Welcome back. So, in the last class, we were
discussing about the different modeling approach
and I have said that what is the advantages
and disadvantages of each modeling approach.
And we have already covered about empirical
correlation and Lagrangian track in the previous
classes. In Lagrangian track is the same whatever
we were solving the Newton’s second law
of motion by taking the different forces into
the account. Whether, it is a buoyancy, it
is a drag it is a gravity all those forces
we are balancing the demo metafilter force
or the momentum integral. So, that is what
we are doing it.
Now, what we are going to discuss now is the
other methods and one of the most kind of
one of the other method are kind of very initial
methods which is being used for the simulation
or for the modeling of the multi-phase flow
is the algebraic slip method. Now, why it
has been used? Because, the major advantage
of this model is that it is computationally
less expensive as we are solving only one
equation for both the phases.
So, there is a continuous or dispersed phase
both the phases we solve only one equation
and we assume that the phase should be interpenetrating
continuum. So, they are kind of inter penetrating
each other and we solve the equation based
on the mixture properties. So, that is the
algebraic slip model. And it is initially
developed and it is computationally very,
very kind of simple or you can say the less
rigorous and computational power requirement
is less, compared to the other model whatever
we are going to discuss.
So, let us begin with the algebraic slip model.
So, what is algebraic slip model? As I said
that it actually depend on the slip velocity
and we solve the algebraic equations. That
is why it is called algebraic slip model because
for slip velocity we solve the algebraic equation.
What we do we solve one set of momentum equation
for the mass average velocity and we track
the volume fraction of each fluid throughout
the domain.
So, what we do? We solve once momentum equation
one continuity equation which is based on
the mass average velocity of the mixture.
And what we do? We solve the volume fraction
of each fluid throughout the domain. So, we
are solving only one phase 1 momentum one
continuity equation based on the mass average
velocity, volume fraction throughout the domain.
So, what we can do we can get the velocity
we can get your mass fraction or mass distribution
and then with the volume fraction we will
get that where the phases are actually present.
So, that is the whole idea of algebraic slip
model.
And because you are using the mass average
velocity, you are just you just need to solve
only one equation for both the phases and
the mass average velocity will depend on the
both the phases. We will see the equation
then it would be more clear. So, what we do
here the major assumption is an empirically
derived relation for the relative velocity
of the phases and that is what I said that
and this equation is the algebraic equation
which is empirically derived and that is the
major assumption.
Because we are going to solve the velocity
based on that what is the relative velocity
and slip velocity in between the 2 phases
and we are going to take the mass average
velocity for the mature equation. And this
relative velocity formula this relative velocity
equation is actually empirically developed
equation. So, your accuracy or your prediction
will depend on the accuracy of this empirically
derived relation and that is the major assumption
of this kind of this modeling method. And
that is why the accuracy of this modeling
method is not very high particularly, when
the dispersed phase concentration is very
high or the slip velocity is very, very high
and we will see that.
So, for turbulent flows again a single set
of turbulent transport equations are solved.
If you have the fluid which is turbulent you
can solve the momentum equation, you can solve
the continuity equation, you can solve the
volume fraction equation and you can solve
the turbulent equation along with it. So,
what will happen you have now we will increase,
number of equation will increase, the computational
time will increase, but still as well or the
model way it gives you the luxury to solve
the turbulent flow also.
This approach is actually very good for the
flow fields where both the phases or generally
flow in the same direction. Please mind this
word generally it is not like you cannot do
the counter current, but generally it works
well when both the phases are flowing in the
co current mode. It means they are following
in the same direction. Why? Because in that
way the slip velocity is actually becomes
lower.
So, in that this is the way over all kind
of you can say that the overall approach which
is being used in the algebraic slip model
one of the initially developed model which
actually starts solving the multi-phase flow
equation from the one dimensional to the 3
dimensional domain, but because the computational
power was not very high the equation was developed
it in this way that you can still solve this
problem of the multi-phase typical multi-phase
problems and get some solution and we will
discuss that what kind of a solution you can
get from it.
Overall, what you do you solve one momentum
equation one continuity equation for based
on the mass average velocity. And then you
solve one equation for the volume fraction
and you solve one equation for turbulent flow
if the flow is turbulent, if the flow is not
turbulent you just solve the 3 equations and
this is the reason that why your computational
time is requirement is relatively lower compared
to other models which we will discuss later
in this course. That is the best part of this
and if you just see the equation I am writing
the equation here, this equation is for the
continuity equation.
If you see that this δρ / δt term is there,
this is the unsteady state mass transfer term
and this is the convective mass transfer term.
So, this is δ(ρux) /δx will be equal to
0, you are getting it from the ui and x i,
it means this will be in all the 3 dimension,
it will be kind of δ by δx δ u by δ x
plus δ v by δ y plus δ v by δz or you
can say that δ ρun upon δx1 plus δρun
δρu2 upon δx2 plus del δu3 upon δx3.
So, you can write it in this form and you
can solve this equation the continuity equation.
You solve the volume fraction transport equation
which was I was take talking about that how
the volume fraction is being transported.
Because you are solving only one equation
and at the end I want that where is my gas?
Where is my liquid? Suppose, if I am talking
about a bubble column, again the same bubble
column liquid is filled. Bubbles are being
passed here, air is being kind of you can
say it is passed and bubble is moving. At
the end, I also want that where is my bubble?
Where is my liquid? What is the fraction of
the bubble? What is the fraction of air fraction?
Or what is the water fraction available at
any domain?
So, to do that, if you do not have that information
then the solution will have very limited application.
So, to get that information we solve a additional
equation here that is the volume fraction
tracking equation and that volume fraction
tracking equation actually differentiate between
the phases that which phase is there; so because
we are solving only one cut momentum equation
and one continuity equation. So, this is sure
actually equation a is gives you the benefit
or give you the luxury to separate the 2 phases
and find it out that where the fractions are
there.
And if you see that volume fraction tracking
equation is very similar to the mass conservation
equation. Instead of the ρ we have replaced
the ρ with α in the volume fraction equation
if you will see here. So, if you see that
equation this is del α by del t and this
a kind of unsteady state for a term for the
volume fraction that how the volume fraction
is changing with the time. And then this is
the convective volume fraction transport term.
So, how the volume fraction is being transported
from one location to other location or you
say the convective motion because of that
how the kind of fraction of the phases is
being transported.
So, this is exactly similar to the continuity
equation the only thing is now, instead of
the mass conservation or mass conservation
you are solving the volume conservation equation
or you can say that you are solving the volume
transport equation instead of the mass transport
equation. So, that is the way it has been
written then we solve one momentum equation
and that momentum equation is actually based
on the mixture property mass average mixture
velocities and we also introduced a drift
velocity inside.
So, that we can understand that what is the
slip between these 2 velocity of 2 phases.
So, what we do? We solve the momentum equation
if you see this is a clear cut or Navier stokes
equation which is I have written in the 2-dimensional
domain. It is an Navier stokes equation and
then you add a extra term which shows that
how the drift taking place between the 2 phases.
So, let us discuss this term this is the unsteady
state acceleration term, this is how the particle
is moving. So, δ δu / δt local acceleration
or any steady state velocity term that it
momentum term you can say a local acceleration
I will say local acceleration. Then this umi
is again based on that the mass average mixture
velocity and then you solve the convective
acceleration term, this is convective term
yeah you solve the convective term. And then
the pressure gradient you solved that what
will be the pressure gradient this is pressure
gradient.
