,
00:39:46.609 --> 00:39:49.570
once you add the equation of motion means
newtons second law of motion once you are
00:39:49.570 --> 00:39:53.660
solving. It will give you the position, and
the velocity as we have done in the particle
00:39:53.660 --> 00:39:57.160
tracking earlier.
Then you calculate the relevant transport
00:39:57.160 --> 00:40:04.059
quantities, bulk properties and analyze the
microstructure. So, that is the basic DEM
00:40:04.059 --> 00:40:08.829
formulation. It means first what you do? You
take a distribution of the solids, you will
00:40:08.829 --> 00:40:13.690
try to find it out that what are the interaction
forces is going to be there what are the attraction
00:40:13.690 --> 00:40:19.599
model which you need to incorporate, how the
energy dissipation will take place, then you
00:40:19.599 --> 00:40:24.619
add newtons second law of motion or equation
of motion into it, and try to find it out
00:40:24.619 --> 00:40:29.400
how the particle position, and velocity is
changing with the time, and with the location.
00:40:29.400 --> 00:40:33.640
And then based on that once you know this
you can calculate the transport quantities
00:40:33.640 --> 00:40:40.140
like you can calculate this velocity. You
can calculate your kind of fluctuation, you
00:40:40.140 --> 00:40:44.329
can calculate your diffusion all those quantities
the transport quantities.
00:40:44.329 --> 00:40:48.670
The bulk properties all those things you can
calculate, because you are tracking the individual
00:40:48.670 --> 00:40:53.000
motion of the discrete phase if any structures
are forming like if suppose some of the group
00:40:53.000 --> 00:40:58.010
of the particle is coming together, you can
get that if the particle Are getting agglomerate,
00:40:58.010 --> 00:41:02.140
you can get that although the structures you
can easily get. So, that is the basic idea
00:41:02.140 --> 00:41:05.079
of the DEM modeling in which there is no fluid
in wall.
00:41:05.079 --> 00:41:09.940
So, the basic flow chart of the DEM modeling.
We have already discussed that is why I am
00:41:09.940 --> 00:41:13.890
going quicker, because most of the things
whatever we have done we have already discussed,
00:41:13.890 --> 00:41:17.400
said while discussing the tracking the path
of a single particle. Now what we are going
00:41:17.400 --> 00:41:22.309
here? We are doing we are solving that equation
along with the fluid phase equation now you
00:41:22.309 --> 00:41:26.799
are going to solve. So, we are going to kind
of see that how the particle is affecting
00:41:26.799 --> 00:41:31.900
the motion of the fluid fluid is affecting
the motion of the particle And how the particle
00:41:31.900 --> 00:41:33.660
collision is affecting the motion of each
other.
00:41:33.660 --> 00:41:38.080
So, again we are solving the 3-way coupling.
In this way in that way if you see this the
00:41:38.080 --> 00:41:43.200
initial position and orientation of the velocity,
first what we do we define a geometry in the
00:41:43.200 --> 00:41:48.730
basic flow chart if I discuss. We define that
what is the geometry of my geometry, how the
00:41:48.730 --> 00:41:52.781
particle is initially fitted. So, suppose
this is a pad bed we define that how the particle
00:41:52.781 --> 00:41:58.309
has been placed inside the column of interest.
Once it is done, what we do? We update the
00:41:58.309 --> 00:42:03.890
particle link list and find a new broken context.
Now, what we do suppose if this is say a granular
00:42:03.890 --> 00:42:08.349
flow, where we are saying that this is my
initial position of the particle, and the
00:42:08.349 --> 00:42:12.750
particle is falling down and filling in a
time . At time t equal to 0, I will give a
00:42:12.750 --> 00:42:16.599
position of the particle. So, earlier all
the particle was there suppose in a hopper.
00:42:16.599 --> 00:42:21.590
At t equal to 1 what I did? I suppose, I put
a small plate and I opened that opening at
00:42:21.590 --> 00:42:25.609
time t equal to 0. So, what will happen? The
particle will start falling and they are filling
00:42:25.609 --> 00:42:30.880
in this bottom tank, what will happen? Because
this moved the particle the position will
00:42:30.880 --> 00:42:33.700
be changed. They will be broken the contacts
will be broken.
00:42:33.700 --> 00:42:37.290
And then what you do because the context has
been broken the particle will start moving.
00:42:37.290 --> 00:42:42.380
Now the particle is start moving you calculate
the force and torque acting on each particle.
00:42:42.380 --> 00:42:45.650
How you can calculate force? You know the
mass of the particle you know the velocity
00:42:45.650 --> 00:42:50.119
dv by dt, you calculate the acceleration,
you calculate the force acting on the particle.
00:42:50.119 --> 00:42:54.450
Once you know the forces, you can also calculate
the toque. So, you calculate the force and
00:42:54.450 --> 00:42:59.700
torque on the particle, then what you do?
You integrate the motion of the motion of
00:42:59.700 --> 00:43:04.349
newtons second law of motion, and then you
calculate that what is the position of the
00:43:04.349 --> 00:43:05.960
particle.
So, we know that how the particle has been
00:43:05.960 --> 00:43:11.230
broken. Now they start moving we use the second
law of motion, newtons second law of motion
00:43:11.230 --> 00:43:15.640
and we try to find it out what will be the
next position of the particle. So, you get
00:43:15.640 --> 00:43:19.910
the next position of the particle we get the
velocity of the particle, even we can get
00:43:19.910 --> 00:43:24.619
the orientation of the particle ok. That if
suppose this is not a spherical particle this
00:43:24.619 --> 00:43:30.301
is a particle which has certain l by d ratio
say 2. You can also get that how it will be
00:43:30.301 --> 00:43:33.890
oriented, because you can calculate the torque
based on the torque you can find it out what
00:43:33.890 --> 00:43:37.529
is the alignment of the particle.
So, you can get that then again what you do
00:43:37.529 --> 00:43:42.770
you accumulate the statistics and calculate
the transport properties like velocity fluctuations
00:43:42.770 --> 00:43:47.650
force field all those things you calculate,
and then what you do you increment it to the
00:43:47.650 --> 00:43:52.400
next time. So, earlier we were doing say t
equal to 0 plus delta t you again do delta
00:43:52.400 --> 00:43:58.520
t plus delta t it means 2 delta t you actually
do that and then again you start you update
00:43:58.520 --> 00:44:03.779
the particle position. You again calculate
the forces, you again include the integrate
00:44:03.779 --> 00:44:08.920
the equations of motion and calculate the
properties like position and velocity then
00:44:08.920 --> 00:44:12.319
again you calculate the transport properties
and again update it.
00:44:12.319 --> 00:44:17.420
So, in this way you keep on doing this calculation,
and this is the basic flow chart and you find
00:44:17.420 --> 00:44:21.840
that how the particles are moving, or dumping
into the next part into the next reservoir.
00:44:21.840 --> 00:44:26.720
That way you can do this discrete particle
simulation DEM simulation to see that how
00:44:26.720 --> 00:44:31.400
the particle is moving. So, that is the flow
chart of the DEM simulation. Now, what we
00:44:31.400 --> 00:44:36.540
do here, actually if you see that this is
actually that we are trying to do the energy
00:44:36.540 --> 00:44:39.690
conservation.
So, what we will solve? We are trying to solve
00:44:39.690 --> 00:44:45.119
the energy conservation of molecular system
energy, and instead of the energy dissipation.
00:44:45.119 --> 00:44:50.250
So, in the molecular dynamics what we do we
solve the energy conservation for the molecular
00:44:50.250 --> 00:44:54.970
system energy and with energy dissipation.
We say that how the energy is being calculated,
00:44:54.970 --> 00:45:00.760
while in the granule at temperature granular
flow what we do we calculate the energy dissipation.
00:45:00.760 --> 00:45:06.069
So, that is the difference in the molecular
dynamics simulation and the DEM simulation.
00:45:06.069 --> 00:45:09.619
In molecular dynamics we solve the energy
conservation equation for the system, and
00:45:09.619 --> 00:45:14.500
here we solve the energy dissipation; which
is becomes a very critical characteristics
00:45:14.500 --> 00:45:19.030
of the granular flow, that how the energy
is being dissipated. And then from there we
00:45:19.030 --> 00:45:25.890
actually find that the realistic of this approximation
we do and we found that how the energy dissipation
00:45:25.890 --> 00:45:31.789
is going through by the collision of the particles.
So, we calculate the energy dissipation energy
00:45:31.789 --> 00:45:37.450
dissipation we say that how much energy will
be lost, because of some realistic approximation
00:45:37.450 --> 00:45:42.569
we use some correlations we will discuss about
the correlation later, we use some correlation
00:45:42.569 --> 00:45:48.220
to see that how this energy is actually dissipating.
And that dissipation generally occurs because
00:45:48.220 --> 00:45:53.779
of the particle collision, and therefore,
the particle collision solution solving is
00:45:53.779 --> 00:45:59.549
the most critical part of any DEM simulation.
And to do that, what we need to do? We have
00:45:59.549 --> 00:46:05.079
to have some model of some basic or kind of
realistic approximation we need to do which
00:46:05.079 --> 00:46:10.339
can solve the particle collisional equation.
And that becomes a heart of any dm simulation,
00:46:10.339 --> 00:46:14.710
that how you are modeling the particle collisional
equation for collision will be heavier.
00:46:14.710 --> 00:46:19.670
And to model that these 2-basic approach has
been used in literature one is hard sphere
00:46:19.670 --> 00:46:23.900
approach one is softest sphere approach. So,
again I am coming back that how though I said
00:46:23.900 --> 00:46:27.600
that very people confused that this is very
close to the molecular dynamics simulation,
00:46:27.600 --> 00:46:31.740
yes, it is close to the molecular dynamics
simulation. If you are just solving the granular
00:46:31.740 --> 00:46:36.340
phase flow, you are not solving the fluid
phase at all. Then it is very close, but the
00:46:36.340 --> 00:46:40.069
molecular simulation the approach is energy
conservation the model.
00:46:40.069 --> 00:46:45.579
Here we model the energy dissipation and we
try to find it out that how the energy dissipation
00:46:45.579 --> 00:46:49.539
is taking place through the collision of the
particles. And based on the collision how
00:46:49.539 --> 00:46:54.430
you are modeling the collision the DEM approach
has been further divided in 2 part. One is
00:46:54.430 --> 00:47:04.900
the hard particles model or we say it hard
sphere approach some book also follow it . I
00:47:04.900 --> 00:47:08.700
say particle because the particle can be of
any shape, but some people say is hardest
00:47:08.700 --> 00:47:13.450
sphere and softest sphere approach. Or you
can say hardest sphere model and softest sphere
00:47:13.450 --> 00:47:16.289
model.
In the hardest sphere model what we do? We
00:47:16.289 --> 00:47:21.250
actually consider the particle to be infinitely
stiff it means they are very stiff they are
00:47:21.250 --> 00:47:26.339
very hard particles. Say, you think about
2 iron particles which are very, very hard
00:47:26.339 --> 00:47:31.880
very stiff. And the model assuming instantaneous
and binary collisions. So, we assume that
00:47:31.880 --> 00:47:35.450
there is 2 particles which are very hard,
very stiff they are not going to deform at
00:47:35.450 --> 00:47:41.640
any cost, they will hit other each other instantaneous
hitting will be there, and the collision will
00:47:41.640 --> 00:47:45.760
be binary.
So, that is the way we model the collisional
00:47:45.760 --> 00:47:50.740
equation by using some operator collisional
operator and that collisional operator will
00:47:50.740 --> 00:47:56.960
be the function of friction normal and tangential
restitution coefficients and pre and post
00:47:56.960 --> 00:48:00.860
collisional velocity and a spins, even the
mass of the particles.
00:48:00.860 --> 00:48:05.650
So, that is the hardest sphere model has been
used we will discuss it then you will understand
00:48:05.650 --> 00:48:11.060
it in in more detail such an approach is approximately
collisional dominating system, where the continuous
00:48:11.060 --> 00:48:15.870
and multiple contexts are not the characteristics.
So, that is the major limitation that the
00:48:15.870 --> 00:48:20.650
continuous collision or multiple contexts
at the same time should not be the characteristic
00:48:20.650 --> 00:48:25.910
of the flow. So, what it says? It says that
suppose what is how it will be going on, I
00:48:25.910 --> 00:48:29.200
have suppose 2 spheres of made of say iron
ore ss.