Now, this is viscous term I have written in
the 2 dimension that is why you are seeing
the 2 thing. This is viscous force or viscous
term, this is the gravity of body force, this
is any extra force any additional force which
is acting on that. Please see that we have
not taken the drag here because we are solving
only one equation. So, we are solving maintain
the equation based on the mixture property,
that is why we are not including the drag
in the momentum equation term.
Now, this is the term which has been extra
internal included. So, that we can get the
individual information of both the phases
and that is very close to whatever the convective
term is there. If you see these 2 terms which
are very close there the convective term the
only difference is that you are now multiplying
it with the αi. So, what you are going to
do you are going to solve all these 3 equations
simultaneously. We are going to solve all
these 3 equations simultaneously to get that
how the individual drift is acting on each
phases? That is why there is a summation time.
If you have 2 phase, you will solve this will
be the 2 term one will be the 4 come say phase
1 and another term will be the same thing
will be phase 2 and you will calculate this
drift velocity. So, this u r is nothing but
is a drift velocity and how to calculate.
So, what is there days more unknowns are now
you have introduced 2 more unknowns one is
which is the mass average mixture velocity.
So, we need to define that what is this and
we need to define your; which is nothing but
the drift velocity. So, we have just modified
the momentum equation we have just added one
extra term, rest of the momentum equation
if you will see it is very close it is similar
to the single-phase flow. Only thing which
is being added is this drift velocity term
which is nothing but the convective acceleration
of individual phases.
How the individual regions are moving and
that is why the α is being introduced here.
So, that we can have the information of the
phase which we are the phase of the interest
or for which we are trying to solve the problem
and then we solve the α overall α volume
and volume fraction equations so that we can
sign that how the volume fraction transport
taking place.
The accuracy or the unknown terms here is
now is still the um and ur that how they should
be defined, how the ur should be defined,
and as I said that the accuracy of these models
depend on the definition of this and ur which
are being defined mostly through the empirical
correlations. How this ρ is being defined?
Because, if you see here also ρ is everywhere
the ρ is the mixture ρ this is also ρm
this is the mixture ρ.
So, mixture ρ is being defined as we have
already discussed during the introduction
phase that the mixture ρ is nothing but volume
fraction of the phase 1, plus volume fraction
of the phase 2. So, that is the way this mixture
model this ρ is being defined, a logistic
model through mixture has been defined, then
we need to define the mass average velocity.
Now, how to define the mass average velocity;
so, mass average velocity has been defined
by this empirical correlation which says that
it is nothing but ρ α one into un. If suppose
it is a 2 phase, if it is a 2 phase 1 and
2 then it will be the density of the first
phase multiplied by the fraction of that phase
which is present at that location for where
you are solving this equation.
If you are solving it numerically you will
what you do you will discretize the whole
domain, if suppose you are doing a numerical
simulation we are going to discretize the
whole domain and you will solve this equation
for each cell for each node point. Depending
upon what approach you are solving you are
solving the cell weight wise all your solving
the node wise you will solve it and then for
each cell you fine to calculate this. So,
what will happen? That α ρ1 is going to
be remain same that is the density of that
phase 1 α one will be changed and it will
be updated through the volume transport equation
for each cell and for each location and for
each time. Un will be the velocity of that
phase at that location which will be calculated
ρ2 is nothing.
But that what is the density of that face
α2 again what is the holdup of volume fraction
of that phase at u2 and then it will be averaged
with the mass average we are calculating the
velocity. So, we have to divide by the total
mass now what will be the total mass per unit
volume if you take then it will be ρ1 into
α1 plus ρ2 into α3 that will be the mass
per unit volume. So, that is the way the mass
average velocity has been calculate it means
calculated in this cases and then you can
use this here un u2 ρ1 and ρ2 or the velocity
and density of each phases respectively.
Then we calculate the drift velocity. So,
how to calculate the if you remember that
the drift velocity term which we have introduced,
which actually multiplied by the α which
shows that the phase information actually
comes from that place the individual phase
information. So, the drift velocity is being
defined by ur drift velocity is the velocity
of that phase minus the velocity of mass average
velocity of the mixture.
So, this is mixture velocity of mass average
mixture velocity and this is the velocity
of that phase that will be the drift velocity
of that phase. So, u and r it means the velocity
drift velocity this is drift velocity of phase
1 of phase 1 will be equal to u n is the velocity
of that phase velocity of phase 1 minus mass
average velocity of mixture.
So, it what information it gives? It gives
that how much fast or how much slow this velocity
phase is moving compared to the mass average
velocity of the mixture. So, you get the individual
information of that particle and this ur you
calculate and then from ur you can get the
un. So, you can get the velocity of each phase
there and then µis being solved with µ 1
α 1 plus µ2 α 2 as we have already discussed
that the accuracy of these equation or µ
this way to calculate the effective viscosity
of the mixture is always questionable.
But this is the one of the best possible way
to calculate the µor you can develop your
µrating or your mutual relation based on
the experimental observation by using some
viscometer, you can calculate that. The best
way of calculating it is this it may not be
the correct way or 100 percent correct way,
but this is the one of the best way to calculate
the viscosity and you can do that with the
µ1 α1 plus µ2 α2.
So, what you will do? Now, if you see this
equation you are almost able to know everything.
Now you can solve this and you can get the
value of velocity for individual phase.
So, that is the way it has been solved and
then what you do you use some empirical correlation
because what we have not included till now
is the drag we have not included somewhere
the drag and we know that if the 2 phases
are flowing together, one phase is in discrete,
then what is going to add the most important
force is going to be the drag.
So, what we need to do we need to include
the drag in our equation then only being able
to see the relative motion between the 2 features
or the slip motion between the 2 phases. So,
the drag is being defined that what we do
is define the relative velocity that what
will be the relative velocity or the slip
velocity and the slip velocity or relative
velocity has been defined with the a bar of
τp prime. So, this is not the real particle
relaxation time this is very close to particle
relaxation time that is why we have defined
it as a τp, but I put a prime. So, that it
is looks little bit different from whatever
the particle relaxation time which we have
discussed earlier.
So, please do not get confused this is the
particle relaxation time, but it is not purely
the particle relaxation time. It is the particle
relaxation time based on the drag, that is
the way we define inside that this the particle
relaxation time or response time based on
drag or if I have been defined there is a
2 term have been introduced one is a and these
all are empirical equation developed equation.
This is a times τp prime and it is being
defined as a overall acceleration which the
particle is seeing.
So, that is being the gravity if the particle
is moving or discrete phase is moving anywhere,
the gravity will be acting what is the convective
acceleration which is going to take a role.
So, this is dot del so, that is the convective
acceleration term and then what is the local
acceleration term δum/δt and is being calculated
is nothing but the mass average velocity.
So, that is the way we calculate the a value
if you know the you can calculate the a value
very easily then this is the τp prime.
Now, again I said that we have not included
the drag earlier. So, we have include the
drag in the slip velocity or relative velocity
and that we are using again question this
is being empirically developed. That is the
τp prime and τp prime is being actually
defined as ρ m minus ρ p, which is the mixture
velocity mixture density minus the ρ of that
phase ρa particle or discrete phase into
multiplied by dp square which is nothing but
the dp square is nothing but the that phase
diameter or discrete phase diameter upon 18
µf into f drag.
So, we multiply it with the drag factor and
this drag factor is being calculated based
on the Schiller Norman correlation for Reynold
number less than thousand and for the Reynold
number greater than thousand it is being used
with 0.0175 into re. Now, you can always use
your own drag you can write a you do for you
can write your own program to use your own
drag whatever you have kind of you want to
use or you feel that it will be best suited
for your system, but this is the way it has
been introduced.