00:48:29.200 --> 00:48:33.900
They are various tests they have very hard
particles it means they are not going to deform,
00:48:33.900 --> 00:48:37.750
they are moving together they will be moving
together they will be interacting if they
00:48:37.750 --> 00:48:41.230
will interacting they will hit each other.
Now once they will hit each other, what will
00:48:41.230 --> 00:48:48.049
happen? The collisional parameters will be
governed or will be found by using first that
00:48:48.049 --> 00:48:51.319
what is the friction between the particles
which is acting once they are interacting
00:48:51.319 --> 00:48:55.170
with each other in presence of some fluid
or something.
00:48:55.170 --> 00:48:59.880
Then what are their normal and continual restitution
coefficient, now what is the restitutional
00:48:59.880 --> 00:49:04.990
coefficient I said that this is e value, if
the collusion is completely elastic the value
00:49:04.990 --> 00:49:10.150
is going to be one, if collisional is completely
inelastic it means after collision the velocity
00:49:10.150 --> 00:49:15.440
becomes 0, the value will be going to 0, or
if this is between the between the elastic
00:49:15.440 --> 00:49:20.049
and inelastic this value will be equal to
between 0 and 1, it means, this co elastic
00:49:20.049 --> 00:49:24.280
the value will be between 0 and 1 bar.
So, you model this you model see the property,
00:49:24.280 --> 00:49:29.040
now this things depends on the properties,
like this properties particle properties what
00:49:29.040 --> 00:49:34.780
is the non-friction normal and tangential
ristitution coefficient is also depends that,
00:49:34.780 --> 00:49:39.529
what is the velocity at which they were approaching
towards each other before the collision what
00:49:39.529 --> 00:49:43.990
is the mass of the particles which you are
interacting with each other before the collision
00:49:43.990 --> 00:49:48.349
if suppose there is no mass change, then before
the collision and after the collision mass
00:49:48.349 --> 00:49:52.390
will going to be the same.
And then what is the velocity of the system
00:49:52.390 --> 00:49:59.040
after the collision? That is the way the collisional
parameters in hard sphere model is being developed
00:49:59.040 --> 00:50:04.010
a hard particle model is being developed.
And generally, these models are used for the
00:50:04.010 --> 00:50:11.930
flow condition there the collisional dominating
flow is there, but the continuous and multiple
00:50:11.930 --> 00:50:15.500
interactions are not the characteristics of
the system. It means, it should not be multi
00:50:15.500 --> 00:50:20.589
particle collision, it should not be multi
particle collision at the same time, it should
00:50:20.589 --> 00:50:24.380
not be that phase this model is being used.
That is called hard sphere model.
00:50:24.380 --> 00:50:29.760
The use of this model because of this limitation
that it should not be the collisional dominating
00:50:29.760 --> 00:50:34.329
flow, or the context should not be a kind
of context dominating flow, this limits the
00:50:34.329 --> 00:50:38.660
application of hard sphere in most of the
chemical engineering equipment of chemical
00:50:38.660 --> 00:50:43.359
engineering reactor which we use like fluidized
bed and all. And most of the time what we
00:50:43.359 --> 00:50:48.789
use is called soft particle model. So, in
the soft particle model what we do? We allow
00:50:48.789 --> 00:50:52.400
as the name suggests the particle is soft.
Now, we are not assuming that infinite stiff
00:50:52.400 --> 00:50:57.349
particle, it is very hard particle you are
not assuming that, we are assuming that particles
00:50:57.349 --> 00:51:02.630
are soft let us say a rubber ball, which can
deform little bit it means if suppose 2 rubber
00:51:02.630 --> 00:51:06.680
ball are interacting with each other, what
will happen like this is a step body. If I
00:51:06.680 --> 00:51:10.880
am interacting there is no deformation it
is not like I am going inside here, but if
00:51:10.880 --> 00:51:14.510
2 rubber ball will be there what they will
do? They will actually change the shape and
00:51:14.510 --> 00:51:19.480
they will allow the overlap. It means one
ball say it was this like this before the
00:51:19.480 --> 00:51:25.880
collision.
During the collision, it may be like this,
00:51:25.880 --> 00:51:30.859
it have a small overlap because of this. So,
what we can do? We can assume this is a spherical
00:51:30.859 --> 00:51:35.230
we can assume this is a physical and I can
say that this is small overlap is possible.
00:51:35.230 --> 00:51:42.240
This is called overlap . This is the particle
Again, and this is after the collision the
00:51:42.240 --> 00:51:46.769
assume both the particle 2 means and we say
that, a small deformation is being accommodated
00:51:46.769 --> 00:51:51.760
overlap is being given. So, that is the major
advantage of this that the soft particle Also
00:51:51.760 --> 00:51:57.380
it can be used, and you assume that you model
this part is that there is a scope is available
00:51:57.380 --> 00:52:02.220
for the particle deformation.
And then, use the particle deformation theory
00:52:02.220 --> 00:52:07.790
to model the contacting in between. So, that
is the softest fear model again we will discuss.
00:52:07.790 --> 00:52:11.900
Now that what how you model the collision
in the hard particle, and you model the collision
00:52:11.900 --> 00:52:16.809
in the soft particle. But this is the basic
approach of the hardest sphere model and the
00:52:16.809 --> 00:52:22.440
softest sphere model. And the hardest sphere
model is basically being used where the multiple
00:52:22.440 --> 00:52:26.190
characteristics multiple contexts are not
the characteristics of the flow. If that is
00:52:26.190 --> 00:52:31.859
not their hardest sphere particle model can
also be used soft sphere model is have a widely
00:52:31.859 --> 00:52:36.930
accepted, it can be used for any system.
Now, what we need to do we need to model the
00:52:36.930 --> 00:52:42.501
contacting. As we said that in DEM the contacting
is the most important part. And that is what
00:52:42.501 --> 00:52:47.329
it is kind of differentiating the dm simulation
with the LaGrange track which we have done
00:52:47.329 --> 00:52:52.510
earlier. Where the single particle was removed,
and we have assumed that the single particle
00:52:52.510 --> 00:52:57.539
motion is not getting the one particle motion
is not getting effect between the other particle,
00:52:57.539 --> 00:53:01.329
and they are not contacting with each other.
So, even there multiple particle flowing there
00:53:01.329 --> 00:53:05.510
in the LaGrange interact whatever we solved
earlier, they are not interacting with each
00:53:05.510 --> 00:53:09.490
other.