So, drag is also being introduced. So, if
you see that very smart way of modeling the
thing in a simplistic solution where you can
solve very few equation, most of the equations
are these equations are algebraic equations.
So, it does not numerically does not require
much time to solve and you have only 3 equations
which are coupled together and you solve it.
So, all these equations we solve together
to get the velocity field to get the volume
fraction field and we have also included the
drag so that we can have the idea about the
slip velocity and relative velocity also.
This is the way we solve the algebraic slip
modern equation, these are the equations which
we solve we solve is simultaneously and we
get the data. The only problem or restriction
is that as I said that, we are what we are
doing we are solving only one equation instead
of the 2 individual equations for individual
phases now we are solving only one equation
and we are assuming that the slip velocity
is not very high. You have to see that there
will be several restrictions of this flow
you cannot use it the accuracy is limited.
Because we have oversimplified our model accuracy
is going to be limited, but it predicts well
under certain conditions and there are certain
condition it is now you cannot use it. So,
the major restrictions which are there for
this model this is not only these restriction,
these are the few restrictions which are very,
very critical and one need to be understandable
need to be careful before using this kind
of approach is first after this the restriction
is the particle relaxation time.
The way we define the particle relaxation
time, I will say that particle in relaxation
time based on drag. So, that there should
be no confusion with the initial particle
relaxation time response time the way we have
defined. So, τp prime into f drag which is
being nothing but the particle reactions in
time ρm minus ρp into dp square upon 18
µf and should be less than 0.001 to 0.01
second, it should be in this region or it
should be less than that.
Why I have given 2 range because some book
says this some book follows this. So, that
is why I have given both the range it should
be definitely less than this. This value lesser
the accuracy of the model will be much higher,
but some books also say it is odds group also
says that it should be even less than 0.01,
it will be working fine that is the way and
if you see that if I know about my system,
I do not need to do the simulation and that
is why we define it in this way that it is
τp prime into f dragged because f drag.
If you want to calculate you have to calculate
the non-number for that you need the sitting
inside everything to need. So, that is why
τp rime into f dragged that way which is
nothing but ρ m minus ρp dp square upon
18 µf, you can calculate all these things
ρm can be calculated.
If you know that what is your lined off density
of individual phases, what is the volume fraction
of the individual phases? If you know the
density of the discrete phase, if you know
that roughly that what should be the size
of your discrete phase particle or discrete
phase bubble. It can be bubble it can be droplet,
it can be particle what which will be you
can calculate the µf if you have again some
idea about the volume fraction if it is less
than 0.001 you can use this model or less
than 0.01 you can use this model.
But above than that is this model applicability
is really low and accuracy will be very, very
low then again there is one limitation again
the other limitation is that one phase should
be continuous and one phase should be dispersed.
If 2 continuous phases are moving together
then it is difficult to model. So, it means
if you want to model a separated flow it is
suppose there is a pipe, where gas and liquid
are flowing this is liquid this is gas. So,
this is liquid this is gas then you cannot
use the algebraic slip model.
So, 1 phase need to be in discrete one or
dispersed 1 phase need to be continuous then
only this model can be use then, if suppose
the phase is disposed you cannot see the interaction
inside the dispersed phase. So, suppose if
there is a bubble and that bubble is rotating
or is changing the shape of it. Say this and
it is happened with the Danckwert’s theory
or surface renewal theory, if you will see
that this is that the bubble actually keep
on changing the shape is the surface stretch
theory. Those kind of interactions if you
want to model which is inside the dispersed
space you cannot model it ok.
Even we are not modeling anything about the
bubble coalescence. So, suppose if the 2 bubbles
are there they are coming and close to each
other and they are being form a bigger bubble.
We are not doing we are not modeling all those
things. So, the interaction inside the dispersed
phase we are not modeling. So, that is again
a limitation because most of the time in the
bus a gas liquid system particularly bubble
column, we have seen that bubble shape bubble
size keep on changing. So, that that the interactions
you cannot actually mind this cannot solve
or cannot model by using the algebraic slip
model. So, that is the one of the limitation.
Another limitation is that the discrete phase
fractions would be less than 10 percent and
this severely limit the application of this.
Now, again there is a debate here some people
say 10 percent, some people says 5 percent.
I am just writing all the numbers I am giving
you the bigger number here say 10 percent
definitely above than 10 percent you cannot
do that because discrete phase interactions
will be very, very high one way coupling will
cannot be used you have to have 2 way, 3 way,
coupling at least or 4 way coupling and that
is why here whatever we are doing we are solving
just one way coupling we are solving the equation
of both the phases it is only troubled with
the drag only.
And that drag is also not the part of the
main equation drag, we are actually taking
into the account in the slip velocity ok.
So, we kind of that is the major limitation
of this that if your discrete phase fraction
is more than 10 percent ASM model algebra
slip model cannot be used. And in most of
the industrial problem industrially relevant
thing kind of reactors or element vessels
the volume fraction of discrete phase is generally
more than 10 percent almost all the time ok.
Other than the few specialized cases and that
is why the accuracy or kind of just the model
prediction is very, limited The applicability
of this model is very limited then again as
I said that it is desirable that both the
phases would flow in the same direction, then
the accuracy of the model is higher. So, in
this model ideally cannot be used for the
counter current flow and for sedimentation
slope it means where the particles are being
settled out and the fluid is moving up.
In this kind of a flow, you cannot use this
model and even in case of the counter current
flow say gas is coming from the top, liquid
is going from the bottom or vice versa gas
is going from the bottom to top and liquid
is coming from the top to the bottom you cannot
use this model the accuracy of this model
will be limited. So, that is the major kind
of you can say restriction. So, I have shown
here the example of a bubble column which
has been solved by using the algebraic slip
model.
And the geometry is there is a column is a
2-dimensional column and what we have done
is a small bit here to show that the column
can be used or the model can be used for even
the 3-dimensional domain this is the geometry
and at the centre we have made a hole of certain
dimensions. I think this dimension is more
around one centimeter and we have passed the
air from this. So, this whole column is filled
with the liquid this whole column is filled
with the liquid and air is passed at the center
of the column by air is this passed through
the center of the column.
So, this is liquid and this is the air is
passed there. So, what will happen? The air
will lift at the center in the downward directions
well the bubble will lift here at this location,
the velocity is very low. So, the bubble fraction
is very low bubble will lift it in this way
liquid will move upward at the center of the
column which you can see here. So, most of
the bubble fraction actually will be at the
center of the column. You can see here the
air fraction which is actually at the center
of the column it is showing here the white
color is actually shows air this is air and
this is liquid or water in this case we have
taken water.
So, you can see the air is flowing at the
center there is no very small amount of the
air or almost no air is there near the wall
and most of the liquid is actually near the
wall. This is the air this is the water you
are able to discriminate the 2 phases very
clearly and that is you are able to do because
you are solving the volume transport equation.
So, though we are solving only one equation,
we are able to discriminate between the phases
that which phase is present at got location.
Entirely, see that there is more air fraction
at the center there is all no air fraction
or very low air fraction near the wall.
We are able to differentiate between the phases
and we are able to do it why because we are
solving the volume transport equation, we
are able to also solve the drift equation
drift velocity and we are coupling it with
the slip velocities too. So, you can solve
that the relative motion then we have also
calculated the velocity of the water why we
have not plotted the velocity of air because
the velocity of the air will be only at the
center end and it will be moving upward.
So, we have also tried to show the velocity
of the water this is the velocity of water
this is water velocity and this simulations
are being done in and it is fluent. And so,
what we have this the air is being passed
because the air is moving up at the center
of the column water will also move up at the
center of the column and you can see that
if I just zoom it, you can see that the velocity
are positive at the center of the column.
If you will see this the velocity is positive
look at these location you can see that the
velocity is positive ok. So, you can see this
and near the wall the velocity is negative
because, what will happen? If you see this
location again that near the wall the velocity
is negative and why it is happening because
if you see positive here, it is negative velocity
negative I do not know whether, you are able
to see it earlier positive and zoom it you
can see this velocities that is here the particle
is moving up and in this location the part
this water is actually going down.
So, here liquid or water is moving up here,
it is going down why it is moving up here
because air is moving up a bubble is moving
up that side it is pushing some of the liquid
with it. And that is why the liquid is also
moving up then at a certain height at this
height what happened that air actually disposed
to the atmosphere and move out of the column
and goes to the atmosphere while liquid have
not that much momentum it has no more buoyancy
force, it will fall down again and then it
will start coming down from this side.
So, you are seeing a simple circulation cell
and we are able to predict that circulation.
So, the ASM model ok. So, these are the simulations
this simulation has already been shown in
the literature we have just repeated the simulation
to see that ok. You are able to get the same
profile this is a standard way to solve the
bubble column. So, if your bubble is the discrete
phase volume fraction is very low say less
than 5 percent or 10 percent you can use this
column to get the information and we have
able to get that those information in a very,
very in a very quick way or relatively faster
way compared to the models which we will discuss
here after. So, that is the advantage.
But the major problem is as I said that the
particle relaxation time multiplied by f drag
should be less than 0.001 or 0.01 depending
upon which group you follow I would recommend
0.001 here and second thing that the flow
should not be counter current and the discrete
phase fraction should be less than 10 percent
for sure. So, lower the discrete phase fraction
better accuracy that is the way it is kind
of the model has been developed is relatively
simpler as we seen the equations and you can
solve that approach. So, that is all about
the algebraic slip model which is being used
some people also called it as a mixture model
or algebraic slip model which is being used
here I have shown here that how-to kind of
calculate a model in this 2 phase flow.
In some of the similar kind of commercial
software’s though if you can want to write
you can write your own course, but if you
are planning to use some commercial software
be aware some of the commercial software does
not allow to use more than 2 phases for ASM
model. It does not mean that ASM model cannot
solve 3 phase problem it can still solve the
3-phase problem you just need to modify the
equations we just need to modify the mixture
velocity, the mixture density mixture viscosity
the way the mass average velocity has been
defined you have to modify those equation
and can easily modified the third component
will be added or third phase will be added.
But most of the commercial software’s actually
restrict the ASM application for 2 phase flow
only. So, kind of do not worry do not be confused
that it can be used only for the 2 phase it
can also be used for the 3 phase definitely
the volume fraction if it will be 3 phase
will be further higher interactions of the
phases will be further higher. So, accuracy
of the productions will be lower.
But mathematically there is no restriction
you can use it theoretically there is no restriction
you can use it ok. So, that is the overall
about the algebraic slip model now we will
move to the next model which is one of the
most commonly used and very popular model
to model the multi-phase flow reactor and
that model name is Eulerian-Eulerian model
as I said earlier also in discussion that
Eulerian-Eulerian model. What we do we assume
all phases to be interpenetrating continuum.
And that is the one of the major assumption
which you do I have already discussed about
that a junction that what interpenetrating
continuum means it means suppose, there is
a solid phase there is a gas phase they are
moving together. You will allow the solid
to go inside the gas you will allow the gas
to go inside the solid physically in the real
world it is not possible, but in the Euler
models you. So, that all your kind of describe
it it in that way.
So, what happens for any small cell whatever
you are solving you will not get that where
is the particle exactly where is the solid
gas exactly, you will get the fraction of
the solid fraction of the gas in that cell
and that is the way the kind of the phases
has been defined with the volume fraction
here again and we will multiply the equations
with the volume fraction for each word term.
So, that we can get that for each velocity
for each cell whatever the velocity or whatever
the stress we are calculating what is the
contribution of each region.
You will see the equations, but that is the
major assumption that you assume that the
particle should be interpolated continuum
there is no discrete nature of the particle
and that is the major limitation also, because
you do that several interaction forces you
cannot solve it or you have to solve by using
the empirically developed equation or you
have to model it by using some equations,
but directly you cannot solve this and that
is limits that kind of use or you can say
the accuracy of this approach.
But still this is the only approach till date
which can be used at all the scale you can
solve the laboratory scale problem you can
solve the pilot plant a scale problem you
can solve the industrial scale problem, but
only the completion time will be higher, but
it is has a potential to do that and that
is why this model is widely used this model
is also being known as a 2 fluid model.
Particularly if you are solving it for both
the fluid phases and you see is called as
a 2 fluid model and why 2 fluid model is being
solved called because both the phases whether
it is a solid or it said any other phase it
is a solid also gas also liquid whether it
is a dispersed stage both the phases you assumed
to be fluid and that is why it is also called
2 fluid model commonly.
So, what we do in this we solve the momentum
equation and continuity equation for each
phases individually. So, what we were doing
earlier we are solving only one momentum one
continuity here you will solve n number of
momentum if you have n number of phase you
will solve n number of momentum equation and
number of continuity equation. So, it means
if I have a 2 phase I will solve 2 continuity
2 momentum for both the in this kind of approach.
So, what is going to happen your computational
power requirement is going to increase. Because
earlier, you were solving only 3. Now, you
are going to solve 4 if you have a turbulent
model you will solve actually 5 you will also
include the turbulence if you are including
the turbulence in both the phases you are
going to solve 6. So, your number of equations
are increased and that is why the computational
time means higher compared to the algebraic
slip model.
But the accuracy is also better compared to
the algebraic slip model because now you are
solving the individual equation for individual
phases has been widely used for the modeling
of fluidized bed risers pneumatic lines hoppers
standby is particle laden flow bubble columns
slurry bubble column all kind of a reactor
this approach has been widely used to model
other than the pure granular flow where only
granules are flowing ok. So, that is the place
where it is kind of it is being not model
or it is decorations will be hampered.
But other than that, it can be used anywhere
even for the flow which is being with the
both the phases are separated then also this
model can be used if they are dispersed then
also this model can be used. So, the applicability
of this model is very wide you can use for
almost all the cases in this model and that
is why the acceptability of this model is
also relatively higher. Because, we are solving
the individual phases equations and we are
coupling those equations are those phase with
the drag and the other forces which will be
required to solve the a higher order coupling.
So, we discuss about the equation, but that
is the way this whole thing is there and it
has been approved used for modeling all kind
of a reactor the discrete phase volume fraction
can be anywhere between 0 to 60 percent. So,
it gives you almost the luxury to solve all
kind of a reactor. So, even if you solve the
pad bed the generally the impact, but the
maximum volume fraction goes in the range
of 60 percent. So, you can solve till the
back bed then you can solve the pad bed to
find that how the velocity distribution is
taking place in the trickle bed also people
have used that this is gives a lot of flexibility.
Now, if you see the restriction of all the
ASM restriction is now being removed actually
and then you can use the drag of whatever
the drag you want for your system. You can
write your UDF or use your own code to use
the drag. So, it does not restrict on your
drag model you can use the kinetic theory
of granular flow or you can based on that
to solve the granule stress in the viscous
regime and it means what it is going to solve
the equation between if you remember, the
closure equation you will solve the drag if
you want to do the find that the interaction
between the mean motion of the solid mean
motion of the gas or I will say mean motion
of the discrete phase and mean motion of the
continuous phase.