But here, we assume in DEM that they are interacting
00:53:09.490 --> 00:53:15.720
with each other, the flow is collisional dominating,
we model the collision, that is what we say
00:53:15.720 --> 00:53:19.789
that we model we first assume the force field
if you go for the dm flow chart. We first
00:53:19.789 --> 00:53:25.319
assume the flow field force field, we model
the contacting, and then we impose the recursion
00:53:25.319 --> 00:53:30.069
of motion to find the new particle position
and velocity, and again we calculate the transport
00:53:30.069 --> 00:53:34.599
properties again we model the flow through
collisions. So, that is the way we solve it.
00:53:34.599 --> 00:53:39.910
So, the collisional equations is modeled properly
need to be modeled properly, and then based
00:53:39.910 --> 00:53:45.660
on that collisional equation model, we define
the approach DEM model in 2 part.
00:53:45.660 --> 00:53:50.050
One is the hard particles model, another one
is the soft particle model. And now what we
00:53:50.050 --> 00:53:55.040
are going to do? We are going to discuss what
is hard particle And what is the soft particle.
00:53:55.040 --> 00:53:59.609
Again, I am not going in detail of this, because
if I cover all the things in great detail
00:53:59.609 --> 00:54:04.020
we will take huge time, you can go through
these equations there are several papers available
00:54:04.020 --> 00:54:08.529
on this, you can go and read the Cundall and
Strack paper you can read the papers from
00:54:08.529 --> 00:54:13.180
DEM papers mostly coming from the Kuipers
group susy group, many other people group
00:54:13.180 --> 00:54:16.880
which is being there.
If you have any problem we can discuss over
00:54:16.880 --> 00:54:21.829
the forum. What is the elementary of the heart
particle model? So, what we have said that
00:54:21.829 --> 00:54:26.160
the 2 infinitely stiff particle Are having
collision with each other. When a particle
00:54:26.160 --> 00:54:31.890
collide, some of the kinetic energy is lost
that is proportional to the initial kinetic
00:54:31.890 --> 00:54:37.160
energy goes into the deforming object.
What is happening to particles is having collision?
00:54:37.160 --> 00:54:41.910
Because one particle And 2 particle, let us
assume one particle is stationary this particle
00:54:41.910 --> 00:54:48.400
say is stationary , and this particle is moving
with a velocity v. Once they will have in
00:54:48.400 --> 00:54:52.549
the collision, then what will happen? The
kinetic energy of this particle will be lost,
00:54:52.549 --> 00:54:54.569
the kinetic energy of this particle will be
lost.
00:54:54.569 --> 00:54:58.680
Because it was say moving with the velocity
v the kinetic energy will be lost, and how
00:54:58.680 --> 00:55:02.920
much it will be lost? That will depend that
it will transfer some of it is kinetic energy
00:55:02.920 --> 00:55:06.950
to this and this particle will start moving
how much loss in the kinetic energy will be
00:55:06.950 --> 00:55:11.509
there. That will be actually equal to the
deforming the object, that how much object
00:55:11.509 --> 00:55:15.180
has been deformed how much objects has been
moved from that location, how much it has
00:55:15.180 --> 00:55:20.109
been deformed opposed from that location.
That is the first elementary that when a particle
00:55:20.109 --> 00:55:24.509
collision take place we say that someone will
lose it is kinetic energy, and that lose in
00:55:24.509 --> 00:55:28.589
the kinetic energy will be proportional to
the deformation of the object.
00:55:28.589 --> 00:55:33.770
Then it means what? If a ball dropped and
hits the ground will not rebound to the same
00:55:33.770 --> 00:55:39.000
height from which it was released. Till the
this is not completely elastic. It means suppose
00:55:39.000 --> 00:55:44.470
this is a ground, we have already did that,
there is a ball, if it goes and hit the ground
00:55:44.470 --> 00:55:48.740
what will happen? The ball will try to come
up. If it is a very rigid particles what will
00:55:48.740 --> 00:55:55.420
happen? Say it started with a height h, it
will achieve a new height which will be say
00:55:55.420 --> 00:56:01.670
h 1, and h 1 will always be less than h. It
will not able to achieve the same height ok.
00:56:01.670 --> 00:56:05.690
Because of some loss in the energy to the
surface, and that loss in the energy what
00:56:05.690 --> 00:56:11.369
it will do it will try to deform this surface.
What we do? We model it we model the collision
00:56:11.369 --> 00:56:15.619
with the coefficient of restitution. The first
approach is there that model the energy loss
00:56:15.619 --> 00:56:21.950
by modeling the coefficient of restitution
which is denoted by a, and e is equal to 1;
00:56:21.950 --> 00:56:28.170
means, perfectly elastic no energy loss. No
energy loss means h 1 will be equal to h e
00:56:28.170 --> 00:56:33.089
equal to 0 complete elastic complete plastic
flow. It means, complete energy loss will
00:56:33.089 --> 00:56:38.210
be there. It means, the actual value will
be actually 0, it will not move anywhere it
00:56:38.210 --> 00:56:42.869
will just loss all it is energy to the surface
and it will stay there.
00:56:42.869 --> 00:56:49.890
And e between 0 and 1, it means what? That
h 1 will be less than h it will lose some
00:56:49.890 --> 00:56:55.410
of it is energy it will never gain that energy
more than that, and it will achieve a height
00:56:55.410 --> 00:57:00.539
which will be smaller than that. The hardest
sphere model we use this approach, we model
00:57:00.539 --> 00:57:04.849
the collision by the coefficient of restitution
through the coefficient of restitution. We
00:57:04.849 --> 00:57:09.869
calculate the value of e, how the is being
calculated? Again, you have might have done
00:57:09.869 --> 00:57:14.940
it in your some undergraduate physics courses.
But let me again revise it. So, what we do?
00:57:14.940 --> 00:57:20.040
The coefficient of correlation is related
to the velocity of the 2 particles which is
00:57:20.040 --> 00:57:24.190
being hitting each other. And let me tell
you it is not only the velocity it is also
00:57:24.190 --> 00:57:28.829
the mass of the particle. We assume to the
velocity, if suppose both the particles are
00:57:28.829 --> 00:57:33.160
made of the same material definitely the mind
the same size mass of the particle is going
00:57:33.160 --> 00:57:36.819
to be the same. If they are not of the same
material, this will be different actually.
00:57:36.819 --> 00:57:42.109
That before there is 2 particle particle A
and particle B, before the collision they
00:57:42.109 --> 00:57:46.839
were at a certain distance, particle A was
moving with the v A particle B was moving
00:57:46.839 --> 00:57:49.160
with a velocity v, v and they are having a
collision.
00:57:49.160 --> 00:57:53.559
What will happen? Once they will have a collision,
they will hit each other say they are hitting
00:57:53.559 --> 00:57:57.049
normally, as I said my hand is saying that
after hitting normally they will go back to
00:57:57.049 --> 00:58:01.420
back they will change their direction they
will go back to back. And suppose if I assume
00:58:01.420 --> 00:58:05.869
that the mass of the particle is not changing,
because there is no deformation of the particle.
00:58:05.869 --> 00:58:11.359
So, mass of the particle is not changing,
velocity is v A prime for the a. Velocity
00:58:11.359 --> 00:58:16.690
is v B prime for B after the collision then
the restitution coefficient is defined as
00:58:16.690 --> 00:58:20.569
v B prime minus v A prime upon v A minus v
B.
00:58:20.569 --> 00:58:26.980
Relative velocity after the collision divided
by relative velocity before the collision,
00:58:26.980 --> 00:58:31.490
and that is what is the restitution coefficient.
That is the ratio is called the restitution
00:58:31.490 --> 00:58:36.869
vision, before collision what was the velocity
after collision what was the velocity that
00:58:36.869 --> 00:58:40.839
relative velocity is being defined with the
restitution coefficient e, if suppose one
00:58:40.839 --> 00:58:45.300
particle is a stationary, say B particle is
stationary B, B value will be 0. That way
00:58:45.300 --> 00:58:48.440
it is being done.
Now, if you see the restitution coefficient
00:58:48.440 --> 00:58:54.119
value, it is it in cm minus v B and v B minus
v A because of the energy transfer. So, that
00:58:54.119 --> 00:58:58.630
restitution coefficient values does not go
in negative, it will always be the positive.
00:58:58.630 --> 00:59:02.599
That is there and if suppose you consider
the mass of the particles is also different
00:59:02.599 --> 00:59:07.049
you want to add the mass of the particle here,
then and you want to find it out that how
00:59:07.049 --> 00:59:11.860
much total loss in kinetic energy will take
place. So, let us assume the mass of the particle
00:59:11.860 --> 00:59:18.500
A is m A velocity is g A, mass of the particle
is m B and let us assume that it is velocity
00:59:18.500 --> 00:59:23.110
is 0, it is a constant to simplify it you
can again derive it for the this equation
00:59:23.110 --> 00:59:28.010
for the particle mv is also moving.
What is the kinetic energy loss? Delta KE
00:59:28.010 --> 00:59:32.900
during the impact so, if they will help. So,
suppose a particle which is moving with a
00:59:32.900 --> 00:59:40.069
velocity v A, mass is m A, it is hitting a
particle And of mass B mv, but the particle
00:59:40.069 --> 00:59:43.530
is a stationary. So, what will happen? It
will lost it is kinetic energy, how much lost
00:59:43.530 --> 00:59:48.589
in kinetic energy is there I think this has
been derived, when your solid mechanics courses,
00:59:48.589 --> 00:59:54.490
you will get that it says m A plus m m A into
m B upon twice and may plus m B, v A square
00:59:54.490 --> 00:59:58.460
into 1 minus c square you can revise this
you can find this formula.
00:59:58.460 --> 01:00:02.109
If you are not able to get it we can derive
which it in detail. We can discuss it over
01:00:02.109 --> 01:00:06.470
the forum many things, I am leaving over the
forum the major region is, these are something
01:00:06.470 --> 01:00:11.160
which we cannot cover in detail in one class
or 2 class, it will require enormous time.
01:00:11.160 --> 01:00:15.080
So, these are the basics which you have already
done in your physics courses or in your mechanics
01:00:15.080 --> 01:00:19.570
courses, you can revise then, if there is
any problem we can discuss over the forum.
01:00:19.570 --> 01:00:24.599
So, that is the way it is being calculated
and kinetic energy loss you can calculate.
01:00:24.599 --> 01:00:28.759
Let how much loss in the kinetic energy will
be there. That is the hardest sphere model
01:00:28.759 --> 01:00:33.520
or hard particle model approach is being used
which is being used in the DEM simulation,
01:00:33.520 --> 01:00:37.820
if you are doing that, your collision is being
modeled this way, you model there is 2 you
01:00:37.820 --> 01:00:41.559
calculate the restitution coefficient you
calculate the kinetic energy loss. And from
01:00:41.559 --> 01:00:45.690
there calculate that how the velocity loss
will be there, and how the pollination of
01:00:45.690 --> 01:00:49.510
the particle will be changing.
In the first sphere field model, what we do?
01:00:49.510 --> 01:00:54.099
What we actually solve the equation for the
gas phase, you solve the equation for the
01:00:54.099 --> 01:00:58.380
solid phase. So now, we are solving about
the Euler LaGrange model, in the Euler LaGrange
01:00:58.380 --> 01:01:04.180
DEM model with the soft sphere approach what
we do? We solve first the equation for the
01:01:04.180 --> 01:01:08.779
gas phase we solve the continuity equation,
we solve the momentum equation for the gas
01:01:08.779 --> 01:01:13.859
phase here, and the continuity equation or
momentum equation is same as whatever we solve
01:01:13.859 --> 01:01:18.039
for the normal cases this is the equation
which we have already discussed.
01:01:18.039 --> 01:01:23.190
And then we solve in the newtons second law
of motion for tracking the velocity of individual
01:01:23.190 --> 01:01:27.609
phase individual particle. So, this is the
equation you are already familiar, we have
01:01:27.609 --> 01:01:33.630
solved this equation for the LaGrange tracking,
the total force m d by dt m into v is equal
01:01:33.630 --> 01:01:38.090
to acting on the particle is equal to the
total force acting on the particle. So, rate
01:01:38.090 --> 01:01:42.300
of change of momentum is equal to the total
force acting on the particle.
01:01:42.300 --> 01:01:47.559
Now, what are the total force acting on the
particle? Earlier we have done drag, gravity,
01:01:47.559 --> 01:01:53.750
buoyancy or any other body forces, say electrical
forces or magnetic forces. Now what we do?