You can take the turbulence equation you can
include the turbulence equation, if you want
to solve the mean motion of the gas range
or the fluid phase to the fluctuating motion
of the fluid phase. You can use the kinetic
theory of gases if you want to solve the interactions
between the mean motion of the solid phase
and subjugating motion of the solid phase
all these things. You can include, you can
solve, the viscous regime flow you can solve
the granularity stress by using the KTGF.
So, that flexibility is also there you can
use that the frictional velocity formula in
the plastic regime can also be used. It means
it can be used not only for the viscous regime
it can also be used for the plastic regime
you can add all the forces. Whatever we have
discussed whether, it was virtual mass force
or added mass force it is also called virtual
mass force lift force all these equations
you can include and you can solve this and
that is why this is the applicability of this
model is enormous.
Because, it gives you the flexibility to do
whatever you want to do the only problem is
that accuracy of all this equations this model
depends on the closure equations or the closure
models which you used to solve these equations
and because these equations are now the number
of variables are more number of equations
are less. You have to use the closure equation
to close that and your accuracy of the solution,
now is going to depend that how good are your
closure equations.
So, how good are your drag force equations
we have discussed, several drag forces for
the same thing for either it is a gas liquid
flow liquid. You can flow a gas solid flow
we have discussed several type of drags. Now,
depending upon how accurately your drag is
able to predict the motion it will the accuracy
of this Euler model will depend on that. How
accurately the kinetic theory of granular
flow models are being used to predict the
fluctuations; that means, the way it will
able to predict the motion or how accurately
your turbulence model. Whatever, you are using
whether it is a k epsilon k omega or any RNG
model whatever the accuracy; however, it is
able to predict the fluctuation motion.
The accuracy of the island model is going
to depend on that. So, that is the way it
has been solved and if you use for the granular
flow regime before I go and discuss the equation.
So, I said that it can be used for plastic
regime it can be used for the viscous regime
for the granular flows. So, generally for
the granular flow 3 regimes are there one
is a elastic regime plastic regime and viscous
regime this things you might have been done
in your solid mechanics course.
But just for a revision I am doing it that
this Euler model can be used for this places
the elastic regimes are generally stagnant.
They do not move till move and the stress
and strain are dependent to each other and
may still clean the collision is modeled with
the elasticity that, how they will move or
the collision is modeled totally through the
elasticity ok. So, that is this kind of a
flow is called elastic regime.
The plastic regime is where the flow is very
slow the stress and strain rate is independent
that does not depend it on that the motion
is not depend on the strain rate. And generally,
the models and when the flow models has been
done through the solid mechanics that how
the solid mechanics problems we solve we solve
in the plastic regime in the viscous regime
the flow is very fast it moves very rapidly
because they are in the viscous regime.
A strain rate are dependent ok. So, if you
change the strain rate your overall mission
will also get changed your collision will
also get changed and the flow is missed mostly
dependent on the kinetic theory.
So, the kinetic theory of granular flow is
being used to model the flow as the flows
are very much collisional dominating and they
are moves very fast and the strain rate is
dependent these are the 3 regimes I am not
going in detail of this. I am just giving
you just trying to revise it you can see it
in your solid mechanics courses that what
are the details of these regimes, but that
is the basic classification of these 3 regimes
and Euler Euler can be used for the plastic
regime and viscous regime.
Now, what I said that in the viscous regime,
but maybe most of the fluid ice bits comes
under the viscous regime what we need we need
a kinetic theory of granular flow to model
the collision to model the flow actually because
the flow is depend on the kinetic theory of
the flow now kinetic theory of granular flow
it is itself is a very vast topic and cannot
be covered here.
So, the domain of this course is not to discuss
the kinetic theory of granular flow in detail
or the derivation of the kinetic theory of
granular flow you can follow any book to see
that, what is the how the kinetic theory of
granular flow has been defined or derived?
And you will see that it is actually nothing
but the kinetic theory of the granular flow
is very similar to the kinetic theory of gases
you take that approach instead of the gases
in gaseous kinetic theory of gases. We assume
the gas molecular to be a point particles
the size is infinitely small here we take
a finite size of the particles.
Now, the moment you take the finite size of
the particle and the particle is being placed
in the fluid which is moving and the particles
are free to move randomly. Anywhere, they
will have the collision and the flow will
be mainly depend on the kinetic transport
it means the transport which is because of
the fluid and again, the another transport
which will be dominating is the transport
because of the particle collision. So, that
is the way the solute particle system is being
defined.
And for this kind of a flow kinetic theory
of granular flow is being used and being modeled
how the kinetic theory of granular flow model
has been described this all a brief description
again I am telling that it is a brief description
it is very difficult to cover it in a small
time it requires abstention time to show you
that how the character theory of grammar flow
has been defined.
So, basically kinetic theory of granular flow
is very close to the Chapman Enskog theory,
it is not very close to the actually the kinetic
theory of the ideal gases or kinetic theory
of gases, but it is very close to the Chapman
Enskog theory where the Chapman Enskog theory.
If you remember your kinetic courses then
it has been might have been told that Chapman
Enskog theory what we do it is for the real
gas.
So, you do not assume that the particle is
a point particle you assume the particle has
certain size and then you try to calculate
that what will be the interaction forces will
be acting on that. So, the Chapman Enskog
and then you define the attraction potential
repulsion potential energy own potential all
those quantities you define.
Kinetic theory of granular flow also is very
close to the Chapman Enskog theory and we
assume that the particle has a finite size
and that is particle finite size we see the
interactions. So, the basically if you see
that the application of kinetic theory for
granular flow has been developed initially
by the Jenkins and savage in 1983, then Lun
et al has further modified it in 1984 ding
and get Gidaspow has done the same in 1990.
So, this is a chronological development of
this; what is based on this is based on the
collisional particle interactions. So, what
we are doing we are seeing that how the particle
will you have a collision and because of that
collision how the motion of mean motion will
be correlated to the fluctuating motion of
the particle. So, that is the whole idea and
the kinetic theory of the underflow.
So, what we do the velocity fluctuation, what
is the assumptions are that the velocity fluctuation
of the solid is much smaller than their mean
velocity. So, that is the one thing which
we should keep in the mind that if you are
using the kinetic theory of the granular flow
your kinetic, it means the velocity fluctuations
in the solid should be much lower compared
to the mean velocity of the solid.
If the flow is completely collisional dominating
and the mean velocity of the solid is much
lower, then the fluctuating velocity of the
solid may be this model will not give you
the much accurate position, this kind of prediction
and why I have said this because some of the
bits. And we will discuss it while we discuss
the binary fluidized bed and all where you
have the distribution of the particle, which
you use may be it may possible that the flow
is actually collisional dominating.
We will discuss those cases, where the flow
you will see that the velocity of fluctuation
velocities are much higher compared to the
mean velocity. So, in these kind of cases
the accuracy of the granular model predictions
the kinetic theory of granular gas granular
flow will be limited then dissipation of the
fluctuation energy due to inelastic deformation.
We do that the dissipation of the fluctuation
energy due to the inelastic deformation that
we take and then dissipation also due to the
friction of the particle with the flow.
So, these all dissipation we take into the
account and we use the Chapman Enskog theory
model to do good you find that, what is the
kinetic theory of granular we develop the
kinetic theory of the granular flow.
So, what is the major thing this is that the
particle fluctuation velocity should be less
than the mean velocity. We calculate the dissipation
of the fluctuation energy due to the inelastic
inferred deformation. So, particle deformation
is also being taken into the account the dissipation
due to the friction of the particles with
the fluid is also being taken into the account.
So, all these things is taken into the account
and based on that a granule phase kinetic
theory of granular flow has been developed
and a quantity has been defined that is called
granular temperature.
So, what we do we take the velocity of particle
is decomposed in the mean velocity and superimposed
it with the fluctuation regime of fluctuation
thing random velocity. So, we assume that
there is a particle which has certain mean
velocity we super imposed that mean velocity
with the flux random fluctuating velocity.
So, that we do then we define a term which
is very analogous to the thermodynamic temperature
of the gas, when we use define the kinetic
theory of gas we define the thermo dynamic
temperature.
And how do we define the thermodynamic temperature?
We define it with the collision between the
particles or collision, between the gas molecules
and we say that the collision between the
gas molecules are the kinetic energy developed
because of that collision is equal to 0.5
ρ v2 and that is equal to nothing but (3/
2) KT where k is the Boltzmann constant and
t is the thermodynamic temperature.
So, from there you define the kinetic theory
of gas this temp analogous to the temperature
you say (3/2) KT is equal to half mv square.
So, you say that this is the kinetic theory
of this who you define the kinetic energy
and you say that this kinetic energy is basically
because of the collision and that is why you
feel the temperature. So, the temperature
you define as 1 upon 3 V square upon K and
if I take it the velocity as a of this velocity
per unit mass if I will take or I will divide
that the temperature in the mass independent
quantity if I define in that way it will be
coming it in this Y phis.
Similar way, we define the granular temperature
which is nothing but it is associated with
the random fluctuation of the fluctuating
velocity of the solid particle in the kinetic
theory of gases just go and revise it. And
if there is any confusion you can write me
I we can try to talk those problems we assume
that the particles or molecules are having
a collision with each other and they are randomly
moving and because of that random motion their
kinetic energy. Because, of this collision
and the random motion the kinetic energy has
increased and that is because of that only
the temperature of the body will increase.
So, the temperature of the body is actually
related to the random motion or the kinetic
energy. So, what will happen if you start
hitting any fluid say gas what will happen
it is temperature will increase if it is temperature
will increase then kinetic energy will increase
number of collision will increase. So, you
can say that increase in the temperature is
being general seen only because of the increase
in the collision.
Similar things we are also doing here in the
kinetic theory of granular slope I am not
deriving it, but what we do we take that mean
velocity of the particle we superimpose it
with the fluctuating velocity then with the
analogous to the thermodynamic temperature.
We define a quantity called granular temperature
and that is actually being defined as a quantity
which is associated with the random flux creation
of the fluctuating velocity of the solid particle.
So, what does it means? That the fluctuation
particle whatever the particle is which is
there what is it is fluctuating velocity and
how the variation in the fluctuating velocity
is taking place. That is the way the granular
temperature has been defined and the source
of the granular temperature or the particle
fluctuation is actually comes from the collision
of the neighborhood particle. The particle
fluctuation source whatever we account it
is a counted based on the collision which
the particle is having with the neighboring
particle.
So, in that way the granular temperature has
been defined and it has been defined as the
exactly same way 2 upon 3 into Ks. So, suppose
if I take it as a Ks which is the kinetic
energy then it will be the temperature will
be nothing but it will be 2 upon 3 of Ks divided
by Ks am not including the Boltzmann constant
here.
So, exactly same way analogous to that this
whatever thermodynamic temperature is being
calculated or defined analogous to that the
granular temperature has been defined which
will be nothing but 2 by 3 Ks and the Ks is
nothing but the kinetic energy due to the
solid velocity fluctuation per unit mass.
So, I did the per unit mass. So, that this
m will be removed and it will be nothing but
half V square that is the case kinetic energy
due to the solid velocity fluctuation per
unit mass exactly same way the kinetic energy
has been defined here that in the kinetic
theory of the gases that is why this granular
temperature and that thermodynamic temperature
are a very analogous quantity.
And the ks which is the kinetic energy of
fluctuation per unit mass it will be defined
as 
the same way the ks will be half it will be
VX prime square plus vy prime square plus
VZ prime square. So, suppose you have 3-dimensional
velocity it will be VX prime is the fluctuating
velocity in x direction. So, solid fluctuating
velocities I will write it let us make it
very solid fluctuating velocity in the x direction
this is solid fluctuating velocity in the
Y direction this is solid fluctuating velocity
in the J direction.
So, what you are saying that how the solid
fluctuations are taking place in all the 3
directions you are adding them and then you
are dividing it by 1 by 2 you are square with
them and then I had actually. So, it will
be actually averaged these all are averaged.
So, what you do suppose in a small cell? If
the particle is there is 10 particles are
there you take the fluctuation particle fluctuating
velocity of all the particles you square them
and then you add them ok.
So, that is the way it has been there. So,
it is a mean fluctuating velocity of in the
x direction this is the mean solid I will
write it here mean also means solid fluctuating
velocity in the x direction means solid fluctuating
velocity in the y direction mean solid fluctuating
velocity in the J direction you take average
all together. So, that is the way that Ks
has been defined and theta is nothing but
2 by 3 of Ks or I can say theta s is nothing
but 1 upon 3 it will be v x prime square v
y prime square VZ prime square.
So, that is the way the granular temperature
can be defined. So, this quantity has been
defined to model the collision or to predict
the collision that, what will be the collision
forces? And how the fluctuation velocities
are being correlated?
So, that is the way the kinetic theory of
granular flow has been defined again this
is a very brief way I have just tried to cover
give the very basics of this.
But what is the difference between the kinetic
theory of the particle kind of gases and the
gas molecules or kinetic theory of granular
flow and kinetic theory of gases the difference
is actually whatever the difference is there
between the gas molecules and the particle.
So, that is the main difference between the
kinetic theory of gas and kinetic theory of
granular flow that, kinetic theory of the
granular flow solid particles are a few order
of magnitude larger compared to the gas molecules.
So, the gas molecules even if you check the
check the Chapman Enskog theory. We assume
to the finite size of the gas molecules, but
the particles are generally few order magnitude
higher. So, suppose a gas molecules even if
you take a finite size generally will be in
the order of any strong here you will get
in the order of micrometer or nano meter even
if you are taking nanometer particles. Then
also, it is one order of magnitude higher
if you are talking about the micrometer level
particle the 4-order magnitude higher.
So, velocity fluctuation of the solids are
much smaller than the mean velocity and that
is most of the time it is true while, in the
case of gases it may not true the fluctuating
velocity may be higher compared to the mean
velocity the kinetic where a part of the solid
fluctuation is anisotropic again this is very
critical most of the in the gas kinetic theory
of the gases we always assume that the fluctuation
are isotropic in nature it means what dx prime
is equal to VY prime will be equal to VZ prime
or you can say VX prime is square VY prime
with square VZ prime is quite extreme anyway
you can define.
This is the isotropic fluctuation, but whatever
VZ in the kinetic theory of the solids that
they are their fluctuation are mostly an isotropic
in nature. So, if they are anisotropic it
means they are not equal then velocity fluctuation
of the solids dissipate into heat rather fast
at the result of inter particle collision.
So, if you are having a collision the particle
actually the energy will dissipate very fast
compared to the way it has been dissipated
in the kinetic theory of gases or with the
gas molecule and granular temperature the
way we have defined is actually the byproduct
of the flow or the collisional dominating
flow or by product of the collisional slope
that is the way we define it.
And that is the difference major difference
the way gas molecules and particles are different
or the way the kinetic theory of gases or
the kinetic theory of granular flow is different.
So, what we do with this what we have done
we have taken we have defined the third order
and this kind of integration or third closure
and that third closure was between the particle
mean motion to the particle, fluctuating motion
which has been defined with the kinetic theory
of granular flow and has been could defined
a term of this granular temperature which
will be actually correlating them.