01:01:53.750 --> 01:01:58.980
We are saying that their particle can affect
they can have a collision and their motion
01:01:58.980 --> 01:02:03.749
can be get affected with each other. So, definitely
we are saying that the rate of change of momentum
01:02:03.749 --> 01:02:09.049
is definitely going to equivalent to the drag,
it will be the body force, it will be the
01:02:09.049 --> 01:02:13.589
pressure force that what is the pressure force
it is there a pressure gradient in which the
01:02:13.589 --> 01:02:16.150
body has been kept because now the fluid is
moving.
01:02:16.150 --> 01:02:20.309
So, what is the pressure gradient the body
has been kept. That is also going to see that
01:02:20.309 --> 01:02:26.890
how the body forces will change, but now what
we are adding? We are adding the contact forces
01:02:26.890 --> 01:02:31.660
that contact forces is f ca we are adding
the van der waal forces which will be acting
01:02:31.660 --> 01:02:36.560
between the particle, we add these 2 forces
extra compared to what the LaGrange track
01:02:36.560 --> 01:02:41.340
we were solving. We added the velocity gradient
or the pressure gradient field in which the
01:02:41.340 --> 01:02:45.690
solid has been also added.
But this is the 2-major thing which we have
01:02:45.690 --> 01:02:50.700
added, and these 4 solve the annular motion,
we solve the annular equation IA is equal
01:02:50.700 --> 01:02:55.970
to d omega by dt equal to ta. Again, the rate
of change of angular momentum is equal to
01:02:55.970 --> 01:02:59.430
the force acting or regular force acting that
is nothing but equal to the torque.
01:02:59.430 --> 01:03:03.579
So, you can calculate the torque, you can
solve even the spin of the particles. So,
01:03:03.579 --> 01:03:08.550
that is the way we are solving the equations
for this. Now what is needed? You needed the
01:03:08.550 --> 01:03:14.349
contact forces fc, you needed the drag force,
you needed the pressure gradient and you needed
01:03:14.349 --> 01:03:18.609
the van der waal forces. Now van der waal
forces, I hope everyone knows how to calculate
01:03:18.609 --> 01:03:23.230
the van der waal forces, you can calculate
the van der waal forces, drag is pretty much
01:03:23.230 --> 01:03:26.460
same which is going to affect the motion of
the fluid also.
01:03:26.460 --> 01:03:30.430
So, drag equations are going to be the same;
which we have already discussed how to use
01:03:30.430 --> 01:03:34.680
the drag we have already solved the problem.
You can add any body force also if you want
01:03:34.680 --> 01:03:39.200
say you can add one more force any other force
which is acting here. Gravitational force
01:03:39.200 --> 01:03:44.210
we already know how to model the only thing
is in the shortest year how to model the contact
01:03:44.210 --> 01:03:49.940
forces. Now as I said in the softest sphere,
the 2 soft particles are being hitting each
01:03:49.940 --> 01:03:53.869
other, and overlap is allowed it means deformation
of the particles is allowed.
01:03:53.869 --> 01:03:58.260
What will happen? Once the particles say is
having collision, and this is again the Cundall
01:03:58.260 --> 01:04:03.269
and Strack paper in 1979 from where this approach
has been taken, that if the 2 particles are
01:04:03.269 --> 01:04:08.580
having collision, what we do? We model the
collision by using the spring dashpot model,
01:04:08.580 --> 01:04:13.200
it means what? If suppose the particle are
hitting each other like this, we allow certain
01:04:13.200 --> 01:04:18.420
overlap we are allowing certain overlaps here.
And we are saying that, these one spring and
01:04:18.420 --> 01:04:22.349
one dashpot is being placed in both normal
and tangential direction.
01:04:22.349 --> 01:04:27.109
So, this is can be hit it it in this way,
this can hit it it in this way. In the tangential
01:04:27.109 --> 01:04:31.589
direction in the normal direction. So, we
are assuming one spring, and one dashpot in
01:04:31.589 --> 01:04:36.720
the normal direction and tangential direction
to model this collision. And that is why it
01:04:36.720 --> 01:04:41.329
is also called a spring dashpot model. Now
why there spring and dashpot model we are
01:04:41.329 --> 01:04:46.549
using, a spring coefficient will tell me about
the restitution coefficient values it means
01:04:46.549 --> 01:04:51.519
hide the spring coefficient better the elasticity
lower the spring coefficient lower the elasticity.
01:04:51.519 --> 01:04:56.359
So, it means the spring coefficient values
are high, it will go back to the same distance,
01:04:56.359 --> 01:05:00.960
it is low it will not go back to the same
distance. The damping is being used because,
01:05:00.960 --> 01:05:04.880
it will lost some energy, and this loss in
the energy will be model through the damping.
01:05:04.880 --> 01:05:10.829
So, again we are using the same approach,
that if I throw a particle on the solid surface
01:05:10.829 --> 01:05:15.210
it will not able to achieve the same place,
because there will be some loss in the energy.
01:05:15.210 --> 01:05:20.019
So, the loss in the energy is being modeled
in the damping way, and that damping is being
01:05:20.019 --> 01:05:24.170
caused mainly because of the fluid which is
being placed in between and also particle
01:05:24.170 --> 01:05:29.170
surface also. That is the way the Cundall
and Strack with soft sphere approach collisional
01:05:29.170 --> 01:05:32.579
things are modeled.
Now, you can see this equation actually if
01:05:32.579 --> 01:05:36.400
you see this is the spring coefficient equation,
typical spring coefficient equation in the
01:05:36.400 --> 01:05:40.990
normal direction, only the damping factor
has been added. That how much it is being
01:05:40.990 --> 01:05:46.089
going to damp the k is the spring coefficient
value epsilon is the displacement. So, if
01:05:46.089 --> 01:05:50.609
you remember your basic spring coefficient
a spring force that is f equal to minus k
01:05:50.609 --> 01:05:55.529
s. So, that is the spring coefficient value,
eta is the damping coefficient, because again
01:05:55.529 --> 01:05:59.789
as I said that it will damp something it will
lose the energy. So, that is why we multiply
01:05:59.789 --> 01:06:03.910
with the eta in this.