So, based on that we again back to the two-fluid
model or Euler Euler model ok; so, please
do not get confused if the name is same Euler
Euler or two fluid model and what we do we
solve the continuity equation. We solve the
momentum equation for the individual features.
So, if suppose you have a 2 phase you solve
the 2 continuity equations say this is for
the phase q the similar thing you will solve
for the phase s.
So, if they have a 2 phase you will solve
it in this way the α is actually the volume
fraction do not get confused this α is actually
whatever we are representing here α is actually
nothing but epsilon. So, it is the volume
fraction inside and then you solve the momentum
equation. So, if you have 2 phases you solve
this momentum equation and you keep or you
solve it twice. So, if the both the phases
are liquid you solve this equation twice.
So, if you see this here again local acceleration
convective acceleration you solve the pressure
gradient term, but now the pressure gradient
is being multiplied with the volume fraction
of that phase it means now you are solving
that what is the pressure difference or what
is the pressure gradient which is the contribution
because of that phase similarly if you see
their local activation also we are solving
we are multiplying with the α it means the
phase fraction then we are multiplying with
the α here again means what this is the contribution
only of that phase.
Then we solve the τ that what is the stress
viscous term we solve this then any Body force
definitely gravity is going to act. So, gravity
on that stage that is why we multiply here
again with the α and then we include a drag.
So, drag is here this is the drag we have
write in the drag we modified generally in
the commercial software in the drag we write
it in this way ks into v as minus d and the
case will be defined with the rest of the
term, whatever the way we have discussed.
So, the drag we defined as half ρ A CD into
slip velocity say (v-u )2. Now, what we do
they define this is as a k and v minus u and
rest of the term will be defined in terms
of the k which will be also correlated with
the ceiling. So, we will discuss that some
of this terms, but that is the way the drag
has been defined which is being acting between
the 2 phases and you are modeling the mean
motion interaction of the 2 phases which is
the part of the equation now which was not
there in the algebraic slip model.
In algebraic slip model, you are defining
the drag you are including the drag in the
slip velocity or in the relative velocity
and that you are being defined with the empirically
developed correlation. Now, here we are using
the model of the drag directly into the equation.
So now, if suppose you have the 2 phases both
the phases are fluid it means say if you are
solving for the gas liquid system fluid does
not mean that this would be the same phase.
So, if you are solving for the gas liquid
system. So, the same equation once you will
solve for the gas phase once we will solve
for the liquid phase what will happen you
have 2 continuity equation 2 momentum equations.
So, that what we saw and then we try to find
it out the closure for the drag if you have
any other interactions forces like virtual
mass force if you want to use the lift force
any other force you can also include it here
to solve this equation. So, that is what in
the 2-fluid model or Euler model we do.
Now, once the flow is solid what we do is
in case of the solid flow in the things becomes
a little bit different. So, in the gas phase
or both the phases which are fluid then the
things are little bit simpler you solve exactly
this equation is a pure damage to equation
or you can say the name is very close to Navier
stokes equation where you just include the
drag force which is the interaction between
the 2 phases you just multiply with the volume
fraction of each phases in that place. So,
each term will be multiplied by the volume
fraction. And so, that if you sum those both
the equation you will get the total forces
which is acting on the body of it in the wholes
kind of domain.
So, that is the way it has been solved continuity
equation is also solved individually for one
individual phases. So, the mass conservation
equation always validated you never see a
problem of the mass loss that your mass conservation
is not being balanced. So, that is the major
advantage, but once the solid comes into the
picture suppose now you are want to solve
for the gas solid or liquid solid this equation
will be modified and some of the term will
be this equation will be modified.
How this equation will be modified? We will
see it here that how this equation will be
modified you know you just enriching this
so that this equation will be more visible.
So, if you see the left id signed I order
term is not being modified because that is
nothing but drag federation term local acceleration
convective acceleration they are exactly same
for the solid phase also and they are just
being multiplied with the volume fraction.
So, this is being this terms has been used
α s term.
Now, we include here if you see there is only
one pressure term here we include one more
pressure and that pressure is the solid pressure
term and we keep this term exactly same αs
into ∇p instead of this ∇p αq or gas
phase say volume fraction. Now, you multiply
with the solid phase volume fraction and we
subtract that solid pressure term we include
that solid pressure term ok.
Now, why the solid pressure term has been
include how it has been defined we will discuss
in the next slide then we use a solid stress
term now this is very critical because now,
what we are going to model? We are going to
model the delta τs and the τs is being defined
for the fluid with the same stress strain
dependent ratio.
So now you are treating the solid as a fluid
you are taking it as a viscous fluid and we
will see that that is why you need to modify
this τs values also and we will see those
things later. So, these 2 terms are very,
critical because you are defining it for the
solid ideally the solid has no viscosity or
you can say that infinitely high viscosity.
So, if I take the solid if I start rubbing
it I will not deform that is not the curve
this my fluid nature this is not the nature
of the solid so, but what we are doing to
do? We are taking that viscous deformation
forces here in they are defining it it in
that way. So, what we need to do we need to
define the µ value of the solid.
Now, µvalue of the solid you cannot get.
So, you have to define some correlation to
find the µvalue of this and that is why now
you will see that though the equations looks
very simple the approach looks very simple
you need to define many quantities. And that
is why the accuracy of this equation will
depend the accuracy of all those models which
we are using to define the solid pressure
term to solid viscosity term the drag force
term all these terms how you are defining
your accuracy is going to be limited with
that.
And if you are going taking the granular temperature
also and what is the terms which we are including
to model the granular temperature. We will
discuss that there certain terms will be needed
this is the solid pressure term.
Which is being defined at it in this way we
will discuss anyway I do not want to kind
of get into it. So, the solid pressure term
that is what this would want to focus that
if you see that for both the fluid phase or
in the solid phase. The equation is same the
way it has been defined is same and it is
nothing but Newton’s law of viscosity.
So, if you see that that is the way it has
been defined the only thing has been done
it has been multiplied by the α. So, that
if you are solving for the gas phase it will
tell you about the gas phase thing if you
are solving for the liquid phase it will tell
you the information about the liquid phase.
So, this is the way it has been defined this
is the if you guys go ahead and open the transport
phenomena book you will see that the τ has
been defined it in this way, there is a region
to define it in this way because the τ is
a symmetric tensor. So, that is why the other
transpose term has been used. So, that you
get the symmetric nature.
The volume change can affect the viscosity
that is why this terms has been used and if
the fluid is incompressible this del dot v
term will be 0 with the continuity equation
anyway. So, this is the way it has been defined
and that is the major limitation because the
solids which cannot have any µ value. Now,
you need to define that µ value and for that
you need to depend on some of the equation
which is being developed and those equation
accuracy will limit the accuracy of all these
models.
This is the as I said that this is nothing
but a drag and the drag is being defined this
is the interface momentum is or drag which
has been defined as α as ρ s as f upon τs.
So, what we have done we have already said
that as I said that in this force we have
actually defined it in this we have put the
drag factor we have defined the α s ρ s
upon τp if you do that you will actually
get this value.
And these are the forces which can be any
additional force as I said it can be any external
force a electrical say you can have magnetic
field it can have a lift force it can be virtual
mass force it can be message history force
any force which you want to include you can
include. So, that is the beauty of this model
it gives you a lot of luxury, but the major
limitation of this model is that for these
things your accuracy will depend based on
the accuracy of these closure equations which
you are using to solve this equation.