And then you also substrate that how much
01:06:03.910 --> 01:06:09.180
is the velocity and the damping coefficient
is there before after moving. So, that is
01:06:09.180 --> 01:06:14.089
the way you calculate the contact forces in
the normal direction. Similarly, you calculate
01:06:14.089 --> 01:06:18.119
the contact forces in the tangential direction
again if you see this is nothing but the same
01:06:18.119 --> 01:06:23.380
thing whatever I have written, it is the k
zeta t into eta only the damping coefficient
01:06:23.380 --> 01:06:27.480
in the tangential we have removed because
now it is moving it it in this way. And then
01:06:27.480 --> 01:06:31.900
to calculate that damping coefficient, actually
in the tangential direction 2 equations are
01:06:31.900 --> 01:06:34.840
being used.
One is being calculated it into the first
01:06:34.840 --> 01:06:40.190
equation is like this way, second equation
is you take the normal forces, you take is
01:06:40.190 --> 01:06:44.589
the coefficient of the friction into the account
multiply the normal forces with the coefficient
01:06:44.589 --> 01:06:49.200
of the friction, and then you find that what
is the f t.
01:06:49.200 --> 01:06:55.140
And then if you see that whatever the ft calculated
by this way, if the Ftab, whatever you have
01:06:55.140 --> 01:07:01.359
calculated is less than the mu times of f
t . Mu times of f and it means this, then
01:07:01.359 --> 01:07:06.440
you use this equation if it is greater than
that you use this equation. It means what
01:07:06.440 --> 01:07:12.069
we are saying that the tangential collision
is going to be lower than the mu times of
01:07:12.069 --> 01:07:16.700
the normal collision. And mu is the coefficient
of friction mu times of this. If it is lower
01:07:16.700 --> 01:07:21.460
or equal then you use this equation.
If this equation is predicting a value which
01:07:21.460 --> 01:07:28.999
is lower or equal then mu times of F and yb.
You use this equation, if not you use this
01:07:28.999 --> 01:07:33.869
equation. So, by this way you model the total
contact forces, van der waal forces we know
01:07:33.869 --> 01:07:39.049
drag we know, and we solve the whole equation
for the particle movement, and we couple it
01:07:39.049 --> 01:07:44.519
with the fluid movement, fluid velocity with
the continuity and the momentum equation for
01:07:44.519 --> 01:07:48.750
the fluid and we found that how the phases
are interacting with each other.
01:07:48.750 --> 01:07:52.480
How the particles are having collision with
each other, and how the collision is changing
01:07:52.480 --> 01:07:56.890
their motion, how the fluid is changing the
motion of the particle all the things you
01:07:56.890 --> 01:08:01.589
can solve. But the problem again what I am
saying again with this approach, this is very
01:08:01.589 --> 01:08:06.970
close to the physics very accurate towards
the real physics. But the problem is the computational
01:08:06.970 --> 01:08:13.150
time. So, if you just assume 50,000 particles
is a very small fluidized bed. Very, very
01:08:13.150 --> 01:08:18.081
small fluidized bed will have a 50,000 particles,
you want to use DEM simulation you have to
01:08:18.081 --> 01:08:28.509
solve 25, 50,000 equations for your you newton
second law of motion equations. For that 50,000
01:08:28.509 --> 01:08:33.350
equation you have to calculate the collisional
parameter just spring dashpot models, and
01:08:33.350 --> 01:08:38.159
then you have to solve 2 equations other.
So, you have to solve enormous number of equations,
01:08:38.159 --> 01:08:44.070
and that will take huge time huge computational
time to model. Sometimes even in the current
01:08:44.070 --> 01:08:49.619
era, if you want to model 50,000 particles
with a very high accuracy, it will take days
01:08:49.619 --> 01:08:55.730
to complete that simulation. So, that is the
major limitation of DEM simulation, but it
01:08:55.730 --> 01:09:00.770
is very close to the real-world problem to
the real physics. And that is why there is
01:09:00.770 --> 01:09:06.600
a scope that you can develop even the collision
model from this place, and if we use the multiple
01:09:06.600 --> 01:09:10.609
multiple kind of modelling approach, this
collisional model which has been developed
01:09:10.609 --> 01:09:15.950
or validated at this scale can be used in
the Euler Euler simulation for having better
01:09:15.950 --> 01:09:19.560
prediction.
So, that is the one way but again DEM model,
01:09:19.560 --> 01:09:24.190
though it is very close to the physics we
are using the basic 6 models. But we are solving
01:09:24.190 --> 01:09:29.339
actually we have to find the value of k t
k d value k n value we have to find the damping
01:09:29.339 --> 01:09:35.109
coefficient values; these values can be calculated
through the youngs modulus experiment there
01:09:35.109 --> 01:09:39.279
are several correlations are there to calculate
all this. But sometimes we use some values,
01:09:39.279 --> 01:09:41.900
and because of that experimental validation
is needed.
01:09:41.900 --> 01:09:47.339
Again, you are using drag force, because you
are using drag force and drag force is actually
01:09:47.339 --> 01:09:52.319
being modeled with lot of empirical correlations.
Again, if you are using those empirical correlation
01:09:52.319 --> 01:09:57.210
Gidaspow Syamlal Wen Yu anything, again you
need experimental validation. So, even at
01:09:57.210 --> 01:10:01.901
the current state of the art DEM cannot be
said that it is 100 percent correct. You need
01:10:01.901 --> 01:10:07.920
to validate it with the experimental data
to find the validity or accuracy of the DEM
01:10:07.920 --> 01:10:11.550
simulations. This is the typical dm simulation
fluidized bed simulations.
01:10:11.550 --> 01:10:15.510
We will discuss the fluidized bed we will
again come back to the same, but if you see
01:10:15.510 --> 01:10:19.810
this is the rate 2 different particles, the
smaller particle smaller density particle
01:10:19.810 --> 01:10:28.250
is blue colored, this is say density rho b,
this is rho d, rho r and rho b is having the
01:10:28.250 --> 01:10:34.081
lower density then rho r. So, blue particle
is smaller color a smaller density, red particle
01:10:34.081 --> 01:10:38.110
is are heavier densities, we are fluidizing
once we are fluidizing this what is happening
01:10:38.110 --> 01:10:41.880
the particles are moving blue particles are
definitely moving is on the top of the bed
01:10:41.880 --> 01:10:45.920
because they are lighter. And they are moving
faster and we are seeing slowly how the red
01:10:45.920 --> 01:10:48.600
particle is diffusing and they are moving
with each other.