In case of the solid, the accuracy will depend
on the model equation or the closure the equations
which we are using to solve the ∇ps and
∇ τs values and also the drag forces.
We will see that, how this individual terms
are being defined. So, shear stress strain
tensor term as I said that it is for the continuous
phase it has been defined and it has been
defined as µ ∇v plus x transpose (2 /3)
µ-k where k is the dilatational viscosity
into ∇ dot v I which actually this term
take care of when the volume change the dilatational
viscosity is there which is caused.
Because of the volume change of the fluid
some of the fluid changes the volume this
term has been included to give the symmetric
nature to you to you are just out insert τis
the second order tensor which are symmetric
to make it symmetric the transpose term has
been included and this is the way it has been
defined now there is a dilation viscosity
I have already discussed that. So, if I try
to solve it in the conventional way the way
we define the τ for the fluid field it is
being defined as minus µ δ v upon δ x into
δ v x upon δ y if it is a τxy ok.
So now if you see that τxy will be equal
to τy x. So, that does give the symmetric
nature I have removed this term because that
this term will be there this is normally this
term identity matrix or normal this tensor
that opponent will be theirs when the normal
tensor will be one for the shear stress term
this value is anyway going to be 0. So, that
is why this value will be defined it in this
way. So, this will be τyx if you write the
τyx it will be also the same µδ vx upon
δ y plus δ v y upon δ x.
So, this gives you the symmetric nature and
that is why this way it has been defined for
the τxx term or the normal component of the
stress you define it in this way this will
be δ v x plus δ v x that is why it becomes
twice of δ vx into 2 by 3 µminus k into
del dot v these terms are you already me knowing
it it has been very discussed in the transport
phenomena. So, I am not touching it much,
but that is the way the shear stress tensor
in the continuous phase has been defined.
So, the first part if you see here it is not
a problem to define the τwhich is the continuous
stage we can easily define that because for
that I can easily find out that what is my
new value ok.
Now, the problem starts once you go for the
solid stress tensor double. So, what happened
the solid stress tensor term we define exactly
same way as the way we have defined the fluid
stress tensor term the only problem in the
solid stress tensor term is now you are having
a µs and that is going to create a problem.
Because solid as I said ideally have no viscosity
or you can say have a in finite viscosity.
So, they do not move they do not deform even
for a very high strain rate ok. So, that is
you can say in a very infinite viscosity or
no viscosity the way you want to tell you
can do that, but the major problem comes with
the µs how you could define that δvs you
can still find because the gradient in the
velocity will be there in the x direction
and the y direction if there is a gradient
you will see that gradient.
The only problem is the µs. So, to define
the µs several people or several researchers
have given many correlations, but some of
the correlations are very critical or very
important I am going to discuss that correlation
only. So, the µs actually which is solid
in viscosity has been defined it is a summation
of the 3-different viscosity ok. So, viscosity
is what it is a resistance to the flow.
So, that approach has been used to define
that what will be the solid shear viscosity.
So, that we are trying to see that what will
be the resistance force will be there if the
solids are moving. So, what we are saying
that that resistance will be mainly because
of the collision it will be because of the
kinetic energy it will be because of the friction
these 3 will be there. So, the µcm has been
defined as a summation of µcollisional µkinetic
µfrictional ok. So, this is frictional.
So, this is kinetic viscosity term frictional
viscosity term and this is collisional viscosity
term 
it has been defined it in this way that µor
total resistance to the flow this comes because
in the stress of the solid will come because
of the collision of viscosity, kinetic viscosity
and the frictional viscosity. In that way,
it has been defined for each viscosity several
equations have been proposed in the literature.
Again, I am saying only those equations which
are most popularly used it does not mean that
you have to limit it yourself till this point.
You can develop that these equations there
are several ways to develop these equations
and this several equations are also available.
So, we are discussing some of the tentative
equations which are widely used. So, collisional
viscosity equation has been defined by 4 by
5 α as ρs ds into here ds is the solid diameter
or diameter of the solid particle g0 is nothing
but the radial distribution function this
is radial distribution function g naught.
We will discuss that this radial distribution
function 
and es is the restitution coefficient it is
restitution coefficient. So, that is the way
it has been defined and if you see that that
is also the function of granular temperature.
So, this theta s is nothing but it is a granular
temperature. So, it says that new collisional
is going to be the proportional to under root
of granular temperature it is going to be
the function of radial distribution function
of the solid it means how the solids are radially
distributed.
What is the restitution coefficient? Restitution
coefficient has been defined that if the particles
are having collision if the pollution is completely
elastic the value of es s or restitution coefficient
will be one if it is completely inelastic
the value will be 0, if it is viscoelastic.
It means, it is not completely elastic it
is not completely in the elastic the value
will be between 0 to one. So, that is the
restitution coefficient it is d prime upon
v it means the column velocity of the particle.
After the collision to the velocity of the
particle, before the collision if this is
a completely elastic both the velocity will
be the same. Then it is value will be one
if it is complete inelastic then after the
collision particle will actually get is stationary.
It will value will be 0 if it is viscoelastic,
the value will be in between this that is
the way the restitution coefficient has been
defined and the way it has been defined is
that that what is the fraction of the solid
present at that place. What is the ρ value
although it means the total momentum it is
going to take place that ρ value will depend?
Then what is the particle size then what is
the radial distribution of the particle. So,
how the particles is being radially distributed?
So, suppose this is their how the particle
radial distribution is there then what is
the restitution coefficient of the particle.
It means after the collision whether it will
be stopped where there will be elastic. It
will be inelastic and then what is the fluctuating
velocity component in the particle and that
is nothing but the granular temperature it
has been included that is the way the collisional
shear viscosity term has been defined.
Similarly, kinetic viscosity term has been
defined again I am telling again and again
that these are not normal equations there
are several equations available do not get
confused that really these equations are there
are other equations are also ever this is
the equation which is given by Shyamlal et
al. and they have also found that the µ kinetic
energy is also actually the function of Θs
then the temperature.
They are also going to see that; what is the
fluctuating velocity of the particle they
are also the function of es restitution coefficient
and g naught that; what is the radial distribution
of the particle. Similarly, I am not detail
of that how these equations are discussed
then the viscosity is also being defined and
it is also in the same way that what is the
frictional viscosity you define with the friction
and you find it out that what is the value
of this.
λs is nothing but the bulk viscosity which
has again defined it it in this way and again
if you see the bulk viscosity is very close
to the wave. We have defined the collisional
shear viscosity these 2-viscosity value, if
you will see other than some constant the
values are exactly same if you see these 2
why it is going to be it means because the
bulk viscosity is mainly because of the collision.
So, only some parameter is being changed and
you get the equation for this also.
Similarly, the solid pressure; so, this is
the way the shear strain model is being developed.
You use all these collisional equations to
model that and you see that how these equations
are kind of going to be implemented in this.
And the accuracy of this shear stress or kind
of all it is stress tensor or solid stress
term will depend on how accurately these models
are working. So, that is the way and if you
see that all this value of g naught we will
discuss it, actually are being empirically
defined or is being calculated based on some
values or some notion like restitution coefficient
values calculating for the solid is very different
it.
So, whether used for the restitution as a
bulk parameter or not, the collision should
be dumped at a trial situation or not this
is all a different question. Altogether, we
will discuss translator, but the way it has
been defined in the next class. What I am
going to do? I am going to discuss about the
solid pressure term also and then we will
see that how the accuracy of the Euler Euler
model will be there? How to use that model?
And then after that we will see the discuss
the limitation of the Euler Euler model. And
then we will use towards the Euler Lagrangian
model or which is also called a discrete phase
model ok.
Thank you.