01:10:48.600 --> 01:10:53.560
So, we are solving this the whole DEM simulation,
we are tracking the motion of all the particles,
01:10:53.560 --> 01:10:58.440
we are solving the interactions between the
blue particles, interaction between the blue
01:10:58.440 --> 01:11:03.070
particles and the red particles, interaction
between the red particles, and interaction
01:11:03.070 --> 01:11:07.770
of these particles with the fluid we are solving
all this together to get this fluid behavior
01:11:07.770 --> 01:11:13.739
of the fluidized bed. But the problem is to
solve this it is computationally very expensive.
01:11:13.739 --> 01:11:18.300
And that is why most of the time we limit
ourselves to a very small dimensions of the
01:11:18.300 --> 01:11:21.320
bed.
So, we call it as a sub bed scale model, we
01:11:21.320 --> 01:11:26.100
can model it for the very smaller bed dimensions.
So, with this our modeling part is over if
01:11:26.100 --> 01:11:30.680
I will try to conclude some of the model I
have not discussed, then I can say the multiphase
01:11:30.680 --> 01:11:36.469
modeling in divided in 2 parts. One is for
the gas liquid flow, one is for the gas solid
01:11:36.469 --> 01:11:39.460
flows.
So, in the gas solid flows if you want for
01:11:39.460 --> 01:11:44.050
the more modeling or more physics, then if
you go from the bottom to top, you will be
01:11:44.050 --> 01:11:49.150
coming very close to the physics, it means
2 fluid model or a multi fluid model will
01:11:49.150 --> 01:11:52.800
have more modeling, as I am going from the
top to bottom it means we are using more modeling
01:11:52.800 --> 01:11:57.909
because we are using more and more assumptions.
Once you move from the bottom to the top you
01:11:57.909 --> 01:12:01.659
use more physics, I have not discussed the
viewing of modeling that is another class
01:12:01.659 --> 01:12:07.090
of model, where you actually do the interface
tracking to see that how the bubble is moving,
01:12:07.090 --> 01:12:12.080
or liquid droplet or bubble is moving inside.
And not discuss that, but if you go through
01:12:12.080 --> 01:12:16.500
this, you can solve the multi fluid model,
you can solve the bubble tracking model you
01:12:16.500 --> 01:12:19.300
track the bubbles, and you can solve the vof
model.
01:12:19.300 --> 01:12:23.610
So, once you go from bottom to the top, your
physics is increased, but computational time
01:12:23.610 --> 01:12:28.070
is also increased, because you are using less
model. Your computational time will be enormous,
01:12:28.070 --> 01:12:32.130
and those models will be restricted to the
sub grade scale. You cannot model the whole
01:12:32.130 --> 01:12:36.630
bubble column by using the vof it is ruled
out. You can model a small section. And similarly,
01:12:36.630 --> 01:12:42.330
you come from the top to the bar this in the
gas solid flow. I will say that it is a multi-fluid
01:12:42.330 --> 01:12:47.270
model with the bubble tracking or you do that.
You use the 2-fluid model, you use the DPM
01:12:47.270 --> 01:12:51.290
model, you go to the lattice Boltzmann model.
Again, lattice Boltzmann model, I have not
01:12:51.290 --> 01:12:55.890
discussed because it is too mathematical and
it is very close towards the molecular simulations.
01:12:55.890 --> 01:13:00.610
I just limited myself in the 2 voiced popular
approach; which has being popularly used in
01:13:00.610 --> 01:13:04.760
the chemical engineering domain; that is,
fluid model medical model or 2 fluid model
01:13:04.760 --> 01:13:09.420
and DPM. Again, you go from the bottom to
the top, you go for more physics, top to the
01:13:09.420 --> 01:13:13.420
bottom more modeling.
So, definitely the computational time increases,
01:13:13.420 --> 01:13:26.880
this is computation increase in computational
time . So, if you go there, the increase in
01:13:26.880 --> 01:13:36.690
the computational time , but increase in the
accuracy too . Based upon whatever the flow
01:13:36.690 --> 01:13:41.860
field you want to use, whatever the system
you have, how much accurate system accurately
01:13:41.860 --> 01:13:46.610
you want to predict your flow, how much kind
of you can compromise with the accuracy what
01:13:46.610 --> 01:13:52.130
label degree of accuracy you need, how much
computationally time you can devote, how much
01:13:52.130 --> 01:13:56.530
of time you can wait for your simulation results,
you can choose one of the models.
01:13:56.530 --> 01:14:00.760
And we have discussed that what is the limitation
of each model what equation you solve in each
01:14:00.760 --> 01:14:06.951
model, and what is the accuracy you can think.
But as of now none of the model is. So, accurate
01:14:06.951 --> 01:14:11.630
that none of the experimental validation is
needed. The experimental validation is must.
01:14:11.630 --> 01:14:17.570
The only thing is if you are using the 2-fluid
model definitely the balance the kind of the
01:14:17.570 --> 01:14:22.719
experimental validations should not be only
at the mean label you should touch to the
01:14:22.719 --> 01:14:26.110
fluctuation velocities too.
Ideally it will be good if you validate all
01:14:26.110 --> 01:14:31.530
your model to the way motion prediction mean
predictions as well as the fluctuation prediction.
01:14:31.530 --> 01:14:36.530
But 2 fluid model it is almost must to do
that. So, that is the whole basics of how
01:14:36.530 --> 01:14:42.150
the modeling is being done for the multi-phase
flow reactors, and now what we are going to
01:14:42.150 --> 01:14:46.290
do from the next classes. We will discuss
about the different multi-phase flow equipments,
01:14:46.290 --> 01:14:52.530
we will see that what are their basic features,
and based on that basic feature, you can see
01:14:52.530 --> 01:14:55.320
that what is the modeling approach you can
use.
01:14:55.320 --> 01:15:00.050
What are the equations? you can solve to find
the features or to model those multi-phase
01:15:00.050 --> 01:15:04.989
flow reactor with. This the the 6th portion
is over, now we will discuss the module 7
01:15:04.989 --> 01:15:20.060
and module 8, which will be mainly based on
the different multi-phase flow reactors.
01:15:20.060 --> 01:15:28.189
Thank you .